New Insight Mathematics 9 5.3

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Chapter 1 Review of Stage 4 This chapter reviews Stage 4 of the Mathematics syllabus and covers the outcomes of Number, Patterns and Algebra, Data, Measurement, and Space and Geometry. After completing this chapter you should be able to: ✓ recognise the properties of special groups of whole numbers and apply a range of strategies to aid computation

✓ collect statistical data using either a census or a sample, and analyse data using measures of location and spread

✓ compare, order and calculate with integers

✓ use formulae and Pythagoras’ theorem in calculating perimeter and area of circles and figures composed of rectangles and triangles

✓ operate with fractions, decimals, percentages, ratios and rates ✓ solve probability problems involving simple events ✓ use letters to reperesent numbers and translate between words and algebraic symbols ✓ create, record, analyse and generalise number patterns using words and algebraic symbols in a variety of ways ✓ use the algebraic symbol system to simplify, expand and factorise simple algebraic expressions ✓ use algebraic techniques to solve linear equations and simple inequalitites ✓ graph and interpret linear relationships on the number plane ✓ construct, read and interpret graphs, tables, charts and statistical information

✓ calculate surface area of rectangular and triangular prisms and volume of right prisms and cylinders ✓ perform calculations of time that involve mixed units ✓ describe and sketch three-dimensional solids including polyhedra, and classify them in terms of their properties ✓ identify and name angles formed by the intersection of straight lines, including those related to transversals on sets of parallel lines, and make use of the relationships between them ✓ classify, construct and determine the properties of triangles and quadrilaterals ✓ identify congruent and similar two-dimensional figures stating the relelvant conditions.

Syllabus references NS4.1, 4.2, 4.3, 4.4, PAS4.1, 4.2, 4.3, 4.4, 4.5, DS4.1, 4.2, MS4.1, 4.2, 4.3, SGS4.1, 4.2, 4.3, 4.4 WM: S4.1–S4.5

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Review of Stage 4 (Chapter 1)

2

A. REVIEW OF NUMBER (NS4.1) Exercise 1A 1

Write the multiples of: a 4 between 11 and 37 b 9 between 8 and 46 c 7 that are less than 78

2

a List the multiples of 2 less than 30. c Write the common multiples of 2 and 6.

3

a Write the factors of 48. c What is the HCF of 48 and 27?

4

Use a factor tree to write the following numbers as a product of prime factors. a 54 b 84 c 144

5

Use the method of division by primes to write the following numbers as a product of prime factors. a 80 b 240 c 600

6

a Write 35 and 60 as a product of prime factors. b Find the HCF of 35 and 60. c Find the LCM of 35 and 60.

7

a Write 140 and 84 as a product of prime factors. b Find the HCF of 140 and 84. c Find the LCM of 140 and 84.

8

Find the HCF and LCM of the following pairs of numbers. Write the numbers as a product of prime factors first. a 84 and 56 b 100 and 75 c 175 and 200

9

Find: a

10

16

b

121

Write 3416 in: a Roman numerals

c

169

b List the multiples of 6 less than 30. d What is the LCM of 2 and 6? b Write the factors of 27.

d

3

125

e

3

216

b base 2

11

Which of the numbers 2, 3, 4, 5, 6, 8, 9, 10, 11, 12 are factors of: a 2628 b 9603 c 233 244?

12

Find 374 × 12.

f

3

1331

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Review of Stage 4 (Chapter 1)

B. REVIEW OF NUMBER (NS4.2) Exercise 1B 1

Plot the following numbers on the number line: a 1, 3, 5, 7 b 8, 1, 6, 2

2

a b c d e

3

Represent the following operations on a number line and hence find the answer. a 2+3 b 2+4−3 c 6×2−7

4

Write the opposite of each of the following statements. a a withdrawal of $15 b an increase of 30 cm in length c a gain of 8 kg in weight d a rise of 15°C in temperature

5

State the combined effect of each of the following. a a deposit of $8 followed by a withdrawal of $7 b a deposit of $8 followed by a withdrawal of $3 c a deposit of $8 followed by a withdrawal of $14

6

If east is the positive direction, write directed numbers for a journey: a 1 km east b 2 km west c 2 km east then 7 km west

7

Plot the following whole numbers on a number line, then write them in ascending order. a −12, −5, −4, −10, −2 b 0, 3, −3, −4, 5

8

Plot the following numbers on a number line then write them in ascending order. a 2--3- , 1 1--3- , −2 2--3- , − 2--3- , 0 b −1.8, −1, 1.8, 2.3, 0

9

Using a number line to help you, insert > or < symbols to make the following statements true. a −3 −4 b −5 5 c −5 0

10

11

12

13

Plot the numbers 2 and 6 on a number line. Write a statement using < to describe the numbers 2 and 6. Write a statement using > to describe the numbers 2 and 6. Write two whole numbers between 2 and 6. Write three other numbers between 2 and 6.

Simplify: a −8 − 5

b −7 − (−4)

c 3 − (−2)

Simplify: a 5 × (−5) × 2

b (−3) × (−5) × 2

c (−2) × (−3) × (−4)

Simplify: a 14 ÷ −2

b −55 ÷ (−5)

c (−72) ÷ (−12)

Simplify: a 19 − 4 × 2 d −54 ÷ 9 × (−3)

b −17 + 5 × 3 e (−7) × 6 ÷ (−2)

c 19 + 2 − 8 f 45 ÷ (−9) × (−2)

3

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4

C. REVIEW OF NUMBER (NS4.3) Exercise 1C 1

Copy the diagram opposite into your 7 -. exercise book and shade ----10

2

In a class of 20 students 1--4- play soccer, 1--5- play netball and the remainder play football. What fraction of the class plays football?

3

Convert

4

Convert 3 5--8- to an improper fraction.

5

155 31 Complete ---------- = -----20

6

Simplify

7

Arrange in descending order: 4--5- ,

8

State the reciprocal of 2 2--3- .

9

Calculate

146 ---------11

to a mixed number.

175 ---------240

3 --8

8 -----15

,

2 --3

of 592 kg.

10

Liam earns $600 per week. He banks 1--5- , spends 2--3- on rent and food, and uses the remaining money for personal use. a How much does Liam bank each week? b How much is spent weekly on rent and food? c What fraction of Liam’s weekly wage is for personal use? d How much is spent on personal use?

11

State the value of 2 in 4.0203.

12

Express 8 +

13

Write 4.2 as a fraction.

14

Write 3 3--8- as a decimal.

15

Express

16

Express 3.85444 correct to the nearest hundredth.

1 --6

3 -----10

+

7 ------------1000

as a decimal.

as a decimal correct to 2 decimal places.

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17

Express 3.5217 correct to the nearest whole number.

18

Simplify:

a 12.6 − 11.8 + 3.84

19

Simplify:

a (2.1 + 3) × (11.9 − 5.9)

20

a Ahmed earns $4.60 per hour. How much does he earn if he works 10 1--2- hours?

b 15.5 ÷ 0.05 + 22.4

c 16.2 ÷ 2 + 5.7 − 1.9

b (10.3 − 8.7) + (0.4 × 9)

b Sylvanna won $1 216 320 in a lottery. She decided to share it equally between eight people. How much did each person receive? 21

Shade 75% of this diagram.

22

Write 48 out of 100 as a percentage.

23

Convert 37% to a fraction.

24

Convert 57% to a decimal.

25

Express 3.8 as a percentage.

26

Express

27

Convert each to percentages and arrange in ascending order: 4--5- , 70%, 0.65

28

Convert

29

Convert 15% to a simplified fraction.

30

Convert 425% to a decimal.

31

Calculate 15% of $360.

32

Find 25% of 48 m.

33

Express 13 kg as a percentage of 52 kg.

34

Increase 100 by 30%.

35

Decrease 320 by 25%.

36

What is a rate?

37

In a class of 28 students there are 13 boys. Write the ratio of boys to girls.

38

Express as a ratio in simplest form: a 12 : 40

39

Find the value of x if 4 : 7 = x : 42.

40

Divide $1000 in the ratio 3 : 5.

41

Divide 46.8 m into the ratio 2 : 3 : 1.

5 --8

as a percentage.

27 ---------100

to a percentage.

b 30 : 108

5

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6

42

If Susan speaks to Clara for 18 m and 53 s, calculate the cost of the phone call if it is 49c/min or part thereof.

43

A car travels at a speed of 97 km/h for 3 1--4- hours. Calculate the distance travelled.

44

A map has a scale of 1 : 300. Convert the following scale distances to real distances. a 11 cm b 15.8 cm

45

The ratio of the size of a model to the size of the real building is 1 : 500. If the model has a height of 25 mm, find the actual height of the building in metres.

25 mm

D. REVIEW OF NUMBER (NS4.4) Exercise 1D 1

A hat contains 1 red, 1 blue, 1 yellow and 1 green ticket. One ticket is chosen. a List the sample space. b What is the probability of selecting the red ticket?

2

Ten cards with the numbers 1 to 10 written on them are shuffled and one card is chosen. a List the sample space. b What is the probability that the selected card has 7 written on it?

3

Copy and complete. Fraction a

Percentage

0.7

b c

Decimal

25% 5 --8

4

A bag contains 4 green, 9 red and 7 blue marbles. One is chosen at random. a How many marbles are in the bag? b How many marbles are red? c What is the probability of selecting a red marble?

5

One card is selected from a normal deck of 52 cards. What is: a P(a diamond) b P(a red card) c P(a king)?

6

a Write a statement describing a probability of 0. b Estimate a percentage probability for the phrase ‘even chance’. c Write a phrase for an event with a probability of success of about 85%.

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7

A die with the numbers 1–6 is rolled once. Describe events that would be: a certain b impossible c of even chance

8

A spinner has five equal-sized sectors coloured green, yellow, orange, brown and white. It is spun once. a What is the probability of white? b What is the probability of any colour except white? c What is the probability of yellow or orange? d What is the probability of any colour except yellow or orange?

E. REVIEW OF ALGEBRA (PAS4.1) Exercise 1E 1

2

3

4

5

6

7

If there are p marbles in each cup, write algebraic expressions for the total number of marbles in each of the following diagrams. a

b

c

d

Simplify: a 6×p e 3×m×m

b g×r f 5×a+3×q

Insert multiplication signs to show the meaning of: a 3p b ab c m2 Simplify: a p+p+p e 0 × 5p

b y+y+y+y+y

c m×5

d 5x 2

c z ×1

d 8×a×b

e 6pq

d 3pq × 1

If m = 3 and n = 4, evaluate: a mn b 5mn

c 7m − 3n

d n2

e 4n 2

Write in fraction form: a t÷2 b g÷r

c r÷g

d 4w ÷ 7

e 3 ÷ 2x

Show the meaning of the following expressions by inserting a division sign. k 4 p 3e a --b ---c --d -----3 m q 4

mn e -------t

7

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8 8

9

10

If p = 4 and q = 5, evaluate: q 24 a --b -----5 p

c

5p -----q

4q d -----p

If p = 7 and q = 3, evaluate: a 3(p + 1) b 4(q − 3) d 5(q − 4) e pq(p − 5) If p = 12 and q = 5, evaluate: p+9 q–3 a -----------b -----------3 2

5p e -----2q

c q(q + 1)

c

26 -----------p+1

p+6 d -----------q+1

3p + 3 e ---------------q+8

F. REVIEW OF ALGEBRA (PAS4.2) Exercise 1F 1

Write down the next three numbers in the sequence: a 2, 7, 12, 17, ___ b 21, 19, 17, 15, ___

2

Using the rules given, find the first four terms of each sequence. a Start with 3 and multiply by 2. b Start with 3, multiply by 2 and then subtract 1.

3

a Complete the table for the above pattern of shapes. Number of pentagons

1

2

3

4

5

Number of matches needed b i

Plot the values in your table as points on a number plane like the one given opposite. ii Describe the geometric pattern formed by the points on the graph. c Write in words a rule to describe the pattern formed. Begin with ‘The number of matches needed is _____.’ d If x = the number of pentagons and y = the number of matches needed, write an algebraic statement for the rule in part c. e How many matches would be needed to build 200 pentagons? 4

1 2 3 4

a Draw the next two shapes in this matchstick pattern.

b Write a sequence for the number of matches needed to make each shape. c How many matches are needed for the 100th shape?

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5

The algebraic rule for a pattern of shapes is p = 3 + 2k where k represents the number of shapes and p represents the number of matches needed to build the shapes. Find the number of matches needed to build: a 5 b 10 c 50 shapes

6

The terms of a number sequence are given by the rule y = 2z + 3 where z represents the position number of the term and y represents the term. Find the first five terms of the sequence.

G. REVIEW OF ALGEBRA (PAS4.3) Exercise 1G 1

2

3

4

5

6

7

8

9

10

11

Simplify: a 9x + 5x

b 7y − y

c 3a 2 + 4a 2

d 9ac − 3ca

Simplify: a 5 × 12n

b −5 × 2a

c 8m × 3

d −5p × −7

Simplify: a 10a ÷ 2

b 12m ÷ −3

c

Simplify: a 4wx + 2y − 5xw − 5y Simplify: 4a a a ------ + --7 7

12m d ----------3

b 6m + 2m − 8m

a a b --- – --3 5

c

Expand each of the following. a a(a − n) b mn(2n − 5) Expand and simplify. a 3(5a + 3) − 4(8 − 4a) Factorise: a mn2 + mn

abc --------a

q --- × q 5

c 4p(3p + 2)

d 6p ÷ 2p d −2p(4y − 2w)

b 3x(y − 4) + 4y (5x − 2) b pq − aq

c 4p − 12d

d 25f − 15

Factorise each by taking out a negative factor. a −3K + 9 b −4p − 12d Define: a pronumeral

b coefficient

If Q = 7 and p = −4, evaluate the following expressions. Qp a 4Q + p b ------c 6p − 5Q 8

d 3(Q − p) + 7p − 8Q

9

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10

12

Complete the value tables. a

m

1

2

3

b L = 3N + 5

4

m+3

N

1

2

3

4

5

L 13

Write an algebraic expression for each: a the product of six and d plus twenty-three b the difference between x and seven multiplied by three. The result is then divided by eight.

H. REVIEW OF ALGEBRA (PAS4.4) Exercise 1H 1

Show each step required to get from the expression 4x + 12 back to x.

2

Solve these questions: a x + 11 = 17 c 4x = 36 e 3y + 18 = 29 g 4d + 8 = 3d − 12 i 3(m + 6) = 2(m − 1)

b d f h j

Solve these equations: 4p a ------ = 6 5

3x + 12 b ------------------- = 12 7

3

x + 9 = −6 −9x = 63 5 − 4p = −47 18 + 7c = 32 − 3c 8(q − 5) = −3(10 + 3q)

4

Is the given value for the pronumeral a solution to the equation? x a 5d + 12 = 28; d = 3 b --- + 7 = 24; x = 3 2--55

5

Write an equation and solve this problem. The sum of a certain number and 23 is 114. What is the number?

6

Solve the following inequalities. a x + 9 ≥ −3

m b ---- < 4 7

I. REVIEW OF ALGEBRA (PAS4.5) Exercise 1I 1

Plot these points on a number plane: A(0, –3), B(–2, –3), C(3, –4), D(–3, 2), E(2, 5)

2

a Plot the points A(–3, 6), B(3, 6) and C(3, 0). b IF ABCD is a square, find the coordinates of the point D.

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3

Use this pattern of matches.

a Copy and complete this table.

Number of patterns 1

2

3

4

5

Number of matches b Write a rule describing the number of matches required to make each pattern. c Using x to represent the number of patterns and y to represent the number of matches, write a set of points describing this information. d Graph these points on the number plane. e Mark in the next two points and write their coordinates. 4

Bulk washing powder is sold for $4.50 per kilogram. The following table shows weight versus cost for various quantities of washing powder. Number of kg

0

1

2

4

5

10

20

Cost ($)

0

4.5

9

18 22.5 45

90

a Using x to represent the number of kilograms and y to represent the cost in dollars, write a set of points describing this information. b Graph these points on the number plane and draw a straight line through them. c Use the graph to find how much 8.5 kg of washing powder would cost. d Use the graph to find how much washing powder could be purchased for $72. 5

Copy and complete the table and draw the graph of y = 2x − 3. Some of the points are provided.

x –2 –1 0 y

6

This graph shows a straight line. a Use the graph to complete this table of values. x –2 –1 0

1

y b Write the rule describing this straight line. The rule is of the form y = ❒x ± ∆.

–5

1

2 1

11

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12

J. REVIEW OF DATA (DS4.1) Exercise 1J 1

This graph shows a student’s pulse rate during exercise. a What is the pulse rate after 3 minutes? b When is the pulse rate 110? c How many times was the pulse rate 90?

2

The sector graph shows favourite activities. a What was the second most popular activity? b What is the fraction of people cycling? c What is the angle at the centre for walking?

3

The step graph shows postage charges. a What is the cost of posting a package weighing 140 g? b What is the heaviest package that can be posted for $2?

4

The travel graph shows Grace’s cycling journey. a When does she rest? b How far does she travel?

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5

Draw a divided bar graph of length 15 cm to represent the results from a survey of 30 year 9 students. Favourite snack

Number of students

Hot chips

15

Hamburger

5

Kebab

1

Chicken

5

Pie

4

6

Draw a travel graph to describe this journey. Jack started from home at 7.00 a.m. and travelled 25 km in one hour. He stopped for 45 min, then continued for another 2 h travelling another 80 km. Jack stopped for 30 min then returned home taking 2 1--2- h.

7

Classify these as nominal or numerical variables. a hair colour b height

8

9

c distance travelled

a Construct a frequency distribution table for this information. Winning margins in a series of soccer matches: 2 2 3 0 0 2 0 4 3 1 0 3 3 3 1 1 2 2 0 0 0 1 2 1 0 4 5 3 0 1 5 1 4 3 4 2 3 0 2 3 1 2 2 0 0 2 3 2 0 4 b Draw a frequency histogram and polygon for this information. c i How many soccer matches were played? ii How many winning margins of 2 were there? iii What does a winning margin of 0 mean?

3 3 0 4 0

5 1 2 0 2

a Draw a frequency distribution table to show the following information: 8 9 5 10 7 6 6 7 5 1 8 6 2 8 6 2 6 4 9 4 7 4 9 2 6 5 9 10 10 1 7 6 2 2 3 1 b How many numbers were 8 or more?

10

Draw a stem-and-leaf plot for the following information. Use stems of 12, 13, 14, 15, 16. 146 145 128 138 161 150 149 142 132 142 150 145 151 131 148 145 132 145 144 153

11

a Draw a column graph for the data in this table. b Draw a dot plot showing this information.

Drink

Frequency

Soft drink

18

Still water

10

Juice

3

Tea/coffee

2

Other

2

13

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Review of Stage 4 (Chapter 1)

14

12

The price of CML shares over a fortnight varied as shown. Day Price ($)

M

T

W

T

F

M

T

W

T

F

7.40

7.20

7.08

7.00

7.35

7.55

7.80

7.25

7.40

7.30

Draw a line graph to show this variation in price.

K. REVIEW OF DATA (DS4.2) Exercise 1K 1

Define the statistical term sample.

2

Would a census or sample be used to investigate the number of people who use a particular brand of toothpaste? Why?

3

Describe the sample you would use if you wanted to gather support for improved skateboard facilities at your local park.

4

For the scores 11, 14, 15, 19, 19, 21, find the: a mean b mode

5

6

For the scores in this stem-and-leaf plot find the: a mean b mode c median d range

c median

d range

Stem

Leaf

2

788

3

001234566

4

1244468

5

3578

6

23

The back-to-back stem-and-leaf plot compares the marks gained by two classes, A and B, in their half yearly Mathematics exam. Class B

Class A

Leaf

Stem

Leaf

21

2

88

6421

3

0356

65310

4

02668

110

5

369

7

6

7

a Find the mean, mode, median and range for each class. b Which class performed better? Explain.

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7

From a school of 800 students, a random sample of 50 students was selected. There were 13 left-handed students in the sample. a What fraction of the sample was left-handed? b Estimate how many of the school population are left-handed.

L. REVIEW OF MEASUREMENT (MS4.1) Exercise 1L 1

Estimate the width of your classroom.

2

Convert to millimetres:

a 0.27 m

b 0.004 km

3

Convert to centimetres:

a 0.34 m

b 0.07 km

4

Calculate the perimeter of a rectangle given a width of 11.9 cm and a length of 26.3 cm.

5

Calculate the perimeter of each shape. All measurements are in centimetres. a

b

6

An octagon has a perimeter of 1012.16 cm. Calculate the length of each side.

7

Find the length of each side marked with a pronumeral, then calculate the perimeter. All measurements are in centimetres. 19.8

8

By counting squares, find the area of the figure.

9

Find the areas of the following rectangles, squares and triangles. a

b

25.3

c

d

15

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16

Review of Stage 4 (Chapter 1)

10

Find the areas of the following composite figures. a

11

b

c

Find the shaded area in these figures. a

b

12

Copy and complete the following conversions. a 5 cm = _____ mm b 800 cm = _____ m c 640 mm = _____ cm d 11.6 m = _____ cm e 43.8 cm = _____ mm f 8400 cm = _____ m g 8 cm2 = _____ mm2 h 7.2 m2 = _____ cm2 i 9000 mm2 = _____ cm2

13

In these triangles

a

b

R

a

i Which side is the hypotenuse? ii Write an expression for Pythagoras’ rule for the triangle. Show whether the following triangles are right angled. a

15

a Find the value of 92. b Calculate the value of

Q

b

c

14

P

b

70 correct to 1 decimal place.

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16

17

Find the length of the hypotenuse correct to 1 decimal place. a

b

Find the length of the third side. a

b 44.2 cm

9.4 cm x cm

21.3 cm

7.2 cm

18

x cm

Find the value of the pronumeral in each of the following, correct to 1 decimal place. a

b 8 cm

12 cm 29.3 cm

10.8 cm

19

Calculate the length of the diagonal of a square with side length 36 cm, correct to 2 decimal places.

20

Name the feature of each circle shown. a

b

21

What fraction of a circle is represented by this sector?

22

Write down the formula for the circumference of a circle when given the diameter.

23

Calculate the circumference of a circle with a diameter of 11.4 cm, correct to 1 decimal place.

24

Write down the formula for the circumference of a circle when given the radius.

25

Calculate the circumference of a circle with a radius of 6.8 cm, correct to 2 decimal places.

108o

17

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18

26

Write down the formula for calculating the area of a circle.

27

Calculate the area of a circle, correct to 1 decimal place given: a radius = 7 cm b diameter = 3.9 cm Calculate the area of this shape correct to 1 decimal place.

28

15.7 cm

M. REVIEW OF MEASUREMENT (MS4.2) Exercise 1M 1

a If ABFE is the top face of the rectangular prism, name the bottom face. b Name the front and back faces. c Name the two sides.

2

a Construct a net of the cube shown.

3.8 cm

b Use the net to calculate the total surface area of the cube. 3

Calculate the surface area of each prism. a

b

15.2 cm

6.8 cm

4

5 cm

3 cm

Construct prisms from the following cross-sections. a

b

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5

Draw the cross-section of each prism if it is cut along the dotted line shown. a

6

b

Calculate the volume of the following solids. a

b

A= 83.4 cm2 41.6 cm

7.4 cm

7

Calculate, to the nearest cm3, the volume of the cylinder shown. 38 cm

15.3 cm

8

Calculate the volume of each solid. a

b 10.3 cm

38.7 cm

7.8 cm 5.8 cm 9.4 cm

23.5 cm

9

Complete the following conversions. a 1 cm3 = ____ mm3 b 1 kL = ____ mL 3 d 1 m = ____ L e 1 m3 = ____ kL

c 1 kL = ____ cm3

N. REVIEW OF MEASUREMENT (MS4.3) Exercise 1N 1

How many hours in two days?

2

Complete the following conversions. a 240 s = ____ min

b 300 min = ____ h

19

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Review of Stage 4 (Chapter 1)

20 3

Convert 210 min to hours and minutes.

4

Calculate:

5

If Sergio caught the bus at 6.35 a.m., what time did he arrive at work if the bus trip took 42 min?

6

High tide is at 5.20 a.m. and low tide is at 9.08 a.m. Calculate the time difference between high and low tide.

7

Convert 2 1--3- h to hours and minutes.

8

Round the following calculator display to the nearest minute: 3° 16° 41.3

a 3 h 35 min + 5 h 48 min

b 3 h 21 min − 1 h 42 min

O. REVIEW OF SPACE AND GEOMETRY (SGS4.1) Exercise 1O 1

Name the common solids that have been combined to make each of the following solids. a

2

b

Sketch the view of each of the following solids from the: a

3

d

i

front

b

Sketch the cross-section when the shapes below are sliced as shown. a

4

c

Draw a possible net for this solid.

b

ii top

iii side

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Review of Stage 4 (Chapter 1)

5

A solid was built from cubes. From the views given below, build the solid and sketch it on isometric grid paper.

6

Name the edges which: a intersect at E b are parallel to AB c are skew to BE

P. REVIEW OF SPACE AND GEOMETRY (SGS4.2) Exercise 1P 1

Name the angles marked with a ★ and a ■ in each of the following. a

C

B

A

b

G F

E

D

E

F

C

D

A

B

2

Draw an interval PQ, 4 cm long. Using P as the vertex and your protractor, draw angles of the following sizes. a 55° b 196° c 228° d 315°

3

Use the diagram shown to classify these angles according to size. a ∠BAC b ∠BCA c ∠BAD d ∠ACB

4

a Describe adjacent complementary angles. b Draw a diagram showing adjacent complementary angles.

5

Find the angles supplementary to: a 135° b 60°

c 72°

d 133°

21

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Review of Stage 4 (Chapter 1)

22 6

Explain why ∠QRS is equal to 125°.

125°

7

From the diagram in question 6, calculate the size of: and give reasons.

8

Calculate the value of the pronumeral and state a reason. a

∠QRU

ii ∠SRT

b

73°

115°

9

i

Find the value of the pronumeral, giving a reason. a

b 134° 50°

34°

125°

10

Are the lines AB and CD parallel? Provide a reason. a

b 75°

128°

52°

Q. REVIEW OF SPACE AND GEOMETRY (SGS4.3) Exercise 1Q 1

Construct a triangle with side lengths 13 cm, 12 cm and 5 cm.

2

Draw an right-angled isosceles triangle.

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Review of Stage 4 (Chapter 1)

3

Find x, giving a reason. a

b 68°

4

142°

Write an equation and solve it to determine the value of each angle.

120° 40°

5

Find the value of the pronumerals, giving reasons for your answers, in alphabetical order (u to z). 43° 52°

6

a Draw a rhombus. b List the properties of a rhombus.

7

Draw a line interval 8 cm long and bisect it using a construction.

8

a Draw a diagram of a convex quadrilateral. b Comment on the diagonals of your diagram.

R. REVIEW OF SPACE AND GEOMETRY (SGS4.4) Exercise 1R 1

Show how these shapes can be divided into two congruent figures. Name the resulting figures. a

2

Divide the rectangle into two congruent figures in two ways.

b

23

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Review of Stage 4 (Chapter 1)

24 3

Each pair of figures is similar. Find the scale factor and the value of the pronumeral. a

b 9 cm

5 cm

24 cm 40 cm

4

In this pair of similar triangles, find the value of the pronumeral. y cm 2.5 cm

7.5 cm 8 cm

5

In ∆STU, which side corresponds to a MN b NL c ML ?

N 14 7

M L

6

The shadow cast by a flagpole is 9 m long. At the same time the shadow of a 30 cm ruler is 40 cm. Calculate the height of the flagpole.

7

Draw a scale drawing of this floor plan of a building with the given dimensions. Use a scale factor of 200.

3m

10 m

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Chapter 2 Indices and Scientific Notation This chapter deals with indices and scientific notation. After completing this chapter you should be able to: ✓ change numbers to index form and vice versa ✓ use the terms base, power, index, exponent ✓ use the index laws to simplify expressions ✓ define and use zero and negative integral indices ✓ define fractional indices for square and cube roots ✓ express very large and very small numbers in scientific notation and vice versa ✓ perform calculations with, and order, numbers expressed in scientific notation.

Syllabus reference NS5.1.1 WM : 5. 1. 1–5. 1. 4, 5. 2 .1 –5 .2 .4

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Diagnostic test

1

When written in index form 5 × 5 × 5 × 5 × 5 × 5 is: A 65

B 56 6

C 555555 2

3

A 4×4

B 4×4×4×4

C 2×2

D 2×2×2×2

9

10

C 625

D 1024

B 214

C 245

D 445

13

When simplified ( 73)5 = B 78

14

C 735

D 735

315 ÷ 35 = B 110

C 33

D 310

15

230 = A 23

8

B 54

25 × 29 =

A 13 7

Which of the following numbers is written in scientific notation? A 49 × 1015

B 3.7 × 1 000 000

C 360 000

D 2.1 × 10–9

When written in scientific notation 6 430 000 becomes: A 6.43 × 104

B 6.43 × 105

C 6.43 × 106

D 6.436

When evaluated using a calculator 4 is:

A 715 6

12

5

A 414 5

D 555555

When written in expanded form 24 is:

A 45 4

11

B 1

C 0

D 2

7–5 is the same as: 1 1 A -----5 B -----7 C –35 5 7

16 D –75

A 7

B 7

7

C 3

B 0.000 032

C 3 200 000

D 320 000

(4.2 × 108) × ( 8 × 1011) = A 3.36 × 1088

B 3.36 × 1089

C 3.36 × 1019

D 3.36 × 1020

(3.9 × 10–12) ÷ ( 5.2 × 106) = A 7.5 × 10–19

B 7.5 × 10–6

C 7.5 × 106

D 7.5 × 1018

When the numbers 3.8 × 10–6, 4.7 × 10–8, 8.9 × 10–8 are written in order from smallest to largest, the answer is: B 4.7 × 10–8, 8.9 × 10–8, 3.8 × 10–6

The meaning of 6 is: 1 1 A 3 B -----2 C 6 D --3 6 3 When written in index form 7 = 1 --3

A 0.000 003 2

A 3.8 × 10–6, 8.9 × 10–8, 4.7 × 10–8

1 --2

3

When written as an ordinary number 3.2 × 10–5 is:

D 3

C 3.8 × 10–6, 4.7 × 10–8, 8.9 × 10–8 D 8.9 × 10–8, 3.8 × 10–6, 4.7 × 10–8 1 --7

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–3

4

5

6

7

8

9, 10

11–13

14–16

A

B

C

D

E

F

G

H

I

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

A. INDEX NOTATION Index notation is a short way of writing the repeated product of numbers, for example 6 × 6 × 6 × 6 × 6 may be written 65. This is read as ‘6 to the power 5’ or ‘6 to the fifth (power)’. The 6 is called the base. It is the number that is being repeated. The 5 is called the power, index or exponent. It tells us how many times the base has been repeated.

Example 1 a Write 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 in index form. b Write the answer to part a in words. c State which number is the i base ii index.

‘In index form’ means ‘using index notation’.

a The number being repeated is 7 and the number of times the 7 is repeated is 9, so 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 = 79. b 7 to the power 9. c i The base is 7. ii The index is 9.

Exercise 2A 1

a Write 4 × 4 × 4 × 4 × 4 in index form. b Write the answer to part a in words. c State which number is the i base

Exponent is another word for index.

ii index.

2

a Write 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 in index form. b Write the answer to part a in words. c State which number is the i base ii index.

3

Write the following products using index notation. a 3×3×3×3×3×3×3 b 2×2×2×2×2×2×2×2×2×2 c 4×4×4 d 5×5×5×5×5×5 e 2×2×2×2

27

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Example 2 For each of the following numbers: i write down the base, i.e. the number that is being repeated ii write down the index, i.e. the number of times the base is repeated iii write the number in expanded form. a 54 b 27 ‘In expanded a i ii iii b i ii iii

4

form’ means as a repeated product.

The base is 5. This is the number being repeated. The index is 4. This is the number of times the base is repeated. Hence 54 = 5 × 5 × 5 × 5 The base is 2. This is the number being repeated. The power is 7. This is the number of times the 2 is repeated. Hence 27 = 2 × 2 × 2 × 2 × 2 × 2 × 2

For each of the following numbers: i write down the base, i.e. the number that is being repeated ii write down the index, i.e. the number of times the base is repeated iii write the number in expanded form. a 35 b 24 c 73 d 96

e 58

Example 3 Use your calculator to evaluate 56. Find the xy key on your calculator. Enter the base, 5. Press the xy key. Enter the power, 6. Press the

= key.

6

5 = 15 625

5

Use your calculator to evaluate: a 210 b 74 4 f 9 g 63

c 35 h 36

d 106 i 85

e 54 j 58

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

B. MULTIPLYING NUMBERS WITH THE SAME BASE Example 1 a Write in expanded form i 32 b Write the answer to part iii in index form. c Is 32 × 34 = 32 + 4 ?

ii 34

a i 3×3 ii 3 × 3 × 3 × 3 6 b 3 because the 3 is repeated 6 times. c Yes, because 2 + 4 = 6.

iii

32 × 34

3×3×3×3×3×3

iii

Exercise 2B 1

a Write in expanded form i 73 b Write the answer to part iii in index form. c Is 73 × 74 = 73 + 4 ?

ii 74

iii 73 × 74

2

a Write in expanded form i 54 b Write the answer to part iii in index form. c Is 54 × 52 = 54 + 2 ?

ii 52

iii 54 × 52

3

a Write in expanded form i 63 b Write the answer to part iii in index form. c Is 63 × 65 = 63 + 5 ?

ii 65

iii 63 × 65

Example 2 Write in index form: a 57 × 54

b 38 × 310

From the answers to part c in the above questions, b 38 × 310 = 38 + 10 = 318 a 57 × 54 = 57 + 4 = 511

When multiplying numbers with the same base, add the indices. 4

Simplify by writing in index form. a 35 × 34 b 27 × 25 f 69 × 64 g 105 × 104

c 72 × 78 h 210 × 210

d 57 × 52 i 520 × 510

e 410 × 46 j 311 × 37

29

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Example 3 a Write in expanded form b Is 56 × 5 = 56 or 57 ?

ii 56 × 5

56

i

a i 56 = 5 × 5 × 5 × 5 × 5 × 5 ii 56 × 5 = 5 × 5 × 5 × 5 × 5 × 5 × 5 b From part ii, the 5 is repeated 7 times. Hence 56 × 5 = 57, i.e. 5 may be written as 51. Hence 56 × 5 = 56 × 51 = 56 + 1 = 57 5

Simplify by writing in index form. a 54 × 5 b 37 × 3

c 29 × 2

d 5 × 58

e 7 × 711

Example 4 a Write in expanded form i 23 × 34 3 4 3+4 b Is 2 × 3 = 6 ? Give a reason.

ii 67

a i 2×2×2×3×3×3×3 ii 6 × 6 × 6 × 6 × 6 × 6 × 6 b No, because the bases are different. 6

a Write in expanded form b Is 35 × 42 = 125 + 2 ? Give a reason.

i

35 × 42

ii 127

7

a Write in expanded form b Is 53 × 25 = 103 + 5 ? Give a reason.

i

53 × 25

ii 108

8

a Write in expanded form b Is 35 × 32 = 95 + 2 ? Give a reason.

i

35 × 32

ii 97

9

a Write in expanded form b Is 23 × 25 = 43 + 5 ? Give a reason.

i

23 × 25

ii 48

10

Write true or false. a 37 × 35 = 312 d 25 × 24 = 29 g 46 × 47 = 442

b 37 × 25 = 612 e 25 × 54 = 109 h 46 × 37 = 1213

c 37 × 35 = 912 f 25 × 24 = 49 i 46 × 47 = 1613

Example 5 Simplify by writing in index form.

11

a 23 × 25 × 24

b 35 × 36 × 33

a 23 × 25 × 24 = 23 + 5 + 4 = 212

b 35 × 36 × 33 = 35 + 6 + 3 = 314

Simplify by writing in index form. a 24 × 26 × 23 b 38 × 33 × 37

c 53 × 57 × 54

d 46 × 43 × 4

e 53 × 54 × 5 × 52

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

C. RAISING A NUMBER TO A POWER Example 1 Simplify by writing in index form a

(32)4

b (75)3

c (24)7

a

(32)4 = (3 × 3)4 or (32)4 = 32 × 32 × 32 × 32 = (3 × 3) × (3 × 3) × (3 × 3) × (3 × 3) = 32 + 2 + 2 + 2 =3×3×3×3×3×3×3×3 = 34 × 2 8 =3 = 38

b (75)3 = 75 × 75 × 75 = 75 + 5 + 5 = 73 × 5 = 715 c

(24)7 = 24 × 7 = 228

When raising a number to a higher power, multiply the indices.

Exercise 2C 1

Simplify by writing in index form. a

(32)3

b (53)2

c (23)4

d (35)3

f

(102)5

g (42)6

h (63)7

i

(38)3

e (74)5 j

(27)10

Example 2 Simplify: a (32)4 × 37 a

2

b (72)3 × (73)4

(32)4 × 37 = 38 × 37 = 315

b (72)3 × (73)4 = 76 × 712 = 718

Simplify: a

(32)3 × 35

e 76 × (72)5 i

(34)5 × (32)4

b (23)4 × 25 f

(23)2 × (24)3

j

(92)5 × (93)4

c (54)3 × 52

d 34 × (33)5

g (52)4 × (53)2

h (74)2 × (73)3

31

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

D. DIVIDING NUMBERS WITH THE SAME BASE Example 1 a Write in expanded form 36 ÷ 32. b Write the answer to part a in index form. c Is 36 ÷ 32 = 36 – 2 ? 6

3 a 36 ÷ 32 = -----2 3 1 1 3×3×3×3×3×3 = --------------------------------------------------31 × 31

b 34 c Yes because 6 – 2 = 4.

=3×3×3×3

Exercise 2D 1

2

3

a Write in expanded form 37 ÷ 32 b Write the answer to part a in index form.

c Is 37 ÷ 32 = 37 – 2 ?

a Write in expanded form 75 ÷ 73 b Write the answer to part a in index form.

c Is 75 ÷ 73 = 75 – 3 ?

a Write in expanded form 28 ÷ 23 b Write the answer to part a in index form.

c Is 28 ÷ 23 = 28 – 3 ?

Example 2 Simplify, writing your answer in index form. a 210 ÷ 27

b 38 ÷ 37

From the answers to part c in the questions above a 210 ÷ 27 = 210 – 7 b 38 ÷ 37 = 38 – 7 = 23 = 31 =3

When dividing numbers with the same base, subtract the indices. 4

Simplify, writing your answer in index form. a 35 ÷ 33 b 28 ÷ 25 c 510 ÷ 54 8 4 8 6 f 6 ÷6 g 2 ÷2 h 311 ÷ 39 k 35 ÷ 31 l 56 ÷ 51 m 27 ÷ 2

d 49 ÷ 45 i 47 ÷ 46 n 104 ÷ 10

e 1012 ÷ 107 j 54 ÷ 53 o 79 ÷ 7

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Example 3 Write true or false. a 65 ÷ 23 = 32

b 36 ÷ 32 = 14 3

3

3

6×6×6×6×6 a 65 ÷ 23 = ----------------------------------------21 × 21 × 21

We can only subtract the indices when the bases are the same.

=3×3×3×6×6 32 = 3 × 3 3 2 ∴ 65 ÷ 2 ≠ 3 The answer is false. 1

1

3 ×3 ×3×3×3×3 b 36 ÷ 32 = -------------------------------------------------------31 × 31 = 34 2

36 ÷ 3 ≠ 1

4

The answer is false. 5

Write true or false. a 58 ÷ 55 = 53 e 65 ÷ 32 = 23

b 67 ÷ 24 = 33 f 26 ÷ 22 = 14

c 59 ÷ 53 = 16 g 39 ÷ 33 = 33

d 410 ÷ 42 = 45 h 107 ÷ 53 = 24

Example 4 Simplify: a 35 × 37 ÷ 38

b (28)3 ÷ 210

a Working from left to right, 35 × 37 ÷ 38 = 35 + 7 ÷ 38 = 312 ÷ 38 = 312 – 8 = 34

c 57 ÷ 53 × 59

b (28)3 ÷ 210 = 28 × 3 ÷ 210 = 224 ÷ 210 = 224 – 10 = 214

c 57 ÷ 53 × 59 = 57 – 3 × 59 = 54 × 59 = 54 + 9 = 513 6

Simplify: a 36 × 34 ÷ 35 e (45)2 ÷ 47 i 210 ÷ 24 × 23 m 23 × 26 × 24 ÷ 27

b f j n

25 × 28 ÷ 26 (33)5 ÷ 37 512 × 52 ÷ 54 (52)3 × (54)2 ÷ 511

c g k o

710 × 78 ÷ 716 58 ÷ 53 × 54 (24)3 ÷ 29 320 ÷ 38 ÷ 37

d h l p

(56)3 ÷ 510 36 ÷ 33 × 34 310 ÷ 38 × 35 725 ÷ (73)5

33

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

E. THE ZERO INDEX Example 1 Complete the following table to find the value of 30. 35 243

34 81

33

32

31

30

27

As we read the first row of numbers, we can see that the powers of the 3 are going down by one. The missing numbers, in the second row, can be found by dividing the number before it by 3. 35

34

33

32

31

30

243

(243 ÷ 3 = )81

(81 ÷ 3 = )27

(27 ÷ 3 = )9

(9 ÷ 3 = )3

(3 ÷ 3 = )1

From the table, 30 = 1.

Exercise 2E 1

2

Complete the following table to find the value of 50. (Divide the second row of numbers by 5.) 55

54

53

3125

625

125

52

51

50

Complete the following table to find the value of 20. 25

24

23

32

16

8

22

21

20

Example 2 a Use the index laws to simplify 74 ÷ 74. b By writing in expanded form, show that 74 ÷ 74 = 1. c Hence show that 70 = 1. a

Using the index laws

74 ÷ 74 = 74 – 4 = 70

b

71× 71× 71× 71 74 ÷ 74 = -------------------------------71 × 71 × 71 × 71 =1

c From parts a and b, 70 = 1

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

3

a Use the index laws to simplify 73 ÷ 73. b By writing in expanded form, show that 73 ÷ 73 = 1. c Hence show that 70 = 1.

4

a Use the index laws to simplify 95 ÷ 95. b By writing in expanded form, show that 95 ÷ 95 = 1. c Hence show that 90 = 1.

5

Use the x y key on your calculator to find the value of: a 70 b 130 c 290 d 5.60

e 31.70

f

5 ( --- )0 8

e 12.90

f

3 ( --- )0 4

From questions 1 to 7, you should have discovered that: Any number raised to the power zero is equal to 1. Without using your calculator, write down the value of: a 40 b 230 c 9550 d 8.670

6

F. NEGATIVE INDICES Example 1 Complete the following table to find the meaning of 3–1, 3–2, 3–3. 35

34

33

243

81

27

32

31

30

3–1

3–2

3–3

The powers of the numbers in the first row are going down by 1. The numbers in the second row can be found by multiplying the number before it by 1--3- . 35

34

33

32

31

30

3–1

243 (243× --13- =)81 (81× --13- =)27 (27× --13- =)9 (9× --13- =)3 (3× --13- =)1 (1× --13- =) --13From the table, we see that: 1 1 3–1 = --- = -----1 3 3

1 1 3–2 = --- = -----2 9 3

1 1 3–3 = ------ = -----3 27 3

Multiplying a number 1 by --- is the same as 3 dividing it by 3.

3–2 ( --13- ×

1 --3

3–3 1 =) --19- ( --19- × --13- =) ----27

35

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Exercise 2F 1

2

1 --2

By multiplying the numbers in the second row by meaning of 2–1, 2–2, 2–3. 25

24

23

32

16

8

22

21

20

2–1

2–2

By multiplying the numbers in the second row by meaning of 10–1, 10–2, 10–3. 105

104

103

100 000

10 000

1000

102

101

1 -----10

complete the following table to find the

2–3

complete the following table to find the 100

Example 2 a Use the index laws to simplify 34 ÷ 36. 1 b By writing in expanded form, show that 34 ÷ 36 = -----2 . 3 1 c Hence show that 3–2 = -----2 . 3 a 34 ÷ 36 = 34–6 a 34 ÷ 36 = 3–2 31 × 31 × 31 × 31 b 34 ÷ 36 = ------------------------------------------------------------31 × 31 × 31 × 31 × 3 × 3 1 = -----------3×3 1 a 34 ÷ 36 = -----2 3 1 c From parts a and b, 3–2 = -----2 3 3

a Use the index laws to simplify 22 ÷ 25. 1 b By writing in expanded form, show that 22 ÷ 25 = -----3 . 2 1 c Hence show that 2–3 = -----3 . 2

4

a Use the index laws to simplify 53 ÷ 57. 1 b By writing in expanded form, show that 53 ÷ 57 = -----4 . 5 1 c Hence show that 5–4 = -----4 . 5

10–1

10–2

10–3

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Example 3 Write down the meaning of:

5

a 9–1

b 6–3

c 7–5

From questions 1 to 4, 1 a 9–1 = -----1 9

1 b 6–3 = -----3 6

1 c 7–5 = -----5 7

Write down the meaning of: a 3–1 b 4–3 –1 f 12 g 9–2 k 2–8 l 5–1

c 2–5 h 6–1 m 10–5

d 8–2 i 7–3 n 5–10

e 5–4 j 3–6 o 4–15

Example 4 Write as a simplified fraction:

6

a 5–2

b 3–5

1 a 5–2 = -----2 5 1 = -----25

1 b 3–5 = -----5 3 1 = ---------- (using a calculator) 243

Write as a simplified fraction. a 3–2 b 2–5 –3 f 6 g 9–2 k 7–3 l 4–4

c 4–3 h 3–4 m 3–6

d 5–4 i 5–5 –1 2 n  ---  5

Example 5 Write using a negative index: 1 a --b 3 1 a -----1 = 3–1 3

1 -----2 3

c

1 -----8 3

1 b -----2 = 3–2 3

c

1 -----8 = 3–8 3

e 2–10 j 2–9 –1 3 o  1 ---  4

37

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

7

Write using a negative index: 1 1 a --b -----2 2 2 f

1 --5

k

1 ------10 3

1 -----8 2 1 h -----3 4 1 m -----5 7 c

1 g -----2 7 1 l --6

1 d -----5 2 1 i -----4 3 1 n -----9 4

1 e -----3 2 1 j -----6 5 1 o -----10

G. FRACTIONAL INDICES Example 1 1 --2 2

a Use the index laws to simplify ( 5 ) . 1 b Find ( 5 )2. --2 c Hence write down the meaning of 5 . a

1 --2

( 5 )2 = 5

1 --- × 2 2

1 --2

b ( 5 )2 = 5

c Since ( 5 )2 = ( 5 )2 1 --2

= 51

then 5 =

5

=5

Exercise 2G 1

1 --2

a Use the index laws to simplify ( 3 )2.

b Find ( 3 )2.

1 --2

c Hence write down the meaning of 3 . 2

1 --2 2

a Use the index laws to simplify ( 7 ) .

b Find ( 7 )2.

1 --2

c Hence write down the meaning of 7 .

Example 2 1 --3

a Use the index laws to simplify ( 7 )3. b Use your calculator to find ( 3 7 )3. 1--3 c Hence write down the meaning of 7 . 1 --3

a ( 7 )3 = 7

1 --- × 3 3

= 71 =7

b ( 3 7 )3 = 7

1 --3

c Since ( 7 )3 = ( 3 7 )3 1 --3

then 7 =

3

7

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

3

1 --3 3

a Use the index laws to simplify ( 5 ) .

b Find ( 3 5 )3.

1 --3

c Hence write down the meaning of 5 . 4

1 --3

a Use the index laws to simplify ( 2 )3.

b Find ( 3 2 )3.

1 --3

c Hence write down the meaning of 2 .

Example 3 Write down the meaning of: a 24

1 --2

b 167

1 --3

From the questions above: 1 --2

a 24 =

5

1 --3

b 167 =

24

3

167

Write down the meaning of: a 29 f

69

1 --3

b 29

1 --3

g 31

1 --2

c 13

1 --2

h 47

1 --3

d 13

1 --2

i

1 --2

195

Example 4 Evaluate: a 81

1 --2

1 --2

a 81 = =

6

81 9

b

64

b

64

1 --3

1 --3

= =

3

64 4

Evaluate: a 49 e 25

1 --2 1 --2

b f

8

1 --3

1000

c 100 1 --3

1 --2

d 27

1 --3

e 27 1 --3

j

1 --3

278

1 --2

39

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Example 5 Write in index form:

7

a

6

a

6 = 6

b 1 --2

Write in index form: a b 3 5

b

c

3

4

d

3

2

3

3

7 7 = 7

1 --3

e

10

f

3

12

H. SCIENTIFIC NOTATION The distance of Mars from the sun is approximately 229 000 000 kilometres. The diameter of the hydrogen atom is 0.000 000 000 025 4 metres. Scientists invented a more convenient method of writing very large and very small numbers like the ones above. It is called scientific notation or standard notation. To write a number in scientific notation, it is written as the product of a number between 1 and 10 and a power of 10.

Example 1 State whether or not the following numbers are written in scientific notation. a 6.7 × 108 d 2.96 × 10–7

b 23 × 105 e 480 000

c 3.65 × 1000

a Yes, as the first number (6.7) is between 1 and 10 and it is multiplied by a power of 10 (108). b No, because the first number (23) is not between 1 and 10. c No, because the second number (1000) is not expressed as a power of 10. d Yes, as the first number (2.96) is between 1 and 10 and it is multiplied by a power of 10 (10–7). e No, as it is not written as a product.

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Exercise 2H 1

State whether or not the following numbers are written in scientific notation. a 5.9 × 106 b 34 × 108 c 8.97 × 10 000 d 5.03 × 10–9 –15 4 e 28 000 f 7 × 10 g 0.85 × 10 h 4.2 × 100 68 i 163 000 000 j 2.006 × 10

2

Copy and complete the following table. 100

101

1

10

102

104 1000

105 1 000 000

Example 2 Write the following numbers in scientific notation. a 5 000 000

b 40 000

a 5 000 000 = 5 × 1 000 000 = 5 × 106 (using the table in question 2 above) b 40 000 = 4 × 10 000 = 4 × 104

3

Write the following numbers in scientific notation. a 3 000 000 b 70 000 c 8000

d 600 000

e 500

Example 3 Write the following numbers in scientific notation.

4

a 5300

b 284 000

a 5300 = 5.3 × 1000 = 5.3 × 103

b 284 000 = 2.84 × 100 000 = 2.84 × 105

Write the following numbers in scientific notation. a 4800 b 392 000 c 64 000

d 2 180 000

e 760

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Example 4 Write the following numbers as ordinary decimal numerals.

5

6

a 6 × 105

b 3.94 × 106

a 6 × 105 = 6 × 100 000 = 600 000

b 3.94 × 106 = 3.94 × 1 000 000 = 3 940 000

Write the following numbers as ordinary decimal numerals. a 3 × 104 b 7 × 103 c 9 × 106 d 4 × 105 f 4.6 × 105 g 6.71 × 103 h 3.9 × 106 i 8.36 × 104

e 8 × 102 j 5.2 × 105

Complete the table. 0.1

0.01

1 -----10

0.0001 1 ------------1000

0.000 001 1 --------------------100 000

Example 5 Write the following numbers in scientific notation. a 0.004

b 0.000 009

a 0.004 = 4 × 0.001 1 = 4 × ------------- (from table in question 6) 1000 1 = 4 × --------3- (from table in question 2) 10 = 4 × 10–3

7

Write the following numbers in scientific notation. a 0.003 b 0.000 007 c 0.0005

b 0.000 009 = 9 × 0.000 001 1 = 9 × -------------------------1 000 000 1 = 9 × --------610 = 9 × 10–6

d 0.000 02

e 0.09

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Example 6 Write as an ordinary decimal numeral.

8

a 5 × 10–2

b 7 × 10–6

1 a 5 × 10–2 = 5 × --------210 1 = 5 × ---------100

1 b 7 × 10–6 = 7 × --------610 1 = 7 × -------------------------1 000 000

= 5 × 0.01

= 7 × 0.000 001

= 0.05

= 0.000 007

Write as an ordinary decimal numeral. a 6 × 10–2 b 3 × 10–6 c 2 × 10–3

d 5 × 10–4

Example 7 Explain the difference between: a 2 × 104 and 24 a 2 × 104 = 2 × 10 000 = 20 000 1 b 2 × 10–4 = 2 × --------410

b 2 × 10–4 and 2–4 and

and

24 = 2 × 2 × 2 × 2 = 16 1 2–4 = -----4 2

1 = 2 × -----------------10 000

1 = -----16

= 2 × 0.000 01

= 0.0625

= 0.000 02

9

Explain the difference between: a 3 × 104 and 34 b 5 × 10–2 and 5–2 –5 –5 d 2 × 10 and 2 e 4 × 106 and 46

c 2 × 103 and 23

e 9 × 10–6

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Example 8 Write 246 000 in scientific notation. Here is a quick method for writing numbers in scientific notation. Step 1: Move the decimal point so that it is positioned between the first and second digits of the number. This always produces a number between 1 and 10. In this case we get 2.46000. Step 2: Count the number of places back to the original position of the decimal point in the number. 2.46000.

Number of places = 5 to the right = +5 This becomes the power of 10.

So 246 000 = 2.46 × 105.

10

Write in scientific notation a 372 000 b 54 000 f 87 500 g 7 698 000

c 2 980 000 h 361 000 000

d 3400 i 8000

e 609 000 j 56 000 000

Example 9 Write 0.000 71 in scientific notation. Step 1: Move the decimal point so that it is positioned between the first and second digits of the number. In this case we get 7.1. Step 2: Count the number of places back to the original position of the decimal point in the number. 0.0007.1

Number of places = 4 to the left = –4 This becomes the power of 10.

So 0.000 71 = 7.1 × 10–4.

11

Write in scientific notation: a 0.000 57 b 0.000 078 e 0.000 801 f 0.000 000 5 i 0.000 09 j 0.000 000 004 9

c 0.0061 g 0.004 39

d 0.000 002 96 h 0.000 002 8

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Example 10 Write 6.48 × 106 as an ordinary number. Reversing the process of example 8: Since the power of 10 is +6, then the decimal point is moved back 6 places to the right, i.e. 6.480000. Hence 6.48 × 106 = 6 480 000

12

Write as an ordinary number: a 7.32 × 106 b 5.2 × 104 e 9.27 × 107 f 6.914 × 104 8 i 2 × 10 j 3.08 × 105

c 5.67 × 105 g 3.275 × 106

d 3.8 × 103 h 7 × 105

Example 11 Write 3.51 × 10–6 as an ordinary number. Reversing the process of example 9: Since the power of 10 is –6, then the decimal point is moved back 6 places to the left, i.e. 0.000 003.51 Hence 3.51 × 10–6 = 0.000 003 51

13

Write as an ordinary number: a 3.98 × 10–6 b 5.3 × 10–4 e 5.9 × 10–6 f 3.07 × 10–7 –5 i 2.71 × 10 j 3.6 × 10–10

c 7.09 × 10–5 g 6 × 10–4

d 8.8 × 10–3 h 3 × 10–6

14

Express the following numbers in scientific notation. a The number of hairs on a person’s head is approximately 129 000. b The distance from the Earth to the Sun is 152 000 000 kilometres. c The diameter of a hydrogen atom is 0.000 000 002 54 centimetres. d The size of the influenza virus is 0.000 000 26 metres. e The average speed of the Earth around the Sun is 107 000 kilometres/hour.

15

Express the following as ordinary numbers. a The distance of Mars from the Earth is 7.83 × 107 kilometres. b The population of China is approximately 1.4 × 109. c A human brain cell is 2.8 × 10–5 metres long. d A microsecond is equivalent to 2.5 × 10–9 hours. e The number of cells in the human body is approximately 1013.

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

I. CALCULATIONS USING SCIENTIFIC NOTATION Example 1 Use the index laws to calculate (leave the answer in scientific notation): a

(3 × 1015) × (6 × 10–7)

a

(3 × 1015) × (6 × 10–7) = (3 × 6) × (1015 × 10–7) = 18 × 1015 + –7 = (1.8 × 101) × 108 = 1.8 × (101 × 108) = 1.8 × 109

b (8 × 10–4) ÷ ( 2 × 106)

b (8 × 10–4) ÷ (2 × 106)

c (5 × 107)3

= (8 ÷ 2) × (10–4 ÷ 106) = 4 × 10–4 – 6 = 4 × 10–10

c (5 × 107)3 = 53 × (107)3 = 125 × 1021 = (1.25 × 102) × 1021 = 1.25 × (102 × 1021) = 1.25 × 1023

Exercise 2I 1

Use the index laws to calculate (leave the answer in scientific notation): b (8 × 1012) × (3 × 109) a (3 × 108) × (4 × 106) 15 –7 c (7 × 10 ) × (6 × 10 ) d (2 × 10–8) × (3 × 10–7) e (5 × 10–9) × (4 × 1020) f (9 × 1016) ÷ (3 × 106) –4 6 g (6 × 10 ) ÷ (2 × 10 ) h (8 × 104) ÷ ( 4 × 1016) i (2 × 105)3 j (7 × 109)2 –6 3 k (3 × 10 ) l (8 × 10–10)2

Remember, one digit before the decimal point.

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Example 2 Use your calculator to evaluate (leave the answer in scientific notation): a (5.3 × 108) × (7.2 × 1011)

b (4.8 × 10–15) ÷ (1.6 × 107)

c (3 × 107)4

d

a Press 5.3

EXP

8 × 7.2

EXP

11

24

( 5.76 × 10 )

=

The display on the calculator will show

or

3.816 × 1020

3.81620

This is the way the calculator displays 3.816 × 1020. (It does not mean 3.81620.) So (5.3 × 108) × (7.2 × 1011) = 3.816 × 1020 b Press 4.8

EXP

–15 ÷ 1.6

The calculator displays

EXP

=

7

or

3 × 10–22

3–22

So (4.8 × 10–15) ÷ (1.6 × 107) = 3 × 10–22 c Press 3

EXP

7

xy 4 =

The calculator displays

d Press

5.76

EXP

The calculator displays

8.1 × 1029

So (3 × 107)4 = 8.1 × 1029 2

√

So

24

18

( 6.25 × 10 )

o

3

19

( 2.7 × 10 )

( 1.369 × 10

n p

3

( 1.25 × 10

3

a Calculate (3.6 × 108) – (4.9 × 107). b Is the answer positive or negative? c Is (3.6 × 108) bigger or smaller than (4.9 × 107)?

4

a Calculate (7.2 × 105) – (2.6 × 108). b Is the answer positive or negative? c Is (7.2 × 105) bigger or smaller than (2.6 × 108)?

5

a Calculate (4.9 × 10–4) – (5.3 × 10–5). b Is the answer positive or negative? c Is (4.9 × 10–4) bigger or smaller than (5.3 × 10–5)?

– 23

– 13

)

)

=

2.4 × 1012

( 5.76 × 10 ) = 2.4 × 1012

Use your calculator to evaluate (leave the answer in scientific notation): a (4.8 × 109) × (3.2 × 1010) b (2.7 × 106) × (9 × 1012) c (3.6 × 1013) × (2.5 × 10–5) d (1.8 × 10–8) × (1.5 × 10–10) 14 7 e (1.2 × 10 ) ÷ (1.5 × 10 ) f (3.6 × 10–12) ÷ (4.8 × 106) g (8 × 1012) ÷ (3.2 × 10–9) h (5.6 × 10–18) ÷ (4 × 10–6) 8 4 j (5.2 × 10–6)2 i (2 × 10 ) k (6 × 1012)3 l (5 × 10–8)5 m

24

47

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Example 3 Write the following numbers in order, from smallest to largest. a 2.5 × 1016, 7.8 × 1014

b 1.9 × 10–8, 4.3 × 10–10

c 4.8 × 106, 7.8 × 106

To compare numbers written in scientific notation: 1

compare the powers of 10

2

if the powers of 10 are the same then compare the first numbers.

a Comparing the powers of 10, 14 < 16, hence 7.8 × 1014 < 2.5 × 1016 (Check: 7.8 × 1014 – 2.5 × 1016 = –2.422 × 1016 < 0 If the difference between two numbers is negative, the first number is smaller than the second.) b Comparing the powers of 10, –10 < –8, hence 4.3 × 10–10 < 1.9 × 10–8 (Check: 4.3 × 10–10 – 1.9 × 10–8 = –1.857 × 10–8 < 0) c The powers of 10 are the same but 4.8 < 7.8, hence 4.8 × 106 < 7.8 × 106 (Check: 4.8 × 106 – 7.8 × 106 = –3 × 106 < 0) 6

Write the following numbers in order, from smallest to largest. a 7.2 × 1015, 4.6 × 1014 b 4.5 × 1016, 3.4 × 1018 c 9.6 × 10–12, 6.8 × 10–9 d 3.8 × 10–6, 7.8 × 10–8 –4 5 e 2.5 × 10 , 7.1 × 10 f 2.9 × 1016, 3 × 1016 g 8.5 × 10–10, 6.4 × 10–10 h 5.9 × 1016, 8.1 × 1014, 2.8 × 1015 –6 –5 –8 i 5 × 10 , 3.9 × 10 , 8.9 × 10 j 6.3 × 106, 7.8 × 10–5, 8.3 × 10–3

7

Light travels at 3 × 105 km/s. How far will it travel in 1 h?

8

The star Alpha Centauri is 4.1 × 1013 km from the Earth. The distance to Altair is 1.5 × 1014 km. Which star is closer to the Earth?

9

The diameter of the hydrogen atom is 2.54 × 10–9 cm. If one million hydrogen atoms could be placed next to each other in a straight line, how long would the line be?

10

The average speed of the Earth around the Sun is approximately 105 km/h. How many days would it take the Earth to travel 9.6 × 108 km?

11

Light travels at 3 × 105 km/s and sound travels at 330 m/s. A timekeeper stands at the end of a straight 100 m running track. After the starter fires the starting gun, how long does it take: a the sight of the smoke to reach her b the sound of the gun to reach her?

12

The area of Australia is approximately 7.7 × 1012 square metres. If the population of Australia in 2010 is expected to be 22 million people, how much land will there be for each head of population?

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

non-calculator activities

Do the following questions without using a calculator. 1

Write in index form: a 9×9×9×9

b 17 × 17 × 17 × 17 × 17 × 17 × 17

2

Write down the base and index of the numbers: a 311 b 59

3

Write in expanded form: b 73 a 86

4

Evaluate: a 25

5

6

7

Simplify, leaving the answer in index form: a 65 × 67 b (34)10 c 816 ÷ 810 Write true or false. a 34 × 35 = 99

b 58 ÷ 54 = 14

b 3

1 --2

c 10

1 --3

b 36

1 --2

c 8

1 --3

Write the following numbers in scientific notation. a 360 000 b 0.000 006 5

10

Write the following as ordinary numbers. a 7 × 105 b 1.2 × 10–4

11

Explain why 2 × 105 ≠ 25.

12

Find the value of (leave answer in scientific notation): a (5 × 108) × (3 × 106) b (9 × 1015) ÷ (3 × 10–8) c (3 × 106)2

13

e 2100 × (25)4

Evaluate: a 2–3

9

d 320 ÷ 3

Write down the meaning of: a 7–4

8

b 34

d

12

( 4 × 10 )

Write the following numbers in order, from smallest to largest: a 6.7 × 1012, 3.5 × 1013 b 4.2 × 10–10, 5.3 × 10–16

d 190

49

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

Language in Mathematics

1

Write in words: a 35

b 82

c 23

d

7

e

2

Write numerical expressions for: a seven squared b four cubed c six to the power five d two to the fourth e the square root of six f the cube root of five

3

Replace the vowels to make words that mean the same as ‘power’. a __nd__x b __xp__n__nt

4

When writing numbers in index notation, what name is given to: a the number that is being repeated b the number of times it is repeated?

5

Explain why 23 × 24 ≠ 47

6

Complete the sentence:

3

9

Scientific notation is also known as _________ notation. 7

Explain how to write a number in scientific notation.

8

Explain why the following numbers are not written in scientific notation. a 34 × 107 b 6.9 × 1 000 000

9

Explain the difference between 3 × 104 and 34.

10

Write down the mathematical meaning and one other meaning of these words. a product b order c base d index

Glossary base

compare

convenient

evaluate

exponent

expression

fractional

index

indices

integral

negative

notation

order

pattern

power

reverse

scientific

simplify

standard

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51

CHECK YOUR SKILLS 1

2

3

4

5

6

7

8

9

When written in index form 7 × 7 × 7 × 7 × 7 × 7 is: A 76 B 67 C 777 777 When written in expanded form 35 is: A 5×5×5 B 5×5×5×5×5 When evaluated using a calculator 74 is: A 74 B 47 34 × 38 = A 912

B 312

When simplified (94)6 = A 910

B 924

520 ÷ 510 = A 110

B 12

3.70 = A 3.7

B 1

2–6 is the same as 1 1 A -----2 B -----6 6 2 When written in index form 6 = A

10

11

12

13

14

6

1 --2

B 62 1 --3

The meaning of 6 is: A 2

C 3×3×3

D 777 7776

D 3×3×3×3×3

✓ C 20 448

D 2401

C 332

D 932

C 946

D 946

C 510

D 52

C 0

D 37

C –12

D –26

1 C 6 × --2

1 D 6 ÷ --2

1 B -----3 C 36 6 Which of the following numbers is written in scientific notation? A 53 × 1018 B 9.2 × 100 000 C 300 000

D 3.7 × 10–45

When written in scientific notation 23 000 000 becomes: B 2.3 × 107 C 2.3 × 108 A 2.3 × 106

D 2.3 × 109

When written as an ordinary number 5.1 × 10–6 is: A 0.000 51 B 0.000 051 C 0.000 005 1 (8.6 × 1018) × (2.5 × 1014) = B 2.15 × 1033 A 2.15 × 1032

C 2.15 × 10252

D

1 --2

D 0.000 000 51

D 2.15 × 10253

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

15

16

(3.6 × 107) ÷ (4.5 × 10–8) = A 8 × 10–14 B 8 × 10–15

C 8 × 1014

D 8 × 1015

When the numbers 2.9 × 10–7, 5.2 × 10–9, 3.8 × 10–9 are written in order from smallest to largest, the answer is: A 2.9 × 10–7, 3.8 × 10–9, 5.2 × 10–9 B 5.2 × 10–9, 3.8 × 10–9, 2.9 × 10–7 –7 –9 –9 C 2.9 × 10 , 5.2 × 10 , 3.8 × 10 D 3.8 × 10–9, 5.2 × 10–9, 2.9 × 10–7

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–3

4

5

6

7

8

9, 10

11–13

14–16

A

B

C

D

E

F

G

H

I

✓

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

REVIEW SET 2A 1

Write in index form: a 2×2×2×2

b 5×5×5×5×5×5×5

2

Write down the base and index of the numbers: a 79 b 310

3

Write in expanded form: a 35

b 76

4

Use a calculator to evaluate: a 38 b 56

5

Write true or false. a 34 × 25 = 69

6

7

8

9

b 86 ÷ 42 = 24

Simplify, leaving the answer in index form. a 512 × 520 b (45)3 c 28 ÷ 26 1 Write down the meaning of: --2 a 4–6 b 3

Evaluate: a 5–3

b 4

1 --2

d 75 × 7

c 15

c 27

e 56 × 57 ÷ 59

1 --3

1 --3

d 190

Write the following numbers in scientific notation. a 23 000 000 b 0.000 05

10

Write the following as ordinary numbers. a 9.8 × 104 b 3.7 × 10–5

11

Explain why 7 × 105 ≠ 75.

12

Use your calculator to find the value of (leave answer in scientific notation): a (3.4 × 104) × (3.5 × 109) b (5.6 × 1010) ÷ (1.4 × 105) c (3 × 109)5

13

d

12

( 2.25 × 10 )

Write the following numbers in order, from smallest to largest. a 3.8 × 1015, 4.6 × 1013 b 7.7 × 10–16, 3.1 × 10–12

53

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

REVIEW SET 2B 1

Write in index form: a 7×7×7×7×7

b 9×9×9×9×9×9×9

2

Write down the base and index of the numbers: a 38 b 52

3

Write in expanded form: a 64

b 73

4

Use a calculator to evaluate: a 29 b 36

5

Simplify, leaving the answer in index form. a 310 × 33 b (72)6 c 410 ÷ 45

6

7

8

9

Write true or false. a 45 × 26 = 811

e (25)4 × 210

b 57 ÷ 53 = 14

1 Write down the meaning of: --2 –5 a 7 b 6

Evaluate: a 6–2

d 612 ÷ 6

b 9

1 --2

c 9

c 8

1 --3

1 --3

d 70

Write the following numbers in scientific notation. a 46 000 b 0.000 3

10

Write the following as ordinary numbers. a 4 × 107 b 1.8 × 10–6

11

Explain why 4 × 106 ≠ 46.

12

Use your calculator to find the value of (leave answer in scientific notation): b (8 × 1010) ÷ (1.6 × 10–5) a (3.9 × 1013) × (4 × 10–5) c (3 × 10–6)5

13

d

( 1.96 × 10

– 10

)

Write the following numbers in order, from smallest to largest: a 4.1 × 109, 5 × 109 b 4.5 × 10–11, 3.1 × 10–15

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

REVIEW SET 2C 1

Write in index form: a 2×2×2×2×2×2×2×2

b 10 × 10 × 10

2

Write down the base and index of the numbers: a 84 b 3–6

3

Write in expanded form: a 62

b 57

4

Use a calculator to evaluate: a 54 b 37

5

Simplify, leaving the answer in index form. a 912 × 97 b (23)5 c 512 ÷ 54

6

7

8

9

Write true or false. a 47 × 34 = 1211

e (28)2 × (210)3

b 158 ÷ 32 = 56

1 Write down the meaning of: --2 –8 a 4 b 3

Evaluate: a 4–3

d 69 × 6

b 4

1 --2

c 2

1 --3

c 64

1 --3

d 40

Write the following numbers in scientific notation. a 17 000 000 000 b 0.000 000 35

10

Write the following as ordinary numbers. a 2.86 × 105 b 3.06 × 10–4

11

Explain why 3 × 104 ≠ 34.

12

Use your calculator to find the value of (leave answer in scientific notation): a (3 × 10–8) × (5.3 × 10–6) b (4.5 × 10–9) ÷ (5 × 10–16) c (7.3 × 1015)2

13

d

3

–7

( 1.25 × 10 )

Write the following numbers in order, from smallest to largest. a 2.94 × 1015, 2.94 × 1016 b 6.5 × 10–14, 1.4 × 10–10

55

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Indices and Scientific Notation (Chapter 2) Syllabus reference NS5.1.1

REVIEW SET 2D 1

Write in index form: a 6×6×6×6×6

b 5×5×5×5×5×5×5×5

2

Write down the base and index of the numbers: a 68 b 3–10

3

Write in expanded form: a 45

b 27

4

Use a calculator to evaluate: a 64 b 212

5

Simplify, leaving the answer in index form. a 520 × 520 b (37)10 c 516 ÷ 512

6

7

8

9

Write true or false. a 24 × 25 = 49

e 330 × (35)3

b 76 ÷ 74 = 12

1 Write down the meaning of: --2 a 6–5 b 34

Evaluate: a 2–5

d 423 ÷ 4

b 100

c 51 1 --2

1 --3

c 125

1 --3

d 130

Write the following numbers in scientific notation. a 205 000 000 b 0.000 356

10

Write the following as ordinary numbers. a 4.21 × 107 b 9 × 10–5

11

Explain why 2 × 106 ≠ 26.

12

Use your calculator to find the value of (leave answer in scientific notation): b (3.9 × 1015) ÷ (6 × 10–8) a (4.5 × 10–6) × (5 × 1030) c (2 × 1018)4

13

d

3

11

( 3.43 × 10 )

Write the following numbers in order, from smallest to largest. a 7.6 × 10–10, 4.7 × 10–10 b 2.4 × 10–11, 3.5 × 10–16

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Chapter 3 Data Representation and Analysis This chapter deals with grouping data to aid analysis, and constructing frequency and cumulative frequency tables and graphs. After completing this chapter you should be able to: ✓ construct a cumulative frequency table for grouped and ungrouped data ✓ construct cumulative frequency histograms and polygons ✓ find the median using a cumulative frequency polygon ✓ group data into class intervals ✓ find the mean using the class centres ✓ find the modal class.

Syllabus reference PAS5.1.1 WM: S5.1.1–S5.1.5

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Diagnostic Test Use the table below to answer questions 1–4. Score

5

6

7

8

9

10

Frequency

4

7

12

17

11

5

1

The mean of the data in the table above is closest to: A 7

B 7.7

C 8

6

The cumulative frequency for the 40–49 class is: A 64

B 44.5

C 26

D impossible to determine without the exact scores 7

An estimate for the median is closest to: A 40–49 class

D 7.5

B 50

C 46 2

The median of the data in the table above is closest to: A 17

3

B 7.7

D 7.5

C 8

D 7.5

The cumulative frequency for the score of 8 is:

8

Which frequency distribution table represents the following scores? 12, 30, 38, 49, 13, 28, 33, 17, 21, 31, 23, 32, 25, 26, 39, 36, 42, 46, 36, 50, 48, 32, 45, 57, 43, 51, 49, 53, 42, 33 A Class

Frequency

10–19

3

20–29

6

30–39

9

40–49

8

50–59

4

Class

Frequency

10–19

3

120

20–29

5

100

30–39

10

40–49

8

50–59

4

A 40 5

C 8

The mode of the data in the table above is closest to: A 7

4

B 7.7

D impossible to determine without the exact scores

B 17

C 8

D 23

The ogive is the: A frequency polygon B frequency histogram C cumulative frequency polygon D cumulative frequency histogram

Cumulative frequency

Here is a cumulative frequency histogram and polygon for questions 6 and 7. Cumulative frequency histogram

80 60 40 20 0 0 10 20 30 40 50 60 70

Class boundaries

B

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

C

Use this table to answer questions 10–12. Class

Frequency

Class

10–19

4

10–16

3

20–29

5

17–23

15

30–39

9

24–30

8

40–49

9

31–37

12

50–59

3

38–44

5

Class

Frequency

10–19

4

20–29

4

30–39

10

40–49

7

50–59

5

D

Class centre

A 27

C 70

C 5

D 10–44

B 10

C 8.6

D impossible to determine without the exact scores The modal class is: A 24–30 C 24–30

The class centre for the 32–38 class is: B 38

B 42

The mean for the data is closest to:

11

12 A 32

fx

The class centre for the 38–44 class is:

10

A 84

9

Frequency

D 35

B 17–23 D 31–37

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1–4

5, 6

7

8

9–12

Section

A

B

C

D

E

A. UNGROUPED DATA This section reviews some important aspects of DS4.2 that are needed for this section.

Example 1 For the scores in this table, find the: a mode

b mean

c median.

Score

8

9

10

11

12

13

14

15

Frequency

3

5

12

18

14

5

2

1

a The mode is 11 as this is the score with the highest frequency of 18. b Add an fx column to find the mean.

☞

59

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

Score (x)

Frequency (f)

fx

8

3

24

9

5

45

10

12

120

11

18

198

12

14

168

13

5

65

14

2

28

15

1

15

∑f = 60

∑fx = 663

∑ means ‘sum of’.

Σfx Mean = -------Σf 663 = ---------60 = 11.05 c To find the median add a cumulative frequency cf column. Score (x)

Frequency (f)

Cumulative frequency (cf)

8

3

3

9

5

3+5=8

10

12

8 + 12 = 20

11

18

20 + 18 = 38

12

14

38 + 14 = 52

13

5

52 + 5 = 57

14

2

57 + 2 = 59

15

1

59 + 1 = 60

There are 60 scores so the middle scores are the 30th and 31st scores. These are both 11. The median is 11.

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

Exercise 3A 1

Add an fx column and find the mean for each frequency distribution table. a

x

f

2

b

x

f

8

17

3

11

4

c

x

f

3

40

2

18

5

41

4

15

19

11

42

8

5

17

20

4

43

9

6

3

21

2

44

10

7

5

22

1

45

7

46

3

47

2

48

2

2

Write the mode for each frequency distribution table in question 1.

3

Add a cumulative frequency column to each table in question 1 and find the median.

4

Use the statistics function of your calculator to find the mean from the tables in question 1.

5

The table shows the number of glasses of water drunk by 9 Orange in a day. Number of glasses

0

1

2

3

4

5

Frequency

10

3

5

2

6

4

a Copy the table and add a cumulative frequency column. b Calculate the median number of glasses of water consumed. c Use the cumulative frequency column to find the number of students who drank: i less than 3 glasses of water ii less than 4 glasses of water d Use your answer from part c i to find the number of students who drank 3 or more glasses of water. e Draw a frequency histogram and polygon for the data. The histogram is like a column graph.

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6

Here is a frequency histogram showing the number of days absent for the 28 students in 9 Blue over a term. Absences 9 8 7

Frequency

6 5 4 3 2 1 0 1

2

3

4

5

6

7

Days absent

a b c d

Here is a frequency histogram showing the number of days that the families of students in Year 9 hired a video in a week. Videos hired 10

8

Frequency

7

Draw a frequency distribution table showing this information. Add a cumulative frequency column to the table. How many students had 2 or less days absent? After 3 days absent a note is sent home. How many notes were sent home? No student receives more than one note per term.

6

4

2

0 1

2

3

4

5

6

7

Number of days

a b c d

Draw a frequency distribution table showing this information. Add a cumulative frequency column to the table. How many students hired less than 2 videos? How many families hired more than 5 videos?

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

Investigation 1 WM: Applying Strategies, Communicating

Graphics calculator exercise This is written for a Casio CFX 9850GB PLUS. 1

Here is a list of the number of hours of sleep on Friday night for each student in 9M3. 8, 8, 7, 6, 8, 10, 10, 9, 8, 6, 4, 5, 5, 8, 8, 8, 9, 7, 9, 5, 3, 4, 8, 6, 9, 8, 8, 5, 5, 3 a Select STAT from the main menu. b Enter the data into list 1. List 1 List 2 List 3 List 4 1 8 To clear old data press 2 F6 to next menu and 3 4 DEL-A the YES . 5

c Select GRPH using F1 . Use F6 to get back to this menu after deleting. d Use SET F6 to give the menu to select the graph type to Hist. Press EXIT to return. e Select GPH 1 F1 to next screen, ignore ‘Set Interval’, press DRAW F6 . The histogram is drawn. Set Interval Start: 3 pitch: 0.778 F6 (DRAW) DRAW

1 VAR

F4

f

Select IVAR F1 to obtain statistics data display.

1–Variable x = ∑x = ∑x2 = xón = -1 xón = n =

6.9 207 1545 1.97230829 2.0060254 30

DRAW

g Note that the mean is 6.9. Scroll down to see the median is 8 and the mode is 8. h Use EXIT to return. i 2

Use F6 to next menu and DEL-A F4 to delete.

Collect information on hours slept and use the graphics calculator to draw the histogram and calculate the mean, mode and median. Discuss your results.

Note: It may be necessary to set the data lists. From the list press CALC F2 then SET F6 . Have the display read as shown.

1Var 1Var 2Var 2Var 2Var

XList Freq XList YList Freq

:List1 :1 :List1 :List2 :1

List1 List2 List3 List4 List5 List6

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

B. CUMULATIVE FREQUENCY DIAGRAMS In stage 4 you drew frequency histograms and polygons. This section involves drawing cumulative frequency histograms and cumulative frequency polygons for grouped data. Ogive is the correct name for cumulative frequency polygon.

Example 1 This table shows the height of Year 9 students in a particular school. The information is shown in groups. Class

Frequency

Cumulative frequency

140–146

1

1

147–153

3

4

154–160

8

12

161–167

12

24

168–174

6

30

175–181

2

32

182–188

2

34

Draw a cumulative frequency histogram and ogive.

Class boundaries are used for the columns.

Height of year 9 students 40 35

Cumulative frequency

64

Use the lowest value as the first number and the class boundaries to form the columns. First draw the cumulative frequency histogram. A histogram has no gaps between the columns. Next join the bottom left corner to the top right corner of each column to form the polygon, called an ogive.

30 25 20 15 10 5 0 140 147 154 161 168 175 182 189

Class boundaries

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

Note: There are three ways of displaying the class values. 1 Write the class boundaries starting from the lowest value. The last value will be the start of the next class after the final class in the table. 2 Use class centres in the middle of the columns. 3 Write the whole class below each column. We will use the first method as it makes finding the median in section C easier.

Exercise 3B 1

For each of the following grouped frequency tables with cf columns, draw a cumulative frequency histogram and ogive. a

c

Class

Frequency

140–146

2

147–153

Cumulative frequency

b

Class

Frequency

2

70–78

6

6

4

6

79–87

7

13

154–160

5

11

88–96

6

19

161–167

15

26

97–105

4

23

168–174

11

37

106–114

4

27

175–181

7

44

115–123

7

34

182–188

8

52

124–132

9

43

Class

Frequency

Cumulative frequency

Class

Frequency

Cumulative frequency

40–49

16

16

20–24

7

7

50–59

13

29

25–29

13

20

60–69

15

44

30–34

6

26

70–79

12

56

35–39

17

43

80–89

21

77

40–44

2

45

90–99

5

82

45–49

8

53

d

Cumulative frequency

65

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

Example 2 Use this cumulative frequency polygon to complete the frequency distribution table. Cumulative frequency 60

Class Cumulative frequency

66

Frequency

50

Cumulative frequency

10–16

40

17–23 30

24–30 20

31–37

10 0 10

38–44 17

24

31

38

45

Score

First read the values from the graph and complete the cumulative frequency column. Class

Frequency

Cumulative frequency

10–16

9

17–23

21

24–30

27

31–37

34

38–44

52

Then subtract the values in the cf column to get the frequencies. Class

Frequency

Cumulative frequency

10–16

9

9

17–23

12

21

24–30

6

27

31–37

7

34

38–44

18

52

21 – 9 = 12 27 – 21 = 6 34 – 27 = 7 52 – 34 = 18

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

Use these cumulative frequency histograms to complete the frequency distribution tables. a

Cumulative frequency

Class

Frequency

45

Cumulative frequency

20–24

40

25–29 Cumulative frequency

35

30–34

30

35–39

25

40–44

20

45–49

15 10 5 0 20

25

30

35

40

45

50

Score

b

Cumulative frequency

Class

50

70–78

45

79–87

40

88–96 Cumulative frequency

2

35

97–105

30

106–114

25

115–123

20 15 10 5 0 70

79

88

97

Score

106

115

124

Frequency

Cumulative frequency

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

c

Cumulative frequency

Class

Frequency

Cumulative frequency

Frequency

Cumulative frequency

45

40–49

40

50–59 Cumulative frequency

35

60–69 30

70–79

25

80–89

20

90–99

15 10 5 0 40

50

60

70

80

90

100

Score

d

Cumulative frequency

Class

80

20–24

70

Cumulative frequency

68

25–29

60

30–34 50

35–39

40

40–44

30

45–49

20 10 0 20

25

30

35

Score

40

45

50

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

Investigation 2 WM: Communicating, Reflecting, Reasoning

Reading level 1

Select a section of a newspaper or magazine article of about 100 words. a Complete a frequency distribution table with the number of letters per word. b Add a cumulative frequency column. c Find the number of words with: i less than 3 letters ii less than 5 letters iii more than 6 letters. d Use a graphics calculator or otherwise calculate the mean, median, and mode number of letters per word. e Draw a frequency histogram. f Draw a cumulative frequency histogram. (You need to enter these values into the graphics calculator.)

2

a Compare your newspaper or magazine article and discuss the summary statistics with others in the class. b Discuss any results of your comparisons.

3

There are various reading level tests. Investigate some of these and how they are used.

C. CALCULATING THE MEDIAN While the median can be calculated for ungrouped data using a cumulative frequency column, only the median class can be found in this way for grouped data. The cumulative frequency polygon, the ogive, can be used to find an estimate for the median of grouped data. To do this, first draw an ogive for the grouped data, draw a line across at the halfway point until it meets the ogive: the value obtained is the estimate for the median.

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

Example 1 Find the median class and estimate the median for the data in this table.

Class

Frequency

Cumulative frequency

1–10

8

8

11–20

19

27

21–30

32

59

31–40

16

75

41–50

25

100

To find the scores in the median class, n + 1 100 + 1 use ------------ = ------------------- = 50.5 2 2 ∴ we look for the 50th and 51st scores. The class containing the median, called the median class, is the 21–30 class because it contains the 50th and 51st scores. Note: To estimate the median we use an ogive or the cumulative frequency polygon. Draw a line across at 50 to the ogive. Then draw the line down to the axis. Estimate the median from the scale median
28.

The columns form the cumulative frequency histogram, and the line graph is the ogive. Remember: The ogive is drawn by joining the top right corner of each column, not the centre of each column.

Usually halve the number of scores.

100 90

Cumulative frequency

70

80 70 60 50 40 30 20 10 0 1

11

21

31

Scores in class intervals

41

51

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

Exercise 3C 1

By drawing a line across at 50, find an estimate for the median of each of these distributions with 100 scores. a b Cumulative frequency Cumulative frequency

120

Cumulative frequency

Cumulative frequency

120 100 80 60 40 20 0 1 11 21 31 41 51 61 71

100 80 60 40 20 0 3 10 17 24 31 38 45 52

Score

Score

c

Cumulative frequency

d Cumulative frequency

120

Cumulative frequency

Cumulative frequency 120

100 80 60 40 20 0 5 14 23 32 41 50 59 68

100 80 60 40 20 0 30 40 50 60 70 80 90 100

Score

Score

These ogives are for distributions with 60 scores. The line to find the estimate for the median is drawn across at 30. Find estimates for the median from these ogives. a

Cumulative frequency

b

Cumulative frequency

70

70

Cumulative frequency

Cumulative frequency

2

60 50 40 30 20 10 0 1 11 21 31 41 51 61

Score

60 50 40 30 20 10 0 5 11 17 23 29 35 41

Score

☞

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

c

d Cumulative frequency

Cumulative frequency 70

Cumulative frequency

Cumulative frequency

70 60 50 40 30 20 10

60 50 40 30 20 10 0 4 11 18 25 32 39 46

0 22 30 38 46 54 62 70

Score

Score

These ogives are for different numbers of scores. By halving the highest number, find an estimate for the median. a

b Cumulative frequency

90

45

80

40

Cumulative frequency

Cumulative frequency

Cumulative frequency

70 60 50 40 30 20 10

35 30 25 20 15 10 5

0 3 12 21 30 39 48 57

0 1 11 21 31 41 51 61

Score

c

Score

d

Cumulative frequency

Cumulative frequency Cumulative frequency

124

Cumulative frequency

3

120 100 80 60 40 20 0 30 45 60 75 90 105 120 135

Score

60 50 40 30 20 10 0 0

9 18 27 36 45

Score

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

4

The results for 60 students in a geography test are given in the table. Score

0–9

10–19

20–29

30–39

40–49

Frequency

2

5

15

31

7

a Add a cumulative frequency column. c Draw a cumulative frequency polygon. 5

6

The management of a small production business carried out a survey on the production level of its employees. The results appear in the table opposite. a Add a cumulative frequency column. b Find the median class. c Draw a cumulative frequency polygon (ogive). d Estimate the median from the ogive. A survey was conducted on the age structure of a small country town, with the following results. a Add a cumulative frequency column. b Find the median class. c Draw a cumulative frequency polygon. d Estimate the median from the ogive.

b Find the median class. d Estimate the median from the polygon. Number of items

Frequency

11–20

3

21–30

17

31–40

27

41–50

5

51–60

3

Age last birthday

Frequency

0–9

165

10–19

112

20–29

103

30–39

129

40–49

94

50–59

85

60–69

73

70–79

22

80–89

9

D. GROUPED DATA In stage 4 stem-and-leaf plots were often used when there was a large spread of data. This section uses groups to collect data into frequency distribution tables.

Example 1 a Organise this data into a grouped frequency distribution table. Use groups 20–29, 30–39 and 40–49. 33 38 23 36 30 41 47 49 35 26 24 34 23 35 41 37 42 40 48 27 35 33 31 42 b Construct a frequency histogram for this data.

☞

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

a

Group

Tally

Frequency

20–29

5

30–39

12

40–49

7 Frequency histogram

b

15

Frequency

10

5

0 20

30

40

50

Class boundaries

Exercise 3D 1

a Complete this frequency distribution table using this data. Group

Tally

Frequency

10–19 20–29 30–39 40–49 10 33 22

18 28 36

23 41 37

35 47 25

37 33 31

19 28 43

42 19 33

48 41 48

16 33 34

20 39 15

b Draw a frequency histogram. 2

a Organise this data into a frequency distribution table using the classes 4–10, 11–17, 18–24, 25–31, 32–38. 17 5 15 6

6 12 21 19

19 15 15 12

23 23 28 18

34 25 17 15

b Draw a frequency histogram.

36 24 15 22

25 33 38 23

11 31 23 31

18 7 37 33

38 19 36 37

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

3

a Organise the data from question 2 into a frequency distribution table using the classes 4–12, 13–21, 22–30, 31–39. b Draw a frequency histogram. c Compare your histogram with that from question 2. Explain any differences.

E. FINDING THE MEAN AND MODE This section revises the method for finding the mean in a frequency distribution table and then extends that to grouped data. When using grouped data, the class with the highest frequency is called the modal class. The mean is calculated by using the class centres. (There will be more on class centres later in the section.)

Example 1 For the following distribution, find: a the mode b the mean

The mode is the score with the highest frequency.

Score

Frequency

19 20 21 22 23 24

5 4 8 11 2 6

a To find the mode look for the highest number in the frequency column, and the mode is the score with that frequency. The highest frequency is 11, ∴ the mode = 22. b To find the mean you need to add an fx column to the frequency distribution table. The fx column stands for frequency (f ) times score (x). This column groups all of the scores of each value and finds the total value for each. Score (x)

Frequency (f )

fx

19 20 21 22 23 24

5 4 8 11 2 6

19 × 5 = 95 20 × 4 = 80 21 × 8 = 168 22 × 11 = 242 23 × 2 = 46 24 × 6 = 144

Totals

∑f = 36

∑fx = 775

‘∑’ means ‘sum of’. It is the Greek letter ‘sigma’.

☞

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

∑ ∑

 fx or  ------------ f 

sum of fx column x = ------------------------------------------------------sum of the f column 775 ∴ x = ---------36 ∴ x = 21.527778 ∴ x = 21.5 (to 1 d.p.)

The mean is the total of the fx column divided by the total of the frequency column i.e. fx _ x = ------------f

∑ ∑

Example 2 Find the mean and mode for this frequency distribution table.

x

35

36

37

38

39

f

8

7

4

9

2

We include an fx column. x

f

fx

35 36 37 38 39

8 7 4 9 2

280 252 148 342 78

∑f = 30

∑fx = 1100

1100 mean = ------------30
36.7 (to 1 d.p.) mode = 38 (since 38 has the highest frequency of 9)

Exercise 3E 1

Copy the following tables, complete the fx column and find the mean. a b c x f x f x 8 9 10 11 12

1 6 5 10 3

121 122 123 124 125

4 11 11 3 1

85 86 87 88 89

f 1 10 8 4 16

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

2

Find the mode for each frequency distribution table in question 1.

Example 3 Find the mean of this data using a scientific calculator.

Score

19

20

21

22

23

24

Frequency

5

4

8

11

2

6

Instructions are for a Sharp calculator. To find the mean, using the statistics function of the calculator: 1 Set the calculator to statistics mode SD by pressing 2ndF

Mode

1 .

2 Make sure the statistics memory is clear by pressing 2ndF DEL. 3 Enter the scores by pressing the score then STO then the frequency and the M+ key. (Do not press = .) 4 When all the scores have been entered, press the appropriate buttons for the x

mean RCL

4.

5 Press the appropriate keys to check that the correct number of scores have been n

entered RCL

0 .

6 For this example the calculator steps are: Mode

2ndF

1

or

Mode

SD

or the appropriate key for your calculator.

to clear the contents of the memory.

2ndF DEL

19 STO

5

M+

20 STO

4

M+

21 STO

8

M+

22 STO 11

M+

23 STO

2

M+

24 STO

6

M+

Always enter the score first, followed by the frequency.

to enter the scores.

x _ Using the keys for the mean RCL , 4 , x = 21.52778. n

Using the keys for the number of scores RCL check that n = 36.

0

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

3

By following the correct calculator steps for your calculator, find the mean for these frequency distribution tables. a b c d x f x f x f x f 63 64 65 66 67 68

4

4 10 2 5 7 3

10 20 30 40 50

8 6 12 8 11

35 40 45 50 55

11 7 5 8 12

840 850 860 870 880 890 900

11 3 7 2 5 4 7

Find the mode for each frequency distribution table in question 3.

Example 4 Find the class centre for the classes: a 1–9

b 20–29

c 21–25

To find the class centre, average the class boundaries. a

1+9 ------------2 10 = -----2 =5 Class centre is 5

5

b

20 + 29 ------------------2

c

49 = -----2 = 24 1--2Class centre is 24 1--2-

Find the class centre for each of these classes. a 11–19 b 20–30 c 34–40 f 35–40 g 50–59 h 14–22

21 + 25 ------------------2 46 = -----2 = 23 Class centre is 23

d 21–30 i 16–24

e 26–30 j 17–24

Example 5 The following information appears in a grouped frequency distribution table. Find: a the mean b the modal class

Class

Frequency

1–10 11–20 21–30 31–40 41–50

8 19 32 16 25

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

To find the mean use the class centres and the frequency. Enter these into the calculator. The assumption is that the scores within a class are evenly distributed throughout the class. You may use this method or the statistics function on your calculator. Class

Class centre (x)

Frequency (f )

fx

= 5.5

8

44.0

1–10

1 + 10 ---------------2

11–20

11 + 20 ------------------2

= 15.5

19

294.5

21–30

21 + 30 ------------------2

= 25.5

32

816.0

31–40

31 + 40 ------------------2

= 35.5

16

568.0

41–50

41 + 50 ------------------2

= 45.5

25

1137.5

∑f = 100

∑fx = 2860.0

Totals

∑ ∑

fx 2860 a Mean = ------------ = ------------- = 28.6 100 f b There is no single score as the mode, but the class with the highest frequency is the modal class. The modal class is 21–30.

The class with the highest frequency is the modal class.

Class centres are equally spaced, just as the classes are.

6

a Complete the frequency distribution table. b Find the modal class. c Find the mean age of the population.

Class

Class centre

Frequency

6–10

8

7

11–15

13

8

16–20

18

16

21–25

12

26–30

4

31–35

3

79

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7

The management of a small business carried out the following survey on the production level of its employees and came up with the following. No. of items produced Frequency

11–20

21–30

31–40

41–50

51–60

3

18

26

6

2

a Draw a frequency distribution table including the class centre column. b Find the mean number of items produced. 8

9

A survey was carried out on the age structure of the population of a country town and the following results were obtained. Find the mean age of the population. Age last birthday

Frequency

0–9

170

10–19

107

20–29

111

30–39

121

40–49

104

50–59

75

60–69

63

70–79

32

80–89

9

The ages of a class completing the ICDL (International Computer Drivers Licence) are listed. 49, 18, 36, 21, 33, 42, 26, 25, 60, 19, 22, 20, 36, 43, 39, 21, 22, 57, 20, 34, 28, 18, 39, 55, 41, 21, 31, 31, 40, 63, 65, 30, 30, 34, 33 a Using class intervals 16–25, 26–35, 36–45, 46–55, 56–65, construct a frequency distribution table and draw a histogram to represent the data. b Using class intervals 16–20, 21–25, 26–30, 31–35, 36–40, 41–45, 46–50, 51–55, 56–60, 61–65, construct a frequency distribution table and draw a histogram. c Compare the two histograms and comment on their shapes. d The next ICDL class has these ages. 17, 35, 43, 54, 18, 33, 38, 22, 26, 36, 28, 37, 39, 30, 31, 44, 41, 47, 51, 41, 55, 53, 37, 40, 62, 48, 56, 54, 58, 63, 47, 65, 38, 47, 53 i Sort these ages into frequency tables using the same classes as parts a and b. ii Draw histograms. iii Compare the two classes. iv Comment on your displays. e Calculate the mean in each of the four groupings. Comment on your answers. f Calculate the modal class for each. Comment on the differences. g Add a cumulative frequency column and find estimates for each of the four medians. Comment on your answers.

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

Example 6 a The mean of 5 scores is 12.2. What is the sum of the scores? b Find x if 10, 7, 3, 6 and x have a mean of 8. a Let S = sum of scores S ∴ --- = 12.2 5

b There are 5 scores. 10 + 7 + 3 + 6 + x ∴ --------------------------------------------- = 8 5 26 + x ∴ --------------- = 8 5 ∴ 26 + x = 40

∴ S = 12.2 × 5 ∴ S = 61

∴ x = 14

i.e. the sum of scores is 61. 10

a The mean of 8 scores is 7.5. What is the sum of the scores? b The mean of 9 scores is 11.6. What is the sum of the scores? c While on an outback safari Bill drove, on average, 262 km per day for a period of 12 days. How far did Bill drive in total while on safari? d The mean monthly sales for a clothing store is $15 467. Calculate the total sales of the store for the year.

11

a b c d e

Find x if 8, 11, 5, 7 and x have a mean of 8. Find x if 3, 15, 7, 9, 11 and x have a mean of 10. Find x if 5, 9, 11, 12, 13, 14, 17 and x have a mean of 12. Find a, given that 3, 0, a, a, 4, a, 6, a and 3 have a mean of 4. Over the complete assessment period, Jenny averaged 35 out of a possible 40 marks for her eight maths tests. However, when checking her files, she could only find 7 of the tests. For these she scored 29, 36, 32, 38, 35, 34 and 39. Can you determine how many marks out of 40 she scored for the eighth test?

Example 7 A cricketer played 12 innings at an average of 38.5, and then scored 12 and 71 in the next two innings. Find the cricketer’s new average. There are 12 scores. Let S = sum of scores S ∴ ------ = 38.5 12 ∴ S = 462 462 + 12 + 71 ∴ new average = -----------------------------------14 545 = ---------14
38.9 12

(There are now 12 + 2 = 14 scores in total.)

a Lili played 14 games of netball and had an average of 16.5 goals per game. In the final two games of the season Lili threw 21 goals and 24 goals. Find Lili’s new average.

☞

81

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b A cricketer played 11 matches and had an average of 23 runs per game. In the last two games she scored 41 and 35 runs. Find her new average. c A tennis player averaged 8 aces per match in her first six matches. In the next three matches she served 6, 11, 13 aces. Find her new average. Extension: 13

A sample of 12 measurements has mean 16.5 and a sample of 15 measurements has mean 18.6. Find the mean of all 27 measurements.

14

15 of 31 measurements are below 10 cm and 12 measurements are above 11 cm. Find the median if the other 4 measurements are 10.1 cm, 10.4 cm, 10.7 cm and 10.9 cm.

15

The mean and median of a set of nine measurements are both 12. If seven of the measurements are 7, 9, 11, 13, 14, 17 and 19, find the other two measurements.

Investigation 3 WM: Reasoning, Communicating, Reflecting

Grouping data 1

A drive-through fast food outlet boasts that the time taken to fill an order is 4 minutes. To support this claim, a survey was done on the times taken to fill the orders of 50 customers on their busiest day. The following results were recorded in minutes. 3.1 2.5 2.9 4.6 5.2 1.9 2.3 6.4 4.1 3.8 3.8 6.2 5.4 1.5 3.4 4.2 2.2 4.4 5.3 1.4 6.7 2.2 1.9 2.4 1.8 3.7 5.5 5.7 6.4 3.4 4.9 3.3 2.9 4.5 5.2 7.3 5.5 3.1 2.2 8.6 4.4 4.9 2.8 5.1 3.9 4.1 6.5 2.6 2.2 4.6 a Put this information into a frequency distribution table using the classes 1.0–1.9, 2.0–2.9, 3.0–3.9, and so on. b Calculate the mean time taken. c Draw an ogive and estimate the median time taken. d Find the modal class. e Do these results support the drive-through outlet’s claim? f Which of these results is most useful when commenting on this claim?

2

Repeat question 1 using the classes 1.4–2.3, 2.4–3.3, 3.4–4.3, 4.4–5.3, and so on.

3

Using the data in question 1: a Calculate the true mean using all the scores. b How does it compare to the means in questions 1 and 2? c Calculate the true median. d How does it compare to the medians in questions 1 and 2?

4

By analysing the answers to questions 1, 2 and 3, how do the class groupings affect: a the mean b the median c the modal class?

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

non-calculator activities

1

17 Write ------ as a mixed numeral. 5

2

Write 45.876 correct to two decimal places.

3

Find the value of 543 × 30.

4

Increase $630 by 10%.

5

If −5 × 2 =

6

How many weeks are there in two years?

7

1 Write 7 – --- as a mixed fraction. 8

8

What is the value of 0.05 × 0.3?

9

341.786 × 10 =

10

÷ 6, what is the missing number?

90% is the same as: 9 1 A -----B --10 9

C 0.9

D both A and C

11

The fraction ------ has a value of 9. What is the missing number? 4

12

What fraction is 3 kg of 9 kg? Give your answer in simplest fraction form.

13

The temperature at 6 p.m. in Jindabyne was 3°C but it fell 2°C each hour for the next 6 hours. What was the temperature at midnight?

14

59.8 Estimate the value of ----------------------- giving your answer as a whole number. 3.2 + 2.8

15

Find the missing number in the box. 32.7057 = 32 + 0.7 + 0.005 +

16

If John earns $150 for 10 hours work, how much will he receive for 4 hours work?

17

Erin knows that 342 × 76 = 25 992. Use this data to find the answer to Erin’s question 25 992 ÷ 342 =

18

What is

19

What is the next number in the sequence: 85, 77, 69, 61, ____

20

What is the average of 30, 33 and 87?

81 + 4 2 ?

83

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

21

3 + 12 ÷ 3 =

22

Katie solved the equation 7x – 6 + 2x – 1 and got the answer x = 1. Is Katie correct?

23

The product of two numbers is 48. If one number is 6, what is the other number?

24

What is the area of a triangle with a base of 6 cm and a perpendicular height of 8 cm?

25

Peter reverses the digits in the number 9564 and subtracts them from the original number. What is the answer?

Language in Mathematics Read the article about Hanna Neumann and answer the questions. Write your answers in complete sentences.

Hanna Neumann (1914–1971) Hanna Neumann was born Hanna von Coemmerer in Berlin. She was the daughter of a historian with teaching qualifications, who was killed in World War 1. As a result her family was very poor, and from the age of 13 she helped support her family by tutoring younger children. Hanna became an extremely capable student and commenced studies at the University of Berlin in 1932. Her main area of study was mathematics but she also had time for physics, philosophy, history and religion. Here she met her husband Bernhard, who in 1933 left for England after deciding that living in Germany under the Nazis had become too dangerous. She secretly travelled to England in 1934 to become engaged and then returned to continue her study in Berlin. Eventually, in 1938 after completing her studies and working as a research student at Gottingen University, she travelled to England and married Bernhard. Hanna worked extremely hard and was recognised as being an excellent teacher. In 1948 she started opening her house in the evenings for others to come and discuss mathematics. She continued to be involved in teaching and studying mathematics, and in 1964 joined Bernhard at the Australian National University in Canberra. In her position at the university, Hanna was recognised as having an enormous capacity for work and a great concern for those whom she taught. She held a number of administrative positions within professional mathematical bodies and travelled delivering lectures. Unfortunately, at the age of 57, on a lecture tour in Canberra she became ill suddenly and died. She is remembered not only for her mathematical ability, but also for the willingness with which she was prepared to devote her time to her teaching and her students.

