Cambridge IGCSE Mathematics - International 0607 — 2024 May/June Paper 6 · Variant 2

This document has 16 pages. Any blank pages are indicated. [Turn over * 0 6 6 4 7 5 5 0 2 7 * Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 Investigation and Modelling (Extended) May/June 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 6) and part B (Questions 7 to 11). ● 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/CB) 332312/4 © UCLES 2024 ,

 

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2 0607/62/M/J/24 © UCLES 2024 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 6) INTEGER TREES (30 marks) You are advised to spend no more than 50 minutes on this part. This investigation looks at patterns when integers are arranged in the shape of a tree. A tree uses consecutive integers starting with 1 at the top. Each tree has a trunk, branches and leaves. The integers always increase going away from the trunk, on both the right and the left. The first branch is always on the right. This tree uses the integers 1 to 6. It has 2 branches with 2 leaves on each branch. Left trunk Right leaf 1 2 3 6 5 4 branch This tree uses the integers 1 to 12. It has 4 branches with 2 leaves on each branch. The leaves furthest from the trunk are the outer leaves. 1 2 3 6 5 4 7 8 9 12 11 10 left outer leaves right outer leaves A is the sum of the left outer leaves. B is the sum of the right outer leaves. T is the sum of the integers in the trunk. Example For the tree with 4 branches, A B T 6 12 18 3 9 12 1 4 7 10 22 = + = = + = = + + + = . * 0019655403102 * ,

 

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3 0607/62/M/J/24 © UCLES 2024 [Turn over In Questions 1 to 4, the number of branches, n, is always even. 1 This tree has 6 branches with 2 leaves on each branch. 1 2 3 6 5 4 7 8 9 12 11 10 13 14 15 18 17 16 Find the value of A, the value of B and the value of T. A = … B = … T = … [3] 2 (a) Complete the table for trees with two leaves on each branch. Use your answers to Question 1 to help you. Number of branches (n) A B T A B - A B T + - 2 6 3 5 3 4 4 18 12 22 6 8 6 9 8 60 16 [3] (b) (i) Use values from the table to give an example to show that . A B n 1 5 - = . … [1] (ii) Write an expression for A B T + - in terms of n. … [1] * 0019655403103 * ,

 

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4 0607/62/M/J/24 © UCLES 2024 3 In this question trees have 3 leaves on each branch. This tree has 6 branches. 1 2 3 4 8 7 6 5 9 10 11 12 16 15 14 13 17 18 19 20 24 23 22 21 (a) Complete the table. Number of branches (n) A B T A B - A B T + - 2 8 4 6 4 6 4 24 16 28 8 12 6 66 18 8 80 [2] (b) (i) Write an expression for A B - in terms of n. … [1] (ii) Write an expression for A B T + - in terms of n. … [1] * 0019655403104 * ,

 

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5 0607/62/M/J/24 © UCLES 2024 [Turn over 4 The table shows values of A B - and A B T + - for trees with 4 leaves on each branch. Number of branches (n) A B - A B T + - 2 5 8 4 10 16 6 15 24 8 20 32 (a) Complete the table below. Use the table above and your answers from Questions 2(b)(ii) and 3(b) to help you. Number of branches Number of leaves on each branch (x) A B - A B T + - n 2 1.5n n 3 n 4 [2] (b) A tree has an even number of branches, n, with x leaves on each branch. Complete the expressions for A B - and A B T + - in terms of n and x. A B - = n 2(…) A B T + - = … [2] * 0019655403105 * ,

 

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6 0607/62/M/J/24 © UCLES 2024 5 In this question trees have an odd number of branches. Trees have more than one branch. This tree has 5 branches with 2 leaves on each branch. 1 2 3 6 5 4 7 8 9 12 11 10 13 14 15 (a) Find the value of A B - and the value of A B T + - for this tree. A B - = … A B T + - = … [2] (b) In Question 4(b) you found expressions for trees with an even number of branches. Do your expressions in Question 4(b) give correct values for this tree? Show how you decide. A B - … … A B T + - … … [2] * 0019655403106 * ,

 

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7 0607/62/M/J/24 © UCLES 2024 [Turn over (c) The table shows some values of A B - and A B T + - for trees with odd numbers of branches. Number of leaves on each branch (x) Number of branches (n) A B - A B T + - 2 3 6 - 6 5 7 12 - 14 3 3 8 - 9 5 12 - 15 7 16 - 21 4 3 10 - 12 5 15 - 20 7 20 - 28 A tree has an odd number of branches, n, with x leaves on each branch. Find expressions for A B - and A B T + - in terms of n and x. A B - = … A B T + - = … [3] * 0019655403107 * ,

 

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8 0607/62/M/J/24 © UCLES 2024 6 In this question a tree can have either an even number or an odd number of branches. A tree can now start with any integer, k. A tree has p leaves on each branch. This is the start of a tree. It shows expressions for the integers on the trunk and the outer leaves only. k … k p + k p 2 1 + + … k p 1 + + (a) An expression for the outer leaf on the second branch is k p 2 1 + + . Show that this statement is correct. [1] (b) Find expressions for the third integer on the trunk and the outer leaf on the third branch. Write your expressions, in their simplest forms, on the tree. [2] * 0019655403108 * ,

