Cambridge IGCSE Mathematics - International 0607 — 2021 Feb/March Paper 6 · Variant 2

This document has 16 pages. Any blank pages are indicated. Cambridge IGCSE™ DC (CE/SW) 207879/1 © UCLES 2021 [Turn over * 6 4 0 7 4 9 5 7 0 3 * CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 Investigation and Modelling (Extended) February/March 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 5) and part B (Questions 6 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 0607/62/F/M/21 © UCLES 2021 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 TO 5) ENCLOSED CUBOIDS (30 marks) You are advised to spend no more than 50 minutes on this part. This task is about the number of white cubes needed to enclose a grey cuboid with integer sides. 1 The diagram shows: • a grey cube with each edge 1 centimetre long (a 1 cm cube) • a view showing how the grey cube is enclosed in a layer of 1 cm white cubes • the resulting cube. First cube Second cube NOT TO SCALE (a) The enclosing layer of white cubes is 1 cm thick. This makes the second cube with each edge 3 cm long. Find the number of white cubes. … [1] (b) The second cube is enclosed in a layer of 1 cm white cubes to make the third cube. There are now 2 layers enclosing the grey cube. Each layer is 1 cm thick. (i) Write down the length of one edge of the third cube. … [1] (ii) Find the number of white cubes in the third cube. … [2]

3 0607/62/F/M/21 © UCLES 2021 [Turn over (c) For any cube made in this way, write down the relationship between the total number of 1 cm cubes and the number of white cubes. … [1] (d) For any cube made in this way, n = the number of enclosing layers L = the length of the outer edge in centimetres W = the number of white cubes. For the first cube, n = 0 and L = 1. For the second cube, n = 1 and L = 3. Complete this table for a grey cube of edge length 1 cm. Number of enclosing layers 0 1 2 3 4 n L 1 3 W 0 728 [5]

4 0607/62/F/M/21 © UCLES 2021 2 A grey cube has edge length 2 cm. This cube will be enclosed in layers of 1 cm white cubes. The diagram shows some of the white cubes in the first layer. NOT TO SCALE (a) This calculation gives the number of white cubes when there is one enclosing layer. 43 - 23 Explain why this calculation works. … … … [2] (b) Complete this table for a grey cube of edge length 2 cm. Number of enclosing layers 0 1 2 3 4 n L 2 10 W 0 56 992 [4]

5 0607/62/F/M/21 © UCLES 2021 [Turn over 3 (a) Complete this table for a grey cube of edge length 3 cm. Number of enclosing layers 0 1 2 3 4 n L 3 W 0 98 316 702 1304 [2] (b) A grey cube has edge length e cm. Write your expressions for W from questions 1(d), 2(b) and 3(a) in this table. Complete the table. Edge length of grey cube W 1 2 3 e [2]

6 0607/62/F/M/21 © UCLES 2021 4 NOT TO SCALE 3 cm 2 cm 5 cm This grey cuboid has edge lengths of 2 cm, 3 cm and 5 cm. (a) Find the number of white cubes when there is one enclosing layer. … [3] (b) A grey cuboid has edge lengths of a cm, b cm and c cm. The cuboid has n enclosing layers. Find the formula for the number of white cubes, W. … [2]

7 0607/62/F/M/21 © UCLES 2021 [Turn over 5 A grey cuboid has edge lengths of k cm, k cm and 2k cm. The grey cuboid has 8 enclosing layers of 1 cm white cubes. The largest face of the resulting cuboid has 546 white cubes. Find the value of k. … [5]

8 0607/62/F/M/21 © UCLES 2021 B MODELLING (QUESTIONS 6 to 10) ENCLOSURES FOR CATS (30 marks) You are advised to spend no more than 50 minutes on this part. Volume, V, of cylinder of radius r, height h. V r h 2 r = Volume, V, of cone of radius r, height h. V r h 3 1 2 r = Volume, V, of sphere of radius r. V r 3 4 3 r = This task looks at modelling the floor area and volume of enclosures for cats. Dorothy has two cats and is going to build an enclosure for them. She considers three designs. 6 Design 1 is a cone and design 2 is a hemisphere. The cone and the hemisphere have the same diameter, d. d d h Hemisphere Cone (a) (i) Show that the volume of the cone is d h 12 2 r . [1]

