Cambridge IGCSE Mathematics (with coursework) 0581 — 2013 Oct/Nov Paper 4 · 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 a pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, 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 130. MATHEMATICS 0581/41 Paper 4 (Extended) October/November 2013 2 hours 30 minutes Candidates answer on the Question Paper. Additional Materials: Electronic calculator Geometrical instruments Tracing paper (optional) UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certifi cate of Secondary Education This document consists of 19 printed pages and 1 blank page. [Turn over IB13 11_0581_41/FP © UCLES 2013 *4113871434* www.XtremePapers.com

2 0581/41/O/N/13 © UCLES 2013 For Examiner′s Use 1 David sells fruit at the market. (a) In one week, David sells 120 kg of tomatoes and 80 kg of grapes. (i) Write 80 kg as a fraction of the total mass of tomatoes and grapes. Give your answer in its lowest terms. Answer(a)(i) … [1] (ii) Write down the ratio mass of tomatoes : mass of grapes. Give your answer in its simplest form. Answer(a)(ii) … : … [1] (b) (i) One day he sells 28 kg of oranges at $1.56 per kilogram. He also sells 35 kg of apples. The total he receives from selling the oranges and the apples is $86.38 . Calculate the price of 1 kilogram of apples. Answer(b)(i) $ … [2] (ii) The price of 1 kilogram of oranges is $1.56 . This is 20% more than the price two weeks ago. Calculate the price two weeks ago. Answer(b)(ii) $ … [3] (c) On another day, David received a total of $667 from all the fruit he sold. The cost of the fruit was $314.20 . David worked for 10 2 1 hours on this day. Calculate David’s rate of profi t in dollars per hour. Answer(c) … dollars/h [2] _____________________________________________________________________________________

3 0581/41/O/N/13 © UCLES 2013 [Turn over For Examiner′s Use 2 Emily cycles along a path for 2 minutes. She starts from rest and accelerates at a constant rate until she reaches a speed of 5 m/s after 40 seconds. She continues cycling at 5 m/s for 60 seconds. She then decelerates at a constant rate until she stops after a further 20 seconds. (a) On the grid, draw a speed-time graph to show Emily’s journey. 5 4 3 2 1 0 10 20 30 40 50 60 Time (seconds) 70 80 90 100 110 120 Speed (m/s) [2] (b) Find Emily’s acceleration. Answer(b) … m/s2 [1] (c) Calculate Emily’s average speed for the journey. Answer(c) … m/s [4] _____________________________________________________________________________________

4 0581/41/O/N/13 © UCLES 2013 For Examiner′s Use 3 h 13 cm 5 cm NOT TO SCALE (a) The diagram shows a cone of radius 5 cm and slant height 13 cm. (i) Calculate the curved surface area of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = πrl.] Answer(a)(i) … cm2 [2] (ii) Calculate the perpendicular height, h, of the cone. Answer(a)(ii) h = … cm [3] (iii) Calculate the volume of the cone. [The volume, V, of a cone with radius r and height h is V = 3 1 πr2h.] Answer(a)(iii) … cm3 [2] (iv) Write your answer to part (a)(iii) in cubic metres. Give your answer in standard form. Answer(a)(iv) … m3 [2]

5 0581/41/O/N/13 © UCLES 2013 [Turn over For Examiner′s Use (b) h 13 cm 5 cm NOT TO SCALE A B O O The cone is now cut along a slant height and it opens out to make the sector AOB of a circle. Calculate angle AOB. Answer(b) Angle AOB = … [4] _____________________________________________________________________________________

6 0581/41/O/N/13 © UCLES 2013 For Examiner′s Use 4 70 m 55 m A B C D 40° 32° NOT TO SCALE The diagram shows a school playground ABCD. ABCD is a trapezium. AB = 55 m, BD = 70 m, angle ABD = 40° and angle BCD = 32°. (a) Calculate AD. Answer(a) AD = … m [4] (b) Calculate BC. Answer(b) BC = … m [4]

