WASC Question Paper

7/27/2019 WASC Question Paper

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General Mathematics Paper 2, May/June 2009

Disclaimer - These questions and answers were taken from WAEC website and I am not in any way

responsible for the correctness of the information or copyright. This is to help intending WAEC

students have a downloadable copies of past WAEC exam questions and solutions.

Question1(a) Given that ( - 5 )( + ) = a + b , find a and b.

(b) If 21 y x 2y 1 = 82 3y, find y.2y + 2

_____________________________________________________________________________________________________

ObservationThe part (a) of this question was well attempted. It involved the expansion of the expression and comparing the

coefficients. It did not pose much problems to the candidates. Candidates were able to expand the expression to get

3 - 5 + - 10 which with a little computation gave a = -7 and b = -4. However many candidates expanded butdid not compare the coefficients.

In part (b) of the question, the fundamental laws of indices were required to solve this problem. They were toexpress the right hand side of the equation in powers of 2 and equate the coefficients of both sides to get y= 1.Most candidates did well in this question but there were a significant number who were not able to apply the laws

correctly.

Question 2(a) If 9 cos x 7 = 1 and 0 x 90, find x.

(b) Given that x is an integer, find the three greatest values of x which

satisfy the inequality 7x < 2x 13.

_____________________________________________________________________________________________________Observation

The part (a) of the question was on trigonometry and required the use of tables. Rearranging the equation results incos x = 0.8889 and from the Mathematical Tables, it showed x = 27.27o.

Most of the candidates scored above average in the question.

In part (b), Candidates were able to solve the expression but were unable to get the three greatest integersrequired. On solving the inequality, 7x 2x < -13 gave x < -23/5. The three greatest integers were -3, -4 and -5.


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Question 3The table shows the number of children per family in a community.

No. of children

0 1 2 3 4 5

No. of families 3 5 7 4 3 2

(a) Find the:

(i) Mode;

(ii) third quartile;

(iii) probability that a family has at least 2 children.

(b) If a pie chart were to be drawn for the data, what would be the sectoral angle representing families with onechild?

_____________________________________________________________________________________________________Observation

The performance of most of the candidates who attempted this question were found to be satisfactory. They wereable to determine the mode which was 2. They obtained the number of at least 2 children to be 16. Therefore the

probability of at least 2 children = = . The number of families with one child = 5 hence the required sectoral

angle = x 360 = 75.However, a good number of the candidates while calculating the 3rd quartile obtained the position of the third

quartile i.e. 3rd quartile = x 24 = 18th position but failed to obtain the third quartile which was 3.


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Question 4(a)Out of 30 candidates applying for a post, 17 have degrees, 15 diplomas and 4 neither degree nor diploma. How

many of them have both?

(b) In triangle PQR, M and N are points on the sides PQ and PR respectively such that MN is parallel to QR. If


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Question 5

In the diagram, ABCDEF is a triangular prism.


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Question 6(a)If log5 = 0.6990, log7 = 0.8451 and log8 = 0.9031, evaluate:

log( ).

(b) For a musical show, x children were present. There were 60 more adults than children. An adult paid D5 and achild D2. If a total of D1280 was collected, calculate the

(i) value of x;

(ii) ratio of the number of children to the number of adults;

(iii) average amount paid per person;

(iv) percentage profit if the organizers spent D720 on the show._____________________________________________________________________________________________

________ObservationCandidates performance in part (a) of the question was said to be satisfactory. They exhibited a good understandingof the theories of logarithms and were able to apply them correctly. They were able to express the figures in terms

of 5, 7 and 8 thus: log ( ) = log( ) which on applying the laws of logarithms gives 4log 7 = 4 x0.8451 = 3.3804.

In the part (b), candidates performance was reportedly fair. A good number of them were able to get the equation

2x + 5(x + 60) = 1280. Solving it gave x = 140. The ratio was given by = 7:10. Average cost = = D3.76. The

percentage profit = x 100 = 77.8%.

Question 7(a) A woman looking out from the window of a building at a height of 30m, observed that the angle of depression of

the top of a flag pole was 44. If the foot of the pole is 25m from the foot of the building and on the samehorizontal ground, find, correct to the nearest whole number, the

(i) angle of depression of the foot of the pole from the woman;

(ii)height of the flag pole.

