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

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

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

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