Cambridge IGCSE Mathematics 0580 — 2022 May/June Paper 2 · Variant 2

1 At noon, the temperature is 4 °C. At midnight, the temperature is - 9 °C. Work out the difference in temperature between noon and midnight. ............................................. °C [1]

Mark scheme: Question Answer Marks Partial Marks 1 13 or – 13 1

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2 Thibault records the number of cars of each colour in a car park. Colour Black White Silver Red Number of cars 8 5 4 3 He draws a pie chart to show this information. Calculate the sector angle for the red cars. ................................................. [2]

Mark scheme: 2 54 2 360 3 M1 for  3 or  360  oe 8  5  4  3 8  5  4  3

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3 Figs cost 43 cents each. Lyra has $5 to buy some figs. Calculate the largest number of figs Lyra can buy and the amount of change, in cents, she receives. ........................... figs and ........................... cents change [3]

Mark scheme: 3 11 27 3 M1 for 500 ÷ 43 oe M1 for 500  their 11  43 oe their 11 must be an integer from 2 to 11

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4 Find the value of 68 # 153 . ................................................. [1]

Mark scheme: 4 102 1

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5 Find the total surface area of a cuboid with length 8 cm, width 6 cm and height 3 cm. .......................................... cm2 [3]

Mark scheme: 5 180 3 M2 for  2   8  6 8 3 3 6  oe or M1 for 8  6 or 8  3 or 3  6

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6 Some cards have either a square, a circle or a triangle drawn on them. Piet chooses one of the cards at random. Complete the table to show the probability of choosing a card with each shape. Shape Square Circle Triangle Probability 0.2 0.32 [2]

Mark scheme: 6 0.48 oe 2 M1 for 1   0.2  0.32  oe

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7 The price of a coat is $126. In a sale, this price is reduced by 18%. Find the sale price of the coat. $ ................................................. [2]

Mark scheme: 7 103.32 cao 2  18  M1 for 126  1  oe    100  or B1 for 22.68

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8 The nth term of a sequence is n 2 + 12 . Find the first three terms of this sequence. ....................... , ....................... , ....................... [2]

Mark scheme: 8 13 16 21 2 B1 for 2 correct terms in correct position or SC1 for 12, 13, 16

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9 North B NOT TO SCALE A The bearing of B from A is 059°. Work out the bearing of A from B. ................................................. [2]

Mark scheme: 9 239 2 M1 for 180 + 59 or 360 – (180 − 59) oe or indicates correct angle on diagram

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2 - 1 10 p = q = e8o e 4o (a) Find (i) p - q , [1] f p (ii) 6p. [1] f p (b) Find p - q . ................................................. [2]

Mark scheme: 10(a)(i) 3 1  4 10(a)(ii)  12  1    48  10(b) 5 2 M1 for (their 3) 2  (their 4) 2 or better

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11 Find the value of p when 6 p # 6 4 = 6 28 . p = ................................................. [1]

Mark scheme: 11 24 1

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12 Annette cycles a distance of 70 km from Midville to Newtown. Leaving Midville, she cycles for 1 hour 30 minutes at a constant speed of 20 km/h and then stops for 30 minutes. She then continues the journey to Newtown at a constant speed of 16 km/h. 80 60 Distance (km) 40 20 0 0 1 2 3 4 5 Time (h) (a) On the grid, draw the distance–time graph for the journey. [3] (b) Calculate the average speed for the whole journey. ........................................ km/h [3]

Mark scheme: 12(a) correct graph 3 B1 for line from (0, 0) to (1.5, 30) B1 for horizontal line from (their 1.5, their 30) for 0.5 hours B1 for a line from (their 2, their 30) ending at distance 70 with a gradient of 16 Provided it fits on the grid and their 30 is <70 12(b) 15.6 or 15.55 to 15.56 3 M2 for 70  (their final time in hours) nfww 70 – their 30 (final time =) 1.5 + 0.5 + 16 or 4.5 or their final time from graph or M1 for 70  any time

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1 513 Without using a calculator, work out 4 - 2 . 8 6 You must show all your working and give your answer as a mixed number in its simplest form. ................................................. [3]

Mark scheme: 13 33 17 1 5 B1 Correct step for dealing with mixed numbers or  8 6 28 6 33k 17 k Allow or 8 k 6 k 99 68 20 M1 Correct method to find common denominator and 32  24 24 24 24 3 20 e.g. 4 and 2 24 24 7 A1 124 cao and correct working

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14 Carlos invests $4540 at a rate of r % per year compound interest. At the end of 10 years he has earned $1328.54 in interest. Calculate the value of r. r = ................................................. [3]

Mark scheme: 14 2.6[0] or 2.600… 3 1328.54  4540 M2 for 10 4540 or M1 for 4540  k 10 = 1328.54 + 4540 for any k If 0 scored SC1 for answer –11.6 or –11.56…