7 0581/41/O/N/13 © UCLES 2013 [Turn over For Examiner′s Use (c) (i) Calculate the area of the playground ABCD. Answer(c)(i) … m2 [3] (ii) An accurate plan of the school playground is to be drawn to a scale of 1: 200 . Calculate the area of the school playground on the plan. Give your answer in cm2. Answer(c)(ii) … cm2 [2] (d) A fence, BD, divides the playground into two areas. Calculate the shortest distance from A to BD. Answer(d) … m [2] _____________________________________________________________________________________

8 0581/41/O/N/13 © UCLES 2013 For Examiner′s Use 5 (a) T y x 0 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 U (i) Draw the refl ection of triangle T in the line y = 5. [2] (ii) Draw the rotation of triangle T about the point (4, 2) through 180°. [2] (iii) Describe fully the single transformation that maps triangle T onto triangle U. Answer(a)(iii) … [3] (iv) Find the 2 × 2 matrix which represents the transformation in part (a)(iii). Answer(a)(iv) f p [2]

9 0581/41/O/N/13 © UCLES 2013 [Turn over For Examiner′s Use (b) P Q O S R p s NOT TO SCALE In the pentagon OPQRS, OP is parallel to RQ and OS is parallel to PQ. PQ = 2OS and OP = 2RQ. O is the origin, = p and = s. Find, in terms of p and s, in their simplest form, (i) the position vector of Q, Answer(b)(i) … [2] (ii) . Answer(b)(ii) = … [2] (c) Explain what your answers in part (b) tell you about the lines OQ and SR. Answer(c) … [1] _____________________________________________________________________________________

10 0581/41/O/N/13 © UCLES 2013 For Examiner′s Use 6 (a) y x 5 4 3 2 1 –1 –2 –3 –4 –5 0 1 –1 –2 –3 2 3 The diagram shows the graph of y = f(x) for –3 Ğ x Ğ 3. (i) Find f(2). Answer(a)(i) … [1] (ii) Solve the equation f(x) = 0. Answer(a)(ii) x = … [1] (iii) Write down the value of the largest integer, k, for which the equation f(x) = k has 3 solutions. Answer(a)(iii) k = … [1] (iv) By drawing a suitable straight line, solve the equation f(x) = x. Answer(a)(iv) x = … or x = … or x = … [3]

11 0581/41/O/N/13 © UCLES 2013 [Turn over For Examiner′s Use (b) g(x) = 1 – 2x h(x) = x2 – 1 (i) Find gh(3). Answer(b)(i) … [2] (ii) Find g–1(x). Answer(b)(ii) g–1(x) = … [2] (iii) Solve the equation h(x) = 3. Answer(b)(iii) x = … or x = … [3] (iv) Solve the equation g(3x) = 2x. Answer(b)(iv) x = … [3] _____________________________________________________________________________________

12 0581/41/O/N/13 © UCLES 2013 For Examiner′s Use 7 120 students are asked to answer a question. The time, t seconds, taken by each student to answer the question is measured. The frequency table shows the results. Time 0 < t Y 10 10 < t Y 20 20 < t Y 30 30 < t Y 40 40 < t Y 50 50 < t Y 60 Frequency 6 44 40 14 10 6 (a) Calculate an estimate of the mean time. Answer(a) … s [4] (b) (i) Complete the cumulative frequency table. Time t Y 10 t Y 20 t Y 30 t Y 40 t Y 50 t Y 60 Cumulative frequency 6 104 120 [2] (ii) On the grid below, draw a cumulative frequency diagram to show this information. 10 20 30 Time (seconds) 40 50 60 120 100 80 60 40 20 0 Cumulative frequency t [3]

13 0581/41/O/N/13 © UCLES 2013 [Turn over For Examiner′s Use (iii) Use your cumulative frequency diagram to fi nd the median, the lower quartile and the 60th percentile. Answer(b)(iii) Median … s Lower quartile … s 60th percentile … s [4] (c) The intervals for the times taken are changed. (i) Use the information in the frequency table on the opposite page to complete this new table. Time 0 < t Y 20 20 < t Y 30 30 < t Y 60 Frequency 40 [2] (ii) On the grid below, complete the histogram to show the information in the new table. One column has already been drawn for you. 10 20 30 Time (seconds) 40 50 60 4 3.5 3 2.5 2 1.5 1 0.5 0 Frequency density t [3] _____________________________________________________________________________________

