The Chief Examiner reported that this question was not attempted by majority of the candidates. The report further stated that those who attempted it performed better in part (b) than in part (a) and those who were able to solve the question in part(a) were those who drew a diagram to illustrate it. However, candidates’ performance in this question was reported to be generally poor.

In part (a), candidates were encouraged to represent the information in a diagram such as

** **

From the diagram, the total surface area of the open tank = (2w x w) + 2(2w x (w-4)) + 2= 8w2 – 24w, where w = width of the tank. Equating this area to the area of the steel place and simplifying, gave the equation w2-3w-180 = 0. Solving this equation gave w = 15m. Therefore the length of the tank was

15 x 2 = 30m.

In part(b) candidates were expected to obtain ∠WXV using the trigonometric ratio tan ∠WXV = which gave ∠WXV = tan-1) = 38.6598. The angle formed by the arc WV at the centre of the circle = 2 x angle at the circumference = 2 x 38.6598 = 77.3196. Therefore length of arc WV = x 2 x x = 9.45 cm correct to 2 decimal places.