Cambridge IGCSE Mathematics - International 0607 — 2015 Oct/Nov Paper 6 · Variant 2

This document consists of 12 printed pages. DC (ST/FD) 104250/3 © UCLES 2015 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 8 7 7 2 0 7 7 2 8 1 * CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 (Extended) October/November 2015 1 hour 30 minutes Candidates answer on the Question Paper. Additional Materials: Graphics calculator READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. Do not use staples, paper clips, glue or correction fluid. You may use an HB pencil for any diagrams or graphs. DO NOT WRITE IN ANY BARCODES. Answer both parts A and B. You must show all relevant working to gain full marks for correct methods, including sketches. In this paper you will also be assessed on your ability to provide full reasons and communicate your mathematics clearly and precisely. At the end of the examination, fasten all your work securely together. The total number of marks for this paper is 40.

2 0607/62/O/N/15 © UCLES 2015 Answer both parts A and B. A INVESTIGATION STARS (20 marks) You are advised to spend no more than 45 minutes on this part. This investigation looks at how stars can be drawn by a robot ant. Each diagram in this investigation is the path that the ant draws by repeating these two steps. • move one unit forward • turn anticlockwise through A° where A < 180 1 (a) A ° A ° The ant repeats the two steps 7 times, going round the polygon until it reaches its starting position. The ant makes 1 complete revolution (360°) to draw this regular 7-sided polygon. Show that A = 51.4, correct to 1 decimal place. (b) The ant draws a regular n-sided polygon. Write down a formula for A in terms of n. …

3 0607/62/O/N/15 © UCLES 2015 [Turn over 2 (a) The diagrams show how the ant draws a 7-pointed star. 1st turn 5th turn 2nd turn 3rd turn A ° A ° A ° A ° A ° 6th turn A ° 7th turn A ° During the 4th turn the ant completes its first revolution. Here is the completed 7-pointed star. To draw a 7-pointed star, the ant must repeat the two steps 7 times, as shown in the diagrams. In doing so, the ant makes 2 complete revolutions (720°). Calculate A. …

4 0607/62/O/N/15 © UCLES 2015 (b) The diagram shows another 7-pointed star. To draw it, the ant repeats the two steps 7 times. (i) In drawing this star, how many complete revolutions does the ant make before it reaches its starting position? … (ii) Explain why A 7 3 360 # = . … … (iii) To draw a 7-pointed star, 7 3 360 # is the largest possible value of A which is less than 180°. Show that 4 complete revolutions do not give a suitable value for A.

5 0607/62/O/N/15 © UCLES 2015 [Turn over 3 The diagram shows the 5-pointed star drawn using the largest possible value of A which is less than 180°. To draw it, the ant repeats the two steps 5 times and makes 2 complete revolutions. Write down the calculation to find A. Give your answer in a similar form to question 2(b)(ii). … 4 (a) Complete the table for stars that are drawn using the largest possible value of A which is less than 180°. The first row gives a polygon and is shown here to help you see the pattern. Number of points on the star Number of revolutions, n, that the ant makes Calculation of A A 3 1 3 1 360 # 120 5 2 144 7 7 360 3 # 154.3 9 11 163.6 (b) Find a formula for A in terms of n. … (c) Find the number of points on the star when A = 172.8 . …

6 0607/62/O/N/15 © UCLES 2015 5 Question 2 shows the only two possible 7-pointed stars. (a) Find the number of possible 11-pointed stars. Write down the number of complete revolutions that the ant makes to draw each of these stars. … (b) Explain why the calculation of A for a 15-pointed star using 6 complete revolutions gives the calculation for the 5-pointed star. … …

7 0607/62/O/N/15 © UCLES 2015 [Turn over (c) Find all the values of A that give a 15-pointed star. …

8 0607/62/O/N/15 © UCLES 2015 B MODELLING BODY MASS (20 marks) You are advised to spend no more than 45 minutes on this part. This task looks at three different models for the approximate body mass of an adult. 1 Here is a simple model to calculate the approximate body mass of an adult. Measure the height in centimetres and subtract 100 to get the approximate mass in kilograms. (a) Find the approximate body mass of an adult who is 1.8 m tall. … (b) Find the height of an adult who has an approximate body mass of 50 kg. … (c) For a height, h metres, the approximate body mass is M kilograms. Find the formula for M in terms of h. … (d) Sketch, showing appropriate scales, the graph of the approximate body mass, M. M 0 h

