Revision Guide.pdf

Revision Guide For Maths 9-1

IGCSE NUMBER First, some important words; know what they mean (get someone to test you): Integer – a whole number. 2, 0 and -17 are integers, but ½, 0.247, √3 and π are not Factor – a whole number that 'goes into' another number exactly. the factors of 6 are 1, 2, 3 and 6 Prime Number – a whole number with exactly two factors, namely 1 and itself. 2, 3, 5, 7 and 11 are prime numbers, but 1, 4, 6, 8 and 9 are not Highest Common Factor (HCF) – the biggest number that goes into two (or more) others. the HCF of 20 and 28 is 4 Lowest Common Multiple (LCM) – the smallest number that two (or more) others go into. the LCM of 20 and 28 is 140 Product – the result of multiplying several numbers together. the product of 2 and 3 is 6 Square Number – a whole number squared (or multiplied by itself). 1, 4, 9 and 16 are square numbers because they are given by 1×1, 2×2, 3×3, 4×4 Surd – a 'nasty' root, one whose decimal goes on for ever without repeating. √2 (= 1.41421356...) and √3 (= 1.7320508...) are surds, but √4 (= 2 exactly) is not Recurring Decimal – one which contains a digit (or block of digits) repeating for ever. 1 /3 = 0.33333... and 1/7 = 0.142857142857... are recurring decimals Significant Figures (sig figs or sf) – start counting from the first non-zero digit. The number 0.000016573 has 5 significant figures Decimal Places (dp) – start counting from the decimal point. The number 0.000016573 has 9 decimal places Indices (plural of index) – little raised numbers representing powers (squared, cubed, etc.). in the example 3², 3 is the base and 2 is the index Standard Form – a number written as A × 10N, where 1 ≤ A < 10 and N is a whole number. 32000 = 3.2 × 104 and 0.00198 = 1.98 × 10-3 when converted to standard form Reciprocal – 'one over' a number, 1 divided by a number; you 'flip' a fraction to do this. the reciprocal of 4 is ¼; the reciprocal of 7/11 is 11/7 Numerator – the top number in a fraction. the numerator of ¾ is 3 Denominator – the bottom number in a fraction. the denominator of ¾ is 4 Common denominator – the LCM of several denominators, used if we add/subtract fractions. to add 2/3 and 1/5, we use a common denominator of 15 Mixed Number or Mixed Fraction – a number consisting of an integer and a fraction. 1½ is a mixed number, while 3/2 is an improper (or top-heavy) fraction Evaluate – work out the value of. to evaluate 3.4 × 1.4², work it out according to BIDMAS = 3.4 × 1.96 = 6.664 Estimate – work out the rough value of (round numbers in the question to 1 sig fig, usually). to estimate 3.9 × 9.1, first round the values and then work out 4 × 9 = 36 Page 2

IGCSE NUMBER ARITHMETIC 

Evaluate means 'work out the value of'. You can usually type the numbers straight into your calculator, but beware!  Faulty Squares: if you type –2² the calculator gives –4 (it should be 4). Get round this by typing (–2)²  Big Divide: Use the

12 is not the same as 1 + 2 ÷ 4 + 5. 45

key to enter big fractions.

Otherwise, insert brackets like this: (1 + 2) ÷ (4 + 5) IGCSE INSIDER INFO: 27% of marks rely on calculator usage. That's a C grade! 

BIDMAS stands for Brackets, Indices, Divide Multiply, Add Subtract. This is the order in which you should carry out calculations. Do anything in brackets first, followed by indices (powers and roots), then ÷ and × together, then + and – together. TJP TOP TIP: Not a lot of people know this, but Divide Multiply are on the same level, and Add Subtract are on the same level. This means that if you have 1 – 2 + 3, you don't do + before – to get 1 – 5 = –4. You just work from left to right to get 1 – 2 + 3 = 2. It's the same with × and ÷; just work from left to right. Your calculator does BIDMAS automatically, but note 'Faulty Squares' and 'Big Divide' mentioned above.



Negative Numbers – remember these rules!



SKILL: Add or subtract negative numbers. Go up/down in a lift, but 'minus a minus' goes up. (Imagine making a + sign out of two adjacent – signs.) TJP TOP TIP: Sending someone naughty out of the class ('subtracting a negative') has a positive effect on the class. Q: Work out A: Total so far:



2 2

+

(-3) – (↓ 3) -1

4 (↓ 4)

– -5

(-5) (↑ 5)

0.

SKILL: Multiply or divide negative numbers. Two minuses make a plus. One minus makes a minus. Q: Work out (-3) × (-4) ÷ (-2): A: First, do the numbers: 3 × 4 ÷ 2 = 6. Then decide if it's + or –: two of the minuses cancel out, leaving –6.

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IGCSE NUMBER ROUNDING AND ESTIMATION 

Rounding numbers We are often asked to round our answers in IGCSE exams. Remember: A number rounds up if the next digit is 5 or above. A number rounds down if the next digit is 4 or below. IGCSE INSIDER INFO: Really important, this one... You can't (usually) lose marks for being too accurate! [The mark scheme will say 'award marks for 3 sig figs or greater'] So: if in doubt, just write down all the digits of your answer. WARNING: The exception to this is when you have just given an accurate answer and then you are asked to round it.



SKILL: Round a number to N decimal places. Start counting digits after the decimal point, then round up/down as appropriate. Q: Round 31.5735189 to 3 decimal places. A: Chop the number off after 3 decimal places: Now round up because a 5 comes next.

31.573 5189 31.574

 Sneaky super-rounding situation: 99.9999 rounded to 2 dps becomes 100.00. (You may have to pad out the number with zeros to give it the right number of dps.) 

SKILL: Round a number to N significant figures. Start counting from the first non-zero digit, then round up/down as appropriate. Q: Round 0.00084631 to 2 significant figures. A: Chop the number off after 2 significant figures: Now round up because a 6 comes next.

0.00084 631 0.00085

 Ever-so-slightly evil example: 137423 rounded to 3 sf becomes 137000, not 137. (You have to pad out the number with enough zeros to make it the right size.) 

SKILL: Estimate the answer to a calculation. Round all the numbers in the question to 1 significant figure, then find the answer. Q: Estimate (3.187 + 6.893) × 0.2096. A: Round to 1 sig fig first: ≈ (3 + 7) × 0.2 = 10 × 0.2 = 2.  The ≈ symbol means “approximately equal to”.  Make sure you don't round 0.2096 down to zero by mistake... It's sig figs, not dp!

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IGCSE NUMBER UPPER AND LOWER BOUNDS 

Upper and Lower Bounds This is roughly the reverse of the previous section: what is the question to my answer? Q: If a number has been rounded to 1.23 (correct to 2 dps), what could it have been originally? A: It can help to sketch out a ruler marked with 1.22, 1.23, 1.24, etc.

1.22

1.24

1.23

Any measurements in the grey region would round to 1.23 rather than to 1.22 or 1.24. This region clearly extends from 1.225 up to 1.235 (not inclusive). So the lower bound is 1.225 and the upper bound is 1.235. Although that upper bound is not inclusive, it's the answer they want you to give. Don't say 1.2349, because we could come along with 1.23499999 and beat that... 

SKILL: Find the lower and upper bounds of a rounded number Sketch the appropriate ruler, then read off 'half way down' and 'half way up'. Q: Find the lower and upper bounds of 3.009cm (correct to 4 sig figs). A: Sketch the ruler:

3.008 Lower bound = 3.0085cm 

3.010

3.009

Upper bound = 3.0095cm

Calculations with Upper and Lower Bounds Q: If a runner covers 200m in 23.3 sec, what is her greatest and least possible speed? A: Well, if the distance is accurate to the nearest metre and the time is accurate to the nearest 0.1 second, we can find the lower and upper bounds for each of these: Distance: Time:

lower bound = 199.5m, lower bound = 23.25 sec,

upper bound = 200.5m upper bound = 23.35 sec

Now using the DST triangle, Speed = Distance/Time. So her greatest possible speed = Max Distance/Min Time = 200.5/23.25 = 8.624 m/s and her least possible speed = Min Distance/Max Time = 199.5/23.35 = 8.544 m/s

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IGCSE NUMBER PRIMES AND FACTORS 

A factor is a whole number that 'goes into' another number exactly. For example, the factors of 10 are 1, 2, 5 and 10.



SKILL: List all the factors of a number. Q: List all the factors of 72. A: Do this in pairs as follows (each pair multiplies to make 72): 1, 72 2, 36 3, 24 4, 18 6, 12 8, 9



A prime number has exactly two factors, namely 1 and itself. The first prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, etc. 4 isn't prime because 2 goes into it; 6 isn't prime because 2 and 3 go into it. 1 isn't a prime number because it has only one factor, not two.



As you might guess, a prime factor is a factor which is a prime number. Every number can be written by multiplying its prime factors together. It's a bit like making compounds out of chemical elements. Instead of water = H2O, we can write 12 = 2²×3



SKILL: Write a number as a product of its prime factors. TJP TOP TIP: Set out your working like this; it's a method similar to LCM and HCF (see below): learn one, get two free! Q: Write 420 as a product of its prime factors. A:

2 2 3 5 7

420 210 105 35 7 1

(2 goes into 420, leaving 210 in the next line) (2 goes into 210, leaving 105) (3 goes into 105, etc.) (7 goes into 7, leaving 1) (No primes go into 1, so we stop)

420 = 2×2×3×5×7  Remember to divide by prime numbers only, working your way up from 2.  If we are asked to write a number using powers of prime factors, use indices: 420 = 2²×3×5×7

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IGCSE NUMBER HCF AND LCM 

HCF stands for Highest Common Factor. It is the largest number that goes into two (or more) others exactly. The HCF of 12 and 21 is 3.



LCM stands for Lowest Common Multiple. It is the smallest number that two (or more) others go into exactly. The LCM of 12 and 21 is 84.



SKILL: Find the HCF and LCM of two numbers. TJP TOP TIP: Set out your working like this; it's similar to finding prime factors (above). Q: Find the LCM and HCF of 45 and 75. A:

3

30

75

(3 goes into 45 and 75, leaving 15 and 25)

5

10

25

(5 goes into 10 and 25)

2

5

(nothing goes into 2 and 5, so we stop)

Now draw an L shape... (shaded here) Multiply all the numbers in the L to get the LCM (geddit?) Multiply the numbers in the left-hand column to get the HCF. LCM = 3×5×2×5 = 150

HCF = 3×5 = 15.

 Remember: it's L for LCM! Q: Find the LCM and HCF of 240 and 768. A:

2

240

768

2

120

384

2

60

192

2

30

96

3

15

48

5

16

(nothing goes into 5 & 16 so we stop)

Now draw an L shape... (shaded here) LCM = 2×2×2×2×3×5×16 = 3840

HCF = 2×2×2×2×3 = 48.

Note: if you have three numbers, the method will work for HCF only. Don't get carried away; you may be asked for just the LCM or just the HCF...

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IGCSE NUMBER INDICES 

The little numbers used for powers (7², 7³, etc.) are called indices (singular is index). The big number underneath (7 in this case) is called the base. There are three laws for combining indices – they only work if the bases match:



Multiply by adding the indices:

74 × 75 = 79

Divide by subtracting the indices:

712 ÷ 72 = 710

Do brackets by multiplying the indices:

(73)5 = 715

LEARN!

There are three further facts to know about indices: 

Negative indices mean '1 over':

7-2 = 1 / 72 = 1/49

Fractional indices are roots:

91/2 = √9 = 3

Zero indices always equal 1:

70 = 1

LEARN!

TJP TOP TIP: Some hints to help you remember these extra facts:  Imagine the '–' of the negative index turning into the division line of 1/...  The number underneath the fraction = roots underneath a tree.  'The Power of Love' (= 0; think tennis) got to No. 1 in the charts. [Huey Lewis and the News, Frankie Goes to Hollywood, Jennifer Rush, Céline Dion, etc.]



SKILL: Simplify expressions involving indices. Use the above laws and facts to do this. Q: Simplify

64 ×69 . 65

64 ×69 = 6 49−5 = 68 . A: 5 6 Q: Simplify 81−1/2 , giving your answer as a fraction. −1/2

A: 81

=

1 = 811/ 2

1 1 = . 9  81

Q: Simplify 163/2 . TJP TOP TIP: With fractional indices such as 3/2, remember it's like a tree: “Power from above, roots down below.” Hint: do the root first to keep the number small, then do the power. 3

A: 163/2 =  16 = 43 = 64 .

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IGCSE NUMBER STANDARD FORM 

Standard Form is a convenient way of writing really big and really small numbers. (These often occur in astronomy and atomic physics, for instance.) For example, 51000 = 5.1 × 104 and 0.00000492 = 4.92 × 10-6. A number is in standard form if it is written as A × 10N, where 1 ≤ A < 10 and N is a whole number (in other words, there is one digit before the decimal point).



SKILL: Convert a number to standard form. TJP TOP TIP:  Move the decimal point until it's after the first non-zero digit.  Count how many times you moved it – this gives the index.  Small numbers (<1) have a negative index.  Big numbers (>1) have a positive index. Q: Convert 299000000 to standard form. A: 299000000. We need to move the decimal point 8 times. The number is big, so the index is 8, not –8. So the answer is 2.99 × 108.



SKILL: Convert a number out of standard form. TJP TOP TIP:  Move the decimal point the number of times given by the index.  If the index is negative, move the decimal point to make a small number.  If the index is positive, move the decimal point to make a big number. Q: Convert 7.4 × 10-3 to a normal number. A: Move the decimal point 3 times to make a small number. 0.0074 The answer is 0.0074.



SKILL: Combine (algebraic) numbers written in standard form. Q: Work out A × 106 + B × 107 (where 1 ≤ A < 10 and 1 ≤ B < 9). A: The answer will be 'something × 107' so we convert both parts to 'something × 107'. (A÷10) × 107 + B × 107 We multiplied 106 by 10, so we have to divide A by 10 to compensate. So the answer is (B + A÷10) × 107. Q: Work out (A × 106) × (B × 107) (where 4 ≤ A < 10 and 4 ≤ B < 10). A: We begin by getting (A × B) × 1013 , but we need to adjust this since A × B > 10. So we divide A × B by 10 and multiply 1013 by 10 to compensate. The answer is (A × B ÷ 10) × 1014. IGCSE INSIDER INFO: IGCSE examiners like these algebraic questions nowadays...

Page 9

IGCSE NUMBER SURDS 

A surd is a root of a whole number which is not a whole number; a 'messy' root. So √2 (= 1.41421356...) and √3 (= 1.7320508...) are surds, but √4 (= 2 exactly) is not. (The word comes from the same root as 'absurd'; in other words, not rational.)



SKILL: Simplify a surd. TJP TOP TIP: Find the biggest square number that goes into the number. You can then square root this part exactly. Q: Simplify √32. A: √32 = √(16 × 2) = √16 × √2 = 4√2



SKILL: Rationalise the denominator of a surd. TJP TOP TIP: If there is a surd on the bottom of a fraction, multiply top & bottom of the fraction by this same surd. This gets the surd on the top instead. Why bother? So that we can add it to other surds (see next skill). 2 . 6

Q: Simplify A: 

2 6 2 6 6 × =  =  . 6 3 6 6

SKILL: Add or subtract surds. TJP TOP TIP: Simplify the surds first! Q: Simplify A:



 9×3

 27





 4×3

 12 −

6 . 3

−

6 3 6 3 × = 3 3  2  3 −  = 3  3 . 3 3  3

SKILL: Multiply or divide surds. TJP TOP TIP: Simplify the surds last! Q: Work out A:

6

×

 42

÷

7 .



7

6×42 =  36 = 6 .  6 ×  42 ÷  7 =

Q: Simplify 3 53− 5 . A: 3 53− 5 = 3×3  3×−5  3×  5 − = 9−5 = 4 .

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 5× 5

(using FOIL)

IGCSE NUMBER FRACTIONS 

The main thing about fraction questions is that you have to show all your working to get any marks at all! They know that your calculator can find the answer for you... First, a quick reminder: TJP TOP TIP: Numerator (Nearer to Neptune) – the top number in a fraction. Denominator (Downstairs) – the bottom number in a fraction.



