Cambridge IGCSE Mathematics - International 0607 — 2024 Oct/Nov Paper 4 · Variant 2

This document has 20 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ DC (CE/CT) 337373/2 © UCLES 2024 CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/42 Paper 4 (Extended) October/November 2024 2 hours 15 minutes You must answer on the question paper. You will need: Geometrical instruments INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● You should use a graphic display calculator where appropriate. ● You may use tracing paper. ● You must show all necessary working clearly and you will be given marks for correct methods, including sketches, even if your answer is incorrect. ● Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. ● For r, use your calculator value. INFORMATION ● The total mark for this paper is 120. ● The number of marks for each question or part question is shown in brackets [ ]. * 4 0 6 3 8 3 8 0 9 6 * ,










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2 0607/42/O/N/24 © UCLES 2024 Formula List For the equation ax bx c 0 2 + + = x a b b ac 2 4 2 ! = - - Curved surface area, A, of cylinder of radius r, height h. r A rh 2 = Curved surface area, A, of cone of radius r, sloping edge l. r A rl = Curved surface area, A, of sphere of radius r. r A r 4 2 = Volume, V, of pyramid, base area A, height h. V Ah 3 1 = Volume, V, of cylinder of radius r, height h. r V r h 2 = Volume, V, of cone of radius r, height h. r V r h 3 1 2 = Volume, V, of sphere of radius r. r V r 3 4 3 = sin sin sin A a B b C c = = cos a b c bc A 2 2 2 2 = + - sin bc A 2 1 Area = A C B c b a * 0000800000002 * ,










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3 0607/42/O/N/24 © UCLES 2024 [Turn over Answer all the questions. 1 (a) The line x y 2 3 12 + = crosses the x-axis at P and the y-axis at Q. Find the coordinates of P and the coordinates of Q. P ( … , … ) Q ( … , … ) [2] (b) (i) A is the point (-2, 5) and B is the point (4, -1). Find the column vector of the translation from point A to point B. f p [1] (ii) Find the coordinates of the mid-point of AB. ( … , … ) [2] (c) Find the magnitude of 9 7 b l. … [2] (d) The line L has gradient 3 and passes through the point (-2, 7). Find the equation of line L. Give your answer in the form y mx c = + . y = … [2] * 0000800000003 * ,










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4 0607/42/O/N/24 © UCLES 2024 2 y x 5 3 0 - 5 - 5 ( ) ( ) f x x x 2 4 = + - (a) On the diagram, sketch the graph of ( ) f y x = for values of x between -5 and 3. [2] (b) On the diagram, draw the asymptote to the graph of ( ) f y x = . [1] (c) Solve the equation ( ) f x 1 =- . … [2] (d) (i) Solve the equation ( ) f x x 1 2 = - . … [2] (ii) Solve the inequality ( ) f x x 1 2 1 - . … [2] * 0000800000004 * ,










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5 0607/42/O/N/24 © UCLES 2024 [Turn over 3 (a) Write 1 25 1 as a percentage. … % [1] (b) In a sale, all prices are reduced by 15%. Lola buys a jacket which has an original price of $64. Calculate the sale price of this jacket. $ … [2] (c) Nina invests $x at a rate of 1.8% per year compound interest. At the end of 3 years the value of this investment is $3375.93, correct to the nearest cent. Calculate the value of x. x = … [2] (d) Olav buys a car for $13 000. Each year the value of the car decreases by 12% of its value in the previous year. Calculate the number of complete years it takes for the value of Olav’s car to first become less than $5000. … [4] * 0000800000005 * ,










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6 0607/42/O/N/24 © UCLES 2024 4 (a) NOT TO SCALE The diagram shows a solid metal shape made from a cylinder and two hemispheres. The radius of the cylinder and of the hemispheres is 3 cm. The length of the cylinder is 15 cm. (i) Show that the total volume of the shape is cm 537 3, correct to 3 significant figures. [3] (ii) The shape is melted and all the metal is used to make 600 identical small cubes. Calculate the side length of one of these cubes. Give your answer in millimetres. … mm [3] * 0000800000006 * ,










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7 0607/42/O/N/24 © UCLES 2024 [Turn over (b) 120° B A O r NOT TO SCALE The diagram shows a sector OAB with radius r cm and centre O. The sector angle is 120°. The shaded segment has an area of . cm 18 4 2. Calculate the length of the arc AB. … cm [5] * 0000800000007 * ,