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

1

a b c d

2

Complete these glossary terms by inserting the vowels. a c__m__l__t__v__ b m__d__ __n c __g__v__ e fr__q__ __ncy f __st__m__t__

Where and when was Hanna Neumann born? What were her main areas of interest? For what is she remembered? Construct a timeline of her life.

d h__st__gr__m

✓

3

Rearrange these words into sentences. The word with the capital letter starts the sentence. a The the is mean average b The the score median is middle c The the calculate centre used mean class is to d class class frequency modal highest with is the The the e frequency name ogive cumulative The another for is polygon f estimate data ogive grouped mean used is to the for The

4

Use every third letter to reveal a message. AJTSEHTBEMIMUTEEDAASNQAMWEORLDPDEVUAISNACDRNM KYEWDDTVINBAZXNATAHIRLPEORCWEETFNGHTBFRSWAAXL QRTTBODQTCIHOAESHSRITPLUNGDDSYFGODGFSDSAETGHA JKTRTIYUSIHTFDISACQSS

Glossary analyse estimate histogram modal class statistics

class centres frequency interpret mode

class intervals frequency table mean ogive

CHECK YOUR SKILLS

cumulative frequency grouped data median polygon

Use the table below to answer questions 1–4. Score

8

9

10

11

12

13

Frequency

8

11

18

12

16

10

1

85

The mean of the data in the table above is closest to: A 10 B 10.5 C 10.6

D 12.3

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

2

3

4

5

The median of the data in the table above is closest to: A 10 B 10.5 C 11

D 12

The mode of the data in the table above is closest to: A 10 B 10.5 C 10.6

D 12.3

The cumulative frequency for the score of 12 is: A 16 B 25

D 65

C 50

Which is used to calculate an estimate for the median of grouped data: A frequency polygon B frequency histogram C cumulative frequency polygon D cumulative frequency histogram

This cumulative frequency histogram and polygon is for questions 6–8. Cumulative frequency 90

Cumulative frequency

80 70 60 50 40 30 20 10 0 0

7 14 21 28 35 42 49

Score

✓

6

The cumulative frequency for the 21–27 class is: A 63 B 27.5 C 23 D impossible to determine without the exact scores

7

The frequency for the 15–21 class is: A 63 B 27.5 C 23 D impossible to determine without the exact scores

8

An estimate for the median is closest to: A 16–23 class B 16 C 20 D impossible to determine without the exact scores

9

Which frequency distribution table represents the following scores? 12, 30, 38, 49, 13, 28, 33, 17, 21, 31, 23, 32, 25, 26, 39, 46, 42, 46, 36, 50, 48, 32, 45, 57, 43, 51, 49, 53, 42, 33

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

A

C

10

B

Class

Frequency

Class

10–19

4

10–19

3

20–29

6

20–29

5

30–39

9

30–39

9

40–49

8

40–49

9

50–59

4

50–59

4

Class

Frequency

10–19

4

10–19

4

20–29

5

20–29

4

30–39

9

30–39

10

40–49

9

40–49

7

50–59

3

50–59

5

Class

Class centre

Frequency

10–18

4

19–27

21

28–36

13

37–45

25

46–54

9

11

D 35

C 13

D 10–54

fx

The class centre for the 28–36 class is: A 64 B 32

12

The mean for the data is closest to: A 34 B 10 C 8.6 D impossible to determine without the exact scores

13

The modal class is: A 19–27

B 28–36

Frequency

C 70

Use this table to answer questions 11–13. Class

Frequency

✓

D

The class centre for the 32–38 class is: A 32 B 38

87

C 37–45

D 46–54

If you have any difficulty with these questions, refer to the examples and questions in the section listed in the table. Question

1–4

5–7

8

9

10–13

Section

A

B

C

D

E

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

REVIEW SET 3A 1

a Find the mean, mode and median for the scores in this table. Score

21

22

23

24

25

26

27

28

Frequency

3

7

4

15

3

18

11

5

b Use a graphics calculator to check your calculations. 2

The table below shows the number of times the students of 9 Red ate take-away for dinner over 39 days. Number of days

Frequency

0–4

12

5–9

9

10–14

5

15–19

3

20–24

1

25–29

2

30–34

1

35–39

1

a Add a cumulative frequency column. b Draw a cumulative frequency histogram and polygon. c What is the modal class? d Calculate the mean. e Find the median using the cumulative frequency column.

REVIEW SET 3B 1

a Find the mean, mode and median for the scores in this table. Score

9

10

11

12

13

14

15

Frequency

23

16

3

8

12

6

1

b Use a graphics calculator to check your calculations.

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

2

a Use this cumulative frequency histogram to complete the frequency distribution table. Cumulative frequency

Class

Cumulative frequency

70

Frequency

20–25

60

26–31

50

32–37

40 30 20 10 0 20 26 32 38 44 50 56 62

Score

b c d e f

How many scores were more than 37? Why can’t you find the number of scores greater than 29? Make an estimate for the median. Calculate the mean. What is the modal class?

REVIEW SET 3C 1

a Find the mean, mode and median for the scores in this table. Score

4

5

6

7

8

9

Frequency

12

18

23

4

11

6

b Use a graphics calculator to check your calculations. 2

The table shows the height of Year 9 Red students. a Add a cumulative frequency column. Class Frequency b Draw a cumulative frequency 140–149 5 histogram and ogive. 150–159 9 c Use the ogive to make an estimate for the median. 160–169 12 d Calculate the mean. 170–179 3 180–189

0

190–199

1

89

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Data Representation and Analysis (Chapter 3) Syllabus reference PAS5.1.1

REVIEW SET 3D 1

a Find the mean, mode and median for the scores in this table. Score

20

21

22

23

24

25

Frequency

6

7

14

3

5

2

b Use a graphics calculator to check your calculations. 2

The histogram shows the number of videos hired in a week by families of a class. a Complete this frequency distribution table. Videos hired

Score

10

Frequency

90

0

8 6

1

4

2

2

3

0 0

1

2

3

4

No. of videos

b c d e f g

5

4 5

Add a cumulative frequency column. How many families hired less than 3 videos in a week? How many families hired more than 3 videos in a week? Calculate the mean number of videos hired. Use the cumulative frequency column to find the median. Find the mode.

Frequency

LEY_bk9_04_finalpp Page 91 Wednesday, January 12, 2005 10:32 AM

Chapter 4 Algebraic Techniques This chapter deals with operations involving algebraic terms including indices. After completing this chapter you should be able to: ✓ add, subtract, multiply and divide algebraic fractions ✓ apply the index laws to simplify algebraic expressions ✓ establish the meaning of the zero index ✓ define indices for square root and cube root ✓ establish the meaning of negative indices ✓ simplify expressions involving fractional and negative indices ✓ remove grouping symbols and simplify by collecting like terms ✓ factorise by determining common factors.

Syllabus reference PAS5.1.1, 5.2.1 W M: S5. 1. 1–S5. 1. 5, S 5 .2 .1 –5 .2 .5

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Diagnostic Test

1

2

3

4

5

6

7

Which of the following is not true? 2x 4x 2x 10x A ------ = -----B ------ = --------3 5 3 15 2x 4x 2x 6x C ------ = -----D ------ = -----3 6 3 9 12a When reduced to its simplest form ---------- = 18a 12 2a A -----B -----18 3a 2 4 C --D --3 6 3x 5x ------ + ------ = 11 11 8x 8x A -----B -----22 11 8x 88x C ---------D --------121 11 2x x ------ + --- = 3 4 3x 3x A -----B -----7 12 11x 11x C --------D --------12 7 3m 4 -------- × ------ = 7 5y 34m 12m A ----------B ----------75y 35y 12my 34my C -------------D -------------35 75 12ab 10 When simplified ------------- × ------ = 5 3a 120ab 8ab A ----------------B ---------15a a 40b C ---------D 8b 5 a b --- ÷ --- = 3 2 ab A -----6 2a C -----3b

6 B -----ab 3b D -----2a

8

9

5xy 3x --------- ÷ ------ = 8 2 5y A -----12 2 15x y C --------------16

12 B -----5y 16 D -------------2 15x y Which of the following does not simplify to t20? A t4 × t5

B t 30 ÷ t 10

C (t 4)5

D t16 × t 4

3 4

10

11

12

(a ) ---------------- = 4 2 a ×a A a2 B a4

A 8a7b 9

B 8a12b 20

C 8ab16

D 8(ab)16

(2m 5)3 = 2m 8 B 8m 8

B 3

17

1 --2

1 --2

D 9

B ( 4y ) C 2y

The meaning of y

1 --3

1 --2

D 4y2

is:

1 D ----3y Which of the following is equivalent 1 to ----3- ? 1 a --3 1 A 3a B -----C a D a –3 3a A

16

C 5

4 y may be written in index form as: A 4y

15

C 2m15 D 8m15

3k0 + 2 = A 2

14

D 16

4a 3b 5 × 2a 4b 4 =

A 13

C a6

1 --3

y

B 3y

C

3

y

3m −2 is equivalent to: 1 3 A –6m B ----------2- C ------29m m

1 D -------3m

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

18

8y 5 ÷ 2y 1 --4

A 19

y

6

–1

=

B

21 1 --4

y

4

C 4y

6

D 4y

4

The highest common factor of 6y 2 and 12y is: A 6

B 3y

C 6y

D 12

–2p 2(5p 2 + 3pq) = A –10p 2 – 6pq

B –10p 2 – 6p 3q

C –10p4 + 6p 3q

D –10p4 − 6p 3q

22

When fully factorised 12ab – 3a + 9a2 = A 3a(4b – 1 + 3a) B 3(4ab – a + 3a2) C a(12b – 3 + 9a) D 3ab(4 – 1 + 3a)

20

When expanded and simplified 3(2m – 1) – (m + 5) = A 5m – 8

B 5m + 2

C 5m – 6

D 5m + 4

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1–8

9, 10

11, 12

13

A

B

C

D

Section

14, 15 16, 17 E

F

18

19, 20 21, 22

G

H

I

A. ALGEBRAIC FRACTIONS Example 1 Complete the following equivalent fractions. 2 ■ a --- = -----3 15

■ 3n b ------ = -----7 14

c

2ab ■ ---------- = -----5 20

2 5 10 a --- × --- = -----3 5 15

3n 2 6n b ------ × --- = -----7 2 14

c

2ab 4 8ab ---------- × --- = ---------5 4 20

Exercise 4A 1

Complete the following equivalent fractions. 7k ■ 2y ■ 4m ■ a ------ = -----b -------- = ---c ------ = -----4 20 5 15 3 6

3t ■ d ------ = ---------10 100

5a ■ e ------ = -----2 12

Example 2 Reduce to its simplest form: 15 a -----20 c

7a -----9a

Dividing the numerator and denominator by the same number is sometimes called cancelling.

8t b -----12 6ab d ---------9a

☞

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

a Dividing the numerator and denominator by 5,

b Dividing numerator and denominator by 4,

3

2

15 15 ------ = --------420 20 3 = --4

8t 8 t ------ = --------312 12 2t = ----3

c Dividing numerator and denominator by a,

d Dividing numerator and denominator by 3 and by a,

1

2 1

7a 7a ------ = --------19a 9a 7 = --9

2

6ab 6 a b ---------- = -------------3 1 9a 9 a 2b = -----3

Reduce to its simplest form: 6x 3m a -----b -------12 9 6a 3p f -----g ---------7a 10p

5t -----20 8x h --------12x

10y d --------15 8ab i ---------4a

c

9b e -----12 12pq j ------------9q

Example 3 Simplify:

3

7 4 a ------ + -----15 15

5x 4x b ------ + -----11 11

7 4 7+4 a ------ + ------ = ------------15 15 15 11 = -----15

5x 4x 5x + 4x b ------ + ------ = ------------------11 11 11 9x = -----11

Simplify: 3x 4x a ------ + -----10 10 9m 6m d -------- – -------10 10

7b 8b b ------ + -----11 11 11k 3k e ---------- – -----12 12

c c

c

7a 2a ------ – -----10 10 7a 2a 7a – 2a ------ – ------ = ------------------10 10 10 5a = -----10 a = --2

2a 3a ------ + -----15 15

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 4 Simplify: 3 2 a --- + --4 3

5n 2n b ------ + -----6 3

3 2 3 3 2 4 a --- + --- = --- × --- + --- × --4 3 4 3 3 4 9 8 = ------ + -----12 12 17 = -----12

4

7m m -------- – ---8 3

5n 2n 5n 2n 2 b ------ + ------ = ------ + ------ × --6 3 6 3 2 5n 4n = ------ + -----6 6 9n = -----6 3n = -----2

5 = 1 ----12

c

c

7m m 7m 3 m 8 -------- – ---- = -------- × --- – ---- × --8 3 8 3 3 8 21m 8m = ----------- – -------24 24 13m = ----------24

Simplify: 2x a ------ + 3 3t f ------ + 10

x --4 2t ----9

5k 3k b ------ + -----6 4 4w w g ------- – -----3 12

7b b ------ – --8 4 3v 4v h ------ + -----2 3 c

3a a d ------ + -----5 10 11e 3e i ---------- – -----10 5

4z 2z e ------ – -----5 3 5x 3x j ------ – -----6 8

Example 5 Simplify:

5

2 5 a --- × --3 9

2a b b ------ × --3 9

2 5 2×5 a --- × --- = -----------3 9 3×9 10 = -----27

2a b 2a × b b ------ × --- = ---------------3 9 3×9 2ab = ---------27

Simplify: m n a ---- × --3 4

k m b --- × ---5 3

c

2p q ------ × --3 5

c c

3h 4 ------ × -------7 5m 3h 4 3h × 4 ------ × -------- = -----------------7 5m 7 × 5m 12h = ----------35m

3a 4 d ------ × -----5 7b

2b 5d e ------ × -----3c 7e

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Example 6 Simplify: 8 15 a --- × -----9 16

8a 15 b ------ × ---------9 16b 1

5

1

8 15 8 15 a --- × ------ = -----3 × --------29 16 9 16 5 = --6 4 1

c

12ab 10 ------------- × -----5 3a

c

5

8a 15 8 a 15 - × -----------b ------ × ---------- = -------3 2 9 16b 16 b 9 5a = -----6b 2

12ab 10 12 a b 10 ------------- × ------ = -----------------× ----------1 1 1 5 3a 3 a 5 8b = -----1 = 8b

6

Simplify: 3m 10n a -------- × ---------5 7 5y 9 f ------ × -----3 2y

2k 6n b ------ × -----9 5 7 3z g ------ × -----2z 14

4w 9z ------- × -----3 8 2ab 6 h ---------- × -----3 2b c

8a d ------ × 5 8mn i -----------9

15b ---------16 15 × -------3m

3t 10 e ----- × -----5 9u 6pq 25 j ---------- × -----5 3q

Example 7 Simplify: 2 5 a --- ÷ --3 8

a 5 b --- ÷ --4 b

To divide by a fraction, we multiply by its reciprocal. 5 8 a The reciprocal of --- is --- , hence 8 5 2 5 2 8 --- ÷ --- = --- × --3 8 3 5 16 = -----15

5 b b The reciprocal of --- is --- , hence b 5 a 5 a b --- ÷ --- = --- × --4 b 4 5 ab = -----20

1 = 1 ----15

7

Simplify: a b a --- ÷ --5 7

w z b ---- ÷ --6 2

c

p 5 --- ÷ --4 q

3 n d ---- ÷ --m 8

a c e --- ÷ --b d

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 8 Simplify: 2a 6b a ------ ÷ -----3 7

b 1

2a 6b 2 a 7 a ------ ÷ ------ = --------- × -------3 3 7 3 6 b 7a = -----9b 8

Simplify: 2x 8y a ------ ÷ -----3 5 16 8 e ------- ÷ ------9w 3w 4xy 2x i --------- ÷ -----3 5

3a 6b b ------ ÷ -----2 7 6k 7k f ------ ÷ -----5 2 9 6 j --------------- ÷ -------10km 5m

5pq 3p ---------- ÷ -----8 2 1

b

5p 10q ------ ÷ ---------3 9 4m 2m g -------- ÷ -------3 5 c

B. THE INDEX LAWS The index laws for numbers were established in chapter 2: 1 When multiplying numbers with the same base, we add the indices. For example, 36 × 34 = 36 + 4 = 310 2 When dividing numbers with the same base, we subtract the indices. For example, 36 ÷ 34 = 36 − 4 = 32 3 When raising a power of a number to a higher power, we multiply the indices. For example, (36)4 = 36 × 4 = 324

If we use letters to represent numbers then the rules can be generalised: am × an = am + n am ÷ an = am − n (a m)n = a mn

1

5pq 3p 5p q 2 ---------- ÷ ------ = -----------× --------14 8 2 3p 8 5q = -----12

7 3 d ------ ÷ --------5v 10v 7 m h -------- ÷ ---2m 8

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Example 1 Show by writing in expanded form that: a m4 × m3 = m7

b m5 ÷ m2 = m3

a m4 × m3 = (m × m × m × m) × (m × m × m) =m×m×m×m×m×m×m = m7

c (m4)3 = m12 1

c (m4)3 = (m × m × m × m) × (m × m × m × m) × (m × m × m × m) =m×m×m×m×m×m×m×m×m×m×m×m = m12

Exercise 4B 1

Show by writing in expanded form that: a m 2 × m4 = m6 b m6 ÷ m 2 = m4

c

(m 2)4 = m8

Example 2 a Use a calculator to evaluate the following expressions when a = 3. i a4 × a3 ii a7 b Does the value of a4 × a3 = the value of a7? a i

a4 × a3 = 34 × 33 = 81 × 27

1

m ×m ×m×m×m b m5 ÷ m2 = ----------------------------------------------------1 1 m ×m m×m×m = -------------------------1 3 =m

ii a7 = 37 = 2187

= 2187 b Yes 2

a Use a calculator to evaluate the following expressions when a = 2. ii a9 i a5 × a4 b Does the value of a5 × a4 = the value of a9?

3

a Use a calculator to evaluate the following expressions when m = 5. i m8 ÷ m2 ii m6 8 b Does the value of m ÷ m2 = the value of m6?

4

a Use a calculator to evaluate the following expressions when n = 3. i (n4)2 ii n 8 b Does the value of (n4)2 = the value of n 8?

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 3 Use the index laws to simplify: a y7 × y3

b y 18 ÷ y 17

a y7 × y3 = y7 + 3

b y 18 ÷ y 17 = y 18

= y 10

c – 17

(b5)3

c (b5)3 = b5 x 3

= y1

= b15

=y

5

6

7

8

Use the index laws to simplify: a m3 × m6 b q8 × q7

c t 10 × t 9

d b15 × b × b 4

e v × v5 × v7

Use the index laws to simplify: a a12 ÷ a10 b x15 ÷ x 5

c w8 ÷ w2

d b6 ÷ b 5

e z 20 ÷ z 19

Use the index laws to simplify: a (b4)2 b (h 5)3

c (k 8)2

d (z10)6

e (n 2)4

Use the index laws to simplify: a m4 × m 2 b x9 ÷ x 6 8 7 f n ÷n g b8 ÷ b

c (b4)6 h (y 5)5

d m 3 × m6 × m4 i t10 × t20 × t

e (v 7)10 j a12 ÷ a6

Example 4 Explain why the index laws cannot be used to simplify: a p3 × q4

b m6 ÷ n4

a p3 × q4 = p × p × p × q × q × q × q = p 3q 4 Since the bases are not the same we cannot simplify further. m×m×m×m×m×m b m6 ÷ n4 = -----------------------------------------------------------n×n×n×n 6 m = ------4n Again, since the bases are different we cannot simplify further.

9

Explain why the index laws cannot be used to simplify: a k5 × m 3 b x9 ÷ y6

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10

Are the following statements true or false? a b4 × b 3 = b7

b m 5 × m 2 = m10

c p 4 × p 5 = p 20

d e 6 × e10 = e16

e a4 × b 5 = ab 9

f

z10 ÷ z 2 = z 8 6 4 p p ----2- = ----q q

g p12 ÷ p 3 = p4

h t8 ÷ t7 = t

k (b7)2 = b14

l

i

w15 ÷ w 3 = w 5

j

(n10)3 = n13

C. APPLYING THE INDEX LAWS Example 1 Simplify: 5

6

5

6

5 4

(a ) b ---------------3 2 a ×a

p ×p a ----------------8 p

5 4 5×4 (a ) a ------------b ---------------= 3 2 3+2 a a ×a

5+6

p ×p p a ----------------= ----------8 8 p p 11

20

p = ------8 p = p11 – 8

a = ------5 a = a20 – 5

= p3

= a15

Exercise 4C 1

Simplify: 5

7

3

x ×x a ---------------6 x 7

6

9

a ×a e ---------------8 2 a ×a k -----------------16 5 k ×k

c

11

f

y ×y -----------------10 8 y ×y

j

(m ) × m

30

i

10

w ×w b --------------------8 w

2 3

16

6

4

k ×k d -----------------8 5 k ×k

10

16

2

x h ---------------3 4 x ×x

14

z ×z g -----------------10 7 z ×z 4 5

5

3 4

k (a ) × (a ) 4

10

a ×a ×a n ------------------------------12 8 a ×a ×a

5 5

y ) m (----------20 y

8

m ×m -------------------10 m

20

30

b ×b ×b o -----------------------------------4 5 (b )

Example 2 Simplify: 4

a 5m × 3m

6

b

7

3

2k × 4k × 3k

5

5 6

l

(t ) ----------10 t

6

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

4

5m × 3m

a

6

4

b

= 5×3×m ×m = 15 × m = 15m

2

6

7

3

2k × 4k × 3k 7

5

3

= 2×4×3×k ×k ×k

4+6

= 24 × k

10

= 24k

5

7+3+5

15

Simplify: 5

a 4m × 3m d 10a

12

7

× 7a

6

b 4

8

g 3z × 4z × 2z

4

6

c

3t × 6t

9

10

f

5b × 6b × b

i

d × 6d × 3d

5p × 2p

e 4w × 6w 3

5

7

h 2q × 5q × 8q

6

8

4

3

4

2

6

4 8

Example 3 Simplify: 8

10

12m a ------------6 3m

b 8

a

10

12m ------------6 3m

b

20a -------------4 16a

12 × m = ------------------6 3×m

8

20 × a = -------------------4 16 × a

10

8

20 a = ------ × ------16 a 4

10

12 m = ------ × ------63 m = 4×m = 4m

3

20a -------------4 16a

6 5 = --- × a 4

2

6

5a = --------4

2

Simplify: 7

6m a ----------23m 10

f

9e ----------66e

12

10a b -------------7 5a 8

2m g ----------36m

10

c

12w --------------8 4w

12

8z d ----------86z

15

6a h -------------10 12a

9

16k e -----------312k

13

i

9t ----------612t

11

j

15b -------------6 20b

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 4 3 5

Simplify ( 2a ) 3 5

( 2a ) 3

3

3

3

= 2a × 2a × 2a × 2a × 2a 3

3

3

3

3

= 2×2×2×2×2×a ×a ×a ×a ×a

3

3 5

5

= 2 × (a ) = 32a

4

15

Simplify: 4 3

b ( 2m )

3 2 5

g (m n )

a ( 3a )

(x y )

f

3 6

c ( 7p )

5 2

4 6 3

h (p q )

2 4

d ( 10k )

7 3 4

i

4 10 2

(a b )

Example 5 Simplify: 2 4

10 8

5 8

a 5m n × 3m n

a

2 4

b

10 8

5 8

5m n × 3m n 2

4

5

7

7 12

= 15m n

12

10

8

5

8

y 12 x = ------ × ------6- × ----28 x y

4

8

=

= 5×3×m ×m ×n ×n = 15 × m × n

12x y ------------------6 2 8x y

b

= 5×m ×n ×3×m ×n 2

12x y ------------------6 2 8x y

3 --2

× x4 × y6 4 6

3x y = -------------2

11 3

e ( 5t ) j

5 2 3

( 2x y )

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

5

Simplify: 3 2

5 3

b 5m n × 2m n

5 8

6 7

d 10x y × 3x y

a 4a b × 2a b

c 3p q × 4p q 10 12

e 2w z

4 5

× 6w z

10 9

4

4 3

6

5 6

f

6a b -------------3 2 4a b 7

15x y g ------------------6 2 5x y

Remember to add indices when multiplying and subtract them when dividing.

12

2k m h ------------------3 6 10k m

11 6

i

6 7

7 8

9a b ----------------8 12a b

j

12m n ------------------6 8 15m n

D. THE ZERO INDEX Example 1 a Use the index laws to simplify a5 ÷ a5. b Hence show that a0 = 1. a Using the index laws, a5 ÷ a5 = a5 − 5 = a0 b But a5 ÷ a5 = 1 (Since any number divided by itself = 1) Hence a0 = 1

Exercise 4D 1

a Use the index laws to simplify a4 ÷ a4.

b Hence show that a0 = 1.

2

a Use the index laws to simplify k 7 ÷ k 7.

b Hence show that k0 = 1.

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Example 2 Evaluate: a x0

b (3x)0

c 3x 0

a x0 = 1

b (3x)0 = 1

c 3x 0 = 3 × x 0 =3×1 =3

3

Evaluate: a y0 f (6z)0 k 3m0 + 1

b (3y)0 g (10m)0 l 9e0 – 3

c 3y 0 h 10m0 m 6p0 + 7

d 4k 0 i 8b0 n 3a0 + 2b0

e 9t 0 j (7q)0 o 6x 0 – 4y 0

E. INDICES FOR SQUARE ROOTS AND CUBE ROOTS Example 1 2

a Show that ( 6 ) = 6. 1 --2 2

b Use the index laws to show that ( 6 ) = 6. 1 --2

c Hence show that a ( 6)

2

6 = 6 . 1 --2

1 --- × 2 2

=

6 ×

=

6×6

= 61

=

36

=6

6

b

( 6 )2 = 6

=6 1 --2

2

c Since ( 6 ) = 6 and ( 6 )2 = 6 then

1 --2

6 =6 .

Exercise 4E 1

c Hence show that 2

1 --2

a Show that ( 5 )2 = 5.

b Use the index laws to show that ( 5 )2 = 5. 1 --2

5 =5 . 1 --2

a Show that ( a )2 = a. c Hence show that

b Use the index laws to show that ( a )2 = a. 1 --2

a =a .

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 2 Write in index form: a

23

a

23 = 23

1 --2

b

t

b

t =t

1 --2

c

7t

d 7 t

c

7t = ( 7t )

1 --2

d 7 t =7×

t

=7× t = 7t

3

Write in index form: a 3 5k e

12 b f 5 k

x c g 6 y

1 --2

1 --2

m 6y

d h

Example 3 Write down the meaning of: a 5

1 --2

b ( 7z )

1 --2

a 5 =

1 --2

1 --2

b ( 7z ) =

5

7z

c

7z

c

7z

1 --2

1 --2

=7× z =7×

1 --2

z

=7 z

4

Write down the meaning of: a 8

1 --2

e ( 3z )

b 13 1 --2

f

1 --2

( 2m )

c p 1 --2

1 --2

g 5k

d q 1 --2

1 --2

h 4t

1 --2

Example 4 3

a Show that ( 3 5 ) = 5. 1 --3

b Use the index laws to simplify ( 5 )3. c Hence show that

3

1 --3

5 =5 .

☞

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3

1 --- × 3 3

3

5 ×

=

3

5×5×5

= 51

=

3

125

=5

3

5 ×

1 --3

a (3 5) =

3

b ( 5 )3 = 5

5

=5 1 --3

3

c Since ( 3 5 ) = 5 and ( 5 )3 = 5 then

5

1 --3

5 =5 . 1 --3

3

a Show that ( 3 6 ) = 6. c Hence show that

6

3

3

b Use the index laws to simplify ( 6 )3. 1 --3

6 =6 . 1 --3

3

a Show that ( 3 a ) = a. c Hence show that

3

b Use the index laws to simplify ( a )3. 1 --3

a =a .

Example 5 Write in index form: a

3

7

a

3

7

=7

1 --3

b

3

e

b

3

e

=e

c

3

4z

c

3

4z

1 --3

d 43 z d 43 z = 4 ×

= ( 4z )

1 --3

=4× z = 4z

7

3

z 1 --3

1 --3

Write in index form: a

3

2

b

3

9

c

3

c

d

3

e

3

5y

f

53 y

g

3

9m

h 93 m

w

Example 6 Write down the meaning of: a 8

1 --3

b n

1 --3

c

( 5x )

1 --3

d 5x

1 --3

☞

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

a 8

1 --3

b n

3

=

8

=

1 --3

3

c n

( 5x ) =

3

1 --3

d 5x

1 --3

=5×x =5×

5x

3

1 --3

x

= 53 x

8

Write down the meaning of: a 12

1 --3

b 35

e ( 6m ) 9

c k

1 --3

f

6m

47

b

p

e 5 x

f

63 x

1 --3

g ( 7v )

d d 1 --3

1 --3

h 7v

1 --3

Write in index form: a

10

1 --3

1 --3

c

3

d 83 p

x

h 33 r

g 4 x

Write down the meaning of: a 5 e

1 --3

b 6

( 5p )

1 --2

f

1 --2

( 6r )

c x 1 --3

1 --3

g ( 5xy )

d t 1 --2

h ( 4pq )

F. NEGATIVE INDICES Example 1 4

a a Use the index laws to simplify ----5- . a 1 –1 c Hence show that a = --- . a 4

4

= a–1 4

4

a b Expand and simplify ----5- . a

1

1

1

1

a a ×a ×a ×a b ----5- = -------------------------------------------------1 1 1 1 a ×a ×a ×a ×a a 1 = --a

a a ----5- = a4 – 5 a

4

a a 1 1 c Since ----5- = a–1 and ----5- = --- then a–1 = --- . a a a a

1 --2 1 --3

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Exercise 4F 3

1

2

3

4

5

3

a a Use the index laws to simplify ----4- . a 1 c Hence show that a–1 = --- . a 3 a a Use the index laws to simplify ----5- . a 1 –2 c Hence show that a = ----2- . a 2 a a Use the index laws to simplify ----5- . a 1 –3 ---c Hence show that a = 3 . a 2 a a Use the index laws to simplify ----6- . a 1 c Hence show that a–4 = ----4- . a a a Use the index laws to simplify ----6- . a 1 c Hence show that a–5 = ----5- . a

a b Expand and simplify ----4- . a 3

a b Expand and simplify ----5- . a 2

a b Expand and simplify ----5- . a 2

a b Expand and simplify ----6- . a a b Expand and simplify ----6- . a

From the above exercises, it can be seen that, in general, 1 a–n = ----n a

Example 2 Write down the meaning of:

6

a k–9

b m–15

1 a k–9 = ----9k

1 b m–15 = -------15 m

Write down the meaning of: a y −2 b k –1

c m–3

Example 3 Write with a negative index: 1 a ----5a

1 b ----7y

1 a ----5- = a–5 a

1 b ----7- = y –7 y

d x –6

e t –10

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

7

Write with a negative index: 1 1 a ----8b ----2k a

c

1 -----11 x

1 d ------14 n

1 e -----20 z

Example 4 Write down the meaning of: a 3m–2

b (3m)–2

a 3m–2 = 3 × m–2

1 b (3m)–2 = ---------------2 ( 3m ) 1 = ----------29m

3 1 = --- × ------21 m 3 = ------2m

8

Write down the meaning of: a 3k –1 b (3k)–1 –5 d (2y) e 3t –4

2y –5 (3t)–4

c f

G. FURTHER USE OF THE INDEX LAWS Example 1 Use the index laws to simplify: 1 --2

1 --2

1 --2

1 --2

a 3y × 2y

1 --3

1 --3

1 --3

1 --3

b 10 n ÷ 5 n 1 --2

a 3y × 2y = 3 × y × 2 × y 1 --2

=3×2× y × y =6× y

1 1 --- + --2 2

= 6 × y1 = 6y

1 --2 1 --2

1 --3

10n b 10 n ÷ 5 n = -----------1 5n

--3

1 --3

10 n = ------ × -----1 5 --3 n =2× n

1 1 --- – --3 3

= 2 × n0 =2×1 =2

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Exercise 4G 1

Use the index laws to simplify: 1 --2

a 2z × 6z d 12 a

1 --3

1 --2

1 --2

b 8p ÷ 2p

÷4a

1 --3

1 --2

1 --3

1 --3

2m × 3m × 4m

c

1 --3

3

k e ---------------1 1 --2

k ×k

--2

Example 2 Use the index laws to simplify: a m–6 × m2

b q–2 ÷ q−7

c (x –3)5

a m–6 × m2 = m–6 + 2

b q−2 ÷ q –7 = q–2 – (−7)

c (x –3)5 = x –3 × 5

= m–4 2

= q5

Use the index laws to simplify: –5 –2 –3 7 a a ×a b y ×y e b i

–6

÷b

2

–2 4

(y )

3

f

w ÷w

j

(t )

–2

= x –15

5

c e ×e g z

–2

–7

÷z

4

d n ×n

–4

h k

5 –4

Example 3 Simplify: a 5m–3 × 6m7

b 4y 7 ÷ 5y –2

a 5m–3 × 6m7 = 5 × 6 × m–3 × m7 = 30 × m–3 + 7 = 30 × m4 = 30m4

c (5y –2)3 = 5y –2 × 5y –2 × 5y –2 = 5 × 5 × 5 × y –2 × y –2 × y –2 3

–2 3

= 5 × (y ) = 125y –6

c (5y –2)3 7

4y b 4y 7 ÷ 5y –2 = ---------–2 5y 7 4 y = --- × -----5 y –2 4 = --- × y 7–(–2) 5 4 = --- × y 9 5 9 4 4y = --- y 9 or -------5 5

–6

–3

÷k

–2

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

3

Simplify: a 10a 5 × 9a –3 e 6p–4 ÷ 2p 2 i (3w –6)2

b 6b–5 × 3b–2 f 3k –4 ÷ 8k –2 j 4n–3 × 3n–4 ÷ 6n–5

c 3v –6 × 2v 2 g (5z –4)3

d 8y 5 ÷ 2y –1 h (2m–3)5

Example 4 State whether the following are true or false. a m3 ÷ m5 = m2

b 3y 0 = 1

1 d 2p–3 = --------32p

c 6k4 ÷ 2k4 = 3

a m3 ÷ m5 = m3 – 5

b 3y 0 = 3 × y 0 =3×1

= m–2 Statement is false.

=3 Statement is false.

4

6 k c 6k 4 ÷ 2k 4 = --- × ----42 k = 3 × k4 – 4

d 2p–3= 2 × p–3 1 = 2 × ----3p 2 = ----3p Statement is false.

= 3 × k0 =3×1 =3 Statement is true.

4

State whether the following are true or false. a 6m 0 = 1

b a4 ÷ a7 = a 3

c

8t 9 ÷ 2t 9 = 4

1 d 3c −2 = --------23c 6 g 5x ÷ x 6 = 5x

e 4k 0 = 4

f

b ÷ b6 = b5

4 h 4y −3 = -----3 y

i

8 (2p −1)3 = ----3p

Example 5 By substituting a = 5, show that a–2 ≠ −2a.

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1 a−2 = ----2a 1 = -----2 5 1 = -----25

–2a = −2 × 5 = –10

Hence a–2 ≠ –2a 5

By substituting a = 3, show that: a a 2 ≠ 2a d a–3 ≠ –3a g a2 – a2 ≠ a0

b a3 ≠ 3a e a2 × a ≠ a2 + a h 5a2 × 3a ≠ 5a2 + 3a

c f

a–2 ≠ –2a a2 + a2 ≠ a4

H. REMOVING GROUPING SYMBOLS Example 1 Expand: a 3(x + 5)

b 4(3y – 2z)

Expand means to write the expression without the grouping symbols.

a 3(x + 5) = (x + 5) + (x + 5) + (x + 5) =x+x+x+5+5+5 =3×x+3×5 = 3x + 15 b 4(3y – 2z) = (3y – 2z) + (3y – 2z) + (3y – 2z) + (3y – 2z) = 3y + 3y + 3y + 3y – 2z – 2z – 2z – 2z = 4 × 3y – 4 × 2z = 12y – 8z

From the examples above we can see that, in general: a × (b + c ) = a × b + a × c that is, to remove the grouping symbols we multiply each term inside them by the number at the front. This result is known as the distributive law.

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 2 Use the distributive law to expand: a 5(2y + 3)

b 7(3y – 4w)

a 5(2y + 3) = 5 × 2y + 5 × 3

b 7(3y – 4w) = 7 × 3y – 7 × 4w

= 10y + 15

= 21y – 28w

Exercise 4H 1

Use the distributive law to expand: a 3(2w + 5) b 6(3z – 2) 2 e 10(x + 6) f 7(ab – 2a 2) i 5(4b + 2a + 3) j 3(5x – 3y – 2z)

c 5(4a + 3b) g 4(m 2 + n 2)

d 2(4x – 3y) h 2(m 3 – 3mn)

Example 3 Expand: a 3w(2y + 4z)

b 2a(3a – 4b)

a 3w(2y + 4z) = 3w × 2y + 3w × 4z

c 4m2(m3 + 2m5)

b 2a(3a – 4b)= 2a × 3a – 2a × 4b = 6a2 – 8ab

= 6wy + 12wz c 4m 2(m 3 + 2m5) = 4m 2 × m3 + 4m 2 × 2m5 = 4m5 + 8m7

2

Expand: a 3a(2b + 4c) e y2(y3 – 4) i 2p5(p2 + 3p3)

b 4x(3x – 2y) f 6x(2y – 5x2) j 5x2(2x3 – 3xy)

c 10k(6k – 4m) g 3k2(2k2 + 5)

d m(m2 + 2) h a3(5a2 – 2)

Example 4 Expand: a –3(2w + 5)

b –2(4a – 3b)

c –(4m + 3n)

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a –3(2w + 5) = –3 × 2w + –3 × 5

b –2(4a – 3b) = –2 × 4a – –2 × 3b

= –6w + –15

= –8a – –6b

= –6w –15

= –8a + 6b

c –(4m + 3n) = –1 × 4m + –1 × 3n = –4m + –3n = –4m – 3n

3

Expand: a –2(y + 3) e –(t + 3) i –(7w + 3)

b –5(a + 2) f –(b + 6) j –(4x – 1)

c –3(w + 4) g –3(2k + 5)

d –4(m – 7) h –2(4m – 5)

Example 5 Expand:

4

a –3a(5a2 + 2ab)

b –n3(2n4 – 5n2p)

a –3a(5a2 + 2ab) = –3a × 5a2 + –3a × 2ab

b –n3(2n4 – 5n2p) = –n3 × 2n4 – –n3 × 5n2p

Expand: a –2a(3a 2 + 2ab) d –y3(4y2 – 3xy)

= –15a3 + –6a2b

= –2n7 – –5n5p

= –15a3 – 6a2b

= –2n7 + 5n5p

b –4x(2x 2 – 3xy) e –3m4(2m 2 + 5mn)

Example 6 Expand and simplify by collecting like terms. a 3(a + 2) + 7

b 3 + 2(3n – 5)

a 3(a + 2) + 7 = 3a + 6 + 7 = 3a + 13

Remember to multiply before adding!

b 3 + 2(3n – 5) = 3 + 6n – 10 = –7 + 6n or 6n – 7

c

–3p 2(3p 2 + 4pq)

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

5

Expand and simplify: a 4(a + 3) + 6 e 6(3z – 1) + 4 i 4 + 3(2w – 4)

b 2(3b – 12) + 12 f 10 + 2(4x + 3) j 16 + 5(4e – 6)

c 3(4w + 2) – 7 g 12 + 2(3b – 5)

d 5(2y – 3) – 2 h 13 + 4(y + 5)

Example 7 Expand and simplify 5 – 2(4y – 3). 5 – 2(4y – 3) = 5 – 8y + 6 = 11 – 8y or – 8y + 11 6

Expand and simplify: a 12 – 2(a + 5) e 20 – 3(2w + 5) i 5 – 3(3 + 4z)

b 8 – 3(y – 2) f 2 – 5(3t – 4) j 3 – 10(1 – 2w)

c 9 – 4(b + 3) g 4 – 3(5x + 2)

d 7 – 2(v – 6) h 10 – 2(3k – 1)

Example 8 Expand and simplify 4(2m – 3) + 3(m – 2). 4(2m – 3) + 3(m – 2) = 8m – 12 + 3m – 6 = 11m – 18 7

Expand and simplify: a 5(2k + 3) + 3(k – 2) c 4(2p – 1) + 2(3p + 5) e 2(5x – 3) + 5(3x – 1) g (6v – 1) + 3(2v – 5) i 7(2a – 3b) + 3(3a + 4b)

b d f h j

2(6m + 7) + 3(m – 1) 3(3a + 2) + 4(a – 3) 3(4y – 2) + (2y + 7) 4(3x + 2y) + 2(5x – 3y) 3a(2a + 6) + 4a(3a – 5)

Example 9 Expand and simplify: a 2(3p + 4q) – 4(2p – 3q)

b 3(4m – 1) – (m + 4)

a 2(3p + 4q) – 4(2p – 3q) = 6p + 8q – 8p + 12q = –2p + 20q or 20q – 2p b 3(4m – 1) – (m + 4) = 12m – 3 – m – 4 = 11m – 7

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8

Expand and simplify: a 3(2k + 5) – 2(k + 3) c 2(6t + 1) – 3(t + 4) e 2(a + 5) – 4(a – 1) g 4(3x + y) – (2x – 7y) i 2q(q – 5) – 4(q – 5)

b d f h j

5(w + 4) – 3(w – 2) 3(5z – 1) – (2z + 5) 5(d – 3) – 3(2d + 1) 3(2a – 3b) – (2a + 3b) 4z(3z + 2) – (z – 1)

I. FACTORISING Use the distributive property to expand 2(3a + 5) = 2 × 3a + 2 × 5 = 6a + 10 Reversing the process, 6a + 10 = 2 × 3a + 2 × 5 = 2 × (3a + 5) = 2(3a + 5) The reverse process to expanding is called factorising.

Note that the 2 is the HCF of 6a and 10.

Example 1 Factorise: a 3y + 12

b 20k – 8

a The HCF of 3y and 12 is 3, hence

b The HCF of 20k and 8 is 4, hence

3y + 12 = 3 × y + 3 × 4

20k – 8 = 4 × 5k – 4 × 2

= 3 × (y + 4)

= 4 × (5k – 2)

= 3(y + 4)

= 4(5k – 2) Note that as 2 is a common factor of 20k and 8, 20k – 8 = 2 × 10k – 2 × 4 = 2(10k – 8) While this is a correct equivalent expression for 20k – 8, it has not been fully factorised. 20k – 8 = 4(5k – 2).

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Exercise 4I 1

Factorise: a 8a + 10 e 4w – 12 i 24k – 18n

b 6x – 4 f 16m – 8 j 16x 2 + 24y2

c 3a + 6b g 12ab + 8

d 5x + 10y h 10m – 20n

Example 2 Factorise: a 3u + 12v + 9w

b 12a – 8b + 20c

a The HCF of 3u, 12v and 9w is 3, hence 3u + 12v + 9w = 3 × u + 3 × 4v + 3 × 3w = 3(u + 4v + 3w) b The HCF of 12a, 8b and 20c is 4, hence 12a – 8b + 20c = 4 × 3a – 4 × 2b + 4 × 5c = 4(3a – 2b + 5c)

2

Factorise: a 5x + 15y + 10z d 12m – 6n – 18r

b 3p – 6q + 9r e 20xy + 50z + 30

c

4a + 12b – 8c

Example 3 Complete the following factorisations and check by expanding. a y 2 + 2y = y (__ + __)

b 12a2 – 8ab = 4a (__ – __)

a Since y 2 + 2y = y × y + y × 2, then

b

y 2 + 2y = y(y + 2) Check: y(y + 2) = y × y + y × 2 = y 2 + 2y

3

Since 12a 2 – 8ab = 4a × 3a – 4a × 2b, then 12a 2 – 8ab = 4a(3a – 2b) Check: 4a(3a – 2b) = 4a × 3a – 4a × 2b = 12a 2 – 8ab

Complete the following factorisations and check by expanding. a y 2 + 7y = y (__ + __) b m 2 – 3m = m (__ – __) c 3mn + 4m = m (__ + __) d 9p – 5pq = p (__ – __) 2 f 2bc – b 2 = b (__ – __) e x + 5xy = x (__ + __)

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g 6m 2 – 3m = 3m (__ – __) i 16pq – 12p 2 = 4p (__ – __)

h 12a 2 – 10ab = 2a (__ – __) j 12k 2 +18k = 6k (__ + __)

Example 4 Complete the following factorisations and check by expanding. a y2 + 7y = __ (y + 7)

b 2m2 – 8m = __ (m – 4)

a The HCF of y2 and 7y is y, hence

b The HCF of 2m2 and 8m is 2m, hence

y2 + 7y = y(y + 7)

2m2 – 8m = 2m(m – 4)

Check: y(y + 7) = y × y + y × 7

Check: 2m(m – 4) = 2m × m – 2m × 4

= y2 + 7y

4

= 2m2 – 8m

Complete the following factorisations and check by expanding. a p2 + 3p = __ (p + 3) b k2 – 2k = __ (k – 2) 2 c 3w + 2w = __ (3w + 2) d 2z2 – z = __ (2z – 1) e 4mn – 3m2 = __ (4n – 3m) f 2x2 + 8x = __ (x + 4) 2 g 4pq – 12p = __ (q – 3p) h 8pq – 12pr = __ (2q – 3r) 2 i 10z – 5z = __ (2z – 1) j 6km – 8m2 = __ (3k – 4m)

Example 5 Explain why the following factorisation is incorrect. 18x2y + 12xy2 = 6xy(3x + 2) 6xy (3x + 2) = 6xy × 3x + 6xy × 2 = 18x2y + 12xy ≠ 18x2y + 12xy2

5

Explain why the following factorisations are incorrect. a y2 + 7y = 7(y2 + y) b 24ab + 8a = 8a(3b – 1) d 15x2y + 12xy 2 = 3xy(5x + 4) c x2 − 6xy = 6x(x − y) e 10ab2 – 5a2b = 5ab(2b2 – a)

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

Example 6 Factorise: a w 2 + 5w

b 3y 2 – 6y

a The HCF of w 2 and 5w is w, hence

c 4a 2 + 8ab b The HCF of 3y 2 and 6y is 3y, hence

w 2 + 5w = w × w + w × 5

3y 2 – 6y = 3y × y – 3y × 2

= w(w + 5)

= 3y(y – 2)

c The HCF of 4a 2 and 8ab is 4a, hence 4a 2 + 8ab = 4a × a + 4a × 2b = 4a(a + 2b)

6

Factorise: a x 2 + 4x e 2k 2 + 4k i 12pq – 18p 2

b y 2 – 7y f 3y 2 – 12y j 16km + 24k 2

c a 2 + ab g 10b 2 + 5ab

d m 2 – 5mn h 9w 2 – 6w

Example 7 Factorise 18ab – 3a + 9a2. The HCF of 18ab, 3a and 9a 2 is 3a, hence 18ab – 3a + 9a 2 = 3a × 6b – 3a × 1 + 3a × 3a = 3a(6b – 1 + 3a)

7

Factorise: a 2ab + 4a + 4a 2 d 5xy – 10x – 5x 2

b 6x 2 + 3x + 9xy e 8k 2 – 6k – 10km

c

2m – 6m 2 + 4mn

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Language in Mathematics

1

Three of the words in the following list have been spelt incorrectly. Find these words and write the correct spelling. reduse, simplify, subistute, aply, numerator, equivalent

2

Complete the following words used in this chapter by replacing the vowels: a f __ ct __ r b z __ r __ c __ nd __ x d __ lg __ br __ __ c e f __ ct __ r __ s __

3

Using an example, explain the meaning of: a reciprocal b highest common factor

4

Write in words:

c

x3

5

Write down the names of the following grouping symbols: a () b []

c

6

7

a

x2

b

x

d

3

x

{}

Match each word with its meaning. a equivalent

A go backwards

b expand

B the same as

c reverse

C a letter used to represent numbers

d evaluate

D remove the grouping symbols

e pronumeral

E find the value of

How many words of three or more letters can you make from the word DISTRIBUTIVE. (No proper names or plurals allowed.)

Glossary algebraic

apply

base

cancel

check

common

cube root

denominator

distributive law

equivalent

evaluate

expand

expression

factor

factorise

grouping symbols

index

indices

negative

numerator

power

reciprocal

reduce

reverse

simplify

square root

substitute

value

zero

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

121

CHECK YOUR SKILLS 1

Which of the following is not true? 3x 9x A ------ = -----5 15

2

3

7

5a C -----81

45a D ---------9

11x B --------15

11x C --------8

3x D -----15

45a B ---------21b

28ab C ------------15

20ab D ------------21

xy C -----6y

x D --6

5x B -----3y

5y C -----3x

xy D -----15

14a B ---------15

35ab C ---------------54

x y --- ÷ --- = 5 3

7ab 5b ---------- ÷ ------ = 9 6

2

54 D ---------------235ab

Which of the following does not simplify to k15? A (k5)3 7

10

5a B -----18

5 3xy When simplified ------ × --------- = 9y 10 15xy 15x A -----------B --------90 90

15 A ---------14a 9

3xy D --------2x

4a 5 ------ × ------ = 3 7b

3x A -----5y 8

✓

3y C -----2

2x x ------ + --- = 5 3

20a A ---------21b 6

3x 6x D ------ = -----5 8

4a a ------ + --- = 9 9

3x A -----8 5

3x 15x C ------ = --------5 25

9xy When reduced to its simplest form --------- = 6x 9y A -----B 3y 6

5a A -----9 4

3x 6x B ------ = -----5 10

B k

10

×k

5

C k

30

÷k

2

3 5

D (k )

9

b ×b ----------------= 2 4 (b ) A b55

B b2

C b8

D 18

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9 8

11

12x y ----------------= 6 2 8x y 3 6

A 4x3y6 12

17

18

1 --2

20

21

22

D 9m7

B 20

C 3

D 8

B ( 9a )

1 --2

C 3a

1 --2

The meaning of 2a is: 3

2a C -----3

B 23 a

2a

1 Which of the following is equivalent to ------2- ? m 1 A -------B m–2 2m

2 C ---m

2c–2 is equivalent to: 2 A ----2c

B −4c

1 C --------22c

3xy -----------2

B 3m–2

1 C --- m2 3

✓ D 9a

2

D 2a

3

D m

6m –3 ÷ 2m–5 = A 3m2

19

C 3m7

1 --3

A 16

B 9m10

9 a may be written in index form as: A 9a

15

D

5y 0 + 3 = A 4

14

9

3x y C -------------2

(3m5)2 = A 3m10

13

3 4

3x y B -------------2

−3k(2k 2 – 4km) =

1 --2

1 D --------24c

1 D --- m–2 3

A –6k 2 + 12km

B

–6k 2 – 12km

C –6k 3 + 12k 2 m

D –6k 3 + 12k 2 m

When expanded and simplified 2w(w – 6) + 3(w – 5) = B 2w2 – 9w – 15 C –7w – 15 A 2w2 – 3w – 15 The highest common factor of 10pq and 12p2 is: A 2p B 2

C p

When fully factorised 4x2 – 6x = A 2x2 (2x2 – 3) B 2(2x2 – 3x)

C x(4x – 6)

D –7w2 – 15

D 2pq

D 2x(2x – 3)

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1–8

9, 10

11, 12

13

Section

A

B

C

D

14, 15 16, 17 E

F

18 G

19, 20 21, 22 H

I

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

REVIEW SET 4A

2

Reduce to simplest form: 12a 9xy a ---------b --------15 6x

3

Simplify: 4a 7a a ------ + -----13 13

2m 5m b -------- + -------3 8

Simplify: a y 10 × y 7 e (5m4 )3

b k 11 ÷ k 5 f 3a 5 b 3 × 2ab 6

c (p7 )2

t ×t d -------------3 4 t ×t

Evaluate: a v0

b 5v 0

c (5v)0

d 2v 0 + 1

5

6

8

9

10

11

12

c

4w 2w ------- × ------5 3

5x y d ------ ÷ --6 3 7

8

Write the meaning of: a x

7

b

3mn ■ ------------ = -----4 20

Complete:

4

a

2x ■ ------ = -----9 18

1

1 --2

b 3x

1 --2

c ( 3x )

Write the meaning of: a z –3

b 2z –3

Simplify: a y –3 × y5 d 6b–2 × 3b7

b e6 ÷ e–2 e 4k–5 ÷ 2k–3

1 --2

State whether the following are true or false. a 4q 0 = 4 b a 5 ÷ a7 = a2 4 –2 d 4b = ----2e n2 × n = n2 + n b Expand: a 5(2v – 4w) b a3 (2a2 + 4a) Expand and simplify: a 4(m – 2) + 3(2m + 5)

b 3(3a – b) – (2a – b)

Factorise: a 8w + 20

b x2 + 9x

d x

1 --3

c (2z)–3

c

(n–4 )5

c

6m5 ÷ 3m5 = 2m

c

–3(4x + 5)

c

4pq – 12q2

e 2x

1 --3

123

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REVIEW SET 4B 3x ■ a ------ = -----7 21

Complete:

2

Reduce to simplest form: 8m 4ab a -------b ---------12 2a

3

Simplify: 2a 7a a ------ + -----9 9

3y 2y b ------ + -----4 3

Simplify: a m14 × m6 e (2m7 )4

b t25 ÷ t5 f 4p3 q7 × 5p4q

c (z6 )4

(b ) d ----------------7 4 b ×b

Evaluate: a s0

b 4s 0

c (4s)0

d 4s 0 − 1

4

5

6

8

9

10

11

12

c

3k 2m ------ × -------5 3

3w 2w d ------- ÷ ------5 3 5 6

Write the meaning of: a c

7

b

2ab ■ ---------- = -----3 18

1

1 --2

b 2c

1 --2

c ( 2c )

Write the meaning of: a e–4

b 3e–4

Simplify: a k –6 × k2 d 5n−3 × 4n8

b m–4 ÷ m–1 e 2a−5 ÷ 4a−8

1 --2

State whether the following are true or false. b b6 ÷ b9 = b–3 a 3w 0 = 1 1 d 2t –2 = -------2e p –2 = –2p 2t Expand: a –10(4p + 3) b m2 (3m5 − m3) Expand and simplify: a 3(2q + 6) − 2(q − 7)

b 2a(2a – 5) – (3a + 1)

Factorise: a 12x − 18

b 2y2 − 7y

d c

1 --3

c (3e)–4

c

(n3)–5

c

a7 ÷ a7 = a

c

–a(2a − 5)

c

4a2 – 3ab + 2a

e ( 2c )

1 --3

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Algebraic Techniques (Chapter 4) Syllabus reference PAS5.1.1, 5.2.1

REVIEW SET 4C

2

Reduce to simplest form: 8w 5mn a ------b -----------24 7n

3

Simplify: 2d 6d a ------ + -----7 7

7b 3b b ------ – -----8 4

Simplify: a p6 × p8 e (3v7 )3

b y16 ÷ y 8 f 3x8 y9 × 6x2y5

c (t5 )6

c ×c d -----------------4 5 (c )

Evaluate: a a0

b 7a0

c (7a)0

d 7a0 + 5

5

6

8

9

10

11

12

c

6x 10y ------ × --------5 9

3w 2w d ------- ÷ ------4 5 12

8

Write the meaning of: a q

7

b

4xy ■ --------- = -----3 12

Complete:

4

a

3h ■ ------ = -----5 20

1

1 --2

b 4q

1 --2

c ( 4q )

1 --2

d q

Write the meaning of: a b −5

b 3b −5

c (3b)−5

Simplify: a d –5 × d –3 e 6m3 ÷ 9m–4

b n –2 ÷ n–3

c (k –2)–3

State whether the following are true or false. a 7h0 = 7 b a ÷ a5 = a−5 3 −4 d 3s = ----4e n2 + n2 = n4 s Expand: a 4(5w + 2x) b –k4 (2k3 − 4k) Expand and simplify: a 2(5t − 1) + 3(2 − 3t)

b 10(x – 2y) – 5(2x − y)

Factorise: a 15n − 20

b 4b2 + 6b

1 --3

c

1 2p5 ÷ 6p5 = --- p 3

c

–(4s − 7)

c

12m2 + 14mn

e 4q

1 --3

d 5a–6 × 3a3

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REVIEW SET 4D

2

Reduce to simplest form: 20h 4pq a ---------b ---------16 6q

3

Simplify: 10k 5k a ---------- – -----11 11

2w w b ------- – ---9 6

Simplify: a y8 × y5 e (9h5 )2

b k16 ÷ k10 f 6a7 b8 × a6b 3

c (p5 )10

c ×c d -----------------4 2 (c )

Evaluate: a w0

b 8w 0

c (8w) 0

d 8w 0 − 9

5

6

8

9

10

c

3a 4b ------ × -----2 5

9x 4x d ------ ÷ -----10 3 3

13

Write the meaning of: a m

7

b

7pq ■ ---------- = -----2 10

Complete:

4

a

5b ■ ------ = -----8 24

1

1 --2

b 6m

1 --2

c ( 6m )

Write the meaning of: a x –2

b 2x –2

Simplify: a y –2 × y –5 d 5z−4 × 3z7

b n–6 ÷ n–8 e 6m−4 ÷ 9m−7

1 --2

State whether the following are true or false. b a9 ÷ a7 = a–2 a 2b0 = 1 7 d 7n–3 = ----3e (2m)2 = 4m2 n Expand: a 2x(6y − 3z) b –m3(3a3 − 4m2)

11

Expand and simplify: a 5(3 + 5n) + 2(4 − 3n) b a(6a + 1) – 2a(a + 1)

12

Factorise: a 24x + 18

b h2 − 8h

d m

1 --3

c (2x)–2

c

(p–3)5

c

2p6 ÷ 10p6 = 2

c

–2(5 − 2a)

c

3y2 + 6y − 9

e ( 6m )

1 --3

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Chapter 5 Rational Numbers This chapter deals with rounding numbers to a specified number of significant figures, expressing recurring decimals as fractions and converting rates from one set of units to another. After completing this chapter you should be able to: ✓ identify significant figures ✓ round numbers to a specified number of significant figures ✓ use the language of estimation appropriately ✓ use symbols for approximation ✓ determine the effect of truncating or rounding during calculations on the accuracy of results ✓ write recurring decimals as fractions ✓ convert rates from one set of units to another.

Syllabus reference NS5.2.1 WM: 5.2.2–5.2.4

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Diagnostic Test

1

The value of the digit 8 in the number 136.832 is:

8

When rounded to the nearest hundred 23 629 is: A 236 B 237 C 23 600 D 23 700

Which of the following numbers are not equal to 1490 when rounded off to 3 significant figures? A 1486.23 B 1493.99 C 1495 D 1485

9

When 59 700 is rounded to the nearest thousand the answer is: A 59 000 B 59 800 C 591 000 D 60 000

When rounded to 2 significant figures, 3.968 is: A 3.97 B 4.0 C 3.9 D 4

10

Which of the following numbers do not have 3 significant figures? A 6.931 B 3.60 C 245 D 35.7

11

Rounding numbers before the last step of a calculation:

A 80 2

3

4

5

6

7

B 8

C

8 -----10

D

8 ---------100

Which of the following numbers are not equal to 52.39 when rounded off, correct to 2 decimal places? A 52.387 B 52.392 C 52.385 D 52.395

A never affects the accuracy of the answer

When a number is rounded off to the nearest 10 the answer is 60.

B often affects the accuracy of the answer

The smallest the number could be is: A 55 B 55.1 C 59 D 59.99

C always affects the accuracy of the answer

Janelle was measured to be 162 cm tall, to the nearest centimetre. Within what range of values does her actual height lie? A 161 ≤ height < 163 cm B 161 < height < 163 cm C 161.5 ≤ height < 162.5 cm D 161.5 < height < 162.5 cm The first significant figure in the number 0.04072 is: A 0

B 4

C 7

D 2

D makes the answer bigger than it should be 12

The recurring decimal 0. 5˙ 2 7˙ is equivalent to: A 0.527777… B 0.527527527… C 0.5272727… D 0.052705270527…

13

When converted to a decimal, 1 2--3- is: A 1.666666667 B 1.666666666 C 1. 6˙ D 1.7

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14

To convert 0. 5˙ 7˙ to a fraction: Step 1: Let n = 0.575757… Step 2: then 100n = 57.575757… Step 3: 19 ------ = -----Step 4 : Hence n = 57 99 33 The missing Step 3 is: A 19n = 57 B 33n = 57 C 99n = 57 D 90n = 57

15

A rate of 18 tonnes/hectare is equivalent to: A 0.18 kg/m2 B 1.8 kg/m2 C 18 kg/m2 D 180 kg/m2

16

Which of the following rates are not equivalent to 6 m/s? A 360 m/min B 21 600 m/h C 21.6 km/h D 216 km/h

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–6

7–10

11

12–14

15, 16

A

B

C

D

E

A. APPROXIMATIONS Consider these situations. 1 $1600 is to be shared between seven people and each share deposited in a bank account. Now, $1600 ÷ 7 = $228.571 428… However it is not possible to deposit this exact amount in a bank account. Since the smallest unit of money we can deposit is a cent, and each share is closer to $228.57 than to $228.58, then we would deposit $228.57 into each person’s account. 2 A piece of timber 2600 mm long has to be cut into three equal lengths. Now, 2600 mm ÷ 3 = 866.666 6… mm However it is not possible to cut a piece of timber exactly this long. Since the smallest unit of measurement we are likely to have on a tape measure is a millimetre, and this length is closer to 867 mm than 866 mm, then we could measure and cut each piece of timber to be 867 mm. In both of these examples we have approximated the result of a calculation to make the answer meaningful.

This process of approximating numbers is also called rounding off.

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Example 1 Write down the value of the digit 7 in each of the following numbers. a 273.6

b 1407.2

c

86.457

d 2764

a 7 tens or 7 × 10 (= 70)

b 7 units or 7 × 1 (= 7)

1 7 c 7 thousandths or 7 × ------------- (= ------------- ) 1000 1000

d 7 hundreds or 7 × 100 (= 700)

Exercise 5A 1

Write down the value of the digit 6 in each of the following numbers. a 465.9 b 2346.1 c 3698 e 16 382 000 f 12.836 g 5.698 i 20 600 j 0.0006

d 6284 h 30.562

Example 2 Round off to the nearest thousand: a 7390

b 24 830

c 46 500

a Round off 7390 to the nearest thousand means: is 7390 closer to 7000 or 8000? 6000

7000

8000

7390 By drawing part of a number line showing thousands and using the digit in the hundreds column (3) to find the approximate position of 7390, we can see that it is closer to 7000, that is 7390
7000, to the nearest thousand. Note that the digit to the right of the thousands column (3) determines to which number it is closer. b Round off 24 830 to the nearest thousand means: is 24 830 closer to 24 000 or 25 000? 23 000

24 000

25 000

24 830 By drawing part of a number line showing thousands and using the digit in the hundreds column (8) to find the approximate position of 24 830, we can see that it is closer to 25 000, that is 24 830
25 000, to the nearest thousand. Note that the digit to the right of the thousands column (8) determines to which number it is closer.

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c Round off 46 500 to the nearest thousand means is 46 500 closer to 46 000 or 47 000? 45 000

46 000

‘By convention’ means everyone agrees to do it this way.

47 000

46 500 By drawing part of a number line showing thousands and using the digit in the hundreds column (5) to find the position of 46 500, we know that it is exactly in the middle of 46 000 and 47 000. By convention, we round off to 47 000, that is 46 500
47 000, to the nearest thousand. To round off a number to the nearest thousand, locate the digit in the thousands column. • If the digit to the right of the thousands column is smaller than 5, retain the thousands digit and replace the digits to the right of it by zeros. • If the digit to the right of the thousands column is bigger than 5, increase the thousands digit by one and replace the digits to the right of it by zeros. This is called rounding up. • By convention, if the digit to the right of the thousands column is equal to 5, round up.

Example 3

The symbols
and ≈ mean ‘approximately equal to’.

Round off to the nearest thousand: a 874 296

b 23 741

c

2520

a The digit in the thousands column is 4. The digit to the right of it is 2, which is smaller than 5. Hence we replace all the digits to the right of 4 with zeros. 874 296
874 000, to the nearest thousand b The digit in the thousands column is 3. The digit to the right of it is 7, which is bigger than 5. Hence we increase the thousands digit by one and replace all the digits to the ‘To the nearest right of it by zeros. thousand’ is known as the level of 23 741
24 000, to the nearest thousand accuracy of the answer. c The digit in the thousands column is 2. The digit to the right of it is equal to 5. Hence we increase the thousands digit by one and replace all the digits to the right of it by zeros. 2520
3000, to the nearest thousand

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2

Round off to the nearest thousand: a 36 800 b 83 500 f 623 490 g 180 524 k 760 l 390

c 524 100 h 6287

d 8299 i 2999

e 18 560 j 400 721

Example 4 Round off: a 2380 to the nearest hundred c 28.5 to the nearest whole number

b 7862 to the nearest ten

Using the method of examples 2 and 3: a Locate the digit in the hundreds column (3). The digit to the right of it is 8, which is bigger than 5. Hence we round up. 2380
2400 to the nearest hundred b Locate the digit in the tens column (6). The digit to the right of it is 2, which is smaller than 5. Hence we round off. 7862
7860 to the nearest ten c Locate the digit in the units column (8). The digit to the right of it is 5. Hence we round up. 28.5
29 to the nearest whole number

3

4

5

6

Round off to the nearest hundred: a 5360 b 16 829 f 240 230 g 3075 k 86 l 29

c 20 421 h 9628

d 849 i 450

e 369 j 147

Round off to the nearest ten: a 674 b 2368 f 28 g 306 k 8 l 4

c 825 h 20 056

d 1056 i 409

e 73 j 1251

Round off to the nearest whole number: a 16.7 b 25.3 c 81.5 f 265.07 g 20.9 h 106.28 k 0.71 l 0.36

d 236.67 i 300.7

e 583.1 j 55.5

Round off the following numbers to the nearest: i 1000 ii 100 iii 10 a 46 783.5 b 28 456.7 c 39 165.5 e 182 678.5

iv 1 d 8462.3

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Example 5 Round off to the nearest 1000: a 39 600

b

99 798

a Locate the digit in the thousands column (9). The digit to the right of it is 6, which is bigger than 5. Hence we increase the thousands digit by one (increasing 39 to 40) and replace all the digits to the right by zeros. 39 600
40 000 to the nearest 1000 b Locate the digit in the thousands column (9). The digit to the right of it is 7, which is bigger than 5. Hence we increase the thousands digit by one (increasing 99 to 100) and replace all the digits to the right by zeros. 99 698
100 000 to the nearest 1000

7

8

9

Round off to the nearest 1000: a 29 700 b 59 854

c 179 500

d 199 870

e 799 500

Round off to the nearest 100: a 3970 b 5963

c 23 950

d 79 980

e 19 965

Round off to the nearest 10: a 698 b 1795

c 32 599

d 99

e 6999

1 decimal place can be abbreviated to 1 d.p.

Example 6 Round off 12.3815 to: a 1 decimal place c 3 decimal places

b 2 decimal places

a Locate the digit in the first column after the decimal point (3). The digit to the right of it is 8, which is bigger than 5. Hence we increase the 3 by one and delete any digits to the right of it. 12.3815
12.4 to 1 d.p. b Locate the digit in the second column after the decimal point (8). The digit to the right of it is 1, which is smaller than 5. Hence we round off at the 8 and delete any digits to the right of it. 12.3815
12.38 to 2 d.p. c Locate the digit in the third column after the decimal point (1). The digit to the right of it is 5. Hence we increase the 1 by one and delete any digits to the right of it. 12.3815
12.382 to 3 d.p.

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10

11

Round off 38.2683 to: a 1 d.p. b 2 d.p.

c 3 d.p.

Round off the following numbers to a 8.4382 b 6.5839 f 21.6029 g 4.0611

i 1 d.p. c 0.8625 h 5.0437

ii 2 d.p. iii 3 d.p. d 0.1864 i 7.0069

e 18.5555 j 3.0002

The last zero must be shown in order to indicate the level of accuracy of our approximation.

Example 7 Round off: a 2.497 correct to 2 decimal places b 19.96 correct to 1 decimal place

a Locate the digit in the second column after the decimal point (9). The digit to the right of it is 7, which is bigger than 5. Hence we increase the 9 by one (increasing 49 to 50) and delete any digits to the right of it. 2.497
2.50 to 2 d.p. b Locate the digit in the first column after the decimal point (9). The digit to the right of it is 6, which is bigger than 5. Hence we increase the 9 by one (increasing 199 to 200) and delete any digits to the right of it. 19.96
20.0 to 1 d.p. 12

Round off: a 3.598 to 2 d.p. e 0.996 to 2 d.p.

b 49.96 to 1 d.p. f 4.8997 to 3 d.p.

c g

2.6895 to 3 d.p. 99.98 to 1 d.p.

d 12.997 to 2 d.p. h 69.995 to 2 d.p.

Example 8 When a number was rounded off to the nearest 10, the answer was 60. a What is the smallest the number could have been? b What is the largest the number could have been? Discuss. c Write a mathematical statement that shows the range of possible numbers. a 55 is halfway between 50 and 60 but by convention is rounded up to 60. This is the smallest the number could have been. b We cannot specify the largest number but we do know that it has to be less than 65 (because 65 would be rounded up to 70). c The number could be equal to 55 or between 55 and 65. Mathematicians write this as: 55 ≤ number < 65.

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13

When a number was rounded off to the nearest 10, the answer was 80. a What is the smallest the number could have been? b What is the largest the number could have been? c Write a mathematical statement that shows the range of possible numbers.

14

When a number was rounded off to the nearest 100, the answer was 400. a What is the smallest the number could have been? b What is the largest the number could have been? c Write a mathematical statement that shows the range of possible numbers.

15

For each of the following: i What is the smallest the number could have been? ii What is the largest the number could have been? iii Write a mathematical statement that shows the range of possible numbers, given that when the number was rounded off: a to the nearest 1000, the answer was 28 000 b to the nearest 1 (whole number), the answer was 43 c to 1 decimal place, the answer was 5.7 d to 2 decimal places, the answer was 6.32

16

Emily was measured to be 163 cm tall, to the nearest centimetre. Within what range of values does her actual height lie?

17

The weight of a can of fruit was measured as 420 g, to the nearest 10 g. Within what range does the actual weight of the can lie?

18

The time taken for Ken to complete the 100 m sprint at the athletics carnival was 12.4 s, to the nearest tenth of a second. Within what range does his actual time lie?

B. SIGNIFICANT FIGURES A method that combines the rounding off techniques of section A of this chapter involves the use of significant figures. The first significant figure in a number is the first digit that is not a zero (reading from left to right).

Example 1 Write down the first significant figure in each of the following numbers. a 3790

b 4.0625

c

0.002 86

The first digit which is not a zero is the a 3 b 4 c 2

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Exercise 5B 1

Write down the first significant figure in each of the following numbers. a 2876 b 5 069 836 c 1.0035 d 0.0791

e 0.000 802

Significant figures may be abbreviated to s.f.

Example 2 Round off 63.75091 correct to: a 1 d 4

b 2 c 3 e 5 significant figures.

a The first significant figure is the 6 which is in the tens column. In this case we are required to round off to the nearest 10. ∴ 63.75091
60 correct to 1 significant figure. b The second significant figure is the 3 which is in the units column. In this case we are required to round off to the nearest 1 (whole number). ∴ 63.75091
64 correct to 2 s.f. c The third significant figure is the 7 which is in the first place after the decimal point. In this case we are required to round off to 1 decimal place. ∴ 63.75091
63.8 correct to 3 s.f. d The fourth significant figure is the 5 which is in the second place after the decimal point. In this case we are required to round off to 2 decimal places. ∴ 63.75091
63.75 correct to 4 s.f. e The fifth significant figure is the 0 which is in the third place after the decimal point. In this case, we are required to round off to 3 decimal places. ∴ 63.75091
63.751 correct to 5 s.f.

2

3

Round off 28.470 58 correct to: a 1 b 2 c 3 e 5 significant figures, giving the level of accuracy of the answer. Round off each of the following numbers to: i 1 ii 2 iii 3 significant figures. a 428.3 b 6238 c 7.819 e 53 689 f 725 600 g 0.039 26 i 6103 j 2005

d 4

d 0.5273 h 0.005 072

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Example 3 Write correct to 3 significant figures: a 249 700

b 629.51

c

0.001 896

d 6.998

a The third significant figure is the 9 in the 1000s column. We need to round off to the nearest 1000. ∴ 249 700
250 000 to 3 s.f. b The third significant figure is in the units column. We need to round off to the nearest whole number. ∴ 629.51
630 to 3 s.f. c The third significant figure is in the fifth place after the decimal point. We need to round off to 5 decimal places. ∴ 0.001 896
0.001 90 to 3 s.f. d The third significant figure is in the second place after the decimal point. We need to round off to 2 decimal places. ∴ 6.998
7.00 to 3 s.f.

4

Write correct to 3 significant figures: a 369 800 b 239.6 f 499.7 g 0.039 98

c 0.005 798 h 0.299 9

d 8.997 i 0.001 999

e 299 700 j 999 900

Example 4 When a number was rounded off to 2 significant figures, the answer was a 430 b 3.7 i What is the smallest the number could have been? ii What is the largest the number could have been? iii Write a mathematical statement that shows the range of possible numbers. a The second significant figure is in the tens column, hence the number has been rounded off to the nearest 10. i 425 is halfway between 420 and 430 but by convention is rounded up to 430. This is the smallest the number could have been. ii We cannot specify the largest number but we do know that it has to be less than 435 (because 435 would be rounded up to 440). iii The number could be equal to 425 or between 425 and 435. Mathematicians write this as: 425 ≤ number < 435

☞

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b The second significant figure is in the first column after the decimal point, hence the number has been rounded off to 1 decimal place. i 3.65 is halfway between 3.6 and 3.7 but by convention is rounded up to 3.7. This is the smallest the number could have been. ii We cannot specify the largest number but we do know that it has to be less than 3.75 (because 3.75 would be rounded up to 3.8). iii The number could be equal to 3.65 or between 3.65 and 3.75. Mathematicians write this as: 3.65 ≤ number < 3.75

5

When a number was rounded off to 2 significant figures the answer was: a 560 b 8.2 c 48 d 0.72 e 37 000 f 0.084 i What is the smallest the number could have been? ii What is the largest the number could have been? iii Write a mathematical statement that shows the range of possible numbers.

6

When a number was rounded off to 3 significant figures the answer was: a 483 b 3.86 c 14 500 d 0.128 e 56.9 f 3210 Write a mathematical statement that shows the range of possible numbers in each case.

7

For each of the following write a mathematical statement that shows the range of possible numbers, given that when rounded to: a 2 s.f. the answer is 300 b 2 s.f. the answer is 3000 c 3 s.f. the answer is 6000 d 3 s.f. the answer is 24 000 e 3 s.f. the answer is 500 000 f 2 s.f. the answer is 0.80 g 3 s.f. the answer is 0.400

Example 5 How many significant figures are there in each of the following numbers? a 294

b 0.3

c

4.20

d 0.0017

e 56 000

a 3 b 1 c 3 d 2 e We are unable to tell precisely. 56 300 rounded to the nearest thousand
56 000; 55 970 rounded to the nearest hundred
56 000; 56 003 rounded to the nearest ten
56 000; 55 999.6 rounded to the nearest whole number
56 000. So there could be 2, 3, 4 or 5 s.f.

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8

How many significant figures are there in each of the following numbers? a 38 b 0.49 c 2896 d 0.075 f 1.800 g 0.0053 h 0.060 i 400 k 23 000 l 8 000 000

e 0.40 j 7000

The number of significant figures indicates the level of accuracy of the measurement.

Example 6

Explain the difference between measurements of: a 5.62 m and 5.620 m

b

2.4 kg and 2.40 kg

a 5.62 m has 3 significant figures. The last significant figure (2) is in the hundredths (of a metre) column. This indicates that the measurement has been made to the nearest hundredth of a metre, i.e. to the nearest centimetre. 5.620 m has 4 significant figures. The last significant figure (0) is in the thousandths (of a metre) column. This indicates that the measurement has been made to the nearest thousandth of a metre, i.e. to the nearest millimetre. So 5.62 m has been measured to the nearest centimetre, and 5.620 has been measured to the nearest millimetre. The second measurement is more accurate. b 2.4 kg has 2 significant figures. The last significant figure (4) is in the tenths (of a kilogram) column. This indicates that the measurement has been made to the nearest tenth of a kilogram, i.e. to the nearest 100 g. 2.40 kg has 3 significant figures. The last significant figure (0) is in the hundredths (of a kilogram) column. This indicates that the measurement has been made to the nearest hundredth of a kilogram, i.e. to the nearest 10 g. So 2.4 kg has been measured to the nearest 100 g, and 2.40 kg has been measured to the nearest 10 g. The second measurement is more accurate.

9

10

Explain the difference between measurements of: a 3.64 m and 3.640 m b c 12 s and 12.0 s d e 23.8 s and 23.80 s f g 1.5 t and 1.50 t h

5.8 kg and 5.80 kg 36 cm and 36.0 cm 7.29 km and 7.290 km 5.83 L and 5.830 L

Working in groups, discuss the level of accuracy of the numbers in the following statements. (You could consider the number of significant figures and how the numbers may have been rounded.) a There were 42 000 people at the concert. b The profit made by the bank was $600 million. c The radius of the Earth is 6400 km. d The distance of Mars from the Sun is 229 000 000 km. e Light travels at 300 000 km/h. f The length of the influenza virus is 0.000 26 mm.

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C. CALCULATIONS AND ROUNDING NUMBERS Exercise 5C 1

a Calculate 12 843 + 678 + 15 632. b Round off the answer to part a, correct to the nearest 100. c Round off to the nearest 100: i 12 853 ii 678 iii 15 692 d Find the sum of the three answers in part c. e Compare the answers to parts b and d. Comment.

2

a Calculate 8085 − 2834 correct to the nearest 10. b Round off to the nearest 10: i 8085 ii 2834 c Find the difference between the answers in part b. d Compare the answers to parts a and c. Comment.

3

a b c d

Multiply 2.341 by 8 and round off the answer to 1 decimal place. Write 2.341 correct to 1 decimal place. Multiply the answer to part b by 8. Compare the answers to parts a and c. Comment.

Truncate to 1 decimal place means to cut off all digits after the first decimal place (without rounding).

4

a Calculate 4.83 × 29.68 and give the answer to 1 decimal place. b Round off each of the numbers in part a to 1 decimal place and then perform the multiplication, giving the answer to 1 decimal place. c Compare the answers to parts a and b. Comment. d Truncate each of the numbers in part a to 1 decimal place and then perform the multiplication, giving the answer to 1 decimal place. e Compare the answers to parts a and d. Comment.

5

a Calculate 32.683 + 16.87 + 25.619. b Round off the answer to part a, correct to 2 significant figures. c Round off to 2 s.f.: i 32.683 ii 16.87 iii 25.619 d Find the sum of the three answers in part c to 2 s.f. e Compare the answers to parts b and d. Comment.

6

a b c d

Multiply 3.5 by 17 and round off the answer to 1 s.f. Write 3.5 and 17, correct to 1 s.f. Multiply the numbers in part b and give the answer to 1 s.f. Compare the answers to parts a and c. Comment.

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7

a Calculate 16.358 ÷ 7.499 giving the answer to 3 s.f. b Find: i 16.358 ii 7.499 correct to 3 s.f. c Divide the answers in b and give the answer to 3 s.f. d Compare the answers to parts a and c. Comment.

8

a Multiply 32.86 by 128.57, giving the answer correct to 3 significant figures. b Round off each of the numbers in part a to 3 significant figures and then perform the multiplication, giving the answer to 3 significant figures. c Compare the answers to a and b. Comment.

9

a Calculate 145 ÷ 784, giving the answer correct to 2 significant figures. b Round off each of the numbers in part a to 2 significant figures and then perform the division, giving the answer to 2 significant figures. c Compare the answers to a and b. Comment.

10

a Find 11 correct to 2 significant figures. b Multiply the answer in part a by itself (i.e. square the answer.) Is the result 11? c Repeat parts a and b using i 3 ii 4 iii 5 iv 6 significant figures.

It should be clear from the above questions that rounding numbers during a calculation often affects the accuracy of the result. So never round off before the last step of a calculation. 11

The value of π is 3.141 592 65 correct to 9 significant figures. Calculate and comment on the accuracy of the following approximations for π , i.e. to how many significant figures is the approximation accurate? 22 ------ to 1, 2, 3, … s.f. and compare with the value a -----(Round off the value of 22 7 7 of π to 1, 2, 3, … s.f.) 355 b ---------113

553 2 c  ----------  312

d

3

31

e

9.87

Example 1 Five metres of rope have to be cut into three equal lengths. Calculate the length of each piece, rounding off to a reasonable level of accuracy. 5 m ÷ 3 = 1.6666… m A reasonable level of accuracy for the result of this calculation could be ‘correct to 4 significant figures’ or ‘correct to 3 decimal places’, as this would be equivalent to ‘correct to the nearest millimetre’, i.e. 5 m ÷ 3
1.667 m. Note that if our tape measure only showed centimetres, then rounding off to 3 significant figures or 2 decimal places (i.e. to the nearest centimetre) would be more appropriate, i.e. 5 m ÷ 3
1.67 m. Hence, several answers are possible, depending on the accuracy of our measuring instrument.

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12

Perform each of the following calculations, rounding off the answer to a reasonable level of accuracy. a 100 mm of cotton has to be cut into three equal pieces. Find the length of each piece. b 100 cm of string has to be cut into three equal pieces. Find the length of each piece. c 100 m of rope has to be cut into three equal pieces. Find the length of each piece. d 2 m of rope has to be cut into three equal pieces. Find the length of each piece. e A running track is 11 m wide. If the track is to be divided into 8 lanes, how wide should each lane be? f There are 2100 families in a country town and the total number of children is 4897. Find the average number of children per family. g A bank announced that its annual profit was $170 million. Calculate its average monthly profit.

D. RECURRING DECIMALS As a decimal

3 --8

= 0.375 and

1 --3

= 0.333 33…

When converted to a decimal, the fraction three places have been filled.

3 --8

terminates. The digits after the decimal point stop after

When the fraction 1--3- is converted to a decimal, the digits after the decimal point keep repeating or recurring. We call this a recurring decimal.

When converted to a decimal, all fractions either terminate or recur.

0.3333… is written 0. 3˙ The dot above the 3 indicates that this digit recurs.

Example 1 Write the following recurring decimals using the dot notation. a 0.4444… d 0.415 415 415…

a

0. 4˙

b 0.4 1˙

b 0.411 11… e 0.415 341 534 153…

c

0. 4˙ 1˙

d

0. 4˙ 1 5˙

The dots are put above the first and last digits of the group of digits that repeat.

c

0.414 141…

e 0. 4˙ 15 3˙

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Rational Numbers (Chapter 5) Syllabus reference NS5.2.1

Exercise 5D 1

Write the following recurring decimals using the dot notation. a 0.7777… b 0.355 55… c 0.282 828… d 0.325 325 325… e 0.678 467 846 784… f 1.4444… g 6.922 22… h 0.494 949… i 0.234 234 234… j 0.033 33… k 0.909 090… l 0.536 666… m 0.217 77…

Example 2 Use your calculator to convert: 5 2 a --b --- to a decimal. 8 3 a By calculating 5 ÷ 8 or using the fraction key

5 --8

= 0.625.

b By calculating 2 ÷ 3 or using the fraction key, the display could show 0.666666666 or 0.666666667, depending on the calculator used. Both are approximations for the recurring decimal 0. 6˙ . In the first case the answer has been truncated (because of the limitations of the calculator display) and, in the second case, the calculator has automatically rounded up to the last decimal place, 2 i.e. --- = 0. 6˙ . 3

2

Convert the following fractions to decimals. a

7 --8

b

5 --9

c

1 --6

d

2 -----11

5 e 1 ----12

f

1 2--3-

g

11 -----18

h

22 -----33

i

-----1 13 22

j

11 -----24

Example 3 Convert the following decimals to fractions:

3

a 0.8

b 0.63

8 4 a 0.8 = ------ = --10 5

b

63 0.63 = ---------100

Convert the following decimals to fractions. a 0.6 b 0.78 c 0.125

c

0.148

148 37 c 0.148 = ------------- = ---------1000 250

d 0.08

e 0.256

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Example 4 Convert 0. 4˙ to a fraction. Let n = 0. 4˙ n = 0.4444… 10n = 4.4444… 9n = 4

then by subtraction

4 n = --9

hence

4 0. 4˙ = --9

i.e.

4

Convert the following recurring decimals to fractions. a

5

Multiplying by 10 moves the decimal point one place to the right.

0. 2˙

b

0. 3˙

c

0. 5˙

d

0. 8˙

e

0. 7˙

e

0. 9˙ 8˙

Convert 0. 9˙ to a fraction. Discuss the result with your class.

Example 5 Convert 0. 5˙ 7˙ to a fraction.

then by subtraction hence i.e.

6

7

Multiplying by 100 makes the decimal parts the same.

n = 0.575757… 100n = 57.575757… 99n = 57 57 19 n = ------ = -----99 33 19 0. 5˙ 7˙ = -----33

Convert the following recurring decimals to fractions. a 0. 4˙ 6˙ b 0. 9˙ 1˙ c 0. 3˙ 0˙

d

0. 6˙ 3˙

Convert the following recurring decimals to fractions (Hint: Multiply by 1000). a 0. 5˙ 8 6˙ b 0. 2˙ 3 9˙ c 0. 8˙ 5 2˙ d 0. 4˙ 2 3˙ e

0. 6˙ 1 5˙

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Rational Numbers (Chapter 5) Syllabus reference NS5.2.1

Example 6 Convert 0.3 5˙ to a fraction. Let n = 0.3 5˙ then and by subtraction hence i.e.

8

n 10n 100n 90n

= = = =

0.35555… 3.5555… 35.5555… 32 32 16 n = ------ = -----90 45 16 ˙ 0.3 5 = -----45

We need to make the decimal parts the same before we subtract.

Convert the following recurring decimals to fractions. a 0.3 8˙ b 0.6 5˙ c 0.9 2˙

d 0.1 6˙

e

0.0 9˙

e

0.04 9˙

Example 7 Convert 0.51 2˙ to a fraction. Let n = 0.51 2˙ then and by subtraction hence i.e.

9

n 100n 1000n 900n

= = = =

0.512222… 51.2222… 512.2222… 461 461 n = ---------900 461 0.51 2˙ = ---------900

Convert the following recurring decimals to fractions. a 0.54 6˙ b 0.72 3˙ c 0.76 2˙

d

0.90 5˙

E. RATES A rate is a comparison between different kinds of quantities. For instance, we could compare distance travelled with the time taken to travel this distance (this well known rate is called speed), mass of fertiliser with the area to be fertilised, cost of washing powder with the mass bought, and so on.

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Example 1 Farmer White spread 1500 kg of fertiliser over 1200 m2 of a field. Write the rate of application of fertiliser in simplified form. We are required to find the amount of fertiliser per (one) square metre of area. 1500 kg 1500 kg per 1200 m2 = ------------- ------21200 m kg kg ------ is usually written 2 = 1.25 ------2m 2 m kg/m . 2 = 1.25 kg/m

Exercise 5E 1

a b c d e

Julie paid $8.60 for a 2.5 kg packet of washing powder. Find the cost per kilogram. Ben was paid $100.48 for 8 hours of work. Calculate his rate of pay (per hour). Kylie typed 660 words in 12 min. How many words per minute can she type? On a journey of 882 km a car used 84 L of petrol. Express the petrol consumption in km/L. A truck travelled 585 km in 6 1--2- h. Calculate the average speed of the truck.

Example 2 Convert a rate of: a 83 cents/metre to cents/centimetre b $8.70/hour to i cents/hour a 83 c/m

83 c = ------ ---1 m 83 c = ---------- ------100 cm c = 0.83 ------cm or 0.83 cents/centimetre

ii cents/minute b i $8.70/h = $ 8.70 ----------1h 870 c = ---------- --- or 870 cents/hour 1 h 870 c ii = ---------- ---------60 min c = 14.5 ---------- or 14.5 cents/minute min

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Rational Numbers (Chapter 5) Syllabus reference NS5.2.1

Example 3 Convert 12 tonnes/hectare to grams/square metre. Write the rate as a fraction then convert.

12 t 12 t/ha = ----------1 ha 12 × 1000 × 1000 g = ---------------------------------------------- ------210000 m = 1200 g/m2 2

Convert: a 91 cents/m to cents/cm c 84 L/h to L/min e 9% p.a. to % per quarter g $50.40/h to cents/min i 6.5 t/ha to g/m2 k $36/L to c/mL

b d f h j l

125 cents/m to cents/mm 15% p.a to % per month $9.60/h to cents/min 18 t/ha to g/m2 $19/kg to c/g 7.5 kg/L to g/mL

3

Convert 72 km/h to

i

km/min

ii m/min

iii m/s

4

Convert 108 km/h to

i

km/min

ii m/min

iii m/s

5

Convert 60 km/h to

i

km/min

ii m/min

iii m/s

6

A 600 litre fish tank was filled in 2 hours. Calculate the rate of flow of water into the tank in: i L/h ii L/min iii mL/s

Example 4 Convert: a 5 m/s to km/h

b 12 g/m2 to kg/ha

a 5 m/s = 5 × 60 m/min

b 12 g/m2 = 12 × 10 000 g/ha 12 × 10000 = ----------------------------- kg/ha 1000

= 5 × 60 × 60 m/h 5 × 60 × 60 = ----------------------------- km/h 1000

= 120 kg/ha

= 18 km/h

7

Convert: a 8 m/s to km/h d 15 g/m2 to kg/ha g 0.5 c/g to $/kg j 0.8 g/mL to kg/L

b 20 m/s to km/h e 8 g/m2 to kg/ha h 12 c/min to $/h

c f I

35 m/s to km/h 25 g/m2 to kg/ha 2 mL/s to L/h

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8

An athlete can run 100 m in 10 s. Calculate his speed in km/h.

9

A car uses 125 L of petrol to travel a distance of 1000 km. Calculate the rate of petrol consumption in: i L/km ii L/100 km iii mL/km

10

A machine fills bottles at the rate of 240 bottles/h. a How many bottles could be filled in 4 1--2- hours? b How long would it take to fill 6000 bottles? c Convert the rate to bottles/min. d How many bottles could be filled in 45 min? e How long would it take to fill 200 bottles?

11

Paint covers walls at the rate of 16 m2/L. a What area could be covered by a 20 L drum? b How much paint would be needed to cover an area of 240 m2? c Convert the rate to m2/mL. d What area could be covered by a i 500 mL ii 200 mL can of paint?

12

A car travels at an average speed of 90 km/h. a How far will the car travel in 5 h? b How long will it take the car to travel 675 km? c Convert the speed to m/s. d If the time it takes to stop the car from this speed is 5.4 s, how far does the car travel before it stops?

13

Water can flow from a tap at the rate of 4.5 L/min. a How much water would flow from the tap in 1 h? b How long would it take to fill a 40 L drum? c Convert the rate to mL/s. d If a plastic bottle is filled in 24 s, what is the capacity of the bottle?

14

A patient in hospital is given a saline solution intravenously at the rate of 50 mL/h. a How much saline solution will the patient receive in 24 h? b If each millilitre contains 12 drops, convert the rate to drops/min.

15

A telecommunications company quotes its rate for mobile phone calls as 15c/call plus 8.5c/30 s (or part thereof). 1 gram = 1000 a Convert the time rate to c/min. milligrams b What is the total cost of a 3-minute phone call? c Use the answer to part b to calculate the average cost per minute of a 3-minute phone call. d How much would it cost to make a call lasting 5 min and 10 s? e Joo-Mee decides that when she rings her friends she cannot afford to spend more than $1 per phone call. What is the maximum time she can spend talking to a friend?

16

Slopon UV protection cream contains 72 g/L of the chemical titanide. Express this concentration in: i g/mL ii mg/mL iii mg/10 mL

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Rational Numbers (Chapter 5) Syllabus reference NS5.2.1

non-calculator activities

1

Round off: a 34 671 to the nearest thousand c 15.187 to the nearest whole number e 7.6351 to 3 significant figures

b 7.296 to 2 decimal places d 2454 to 2 significant figures

2

How many significant figures are there in each of the following numbers? a 2.34 b 0.0087 c 40 000

3

Write the following using dot notation. a 0.4444… b 0.525 252… e 0.213 213 213…

4

By division, convert to a decimal:

5

Convert 0. 3˙ 7˙ to a fraction.

6

Convert:

a 36 km/h to m/s

a

c 0.566 66…

3 --8

b

d 0.576 666…

4 --9

b 1.5 g/mL to kg/L.

Language in Mathematics 1

The value of the digit 7 in the number 13.475 is: A 7 tens B 7 tenths C 7 hundreds

D 7 hundredths

2

Explain the difference between the measurements 9 seconds and 9.0 seconds.

3

Explain what is meant by the term ‘level of accuracy’ of an answer.

4

Explain the difference between the words affect and effect.

5

Complete the following words by replacing the missing vowels: a fr__ct__ __n b d__c__m__l c s__m d pr__d__ct e t__rm__n__t__

6

Write in your own words the meanings of: recurring, convert, retain, truncate, convention.

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7

Three of the words in the following list are spelt incorrectly. Find these words and write the correct spelling: significent, repeating, deleet, figgers, accuracy.

8

How many words of three or more letters can you make from the word APPROXIMATION. (No proper names or plurals allowed.)

Glossary

✓

accuracy

affect

approximate

calculation

convention

convert

decimal

delete

digit

effect

equivalent

figure

fraction

identify

product

rate

recurring

repeating

retain

round off

significant

sum

terminate

truncate

CHECK YOUR SKILLS 1

The value of the digit 3 in the number 156.832 is: A 30 B 3

2

When rounded to the nearest thousand 23 629 is: A 23 B 24

3

3 C -----10 C 23 000

When 7982 is rounded to the nearest hundred the answer is: A 7900 B 7990 C 7980

notation

3 D ---------100

D 24 000

D 8000

4

Which of the following numbers are not equal to 36.5, when rounded off correct to 1 decimal place? A 36.48 B 36.54 C 36.55 D 36.50

5

When a number is rounded off to the nearest 10 the answer is 70. The smallest the number could be is: A 69.99 B 69 C 65.01 D 65

6

The mass of a soup can was given as 380 g to the nearest 10 g. The actual mass of the can lies in the range: A 370 ≤ mass < 390 g B 370 < mass < 390 g C 375 ≤ mass < 385 g D 375 < mass < 385 g

7

The first significant figure in the number 0.005064 is: A 0 B 5 C 6

D 4

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Rational Numbers (Chapter 5) Syllabus reference NS5.2.1

8

Which of the following numbers are not equal to 4600 when rounded off to 2 significant figures? A 4639 B 4608 C 4550 D 4650

9

When rounded to 3 significant figures, 4.5976 is: A 4.597 B 4.598

10

C 4.59

Rounding numbers before the last step of a calculation: A never affects the accuracy of the answer B often affects the accuracy of the answer C always affects the accuracy of the answer D makes the answer bigger than it should be

12

The recurring decimal 0.6 4˙ is equivalent to: A 0.6444… B 0.646 464…

13

14

When converted to a decimal, 1 7--9- is: A 1.777777778 B 1.777777777

16

B 11n = 63

A rate of 13 tonnes/hectare is equivalent to: A 0.13 kg/m2 B 1.3 kg/m2

D 0.034

✓

C 0.064 44…

D 0.064 064 064…

C 1.8

D 1. 7˙

C 99n = 63

D 90n = 63

C 13 kg/m2

D 130 kg/m2

To convert 0. 6˙ 3˙ to a fraction: Step 1 Let n = 0.636363… Step 2 then 100n = 63.636363… Step 3 63 7 Step 4 hence n = ------ = -----99 11 The missing Step 3 is: A 7n = 63

15

D 4.60

Which of the following numbers do not have 2 significant figures? A 7.29 B 5.0 C 36

11

Which of the following rates are not equivalent to 5 m/s? A 300 m/min B 18 000 m/h C 18 km/h

D 180 km/h

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

151

1–6

7–10

11

12–14

15, 16

A

B

C

D

E

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Rational Numbers (Chapter 5) Syllabus reference 5.2.1

REVIEW SET 5A 1

2

3

Write down the value of the digit 5 in each of the following numbers. a 253.6 b 1405.2 c 86.457

d 2564

Round off: a 2470 to the nearest hundred

b 7926 to the nearest ten

c 33.5 to the nearest whole number

d 49 900 to the nearest thousand

Round off 8.4625 to: a 1 decimal place

b 2 decimal places

c 3 decimal places

4

Round off:

a 3.497 correct to 2 decimal places

b 19.98 correct to 1 decimal place

5

When a number was rounded off to the nearest 10, the answer was 40. a What is the smallest the number could have been? b What is the largest the number could have been? Discuss. c Write a mathematical statement that shows the range of possible numbers.

6

Write down the first significant figure in each of the following numbers. a 3790

7

10

b 2

c 3

d 4

b 0.6

c 8.20

d

0.0032

Explain the difference between measurements of 3.65 m and 3.650 m. Write the following recurring decimals using the dot notation. a 0.2222…

b 0.422 22…

d

e 0.425 342 534 253…

0.425 425 425…

11

Use your calculator to convert to a decimal:

12

Convert the following decimals to fractions: a 0.6

13

e 5 significant figures

How many significant figures are there in each of the following numbers? a 795

9

c 0.002 86

Round off 48.350 92 correct to: a 1

8

b 4.0625

Convert to a fraction:

b

0.73 a 0. 4˙

c 0.424 242…

a

3 --8

c 0.138 b 0. 4˙ 1˙

b 1 2--3-

e 28 000

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Rational Numbers (Chapter 5) Syllabus reference 5.2.1

14

Convert 16 tonnes/hectare to grams/square metre.

15

Convert 1.2 L/km to L/100 km.

REVIEW SET 5B 1

Write down the value of the digit 6 in each of the following numbers. a

2

3

253.6

b

1607.2

c

83.456

d

2564

Round off: a

13 560 to the nearest hundred

b 4063 to the nearest ten

c

147.55 to the nearest whole number

d 99 900 to the nearest thousand

Round off 1.5607 to: a 1 decimal place

b 2 decimal places

c 3 decimal places

4

Round off:

a 2.695 correct to 2 decimal places

5

Susan was measured to be 164 cm tall, to the nearest centimetre. Within what range of values does her actual height lie?

6

Write down the first significant figure in each of the following numbers. a 24 560

7

b 15.0715

10

11

0.005 09

b 539.53

c 0.002 397

d 1.998

How many significant figures are there in each of the following numbers? a 37

9

c

Write correct to 3 significant figures: a 365 400

8

b 39.96 correct to 1 decimal place

b 1.3

c 17.90

d

0.0008

e 4000

Explain the difference between measurements of 1.85 kg and 1.850 kg. Write the following recurring decimals using dot notation. a

0.3333

b 0.366 66…

e

0.567 856 785 678…

Use your calculator to convert to a decimal:

c 0.282 828…

a

3 -----16

d

0.314 314 314…

b 1 7--9-

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Rational Numbers (Chapter 5) Syllabus reference 5.2.1

12

Convert the following decimals to fractions: a

0.3

b

13

Convert to a fraction: a 0. 8˙

14

Convert 36 km/h to:

15

Convert 5 mL/s to L/h.

0.81

c 0.267 b

i

km/min

0. 4˙ 9˙

ii m/min

iii m/s

REVIEW SET 5C 1

Write down the value of the digit 8 in each of the following numbers. a

2

3

318.6

b 36.8

c 23.487

d

Round off: a 13 827 to the nearest hundred

b

c

d 89 600 to the nearest thousand

24.09 to the nearest whole number

765 to the nearest ten

Round off 13.0652 to: a 1 decimal place

b 2 decimal places

c 3 decimal places

4

Round off:

5

Write down the first significant figure in each of the following numbers. a

6

a 4.196 correct to 2 decimal places

3790

b 4.0625

b

20.95 correct to 1 decimal place

c 0.002 86

Round off 17.6308 correct to: a 1

7

8567

b 2

c 3

d 4

e 5 significant figures

When a number was rounded off to 2 significant figures the answer was 430. a What is the smallest the number could have been? b What is the largest the number could have been? Discuss. c Write a mathematical statement that shows the range of possible numbers.

8

How many significant figures are there in each of the following numbers? a 795

b 0.6

c

8.20

d 0.0032

e

28 000

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Rational Numbers (Chapter 5) Syllabus reference 5.2.1

9 10

Explain the difference between measurements of 6 cm and 6.0 cm. Write the following recurring decimals using dot notation. a 0.1111….

b 0.377 77…

c 0.929 292…

d 0.637 637 637…

e 0.423 333… 11

Use your calculator to convert to a decimal:

12

Convert these decimals to fractions:

13

Convert to a fraction:

14

Convert 72 L/h to mL/s.

15

Convert 2 cents/min to $/day.

a

a

a

0.2

0. 8˙

11 -----40

-----b 1 11 12

b

0.58

c

0.125

b 0. 6˙ 3 5˙

REVIEW SET 5D 1

Write down the value of the digit 3 in each of the following numbers. a 12.03

2

3

b 3568

c 93.257

d 530.8

Round off: a 78 463 to the nearest hundred

b 509 to the nearest ten

c 34.08 to the nearest whole number

d 499 500 to the nearest thousand

Round off 2.0566 to: a 1 decimal place

b 2 decimal places

4

Round off:

5

The weight of a can of tomatoes was measured as 240 g, to the nearest 10 g. Within what range does the actual weight of the can lie?

6

Write down the first significant figure in each of the following numbers. a 135 700

7

a 3.198 correct to 2 decimal places

c 3 decimal places

b 0.0063

b 59.97 correct to 1 decimal place

c 5.0084

Write correct to 3 significant figures: a 24 671

b

67.835

c 0.050 67

d 2.995

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8

How many significant figures are there in each of the following numbers? a

9 10

462

b

0.5

c

3.60

d 0.000 93

e

600

Explain the difference between measurements of 12 seconds and 12.0 seconds. Write the following recurring decimals using dot notation. a 0.9999…

b 0.455 55…

c

0.858 585…

d 0.726 666…

e 0.436 436 436… 11

Use your calculator to convert to a decimal:

a

17 -----80

12

Convert the following decimals to fractions:

a

0.8

13

Convert to a fraction:

a 0. 6˙

14

Convert 90 km/h to:

i

15

Convert 0.75 kg/m2 to tonnes/hectare.

km/min

5 b 1 ----12

b 0.96

b 0.3 9˙ ii m/min

iii m/s

c 0.545

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Chapter 6 Perimeter and Area This chapter deals with the use of formulas to find the area of quadrilaterals, and the perimeter and area of composite figures. At the end of this chapter you should be able to: ✓ develop and use formulas to find the area of quadrilaterals ✓ calculate the area and perimeter of composite figures including quadrants and semicircles ✓ calculate the perimeter and area of sectors and composite figures involving sectors.

Syllabus reference MS5.1.1, 5.2.1 WM: S5.2.1–S5.2.5

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

Diagnostic Test

1

The formula A =

1 --2

h(a + b) could be used

to find the area of a: A kite

B trapezium

C rhombus

D all of these

5

A 30.38 cm2

B 15.19 cm2

C 11.1 cm2

D 22.2 cm2

The area of this quadrilateral is closest to: B

2

5.3 cm

9.8 cm

3

6 cm 20 cm

>

D

A AC = 20 cm

2.2 cm

2

C

4 cm

The area of this trapezium is:

> 6

A 16.61 cm

B 33.22 cm2

C 57.134 cm2

D 114.268 cm2

A 480 cm2

B 200 cm2

C 300 cm2

D 100 cm2

The perimeter of this semicircular garden is:

The area of this kite is: 5.2 m

3.8 cm

7

A 9.3 m

B 21.5 m

C 13.4 m

D 4.08 m

The perimeter of this shape is:

8.5 cm

B 24.6 cm

2

D 16.15 cm2

A 32.3 cm C 12.3 cm 4

7.1 m

2

2

1.8 m

The area of this rhombus is closest to:

8

4.9 cm 6.2 cm

A 19.2 m

B 20.6 m

C 23.5 m

D 12.78 m

Simone has 50 m of garden edging. The radius of the circular garden she can enclose with the edging is: A 15.9 m

B 7.96 m

C 12.5 m

D 157 m

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

9

10

The perimeter of a rhombus with diagonals 4.1 cm and 6.3 cm is closest to:

13

The perimeter of this sector is closest to: A 7.7 cm

A 3.76 cm

B 7.5 cm

B 23.7 cm

C 15 cm

D 30 cm

C 440 cm

55°

D 25.8 cm

The area of this figure is closest to:

8 cm 8.2 cm

14 5.8 cm

The area of this sector is closest to: A 212 cm2 B 248 cm2

3.4 cm

C 59.6 cm2

11

A 161.7 cm2

B 59.12 cm2

C 36 cm2

D 34.8 cm2

75°

D 1350 cm2

18 cm

The shaded area is closest to:

10.1 m

Use this diagram to answer questions 15 and 16.

