2008 - WAEC Mathematics Past Questions and Answers - page 1

x% of 240 = 12

(\frac{x}{100} \times 240 = 12)

x = (\frac{12 \times 100}{240})

x = 5

(\frac{(3.2)^2 - (4.8)^2}{3.2 + 1.8} = \frac{(3.2 - 4.8)(3.2 + 4.8)}{(3.2 + 4.8)})

= 3.2 - 4.8

= -1.60

(\sqrt{50} + \frac{10}}{\sqrt{2}} = (\frac{\sqrt{50}}{1} + \sqrt{10}{\sqrt{2}})

= (\frac{\sqrt{50 \times 2} + 10}{\sqrt{2}})

= (\frac{\sqrt{100} + 10}{\sqrt{2}})

= (\frac{10 + 10}{\sqrt{2} = \frac{20}{\sqrt{2}})

= (\frac{20}{\sqrt{2}}) \times \frac{\sqrt{2}}{\sqrt{2}})

= (\frac{20\sqrt{2}}{2})

= 10(\sqrt{2})

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

where r = r% p.a; I = 0.36p

0.36p = (\frac{P \times r \times 4}{100})

(\frac{0.36 \times 100}{4}) = r

r = 9

Cost price, c.p = (\frac{100}{5}) x N350 = N7000

Selling price, s.p = (\frac{100}{5}) x N290 = N7250

%Gain = (\frac{S.p - C.p}{C.p}) x 100%

= (\frac{7250 - 7000}{7000}) x 100% = (\frac{250 \times 100}{7000})

= 3.6% (approx.)

p - 2g + 1 = g + 3p.........(1)

p - 2 = 0 .........(2)

From (2), p = 2; put p = 2 into (1);

2 - 2g + 1 = g + 3(2)

3 - 2g = g + 6

-2g - g = 6 - 3

-3g = 3

g = (\frac{3}{-3})

g = -1

(\frac{\frac{1}{x} + \frac{1}{y}}{x + y}) = (\frac{\frac{y + x}{xy}}{x + y})

= (\frac{x + y}{xy})

= (\frac{x + y}{xy} \times \frac{1}{x + y})

= (\frac{1}{xy})

3(\sqrt{27x^3y^9}) = 3(\sqrt{27} \times 3\sqrt{3^3} \times 3\sqrt{y^9})

= 3 (\times x \times y^3)

= 3xy³

Given; x = 2; y = (\frac{-1}{4})

= (\frac{x^2y - 2xy}{5})

= (\frac{2^2(\frac{-1}{4}) - 2(2)(\frac{-1}{4})}{5})

= (\frac{4(\frac{-1}{4}) + 4(\frac{-1}{4})}{5})

= (\frac{1 + 1}{5})

= (\frac{0}{5})

= 0

5y² + 2ay - 3a² = 5y² + 5ay - 3a²

= 5y(y + a) - 3a(y + a)

= (y + a)(5y - 3a)