SKILL: Cancel a fraction down to its lowest terms. Q: Cancel

16 down to a fraction in its lowest terms. 24

A: Find the biggest number that goes into 16 and 24 (the HCF). This is 8. 2 Now divide top and bottom by 8 to get . 3 

SKILL: Convert a fraction to an equivalent fraction with a different denominator. (We often need to do this when adding or subtracting fractions.) Q: Write

3 x as a fraction in the form . 5 35

A: Whatever we do to the denominator, we must also do to the numerator. Here, we need to multiply the denominator by 7, turning 5 into 35. Now do the same to the numerator: 3×7 = 21. 21 So the answer is . 35 

SKILL: Convert a top-heavy to a mixed fraction. Q: Convert

23 to a mixed fraction. 5

A: How many times does 5 go into 23? 4 times. What's the remainder when we divide 23 by 5? Remainder is 3. 3 Answer is 4 . 5 

SKILL: Convert a mixed to a top-heavy fraction. Q: Convert 5

2 to a top-heavy fraction. 7

A: Multiply the whole number by the denominator. 5×7 = 35. Add the numerator. 35 + 2 = 37. 37 Answer is . 7

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IGCSE NUMBER 

SKILL: Multiply fractions.  Convert any mixed fractions to top-heavy first.  Cancel down (zap) any numbers with common factors on the top and the bottom.  Then work out top × top and bottom × bottom.  If your answer is a top-heavy fraction, convert it to a mixed fraction if required. 2 5 Q: Evaluate 4 × , showing all your working. 3 7 2 5 14 5 2 5 × = × (cancelling down) A: 4 × = 3 7 3 7 3 1 10 1 = = 3 . 3 3



SKILL: Divide fractions.  Convert any mixed fractions to top-heavy first.  Turn the second fraction upside-down and then multiply (as above). Remember KFC : Keep the first fraction, Flip the second fraction, Change the ÷ to ×. 3 5 Q: Evaluate 2 ÷1 , showing all your working. 4 8 3 5 11 13 11 8 11 2 ÷ = × = × A: 2 ÷1 = (cancelling down) 4 8 4 8 4 13 1 13 22 9 = = 1 13 13



SKILL: Add and subtract fractions. You don’t have to make top-heavy fractions in this case…  Add/subtract any whole numbers first.  Put both fractions over a common denominator.  Add/subtract the tops of the fractions – the denominator stays the same.  Combine with the whole number part to finish. Q: Evaluate 2

3 5  1 , showing all your working. 4 8

3 5 6 5 11 3  =  = = 1 4 8 8 8 8 8 3 3 So the answer is 3  1 = 4 . 8 8

A: 2  1 = 3

Q: Evaluate 4

and

2 3 − 1 , showing all your working. 7 5

2 3 10 21 11 − = − = − 7 5 35 35 35 11 24 = 2 So the answer is 3  − . 35 35

A: 4 − 1 = 3

and

Page 12

IGCSE NUMBER RECURRING DECIMALS 

A recurring decimal is one which contains a digit (or block of digits) repeating for ever. Any recurring decimal can be written as a fraction; your calculator will do this, but once again we have to show all our working to get any marks.



SKILL: Convert a recurring decimal to a fraction. TJP TOP TIP: Set out your working like this. The idea is to cunningly cancel out all those repeating digits... Q: Convert 0.433333... to a fraction in its lowest terms. A: We start by writing

F = 0.433333...

If we multiply by 10 we get

10F = 4.333333...

Now subtract (bottom – top):

9F = 3.9

(we cancel all those '3's)

Rearrange to get

F = 3.9 / 9

But that isn't a fraction; adjust it:

F = 39 / 90

So the fraction in its lowest terms is

13 . 30

Q: Convert 0.45454545... to a fraction in its lowest terms. A: We start by writing

F = 0.45454545...

If we multiply by 100 we get

100F = 45.45454545...

Now subtract (bottom – top):

99F = 45

(zap the pesky decimals)

Rearrange to get

F = 45 / 99

Now cancel down:

F = 5 / 11

So the fraction in its lowest terms is

5 . 11

How do we know whether to multiply by 10 or by 100, etc.? If a single digit repeats, multiply by 10. If two digits repeat, multiply by 100. If three digits repeat, multiply by 1000. (Spot the pattern...) We can also deal with algebraic questions as follows: Q: Convert 0.abcabcabcabc... to a fraction in terms of a, b and c. A: We write Multiply through by 1000: Subtract (bottom – top) Therefore

F=

0.abcabcabcabc...

1000F = abc.abcabcabcabc... 999F = abc F= Page 13

abc . 999

IGCSE NUMBER PERCENTAGES 

‘Percent’ is simply Latin for ‘out of a hundred’. (It's also written as %.) Remember this and there should be no problem.



SKILL: Convert between percent, fractions and decimals. % to Fraction:

Write as a fraction out of 100, then cancel down. 35% = 35/100 = 7/20

Fraction to %:

Multiply the fraction by 100. 7/8 = 7/8 of 100% = (7÷8)×100% = 87.5%

% to Decimal:

Move the decimal point twice to make the number smaller (÷100). 27% = 27÷100 = 0.27

Decimal to %:

Move the decimal point twice to make the number bigger (×100). 0.07 = 0.07×100% = 7%

Remember: percentages are 100× bigger than decimals. 

SKILL: Find a certain percentage of a number. TJP TOP TIP: Write the % as a fraction over 100, and multiply by it. Q: Find 23% of £120. A: 120×(23/100) = £27.60.



SKILL: Increase or decrease a number by a certain percentage. Work out the percentage amount (see above) then add it to the original number to increase or subtract it to decrease. Q: Increase £32 by 40%. A: 40% of £32 = 32×(40/100) = £12.80. Now add it on! £32 + £12.80 = £44.80. Q: Decrease 70kg by 6%. A: 6% of 70kg = 70×(6/100) = 4.2kg. Now subtract it! 70 – 4.2 = 65.8kg. TJP TOP TIP: A more powerful way is to think about how many percent we'll have altogether at the end, including the original amount (100%). Q: Increase £32 by 40%. A: At the end, we'll have 100 + 40 = 140%. So we find 32×(140/100) = £44.80. Q: Decrease 70kg by 6%. A: At the end, we'll have 100 – 6 = 94%. So we find 70×(94/100) = 65.8kg. Page 14

IGCSE NUMBER 

SKILL: Find a percentage change. Percentage Change =

Change ×100 Original Amount

Learn!

(Use the original amount, not the final amount!) Q: A house falls in price from £250,000 to £210,000; find the percentage change. A: The change is £40,000, so

40000 ×100 = 16% fall. 250000

Q: The price of a litre of milk increases from 80p to 86p; find the percentage change. A: The change is 6p, so 

6 ×100 = 7.5% increase. 80

SKILL: Solve a reverse percentage problem. For instance, you are told the sale price of a coat and you need to find the original price. TJP TOP TIP: This is a classic danger area! If 20% was taken off, don't just add 20%; this is wrong! Instead, learn the method below. Q: A coat costs £64 in a sale after being reduced by 20%. Find the original price. A: If 20% was taken off, £64 must be 100 – 20 = 80% of the original price. £64 = (÷80) £0.80 = (×100) £80 =

80% (÷80) 1% (×100) 100%

(get 1%) (and now get 100%)

So the original price was £80. Q: Becki sells a camera for £170, making a 25% profit. How much did she buy the camera for originally? A: If 25% profit was added, £170 must be 100 + 25 = 125% of the original price. £170 = (÷125) £1.36 = (×100) £136 =

125% (÷125) 1% (×100) 100%

(get 1%) (and now get 100%)

So the original price was £136.

Page 15

IGCSE NUMBER COMPOUND INTEREST AND DEPRECIATION 

If you put money in a savings account, you generally get some interest each year; this means you receive extra money on top of your original amount. Yippee! And that's not all; if you leave your money there for several years, you get interest on your interest, too... Yippee squared! Anyway, this is called compound interest. IGCSE INSIDER INFO: Compound interest is a brand new topic this year (2011) so it's quite likely they'll be keen to include a question on it...



SKILL: Solve a compound interest problem. Q: Will puts £500 into a savings account paying 5% compound interest each year. How much money does he have after 10 years? A: Every year, the money is multiplied by 100 + 5 = 105% = (105/100). This consists of the original money (100%) plus the interest (5%). So after 10 years, there will be 500 × (105/100) 10 = £814.45. Q: Mel saves £200 for 5 years at a compound interest rate of 4%. How much interest does she receive in this time? A: Every year, the money is multiplied by 100 + 4 = 104% = (104/100). So after 5 years, there will be 200 × (104/100) 5 = £243.33. But the interest earned is 243.33 – 200 = £43.33.



SKILL: Solve a depreciation problem. When the value of something decreases over time, this is called depreciation. Q: Ed buys a laptop for £400 but it loses 15% of its value each year. How much is it worth after 3 years? A: Every year, the value is multiplied by 100 – 15 = 85% = (85/100). So after 3 years the laptop is worth 400 × (85/100)3 = £245.65. If you prefer, you can work with decimals instead of fractions. For example, 400 × 0.853 = £245.65.

Page 16

IGCSE NUMBER RATIOS There are two types of ratio questions: sharing out and increase/decrease. 

SKILL: Solve a 'sharing' ratio problem. TJP TOP TIP: Add up the number of 'shares', and divide the amount to be shared by this number to get the size of one share. Then answer the question... Q: £90 is to be shared between Bella, Tori and Jack in the ratio 5:3:2. How much does Tori receive? A: There are 5 + 3 + 2 = 10 shares altogether. Each share is worth 90 ÷ 10 = £9. Tori gets 3 shares, so she receives 3 × 9 = £27.



SKILL: Solve an 'increase/decrease' ratio problem. TJP TOP TIP: Scale all the quantities by the given ratio (turn it into a fraction). Get the fraction the right way up by considering whether the quantities should become larger or smaller. Q: A recipe to serve 4 people requires 500g of flour, 2 eggs and 200ml of milk. Adapt it to serve 10 people. A: We have to increase all quantities in the ratio 4:10, so multiply through by (10/4) to make the quantities bigger. 500 × (10/4) = 1250g of flour, 2 × (10/4) = 5 eggs, 200 × (10/4) = 500ml of milk.



Converting units We may be asked to convert between currencies, between units of length, area and volume, or between units of time. This is another form of increase/decrease ratio.



SKILL: Convert between units. Q: If £1.00 = $1.60 and $1.00 = 83 Japanese Yen, convert £7.99 to Japanese Yen. A: Convert £7.99 to $ by multiplying by (1.60/1.00): 7.99 × (1.60/1.00) = $12.784 Convert $12.784 to Yen by multiplying by (83/1): 12.784 × (83/1) = 1061 Yen. Q: Convert a speed of 10m/s to km/h. A: There are 1000m in 1km. To convert m to km, divide by 1000:

10m/s = 0.01km/s

There are 3600 seconds in 1 hour. You travel further in 1 hour than in 1 second. So we now multiply by 3600: 0.01km/s = 36km/h. Q: Convert an area of 3m² to cm². A: There are 100² cm² in 1m² (area is squarier) so answer is 3 × 100² = 30000cm². Page 17

IGCSE NUMBER SETS 

A set is a collection of 'objects' containing no duplicates. The objects in the collection can be listed between curly brackets, for instance: A = {1, 3, 5, 7, 9} We can also define a set using a rule, such as: B = {even numbers from 0 to 10 inclusive} C = {n² : 1 ≤ n ≤ 10} (this means the square numbers from 1 to 100) Some important words and symbols:



Element = an object belonging to a set. Set {a, b, c} contains the elements a, b and c. We usually use BIG letters for sets and little letters for elements. Ԑ = Universal Set. This is the 'world' for the current question, e.g. Ԑ = {All animals}. ∅ = Empty Set = { }. This is the set with nothing in it. A' = Not A = the complement of A. A' contains every element that isn't in A. A ∩ B = A intersection B = the 'overlap' of A and B. A ∩ B contains all elements which are in A and in B. A ∪ B = A union B = A combined with B. A ∪ B contains all elements which are in A or in B (or in both). a ∈ B = a is an element in set B = a is a member of B. 1 ∈ {odd numbers} A ⊂ B = A is a subset of B = set A is contained inside set B. If A = {2, 3, 4} and B = {1, 2, 3, 4, 5, 6} then A ⊂ B. A ⊄ B = A is not a subset of B = set A is not contained inside set B. If A = {2, 3, 7} and B = {1, 2, 3, 4, 5, 6} then A ⊄ B. n(A) = number of elements in A. If A = {a, b, c, d} then n(A) = 4. If B = ∅ then n(B) = 0. TJP TOP TIP: Here are some ways to remember these symbols: Ԑ = Ԑverything (Ԑ is BIG). ∈ = ∈lement (∈ is small). ∩ = i∩tersection = a∩d = the overlap because ∩ goes over the top. ∪ = ∪nion = a mixing bowl ∪ that you chuck everything into. A ⊂ B is like A < B so A is contained in B.

Page 18

Learn!

IGCSE NUMBER 

SKILL: List the elements of a set. Memorise all the symbols from the previous page otherwise you'll just be guessing... Q: If Ԑ = {integers between 1 and 10}, A = {even numbers} and B = {prime numbers}, work out: (i) A ∩ B

(ii) A'

A: (i) A ∩ B = {2} (ii) A' = {1, 3, 5, 7, 9}

(iii) A ∪ B'

(iv) (A ∪ B)'

(v) n(B)

2 is the only number which is even and prime. These are the non-even (odd) numbers between 1 and 10.

(iii) A ∪ B' = {1, 2, 4, 6, 8, 9, 10}

Even numbers or non-prime numbers.

(iv) (A ∪ B)' = {1, 9}

Only 1 and 9 are not (even or prime).

(v) n(B) = 4

There are 4 prime numbers (2, 3, 5, 7) between 1 and 10.

Q: If Ԑ = {animals}, A = {adult animals}, B = {black animals}, C = {cats}, D = {dogs}, and P = {my pets}, describe the following in English: (i) Max ∈ B ∩ C (iii) Ellie ∈ (C ∪ D)' (v) P ⊂ A ∩ B' ∩ C A: (i) Max ∈ B ∩ C

(ii) Sam ∈ A' ∩ D (iv) C ∩ D = ∅ (vi) n(P) = 1 Max is a black cat.

(ii) Sam ∈ A' ∩ D

Sam is a puppy [or a dog which isn't an adult].

(iii) Ellie ∈ (C ∪ D)'

Ellie isn't a cat or a dog (but is some other animal).

(iv) C ∩ D = ∅

No animal is a cat and a dog [there are no cats that are dogs].

(v) P ⊂ A ∩ B' ∩ C

My pets are all adult non-black cats.

(vi) n(P) = 1

I have one pet.

Q: If Ԑ = {quadrilaterals}, A = {shapes with one or two pairs of parallel sides}, B = {shapes with all angles equal to 90°}, C = {shapes whose diagonals are equal}, decide whether the following are true or false. (i) trapezium ∈ A A: (i) trapezium ∈ A

(ii) A ⊂ B

(iii) B ⊂ C

(iv) B' ∩ C = ∅

True; a trapezium has one pair of parallel sides.

(ii) A ⊂ B

False; for instance, a parallelogram need not have 90° angles.

(iii) B ⊂ C

True; all squares or rectangles have equal diagonals.

(iv) B' ∩ C = ∅

False; if two equal diagonals don't cross at their midpoints, the angles aren't all 90°. For instance, this (flat) shape...

Page 19

IGCSE NUMBER VENN DIAGRAMS 

We can show sets graphically using a Venn diagram. Here is an example: Ԑ B

A a, e

b, c, h

f

d, g

From this diagram we can see that: A = {a, b, c, e, h} A' = {d, f, g} A ∩ B = {b, c, h} A' ∩ B = {f} A' ∪ B' = {a, d, e, f, g} (A ∩ B)' = {a, d, e, f, g} n(A) = 5 f∈B 

(everything inside ellipse A) (everything outside ellipse A) (the overlap of A and B; anything inside A and B) (anything outside A and also inside B) (anything outside A or anything outside B) (anything not in the overlap of A and B) (there are five elements inside A) (f is an element of B; f is a member of B)

SKILL: Shade and interpret Venn diagrams. Q: Shade the following regions in the Venn diagrams: (i) A ∪ B

(ii) A' ∩ B ∩ C

A: Ԑ

Ԑ A

B

A

B

C Anything in A or in B.

Anything not in A and in B and in C.

TJP TOP TIP: If you find this difficult, turn the symbols into words, then go through each region of the Venn diagram in turn to check if it 'qualifies'; shade it if it does.

