Cambridge IGCSE Mathematics - International 0607 — 2020 May/June Paper 6 · Variant 1

Cambridge IGCSE™ DC (LK/SG) 187928/1 © UCLES 2020 [Turn over This document has 12 pages. Blank pages are indicated. CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 Investigation and Modelling (Extended) May/June 2020 1 hour 40 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer both part A (Questions 1 to 8) and part B (Questions 9 to 12). ● 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 [ ]. * 7 4 5 7 1 1 5 1 7 9 *

2 0607/61/M/J/20 © UCLES 2020 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 8) COMBINING TRIANGLE NUMBERS (30 marks) You are advised to spend no more than 50 minutes on this part. This investigation looks at results when adding, subtracting and multiplying triangle numbers. Here is a table of the first 5 triangle numbers, T1 to T5. T1 T2 T3 T4 T5 1 3 6 10 15 1 Find the next triangle number. … [2] 2 Complete the table for subtracting consecutive triangle numbers. T1 1 T2 - T1 2 T3 - T2 T4 - T3 T5 - T4 5 T6 - T5 Tn-2 - Tn-3 Tn-1 - Tn-2 Tn - Tn-1 [2]

3 0607/61/M/J/20 © UCLES 2020 [Turn over 3 Complete the table for adding two consecutive triangle numbers. T1 1 T2 + T1 4 T3 + T2 9 T4 + T3 T5 + T4 T6 + T5 Tn + Tn-1 [2] 4 Use the last row of the table in Question 2 to complete equation 1 . T T n n 1 - = - … Use the last row of the table in Question 3 to complete equation 2 . T T n n 1 = + - … (a) By adding equations 1 and 2 together show that T n n 2 n 2 = + . [1] (b) (i) By multiplying equations 1 and 2 together, find a result about the squares of consecutive triangle numbers. [2] (ii) Give a numerical example of this result. [2]

4 0607/61/M/J/20 © UCLES 2020 5 The sum of two different triangle numbers sometimes equals another triangle number. When this happens, we have a triangle triple. Example • Start with the triangle number T 6 3 = • From the table in Question 2 T T 6 6 5 - = So T T T 6 5 3 - = • Rearrange the equation T T T 3 5 6 + = • The triangle triple is then (3, 5, 6) The three different numbers must be written in order of increasing size. (a) Start with triangle number T 15 5 = and complete the method of the Example to find another triangle triple. T15 - … = … So … - … = T5 T5 + … = … The triangle triple is ( 5, … , … ) [3] (b) In the table, each row is a triangle triple. Use part (a) and any patterns you notice to complete the table. Triangle triple 3 5 6 4 5 6 7 [3]

5 0607/61/M/J/20 © UCLES 2020 [Turn over 6 (a) When you add the last two rows in the table in Question 2, you get an expression for T T n n 2 - - . This is the difference between triangle numbers that are two apart. Give this expression in its simplest form. … [1] (b) (i) Find n when T T 15 n n 2 - = - . … [1] (ii) T 15 5 = Use your answer to part (i) to find a triangle triple where • the smallest number is 5 • the difference between the other two numbers is 2. ( 5, … , … ) [2]

6 0607/61/M/J/20 © UCLES 2020 7 (a) By adding rows in the table in Question 2, show that T T n 3 3 n n 3 - = - - . This is the difference between triangle numbers that are three apart. [1] (b) T 105 14 = Use part (a) to find a triangle triple where • the smallest number is 14 • the difference between the other two numbers is 3. ( 14, … , … ) [3]

7 0607/61/M/J/20 © UCLES 2020 [Turn over 8 Find all the triangle triples where the smallest number is 14. [5]