4.8 cm

3m 2

A 248.1 m

B 62 m

2

D 46.8 m2

C 88.2 m 12

25o

2

1.2 m

A farmer fertilises a paddock consisting of a rectangle and a semicircle, shown below. The fertiliser is spread at the rate of 2.3 kilograms per square metre. The amount of fertiliser the farmer needs is closest to:

15 m

15

16 28 m

A 2.6 t

B 1.2 t

C 250 kg

D 966 kg

The perimeter of the shape is closest to: A 29.2 m

B 9.1 m

C 12.7 m

D 9.7 m

The area of the shape is closest to: A 5.56 m2

B 4.25 m2

C 1.96 m2

D 10.36 m2

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–5

6–9

10, 11

12

13–16

B

C

D

E

F

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A. AREA AND PERIMETER REVIEW This section reviews conversions, area and perimeter from Stage 4.

Linear conversions 1 km = 1000 m

1 m = 100 cm

1 cm = 10 mm

Example 1 Convert: a 50 mm to cm

b 8.6 km to m

a

b

50 mm 50 = ------ cm 10 = 5 cm

8.6 km 8.6 × 1000 m = 8600 m

Area conversions When converting area units, which are square units, the linear conversion must be squared. Since 10 mm = 1 cm then 102 mm2 = 12 cm2

(squaring both sides)

∴ 100 mm2 = 1 cm2 The hectare (ha) is a special unit of area: 1 ha = 10 000 m2.

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

Example 2 Convert: a 25 cm2 to mm2 a

b 2000 cm2 to m2

1 cm = 10 mm ∴ 1 cm2 = 102 mm2 ∴ 1 cm2 = 100 mm2 ∴ 25 cm2 = 2500 mm2 2

b

100 cm = 1 m 1002 cm2 = 12 m2 ∴ 10 000 cm2 = 1 m2 ∴ 2000 cm2 = 0.2 m2 (dividing both sides by 5)

Exercise 6A 1

2

Convert: a 21 cm to m d 4 cm to mm g 200 mm to cm j 8.3 cm to mm m 0.05 km to cm

b e h k n

180 mm to cm 2.3 m to cm 280 cm to m 6.3 km to m 3.2 m to mm

c f i l o

3500 m to km 1.8 km to m 5.2 m to cm 0.03 m to cm 83 000 cm to km

Convert: a 4 cm2 to mm2 d 32 km2 to m2 g 5 ha to m2

b 31 m2 to cm2 e 40 000 cm2 to m2 h 7.3 ha to m2

c f i

5.3 m2 to mm2 7 000 000 mm2 to m2 42 000 m2 to ha

When converting square units, square the conversion first.

3

Convert: a 15 cm2 to mm2 c 32 000 cm2 to m2 e 235 m2 to cm2 g 7.82 m2 to ha i 23 km2 to ha k 5.2 m2 to cm2

b d f h j l

15 ha to m2 3280 mm2 to cm2 36.5 ha to m2 3 654 200 cm2 to ha 0.004 2 ha to cm2 0.002 m2 to mm2

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Example 3

The perimeter of a circle is called the circumference.

Find the circumference of this circle to 1 decimal place. C = 2πr C = 2 × π × 8.2 = 51.5221... = 51.5 cm

4

8.2 cm

Find the circumference of these circles, to 1 decimal place if necessary. a

Radius is half the diameter.

b

0.4 km

5.1 cm

c

d 12.6 cm 48 cm

Example 4 Find the area of this rectangle. A = lb = 3.4 × 6.8 cm2 = 23.12 cm2

3.4 cm

6.8 cm

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

5

Find the area of each rectangle. a

b

c

d

12 m 16 cm 8 cm

12 m

15 m 4 cm 5 cm

Example 5 Find the area of each triangle. a

b 3m

8m

5m

2.5 cm

c

6 cm 12 m

a A = 1--2- bh

b A = 1--2- bh

c A = 1--2- bh

=

=

=

1 --2

× 12 × 5 m2

= 30 m2

6

× 8 × 3 m2

1 --2

= 12 m2

1 --2

× 2.5 × 6 cm2

= 7.5 cm2

Find the area of each triangle. a

b

c

d

7m 4 cm

4m

40 m

8 cm

2 cm 9 cm

32 m

Example 6 Find the area of the parallelogram. A = bh = 10 × 5 cm2 = 50 cm2

5 cm

10 cm

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

7

Find the area of each parallelogram. a b

c 6 cm

3 cm

10 cm 6 cm 12 cm

8 cm

Example 7 Find the area of these circles correct to 1 decimal place. a

b 12 cm

5 cm

a Area = πr2 = π × 5 × 5 cm2 = 25π cm2
78.5 cm2

8

b Area = πr2 = π × 6 × 6 m2 = 36π m2
113.1 m2

Find the area of each circle correct to 2 decimal places. a b c 8 cm 14 m

d

e

2 6.

cm

8.5 cm

f

15.3 m

1.26

km

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

Investigation 1 WM: Applying Strategies

Formulas for area In Stage 4 you learnt that the area of a triangle is: A = 1--2- b × h Use this formula to find an expression for the area of a rhombus and a kite. 1

Rhombus Use this diagram to find an expression for the area of a rhombus with diagonals x and y units in length. y y 2

x

y 2

x

2

Kite The formula for the area of a kite is the same as that of a rhombus. Compare this derivation with yours from question 1.

1 2

Divide into two triangles. y cm

y cm

x cm

Area =

1 --2 1 --4 1 --2

×

( 1--2-

y) × x +

1 --2

×

( 1--2-

y) × x

1 2

y cm

xy + xy B. AREAx cmOF SPECIAL =QUADRILATERALS 1 --4

= xy

From Investigation 1, the following formulas have been developed. Use the formula for the area of a triangle to find an expression for the area of a trapezium.

Rhombus Trapezium 3 Use these diagrams to find an expression for the area of a trapezium. a

a

t

cu

h

b

b

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B. AREA OF SPECIAL QUADRILATERALS From Investigation 1, we have developed the following formulas.

Rhombus y

x

Area =

1 -2 1 -2

× product of the lengths of the diagonals

A = xy, where x and y are the lengths of the diagonals

Kite y cm

Area

A=

1 -2

xy

x cm

Trapezium a

A = 1--2- ah + 1--2- bh ∴ A = 1--2- h (a + b)

height

a+b or A =  ------------ h  2  b

Note the ‘height’ is the perpendicular distance between the two parallel sides. Sometimes it is a side but usually it is not.

Example 1 Find the area of the rhombus with diagonals of length 5 cm and 7 cm. Area = 1--2- xy =

1 --2

× 5 × 7 cm2

= 17.5 cm2

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

Exercise 6B 1

Find the area of each rhombus. a b

c 4.3 cm 6.3 m

4 mm 12 mm

7.2 cm

9.5 m

Example 2 Find the area of this kite. Area = 1--2- xy =

1 --2

5 cm

×5×8

8 cm

= 20 cm2

Find the area of each kite. a

b

4.8 m

6 cm

c

15 cm

9 cm

2

11.6 m

4.3 cm

Example 3 Find the area of this trapezium. First identify the height then use the formula. Area = 1--2- h(a + b) =

1 --2

× 4(11 + 16)

= 54 m2

11 m 4m 16 m

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

3

Find the area of each trapezium. Identify the height first. a

The height is perpendicular to the parallel sides.

b

6m 4m

6 cm

3 cm

7 cm

10 m

c

d

16 cm 7 cm

35 mm

50 mm

12 m 28 mm

Example 4 Find the area of this quadrilateral.

Q 3 cm

Area = Area of triangle A + Area of triangle B =

1 --2

× 14 × 3 +

1 --2

A

× 14 × 5

5 cm 14 cm B

= 56 cm2 P

4

Find these areas. a

b

4 cm P

c

4 cm PQ = 8 cm

RS = 13 m

11 m

Use the correct formula to find the area of these quadrilaterals. a b c 10 m

8 cm

8 mm

12 cm

5

8m

6m

15 mm 7m

TU = 28 m

S

Q 1 cm

T

5 cm

R

U

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

d

e

f

3 km 11.3 m

18 m 15 km

7.5 m

8 km 8.5 m 4.1 m

C. PERIMETER OF COMPOSITE FIGURES This section involves finding the perimeter of composite figures, and the solution of worded problems involving perimeter. Composite figures are those made up of more than one plane shape, including curved shapes.

Example 1 Find the perimeter of these figures. a

b 6 cm

40 m

a Perimeter Divide by 2 because it is a = (2πr ÷ 2) + 6 semicircle. =π×3+6 = 3π + 6
15.4 cm (1 d.p.) b Perimeter = circumference of circle + 2(length of straight side) = π × 40 + 2 × 40 = 40π + 80
205.7 m Two semicircles make a circle.

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Exercise 6C 1

Find the perimeter of these, giving answers to one decimal place. a b 10 cm 8 cm

c

d

20 m

20 m

20 m

18 m

Example 2

A quarter of a circle, so divide by 4.

Find the perimeter of these figures. a

b

8 cm

a

2

Perimeter = (2πr ÷ 4) + 2 × 8 = (2π × 8 ÷ 4) + 2 × 8 = 28.6 (1 d.p.)

12 m

b

Perimeter = (2πr ÷ 4) + 4 × 12 = (2 × π × 12 ÷ 4) + 4 × 12 = 66.8 (1 d.p.)

Find the perimeter of these, correct to one decimal place. a b c

d 20 cm

18 mm

10 cm

6 cm 10 m

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3

Find the perimeter of the following (giving your answer correct to 4 significant figures). a b c

4 cm 6 cm 4m

4

5

a i Find the lengths of the paths from A to B along the 4 small semicircles, and along the larger semicircle (to 2 d.p.). ii Which is shorter? b What is the difference between the lengths of the two paths?

A

B 8m

Find the perimeter correct to 2 decimal places (all measurements are in cm). a b c 10

8

11

15 5

7

A farmer decides to fence a 400 m by 350 m paddock with a 4-strand wire fence. Find the total cost of the wire required given that single strand wire costs 12.4 cents per metre.

6

7

Find the total length of string used to tie a box as illustrated. An extra 15 cm is required for the knot and bow.

10 cm

20 cm 15 cm

8 2m

The framing of a toolshed consists of square galvanised tubing which costs $4.65 a metre. Find the total cost of the tubing necessary to make the framing of the shed opposite.

4m 3m

9

A garden consists of six rectangular-shaped 8 m by 7 m garden beds and a 2 m wide path surrounding them as shown. Jarrah timber strips are used to surround each bed and the whole garden area. Find the total length of jarrah required.

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Example 3 Izat ran around a circular track. He ran 500 m. Find the radius of the track. C = 2πr r

500 2×π×r ------------ = --------------------2×π 2×π 500 ∴ r = ----------------(2 × π) = 79.6 cm (to 1 d.p.)

10

A circular plate has circumference 50 cm. Find the radius correct to 1 d.p.

11

Georgette wants a circular track with a circumference of 200 m. Find the radius of the track correct to 1 d.p.

12

A satellite has a circular orbit 800 km above the Earth’s surface. a If the radius of the Earth is 6400 km, find the radius of the orbit of the satellite. b Find the circumference of the satellite’s orbit. c If the satellite makes one orbit in a day, find the speed of the satellite.

13

A bicycle wheel has diameter of 0.6 m. Through how many complete revolutions must the wheel turn during a 100 km trip?

14

A newspaper company decides to place a plastic wrapper around its newspapers. Each wrapper is 50% longer than the circumference of the rolled-up paper, and the average diameter of a paper is 5 cm. Find the number of kilometres of wrapper required to wrap the 275 000 newspapers produced daily. Give your answer correct to 2 d.p.

15

A rhombus has diagonals 24 cm and 10 cm as shown. a Calculate the length of the side of the rhombus. b Calculate the perimeter of the rhombus. Remember Pythagoras.

24 cm

10 cm

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

16

A kite has diagonals as shown. a Calculate the length of one short side of the kite. b Calculate the length of one long side and hence the perimeter.

12 cm 3 cm

10 cm

D. COMPOSITE AREAS Figures that cannot have their areas calculated using one formula are called composite areas. The area of a composite figure can be calculated by dividing it into identifiable shapes, then adding or subtracting the area of these shapes to find the total area.

Example 1 Find the shaded area. The area is found by adding the area of the rectangle and the triangle. Total area = area of triangle + area of rectangle 1 = --- × 14 × 3 + 8 × 14 2 = 133 cm2

11 cm

8 cm

14 cm

Example 2 Find the area of this shape.

Divide the figure into two rectangles and find any unknown side lengths. Area = 18 × 8 + 7 × 8 = 200 cm2 8 cm

18 cm

7 cm 8 cm 15 cm

8 cm

18 cm

7 cm 8 cm 15 cm

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

Exercise 6D 1

Find the area of each shape (correct to 2 d.p. where necessary). All angles are right angles. a

3 cm

b

c 2 cm

5 cm

15 cm

8 cm 8 cm

4 cm

16 cm

7 cm

5 cm

9 cm 6 cm

d

10 cm

e

f 7 cm

3 cm

10 cm 11 cm 9m

15 cm

13 cm

12 m

Example 3 Find the shaded area. 5 cm

The shaded area is found by calculating the total area and then subtracting the unshaded area. Area = area large circle − area small circle = π × 72 − π × 52 cm2 = 49π − 25π cm2 = 24π cm2
75.40 cm2

2

2 cm

Find the shaded areas (correct to 2 d.p. where necessary). a

b

c 1m

3m

4 cm

10 m

8 cm

2m 4m

5m

17 m

7 cm

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

d

e

f

23 cm

P

Q

27 cm 18 cm

8 cm 6 cm

R

S

10 cm

7 cm

11 cm

10 cm PR = 5 cm, SQ = 7 cm

3

Find the areas of the shaded regions (answer to 1 d.p.). a b

c

1 cm

4 cm

3 cm 100 m 5 cm

5 cm

d

9 cm

e

f 8 cm

8 cm 3 cm 11 cm

5 cm 17 cm 8 cm 8 cm

E. AREA APPLICATIONS This section involves practical problems using area.

Exercise 6E 1

Calculate the cost of carpeting a rectangular room 4.8 m long and 7.3 m wide, if carpet costs $72.95 a square metre.

2

The diagram shows the floor plan of a conference room. a Calculate the area of the conference room. b Calculate the cost of tiling the floor of the conference room if the tiling costs $32.80 per square metre. Scale 1 cm : 5 m

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3

A family room is in the shape of a rectangle with semicircles at either end as shown 4m in the diagram. a Calculate the area of the family room. 6m b Tiles cost $45 per square metre, and laying them costs $35 per square metre. Find the total cost of tiling the family room.

4

Concreters charge $18.90 per square metre. Calculate the cost of concreting the area shown.

8m 4.2 m 7.6 m

ro

3 -----10

of the area is used for pasture lands and the homestead. How many hectares are used for these purposes? e Calculate the value of the hobby farm if each square metre is valued at $7.20.

M

d

ad

A small triangular hobby farm is situated along a main road. a Calculate the area (in m2) of the farm. b How many hectares are there in this farm? c How many hectares are used for growing crops, if 7 ------ of the area of the farm is used for crop growing? 10

ain

5

Hobby farm

850 m

620 m

1 ha = 10 000 m2

6

A 2 m wide path is placed around a circular pond of diameter 4 m. Find the area of the path correct to the nearest whole number.

4m

2m

7

A garden bed is in the shape of a quadrant of a circle, radius 3.5 m. A path 1 m wide is to be built around the curved boundary only. Find the area of the path.

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

8

A farmer wants to spread 200 kg of superphosphate per hectare. What weight of superphosphate is required to fertilise this paddock? Give your answer in tonnes.

400 m 200 m

Example 1 A circle has the same area as a square with sides 10 cm. Find its radius.

r cm 10 cm 10 cm

Area of circle = πr 2 cm2 Area of square = 10 × 10 cm2 = 100 cm2 ∴ πr 2 = 100 100 ∴ r 2 = ---------π ∴ r 2
31.83 ∴ r
31.83 (r is positive) ∴ r
5.642 cm

9

A rectangle is 12 cm by 8 cm. a Find the area of the rectangle. b If the length of the rectangle is increased by 3 cm, find the width if the area remains the same.

10

A rectangle is 12 cm by 8 cm. If the length of the rectangle is increased by 4 cm, how must the breadth be varied so that the area remains the same?

11

A 10 cm by 16 cm rectangle has the same perimeter as a square. Which figure has the greater area? By how much?

12

A circle has the same area as a rectangle 15 cm × 7 cm. a Find the area of the rectangle. b Find the radius of the circle (to the nearest hundredth). c Which figure has the larger perimeter?

F. AREA AND PERIMETER INVOLVING SECTORS Earlier in this chapter we found the perimeters and areas of semicircles and quadrants. In this section we will find the areas and perimeters of sectors. To find the perimeter and area of sectors the fraction of the whole circle must be found first.

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Example 1 Calculate the perimeter of this sector. 30 The sector is ---------- of a circle. 360 30 ∴ The curved length = ---------- × 2 × π × r 360 30 = ---------- × 2 × π × 8 360 = 4.189 cm ∴ Perimeter = 4.189 + 8 + 8 = 20.2 cm (to 1 d.p.)

30° 8 cm

Always divide the angle by 360.

Exercise 6F 1

Find the fraction of a circle represented by these sectors. a b

4 cm

c

5 cm 12 cm 120°

60°

2

20°

Calculate the curved length and hence the perimeter of the sectors in question 1.

Example 2 Calculate the area of this sector. 50 The sector is ---------- of a circle. 360 50 ∴ Area = ---------- × πr2 360 50 = ---------- × π × (12.3)2 360 = 66.0 cm2 (to 1 d.p.)

50° 12.3 cm

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

3

Calculate the area of the sectors in question 1.

4

Calculate the perimeter and area of these figures. a

b 5m

c

320°

100°

d

200°

57 m

115 m

10°

15.7 m

Example 3 Calculate the perimeter of this figure. 4m

70 Curved length = ---------- × 2 × π × 4 360 = 4.887 m

70°

Perimeter = 4.887 + 1.5 + 4 + 1.5 + 4 = 15.9 m (to 1 d.p.)

5

1.5 m

Calculate the perimeter of these figures. a

b 8m

5m

40° 50°

5.29 m

5.1 m 1.8 m 8.4 m

c

d 53 cm 3.7 cm 70° 18 cm 50 cm

37° 0.65 m

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

Example 4 Calculate the area of this figure. 55 Area = ---------- × πr2 + lb 360 55 Area = ---------- × π × (135)2 + 81 × 135 360

55°

= 19 682.4 cm2 (to 1 d.p.)

81 cm 135 cm

6

Calculate the area of the figures in question 5.

7

Calculate the area and perimeter of this figure made of semicircles.

24 cm

8

Calculate the area of this figure.

30° 3m 50°

50°

8m 40°

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

Language in Mathematics

Johann Kepler (1571–1630) Johann Kepler was born in the German town of Wurttemberg. Although small and suffering from ill health, he was recognised as being intelligent. With a scholarship he was able to attend the University of Tubingen, where he studied first for the Lutheran ministry and then science. He studied under a master in astronomy who believed in, and taught, the Copernican theory that the Earth rotated around its own axis, and also about the Sun. Kepler taught mathematics in Graz from 1594. In 1600 he went to Prague and became assistant to Tycho Brahe, an important astronomer. After Brahe’s death, Kepler succeeded him as astronomer and mathematician to the emperor. Kepler had access to Brahe’s extensive records of observations and calculations. With his belief in the Copernican theory, he became one of the founders of modern astronomy. He developed three fundamental laws of planetary motion, now known as Kepler’s Laws, in 1609. These proposed, among other things, that the Sun was at the centre of our planetary system, and that the orbits of the planets were elliptical rather than circular. Sixty years later these laws helped Newton develop his Universal Law of Gravitation. Kepler also suggested that tides are caused by the Moon’s gravitational pull on the seas. He produced tables giving the positions of the Sun, Moon and planets, which were used for about 100 years. In 1611 he proposed an improved refracting telescope, and later suggested a reflecting telescope developed by Newton. 1

a b c d e f

How old was Kepler when he died? When and where did Kepler teach Mathematics? Describe the development of Kepler’s ideas concerning planetary motion. Research Kepler’s three laws. For how long were his tables of positions used? How are tides formed?

2

Insert the vowels in these glossary terms. a c __ rcl __ b q __ __ dr __ l __ t __r __ l c c __ mp __s __ t __ d rh __ mb __ s

☞

181

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

e k __ t __ g tr __ p __ z __ __ m

f s __ ct __ r h tr __ __ ngl __

3

Rearrange these words to form a sentence. a circle a semicircle A half is b a is of quarter quadrant A circle c are shapes calculated Composite dividing by area up the d may way than Composite more in areas one be found

4

Use every third letter to find the sentence. WDTRFHTGEHYAUJRNHEGBAVFOEDFSWAAZRDFHHJOLPMOE BQAUZDSFYOIJRBWAQAKCGIHJTIIEOPILLSGFHDEASKLAXFV BTHQHSOEYAPEFRHKOIPDNMUAECSDTCGOHNFBETWXHAUEI ODAGIBHAJKGNHODSNWEADFLTYS

Glossary area diameter perimeter rhombus triangle

circle formula quadrant sector

circumference kite quadrilateral semicircle

CHECK YOUR SKILLS 1

The formula A = 1--2- xy could be used to find the area of a: A parallelogram

2

B trapezium

C rhombus

The area of this trapezium is closest to: A 15.2 cm2 B 30.4 cm2 C 56.202 cm2 D 28.101 cm2

✓ composite figure parallelogram radius trapezium

5.1 cm

3.8 cm

2.9 cm

3

0.9 m

1.3 m

The area of this kite is: A 1.17 m2 B 2.34 m2 C 0.585 m2 D 2.2 m2

D all of these

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

4

The area of this rhombus is: A 107.01 cm2 B 13.37625 cm2 C 26.7525 cm2 D 53.505 cm2

8.7 cm 12.3 cm

5 10 m 4m

4m 20 m

6

The perimeter of this semicircular garden is closest to: A 7.2 m B 8.8 m C 4.4 m D 8.96 m

7 1.6 m

The shaded area is: A 174.9 cm2 B 99.5 cm2 C 225.1 cm2 D 300.5 cm2

2.8 m

✓

The perimeter of this shape is: A 18.05 m B 16.45 m C 10.5 m D 8.9 m

8

Tiarne has 35 m of garden edging. The radius of the circular garden she can enclose with this edging is: A 11.1 m B 5.6 m C 220 m D 3848 m

9

The perimeter of a rhombus with diagonals 5.2 cm and 8.6 cm is closest to: A 3.4 cm B 22.36 cm C 5 cm D 20.1 cm

10

The area of this figure is closest to: A 7.84 m2 B 14 m2 C 5.88 m2 D 9.8 m2

11

4.8 cm

183

1.5 cm

1.4 m

The shaded area is closest to: A 25.16 cm2 B 65.3 cm2 C 29.7 cm2 D 14.8 cm2

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

12

A farmer fertilises the paddock consisting of a rectangle and a semicircle as shown. The fertiliser is spread at the rate of 4.5 kilograms per square metre. The amount of fertiliser the farmer needs is closest to: A 10 t B 7.875 t C 12.2 t D 2.231 t

35 m

50 m

The perimeter of this sector is closest to: A 22 cm B 29 cm C 46 cm D 70 cm

13

105°

12 cm

14

The area of this sector is closest to: A 3.87 cm2 B 7.6 cm2 C 15.6 cm2 D 222.3 cm2

57° 3.9 cm

Use this diagram to answer questions 15 and 16. 15

16

4.2 m

The perimeter of the shape is closest to: A 67.42 m B 18.99 m C 14.8 m D 2.7855 m

38°

✓ 1.8 m

The area of the shape is closest to: A 5.85 m2 B 9.02 m2 2 C 13.41 m D 14.8 m2

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–5

6–9

10, 11

12

13–16

B

C

D

E

F

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

REVIEW SET 6A 1

2

Copy and complete. a 85 cm = ___ m

b 15 000 m2 = ___ ha

Find the perimeter of these shapes. a b

c

3.5 km = ___ m

c

d 8 cm 4 cm

3m

4 cm

3 cm 10 cm

18 m

7m

3

a A rectangular field 110 m by 75 m is to be fenced. Find the total length of fencing required. b Find the perimeter of a right-angled triangle with hypotenuse length 26 cm and one other side length 10 cm.

4

Find the perimeter correct to 1 decimal place. a b 15 cm 13 cm

8 cm

5 cm 24 cm

5 cm

5

A satellite has a circular orbit 700 km above the surface of the Earth. If the radius of the Earth is 6400 km, how far does the satellite travel in one orbit?

6

Find the area of these shapes. a

b

10 m

14 cm

5m

7 cm 5 cm 15 cm

6m

7

Find the shaded area. a

b

c

40°

8 cm

8 cm 3.2 m 4.8 m

10 cm

15 cm

185

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

REVIEW SET 6B 1

2

Copy and complete. a 4.28 ha = ___ m2

b 3 cm2 = ___ mm2

Find the perimeter of these shapes. a b

c

4300 cm = ___ m

c

d 25 m

5 cm

7 cm

10 cm

15 m

12 cm

12 cm

12 m

3

a A rectangular swimming pool is 20 m by 10 m and is surrounded by a path 2 m wide. What is the perimeter around the outside edge of the path? b Find the perimeter of a rhombus with diagonals 12 cm and 16 cm.

4

Find the perimeter correct to 1 decimal place. a b

12 cm

6 cm

15 cm

5

A machine makes circular plates with circumference 60 cm. Find the diameter of the plate correct to 1 decimal place.

6

Find the area of these shapes. a

b

c

7.3 cm

22.4 cm

80°

4.2 cm 5.3 m 6.5 cm

7

18.3 cm

Find the shaded area. a 1m 10 m

15 m

b 12 cm

5 cm

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

REVIEW SET 6C 1

Calculate the perimeter of each shape. a b

7 cm

11 cm

15 cm

8 cm 5 cm

2

3

Write the formula for the area of the following shapes. a rectangle b rhombus c Calculate the area of each shape. a

b

trapezium

17 cm 3 cm

18 cm 7 cm 6 cm

25 cm

9 cm

4

Farmer Smith has a rectangular paddock that is 408 m wide and 673 m in length. Calculate the cost of fencing the paddock if fencing costs $8.53 per m.

5

Karl wishes to cut a triangle from a rectangular piece of wood, as shown. Calculate: a the area of the triangle if the base is 52 cm and the height is 64 cm b the area remaining after the triangle is removed

1.2 m

64 cm 52 cm 2.3 m

6

Find the area and perimeter of these. a b 4.5 m 75°

4.3 m

8.2 m

6m

187

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Perimeter and Area (Chapter 6) Syllabus reference MS5.1.1, 5.2.1

REVIEW SET 6D 1

Calculate the perimeter of each shape. a

b

12 cm 15 cm

10 cm

24 cm 6 cm 12 cm

2

3

Write the formula for the area of these shapes. a trapezium b kite

c parallelogram

Calculate the area of each shape. a

b

29 cm 16 cm

17 cm

35 cm

33 cm

4

Crystal walks around her block three times each morning. If the block is 450 m by 384 m, calculate the distance that she walks each morning. Express your answer in kilometres.

5

Deborah’s lounge room is shown opposite. Calculate the cost of carpeting the lounge room if the carpet costs $119.80 per square metre.

6

Find the perimeter of each. a

3.4 m

2.3 m

2.7 m 1.6 m

b 2.1 m

8.3 m

2.1 m

7.2 cm 4.5 m

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Chapter 7 Coordinate Geometry This chapter deals with operations on the number plane. After completing this chapter you should be able to: ✓ determine the midpoint of an interval from a diagram ✓ use Pythagoras’ rule to find the length of a line interval drawn between two points ✓ find the gradient of an interval using a diagram ✓ determine whether a line has positive or negative slope ✓ find the gradient of a line using a right-angled triangle ✓ draw graphs of horizontal and vertical lines ✓ graph a variety of linear relationships ✓ graph simple non-linear relationships.

Syllabus reference PAS5.1.2 WM: S5.1.1–S5.1.5

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

Diagnostic Test 1

2

3

The midpoint of the join of (4, 3) and (10, 3) is: A (14, 6)

B (6, 3)

C (7, 3)

D (6, 6)

The gradient of the join of X(–5, –1) and Y(3, 5) is: 3 3 4 4 A + --B − --C + --D − --4 4 3 3

9

The gradient of this line is:

The midpoint of the join of (−3, −5) and (−3, 11) is: A (−6, 6)

B (−3, 6)

C (−6, 3)

D (−3, 3)

6 A --7 7 B --6 6 C − --7 7 D − --6

The midpoint of the join of (1, 4) and (8, 0) is: A (9, 4) C

4

8

(4 1--2-

B (7, 4) D (3 1--2- , 2)

, 2)

10

The gradient of the join of A(−5, 9) and B(7, 5) is: 1 1 A + --B − --C +3 D −3 3 3

11

Find the gradient of this line.

The midpoint of the join of (−5, 1) and (7, –5) is: A (1, −2)

B (2, −4)

C (6, 6)

D (6, 3) 4 y

5

6

7

The distance between the points (7, 1) and (2, 9) is: A

89

B

181

C

39

D

13

The distance between the points (−5, 7) and (4, −5) is:

3 2 1 0 −5 −4 −3 −2 −1 0 1 −1 −2

63

B

225

−4

C

5

D

21

−5 −6

The slope of MN is:

−7 −8

N

B −3 4 C --7 7 D --4

x 4 5

−3

A

A +3

2 3

−9

A +2 M

B −2

1 C + --2

1 D − --2

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

12

The equation of this line is:

B y

4 y 3

5 4

2

3

1 0 −5 −4 −3 −2 −1 0 1

13

2 3

A y=3

B x=3

C y = 3x

D x = 3y

2

x 4 5

1 0 −4 −3 −2 −1 0 1 −1 −2

−3

0

x 4

−3

By using this table, the graph of y = 2x −1 is: x

2 3

−4 −5 −6

3

y C A

y

y

6 5

7 6 5

4 3

4

2

3

1 0 −4 −3 −2 −1 0 1 −1 −2

2 1 0 −4 −3 −2 −1 0 1 −1 −2

2 3

x 4

2 3

x 4

−3 −4

−3

−5

−4

−6

−5 −6 −7 −8

D y 1 0 −4 −3 −2 −1 0 1 −1 −2 −3

2 3

x 4

195

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

14

By completing this table of values for each equation, the equation of this graph is: x

−3

0

15

The line containing the point (2, −2) is: A y = 3x − 8

B y = −3x − 3

C y = 4x + 14

D y = −4x − 11

3 16

The equation of this graph is:

y 9 y 8 7

y 9 8

6 5

7 6 5

4

4

2

3

3 2 1 0 −4 −3 −2 −1 0 1 −1 −2

2 3

x 4

1 0 −4 −3 −2 −1 0 1 −1

x 2 3 4

A y = 3x

B y = 2x

C y = x2

D y = x2 + 1

−3 −4 −5 −6 −7 −8 −9

A y = 3x − 1

B y = −3x + 1

C y = 4x − 1

D y = −4x − 1

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–4

5, 6

7, 8

9–11

12

13–15

16

A

B

C

D

E

F

G

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

A. MIDPOINT The midpoint of an interval is the point halfway between the end points of the interval.

S

T

L Midpoint M

B A

Example 1 Plot each pair of points, join them with a straight line and find the coordinates of the midpoint. a (3, 2) and (9, 2) a

b (5, 1) and (5, 7) The length of this horizontal line interval is 6 units. The midpoint then is 3 units from either end.

Midpoint

(3, 2)

(9, 2)

Both points and the midpoint have y-ordinate of 2. The coordinates of the midpoint are (6, 2).

(5, 9)

b Midpoint

(5, 1)

The length of this vertical line interval is 8 units. The midpoint then is 4 units from either end. Both points and the midpoint have x-ordinate of 5. The coordinates of the midpoint are (5, 5).

Exercise 7A 1

Plot the following pairs of points, join them with a horizontal line and find the coordinates of the midpoint. a (1, 4) and (9, 4) b (2, 3) and (12, 3) c (3, 6) and (7, 6)

2

Plot the following pairs of points, join them with a vertical line and find the coordinates of the midpoint. a (2, 1) and (2, 11) b (5, 3) and (5, 7) c (3, 4) and (3, 8)

197

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

Example 2 Plot each pair of points, join them and find the midpoint. a (–5, –2) and (8, –2) a 5

b (1, –5) and (1, 7) The length of the interval is 5 + 8 = 13 units. The midpoint is 13 ÷ 2 = 6 1--2units from either end.

y

4 3

−5 + 6 1--2- = 1 1--2-

2 1 0 −5 −4 −3 −2 −1 0 1 −1 −2 (−5, −2) −3

x 2 3

4

5 6

7 8

or 8 − 6 1--2- = 1 1--2∴ the midpoint is (1 1--2- , −2)

(8, −2)

−4 −5 y 7

b

(1, 7)

The length of the interval is 5 + 7 =12 units. The midpoint is 12 ÷ 2 = 6 units from either end.

6 5 4 3 2 1 0 −5 −4 −3 −2 −1 0 1 2 3 −1 −2

x 4

−5 + 6 = 1 or 7 − 6 = 1 ∴ the midpoint is (1, 1)

5

−3 −4 −5

3

(1, −5)

Plot the following pairs of points and find the coordinates of the midpoint of these horizontal and vertical line intervals. a (1, 2) and (1, 4) b (−2, 1) and (−2, 5) c (3, −1) and (3, 3) d (5, 1) and (3, 1) e (5, 2) and (−3, 2) f (−4, −3) and (−4, 5) g (0, 0) and (0, 9) h (0, 0) and (3, 0) i (7, 14) and (−3, 14)

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

Example 3 Plot the following pairs of points and find the coordinates of these oblique lines. b (−2, −5) and (3, −1)

a (1, 4) and (5, 9) a i

An oblique line is neither vertical nor horizontal.

Plot the points and join with a straight line.

y 9

(5, 9)

8 7 6 5 4 3

(1, 4)

(5, 4)

2 1 1

2 3

4 5 6

x

7

ii Draw vertical and horizontal lines to make a right-angled triangle. iii Write the coordinates of the third vertex. iv Find the midpoint of each interval.

y 9

(5, 9)

8 7

(5, 6 12 )

6 5 4

3 (1, 4) (3, 4) (5, 4) 2 1

x 1

2 3

4

5

6 7

v The midpoint is (3, 6 1--2- ). b Use the same steps as in part a. length is 5 units 5 ÷ 2 = 2 1--2∴ x ordinate is --12−2 + 2 --12- = --121 ∴ midpoint of interval is ( --2- , −5) height is −1 − −5 = 4 units 4÷2=2 ∴ y ordinate is −3 –1 –2 = –3 ∴ midpoint of the line interval is ( --12- , −3)

y1

–2 –1 1

2 3 4

(3, −1)

2

(3, −3)

3 4 (–2, –5)

5 6

5

( 1 , −5) (3, −5) 2

x

199

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

4

Plot the following pairs of points and find the coordinates of the midpoint of these oblique lines. a (5, 2) and (1, 4) b (2, 0) and (0, 8) c (3, −1) and (1, −5) d (−2, 5) and (2, −5) e (2, −1) and (−1, 3) f (5, 7) and (−3, −1) g (−2, 3) and (−5, 1) h (−4, −4) and (−1, 1) i (−2, −3) and (2, 3)

Example 4 Find the midpoint of the line segment joining A(−5, 2) and B(7, −3). Since the midpoint M is halfway between A and B, then the x-coordinate of M will be halfway between the x-coordinates of A and B:

y 4 3

A(−5, 2)

Similarly, the y-coordinate of M will be halfway between the y-coordinates of A and B: 2 + ( –3 ) i.e. y-coordinate of M = --------------------- = − 1--22

Midpoint

2

–5+7 i.e. x-coordinate of M = ----------------- = 1 2

1 −5 −4 −3 −2 −1 −1

x 2

1

3

4

5

6

7

−2 −3 −4

B(7, −3)

∴ the coordinates of the midpoint M are (1, − 1--2- ) 5

Find the midpoint of the line segment joining the following pairs of points. a (5, −8) and (3, −3) b (−2, −2) and (6, −3) c (0, 6) and (6, 0) d (15, 27) and (17, 3) e (51, −12) and (−36, 11) f (0, 0) and (−7, −11)

B. DISTANCE BETWEEN TWO POINTS Pythagoras’ rule can be used to find the length of the line interval joining two points.

Example 1 Plot the points (1, 4) and (8, 6) on the number plane. Find the distance between the two points. First draw in a right-angled triangle. Find the length of each side: • the vertical side = 6 − 4 = 2 units • the horizontal side = 8 − 1 = 7 units Use Pythagoras’ rule: c2 = a2 + b2 = 22 + 72 c 2 = 53 c = 53 = 7.28 units (to 2 d.p.)

8 y 7 6 5 4 3

(8, 6) (1, 4)

2

7

2 1 0

x 0 1

2 3

4 5

6 7 8

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

Exercise 7B 1

Plot the following pairs of points and find the distance between them. a (2, 3) and (5, 7) b (5, 3) and (8, 6) c (8, 7) and (3, 3) d (1, 9) and (7, 2) e (2, 8) and (7, 5) f (0, 0) and (5, 7)

Example 2 Find the distance between the points (–5, 8) and (4, –2). i

Plot the points and draw in the right-angled triangle. ii Find the length of each side. iii Use Pythagoras’ rule: c2 = a2 + b2 c 2 = 102 + 92 10 2 c = 181 (2 + 8 = 10) c = 181 c = 13.45 (to 2 d.p.)

(8, 6)

8 y 7 6 5 4 3 (1, 4) 2 1 0

−6 −5 −4 −3 −2 −1 0 1 –1 –2 9 –3 (5 + 4 = 9)

2

Find the distance between each pair of points. a (−5, 3) and (6, 2) b (−2, −5) and (3, 7) d (−7, 0) and (5, −4) e (−8, −3) and (0, 0)

x 2 3

4 5

6

c (−4, −5) and (5, −1) f (4, 0) and (0, −3)

C. SLOPE (GRADIENT) We use the words slope or gradient when talking about the degree of steepness of a line or a line segment. Horizontal lines have no slope. This line is very steep. It must therefore have a large slope.

To compare the slope of different lines we use the ratio of vertical rise to horizontal run. vertical rise vertical rise slope = -----------------------------horizontal run

horizontal run

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

Here are some varying slopes. House roof

Leaning Tower of Pisa

Escalator

2m 8m

56 m

4m 10 m

slope = =

4m

4 slope = ------

2 --8 1 --4

=

------slope = 56 4

10 2 --5

= 14

For a horizontal line the vertical rise is 0, therefore the slope is 0. When line segments are drawn on graph paper, we can easily determine the slope of the line segments by drawing horizontal and vertical lines to complete a right-angled triangle.

Example 1 Find the slope of AB.

Draw in a right-angled triangle, the same as when calculating distance. Slope of AB vertical rise = -----------------------------------horizontal run =

2 --5

2 5

Exercise 7C 1

Find the slope of the following. a b

c 10 m

150

4m

3m 4m

1000 m an uphill road

6m

a barn roof

a slippery-dip

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

2

In each of the diagrams, draw a right-angled triangle and find the gradient using vertical rise gradient = ------------------------------------ . horizontal run a

b B

B

A

A

c

B

B

d

A A

Example 2 Find the gradient of the line passing through C(−4, −2) and D(3, 2). Plot the points and draw the right-angled triangle showing side lengths. rise gradient = --------run =

3

4 --7

Find the gradient of the line passing through each pair of points. a C(–5, –2) and D(4, 5) b A(–3, –1) and B(5, 2) c C(–5, 3) and P(7, 7) d M(1, –5) and N(2, 6)

203

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

Investigation 1 WM: Communicating, Reflecting

Varying the slope Copy and complete the table.

1 N

Line segment

x-run

y-rise

Slope

L

AB

J

CD H F

M I K

G

GH IJ

D

E C A

2

EF

B

KL MN

Copy and complete: a The slope of a horizontal line is ____. b The slope of a vertical line is ____. c As the line segments become steeper, their slopes ____.

D. POSITIVE AND NEGATIVE GRADIENTS Negative slopes In the figure opposite Line 1 and Line 2 are parallel, and each of them has the same slope of 2. Line 3 is not parallel to Lines 1 and 2, yet it has the same degree of steepness. Line 1

Line 2

Line 3

We say that Lines 1 and 2 are forward sloping whereas Line 3 is backward sloping. As we go from left to right on Line 1 we are going uphill, whereas on Line 3 we are going downhill.

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

Example 1 Determine the slope of AB and CD in the following.

B

C

D

A

B C 6

5

A 4

The slope of AB is positive (uphill) rise ∴ slope AB = + --------run +6 = -----4 = +1 1--2-

2 D

The slope of CD is negative (downhill) rise ∴ slope CD = – --------run 5 = – --2 = –2 1--2-

Determine positive or negative slope first.

205

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

Exercise 7D 1

Find the gradient of each line. Determine positive or negative slope first. a

b

c

T C

B

A

d

D

e

E

S

X Y

N

D f

g

T

V M 2

Determine the slope of: a OA b OB c OC d OD e OE f OF g OG

G

F E

D

C B O

A

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

3

Determine the slope of: a AP b AQ c AR d AS e AT f AU

P R

Q

T

S

U

4

U S Q R

P

V

T

O

Imagine you are walking across the countryside from O to W (i.e. from left to right). a When are you going uphill? b When are you going downhill? c Where is the steepest positive slope? d Where is the steepest negative slope? e Where is the slope 0? f When is the slope not zero but least?

Example 2 a Plot the points A(–3, 5) and B(7, 2). b Find the gradient of the lines though A and B. a

Plot the points and draw the line. (–3, 5) A

y

B (7, 2) 10

b

x

Find the side lengths of the triangle. The slope is negative as it is ‘downhill’. rise Gradient = --------run 3 = – -----10

W

207

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

5

Plot the following pairs of points and determine the gradient of the line passing through the two points. a A(–4, 6) and B(7, 2) b C(–4, –1) and D(5, 3) c P(1, 3) and Q(–4, –1) d R(0, 0) and S(5, 3) e M(5, 3) and N(–5, 2) f S(–3, –2) and T(4, –6)

Example 3 Find the gradient of this line.

y 4 3 2 1 x –2 –1

1

2 3

–2 –3

First choose any two points on the line and draw in a right-angled triangle. The slope is positive ‘uphill’. rise Gradient = --------run 6 = + --5

6

By choosing two points on each line, find the gradient of these straight lines. a b

4

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

c

d

Example 4 Find the gradient of the given line.

Draw a right-angled triangle, labelling the rise and run. rise Gradient = --------run 3 = --6 1 = --2

7

Find the gradient of the following lines. a

b 3

c

209

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

d

e

f

–3

8

Find the slope of each of these graphs. Be careful as the scales are not the same. a

b

c

d

e

f

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

Investigation 2 WM: Reasoning, Communicating

The slope of a line 1

Copy and complete the table

Line segment

x-run

y-rise

y-rise ----------------x-run

BC

2

1

1 --2

DE AC BE AE AF 2

State, in sentence form, any conclusions drawn from the graph and table.

Investigation 3 WM: Reflecting, Communicating

Relating gradient and the tangent ratio 1

Plot the points A (1, 2) and B (5, 9).

2

Form a right-angled triangle and write the lengths of the horizontal and vertical lines.

3

Find the gradient of AB.

4

Label the angle at A as θ.

5

With respect to θ, label the sides as opposite, adjacent and hypotenuse.

6

Write an expression for tan θ.

7

Compare tan θ and the gradient.

8

Explain the result from question 7.

9

Calculate the angle equal to the slope of the line with the x-axis.

10

Calculate the angle for the gradient of the join of the points in Exercise 12D Question 5.

11

Copy and complete: The gradient of a line is equal to ___ θ, where θ is the angle made by the line and the ___ axis.

211

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

E. LINES PARALLEL TO THE AXES In section A of this chapter, we found the midpoint of horizontal and vertical lines. In the first example, the points (3, 2) and (9, 2) were joined to give a midpoint of (6, 2). By noticing that all points have a y-ordinate of 2, the equation of that horizontal line must be y = 2. The points (5, 1) and (5, 9) were joined by a vertical line, and the midpoint was found to be (5, 5). All three points have a x-ordinate of 5. This shows that the equation of the vertical line is x = 5. Horizontal lines are of the form y = a where a is a positive or negative number. Vertical lines are of the form x = b where b is a positive or negative number.

Example 1 Graph these lines: a x=3 a x=3

b y = –2

This should be a vertical line. Use this table of values to check.

x

3

3

3

3

y

–1

0

1

2

All x values are 3.

Plotting the points gives:

b y = –2

This should be a horizontal line. Use this table of values to check.

x

–2

–1

0

1

2

y

–2

–2

–2

–2

–2

All y values are –2.

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Plotting the points gives:

Exercise 7E Graph these horizontal and vertical lines. a y=3 b x=1 e x=5 f y=8

1

c x = –2 g y = –3

d y = –4 h x=7

2

a List the equations from question 1 that represent horizontal lines. b Write the coordinates of the point where each of these lines cuts the y-axis.

3

a List the equations from question 1 that represent vertical lines. b Write the coordinates of the point where each of these lines cuts the x-axis.

4

a Graph these horizontal lines on the sane number plane: y = –2, y = –1, y = 1, y = 2 b The x-axis is a horizontal line. The equation of the x-axis is y = ___. c Explain your answer in part b.

5

a Graph these vertical lines on the same number plane: x = –2, x = –1, x = 1, x = 2 b Write the equation of the y-axis. c Explain your answer to part b.

F. GRAPHING LINEAR RELATIONSHIPS Consider all points (x, y) in which y = 2x. We can write down any number of ordered pairs that satisfy this rule. For example, (1, 2), (2, 4), (3, 6), (–2, –4), (0.7, 1.4). To visualise all the points that satisfy the equation y = 2x, we draw a graph of all points near the origin, O. Often we find that a table of values is useful. For y = 2x, a table of values is: x

–3

–2

–1

0

1

2

3

y

–6

–4

–2

0

2

4

6

Graphing these seven points only, we get the graph in Figure 1. However, not just the integers satisfy this rule. Figure 2 on the next page is the plot of all points with x-values 0.25 units apart.

☞

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In fact, if we imagine ordered pairs where the x-values are the complete set of real numbers, we would obtain the complete line as in Figure 3.

Example 1 Draw the graph of the lines with these equations. a y=x+2

b y = –2x

c y= x–1

When you choose values for the table you can choose any x value you like. a y=x+2 x

–3

–2

–1

0

1

2

3

y

–1

0

1

2

3

4

5

b y = –2x x

–2

–1

0

1

y

4

2

0

–2

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

c y=

1 --2

x–1

x

–3

–2

–1

0

1

2

3

y

–2 1--2-

–2

–1 1--2-

–1

– 1--2-

0

1 --2

Note: When graphing the lines shown above, the line is extended past the plotted points with an arrow on each end to show that it continues in both directions. Write the equation on the line.

Exercise 7F In each of the following, copy and complete the table of ordered pairs.

1

a y=x x

b y=x–2 1

2

3

4

5

x

y

0

1

2

3

–1

0

1

2

–1

0

1

2

–1

0

1

2

y d y=5–x

c y = –x x

–2

–1

0

1

2

x

y

–2

y

e y = 2x + 1 x

–2

f –1

0

1

2

y

y = – 1--2- x x

–2

y

g y = 8 – 2x x

–2

y

2

–1

h y = 1 – 3x –1

0

1

2

x

–2

y

i

Copy and complete the following tables using the rule provided.

ii Plot each set of ordered pairs on separate axes and draw the straight line through the points.

☞

215

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a y=x+1 x

y=x+3

b

–2

–1

0

1

2

x

y

–2

0

2

4

–1

0

1

2

–1

0

1

2

y

c y = –x – 1 x

–4

y=4–x

d –2

0

2

x

4

–2

y

y e y = 2x – 2 x

–2

y = 3 – 2x

f –1

0

1

x

2

–2

y

y

3

–4

For the straight lines in questions 1 and 2: a Write a list of the equations with a positive gradient. b Write a list of the equations with a negative gradient. c What is the difference between these groups of equations? d Without drawing, state whether each of these equations has positive or negative slope. i y = 2x – 1 ii y = –3x + 4 iii y = 5 – 7x iv y = 3 + 2x v y = 7x – 1 vi y = –5x + 2

Example 2 By using this table of values draw graphs of: x

–3

0

3

y

a

x+y=7

When x = –3 –3 + y = 7 y =7+3 y = 10

a

x+y=7

b

x–y=3

x

–3

0

3

y

10

7

4

When x = 0 0+y=7 y =7

When x = 3 3+y=7 y =7–3 =4

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

b x–y=3

When x = –3 –3 – y = 3 –y = 3 + 3 –y = 6 y = –6

4

x

–3

0

3

y

–6

–3

0

When x = 0 x–y =3 0–y =3 y = –3

When x = 3 x–y =3 3–y=3 –y = 0 y =0

By using a table of values, draw on separate number planes the graphs of:

x

–3

0

3

y a y=x–4

b y=x+4

e y = 1--2- x

f j

i

x–y=8

m 3y – 2x = –12

c y = 2x

d y=1–x

2x + 3 y = ---------------4

g x + y = –3

h x+y=1

x–y=6

k y = –4 + x

l

3x n y = ------ – 2 2

o 2x – 3y = 6

p 5x + 3y = 30

x+y=8

Example 3 Does the point (2, 2) lie on the line y = 2x – 1? Draw a table of values for y = 2x – 1. x

–2

–1

0

1

y

–5

–3

–1

1

Plot these points and draw the line y = 2x – 1. Plot the point (2, 2). The point (2, 2) does not lie on the line.

5

Using the graphs you drew in question 4, answer the following questions. a Does the point (1, –3) lie on the line y = x – 4? b Does the point (–1, 2) lie on the line y = x + 4? c Does the point (0, 2) lie on the line y = 2x?

☞

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d Does the point (3, –2) lie on the line y = 1 – x? e Does the point (4, 1) lie on the line y = 1--2- x? 2x + 3 Does the point (0, 2) lie on the line y = ---------------- ? 4 g Does the point (–2, –1) lie on the line x + y = –3? h Does the point (6, –2) lie on the line x – y = 8? i Does the point (2, –4) lie on the line x – y = 6? f

6

When making up a table of values from rules, I unfortunately mix them up. Can you sort out which graph belongs to which table of values? a A y = 2x x –4 –2 0 2 4 y

–2

–1

0

1

2

b x

0

1

2

3

4

y

0

2

4

6

8

x

–4

–2

0

1

3

y

–2

0

2

3

5

x

–2

0

1

2

3

y

4

2

1

0

–2

x

–3

–2

–1

0

1

y

3

2

1

0

–1

B y = –x

c C y=

1 --2

x

d D y=x+2

e E y=2–x

Example 4 You can check if a point lies on a line without drawing it. By substituting the x value of the point (13, 25) and finding the corresponding y value, decide if the point in brackets lies on the line y = 2x – 1. The x value is 13.

As y = 2x - 1 y = 2 (13) - 1 y = 25 Since the y values are equal the point lies on the line.

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7

By substituting the x-value of the point given in brackets and finding the corresponding y-value, decide if the point in brackets lies on the given line. a y=4–x (2, 2) b y=x+4 (1, 5) c y = 2x (3, 8) d y=1–x (5, 4) e y = 1--2- x (3, 6) f y = 2x + 3 (–2, –1) g y = 2x – 3 (–4, –11) h y = 3 – 2x (–5, 4) i y = 3x – 2 (10, 28)

8

Find five points that lie on these lines. a y=x+3 b 2x + y = 5

c 3x – 2y = 6

Investigation 4 WM: Applying Strategies, Communicating

Graphics calculator 1

These instructions are for a CASIO CFX9850GB PLUS. a Select GRAPH from the MAIN MENU. b Enter the equation y = 2x + 3 by pressing x,θ,T + 3 then EXE 2 c Press F6 to DRAW. d The graph should appear on the screen. If the scale of the axes needs adjustment V–Window

press

F3

and adjust as needed. Press EXIT to return.

e Press F2 F1 to delete any graphs after choosing the equations. f To have more than one graph at a time on the screen omit instruction e. 2

a Draw these graphs on the same screen. (Use different colours if you can.) y = 2x +1, y = 2x – 1 and y = 2x + 3 b What observation can you make?

3

a Graph: y = –3x – 1, y = –3x and y = –3x + 2. b Comment on these graphs.

4

a Graph: y = x + 2, y = x and y = x – 5. b Comment on these graphs.

5

a Graph: y = 3x – 1, y = 2x – 1 and y = –4x – 1. b What comment can be made?

6

a Draw the graphs of y = x 2, y = x 2 + 1 and y = x 2 – 2 on the same axes. b Comment on these graphs.

☞

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

7

a Graph y = 2 x, y = 3 x and y = 5 x on the same set of axes. b Comment on the similarities in these graphs.

8

Draw the graphs from the exercises in sections F and G.

9

Make up some of your own.

G. NON-LINEAR RELATIONSHIPS Straight lines are one type of relationship that can be graphed. There are many relationships that, when graphed, are not straight lines. This section examines some of these. Note that graphics calculators are an excellent tool in this section. The parabola is the name given to the graph relating y to x 2. The simplest parabola is y = x 2. The exponential graph has x as a power. An example is y = 2x.

Exercise 7G 1

a Complete this table of values for y = x2. x

–4

–3

–2

– --12-

–1

1 --2

0

1

2

3

4

y

b Plot these points and draw a smooth curve through them. 2

By using a table with the same values as question 1, or a graphics calculator, graph on the same number plane: a y = x2, y = 2x2, y =

1 2 --- x 2

b y = x2, y = 3x2, y = 4x2, y =

1 2 --- x 4

c y = x2, y = x2 + 1, y = x2 + 2, y = x2 – 1 3

Write some observations about each of the four groups of parabolas in question 2.

Example 1 Complete this table and sketch y = 3x. x y

–3

–2

–1

0

1

2

3

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

Use the Xy button on the calculator to find the values. x

–3

–2

–1

0

1

2

3

y

0.04

0.1

0.3

1

3

9

27

x

Extend the graph past the end points.

4

a By using the table, or a graphics calculator, graph on the same number plane: y = 2x, y = 3x and y = 4x. x

–3

–2

–1

0

1

2

3

y b What do you notice about all three graphs?

Language in Mathematics Pierre de Fermat

(1601–1665)

Pierre de Fermat was born in Beaumont-de-Lomagne in France, near the border with Spain. He studied Latin and Greek literature, ancient science, mathematics and modern languages at the University of Toulouse, but his main purpose was to study law. In 1629 Fermat studied the work of Appollonius, a geometer of ancient Greece, and discovered for himself that loci or sets of points could be studied using coordinates and algebra. His work ‘Introduction to Loci’ was not published for another fifty years, and together with ‘La Geometrie’ by Descartes, formed the basis of Cartesian geometry.

☞

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In 1631 Fermat received his degree in law, was later awarded the status of a minor nobleman, and in 1648 became King’s Councillor, Fermat was a man of great integrity who worked hard. He remained aloof from matters outside his own jurisdiction, and pursued his great interest in mathematics. He worked with Pascal on the theory of probability and the principles of permutations and combinations. He worked on a variety of equations and curves and the Archimedean spiral. In 1657 he wrote ‘Concerning the Comparison of Curved and Straight Lines’ which was published during his lifetime. Fermat died in 1665. He was acknowledged master of mathematics in France at the time, but his fame would have been greater if he had published more of his work while he was alive. He became known as the founder of the modem theory of numbers. In mid-1993, one of the most famous unsolved problems in mathematics, Fermat’s Last Theorem was solved by Andrew Wiles of Princeton University (USA). Wiles made the final breakthrough after 350 years of searching by many famous mathematicians (both amateur and professional). Wiles is a former student and collaborator of Australian Mathematician John Coates. Fermat’s Last Theorem is a simple assertion which he wrote in the margin of a mathematics book, but which he never proved, although he claimed he could. The theorem is: The equation xn + yn = zn, when the exponent n is greater than 2, has no solutions in positive integers. Wiles’ work establishes a whole new mathematical theory, proposed and developed over the last 60 years by the finest mathematical minds of the 20th century. 1

Read the article about Pierre de Fermat and answer the questions. a How many years was Fermat alive? b List four of Fermat’s achievements c How many publications did Fermat have in his lifetime? d Why was Fermat not as famous as he could have been? e What was the only article published by Fermat in his lifetime?

2

Complete these glossary words by inserting the vowels. a v__rt__c__l b h__r__z__nt__l

3

c

gr__d__ __nt

d d__st__nc__

e

__bl__q__ __

f

g

l__n__ __r

sl__p__

Rearrange these sentences, the first word has a capital letter. a The equation y-axis x = 0 has the b has The x-axis y = 0 equation the c positive uphill gradient slope An is a d in is a midpoint middle The the of line e a the If gradient is goes downhill negative line f interval Pythagoras’ found The is using an length of theorem

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

4

Use every third letter to reveal the message.

AWIERNGHCUJOIKOOPRLKDNGISENRTAASTXCEVFGBGENHOMJM NHEWETRTRGFYDCAESNEDIFGNJKTOPELKRJHVNBAGFLCDJESO WAISDNFGSHJTKMWNBOVCPASODFIGHNBGTVFSCDADENDSDSEH WSAQASASLASEADNFGGHJTKLHIFADFMGBIHJDKMPDEOSCIFJN MTYAAXINKEDSJSMWLZFOGHPJKE

Glossary coordinates horizontal negative positive triangle

decimal place interval non-linear Pythagoras vertex

endpoint length number plane relationships vertical

✓ gradient linear oblique right-angled triangle

halfway midpoint operations slope

CHECK YOUR SKILLS 1

2

3

4

5

6

7

223

The midpoint of the join of (3, 5) and (9, 5) is: A (12, 5) B (6, 5)

C (6, 10)

D (6, 0)

The midpoint of the join of (2, –4) and (2, 10) is: A (2, 3) B (2, 6)

C (0, 6)

D (4, 6)

The midpoint of the join of (–1, 3) and (9, 1) is: A (8, 2) B (5, 2)

C (8, 4)

D (4, 2)

The midpoint of the join of (–6, 3) and (5, –7) is: A (–1, –4) B (– 1--2- , –2)

C ( 1--2- , 2)

D (–5 1--2- , 5)

The distance between the points (5, 3) and (0, 10) is: A 12 B 144 C

24

D

74

The distance between the points (–5, 8) and (6, –5) is: A 290 B 48 C

170

D

2

The slope of AB is: 7 A --5 –7 B -----–5 5 C --7 –5 D -----–7

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

8

The gradient of the join of A(–7, –1) and B(3, 6) is: 10 A -----7

9

–10 B --------7

7 C -----10

The gradient of this line interval is: A 7 B –7 C 1 D –1

10

11

The gradient of the join of A(–5, 7) and B(3, 5) is: 1 –1 A --B -----4 4 Find the gradient of this line. A +1 B –1 C +8 D –8

12

–7 D -----10

The equation of this line is: A y=2 B x=2 C y = 2x D x = 2y

C 4

✓ D –4

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

13

225

By using this table the graph of y = 2x +1 is: x

–3

0

3

y A

B

C

14

D

x

–3

0

3

y

A y = 3x – 1 B y = –3x + 1 C y = 4x – 1 D y = –4x – 1

15

✓

By completing this table of values for each equation, the equation of this graph is:

The line containing the point (–4, 9) is: A y = 3x – 8 B y = –3x – 3

C y = 4x + 14

D y = –4x – 11

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16

The equation of this graph is: A y = 3x B y = 2x C y = x2 D y = x2 + 1

x

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1–4

5, 6

7, 8

9–11

12

13–15

16

Section

A

B

C

D

E

F

G

REVIEW SET 7A 1

By drawing a diagram and plotting the points, find the midpoint of the join of: a (4, 3) and (10, 3) b (2, 5) and (2, 9) c (4, 1) and (8, 6) d (–4, 3) and (10, –1)

2

Using Pythagoras’ theorem, find the distance between these pairs of points: a (4, 3) and (10, 7) b (–2, 5) and (2, 3)

3

In this diagram draw a right-angled triangle and find the gradient of the line.

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

4

Find the gradient of the line passing through these pairs of points. a (–5, –2) and (6, 3) b (–3, 6) and (7, 1)

5

Find the gradient of this line.

6

Draw a sketch of each of these lines. a y=3

b y = –4

e x+y=5

f

c y=x+3

d y = 4 – 3x

x–y=2

7

Does the point (4, –3) lie on the line y = 2x – 11? Explain your answer.

8

Draw a neat sketch of the relation y = x2.

REVIEW SET 7B 1

Use these diagrams to find the midpoint of each line interval. a b

2

a Find the distance between the two points in question 1 part b. b Find the distance between the points (1, 8) and (5, 1).

3

a Find the gradient of the line in question 1 part b. b Find the gradient of the line passing through the points (–4, –2) and (2, 3).

227

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

4

Find the gradient of this line.

5

Complete this table for each relation and draw a sketch of each of the lines. a y=x+4 b y = 2 – 3x x –3 0 3 c x + y = –8 d x–y=0 y

6

Does the point (–1, –3) lie on the line x – y = 4? Explain your answer.

7

a Complete this table for the relation y = 2x. b Draw a neat sketch of the relation y = 2x.

x

–2

–1

0

y

REVIEW SET 7C 1

By drawing a diagram and plotting the points, find the midpoint of the join of: a (5, 3) and (11, 3) b (2, 1) and (2, 9) c (3, 1) and (8, 4) d (–5, 3) and (10, –4)

2

Using Pythagoras’ theorem, find the distance between these pairs of points: a (4, 3) and (9, 8) b (–2, 7) and (3, 3)

3

In this diagram draw a right-angled triangle and find the gradient of the line.

4

Find the gradient of the line passing through these pairs of points: a (–7, –2) and (6, 3) b (–3, 6) and (8, 3)

1

2

3

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Coordinate Geometry (Chapter 7) Syllabus reference PAS5.1.2

5

Find the gradient of this line.

6

Draw a sketch of each of these lines: a y = –5 c y=x–2 e y = 2x + 1 g x–y=6

b x=2 d y = 5 – 4x f x + y = –1

7

Does the point (2, 5) lie on the line y = –3x – 1? Explain your answer.

8

Draw neat sketches of these relations. a y = x2 + 2 b y = 3x

REVIEW SET 7D 1

Use these diagrams to find the midpoint of each line interval. a

2

b

a Find the distance between the two points in question 1 part b. b Find the distance between the points (3, 8) and (5, 4).

229

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3

a Find the gradient of the line in question 1 part b. b Find the gradient of the line passing through the points (–7, –2) and (2, 1).

4

Find the gradient of this line.

5

Complete this table for each relation and draw a sketch of each of the lines. x y

–3

0

3

a y=3–x c x + y = –1

b y = 2x – 3 d x–y=6

6

Does the point (–1, 5) lie on the line x + y = 4? Explain your answer.

7

a Complete this table for the relation y = 3x. b Draw a neat sketch of the relation y = 3x.

x y

–2

–1

0

1

2

3

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Chapter 8 Polygons This chapter deals with the development and application related to the angle sum of interior and exterior angles for convex polygons. After completing this chapter you should be able to: ✓ find the interior angle sum of polygons ✓ establish the result for the sum of the exterior angles of a polygon ✓ apply angle sum results to find unknown angles.

Syllabus reference SGS5.2.1 WM: S5.2.1–S5.2.4

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Polygons (Chapter 8) Syllabus reference SGS5.2.1

Diagnostic Test

1

The angle sum of a seven-sided polygon is:

5

A 540° B 720° C 900° D 1080° 2

3

4

A six-sided polygon is called a: A pentagon

B octagon

C decagon

D hexagon

6

A regular polygon with 20 sides has interior angle size of: A 3240°

B 162°

C 3600°

D 18°

A pentagon has one right angle and the other angles are all equal. The size of the equal angles is: A 540°

B 450°

C 112.5°

D 90°

The exterior angle of a 15-sided regular polygon is: A 24°

B 156°

C 360°

D 2340°

The value of x in this diagram is:

An irregular polygon has one exterior angle of 100° and all others 10°. The number of sides of this polygon is:

A 25°

A 26

7

B 27

C 36

D 37

B 65° C 205°

120°

85°

D 115° x°

130°

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–5

6, 7

A

B

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Polygons (Chapter 8) Syllabus reference SGS5.2.1

A. INTERIOR ANGLE SUM OF A POLYGON A polygon is a plane shape with straight sides. Polygons are named according to the number of sides. The simplest polygon is a triangle, which has an interior angle sum of 180°. A quadrilateral has four sides and an angle sum of 360°.

Example 1 Find the sum of the interior angles of a heptagon. A heptagon is a polygon with seven sides. Choose one vertex and draw all the diagonals from that vertex. There are five triangles in the heptagon. The angle sum of each triangle is 180°. ∴ interior angle sum = 5 × 180° = 900°

5

4

A

3 2

1

Exercise 8A 1

a Draw any pentagon (5-sided polygon) and label one of its vertices A. b Draw in the diagonals from A. c Find the interior angle sum of the pentagon.

2

Repeat question 1 with different polygons, drawing diagonals from one vertex only. Copy and complete the table opposite. Example 1 shows this for the heptagon.

Polygon

Number of sides

Number of triangles

Angle sum of polygon

quadrilateral

4

2

2 × 180° = 360°

7

5

pentagon hexagon heptagon octagon

5 × 180° = 900°

nonagon decagon n-gon 3

Copy and complete the following statement: The sum of the measure of the interior angles of any n-sided polygon is ______ × 180°.

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Example 2

A regular polygon has all sides and interior angles equal.

Find the size of each interior angle in a regular dodecagon. A dodecagon has 12 sides. The angle sum = (12 − 2) × 180° = 10 × 180° = 1800° As there are 12 equal angles, the size of each is 1800° ÷ 12 = 150°.

4

Find the size of each interior angle in these regular polygons. a pentagon b hexagon c heptagon d octagon e nonagon f decagon

5

A regular polygon has 24 sides. a Find the sum of the interior angles.

b Find the size of each interior angle.

Example 3 Find x giving a reason.

110°

The polygon is a pentagon. The angle sum = (5 − 2) × 180° = 540° ∴ x + 120 + 130 + 110 + 100 = 540 (angle sum of a pentagon) x + 460 = 540 x = 540 − 460 x = 80

6

Find the value of x in these pentagons. Give a reason. a b 80°

120°

100° 130°

x°

c 100°

70°

115° 160°

150°

130°

140° 140° 100°

108°

100° x°

x°

x°

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Polygons (Chapter 8) Syllabus reference SGS5.2.1

7

a Find the angle sum of a hexagon. b Find the value of x in these hexagons. i ii

iii

120°

138°

150° 150°

x°

147°

x° 97°

130° 160°

155°

170°

110°

118°

130° x°

110°

Example 4 Find x, giving brief reasons.

x°

The pentagon has 5 sides. ∴ the sum of interior angles is 3 × 180° = 540° ∴ x + x + x + 132 + 90 = 540 ∴ 3x + 222 = 540 ∴ 3x = 318 ∴ x = 106

8

Find x, giving brief reasons. a

x°

132°

x°

b

c x° 140°

120° x°

130°

125°

100°

160°

x°

d

80°

70°

e

f 2x° 2x°

150°

110°

2x°

2x° 120° 120° 2x°

x°

x° x°

x°

☞

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g

h x°

3x° 6x° 3x°

2x°

9

10

i x°

x°

x°

60°

x°

x°

x°

x°

x°

2x°

3x°

x°

x°

x°

x°

A pentagon has three right angles and two other equal angles. Find the size of each of the two equal angles. A hexagon has two right angles and all other angles equal. Find the size of each of the equal angles.

Investigation 1 WM: Applying Strategies, Communicating

Exterior angle sum 1

The five exterior angles of a pentagon are shown. As this is a regular pentagon all exterior angles are equal. a Measure the exterior angles. b Find the sum of the five exterior angles.

A non-regular pentagon is shown opposite. a Measure the five exterior angles. b Find the sum of the exterior angles.

2

3

a Draw a hexagon. c Find the sum of the exterior angles.

b Measure the exterior angles.

4

a Draw an octagon. c Find the sum of the exterior angles.

b Measure the exterior angles.

5

Copy and complete: The sum of the exterior angles of any polygon is ______.

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Polygons (Chapter 8) Syllabus reference SGS5.2.1

6 j

d i

e

f c h g b

a

Copy and complete: a i (a + f ) = ______ ii (b + g ) = ______ b The sum of the exterior and interior angle is (a + f ) + (b + g ) + (c + h ) + (d + i ) + (e + j ) = 5 × ______ c The sum of the interior angles f + g + h + i + j = 3 × ______ d The sum of the exterior angles is ______.

B. EXTERIOR ANGLES OF POLYGONS When the sides of a polygon are extended, the exterior angles are formed. From investigation 1, the angle sum of the exterior angles of any polygon is 360°.

Example 1 a Find the size of each exterior angle of a regular decagon. b Hence, find the size of each interior angle. c Find the angle sum of a decagon.

A decagon has 10 sides.

a The exterior angle sum of any polygon is 360°. 360 ∴ Exterior angle = ---------10 = 36° b The interior and exterior angles make a straight line. ∴ Interior angles = 180° − 36° = 144° c Angle sum = 144° × 10 = 1440°

Exercise 8B 1

a Find the size of each exterior angle of a regular octagon. b Use the exterior angle to find the size of the interior angle, giving a reason. c Hence, find the angle sum of an octagon.

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2

a Find the size of each exterior angle of a regular 20-sided polygon. b Find the size of each interior angle. c Hence, find the angle sum of a 20-sided polygon.

Example 2 a A regular polygon has exterior angles measuring 12°. Find the number of sides. b An irregular polygon has one exterior angle of 80° and all the others are 7°. How many sides in this polygon? 360° a Exterior angle = -----------where n is the number of sides n 360° ∴ n = ------------------------angle size 360° = -----------12 = 30 ∴ The polygon has 30 sides. b Since one angle is 80° the others add to 360° − 80° = 280° 280° Then ------------ = 40 7° ∴ The polygon has 41 sides.

Why not 40 sides?

3

A regular polygon has each exterior angle measuring 15°. Find the number of sides of the polygon.

4

Find the number of sides in regular polygons with exterior angles of: a 10° b 18° c 24°

d 90°

5

An irregular polygon has one exterior angle 100° and all others 13°. How many sides?

6

Find the number of sides in irregular polygons with: a one exterior angle of 60° and the rest 10° b one exterior angle of 120° and the rest 12° c one exterior angle of 45° and the rest 15°

7

An irregular polygon has two exterior angles twice the size of the rest. If the other exterior angles are 15°, find the number of sides.

8

An irregular polygon has two exterior angles twice the size of the rest. If the other exterior angles are 20°, find the number of sides.

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Polygons (Chapter 8) Syllabus reference SGS5.2.1

Investigation 2 WM: Applying Strategies, Investigating, Communicating

Tessellation A tessellation is a pattern of shapes that fit exactly together. A regular tessellation is made up of regular polygons of one type and size only. You will need: square grid paper, triangular grid paper. 1

A regular hexagon will tessellate as shown in this diagram. a What is the size of each interior angle of a regular hexagon? b Look at the circled part of the tessellation. How many hexagons meet at the point? c What is the sum of the angles at the point circled? d Copy and complete: A regular polygon will tessellate if the size of the interior angles divides exactly into _____.

2

The size of the interior angles of a regular pentagon is 108°. Explain why a regular pentagon will not tessellate.

3

a Copy and complete this table. b Which of these eight regular polygons will tessellate? Give reasons.

Regular polygon

Number or sides

Triangle Square Pentagon Hexagon Heptagon Octagon Nonagon Decagon

3 4 5 6 7 8 9 10

Angle sum = 180(n − 2)

Interior angle size = angle sum ÷ n

540° 720°

108° 120°

4

a On grid paper draw a tessellation of a square with side length 2 units. b Draw a different tessellation using the same square.

5

a On a sheet of triangular grid paper draw a tessellation of equilateral triangles with side length 2 units. b Using a different colour show that this is a tessellation of regular hexagons. c What other plane shapes can be seen in the tessellation?

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Polygons (Chapter 8) Syllabus reference SGS5.2.1

Investigation 3 WM: Applying Strategies, Investigating

Archimedes Archimedes used polygons inside and outside a circle of given radius to give an approximation for π. C Given that C = πd, then π = ---- . D This software investigation uses Archimedes theory to find an estimate for π by drawing a regular polygon inside a circle of constant radius (inscribed) and another polygon with the same number of sides outside, but touching, the circle (circumscribed). The estimate of π is found by dividing the perimeters of the polygons by the radius of the circle. perimeter of inscribed polygon perimeter of circumscribed polygon ------------------------------------------------------------------------------- ≤ π ≤ -------------------------------------------------------------------------------------------radius radius

Investigation 4 WM: Reasoning

Circles and doughnuts 1

Use a pencil to draw a starter circle of radius 3 cm, say, and use your protractor to divide the circle into equal angles of 10°.

2

At each of the 36 points around the starter circle, as centre, draw a circle of radius 2 cm.

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Polygons (Chapter 8) Syllabus reference SGS5.2.1

3

There are various ways in which you can colour your doughnut. Experiment with different colour patterns. Three different ones are shown here.

Can you find other ways to colour your doughnut?

Language in Mathematics

1

Rearrange these words to form sentences relating to this chapter. a sides A has five pentagon b has sides equal A regular all polygon c A eight octagon sides polygon an is with d sum 360° is The polygon exterior a of angle

2

Use every third letter to reveal the message. EDTTHOFVFBNIWANSDDTTTUOHPLEJQIAVNCETRGEYJRMZIA EOTERITAOUNTFGESLFGEBNSCWUQAMZSOAEFTRAUOPITOH JLLPYOIGUCOSENQASQUUTFBWDTERRCVANKCLOTZATQWWE ROTFFVIROWOASMCGTBYHJKELONPQUASMASBDEERFRTIOY TFFVSBHITGDJKELOSPIAYTNEWDASMDCUFVLGBTHNIJMPKL LPOYYTBREYWSOQANZSEDRHFTUGYNHUDJIRKOELMDHFADA NQEDTUEOPIRZGCHHJOTPNYQRDTSEAEGASRFGEHJEUYS

Glossary angle sum exterior nonagon regular

concave heptagon pentagon tessellate

convex hexagon polygon tessellation

decagon interior quadrilateral

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CHECK YOUR SKILLS 1

2

3

4

The angle sum of an eight-sided polygon is: A 540° B 720°

C 900°

A ten-sided polygon is called a: A pentagon B octagon

C decagon

A regular polygon with 24 sides has interior angle size of: A 3960° B 165° C 4320° The value of x in this figure is: A 145° B 235° C 325° D 35°

140° 165°

x°

D 1080°

✓ D hexagon

D 180°

5

A heptagon has three right angles and the other angles are all equal. The size of the equal angles is: A 900° B 112.5° C 157.5° D 247.5°

6

The exterior angle of a 30-sided regular polygon is: A 12° B 168° C 360°

7

D 5040°

An irregular polygon has one exterior angle of 100° and all others 13°. The number of sides of this polygon is: A 20 B 21 C 30 D 113

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–5

6, 7

A

B

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Polygons (Chapter 8) Syllabus reference SGS5.2.1

REVIEW SET 8A 1

a Draw a hexagon. c Find the angle sum of a hexagon.

2

Find x giving a reason. a

b Draw in all diagonals from one vertex.

b

120° 100°

150° x°

130°

x°

130° x°

3

4

100°

130°

For a regular octagon: a What is the size of each interior angle?

b What is the sum of the exterior angles?

A regular polygon has exterior angles 10°. a How many sides does it have?

b What is the sum of the interior angles?

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Polygons (Chapter 8) Syllabus reference SGS5.2.1

REVIEW SET 8B 1

a Find the angle sum of a dodecagon (12 sides). b Find the size of each interior angle of a regular dodecagon.

2

Find x, giving a reason. a

b

135°

140°

148°

100°

x° x°

150° 100° x°

3

A nine-sided figure has three right angles and all other angles equal. What is the measure of each of these equal angles?

4

a Find the size of each exterior angle of a 25-sided regular polygon. b Find the size of each interior angle. c Hence, find the angle sum of a 25-sided polygon.

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Polygons (Chapter 8) Syllabus reference SGS5.2.1

REVIEW SET 8C 1

2

A regular polygon has 36 sides. a Find the sum of the interior angles.

b Find the size of each interior angle.

Find x, giving a reason. a

b 3x°

120°

115°

x°

2x°

150° 168° x° x°

135°

2x°

3

A regular pentagon has two right angles and all other angles equal. Find the size of the equal angles.

4

Find the size of the exterior angle in a regular 30-sided figure.

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Polygons (Chapter 8) Syllabus reference SGS5.2.1

REVIEW SET 8D 1

a Draw a pentagon. c What is the angle sum of a pentagon?

2

Find x giving a reason. a

b Draw all diagonals from one vertex.

b x°

130° 150°

155°

x°

130°

x° 120°

100°

170°

3

A hexagon has three right angles and all other angles equal. Find the size of the equal angles.

4

Find the size of the exterior angle in a regular octagon.

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Chapter 9 Probability This chapter deals with relative frequencies and theoretical probabilities. After completing this chapter you should be able to: ✓ conduct experiments to determine the relative frequency of an event ✓ estimate the probability of an event from experimental data ✓ express the probability of an event using the probability definition ✓ calculate probabilities for simple events ✓ simulate events using random number generators.

Syllabus reference NS5.1.3 WM: S5.1.1–S5.1.5

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Probability (Chapter 9) Syllabus reference NS5.1.3

Diagnostic Test Use this table for questions 1 and 2. A tennis racquet is spun 100 times. It has rough on one side and smooth on the other. The table shows the results. Outcome

Frequency

Rough

81

Smooth

19

1

The relative frequency for ‘rough’ is: 81 19 81 A 81 B ---------- C -----D -----100 81 19

2

Based on the table, the probability that a spin of this racquet will result in ‘smooth’ is: 19 81 A 81 B 19 C ---------- D ---------100 100

3

4

5

6

Relative frequency

7

A 52 card pack is shuffled and one card is dealt. The probability that it is a diamond is: 1 1 A --B -----C 52 D 13 4 52

8

Which statement is true? A There are 8 teams in our netball competition so our probability of winning the competition is 1--8- .

A normal six-sided die is thrown once. The probability of getting a 4 is: 1 4 1 A --B --C --D even 4 6 6

B A coin has been tossed 8 times and all 8 have been heads, therefore the next toss must be tails. C Traffic lights can be red, amber or green. Therefore the probability the light is green is 1--3- .

A hat contains 1 red, 5 white and 7 blue tickets. A ticket is selected at random from the hat. The probability that the ticket is white is: 5 1 7 A 5 B -----C -----D -----13 13 13 A raffle has 100 tickets numbered from 1 to 100. The probability that the ticket selected is a number between 8 and 15 inclusive is: 1 8 15 23 A ---------- B ---------- C ---------- D ---------100 100 100 100 A poker die has faces A, K, Q, J, 10, 9 and is rolled once. The probability of getting a K or a J is: 1 2 4 1 A --B --C --D --6 6 6 4

Percentage

D There are 24 horses in the Slipper Handicap horse race. Therefore the probability that I will draw the favourite out of a hat containing the 1 -. names of all the horses is ----24 9

A netball shooter has a 60% chance of scoring a goal from just inside the goal circle. To simulate the number of goals from 50 shots at goal a table of random numbers can be used by: A assigning the digits 0, 1, 2, 3, 4, 5, 6 to scoring a goal B assigning the digits 0, 6 to missing C assigning the digits 1, 2, 3, 4, 5 to scoring a goal D assigning the digits 0, 1, 2, 3 to missing

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Probability (Chapter 9) Syllabus reference NS5.1.3

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1, 2

3–7

–

8

9

A

B

C

D

E

Investigation 1 WM: Communicating, Applying Strategies, Reasoning

Probability experiments A normal coin has two sides, heads and tails. The theoretical probability of each event is 1--2- . This experiment examines the probability of heads and tails. 1

a Toss a coin 50 times and complete this table. Result

Tally

Frequency

Fraction of total

Heads Tails b c d e

Toss the coin another 50 times and complete a second table. Compare the results in the two tables and comment on any differences. Combine your two tables and compare the results of the new table out of 100 trials. Combine the tables of nine other people with yours. Comment on the fractions out of 1000 for each of the two events. f The fraction for each should be 1--2- . Comment on the difference in results from the 50 trials versus the 1000 trials. g What would you expect if this experiment was repeated 1 000 000 times? h Comment on the statement ‘The more times the coin is tossed, the closer the fraction of heads and tails gets to 1--2- each.’ 2

Chircop tossed a coin 100 times: heads came up 53 times and tails came up 47 times. 47 - . Is she correct? Comment. a Chircop concluded that the probability of tails is --------100 b Should Chircop expect the same results if she repeats her experiment of tossing the coin 100 times? Explain. c If Chircop tossed the coin 1 000 000 times, how many tails would she expect? Explain.

☞

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3

When a die is thrown there are six equally likely outcomes: 1, 2, 3, 4, 5 or 6. a Throw a die 120 times Result Tally Frequency Fraction and complete the table. (Probability) 1 2 3 4 5 6

b How many of each number did you expect? c Explain any differences Total between your expected values and the results. d Combine your results with the class and comment. 4

120

Rachel rolled a normal six-sided die twelve times. She did not throw a 6. Rachel concluded that the probability of obtaining a 6 was 0. Why is she wrong? How many 6s would be expected in twelve throws of the die?

Research Assignment WM: Communicating, Reasoning

Chance statements 1

Collect statements involving the use of ‘chance’ language in the media.

2

Organise these statements from most likely to least likely to occur.

3

Assign each ‘event’ with a probability between 0 and 1.

4

Comment on the statements.

A. THEORETICAL PROBABILITY VERSUS EXPERIMENTAL RESULTS From investigation 1 you would have concluded that the more trials conducted the closer the experimental results are to the theoretical results. In fact, theoretical probability only makes predictions for overall results in the long term. With experiments of a small number of trials, there may be little or no correlation between expected results and experimental results. In some instances it is either very complicated or impossible to calculate the theoretical probability of an event. In this case the relative frequency gives an estimate for the probability.

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Probability (Chapter 9) Syllabus reference NS5.1.3

Example 1 A cylindrical can is tossed 200 times. The number of times it landed on its side and on an end were recorded in this table. Outcome

Frequency

Top end Bottom end Side

Relative frequency

Percentage

36 38 126

a Complete the table. b In future tosses of the can estimate the probability that it will land on its side. a

Outcome

Frequency

Top end

36

Bottom end

38

Side

126

Relative frequency 36 ---------200 38 ---------200 126 ---------200

Percentage 18% 19% 63%

b P(lands on side) = 63%

Exercise 9A 1

A tennis racquet is spun 200 times. It has rough on one side and smooth on the other. The table shows the results. Outcome Rough Smooth

Frequency 86 114

Relative frequency

Percentage

a Copy and complete the table. b Estimate the probability that a spin will result in ‘rough’. 2

Four hundred car salespeople were randomly selected and asked the country in which their car was manufactured. The results are shown in the table. Country Australia Japan Korea Germany Other

Frequency 146 128 56 48 22

Relative frequency

Percentage

a Copy and complete the table. b Use this data to estimate the probability that a salesperson chosen at random will have a car manufactured in Australia. c Estimate the probability that the salesperson’s car is manufactured in Germany.

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Probability (Chapter 9) Syllabus reference NS5.1.3

3

At a state wide teachers’ conference the colour of each teacher’s car was recorded; the results appear in the table. There were 200 teachers’ cars. Colour White

Number 53

Red

48

Blue

27

Green

25

Yellow

21

Black

12

Silver

8

Other

6

Relative frequency 53 ---------200 48 ---------200

Percentage 26.5%

a Copy the table and complete the relative frequency and percentage columns. b Using these results, how many red teachers’ cars would you expect in a school with 50 teachers? Explain. c Complete a similar table for the teachers in your school. How do your school’s results compare? Explain any differences. 4

In a year, a restaurant served 4754 bottles of wine. Of these 86 were returned because the wine was faulty. Based on this information, estimate the probability that a bottle of wine from this restaurant is faulty when opened.

5

In a survey of 5000 marriages it was found that 1285 ended in divorce. Matt and Diane are getting married. Based on this information, what is the approximate probability that their marriage will end in divorce?

B. THEORETICAL PROBABILITY In stage 4 we had the definition of the probability of an event A occurring. It was: number of favourable outcomes P(A) = ------------------------------------------------------------------------n where n is the total number in the sample space. In stage 4 the following probability properties were developed: • The probability of an event occurring is between 0 and 1. • If P(A) = 0 then the event A is impossible. • If P(A) = 1 then the event A is a certainty. • P(the event A does not occur) = 1 − P(A).

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Probability (Chapter 9) Syllabus reference NS5.1.3

Example 1 A spinner is made from a regular pentagon with equal sections containing the numbers 1, 2, 3, 4, 5. a b c d

List the sample space. Find the probability of spinning a 3. Find the probability of spinning an odd number. Find the probability of not spinning a 3.

1

2 3

5 4

a The five numbers 1, 2, 3, 4, 5 comprise the sample space ∴ n = 5. number of 3s 1 b P(3) = --------------------------------------------------------------- = --number in sample space 5 number of odd numbers 3 c P(odd number) = --------------------------------------------------------------- = --number in sample space 5 d The complementary event to ‘spinning a 3’ is ‘not spinning a 3’. ∴ P(not a 3) = 1 – P(3) Checking: = 1 – 1--5The numbers that are not 3 are 1, 2, 4, 5 4 = --5∴ P(not a 3) = 4--5- .

Exercise 9B 1

A square spinner has the numbers 1, 2, 3, 4 in equal sections. For one spin, determine the probability of getting a: a 1 b 3 c 5 d number less than 5 e number other than 1

2

A spinner in the shape of a regular hexagon has equal sections marked 1 to 6. For one spin, determine the probability of getting a: a 6 b 4 c 3 d odd number less than 6 e even number less than 7 f number less than 7 g 8 h number other than 4

3

A spinner is made from a regular decagon with 10 equal sections containing the numbers 1 to 10. a List the sample space. b Find the probability of spinning a 5. c Find the probability of spinning an even number. d Find the probability of spinning an odd number. e Find the probability of not spinning a 10. f Find the probability of spinning an 11.

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Probability (Chapter 9) Syllabus reference NS5.1.3

4

A bag contains 3 red, 4 white and 5 blue tickets. A ticket is selected at random from the bag. Determine the probability that the ticket is: a red b white c blue d green e not red f not white g red or blue h not red or white i red, white or blue

5

Adam has spread out a pack of playing cards. He picks a card at random. Hearts and diamonds are red. Clubs and spades are black. Picture cards are King, Queen, Jack.

P(spade) means the probability of selecting a spade.

Find: a P(spade) d P(a black 10) g P(a 5 or a 6)

b P(4 of hearts) e P(ace) h P(green card)

c f i

P(club) P(picture card) P(not a picture card)

6

List the sample space for: A card chosen at a tossing a coin random means each card has the same b the sexes of a 2-child family chance of being selected. c tossing 2 coins at the same time d rolling a die e the order in which 3 people can stand in a line

7

A fair die is rolled. Determine the probability of getting: a a 2 or a 3 b a positive integer c a result greater than 4 d a non-6 e a7

8

A poker die has faces A, K, Q, J, 10 and 9, and is rolled once. Determine the probability of getting: a an A b a number c an A or a number

9

A symmetrical octahedral (8-sided) die has numbers 1 to 8 marked on its faces. It is rolled once. Determine the probability of throwing: a a2 b a number less than 4 c a number less than 1 d a number between 0 and 9

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Probability (Chapter 9) Syllabus reference NS5.1.3

10

A regular pentagonal (5-sided) spinner has the numbers 1 to 5 marked on its partitions. Determine the probability that after a spin the result will be: a an even number b a prime c a factor of 6

11

A bag contains 3 red and 7 blue buttons, and one is randomly selected from the bag. Determine the probability that the button is: a red b blue c red or blue d green

12

The $2 lottery has 100 000 tickets. Find the probability of: a winning first prize with one ticket b not winning first prize with 50 tickets

13

One ticket is chosen in a lottery consisting of 100 tickets numbered 1 to 100, and the choice is made randomly. Determine the probability that the ticket is: a a two-digit number b a multiple of 12 c a multiple of 7 or 11

14

Determine the probability that a person randomly selected in the street has his (or her) birthday in September. (Don’t forget leap years.)

15

Many games require a 6 to start, when rolling a normal die. a What is the probability of starting on the first roll of the die? b Kristie says 3 is her lucky number. Is she more likely to start first roll if a 3 is required instead of a 6?

Investigation 2 WM: Reasoning, Communicating, Applying Strategies

A pair of dice From the illustration, we can clearly see that when two dice are rolled there are 36 possible outcomes. Of these {1, 3}, {2, 2} and {3, 1} give us a sum of 4. Consider the following question: What is the likelihood of the sum of the numbers being 4, when two dice are rolled?’

☞

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Probability (Chapter 9) Syllabus reference NS5.1.3

The answer to this question lies in the fact that 3 out of the possible 36 outcomes give a sum of 4. Hence the probability of a sum of 4 when two dice are rolled is

3 -----36

(or 8.333%).

1

Using the illustration 2 3 4 5 6 7 8 9 10 11 12 Sum above, copy and complete the table. 3 1 ----------Fraction of total 36 36 The figures listed in this table are our expected results for the experiment of determining the likelihood of a particular sum when two dice are rolled.

2

Now toss two dice 200 times and record in a table the sum of the two numbers for each toss. Copy and complete the table given opposite.

3

Pool as much data as you can with other students and find the overall percentage of each sum. Make a comparison between the results you have obtained and the expected results from the previous table.

Sum

Tally

Frequency

Fraction of total

2 3 4 5 6 7 8 9 10 11 12 Total

200

C. CALCULATING PROBABILITIES This section involves two groupwork activities designed to promote the understanding of probability concepts relating to theoretical and calculated probabilities.

Groupwork 1 1

Each person designs their own spinner with four colours. Make the spinner so that all colours are not equally likely but do not tell the probabilities of each colour occurring.

2

Using the spinner you made, conduct a trial of 100 spins to see how closely the experimental results match the probabilities.

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Probability (Chapter 9) Syllabus reference NS5.1.3

3

a Working in pairs, use another person’s spinner. One person spins, the other person records the results. The person recording cannot look at the spinner. b The recorder decides on the number of trials and must estimate the probabilities of each of the colours after the experiment. c Compare your results and explain them. d Swap over so you are recording and spinning with another spinner, and repeat parts a, b and c.

4

What can you conclude about the number of trials compared with the accuracy of the results?

Groupwork 2 1

a Using between 20 and 52 cards, each person designs a deck with their own probabilities for drawing a red or a black card. b Swap decks around the group. To experimentally find the probabilities of red or black cards, shuffle the deck then select a card and record its colour. Replace the card, shuffle, then select another card. c The challenge is to be the first to correctly identify the probabilities. You may not guess and you are only permitted one answer. You must decide how many trials are enough.

2

The teacher prepares a number of bags with counters or marbles of two different colours in each bag. Only the teacher knows the number of each colour. a Students in their groups draw a counter or a marble, note its colour and replace it. b They repeat this process until they are ready to answer with the probability of each colour. Each group is allowed only one answer. c The first correct group wins. d Compare the experimental results of all groups. Does this provide a better estimate?

Investigation 3 WM: Applying Strategies, Reasoning, Communicating

Rectangular spinners This investigation is suitable for four students. A rectangular spinner can be made from cardboard using a match or toothpick for its spindle. You are to investigate the chances of the spinner finishing on blackened edges on opposite sides of the spinner. The other two sides are to be 4 cm long.

☞

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1

Make eight different spinners for, say, x = 1 cm, 2 cm, 2.5 cm, 3 cm, 3.5 cm, 4 cm, 5 cm, 6 cm or other values of your own choosing.

4 cm x cm

2

Each spinner is to be twirled several hundred times so that an estimate of the probability of getting a black edge results for each of the spinners. Divide this task between the members of your group.

3

Collect the data and complete a table like the one given below. x

1

2

2.5

3

3.5

4

5

6

P(black) estimate 4

Graph P(black) against x. Write about the shape of your graph and explain how it could be used to estimate P(black) for various values of x.

5

From your graph estimate: a x, when P(black) = 0.1, 0.4, 0.8

b P(black) if x = 4.5

D. EQUALLY LIKELY EVENTS Equally likely events are events that have the same (equal) chance of occurring. For example, when a coin is tossed, heads and tails are equally likely events. When selected at random, they have an equal chance of occurring. Sometimes events are not equally likely.

Example 1 Comment on these statements: a My family has four boys so the next baby born will be a boy. b There are ten teams in the football competition, therefore the probability that my 1 -. team will win the competition is ----10 a The fact that four boys were born is a coincidence. The next baby has approximately an equal chance of being a boy or a girl. b Some teams are better than others, so the probability of each team winning is not 1 - , if it is poor the equal. If your team is good the probability will be greater than ----10 probability could be very close to 0.

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Probability (Chapter 9) Syllabus reference NS5.1.3

Exercise 9D 1

Comment on these statements: a There are 26 letters in the English alphabet. Therefore the probability that a person’s 1 -. name starts with Z is ----26 b There are 12 teams in my netball competition. Therefore the probability that my team will 1 -. win the competition is ----12 c There are 128 players in the main draw of Wimbledon. Therefore the probability of picking 1 -. the winner at the start of the tournament is --------128 d Traffic lights can be red, amber or green. Therefore the probability that a particular traffic light is red is 1--3- . 1 -. e There are 52 cards in a pack. Therefore the probability of selecting an ace is ----13 f I need a 6 on a normal die to start a game. Therefore the probability that I will start first go is 1--6- . g There are 24 horses in the Melbourne Cup field. Therefore the probability of selecting the 1 -. winner from a list of their names is ----24 h A letter is chosen from the word INSIGHT. Therefore the probability that it is an I is 2--7- .

2

What assumptions are made in these statements? a 30% of the population do not work. Therefore the probability of being unemployed is 0.3. b I asked five people leaving the corner store what drink they bought. Four of them bought orange juice and one bought lemonade. Therefore the probability that someone buys orange juice is 4--5- . c Of the thirty students in 9 Red, only three watch the evening news. Therefore the 1 -. probability that someone watches the evening news is ----10

Investigation 4 WM: Applying Strategies, Reasoning

Instant money and bingo Many sporting clubs and service clubs use instant money or bingo type tickets to raise money. We will consider an instant money type of game where amounts of $1, $2, $5, $10 and $25 can be won. A ticket will show five different amounts, and if three of them are the same then that ticket will win.

$5 $2 $1 $25 $10

The second ticket wins $2 because three $2 symbols appear.

$2 $2 $10 $5 $2

Losing ticket

Winning ticket

The social club is raising money for new clubrooms. They decide to sell 25 000 of these tickets at 20 cents each. They will distribute $2500 in prizes. They decide to print: 5 prizes of $25 25 prizes of $10 125 prizes of $5 250 prizes of $2 and the remaining prize money is in $1 tickets.

☞

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1

Determine how many prizes of $1 will have to be produced so $2500 is given in prizes.

2

When all tickets have been sold, how much money does the social club expect to make?

3

Determine the probability that when buying one ticket you will win: a a $25 prize b a $10 prize c a $5 prize d a $2 prize e a $1 prize f any prize

Investigation 5 WM: Applying Strategies, Reasoning, Communicating

Roulette One game played extensively in casinos is roulette. The game consists of a horizontal rotating wheel containing 38 equal slots and a steel marble that will spin into one of them. The slots are numbered 00, 0, 1, 2, 3, 4, 5, … up to 36. The slots are red or black for the numbers 1 to 36. 0 and 00 are usually green. (00 appears on the American version.) Gamblers place their betting chips on the table as shown (International layout). They can bet on red or black or single numbers or combinations of numbers.

a b c d e f g h

i

Result

$ won

Individual number 5 (pleine) Two adjacent numbers (cheval) Three numbers in a row (transversale pl.) Four numbers in a square (carre) First four numbers (0, 1, 2, 3) Two rows, six numbers (transversale 6) A vertical row of 12 numbers (colonne) First twelve numbers 1–12 (12 p.) Second twelve numbers 13–24 (12 m.) Third twelve numbers 25–36 (12 d.) All even numbers 2, 4, 6, 8 ... (pair) All odd numbers 1, 3, 5, 7... (impair) All red numbers 1, 3, 5, 7 ... (rouge) All black numbers 2, 4, 6, 8, 10, 11, ... (noir) All numbers 1–18 (manque) All numbers 19–36 (passe)

35 17 11 8 8 5 2 2 2 2 1 1 1 1 1 1

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Probability (Chapter 9) Syllabus reference NS5.1.3

In the result table, the $ won shows how much can be won from a $1 bet. A loss results in the casino keeping your $1. For example: • a win on number 17 pays $35 (and you also get back your original bet of $1) • a win on b above pays $17 if an 8 or 9 occurs. (Assume that we use the table as shown, i.e. no 00 slot.) 1

If you bet $10 on number 23 and win, what will be your return?

2

If you bet $1 on each number, how much does it cost you and what will be your return?

3

From your answer to question 2 determine what percentage profit the casino expects to make.

4

What profit does the casino hope to make on coloured bets?

5

Your expected gain can be calculated using: Expected gain = Possible $ winnings × P(winning) − Amount bet × P(losing) and a negative answer means an expected loss. Copy and complete the following table assuming that you bet $1 each time. Type of bet

Possible $ winnings

P(winning)

P(losing)

Expected $ gain or loss

5

6 -----37

31 -----37

−0.027

a b c d e f g h i 6

Is the game of roulette ‘fair’?

7

What type of bet (if any) is ‘best’ to use in roulette?

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Investigation 6 WM: Applying Strategies, Reasoning

Concealed number tickets Many clubs have ticket machines that contain sets of consecutive numbers from 0001, 0002, 0003, … up 2000. These ‘games’ are relatively inexpensive to play and are used as fundraisers for the club (e.g. football club, golf club). Tickets are ejected at random at a cost of 20 cents each. A small cardboard cover is removed to reveal the concealed number. Suppose a golf club can buy golf balls as prizes for $2.50 each and a set of 2000 tickets for the machine at $30. One club shows the following winners table. Winning numbers 777 1000, 2000 any multiple of 25

Prize

Cover removed

4 golf balls 2 golf balls 1 golf ball

1

If all tickets are sold, how many balls are paid out as prizes?

2

Determine the total cost to the club for a complete round of 2000 tickets going through the machine.

3

What percentage profit is made by the club?

4

If you purchase one ticket, what is your chance of winning at least one ball?

5

What is your $ expectation for the playing of one game, i.e. purchasing one ticket.

6

Design your own concealed ticket game for a club or group so that the customer receives an expected payout of around 80% (and certainly not less than 80%).

E. RANDOM NUMBER GENERATORS A simulation is a simple mathematical model that allows us to investigate a more complicated, time-consuming or costly situation. In some cases we can estimate the probability of events that are very difficult to determine using the formal rules of probability. This section uses tables of random numbers to model situations.

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Probability (Chapter 9) Syllabus reference NS5.1.3

This table of random numbers is to be used in this exercise. 0 8 2 1 3 2 9 7 8 4 7 6 5 6 1 4 6 6 3 5 1 0 1 9 5 1 0 0 1 2 4 9 3 3 8 4 0 6 3 3 1 4 5 8 0 2 9 6 2 5 7 2 3 9 9 6 7 5 5 3 6 1 9 6 0 9 3 0 6 4 8 6 7 9 8 6 7 1 5 5 5 2 9 3 1 5 8 0 0 2 1 7 6 0 0 4 9 3 3 3 4 7 5 4 5 8 6 4 0 6 9 3 3 0 1 7 9 2 6 3 0 5 1 5 1 4 8 2 6 8 4 4 7 2 9 8 8 1 9 5 4 0 6 9 3 4 1 2 7 0 2 5 5 3 4 4 9 2 0 3 4 4 7 7 4 7 4 0 4 5 9 3 4 6 8 2 4 4 7 2 3 8 2 5 5 3 5 7 9 9 7 4 5 3 9 4 2 7 9 6 9 0 3 1 4 9 5 9 2 9 9 1 8 3 3 3 2 5 1 4 5 6 8 2 7 0 9 2 2 0 9 3 4 6 5 6 8 1 3 7 7 1 1 9 8 8 5 0 5 5 7 1 7 9 7 8 3 6 9 2 3 3 5 2 3 4 1 6 1 7 6 5 6 8 1 3 7 7 1 1 9 8 8 5 0 5 5 7 1 7 9 7 8 3 6 7 4 6 3 1 0 3 0 7 8 6 4 6 7 8 4 9 8 9 6 8 0 2 9 2 0 8 9 4 3 1 1 7 0 5 0 1 3 9 1 7 0 5 8 7 6 4 6 4 8 9 4 1 0 6 3 0 3 9 0 0 3 8 9 0 5 9 4 6 5 7 1 3 9 8

Example 1 A football goalkicker has a 60% chance of kicking a goal from the sideline. a Simulate five kicks, using a table of random numbers, to find the number of goals he would score by: i assigning digits to represent a goal and a miss ii selecting five random one-digit numbers from the table b Repeat the simulation to estimate the number of goals he would kick from 50 attempts. c Compare your results with the expected probability. a i

Let each digit in the table represent a kick. Let the six digits 0, 1, 2, 3, 4 and 5 signify a goal, and the four digits 6, 7, 8, 9 signify a miss. Six out of ten digits or 60% represent a goal, and four out of ten digits or 40% represent a miss. ii Choosing any place in the table as a starting point, read off five digits representing the five kicks at goal. If the starting point is the third row with the seventh digit in, shown in bold, then the five digits are 5, 2, 9, 3, 1. This represents four goals and one miss. b Starting at the same place the fifty digits are: 52931580021760049333475458640693301792630 515148268 representing 34 goals and 16 misses. c The expected number of goals from 50 kicks would be 30; from the simulation it is 34. Note: • Any six of the digits 0–9 could have been chosen to represent a goal. • Starting at a different place may give different results. • A calculator or spreadsheet could be used to generate random numbers. If it is possible or practical to use technology to generate the numbers, then use the technology rather than the table for questions 3–5. This spreadsheet provides a random number generator.

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Exercise 9E 1

Repeat the question from the example starting with the first digit in the table.

2

The probability that a particular soccer striker will score from a penalty is 90%. Simulate 50 kicks, using a table of random numbers, to find the number of goals he would score by: a i assigning digits to represent a goal and a miss ii selecting fifty random one-digit numbers from the table starting with the first number in the fifth row. b Compare your results with the expected number of 45 goals. c Repeat, starting from a different position in the table, and compare the results.

3

The probability that a netball shooter will score a goal is 70%. Simulate 50 shots at goal, using a table of random numbers, to find the number of goals she would score by: a i assigning digits to represent a goal and a miss ii selecting fifty random one-digit numbers from the table starting with the first number in the fifth row. b Compare your results with the expected number of 35 goals. c Repeat, starting from a different position in the table, and compare the results.

4

A plant seedling has an 80% chance of surviving the first six weeks after planting. Simulate the survival of 100 plants using a table of random numbers by: a i assigning digits to represent the survival and death of the seedling ii selecting 100 random one-digit numbers from the table. b Compare your results with the expected number of 80 survivals.

5

The chance of a couple having a male child or a female child are each 1--2- . Simulate the number of boys and girls in a family of four children, using a table of random numbers, by: a i assigning even numbers to represent boys and odd numbers to represent girls ii selecting four one-digit numbers from the table and recording the number of boys and girls. b Repeat the process until you have results for ten families. c Combine the results of the class. d Find the mean number of boys in a family of four, using the whole class results. e From your results, what is the probability in a family of four children that there are 0, 1, 2, 3 or 4 boys?