Page 20

IGCSE ALGEBRA CONTENTS Page

Topic

2

Algebra Words

3

Algebra Basics

4

Tips for Writing Algebra

4

Expressions, Formula, Equations and Inequalities

5

Wordy Questions

5

Evaluating Expressions and Formulae

6

Brackets: Basic Expanding and Factorising

7

Collecting Like Terms

7

Simplifying Indices

8

Solving Linear Equations

9

Solving Linear Inequalities

10

Brackets: Expanding Quadratics

11

Brackets: Factorising Quadratics

12

Solving Quadratic Equations

13

Solving Quadratic Inequalities

14

Algebraic Fractions

15

Solving Linear Simultaneous Equations

16

Solving Quadratic Simultaneous Equations

17

Rearranging Formulae

18-19

Functions

20

Proportion

21

Sequences

22-23

Differentiation

Page 24

IGCSE ALGEBRA First, some important words; know what they mean (get someone to test you): Expression – a fragment of algebra with no '=' sign. 3 x5 is an expression, and so is  r 2 . Formula – tells you the connection between two or more quantities. 2 A =  r is a formula giving the area of a circle in terms of its radius. Equation – a mathematical statement with an '=' sign; it is only true for certain values. 3 x5 = 14 is an equation which is only true when x = 3 . Inequality – a mathematical statement with a '<', '>', '≤' or '≥' sign. 3 x5  14 is an inequality which is only true when x  3 . Expand – get rid of brackets by multiplying out in full (opposite of Factorise). a) Expand 3 2 x−5 y Answer 6 x−15 y b) Expand  x−5 x2 Answer x 2−3 x−10 Factorise – put into brackets (opposite of Expand). a) Factorise 6 xy−4 x 2 Answer 2 x  3 y−2 x  2 b) Factorise x 5 x 4 Answer  x1 x4 Simplify – gather together any matching bits. a) Simplify 3 x4 y−5 x6 y Answer −2 x10 y 3 2 4 b) Simplify 4 x y ×3 x y Answer 12 x 4 y 6 Solve – work out what number(s) make an equation true. a) Solve 3 x−7 = 26 Answer 3 x = 33 so x = 11 b) Solve  x2 x−9 = 0 Answer x = −2 or x = 9 Term – a 'bit' of an equation or an expression separated by + or – signs. The second term of 3 x−4 y5 z is −4 y . Coefficient – the number part of a term. The coefficient of x in 2 x 2 −7 x9 is –7. Linear – something with x in it (but no higher powers or roots). 3 x−5 is a linear expression. Quadratic – something with x² in it (but no higher powers or roots). 2 3 x 7 x −8 is a quadratic expression. Function – a mathematical rule for changing an input number into an output number. f  x =2 x 1 takes an input number, doubles it and adds 1 to give an output number. Domain – the set of all numbers that can go into a function (domaIN). The domain of g  x=  x−4 is x≥4 because we can't square root a negative number. Range – the set of all number that can come out of a function. The range of g  x=  x−4 is g  x≥0 because the numbers coming out are 0 or above. Inverse function – a mathematical rule that 'undoes' a given function (reverse flowchart). f −1  x = x−1/2 is the inverse function of f  x =2 x 1 . Differentiate – find a formula for the gradient (the derivative or dy/dx) of a given curve. dy 4 2 If y = 4 x 5−2 x 33 x −5 , then dx = 20 x −6 x 3 . Turning point (maximum or minimum) – a point on a curve having zero gradient. dy The curve y = x 2 3 has a minimum at (0, 3) since dx = 2 x = 0 at x = 0 . Page 2

IGCSE ALGEBRA ALGEBRA BASICS 

Algebra is a branch of mathematics where letters are used instead of numbers. Why? (i) we want a formula that works for any value of radius (let's say), so we call it r . (ii) we don't yet know the number we want (we're solving an equation to find x ). (iii) we're plotting a graph where x & y are always changing, not fixed numbers. You may not have thought about this much, but certain letters are used for certain things. Don't bother learning this list off by heart, but it might be useful as a reference.

a ,b , c , etc. are numbers which are fixed for a particular question (constants). Write down the values of a ,b and c where a x 2b xc = 0 . a ,b , c are also the unknown sides of a triangle, whether right-angled or not. Find the hypotenuse c using Pythagoras' Theorem a 2b2 = c2 . A , B , C are the angles opposite sides a ,b , c in a non right-angled triangle. c is also the y-intercept of a straight line. Find c if y = 2 xc passes through the point (1, 8). f , g , h are functions. Find the inverse function of f  x  = 4 x−5 .

h can also be height. 1 2 Evaluate 3  r h to find the volume of the cone. l is usually length. Curved surface area =  r l . m can be the gradient of a straight line, or it can be mass. Express in the form y = m xc .

n is a variable, unknown whole number. Find the 100th term of the sequence t n = n21 . r is usually radius. Work out the value of  r 2 . s can be distance. 1 If s = 2 uv t , rearrange this formula to make u the subject. t is usually time. Sketch the graph for 0≤t≤10 . u , v can be speed or velocity. Make m the subject in I = mv−mu .

w is usually width. Express the perimeter in terms of l and w . x , y , z are used for co-ordinates as well as for unknown numbers in equations. Plot the line y = 3 x5 ; solve the equation x 24 x3 = 0 .

CAPITAL LETTERS are often used for unknown length-based quantities such as: A = area, C = circumference, L = length, P = perimeter, V = volume.  ,  ,  ,  are Greek letters (theta, phi, alpha, beta) used for unknown angles. Find the value of angle  , showing all your working.

Page 3

IGCSE ALGEBRA TIPS FOR WRITING ALGEBRA 

We write 2 x instead of 2×x to mean 'two lots of x ' or x x . We leave out the times symbol because it might get confused with x .



We write x 2 instead of x× x or x x to mean ' x times itself'. Using indices is clearer and quicker once we get to x 9 instead of x x x x x x x x x …



If you're multiplying a whole load of numbers and letters, remember: Put the number first, followed by the letters in alphabetical order. This makes it easier to see if two terms match (contain the same letters). So we'd write 3 x 2 y z rather than z y x 2 3 .



We hardly ever use the division symbol '÷'; instead we use a division bar,



An '=' sign in algebra is not an instruction to work out the answer (like on a calculator). '=' means 'is the same as' or 'balances', not 'makes' or 'write the answer here'.

x . y

Never write '=' between two things that aren't the same! 22=43=7 is wrong! 

Use brackets to 'over-ride' BIDMAS. If you want to add a and b and then double them, you can write 2 ab . (If you simply wrote 2 ab , this would not be right according to BIDMAS.)

EXPRESSIONS, FORMULAE, EQUATIONS AND INEQUALITIES These are all 'bits' of algebra, but we need to know the difference between them. IGCSE INSIDER INFO: If you are asked for an expression and you give a formula or equation instead (or the other way round) you will lose marks! 

Expression: a fragment of algebra with no '=' sign. 3 x5 is an expression, and so is  r 2 .



Formula: tells you the connection between two or more quantities; it has an '=' sign. 2 A =  r is a formula giving the area of a circle in terms of its radius.



Equation: a mathematical statement with an '=' sign; it is only true for certain values. 3 x5 = 14 is an equation which is only true when x = 3 . TJP TOP TIP: EQUAtion has EQUAls in it.



Inequality: a mathematical statement with a '<', '>', '≤' or '≥' sign. 3 x5  14 is an inequality which is only true when x  3 . TJP TOP TIP: The wider end of the inequality goes with the bigger number. (Personally, I don't do crocodiles, but if they work for you...)

Page 4

IGCSE ALGEBRA WORDY QUESTIONS 

We may have to take a situation described in English and 'translate' it into algebra. TJP TOP TIP: If in doubt, pretend that the letters are numbers and think about what you'd do with the numbers. Then swap the numbers for letters.



SKILL: convert a wordy question into algebra. Q: If apples cost 20p each and bananas cost 15p each, how much do a apples and b bananas cost? A: [Suppose we had 2 apples and 3 bananas. We'd work out 20×215×3 .] So a apples and b bananas would cost 20 a15b pence. Q: Farmer Chris has g geese and h horses. How many feet do they have? A: [If we had 5 geese (each with 2 feet) and 6 horses (each with 4 feet), they would have 2×54×6 feet.] So g geese and h horses have 2 g 4 h feet. Q: If a square of side x has an area equal to its perimeter, find the value of x . A: [If a square has side 3, its area is 3² and its perimeter is 3+3+3+3.] If a square has side x , its area = x 2 and its perimeter = x xx x=4 x , so we need to solve x 2=4 x (which we'll come to later on).

EVALUATING EXPRESSIONS AND FORMULAE 

A complicated name for a simple idea with easy marks... All you need to do is replace all the letters with numbers and work out the answer!



SKILL: Evaluate an expression or a formula. Remember to use BIDMAS and your calculator correctly (see NUMBER guide). Q: Work out the value of A: Substitute to get

x2 if x=−6, y=−3, z=12 .  yz 

−62 36 = = 4. −312 9

NB a negative number squared must be positive, even if your calculator says it's not. Also, with a division bar you need to work out the whole top line and the whole bottom line first, and then divide them. Q: Evaluate V =

4  r 3 where r =5.73 , giving your answer correct to 2 decimal places. 3

A: Substitute to get V =

4 ×5.733 = 788.05 . 3

Page 5

IGCSE ALGEBRA DEALING WITH BRACKETS (I): BASIC EXPANDING AND FACTORISING We need to be good at taking things out of brackets and putting them back into brackets. 

SKILL: Multiply out (expand) a single term by a bracket. Multiply each term inside the bracket by whatever is in front of the bracket. Q: Expand −34 x −5 . A: −34 x −5 = −3×4 x−3×−5 = −12 x15 .

(NB: last term is 15 , not −15 .)

Q: Multiply out 5 a 2 b 2 a−3b4 a b  . A: 5 a 2 b 2 a−3b4 a b  = 5 a 2 b×2 a5 a 2 b×−3 b5 a 2 b×4 a b = 10 a 3 b−15 a 2 b 2+20 a 3 b2 

SKILL: Factorise into a single bracket.  Find the HCF of all the terms [the biggest number and letter(s) that goes into them].  Write down this HCF in front of the brackets.  Divide all the original terms by this HCF and put them in the brackets. TJP TOP TIP: You can always check your answer by expanding the brackets again... If you've done it right, the terms in the bracket should have nothing in common. Q: Factorise 18 a−27 . A: The HCF is 9 so we get 9 2 a−3 [Check: 9 2 a−3 = 9×2 a9×−3 = 18 a−27 . Correct!] Q: Factorise 8 x y 2 20 x 2 . A: The HCF is 4 x so we get 4 x 2 y 25 x  . TJP TOP TIP: With more complicated questions, you can set out your working as for HCF and LCM (see NUMBER guide). You get the HCF by multiplying what's in the left-hand column, and the brackets contain the terms in the bottom row. See below: Q: Factorise 70 a 3 b5 c 2−28 a b3 c 342 a 2 b 4 c4 A:

2 7 a b3 2 c

3

5

2

3

3

2

4 4

70 a b c −28 a b c 42 a b c 35 a 3 b 5 c 2−14 a b 3 c 321 a 2 b 4 c 4 3 5 2 3 3 2 4 4 5 a b c −2 a b c 3 a b c 5 a 2 b 5 c2 −2 b3 c 3 3 a b 4 c 4 2 2 2 3 4 5 a b c −2 c 3 a b c 2

5a b

2

−2 c

3 a b c

a is the biggest factor of a 3 , a , a 2 b3 is the biggest factor of b5 , b3 , b 4 2 2 3 4 c is the biggest factor of c , c , c

2

HCF = 2×7 a b3 c 2 = 14 a b 3 c2

(from the left-hand column)

Answer: 14 a b3 c 2 5 a 2 b 2−2 c3 a b c 2 

(using the bottom row)

[Check: the terms in brackets have no common factors.]

Page 6

IGCSE ALGEBRA COLLECTING LIKE TERMS 

In the course of a question, you may need to gather matching 'bits' together. This is called collecting like terms. TJP TOP TIP: Matching terms must have exactly the same letters and powers, but the numbers in front don't matter. Very simply, 2 apples + 3 bananas + 5 apples + 8 bananas = 7 apples + 11 bananas. Add up all the apples, then add up all the bananas.



SKILL: Collect like terms. Q: Simplify 3 a4 b−5 a−8b . A: 3 a4 b−5 a−8b = −2 a−4 b . [Go through once adding the a terms, then go through again adding the b terms.] Q: Simplify x 23 x4 x12 . A: x 23 x4 x12 = x 2 7 x12 . [Only the x terms match; x is different from x 2 , so don't group them together.] Q: Expand and simplify 3 x 2 y −3 x−2 y . A: This is what I call GROTBAG: Get Rid Of The Brackets And Group. 3 x 2 y −3 x−2 y = 3 x6 y −3 x6 y (The last term is 6 y not −6 y because −3×−2 y  = 6 y .) = 12 y (The x terms cancel out.)

SIMPLIFYING INDICES 

Another way of tidying up algebra in questions is by simplifying indices. We use the three laws of indices:



4

5

x ×x = x

Divide by subtracting the indices:

y 12÷ y 2 = y 10

Do brackets by multiplying the indices:

 z 3 5 = z 15

Any numbers should be worked out alongside the indices. 

9

Multiply by adding the indices:

SKILL: Simplify algebraic indices. Q: Simplify 3

A:

2

3 x3 y 2×4 x y 4 . 6 x2 y5 4

3 x y ×4 x y 3×4 = × x 31−2 y 24−5 = 2 x 2 y 2 5 6 6x y

Page 7

LEARN!

IGCSE ALGEBRA SOLVING LINEAR EQUATIONS 

A linear equation is one with only numbers and x terms (no higher powers or roots). 'Solve' means find the value(s) of x that makes the equation true. Remember to do exactly the same thing to each side of the equation to keep them equal. IGCSE INSIDER INFO: If you show no working but get the right answer, you get no marks! So even if you can do it all in your head, write down every step. TJP TOP TIP: To solve an equation, first of all group the x terms together if required. Then 'undo' the equation, one step at a time, to get x=... . The last thing to happen (by BIDMAS) is the first thing to undo.



SKILL: Solve a linear equation. Q: Solve 3 x7 = 19 . A: 3 x7 = 19 −7 −7 3x = 12 ÷3 ÷3 x

Last thing to happen is 7 . First, undo 7 with −7 . Then undo ×3 with ÷3

= 4

Q: Solve 3 x7 = x−9 . A: 3 x7 = x−9 −x −x  2 x 7 = −9 −7 −7 2x = −16 ÷2  ÷2 x

First, group the x terms together. Subtract the smaller of 3 x and x , to keep things positive. From here on it's like the previous example.

= −8

If the equation contains fractions, get a common denominator before proceeding. Q: Solve A:

2 x1 3 x−4 = . 3 5

2 x1 3 x−4 = 3 5 10 x5 9 x−12 = 15 15 10 x5 = 9 x−12 −9 x −9 x x5 = −12 −5 −5

x

First, get a common denominator. Now the numerators must be equal. Group the x terms together like before.

= −17

Page 8

IGCSE ALGEBRA SOLVING LINEAR INEQUALITIES 

A linear inequality has a '<', '>', '≤' or '≥' sign instead of an '='. Just treat it exactly like an equation, with one important exception: TJP TOP TIP: Flip the inequality if you multiply or divide by a negative number. This is because you are flipping the number line, so 'less than' becomes 'greater than'.



SKILL: Solve a linear inequality. Q: Solve 5 x−4  11 . A: 5 x−4  11 4 4 5x  15 ÷5 ÷5 x  3

Do the same as for an equation.

Q: Solve −4 x7 ≥ 23 . A: −4 x7 ≥ 23 −7 −7 −4 x ≥ 16 ÷−4 ÷−4 x ≤ −4 

Flip the inequality because we're dividing by –4.