 

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9 0607/62/M/J/24 © UCLES 2024 [Turn over (c) (i) Find expressions for A B T + - when a tree has 2 branches, 3 branches and 4 branches. Give each expression in its simplest form. … … … [3] (ii) Find an expression for A B T + - when a tree has n branches. … [1] * 0019655403209 * ,

 

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10 0607/62/M/J/24 © UCLES 2024 The modelling task starts on the next page. * 0019655403210 * ,

 

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11 0607/62/M/J/24 © UCLES 2024 [Turn over B MODELLING (QUESTIONS 7 to 11) CARBON FOOTPRINT OF AIR TRAVEL (30 marks) You are advised to spend no more than 50 minutes on this part. This task looks at modelling the personal carbon footprint of someone travelling by air. Carbon footprint is the amount of carbon dioxide that an activity puts into the atmosphere. Nikita wants to estimate her carbon footprint when she travels by air. Her carbon footprint depends on the amount of fuel used by the aircraft, the length of the journey and the number of people on the aircraft. Average fuel use is the amount of fuel, in kilograms, used for each kilometre travelled. 7 An aircraft uses T kg of fuel for a journey of length L km. A formula for the average fuel use, F kg/km, is F L T = . An aircraft makes a journey of 850 km and uses 4760 kg of fuel. Find the average fuel use for this journey. … [2] * 0019655403211 * ,

 
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12 0607/62/M/J/24 © UCLES 2024 8 Nikita finds this data for an aircraft making journeys of different lengths. Journey length (L km) 260 400 520 740 950 1080 1200 1460 Average fuel use (F kg/km) 10.6 7.5 6.2 5.8 5.0 4.6 4.5 4.3 She plots the data on a grid. 0 200 400 600 800 Journey length (km) Average fuel use (kg/km) 1000 1200 1400 1600 L F 0 2 4 6 8 10 12 (a) What happens to the average fuel use as the journey length increases? … [1] (b) Nikita thinks a model for the data is F a L b = + where a and b are constants. (i) Use the values of F from the table for L 400 = and L 00 12 = to write down two equations in terms of a and b. … … [2] * 0019655403212 * ,

 
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13 0607/62/M/J/24 © UCLES 2024 [Turn over (ii) Solve your equations to find the value of a and the value of b and write down the model. a = … b = … Model … [3] 9 The model gives these values of F, correct to one decimal place. Journey length (L km) 200 400 600 800 1000 1200 1400 1600 Average fuel use (F kg/km) 12.0 7.5 6.0 5.3 4.8 4.5 4.3 4.1 (a) Plot these values on the grid in Question 8 and draw the graph of the model. [2] (b) How well does the model fit Nikita’s data? … [1] (c) The aircraft makes a journey of 10 000 km. Find the average fuel use that the model gives. … [2] (d) Is your answer to part (c) valid? Give a reason. … [1] * 0019655403213 * ,

 
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14 0607/62/M/J/24 © UCLES 2024 10 The amount of carbon dioxide given out by an aircraft depends on the amount of fuel used. A kilogram of aircraft fuel makes 3200 grams of carbon dioxide. (a) The aircraft makes a journey of 800 km and uses 5.3 kg of fuel per km. Work out the amount of carbon dioxide, in grams, that this aircraft gives out for each kilometre of the journey. … [2] (b) The personal carbon footprint, C, for an aircraft passenger is the amount of carbon dioxide, in grams, that the aircraft gives out per kilometre divided by the number of passengers. Nikita assumes that this aircraft has 200 passengers. (i) Use your model for F in Question 8(b)(ii) to show that a model for C for a passenger on this aircraft with a journey length L is C L 48 28800 = + . [2] (ii) Sketch the graph of this model for 200 G L G 1600. 0 200 Journey length (km) Personal carbon footprint (g/passenger km) 1600 L C 0 [3] * 0019655403214 * ,

 
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15 0607/62/M/J/24 © UCLES 2024 (c) Nikita travels 900 km on this aircraft. Use the model in part (b)(i) to calculate the amount of carbon dioxide, in kg, that the aircraft gives out for each passenger for the whole journey. … [3] 11 Nikita wants to compare her carbon footprint for air travel and train travel. (a) She finds that the personal carbon footprint, C, for train travel is 40 g/km for each passenger for any length of journey. Sketch the graph of the model for C for train travel on the axes in Question 10(b)(ii). [2] (b) One train company claims: When you travel by train, your carbon footprint will never be more than 30% of your carbon footprint for the same journey by air. For what lengths of journey is this claim correct? … [4] * 0019655403215 * ,

 
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16 0607/62/M/J/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 * 0019655403216 * ,