9 0607/62/F/M/21 © UCLES 2021 [Turn over (ii) Find the volume of the hemisphere. Write your answer in its simplest form, in terms of r and d. … [2] (b) Write h in terms of d when the volume of the cone is equal to the volume of the hemisphere. … [1] (c) (i) Find the volume of the cone when h d 2 = . Write your answer in its simplest form, in terms of r and d. … [2] (ii) Which is smaller, the volume of the cone in part (c)(i) or the volume of the hemisphere? Show how you decide. … [1]

10 0607/62/F/M/21 © UCLES 2021 Information about cat enclosures Minimum floor sleep area per cat 0.5 m2 Minimum floor exercise area per cat 2.8 m2 Minimum extra floor area per cat 1 m2 Highest point above floor must be 2 m or more 7 (a) Show that the total minimum floor area for two cats is 8.6 m2. [1] (b) Calculate the diameter that gives this minimum floor area for both the cone and the hemisphere. … [2] (c) Explain why a hemisphere with this minimum floor area is not a suitable enclosure. … … [1]

11 0607/62/F/M/21 © UCLES 2021 [Turn over 8 Dorothy decides on a volume of 10 m3 for the enclosure for the two cats. (a) Use question 6(a)(i) to find a model for h, in terms of r and d, for the cone enclosure. … [2] (b) Sketch your model for h for d 0 12 1 G . 0 Diameter Height 12 12 d h [2] (c) Find the total floor area of the enclosure when the height of the cone is 2 m. … [3]

12 0607/62/F/M/21 © UCLES 2021 9 Design 3 is a half-cylinder. It has diameter d and length w. Half-cylinder d w (a) Use the information on page 10 to write down the smallest value of d. … [1] (b) Show that the volume of the half-cylinder is d w 8 2 r . [2] (c) (i) Show that a model for w for a half-cylinder enclosure that has a volume of 10 m3 is w d 80 2 r = . [1]

13 0607/62/F/M/21 © UCLES 2021 (ii) Sketch the model for w for d 0 12 1 G . Length w 0 Diameter 12 12 d [1] (d) (i) The total floor area is 8.6 m2. Write down a model for w in terms of d. … [1] (ii) On the axes above sketch your model from part (d)(i). [2] (e) Find the height of the half-cylinder enclosure that fits both models. … [3] 10 Write down which enclosure of volume 10 m3, cone or half-cylinder, you think Dorothy should choose for her two cats. Explain your choice. … … [1]

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16 0607/62/F/M/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 9 printed pages. © UCLES 2021 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 (Extended) March 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 March 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 March 2021 © UCLES 2021 Page 2 of 9 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 March 2021 © UCLES 2021 Page 3 of 9 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 March 2021 © UCLES 2021 Page 4 of 9 Question Answer Marks Partial Marks A INVESTIGATION ENCLOSED CUBOIDS 1(a) 26 1 1(b)(i) 5 1 1(b)(ii) 53 – 1 oe or 4 [layers] ×25 + [1 layer] 24 oe or their 26 + 6[faces]×32 + 12[edges]×3 + 8[vertices] oe or their 26 + 53 – 33 oe C1 124 1 1(c) Total number of cubes = number of white cubes + 1 oe 1 1(d) 73 – 1 oe or at least three differences of 2 C1 1 2 3 4 n 3 5 7 9 2n + 1 oe 26 124 342 728 (2n+1)3−1 oe isw 4 B1 for 7 and 9 B1 for 342 B1 for 2n + 1 oe B1FT for (their (2n + 1))3 – 1 oe 2(a) 43 is the [total] volume of the cube [in cm3] 23 is the volume of the grey cube [in cm3] [The difference is the number of white cubes oe] OR 4 [cm] is the outer cube edge length oe 2 [cm] is the grey cube edge length oe Cubing edge length gives volume oe [The difference is the number of white cubes oe] 2 B1 for each OR B1 for two correct statements 2(b) 63 – 23 oe or 83 – 23 oe C1 n 1 2 3 4 n L 4 6 8 10 2n + 2 oe W 56 208 504 992 (2n+2)3−23 isw oe 3 B1 for 2nd row correct B1 for 208 and 504 B1FT for (their (2n + 2))3 – 23 oe 3(a) 0 1 2 3 4 n 3 5 7 9 11 2n + 3 oe 0 98 316 702 1304 (2n + 3)3 − 33 oe 2 B1 for row 2 correct B1FT for (their (2n + 3))3 – 33 oe

