Cambridge IGCSE Mathematics - International 0607 — 2025 Feb/March Paper 6 · Variant 2

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 Investigation and Modelling (Extended) February/March 2025 1 hour 30 minutes You must answer on the question paper. No additional materials are needed. 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, 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 50. ● The number of marks for each question or part question is shown in brackets [ ]. * 7 2 5 0 0 7 8 5 5 5 * DC (PQ/CT) 346897/1 © UCLES 2025 ,










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2 0607/62/F/M/25 © UCLES 2025 Section A INVESTIGATION REGULAR POLYGONS You are advised to spend no longer than 45 minutes on this section. In this task you will investigate fitting 3 regular polygons together at a point. All the polygons have sides of the same length. The 3 polygons fit together exactly with one corner from each polygon meeting at a point with no gaps or overlaps. 1 The diagram shows a square and 2 regular octagons fitting together at a point. This is the formula for the size of each interior angle, a°, of a regular polygon with n sides. a n 180 360 = - (a) Show that the size of each interior angle of a regular octagon is 135°. [1] (b) Use angles to show that 2 regular octagons and a square fit together at a point. [2] * 0000800000002 * ,










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3 0607/62/F/M/25 © UCLES 2025 [Turn over 2 Use the formula in Question 1 to complete the table. Number of sides, n 5 6 8 9 10 Interior angle, a° 135 144 [3] 3 Rearrange the formula a n 180 360 = - to show that n a 180 360 = - . [2] * 0000800000003 * ,










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4 0607/62/F/M/25 © UCLES 2025 4 The diagram shows regular polygons A and B and part of regular polygon C. NOT TO SCALE A B C A is a regular 10-sided polygon. B is an equilateral triangle. C is a different regular polygon. Work out the number of sides of polygon C. … [3] * 0000800000004 * ,










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5 0607/62/F/M/25 © UCLES 2025 [Turn over 5 Charlie finds other ways to fit 3 regular polygons together at a point. The first polygon has p sides. The second polygon has q sides. The third polygon has r sides. (a) Consider the angles around a point and use the formula a n 180 360 = - to show that p q r 1 1 1 2 1 + + = . [3] (b) p q r G G (i) When p 6 = , use p q r 1 1 1 2 1 + + = to find the value of q and the value of r. q = … r = … [1] (ii) p cannot be greater than 6. Give an example to show this. [1] * 0000800000005 * ,










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6 0607/62/F/M/25 © UCLES 2025 6 In this question, the value of p is 3 and p q r G G . Charlie wants to find all the possible ways to fit 3 regular polygons at a point. (a) Use p q r 1 1 1 2 1 + + = to show that q r 1 1 1 6 + = . [1] (b) q and r are positive integers. Give a reason why q 7 H . [1] (c) Complete the table. Number of sides of first polygon, p Number of sides of second polygon, q Number of sides of third polygon, r 3 7 3 8 24 3 9 3 10 3 12 12 [3] * 0000800000006 * ,










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7 0607/62/F/M/25 © UCLES 2025 [Turn over 7 Find all the possible ways to fit a square and 2 other regular polygons at a point. … [4] * 0000800000007 * ,










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8 0607/62/ F/M/25 © UCLES 2025 Section B MODELLING STREET LIGHTS You are advised to spend no longer than 45 minutes on this section. In this task you will be modelling how to place lights on paths. In this task: • paths are straight • lights are shown as large dots • lights shine on an area that is a circle called the circle of light • lights must shine on all parts of the path • lights are placed as far apart as possible. In this task you will find Pythagoras’ theorem useful. a c b a b c 2 2 2 + = 8 This path has a width of 3 m. The radius of each circle of light is 5 m. The distance along the path between the lights is d metres. (a) The lights are placed on the same side of the path as shown. circle of light light path NOT TO SCALE dm 3m 5m Use Pythagoras’ theorem to find the value of d. … [3] * 0000800000008 * ,