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Probability (Chapter 9) Syllabus reference NS5.1.3

non-calculator activities

1

2.5 − 1.7 =

2

3

Find the value of 2820 ÷ 20.

4

5

Five more than three times the sum of 4 and 6?

6

7 9 10

11

1 Write 8 − --- as a mixed fraction? 3 35.6543 × 1000 =

8

7 Change ------ to a percentage. 20 Find 20% of $180. 1 How many minutes in 3 --- hours? 2 What is the value of (0.4)2?

Consider the 4 in the number 13 746 892. Its value would be: A 4 B 4000 C 40 000

D 400 000

Which of the following numbers is closest to 7? A 6.9 B 7.01

D 7.1

C 6.93

12

What fraction is 4 days of 1 fortnight? Give your answer in simplest fraction form.

13

An urn contains 9 blue and 7 red marbles. If I withdraw one marble from the urn, what is the probability that it will be blue?

15

20.4 Estimate the value of ----------------------- giving your answer as a whole number. 2.1 + 1.8 Write the number for 200 + 3 + 0.02 + 0.004.

16

If John earns $140 for 7 hours work, how much will he receive for 5 hours work?

17

Rachel knows that 225 × 31 = 6975. Use this data to find the answer to Rachel’s question: 69 750 000 ÷ 31 =

18

What is

19

What is the next number in the sequence: 1, 4, 9, 16, 25, _____ ?

20

The average of ten numbers is 3. What is the total of all ten scores?

21

5+7×2+1=

23

2 1 --- + --- = 3 6

14

25

2

36 + 4 ?

12 345 km = _____ metres.

22 Ken has finished reading 25% of a 40 page book. How many pages are left to read? 24 What is the length of the side x in the diagram?

x cm 6 cm 8 cm

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Language in Mathematics Blaise Pascal (1623–1662) Blaise Pascal was born in Clermont-Ferrand, France, in 1623. Pascal’s father was a judge in the tax court and his mother died when he was only three years old. He began showing a great insight and understanding of mathematics at a relatively young age. When he was only 16, Pascal wrote a paper on conic sections, the study of shapes obtained when a right circular cone is cut at various angles. Before he turned 20, Pascal developed a calculating machine (patented 1647) to assist his father with tax calculations. This machine was, in a way, the first digital calculator. Pascal conducted experiments with his father on vacuums, and in 1647 described the effects of air pressure on tubes of mercury, which led to the construction of barometers. There appears very strong evidence that the theory and study of probability originated in the gambling halls of France where players had little idea of odds or percentages. However, Pascal is one of the main theorists in the development of the modem theory of probability as it is known today. Pascal is remembered not only for his contribution to mathematics, but also for his involvement in physics and religious philosophy. His family originally held strong Catholic beliefs, but as a result of the illness of his father he came into contact with people with an even stricter moral approach to religion. As a result, Pascal became extremely interested in Christian beliefs and ethics, and produced a large number of papers on these subjects. Despite his great contributions to mathematical and religious thinking, the one criticism of Pascal is that he was probably too concrete a thinker—some emotional problems he appeared to analyse like a geometry problem. However, this minor criticism is far outweighed by the enormous contributions he made in other areas. 1

a b c d

How many years did Pascal live? Why is Pascal remembered? What criticism is made of Pascal? Why did Pascal become interested in Christian beliefs?

2

Replace the vowels in these glossary terms. a ch __ nc __ c pr __ b __ b __ l __ ty e pr __ d __ ct __ __ n

b __ v __ nt d r __ nd __ m f th __ __ r __ t __ c __ l

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Probability (Chapter 9) Syllabus reference NS5.1.3

3

Rearrange these words to form a sentence: a used estimate Relative to frequencies probability are b are number simulate to used Random generators events

4

Use every third letter to find the sentence. DFPGYRTROEWBDFAGHBBVICDLXSIZATQAYWSEEESRTTYUII OMPLAKJTKMEMRSFGBDSEERCGHOSCMEFERTMYGOVDRWAE SDSRFTVFACSBAQLWEERTAGHSJHTEDHWSEESNDSUCDMVG BMIEOURQWOESFAGTHPREMIHAAQRLCISPDIEFNHJCTRRAA ESVABNSHYEFGS

Glossary certainty event experimental probability impossible random simulate

chance equally likely events fair prediction random number generator theoretical

CHECK YOUR SKILLS

Use this table for questions 1 and 2. Three hundred car salespeople were randomly selected and asked the country in which their car was manufactured. The results are shown in the table.

✓ complementary event experimental data favourable outcomes probability relative frequency trial

Country

2

3

The relative frequency for Korea is: 42 A 42 B ---------100

Frequency

Australia

125

Japan

103

Korea

42

Germany

24

Other 1

267

42 C ---------300

Based on the table, the probability that a car will be Australian is: 125 125 100 A ---------B ---------C ---------100 300 125 A normal six-sided die is thrown once. The probability of getting a 1 is: 1 4 1 A --B --C --4 6 6

6

42 D ---------125 125 D ---------175

D even

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4

A hat contains 7 red, 3 white and 1 blue ticket. A ticket is selected at random from the hat. The probability that the ticket is red is: 7 1 7 A 7 B -----C -----D -----11 11 10

5

A raffle has 50 tickets numbered from 1 to 50. The probability that the ticket selected is the numbers 7 or 13 is: 2 1 7 13 A -----B -----C -----D -----50 50 50 50

6

A poker die has faces A, K, Q, J, 10, 9, and is rolled once. The probability of getting an A or a 10 or a 9 is: 1 2 3 1 A --B --C --D --6 6 6 3

7

A 52-card pack is shuffled and one card is dealt. The probability that it is a diamond is: 1 2 13 4 A -----B -----C -----D -----52 52 52 52

8

Which statement is true? A There are 12 teams in our netball competition, so our probability of winning the 1 -. competition is ----12 B A coin has been tossed 5 times and all 5 have been heads, therefore the next toss must be tails. C A die has six sides with the numbers 1 to 6 on them. The chance that I will get a 6 to start a game is 1--6- . D There are 18 horses in the Slipper Handicap horse race. Therefore, the probability that 1 -. the favourite will win is ----18

9

A plant seedling has an 80% chance of surviving the first six weeks after planting. To simulate the survival of 100 plants a table of random numbers can be used by: A assigning the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 to survival B assigning the digits 1, 8, 0 to not surviving C assigning the digits 1, 2, 3, 4, 5, 6, 7, 8 to survival D assigning the digits 0, 1, 2 to not surviving

✓

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1, 2

3–7

–

A

B

C

8

9

D

E

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Probability (Chapter 9) Syllabus reference NS5.1.3

REVIEW SET 9A 1

A tennis racquet is spun 200 times. It has rough on one side and smooth on the other. The table shows the results. Outcome

Frequency

Rough

166

Smooth

34

Relative frequency

Percentage

a Copy and complete the table. b Estimate the probability that a spin will result in ‘rough’. 2

A normal six-sided die is thrown once. Determine the probability of getting: a a4 b a5 c an even number d a 1 or a 6

3

Comment on the statement: ‘There are 10 teams in our netball competition, so the probability 1 - ’. that our team will win the competition is ----10

4

The probability that a particular super-12 kicker will score from a penalty is 80%. Simulate 50 kicks, using a table of random numbers, to find the number of goals he would score by: a i assigning digits to represent a goal and a miss ii selecting fifty random one-digit numbers from the table starting with the first number in the fifth row. b Compare your results with the expected number of 40 goals. c Repeat, starting from a different position in the table, and compare the results.

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REVIEW SET 9B 1

Kim rolled a normal six-sided die six times. She did not throw a 6. Kim concluded that the probability of obtaining a 6 was 0. Why is she wrong? How many 6s would be expected in six throws of the die?

2

A hat contains 3 red, 5 white and 9 blue tickets. A ticket is selected at random from the hat. Determine the probability that the ticket is: a red b blue c white d not blue

3

A raffle has 100 tickets numbered from 1 to 100. Determine the probability that the ticket selected is: a number 12 b a number less than 10 c a number between 19 and 31

4

Comment on the statement: ‘Traffic lights can be red, amber or green. Therefore the probability the light is green is 1--3- .’

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Probability (Chapter 9) Syllabus reference NS5.1.3

REVIEW SET 9C 1

Four hundred car salespeople were randomly selected and asked the country in which their car was manufactured. The results are shown in the table. Country

Frequency

Australia

195

Japan

103

Korea

62

Germany

34

Other

Relative frequency

Perecentage

6

a Copy and complete the table. b Use this data to estimate the probability that a salesperson chosen at random will have a car manufactured in Australia. c Estimate the probability that the salesperson’s car is manufactured in Japan. 2

A poker die has faces A, K, Q, J, 10, 9 and is rolled once. Determine the probability of getting: a aK b a number c an A and a J

3

Comment on the statement: ‘I need a 6 to start a game. Therefore, the probability that I will start on my first roll is 1--6- .’

4

A plant seedling has a 90% chance of surviving the first six weeks after planting. Simulate the survival of 100 plants using a table of random numbers by: a i assigning digits to represent the survival and death of the seedling ii selecting 100 random one-digit numbers from the table. b Compare your results with the expected number of 90 survivals.

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REVIEW SET 9D 1

In a year, a restaurant served 5188 bottles of wine. Of these 74 were returned because the wine was faulty. Based on this information, estimate the probability that a bottle of wine from this restaurant is faulty when opened.

2

A 52 -card pack is shuffled and one card is dealt. Determine the probability that it is: a an ace b a red card c a heart d a picture card

3

Matthew and Melissa each tossed a coin 100 times. Matthew counted 53 tails and so stated 53 - . Melissa counted 46 tails and stated that the probability that the probability of tails is --------100 46 - . Who is correct? Explain. of tails is --------100

4

Comment on the statement: ‘There are 24 horses in the Slipper handicap horse race. Therefore, the probability that the 1 - .’ favourite will win is ----24

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Chapter 10 Algebraic Techniques This chapter deals with simplifying expressions, expanding binomial products, and factorising quadratic expressions. After completing this chapter you should be able to: ✓ simplify algebraic expressions, including those involving fractions ✓ generate, evaluate and expand quadratic expressions ✓ recognise perfect squares and complete the square ✓ factorise expressions ✓ simplify expressions involving algebraic fractions ✓ generate and describe quadratic expressions.

Syllabus reference PAS5.3.1 WM: S5.3.1–S5.3.5

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Diagnostic Test 1

15a – a = A 15

2

10 B 14

C 7a – 2b

4

5

6

7

8

9

D 14a

3a – 2b + 4a = A –a – 2b

3

C 15a

B –a + 2b

11

2

D 5a b

5 – (2x – 3) = A 8 – 2x

B 2 – 2x

C –2x – 2

D 5+x

12

B y 2 – 15y

C 3y 2 – 15y

D 3y 2 – y

t t --- – --- = 3 5 A 0 2t C -----15 a --- + a = 3 2a A -----3 a2 C ----3

A 3x 2 + 12x – 2

B –3x 2 + 12x – 2

C –3x 2 + 3x + 2

D –3x2 + 2x + 2

(x – 3)2 = A x2 – 9

B x2 + 9

C x 2 – 6x + 9

D x 2 + 6x – 9

(5 – 3x)2 = A 25 – 3x 2

y(y – 7) – 2y(4 – y) = A –y 2 – 15y

7x – (3x – 1)(x + 2) =

B 25 – 9x 2

C 25 – 15x + 9x 2 D 25 – 30x + 9x 2 13

(5x – 1)(5x + 1) = A 25x 2 – 1

B 25x 2 + 1

C 25x 2 – 10x + 1 D 25x 2 + 10x – 1 B –2 D 2t

4a B -----3 2a 2 D --------3

2y 3x 3y – 4x – ------ + ------ = 3 4 – 5y + 5x 28y – 39x A ----------------------B ------------------------12 12 7y – 13x y–x C ---------------------D ----------12 12

14

15

16

)2 = x 2 – 8x + ∆ then:

If (x – A

= 8; ∆ = 64

B

= 4; ∆ = 16

C

= 2; ∆ = 4

D

=

When factorised, pq – 5p 2 = A –4p 2

B p(q – 5p)

C pq(1 – 5p)

D cannot be factorised

When factorised, 4x + 4t – x 2 – xt = A (4 – x)(x + t)

B (4 – x)(4 – t)

C x(4 – x – t) + 4t

D 3x – 3t

(x – 3)(x + 5) = A x 2 – 15

B x 2 + 2x – 15

C 2x + 2

D x 2 – 2x – 15

17

When factorised, m 2 – 4n 2 = A (m – 4n)(m + 4n) B (m – 4n)2

(3x – 5)(2x + 3) = A 6x 2 – x – 15

B 6x 2 + 5x – 15

C 6x 2 – x – 2

D 6x 2 + 5x – 2

8; ∆ = 8

C (m – 2n)2 D (m + 2n)(m – 2n)

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18

When factorised, 81x 2 – 16y 2 =

23

A (9x – 2y)(9x + 2y)

A (4x – 1)(3x + 2)

B (9x – 4y)(9x + 4y)

B (12x + 1)(x – 2)

C (3x – 2y)(3x + 2y)

C (4x + 1)(3x – 2)

2

D (9x – 4y) 19

20

21

D (12x – 1)(x + 2)

When factorised, x 2 – 18x + 81 =

24

When factorised, x 3 + x 2 + 12x =

A (x + 9)(x – 9)

B (x + 9)2

A x(x 2 + x + 12)

C (x – 18)2

D (x – 9)2

B x(x – 3)(x + 4) C x(x + 12)(x + 1)

When factorised, 16x 2 – 40x + 25 = A (16x – 25)2

B (16x + 25)2

C (4x – 5)2

D (4x + 5)2

D x 2(x + 1) + 12x 2

25

When factorised, x 2 – 7x + 10 = A (x – 5)(x – 2)

B (x + 5)(x + 2)

C (x – 10)(x – 1) D (x + 10)(x + 1) 22

When factorised, 12x 2 + 5x – 2 =

26

When factorised, 3x 2 + x – 10 = A (3x + 1)(x – 10)

x + 5x – 14 = When simplified, -----------------------------2x + 14 B x(x + 3) A x 2 + 3x – 2C x–1 D x----------2 3 5 - – ------------- = When simplified, ----------------2 2 x + 2x x – 4 –2 ( x + 3 ) A ------------------------------------x(x + 2)(x – 2) 2

B (3x + 5)(x + 2)

– 2x – 10x – 12 B -----------------------------------------2 2 ( x + 2x ) ( x – 4 ) –2 --------------C 2x – 4

C (3x – 1)(x + 10) D (3x – 5)(x + 2)

–2 D -----------------------------2 2x + 2x – 4 If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–4

5–7

8–10

11–14

15, 16

17, 18

19, 20

21

22, 23

24

25, 26

–

A

B

C

D

E

F

G

H

I

J

K

L

A. SIMPLIFYING EXPRESSIONS (REVIEW) Example 1 Simplify, where possible, by collecting like terms. a 3a + 4a b 11b – b c 2ab + 3ab

d 3x 2 + 2x

☞

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a 3a + 4a = 7a

.

b 11b – b = 11b – 1b = 10b d 3x 2 + 2x is in simplest form (x 2 and x are unlike terms)

c 2ab + 3ab = 5ab

Exercise 10A 1

Simplify, where possible, by collecting like terms. a 4a + 5a b 3x + 2x e 2x + x f x + 3x i a2 + a2 j 7x + 3 m 17x – x n x2 + x q 3b 2 – b 2 r 2ba + 3ab u xy + 2yx v 3p – 2p

c g k o s w

5x – 2x 3x – 2x x 2 + 10x 2 7b + b 11n – 11n 15b – 8b

d h l p t x

b+b 3x – x 17x – 7 7b – b 3ab + ba 7bca – 5abc

Example 2

2

3

Simplify, where possible, by collecting like terms. a 6d + 3d + 5 b 4x + 5 + 2

c 3x – x 2 + 4x

a 6d + 3d + 5 = 9d + 5

c 3x – x 2 + 4x = 7x – x 2

b 4x + 5 + 2 = 4x + 7

Simplify, by collecting like terms. a 6x – 3x + 7x b 7y – 2y – 3y d 7ab – 2ab + 11ab e 4x 2y + 13x 2y – 8x 2y g 11xy – 5xy – 6xy h 6ab – 3ab – 2ab + 7ab

c 9a 2 + 13a 2 – 17a 2 f 7q + q + 4q – 10q i 4p 2q + 3p 2q – 5p 2q – p 2q

Simplify, where possible, by collecting like terms. a 5x + 4x – 2 b m+7+4 d 2y + y + 3 e p + 3p – 5 g 6t + 4 – 3t h 16n – 16 + 5n j 7k + k – 8 k 4a2 – a2 – 7a 2 2 m 2m – m + 5n n 3x + 2y + 5x 2 2 p 2ab + b + 2b q 6x + 4x – 10 t 5 + 2a – 1 s n + n + 2n 2 v 5mn – 8m – 4mn w a2b + 2ab 2 + 4a 2b

c f i l o r u x

x+6+x 7 + 4x – x p 2 + p + 4p 2 5cd + 2dc – 2 3a + 2b – a 6a + 3a – ab x – x 2 + 2x 5x 2 – 3x 2 + 6x 3

Example 3 Simplify, where possible, by collecting like terms. a 4x – 7x – 5 + x b 5 – 5a – a + 7

c x – 5 – 2x – 1

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a 4x – 7x – 5 + x = –2x – 5 (as 4 – 7 + 1 = –2)

4

5

6

b 5 – 5a – a + 7 = 5 + 7 – 5a – a = 12 – 6a

c x – 5 – 2x – 1 = x – 2x – 5 – 1 = –x – 6

Simplify, where possible, by collecting like terms. a 4x – 7x b –4x – 7x d 6d – d e –6d – d g 4n – 11n h –4n + 11n j 3a + 2 – 6a k 5–d–8 m –3g – (–g) n 5a – (–a) – 2a

c f i l o

–4x + 7x –6d + d –4n – 11n x – (–2x) 4ac – 5ca

Simplify, where possible, by collecting like terms. a a + 3 + 2a + 7 b 5 + 2a + 3 + 4a d 2a + 3b + 3a + b e 3a2 + a + a2 + 2a g 3 + 6y + 1 + 2y h n 2 + n – n + n2 j x2 + 2x – x2 + 5x k 3x – 5x – 3 + 2x m 7 – 5x – 7x + 3 n 3p + 7p – 8 + 4p p 8 – 7x – 5 + 3 q x – 8 – 7x – 5

c f i l o r

3a + 2 + a + 4 ab + b2 + 2ab + 2b2 18c + 5 – 4 – 11c 7 – 5p + 3p – 12 3x + 7x – 2 – x x2 + x + 2 – 5x

Simplify, where possible, by collecting like terms. a –8l – 4 + 3l – 6 b x2 + 2x – 5x – x2 d ab + b – 2ab – 3b e –x – 5 – 2x – 3 g 3a – 2 – a + 3 – a h a2b + a2b – 3a2b + 7b j 4p5 + 5p4 – p5 – 6p4 k 8m 2 – 5n2 + 3n2 – 4m2

c f i l

3x – 2y – (–x) + y 5t – (–t) + 6 – t 5d – c + d – 2c + 2 6s2t + 5s2 – 8s2t – 9s2

Example 4 Find an expression for the perimeter of this rectangle.

x+2 x

Perimeter = (x + 2) + x + (x + 2) + x = 4x + 4

7

Write a simplified expression for the perimeter of the following shapes. a b c

x x+5

x–3 x

y y+3

☞

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d

e

f

y

x+1

x

x–1 x

x+4

Example 5 A rectangular garden has one side 4 metres longer than the other. Write two different expressions for its perimeter. Method 1: Let the shorter side be x metres, then the longer side is (x + 4) metres. The perimeter is given by P = x + (x + 4) + x + (x + 4) = 4x + 8 metres

Method 2: Let the longer side be y metres, then the shorter side is (y – 4) metres. The perimeter is given by P = y + (y – 4) + y + (y – 4) = 4y – 8 metres

(y – 4) m

xm ym

(x + 4) m

8

Write down two expressions for the perimeter of a rectangular garden with one side a 2 m longer than the other c 5 m shorter than the other

b 3 m longer than the other

9

A triangular fence has the longest side 4 metres longer than the second side which is 3 metres longer than the smallest side. Write three expressions for the perimeter.

10

A rectangular garden has one side 1 metre longer than the other. Write two expressions for its area.

Example 6 Expand and simplify: a 2 – (3 – 4x)

b 4 + 3(x – 5)

a

b

2 – (3 – 4x) = 2 – 3 + 4x = –1 + 4x = 4x – 1

4 + 3(x – 5) = 4 + 3x – 15 = 3x – 11

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11

Expand and simplify. a 6 – (3x + 5) d 6x – (3 – 4x) g 3x – (2x – 1) j 3(2 – x) – 11x

b e h k

4 – 2(1 – 2x) 5x – 3(2x + 1) 5 – 3(x – 2) 2(x – 3) + 1

c f i l

19 – (5x + 8) 7 – 5(1 – 2x) 7 – 6(3x – 2) 3(x – 2) – 9x

Example 7

12

13

Expand and simplify. a 4(x + 2) + 3(5 – x)

b x(2 – x) – 3(4 – 3x)

a

b

4(x + 2) + 3(5 – x) = 4x + 8 + 15 – 3x = 4x – 3x + 8 + 15 = x + 23

Expand and simplify. a 2(x + 3) + 3(x – 1) d 2(1 – x) – 3(x – 1) g n(n + 2) + n(2n + 1) j x(x + 7) – 3x(2 – x) m 2x(x + 3) – 5(5 + x) p –2x(1 – x) + 3(x – 4)

b e h k n q

x(2 – x) – 3(4 – 3x) = 2x – x 2 – 12 + 9x = 11x – x 2 – 12

3(y + 1) + 2(y + 3) d(d + 1) + d(d – 1) n(n + 2) – n(2n + 1) a(b + c) – b(c + a) –(x – 3) – 2(2 – x) –2(x + 3) – 5x(x + 1)

c f i l o r

Write an expression for the area of the following shapes in: i factorised form ii expanded form. a b c

4x

3 x+5

2(p + 1) – 3(p – 2) d(d + 3) – d(d – 4) 4(2 + 3x) + 3(x + 5) 2x(x + 1) + 3(x + 2) 4x(1 – x) + 2(x + 6) –x – 5 – 8(x + 1)

d

3x + 1

2x + 7

x+y 6

2y

B. EXPRESSIONS INVOLVING FRACTIONS To add (or subtract) two or more algebraic fractions, we must first form a common denominator and then add (or subtract) the numerators. For example, just as

2 3 -- + -3 4 2 4 3 3 = -- × -- + -- × -3 4 4 3 8+9 = -----------12 17 = ----12

(as LCD is 12)

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x 5x to find -- + ----- we find the LCD and proceed in the 3 4 same way as for ordinary fractions. x 5x So, -- + ----3 4 x 4 5x 3 = -- × -- + ----- × -3 4 4 3

LCD means lowest common denominator.

4x 15x = ----- + -------12 12 19x = -------12

Example 1 Simplify: x 3x a --- + -----2 4 a

a 2a b --- – -----3 5

x 3x --- + -----2 4 x 2 3x = --- × --- + -----2 2 4

b

a 2a --- – -----3 5 a 5 2a 3 = --- × --- – ------ × --3 5 5 3 5a – 6a = ------------------15 –a = -----15 a = – -----15

2x + 3x = ------------------4 5x = -----4

Exercise 10B 1

Simplify by writing as a single fraction. a --a- + --a2 3

bb --b- – ----5 10

e --b- + --b3 4

f

5t--t- – ---3 9

g m ---- + 2m -------7 21

j

3r- + --r----7 5

k

i

2p p ------ – -----5 15

3x- – --xm ----8 2

9r- – ----3rn ----16 4

c

--c- + 3c -----4 2

6x ------ – 2x -----7 3

o 3m -------- + m ---11 5

d --x- – --x7 2 h 5d ------ – --d6 3 l

4m -------- – m ---9 3

p 8d ------ – 7d -----5 4

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Example 2 Write as a single fraction: a --b- + 1 3 a

b

--b- + 1 3

b

--a- – a 4 --= --a- – a × 4 4 4 a – 4a = --------------4 – 3a = ---------4

= --b- + 3 --3 3 +3 = b -----------3

2

--a- – a 4

Write as a single fraction: x a --- + 1 2 x e --- – 4 2 a i --- + a 3 x m --- – 3x 3

y b --- – 1 3 x f 2 – --5 a j --- – a 2 m m m n ---- + ---- + ---2 3 6

a --- + 2 2 a g a + --2 x k x + --7 a a a o --- – --- + --2 3 6

c

b d --- – 3 4 b h b + --3 x 2x + --2 x x x p --- – --- + --4 3 5 l

Example 3 Simplify: 2x 9a 2a – 5x + ------ – -----3 5 2x 9a 2a – 5x + ------ – -----3 5 15 15 2x 5 9a 3 = 2a × ------ – 5x × ------ + ------ × --- – ------ × --15 3 5 5 3 15 30a 75x 10x 27a = ---------- – --------- + --------- – ---------15 15 15 15 3a – 65x = ---------------------15 3

Simplify: x a 3x – 2a + 3a ------ – --4 2 7p 4r d ------ – ------ + 2p – r 3 5 3t 5m g ----- – -------- + m ---- – 3t 4 3 4

b e h

3p 3m -------- – 2p + ------ – 4m 2 5 6m n -------- – --- + 4m – 3n 5 2 5r ------ – 3m -------- + 3r ------ – 7m -------7 4 2 2

c f i

4a – 7b + 3b ------ – 4a -----4 5 7a 3b 4a – 5b – ------ + -----4 5 7x 2y ------ – ------ + 3x – y 3 5

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C. QUADRATIC EXPRESSIONS Quadratic expressions are of the form ax 2 + bx + c where x is a variable, and a, b and c are constants, with a ≠ 0.

Example 1 A rectangle has length 4 cm more than its breadth. Write an expression for the area. Let the breadth be x cm. Then the length is (x + 4) cm. Area = length × breadth ∴ A = (x + 4) × x ∴ A = x(x + 4) ∴ A = x 2 + 4x cm2

x+4 x

Exercise 10C 1

A rectangle has length 6 cm more than its breadth. Write an expression for the area.

2

A rectangle has length 5 cm more than its breadth. Write an expression for the area.

3

A rectangle has breadth 3 cm less than its length. Write an expression for the area.

4

A triangle has perpendicular height 3 cm more than its base length. Write an expression for the area.

5

What is an expression for the area of the rectangle with sides (x + 3) and (x + 1)?

6

What is an expression for the area of the triangle with base (x + 7) and altitude (x + 2)?

Example 2 Expand and simplify the expression for the area of a rectangle with sides (x + 3) and (x + 2). x

+3

x

x2

+3x

+2

+2x

+6

Area = (x + 3)(x + 2) A = x(x + 2) + 3(x + 2) = x 2 + 2x + 3x + 6 = x 2 + 5x + 6

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Example 3 Expand and simplify: a (x + 3)(x – 5)

b (x – 2)(x – 1)

c (x – 5)(x + 7)

a

b

c

x + 3)(x – 5) = x(x – 5) + 3(x – 5) = x 2 – 5x + 3x – 15 = x 2 – 2x – 15

(x – 2)(x – 1) = x(x – 1) – 2(x – 1) = x 2 – x – 2x + 2 = x 2 – 3x + 2

(x – 5)(x + 7) = x(x + 7) – 5(x + 7) = x 2 + 7x – 5x – 35 = x 2 + 2x – 35

7

Expand and simplify the expression for the area of a rectangle with sides: a (x + 5) and (x + 1) b (x + 7) and (x + 2) c (x + 3) and (x + 8)

8

Expand and simplify: a (x + 4)(x + 2) d (x + 4)(x – 3) g (x – 10)(x + 3) j (x – 3)(x – 2)

b e h k

(x – 3)(x + 2) (x – 5)(x + 3) (x – 4)(x – 7) (x – 6)(x – 9)

c f i l

(x + 6)(x – 2) (x + 9)(x – 5) (x + 7)(x – 1) (x – 12)(x – 3)

Example 4 If x = 3 and y = –2, find the value of: a 3x 2 – 2x + 5 b (x + 2)(y + 3) a

9

2

3x – 2x + 5 = 3(3)2 – 2(3) + 5 = 26

b

(x + 2)(y + 3) = (3 + 2)(–2 + 3) = (5)(1) =5

If x = 4 and y = –3, find the value of: a 2x 2 – 3x + 1 b 4x 2 + 2x – 1 2 d 4y – 7y – 2 e 5x 2 – 7x + 1 g (x – 5)(x + 2) h (y – 3)(y + 5) j (4x + 1)(3y – 2) k (5x – 2)(3x + 1)

c (3x – 2)(x + 5) c

(3x – 2)(x + 5) = (3(3) – 2)(3 + 5) = (7)(8) = 56

c f i l

y 2 – 3y + 5 (x + 3)(y – 2) (3x – 7)(2x + 1) (8y – 2)(y + 1)

Example 5 Expand and simplify: a (2x – 5)(x + 3) a

b (5x – 7)(2x + 5)

c (3x + 2)(2x – 1)

(2x – 5)(x + 3) b (5x – 7)(2x + 5) c (3x + 2)(2x – 1) = 2x(x + 3) – 5(x + 3) = 5x(2x + 5) – 7(2x + 5) = 3x(2x – 1) + 2(2x – 1) = 2x 2 + 6x – 5x – 15 = 10x 2 + 25x – 14x – 35 = 6x 2 – 3x + 4x – 2 = 10x 2 + 11x – 35 = 6x 2 + x – 2 = 2x 2 + x – 15

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10

11

Expand and simplify: a (2x + 3)(x – 1) d (3x + 2)(3x – 5) g (4x + 1)(3x – 1) j (7x – 2)(7x + 2)

b e h k

(2x – 5)(x – 8) (2x + 9)(3x – 2) (2x – 5)(3x – 2) (5x – 3)(2x – 5)

c f i l

(3x – 5)(x – 5) (4x – 5)(4x + 5) (2x + 3)(4x – 5) (4x + 1)(3x – 5)

Expand and simplify: a 3x + (x – 5)(x + 2) d (x + 5)(x + 2) – 5x g 4 – (x – 3)(x + 2)

b e h

6x + (2x – 1)(3x + 4) (3x – 7)(x – 2) + 5x 7x – (2x + 1)(x – 5)

c f i

(x + 2)(x – 7) – 2x (4x – 2)(x + 3) – 4x 2 + 2 6x – (x – 2)(2x – 3)

c

12x 2 + 25x – 14

Investigation 1 WM: Applying Strategies, Reasoning

Using substitution Rachel and Diana have different answers to a question. Rachel has (3x – 2)(4x + 7) = 12x 2 + 13x – 14 and Diana has (3x – 2)(4x + 7) = 12x 2 + 25x – 14. 1

Substitute x = 1 and evaluate the three expressions: a (3x – 2)(4x + 7) b 12x 2 + 13x – 14

2

Who is correct, Rachel or Diana?

3

Robert substitutes x = 0 into the three expressions and says that both Rachel and Diana are correct. a Evaluate each expression when x = 0. b Explain the flaw in Robert’s reasoning.

4

How can you ensure your substitution will work?

D. BINOMIAL PRODUCTS

a

Perfect squares expansion (a + b)2 = (a + b)(a + b)

b

a

a2

ab

b

ab

b2

= a(a + b) + b(a + b) = a 2 + ab + ba + b 2 = a 2 + 2ab + b 2

So,

(a + b) 2 = a 2 + 2ab + b 2

and similarly

(a – b) 2 = a 2 – 2ab + b 2

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Difference of two squares expansion (a + b)(a – b) = a(a – b) + b(a – b) = a 2 – ab + ba – b 2 = a2 – b2

(a + b)(a – b) = a 2 – b 2

So,

Example 1 Expand and simplify: a (x + 5)2 a

(x + 5)2 = x 2 + 2 × x × 5 + 52 = x 2 + 10x + 25

b (x – 3)2 b

c (4x – 5)2

(x – 3)2 = x2 – 2 × x × 3 + 32 = x 2 – 6x + 9

c

(4x – 5)2 = (4x)2 – 2(4x)(5) + (5)2 = 16x 2 – 40x + 25

Exercise 10D 1

2

3

Expand the following perfect squares. a (x + 2)2 b (x + 6)2 e (2x + 3)2 f (4a + 5)2 2 i (4 + 3x) j (1 + 2x)2

c (y + 10)2 g (5x + 4)2 k (5 + 2x)2

d (3x + 1)2 h (3y + 2)2 l (4 + 5x)2

Expand the following perfect squares. a (x – 2)2 b (x – 6)2 2 e (2x – 3) f (5a – 4)2 i (5 – 2x)2 j (1 – 3x)2

c (y – 9)2 g (3x – 4)2 k (5 – 3x)2

d (3x – 1)2 h (4y – 1)2 l (4 – 2x)2

Expand these perfect squares. a (x + 5)2 b (x – 5)2 2 e (3x – 5) f (4x + 3)2

c (2x – 7)2 g (5 + 2x)2

d (2x + 7)2 h (3 – 7x)2

Example 2 Expand the following using the difference of two squares. a (x + 5)(x – 5) b (2x – 3)(2x + 3) a

4

(x + 5)(x – 5) = x 2 – 52 = x 2 – 25

b

(2x – 3)(2x + 3) = (2x)2 – (3)2 = 4x2 – 9

Expand and simplify using the difference of two squares. a (x + 3)(x – 3) b (x + 4)(x – 4) c (x + 6)(x – 6) d (x – 10)(x + 10) e (x + 1)(x – 1) f (2x – 5)(2x + 5) g (3x – 2)(3x + 2) h (5x + 1)(5x – 1) i (7x + 8)(7x – 8)

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5

Expand and simplify using one of the rules. a (x – 10)2 b (x + 7)2 d (x + 11)(x – 11) e (x + 4)(x + 4) g (x – 7)2 h (x – 12)(x + 12)

c (x + 8)(x – 8) f (x – 7)(x – 7) i (x + 12)(x + 12)

Example 3 Complete the following expressions. a (x + 5)2 = x 2 + __ x + __ b (x – __)2 = x 2 – 8x + __

c (y + __)2 = y 2 + 5y + __

These expressions are based on the perfect square expansions (x + y)2 = x 2 + 2xy + y 2 and (x – y)2 = x 2 – 2xy + y 2. a (x + 5)2 = x 2 + __ x + __ The coefficient of x is the ‘twice the product’ term, i.e. 2 × 1 × 5 = 10. The constant term is the second term squared, i.e. 52 = 25. ∴ (x + 5)2 = x 2 + 10x + 25 b

(x – __)2 = x 2 – 8x + __ The coefficient of x must be halved, i.e. –8 ÷ 2 = –4. The constant term is then (–4)2 = 16 ∴ (x – 4)2 = x 2 – 8x + 16

c (y + __)2 = y 2 + 5y + __ 5 The coefficient of y must be halved, i.e. 5 ÷ 2 = --- . 2 5 2 25 The constant term is then  --- = ----- 2 4 25 2 5 2 ∴ (y + --2- ) = y + 5y + -----4

6

Copy and complete the following expressions. a (x + 3)2 = x 2 + __x + __ b (x – 7)2 = x 2 – 6x + __ 2 2 e (x – __)2 = x 2 – 10x + __ d (x – __) = x – 6x + __ g (x + __)2 = x 2 __ __ + 49 h (x + __)2 = x 2 + 18x + __

c (x – 2)2 = x 2 __ __ + __ f (x + __)2 = x2 + 12x + __

Example 4 What number needs to be added to complete the square? a x2 + 4x b x2 – 7x a x 2 + 4x x2 + 4x + __ = (x + __)2 constant = ( 4--2- )2 = 4 ∴ add 4

b x2 – 7x x2 – 7x + __ = (x – __)2 constant = ( – 7--- )2 = 2

49 ∴ add -----4

49 -----4

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7

What number needs to be added to complete the square in the following expressions? a x 2 + 6x b x 2 + 10x c x 2 – 8x d x 2 – 4x e x 2 + 12x f x 2 – 18x g x 2 + 7x h x 2 + 15x 2 2 2 i x – 3x j x – 9x k x –x l x2 + x

E. COMMON FACTORS (REVIEW) Factorisation is the reverse process of expansion. As 3(x + 2) = 3x + 6, the factorisation of 3x + 6 = 3(x + 2). Remove the highest common factor.

Example 1 Factorise fully by removing the HCF. a 10x + 5 b 4x 2 – 2x a

10x + 5 = 5(2x + 1)

b

4x 2 – 2x = 2x(2x – 1)

c p 2q – q 2p c

p 2q – q 2p = pq(p – q)

Exercise 10E 1

2

Factorise fully: a 3a – 3b c pq – qr e 4x 2 + x g pq – 3q 2 i 6x2y – 18xy 2 k 9x 2y + 27xy

b d f h j l

5m + 10n x 2 – 5x 15x + 3x2 2πR – 2πr 28p 2a – 21pa 3pqr – 15p 2q

Factorise fully, removing the negative factor. a –3a – 3b b –4x 2 – 2x c –8a + 4b d –4 – 8b e –3 – x f –18x 2 + 9x

Check factorising by expanding your answer.

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Example 2 Factorise fully: a 6R + xR + yR c 4(x – 1) + y(x – 1)

b 9x + 18xy + 12x 2 d x(y + 3) – (y + 3)

a

b

6R + xR + yR = R(6 + x + y) c 4(x – 1) + y(x – 1) = (x – 1)(4 + y)

3

Take care when removing a negative sign.

9x + 18xy + 12x 2 = 3x(3 + 6y + 4x) d x(y + 3) – (y + 3) = (y + 3)(x – 1)

Factorise fully: a 6B + aB + cB b 4R – xR + yR 2 d 8x – 24xy + 16xyz e 4(x – 2) + y(x – 2) g a(x + 1) + 3(x + 1) h x(x – 4) – (x – 4) j x(a + 1) – (a + 1) – y(a + 1)

c 6x + 14xy – 3xz f 3(x – 1) + y(x – 1) i 3(p – 3) + x(p – 3) + y(p – 3)

Example 3

4

Factorise fully: a 3x + 6 + xy + 2y

b 4x – 4 + xz – z

a

b

3x + 6 + xy + 2y = 3(x + 2) + y(x + 2) = (x + 2)(3 + y)

Factorise fully: a 4x + 2 + 2x 2 + x d xy – 2y + 4x – 8 g xy + 3x – 2y – 6 j 3x + 3t – x 2 – xt m 4 + 4y – 3x – 3xy

b e h k n

4x – 4 + xz – z = 4(x – 1) + z(x – 1) = (x – 1)(4 + z)

3x – 3 + xz – z x 2 – 7x + xy – 7y 2xy – 8x + 5y – 20 3a + ac – 3b – bc 6a – 5ay + 6b – 5by

c f i l o

xy + 5y + 3x + 15 4x – x 2 + 4y – xy 3xy – 7y + 12x – 28 3x 2 + 3xy – 2x – 2y 4p – 3p 2 – 4q + 3pq

F. DIFFERENCE OF TWO SQUARES FACTORISATION Since (x + y)(x – y) = x 2 – y 2, the factorisation of x 2 – y 2 is (x + y)(x – y). This is the difference of two squares factorisation, and can be shown in a diagram.

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y

x

x (x – y)

(x – y)

x x

y

y

y

(x – y)

Area = x 2 – y 2

So, x 2 – y 2 = (x + y)(x – y)

Area = (x + y)(x – y)

Example 1 Factorise: a x2 – 9 a

b 4y 2 – 9

x2 – 9 = (x + 3)(x – 3)

b

c 4x 2 – 25

4y 2 – 9 = (2y + 3)(2y – 3)

c

4x 2 – 25 = (2x + 5)(2x – 5)

Exercise 10F 1

Factorise fully: a x2 – 4 e c 2 – 25 i 4x 2 – 1 m 25y 2 – 16x 2

b f j n

y2 – 9 x2 – y2 9x 2 – 4 100x 2 – 81y 2

c g k o

z 2 – 16 a2 – c 2 9x 2 – 1 64a2 – 25b 2

Example 2 Evaluate using the difference of two squares factorisation. b (4.8)2 – (2.8)2 a 1012 – 992 a

1012 – 992 = (101 + 99)(101 – 99) = (200)(2) = 400

b

(4.8)2 – (2.8)2 = (4.8 + 2.8)(4.8 – 2.8) = (7.6)(2) = 15.2

2

Evaluate using the difference of two squares factorisation. a 3012 – 2992 b 2012 – 1992 c 1052 – 952 2 2 2 2 d (3.5) – (2.5) e (9.4) – (9.3) f 8562 – 8552

3

Factorise: a x4 – y4

b 16a 4 – 81b 4

c (p – q)2 – (p + q)2

d h l p

a2 – 9 m 2 – n2 16y 2 – 9 121r 2 – 9t 2

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G. PERFECT SQUARES FACTORISATION Since (x + y)2 = (x + y)(x + y) = x 2 + 2xy + y 2, the factorisation of x 2 + 2xy + y 2 is (x + y)2. (x – y)2

Similarly

= (x – y)(x – y) = x 2 – 2xy + y 2 So the factorisation of x 2 – 2xy + y 2 is (x – y)2. The sign of the coefficient of x is the sign inside the bracket.

The coefficient of x must be double this number.

Example 1 Factorise: a x 2 + 4x + 4 a

x 2 + 4x + 4 = (x + 2)2

b x 2 – 6x + 9 b

x 2 – 6x + 9 = (x – 3)2

Exercise 10G 1

Factorise: a x 2 + 10x + 25 d x 2 – 16x + 64 g y 2 – 6y + 9 j p 2 – 8p + 16

b e h k

x 2 – 20x + 100 x 2 – 14x + 49 y 2 + 2y + 1 m2 + 6m + 9

c f i l

x 2 + 18x + 81 x 2 + 22x + 121 x 2 – 24x + 144 a2 – 10a + 25

Example 2 Factorise: a 4x 2 + 12x + 9

b

9x 2 – 30x + 25

1 Square root the coefficient of x 2. 2 Square root the constant term. (Note it must be positive.) 3 Check that the product of these multiplied by 2 is the coefficient of x. a 4x 2 + 12x + 9 4 = 2, 9 = 3 2 × 3 × 2 = 12 then 4x 2 + 12x + 9 = (2x + 3)2

b

9x 2 – 30x + 25 9 = 3, 25 = 5 2 × 3 × 5 = 30 then 9x 2 – 30x + 25 = (3x – 5)2

Square root the constant term first.

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Factorise: a 4x 2 + 20x + 25 d 9x 2 – 48x + 64 g 49x 2 + 140x + 100 j 49x 2 – 42x + 9

2

b e h k

25x 2 + 60x + 36 121x 2 – 132x + 36 25x 2 – 20x + 4 9x2 – 30x + 25

16x 2 – 72x + 81 81x 2 + 90x + 25 4x 2 + 44x + 121 100x2 – 180x + 81

c f i l

Example 3 Factorise: a x 2 + 2xy + y 2 These are both perfect squares. a x 2 + 2xy + y 2 = (x + y)2

3

Factorise: a p 2 + 2pq + q 2 d d 2 – 2dp + p 2

b

a 2 – 2ab + b 2

b

a 2 – 2ab + b 2 = (a – b)2

b m 2 – 2mn + n 2 e n 2 – 2nt + t 2

c r 2 + 2rt + t 2 f r 2 + 2ry + y 2

H. QUADRATIC TRINOMIALS Expand

(x + a)(x + b) = x(x + b) + a(x + b) = x 2 + bx + ax + ab = x 2 + (a + b)x + ab the sum of the numbers

The sum of the numbers is the coefficient of x and the product of the numbers is the constant term.

the product of the numbers

So, the factorisation of x 2 + (a + b)x + ab = (x + a)(x + b).

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Example 1 Factorise: a x 2 + 5x + 6

b x 2 + 7x + 10

c x 2 + 7x + 12

a x 2 + 5x + 6 Two numbers that add to give 5 and whose product is 6 are 3 and 2. ∴ x 2 + 5x + 6 = (x + 3)(x + 2) b x 2 + 7x + 10 Two numbers that add to give 7 and whose product is 10 are 5 and 2. ∴ x 2 + 7x + 10 = (x + 5)(x + 2) c x 2 + 7x + 12 Two numbers that add to give 7 and whose product is 12 are 4 and 3. ∴ x 2 + 7x + 12 = (x + 4)(x + 3)

Exercise 10H 1

Factorise: a x 2 + 8x + 7 e x 2 + 10x + 24 i x 2 + 9x + 20

b x 2 + 8x + 12 f x 2 + 13x + 30 j x 2 + 9x + 18

c x 2 + 13x + 12 g x 2 + 11x + 30 k x 2 + 19x + 18

d x 2 + 10x + 9 h x 2 + 12x + 20 l x 2 + 13x + 42

Example 2 Factorise: a x 2 – 4x + 3

b

x 2 – 8x + 12

a x 2 – 4x + 3 Two numbers whose sum is –4 and product is 3 are –3 and –1. ∴ x 2 – 4x + 3 = (x – 3)(x – 1) b x 2 – 8x + 12 Two numbers whose sum is –8 and product is 12 are –6 and –2. ∴ x 2 – 8x + 12 = (x – 6)(x – 2) 2

Factorise: a x 2 – 6x + 5 e x 2 – 9x + 8 i x 2 – 16x + 15

b x 2 – 8x + 7 f x 2 – 7x + 10 j x 2 – 9x + 14

c x 2 – 12x + 11 g x 2 – 11x + 10 k x 2 – 15x + 14

d x 2 – 6x + 8 h x 2 – 8x + 15 l x 2 – 11x + 24

Example 3 Factorise: a x 2 – 3x – 10

b

x2 + x – 6

a x 2 – 3x – 10 Two numbers whose sum is –3 and product is –10 are –5 and 2. ∴ x 2 – 3x – 10 = (x – 5)(x + 2). b x2 + x – 6 Two numbers whose sum is 1 and product is –6 are 3 and –2. 2 ∴ x + x – 6 = (x + 3)(x – 2).

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3

4

Factorise: a x 2 + 7x – 8 e x 2 + 4x – 12 i x 2 + 4x – 21

b x 2 + 3x – 10 f x 2 – 11x – 12 j x 2 – 20x – 21

c x2 + x – 2 g x 2 – 5x – 24 k x 2 + 17x – 60

d x 2 + x – 42 h x 2 + 5x – 24 l x 2 + 3x – 54

Factorise: a x 2 + 19x + 18 e x 2 + 53x – 54 i x 2 – 30x – 64

b x 2 – 7x – 18 f x 2 – 25x – 54 j x 2 + 2x – 35

c x 2 + 17x – 18 g x 2 – 16x + 64 k x 2 + 7x – 30

d x 2 + 15x + 54 h x 2 + 12x – 64 l x 2 – 15x + 50

I. FURTHER QUADRATIC TRINOMIALS There are several methods for factorising trinomials of the form ax2 + bx + c where a ≠ 1. One of these methods is given in the following example.

Example 1 Factorise: a 2x 2 + x – 3

b

a 2x 2 + x – 3 ( 2x + 3 ) ( 2x – 2 ) = -----------------------------------------2

3x 2 + 16x + 5

c 5x 2 + 13x – 6

2 × –3 = –6 ∴ need two numbers with a product of –6 and a sum of +1. These are 3 and –2.

= (2x + 3)(x – 1) b 3x 2 + 16x + 5 ( 3x + 15 ) ( 3x + 1 ) = --------------------------------------------3

3 × 5 = 15 ∴ need two numbers with a product of 15 and a sum of 16. These are 15 and 1.

= (x + 5)(3x + 1) c 5x 2 + 13x – 6 ( 5x + 15 ) ( 5x – 2 ) = --------------------------------------------5

5 × –6 = –30 ∴ need two numbers with a product of –30 and a sum of +13. These are 15 and –2.

= (x + 3)(5x – 2) Place the coefficient of x2 together with x at the beginning of each bracket and divide the whole expression by this coefficient to maintain equality.

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Exercise 10I 1

2

3

Fully factorise: a 2x 2 + 5x + 3 e 2x 2 + 7x + 5 i 5x 2 – 14x – 3

b 2x 2 – 9x – 5 f 2x 2 + 3x – 2 j 5x 2 + 2x – 3

c 3x 2 + 5x – 2 g 7x 2 + 9x + 2 k 5x 2 – 8x + 3

Fully factorise: a 2x 2 + 5x – 12 e 3x 2 + 13x + 4 i 3x 2 + 10x – 8

b 3x 2 – 7x – 6 f 3x 2 – 17x + 10 j 2x 2 + 17x – 9

c g k

Fully factorise: a 2x 2 + 9x – 35 d 3x 2 – x – 2 g 11x 2 – 52x – 15

3x 2 + 7x + 4 3x 2 + 8x + 4 2x 2 + 9x – 18

b 3x 2 + 5x – 12 e 5x 2 – 29x + 20 h 7x 2 – 61x + 40

d 3x 2 – 5x – 2 h 2x 2 + 3x – 5 l 11x 2 – 9x – 2

d 2x 2 – 3x – 9 h 5x 2 – 13x – 6 l 2x 2 + 11x – 21

c 5x 2 – 8x + 3 f 7x 2 + 15x + 2 i 5x 2 – 52x + 63

Example 2 Factorise: a 6x 2 – 13x – 5

b 12x 2 – 5x – 2

a 6x 2 – 13x – 5 ( 6x – 15 ) ( 6x + 2 ) = --------------------------------------------6 3 ( 2x – 5 )2 ( 3x + 1 ) = ------------------------------------------------6 = (2x – 5)(3x + 1) b 12x 2 – 5x – 2 ( 12x – 8 ) ( 12x + 3 ) = -----------------------------------------------12 4 ( 3x – 2 )3 ( 4x + 1 ) = ------------------------------------------------12 = (3x – 2)(4x + 1)

4

5

Fully factorise: a 8x 2 + 14x + 3 e 6x 2 + 19x + 3 i 4x 2 + 4x + 1 Fully factorise: a 6x 2 – 7x – 3 d 12x 2 – 23x + 5 g 10x2 + 19x – 15

6 × –5 = 30 ∴ need two numbers with a product of –30 and a sum of –13. These are –15 and 2. (factorise each bracket) (cancel) 12 × –2 = –24 ∴ need two numbers with a product of –24 and a sum of –5. These are –8 and 3. (factorise each bracket) (cancel)

b 15x 2 + x – 2 f 10x 2 + 17x + 3 j 10x 2 + x – 2

b 4x 2 – 23x + 15 e 12x 2 – 7x – 10 h 20x 2 – 31x – 7

c 21x 2 + 17x + 2 g 14x 2 + 37x + 5 k 9x 2 – 12x + 4

d h l

6x 2 + 5x + 1 21x 2 – 62x – 3 3x 2 + 14x + 8

c 9x 2 – 6x – 8 f 12x 2 – 79x – 35 i 18x 2 + 19x – 12

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J. MISCELLANEOUS FACTORISATION

Always look for common factor first.

We suggest you use the following order of factorising. Follow these steps: 1 common factor 2 difference of two squares 3 quadratic trinomial 4 grouping in pairs

Example 1 Factorise fully: a 3x 2 – 12 a

b 2x 2 – 10x + 12

3x 2 – 12 = 3(x 2 – 4) = 3(x + 2)(x – 2)

b

c x 4 – 9x 2

2x 2 – 10x + 12 = 2(x 2 – 5x + 6) = 2(x – 3)(x – 2)

c

x 4 – 9x 2 = x 2(x 2 – 9) = x 2(x + 3)(x – 3)

Exercise 10J 1

2

3

Fully factorise: a 3x 2 + 2x d 3b 2 – 75 g x 2 – 8x – 9 j 2g 2 – 12g – 110 m 12 – 11x – x 2 p x4 – x2 s a 3b 2 – ab 2 v 9x 4 – 4x 2

b e h k n q t w

x 2 – 81 2x 2 – 32 d 2 + 6d – 7 4a 2 – 9d 2 5a 2 – 5a – 10 d 4 + 2d 3 – 3d 2 x2 – x – 6 x 2 + 8x – 9

c f i l o r u x

2p 2 + 8 n 4 – 4n 2 3x2 – 108 4t + 8t 2 2c 2 – 8c + 6 b 2+ 3b – 28 x 3 + 4x 2 + 4x –2a 2 – 12a – 18

Fully factorise: a 14 – x 2 – 5x d 18x – 2x 3 g 4x 2 – 2x 3 – 2x j (x + 2)2 – 4 m (x + 1)a + (x + 1)b p x(x + 2) + 3(x + 2)

b e h k n q

x 2 + 14x + 49 ab + ac – 2a x 3y – 4xy 4x 4 – 64 x4 – a4 x 3 + x2 + x + 1

c f i l o r

4a3 – 4ab 2 a 2b 2 – 2ab (a + b)2 – 9 (x – 2)y – (x – 2)z (x – y)a + (x – y) x 3 + 2x 2 + x + 2

Where possible, fully factorise the following miscellaneous expressions. b 4x 2 – 1 c 5x 2 – 15x a 3x 2 + 9x

☞

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d g j m p s v 4

3x – 5x 2 x 3 + 2x 2 3x 2 – 12 x2 – x – 6 x 2 – 16x + 39 9x – 18x 2 13x 2 – 52x

e h k n q t w

x 2 + 3x – 40 x2 – 9 3x 3 + 6x 2 4x 2 + 8x 7x 2 – 21x 8x 2 – 12x x 2 + 2x – 3

f i l o r u x

x 2 – 16 3x 3 + 6x x 2 + 10x + 25 9x 2 – 25 2x 2 – 50 4x 2 + 4x – 3 x 3 – 9x

Where possible, fully factorise the following miscellaneous expressions. a x3 + x2 + x b x 2 – 17x – 60 c 3x 2 – 27 d x 2 – 2x – 8 e x 2 + 4x + 4 f 6x 2 + 5x – 6 2 2 g x – 5x + 6 h 36x + 25 i 4x 2 – 8x – 60 j 3x 2 – 42x + 99 k x 2 + 11x + 30 l 49x 2 – 1 2 2 m x – 7x + 12 n x + 6x – 16 o x 2 – 5x – 24 p x 2 – 8x + 16 q x 2 – 9x + 14 r x 2 + 13x + 36 2 2 s x – 9x – 36 t x + 7x – 18 u x 2 – 10x + 25 v 3x 2 + 6x – 72 w 4x 2 – 4x – 48 x (2x + 1)2 – 9

K. FACTORISING MORE COMPLEX EXPRESSIONS To simplify algebraic expressions with numerators and denominators, first factorise all expressions fully. Then simplify the expression by cancelling as appropriate.

Example 1 Factorise and simplify: x2 – x – 6 a ----------------------x–3 a

b

x2 – x – 6 ----------------------x–3 (x – 3)(x + 2) = ---------------------------------x–3

b

=x+2

x 2 – 16 ------------------8x – 32 x 2 – 16 ------------------8x – 32 (x + 4)(x – 4) = ---------------------------------8(x – 4) x+4 = -----------8

Exercise 10K 1

Factorise and simplify: x 2 + 2x a ----------------2 x –4 2x 2 + 6x – 8 e -----------------------------2 x – x – 20

3x + x 2 b ----------------9 – x2 f

x 2 + 6x + 9 ----------------------------x 2 – 5x – 24

3x 2 – 9x c -------------------------x 2 – 2x – 3

x 2 + 2x + 1 d --------------------------x2 – 1

3x 2 – 12 g -----------------------------214 – 5x – x

2x 2 + 6x – 36 h ---------------------------------212 + 8x – 4x

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Example 2 Factorise and simplify: 2y – 4 10y a --------------- × ---------------5 y 2 – 2y a

2y – 4 10y --------------- × ---------------5 y 2 – 2y

b

b

2 2(y – 2) 10y = -------------------- × ------------------y – 2) 5 y ( 1

x 2 – 9 x 2 + 4x + 3 -------------- ÷ --------------------------x + 5 x 2 + 6x + 5 x 2 – 9 x 2 + 4x + 3 -------------- ÷ --------------------------x + 5 x 2 + 6x + 5 (x + 3)(x – 3) (x + 3)(x + 1) = ---------------------------------- ÷ ----------------------------------(x + 5) (x + 5)(x + 1) (x + 3)(x – 3) (x + 5)(x + 1) = ---------------------------------- × ----------------------------------(x + 5) (x + 3)(x + 1)

=4

=x–3

2

Factorise and simplify: 3x – 6 14x a --------------- × ----------------2 7 x – 2x 2 2 x – 16 x + 5x + 4 c ------------------- ÷ --------------------------4x – 16 2x + 2

2

b

2

d

2

7x + 7 x +x–6 - × ----------------------e ---------------------2 x – x – 2 5x + 15 2 x –4 x+2 - ÷ -------------------------g -------------------------2 2 x + 2x – 8 x + 3x – 4

2x + 10x 6 ------------------------- × ------------------4x 3x + 15

f h

x + 8x + 15 4x + 12 ------------------------------ ÷ -----------------2 2 x – 25 x – 5x 2 2 2x – 10x x – 2x – 15 -----------------------÷ ----------------------------2 2 3x – 9x x –9 2 x + 2x 3x – 15 ------------------ × ----------------------------2 x+4 x – 3x – 10

Only cancel terms in brackets or in front of brackets

Example 3 Factorise and simplify: 3 2 - – ------------a ----------------2 2 x + 2x x – 4

b

3 4 ----------------------------- + ------------------2 2x – 10 x – 3x – 10

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a

2 3 ----------------- – ------------2 2 x + 2x x – 4 2 3 = -------------------- – ---------------------------------x(x + 2) (x – 2)(x + 2)

4 3 ----------------------------- + ------------------2 2x – 10 x – 3x – 10 4 3 = ---------------------------------- + -------------------(x – 5)(x + 2) 2(x – 5)

b

2 x–2 3 x = --------------------  ------------ – ----------------------------------  -- x ( x + 2 )  x – 2 ( x – 2 ) ( x + 2 )  x

4 2 3 x+2 = ----------------------------------  --- + --------------------  ------------ ( x – 5 ) ( x + 2 )  2 2 ( x – 5 )  x + 2

2 ( x – 2 ) – 3x = ------------------------------------x(x + 2)(x – 2)

8 + 3(x + 2) = -------------------------------------2(x – 5)(x + 2)

2x – 4 – 3x = ------------------------------------x(x + 2)(x – 2)

3x + 14 = ------------------------------------x(x – 5)(x + 2)

–x–4 = ------------------------------------x(x + 2)(x – 2) 3

Factorise and simplify: 3 4 a ----------------- – ------------2 2 x + 3x x – 9 2 3 c ----------------------------- + ----------------2 2 x – 4x – 12 x – 6x 2 1 e ----------------– -----------------------------2 2 x – 2x x + 5x – 14 1 3 g ----------------- + -------------------------2 2 x – 25 x – 6x + 5

2 5 b ----------------- + ------------------2 x – 16 4x – 16 4 2 d ------------------- – -------------------------3x – 12 x 2 – 3x – 4 7 3 -----------------------------+ ----------------2 2 x – 7x + 10 x – 2x 4 2 h ----------------– ------------2 2 x – 3x x – 9 f

L. QUADRATIC RELATIONSHIPS Many graphs are not straight lines. We see curved graphs in many real-life situations. One of these graphs, a parabolic graph, occurs when we have a quadratic relationship. The parabola is the basic shape used in headlight and torch reflectors, solar furnaces and satellite dishes. We observe quadratic relationships in geometric patterns.

Example 1 Examine this pattern of squares and determine the relationship between the width of each figure (W) and the total number of squares (S). W

W

W

W

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a Complete the table, drawing extra diagrams if necessary.

W

b Graph S versus W on a number plane.

S

1

2

3

4

5

c State the relationship between S and W as an algebraic formula. a W

1

2

3

4

5

S

2

5

10

17

26

Graph S vs W means W on the horizontal axis and S on the vertical axis.

b 25

S

20 15 10 5

W 1 2

3

4 5 6

7 8

c S = W2 + 1

The class could discuss the following questions. 1

What happens for values of W that are not integers, e.g. W = 1.5? What is the value of S?

2

Can W have values less than 1, e.g. W = 0.5?

3

Can W have value zero?

4

What happens as W gets larger?

Exercise 10L 1

Examine this pattern of squares and determine the relationship between the width of each diagram (W) and the total number of squares (S). a Copy and complete the given table, drawing extra diagrams if necessary. b Graph S vs W on a number plane. c State the relationship between S and W as an algebraic formula.

2

Repeat question 1 for each of these patterns. a W W

W

1

2

3

4

5

S

W

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b

W

W

W

Example 2 A rectangle has perimeter 24 cm. a Complete this table comparing length (l), breadth (b) and area (A). l

1

2

3

4

5

6

7

8

9

10

11

b A b c d e

Graph l vs A for this rectangle. What is the maximum area? When does it occur? Can l be 13 or 0? Find the value of A when l = 3.5.

a

p = 2l + 2b i.e. 2l + 2b = 24 2(l + b) = 24 l + b = 12

∴ b = 12 – l If l = 1 b = 12 – 1 ∴ b = 11

l

1

2

3

4

5

6

7

8

9

10

11

b

11

10

9

8

7

6

5

4

3

2

1

A

11

20

27

32

35

36

35

32

27

20

11

b c Maximum area = 36 cm2 when l = 6 and b = 6, i.e. a square. d If l = 13 then b = –1, and it is not possible to have a side length of –1. ∴ l ≠ 13. If l = 0 then the side would be zero, i.e. no rectangle at all. e If l = 3.5, b = 12 – 3.5 = 8.5, A = 3.5 × 8.5 = 29.75 cm2.

A 40 36 30 20 10 l 2

4

6

8

10

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3

4

A rectangle has perimeter 20 cm. a Complete this table comparing length, breadth and area. b Graph l vs A for this rectangle. c Find the value of A when l = 3.5.

l

1

2

3

4

5

6

7

8

9

b A

Repeat question 3 using rectangles with the following perimeters. i 18 cm ii 28 cm iii 22 cm iv 26 cm

v 30 cm

Example 3 a Find an equation that describes y in terms of x for these tables of values. b Draw a graph of y vs x for each table of values. i x –5 –4 –3 –2 –1 0 1 2 3 4

5

y

27

18

11

6

3

2

3

6

11

18

27

x

–5

–4

–3

–2

–1

0

1

2

3

4

5

y

29

20

13

8

5

4

5

8

13

20

29

2

4

6

2

4

6

ii

i

a

The values are not increasing by a constant value but starting from x = 0 they increase by 1 then 3 then 5 then 7 and so on. ∴ look for a squaring pattern: If x = 1 then y = 3, y = (1)2 + 2 If x = 2 then y = 6, y = (2)2 + 2 If x = 3 then y = 11, y = (3)2 + 2 ∴ y = x2 + 2

b

y 35 30 25 20 15 10 5 –6

ii a

Look for a squaring pattern: If x = 1 then y = 5, y = (1)2 + 4 If x = 2 then y = 8, y = (2)2 + 4 If x = 3 then y = 13, y = (3)2 + 4 ∴ y = x2 + 4

x

–2

–4

y 35

b

30 25 20 15 10 5

x –6

5

–2

–4

Find an equation which describes y in terms of x for these tables of values. a x

–5

–4

–3

–2

–1

0

1

2

3

4

5

y

26

17

10

5

2

1

2

5

10

17

26

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b x

–5

–4

–3

–2

–1

0

1

2

3

4

5

y

30

21

14

9

6

5

6

9

14

21

30

x

–5

–4

–3

–2

–1

0

1

2

3

4

5

y

24

15

8

3

0

–1

0

3

8

15

24

x

–5

–4

–3

–2

–1

0

1

2

3

4

5

y

23

14

7

2

–1

–2

–1

2

7

14

23

x

–5

–4

–3

–2

–1

0

1

2

3

4

5

y

28

19

12

7

4

3

4

7

12

19

28

c

d

e

6

Draw a graph of y versus x for each of the tables in question 5.

7

The perimeter of a rectangle is 50 cm. a Find an expression for the area, A cm2, of the rectangle. b Sketch the graph of A against l, the length of the rectangle.

8

The perimeter of a rectangle is 38 cm. a Find an expression for the area, A cm2, of the rectangle. b Sketch the graph of A against l, the length of the rectangle.

non-calculator activities

1

Evaluate 8 – 2.73.

2

Meat pies are $2.80. On Sundays they cost 10% more. How much is a pie on Sunday?

3

Convert 4.2 m to cm.

4

1 Evaluate  1 --- .  4

5

Insert grouping symbols to make this statement true: 10 – 3 + 4 × 3 = 33

6

The temperature at dusk was 3°C. The temperature dropped 0.5°C per hour. What was the temperature after 9 hours?

2

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Algebraic Techniques (Chapter 10) Syllabus reference PAS5.3.1

7

If 512 × 6284 = 3 217 408, what is 5.12 × 62.84?

8

Solve 5 – 3p = 7.

9

A bag contains 6 white, 5 green and 2 blue disks. What is the probability of not selecting a green disk?

10

Sarah drives 350 km and uses 28 L of petrol. Express this rate in L per 100 km.

11

The mode of the data shown in this stem-and-leaf plot is 43. What is the value of
? Stem

Leaf

3

7, 8, 9

4

0, 2, 3,
, 5, 6, 9

5

0, 2

12

The 3-digit number 4
1 is divisible by 9. What is the missing digit?

13

Between which consecutive whole numbers is the square root of 53?

14

1 – 3--Find -----------4- . 1 + 3--4

15

Andreas collects 15 cans of food from each of 8 friends. He then gives six cans to each home room class. How many home room classes are there?

Language in Mathematics Read the following biography and answer the questions below.

Sir William Rowan Hamilton (1805–1865) William Rowan Hamilton was born in Dublin in 1805, the son of a solicitor. His ability was evident at an early age, as by the age of 13 he had managed to learn thirteen different languages. This mastery of languages helped him to become one of the few great mathematicians with the facility for involved mental calculations. In 1823 Hamilton entered Trinity College in Dublin and was appointed to the post of Andrew’s Professor of Astronomy and Royal Astronomer to Ireland while still an undergraduate. In 1827 he moved to Dunsink Observatory just outside Dublin. Hamilton continued his work in physics, astronomy and algebra and in 1835 was knighted for his

303

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contribution to science. In the same year he discovered quaternions—a very important step in the development of modern algebra. He continued to work on quaternion theory and spent the last 20 years of his life as a recluse. The results of his research The Elements of Quaternions were published after his death in 1866. He was honoured by many foreign academics for his contributions in each of his fields of interest. 1

a b c d e

How old would Hamilton be in 2007? For what was he knighted? When? How old was Hamilton when he moved to Dublin? What were Hamilton’s main interests? What is a recluse?

2

Rearrange these words to form sentences. a opposite Expanding is factorising of the b highest Always factor factorise common possible the c bracket bracket the the by in outside Multiply to term all terms the the

3

Use every third letter to find a sentence. HKTQTOFGCYUHFDESACEEKRRATLFACABTCIETQWOOPRDFI ZTSNHAZYSKNOJTLPBUACTBEIRSOAANIOYPRODTUGTCHYAE RNAXEVTINHTMOHQQEAERFEEGEXTTPHJAAANCVDBGYEDOY HUKHRARATINOISBKWCDEXSRWEORGRHJSKMUNBBGFSDQTE HIOLTFXUFGTUUEEUANJNADUTHMUJBWQEASRCTIBUNNITOP OFDTGHHTEEAEQSHUNYE FGSJKTERITHOASNFGAYUNIODLL TASHCFEVYANUNMISOLWWCEEVRTNTYJOIESASEDFEGHIJKF LOTTRHQAEZXYCVABNRMKEHJEGFQSAUQWAERL

4

Give an algebraic example of five glossary terms.

Glossary algebra expand perfect square trinomial

binomial products expression quadratic expressions

common factor grouping in pairs solve

difference of two squares factorise substitution

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Algebraic Techniques (Chapter 10) Syllabus reference PAS5.3.1

305

CHECK YOUR SKILLS 1

2

3

4

5

6

7

8

9

10

11

12

13

8z – z = A 8

B 7

C 7z

4x – 3y + 5x – 2y = A 9x – 5y

B –x – 5y

C –x – y

D 9x + 5y

2x – (4 – 3x) = A –x – 4

B –4 + x

C 5x + 4

D 5x – 4

C 5m 2 + 25m

D –3m 2 + 25m

m(m + 3) – 4m(7 – m) = A –3m 2 – 25m B 5m 2 – 25m p 3p --- – ------ = 4 7 – 2p A ---------–3 t t + --- = 4 5t A ----4 3x 3p 4p – ------ + ------ – 2x = 4 5 7p 5x A ------ – -----5 4

✓

– 5p B ---------28

C

2t B ----4

C 5t

7p – 5x B ------------------20

C

(x + 7)(x – 5) = A x 2 – 35

B x 2 + 2x – 35

C x 2 + 2x + 2

D 2x + 2

(4x – 3)(2x + 5) = A 8x 2 + 14x – 15

B 8x 2 + 6x – 15

C 6x + 2

D 8x 2 + 26x – 15

4x – (2x + 1)(x – 3) = A –2x 2 – x – 3

B –2x 2 – x + 3

C –2x 2 + 9x + 3

D –2x 2 + 9x – 3

(y – 5)2 = A y 2 – 25

B y 2 + 25

C y 2 – 5x + 25

D y 2 – 10x + 25

(5x – 3)2 = A 25x 2 – 9

B 5x 2 – 9

C 25x 2 – 30x + 9

D 25x 2 – 30x – 9

(3x – 2)(3x + 2) = A 3x – 2

B

9x2 – 4

C

– 2p ---------28

D 8z

92p + 55x -------------------------20

9x 2 – 12x – 4

4p D -----28

3t D ----4

92p – 55x D -------------------------20

D 9x 2 – 12x + 4

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14

15

16

17

18

19

20

21

22

23

24

If (x –
)2 = x 2 – ∆x + 16 then: A
= 4, ∆ = 8 B
= 16, ∆ = 8

C
= 256, ∆ = 128 D
= 16, ∆ = 4

When factorised, 3x 2 – 6x = A –3x B –3x 3

C

3x(x – 2)

D 3(x 2 – 2x)

When factorised, px + 3q – 3p – qx = A (p – q)(x – 3) B pq(3 – x)

C

(p – 3)(x – q)

D cannot be factorised

When factorised, x 2 – y 2 = A (x – y)(x + y) B (x – y)2

C

(x + y)2

When factorised, 4p 2 – 25q 2 = A (2p – 5q)2 B (2p – 5q)(2p + 5q)

C

(4p – 25q)(4p + 25q) D (4p – 25q)2

When factorised, x2 – 10x + 25 = A (x + 5)(x – 5) B (x + 5)2

C

(x – 10)2

When factorised, 81x2 – 198x + 121 = A (9x + 11)(9x – 11) B (9x – 11)2

C

(81x + 11)2

When factorised, x 2 + 4x – 21 = A (x + 7)(x – 3) B (x – 7)(x + 4)

C

(x + 21)(x – 1)

D (x – 21)(x + 1)

When factorised, 5x 2 – 7x – 6 = A (5x – 3)(x + 2) B (5x + 1)(x – 6)

C

(5x + 3)(x – 2)

D (5x – 1)(x + 6)

When factorised, 8x 2 + 5x – 3 = A (4x – 1)(2x + 3) B (8x + 3)(x – 1)

C

(4x + 1)(2x – 3)

D (8x – 3)(x + 1)

When factorised, 16x 2 – 4 = A (4x – 2)(4x + 2) B 4(2x + 1)(2x – 1)

C

4(4x2 – 1)

D 4(x + 1)(x – 1)

C

x2 – 9x + 20

2

25

26

x – 12x + 35 When simplified ---------------------------------- = 3x – 15 2 x – 4x + 7 x–7 A --------------------------B -----------–3 3 7 2 When simplified, -----------------------------– ----------------= 2 2 x – 7x + 10 x – 5x 2

5 A ---------------------------------------------------------2 2 ( x – 7x + 10 ) ( x – 5x )

7x – 33x – 4 B ------------------------------------x(x – 5)(x – 2)

✓ D 2(x – y)

D (x – 5)2

D (81x – 121)2

D

x–5 -----------3

2

5x + 4 C ------------------------------------x(x – 5)(x – 2)

D

7x – 37x – 4 ------------------------------------x(x – 5)(x – 2)

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–4

5–7

8–10

11–14

15, 16

17, 18

19, 20

21

22, 23

24

25, 26

–

A

B

C

D

E

F

G

H

I

J

K

L

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Algebraic Techniques (Chapter 10) Syllabus reference PAS5.3.1

REVIEW SET 10A 1

2

Expand and simplify: a 4 – (5x + 3)

c n(n + 2) – 2n(n – 1)

a b a + --3

2x 2a c 2a – 3x + ------ – -----5 3

Write as a single fraction: 3p 2p a ------ – -----7 5

3

b 3(x – 5) – 2

Expand and simplify if possible. a (x + 5)(x – 3) b (2x – 7)(3x – 8) d (3x – 4)2 e (x – 2)(x + 2)

c (x – 8)2 f (3x – 5)(3x + 5)

4

a If x = 3, evaluate 3x2 – 5x + 8. b Which number must be added to complete the square for: i x 2 – 6x ii x 2 + 11x?

5

Fully factorise: a 7x + 14 d 4y 2 – 25 g x 4 – 16x2

6

b 8x 2y – 20xy e x 2 + 7x – 8 h –2x 2 + 2x + 24

c 3x – 9 + xy – 3y f 3x 2 + 11x – 4 i 3x + 6y – 9z

Factorise and simplify: 2

x +x–6 a ----------------------x–2

2

x –9 2x + 4 b --------------------- × ---------------------4x + 12 x 2 – x – 6

c

2 3 ----------------+ ----------------2 2 x – 4x x – 16

307

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REVIEW SET 10B 1

2

Expand and simplify: a –3 – 2(3x – 1)

c p(p + 7) – 3p(4 – p)

x b --- + 2x 5

a 2x c 3x – --- + ------ – 3a 4 3

Write as a single fraction: 4x 2x a ------ – -----3 5

3

b 4(2x – 3) – 5x

Expand and simplify if possible. a (x + 2)(x – 11) b (3x – 8)(4x – 3) 2 d (8x – 5) e (x – 3)(x + 3)

c (x + 4)2 f (5x – 3)(5x + 3)

4

a If x = –2, evaluate 2x 2 – 9x + 5. b What number must be added to complete the square for: i x 2 + 10x ii x 2 – 9x?

5

Factorise fully: a 3a – 9 d 16y 2 – 25 g 9x 4 – 16x 2

b 12xy + 18x 2 e x 2 – 3x + 2 h x 3 + 3x 2 + x + 3

Factorise and simplify: 2 x –9 a ------------------6x – 18

x – 3x – 4 x – 4x – 5 b --------------------------- ÷ -------------------------2 8x – 32 x – 25

6

2

c 2xy – 6x + 7y – 21 f 6x 2 – 7x – 5 i 3x + 2xy – 4xz

2

c

1 3 ------------------------------ – ------------2 2 x + 7x + 10 x – 4

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Algebraic Techniques (Chapter 10) Syllabus reference PAS5.3.1

REVIEW SET 10C 1

2

3

Expand and simplify: a 4 – (3x – 2)

b 2(x + 3) – 4

c y(y + 2) – 3y(y + 4)

Write as a single fraction: p 4x 2x a ------ – -----b --- + p 3 3 5

x 3y c 3x – 4y + --- – -----3 4

Expand and simplify, if possible. a (x – 3)(x + 4) b (2x – 5)(3x – 7) d (4p – 5)2 e (x – 3)(x + 3)

c (x – 4)2 f (4x – 3)(4x + 3)

4

a Use x = 3 to show that 6x 2 – x – 2 = (3x – 2)(2x + 1). b Copy and complete: x 2 – 4x +
= (x – 2)2

5

Factorise: a 5x + 10 d 9x 2 – 100 g x 3 – 16x

b 3x 2 – 6xy e x 2 – x – 12 h –3x 2 – 21x + 24

Simplify: 2 x – x – 20 a -------------------------x–5

x –4 3x – 15x b ----------------------× ----------------------------2 2 x + x – 6 x – 3x – 10

6

2

c 2x – 4 + 3xy – 6y f 3x 2 + 16x + 5 i 12x – 9p + 6z

2

c

4 5 ------------- – ----------------2 2 x – 9 x + 3x

309

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Algebraic Techniques (Chapter 10) Syllabus reference PAS5.3.1

REVIEW SET 10D 1

2

3

Expand and simplify: a 2x – 2(4x – 7)

b 3(7 – 4x) – 3x

Write as a single fraction: 3z 4z a ------ – -----b 7 5

c m(m – 4) – 3m(6 – m)

r --- + 2r 4

2p 5x c 2x – 4p – ------ + -----3 2

Expand and simplify, if possible. a (x – 2)(x + 11) b (3x – 7)(4x + 2) d (6z – 5)2 e (x – 9)(x + 9)

c (y + 3)2 f (6x – 5)(6x + 5)

4

a Use x = –1 to show that 10x 2 – x – 3 = (5x – 3)(2x + 1). b Copy and complete: x 2 +
+ 9 = (x – 3)2

5

Factorise: a 3x – 15 d 16x 2 – 25 g 6x 2 – 11x – 10

6

b 12xy – 8yz e x 2 + 4x – 21 h 2x 3 – 18x

c xp + 2x – yp – 2y f 5x 2 + 7x – 6 i –15x – 20y + 10z

Simplify: 2

x + 5x – 14 a -----------------------------2x + 14

2

b

2

x – 25 x – 4x – 5 ------------------------ ÷ -------------------------2 2 3x + 15x x +x

2 3 c ----------------– ----------------2 2 x – 5x x – 25

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Chapter 11 Consumer Arithmetic This chapter deals with solving consumer arithmetic problems involving earning and spending money, simple interest and loans. After completing this chapter you should be able to: ✓ calculate weekly, fortnightly, monthly and yearly earnings for various types of income ✓ calculate net income after considering common deductions ✓ calculate simple interest using the formula ✓ apply the simple interest formula to problems involving investing money ✓ calculate and compare the cost of purchasing goods by different means ✓ calculate a ‘best buy’.

Syllabus reference NS5.1.2 WM: S5.1.1–S5.1.5

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Consumer Arithmetic (Chapter 11) Syllabus reference NS5.1.2

Diagnostic test 1

Carol earns $397.20 per week. This is equivalent to a yearly salary of: A $19 860 B $20 654.40 C $10 327.20 D $19 065.60

2

A salary of $41 808 p.a. is not equivalent to:

7

A $804 per week B $3216 per fortnight

Alex is a real estate agent. He charges the following commission for selling home units: 3% of the first $150 000 and 1.5% for the remainder of the selling price. His commission for selling a home unit for $210 000 would be: A $6300

B $3150

C $4500

D $5400

8

C $3484 per month

4

5

6

Sam earns $360 per week. This is equivalent to a monthly income of: A $1440

B $1260

C $1560

D $1594.29

B $831.60

C $920.70

D $950.40

9

Bettina earns $680 per week. She is entitled to 4 weeks annual recreation leave and receives an additional holiday loading of 17.5%. Her total holiday pay for 4 weeks is: A $2720

B $2839

C $3196

D $476

Deborah is paid $0.48 for each pair of shorts that she sews. If she can sew an average of 12 pairs of shorts per hour and she works a 38-hour week, then her average weekly earnings are:

Sun.

$14.38

$17.98

$21.57

The table shows the award wages for a kitchen hand employed as a casual. The wages of a casual kitchen hand who works 10 hours Monday to Friday, 4 hours on Saturday plus 5 hours on Sunday is:

Chan works a 36-hour week and is paid $19.80 per hour. His total wages for a week in which he works an additional 4 hours at time-and-a-half and 3 hours at double time is: A $712.80

Sat.

Kitchen hand

D $10 452 per quarter 3

Mon.–Fri.

10

11

A $273.22

B $323.57

C $305.62

D $337.93

Simon earns $465 per week. The deductions from his salary each week are tax $140, superannuation $32, health insurance $36.80. His net pay for the week is: A $256.20

B $673.80

C $320.20

D $536.20

The simple interest on $2490 at 4.5% p.a. for 5 years is: A $112.05

B $2602.05

C $560.25

D $3050.25

Michelle invested $3000 for 4 years and earned $780 in interest. The annual rate of interest was:

A $5.76

B $18.24

A 26%

B 6.5%

C $218.88

D $1094.40

C 1.04%

D 4%

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Consumer Arithmetic (Chapter 11) Syllabus reference NS5.1.2

12

13

A camera store offers a discount of 12% for paying cash. The cash price of a camera marked as $499 is:

The following conditions for the deferred payment scheme apply: i

Pay nothing for 12 months.

A $439.10

B $439.12

ii

C $59 90

D $59.88

Balance plus interest to be repaid by equal monthly instalments over the two years following the interest free period.

iii

Simple interest of 16% p.a. is charged for the 3-year period of the agreement.

The method of purchasing goods by which a deposit is paid and the balance is paid off over a short period of time with no interest charged, but the goods cannot be taken until full payment has been made is called: A time payment

The total amount you would have to pay for the television under this scheme is:

B hire purchase

C deferred payment D lay-by 14

15

16

A refrigerator costing $1895 can be bought on terms for $295 deposit and 24 monthly instalments of $84. The total cost of buying the refrigerator on terms would be:

17

A $1498

B $1737.68

C $1977.36

D $2217.04

Using the table on page 344, the monthly repayment on a loan of $16 500 over 3 1--2- years at 12% p.a. is, to the nearest cent:

A $1895

B $2016

A $468.41

B $548.04

C $2311

D $2190

C $639.34

D $483.05

A television set costing $1289 can be bought on the following terms: deposit $289 and the balance to be repaid over 3 years by equal monthly instalments. Simple interest is charged at 11% p.a. If the TV is bought on these terms, the monthly repayment would be:

18

A $36.94

B $47.62

19

C $39.59

D $30.83

Which of the following is the best value? A 350 mL for $1.40 B 750 mL for $2.85 C 2 L for $6.40 D 5 L for $16.50

A television set is advertised as follows:

The price of a TV including GST is $495. The amount of GST included is: A $49.50

B $45

C $445.50

D $450

98sit $N1o 4 depo No repayments for 12 months (conditions apply)

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–3

4, 5

6

7

8

9

10, 11

12

13

14, 15

16

17

18

19

B

C

D

E

F

G

I

J

L

M

N

O

P

Q

313

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Consumer Arithmetic (Chapter 11) Syllabus reference NS5.1.2

A. EARNING AN INCOME There are a number of different ways in which people are paid for providing their labour, knowledge, skills and services. If people work for themselves they charge a fee, some people rely on income from investments, but most people work for an employer. By research and discussion, complete the table below that shows the ways people are paid when they work for an employer.

Exercise 11A Earning an income Method of payment

Description

Salary

A fixed amount per year usually paid weekly or fortnightly.

Wages

Based on an hourly rate for an agreed number of hours per week. Usually paid weekly or fortnightly.

Commission

A percentage of the value of goods or services sold is paid. Sometimes a low wage, called a retainer, is paid in addition to this.

Piecework

A fixed amount for each item produced or completed.

Fee

A fixed amount for a service provided.

Casual

A fixed hourly rate for the number of hours worked.

Examples of occupations

Advantanges/ Disadvantages

B. SALARIES AND WAGES Example 1 Georgina earns a salary of $670.85 per week. How much does she earn per: a fortnight b year? a Fortnightly salary = $670.85 × 2 = $1341.70 b Yearly salary = $670.85 × 52 = $34 884.20

(using 1 fortnight = 2 weeks) (using 1 year = 52 weeks)

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Consumer Arithmetic (Chapter 11) Syllabus reference NS5.1.2

Exercise 11B 1

Convert the following weekly salaries into the equivalent salary per: i fortnight ii year a $457 b $1025.60 c $1378.94

Example 2 Harry earns a salary of $48 600 p.a. How much does he earn per: a week

b fortnight

p.a. is short for per annum, which means per year.

c month?

a Weekly salary = $48 600 ÷ 52 (using 1 year = 52 weeks) = $934.62 to the nearest cent b Fortnightly salary = $48 600 ÷ 26 (using 1 year = 26 fortnights) = $1869.23 to the nearest cent c Monthly salary = $48 600 ÷ 12 (using 1 year = 12 months) = $4050

2

Convert the following yearly salaries into the equivalent salary per: i week ii fortnight iii month a $52 400 b $95 370 c $82 900

3

Convert the annual salaries shown in the advertisements below to the equivalent: i weekly ii fortnightly iii monthly salaries a b Fashion Girl's Surfwear Designer $80K Exciting position for the right person. Ph 9444 222

c Cleaner/Housekeeper $40K Rare opportunity to work in fine home. Ph 9666 000

Foreman $110K Experienced foreman required for city project. Ph 9333 000

$80K is a short way of indicating $80 000.

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Example 3 Brad earns $288 per week. What is his monthly salary? 1 month ≠ 4 weeks, so we must calculate Brad’s yearly salary first. Yearly salary = $288 × 52 = $14 976 Monthly salary = $14 976 ÷ 12 = $1248 4

Convert the following weekly salaries into monthly salaries: a $225 b $196 c $674

Example 4 Bruno earns $3600 per month. What is his equivalent weekly salary? Again, we must calculate the yearly salary first. Yearly salary = $3600 × 12 = $43 200 Weekly salary = $43 200 ÷ 52 = $830.77 to the nearest cent 5

Convert the following monthly salaries to the equivalent weekly salaries. a $4200 b $5635 c $3599

6

Scott earns $68 840 p.a., Lisa earns $1350 per week and Paula earns $5700 per month. Who earns the most?

Example 5 Ella works a 35-hour week and is paid $23.86 per hour. What are her weekly wages? Weekly wages = 35 × $23.86 = $835.10 7

Calculate the weekly wages for a person who works a 35-hour week and is paid: a $18.90/h b $26.48/h c $84.50/h

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Example 6 Yoshi earns $1389.50 for working a 35-hour week. What is his hourly rate of pay? Hourly rate = $1389.50 ÷ 35 = $39.70 8

Calculate the hourly rate of pay for a person who works a 35-hour week and is paid weekly wages of: a $994 b $847 c $626.50

Example 7 Sophie works a 38-hour week and is paid $28.75 per hour. How much does she earn in a: a week

b fortnight

c year?

a Weekly wages = 38 × $28.75 = $1092.50 b Fortnightly wages = $1092.50 × 2 = $2185 c Yearly wages = $1092.50 × 52 = $56 810

9

How much does a person earn in a i week ii fortnight iii year if the person works a 38-hour week and is paid: a $43/h b $52.90/h c $75.30/h

Example 8

Remember that 1 month is not equal to 4 weeks!

Fiona earns $16.80 per hour and works a 36-hour week. What are her average monthly wages? Weekly wages = $16.80 × 36 = $604.80 Yearly wages = $604.80 × 52 = $31 449.60 Monthly wages = $31 449.60 ÷ 12 = $2620.80

10

What are the average monthly wages for a person who works a 36-hour week and earns: a $18.20/h b $32.90/h c $76.50/h

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C. ADDITIONAL PAYMENTS Example 1 Ben normally works a 35-hour week and is paid $18.90 per hour. Calculate his total wages for a week in which he works an additional 5 hours overtime at time-and-a-half. Full-time employees who earn wages are expected to work a minimum number of hours each day, or each week, as negotiated in their workplace agreement. Overtime is paid to people who work hours in addition to those required by their workplace agreement and it is paid at a higher rate. The most common rates of overtime payment are: a ‘time-and-a-half’, i.e. the employee is paid at 1 1--2- times the normal hourly rate of pay, e.g. if the normal rate of pay is $20/hour then the employee would be paid ($20 × 1 1--2- =) $30/hour at time-and-a-half. b ‘double time’, i.e. the employee is paid double the normal rate of pay, e.g. if the normal rate of pay is $20/hour then the employee would be paid ($20 × 2 =) $40/hour at double time. Normal pay = $18.90 × 35 = $661.50 Overtime = ($18.90 × 1.5) × 5 = $141.75 Total wages = $661.50 + $141.75 = $803.25

Exercise 11C 1

Dianne normally works a 35-hour week and is paid $23.40 per hour. Calculate her total wages for a week in which she works an additional 4 hours at time-and-a-half.

2

Rebecca normally works a 36-hour week and is paid $17.20 per hour. Calculate her total wages for a week in which she works an additional 3 hours at time-and-a-half.

3

Tim is paid $18.60 per hour for a normal 35-hour week and time-and-a-half for any extra hours worked. How much would he earn for a week in which he worked 40 hours?

Example 2 Ringo normally works a 35-hour week and is paid $36.15 per hour. Calculate his total wages for a week in which he works an additional 5 hours at time-and-ahalf and 3 hours at double time. Normal pay = $36.15 × 35 = $1265.25 Overtime = ($36.15 × 1.5) × 5 + ($36.15 × 2) × 3 = $488.03 Total wages = $1265.25 + $488.03 = $1753.28

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4

Jarrod normally works a 35-hour week and is paid $31.20 per hour. Calculate his total wages for a week in which he works an additional 5 hours at time-and-a-half and 3 hours at double time.

5

Martha normally works a 35-hour week and is paid $28.60 per hour. Calculate her total wages for a week in which she works an additional 8 hours at time-and-a-half and 5 hours at double time.

6

Dana normally works a 38-hour week and is paid $36.15 per hour. Calculate her total wages for a week in which she works an additional 4 hours at time-and-a-half and 2 hours at double time.

7

Erin is paid $24.70 per hour. She is paid the normal rate for the first 7 hours worked each day, time-and-a-half for the next 2 hours and double time thereafter. Calculate her total wages for a day on which she worked: a 8 hours b 9 hours c 10 hours

8

Rob is paid $26.30 per hour. He is paid the normal rate for the first 6 hours worked each day, time-and-a-half for the next 2 hours and double time thereafter. Calculate his total wages for a day on which he worked a 8 hours b 9 hours c 10 hours

Example 3 Don works a 35-hour week and is paid time-and-a-half for any extra hours worked. One week he worked 5 hours overtime and was paid $969. What is his hourly rate of pay? Let the hourly rate of pay be $y, then y × 35 + y × 1.5 × 5 = 969 35y + 7.5y = 969 42.5y = 969 969 y = ----------42.5 = 22.8 i.e. Don earns $22.80 per hour.

9

Angela works a 35-hour week and is paid time-and-a-half for any extra hours worked. One week she worked 4 hours overtime and was paid $1102.90. What is her hourly rate of pay?

10

Daniel works a 35-hour week and is paid time-and-a-half for any extra hours worked. One week he worked 7 hours overtime and was paid $982.80. What is his hourly rate of pay?

11

Pete works his normal 35-hour week plus 4 hours overtime at time-and-a-half and 3 hours at double time. He was paid $1576.38. What is his hourly rate of pay?

12

Glenda is paid $17.90 per hour for working a 35-hour week and time-and-a-half for any extra hours worked. One week she was paid $733.90. How much overtime did she do?

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Example 4 Paul works for a builder and earns $66 000 per year. At the end of the year the builder decides to pay Paul a bonus equal to one month’s salary. Calculate Paul’s bonus. A bonus is an extra payment made to employees, often as a reward for meeting deadlines, exceeding profit targets, producing a high quantity of work etc. Bonus = $66 000 ÷ 12 = $5500 13

Jenni works as a secretary and earns $58 600 per year. At the end of the year her employer pays her a bonus of one month’s salary. Calculate Jenni’s bonus.

14

Abdul is paid $23.50 per hour and works a normal 35-hour week. At the end of the year his employer pays him a bonus of 5% of his yearly wages. Calculate Abdul’s bonus.

15

A company made a profit of $194 000 for the year. The owner decided to share 60% of the profit between her 80 employees as a bonus. Calculate the bonus paid to each of the employees.

16

For completing a project ahead of schedule, each member of the project team was given a bonus of 3% of the after-tax profit made. Calculate the bonus paid to each member of the team if the after-tax profit was $120 000.

17

An engineering design company decided to pay its 12 employees an equal share of 20% of the profit on a special project, as a bonus. If each of the 12 employees received a bonus of $1300, how much profit did the company make on this project?

18

Mark’s total income for the year was $57 337.28. This included a bonus of one month’s salary. What is Mark’s normal annual income (i.e. his income without any bonus)?

Example 5 Tanya earns $810 per week. She is entitled to 4 weeks annual recreation leave and receives an additional holiday loading of 17.5%. Calculate Tanya’s: a holiday loading b total pay for this holiday period. Holiday loading (leave loading) is an extra payment given to employees when they take their annual recreation leave. It is usually calculated as 17.5% of 4 weeks normal salary or wages. a Normal pay for 4 weeks = $810 × 4 = $3240 Holiday loading = 17.5% of $3240 = 0.175 × $3240 = $567

b Holiday pay = $3240 + $567 = $3807 or Holiday pay = 117.5% of 4 weeks pay or Holiday pay = 1.175 × 4 × $810 or Holiday pay = $3807

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19

Kylie earns $760 per week. She is entitled to 4 weeks annual leave and receives an additional holiday loading of 17.5%. Calculate Kylie’s: a holiday loading b total pay for this holiday period

20

Vinh works a normal 35-hour week and is paid $17.90 per hour. He is entitled to 4 weeks annual leave and receives an additional holiday loading of 17.5%. Calculate Vinh’s: a holiday loading b total pay for this holiday period

21

Sunny earns $1230 per fortnight. She is entitled to 4 weeks annual leave and receives an additional holiday loading of 17.5%. Calculate Sunny’s total pay for this holiday period.

22

Wesley earns $43 940 per year. He is entitled to 4 weeks annual leave and receives an additional holiday loading of 17.5%. Calculate Wesley’s total pay for this holiday period.

Example 6 Zoe works as a receptionist. She is entitled to 4 weeks annual leave and receives a holiday loading of 17.5%. One year her total holiday pay was $3092.60. What is Zoe’s weekly salary? Let the weekly salary be $z, then 117.5% of 4 × z = 3092.60 i.e. 1.175 × 4 × z = 3092.60 4.7 × z = 3092.6 3092.6 z = -----------------4.7 = 658 i.e. Zoe earns $658 per week.

23

Tiffany is entitled to 4 weeks annual leave and receives a holiday loading of 17.5%. One year her total holiday pay was $4512. What is Tiffany’s weekly salary?

24

Bin is entitled to 4 weeks annual leave and receives a holiday loading of 17.5%. One year his total holiday pay was $4812.80. Calculate his holiday loading.

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Investigation 1 WM: Reasoning, Applying Strategies

Calculating total pay Here is a spreadsheet to calculate the total pay for the employees of a factory. 1 2 3 4 5 6 7 8

A Employee

B Rate ($/h)

C Normal time (h)

24.72 18.94 23.65 26.36 16.78 15.43

36 36 36 35 35 40

Bill Sue Alan Gillian Natasha Eric

D Overtime (h) time-and-a-half 8 6 4 5

E Overtime (h) double time 4 1 2 3

1

1

F Total pay

1

Copy the spreadsheet.

2

In Cell F3 type the formula ‘= (C3 + D3*1.5 + E3*2) × B3’.

3

To find the total pay for the other employees: • Highlight cells F3 to F8. • Go to Edit. • Select Fill Down.

4

Add some more employees, put in their rate and the number of hours worked. Calculate their total pay.

D. PIECEWORK Piecework is a method of earning money in which the employee is paid for the number of items (pieces) produced or completed.

Example 1 Peta works at home sewing children’s tops. She is paid $3.20 for each top she produces. How much does she earn for a week in which she produces 120 tops? Income = 120 × $3.20 = $384

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Exercise 11D 1

Kerry is paid $0.83 per item for ironing shirts in a factory. How much does she earn for a week in which she irons 240 shirts?

2

Patricia earns $1.03 for each dress she finishes in a clothing factory. If, on average, she can finish 7 dresses per hour and she works a 35-hour week, what are her average weekly earnings?

3

Joe works for a men’s hairdresser and is paid $9 for each haircut he does. If he averages 16 haircuts per day for 6 days, how much does he earn?

4

Terry has a job assembling door locks. One week he assembles 450 locks and is paid $216. How much is he paid for assembling each lock?

5

Wayne works for Sparkler Lighting Co. assembling lamps. He is paid the following daily piecework rates: up to 50 lamps $1.45 /lamp No. assembled for each lamp over 50 and up to 70 $1.60 /lamp 55 for each lamp over 70 $1.90 /lamp Mon. Tues. 48 Here is Wayne’s work card for the week. Wed. 62 Calculate Wayne’s earnings for the week. Thurs. 76 Fri. 52

E. COMMISSION Commission is a method of earning income by which the employee is paid a percentage of the value of their sales.

Example 1 Georgia works as a salesperson and is paid a commission of 6% of the value of her sales. If Georgia sells $12 000 worth of goods one week, what is her commission? Commission = 6% of $12 000 = 0.06 × $12 000 = $720

Exercise 11E 1

A real estate agent charges a commission of 1.5% of the value of any house he sells. Calculate how much he will earn if he sells a house for: a $660 000 b $320 000 c $980 000

2

Joanne has a part-time job selling cosmetics. She is paid a commission of 18% of all sales. Calculate how much she earns in one month if her sales are: a $9000 b $5400 c $2300

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3

Daina is a stockbroker. She receives a commission of 2.5% of the selling price of any shares that she sells. How much commission would she earn for selling shares worth: a $15 000 b $27 000 c $243 000?

The commission earned for buying or selling shares is called brokerage.

Example 2

‘In excess of’ means ‘more than’.

Steve has a job selling clothing. He earns a commission of 19.5% of all weekly sales in excess of $5000. How much commission does he earn on sales of: a $4800

b $8650?

a As Steve’s sales are not more than $5000 he earns no commission. b Excess = $8650 − $5000 = $3650 Commission = 0.195 × $3650 = $711.75

4

Fiona has a job selling cleaning equipment. She earns a commission of 17.5% of all weekly sales in excess of $10 000. How much commission does she earn on weekly sales of: a $8000 b $12 000 c $24 000?

Example 3 Carol sells internet plans. She is paid the following rates of commission: • 1.5% of the first $20 000 worth of sales, • 2.5% of any sales above $20 000. Calculate how much she earns in a week in which her sales are: a $16 000

b $24 000

a Commission = 1.5% of $16 000 = 0.015 × $16 000 = $240 b Commission = 1.5% of $20 000 + 2.5% of ($24 000 − $20 000) = 0.015 × $20 000 + 0.025 × $4000 = $400

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5

Harry sells internet plans. He is paid the following rates of commission: • 2% of the first $20 000 worth of sales, • 3.5% of any sales above $20 000. Calculate how much he earns in a week in which his sales are: a $13 000 b $19 990 c $20 000 d $25 000

6

Zane has a job selling advertising. He is paid a commission on his weekly sales as follows: • 1% for the first $50 000 worth of sales, • 2% of the next $30 000 worth of sales and • 6% of the value of any remaining sales.

e $38 000

‘Rate’ means the percentage rate of commission.

How much would Zane earn in a week in which his sales are: a $46 000 b $50 000 c $65 000 d $80 000 e $92 000 7

Marie is a real estate agent. She charges the following commission for selling home units: • 3% for the first $160 000 of the selling price of the unit, • 2% for the next $50 000 and • 1.5% for the remainder of the selling price. Calculate how much Marie would earn for selling a unit for: a $150 000 b $180 000 c $210 000 d $280 000

e $360 000

Example 4 Chad sells washing machines. He is paid a fixed wage of $200 per week plus a commission of 3% of sales. How much does he earn in a week in which his sales are $5480? Commission = 0.03 × $5480 = $164.40 Weekly earnings = Retainer + Commission = $200 + $164.40 = $364.40

The fixed part of Chad’s wages ($200) is called a retainer.

8

Therese sells printers for computers. She is paid a retainer of $250 per week plus a commission of 4% of her sales. How much does she earn in a week in which she sells $14 970 worth of printers?

9

Michael works for a bookseller. He is paid a retainer of $280 per week plus a commission of 2% of sales. How much does he earn in a week in which his sales are: a $7650 b $3000 c $12 000 d $4700 e $8260?

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10

Complete the following table to show the weekly earnings of the sales team for a pharmaceutical company. Employee R. Roberts H. Low J. Thum K. Trau G. Flood

Retainer

Rate

Sales

Commission

$200 $150 $100 nil nil

2% 5% 6% 8% 9%

$4 200 $8 600 $10 450 $12 900 $15 360

$84

Weekly earnings

11

Jacqueline works as a sales representative for a hardware company. She is paid a retainer of $250 per week plus a commission of 3% of any sales in excess of $6000. How much would she earn in a week in which her sales were: a $4500 b $6400 c $7200 d $8430 e $10 960?

12

Hassan gets a job as a salesperson with a mobile phone company. He is offered two methods of weekly payment: A Retainer of $200 plus commission of 3%, or B No retainer, commission of 8%. a How much would Hassan earn, using each method, if his weekly sales were: i $0 ii $3000 iii $4000 iv $5000 v $10 000? b Which method of payment would you advise Hassan to choose? Give reasons.

13

Phillipa works as a salesperson and is paid a commission of 5% of sales. If Phillipa earns a commission of $821 in one week, what was the value of the goods that she sold?

14

One week Alex sells two cars costing $32 000 each. If he was paid a commission of $1280, what is the rate of commission that he is paid?

15

Joe is paid a retainer plus a commission of 4% of sales. If he receives $980 for selling $18 000 worth of goods, what is the retainer that he is paid?

16

Sally is paid a retainer of $220 per week plus a commission of 3% of sales. One week she earned $478. What was the value of the goods that she sold?

17

Sasha is paid a retainer of $180 per week plus a commission of 6% of sales in excess of $5000. One week he earned $858, what was the value of the goods that he sold?

18

Jacques is paid the following commission for selling computers: 1.5% of the first $25 000 worth of sales 2.5% of any sales in excess of $25 000. One week he earns a commission of $750. What was the value of the computers he sold in this week?

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F. CASUAL AND PART-TIME JOBS Casual and part-time workers are paid for the number of hours worked. The hourly rate is higher than for full-time workers because they may not be entitled to benefits such as holiday leave and sick leave. They may also be paid special rates for weekends and public holidays. The difference between casual and part-time workers is dependent on the number of hours worked.

Example 1 The table below shows part of an award agreement for tradespersons. Tradespersons Bricklayer Carpenter Painter Sign writer Roof tiler

Full-time $ per hour

Casual $ per hour

15.90 16.08 15.63 15.89 15.79

19.08 19.30 18.76 19.07 18.95

a Tom is a full-time bricklayer who works a 35-hour week. Calculate his normal weekly wages. b Bob is a qualified bricklayer who is employed as a casual for 35 hours one week. How much more than Tom does Bob earn? a Tom’s wages = $15.90 × 35 = $556.50 b Bob’s wages = $19.08 × 35 = $667.80 $667.80 − $556.50 = $111.30 i.e. Bob earns $111.30 more than Tom for this week.

Exercise 11F Use the table in example 1 above to do questions 1–4. 1

Emma is a sign-writer who does casual work. She works the following hours one week: Monday 3 hours, Tuesday 4 hours, Wednesday 3 hours, Friday 5 hours. How much does she earn?

2

a Matt is a full-time painter who works a 35-hour week. Calculate his normal weekly wages. b During a busy period he employs a casual to work with him for the 35 hours. How much extra does the casual earn for the week’s work?

3

Jack is a full-time roof tiler. Due to extra demand he employs two casuals to help him for 7 hours on each of 3 days. What is the wages bill for the two casuals?

4

Grant is a carpenter who is employed as a casual. One week he earns $463.20. How many hours did he work?

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Example 2 The table below shows part of the Restaurant Employees Award. Casual $ per hour

Kitchen hand Waiter Grill cook Grade 6 chef

Mon.–Fri.

Sat.

Sun.

