Cambridge IGCSE Mathematics (with coursework) 0581 — 2010 Oct/Nov Paper 4 · Variant 3

This document consists of 19 printed pages and 1 blank page. IB10 11_0581_43/RP © UCLES 2010 [Turn over *6814001826* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MATHEMATICS 0581/43 Paper 4 (Extended) October/November 2010 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 2010 0581/43/O/N/10 For Examiner's Use 1 Thomas, Ursula and Vanessa share $200 in the ratio Thomas : Ursula : Vanessa = 3 : 2 : 5. (a) Show that Thomas receives $60 and Ursula receives $40. Answer(a) [2] (b) Thomas buys a book for $21. What percentage of his $60 does Thomas have left? Answer(b) % [2] (c) Ursula buys a computer game for $36.80 in a sale. The sale price is 20% less than the original price. Calculate the original price of the computer game. Answer(c) $ [3] (d) Vanessa buys some books and some pencils. Each book costs $12 more than each pencil. The total cost of 5 books and 2 pencils is $64.20. Find the cost of one pencil. Answer(d) $ [3]

3 © UCLES 2010 0581/43/O/N/10 [Turn over For Examiner's Use 2 S R Q P 85° 40° 4 km 7 km 4.5 km NOT TO SCALE The diagram shows five straight roads. PQ = 4.5 km, QR = 4 km and PR = 7 km. Angle RPS = 40° and angle PSR = 85°. (a) Calculate angle PQR and show that it rounds to 110.7°. Answer(a) [4] (b) Calculate the length of the road RS and show that it rounds to 4.52 km. Answer(b) [3] (c) Calculate the area of the quadrilateral PQRS. [Use the value of 110.7° for angle PQR and the value of 4.52 km for RS.] Answer(c) km2 [5]

4 © UCLES 2010 0581/43/O/N/10 For Examiner's Use 3 (a) Expand the brackets and simplify. )1 ( 4 ) 3 ( − + + x x x x Answer(a) [2] (b) Simplify 3 3) 3 ( x . Answer(b) [2] (c) Factorise the following completely. (i) 14 7 14 7 x x + Answer(c)(i) [2] (ii) aw ay xw xy 2 2 + + + Answer(c)(ii) [2] (iii) 49 4 2 − x Answer(c)(iii) [1]

5 © UCLES 2010 0581/43/O/N/10 [Turn over For Examiner's Use (d) Solve the equation. 0 1 5 2 2 = + + x x Show all your working and give your answers correct to 2 decimal places. Answer(d) x = or x = [4]

6 © UCLES 2010 0581/43/O/N/10 For Examiner's Use 4 (a) A =       5 4 3 2 B =       7 2 C = ( ) 2 1 Find the following matrices. (i) AB Answer(a)(i) [2] (ii) CB Answer(a)(ii) [2] (iii) A-1, the inverse of A Answer(a)(iii) [2] (b) Describe fully the single transformation represented by the matrix       1 _ 0 0 1 . Answer(b) [2] (c) Find the 2 by 2 matrix that represents an anticlockwise rotation of 90° about the origin. Answer(c)             [2]

7 © UCLES 2010 0581/43/O/N/10 [Turn over For Examiner's Use 5 D C B A The diagram shows an area of land ABCD used for a shop, a car park and gardens. (a) Using a straight edge and compasses only, construct (i) the locus of points equidistant from C and from D, [2] (ii) the locus of points equidistant from AD and from AB. [2] (b) The shop is on the land nearer to D than to C and nearer to AD than to AB. Write the word SHOP in this region on the diagram. [1] (c) (i) The scale of the diagram is 1 centimetre to 20 metres. The gardens are the part of the land less than 100 m from B. Draw the boundary for the gardens. [1] (ii) The car park is the part of the land not used for the shop and not used for the gardens. Shade the car park region on the diagram. [1]

