Cambridge IGCSE Mathematics - International 0607 — 2020 Oct/Nov Paper 6 · Variant 3

Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/63 Paper 6 Investigation and Modelling (Extended) October/November 2020 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 8). ● 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 [ ]. DC (LK/CB) 189359/1 © UCLES 2020 [Turn over This document has 20 pages. Blank pages are indicated. * 8 1 4 0 9 9 6 7 1 9 *

2 0607/63/O/N/20 © UCLES 2020 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 TO 4) AREAS OF POLYGONS INSIDE AND OUTSIDE A CIRCLE (30 marks) You are advised to spend no more than 50 minutes on this part. This investigation looks at the areas of polygons drawn inside and outside a circle of radius 10 cm. An inscribed polygon is a polygon in which all the vertices lie on a circle. This is an inscribed square. A circumscribed polygon is a polygon in which each side is a tangent to a circle. This is a circumscribed square. You may find some of these formulas useful. Area, A, of circle, radius r A r2 r = Area, A, of triangle, base b, height h A bh 2 1 = In a right-angled triangle, hypotenuse opposite sini = , cos hypotenuse adjacent i = , opposite tan adjacent i = .

3 0607/63/O/N/20 © UCLES 2020 [Turn over 1 (a) NOT TO SCALE 10 cm O A square circumscribes a circle, centre O, radius 10 cm. Work out the area of the square. … [1]

4 0607/63/O/N/20 © UCLES 2020 (b) NOT TO SCALE 10 cm O A square is inscribed in a circle, centre O, radius 10 cm. Work out the area of the square. … [2] (c) Show that the area of a circle, radius 10 cm, is cm 100 2 r . [1] (d) Area of inscribed square 1 Area of circle 1 Area of circumscribed square Use this statement to complete the inequality below. … 1 r 1 … [1]

5 0607/63/O/N/20 © UCLES 2020 [Turn over 2 (a) NOT TO SCALE 10 cm O A regular hexagon is inscribed in a circle, centre O, radius 10 cm. Find the area of the hexagon. … [3]

6 0607/63/O/N/20 © UCLES 2020 (b) (i) NOT TO SCALE 10 cm An equilateral triangle has height 10 cm. Find the area of the triangle. … [3] (ii) NOT TO SCALE O 10 cm A regular hexagon circumscribes a circle, centre O, radius 10 cm. Using your answer to part (i), find the area of the hexagon. … [2]

7 0607/63/O/N/20 © UCLES 2020 [Turn over (c) (i) Use Question 1(c), Question 2(a) and Question 2(b)(ii) to complete the inequality. … 1 r 1 … [1] (ii) Give a geometric reason why the range in the inequality in Question 2(c)(i) is smaller than the range in the inequality in Question 1(d). … … [1]

8 0607/63/O/N/20 © UCLES 2020 3 (a) NOT TO SCALE O 10 cm A regular 12-sided polygon is inscribed in a circle, centre O, radius 10 cm. Find the area of this polygon. … [2]

9 0607/63/O/N/20 © UCLES 2020 [Turn over (b) NOT TO SCALE O 10 cm A regular 12-sided polygon circumscribes a circle, centre O, radius 10 cm. Find the area of this polygon. … [3] (c) Use the answers to part (a) and part (b) to complete the inequality. … 1 r 1 … [1]

10 0607/63/O/N/20 © UCLES 2020 4 (a) (i) Show that a formula for the area, cm A 2, of a regular polygon with n sides inscribed in a circle, radius 10 cm, is sin A n n 50 360 ° = b l . [2] (ii) Show that a formula for the area, cm B 2, of a regular polygon with n sides that circumscribes a circle, radius 10 cm, is tan B n n 100 180 ° = b l . [2]

11 0607/63/O/N/20 © UCLES 2020 [Turn over (b) (i) Work out the area of a regular polygon with 100 sides that is inscribed in a circle, radius 10 cm. Give your answer correct to 4 significant figures. … [2] (ii) Work out the area of a regular polygon with 100 sides that circumscribes a circle, radius 10 cm. Give your answer correct to 4 significant figures. … [2] (c) Use your answers to part (b) to explain how you can find the value of r correct to 3 significant figures. … … [1]

