Cambridge IGCSE Mathematics - International 0607 — 2024 Oct/Nov Paper 6 · Variant 1

This document has 16 pages. Any blank pages are indicated. [Turn over * 8 5 7 1 6 1 9 1 5 4 * Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 Investigation and Modelling (Extended) October/November 2024 1 hour 40 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer both part A (Questions 1 to 3) and part B (Questions 4 to 7). ● 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, including sketches, to gain full marks for correct methods. ● In this paper you will be awarded marks for providing full reasons, examples and steps in your working to communicate your mathematics clearly and precisely. INFORMATION ● The total mark for this paper is 60. ● The number of marks for each question or part question is shown in brackets [ ]. DC (DE/CT) 337380/2 © UCLES 2024 ,










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2 0607/61/O/N/24 © UCLES 2024 The investigation starts on the next page. * 0000800000002 * ,










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3 0607/61/O/N/24 © UCLES 2024 [Turn over Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 TO 3) HOUSE OF CARDS (30 marks) You are advised to spend no more than 50 minutes on this part. This investigation looks at the number of cards in a house of cards. The diagram shows a house of cards with three rows. Rows are counted down from the top of the house. In this investigation is a horizontal card and are diagonal cards. Example 1 This house of cards has 3 rows of cards. Row 1 Row 2 Row 3 Example 2 This house of cards has 5 rows of cards. Row 1 Row 2 Row 3 Row 4 Row 5 * 0000800000003 * ,










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4 0607/61/O/N/24 © UCLES 2024 1 This is Row 1 in a house of cards. There are 0 horizontal cards. There are 2 diagonal cards. There are 2 cards in total. This is Row 2 in a house of cards. There is 1 horizontal card. There are 4 diagonal cards. There are 5 cards in total. This is Row 3 in a house of cards. (a) Complete the table. Row (n) Number of horizontal cards Number of diagonal cards Total number of cards 1 0 2 2 2 1 4 5 3 4 5 [3] (b) Find an expression, in terms of n, for the total number of cards in Row n. … [3] * 0000800000004 * ,










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5 0607/61/O/N/24 © UCLES 2024 [Turn over (c) The total number of cards in Row p is 368. Work out how many diagonal cards are in Row p. … [3] * 0000800000005 * ,










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6 0607/61/O/N/24 © UCLES 2024 2 The house number is the number of rows in the house. This is House 1. There are 0 horizontal cards. There are 2 diagonal cards. There are 2 cards in total. This is House 2. There is 1 horizontal card. There are 6 diagonal cards. There are 7 cards in total. This is House 3. (a) Complete the table. You may use the grid to help you. House (h) Number of horizontal cards Number of diagonal cards Total number of cards 1 0 2 2 2 1 6 7 3 4 5 [4] * 0000800000006 * ,










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7 0607/61/O/N/24 © UCLES 2024 [Turn over (b) Find an expression, in terms of h, for the number of diagonal cards in House h. … [2] (c) This is an expression for the number of horizontal cards in House h. . ( ) h h 0 5 1 - Use this expression and your answer from part (b) to find an expression for the total number of cards in House h. Give your answer in its simplest form. … [2] (d) The total number of cards in House k is 737. Find the number of rows in House k. … [3] * 0000800000007 * ,










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8 0607/61/O/N/24 © UCLES 2024 3 The investigation now looks at the total number of cards in a sequence of houses of cards. This is the first diagram in the sequence of houses. There is 1 house. There are 0 horizontal cards. There are 2 diagonal cards. There are 2 cards in total. This is the second diagram in the sequence of houses. There are 2 houses. There is 1 horizontal card. There are 8 diagonal cards. There are 9 cards in total. This is the third diagram in the sequence of houses. There are 3 houses. There are 4 horizontal cards. There are 20 diagonal cards. There are 24 cards in total. (a) Complete the table. You may use the table in Question 2(a) to help you. Total number of houses (t) Number of horizontal cards (H ) Number of diagonal cards Total number of cards 1 0 2 2 2 1 8 9 3 4 20 24 4 5 [2] * 0000800000008 * ,










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9 0607/61/O/N/24 © UCLES 2024 [Turn over (b) This is a formula for the number of horizontal cards, H, in a sequence of t houses of cards. ( )( ) H t t a t a 6 1 = + - , where a is a positive constant. Find the value of a and write down the formula. a = … H = … [3] (c) The nth diagram, with n houses, in the sequence of houses has 2925 horizontal cards. Use part (b) and Question 2(c) to find the total number of cards in the last house in the diagram. … [5] * 0000800000009 * ,