 
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This document consists of 8 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 (Extended) May/June 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 May/June 2024 series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 2 of 8 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 May/June 2024 © Cambridge University Press & Assessment 2024 Page 3 of 8 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. 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 May/June 2024 © Cambridge University Press & Assessment 2024 Page 4 of 8 Question Answer Marks Partial Marks 1 36 27 51 2 B1 for two correct 6 + 12 + 18 or 3 + 9 + 15 or 1 + 4 + 7 + 10 + 13 + 16 C1 2(a) 19 20 21 24 23 22 or three common differences seen in either of last two columns or 36 + 27 – 51 [=12] or 60 – 12 = 48 C1 n A B T A – B A + B – T 2 6 3 5 3 4 4 18 12 22 6 8 6 36 27 51 9 12 8 60 48 92 12 16 2 B1 for two correct 2(b)(i) 1.5×2 = 3 or 1.5×4 = 6 or 1.5×6 = 9 or 1.5×8 = 12 1 2(b)(ii) 2n oe 1 3(a) n A B T A – B A + B – T 2 8 4 6 4 6 4 24 16 28 8 12 6 48 36 66 12 18 8 80 64 120 16 24 2 B1 for three correct 3(b)(i) 2n oe 1 3(b)(ii) 3n oe 1 4(a) branches leaves A – B A + B – T n 2 1.5n 2n n 3 2n 3n n 4 2.5n oe 4n oe 2 B1 for each value third row

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 5 of 8 Question Answer Marks Partial Marks 4(b) x + 1 oe xn oe 2 B1 for each 5(a) A = 18 B = 27 T = 35 C1 – 9 10 1 5(b) No oe and 5 (2 1) 2   = 7.5 or ‘is not negative’ 1 FT their expression in x only in 4(b) Yes and 2 × 5 = 10 1 5(c) Correct formula for A – B for a particular x or n stated C1 [A – B = ] ( 1)( 1) 2   x n oe [A + B – T = ] xn 2 B1 for one correct 6(a) k + p + 1 + p or p + k + p + 1 1 6(b) k + 2p + 2 oe k + 3p + 2 oe in correct places on tree 2 B1 for each 6(c)(i) Two from the expressions for 2, 3 and 4 branches (2) (k + 2p + 1) + (k + p) – (2k + p + 1) (3) (k + 2p + 1) + (2k + 4p + 2) – (3k + 3p + 3) (4) (2k + 6p + 4) + (2k + 4p + 2) – (4k + 6p +6) C2 C1 for one or for k+4p+3 and k+3p+3 on 4th line of tree 2p 3p 4p oe 1 Dependent on C2 6(c)(ii) np oe 1 7 5.6 1 kg/km or kg km–1 C1 8(a) decreases oe 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 6 of 8 Question Answer Marks Partial Marks 8(b)(i) 7.5 = a + 400 b oe isw 4.5 = a + 1200 b oe isw 2 B1 for one correct 8(b)(ii) 3 = 600 b or 2 1200 b C2 C1 for a negative sign on either side of this equation [a =] 3 [b =] 1800 F = 3 + 1800 L 1 Alternative Eliminating a or equating coefficients of b or removing fractions and equating coefficients of a or sketch of intersecting straight lines with negative gradients (C1) FT their equations of the form k = a + ...00 b Simplified equation with one variable on one side nfww or equations of lines and intersection marked on the sketch. (C1) Only if first C1 awarded [a =] 3 [b =] 1800 F = 3 + 1800 L (1) 9(a) Accurate plots 1 Smooth curve through their plotted points 1 9(b) [very] good fit oe 1 9(c) 3 + 1800 10000 C1 FT their 8(b)(ii) 3.18 or 3.2 kg/km 1 FT their 8(b)(ii) if answer positive 9(d) No, [too far] outside given data oe 1 10(a) 5.3 × 3200 C1 16 960 g 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 7 of 8 Question Answer Marks Partial Marks 10(b)(i) 3200 200 × 1800 3        L leading to 28800 48   C L 2 B1 for their F × 3200 or their F  200 or 3200 200 = 16 or × 3200 200 seen 10(b)(ii) Correct sketch from approximately 200 to 1600 for L and approximately 200 to 70 for C 2 B1 for correct shape Scale on C axis implying approx. (200, 192) or (1600, 66) C1 10(c) 28800 48 900        × 900 or 28800 48 900        ÷ 1000 or 80[g] = 0.08 kg C2 C1 for 28800 48 900        72 1 11(a) Horizontal line below the curve 1 40 marked on C-axis or (200, 40) marked or line segment labelled C = 40 C1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 8 of 8 Question Answer Marks Partial Marks 11(b) 40 0.3 oe or 28800 0.3 48        L or 40 28800 48        L soi 1 If 0 out of 3 scored, SC1 for 3150. Sketch of 28800 0.3 48         C L with C = 40 on one diagram or sketch of C = 40 0.3 with 28800 48   C L on one diagram or 28800 0.3 48        L or 8640 14.4  L = 40 and isolating Ls on one side or L 28800 0.3 48        L 337 40.038 338 39.962 or L 28800 48 L  337 133.46 338 133.21 or L 40 28800 48        L 337 0.2997 [29.97%] 338 0.3003 [30.03%] C1 337.5 or 337 or 338 1 L ⩽ 337.5 km 1dep FT their positive value for L dependent on C1