Cambridge IGCSE Mathematics - International 0607 — 2022 Oct/Nov Paper 6 · Variant 2

This document has 12 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ DC (LK/FC) 303211/3 © UCLES 2022 CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 Investigation and Modelling (Extended) October/November 2022 1 hour 40 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer both part A (Questions 1 to 6) and part B (Questions 7 to 10). ● 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 7 7 4 4 7 7 4 5 3 *

2 0607/62/O/N/22 © UCLES 2022 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 6) TWO-STEP SEQUENCES (30 marks) You are advised to spend no more than 50 minutes on this part. This investigation looks at two-step sequences. These are sequences which use two steps to get from one term to the next. The first term in every sequence is 1. The two steps are: • multiply by a given number • then add a given number. 1 In this question the two steps are: • multiply by 2 • then add 1. 1st term = 1 2nd term = 1st term # 2 + 1 = 1 # 2 + 1 = 3 3rd term = 2nd term # 2 + 1 = 3 # 2 + 1 = 7 4th term = 3rd term # 2 + 1 = 7 # 2 + 1 = 15 (a) Work out the 5th term of this sequence. 1, 3, 7, 15, … [2] (b) The nth term of another sequence is 2n. Calculate the 2nd, 3rd and 4th terms of this sequence. 2, … , … , … , 32 [1] (c) Look at your answers to part (a) and part (b). Write down an expression, in terms of n, for the nth term of the sequence in part (a). … [1]

3 0607/62/O/N/22 © UCLES 2022 [Turn over 2 In this question the two steps are: • multiply by 2 • then add 3. The first term is 1. (a) Work out the 2nd, 3rd and 4th terms of this sequence. 1 , … , … , … , 61 [2] (b) The nth term of this sequence is a b 2n # + . (i) Substituting n 1 = , to get the first term of the sequence, gives the equation a b 2 1 + = . Substitute another value for n to make another equation in terms of a and b. … [1] (ii) Solve the simultaneous equations in part (i) to show that the nth term of the sequence is 2 2 3 n # - . [2] 3 In this question the two steps are: • multiply by 2 • then add 5. The first term is 1. The expression for the nth term is 3 2 5 n # - . Show that this expression gives the correct value for the 4th term of this sequence. [3]

4 0607/62/O/N/22 © UCLES 2022 4 In this question the two steps are always: • multiply by 2 • then add k. The first term is 1. (a) Complete the table. Use your answer to Question 1(c) and any patterns you notice. Steps to get the next term Expression for the nth term Multiply by 2, then add 1 … Multiply by 2, then add 3 2 2 3 n # - Multiply by 2, then add 5 2 3 5 n # - Multiply by 2, then add 7 … Multiply by 2, then add … … 9 - [2] (b) An expression for the nth term of this sequence is a b 2n # + . Find expressions for a and b in terms of k. Write down the expression for the nth term of the sequence. a = … b = … nth term = … [3]

5 0607/62/O/N/22 © UCLES 2022 [Turn over (c) The 5th term of a sequence using the nth term in part (b) is 286. Complete the two steps. • multiply by 2 • then add … [3] 5 In this question the two steps are: • multiply by 3 • then add 2. The expression for the nth term is a b 3( ) n 1 # + - . (a) The first term is 1. (i) Find the value of the second term of the sequence. … [1] (ii) Use the first two terms to write two equations in terms of a and b. … … [2] (b) Find the value of a and the value of b. a = … b = … [3]

6 0607/62/O/N/22 © UCLES 2022 6 (a) Complete the table. Use your answer to Question 5(b) and any patterns you notice. Steps to get the next term Expression for the nth term Multiply by 2, then add 1 2 2 1 ( ) n 1 # - - Multiply by 3, then add 2 Multiply by 4, then add 3 Multiply by 5, then add 4 Multiply by 6, then add 5 2 1 6( ) n 1 # - - [1] (b) For the sequence in the last row of the table, the first term has the value 1 and the second term has the value 11. Find which term has its value closest to 20 000 000. … [3]

