Cambridge IGCSE Mathematics - International 0607 — 2021 Oct/Nov Paper 6 · Variant 2

This document has 12 pages. Cambridge IGCSE™ * 0 0 9 3 8 7 1 8 8 1 * DC (PQ/CGW) 215051/4 © UCLES 2021 [Turn over CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 Investigation and Modelling (Extended) October/November 2021 1 hour 40 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer both part A (Questions 1 to 4) and part B (Questions 5 to 7). ● 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 0607/62/O/N/21 © UCLES 2021 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 4) CONNECTING DOTS (30 marks) You are advised to spend no more than 50 minutes on this part. This investigation looks at the number of ways of connecting dots using straight lines. This diagram shows 1 dot. There is 1 row and 1 column. This is a 1 by 1 diagram. There are no connections to other dots. This diagram shows 4 dots. There are 2 rows and 2 columns. This is a 2 by 2 diagram. There are 6 ways to join 2 dots. These are: • 2 vertical connectors (solid lines) • 2 horizontal connectors (solid lines) • 1 up diagonal connector (dashed line) • 1 down diagonal connector (dashed line). 1 (a) This is a 3 by 3 diagram. The diagram shows: • 6 horizontal connectors • 4 up diagonal connectors. Each connector joins 2 dots. Write down the number of vertical connectors and the number of down diagonal connectors that join 2 dots. Vertical … Down diagonal … [2]

3 0607/62/O/N/21 © UCLES 2021 [Turn over (b) Complete the table for the numbers of connectors that join 2 dots. Use part (a) and any patterns you notice. You may use the square dotty paper on page 12 for diagrams. Numbers of connectors that join 2 dots Horizontal Vertical Up diagonal Down diagonal Total Size of diagram (n by n) 1 by 1 0 0 0 0 0 2 by 2 2 2 1 1 6 3 by 3 6 4 4 by 4 9 5 by 5 20 6 by 6 110 [3] (c) In an n by n diagram there are n rows and n columns. (i) Find an expression, in terms of n, for the number of up diagonal connectors that join 2 dots on an n by n diagram. … [2] (ii) Find an expression, in terms of n, for the number of horizontal connectors that join 2 dots on an n by n diagram. … [3] (iii) Use your answers in part (i) and part (ii) to find an expression for the total number of connectors that join 2 dots. Do not simplify your expression. … [1]

4 0607/62/O/N/21 © UCLES 2021 2 This is a 3 by 3 diagram. There are 8 ways to join 3 dots together. These are: • 3 vertical connectors • 3 horizontal connectors • 1 up diagonal connector • 1 down diagonal connector. (a) This is a 4 by 4 diagram. Find the number of horizontal, vertical, up diagonal and down diagonal connectors that join 3 dots. Two horizontal connectors have been drawn for you. Horizontal … Vertical … Up diagonal … Down diagonal … [2]

5 0607/62/O/N/21 © UCLES 2021 [Turn over (b) Complete the table for the numbers of connectors that join 3 dots. Use your answers to part (a) and any patterns you notice. You may use the square dotty paper on page 12 for diagrams. Numbers of connectors that join 3 dots Horizontal Vertical Up diagonal Down diagonal Total Size of diagram (n by n) 2 by 2 0 0 0 0 0 3 by 3 3 3 1 1 8 4 by 4 5 by 5 15 6 by 6 80 [2] (c) (i) This is an expression for the number of up diagonal connectors that join 3 dots on an n by n diagram. ( ) n 2 2 - Work out the number of up diagonal connectors that join 3 dots on a 20 by 20 diagram. … [1] (ii) This is an expression for the number of horizontal connectors that join 3 dots on an n by n diagram. n an 2 + Find the value of a and write down the expression. … [3] (d) Find an expression, in terms of n, for the total number of connectors that join 3 dots on an n by n diagram. Do not simplify your expression. … [1]

6 0607/62/O/N/21 © UCLES 2021 3 (a) Complete the table for the numbers of connectors that join 4 dots. Numbers of connectors that join 4 dots Horizontal Vertical Up diagonal Down diagonal Total Size of diagram (n by n) 3 by 3 0 0 0 0 0 4 by 4 10 5 by 5 10 6 by 6 54 n by n [4] (b) Find an expression, in terms of n, for the total number of connectors that join 4 dots on an n by n diagram. Do not simplify your answer. … [1] 4 (a) This is an expression for the total number of connectors that join m dots on an n by n diagram. ( ) ( ) n n k n k 2 2 2 - + - What is the relationship between k and m? … [1]

