Cambridge IGCSE Mathematics 0580 — 2014 May/June Paper 2 · Variant 1

31 Use your calculator to work out + 2–1. 4 Give your answer correct to 2 decimal places. Answer ................................................ [2] __________________________________________________________________________________________

Mark scheme: Question Answers Mark Part Marks 33 11 1 1.37 2 B1 for 0.866… or or 0.5 or 2 2 or B1 for 1.366… as final answer 1 2 36 f b

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2 x 2 2 y = 2 + x 2 Find the value of y when x = 6. Give your answer as a mixed number in its simplest form. Answer y = ................................................ [2] __________________________________________________________________________________________

Mark scheme: 1 2 36 2 18 2 M1 for + or better 18 36 2 n f

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3 Solve the equation. n - 8 = 11 2 Answer n = ................................................ [2] __________________________________________________________________________________________

Mark scheme: n 3 30 2 M1 for n – 8 = 22 or = 15 2

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4.8 # 1.98276 4 p = 16.83 (a) In the spaces provided, write each number in this calculation correct to 1 signifi cant fi gure. Answer(a) ............ × ............ ............ [1] (b) Use your answer to part (a) to estimate the value of p. Answer(b) ................................................ [1] __________________________________________________________________________________________

Mark scheme: 4 (a) 5× 2 1 20 (b) 1 1 0.5 or cao 2

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5 Write the following in order of size, smallest fi rst. 0.52 0.5 0.53 3 0.5 Answer .................. < .................. < .................. < .................. [2] __________________________________________________________________________________________

Mark scheme: 5 0.53 0.52 0.5 3 5.0 2 B1 for 0.25 , 0.125 and 0.793… seen or for three in correct order

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6 Carlo changed 800 euros (€) into dollars for his holiday when the exchange rate was €1 = $1.50 . His holiday was then cancelled. He changed all his dollars back into euros and he received €750. Find the new exchange rate. Answer €1 = $ ................................................. [3] __________________________________________________________________________________________

Mark scheme: 6 1.6[0] 3 M1 for 800 × 1.5 and M1 for their 1200 ÷ 750

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7 Make x the subject of the formula. y = (x – 4)2 + 6 Answer x = ................................................ [3] __________________________________________________________________________________________

Mark scheme: 7 4 ± y − 6 3 M1 for their 6 moved correctly M1 for their √ taken correctly M1 for their 4 moved correctly

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

Mark scheme: 8 2 3 B1 for common denominator x (x +1) seen M1 for 2 (x + 1) – 2x oe or better x ( x + )1

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9 A bus company in Dubai has the following operating times. Starting Finishing Day time time Saturday 06 00 24 00 Sunday 06 00 24 00 Monday 06 00 24 00 Tuesday 06 00 24 00 Wednesday 06 00 24 00 Thursday 06 00 24 00 Friday 13 00 24 00 (a) Calculate the total number of hours that the bus company operates in one week. Answer(a) ............................................. h [3] (b) Write the starting time on Friday in the 12-hour clock. Answer(b) ................................................ [1] __________________________________________________________________________________________

Mark scheme: 9 (a) 119 3 M2 for 18 × 6 + 11 oe or B1 for 18 or 11 or 108 (b) [0] 1 [00] pm cao 1

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10 Factorise completely. (a) ax + ay + bx + by Answer(a) ................................................ [2] (b) 3(x – 1)2 + (x – 1) Answer(b) ................................................ [2] __________________________________________________________________________________________

Mark scheme: 10 (a) (a + b)(x + y) 2 B1 for a(x + y) + b(x + y) or x(a + b) + y(a + b) (b) (x – 1)(3x – 2) 2 B1 for (x – 1)(3(x – 1) + 1) If B0 then SC1 for (x + a)(3x + b) where 3a + b = – 5 or ab = 2 or 3(x – 1)(x – ⅔) IGCSE – May/June 2014 0580 21 8 2 + 2 2 − 9 2

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11 A triangle has sides of length 2 cm, 8 cm and 9 cm. Calculate the value of the largest angle in this triangle. Answer ................................................ [4] __________________________________________________________________________________________

Mark scheme: 8 + 2 9 11 113.9 to 114.0 4 M2 for [cos =] 2 × 8 × 2 or M1 for 92 = 82 + 22 – 2 × 8 × 2 × cos x 13 A1 for −0.406 or −0.4063 to −0.4062 or − 32 If 0 scored SC2 for 54.3[1…] or 11.7 or 11.71 to 11.72 9 2 + 2 2 − 8 2 SC1 for [cos =] or 2 × 9 × 2 9 2 + 8 2 − 2 2 [cos =] 2 × 9 × 8 10 9

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12 p = 4 × 105 q = 5 × 104 Find, giving your answer in standard form, (a) pq, Answer(a) ................................................ [2] q (b) . p Answer(b) ................................................ [2] __________________________________________________________________________________________

