Question 4The diagram shows a cone with slant height 10.5 cm. If the curved surface area of the cone is 115.5 cm²;, calculate, correct to 3 significant figures, the:
(a)base radius, r;
(b)height, h;
(c)volume of the cone. [Taken= 22]
7
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Observation
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.
Question 5
Two fair dice are thrown.
M is the event described by "the sum of the scores is 10" and
N is the event described by "the difference between the scores is 3".
(a) Write out the elements of M and N.
(b) Find the probability of M or N.
(c) Are M and N mutually exclusive? Give reasons.
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Observation
This question was reported to be popular and well attempted by majority of the candidates. However, a few of them did not list the elements of M and N as required but drew the sample space when 2 dice are thrown. Some of them who listed these elements did not do so completely. Majority of them were reported not to be able to state correctly the condition for which events are mutually exclusive. Candidates were expected to show that M = { (4,6) , (5,5), (6,4}. N = { (1,4), (2,5), (3,6), (4,1),
(5,2), (6,3) }. Probability of M = 3/36 = 1/12. Probability of N = 6/36 = 1/6.
Probability of Nor M:1/12 + 1/6 = 1/4. M and N are mutually exclusive because no element is common to both sets. Thus the two events cannot happen at the same time i.e. n(M n N) = O. It was also reported that while listing the elements of N, some candidates would list (1,4) but not (4,1).
(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 km²,
(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 cm². 20,000 em on the ground = 1 cm on the map, hence 400,000,000 cm² on the ground = 1 cm2850,000,000,000 = 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
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