Cambridge IGCSE Mathematics - International 0607 — 2019 Oct/Nov Paper 6 · Variant 3

This document consists of 12 printed pages. DC (CE/FC) 170794/3 © UCLES 2019 [Turn over * 4 8 2 1 2 0 3 5 5 7 * CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/63 Paper 6 (Extended) October/November 2019 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 6) and B (Questions 7 to 9). You must show all 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 to 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. Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education

2 0607/63/O/N/19 © UCLES 2019 The Investigation starts on the next page.

3 0607/63/O/N/19 © UCLES 2019 [Turn over Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 6) REMAINDERS (20 marks) You are advised to spend no more than 45 minutes on this part. This investigation is about the remainder when one positive integer is divided by another. Example 8 38 0 ' = , with a remainder of 8. This can be written as [ ] 8 38 8 R ' = . 1 Find (a) [ ] 13 5 R ' , … (b) [ ] 5 13 R ' . … 2 Show that [ ] 4000 19 10 R ' = . 3 x is a factor of 20. Show that [ ]x 20 0 R ' = . 4 For a positive integer, n, write down the largest and smallest values of [ ] n 100 R ' . Largest … Smallest …

4 0607/63/O/N/19 © UCLES 2019 5 (a) Complete this table of values of [ ] a b R ' . b 1 2 3 4 5 6 a 1 0 1 1 1 1 1 2 0 3 0 4 0 5 0 6 0 (b) [ ] [ ] 10 10 10 9 9 R R ' ' + = [ ] [ ] 11 11 11 7 7 R R ' ' + = These examples suggest that [ ] [ ] x x x y y R R ' ' + = . Use values from the table to show one example that this is not always true. (c) The remainder when a positive integer, n, is divided by 100 is [ ] n 100 R ' . Explain why dividing [ ] n 100 R ' by 100 gives the same remainder. … …

5 0607/63/O/N/19 © UCLES 2019 [Turn over 6 In this question x, y and z are positive integers. (a) When [ ] x z a R ' = and [ ] y z b R ' = , then [ ] [ ] xy z ab z R R ' ' = . (i) Check that this is true when x = 2, y = 8 and z = 5. (ii) Using algebra, show that [ ] [ ] x z a z R R 2 2 ' ' = . (iii) Use the result in part(a)(ii) to show that [ ] 76 7 1 R 2 ' = .

6 0607/63/O/N/19 © UCLES 2019 (b) From part (a) [ ] [ ] xy z ab z R R ' ' = and [ ] [ ] x z a z R R 2 2 ' ' = . (i) Use patterns to help you complete the table. [ ] 7919 13 R ' = 2 [ ] [ ] 7919 13 2 13 R R 2 2 ' ' = [ ] 4 13 R ' = 4 = [ ] [ ] 7919 13 13 4 R R 2 4 ' ' = [ ] 13 16 R ' = 3 = [ ] [ ] 7919 13 13 3 R R 8 2 ' ' = [ ] 13 9 R ' = 9 = [ ] R = (ii) 79197is over 20 digits long. The following example shows how to calculate [ ] 7919 13 R 7 ' . 7919 7919 7919 7919 7 4 2 1 # # = [ ] [( ) ] 7919 13 3 4 2 13 R R 7 ' ' # # = [ ] 24 13 R ' = 11 = Find [ ] 7919 13 R 11 ' . …

7 0607/63/O/N/19 © UCLES 2019 [Turn over (iii) Is 13 a factor of 79199? Show how you decide. (c) Work out whether 7 is a factor of 7919 5 64 + .

8 0607/63/O/N/19 © UCLES 2019 B MODELLING (QUESTIONS 7 to 9) ORBITING SATELLITES (20 marks) You are advised to spend no more than 45 minutes on this part. This investigation is about satellites orbiting the Earth at a height, h kilometres, above the ground. h 7 The scatter diagram shows the heights and the orbit times, t minutes, for 9 satellites making circular orbits around the Earth. 200 90 95 100 105 Time (minutes) Height (km) 110 115 t h 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 (a) (i) The mean of these satellite heights is 770 km and the mean orbit time is 100 minutes. Plot this point. (ii) Draw a line of best fit. (iii) Use your line of best fit to find a straight line model (Model A) connecting t and h. Give your answer in the form . t mh c = +

9 0607/63/O/N/19 © UCLES 2019 [Turn over (b) Another model (Model B) connecting t and h is . ( ) t h 1 659 10 6370 4 3 # # = + - . The satellite NORAD 40730 has a circular orbit. It takes 728.9 minutes to orbit the Earth at a height of 20 450 km. Find which of the two models gives an orbit time closer to the actual orbit time. (c) Communication satellites need an orbit time of 1440 minutes. Use Model B to find the height which gives an orbit time of 1440 minutes.

10 0607/63/O/N/19 © UCLES 2019 8 Some satellites do not have circular orbits. 114 027 km 7079 km NOT TO SCALE The height of satellite NORAD 25989 is between 7079 km and 114 027 km. It has an orbit time of 2872 minutes. Use h as the mean of these two heights to calculate the orbit time using Model B. . ( ) t h 1 659 10 6370 4 3 # # = + - How does this time compare with the actual orbit time?

