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NOVEMBER/DECEMBER 2010 WASSCECHIEF EXAMINER'S REPORT GENERAL MATHEMATICS 2 GENERAL COMMENTS ON THE PAPER AS A WHOLE The Chief Examiner reported that the standard and outlay of the paper compared favourably with those of the previous years. According to the report, the wordings were clear, unambiguous and covered a wide range of topics. All the questions were drawn from within the ambit of the syllabus and they tested the general knowledge of quite a number of concepts and definitions as well as candidates' ability to apply these basic concepts and principles in solving problems. The marking scheme was reportedly appropriately drawn, generous in the allocation of marks to correct points and was flexible enough to adequately cover expected alternatives. Candidates' performance was reported to be comparable with those of the previous years. The report further stated that a few centres recorded good performance by the candidates as well as reduced rate of examination malpractice. However, most candidates failed in the application of basic concepts and principles. The overall performance of the candidates was generally reported not to be different from those of the previous years. SUMMARY OF CANDIDATES' STRENGTHS Candidates' performance was commended in the following areas: 1. Number bases 2. Solving linear inequalities 3. Venn diagrams 4. Completing tables of values and drawing of graphs 5. Probability 282 . 6. did not write answers correct to the required degree of accuracies.Candidates should read the instructions carefully before answering the questions so as to know the demands of the question. majority of them were not able to find the vertical distance between the center of the earth and the center of the small circle on which P and Q lie. 2. Others could not interpret given problems into mathematical statements and solve them correctly. Although the question on Longitude and Latitude was reportedly well attempted by majority of the candidates.. 7. Some of them also. 3. 4. Majority of the candidates were also reported not to be able to recall the properties of a Rhombus. · SUGGESTED REMEDIES 1. Arithmetic Progression 5.Candidates are encourage to adequately cover the syllabus and solve a lot of practice problems while preparing for the examination.Solving problems on mensuration in three dimensions. Representing given information on diagrams. 283 . 2.Statistics: many candidates did not draw the cumulative frequency curve using the upper class boundaries.Everyday Arithmetic as in questions 6b and llb 3.Geometry: Questions on this topic were reported to be avoided by majority of the candidates and the few who attempted them performed poorly . Reading and drawing inferences from graphs 7. SUMMARY OF CANDIDATES' WEAKNESSES Many candidates were reported not to adhere to the rubrics of the paper. Other areas where candidates' weaknesses were observed are: 1.Candidates are encouraged to draw necessary diagrams that would help them in solving problems in areas requiring diagrams. Candidates are encouraged to avail themselves of past WASSCE General mathematics question papers while preparing for the examination D E TA IL E D C O M M E NON EA C H Q U E STIO N TS QUESTION 1 a) Two children shared an amount of money in the ratio 3 If the smaller share was b) GH¢ 25. the probability of both red = 2. what is the probability that both balls are red? Majority of the candidates were able to find the sum of the ratios correctly but forming the equation to find the amount shared posed a problem to some of them. Therefore. Thus. 6. how much was shared between them? 4 5 (b) A box contains 5 red.and the second being red as ~.88 B In part (b) Majority of the candidates performed well in this question however.4. Hence sum of ratio is 23. a few of them interpreted the question as sum of probability instead of product of probability. If the amount shared between them was y. 3 green and 4 blue balls of the same size. They were able to obtain the probability of the first being red as 2. mathematical and statistical tables published for WAEC. Some others were not able to identify the smaller ratio. If a boy picks two balls from the box one after the other without replacement.Teachers are encouraged to adopt teaching techniques that would demystify the subject 7.00. Y = GH¢ . 