2012 - WAEC Mathematics Past Questions and Answers - page 1

(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{2} \times 3\sqrt{2}})

= (\frac{\sqrt{5} \times 2\sqrt{6}}{\sqrt{2} \times \sqrt{3}})

= (\frac{\sqrt{5} \times 2 \sqrt{6}}{\sqrt{6}})

= 2(\sqrt{5})

Divide cash enrollment by their common factors then the answer of each division gives the required ratio. i.e. x = (\frac{1050}{70}) = 15

and y = (\frac{1190}{70}) = 17

ratio of enrollment of x to y = 15:17

T₄ = 5 + (4 - 1)²; where n = 4

= 5 + (3)² = 5 + 9

= 14

T = 5 + (6 - 1)²

where n = 6

= 5 + (5)²

= 5 + 25 = 30

T⁴ + T⁶ = 14 - 30

= -16

(\frac{720 \times 1000}{8 \times 60 \times 60}) = (\frac{20 \times 10}{8})

= (\frac{200}{8})

= 25 m/s

A = 3,500, P = N2,500

A = P + I

But I = N3,500 - N2,500

I = N1,000

I = (\frac{PRT}{100})

N1,000 = (\frac{2.500 \times R \times 4}{100})

1000 = 100R

R = (\frac{1000}{100})

= 10%

(\frac{1}{8} + \frac{2}{3x} = \frac{1}{3})

= (\frac{1}{2})

(\frac{5}{3x} = \frac{1}{3})

3x = 15

x = (\frac{15}{3})

= 5

(\frac{54k^2 - 6}{3k + 1} = \frac{6(9k^2 - 1)}{3k + 1})

= (\frac{6(3k + 1) - (3k - 1)}{3k + 1})

= 6(3k - 1)

q = (\frac{3p}{r} + \frac{s}{2})

q = (\frac{6p + rs}{2r})

6p + rs = 2qr

6p = 2qr - rs

p = (\frac{2qr - rs}{6})