WAEC 2018 · Paper 2 · Q1

(a)
Evaluate without using a calculator: $\left(\frac14 \times 9\frac17 + \frac25\left(\frac23 + \frac34\right)\right) \div \left(\frac25 - \frac14\right)$ .
(b)
A hunter walked $250\text{ m}$ from point $P$ to $Q$ on a bearing of $042^\circ$ . Calculate, correct to the nearest metre, the vertical distance he has moved (i.e. how far north).

Worked solution (try it first)

(a)

Work inside the big bracket first.
$\frac14 \times 9\frac17 = \frac14 \times \frac{64}{7}$
$= \frac{16}{7}$ .
$\frac23 + \frac34 = \frac{17}{12}$ , and $\frac25 \times \frac{17}{12} = \frac{17}{30}$ .
So the bracket is $\frac{16}{7} + \frac{17}{30} = \frac{480 + 119}{210}$
$= \frac{599}{210}$ .
The divisor is $\frac25 - \frac14 = \frac{3}{20}$ .
So the value is $\frac{599}{210} \times \frac{20}{3} = \frac{1198}{63}$
$= 19\frac{1}{63}$ .

(b)

Draw north at $P$ and the path $PQ$ , 250 m on $042^\circ$ .
Drop a perpendicular from $Q$ to the north line.
The distance moved north is the side next to the $42^\circ$ angle, with $PQ$ as the hypotenuse: $250\cos 42^\circ = 250 \times 0.7431$
$\approx 186$ m.

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