Cambridge IGCSE Mathematics - International 0607 — 2014 Oct/Nov Paper 6 · Variant 3
0607/63/O/N/14 · 40 marks · ≈45 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper12 pages












Mark scheme5 pages
Answers below. Sit the paper first if you are practising.





Paper as text
Question paper, page 1
This document consists of 12 printed pages. IB14 11_0607_63/5RP © UCLES 2014 [Turn over *2009383695* Cambridge International Examinations Cambridge International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/63 Paper 6 (Extended) October/November 2014 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.
Question paper, page 2
2 © UCLES 2014 0607/63/O/N/14 Answer both parts A and B. A INVESTIGATION THE END RESULT (20 marks) You are advised to spend no more than 45 minutes on this part. This investigation looks at some relationships between unit fractions. A unit fraction is a fraction which has a numerator of 1. 1 The difference between the unit fractions 4 1 and 5 1 is 20 1 20 4 5 5 1 4 1 = − = − . The product of the unit fractions 4 1 and 5 1 is 20 1 5 1 4 1 = × . The difference and the product of the unit fractions 4 1 and 5 1 are the same. (a) (i) Find the difference and the product of the unit fractions 3 1 and 4 1 . difference = … product = … (ii) Find the difference and the product of the unit fractions 3 1 and 5 1 . difference = … product = …
Question paper, page 3
3 © UCLES 2014 0607/63/O/N/14 [Turn over (iii) Find another pair of unit fractions whose difference and product are the same. … and … (b) a and b are positive integers where a < b. Write as a single algebraic fraction (i) b a 1 1 − , … (ii) b a 1 1 × . … (c) A pair of unit fractions has the same difference and product. Complete the following statement. When the larger fraction is n 1 , the smaller fraction is …
Question paper, page 4
4 © UCLES 2014 0607/63/O/N/14 2 The sum of the unit fractions 2 1 and 4 1 is 4 3 8 6 8 2 4 4 1 2 1 = = + = + . The numbers 3 and 4 form two sides of a right-angled triangle. 3 4 NOT TO SCALE The hypotenuse of this triangle is 5 25 4 3 2 2 = = + . The set of integers (3, 4, 5) is called a Pythagorean Triple. (a) (i) Find the sum of the unit fractions 5 1 and 7 1 . … (ii) Using your answer to part (a)(i), find a Pythagorean Triple that may be formed from the sum of 5 1 and 7 1 . ( … , … , … ) (iii) Find the sum of 9 1 and 11 1 . … (iv) Can a Pythagorean Triple be formed from your answer to part (a)(iii)? Explain your answer. …
Question paper, page 5
5 © UCLES 2014 0607/63/O/N/14 [Turn over (b) For some values of p and q, the sum of the unit fractions p 1 and q 1 can be used to form a Pythagorean Triple. (i) Write as a single algebraic fraction q p 1 1 + . … (ii) Complete the Pythagorean Triple in terms of p and q. ( … , … , pq + 2) (iii) By applying Pythagoras’ Theorem to your answer to part (b)(ii), show that p2 + q2 = 2pq + 4.
Question paper, page 6
6 © UCLES 2014 0607/63/O/N/14 (iv) Find the two algebraic relationships between p and q so that q p 1 1 + can be used to form a Pythagorean Triple. … …
Question paper, page 7
7 © UCLES 2014 0607/63/O/N/14 [Turn over PART B STARTS ON PAGE 8.
Question paper, page 8
8 © UCLES 2014 0607/63/O/N/14 B MODELLING RESCUE MISSION (20 marks) You are advised to spend no more than 45 minutes on this part. A rescue team is going to fly to a disaster area. There must be at least 35 people in the team and they need at least 80 tonnes of equipment. The objective of this task is to find the best way to organise the rescue mission. Two planes, X and Y, are available to take the rescue team and their equipment. Both planes must be back at the starting airport in less than 24 hours. The maximum loads, flight times and total cost for each plane are shown in the table. Plane Maximum number of people per flight Maximum mass of equipment (tonnes) Maximum return flight time (hours) Cost for return flight ($ thousand) X 5 10 3 40 Y 7 20 4 65 Let x be the number of return flights made by plane X and y be the number of return flights made by plane Y. The rescue mission is to be modelled by five inequalities. 1 (a) (i) The total mass, in tonnes, of the equipment carried by the two planes is 10x + 20y. Explain why the mass of the equipment to be carried by the two planes can be shown by the inequality 10x + 20y [ 80. … (ii) Find, in terms of x and y, an inequality to show the total number of people carried by the two planes. … (iii) Find, in terms of x and y, an inequality to show the total flight time of the two planes. …
Question paper, page 9
9 © UCLES 2014 0607/63/O/N/14 [Turn over (b) Complete each inequality below to show (i) the greatest number of flights that plane X would be able to make, 0 Y x Y … (ii) the greatest number of flights that plane Y would be able to make. 0 Y y Y … (c) Write an expression, in terms of x and y, for the total cost of the flights, in thousands of dollars. …
