MATHEMATICS (SYLLABUS D)mes.intnet.mu/English/Documents/Examinations... · heights of the two bottles. A good number of candidates, however, gave 1:4 as their answer. In part (b),

Cambridge General Certificate of Education Ordinary Level 4029 Mathematics (Syllabus D) November 2014

Principal Examiner Report for Teachers

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MATHEMATICS (SYLLABUS D)

Paper 4029/01

Paper 1

General comments The standard of the question paper was comparable to that of the papers set in recent years. It provided candidates with the opportunity to demonstrate their ability in mathematics. A good number of the questions were relatively simple and straight-forward. Others proved stimulating to many and were successfully tackled by a handful of candidates. There was no evidence this year that more candidates were rushing to finish the paper than in previous years. Comment on specific questions Question 1 Both parts of this question were generally well-answered. In part (a), a small number of candidates added the cost of one drink only to the cost of the cake. In part (b), most of the mistakes observed were computational in nature. Answers: (a) 256; (b) 2 hours 35 minutes. Question 2 Part (a) was successfully answered by a large number of candidates. However, some candidates correctly obtained 39 minutes but forgot to subsequently add 20 minutes to the 39 minutes. Many candidates also mistook the time to cook 500 g of meat to be equal to 33 minutes (13 + 20) and, therefore, multiplied 33 by 3 to obtain 99 minutes as their answer. Part (b) was not successfully attempted by many candidates. It seemed that many were unsure as to how to

tackle the question. A significant number of candidates obtained the incomplete answer 500

13M.

Answers: (a) 59; (b) 20500

13MT += .

Question 3 Part (a) was not well answered. Quite a large proportion of the candidates encountered problems while dealing with the negative scores. In particular, they could not arrange the scores in ascending order and thus could not find the median. A persistent mistake was to give the rank of the median as answer. Candidates were more successful in part (b). Common mistakes occurred as a result of either mistakes on adding the scores or neglecting to divide the sum of the scores obtained by 10. Answer: (a) –0.5; (b) 0.1.



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A B

C

Question 4 Part (a) was generally well answered by the vast majority of candidates. A few candidates, however, made

computational mistakes while evaluating f (2

1).

Part (b) was also well-answered in general with only a small number of candidates not attempting the question.

Answers: (a) –5; (b) 2

6+x

.

Question 5 Part (a) was successfully answered by a large number of candidates. A few unsuccessful attempts resulted from candidates truncating the value as their answer. A minority of candidates did not seem to appreciate the concept of significant figures and gave their answer as 1234.567.

Part (b) was less well answered than part (a). A few candidates rightly approximated π

28 to 9 and obtained

the correct answer. A common mistake made by others was to approximate 28 to 25 and π to 4 (the nearest square numbers) respectively to get 2.5 as their final answer. Answers: (a) 1200; (b) 3. Question 6 A range of wrong answers were recorded in part (a). Many candidates seemed to have shaded regions of the Venn diagram without reference to the inequalities. Candidates’ performance in part (b) was no better than in part (a). The majority of candidates could not understand what set Q was and, consequently, could not find the number of elements in Q. Some candidates correctly listed the elements of Q and provided the list as their answer, overlooking the fact that they were required to give the number of elements in Q instead. Answers: (a) (b) 3. Question 7 The vast majority of candidates correctly converted 90 km to 90 000 metres and 1 hour to 3600 seconds respectively but a fair proportion of them subsequently made mistakes in dividing 90 000 by 3600. A few others divided 3600 by 90 000 instead. Answer: 25 Question 8 This question was successfully attempted by the higher scoring candidates. In part (a), they correctly reasoned that they would need to take the square roots of the ratio of the areas to obtain the ratio of the heights of the two bottles. A good number of candidates, however, gave 1:4 as their answer. In part (b), many of the candidates who did not score in part (a) earned credit for recognising that they had to raise the answer they obtained in part (a) to the power of 3. Answers: (a) 1:2; (b) 1:8.



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Question 9 This question was fairly well attempted by candidates. In part (a), a frequent wrong answer noted was 54.3. Another common mistake was to obtain 53.8 upon subtracting 0.5 from 54.3. Candidates were more successful in attempting part (b) as they rightly recognised they had to divide d + 0.5

by their answer to part (a). A good number of candidates, however, gave 54.25

0.5 -d as their answer neglecting

the fact that the average speed was directly proportional to the distance run. A handful of candidates also

gave 54.25

.5d as answer to part (b).