14 0581/41/O/N/13 © UCLES 2013 For Examiner′s Use 8 (a) Solve the equation 8x2 – 11x – 11 = 0. Show all your working and give your answers correct to 2 decimal places. Answer(a) x = … or x = … [4] (b) y varies directly as the square root of x. y = 18 when x = 9. Find y when x = 484. Answer(b) y = … [3]

15 0581/41/O/N/13 © UCLES 2013 [Turn over For Examiner′s Use (c) Sara spends $x on pens which cost $2.50 each. She also spends $(x – 14.50) on pencils which cost $0.50 each. The total of the number of pens and the number of pencils is 19. Write down and solve an equation in x. Answer(c) x = … [6] _____________________________________________________________________________________

16 0581/41/O/N/13 © UCLES 2013 For Examiner′s Use 9 y x 1 2 3 4 5 6 7 8 9 10 5 4 3 2 1 0 L2 L3 L1 R (a) Find the equations of the lines L1, L2 and L3. Answer(a) L1 … L2 … L3 … [5] (b) Write down the three inequalities that defi ne the shaded region, R. Answer(b) … … … [3]

17 0581/41/O/N/13 © UCLES 2013 [Turn over For Examiner′s Use (c) A gardener buys x bushes and y trees. The cost of a bush is $30 and the cost of a tree is $200. The shaded region R shows the only possible numbers of bushes and trees the gardener can buy. (i) Find the number of bushes and the number of trees when the total cost is $720. Answer(c)(i) … bushes … trees [2] (ii) Find the number of bushes and the number of trees which give the greatest possible total cost. Write down this greatest possible total cost. Answer(c)(ii) … bushes … trees Greatest possible total cost = $ … [3] _____________________________________________________________________________________

18 0581/41/O/N/13 © UCLES 2013 For Examiner′s Use 10 (a) 1 = 1 1 + 2 = 3 1 + 2 + 3 = 6 1 + 2 + 3 + 4 = 10 (i) Write down the next line of this pattern. Answer(a)(i) … [1] (ii) The sum of the fi rst n integers is k n (n + 1). Show that k = 2. Answer(a)(ii) [2] (iii) Find the sum of the fi rst 60 integers. Answer(a)(iii) … [1] (iv) Find n when the sum of the fi rst n integers is 465. Answer(a)(iv) n = … [2] (v) 1 + 2 + 3 + 4 + … + x = ( 8)( 7) n n 2 - - Write x in terms of n. Answer(a)(v) x = … [1]

19 0581/41/O/N/13 © UCLES 2013 [Turn over For Examiner′s Use (b) 13 = 1 13 + 23 = 9 13 + 23 + 33 = 36 13 + 23 + 33 + 43 = 100 (i) Complete the statement. 13 + 23 + 33 + 43 + 53 = … = (…)2 [2] (ii) The sum of the fi rst n integers is 2 n (n + 1). Find an expression, in terms of n, for the sum of the fi rst n cubes. Answer(b)(ii) … [1] (iii) Find the sum of the fi rst 19 cubes. Answer(b)(iii) … [2] _____________________________________________________________________________________

20 0581/41/O/N/13 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. BLANK PAGE © UCLES 2013

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2013 series 0581 MATHEMATICS 0581/41 Paper 4 (Extended), maximum raw mark 130 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 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. www.XtremePapers.com

Page 2 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0581 41 © Cambridge International Examinations 2013 Abbreviations cao correct answer only cso correct solution only dep dependent ft follow through after error isw ignore subsequent working oe or equivalent SC Special Case www without wrong working art anything rounding to soi seen or implied Qu Answers Mark Part Marks 1 (a) (i) 5 2 cao 1 (ii) 3 : 2 cao 1 (b) (i) 1.22 2 M1 for 86.38 – 28 × 1.56 (ii) 1.3 [0] nfww 3 M2 for 1.56 ÷ 1.2 oe or M1 for 1.56 = 120% soi (c) 33.6[0] 2 M1 for (667 – 314.2) ÷ 10.5 oe 2 (a) 3 correct lines on grid (0, 0) to (40, 5) (40, 5) to (100, 5) (100, 5) to (120, 0) 2 Allow good freehand SC1FT for 2 lines correct, FT from an incorrect line (b) 40 5 oe 1 (c) 3.75 4 M2 for 0.5 × 40 × 5 + 60 × 5 + 0.5 × 20 × 5 oe [450] or M1 for evidence of a relevant area = distance and M1dep their area (or distance) ÷ 120