14.38 14.92 15.73 18.69

17.98 18.65 19.66 23.36

21.57 22.38 23.59 28.03

Calcuate the wages of a casual waiter who works 8 hours from Monday to Friday, 6 hours on Saturday and 4 hours on Sunday. Wages = 8 × $14.92 + 6 × $18.65 + 4 × $22.38 = $320.78

Use the table in example 2 to do questions 5–8. 5

Emily is employed as a casual kitchen hand for 3 hours per day for each day Monday to Friday. Calculate her wages.

6

Trent is a grade 6 chef who works as a casual on Saturday for 6 hours and Sunday for 6 hours. Calculate his wages.

7

Calculate the wages of a casual grill cook who works the following hours: M

T

W

3 8

T

F

S

S

3

4

6

3

Con is employed as a casual kitchen hand. One week he worked 3 hours on Saturday, 4 hours on Sunday and the remaining hours were all worked in the period Monday to Friday. If he received $226.50 for the week’s work, how many hours did he work from Monday to Friday?

The table below shows the Restaurant Award rates for Grade 1 juniors. Juniors: Casual $ per hour Age (years)

Mon.–Fri.

Sat.

Sun.

17 18 19 20

8.92 10.07 11.51 12.95

11.15 12.58 14.38 16.18

13.38 15.10 17.26 19.42

Use the table above to answer questions 9–12.

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9

How much would a 17-year-old casual employee earn for working: a 8 hours from Monday to Friday b 4 hours on Saturday c 6 hours on Sunday d 6 hours Monday to Friday and 5 hours on Saturday e 10 hours Monday to Friday, 4 hours on Saturday and 3 hours on Sunday?

10

a Ben is 18 years old and does casual work in a coffee shop. How much does he earn for working 4 hours on Saturday and 6 hours on Sunday? b Lara is 20 years old; how much would she earn for working the same hours as Ben?

11

Sarah and Ella work in a café. Sarah is 17 years old and Ella is 18 years old. One week they both work the same shifts, as shown below. M

T

W

T

F

S

S

3

3

–

4

4

5

3

How much more than Sarah does Ella earn for this week? 12

Jenny is 19 years old and does casual work as a waitress on Friday, Saturday and Sunday nights. One week she worked 5 hours on Friday night, 6 hours on Saturday night and was paid $204.24 for the week. How many hours did she work on Sunday night?

G. NET EARNINGS The total amount earned by an employee is called gross income. However this is not the amount of money that the employee actually takes home because deductions are made. The most common deductions are federal income tax, health insurance and superannuation. The amount actually received, after deductions, is called net earnings or take-home pay. Net Earnings = Gross Income − Deductions

Example 1 Julie earns $890 per week. The deductions from her salary each week are: tax $265.83, health insurance $43.59 and superannuation $42.70. Calculate her net earnings each week. Total deductions = $265.83 + $43.59 + $42.70 = $352.12 Net Earnings = $890 − $352.12 = $537.88

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Exercise 11G 1

Patricia earns $940 per week. The deductions from her salary each week are: tax $280.78, health insurance $37.62 and superannuation $56.40. Calculate her net earnings each week.

2

David earns $760 per week. The deductions from his salary each week are: tax $212.80, health insurance $32.40 and superannuation $30.40. Calculate his net earnings each week.

3

Sue earns $43.80 per hour and works a 36-hour week. The deductions from her wages each week are: tax $536.11, superannuation $94.61 and health insurance $51.25. She pays union fees of $7.60 and also has $50 per week paid directly into an investment account. Calculate her take-home pay each week.

4

Yuchen earns $63.70 per hour and works a 38-hour week. The weekly deductions from her wages include: tax $871.42, superannuation $217.85 and health insurance $44.90. She also pays $5 per week to her favourite charity and has $70 per week paid into a special savings account. Calculate her take-home pay.

5

James’s gross salary is $1230 per week. His employer deducts 31% of his gross earnings for tax and he contributes 9% of his gross income into a superannuation fund. His health fund contributions are $39.99 and professional association fees are $13.20 per week. Calculate his net weekly earnings.

H. BUDGETS A budget is a financial plan for the future. It is a means by which you can save for future purchases and avoid over-spending. To prepare a budget you need to determine your expected income and estimate your expected expenses. Your income needs to be larger than your expenses if you are to live within your means. To prepare a budget, for a given time period (e.g. week, month, year): • calculate your total income • estimate your total expenses • calculate income minus expenses • adjust income or expenses if necessary.

Example 1 Karen has just started work and still lives at home with her parents. Her weekly take-home salary is $480. Each week she pays $110 for board, $49 for fares and $35 for lunches. She spends $120 per week on entertainment, $95 per fortnight on personal items and $330 per month on clothes. a Prepare an annual budget for Karen. b Karen wants to go on an overseas holiday in 3 years time. The cost of the holiday is $7899. Determine whether or not Karen will be able to take her holiday. c If Karen will not have sufficient money to take her holiday, how could she adjust her budget so that she would be able to afford the holiday?

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Salary ($480 × 52) = $24 960 Board ($110 × 52) = $5720 Fares ($49 × 52) = $2548 Lunches ($35 × 52) = $1820 Entertainment ($120 × 52) = $6240 Personal items ($95 × 26) = $2470 Clothes ($330 × 12) = $3960 Total = $22 758 Income − Expenses = $2202 Karen has an excess of income over expenses so she will be able to live satisfactorily on this budget and save some money. b If Karen saves all her money, then in 3 years she will have $2202 × 3 = $6606. She is $7899 − $6606 = $1293 short of her target. c Karen must either increase her income or decrease her expenses by at least ($1293 ÷ 3 =) $431 per year. She could increase her income by finding employment with a higher salary or getting a second job. She could decrease her expenses by, for example, reducing her spending on clothes to $290 per month. She would then save ($40 × 12 =) $480 per year on expenses. Or, if she reduced her spending on entertainment to $110 per week, she would save ($10 × 52 =) $520 per year. She would then be able to afford to take the holiday. a Income Expenses

Exercise 11H 1

Naomi lives at home with her parents. Her weekly take-home salary is $590. Each week she pays $100 for board, $53 for fares and $42 for lunches. She spends $150 per week on entertainment, $84 per fortnight on personal items, $380 per month on clothes. a Prepare an annual budget for Naomi. b Naomi wants to go on an overseas holiday in 3 years time. The cost of the holiday is $12 550. Determine whether or not Naomi will be able to take her holiday.

2

Matthew’s net earnings are $540 per week. He shares a house for which he pays $120 per week rent. Each week he spends $110 on food, $145 on entertainment and $25 on personal items. The loan repayments on his car are $380 per month. He spends $45 per week on petrol and the six-monthly service is $380. Annual registration and insurance amount to $1148. His mobile phone costs him $24 per month. a Prepare an annual budget for Matthew. b How could Matthew adjust his budget so that he can live within his means?

3

George is a full-time TAFE student. He receives an allowance of $320 per fortnight from the government and averages earnings of $120 per week from his part-time job. His expenses are: rent $320 per month, food $90 per week, phone $110 per quarter, entertainment $70 per week, books $350 per year. a Prepare an annual budget for George. b George has already saved the money to buy a car. He estimates that it would cost him $30 per week for petrol, $80 per month for maintenance and $840 per year for registration and insurance. Can George afford to own and run a car?

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4

Kate and Robert want to prepare a budget for next year and have gathered the following information. Income: Kate’s take-home pay is $490 per week and Robert clears $380 per week. Interest of $230 from investments is expected in February and August of next year. Expenses: Home loan repayments, $980 per month Food, $160 per week Electricity, $480 each quarter Telephone, $110 per month Council rates, $340 each quarter Water rates, $186 per quarter Car registration and insurance, $780 per year Comprehensive car insurance, $810 per year Car loan repayments, $108 per week Car running expenses, average $190 per month Clothing, average $350 per month Personal items, $45 per week a Prepare an annual budget for Kate and Robert. b In order to reduce the cost of their loans, Kate and Robert wish to increase their loan repayments. Can they afford to do this? What advice would you give Kate and Robert?

I. SIMPLE INTEREST When investing money in a financial institution, such as a bank, the bank pays for the use of your money. This payment by the bank is called interest and is calculated as a percentage of the amount invested. Similarly, when you borrow money a charge is made for the use of the bank’s money. This charge is also called interest and is calculated as a percentage of the amount borrowed. There are two methods of calculating the interest: simple interest and compound interest. If the interest is calculated as a fixed percentage of the original amount invested (or borrowed), then it is called simple interest.

Example 1 Calculate the simple interest if $8000 is invested for 3 years at 4.5% p.a. Interest for 1 year = 4.5% of $8000 = 0.045 × $8000 = $360 Interest for 3 years = $360 × 3 = $1080 If $P is invested for T years at r % p.a. then the simple interest, I, can be found using the formula: I = PRT

r where P is called the principal, R is called the interest rate p.a.  R = -------- and T is the time in years.  100

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Example 2 Use the simple interest formula to calculate the simple interest earned on an investment of $10 800 at 3.9% p.a. for 5 years. I = PRT = $10 800 × 0.039 × 5 = $2106

Exercise 11I 1

Calculate the simple interest if $7000 is invested for 2 years at 5% p.a.

2

Calculate the simple interest if $12 000 is invested for 4 years at 3% p.a.

3

Complete the following table. Principal

Annual interest rate

Time invested (years)

$5 800 $15 000 $24 000 $6 500 $18 000 $9 300 $6 000

7% 3.5% 4.5% 5% 2.8% 3.4% 3%

4 3 5 6 2 4 3

Simple interest

Example 3 Calculate the amount to which $7000 will grow in 3 years if invested at 6.5% p.a. simple interest. Interest = $7000 × 0.065 × 3 = $1365 Amount after 3 years = $7000 + $1365 = $8365

4

Calculate the amount to which $9000 will grow in 3 years if invested at 6.5% p.a. simple interest.

5

Calculate the amount to which $20 000 will grow in 5 years if invested at 4% p.a. simple interest.

6

If I invest $13 500 at 7.4% p.a. simple interest, how much will I have in 4 years time?

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Example 4 Calculate the simple interest earned on $6000 at 8% p.a. for 16 months. 16 The number of years the money is invested = ------ , hence 12 16 Interest = $6000 × 0.08 × -----12 = $640 7

Calculate the simple interest earned on the following investments: a $5000 at 9% p.a. for 18 months b $7000 at 8% p.a. for 15 months c $12 500 at 10% p.a. for 9 months d $3800 at 12% p.a. for 27 months e $24 000 at 7.8% p.a. for 45 months

Example 5 Rene invested $4700 at 6% p.a. simple interest and earned $1128 in interest. For how long did he invest his money? Interest for 1 year = 0.06 × $4700 = $282 No. of years invested = $1128 ÷ $282 =4 i.e. Rene invested his money for 4 years. 8

Harry invested $13 000 at 6% p.a. simple interest and earned $4680 in interest. For how long did he invest his money?

9

Joy invested $2800 at 3.5% p.a. simple interest and earned $490 in interest. For how long did she invest her money?

Example 6 Colin invested $4000 for 5 years and earned $700 in interest. What was the annual rate of simple interest? Interest for 1 year = $700 ÷ 5 = $140

140 Annual interest rate = ------------- × 100% 4000 = 3.5%

10

Kim invested $6000 for 5 years and earned $2100 in interest. What was the annual rate of simple interest?

11

Lauren invested $17 000 for 4 years and earned $3128 in interest. What was the annual rate of simple interest?

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J. PURCHASING GOODS BY CASH When paying cash to purchase goods, the cost is rounded to the nearest 5 cents.

Example 1 How much would you actually pay in cash to purchase goods that cost: a 87 cents

b $2.43

c $2.99?

Rounding off to the nearest 5 cents, you would pay a 85 cents b $2.45

c $3.00

Exercise 11J 1

2

How much would you actually pay in cash to purchase goods that cost: a 76c b $5.28 c $2.79 d $7.31 e $3.97 g $16.23 h $21.99 i $54.85 j $39.14 k $17.36 Calculate the change given when: a $10 is offered to pay for goods worth $5.83 c $10 is offered to pay for goods worth $8.22 e $20 is offered to pay for goods worth $18.36 g $50 is offered to pay for goods worth $28.57

b d f h

f l

$8.52 $69.98

$10 is offered to pay for goods worth $4.99 $20 is offered to pay for goods worth $12.84 $20 is offered to pay for goods worth $6.01 $50 is offered to pay for goods worth $48.19

Example 2 An electrical store offers a discount of 12% for cash purchases. Find the cash price of a television set marked as $799. Discount = 12% of $799 or Price = 88% of $799 = 0.12 × $799 = 0.88 × $799 = $95.88 Price = $799 − $95.88 = $703.12 = $703.12 Rounding the discounted price to the nearest 5 cents, the cash price is $703.10.

3

An electrical store offers a discount of 12% for cash purchases. Find the cash price of a sound system marked at $479.

4

A builders’ hardware store offers a discount of 6% for cash purchases. Find the cash price for goods worth: a $147 b $463 c $224 d $180.56 e $68.99

5

List some advantages and disadvantages of using cash to purchase goods.

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K. USING CREDIT CARDS A credit card is a convenient method for purchasing and paying for goods. You can pay for the goods later, you don’t need to carry large amounts of cash, you can take advantage of sales, and a monthly statement of purchases is provided. The financial institution issuing the card charges an annual fee and if the balance owing at the end of each month is paid within the interest-free period (which varies from 0 to 55 days) then no further costs are involved. However, if any balance is owing after the interest-free period has finished then there is an initial charge equal to one month’s interest on the balance outstanding and, in addition, an interest charge compounded daily from the end of the interest free period. There is a minimum payment that must be made each month. It is also possible to obtain cash advances up to a certain limit. In this case interest is charged daily from the time the cash is withdrawn.

Exercise 11K

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Use the statement on the previous page to answer the following questions. 1

What is the: a date the statement period starts and ends c daily interest rate for purchases e available credit g minimum payment that must be made

b annual interest rate for purchases d credit limit f date by which payment must be made

2

Calculate the: a total purchases made b total credits (CR) c Opening Balance + Purchases + Financial Institution Tax − Credits. Is this the closing balance?

3

What percentage is the minimum repayment due of the closing balance?

4

List some advantages and disadvantages of using credit cards to purchase goods.

L. LAY-BY Some retail stores allow customers to purchase goods by a method called lay-by. Under a lay-by agreement a deposit is paid and the goods are put aside. The remainder of the cost price must be paid off within a given period of time. The customer cannot collect the goods until the balance is completely repaid, but no interest is charged.

Example 1 Nick decides to lay-by a tool set costing $849 and pays a deposit of $100. Over the next 3 months he makes repayments of $150, $85, $90, $160, $120 and $70. How much more does he have to repay to be able to collect the tool set? Total amount repaid = $100 + $150 + $85 + $90 + $160 + $120 + $70 = $775 Balance = $849 − $775 = $74 Nick still has $74 to pay before he can collect the tool set.

Exercise 11L 1

Ben decides to lay-by an electric saw costing $569 and pays a deposit of $120.Over the next 3 months he makes repayments of $60, $45, $90, $70, $70 and $80. How much more does he have to repay to be able to collect the saw?

2

Zoe lay-bys a bike costing $225 for her son’s birthday and pays a deposit of $60. Each week for the next 5 weeks she makes payments of $25. How much more does she have to repay to be able to collect the bike?

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Example 2 Isabella lay-bys a dress costing $324 and pays a deposit of $50. She wants to collect the dress in approximately 3 months time to wear to a wedding. If she pays off the balance by making 3 equal monthly instalments, calculate: a the balance to be repaid

b the amount of each monthly instalment

a Balance = $324 − $50 = $174

b Monthly instalment = $174 ÷ 3 = $58

3

Martine lay-bys a dress costing $485. She pays a deposit of $80 and pays off the balance by making 4 equal monthly instalments. Calculate: a the balance to be repaid b the amount of each monthly instalment

4

Yvonne lay-bys a new electric oven costing $778. She is required to pay a 10% deposit and repay the balance by 12 equal weekly instalments. Calculate the: a deposit b balance to be repaid c amount of each weekly instalment

5

Josh lay-bys a new DVD player costing $456. He is required to pay a 15% deposit and repay the balance by 6 equal fortnightly instalments. Calculate the: a deposit b balance to be repaid c amount of each fortnightly instalment

6

List some advantages and disadvantages of using the lay-by method to purchase goods.

M. BUYING ON TERMS

Buying on terms is sometimes called hire purchase.

When an item is bought on terms a deposit is paid and the item is received immediately. The balance of the price is borrowed and this balance plus simple interest is repaid by equal instalments over a fixed term.

Example 1 A refrigerator costing $2998 can be bought on terms for $299 deposit and 24 monthly instalments of $139.45. a Calculate the total cost of buying the refrigerator on terms. b How much would you save by paying cash? a Total cost = $299 + 24 × $139.45 = $3645.80 b Amount saved by paying cash = $3645.80 − $2998 = $647.80

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Exercise 11M 1

A refrigerator costing $2599 can be bought on terms for $399 deposit and 24 monthly instalments of $115.50. a Calculate the total cost of buying the refrigerator on terms. b How much would you save by paying cash?

2

A laptop computer costing $2298 can be bought on terms for $229 deposit and 18 monthly repayments of $135.60. a Calculate the total cost of buying the computer on terms. b How much would you save by paying cash?

3

A home theatre system costing $1598 can be bought on the following terms: 10% deposit and 48 weekly instalments of $37.15. a Calculate the total cost of buying the system on terms. b How much would you save by paying cash?

4

A hi-fi sound system costing $879 can be bought on the following terms: deposit 15% and 26 fortnightly repayments of $39.98. a Calculate the total cost of buying the system on terms. b How much would you save by paying cash?

Example 2 A computer costing $3498 can be bought on terms for $300 deposit and 36 monthly repayments of $124.17. a b c d e

Calculate the total cost of buying the computer on terms. Find the total amount of interest charged. Calculate the amount of interest paid annually. What was the amount of money borrowed? Calculate the annual rate of interest charged.

a Total cost = $300 + 36 × $124.17 b Interest = $4770.12 − $3498 = $4770.12 = $1272.12 c Annual interest = $1272.12 ÷ 3 = $424.04 d Amount borrowed = Balance owing after paying the deposit = $3498 − $300 = $3198 e Annual interest rate = =

annual interest × 100% amount borrowed $424.04 × 100% $3198

= 13.3%

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5

A camera costing $1499 can be bought on terms for $200 deposit and 24 monthly repayments of $74.69. a Calculate the total cost of buying the camera on terms. b Find the total amount of interest charged. c Calculate the amount of interest paid annually. d What was the amount of money borrowed? e Calculate the annual rate of interest charged.

6

A television costing $5890 can be bought on terms for $300 deposit and 36 monthly repayments of $199.53. a Calculate the total cost of buying the computer on terms. b Find the total amount of interest charged. c Calculate the amount of interest paid annually. d What was the amount of money borrowed? e Calculate the annual rate of interest charged.

7

A dining room suite was advertised for $5990 or $500 deposit and 48 monthly repayments of $187.58. a Calculate the total cost of buying the dining room suite on terms. b Find the total amount of interest charged. c Calculate the amount of interest paid annually. d What was the amount of money borrowed? e Calculate the annual rate of interest charged.

Example 3 A wide-screen plasma TV set can be bought for $7998 cash or on the following terms: deposit $799, the balance to be repaid over 2 years by 24 equal monthly repayments. Simple interest is charged on the balance at 12% p.a. If the TV is bought on terms calculate: a the balance owing after the deposit is paid b the interest charged on the balance owing c the monthly repayment a Balance owing = $7998 − $799 b Interest = $7199 × 0.12 × 2 = $7199 = $1727.76 c Balance owing + Interest = $7199 + $1727.76 = $8926.76 Monthly repayment = $8926.76 ÷ 24 = $371.95 (to the nearest cent) 8

Peter buys a second-hand car advertised for $9600 on the following terms: deposit $2000, the balance to be repaid over 2 years by equal monthly repayments. Simple interest is charged at 12% p.a. Calculate: a the balance owing b the interest charged on the balance owing c the monthly repayment

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9

Angela buys a car advertised for $12 900 on the following terms: deposit $3000, the balance to be repaid over 3 years by equal monthly repayments. Simple interest is charged at 9% p.a. Calculate: a the balance owing b the interest charged on the balance owing c the monthly repayment

10

Adrienne buys a washing machine advertised for $4990 on the following terms: deposit 10% and the balance repaid over 2 years by equal monthly repayments. Simple interest is charged at 15% p.a. Calculate: a the deposit b the balance owing c the interest charged on the balance owing d the monthly repayment

11

Robin buys a new car advertised for $19 900 on the following terms: deposit 15%, the balance to be repaid over 4 years by equal monthly repayments. Simple interest is charged at 11.9% p.a. Calculate: a the deposit b the balance owing c the interest charged on the balance owing d the monthly repayment

12

List some advantages and disadvantages of purchasing goods on terms.

Investigation 2 WM: Applying Strategies, Reasoning

Monthly repayments The spreadsheet below calculates the monthly repayment when an item is bought on terms. A 1 Item

B

C

D

E

F

G

H

Cash Deposit Interest Repayment Balance Interest on Monthly price ($) rate (% p.a.) period (years) owing balance Repayment

2 Computer

2998

298

12

2

3 TV

1899

189

15

2

4 Furniture

4672

250

11.6

3

5 Air conditioner

7659

1000

14.2

4

6 Refrigerator

3628

628

9.9

3

1

Copy the spreadsheet.

2

In Cell F2 type the formula ‘= B2 − C2’. This is the balance owing after the deposit is paid.

3

In Cell G2 type the formula ‘= F2*D2/100*E2’. This is the amount of interest charged on the balance.

☞

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4

In Cell H2 type the formula ‘= (F2 + G2)/(E2*12)’. This is the monthly repayment.

5

If the computer was paid off over 3 years instead of 2: a What would be the monthly repayment? b How much more interest would be paid? (Hint: Change repayment period to 3 and use the arrow key to move right.)

6

Try changing the repayment period and/or the interest rate for the other items to investigate the effect on the monthly repayment and the amount of interest paid.

7

Use some advertisements from newspapers or magazines to check the advertised monthly repayment for several items. If there is a difference, investigate for hidden charges.

N. DEFERRED PAYMENT Many advertisements make statements such as ‘No repayments for 12 months’ or ‘Pay nothing until next June’. Under these arrangements the goods may be taken immediately the finance contract is approved and no payment needs to be made for the agreed period of time. This type of financial arrangement is known as a deferred payment scheme.

Example 1 Michael sees a television set advertised as shown opposite. When Michael approaches the retailer to buy the television, he is given the conditions opposite for the deferred payment scheme. a Calculate the total amount Michael would have to pay for the television under this scheme. b Calculate the monthly instalments.

$2498

NO DEPOSIT NO DEPOSIT NO REPAYMENTS NO REPAYMENTS FOR 1212 MONTHS FOR MONTHS (Conditions apply.)

Conditions: (i) Pay nothing for 12 months. (ii) Balance plus interest to be repaid by equal monthly instalments over the two years following the interest-free period. (iii) Simple interest of 16% p.a. is charged for the 3-year period of the agreement. (iv) Establishment fee of $110. (v) Account service fee of $2.95 per month for the 3-year period of the agreement.

a Michael must pay interest on $2498 for 3 years. Interest = $2498 × 0.16 × 3 = $1199.04 Total cost = Price + Interest + Establishment fee + Account service fee = $2498 + $1199.04 + $110 + $2.95 × 36 = $3913.24 b Michael has to repay this amount by 24 equal instalments. Monthly instalment = $3913.24 ÷ 24 = $163.05

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Exercise 11N 1

$2999

A computer is advertised as shown opposite. a Calculate the total amount you would have to pay for the computer under this scheme. b Calculate the monthly instalments.

NO DEPOSIT NO REPAYMENTS FOR 12 MONTHS (Conditions apply.)

Conditions: (i) Pay nothing for 12 months. (ii) Balance plus interest to be repaid by equal monthly instalments over the two years following the interest-free period. (iii) Simple interest of 16% p.a. is charged for the 3-year period of the agreement. (iv) Establishment fee of $110. (v) Account service fee of $2.95 per month for the 3-year period of the agreement.

2

A second-hand car is advertised as follows.

$8599 No Deposit No Repayments for 12 Months

a Calculate the total amount you would have to pay for the car under this scheme. b Calculate the monthly instalments.

(Conditions apply.) Conditions: (i) Pay nothing for 12 months. (ii) Balance plus interest to be repaid by equal monthly instalments over the three years following the interest-free period. (iii) Simple interest of 15% p.a. is charged for the 4-year period of the agreement. (iv) Establishment fee of $125. (v) Account service fee of $2.55 per month for the 4-year period of the agreement.

3

A sofa bed is advertised for $1598 with no deposit and no repayments for 6 months. The conditions of the agreement are: (i) Pay nothing for 6 months. (ii) Balance plus interest to be repaid by equal monthly instalments over the 12 months following the interest free period. (iii) Simple interest of 1.5% per month is charged for the 18 months of the agreement. (iv) Establishment fee of $135. (v) Account service fee of $2.85 per month, for the 18 month period of the agreement a Calculate the total amount you would have to pay for the bed under this scheme. b Calculate the monthly instalments.

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O. LOANS To purchase expensive items some people prefer to organise a personal loan through a bank, credit union or other financial institution. Personal loans often cost less than other forms of payment, such as buying on terms or deferred payment schemes, because of lower interest rates and lower fees. Banks have tables from which the monthly repayments can be determined. For example, the table below shows the monthly repayments of principal plus interest on each $1000 borrowed for various interest rates. Monthly repayments on each $1000 borrowed Annual interest rate Loan term (months)

10.0%

10.5%

11.0%

11.5%

12.0%

12.5%

13.0%

13.5%

14.0%

12

87.9159 88.1486 88.3817 88.6151 88.8488 89.0829 89.3173 89.5520 89.7871

18

60.0571 60.2876 60.5185 60.7500 60.9820 61.2146 61.4476 61.6811 61.9152

24

46.1449 46.3760 46.6078 46.8403 47.0735 47.3073 47.5418 47.7770 48.0129

30

37.8114 38.0443 38.2781 38.5127 38.7481 38.9844 39.2215 39.4595 39.6984

36

32.2672 32.5204 32.7387 32.9760 33.2143 33.4536 33.6940 33.9353 34.1776

42

28.3168 28.5547 28.7939 29.0342 29.2756 29.5183 29.7621 30.0071 30.2532

48

25.3626 25.6034 25.8455 26.0890 26.3338 26.5800 26.8275 27.0763 27.3265

54

23.0724 23.3162 23.5615 23.8083 24.0566 24.3064 24.5577 24.8104 25.0647

60

21.2470 21.4939 21.7424 21.9926 22.2444 22.4979 22.7531 23.0098 23.2683

Example 1 Use the table above to calculate the monthly repayments on a loan of $8300 for 4 years at 13%. From the table, the monthly repayment for each $1000 borowed = $26.8275 Monthly repayments for $8300 = $26.8275 × 8.3 = $222.67 to nearest cent

Exercise 11O 1

Use the table above to calculate the monthly repayments on a loan of $9000 for 3 years at 12%.

2

Use the table above to calculate the monthly repayments on loans of: a $85 000 for 2 1--2- years at 10.5% b $67 000 for 5 years at 11% c $14 600 for 3 1--2- years at 13.5% e $12 450 for 42 months at 11.5%

d $16 000 for 54 months at 14%

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Example 2 a Use the table to calculate the monthly repayments on a loan of $6900 for 3 years at 12.5%. b Calculate the total cost of the loan if there is a loan application fee of $180. a Monthly repayment = $33.4536 × 6.9 = $230.83 b Total cost of loan = $230.83 × 36 + $180 = $8489.87

3

a Use the table to calculate the monthly repayments on a loan of $7000 for 3 years at 11%. b Calculate the total cost of the loan if there is a loan application fee of $180.

4

a Use the table to calculate the monthly repayments on a loan of $15 500 for 4 years at 13.5%. b Calculate the total cost of the loan if there is a loan application fee of $250.

5

Calculate the cost of the following loans. (Calculate the monthly repayment first.) a $5000 for 3 1--2- years at 12%, loan application fee $300 b $12 000 for 4 1--2- years at 14%, loan application fee $200 c $8500 for 2 years at 10.5%, loan application fee $260 d $9400 for 60 months at 11.5%, loan application fee $210 e $18 000 for 42 months at 10%, loan application fee $190

Use the table given for questions 6–8. 6

Terry borrowed $20 000 at 11.0% p.a. His monthly repayments were $654.77. Over what period of time did he borrow the money?

7

Natasha borrowed $18 000 over 4 1--2- years. The monthly repayments were $446.59. What was the interest rate charged?

8

Bill took out a loan over 3 years at 12.5% p.a. His monthly repayments were $802.89. How much money did Bill borrow?

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P. COMPARING PRICES Example 1 Which is the better buy, 3 kg of apples for $13.38 or 5 kg for $21.60? Method 1

Method 2

Find the cost per kilogram: 3 kg for $13.38 = ($13.38 ÷ 3) per kg = $4.46 per kg 5 kg for $21.60 = ($21.60 ÷ 5 ) per kg = $4.32 per kg 5 kg for $21.60 is the better buy because the cost per kilogram is cheaper. Find the amount of apples per dollar: 3 kg for $13.38 = (3 ÷ 13.38) kg for $1 = 0.224 kg for $1 5 kg for $21.60 = (5 ÷ 21.6) kg for $1 = 0.231kg for $1 5 kg for $21.60 is the better buy because you get more apples per dollar.

Exercise 11P 1

Which is the better buy: a 3 kg of oranges for $4.00 or 5 kg for $6.60? b 2 kg of meat for $15.96 or 3 kg for $23.97? c 1.5 litres of soft drink for $2.70 or 2 litres for $3.50? d a 350 g packet of cereal for $2.20 or a 575 g packet for $3.69? e a 150 mL bottle of sauce for $2.29 or a 750 mL bottle for $10.99?

2

Which of the following is the best value? a Tuna: 95 g tin for $1.33, 185 g tin for $2.17, 425 g tin for $3.50 b Cordial: 750 mL for $1.12, 2 L for $3.10, 4 L for $6.06 c Sandwich spread: 115 g jar for $1.43, 175 g jar for $1.96, 235 g jar for $2.49 d Chocolate: 55 g block for $0.98, 250 g block for $2.58, 375 g block for $3.94 e Milk Flavouring: 375 g tin for $2.69, 750 g tin for $5.29, 1.25 kg tin for $8.28

3

Harry’s Car Hire charges $28 per day with no limit on the number of kilometres travelled to hire a new Toyota Corolla. Ray’s Car Rental charges $20 per day plus 8 cents per kilometre travelled to rent the same car. Which company is cheaper if you are likely to travel, each day: a 80 km b 100 km c 150 km?

4

To hire a new Holden Commodore each day, Bob’s Rentals charges $45 plus 7 cents per kilometre travelled. Sophie’s Rentals charges $48 per day plus 5 cents per kilometre travelled. Which company is cheaper if you are likely to travel, each day: a 100 km b 300 km c 150 km?

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5

On savings accounts, Bob’s Bank charges a management fee of $5.50 per month. The first five transactions are free and then a fee of 28 cents per transaction is charged. Bill’s Bank charges a monthly management fee of $6.00 plus 9 cents per transaction. Which bank is cheaper to use if your average number of monthly transactions is: a 5 b 10 c 15?

6

The Mobile Phone Company offers two plans. Plan A has a connection fee of $12 per month and calls cost 21 cents/30 seconds. Plan B has a connection fee of $15 plus call charges of 16 cents/30 seconds. Which plan would be cheaper, and by how much, if your expected calls per month were: a 20 minutes b 30 minutes c 50 minutes?

7

Terry’s Telecommunications has two mobile phone plans. The Starnet Plan has a monthly fee of $25, 50 free calls then 45 cents/call. The Supernet Plan has a monthly fee of $35, 100 free calls and then 35 cents/call. Which plan is cheaper, and by how much, if the expected number of calls per month total: a 50 b 100 c 200?

Q. GOODS AND SERVICES TAX (GST) A federal tax, known as the GST, is applied to most goods and services in Australia. It is calculated at the rate of 10% of the purchase price of the goods or services. The price excluding the GST (i.e. the price before the GST is added) is written ‘price excluding GST’ and the price including the GST (i.e. the price after the GST is added) is written ‘price including GST’.

Example 1 Calculate the GST and the price including GST on a camera with a listed price of $710, price excluding GST. GST = 10% of $710 = 0.1 × $710 = $71 Price including GST = $710 + $71 = $781

Exercise 11Q 1

Calculate the GST and the price including GST on the following items: a microwave oven $440, price excluding GST b computer $3690, price excluding GST c TV repairs $258, price excluding GST d DVD player $397, price excluding GST e plumber’s bill for services $1800, price excluding GST

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Example 2 Calculate the price including GST on a mobile phone listed as $299, price excluding GST. Price including GST = list price + 10% of the list price = 110% of the list price = 1.10 × $299 = $328.90

2

Calculate the price including GST on the following items whose price, excluding GST, is given. a car battery $95, price excluding GST b ticket to Rugby Final $225, price excluding GST c bottle of wine $17, price excluding GST d printer repairs $336, price excluding GST e electrician’s bill $457, price excluding GST

Example 3 Calculate the GST included on a television set advertised for $899, price including GST. Price excluding GST + GST = $899 i.e. price excl. GST + 10% of price excl. GST = $899 i.e. 110% of price excl. GST = $899 1.1 × price excl. GST = $899 price excl. GST = $899 ÷ 1.1 = $817.27 GST = $899 − $817.27 = $81.73

3

Calculate the GST included in the price of the following items: a TV $1189, price including GST b lounge suite $4970, price including GST c BBQ chicken $10.89, price including GST d perfume $148, price including GST e dress $124, price including GST

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Example 4 Calculate the GST included on a television set advertised for $899, price including GST. Note that example 3 above could have been calculated as follows. 110% of price excl. GST = $899 11 i.e. ------ × price excl. GST = $899 10 11
price excl. GST = $899 ÷ -----10 10 = ------ × $899 11 1 Hence GST = ------ × $899 11 = $81.73 This leads to the ‘GST Rule of Thumb’ which 1 states that GST = ------ of price including GST. 11

4

Use the GST Rule of Thumb to check your answers to question 3.

5

Find the missing amounts in the following. a

Tax Invoice Services rendered = $850 GST = Total including GST =

c Tax Invoice Services rendered = GST = $48.80 Total including GST = $536.80

b

Tax Invoice Taxable items Shirt $69.95 Tie $29.95 Total including GST = $99.90 GST included in total =

d Tax Invoice Taxable items 5 CDs @ $32.90 including GST = GST included in total =

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non-calculator activities 1

David earns $300 per week. How much does he earn per: a fortnight b year c month?

2

Convert a salary of $43 800 p.a. to the equivalent salary per month.

3

Alice works a 35-hour week and is paid $20 per hour. How much does she earn for a week in which she works an additional 4 hours at time-and-a-half and 1 hour at double time?

4

Katya earns $1.20 for each lamp she makes. If on average she can finish 10 lamps per hour and she works a 36-hour week, calculate her average weekly earnings.

5

Maria sells household cleaners. She is paid a commission of 5% of sales. How much does she earn in a week in which her sales are $8000?

6

Jack’s gross weekly income is $847. The deductions from his salary each week are: tax $206, superannuation $51.60 and health insurance $26.53. Calculate his net earnings each week.

7

A sports goods store offers a discount of 10% for cash purchases. Find the cash price of a basketball marked as $89.

8

Alex lay-bys a tool set costing $638 by paying a deposit of $125. Over the next 3 months he makes repayments of $100, $120 and $185. How much more does he have to repay in order to collect the tool set?

9

A sound system can be bought for $589 cash or on the following terms: deposit $189 and 24 equal monthly repayments of $23. a What is the total cost of the sound system if it is bought on terms? b How much interest would be paid?

10

Calculate the GST included in the price of a DVD player costing $187, price including GST.

Language in Mathematics

1

List four different ways in which people are paid for providing their labour or services.

2

Explain the meaning of: a overtime

b bonus

c holiday loading

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3

Use the following words or phrases in a sentence: gross income, deductions, net earnings

4

What is the difference in meaning between the words principal and principle?

5

The following words have a mathematical meaning as well as other meanings in ordinary English. Use a dictionary to complete the table. Word

Mathematical meaning

351

Other meaning

credit deposit balance 6

7

✓

Complete the following words from this chapter by replacing the vowels. a f — rtn — ghtly b r—t——n—r c b — dg — t

d d — sc — — nt

Three of the words in the following list have been spelt incorrectly. Rewrite them with the correct spelling: peacework, purchase, cash, survice, loan, invesment.

Glossary balance cash deductions excluding gross income including labour overtime repayment simple interest

bonus commission deferred payment expenses GST income lay-by piecework retainer take-home pay

budget compare deposit flat interest holiday loading instalment loan principal salary time-and-a-half

CHECK YOUR SKILLS

buying on terms credit discount fortnight hire purchase investment net earnings purchase service wages

1

Samantha earns $326.80 per week. This is equivalent to a yearly salary of: A $15 686.40 B $16 340 C $16 993.60 D $17 320.40

2

A salary of $33 228 p.a. is not equivalent to: A $639 per week B $2556 per fortnight C $2769 per month D $8307 per quarter

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3

Garry earns $342 per week. This is equivalent to a monthly income of: A $1368 B $1470.60 C $1482

D $1539

4

Sally works a 36-hour week and is paid $14.80 per hour. Her total wages for a week in which she works an additional 5 hours at time-and-a-half and 3 hours at double time is: A $732.60 B $651.20 C $710.40 D $769.60

5

Bianca earns $560 per week. She is entitled to 4 weeks annual recreation leave and receives an additional holiday loading of 17.5%. Her total holiday pay for 4 weeks is: A $2240 B $392 C $2338 D $2632

6

David is paid $0.37 for each tree that he plants. If he can plant an average of 18 trees per hour and he works a 36-hour week, then his average weekly earnings are: A $6.66 B $13.32 C $239.76 D $479.52

7

Tony is a real estate agent. He charges the following commission for selling home units: 3% of the first $150 000 and 1.5% for the remainder of the selling price. His commission for selling a home unit for $220 000 would be: A $6600 B $5550 C $3300 D $9900 Casual $ per hour

8 Waiter

Mon.–Fri.

Sat.

Sun.

14.92

18.65

22.38

✓

The table shows the award wages for a waiter employed as a casual. The wages of a casual waiter who works 10 hours Monday to Friday, 4 hours on Saturday plus 5 hours on Sunday is: A $335.70 B $283.48 C $317.05 D $350.62 9

10

Stephen earns $487 per week. The deductions from his salary each week are tax $139, superannuation $42, and health insurance $31.80. His net pay for the week is: A $699.80 B $421.80 C $358.20 D $274.20 The simple interest on $3480 at 5.5% p.a. for 4 years is: A $7656 B $191.40 C $765.60

D $4245.60

11

Michelle invested $5000 for 3 years and earned $825 in interest. The annual rate of interest was: A 5.5% B 16.5% C 33.3% D 3.33%

12

A camera store offers a discount of 12% for paying cash. The cash price of a camera marked as $459 is: A $55.08 B $55.10 C $403.92 D $403.90

13

The method of purchasing goods by which a deposit is paid, the balance is paid off over a short period of time, no interest is charged but the goods cannot be taken until full payment has been made is called: A time payment B hire purchase C deferred payment D lay-by

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14

A refrigerator costing $1395 can be bought on terms for $295 deposit and 24 monthly instalments of $61. The total cost of buying the refrigerator on terms would be: A $2859 B $1759 C $1464 D $2564

15

A television set costing $1089 can be bought on the following terms: deposit $289 and the balance to be repaid over 2 years by equal monthly instalments. Simple interest is charged at 13% p.a. If the TV is bought on these terms, the monthly repayment would be: A $42 B $57.17 C $37.67 D $51.27

16

A television set is advertised as shown opposite. The total amount you would have to pay for the television under this scheme is: A $1598 B $1853.68 C $2365.04 D $2109.36

$1598

✓

NO DEPOSIT NO REPAYMENTS FOR 12 MONTHS (Conditions apply.)

Conditions: (i) Pay nothing for 12 months. (ii) Balance plus interest to be repaid by equal monthly instalments over the two years following the interest free period. (iii) Simple interest of 16% p.a. is charged for the 3-year period of the agreement.

17

Using the table on page 344, the monthly repayment on a loan of $24 000 over 2 1--2- years at 11.5% is, to the nearest cent: A $38.51 B $924.30 C $696.82 D $1124.17

18

Which of the following is the best value? A 350 mL for $2.80 B 750 mL for $5.25

19

C 2 L for $15.00

D 5 L for $39

The price of a TV, including GST, is $583. The amount of GST included is: A $58.30 B $53 C $524.70

D $530

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–3 4, 5 B

C

6

7

8

9

D

E

F

G

353

10, 11 12 I

J

13 14, 15 16

17

18

19

L

O

P

Q

M

N

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REVIEW SET 11A 1

David earns $463.90 per week. How much does he earn per: a fortnight b year c month?

2

Convert a salary of $56 000 p.a. to the equivalent salary per: a week b fortnight c month

3

Alice works a 35-hour week and is paid $18.70 per hour. How much does she earn for a week in which she works an additional 4 hours at time-and-a-half and 3 hours at double time?

4

Michelle works a 35-hour week and is paid time-and-a-half for any extra hours worked. One week she worked 4 hours overtime and was paid $746.32. What is her hourly rate of pay?

5

Travis earns $560 per week. He is entitled to 4 weeks annual leave and receives an additional holiday loading of 17.5%. Calculate his total pay for this holiday period.

6

Sharon is entitled to 4 weeks annual leave and receives a holiday loading of 17.5%. One year her total holiday pay was $2641.40. What is Sharon’s weekly salary?

7

Nerida earns $0.98 for each dress she finishes in a clothing factory. If on average she can finish 12 dresses per hour and she works 8 hours per day for 4 days, calculate her average weekly earnings.

8

Cass sells computers. She is paid a retainer of $220 per week plus a commission of 2% of sales. How much does she earn in a week in which her sales are $12 800?

9

Jim is paid a retainer plus a commission of 4% of sales. If he receives $800 for selling $13 000 worth of goods, what is the retainer that he is paid?

10

Sam works as a casual in a fruit shop. He gets paid $11.60 for any hours worked from Monday to Friday, $12.90 per hour for Saturdays and $13.60 for Sundays. Calculate how much he earns for a week in which he works 6 hours between Monday and Friday, 5 hours on Saturday and 4 hours on Sunday.

11

Jack’s gross weekly income is $768 per week. The deductions from his salary each week are: tax $224, superannuation $38.40 and health insurance $33.76. Calculate his net earnings each week.

12

Calculate the simple interest on $3600 if invested at 9% p.a. for: a 4 years b 20 months

13

A sports goods store offers a discount of 16% for cash purchases. Find the cash price of a pair of running shoes marked as $179.

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14

List two advantages and two disadvantages of using a credit card to purchase goods.

15

Melanie lay-bys a swing set costing $524 by paying a deposit of $150. Over the next 3 months she makes repayments of $100, $120 and $85. How much more does she have to repay in order to collect the swing set?

16

A car costing $10 999 can be bought on the following terms: deposit $3000, the balance to be repaid over 4 years by 48 equal monthly repayments. Simple interest is charged on the balance at 12% p.a. Calculate: a the balance owing b the interest charged on the balance owing c the monthly repayment

17

Use the table on page 344 to calculate the monthly repayments on a loan of $7800 for 3 1--2- years at 12.5% p.a.

18

Terry borrowed $20000 at 11.5% p.a. His monthly repayments were $770.25. Over what period of time did he borrow the money? (Use the table on page 344.)

19

Which is the better buy: 3 kg of tomatoes for $8.97 or 5 kg for $14.80?

20

Calculate the GST included in the price of a bottle of wine costing $18, price including GST.

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REVIEW SET 11B 1

Dan earns $368.54 per week. How much does he earn per: a fortnight b year c month?

2

Convert a salary of $45 600 p.a. to the equivalent salary per: a week b fortnight c month

3

Olivia works a 36-hour week and is paid $21.36 per hour. How much does she earn for a week in which she works an additional 6 hours at time-and-a-half and 2 hours at double time?

4

Stephanie is paid $21.30 per hour for working a 35-hour week and time-and-a-half for any extra hours worked. One week she was paid $873.30. How much overtime did she do?

5

Terry earns $680 per week. He is entitled to 4 weeks annual leave and receives an additional holiday loading of 17.5%. Calculate his total pay for this holiday period.

6

Nick is entitled to 4 weeks annual leave and receives a holiday loading of 17.5%. One year his total holiday pay was $3741.20. Calculate his holiday loading.

7

Joanne sews buttons on shirts in a clothing factory. She is paid $0.38 per shirt. Calculate her income for a week in which she completed the following number of shirts: Mon 165, Tues 189, Wed 212, Thurs 194, Fri 176.

8

Benita sells printers. She is paid a retainer of $260 per week plus a commission of 1.5% of sales. How much does she earn in a week in which her sales are $22 400?

9

Sally is paid a retainer of $220 per week plus a commission of 3% of sales. One week she earned $598. What was the value of the goods that she sold?

10

Dennis works as a casual in a coffee shop. He gets paid $10.90 for any hours worked from Monday to Friday, $13.64 per hour for Saturdays and $14.28 for Sundays. Calculate how much he earns for a week in which he works 10 hours between Monday and Friday, 4 hours on Saturday and 6 hours on Sunday.

11

John’s gross weekly income is $683 per week. The deductions from his salary each week are: tax $216, superannuation $36.78, health insurance $41.20 and savings $50. Calculate his take-home pay each week.

12

Calculate the simple interest on $18 000 if it is invested at 6% p.a. for: a 3 years b 15 months

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13

An electrical goods store offers a discount of 14% for cash purchases. Find the cash price of a toaster marked as $89.

14

List the advantages and disadvantages of using a lay-by to purchase goods.

15

An outdoor furniture setting costing $1889 can be bought on terms for $300 deposit and 24 monthly instalments of $90.04. a Calculate the cost of buying the furniture on terms. b How much interest is paid?

16

A washing machine costing $1655 can be bought on the following terms: deposit $200, the balance to be repaid over 2 years by 24 equal monthly repayments. Simple interest is charged on the balance at 15% p.a. Calculate: a the balance owing b the interest charged on the balance owing c the monthly repayment.

17

Use the table on page 344 to calculate the monthly repayments on a loan of $12 000 for 5 years at 10.5% p.a.

18

Sam borrowed $24000 over 4 years. The monthly repayments were $614.48. What was the interest rate charged? (Use the table on page 344.)

19

A-One Car Hire Co. charges $34 per day with unlimited kilometres to rent a new Corolla. B-One Car Rentals charges $26 per day plus 6 cents per kilometre travelled. Which company is cheaper if you are likely to travel each day: a 60 km b 100 km c 150 km?

20

Calculate the GST included in the price of a pair of shoes costing $128, price including GST.

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REVIEW SET 11C 1

Convert a salary of $36 000 p.a. to the equivalent salary per: a week b fortnight c month

2

Convert a salary of $365 per week to the equivalent monthly salary.

3

Alice works a 38-hour week and is paid $19.20 per hour. How much does she earn for a week in which she works an additional 3 hours at time-and-a-half and 1 hour at double time?

4

Kim works a 35-hour week and is paid time-and-a-half for any extra hours worked. One week she worked 5 hours overtime and was paid $1140.70. What is her hourly rate of pay?

5

Kelly earns $632 per week. She is entitled to 4 weeks annual leave and receives an additional holiday loading of 17.5%. Calculate her total pay for this holiday period.

6

Karen is entitled to 4 weeks annual leave and receives a holiday loading of 17.5%. One year her total holiday pay was $5931.40. What is Karen’s weekly salary?

7

Peta earns $538 per week. At the end of the year her employer pays her a bonus of 5% of her annual salary. Calculate Peta’s bonus.

8

Cameron sells real estate. He charges the following commission for selling home units: 3% of the first $150 00 2% of the next $50000 1% of the remainder of the selling price. Calculate how much Cameron would earn for selling a home unit for: a $145 000 b $185 000 c $220 000

9

Mick is paid a retainer plus a commission of 7% of sales. If he receives $992 for selling $9600 worth of goods, what is the retainer that he is paid?

10

James works as a casual in a bar. He gets paid $15.20 for any hours worked from Monday to Friday, $17.68 per hour for Saturdays and $19.32 for Sundays. Calculate how much he earns for a week in which he works 12 hours between Monday and Friday, 6 hours on Saturday and 6 hours on Sunday.

11

Josh’s gross weekly income is $940 per week. The deductions from his salary each week are: tax $312, superannuation $56.30 and health insurance $41.22. Calculate his net earnings each week.

12

Calculate the simple interest on $13 000 if invested at 6% p.a. for: a 5 years b 21 months

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13

A sports goods store offers a discount of 18% for cash purchases. Find the cash price of a tennis racquet marked as $279.

14

List the advantages and disadvantages of using cash to purchase goods.

15

Sylvie lay-bys a dress costing $465 by paying a deposit of 10%. Over the next 4 weeks she makes repayments totalling $320. How much more does she have to repay in order to collect the dress?

16

A car costing $12 000 can be bought on the following terms: deposit $2000, the balance to be repaid over 3 years by 36 equal monthly repayments. Simple interest is charged on the balance at 8% p.a. Calculate: a the balance owing b the interest charged on the balance owing c the monthly repayment

17

Use the table on page 344 to calculate the monthly repayments on a loan of $24 000 for 4 years at 13% p.a.

18

Lenny borrowed $35 000 at 10.5% p.a. His monthly repayments were $999.41. Over what period of time did he borrow the money? (Use the table on page 344.)

19

Which is the best value? Chocolate: 55 g block for $1.10, 250 g block for $4.88, 375 g block for $7.35.

20

Calculate the GST and the price including GST on a pair of boots costing $498, price excluding GST.

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REVIEW SET 11D 1

Convert a salary of $56 000 p.a. to the equivalent salary per: a week b fortnight c month

2

Holly earns $528 per week. How much is this per month?

3

Alice works a 35-hour week and is paid $24.10 per hour. How much does she earn for a week in which she works an additional 5 hours at time-and-a-half and 4 hours at double time?

4

Gayatri is paid $36.90 per hour for working a 35-hour week and time-and-a-half for any extra hours worked. One week she was paid $1734.30. How much overtime did she work?

5

Ken earns $720 per week. He is entitled to 4 weeks annual leave and receives an additional holiday loading of 17.5%. Calculate his total pay for this holiday period.

6

Ray is entitled to 4 weeks annual leave and receives a holiday loading of 17.5%. One year his total holiday pay was $4091.82. Calculate his holiday loading.

7

Isabella earns $0.71 for each part she builds in a factory that produces electrical appliances. If on average she can finish 15 parts per hour and she works 6 hours per day for 5 days, calculate her average weekly earnings.

8

Kate sells mobile phone plans. She is paid a retainer of $180 per week plus a commission of 6% of sales. How much does she earn in a week in which her sales are $9200?

9

Olivia is paid a retainer of $250 per week plus a commission of 6% of sales. One week she earned $768.40. What was the value of the goods that she sold?

10

Ann works as a casual in a cafe. She gets paid $12.34 for any hours worked from Monday to Friday, $13.85 per hour for Saturdays and $15.98 for Sundays. Calculate how much she earns for a week in which she works 8 hours between Monday and Friday, 6 hours on Saturday and 3 hours on Sunday.

11

Phil’s’s gross weekly income is $895 per week. The deductions from his salary each week are: tax $291, superannuation $42.81 and health insurance $38.26. He also deposits $100 a week into a special savings account and has $10 per week donated directly to a charity. Calculate his take-home pay each week.

12

Calculate the simple interest on $11 400 if invested at 8% p.a. for: a 3 years b 15 months

13

A store offers a discount of 12% for cash purchases. Find the cash price of a pair of sun glasses marked as $189.

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14

List the advantages and disadvantages of using a deferred payment option to purchase goods.

15

A car costing $10 999 can be bought on the following terms: deposit $3000, the balance to be repaid over 4 years by 48 equal monthly repayments. Simple interest is charged on the balance at 12% p.a. Calculate: a the balance owing b the interest charged on the balance owing c the monthly repayment No Deposit

$3999

16

A computer is advertised as shown opposite. a Calculate the total amount you would have to pay for the computer under this scheme. b Calculate the monthly instalments.

No Repayments for 12 months (Conditions apply.)

Conditions: (i) Pay nothing for 12 months. (ii) Balance plus interest to be repaid by equal montly instalments over the two years following the interest free period. (iii) Simple interest of 15% p.a. is charged for the 3-year period of the agreement. (iv) Establishment fee of $110.

17

Use the table on page 344 to calculate the monthly repayments on a loan of $15 500 for 4 1--2- years at 14% p.a.

18

Will borrowed $28000 over 5 years. The monthly repayments were $651.51. What was the interest rate charged? (Use the table on page 344.)

19

On savings accounts, Bob’s Bank charges a management fee of $5.50 per month. The first 5 transactions are free and then a fee of 26 cents per transaction is charged. Bill’s Bank charges a monthly management fee of $7.00 plus 9 cents per transaction. Which bank is cheaper to use if your average number of monthly transactions is: a 10 b 15 c 20?

20

Calculate the GST in the price of a cooked chicken costing $9.90, price including GST.

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Chapter 12 Right-angled Trigonometry This chapter deals with the solution of right-angled triangles. After completing this chapter you should be able to: ✓ identify and label sides of a right-angled triangle ✓ define sine, cosine and tangent ratios ✓ use a calculator to find trigonometric ratios and angles ✓ use trigonometry to find unknown sides and angles in right-angled triangles ✓ solve problems involving trigonometry and angles of elevation and depression.

Syllabus reference MS5.1.2 WM: S5.1.1–S5.1.5

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Diagnostic test 1

2

3

4

5

The side opposite angle A in this triangle is: A A AD B AR C DR R D D the hypotenuse

8

The hypotenuse in this triangle is: A MT B MV C TV T D m

9 M

V

The expression for sin θ in this triangle is: r A t

b B --t

t C r

t D --b

r

AD A -------DR

DR B -------- A AD

AR C -------AD

DR D -------AR D

11

The value of θ in the triangle is closest to: A 52° 23 cm B 38° C 1° θ 18.2 cm D 51°

12

The value of θ in the triangle is closest to: A 62° θ B 28° 1.9 m C 33° 3.5 m D 57°

13

The value of θ in the triangle is closest to: A 68° θ B 70° 16.2 cm C 32° 43.9 cm D 22°

b

The value of cos 53° is closest to: A 53 B 0.6018 C 0.8192 D –0.9848

6

The value of θ in the expression tan θ = 3.466 is closest to: A 74° B 0.0606 C 19° D 3

7

The value of angle A in the expression 11.5 cos A = ----------- is closest to: 25 A 0.9799 B 0.99996 C 62° D 63°

The value of x is closest to: A 27.2 14 cm B 12 C 23.3 59° x cm D 7.2

The value of x is closest to: A 375 72° B 40 x mm C 128 122 mm D 116

θ

θ

x mm

10

t

The expression for tan θ in this triangle is:

The value of x is closest to: A 14.3 32.6 mm B 74.4 C 29.3 26° D 15.9

R

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14

The diagram shows that the angle of elevation of the top of a cliff from a boat 1500 m out to sea is 6°. The height of the cliff above the boat is closest to: A B C D

150 m 14 272 m 160 m 1492 m

15

The angle of depression from the top of a cliff 300 m above sea-level to a boat is 76°. The distance of the boat from the cliff is closest to: A 1203 m 76° B 75 m 300 m C 73 m Dm D 1240 m

hm 6° 1500 m

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–4

5–7

8–10

11–13

14, 15

A

B

C

D

E

Trigonometry is a branch of mathematics that combines arithmetic, algebra and geometry. The word trigonometry is derived from Greek and means triangle measurement. The study of trigonometry enables us to compare similar triangles so that lengths that are difficult or impossible to measure directly can be calculated. Greek, Persian and Hindu astronomers first developed trigonometry around 200 BC. Hipparchus is credited with being the originator of the science at that time. Today trigonometry is used by astronomers, architects, surveyors, engineers and navigators of both planes and ships.

Investigation 1 WM: Reasoning, Applying Strategies, Communicating

Ratios of sides 1

Here are three right-angled triangles. They are all equiangular. a b c C

B

A

C 40°

A C

40°

B

B

40°

Measure the lengths of all sides to the nearest millimetre.

A

☞

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

2

Copy and complete this table. Triangle

AB

BC

AC -------BC

AC

AC -------AB

BC -------AB

a b c

3

Compare your answers in the last three columns. What do you notice?

4

Draw a triangle ABC with base AB of length 6 cm, angle A of measure 30° and angle B of measure 90°.

5

Draw a second triangle ABC with AB of length 10 cm and angles A and B as before.

6

Draw a third triangle ABC with AB of length of your choosing and again angles A and B are the same as before.

7

Clearly the triangles are equiangular. Measure the lengths of the unknown sides and complete a table like the one below. Triangle

AB

a

6 cm

b

10 cm

BC

AC -------AB

AC

BC -------AB

c 8

What do you notice about the answers in the last two columns?

9

a Draw this diagram to scale. G F E

A

40° 6 cm

B 4 cm

C

5 cm

D

☞

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

b

Copy and complete this table: Triangle ABE

AB

EB

AE

EB -------AB

AB -------AE

Triangle ACF

AC

FC

AF

FC ------AC

AC ------AF

Triangle ADG

AD

GD

AG

GD -------AD

AD -------AG

c

What do you notice about the answers in the last two columns?

d

Write a paragraph describing your results in this investigation.

A. DEFINING TRIGONOMETRIC RATIOS For convenience, we give special names to the sides of a right-angled triangle. The side opposite the right angle is known as the hypotenuse. It is the longest side of the right-angled triangle. In this triangle the hypotenuse is AC. Theta (θ) and phi (φ) are Greek letters often used for angles.

hypotenuse

A

θ

φ

C

B

The side BC is opposite the angle at A (angle θ). The side AB is adjacent or next to the angle at A (angle θ). If we look at the angle at C (angle φ), AB is now the opposite side, and BC is the adjacent side.

371

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

Example 1 In this triangle, what is the: a hypotenuse b side opposite angle P c side adjacent to angle P d side opposite angle R e side adjacent to angle R?

Q

P

a The hypotenuse is PR. c PQ is the side adjacent to angle P. e QR is the side adjacent to angle R.

R

b QR is the side opposite angle P. d PQ is the side opposite angle R.

Remember how to name sides R

Label the side opposite T as t, R as r, and S as s.

t S

T

r

Exercise 12A

In the diagrams below, find: i the hypotenuse ii the side opposite the angle marked θ iii the side adjacent to the angle marked θ a b c A

P

θ

C

B

X

θ

R

d

m

e

Q

θ

1

s

Y

f θ

l

r

p

E

g

h z x

n θ

θ

q

D

i

U θ

θ

V

j

w

T

k

l

S

h

g

θ

k m

θ k

θ

l

U

T

v

u

y

θ

F

Z

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

2

The triangle opposite has hypotenuse of length a units and other sides of length b and c units. θ and ø are the two acute angles.

φ

a

b

Find the: a side opposite θ b side opposite ø c side adjacent to θ d side adjacent to ø

θ

c

Example 2 Using the given triangle, write expressions to q complete the table.

θ

p r

opposite ---------------------adjacent

opposite ----------------------------hypotenuse

adjacent ----------------------------hypotenuse

θ

The hypotenuse is p, the side opposite θ is q and the side adjacent to θ is r. θ

3

opposite ---------------------adjacent

opposite ----------------------------hypotenuse

q --r

q --p

adjacent ----------------------------hypotenuse r --p

Complete this table for the triangles in question 1. θ

opposite ---------------------adjacent

opposite ----------------------------hypotenuse

adjacent ----------------------------hypotenuse

The ratios from question 3 are given names. ˙˙ opposite opposite The ratio ------------------ is the tangent of the angle θ and is abbreviated to tan θ = -----------------. adjacent adjacent opposite opposite The ratio ------------------------- is the sine of the angle θ and is abbreviated to sin θ = ------------------------- . hypotenuse hypotenuse adjacent adjacent The ratio ------------------------- is the cosine of the angle θ and is abbreviated to cos θ = -------------------------. hypotenuse hypotenuse The trigonometric ratios can be remembered using a mnemonic. SOH sin θ

CAH opp = -------hyp

cos θ

TOA adj = ------hyp

tan θ

opp = -------adj

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

Example 3 In the triangle ABC, find tan θ, cos θ, and sin θ. B

θ

C

tan θ

4

opp = --------adj BC = ------AC

opp = --------hyp BC = ------AB

sin θ

i tan θ

In the following diagrams, find: a

b

U

cos θ

ii sin θ

adj = --------hyp AC = ------AB

iii cos θ c

a

θ T

θ

c

b

S

A

s u

t

θ

d

e

f

Z

R

θ

θ

Y

X

5

i

h

Find:

T

j

i sin A

ii cos A

a

b

iii tan A T

θ

S

in these diagrams. V

c

A

C

B

A

A

P

R

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

B. TRIGONOMETRIC RATIOS AND THE CALCULATOR This section uses the calculator to find approximations for trigonometric ratios and angles from decimals.

Example 1 Find, correct to 4 decimal places: a sin 41° b cos 78° a sin 41°
0.6561

b cos 78°
0.2079

sin 41 =

cos

c

tan 15°

c

tan 15°
0.2679

78 =

tan

15 =

Exercise 12B 1

Find, correct to 4 decimal places: a sin 24° b cos 65° e cos 81° f tan 5° i sin 54° j sin 85°

c tan 35° g tan 75° k tan 55°

d sin 14° h cos 38° l cos 11°

Example 2 Find the value of θ, correct to the nearest degree if: a sin θ = 0.4718 b tan θ = 3.624

cos θ = 0.7

c

a sin θ = 0.4718
28.151202
28° b tan θ = 3.624
74.57378579
75° c cos θ = 0.7
45.572996
46° 2

-1

sin

0.4718

Check your calculator steps.

tan-1 3.624

cos--1 0.7

Find the value of θ correct to the nearest degree. a sin θ = 0.2431 b cos θ = 0.1251 d cos θ = 0.4 e tan θ = 0.5 g tan θ = 0.041 h sin θ = 0.552 j sin θ = 0.3004 k tan θ = 25.3715 m cos θ = 0.5484 n sin θ = 0.7474

c f i l o

tan θ = 3.2415 sin θ = 0.7 cos θ = 0.044 cos θ = 0.8844 tan θ = 0.3333

375

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

Example 3 Find the value of ∠A correct to the nearest degree 5 a sin A = --7

7.3 b cos A = -------12

c

25 tan A = -------8.7

5 a sin A = --7 A
45.5846914
46°

sin-1

( 5 ÷ 7 )

=

The fraction button may be used if both numerator and denominator are whole numbers.

7.3 b cos A = -------12 A
52.53091057
53°

cos-1

( 7.3 ÷ 12 )

=

tan-1

( 25 ÷ 8.7 )

=

25 c tan A = -------8.7 A
70.8121034
71°

With DAL calculators enter the trig ratio first.

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

3

4

Find angle B to the nearest degree. 17 a tan B = -----5

8 b sin B = -----11

12 c cos B = -----17

8.3 d sin B = -------15

7.1 e cos B = ----------11.2

8.4 f tan B = -------3.9

0.7 g cos B = -------1.2

6.3 h tan B = -------7.1

0.05 i sin B = ----------0.13

a If cos E = 0.52, find the size of ∠E. b If sin P = 0.1352, find the size of ∠P. c If tan R = 5.31, find the size of ∠R. 11.3 d If cos M = ----------- , find the size of ∠M. 15.8

Investigation 2 WM: Reasoning, Communicating

Comparing ratios 1

Use your calculator to complete the table below correct to three decimal places. a What do you notice about the answers in the tan θ column compared with the sin θ -------------- column? cos θ b Based on your answer to part a, sin θ complete -------------- = cos θ c From the table above, between which values do: i sin θ ii cos θ iii tan θ lie?

θ 0° 10° 20° 30° 40° . . . 90°

sin θ

cos θ

tan θ

sin θ cos θ -------------- -------------cos θ sin θ

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

C. USING TRIGONOMETRY TO FIND SIDES Example 1 Use the sine ratio to find the value of x correct to one decimal place. a

b xm

20 cm x cm 27°

68° 50 m

a

b xm

20 cm x cm 27°

68° 50 m

opp sin θ = --------hyp

opp sin θ = --------hyp

x sin 27° = -----20

x sin 68° = -----50

∴ x = 20 sin 27°

x = 50 sin 68°

x
9.1 20

×

sin

x
46.4 27 =

50

×

sin

68 =

Exercise 12C 1

Use the sine ratio to find the value of the unknown correct to one decimal place. a b c t cm a cm 68°

18 cm x cm 37°

d

43°

3 cm

e

16.5 cm

f

48° 11.7 cm

x cm

y cm 35°

15.2 cm

xm 43° 2.5 m

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

g

h

x cm

i

67°

zm 73°

16.2 cm 53°

y mm

11.5 mm

13.1 m

Example 2 Use the sine ratio to find the value of the unknown correct to one decimal place. a

b ym

4m

18°

6.2 m xm

65°

opp sin θ = --------hyp

a

6.2 sin 18° = -------y

4 sin 65° = --x ∴ x sin 65° = 4

4 ÷

2

opp sin θ = --------hyp

b

∴ y sin 18° = 6.2

4 ∴ x = ------------------sin 65°

6.2 ∴ y = ------------------sin 18°

∴ x
4.4

∴ y
20.1

sin 65 =

6.2 ÷

sin 18 =

Use the sine ratio to find the value of the unknown correct to one decimal place. a

b ym

27° 8 cm

c

14°

x cm

6.2 mm

43°

d

e

f

53.2 cm 28° 41°

z cm

a mm

11.2 m

ym

115 mm

a mm 48°

6.2 m

379

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

3

Find the value of x, giving your answer correct to one decimal place. a b c 10 cm

x cm

50°

xm

x cm

35°

72° 50 cm

2m

d

e

f

4.3 m 173 cm

x cm

x cm

xm

43° 30°

100 cm 60°

Example 3 Use the cosine ratio to find the value of x correct to one decimal place. a b 23 m 52 cm 32°

51°

xm

a

xm

adj cos 32° = --------hyp

23 m

x ∴ cos 32° = -----23 ∴ x = 23 cos 32
19.5 b

52 cos 51° = -----x 52 ∴ x = -------------------cos 51°
82.6

xm

( 23

×

32° cos

( 52 ÷

32 )

=

52 cm is opposite.

52 cm 51°

x is adjacent.

xm cos

51 )

=

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

4

Use the cosine ratio to find, correct to one decimal, the value of x. a

b

c

x km 66°

12 m x cm

8 cm

200 km

50°

33° xm

d

e

x cm

6 cm

f

x cm

48°

h

xm

i

24°

x cm

3m

119 mm

16.8 cm

64°

g

53°

x mm 71°

xm

25.2 cm

48° 16.2 m

5

Find the length of the hypotenuse using the cosine rule. Give answers correct to one decimal place. a

b

53 cm 71°

c 3.8 m

143 mm 39°

25°

x cm a mm

ym

d

e

f

21°

310 mm 39 cm

3.2 m 43°

6

When finding the hypotenuse you will divide by the angle.

xm

z mm 31°

a cm

Use the cosine ratio to find, correct to one decimal place, the value of x. a b c xm 15°

43 cm 40°

x cm

15 m x cm

60° 8.64 cm

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

d

e

f 16°

x cm

x cm 15 cm

15.8 cm 40° 11.7 cm

x cm

41°

Example 4 Use the tangent ratio to find the value of x correct to one decimal place. a

16 cm

b xm x cm 31°

53°

8m

opp tan 31° = --------adj

a

opp tan 53° = --------adj

b

16 ∴ tan 53° = -----x

x ∴ tan 31° = --8

16 x = -------------------tan 53°

∴ x = 8 tan 31° x = 4.8 8

7

×

tan

x = 12.1

31 =

16

÷

tan

53 =

Use the tangent ratio to find the value of x, correct to one decimal place. a

b

c x cm

xm

23°

12 cm

34°

x km

40°

20 km

4m

d

e

f

58° x cm 49° 18.7 cm

x mm

91.3 cm

x cm

61° 210 mm

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

8

Use the tangent ratio to find the value of x, correct to one decimal place. a b c 16.5 cm

x mm

18°

xm

43° 214 mm

x cm 67° 4.3 m

d

e

f x cm

24 m

42° x km

16.9 km

40 cm 58°

9

48° xm

Use the tangent ratio to find the value of x, correct to one decimal place. a

b

c

52.9 cm

15.3 cm 28° 137 mm

x cm

x cm 41° 75° x mm

10

Use one of the sine, cosine or tangent ratios to find the value of the pronumeral, correct to one decimal place. a

b

15 cm

x cm

d

c

29.3 cm 41°

x m 40° 11°

e

14.2 m

xm

18 m

f 17°

48.3 mm x cm

14.3 m 28°

16°

g

y mm

h

i 8.3 cm 50°

15°

x cm 83 cm

x cm

18.3 m 21° xm

xm

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

D. USING TRIGONOMETRY TO FIND ANGLES Example 1 Use the sine ratio to find the value of θ correct to the nearest degree.

θ

40 cm

28 cm

opp sin θ = --------hyp

θ

40 cm

∴ θ
44.43° ∴ θ
44°

sin

( 28 ÷ 40

)

=

28 cm

Exercise 12D 1

Using the sine ratio, find to the nearest degree, the value of θ. a

b

c θ

32 cm

8m

5m

8.7 km

θ θ

d

3.5 km

50 cm

e

f

11 cm θ 16 cm

θ

g

15 cm

θ 22.3 cm 33.6 cm

28 cm

h θ

423 mm

i 6.25 m

θ 0.81 m

1.2 m θ

312 mm

2

4.37 m R

RQ is half as long as PR. Using the sine ratio, find the measure of angle RPQ to the nearest degree.

P

Q

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

Example 2 Use the cosine ratio to find the value of θ correct to the nearest degree.

Q

12 m

R

∴ cos θ = 12 m

3

P

4m

adj cos θ = --------hyp

Q

R θ4m

θ

4 -----12

cos

( 4 ÷ 12 )

=

∴ θ
70.528…
71°

P

Using the cosine ratio, find to the nearest degree, the measure of the unknown angle. a b c 18 km

θ 10 cm

9m

θ 4 cm

d

12 m

θ

10 cm θ

e

12 km

f 14 cm

21 cm

18 cm

11.3 cm θ 41.2 cm

θ

g

h

i

671 mm θ 258 mm

4

2.4 m

7.62 m θ

θ 0.92 m

3.47 m

AC is three times longer than BC. Using the cosine ratio, find the measure of angle BCA to the nearest degree.

A

B

C

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

Example 3 Use the tangent ratio to find the value of θ correct to the nearest degree.

R 5 cm θ

P

opp tan θ = --------adj

R

∴ tan θ =

5 cm P

5

7 cm

Q

7 cm

5 --7

tan

( 5 ÷ 7

)

=

∴ θ
36°

Q

Using the tangent ratio, find the measure of the angle marked ø to the nearest degree. a b c 12 cm φ

52 cm 26 cm

4.2 m φ

φ

d

18.1 cm 8.35 m

e 12 cm

f

10.8 cm

φ 12.1 cm

18 cm

8 cm φ

φ

g

7 cm

h 8.9 cm φ

i

7.13 m φ

235 mm

15.6 cm

118 mm

9.26 m

φ

6

Use one of the sine, cosine or tangent ratios to find the unknown angle to the nearest degree. a b c 8.2 m 11 m

6.2 cm

6.3 m

18 m θ

θ 14.9 cm

θ

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

d

e 46.9 cm

θ

θ 41 cm

g

f

h

θ

i θ 3.1 m

18.3 cm

11.1 cm

13.9 m

4.3 cm θ

θ

6.2 cm

1.9 cm

11.8 m

16.8 cm 1.9 m

j

k

14.2 cm

l 1200 mm

16.1 cm θ

θ

θ

1.1 m 4300 mm

3.2 m

E. ANGLES OF ELEVATION AND DEPRESSION When an object is higher than an observer, the angle of elevation is the angle from the horizontal up to the object. object

observer

angle of elevation

horizontal

angle of depression

object

When an object is lower than an observer, the angle of depression is the angle from the horizontal down to the object.

387

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

Example 1 The diagram shows the angle of elevation of the top of a flagpole, as observed from a point 15 m from its base, is 63°. Find the height of the flagpole.

hm 63° 15 m

opp tan θ = --------adj h tan 63° = -----15 h = 15 tan 63° h
29.4 m

Exercise 12E 1

The diagram shows the angle of elevation of the top of a flagpole, as observed from a point 20 m from its base, is 48°. Find the height of the flagpole.

hm 48° 20 m

The top of a tree, when viewed 50 m from its base, has an angle of elevation of 23°. Find the height of the tree.

2 hm 23° 50 m

3

A person is 250 m from a cliff. The angle of elevation of the to of the cliff is 61°. Find the height of the cliff. hm 61° 250 m

Example 2

The angle of depression from the top of a cliff 180 m above sea level to a boat is 48°. How far is the boat from the base of the cliff? 48° 180 m

dm

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

In the diagram the lines are parallel as shown. 48°

Alternate angles are equal.

180 m 48° dm

This means that the angle in the bottom corner is also 48°. 180 Now tan 48° = ---------d 180 ∴ d = -----------------tan 48° d = 162 m (to nearest metre)

4

The angle of depression from the top of a cliff 200 m above sea level to a boat is 57°. How far is the boat from the base of the cliff?

57° 200 m

dm

5

33°

When looking down to a person standing in a park 180 m from the base of a building, the angle of depression is 33°. How high is the h m building?

180 m

Example 3 A ladder leaning against a vertical wall reaches 3.5 m up the wall and makes an angle of 55° with the ground. Determine the length of the ladder. 3.5 sin 55° = -------x ∴ x sin 55° = 3.5 3.5 m

x

3.5 ∴ x = -----------------sin 55°

55°

∴ x
4.273 ∴ the ladder is 4.273 m long.

( 3.5 ÷

sin

55 =

)

389

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

Example 4 Determine the length of the roofing beam required to support the roof of pitch 14° as shown in the diagram.

8.2 m 14°

x cos 14° = -------8.2

8.2 m xm 14°

xm

∴ x = 8.2 cos 14° ( 8.2 ∴ x
7.956

×

cos 14 =

)

∴ the length of beam is 2 × 7.956
15.91 m 6

The diagonal of a rectangle is 12 cm in length and the longer side is 9 cm. Determine the measure of the angle between the diagonal and the shorter side. Answer to the nearest degree.

12 cm θ

9 cm

A see-saw has length 5.2 m. When one end is resting on the ground it makes an angle of 23° with the ground. Find the height of the other end above ground level.

7 23°

8

The diagonal of a rectangle is 13.5 cm in length and the angle between the diagonal and the longer side is 23°. Find the length of the rectangle.

9 6.4 m

13.5 cm 23°

A 6.4 m long ladder leaning against a wall has its base 3.6 m from the foot of the wall. Find the angle between the ladder and the ground.

3.6 m

10

A rectangle has sides of length 12 cm and 8 cm. Determine the measure of the angles between the diagonal and each of the sides, correct to the nearest degree.

12 cm 8 cm

11 1m

72° 25 m

A young boy, with eyes 1 m above ground level, stands 25 m from the base of a tall building. If he looks up to the top of the building at an angle of elevation of 72°, find the height of the building.

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

12

When the sun is at an angle of elevation of 63°, a tree casts a shadow of length 6.2 m. Find the height of the tree. 63° 6.2 m

13

A boat has an anchor rope of length 55 m. Due to the ocean current, the boat drifts so that the rope makes an angle of 63° with the surface of the water. Find the depth of the water at the position where the anchor lies on the bottom.

14

A ski slope falls 115 m over a 415 m run. What is the angle of depression from the top of the slope?

θ 415 m

15

Determine the length of roofing beam, l, required to support a roof of pitch 16° as shown.

6.7 m 16° l

Find half first.

Language in Mathematics

1

2

115 m

Add vowels to complete these words. a tr__g__n__m__try b d __ngl__ __f d__pr__ss__ __n e g s__n__ h

__pp__s__t__ c__s__n__ __dj__c__nt

c hyp__t__n__s__ f t__ng__nt

Rearrange these words to form a sentence, the first word has a capital letter. a measured degrees are Angles in b tangent all cosine ratios and Sine are c of the depression down Look for angle d opposite hypotenuse The angle right is the e side Cosine the the divided adjacent hypotenuse by is

391

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392

Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

3

Use every third letter to reveal a sentence about trigonometry.

AATERHFGEHHONBRVFIAWGEEISSNDROTYFHJTOPHMAEASWWEO RERFRDYGCUJOSESFGINJNOIEPOIOISASFXCRFGOR EMWQTZCHV GENHWKOOLIRJYDHTSDECAQOSWMDEPRFLTHEUJMIKEOPNNGTR DOGAFJKSWEIGYNTHEDESTYOASTXDHFVABNTMKTLOHIUEUICYU OYTSRTITENEREWQOSDFFTAASNDEARGNHJGKILOLEEFEDFQEFU SDACFLVGSBHTDWHASEDGSJNITVNUREBHOPQFJLNOOICVNIGEH JTKLYOMUYITRNEWUQASZSTERHFVEBGAYHNJUGIKLGFE 4

Investigate the origin of the terms sine, cosine, tangent and trigonometry. Write a report.

Andrey Nikolayevich Kolmogorov

(1903–1987)

Andrey Nikolayevich Kolmogorov was born in 1903 in Tambov, Russia. At the age of 17 he enrolled in Moscow State University where his initial interest was in ancient Russian arts; an interest he maintained throughout his life. He began his first productive mathematical research in 1921 with research on trigonometrical series and operations on sets. In the following years he made considerable contributions to the areas of differentiation, integration and measurable sets. He continued to expand his fields of interest to include mathematical logic. In 1925 he graduated, was appointed a research associate and began to work in the field of probability theory. He later used this work to study the motion of the planets and the turbulent flow of air from a jet engine. After being appointed a professor of the university in 1931 and subsequently a director of the Institute of Mathematics, he continued to work in the field of stochastic processes and probability theory. He extended this theory to incorporate what are now known as Markov processes, related the theory to physics and the areas of Brownian motion and diffusion. He also developed two systems of equations (that now have his name), which describe Markov processes. Throughout his research, Kolmogorov was surrounded by young mathematicians who wished to learn. In his later years he became interested in the mathematical education of school-children and was appointed chairman of the Commission for Mathematical Education in the USSR. He was recognised as the 20th century’s most influential Soviet Mathematician, both in his own country and abroad. He received a number of prizes and was elected a member of numerous scientific academies as well as holding honorary doctorates from Paris, Stockholm and Warsaw universities. 5

a b c d

How old was Kolmogorov when he died? What did he research after graduating? What are the Markov processes? What interested Kolmogorov in his later years?

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

393

Glossary adjacent approximation opposite ratio theta

alternate angles cosine parallel right angled triangle

angle of depression degree phi sine trigonometry

CHECK YOUR SKILLS

1

The side opposite angle B in this triangle is: A MB C TB

2

L

3

4

M

The hypotenuse in this triangle is:

N

A LM C MN

R

AD A -------AR AR C -------AD

DR B -------AD DR D -------AR

The value of cos 28° is closest to:

g

x

B 0.8829

C 0.1392

D 0.1736

The value of θ in the expression tan θ = 4.29 is closest to: A 77°

7

θ

The expression for tan θ in this triangle is:

A 28 6

T

B

a

g B --a g D --x

θ

D

5

B LN D m

The expression for sin θ in this triangle is:

A

✓ M

B BT D the hypotenuse

a A --g a C --x

angle of elevation hypotenuse pitch tangent

B 0.075

C 13°

0.56 The value of angle A in the expression sin A = ----------- is closest to: 1.8 A 53° B 0.0098 C 34°

D 4

D 18°

LEY_bk953_12_2ndpp Page 394 Wednesday, January 12, 2005 12:04 PM

394

Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

The value of x is closest to:

8

A 82 C 47

67 mm

B 55 D 96

35° x mm

The value of x is closest to: A 151 B 160 C 49 D 55

9 x cm

52 cm 71°

10

The value of x is closest to:

✓ x mm 44°

A 82 C 41

B 57 D 85

The value of θ in the triangle is closest to: A 36° B 54° C 47° D 43°

11 15 cm θ 20.6 cm

12

59 mm

The value of θ in the triangle is closest to: A 23° B 67° C 26° D 64°

12 m

θ

5.2 m

13 61 cm

14

15

θ

38 cm

The value of θ in the triangle is closest to: A 38° B 51° C 32° D 58°

The diagram shows that the angle of elevation of the top of a cliff from a boat 1000 m out to sea is 4°. The height of the cliff above the boat is closest to: A 23 m B 1000 m C 14 300 m D 70 m

hm

4° 1000 m

The angle of depression from the top of a cliff 200 m above sea level to a boat is 67°. The distance of the boat from the cliff is closest to: A 85 m B 471 m C 78 m D 184 m

67°

200 m

Dm

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1–4

5–7

8–10

11–13

14, 15

Section

A

B

C

D

E

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

REVIEW SET 12A 1

Write down expressions for sin α, cos α, tan α, sin β, cos β and tan β in each of the following. a b c d

x

2

l

z

α

y

m

β

α

t

β u

n

α

β

β

b

a

r

α c

Find the length of the side marked x correct to one decimal place. a

b

c 11.2 cm

x cm

xm

18 cm 63°

38°

3

21 m

14°

x cm

Find the value of θ correct to the nearest degree. a

b

8.7 cm θ

c 15 cm

θ 12.8 cm 65 m

4

108 m

θ

28 cm

Solve the following problems involving trigonometry. a The shadow of a tree is 40 m in length and the angle of elevation from the end of the shadow to the tree top is 33°. 1 - of a metre. Find the height of the tree to the nearest ----10 33° 40 m

b 80 m

A kite string is pinned to the ground. The string makes an angle of 55° to the ground and is 80 m long. How high is the kite above ground level?

55°

c Find the measure of all angles of a triangle that has sides 3 cm, 4 cm and 5 cm.

395

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396

Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

REVIEW SET 12B 1

Write down expressions for sin θ, cos θ and tan θ in each of the following. a b c d y

q

p x

z

θ

θ

2

V

S

B

θ

r

θ

A

C

T

Find the length of the side marked x correct to 3 significant figures. a b c 33.1 cm x cm

3

75°

18.3 cm 73°

58° 16 cm

Find the value of θ correct to the nearest degree. a

11.5 cm θ

b

c 11.9 cm

17.6 m 18.3 cm

4

x cm

x cm

15.3 m θ

Solve these problems using trigonometry. a The angle of elevation of the top of a tree from a point 15 m from the base of a tree is 38°. Find the height of the tree.

28.3 cm

θ

Hm 38° 15 m

b An aeroplane takes off at a constant angle of 20°. When it has flown 1000 m, what is its altitude to the nearest metre? c A ladder is 5 m long and makes an angle of 75° with the ground. How far up the wall does it reach (to the nearest 10 cm)?

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Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

REVIEW SET 12C 1

Write down expressions for sin α, cos α, tan α, sin β, cos β and tan β in each of the following. a b c d p α

α

β

a

b

r

q

β

α

x

c

2

d

z

y

e β

α

β

f

Find the length of the side marked x to 3 significant figures. a b c

d

8m

x cm

48° 10 cm

3

xm 54°

20 cm

x cm

x km

57°

48°

Find, correct to the nearest degree, the value of θ. a b

9 km

c

θ

12 m 3 cm

9m θ

9 km

5 km θ

7 cm

4

Solve the following problems using trigonometry. a

Find all the sides and angles in this triangle.

57°

36 m

b The diagram shows that the angle of elevation of the top of a cliff from a boat 100 m out to sea is 8°. Calculate the height of the cliff above the boat. 50°

c 150 m

Dm

Hm 8° 100 m

The angle of depression from the top of a cliff 150 m above sea level to a boat is 50°. Calculate the distance of the boat from the cliff.

397

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398

Right-angled Trigonometry (Chapter 12) Syllabus reference MS5.1.2

REVIEW SET 12D 1

Write expressions for sin θ, cos θ and tan θ in the following diagrams. a b c d P

b

a

T

s

θ r

θ

c

2

R

θ

M

T L

t θ

Find the length of the side marked x to one decimal place. a b c

d x cm

3.2 m

xm

x cm

68°

5.1 cm

28°

3

6.2 cm 34°

4.1 cm 53°

x cm

Find, correct to the nearest degree, the value of θ. a

40 km

20 m

b

c

θ

25 cm

23 km θ

4

θ

48 m

7 cm

Solve the following problems using trigonometry. a To measure the width of a river a surveyor finds a point B directly opposite a landmark T, such as a tree, on the bank on the other side of the river. He then moves 20 m along the bank at right angles to BT to a point A. With a theodolite he measures angle BAT as 66°. Calculate the width of the river to the nearest metre.

T xm 66° A

20 m

B

b An isosceles triangle has sides 7 cm, 7 cm and 8 cm long. Find the measure of the base angles of the triangle to the nearest minute. c From the top of a vertical cliff 50 m high, the angle of depression of a boat straight out to sea is 15°. How far is the boat from the foot of the cliff, to the nearest metre?

15° 50 m

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Chapter 13 Equations and Inequalities This chapter deals with the solution of linear equations and inequalities, and the solution of simple quadratic equations. After completing this chapter you should be able to: ✓ solve linear equations ✓ solve word problems using linear equations ✓ substitute into formulas and solve ✓ explain why a particular value could be a solution to an equation ✓ solve simple quadratic equations ✓ solve linear inequalities.

Syllabus reference PAS5.2.2 WM: S5.2.2–S5.2.4

(not including simultaneous equations)

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400

Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

Diagnostic Test 1

The solution to 5x − 3 = −13 is: A x = −2 C x=

2

B x=5

-----− 16 5

5

10

D x = −1

-----B − 14 5

C

17 -----2

D 6

A man is currently three times as old as his son. In 12 years from now he will be twice as old as his son will be then. An equation to find the son’s age now is: B 2(x + 12) = 3x + 12 C x + 12 = 3x

A x = −1

B x = −5

D 3(x + 12) = 2x + 12

C x=5

D x=1

11

A solution to x2 = 16 is:

x = 3 is not a solution of:

A x=8

B x = 32

A 5x + 2 = 26 − 3x

C x = −4

D x = 256

D 5x − 3 = 4x

12

13

C 3x = 2 2 --3

5x + 2 The solution to ---------------- = −4 is: 3 -----A x = − 6--5B x = − 10 3 -----C x = − 14 5

B 8x = −15 C x= D x=

15 -----8 1 7--8-

14

D x=2

The line without an error in the solution 4x x 1 of ------ – --- = – --- is: 5 3 2 A 12x − 5x = −15

5 --2

C x = 2.4

B 3x + 1 = 3 D x=

A solution to 4x2 = 25 is: A x=

The first line with an error in solving 5(x + 1) − 2(x − 2) = 3 is: A 5x + 5 − 2x − 4 = 3

8

C 5

The solution to 15 − 3x = −5 + x is:

C 2x − 1 = 7

7

B 4

A x + 12 = 2(3x + 12)

D 4

B 4 − 3x = 2x − 11

6

When half a number is added to one-third of a number, the answer is 5. The number is: A 3

If y = 3 − 5x and y = −17 then x is: A 88

4

D x=2

x The solution to --- – 2 = 1 is: 3 A x=3 B x=9 C x=1

3

9

15

-----B x = − 25 4

D x=

5 --4

The area A of a circle is given by A = πr 2. The value of r when A = 100 is closest to: A 17.7

B 31.8

C 10

D 5.6

In symbols ‘four times a number plus eight is always less than 90’ is: A 4 × 8 < 90

B 4n + 8 ≤ 90

C 4n + 8 > 90

D 4n + 8 < 90

A solution to 4 – 3x < 10 is: A x = −4

B x = −3

C x = −2

D x = −1

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

16

The number that is not a solution of 2x + 1 ---------------- ≤ 5 is: 2 A 5 B 4 C 3 D 2

A solution to 5 − 2x > 11 + 3x is:

17

A x = 16

-----B x = − 16 5

C x = − 6--5-

D x = − 4--5-

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–3

4–6

7, 8

9, 10

11, 12

13

14, 15

16, 17

A

B

C

D

E

F

G

H

A. LINEAR EQUATIONS Linear equations are equations of the form (or can be simplified to the form) ax + b = 0, where a and b are constants and x is the unknown (or variable).

Example 1 Solve: a 7x − 9 = −5 a

b 17 = 8 – 4x

7x − 9 = −5 ∴ 7x − 9 + 9 = −5 + 9 ∴ 7x = 4 7x 4 ∴ ------ = --7 7 ∴x =

b (+9) (÷7)

17 = 8 – 4x ∴ 17 – 8 = 8 – 8 – 4x ∴ 9 = −4x 9 – 4x ∴ ------ = --------–4 –4

(–8) (÷ − 4)

∴ − 9--4- = x

4 --7

∴ x = − 9--4∴ x = −2 1--4-

Exercise 13A 1

Solve for x : a x + 3 = 10 e 5x + 8 = 2 i 6 + 7x = −2 m 6 − x = −5 q 3 − 7x = −2 u 8 = 3 − 2x

b f j n r v

3x = −9 4x − 9 = 1 5 = 3x + 7 −4x = 15 17 − 2x = −1 6 = −1 − 7x

c g k o s w

3x + 6 = 0 8x – 6 = 10 6x – 7 = −1 3 – 2x = 7 11 = 3 – 2x −15 = 3 – 6x

d h l p t x

3x − 4 = −6 3x + 6 = 7 −1 = 2x + 6 5 − 4x = −7 15 − 2x = −1 11 = −4 – 3x

401

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402

Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

Example 2 Solve: m ---- – 5 = – 2 3 m ---- – 5 = – 2 3 m ∴ ---- − 5 + 5 = −2 + 5 3

(adding 5 to both sides)

m ∴ ---- = 3 3 m ∴ ---- × 3 = 3 × 3 3

(multiplying both sides by 3)

∴m=9

2

3

Solve for x: x a --- + 3 = 8 2 x d --- + 3 = −4 6

x b --- − 1 = 4 3 x e --- − 2 = 4 7

c f

x --- + 2 = −3 5 x ------ − 6 = −1 10

Check the given solution by substitution and say whether or not it is correct. a 2x + 8 = 15 (x = 7) b 7 + 5x = 9 (x = 2) x c −15 = 6 − 7x (x = 3) d --- − 3 = 6 (x = 9--5- ) 5

Example 3 If y = 5x − 3 find x when y = −18. y = 5x − 3

4

−18 = 5x − 3

(substitute the value for y )

−15 = 5x

(add 3 to both sides)

−3 = x ∴ x = −3

(divide by 5)

a If y = 3x − 5, find x when y = 5

b If y = 4x + 2, find x when y = 11

c If y = 7 − 5x, find x when y = 0

d If y = 4 − 3x, find x when y = −3

e If y = 5 − 7x, find x when y = −5

f

If y = 3x − 5, find x when y = 8

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

B. PRONUMERALS ON BOTH SIDES OF AN EQUATION When you are solving equations with pronumerals on both sides, as well as adding and subtracting numbers from both sides, you may have to add and subtract pronumerals from both sides. The first step to adding or subtracting pronumerals is to move them to one side. It does not matter which side. Next add or subtract to move the numbers to the other side of the equation.

Example 1 Solve: a 5x + 2 = 3x − 5 a

b 15 − 2x = 11 + x

5x + 2 = 3x − 5 ∴ 5x + 2 − 3x = 3x − 3x − 5 (− 3x) ∴ 2x + 2 = −5 ∴ 2x = −7 (− 2) ∴ x = − 7--2(÷ 2) ∴ x = −3 1--2-

b

15 − 2x = 11 + x ∴ 15 − 2x + 2x = 11 + x + 2x ∴ 15 = 11 + 3x ∴ 15 − 11 = 11 − 11 + 3x ∴ 4 = 3x

Always do the same thing to both sides.

∴ 4--3- = x ∴ x = 1 1--3-

Exercise 13B 1

2

Solve the following equations with integer solutions. a 5x + 2 = 2x + 14 b 3x + 7 = 11 − x d 3x − 4 = 5x − 2 e 3−x=x+7 g 2x − 3 = x + 6 h 5x − 9 = 1 + 6x

c 5 + x = 8 − 2x f 4 − 2x = 3 − x i 3x − 5 = 7 − x

Solve: a 8x + 7 = 4x − 2 d x − 3 = 5x + 7 g 2x + 5 = 9 − 2x j 5a + 3 = a − 1 m 11a − 7 = 5a + 12

c f i l o

b e h k n

7x + 3 = 2x + 7 3 + x = 17 + 4x 3x − 5 = 5x + 9 4 − 3s = 2s + 17 3y − 5 = −14 − 2y

5 + 2x = 11 − x 15 − 3x = 2 − x 5 − 7x = 3x + 2 9x − 4 = 3 + 4x 7p = 15 − 3p

(+ 2x) (− 11) (÷ 3)

403

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404

Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

Example 2 By substituting, check the solutions to the following equations. a 2x − 5 = 10 − 3x a

3

(x = 3)

Is 2x − 5 = 10 − 3x ? Is 2(3) − 5 = 10 − 3(3) 1=1 ∴ x = 3 is the solution

b 5x + 2 = 2x − 7 b

(x = 2)

Is 5x + 2 = 2x − 7 ? Is 5(2) + 2 = 2(2) − 7 12 ≠ −3 ∴ x = 2 is not the solution

By substituting, check the solutions of the following equations. a 3x + 9 = 4 + 2x (x = 1) b 9a + 2 = 7a − 4 c 7a − 5 = 3 − a (a = 2) d 15 − 2x = 6 + x 5 e 2x − 3 = 7 − 4x (x = --3- ) f 5x − 7 = 3 + x

Example 3 Solve: a 5(x + 1) − 2(x − 2) = 7 a 5(x + 1) − 2(x − 2) = 7 ∴ 5x + 5 − 2x + 4 = 7 ∴ 3x + 9 = 7 ∴ 3x + 9 − 9 = 7 − 9 ∴ 3x = −2 ∴ x = − 2--3b

3(x + 1) = 5x + 3(2x − 1) 3x + 3 = 5x + 6x − 3 3x + 3 = 11x − 3 3x − 3x + 3 = 11x − 3x − 3 3 = 8x − 3 6 = 8x 6 8x --- = -----8 8 3 --4

=x

x=

3 --4

b 3(x + 1) = 5x + 3(2x − 1) (expanding the brackets) (collecting ‘like’ terms) (subtracting 9 from both sides)

(expanding the brackets) (collecting ‘like’ terms) (subtracting 3x from both sides) (adding 3 to both sides) The number and its sign in front of the brackets are multiplied by each term within the brackets.

(a = −3) (x = 3) (x = 3 1--2- )

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

4

5

Solve for x given that all answers are integers. a 3(x + 1) − 2(x − 4) = 13 c 4(x − 5) + 5(x + 1) = 12 e 4(x − 2) = 3x + 4(x − 2) g 4 − x = 2 − 3(x + 2)

b d f h

2(x − 5) + 3(x + 2) = −9 2(x − 1) = 3(x + 5) − 22 2(x − 1) = 4(2x + 1) − 9x 6 − 2(x + 5) = 2(2x − 1) −5x

Solve for x : a 2(x + 1) − 1 = 8 c 3(x + 2) − 7 = 11 e 4(2x − 1) + 7 = 0 g 3 − 2(x + 1) = −4 i 5x − 4(4 − x) = x + 1 k 2(x − 1) = 1 − (3 − x)

b d f h j l

5(1 − 3x) = −4 2(x + 1) + 3(x − 1) = 6 11 − 2(x − 1) = 7 7 − (2 – x) = 2x 3 − x = 5 − 2(x + 1) x + 7(4 − x) = 2x + 3(x − 1)

Example 4 If y = 3 − 5(x + 4), find x when y = −32. y = 3 − 5(x + 4) −32 = 3 − 5(x + 4) −32 = 3 − 5x − 20 −32 = −17 − 5x −32 + 17 = −17 + 17 − 5x −15 = −5x 3=x ∴x=3

6

a b c d e f g h

(substituting y = −32) (collecting like terms) (add 17 to both sides) (dividing by −5)

If y = 7 − 3(x + 2), find x when y = −5. If y = 5 − 4(x − 3), find x when y = 37. If y = 4 − 5(2x − 5), find x when y = 12. If y = 14 − 3(2x − 8), find x when y = 0. If y = 3x − 2(5x + 1), find x when y = −16. If y = 4x − 3(5 − 2x), find x when y = 8. If y = 3(2x − 1) − 4(x + 2), find x when y = −3. If y = 4(1 − 3x) − 2(1 – x), find x when y = 2.

405

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406

Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

Example 5 The equation 3(x − 4) = 5 – (2x + 3) has solution x =

14 -----5

.

Change one term or sign to make the solution x = 3. 3(x − 4) = 5 − (2x + 3) For x = 3 to be a solution, both sides of the equation must be equal. LHS = 3(x − 4) RHS = 5 − (2x + 3) = 3(3 − 4) = 5 − (2(3) + 3) = 3(−1) = 5 − (9) = −3 =−4 To make LHS = RHS, we add 1 to the RHS. ∴ the equation becomes 3(x – 4) = 6 – (2x + 3) where the 5 becomes a 6.

7

Change one term or sign so that each of the following equations has solution x = 2. a 2(x − 3) = 1 − (2x − 5) b 3(x − 1) = 4 − (3x − 2) c 5(2x + 3) = 2 − (5x + 1) d 6x − 5 = 5(3x − 1)

8

Write three equations of your own (with at least four terms) that have solution x = 3.

Investigation 1 WM: Reasoning, Applying Strategies

Spreadsheet Use a spreadsheet program to solve the equation 3x − 7 = 9 − x given that the solution is an integer. The spreadsheet needs to have three columns labelled x, 3x − 7, and 9 − x. In cell A1 enter x, in cell B1 enter 3x − 7, in cell C1 enter 9 – x, in cell A2 enter 0, in cell B2 enter = A2 * 3 − 7, in cell C2 enter = 9 − A2.

x

3x − 7

9−x

0

–7

9

1

–4

8

2

–1

7

3

2

6

It can be seen that x = 4 is the solution.

4

5

5

x = 4 gives the same value for both expressions.

5

8

4

6

11

3

Use the fill down command to find the values for each side of the equation. The answer is the value of x that gives the same value in each column.

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

1

Change the spreadsheet to solve the following equations with integer solutions. a 5x − 3 = 53 − 2x b 3x + 5 = 35 − 2x c 19 − 2x = 7x − 44 d 6x + 11 = 41 − 4x e 3x − 17 = 33 − 7x f 9x + 15 = 79 − 7x

2

Change the spreadsheet to solve the following equations with negative integer solutions. a 4x − 3 = 13 + 6x b 7x − 3 = −25 + 5x c 8 − 7x = 2 − 8x d 3 − 5x = 39 − 3x

3

Explain how to modify the spreadsheet to solve equations that do not have integer solutions.

4

Solve: a 3x − 7 = 6 − 9x

b 5x + 23 = 2x − 8

c 8 − 7x = 4x + 59

C. EQUATIONS WITH FRACTIONS To solve equations involving algebraic fractions, multiply both sides by the lowest common denominator.

Example 1 Solve for x: x 2 a --- = --3 5 a

x 2 --- = --3 5 x 2 ∴ 15 × --- = --- × 15 3 5 ∴ 5x = 6 6 ∴ x = --5

4 x b --- = --7 3 b

4 x --- = --7 3 4 x ∴ 21 × --- = --- × 21 7 3 ∴ 12 = 7x 12 ∴ x = -----7

Our first step is to eliminate the fraction by multiplying both sides by the LCD.

c c

x --- = 4 2 x --- = 4 2 x ∴ 2 × --- = 4 × 2 2 ∴x=8

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Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

Example 2 Solve for x: 4x + 3 a ---------------- = – 2 5 a

b

4x + 3 –2 ---------------- = -----5 1 4x + 3 –2 5 × ---------------- = ------ × 5 5 1 ∴ 4x + 3 = −2 × 5 ∴ 4x + 3 = −10 ∴ 4x + 3 − 3 = −10 − 3 ∴ 4x = −13 -----∴ x = − 13 4

1 --3

(2x − 1) = −4 1 --3

b

(2x − 1) = −4

Do the same to both sides.

3 × 1--3- (2x − 1) = −4 × 3 2x − 1 = −12 2x − 1 + 1 = −12 + 1 2x = −11

(subtracting 3 from both sides)

-----∴ x = − 11 2

(dividing both sides by 4)

Exercise 13C 1

Solve for x: x a --- = 5 3 x d --- – 2 = – 1 4 2x + 7 g ---------------- = 0 3 1 – 2x j --------------- = 3 2

2x b ------ = – 4 5 x–1 e ------------ = 6 2

c f

h

1 --2

(3x + 1) = −1

i

k

1 --5

(4 − 3x) = −1

l

x --- + 1 = – 5 2 x+5 ------------ = – 1 3 1 + 2x ---------------- = 6 7 1 --4

(5 − 2x) = −2

Example 3 Solve for x: 3x + 1 a ---------------- = 2 3 a

3x + 1 ---------------- = 2 3 3x + 1 ∴ 3  ---------------- = 2 × 3  3  ∴ 3x + 1 = 6 ∴ 3x = 5 x = 5--3-

3x – 1 2x b --------------- = -----5 7 b

3x – 1 2x --------------- = -----5 7 3x – 1 2x 35  --------------- =  ------ × 35  5   7 ∴ 7(3x − 1) = 5(2x) ∴ 21x − 7 = 10x ∴ 21x − 10x − 7 = 10x − 10x ∴ 11x − 7 = 0 ∴ 11x = 7 7 ∴ x = ----11

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

2

Solve for x : 2x + 1 1 a ---------------- = --3 2 c

4x + 1 b ---------------- = 2 5

3x – 2 --------------- = −2 4

When multiplying both sides of an equation by the same number we sometimes need to use brackets.

2x + 1 d ---------------- = 3 4

3x + 1 e ---------------- = 3 2

f

3x + 2 x – 1 ---------------- = -----------5 4

1–x x+2 g ------------ = -----------2 3

x+1 x h ------------ = --2 3

i

2x – 1 3x --------------- = -----7 5

j

x + 1 2x – 3 ------------ = --------------2 3

k

2x + 5 ---------------- = x + 4 3

l

2x + 7 ---------------- = x − 5 3

When either the LHS or RHS of a fractional equation has more than one term, we solve it by multiplying both sides of the equation by the lowest common denominator (LCD).

Example 4 Solve for x: 2x x a ------ – --- = 5 3 2

b

2x x a ------ – --- = 5 has LCD of 6 3 2 2x x ∴ 6  ------ − 6  --- = 6 × 5  3  2 ∴ 2(2x) − 3(x) = 30 ∴ 4x − 3x = 30 ∴ x = 30

3

Solve for x : x x a --- + --- = 2 2 5 x 5x e --- + ------ = 14 3 6

x 2x 5 b --- − ------ = --2 3 6 x 3x f --- + ------ = 6 1--23 4

x 3x --- − 3 = -----5 8

x 3x b --- − 3 = -----has LCD of 40 5 8 x 3x ∴ 40  --- − 40 × 3 = 40  ------  5  8

Multiply all terms by the LCD.

∴ 8(x) − 120 = 5(3x) ∴ 8x − 120 = 15x ∴ −120 = 15x − 8x ∴ −120 = 7x ---------∴ x = − 120 7

3x x ------ − --- = 11 2 8 2x x g ------ − --- = −2 5 2 c

x x d --- + --- = 5 2 4 x 7x h --- − 2 = -----3 12

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Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

Investigation 2 WM: Applying Strategies

Equation solver 1

Use a graphics calculator to solve the equation 3x + 8 = 6 − 2x. Instructions for a Casio CFX9850 GB Plus follow. 1 Select EQUA from MAIN MENU. Use brackets to 2 Select type solver by pressing F3. enter fraction 3 Enter equation parts. 3 x,θ,T

8

= . 6

2

x,θ,T

4 Press EXE to store. 5 Press F6 to solve. 6 Use F2 to delete the equation. 2

3

Solve: a 4x − 5 = 7 − 3x c 6 − 5x = 3 + 2x 4 – 5x e --------------- = 8 3

b 6 − 2x = 5x + 3 d 4(x + 3) = 7(4 − 3x) 6 – 2x 4x + 1 f --------------- + ---------------- = 3 5 2

Solve other equations.