SKILL: Show an inequality on a number line. TJP TOP TIP: Think 'more ink/less ink' If it's '≤' or '≥', use a solid blob ● (symbols using more ink). If it's '<' or '>', use a hollow blob ○ (symbols using less ink). Also, the arrowhead points the same way as the '<' or '>'. Q: Show x  3 on a number line. A: 3 Q: Show x ≥ 1 on a number line. A: 1 Q: Show 1 ≤ x  3 on a number line. A: 1

3

Page 9

IGCSE ALGEBRA DEALING WITH BRACKETS (II): EXPANDING QUADRATICS 

To multiply a bracket by a bracket, use FOIL (or Smiley Face). FOIL stands for First, Outer, Inner, Last. This means: Multiply the First terms in each bracket,

 x2 x3 → x 2

Multiply the Outer two terms in the brackets,

 x2 x3 → 3 x

Multiply the Inner two terms in the brackets,

 x2 x3 → 2 x

Multiply the Last terms in each bracket.

 x2 x3 → 6

Then add these bits together to get the final answer: x 25 x 6 . In other words, use MAAD (Multiply Across, Add Down) on your FOIL. If you prefer, Smiley Face does the same thing like this:  x2 x3

[The curves tell you what bits to multiply together.] [The eyes are just for fun...]



SKILL: Expand brackets using FOIL. Q: Expand  x3 x−5 . A:

F: O: I: L:

x× x = x×−5 3× x = 3×−5

2

x = −5 x 3x = −15

Answer: x 23 x−5 x−15 = x 2−2 x−15

Q: Expand 4 x 53 x−2 . A:

F: O: I: L:

4 x×3 x = 12 x 2 4 x×−2 = −8 x 5×3 x = 15 x 5×−2 = −10

Answer: 12 x 2−8 x15 x−10 = 12 x 27 x −10

Q: Work out  x−42 . A: Warning! This is FOIL in disguise! First, rewrite it as  x−42 =  x−4 x−4 . F: O: I: L:

2

x× x = x x×−4 = −4 x −4×x = −4 x −4×−4 = 16

Answer: x 2−4 x−4 x16 = x 2−8 x 16

Page 10

IGCSE ALGEBRA DEALING WITH BRACKETS (II): FACTORISING QUADRATICS 

To factorise a quadratic expression into brackets, use the 'Anti-FOIL' methods below:



SKILL: Factorise a simple quadratic into brackets (one lot of x 2 ). Q: Factorise x 27 x12 . A: Find two numbers that add to make 7 and multiply to make 12. [Start with the 'multiply' part to cut down your options. Could be 1 & 12, 2 & 6 or 3 & 4.] The required numbers are 3 and 4. Answer:  x3 x4 Q: Factorise x 22 x−8 . A: Find two numbers that add to make 2 and multiply to make –8. [Could be 1 & 8 or 2 & 4, then decide about minus signs.] The required numbers are –2 and 4. [We need one +ve and one –ve number to multiply to make –8. Since they add to +2, the +ve number must 'beat' the –ve one, so it's –2 & 4, not –4 & 2.] Answer:  x−2 x4 IGCSE INSIDER INFO: The next type of question often comes up – don't be caught out! Q: Factorise x 2−9 x . A: Find two numbers that add to –9 and multiply to 0 (you can write it as x 2−9 x0 ). These numbers are 0 and –9. Answer:  x−0 x−9 = x  x−9 . Or, note that the terms have an x in common, so the answer is immediately x  x−9 . Q: Factorise x 2−9 .

[Note: This is called Difference Between Two Squares]

A: Find two numbers that add to 0 (there are no lots of x ) and multiply to –9. These numbers are –3 and 3. Answer:  x−3 x3 

If the coefficient of x 2 is greater than 1, we use a slightly 'tweaked' method (Fairbrother).



SKILL: Factorise a harder quadratic into brackets (more than one lot of x 2 ). Q: Factorise 12 x 2− x−6 . A: Find two numbers that add up to –1 and multiply to make 12×(–6) = –72. [Could be 1 & 72, 2 & 36, 3 & 24, 4 & 18, 6 & 12 or 8 & 9.] The required numbers are –9 and 8. Start by writing 12 x−912 x8 and then divide by common factors 3 and 4 to give Answer: 4 x −33 x 2

Page 11

IGCSE ALGEBRA SOLVING QUADRATIC EQUATIONS 

There are two totally different methods for solving quadratic equations (which are equations with an x 2 in them, and no higher powers). Fortunately, there is a cunning way to decide which method to use... IGCSE INSIDER INFO: If you are asked to round your answer (to 3 sig figs, 2 dp), use the Quadratic Formula. If you don't have to round your answer, factorise the quadratic into two brackets. Whichever you use, you must rearrange your equation to get '= 0' on the right hand side.



SKILL: Solve a quadratic equation using the formula. The quadratic formula is printed inside the front cover of your examination paper. If a x 2b xc = 0 then x =

−b± b 2−4 ac 2a

The ± symbol is a sort of 'buy one, get one free'. You work the formula out once with a +, then you do it all again with a –. Yes, there are usually two answers to a quadratic equation... [And equations with x 3 in them can have up to three answers, but they're not in IGCSE.] TJP TOP TIP: Beware Faulty Squares: your calculator may claim that −32 = −9 , which is wrong. A negative number squared is always positive. I also recommend using the

key on the calculator for working this out (Big Divide).

Be exceptionally careful about which numbers are positive and which are negative... Q: Solve 4 x 2−3 x−5 = 0 , giving your answers to 3 significant figures. A: Here, a=4, b=−3, c=−5 so x = So x = 

−−3±−32−4×4×−5 3± 9−−80 = 2×4 8

3 89 3− 89 = 1.55 or x = = −0.804 . 8 8

SKILL: Solve a quadratic equation by factorising. Simply put the quadratic into brackets, then read off the solutions. Q: Solve x 2−3 x−28 = 0 . A: Find two numbers that add to –3 and multiply to –28 and hence rewrite this as  x−7 x4 = 0 . The solutions are found by considering what number would make each bracket zero. So x = 7 or x = −4 .

Page 12

IGCSE ALGEBRA SOLVING QUADRATIC INEQUALITIES 

If you are asked to solve a quadratic inequality, it won't usually need the quadratic formula – it'll tend to be fairly simple to work out. BUT there's a catch! Once you have your two numbers, you have to decide if it's a sandwich or an anti-sandwich. Here's an example to explain...



SKILL: Solve a quadratic inequality. Q: Solve the inequality x 2 ≤ 25 . A: It's tempting to square root both sides to get x ≤ 5 . While this is true, it's not the whole answer. For example, if x = −10 (which is certainly less than 5), then x 2 = 100 (which is not less than 25). So this value of x doesn't work. In fact the answer is −5 ≤ x ≤ 5 , which is a sandwich ( x is sandwiched between –5 and 5). On the number line, this looks like -5

5

Here's another example. Q: Solve  x−2 x−5  0 . A: Start by solving the matching equation  x−2 x−5 = 0 . The solutions are x = 2 and x = 5 . Back to the inequality; is it true for values of x between 2 and 5 or outside 2 and 5? Do an 'experiment' to find out. If we choose x = 3 (which is between 2 and 5), then 3−23−5 = −2  0 which doesn't work. So we need an anti-sandwich, x  2 or x  5 . On the number line, this looks like

2

5

TJP TOP TIP: If the inequality is ' x 2 bits < (or ≤) number', it's a sandwich. Imagine the x 2 eating the sandwich with its < (or ≤) mouth... Otherwise it's an anti-sandwich.

Page 13

IGCSE ALGEBRA ALGEBRAIC FRACTIONS 

Algebraic fractions are simply fractions which contain letters instead of/as well as numbers. We use the same standard rules as before for adding, multiplying, etc.



SKILL: Simplify algebraic fractions. TJP TOP TIP: Factorise everything in sight! Then cancel down any matching bits. Q: Simplify

3 x12 . 2 x8

A: Factorise and then cancel down:

3 x4 3 x12 3 = = . 2 x8 2 x 4 2

x 2−x −12 Q: Simplify 2 . x −2 x−8 A: Factorise and then cancel down: 

 x −4 x3 x 2−x −12 x3 = = . 2  x−4 x2 x2 x −2 x−8

SKILL: Add or subtract algebraic fractions. Just get a common denominator first and then add/subtract the numerators. Q: Work out

x−3 x1 − . 2 5

A: Common denominator is 10: Q: Work out A: 

x−3 x1 5 x−15 2 x2 3 x−17 − = − = . 2 5 10 10 10

1 1  . x x 1

1 1 x1 x 2 x1  =  = . x x 1 x  x1 x  x1 x  x1

SKILL: Solve equations involving algebraic fractions. Get a common denominator, then 'zap' it (multiply through by it). Q: Solve A:

1 1 x  = . 3 x 1 3

x  x 1 x1 3  = 3 x1 3  x1 3  x1

[now multiply through by 3 x 1 ]

 x1  3 = x  x1 = x 2 x 4 = x2 x = ±2

Page 14

IGCSE ALGEBRA SOLVING LINEAR SIMULTANEOUS EQUATIONS 

Linear simultaneous equations are two (or more) linear equations that must be true at the same time. 'Linear' means there are no pesky x 2 terms (or roots, etc.). TJP TOP TIP: To solve linear simultaneous equations:  Match up the number of x or y first (multiply through by a number if required).  SSS? This means 'Same Sign? Subtract!'. If the matching terms are both +ve (or both –ve) then subtract one equation from the other. Otherwise add the equations.  Solve this new equation, then go back and find the other letter.



SKILL: Solve linear simultaneous equations. Q: Solve

2 x3 y = 13 . 4 x11 y = 41

A: Match up the number of x by doubling the first equation. 4 x + 6 y = 26 4 x + 11 y = 41 SSS? Yes! The matching x terms are both positive. So subtract the top equation from the bottom one (to keep things positive): 5 y = 15 y = 3 Now go back and find x ; substitute for y in the easiest original equation. 2 x 3×3 = 2 x9 = 13 2x = 4 x = 2 So x = 2 and y = 3 . [You can check these values in the original equations.] Q: Solve

3 x−4 y = −12 . 5 x6 y = −1

A: Match up the number of y by tripling the first equation and doubling the second one. 9 x − 12 y = −36 10 x + 12 y = −2 SSS? No! The matching y terms have opposite signs. So add the two equations: 19 x = −38 x = −2 Now go back and find y ; substitute for x in the easiest original equation. 5×−26 y = −106 y = −1 6y = 9 y = 1.5 So x = −2 and y = 1.5 . TJP TOP TIP: If you get horrible answers (not a whole number or a simple fraction), you are probably wrong... And you can always check your answers anyway.

Page 15

IGCSE ALGEBRA SOLVING QUADRATIC SIMULTANEOUS EQUATIONS 

Quadratic simultaneous equations are two (or more) equations that must be true at the same time. 'Quadratic' means there are x 2 or y 2 terms in there. TJP TOP TIP: To solve quadratic simultaneous equations:  Match up the number of x or y first, if possible. [If not, read on...]  SSS? This means 'Same Sign? Subtract!'. If the matching terms are both +ve (or both –ve) then subtract one equation from the other. Otherwise add the equations.  Solve this new quadratic equation, then go back and find the other letter.



SKILL: Solve simple quadratic simultaneous equations. Q: Solve

y = 2 x3 . y = x2

A: The y terms already match! This often happens in these questions. SSS? Yes! The matching y terms are both positive. So subtract the top equation from the bottom one (to keep the x 2 term positive): y− y = x 2−2 x3 0 = x 2−2 x−3 . Now factorise this to solve: 0 =  x −3 x1 So x = 3 or x = −1 . Now go back and find y ; substitute for x in the easiest original equation, y = x 2 . If x = 3 , then y = 32 = 9 . If x = −1 , then y = −12 = 1 . 

SKILL: Solve harder quadratic simultaneous equations. If we can't match up x or y , we have to substitute instead... Q: Solve

2 x y = 1 . 2 2 x y = 2

A: Rearrange the first equation to get y = 1−2 x . Now substitute this into the second equation: 2 2 x 1−2 x  = 2 x 21−4 x4 x 2 = 2 2 Put the quadratic equal to zero. 5 x −4 x1 = 2 2 Now for a spot of Fairbrother... 5 x −4 x−1 = 0 Find two numbers which add to –4 and multiply to 5×(–1) = –5. These are –5 and 1, so we get: 5 x−55 x 1 = 0

or

 x−15 x1 = 0 (cancelling down)

If x = 1 , then y = 1−2×1 = −1 . 1 1 2 If x = − 5 , then y = 1−2×− 5 = 1 5 .

Page 16

IGCSE ALGEBRA REARRANGING FORMULAE 

We may be asked to rearrange a formula to get a different letter on the left-hand side.



SKILL: Rearrange a formula where the letter appears once. TJP TOP TIP: Draw up a flowchart showing what happens to this letter, then reverse it. This is a really powerful and easy method! You can use it for inverse functions, too. Q: Rearrange A =  r 2 to make r the subject. A: Draw up a flowchart: 2

r 

 ×  A

Start with r , square it, × , answer A.

Now reverse it:

  ÷  A

r 

Start with A , ÷ , square root it, answer r .

Read off the answer: r =

 A÷

Q: Rearrange t = A: Flowchart: 2

m 



b m2 a  to make m the subject. e

 a  ×b  ÷e 

  t

Reverse it:

  −a  ÷b  ×e 

m 

2

 t

Read off: m =



e t2 −a . b

Make sure that BIDMAS matches your flowchart so things get done in the right order. 

SKILL: Rearrange a formula where the letter appears twice. If the letter appears twice, first group them together in one place! Remember: Expand, Group, Factorise. Q: Rearrange

yx = C to make y the subject. y−x

A: yx = C  y−x 

Multiply both sides by the denominator

yx = Cy−Cx

Expand the brackets

xCx = Cy− y

Group the y terms on one side

x 1C  = y C −1

Factorise (put back into brackets again)

x C1 = y C −1

Divide to finish

Page 17

IGCSE ALGEBRA FUNCTIONS 

A function is a mathematical 'black box' that does something to whatever you put into it. There are two ways of writing a function, but they are both the same: f  x  = (expression) f : x  (expression) or



SKILL: Evaluate a function. Q: If f  x  = 2 x 2 3 , find f 4  , f −1 , f a in turn. A: f 4  = 2×42 3 = 2×163 = 35 f −1 = 2×−123 = 2×13 = 5 2

(Remember BIDMAS!) Replace the x with a ; nothing else to do.

f a  = 2 a 3 

Simply replace the x with 4.

SKILL: Find the domain and range of a function. TJP TOP TIP: Some terminology: Domain = the allowed numbers that go in to a function. Range = the numbers that come out of a function. Numbers not allowed in the domain almost always involve:  Square roots of negative numbers  Division by zero The range is the 'height' of the graph; the y-values it can take. Q: Find the domain of f  x  =

 x5 .

A: We can't square root a negative number (try it on your calculator!) so we need x5 ≥ 0 Answer: Domain is x ≥ −5 . Q: Which values of x must be excluded from the domain of g : x 

1 ? x −25 2

A: We can't divide by zero so we must exclude any values where 2 x −25 = 0 . Answer: Exclude x = 5, x = −5 from the domain. Q: Find the range of h  x  =  x −22−7 . A: It's tempting to multiply this out, but don't do it! Consider how this graph has been translated: right 2 and down 7. It thus has a minimum at (2, –7), so the height of the graph is always –7 or above. Answer: Range is h  x  ≥ −7 . Q: What is the range of f : x  4 x 1 when the domain is 0≤x≤10 ? A: Here, the domain is restricted to x values from 0 to 10. This is a straight-line graph, so its range (the possible y values) go from 4×01 = 1 up to 4×101 = 41 . Answer: Range is 1 ≤ f ≤ 41 . Page 18

IGCSE ALGEBRA 

The inverse of a function 'undoes' the original function; it's a flowchart put into reverse. The inverse of f  x  is written as f −1  x  .



SKILL: Find the inverse of a function. If x occurs only once, do a flow chart and reverse it. Q: If f  x  = 2  x 3−17 , work out f −1  x  . A: Flowchart:

x  −1

3

Reverse it:

f

Answer:

f −1  x  =

 −17  ×2  f  x

x 

3  17  ÷2  x

 3



x 17 . 2

If x occurs more than once, do the following steps:  Write the function as y = ...  Change all x into y and y into x  Rearrange to make y the subject; the right-hand side is the inverse function. Q: Find the inverse function of g  x =

x . x−2

x y becomes x = . x−2 y−2 x  y−2 = y xy−2 x = y xy− y = 2 x y  x−1 = 2 x 2x 2x −1 y = so g  x  = . x−1 x−1

A: y =

If f  x  = f −1  x  , then the function is self inverse; it 'undoes' itself. E.g. f  x  = 6− x . 

