Cambridge IGCSE Mathematics - International 0607 — 2021 May/June Paper 6 · Variant 1

This document has 16 pages. Any blank pages are indicated. Cambridge IGCSE™ DC (CE/JG) 207497/3 © UCLES 2021 [Turn over CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 Investigation and Modelling (Extended) May/June 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 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 [ ]. * 7 3 2 6 3 4 3 8 5 4 *

2 0607/61/M/J/21 © UCLES 2021 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 TO 6) ROLLING SQUARE (30 marks) You are advised to spend no more than 50 minutes on this part. This investigation looks at the path of a point on a square as it rolls along the x-axis. A square of side 1 cm rolls along the x-axis. One roll is a turn of 90° clockwise about its bottom right corner. 0.5 1 1.5 0.5 1 1.5 Diagram 1 Starting position 2 2.5 3 y x 0 0 0.5 1 1.5 0.5 1 1.5 Diagram 2 After one roll 2 2.5 3 y x Position 1 Centre of rotation Position 1 Position 2 Centre of rotation Diagram 1 shows the square in Position 1. One side of the square is bold to help show the rotation. The centre of the square is (0.5, 0.5). Diagram 2 shows the square rolled 90° clockwise about (1, 0) to Position 2. 1 To get to Position 3 the square rolls 90° clockwise about (2, 0). To get to Position 4 the square then rolls 90° clockwise about (3, 0). (a) On the diagram below, draw the square in Position 4, Position 5 and Position 6. 0 1 2 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 y x [1]

3 0607/61/M/J/21 © UCLES 2021 [Turn over (b) Complete this table to show the x-coordinate of the centre of the square in each position. You may use the diagram on page 2 to help you. Position (n) 1 2 3 4 5 6 n x-coordinate 0.5 1.5 2.5 [2] (c) Find the x-coordinate of the centre of the square in Position 92. … [2] 2 The side of the square is now 2 cm. 1 2 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 y x 0 The square rolls along the x-axis in the same way as in Question 1. (a) Complete the table of x-coordinates of the centre of the square in different positions. Position (n) 1 2 3 4 5 6 n x-coordinate 1 3 [3] (b) Find the coordinates of the centre of the square in Position 35. (… , …) [1]

4 0607/61/M/J/21 © UCLES 2021 3 The side of the square is now 3 cm. Complete the table of x-coordinates of the centre of the square in different positions. You may use the diagram below to help you. Position (n) 1 2 3 4 5 6 n x-coordinate 1.5 1 2 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 y x 0 [3] 4 Write your expressions from Questions 1(b), 2(a) and 3 in the table below. Complete the table using any patterns you notice. You may use the grid on page 5 to help you. Side of square (w cm) x-coordinate in Position n 1 2 3 4 5 w

5 0607/61/M/J/21 © UCLES 2021 [Turn over 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 y x 0 [5] 5 A square of side w cm rolls from Position 1 to Position 120. At Position 120, the x-coordinate of the centre of the square is 2151. Find the value of w. … [3]

6 0607/61/M/J/21 © UCLES 2021 6 A square of side 2 cm rolls along the x-axis. A 1 2 3 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 y x Position 1 Position 2 0 (a) The table shows the x-coordinate of the point A for each position. Complete the table. Position 1 2 3 4 5 6 7 8 9 10 11 12 13 x-coordinate 0 4 6 8 12 14 20 22 24 [1] (b) In the row of x-coordinates there are four sequences. For Positions 4, 8, 12, ... the expression for the position is 4a, where a is a positive integer. Complete the table. a 1 2 3 4 5 a Position (4a) 4 8 12 16 20 4a x-coordinate 14 22 8a - … Position (4a – 1) 3 7 11 15 19 4a – 1 x-coordinate 6 … Position (4a – 2) 2 6 10 14 18 4a – 2 x-coordinate 4 12 20 … Position (4a – 3) 1 4a – 3 x-coordinate 0 … [5]

7 0607/61/M/J/21 © UCLES 2021 [Turn over (c) The 2 cm square rolls to Position 523. Use part (b) to help you find the coordinates of point A. (… , …) [4]