8 © UCLES 2010 0581/43/O/N/10 For Examiner's Use 6 Sacha either walks or cycles to school. On any day, the probability that he walks to school is 5 3 . (a) (i) A school term has 55 days. Work out the expected number of days Sacha walks to school. Answer(a)(i) [1] (ii) Calculate the probability that Sacha walks to school on the first 5 days of the term. Answer(a)(ii) [2] (b) When Sacha walks to school, the probability that he is late is 4 1 . When he cycles to school, the probability that he is late is 8 1 . (i) Complete the tree diagram by writing the probabilities in the four spaces provided. 3 5 … 1 4 … walks cycles late not late … … late not late [3]

9 © UCLES 2010 0581/43/O/N/10 [Turn over For Examiner's Use (ii) Calculate the probability that Sacha cycles to school and is late. Answer(b)(ii) [2] (iii) Calculate the probability that Sacha is late to school. Answer(b)(iii) [2]

10 © UCLES 2010 0581/43/O/N/10 For Examiner's Use 7 (a) Complete the table for the function f(x) = 1 10 3 + x . x –4 –3 –2 –1 0 1 2 3 f(x) –1.7 0.2 0.9 1 1.1 1.8 [2] (b) On the grid, draw the graph of y = f(x) for –4 Y x Y=3. y x 4 3 2 1 –1 –2 –3 –4 –5 –6 –7 –8 –3 –4 –2 –1 1 0 2 3 [4] (c) Complete the table for the function g(x) = x 4 , 0 ≠ x . x –4 –3 –2 –1 1 2 3 g(x) –1 –1.3 2 1.3 [2]

11 © UCLES 2010 0581/43/O/N/10 [Turn over For Examiner's Use (d) On the grid, draw the graph of y = g(x) for –4 Y x Y –1 and 1 Y x Y 3. [3] (e) (i) Use your graphs to solve the equation x x 4 1 10 3 = + . Answer(e)(i) x = or x = [2] (ii) The equation x x 4 1 10 3 = + can be written as 0 4 = + + b ax x . Find the values of a and b. Answer(e)(ii) a = b = [2]

12 © UCLES 2010 0581/43/O/N/10 For Examiner's Use 8 3 cm 12 cm NOT TO SCALE The diagram shows a solid made up of a hemisphere and a cylinder. The radius of both the cylinder and the hemisphere is 3 cm. The length of the cylinder is 12 cm. (a) (i) Calculate the volume of the solid. [ The volume, V, of a sphere with radius r is 3 3 4 r V π = .] Answer(a)(i) cm3 [4] (ii) The solid is made of steel and 1 cm3 of steel has a mass of 7.9 g. Calculate the mass of the solid. Give your answer in kilograms. Answer(a)(ii) kg [2]

13 © UCLES 2010 0581/43/O/N/10 [Turn over For Examiner's Use (iii) The solid fits into a box in the shape of a cuboid, 15 cm by 6 cm by 6 cm. Calculate the volume of the box not occupied by the solid. Answer(a)(iii) cm3 [2] (b) (i) Calculate the total surface area of the solid. You must show your working. [ The surface area, A, of a sphere with radius r is 2 4 r A π = .] Answer(b)(i) cm2 [5] (ii) The surface of the solid is painted. The cost of the paint is $0.09 per millilitre. One millilitre of paint covers an area of 8 cm2. Calculate the cost of painting the solid. Answer(b)(ii) $ [2]

14 © UCLES 2010 0581/43/O/N/10 For Examiner's Use 9 (a) 3 2 1 –1 –2 –3 –4 0 –1 1 2 3 4 5 –2 –3 –4 C B A y x The points A (5, 3), B (1, –4) and C (–4, –2) are shown in the diagram. (i) Write as a column vector. Answer(a)(i) =             [1] (ii) Find – as a single column vector. Answer(a)(ii)             [2] (iii) Complete the following statement. – = [1] (iv) Calculate . Answer(a)(iv) [2]