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8 0607/42/O/N/24 © UCLES 2024 5 (a) v u as 2 2 2 = - (i) Find the value of v when u = 7, a = 1.5 and s = 10. v = … [2] (ii) Rearrange the formula to write u in terms of v, a and s. u = … [2] (b) Complete the table for sequences A, B and C. Sequence 1st term 2nd term 3rd term 4th term 5th term nth term A 11 8 5 2 B 3 8 15 n n 2 2 + C 4 1 4 1 16 1 [8] * 0000800000008 * ,










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9 0607/42/O/N/24 © UCLES 2024 [Turn over 6 y – 4 – 3 – 2 – 1 1 2 3 4 5 – 1 1 2 3 4 5 – 2 – 3 – 4 – 5 0 x T Q P (a) Describe fully the single transformation that maps (i) triangle T onto triangle P … … [3] (ii) triangle T onto triangle Q. … … [3] (b) Stretch triangle T by factor 2 with invariant line x = 5. [2] * 0000800000009 * ,










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10 0607/42/O/N/24 © UCLES 2024 7 (a) x 3 1 1 G - Show this inequality on the number line. -5 -4 -3 -2 -1 0 1 2 x [2] (b) (i) Solve the inequality x 7 2 1 5 1 G - + . … [2] (ii) Write down the integers that satisfy the inequality x 7 2 1 5 1 G - + . … [2] (c) Solve. ( ) ( ) x x 7 3 3 2 1 1 - - + = x = … [3] * 0000800000010 * ,










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11 0607/42/O/N/24 © UCLES 2024 [Turn over (d) y y 1 5 8 1 + + = (i) Show that y y 4 5 0 2 - - = . [3] (ii) Solve by factorisation. y y 4 5 0 2 - - = y = … or y = … [3] * 0000800000011 * ,









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12 0607/42/O/N/24 © UCLES 2024 8 NOT TO SCALE 50° A B C D 12 m 16 m 7 m The diagram shows four points, A, B, C and D, on horizontal ground. (a) Calculate AD. AD = … m [2] (b) Calculate angle DAB. Angle DAB = … [2] (c) Calculate CD. CD = … m [3] * 0000800000012 * ,









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13 0607/42/O/N/24 © UCLES 2024 [Turn over (d) The point P lies on BC and is the nearest point to D. Calculate BP. BP = … m [3] (e) D is due north of B. Calculate the bearing of C from A. … [5] * 0000800000013 * ,









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14 0607/42/O/N/24 © UCLES 2024 9 ( ) f x x 1 3 = + ( ) g tan x x = ( ) h x 3 1 x = + (a) Find ( ) f 2 - . … [1] (b) Find the exact value of g(120). … [1] (c) On the diagram, sketch the graph of ( ) g y x = for values of x between 0° and 180°. y x 180° 90° 0° [2] (d) Find x when ( ) h x 82 = . x = … [2] * 0000800000014 * ,









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15 0607/42/O/N/24 © UCLES 2024 [Turn over (e) Find ( ) h x 1 - . ( ) h x 1 - = … [2] (f) Simplify fully. ( ) ( ) f f x x 1 - Give your answer as a single fraction. … [3] * 0000800000015 * ,









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16 0607/42/O/N/24 © UCLES 2024 10 (a) Zola measures the height of each of 100 plants in her garden. The table shows her results. Height (h cm) h 0 10 1 G h 10 15 1 G h 0 15 2 1 G h 20 30 1 G h 30 0 6 1 G Frequency 13 21 25 19 22 (i) Calculate an estimate of the mean. … cm [2] (ii) One of the plants is chosen at random. Find the probability that the plant has a height greater than 15 cm. … [1] (iii) Two of the 100 plants are chosen at random without replacement. Find the probability that one plant has a height of 15 cm or less and one has a height greater than 30 cm. … [3] * 0000800000016 * ,









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17 0607/42/O/N/24 © UCLES 2024 [Turn over (b) 50 students are asked if they like football (F ) and if they like swimming (S ). 3 do not like football and do not like swimming. 38 like football. 16 like swimming. (i) Complete the Venn diagram. F S U 3 … … … [2] (ii) Write down the number of students who like football and swimming. … [1] (iii) One of the 50 students is chosen at random. Find the probability that this student likes football or swimming but not both. … [1] (iv) Two of the students who like swimming are chosen at random. Find the probability that they both like football. … [2] * 0000800000017 * ,









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18 0607/42/O/N/24 © UCLES 2024 11 (a) a is a positive integer. Rationalise the denominator. a 3 1 - … [2] (b) ( )( ) g h h g p q 3 3 3 + - = + Find p and q in terms of g and h. p = … q = … [3] * 0000800000018 * ,