D. PRACTICAL EQUATIONS Example 1 If twice a certain number is subtracted from 11, the result is 4 more than the number. Find the number. Let x be the number, ∴ 2x is twice the number and 11 − 2x is twice the number subtracted from 11. Also x + 4 is 4 more than the number. Thus 11 − 2x = x + 4 ∴ 11 = 3x + 4 ∴7 = 3x ∴ --73- = x ∴ the number is 7--3- (or 2 1--3- )

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

Exercise 13D 1

Solve the following problems. a When a number is trebled and then increased by 7, the answer is 19. Find the number. b When a number is subtracted from 11, the result is 5 more than the number. Find the number. c When a number is decreased by 3 and the result is doubled, the answer is equal to the original number. Find the number. d When half a number is added to one-third of a number, the answer is 30. Find the number.

Write the statement as an equation and then solve the equation.

Example 2 The sum of three consecutive even numbers is 132. Find the smallest one. Let x be the smallest even number. ∴ (x + 2) and (x + 4) are the other two even numbers. Now x + (x + 2) + (x + 4) = 132 ∴ 3x + 6 = 132 ∴ 3x = 126 ∴ x = 42 Thus 42 is the smallest even number.

2

Consecutive integers are whole numbers that follow one another.

a If two consecutive integers have a sum of 127, find the numbers. b If three consecutive integers add to 27, find the smallest of them. c Four consecutive integers have a sum of −6. Find the largest of them.

Example 3 If five more than a number is one more than twice the number, find the number. Let n be the number, then 5 + n = 2n + 1 4 + n = 2n 4=n ∴ the number is 4. 3

(subtracting 1) (subtracting n)

Solve the following problems. a If seven more than a number is three more than twice the number, find the number. b If four more than a number is eight more than three times the number, find the number. c If six more than twice a number is four more than four times the number, find the number. d If nine less than five times a number is three less than twice the number, find the number.

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Example 4 The sum of two numbers is 14. When one number is added to twice the other the result is 25. Find the numbers. Let n be one number. ∴ Since the numbers add up to 14, the other number is (14 − n), and ∴ n + 2(14 − n) = 25 ∴ n + 28 − 2n = 25 ∴ −n + 28 = 25 ∴ −n = −3 (subtracting 28 from both sides) ∴n=3 and since 14 − n = 14 − 3 = 11, the numbers are 3 and 11. 4

Solve the following problems. a The sum of two numbers is 10. When one number is added to twice the other, the result is 16. Find the numbers. b The sum of two numbers is 12. When one number is subtracted from three times the other, the result is 4. Find the numbers. c Two numbers differ by 2. Twice the smaller number is added to the larger number and the result is 14. Find the numbers. d Three consecutive even integers are such that the sum of the smaller two is equal to six more than the largest one. Find the integers. e Three consecutive integers are such that three times the sum of the larger pair is equal to five times the sum of the smaller pair. Find the numbers.

Example 5 Apples cost 13 cents each and oranges cost 11 cents each. If I buy 5 more apples than oranges, and the total cost of the apples and the oranges is $2.33, how many oranges and apples did I buy? Let x be the number of oranges bought. Type Number apples x+5 oranges x

Cost/unit 13 cents 11 cents Total

Thus 13(x + 5) + 11x = 233 Setting up a table ∴ 13x + 65 + 11x = 233 may help in solving money problems. ∴ 24x + 65 = 233 ∴ 24x = 168 ∴x=7 ∴ 7 oranges and 12 apples were bought.

Total value 13(x + 5) cents 11x cents 233 cents

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

5

10 5

Solve these problems using a table to assist you. a I have 30 coins in my pocket. The coins are either 5-cent or 10-cent pieces and their total value is $2.10. How many 5-cent coins do I have? b Bananas cost 12 cents each and grapefruit cost 23 cents each. If I buy 3 more bananas than grapefruit, the total cost will be $2.81. How many bananas do I buy? c If I have 10-cent and 20-cent coins only with a total value of $6.90 and I have 36 coins altogether, how many 10-cent coins do I have? d Joe has a collection of 5-cent and 20-cent pieces that have a total value of $3.85. He has two less 20-cent coins than 5-cent coins. How many 5-cent coins has he? e I wish to blend brand A coffee at $7 a kilogram with brand B coffee at $11 a kilogram. If the total weight of the mixture is 10 kg and the total cost of blended mixture is $86, how many kilograms of A are mixed with B?

Example 6 At the moment Jack is 5 years older than Mim. In 7 years Mim’s age will be threequarters of Jack’s age. How old are they at present?

Age now

Age in 7 years

Mim

x yr

(x + 7) yr

Jack

(x + 5) yr

(x + 12) yr

Thus x + 7 = 3--4- (x + 12) ∴ 4(x + 7) = 3(x + 12) (multiplying both sides by 4) ∴ 4x + 28 = 3x + 36 ∴ x + 28 = 36 (subtract 3x from both sides) ∴x=8 ∴ Mim is 8 years old and Jack is (x + 5) = 13 years old. 6

Solve the following problems. a A man is currently 3 times as old as his son. In 11 years from now he will be twice as old as his son will be then. How old is his son now? b At present Guido is 8 years older than Bob. If Guido was one year younger his age would be double Bob’s age. How old is Bob? c In 5 years’ time Pam will be twice as old as Sam was two years ago. Pam is 8 years older than Sam. How old is Sam? d The sum of Peter’s and Susan’s ages is 20 years. If Peter’s age is doubled, it will be five more years than Susan’s age trebled. How old is Susan?

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Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

Investigation 3 WM: Reasoning, Applying Strategies, Communicating

Square numbers 1

Find 3 × 3 and −3 × −3. Compare the answers.

2

Find: a 4 × 4 and −4 × −4 c −6 × −6 and 6 × 6

b 5 × 5 and −5 × −5 d −10 × −10 and 10 × 10

3

What do you notice about the answers to each part in question 2?

4

The solution to the equation x2 = 49 is found by finding a number that when multiplied by itself gives 49. What are the two answers?

5

The solution to x2 = 64 is x = 8 or −8. Explain why there are two answers.

6

Are there always two answers to x2 = c where c is a number? Explain.

E. SIMPLE QUADRATIC EQUATIONS From investigation 3 there are two solutions to the equation x2 = c This equation is called a quadratic equation because the variable x has a power of 2.

Example 1 Solve: a x2 = 25

b x2 = 169

a x2 = 25 x = ± 25 (square root both sides) x = ±5

b x2 = 169 x = ± 169 x = ±13

± is the symbol for plus or minus.

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

Exercise 13E 1

Solve: a x2 = 9 f x 2 = 121

b x 2 = 16 g x 2 = 36

c x 2 = 64 h x 2 = 81

d x 2 = 144 i x 2 = 100

e x 2 = 49 j x2 = 4

Example 2 Solve:

2

a x 2 = 10

b x 2 = 43

a x 2 = 10 x = ± 10 x = ±3.16 (2 d.p.)

b x 2 = 43 x = ± 43 x = ±6.56 (2 d.p.)

Solve, giving answers to 2 decimal places. a x 2 = 12 b x 2 = 51 c x 2 = 19 f x 2 = 28 g x 2 = 68 h x 2 = 91

d x 2 = 47 i x 2 = 193

e x 2 = 83 j x 2 = 200

Example 3 Quadratic equations have two answers.

Solve: a 5x 2 = 80 a

5x 2 = 80 2

5x 80 --------- = -----5 5 2 x = 16

3

b 3x 2 = 75 b 3x 2 = 75 2

3x 75 --------- = -----3 3 2 x = 25

x = ± 16

x = ± 25

x = ±4

x = ±5

Solve: a 2x 2 = 18 e 7x 2 = 175

b 5x 2 = 180 f 10x 2 = 160

c 8x 2 = 72 g 7x 2 = 252

d 3x 2 = 48 h 6x 2 = 294

Example 4 Solve: a 9x 2 = 25

b 81x 2 = 49

☞

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a

9x 2 = 25

b

2

81x 2 = 49 2

9x 25 --------- = -----9 9 25 x 2 = -----9

81x 49 ------------ = -----81 81 49 x 2 = -----81

25 x = ± ---------9 5 x = ± --3 4

Solve: a 4x 2 = 49 e 49x 2 = 144

49 x = ± ---------81 7 x = ± --9

b 100x 2 = 81 f 121x 2 = 64

c 25x 2 = 16 g 144x 2 = 49

d 81x 2 = 16 h 81x 2 = 100

Example 5 Solve: a

3x 2 = 21

b

5x 2 = 12

a

3x 2 = 21

b

5x 2 = 12

2

2

3x 21 --------- = -----3 3

5x 12 --------- = -----5 5 12 2 x = -----5

x2 = 7

5

x=± 7

12 x = ± -----5

x = ±2.65 (2 d.p.)

x = ±1.55 (2 d.p.)

Solve, giving answers to two decimal places if necessary. a 7x 2 = 56 b 4x 2 = 12 2 d 13x = 47 e 7x 2 = 18

Example 6 Solve 4x2 − 5 = 20 4x 2 − 5 = 20

(add 5 to both sides)

2

4x = 25 25 x 2 = -----4 x=± 5 x = ± --2

25 -----4

c f

11x 2 = 66 5x 2 = 23

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

6

Solve, giving answers to two decimal places if necessary. a 9x 2 − 8 = 56 b 4x 2 + 3 = 52 c 81x 2 + 7 = 107

d 49x 2 − 20 = 5

Investigation 4 WM: Applying Strategies, Communicating, Reflecting

Square roots Suppose the square root key is the only key on your calculator that does not work. How can 5 be determined? Here’s a method. Step 1: Step 2: Step 3: Step 4:

Make an estimate of

5 Find the average of a and --- , and call it b. a 5 Find the average of b and --- , and call it c. b And so on.

For example, suppose ∴b= ∴c=

5 and call it a.

5
2 i.e. a = 2

1 5 --- (2 + --- ) = 2.25 2 2 1 5 --- (2.25 + ----------- )
2 2.25

2.2361

∴ d = _____ ∴ e = _____ ∴ f = _____ 1

Find d, e and f in the above example.

2

Find

3

Start with a = 3 and find b, c, d, e and f. Does

4

Can you explain why the method works? 5 is a solution of the equation x 2 = 5, and notice that as the process continues Hint: 5 b
c, c
d, d
e, …, i.e. x
1--2- (x + --- ). x

5

Explain how you would modify the above method to calculate: a 7 b k

5 from your calculator. Does the method appear to work? 5 result?

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Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

F. SUBSTITUTION INTO FORMULAS A formula is an equation that connects two or more variables.

It is normal for one of the variables to be expressed in terms of the other(s). The subject of the formula is the variable that is written in terms of the other variables.

Formula substitution If a formula contains two or more variables and we know the value of all but one of them, we can use the formula to find the value of the unknown variable. Follow the method below. 1

Write down the formula.

2

State the values of the known variables.

3

Substitute into the formula to find an equation with one variable.

4

Solve the equation for the unknown variable.

Example 1 The area of a triangle is given by A = 1--- bh, where A is the area, b is the base of 2 the triangle, and h the height. a Find the area of a triangle with b = 8 m and h = 7 m. b Find the height when A = 30 cm2 and b = 5 cm. a A = 1--2- bh A=

1 --2

b

×8×7

A = 1--2- bh 30 =

= 28 2

The area is 28 m .

1 --2

×5×h

60 = 5h 60 5h ------ = -----5 5 ∴ h = 12 The height is 12 cm.

Exercise 13F 1

The area of a rectangle is given by A = lb, where A is the area, l the length and b the breadth. a Find the area of a rectangle with length 16 cm and breadth 5 cm. b Find the length of a rectangle with area of 30 cm2 and breadth 5 cm.

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

2

The formula for finding the circumference (perimeter) C of a circle of diameter d is C = πd. Find: a the circumference of a circle of diameter 11.4 cm b the diameter of a circle with circumference 250 cm

d

3

The formula for calculating the circumference C of a circle of radius r is C = 2πr. Find: a the circumference of a circle of radius 8.6 cm b the radius of a circle of circumference 100 m

4

When a car travels a distance of d km in time t h, the average speed, s kmph, for the journey d is given by the formula s = --- . Find: t a the average speed of a car that travels 200 km in 2 h b the distance travelled by a car in 3 1--4- h if its average speed is 80 kmph c the time taken for a car to travel 865 km at an average speed of 110 kmph

Example 2 When a stone is dropped down a well the total distance fallen, D metres, is given by the formula D = 1--- gt 2, 2 where t is the time of fall (in seconds) and g is the gravitational constant of 9.8 m s−1. Find:

D

a the distance fallen after 5 s 1 - th s) taken for the stone to b the time (to the nearest --------100 fall 100 m a

D = 1--2- gt 2 where g = 9.8 m s−1 and t = 5 s

∴D=

1 --2

× 9.8 × 52 m

Calculator:

0.5 ∴ D = 122.5 m ∴ the stone has fallen a distance of 122.5 metres. b D = 1--2- gt 2 where D = 100 m, and g = 9.8 m s−1 ∴ 100 = 1--2- × 9.8 × t 2 ∴ 100 = 4.9t 2 100 ∴ ---------- = t 2 4.9 100 ∴ t = ---------- (as t > 0) 4.9 ∴ t
4.5175 … ∴ the time taken is 4.52 seconds.

5

×

9.8

×

5

=

冑

x = 2

Calculator: 100

÷

4.9

When a cricket ball is dropped from the top of a building the total distance fallen is given by the formula D = 1--2- gt 2 where D is the distance in metres and t is the time taken in seconds. Find, given that g
9.8 m s−1, a the total distance fallen in the first 3 s of fall b the height of the building when the time of fall to hit the ground is 5.13 s

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6

A circle’s area A is given by A = πr 2 where r is its radius length. Find: a the area of a circle of radius 5.6 cm b the radius of a circular swimming pool that must have an area of 200 m2

7

A cylinder of radius r and height h has volume given by V = πr 2 h. Find: a the volume of a cylindrical tin can of radius 12 cm and height 17.5 cm b the height of a cylinder of radius 4 cm given that its volume is 80 cm3 c the radius of copper wire of volume 100 cm3 and length 2 km

8 h

D

Earth

9

h

The formula D
3.56 h gives the approximate distance (D km) to the horizon that can be seen by a person with eye level h metres above the level of the sea. Find: a the distance of the horizon when a person’s eye level is 10 m above sea-level b how far above sea-level a person’s eye must be if the person wishes to see the horizon at a distance of 30 km

The formula for calculating the total surface area of a sphere of radius r is given by A = 4πr 2. Find: a the total surface area of a sphere of radius 6.9 cm b the radius of a spherical balloon that is to have a surface area of 1 m2. (Answer in cm.)

10

point of support

l

. 11

r

pendulum

r

The time taken for one complete swing of a simple pendulum is given by T = 1--5- l where l is the length of the pendulum (in cm) and T is the time (called the period) in seconds. Find: a the time for one complete swing of the pendulum if its length is 50 cm b the length of a pendulum if it is to have a period of exactly 1 s

To find the area of a triangle with sides a, b and c units long we find s, its semi-perimeter, using the formula a+b+c s = ---------------------- , and then use A = s ( s – a ) ( s – b ) ( s – c ) . 2 A triangle has sides of length 5 cm, 6 cm and 7 cm. Find its semi-perimeter and hence its area.

a

c

b

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

12

The formula for energy given speed and mass is E = 1--2- mv 2. a Calculate E if m = 100 and v = 5. b Calculate m given E = 1400 and v = 18. c Calculate v if E = 5000 and m = 200. Note v > 0.

13

The volume of a sphere radius r is given by V = 4--3- πr 3. a Find V when r = 2.5. b Find r when V = 1000.

14

The surface area of a closed cylinder is given by A = 2πr 2 + 2πrh. Find h when A = 1800 and r = 3.

15

y2 – y1 -. The formula for gradient, m, is given by m = --------------x – x 2 1 Find: a y2 when x1 = 5, x2 = 3, y1 = 2 and m = −2 b x1 when x2 = 7, y1 = −3, y2 = 4 and m = 1

G. INEQUALITIES These symbols are convenient devices for representing inequalities. Symbol

Meaning

<

less than

>

greater than

≤

less than or equal to

≥

greater than or equal to

≠

not equal to

Example 1 Write in algebraic form: a Three times a number is always smaller than ten. b Twice a number is larger than or equal to eight. c Four more than three times a number is greater than fifteen. Let the unknown number be x. Then, a 3x < 10 b 2x ≥ 8 c 3x + 4 > 15

You may need −− the 3√ button.

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Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

Exercise 13G 1

Write in algebraic form: a Four times a number is always smaller than thirteen. b Five times a number is always greater than fifty. c Ten times a number is greater than or equal to eighty. d Two more than three times a number is less than thirty-five. e Three more than four times a number is more than seventy. f Eight less than a number is less than or equal to thirty. g Thirteen less than twice a number is greater than twenty. h Nine less than twenty times a number is more than ten. i Ten more than half a number is less than nineteen. j Five more than one-third of a number is greater than six.

Example 2 a Is x = 3 a solution to the inequality 3x – 2 > 5? b Is x = −5 a solution to the inequality 5 – 2x ≤ 6? c Is x = 7 a solution to the inequality 3 – 5x > 2x + 1? Check by substituting. a 3x – 2 > 5 Is 3(3) – 2 > 5? Is 9 – 2 > 5? 7 >5 which is true ∴ x = 3 is a solution

2

b

5 – 2x ≤ 6 c 3 – 5x > 2x + 1 Is 5 – 2(−5) ≤ 6? Is 3 – 5(7) > 2(7) + 1? Is 5 + 10 ≤ 6? Is − 32 > 15? 15 ≤ 6 which is false which is false ∴ x = −5 is not a solution ∴ x = 7 is not a solution

Check if the value in brackets is a solution to the inequality. (x = 4)

b 3 – 4x ≥ 13

(x = −5)

c 5x + 1 < 11 (x = 2)

x d --- + 3 > 0 2 3x – 1 f --------------- ≤ 3 4 x–3 h ------------ <7 2

(x = 10)

a 3x – 2 > 7

e 7x – 8 ≥ 5

(x = 10)

g 7 – 4x ≤ 3

(x = −2)

(x = 0)

(x = −3)

j

3x + 7 < 11 – 2x

(x = 3)

k 4x + 1 ≥ 6

(x = 1)

l

7x + 3 < 5 – x

(x = 3)

m 2>5–x

(x = −3)

n 3(x + 1) – 2 > 5 – 5(1 – x)

i

x+6>0

(x = −7)

(x = −1)

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

Example 3 Find a solution to these inequalities using the guess-and-check method. a 5 – 7x ≥ 10

b 3x – 5 < 8 – 5x

a Try x = 0 5 – 7(0) ≥ 10 5 ≥ 10 false ∴ x = 0 is not a solution Try x = 5 5 – 7(5) ≥ 10 − 30 ≥ 10 false ∴ x = 5 is not a solution Try x = −10 5 – 7(−10) ≥ 10 75 ≥ 10 true ∴ x = −10 is a solution

b Try x = 4 3(4) – 5 < 8 – 5(4) 7 < −12 false ∴ x = 4 is not a solution Try x = 0 3(0) – 5 < 8 – 5(0) − 5 < 8 true ∴ x = 0 is a solution

3

Find three solutions to each of the following inequalities, using the guess-and-check method. 4–x a 3x ≤ −9 b ------------ ≥ −3 c −2x ≥ 5 5 d 4 + 3x ≥ 7 + 2x e 5 – 6x ≤ − 7 f 3x – 4 > 5x – 2 x g 2>4–x h 4 – 2x ≥ 3 – x i --- > −2 3

4

How many solutions are there to the inequality: a 5x – 12 > 3 b 5 – 2x ≤ 3 + 4x ?

Investigation 5 WM: Reflecting, Reasoning

Inequalities Example 1 Change both sides of the inequality 5 < 7 by the operation [x(−3)] and insert a sign to make the new inequality true. 5<7 − 15 –21 Thus 15 > −21

(multiply each number by –3) (insert the inequality sign)

☞

423

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Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

1

Both sides of the inequality 6 > 4 are changed by the operation in brackets. Insert > or < signs to make the new inequality true. a 12 8 [×2] b 3 2 [÷2] c 9 7 [+3] d 1 –1 [−5] e –12 –8 [×(−2)] f –3 –2 [÷(−2)] g 4 2 [+2] h 8 6 [−(−2)]

2

Both sides of the inequality –8 < 4 are changed by the operation in brackets. Insert < or > signs to make the new inequality true. a –2 1 [÷4] b –24 12 [×3] c 1 13 [+9] d –15 –3 [−7] e 4 –2 [÷(−2)] f 24 –12 [×(−3)] g 2 –2 [÷(−4)] h –12 0 [+(−4)]

3

Using your answers to questions 1 and 2 write a set of rules for working with inequalities.

4

Inequalities may be written in equation form, as a number line graph, or in words. Copy and complete this table. Inequality x >3 x ≤3 –4 < x ≤ 2 x < 0 or x > 4

Graph –1

–3

0

–2

1

–1

Description

2

0

3

1

2

4

3

4

5

all numbers greater than 3

5

all numbers less than or equal to 3

–4 –3 –2 –1

0

1

2

3

4

5

all numbers between –4 and 2, including 2

–3 –2 –1

1

2

3

4

5

6

all numbers greater than 4 or less than 0

0

x ≥ –2 –3

–2

–1

0

1

2

3

all numbers between 3 and 8, including 3

–1

–8

0

–7

1

2

–6

3

–5

4

–4

5

6

–3

7

–2

all numbers greater than or equal to –8 x ≤ –3 or x ≥2

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

–8 –7 –6 –5 –4 –3 –2 –1 0

1

all numbers less than 0 all numbers less than 2 or greater than 3 4>x 5≤ x –5 –4 –3 –2 –1

5

a b c d

0

1

2

3

What does the filled in dot mean? What does the open dot mean? Can x < 0 and x > 4 both be true at the same time? Why does 0 < x > 4 not make sense?

H. SOLVING INEQUALITIES Inequalities are algebraic sentences containing at least one of the symbols >, <, ≥ or ≤.

For example, 3x + 2 ≥ −7 is an inequality. To solve inequalities, we handle the situation as if they are ordinary equations and carry out the same operation to each side of the inequality. 3x + 2 Left Hand Side (LHS)

≥

−7 Right Hand Side (RHS)

In investigation 5 the rules below were discovered.

Rules

Rules for inequalities • •

Note:

The reverse of > is <. The reverse of ≥ is ≤.

If we add or subtract from both sides of an inequality the inequality sign is maintained. If we multiply or divide both sides of an inequality by: a positive number, we maintain the same inequality sign a negative number, we reverse the inequality sign.

425

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426

Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

For example, if 3x ≥ 12 then x ≥ 4

(dividing both sides by 3)

but, if –3x ≥ 18 then x ≤ −6

(dividing both sides by –3)

Example 1 Solve for x, graphing the solution on the number line. b 3 – 4x ≤ 15

a 3x – 16 > 8 a

3x – 16 > 8 ∴ 3x > 24 ∴ x >8 5

b

6

c 2 > 5 – 2x

(adding 16 to both sides) (dividing both sides by 3) 7

8

3 – 4x ≤ 15 ∴ – 4x ≤ 12 -----∴ x ≥ 12 –4

9

10

11

(subtracting 3 from both sides) (÷ both sides by –4 and reversing the inequality)

∴ x ≥ –3 –6

c

–5

–4

–3

2 > 5 – 2x 2 – 5 > 5 – 5 – 2x –3 > –2x –3 ------ < x –2 x>

–1

0

(subtracting 5 from both sides) (dividing by –2 and reversing the inequality)

3 --2

–2

–2

(changing the x to the LHS) –1

0

1

2

3

4

Exercise 13H 1

Solve for x and graph the solutions on the number line. a x+3>5

b x–2≤3

e 2x ≤ − 12

f

x–3>−4

j

3x ≤ − 9

i

x+5>0

c 3x ≥ 18 x g --- > − 4 4 k −2x ≥ 6

x d --- < −2 3 h x + 4 < −3 l

−5x < 10

m 2x + 1 > 13

n 3x – 2 ≤ 7

o 4x + 1 ≥ 6

p 3–x<0

q 5 – 2x ≥ 0

r 2>5–x

s 3 – 4x ≥ 13

t

5 – 6x ≤ − 7

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

2

Solve for x and graph the solutions on the number line. x x a --- > −3 b ------ ≤ 4 2 –3 2x + 1 3x – 1 d ---------------- ≥ 5 e --------------- ≤ 3 2 4 x–3 4–x g ------------ < 7 h ------------ ≥ −2 –2 5

c f i

x --- + 3 > 0 2 1 --- (3 – 2x) ≥ −1 2 3 – 5x --------------- > −1 –4

Extension

Example 2 Solve for x: a 3 + 2x < 13 – 3x a

3

4

b 5 – 3x ≥ x + 7

3 + 2x < 13 – 3x ∴ 3 + 2x + 3x < 13 – 3x + 3x ∴ 3 + 5x < 13 ∴ 3 – 3 + 5x < 13 – 3 ∴ 5x < 10 ∴ x<2

b

5 – 3x ≥ x + 7 ∴ 5 – 3x – x ≥ x – x + 7 ∴ 5 – 4x ≥ 7 ∴ 5 – 5 – 4x ≥ 7 – 5 ∴ −4x ≥ 2 ∴ x ≤ − 1--2

Solve for x. a 4 + 3x ≥ 5 + x c 3x – 4 > 5x – 2 e 3x + 7 < 11 – 2x g 3x + 2 > x – 5 i 5 – 2x ≥ x + 4 k 3(x – 1) > x + 2 m 3x – 2 > 2(x – 1) + 5x o 5 – (x + 2) ≤ 2(2x – 1) q 3 – x ≥ 5 – 2(x + 1)

b d f h j l n p r

Solve for x. 3+x 2–x a ------------ ≥ -----------4 3 7x – 5 3x – 1 c --------------- ≤ --------------5 4 5x – 3 4 – x e --------------- > -----------2 3 3 – 2x 7x + 1 ----------------------------g < 4 5 11 – 4x i ------------------- – 3 > 5 7 4x – 3 7x – 5 k --------------- + --------------- ≥ – 1 2 3

6x – 5 b --------------2 4x + 3 d ---------------3 5 – 3x f --------------7 9 – 2x --------------h 3 2x – 7 j --------------4 8 – 3x l --------------2

3 + 3x < 13 + x x – 3 ≤ 5x + 7 4 – 2x ≥ 3 – x 2x – 3 < 5 – 7x 7 – 3x ≤ 5 – x 5 – 2x ≤ 2(x + 2) 1 – (x – 3) ≥ 2(x + 5) – 1 3(x + 1) – 2 > 5 – 2(x – 1) x + 7(4 – x) < 2 – 5(1 – x)

3x + 1 < ---------------4 1–x > -----------5 2+x ≤ -----------4 2x + 1 --------------≥ 4 –3<7 4x + 1 – ---------------- ≤ 1 3

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Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

non-calculator activities

1

21 Write ------ as a mixed numeral, 2

2

Write 134.7652 correct to 2 decimal places.

3

Find the value of 2341 × 20.

4

Increase $100 by 10%

5

If −20 ÷ 5 =

6

How many fortnights are there in two years?

7

1 Write 5 − --- as a mixed fraction. 5

8

What is the value of 0.6 × 0.3?

9

2367.5 × 10 =

10

× 2, what is the missing number?

30% is the same as: 1 A --3

3 B -----10

C 0.03

D none of these

11

The fraction ------ has a value of 7. What is the missing number? 5

12

What fraction is 3 kg of 9 kg? Give your answer in simplest fraction form.

13

The temperature at 6 p.m. at Perisher was 4°C but it fell 2°C each hour for the next 6 hours. What was the temperature at midnight?

14

9.9 Estimate the value of ----------------------- giving your answer as a whole number. 3.1 + 1.8

15

Find the missing number in the box. 143.6508 = 143 + 0.6 + 0.05 +

16

If Kritizia earns $250 for 10 hours work, how much will she receive for 3 hours work?

17

Jade knows that 26 × 156 = 4056. Use this data to find the answer to Jade’s question 4056 ÷ 156 = 2

18

What is

25 + 4 ?

19

What is the next number in the sequence 2, 10, 18, 26, __?

LEY_bk953_13_2ndpp Page 429 Wednesday, January 12, 2005 12:05 PM

Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

20

What is the average of 8, 10 and 12?

21

2+6×3=

22

Stuart solved the equation 2x + 2 = x + 5 and got the answer x = 3. Is Stuart correct?

23

The product of two numbers is 12. If one number is 2, what is the other number?

24

What is the area of a triangle with a base of 12 cm and a perpendicular height of 10 cm?

25

Jasmine reverses the digits in the number 4683 and subtracts them from the original number. What is the answer?

Language in Mathematics

Read this article and answer the questions.

Paul Erdos (1913−

)

Paul Erdos was born in Budapest, Hungary, in 1913. He is known as one of the greatest modern-day mathematicians as a result of his considerable contribution in the areas of algebra, analysis, combinatorics, geometry, topology, number theory and graph theory. By the age of three he was able to mentally multiply three-digit numbers. With his mother’s encouragement Erdos read many scientific books from an early age and developed an interest in mathematics, which he pursued. When he began to travel and present papers, his mother went with him. Erdos was extremely depressed when she died. Erods has continued to work in mathematics and travels around the world talking at universities and meeting other mathematicians. He has written many scientific papers of great importance and is renowned for being able to solve problems that confuse others. The ‘Prime Number Theorem’ was stated as conjecture in 1792, and it was Erdos along with Selberg who gave the first non-integral calculus proof in 1948–49. Despite his fame, Erdos is known for his cosmopolitan lifestyle and stimulating conversation as much as for his mathematical contributions. He has very little personal property as he places no value on material goods. With no family commitments, he is able to give all his time to pursuing his own interests in mathematics and helping others to develop their understanding and knowledge of the subject.

429

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Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

1

a b c d e

2

Complete these glossary terms using vowels.

3

How old would Paul Erdos be today? When could he mentally multiply three-digit numbers? What does Paul Erdos do with his time? For what is he most famous? Write the three most important points from the article.

a l __ n __ __ r

b p r __ n __ m __ r __ l

c q __ __ dr __ t __ c

d s __ l __ t __ __ n

e s __ b s t __ t __ t __

f

v __ r __ __ bl __

Write a paragraph describing how a linear equation is solved.

Glossary consecutive

equation

formulae

formulas

fractions

inequalities

like terms

linear

lowest common denominator (LCD)

pronumeral

quadratic equation

solution

solve

substitute

unknown

value

variable

subject

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Equations and Inequalties (Chapter 13) Syllabus reference PAS5.2.2

431

CHECK YOUR SKILLS 1

The solution to 3x – 5 = −6 is: A x=

2

3

–1 -----3

6

7

8

C x=

B

–1 -----2

C

The solution to 8 – 5x = 2x + 3 is: 7 --5

B x=

5 --7

x = −2 is not a solution of: A 3x + 5 = 3 + 2x C 4x + 1 = 7 − 5x

– 11 --------3

1 --2

D x=

11 -----3

✓

C x = 25

If y = 7 – 2x and y = −8, then x is:

A x= 5

1 --3

x The solution to --- + 3 = 2 is: 5 1 A x = –----B x=1 5

A 7 1--24

B x=

C x=

D x = −5

D − 7 1--2-

5 --3

D x=

–5 -----3

D x=

5 -----12

B 7 – 2x = 3x + 17 D 6x + 4 = 2x – 4

x – 3 4 – 2x The first line with an error in solving ------------ – --------------- = 2 is: 2 3 A 3(x − 3) – 2(4 − 2x) = 12 B 3x – 9 – 8 – 4x = 12 C −x – 17 = 12 D x = −29 6 – 5x The solution to --------------- = −2 is: 3 A x = 2 2--5B x=0

C x=

8 --5

3x x + 1 – 1 The line without an error in the solution of ------ – ------------ = ------ is: 2 3 5 A 15 (3x) – 10 (x + 1) = −1 B 45x – 10x + 10 = −1 C 35x = −11 -----D x = − 35 11

9

When one-third of a number is added to twice a number, the answer is 10. The number is: -----A 1 3--7B 10 C 3 1--3D 4 2--77

10

A woman is currently twice as old as her son. In 5 years’ time the sum of their ages will be 70 years. An equation to find the son’s age now is: A n + 2n = 5 B n + 2n = 70 C (n + 5) = 7 + (2n + 5) D (n + 5) + (2n + 5) = 70

11

A solution to x2 = 64 is: A x=4

B x=8

C x = 32

D x = 128

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432

Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

12

A solution to 9x2 = 49 is: 7 A x = –----B x= 9

C x=

–7 -----3

D x=

49 -----9

− + 49--------9

13

The surface area of a sphere is given by A = 4πr2. The value of r when A = 200 is closest to: A 4 B 14 C 50 D 64

14

In symbols ‘five times a number minus three is always greater than the number plus one’ is: A 5n – 3 >1 B 5(n−3) > n + 1 C 3 – 5n > 1 D 5n – 3 > n + 1

15

A solution to 6 – 7x > −8 is: A x=1 B x=5

16

17

C x=3

4 – 3x The number that is not a solution of --------------- > −1 is: 5 A 3 B 2 C 1 A solution to 7 – 3x < 4 + 2x is: 3 A x = –----B x=0 5

C x=

✓ D x=4

D 0

3 --5

D x=1

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question Section

1–3

4–6

7, 8

9, 10

11, 12

A

B

C

D

E

13

14, 15

16, 17

F

G

H

LEY_bk953_13_2ndpp Page 433 Wednesday, January 12, 2005 12:05 PM

Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

REVIEW SET 13A 1

Solve for x: a 2x + 5 = −3

b 4x = 2x + 11

c

x d --- + 9 = −3 7

e 2 – x = 3x + 7

f

2

4 – 2x If x = −3, find y if y = --------------- . 5

3

If y = 3(2x – 1) – (x – 5), find x when y = 7.

4

Solve for x: a 3(2x + 1) = 4 d 3(x – 2) + 5(x + 1) = 15 2x – 1 g --------------- = 4 3

b 2(3 – x) = 3(x + 5) x 3 e --- = --5 2 x x h --- – --- = 4 3 6

x 1 --- = --3 6 x–5 ------------ = −7 4

c 4 – 3(2 – x) = 7 (2x + 1) = −3

f

1 --2

i

x 3 x + --- = --3 4

5

Solve the following problems. a Five times a certain number is equal to 12 more than the number. Find the number. b When three consecutive integers are added the result is 48. Find the largest integer. c One side of a rectangle is 2 cm shorter than the other side and its perimeter is 96 cm. Find the length of the longer side. d Peter has 48 coins. He has three times as many 10-cent coins as 20-cent coins and the remainder are 5-cent coins. Determine the number of each coin type if their total value is $5.10.

6

The period of a pendulum (the time for one complete swing) is given by T = where l cm is the length of the pendulum. Find: a the period if the pendulum has length 74 cm b the length of the pendulum if its period is 2.5 seconds.

7

Solve: a x+5<3 2x – 3 d --------------- > 1 4

x b --- ≥ −3 2 4 – 2x e --------------- ≥ 3 5

1 --5

c 4x + 1 ≥ 2 f

3 – 2x < 14 + 5x

l seconds,

433

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Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

REVIEW SET 13B 1

Solve for x: a 3x – 1 = 2

b 6x = 11 – 3x

c

x d --- – 3 = −6 5

e 4 – 3x = 2x + 1

f

2

5a – 4 If a = −2, find t if t = ---------------- . 3

3

If y = 2(3x – 4) – 5(3 – x), find x when y = −3.

4

Solve for x: a 2(4x + 1) = 3 d 4(x – 5) + 3(2x + 1) = 3 3x – 5 g --------------- = 8 2

b 3(5 – 2x) = 5(x – 3) x 5 e --- = --7 3 x x h --- – --- = −4 5 2

2x ------ = 5 3 x–3 ------------ = 2 4

c 5 – 2(5 – 2x) = 3 (4 – 5x) = −2

f

1 --3

i

x 2 x – --- = --4 3

5

Solve these problems. a Four times a number is equal to 15 more than the number. Find the number. b When four consecutive integers are added, the result is 54. Find the smallest integer.

6

The surface area of a sphere is given by A = 4πr2. Find r when A = 400.

7

Solve: a 2x ≥ −8 d 4 – 3x < 5

b x – 5 < −3 3x – 1 e --------------- ≥ 2 4

c −8x > 16 6–x f ------------ > 4 2

LEY_bk953_13_2ndpp Page 435 Wednesday, January 12, 2005 12:05 PM

Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

REVIEW SET 13C 1

2

Solve for x: a 4x + 5 = 12

b 5 – 3x = −7

c

x d --- – 5 = 7 3

e 3x – 2 = x + 6

f

2x – 3 If y = --------------- find: 5 1 a y when x = --2

x --- = −4 3 2x + 5 ---------------- = −1 4

b x when y = 3

3

If y = 5 – 2(x – 3), find x when y = 3.

4

Solve: a −3(2a + 5) = 15 c 3(2p – 3) = 4(p + 1) x 4 e --- = --2 3 2x + 3 3x – 2 g ---------------- = --------------4 5 x+1 1–x i ------------ + x = -----------5 2

b 2(3a – 4) = a + 9 d x + 5(3 – x) = 2x + 4(x – 1) f

1 --3

(2x + 1) = −4

2x + 3 3x + 1 h ---------------- – ---------------- = 2 3 4

5

Solve the following problems. a I think of a number, double it and add 7. The answer is –4. Find the number. b An apple costs twice as much as a banana. The two of them cost me a total of 48 cents. Find the cost of the apple. c If a number is increased by 5 and then trebled, the result is six more than two-thirds of the number. Find the number. d An athlete can run at 15 kmph and cycle at 48 kmph. In his total training he runs and cycles the same distance. If his training time is 3 hours and 9 minutes, what distance does he cycle and run?

6

The value $V, of precious opal is given by the formula V = 20w2 dollars, where w is the weight in grams. a Find the value of an opal weighing 23.8 grams. b Find the weight of an opal valued at $2300.

7

Solve: –x a ------ > 2 3 x+1 d ------------ < 2 3

b 6–x<1

c 4x + 1 ≥ 7

4 – 3x e --------------- ≥ 1 5

f

5 – 3x ≤ 2(1 – x)

435

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436

Equations and Inequalities (Chapter 13) Syllabus reference PAS5.2.2

REVIEW SET 13D 1

2

Solve for x: a 3x – 2 = 5 x d --- + 3 = −2 7 3x – 5 If y = --------------- find: 3 a y when x =

b 4 – 7x = 3

c

e 4x – 7 = 3x + 5

f

b x when y =

1 --3

3

If y = 7 – 5(4 – x), find x when y = 0.

4

Solve: a −2(4x + 3) = 15 d x – 5(2x + 1) – 3 = 4 3 – 4x x + 5 g --------------- = -----------2 5

b 4(3t + 1) = t – 2 4 x e --- = --3 4 3x – 4 2 – 4x h --------------- – --------------- = 1 5 3

x --- = −4 5 3x – 2 --------------- = 2 5

7 --3

c 4(3p – 7) = 5(1 – p) (3x – 1) = −5

f

1 --4

i

x+1 3–x ------------ + x = ------------ + 1 2 3

5

Solve these problems. a When 12 is added to twice a number, the answer is 8. Find the number. b If a number is decreased by 3 then multiplied by 4, the result is 7 more than five times the number. Find the number.

6

The velocity of an object is given by v 2 = u 2 + 2as. a Find v when u = 10, a = −5 and s = 3. b Find u when v = 20, a = 3 and s = 12.

7

Solve: a x–4>3 x d --- + 2 ≥ 0 3

b −3x > 9 4 – 3x e --------------- ≥ 1 2

c 4 – 3x ≤ 14 f

3 – x ≥ 2 – 4(x – 1)

LEY_bk9_c14_finalpp Page 437 Wednesday, January 12, 2005 11:47 AM

Chapter 14 Straight Lines and Regions This chapter deals with the use and application of various standard forms of the equation of the straight line and the graphing of regions on the number plane. After completing this chapter you should be able to: ✓ find the equation of a straight line ✓ rearrange equations of straight lines into various forms ✓ sketch straight lines given their intercepts ✓ demonstrate that two lines are perpendicular if the product of their gradients is –1 ✓ find the equation of a line parallel or perpendicular to a given line ✓ graph a variety of regions of the number plane involving straight lines.

Syllabus reference PAS5.3.3 WM: S5.3.1–S5.3.5

LEY_bk9_c14_finalpp Page 438 Wednesday, January 12, 2005 11:47 AM

438

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Diagnostic Test 1

2

3

4

The equation of the line passing through (–2, 3) with gradient 2 is: A y = 2x + 3

B y = 2x – 7

C y = 2x – 2

D y = 2x + 7

The equation of the line passing through K (–3, 2) and L (4, –5) is:

7

When written in gradient–intercept form the equation 3x – 2y + 8 = 0 is: A 2y = 3x + 4 3 C y = --- x + 4 2

8

B 3x – 2y = –8 3 D y = --- x + 8 2

The x and y intercepts of 3x – 4y – 12 = 0 are:

A y = –x – 1

B y=x+5

A 4 and –3

B 3 and –4

C y = –3x – 7

D y = –x + 1

C –4 and 3

D –3 and 4

The equation of the line cutting the y-axis at 4 and the x-axis at –5 is: 4 --- x + 4 A y= 4 B y = – --- x + 4 5 5 5 5 C y = --- x + 4 D y = – --- x + 4 4 4

9

10

The equation of this line is:

The line perpendicular to 3x – 5y + 7 = 0 is: A 3x + 5y – 2 = 0 B 3x – 5y – 7 = 0 5 C –3x – 5y + 7 = 0D y = – --- x + 2 3

y 3 2

11

1 –3 –2 –1

The line parallel to y = –2x + 1 is: --- x + 1 --- x + 1 A y= 1 B y= –1 2 2 C y = –2x + 3 D y = –2x + 1

x 1

–1

2

3

–2

--- x – 1 A y= 2 3

3 B y = --- x – 1 2

C y = –x + 3

D y = 2x – 1

12

The triangle ABC with vertices A (–1, 0), B (1, 4) and C (5, 2) is: A scalene

B isosceles

C equilateral

D right angled

The region defined by y < 3 is: A y 4

5

6

The value of a if (a, –2) lies on the line y = 3x – 2 is:

3

A 0

B 1

1

C 2

D 3

--- x – 2 In general form y = 1 --- is: 2 3 A 6y – 3x + 4 = 0 B 3x – 6y – 4 = 0 --- x – y – 2 --- = 0 C 3x + 2y – 2 = 0 D 1 2 3

2

–3 –2 2 –1 1

x –1 –2 –3

1

2

3 4

LEY_bk9_c14_finalpp Page 439 Wednesday, January 12, 2005 11:47 AM

Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

B

14 y

The region defined by 3x – y < 9 is: A

4

2 y

3

x

2

–4 –3 –2 2 –1 1

1 –3 –2 2 –1 1

–1

x –1

1

2

2

3 4

–2 –3

3 4

–4

–2 –3

C

1

–5 5 –6 –7

y

–8 –9

4 3 2 1 –3 –2 2 –1 1

B

x –1

1

2

2 y

3 4

x –4 –3 –2 2 –1 1

–2 –3

3

–5 5 –6

2

–7

1

–8 –9

x 1

2

3 4

–2 –3

13

3 4

–4

y 4

–1

2

–2 –3

D

–3 –2 2 –1 1

–1

1

C

2 y

This region is defined by:

x –4 –3 –2 2 –1 1

y 4

–2 –3

3 2

–4

1 –4 –3 3 –2 2 –1 1

–1

x –1

1

2

3 4

–7

–2 –3

A x ≤ 3 and y < 2

–5 5 –6 –8 –9

B x > –3 and y < 2

C x < –3 and y ≤ 2 D x ≥ –3 and y > 2

1

2

3 4

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

D

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1–5

6–8

9, 10

11

12, 13

14

Section

A and B

C

D

E

F

G

A. EQUATIONS OF LINES The equation of a line is the simplest relationship that connects the x and y values for every point on the line.

For example, this line contains infinitely many points.

y

3

Some of these are (–1, 0), (0, 1), (1, 2), (2, 3), (3, 4), (4, 5).

2 1 –3 –2 –1

x –1 –2 –3

1

2

3 4

Notice that the y-coordinate is always 1 more than the x-coordinate. This is summarised by writing y = x + 1. Thus y = x + 1 is the equation of all points on this line.

–4 y

Similarly, this line contains the points (5, –1), (4, 0), (3, 1), (2, 2), (1, 3), (0, 4), (–1, 5) and so on.

4

Notice that the sum of the x and y coordinates is always 4.

2

3 1

i.e. x + y = 4 Thus the equation of the line is x + y = 4.

–2 –1

x –1 –2

1

2

3 4

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

The equation of a line The equation of a line can be determined if we know: • the gradient of the line, and • the coordinates of a point on the line.

If a straight line has gradient m and passes through the point with coordinates (x1, y1), y–y x – x1

then its equation is y – y1 = m(x – x1) or -----------1 = m.

Proof:

Suppose P (x, y) is any point on the line with gradient m.

y P(x2,y2) Q(x1,y1) x

Equating the slopes gives y – y1 ------------ = m x – x1 or y – y1 = m(x – x1).

Example 1 Find the equation of the line through (–1, 2) having a gradient of 4. The equation of the line is y – y1 = m(x – x1) where (x1, y1) = (–1, 2) and gradient m = 4. y – 2 = 4(x – (–1)) y – 2 = 4(x + 1) y – 2 = 4x + 4 y = 4x + 6

Exercise 14A 1

Find the equation of the line through: a (2, –3) with gradient 3

b (–4, 2) with gradient –2

c (7, 4) with gradient – 1--2-

d (2, –5) with gradient 5

e (3, 7) with gradient

1 --3

g (3, 4) with gradient –1 1--2i

( – 1--2

,

1 1--2-

) with gradient 4

f

(–5, –2) with gradient – 1--2-

h (0, 0) with gradient 6 j

(5, 4) with gradient 0

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Example 2 Find the equation of the line that passes through the points A(–1, 5) and B(2, –3). First we find the gradient of AB using y2 – y1 m = --------------y2 – x1

[use (–1, 5) for (x1, y1) and (2, –3) for (x2, y2)]

–3–5 ∴ m = -------------------2 – ( –1 ) 8 ∴ m = – --3 Second, find the equation of the line using y – y1 = m ( x – x1 ) 8 ∴ y – 5 = – --- ( x – ( – 1 ) ) 3 8 ∴ y – 5 = – --- ( x + 1 ) 3 8 8 ∴ y = – --- x – --- + 5 3 3 8 7 ∴ y = – --- x + --3 3

2

1 8 or y = – --- x + 2 --3 3

Find the equation of the line passing through the following pairs of points. a A(2, 3) and B(4, 7) b A(0, 2) and B(–2, 4) c A(–1, –3) and B(5, –5) d A(6, 3) and B(4, 1) e A(5, –2) and B(2, –5) f P(0, 0) and Q(3, 5) g P(–3, 5) and Q(1, –2) h L(–3, –2) and M(0, 4) i X(2, 2) and Y(–3, –1) j X(0, 6) and Y(–4, 0)

Example 3 Find the equation of the line that cuts the y-axis at 3 and the x-axis at –2. The y-axis is cut when x = 0, i.e. (0, 3) is one point. The x-axis is cut when y = 0, i.e. (–2, 0) is the other point.

B(2, –3) could have been used as (x1, y1 ), giving the same equation.

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

y2 – y1 Gradient, m = --------------x2 – x1

[using (x1, y1) = (0, 3) and (x2, y2) = (–2, 0)]

0–3 = ----------------–2–0 –3 = -----–2 3 = --2 But, y – y1 = m(x – x1) 3 ∴ y – 3 = --- (x – 0) 2 3 ∴ y – 3 = --- x 2 3 i.e. y = --- x + 3 2

Find the equation of the line a cutting the y-axis at –4 and the x-axis at 2 b cutting the y-axis at 7 and the x-axis at –2 c cutting the y-axis at –5 and the x-axis at –3 d with y-intercept –3 and x-intercept 2

3

B. THE GRADIENT–INTERCEPT FORM If a straight line has gradient m, and y-intercept b, then it has equation y = mx + b.

Proof: Since the y-intercept is b, the point (0, b) lies on the line.

y

Thus the equation is y – b = m(x – 0) i.e. y = mx + b b

x

Example 1 Find the equation of the line with gradient –3 and y-intercept 2. Since m = –3 and b = 2, the equation is y = –3x + 2.

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Exercise 14B 1

Find the equation of the line with: a gradient 2 and y-intercept 7

b

c gradient –3 and y-intercept –1 2 e gradient – --- and y-intercept 6 3

gradient 4 and y-intercept –6 1 gradient – --- and y-intercept 2 2 gradient 1 and y-intercept 0

d f

Example 2 Find the equation of the line below. (0, 2) and (4, 3) lie on the line 3–2 1 ∴ gradient is = ------------ = --4–0 4 1 i.e. m = --- and b = 2 4 1 ∴ equation is y = --- x + 2 4

y 4 3

(4, 3)

2 1

x 1

2

2

3

4

Find the equations of the following lines. a b

c y

y

y

4

4

4 3

3

(2, 3)

2

3

(3, 3)

2

2

1

1

1

x

x 1

2

3

4

1

–1

2

3

x –1

4

1

2

3

4

–1

d

e

f

y 3

2

4

2

1

3

1

x x

(–2, 3) 2

1

–4 –3 –2 –1

1

x –2 –1

y

y

5

1

2

3

4

2

3

–3 –2 –1

1

2

3

4

5

–1

–1

–2

–2

–3

(4, –3)

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Find the equation connecting the variables given. a b

3

M

c G

y

4

3

4

3

2

3

2

slope 13

1 1

2

3

x 1

–2 –1

t –2 –1

2

1 2

3

4

1

5

s

–1

4

–1

2

3

4

–1

–2

d

1

e M

F 5

2 1

4

g 1

–1

2

3

4

3

6

5

–1

2

–2

1 –1

(10, 2) x2 1

2

3

4

5

6

7

8

9

10

f p

√n 1

2

3

4

5

6

–1 –2 –3

(6, 3)

–4

Check the variable on each axis.

–5 –6

An alternative method to find the equation of a line is to use y = mx + b.

Example 3 Use y = mx + b to find the equation of the line passing through (–3, 5) with gradient 3. y = mx + b i.e. y = 3x + b We substitute x ∴5 ∴5 ∴ 14 i.e. y

= = = = =

– 3 and y = 5 3 ( –3 ) + b –9+b b 3x + 14

[as the gradient m = 3] [as (–3, 5) lies on the line]

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

4

Find the equation of the line through a (3, –4) with gradient 2 c (8, –2) with gradient –3 e (3, 0) with gradient 5 g (0, 5) with gradient – 1--3i (9, 9) with gradient 1--3-

b d f h j

(5, –2) with gradient 1 (–4, –4) with gradient –2 (6, 8) with gradient 1--2(0, 0) with gradient 8 (6, 2) with gradient 0

Example 4 Find a given that (a, –1) lies on the line y = 2x + 3. Substitute x = a and y = –1 into the equation y = 2x + 3. ∴ –1 = 2(a) + 3 ∴ –1 = 2a + 3 ∴ –4 = 2a ∴ a = –2

5

Find a given that each point below lies on the line with the given equation. a (a, 3) y = 2x – 1 b (a, –2) y=4–x c (4, a) y= x+3 d (–2, a) y = 1 – 3x

Example 5 Draw the graph of the line with equation y = –2x + 1. Method 1 Table of values: x

–2

0

2

y

5

1

–3

Method 2 –2 y-intercept is 1 and gradient is -----1 ∴ start at 1 on y-axis, x-step of 1, then y-step of –2.

y

y y= –2x +1

5

5

4

4

3

3

2

2 1

1 –4 –3 –2 –1

–2

x –1 –2 –3

1

2

3 4

–4 –3 –2 –1

–1 –2 –3

1

x 2

3 4

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

6

Draw the graph of the lines with the following equations. 1 a y = --- x + 2 b y = 2x + 1 2 1 d y = –3x + 2 e y = – --- x 2 3 2 g y = --- x h y = --- x + 2 2 3

c y = –x + 3 f y = –2x – 2 3 i y = – --- x – 1 4

C. GENERAL FORM OF A STRAIGHT LINE EQUATION The general form of the equation of a straight line is Ax + By + C = 0 where A, B and C are all integers and A ≥ 0.

Example 1 Write, in general form, the equations a y = 3x – 1

b

a

b

y = 3x – 1 0 = 3x – y – 1 (subtracting y) (swapping sides) 3x – y – 1 = 0

2 y = – --- x + 4 3 2 y = – --- x + 4 3 2 0 = – --- x – y + 4 3 0 = – 2x – 3y + 12 – 2x – 3y + 12 = 0 2x + 3y – 12 = 0

(subtracting y) (multiplying by 3) (swapping sides) (multiplying by –1 so that A ≥ 0)

Exercise 14C 1

Write the following equations in general form. a y = 2x + 1

b y = 5x – 2

d y = –2x – 5

e y = –3x + 4

1 g y = – --- x – 5 2 2 1 j --- y = --- x + 1 3 4

2 h y = – --- x – 3 3 1 3 k y = --- x – --2 4

c y = 2x + 5 1 f y = --- x + 2 2 3 2 i y = – --- x – --4 3 1 1 l --- y = x + --5 2

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Example 2 Rewrite the equation 4x – 3y – 12 = 0 in the form y = mx + b and hence find the gradient and y-intercept. 4x – 3y – 12 = 0 4x – 12 = 3y (adding 3y to both sides) 3y = 4x – 12 (swapping sides) 4 (dividing by 3) y = --- x – 4 3 4 ∴ gradient is --- m = and y-intercept is –4. 3

2

Rewrite the following equations in y = mx + b form and hence find the gradient and y-intercept. a x + 2y – 4 = 0 b 3x + 2y – 24 = 0 c 2x – y + 4 = 0 d 4x – 2y – 6 = 0 e 5x + 2y + 10 = 0 f 3x + 2y – 8 = 0 g 4x – y – 6 = 0 h 3x – 2y + 17 = 0 i 8x – 2y – 7 = 0

3

Find b if each point below lies on the line with the given equation. a (2, b) x + 2y = –4 b (–1, b) 3x – 4y = 6 c (b, 4) 5x + 2y = 1 d (b, –3) 4x – y = 8

Example 3 Find the x- and y-intercepts of the line with equation 4x – 3y – 12 = 0. x-intercept when y = 0

y-intercept when x = 0

4x – 3(0) – 12 = 0

4(0) – 3y – 12 = 0

4x – 12 = 0

–3y – 12 = 0

4x = 12

–3y = 12

x =3 i.e. x-intercept is 3 4

y = –4 i.e. y-intercept is –4

Find the x- and y-intercepts of the following lines. a x + 2y – 8 = 0 b 3x + 2y – 12 = 0 d 3x – 2y – 36 = 0 e 5x + 2y + 20 = 0 g 4x – y – 5 = 0 h 3x – 2y + 15 = 0

c 2x – y + 6 = 0 f 3x + 2y – 5 = 0 i 9x – 2y – 5 = 0

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Example 4 Draw the graph of the line with equation 5x – 3y – 15 = 0. Find the x-and y-intercepts If x = 0

5(0) – 3y – 15 = 0

If y = 0

5x – 3(0) – 15 = 0

–3y – 15 = 0

5x – 15 = 0

–3y = 15

5x = 15

y = –5 ∴ y-intercept is –5

x=3 ∴ x-intercept is 3

Use a third point as a check. Try x = 1.

y

5(1) – 3y – 15 = 0

3

5 – 3y – 15 = 0

2 1

–3y – 10 = 0 –3y = 10 10 y = – -----3 10 ∴ (1, – ------ ) is on the line. 3

–4 –3 –2 –1

x –1

1

2

3 4

–2 –3 –4 –5

5

Draw the graph of the line with equation a x + 2y – 8 = 0 b 3x – y – 6 = 0 d 4x + 3y – 8 = 0 e x+y–5=0 g 3x – 4y – 12 = 0 h 5x + 2y + 10 = 0

c 2x – 3y – 4 = 0 f x–y+5=0 i x – 2y = 0

Investigation 1 WM: Reasoning, Communicating, Applying Strategies

Graphs of lines 1

a On the same number plane draw the straight lines with equations y = 2x, y = 2x + 1, y = 2x – 3, y = 3x + 1 b What do you notice about these lines?

☞

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

2

a On the same number plane draw the straight lines with equations y = –3x + 1, 3x + y – 2 = 0, 6x + 2y + 3 = 0. b What do you notice about these three lines? c Rewrite the second two lines in y = mx + b form and find their gradients.

3

Copy and complete: Straight lines are _____ if their _____ are equal.

4

a On the same number plane draw the straight lines with equations 1 y = 2x + 1 and y = – --- x + 2. 2 b What do you notice about these lines?

5

a On the same number plane draw the straight lines with equations y = 3x – 1 and x + 3y – 6 = 0. b What do you notice about these lines? c Rewrite x + 3y – 6 = 0 in y = mx + b form and find the gradient.

6

Copy and complete: Straight lines are ____ if the product of their gradient is ____.

D. PARALLEL AND PERPENDICULAR LINES From Investigation 1: If two straight lines have gradients m1 and m2 then: •

the lines are parallel if m1 = m2

•

the lines are perpendicular if m1 × m2 = –1.

Example 1 a Find the equation of the line parallel to the line y = –5x – 7 passing through the point (2, 3). b Find the equation of the line parallel to the line 3x – 6y + 8 = 0 passing through the point (–1, –2).

Use a graphics calculator.

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

a The gradient is –5, since the lines are parallel. i.e. equation is

y ∴3 ∴3 ∴ 13 y

= = = = =

– 5x + b – 5 ( 2 ) + b (substituting (2, 3)) – 10 + b b – 5x + 13 is the equation of the line.

b 3x – 6y + 8 = 0 must be rearranged into y = mx + b form to find the gradient. 3x – 6y + 8 = 0 ∴ 3x + 8 = 6y 3x 8 ∴ ------ + --- = y 6 6 1 4 ∴ y = --- x + --2 3

1 So gradient = --- . 2 1 Use y – y1 = m(x – x1) with m = --- and (–1, –2) for (x1, y1) 2 1 ∴ y – (–2) = --- (x –(–1)) 2 1 1 1 ∴ y + 2 = --- (x + 1) = --- x + --2 2 2 1 1 y = --- x – 1 --2- is the equation of the line. 2

Exercise 14D 1

2

Which of the following pairs of lines are parallel? a y = 3x + 1, y = 3x – 5 b y = 2x – 1, c y = 5x + 3, y = 3x + 5 d y = 4x – 3, e y = 2x – 5, 2x – y + 4 = 0 f y = –x – 5, g 4x – 3y + 5 = 0, 3x + 4y + 2 = 0 h 2x + 3y – 2 = 0,

y = 2x y = 4 – 3x x – 2y + 3 = 0 2x + 3y – 5 = 0

Find the equation of the line: a parallel to the line y = 2x – 5 passing through the point (1, 4) b parallel to the line y = –7x – 2 passing through the point (–5, –2) 1 c passing through the origin parallel to y = – --- x + 5 2 d parallel to the line 5x – 7y + 3 = 0 passing through the point (2, –3) e passing through (–1, –3), parallel to the line passing through the points (1, 5) and (3, 6) f

with y-intercept –2, that is parallel to the line segment joining (–7, 5) and the origin

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Example 2 a Is the line y = 3x – 5 perpendicular to the line 2x + 6y + 9 = 0? b What is the gradient of the line perpendicular to the line 5x – 2y + 4 = 0? c Find the equation of the line passing through (6, –3) perpendicular to the line 2 y = – --- x + 4. 3 a y = –3x – 5 has gradient m1 = 3 We rearrange 2x + 6y + 9 = 0 to find its gradient. 6y = – 2x – 9 – 2x 9 ∴ y = --------- – --6 6 1 1 3 ∴ y = – --- x – ---, and so had gradient m2 = – --3 3 2 1 Now m1 × m2 = 3 × – --- = –1 ∴ the lines are perpendicular. 3 b Rearrange 5x – 2y + 4 = 0 into y = mx + b form. 5x – 2y + 4 = 0 ∴ 5x + 4 = 2y 5x 4 ∴ ------ + --- = y 2 2 5 y = --- x + 2 2 2 ∴ the gradient of the perpendicular line is – --5 2 2 c y = – --- x + 4 has gradient m1 = – --3 3 3 ∴ the gradient of the perpendicular line is --2 3 ∴ y = --- x + b and so we substitute (6, –3) 2 3 ∴ –3 = --- (6) + b 2

5 2 (since --- × – --- = –1) 2 5 2 3 (since – --- × --- = –1) 3 2

∴ –3 = 9 + b –12 = b 3 ∴ y = --- x – 12 is the equation. 2 3

Which of the following pairs of lines are perpendicular? a y = 2x – 5, y = 2x b y = –3x + 5, c y = 4x + 7,

y = –4x + 3

e 3x – 4y + 5 = 0, 4x + 3y – 2 = 0

1 y = --- x – 2 3 2 5 d y = --- x – 1, y = – --- x – 1 5 2 f 7x + 5y + 3 = 0, 5y – 7x = 0

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

4

5

Find the gradients of the lines perpendicular to the given lines. 3 a y = – --- x + 2 b y = 7x – 2 5 4 d y = – --- x – 7 e 3x – 2y + 1 = 0 5

3 c y = --- x – 5 2 f

5x – 7y + 7 = 0

Find the equation of the line: a perpendicular to y = 5x – 2 passing through (–3, 2) 1 b perpendicular to y = – --- x + 7 passing through (0, 5) 4 3 1 c passing through the origin perpendicular to y = --- x + --4 3 d passing through (–2, 5) perpendicular to 3x – 4y + 12 = 0 e with y-intercept 7 perpendicular to 5x + 2y – 7 = 0 f

passing through (–2, –5), perpendicular to the line segment joining (2, 3) and (5, –3)

g passing through the origin, perpendicular to the line segment joining (3, 0) and (0, –5) 6

1 a Find the equations of five lines that are perpendicular to y = --- x – 3. 2 b Find the equations of five lines that are parallel to y = 3x – 2. c Write the equation of all lines parallel to 3x – 5y + 6 = 0. d Write the equation of all lines perpendicular to 5x – 3y + 7 = 0.

7

a Find the equations of three lines that are perpendicular to 3x – 4y + 8 = 0. b Write the equations of these lines in general form. c What do you notice?

8

a Find the equations of three lines that are parallel to 2x – 7y + 3 = 0. b Write the equations of these lines in general form. c What do you notice?

9

Write the equation of all lines: a parallel to 7x – 5y + 6 = 0

b perpendicular to 7x – 5y + 6 = 0

Investigation 2 WM: Communicating, Reasoning

Transformation of lines 1

a On the same number plane, sketch the graphs of y = 3x and y = 3x + 1. b Which transformation, rotation, reflection or translation, would be used to superimpose the graph of y = 3x over that of y = 3x + 1? c Describe a transformation that would superimpose the graph of y = 3x over the graph of y = 3x – 5.

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

2

3

a On the same number plane, sketch the graphs of y = 2x and y = –2x. b Describe a transformation that would superimpose the graph of y = 2x on the graph of y = –2x. 1 a On the same number plane, sketch the graphs of y = 3x and y = – --- x. 3 b Describe a transformation that would superimpose the graph of y = 3x over the graph of 1 y = – --- x. 3

4

Describe a transformation that would superimpose the graph of y = 4x over the graph: 1 a y = –4x b y = 4x – 7 c y = – --- x 4

5

Describe the transformations needed to map: 1 1 a y = 3x to y = –3x to y = 5 – 3x b y = –4x to y = --- x to y = --- x – 2 4 4 1 1 1 c y = --- x to y = – --- x to y = 5 – --- x d y = 3x – 5 to y = 3x to y = –3x 2 2 2 Use a graphics calculator here.

E. FURTHER COORDINATE GEOMETRY This section uses some coordinate geometry aspects from Chapter 7 in conjunction with aspects of this chapter. Remember x1 + x2 y1 + y2 Midpoint formula:  ---------------, ---------------   2 2  Distance formula: d =

2

( x2 – x1 ) + ( y2 – y1 )

2

Exercise 14E 1

Line segments AB and CD bisect each other at T. Find the coordinates of C.

A(5, 3)

C T

B(–3, –1)

D(2, 0)

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

2

A(3, 2)

D

B(8, 4)

ABCD is a parallelogram. Use the fact that the diagonals of the parallelogram bisect each other to find the coordinates of D.

C(6, –1)

3

Triangle ABC has A(–1, 4), B(2, –1) and C(5, 2) as vertices. Find the length of the line segment from A to the midpoint of BC.

4

For the points P(2, 3), Q(0, 0), R(7, 4), S(a, –1), find a if: a PQ is parallel to RS b PR is parallel to QS c PQ is perpendicular to RS d PR is perpendicular to QS

5

The triangle ABC has vertices A(–2, 0), B(2, 1) and C(1, –3). a Find the length of the sides AB, BC and AC using the distance formula. b Classify ∆ABC as scalene, isosceles or equilateral. Give a reason.

6

The triangle PQR has vertices P(1, 0), Q(3, 1) and R(7, 3). a Find the lengths of each of the sides. b Classify ∆PQR as scalene, isosceles or equilateral. Give a reason.

7

Classify ∆LMN with vertices L(–2, –1), M(0, 3) and N(4, 1) as scalene, isosceles or equilateral. Give a reason.

8

The triangle XYZ has vertices X(1, 2), Y(2, 5) and Z(4, 1). a Find the length of each of the sides XY, YZ and XZ. b Use Pythagoras’ rule to decide if ∆XYZ is right angled. Give a reason. c Find the gradient of the sides XY, YZ and XZ. d Use the gradients to decide if ∆XYZ is right angled. Give a reason.

9 10

Is the triangle with the vertices D(–2, –1), E(1, –1) and F(–2, 3) right angled? Give a reason. The triangle PQR has vertices P(–2, 5), Q(3, –1) and R(–4, –7). a Find the coordinates of S, the midpoint of PQ. b Find the coordinates of T, the midpoint of PR. c Show that the length of QR is twice the length of ST. d Show that ST is parallel to QR.

11

The quadrilateral PQRS has vertices P(2, 4), Q(5, 1), R(–1, –2) and S(–4, 1). a Prove that PQRS is a parallelogram by showing that: i the opposite sides are equal ii the opposite sides are parallel iii the diagonals bisect each other iv one pair of opposite sides are equal and parallel.

☞

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

b Is PQRS a rectangle? Explain. c Is PQRS a rhombus? Explain. 12

ABC is a triangle with vertices A(–4, 4), B(3, 2) and C(–2, –3). a Show that ∆ABC is isosceles with AC = AB. b Find the midpoint D of the base BC. c Find the length of AD. d Prove that AD is perpendicular to BC.

13

Triangle XYZ has vertices X(–1, 3), Y(–4, 1) and Z(3, –3). a Show that ∆XYZ is right angled. Do this in two ways. b Find the equation of the perpendicular bisector of XY. c Find the equation of the side YZ. d Show that the point of intersection of the perpendicular bisector of XY and the side YZ is (– 1--2- , –1). e Show that the perpendicular bisector of XZ also passes through the point (– 1--2- , –1). f

14

Explain why (– 1--2- , –1) is the centre of a circle with X, Y and Z on its circumference. What names are given to this point and the circle?

The triangle STU has vertices S(–2, 5), T(2, 3) and U(–4, –1). a Find the equation of the perpendicular bisectors of each side. b Hence, find the circumcentre of ∆STU.

15

The quadrilateral WXYZ has vertices W(–2, –3), X(–4, 4), Y(3, 2) and Z(2, –4). The points A,B, C, D are the midpoints of each of the sides WX, XY, YZ and ZW. a Find the coordinates of A, B, C and D. b What type of quadrilateral is ABCD?

16

Find the area of the triangle enclosed by the lines y = 0, y = 2x and x + y = 6.

F. INEQUALITY GRAPHS USING LINES PARALLEL TO THE AXES y

We know that the solution to the inequality x ≥ 1 can be drawn on the number line as shown.

4 3

x=1

2 –3 –2 –1

0

1 2

3 4

The graph of x = 1 on the number plane is a line parallel to the y-axis passing through x = 1 as shown.

1 –4 –3 –2 –1

x –1 –2 –3 –4

1 2

3 4

5

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Combining these two graphs gives the graph of x ≥ 1 on the number plane. It includes all x-values to the right of x = 1.

4

This is shown by shading the half-plane to the right of the line x = 1.

3

y

2 1 This line is the boundary of the region.

–4 –3 –2 2 –1

x –1

1 2

3 4

5

–2 –3 –4

The graph of the inequality x ≥ 1 is shown opposite.

y 4

The dotted line indicates that x = 1 is not included, which is similar to the open circle on number lines.

3 2 1 –4 –3 –2 2 –1

x –1

1 2

3 4 5

–2 –3 –4

Example 1 Sketch the regions defined by: a x≥2

b

y<3

☞

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

a x≥2

b

y<3

y

y

4

4

3

3

2

2 1

1 –4 –3 –2 2 –1

x

x –1

1

2

3 4

–4 4 –3 –2 2 –1 1

5

–1

–2 –3

–2

–4

–4 4

1

2

3 4

–3

Line is parallel to the y-axis. The line is solid and included in the shaded area

Line is parallel to the x-axis. Since y < 3 the shading is below the line which is not solid.

Exercise 14F 1

Sketch the following regions. a x≥2 d y≤4 g y > –5 1 j x < –1 --4

b e h k

x ≤ –1 x>3 y<7 3 y ≥ --2

c f i l

y ≥ –2 x<1 x ≥ –4 y<1

Example 2 Sketch the region defined by the intersection of: a x < 3 and y ≥ 1 b x ≥ 1 and y ≤ 2 y

a Sketch both graphs on the same number plane.

4 3 y≥1 –4 –3 –2 2 –1

The solution is the region that is shaded by both graphs as shown opposite.

2 1

x –1 –2 –3 –4

1

2

3 4

x<3

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Note: If one or more graphs are dotted lines then the point of intersection is not included.

4

y

3 An open circle shows that the point (3, 1) is not included.

2

y≥1

1

–4 –3 –2 2 –1

x –1

1

–2 –3

2

3 4

x<3

–4

b Sketch both graphs on the same number plane. 4

y x≥1

3 y≤2

2 1

–4 –3 –2 2 –1

x –1

1

2

3 4

–2 –3 –4

The solution is the region that is shaded by both graphs.

4

y x≥1

3

Note: The point (1, 2) is included because both graphs are solid lines.

y≤2 –4 –3 –2 2 –1

2 1

x –1

1

2

3 4

–2 –3 –4

2

3

Sketch the regions defined by the intersection of these regions. a x ≥ 2 and y ≤ 2 b x > 1 and y ≥ 3 d x > –3 and y < –1 e x ≥ 4 and y ≤ –5 Shade the region defined by the intersection of these regions. a x ≥ –2 and x < 4 b y > –3 and y ≤ 2 c x ≥ –4 and x < 1 d y ≥ 0 and y < 4

c x < –1 and y > 3 f x ≤ 2 and y > 4

459

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

4

a Shade the region defined by x ≥ –3, y ≤ 2, x ≤ 1 and y ≥ –2. b Describe the region.

5

Write inequalities that would result in a region that is a rectangle 4 × 6 units on the number plane.

G. REGIONS IN THE NUMBER PLANE In the previous section all the lines were parallel to one of the axes. In this section regions involving any straight line are considered.

Example 1 Sketch the region defined by: a x+y≤3

b 2x – y > 6

a First, sketch the graph of the line x + y = 3. Here the x intercept is 3 and the y intercept is 3. 4

y

Draw an unbroken line because the inequality contains an equals sign.

3 2 1 –4 –3 –2 –1

x –1

1

2

3 4

–2 –3 –4

Second, decide which side of the line to shade. In the previous section this was easy, but here it may not be as clear. To decide where to shade, choose a point on one side of the line and test to see if it satisfies the inequality.

4

y

3

To decide where to shade, choose a point on one side of the line and test to see if it satisfies the inequality.

x

2

+ y ≤

1

3

For example, test (0, 0). x+y ≤3 0+0≤3 0 ≤ 3 is true. ∴ shade on the side of the line containing (0, 0).

–4 –3 –2 –1

–1 –2 –3 –4

1

2

3 4

x

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Always test (0, 0) when possible as it is easy to substitute.

b Step 1: 2x – y = 6 Sketch 2x – y = 6

Step 2: Test (0, 0) 2x – y > 6

x intercept when y = 0 is x = 3

2(0) – 0 > 6

y intercept when x = 0 is y = –6

0 > 6 is false ∴ shade the opposite side of the line to (0, 0).

y 4 3

x 2

>6

2

3 4 5

–2 –3

2x

–1

1

–y

3

1 –5 –4 –3 –2 –1

y

4

2

1 –5 –4 –3 –2 –1

x –1

–5

–2 –3

–6

–4

–4

1

2

3 4 5

–5

The line is dotted as the inequality does not contain an equals sign.

–6

Exercise 14G 1

2

Sketch these regions on the number plane. a x+y≤2 b x+y≥4 e x–y≥5 f x – y ≤ –3

c x+y>3 g y–x<3

d x + y < –2 h y–x>2

Sketch these regions on the number plane. a 2x – y ≤ 4 b 2x – 3y > 6 e y – 2x < –4 f 4x – y < 3

c 4x – y > –4 g 6x – 5y ≥ 15

d 3x + 5y ≤ 15 h 3x + 5y ≥ 15

461

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Example 2 Sketch the region defined by: b y ≤ 4x

a y > 3x a y > 3x

Step 1: Sketch y = 3x using a dotted line y

4

4

y = 3x

3

y = 3x

3 2

2

1

1 –4 –3 –2 –1

y

x

x –1

1

2

–4 –3 –2 –1

3 4

–1

–2 –3

–2 –3

–4

–4

1

2

3 4

Step 2: Choose a point on one side of the line, say (1, 0). Any point may be used.

(0, 0) cannot be used as the line passes through the origin.

Test (1, 0) y > 3x 0 > 3(1) 0 > 3 is false

∴ shade the opposite side of the line to (1, 0) b y ≤ 4x Step 1: Sketch y = 4x using an unbroken line. 4

y y = 4x

3

Step 2: Use the point (0, 2) to test.

2 1 –4 –3 –2 –1

x –1 –2 –3 –4

1

2

3 4

y ≤ 4x

4

2 ≤ 4(0)

3

2 ≤ 0 is false

2

∴ shade the opposite side of the line to –4 –3 –2 –1 (0, 2).

y

1

x –1 –2 –3 –4

1

2

3 4

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

3

Sketch these regions on the number plane. a y ≥ 2x b y < 3x e y > 5x f y > –x

d y≥x h y ≤ –3x

c y<x g y ≤ –2x

Example 3 Write an equation to describe the region. a 4

b

y

6 5 y 4

3 2

3

1 –4 –3 –2 –1

(2, 6)

x –1

1

2

2

3 4

1 x

–2 –3

–4 –3 –2 –1

–4

–1

1

2

3 4

–2 –3 –4 –5 –6

a Step 1: Find the equation of the line. rise m = --------y intercept is 2. run 2 = + --3 2 ∴ equation is y = --- x + 2 3 ∴ 3y = 2x + 6

b Step 1: Find the equation of the line rise m = --------- y intercept is 0. run 6 = --2 =3 ∴ equation is y = 3x

3y – 2x = 6 Step 2: Substitute a point in the region, not on the line, e.g. (0, 0). 3y – 2x = 6 Test 3(0) – 2(0) = 6 0 = 6 is false

Step 2: Substitute a point in the region, e.g. (1, 0). y = 3x Test 0 = 3(1) 0 = 3 is false

Use an inequality to make the statement true, e.g. use <.

Use an inequality to make the statement true, e.g. use <.

∴ inequality is 3y – 2x < 6

∴ inequality is y < 3x

463

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

4

Write equations to describe these regions. a 4

b

y

4

3

3

2

2

1 –4 –3 –2 –1

1

x –1

1

2

y

–4 –3 –2 2 –1

3 4

x –1

–2 –3

–2 –3

–4

–4

c

1

2

3 4

d 4

y

4

3

3

2

2

1 –4 –3 –2 –1

1

x –1

1

2

y

3 4

–5 5 –4 –3 –2 2 –1

x –1

–2 –3

–2 –3

–4

–4

1

2

3 4

5

–5

e

f y (–2, 2 8)

8

4

7

3

6

2

5

1

4

–5 5 –4 –3 –2 2 –1

(4, 2) x

3

–1

2

–2 –3

1 –4 –3 –2 –1

y

x –1 –2 –3 –4

1

2

3 4

–4 –5

1

2

3 4

5

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

non-calculator activities 1

Holly spends $14.25. What is her change from $50?

2

Evaluate 8 – 2.63.

3

Convert 5.23 cm to mm.

4

Peter buys a pair of jeans marked at $75.00. If he received 10% discount, how much does he pay for the jeans?

5

Red and blue lollies are sold in the ratio 2 : 5. If there are 16 red lollies in a bag, how many are blue?

6

Solve 5 – 2x = 9.

7

Explain the difference between gross income and taxable income.

8

Factorise x2 – 11x + 24.

9

Find d in this triangle. 5 cm

13 cm

d cm

10

Al works in a shoe store and notes that the mode female shoe size sold is 7 1--2- . Explain the term mode.

11

What is the probability of rolling a 6 with a normal six-sided die?

12

1 – 1--What is the value of -----------51- ? 1 + --5-

13

My car uses 15 L of petrol per 100 km. Find the number of litres of petrol needed to travel 350 km.

14

Between which two consecutive whole numbers is the square root of 67.3?

15

If 153 × 4785 = 732 105, find the value of 732.105 ÷ 478.5.

465

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

Language in Mathematics

Pierre de Fermat

(1601–1665)

Pierre de Fermat was born in Beaumont-de-Lomagne in France, near the border with Spain. He studied Latin and Greek literature, ancient science, mathematics and modern languages at the University of Toulouse, but his main purpose was to study law. In 1629 Fermat studied the work of Appollonius, a geometer of ancient Greece, and discovered for himself that loci or sets of points could be studied using coordinates and algebra. His work Introduction to Loci was not published for another fifty years, and together with La Geometrie by Descartes, formed the basis of Cartesian geometry. In 1631 Fermat received his degree in law, was later awarded the status of a minor nobleman, and in 1648 became King’s Councillor, Fermat was a man of great integrity who worked hard. He remained aloof from matters outside his own jurisdiction, and pursued his great interest in mathematics. He worked with Pascal on the theory of probability and the principles of permutations and combinations. He worked on a variety of equations and curves and the Archimedean spiral. In 1657 he wrote ‘Concerning the Comparison of Curved and Straight Lines’ which was published during his lifetime. Fermat died in 1665. He was acknowledged master of mathematics in France at the time, but his fame would have been greater if he had published more of his work while he was alive. He became known as the founder of the modem theory of numbers. In mid-1993, one of the most famous unsolved problems in mathematics, Fermat’s Last Theorem was solved by Andrew Wiles of Princeton University (USA). Wiles made the final breakthrough after 350 years of searching by many famous mathematicians (both amateur and professional). Wiles is a former student and collaborator of Australian Mathematician John Coates. Fermat’s Last Theorem is a simple assertion which he wrote in the margin of a mathematics book, but which he never proved, although he claimed he could. The theorem is: The equation x n + y n = z n, when the exponent n is greater than 2, has no solutions in positive integers. Wiles’ work establishes a whole new mathematical theory, proposed and developed over the last 60 years by the finest mathematical minds of the 20th century. 1

Answer these questions based on the biography of Pierre de Fermat. a How old was Fermat when he studied Appollonius?

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

467

b When was his work Introduction to Loci published? What was it about? c Who solved Fermat’s last theorem? d Explain Fermat’s last theorem. Research this theorem and write a one-page report. 2

Write an explanation of each glossary term.

3

Use every third letter to find a sentence. DFTEDWVGOHULIOIKJNHGEFDSASAERRTGEBNPMIEKJRGTP QAEZSNWEDRFIDCCSHUKOLPLADURIOIDEFASTFVHBKENJPM HRKIOLODPOUTUCXYTTTORRFEETDDHFFEXFIVHRNPGOIRYT AEDDSVIHUEKANEDTASSCFIVGSFDNESEAEGAEAAETDRIFGV HNEMJOTGNAXEHUAJKNUPDZCTVBHMKELPYIOAUYRUIEIJPH GAIERETARELETLRTEASLCVIBHFNGTDSHFGEHNIMJRIKGUY RTRAEDDEDISWEADNASTCVSBNANNRMMESSERRQAAUSSAQWL

Glossary boundary gradient–intercept form linear relationship product translation

equation inequality parallel reflection y-intercept

CHECK YOUR SKILLS 1

2

3

4

✓

general form inequation perpendicular region

gradient line point sketch

The equation of the line passing through (–2, –6) with gradient 2 is: A y = 2x + 3 B y = 2x – 7 C y = 2x – 2

D y = 2x + 7

The equation of the line passing through A(–4, 5) and B(2, –13) is: A y = –x – 1 B y=x+5 C y = –3x – 7

D y = –x + 1

The equation of the line cutting the y axis at –4 and the x axis at +5 is: 4 4 5x A y = --- x – 4 B y = – --- x – 4 C y = ------ – 4 5 5 4

5x D y = – ------ – 4 4

The equation of this line is: A y = 3x – 3 B y = –3x – 3 C y = –x + 3 D y = 2x – 3

y 3 2 1 –3 –2 –1

x –1 –2 –3

1

2

3

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

5

6

The value of a if (a, 1) lies on the line y = 3x – 2 is: A 0 B 1 C 2

D 3

1 In general form y = --- x – 2 A 6y – 3x + 8 = 0

D

4 --- is: 3 B 3x – 6y – 8 = 0

C 3x + 2y – 4 = 0

1 4 --- x – y – --- = 0 2 3

7

When written in gradient–intercept form, the equation 3x + 2y + 8 = 0 is: 3x 3x A 2y = 3x + 4 B 3x – 2y = –8 C y = – ------ + 8 D y = – ------ – 4 2 2

8

The x- and y-intercepts of 4x + 3y – 12 = 0 are: A 4 and –3 B 3 and 4

9

10

11

12

The line parallel to y = –3x + 1 is: 1 1 A y = --- x + 1 B y = – --- x + 1 3 3 The line perpendicular to 3x + 5y + 7 = 0 is: A 3x + 5y – 2 = 0 B 3x – 5y – 7 = 0

C –4 and 3

✓ D –3 and 4

C y = –3x + 3

D y = –x + 3

C 3x + 5y + 7 = 0

5x D y = ------ + 2 3

The triangle ABC with vertices A(2, 0), B(1, 4) and C(6, 3) is: A scalene B isosceles C equilateral The region defined by y ≤ 3 is: A 4

B

y

4

3

3

2

2

1 –4 –3 3 –2 2 –1 1

1

x –1

1

2

–4 –3 3 –2 2 –1 1

3 4

x

1

2

1

2

–1 –2 –3

–2 –3

C

D 4

y

4

3

3

2

2

1 –4 –3 3 –2 2 –1 1

y

–1 –2 –3

2

3 4

y

3 4

1

x 1

D right angled

–4 –3 3 –2 2 –1 1

x –1 –2 –3

3 4

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

13

This region is defined by: A x ≤ –3 and y ≤ 2 B x > –3 and y ≤ 2 C x < –3 and y ≤ 3 D x ≥ –3 and y ≥ 2

4

469

y

3 2 1 –4 –3 3 –2 2 –1 1

x –1

1

2

3 4

–2 –3

14

The region defined by 3x – y ≤ 6 is: A

B

3

2

2

1 –4 –3 –2 –1 1

1

x –1

1

2

3 4

–4 –3 –2 –1 1

–2 –3 3

x

–1

1

2

3 4

1

2

3 4

–2 –3 3

–4

–4

–5

–5

–6

–6

–7

–7

C

D y

y

3

3

2

2

1 –4 –3 –2 –1 1

✓ y

y 3

1

x –1

1

2

3 4

–4 –3 –2 –1 1

–2 –3 3

x

–1 –2 –3 3

–4

–4

–5

–5

–6

–6

–7

–7 –8

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1–5

6–8

9, 10

11

12,13

14

Section

A and B

C

D

E

F

G

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

REVIEW SET 14A 1

For the points A(3, –1) and B(–5, 0) find: a the gradient of AB b the equation of the line through A and B c the x- and y-intercepts of the line AB d the equation of the line perpendicular to AB passing through A e the equation of the line parallel to AB passing through the point (3, 0)

2

a Write 3y = 4x – 7 in general form. b Find the equation of the line passing through (–4, 7) and (2, 6) in general form. c Sketch the line 3x – 4y + 12 = 0.

3

a Find the fourth vertex of the parallelogram ABCD for A(–7, 11), B(6, 5) and C(3, 8). b Find the equation of the line from A(7, 2) to the midpoint of the line segment joining B(–6, 4) to C(3, –1).

4

a Find k if 2x + ky = 5 is perpendicular to x – 3y = 11. b A(–1, 3), B(2, 4) and C(t, –1) are collinear. Find t.

5

Write the equation of each inequality. a b 4

y

y

4

3

3

2

2 1

1 –4 –3 –2 –1

x

x –1

1

2

3 4

–6 6 –5 5 –4 –3 –2 –1

–1

–2 –3

–2 –3 3

–4

–4

1

2

3 4

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

REVIEW SET 14B 1

Find the equation of the line passing through M(–4, 2) and N(3, 7).

2

Find the equation of the line passing through (–4, 5) perpendicular to 3x – 2y + 7 = 0.

3

2 Write y = – --- x – 4 in general form. 3

4

Find the equation of the perpendicular bisector of the join of (–2, 3) and (4, 5).

5

Sketch these regions: a y < –1

b

x–y>6

c 2x – 3y ≤ 12.

REVIEW SET 14C 1

For the points P(–6, 3) and Q(–1, –5) find: a the gradient of PQ b the equation of the line through P and Q c the x- and y-intercepts of the line PQ d the equation of the line perpendicular to PQ passing through Q e the equation of the line parallel to PQ passing through the origin

2

a Write 4x – 7y + 8 = 0 in gradient–intercept form. b Find the equation of the line passing through (4, –2) and the origin in general form. c Sketch the line 6x – 7y + 9 = 0.

3

a Determine z if 2x + 5y = 11 and zx – 3y = 9 are perpendicular. b A(–2, 4), B(3, 7) and D(–1, d) lie on the same straight line. Find the value of d. c Find the coordinates of D for parallelogram ABCD if A(–2, 3), B(1, 7) and C(5, 1) are vertices.

4

Sketch x – 3y = 7.

5

Sketch these regions: a x ≤ –1

b 2x – y > 4

471

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Straight Lines and Regions (Chapter 14) Syllabus reference PAS5.3.3

REVIEW SET 14D 1

a Find the equation of the line with gradient 3 passing through (–2, 4). b Find the equation of the line parallel to 3x – 5y = 8 passing through the origin.

2

3x Write y = – ------ + 2 in general form. 4

3

Find the area of the triangle formed when 2x – 3y = 6 cuts the coordinate axes.

4

Does (1, 4) lie on the line 3x – 5y + 2 = 0? Explain.

5

Sketch: a the intersection of y ≥ –2 and x < –1 b 3x – y > 6 c y ≤ –4x

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Chapter 15 Surds and Indices This chapter deals with defining the system of real numbers, distinguishing between rational and irrational numbers, performing operations with surds, using integers and fractions for index notation, and converting between surd and index form. After completing this chapter you should be able to: ✓ define real numbers and distinguish between rational and irrational numbers ✓ simplify expressions involving surds ✓ expand expressions involving surds ✓ rationalise the denominators of simple surds ✓ use the index laws to define fractional indices ✓ translate expressions in surd form to expressions in index form and vice versa ✓ evaluate numerical expressions involving fractional indices ✓ use the calculator to evaluate fractional powers of numbers ✓ evaluate a fraction raised to the power of –1 ✓ prove some general properties of real numbers.

Syllabus reference NS5.3.1 WM: S5.3.1–S5.3.5

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Diagnostic Test 1

Which of the following statements is not correct?

10

A the square root of 4 is 2 or –2

2

3

B

4 = 2

D

–4 = –2

4

12

B 225

C 15

In its simplest form,

Written in the form

6

7

8

30

40 ---------- = 5 A 8

B

13

32 =

180

10 – 2 6

2( 5 – 2 3) =

150 D

900 14

1 B --8

C 2 2 D

D 2 1--9-

B 10 10 – 2 5

C 8 5

D 2 15 + 6 10 17 B 4 2

C 2 3+ 2

D 2 5

C

10 – 4 3

D 10 – 2 6

2

( 5 + 2) = B 7 + 2 10 14

D 2 7

Expressed with a rational denominator,

15 ---------- C 6

B

5 ------3

D

10 ---------6

1 --2

10 = 1 B --5

In index form, B k

1 10 D ---------10

C 4

4 --3

3

k = C k

3 --4

D k7

2 --3

When evaluated, 27 = A 18

18 + 2 = 20

B

A k12 16

A 8 15

10 – 2 3

A 5

35 15

C 4 1--3-

A

5 ----------- = 2 3 5 A ------6

4 10 – 2 5 + 6 10 =

A

D

C

D 9 5

n, 5 6 =

180 C

1 4 --- = 9 37 A ---------- B 2 1--33

9

C 10 30

A 7

(3 5 )2 =

A

B 30 2

11

A 2 8 B 8 2 C 16 2 D 4 2 5

A 30 6

C – 4 = –2

Which of the following is not a rational number? 9 A 11 B ------ C 2 3--4D –4.7 16 A 45

2 3×5 6 =

3 ---  5

B 9

C 6

D 40.5

3 B – --5

5 C – --3

D

–1

A 1 2--3-

= 1 -----15

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1

2

3–7

8, 9

10–12

13

14–16

17

Section

A

B

C

D

E

F

G

H

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A. THE SQUARE ROOT OF A NUMBER The square root of a number, x, is that number which when multiplied by itself is equal to x. For example, the square root of 9 is 3 or –3, since 32 = 9 and (–3)2 = 9. x is the positive square root of x, e.g.,

9 = 3. – x is then the negative square root of x.

Example 1 Write down: a the square root of 81

b

81

c – 81

a the square root of 81 = 9 or –9

b

81 = 9

c – 81 = –9

Exercise 15A 1

Write down: a the square root of 4 d the square root of 25 g the square root of 49 j the square root of 64

4 25 49 64

b e h k

c f i l

– – – –

4 25 49 64

Example 2 Find, where possible: a

9

a

9 =3

b

– 9

c b

–9

d

0

– 9 = –3

c Since there is no number which multiplied by itself is equal to –9, it is not possible to find – 9 . We say that – 9 is undefined. d The square root of 0 is 0, since 0 0 = 0. Zero is neither positive nor negative but we define 0 = 0. 2

Find, where possible: a 4 b – 4 f – 16 g – 16 k – 25 l – 81

c h m

From the above examples: •

x is undefined for x < 0

•

x = 0 for x = 0

•

x is the positive square root of x when x > 0

–4 36 – 64

d 0 i – 36 n – 100

e j o

16 – 36 –1

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B. THE REAL NUMBER SYSTEM Real numbers are those that can be represented by points on a number line. Real numbers are either rational or irrational. a A rational number is one that can be expressed as the ratio - of two integers, where b ≠ 0. b An irrational number is a real number that is not rational.

Example 1 Show that the following are rational numbers: 3 a 2 --b 0.637 c 3 4

d 0. 4˙

e –3.1

3 11 a 2 --- = -----4 4 a 3 This is in the form --- , where a and b are integers, hence 2 --- is a rational number. b 4 637 b 0.637 = ------------1000 a This is in the form --- , where a and b are integers, hence 0.637 is a rational number. b 3 c 3 = --1 a This is in the form --- , where a and b are integers, hence 3 is a rational number. b 4 d 0. 4˙ = --- (You learnt how to convert recurring decimals into fractions in chapter 5.) 9 a This is in the form --- , where a and b are integers, hence 0. 4˙ . is a rational number. b 1 e – 3.1 = – 3 ----10 31 = – -----10 a This is in the form --- , where a and b are integers, hence –3.1 is a rational number. b From Example 1 we note that: Any terminating or recurring decimal is a rational number.

Exercise 15B 1

2

a Show that the following are rational numbers by expressing them in the form --- . b a 4 2--3b 0.91 c 5 d 0. 7˙ e –5 1--2f 4 16 --g 0. 5˙ 3˙ h –2.6 i j k 30% l 9

2.84

Convert the following rational numbers to decimals. 3 5 -----a --b 1 5--8c 4 2--3d 5 12 g 6.5% h 17 2--3- %

69%

e

1 --7

f

7.3%

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From question 2 we note that: When a rational number is converted to a decimal, the decimal either terminates or recurs.

Example 2 Convert the following real numbers to a decimal and discuss whether they are rational or irrational. a

2

b

5

Using a calculator a 2 = 1.414 213 562… b 5 = 2.236 067 978… Since neither decimal terminates or recurs (although we have only been able to show an answer to 9 decimal places, the limit of the calculator display) these numbers cannot be expressed as the ratio of two integers and hence are not rational. They are irrational numbers.

Example 3 Determine whether the following real numbers are rational or irrational. 16 a 6 b -----49 a

b

3

6 = 2.449 897 43… Since the decimal neither terminates nor recurs, it cannot be expressed as the ratio of two integers, hence 6 is an irrational number. 4 4 16 16 4 ------ = --- (since --- × --- = ------ ) 7 7 49 49 7 a This is in the form --- , where a and b are integers, hence b

16 ------ is a rational number. 49

Determine whether the following real numbers are rational or irrational 4 a 8 b c d -----9 11 25

e

5 -----16

Example 4 Using a calculator statements: a

6 = 2.44

6 = 2.449 489 743… Write true or false for the following b

6 = 2.449

c

6 = 2.4494

☞

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a 2.442 = 5.9536 The statement is false. b 2.4492 = 5.997 601 The statement is false. c 2.44942 = 5.999 560 36 The statement is false. Because 6 is irrational its exact value cannot be written as a decimal. The values given in parts a, b and c are rational approximations for 6 . 4

Using a calculator a 2 = 1.41

5

Write rational approximations, correct to 3 decimal places, for: a 11 b 15 c 37 d 99

6

2 = 1.414 213 562… Write true or false for the following statements: b 2 = 1.414 c 2 = 1.4142

e

151

a Using a ruler and set square, copy the diagram below into your exercise book. P1 1 –3

–2

–1

O 0

1 Q 2

3

b Use Pythagoras’ theorem to calculate the length of the interval OP1. c Using a pair of compasses as shown, accurately mark the position of line. 7

2 on the number

a Extend the diagram in question 1 as shown below. P2 1 P1 2 –3

–2

–1

O 0

1 1 Q 2

3

b Calculate the length of OP2. c Mark the position of 3 on the number line. 8

Extend the diagram in question 7 to show the position on a number line of

4,

5,

6, …

Extension: Proof that

2 is irrational

In example 2 we cannot be certain that the decimal form of 2 does not terminate or recur after some large number of decimal places, hence it is not a proof that 2 is irrational. Work through the following proof with your teacher. a Assume that 2 is rational. That is, assume that 2 can be written in the form -- , where a and b are integers b and the fraction is written in its simplest form (i.e. a and b have no common factors). a If 2 = -b 2

then, squaring both sides and

a 2 = ----2b a 2 = 2b 2 …. (1)

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Hence a 2 is even (any multiple of 2 is even) and therefore a is even. If a is even, then a may be written in the form 2k, where k is an integer.

Substituting a 2 = 4k 2 into (1),

i.e. a a2 4k 2 b2

= 2k = 4k 2 = 2b 2 = 2k 2

Hence b 2 is even and therefore b is even. a But if a and b are both even then the fraction -- cannot be in its simplest form, which contradicts our original b statement. a Therefore 2 cannot be written in the form -- where a and b are integers with no common factor, i.e. 2 is b not rational.

C. PROPERTIES OF SURDS In section B we distinguished between rational and irrational numbers. The set of irrational numbers contains numbers such as 2 , 3 2 , π etc. Irrational numbers that contain the radical sign are called surds. When working with surds we may use the following properties: If x > 0 and y > 0, 2

1 ( x) = x =

x

A set is a group of objects (numbers, letters, names …).

2

xy = x × y x x - = -----y y

2 3

Example 1 Simplify: 2

a

( 5)

a

5 is the positive number which multiplied by itself is equal to 5. 2 Hence ( 5 ) = 5

c (3 5 )2 = 3 5 × 3 5 = 9 × ( 5 )2 =9×5 = 45

b

5

2

c (3 5) b

2

5 = 25 =5

2

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Surds and Indices (Chapter 15) Syllabus reference NS5.3.1

Exercise 15C 1

Simplify: a ( 11 )2 f (3 2 )2

2

c ( 8 )2 h (5 2 )2

b 11 g (2 3 )2

2

d 8 i (10 7 )2

e ( 6 )2 j (4 5 )2

Example 2 Simplify: a

5× 3

Using property 2 from above: a 5× 3 = 5×3 = 15

2

Simplify: 2× 7 a d 7 × 11 g 7 × 5 × 10

b e

b

6× 7

b

6× 7 = =

6×7 42

3 × 10 13 × 17

c f

5× 2 3× 2× 5

Example 3 Simplify: a

28

b

45

Using property 2: a 28 = 2 × 14 or 4 × 7 = 2 × 14 or 4 × 7 Since 4 is a perfect square, 4 simplifies to 2 and we choose the second product. i.e. 28 = 4 × 7 = 2× 7 = 2 7 b We look for factors of 45, one of which is a perfect square. 45 = 9 × 5 = 3 5

3

Simplify: a 12 f 90 k 24

b g l

20 50 32

c h m

18 75 48

d i n

27 200 72

e j o

8 98 128

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Example 4 Simplify: a

3 × 12

b

2× 6

a

3 × 12 = 36 = 6

b

2× 6 =

12 4× 3

=

= 2 3 4

Simplify: a 8× 2 f 14 × 2

b g

2 × 32 8× 5

c h

5 × 20 15 × 3

d i

2 × 10 15 × 5

10 × 5 3× 8

e j

Example 5 Express 3 5 in the form 3 5 = =

5

n.