Question paper, page 10
10 © UCLES 2014 0607/63/O/N/14 2 (a) On the grid, find the region satisfied by the inequalities from question 1(a) and question 1(b). Shade the unwanted region. 1 0 2 3 4 5 6 7 8 9 10 10 9 8 7 6 5 4 3 2 1 y x (b) Two of the lines cross when x = 3 2 4 and y = 3 2 1 . Explain why these values should not be used to find the minimum cost. …
Question paper, page 11
11 © UCLES 2014 0607/63/O/N/14 [Turn over (c) When x = 0 and y = 5, the cost of the rescue mission is 325 thousand dollars. Using your diagram, find the minimum cost of the rescue mission. Write down the number of times each plane must fly. plane X … plane Y … minimum cost = … thousand dollars 3 Plane X is not available to make more than 4 return flights. How many times should each plane be used to minimise the cost? What would be the increase in cost compared to your solution in question 2? plane X … plane Y … increase in cost = … thousand dollars 4 Compare your solutions to question 2(c) and question 3. Which is the better solution? Explain your answer. … … …
Question paper, page 12
12 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. 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 2014 0607/63/O/N/14 5 A third plane, Z, is available to assist with the mission. This plane must also be back at the starting airport in less than 24 hours. The maximum loads, flight times and total cost for each plane are shown in the table. Plane Maximum number of people per flight Maximum mass of equipment (tonnes) Maximum return flight time (hours) Cost for return flight ($ thousand) X 5 10 3 40 Y 7 20 4 65 Z 4 8 2 50 Let x be the number of return flights made by plane X, y be the number of return flights made by plane Y, and z be the number of return flights made by plane Z. This rescue mission is to be modelled by six inequalities. (a) Find the six inequalities and an expression for the cost in terms of x, y and z. Inequalities … … … … … … Cost … (b) Explain why the method of solution used in question 2 is not appropriate for this model. … …
Mark scheme, page 1
® 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 2014 series 0607 CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/63 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 2014 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.
Mark scheme, page 2
Page 2 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0607 63 © Cambridge International Examinations 2014 A INVESTIGATION THE END RESULT 1 (a) (i) (ii) (iii) (b) (i) (ii) (c) 1 oe 12 1 oe 12 2 oe 15 1 oe 15 Any correct pair oe b a ab − 1 oe ab 1 1 + n 1 1 1 1 1 1 1 1 C opportunity C opportunity Any pair of fractions whose denominators are n and n + 1 for any integer n > 0 Not 1 4 and 1 5 Not 1 3 and 1 4
Mark scheme, page 3
Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0607 63 © Cambridge International Examinations 2014 2 (a) (i) (ii) (iii) (iv) (b) (i) (ii) (iii) (iv) 12 oe 35 12, 35, 37 oe 99 20 Yes oe and correct reason isw p q pq + p + q, pq, [pq + 2] ( ) ( ) ( )2 2 2 2 pq their q p their pq + + = + 2 2 2 2 2 2 2 4 4 q p pq q p pq q p + + + = + + Correct further step leading to given answer 2 oe q p = + 2 oe q p = − 1 1 1 1FT 1 1 1 1 1 1 1 C opportunity C opportunity C opportunity FT their 99 20 e.g. 2 2 20 99 101 + = May be unsimplified Communication seen in at least two of 1a(i), 1a(ii), 2(a)(i), 2a(ii) or 2(a)(iii) 1
Mark scheme, page 4
Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0607 63 © Cambridge International Examinations 2014 B MODELLING RESCUE MISSION 1 (a) (i) (ii) (iii) (b) (i) (ii) (c) Maximums are 10 and 20 and minimum in total is 80 y x 7 5 + > 35 24 4 3 < + y x 7 5 y x 65 40 + 1 1 1 1 1 1 If 0 scored in (i) and (ii), SC1 for 8 and 6 2 (a) (b) (c) Line from (0, 4) to (8, 0) Line from (0, 5) to (7, 0) Line from (0, 6) to (8, 0) Line y = their 5 and line x = their 7 Correct region [They are] fractions oe 305 1 6 1 1FT 1FT 1FT 1 1 2 B1 for at least 2 correct C opportunity 3 10 3 3 1 4 Identify one solution using any valid comparison of time and cost. 1FT e.g. An extra $10 000 will reduce the time by one hour
Mark scheme, page 5
Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0607 63 © Cambridge International Examinations 2014 5 (a) (b) 35 4 7 5 > z y x + + 80 8 20 10 > z y x + + oe 24 2 4 3 < + + z y x 7 0 < < x 5 0 < < y 11 0 < < z z y x 50 65 40 + + e.g. [The graph used in part 2 is] 2 [dimensional; the problem is now] 3 [dimensional]. oe www 2 1 FT their x and y from 1(b) B1 for any 5 correct 2 not 3 variables All statements must be valid Communication in 2(c) 1
What you needed in this session
Cambridge’s own grade thresholds for 2014 Oct/Nov, Paper 6 · Variant 3. A higher threshold means an easier paper — the bar moves with how the cohort did.