Cambridge IGCSE Mathematics 0580 — 2010 Oct/Nov Paper 4 · Variant 3

1 Thomas, Ursula and Vanessa share $200 in the ratio For Examiner's Thomas : Ursula : Vanessa = 3 : 2 : 5. Use (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]

Mark scheme: Qu. Answers Mark Part Marks 1 (a) 200 ÷ 10 × 3 oe M1 200 ÷ 10 × 2 oe M1 39 (b) 65 2 M1 for × 100 oe 35 is M0 60 (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 4 2 + 4 5 2 7 2

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2 For R Examiner's 4 km Q Use NOT TO SCALE 7 km 4.5 km 85° S 40° P 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]

Mark scheme: 4 + 5.4 − 7 2 (a) (cosQ =) o.e. M2 M1 for 72 = 42 + 4.52 – 2 × 4 × 4.5 × cos(Q) 2 × 4 × 5.4 110.74…. E2 If E0 then A1 for – 0.354(1….) 7 sin 40 RS 7 (b) ( RS = ) M2 M1 for = o.e. sin 85 sin 40 sin 85 4.516 … E1 Can be implied by second M (c) Angle R = 55° B1 (May be seen on diagram) 0.5 × 7 × 4.52 × sin(their 55) o.e. M1 (12.95 – 13.0) their 55 is (180 – 40 – 85) 0.5 × 4 × 4.5 × sin110.7 o.e. M1 (8.418 – 8.42) (s = 7.75) Triangle PRS + Triangle PQR M1 Dependent on M1, M1 21.4 (21.36 – 21.42) A1 www 5 IGCSE – October/November 2010 0580 43 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) 2 M1 for any correct partially factorised expression or 7x7(1 + ...) (ii) (y + w)(x + 2a) 2 M1 for x(y + w) + 2a(y + w) or y(x + 2a) + w(x + 2a) (iii) (2x + 7)(2x – 7) 1 2 ( )

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6 Sacha either walks or cycles to school. For 3 Examiner's On any day, the probability that he walks to school is . Use 5 (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] 1 (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 (i) Complete the tree diagram by writing the probabilities in the four spaces provided. 1 late 4 3 walks 5 not late .......... .......... late .......... cycles not late .......... [3] (ii) Calculate the probability that Sacha cycles to school and is late. For Examiner's Use Answer(b)(ii) [2] (iii) Calculate the probability that Sacha is late to school. Answer(b)(iii) [2]

Mark scheme: 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 5 243   (ii) (0.07776) 2 Accept 0.0778. M1 for  3 oe 3125  5  2 3 1 7 2 3 1 7 (b) (i) , , , 3 B1 for and B1 for B1 for 5 4 8 8 5 4 8 8 1 2 1 (ii) (0.05) cao 2 M1 for their × their 20 5 8 1 3 3 1 (iii) (0.2) ft 2ft ft + their (b)(ii) or M1 for × 5 20 5 4

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x 3 For 7 (a) Complete the table for the function f(x) = + 1 . Examiner's 10 Use 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 4 3 2 1 x –4 –3 –2 –1 0 1 2 3 –1 –2 –3 –4 –5 –6 –7 –8 [4] 4 (c) Complete the table for the function g(x) = , x ≠ 0 . x x –4 –3 –2 –1 1 2 3 g(x) –1 –1.3 2 1.3 [2] (d) On the grid, draw the graph of y = g(x) for –4 Y x Y –1 and 1 Y x Y 3. [3] For Examiner's Use x 3 4 (e) (i) Use your graphs to solve the equation + 1 = . 10 x Answer(e)(i) x = or x = [2] x 3 4 4 (ii) The equation + 1 = can be written as x + ax + b = 0 . 10 x Find the values of a and b. Answer(e)(ii) a = b = [2]

Mark scheme: 7 (a) – 5.4 1 3.7 1 (b) 8 points correctly plotted ft P3 P3ft their table. P2ft for 6 or 7 points. P1ft for 4 or 5 points Smooth cubic curve through all 8 C1 Only ft points if shape not affected. points (c) –2, –4, 4 2 B1 for 2 correct (d) 7 points correctly plotted ft P2 P2ft P1ft for 5 or 6 points Two separate smooth branches of C1 Must pass through all 7 points, only ft if shape rectangular hyperbola not affected and no contact with either axis. (e) (i) –2.9 Y x Y– 2.8 1 Not with y coordinates 2.05 Y x Y 2.15 1 (ii) a = 10 1 b = –40 1 IGCSE – October/November 2010 0580 43 2

