Cambridge IGCSE Mathematics - International 0607 — 2021 Oct/Nov Paper 6 · Variant 3

This document has 12 pages. Cambridge IGCSE™ * 4 4 2 7 9 2 6 5 9 0 * DC (KS/SG) 212585/1 © UCLES 2021 [Turn over CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/63 Paper 6 Investigation and Modelling (Extended) October/November 2021 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 6). ● 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/O/N/21 © UCLES 2021 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 3) GIRARD’S SUMS (30 marks) You are advised to spend no more than 50 minutes on this part. Albert Girard, a 17th century French mathematician, investigated numbers, N, that can be written as the sum of two squares, a b 2 2 + . This task is about these numbers. For this task, a and b are integers where a 0 H and b 0 H . 1 (a) Complete the table. a a2 b b2 N a b 2 2 = + N 4 ' 2 4 6 36 40 10 remainder 0 18 10 106 remainder 0 28 16 256 remainder 0 4 64 20 remainder 0 144 196 85 remainder 0 20 400 884 221 remainder 0 0 0 900 225 remainder 0 [5] (b) (i) When a 2 = and b 4 = then N k 4 = , so N is a multiple of 4. Find the value of k. … [2]

3 0607/63/O/N/21 © UCLES 2021 [Turn over (ii) The values of a and b in the table are all even numbers. When a m 2 = and b n 2 = then N k 4 = . Find an expression for k in terms of m and n. … [3] (c) Not all multiples of 4 can be written as the sum of two square numbers. Show that there are no values of a and b that give k 11 = . [2]

4 0607/63/O/N/21 © UCLES 2021 2 (a) Complete the table. a a2 b b2 N a b 2 2 = + N 4 ' 7 49 5 25 74 18 remainder 2 21 19 802 200 remainder 2 17 289 914 remainder 2 49 170 remainder 2 1 1 remainder [4] (b) When a is an odd number, a n 2 1 = - . (i) Use algebra to explain why, when a is an odd number, a 4 2 ' has a remainder of 1. … [3] (ii) Explain why, for the values in the table in part (a), N is always k 4 2 + . … … [2]

5 0607/63/O/N/21 © UCLES 2021 [Turn over (c) When a and b are both odd, , N k 4 2 = + so N is a multiple of 4 plus 2. Not all multiples of 4 plus 2 can be written as the sum of two square numbers. Find all the values of k from 1 to 9 where N a b 2 2 = + . … [3]

6 0607/63/O/N/21 © UCLES 2021 3 The values of N that can be written as the sum of two square numbers are of the form , k r 4 + where the remainder r is a constant. (a) Explain why r can be 0, 1 or 2 but cannot be 3. [3] (b) N a b 2 2 = + Find all the values of N, where , N 10 30 1 1 that are of the form k 4 1 + . … [3]

7 0607/63/O/N/21 © UCLES 2021 [Turn over THE MODELLING TASK STARTS ON PAGE 8.

8 0607/63/O/N/21 © UCLES 2021 B MODELLING (QUESTIONS 4 to 6) PRODUCTION BOUNDARIES (30 marks) You are advised to spend no more than 50 minutes on this part. This task is about the number of computer tablets and mobile phones a company makes and sells. The company owns two factories, A and B. Factory A makes A-tablets and A-phones. Factory B makes B-tablets and B-phones. A production boundary is a curve or line. Points on the curve or line are the maximum numbers of the two items a factory can make when all resources are used. It is the boundary of the region which shows all the combinations of the two items a factory can make. 4 Factory A makes t A-tablets and p A-phones each day. The manager of factory A uses the model p t 9000 1000 2 = - where , t 0 H as the production boundary for the output of A-tablets and A-phones. (a) On the axes below, sketch this model. 10 000 p t 4000 0 Number of A-phones Number of A-tablets [2] (b) When factory A makes 9000 A-phones it cannot make any A-tablets. Write down the maximum number of A-tablets it can make when it does not make any A-phones. … [1] (c) On Monday, factory A makes 1000 A-tablets. On Tuesday, factory A makes 1500 A-tablets. Find the decrease in the maximum number of A-phones it can make from Monday to Tuesday. … [3]