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19 0607/42/O/N/24 © UCLES 2024 (c) (i) d is an integer. Work out 3 10 3 10 d d 2 # # + - , giving your answer in standard form. … [2] (ii) Find 8 10 3 # 2000. Give your answer in standard form. … [2] * 0000800000019 * ,









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20 0607/42/O/N/24 © UCLES 2024 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. BLANK PAGE * 0000800000020 * ,










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This document consists of 9 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/42 Paper 4 (Extended) October/November 2024 MARK SCHEME Maximum Mark: 120 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the October/November 2024 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 2 of 9 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors. GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 3 of 9 Mathematics-Specific Marking Principles 1 Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required then no marks will be awarded for a scale drawing. 2 Unless specified in the question, non-integer answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the degree of accuracy is not affected. 3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points. 4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). 5 Where a candidate has misread a number or sign in the question and used that value consistently throughout, provided that number does not alter the difficulty or the method required, award all marks earned and deduct just 1 A or B mark for the misread. 6 Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 4 of 9 MARK SCHEME NOTES The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark M Method marks, awarded for a valid method applied to the problem. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. For accuracy marks to be given, the associated Method mark must be earned or implied. B Mark for a correct result or statement independent of Method marks. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’ is used to indicate that a particular M or B mark is dependent on an earlier mark in the scheme. Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working nfww not from wrong working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied Question Answer Marks Partial Marks 1(a) [P] (6, 0) [Q] (0, 4) 2 B1 for each or for correct answers in reverse 1(b)(i) 6 6     −   1 1(b)(ii) (1, 2) 2 B1 for each coordinate 1(c) 11.4 or 11.40... 2 M1 for 72 + 92 oe 1(d) 3x + 13 2 B1 for y = 3x + k, k ≠ 0 or M1 for 7 = 3(–2) + c or 7 3( ( 2)) − = −− y x

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 5 of 9 Question Answer Marks Partial Marks 2(a) Correct sketch 2 B1 each branch Max 1 mark if branches joined or if significant overlap or gap 2(b) x = –2 drawn 1 Vertical line between branches 2(c) –3, 2 2 B1 for each or B1 for (–3, –1) and (2, –1) as final answer 2(d)(i) 1.09 or 1.094 to 1.095 2 B1 for correct sketch of y = 2 1 − x added to diagram or if y co-ordinate also given 2(d)(ii) x < –2, x > their positive 1.09 2 B1 for each 3(a) 104 1 3(b) 54.4[0] 2 M1 for 15 64 64 100 −  oe or B1 for 9.60 3(c) 3200 2 M1 for 3 1.8 1 3375.93 100    + =     x oe 3(d) 8 nfww 4 B3 for 7.47 or 7.474 to 7.475 or M3 for 12 5000 log 1 log 100 13000     − =         n oe or good sketch indicating value between 7 and 8 or correct trials reaching 7 and 8 or M2 for 12 5000 1 100 13000   − =     n oe or suitable graph with n > 1 or at least 3 correct trials or M1 for 12 13000 1 5000 100    − =     n oe soi by at least 2 trials with n > 1 2 0 -2 -4 4 2 0 -2 -4 2 0 -2 -4 4 2 0 -2 -4 2 0 -2 -4 4 2 0 -2 -4 2 0 -2 -4 4 2 0 -2 -4

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 6 of 9 Question Answer Marks Partial Marks 4(a)(i) 3 2 2 π 3 3   oe M1 2 π 3 15   M1 537.2 to 537.3 A1 4(a)(ii) 9.64 or 9.636 to 9.639 3 M2 for 3 537 600 or M1 for 537 600 4(b) 11.5 or 11.46... 5 M3 for 2 2 120 1 sin120 18.4 360 2    −   = r r oe OR M1 for 2 120 360   r oe M1 for 2 1 sin120 2   r oe AND M1 for 120 2 360   their r oe 5(a)(i) 4.36 or 4.358 to 4.359 2 M1 for 72 – 2(1.5)(10) or better 5(a)(ii) 2 2 + v as final answer 2 M1 for correct rearrangement making u2 the subject of the equation M1 for correct square root of their expression Max 1 mark if final answer incorrect 5(b) A –1 14 – 3n oe 3 B1 B2 or B1 for k – 3n for integer k or 14 – kn ( ) 0  k B 24, 35 2 B1 for each C 1 64 2 4 −n oe 3 B1 B2 or B1 for (2 ) 4 or 2 k k ( ) 0  k 6(a)(i) Rotation 90˚ clockwise oe [centre] (3, 2) 3 B1 for each

