Cambridge IGCSE Mathematics - International 0607 — 2017 May/June Paper 6 · Variant 2

* 3 0 0 6 0 2 3 6 3 5 * This document consists of 12 printed pages. DC (LK/SG) 133976/4 © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 (Extended) May/June 2017 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 and B. 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 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 0607/62/M/J/17 © UCLES 2017 Answer both parts A and B. A INVESTIGATION NUMBER STEMS (20 marks) You are advised to spend no more than 45 minutes on this part. This investigation is about finding numbers that have the same Number Stem. The possible Number Stems are the nine integers from 1 to 9. Here is how to calculate the Number Stem of a number. Step 1 Add the digits of the number to get a total. Step 2 If the total is 9 or less, STOP. Otherwise, add the digits of the total. Step 3 Repeat Step 2. Examples Number 124 Number 893 Step 1 1 + 2 + 4 = 7 Step 1 8 + 9 + 3 = 20 Step 2 STOP Step 2 2 + 0 = 2 Number Stem is 7. Step 3 STOP Number Stem is 2. 1 (a) Complete the tables to show the Number Stems for these multiples of 3 and 12. Multiple of 3 3 6 9 12 15 18 21 24 27 30 Number Stem 3 6 9 3 9 3 6 9 3 Multiple of 12 12 24 36 48 60 72 84 96 108 120 Number Stem 3 6 9 (b) (i) Complete the sequence to show the first four numbers greater than 3 that have a Number Stem of 3. 12, 21, 30, … (ii) Find the nth term of the sequence in part (b)(i). …

3 0607/62/M/J/17 © UCLES 2017 [Turn over (iii) Find the 87th number greater than 3 that has a Number Stem of 3. … (c) (i) Complete this table. Number Number Stem Calculation Answer 3 3 3 ÷ 9 0 remainder 3 19 1 19 ÷ 9 2 remainder 1 22 … 22 ÷ … 2 remainder 4 35 … 35 ÷ 9 … remainder … … 7 … ÷ 9 … remainder … (ii) A number, that is not a multiple of 9, is divided by 9. What is the connection between its Number Stem and the remainder? … (iii) Using your answer to part (c)(ii) write down the remainder when 104 020 100 is divided by 9. …

4 0607/62/M/J/17 © UCLES 2017 2 The sequence shows the first three numbers greater than 2 with a Number Stem of 2. 11, 20, 29, … (a) Write down the next two numbers of the sequence. … , … (b) Find the nth term of this sequence. … (c) Using your answer to part (b), find the largest number less than 10 000 that has a Number Stem of 2. … 3 The integer k is a Number Stem. (a) Write down, in terms of k, the first four numbers greater than k with a Number Stem of k. … , … , … , … (b) Write down, in terms of n and k, the nth term for the sequence of numbers greater than k with a Number Stem of k. …

5 0607/62/M/J/17 © UCLES 2017 [Turn over 4 (a) Complete this table. Calculation Answer 7 ÷ 12 0 remainder 7 15 ÷ 12 1 remainder … 23 ÷ 12 … remainder … (b) An integer, that is not a multiple of 12, has remainder f when it is divided by 12. Find, in terms of n and f, the nth term for the sequence of numbers greater than f with a remainder of f. … (c) Show that f + 10 cannot be a term of the sequence of numbers greater than f with a remainder of f.

6 0607/62/M/J/17 © UCLES 2017 B MODELLING ELEVATORS (20 marks) You are advised to spend no more than 45 minutes on this part. This task is about the mass an elevator carries and the time it takes to move between floors. EasyUp is a company that makes elevators. For each type of elevator, the company uses two mathematical models. Model 1 The company models the masses of the passengers using the elevator. Model 2 The company models the time it takes the elevator to move between floors. 1 The EasyUp-5 elevator carries a maximum of 5 passengers. (a) For Model 1, the company estimates that • 10 2 of the passengers have a mass of 50 kg • 10 4 of the passengers have a mass of 70 kg • 10 4 of the passengers have a mass of 85 kg. From the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 • 0 and 1 give a mass of 50 kg • 2, 3, 4 and 5 give a mass of 70 kg • 6, 7, 8 and 9 give a mass of 85 kg.