9 0607/63/O/N/21 © UCLES 2021 [Turn over (d) (i) On Wednesday, factory A makes 5000 A-phones. Use your graph from part (a) to explain why it is not possible for it to make 2500 A-tablets on Wednesday. … … [1] (ii) On the graph in part (a) shade the region that represents the numbers of A-phones and A-tablets that factory A can make. [1] (e) The company sells all the A-phones and A-tablets that factory A makes each day. The company makes $160 profit for each A-tablet and $100 profit for each A-phone it sells. The greatest possible daily profit at factory A is $964 000. (i) Write down a linear equation for this profit in terms of p and t. Give your answer in the form p mt c = + . … [2] (ii) Find the number of A-tablets and A-phones that factory A should sell in order to make a profit of $964 000. t = … p = … [3]

10 0607/63/O/N/21 © UCLES 2021 5 Factory B makes t B-tablets and p B-phones. The table shows the maximum numbers of B-phones that factory B can make each day for some numbers of B-tablets. Number of B-tablets t Number of B-phones p 1000 8000 2000 6000 3000 4000 4000 2000 As the number of B-tablets increases, the number of B-phones decreases at a constant rate. (a) (i) Draw the production boundary for factory B on the axes below. p t 0 [2] (ii) Find the equation which models this production boundary, giving p as a function of t. … [2]

11 0607/63/O/N/21 © UCLES 2021 [Turn over (iii) Factory B makes at least 1000 B-tablets but no more than 4000 B-tablets each day. Write down the domain of the model in part (a)(ii). … [1] (b) The company sells all the B-tablets and B-phones factory B makes each day. The company makes $200 profit for each B-tablet and $190 profit for each B-phone it sells. Each day, the manager of factory B expects to make the greatest possible profit. (i) Find the greatest possible profit each day. … [3] (ii) One day factory B has to make 2500 B-tablets. On this day the profit is 73.3% of the greatest possible profit. Work out the number of B-phones factory B makes on this day. … [4] Question 6 is printed on the next page.

12 0607/63/O/N/21 © UCLES 2021 6 The company puts new machinery to make phones in factory A and factory B. Factory A can now make double the number of A-phones. Factory B can now make 10% more B-phones. All other conditions remain the same. (a) Complete the following models for the production boundaries at each factory after the changes. Use the models in Question 4 and Question 5(a). Factory A: p = … for t 0 H Factory B: . p t 2 2 11000 = - + for … G t G … [2] (b) After the changes, the greatest possible profit made each day by factory A is $1 830 000. Find the total greatest possible profit made each day by the company. … [3] 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 the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge.

This document consists of 8 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/63 Paper 6 (Extended) October/November 2021 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 2021 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 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 descriptors 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 October/November 2021 © UCLES 2021 Page 3 of 8 Maths-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, 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 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 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 October/November 2021 © UCLES 2021 Page 4 of 8 Question Answer Marks Partial Marks A INVESTIGATION GIRARD’S SUMS 1(a) a a2 b b2 N = a2 + b2 4 N Rem 2 4 6 36 40 10 0 18 324 10 100 424 106 0 28 784 16 256 1040 260 0 4 16 8 64 80 20 0 12 144 14 196 340 85 0 20 400 22 484 884 221 0 30 900 0 0 900 225 0 5 B4 for 13, 14 or 15 correct cells or B3 for 10, 11 or 12 correct cells or B2 for 7, 8 or 9 correct cells or B1 for 4, 5 or 6 correct cells 1(b)(i) 4 + 16 = 20 or 22 + 42 = 20 C1 [k =] 5 1 1(b)(ii) 2 2 (2 ) (2 ) m n + C1 [ ] 2 2 4 4 4 k m n = + 1 2 2 m n + 1 1(c) Shows, using an exhaustive list, that there are no values of 2 2 + m n that give 11 by adding square numbers or Shows, using an exhaustive list, that there are no values of 2 m or 2 n that when subtracted from 11 give a square number OR Shows, using an exhaustive list, that adding square numbers that are multiples of 4 and less than 44 does not give 44 or Shows, using an exhaustive list, that subtracting square numbers that are multiples of 4 and less than 44, from 44, does not produce a square number that is even and a multiple of 4 2 B1 for finding one correct example

