Cambridge IGCSE Mathematics - International 0607 — 2024 May/June Paper 6 · Variant 3

This document has 12 pages. [Turn over Cambridge IGCSE™ * 7 3 0 7 6 8 4 1 9 6 * DC (DE/JG) 332316/2 © UCLES 2024 CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/63 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 10). ● 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 [ ].

2 0607/63/M/J/24 © UCLES 2024 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 6) CHANGING THE ORDER OF OPERATIONS (30 marks) You are advised to spend no more than 50 minutes on this part. This investigation is about the results of two calculations when you change the order of operations. Each calculation uses three integers, x, y and z. Calculation 1 Add y and z. Multiply the answer by x. Calculation 2 Multiply y and z. Add the answer to x. Example x 3 = y 5 = z 7 = Calculation 1 y z 5 7 12 + = + = x 12 12 3 36 # # = = 36 Result = Calculation 2 y z 5 7 35 # # = = x 35 35 3 38 + = + = 38 Result = 1 Complete the table. x y z Result of calculation 1 Result of calculation 2 4 7 5 48 39 4 7 6 46 4 7 7 56 4 7 8 5 9 7 80 68 5 9 8 85 5 9 9 86 5 9 10 [4]

3 0607/63/M/J/24 © UCLES 2024 [Turn over 2 Show that when x 3 = , y 4 = and z 9 = the two calculations give the same result. [2] 3 y 5 = and z 6 = . Find the value of x when the two calculations give the same result. … [3]

4 0607/63/M/J/24 © UCLES 2024 4 (a) Complete the table. x y z Result of calculation 1 Result of calculation 2 6 11 12 138 138 7 13 14 189 8 15 16 248 9 17 18 [2] (b) Use the table in part (a) to write y and z in terms of x. y = … z = … [2] (c) The results of calculation 1 and calculation 2 are always the same. Use your expressions for y and z in part (b) to show that this statement is true. [3]

5 0607/63/M/J/24 © UCLES 2024 [Turn over 5 In this question x, y and z are consecutive, for example 15, 16 and 17. (a) Write y and z in terms of x. y = … z = … [1] (b) The results of calculation 1 and calculation 2 are the same. Show that x x 2 0 2 - - = . [4] (c) The results of calculation 1 and calculation 2 are the same. x, y and z are consecutive. Find two sets of values for x, y and z. … [3]

6 0607/63/M/J/24 © UCLES 2024 6 When the result of calculation 1 is the same as the result of calculation 2, ( ) x y z x yz + = + . (a) ( ) ( )( ) x x y x z x 1 - = - - Expand the brackets and rearrange to show that this becomes ( ) x y z x yz = + + . [2] (b) The results of calculation 1 and calculation 2 are the same. x 6 = and z y 2 . Find all the possible values for x, y and z. … [4]

7 0607/63/M/J/24 © UCLES 2024 [Turn over B MODELLING (QUESTIONS 7 to 10) FARMERS’ FIELDS (30 marks) You are advised to spend no more than 50 minutes on this part. This task looks at the area of a field inside a fence. 7 A farmer uses 600 metres of fencing to make a rectangular field. The length of the field is x metres. (a) The area of the field is A square metres. Show that a model for A, in terms of x, is ( ) A x x 300 = - . [2] (b) Find the range of possible lengths of the field. … [1] (c) Sketch the graph of the model ( ) A x x 300 = - . A x [3]

8 0607/63/M/J/24 © UCLES 2024 (d) Find the maximum area of the field and the length which gives this maximum area. Area = … Length = … [2] 8 The farmer uses a wall as one side of the field. The farmer uses 500 metres of fencing for this field. fence fence fence wall x metres (a) The length of the field is x metres. Find a model for the area of the field, m B 2. … [3]

9 0607/63/M/J/24 © UCLES 2024 [Turn over (b) Which model gives the greater maximum area of the field, the model for A or the model for B? Find the difference between these two maximum areas. Model … Difference … [3]

10 0607/63/M/J/24 © UCLES 2024 9 The farmer uses the model ( ) A x x 300 = - . These are the costs for making the field: • grass seed $0.27 per square metre • fertiliser $0.08 per square metre • a total of $7000 for fencing and labour. (a) The cost to make a rectangular field with length x metres is $C. Show that the model for C is . C x x 105 0 35 7000 2 = - + . [3] (b) Use your answer to Question 7(d) to find the cost of making the field with maximum area. … [2] (c) The farmer keeps sheep in a field. Each sheep has an area of 450 square metres. The farmer is given $5 for each sheep. He uses the money that he receives for the sheep to reduce his costs for making the field. (i) Use this information to change the model in part (a). … [2]

11 0607/63/M/J/24 © UCLES 2024 [Turn over (ii) Find the cost for the field with the maximum area. … [2] 10 The farmer makes a rectangular field. The length of the diagonal is 200 metres. w 200 metres (a) Show that a model for the area, m A 2, of this new field is ( )( ) sin cos A w w 40000 = . [2] (b) Sketch the graph of the model. A w [2] Question 10(c)(i) and (ii) are printed on the next page.

