Igcse Maths Ce S Report P3h And 4h Nov 05 Final

IGCSE MATHEMATICS 4400, NOVEMBER 2005 CHIEF EXAMINER’S REPORT Paper 3H General Comments This paper gave candidates the opportunity to demonstrate positive achievement, which was taken by the majority. Most of the questions proved accessible and many had a high success rate. The only exceptions to this were Questions 3 (kite construction), 15(a) (inequality) and 20 (Venn Diagram). Methods were generally well explained with working clearly shown. Question 1 There were few errors with the calculation in part (a), although 0.17037… ⎛ ⎛ 9.7 ⎞⎞ + 1.2 ⎟ ⎟ appeared occasionally. Most (2.6 − 9.8 ÷ 2.7 + 1.2) and – 2.229 ⎜ 2.6 − ⎜ ⎝ 2.7 ⎠⎠ ⎝ candidates scored the mark for rounding in part (b) but a minority gave 0.09 or 0.1. Question 2 Many correct lines were seen but, surprisingly often, an inappropriate choice of scale on the y-axis resulted in the graph going off the grid at x = -1. Question 3 There was wide variation between centres but, in general, this question proved difficult and was badly answered, even by many of the best candidates. Some misinterpreted the information given, misunderstanding the term “shorter diagonal”, while others did not appreciate the meaning of “construct” and ignored the instruction to “show all construction lines”. A drawing within the tolerance but with no visible construction scored 1 mark out of 4. The same mark was awarded for a drawing within the tolerance to which an assortment of spurious or irrelevant arcs had been added. A preliminary sketch is often helpful in questions of this type. Question 4 Both parts were well answered, the most common error being the evaluation of 0 x 1 as 1 in the calculation of the mean in part (a). Question 5 Many gained full marks. If marks were lost, it was usually for inadequate reasons, especially giving just “parallel lines” as the reason that angle ACD is 18°. “Alternate angles” was required. “Z angles” was also accepted but will not be in future. A minority of candidates wrote mini essays, which should be discouraged. Working with concise reasons alongside the related line in the working is much easier to follow. Question 6 The calculation was usually performed accurately, although occasional sign errors led to answers of -22.8. Question 7 Direct proportion was well understood and errors were rare.

Question 8 “N =” was sometimes omitted in the first part but, otherwise, few marks were lost. The second part proved much more demanding, although a substantial minority PH gave a correct formula in a simplified form such as N = . Any correct formula, 12 1 PH even if unsimplified, N = 2 , for example, scored full marks. Of incorrect 6 PH formulae, N = PH and N = were the most popular. 6 Question 9 Both transformations were usually correct. The only regular errors were a clockwise rotation in part (a) and a wrongly positioned image in part (b), but both of these were rare. Question 10 This question had a high success rate. In the first part, 26% of 85 (22.1) was sometimes calculated but not subtracted from 85 while, in the second part, the ⎛ 48.1 ⎞ most common wrong answer was 185 ⎜ ⎟. ⎝ 0.26 ⎠ Question 11 The vast majority of candidates solved the equation correctly. Question 12 Part (a)(i) was almost always correct but, in part (a)(ii), a substantial minority gave the cumulative frequency for an age of 54 years as the answer, failing to subtract it from 100. Errors were rare in part (b), but the most likely ones were finding the upper and lower quartiles but not their difference, finding the median or giving an answer of 50 (75 – 25). Question 13 There were many completely correct diagrams, the quality of answers being noticeably centre dependent. It was not surprising that the line x + 2 y = 6 posed problems for some candidates or that the line y = 1 was sometimes drawn instead of the line x = 1. It was, however, surprising that the line y = x caused just as many difficulties, regularly being interpreted as the x-axis or the line y = -x. Question 14 Most candidates gave a complete method in the first part, usually by using Pythagoras’ Theorem to find the length of AC, subtracting 8 cm and then dividing the result by 2. In the second part, several methods were used successfully but the most popular one was to use the Cosine Rule in triangle APD. This was often done correctly but it was not uncommon for 82 + 1.66 2 − 2 × 8 × 1.66 cos 45° to be evaluated as (82 + 1.66 2 − 2 × 8 × 1.66)cos 45° . Some candidates, having evaluated the expression correctly, then failed to find its square root. Another regular error was to treat triangle APD as right-angled.