Mark scheme: 12 (a) 2 × 1010 2 B1 for 20 × 109 or 20 000 000 000 (b) 1.25 × 10−1 2 B1 for 0.125 oe

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13 C NOT TO SCALE O B 58° D 23° A A, B, C and D lie on a circle centre O. Angle ABC = 58° and angle CAD = 23°. Calculate (a) angle OCA, Answer(a) Angle OCA = ................................................ [2] (b) angle DCA. Answer(b) Angle DCA = ................................................ [2] __________________________________________________________________________________________

Mark scheme: 13 (a) 32 2 B1 for AOC = 116 (b) 35 2 B1 for CDA = 122

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14 A(5, 10) NOT TO SCALE B(13, –2) A(5, 10) and B(13, –2) are two points on the line AB. 2 The perpendicular bisector of the line AB has gradient . 3 Find the equation of the perpendicular bisector of AB. Answer ................................................ [4] __________________________________________________________________________________________

Mark scheme: 14 2 4 B1 for (9, 4) y = x − 2 oe and 3 2 M2 for y = kx − 2 (k ≠ 0) or y = x + k (k ≠ 0) or 3 2 x − 2 3 2 2 or M1 for y = x or x + k (k ≠ 0) 3 3

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15 Solve the inequality for positive integer values of x. 21 + x > x + 1 5 Answer ................................................ [4] __________________________________________________________________________________________ 1

Mark scheme: 15 [0], 1, 2, 3 4 M1 for moving the 5 correctly M1 for collecting their terms A1 for a correct inequality for x eg [0 ≤ ] x < 4 12

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16 (a) (224) 2 = p4 Find the value of p. Answer(a) p = ................................................ [2] q2 + q 2 (b) Simplify 1 1 . q 4 # q 4 Answer(b) ................................................ [3] __________________________________________________________________________________________

Mark scheme: 16 (a) 8 2 B1 for 212 or 4096 3 3 (b) 2 q 2 3 B2 for kq 2 as the answer or 1 B1 for 2q2 and B1 for q 2 oe nfww

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17 Thailand NOT TO Hong SCALE Kong Singapore 150° Malaysia A travel brochure has 72 holidays in four different countries. The pie chart shows this information. (a) There are 24 holidays in Thailand. Show that the sector angle for Thailand is 120°. Answer(a) [2] (b) The sector angle for Malaysia is 150°. The sector angle for Singapore is twice the sector angle for Hong Kong. Calculate the number of holidays in Hong Kong. Answer(b) ................................................ [3] __________________________________________________________________________________________

Mark scheme: 17 (a) correct working 2 M1 for 1 holiday = 5 or 360 ÷ 72 = 5 and B1 for 24 × 5 [= 120] or 24 M2 for × 360 [=120] oe 72 (b) 6 nfww 3 M1 for 150 + 120 + x + 2x = 360 oe A1 for 30 identified as the required angle 1 1 10 4

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18 NOT TO SCALE 10 cm 4 cm A solid cone has base radius 4 cm and height 10 cm. A mathematically similar cone is removed from the top as shown in the diagram. 1 The volume of the cone that is removed is of the volume of the original cone. 8 (a) Explain why the cone that is removed has radius 2 cm and height 5 cm. Answer(a) [2] (b) Calculate the volume of the remaining solid. 1 [The volume, V, of a cone with radius r and height h is V = πr 2h.] 3 Answer(b) ......................................... cm3 [4] __________________________________________________________________________________________ Question 19 is printed on the next page.

Mark scheme: 10 4 1 1 = = 2 or 3 8 = 2 AND = 5 oe and18 (a) correct working 2 B2 for 3 8 2 2 2 oe or 1 1 1 = ( 3) or 3 8 or 8 = 23 or B1 for 3 8 8 2 IGCSE – May/June 2014 0580 21 7 1 2 (b) 147 or 146.5 to 146.6… 4 M3 for × × π × 4 × 10 8 3 or 1 2 M1 for × π × 4 × 10 3 and 1 2 M1 for × π × 2 × 5 3 and M1 for subtracting their volumes 1 i

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19 E D C NOT TO SCALE 8 cm 60° 30° A B 8 cm The diagram shows a rectangle ABCE. D lies on EC. DAB is a sector of a circle radius 8 cm and sector angle 30°. Calculate the area of the shaded region. Answer ......................................... cm2 [7]

Mark scheme: 1 19 1.38 or 1.39 or 1.384 to 1.389 7 M3 [Area ∆ =] × 8 cos 60 × 8 sin 60 2 or M1 for [ AE =] 8cos 60 and M1 for [ ED] = 8sin 60 and 30 2 M1 for Area sector × π × 8 360 and M1 for Area rectangle = 8 × 8cos60 or 8 × 4 M1 for their 32 – (their 13.86 + their 16.76) or better

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