(b)


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In the diagram, O is the centre of the circle,


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Question 9Simplify x + 2 - x + 3

x - 2 x - 1

(a)The graph of the equation y = A x2 + B x + C passes through the points (0, 0), (1, 4) and (2, 10). Find the:(i) value of C;

(ii) values of A and B;

(iii) co-ordinates of the other point where the graph cuts the x-axis.

_____________________________________________________________________________________________________Observation

The part (a) of this question attracted majority of the candidates. However, many of them were unable to get theL.C.M. A few others cross multiplied both sides of the minus sign. They were expected to use (x - 2)(x - 1) as L.C.Mand obtained

(x+2(x-1) (x+3)(x-2) = 4(x - 2) (x 1) (x - 2)(x - 1)

In part (b), candidates performance was fair. However, majority of the candidates appeared not to be familiar with

coordinate of points on a graph. To find point C, candidates were expected to use x = 0 and y = 0 to get c =0. Similarly, substituting 1 for x and 4 for y as well as 2 for x and 10 for y gives two equations in A and B i.e. A + B =4; 4A + 2B = 10. Solving these equations simultaneously gave A = 1, B = 3, which gave the equation as y = x2 + 3x.

To get the 2nd point of intersection with the x axis, put y = 0, in the equation to get x2 + 3x = 0. Solving thisequation gave x = -3, Thus the required point = (-3, 0)

Question 10(a)Using ruler and a pair of compasses only, construct:

(i) quadrilateral PQRS such that /PQ/ = 10cm, /QR/ = 8cm,/PS/ = 6cm, PQR = 60 and QPS = 75;

(ii) the locus 11 of points equidistant from QR and RS;(iii) locus 12 of points equidistant from R and S.

(b) Measure /RS/_____________________________________________________________________________________________

______

ObservationAs usual, majority of the candidates did not attempt this question. However, the few who attempted the questionperformed well. They were able to construct the required angles and line segments.


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Question 11(a) A circle is inscribed in a square. If the sum of the perimeter of the square and the circumference of the circle is

100 cm, calculate the radius of the circle. [Take p = ]

(b) A rope 60cm long is made to form a rectangle. If the length is 4 times its breadth, calculate, correct to onedecimal place, the

(i) length;

(ii) diagonal

of the rectangle.

_____________________________________________________________________________________________________

ObservationIn part (a), majority of the candidates could not translate the question into a meaningful diagram. Others who wereable to draw it took the length of the square to be r instead of 2r hence they lost some marks. Perimeter of the

square = 8r, circumference of the circle = 2pr. Thus 2pr + 8r = 100 i.e. + 8r = 100 r = 7cm.

It was reported that majority of the candidates did not attempt the part (b). Some of them who attempted it couldnot manipulate the algebraic expressions involved. Since the length was said to be four times the breadth, it meansthat if the breadth = b, then length (L) = 4b. The perimeter was given as 60cm.Therefore, perimeter= 2(L + b) = 2(4b+b) = 10b = 60. Solving this gave b = 6cm. Hence, length (L) = 6 x 4 = 24cm. Diagonal = 242+62) = 24.7cm.


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Question 12(a)Copy and complete the table of values for y = sin x + 2 cos x, correct to one decimal place.

X 0 30 60 90 120 150 180 210 240

Y 2.2 -1.2 -2.0 -1.9

(b) Using a scale of 2 cm to 30 on the x axis and 2 cm to 0.5 units

on the y-axis, draw the graph of y = sin x + 2cos x for 0 x 240 (c) Use your graph to solve the equation:

(i) sin x + 2cos x = 0;(ii) sin x = 2.1 2cos x.

(d) From the graph, find y when x = 171._____________________________________________________________________________________________

________

ObservationThis was another question where candidates performance was reported to be very poor. Majority of them couldnot complete the table of values correctly. The few who drew the graph correctly could not read from it. More

ork should be done by teachers on this area of the syllabus. The expected the table is as shown:

X 0 30 66 90 120 130 180 210 240

Y 2.0 2.2 1.9 1.0 -0.1 -1.2 -2.0 -2.2 -1.9

They were also expected to use the given scale and the table to draw the graph, locate the line y = 2.1 and use it tosolve the equation, sin x = 2.1 2cos x, to obtain x = 9, 45. When x = 171, y = 1.75.