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10 0607/61/O/N/24 © UCLES 2024 B MODELLING (QUESTIONS 4 TO 7) AGE AND FASTEST TIMES (30 marks) You are advised to spend no more than 50 minutes on this part. This task looks at how age affects the fastest recorded times for athletes to run 100 metres. 4 The table shows the age of athletes running 100 m and their fastest times. Age (x years) 33 34 37 40 45 50 55 60 65 70 76 Time ( y seconds) 9.74 9.80 9.87 9.93 10.72 10.88 11.30 11.70 12.31 12.77 13.25 (a) Complete the scatter diagram to show the results. The first seven points have been plotted for you. Time (seconds) 13 12.5 13.5 12 11.5 11 10.5 10 9.5 30 40 50 60 70 35 45 55 65 75 Age (years) 80 y x [2] * 0000800000010 * ,










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11 0607/61/O/N/24 © UCLES 2024 [Turn over (b) A straight line through the points (38, 10) and (74, 13) models the data. (i) On the grid, draw the model. [1] (ii) Find the equation of the model. … [3] (iii) The fastest time for a certain age is 12 seconds. Use the model to find this age. … [2] (iv) The fastest recorded time to run 100 m is 9.58 seconds. Comment on the validity of the model for an athlete aged 20 years. … [2] * 0000800000011 * ,









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12 0607/61/O/N/24 © UCLES 2024 5 For athletes younger than 20 years there is a different model for the fastest time to run 100 m. This is a graph of the model. Time (seconds) 14 15 13 12 11 10 5 7 9 11 13 Age (years) 15 6 8 10 12 14 16 17 y x (a) An athlete aged 13 years runs 100 m. Use the graph to write down the fastest time for this age. … [2] (b) The model for the fastest times for athletes younger than 20 years is y c x 268 .0 0139 # = + where c is a constant. Use your answer to part (a) to find the value of c correct to the nearest integer. Write down the model. c = … y = … [3] * 0000800000012 * ,









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13 0607/61/O/N/24 © UCLES 2024 [Turn over (c) 10.0 seconds is the fastest time for a certain age that is below 20 years. Using your model in part (b), solve an equation to show that this age is 17 years. [4] * 0000800000013 * ,









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14 0607/61/O/N/24 © UCLES 2024 6 For athletes aged from 82 years to 105 years there is a different model for the fastest time to run 100 m. This model is . . y x x 0 0381 6 23 269 2 = - + . (a) On the axes, sketch the graph of the model for x 82 105 G G . y x 105 82 10 [3] (b) The fastest time for an athlete aged 100 years to run 100 m is 26.99 seconds. Find the difference between this time and the time that the model predicts. … [2] (c) 18.32 seconds is the fastest time for a certain age between 82 years and 105 years. Use the model to find this age. … [1] * 0000800000014 * ,









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15 0607/61/O/N/24 © UCLES 2024 7 For any age, the fastest recorded time to run 100 m is 9.58 seconds. Use each model to find the possible ages of the athlete who ran this fastest time. [5] * 0000800000015 * ,









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16 0607/61/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 * 0000800000016 * ,









