Cambridge IGCSE Mathematics (with coursework) 0581 — 2011 May/June Paper 4 · Variant 1

This document consists of 16 printed pages. IB11 06_0581_41/RP © UCLES 2011 [Turn over *3153513379* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MATHEMATICS 0581/41 Paper 4 (Extended) May/June 2011 2 hours 30 minutes Candidates answer on the Question Paper. Additional Materials: Electronic calculator Geometrical instruments Mathematical tables (optional) Tracing paper (optional) 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 fluid. 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 specified in the question, and if the answer is not exact, give the answer to three significant figures. 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. www.XtremePapers.com

2 © UCLES 2011 0581/41/M/J/11 For Examiner's Use 1 A school has a sponsored swim in summer and a sponsored walk in winter. In 2010, the school raised a total of $1380. The ratio of the money raised in summer : winter = 62 : 53. (a) (i) Show clearly that $744 was raised by the swim in summer. Answer (a)(i) [1] (ii) Alesha’s swim raised $54.10. Write this as a percentage of $744. Answer(a)(ii) %[1] (iii) Bryan’s swim raised $31.50. He received 75 cents for each length of the pool which he swam. Calculate the number of lengths Bryan swam. Answer(a)(iii) [2] (b) The route for the sponsored walk in winter is triangular. North B A C 110° NOT TO SCALE (i) Senior students start at A, walk North to B, then walk on a bearing 110° to C. They then return to A. AB = BC. Calculate the bearing of A from C. Answer(b)(i) [3]

3 © UCLES 2011 0581/41/M/J/11 [Turn over For Examiner's Use (ii) North B A C 110° 110° NOT TO SCALE 4 km AB = BC = 6 km. Junior students follow a similar path but they only walk 4 km North from A, then 4 km on a bearing 110° before returning to A. Senior students walk a total of 18.9 km. Calculate the distance walked by junior students. Answer(b)(ii) km [3] (c) The total amount, $1380, raised in 2010 was 8% less than the total amount raised in 2009. Calculate the total amount raised in 2009. Answer(c) $ [3]

4 © UCLES 2011 0581/41/M/J/11 For Examiner's Use 2 In this question give all your answers as fractions. The probability that it rains on Monday is 5 3 . If it rains on Monday, the probability that it rains on Tuesday is 7 4 . If it does not rain on Monday, the probability that it rains on Tuesday is 7 5 . (a) Complete the tree diagram. Monday Tuesday Rain No rain Rain No rain No rain Rain [3] (b) Find the probability that it rains (i) on both days, Answer(b)(i) [2] (ii) on Monday but not on Tuesday, Answer(b)(ii) [2] (iii) on only one of the two days. Answer(b)(iii) [2] (c) If it does not rain on Monday and it does not rain on Tuesday, the probability that it does not rain on Wednesday is 4 1 . Calculate the probability that it rains on at least one of the three days. Answer(c) [3]

5 © UCLES 2011 0581/41/M/J/11 [Turn over For Examiner's Use 3 (a) p varies inversely as (m + 1). When p = 4, m = 8. Find the value of p when m = 11. Answer(a) p = [3] (b) (i) Factorise x2 – 25. Answer(b)(i) [1] (ii) Simplify 25 5 11 2 2 2 − + + x x x . Answer(b)(ii) [3] (c) Solve the inequality 5(x – 4) I 3(12 – x). Answer(c) [3]

6 © UCLES 2011 0581/41/M/J/11 For Examiner's Use 4 (a) H F G 12 cm 14 cm 6 cm NOT TO SCALE The diagram shows triangle FGH, with FG = 14 cm, GH = 12 cm and FH = 6 cm. (i) Calculate the size of angle HFG. Answer(a)(i) Angle HFG = [4] (ii) Calculate the area of triangle FGH. Answer(a)(ii) cm2 [2]

7 © UCLES 2011 0581/41/M/J/11 [Turn over For Examiner's Use (b) R P Q 18 cm 12 cm 117° NOT TO SCALE The diagram shows triangle PQR, with RP = 12 cm, RQ = 18 cm and angle RPQ = 117°. Calculate the size of angle RQP. Answer(b) Angle RQP = [3]