Answers: (a) 54.25; (b) 54.25

0.5+d.

Question 10 This was a generally well-answered question in which a very large number of correct answers were recorded. Occasionally, candidates made computational mistakes but used a correct method. A small number of candidates overlooked the term ‘inversely’ and worked out the question by taking y to be directly proportional to x. Answer: 12 Question 11 This was a question that was successfully attempted by many candidates although a significant number of them found part (c) of the question quite demanding. In general, candidates gave an expression in n without verifying if it fitted the data given. Some candidates did not attempt the question.

Answers: (a) 1; (b) 41, 40, 81; (c) ( )212 +n .

Question 12 This question was well answered by the vast majority of candidates. The main mistake noted in part (a) was to write 10

4 instead of 10

–4.

In part (b), a large number of candidates carried out the division properly but a few of them did not give their answer in standard form. Many also wrote 0.6 × 10

1 which they obtained by adding the indices (–5 + 6)

rather than subtracting them (–5 – 6).

Answers: (a) 4

105.67−

× ; (b) 12

106−

× .

Question 13 Many candidates fared well in part (a). Some candidates, however, obtained 140% from which they

subsequently subtracted 100. Less successful responses evaluated 1007

5× .

Very few candidates used the answer which they obtained in part (a) to answer part (b). Among those who did so, many managed to find that AD = 4.2 cm but some forgot to subtract 3 from AD to find the length of ED. Most of the candidates who attempted the question used the properties of similar triangles and arrived at

AD = 5

21 cm. A very large proportion then evaluated

5

21 as 4.5 cm instead of 4.2 cm. A few unsuccessful

attempts stemmed from using the ratio of the sides of the two triangles which were not corresponding sides. Answers: (a) 140; (b) 1.2 .



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Question 14 This question proved to be challenging to the vast majority of candidates. Very few candidates realised that the radius of the sector was the slant height of the cone. Most of the candidates thus took the value of the radius of the base of the cone to be 6 cm. A minority took the value of r to be equal to the vertical height of the cone, which they calculated as being equal to 8 cm by applying Pythagoras’ theorem. Part (b) was also not successfully attempted. Very few candidates equated the length of the arc to the circumference of the base of the cone, or the area of the sector to the area of the base of the cone. Answers: (a) 10; (b) 216. Question 15 This proved to be another challenging question for candidates. In part (a), most knew how to apply the formula to calculate the volume of a sphere. A large number of candidates, however, evaluated 3

3 as 9 and obtained 240 as their answer. Some candidates found the

volume of a single sphere and thus obtained 36 as the value for k.

In part (b), very few candidates recognised that they had to equate ( ) h62

π to π720 to obtain the correct

answer. A large number of candidates started with an assumed value for the height of the water (often 60 cm) with the spheres placed inside the cylinder, found the volume of the water and the spheres using this

value, subtracted π720 from their answer and then proceeded to find the new height of the water. Most

made errors while using this lengthy method but a few arrived at the correct answer. Where candidates started with h as the initial height, the algebra usually included mistakes. Answers: (a) 720; (b) 20. Question 16 This question was not well answered in general. A very large number of candidates did not know how to tackle parts (a) and (b). They arbitrarily wrote any column vector as an answer. A small minority knew the matrices representing the transformations and used them successfully. A few others successfully tried to find the images of AB on the diagram. The overall performance in part (c) was better than in parts (a) and (b) with the majority of the candidates obtaining the correct answer.

Answer: (a)

−

−

3

4 ; (b)

−

−

4

3 ; (c) 5.

Question 17 This was successfully attempted by almost all the candidates. In part (a), a few stumbled on the evaluation of p0.

In part (b), a minority determined the cube root of only one of the two terms given.

Answers: (a) 35−p ; (b)

23x .



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Question 18 This question was relatively easy on the whole. In part (a), a few cases of incomplete factorisation were noted.

In part (b), common mistakes observed were to give the answer as ( )23 cb − or ( )( )cbcb +− 99 .

In part (c), most of the candidates knew they had to group the terms in pairs but many had trouble to work with the negative signs. To counteract this difficulty, some candidates decided to replace the negative signs by positive signs.

Answers: (a) ( )aa 414 − ; (b) ( )( )cbcb +− 33 ; (c) ( )( )yxx −+ 5 .