15 © UCLES 2010 0581/43/O/N/10 [Turn over For Examiner's Use (b) A B D C M u t NOT TO SCALE ABCD is a trapezium with DC parallel to AB and DC = 2 1 AB. M is the midpoint of BC. = t and = u. Find the following vectors in terms of t and / or u. Give each answer in its simplest form. (i) Answer(b)(i) = [1] (ii) Answer(b)(ii) = [2] (iii) Answer(b)(iii) = [2]

16 © UCLES 2010 0581/43/O/N/10 For Examiner's Use 10 (a) For a set of six integers, the mode is 8, the median is 9 and the mean is 10. The smallest integer is greater than 6 and the largest integer is 16. Find the two possible sets of six integers. Answer(a) First set , , , , , Second set , , , , , [5] (b) One day Ahmed sells 160 oranges. He records the mass of each orange. The results are shown in the table. Mass (m grams) 50 < m Y 80 80 < m Y 90 90 < m Y 100 100 < m Y 120 120 < m Y 150 Frequency 30 35 40 40 15 (i) Calculate an estimate of the mean mass of the 160 oranges. Answer(b)(i) g [4]

17 © UCLES 2010 0581/43/O/N/10 [Turn over For Examiner's Use (ii) On the grid, complete the histogram to show the information in the table. 5 4 3 2 1 0 50 60 70 80 90 100 110 120 130 140 150 m Frequency density Mass (grams) [4] Question 11 is printed on the next page.

18 © UCLES 2010 0581/43/O/N/10 For Examiner's Use 11 Diagram 1 Diagram 2 Diagram 3 Diagram 4 The first four Diagrams in a sequence are shown above. Each Diagram is made from dots and one centimetre lines. The area of each small square is 1 cm2. (a) Complete the table for Diagrams 5 and 6. Diagram 1 2 3 4 5 6 Area (cm2) 2 6 12 20 Number of dots 6 12 20 30 Number of one centimetre lines 7 17 31 49 [4] (b) The area of Diagram n is )1 ( + n n cm2. (i) Find the area of Diagram 50. Answer(b)(i) cm2 [1] (ii) Which Diagram has an area of 930 cm2? Answer(b)(ii) [1] (c) Find, in terms of n, the number of dots in Diagram n. Answer(c) [1]

19 © UCLES 2010 0581/43/O/N/10 For Examiner's Use (d) The number of one centimetre lines in Diagram n is 1 2 2 + + pn n . (i) Show that p = 4. Answer(d)(i) [2] (ii) Find the number of one centimetre lines in Diagram 10. Answer(d)(ii) [1] (iii) Which Diagram has 337 one centimetre lines? Answer(d)(iii) [3] (e) For each Diagram, the number of squares of area 1 cm2 is A, the number of dots is D and the number of one centimetre lines is L. Find a connection between A, D and L that is true for each Diagram. Answer(e) [1]

20 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 2010 0581/43/O/N/10 BLANK PAGE

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 0581 MATHEMATICS 0581/43 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. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 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 2010 0581 43 © UCLES 2010 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) 200 ÷ 10 × 3 oe 200 ÷ 10 × 2 oe M1 M1 (b) 65 2 M1 for 100 60 39 × oe 35 is M0 (c) 46 3 M2 for 36.80 ÷ 0.8 oe or M1 for 80% = 36.80 oe (d) 0.6(0) 3 M2 for 5(x + 12) + 2x = 64.2 oe or (64.2 – 5 × 12) ÷ 7 or 5x + 2(x – 12) = 64.2 oe or (64.2 + 2 × 12) ÷ 7 or M1 for y = x + 12 and 5y + 2x = 64.2 or y = x – 12 and 5x + 2y = 64.2 After M0, SC1 for k(x ± 12) seen 2 (a) (cosQ =) 5.4 4 2 7 5.4 4 2 2 2 × × − + o.e. 110.74…. M2 E2 M1 for 72 = 42 + 4.52 – 2 × 4 × 4.5 × cos(Q) If E0 then A1 for – 0.354(1….) (b) 85 sin 40 sin 7 ) ( = RS M2 M1 for 85 sin 7 40 sin = RS o.e. 4.516 … E1 Can be implied by second M (c) Angle R = 55° 0.5 × 7 × 4.52 × sin(their 55) o.e. 0.5 × 4 × 4.5 × sin110.7 o.e. Triangle PRS + Triangle PQR 21.4 (21.36 – 21.42) B1 M1 M1 M1 A1 (May be seen on diagram) (12.95 – 13.0) their 55 is (180 – 40 – 85) (8.418 – 8.42) (s = 7.75) Dependent on M1, M1 www 5