A composite function is what you get when you put one function 'inside' another function. WARNING: This is not the same as multiplying the functions! TJP TOP TIP: Always write in the invisible brackets: fg  x really means f  g  x . Then work from the inside outwards.



SKILL: Work out composite functions. Q: If f  x  = x 2 and g  x = x2 , find fg  x and gf  x  . A: fg  x = f  g  x = f  x2 =  x22 = x 24 x4 gf  x  = g  f  x  = g  x2  = x 22

Put in the invisible brackets! Start from the inside, replacing g  x with x2 . Then put x2 through the 'squaring' function, f  x  . Put in the invisible brackets! Start from the inside, replacing f  x  with x 2 . Then put x 2 through the '+2' machine, g  x .

Sneaky trick question: f f −1  x  = x because f −1  x  does the opposite of f  x  . Page 19

IGCSE ALGEBRA PROPORTION 

Two quantities are proportional if they change so that one of them is always a fixed multiple of the other. This means that if you double one quantity, the other one is doubled, too. There are several words and symbols meaning exactly the same thing: A varies as B;

A is (directly) proportional to B;

A ∝ B;

A = kB

There are five more possibilities listed in the syllabus, namely: A is proportional to (or varies as) the square of B A is proportional to (or varies as) the cube of B A is inversely proportional to (or varies inversely as) B A is inversely proportional to (or varies inversely as) the square of B A is proportional to (or varies as) the square root of B 

A ∝ B² A ∝ B³ A ∝ 1/B A ∝ 1/B² A ∝ √B

SKILL: Solve proportion problems. Good news: proportion questions are remarkably predictable...  Write down the proportion relation and swap the '∝' for '= k×'  Use the given data to get the value of k and write down the master formula.  Use this formula forwards.  Use this formula in reverse. Q: If p varies as the square of q , and p = 20 when q = 2 , find (a) p in terms of q

(b) p when q = 10

(c) q when p = 605

A: (a) Here, p ∝ q2 is rewritten as p = kq2 . Then substitute p = 20, q = 2 to get 20 = k ×22 , so k = 5 . Master formula is: p = 5 q 2 . (b) When q = 10, p = 5×102 = 5×100 = 500 . (c) When p = 605, 605 = 5 q 2 so 121 = q2 . Therefore q = ±11 . Q: If y is inversely proportional to x and y = 4 when x = 3 , find (a) y in terms of x A: (a) We rewrite y ∝

(b) y when x = 2

(c) x when y = 3

1 1 as y = k × . x x

Then substitute x = 3, y = 4 to get 4 = k× Master formula is y = 12×

1 , so k = 12 . 3

1 12 (or y = if you prefer). x x

(b) When x = 2, y = 12×

1 = 6. 2

(c) When y = 3, 3 = 12×

1 3 1 1 = = so . Therefore x = 4 . x 12 4 x

Page 20

IGCSE ALGEBRA SEQUENCES 

A sequence is a list of numbers, usually following some pattern. Imagine a row of boxes, numbered 1, 2, 3, etc., and each box contains a number in the sequence: 4 1

7 2

10

?

3

100

Mathematically we say that t1 = 4, t2 = 7, t3 = 10, etc. What is in box 100? We could just draw all the boxes and count all the way up to 100… Or we could find a formula for what is in box n, where n stands for n-ything. 3n+1

In this case, t n = 3 n1 n

t n means ‘what’s in the nth box’ or ‘the nth term’ (the nth number in the list). Sometimes you will see u n or another letter used, but it’s exactly the same idea.

Now to find what’s in box no. 100, we simply put n = 100 into the formula. So t 100 = 3 n1 = 3×1001 = 301 . 

SKILL: Find a formula for a linear sequence. TJP TOP TIP: Find two things from your sequence:  the gap between next-door numbers (–ve if they go down, +ve if they go up)  the 0th term; the number that would come before the first number in the list. Then tn = Gap×n + 0th term LEARN!!!

[only works if the gap is fixed]

Q: Find a formula for the sequence 7, 5, 3, 1, -1, … A: Gap = –2 [numbers go down by 2] 0th term = 9 [would come before the 7] So t n = −2 n9 . 

Some well-known sequences – learn to recognise them! 1, 4, 9, 16, 25, … 1, 3, 6, 10, 15, … 1, 8, 27, 64, 125, … 2, 3, 5, 7, 11, 13, … 1, 1, 2, 3, 5, 8, 13, …

Square numbers [ t n = n2 ] 1 Triangle numbers [ t n = 2 n n1 ] Cube numbers [ t n = n3 ] Prime numbers [no pattern to the gaps] Fibonacci numbers [add the previous two to get the next one]

Page 21

IGCSE ALGEBRA DIFFERENTIATION 

Differentiation is all about finding the gradient (slope) of a curve. The gradient will change as we move along the curve, so the gradient is a formula involving x .

TJP TOP TIP: Lots of different words and symbols, all meaning the same thing: Find the gradient;

differentiate;

find the derivative;

find dy/dx;

find f'(x)

Find the gradient formula by doing this to each term: x n ⇒ n x n−1 In English; multiply by the power, then decrease the power by 1. [Also, x terms go to numbers, numbers disappear.] 

SKILL: Differentiate an expression or a function. Q: Differentiate 4 x5 −6 x 29 x−13 . A: 5×4 x 5−1−2×6 x 2−19 = 20 x 4 −12 x9 If we have '1 over' terms, rewrite them using negative indices, then differentiate. Q: If y =

1 4 dy − , find . 3 x dx x

A: Rewrite y = x −3−4 x−1 , so dy −3 4 −3−1 −1−1 −4 −2 = −3×x −−1×4 x = −3 x 4 x = 4  2 . dx x x 

Where a graph reaches a maximum (top of a hill) or a minimum (bottom of a valley), the gradient must be zero ('cos we're not going uphill or downhill – see the graph above). Maximum and minimum points are called turning points, and we can find them by working out the derivative, setting it equal to zero and solving this equation. To decide whether we have a maximum or a minimum point: (a) use the graph's shape; x 2 graphs have a minimum, −x 2 graphs have a maximum, (b) with two turning points ( x 3 graphs), the higher one is the maximum.



SKILL: Find the turning points of a graph. Q: Find and identify any turning points of the graph y = x 3−6 x 210 . A:

dy = 3 x 2−12 x = 3 x ( x−4) = 0 at any turning points. dx At x = 0 , y = 0 3−6×0210 = 10 At x = 4 , y = 43−6×4 210 = −22 (0, 10) is a maximum and (4, –22) is a minimum (it's lower).

Page 22

IGCSE ALGEBRA 

SKILL: Find where the gradient takes a certain value. Q: Find the co-ordinates of the point where the graph y = −x 2+5 x+8 has a gradient of –3. A: Get the gradient and set it equal to –3. dy = −2 x+5 = −3 dx −2 x = −8 x = 4 When x = 4, y = −42+5×4+8 = −16+20+8 = 12 . So the point is (4, 12).



Kinematics is the study of the motion of objects – it comes from the same origin as 'cinema' (moving pictures) and 'kinetic energy' (motion energy). TJP TOP TIP: Kinematics questions need the following facts: dx v = Velocity = derivative of position. dt dv a = Acceleration = derivative of velocity. dt



SKILL: Solve a kinematics question. Q: If a ball is thrown in the air so that its height is given by h = 5 t−5t 2 metres, find (i) its velocity and (ii) its acceleration after t seconds. Hence find its maximum height. A: The position is given by h , so we differentiate this once to get v and again to get a . (i) v =

dh = 5−10 t m/s dt

(ii) a =

dv = −10 m/s². dt

To find its maximum height, note that its velocity v = 5−10 t is zero when t = 0.5 . [It reaches the top when its velocity drops to zero.] 1 1 2 At this time, the height h = 5× 2 − 5×( 2 ) = 1.25 metres.

Page 23

IGCSE STATS & PROBABILITY CONTENTS Page

Topic

2

Statistics and Probability Words

3-4

Probability

5

Simple Tree Diagrams

6

Tree Diagrams for Conditional Probability

7

Mean, Median, Quartiles, Mode and Range from a List

8

Mean, Median, Mode and Range from a Table

9

Mean, Median Class, Modal Class and Range from a Grouped Table

10

Cumulative Frequency

11

Histograms

Page 12

IGCSE STATS & PROBABILITY First, some important words; know what they mean (get someone to test you): Mean – the sum of the data values divided by the number of items. The mean of 1, 2, 2, 3, 4, 6 is (1+2+2+3+4+6) ÷ 6 = 18 ÷ 6 = 3 Median – the middle data value when all the numbers are listed in order. The median of 1, 2, 2, 3, 4, 6 is (2+3) ÷ 2 = 5 ÷ 2 = 2.5 Mode – the most common data value. The mode of 1, 2, 2, 3, 4, 6 is 2 Range – the largest data value minus the smallest data value. The range of 1, 2, 2, 3, 4, 6 is 6 – 1 = 5 Lower Quartile – the data value that is one quarter of the way up from the lowest value. The lower quartile of 1, 2, 2, 3, 4, 6 is 2 Upper Quartile – the data value that is three quarters of the way up from the lowest value. The upper quartile of 1, 2, 2, 3, 4, 6 is 4 Interquartile Range = Upper Quartile – Lower Quartile. The interquartile range of 1, 2, 2, 3, 4, 6 is 4 – 2 = 2 Frequency – the number of times that an event occurs. If a roll a die 20 times and get the number three on four occasions, the frequency is 4. Cumulative Frequency – the running total of the frequency values. These running totals are often then plotted as an S-shaped curve. We can then read off the median and the quartiles. Grouped Frequency Table – a table where data values are grouped in 'bins'. A survey might record the number of people aged 0-4, 5-9, 10-19, etc. Class Width – the width of a 'bin' in a grouped frequency table. Frequency Density – the frequency divided by the class width, used in histograms. Histogram – a chart plotting frequency density on the y axis with classes along the x axis. It is like a bar chart but corrected for the misleading effect of having differing bin widths. Probability – the chance of an event happening. Probability is always a number between 0 (impossible) and 1 (certain). Outcome – the result of an event. If a coin is tossed, the possible outcomes are Heads and Tails. Expected Number – the number of times you would expect an outcome to occur. If I roll a die 100 times and the chance of rolling a '3' is 0.2, I expect to get 20 '3's. Mutually Exclusive Events – events that cannot both/all happen at the same time. If you roll a die, getting a 1 or getting a 2 are mutually exclusive (can't happen together). But having a beard or wearing glasses are not mutually exclusive (can happen together). Independent Events – events that do not affect one another's outcomes. If a coin is tossed twice, the outcomes are independent – there is no 'memory effect'. But if sweets are picked from a bag and eaten, successive events are not independent. Tree diagram – a way of showing all the possible outcomes when two or more events occur, along with their probabilities. The diagram branches repeatedly, like a tree (on its side).

Page 2

IGCSE STATS & PROBABILITY PROBABILITY 

Probability means the chance of an event happening. It is always given as a number between 0 and 1, with 0 = impossible and 1 = certain. You can give probabilities as decimals or fractions, but never as percentages.



The outcome is the result of an event. If you consider all the possible outcomes of an event, their probabilities must add to 1. This is because it is certain that one of the outcomes will happen (we just don't know which one).



SKILL: Complete a table of probabilities listing all outcomes. Q: If a spinner is numbered 1, 2, 3, 4 and 5 and it lands on these numbers with the following probabilities, complete the table. Spinner

1

2

3

4

Probability

0.1

0.2

0.1

0.05

5

A: Since the probabilities must add up to 1, the missing number is 1 – 0.1 – 0.2 – 0.1 – 0.05 = 0.55. If the probability of an event happening is p, the probability of it not happening is 1 – p. This is because something either happens or it doesn't – there's no other option. 

Events are mutually exclusive if they can't both/all happen at once. An example is getting Heads or getting Tails when you flip a coin; you can't get both. We combine the probabilities of mutually exclusive events by adding them. TJP TOP TIP: Remember ADD-OR (as in 'we add-or statistics').



SKILL: Combine probabilities using the OR rule. Q: For the spinner mentioned in the previous question, find the probability of getting a 2 or a 3. A: Getting a 2 and getting a 3 are mutually exclusive, so we add the probabilities. Prob(getting 2 or 3) = 0.2 + 0.1 = 0.3. Q: A pupil is picked at random from a class. The probability of picking someone wearing glasses is 0.3 and the probability of picking a girl is 0.5. Explain why the probability of picking a girl or someone wearing glasses is not 0.8. A: Wearing glasses and being a girl are not mutually exclusive; there could be one or more girls who wear glasses. So we can't just add the probabilities; we'd be counting any girls with glasses twice.

Page 3

IGCSE STATS & PROBABILITY 

Two events A and B are independent if they have no effect on one another. In this case, Prob(A and B) = Prob(A) × Prob(B) This is the multiplication law for independent events.



SKILL: Combine probabilities using the AND rule. Q: The probability of spinning a 2 is 0.2 and the probability of picking a red ball out of a bag is 0.3. Find the probability of spinning a 2 and picking a red ball. A: These are independent events; they don't affect each other. So we multiply: Prob(2 on spinner and red ball) = 0.2 × 0.3 = 0.06.



SKILL: List all possible outcomes to solve probability questions. To list all the possible outcomes, you need to be systematic. Here are two common examples: Q: Three fair coins are tossed. List all the possible outcomes. Hence find the probability of getting three tails. A:

HHH HHT HTH HTT

THH THT

TTH

TTT

There are 8 possible outcomes (spot the pattern...) So the probability of getting TTT is 1/8. Q: Two spinners are marked with numbers from 1 to 4. Draw a table to show all the possible outcomes. If each number is equally likely, find the probability of getting a total of 6. A:

1

2

3

4

1

.

.

.

.

2

.

.

.

x

3

.

.

x

.

4

.

x

.

.

There are 16 possible outcomes (4 × 4), of which the 3 marked ones add up to 6. So the probability of getting a total of 6 is 3/16. 

The expected number or expected frequency is the number of times you would expect an event to happen. Simply multiply the probability by the number of trials.



SKILL: Find the expected frequency. Q: The probability of getting a '3' when rolling a die is 0.15. How many times would you expect to get a 3 is you roll it 200 times? A: 0.15 × 200 = 30 times.

Page 4

IGCSE STATS & PROBABILITY SIMPLE TREE DIAGRAMS 

Probability problems are often tackled using a tree diagram. This looks like a tree on its side, where the branches show the different outcomes along with their probabilities. The first set of branches from the left correspond to the first event to happen. The next set of branches coming off these correspond to the second event, etc. TJP TOP TIP: Remember MAAD for tree diagrams: Multiply Across, Add Down. This refers to how to combine the probabilities marked on the branches.



SKILL: Construct a tree diagram and use it to answer a probability question. Q: The probability of Esmee rolling a six on a fair die is 1/6. a) Draw a tree diagram to show the possible outcomes when she rolls this die twice. b) Use this tree diagram to find the probability of Esmee rolling: i) two sixes ii) exactly one six A: a) Draw the tree diagram. Second roll

1/6

First roll

6

1/6

5/6 1/6

5/6

Not 6 6

Not 6 5/6

b) i) P(two sixes) =

6

Not 6

1 1 1 × = 6 6 36

ii) P(exactly one six) =

1 5 5 1 5 5 10 5 × + × = + = = 6 6 6 6 36 36 36 18

[This could be '6' followed by 'not 6', or 'not 6' followed by '6'.] Note: Tree diagrams can have three or more branches at each stage, and three or more stages. But if you end up drawing hundreds of branches, there's probably an easier way...

Page 5

IGCSE STATS & PROBABILITY TREE DIAGRAMS FOR CONDITIONAL PROBABILITY 

Read the question carefully to see if it says objects are picked without replacement. If so, the probabilities will change after each selection. This is an example of conditional probability, where the probabilities depend on what has already happened.



SKILL: Use a tree diagram to answer a conditional probability question. Q: A bag contains 3 stoats and 2 weasels; two animals are then picked at random without replacement. Use a tree diagram to calculate: a) Prob(2 stoats) b) Prob(1 stoat and 1 weasel, in either order) c) Prob(at least one weasel) A: First, draw the tree diagram.