7 0607/62/M/J/17 © UCLES 2017 [Turn over (i) Numbers are chosen at random from 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. Each number models the mass of a passenger. The 5 passengers are A, B, C, D and E. Here is a random number table, arranged in groups of 5 numbers. 1 6 8 5 2 4 7 9 1 1 6 9 8 7 4 9 1 4 6 8 4 2 1 9 0 8 2 8 8 0 8 9 4 7 6 7 4 1 6 5 1 0 4 7 6 6 0 0 4 5 Use the last four rows of the random number table to complete the table of trials below. The first six trials have been completed for you. Mass of passenger (kg) Total mass (kg) A B C D E Trial 1 50 85 85 70 70 360 Trial 2 70 85 85 50 50 340 Trial 3 85 85 85 85 70 410 Trial 4 85 50 70 85 85 375 Trial 5 70 70 50 85 50 325 Trial 6 85 70 85 85 50 375 Trial 7 85 85 Trial 8 Trial 9 Trial 10 (ii) The EasyUp-5 elevator carries a maximum total mass of 400 kg. Use the results of trials 1 to 10 to work out the relative frequency that the total mass will be more than 400 kg. …

8 0607/62/M/J/17 © UCLES 2017 (b) For Model 2, the diagram below is a distance-time graph for the EasyUp-5 elevator. The graph modelling the movement is y = f(t), where f(t) is the number of floors above or below the ground floor (floor 0) at time t seconds. y t 5 10 15 20 4 3 2 1 0 –1 –2 Ground floor (i) The graph shows the elevator starting one floor below the ground floor. At which floor does it stop? … (ii) Between which two floors does the elevator have the greatest average speed? … and … (iii) Find the average time it takes the elevator to move from one floor to the next. …

9 0607/62/M/J/17 © UCLES 2017 [Turn over 2 The EasyUp-3 elevator carries a maximum of 3 passengers. The maximum total mass is 240 kg. (a) For Model 1, the company now uses different proportions of passengers for each mass, as shown in the table. A number is chosen from 0, 1, 2, 3, 4, 5, 6 and 7 to give the mass. For example, the numbers 1, 2, 3, 4 and 5 each give the mass 70 kg. (i) Complete the table below. Proportion of passengers 8 5 8 2 Amount of numbers 5 Numbers 0 1, 2, 3, 4, 5 Mass of passenger (kg) 50 70 85 (ii) Random numbers model the masses of the 3 passengers, A, B and C. Here is a random number table, arranged in groups of four numbers. Three random numbers are needed for each trial. The numbers 8 and 9 are not used. Cross out the numbers 8 and 9 in the table. When four numbers remain in a row, cross out the last number. 8 2 1 5 1 6 3 3 6 7 0 5 0 9 1 5 2 0 8 6 1 0 1 1 3 4 8 2 9 0 4 3 Use the last four rows of the random number table to complete the table of trials below. The first four trials have been completed for you. Mass of passenger (kg) Total mass (kg) A B C Trial 1 70 70 70 210 Trial 2 70 85 70 225 Trial 3 85 85 50 220 Trial 4 50 70 70 190 Trial 5 Trial 6 Trial 7 Trial 8

10 0607/62/M/J/17 © UCLES 2017 (b) For Model 2, the diagram below is a distance-time graph for the EasyUp-3 elevator. The graph modelling the movement is y = h(t), where h(t) is the number of floors above or below the ground floor (floor 0) at time t seconds. 15 20 10 5 1 0 –1 y t Ground floor (i) The graph shows that the EasyUp-3 elevator starts to move from floor 1. Find the average time it takes the elevator to move from one floor to the next. … (ii) ( ) h cos t kt = . Find the value of the integer k. … 3 The mass carried by an elevator is x kilograms. EasyUp say that an elevator is well designed when • the probability that x is less than the maximum total mass is greater than 0.95 and • it takes at most 5 seconds on average to move between floors. Explain whether (a) the EasyUp-5 elevator is well designed, … … (b) the EasyUp-3 elevator is well designed. … …

11 0607/62/M/J/17 © UCLES 2017 [Turn over 4 Write down two ways to improve Model 1 in question 1(a). … … … … 5 The EasyUp-n elevator carries a maximum of n passengers. The maximum total mass is 80n kilograms. (a) For Model 1, the company uses different proportions of passengers for each mass, as shown in the table. A number is chosen at random from m integers to give the mass. (i) Complete the table. Proportion of passengers m 2 m m 3 - Amount of numbers Mass of passenger (kg) 50 70 85 (ii) Explain why m 4 H . … … Question 5(b) and question 5(c) are printed on the next page.

12 0607/62/M/J/17 © UCLES 2017 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. (b) For Model 2, the distance-time graph for the EasyUp-7 elevator is modelled by ( ) h y t2 =- . (i) Use your answer to question 2 (b)(ii) to write down the equation of this distance-time graph. … (ii) On the grid below, sketch the graph of ( ) h y t2 =- , for t 0 10 G G . 10 5 0 1 2 –1 –2 y t (c) The mass carried by an elevator is x kilograms. The probability that x is less than the maximum mass is 0.99 . Use this information and your graph in part (b)(ii) to explain why the EasyUp-7 elevator is well designed. … …

® IGCSE is a registered trademark. This document consists of 7 printed pages. © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 (Extended) May/June 2017 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 will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 2 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

