(a) The scale of a map is 1:20,000. Calculate the area, in square centimetres, on the map of a forest reserve which covers 85 km",
(b) A rectangular playing field is 18 m wide. It is surrounded by a path 6m wide such that its area is equal to the area of the path. Calculate the length of the field.
(c]
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Observation
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,OOOcm2•
85km2 = 850,000,000,000 em". 20,000 em on the ground = 1 cm on the map, hence 400,000,000 em2 on the ground = 1 cm22 on the ground is equivalent to 850,000,000,000 = 2125 cm2 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 22x 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
Question 8
Using ruler and a pair of compasses only,
(a) construct
(i) a quadrilateral PQRS with IPSI :: 6 cm, LRSP:: 90°,
IRSI = 9 ern, IQRI = 8.4 cm and IPQI :: 5.4 cm;
(ii) the bisectors of LRSP and LSPQ to meet at X;
(iii) The perpendicular XTto meet PS at T.
(b) Measure IXT/'
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Observation
This question on geometrical construction was reported to be quite unpopular among the candidates. Very few of them attempted it and their performance was said to be poor. A good number of them measured the angles instead of constructing them. Others could not construct a perpendicular from a given point to a given line segment. Teachers are encouraged to emphasize this area of the syllabus.
Question 9
In the diagram, /AB/ = 8 km, /BC/ = 13 km, the bearing of A from B is 310° and the bearing of B from C is 230°. Calculate, correct to 3 significant figures,
(a) the distance AC;
(b) the bearing of C from A;
(c) how far east of B, C is
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Observation
This question was reported to have been attempted by majority of the candidates and their performance was described as satisfactory. However, most of them were reported not to calculate LABC correctly hence got wrong answers even though they were able to apply the cosine rule correctly to their wrong values. Others were not able to determine the required bearing correctly. The expected responses were as follows:
LASC = 100°. Therefore /AC/ = 82 + 132 - 2 (8)(13)cos100 which gave / AC/ = 16.4 km.
sin (LCAS) = sin100. Hence, sin (LCAB) = 13Sin100
13 16.4 16.4
Simplifying gave LCAB = 51.32°. Bearing of C from A = 180 - (50 + 51.32) = 079°.
If the distance of C east of B = BD, then BO = BC cas 40° = 13 x cas 40° = 9.96 km.
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