Page 3 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2010 0581 43 © UCLES 2010 3 (a) 5x2 – x or x(5x – 1) 2 M1 for x2 + 3x or 4x2 – 4x correct (b) 27x9 2 B1 for 27 or for x9 (c) (i) 7x7(1 + 2x7) (ii) (y + w)(x + 2a) (iii) (2x + 7)(2x – 7) 2 2 1 M1 for any correct partially factorised expression or 7x7(1 + ...) M1 for x(y + w) + 2a(y + w) or y(x + 2a) + w(x + 2a) (d) ( ) ) 2 ( 2 )1( 2 4 5 5 2 − ± − oe 2 In square root B1 for 5² – 4(2)(1) or better (17) If in form r q p + or r q p − B1 for p = – 5 and r = 2(2) –2.28 –0.22 1 1 SC1 for –2.3 or –2.281 to –2.280 and –0.2 or –0.220 to –0.219 4 (a) (i)     43 25 1 1 If 0, 0 then SC1 for 25 and 43 seen (ii) (16) 2 B1 for 16 without brackets (iii)     − − − 2 4 3 5 2 1 isw or       − − 1 2 2 3 2 5 2 B1 for determinant = –2 or B1 for     − − 2 4 3 5 k (b) Reflection only x-axis oe 1 1 If more than one transformation given – no marks available independent (c)     − 0 1 1 0 2 B1 for one correct column

Page 4 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2010 0581 43 © UCLES 2010 5 (a) (i) Accurate perpendicular bisector, with 2 pairs of arcs, of CD. (ii) Accurate angle bisector, with two pairs of arcs, of angle A. 2 2 SC1 if accurate without arcs. SC1 if accurate without arcs. (b) SHOP written in correct region S1 Dependent on at least SC1 in (i) and (ii) and intersection (c) (i) Arc, centre B, radius 5cm, reaching across ABCD. (ii) Area outside their arc centre B and outside SHOP shaded 1 1ft Allow good freehand dep on S1 6 Accept fraction, %, dec equivalents (3sf or better) throughout but not ratio or words i.s.w. incorrect cancelling/conversion to other forms Pen –1 once for 2 sf answers (a) (i) 33 1 (ii) 3125 243 (0.07776) 2 Accept 0.0778. M1 for 5 5 3      oe (b) (i) 5 2 , 4 3 , 8 1 , 8 7 3 B1 for 5 2 and 4 3 B1 for 8 1 B1 for 8 7 (ii) 20 1 (0.05) cao 2 M1 for their 5 2 × their 8 1 (iii) 5 1 (0.2) ft 2ft ft 20 3 + their (b)(ii) or M1 for 4 1 5 3 × 7 (a) – 5.4 3.7 1 1 (b) 8 points correctly plotted ft Smooth cubic curve through all 8 points P3 C1 P3ft their table. P2ft for 6 or 7 points. P1ft for 4 or 5 points Only ft points if shape not affected. (c) –2, –4, 4 2 B1 for 2 correct (d) 7 points correctly plotted ft Two separate smooth branches of rectangular hyperbola P2 C1 P2ft P1ft for 5 or 6 points Must pass through all 7 points, only ft if shape not affected and no contact with either axis. (e) (i) –2.9 Y x Y– 2.8 2.05 Y x Y 2.15 (ii) a = 10 b = –40 1 1 1 1 Not with y coordinates

