Cambridge IGCSE Mathematics - International 0607 — 2018 May/June Paper 6 · Variant 1

* 5 5 7 2 3 8 5 1 7 2 * This document consists of 12 printed pages. DC (LK/SG) 153611/3 © UCLES 2018 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 (Extended) May/June 2018 1 hour 30 minutes Candidates answer on the Question Paper. Additional Materials: Graphics Calculator READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. Do not use staples, paper clips, glue or correction fluid. You may use an HB pencil for any diagrams or graphs. DO NOT WRITE IN ANY BARCODES. Answer both parts A (Questions 1 to 4) and B (Questions 5 to 9). You must show all the relevant working to gain full marks for correct methods, including sketches. In this paper you will also be assessed on your ability to provide full reasons and communicate your mathematics clearly and precisely. At the end of the examination, fasten all your work securely together. The total number of marks for this paper is 40.

2 0607/61/M/J/18 © UCLES 2018 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 4) LARGEST PRODUCTS (20 marks) You are advised to spend no more than 45 minutes on this part. This investigation looks at finding the largest product when two or more positive integers have a given sum. For the positive integers 2 and 5 • the sum 2 + 5 is 7 • the product 2 × 5 is 10. 1 (a) (i) Complete the table for all the different pairs of positive integers that have a sum of 8. Integers Sum Product 1 8 2 8 3 5 8 4 8 16 Write down the calculation that gives the largest product. … (ii) Find the largest product of two positive integers that have a sum of 20. … Write down the calculation that gives this product. …

3 0607/61/M/J/18 © UCLES 2018 [Turn over (b) (i) Complete this table for all the different pairs of positive integers that have a sum of 9. Note that 2 and 7 is the same pair as 7 and 2. Integers Sum Product 1 9 9 9 9 Write down the calculation that gives the largest product. … (ii) Find the largest product for two positive integers that have a sum of 21. … Write down the two positive integers that give this product. …

4 0607/61/M/J/18 © UCLES 2018 2 Tom wants to find the set of positive integers that have a sum of 19 and give the largest product. He uses the ideas from question 1 and the following method. • Write 19 in the top box of the diagram. • Write the two positive integers that give the largest product in the two boxes on the next row. • Complete the remaining boxes in the same way. (a) Complete the diagram. 19 9 2 2 2 3 2 3 2 3 The largest product using Tom’s method is 2 × 2 × 2 × 3 × 2 × 3 × 2 × 3 = 25 × 33 = 864. (b) Explain why Tom does not replace the 2s and the 3s to make another row in the diagram. … …

5 0607/61/M/J/18 © UCLES 2018 [Turn over (c) Complete the diagram and find the largest product that Tom’s method gives for positive integers that have a sum of 22. 22 …

6 0607/61/M/J/18 © UCLES 2018 3 In any set of positive integers, the replacement of a set of three 2s with a set of two 3s will not change the sum. (a) Explain why this is correct. … (b) Explain why this replacement will always increase the product of a set of positive integers. … … (c) Here is Tom’s result from question 2(a). 2 × 2 × 2 × 3 × 2 × 3 × 2 × 3 = 25 × 33 = 864 (i) Use the replacement to calculate the largest product of a set of positive integers that have a sum of 19. … (ii) Write your answer to part (i) as a product of powers of 2 and 3. …

7 0607/61/M/J/18 © UCLES 2018 [Turn over 4 (a) The sum of a set of positive integers is N. After replacing each set of three 2s in Tom’s method with two 3s, the largest product will be 2x × 3y. (i) Find, in terms of x and y, a formula for the sum N. … (ii) Explain why x can only be 0, 1 or 2. … … (b) Use part (a) to find the largest product of a set of positive integers that have a sum of (i) 60, … (ii) 62. … (c) 6 377 292 is the largest product of a set of positive integers that have a sum of N. Find the value of N. …

8 0607/61/M/J/18 © UCLES 2018 B MODELLING (QUESTIONS 5 to 9) COUNTING PRIME NUMBERS (20 marks) You are advised to spend no more than 45 minutes on this part. Prime numbers have only two factors: 1 and the number itself. In this task you will use models for the number of primes in a given range. The function P(x) gives the number of primes that are less than x. For example, P(10) = 4 because there are exactly four primes that are less than 10. These four primes are 2, 3, 5 and 7. 5 (a) Show that P(20) = 8. (b) P(40) = 12. Find P(50). … (c) P(100) = 25. Find P(90). …

9 0607/61/M/J/18 © UCLES 2018 [Turn over 6 Use your answers to question 5 to complete the table. x 10 20 40 50 90 100 140 200 P(x) 4 8 12 25 34 46 On the grid plot this information. The first three points have been plotted for you. 50 40 30 20 10 0 0 20 40 60 80 100 120 140 160 180 200 x P(x)

10 0607/61/M/J/18 © UCLES 2018 7 A model, L(x), for the number of primes less than x, is a straight line that passes through the points (40, 12) and (140, 34). (a) Find the equation of the line. … (b) Use your model to estimate the number of primes that are less than 1000. … 8 A quadratic model, Q(x), for the number of primes that are less than x, is Q(x) = kx – 0.001x2 . (a) The model estimates that there are 34 primes that are less than 140. Find the value of k, giving your answer correct to 1 decimal place. … (b) The model is quite accurate for the number of primes when x 1 200. Why is the model unsuitable for the number of primes when x 2 200? … …

11 0607/61/M/J/18 © UCLES 2018 [Turn over 9 A very good model for the number of primes less than x, where x is less than 1000, is ( ) ( ) N log x x x 2 10 = . (a) Use the model to find how many primes there are that are (i) less than 400, … (ii) less than 800. … (b) Use your answers to part (a) to complete the table. x 0 200 1 G x 200 400 1 G x 400 00 6 1 G x 600 00 8 1 G Number of primes 44 31 Question 9(c) and 9(d) are printed on the next page.