9× 5 45

Express in the form n . a 3 2 b 2 3

c 4 5

d

5 2

e

10 7

Example 6 Simplify: 16 a -----25

b

Using property 3: 16 16 a ------ = ---------25 25 4 = --5

6

Simplify: 9 a -----16 f

21 -----9

b

11 -----25

21 ---  4

c

11 11 ------ = ---------25 25 11 = ---------5

c 21 --- =  4

9 --4

1 3 = --- = 1 --2 2

b

9 -----25

c

17 -----25

d

5 -----16

e

11 -----16

g

61 ---  4

h

17 ---  9

i

13 ---  4

j

25 ---  9

481

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Example 7 Simplify: 12 a ---------3

b

15 ---------5

b

15 ---------- = 5

c

40 ---------5

c

40 ---------- = 5

Using property 3: 12 a ---------- = 3

12 -----3

= 4 = 2 7

3

=

=

40 -----5 8

= 2 2

Simplify: 24 a ---------6

18 b ---------2

c

30 ---------6

24 g ---------3

24 h ---------2

f 8

15 -----5

18 ---------3

20 d ---------10 i

24 e ---------8

32 ---------2

j

54 ---------3

Determine whether the following statements are true or false. Give a reason. 2

a ( 5) = 5 e

20 × 5 = 10

b 3× 7 = f

12 ---------- = 2

2

c (4 2) = 8

21 6

d

18 = 2 3

1 1 1 --- = 1 --4 2

g

D. ADDITION AND SUBTRACTION OF SURDS Example 1 Simplify: a 3 2+5 2

b

8 5–2 5

a 3 2+5 2 = 8 2

b

8 5–2 5 = 6 5

Exercise 15D 1

Simplify: a 5 3+4 3

b 7 11 + 6 11

c 7 5–3 5

e 6 3+4 3+5 3

f

8 3+5 3–7 3

g 15 6 – 3 6 – 4 6

h

i

5 10 – 7 10

j

8 5–3 5+2 5

3 5–8 5+2 5

d 9 7–4 7

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Example 2 Simplify: a 4 2+5 3+3 2

b

7 6+4 7–3 6–5 7

b

7 6+4 7–3 6–5 7 = 4 6– 7

Collect like surds. a 4 2+5 3+3 2 = 7 2+5 3

2

Simplify: a 5 2+4 3+6 2

b 7 3+4 5+3 5

c 6 2+5 2+2 3

d 5 7 – 2 10 + 3 7

e 6 5 + 2 11 – 3 5

f

7 10 – 4 6 – 6 10

g 5 2+6 3–3 2+ 3

h

6 3+2 7–5 3–4 7

i

j

4 11 – 3 10 – 6 11 – 2 10

10 5 – 4 3 – 5 3 + 2 5

Example 3 Simplify: a

18 + 2

b

50 – 18

a

18 + 2 = 3 2 + 2

b

50 – 18 = 5 2 – 3 2 = 2 2

= 4 2

3

4

5

Simplify: a

18 + 4 2

b

12 + 5 3

c

20 – 2 5

e

45 + 20

f

54 – 24

g

48 – 12

h

24 + 3 6 – 4 6

i

j

75 – 48 – 27

6 2 – 18 – 2 2

d 6 2– 8

Simplify: a 5 6 + 24 – 3 5

b

d

e 7 5 – 20 + 45 – 5 6

28 + 27 + 63 + 12

50 + 6 3 – 8

c 8 10 + 4 5 – 90

Determine whether the following statements are true or false. Give reasons. a

5+ 3 =

8

d 4 3 + 2 7 = 6 10

b 3 5–2 5 = 1 e 6 2 =

72

c

24 – 6 =

6

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E. MULTIPLICATION OF SURDS Example 1 Simplify: a 4×3 7

b 3 5×2 3

c

6 2×3 8

d 5 8×4 3

b 3 5 × 2 3 = (3 × 2) × ( 5 × 3)

a 4×3 7 = 4×3× 7

= 6 15

= 12 7

d 5 8 × 4 3 = (5 × 4) × ( 8 × 3)

c 6 2 × 3 8 = (6 × 3) × ( 2 × 8)

= 20 × 24

= 18 × 16 = 18 × 4 = 72

= 20 × 2 6 = 40 6

Exercise 15E 1

Simplify: a 5×2 3

b 2×6 2

e 3 3×6 2

f j

6 2 × 3 18

i

m 5 3×2 8

c 4 7 × 10

d 2 6×5

10 5 × 3 7

g 6 7×7 6

h 5 12 × 2 3

2 32 × 5 2

k

o 6 6×2 2

5 × 3 20

l

3 6×4 3

p 3 10 × 2

Example 2 Simplify: a 3( 7 + 2 6) a

2( 7 – 3 5)

b

3( 7 + 2 6) = 3 × 7 + 3 × 2 6

b

4 3(2 5 + 3 2)

c

2( 7 – 3 5) = =

= 3 7+6 6

2× 7– 2×3 5 14 – 3 10

c 4 3 (2 5 + 3 2) = 4 3 × 2 5 + 4 3 × 3 2 = 8 15 + 12 6

2

Simplify: a 5( 6 + 3) d

3( 5 + 2)

g 3 2( 7 – 4 3) j

2 10 ( 6 2 – 3 10 )

b 2( 5 + 4 3)

c 4 ( 2 10 – 3 2 )

5( 6 – 4 3)

f

2(3 5 + 2 7)

h 4 3 ( 2 5 + 5 10 )

i

3 5(2 3 + 4 5)

k 5 2(3 5 – 2 6)

l

3 6(4 3 + 2 8)

e

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Example 3 Simplify: a ( 5 + 3)( 5 – 4)

b (2 3 – 2 5)(4 3 + 5)

a ( 5 + 3)( 5 – 4) =

( 7 + 3)

c

2

5( 5 – 4) + 3( 5 – 4)

= 5 – 4 5 + 3 5 – 12 = –7 – b

5

(2 3 – 2 5)(4 3 + 5) = 2 3(4 3 + 5) – 2 5(4 3 + 5) = 24 + 2 15 – 8 15 – 10 = 14 – 6 15

c

2

2

( 7 + 3) = ( 7) + 2 × 7 × 3 + ( 3)

2

= 7 + 2 21 + 3 = 10 + 2 21

3

Simplify: a ( 3 + 5)( 3 + 2)

b ( 7 + 3)( 7 – 4)

c ( 10 – 6 ) ( 10 – 1 )

d ( 5 – 3)( 5 + 3)

e (2 3 + 5)( 3 + 1)

f

(3 2 + 4)(2 2 – 7)

g (2 5 – 3)(4 5 + 3)

h (5 2 – 2 3)(3 2 + 4 3)

i

(3 7 + 5 2)( 7 – 4 2)

k (3 5 + 4 2)(3 5 – 3 2)

l

(2 7 + 5)(2 7 – 5)

j

(2 6 – 4 3)(3 6 – 3)

m (4 5 – 3 2)(4 5 + 3 2) n ( 3 + 6) p ( 10 – 5 ) 4

2

2

o ( 5 + 2)

q (2 5 + 3)

2

2

r (3 7 – 2 5)

2

Determine whether the following statements are true or false. Give reasons. a 2 3×3 2 = 6 6 d

7( 2 – 3) =

14 – 3

b 4 × 3 5 = 12 60

c 3 12 × 5 3 = 90

2

e ( 5 + 3) = 8

F. RATIONALISING THE DENOMINATOR OF A SURD Example 1 Rationalise the denominator of: 1 5 a ------b ------7 7

c

3 ------7

d

5 3 ----------7

☞

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Rationalise the denominator of a fraction means to convert it to an equivalent fraction with a rational denominator. Since a × a = a , we proceed as follows. 5 7 5 b ------- = ------- × ------7 7 7

1 7 1 a ------ = ------- × ------7 7 7

c

7 = ------7

5 7 = ----------7

3 7 3 ------- = ------- × ------7 7 7

5 3 7 5 3 d ----------- = ----------- × ------7 7 7 5 21 = -------------7

21 = ---------7

Exercise 15F 1

Rationalise the denominator of: 1 a ------5

1 b ------3

c

f

8 ------7

3 g ------5

7 h ------3

i

k

3 2 ----------5

l

3 7 m ----------6

5 5 n ----------2

4 3 ----------10

1 ---------10

5 d ------2

3 e ------6

11 ---------6

j

2 ------7

4 10 o -------------3

Example 2 Rationalise the denominator of: 1 5 a ----------b ----------4 7 4 7

c

1 7 1 a ---------- = ----------- × ------4 7 7 4 7

c

b

5 7 5 ----------- = ----------- × ------4 7 7 4 7

7 = ------28

2

5 3 ----------4 7 5 3 7 5 3 ----------- = ----------- × ------4 7 7 4 7

5 7 = ----------28

5 21 = -------------28

Rationalise the denominator of: a

1 ----------2 3

b

1 ----------3 5

c

4 ----------5 2

d

8 ----------3 7

e

5 ----------3 2

f

10 ----------4 3

g

5 7 -------------3 10

h

6 2 ----------5 3

i

3 5 -------------2 11

j

2 7 ----------5 6

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Extension

Example 3 a Expand and simplify ( 5 + 3 ) ( 5 – 3 ) . 1 b Hence rationalise the denominator of i --------------------5+ 3

1 ii --------------------5– 3

a ( 5 + 3 ) ( 5 – 3 ) = 5 – 15 + 15 – 3 = 2

b i

1 5– 3 1 --------------------- = --------------------- × --------------------5+ 3 5– 3 5+ 3

ii

1 5+ 3 1 --------------------- = --------------------- × --------------------5– 3 5+ 3 5– 3 5+ 3 = --------------------2

5– 3 = --------------------2 3

a Expand and simplify ( 7 + 2 ) ( 7 – 2 ) . 1 i -------------------------( 7 + 2) a Expand and simplify ( 10 + 3 ) ( 10 – 3 ) . 1 b Hence rationalise the denominator of i ----------------------------( 10 + 3 ) a Expand and simplify ( 5 + 2 ) ( 5 – 2 ) . 1 b Hence rationalise the denominator of i ---------------------( 5 + 2) a Expand and simplify ( 6 + 3 ) ( 6 – 3 ) . 5 b Hence rationalise the denominator of i ---------------------(6 + 3) a Expand and simplify ( 2 3 + 2 ) ( 2 3 – 2 ) . 5 b Hence rationalise the denominator of i -----------------------------(2 3 + 2) a Expand and simplify ( 3 2 + 5 ) ( 3 2 – 5 ) . ( 2 + 3) b Hence rationalise the denominator of i ------------------------(3 2 + 5) Rationalise the denominator of: b Hence rationalise the denominator of

4

5

6

7

8

9

Remember that ˙2 2 (a + b)(a – b) = a – b

1 a --------------------( 6 – 2)

3 b ----------------------------( 10 + 2 )

( 10 – 2 ) d ------------------------10 + 2

(2 5 + 3 2) e --------------------------------(2 5 – 3 2)

1 ii ------------------------( 7 – 2) 1 ii -----------------------10 – 3 1 ii --------------------( 5 – 2) 10 ii --------------------(6 – 3) 7 ii ----------------------------(2 3 – 2) ii

( 5 + 1) c ---------------------3– 2

(3 2 – 4) ------------------------(3 2 – 5)

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G. FRACTIONAL INDICES Example 1 1 4

 --4- a Use the index laws to simplify  a  .   4

b Simplify ( 4 a ) . a

1 4

 --4- a   

= a = a

c b

1 --- × 4 4

1 --4

Hence show that a = 4

(4 a) =

4

4

a.

a×4 a×4 a×4 a

= a 1

= a 1 4

 --4- c Since  a   

1 --4

4

= ( 4 a ) = a , then a =

4

a

Exercise 15G 1 2

1

 --2- a Use the index laws to simplify  a  . 2   b Simplify ( a ) . 1--2 c Hence show that a = a . 1 3

2

 --3- a Use the index laws to simplify  a  . 3   b Simplify ( 3 a ) . --13 c Hence show that a = 3 a .

Example 2 Write in surd form: a k a

3

1 --5 1 --5

k =

5

a

b

z

b

z

1 -----10 1 -----10

=

10

z

Write in surd form: a m

1 --4

e 17

1 --6

b y f

1 --2

25

c p 1 --4

1 --3

g 62

d t 1 --3

1 --8

h a

1 --n

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Example 3 Evaluate: a 25

1 --2

25 = 25 = 5

a

4

1 --4

b

256

b

256 = 4 256 = 4 ( since 4 × 4 × 4 × 4 = 256 )

1 --4

Evaluate: a 49 f

5

1 --2

8

1 --2

b 27

1 --3

1 --3

h 121

c 1 --2

i

625 81

1 --4

d 32

1 --4

j

64

1 --5

e 1000 000

1 --3

k 1

1 -----10

Use your calculator to evaluate, correct to 2 decimal places: a 5

1 --3

b 2

1 --4

c 298

1 --5

d 41

1 --2

e 831

Example 4 Write in index form: a a

6

3

3

7 7 = 7

1 --3

Write in index form: a 10 6 e 56

b f

3 3

b

5

m

b

5

m = m

1 --5

c g

7 y

Example 5 Write in index form: a ( m)

3

b

m

3 1 --3 2

1 3

a

 --2- ( m) = m    3

= m = m

1 --- × 3 2 3 --2

b

3

m = (m ) = m = m

1 3 × --2 3 --2

4 5

5 k

d h

5

38 m

1 --6

1 --6

489

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490

Surds and Indices (Chapter 15) Syllabus reference NS5.3.1

7

Write in index form: 5 a ( m) b f

a

5

m 4

g

5

c (3 m)

3

h

k

3

w

2

4

d

3

m

i

5

y

2

e ( a)

3

j

(4 t)

Example 6 Write down the meaning of: a k

3 --4

b

 --4- k = k   

a

1 --3 4

1 3

3 --4

3

8

4

k

5 --3 1 5

5 --3

1 --5 3

 --3- b w = w   

or ( k )

= ( 4 k ) or

w

3

= (3 w)

or ( w ) 5

or

3

w

5

Write down the meaning of: a

k

2 --3

b

m

4 --3

c

t

3 --2

d

a

3 --5

e

w

Example 7 Evaluate: a 8

5 --3

5 ---

a 83 = ( 3 8 )

5

25

b

25 = ( 25 )

3 --2

= 5 = 125

Evaluate: a e

10

3

3

5

= 2 = 32

9

3 --2

b

9 4

3 --2

b

5 --2

f

8 8

2 --3

c 16

4 --3

g 49

3 --4

d 27

3 --2

h 32

Use your calculator to evaluate: 4 ---

a 729 3

3 ---

b 16 807 5

13 ------

c 1024 10

4 --3 3 --5

5 --2

5

7

LEY_bk9_15_finalpps Page 491 Wednesday, January 12, 2005 12:08 PM

Surds and Indices (Chapter 15) Syllabus reference NS5.3.1

H. SOME GENERAL PROPERTIES OF REAL NUMBERS Exercise 15H 1

Use the fraction key on your calculator to evaluate (as a fraction): a

2

5 ---  2

–1

b

6 ---  5

–1

c

9 ---  7

–1

d

2 ---  3

–1

e

3 ---  4

–1

a –1 b Show that  --- = --- .  b a

3

If a and b are real numbers, determine whether the following statements are true or false. If false give a counter example. A counter example (Hint: Try several different pairs of real is an example that numbers to test the truth of the statements.) demonstrates the statement is false. a i a + b is a real number ii a – b is a real number iii a × b is a real number iv a ÷ b is a real number b i a+b=b+a ii a – b = b – a iii a × b = b × a iv) a ÷ b = b ÷ a c i (a + b) + c = a + (b + c) ii (a – b) – c = a – (b – c) iii (a × b) × c = a × (b × c) iv (a ÷ b) ÷ c = a ÷ (b ÷ c) d i a×0=0 ii a + 0 = a iii a × 1 = a iv a ÷ a = 1

4

If m and n are rational numbers, determine whether the following statements are true or false. If false give a counter example. a m + n is always rational b m – n is always rational c m × n is always rational d m ÷ n is always rational

5

Find a pair of surds which satisfy the condition: a the product of the surds is irrational b the product is rational c the quotient is irrational d the quotient is rational

6

a Write down three consecutive integers starting with y. b Hence show that the sum of any three consecutive integers is divisible by 3.

7

a Show that any even number can be written in the form 2k, where k is an integer. b Show that any odd number can be written in the form 2k + 1, where k is an integer. c Hence prove the following properties of numbers: i the sum of any two even numbers is even ii the sum of any two odd numbers is even iii the sum of an even number and an odd number is odd iv the product of two even numbers is even v the product of an odd number and an even number is even vi the product of two odd numbers is odd.

491

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492

Surds and Indices (Chapter 15) Syllabus reference NS5.3.1

Language in Mathematics 1

a Explain, in words, how to calculate: i the square of a number ii the square root of a number. b Write down the: i square of 9 ii square root(s) of 9. c Which of the following numbers are perfect squares: 1, 4, 12, 25, 27, 200? d Write in words:

i

m2

a real number

ii

m

iii – m

b rational number

1 ---

iv m 2

2

What is a:

c irrational number?

3

Use words from the list below to complete the following statements. radical, terminating, false, rationalising, recurring a Every rational number can be expressed as a ___ or ___ decimal. b Converting the denominator of a fraction into a rational numbers is called ___ the denominator. c The mathematical sign is called the ___ sign. d A counter example is an example which demonstrates that a statement is ___ .

4

Write down an example of three consecutive integers.

5

a Find the i sum and b What is the quotient when 13 is divided by 5?

6

Use a dictionary to write down two meanings of each of the following words: general, property, index

7

Three of the words in the following list have been spelt incorrectly. Rewrite them with the correct spelling. ratio, convurt, indixes, recur, fractional, intejer

ii product of 5 and 9.

Glossary approximation fractional integer property rational recurring

consecutive general irrational quotient rationalise square

convert index perfect square radical sign real square root

counter example indices product ratio recur surd

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Surds and Indices (Chapter 15) Syllabus reference NS5.3.1

493

CHECK YOUR SKILLS 1

2

Which of the following statements is not correct? A the square root of 25 is 5 or –5 B 25 = 5 Which of the following is not a rational number? A

3

4

5

6

10

11

12

2

B

9 -----16

C 3 3--4-

= B 36

In its simplest form, A 10 8 Written in the form A 14 50 ---------- = 5 A 10

80 = B 8 10 n, 2 7 = B 28

B

10

C 6

✓ D

– 25 = –5

D –21

D

6

C 2 20

D 4 5

C

D

98

196

C 2 5

D 5 2

B 3 1--2-

C 3 1--4-

D 9 1--2-

6 5–2 3+3 5 = A 7 7

B 9 5–2 3

C 7 2

D 4 2+3 5

12 + 27 = A 39

B 5 3

C 5 6

D 3 5

4 3×5 2 = A 20 6

B 60 2

C 5 24

D 120

5(3 2 – 3) = A 3 10 – 3 5

B

C 3 10 – 15

D 3 10 – 3 5

C 5 – 2 14

D 9 – 2 14

1 9 --4 A

9

17

(2 3) A 12

7

8

C – 25 = –5

37 ---------2

30 – 15

2

( 7 – 2) = A 5

B 25

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494

Surds and Indices (Chapter 15) Syllabus reference NS5.3.1

13

3 2 Expressed with a rational denominator, ----------- = 2 5 3 2 3 2 A ----------B ----------10 5

C

3 10 -------------10

C

12

D

6 -------------2 10

1 ---

14

15

12 2 = A 6 In index form, A k

16

1 B --6 3

4

k =

12

B k

When evaluated, 16

3 --4

4 ---  3

4 --3

C k

3 --4

B 21 1--3-

C 8

1 B -----12

3 C --4

–1

1 D --8

–1

=

4 A – --3

✓ D k

=

A 12 17

1 D ---------144

3 D – --4

If you have any difficulty with these questions, refer to the examples and questions in the sections listed in the table. Question

1

2

3–7

8, 9

10–12

13

Section

A

B

C

D

E

F

14–16

17

G

H

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Surds and Indices (Chapter 15) Syllabus reference NS5.3.1

REVIEW SET 15A 1

2

3

4

Find, where possible: a the square root of 16 b

a ( 2)

2

b (5 2)

Express in simplest form: a

7

Express in the form

8

Simplify: a

2

18

2 × 32 = 8

c b

a 2 5

16 -----25

e

1 --3

0

d 3 1--2-

b

17 -----25

f

20 ---------5

6× 7 20

e

3 × 12

c

d

1 2 --4

c

4 1 --9

d

24 g ---------3 15

c

45 = 5 3

20 ---------- = 2

5

f

1 1 4 --- = 2 --4 2

b 8 2–6 6+4 6

c

Write true or false. a 6+ 3 = 9

b

c 5 7–4 7 =

Simplify: a 2 8×5 2

b 6 8×2 3

27 – 6 =

21

50 – 18

c 4 × 5 10

f ( 6 + 3)( 6 – 2)

Rationalise the denominator of: 1 a ---------10

8× 3

b 5 2

b 3 5 = e

16 -----9

d 3 5×2 7

Simplify: a 5 3+7 3–2 3

e 3 2(2 5 – 3 3) 13

n:

15 e ---------3 Write true or false. 2 a 7 = 7 d

12

– 16

Determine whether the following real numbers are rational or irrational. 3 a –3 b 5 c 9 d --4

6

11

d

Convert the following rational numbers to decimals. 3 a --b 23% c 8

Simplify:

10

c – 16

a Show that the following are rational numbers by expressing them in the form --- . b a 3 1--4b 5 c 0.83 d 0. 5˙

5

9

16

b

g (3 7 + 4)

3 -------------2 10

d 2 5 + 20

7

d 2

5( 7 + 2)

495

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496

Surds and Indices (Chapter 15) Syllabus reference NS5.3.1

14

15

a Expand and simplify ( 2 + 1)( 2 – 1). 1 b Hence rationalise the denominator of ----------------- . 2+1 Write as a surd: a k

16

17

1 --3

Write in index form: a z

k

b

3

1 --5

c k

y

3 --4

c

d k

m

2 --3

5

Evaluate: a 25

18

b

1 --2

b 8

5 Write as a fraction  ---  4

4 --3

c 343

2 --3

–1

REVIEW SET 15B 1

2

Find, where possible: a the square root of 36

b

36

c – 36

d

– 36

Convert the following rational numbers to decimals. 5 a --b 49% c 8

4

Determine whether the following real numbers are rational or irrational. 5 a –5.2 b c d 7 --36 8

6

7

0

a Show that the following are rational numbers, by expressing them in the form --- . b 1 a 3 --b 4 c 0.27 d 0. 3˙ 3

3

5

e

Simplify: 2 a ( 3)

b (5 3)

Express in simplest form: a b 8 Express in the form a 2 7

2

45

n: b 4 3

2 --3

1 d 4 --2

e

16 -----9

c

5× 6

d 2 7×6 2

c

8× 2

d

8× 6

LEY_bk9_15_finalpps Page 497 Wednesday, January 12, 2005 12:08 PM

Surds and Indices (Chapter 15) Syllabus reference NS5.3.1

8

9

10

11

12

13

14

15

Simplify: 9 a --4 18 e ---------6

17

c

1 3 --4

d

2

a

10 = 10

b 5 3 =

15

c

54 = 6 3

d

2 × 18 = 6

e

20 ---------- = 4

5

f

1 1 9 --- = 3 --4 2

Simplify: a 4 2 + 5 2 – 10 2

b 5 3–6 5–4 5

Write true or false. a 7 + 5 = 12

b

27 – 12 =

Simplify: a 2 12 × 5 3

b

e 2 5(2 3 – 3 5)

f ( 7 + 3)( 7 – 2 3)

c

45 – 20

15

4 6×2 3

d 3 6 + 24

c 6 6–5 6 = 1

c 3×7 2 g (2 6 – 5)

d

3 ( 10 – 3 )

2

Rationalise the denominator of: 1 2 a ------b ----------3 5 5 a Expand and simplify ( 3 + 2 ) ( 3 – 2 ) . 1 b Hence rationalise the denominator of --------------------- . 3– 2 Write as a surd: 1 --2

b n

Write in index form: a m

1 --4

b

c n

3

w

4 --3

d n

c

t

3

Evaluate: a 49

18

f

7 1 --9 24 g ---------2

11 -----4 24 ---------6

Write true or false.

a n 16

b

1 --2

3 –1 Write as a fraction  --- .  5

b 16

3 --4

c 324

3 --2

3 --5

497

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498

Surds and Indices (Chapter 15) Syllabus reference NS5.3.1

REVIEW SET 15C 1

2

Find, where possible: a the square root of 9 b

9

c – 9

d

–9

e

a Show that the following are rational numbers by expressing them in the form --- . b 7 a 1 --b 2 c 0.314 d 0. 6˙ 8

3

Convert the following rational numbers to decimals. 1 a --b 137% c 8

4

Determine whether the following real numbers are rational or irrational. 7 a –17 b 11 c 100 d --9

5

6

7

8

9

10

11

12

0

Simplify: 2 a ( 7)

b (2 7)

Express in simplest form: a 60 b Express in the form a 10 2

2

54

5 --9

2 d 5 --3

e

c

5 × 11

d 3 6×2 5

c

20 × 5

d

24 × 2

d

5 1 -----16

n:

Simplify: 9 a -----16 30 e ---------10 Write true or false. 2

b 3 7

b f

9 1 -----16 48 g ---------6

17 -----16 27 ---------3

c

a

3 = 3

b 4 3 =

12

c

63 = 7 3

d

27 × 3 = 9

e

40 ---------- = 4

10

f

4 2 1 --- = 1 --9 3

Simplify: a 2 5+8 5– 5 Write true or false. a 10 + 10 = 20 Simplify: a 2 8×5 8 e 5 2(3 5 – 2 2)

16 -----25

b 6 3–5 3+2 7

b

12 – 3 = 3

b 4 8×2 6 f ( 10 + 5 ) ( 10 – 6 )

c

d

32 – 18

3 + 27

c 9 5–8 5 = 1 c 3×2 7 2 g ( 5 + 2 3)

d

3( 6 + 5)

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Surds and Indices (Chapter 15) Syllabus reference NS5.3.1

13

14

15

Rationalise the denominator of: 2 1 a ------b ----------3 7 7 a Expand and simplify ( 5 + 2 ) ( 5 – 2 ) . 1 b Hence rationalise the denominator of ----------------5+2 Write as a surd: a w

16

17

1 --3

Write in index form: a n

b

c w

4

3 --2

d w

z

3

c

n

2 --3

2

Evaluate: a 16

18

b w

1 --6

1 --2

b 25

3 --2

c 9261

4 --3

3 –1 Write as a fraction  --- .  8

REVIEW SET 15D 1

2

Find, where possible: a the square root of 81 b

81

c – 81

d

– 81

Convert the following rational numbers to decimals. 13 a ---------b 174% c 100

4

Determine whether the following real numbers are rational or irrational. 17 a –7 b c d 15 36 -----36

6

7

0

a Show that the following are rational numbers by expressing them in the form --- . b 7 a 4 --b –6 c 0.09 d 0. 2˙ 8

3

5

e

Simplify: 2 a ( 11 )

b ( 3 11 )

Express in simplest form: a b 12 Express in the form a 6 2

2

48

n: b 3 3

5 --6

3 d 3 --4

e

25 -----36

c

8× 7

d 4 2×9 5

c

2 × 18

d

21 × 3

499

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500

Surds and Indices (Chapter 15) Syllabus reference NS5.3.1

8

Simplify: 25 a -----36 21 ---------3 Write true or false. e

9

10

11

12

14

15

17

f

45 ---------5

g

2

13 1 -----36

d

14

c

72 = 2 6

d

20 × 5 = 10

48 e ---------- = 8

6

f

1 1 6 --- = 2 --4 2

Simplify: a 6 3+2 3–5 3–3 3 b 5 3–2 6+3 6–8 3 Write true or false. a 1+ 5 = 6 Simplify: a 5 7×2 7

b

20 – 10 =

b 2 15 × 3 3 f

17 1 -----36

60 ---------5

b 2 7 =

c

50 – 8

d 5 3 + 27

c 2 11 – 11 = 1

10

c 7×2 3

( 3 + 2 2)(2 3 – 2)

2 ( 10 + 6 )

d

g ( 5 + 3 7)

2

Rationalise the denominator of: 1 2 5 a ------b ----------5 3 3 a Expand and simplify ( 5 + 2 3 ) ( 5 – 2 3 ) . 1 b Hence rationalise the denominator of -------------------- . 5–2 3 Write as a surd: 1 --2

b z

Write in index form: a y

1 --3

b

c z

3

y

2

5 --2

d z

c

4

y

3

Evaluate: a 8

18

c

5 = 5

a z 16

29 -----36

a

e 3 5(2 3 – 5) 13

b

1 --3

1 Write as a fraction  1 ---  2

b 4 –1

.

5 --2

c 1296

3 --4

5 --3

LEY_bk953_answers_finalpp Page 504 Thursday, January 13, 2005 3:23 PM

504

Answers

ANSWERS

5 a c

deposit of $1 debt of $6

CHAPTER 1

6 a

+ 1 km

Exercise 1A

7 a

1 a b c

12, 16, 20, 24, 28, 32, 36 9, 18, 27, 36, 45 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77

–12 –11–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0

2 a

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28 6, 12, 18, 24 c 6, 12, 18, 24 d 6

b 3 a b

1, 2, 3, 4, 6, 8, 12, 16, 24, 48 1, 3, 9, 27 c 3

4 a

54 = 2 × 33

c 5 a c

b

84 = 22 × 3 × 7

b

240 = 24 × 3 × 5

144 = 24 × 32 80 = 24 × 5

– 2 km

–5 –4 –3 –2 –1 0 –2 2--3- , – 2--3- , 0,

8 a 9

b

a

2 --3

deposit of $5

, 1 1--3-

b

c

1

2

– 5 km

3

4

35 = 5 × 7, 60 = 22 × 3 × 5 5 c 420

7 a b

140 = 2 × 5 × 7, 84 = 2 × 3 × 7 28 c 420

8 a

84 = 22 × 3 × 7, 56 = 23 × 7, HCF = 28, LCM = 168

–1.8, –1, 0, 1.8, 2.3

–3 > –4

b

–5 < 5

c

–5 < 0

10 a

–13

b

–3

c

5

11 a

–50

b

30

c

–24

12 a

–7

b

11

c

6

13 a d

11 18

b e

–2 21

c f

13 10

2

Exercise 1C 1

100 = 22 × 52, 75 = 3 × 52, HCF = 25, LCM = 300

c

175 = 52 × 7, 200 = 23 × 52, HCF = 25, LCM = 1400 4 5

Shade any 14 squares.

2

b

b e

11 6

c f

10 a

MMMCDXVI

11 a c

2, 3, 4, 6, 9, 12 b 2, 3, 4, 6, 9, 11, 12

b

2

11 -----20

3 13

5 100

13 11

8

6

3 --8

3 -----11

35 -----48

4

29 -----8

7

4 --5

,

10 a

3, 9, 11

$120

b

$400

c

2 -----15

d

12 8.307

13 4 1--5-

14 3.375 17 4

16 3.85

18 a

4.64

b

1 a

19 a

30.6

b

5.2

20 a

$48.30

b

$152 040

0

1

2 3

4 5

6

7

0

1

2 3

4 5

6

7

0

1

2 3

4 5

6

7

2<6

c

6>2 b

3

d

8 -----15

$80

11 hundredths

15 0.17

2 a

,

9 222 kg

Exercise 1B

b

2 --3

110 101 011 000

12 4488

b

5

600 = 23 × 3 × 52

6 a b

9 a d

b

b

3, 4

8

e c

332.4

c

11.9

21

3 1--2-

3 a

5

5

4 a b c

deposit of $15 decrease of 30 cm in length loss of 8 kg d fall of 15°C

,

4 1--2-

,5

37 ---------100

22 48%

23

25 380%

26 62.5%

27 0.65, 70%,

24 0.57

4 --5

28 27%

29

3 -----20

30 4.25

31 $54

32 12 m

33 25%

LEY_bk953_answers_finalpp Page 505 Thursday, January 13, 2005 3:23 PM

Answers

34 $130

35 $240

Exercise 1F

36 A comparison between 2 different quantities, e.g. 30 km/h.

1 a

22, 27, 32

b

13, 11, 9

2 a

3, 6, 12, 24

b

3, 5, 9, 17

37 13 : 15 38 a

3 : 10

b

39 x = 24

3 a

5 : 18

40 $375, $625 b

42 $9.31 44 a

1

2

No. of matches

5

10 15 20 25

43 315.25 km

3300 cm

b

4740 cm

45 12.5 m

(red, blue, green, yellow)

b

1 --4

2 a

(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

b

1 -----10

3 a

7 -----10

c

, 70%

b

1 --4

, 0.25

4 a

20

b

9

c

9 -----20

5 a

1 --4

b

1 --2

c

1 -----13

6 a b

tossing a coin and getting a 1 50% c high probability

7 a b c

getting a number from 1 to 6 getting a 7 getting an even number

8 a

1 --5

4 --5

c

2 --5

3 --5

d

5

20 15 10 5

1 2 3 4 5 No. of pentagons ii The number of matches increases by 5, and points lie in a straight line. c The number of matches needed is five times the number of pentagons. d y = 5 × x, y = 5x e 1000 4 a

0.625, 62.5%

b

4

25

Exercise 1D 1 a

3

i

No. of matches

41 15.6, 23.4, 7.8

No. of pentagons

, b Multiply the shape number by 5 and add 1. c 501 5 a 13 b 23 c 103 6 5, 7, 9, 11, 13 Exercise 1G 1 a

14x

b

6y

c

7a 2

d

6ac

Exercise 1E

2 a

60n

b

–10a

c

24m

d

35p

1 a

p+1

3 a

5a

b

–4m

c

bc

d

4m

2 a

6p

4 a

–wx – 3y

b

5 a

5a -----7

2a -----15

c

q2 -----5

d

3

6 a

a2 – an

d 3 a d

b

3p b

c

2p + 2 d

gr

c 2

3p + 3

5m

8ab

e

3m

f

5a + 3q

3×p 5×x×x

b e

a×b 6×p×q

c

m×m c

b

b

2mn2 – 5mn

2

12p + 8p

d

–8py + 4pw

b

23xy – 12x – 8y

4 a

3p

b

5y

c

z

d

3pq e

0

5 a

12

b

60

c

9

d

16

64

7 a

31a – 23

3 -----2x

8 a

mn(n + 1)

t --2

6 a 7 a d

g --r

b

r --g

c

k÷3 3e ÷ 4

b e

e

4w ------7

d

4÷m mn ÷ t

c

e p÷q

a

1

b

6

c

4

d

5

e

2

9

a

24

b

0

c

12

d

–5

e

42

7

b

1

c

2

d

3

e

3

10 a

c

4(p – 3d)

9 a

–3(k – 3)

10 a

8

b 11 a c

0

b

b

q(p – a)

d

5(5f – 3)

–4(p + 3d)

A pronumeral is a letter that stands for a number. A coefficient is a number in front of a pronumeral. 24

b

–59

d

– 28 –7 --------- = -----8 2 –51

505

LEY_bk953_answers_finalpp Page 506 Thursday, January 13, 2005 3:23 PM

506

Answers

Exercise 1G continued

b

13 a

m

1

2

3

4

m÷3

1 --3

2 --3

1

4 --3

N

1

2

3

4

L

8

11 14 17 20

6d + 23

3 a

b

5

b

(–3, 0)

Exercise 1H 1 –12, ÷ 4 2 a x=6 c x=9 11 e y = -----3 g d = –20

b d

x = –15 x = –7

f

p = 13

m = –20

j

15 p = -----2 4 a No 5 x + 23 = 114, x = 91 6 a x ≥ –12

b

14 7 c = ------ = --10 5 10 q = -----17 x = 24

b

No

b

m < 28

i

h

3 a

D (–3, 2)

A

y

6

–3 –2 –1 0 –1 –2

5

No. of matches

1

3

5

7

9

14 12 10 8 6 4

2 3 4 5 6 7 No. of pattern (6, 11), (7, 13) (0, 0), (1, 4.5), (2, 9), (4, 18), (5, 22.5), (10, 45), (20, 90)

e 4 a b

Cost for various quantities of washing powder 100 95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5

x

B

5 4 3 2 1

4

1

E (2, 5)

–5 –4 –3 –2 –1 1 2 3 4 5 –1 –2 –3 A (0, –3) C (3, –4) B (–2, –3) –4 –5 2 a

3

Cost ($)

y

5 4 3 2 1

2

2

Exercise 1I 1

1

Number of matches equals the number of patterns multiplied by two minus one. y = 2x – 1

c d

3(x – 7) -------------------8

No. of patterns

No. of matches

12 a

b

C x 1 2 3 4 5

2

c 5

4

6

$38.25

8

10 12 14 No. of kg

d

x

–2 –1

0

1

2

y

–7 –5 –3 –1

1

16 kg

16

18

20

LEY_bk953_answers_finalpp Page 507 Thursday, January 13, 2005 3:23 PM

Answers

5

y

8 a

Score

3 2 1 –3 –2 –1 0 –1

2 3x

1

–2 –3

Tally |||| |||| |||| |

1

|||| ||||

2

|||| |||| ||||

14

3

|||| |||| ||

12

4

|||| |

6

5

|||

3

b

–5

18

–8

16

x

–2 –1

0

1

2

y

0

6

9

12

3

Frequency

–7

10 8

y = 3x + 6

6 4

1 a b

about 90 beats per minute 5 minutes c twice

2 a

jogging

3 a 4 a

$4 2–3 p.m.

b

1 -----18

c b b

2 0

200°

50 g 40 km

c

i 60

2 3 Score

4

5

iii The game was a draw.

Kebab

Hot chips

Distance from home (km)

1

ii 14

5 Favourite snacks

7 a c

60

14 12

Exercise 1J

6

9

Winning margins in a series of soccer matches

–6

b

16

Total

–4

6 a

Frequency

0

Hamburger

Jack’s journey

9 a

Chicken

Score

Tally

Pie

Frequency

125

1

|||

3

100

2

||||

5

75 50

3

|

1

4

|||

3

5

|||

3

6

|||| ||

7

7

||||

4

8

|||

3

9

||||

4

10

|||

3

25 . . . . m m m m m m m m a. a. a. a. a. p. p. p. 7 8 9 10 11 12 1 2 Time .

nominal numerical

.

.

.

b

numerical

Total b

10

36

507

LEY_bk953_answers_finalpp Page 508 Thursday, January 13, 2005 3:23 PM

508

Answers

Exercise 1J continued 10

Stem

3 You would use a sample of people that like or do skateboarding. 4 a 16.5 b 19 c 17 d 10 5 a 41.48 b 44 c 41 d 36 6 a Class A: mode = 28 and 46, range = 39, mean = 43.13, median = 42. Class B: mode = 51, range = 46, mean = 40.67, median = 41. b Class A

Leaf

12 13 14 15 16

8 1228 2245555689 0013 1

11 a

Choice of drink 20 18

1 2 3 4 5 6 7

14 12 10 8 6 4

Student to answer. a 270 mm b 4000 mm a 34 cm b 7000 m 76.4 cm a 174.7 cm b 102.94 cm 126.52 cm x = 9.12 cm, y = 12.6 cm, w = 7.2 cm, z = 12.6 cm, perimeter = 90.2 m 180 mm2

9 a

er

nk

at

i

r td

f

So

lw

il

St

ce

i

Ju

er

ee

ff

o /c

c 11 a 12 a c e

Soft drink Still water Juice Tea/coffee Other

g i 13 a 14 15 16 17 18 19 20

Price of CML shares

Price ($)

10 m

10 a

Frequency

12

2

c

th

O

a Te Drink

Drink

7.80 7.75 7.70 7.65 7.60 7.55 7.50 7.45 7.40 7.35 7.30 7.25 7.20 7.15 7.10 7.05 7.00

21

104 cm2

W

T

F M Day

T

W

T

F

49 cm2

d

44 cm2

b

5400 mm2

22 m2 50 mm 64 cm 438 mm

b b d f

2100 cm2 8m 1160 cm 84 m

800 mm2

h

72 000 cm2

243 m

2

90 cm

ii b2 = a2 + c2

i b

b i PR a No a 81 a 7.5 cm a 6.0 cm a 14.4 cm 50.91 cm a diameter 3 -----10

27 a T

b

2

24 C = 2 × π × r M

208

8 29 units2

2

b

b

Exercise 1L

16

Frequency

13 -----50

7 a

2

153.9 cm

ii b b b b b

PR2 = QR2 + PQ2 Yes 8.4 17.1 cm 38.7 cm 27.2 cm

b

chord

22 C = πd

23 35.8 cm

25 42.73 cm

26 A = πr 2

b

11.9 cm2

2

28 96.8 cm

Exercise 1K

Exercise 1M

1 Only a part of the whole population. 2 Sample, as a census is too costly and time consuming.

1 a c

DCGH AEHD, BFGC

b

ABCD, EFGH

LEY_bk953_answers_finalpp Page 509 Thursday, January 13, 2005 3:23 PM

Answers

2 a

b

86.64 cm2

4

5

3.8 cm 3 a 4 a

338.72 cm2

b b

492 cm2 6 a

FE, DE, BE

b

FE

c

FD, CA

Exercise 1P

5 a

6 a

b

3163.5 cm3

b

3469.44 cm3

7 27 946 cm3 8 a

3546.86 cm3

b

793.27 cm3

9 a

1000 mm3

b

1 000 000 mL

1 000 000 cm3 1 kL

d

1000 L

c e

Exercise 1N 1 2 3 4 5 7

48 hours a 4 min 3 h 30 min a 9 h 23 min 7.17 a.m. 2 h 20 min

b

5 hours

b 1 h 39 min 6 3 h 48 min 8 3°17′

Exercise 1O 1 a b c d 2 a

b

3 a

rectangular prisms rectangular prism and triangular prism cylinder and cone cube and square based pyramid i ii iii

i

ii

b

iii

1 a ∠EBD, ∠GEB, alternate b ∠EJO, ∠NOR, corresponding 2 Student to answer. 3 a acute b right-angled c obtuse d right-angled 4 a Two or more angles that share a common vertex and arm and add to give 90°. b

5 a 45° b 120° c 108° d 47° 6 vertically opposite angle 7 i 55° straight angle ii 55° vertically opposite ∠QRU 8 a k = 65°, co-interior angles, supplementary, parallel lines b n = 107°, straight line is 180° 9 a x = 83° vertically opposite b x = 142° angles at a point 10 a Yes b Yes, corresponding angles equal and parallel to a third line. Exercise 1Q 1 Student to answer. 2

3 a x = 56°, base angles of isosceles ∆ b x = 56°, exterior angle of ∆ 4 x = 180 − 120 − 40 − 10, x = 10°, angle sum of ∆ 5 u = 38° (supplementary angles), v = 99° (angle sum of ∆), w = 85° (angle sum of ∆), x = 85° (vertically opposite w), y = 81° (supplementary angles), z = 14° (angle sum of ∆) 6 a b • 2 axes of symmetry • 4 sides equal • diagonals bisect each other at right angles • opposite angles equal • opposite sides parallel

509

LEY_bk953_answers_finalpp Page 510 Thursday, January 13, 2005 3:23 PM

510

Answers

Exercise 2B

Exercise 1Q continued 7 Student to answer. 8 a b

b 77 c Yes 2 a i 5×5×5×5 ii 5 × 5 iii 5 × 5 × 5 × 5 × 5 × 5

Exercise 1R 1 a

i 7×7×7 ii 7 × 7 × 7 × 7 iii 7 × 7 × 7 × 7 × 7 × 7 × 7

1 a 1 diagonal inside 1 diagonal outside

56 c Yes i 6×6×6 ii 6 × 6 × 6 × 6 × 6 iii 6 × 6 × 6 × 6 × 6 × 6 × 6 × 6

b 3 a

b

1 right-angled triangle

trapesium

b

68

c

Yes

4 a

39

b

212

c

710

d

59

e

416

613

g

109

h

220

i

530

j

318

f

2

5

b 10 a f

5 b 3 c 2 d 5 e 712 i 3×3×3×3×3×4×4 ii 12 × 12 × 12 × 12 × 12 × 12 × 12 No, bases are different. i 5×5×5×2×2×2×2×2 ii 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 No, bases are different. i 3×3×3×3×3×3×3 ii 9 × 9 × 9 × 9 × 9 × 9 × 9 No, bases are different. i 2×2×2×2×2×2×2×2 ii 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 No, bases are different. T b F c F d T e F F g T h F i F

11 a

213

5 a 6 a b 7 a

and

b

3 --8

3 a 4 5 6 7

,x=9

b

1 --5

,x=

y = 24 a ST b SU 6.75 m Student to answer.

c

9 --5

, y = 20 TU

CHAPTER 2 Diagnostic Test 1 5 9 13

B A C B

2 6 10 14

45

a

9

b a

10

9

b

318

c

514

d

410

e

510

b

56

c

212

d

315

e

720

412

h

621

i

324

c

14

Exercise 2C D D B D

3 7 11 15

D B D A

4 8 12 16

B A C B

c

i 4

1 a f 2 a f

Exercise 2A 1 a

8

8

36

1010 g 11

3

18

2

b g

17

2

14

5

h

5

17

7

d i

j

270

19

e

716

28

j

922

3 3

Exercise 2D b

4 to the power 5

ii 5

2 a 109 b 10 to the power of 9 or to the ninth c i 10 ii 9 3 a 37 b 210 4 a i 3 ii 5 b i 2 ii 4 c i 7 ii 3 d i 9 ii 6 e i 5 ii 8 5 a 1024 b d 1 000 000 e g 216 h j 390 625

c 43 d 56 e 24 iii 3 × 3 × 3 × 3 × 3 iii 2 × 2 × 2 × 2 iii 7 × 7 × 7 iii 9 × 9 × 9 × 9 × 9 × 9 iii 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 2401 c 243 625 f 6561 729 i 32 768

1 a b 2 a b 3 a

3×3×3×3×3×3×3 ------------------------------------------------------------- = 3 × 3 × 3 × 3 × 3 3×3 35

c

Yes

7×7×7×7×7 ----------------------------------------- = 7 × 7 7×7×7 72

c

Yes

2×2×2×2×2×2×2×2 ---------------------------------------------------------------------- = 2 × 2 × 2 × 2 × 2 2×2×2 b 4 a f k

25 32 64 34

b g l

23 22 55

c c 56 h 32 m 26

Yes d 44 i 41 n 103

e j o

105 51 78

LEY_bk953_answers_finalpp Page 511 Thursday, January 13, 2005 3:23 PM

Answers

5 a e 6 a r i m

T F 35 43 29 26

b f b f j n

F F 27 38 510 53

c g c g k o

F F 72 59 23 35

d h d h l p

F F 58 37 37 710

1 m --------10 5

55

54

3125 625

53

52

51

50

125

25

5

1

j

1 -----36

k

1 -----28

n

1 -------5 10

o

1 -------4 15

1 -----51

l

1 --9

b

1 -----32

c

1 -----64

d

1 ---------625

e

1 ------------1024

f

1 ---------216

g

1 -----81

h

1 -----81

i

1 ------------3125

6 a

Exercise 2E 1

1 -----73

i

25

24

23

22

21

20

j

1 ---------512

k

1 ---------343

l

1 ---------256

32

16

8

4

2

1

1 m ---------729

n

1 2 --2

o

4 --7

2

1

73 – 3 = 70

3 a

70 = 1

c

5–5

4 a

9

0

=9

b

1

1

1

1

1

1

1

9 =1

5 a d 6 a d

1 1 1 1

b e b e

1

1

1

4

2

2

32 –1

2

2

16

1

1

1 1 1 1

8

2

2

4

c f c f

1

0

1

1

2

2

2

–2

2

2

1 --2

1

–3

2

1 --4

104

103 102 101 100 10–1 10–2 10–3 1 -----10

1 ---------100

1 ------------1000

1 1 = ------------- = --------1000 10 3

c 4 a c 5 a e

3–4 –5

m 7

22 – 5 = 2–3

b

1 2–3 = -----23

1

2×2 1 ----------------------------------------- = -----3 2×2×2×2×2 2 1 1

f

1 --------12 1

d

2–5

f

5

g

–2

7

h

4–3

j

5–6

k

3–10

l

6–1

o

–1

–9

4

10

1 a

3

b

3

c

3

2 a

7

b

7

c

7

3 a

5

b

5

c

3

5

c

3

2

c

3

13

27

f

3

69

47

i

3

195

c f

10 10

c

43

f

12 3

4

a

2

5

a

3

b

d

13

e

g

31

h

j a d

7 3

7

a

52

d

23

2

b

29

6

1 a f 2

1

1

29 3

278

1 --1 ---

g

1 -----92

h

1 -----61

Yes

b

Yes 0

b e

2 5

b

32

e

10 2

No

g 1

No 2

10 10 10

1

5×5×5 1 53 – 7 = 5–4 b ------------------------------------------------------------- = -----4 5×5×5×5×5×5×5 5 1 1 1 1 –4 5 = -----54 1 1 1 1 -----b -----c -----d -----31 43 25 82 1 -----54

2–8

n

2–2

1 --1 ---

1 --1 ---

Exercise 2H 1

3 a

i

1 --8

1 1 1 10–1 = --------- , 10–2 = ---------- = --------- , 100 10 1 10 2 10

2

c

–1

b

Exercise 2G

1 1 1 1

–1

100 000 10 000 1000 100 10 1

–3

e

–3

1 1 1 1 1 = ------ , 2–2 = --- = ------ , 2–3 = --- = -----4 8 21 22 23

105

2

3

2–1

1

Exercise 2F 5

7 a

9×9×9×9×9 ----------------------------------------- = 1 9×9×9×9×9

0

c

1

7×7×7 ---------------------- = 1 7×7×7

b

3

10

c

No

d

Yes

e

No

h

No

i

No

j

Yes

4

5

10

106

10

1 10 100 1000 10 000 100 000 1 000 000 3 a d 4 a

3 × 106

b

7 × 104

6 × 10

e

5 × 10

4.8 × 103

b

3.92 × 105

5

c

8 × 103

2

c

6.4 × 10

d

2.18 × 106

e 5 a

7.6 × 10 30 000

b

7000

4 2

511

LEY_bk953_answers_finalpp Page 512 Thursday, January 13, 2005 3:23 PM

512

Answers

Exercise 2H continued c e g i 6

9 000 000 800 6710 83 600

d f h j

400 000 460 000 3 900 000 520 000

1 ---------100

1 ------------1000

1 -----------------10 000

1 --------------------100 000

7 a

3 × 10

d 8 a c e

2 × 10–5 e 0.06 0.002 0.000 009

9 a

3 × 104 = 30 000, 34 = 81

–3

b

7 × 10

–6

c

5 × 10–2 = 0.05, 5–2 =

c

2 × 103 = 2000, 23 = 8

d

2 × 10–5 = 0.000 02, 2–5 =

e

4 × 106 = 4 000 000, 46 = 4096

10 a

3.72 × 10

c

2.98 × 10

e

6.09 × 10

g

7.698 × 10

i 11 a

1 -------------------------1 000 000

5 × 10

–4

9 × 10–2 b 0.000 003 d 0.0005

b

6 × 10–15

e

2.0 × 1012 f

3 × 1010

g

3 × 10–10

h

2 × 10–12

i

8 × 1015

j

4.9 × 10

k

2.7 × 10

l

6.4 × 10–19

–17

b

2.43 × 1019

c

9 × 10

d

2.7 × 10–18

e

8 × 10

f

7.5 × 10–19

g

2.5 × 1021

h

1.4 × 10–12

i

1.6 × 10

j

2.704 × 10–11

k

2.16 × 1038

l

3.125 × 10–37

n

3.7 × 10–12

p

5 × 10–5

8 6

33

m 2.5 × 10

9

o

1 -----25

1 -----32

3 × 106

3 a c

3.11 × 10 Bigger

b

Positive

4 a c

–2.5928 × 108 Smaller

b

Negative

5 a c

4.37 × 10–4 Bigger

b

Positive

6 a

4.6 × 1014 < 7.2 × 1015

b

4.5 × 1016 < 3.4 × 1018

c

9.6 × 10–12 < 6.8 × 10–9

8

b

5.4 × 10

d

3.4 × 10

f

8.75 × 10

d

h

3.61 × 10

7.8 × 10–8 < 3.8 × 10–6

e

8.0 × 10

j

5.6 × 10

2.5 × 10–4 < 7.1 × 105

f

5.7 × 10–4

7.8 × 10–5

2.9 × 1016 < 3 × 1016

b

g

6.4 × 10–10 < 8.5 × 10–10

5 6 5 6

3

4 3 4 8

7

c

6.1 × 10

d

2.96 × 10

h

8.01 × 10–4

f

5 × 10–7

8.1 × 1014 < 2.8 × 1015 < 5.9 × 1016

e

i

8.9 × 10–8 < 5 × 10–6 < 3.9 × 10–5

g

j

7.8 × 10–5 < 8.3 × 10–3 < 6.3 × 106

–3

–6

4.39 × 10

h

2.8 × 10

i 12 a c e g i 13 a c e g i

9 × 10–5 7 320 000 567 000 92 700 000 3 275 000 200 000 000 0.000 003 98 0.000 070 9 0.000 005 9 0.000 6 0.000 027 1

j b d f h j b d f h j

4.9 × 10–9 52 000 3800 69 140 700 000 308 000 0.000 53 0.008 8 0.000 000 307 0.000 003 0.000 000 000 36

14 a

1.29 × 105

b

1.52 × 108 km

d

2.6 × 10–7 m

c e 15 a c d e

–3

2.54 × 10

–9

cm

–6

1.07 × 105 km/h 78 300 000 km b 1 400 000 000 0.000 028 m 0.000 000 002 5 hours 10 000 000 000 000

7 1.08 × 109 km 8 Alpha Centauri 9 2.54 × 10–3 cm or 0.002 54 cm 10 400 days 11 a 0.000 000 3˙ s b

1.2 × 1015 b

2.4 × 1022 c

4.2 × 109

0.3˙ 0˙ s

2

12 350 000 m /person Non-calculator Activities 1 a 2 a b 3 a 4 a

b 177 94 base = 3, index = 11 base = 5, index = 9 8×8×8×8×8×8 32 b 81

5 a

612

d 6 a 7 a

19

3 F 1 -----74

8 a

1 --8

9 a

3.6 × 10

Exercise 2I 1 a

19

1.536 × 1020

2 a

0.1 0.01 0.001 0.0001 0.000 01 0.000 001 1 -----10

d

b e b b b 5

6

340

b

7×7×7

c

86

c

3

10

d

1

120

2 F

3 c

2

b

6.5 × 10

–6

LEY_bk953_answers_finalpp Page 513 Thursday, January 13, 2005 3:23 PM

Answers

10 a

700 000

b 5

1.5 × 1015

12 a

9 × 10

12

c

b

3 × 1023

11 7 × 105 = 700 000, 75 = 16 807

d

2 × 10

12 a

1.19 × 1014

b

c

2.43 × 1047

d

6

6.7 × 1012 < 3.5 × 1013

13 a

5.3 × 10

–16

b

< 4.2 × 10

–10

13 a b

Language in Mathematics three to the power of five eight squared two cubed the square root of seven the cube root of nine

2 a

72

b

43

d 3 a 4 a

24 index base

e b b

6 f 35 exponent power, index or exponent

65

c

5 23 × 24 = 27 = 128; 47 = 16 384 6 Scientific notation is also known as index notation. 7 Scientific notation is a way of writing the repeated product of numbers,

9 3 × 104 = 30 000 while 34 = 3 × 3 × 3 × 3 = 81. 10 Student to answer. Check Your Skills 2 6 10 14

D C C B

3 7 11 15

D B D C

4 8 12 16

B B B D

Review Set 2A 1 a 2 a b 3 a 4 a 5 a

24 b 57 base = 7, index = 9 base = 3, index = 10 3×3×3×3×3 b 7×7×7×7×7×7 6561 b 15 625 F b F 32

6 a

5

7 a

1 -----46

8 a

1 ---------125

b

15

4

b

c

2

2

d

3

2

6

7 c

c

3

< 3.8 × 10

7.7 × 10–16 < 3.1 × 10–12

b 97 75 base = 3, index = 8 base = 5, index = 2 6×6×6×6 b 512 b 729

5 a 6 a

313 F

7 a

1 -----75

8 a

1 -----36

5

15

d

1

b

712 c 45 b F b b

7×7×7 d

611

c

6

3

c

2

230

e 3

9

d

1

3 × 10–4 b 0.000 001 8

11 4 × 106 = 4 000 000, 46 = 4096 12 a c 13 a b

1.56 × 109 b

5 × 1015

2.43 × 10–28

d

1.4 × 10–5

4.1 × 10 < 5 × 10 9

9

3.1 × 10–15 < 4.5 × 10–11

Review Set 2C 1 a 2 a b 3 a 4 a

b 103 28 base = 8, index = 4 base = 3, index = –6 6×6 b 5×5×5×5×5×5×5 625 b 2187

5 a 6 a

919 F

7 a

1 -----48

8 a

1 -----64

b

215 c 58 b F b b

2

d

610

c

3 c

4

3

12 a

1.59 × 10–13

b

9 × 106

c

5.329 × 1031

d

5 × 10–3

13 a

2

d

11 3 × 104 = 30 000, 34 = 81

2.95 × 10

15

< 2.94 × 1016

6.5 × 10–14 < 1.4 × 10–10

246

e

b 3.5 × 10–7 b 0.000 306

9 a 1.7 × 1010 10 a 286 000

4

e 3

1 a 2 a b 3 a 4 a

b b

1.5 × 106 15

9 a 4.6 × 104 b 10 a 40 000 000

e.g. 3 × 3 × 3 × 3 = 34. Write down the base (3), i.e. the number being repeated, and then the index, i.e. the number of times the base is repeated. 8 a The number is not between 1 and 10. b The second number is not written as a power of 10.

A B A C

4.6 × 10

13

4 × 105

Review Set 2B

1 a b c d e

1 5 9 13

b 5 × 10–5 b 0.000 037

9 a 2.3 × 107 10 a 98 000

0.000 12

11 2 × 10 = 200 000, 2 = 32 5

1

513

LEY_bk953_answers_finalpp Page 514 Thursday, January 13, 2005 3:23 PM

Answers

Review Set 2D

3 a

5

b 5 6 base = 6, index = 8 base = 3, index = –10 4×4×4×4×4 2×2×2×2×2×2×2 1296 b 4096

1 a 2 a b 3 a b 4 a

40

5 a 6 a

5 F

7 a

1 -----65

8 a

1 -----32

cf

8

70

b

3

4

c 5 b F b

d

22

4

45

e

3

b

3

c

3

2

8

6

34

19

14

51

23

23

54

25

33

59

26

40 45

51

47 b

10

c

5

d

1

b 3.56 × 10–4 b 0.000 09

2.25 × 1025

b

6.5 × 1022

1.6 × 1073

d

7 × 103

13 a

4.7 × 10–10 < 7.6 × 10–10

b

3.5 × 10–16 < 2.4 × 10–11

4 a 5 a


4.2

Diagnostic Test

e

2 C 6 C 10 B

3 7 11

C C A

fx

43.6

cf

b c

0

10

10

1

3

13

2

5

18

3

2

20

4

6

26

5

4

30

Total

30

2 i 18 ii 20 12

d

10 9

b

fx

c

fx

7 6

51

80

33

90

164

60

209

336

85

80

387

18

42

440

3 2

35

22

315

1

138 96 b

19

c

44

5 4

0

94

5

c

8

16

2 a

19

Water drunk by 9 Orange in a day

4 A 8 B 12 B

Exercise 3A 1 a

b

No. of glasses Freq.

CHAPTER 3

1 B 5 C 9 D

cf

8

11 2 × 106 = 2 000 000, 26 = 64 c

c

43

34

9 a 2.05 × 108 10 a 42 100 000 12 a

cf

19

Frequency

514

6 a&b

1 2 3 4 5 No. of glasses

Days absent

Freq.

cf

1

3

3

2

6

9

3

8

17

4

3

20

5

1

21

6

2

23

7

5

28

Total

28

c d

9 19

LEY_bk953_answers_finalpp Page 515 Thursday, January 13, 2005 3:23 PM

Answers

c Number of days Freq.

cf

1

8

8

2

5

13

3

3

16

4

4

20

5

6

26

6

3

29

7

1

30

Total

30

c d

Cumulative frequency 90

8 4

1 a

52 37

Cumulative frequency

44

40 29

30

11 6 2

43

19

2 a

13

88

70 80 Class

97 106 115 124 133 Class

b

90 100

53 43

45

26 20 7 25

30

35 40 Class

27

6 79

55 50 45 40 35 30 25 20 15 10 5 20

34

60

Cumulative frequency

26

23

50

d

Cumulative frequency

70

50

40

44

140 147 154 161 168 175 182 189 Class

45 40 35 30 25 20 15 10 5

56

60

10

Cumulative frequency

Cumulative frequency

Cumulative frequency

b

70

20 16

Exercise 3B

55 50 45 40 35 30 25 20 15 10 5

82 77

80

Cumulative frequency

7 a&b

45

50

Cumulative frequency

Class

Frequency

20–24

8

8

25–29

4

12

30–34

9

21

35–39

12

33

40–44

5

38

45–49

2

40

Class

Frequency

Cumulative frequency

70–78

6

6

79–87

10

16

88–96

11

27

97–105

4

31

106–114

8

39

115–123

6

45

515

LEY_bk953_answers_finalpp Page 516 Thursday, January 13, 2005 3:23 PM

Answers

Exercise 3B continued

5 a

Class

Frequency

Cumulative frequency

40–49

4

4

50–59

7

11

60–69

5

16

70–79

8

24

80–89

6

30

90–99

9

39

Class

Frequency

Cumulative frequency

20–24

9

9

25–29

12

21

30–34

7

28

d

35–39

4

32

40–44

20

52

45–49

16

68 d

26 34

b b

27 18

c c

44 56

d d

65 17

3 a

27

b

28

c

83

d

24

55 50 45 40 35 30 25 20 15 10 5

6 a

cf

51

60

b

30–39

277

cf

380

2 5 15 31 7 60

2 7 22 53 60

509 603 688 761 783 792

Cumulative frequency 60 53

50 40 30 22 20

Age last birthday

c

Cumulative frequency

30–39

33

31 41 Class

165

Frequency

10

21

about 33 or 34

0–9 10–19 20–29 30–39 40–49 Total

0 d

Cumulative frequency

Score

60

31–40

3 20 47 52 55

11

1 a 2 a

b c

b

c

Exercise 3C

4 a

cf

Cumulative frequency

c

Cumulative frequency

516

800 700 600 500 400 300 200 100

7

2 10

20 30 Class

40

49

0 10 20 30 40 50 60 70 80 90 Age range d

31

LEY_bk953_answers_finalpp Page 517 Thursday, January 13, 2005 3:23 PM

Answers

Exercise 3D 1 a

c

Group

Tally

10–19

|||| |

20–29

|||| |

30–39

|||| |||| |

40–49

|||| ||

b

Frequency 6

Exercise 3E

6

1 a

11 7

Frequency

Frequency histogram 15 11

10 6

7

6

c

Classes

Tally

Frequency

4–10 11–17 18–24 25–31

|||| |||| |||| |||| |||| || ||||

4 10 12 5

32–38 b

|||| ||||

9

Total

40

Frequency

10

10

12 9 5

4

5 4

3 a

2 3 4 5

Frequency histogram 15

11

18 25 Classes

32

Tally

Frequency

|||| || |||| |||| ||| |||| ||||

7 13 9

|||| |||| |

11

Total

40

Frequency

9

10

b 7 a

11

5 13

22 31 Classes

8 9 10 11 12 T

1 6 5 10 3 25

8 54 50 110 36 258

39

x

f

fx

121 122 123 124 125 T

4 11 11 3 1 30

484 1342 1353 372 125 3676

Mean = 122.5

x

f

fx

85 86 87 88 89 T

1 10 8 4 16 39

85 860 696 352 1424 3417

20.5

7

4

b

j

b

13

fx

Mean = 87.6 11 b 122 and 123 65.3 b 31.8 c 45.3 64 b 30 c 55 15 b 25 c 25.5 e 28 f 54.5 h 18 i

38

4–12 13–21 22–30

15

f

a a a a d g

6 a

Classes

31–39

x

Mean = 10.32

5 10 20 30 40 50 Class boundaries

2 a

Scores are not so spread out but the classes are closer in number.

b

c 89 d 867 d 840 37 37.5 20

Class

Class centre

Frequency

6–10 11–15 16–20 21–25 26–30

8 13 18 23 28

7 8 16 12 4

31–35

33

3

16–20

c

18.7

No. of items

Class centre

Frequency

11–20 21–30 31–40 41–50

15.5 25.5 35.5 45.5

3 18 26 6

51–60

55.5

2

111

55

1813.5

32.95

fx 46.5 459 923 273

517

LEY_bk953_answers_finalpp Page 518 Thursday, January 13, 2005 3:23 PM

518

Answers

Exercise 3E continued

d

8 31.73 ≈ 32 Tally

c.c. Frequency

cf

11 10 8 2

11 21 29 31

4

35

16–25 26–35 36–45 46–55

|||| |||| | |||| |||| |||| ||| ||

20.5 30.5 40.5 50.5

56–65

||||

60.5

Class intervals

Tally

Class centre

Freq.

cf

16–25 26–35 36–45 46–55 56–65

||| |||| | |||| |||| | |||| |||| ||||

20.5 30.5 40.5 50.5 60.5

3 6 11 10 5

3 9 20 30 35

b

Frequency

20

Ages of a ICDL class

Class Tally C.c. intervals

15 11 10

10

8 4

5 0

b

a

2 16

26 36 46 56 Class intervals

Class Tally C.c. Frequency intervals 16–20 21–25 26–30 31–35 36–40 41–45 46–50 51–55 56–60 61–65

|||| |||| | |||| |||| | |||| ||| | | || ||

18 23 28 33 38 43 48 53 58

5 6 4 6 5 3 1 1 2

63

65

cf 5 11 15 21 26 29 30 31 33

2

35

f

cf

16–20 21–25 26–30 31–35 36–40 41–45 46–50 51–55 56–60

|| | ||| ||| |||| || |||| |||| |||| | ||

18 23 28 33 38 43 48 53 58

2 1 3 3 7 4 4 6 2

2 3 6 9 16 20 24 30 32

61–65

|||

63

3

35

ii

Histogram 1: Age of next ICDL class 12

11 10

10

Frequency

9 a

Class intervals

i

8 6

6

5

4

3

Frequency

2

6 4 2

5

Ages of a ICDL class 6 6 5 4 3 1

1

0

2

2

0 16 21 26 31 36 41 46 51 56 61 65 Class intervals c

The first gives a downward trend while the second one is more even in height.

16

26 36 46 56 Class interval

66

LEY_bk953_answers_finalpp Page 519 Thursday, January 13, 2005 3:23 PM

Answers

g ii

Age of next ICDL class

Frequency

8

7 6

6 4

4

3

4

3

2

2

1

0

16 21 26 31 36 41 46 51 56 Class interval

iii The first class has a greater number of younger students. iv The second histogram shows that there are two classes that are bigger than the others, while the first shows them as about the same. e f g 10 a c 11 a d 12 a c 13 a

i 34.2 ii 34.3 iii 42.8 iv 42.7 i 16–26 ii 21–25, 31–35 iii36–45 iv 36–40 i 33 ii 33 iii 41 iv 41 60 b 104.4 3144 km d $185 604 9 b 15 c 15 11 e 37 17.25 b 25.3 (1 d.p.) 8.7 (1 d.p.) 17.7 (1 d.p.) 14 10.1 15 12 and 6

f

The grouped data is used to estimate mean for the ogive. 4 The mean, mode and median are central to the study of statistics. Check Your Skills 1 5 9 13

C C B B

1 a b 2 a

3 16 290

5 –60

6

6 7--8-

7

8

0.015

104

9 3417.86

10 D

11 36

12

1 --3

13 –9°C

14 10

15 0.0007

16 $60

17 76

18 25

19 53

20 50

21 5

22 Yes

23 8

24 24 cm2

d e

cf

26 29 30 32 33 34 b

Language in Mathematics cumulative b median c ogive histogram e frequency f estimate The mean is the average. The median is the middle score. The centre is used to calculate the mean class. The modal class is the class with the highest frequency. The ogive is another name for cumulative frequency polygon.

mean = 24.98, mode = 26, median = 26 Student to answer.

21

25 4905

2 a d 3 a b c

4 D 8 B 12 A

12

Cumulative frequency

2 45.88

4 $693

3 A 7 C 11 B

Review Set 3A

Non-calculator Activities 1 3 2--5-

2 C 6 A 10 D

No. of time students of 9 Red ate take-away 33 34 32 29 30

34 32 30 28 26 26 24 21 22 20 18 16 14 12 12 10 8 6 4 2 0

c

0–4

5 10 15 20 25 30 35 40 No. of days d 10.1 e 8

Review Set 3B 1 a b

mean = 10.88, mode = 9, median = 10 Student to answer.

519

LEY_bk953_answers_finalpp Page 520 Thursday, January 13, 2005 3:23 PM

Answers

Review Set 3B continued 2 a

b c d

2 a&b

b 2 a

0

8

8

20–25

5

1

9

17

26–31

13

2

7

24

32–37

12

3

2

26

38–43

14

4

1

27

44–49

9

5

3

30

50–55

5

56–62

2

c

mean = 6.03, mode = 6, median = 6 Student to answer. cf 14

d

4

e

1.6

f

1

g

1

CHAPTER 4 Diagnostic Test 1 5 9 13 17 21

A B A C C C

2 6 10 14 18 22

C D C A C A

3 7 11 15 19

B C A C D

4 8 12 16 20

C A D D A

1 a

6y -----15

b

8m -------- c 6

35k ---------- d 20

30t ---------- e 100

30a ---------12

2 a

x --2

b

m ---3

c

t --4

d

2y -----3

e

3b -----4

f

6 --7

g

3 -----10

h

2 --3

i

2b

j

4p -----3

3 a

7x -----10

b

15b ---------- c 11

a --3

d

3m -------- e 10

2k -----3

4 a

11x --------- b 12

19k ---------- c 12

5b -----8

d

7a -----10

e

2z -----15

f

47t --------- g 90

5w ------4

17v --------- i 6

e --2

j

11x --------24

c

2pq ---------15

26 29 29 30 b

Height of Year 9 Red students 29

30

29

30

26 25 20 15 10 5

5

163

d

160

b

km -------15

d

12a ---------35b

e

10bd ------------21ce

6 a

6mn -----------7

b

4km ----------15

c

3wz ----------2

d

3ab ---------2

e

2t -----3u

f

7 1--2-

g

3 --4

h

2a

i

40n ---------9

j

10p

Review Set 3D mean = 22, mode = 22, median = 22 Student to answer.

h

mn -------12

5 a

14

140 150 160 170 180 190 200 Class

1 a b

24

Exercise 4A

5

c

cf

Frequency

Review Set 3C 1 a

Score Frequency

Class

28 It is within a group and individual scores are not known in this case. 39 e 37.7 f 38–43

Cumulative frequency

520

7 a

7a -----5b

b

w -----3z

c

pq -----20

d

24 -------- e mn

ad -----bc

8 a

5x --------- b 12y

7a -----4b

c

3p -----2q

d

4 2--3-

2 --3

12 -----35

3 1--3-

h

28 ------m2

i

10y --------- j 3

f

g

e

3 -----4k

LEY_bk953_answers_finalpp Page 521 Thursday, January 13, 2005 3:23 PM

521

Answers

Exercise 4E

Exercise 4B 1 2 3 4

1–2 Student to answer.

Student to answer. a i 512 ii 512 a i 15 625 ii 15 625 a i 6561 ii 6561

b b b

Yes Yes Yes

m9

b

q15

c

t 19

d

b 20

e

v 13

6 a

a

2

b

10

x

c

w

6

d

b

e

z

7 a

b8

b

h15

c

k16

d

z 60

e

n8

8 a

6

b

3

c

b

24

n g b h y Bases are different. Bases are different. T b F c F T g F h T T l F

25

f 9 a b 10 a f k

m

x

7

d i

d i

m t

31

T F

70

e

v

j

a6

e j

F F

Exercise 4C x

6

f

y

2

k

a32

1 a

5

b

w

2

c

m

7

g

z

h

l

t 20

m y5

d

x

3

e

a

9

k k

j

m11

n

a5

o

b40

2 a

12m12

b

10p10

c

18t12

d

70a16

e

24w 19

f

30b9

g

17

24z

i

18d18

3 a

2m5

b

2a5

c

3w 2

d

4z 4 --------3

e

4k 6 --------3

f

3e 4 --------2

g

m5 ------3

h

a5 -----2

i

3t 7 -------4

j

3b 5 --------4

h

18

80q

12 2 c

5k 2 g

6y 2 h

1 ---

4 a

1 ---

1 ---

13

1 ---

23

d

w3

g

c

p

d

q

g

5 k

h

4 t

1 ---

b

93

e

( 5y ) 3

( 9m ) 3

h

9m 3

8 a

3

12

b

3

d

3

d

e

3

g

3

7v

h

73 v

1 ---

1 ---

9 a

47

1 --2 1 ---

e

5x 2 3

10 a e

b

p

f b

5p

f

1 ---

c

c3

f

5y 3

35

c

3

6m

f

63 m

1 ---

1 ---

1 ---

1 --2

1 ---

c

x3

6x 3

g

4x 2

6

c

6r

g

1 ---

5

1 ---

( 5k ) 2

e

m2

( 6y ) 2

e f 3z 2m 5–6 Student to answer. 7 a

1 ---

d

x2

1 ---

b

8

3

i

1 ---

b

32

f

5 a

13

1 ---

3 a

3

k

1 ---

d

8p 3

h

3r 3

x

d

t

5xy

h

1 ---

3

1 ---

3

Exercise 4F 1–5 Student to answer. 6 a

1 -----y2

b

1 --k

c

1 ------m3

b

k –2

c

x –11 d

1 ----x6

e

1 ------t 10

n –14 e

z –20

d

27a12

b

64m18

c

49p10

7 a

a–8

d

10 000k 8

e

125t 33

f

x15y 10

8 a

b

m12n18

h

p 28q12

i

a 8b 20

1 -----3k

c

g

3 --k

2 -----y5

1 ------------32y 5

e

3 ---t4

f

1 -----------81t 4

b

4

c

24m

4 a

15 6

d

j

8x y

5 a

8a8b 5

b

10m10n8

c

12p11q15

30x10y 4

e

12w14z17

f

3a 2 b 4 ---------------2

d g j

3x4y 7

h

k4m6

------------5

i

3a 3 b 5

---------------4

4m -------5

1–2 Student to answer. 3 a 1 b 1 c 3 f 1 g 1 h 10 k 4 l 6 m 13

Exercise 4G 1 a d

d i n

4 8 5

e j o

9 1 2

12z 3

2 a

a–7

e

–8

b

2

e b f

k

y4 w

5

c

e–2

d

n

g

z

2

h

k –4

d

4y 6

–8

j

t

3 a

90a2

b

18b–7

c

6v –4

e

3p–6

f

3k –2 -----------8

g

125z –12 h

i

Exercise 4D

4pq

y

–20

i 9w –12 j 2n–2 4 a F b F c f F g F h 5 Student to answer.

T T

d i

F T

32m –15

e

T

LEY_bk953_answers_finalpp Page 522 Thursday, January 13, 2005 3:23 PM

522

Answers

Exercise 4H 1 a c

6w + 15 20a + 15b

b d

18z – 12 8x – 6y

e

10x 2 + 60

f

7ab – 14a 2

4m + 4n 20b + 10a + 15

h j

2m3 – 6mn 15x – 9y – 6z

6ab + 12ac

b

12x 2 – 8xy

c

60k2 – 40km

d

m3 + 2m

e

y 5 – 4y 2

f

12xy – 30x 3

h

5a5 – 2a3

g i 2 a

2

2

4

2

g

6k + 15k

i 3 a d g j

2p7 + 6p8 –2y – 6 –4m + 28 –6k – 15 –4x + 1

4 a

–6a3 – 4a2b

c

b e h

j 10x5 – 15x3y –5a – 10 c –3w – 12 –t – 3 f –b – 6 –8m + 10 i –7w – 3

–9p4 – 12p3q 6

b

–8x3 + 12x2y

d

–4y 5 + 3xy 4

5

e 5 a d g j 6 a d g j 7 a d g

–6m – 15m n 4a + 18 b 10y – 17 e 2 + 6b h –14 + 20e 2 – 2a b 19 – 2v e –2 – 15x h –7 + 20w 13k + 9 b 13a – 6 e 12v – 16 h

j 8 a c e g

18a2 – 2a 4k + 9 9t – 10 –2a + 14 10x + 11y

b d f h

2w + 26 13z – 8 –d – 18 4a – 12b

2q 2 – 14q + 20

j

12z 2 + 7z + 1

i

6b – 12 18z – 2 33 + 4y

c f i

12w – 1 16 + 8x –8 + 6w

14 – 3y 5 – 6w 12 – 6k

c f i

–3 – 4b 22 – 15t –4 – 12z

15m + 11 25x – 11 22x + 2y

c f i

14p + 6 14y + 1 23a – 9b

4 a p(p + 3) c w(3w + 2) e m(4n – 3m) g 4p(q – 3p) i 5z(2z – 1) 5 Student to answer. 6 a x(x + 4) c a(a + b) e 2k(k + 2) g 5b(2b + a) i 6p(2q – 3p) 7 a 2a(b + 2 + 2a) c 2m(1 – 3m + 2n) e 2k(4k – 3 – 5m)

2(4a + 5) 3(a + 2b) 4(w – 3) 4(3ab + 2)

b d f h

2(3x – 2) 5(x + 2y) 8(2m – 1) 10(m – 2n)

i 2 a c e 3 a c e g i

6(4k – 3n) 5(x + 3y + 2z) 4(a + 3b – 2c) 10(2xy + 5z + 3) y(y + 7) m(3n + 4) x(x + 5y) 3m(2m – 1) 4p(4q – 3p)

j b d

8(2x 2 + 3y 2) 3(p – 2q + 3r) 6(2m – n – 3r)

b d f h j

m(m – 3) p(9 – 5q) b(2c – b) 2a(6a – 5b) 6k(2k + 3)

k(k – 2) z(2z – 1) 2x(x + 4) 4p(2q – 3r) 2m(3k – 4m)

b y(y – 7) d m(m – 5n) f 3y(y – 4) h 3w(3w – 2) j 8k(2m + 3k) b 3x(2x + 1 + 3y) d 5x(y – 2 – x)

Language in Mathematics 1 reduce, substitute, apply 2 a a factor b zero c index d algebraic e factorise 4 a x squared (to the power of 2) b square root of x c x cubed (to the power of 3) d cube root of x 5 a parentheses b brackets c braces Check Your Skills 1 5 9 13 17 21

D A C D A A

2 6 10 14 18 22

C D C A A D

3 7 11 15 19

A A B B C

4 8 12 16 20

B B B B B

Review Set 4A 1 a

4x -----18

b

15mn --------------20

2 a

4a -----5

b

3y -----2

3 a

11a ---------13

b

31m ----------24

4 a

y 17

d 5 a

8

Exercise 4I 1 a c e g

b d f h j

t 1

e 5

b

6 a

x

7 a

1 ----z3

8 a 9 a

y2 T

10 a

b

b

b b

k6 12

125m c

3 x c b e8 F

10v – 20w b

8w 2 ---------15

c

2 ----z3 c c

c

p14

f

6a6b9 d 3

1

3x d

3

c n –20 d F d

5x -----2y

d

x

e 1 --------8z 3

18b5 e T e

2a5 + 4a4 c

23 x

2k –2 F

–12x – 15

LEY_bk953_answers_finalpp Page 523 Thursday, January 13, 2005 3:23 PM

Answers

11 a 12 a

10m + 7 b 4(2w + 5) b

Review Set 4D

7a – 2b x(x + 9)

c

4q(p – 3q)

Review Set 4B

1 a

15b ---------24

b

35pq ------------10

1 a

9x -----21

b

12ab ------------18

2 a

5h -----4

b

2p -----3

2 a

2m -------3

b

2b

3 a

5k -----11

b

w -----18

c

6ab ---------5

d

27 -----40

3 a

a

b

17y --------12

c

2km ----------5

d

9 -----10

4 a

y13

b

k6

c

p50

d

c8

4 a

m20

b

t 20

c

z 24

d

b19

e 5 a

81h10 1

f b

6a13b11 8 c

1

d

–1

6 a

m

b

6 m

c

1

d

3

m

e

3

b

2 ----x2

e 5 a

16m 1

6 a

c

d

3

28

f b

c

7 8

20p q 4 b

2 c

e

3

1 ----e4

8 a

k –4

b

m–3 c

n–15 d

20n5 e

1 --2

9 a

F

b

T

F

d

F

F

3m7 – m5

c

10 a

3 ----e4

c

c

–40p – 30 b

1 -----------81e 4

e

a3

–2a 2 + 5a

2

11 a 12 a c

4q + 32 b 4a – 13a – 1 6(2x – 3) b y(2y – 7) a(4a – 3b + 2)

12h ---------20

2 a

w ---3

3 a

8d -----7

4 a

p14

b

y8

c

t 30

d 5 a

1 1

e 7

27v 21 c 1

f

18x10y 14 d 12

b

4 q

c

e

43

d

b

b q

3

q

b

16xy -----------12

b

5m -------7

b --8

c

4xy --------3

1 7--8-

d

c

p–15 d

9 a

F

b

F

c

F

d

15z3 e T 3

e

2m 3 ----------3 T

10 a c

12xy – 6xz –10 + 4a

b

–3a m + 4m5

11 a 12 a

23 + 19n 6(4x + 3)

b b

4a 2 – a h(h – 8)

D B B B

4 8 12 16

D C B D

c

600

3

3(y 2 + 2y – 3)

1 ---------------243b 5

8 a

d –8

b

n

c

k6

15a–3

e

10 a 20w + 8x b 11 a t + 4 b 12 a 5(3n – 4) c 2m(6m + 7n)

C A B C

1 a d

q c

F

1 5 9 13

g

3 -----b5

b

n2

4q

b

T

b

2m 7 ----------3 c F 7

d

T 5

e

2 6 10 14

C C A C

3 7 11 15

Exercise 5A

1 -----b5

9 a

y –7

Diagnostic Test

7 a

d

8 a

1 --------4x 2

c

CHAPTER 5

1 a

6 a

7 a

c

Review Set 4C

6m

1 ----x2

2c

7 a

b

3

d

2c

c

6m

c

F

–2k + 4k c –4s + 7 –15y b 2b(2b + 3)

2 a d g j 3 a d g j 4 a d g j

60

b

6

6000

e

6 000 000 f

6 -----h 10 37 000 8000 181 000 401 000 5400 800 3100 100 670 1060 310 1250

6 ---------i 600 100 b 84 000 c e 19 000 f h 6000 i k 1000 l b 16 800 c e 400 f h 9600 i k 100 l b 2370 c e 70 f h 20 060 i k 10 l

6 ------------1000 6 -----------------10 000 524 000 623 000 3000 0 20 400 240 200 500 0 830 30 410 0 j

523

LEY_bk953_answers_finalpp Page 524 Thursday, January 13, 2005 3:23 PM

524

Answers

Exercise 5A continued 5 a 17 b 25 c 82 d 237 e 583 f 265 g 21 h 106 i 301 j 56 k 1 l 0 6 a i 47 000 ii 46 800 iii 46 780 iv 46 784 b i 28 000 ii 28 500 iii 28 460 iv 28 457 c i 39 000 ii 39 200 iii 39 170 iv 39 166 d i 8000 ii 8500 iii 8460 iv 8462 e i 183 000 ii 182 700 iii 182 680 iv 182 679 7 a 30 000 b 60 000 c 180 000 d 200 000 e 800 000 8 a 4000 b 6000 c 24 000 d 80 000 e 20 000 9 a 700 b 1800 c 32 600 d 100 e 7000 10 a 38.3 b 38.27 c 38.268 11 a i 8.4 ii 8.44 iii 8.438 b i 6.6 ii 6.58 iii 6.584 c i 0.9 ii 0.86 iii 0.863 d i 0.2 ii 0.19 iii 0.186 e i 18.6 ii 18.56 iii 18.556 f i 21.6 ii 21.60 iii 21.603 g i 4.1 ii 4.06 iii 4.061 h i 5.0 ii 5.04 iii 5.044 i i 7.0 ii 7.00 iii 7.007 j i 3.0 ii 3.00 iii 3.000 12 a 3.60 b 50.0 c 2.690 d 13.00 e 1.00 f 4.900 g 100.0 h 70.00 13 a 75 b < 85 c 75
no. < 85 14 a 350 b < 450 c 350
no. < 450 15 a i 27 500 ii < 28 500 iii 27 500
no. < 28 500 b i 42.5 ii < 43.5 iii 42.5
no. < 43.5 c i 5.65 ii < 5.75 iii 5.65
no. < 5.75 d i 6.315 ii < 6.325 iii 6.315
no. < 6.325 16 162.5
height < 163.5 cm 17 415
weight < 425 g 18 12.35
time < 12.45 s Exercise 5B 1 a

2

b

5

c

1

d

7

e

8

2 a d 3 a b c d e f g h i j 4 a d g j 5 a

6

7

8

9

30 b 28 c 28.5 28.47 e 28.471 i 400 ii 430 iii 428 i 6000 ii 6200 iii 6240 i 8 ii 7.8 iii 7.82 i 0.5 ii 0.53 iii 0.527 i 50 000 ii 54 000 iii 53 700 i 700 000 ii 730 000 iii 726 000 i 0.04 ii 0.039 iii 0.0393 i 0.005 ii 0.0051 iii 0.005 07 i 6000 ii 6100 iii 6100 i 2000 ii 2000 iii 2010 370 000 b 240 c 0.005 80 9.00 e 300 000 f 500 0.0400 h 0.300 i 0.002 00 1 000 000 i 555 ii < 565 iii 555
no. < 565 b i 8.15 ii < 8.25 iii 8.15
no. < 8.25 c i 47.5 ii < 48.5 iii 47.5
no. < 48.5 d i 0.715 ii < 0.725 iii 0.715
no. < 0.725 e i 36 500 ii < 37 500 iii 36 500
no. < 37 500 f i 0.0835 ii < 0.0845 iii 0.0835
no. < 0.0845 a 482.5
no. < 483.5 b 3.855
no. < 3.865 c 14 450
no. < 14 550 d 0.1275
no. < 0.1285 e 56.85
no. < 56.95 f 3205
no. < 3215 a 295
no. < 305 b 2950
no. < 3050 c 5995
no. < 6005 d 23 950
no. < 24 050 e 499 500
no. < 500 500 f 0.795
no. < 0.805 g 0.3995
no. < 0.4005 a 2 b 2 c 4 d 2 e 2 f 4 g 2 h 2 i 1, 2 or 3 j 1, 2, 3 or 4 k 2, 3, 4 or 5 l 1, 2, 3, 4, 5, 6 or 7 a 3.64 m is the total nearest cm, 3.640 to the nearest mm b 5.8 kg is to the nearest 100 g, 5.80 kg to the nearest 10 g c 12 s is to the nearest second, 12.0 s to the nearest tenth of a second d 36 cm is to the nearest cm, 36.0 cm is to the nearest mm

LEY_bk953_answers_finalpp Page 525 Thursday, January 13, 2005 3:23 PM

Answers

e

23.8 s is to the nearest tenth of a second, 23.80 to the nearest hundredth of a second f 7.29 km is to the nearest 10 m, 7.290 km to the nearest m g 1.5 t is to the nearest 100 kg, 1.50 t to the nearest 10 kg h 5.83 L is to the nearest 10 mL, 5.830 L to the nearest mL 10 Student to answer. Exercise 5C 1 a c d 2 a b d 3 a d 4 a c e 5 a c d 6 a d 7 a b c 8 a 9 a 10 a c

29 153 b 29 200 i 12 900 ii 700 iii 15 700 29 300 e Differ by 100. 5250 i 8090 ii 2830 c 5260 Differ by 10. 18.7 b 2.3 c 18.4 Differ by 0.3. 143.4 b 142.6 Differ by 1.2. d 142.1 Differ by 1.3. 75.172 b 75 i 33 ii 17 iii 26 76 e Differ by 1. 60 b 4, 20 c 80 Differ by 20. 2.18 i 16.4 ii 7.50 2.19 d Differ by 0.01. 4220 b 4240 c Differ by 20. 0.18 b 0.19 c Differ by 0.01. 3.3 b 10.89 i 3.32, 11.0224 ii 3.317, 11.002 … iii 3.3166, 10.9998 … iv 3.31662, 10.99996 … 11 a 3 s.f. b 7 s.f. c 4 s.f. d 3 s.f. e 4 s.f. 12 Student to answer. Exercise 5D 0.7˙

b

0.35˙

c

0.2˙ 8˙

d

0.3˙ 25˙

e

0.6˙ 784˙

f

1.4˙

g

6.92˙

h

0.4˙ 9˙

i

0.2˙ 34˙

j

0.03˙

k

0.9˙ 0˙

l

0.536˙

1 a

m 0.217˙ 2 a d

0.875

b

0.5˙

c

0.16˙

0.1˙ 8˙

e

1.416˙

e

1.6˙

g

0.61˙

j

0.4583˙

h

0.6˙

i

1.59˙ 0˙

3 a

3 --5

b

39 -----50

c

1 --8

d

2 -----25

e

32 ---------125

4 a

2 --9

b

1 --3

c

5 --9

d

8 --9

e

7 --9

b

91 -----99

c

10 -----33

d

7 -----11

e

98 -----99

5

0.9˙ = 1

6 a

46 -----99

7 a

586 ---------- b 999

239 ---------- c 999

284 ---------- d 333

47 ---------- e 111

205 ---------333

8 a

7 -----18

b

59 -----90

83 -----90

1 --6

e

1 -----10

9 a

41 -----75

b

217 ---------- c 300

163 ---------- e 180

1 -----20

c

Exercise 5E 1 a $3.44/kg c 55 words/min e 90 km/h 2 a 0.91 c/cm c 1.4 L/min e 2.25% per quarter g i k 3 i iii 4 i iii 5 i

8 9 10

343 ---------- d 450

b d

$12.56/h 10.5 km/L

b d f

0.125 c/mm 1.25% per month 16 c/min

84 c/min

h

1800 g/m2

650 g/m2 3.6 c/mL 1.2 km/min 20 m/s 1.8 km/min 30 m/s 1 km/min

j 1.9 c/g l 7.5 g/mL ii 1200 m/min ii 1800 m/min ii 1000 m/min

iii 16.6˙ m/s 6 i 300 L/h 7

d

ii 5 L/min

iii 83.3˙ mL/s a 28.8 km/h b 72 km/h c 126 km/h d 150 kg/ha e 80 kg/ha f 250 kg/ha g $5/kg h $7.20/h i 7.2 L/h j 0.8 kg/L 36 km/h i 0.125 L/km ii 12.5 L/100 km iii 125 mL/km a 1080 b 25 h c 4 bottles/min d 180 e 50 min

11 a d 12 a c 13 a c 14 a

320 m2 b 2

i 8m 450 km 25 m/s 270 L 75 mL/s 1200 mL

15 L

c

0.016 m2/mL

2

ii 3.2 m b d b d b

7.5 h 135 m 8 h 53 min 1800 mL 10 drops/min

525

LEY_bk953_answers_finalpp Page 526 Thursday, January 13, 2005 3:23 PM

526

Answers

Exercise 5E continued 15 a 17c/min b 66c c 22c/min d 108.5c e 5 min 16 i 0.072 g/mL ii 72 mg/mL iii 720 mg/10 mL Non-calculator Activities 1 a d 2 a

35 000 2500 3 b

3 a d 4 a

b e 2

7.30 7.64 c

0.4˙

b

0.5˙ 2˙

0.576˙

e

0.2˙ 13˙

b

0.4˙

0.375

c

15

1, 2, 3, 4 or 5 c

0.56˙

15 120 L/100 km Review Set 5B

2 3 4 5 6 7 8 9

37 5 -----99 6 a 10 m/s

b

6 6 -----b 600 c ------------- d 60 1000 10 a 13 600 b 4060 c 148 d 100 000 a 1.6 b 1.56 c 1.561 a 2.70 b 40.0 163.5
height < 164.5 cm a 2 b 1 c 5 a 365 000 b 540 c 0.002 40 d 2.00 a 2 b 2 c 4 d 1 e 1, 2, 3 or 4 1.85 kg is to the nearest 10 g, 1.850 to the nearest gram.

1 a

0.3˙

b

0.36˙

d

0.3˙ 14˙

e

0.5˙ 678˙

11 a

0.1875

b

1.7˙

12 a

3 -----10

b

81 ---------100

10 a

1.5 kg/L

Language in Mathematics 1 D 5 a fraction d product

b e

decimal terminate

c

sum

Check Your Skills 1 5 9 13

D D D D

2 6 10 14

50

3 7 11 15

D B B B

4 8 12 16

b

5

c

5 ---------100 34 8.463

0.2˙

b

0.42˙

d

0.4˙ 25˙

e

0.4˙ 253˙

11 a

0.375

b

1.6˙

12 a

3 --5

b

73 ---------100

4 --9

b

14 1600 g/m2

267 ------------1000

8 49 --b -----9 99 14 i 0.6 km/min ii 600 m/min iii 10 m/s 15 18 L/h

C D A D

d

500

a 2500 b 7930 c d 50 000 a 8.5 b 8.46 c a 3.50 b 20.0 a 35 b < 45 c 35
no. < 45 a 3 b 4 c 2 a 50 b 48 c 48.4 d 48.35 e 48.351 8 a 3 b 1 c 3 d 2 e 2, 3, 4 or 5 9 3.65 m is to the nearest cm, 3.650 to the nearest mm.

13 a

c

Review Set 5C

2 3 4 5 6 7

10 a

0.2˙ 8˙

13 a

D C A C

Review Set 5A 1 a

c

41 -----99

c

0.4˙ 2˙

8 8 -----c ---------d 8000 100 10 a 13 800 b 770 c 24 d 90 000 a 13.1 b 13.07 c 13.065 a 4.20 b 21.0 a 3 b 4 c 2 a 20 b 18 c 17.6 d 17.63 e 17.631 a 425 b < 435 c 425
no. < 435 a 3 b 1 c 3 d 2 e 2, 3, 4 or 5 6 cm is to the nearest cm, 6.0 cm to the nearest mm.

1 a 2 3 4 5 6 7 8 9

8

0.1˙

b

0.37˙

d

0.6˙ 37˙

e

0.423˙

11 a

0.275

b

1.916˙

12 a

1 --5

b

29 -----50

13 a

8 --9

b

635 ---------999

10 a

c

69 ---------500

b

c

0.9˙ 2˙

c

1 --8

LEY_bk953_answers_finalpp Page 527 Thursday, January 13, 2005 3:23 PM

Answers

14 20 mL/s 15 $28.80/day

4 a c

32.0 cm 150.8 cm

b d

2.5 km 39.6 cm

Review Set 5D

5 a

32 cm2

b

180 m2

c

80 cm2

d

144 m2

3 ---------b 3000 c 3 d 30 100 a 78 500 b 510 c 34 d 500 000 a 2.1 b 2.06 c 2.057 a 3.20 b 60.0 235 g
weight < 245 g a 1 b 6 c 5 a 24 700 b 67.8 c 0.0507 d 3.00 a 3 b 1 c 3 d 2 e 1, 2 or 3 12 s is to the nearest second, 12.0 to the nearest tenth of a second.

1 a 2 3 4 5 6 7 8 9

10 a d

0.9˙

b

0.45˙

0.726˙

e

0.4˙ 36˙

11 a

0.2125

b

1.416˙

12 a

4 --5

b

24 -----25

2 13 a --b 3 14 i 1.5 km/min iii 25 m/s 15 7.5 t/ha

c

0.8˙ 5˙

c

109 ---------200

2 --5 ii 1500 m/min

2 6 10 14

A C B A

3 7 11 15

D C B D

4 8 12 16

B B B A

Exercise 6A 1 a d g j m 2 a d g 3 a

0.21 m 40 mm 20 cm 83 mm 5000 cm 400 32 000 000 50 000

b e h k n b e h

2

1500 mm

18 cm 230 cm 2.8 m 6300 m 3200 mm 310 000 4 73 000 b

c f i l o c f i

3.5 km 1800 m 520 cm 3 cm 0.83 km 5 300 000 7 4.2 2

150 000 m

c

3.2 m2

d

32.8 m2

e g

2 350 000 cm2 0.000 782 ha

f h

365 000 m2 0.036 542 ha

i

2300 ha

j

420 000 cm2

l

2000 mm2

k

2

52 000 cm

36 cm2

7 a

24 cm2

2

b

4 cm

b

36 cm2

2

2

c

640 m2

c

120 cm2

c

56.75 cm2

8 a

201.06 cm b

153.94 m

d

30.19 cm2 e

183.85 m2 f

1.25 km2

29.925 m2 c

15.48 cm2

Exercise 6B 1 a

24 mm2

2 a

45 cm

3 a

2

d

b

2

b

32 m

2

27.84 m 2

c

19.35 cm2

b

30 cm

c

98 cm2

b

58.5 cm2

c

266 m2

b

48 cm2

c

60 mm2

1190 mm2

4 a

20 cm2

5 a

51 m2 2

82.5 km

2

e

108.375 m f

23.165 m2

b

25.7 cm

c

102.8 m

b

64.3 mm

c

55.7 m

b

21.42 cm

c

57.1 cm

Exercise 6C 1 a d 2 a d 3 a

20.6 cm 86.3 m 21.4 cm 75.7 cm 14.28 m

2

Diagnostic Test B D C B

14 m

d

d

CHAPTER 6 1 5 9 13

6 a

2

2

4 a i & ii 4πm = 12.57 m b No difference. 5 a 67.99 cm b 45.70 cm c 37.85 cm 6 $744 7 125 cm 8 $237.15 9 152 m 10 8.0 cm 11 31.8 m 12 a 7200 km b 45 239 km c 1885 km/h 13 53 052 14 64.80 km 15 a 26 cm b 104 cm 16 a 6.7 cm b 11.7 cm, 36.8 cm Exercise 6D 1 a d 2 a

168 cm2

b

100.5 cm2 e 2

33 m

b 2

118 cm2

c

126.85 m2 f 2

50 m

2

38 cm2

f

54.54 cm2

265.5 cm

e

52.5 cm

3 a

12.6 cm2

b

17 854 m2 c

d

301.6 cm

Exercise 6E 1 $2556.17

e

2

64.7 cm

117 cm2

c

d

2

65 cm2

f

16.9 cm2 38.0 cm2

527

LEY_bk953_answers_finalpp Page 528 Thursday, January 13, 2005 3:23 PM

528

Answers

400 m2

b

$13 120.00

9 D 13 C

3 a 36.6 m2 4 $892.08

b

$2925.31

Review Set 6A

b d

26.35 ha 7.905 ha

Exercise 6E continued 2 a

5 a c e

263 500 m2 18.445 ha $1 897 200

6 38 m2 7 6.3 m2 8 1.914 t 9 a 96 cm2 b 6.4 cm 10 Decrease breadth to 6 cm.

10 A 14 B

11 D 15 C

12 A 16 C

1 a 85 cm = 0.85 m b 15 000 m2 = 1.5 ha c 3.5 km = 3500 m 2 a 20 m b 46.3 m c 12 cm d 25 cm 3 a 370 m b 60 cm 4 a 51.7 cm b 73.5 cm 5
44 611 km 6 a 7 a

40 m2 2

23.4 m

b

147 cm2

b

52 cm2

c

126.4 cm2

c

30.86 m

c

19.6 m2

2

11 Square; by 9 cm 12 a c

105 cm2 Rectangle


5.78 cm

b

Review Set 6B 1 a

Exercise 6F

4.28 ha = 42 800 m2

b 3 cm2 = 300 mm2 c 4300 cm = 43 m a 30 cm b 44 cm d 97.1 m a 76 m b 40 cm a 34.3 cm b 34.7 cm 19.1 cm

2 a c

60 120 1 ---------- = --- b ---------- = 360 360 6 4.2 cm, 12.2 cm b 4.2 cm, 28.2 cm

3 a

8.4 cm2

4 a

P = 58.8 m, A = 215 m2

6 a

13.65 cm2 b

326.2 m2

b

P = 37.9 m, A = 69.8 m2

7 a

46 m2

b

102.7 cm2

c

P = 123.9 m, A = 283.5 m2

Review Set 6C b

66 cm 1 A = --- xy 2

1 a

b

1 20 1 --- c ---------- = -----3 360 18 10.5 cm, 20.5 cm

26.2 cm2

c

d 5 a c

P = 631.4 m, A = 23 082 m2 17.96 m b 32.4 m 185.8 m d 11.1 m

6 a

19.9 m2

25.1 cm2

2 3 4 5

1 a

20 cm

7 A = 169.64 cm , P = 75.4 cm

A = lb b 1 -c A = h(a + b) 2 b 132 cm2 3 a 129 cm2 4 $18 441.86

8 65.6 m2

5 a

1664 cm2

6 a

P = 29.7 m, A = 61.7 m2

b

P = 23.9 m, A = 26.8 m2

c

2

2166 cm

b

64.2 m2

d

6.8 m2

2

Language in Mathematics 2 a c e g 3 a b c

circle b quadrilateral composite d rhombus kite f sector trapezium h triangle A semicircle is half a circle. A quadrant is quarter of a circle. Composite shapes are calculated by dividing up the area. d Composite areas may be found in more than one way. 4 The area of a rhombus or a kite is half the product of the diagonals. Check Your Skills 1 C 5 A

2 A 6 A

3 C 7 D

4 D 8 B

2 a

b

2.5936 m2

Review Set 6D 1 a 2 a c 3 4 5 6

32 cm b 69 cm 1 1 A = --- h(a + b) b A = --- xy 2 2 A=B×H

a 280 cm2 5.004 km $419.30 a 25.7 cm

b

759 cm2

b

36.4 m

LEY_bk953_answers_finalpp Page 529 Thursday, January 13, 2005 3:23 PM

Answers

Cumulative Review: chapters 2–6 1 a b c d e f g h 2 a b c d e 3 a b

c

i 66 ii 511 i base = 7, index = 8 ii base = 4, index = –11 i 4×4×4×4×4×4 ii 3 × 3 × 3 × 3 × 3 × 3 × 3 i 625 ii 2048 20 ii 384 iii 53 i 5 24 48 iv 4 v 3 i False ii False 1 i -----ii 24 iii 3 41 75 1 i -----81

ii 8

iii 5

i

ii

b

w i ---4

ii 9m

c

5d i -----7

ii

xy iii ----3

15 iv -----16 ii y12

e

v 27v 15 i 1 ii 7

f

i

ii 9 t

iii

g

1 i ----c5

4 ii ----c5

1 iii -------------( 4c ) 5

h

i d–7

ii n3

iii k10

iv 15a–6

v

28 5 a

30

b c

31

Height of Year 9 students

ii

35

6 a

30

b

25

c d e f

20 15 10

g

5 h

140 150 160 170 180 190 200 Class iii 160–169 iv 163

4b b ------ = --8 2

i p12

i

30

20xy -----------15

d

t

cf 12

ii 38 iii grouped data iv 32–37 v 37.5 vi 32–37

5 15 27 20 8 8 2

15 h i ----------25

4 a

5

Cumulative frequency

Class Frequency 20–25 26–31 32–37 38–43 44–49 50–55 56–61

iv 1

ii 7.4 × 10–5 i 3.25 × 108 i 62 100 000 ii 0.000 03 6 4 × 10 = 4 000 000 but 46 = 4 × 4 × 4 × 4 × 4 × 4 = 4096 ii 4.86 × 1022 i 1.95 × 1025 iii 4.10 × 1075 iv 7 × 105 –16 7.5 × 10 , 2.4 × 10–11, 6.7 × 10–10, 7.6 × 10–10 i mean = 6.05, mode = 6, median = 6 ii Student to answer.

i

i

iii t 35

vi 21x10y 15 iii 1 iv 12

2m 5 ----------3 ii False v False

i True iv True

i 35w + 14x iii –4s + 8 i 11t + 2 i 5(5n – 4) iii 4m(3m + 4n) i 300

ii 30

i 13 500 iii 780 i 24.1 i 4.2 i 4 i 80 iv 81.46

iv c10

ii ii ii ii v

9t

iii False

ii –2k7 + 3k 5 ii 2x – 16y ii 2b(4b + 3)

3 iii ---------100 ii 41 iv 80 000 24.07 iii 20.97 2 iii 81 iii 81.461 ii 744.9˙

iv 3000

24.071 8 81.5

i 735 iii 735
number < 745 i 4 ii 2 iii 4 iv 3 v 2, 3, 4 or 5 15 s is to the nearest second but 15.0 s is to the nearest tenth of a second. 15.0 s is more accurate.

529

LEY_bk953_answers_finalpp Page 530 Thursday, January 13, 2005 3:23 PM

530

Answers

Cumulative Review: chapters 2–6 continued i 0.8˙

ii 0.43˙

iv 0.723˙

v 0.4˙ 36˙

b

i 0.2375

ii 1.583˙

c

6 3 i ------ = --10 5

7 a

iii 0.8˙ 5˙

92 23 ii ---------- = -----100 25

e

2 i --3 i 2 km/min

f

iii 33.3˙ m/s 8.5 t/ha

24 ii -----45 ii 2000 m/min

CHAPTER 7 Diagnostic Test 1 5 9 13

C A C B

2 6 10 14

D B B D

3 7 11 15

C C A A

4 8 12 16

A A A B

Exercise 7A 1 a 2 a 3 a d

(5, 4) (2, 6) (1, 3) (4, 1)

b b b e

(7, 3) (5, 5) (–2, 3) (1, 2)

c c c f

(5, 6) (3, 6) (3, 1) (–4, 1)

(0, 4 1--2- )

h

(1 1--2- , 0)

i

(2, 14)

4 a

(3, 3)

b

(1, 4)

c

(2, –3)

d

(0, 0)

e

( 1--2- , 1)

f

(1, 3)

g

(–3 1--2- , 2)

h

(–2 1--2- , –1 1--2- ) i

(0, 0)

5 a

(4, –5 --12- )

b

(2, –2 --12- )

c

(3, 3)

d

(16, 15)

e

(7 --12- , – --12- )

f

(–3 --12- , –5 --12- )

g

2

d

5 --3

3 --8

c

1 --3

d

11

d

+2

d

7 --6

3 --4

b

3 --6

3 a

7 --9

b

=

1 --2

Exercise 7D

544 68 iii ------------- = --------1000 125 d

c

2 a

1 a

+2

b

– 4--3-

c

–4

e

– 2--5-

f

5 --3

g

– 3--4-

2 a

0

b

1 --5

c

3 --5

e

7 --3

f

9 --2

g

undefined

3 a

0

d 4 a c

–1 e –2 OP, PQ, RS, TU b TU d ST e

f –9 QR, ST, UV VW f PQ

5 a

4 – ----11

4 --5

– --13-

b

b

4 --9

c

d

3 --5

6 a

4 --3

7 a

1

b

4 --3

c

– 2--3-

d

1 --2

e

2

f

– --14-

8 a

2

b

13

c

–6

e

5 -----12

f

11 -----40

d

1 -----10

e b

1 – --------125

– --43-

c

4 --5

y

1 a

y

b

3 x y

1

5

b

18 = 4.24

d

85 = 9.22

e

34 = 5.83

f

74 = 8.60

122 = 11.05

b

c

97 = 9.85

d

e

73 = 8.54

f

y

d

x x

–2

e

y

5

x

x

13 160 = 12.65

g

y

5

h

1 --5

c

3 --4

y

x –3

b

y

f

8

Exercise 7C 3 -----20

x

–4

41 = 6.40

1 a

– --43-

d

Exercise 7E

c

c

2 a

– 4--7-

f

Exercise 7B 1 a

– --23-

c

2 a

y = 3, y = –4, y = 8, y = –3

7

x

LEY_bk953_answers_finalpp Page 531 Thursday, January 13, 2005 3:23 PM

Answers

b 3 a b 4 a

b

(0, 3), (0, –4), (0, 8), (0, –3) x = 1, x = –2, x = 5, x = 7 (1, 0), (–2, 0), (5, 0), (7, 0) y b y=0 c between y = –1 and 2 y=1 1 x –1 –2 y

5 a

–2 –1

b c

y=x+3 x

–2

–1

0

1

2

y

–1

1

3

5

7

y=x+3

x

c

x=0 between x = –1 and x=1

y = –x – 1 x

–2

–1

0

1

2

y

3

1

–1

–3

–5

y

1 2

y = –x – 1

b

c

d

e

f

g

h

2 a

x

–1

Exercise 7F 1 a

y 3

x

1

2

3

4

5

y

1

2

3

4

5

x

–1

0

1

2

3

x

–2

–1

0

1

2

y

–3

–2

–1

0

1

y

6

5

4

3

2

x

–2

–1

0

1

2

y

2

1

0

–1

–2

x

–2

–1

0

1

2

y

7

6

5

4

3

x

–2

–1

0

1

2

y

–3

–1

1

3

5

x

–2

–1

0

2

y

1

1 --2

0

1 – --1-

–1

x

–2

–1

0

1

2

y

12

10

8

6

4

x

–2

–1

0

1

2

y

7

4

1

–2

–5

2

d

y=4–x

y 4

x e

–2

–1

0

1

2

y

–1

0

1

2

3

y = 2x – 2 x

–2

–1

0

1

2

y

–6

–4

–2

0

2

y y = 2x – 2 1 x

–2

f

y=x+1 x

y=4–x

y = 3 – 2x x

–2

–1

0

1

2

y

7

5

3

1

–1

y 3

y

x y = 3 – 2x

1 x y=x+1

3 a

y = x, y = x – 2, y = 2x + 1, y = x + 1, y = x + 3, y = x – 1, y = 2x – 2

b

y = –x, y = 5 – x, y = – 1--2- x, y = 8 – 2x, y = 1 – 3x, y = 4 – x, y = 3 – 2x

531

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532

Answers

Exercise 7F continued c The coefficient of x is positive for a positive gradient and the coefficient of x is negative for a negative gradient. d i positive ii negative iii negative iv positive v positive vi negative 4 a b y y y=x+4 4

4 x –4

y

c

x

–4

y=x–4 d

y 1

y = 2x x

x

y=1–x 1

y y=

f x

5 a Yes b No d Yes e No g Yes h Yes 6 a C b A c D 7 a Yes b Yes d No e No g Yes h No 8 Student to answer.