12 0607/63/O/N/20 © UCLES 2020 B MODELLING (QUESTIONS 5 TO 8) MODELLING CONTAINERS (30 marks) You are advised to spend no more than 50 minutes on this part. Olivia wants to design a closed container with a volume of cm 1000 3 and minimum surface area. 5 Olivia uses a square-based cuboid to model the container. NOT TO SCALE h cm x cm x cm (a) (i) Write down a formula for the volume of the cuboid, cm V 3, in terms of x and h. … [1] (ii) Find a formula for the surface area, cm S 2, of the cuboid, in terms of x and h. Give your answer in its simplest form. … [2] (b) (i) V 1000 = . Write h in terms of x. … [1] (ii) Show that S x x 2 4000 2 = + . [1]

13 0607/63/O/N/20 © UCLES 2020 [Turn over (iii) Work out the value of S when x 25 = . … [1] (c) Sketch the graph of S x x 2 4000 2 = + for x 0 25 1 G . 0 0 S x 25 [3] (d) (i) Find the minimum surface area of the cuboid. … [1] (ii) Describe the container that gives the minimum surface area for Olivia’s model. … … [2]

14 0607/63/O/N/20 © UCLES 2020 6 Volume, V, of a cylinder of radius r, height h V r h 2 r = Curved surface area, A, of a cylinder of radius r, height h rh A 2r = Olivia now uses a cylinder to model the container. NOT TO SCALE h cm r cm The total surface area of this model is cm T 2. (a) V 1000 = . Show that T r r 2 2000 2 r = + . [3] (b) (i) Find the minimum surface area of the cylinder. … [2]

15 0607/63/O/N/20 © UCLES 2020 [Turn over (ii) Find the dimensions of the cylinder with the minimum surface area. r = … h = … [2]

16 0607/63/O/N/20 © UCLES 2020 7 Volume, V, of a pyramid, base area A, height h V h A 3 1 = Olivia now uses a square-based pyramid to model the container. NOT TO SCALE x cm O C D A B E h cm The pyramid, OABCD, has a square base of side x cm and height h cm. The vertex of the pyramid, O, is directly above the centre of the square base. E is the mid-point of BC. (a) Find an expression for OE in terms of h and x. … [2]

17 0607/63/O/N/20 © UCLES 2020 [Turn over (b) The total surface area of this model is cm P 2. V 1000 = . Show that P x x x 36000000 2 6 = + + . [4]

18 0607/63/O/N/20 © UCLES 2020 (c) (i) Find the minimum surface area of the pyramid. … [2] (ii) Find the dimensions of the pyramid with the minimum surface area. x = … h = … [2]

19 0607/63/O/N/20 © UCLES 2020 8 Olivia recommends the container with the smallest surface area to a company. Give a geometric reason why the company might not accept Olivia’s recommendation. Olivia recommends the … Geometric reason … … [1]

20 0607/63/O/N/20 © UCLES 2020 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 2020 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/63 Paper 6 (Extended) October/November 2020 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 2020 series for most Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 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.

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 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