11 0607/63/O/N/19 © UCLES 2019 [Turn over 9 Some satellites eventually fall back to the Earth. A model for the descent time of a satellite, D days, with an initial height, h kilometres, is D h 5 10 . 24 9 64 # = - ` j . (a) Sketch the graph of this model for h 400 600 G G . D h 400 0 3000 600 (b) The first satellite, Sputnik 1, had an initial height of 577 km and a descent time of 92 days. Compare this descent time with the descent time given by the model. Question 9(c) is printed on the next page.

12 0607/63/O/N/19 © UCLES 2019 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. (c) Radiation from the Sun affects the descent time, D days. This radiation, s, is measured in solar flux units (SFU). The table shows the descent times, in days, for different values of h and s. Radiation (s SFU) 20 40 60 80 100 120 140 160 Initial height (h km) 500 553 350 233 163 118 88 68 53 475 298 195 134 96 71 54 43 34 450 158 107 76 56 43 33 27 22 425 82 58 42 32 25 20 16 14 400 42 31 23 18 15 12 10 9 (i) Write down the descent time of a satellite with an initial height of 450 km when the radiation is 100 SFU. … (ii) What does the table show about the effect of radiation on the descent time of a satellite? … … (iii) A satellite has an initial height of 500 km. D = ks1.1 D = k – s D = ks0.9 D = ks–1 where k is a constant. Which one of these equations is the best model connecting descent time and radiation? Give a reason for your answer. … …

This document consists of 6 printed pages. © UCLES 2019 [Turn over Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/63 Paper 6 (Extended) October/November 2019 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 October/November 2019 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 2019 © UCLES 2019 Page 2 of 6 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 2019 © UCLES 2019 Page 3 of 6 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 2019 © UCLES 2019 Page 4 of 6 Question Answer Marks Partial Marks A INVESTIGATION REMAINDERS 1(a) 3 1 1(b) 5 1 2 4000 – (210 × 19) or 4000 ÷ 19 with 210 R 10 or 0.5263… × 19 = …… rounds to 10 or 4000 – 10 = 3990, 3990 ÷ 19 = 210 1 3 20 divided by at least 5 of the 6 factors of 20 giving the answer and R = 0 for each case 1 4 [Largest] 99 [Smallest] 0 1 5(a) 20 correct entries b 1 2 3 4 5 6 a 1 0 1 1 1 1 1 2 0 0 2 2 2 2 3 0 1 0 3 3 3 4 0 0 1 0 4 4 5 0 1 2 1 0 5 6 0 0 0 2 1 0 1 5(b) A correct counter example, using numbers 1 to 6, showing sum of two remainders 1 5(c) [The] remainder [is] less than 100 oe 1 6(a)(i) [a = ] = 2 [b = ] = 3 R[16 ÷ 5] = 1 oe R[6 ÷ 5] = 1 oe 2 B1 for 2 statements correct 6(a)(ii) x = y and a = b 1 6(a)(iii) R[76 ÷ 7] = 6 and R[62 ÷ 7] = 1 1

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 5 of 6 Question Answer Marks Partial Marks 6(b)(i) R[791916 ÷ 13] [=] R[92 ÷ 13] [=] R[81 ÷ 13] [=] 3 2 B1 for [92 ÷ 13] 6(b)(ii) 7 2 M1 for 7919a × 7919b × … where • only 7919 used • a, b, … are 1 or 2 or 4 or 8 • a + b + … = 11 6(b)(iii) [No] R[18 ÷ 13] = a value or ≠ 0 oe 1 6(c) R[791964 ÷ 7] = 2 2 + 5 = 7 oe leading to ‘yes’ oe 2 B1 for remainder 2 linked to 791964 Communication: seen in one of the following questions 1 6(a)(i) R[2 ÷ 5] and R[8 ÷ 5] seen 6(b)(ii) 72 ÷ 13 or 384 ÷ 13 or 2048 ÷ 13 6(c) R[7919 ÷ 7] = 2

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2019 © UCLES 2019 Page 6 of 6 Question Answer Marks Partial Marks B MODELLING ORBITING SATELLITES 7(a)(i) Point (770, 100) plotted 1 7(a)(ii) Correct straight line through (770, 100) 1 7(a)(iii) [t =] (0.015 to 0.025) h + (80 to 89) 2 B1 for either gradient or y-intercept correct 7(b) [Model A: t =] 300 to 650 [Model B: t =] 728 to 729 [Model B] 2 B1 for correct model B time 7(c) 35 800 to 35 900 2 B1 for one of 8 670 000 to 8 680 000 or 7.53… × 1013 or 42 200 to 42 240 8 2872.1 to 2872.2 and very close oe 3 B2 for 2872.1 to 2872.2 or B1 for 60 553 FT their 2872. … for comment 9(a) Correct sketch 1 9(b) 2070 to 2074 and not close oe 2 B1 for 2070 to 2074 9(c)(i) 43 1 9(c)(ii) Increasing radiation decreases descent time oe 1 9(c)(iii) D = ks–1 only indicated and valid reason 2 B1 for indicating D = ks–1 only Communication: seen in four of the following questions 2 C1 for two or three opportunities seen 7(b) ( ) 3 20450 6370 + seen 7(c) ‘km’ seen as units 8 114027 7079 2 + seen 9(b) 24 9.64 5 10 577 − × × or ( )[ ] 24 9.64 5 10 577 − × × seen 9(c)(i) ‘Days’ seen as units