WAEC Mathematics Past Questions & Answers - Page 191

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

\(\frac{11}{2}\) ÷ \(\frac{33}{2}\) = \(\frac{11}{2}\) * \(\frac{2}{33}\)

  = \(\frac{11}{33}\) 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) \(=\frac{Number\hspace{1mm} of\hspace{1mm} girls}{Number \hspace{1mm} of \hspace{1mm} boys \hspace{1mm} and \hspace{1mm} girls\hspace{1mm} in \hspace{1mm} class}\\
= \frac{12}{m+12}\)

The sum of angles at point = 360^o
(x+10)^o + (4x+50)^o + 20^o + 3x^o + 2x^o = 360^o
10x + 80^o = 360^o
10x = 360^o - 80^o = 280^o
\(\frac{280}{10}=28^{\circ}\)

\(\frac{PR}{QR} = \frac{QR}{QY} = \frac{QP}{YR}\) (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 = \(\frac{x}{1}-\left(\frac{x}{4}+\frac{x}{3}\right)\\
\frac{3x+4x}{12}; = \frac{12x-7x}{12} = \frac{5x}{12} = \frac{5}{12}\)