WAEC 2015 · Paper 2 · Q11

(a)
Make $m$ the subject of the relation $h = \dfrac{mt}{d(m + p)}$ .
(b)
In the diagram, $WY$ and $WZ$ are straight lines; $O$ is the centre of circle $WXM$ and $\angle XWM = 48^\circ$ . Calculate the value of $\angle WYZ$ .
(c)
An operation $\oplus$ is defined on the set $X = \{1, 3, 5, 6\}$ by $m \oplus n = m + n + 2 \pmod 7$ , where $m, n \in X$ . (i) Draw a table for the operation. (ii) Using the table, find the truth set of: I. $3 \oplus n = 3$ ; II. $n \oplus n = 3$ .
Model answer
(i) Work out $m + n + 2$ and take the remainder on dividing by 7:

$\oplus$ 1 3 5 6

1 4 6 1 2

3 6 1 3 4

5 1 3 5 6

6 2 4 6 0

(ii) I. In the row for 3, the entry 3 is under $n = 5$ : truth set $\{5\}$ . II. The diagonal ( $n \oplus n$ ) reads $4, 1, 5, 0$ and never 3: truth set $\{\}$ (empty).

$\oplus$	1	3	5	6
1	4	6	1	2
3	6	1	3	4
5	1	3	5	6
6	2	4	6	0

Worked solution (try it first)

Join $XM$ .
$WM$ passes through the centre $O$ , so it is a diameter and $\angle WXM = 90^\circ$ (angle in a semicircle).
In $\triangle WXM$ : $\angle WMX = 180^\circ - 90^\circ - 48^\circ$
$= 42^\circ$ (angles in a triangle).
$XYZM$ is a cyclic quadrilateral and $W, M, Z$ are in a straight line, so $\angle WMX$ is its exterior angle at $M$ .
An exterior angle of a cyclic quadrilateral equals the interior opposite angle: $\angle XYZ = 42^\circ$ .
Hence $\angle WYZ = 42^\circ$ .

$\oplus$	1	3	5	6
1	4	6	1	2
3	6	1	3	4
5	1	3	5	6
6	2	4	6	0

For I, look along the row for 3: the entry 3 appears only in the column for 5, so the truth set is $\{5\}$ .
For II, look at the diagonal, where $n \oplus n$ gives $4, 1, 5, 0$ .
None of them is 3, so the truth set is the empty set, $\{\}$ or $\varnothing$ .