3/5

2/5

2/4

S

2/4

W

3/4

S

1/4

W

S

W

[The top right probability is 2/4 because if we remove a stoat, there are now only 2 stoats left out of 4 animals still in the bag.] Now use the tree diagram to answer the questions. a) P(2 stoats) =

3 2 6 3 × = = 5 4 20 10

b) P(1 stoat and 1 weasel) =

3 2 2 3 6 6 12 3 ×  × =  = = 5 4 5 4 20 20 20 5

[Note: this could be Stoat then Weasel, or Weasel then Stoat.] c) P(at least one weasel) = 1 – P(no weasels) = 1 −

3 2 6 14 7 × = 1 − = = 5 4 20 20 10

TJP TOP TIP: If it's a 'one of each' question, remember that there is more than one way to get this on your tree diagram. For example, 'A then B' or 'B then A'.

Page 6

IGCSE STATS & PROBABILITY MEAN, MEDIAN, QUARTILES, MODE AND RANGE FROM A LIST 

Mean = total of all data values ÷ total number of items. It's sensitive to any 'freak results' that are unusually high or low.



Median = middle data value when sorted into increasing order. If there are two middle data values, take their mean. The median is not sensitive to 'freak results'.



Lower Quartile = the median of the bottom half of the list. It's the value ¼ of the way up the list.



Upper Quartile = the median of the top half of the list. It's the value ¾ of the way up the list.



Interquartile Range = upper quartile – lower quartile. It indicates how spread out the data values are.



Mode = most common data value. If there are two most common values, the distribution is bimodal.



Range = highest value – lowest value.



SKILL: Find the mean, median, quartiles, mode and range from a list of data. Q: Find the quartiles and median of 4, 5, 6, 8, 10, 13, 15, 16, 19. A: The median is the middle number, 10. The lower quartile is the median of the bottom half 4, 5, 6, 8 which is 5.5. The upper quartile is the median of the top half 13, 15, 16, 19 which is 15.5. The interquartile range = 15.5 – 5.5 = 10. [Note: don't include the middle number 10 in the bottom or top half of the list.] Q: Find the mean, median, quartiles, mode and range of 1, 3, 3, 3, 4, 5, 6, 7, 10, 11. A: Mean = (1+3+3+3+4+5+6+7+10+11) ÷ 10 = 5.3 Median = (4+5) ÷ 2 = 4.5 Lower Quartile = 3 Upper Quartile = 7 Interquartile Range = 7 – 3 = 4 Mode = 3 Range = 11 – 1 = 10 Q: Three numbers are 6, x and 2x (with x>6). Show that the mean is 2 greater than the median. A: The median is x. The mean is (6 + x + 2x) ÷ 3 = (6 + 3x) ÷ 3 = 2 + x. So the mean (x + 2) is 2 greater than the median (x).

Page 7

IGCSE STATS & PROBABILITY MEAN, MEDIAN, MODE AND RANGE FROM A TABLE 

To work out these quantities if the data values are listed in a table, do the following.



SKILL: Find the mean, median, mode and range from a table. Q: Find the mean, median, mode and range of the following data. length

frequency

1

3

2

2

3

6

4

5

5

4

A: Add a column to the table for length × frequency. length

frequency

length × frequency

1

3

3

2

2

4

3

6

18

4

5

20

5

4

20

20

65

Mean = (1×3 + 2×2 + 3×6 + 4×5 + 5×4) ÷ (3 + 2 + 5 + 6 + 4) = 65 ÷ 20 = 3.25. [Use MAAD – Multiply Across, Add Down – on the table.] Median = the length category containing the middle (two) items when listed in order. There are 20 items altogether, so we need the 10 th and 11th items. Count down from the top: Items 1-3 have length 1; Items 4-5 have length 2; Items 6-11 have length 3. So the median is 3. Mode is the length category with the most (the biggest frequency) = 3. Range = biggest length – smallest length = 5 – 1 = 4. TJP TOP TIP: To find the position of the median, do the mean of the first and last positions. So in a list of 123 items, the median is at position (1 + 123) ÷ 2 = 62. If this position is 'X and a half', the two middle numbers are at X and X+1.

Page 8

IGCSE STATS & PROBABILITY MEAN, MEDIAN CLASS, MODAL CLASS AND RANGE FROM A GROUPED TABLE 

If you need to work out these quantities from a grouped table (where data values are grouped into 'bins' so we don't know their exact values any more): ● Find the mean of grouped data using the middle value of each class. ● Find the class containing the median (see previous page). ● The modal class is the group or class with the most (the highest frequency). ● The range is the upper limit of the highest group (class) minus the lower limit of the lowest group (class).



SKILL: Find the mean, median class, modal class and range from a grouped table. Q: Find the mean, median class, modal class and range of the following grouped data. height (cm)

frequency

101-120 121-130 131-140 141-150 151-160 161-170 171-190

1 3 5 7 4 2 1

A: Add two columns to the table, for the midpoint of the class and for freq × midpoint. height (cm)

frequency

midpoint

freq × midpoint

101-120 121-130 131-140 141-150 151-160 161-170 171-190

1 3 5 7 4 2 1

110.5 125.5 135.5 145.5 155.5 165.5 180.5

110.5 376.5 677.5 1018.5 622 331 180.5

23

3316.5

Mean = 3316.5 ÷ 23 = 144 (midpoint MAADness...) This is just an estimate because we don't know the exact data values. Median class = the class containing the middle item, no. 12 in the list. The 12th item occurs in the 141-150 class. Modal class = 141-150 because it has the highest frequency. Range = 190 – 101 = 89.

Page 9

IGCSE STATS & PROBABILITY CUMULATIVE FREQUENCY 

To find the median and quartiles accurately from grouped data, it is helpful to draw a cumulative frequency graph and read off the values from it. In a cumulative frequency graph, we find the running total of the frequencies. We then plot this against the upper end of each class interval to show how many data values there are up to a particular limit.



SKILL: Plot a cumulative frequency curve and find the median and quartiles. Q: Plot a cumulative frequency curve from this table and find the median and quartiles. Value x

Frequency f

0-20

12

20-30

20

30-60

15

60-100

25

A: First work out the cumulative frequency (running total). Value x

Frequency f

Cumulative Freq

0-20

12

12

20-30

20

32

30-60

15

47

60-100

25

72

We now plot points at (0, 0) (20, 12) (30, 32) (60, 47) (100, 72) and draw a smooth curve through these points to give that classic S-shaped curve. Cumulative 72 frequency 54 36 18 0

LQ Median UQ

x

Then we can read off the Median and the Upper and Lower Quartiles off the x-axis. (No actual numbers here this time, but there will be in the exam.) The Interquartile Range = Upper Quartile – Lower Quartile.

Page 10

IGCSE STATS & PROBABILITY HISTOGRAMS 

Histograms are a bit like bar charts except that we must plot frequency density = frequency ÷ class width, not the frequency. This means that the area of each bar is equal to the frequency. Also remember that the bars should not have gaps between them. TJP TOP TIP: Two utterly essentially vitally important facts for histograms: ● Always plot frequency density (= frequency ÷ width). Hint: think alphabetically... frequency comes before width. ● The frequency is given by the area of each bar, not the height.



SKILL: Plot a histogram from a grouped frequency table. Q: Display the following data on a histogram. Value x

Frequency f

0-20

12

20-30

20

30-60

15

60-100

25

A: We must begin by working out the frequency density = frequency ÷ width. Draw an extra column (or row) on the table if necessary. Value x

Frequency f

Freq density

0-20

12

12÷20 = 0.6

20-30

20

20÷10 = 2.0

30-60

15

15÷30 = 0.5

60-100

25

25÷40 = 0.625

Freq density 2 1 0

0

20

30

60

100

x

TJP TOP TIP: Sometimes they give us a bar that is already drawn on the histogram as well as featuring in the table. Use this known bar to label the y-axis correctly. To read values off the histogram, remember that the area of a bar gives the frequency.

Page 11

IGCSE GRAPHS CONTENTS Page

Topic

2

Graphs and Co-ordinates

3

Straight Lines

4-5

Plotting and Recognising y = mx + c

6

Other Straight Line Equations

6

Parallel Lines

7-8

Graphing Inequalities, Shading Regions

9

Conversion Graphs

9

Distance-Time Graphs

10

Plotting Curves: Table of Values

11

Gallery of Graphs

12

Intersection of Two Graphs

13

Gradients of Curves

14

Functions with Graphs

15

Transformations

16

Translations

17

Enlargements

18

Reflections

19

Rotations

Page 20

IGCSE GRAPHS GRAPHS AND CO-ORDINATES 

A graph is a set of points displayed on a grid. This grid usually has a horizontal axis and a vertical axis (plural axes), each marked with numbers to help locate points on the graph. The position of a point on a graph is given by its (x, y) co-ordinates. This position is always measured from the origin, (0, 0) where the axes cross. TJP TOP TIP: x is a cross (=across), and y to the sky. Right and Up are the positive directions. Left and Down are the negative directions. y axis y 6

A X

5 4 3 2 1

-6

-5

-4

-3

-2

-1 0 -1 -2 -3

1

2

3

4

5

6x

x axis

B X

-4 -5 -6 Point A is at (-3, 5) and point B is at (1, -3). It is best to plot points by marking an X using a pencil (just like voting, really...) An X marks the exact point more accurately than a blob (if your pencil is sharp). An X also shows up better on graph paper than a + or a dot. Once you have plotted your points, you may be asked to join them to make a polygon or a straight line, or else draw a smooth curve passing through them. Once again, do this using a sharp pencil – just in case you need to change anything. If you have a graph using different letters, for example (t, v), then the first letter goes across (like x) and the second letter goes up (like y).

Page 2

IGCSE GRAPHS STRAIGHT LINES 

A straight line can be drawn through any two points plotted on a graph. Mathematically, a line can go on for ever, but we can slice it off if necessary to make a line segment instead.



SKILL: Find the midpoint, length and gradient of a line segment. The midpoint is the average of the co-ordinates of the end points. The length is found using Pythagoras with the 'rise' and the 'run'. The gradient (or the slope) is given by 'rise over run'. [rise is the change in y along the line, run is the change in x.] Q: Find the midpoint, length and gradient of the line segment AB. y 6

A RUN = 4 X 5 4 3 2

RISE = -8

1 -6

-5

-4

-3

-2

-1 0 -1 -2 -3

1

2

3

4

5

B X

-4 -5 -6 A: Point A is (-3, 5) and point B is (1, -3) so: Midpoint = Length =



 4 2−82

Gradient =



−31 5−3 , = −1, 1 . 2 2 =

 1664 =  80 = 8.94 .

rise −8 = = −2 . run 4

Remember: ● A line sloping up to the right has a positive gradient. ● A line sloping down to the right has a negative gradient. ● Steep lines have big gradients. ● Shallow lines have small gradients.

Page 3

6x

IGCSE GRAPHS STRAIGHT LINE EQUATION: y = mx + c 

The equation of a straight line can almost always be written as y = mx + c where m is the gradient and c is the y-intercept. The gradient is 'rise over run' (how far the graph goes up for every step to the right). The y-intercept is where the graph cuts the y-axis.



SKILL: Find the equation of a straight line. Q: Find the equation of this line. y 6 5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6

1

2

3

4 5

6x

A: The line cuts the y-axis at y = –2 so c = –2. The gradient m = 3 because the 'stairs' go up in steps of 3.

So y = 3x – 2.

Q: Find the equation of this line. y 6 5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6

1

2

3

4 5

6x

A: The line cuts the y-axis at y = 3 so c = 3. The gradient m = –¼ because the 'stairs' go down in steps of ¼. How can we tell it's ¼? Well, it takes 4 steps to go down a whole square. So y = –¼x + 3. Page 4

IGCSE GRAPHS 

SKILL: Plot a straight line (using y = mx + c). TJP TOP TIP: Start at the y-intercept, then go right 1, up m (where m = gradient). Always go right 1. If m is negative, you go down instead of up. Q: Plot y = –x + 3 A: The y-intercept is 3, so start at 3 on the y-axis. Then go right 1, down 1 (because there is –1 lot of x). y 6 5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6

1

2

3

4 5

6x

y = -x + 3

Q: Plot y = 4 and y = x on the same axes. A: The line y = 4 has a y-intercept of 4 and zero gradient (no x at all). So start at 4 on the y axis and go right 1, up 0 (in other words, horizontal). The line y = x has a y-intercept of 0 and a gradient of 1 (1 lot of x). So start at 0 in the y axis and go right 1, up 1. y=x

y 6 5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6

y=4

1

2

3

4 5

Page 5

6x

IGCSE GRAPHS STRAIGHT LINES: OTHER EQUATIONS 

We may come across two other forms of straight line equation. A vertical line has no y-intercept and an infinite gradient, so it cannot be written as y = mx + c. Instead we write x = c where c is the x-intercept. So x = 7 is a vertical line cutting the x-axis at 7. Also, the form ax + by = c can be used to represent any straight line. We can either rearrange it to get y = … or else use the trick shown below:



SKILL: Plot a straight line (using other equations). Q: Plot the lines x = –4 and 2x + 3y = 6 on the same axes. A: First, x = –4 is a vertical line cutting the x axis at –4. To plot 2x + 3y = 6, first set x = 0 and find y. This gives the point (0, 2).

y = 6÷3 = 2.

Now set y = 0 and find x. This gives the point (3, 0).

x = 6÷2 = 3.

Finally draw a straight line through these two points. x = -4 y 6 5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6

1

2

3

4 5

6x 2x + 3y = 6

PARALLEL LINES 

Parallel lines have the same gradient. Q: Find the line parallel to y = 2x – 7 that passes through the point (3, 2) A: Our new line must have the same gradient, but we don't know its y-intercept. y = 2x + c Now find c by substituting the values x=3 and y=2 into the equation. 2 = 2×3 + c so c=2–6=–4 Answer: y = 2x – 4 Page 6

IGCSE GRAPHS GRAPHING INEQUALITIES AND SHADING REGIONS 

If we are given an inequality such as y > 2x + 3, this corresponds to a region on a graph, not a line. Namely, all the points above the line y = 2x + 3. First mark the boundary line by changing the inequality to an equation. Then decide whether we want the region above/below the line or left/right of the line. TJP TOP TIP: The 'more ink/less ink' rule works again: ● A < or > inequality is plotted with a dotted line. ● A ≤ or ≥ inequality is plotted with a solid line.



SKILL: Shade the region satisfying one or more inequalities. Q: Shade the region defined by y > 2x + 3, x < 1 and y > –x – 2. A: First, mark the dotted lines y = 2x + 3, x = 1 and y = –x – 2 (less ink with < and >). [See pages 5 and 6 for how to plot a straight line.] Then decide if we want the region above/below or left/right. 'y > …' is above the line, 'x < …' is left of the line. It is helpful to use little arrows to mark above/below or left/right. You then want the region that all the arrows are pointing into. Shade this region y 6 5 4 3 2 1 -6

-5

-4

-3

-2

-1 0 -1

1

2

3

4

5

6 x

-2 -3 -4 -5 -6 y = 2x + 3

y = -x - 2

x = -1

Note: sometimes you may be asked to shade the unwanted region instead.

Page 7

IGCSE GRAPHS 

SKILL: Find the inequalities which define a given region. Q: List the inequalities which define the shaded region. y 6 5 4 3 2 1 -6

-5

-4

-3

-2

-1 0 -1

1

2

3

4

5

6 x

-2 -3 -4 -5 -6

A: First, get the equations of the boundary lines: [See page 4 for how to read off the equation of a straight line.] Here, y = 4, y = x – 2 and y = –½x + 2 Now decide on the inequalities: All the lines are solid, so use ≤ or ≥. Below the y = 4 line, so y ≤ 4. Above the y = x – 2 line, so y ≥ x – 2. Above the y = –½x + 2, so y ≥ –½x + 2.

Page 8

IGCSE GRAPHS CONVERSION GRAPHS 

You may get questions where you use a graph to convert one quantity into another. Common examples are currency exchange rates or converting units. Simply read off the values from the graph, but make sure you use the right axis.



SKILL: Use a conversion graph. Q: Use this conversion graph to change 50°C to Fahrenheit. A:

°F 240 220 200 180 160 140 120 100 80 60 40 20 0

10 20 30 40 50 60 70 80 90 100110 120 °C

Answer is approximately 120°F. [You are always allowed a little leeway.]

