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

1 Examiner's Use For the diagram, write down (a) the order of rotational symmetry, Answer(a) [1] (b) the number of lines of symmetry. Answer(b) [1]

Mark scheme: Qu. Answers Mark Part Marks 1 (a) 1 1 Allow none (b) 1 1

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2 Calculate 3sin120° − 4(sin120°)3. Answer [2]

Mark scheme: 2 0 2 M1 4sin3120 evaluated and rounding to 2.6 or 3 3 better (2.598…) or 2 3 2

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3 Write the following in order of size, smallest first. 2 3 2 − 3 3 2 − 3 2 Answer < < < [2]

Mark scheme: 3 2 3 2 – 3 , 2 – , , 3 2 M1 correct decimals seen 2 3 15a + 32

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3a 4 Examiner's 4 Write as a single fraction + . Use 8 5 Answer [2]

Mark scheme: 15a + 32 4 oe 2 B1 15a + 32 seen 40 15 a 32 or SC1 + on answer line 40 40

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5 Write 28 × 82 × 4-2 in the form 2n. Answer [2]

Mark scheme: 5 210 2 M1 26 or 2–4 seen

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6 Change 64 square metres into square millimetres. Give your answer in standard form. Answer mm2 [2]

Mark scheme: 6 6.4 × 107 2 M1 64 × 1002 × 102 or 64 000 000 oe

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7 A B C The shaded area in the diagram shows the set (A ∩C ) ∩B'. Write down the set shown by the shaded area in each diagram below. A B A B C C [2]

Mark scheme: 7 (A ∪ B ∪ C)' 1 or A' ∩ B' ∩ C' or A' ∩ (B ∪ C)' (A ∪ C)' ∩ B 1 or A' ∩ C' ∩ B

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8 Examiner's y Use 1 y = sinx 0.5 0 x 30° 60° 90° 120° 150° 180° –0.5 y = cosx –1 The diagram shows accurate graphs of y = sinx and y = cosx for 0° Y x Y 180°. Use the graph to solve the equations (a) sinx – cosx = 0, Answer(a) x = [1] (b) sinx – cosx = 0.5. Answer(b) x = [2]

Mark scheme: 8 (a) 43 to 47 1 (b) 64 to 68 2 SC1 23 to 27

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9 A fence is made from 32 identical pieces of wood, each of length 2 metres correct to the nearest centimetre. Calculate the lower bound for the total length of the wood used to make this fence. Write down your full calculator display. Answer m [3]

Mark scheme: 9 63.84 cao 3 M1 figs 1995 M1 32 × their lower bound 3

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10 Make x the subject of the formula. Examiner's Use x + 3 P = x Answer x = [4]

Mark scheme: 3 10 x = 4 M1 for each of the four moves completed P − 1 correctly

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11 NOT TO SCALE O Q C 6 cm 4 cm P T Two circles, centres O and C, of radius 6 cm and 4 cm respectively, touch at Q. PT is a tangent to both circles. (a) Write down the distance OC. Answer(a) OC = cm [1] (b) Calculate the distance PT. Answer(b) PT = cm [3]

Mark scheme: 11 (a) 10(.0..) 1 (b) 9.80 3 M2 √((a)2 – 22) or M1 PT2 + 22 = (a)2 h

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12 The diagram represents the ski lift in Queenstown New Zealand. Examiner's Use T NOT TO SCALE 730 m h 37.1° B (a) The length of the cable from the bottom, B, to the top, T, is 730 metres. The angle of elevation of T from B is 37.1°. Calculate the change in altitude, h metres, from the bottom to the top. Answer(a) m [2] (b) The lift travels along the cable at 3.65 metres per second. Calculate how long it takes to travel from B to T. Give your answer in minutes and seconds. Answer(b) min s [2]

Mark scheme: h 12 (a) 440 2 M1 sin 37.1 or cos 52.9 = oe 730 730 (b) 3 min 20 sec 2 M1 .365  6x 3   6x 3 

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13 Examiner's  6 − 3  x  Use M =    .  4 5  1  (a) Find the matrix M. Answer(a) M = [2] (b) Simplify ( x 1 ) M. Answer(b) [2]

Mark scheme: 13 (a)  6 x − 3  but not  6 x − 3  2 B1 6x – 3 or B1 4x + 5 in a (2 × 1) matrix on  4 x + 5   4 x ( + )5  answer line (b) (6x2 + x + 5) cao 2 M1 any 1 × 1 matrix in answer space

