General Mathematics Paper 2,May/June. 2014
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Main
Weakness/Remedies
Strength

Question 10
1. Solve: ( x – 2)(x – 3)  = 12

(b)

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Observation

In the diagram, M and N are the centres of two circles of equal radii 7 cm.  The circles intercept at P and Q.  If ∠PMQ = ∠PNQ = 60o, calculate, correct to the nearest whole number, the area of the shaded portion.  [ Take

Part (a) of this question was reported to be well attempted by majority of the candidates. They were reported to have correctly expanded the bracket and simplified to obtain 2Solving this quadratic equation gave .

In part (b), majority of the candidates were reported not to attempt the question. Most of them answered only part (a). The performance of majority of those who attempted it was also reported to be poor. Majority of the candidates were reported not to have recognized the shaded area as two segments from the two circles.  Candidates were expected to obtain the area of the segment in each circle and add them.  Since the circles were equal, they would only obtain the area for one and multiply it by 2 to obtain the area of the shaded portion as follows

Area of sector cm2 or 25.667cm2.
Area of triangle NPQ=22 or cm2.  Therefore, area of shaded portion 22)2, correct to the nearest whole number.