WAEC 2010 · Paper 2 · Q4

(a)
Find the value of $x$ if $x3_{\text{five}} - 14_{\text{five}} = 2x_{\text{five}}$ .
(b)
The diagram is a circle passing through the points $A$ , $B$ , $C$ and $D$ such that $AC$ and $BD$ meet at a point $E$ inside the circle. If $\angle DAC = 27^\circ$ , $\angle ABD = 54^\circ$ and $\angle ACB = 63^\circ$ , find: (i) $\angle CAB$ ; (ii) $\angle AEB$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Write each number in base ten: $x3_{\text{five}} = 5x + 3$ , $14_{\text{five}} = 5 + 4 = 9$ and $2x_{\text{five}} = 10 + x$ .
So $(5x + 3) - 9 = 10 + x$ .
Simplify: $5x - 6 = 10 + x$ , so $4x = 16$ and $x = 4$ .

(b)(i)

Angles in the same segment are equal: $\angle DBC = \angle DAC = 27^\circ$ (both stand on arc $DC$ ).
The angles of triangle $ABC$ add up to $180^\circ$ , and $\angle ABC = 54^\circ + 27^\circ$
$= 81^\circ$ .
So $\angle CAB = 180^\circ - 81^\circ - 63^\circ$
$= 36^\circ$ .

(ii)

In triangle $ABE$ : $\angle AEB = 180^\circ - 36^\circ - 54^\circ$
$= 90^\circ$ .

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