5. Teachers are encouraged to emphasize these areas of weaknesses during instruction.00.x 25 = GH¢ 71.Candidates are encouraged to familiarize themselves with/acquire necessary materials such as mathematical sets.x1~1= 2-. Candidates were expected to convert the fraction to whole numbers by multiplying by their LCM which was 20 to obtain the ratio 15 :8. Working with fractions also posed some problems for some of the candidates. then -y B 23 23 = GH¢ 25. 2 1 1 2 1 1 33 284 . for use while preparing for the examinations and during the examination. etc. Candidates were expected to show that if the price of the apple before the increase was N x. (b) When the price of an apple increased by N 5. The new price = N (15 + 5) = N 20.00. 18(x + 5) . It was also reported that many candidates did not recall that the sign of the inequality has to change when multiplying or dividing with a negative number. They were also able to represent this solution on the number line. (ii) Illustrate the solution on a number line.00. correct to three significant figures.~ ex + 2) ::. Some candidates could not clear the fractions correctly while others changed the sign of inequality to equality. 285 . sin x = .3. obtained x ? -2.00. • ~I I I I -3 -2 -1 0 II 12 I• 3 In part (b). then the price after the increase became N (x + 5).00 more than 20 apples cost before the increase. Find the new price of an apple. majority of the candidates were reported to have performed poorly in this question. (ii) Calculate. Therefore. Majority of the candidates reportedly performed well in this question. QUESTION 3 (a) In a right angled triangle.Evaluate 5(cos x? .20x = N 60. 1 + x. Hence they could not solve them correctly. (b) The angle of elevation of the top of a vertical pole from a point 63m east of the base of the pole is 30°.. Solving this equation gave the value of N x as N 15. the angle of elevation of the top is 60° (i) Draw a sketch diagram to illustrate the information. the distance of the second point from the base of the pole. The report stated that majority of the candidates were not able to translate the given word problem into equations. However.00f 18 apples cost N 60. They were able to clear the fractions by multiplying the inequality by the LCM which was 6 and after bringing like terms together and slrnplifylnq. From another point due west of the pole.QUESTION 2 (a) (i) Solve the inequality: i x . If LDAC= 27°.14five = 2X five • (b) The diagram is a circle passing through the points A. 286 . Cand Dsuch that ACand BD meet at a point Einside the circle. In part (b). (ii) LAEB.Part (a) of this question was well attempted by majority of the candidates. Teachers are encouraged to emphasize the need for diagrams during instruction. find: to LCAB. {fB/ = xtan 60° = 63tan 30 • This implied that x = 63tan30 = 21. They were able to apply the Pythagoras theorem correctly to find the length of the third side of the triangle final answer as ~. LABD= 54° and LACB= 63°. the report stated that majority of the candidates could not interpret the question correctly which showed poor understanding of the topic.0 m tan 60 QUESTION 4 (a) Find the value of X if X3five . B. They were unable to draw the diagram correctly and this also affected their being able to answer the question correctly. Substituting this value of cos x in the given expression and simplifying gave the L x B 63 m V M From the diagram. T =4 and cos x as ~. . They wrote -5 + 4 instead of -(5 + 4) when converting -14five to base 10.LABD.432m3• Volume of cubical block (0. 1800 . Converting the given expression to base ten gave 5X + 3 .Part (a) of this question was reported to be very popular among the candidates and majority of them performed very well in it.270 ./72+ 4.LBCA(sum of angles of a triangle). b = breadth (width) and h = height.432 + 0. However. Majority of the candidates could not recall the appropriate theorems that would have enabled them to solve the question correctly.557 m = 0. 1.Therefore.4) = = = = = 0.4m. Volume of the tank before the block was lowered into it = (1. correct to two Significant figures.557m3• New height of water 0.52 = 8.LDBC.5m)3 0. = = Part (b) was reported to be poorly attempted by majority of the candidates.