Cambridge IGCSE Mathematics - International 0607 — 2010 May/June Paper 6 · Variant 1

This document consists of 7 printed pages and 1 blank page. IB10 06_0607_06/5RP © UCLES 2010 [Turn over *0661845285* *0661845285* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/06 Paper 6 (Extended) May/June 2010 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, highlighters, glue or correction fluid. You may use a pencil for any diagrams or graphs. Answer both parts A and B. You are advised to spend 45 minutes on Part A and 45 minutes on Part 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 the marks for this paper is 40. www.XtremePapers.com

2 © UCLES 2010 0607/06/M/J/10 For Examiner's Use Answer both parts A and B. A. INVESTIGATION FERMAT’S LITTLE THEOREM (20 marks) You are advised to spend 45 minutes on part A. The division 46 ÷ 5 gives 9 with a remainder of 1. A method for finding the remainder is 46 ÷ 5 = 9.2 Because 9 × 5 = 45, the remainder is 46 – 45 = 1. The division 921 ÷ 7 gives a remainder of 4. A method for finding the remainder is 921 ÷ 7 = 131.571…. Because 131 × 7 = 917, the remainder is 921 – 917 = 4. The division 211 ÷ 13 gives a remainder of 7. A method for finding the remainder is 211 ÷ 13 = 157.5384…. Because 157 × 13 = 2041, the remainder is 211 – 2041 = 7. 1 Find the remainder in these divisions. (a) 1234 ÷ 7 (b) 29 ÷ 9

3 © UCLES 2010 0607/06/M/J/10 [Turn over For Examiner's Use 2 In 1640 the French mathematician Fermat found something interesting about the remainder when dividing by a prime number. Some of his results are shown in the table below. Prime Division Remainder Division Remainder Division Remainder 3 23 ÷ 3 5 25 ÷ 5 35 ÷ 5 45 ÷ 5 7 27 ÷ 7 2 37 ÷ 7 47 ÷ 7 11 211 ÷ 11 Complete the unshaded boxes in this table. You may use the space below to show any working. 3 Use the patterns you have found in your table to complete the following statements. (a) 711 ÷ has a remainder of . (b) 817 ÷ has a remainder of . 4 From the table 27 ÷ 7 has a remainder of 2. So 27 – 2 has a prime factor of 7. Because 27 – 2 = 2(26 – 1) and 7 is not a factor of 2, 26 – 1 has a prime factor of 7. (a) Complete the following statements to show why 512 – 1 has a prime factor of 13. has a remainder of 5. So 513 – 5 has a prime factor of . Because 513 – 5 = and is not a factor of 5, 512 – 1 has a prime factor of 13. (b) Write down a prime factor of 816 – 1. Part A continues on page 4.

4 © UCLES 2010 0607/06/M/J/10 For Examiner's Use 5 Complete the general statement below. a p-1 – 1 has a prime factor of . This is known as Fermat’s Little Theorem. 6 Fermat noted that p must not be a factor of a. Give an example to show the result is not true when p is a factor of a. 7 By writing 724 – 1 in different ways, you can use Fermat’s Little Theorem to find prime factors. Examples: 724 – 1 = ( ) 1 24 7 – 1 = ( ) 2 1 24 7 − – 1 Using Fermat’s Little Theorem with p = 2, 724 – 1 has a prime factor of 2. 724 – 1 = ( ) 3 8 7 – 1 = ( ) 4 1 8 7 − – 1 Fermat’s Little Theorem with p = 4 cannot be used because 4 is not a prime number. Use the above method to find as many prime factors as you can of 724 – 1. Remember: p must not be a factor of a. Show all your working.

5 © UCLES 2010 0607/06/M/J/10 [Turn over For Examiner's Use BLANK PAGE Part B starts on page 6.

6 © UCLES 2010 0607/06/M/J/10 For Examiner's Use B. MODELLING CHANGE OF AVERAGE SPEED (20 marks) You are advised to spend 45 minutes on part B. Sam makes a journey in two stages. In Stage 1, he cycles at 10 km/h for one hour. In Stage 2, he cycles at 20 km/h. 1 In Stage 2, Sam cycles for 30 minutes. (a) Find the total distance he travels. km (b) Show that his average speed for the whole journey is 13.3 km/h. 2 If, in Stage 2, he cycles for 15 minutes, show that his average speed is now 12 km/h. 3 If, in Stage 2, he cycles for 12 minutes, find his average speed. km/h 4 (a) In Stage 2, Sam cycles for x minutes. Write down a formula for S, the average speed, in km/h. (b) Your formula is a mathematical model for the average speed. Show that it simplifies to S = 600 + 20 60 + x x .