7 0607/62/O/N/21 © UCLES 2021 [Turn over (b) (i) Use part (a) to show that when n 5 = and m 2 = there is a total of 72 connectors. [1] (ii) Find all the possible values for n and m that give a total of 72 connectors. … [3]

8 0607/62/O/N/21 © UCLES 2021 B MODELLING (QUESTIONS 5 to 7) BREEDING DEER (30 marks) You are advised to spend no more than 50 minutes on this part. This task looks at modelling the population of a herd of deer. In this herd there are equal numbers of males and females. Each adult female deer gives birth to a male and a female fawn (baby deer) every year. At the start of the first year a farmer has 20 adult deer. There are 10 males and 10 females. At the end of each year all the deer are adult. 5 (a) (i) At the end of the first year the farmer sends 20% of the 20 deer to other farmers. He sends an equal number of male and female deer. Show that at the start of the second year he has 36 deer in his herd. [2] (ii) Pn = the number of deer in the herd at the start of year n. Pn 1 = + the number of deer in the herd at the start of year n 1 + . At the end of each year the farmer sends 20% of Pn to other farmers. He always sends an equal number of male and female deer. Show that . P P 1 8 n n 1 = + . [2] (b) (i) When the farmer finds Pn 1 + he rounds the value to the nearest even integer. The table shows the number of deer in the herd, Pn, at the start of year n. Use part (a)(ii) to complete the table. Year (n) 1 2 3 4 5 6 7 8 Number in herd ( ) Pn 20 36 116 374 [4]

9 0607/62/O/N/21 © UCLES 2021 [Turn over (ii) Use your answers to part (b)(i) to plot the four missing points. 200 0 1 0 400 600 800 1000 1200 1400 2 3 4 Year Number in herd 5 6 7 8 P n [2] (c) The farmer models the data using P abn 1 = - . Use the first two years in the table in part (b)(i) to find the value of a and the value of b and write down the model. … [3]

10 0607/62/O/N/21 © UCLES 2021 6 The farmer now models the data in the table on page 8 using ( ) ( ) P a n b n c 1 1 2 = - + - + . (a) (i) Use the first year in the table in Question 5(b)(i) to show that c 20 = . [1] (ii) Use year 2 and year 4 in the table in Question 5(b)(i) to write down a pair of simultaneous equations in terms of a and b. … … [2] (b) (i) Solve your simultaneous equations from part (a)(ii) and write down the model. … [4] (ii) Sketch your model on the grid in Question 5(b)(ii) for n 1 8 G G . [2] (c) Is this a suitable model for the number of deer in the herd? Give one reason for your answer. … … [1]

11 0607/62/O/N/21 © UCLES 2021 [Turn over 7 The farmer now models the data in the table on page 8 using ( ) P a n 1 b = - . (a) Use the data for year 2 and year 4 in the table in Question 5(b)(i) to (i) find a, … [2] (ii) find b and write down the model. … [3] (b) Give two reasons why the model in part (a) is not suitable for the number of deer in the herd. 1 … … 2 … … [2]

12 0607/62/O/N/21 © UCLES 2021 This square dotty paper may be used for your diagrams in the investigation. 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.

This document consists of 8 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 (Extended) October/November 2021 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 2021 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 2021 © UCLES 2021 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 2021 © UCLES 2021 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 2021 © UCLES 2021 Page 4 of 8 Question Answer Marks Partial Marks A INVESTIGATION CONNECTING DOTS 1(a) (Vertical) 6 (Down diagonal) 4 2 B1 for each 1(b) H V UD DD T 1 × 1 0 0 0 0 0 2 × 2 2 2 1 1 6 3 × 3 6 6 4 4 20 4 × 4 12 12 9 9 42 5 × 5 20 20 16 16 72 6 × 6 30 30 25 25 110 3 B1 for each row If 0 scored, B1 for all up diagonals or all down diagonals correct 1(c)(i) Reference to square numbers in words or in working or at least one correct numerical example or at least 3 second differences of 2 C1 (n – 1)2 oe isw 1 1(c)(ii) n rows/columns and ( )1 − n connectors or at least 3 second differences of 2 and working from e.g. simultaneous equations to find −n or at least 3 correct numerical examples C2 C1 for n rows or n columns or (n – 1) connectors or at least 3 second differences of 2 or working from e.g. simultaneous equations to find –n or at least 1 correct numerical example n(n – 1) oe 1 1(c)(iii) 2(n – 1)2 + 2n(n – 1) oe isw 1 FT 2 × their (c)(i) + 2 × their (c)(ii) 2(a) [horizontal] 8 [vertical] 8 [up diagonal] 4 [down diagonal] 4 2 B1 for 1, 2 or 3 correct

