2010 - WAEC Mathematics Past Questions and Answers - page 2

3(x + 1) \(\leq\) 5(x + 2) + 15

3x + 3 \(\leq\) 5x + 10 + 15

3x - 5x \(\geq\) 10 + 15 - 3

-2x \(\geq\) 22

x \(\leq\) \(\frac{-22}{2}\)

x \(\leq\) -11

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})