Cambridge IGCSE Mathematics 0580 — 2010 May/June Paper 2 · Variant 3

1 During one week in April, in Quebec, the daily minimum temperatures were Examiner's Use –5°C, –1°C, 3°C, 2°C, –2°C, 0°C, 6°C. Write down (a) the lowest of these temperatures, Answer(a) °C [1] (b) the range of these temperatures. Answer(b) °C [1]

Mark scheme: Qu. Answers Mark Part Marks 1 (a) –5 1 (b) 11 1 53

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53 2 23 48% 4.80 11 Write the numbers in order of size with the largest first. Answer K= K K [2]

Mark scheme: 53 2 > .480 > 23 > 48 % 2 M1 for decimals seen 11 4.7958… 0.48 (4.80) 4.81(…)

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3 Ricardo changed $600 into pounds (£) when the exchange rate was $1 = £0.60. He later changed all the pounds back into dollars when the exchange rate was $1 = £0.72. How many dollars did he receive? Answer $ [2]

Mark scheme: 3 500 2 M1 for 600 × 0.6 ÷ 0.72 seen

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4 The maximum speed of a car is 252 km/h. Change this speed into metres per second. Answer m/s [2]

Mark scheme: 4 70 2 M1 for 252 × 1000 ÷ 60 ÷ 60 oe

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5 Amalie makes a profit of 20% when she sells a shirt for $21.60. Examiner's Use Calculate how much Amalie paid for the shirt. Answer $ [2]

Mark scheme: 5 18 2 M1 for 21.6 ÷ 1.2 oe 8

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6 3x × 94 = 3n. Find n in terms of x. Answer n = [2]

Mark scheme: 6 x + 8 2 M1 38 seen

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7 Shade the required regions in the Venn diagrams below. A B A B C C [2]

Mark scheme: 7 2 B1 for one correct Venn diagram 5 x − 3

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8 Write as a single fraction in its simplest form x x − 1 + . 3 2 Answer [2]

Mark scheme: 5x 3 8 2 B1 for 5x – 3 seen 6 5 3 SC1 x − on answer line 6 6 5 k

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9 1 second = 106 microseconds. Examiner's Use Change 3 × 1013 microseconds into minutes. Give your answer in standard form. Answer min [2]

Mark scheme: 9 5(.00) × 105 2 SC1 for 5 × 10k or 500 000 on answer line

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10 The length of each side of an equilateral triangle is 74 mm, correct to the nearest millimetre. Calculate the smallest possible perimeter of the triangle. Answer mm [2]

Mark scheme: 10 220.5 cao 2 M1 for 73.5 seen h 55

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11 P cliff NOT TO SCALE beach A F 55 m The diagram shows a point P at the top of a cliff. The point F is on the beach and vertically below P. The point A is 55m from F, along the horizontal beach. The angle of elevation of P from A is 17°. Calculate PF, the height of the cliff. Answer PF = m [3]

Mark scheme: h 55 11 16.8 3 M2 tan17 = or tan73 = 55 h 55 h or M1 tan17 = or tan73 = if angle seen in h 55 wrong place at P 2 2 2

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12 Expand and simplify 2(x – 3)2 – (2x – 3)2. Examiner's Use Answer [3]

Mark scheme: 12 9 – 2x2 3 B1 for x2 – 3x – 3x + 9 or 2x2 – 6x – 6x + 18 B1 for 4x2 – 6x – 6x + 9 or –4x2 + 6x + 6x – 9

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13 (a) Write down the number of lines of symmetry for the diagram below. Answer(a) [1] (b) Write down the order of rotational symmetry for the diagram below. Answer(b) [1] (c) The diagram shows a cuboid which has no square faces. Draw one of the planes of symmetry of the cuboid on the diagram. [1]

Mark scheme: 13 (a) 0 1 (b) 2 1 (c) plane across centre of shape 1 Three possibilities

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y Examiner's 14 Solve the equation 3( y − 4 ) + = 9 . Use 2 Answer y = [3]

Mark scheme: 14 6 3 M1 for one correct first step which leads towards simplifying y 3 y − 12 + = 9 2 or 6(y – 4) + y = 18 y or y – 4 + = 3 6 M1 correctly collecting their terms to py = q IGCSE – May/June 2010 0580 23