Question 13(a)How many numbers between 75 and 500 are divisible by 7?

(b) The 8th term of an Arithmetic progression (A.P.) is 5 times the third term while the 7th term is 9 greater thanthe 4th term. Write the first five terms of the A.P.

_____________________________________________________________________________________________________

ObservationThis question posed quite some problems for a good number of candidates who attempted it. In the part (a), somecandidates could not recognise the problem, as an A.P. problem. A good number of them counted these numbersout and thereby wasted time. They were expected to recognize the first number divisible by 7 as the first term ofthe A.P. (i.e. 77) and the last number (497) as the last term and substitute these values in the formula Tn = a+ (n-1)d (where Tn = value of the nth term, a = first term, d = common difference and n = required term) i.e. a = 77, d =7, Tn = 497 and after a little computation yields n, the required number to be equal to 61.In part (b) some candidates were able to compare the 3rd term with the 8th term to obtain the equation a + 7d = 5(a+ 2d) but could not get the second equation a + 6d = a + 3d + 9. Some others were able to obtain d = 3 by solvingthe 2nd equation correctly but could not obtain the value of a nor the required 5 terms. From the second equation,bringing the like terms together gave 3d = 9. Hence, d = 3.Similarly, from the first equation, 4a = - 3d. Substituting3 for d in the equation gave 4a = -3(3) = -9. Therefore, a = - 9/4. The first .five terms of the AP were -9/4, , 15/4,

22/4, 39/4.


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General Mathematics Paper 2, May/June 2010




Question1A = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x: 3


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Question 3(a) In the diagram, L PQR = 125, LQRS = r, LRST = 800 and LSTU = 44. Calculate the value of r

b) .

In the diagram, TS is a tangent to the circle at A.ABI ICE, LAEC = sx", LADB = 60 and LTAE = xo. Find the value ofx",

_____________________________________________________________________________________________________

ObservationPart (a) of this question required that candidates drew straight lines, one each passing through points Sand R and

parallel to PQ and UT as shown in the diagram below but the report stated that majority of the candidates didnot.From the diagram, LMQR = LQRN = 180 -125 = 55 ( alternate angles), LVST = 44 (alternate angle to LSTU).

Therefore LNRS = LVSR = 80 - 44 = 360Hence, r = 55 + 36 = 91

In part (b), candidates' performance was reported to be worse than part (a). Candidates were reported to haveexhibited poor understanding of circle theorems. Teachers were encouraged to do a lot of work in this area.

From the diagram, LBDA = LBAS = 600 (angles in the alternate segment). LBAE = 180 - 5x (adjacent angles on atransversal). Therefore LBAS + LBAE + LEAT = 60 + 180 - 5x + x = 180 (angles on a straight line). Solving this simple

equation gave x = 150. Candidates' performance this question was described as poor.


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Question 4

The diagram shows a cone with slant height 10.5 cm. If the curved surface area of the cone is 115.5 cm;, calculate,correct to 3 significant figures, the:

(a)base radius, r;(b)height, h;

(c)volume of the cone. [Taken= 22]7

_____________________________________________________________________________________________________

ObservationThis 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 ofthem obtained full marks. The curved surface area,

iu! = 115.5cm, slant height,l= 10.scm, n = ll, hence r = 115.57nl

= 11.5 x 7 = 3.S0cm. r2 + h2 = (10.5}2 => h = .J(10.5)2 - (3.5)2 = 9.90cm22 x 10.5

Volume of cone = .1 x 22 x 3.5 x 9.899 == 127 cm3 7 1

A good number of the candidates were however reported to have lost some marks, especially the A marks, becausethey did not give their answers in 3 significant figures as required by the question.

Question 5Two fair dice are thrown.

M is the event described by "the sum of the scores is 10" andNis 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 Nmutually exclusive? Give reasons._____________________________________________________________________________________________

________

ObservationThis question was reported to be popular and well attempted by majority of the candidates. However, a few of themdid not list the elements of M and N as required but drew the sample space when 2 dice are thrown. Some of themho listed these elements did not do so completely. Majority of them were reported not to be able to state correctlythe 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 theelements of N, some candidates would list (1,4) but not (4,1).