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8 For Examiner's NOT TO Use SCALE 3 cm 12 cm 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. 4 3 [ The volume, V, of a sphere with radius r is V = πr .] 3 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] (iii) The solid fits into a box in the shape of a cuboid, 15 cm by 6 cm by 6 cm. For Calculate the volume of the box not occupied by the solid. Examiner's Use 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 A = 4πr 2 .] 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]

Mark scheme: 2 8 (a) (i) 396 (395.6 – 396) 4 M1 for × π × 33 and M1 (independent) for 3 π × 32 × 12, M1 (dependent on M2) for adding 126 π implies M3 (ii) 3.13 (3.125 – 3.128….) ft 2ft ft their (i) × 7.9 ÷ 1000 . M1 for × 7.9 soi by figs 313 or 3125 – 3128… (iii) 144 (144 – 144.4) ft 2ft ft 15 × 6 × 6 – their (a)(i) M1 for 6 × 6 × 15 oe (b) (i) 311 (310.8 – 311.1) 5 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) (ii) 3.50 (3.496 to 3.50) ft 2ft ft their (b)(i) × 0.01125 M1 for their (b)(i) ÷ 8 and × figs 9 implied by figs 3496 to 350 9 

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9 (a) For y Examiner's Use 3 A 2 1 x –4 –3 –2 –1 0 1 2 3 4 5 –1 C –2 –3 B –4 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] (b) For u Examiner's D C Use NOT TO SCALE t M A B 1 ABCD is a trapezium with DC parallel to AB and DC = AB. 2 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]

Mark scheme: 9 (a) (i)  9  1  5  (ii)  4  1 If 0, SC1 for CB =  5  seen  7  1  − 2  (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 1 1 (ii) (t – u ) oe 2 M1 for (their BA + AD + DC ) or equivalent 2 2 correct route for BM , along obtainable vectors in terms of t and u or M1 for correct unsimplified answer 3 1 (iii) u + t oe ft 2ft ft their (i) + their (ii) simplified 2 2 or t + u – their (b)(ii) simplified M1 for correct (or ft) unsimplified (i) + (ii) or t + u – their (b)(ii) IGCSE – October/November 2010 0580 43

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10 (a) For a set of six integers, the mode is 8, the median is 9 and the mean is 10. For Examiner's The smallest integer is greater than 6 and the largest integer is 16. Use 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] (ii) On the grid, complete the histogram to show the information in the table. For Examiner's Use 5 4 3 Frequency density 2 1 0 m 50 60 70 80 90 100 110 120 130 140 150 Mass (grams) [4] Question 11 is printed on the next page.

Mark scheme: 10 (a) 7, 8, 8, 10, 11, 16 5 Mark answer spaces only or clearly indicated and 8, 8, 8, 10, 10, 16 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 + 4 M1 for mid-values soi (allow 1 error/omission) with x in correct 40 × 110 + 15 × 135) ÷ 160 and M1 for use of ∑fx interval including both boundaries allow one further error/omission and M1 (dependent on second M) for ÷ 160 94.7 (94.68 – 94.69) www 4 (ii) Heights of 4, 2, 0.5 with correct 4 B3 for 2 correct interval widths or B2 for 1 correct or B1 for all three freq. densities correct but no/incorrect graph

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11 For Examiner's Use 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 n ( n + )1 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] (d) The number of one centimetre lines in Diagram n is 2 n 2 + pn + 1 . For Examiner's Use (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]

Mark scheme: 11 (a) 30 42 4 B3 for 2 correct rows 42 56 or B2 for 1 correct row 71 97 or B1 for any term in column 5 correct (b) (i) 2550 1 (ii) 30 1 (c) (n + 1)(n + 2) oe final ans 1 (d) (i) 2n2 + pn + 1 = t 2 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 Correct solution shown with 1 intermediate step to p = 4 E1 OR Use p = 4 to get 2n² + 4n + 1 = 31 and simplifies to 3 term eqn M1 Solve correctly to get n = 3 E1 OR both 2 × 9 + 4 × 3 + 1 (= 31) M1 Conclusion e.g. 31 = 31 E1 with one part evaluated OR n(n + 1) + (n + 1)(n + 2) – 1 Correct simplification to 2n2 + 4n + 1 E1 or better M1 (ii) 241 1 (iii) 12 3 M1 for 2n2 + 4n + 1 = 337 and M1 for (n – 12)(n +14) or correct expression for n using formula (e) L = A + D − 1 oe 1

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