7 0607/62/O/N/22 © UCLES 2022 [Turn over B MODELLING (QUESTIONS 7 to 10) DRIVING TO MY PLACE OF WORK (30 marks) You are advised to spend no more than 50 minutes on this part. This task looks at a model for the time that I take to drive from my home to my place of work. I live 20 km from my place of work. When I leave my home at 7.00 am, I drive at an average speed of 50 km/h. 7 (a) Calculate the time, in minutes, to drive to work when I leave home at 7.00 am. … [3] (b) The time that it takes me to drive to work is m minutes. Find, in its simplest form, a model for m when my average speed is v km/h. … [1]

8 0607/62/O/N/22 © UCLES 2022 8 When I leave home after 7.00 am, there is more traffic, and my average speed is less than 50 km/h. My average speed decreases steadily by 1 km/h for every 2 minutes after 7.00 am that I leave home. For example, when I leave at 6 minutes after 7.00 am, my average speed is 3 km/h less, which is 47 km/h. (a) I leave home at 7.40 am. (i) Find my average speed. … [2] (ii) Show that the time to drive to work is 40 minutes. [1] (b) I leave home x minutes after 7.00 am. Show that a model for the time, T minutes, to drive to work is T x 100 2400 = - . [2]

9 0607/62/O/N/22 © UCLES 2022 [Turn over (c) Sketch the graph of the model T x 100 2400 = - for x 0 90 G G . 0 90 x T 0 [2] (d) I do not want to drive for more than 30 minutes. Find the latest time that I should leave home. … [2] (e) I must be at work by 9.00 am. One day I oversleep and leave home at 8.35 am. (i) Use the model to find how late I will be for work. Give your answer in hours and minutes. … [3] (ii) Make a statement about the suitability of the model. … [1]

10 0607/62/O/N/22 © UCLES 2022 9 I leave home x minutes after 7.00 am. (a) Explain why a model for A, the number of minutes after 7.00 am when I arrive at work, is x A x 100 2400 = + - . … [1] (b) I must be at work by 9.00 am, which is two hours after 7.00 am. So my maximum value of A is 120. (i) Show that, for this maximum value of A, x is a solution to the equation x x 220 9600 0 2 - + = . [3] (ii) Find this value of x. … [3] (iii) Find the latest time that I can leave home to arrive at work on time. … [1]

11 0607/62/O/N/22 © UCLES 2022 10 I move to a new home and now live d km from my work. When I leave my new home at 7.00 am, my average speed is v km/h. As before, my average speed decreases steadily by 1 km/h for every 2 minutes after 7.00 am that I leave home. (a) I leave my new home x minutes after 7.00 am. Show that a model for the time, T minutes, to drive to work is T v x d 2 120 = - . [2] (b) I want to leave my new home at 7.30 am and arrive at work at 9.00 am. Find a model for v in terms of d. … [3]

12 0607/62/O/N/22 © UCLES 2022 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 Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. BLANK PAGE

This document consists of 8 printed pages. © UCLES 2022 [Turn over Cambridge IGCSE™ MATHEMATICS 0607/62 Paper 6 Extended October/November 2022 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 2022 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 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.

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 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