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Question 6(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]

_____________________________________________________________________________________________________Observation

Candidates' performance in part (a) was reported to be poor. A good number of them were reported not to haveattempted 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,OOOcm285km2 = 850,000,000,000 cm. 20,000 em on the ground = 1 cm on the map, hence 400,000,000 cm on the ground

= 1 cm2850,000,000,000 = 2125 cm2 on themap. 400,000,000

In part (b), candidates' performance was said to be better than it was in part (a). However, many candidates did notdraw 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 as360 - 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 cm2360 7 2 2

This meant that 360 - x = 360 x 2 x 27.5 from where we obtain x = 103 to the nearest degree77


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Question 7(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]

_____________________________________________________________________________________________________Observation

Candidates' performance in part (a) was reported to be poor. A good number of them were reported not to haveattempted 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,OOOcm285km2 = 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 themap. 400,000,000

In part (b), candidates' performance was said to be better than it was in part (a). However, many candidates did notdraw 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 as360 - 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 cm2360 7 2 2

This meant that 360 - x = 360 x 2 x 27.5 from where we obtain x = 103 to the nearest degree77


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Question 8Using 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 ofLRSPand LSPQto meet atX;

(iii) The perpendicularXTto meet PS at T.

(b) Measure IXT/'

_____________________________________________________________________________________________________Observation

This question on geometrical construction was reported to be quite unpopular among the candidates. Very few ofthem 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_____________________________________________________________________________________________

________ObservationThis 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 wronganswers 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) = 13Sin10013 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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Question 10(a) Copy and complete the table of values for the relation V = -X +X+ 2 for -3 x 3.

(b) Using scales of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the v-axis, draw a graph of the relationy = -X + X + 2.

(c) From the graph, find the:(i)Minimum value of y;

(ii)Roots of the equation X -x-2 = 0;(iii)Gradient of the curve at x = -0.5.

_____________________________________________________________________________________________________Observation

This question was reported to have been attempted by majority of the candidates who were able to complete thetable of values and plotted the graph correctly. However, majority ofthe candidates were reported not to draw the

tangent to the curve as required. Hence they were unable to determine the gradient of the curve atx = -0.5.

Question 11In the diagram, L.PTQ= L.PSR = 900,/PQ/ = 10 ern,/PS/ = 14.4 cm and/TQ/ = 6 cm.

Calculate the area of quadrilateral QRST.(b) Two opposite sides of a square are each decreased by 10% while the other two are each increased by 15% to

form a rectangle. Find the ratio of the area of the rectangle to that of the square._____________________________________________________________________________________________

________

ObservationThis question was reported not to be very popular and candidates' performance was described as not satisfactory.

Majority of the candidates were reported not to apply the concept of similar triangles correctly. Others did notrecognize the quadrilateral as a trapezium and so failed to use the correct formula when finding its area. Part (b) of

the question was reported not to have been done satisfactorily either. Candidates were expected to show that:/PT/ = ..../102 - 62 = 8 cm. m: =!J2L i.e ~ = 14.4.Hence, /SR/ = 10.8 cm.

/TO/ /SR/ 6 /SR/Area of quadrilateral QRST = Yz (6 + 10.8) x 6.4 = 53.76 cm2. Some candidates were reported to have subtractedthe area of triangle PQT from triangle PRS. This was also in order. In part (b) if the side of the square was y, then

new breadth = 90 x y= 0.9y.100

New length = 115 x Y = 1.15y. New area = 1.15yx 0.9y= 1.035/.100

Hence, ratio = 1.035y : y = 1.035 : 1 or 207:200 .


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Question 12The frequency distribution of the weight of 100 participants in a high jump competition is as

shown below:

Weight (kg) 20-29 30 - 39 40 - 49 SO - 59 60 - 69 70-79Number of

pa rtici pa nts 10 18 22 25 16 9

(a) Construct the cumulative frequency table.(b) Draw the cumulative frequency curve.

(c) From the curve, estimate the:(i) median;

(ii) semi-interquartile range;(iii) probability that a participant chosen at random weighs at least 60 kg.

