Cambridge IGCSE Mathematics - International 0607 — 2024 May/June Paper 5 · Variant 3

This document has 8 pages. [Turn over * 1 9 6 0 0 2 2 8 6 5 * Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/53 Paper 5 Investigation (Core) May/June 2024 1 hour 10 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 36. ● The number of marks for each question or part question is shown in brackets [ ]. DC (DE) 332310/1 © UCLES 2024

2 0607/53/M/J/24 © UCLES 2024 Answer all the questions. INVESTIGATION CHANGING THE ORDER OF OPERATIONS 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 In this question x 2 = , y 4 = , z 6 = . Complete calculation 1 and calculation 2. Calculation 1 y z 4 6 + = + = … … x # = … 2 # = … Result = … Calculation 2 y z 4 6 # # = = … … x = + … 2 = + … Result = … [4]

3 0607/53/M/J/24 © UCLES 2024 [Turn over 2 (a) Work out the results of calculation 1 and calculation 2 when x 3 = , y 4 = , z 9 = . Result of calculation 1: … Result of calculation 2: … [2] (b) What do you notice about your results in part (a)? … [1]

4 0607/53/M/J/24 © UCLES 2024 3 In this question x and z do not change in each part. (a) In this part x 2 = and z 4 = . Complete the table. Calculation 1 Calculation 2 x y z y z + x result # = y z # x result + = 2 1 4 5 4 2 2 4 12 10 2 3 4 [3] (b) In this part x 3 = and z 6 = . y increases by 1 each time. Continue the table until the two results are the same. You may not need all the rows. x y z y z + x result # = y z # x result = + 3 1 6 7 21 6 9 3 2 6 8 24 12 15 3 3 6 [3]

5 0607/53/M/J/24 © UCLES 2024 [Turn over (c) In this part x 4 = and z 8 = . Find the value of y that makes the results of calculation 1 and calculation 2 the same. Use this table to help you. Calculation 1 Calculation 2 x y z y z + x result # = y z # x result = + y = … [3]

6 0607/53/M/J/24 © UCLES 2024 4 The results of calculation 1 and calculation 2 are the same. (a) Complete the table. Use your answers to Question 3 and any patterns you notice to help you. x y z 2 4 3 6 4 8 5 6 17 [3] (b) Find expressions in terms of x for y and for z. y = … z = … [2] 5 In this question x 1 = and y 1 = . (a) Write down expressions for calculation 1 and for calculation 2 in terms of z. Show that these results are the same. [2] (b) What do your results in part (a) tell you about the value of z? … [1]

7 0607/53/M/J/24 © UCLES 2024 [Turn over 6 In this question x, y and z are consecutive. For example 15, 16 and 17. (a) Complete the table. x y z 15 16 17 12 20 [1] (b) Write y and z in terms of x. y = … z = … [1] (c) Show that the result of calculation 1 is x x 2 3 2 + . [2] (d) Find the result of calculation 2 in terms of x. Give your answer in its simplest form. … [3] (e) The results of calculation 1 and calculation 2 are the same. Show that x x 2 2 = + . [1] Question 6(f) is printed on the next page.

8 0607/53/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. (f) The results of calculation 1 and calculation 2 are the same. x, y and z are consecutive. x, y and z are all between 5 - and 5. Find two sets of values for x, y and z. … [4]

This document consists of 5 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/53 Paper 5 (Core) May/June 2024 MARK SCHEME Maximum Mark: 36 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.

0607/53 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 2 of 5 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.

0607/53 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 3 of 5 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

0607/53 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 4 of 5 Question Answer Marks Partial Marks 1 (4 + 6 = )10 (10 × x = 10 × 2 =) 20 (4 × 6 =) 24 (24 + x = 24 + 2) = 26 4 B1 for each answer FT their 10  2 FT their 24 + 2 2(a) 4 + 9 (=13) and their 13  3 or 4  9 (= 36) and their 36 + 3 C1 39 39 1 2(b) Results are the same oe 1 FT their answers to 2(a) 3(a) 5 10 4 6 6 12 8 10 7 14 12 14 3 B2 for four cells correct or B1 for two cells correct 3(b) 7 21 6 9 8 24 12 15 9 27 18 21 10 30 24 27 11 33 30 33 3 B2 for eight cells correct or B1 for four cells correct 3(c) 2 correct different rows in table C1 7 2 B1 for 60 seen correctly in y = 7 row or for both results 60 linked to 7 4(a) x y z 2 3 4 3 5 6 4 7 8 5 9 10 6 11 12 9 17 18 3 B2 for four entries correct or B1 for two entries correct If 0 or 1 scored and y numbers consecutive, SC1 for 17 – their 11 added to x = 6 and then correctly doubled to get z (in last row)

0607/53 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 5 of 5 Question Answer Marks Partial Marks 4(b) (y =) 2x – 1 oe (z =) 2x oe 2 B1 for each 5(a) (1 + z) × 1 1 × z + 1 1 1 + z and z + 1 isw 1 5(b) z can be any integer oe 1 6(a) x y z 15 16 17 12 13 14 18 19 20 1 6(b) (y =) x + 1 oe (z =) x + 2 oe 1 6(c) [x] (x + 1 + x + 2) 1 x2 + x + x2 + 2x or x (2x + 3) leading to 2x2 + 3x 1 6(d) (x + 1)(x + 2) [+ x] C1 FT their expressions in 6(b) if binomial x2 + x + 2x + 2 [+ x] C1 FT correct expansion of their brackets x2 + 4x + 2 1 6(e) 2x2 + 3x = x2 + 4x + 2 leading to x2 = x + 2 1 6(f) x = 2, y = 3, z = 4 x = –1, y = 0, z = 1 2 B1 for each set Correct sketch of x2 and x + 2 or of x2 – x – 2 AND both intersections marked OR Correct trials for x = 2 and x = –1 C2 C1 for correct sketch without intersections OR C1 for one correct trial for either x = 2 or x = –1