Cambridge IGCSE Mathematics - International 0607 — 2022 Oct/Nov Paper 5 · Variant 3

This document has 8 pages. [Turn over Cambridge IGCSE™ DC (CJ) 316794 © UCLES 2022 CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/53 Paper 5 Investigation (Core) October/November 2022 1 hour 10 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer all questions. ● 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 36. ● The number of marks for each question or part question is shown in brackets [ ]. * 0 5 4 1 1 1 8 2 8 1 *

2 0607/53/O/N/22 © UCLES 2022 Answer all the questions. INVESTIGATION TWO-STEP SEQUENCES 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. … [2] (b) The 3rd term of this sequence is 7. You can write 7 as 2 1 3 - . Complete the table. 1st term 1 2 1 1 - 2nd term 3 2 1 2 - 3rd term 7 2 1 3 - 4th term 15 5th term [1]

3 0607/53/O/N/22 © UCLES 2022 [Turn over (c) Calculate the 20th term of this sequence. Write down all the digits shown on your calculator. … [2] (d) (i) Use the last column in the table to write down an expression for the nth term of this sequence. … [1] (ii) Show that your expression gives the correct value for the 6th term of this sequence. [2]

4 0607/53/O/N/22 © UCLES 2022 2 In this question the two steps are: • multiply by 3 • then add 4. The first term is 1. (a) Calculate the 2nd, 3rd and 4th terms of this sequence. 1, … , … , … [3] (b) Complete the table. 1st term 1 3 2 1 - 2nd term 3 2 2 - 3rd term 4th term 5th term 241 35 - [2] (c) Write down an expression for the nth term of this sequence. … [1]

5 0607/53/O/N/22 © UCLES 2022 [Turn over 3 In this question the two steps are: • multiply by 4 • then add 9. The first term is 1. Show that the expression for the nth term, 4 3 n - , gives the correct value for the 3rd term of this sequence. [4]

6 0607/53/O/N/22 © UCLES 2022 4 (a) Copy your results from Question 1(d)(i) and Question 2(c) into the table. Use any patterns you notice to complete the table. Steps to get the next term Expression for the nth term Question 1(d)(i) Multiply by 2, then add 1 … Question 2(c) Multiply by 3, then add 4 … Multiply by 4, then add 9 3 4n - Multiply by … , then add 16 5n - … Multiply by 6, then add … … 5 - Multiply by 7, then add 36 … Multiply by … , then add … 8 7 n - [4] (b) A sequence has the two steps that you found in the last row of the table. Show that the expression for the nth term gives the correct value for the 3rd term of this sequence. [3]

7 0607/53/O/N/22 © UCLES 2022 [Turn over (c) The nth term of a two-step sequence is 22 21 n - . Find the two steps. • … • … [2] (d) In a two-step sequence the steps are: • multiply by 11 • then add 100. The first term is 1. (i) Find the value of the term nearest to 20 000 000. Write down all the digits shown on your calculator. … [2] (ii) Which term in the sequence is your answer to part (i)? … [1] Question 5 is printed on the next page.

8 0607/53/O/N/22 © UCLES 2022 5 In this question the steps in Question 1 are in the reverse order. The two steps are now: • add 1 • then multiply by 2. (a) The first term is 1. The second term is 4. Calculate the 3rd, 4th and 5th terms. 1, 4, … , … , … [2] (b) This two-step sequence has nth term equal to a 2 2 n # - . (i) The first term is 1. Use this to find the value of a. … [2] (ii) Use part (i) to show that the expression for the nth term gives the correct value for the 3rd term of this sequence. [2] 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.

This document consists of 5 printed pages. © UCLES 2022 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/53 Paper 5 (Core) October/November 2022 MARK SCHEME Maximum Mark: 36 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.

0607/53 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 2 of 5 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.

0607/53 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 3 of 5 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

0607/53 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 4 of 5 Question Answer Marks Partial Marks 1(a) 15 × 2 + 1 C1 31 1 1(b) 24 – 1 25 – 1 1 1(c) 220 – 1 C1 1 048 575 1 1(d)(i) 2n – 1 oe 1 1(d)(ii) 31 × 2 + 1 = 63 1 26 – 1 = 63 1 2(a) 1 × 3 + 4 = 7 or 7 × 3 + 4 = 25 or 25 × 3 + 4 = 79 C1 7, 25, 79 2 B1 for 7 B1FT for one correct calculation using their 7 or their 25 2(b) 33 – 2 34 – 2 [35 – ]2 2 B1 for two correct expressions 2(c) 3n – 2 1 3 1 × 4 + 9 = 13 C1 13 × 4 + 9 = 61 2 M1 for their13 × 4 + 9 43 – 3 = 61 1 4(a) Multiply by 5, add 16 5n – 4 Multiply by 6, add 25 6n – 5 Multiply by 7, add 36 7n – 6 Multiply by 8, add 49 8n – 7 4 B1 for 5 and 8 B1 for 25 and 49 B1 for 6n and 7n B1 for 4 and – 6 4(b) 1 × 8 + 49 = 57 57 × 8 + 49 = 505 2 FT first statement with their 8 and their 49 B1 for each statement or B1 for 1 × 8 + 49 = 57 × 8 + 49 = 505 83 – 7 = 505 1

0607/53 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2022 © UCLES 2022 Page 5 of 5 Question Answer Marks Partial Marks 4(c) multiply by 22 [then] add 441 2 B1 for each 4(d)(i) Correct use of 11n – 10 seen once or [1,] 111, 1321, 14 631, 161 041 C1 19 487 161 1 4(d)(ii) 7[th] 1 5(a) ( 4 + 1) × 2 or 5 × 2 and (10 + 1) × 2 or 11 × 2 and (22 + 1) × 2 or 23 × 2 C1 FT correct calculation of next term 10, 22, 46 1 5(b)(i) 2a – 2 = 1 C1 a = 1.5 oe 1 5(b)(ii) 1.5 × 23 – 2 = 10 2 M1FT for their 1.5 in the substitution.