_____________________________________________________________________________________________________

ObservationThis question was reported to be attempted by majority of the candidates. According to the report, candidates'

performance in this question was fair. Majority of them were reported not to have read from their Ogives correctly.Others did not draw the Ogive using class boundaries.

The median was 49.5 while the first quartile(Ql) was 37.8 and the third Quartile (~)59.5.Hence, the semi-interquartile range Q1-Q1 = 59.5 - 32.8 = 10.85 1.

2 2Number of participants who weighed at least 60kg = 25. Therefore probability of choosing a

participant who weighed at least 60kg = 25= -.1 .100 4

Question 13(a)The third term of a Geometric Progression (G.P) is 24 and its seventh term is 4(20/27) .Find Its irst term.

(b)Given that y varies directly as x and inversely as the square of z. Ify= 4, whenx= 3 and z = 1, find y whenx= 3and z = 2.

_____________________________________________________________________________________________________Observation

This question was reported to have been attempted by majority of the candidates and their performance wascommended. However in part (a), a few of them divided the indices instead of subtracting them. A few others were

not able to manipulate the fraction involved. Part (b) was also well attempted and majority of them obtained fullmarks. The expected responses were as follows:


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General Mathematics Paper 2, Nov/Dec 2009




Question1Simplify, without using tables or calculator

1. 31/3 11/4 4/9;2. 2 + 96 - 4 (6 -1)2 and express your answer in the form m + n6 where

m and n are real numbers._____________________________________________________________________________________________

________

ObservationThe part (a) of the question required the use of BODMAS rule, finding the LCM and simplifying. Most candidateswere reported to have followed this procedure and were able to manipulate the fractions well. However, some

candidates misinterpreted to mean + . Others subtracted first before dividing. 31/3 11/4 4/9 = 10/3 x 4/5 4/9 = 8/3 4/9 = 20/9 = 22/9

In part (b), many candidates reportedly found the expansion of (6 -1)2quite challenging. Others could not reduce 96 to its lowest form. The expansion of (6 -1)2 = (6 -1) (6 -1) = 6 -

6 -6 + 1 = 7 - 26. 96 = 6 x 16 = 46Therefore 2 + 96 4(6 -1)2 = 2 + 46 4(7-26) = 2 + 46 28 + 86 which after a little computation resulted in -

26 + 126.

Question2(a) Solve the equation 4x-1 - 3x-1 = 5-2x

3 2 4(b) From a shop, Kofi bought 2 singlets and 3 shirts for GH31.00 while Kwasi bought 3 singlets and 2shirts for GH29.00. How much will Yaw pay for one singlet and one shirt he bought from the same shop?_____________________________________________________________________________________________

________

ObservationAccording to the chief examiner, part (a) of this question was well attempted by majority of the candidates. Theywere able to find the LCM of the denominators and cross multiply both sides of the equality sign by the LCM which

was 12. Simplifying gave 4(x-1) - 6(3x 1) -3(5-2x) which further gave x = 31/4. In part (b), it was reported that thequestion was attempted by majority of the candidates who were able to form the simultaneous equation 2x + 3y =

31; 3x+2y = 29, solve these equation simultaneously to obtain x = 5 and y = 7. However, some of the candidates didnot add these values to obtain the required price which was GH12.00.


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Question3The probability that a malaria patient (M) survives when administered with a newly discovered drug is 0.27and the probability that a thyphoid patient (T) survives when injected with another newly discovered drug is0.85. What is the probability that

either of the two patients survives?neither of two patients survive?

at least one of the two patients survive?

Give your answers correct to 2 significant figures._____________________________________________________________________________________________

________

ObservationIt was reported that the performance of candidates in this question was not commendable. They found it difficult;hence they scored very low marks. They were expected to show that if P(M) = probability of the malaria patient

surviving = 0.27, probability of not surviving, P(M) = 1- 0.27 = 0.73. Similarly, if P(T) = probability of the thypoidpatient surviving = 0.85, then probability of not surviving = P(T1) = 1-P(T) = 1- 0.85 = 0.15. Probability of either

surviving = (0.27 x 0.15) + (0.85 x 0.73) = 0.66. Probability of neither surviving =

(0.73 x 0.15) = 0.11. Probability of at least one surviving = 1- 0.11 = 0.89.