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15 Find the highest common factor (HCF) of 12a 3 b and 20a 2 b 2 . ................................................. [2]

Mark scheme: 15 4a 2 b final answer 2 M1 for two correct parts out of three from 4, a2 and b in final answer

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16 The Venn diagram shows the number of students in a class of 40 who study physics (P), mathematics (M) and geography (G). P M 2 4 11 9 5 3 6 G (a) Use set notation to describe the shaded region. ................................................. [1] (b) Find n (( P + G ) , M l ) . ................................................. [1] (c) A student is chosen at random from those studying geography. Find the probability that this student also studies physics or mathematics but not both. ................................................. [2]

Mark scheme: 16(a)  M  G   P  1 16(b) 22 1 16(c) 8 2 oe M1 for 23 k k 8 3  5 or or or c ≠ 1 23 3  9  5  6 c c or for 8 and 23 identified

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17 (a) Sketch the graph of y = sin x for 0° G x G 360° . y 1 O x 360 – 1 [2] (b) Solve the equation 3 sinx + 1 = 0 for 0° G x G 360° . x = ................. or x = ................... [3]

Mark scheme: 17(a) 2 Correct sketch to go through B1 for correct sine curve shape through the origin (0, 0), (180, 0) and (360, 0) 17(b) 199.5 or 199.47… 3 B2 for one correct and 340.5 or 340.52 to 340.53… 1 or M1 for sin x =  oe 3 If 0 scored SC1 for two reflex angles with sum of 540 or two non-reflex angles with sum of 180

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18 (a) y is directly proportional to the cube root of ( x + 1) . When x = 7 , y = 1. Find the value of y when x = 124 . y = ................................................ [3] (b) F is inversely proportional to the square of d. Explain what happens to F when d is halved. ..................................................................................................................................................... [1]

Mark scheme: 18(a) 2.5 3 M1 for y  k  3 x  1 M1 for y  theirk  3 124  1 18(b) multiplied by 4 oe 1

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4 x19 f ( )x = 7 x - 8 g ( )x = + 5 h ( )x = 2 + 1 x (a) Find f -1 ( )x . f -1 ( )x = ................................................ [2] 1 (b) Find the value of x when h ( )x = g e 3 o. x = ................................................ [2]

Mark scheme: 19(a) x 8 2 y 8 final answer M1 for x  7 y  8 or y  8  7 x or  x  7 7 7 19(b) 4 2 1 M1 for 4 ÷ + 5 oe or better 3

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20 Factorise completely. (a) 2m + 3p - 8km - 12kp ................................................. [2] (b) 5x 2 - 20y 2 ................................................. [3]

Mark scheme: 20(a)  2 m  3 p 1  4 k  final 2 B1 for 2 m  3 p  4 k  2 m  3 p  or better answer or 2 m 1  4 k   3 p 1  4 k  or correct answer seen and spoilt 20(b) 5  x  2 y  x  2 y  final 3 B2 for (5x – 10y)(x + 2y) or (x – 2y)(5x + 10y) or correct answer seen then spoilt answer or B1 for 5 x 2  4 y 2   or for  x  2 y  x  2 y 

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21 The nth term of a sequence is an 2 + bn - 4 . The first term is - 3 and the second term is 2. Find the value of a and the value of b. a = ................................................ b = ................................................ [5]

Mark scheme: 21 [a =] 2 5 M2 for correct method to find two simultaneous [b =] − 1 equations e.g. two from a  21 b 1 4 3 a  2 2  b  2  4  2 3a + b = 2 – – 3 or M1 for 1 correct equation M1 for correctly eliminating one variable for their simultaneous equations A1 for a = 2 A1 for b = − 1

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22 A D B NOT TO x SCALE y O 3 4 OA = x , OB = y and OD = x + y . 7 7 Calculate the ratio AD : DB. ........................ : ....................... [2]

Mark scheme: 22 4 : 3 oe 2 M1 for  4 4  3 3 AD  x  y oe or DB  x  y oe 7 7 7 7

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23 6.2 cm NOT TO SCALE l The diagram shows a solid metal shape made from a cone and a hemisphere, both with radius 6.2 cm. The total surface area of the solid shape is 600 cm2. Calculate the slant height, l, of the cone. [The surface area, A, of a sphere with radius r is A = 4rr 2 .] [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] l = ............................................ cm [4]

Mark scheme: 23 18.4 or 18.40… 4 1 2 600  4  6.2 2 M3 for oe 6.2  or M2 for 1 2 4  6.2   6.2 l 600 oe 2 600  4  6.2 2 or or better 6.2  1 2 or M1 for 4  6.2 or  6.2  l  2

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