Page 3 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0581 41 © Cambridge International Examinations 2013 Qu Answers Mark Part Marks 3 (a) (i) 204 or 204.2 to 204.23 2 M1 for 13 5× × π implied by answer in range 204.1 to 204.3 (ii) 12 cao 3 M2 for 2 2 5 13 − or states 5, 12, 13 triangle or M1 for 132 = 52 + h2 or better (iii) 314 or 314.1 to 314.2 2 M1 for × × × 2 5 3 1 π their (a) (ii) implied by answer in range 314 to 314.3 (iv) 3.14 × 10–4 or 3.141 to 3.142 × 10–4 2FT FT their (a) (iii) ÷ 1003 correctly evaluated and given in standard form to 3 sig figs or better or M1 FT for their (a) (iii) ÷ 1003 or SC1 for conversion of their m3 into standard form only if negative power (b) 138 or 138.3 to 138.5 4 M3 for 360 26 10 × π π oe or 360 13 13 5 2 × × × × π π (i) (a) ortheir oe or M2 for a correct fraction without × 360 or M1 for 13 2× × π oe [81.6 to 81.8] seen or 2 13 × π oe [530.6 to 531.2] seen 4 (a) 45.[0] or 45.01 to 45.02 nfww 4 M2 for 552 + 702 – 2.55.70 cos 40 or M1 for correct implicit equation A1 for 2026. …. (b) 84.9 or 84.90 to 84.92 4 B1 for angle BDC = 40 soi M2 for 32 sin ) 40 ( sin 70 their or M1 for correct implicit equation (c) (i) 4060 or 4063 to 4064 nfww 3 M2 for 2 1 ( ) 40 sin 70 × 55 + 2 1 ( )) 32 40 180 ( sin ) ( 70 − − × their b their oe or M1 for correct method for one of the triangle areas (ii) 1020 or 1015 to 1016 2FT FT their (c) (i) ÷ 4 oe correctly evaluated or M1 their (c) (i) ÷ figs 4 oe (d) 35.4 or 35.35… nfww 2 M1 for sin 40 = 55 distance or better or for 2 1 (55 × 70 sin 40) = (70 × distance) ÷ 2 or better

Page 4 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0581 41 © Cambridge International Examinations 2013 Qu Answers Mark Part Marks 5 (a) (i) Correct reflection to (4, 8) (2, 9) (4, 9) 2 SC1 for reflection in line x = 5 or reflection in y = k Ignore additional triangles (ii) Correct rotation to (4, 2), (4, 3) (6, 3) 2 SC1 for rotation 180˚ with incorrect centre Ignore additional triangles (iii) Shear, x-axis oe invariant, [factor] 2 3 B1 each (independent) (iv)       1 0 2 1 2FT FT their shear factor B1FT for one correct column or row in 2 by 2 matrix but not identity matrix or SC1FT for       1 2 0 1 (b) (i) p + 2s final answer 2 M1 for recognising OQ as position vector soi (ii) s + 2 1 p final answer 2 B1 for s + kp or ks + 2 1 p or correct route (k ≠ 0) (c) parallel and OQ = 2SR oe 1 6 (a) (i) 1.4 to 1.6 1 (ii) 1.15 to 1.25 1 (iii) – 1 1 (iv) – 2.25 to – 2.1 – 0.9 to – 0.75 2.2 to 2.35 3 B2 for 2 correct or B1 for one correct or B1 for y = x drawn ruled to cut curve 3 times (b) (i) – 15 2 B1 for [h(3) =] 8 seen or M1 for 1 – 2(x2 – 1) or better (ii) 2 1 x − or 2 2 1 x − oe final answer 2 M1 for2 x = 1 – y or x = 1–2y or better (iii) – 2, 2 3 M1 for x2 – 1 = 3 or better B1 for one answer (iv) 8 1 oe nfww 3 M2 for 8x = 1 or 8x – 1 = 0 or M1 for 1 – 2(3x) [= 2x]