(5 + 4) 10 + X. given by 4 x length of one side = 4 x 8. Simplifying this given expression gave X 4. majority of the candidates who attempted this question performed well in it however.9 x 0. Many candidates demonstrated poor knowledge of these properties. i.9 287 . many of them who did not convert the dimensions to the same unit did not get the full marks for this question.630 = 360• Angle AEB = 270 + 630 = 900• QUESTION 5 (a) The diagonals of a rhombus are 14em and gem.0.2 XO. the perimeter of the rhombus. as x = . Difference in height = (0.id i s a c o where I = length. (b) The cross section of a rectangular tank measures l.125)m3 = 0.5157 . If a cubical block of side 50em is lowered into the tank. Part(a) of the question required the knowledge of the properties of the rhombus.4)m3 = 0.125m3• Therefore total volume (0. Candidates were expected to divide each diagonal by 2 and apply the Pythagoras theorem to obtain the length of the rhombus. In part (b). Calculate.540 .2 x 0.5157m.e. It contains water to a depth of O.9m. It has equal sides and the diagonals intersect at right angles to each other . Candidates were expected to recall that LDBC= 270 (angles in the same segment as LCAD. the rise in the water level (in metres).. it was observed that some candidates failed to subtract correctly.12m..2m by 0. correct to the nearest centimetre. calculate.32 = 33 cm to the nearest centimetre.32 cm. x. correct to 2 significant figures. Perimeter of the rhombus is . Candidates were expected to recall that volume of a cube = 13 w h i l e t h a t o f I x buXb h. LCAB= 1800 . QUESTION 6 (a) In a class of 50 students. Hence the cost price = ~ ~ ':00 x100 = 125% 92 which gave N 720.00. They were expected to show that if 8% discount was given.» .15 = 15. find the (i) cost price of the article. the marked price = N 828 x100 = N 900. 15 offered History and Geography while 3 did not offer any of the two subjects. If the article was initially marked to gain 25%. Their responses were reported to indicate candidates' poor understanding of transactions involving discounts. (ii) Find the number of candidates that offered: (A) History only.00 is equal to 92% of the marked price. They were able to draw the required Venn diagram correctly and were also able to determine the number who offered History only and Geography only correctly.00.3 = 17 In part (b). = 50 . (i) Represent the information on a Venn diagram.. candidates' performance was reported to be poor. Candidates were expected to draw the Venn diagram thus: n(U) = 50 H = 3 . then N 828. n(H) only n(G only) = 30 . Therefore. 125 ._ _ ~ From the diagram. (b) A trader sold an article at a discount of 8% for N 828.00. (B) Geography only. (ii) discount allowed. The discount allowed = N (900 . Since it was initially marked to gain 25%.30 .00.828) = N 72. Part (a) was reported to be well attempted by majority of the candidates. 30 offered History. it implied that N 900 = 125% of the cost price. 288 . They were expected to complete the table as follows: The range of values of x for which y is negative can be gotten from the graph by noting the x-coordinates of the part of the graph that fell below the x-axls. reading and drawing inferences from the graph posed some problems to candidates. (b) (i) construct I. draw the graph of the relation y 2 = ~x(x . the range was < x < 6. The minimum value obtained from the graph was -4.35 and 7. 2 The Chief Examiner reported that this question was attempted by majority of the candidates. (ii) minimum value of y. and 2cm to 2 units on the y-axis. (ii) locate M.6) for -2 ~ x ~ 8. 2 (b) Using scales of 2cm to 1 unit on the x . (iii) roots of the equation ~x(x .QUESTION 7 (a) Copy and complete the following table for the relation y = ~x(x . In this case.5 while the roots of the equation was x = -1.6) = 5. o QUESTION 8 Using ruler and a pair of compasses only: (a) construct a triangle PQRwith /PQ/ = lOcm. LQPR = 90° and LPQR = 30°. the locus of all points equidistant from PR and QR. 289 . Finding the roots of the equation V2 x(x . While only a few candidates were reported not to have plotted the points correctly.axis. the point where / intersects with p o. majority of them were reported not to have determined the required range of values correctly.6) for -2 ~ x ~ 8.6) = 5 also posed