Quantitative Notes: Learnfast Review And Tutorial Hub – Nmat Reviewer

3

Quantitative Notes

SPECIAL PRODUCTS

LAWS OF EXPONENTS

(x + y) 3 = x 3 + 3x 2y + 3xy 2 + y 3 (x – y) 3 = x 3 – 3x 2y + 3xy 2 – y 3

Given x, y in real numbers, m and n in integers: Note: rules also apply if m, n are rational. a. 𝑥 𝑚 𝑥 𝑛

= 𝑥 𝑚+𝑛

b.

𝑥𝑚 𝑥𝑛

c. (𝑥 𝑚 )𝑛

= 𝑥 𝑚𝑛

d.

(𝑥𝑦)𝑛 = 𝑥 𝑛 𝑦 𝑛

f.

𝑥 −𝑛 =

e. (

𝑥 𝑛 ) 𝑦

=

𝑥𝑛 𝑦𝑛

note: y ≠ 0

= 𝑥 𝑚−𝑛

Rule 2 :

1

Rule 3:

√𝑥 = 𝑥 𝑛

1 𝑥𝑛

𝑛

𝑛

𝑛 √𝑥𝑦 = √𝑥 √𝑦

Rule 4:

𝑛

𝑥

√𝑦 =

Mode is the most frequently seen elements in as set. Example: Ronald’s grades in Math and Science for 2 nd semester are shown below. Math: 82, 80, 83, 85, 85, 87, 85, 89, 90, 85, 86 Science: 86, 90, 89, 87, 85, 84, 89, 89, 82, 82 Find the mean, median and mode of each subject.

𝑛

√𝑥 √𝑦

𝑛

𝑛

𝑚

√𝑥 𝑚 = 𝑥 𝑛

Since a logarithm is simply an exponent which is just being written down on the line, we expect the logarithm laws to work the same as the rules for exponents, and luckily, they do.

Logarithmic

𝑏𝑚 × 𝑏𝑛 = 𝑏𝑚+𝑛 𝑏𝑚 ÷ 𝑏𝑛 = 𝑏𝑚−𝑛

𝑙𝑜𝑔𝑏 𝑥𝑦 = 𝑙𝑜𝑔𝑏 𝑥 + 𝑙𝑜𝑔𝑏 𝑦 𝑥 𝑙𝑜𝑔𝑏 ( ) = 𝑙𝑜𝑔𝑏 𝑥 − 𝑙𝑜𝑔𝑏 𝑦

(𝑏𝑚 )𝑛 = 𝑏𝑚𝑛 𝑏0 = 1

𝑙𝑜𝑔𝑏 (𝑥 ) = 𝑛𝑙𝑜𝑔𝑏 𝑥 𝑙𝑜𝑔𝑏 (1) = 0

𝑦 𝑛

Note: On our calculators, “log” (without any base) is taken to mean “log base 10”. So, for example “log 7” means “log107”.

FACTORING METHODS a(x + y) = ax + ay (x + y)(x – y) = x 2 – y 2 (x + y) 2 = x 2 + 2xy + y 2 (x – y) 2 = x 2 – 2xy + y

3

Median is the middle of all elements or data in a set arranges in descending or ascending order.

LAWS OF LOGARITHMS

Exponential

(x – y)(x 2 + xy + y 2) = x 3 – y

Mean is the average of the data in a set. It is the sum of all data/elements by the number of data/elements.

Given x, y are real numbers, n > 0, n an integer, and x > 0, y > 0, if n is even. 𝑛

3

MEAN, MEDIAN AND MODE

LAWS OF RADICALS (RATIONAL EXPONENTS)

Rule 1 :

(x + y) (x 2 – xy + y2) = x 3 + y

Solution: 1. For Math Mean = 82+80+83+85+85+87+85+89+90+85+86 11 = 85.18 Median = 85 Arrange the grades in descending order: 90, 89, 87, 86, 85, 85, 85, 85, 83, 82, 80. Since there are 11 grades, the middle is in the 6th place, which is 85. Therefore the median is 85. Mode is 85, since 85 appears the most number of times that the others. 2. For Science Mean = 86+90+89+87+85+84+89+89+82+82 10 =86.3 Median = 86.5 Arranging the grades in descending order we have: 90, 89, 89, 89, 87, 86, 85, 84, 82, 82. Since the number of grades is 10 (even), the middle is in the 5th and 6th places. Therefore, the median is the average of 87 and 86 which is 86.5. Mode is 89, since 89 appears the most number of times than the others.

2

Factorial Notation n! = n(n-1)(n-2)…2.1, where n is a positive integer 0! = 1 Example: 4! = 4*3*2*1 = 24

LEARNFAST REVIEW AND TUTORIAL HUB – NMAT REVIEWER

4 Permutation of n distinct objects Permutation is an ordered arrangement of objects in a set. The number of permutations of n distinct objects taken r at a time is nPr

Example: What is the probability of getting an even number when a die is rolled?

𝑛!

= P(n, r) = n(n-1)(n-2)…(n-r+1)= (𝑛−𝑟)!

Solution: Possible outcomes = {1,2,3,4,5,6} Favorable outcomes = {2,4,6} 3 1 P(even number) = =

Example: How many ways can a 3-digt number be arranged to form a bank password?

6

Solution: There are 10 digits taken 3 at a time P(n, r) = P(10,3)

10! (10−3)!

=

10! 7!

=

10.9.8.7! 7!

= 720

Permutation of n objects non-distinct objects The permutations of n objects taken n at a time, in which q are alike, r are alike and so on is P(n,r) =

Solution: Since the coin consists of one head and one tail, the probability of getting a head is P = ½

𝑛! 𝑞!𝑟!

Deck of Cards 4 suits: diamond, heart, spade, club Face Cards: King, Queen, Jack Numbered Cards: 2, 3, 4, 5, 6, 7, 8, 9, 10 Lettered Card: Ace Number of cards in a deck: 52 Number of Colors: 2 (red and black) Example: In a deck of card, what is the probability of getting an Ace or black card?

Example: How many permutations are there in the word LOBBY?

Solution: P(Ace or black) = P(Ace) + P(black) – P(Ace and black) = 4 26 2 28 7 + − = =

Solution: Letter B = 2 Letter L = 1 Letter O = 1 Letter Y = 1 5! 5.4.3.2! P(n,r) = = = 60 ways 2!1!1!1!

2

52

52

52

52

13

2!

Circular Permutation The permutations of n objects taken n at a time arranged in a circular position. P = (n-1) Example: How many ways can 5 people sit around a round table? Solution: P = (n-1)! P = (5-1)! = 4! = 24 ways Combination A combination is an unordered arrangement of objects in a set. The number of combinations of n objects taken r at a time is 𝑛! C(n,r) = 𝑟!(𝑛−𝑟)!

Example: How many combinations can be made on 6 girls taken 2 at a time to appear in a variety show? Solution: C(6,2)=

6! 2!(6−2)!

=

6! 2!4!

=

6.5.4! 2.1.4!

= 15

Probability If an event can happen in m many ways and may fail in n 𝑚 ways, then the probability that it will happen is P = and the probability that it will fail is Q = P+Q=1

𝑛 𝑚+𝑛

𝑚+𝑛

and such that

Note: P(EVENT) = # of favourable outcomes / # of possible outcomes. Probability is usually expressed in percent or fraction 0 < P(EVENT) < 1 Example: In tossing a coin, what is the probability that the head, H, will appear?

LEARNFAST REVIEW AND TUTORIAL HUB – NMAT REVIEWER