a)

i.

P and R are subsets of W ⇒ P and R elements must be found in W.

W={natural numbers greater or equal 3 but less than 21} ⇒ W={3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}

P={multiples of 3} ⇒ P={3,6,9,12,15,18}

R={factors of 18} ⇒ R ={1,2,3,6,9,18}

ii.

α

The complement of a set is the set that includes all the elements of the universal set that are not present in the given set.

P^' read as P complement

⇒ = P^' = {4,5,7,8,10,11,13,14,16,17,19,20}

β

(P ∩ R)^' read as P ∩ R complement

First find the elements of P ∩ R and later find its complement (Elements in W but not in P ∩ R)

Intersection of two sets is the list of elements which are found in both sets

⇒ (P ∩ R) = {3,6,9,18}

⇒ (P ∩ R)^' = {4,5,7,8,10,11,12,13,14,15,16,17,19,20}

b)

Concept

A^b/_c = (Axc+b)/c

⇒ 3¹/₁₅ = (3x15+1)/15 = 46/15

⇒ 2¹/₃ = (2x3+1)/3 = 7/3

⇒ 2²/₁₅ = (2x15+2)/15 = 32/15

⇒ 1²/₃ = (1x3+2)/3 = 5/3

Applying BODMAS, the bracket must be solved first

⇒ (46/15 - 7/3) ÷ (32/15 + 5/3) = ([46-7x5]/15) ÷ ([32+5x5]/15)

([46-7x5]/15) ÷ ([32+5x5]/15) = ([46-35]/15) ÷ ([32+25]/15) = 11/15 ÷ 57/15

Applying a/b ÷ c/d = a/b x d/c (Thus d moves to the top and c down and the division changes to multiplication)

⇒ 11/15 ÷ 57/15 = 11/15 x 15/57 = 11/57 (The 15s cancel each other)

c)

Method I

Sum of all fractions add up to 1

⇒ fraction of students who speak French + Ga + Twi = 1

French = ⅙

Ga = ⅔

⇒ ⅙ + ⅔ + Twi = 1

⇒ fraction of students who speak Twi = 1 - (⅙ + ⅔)

Twi = 1 - ([1+2x2]/6) = 1 -5/6 = (1x6 - 5)/6 = ⅙

Applying the concept, value of a fraction = fraction x total number

Let x = the total number of students in the school

The total number of students who speak Twi is known, 60(value)

⇒ fraction of students who spean Twi x total number of students = 60

⇒ ⅙ x x = 60

Get rid of the denominator (6) by multiplying each side of the equation by the LCM (6) of the denominators

⇒ 6 x ⅙ x x = 60x6

⇒ x = 360

∴ there are 360 students in the school

Method II

Let x = the total number of students in the school

Applying the concept, value of a fraction = fraction x total number

⇒ No. of students who speak French = ⅙ x x

⇒ No. of students who speak Ga = ⅔ x x

The sum of No. of students who speak French + Ga + Twi = x x (Total number of students in the school)

⇒ ⅙ x x + ⅔ x x + 60 = x

Solve for the value of x

Multiply both sides of the equation by the LCM of the denominators (6)

6 x ⅙ x x + 6 x ⅔ x x + 60x6 = x x 6

x+4x+360 = 6x

5x+360 = 6x

360 = 6x - 5x

360 = x

∴ there are 360 students in the school