DISTANCE-TIME GRAPHS 

Distance-time graphs always have time along the horizontal axis. The gradient then gives the velocity because rise ÷ run = distance ÷ time. Remember the



D triangle [Dauntsey's School Triangle]. S ∣ T

SKILL: Draw and interpret a distance-time graph. Q: Use this graph to find the total distance travelled and the maximum speed. Dist (m) 60 50 40 30 20 10 0

10

20

30

40

50

60 70

80

90

100 110 120 Time (sec)

A: The total distance travelled is 60 (forwards) + 30 (back again) = 90 m. The maximum speed is between 60 and 80 seconds: speed = 40÷20 = 2 m/s. Page 9

IGCSE GRAPHS PLOTTING CURVES: TABLE OF VALUES 

You may be asked to plot a curve rather than a straight line. Use a table of values; there is a clever trick to get your calculator to do this for you! It is worth knowing the sort of graphs you can be asked to plot: IGCSE INSIDER INFO: The syllabus says that all graphs to be plotted must be either y = A x 3+B x 2+C x+D where A , B , C , D are integers (and could be zero) or E F y = A x 3+B x 2+C x+D+ + 2 where a minimum of three of A , B , C , D , E , F x x are zero (so there will never be more than three terms in the function). Also, x and y may be replaced by other letters. Here are some examples listed in the syllabus itself: 3 3 2 3 y = x ; y = 3 x −2 x +5 x−4 ; y = 2 x −6 x+2 ; V = 60 w(60−w)

y = 

2 1 1 2 3 x −5 W = 5 ; y = 2 x +3 x+ ; y = ; 2 x x x d

SKILL: Fill in a table of values for plotting a curve. TJP TOP TIP: Use your calculator to do all the hard work! (This works for the Casio FX-83 and FX-85; consult your manual for other models.) ● Press MODE and then 3 . ● Enter your function using ALPHA

) to enter x.

● Start? Enter the first x value in the table printed in the question. ● End? Enter the last x value in the table. ● Step? Enter the gap between the x values in the table. ● You will now see a table of x, y values on the screen. The y values are labelled f(x). Move up and down using the cursor keys, and copy the y values into your table. [Ignore the very left-hand column – these numbers are not required.] ● Press MODE and then 1 at the end to return to the normal calculator mode. Q: Complete the following table of values when y = 3x³ + 2 – 18/x²

x

1

2

3

4

5

y

A: Enter the function, set Start = 1, End = 5 and Step = 1. Now read off the values and write them in the table: x

1

2

3

4

5

y

-13

21.5

81

192.875

376.28

TJP TOP TIP: Always place your hand inside the curve when you draw the curve.

Page 10

IGCSE GRAPHS A GALLERY OF GRAPHS 

You are expected to be able to sketch some common graphs – learn the following!

y = x³

y = x²

This is called a parabola.

This is a cubic graph; it goes flat at x=0.

y = 1/x

y = 1/x²

This is a reciprocal graph. It diverges to infinity at x=0.

This graph also diverges to infinity at x=0.

TJP TOP TIP: Note the symmetry of these graphs. Graphs with even powers of x have line symmetry in the y axis. Graphs with odd powers of x have rotational symmetry order 2 about the origin.

Page 11

IGCSE GRAPHS INTERSECTION OF TWO GRAPHS 

As well as solving two equations simultaneously, we may be asked to solve them graphically. This means plotting the two lines/curves and reading off where they cross.



SKILL: Solve two equations graphically. Q: Use the graphs of y = x 3 and y = 5 x to solve x 3 = 5 x . A:

y 30

y = x³

20

y =5x

10 -3

-2

-1

0

1

2

3x

-10 -20 -30 The graphs cross at x = –2.3, x = 0 and x = 2.3. [You are allowed a margin for error.] 

Sometimes we need to find a straight line to plot in order to solve an equation graphically. TJP TOP TIP: Subtract the equation to be solved from the curve that's plotted. This is explained in the example below.



SKILL: Choose which straight line to plot to solve an equation graphically. Q: What straight line should be plotted with y = x 2 +4 x −7 if we want to solve 2 x +2 x+3 = 0 . A: Subtract the equation to solve from the curve are given. 2

y = x +4 x −7 minus 2 0 = x +2 x+3 equals y = 2 x−10 So we should plot the straight line y = 2 x−10 .

Page 12

IGCSE GRAPHS GRADIENTS OF CURVES 

To find the gradient of a curve at a point from its graph: ● draw the tangent, ● work out rise ÷ run. Warning: first make sure that you aren't supposed to use dy/dx to find the gradient.



SKILL: Find the gradient of a curve at a given point. Q: Find the gradient of this curve at x = 3. y 6 5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6

1

2

3

4 5

6x

A: Draw the tangent touching the curve at x = 3, then work out its rise ÷ run. y 6 5 Tangent 4 3 2 Rise = 11 1 -6 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6

1

2

3

4 5

6x

Run = 8

Gradient = Rise ÷ Run = 11 ÷ 8 ≈ 1.4. [Anything close to the right answer will be fine.] TJP TOP TIP: Make the rise ÷ run triangle as big as possible for maximum accuracy. Remember: it's a negative gradient if the tangent slopes down to the right.

Page 13

IGCSE GRAPHS FUNCTIONS WITH GRAPHS 

A function question may often use a graph instead of a formula. All you need to know is that the domain (what goes in) is on the x axis and the range is on the y axis.



SKILL: Use a graph to answer questions on functions. Q: Here is a graph of f(x) and g(x). Use it to find the following: (a) f(3) (b) fg(1) (c) solve f(x) = 2 (d) solve f(x) = g(x) y 6 5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 -1 -2 -3 -4 -5 -6

y = f(x)

1

2

3

4 5

6x

y = g(x)

A: (a) f(3) = 3. (b) fg(1) = f(g(1)) = f(0) = -3. (c) f(x) = 2 at x = -3.5 and x = 2.5. (d) f(x) = g(x) where the graphs cross, at x = 1.5.

Page 14

IGCSE GRAPHS TRANSFORMATIONS 

A transformation is a way of changing the position, shape or size of an object. There are many different types of transformations, and most of them correspond to the different ways you can alter a graphics object on a computer screen. However, we only need to know four transformations for IGCSE. These are: ● Translation: a move or a shift ● Enlargement: a grow or a shrink ● Reflection: a flip ● Rotation: a spin For each of these transformations, we need to have certain pieces of information. ● Translation: the movement in the x and y directions as a column vector. Translate by

(−23 ) means 'move right 3 and down 2'. 3 -2

● Enlargement: the co-ordinates of the centre of enlargement, and the scale factor. Enlarge by factor 3, centre (2, 5) means 'make it three times bigger, keeping the point (2, 5) fixed'. x (2, 5)

● Reflection: the line of reflection (or its equation). Reflect in the line y = 0 means 'use the x-axis (y=0) as a mirror line'. y=0

● Rotation: the co-ordinates of the centre of reflection, and the angle. Rotate by +90° about (-2, -5) means 'turn the object through 90° anticlockwise about the point (-2, -5).

x (-2, -5) Note: a positive angle is taken to be anticlockwise, and a negative angle is clockwise.

Page 15

IGCSE GRAPHS TRANSLATIONS 

SKILL: Translate a shape by a given amount. Q: Translate the shape A by

(−42 )

A: Move it 2 right and 4 down.

and label the new shape B.

A

Count between corresponding corners!

y 6 5 4 3 2 1

A

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 -2 -3 -4 -5 -6 

y 6 5 4 3 2 1

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x B -1 -2 -3 -4 -5 -6

SKILL: Find a translation vector. Q: Find the translation that moves A onto B. y 6 5 4 3 2 1

A

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 -2 -3 B -4 -5 -6

A: Pick a matching corner on both shapes and count the squares. y 6 5 4 A 3 2 1 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 -2 -3 B -4 -5 -6

Translation of

Page 16

. (−6 −7)

IGCSE GRAPHS ENLARGEMENTS 

SKILL: Enlarge a shape by a given scale factor from a given centre. Q: Enlarge shape A by factor 2, centre (-6, -6) and label the new shape B.

A: Draw guidelines out from (-6, -6) through the corners of A, and double the lengths. Count squares to be more accurate: for top corner, right 2 up 6 becomes right 4 up 12.

y

y 6 5 4 3 2 1

6 5 4 3 B2 1

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 -2 A -3 -4 -5 -6 

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 -2 A -3 -4 -5 -6

SKILL: Identify an enlargement (scale factor and centre). Q: Find the enlargement that turns shape A into shape B. y 6 5 4 3 2 1

A: Draw straight lines through matching points on the two shapes. Scale factor = new height ÷ old height. y 6 5 4 3 2 1

A

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 B -2 -3 -4 -5 -6

Page 17

A

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 B -2 -3 -4 -5 -6 Centre of enlargement is (-4, -3), Scale factor is ⅓.

IGCSE GRAPHS REFLECTIONS 

SKILL: Reflect a shape in a given line. Q: Reflect shape A in the dotted line and label the new shape B. y 6 5 4 3 A 2 1

A: Count (diagonally) to the mirror line, then count the same distance beyond. y 6 5 4 3 A 2 1

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 -2 -3 -4 -5 -6 

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x B -1 -2 -3 -4 -5 -6

SKILL: Find a line of reflection. Q: Give the equation of the line of reflection.

A: Join matching points on the two shapes; the mirror line is exactly half way between the shapes. y 6 5 4 3 2 1

y 6 5 4 3 2 1 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 -2 -3 -4 -5 -6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 -2 -3 -4 -5 -6 The line of reflection is y = x + 1.

Page 18

IGCSE GRAPHS ROTATIONS 

SKILL: Rotate a shape by a given angle around a given point. Q: Rotate A by 90° about (3,1) and label the new shape B. y 6 5 4 A 3 2 1

A: Draw an 'L' from the centre of rotation to the shape, and rotate the L. y 6 5 4 3B 2 1

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 -2 -3 -4 -5 -6

A X

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 -2 -3 -4 -5 -6

IGCSE INSIDER INFO: Ask for tracing paper in the exam to help you do rotations. 

SKILL: Identify a rotation (angle and centre). Q: Find the angle and centre of rotation that moves shape A onto shape B. y 6 5 4 3 2 1

A: Join matching points with a '+' shape. These focus to the centre of rotation. y 6 5 4 3 2 X1

A

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 -2 -3 B -4 -5 -6

A

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x -1 -2 -3 B -4 -5 -6 Rotation angle is –90° (90° clockwise), Centre of rotation is (-1, 1).

Page 19

IGCSE SHAPE & SPACE CONTENTS Page

Topic

2-3

Angle and Polygon Words

4-5

Angle and Polygon Facts

6-7

Construction

8

Parts of Circles

9-10

Circle Theorems

11

Flowchart for Triangle Formulae

12

Pythagoras

13-14

Trigonometry (SOHCAHTOA)

14

3-D Problems

15

Sine Rule

16

Cosine Rule

17

Area of a Triangle

18

Perimeter and Area

19

Surface Area and Volume

20

Similarity and Enlargement

21

Metric Units

22-23

Vectors

Page 24

IGCSE SHAPE & SPACE First, some important words; know what they mean (get someone to test you): Acute angle: less than 90°.

Right angle: exactly 90°.

Obtuse angle: between 90° and 180°.

Reflex angle: greater than 180°.

Parallel: two or more lines going in the same direction (with the same gradient).

Perpendicular: two lines at 90° to each other.

Polygon: a closed 2-D shape with straight sides.

Regular Polygon: a polygon with all angles equal and all sides equal.

Equilateral triangle: three equal angles and three equal sides.

Isosceles triangle: two equal angles and two equal sides.

Right-angled triangle: one angle of 90°.

Scalene triangle: no special angles or sides.

Page 2

IGCSE SHAPE & SPACE Quadrilateral: a four-sided 2-D shape.

Cyclic quadrilateral: a four-sided shape whose corners lie on a circle.

Trapezium: a shape with one pair of parallel sides.

Parallelogram: a shape with two pairs of parallel sides.

Rhombus: a parallelogram which has all sides the same length (a diamond).

Kite: a shape with a line of symmetry passing through two opposite corners.

Interior angle: an angle inside a corner of a polygon.

Exterior angle: the angle the line turns through at each corner (the 'turtle' angle).

Angle of elevation: the angle above the horizontal.

Angle of depression: the angle below the horizontal.

Page 3

IGCSE SHAPE & SPACE ANGLES 

Angles are measured in degrees (°) so that there are 360 degrees in a full circle. Why 360? Well, it's probably to do with the number of days in a year, combined with the fact that lots of numbers go into 360 exactly (it has many factors). Here are some more important angle facts: Angles on a straight line add up to 180°.

Angles in a triangle (interior angles) add up to 180°.

b

a

c

c

a a + b + c = 180

a + b + c = 180

Exterior angle in a triangle is the sum of the other two interior angles. a

b

Angles in a quadrilateral add up to 360°.

a

b

b d

c a+b=c Exterior angles in a polygon add up to 360°, always.

c

a + b + c + d = 360 Interior angles in an n-sided polygon add up to 180(n – 2)°.

If you walk around the shape once, you turn through a total angle of 360°.

You can divide an n-sided polygon into n – 2 triangles, each 'worth' 180°.

Exterior angle of a regular polygon with n sides is 360÷n°.

Interior angle of a regular polygon with n sides is 180(n – 2)÷n°.

Page 4

IGCSE SHAPE & SPACE Alternate angles are equal.

Corresponding angles are equal. c

a a

c

HB HELPFUL HINT: This is like an A and and A joined together, spelling Alternate Angle.

TJP TOP TIP: These are 'F'-ing Corresponding Angles.

A

A

F

Vertically opposite angles are equal.

v

v

TJP TOP TIP: This is a V and a V (for Vertical), Opposite one another.

IGCSE INSIDER INFO: You must use the proper names: 'Z angle', 'F angle' and 'X angle' will score zero marks.

V

V



A Bearing is simply an angle measured clockwise from North. Bearings are often used in navigation where a compass is used to find the direction of North. The bearing of A from B means we are at B, going towards A. The bearing of B from A is 180° different from the bearing of A from B. N N 072°

A

B

Page 5

252° = 72 + 180

IGCSE SHAPE & SPACE CONSTRUCTION 

You are expected to be able to draw and measure:  a straight line to the nearest mm  an angle to the nearest degree so bring a sharp pencil, a sharpener, an eraser, a ruler and a protractor. You will also need a pair of compasses for the following methods:



SKILL: Find the perpendicular bisector of a line segment AB (without measuring).  Set your compasses to a little over half way between the line ends A and B.  Draw arcs with the compass point on A and then on B.  Draw a straight line through the points where these arcs cross one another.

B

A



SKILL: Bisect an angle (without a protractor).  Set your compasses to a fixed radius, just shorter than the lines making the angle.  Put the compass point on the angle corner O and draw an arc cutting each line at A, B.  Then draw arcs with the compass point on A and B in turn to make a rhombus OACB.  Draw a line through OC; this bisects the angle. A C

O B

Page 6

IGCSE SHAPE & SPACE 

SKILL: Construct triangles and other 2-D shapes (compasses and straightedge). Equilateral Triangle:  Given a baseline AB, set your compasses to this distance.  Draw two arcs centred on points A and B so that the arcs cross at point C.  ABC then forms an equilateral triangle. C

B

A

Regular Hexagon:  Set your compasses to the required side length.  Draw a circle centre O.  Use the compasses to mark off the side length all around the circumference.  Join the marks to construct the hexagon ABCDEF. C

D

B

A

O

E

F

Square:  Given a baseline AB, construct a perpendicular line at A (see previous page).  Use your compasses to mark off the distance AB up this line to give C.  Using this same radius, now draw arcs centred on C and B, crossing at D. C

D

A

B

Page 7

IGCSE SHAPE & SPACE PARTS OF CIRCLES 

Learn all the following words for parts of circles...

DIAM E

RA DIU S

TANG EN T Just t ouche s the

TER

CIRCUMFERENCE

circle

Radius < Diameter < Circumference, just like the length of the words...

Think gu

te la OR T IN N co M ME cho G . SE rry's e.. g Te ran ink o Th

CHOR

D

itar st rings.. .

MAJOR SEGMENT

ARC MINOR SECTOR

Like an arch...

MAJOR SECTOR

Page 8

IGCSE SHAPE & SPACE CIRCLE THEOREMS

O

O θ

θ

Radius meets tangent at 90°.

Isosceles triangle (with two corners on the circumference and one at the centre).

θ

θ O

θ

2θ

Angle at the centre is twice the angle at the circumference.