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 5 of 8 Question Answer Marks Partial Marks 2(b) H V RD LD T 2 × 2 0 0 0 0 0 3 × 3 3 3 1 1 8 4 × 4 8 8 4 4 24 5 × 5 15 15 9 9 48 6 × 6 24 24 16 16 80 2 B1 for each row 2(c)(i) 324 1 2(c)(ii) n rows/columns × ( ) 2 − n connectors or form a correct equation in a and use a correct step towards solving this equation or at least 3 correct numerical examples C2 C1 for n rows or n columns or ( ) 2 − n connectors or form a correct equation in a or at least 1 correct numerical example [a =] −2 and n2 − 2n 1 2(d) 2(n – 2)2 + 2(n2 – 2n) oe isw 1 FT 2 × (n – 2)2 + 2 × their 2(c)(ii) 3(a) H V UD DD T 3 × 3 0 0 0 0 0 4 × 4 4 4 1 1 10 5 × 5 10 10 4 4 28 6 × 6 18 18 9 9 54 n × n (n2 – 3n) (n2 – 3n) (n – 3)2 (n – 3)2 4 B2 for 3 numerical rows correct or B1 for 2 numerical rows correct and B1 for (n2 – 3n) in H and V and B1 for (n – 3)2 in UD and DD 3(b) 2(n2 – 3n) + 2(n – 3)2 oe 1 FT 2 × their (n2 – 3n) + 2 × their (n – 3)2 seen in 3(a), could be simplified 4(a) 1 = − k m oe or m is one more than k oe 1 4(b)(i) 2 × 5(5 – 1) + 2(5 – 1)2 leading to 72 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 6 of 8 Question Answer Marks Partial Marks 4(b)(ii) n = 9, m = 7 n = 16, m = 15 n = 35, m =35 [n = 5, m = 2] 3 B1 for each B MODELLING BREEDING DEER 5(a)(i) 20 20 4 100 × = oe M1 Calculations showing 20 + 20 oe and – 4 = 36 nfww A1 5(a)(ii) + Pn or– 0.2 Pn or 36 20 1 Pn + Pn – 0.2 Pn or 1.8 Pn from 36 20 1 5(b)(i) Calculation e.g. 36 × 1.8 oe C1 n 1 2 3 4 5 6 7 8 nP 20 36 64 116 208 374 674 1214 3 B2 for 3 correct or B1 for 1 or 2 correct If 0 scored SC1 for 1 of 64.8, 65, 208.8, 209, 673.2, 673, 1213.2, 1213 in correct place in table 5(b)(ii) Four correct plots 2 B1 for 2 or 3 correct 5(c) 20 = ab0 or [ ]1 36 = ab or [ ]1 36 = × their a b C1 1 20 1.8 − = × n P 2 B1 for either 20 or 1.8 If 0 scored SC1 for a = 20 and b = 1.8 6(a)(i) 20 = a(1 – 1)2 + b(1 – 1) + c and c = 20 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 7 of 8 Question Answer Marks Partial Marks 6(a)(ii) a + b + 20 = 36 oe 9 3 20 116 + + = a b oe isw 2 B1 for each 6(b)(i) Correct multiplication or division of one of their equations or Correct rearrangement of one of their equations or correct sketch of one of their equations C1 FT their linear 6(a)(ii) Correct subtraction of their equations or Correct substitution of their equation plus one correct step or correct sketch of second of their equations C1 FT P = 8(n – 1)2 + 8(n – 1) + 20 oe 2 B1 for either a or b correct If 0 scored SC1 for a = 8 and b = 8 6(b)(ii) Correct sketch 2 B2 for increasing continuous curve below plotted points for n > 4 and finishing between 400 and 600 at n = 8 or B1 for increasing continuous curve below plotted points for n > 5 and finishing between 300 and 600 at n = 8 or B1 for plotting points correctly without joining 6(c) No, with justification referring to n > 4 1 7(a)(i) 36 = a(2 – 1)b C1 [a =] 36 1 7(a)(ii) 116 = their 36 × (4 – 1)b C1 FT their 7(a)(i) 116 log log3 36 ÷ their or log3 116 36       their oe C1 or suitable sketch P = 36(n – 1)1.07 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2021 © UCLES 2021 Page 8 of 8 Question Answer Marks Partial Marks 7(b) Suitable sketch or calculation C1 Two appropriate reasons e.g. zero deer at start (after 1 year) – starts at (1, 0) [almost] linear most values don’t match only correct for some years [2 and 4] gradient less steep/different [after n = 4] 1