This question was reported as one of the questions where candidates performed very well. They showed good understanding of mensuration of right circular cones as they were able to apply relevant formulae and majority of them obtained full marks. The curved surface area,
iu! = 115.5cm², slant height,l= 10.scm, n = ll, hence r = 115.5
                                                                                                     7nl

= 11.5 x 7 = 3.S0cm. r2 + h2 = (10.5}2 => h = .J(10.5)2 - (3.5)2 = 9.90cm
   22 x 10.5
Volume of cone = .1 x 22 x 3.5 x 9.899 == 127 cm²
                                                     3 7                 1
A good number of the candidates were however reported to have lost some marks, especially the A marks, because they did not give their answers in 3 significant figures as required by the question.