Question 19 Part (a) was successfully attempted by many. A small number of candidates did not appreciate the concept of rotational symmetry and often gave the answer 1. Few were able to find all the remaining angles of the hexagon in part (b) but many could find 90° as one of the required angles. A good number of candidates were also able to find 150° as another angle of the hexagon but did not realise that there were two such angles. The last pair of angles which was equal to 135° was most difficult to find. Candidates in general made mistakes in using either 360° or 540° as the sum of the interior angles of the hexagon. Even those who used 720° as the sum of the interior angles forgot, in many cases, to subtract the given angle of 60° in their attempt to find the missing angles. A handful of candidates subtracted 60° from 720° and divided the result by 5, giving 132° as the size of all the five remaining angles of the hexagon. Answer: (a) 4 (b) 90, 150, 150, 135, 135 Question 20 Candidates who recalled how to use the angle properties of circles worked out the question without any major difficulty. Others used any angle properties they could remember, appropriate or not, and very often obtained answers which would have been correct to other parts of the question. Answers: (a) 68; (b) 44; (c) 112; (d) 44. Question 21 The majority of the candidates did not seem to have understood the requirements of the question and correct answers to part (a) were rare. Many candidates simply duplicated the second part of the diagram given in the empty space provided in their attempt to complete the tree diagram. In some cases, the probabilities on the pairs of branches did not add up to 1. Correct answers to part (b)(i) were more frequent as candidates had to use the information already provided in the diagram. A significant number of candidates added the probability that the first ball is blue to the probability that the second ball is also blue, instead of multiplying the two probabilities. Part (b)(ii) was less well answered as many candidates omitted one possible outcome from their calculations. Others multiplied probabilities which had to be added, or added probabilities which had to be multiplied. A few candidates simply subtracted their answer to part (b)(i) from 1. Some of the answers given were greater than 1.

Answers: (a) (b)(i) 10

1; (ii)

50

17.

blue

red



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Question 22 Many attempted this question with success. A very large number of candidates correctly determined the gradient in part (a). In part (b), many could find the speed at t = 9 through direct proportion. In part (c), the majority of the candidates knew they had to find the area under the graph but a few left arithmetical mistakes along the way. Others sometimes misread the values given on the graph and used those to calculate the distance travelled from t = 0 to t = 60. Answers: (a) 1.2; (b) 3.6; (c) 480. Question 23 Part (a) was generally well-answered on the whole. Only a few candidates failed to replace x by 8 in the equation 2y = 12 + x. In part (b), since the equations of the lines were already given in the question, many correct answers were seen. However a good number of candidates, nevertheless, gave the wrong inequality signs in their answers. There were few correct answers to part (c). A minority thought of finding the coordinates of all the three vertices of the triangle ABC to reach the answer. In all the other cases, the answers given were the result of apparent guesswork. Answers: (a) (8, 10); (b) x > 8, 2y > 12 + x; (c) (9, 11). Question 24 There were many good answers to part (a) but there were also numerous cases of candidates reading the wrong scale of their protractor. A handful of candidates found the length of AB instead of finding the bearing of B from A. Part (b) was quite well answered on the whole. A considerable number of candidates indicated they knew which loci to draw but drew an incomplete circle. Quite a high number of candidates drew perpendicular bisectors which did not intersect their circle at two points. A few others attempted to complete a possible triangle without drawing any loci. Answer: (a) 137° to 139° Question 25 Part (a) was quite straightforward for the vast majority of the candidates. Quite a high number of candidates were did not handle the negative numbers correctly, however. A few others interchanged the x- and the y- coordinates. Part (b) was also found relatively straightfoward. A few candidates forgot to include the negative sign. Nevertheless, a significant number of candidates divided the difference in x-coordinates by the difference in y-coordinates. Part (c)(i) proved more demanding. Some tried to use column vectors and a few succeeded. Part (c)(ii) appeared difficult for most candidates. There were many attempts at finding the lengths of PM and MR, with little success.

Answers: (a)

− ,12

1; (b)

7

6− ; (c)(i) (10, –8); (ii)

3

1.



