The chief examiner reported that part (a) of the question was poorly handled by majority of the candidates who attempted it.  Many of the candidates used three consecutive whole numbers instead of three positive numbers that are in A.P. A few others attempted it by trial and error.  Candidates were expected to show that since the numbers are in A.P., they can be written as a, a + d, a+2d.  Their sum being 21 implied that a+a+d+a+2d = 3a+3d = 21.  Hence, a+d = 7 i.e. the middle number is 7.  The sum of their squares = a2+ (a+d)2 = (a+2d)2 = 3a2+6ad+5d2 = 155.  Substituting 7-d for a and solving resulted in d=2.  Hence, the numbers were 5, 7, 9.

The part (b) of the question demanded the use of the formula for finding the total surface area of a sphere = 4pr2.  Majority of the candidates applied it well and obtained full marks but there were a few others who rather used the formula for finding the volume of the sphere.  Since the total surface area was 154 cm2, it implied that 4 x 22/7 x r2 = 154 i.e. r2 = 49/4.  This gave r = 7/2 or 3.5cm.