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9 0607/62/ F/M/25 © UCLES 2025 [Turn over (b) The lights are placed on both sides of the path as shown. circle of light light path NOT TO SCALE dm 3m 5m Show that the value of d is now 9. [1] (c) Lights are placed on a long path. Which arrangement, part (a) or part (b), uses the smaller number of lights? Give a reason for your answer. … [1] * 0000800000009 * ,










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10 0607/62/F/M/25 © UCLES 2025 9 A different path has a width of 4 m. The radius of each circle of light is 5 m. Calculate the distance along the path between lights (a) when the lights are all on the same side of the path … [2] (b) when the lights are on both sides of the path. … [2] * 0000800000010 * ,










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11 0607/62/ F/M/25 © UCLES 2025 [Turn over 10 circle of light light path dm wm rm For this path: • the width of the path is w metres • the radius of the circle of light is r metres • lights are on both sides of the path. (a) A model for d in terms of r and w is d r r w 2 2 = + - . Draw and label a diagram to show this is correct. [2] (b) Find the value of d when w 6 = and . r 6 8 = . … [2] * 0000800000011 * ,









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12 0607/62/ F/M/25 © UCLES 2025 11 bulb circle of light r x h The diagram shows a light with height h metres and its circle of light. A model for the intensity of light, L, at distance x metres from the bulb is L x B 2 = . B is a number called the brightness index of the bulb. When B is larger the light is brighter. The intensity of light on the circumference of the circle of light must be 0.05 units. (a) Show that r B h 20 2 = - . [3] * 0000800000012 * ,









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13 0607/62/F/M/25 © UCLES 2025 [Turn over (b) The height of the light is 6 m. The radius of the circle of light is 10 m. Find the value of B. … [3] (c) The height of the lights is h metres. Change the model d r r w 2 2 = + - to find a model for d in terms of B, h and w. … [1] * 0000800000013 * ,









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14 0607/62/F/M/25 © UCLES 2025 12 The height of the lights is 3.0 m. The lights are on both sides of the path. The brightness index is 3.5 . The intensity of light on the circumference of the circle of light is still 0.05 units. Find the greatest possible width for the path. … [5] * 0000800000014 * ,









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15 0607/62/F/M/25 © UCLES 2025 BLANK PAGE * 0000800000015 * ,









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16 0607/62/F/M/25 © UCLES 2025 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 2025 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 (Extended) February/March 2025 MARK SCHEME Maximum Mark: 50 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 February/March 2025 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 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.

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 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.

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 4 of 10 Annotations guidance for centres Examiners use a system of annotations as a shorthand for communicating their marking decisions to one another. Examiners are trained during the standardisation process on how and when to use annotations. The purpose of annotations is to inform the standardisation and monitoring processes and guide the supervising examiners when they are checking the work of examiners within their team. The meaning of annotations and how they are used is specific to each component and is understood by all examiners who mark the component. We publish annotations in our mark schemes to help centres understand the annotations they may see on copies of scripts. Note that there may not be a direct correlation between the number of annotations on a script and the mark awarded. Similarly, the use of an annotation may not be an indication of the quality of the response. The annotations listed below were available to examiners marking this component in this series. Annotations Annotation Meaning More information required Accuracy mark awarded zero Accuracy mark awarded one Accuracy mark awarded two Accuracy mark awarded three Independent mark awarded zero Independent mark awarded one Independent mark awarded two Benefit of the doubt Communication mark Incorrect point Follow through Highlighter Highlight a key point in the working Ignore subsequent work Method mark awarded zero Method mark awarded one Method mark awarded two Misread Omission

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 5 of 10 Annotation Meaning Off-page comment Allows comments to be entered at the bottom of the RM marking window and then displayed when the associated question item is navigated to. On-page comment Allows comments to be entered in speech bubbles on the candidate response. Premature rounding/approximation Special case Indicates that work/page has been seen Transcription error Correct point Not from wrong working 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