8 0607/61/M/J/21 © UCLES 2021 B MODELLING (QUESTIONS 7 TO 10) WIND TURBINES (30 marks) You are advised to spend no more than 50 minutes on this part. This task looks at the use of wind turbines to generate electricity. Area, A, of circle, radius r. A = πr2 Circumference, C, of circle, radius r. C = 2πr 7 This is the front view and the side view of a wind turbine. Blade length Height of tower Area covered by blade Side view Wind turbines with longer blades generate more electrical power than wind turbines with shorter blades. Power is measured in kilowatts (kW). A wind turbine has blades that are 27 m long and a tower of height 80 m. (a) Find the greatest and least height above the ground for the tip of a blade as it turns. Greatest height … Least height … [2]

9 0607/61/M/J/21 © UCLES 2021 [Turn over (b) An international soccer pitch is a rectangle measuring 70 m by 105 m. (i) Find the area covered by the blades of this wind turbine. Write your answer as a percentage of the area of the international soccer pitch. … [3] (ii) New wind turbines have blades that are 107 m long. Find the area covered by these blades as a percentage of the area of the international soccer pitch. … [2]

10 0607/61/M/J/21 © UCLES 2021 8 The amount of power generated depends on wind speed as well as the area covered. This table shows the power (in kW) for blades of different lengths at a fixed wind speed. Blade length (b metres) 27 33 40 44 48 54 64 72 80 Power (P kW) 225 300 500 600 750 1000 1500 2000 2500 (a) Plot the last three points on this graph. The first six points have been plotted for you. 0 500 0 1000 1500 2000 2500 10 20 30 40 50 Power (kW) Blade length (m) 60 70 80 P b [1]

11 0607/61/M/J/21 © UCLES 2021 [Turn over (b) A model for the power, P kW, is P = cb2, where b is the length of the blade in metres and c is a constant. Use the information to find a value for c and write down the model. … [2] (c) Another wind turbine generates 1200 kW. Use your model to find the length of its blade. … [2]

12 0607/61/M/J/21 © UCLES 2021 9 (a) A blade rotates through 30° every second. (i) Find the time it takes to complete a full turn and the number of complete turns it makes in a minute. Time = … Number of turns = … [3] (ii) Different parts of the blade travel through air at different speeds. 27 m Show that the speed of the tip of this blade, with length 27 m, is 14.1 m/s, correct to 3 significant figures. [3]

13 0607/61/M/J/21 © UCLES 2021 [Turn over (b) The blade with length 27 m now rotates through 40° every second. Find the new speed of the blade tip in m/s. … [2] (c) A blade turns through t degrees every second. The length of the blade is L metres. Write a model for the speed, S m/s, of the blade tip in terms of π, t and L. Give your answer in its simplest form. … [2] (d) The maximum speed for a blade tip is 72 m/s. Find the maximum speed of rotation, in degrees per second, for a blade with length 107 m. … [3]

14 0607/61/M/J/21 © UCLES 2021 10 Wind enters a turbine at a speed of u m/s. The wind leaves the turbine at a reduced speed of v m/s. u v x is the fraction that v is of u, so x u v = . A model for the efficiency, E, of the wind turbine is ( )( ) E x x 2 1 1 2 = - + . (a) What can you say about the wind speeds v and u if the efficiency, E, is zero? … [1]

15 0607/61/M/J/21 © UCLES 2021 (b) Sketch the graph of the model for E for x 0 1 G G . 0 E x 1 1 [2] (c) Find the value of x that gives maximum efficiency. … [1] (d) Find the greatest value for E. Give your answer as a percentage. … [1]

16 0607/61/M/J/21 © UCLES 2021 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. BLANK PAGE