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 5 of 8 Question Answer Marks Partial Marks 2(a) a a2 b b2 N = a2 + b2 4 N Rem 7 49 5 25 74 18 2 21 441 19 361 802 200 2 17 289 25 625 914 228 2 11 121 7 49 170 42 2 1 1 1 1 2 0 2 3 B2 for 9, 10, 11, 12 or 13 cells correct or B1 for 6, 7 or 8 cells correct A relevant calculation C1 2(b)(i) 4n2 − 2n − 2n + 1 C1 4(n2 − n) + 1 or 2 4 4 1 4 − + n n 1 Correct statement about multiple or factor of 4 with remainder 1 1 2(b)(ii) Shows, using an exhaustive list that each of the five values of N is a multiple of 4 + 2 OR Valid explanation e.g. Both a and b are odd numbers, therefore a2 and b2 are both multiples of 4 remainder 1 so the total is a multiple of 4 remainder 2 2 B1 for two correct examples OR B1 for a partially correct explanation e.g. both are multiples of 4 remainder 1 so the total is a multiple of 4 remainder 2 2(c) Four correct values of 4k + 2 seen from k 1 2 3 4 5 6 7 8 9 4k + 2 6 10 14 18 22 26 30 34 38 C1 Finds at least two of the sums of two odd squares using 1, 9, 25: 12 + 12 = 2, 12 + 32 = 10, 32 + 32 = 18, 12 + 52=26, 32 + 52 = 34 C1 Three from 2, 4, 6, 8 1

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 6 of 8 Question Answer Marks Partial Marks 3(a) (even)2 + (even)2 gives remainder 0 (odd)2 + (odd)2 gives remainder 2 C1 [Remaining case] (odd)2 + (even)2 gives remainder 1 and These are the only possible cases so r cannot be 3 oe 2 B1 for (odd)2 + (even)2 gives remainder 1 3(b) Two from 13 = 32 + 22 oe 17 = 42 + 12 oe 18 = 32 + 32 oe 20 = 22 + 42 oe 25 = 42 + 32 oe 26 = 12 + 52 oe 29 = 52 + 22 oe C1 13, 17, 25, 29 2 B1 for any two correct B MODELLING PRODUCTION BOUNDARIES 4(a) Correct sketch 4000 10000 0 9000 3000 t p 1 correct shape 3000 and 9000 as intercepts indicated on graph C1 4(b) 3000 1 4(c) their 8000 – their 6750 or 2 1000 9000 1000 − or 2 1500 9000 1000 − or vertical lines at 1000 and 1500 on sketch C1 8000 or 6750 1 1250 1 4(d)(i) Valid explanation using graph e.g. The maximum number of A-tablets would be 2000 or It is a point beyond the curve 1

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 7 of 8 Question Answer Marks Partial Marks 4(d)(ii) Correct region shaded 4000 10000 0 9000 3000 t p 1 FT their sketch 4(e)(i) 160t + 100p = 964 000 oe C1 p = −1.6t + 9640 1 4(e)(ii) sketch of graphs or suitable line drawn on graph in 4(a) or equating models: 9640 – 1.6t = 9000 − 2 1000 t oe C1 [t =] 800 [p =] 8360 2 B1 for each 5(a)(i) Correct ruled graph 5000 10000 0 p t 1 5000 and 10 000 as intercepts indicated on graph C1 5(a)(ii) p = −2t + 10 000 oe 2 B1 for −2t + 10 000 or [p =] −2t + c, c ≠ 0 or [p =] mt + 10 000, m ≠ 0 5(a)(iii) 1000 ⩽ t ⩽ 4000 1

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 8 of 8 Question Answer Marks Partial Marks 5(b)(i) A correct valid calculation e.g. 1000 × 200 + 8000 × 190 C1 1 720 000 1 For $ used with their numerical answer C1 5(b)(ii) 4004 nfww 4 M1 for 0.733 × their 1 720 000 or 1 260 760 seen M1 for 2500 × 200 or 500 000 seen M1 for 1260760 500000 190 − their their soi 6(a) A: 2 2 9000 1000   = −     t p oe where t ⩾ 0 1 B: p = −2.2t + 11 000 for 1000 ⩽ t ⩽ 4000 1 6(b) (−2.2 × 1000) + 11 000 or 1.1 × 8000 or suitable sketch C1 [1 830 000 +] 200 × 1000 + 190 × 8800 C1 3 702 000 1