Cambridge IGCSE Mathematics - International 0607 — 2011 Oct/Nov Paper 6 · Variant 1

This document consists of 12 printed pages. IB11 11_0607_06/3RP © UCLES 2011 [Turn over *1866204662* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/06 Paper 6 (Extended) October/November 2011 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. 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. www.XtremePapers.com

2 © UCLES 2011 0607/06/O/N/11 For Examiner's Use Answer both parts A and B. A INVESTIGATION MAXIMISING THE PERIMETER (20 marks) Identical shapes can be joined to make larger shapes. 1 Equilateral triangles of side 1 cm may be joined edge to edge, for example but not like this. (a) The diagram below shows a shape made of 4 equilateral triangles and a shape made of 5 equilateral triangles. Draw a different shape made of 4 equilateral triangles and a different shape made of 5 equilateral triangles. (b) (i) The diagram below shows a shape, made of 6 equilateral triangles, with a perimeter of 6 cm. Draw a different shape, made of 6 equilateral triangles, with a perimeter greater than 6 cm.

3 © UCLES 2011 0607/06/O/N/11 [Turn over For Examiner's Use (ii) The diagram below shows a shape, made of 7 equilateral triangles, with a perimeter of 7 cm. Draw a different shape, made of 7 equilateral triangles, with a perimeter greater than 7 cm. (c) (i) This table shows the greatest possible perimeters for shapes made of equilateral triangles. Complete the table. Number of equilateral triangles 2 3 4 5 6 7 8 Greatest perimeter (cm) 4 10 You may use the grid below to help you.

4 © UCLES 2011 0607/06/O/N/11 For Examiner's Use (ii) Write down the greatest perimeter for a shape made of 20 equilateral triangles. cm (iii) How many equilateral triangles make the shape when the greatest perimeter is 32 cm? (d) Write down an expression, in terms of x, for the greatest perimeter for a shape made of x equilateral triangles. 2 Squares of side 1 cm may be joined edge to edge, for example but not like this. (a) Find the greatest perimeter for a shape made of 6 squares. cm You may use the grid opposite to help you.

5 © UCLES 2011 0607/06/O/N/11 [Turn over For Examiner's Use (b) (i) Complete this table. Number of squares 2 3 4 5 6 7 8 9 10 Greatest perimeter (cm) 6 12 22 (ii) Write down the greatest perimeter for a shape made of 17 squares. cm (iii) How many squares make the shape when the greatest perimeter is 32 cm? (c) Write down an expression, in terms of x, for the greatest perimeter for a shape made of x squares.

6 © UCLES 2011 0607/06/O/N/11 For Examiner's Use 3 (a) This table shows the greatest perimeters for shapes made of regular hexagons of side 1 cm. Complete the table. Number of regular hexagons 2 3 4 5 6 Greatest perimeter (cm) 26 (b) Write down an expression, in terms of x, for the greatest perimeter for a shape made of x regular hexagons. 4 Find an expression, in terms of x, for the greatest perimeter for a shape made of x regular octagons.

7 © UCLES 2011 0607/06/O/N/11 [Turn over For Examiner's Use 5 (a) Write down an expression, in terms of x and y, for the greatest perimeter for a shape made of x regular polygons each with y sides. (b) The greatest perimeter for a shape made of x regular polygons, each with y sides is 26 cm. Find three possible pairs of values of x and y. x = y = x = y = x = y =

8 © UCLES 2011 0607/06/O/N/11 For Examiner's Use B MODELLING COVERING CAKES (20 marks) Different shaped cakes are made each with a volume of 4000 cm3. The top and sides of each cake are covered in chocolate. 1 A square-based cake measures x cm by x cm by y cm, as shown in the diagram. x cm x cm y cm (a) Show that 2 4000 y x = . (b) The area covered in chocolate is S cm2. By finding an expression for S in terms of x and y show that 2 16000 S x x = + .