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2010 0581 43 © UCLES 2010 8 (a) (i) 396 (395.6 – 396) 4 M1 for 3 2 × π × 33 and M1 (independent) for π × 32 × 12, M1 (dependent on M2) for adding 126 π implies M3 (ii) 3.13 (3.125 – 3.128….) ft (iii) 144 (144 – 144.4) ft 2ft 2ft ft their (i) × 7.9 ÷ 1000 . M1 for × 7.9 soi by figs 313 or 3125 – 3128… ft 15 × 6 × 6 – their (a)(i) M1 for 6 × 6 × 15 oe (b) (i) 311 (310.8 – 311.1) (ii) 3.50 (3.496 to 3.50) ft 5 2ft M1 for 2 × π × 32 and M1 (independent) for π × 6 × 12 and M1 for π × 32, M1 (dependent on M3) for adding. (99π implies M4) ft their (b)(i) × 0.01125 M1 for their (b)(i) ÷ 8 and × figs 9 implied by figs 3496 to 350 9 (a) (i)     5 9 1 (ii)     7 4 1 1 If 0, SC1 for     − = 2 5 CB seen (iii) BA or – AB 1 BA not indicated as a vector is not enough. (iv) 10.3 (10.29 – 10.30) 2 M1 for (their 9)2 + (their 5)2 (b) (i) 2u 1 (ii) 2 1 (t – u ) oe 2 M1 for 2 1 (their DC AD BA + + ) or equivalent correct route for BM , along obtainable vectors in terms of t and u or M1 for correct unsimplified answer (iii) 2 3 u + 2 1 t oe ft 2ft ft their (i) + their (ii) simplified or t + u – their (b)(ii) simplified M1 for correct (or ft) unsimplified (i) + (ii) or t + u – their (b)(ii)

Page 6 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2010 0581 43 © UCLES 2010 10 (a) 7, 8, 8, 10, 11, 16 and 8, 8, 8, 10, 10, 16 5 Mark answer spaces only or clearly indicated lists. Allow numbers in any order but must be lists of 6 integers B4 for either correct list If not B4 then B1 for a series with mode 8 and B1 for a series with median 9 and B1 for a series with sum 60 (b) (i) (30 × 65 + 35 × 85 + 40 × 95 + 40 × 110 + 15 × 135) ÷ 160 4 M1 for mid-values soi (allow 1 error/omission) and M1 for use of ∑fx with x in correct interval including both boundaries allow one further error/omission and M1 (dependent on second M) for ÷ 160 94.7 (94.68 – 94.69) (ii) Heights of 4, 2, 0.5 with correct interval widths 4 www 4 B3 for 2 correct or B2 for 1 correct or B1 for all three freq. densities correct but no/incorrect graph 11 (a) 30 42 42 56 71 97 4 B3 for 2 correct rows or B2 for 1 correct row or B1 for any term in column 5 correct (b) (i) 2550 (ii) 30 1 1 (c) (n + 1)(n + 2) oe final ans 1 (d) (i) 2n2 + pn + 1 = t Uses a value of n up to 6 and a matching t from the table e.g. puts n = 3 and t = 31 2 × 3² + 3p + 1 = 31 M1 OR Use p = 4 to get 2n² + 4n + 1 = 31 and simplifies to 3 term eqn M1 OR both 2 × 9 + 4 × 3 + 1 (= 31) M1 with one part evaluated OR n(n + 1) + (n + 1)(n + 2) – 1 or better M1 (ii) 241 (iii) 12 2 1 3 Correct solution shown with 1 intermediate step to p = 4 E1 Solve correctly to get n = 3 E1 Conclusion e.g. 31 = 31 E1 Correct simplification to 2n2 + 4n + 1 E1 M1 for 2n2 + 4n + 1 = 337 and M1 for (n – 12)(n +14) or correct expression for n using formula (e) 1 − + = D A L oe 1