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Question5(a)

In the diagram, the two circles intersect at X and Y. The centre of the smaller circle is on the circumferenceof the bigger circle. A and B are any two points on the major arcs, one on each circle. Find an equationconnecting a and b.

(b)

In the diagram


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Question6(a) By how much is 110002 greater than or les than 1112 x 112?(b) A shopkeeper has in stock 20 televisions set percent. He sells 18 of them at a profit of 15 percent andthe remaining two at a loss of 5 percentage. Find his profit on the 20 sets._____________________________________________________________________________________________

________

ObservationThe report also stated that although majority of the candidates attempted part (a) of the question and performedwell in it, it was observed that majority of them converted first to base ten, solved it and converted back to base

2. Very few candidates were reported to have worked in base 2. 1112 = 112 = 101012.110002 101012 = 112.

In part (b), majority of the candidates were reported to have found the question quite challenging. They wereexpected to show that if p is the cost price of a television set, then the total sales for 18 sets = 115 x 18p =

20.7p. Similarly, total 100 sales for 2 sets = 95/100 x 2p = 1.9p. Therefore total sales on the 20 sets x 20.7p + 1.9p= 22.6p. Total cost price on the 20 sets = 20p. Hence, percentage profit

= 22.6p 20p x 100 = 13%.20p 1


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Question8The area of a rectangular football field is 7200m2 while its perimeter is 360m. calculate the:

1. dimensions of the field;2. cost of clearing the field at N6.50 per square meter, leaving a margin of 2m wide along the longer

sides;3. percentage of the part not cleared.

_____________________________________________________________________________________________________Observation

It was reported that majority of the candidates who attempted this question were able to derive the two equationsi.e. if x = length of field and y = breadth, then

xy = 7200; x + y = 180. They were able to solve these equations and also obtained the dimensions as 60m and120m. However, attempt on the parts (b) and (c) was poorly handed by some of the candidates. Many took the

margin to be all round the field but the question specified along the longer sides. Thus cost of clearing the field =N6.50 x 56 x 120 = N43,680.00. The percentage of the part not cleared

= 7200 (120 x 56) x 100 = 62/3 or 6.67%7200

Candidates, who sketched the diagram, were reported to have performed well in the question.

Question9In the diagram, a ladder LN 10m long, rests on a wall 4.5m high such that 2.5m of it projects beyond the wall.

1. Calculate, correct to one decimal place, the angle which the ladder makes with the ground.2. How high above the ground is the upper end of the ladder?

If the foot of the ladder is moved by 2m further away from the wall, calculate, correct to the nearest degree,the angle which the ladder makes with the ground

_____________________________________________________________________________________________________Observation

Candidates attempt on this question was fair especially parts (a) and (b). Many of them missed the (c) part becausethey did not quite understand what would happen as the ladder was moved by 2m further away from the wall. If the

foot of ladder is moved a further 2m away from the wall, LP becomes (6 + 2)m = 8m, tanx = 4.5/8 and from thetables, x = 29, to the nearest degree


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Question10In a class of 200 students, 70 offered Physics, 90 Chemistry, 100 Mathematics while 24 did not offer any ofthe three subjects. Twenty three (23) students offered Physics and Chemistry, 41 Chemistry and Mathematics

hile 8 offered all three subjects.

Draw a Venn diagram to illustrate the information.Find the probability that a student selected at random from the class offered:o Physics only;o Exactly two of the subjects.

_____________________________________________________________________________________________________Observation

This question was popular and the candidates did well in it. Though many were able to draw the correct Venndiagram, they could not determine the correct number of students who offered physics and those who offered

exactly two of the subjects. Hence, they could not find the required probabilities. They were expected to denotethe number of those who offered Physics and Chemistry only by a variable say x, equate the entries to 200 to get

100 + x + 49 x + 47 x + 24 = 200, from where x = 20. Number who offered Physics only = 47 20 = 27. Thereforeprobability of those who offered Physics only = 27/200.

Number of those who offered exactly 2 subjects = 15 + 20 + 33 = 68. Probability of those who offered exactly 2

subjects = 68/200 = 17/50

Question11Using ruler and a pair of compasses only, construct a quadrilateral PQRS in which /QR/ = 6cm,


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Question12

In the diagram, /PR/ = /RQ/, /RS/ = 10cm

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