9 © UCLES 2011 0607/06/O/N/11 [Turn over For Examiner's Use (c) Sketch the graph of S against x for 2 Y x Y 40 and 0 Y S Y 10000 on the axes below. S 0 x 40 10000 (d) Find the minimum surface area to be covered in chocolate. Write down the values of x and y. minimum surface area = cm2 x = y =

10 © UCLES 2011 0607/06/O/N/11 For Examiner's Use 2 A circular-based (cylindrical) cake has a radius of x cm and a height of y cm. The area to be covered in chocolate is S cm2 and the volume of the cake is 4000 cm3. y cm x cm (a) Show that 2 8000 S x x π = + . (b) Sketch the graph of S against x for 1 Y x Y 20 and 0 Y S Y 10000 on the axes below. S 0 x 20 10000

11 © UCLES 2011 0607/06/O/N/11 [Turn over For Examiner's Use (c) Find the minimum surface area to be covered in chocolate. Write down the values of x and y. minimum surface area = cm2 x = y = 3 2 16000 S x x = + and 2 8000 S x x π = + are models for the amount of chocolate required to cover the top and sides of each cake. (a) Explain how you could use these models for surface area to find the volume of chocolate required. (b) Comment on whether the models give realistic results for the volume of chocolate. Question 4 is printed on the next page

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. 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 2011 0607/06/O/N/11 For Examiner's Use 4 For a cake with minimum surface area, bakers use the following rule: Test this rule on both cakes. Show your working. There is twice as much chocolate on the sides as on the top.

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2011 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. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the October/November 2011 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 – October/November 2011 0607 06 © University of Cambridge International Examinations 2011 A INVESTIGATION MAXIMISING THE PERIMETER 1 (a) 4 joined equilateral triangles (not in row) 5 joined equilateral triangles (not in row) e.g. 1 Shapes may not be rotations or reflections of those given 1 for both a 4 triangle and a 5 triangle diagram (b) (i) 6 joined equilateral triangles with a perimeter > 6 e.g. or 1 (ii) 7 joined equilateral triangles with a perimeter > 7 1 (c) (i) Number of equilateral triangles 2 3 4 5 6 7 8 Greatest perimeter (cm) 4 5 6 7 8 9 10 1 C –1 any error or omission C opportunity (ii) 22 (cm) 1 (iii) 30 (triangles) 1 (d) x + 2 oe 1 Not x = y = – 1 mark once only 2 (a) 14 (cm) 1 C C opportunity (b) (i) Number of squares 2 3 4 5 6 7 8 9 10 Greatest perimeter(cm) 6 8 10 12 14 16 18 20 22 1 –1 any error or omission (ii) 36 (cm) 1 (iii) 15 (squares) 1 (c) 2x + 2 oe 1

Page 3 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0607 06 © University of Cambridge International Examinations 2011 3 (a) Number of regular hexagons 2 3 4 5 6 Greatest perimeter (cm) 10 14 18 22 26 1 C –1 any error or omission C opportunity (b) 4x + 2 oe 1 4 6x + 2 oe 1 5 (a) (y – 2)x + 2 oe 2 1 for y – 2 seen (b) x = 24, y = 3 x = 12, y = 4 x = 8, y = 5 x = 6, y = 6 x = 4, y = 8 x = 3, y = 10 x = 2, y = 14 x = 1, y = 26 2FT C ft their part (a) 1 for one or two correct pairs C opportunity C1 1 for two C opportunities seen [Total: 20]

Page 4 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0607 06 © University of Cambridge International Examinations 2011 B MODELLING COVERING CAKES 1 (a) Volume = x × x ×y oe e.g. V = x²y y = 4000 x² 1 C C opportunity (b) S = x² + 4xy oe S = x² + 4x(4000) x² S = x² + 16000 x 1 1 (c) correct sketch 1 C C opportunity (d) (minimum surface area =) 1200 (cm2) (x =) 20 (y =) 10 1 1 1 2 (a) V = πx2y (= 4000) S = πx2 + 2πxy y = 4000 or πxy = 4000 oe πx2 x S = πx2 + 2πx 4000 πx2 S = πx2 + 8000 x 1 1 1 1 C C opportunity (b) correct sketch 1 C C opportunity (c) (minimum surface area =) 1110 (cm2) or better (1107.162…) (x =) 11 or better (10.8385…) (y =) 11 or better (10.8385…) 1 1 1 3 (a) Multiply by thickness 1 explanation (b) Not uniform thickness or Missing elements of volume 1 comment

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2011 0607 06 © University of Cambridge International Examinations 2011 4 Square based: top = 400 cm2 : sides = 800 cm2 Circular based: Top = 369(.05..) cm2 : sides = 738(.1..) cm2 Yes, both in ratio – top : sides = 1 : 2 1 1 CFT for areas for areas C opportunity for statement that FT their areas C1 1 for two opportunities seen [Total 20]