Candidates' performance in part (a) was reported to be poor. A good number of them were
reported not to have attempted this part of the question. They were expected to recall that
lkm = 100,000 ern, therefore 1km2 = 1km x 1km = 100,000 x 100,000 = 10,000 000,OOOcm^{2}•

85km^{2} = 850,000,000,000 em". 20,000 em on the ground = 1 cm on the map, hence
400,000,000 em^{2} on the ground = 1 cm^{2}2 on the
ground is equivalent to __850,000,000,000__ = 2125 cm^{2} on the map.
400,000,000

In part (b), candidates' performance was said to be better than it was in part (a). However,
many candidates did not draw the diagram correctly and so were not able to solve the problem.

From the diagram, area of path = 2(30 x 6) + 12a = 360 + 12a. Area of field = 18a. Since they are
equal, 18a = 12a + 360. This gave a = 60 m.

In part (c), candidates' performance was described as fair. However, some candidates did not
see the reflex angle as 360 - x, hence, did not subtract their final answer from 360° when they
had calculated the value of the reflex angle. Here, __360 - x__ x __22__x __Z x Z__ = 27.5 cm2

360 7 2 2

This meant that 360 - x =

__360 x 2 x 27.5__ from where we obtain x = 103° to the nearest degree

77