WAEC 2018 · Paper 2 · Q4

(a)
In the diagram, $QOS$ is a diameter, $\angle RQS = x^\circ$ and $\angle QST = (3x + 15)^\circ$ , where $RST$ is a straight line. Find: (i) the value of $x$ ; (ii) $\angle RSQ$ .
Separate values with commas, e.g. 3, −2
(b)
If $2N4_{\text{seven}} = 15N_{\text{nine}}$ , find the value of $N$ .

Worked solution (try it first)

(a)(i)

$QOS$ is a diameter, so $\angle QRS = 90^\circ$ (angle in a semicircle).
$\angle QST$ is an exterior angle of triangle $QRS$ , so it equals the sum of the two opposite interior angles: $3x + 15 = x + 90$ .
So $2x = 75$ and $x = 37.5$ .

(ii)

$\angle QST = 3(37.5) + 15 = 127.5^\circ$ .
$RST$ is a straight line: $\angle RSQ = 180^\circ - 127.5^\circ$
$= 52.5^\circ$ .

(b)

Change both to base ten: $2N4_{\text{seven}} = 2 \times 49 + 7N + 4$
$= 102 + 7N$ and $15N_{\text{nine}} = 81 + 45 + N = 126 + N$ .
So $102 + 7N = 126 + N$ , $6N = 24$ and $N = 4$ .
(A digit 4 is allowed in both base seven and base nine.)

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