This document consists of 7 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 (Extended) May/June 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 May/June 2021 series for most Cambridge IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 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 2021 © UCLES 2021 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 2021 © UCLES 2021 Page 4 of 7 Question Answer Marks Partial Marks A INVESTIGATION ROLLING SQUARE 1(a) Three correct squares in positions 4, 5 and 6 1 1(b) 3.5 4.5 5.5 n – 0.5 oe 2 B1 for 3.5 4.5 5.5 B1 for n – 0.5 oe 1(c) Correct substitution of 92 in their expression or 92 – 0.5 C1 FT their expression in n 91.5 1 FT their expression in n 2(a) 5 7 9 11 2 1 − n oe 2 B1 for 5 7 9 11 B1 for 2 1 − n oe At least 3 further squares on diagram or at least 3 further centres marked or at least 3 differences of 2 seen C1 2(b) (69, 1) oe 1 3 4.5 7.5 10.5 13.5 16.5 3 1.5 n − oe 2 B1 for 4.5 7.5 10.5 13.5 16.5 B1 for 3 1.5 − n oe At least 3 squares on diagram or at least 3 centres marked or at least 3 differences of 3 seen C1 4 4 2 − n oe 1 5 2.5 − n oe 1 2 −w wn oe 2 B1 for wn oe or 2 −w oe At least 3 differences seen for the term in n or for the constant or for the expression C1 5 their 2   −     w wn = 2151 C1 FT their expression from Q4 Simplifies to a single term in w C1 FT their equation of the form wn +aw = 2151 where a is a constant 18 nfww 1 mark final answer 6(a) 6 14 16 22 1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 5 of 7 Question Answer Marks Partial Marks 6(b) 1 2 3 4 5 4 8 12 16 20 6 14 22 30 38 8a – 2 3 7 11 15 21 6 14 22 30 38 8a – 2 2 6 10 14 18 4 12 20 28 36 8a – 4 1 5 9 13 17 0 8 16 24 32 8a − 8 5 B1 for 30, 38, and 30, 38, and 28, 36 correctly placed B1 for 8a – 2 correctly placed twice B1 for 8a – 4 oe correctly placed B1 for 5, 9, 13, 17 and 32 in bottom row B1 for 8a – 8 oe correctly placed 6(c) Method 1: Using algebra 523 4 or 523 2 4 + or 523 3 4 + or 523 1 4 + or 4a – 1 = 523 or 4(131) – 1 = 523 or 523 3 4 − C1 (their 131) × 8 – 2 or (their130) × 8 – 2 + 8 or (their130) × 8 – 2 C1 FT their 8a – 2 from (b) providing of the form ka + c where k and c are non- zero constants (1046, 0) 2 B1 for each coordinate Method 2: Using patterns Identifies the correct row in the table as 4a – 1 and states 2 × 523 C2 C1 for identifying the correct row in the table as 4a – 1 (1046, 0) 2 B1 for each coordinate

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 6 of 7 Question Answer Marks Partial Marks B MODELLING WIND TURBINES 7(a) 80 + 27 or 80 – 27 or m[etres] on answer line[s] C1 [Greatest] 107 [Least] 53 1 7(b)(i) 2 27 π × or 70 105 × C1 FT their27 if seen in (a) 31.2 or 31.15 to 31.16 2 M1 for [ ] 2 27 100 70 105 π × × × their oe If 0 scored, SC1 for 10.4 or 10.38 to 10.39 7(b)(ii) 489 or awrt 489.4 2 M1 for [ ] 2 π 107 100 (70 105) × × × their oe 8(a) 3 points correctly plotted 1 8(b) Correct substitution of P and b from the table into P = cb2 or [ ] 2 = P c b C1 P = (0.275 to 0.391)b2 nfww 1 8(c) 1200÷their c oe or 2 1200 = b their c C1 55.4 to 66.1 1 FT their (b) if possible 9(a)(i) 12 or 0.2 and 5 2 B1 for each Correct units of time seen on answer line or 360 30 or 60 12 or 30 60 360 × oe C1 9(a)(ii) (2 π 27) 12 × × ÷ their or 30 2 360 × × π× 27 oe 2 may be in steps B1 for 2 π 27 × × or 54π soi 14.13 to 14.14 1 9(b) 18.8 or 18.84 to 18.85 2 M1 for 40 2 27 360 × × π× oe or 6π 9(c) 180 π = tL S or other fully-simplified form 2 M1 for 180 πtL or equivalent unsimplified form

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2021 © UCLES 2021 Page 7 of 7 Question Answer Marks Partial Marks 9(d) Method 1: Substitutes then changes the subject 38.54 to 38.6 or 38.5 2 M1 for substitution of S = 72 and L = 107 in 180 π = tL S oe Correct change of subject of 72 180 107π = t to t C1 Method 2: Changes the subject then substitutes 38.54 to 38.6 or 38.5 2 M1 for changing the subject of 180 π = tL S oe to t Substitution of S = 72 and L = 107 C1 FT their rearrangement Method 3: Finds the number of seconds and then the number of degrees per second 38.54 to 38.6 or 38.5 2 M1 for 2π(107) ÷ 72 oe 360 ÷ (2π(107) ÷ 72) oe C1 10(a) They are equal oe 1 10(b) Correct sketch 1 1 E x 0 2 B1 for correct shape B1 for a curve meeting E-axis at approx 0.5 and x-axis at 1 10(c) 0.33[3…] 1 10(d) 59.2 to 59.3 1