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Question 26 There were many correct answers to part (a), although there was a fair number of candidates who mistook k for the determinant and gave 7 as answer. There were also many correct answers to part (b) except in a few cases where careless arithmetic mistakes were left. There were a pleasing number of correct answers to part (c) recorded with many candidates post-multiplying by the inverse of matrix A. There were also many others who attempted the post-multiplication but made mistakes in multiplying the matrices. A few candidates replaced Y by (a, b) and obtained a pair of equations which they solved simultaneously, resulting in a range of varied answers. The method being perfectly valid, the candidates earned partial credit.

Answers: (a) 7

1; (b)

−−

02

41; (c) ( )02 .



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MATHEMATICS (SYLLABUS D)

Paper 4029/02

Paper 2

General comments The standard of numeracy and algebraic manipulation was generally good. There was no sign that candidates had difficulties in completing the paper in the allocated time. The need to work to the required level of accuracy has been emphasised in previous reports. Unless stated otherwise within a specific question, three figure accuracy is required. This means that candidates should use at least four figure accuracy in intermediate workings, including cases where answers are used in subsequent parts of the question. There were rare cases of the use of grad or rad in trigonometric questions. Unlike previous years, more successful attempts were seen in Section B than in Section A. Questions 1(b), 3(c), 3(d), 6(a) and 8 were highly scoring whereas questions 2(a)(ii), 2(a)(iii), 2(b), 5, 7(a), 9(a)(iii), 9(b)(ii), 10(e) and 11(b)(iv)(b) proved to be difficult for many candidates. Comments on specific questions Section A

Question 1 This question was relatively well answered by many candidates. (a) (i) Most candidates answered this part correctly. (ii) The amount deducted for tax was frequently seen but candidates were confused while calculating

the weekly salary. (iii) Many candidates calculated 25% or 75% of Rs 937500, overlooking the fact that Rs 937500

represented a percentage of 125%. (b) (i) This part was generally well answered by most candidates. (ii) Many candidates correctly attempted this part. (iii) Most candidates who found the ratio between Indian Rupees and Swiss Francs were able to obtain

the correct answer. Answers: (a)(i) 30; (ii) 15252.40; (iii) 750000; (b)(i) 65407.50; (ii) 294.12; (iii) 877.19. Question 2 Few candidates scored full marks in this question. (a) (i) The correct answer was obtained by most candidates. (ii) The correct angle was seen but the explanation given by candidates was often incorrect. (iii) In general, candidates did not provide a correct reason.



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(b) Only a few candidates attempted this part successfully. Answers: (a)(i) 23; (ii) 90 with reason. Question 3 This question was generally well attempted, except for part (b).

(a) (i) Quite a large number of candidates obtained 16

1.

(b) Few candidates obtained the correct answer. The wrong answer 256

21 was frequently seen.

(c) (i) This part was generally well answered. (ii) This part was correctly answered by many candidates. (d) Many candidates deduced the answer using sequence.

Answers: (a) 16

1; (b)

256

42; (c)(i) 26; (ii) m = 5 and n = –3; (d) p = 17.

Question 4 This question was well attempted by many candidates. (a) (i) Common mistakes noted in this part were to obtain 210 as the final answer and to calculate

( )10733

1××× instead of calculating ( )1073

2

1××× .

(ii) Some candidates mistook the length of the hypotenuse to be 10 and calculated the total surface

area using that value. (b) (i) Good attempts were made but the answers provided were often inaccurate. (ii) A common mistake made by candidates was to give their answer as 0.28 m. Answers: (a)(i) 105; (ii) 197.2; (b)(i) 0.845; (ii) 0.280 . Question 5 A significant number of candidates answered this question incorrectly. (a) Many candidates were not able to distinguish between radius and diameter. (b) Candidates were able to identify the total area as the sum of the area of a circle and that of a

rectangle. (c) (i) A significant number of candidates measured the angle TOA from the diagram and used it to

calculate the length of the arc TA, instead of recognising that TA was equal to the difference between the perimeter of the outer track and that of the inner track.

(ii) This part was poorly attempted. Answers: (a) 63.7; (b) 9550 or 9560; (c)(i) 18.8 – 19.0; (ii) 31°.