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 7 of 9 Question Answer Marks Partial Marks 6(a)(ii) Enlargement [scale factor] –2 [centre] (2, 0) 3 B1 for each 6(b) Image at (1, 1), (1, 2), (–3, 1) 2 B1 for stretch factor 2 with invariant line x = k or y = 5 7(a) 2 B1 for one circle correct at end of the line or if correct circles are not joined by a line 7(b)(i) –4 ≤ x < 2 2 B1 for –4 ≤ x or for x < 2 or for –8 ≤ 2x < 4 7(b)(ii) –4, –3, –2, –1, 0, 1 2 FT their (i) B1 if one omission or extra 7(c) 25 3 M1 for correct expansion M1 for correctly collecting their like terms 7(d)(i) 5 8 ( 5) + + = + y y y y oe M1 Correct removal of fractions 2 5 8 5 + + = + y y y y oe M1 Correct removal of brackets Leading to 2 4 5 0 − − = y y A1 No errors or omissions 7(d)(ii) ( 5)( 1) − + y y M2 B1 for ( )( ) + + y a y b with ab = –5 or a + b = –4 or for ( 1) 5( 1) or for ( 5) [1]( 5) y y y y y y + − + − + − –1, 5 A1 strict FT their factors 8(a) 13.9 or 13.89... 2 M1 for 72 + 122 8(b) 59.7 or 59.74... 2 M1 for 12 tan 7 = oe 8(c) 12.4 or 12.37 to 12.38 3 M2 for 2 2 12 16 2 12 16cos50 + −  OR M1 for 2 2 12 16 2 12 16cos50 + −  A1 for 153 or 153.1 to 153.2 8(d) 7.71 or 7.713... 3 M2 for 12cos50 oe or M1 for recognising shortest distance –3 1

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 8 of 9 Question Answer Marks Partial Marks 8(e) [0]61.9 or [0]61.89 to 61.91… 5 M1 for 2 2 2 [ ]7 16 2 7 16cos(90 50) = + −  + AC M2 for 16sin(90 50) [sin ] + = CAB their AC or M1 for 16 sin(90 50) sin( ) = + theirAC CAB M1 for 90 – theirCAB oe or SC1 for final answer 242 or 241.89 to 241.91… Or Line AB produced to E such that AEC = o 90 M1 for CE 16sin(90 50) = − M1 for BE 16cos(90 50) = − M1 for tan 7 = + theirCE CAE theirBE M1 for 90 – their CAE 9(a) –7 1 9(b) 3 − 1 9(c) Correct sketch 2 B1 for each branch. Max B1 if branches joined or if too much overlap or too large a gap 9(d) 4 2 M1 for 3 1 82 + = x or better 9(e) 3 log ( 1) − x or log( 1) log3 − x final answer 2 M1 for 1 3 or for 3 1 −= = + x y y x or 3 log 1 − x or log 1 log3 − x 150 100 50 0 10 5 0 -5 -10 150 100 50 0 10 5 0 -5 -10 150 100 50 0 10 5 0 -5 -10 150 100 50 0 10 5 0 -5 -10

0607/42 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 9 of 9 Question Answer Marks Partial Marks 9(f) 6 3 3 2 1 + + x x x or 3 3 3 ( 2) 1 + + x x x final answer 3 M2 for 3 2 3 ( 1) 1 1 + − + x x oe or better or M1 for 3 3 1 1 1 + − + x x 10(a)(i) 22.3 2 M1 for at least 4 mid values soi 10(a)(ii) 33 50 oe 1 10(a)(iii) 34 225 oe 3 M2 for 34 22 22 34 100 99 100 99  +  oe or M1 for one product 10(b)(i) 31, 7, 9 correctly placed 2 B1 for 7 in intersection 10(b)(ii) 7 1 FT their Venn diagram 10(b)(iii) 4 5 oe 1 FT their Venn diagram 10(b)(iv) 7 40 2 M1 for 7 7 1 16 15 −  their their 11(a) 3( 1) 1 + − a a or 3 3 1 + − a a 2 M1 for 1 1 +  + a a 11(b) [p =] –2gh [q =] 2 2 − h g 3 B2 for one correct or B1 for three terms correct in 2 2 3 3 3 − + − gh g h gh 11(c)(i) 3.03 10  d 2 B1 for figs 303 or M1 for [0].03 10  d or 2 300 10 −  d 11(c)(ii) 9.28 or 9.283... × 10666 2 B1 for figs 928 or 9283... or M1 for 800 × 1998 10 or 0.8 × 2001 10 or 666 x 10 k with 9.284 9.35...   k