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 4 of 7 Question Answer Marks Partial Marks A INVESTIGATION AREAS OF POLYGONS INSIDE AND OUTSIDE A CIRCLE 1(a) 400 1 1(b) 200 2 M1 for 0.5 × 10 × 10 or 102 + 102 or 2 × 10 × 10 1(c) 102π or π × 10 × 10 1 1(d) 2 < π < 4 1 FT their 1(b) and their 1(a) 2(a) 260 or 259.8[...] 3 M2 for 6 × 0.5 × 10 × 10 × sin 60 or 2 2 0.5 10 10 5 × × − oe or M1 for 0.5 × 10 × 10 × sin 60 or (102 – 52) 2(b)(i) 57.7 or 57.5 to 57.8 3 M1 for 10 ÷ (0.5 × base) = tan 60 or correct use of equivalent trig methods or base2 = 102 + (0.5 × base)2 oe M1 for 0.5 × their base × 10 oe or 0.5 × their base × their base × sin 60 2(b)(ii) 6 × their (b)(i) oe or base 1 0 6 2 their × × oe C1 346 or 345 to 346.8 B1 FT their (b)(i) 2(c)(i) 2.6[0] < π < 3.46 1 FT their 2(a) and their 2(b)(ii) 2(c)(ii) Correct explanation relating areas of polygons and the circle 1 e.g. Because the areas of the hexagons are closer to the area of the circle e.g. Less space between the circle and polygons 3(a) Correct method for area of one triangle with apex at centre C1 300 B1 Not from rounding 3(b) Correct trig method to find base or slant height of triangle with apex at centre C1 322 or 321.[5...] 2 M1 for correct method to find area of one triangle e.g. 0.5 × 10 × 10 × tan (their 15) 3(c) 3 < π < 3.22 or 3 < π < 3.21[5...] 1 FT their 3(a) and their 3(b)

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 5 of 7 Question Answer Marks Partial Marks 4(a)(i) n × 0.5 × 10 × 10 × 360 sin n leading to 360 50 sin n n 2 M1 for area of [one] triangle = 0.5 × 10 × 10 × 360 sin n 4(a)(ii) n × 0.5 × 10 × 2 × 360 10tan 2n leading to 180 100 tan n n 2 M1 for base of triangle = 2 × 360 10tan 2n 4(b)(i) 360 50 100 sin 100 × × C1 314.0 B1 4(b)(ii) 180 100 100 tan100 × × C1 314.3 B1 4(c) Correct explanation 1 e.g. 314.0 < 100 π < 314.3 so 3.140 < π < 3.143 so π = 3.14 to 3sf or Divide both areas by 100 and round the answers to 4 significant figures. The value of π is the first 3 figures. B MODELLING MODELLING CONTAINERS 5(a)(i) V = x2h oe 1 5(a)(ii) S = 2x2 + 4xh or S = 2x(x + 2h) 2 B1 for 2x2 or 4xh or correct expression not fully simplified or incorrectly simplified 5(b)(i) [ ] 2 1000 h x = 1 5(b)(ii) 2 2 1000 = 2 + 4 S x x x × × leading to 2 4000 = 2 + S x x 1 5(b)(iii) 1410 1

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 6 of 7 Question Answer Marks Partial Marks 5(c) Correct sketch B2 B1 for correct shape (dependent on vertical axis) B1 for correct endpoints and not crossing axes scale on vertical axis C1 Indicating minimum at approximately 500 5(d)(i) 600 1 5(d)(ii) Cube 1 [side] 10 cm 1 6(a) 2 = 2 + 2π T r rh π oe M1 2 1000 = π h r oe M1 2 2 1000 = 2π + 2π × π T r r r leading to 2 2000 2π + T r r = A1 6(b)(i) Correct sketch or 3 correct trials C1 554 or 553.5 to 553.6 B1 6b(ii) r = 5.42 or 5.419... h = 10.8 or 10.83 to 10.84 2 FT their (b)(i) B1 for each 7(a) sketch of right-angled triangle with short sides labelled 2 x and h C1 2 2 + 4 x h oe B1

0607/63 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2020 © UCLES 2020 Page 7 of 7 Question Answer Marks Partial Marks 7(b) 2 3000 = h x oe M1 [ ] 2 1 + 4× × × 2 P x x theirOE = oe M1 [P =] 2 2 4 1 9000000 + 4× × × + 2 4 x x x x oe M1 leading to 6 2 + 36000000 = + x P x x oe A1 7(c)(i) Correct sketch or 3 correct trials C1 660 or 660.3 to 660.4 B1 7c(ii) x = 12.8 or 12.84 to 12.85 1 h = 18.2 or 18.17[...] 1 FT their x 8 Cylinder and correct reason 1 e.g. the nets do not tessellate (wasted material in production) 6 2 4 + 36000000 [ =] + 2 4 x P x x x