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 6 of 10 Question Answer Marks Partial Marks 1(a) 360 180 8 − 135 = 1 1(b) 135 + 135 + 90 = 360 oe 2 B1 for 135 + 135 + 90 or 360 seen 2 5 6 8 9 10 108 120 135 140 144 2 B1 for two correct 360 180 5 or 6 or 9 − C1 3 Isolate term in n e.g. 360 180 = −a n 1 360 (180 ) = − n a or 1 360 180 = − n a or 1 180 360 − = a n leading to 360 180 = − n a 1 4 15 1 360 – 144 – 60 OR 60 – 36 and either 360 or 10 C1 360 180 156 − OR 360 24 their C1 FT their 156 provided < 180 5(a) 360 360 360 180 180 180 360       − + − + − =             p q r 1 Collect numerical terms or divide by or factor out 180 or divide by or factor out 360 1 Leading to 1 1 1 1 2 + + = p q r 1 5(b)(i) q = 6 r = 6 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 7 of 10 Question Answer Marks Partial Marks 5(b)(ii) Values of p, q and r with p q r and p > 6 and 1 1 1 k p q r + + = where 1 2 k  e.g. 1 1 1 7 7 7 + + = 3 7 1 6(a) 1 1 1 1 1 2 3 6 + = − = q r 1 6(b) 1 1 1 6 = − q r , so 1 q is less than 1 6 , so q is more than 6 oe 1 6(c) p q r 3 7 42 3 8 24 3 9 18 3 10 15 3 12 12 2 B1 for one correct value 1 1 1 6 7 42 − = or 1 1 1 6 9 18 − = C1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 8 of 10 Question Answer Marks Partial Marks 7 [4,] 5, 20 [4,] 6, 12 [4,] 8, 8 and no others OR [The angles are] [90,] 108, 162 [90,] 120, 150 [90,] 135, 135 and no others 2 B1 for one, ignoring incorrect sets or correct values of q and r with incorrect value for p Fraction Method 1 1 1 4 + = q r C1 1 1 4 5 or 6 or 8 − or 4 4 = − q r q where q = 5 or 6 or 8 C1 OR Angle Method Interior 270 – 108 = 162 or 360 – 90 – 108 = 162 or 270 – 120 = 150 or 360 – 90 – 120 = 150 or 270 – 135 = 135 or 360 – 90 – 135 = 135 oe Exterior 360 5 or 6 or 8 oe (C1) 360 180 150 − or 360 180 162 − or 360 180 120 − 360 ) ( 5 or 6 or 8 1 80 their − or 360 18 or 360 30 or 360 45 oe (C1) 8(a) 8 1 2 2 5 3 − may be seen in stages C2 C1 for 2 2 2 3 5 + = x oe or 52 – 32 or square root seen 8(b) 4 + 5 = 9 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 9 of 10 Question Answer Marks Partial Marks 8(c) part (b) because 9 is bigger than 8 oe 1 9(a) 6 m 1 2 2 5 4 − may be seen in stages C1 9(b) 8 1 m C1 10(a) Correct diagram 2 B1 for diagram with one correct label for r or 2 2 − r w 10(b) 10 1 2 2 6.8 6.8 6 + − C1 11(a) x2 = r2 + h2 oe 1 2 0.05 = B x or B = 0.05x2 1 leading to 2 20 = − r B h 1 11(b) 6.8 2 M1 for 20B = 136 or for 0.05 = 1 36 B or 0.05 × 136 oe or for 11.672 used correctly for 136 2 10 20 6 = − B oe or x2 or r2 + h2 or 20B = 102 + 62 oe C1 e.g. 2 2 10 20 6 = − B 11(c) 2 2 2 20 20 = − + − − d B h B h w isw 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED February/March 2025 © Cambridge University Press & Assessment 2025 Page 10 of 10 Question Answer Marks Partial Marks 12 m C1 7.8 2 B1 for 61 or 61.05… to 61.1 seen or 7.8[1] seen not as answer 2 2 2 20 3.5 3 20 3.5 3  − +  − −w or 2 3.5 3 0.05 − C1 FT their 11(c) answer w2 ⩽ 61 or 61 w oe e.g. in words C1