4

1

1 --4

0

1 --2 1 --4

1

2

3

1

4

9 16

y

4 x

–4 2 a

1 x

1

y

y = 2x2 y = x2 y = 12 x2

x+y=1 y

i

B

y = x2

y

–3

No No Yes e No Yes Yes

x

x

–3

c f i E c f i

d

x –4 –3 –2 –1 – 1--2- 0

b

– 32

x + y = –3

x

Exercise 7G

y

h

y

6

2x – 3y = 6

y = 2x 4+ 3 g

5x + 3y = 30

3 x

–2

y 16 9

3 4

x 2

y

p

10

1 a e

y

o

y

j

x –8 k

8 x

–6

x–y=8 l

y

6 x

b

x–y=6

y

y = 4x2 y = 3x2 y = x2

y

y = 14 x2

8 –4 m

4 x

y

–4

8 x x+y=8

y = –4 + x n 6 x

3y – 2x = –12

x c

y

y

–2

y=

y = x2

x

4 3

2 3x 2

–2

y = x2 + 2 y = x2 + 1 y = x2 – 1

1 x –1

4

LEY_bk953_answers_finalpp Page 533 Thursday, January 13, 2005 3:23 PM

Answers

3 Student to answer. 4 a y

y

6 a y = 4x

3

x

y = 3x

1

c

3

4

Language in Mathematics

y

e

1 a b

64 years Loci and set work, received a law degree and became King’s Councillor. c One d As many of his achievements were published after he died. e ‘Concerning the Comparison of Curved and Straight Lines.’ 2 a vertical b horizontal c gradient d distance e oblique f slope 3 a The y axis has the equation x = 0. b The x axis has the equation y = 0. c An uphill gradient is a positive slope. d The midpoint of a line is in the middle. e If a line goes downhill the gradient is negative. f The length of an interval is found using Pythagoras’ theorem. 4 In coordinate geometry an interval joins two points and has length, a midpoint and slope. Check your Skills 2 6 10 14

A A B B

3 7 11 15

D C A B

4 8 12 16

(7, 3)

b

(2, 7)

2 a

52

b

20

3

c

(6, 3 1--2- ) d

5 -----b 11 5 y = –2x – 2

y

–2

x–y=2

Review Set 7B 1 a

(3 1--2- , 3)

b

(3, 5 1--2- )

2 a

63

b

65

3 a 4

7 --4

b

5 --6

1 – --3 y

b

y

–4 c

2 x y=x+4 y

x + y = –8

d

x

y

x

–8

6 No since –1 – (–3) = 2 not 4. 7 a

y = 2 – 3x 2 3

–8 – –1 --2

2 x

x

4

(3, 1)

x

7 When x = 4, y = –3 ∴ Yes 8 y y = x2

3 --5

4 a

f

5 x x+y=5

Review Set 7A 1 a

4 3

5

5 a

B C B C

y = 4 – 3x

x y=x+3

–3 All pass through (0, 1).

B D D C

y

d

y

x

1 5 9 13

–4

x

y = 2x

b

y

b

1 1 --- , --- , 1, 2, 4, 8 4 2

x x–y=0

533

LEY_bk953_answers_finalpp Page 534 Thursday, January 13, 2005 3:23 PM

534

Answers

Review Set 7D

Review Set 7B continued b

y

y = 2x

c 2 a

(8, 3) ,

2 1--2-

)

50

b

(2, 5)

5

d

2 1--2-

b

,

– 1--2-

2 a

85

b

20

6 – --7 y

b

y

b

3

x –3 y = 2x – 3

3 x y=3–x

41 y

c 5 -----13

1 --3

b

)

y

d x + y = –1 x

–1

3 – -----11

–6

–1

5 --2

6 a

(5 --12- , 5)

3

4 3 – --7 4 a

b

4 –1 5 a

Review Set 7C (5 1--2-

(1, 2)

3 a

x

1 a

1 a

6 Yes, since –1 + 5 = 4 y

b

y

7 a x=2

x 2 –5 c

x –2 –1 1 --9

y

x

1 --3

y

b

y = –5

1

2

3

1

3

9

27

y = 3x

1

y

d

y

0

x 5

y = 5 – 4x

2 x –2 y = x – 2

5 4

x

Chapter 8 Diagnostic Test

e

f

y

y

x + y = –1

1

x

–1 – 12 g

x y = 2x + 1

1 C 5 C

2 D 6 A

3 B 7 B

4 D

Exercise 8A

–1

1 b

A

y

–6

x–y=6

7 No, since –3(2) – 1 = –7 not 5. 8 a b y

2

c

6 x

y = x2 + 2 x

y

y = 3x

x

540°

A

A

6 x x–y=6

LEY_bk953_answers_finalpp Page 535 Thursday, January 13, 2005 3:23 PM

Answers

2

Polygon

Number Number of Angle sum of of sides triangles polygon

quadrilateral

4

2

360°

Investigation 2 1 a c

120° b 3 hexagons, 360° A regular polygon will tessellate if the size of the interior angles divides exactly into 360°. 2 108° does not divide exactly into 360°.

pentagon

5

3

540°

hexagon

6

4

720°

heptagon

7

5

900°

octagon

8

6

1080°

triangle

3

180

60

3 a

Interior angle size

9

7

1260°

square

4

360

90

decagon

10

8

1440°

pentagon

5

540

108

n-gon

n

(n – 2)

(n – 2) × 180°

108°

b

120°

c

128 4--7-

°

d 135° e 140° f 144° a 3960° b 165° a 80° b 70° c 77° a 720° b i 50° ii 70° iii 65° a x = 50 (angle sum of quadrilateral is 360°) b x = 165 (angle sum of pentagon is 540°) c x = 120 (angle sum of hexagon is 720°) d x = 20 (angle sum of quadrilateral is 360°) e x = 120 (angle sum of hexagon is 720°) f x = 60 (angle sum of pentagon is 540°) g x = 40 (angle sum of hexagon is 720°) h x = 125 (angle sum of septagon is 900°) i x = 135 (angle sum of octagon is 1080°) 9 135° 10 135°

5 6 7 8

Investigation 1 1 2 3 4 5 6

Number Angle of sides sum

nonagon

3 (n – 2) 4 a

Regular polygon

a All 72° b 360° c 360° c 360° 360° a i 180° b 180°

6

720

120

7

900

128 --47-

octagon

8

1080

135

nonagon

9

1260

140

decagon

10

1440

144

Triangle, square, hexagon.

Language in Mathematics 1 a A pentagon has five sides. b A regular polygon has all sides equal. c A polygon with eight sides is an octagon. d The exterior angle sum of a polygon is 360°. 2 To find the interior angle sum of a polygon subtract two from the number of sides and multiply by one hundred and eighty degrees. Check Your Skills 1 D 5 C

2 C 6 A

3 B 7 B

4 A

Review Set 8A b

360°

ii 180° c 180°

1 a

d

360°

Exercise 8B 1 a 45° c 1080° 2 a 18° 3 24 sides 4 a 36 sides d 4 sides 5 21 sides 6 a 31 sides 7 22 sides 8 16 sides

b

hexagon heptagon

b

135°, straight line is 180°

b

162°

c

3240°

b

20 sides

c

15 sides

b

21 sides

c

22 sides

2 a b 3 a 4 a

b

c 720°

x = 100 (angle sum of a hexagon is 720°) x = 105 (angle sum of a pentagon is 540°) 135° b 360° 36 b 6120°

Review Set 8B 1 a 1800° b 150° 2 a x = 67 (angle sum of a pentagon is 540°) b x = 120 (angle sum of a hexagon is 720°) 3 165° 4 a 14.4° b 165.6° c 4140° Review Set 8C 1 a 2 a b

6120° b 170° x = 32 (angle sum of a hexagon is 720°) x = 60 (angle sum of a pentagon is 540°)

535

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536

Answers

Review Set 8C continued 3 120° 4 12° Review Set 8D 1 a

b

c 540°

2 a x = 40 (angle sum of a pentagon is 540°) b x = 87.5 (angle sum of a hexagon is 720°) 3 150° 4 45°

Diagnostic Test

b 2 a

b

2 C 6 B

3 C 7 A

3 a

b

Outcome Frequency

4 B 8 D

Relative Percentage frequency

Rough

86

86/200

43%

Smooth

114

114/200

57%

43 ---------100 Country Frequency

Relative Percentage frequency

Australia

146

146/400

36.5%

Japan

128

128/400

32%

Korea

56

56/400

14%

Germany

48

48/400

12%

Other

22

22/400

5 --12- %

36.5% Colour

c

1285 257 ------------- = ------------5000 1000

Exercise 9B 1 a

1 --4

b

1 --4

c

0

d

1

e

3 --4

2 a

1 --6

b

1 --6

c

1 --6

d

1 --2

e

1 --2

f

1

9 -----10

f

0

3 a

Number

Relative Percentage frequency

53

53/200

26.5

Red

48

48/200

24

Blue

27

27/200

13.5

Green

25

25/200

12.5

Yellow

21

21/200

10.5

Black

12

12/200

6

Silver

8

8/200

4

Other

6

6/200

3

c

Student to answer.

5 --6 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 1 1 1 ------ c --d --e 10 2 2 g

0

h

3 1 ------ = --12 4

b

4 1 ------ = --12 3

c

5 -----12

d

0

e

9 3 ------ = --12 4

f

8 2 ------ = --12 3

g

8 2 ------ = --12 3

h

5 -----12

i

1

5 a f 6 a c e 7 a

1 --4

1 -----52

b

c

1 --4

d

1 -----26

1 -----13

e

3 2 10 ------ g ------ h 0 i -----13 13 13 H, T b M, F HH, HT, TH, TT d 1, 2, 3, 4, 5, 6 ABC, ACB, BAC, BCA, CBA, CAB 1 1 5 --b 1 c --d --e 0 3 3 6

8 a

1 --6

b

2 1 --- = --- c 6 3

1 --2

9 a

1 --8

b

3 --8

0

c

10 a

2 --5

b

4 --5

c

3 --5

11 a

3 -----10

b

7 -----10

c

1

12 a

1 ---------------------100 000

b

950000 19 ---------------------- = -----100 000 20

13 a

9 -----10

b

8 2 ---------- = ------ c 100 25

12%

White

12

5

4 a

Exercise 9A 1 a

86 43 ------------- = ------------4754 2377

b

CHAPTER 9

1 B 5 B 9 D

4

d

1

d

0

22 ---------100

1463 30 30 14 In days ---------- × 3 + ---------- = ------------4453 365 366 15 a

1 --6

b

No

Exercise 9D 1 a

False, few names start with z ∴ not

1 -----26

.

LEY_bk953_answers_finalpp Page 537 Thursday, January 13, 2005 3:23 PM

Answers

b c d e

False, some teams are better than others. False, some players are better than others. False, some colours stay on longer than others. 1 -. True, 4 aces from 52 is ----13

f

True, P(6) =

g

True, if you know nothing about the horses and just pick one. However, the probability of 1 -. a particular horse winning is not ----24 True. It assumes that people who don’t work are unemployed. They may be retired etc. The sample space is too small. The students in 9 Red are not representative of the sample.

h 2 a b c

1 --6

.

Exercise 9E 1 5 kicks : 4 goals and 1 miss 50 kicks : 34 goals and 16 misses 2 a i Digits 0, 1, 2, 3, 4, 5, 6, 7, 8 signify a goal and the digit 9 is a miss. ii 45 goals, 5 misses b 45 goals c Student to answer. 3 a i 0, 1, 2, 3, 4, 5, 6 represents a goal 7, 8, 9 represents a miss ii 35 goals b same c Student to answer. 4 a i 0, 1, 2, 3, 4, 5, 6, 7 represent survival 8, 9 represent non-survival ii Student to answer. b Student to answer. 5 Student to answer.

0.8 2 $36 5 2 7 --38 C 11 9 -----14 16 $100 17 36 20 30 23 12 345 000

13 16 19 22 25

35% 35 0.16 B 5

b

1 C 5 A 9 C

2 250 000 18 22 30 21 20 5 --24 10 6

chance b event c probability random e prediction f theoretical Relative frequencies are used to estimate probability. Random number generators are used to simulate events.

2 B 6 C

3 C 7 C

4 B 8 C

Review Set 9A 1 a

b

Relative frequency Percentage 166/200

83%

34/200

17%

83%

1 1 1 1 --b --c --d --6 6 2 3 3 False, not all teams are equal in ability. 4 a i 0, 1, 2, 3, 4, 5, 6, 7 represent a goal, 8, 9 a miss ii 43 b overestimate c Student to answer. 2 a

Review Set 9B 1 Wrong, expect 1 2 a

3 -----17

b

9 -----17

c

5 ------ p 17

d

8 -----17

1 9 11 ---------b ---------c ---------100 100 100 4 Some colours are showing for longer times than others. 3 a

Review Set 9C

3 141 6 210 minutes 9 35 654.3 2 12 --7 15 203.024

Language in Mathematics 2 a d 3 a

Check Your Skills

1 a

Non-calculator Activities 1 4 7 10

4 Probability estimates become more stable as the number of trials increases.

Country Frequency Australia

195

195/400

Japan

103

103/400

25.75%

Korea

62

62/400

15.5%

Germany

34

34/400

8.5%

6

6/400

1.5%

Other b

Relative Percentage frequency

195 ---------400

c

48.75%

103 ---------400

1 1 1 --b --c --6 3 3 3 True 4 a i 0, 1, 2, 3, 4, 5, 6, 7, 8 represent survival 9 represents no survival ii Student to answer. b Student to answer. 2 a

537

LEY_bk953_answers_finalpp Page 538 Thursday, January 13, 2005 3:23 PM

538

Answers

Review Set 9D 1

r 6 a d

74 37 ------------- = ------------5188 2594

1 1 1 3 -----b --c --d -----13 2 4 13 3 Neither and both as a small number of trials is

g

2 a

50 ---------100

not accurate. Theoretically

7 8

CHAPTER 10 9

Diagnostic Test 1 5 9 13 17 21 25

D C A A D A D

2 6 10 14 18 22 26

C B D B B D A

3 7 11 15 19 23

A B C B D A

4 8 12 16 20 24

C B D A C B

10 11

12

a d

9a 2b

b e

5x 3x

c f

3x 4x

g

x

h

2x

i

2a2

j

7x + 3

k

11x2

l

17x – 7

m 16x

n

x2+ x

o

8b

p s v

q 2b2 t 4ab w 7b

r u x

5ab 3xy 2abc

2 a e 3 a e

6b 0 p 10x b

2y

c

5a2

d

2

i

5p2 + p

j

l

7cd – 2

m m2 + 5n

o

2a + 2b

8k – 8 2

3a2 – 7a

n

8x + 2y

q

10x – 10

2

t

4 + 2a

s

2n + 2n

u

3x – x2

v

mn – 8m

w 4 a e i m 5 a

k

2ab + 3b

9a – ab

5a2b + 2ab2 x 2x2 + 6x3 –3x b –11x c 3x –7d f –5d g –7n –15n j –3a + 2 k –3 – d –2g n 4a o –ac 3a + 10 b 6a + 8 c 4a + 6 2

2

e

4a + 3a

f

h l o

i 2n2 –2p – 5 9x – 2

7c + 1 j 7x m 10 – 12x n p 6 – 7x q

3ab + 3b

g

4

5d 7n 3x

i

6d – 3c + 2

2

5 – 5x

e

2d2

f

7d

g

2

3n + 3n

h

–n2 + n

i

15x + 23

j

4x2 + x

k

ac – bc

l

2x2 + 5x + 6

n

x–1

–4x2 + 6x + 12

p

2x2 + x – 12

q 13 a

–5x2 – 7x – 6 i 3(x + 5)

r ii

–9x – 13 3x + 15

b c

i 4x(2x + 7) i 3(3x + 1)

ii ii

8x2 + 28x 9x + 3

d

i y(x + y)

ii

xy + y2

Exercise 10B 1 a

5a -----6

b

b -----10

c

7c -----4

d

5x – -----14

e

7b -----12

f

2t ----9

g

5m -------21

h

d --2

i

p --3

j

22r --------35

k

4x -----21

l

m ---9

n

3r – -----16

o

26m ----------55

p

3d – -----20

c

a+4 -----------2

d

b – 12 --------------2

x m – --8 d h l

2

4x – y 5t + 6

d

o

16ab

p

r

–a2b + 7b

h

m 2x2 + x – 25

h 8ab i p2q 2x + 6 d 3y + 3 3t + 4 h 21n – 16

9x y f 2q g 0 9x – 2 b m + 11 c 4p – 5 f 7 + 3x g

a+1

c f

k 4m – 2n l –2s2t – 4s2 j 3p – p a 4x + 10 b 4x – 6 c 4y + 6 d 3x – 1 e 3y f 4x + 5 Let x m be the length of the longer side, let y m be the length of the shorter side. a 4x + 4; 4y – 8 b 4x + 6; 4y – 6 c 4x + 10; 4y – 10 x m is the length of the shorter side ∴ 3x + 10 y m is the length of the middle side ∴ 3y + 1 z m is the length of the longest side ∴ 3z – 11 x(x + 1), y(y – 1). a 1 – 3x b 4x + 2 c 11 – 5x d 10x – 3 e –3 – x f 10x + 2 g x+1 h 11 – 3x i 19 – 18x j 16 – 14x k 2x – 5 l –6x –6 a 5x + 3 b 5y + 9 c –p + 8

Exercise 10A 1

–3x –3x – 8

5

.

4 False, not all of equal ability.

x2 – 4x + 2 –5l – 10 b –ab – 2b e

2 a

x+2 -----------2

b

y–3 -----------3

e

x–8 -----------2

f

10 – x --------------- g 5

3a -----2

h

4b -----3

8y + 2

i

4a -----3

j

–a -----2

k

8x -----7

l

5x -----2

k –3 14p – 8 –6x – 13

– 8x m --------3

n

m

o

0

p

7x -----60

d

5a + 4b

LEY_bk953_answers_finalpp Page 539 Thursday, January 13, 2005 3:23 PM

Answers

3 a c

11x – 5a ---------------------4

b

64a – 125p ----------------------------20

d

– 25m – 14p -------------------------------10

c

y2 – 18y + 81

d

9x2 – 6x + 1

e

4x2 – 12x + 9

f

25a2 – 40a + 16

65p – 27r ------------------------15

g

9x2 – 24x + 16

e

52m – 35n ---------------------------10

f

45a – 88b -------------------------20

g

– 3t – 17m ---------------------------12

h

62r – 119m -----------------------------28

80x – 21y ------------------------15

i

4 6

2 x2 + 5x

1 --2

(x2 + 3x)

1 --2

(x + 7)(x + 2)

16y2 – 8y + 1

i

25 – 20x + 4x

j

1 – 6x + 9x2

k

25 – 30x + 9x2

l

16 – 16x + 4x2

x2 + 10x + 25

b

x2 – 10x + 25

3 a c

4x – 28x + 49

d

4x2 + 28x + 49

e

9x2 – 30x + 25

f

16x2 + 24x + 9

g

25 + 20x + 4x2

h

9 – 42x + 49x2

3 x2 – 3x

5 (x + 3)(x + 1)

b

x – 16

c

x2 – 36

d

x2 – 100

e

x2 – 1

f

4x2 – 25

g

9x2 – 4

h

25x2 – 1

i

49x2 – 64

5 a 2

7 a

x + 6x + 5

b

2

x + 9x + 14

2

x + 11x + 24

c

8 a x2 + 6x + 8

c x2 + 4x – 12

2

2

2

2

x – 20x + 100 b x + 14x + 49 c x2 – 64

d

x2 – 121

e x2 + 8x + 16

g

x2 – 14x + 49

h x2 – 144 i x2 + 24x + 144

6 a

b x2 – x – 6

2

x –9

4 a

Exercise 10C 1 x2 + 6x

h

2

2

x + 6x + 9 b x2 – 14x + 49 c x2 – 4x + 4

d

(x – 3)2 = x2 – 6x + 9

e

(x – 5)2 = x2 – 10x + 25

g x – 7x – 30 h x – 11x + 28 i x + 6x – 7

f

(x + 6)2 = x2 + 12x + 36

j x2 – 5x + 6 k x2 – 15x + 54 l x2 – 15x – 36 9 a 21 b 71 c 23 d 55 e 61 f –35 g –6 h –12 i 45 j –187 k 234 l 52

g

(x + 7)2 = x2 + 14x + 49

2

d x + x – 12

2

e x – 2x – 15

2

2

f x + 4x – 45

2

2

2x2 + x – 3

b

2x2 – 21x + 40

c

3x2 – 20x + 25

d

9x2 – 9x – 10

e

6x2 + 23x – 18

f

16x2 – 25

10 a

g

12x – x – 1

h

6x2 – 19x + 10

i

8x2 + 2x – 15

j

49x2 – 4

k

10x2 – 31x + 15

l

12x2 – 17x – 5

x – 10

b

6x2 + 11x – 4

c

x2 – 7x – 14

d

x2 + 2x + 10

e

3x2 – 8x + 14

f

10x – 4

g

–x2 + x + 10

h

–2x2 + 16x + 5

i

–2x2 + 13x – 6

11 a

2

2

Exercise 10D 1 a

x2 + 4x + 4 2

b

x2 + 12x + 36

c

y + 20y + 100

d

9x2 + 6x + 1

e

4x2 + 12x + 9

f

16a2 + 40a + 25

g

25x2 + 40x + 16

h

9y2 + 6y + 1

i

2

16 + 24x + 9x

j

1 + 4x + 4x2

k

25 + 20x + 4x2

l

16 + 40x + 25x2

x2 – 4x + 4

b

x2 – 12x + 36

2 a

h 7 a

f x2 – 49

(x + 9)2 = x2 + 18x + 81 9 b 25 c 16 d

f

81

g

49 -----4

k

1 --4

l

1 --4

h

225 ---------4

i

4

e

36

9 --4

j

81 -----4

Exercise 10E 1 a d g j 2 a d 3 a c e g i 4 a c e g i k m o

3(a – b) b 5(m + 2n) c q(p – r) x(x – 5) e x(4x + 1) f 3x(5 + x) q(p – 3q) h 2π(R – r) i 6xy(x – 3y) 7ap(4p – 3) k 9xy(x + 3) l 3pq(r – 5p) –3(a + b) b –2x(2x + 1) c –4(2a – b) –4(1 + 2b) e –(3 + x) f –9x(2x – 1) B(6 + a + c) b R(4 – x + y) x(6 + 14y – 3z) d 8x(x – 3y + 2yz) (x – 2)(4 + y) f (x – 1)(3 + y) (x + 1)(a + 3) h (x – 4)(x – 1) (p – 3)(3 + x + y) j (a + 1)(x – 1 – y) (2x + 1)(2 + x) b (x – 1)(3 + z) (x + 5)(y + 3) d (x – 2)(y + 4) (x – 7)(x + y) f (4 – x)(x + y) (y + 3)(x – 2) h (y – 4)(2x + 5) (3x – 7)(y + 4) j (3 – x)(x + t) (a – b)(3 + c) l (3x – 2)(x + y) (4 – 3x)(1 + y) n (6 – 5y)(a + b) (4 – 3p)(p – q)

539

LEY_bk953_answers_finalpp Page 540 Thursday, January 13, 2005 3:23 PM

540

Answers

Exercise 10F 1 a c e g i k m o 2 a d

(x – 2)(x + 2) b (y + 3)(y – 3) (z – 4)(z + 4) d (a – 3)(a + 3) (c + 5)(c – 5) f (x + y)(x – y) (a – c)(a + c) h (m – n)(m + n) (2x – 1)(2x + 1) j (3x – 2)(3x + 2) (3x + 1)(3x – 1) l (4y – 3)(4y + 3) (5y – 4x)(5y + 4x) n (10x – 9y)(10x + 9y) (8a – 5b)(8a + 5b) p (11r – 3t)(11r + 3t) 1200 b 800 c 2000 6 e 1.87 f 1711

3 a

(x – y)(x + y)(x2 + y2)

b c

(2a – 3b)(2a + 3b)(4a2 + 9b2) –4pq

Exercise 10G 1 a

(x + 5)2

b

(x – 10)2

c

(x + 9)2

d

(x – 8)2

e

(x – 7)2

f

(x + 11)2

g

(y – 3)2

h

(y + 1)2

i

(x – 12)2

2

1 a c e g i k 2 a c e g i k 3 a c e g i 4 a c e g

k

(m + 3)

l

(a – 5)2

2 a

(2x + 5)2

b

(5x + 6)2

c

(4x – 9)2

d

(3x + 8)2

e

(11x – 6)2 f

(9x + 5)2

g

(7x + 10)2 h

(5x – 2)2

(2x + 11)2

j

(7x – 3)2

k

(3x + 5)2 l

(10x – 9)2

3 a

(p + q)2

b

(m – n)2

c

(r + t)2

i

d

(d – p)2

e

(n – t)2

f

(r + y)2

k 5 a c e g i

i

Exercise 10H 1 a c e g i k 2 a c e g i k 3 a c e g i k 4 a c e

(x + 7)(x + 1) (x + 12)(x + 1) (x + 6)(x + 4) (x + 6)(x + 5) (x + 5)(x + 4) (x + 18)(x + 1) (x – 5)(x – 1) (x – 11)(x – 1) (x – 8)(x – 1) (x – 10)(x – 1) (x – 15)(x – 1) (x – 14)(x – 1) (x + 8)(x – 1) (x + 2)(x – 1) (x + 6)(x – 2) (x – 8)(x + 3) (x + 7)(x – 3) (x + 20)(x – 3) (x + 18)(x + 1) (x + 18)(x – 1) (x + 54)(x – 1)

b d f h j l b d f h j l b d f h j l b d f

(x + 6)(x + 2) (x + 9)(x + 1) (x + 10)(x + 3) (x + 10)(x + 2) (x + 6)(x + 3) (x + 7)(x + 6) (x – 7)(x – 1) (x – 4)(x – 2) (x – 5)(x – 2) (x – 5)(x – 3) (x – 7)(x – 2) (x – 8)(x – 3) (x + 5)(x – 2) (x – 7)(x + 6) (x – 12)(x + 1) (x + 8)(x – 3) (x – 21)(x + 1) (x + 9)(x – 6) (x – 9)(x + 2) (x + 9)(x + 6) (x – 27)(x + 2)

(x – 8)(x – 8) (x – 32)(x + 2) (x – 3)(x + 10)

h j l

(x + 16)(x – 4) (x – 5)(x + 7) (x – 5)(x – 10)

(2x + 3)(x + 1) (3x – 1)(x + 2) (2x + 5)(x + 1) (7x + 2)(x + 1) (5x + 1)(x – 3) (5x – 3)(x – 1) (2x – 3)(x + 4) (3x + 4)(x + 1) (3x + 1)(x + 4) (3x + 2)(x + 2) (3x – 2)(x + 4) (2x – 3)(x + 6) (2x – 5)(x + 7) (5x – 3)(x – 1) (5x – 4)(x – 5) (11x – 3)(x + 5) (5x – 7)(x – 9) (4x + 1)(2x + 3) (7x + 1)(3x + 2) (6x + 1)(x + 3) (7x + 1)(2x + 5)

b d f h j l b d f h j l b d f h

(2x + 1)(x – 5) (3x + 1)(x – 2) (2x – 1)(x + 2) (2x + 5)(x + 1) (5x – 3)(x + 1) (11x + 2)(x – 1) (3x + 2)(x – 3) (2x + 3)(x – 3) (3x – 2)(x – 5) (5x + 2)(x – 3) (2x – 1)(x + 9) (2x – 3)(x + 7) (3x – 4)(x + 3) (3x + 2)(x – 1) (7x + 1)(x + 2) (7x – 5)(x – 8)

b d f h

(5x + 2)(3x – 1) (3x + 1)(2x + 1) (5x + 1)(2x + 3) (21x + 1)(x – 3)

(2x + 1)2

Exercise 10I

(p – 4)

j

2

g i k

j

(5x – 2)(2x + 1)

2

l b d f h

(3x + 2)(x + 4) (4x – 3)(x – 5) (4x – 1)(3x – 5) (12x + 5)(x – 7) (4x – 7)(5x + 1)

x(3x + 2)

b

(x + 9)(x – 9)

(3x – 2) (2x – 3)(3x + 1) (3x – 4)(3x + 2) (3x + 2)(4x – 5) (5x – 3)(2x + 5) (9x – 4)(2x + 3)

Exercise 10J 1 a

2

c

2(p + 4)

d

3(b + 5)(b – 5)

e g i k m

2(x – 4)(x + 4) (x – 9)(x + 1) 3(x – 6)(x + 6) (2a – 3d)(2a + 3d) –1(x + 12)(x – 1)

f h j l n

n2(n + 2)(n – 2) (d + 7)(d – 1) 2(g – 11)(g + 5) 4t(1 + 2t) 5(a – 2)(a + 1)

o

2(c – 3)(c – 1)

p

x2(x + 1)(x – 1)

q

d2(d + 3)(d – 1)

r

(b + 7)(b – 4)

s

ab2(a + 1)(a – 1)

t

(x – 3)(x + 2)

u

x(x + 2)2

v

x2(3x – 2)(3x + 2)

w (x + 9)(x – 1)

x

–2(a + 3)2

b d f

(x + 7)2 2x(3 – x)(3 + x) ab(ab – 2)

2 a c e

–(x + 7)(x – 2) 4a(a + b)(a – b) a(b + c – 2)

LEY_bk953_answers_finalpp Page 541 Thursday, January 13, 2005 3:23 PM

Answers

g i

–2x(x – 1)2 h xy(x + 2)(x – 2) (a + b – 3)(a + b + 3) j x(x – 4)

k

4(x2 + 4)(x + 2)(x – 2) l (x – 2)(y – z)

m (x + 1)(a + b) o (x – y)(a + 1) q 3 a c e

(x + 1)(x2 + 1) 3x(x + 3) 5x(x – 3) (x + 8)(x – 5)

2

g

4x + 14 2x + 12 ---------------------------------------------------- h ------------------------------------(x + 5)(x – 5)(x – 1) x(x – 3)(x + 3)

Exercise 10L

2

n (x + a )(x + a)(x – a) p (x + 2)(x + 3) r b d f

(x2 + 1)(x + 2) (2x + 1)(2x – 1) x(3 – 5x) (x + 4)(x – 4)

1 a

W

1

2

3

4

5

S

1

4

9

16

25

b

x (x + 2)

h

(x + 3)(x – 3)

20

i

3x(x2 + 2)

j

3(x + 2)(x – 2)

l n p r t v

2

(x + 5) 4x(x + 2) (x – 13)(x – 3) 2(x + 5)(x – 5) 4x(2x – 3) 13x(x – 4)

15 10

x

x2(x – 9)

x(x + x + 1) 3(x + 3) (x – 3)

b d

(x – 12)(x + 5) (x – 4)(x + 2)

e

(x + 2)2

f

(3x – 2)(2x + 3)

30

g i k m

(x – 3)(x – 2) 4(x – 5)(x + 3) (x + 6)(x + 5) (x – 4)(x – 3)

h j l n

36x2 + 25 3(x – 11)(x – 3) (7x + 1)(7x – 1) (x + 8)(x – 2)

25

o q s

(x – 8)(x + 3) (x – 7)(x – 2) (x – 12)(x + 3)

p r t

(x – 4)2 (x + 4)(x + 9) (x + 9)(x – 2)

u (x – 5)2 w 4(x – 4)(x + 3)

v x

3(x + 6)(x – 4) 2(x – 1)(2x + 3)

k m o q s u

2

3x (x + 2) (x – 3)(x + 2) (3x – 5)(3x + 5) 7x(x – 3) 9x(1 – 2x) (2x – 1)(2x + 3)

w (x + 3)(x – 1) 4 a c

2

5

x -----------x–2

b

x -----------3–x

c

3x -----------x+1

d

x+1 -----------x–1

e

2(x – 1) -------------------x–5

f

x+3 -----------x–8

g

–3 ( x + 2 ) -----------------------x+7

h

–( x + 6 ) --------------------2(x + 1)

2 a

6

b

1

c

1 --2

d

x --4

e

7 --5

f

2 --3

g

x–1

h

3x -----------x+4

W

1 2 a

2 3 4 5

i W

1

2

3

4

5

S

3

6

11

18

27

ii

iiiS = W 2 + 2

S

20 15 10 5 0 b

W

1

2 3 4 5

i

Exercise 10K 1 a

S = W2

25

g

2

c

A

W

1

2

3

4

5

S

4

7

12

19

28

ii

iii S = W 2 + 3

S

30 25 20

3 a

–x–9 ------------------------------------x(x + 3)(x – 3)

b

5x + 28 ------------------------------------x(x + 4)(x – 4)

c

5x + 6 ------------------------------------x(x – 6)(x + 2)

d

4x – 2 -------------------------------------3(x – 4)(x + 1)

e

x + 14 ------------------------------------x(x – 2)(x + 7)

f

5 ( 2x – 3 ) ------------------------------------x(x – 5)(x – 2)

15 10 5 0

W

1

2 3 4 5

3 a l

1

2

3

4

5

6

7

8

9

b

9

8

7

6

5

4

3

2

1

A

9

16

21

24

25

24

21

16

9

541

LEY_bk953_answers_finalpp Page 542 Thursday, January 13, 2005 3:23 PM

542

Answers

iv a

Exercise 10L continued b A 25

20 15 10

l

1

2

3

4

5

6

7

8

9

10

11

12

b

12

11

10

9

8

7

6

5

4

3

2

1

A

12

22

30

36

40

42

42

40

36

30

22

12

b

5

l

30 20

22 3--4-

4 i

a

l

1

2

3

4

5

6

7

10

8

b

8

7

6

5

4

3

2

1

A

8

14

18

20

20

18

14

8

l

0 v

b

A

20 15 10

1

5

6

7

8

9

10 11 12 13 14

b

14 13 12 11 10

9

8

7

6

5

A

14 26 36 44 50 54 56 56 54 50 44 36 26 14

7

ii a l

1

2

3

4

5

6

7

8

9

10

11

12

13

b

13

12

11

10

9

8

7

6

5

4

3

2

1

A

13

24

33

40

45

48

49

48

45

40

33

24

13

50 40 30 20

3

4

c

3

2

40.25

l

2

4 6 8 10 12 14

5 a

y = x2 + 1

b

y = x2 + 5

d 6 a

y = x2 – 2

e

y = x2 + 3 b

y

c

y = x2 – 1

y 5

c 36.75

A

4

A

20 0 0

8

15 3--4-

b

2

60 40

l

2 3 4 5 6

4 6 8 10 12

l

b

1

2

a

5

c

33.25

40

1 2 3 4 5 6 7 8 9 c

c

A

50

1 x c

10

x d

y

y

l

0 2 4 6 8 10 12 14 iii a

x

l

1

2

3

4

5

6

7

8

9

10

b

10

9

8

7

6

5

4

3

2

1

A

10

18

24

28

30

30

28

24

18

10

–2 e

b

c 26.25

A

30 20 10 0

x

–1

y

3 l

0

2

4 6 8 10

x

1

LEY_bk953_answers_finalpp Page 543 Thursday, January 13, 2005 3:23 PM

Answers

A = 25x – x 2

7 a

b

A = 25l – l 2

x 2 + 2x – 15

b

6x 2 – 37x + 56

c

x 2 – 16x + 64

d

9x 2 – 24x + 16

e

x2 – 4

f

9x2 – 25

3 a

A 156.25

4 a 5 a d

l 12.5 A = 19x – x 2

8 a

b

25

f h

A = 19l – l 2

A

6 a

121 ---------4 7(x + 2) b 4xy(2x – 5) c (x – 3)(3 + y) (2y – 5)(2y + 5) e (x + 8)(x – 1) 20

b

i 9

(3x – 1)(x + 4) –2(x – 4)(x + 3) x+3

ii

g i

5x + 8 c ------------------------------------x(x – 4)(x + 4)

1 --2

b

90.25

x2(x + 4)(x – 4) 3(x + 2y – 3z)

Review Set 10B l 9.5

19

Non-calculator Activities 1

5.27

4

25 -----16

2

$3.08

5

(10 – 3 + 4) × 3 = 33

6

–1.5ºC

7

321.7408 8

9

8 -----13

10 8 L/100 km 11 3

=

9 1 ----16

12 4

3

13 7 and 8

420 cm

P=–

14

1 a

–6x – 1

b

3x – 2

c

4p2 – 5p

2 a

8x -----15

b

11 -----5

c

44x – 39a ------------------------12

3 a

x 2 – 9x – 22

b

12x 2 – 41x + 24

c

x 2 + 8x + 16

d

64x 2 – 80x + 25

f

25x 2 – 9

e 2 --3

1 --7

15 20

2 a b

x –9

4 a

31

5 a c e

3(a – 3) (y – 3)(2x + 7) (x – 2)(x – 1)

g i

Language in Mathematics

2

b

i 25 b d f

81 -----4 6x(2y + 3x) (4y + 5)(4y – 5) (3x – 5)(2x + 1) ii

x 2(3x – 4)(3x + 2) h x(3 + 2y – 4z)

(x 2 + 1)(x + 3)

x+3 -----------6

x+5 -----------8

Expanding is the opposite of factorising. Always factorise the highest possible common factor. c Multiply all the terms in the bracket by the term outside the bracket. 3 To check a factor is a solution you can either expand your answer or substitute a number into the question and the answer to see if they are equal.

1 a

–3x + 6

b

2x + 2

c

–2y 2 – 4y

2 a

26x --------15

b

4p -----3

c

40x – 57y ------------------------12

Check Your Skills

3 a

x 2 + x – 12

b

6x 2 – 29x + 35

c

x 2 – 8x + 16

d

16p 2 – 40p + 25

e

x2 – 9

f

16x 2 – 9

1 5 9 13 17 21 25

C B A B A A B

2 6 10 14 18 22 26

A A C A B C C

3 7 11 15 19 23

D D D C D D

4 8 12 16 20 24

B B C A B B

Review Set 10A 1 a

–5x + 1

2 a

p -----35

b

3x – 17

b

4a -----3

c

–n2 + 4n

c

20a – 39x ------------------------15

6 a c

b

– 2x – 17 ---------------------------------------------------(x + 5)(x + 2)(x – 2)

Review Set 10C

4 a 5 a c e g i 6 a c

Both expressions equal 49. b =4 5(x + 2) b 3x(x – 2y) (x – 2)(2 + 3y) d (3x + 10)(3x – 10) (x + 3)(x – 4) f (3x + 1)(x + 5) x(x + 4)(x – 4) h –3(x + 8)(x – 1) 3(4x – 3p + 2z) 3x (x + 4) b -----------x+3 –x+3 ------------------------------------x(x + 3)(x – 3)

543

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Answers

Review Set 10D 1 a

– 6x + 14

2 a

– 13z ------------35

2

b

21 – 15x

c

4m – 22m

b

9r -----4

c

27x – 28p -------------------------6

x 2 + 9x – 22

b

12x 2 – 22x – 14

c

y 2 + 6y + 9

d

36z 2 – 60z + 25

e

x 2 – 81

f

36x 2 – 25

3 a

4 a 5 a c e g i

Both expressions equal 8. b = –6x 3(x – 5) b 4y(3x – 2z) (x – y)(p + 2) d (4x – 5)(4x + 5) (x – 3)(x + 7) f (5x – 3)(x + 2) (3x – 2)(2x + 5) h 2x(x + 3)(x – 3) – 5(3x + 4y – 2z)

6 a

x–2 -----------2

b

1 --3

10 – x ------------------------------------x(x – 5)(x + 5)

c

CHAPTER 11 Diagnostic Test 1 5 9 13 17

B C A D D

2 6 10 14 18

B C C C C

3 7 11 15 19

C D B A B

4 8 12 16

D B A D

Exercise 11A Student to answer. Exercise 11B 1 a b c 2 a

3

4 5 6 7

i $914 ii $23 764 i $2051.20 ii $53 331.20 i $2757.88 ii $71 704.88 i $1007.69 ii $2015.38 iii $4366.67 b i $1834.04 ii $3668.08 iii $7947.50 c i $1594.23 ii $3188.46 iii $6908.33 a i $1538.46 ii $3076.92 iii $6666.67 b i $2115.38 ii $4230.77 iii $9166.67 c i $769.23 ii $1538.46 iii $3333.33 a $975 b $$849.33 c $2920.67 a $969.23 b $1300.38 c $830.54 Lisa ($70 200 p.a.) a $661.50 b $926.80 c $2957.50

8 9

a a b c

10 a

$28.40 b $24.20 c $17.90 i $1634 ii $3268 iii $84 968 i $2010.20 ii 4020.40 iii $104 530.40 i $2861.40 ii $5722.80 iii $148 792.80 $2839.20 b $5132.40 c $11 934

Exercise 11C 1 3 5 7 8 9 12 15 18 20 21 23

$959.40 $790.50 $1630.20 a $209.95 a $236.70 $26.90 4 hours $1455 $52 926.72 a $438.55 $2890.50 $960

2 $696.60 4 $1513.20 6 $1735.20 b $247 c $296.40 b $289.30 c $341.90 10 $21.60 11 $33.54 13 $4883.33 14 $2138.50 16 $3600 17 $78 000 19 a $532 b $3572 b $2944.55 22 $3971.50 24 $716.80

Exercise 11D 1 $199.20 4 $0.48

2 $252.35 5 $433.40

3 $864

b b b b b e b e b e

$4800 $972 $675 $350 $399.80 $1030 $500 $1820 $5200 $8050

c c c c c

$14 700 $414 $6075 $2450 $400

c

$800

c

$5800

b e

$340 $445.20

c

$520

Exercise 11E 1 2 3 4 5

a a a a a d 6 a d 7 a

$9900 $1620 $375 Nil $260 $575 $460 $1100 $4500 d $6850 8 $848.80 9 a $433 d $374 10

11 a d 12 a

Commission

Weekly earnings

$84

$284

$430

$580

$627

$727

$1032

$1032

$1382.40

$1382.40

$250 b $262 c $286 $322.90 e $398.80 i A $200, B Nil ii A $290, B $240 iii A $320, B $320 iv A $350, B $400

LEY_bk953_answers_finalpp Page 545 Thursday, January 13, 2005 3:23 PM

Answers

v A $500, B $800 If sales are more than $4000 per week method B is better. 13 $16 420 14 2% 15 $260 16 $8600 17 $16 300 18 $40 000 b

Exercise 11F 1 2 3 6 9

$286.05 a $547.05 $795.90 $308.34 a $71.36 d $109.27 10 a $140.92

b 4 7 b e

$109.55 24 hours 5 $215.70 $346.03 8 6 hours $44.60 c $80.28 $173.94 b $181.24 12 3 1--2- hours

11 $28.41 Exercise 11G 1 $565.20 4 $1211.43

2 $484.40 5 $684.81

3 $837.23

Exercise 11H 1 a b 2 a b 3 a b 4 a b

Income = $30 680, Expenses = $24 684, Income – Expenses = $5996 p.a. 3 × $5996 – $12 550 = $5438. Naomi will be able to take her holiday. Income = $28 080, Expenses = $29 896, Income – Expenses = –$1816 p.a. He needs to reduce his expenses by approximately $35 per week. Income = $14 560, Expenses = $12 950, Income – Expenses = $1610 p.a. Cost of car = $3360 p.a.; No. Income = $45 700, Expenses = $41 450, Income – Expenses = $4250 p.a. Yes

Exercise 11I 1 $700 3

2 $1440

Simple interest $1624 $1575 $5400 $1950 $1008 $1264.80 $540

4 $10 755 7 a $675 d $1026 8 6 years 10 7% p.a.

5 b e 9

$24 000 6 $17 496 $700 c $937.50 $7020 5 years 11 4.6% p.a.

Exercise 11J 1 a 75c b d $7.30 e g $16.25 h j $39.15 k 2 a $4.15 b d $7.15 e g $21.45 h 3 $421.50 4 a $138.20 b d $169.75 e 5 Student to answer.

$5.30 $3.95 $22.00 $17.35 $5.00 $1.65 $1.80

c f i l c f

$2.80 $8.50 $54.85 $70.00 $1.80 $14.00

$435.20 $64.85

c

$210.55

Exercise 11K 1 a 13 Dec., 10 Jan. b 15.95% c 0.04369% d $10 000 e $7674 f 4 Feb. g $34 2 a $2368.72 b $4290.76 c $2326.26; Yes 3 1.46% 4 Student to answer. Exercise 11L 1 3 4 5 6

$34 2 a $405 b a $77.80 b a $68.40 b Student to answer.

$40 $101.25 $700.20 $387.60

c c

$58.35 $64.60

Exercise 11M 1 2 3 4 5

a a a a a d a d a d a a

$3171 b $572 $2669.80 b $371.80 $1943 b $345 $1171.33 b $292.33 $1992.56 b $493.56 c $246.78 $1299 e 19% 6 $7483.08 b $1593.08 c $531.03 $5590 e 9.5% 7 9503.84 b 3513.84 c 878.46 5490 e 16% 8 $7600 b $1824 c $392.67 9 $9900 b $2673 c $349.25 10 a $499 b $4491 c $1347.30 d $243.26 11 a $2985 b $16 915 c $8051.54 d $520.14 12 Student to answer. Exercise 11N 1 a 2 a 3 a

$4654.72 $14 005.80 $2215.76

b b b

$193.95 $389.05 $184.65

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Answers

Check Your Skills

Exercise 11O 1 $298.93 2 a $3233.77 d $401.04 3 a $229.17 4 a $419.68 5 a $6447.96 d $12 613.80 6 3 years

b e b b b e 7

$1456.74 c $438.10 $361.48 $8430.15 $20 394.77 $16 442.12 c $9720.80 $21 597.40 13.5% 8 $24 000

Exercise 11P 1 a c e 2 a c e 3 a c 4 a c 5 a c 6 a 7 a b c

5 kg for $6.60 b 2 kg for $15.96 2 L for $3.50 d 350 g for $2.20 750 mL for $10.99 425 g for $3.50 b 750 mL for $1.12 235 g for $2.49 d 250 g for $2.58 1.25 kg for $8.28 Ray’s b Both the same. Harry’s Bob’s b Sophie’s Both the same. Bob’s b Both the same. Bill’s A b Both the same. c B Starnet by $10 Supernet by $12.50 Supernet by $22.50

Exercise 11Q 1 a $44, $484 b $369, $4059 c $25.80, $283.80 d $39.70, $436.70 e $180, $1980 2 a $104.50 b $247.50 c $18.70 d $369.60 e $502.70 3 a $108.09 b $451.82 c $0.99 d $13.45 e $11.27 4 Student to answer. 5 a $85, $935 b $9.08 c $488 d $164.50, $14.95 Non-calculator Activities 1 2 5 8 9 10

a $600 $3650 $400 $108 a $741 $17

b $15 600 3 $860 6 $562.87 b

c $1300 4 $432 7 $80.10

$152

Language in Mathematics 1–5 Student to answer. 6 a fortnightly b retainer c budget d discount 7 piecework, service, investment

1 5 9 13 17

C D D D B

2 6 10 14 18

B C C B B

3 7 11 15 19

C B A A B

4 8 12 16

A A D C

Review Set 11A 1 2 3 6 9 12 13 15 16

a $927.80 a $1076.92 $878.90 $562 $280 a $1296 $150.35 $69 a $7999

17 $230.24

b $24 122.80 c $2010.23 b $2153.85 c $4666.67 4 $18.20 5 $2632 7 $376.32 8 $476 10 $188.50 11 $471.84 b $540 14 Student to answer. b

$3839.52 c

18

2 1--2-

$246.64

years

19 5 kg for $14.80 20 $1.64 Review Set 11B 1 2 3 6 8 10 12 13 15 16 17 19 20

a $737.08 a $876.92 $1046.64 $557.20 $596 $249.24 a $3240 $76.55 a $2460.96 a $1455 $257.93 a B-One $11.64

b $19 164.08 c $1597.01 b $1753.85 c $3800 4 4 hours 5 $3196 7 $355.68 9 $12 600 11 $399.02 b $1350 14 Student to answer. b $571.96 b $436.50 c $78.81 18 10.5% b B-One c A-One

Review Set 11C 1 2 5 8 9 12 13 15 16

a $692.31 $1581.67 $2970.40 a $4350 $320 a $3900 $228.80 $98.50 a $10 000

17 $643.86

b $1384.62 c $3000 3 $854.40 4 $26.84 6 $1262 7 $1398.80 b $5200 c $5700 10 $404.40 11 $530.48 b $1365 14 Student to answer. b 18

$2400 3 1--2-

c

$344.44

c

$4666.67

years

19 250 g for $4.88 20 $49.80, $547.80 Review Set 11D 1 a

$1076.92

b

$2153.85

LEY_bk953_answers_finalpp Page 547 Thursday, January 13, 2005 3:23 PM

Answers

2 5 7 10 12 13 15 16 17 19 20

$2288 $3384 $319.50 $229.76 a $2736 $166.32 a $7999 a $5908.55 $388.50 a Bob’s $0.90

3 $1217.05 4 8 hours 6 $609.42 8 $732 9 $8640 11 $412.93 b $1140 14 Student to answer. b $3839.52 c $246.64 b $246.19 18 14% b Bob’s c Bill’s

y

ii

y = 3x

1 x k

y

i

y = x2

x

Cumulative Review: chapters 7–11 1 a b

y

ii

i (4 1--2- , 5)

ii (3, 5)

i (6, 3)

ii (2, 6 1--2- )

iii (6 1--2- , 2 1--2- )

iv ( 5--2- , – 3--2- )

e

–1

i d =

52 = 7.2

ii d =

58 = 7.6

d

i d =

50 = 7.1

ii d =

41 = 6.4

6 i --- = 1 1--2 4 f –3 y h i

ii g ii

6 2 --- = --9 3 2 y

2 a d g

1260° 135° 26 sides

y

5 3

x + y = –3 x

–3 –3 y

i

Yes, since –1 –(–3) = 2.

j

i

x –2 –1 y

1 --9

1 --3

0

1

2

3

1

3

9

27

165° 18° 46.3 cm

k

Perimeter = 177.5 m, Area = 1593.6 m2 Relative Percentage frequency

Outcome Frequency Heads

x

92 ---------100 108 ---------200

92

Tails

108

46% 54%

b

Based on table 46% or 0.46.

c

i

d

All numbers have equal chance. 3.

e

i P(R) =

f g

2x – y = 0 x

c f i

169.6 cm2

1 --6

ii

4 a c

1 --6

iii

4 -----17

iii P(W) = iv

6 150° 4474.5 kg

y = 5 – 3x

5 x y=x+2

b e h

j

3 a

2

iii

1 x

–1

c

–2

y = x2 – 1

3 --6

=

1 --2

iv

ii P(B) = 6 -----17

2 --6

=

1 --3

7 -----17

iv P(not blue) =

10 -----17

Not all teams are of equal ability. i Kick will be a goal 0, 1, 2, 3, 4, 5. Kick will miss 6, 7, 8, 9. ii 7 0 2 5 5 3 4 4 9 2 0 3 4 4 7 7 4 7 4 0 4 5 9346824472382553579974 539427 iii 34 goals compared with 30 expected iv Student to answer. 11y 2 – 8y

b

35x – 54y ------------------------10

i x 2 – 3x – 15

ii 6x 2 – 29x + 28

iii x 2 + 2x + 1

iv 9r 2 – 24rp + 16p 2

2

v x –4

vi 9x 2 – 1

d x 2 – 4x + 4 = (x – 2)2 e i 3(x – 3) ii 4x(x – 3y)

547

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Answers

Cumulative Review: chapters 7–11 continued iii (x – 3)(2 – 3y) v (x – 5)(x – 2)

iv (4x – 5)(4x + 5) vi (2x – 5)(3x + 1)

2 i --x 5 a i $1288.46 ii $2576.92 iii $5583.33 b $2192.67 c $914.85 d $3854 e $406.70 f $650 g $285.40 h $332.93 i i $3528 ii $1176 j $214.14 k i $13 999 ii $6719.52 iii $431.64 l i $5328.55 ii $222.02 m i Bob’s by $2.25 ii Bob’s by $1.45 iii Bob’s by 65 cents n 80 cents GST

opposite adjacent ------------------------------- ------------------------------hypotenuse hypotenuse

Q

opposite -----------------------adjacent

vi

ED -------FD

ED -------EF

FD ------EF

vii

x -z

x -y

z -y

viii

TV ------UT

TV ------UV

UT ------UV

ix

u ---w

u --v

w ---v

x

SU ------ST

SU ------TU

ST ------TU

xi

h --k

h --g

k --g

xii

k --l

k ---m

l ---m

3

f

CHAPTER 12

1 5 9 13

C B D D

2 6 10 14

A A B C

3 7 11 15

C D B B

4 A 8 A 12 A

Exercise 12A 1 a b c d e f g h i j k l 2 a 3

i AB i RQ i XZ i r i n i EF i y i UV i v i TU i g i m b

UT i ------TS

ii

UT ------US

iii

TS ------US

b

a i --b

ii

a --c

iii

b --c

c

t i -s

ii

t --u

iii

s --u

d

XY i ------XZ

ii

XY ------YZ

iii

XZ ------YZ

e

h i --j

ii

h --i

iii

j --i

f

ST i ------RT

ii

ST ------RS

iii

RT ------RS

5 a

BC i ------CA

ii

BA ------AC

iii

CB ------BA

b

TV i ------AV

ii

AT ------AV

iii

TV ------TA

c

PR i ------AP

ii

AR ------AP

iii

PR ------AR

b e h k b e h k n b e h 8°

0.4226 0.1564 0.7880 1.4281 83° 27° 34° 88° 48° 47° 51° 42° c 79°

4 a

Diagnostic Test

b

ii ii ii ii ii ii ii ii ii ii ii ii c

BC PR XY p m ED x TV u SU h k

iii iii iii iii iii iii iii iii iii iii iii iii c

c

AC PQ ZY q l FD z UT w ST k l d b

Q

opposite -----------------------adjacent

i

BC ------AC

BC ------AB

AC ------AB

ii

PR -------PQ

PR -------RQ

PQ -------RQ

iii

XY ------ZY

XY ------XZ

ZY ------XZ

iv

p --q

p --r

q --r

v

m ---l

m ---n

l --n

opposite adjacent ------------------------------- ------------------------------hypotenuse hypotenuse

Exercise 12B 1 a d g j 2 a d g j m 3 a d g 4 a

0.4067 0.2419 3.7321 0.9962 14° 66° 2° 17° 57° 74° 34° 54° 59° b

c f i l c f i l o c f i

0.7002 0.0875 0.8090 0.9816 73° 44° 87° 28° 18° 45° 65° 23° d 44°

LEY_bk953_answers_finalpp Page 549 Thursday, January 13, 2005 3:23 PM

Answers

Exercise 12C 1 a d g 2 a d 3 a d 4 a d g 5 a d 6 a d 7 a d 8 a d 9 a 10 a d g

e

10.8 8.7 12.9 17.6 81.1 5.7 8.6 5.1 4.0 1.8 162.8 4.4 56.1 11.3 2.7 21.5 7.0 15.0 8.1 2.9 49.5 309.8

b e h b e b e b e h b e b e b e b e b b e h

2.8 8.7 10.6 16.4 13.2 2.6 199.8 10.1 7.4 23.0 4.2 332.1 14.5 12.2 10.1 146.1 13.2 44.4 36.7 13.8 102.9 12.9

c f i c f c f c f i c f c f c f c f c c f i

Cosine is the adjacent side divided by the hypotenuse. 3 The origin of the word cosine is from the words complement sine so that the cosine of an angle equals the sine of ninety minus the angle.

11.3 1.7 12.5 25.6 154.7 47.6 68.2 81.3 38.7 10.8 184.0 45.5 17.3 12.1 8.5 378.9 229.5 15.2 60.9 25.5 4.2 19.6

Check Your Skills 1 5 9 13

D B C B

1 a

b

c

40° 48°

c h

24° 44°

d i

43° 42°

e

b g

41° 67°

c h

48° 63°

d i

56° 67°

e

48°

b g b g l

27° 60° 25° 53° 20°

c h c h

56° 63° 52° 76°

d i d i

56° 52° 61° 38°

e

21°

e j

50° 41°

2 21.2 m 5 117 m 8 12.4 cm 11 78 m 14 16°

3 6 9 12 15

4 B 8 B 12 C

y z y z sin α = -- , cos α = -- , tan α = -- , sin β = -- , x x z x

l m l m sin α = --- , cos α = ---- , tan α = ---- , sin β = ---- , n n m n

32° d

451 m 49° 56° 12.2 m 12.88 m

trigonometry b opposite hypotenuse d angle of depression cosine f tangent sine h adjacent Angles are measured in degrees. Sine, cosine and tangent are all ratios. Look down for the angle of depression. The hypotenuse is opposite the right angle.

t u t u sin α = - , cos α = --- , tan α = --- , sin β = --- , r r u r t u cos β = - , tan β = --r t b c b c sin α = --- , cos α = --- , tan α = --- , sin β = --- , a a c a b c cos β = --- , tan β = --a b

2 a 3 a 4 a

11.1 b 31° b 26.0 m b

12.6 c 47° c 65.5 m c

84.2 32° 90°, 53°, 37°

Review Set 12B 1 a

y x y sin θ = -- , cos θ = -- , tan θ = -z z x

b

r p r sin θ = --- , cos θ = --- , tan θ = --q q p

c

AB AC AB sin θ = ------- , cos θ = ------- , tan θ = ------BC BC AC

d

SV VT SV sin θ = ------- , cos θ = ------ , tan θ = ------ST ST VT

Language in Mathematics 1 a c e g 2 a b c d

B D A A

l m cos β = --- , tan β = ---n l

b g

22.2 m 130 m 2m 34° and 56° 49 m

3 7 11 15

y z cos β = -- , tan β = -x y

Exercise 12E 1 4 7 10 13

A A D D

Review Set 12A

Exercise 12D 1 a 39° f 42° 2 30° 3 a 66° f 74° 4 71° 5 a 63° f 42° 6 a 52° f 18° k 74°

2 6 10 14

2 a 3 a 4 a

70.7 51° 11.7 m

b b b

31.7 49° 342 m

c c c

25.6 25° 4.8 m

Review Set 12C 1 a

b c b c sin α = --- , cos α = --- , tan α = --- , sin β = --- , a a c a b c cos β = --- , tan β = --a b

549

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550

Answers

Review Set 12C continued b

m x = 11

n

q p q p sin α = --- , cos α = --- , tan α = --- , sin β = --- , r r p r

-----x = – 15 4

p

x=3

q

x=

q p cos β = --- , tan β = --r q

s

x = –4

t

v 2 a d 3 a

x = –1 x = 10 x = –42 No b

w x=3 x b x = 15 c e x = 42 f No c Yes

4 a

x=

10 -----3

b

x=

9 --4

c

x=

7 --5

d

x=

7 --3

e

x=

10 -----7

f

x=

13 -----3

c f i

x=1 x=1 x=3

c

x=2

x y x y sin α = -- , cos α = -- , tan α = -- , sin β = -- , z z y z

c

x y cos β = -- , tan β = -z x e d e d sin α = --- , cos α = --- , tan α = --- , sin β = --- , f f d f

d

e d cos β = --- , tan β = --f e 17.0 b 14.9 cm 8.10 km 23° b 41°

2 a d 3 a 4 a

c

5.20 m

c

61°

33° 19.6 m

b

14 m

36 m c

126 m

a c a sin θ = --- , cos θ = --- , tan θ = --b b c

b

RT PR RT sin θ = ------- , cos θ = ------- , tan θ = ------PT PT PR

c

LM TM LM sin θ = -------- , cos θ = -------- , tan θ = -------LT LT TM s r s sin θ = -- , cos θ = - , tan θ = -t t r 1.5 b 10.9 c 6.8 d 55° b 23° c 16° 44.9 m b 55° c 187 m

d 2 a 3 a 4 a

3.5

2 6 10 14

B A B D

3 7 11 15

D C C D

4 8 12 16

x=8

u

x = – 5--2x = –5 x = –25 x = 50 d No

x=1 x = –2 x = –10

2 a

x = – 9--4-

b

x=

d

x = – 5--2-

e

-----x = – 14 3

f

x=

13 -----2

g

x=1

h

x = –7

i

x=

3 -----10

j

a = –1

k

-----s = – 13 5

l

x=

7 --5

19 -----6

n

y = – 9--5-

o

p=

3 --2

3 a d 4 a d g

No Yes x=2 x=5 x = –4

b e b e h

Yes Yes x = –1 x=0 x = –2

c f c f

No No x=3 x=2

5 a

x=

7 --2

b

x=

c

x=4

d

x=

7 --5

e

x = – --38-

f

x=3

g

x=

5 --2

h

x=5

i

x=

17 -----8

j

x=0

k

x=0

l

x=

31 -----11

6 a

x=2

b

x = –5

c

x=

17 -----10

e

x=2

f

x=

23 -----10

x=

19 -----3

4 --5

3 --5

g x=4 h x=0 7 a 2(x – 3) = –3 – (2x – 5) b 3(x – 1) = 7 – (3x – 2) c 5(2x + 3) = 46 – (5x + 1) d 6x + 13 = 5(3x – 1) 8 Student to answer.

C D A A

Exercise 13A 1 a

x=9

b e h

d

Diagnostic Test A C D D B

r

x=4 x = –1 x=9

CHAPTER 13

1 5 9 13 17

5 --7

1 a d g

m x=

Review Set 12D 1 a

x = –2

Exercise 13B

30.2 m

57°

o

Exercise 13C

x=7

b

x = –3

c

x = –2

d

x = – --23-

e

x = – --65-

f

x=

g

x=2

h

x=

1 --3

i

j

x = – 2--3-

k

x=1

l

1 a d

x = 15 x=4

b e

x = –10 x = 13

c f

x = –12 x = –8

x = – 8--7-

g

x = – 7--2-

h

x = –1

i

x=

41 -----2

x = – 7--2-

j

x = – 5--2-

k

x=3

l

x=

13 -----2

5 --2

LEY_bk953_answers_finalpp Page 551 Thursday, January 13, 2005 3:23 PM

Answers

2 a

x=

1 --4

b

x=

9 --4

c

5 --3

f

-----x = – 13 7

i

5 x = – ----11

d

x = 5 1--2-

e

x=

g

x = – 1--5-

h

x = –3

j 3 a d g

x=9

x = –2

k

x = –7

l

x = 22

x=

20 -----7

b

x = –5

c

x=8

x=

20 -----3

e

x = 12

f

x=6

x = 20

h

x = –8

Exercise 13D 1 2 3 4

a a a a d 5 a b e 6 a

4 b 3 c 6 63, 64 b 8 c 4 b –2 c 1 6, 4 b 4, 8 c 8, 10, 12 e 1, 2, 3 12 × 10 cents, 18 × 5 cents 10 c 3 d 17 6 kg brand A, 4 kg brand B 11 b 7 c 17

d 36 0 d 2 4, 6

d

7

1 a d g j 2 a d g j 3 a d g

x = ±3 x = ±12 x = ±6 x = ±2 ±3.46 ±6.86 ±8.25 ±14.14 x = ±3 x = ±4 x = ±6

b e h

x = ±4 x = ±7 x = ±9

c f i

x = ±8 x = ±11 x = ±10

b e h

±7.14 ±9.11 ±9.54

c f i

±4.36 ±5.29 ±13.89

b e h

x = ±6 x = ±5 x = ±7

c f

x = ±3 x = ±4

4 a

x = ± --72-

b

9 x = ± ----10

c

x = ± --45-

d

x = ± --49-

e

12 x = ± ----7

f

8 x = ± ----11

g

7 x = ± ----12

h

-----x = ± 10 9

5 a d

x = ±2.83 x = ±1.90

b e

x = ±1.73 x = ±1.60

6 a

x=±

8 --3

b

x=±

7 --2

-----x = ± 10 9

d

x=±

5 --7

c

7 a c 8 a

98.5 cm2 (1 d.p.) 3

7916.8 cm 0.0126 cm 11.3 km (1 d.p.)

9 a 598.3 cm2 10 a 1.4 seconds

x = ±2.45 x = ±2.14

b

7.98 m (1 d.p.)

b

1.6 m (1 d.p.)

b

71 metres

b b

28.2 cm 25 cm

9, 14.7 cm2 (1 d.p.) a 1250 b 8.64 a 65.4 b 92.5 a yz = 6 b

11 12 13 14 15

c

7.07

6.2 x1 = 0

Exercise 13G 1 a c e g

4x < 13 10x  80 4x + 3 > 70 2x – 13 > 20

b d f h

x --- + 10 < 19 2 2 a Yes b Yes e Yes f Yes i Yes j No m No n Yes 3 Student to answer. 4 a Infinitely many. i

Exercise 13E

c f

6 a

5x > 50 3x + 2 < 35 x – 8
30 20x – 9 > 10

c g k

x --- + 5 > 6 3 No d No h No l

b

Infinitely many.

j

Exercise 13H 1 a

x>2

b

x
5

c

x6

d

x < –6

e

x
–6

f

x > –1

g

x > –16

h

x < –7

i

x > –5

j

x
–3

k

x
–3

x

0

2

0

5

0

6

–6

0

x x

–6

0

–1

0

–16

x x x x

0

–7

0

–5

0

–3

0

–3

0

x

Exercise 13F 1 2 3 4

a a a a c 5 a

80 cm2 b 6 cm 35.81 cm (2 d.p.) b 54.0 cm (1 d.p.) b 100 km/h b 7 hours 52 minutes 44.1 m b

79.58 cm (2 d.p.) 15.9 m (1 d.p.) 260 km about 129 m

x x x

Yes Yes No

551

LEY_bk953_answers_finalpp Page 552 Thursday, January 13, 2005 3:23 PM

552

Answers

Exercise 13H continued l

x > –2

–2

m x>6 n

x
3

o

x

p

5 --4

x>3

q

x


r

x>3

s

x
– --52-

p

6

0

3

0

5 4

0

x x x

3 –5

t

x2

2 a

x > –6

b

x  –12

c

x > –6

d

x

9 --2

e

x


13 -----3

f

x


5 --2

g

x > –11

h

x
14

i

x>

0

2

–6

0

x

0

0

9 2

x

–11

x

5 3

x>1

f

x

g

x>

11 -----38

h

x


33 -----14

i

-----x < – 45 4

j

x<

33 -----2

k

x

1 --2

l

x

16 -----17

1 10 1--2-

2 134.77

3 46 820

4 $110

5 –2

6 52

7 4 4--5-

8 0.18

9 23 675

10 B

11 35

12

1 --3

13 16 19 22 25

14 17 20 23

15 18 21 24

0.000 8 21 20 60 cm2

–8°C $75 34 Yes 819

2 26 10 6

pronumeral solution variable

A C D A D

2 6 10 14

D B D D

3 7 11 15

A A B A

4 8 12 16

B C B A

x = –4

b

x=

11 -----2

c

x=

1 --2

x = –84

e

x = – 5--4-

f

x = –23

1 --6

b

x = – 9--5-

c

x=3

4 a

x

d

x=2

e

x = 7 --12-

f

x = – --72-

g

x = 6 1--2-

h

x = 24

i

x=

x

1 --2

b

x<5

c

x < –1

d

x

–5 -----2

e

x<

4 --5

f

x
1

g

x>

h

x<

8 --9

i

x


1 --3

j

x>1

k

x>

5 --2

l

x

1 --4

m x<0

n

x


o

x1

–5 -----3

e

x

3 a

–7 -----2

6 -----19

-----x > – 12 23

d

x

0

5

15 -----13

2 2 3 x=1

14

–1

x


d

1 a

x

0

0

c

x<

Review Set 13A

5 2

0

11 -----9

31 -----11

b

1 5 9 13 17

x

13 3

0

x>

Check Your Skills

x

0

–6

r

Language in Mathematics 2 a linear b c quadratic d e substitution f

x

–12

– 1--5-

x

0

2

x0

Non-calculator Activities

x

5 2

0

q

x  – 1--7-

x

3

6 --5

4 a

x

0

0

5 --2

x

0

x>

x=

9 -----16

5 a d 6 a

x=3 b 17 c 25 cm 9 × 20 cents, 27 × 10 cents, 12 × 5 cents 1.72 s b 156.25

7 a

x < –2

d

b

x  –6

c

x

e

11 x
– ----2

f

11 x > – ----7

x=1

b

x=

11 -----9

c

x=

x = –15

e

x=

3 --5

f

x = 11

x>

7 --2

1 --4

Review Set 13B 1 a d

15 -----2

LEY_bk953_answers_finalpp Page 553 Thursday, January 13, 2005 3:23 PM

Answers

3 x=

20 -----11

1 --8

b

x=

30 -----11

c

x=2

5 A 9 C 13 C

x=2

e

x=

35 -----3

f

x=2

Exercise 14A

x=

40 -----3

i

8 --9

-----2 – 14 3

4 a

x=

d g

x=7

h

5 a Number is 5. b The smallest integer is 12. 6 5.64 7 a x  –4 b x<2 c x > – --13-

d

e

x3

f

x=

x < –2 x < –2

Review Set 13C 1 a

x=

f

y = – 1--2- x – 4 1--2-

g

y = – 3--2- x + 8 1--2-

h

y = 6x

i

y = 4x + 3 1--2-

j

y=4

y = 2x – 1

b

y = –x + 2

d

y=x–3

f

y=

h

y = 2x + 4

j

y=

3 --2

x+6

y = 2x – 4

b

y=

7 --2

x+7

y = – 5--3- x – 5

d

y=

3 --2

x–3

y=x–7

g

y = – --74- x –

i

y=

a=

17 -----5

c

p=

8 --3

f

-----x = – 13 2

d

x=

19 -----10

e

x=

g

x=

23 -----2

h

x = –15

i

x=

5 a

-----– 11 2

b

32 cents

c

-----– 27 7

d 6 a

36 km $11 328.80

7 a

x < –6

13 -----2

3 -----17

3 a c

1 a b

d

10.72 g

x>5

c

x

e

x


– 1--3-

f

x>3

b

x=

1 --7

x = –35

e

x = 12

2 a

y = – 4--3-

b

x=4

3 x=

13 -----5

3 --2

Review Set 13D

2 a d 3 a

c

x = –20

f

x=4

d f 4 a

21 x = – ----8

b

6 t = – ----11

d

x = – --43-

e

x=

g

x=

5 -----22

h

x=

33 -----17

c

p=

16 -----3

f

19 x = – ----3

37 -----29

i

x=

x+

1 --4

4 --5

9 -----11

y=

– 1--2-

x+2e

y=x+1 y=

1 --2

b

b

x+4 e

y=

– 2--3-

y=

4 --3

N=

1 --2

M = –g + 2 e

F=

1 -----10

x+5

g

y = – 1--3- x + 5 h

y = 8x

y=

1 --3

x+6

d

7

j 5 a 6 a

y=2 2

b

6

c

x


f

x

1 –4 –3 –2 –1 0 –1 –2

D

y = –3x + 22 1 --2

e

4

c

y=

x  –6

B

x2 + 1

y = 5x – 15 f

-----x  – 10 3

3

G = – 1--2- S + 2

y = –2x – 12e

c

A

y=x

d

x < –3

2

x–1 c

1 P = – --- n – 2 6 y = 2x – 10 b y = x – 7

b

D

x+6f

y = –3x – 1

y = – --14- x – 2

x>7

1

c

y = –2x – 2 f

7 a

Diagnostic Test

x

y = – --23- x + 2

y 3 2

CHAPTER 14

5 --3

x–1 c

t+2 b

M=

1 --3

y = 4x – 6

The number is –2. ±8.4

2 --3

The number is –19. ±18.1

y = 2x + 7

5 a 6 a

d

b b

3 --5

x–

3 1--3-

Exercise 14B

b

7 --3

– 1--3-

e

b

x<5

x+6

x = – --92-

3 x=4 a = –5

1 --3

f

x=9

4 a

y=

y=

b

d

e

c

g = – 2--5-

x=

y = 5x – 15

x = –12

2 a

1 a

y = –2x – 6

d

c

x=4

d

b

y = – 1--2- x + 7 1--2-

x=4

e

8 A 12 D

y = 3x – 9

b

x = 36

7 C 11 B

c

2 a

7 --4

d

4 a

1 a

6 B 10 A 14 D

i 5

(2, 3)

1 2 3

4 x

553

LEY_bk953_answers_finalpp Page 554 Thursday, January 13, 2005 3:23 PM

554

Answers

Exercise 14B continued b

g

y 3 2

(1, 3)

1 2 3

–4 –3 –2 –1 0 –1

4 x

–2 c

h

y 4

1 1 2 3

–4 –3 –2 –1 0 –1

4 x

i

(–2, 1)

–4 3

2 3

1 2 3

4 x

1

–4 –3 –2 –1 0 –1

4 x

–2 Exercise 14C

1 1 2 3

4 x

–2 f

1 2 3

2

y 3 2

–4 –3 –2 –1 0 –1

4 x

y 3

–2 e

1 2 3

–2

y 3 2

–4 –3 –2 –1 0 –1

(3,4)

2

–2

1

4 x

3

1

d

1 2 3

–2

y 3 2

–4 –3 –2 –1 0 –1

(2, 3)

1

1 –4 –3 –2 –1 0 –1

y 3 2

1 a c e g i k

2x – y + 1 = 0 2x – y + 5 = 0 3x + y – 4 = 0 x + 2y + 10 = 0 9x + 12y + 8 = 0 2x – 4y – 3 = 0

b d f h j l

5x – y – 2 = 0 2x + y + 5 = 0 x – 2y + 4 = 0 2x + 3y + 9 = 0 3x – 8y + 12 = 0 10x – 2y + 5 = 0

2 a

y = – 1--2- x + 2, m = – 1--2- , b = 2

y 3 2

b

y = – 3--2- x + 12, m = – 3--2- , b = 12

c d

y = 2x + 4, m = 2, b = 4 y = 2x – 6, m = 2, b = –3

1

e

y = – --52- x – 5, m = – --52- , b = –5

f

y = – 3--2- x + 4, m = – 3--2- , b = 4

g

y = 4x – 6, m = 4, n = –6

h

y=

i

y = 4x – 3 1--2- , m = 4, b = –3 1--2-

–4 –3 –2 –1 0 –1 –2

1 2 3

4 x

3 --2

x + 8 1--2- , m =

3 --2

, b = 8 1--2-

3 a

–3

b

– 9--4-

c

– 7--5-

4 a

8, 4

b

4, 6

c

–3, 6

d

5 --4

LEY_bk953_answers_finalpp Page 555 Thursday, January 13, 2005 3:23 PM

Answers

d

12, –18

e

–4, –10

f

5 --3

,

g

5 --4

h

–5, 7 1--2-

i

5 --9

, –2 1--2-

5 a

, –5

5 --2

e

x+y=5 y 8

x + 2y – 8 = 0

6 y 6

2

4

–8 –6 –4 –2 0 –2

2 4 6

8 x

–4 f

3x – y – 6 = 0

y 8

4

6

2

4

–4

2 5 –6 –4 –2 0 –2

–6

–4

2 4 6

8 x

g

2x – 3y – 4 = 0

–2 d

–4 h

y 8

–6 –4 –2 0 –2

2 4 6 x –3

5x + 2y + 10 = 0

2

y x

–4 –2 0 –2

6 2

2 4 x

3x – 4y – 12 = 0

–4 –2 0 –2

1 2 3 x – 43

4x + 3y – 8 = 0

4

5

y

y

–3 –2 –1 0 –1

x–y+5=0

y 6

–8 –6 –4 –2 0 –2

c

5 2 4 6 x

–6 –4 –2 0 –2

2

b

5

4

8 3

–4

2 –5

–6

2 4 6 x i

x – 2y = 0

–4

y

3 2

(2, 1)

1 x

–2 –1 0 –1 –2

1 2 3

555

LEY_bk953_answers_finalpp Page 556 Thursday, January 13, 2005 3:23 PM

556

Answers

Exercise 14D

11 a

1 a, b, e, h 2 a

y = 2x + 2

d

5x – 7y – 31 = 0

f

y=

– 5--7-

e

1 --2

y=

x–

–

– 4--3-

7 --5

c

y=

e

2x – 5y + 35 = 0

6 a c 7 b c

– 3--5-

1 --7

x

– --23-

c

– --23-

e

y=

iv dPQ =

– --75-

f b

y = 4x + 5

d

4x + 3y – 7 = 0

f

y=

x–4 b

x

Exercise 14E 1

b c 12 a

1 --2

y = –2x + b b y = 3x + b 3x – 5y + c = 0 d 3x + 5y + c = 0 4x + 3y + c = 0 Co-efficients of x and y are interchanged and sign of y is changed. 2x – 7y + c = 0 Same co-efficients of x and y. 7x – 5y + c = 0 b 5x + 7y + c = 0

8 b c 9 a

(0, 2)

d 13 b e

2

(1, –3)

3

c

14 --12-

130 -------------2

11 -----3

5 a b

dAB = 17 , dBC = 17 , dAC = isosceles since AB = BC

6 a b

dPQ = 5 , dQR = 20 , dPR = points are collinear

b

–5

d

f 14 a b


5.7 1 --5

18

dAB = dAC = D=

( 1--2-

dQR =

d

mST = mQR =

85 , dST = 6 --7

85 -----4

b

1 --2

c

AD =

81 -----2

(–1 2--7- , 1 3--7- ) A(–3,

1 --2

), B(– --12- , 3), C(2 --12- , –1), D(0, –3 --12- )

parallelogram

16 A = 12 units2

Exercise 14F 1 a

x ≥2

y 3

2

dQR

1 2 3 x

–2 –3 b

T = (–3, –1) =

)

2x – 3y – 5 = 0 circumcentre, circumcircle 2x – y + 4 = 0, 3x + 2y + 1 = 0, x + 3y – 3 = 0

10

d Right-angled, as mXY × xmXZ = –1 9 EFD is right-angled at D (3-4-5 triangle).

c

43 so ABC is isosceles

–3 –2 –1 0 –1

mXY = 3, mYZ = –2, mXZ = – 1--3-

b

,

– 1--2-

1

(XY)2 + (XZ)2 = (YZ)2.

S = ( 1--2- , 2)

18 , mPQ = –1,

mAD = –1, mBC = 1, so AD ⊥ BC 6x + 4y + 7 = 0 c 4x + 7y + 9 = 0 perpendicular bisector of XZ is

45

dXY = 10 , dYZ = 20 , dXZ = XYZ is right-angled as

10 a

18 , dRS =

mRS = –1 so PQ || RS  opposite sides equal and parallel PQRS ≠ rectangle since PQ is not ⊥ QR PQRS ≠ rhombus since PQ ≠ QR

7 dLM = 20 , dMN = 20 , dLN = 40  LMN is isosceles as LM = MN

c

1 --2

2x – 3y – 5 = 0; (– 1--2- , –1) lies on

15 a

4 a

8 a b

, mRS = –1, mPS =

iii midPR = ( 1--2- , 1), midQS = ( 1--2- , 1) b

y = – 1--5- x +

g

1 --2

so diagonals bisect each other

5 --4

5 a

18 ,

 QR || PS and PQ || RS

x–2

5 --3

d

45 , dRS =

45  QR = PS and PQ = RS

ii mPQ = –1, mQR =

2 1--2-

3 b, d, e 4 a

18 , dQR =

dPS =

y = –7x – 37 c y = – 1--2- x

b

i dPQ =

x≤–1

y 3

2 1 –3 –2 –1 0 –1 –2 –3

1 2 3 x

LEY_bk953_answers_finalpp Page 557 Thursday, January 13, 2005 3:23 PM

Answers

c

y ≥–2

y 3

y ≤4

y >–5

2

1

1 –3 –2 –1 0 –1

1 2 3 x

–2

–2

–3

–3 –5 h

y <7

2 1 –3 –2 –1 0 –1

x>3

1 2 3 x

–4 y 4

3

e

y 3

2

–3 –2 –1 0 –1

d

g

y 7

6 5

1 2 3 x

4

–2

3

–3

2 1

y 3

–3 –2 –1 0 –1

2

–2

1 –3 –2 –1 0 –1

1 2 3 x

–3 1 2 3 x

x ≥–4

–2

y 3

–3

2

i

1 f

x<1

y 3

–4 –3 –2 –1 0 –1

2

–2

1

–3

–3 –2 –1 0 –1

1 2 3 x

1 2 3 x j

x < – 1 1--4-

–2

y 3

–3

2 1

–114

–3 –2 –1 0 –1 –2 –3

1 2 3 x

557

LEY_bk953_answers_finalpp Page 558 Thursday, January 13, 2005 3:23 PM

558

Answers

Exercise 14F continued k

y ≥

3 --2

c y 3

2

2

1

1

–3 –2 –1 0 –1

–3 –2 –1 0 –1

1 2 3 x

–3

–3 y <1

d

y 3

2 1

1

–3 –2 –1 0 –1

1 2 3 x

–3

–3 e

y 3

1

1

b

y 3

2

2

–3 –2 –1 0 –1

1 2 3 x

–2

–2

2 a

y 3

2

–3 –2 –1 0 –1

1 2 3 x

–2

–2

l

y 3

–3 –2 –1 0 –1

1 2 3 x

1 2 3 4 5 6 x

–2

–2

–3

–3

–4 –5

y 3

f

2

y 4

1

3

–3 –2 –1 0 –1 –2 –3

1 2 3 x

2 1 –3 –2 –1 0 –1 –2 –3

1 2 3 x

LEY_bk953_answers_finalpp Page 559 Thursday, January 13, 2005 3:23 PM

Answers

3 a

4 a

y 4

y 3

3

2

2

1

1 –3 –2 –1 0 –1

–3 –2 –1 0 –1

1 2 3 4 x

1 2 3 x

–2

–2

–3

–3 b

b A square of side length 4 units. 5 Student to answer.

y 3

Exercise 14G

2

1

a

1 –3 –2 –1 0 –1

1 2 3 4 x

y 4

3 2

–2

1

–3 –3 –2 –1 0 –1

–4 c

–2

y 4

–3

3

b

y 4

2 1 –5 –4 –3 –2 –1 0 –1

3 2

1 2 3 x

1

–2

–4 –3 –2 –1 0 –1

–3 d

1 2 3 4 x

–2

y 4

–3 –4

3 2

c

1 –4 –3 –2 –1 0 –1

1 2 3 x

1 2 3 4 x

y 3

2 1

–2 –3

–3 –2 –1 0 –1 –2 –3

1 2 3 x

559

LEY_bk953_answers_finalpp Page 560 Thursday, January 13, 2005 3:23 PM

560

Answers

h

Exercise 14G continued d

y 3

2

2

1

1

–3 –2 –1 0 –1

–3 –2 –1 0 –1

1 2 3 x

–3

–3

2

a

y 3

2 1

1

–3 –2 –1 0 –1 1 2 3 4 5 6 x

–3

–3

–4 b

–5 –6

y 3

2 1

y 4

3

–3 –2 –1 0 –1

2

–2

1

–3

–4 –3 –2 –1 0 –1

1 2 3 4 x

–2

c

y 5

3 2

y 4

1

3 2

–3 –2 –1 0 –1

1

–2

–2

1 2 3 4 x

4

–3

–4 –3 –2 –1 0 –1

1 2 3 4 x

–2

–2 –4

g

y 3

2

–4 –3 –2 –1 0 –1

f

1 2 3 x

–2

–2

e

y 3

–3 1 2 x

1 2 3 x

LEY_bk953_answers_finalpp Page 561 Thursday, January 13, 2005 3:23 PM

Answers

d

h

y 4

3

3

2

2

1

1

–3 –2 –1 0 –1

–2 –1 0 –1

1 2 3 4 5 6 x

–2 3

a

y 3

1

1

–3 –2 –1 0 –1

1 2 3 4 x

–3

–3

b

–4

y 3

2

y 3

1

2

–3 –2 –1 0 –1

1 –3 –2 –1 0 –1

1 2 3 x

–2

–2

f

y 3

2

2

–3 –2 –1 0 –1

1 2 3 4 5 6 x

–2

–3 e

y 4

1 2 3 x

–2

1 2 3 x

–3

–2

c

–3 –4

y 3

2 g

1

y 3

2

–3 –2 –1 0 –1

1

–2

–3 –2 –1 0 –1 –2 –3 –4

1 2 3 x

–3

1 2 3 4 x d

y 3

2 1 –3 –2 –1 0 –1 –2 –3

1 2 3 x

561

LEY_bk953_answers_finalpp Page 562 Thursday, January 13, 2005 3:23 PM

562

Answers

Non-calculator Activities

Exercise 14G continued e

1 4 7

y 5

4

$35.75 2 5.37 3 52.3 mm $67.50 5 40 6 x = –2 Gross income is total income. Taxable income is gross income minus tax deductions. (x – 8)(x – 3) 9 d = 12

3

8

2

10 Most common shoe size is 7 1--2- .

1

1 --6

11

–3 –2 –1 0 –1

1 2 3 x

12

2 --3

14 8 and 9

13 52.5 L 15 1.53

–2

Language in Mathematics

–3

3

Two lines are perpendicular if the product of their gradients is negative one and they are parallel if their gradients are equal.

–4 f

y 3

Check Your Skills 1 5 9 13

2 1 –3 –2 –1 0 –1

1 2 3 x

C B C A

1 a

–3 y 3

1

– 1--8-

b

4 A 8 B 12 A

y = – 1--8- x –

5 --8

y-intercept = – 5--8- , x-intercept = –5

d

y = 8x – 25

e

y = – 1--8- x +

4x – 3y – 7 = 0 3x – 4y + 12 = 0

b

x + 6y – 38 = 0

3 --8

y 3

1 2 3 x

2

–2

1

–3 h

3 B 7 D 11 A

c

2 a c

2

–3 –2 –1 0 –1

C B B A

Review Set 14A

–2

g

2 6 10 14

–4 –3 –2 –1 0 –1

y 3

1 2 x

2

3 a

(–10, 14)

b

x – 17y + 27 = 0

1

4 a

2 --3

b

t = –13

5 a

y <4

b

x – 3y ≥ – 6

–3 –2 –1 0 –1

1 2 3 x

Review Set 14B

–2

1 5x – 7y + 34 = 0 3 2x + 3y + 12 = 0

–3 –4 4 a

3x + y ≤ 3 b

y–x ≥2

c

x–y ≤3

d

2y – x ≤ 4 e

y ≤ – 4x

f

y ≥

1 --2

x

2 3x + 2y + 2 = 0 4 y = –3x + 7

LEY_bk953_answers_finalpp Page 563 Thursday, January 13, 2005 3:23 PM

Answers

5 a

c

y 3

2

2

1

1

–3 –2 –1 0 –1

b

–3 –2 –1 0 –1

1 2 3 x

–2

–2

–3

–3 3 a

y 3

4

-----– 15 2

1

1 2 3 4 5 6 x

–2 –1 0 –1

–3

–2

–4

–3 5 a

–6

2 1

2

–3 –2 –1 0 –1

1

–3 b

y 3

–3 –4

2

–5

1 –3 –2 –1 0 –1

Review Set 14C 8x b y = – ------ – 6 3--5- ; 8x + 5y + 33 = 0 5

–3

3 ------ , y-intercept –6 --x-intercept – 33 8 5

d

e

8x y = – -----5

Review Set 14D

b

x + 2y = 0

1 a

y=

4 --7

x+

8 --7

1 2 3 x

–2

c

5x – 8y – 35 = 0

1 2 3 x

–2

1 2 3 4 5 6 7 y

–2

2 a

1 2 3 4 5 6 7 x

y 3

x 3

– 8--5-

(2, –3)

y

–5

1 a

c

2

–2

–3 –2 –1 0 –1

1 2 x

x – 3y = 7

1

c

9 7

4 3--5-

b

2

–3 –2 –1 0 –1

6x – 7y + 9 = 0

y 3

–4

y = 3x + 10

b

3x – 5y = 0

2 3x + 4y – 8 = 0 3 3 units2 4 No, as 3(1) – 5(4) + 2 ≠ 0.

563

LEY_bk953_answers_finalpp Page 564 Thursday, January 13, 2005 3:23 PM

564

Answers

2 a d g j m

b e h k n

–2 4 6 –5 –10

c f i l o

Undefined –4 –6 Undefined Undefined

14 -----3

b

91 ---------100

c

5 --1

d

7 --9

e

– 11 --------2

f

-----2 21 25

g

53 -----99

h

–2 3--5-

I

4

y 3

j

2 --3

k

3 -----10

l

3 7 ----10

2

2 a

0.6

b

1.625

c

4. 6˙

1

d

0.41 6˙

e

0.142 85 7˙ f

0.06 5˙ irrational rational False 3.317 9.950

h b e b b e

0.17 6˙ rational irrational False 3.873 12.288

Review Set 14D continued 5 a

y 3

2 1 –3 –2 –1 0 –1

Exercise 15B

1 2 3 x

1 a

–2 –3 b

–3 –2 –1 0 –1

g 3 a d 4 a 5 a d

1 2 3 x

–2 –3 –4 –5

c

(–1, 4)

–6

6 b

–7

Exercise 15C

A C B C

2 or –2 b 5 f –7 j

False 6.083

c f i

8 18 700

3

2 a

14

b

30

c

10

d

77

e

221

f

30

g

350

11 5 50

–2

3 a

2 3

b

2 5

c

3 2

–3

d

3 3

e

2 2

f

3 10

g

5 2

h

5 3

i

10 2

j

78

k

2 6

l

4 2

m 4 3 4 a 4

n b

6 2 8

o c

8 2 10

3 7 11 15

A A B C

4 8 12 16

D B B B

Exercise 15A 1 a e i

c c

1

Diagnostic Test 2 6 10 14

b e h

irrational

2

1 2 3 x

11 8 12 80

7 b

c

3

CHAPTER 15

D C B B A

2

0.69

1 a d g j

y 4

–3 –2 –1 0 –1

1 5 9 13 17

2 0 Undefined Undefined Undefined

2 c –5 g 8 or –8 k

–2 d 7 or –7 h 8 l

5 or –5 7 –8

d

2 5

e

5 2

f

2 7

g

2 10

h

3 5

i

5 3

j

2 6

5 a

18

b

12

c

80

d

50

e

700

6 a

3 --4

b

3 --5

c

17 ---------5

LEY_bk953_answers_finalpp Page 565 Thursday, January 13, 2005 3:23 PM

Answers

d

5 ------4

e

11 ---------4

f

21 ---------3

e

30 – 4 15

f

3 10 + 2 14

g

5 --2

h

4 --3

i

7 ------2

g

3 14 – 12 6

h

8 15 + 20 30

j

23 ---------3

i

6 15 + 60

j

24 5 – 60

k

15 10 – 20 3

l

36 2 + 24 3

3 a

3 + 7 3 + 10

b

–5 – 7

c

16 – 7 10

d

–4

e

11 + 7 3

f

–16 – 13 2

g

31 – 6 5

h

6 – 14 6

i

– 19 – 7 14

j

48 – 42 2

k

21 + 3 10

l

23

m 62

n

39 + 12 3

o

p

15 – 2 50

r

83 – 12 35 d F e

7 a

2

b

3

c

d

2

e

3

f

g

2 2

h

2 3

i

j 8 a d g

3 2 T F F

b e

F T

c f

6 5 4

F F

Exercise 15D 1 a

9 3

b

13 11

c

4 5

d

5 7

e

15 3

f

6 3

g

8 6

h

7 5

i

–2 10

j

–3 5

2 a

11 2 + 4 3

b

7 3 +7 5

c

11 2 + 2 3

d

8 7 – 2 10

e

3 5 + 2 11

f

10 – 4 6

g

2 2 +7 3

h

3 –2 7

i

12 5 – 9 3

j

7 2

b

7 3

c

d

4 2

e

5 5

f

6

g

2 3

h

6

i

2

j

–2 3

4 a c e 5 a

0

7 6 –3 5

b

3 2 +6 3

5 10 + 4 5

d

5 7 +5 3

8 5 –5 6 F b F

c

T

d

F

e

T

1 a

10 3

b

12 2

c

40 7

d

10 6

e

18 6

f

30 35

g

42 42

h

60

i

108

j

80

k

30

l

36 2

m 20 6

n

24 3

o

6 5

c

5 6 +5 3

b

8 10 – 12 2

d

6

T

F

5 ------5

b

3 ------3

c

10 ---------10

d

5 2 ----------2

e

6 ------2

f

8 7 ----------7

g

15 ---------5

h

21 ---------3

i

66 ---------6

j

14 ---------7

k

3 10 -------------5

l

2 30 -------------5

m

42 ---------2

n

5 10 -------------2

o

4 30 -------------3

3 ------6

b

5 ------15

c

2 2 ----------5

d

8 7 ----------21

e

10 ---------6

f

30 ---------12

g

70 ---------6

h

6 6 ----------15

i

3 55 -------------22

j

42 ---------15

3 a

5

b

7– 2 i --------------------5

ii

7+ 2 --------------------5

4 a

7

b

10 – 3 i -----------------------7

ii

10 + 3 -----------------------7

5 a

1

b

i

6 a

33

b

30 – 5 3 i ----------------------33

7 a

10

b

i

2 15 – 10 ------------------------------10

5 –2

ii

ii

5 +2

60 + 10 3 ii --------------------------33

2 21 – 14 ------------------------------10

8 a

–7

b

–9+4 2 i ------------------------–7

9 a

6+2 ----------------- b 2

3 10 – 3 2 ------------------------------8

2 5 +8 3 15 +

c

1 a

2 a

Exercise 15E

2 a

29 + 12 5 T b F

Exercise 15F

–5 10 – 2 11

3 a

q 4 a

7 + 2 10

–2+3 2 ii ------------------------–7

565

LEY_bk953_answers_finalpp Page 566 Thursday, January 13, 2005 3:23 PM

566

Answers

4 a

Exercise 15F continued c

15 + 10 + 3 + 2 -------------------------------------------------------1

d

5 – 10 -------------------2

c 2 a c

a

6 a b 7 a

Since a

1 --(a 2 )2

b

a = a, then a

1 ---

m

b

e 4 a e j 5 a e

6

17

7 10 4 1.71 3.07

f b f k b

3 2 1 1.19

6 a

10 2

b

73

1 ---

56 m

e

a

i

y

8 a

3

1 --6

5 --2

b

5 --2

3 --5

k

f

2

3

p

d

8

t

25

g c h

3

62

h d i

n

a

5 --2

m

f

a

j

t

b

3

c

3.12

c

54

c

5 --2

g

k

1 --5

m k

d

6.40

d

38 5

h

2 --3

3 --4

m

d

m

h

w

t

3

d

5

8 343

Language in Mathematics 1 a

i i

Multiplying the number by itself Thinking of a number which when multiplied by itself gives a particular number, e.g. 25 = 5 × 5. i 81 ii 3 or –3 1, 4, 25 i m squared ii square root of m iii the negative square root of m iv the square root of m Real numbers are those that can be represented by points on the number line. They can be rational or irrational. Rational numbers can be expressed as a a fraction --- , b
0. b Irrational is a real number that is not rational. terminating, recurring rationalising radical d false y, y + 1, y + 2 i 14 ii 45

1 --2

b c d

2 --3

4 --3

a

c

d 81 h 8 8192

Exercise 15H 2 --5

b

5 --6

d

3 --2

e

4 --3

2 no answer 3 a i T

c

7 --9

3

b

i F i F i T

ii T

ii F ii F ii T

c 3 a b c 4 a 5 a

2 3--5-

b iii T

iii T iii T iii T

6 Student to answer.

7 convert, indices, integer Check Your Skills

a iv F  --- is undefined  0 b c d

4k 2k + 2k = ------ ∴ even 2

2

1 ---

1 ---

c

4 c 16 g b 343

1 a

25 × 4 = 10

4k + 2k v (2k + 1) × 2k = ---------------------- = even 2

2 3

5

e w 9 a 27 b e 32 f 10 a 6561

T

2

2 a

4

b

6

d

4k iv 2k × 2k = --------- ∴ even 2

7 --4

m

T

vi (2k + 1)(2k + 1) = 4k 2 + 4k + 1 ∴ odd

g

5 --2

5 11

c

4k + 2 ii 2k + 1 + 2k + 1 = ---------------- ∴ even 2 iii 2k + 2k + 1 = 4k + 1∴ odd

a.

c

1 ---

y

3

y 4

i

a.

1 ---

4

7 a

=

Since (a 3 )3 = ( 3 a )3 = a, then a 3 =

3 a

e

c

1 --a2

T

7 16 d ---------- = 2 ( 7 ) ÷ 3 = ------3 4 y, y + 1, y + 2 y + y + 1 + y + 2 = 3y + 3 = (y + 1) no answer b no answer

c

e 16 + 6 10

b

b 2× 3 =

5 a

Exercise 15G 1 a

T

iv F iv F iv T

1 5 9 12

D B B A

2 6 10 13

A B A C

3 7 11 14

A 4 D A 8 B A or D C 15 B

LEY_bk953_answers_finalpp Page 567 Thursday, January 13, 2005 3:23 PM

Answers

16 C

17 C

Review Set 15A 4 or –4 undefined

b e

2 a

12 -----4

5 --1

3 a 4 a d

0.375 b rational irrational

0.23 c 0.3 b irrational c e rational

d 3.5 rational

5 a

2

b

50

c

d

6 35

6 a

3 2

b

2 5

c

d

2 6

7 a

20 4 --5

8 a e 9 a e

5 T T

10 a

4 0

c

e

1 a d

b

c

83 ---------100

6

9 5 --9

17 ---------5

c

3 --2

f b f

2 F F

g c

2 2 F

13 ---------2

d

d

b

F

c

T

12 a

40

b

24 6

c

20 10

f

5

d

F

b

F

5 6 c

12 a

60

b

24 2

e

4 15 – 30 f

d

30 – 3

g

49 – 20 6 10 ---------15

14 a

1

b

3+ 2 --------------------1

15 a

n

16 a 17 a

m2 7

2 a

15 -----8

14 a

1

b

2–1 ---------------1

15 a

3

3 a 4 a c e

16 a 17 a

z2 5

18

1 ---

1 ---

b b

y3 16

c c

k

2

5 ---

m2 49

4 --5

b e

2 a

10 -----3

4 --1

c

27 ---------100

3 a 4 a c e

0.625 b rational rational rational

0.49

c b d

0. 6˙ d irrational irrational

5 a

3

b

75

c

6 a

2 2

b

3 5

c

28

b

6 0

c

b

4 48

5

d

n

3

3 ---

w3 8

c c

t2 5832

3 or –3 b undefined e

3 0

c

–3

157 ---------500

0.125 b rational rational rational

1.37

c b d

0. 5˙ d irrational irrational

5 a

7

b

28

c

6 a

2 15

b

3 6

c

b

17 ---------4

c

f

3

g 2 2 c F

1 --3

4.5

9

b

3 a e

T F

b f

10 a 11 a

9 5 F

b

12 a

80

d

12 14

d

3 2 +

d

4 3

f

–20 –

13 a

7 ------7

F T b b 15

10

d

55 10

b

200 3 --4

e

30

1 ---

b b

4

c

–6 d

n

2 --1

8 a

6 or –6 undefined

7 a

a d

7 a

Review Set 15B 1 a d

3

Review Set 15C

30 ---------20

3

c

n

21

5 --3

18

d

4

b

1 ---

b

3

1–

b

10 ---------10

k

21 2

5 ------5

13 a

4

c

T

13 a

1

c

T

c 11 a

79 + 24 7

k

d

5 3 – 10 5

g

5

13 ---------2

b

6

b

g 2 3 c F

F F

d

– 2

6 10 – 9 6

k

2 b f

4 --3

10 a

e

10

f

c

T F

8 2 –2 6 c 2 2

4 5 F

35 +

11 ---------2

T

d 11 a

d

3

b

a e

50

b

b

–4 d

42

b

10 3

3 --2

8 a

2 --3

5.6

d

6 30

d

4 3

d

21 ---------4

d

T

63 5 --4

3 +2 7 c 2 d F c F 32 3

c

4 3

6 7

e

15 10 – 20

g

17 + 4 15

b

14 ---------21

567

LEY_bk953_answers_finalpp Page 568 Thursday, January 13, 2005 3:23 PM

568

Answers

Cumulative Review: chapters 12–15 1 a AY, AR, RY; a, r, y b RT or TR; o

Review Set 15C continued 14 a

1

15 a

3

w w

c

3

1 --n2

16 a 17 a

5–2 ---------------1

b

6

w

d

3

w

1 --z4

b b

4

b

c 2

c c

125

194 481

1 a d

–9 or 9 undefined

b e

2 a

39 -----8

– 6--1-

c

9 ---------100

3 a 4 a c e

0.13 b rational rational rational

1.74

c b d

0.8 3˙ d irrational irrational

3.75

5 a

11

b

99

c

2 14

d

36 10

6 a

2 3

b

4 3

c

6

d

3 7

7 a

72

b

5 --6

e 9

d

7 a e

T F

9 0

c

b c

f

3

g 2 3 c F

F T

2 --9

a iii cos θ = DR -------- = --AR d e i 0.6018 ii 0.9455 f i 74° ii 33° iv 63° g i x = 13.3 ii iii x = 80.8 iv h i 69° ii iii 35° iv i 281 m j 2 a

7 --6

b

c 11 a

3 2 F

d b

F

12 a

70

b

18 5

d

53 ---------6

d

8 3 c c

c

i x = – 3--2-

2 5 +2 3

e

6 15 – 15

f

2+3 6

g

68 + 6 35

3 ------3

b

2 15 -------------15

b

5+2 3 -------------------–7

z

16 a 17 a 18 a

1 --y2

2

b

3

b b

c

z 2 --y3

32

z

5

c c

x=1

vi x = – 1--3-

ii t =

8 --7

iii r =

e

f

i t>9

ii x
– 8--5-

iv x 

b

40 -----3

v

4075

x

3 --y4

216

iii x < 10

19 -----3

y = – --35- x + 3 --15i ii

y = 3x – 1 2x + 5y – 29 = 0

c 8x + 6y – 3 = 0 d y =

d

34 -----11

3 --5

–3

f

15 a

iii x = –6

d

3 a

x = 16.1 x = 50.1 52° 69° 169 m

23 ii x = ------ = 11 --122

1 --4

i y=

F 14 3

v

iii 0.5095 iii 70°

7 --5

ii x =

b

T

6 –3 3

d

–7

5 --4

iv x =

0

14 a

i x=

iv m = 35

27

29 ---------6

b f

–9 d

b

10 a

13 a

r i tan θ = AD -------- = --DR a r ii sin θ = AD -------- = --AR d

Review Set 15D

8 a

a ii cos θ = ---m

t iii tan θ = --a

2 --n3

8 --3

18

t i sin θ = ---m

y

i

3 --7

ii

x+

8 --7

e y = 1--2- x –

y 4

3

z

5

x

–3 iii

y

2 --3

4 x –4

x

7 --2

LEY_bk953_answers_finalpp Page 569 Thursday, January 13, 2005 3:23 PM

Answers

y

iv

4x –3

7 b i --4

ii

4 --1

2 iv --3

c

d

irrational

e

i 5

ii 63

f

3 6

g

i

h

10 5

4 a 3

300 ii

iii 15 7 7 ------7

j

i

k

i 4

l

i

3

iv

3

w2

m i

1.47

n

w

1 --2

31 iii ---------100

45

i

i

80 ii 48 2 42 + 35

iv ii

14 ---------21

ii

5–1 ---------------4

ii

6

ii

z

w

1 --4

v

30 3 – 20

iii

2

iii n

w3

2 --3

569