WAEC 2025 · Paper 2 · Q4✱✱

(a)
A pack of 52 playing cards is shuffled and one card is drawn. Find the probability that it is either a five or a red nine.
(b)
The bearing of $Q$ from $P$ is $150^\circ$ and the bearing of $R$ from $Q$ is $060^\circ$ . If $|PQ| = 5\text{ m}$ and $|QR| = 3\text{ m}$ , find the bearing of $R$ from $P$ , correct to the nearest degree.

Worked solution (try it first)

There are four fives and two red nines (hearts and diamonds), with no card counted twice: 6 cards.
Probability $= \frac{6}{52} = \frac{3}{26}$ .

Break each leg into east and north parts.
$P$ to $Q$ on $150^\circ$ , 5 m: east $5\sin 150^\circ = 2.5$ , north $5\cos 150^\circ \approx -4.330$ (so 4.330 south).
$Q$ to $R$ on $060^\circ$ , 3 m: east $3\sin 60^\circ \approx 2.598$ , north $3\cos 60^\circ = 1.5$ .
Altogether, $R$ is $5.098$ m east and $2.830$ m south of $P$ .
That direction is south-east of $P$ , with angle $\tan^{-1}\frac{2.830}{5.098} \approx 29.0^\circ$ below due east.
Bearing of $R$ from $P$ $= 090^\circ + 29.0^\circ$
$\approx 119^\circ$ .