12 0607/61/M/J/18 © UCLES 2018 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 International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. (c) The model is still a good model for x 1000 H . (i) A number is chosen at random from the numbers 1 to 10 000. Use the model to estimate the probability that it is prime. … (ii) A number is chosen at random from numbers in the range 1 to x. Find, in its simplest form, a model for the probability that it is prime. … (d) Use the model ( ) ( ) log x x x 2 N 10 = to estimate the 100th prime. …

IGCSE™ is a registered trademark. This document consists of 7 printed pages. © UCLES 2018 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 (Extended) May/June 2018 MARK SCHEME Maximum Mark: 40 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 2018 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 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 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.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 3 of 7 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

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 4 of 7 Question Answer Marks Partial Marks A INVESTIGATION LARGEST PRODUCTS 1(a)(i) 1 7 8 7 2 6 8 12 3 5 8 15 4 4 8 16 4 × 4 1 1(a)(ii) 100 10 × 10 1 1(b)(i) 1 8 9 8 2 7 9 14 3 6 9 18 4 5 9 20 4 × 5 or 5 × 4 1 1(b)(ii) 110 10 11 1 2(a) 10 4 5 5 5 1 2(b) Valid explanation 1 2(c) 2916 nfww 2 B1 for 2 3 3 3 2 3 3 3 nfww If 0 scored, SC1 for correct product of their bottom row of 2s and 3s C opportunity 3(a) 2 + 2 + 2 = 3 + 3 [= 6] oe 1 3(b) 2 × 2 × 2 [8] < 3 × 3 [9] oe 1 3(c)(i) 972 1 3(c)(ii) 22 × 35 1 4(a)(i) [N = ] 2x + 3y 1 C opportunity 4(a)(ii) No more sets of three 2s to change when x < 3 oe 1 4(b)(i) 3 486 784 401 2 B1 for x = 0, y = 20 soi or 34 × 224 = 34 × 316 oe seen C opportunity

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 5 of 7 Question Answer Marks Partial Marks 4(b)(ii) 6 973 568 802 1 FT their (b)(i) × 2 C opportunity 4(c) 43 nfww 2 B1 for [22] × 313 or [4] × 313 soi or 3 log 1594 323 oe or x = 2, y = 13 soi C opportunity Communication: Seen in two of the following questions 1 2(c) Showing how their 2916 was calculated e.g. 36 × 22 or 3 × 3 × 3 × 3 × 3 × 3 × 2 × 2 4(a)(i) N = an expression in x and/or y 4(b)(i) 3y = 60 or N = [0] + 3(20) or 60 = [0] + 3(20) oe 4(b)(ii) Solution to 2x + 3y = 62 shown by x = 1, y = 20 or 2 + 3 × 20 = 62 or 2 + 60 = 62 or N = 2 + 60 oe 4(c) 6 377 292 ÷ 4 or 1594323 log log3 or 3 log 1594 323 or 6 377 292 ÷ 3 repeatedly reducing to 4 oe or correct method for prime factorisation of 6 377 292

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 6 of 7 Question Answer Marks Partial Marks B MODELLING COUNTING PRIME NUMBERS 5(a) 2, 3, 5, 7, 11, 13, 17, 19 1 5(b) 15 nfww 1 C opportunity 5(c) 24 nfww 1 C opportunity 6 Table completed Correct points plotted 1 FT their table 7(a) L(x) = 0.22x + 3.2 oe 2 M1 for 0.22 or 34 12 140 40 − − oe soi or for correctly setting up and solving a pair of simultaneous equations to find 0.22 or 3.2 C opportunity 7(b) 223 1 FT their y = mx + c with their m between 0 and 1 and their c ≠ 0 C opportunity 8(a) [0].4 2 M1 for 140k – 0.001(1402) = 34 or better C opportunity 8(b) There are more primes but the graph or function value decreases oe 1 9(a)(i) 77 1 Condone 76 9(a)(ii) 138 1 Condone 137 If 0 scored in (a)(i) and (a)(ii) then SC1 for 76.8[6…] or 76.9 in (a)(i) and 137.7[8…] or 137.8 in (a)(ii) 9(b) 33 30 1 STRICT FT For 33, FT their (a)(i) – 44 provided integers For 30, FT their (a)(ii) – their (a)(i) – 31 provided integers C opportunity 9(c)(i) 0.125 oe 2 B1 for 1250 9(c)(ii) 2 1 1 or 2log log x x 1 C opportunity

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2018 © UCLES 2018 Page 7 of 7 Question Answer Marks Partial Marks 9(d) 547 2 M1 100 2log x x = soi C opportunity Communication: Seen in five of the following questions 2 1 for communication seen in three questions 5(b) 41, 43, 47 indicated 5(c) 97 indicated 7(a) Substitution of (40, 12) or (140, 34) in y = their 0.22x + k or y – b = their 0.22(x – a) oe or method for finding the second unknown if using simultaneous equations 7(b) Substitution of 1000 seen 8(a) Correct intermediate step after initial substitution 9(b) Relevant subtraction seen 9(c)(ii) Correct unsimplified form 9(d) Sketch of a pair of relevant graphs seen or three improving trial and improvement steps