Question 15 Hardly any candidates solved the inequality completely in part (a), x ≤ 2 appearing much more often than −2 ≤ x ≤ 2 . In part (b), full marks were awarded to candidates who correctly represented the solution set to their answer to part (a), even if it was incorrect. Question 16 Both parts were well answered, the only regular error being in part (a), where an answer of 84° was quite popular with “angles in the same segment” as the reason. Question 17 The quality of answers was particularly centre dependent. Candidates with any knowledge of functions were able to answer part (a) correctly but, in part (b), it was not uncommon to see one of the solutions omitted, usually x = -1. Part (c) proved difficult, many candidates just finding f(1). In part (c), many drew a tangent at the vertical difference point (-1, 6) but, in finding , either failed to take account of horizontal difference the scale on the x-axis or did not appreciate that the gradient was negative. The 1

coordinates (-1, 6) were sometimes used to obtain answers like -6 and − 6 . In part (e), many did not understand what was required and it was not unusual to see the x values -2 and 4 simply inserted in the inequality. Question 18 There were many correct solutions. The most common mistake made by candidates who used a correct method was with the inner radius, usually taking it as 4.8 cm. It was also not unusual to see diameters used instead of radii in the formula for the volume of a sphere. Some did not appreciate the 3-dimensional nature of the question and used πr 2 . A few used 4πr 2 or found the volume of a prism of length 0.4 cm with the given annulus as its cross-section. P

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Question 19 Both parts were well answered, especially the first, but in the second part, the probability that Gill walks to school on both days was regularly omitted. Question 20 This question was poorly answered and completely correct solutions were rare. Whether this was because of problems with the notation or because of unfamiliarity with this style of question was not obvious. Two frequent errors were 5, which is n((P ∪ Q )′) , for part (b)(ii) and Ø, instead of 0, for part (b)(iii). Question 21 Many candidates solved the simultaneous equations correctly, usually by eliminating y and solving 3x 2 − 2 x − 5 = 0 , generally by factorisation but sometimes using the y −5 for x was less popular and more algebraically demanding formula. Substituting 2 but was often completed successfully. A sizeable minority found the x values correctly but did not go on to find the y values.

Paper 4H General Comments Most candidates appeared to have been prepared adequately for this paper. Presentation was good on the whole, with few candidates omitting working. The algebraic questions were well answered on the whole. A large proportion of candidates lost marks by rounding prematurely, especially in question 21 and also in question 9. Question 1 (a) Most candidates answered this part correctly, although a few gave extra numbers such as 9 or 17. (b) This part was also well answered. Some candidates failed to understand the intersection notation. A few probably understood it, but gave the answer “0” which did not score the mark. (c) This was generally well answered, although a few wrote “yes” and then correctly explained why 11 is not a member of B. Presumably they understood ∈ to mean ∉. Question 2 Both parts were well answered by almost all candidates. A few multiplied instead of dividing. In part (b), candidates who scaled up all three parts of the ratio and then added were generally less successful than those who added 1 + 2 + 5 = 8 and then scaled the 200 up. Question 3 This question was very well answered . Some very occasional errors were (a) (x – 5) (x – 5), (b) 2x 2 + 3y and (c) x 2 + 6x – 8. P

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Question 4 (a) A few candidates misread the scale, giving 22 and 82. (b) The answer “at his destination” was occasionally seen, but gained no mark. (c) Some candidates confused distance and speed, giving the answer 40 to 60. (d) Many candidates did not convert to hours. Some misread the graph, giving 6/20 or 4.4/20. (e) A few candidates attempted a distance/speed calculation. Question 5 (a) Common incorrect answers were 140°, 40° and 50°. (b) Most candidates attempted a perpendicular bisector, but many either did not attempt a construction or drew only one pair of arcs. (c) Few candidates answered this correctly. Common answers were 60° and 240°. Question 6 (a) Most candidates gave a partially correct answer, omitting the centre and/or the sense. A few gave a rotation and translation, gaining no marks. Some lost the mark for “rotation” by using the term “turn” instead. (b) The responses to this part were pleasing. The most commonly incorrectly answered sub-part was (iv). This suggests that the term “congruent” was unfamiliar to some candidates. Question 7 Most candidates answered both parts correctly. In part (b), some added correctly and then subtracted from 1. Others added 0.25 + 0.25. A few multiplied.

Question 8 Responses to this question gave less evidence than usual of lack of understanding. A few candidates made the traditional errors of ∑f/4 and ∑x/∑f. However, more common errors were the use of UCBs or incorrect midpoints. Question 9 Many candidates made incorrect statements such as x = 4.8cos68° or xsin68° = 4.8. A few started correctly with cos68° = 4.8/x, but rearranged incorrectly. Some candidates wasted time by finding the third side and then used Pythagoras’ theorem. Premature rounding sometimes lost a mark. Question 10 (a) This part was generally answered correctly. It is noteworthy that parts (b) and (c) can be answered by mechanical use of the calculator. Many candidates unnecessarily converted the given data into normal format. This gave extra scope for errors to creep in. (b) A few candidates could not handle the notation. Some gave a wrong power. Many gave a correct answer but not expressed in standard form as required by the question. (c) Candidates answered this part well. although some failed to give their answer in standard form and therefore lost the only available mark. Question 11 (a) Most candidates gave the correct equation in some form. A few omitted the squares; others made a sign error. (b) Algebraic rearrangement was generally correct, though a few candidates tried to solve rather than to derive the equation. (c) Substitution into the formula was generally correct although some candidates omitted essential brackets or made sign errors. Many candidates ignored the instruction to give answers to one decimal place. Others rounded 3.449 incorrectly to 3.5. Question 12 (a) (a)(i) This part was well answered, but in (ii) many candidates gave an explanation of their method for finding the gradient rather than an interpretation. Some gave an interpretation involving x and y or squares on the graph rather than referring to the context. (b) Most candidates gave the correct equation. (c) Many found the correct answer by trial and improvement. Others drew a graph for the second shop. A few solved the equation 3 + 1.5x = 2x + 1. A significant minority misinterpreted the information given in (c) to mean 3x + 1.5. A common error was to find the correct values (x = 4 and y = 9) but then to give the answer 9.