8 © UCLES 2011 0581/41/M/J/11 For Examiner's Use 5 4 3 2 1 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 –1 1 0 2 3 4 5 6 7 y x A B C (a) On the grid above, draw the image of (i) shape A after translation by the vector       − − 2 3 , [2] (ii) shape A after reflection in the line x = −1 . [2] (b) Describe fully the single transformation which maps (i) shape A onto shape B, Answer(b)(i) [3] (ii) shape A onto shape C. Answer(b)(ii) [3] (c) Find the matrix representing the transformation which maps shape A onto shape B. Answer(c)       [2] (d) Describe fully the single transformation represented by the matrix       − − 1 0 0 1 . Answer(d) [3]

9 © UCLES 2011 0581/41/M/J/11 [Turn over For Examiner's Use 6 A C D F B E 36 cm 19 cm 14 cm NOT TO SCALE In the diagram, ABCDEF is a prism of length 36 cm. The cross-section ABC is a right-angled triangle. AB = 19 cm and AC = 14 cm. Calculate (a) the length BC, Answer(a) BC = cm [2] (b) the total surface area of the prism, Answer(b) cm2 [4] (c) the volume of the prism, Answer(c) cm3 [2] (d) the length CE, Answer(d) CE = cm [2] (e) the angle between the line CE and the base ABED. Answer(e) [3]

10 © UCLES 2011 0581/41/M/J/11 For Examiner's Use 7 (a) Complete the table of values for the equation y = 2 4 x , x ≠ 0. x O4 O3 O2 O1 O0.6 0.6 1 2 3 4 y 0.25 0.44 11.11 4.00 0.44 [3] (b) On the grid, draw the graph of y = 2 4 x for O4 Y x Y O0.6 and 0.6 Y x Y 4 . –4 –3 –2 –1 1 0 2 3 4 12 11 10 9 8 7 6 5 4 3 2 1 –1 –2 y x [5]

11 © UCLES 2011 0581/41/M/J/11 [Turn over For Examiner's Use (c) Use your graph to solve the equation 2 4 x = 6 . Answer(c)x = or x = [2] (d) By drawing a suitable tangent, estimate the gradient of the graph where x = 1.5. Answer(d) [3] (e) (i) The equation 2 4 x O x + 2 = 0 can be solved by finding the intersection of the graph of y = 2 4 x and a straight line. Write down the equation of this straight line. Answer(e)(i) [1] (ii) On the grid, draw the straight line from your answer to part (e)(i). [2] (iii) Use your graphs to solve the equation 2 4 x O x + 2 = 0. Answer(e)(iii) x = [1]

12 © UCLES 2011 0581/41/M/J/11 For Examiner's Use 8 The table below shows the marks scored by a group of students in a test. Mark 11 12 13 14 15 16 17 18 Frequency 10 8 16 11 7 8 6 9 (a) Find the mean, median and mode. Answer(a) mean = median = mode = [6] (b) The table below shows the time (t minutes) taken by the students to complete the test. Time (t) 0 I=t Y=10 10 I=t Y=20 20 I=t Y=30 30 I=t Y=40 40 I=t Y=50 50 I=t Y=60 Frequency 2 19 16 14 15 9 (i) Cara rearranges this information into a new table. Complete her table. Time (t) 0 I=t Y=20 20 I=t Y=40 40 I=t Y=50 50 I=t Y=60 Frequency 9 [2] (ii) Cara wants to draw a histogram to show the information in part (b)(i). Complete the table below to show the interval widths and the frequency densities. 0 I=t Y=20 20 I=t Y=40 40 I=t Y=50 50 I=t Y=60 Interval width 10 Frequency density 0.9 [3]

13 © UCLES 2011 0581/41/M/J/11 [Turn over For Examiner's Use (c) Some of the students were asked how much time they spent revising for the test. 10 students revised for 2.5 hours, 12 students revised for 3 hours and n students revised for 4 hours. The mean time that these students spent revising was 3.1 hours. Find n. Show all your working. Answer(c) n = [4]

14 © UCLES 2011 0581/41/M/J/11 For Examiner's Use 9 Peter wants to plant x plum trees and y apple trees. He wants at least 3 plum trees and at least 2 apple trees. (a) Write down one inequality in x and one inequality in y to represent these conditions. Answer(a) , [2] (b) There is space on his land for no more than 9 trees. Write down an inequality in x and y to represent this condition. Answer(b) [1] (c) Plum trees cost $6 and apple trees cost $14. Peter wants to spend no more than $84. Write down an inequality in x and y, and show that it simplifies to 3x + 7y Y 42. Answer(c) [1]

15 © UCLES 2011 0581/41/M/J/11 [Turn over For Examiner's Use (d) On the grid, draw four lines to show the four inequalities and shade the unwanted regions. 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 y x [7] (e) Calculate the smallest cost when Peter buys a total of 9 trees. Answer(e) $ [2] Question 10 is printed on the next page.

