2010 - WAEC Mathematics Past Questions and Answers - page 2

Volume of water in tank = l x b x h = 180 litres

but 1 litre = 1000cm²

250 x 120 x hw = 180 x 1000

hw = \(\frac{180 \times 1000}{250 \times 120}\)

= 6cm

\(\frac{5^{n +3}}{25^{2n -2}}\) = 5^o

\(\frac{5^{n + 3}}{5^{2(2n - 3)}}\) = 5^o

n + 3 - 4n + 6 = 0

-3n + 9 = 0

-3n = -9

n = \(\frac{-9}{-3}\)

n = 3

pr. (More than one pet)

= \(\frac{\text{No. of students with > 1 pet}}{\text{total no. of students}}\)

= \(\frac{5 + 10 + 3}{30}\)

= \(\frac{18}{30}\)

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

2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\)

= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} + \frac{3}{\sqrt{9 \times 3}}\)

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

= 2\(\sqrt{3} = 6 \frac{\sqrt{3}}{3} + \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)

= 2\(\sqrt{3} - 2\sqrt{3} + \frac{\sqrt{3}}{3}\)

= \(\frac{\sqrt{3}}{3} = \frac{1}{3} \sqrt{3}\)

sin 3y = cos 2y, but sin \(\theta\) = cos(90 - \(\theta\))

sin 3y = cos(90 - 3y)

cos(90 - 3y) = cos 2y

90 - 3y = 2y

5y = 90

y = \(\frac{90}{5}\)

y = 18^o

l = x; b = x - 1

perimeter = 2(l + b) = 16

l + b = \(\frac{16}{2}\) = 8

l + b = 8

x + x - 1 = 8

2x = 8 + 1

2x = 9

x = \(\frac{9}{2}\)cm

x = 4\(\frac{1}{2}\)

Given tan x = 1

x = tan^-1(1)

x = 45^o

Now, \(\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}\)

= \(\frac{1 - \frac{1}{2}}{\frac{1}{2}}\)

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

= \(\frac{1}{2} \times \frac{1}{\sqrt{2}}\)

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

perimeter = 2(l + b) = 32

l + b = \(\frac{32}{2}\)

l + b = 16

b = 16 - 1.......(1)

Area = l + b = 39

lb = 39 .....(2)

put (1) into (2)

l(16 - 1) = 39

16l - l² = 39

l² = 13l - 3l + 39 = 0

l(l - 13) - 3(1 - 13) = 0

(l - 3)(l + 130 = 0

l - 3 = 0 or l - 13 = 0

l = 3cm or l = 13cm; The length in 13cm

\(\frac{2}{2 + x} + \frac{2}{2 - x}\)

\(\frac{2(2 - x) + 2(2 + x)}{(2 + x)(2 - x)} = \frac{4 - 2x + 4 + 2x}{4 - 2x + 2x - x^2}\)

= \(\frac{8}{4 - x62}\)