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 5 of 9 Question Answer Marks Partial Marks 3(b) Edge length of grey cube W 1 (2n + 1)3 − 1 2 (2n + 2)3 − 23 3 (2n + 3)3 − 33 e (2n + e)3 – e3 oe 2 B1 for (2n + e)3 oe or for – e3 4(a) 4 × 5 × 7 – 2 × 3 × 5 or 2(5 × 3 + 5 × 2 + 3 × 2)[faces] + 4(5 + 3 + 2)[edges] + 8[corners] C2 C1 for 4 × 5 × 7 or 4, 5 and 7 are edge lengths soi or for 2(5 × 3 + 5 × 2 + 3 × 2) + 4(5 + 3 + 2) 110 1 4(b) W = (a + 2n)(b + 2n)(c + 2n) – abc isw 2 B1 for a + 2n or b + 2n or c + 2n or for (a + 2)(b + 2)(c + 2) – abc or for using any stated numerical n

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 6 of 9 Question Answer Marks Partial Marks 5 2k + 16 and k + 16 seen C1 (2k + 16)(k + 16) = 546 C1 FT their length for (2k + a) and (k + a) Method 1 Sketch of quadratic with exactly one positive intersection with horizontal line for 546 or 273. C2 OR Method 2 Correct expansion of their equation if of the form (ak + b)(ck + d) = 546. C1 Method of solving their quadratic equation with non- zero term in k C1FT Dependent on first C1. Sketch of quadratic with exactly one positive intersection with horizontal axis or horizontal line. or factorisation or substitution into the quadratic formula 5 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 7 of 9 Question Answer Marks Partial Marks B MODELLING ENCLOSURES FOR CATS 6(a)(i) 2 1 π 3 2 d h       or 2 1 π 3 4 d h 1 6(a)(ii) [V =] 3 π 12 d isw 2 B1 for a correct unsimplified form e.g. 3 1 4 π 2 3 2 d   ×     oe 6(b) [h =] d 1 FT their (a)(ii) 6(c)(i) [V =] 2 π [ ] 12 2 d d × oe C1 [V =] 3 π 24 d 1 6(c)(ii) Cone with valid justification e.g. denominator for cone is greater or 1 12 or 2 24 > 1 24 oe or 0.83 > 0.042 (or 0.0416) or ratio = 1:2 oe 1 FT their 1 12 in part (a)(ii) > their 1 24 in part (c)(i) only if numerators for volume the same. 7(a) 0.5 + 2.8 + 1 [= 4.3] oe and [4.3] × 2 [= 8.6] oe 1 7(b) 8.6 π r = oe or d2 = 8.6 4 π × or better C1 3.31 1 If C0 and 0 scored, SC1 for 2.34 7(c) Height 1.65 is less than 2[m] or 1.65 less than highest point oe 1 FT 1 2 their 3.31 if less than 2. 8(a) 2 π 12 d h = 10 oe C1 h = 2 120 πd 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 8 of 9 Question Answer Marks Partial Marks 8(b) Correct sketch 2 B1 for correct shape, not cutting the d-axis but < 0.6 at d = 12 B1 for correct shape to the right of d = 1.2 8(c) 2 1 2 10 3 π r × = OR π d2 = 60 oe or π (2r)2 = 60 and 2 π 4 d × or 4 π r2 = 60 OR [d=] 120 2π oe or r = 15 π oe and 2 120 2π π 2     ×    or better OR [d=] 4.37[0…] or r = 2.185 or 2.19 and 2 4.37 π 2 their   ×    oe or π×(their 2.185)2 C2 C1 2 1 2 3 × π r C1 for each C1 for each C1 for each 15 1 9(a) 4 1 9(b) Using 2 2 r h π with correct substitution of 2 d and w leading to 2 8 d w π 2 B1 for using 2 2 r h π or correct substitution of 2 d and w into cylinder formula 9(c)(i) 2 8 d w π = 10 and πd2w = 80 leading to w = 2 80 πd 1

0607/62 Cambridge IGCSE – Mark Scheme PUBLISHED March 2021 © UCLES 2021 Page 9 of 9 Question Answer Marks Partial Marks 9(c)(ii) Correct sketch 1 Correct shape that does not cross either axis 9(d)(i) w = 8.6 d 1 9(d)(ii) Correct sketch 2 B1 for correct shape, not touching either axis B1 dep for one point of intersection, If 0 scored, SC1 for one point of intersection and a curve with only one axis as an asymptote. 9(e) Intersection marked on graph, or use of intersection point stated. or 2 80 πd = their 8.6 d or better C1 [d =] 2.96[…] or 80 8.6π oe 1 1.48[…] 1 FT their (d) 10 Choice with valid reason e.g. Cone because half-cylinder is not high enough. Half-cylinder has space that is easier to use. 1 FT their 9(e)