16 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 0581/41/M/J/11 For Examiner's Use 10 The first and the nth terms of sequences A, B and C are shown in the table below. (a) Complete the table for each sequence. 1st term 2nd term 3rd term 4th term 5th term nth term Sequence A 1 n3 Sequence B 4 4n Sequence C 4 (n + 1)2 [5] (b) Find (i) the 8th term of sequence A, Answer(b)(i) [1] (ii) the 12th term of sequence C. Answer(b)(ii) [1] (c) (i) Which term in sequence A is equal to 15 625? Answer(c)(i) [1] (ii) Which term in sequence C is equal to 10 000? Answer(c)(ii) [1] (d) The first four terms of sequences D and E are shown in the table below. Use the results from part (a) to find the 5th and the nth terms of the sequences D and E. 1st term 2nd term 3rd term 4th term 5th term nth term Sequence D 5 16 39 80 Sequence E 0 1 4 9 [4]

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2011 question paper for the guidance of teachers 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 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 May/June 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 – May/June 2011 0581 41 © University of Cambridge International Examinations 2011 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) 62 53 62 1380 × + 1 Allow 115 for 62 + 53 (ii) 7.27 (7.271 to 7.272) 1 (iii) 42 2 M1 for 75 3150 oe (b) (i) 235 3 B2 for angle ACS = 55 or angle ACN = 125 B1 for 55 seen (ii) 12.6 (12.58 to 12.59) 3 M2 for 9. 18 6 4 × or 55 cos 4 2 4 4 × × + + or 35 sin 4 2 4 4 × × + + oe (M1 for 6 4 soi or 55 cos 4 2 × × or 35 sin 4 2 × × soi oe) (c) 1500 3 M2 for 08 .0 1 1380 − oe (M1 for recognition that 92% = 1380)

Page 3 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0581 41 © University of Cambridge International Examinations 2011 2 (a) Monday 5 3 , 5 2 1 Tuesday 7 4 , 7 3 1 7 5 , 7 2 1 (b) (i) 35 12 oe cao 2 M1 5 3 × 7 4 ft their tree (ii) 35 9 oe cao 2 M1 5 3 × 7 3 ft their tree (iii) 35 19 oe 2 ft ft their (b)(ii) + 35 10 ft their tree throughout (iii) M1 for 5 2 × 7 5 + their (b)(ii) or 7 2 5 2 7 4 5 3 1 × − × − (c) 35 34 oe cao 3 ft their tree throughout (iv) M2 for 1 –       − = × × 35 1 1 4 1 7 2 5 2 (M1 for      = × × 35 1 4 1 7 2 5 2 ) or M2 for 4 3 7 2 5 2 7 5 5 2 5 3 × × + × + (M1 for any two of these) 3 (a) 3 www 3 M1 for ( )1 + = m k p oe A1 for k = 36 or M2 for 4 × 9 = p × 12 oe (b) (i) (x + 5)(x – 5) 1 (ii) ( ) ) 5 ( 1 2 − + x x final answer 3 B2 for factors ( )( ) 5 1 2 + + x x or SC2 for final answer 5 2 1 − + x x (B1 for ( )( ) b x a x + + 2 where ab = 5 or 2b + a = 11 or SC1 for ) 5 )( ( 2 1 + + x x ) (c) x < 7 oe final answer 3 M2 for 8x * 56 where * is inequality or = sign (B1 for 5x – 20 or 36 − 3x)

