(i)

Let n = number of individuals who like basketball.

F = {Individuals who like football}

B = {Individuals who like basketball}

Number of individuals who like football = 3 x n = 3n

Number of individuals who like both games = $\frac{1}{3}$ x n = $\frac{\mathrm{n}}{3}$

(ii)

Coverying F

3n + n - $\frac{\mathrm{n}}{3}$ + 139 = 1382

Get rid of the fraction by multiplying both sides by the denominator 3.

3 x 3n + 3 x n - $\frac{\mathrm{n}}{3}$ x 3 + 139 x 3 = 1382 x 3

9n + 3n - n + 417 = 4146

11n = 4146 - 417

11n = 3729

Divide both sides by 11

n = $\frac{3729}{11}$ = 339

(a)

Let U = {Artistes}
C = {artistes who preferred Cymbal}
S = {artistes who preferred Saxophone}
B = {artistes who preferred Bongo}
x = number of artistes who preferred Cymbal and Bongo only
y = number of artistes who preferred Saxophone and Bongo only

(b)

(i)

Let U = {People}

n(U) = 100%

Let T = {People who owned Television}

n(T) = 68%

Let R = {People who owned Radio}

n(R) = 72%

Let x = Percentage who owned both Television and Radio

68% - x% + x% + 72% - x% + 12% = 100%

68 - x + x + 72 - x + 12 = 100

152 - x = 100

152 - 100 = x

52 = x

(ii)

16% of the population owned Television only

(a)

Let U = {Officers}

n(U) = 50

Let B = {Officers who chose beans}

n(B) = 21

Let R = {Officers who choose rice}

n(R) = 18

x = number of officers who chose beans and plantain only

y = number of officers who chose beans and rice only

z = number of officers who chose rice and plantain only

a = number of officers who chose none of the three kinds of food

Covering the rice (R) region

18 + 3 + x + 9 + a = 50

x + a + 30 = 50

x + a = 50 - 30

x + a = 20 ------ eqn 1

Covering the beans (B) region

21 + a + 2 + z + 9 = 50

a + z + 32 = 50

a + z = 50 - 32

a + z = 18 ------ eqn 2

Covering the plantain (P) region

24 + 3 + y + 2 + a = 50

y + a + 29 = 50

y + a = 50 - 29

y + a = 21 -------- eqn 3

Sum of the beans (B) region should give 21.

3 + y + 5 + x = 21

y + x + 8 = 21

y + x = 21 - 8

y + x = 13 ------- eqn 4

Get rid of a by subtracting --- eqn 1 from ---- eqn 3

y - x = 1 -------- eqn 5

Solve ---- eqn 4 and ---- eqn 5 simultaneously.

---- eqn 4 + ----- eqn 5

2y = 14

Divide both sides by 2

y = $\frac{14}{2}$ = 7

Substitute y = 7 into ---- eqn 5

7 - x = 1

7 - 1 = x

6 = x

x = 6

Substitute x = 6 into ------ eqn 1

6 + a = 20

a = 20 - 6

a = 14

Substituting a = 14 into --- eqn 2

14 + z = 18

z = 18 - 14

z = 4

Redraw the Venn diagram and substitute the values of the variables

(b)

(i)

Let U = {Students in the class}

x = {Students who like football and cricket only

n(Football only) = 24 - (x + 6 + 5)

n(Football only) = 24 - (x + 11)

n(Football only) = 24 - x - 11

n(Football only) = 13 - x

n(Cricket only) = 18 - (x + 5 + 7)

n(Cricket only) = 18 - (x + 12)

n(Cricket only) = 18 - x - 12

n(Cricket only) = 6 - x

(ii)