Cambridge IGCSE Mathematics 0580 — 2012 May/June Paper 1 · Variant 3

2 For 1 Write as a percentage. Examiner's 5 Use Answer %[1]

Mark scheme: Qu. Answers Mark Part Marks 1 40 1

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2 Change 5.2 square metres into square centimetres. Answer cm2 [1]

Mark scheme: 2 52 000 1

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3 Mohinder changes $240 into Rupees. The exchange rate is $1 = 46.2875 Rupees. Calculate how many Rupees he receives. Answer Rupees [1]

Mark scheme: 3 11 109 1

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4 (a) Write down the next prime number after 47. Answer(a) [1] (b) Write down the next square number after 49. Answer(b) [1]

Mark scheme: 4 (a) 53 1 (b) 64 1

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5 I K = For Examiner's Choose one of these symbols to make each statement correct. Use (a) −15 −5 [1] (b) (−5)2 25 [1]

Mark scheme: 5 (a) < 1 (b) = 1 750 × 2 × 8 f

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6 Hans invests $750 for 8 years at a rate of 2% per year simple interest. Calculate the interest Hans receives. Answer $ [2]

Mark scheme: 750 × 2 × 8 6 120 2 M1 for 100 oe seen or SC1 870 as final answer

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7 A 73° B NOT TO 120° SCALE 82° x° C D E The diagram shows a quadrilateral ABCD. CDE is a straight line. Calculate the value of x. Answer x = [2]

Mark scheme: 7 95 2 B1 for 85 seen or M1 x = 180 – ‘their angle ADC’, if it is clearly seen

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8 Work out For Examiner's Use  5   6  (a)   −   ,  3   −2      Answer(a)   [1]      3 (b) 5   .  −4      Answer(b)   [1]    

Mark scheme: 8 (a) − 1  1    5  (b)  15  1    −20 

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9 Simplify (a) a0 , Answer(a) [1] (b) b3 × b-5 . Answer(b) [1]

Mark scheme: 9 (a) 1 1 (b) b–2 1 accept 21 b

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10 During her holiday, Hannah rents a bike. She pays a fixed cost of $8 and then a cost of $4.50 per day. Hannah pays with a $50 note and receives $10.50 change. Calculate for how many days Hannah rents the bike. Answer days [3]

Mark scheme: 10 7 cao 3 B1 for 39.5(0) or 31.5(0) or 42 M1 for (their 39.5 – 8) ÷ 4.5 or (their 42 – 10.5) ÷ 4.5

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11 For E Examiner's D F Use b° 52° NOT TO SCALE a° A C B In the diagram lines AC and DF are parallel and AE = EB. Angle AEB = 52°. (a) Write down the mathematical name for triangle AEB. Answer(a) [1] (b) Work out the value of a. Answer(b) a = [1] (c) Explain why a = b. Answer(c) [1]

Mark scheme: 11 (a) isosceles 1 (b) 64 1 (c) alternate (angle) 1 accept z angle IGCSE – May/June 2012 0580 13

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12 Solve the simultaneous equations. 4x + y = 18 5x + 3y = 19 Answer x = y = [3]

Mark scheme: 12 [x =] 5, [y =] –2 3 M1 for consistent multiply and add/subtract as appropriate. Allow computational errors. Other methods allowed. A1 for correct x or y. 4

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13 (a) Write 0.000 64 in standard form. For Examiner's Use Answer(a) [1] (b) Calculate, writing the answer in standard form. 7 8.18 × 10 4 5.84 × 10 Answer(b) [2]

Mark scheme: 13 (a) 6.4 × 10–4 1 (b) 1.4 × 103 2 M1 for 1400 or answer rounding to 1401 or 1.4 × 10k

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14 7 3 8 2 5 1 5 3 4 6 2 3 For the numbers above work out the (a) mode, Answer(a) [1] (b) median, Answer(b) [2] (c) range. Answer(c) [1]

Mark scheme: 14 (a) 3 1 (b) 3.5 2 M1 for at least 7 numbers in order and an attempt to find the middle number (c) 7 1

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15 Without using your calculator, work out the following. For Show all the steps of your working and give each answer as a fraction in its simplest form. Examiner's Use 11 1 (a) − 12 3 Answer(a) [2] 1 11 (b) ÷ 4 13 Answer(b) [2]

Mark scheme: 15 (a) 11 4 M1 correct use of a common denominator − oe 2 12 12 7 cao ww 0 A1 12 1 13 (b) × oe 2 M1 inversion and operation change 4 11 13 cao ww 0 44 A1 21

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16 (a) Solve the equation 5(x − 3) = 21 . Answer(a) x = [2] (b) Make x the subject of the equation y = 3x − 2 . Answer(b) x = [2]

Mark scheme: 16 (a) 7.2 oe 2 M1 for 5x – 15 = 21 or x – 3 = 215 y + 2 (b) [x =] 3 2 M1 for 3x = y + 2 or – 3x = –2 – y

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17 For 12 cm Examiner's Use 4 cm NOT TO 12 cm 18 cm 18 cm SCALE 34 cm For the shape above, work out (a) the perimeter, Answer(a) cm [2] (b) the area. Answer(b) cm2 [2]

Mark scheme: 17 (a) 112 2 M1 Attempt to add 6 given and their 2 sides (b) 564 2 M1 for 18 × 34 – 12 × 4 : (612 – 48) or (18 × 12) + (14 × 12) + (10 × 18) or (4 × 12) + (10 × 4) + (34 × 14)

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18 (a) Find the value of 7p – 3q when p = 8 and q = O5 . Answer(a) [2] (b) Factorise completely. 3uv + 9vw Answer(b) [2]

Mark scheme: 18 (a) 71 2 M1 for 7×8 – 3×–5 or B1 56 and –15 (b) 3v (u + 3w) final answer 2 B1 for 3(uv + 3vw) or v (3u + 9w) As final answer

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19 For North Examiner's Use North B 26 km NOT TO SCALE 74 km A 28° C (a) Work out the bearing of A from C. Answer(a) [2] (b) Calculate the distance AB. Answer(b) km [2]

Mark scheme: 19 (a) 332 2 M1 for BCA = 28. Or 360 – 28 or 152 marked correctly at C or 180 + 152 (b) 78.4 2 M1 for AB2 = 742 + 262 or better

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20 (a) Colin has some seeds. For The probability a seed will grow is 0.85 . Examiner's Use Find the probability that a seed will not grow. Answer(a) [1] (b) Richard grows flowers. Some of his flowers are chosen at random. The colours are recorded in the table below. Colour of Relative Frequency flower Frequency Red 20 0.16 Blue 15 Yellow 35 Other 55 (i) Complete the table to show the relative frequency of each colour. [2] (ii) Richard grows 800 flowers in total. Estimate how many of these flowers are red. Answer(b)(ii) [2]

Mark scheme: 20 (a) [0].15 oe 1 (b) (i) 0.12, 0.28, 0.44 oe 2 M1 for division of 15, 35 or 55 by their 125 Or B1 for 1 correct (ii) 128 2 M1 for 800 × [0].16

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