some problems to these candidates. (c) Use the graph to find the: (i) range of values of x for which y is negative. Majority of those who attempted it were reported to have completed the table of values and drew the graph correctly.35. However. (a) calculate. (ii) calculate the area of the circler correct to one decimal place. They were reported to have constructed the triangle correctly.Its If IABI ::: 6em. I pyramid with a rectangular base vertical height is OG. 290 . Very few candidates attempted it. majority of them were unable to obtain the required locus hence could not go further. (iii) angle between the triangle OAB and the base ABCD. Generally. However.(c) (i) with M as centre and radius MP. They could not calculate angles between edges and faces in three dimensional problems correctly. it was observed that candidates performed better in part (a)(i) and (b) of this question than in part (a)(ii) and (iii). (b) find the volume of the pyramid. BC/ = 8em and each slant edge is 13em. there appeared to be an improvement in the performance of the candidates. However. 7 22 The reported stated that although this question was quite unpopular among the candidates. QUEsnON9 ~----~--------~C A OABCDis a right ABCD. this question was quite unpopular among the candidates. [Take 1 C = . correct to one decimal placer the (i) vertical height IOGI. candidates' performance was reportedly very poor. (ii) angle between a slant edge and the base ABCD. According to the report. draw a circle.]. Candidates are also encouraged to draw sketches before they began to construct as this will serve as a guide to what you are expected to do. Teachers are encouraged not to relent in their effort at teaching this topic. . Similarly. Then tan a the value of a as 71. then the angle between faces OAB and ABCD 15) = 13 cos(LOAG). IO G /= vl144 = 12cm. However. a good number of the candidates used class mid-points or upper class limit rather than upper class boundaries to draw the curve. Therefore.59 40 60 .29 60 30 -39 72 40 . (c) If the minimum mark for distinction is 75%.4°.1 e = 67. (b) Use your graph to estimate the semi-interquartile range.e.es are drawn using upper class boundaries against the cumulative frequencies. The report also stated that majority of the candidates could not obtain the quartiles from the graph hence could not calculate the semiinterquartile range as required. i. = 2 2 IO G l+ IA G l= IA O2/• This implies that IO G2/ = /A O 2/ . ~ AO = cos. If X is the mid point of AB.From the diagram. Angle between a slant edge and the base ABCD is LOAG which can be obtained as follows: //A G // cos(LOAG).69 25 70 -79 10 80 -89 5 candidates (a) Draw a cumulative frequency curve for the distribution. Candidates should know that cumulative frequency curv. how many candidates passed with distinction? This question was reportedly very popular among the candidates and their performance was described as fair.49 80 50 . LOAG = can be obtained as follows: Let the angle be a. Marks (%) Number of 0-9 20 10 . The volume of the pyramid = !(base area x height) = ! x 6 x 8 x 12 = /O G = 12 .19 48 20 .52 = 144. The inability of candidates to read from their curves was very noticeable 291 ./AGl = 132 ./ /A B P + /B C)P= V 2(. IA G I = lh/A C I V 2(.J6 + 8 ) = 5cm.Simplifying this gave // /GX 4 3 3 = 192cm • 3 QUESTION 10 The table gives the distribution of marks for 360 candidates who sat for an examination... Majority of the candidates were reported to have constructed the cumulative frequency table correctly.6°. Bearing of Q from P = 270° + 66° = 336°. Q 3km H 3km p The distance between the two ships could be obtained using either the sine or the cosine rule. Therefore.420 = 480. According to the report. (il) Find the bearing of ship Q from ship P . They were expected to relate the time-taken during each part of the journey with the total time for the whole journey. (i) Find the distance between the two ships. it was reported that although this question was attempted by majority of the candidates.48 In part (b). Candidates performed better in part (a) than in part (b). -60 = 4hrs. find v.44km. From the diagram.e. he whole i d expecte average speetime taken for t e whole Journey as to lnd t e lime taken when travelling with 75 krn/h = 300 total time taken total distance