WAEC 2014 · Paper 2 · Q9

(a)
In the Venn diagram, $P$ , $Q$ and $R$ are subsets of the universal set $U$ . The regions are: $P$ only $16 - 2x$ , $P \cap Q$ only $5x$ , $Q$ only $6 + x$ , $P \cap R$ only $8x$ , $Q \cap R$ only $7x$ , $P \cap Q \cap R$ $4x$ , $R$ only $19 - 3x$ , and $4$ outside. If $n(U) = 125$ , find: (i) the value of $x$ ; (ii) $n[(P \cup Q) \cap R']$ .
Separate values with commas, e.g. 3, −2
(b)
In the diagram, $O$ is the centre of the circle, $XOW$ is a diameter, $WX$ is parallel to $YZ$ , and the lines $WY$ and $OZ$ meet at $E$ . If $\angle WXY = 50^\circ$ , find the value of: (i) $\angle WXZ$ ; (ii) $\angle YEZ$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

The eight regions together hold all 125 elements: $(16 - 2x) + 5x + (6 + x) + 8x + 7x + 4x + (19 - 3x) + 4 = 125$ .
Collect the numbers and the $x$ terms: $45 + 20x = 125$ , so $20x = 80$ and $x = 4$ .

$(P \cup Q) \cap R'$ is the part of $P$ or $Q$ that is outside $R$ : the regions $P$ only, $P \cap Q$ only and $Q$ only.
With $x = 4$ these hold $16 - 8 = 8$ , $5 \times 4 = 20$ and $6 + 4 = 10$ , so $n[(P \cup Q) \cap R'] = 8 + 20 + 10 = 38$ .

$XOW$ is a diameter, so $\angle XYW = 90^\circ$ (angle in a semicircle).
In $\triangle WXY$ : $\angle XWY = 180^\circ - 90^\circ - 50^\circ$
$= 40^\circ$ .
$WX \parallel YZ$ , so $\angle WYZ = \angle XWY = 40^\circ$ (alternate angles).
$\angle WXZ$ and $\angle WYZ$ stand on the same arc $WZ$ , so $\angle WXZ = 40^\circ$ (angles in the same segment).

The angle at the centre is twice the angle at the circumference on the same arc: $\angle WOZ = 2 \times \angle WYZ$
$= 80^\circ$ .
In $\triangle OEW$ : $\angle OEW = 180^\circ - 80^\circ - 40^\circ$
$= 60^\circ$ .
$\angle YEZ$ and $\angle OEW$ are vertically opposite, so $\angle YEZ = 60^\circ$ .