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Question 6 This question was generally well attempted. (a) Misuse of the protractor, and both angles drawn as acute, were frequently seen. (b) Candidates were able to use the required scale correctly. Answers: (b) 120 – 125 Section B

Question 7 Few candidates attempted this question. (a) (i) This part proved to be challenging for the majority of candidates. In general, candidates were not

able to show that DE was parallel to BC. A number of irrelevant and incorrect facts were provided. (ii) Many candidates did not realise that triangles ADE and triangles ABC were similar. The ratio 3 : 5

was commonly seen. (b) (i) Most candidates were able to calculate the vertices of triangle B using matrix multiplication and the

correct triangle was drawn. (ii) Many candidates interpreted the transformation S1 as being an enlargement instead of a stretch. (iii) Candidates realised that the required matrix would be obtained by carrying out the product of the

two 2 × 2 matrices, S2 and S1, but some calculated S1S2 instead of S2S1. (iv) Candidates were not able to understand that the inverse of the matrix in (b)(iii) was required. Many

incorrect attempts were made in this part.

Answers: (a)(ii) 9 : 25; (b)(i) Triangle with vertices (6, 1), (10, 1) and (10, 4); (ii) Stretch; (iii)

12

02;

(iv)

− 11

02

1

.

Question 8 This question was well answered on the whole. (a) (i) This part was successfully attempted by most of the candidates. (ii) Candidates made a good attempt to make h the subject of formula. Algebraic errors were seen in

the removal of the square root. Answers with π2

2T

were often seen.

(b) Many candidates successfully attempted this part. A significant number of candidates obtained 42

or 16 as the value of p as a result of leaving a mistake either in expanding the brackets or in grouping common terms.

(c) Candidates were able to apply LCM to express the left-hand side of the equation as a single

fraction. A few candidates wrongly removed the denominators by cross multiplication. In addition, the value 0 × 12 = 12 was commonly seen.



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(d) This part was generally well answered and the correct use of the quadratic formula was noted. However a few candidates tried to solve the quadratic equation by factorisation.

Answers: (a)(i) 2.24 (ii) 2

2

4π

gT (b) 14 (c) –5.5 (d) –0.41 and –3.26

Question 9 This question was fairly well attempted by the candidates.

(a) (i) Many candidates were able to apply the formula Cabsin2

1to show the given area.

(ii) This part was also well attempted by candidates. (iii) Candidates were generally unable to obtain the correct angle as they often mistook half of the area

of the parallelogram to be equal to the area of triangle ABC. (b) (i) This part was successfully answered by most candidates. (ii) Candidates made good attempts to find angle CAM but were not able to obtain the correct value as

algebraic errors were seen in making cos CAM the subject of the formula. In addition, some candidates stopped at angle CAM, neglecting to subtract 30º from it.

Answers: (a)(ii) 39.1º or 39.2º; (iii) 136.3º; (b)(i) 6.16; (ii) 41.4 . Question 10 This question was attempted by the majority of candidates. (a) Candidates were able to complete the table correctly. (b) The plotting of points and drawing of a smooth curve were generally well done. However, cases

where a wrong scale or a non-uniform scale was used were noted. (c) Drawing of the tangent at (3, – 4) was quite straightforward but some drew chords or left ‘daylight’

between the tangent and the curve. (d) (i) This part was well answered. (ii) Candidates did not realise that the required answers would be given by finding the x-coordinates of

the points where the line y = 4 intersects the curve y = x2 – 4x – 1.

(e) Candidates did not seem to know how to use the curve drawn in part (b) to solve the quadratic

equation. However, many candidates drew a new curve and attempted to read values where the new curve intersected the x-axis.

Answers: (a) 11 11; (c) 2; (d)(i) –5; (ii) a = –1 and b = 5; (e) 0.6 and 3.4 . Question 11 This question was quite popular and many candidates attempted parts (a) and (b)(i). (a) This part was generally well answered by many candidates. However, there were attempts made at

finding the frequency densities by using a fixed class width of 10. (b) (i) This part was well attempted by the majority of candidates. A few candidates were inaccurate in

plotting the points to draw the cumulative frequency curve. (ii)(a) Many candidates took the total frequency to be 250 instead of 240 while finding the median.



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(b) Many candidates used 250 as the total frequency to calculate the lower quartile and the upper quartile.

(iii) This part was well answered by most candidates but a few gave values from the cumulative

frequency table. (iv)(a) This part was fairly well answered. (b) It seemed that candidates did not understand the meaning of percentile. This part was often left

unattempted. Answers: (a) Correct histogram; (b)(i) Correct cumulative frequency curve; (ii)(a) 195; (b) 72 to 88; (iii) 50 78 72 32 4; (iv)(a) 36; (b) 8.