7 © UCLES 2010 0607/06/M/J/10 [Turn over For Examiner's Use 5 Use the model to find Sam’s average speed if, in Stage 2, he cycles for 13 minutes. km/h 6 On the axes below, sketch the graph of S against x for 0 Y x Y 500. Show any important details on your sketch. S x 0 0 500 7 Find how many minutes Sam must cycle in Stage 2 to have an average speed of 13 km/h. min Part B continues on page 8.

8 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. University of 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. © UCLES 2010 0607/06/M/J/10 For Examiner's Use 8 In Stage 2, instead of 20 km/h, Sam cycles at y km/h for x minutes. (a) Modify the model in question 4(b). (b) Find y if, after 24 minutes in Stage 2, Sam’s average speed is 8 km/h. y = (c) On the axes below sketch the graph of S against x if he stops for x minutes at the end of Stage 1. S x 0 0

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2010 question paper for the guidance of teachers 0607 CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/06 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.XtremePapers.com

Page 2 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2010 0607 06 © UCLES 2010 M marks are given for a correct method. A marks are given for an accurate answer following a correct method. B marks are given for a correct statement or step. D marks are given for a clear and appropriately accurate drawing. P marks are given for accurate plotting of points. E marks are given for correctly explaining or establishing a given result. C marks are given for clear communication. Abbreviations cao correct answer only cso correct solution only ft follow through oe or equivalent soi seen or implied ww without working www without wrong working

Page 3 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2010 0607 06 © UCLES 2010 Question Answer Mark Notes Comments A 1 (a) (b) 2 8 1 1 B1 B1 Communication mark possible for complete method shown. 2 Prime Division Remainder Division Remainder Division Remainder 3 23 ÷ 3 2 5 25 ÷ 5 2 35 ÷ 5 3 45 ÷ 5 4 7 27 ÷ 7 2 37÷7 3 47 ÷ 7 4 11 211 ÷ 11 2 311 ÷ 11 3 411 ÷ 11 4 3 Deduct 1 2 for each error or omission and round down B3 Ignore extra entries 3 (a) (b) 11 7 17 8 1 1 B1 B1 4 (a) (b) 513 ÷ 13 13 5(512 − 1) 13 17 4 1 B1 B1 B1 + B1 B1 Accept 3, 5, 7, 13, 97, 241, 257, 653 5 p 1 B1 Accept (p − 1) + 1 or p − 1 + 1 6 Expression with p prime and a factor of a For example 105–1 – 1 or 104 – 1 Evaluation and comment that p is not a factor 2 B1 R1 Ignore extra expressions 7 724 – 1 = [( ) 2 12 7 – 1] = ( ) 1 3 12 7 − – 1 so 3 is prime factor 724 – 1 = [( ) 4 6 7 – 1] = ( ) 1 5 6 7 −– 1 so 5 is prime factor 724 – 1 = [( ) 12 2 7 – 1] = ( ) 1 13 2 7 − – 1 so 13 is prime factor 4 M1 A1 B1 B1 Apply to one correct answer Apply to other correct answers deducting one for each incorrect Further prime factors are 19,43,73,181,193,409,1201 1 C1 Communication seen in question 1

Page 4 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2010 0607 06 © UCLES 2010 Question Answer Mark Notes Comments B 1 (a) 20 1 B1 (b) 2 1 1 20 oe 1 R1 Averaging speeds possible 2 4 1 1 5 10 + oe 2 R1 15 ÷1.25 with time in any form R1 for 15 and 4 1 1 shown in working Accept 12 × 1.25 = 15 3 11.6 to 11.7(km/h) 2 M1 5 1 1 4 10 + oe Ignore extra methods Communication mark possible but not for model or 72 840 4 (a) 10 20 60 1 60 x x + × + oe for numerator 2 B1 for numerator or denominator seen (b) Evidence of either multiplying top and bottom by 60 or common denominators of 60 oe. 1 R1 5 11.7 to 11.8(km/h) 1 B1 Communication mark (can be evidence of substitution) 6 2 G1 correct shape G1 start at (0, 10) 7 26 or better 2 M1 Sketch showing intersection of graphs M1 600 + 20x = 13(60 + x) Communication mark for complete correct method shown or described. Reverse substitution statement does not gain communication

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2010 0607 06 © UCLES 2010 8 (a) (S =) x yx + + 60 600 oe 1 B1 (b) 3 2 M1 24 60 24 600 + + y = 8 soi A1ft for at least same level of difficulty Communication mark (c) 2 G1 decreasing from a point on the y-axis G1 x-axis asymptote 1 C1 Communication seen in two of questions 3, 5, 7, 8(b)