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This document consists of 10 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 (Extended) October/November 2024 MARK SCHEME Maximum Mark: 60 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.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 2 of 10 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.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 3 of 10 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.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 4 of 10 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) n Horizontal Diagonal Total 1 0 2 2 2 1 4 5 3 2 6 8 4 3 8 11 5 4 10 14 3 B1 for each correct row If 0 scored, SC1 for last column = sum of other two in all rows 1(b) 3n – 1 oe 2 B1 for 3n oe or kn – 1 (k ≠ 0) three differences of 3 for total cards or n – 1 + 2n or C1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 5 of 10 Question Answer Marks Partial Marks 1(c) 3p – 1 = 368 or (368 + 1) ÷ 3 C1 FT their (3n – 1) FT correct inverse numerical calculation from their (3n – 1) 246 2 B1 for 123 or B1FT their p × 2 correctly evaluated as final answer or 368 – (their p – 1) oe correctly evaluated as final answer 2(a) h Horizontal Diagonal Total 1 0 2 2 2 1 6 7 3 3 12 15 4 6 20 26 5 10 30 40 3 B1 for each correct row If 0 scored, SC1 for last column = sum of other two in all rows House 4 or House 5 correctly drawn or 3 differences from 1, 2, 3, 4 in horizontal column or 3 differences from 4, 6, 8, 10 in diagonal column or 3 differences from 5, 8, 11, 14 in total column C1 2(b) h2 + h or h(h +1) oe 2 B1 for a quadratic expression or for bad form e.g. h + 1 × h 2(c) ( ) 0.5 1 − h h + h2 + h oe C1 FT their (h2 + h) 1.5h2 + 0.5h oe isw 1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 6 of 10 Question Answer Marks Partial Marks 2(d) 1.5k2 + 0.5k = 737 oe or correct sketch of y = 1.5k2 + 0.5k oe C1 FT their (1.5k2 + 0.5k) for both equation and sketch 2 0.5 0.5 4 1.5 737 2 1.5 −  −  −  or better or (3k + 67)(k – 22) or (1.5k – 33)(k + 22.3 ) oe or line drawn at y = 737 on sketch or solution indicated at y = 0 on appropriate sketch or 3 correct trials of integer k > 10 in 1.5k2 + 0.5k or triangle drawn with at least 13 rows with 3 totals correct from rows > 10 C1 22 cao 1 3(a) t Horizontal H Diagonal Total 1 0 2 2 2 1 8 9 3 4 20 24 4 10 40 50 5 20 70 90 2 B1 for one correct row If 0 scored, SC1 for last column = sum of other two in both rows 3(b) Substitution of any t, H pair from the table into given equation or Method of differences with two third differences of 1 and 6a = 1 C1 FT their H for given t (a =) 1 ( ) ( )( ) 1 1 1 6 = + − H t t t oe 2 M1 for first correct step to solve for a or B1 for a = 1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 7 of 10 Question Answer Marks Partial Marks 3(c) Correct sketch of ( )( ) 1 1 1 6 = + − H t t t and line H = 2925 oe or 3 2925 6  rounded to 26 or 3 correct trials of integer t > 19 in their ( )( ) 1 6 = + − H t t a t a C2 FT ( )( ) 1 6 = + − H t t a t a for their value of a C1 FT for correct sketch without correct straight line C1 for 3 2925 6  C1 for 1 correct trial of integer t > 19 in their ( )( ) 1 6 = + − H t t a t a Substitution of t = 26 into 1.5h2 + 0.5h C1 FT their 26 and their (1.5h2 + 0.5h) in 2(c) 1027 2 B1 for [t =] 26

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 8 of 10 Question Answer Marks Partial Marks Modelling 4(a) 4 points correctly plotted 2 B1 for 2 points correctly plotted 4(b)(i) Correct ruled line drawn 1 4(b)(ii) y = 1 12 x + 41 6 oe 3 M1 for 13 10 74 38 − − oe M1FT for substituting (38, 10) or (74, 13) or (56, 11.5) into y = (their m) x + c oe If 0 scored SC2 for y = 0.0836x + 6.82 or SC1 for y = 0.08x + 6.8 4(b)(iii) For horizontal line on graph at y = 12 or substitution of y = 12 into y = 1 12 x + 41 6 C1 FT their y = 1 12 x + 41 6 62 1 4(b)(iv) Not valid as outside the range of the given data oe 2 B1 for outside the range of the given data/model OR [y =] 1 12 × 20 + 41 6 or sketch graph and indication of (20, 8.5) seen or 9.58 = 1 12 x + 41 6 or horizontal line drawn at 9.58 on graph C1 FT their equation Not valid and [9.58] is slower/worse than model oe or [9.58] is the time for someone older than 20 oe or The answer is not close enough oe 1 5(a) 11 1 Seconds or s C1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 9 of 10 Question Answer Marks Partial Marks 5(b) 0.0139 11 268 13 = +  c C1 FT 11 their (c =) –248 0.0139 ( )268 248 = −  y x 2 M1FT for 0.0139 11 268 13 − =  their c or 0.0139 1 1 13 their = 0.0139 268 13 + c or B1 for c = –248 5(c) Correct sketch of model and horizontal line at y =10 showing intersection or ( ) 0.0139 10 268 248 − = − x or better M2 FT their –248 M1 for correct sketch of model or  0.0139 10 268 248 = + − x [x =] 16.8 to 17.2 leading to 17 A2 A marks dep M1 A1 for 16.8 to 17.2 6(a) Correct sketch 2 B1 for positive quadratic starting above the x axis with continuous positive gradient B1dep for curve reaching at least x = 100 and touching y axis Correct vertical scale indicated C1 6(b) Intersection marked on sketch in (a) at approx. x = 100 or 27 [seconds] or 0.0381× 1002 – 6.23 × 100 + 269 C1 0.01 seconds 1 6(c) 92 years 1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2024 © Cambridge University Press & Assessment 2024 Page 10 of 10 Question Answer Marks Partial Marks 7 32 or 33 1 Sketch or line on graph in 4(a) or 9.58 = 1 12 x + 41 6 oe C1 FT their model in 4(b)(ii) 19 1 Sketch or 9.58 = 0.0139 268 248 − x C1 FT their model in 5(b) There is no age for model in Q6 oe 1