Angles in the same segment are equal.

θ O Φ

Angle in a semicircle = 90°.

Opposite angles in a cyclic quadrilateral add up to 180°. θ+φ = 180°. Page 9

IGCSE SHAPE & SPACE

C

θ A

B D θ

Alternate segment theorem. Marked angles are equal.

Intersecting Chords (I). A×B = C×D HB HELPFUL HINT: The lines make a × shape, so we × the lengths together.

B

B

D A

A C

C

Intersecting Chords (II). A×B = C×D

Intersecting Chords (III). A×B = C²

IGCSE INSIDER INFO: You have to give a reason for each step when finding the unknown angles in a question. Use the official names above, not 'Star Trek', etc.

Page 10

IGCSE SHAPE & SPACE FORMULAE FOR TRIANGLES Which formula do I use to find angles or sides? Here's a flow chart to help... START

Is it a RAT (Right-Angled Triangle)?

NO

YES

Does the question involve only sides (no angles)?

Do we know an oppoSite pair (a side and an angle)?

NO

YES

NO

YES

Use SOHCAHTOA Use SINE RULE

Use PYTHAGORAS

Use COSINE RULE

All these formulae (except for one) are given inside the front cover of your IGCSE paper.

ANATOMY OF A RAT 

Learn these special names for the sides of a right-angled triangle.

OT

θ

OPPOSITE

P HY

E US N E

ADJACENT The Hypotenuse is always opposite the right angle (Pythagoras and SOHCAHTOA) and it is the longest side. The Opposite is opposite the given angle (SOHCAHTOA). The Adjacent is next to (adjacent to) the given angle (SOHCAHTOA).

Page 11

IGCSE SHAPE & SPACE PYTHAGORAS 

If we know two of the three sides of a RAT (Right-Angled Triangle), we can find the missing side using Pythagoras' Theorem.



SKILL: Find an unknown side of a RAT using Pythagoras. c b 2

2

2

Use a b = c

where c 2 is the hypotenuse. a TJP TOP TIP: Go ahead and relabel the triangle if you like, to get c in the right place on the hypotenuse. Just remember to change back to the correct letter at the end. Q: Find the length of side x in this triangle. x cm

5 cm A: Use a 2b2 = c2 substituting to get 5212 2 = x 2 2 25144 = 169 = x So x =

12 cm

169 = 13 ; the side has length 13 cm. 30

Q: Find the value of b . b

34 A: b 2302 = 342 2 b 900 = 1156 2 b = 1156−900 = 256 So b =

 256 = 16 .

Q: Find the height of an equilateral triangle of side 5 m. 5m hm

A: Hint: split it into two RATs! 2 2 2 2.5 h = 5 6.25h2 = 25 2 h = 25−6.25 = 18.75 So h =

5m

2.5 m

 18.75 = 4.33 . The height is 4.33 m (3 sig figs). Page 12

2.5 m

IGCSE SHAPE & SPACE TRIGONOMETRY (SOHCAHTOA) 

sin  =

Opp Hyp

cos =

Adj Hyp

P HY

tan  =

Opp Adj

θ

OT

E US N E

OPPOSITE



Trigonometry gives the connection between the sides and the angles of a RAT.

ADJACENT

Learn these triangles

O

(they spell SOHCAHTOA) S

A H

C

O H

T

A

Decide which SOHCAHTOA triangle to use by seeing which side does not feature in the question (as either a number or an unknown letter). Then cover up the letter you are trying to find; the remaining two letters give the formula you need. For example, to find H in the SOH triangle, we get O ÷ S. Two letters on the bottom line are multiplied; a letter on top is divided by one underneath. 

SKILL: Find an unknown side of a RAT using SOHCAHTOA. Q: Find the length x . 11 x 27° A: The Adjacent side does not feature in this question, so use SOH (no Adj). We are finding the Opposite side, so from the SOH triangle, Opp = Sin × Hyp. x = sin 27×11 = 4.99 (3 sig figs). y Q: Find the length y .

31° 8

A: The Hypotenuse does not feature in this question, so use TOA (no Hyp). We are finding the Adjacent side, so from the TOA triangle, Adj = Opp ÷ Tan y = 8÷tan 31 = 13.3 (3 sig figs).

Page 13

IGCSE SHAPE & SPACE 

SKILL: Find an unknown angle of a RAT using SOHCAHTOA. TJP TOP TIP: Always use the SHIFT button to find an angle. This gives the inverse sin, cos or tan, written as sin −1 , cos−1 , tan−1 . If you get error on your calculator, you have got your numbers the wrong way round. Q: Find the angle θ in this triangle.

57 θ° 49

A: The Opposite does not feature in this question, so use CAH (no Opp). We are finding Cos, so from the CAH triangle, Cos = Adj ÷ Hyp −1  = cos 49÷57 = 30.7 ° (1 dp).

THREE-DIMENSIONAL PROBLEMS 

We may get a 3-D question which we have to break down into 2-D RATs in order to find a mystery length or angle.



SKILL: Use Pythagoras and Trigonometry to solve a 3-D problem. D

Q: Find the long diagonal AD of this cuboid, and also angle DAC.

7

A

20

2

2

AC = 20 6 = 436

c

A

20

c

2

AD = 4367 = 485 AD =

6 B D

Then find AD using this triangle: 2

B

C

C

A: Find diagonal AC first, using this triangle: 2

6

 485 = 22.0 (3 sig figs)

7

A C √436 To find DAC (the angle at A), we could use SOH, CAH or TOA; we know all the sides. Here we'll use TOA. DAC = tan−1 7÷ 436 = 18.5 ° (1 dp).

Page 14

IGCSE SHAPE & SPACE SINE RULE 

If we know the values of an opposite pair (a side and the angle opposite it) in a non right-angled triangle (non-RAT), we use the Sine Rule. a b c = = sin A sin B sin C

TJP TOP TIP: OppoSite pair ⇨Sine Rule. IGCSE INSIDER INFO: You can flip the printed formula upside-down to get the unknown quantity on the top line – this makes life much easier.

a B°

Note: little letters are sides, CAPITAL letters are angles, and a is opposite A, b opposite B, etc. 

C° c

b A°

SKILL: Use the Sine Rule to find an unknown side. Q: Find the value of side b in this triangle.

17 35°

b 67°

A: First, label the sides and angles with a , A and B (we don't need c and C here). 17 b = sin 67 sin 35 17 ×sin 35 = b sin 67

a=17 B=35°

A=67°

b = 10.6 (3 sig figs)



b

SKILL: Use the Sine Rule to find an unknown angle. Q: Find the value of angle C in this triangle. 13 19°

8 C°

A: Label the other sides and angles involved in the question. Then use the flipped version of the formula to get the angle on the top. [Note: we could have used b and B instead of a and A.] sin 19 sin C = 8 13 sin 19 ×13 = sin C 8 sin C = 0.529048

c=13 A=19°

C = sin−1 0.529048 = 31.9° (1 dp).

Page 15

a=8 C°

IGCSE SHAPE & SPACE COSINE RULE 

If we don't know the values of an opposite pair (a side and the angle opposite it) in a non right-angled triangle (non-RAT), we use the Cosine Rule. TJP TOP TIP: No Sine Rule ⇨ Cosine Rule.

2

Be careful with BIDMAS when working out this formula. Don't press '=' until the end...

2

a

C°

B°

And remember to square root!



2

a = b c −2 b c cos A

c

b A°

SKILL: Use the Cosine Rule to find an unknown side. TJP TOP TIP: Relabel the triangle so that the side you are finding is called a . 'Cos that's what is on the left hand side of the formula... Q: Find side y .

y 2.6 35° 6.2

A: Relabel the triangle with a = y , and put angle A opposite it. It doesn't matter where you put b and c , by the way... 2

2

2

a = 2.6 6.2 −2×2.6×6.2 cos 35 2

a = 6.7638.44−32.24 cos 35

a=y b=2.6

a 2 = 18.79

A=35°

a = y = 4.33 (3 sig figs). 

c=6.2

SKILL: Use the Cosine Rule to find an unknown angle. IGCSE INSIDER INFO: LEARN THIS FORMULA! It's not given in the front of the exam paper.

cos A =

b2c 2−a 2 2 bc

Q: Find angle A. 2

a=12 2

4 13 −12 A: cos A = 2×4×13 cos A =

2

b=4 A° c=13

41 = 0.39423 104

A = cos−1 0.39423 = 66.8 ° .

Page 16

IGCSE SHAPE & SPACE AREA OF A TRIANGLE 

The area of a triangle is ½ base × height, but if we have a non right-angled triangle the height may not be obvious. Fortunately, we are given a formula to work out the area.

Area = A =

1 2

a b sin C

TJP TOP TIP: Relabel the triangle so that the angle is called C, sandwiched between sides a and b .



SKILL: Find the area of a non right-angled triangle. Q: Calculate the area of this triangle. 66 cm

41 cm

33°

59°

A: A sneaky question; we first have to fill in the missing angle at the top, which is sandwiched between the two given sides. This angle is 180 – 33 – 59 = 88°.

A =

1 2

a=66 cm

×66×41×sin 88

33°

A = 1352.18 So the area is 1350 cm² (3 sig fig).

Q: If this triangle has an area of 132 cm², find the value of x .

a=x C=42° b=2x A: Substitute into the area formula:

A =

1 2

x×2 x sin 42 = 132

x 2 sin 42 = 132 x2 =

132 sin 42

= 197.27

x = 14.0 cm (3 sig figs).

Page 17

C=88°

b=41 cm 59°

IGCSE SHAPE & SPACE PERIMETER AND AREA 

The perimeter of a shape is the distance all the way around it (imagine walking round it). A circle's perimeter is called the circumference; it gets its own special name. Triangle  Perimeter = sum of the sides  Area = ½ base × vertical height Rectangle  Perimeter = 2 × base + 2 × height  Area = base × height Parallelogram  Perimeter = 2 × base + 2 × slope height  Area = base × vertical height Trapezium  Perimeter = sum of the sides  Area = mean width × vertical height Circle  Circumference = 2 π r  Area = π r² TJP TOP TIP: Area is squarier!



SKILL: Find the perimeter of a sector of a circle. Hint: length of arc = fraction of circle × circumference Perimeter = radius + radius + arc = 5 + 5 + (130/360) × 2 π × 5

130°

= 21.3446 5 cm

= 21.3 cm (3 sig figs) 

SKILL: Find the area of a sector of a circle. Hint: area of sector = fraction of circle × area of circle Area = (130/360) × π × 5²

130°

= 28.3616 = 28.4 cm² (3 sig figs)

5 cm

Page 18

IGCSE SHAPE & SPACE SURFACE AREA AND VOLUME Face – a flat side of a 3-D shape. Edge – a line joining two corners of a 3-D shape. Vertex – a corner of a 3-D shape. 

The surface area is the total area of the outside of a 3-D shape. If you are asked for the curved surface area, just leave out any flat faces. [The formulae you are given include only the curved surface area for cones/cylinders.] Cuboid  Surface area = 2 abbcca   Volume = abc

c b

a Prism  Surface area = sum of rectangles + 2 ends  Volume = area of end × length

r

Cylinder  Surface area = 2  r h2  r 2  Volume =  r 2 h

h

Note: h may be the length if the cylinder is lying on its side. Cone  Surface area =  r l r 2 1 2  Volume = 3  r h

l

h

Note: l , h , r obey l 2 = h 2r 2 where l is the slant height and h is the vertical height.

r

Sphere  Surface area = 4  r 2 4 3  Volume = 3  r r

To find the volume or curved surface area of a hemisphere (a half-sphere), simply work it out for a whole sphere and then halve your answer.

Page 19

IGCSE SHAPE & SPACE SIMILARITY AND ENLARGEMENT 

Shapes are similar if one is an enlargement of the other (so that all its lengths are multiplied by the same amount). The scale factor = new length ÷ old length. Shapes are congruent if they are the same shape and size (so that you could fit one on top of the other). If we enlarge a shape on a special 3-D photocopier, what happens to the lengths, the areas and the volumes? Here is a clue: set the enlargement to double all lengths, and then see what happens.

The area of each face is multiplied by 4 = 2²

(area is squarier)

The volume of the cube is multiplied by 8 = 2³ (volume is 3-D) But the angles in the corners stay at 90°

(don't change the angles!)

In general use LAV to work out scale factors. (This stands for Length Area Volume) Always start by finding the scale factor N. 

L: ×N A: ×N² V: ×N³

SKILL: Solve a problem involving similar shapes. Q: Two shapes are similar; find the missing lengths a and b. 12.5 cm b cm 2 cm

a cm

4 cm

10 cm

A: Use two matching numbers to get the scale factor = 10 ÷ 4 = 2.5. Now use it to find a = 2 × 2.5 = 5 cm and b = 12.5 ÷ 2.5 = 5 cm. Q: A cylinder is enlarged by scale factor 3. Find the new length, area and volume. L = 7cm A = 40cm² V = 100cm³ A: New L = 7 × 3 = 21 cm New A = 40 × 3² = 40 × 9 = 360 cm² New V = 100 × 3³ = 100 × 27 = 2700 cm³

Page 20

IGCSE SHAPE & SPACE METRIC UNITS You need to know all these common units of length and volume. 10 mm = 1 cm 1000 mm = 1 metre 100 cm = 1 metre 1000 m = 1 km milli = 1 thousandth

centi = 1 hundredth

kilo = one thousand

SKILL: Convert between area/volume measures. Q: Convert 3 m² to cm². A: Each square metre has 100×100 = 100² square centimetres in it (not just 100). 100 squares 100 squares

So the answer is 3×100² = 30,000 cm².

Q: How many cubic millimetres are there in one cubic kilometre? A: Along one edge of this giant cube, there are 1000×1000 = 106 millimetres in a kilometre.

So there are 106 × 106 × 106 = 1018 mm³ in one km³.

1 million

m

ill io n

1 million



1000 ml = 1 litre 1 ml = 1 cm³ 1 litre = 1000 cm³

1



Q: How many litres of fruit juice would it take to fill a swimming pool 10m by 50m by 3m? A: The pool has a volume of 10 × 50 × 3 = 1500 m³.

This converts to 1500 × 100³ = 1,500,000,000 cm³ Since 1 litre = 1000 cm³, the pool will hold 1,500,000 litres of juice. Page 21

IGCSE SHAPE & SPACE VECTORS 

A scalar has size (an ordinary number), e.g. time, mass, speed, distance. A vector has size and direction, e.g. force, weight, velocity, displacement. The size (or length) of a vector is called its magnitude or modulus. A vector quantity is shown by a bold letter a (in print), or underlined a if handwritten. The same letter 'a' written normally means the modulus or length of the vector a. If we have a vector going from A to B, we can write this as  AB . We can show the direction of a vector in several different ways: a

 by drawing a diagram:

 by giving an angle or bearing: 10km on a bearing of 327°.  by giving the x and y amounts in brackets (a column vector): c =

−68  .

If you multiply a vector by a scalar (ordinary number) you just make the vector longer or shorter. For example, doubling a vector makes it twice as long (in the same direction). The resultant is the result of adding or subtracting two or more vectors. Use Pythagoras to find the modulus of a column vector. 

SKILL: Solve problems involving column vectors.

  and b = −77 , find (i) 3 a (ii) ab (iii) the modulus of a . 2 6 2 −7 −5 A: (i) 3 a = 3   =   (ii) ab =    =   5 15 5 7 12 2 Q: If a = 5

(iii) modulus of a = 

 2 252

=

 29 = 5.39 (3 sig figs)

SKILL: Solve geometrical vector questions. Q: Use the grid below to write vector expressions for (i)  AC (ii)  HG (iii)  KF (iv)  DI Hint: you can only move in the a and b directions, and backwards is negative. A E

B F

C G

D H

b I

a

J

A: (i)  AC = 2 a

K

L

(ii)  HG = −a

(iii)  KF = −ab (iv)  DI = −3 a−2 b

Page 22

IGCSE SHAPE & SPACE TJP TOP TIP: Two vectors are parallel if one is a multiple of the other. 

SKILL: Use vectors to prove geometrical results. Q: If M is the midpoint of AB and N is the midpoint of AC, find  BC and hence MN and  state two facts about these vectors. A b a N M C B A:  BC = −2 a2 b . MN = −ab and  Therefore  BC = 2 MN , so:  (i) BC is twice as long as  MN ,   (ii) BC is parallel to MN .

Page 23