9 0607/62/O/N/15 © UCLES 2015 [Turn over 2 In the 19th century, the Belgian mathematician Quetelet suggested a different model. In his model the approximate body mass of an adult, M kg, varies as the square of the height, h metres. For an adult with height 2 m the approximate body mass is 88 kg. (a) Show that Quetelet’s model can be written as M = 22h2. (b) Show that an adult with height 1.5 m has an approximate body mass of 49.5 kg. (c) An adult has an approximate body mass of 77 kg. Calculate the height of this adult. …

10 0607/62/O/N/15 © UCLES 2015 3 (a) For some adults the simple model and Quetelet’s model give the same approximate body mass. Find the height of these adults. … (b) For most adults, which model gives a larger approximate body mass? Explain your answer. … …

11 0607/62/O/N/15 © UCLES 2015 [Turn over 4 Quetelet’s model gives an approximate body mass which is too small for taller people. A modern model uses M = khn and the following information. The approximate body mass of an adult is 78 kg for a height of 1.84 m and 50 kg for a height of 1.54 m. (a) Write down two equations in terms of k and n. … … (b) Show that 1.56 = 1.195n, correct to 3 decimal places. (c) Show that n = 2.5, correct to one decimal place. (d) Using n = 2.5, calculate the value of k, correct to the nearest integer. … Questions 4(e) and 5 are printed on the next page.

12 0607/62/O/N/15 © UCLES 2015 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 International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. (e) Sketch the graph of the modern model. M 0 h 5 Above which height does the modern model give a greater approximate body mass than the Quetelet model? …

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME for the October/November 2015 series 0607 CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/62 Paper 6 (Extended), maximum raw mark 40 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 will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2015 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0607 62 © Cambridge International Examinations 2015 Abbreviations cao correct answer only dep dependent FT follow through after error isw ignore subsequent working oe or equivalent rot rounded or truncated SC Special Case nfww not from wrong working soi seen or implied A INVESTIGATION STARS Question Answer Mark Part Marks 1 (a) 360 ÷ 7 oe 1 (b) [A =] n 360 oe 1 2 (a) 102.85… to 102.9 or 103 2 M1 for 7 720 oe (b) (i) 3 1 (ii) 3 revolutions oe and 7 angles oe 1 (iii) 7 360 4× > 180 oe 1 3 5 360 2× or equivalent calculation 1 4 (a) 3 1 360 3 1 × 120 5 2 360 5 2 × 144 7 3 360 7 3 × 154.3 9 4 360 9 4 × 160 11 5 360 11 5 × 163.6 2 B1 for 5 correct cells (b) 1 2 360 ] [ + = n n A oe 1

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0607 62 © Cambridge International Examinations 2015 Question Answer Mark Part Marks (c) 25 3 B2 for [n=] 12 soi or M1FT for their 1 2 360 + n n = 172.8 C opportunities 5 (a) [1], 2, 3, 4, 5 2 Accept in suitable calculations e.g. 360 11 2 × Deduct 1 for extras and 1 for each omission If 0 scored SC1 for 4 or 5 with no working (b) 5 2 15 6 = soi 1 (c) 48, 96, 168 cao 2 B1 for two correct values of A only or B1 for three correct values plus extras less than 180° or B1 for 2, 4 and 7 [revolutions] soi C opportunity Communication seen in one of 4(c) (two possible places) or 5(c) 1

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0607 62 © Cambridge International Examinations 2015 B MODELLING BODY MASS Question Answer Mark Part Marks 1 (a) 80[kg] 1 (b) 1.5[m] or 150cm 1 (c) [M =] 100h – 100 oe seen 1 (d) Straight line with positive gradient approx through (1.5, 50) and (2, 100) 1 1 C opportunity 2 (a) 2 M kh = or 2 M h ∝ 88 = k × (22 or 4) 1 1 If 0 scored SC1 for 88 = 22 × 4 oe C opportunity (b) 22 × 1.52 [= 49.5] oe 1 (c) 1.87[m] or 187cm 1 Condone 1.9[m] but not 190cm C opportunity 3 (a) 1.485 to 1.49 [m] or 148.5 to 149 cm 1 Condone 3.06 as a second answer (b) Simple (100h – 100) and correct conclusion 1 C opportunity 4 (a) 78 = k 1.84n isw 50 = k 1.54n isw 1 (b) n n k k 54 .1 84 .1 50 78 = 1 (c) 195 .1 log 56 .1 log or log1.195 1.56 1 (d) 17 2 M1 for 78 = k × 1.842.5 or 50 = k × 1.542.5 or B1 for 16.98 to 16.99 C opportunity (e) exponential curve 1 C opportunity 5 1.67[…] or 1.68 [m] 1FT FT their 17 rot to at least 2dp C opportunity Communication seen in four of 1(d), 2(a), 2(c), 3(a), 4(d), 4(e) or 5 2 1 mark if seen in two