1993 - WAEC Mathematics Past Questions and Answers - page 1

S = {1, 2, 3, 4, 5, 6};  T = {2, 4, 5, 7}; R = {1, 4, 5}

(S∩T) ∪ R = {2, 4, 5}  ∪ {1, 4, 5}

= {1, 2, 4, 5}

$\frac{3}{4}÷1\frac{1}{4}\times (1\frac{1}{2}-\frac{2}{3})$

$\frac{3}{4}÷\frac{5}{4}\times (\frac{9-4}{6})$

= $\frac{3}{4}\times \frac{4}{5}\times \frac{5}{6}$

= $\frac{1}{2}$

3m + 3 > 9

3m > 9 - 3

3m > 6

m > 2

${89}_{10}$

${89}_{10}={1011001}_2$

when n = 2

(-1)${}^{n-2}$ 2${}^{n+1}$ = 2

When n = 3

(-1)${}^{n-2}$ 2${}^{n+1}$ = -4

Sum = 2 - 4 = -2

$\frac{4^{-\frac{1}{2}}\times {16}^{\frac{3}{4}}}{4^{\frac{1}{2}}}$

= $\frac{{16}^{\frac{3}{4}}}{4^{\frac{1}{2}}\times 4^{\frac{1}{2}}}$

= $\frac{(2^4{)}^{\frac{3}{4}}}{4^{\frac{1}{2}}\times 4^{\frac{1}{2}}}$

= $\frac{2^3}{4}$

= 2

$\frac{\log \sqrt{27}}{\log 81}$

= $\frac{\log \sqrt{3^3}}{\log 3^4}$

= $\frac{\log 3^{\frac{3}{2}}}{\log 3^4}$

= $\frac{\frac{3}{2}\log 3}{4\log 3}$

= $\frac{\frac{3}{2}}{4}$

= $\frac{3}{8}$

2s${}^2$ - 3st - 2t${}^2$

= 2s${}^2$ - 4st + st - 2t${}^2$

= 2s(s - 2t) + t(s - 2t)

= (2s + t)(s - 2t)

x${}^2$ - 2x - 3 = 0

x${}^2$ - 3x + x - 3 = 0

x(x - 3) + 1(x - 3) = 0

(x + 1)(x - 3) = 0

x = (-1, 3)