2013 - WAEC Mathematics Past Questions and Answers - page 2

(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}})

substitute x = 64 and y = 27

(\frac{64^{\frac{1}{2}} - 27^{\frac{1}{3}}}{27 - 64^{\frac{2}{3}}} = \frac{\sqrt{64} - 3\sqrt{27}}{27 - (3\sqrt{64})^2})

= (\frac{8 - 3}{27 - 16})

= (\frac{5}{11})

(\frac{1}{2})x + 2y = 3......(i)(multiply by 2)

(\frac{3}{2})x - 2y = 1......(ii)(multiply by 2)

x + 4y = 6......(iii)

3x - 4y = 2.....(iv) add (iii) and (iv)

4x = 8, x = (\frac{8}{4}) = 2

substitute x = 2 into equation (iii)

x + 4y = 6

2 + 4y = 6

4y = 6 - 2

4y = 4

y = (\frac{4}{4})

= 1(x + y)

2 + 1 = 3

p^{(\frac{1}{3})} = (\frac{3\sqrt{q}}{r})(cross multiply)

3(\sqrt{q}) = r x 3(\frac{\sqrt{q}}{r})(cross multiply)

3(\sqrt{q}) = r x 3(\sqrt{p}) cube root both side

q = 3(\sqrt{r}) x p

q = r^{(\frac{1}{3})}p = pr^{(\frac{1}{3})}

From (\bigtriangleup) OMQ find /MQ/ by Pythagoras OQ² = OM² + MQ²

5² = 2² + MQ²

25 = 4 + MQ²

25 - 4 = MQ²

21 - MQ²

MQ² = 21

MQ² = (\sqrt{21})

Length of chord = 2 x (\sqrt{21}) = 2(\sqrt{21})cm

Sin x = (\frac{5}{13})

0^o (\leq) x (\leq) 90^o, (cos x - tan x)

AC² = AB² + BC²

13² = 5² + BC²

169 - 25 + BC²

169 - 25 = BC²

144 = BC²

Cos x = (\frac{Adj}{Hyp}) = (\frac{12}{13})

BC = (\sqrt{144})

BC = 12

tan x = (\frac{opp}{adj} = \frac{5}{12})

BC = 12

cos x - tan x = (\frac{12}{13} - \frac{5}{12})

(\frac{144 - 65}{156} = \frac{79}{156})

Tan 50^o = (\frac{h}{6})

h = 6 tan 50

= 6 x 1.1917

= 7.1505

= 7.15

The probability not scoring 85% is = 1 - pro(scoring 85%)

= 1 - (\frac{85}{100}) = 1 - 0.85

= 0.15

not scoring would be less than 1

2 log x (3(\frac{3}{8})) = 6(divide both sides by 2)

(\frac{2 log_x(3 \frac{3}{8})}{2} = \frac{6}{2})

(\log_x \frac{27}{8} = 3)

(\frac{27}{8} = x^3)

(x^3 = \frac{27}{8})

x = (\sqrt{\frac{27}{8}})

= ((\frac{27}{8}))^{(\frac{1}{3})}

= (\frac{(3^3)^{\frac{1}{3}}}{(2^3)^{\frac{1}{3}}})

= (\frac{3}{2})