Question 13 (a) A correct tree was frequently seen. However a large number of candidates appeared not to understand the concept of a tree diagram. Many drew only one branch following from each of the two given lower branches (probably trying to follow the “pattern” in the given branches). Others only gave pairs of branches instead of triples. Some correctly labelled their pairs “Red” and “Blue or Green”, but this failed to follow the instruction “ …showing all the probabilities”. A few candidates drew a complete tree for the “without replacement” case. These lost a mark in (a) but could gain all the marks in parts (b) and (c). Some candidates wrote the relevant product at the end of each branch. It is worth noting that a probability tree diagram is complete without these. (b) Many candidates answered this part correctly. Others omitted one of the three possibilities. Some multiplied incorrectly (e.g. 61 × 61 = 62 ). A significant minority of candidates did not multiply at all. Some gave answers greater than 1. (c) Few candidates used the quick method ( 65 × 65 ). Most tried the long method P(BB, BG, GB, GG), but included only three of the four cases. Some incorrectly found 1 - P(RR). Some tried 1 – P(at least one R) but generally missed at least one case. As in part (b), some candidates did not multiply at all and others gave answers greater than 1. Question 14 Most candidates divided upper bound by upper bound. Some divided 1000 by 2.5 and gave the upper bound of the result. A few attempted the correct method, but many of these failed to find the correct bound for one or both figures. Common errors were 1004, 1010, 1040, 1050, 2.44 and 2.4. Of the very few who used the correct method, some rounded 410.2 down instead of up. Question 15 A small minority of candidates showed no familiarity with calculus. (a) Most candidates were able to differentiate correctly, although a few gave 3x – 12 or 3x 2 – 12x. (b) The key fact is that x = 0. This was missed by many candidates. Some gave “3”, presumably because of confusion with y = mx + c. Some substituted x = 0 into y rather than into the derivative. Many candidates performed what, to them, is always the next step after finding a derivative, namely to put it equal to 0 and solve. (c) Most candidates understood that the derivative needs to be equated with 0. Many achieved at least one correct answer, although some omitted x = -2. A few candidates created their own (inadequate) method. They evaluated y for all the relevant integer values of x and chose the two which were clearly a peak and a trough. Others looked at the diagram and guessed the coordinates of A and C. P

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Question 16 Few candidates understood the external case of the intersecting chords theorem. +10) . Common errors were 11x = 109 , 11x = 109 , 10 x x = 11 x 9, x + 10 = 11 + 9 and 109 = ( x20 Some attempted to use Pythagoras’ theorem or trigonometry. Question 17 Some candidates appeared not to understand the idea of a range of a function. (a) Many found 1 and –1, although some then subtracted one from the other. Perhaps this was due to confusion with statistical “range”. (b) Some candidates found the range of g correctly, but then gave this as the answer or gave 0° to 90°. Answers like 1 to 89 were not uncommon.

Question 18 A few candidates resorted to decimals in this question. (a) Some candidates correctly obtained 2 + 2 2 but went on to give an answer of 4 2. (b) (b) 9 2 and 9 were much in evidence. Some candidates multiplied numerator

and denominator by

2 only.

Question 19 There was much confusion between frequency and frequency density. Many candidates calculated 24 ÷ 6 = 4.8 and attempted to use this figure in various ways. Some candidates attempted to find the heights and then added these. Others attempted to find the scale (usually incorrectly) and then used it correctly to find the areas of the blocks. Surprisingly few candidates attempted the easiest method which was to find the area of one square and multiply by the number of squares in the histogram. Question 20

⎛ 2⎞ ⎛ 5 ⎞

⎟ or vice versa. The vast majority of candidates subtracted ⎜⎜ ⎟⎟ − ⎜⎜ − 4⎟ 3 ⎝ ⎠ ⎝

⎠

Candidates who drew a sketch were more likely to obtain the correct answer. Question 21 Many successful attempts were seen, although a large number of able candidates lost a mark through premature rounding. Some candidates unnecessarily calculated VA, but went on the find the correct answer. Some candidates incorrectly thought that tan VAM = VM/AC. Some thought that AM = 5 or that VA = 17.

MATHEMATICS 4400 PAPERS 3H & 4H, GRADE BOUNDARIES Higher Tier Grade

A*

A

B

C

D

E

Lowest mark for award of grade

79

61

43

26

15

9

Note: Grade boundaries may vary from year to year and from subject to subject, depending on the demands of the question paper.