8 0607/61/M/J/20 © UCLES 2020 B MODELLING (QUESTIONS 9 to 12) SPEED OF PLANETS (30 marks) You are advised to spend no more than 50 minutes on this part. This task looks at models for the distance of a planet from the Sun and the time it takes to travel once round the Sun. The task uses these models to find a model for the speed of a planet. Astronomers use astronomical units, au, to measure distance in space. 1 au = the distance from the Sun to the Earth. 9 In the 18th century, the German astronomer Bode numbered the planets Venus to Neptune from 0 to 7. The table shows his numbers and the distance from the Sun to each planet. Bode’s number, n Planet Distance from Sun, R au Bode’s estimates (au) Mercury 0.39 0 Venus 0.72 0.7 1 Earth 1.00 2 Mars 1.52 3 unknown 4 Jupiter 5.20 5 Saturn 9.55 6 Uranus 19.22 19.6 7 Neptune 30.11 Bode estimated the distance from the Sun to Venus as 0.7 au. After that, for each planet, he used the following rule to estimate the distance of the next planet in the table. Double the estimate for the distance and subtract 0.4 (a) Complete the table for Bode’s estimates. [3] (b) Bode’s rule gives a good model for the planets Venus to Uranus. Work out Bode’s estimate for Neptune. Is it a good estimate? … [2]

9 0607/61/M/J/20 © UCLES 2020 [Turn over (c) Find an estimate for Mercury using Bode’s rule. … [2] 10 Bode’s rule requires doubling each time. So a possible model for the distance of a planet from the Sun, R au, is a b R 2n # = + where n is Bode’s number and a and b are constants that transform the graph of R 2n = . (a) Write down the two types of transformation used. … , … [2] (b) Using the information for Earth and Jupiter on page 8, write two equations in a and b. … … [2] (c) Show how solving these simultaneous equations gives . . R 0 3 2 0 4 n # = + . [2]

10 0607/61/M/J/20 © UCLES 2020 11 The table shows the time, T years, it takes a planet to go once round the Sun. The table includes the ‘dwarf planet’ Ceres which Bode’s rule predicts. Planet R (au) T (years) Mercury 0.39 0.24 Venus 0.72 0.62 Earth 1.00 1.00 Mars 1.52 1.88 Ceres 2.77 4.60 Jupiter 5.20 11.86 Saturn 9.55 29.46 Uranus 19.22 84.01 Neptune 30.11 164.80 A possible model for T is T k R p # = where k and p are constants. (a) (i) Use the information in the table for Earth to find k. … [2] (ii) Using the result from part (i) and the information in the table for Jupiter, find p correct to 1 decimal place. … [3] (b) Write down the model for T in terms of R. Is this a good model for the time that it takes Mercury to go once round the Sun? Show how you decide. [2]

11 0607/61/M/J/20 © UCLES 2020 [Turn over 12 (a) Assume that planets travel in a circle with the Sun at the centre. Use your model in Question 11(b) and . . R 0 3 2 0 4 n # = + to show that a model for the average speed of a planet is r . . S 0 3 2 0 4 2 n # = + , where n is Bode’s number and S is measured in au/year. [3] (b) Sketch the graph of S for n 0 5 G G . 0 0 5 n S [2] Questions 12(c) and 12(d) are printed on the next page.

12 0607/61/M/J/20 © UCLES 2020 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) The graph in part (b) is approximately a straight line. Find a linear model for S, in terms of n, by finding the equation of this straight line. Write the numbers in your model correct to 1 decimal place. … [3] (d) Bode’s number for Neptune is 7. Show that your model does not give a sensible answer for the speed of Neptune. [2]

This document consists of 7 printed pages. © UCLES 2020 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 (Extended) May/June 2020 MARK SCHEME Maximum Mark: 60 Published Students did not sit exam papers in the June 2020 series due to the Covid-19 global pandemic. This mark scheme is published to support teachers and students and should be read together with the question paper. It shows the requirements of the exam. The answer column of the mark scheme shows the proposed basis on which Examiners would award marks for this exam. Where appropriate, this column also provides the most likely acceptable alternative responses expected from students. Examiners usually review the mark scheme after they have seen student responses and update the mark scheme if appropriate. In the June series, Examiners were unable to consider the acceptability of alternative responses, as there were no student responses to consider. Mark schemes should usually be read together with the Principal Examiner Report for Teachers. However, because students did not sit exam papers, there is no Principal Examiner Report for Teachers for the June 2020 series. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the June 2020 series for most Cambridge IGCSE™ and Cambridge International A & AS Level components, and some Cambridge O Level components.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 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 2020 © UCLES 2020 Page 3 of 7 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

