Cambridge IGCSE Mathematics (with coursework) 0581 — 2014 Oct/Nov Paper 2 · Variant 1

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. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fl uid. DO NOT WRITE IN ANY BARCODES. Answer all questions. If working is needed for any question it must be shown below that question. Electronic calculators should be used. If the degree of accuracy is not specifi ed in the question, and if the answer is not exact, give the answer to three signifi cant fi gures. Give answers in degrees to one decimal place. For π, use either your calculator value or 3.142. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total of the marks for this paper is 70. MATHEMATICS 0581/21 Paper 2 (Extended) October/November 2014 1 hour 30 minutes Candidates answer on the Question Paper. Additional Materials: Electronic calculator Geometrical instruments Tracing paper (optional) Cambridge International Examinations Cambridge International General Certifi cate of Secondary Education This document consists of 12 printed pages. [Turn over IB14 11_0581_21/FP © UCLES 2014 *9526088866* PAPA CAMBRIDGE

2 0581/21/O/N/14 © UCLES 2014 1 Use your calculator to fi nd the value of 1.357. Give your answer correct to 5 signifi cant fi gures. Answer … [2] __________________________________________________________________________________________ 2 Write the following in order of size, smallest fi rst. π 3.14 7 22 3.142 3 Answer … < … < … < … < … [2] smallest __________________________________________________________________________________________ 3 Write down the letters in the word above that have (a) exactly one line of symmetry, Answer(a) … [1] (b) rotational symmetry of order 2. Answer(b) … [1] __________________________________________________________________________________________

3 0581/21/O/N/14 © UCLES 2014 [Turn over 4 Cheryl recorded the midday temperatures in Seoul for one week in January. Day Mon Tue Wed Thu Fri Sat Sun Temperature (°C) –4 –5 –3 –11 –8 –3 –1 (a) Write down the mode. Answer(a) … °C [1] (b) On how many days was the temperature lower than the mode? Answer(b) … [1] __________________________________________________________________________________________ 5 Without using a calculator, work out 4 1 + 6 1 . Write down all the steps in your working and give your answer as a fraction in its simplest form. Answer … [2] __________________________________________________________________________________________ 6 Write 15.0782 correct to (a) one decimal place, Answer(a) … [1] (b) the nearest 10. Answer(b) … [1] __________________________________________________________________________________________

4 0581/21/O/N/14 © UCLES 2014 7 The population of Dubai at the end of 2012 was 2.1 million. This was predicted to increase at a rate of 6% each year. Calculate the predicted population of Dubai at the end of 2015. Answer … million [3] __________________________________________________________________________________________ 8 On a ship, the price of a gift is 24 euros (€) or $30. What is the difference in the price on a day when the exchange rate is €1 = $1.2378? Give your answer in dollars, correct to the nearest cent. Answer $ … [3] __________________________________________________________________________________________

5 0581/21/O/N/14 © UCLES 2014 [Turn over 9 (a) Write 2.8 × 102 as an ordinary number. Answer(a) … [1] (b) Work out 2.5 × 108 × 2 × 10–2. Give your answer in standard form. Answer(b) … [2] __________________________________________________________________________________________ 10 Solve the equation. x 5 x + = 3 7 Answer x = … [3] __________________________________________________________________________________________

6 0581/21/O/N/14 © UCLES 2014 11 (a) Simplify x8 ÷ x2. Answer(a) … [1] (b) Simplify 6 1 27 x 3 ` j . Answer(b) … [2] __________________________________________________________________________________________ 12 Solve the simultaneous equations. 0.4x – 5y = 27 2x + 0.2y = 9 Answer x = … y = … [3] __________________________________________________________________________________________ 13 y varies directly with 5 x + . y = 4 when x = –1. Find y when x = 11. Answer y = … [3] __________________________________________________________________________________________

7 0581/21/O/N/14 © UCLES 2014 [Turn over 14 A = 2 1 8 4 f p Work out A2 – 4A. Answer f p [3] __________________________________________________________________________________________ 15 A B C (a) Using compasses and straight edge only, construct (i) the perpendicular bisector of AC, [2] (ii) the bisector of angle ACB. [2] (b) Shade the region inside the triangle which is ● nearer to A than to C and ● nearer to AC than to BC. [1] __________________________________________________________________________________________

8 0581/21/O/N/14 © UCLES 2014 16 A helicopter fl ies from its base B to deliver supplies to two oil rigs at C and D. C is 6 km due east of B and the distance from C to D is 8 km. D is on a bearing of 120° from B. North North 6 km 8 km B C D NOT TO SCALE 120° Find the bearing of D from C. Answer … [5] __________________________________________________________________________________________

9 0581/21/O/N/14 © UCLES 2014 [Turn over 17 The diagram shows a child’s toy. 5 cm 8 cm NOT TO SCALE The shape of the toy is a cylinder of radius 5 cm and height 8 cm on top of a hemisphere of radius 5 cm. Calculate the volume of the toy. [The volume, V, of a sphere with radius r is V = 3 4 πr 3.] Answer … cm3 [5] __________________________________________________________________________________________

10 0581/21/O/N/14 © UCLES 2014 18 If it rains today the probability that it will rain tomorrow is 0.4 . If it does not rain today the probability that it will rain tomorrow is 0.2 . On Sunday it rained. (a) Complete the tree diagram for Monday and Tuesday. Rain No rain 0.6 0.4 Rain Monday Tuesday No rain … 0.4 Rain No rain … … [2] (b) Find the probability that it rains on at least one of the two days shown in the tree diagram. Answer(b) … [3] __________________________________________________________________________________________