12 0607/63/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. (c) (i) Find the maximum area of this field. … [1] (ii) When the area is the maximum, find the length of the fence. … [2]

This document consists of 7 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/63 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.

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 2 of 7 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.

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 3 of 7 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

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 4 of 7 Question Answer Marks Partial Marks 1 One calculation 1 and one calculation 2 written out C1 e.g. (7 + 6) × 4 = 52 and (7 × 6) + 4 = 46 x y z Result of calculation 1 Result of calculation 2 4 7 5 48 39 4 7 6 52 46 4 7 7 56 53 4 7 8 60 60 5 9 7 80 68 5 9 8 85 77 5 9 9 90 86 5 9 10 95 95 3 B2 six cells correct or B1 three cells correct 2 4 + 9 = 13, 13 × 3 = 39 4 × 9 = 36, 36 + 3 = 39 2 B1 for each or B1 for 4 + 9 = 13 × 3 = 39 and 4 × 9 = 36 + 3 = 39 3 (5 + 6) × x or (5 × 6) + x or 11x or 30 + x or showing both calculations work for 3 (5 + 6) × 3 and (5 × 6) + 3 C1 11x = 30 + x or 10x = 30 or showing both calculations = 33 (5 + 6) × 3 = 33 and (5 × 6) + 3 = 33 C1 3 1 4(a) x y z Result of calculation 1 Result of calculation 2 6 11 12 138 138 7 13 14 189 189 8 15 16 248 248 9 17 18 315 315 2 B1 for 315 seen correctly placed once 4(b) (y =) 2x – 1 oe (z =) 2x oe 2 B1 for each

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 5 of 7 Question Answer Marks Partial Marks 4(c) (2x – 1 + 2x) x or 2x(2x – 1) + x 1 FT their expressions in x for y and z (4x – 1) x or 2x2 – x + 2x2 or 4x2 – 2x + x M1 Correct equation simplified to 4x2 – x twice A1 5(a) (y =) x + 1 oe (z =) x + 2 oe 1 5(b)      1 2 x x x     and    1 2 x x x    1 FT their binomial expressions in x for y and z 2x2 + 3x and x2 + 3x + 2 [ +x ] 2 FT their y and z of form ax + b (b ≠0) B1 for one correct expansion 2x2 + 3x = x2 + 4x + 2 leading to x2 – x – 2 = 0 1 5(c) x = 2, y = 3, z = 4 x = –1, y = 0, z = 1 1 (x + 1)(x – 2) or     2 1 1 4 1 2 2 1      or better, but with square root or sketch of x2–x–2 showing both x intercepts appropriately or correct trials for x = 2 and x = –1 C2 C1 for brackets that would give – x or –2 or for correct formula with one wrong sign or for quadratic with wrong or missing intercepts or correct trial for x = 2 or x = –1 6(a) x2 – x = yz – xy – xz +x 2 M1 correct expansion of both brackets Correct rearrangement leading to given answer A1 6(b) 6 × 5 = (y – 6)(z – 6) C1 Two pairs from 1 30 2 15 3 10 5 6 C1 6, 7, 36 6, 8, 21 6, 9, 16 6, 11, 12 2 B1 for one correct set other than 6,11,12

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 6 of 7 Question Answer Marks Partial Marks 7(a) [Area =] length × breadth and 600 2 2 x A x    or 600 2 A x x          2 B1 for 600 – 2x or 600 300 2  or clearly stated that the other sides are 300 – x B1 for [A =] length × breadth 7(b) 0 < length < 300 [m] 1 7(c) Correct sketch 1 0 and 300 labelled Suitable scale on A axis C2 C1 for 0 and 300 labelled or for suitable scale on A axis or coordinates of maximum labelled 7(d) [Area =] 22 500 [Length =] 150 1 m2 and m C1 8(a) B = an expression in x C1   500 2 x B x    oe 2 B1 for 500 2 x  seen 8(b) Sketch of   500 2 x B x    with x-intercepts 0 and 500 or substituting x = 250 C1 31 250 or 31200 or 31300 nfww 1 Model B has greater area by 8750 [m2] 1 FT their areas if maximum for B given 9(a) 0.35x (300 – x) or 0.35(300x – x2) or e.g. 81x – 0.27x2 + 24x – 0.08x2 2 B1 for 0.35x or 0.27x (300 – x) or 81x – 0.27x2 or 0.08x (300 – x) or 24x – 0.08x2 Correct completion to C = 105x – 0.35x2 + 7000 1

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 7 of 7 Question Answer Marks Partial Marks 9(b) Substituting their 150 into C and including $ on answer line C1 14 875 1 FT their 150 9(c)(i)   2 300 105 35 7000 5 oe 450 x x x x      2 B1 for   300 5 450 x x   or 5 450 A  oe 9(c)(ii) Substitutes their 150 into their   300 5 450 x x   or their 5 450 A  C1 $14 625 1 10(a) A = 200cos w × 200sin w = 40 000sin wcos w 2 B1 for 200cos w or 200sin w seen 10(b) Correct sketch 1 0 and 90 labelled C1 10(c)(i) 20 000 [m2] 1 10(c)(ii) 566 m 1 200sin 45 or 200cos 45 or 20000 their C1