WAEC Mathematics Past Questions & Answers - Page 190

$7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div16\frac{1}{2}$; $\frac{15}{2} - \left(\frac{5}{2}+\frac{3}{1}\right)\div \frac{33}{2}$
$\frac{15}{2} - \left(\frac{5+6}{2}\right)\div \frac{33}{2}$; $\frac{15}{2} - \frac{11}{2} \div \frac{33}{2}$

$\frac{11}{2}$ ÷ $\frac{33}{2}$ = $\frac{11}{2}$ * $\frac{2}{33}$

  = $\frac{11}{33}$ or $\frac{1}{3}$ (when simplified)

$\frac{15}{2}$ - $\frac{1}{3}$ = $\frac{43}{6}$

7.166

The nearest whole number is 7

Prob. (a girl) $=\frac{Number\hspace{1mm} of\hspace{1mm} girls}{Number \hspace{1mm} of \hspace{1mm} boys \hspace{1mm} and \hspace{1mm} girls\hspace{1mm} in \hspace{1mm} class}\\ = \frac{12}{m+12}$

$\frac{PR}{QR} = \frac{QR}{QY} = \frac{QP}{YR}$ (SSS Congruence)

$\frac{QR}{4} = \frac{8}{5}$

$QR = \frac{4 \times 8}{5}$

= 6.4 cm

Let the girls pocket money be rep. by x. The amount spent on books = $\frac{1}{4}of\hspace{1mm}x = \frac{x}{4}$
The amount spent on dress $=\frac{1}{3} of \hspace{1mm}x=\frac{x}{3}$
∴The fraction that remains = $\frac{x}{1}-\left(\frac{x}{4}+\frac{x}{3}\right)\\ \frac{3x+4x}{12}; = \frac{12x-7x}{12} = \frac{5x}{12} = \frac{5}{12}$

$\frac{3}{4}t+\frac{1}{3}(21-t) = 11; \frac{3t}{4} + \frac{7}{1} - \frac{t}{3} = \frac{11}{1}\\ \frac{3 \times 3t + 7\times 12 – 4 \times t = 11 \times 12}{12}\\ 9t + 84 – 4t = 132; 5t = 132 – 84\\ 5t = 48; t = \frac{48}{5} = 9\frac{3}{5}$