The Chief Examiner reported that part (a) of this question was attempted by majority of the candidates and they performed well in it. Their performance was reported to be better in part (a) than in part (b) which majority of them did not attempt.
In part (a), majority of the candidates were reported to have added the elements in the universal set,(U), and equated it to 125 to obtain .Solving this equation gave.Number of elements in 1, n1), . Some candidates were reported not to add 4 which was outside the three intersecting sets and this give them a wrong value of . Others were also reported not to have understood R1, the compliment of R.
In part (b), candidates were reported to have exhibited poor knowledge of circle theorems. They were expected to respond as follows:-
∠WXY = (given) and∠XYW Therefore ∠XWY = (90 – 50)oSince WX was parallel to YZ, then ∠WYZ = 40o (alternate angle to ∠XWY). Angle WOZ = 2 ∠WYZ = 80o (angle that an arc of a circle subtends at the centre of the circle is twice the angle it subtends at the remaining part of the circumference). Therefore, ∠OEW = (180 – 80 – 40)o = 60o. ∠YEZ = ∠OEW = 60o (vertically opposite angles).
.Solving this equation gave
.Number of elements in 
1, n
1), 
. Some candidates were reported not to add 4 which was outside the three intersecting sets and this give them a wrong value of 
. Others were also reported not to have understood R1, the compliment of R.
(given) and∠XYW 
Therefore ∠XWY = (90 – 50)o
Since WX was parallel to YZ, then ∠WYZ = 40o (alternate angle to ∠XWY). Angle WOZ = 2 
∠WYZ = 80o (angle that an arc of a circle subtends at the centre of the circle is twice the angle it subtends at the remaining part of the circumference). Therefore, ∠OEW = (180 – 80 – 40)o = 60o. ∠YEZ = ∠OEW = 60o (vertically opposite angles).