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15 G g NOT TO SCALE O N h H In triangle OGH, the ratio GN : NH = 3 : 1. = g and = h. Find the following in terms of g and h, giving your answers in their simplest form. (a) Answer(a) = [1] (b) Answer(b) = [2]

Mark scheme: 15 (a) g – h 1 1 3 1 (b) g + h 2 M1 for OH + HN or h + (a) 4 4 4 3 OG + GN or g – (a) 4 5A 5 A − 2 r 3

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r ( y + 2 ) Examiner's 16 Make y the subject of the formula. A = Use 5 Answer y = [3]

Mark scheme: 5 A 5 A 2 r 16 − 2 or 3 M1 for correctly multiplying by 5 r r M1 for correctly dividing by r M1 for correct subtraction in any order 40 2

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17 K NOT TO SCALE 40° O 5.6 cm L OKL is a sector of a circle, centre O, radius 5.6 cm. Angle KOL = 40°. Calculate (a) the area of the sector, Answer(a) cm2 [2] (b) the perimeter of the sector. Answer(b) cm [2]

Mark scheme: 40 2 17 (a) 10.9 2 M1 for × π × 6.5 360 40 (b) 15.1 2 M1 for × π × 2 × 6.5 (= 3.91..) 360

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18 f(x) = x2 + 2 g(x) = (x + 2)2 h(x) = 3x – 5 Examiner's Use Find (a) gf(–2), Answer(a) [2] (b) h –1(22) . Answer(b) [2]

Mark scheme: 18 (a) 64 2 B1 for evidence of f(–2) = 6 x + 5 (b) 9 2 M1 for 3x – 5 = 22 or seen 3 3

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19 v 3 NOT TO SCALE Speed (m / s) t 0 4 15 Time (seconds) The diagram shows the speed-time graph for 15 seconds of the journey of a cyclist. (a) Calculate the acceleration of the cyclist during the first 4 seconds. Answer(a) m/s2 [1] (b) Calculate the average speed for the first 15 seconds. Answer(b) m/s [3]

Mark scheme: 3 19 (a) or 0.75 1 4 (b) 2.6 3 M1 for finding the area under the graph or M1 for their 39 ÷ 15

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20 Examiner's y Use 4 3 2 1 x 0 1 2 3 4 Find the three inequalities which define the shaded region on the grid. Answer [5]

Mark scheme: 20 x [ 0 1 L1 x R 0 1 1 y [ x oe 2 L1 y R x 2 2 x + y Y 4 oe 2 L1 x + y R 4 where R is any one of = < > Y [ B2 all inequalities correct or B1 2 correct sin 140

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21 Examiner's North Use North Q NOT TO 140° SCALE 50 m North R 100 m P The diagram shows three points P, Q and R on horizontal ground. PQ = 50 m, PR = 100 m and angle PQR = 140°. (a) Calculate angle PRQ. Answer(a) Angle PRQ = [3] (b) The bearing of R from Q is 100°. Find the bearing of P from R. Answer(b) [2]

Mark scheme: sin 140 21 (a) 18.7 3 M2 for sin R = 50 × (= 0.3219…) 100 sin R sin 140 or M1 for = oe 50 100 (b) 261(.3) 2ft M1 360 – 80 – their (a)

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22 Examiner's B Use A C The diagram shows a farmer’s field ABC. The farmer decides to grow potatoes in the region of the field which is • nearer to A than to C and • nearer to AB than to AC. Using a straight edge and compasses only, construct two loci accurately and shade this region on the diagram. [5] Question 23 is printed on the next page.

Mark scheme: 22 Perpendicular bisector of AC 2 B1 accurate line B1 two pairs of correct construction arcs Bisector of angle A 2 B1 accurate line B1 two pairs of correct construction arcs Shaded region inside triangle and 1 B1 dep on first B1 being scored for both lines to left of perp bisector of AC and above bisector of angle A ( )

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3 −1 Examiner's23 A = (1 4 ) B = Use ( −2 2 ) Find (a) AB, Answer(a) AB = [2] (b) the inverse matrix B–1 , Answer(b) B–1 = [2] (c) BB–1. Answer(c) BB–1 = [1]

Mark scheme: 23 (a) (− 5 7 ) 2 B1 either correct in a (1 × 2) matrix 1  2 1   2 1  (b) oe 2 M1 for seen or 2 × 3 – –1 × –2 ( = 4) 2 3  4  2 3    1 0  (c)  0 1  or I cao 1

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