WAEC 2011 · Paper 2 · Q13

(a)
$P$ varies directly as $Q$ and inversely as the square of $R$ . If $P = 1$ when $Q = 8$ and $R = 2$ , find the value of $Q$ when $P = 3$ and $R = 5$ .
(b)
An aeroplane flies from town $A(20^\circ\text{N}, 60^\circ\text{E})$ to town $B(20^\circ\text{N}, 20^\circ\text{E})$ . (i) If the journey takes 6 hours, calculate, correct to 3 significant figures, the average speed of the aeroplane. (ii) If it then flies due north from town $B$ to town $C$ , $420\text{ km}$ away, calculate, correct to the nearest degree, the latitude of town $C$ . $[\text{Take radius of the earth} = 6400\text{ km and }\pi = 3.142]$
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

$P$ varies directly as $Q$ and inversely as $R^2$ : $P = \frac{kQ}{R^2}$ .
With $P = 1$ , $Q = 8$ , $R = 2$ : $1 = \frac{8k}{4}$ , so $k = \frac12$ and $P = \frac{Q}{2R^2}$ .
When $P = 3$ and $R = 5$ : $3 = \frac{Q}{50}$ , so $Q = 150$ .

$A$ and $B$ are both on latitude $20^\circ$ N, and the difference in longitude is $60^\circ - 20^\circ = 40^\circ$ .
Distance along the parallel $= \frac{40}{360} \times 2 \times 3.142 \times 6400\cos 20^\circ$
$\approx 4199.1$ km.
Average speed $= \frac{4199.1}{6} \approx 700$ km/h.

Due north is along a meridian, a great circle of radius 6400 km.
If the latitude changes by $y^\circ$ : $\frac{y}{360} \times 2 \times 3.142 \times 6400 = 420$ , so $y \approx 3.76^\circ$ .
$C$ is at $20^\circ + 3.76^\circ \approx 24^\circ$ N.