11 0581/21/O/N/14 © UCLES 2014 [Turn over 19 A B C D NOT TO SCALE Two circles, centres A and B, are each of radius 8 cm and intersect at C and D. Each circle passes through the centre of the other circle. (a) Explain why angle CBD is 120°. Answer(a) [1] (b) For the circle, centre B, fi nd the area of the sector BCD. Answer(b) … cm2 [2] (c) (i) Find the area of the shaded segment CAD. Answer(c)(i) … cm2 [3] (ii) Find the area of overlap of the two circles. Answer(c)(ii) … cm2 [1] __________________________________________________________________________________________ Question 20 is printed on the next page. NOT TO SCALE A B C D 120° 8 cm 8 cm NOT TO SCALE A B C D

12 0581/21/O/N/14 © UCLES 2014 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. 20 f(x) = 3x – 2 g(x) = 1 x + 2 , x ≠ –1 (a) Find gf(2). Answer(a) … [2] (b) Solve g(x) = 10. Answer(b) x = … [2] (c) Simplify. f(2x) – f(x + 2) Answer(c) … [3]

® 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 0581 MATHEMATICS 0581/21 Paper 2 (Extended), maximum raw mark 70 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. PAPA CAMBRIDGE

Page 2 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0581 21 © Cambridge International Examinations 2014 Abbreviations cao correct answer only dep dependent FT follow through after error isw ignore subsequent working oe or equivalent SC Special Case nfww not from wrong working soi seen or implied Qu. Answers Mark Part Marks 1 8.1722 cao 2 B1 for 8.17 or 8.172 or 8.1721 or 8.17215… 2 3 3.14 π 3.142 7 22 2 B1 for 3.141[5...] to 3.1416 and 3.1428 to 3.1429 or 3.143 seen or SC1 for 4 in correct order 3 (a) (b) E B A cao Z cao 1 1 4 (a) (b) –3 4 1 1FT FT their numerical mode 5 12 3 and 12 2 12 5 cao M1 A1 Equivalent denominators can be used, working must be shown. 6 (a) (b) 15.1 cao 20 cao 1 1 7 2.5[0] or 2.501… nfww 3 M2 for ( ) 3 100 6 1 1.2 + × oe or M1 for ( ) n 100 6 1 1.2 + × oe where 2 ≥ n or for figs ( ) 3 100 6 1 21 + × oe 8 0.29 cao 3 M2 for ( ) 2378 .1 24 30 × − or ( ) 30 2378 .1 24 − × or M1 for 2378 .1 24× 9 (a) (b) 280 5 × 106 1 2 B1 for 5 000 000 oe or B1 for answer 6 10 × k or k 10 5×

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0581 21 © Cambridge International Examinations 2014 10 3.75 oe 3 M2 for x x 3 7 5 3 − = × oe or M1 for ( ) x x 7 5 3 = + or x x 3 7 5 = + or 3 7 5 1 = + x or better 11 (a) (b) x6 3 2 x 1 2 B1 for answer kx2 or 3 k x or 3 1 12 5 5 − nfww 3 M1 for correctly eliminating one variable A1 for x = 5 A1 for y = −5 If zero scored SC1 for correct substitution and evaluation to find the other variable 13 [±] 8 nfww 3 M1 for 5 + = x k y A1 for k = [±] 2 or M2 for 5 11 5 1 4 + = + − y oe 14       8 2 16 4 3 M2 for       24 6 48 12 and       16 4 32 8 or M1 for       24 6 48 12 or for       16 4 32 8 15 (a) (i) (ii) (b) 2 2 1 B2 for correct ruled bisector with correct arcs or B1 for correct bisector with no/incorrect arcs B2 for correct ruled bisector with correct arcs or B1 for correct bisector with no/incorrect arcs correct shading 16 142 or 142.0… 5 B1 for CBD = 30 M2 for [ ] 8 sin 6 sin theirB D × = oe or M1 for ) 30 sin( 8 sin 6 their D = oe A1 for [D =] 22 or 22.0 or 22.02… B1FT for 90 + (their30 + their22) evaluated correctly for their final answer or for 360 – 90 – theirBCD evaluated correctly for their final answer

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2014 0581 21 © Cambridge International Examinations 2014 17 890 or 890.1 to 890.2… 5 M4 for 8 5 π 5 π 3 4 2 1 2 3 × × +       × × × or M3 for       × × × 3 5 π 3 4 2 1 and 8 5 π 2 × × or M2 for       × × × 3 5 π 3 4 2 1 or 8 5 π 2 × × or M1 for 3 5 π 3 4 × × 18 (a) 8.0 2.0 6.0 in correct places 2 B1 for 0.6 in correct place B1 for 0.2 and 0.8 in correct places (b) 0.52 oe nfww 3 M2FT for 1 – (their 0.6 × their 0.8) oe or M1FT for a correct product from their tree in (a) 19 (a) (b) (c) (i) (ii) CBA and BDA are equilateral oe 67[.0] or 67.02 to 67.03 39.3 or 39.28 to 39.33 78.6 or 78.7 or 78.56 to 78.66 1 2 3 1FT M1 for 2 360 120 8 π × × oe M2FT for ( ) 120 sin 82 2 1 × × − b their oe or M1 for 120 sin 82 2 1 × × oe FT 2 × their(c)(i) correctly evaluated 20 (a) (b) (c) 0.4 or 5 2 –0.8 or 5 4 − 6 3 − x or ( ) 2 3 − x nfww 2 2 3 B1 for [f(2) =] 4 or M1 for ( ) 1 2 3 2 + − x or better M1 for ( )1 10 2 + = x or better M2 for ( ) ( ) ( ) 2 2 3 2 2 3 − + − − x x or M1 for ( ) [ ] ( ) 2 2 3 2 f − = x x or ( ) [ ] ( ) 2 2 3 2 f − + = + x x