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 4 of 8 Question Answer Marks Partial Marks 1(a) 15  2 + 1 C1 31 1 1(b) 4, 8, 16 1 1(c) 2n – 1 1 2(a) 5, 13, 29 2 B1 for 5 or B1FT for one correct calculation using their 5 or their 13 2(b)(i) Correct substitution for n to make correct equation 1 FT their terms in (a) 2(b)(ii) Correct equation isolating a or b 1 FT their part (b)(i) Correctly substituting a = 2 in an equation in a and b, leading to b = – 3 or correctly substituting b = – 3 leading to a = 2. 1 3 [1  2 + 5 =]7 and 7  2 + 5 = 19 and 19  2 + 5 = 43 2 B1 for 7, 19, 43, or the second statement, or the third statement. 3  24 – 5 = 43 1 4(a) Steps Expression Multiply by 2, then add 1 2n – 1 Multiply by 2, then add 3 2  2n – 3 Multiply by 2, then add 5 3  2n – 5 Multiply by 2, then add 7 4  2n – 7 Multiply by 2, then add 9 5  2n – 9 2 B1 for 9 in first column B1 for 4  2n – 7 and 5  2n – 9 4(b) [a =] 1 2 2 + k or 1 2 + k oe 2 B1 for 2 k or 2 k subsumed in the answer e.g. 1 2 − k [b =] – k 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 5 of 8 Question Answer Marks Partial Marks 4(c) 5 1 2 286 2 k k +  − =     C1 FT their a  32 + their b where both a and b are in terms of k only 16k + 16 – k = 286 oe C1 FT their a and their b where both a and b are in terms of k only 18 1 5(a)(i) 5 1 5(a)(ii) a  3(1–1) + b = 1 or a + b = 1 oe a  3(2–1) + b = 5 or 3a + b = 5 oe 2 B1 for each FT their 5 5(b) Correctly eliminating one variable C1 FT their equations in (a)(ii) [a =] 2 and [b =] – 1 2 B1 for each FT their (a)(ii) equations 6(a) 2  4n – 1 – 1 2  5n – 1 – 1 1 6(b) 2  6(10–1) – 1 = 20 155 391 and 2  6(9–1) – 1 = 3 359 231 OR 6n–1 = 10 000 000.5 and log10 000 000.5 = 9 log6 oe or log6 10 000 000.5 = 9 oe or 69 =10 077 696 or log10000000.5 log6 log6 + = 10 or 9.99… OR Correct sketch of graphs C2 C1 for 2  6(10–1) – 1 = 20 155 391 OR C1 for one statement OR C1 for correct sketch of y = 2 × 6n – 1 10[th] 1 If 0 scored SC1 for n = 9.99[…] seen 7(a) 20 60 50  soi oe C2 C1 for 20 50 or  60 or 50 60 24 minutes 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 6 of 8 Question Answer Marks Partial Marks 7(b) m = 1200 v 1 8(a)(i) 20 seen C1 30 km/h 1 FT 50 – their 20 8(a)(ii) 20 60 30  = 40 oe or 20 2 hour oe 40 30 3 = = [mins] 1 8(b) 50 – 2 x seen 1 1200 50 2 −x leading to 2400 100 −x 1 8(c) Correct sketch 1 Increasing curve from T axis not touching or crossing x axis T intercept marked as 24 C1 8(d) T = 30 drawn on the graph or 2400 100 −x = 30 C1 7.20 oe 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 7 of 8 Question Answer Marks Partial Marks 8(e)(i) 2400 100 95 − or correct sketches of T = 2400 100 −x and x = 95 or x = 95 drawn on graph in 8(c) C1 7h 35 min 2 FT their 95 if between 60 and 100 B1 for 8 [hours] or 480 [minutes] 8(e)(ii) Not useful for leaving home very late oe or unsuitable for 100 minutes [after 7am] oe 1 9(a) number of minutes after 07.00 + time to drive to work oe 1 9(b)(i) x + 2400 100 −x = 120 1 Correct elimination of fractions 1 Correct expansion of brackets leading to x2 – 220x + 9600 = 0 nfww 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 8 of 8 Question Answer Marks Partial Marks 9(b)(ii) Sketch of y = x2 – 220x + 9600 with left positive x intercept clearly marked OR (x – 60)(x – 160) OR ( ) ( )  2 – –220 220 4 1 9600 2  − −  = 60, 160 OR x – 110 = ± 2500 leading to 60, 160 OR A minimum of two trials, one of which is 60. C2 C1 for sketch of y = x2 – 220x + 9600 OR for (x + a)(x + b) where ab = 9600 or a + b = –220 or 60 and 160 with wrong signs OR for ( ) ( )  2 – –220 220 4 1 9600 2  − −  oe OR (x – 110)2 = 2500 OR for 602 – 220  60 + 9600 = 0 60 1 9(b)(iii) 8am oe 1 10(a) v – 2 x seen for speed 1 60 2  − d x v oe leading to 120 2 − d v x 1 10(b) 120 2 30 − d v = 90 oe C1 Correct next step towards isolating v C1 v = 2 15 3 + d oe isw 1