Page 4 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0581 41 © University of Cambridge International Examinations 2011 4 (a) (i) (cos (HFG)) = 14 6 2 12 14 6 2 2 2 × × − + M2 M1 for implicit form 58.4 (58.41…) A2 A1 for 0.5238… (ii) 0.5 × 6 × 14 × sin (their 58.4) oe 35.8 or 35.77 to 35.78 M1 A1ft ft their (i) Correct or ft their (i) (b) (sin (RQP)) = 18 12 ) 117 sin( × 36.4 or 36.44... M2 A1 M1 for implicit form 5 (a) (i) Correct translation (see diagram) 2 SC1 for translation by      − k 3 or by       −2 k (ii) Correct reflection (see diagram) 2 SC1 for reflection in y = −1 (b) (i) Stretch, (factor) 3, y-axis or x = 0 invariant 1 1 1 (ii) Rotation 90° clockwise (1, − 1) 1 1 1 Accept −90° (c) (i)       1 0 0 3 ft from (b)(i) 2 ft SC1 for       3 0 0 1 (ft from (b)(i)) or       1 0 0 k with k algebraic or numeric but ≠ 1 or 0 (ii) Rotation, 180° Origin 1 1 1 Accept O or (0,0) 6 (a) 23.6 (23.60…) 2 M1 for 142 + 192 (b) 2300 or 2303 to 2304 cao 4 M3 for 2 × ½ × 14 × 19 + 14 × 36 + 19 × 36 + their BC × 36 M2 for 4 of these added M1 for ½ × 14 × 19 (c) 4788 or 4790 cao 2 M1 their triangle area × 36 (d) 43(.0) or 43.04 to 43.05 cao 2 M1 for (their (a))2 + 362 or 362 + 192 + 142 (e) 18.9° to 19.02° cao 3 M2 for inv sin       CE their 14 or inv tan       + 2 2 36 19 14 or inv cos         + CE their 36 19 2 2 or complete longer methods (M1 for clearly identifying angle CEA)

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0581 41 © University of Cambridge International Examinations 2011 7 (a) 1(.00) 4(.00) 11.1(1) 1(.00) 0.25 3 B2 for 4 correct, B1 for 3 correct (b) 10 points plotted Correct shaped curve through 10 points (condone 2 points slightly missed) 2 separate curves not crossing x-axis and not touching or crossing y-axis P3 ft C1 ft B1 B2 for 8 or 9 points correct ft B1 for 6 or 7 points correct ft ft their points if shape correct – ignore anything between – 0.6 and 0.6 Independent (c) −0.85 to – 0.75 cao 0.75 to 0.85 cao 1 1 (d) Tangent drawn (ruled) at x = 1.5 – 3 to −2 T1 2 Allow slight daylight Dep on T1 M1 evidence rise/run dependent on tangent SC1 for answer in range 2 to 3 Answer implies M but not the T mark (e) (i) y = x − 2 oe 1 (ii) line ruled to cross curve 2 ft Dependent on (i) in form y = mx + c, m ≠ 0, c ≠ 0 B1 for gradient ft or y intercept ft but again to cross curve at all possible points (iii) 2.5 to 2.7 cao 1 Dependent on (e)(i) correct 8 14.2 14 13 3 2 1 M1 for Σ fx (10 × 11 + 8 × 12 + 16 × 13 + 11 × 14 + 7 × 15 + 8 × 16 + 6 × 17 + 9 × 18 ) (1065) (allow one error or omission) M1dep for ÷ Σf (10 + 8 + 16 + 11 + 7 + 8 + 6 + 9) (75) (allow one further error or omission) M1 for 37th, 37.5th or 38th seen (b) (i) 21, 30, 15 2 B1 for 2 correct (ii) 20 20 10 (10) 1.05 1.5 1.5 (0.9) 3 1, 1, 1 for each correct vertical pair (c) )1.3 ( 12 10 4 3 12 5.2 10 = + + + × + × n n M2 M1 for either numerator or denominator seen multiplying across and collecting terms (n =) 8 www 4 M1 A1 dep on linear numerator and denominator their (68.2 − 25 − 36) = their (4 − 3.1) × n

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – May/June 2011 0581 41 © University of Cambridge International Examinations 2011 9 (a) x [=3 y [=2 1, 1 (b) x + y Y 9 1 (c) 6x + 14y Y 84 1 (d) x = 3 y = 2 9 = + y x Line from (0, 6) to (14, 0) Correct quadrilateral unshaded or clearly indicated 1, 1 2 2 1 Accept clear and freehand lines long enough to define the correct quadrilateral SC1 for line through (0, 9) or (9, 0) B1 for through (0, 6) or (14, 0) (e) $ 70 2 B1 for considering (7, 2) 10 (a) (A 1) 8 27 64 125 (B 4) 8 12 16 20 (C 4) 9 16 25 36 2 1 2 B1 for 3 correct B1 for 3 correct (b) 512 169 1 1 (c) 25 99 1 1 (d) 145 n3 + 4n oe 16 (n + 1)2 – 4n oe but isw 1, 1 1, 1 Likely oe is (n – 1)2