covered d = 600 = 1 0 h rs. 300 = 6. v = 300 = 50km/h. Recall that LQPH = 66°. candidates' performance was reported to be poor. This implies that LHQP= LQPH= the sine rule. i. Using = 2. Another ship Q is also 3 km from the harbour but on a bearing of 042° from the harbour. only those candidates who sketched the diagram correctly performed well on the question.QUESTION 11 (a) A ship Pis 3km due east of a harbour. sin66° 3 = Sjin48 0 •Therefore. time taken when travelling 75 with v km/h = (10 .4) hrs = 6hrs. In part (a). sm66 = 660. Candidates were . Hence. v 6 29 2 . their apparent difficulty was in recalling the fact that average speed = tota l distance . (b) A motorist travelled 300 km at an average speed of 75 km/h and returned at an average speed of v km/b. PQI /PQ/ = 3~in4~O Therefore. LQHP = 900 . If his average speed for the whole journey is 60 km/h. 180. correct to the nearest km. 360 QUESTION 13 (a) The second. According to the report majority of the candidates did not draw the correct diagram. (b) length of the minor arc PQ. Length of minor arc PQ = -x 2 x . correct to the nearest 10km. long 18°W) and Q(lat 400N. . (ii) first term. Calculate.P. long 78OW) are two cities on the surface of the earth. Teachers are encouraged to stress the need for drawing appropriate diagrams that will aid the candidate in answering such questions.4 = 5100 m to t e nearest .) are x .1. x + 1 and 7 respectively.x 4902. fourth and sixth terms of an Arithmetic Progression (A. [Take tt = 2 72and radius of the earth= 6400 km] This question was reportedly attempted by majority of the candidates and their performance was described as fair. Find the: (i) common difference. I_ -+ ____ ___ ~ I -. Majority of the candidates could not find the vertical distance between the centre of the earth and the centre of the small circle on which P and Q lie. (iii) value of x. (c) vertical distance between the centre of the earth and the centre of the small circle on which P and Q lie.6428 = 4114km to the nearest km. correct to the nearest 100 km. Length of the distance I O X=I R sin400 = 6400 x 0. Candidates were also reported to have performed better in parts (a) and (b) than in part (c).4 km = 4900km to the nearest 10 km. 60 22 7 k h 100 km. the: (a) radius of the parallel of latitude on which P and Q lie.-. Others did not draw the diagram at all.q a r ~ -E u to Radius of the parallel of latitude on which P and Q lie = R cos 40° = 4902.QUESTION 12 P(lat 400N. 293 . then V4 x ~ X 22 X r3 = 6000cm3• Solving this equation gave r = 17. Substituting 2 for a and 1 for d in T2' 3 = x . In part (a).1 = a + d. a +5d = 7. This implied that x . n= the term sought.P. Majority of them took the sphere to be filled when 6 litres of water was poured into it instead of one-quarter filled as stated in the question. Simplifying gave a = 2. This the report attributed to their inability to manipulate the algebraic expressions involved. Calculate.8 cm correct to 3 significant figures. candidates' performance was reported to be poorer than it was in part (a). its diameter.l)d.e.T2 = 2d = [(x + 1) .9 em. In part (b).(x . d = common difference and Tn = value of the term sought. Hence diameter = 2 x r = 35.(b) A spherical bowl of radius r em is one-quarter full when 6 litres of water is poured into it. The Chief Examiner reported that responses were not received from candidates for questions 14 and 15 as these questions were meant for candidates in Ghana. Candidates were expected to recall the nth term of an A. they could not obtain the three equations involved. T4 = a + 3d ~ x + 1 = a + 3d. as Tn = a + (n .1)] = 2. [Take tt = 272]. correct to three significant figures. i. Sierra Leone and The Gambia. T4 . 3 7 294 . This question was also reported to being very popular among the candidates.1~ x = 4. hence they lost some marks. T2 = a + d. Since the sphere was one-quarter full. T 6 = 7(given). where a = first term. Majority of the candidates did not get the required answer mainly because they did not read the question carefully to understand what was required. However majority of the candidates were reported to have performed poorly in the question. Others did not convert the unit from litres to em' before solving. Hence 2d = 2 ~ d = 1. 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