WAEC Mathematics Past Questions & Answers - Page 6

Total numbers from 40 to 50 inclusive = 11 (40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50).

Prime numbers between 40 and 50 (inclusive) are 41, 43, and 47.

There are 3 prime numbers in this range.

therefore, Pr(prime numbers) = $\frac{3}{11}$

Converting each number to base 10
$110_{two} = 1 × 2^2 + 1 × 2^1 + 0 × 2^0$
= 1 × 4 + 1 × 2 + 0 × 1
= 4 + 2 + 0
= $6_{ten}$
$31_{eight} = 3 × 8^1 + 1 × 8^0$
= 3 × 8 + 1 × 1
= 24 + 1
= $25_{ten}$
$42_{five} = 4 × 5^1 + 2 × 5^0$
= 4 × 5 + 2 × 1
= 20 + 2
= $22_{ten}$
Hence, $31_{eight} > 42_{five} > 110_{two}$
In ascending order, $110_{two}, 42_{five}, 31_{eight}$

given that diameter = 140cm ⇒radius = 70cm, volume of cylinder = 200litres = $200,000cm^3$

volume of cylinder = base area times height = $\pi r^2h$

$200,000cm^3 = \frac{22}{7} \times 70 \times 70 \times h$

$200,000 = \frac{107,800h}{7}$

cross multiply

1,400,000 = 107,800h 

$h = \frac{1,400,000}{107,800}$

   = 13cm (to the nearest cm).

$y = \frac{ax^3 - b}{3z}$

cross multiply

$ax^3 - b$ = 3yz

$ax^3$ = 3yz + b 

divide both sides by a 

$x^3 = \frac{3yz + b}{a}$

take cube root of both sides

therefore, x = $\sqrt[3] \frac{3yz + b}{a}$

m:n = $2\frac{1}{3} : 1\frac{1}{5}$ = m : n = $\frac{7}{3} : \frac{6}{5}$

$\frac{7}{3} : \frac{6}{5}$ =  $\frac{7}{3} \div \frac{6}{5}$

$\frac{m}{n}$ =  $\frac{7}{3} \times \frac{5}{6}$

$\frac{m}{n}$ =  $\frac{35}{18}$ = m =  $\frac{35n}{18}$

n : q = $1\frac{1}{2} : 1\frac{1}{3}$ =  $\frac{3}{2} : \frac{4}{3}$

$\frac{n}{q}$ =   $\frac{3}{2} \times\frac{3}{4}$

$\frac{n}{q}$ =  $\frac{9}{8}$ = q =  $\frac{8n}{9}$ 

q : m =   $\frac{8n}{9}$ :   $\frac{35n}{18}$

$\frac{q}{m}$ =  $\frac{8n}{9} \div \frac{35n}{18}$

$\frac{q}{m}$ =  $\frac{8n}{9}\times\frac{18}{35n}$

 =$\frac{q}{m} = \frac{16}{35}$ = q : m = 16 : 35