Page 5 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0581 41 © Cambridge International Examinations 2013 Qu Answers Mark Part Marks 7 (a) 24.7 or 24.66 to 24.67 4 M1 for midpoints soi (condone 1 error or omission) (5, 15, 25, 35, 45, 55) and M1 for use of ∑fx with x in correct interval including both boundaries (condone 1 further error or omission) and M1 (dependent on second M) for ∑fx ÷ 120 (b) (i) 50, 90, 114 2 B1 for 2 correct (ii) Correct curve or ruled polygon 3 Ignore section to left of t = 10 B1 for 6 correct horizontal plots and B1FT for 6 correct vertical plots If 0 scored SC1 for 5 out of 6 correct plots and B1FT for curve or polygon through at least 5 of their points dep on an increasing curve/polygon that reaches 120 vertically (iii) 21.5 to 23 15 to 16.5 24 to 26 4 B1 B1 B2 or B1 for 72 or 72.6 seen (c) (i) 50, 30 2 B1 each (ii) Correct histogram 3FT B1 for blocks of widths 0 – 20, 30 – 60 (no gaps) B1FT for block of height 2.5 or their 50 ÷ 20 and B1FT for block of height 1 or their 30 ÷ 30

Page 6 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0581 41 © Cambridge International Examinations 2013 Qu Answers Mark Part Marks 8 (a) ( ) ( )( ) 11 8 4 11 2 − − − or better p = –(– 11), r = 2(8) or better – 0.67, 2.05 final answers B1 B1 B1B1 Seen anywhere or for 2 16 11      − x Must be in the form r q p + or r q p − or B1 for 16 11 16 11 8 11 2 +       + SC1 for – 0.7 or – 0.672 to – 0.671 and 2.0 or 2.046 to 2.047 or answers 0.67 and – 2.05 (b) 132 3 M1 for x k y = oe or ky x = oe A1 for k = 6 oe or better or for k = 0.1666 to 0.167 [k = 6 implies M1A1] oe (c) 20 with supporting algebraic working 6 B2 for 19 5.0 5. 14 5.2 = − + x x oe or B1 for 5.2 x or 5. 5. 14 − x M1dep on B2 for first completed correct move to clear both fractions M1 for second completed correct move to collect terms in x to a single term M1 for third completed correct move to collect numeric term[s] leading to ax = b SC1 for 20 with no algebraic working 9 (a) y = 2 oe y = 2x oe y = – 2 1 x + 5 oe 1 2 2 M1 for y = kx, 0 ≠ k or gradient 2 soi M1 for gradient – ½ soi or y = kx + 5oe or x + 2y = k 0 ≠ k oe If L2 and L3 both correct but interchanged then SC3 (b) y ≥ 2 oe y ≤ 2x oe y ≤ – 2 1 x + 5 oe 3 B1 for each correct inequality, allow in any order After 0 scored, SC1 for all inequalities reversed (c) (i) 4 [bushes], 3 [trees] 2 M1 for any correct trial using integer coordinates in region or 30x + 200y = 720 seen (ii) 2 [bushes], 4 [trees] 860 2 1 M1 for any correct trial using integer coordinates in region

Page 7 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0581 41 © Cambridge International Examinations 2013 Qu Answers Mark Part Marks 10 (a) (i) 1 + 2 + 3 + 4 + 5 = 15 1 (ii) Correct substitution equating to sum e.g. ( ) 3 1 2 2 = + k and k = 2 stated with no errors seen 2 M1 for using a value of n in ( ) k n n 1 + e.g. ( ) 3 1 2 2 = + k or for a verification using k = 2 e.g. ( ) 3 2 1 2 2 = + (iii) 1830 1 (iv) 30 2 M1 for ( ) 2 1 + n n = 465 or better (v) n – 8 1 (b) (i) 225, 15 2 B1 either (ii) ( ) 4 1 2 2 + n n oe 1 (iii) 36100 2 M1 for ( ) 4 1 19 19 2 2 + oe or 1902