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 4 of 7 Question Answer Marks Partial Marks A INVESTIGATION COMBINING TRIANGLE NUMBERS 1 15 + 6 C1 21 B1 2 T1 1 T2 – T1 2 T3 – T2 3 T4 – T3 4 T5 – T4 5 T6 – T5 6 Tn–2 – Tn–3 n – 2 Tn–1 – Tn–2 n – 1 Tn – Tn–1 n 2 B1 for 3, 4 and 6 B1 for n, n – 1, n – 2 3 T1 1 T2 + T1 4 T3 + T2 9 T4 + T3 16 T5 + T4 25 T6 + T5 36 Tn + Tn–1 n2 2 B1 for 16, 25 and 36 B1 for n2 oe 4(a) 2Tn = n2 + n 1 4(b)(i) Tn2 – Tn–12 = n3 2 B1 for each side 4(b)(ii) Correct numerical answer e.g. 32 – 12 = 8 = 23 2 B1 for, e.g. 32 – 12 = 8 or 23 = 8 or 32 – 12 = 23

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 5 of 7 Question Answer Marks Partial Marks 5(a) T15 – T14 = 15 B1 T15 – T14 = T5 T5 + T14 = T15 B1 (5, 14, 15) B1 5(b) 3 5 6 4 9 10 5 14 15 6 20 21 7 27 28 B2 B1 for one column or one row Showing differences or using the method at least once or last column are triangle numbers C1 6(a) 2n – 1 1 6(b)(i) 8 1 FT their 2n – 1 6(b)(ii) T8 – T6 or 36 – 21 = 15 or T5 C1 (5, 6, 8) B1 7(a) Adding last three rows soi or n + n – 1 + n – 2 1 7(b) 3n – 3 = 105 C1 n = 36 B1 (14, 33, 36) B1 8 Two correct linear equations solved for correct n other than 3n – 3 = 105 C2 C1 for one of 2n – 1 = 105 and n = 53 4n – 6 = 105 and n = 27.75 5n – 10 = 105 and n = 23 6n – 15 = 105 and n = 20 7n – 21 = 105 and n = 18 (14, 104, 105) (14, 51, 53) [(14, 33, 36)] (14, 18, 23) B3 B1 for each of (14, 104, 105) (14, 51, 53) (14, 18, 23)

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 6 of 7 Question Answer Marks Partial Marks B MODELLING SPEED OF PLANETS 9(a) Earth 1[.0] Mars 1.6 unknown 2.8 Jupiter 5.2 Saturn 10[.0] 3 B2FT for one error followed through correctly B1FT for two errors followed through correctly 9(b) 38.8 No oe 2 B1 for 38.8 B1FT correct conclusion from their 38.8 9(c) Add 0.4 and divide by 2 oe or 2x – 0.4 = 0.7 oe C1 0.55 B1 10(a) Translation Stretch 2 B1 for each 10(b) 2a + b = 1 oe 16a + b = 5.2 oe 2 B1 for each 10(c) 14a = 4.2 leading to a = 0.3 C1 2 × their 0.3 + b = 1 or 16 × their 0.3 + b = 5.2 leading to b = 0.4 C1 OR 7b = 0.28 leading to b = 0.4 C1 2a + their 0.4 = 1 or 16a + their 0.4 = 5.2 leading to a = 0.3 C1 11(a)(i) 1 = k × 1p C1 [k =] 1 B1 11(a)(ii) 1.5 3 M1 for 11.86 = 5.2p M1 for log11.86 log5.2 B1 for 1.5

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 7 of 7 Question Answer Marks Partial Marks 11(b) [T = R1.5] 0.391.5 = 0.24[…] B1 FT the value of 0.39their1.5 Yes oe B1 FT correct conclusion from their 0.24 12(a) S = 1.5 0.5 distance 2π 2π time R R R = = 3 C1 for S = distance time oe C1 for distance = 2πR soi 12(b) Correct sketch B1 Vertical scale implying intercept between 7 and 8. C1 12(c) rise run calculation C1 S =(–1.3 to –1.0)n + (7.3 to 7.7) B2 B1 for gradient –1.3 to –1 B1 for intercept 7.3 to 7.7 12(d) their gradient × 7 + their intercept = negative answer 2 B1FT for their negative gradient × 7 + their intercept B1 for a negative answer 5 5 10 n S