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 3 of 7 Question Answer Marks Part Marks A INVESTIGATION NUMBER STEMS 1(a) 15 6 48 60 72 84 96 108 120 3 6 9 3 6 9 3 1 1(b)(i) 39 1 1(b)(ii) 9n + 3 oe 2 B1 for 9n + a oe 1(b)(iii) 786 1 FT their (9n + 3) C opportunity 1(c)(i) 4 22 ÷ 9 2 remainder 4 8 35 ÷ 9 3 remainder 8 7 k ÷ 9 j remainder 7 2 B1 for 5, 6 or 7 correct k is any integer with a number stem of 7 j is the integer part of 9 their k 1(c)(ii) [They are the] same oe 1 1(c)(iii) 8 1 Answer found from division scores 0. 2(a) 38, 47 1 2(b) 9n + 2 oe 1 C opportunity 2(c) 9992 2 B1FT for [n=]1110[. …..] C opportunity 3(a) k + 9, k + 18, k + 27, k + 36 oe 1 3(b) 9n + k oe 1 SC1 for 9n + k – 9 oe from an answer of k, k + 9, k + 18, k + 27 in part (a) 4(a) 7 ÷ 12 0 remainder 7 15 ÷ 12 1 remainder 3 23 ÷ 12 1 remainder 11 1 4(b) 12n + f oe 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 4 of 7 Question Answer Marks Part Marks 4(c) 12n + f = f + 10 M1 FT their (12n + f ) = f + 10 soi 12n = 10 and leading to n is not an integer oe A1 SC2 f + 10 is smaller than any term in the sequence f + 12, f + 24… or SC1 if f + 12 not explicitly stated Communication: Seen in two of the following questions 1 1(b)(iii) their (9 × 87 + 3) seen 2(b) At least two differences of 9 seen (may be in Q2 stem or in part(a)) or “The sequence is 1 less than the previous sequence” oe 2(c) their (9n + 2) * 10 000, where * is = or < or ⩽ or two trials of the form 9 × n + 2 with 1000 ⩽ n ⩽ 1200 substituted and number found. or two trials of the form 999N, N a single digit, and correct number stems calculated.

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 5 of 7 Question Answer Marks Part Marks B MODELLING ELEVATORS 1(a)(i) Trial 7 85 85 70 85 85 410 Trial 8 85 70 50 85 70 360 Trial 9 50 50 70 85 85 340 Trial 10 85 50 50 70 70 325 2 B1 for any correct row 1(a)(ii) 2 10 oe 1 FT their completed table 1(b)(i) 3 1 1(b)(ii) 0 and 2 oe 1 Allow 0 and 1 or ground and first or 1 and 2 or ground and second 1(b)(iii) 5 1 C opportunity 2(a)(i) 1 8 oe 1 2 6, 7 1 2(a)(ii) Trial 5 70 50 85 205 Trial 6 70 50 70 190 Trial 7 70 70 70 210 Trial 8 50 70 70 190 1 2(b)(i) 10 1 C opportunity 2(b)(ii) 9 1 C opportunity 3(a) No, and the probability [of less than the maximum] is 0.8 oe or No, and the probability [of more than the maximum] is more than 0.05 oe 1 FT 1 – their 2 10 in 1(a)(ii) 3(b) No, and EasyUp-3 takes 10 seconds [to move between floors] oe 1 FT their 10 in 2(b)(i) Accept “more than 5” instead of 10. If 0 scored in (a) and (b), SC1 for both explanations correct. 4 Increase the number of trials oe Increase the number of masses oe 2 B1 for each

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 6 of 7 Question Answer Marks Part Marks 5(a)(i) 1 m 1 2 m – 3 1 C opportunity 5(a)(ii) Valid comment 1 If m is less than 4 the proportion [with a mass of 85] is 0 [or negative] oe Comment about the number of passengers on its own scores 0. 5(b)(i) [ ] cos( 2 ) y their k t = − × 1 Expect y = –cos18t 5(b)(ii) 1 FT their cosine equation if the graph fits on the axes. 5(c) It moves [between floors] at [an average of] their 5 seconds [per floor] oe and “Probability [that x is less than the max] > 0.95” oe 1 FT their cosine graph. Communication: Seen in two of the following questions 1 1(b)(iii) 4 floors in 20 seconds or 0.2 oe floors in 1 second or 6.3 3.7 4 6 4 + + + or similar values with one decimal place or 20 4 but not if 6 4 4 6 4 + + + oe seen 2(b)(i) seconds in final answer 2(b)(ii) 40 is 360° or 360 40 or 360 40 9 = or 10 is 90° etc. as above or 20 is 180° etc. as above 5 10 -2 -1 1 2 0

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 7 of 7 Question Answer Marks Part Marks 5(a)(i) 3 2 ? 1 m m m − + + = oe or 2 3 2 3 1 m m m m m m m m m − − − + − − = = oe or m – 3 + 2 + 1 = m oe or 3 2 1 2 m m m −+ − = or unsimplified form for 1 in the table: m – 2 – (m – 3) oe