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14 y 10 8 6 4 2 x –4 –2 0 2 4 6 8 10 12 14 16 18 By shading the unwanted regions of the grid above, find and label the region R which satisfies the following four inequalities. y [ 2 x + y [ 6 y Y x + 4 x + 2y Y 18 [4]

Mark scheme: 14 4 Mark the position of the letter R (or the worst unshaded region if R is missing) as follows IGCSE – May/June 2010 0580 22

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15 Examiner's Use y NOT TO SCALE 4 A y = 4 B x 0 2x + y = 8 3x + y = 18 (a) The line y = 4 meets the line 2x + y = 8 at the point A. Find the co-ordinates of A. Answer(a) A ( , ) [1] (b) The line 3x + y = 18 meets the x axis at the point B. Find the co-ordinates of B. Answer(b) B ( , ) [1] (c) (i) Find the co-ordinates of the mid-point M of the line joining A to B. Answer(c)(i) M ( , ) [1] (ii) Find the equation of the line through M parallel to 3x + y = 18. Answer(c)(ii) [2]

Mark scheme: 15 (a) (2, 4) 1 (b) (6, 0) 1 (c) (i) (4, 2) ft 1ft From (a) and (b) (ii) y = – 3x + 14 oe 2 M1 sub their (c)(i) into y = –3x + c oe 1

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16 The graphs show the speeds of two cyclists, Alonso and Boris. Examiner's Use Alonso accelerated to 10 m/s, travelled at a steady speed and then slowed to a stop. 10 Speed Alonso (m / s) 0 2 4 6 8 10 12 14 16 Time (seconds) Boris accelerated to his maximum speed, v m/s, and then slowed to a stop. v NOT TO SCALE Speed (m / s) Boris 0 16 Time (seconds) Both cyclists travelled the same distance in the 16 seconds. Calculate the maximum speed for Boris. Show all your working. Answer m/s [5]

Mark scheme: 1 16 16 or 16.3 5 M1 finding the area under graph A1 130 4 1 M1 × 16 × v 2 M1 equating and solving

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17 Examiner's Use NOT TO Q R SCALE P 8 m S O W T V U The diagram shows the junction of four paths. In the junction there is a circular area covered in grass. This circle has centre O and radius 8 m. (a) Calculate the area of grass. Answer(a) m2 [2] (b) Q NOT TO SCALE 12 m 45° P O The arc PQ and the other three identical arcs, RS, TU and VW are each part of a circle, centre O, radius 12m. The angle POQ is 45°. The arcs PQ, RS, TU, VW and the circumference of the circle in part(a) are painted white. Calculate the total length painted white. Answer(b) m [4]

Mark scheme: 17 (a) 201 2 M1 π × 82 45 (b) 87.9 or 88.0 4 M1 × 2 × π × 12 ….. d 360 M1 2 × π × 8 ……………..e M1 ft for their (4d + e) which must come from multiples of π SC2 43.9 or 44.0

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18 (a) f(x) = 1 – 2x. Examiner's Use (i) Find f(-5). Answer(a)(i) [1] (ii) g(x) = 3x – 2. Find gf(x). Simplify your answer. Answer(a)(ii) [2] (b) h(x) = x2 – 5x – 11. Solve h(x) = 0. Show all your working and give your answer correct to 2 decimal places. Answer(b) x = or x = [4] Question 19 is printed on the next page.

Mark scheme: 18 (a) (i) 11 1 (ii) 1 – 6x 2 M1 3(1 – 2x) – 2 5 ± k (b) –1.65, 6.65 4 M1 M1 √[(–5)2 – 4 × 1 × (–11)] 2 or better A1 A1

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19 The braking distance, d metres, for Alex’s car travelling at v km/h is given by the formula Examiner's Use 200d = v(v + 40). (a) Calculate the missing values in the table. v 0 20 40 60 80 100 120 (km/h) d 0 16 48 96 (metres) [2] (b) On the grid below, draw the graph of 200d = v(v + 40) for 0 Y v Y 120. d 100 90 80 70 60 Distance 50 (metres) 40 30 20 10 v 0 10 20 30 40 50 60 70 80 90 100 110 120 Speed (km / h) [3] (c) Find the braking distance when the car is travelling at 110 km/h. Answer(c) m [1] (d) Find the speed of the car when the braking distance is 80 m. Answer(d) km/h [1]

Mark scheme: 19 (a) 6, 30, 70 2 B1 for 2 correct (b) graph 3 P2 7 plots correct from table P1 5 or 6 plots correct from table C1 smooth curve through the points in the given range within one small square of the plots or the correct position (c) 82.5 or ft ±1 1ft (d) 108 or ft ±1 1ft

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