Cambridge IGCSE Mathematics 0580 — 2024 May/June Paper 2 · Variant 1

1 y 8 7 6 A 5 4 3 C 2 1 B x – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 – 1 – 2 The diagram shows two sides of a parallelogram ABCD. Find the coordinates of point D. ( ...................... , ...................... ) [2]

Mark scheme: Question Answer Marks Partial Marks 1 (–3, 7) 2 B1 for correct diagram or correct coordinates for their point D or for   3, k  or  k , 7 

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2 Geetha has a box of toys. She picks a toy at random from the box. The probability that she picks a wooden toy is 0.6 . (a) Work out the probability that she does not pick a wooden toy. ................................................. [1] (b) The box contains three types of toys, wooden, plastic or metal. Type of toy Wooden Plastic Metal Number of toys 14 14 Probability 0.6 Complete the table. [2]

Mark scheme: 2(a) 0.4 oe 1 2(b) 42 2 B1 for 42 0.2 0.2 B1 for 0.2 and 0.2 If B0 scored SC1 for their two probabilities being half their (a)

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3 The table shows some information about two sequences. nth term 5th term Sequence A 60 – 4n Sequence B n2 – 300 (a) Complete the table. [2] (b) Find the smallest positive number in sequence B. ................................................. [2]

Mark scheme: 3(a) 40 2 B1 for each –275 3(b) 24 2 B1 for 324 or 289 or 300 or 17.3…

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4 Find the greatest odd number that is a factor of 140 and a factor of 210. ................................................. [2]

Mark scheme: 4 35 2 B1 for answer 5, 7 or 70 or M1 for 2 × 2 × 5 × 7 and 2 × 3 × 5 × 7 or two correct factor trees or tables or 5 × 7 × k seen

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5 Calculate. (a) 3 343 - 40.96 ................................................. [1] (b) (192 + 4 # 16 ) .125 ................................................. [1]

Mark scheme: 5(a) 3 1 0.6 or 5 5(b) 1024 1

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6 D A 40° C E B NOT TO SCALE x° The diagram shows 5 kites that are congruent to kite ABCD. Each kite is joined to the next kite along one edge. Angle DAB = 40° and DCE is a straight line. Find the value of x. x = ................................................ [3]

Mark scheme: 6 145 3 M1 for 180 ÷ 6 or any angle congruent to BCD = 30 M1 for (360 – 40 – their 30) ÷ 2 oe

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7 12.8 cm J L NOT TO SCALE 12.8 cm K The diagram shows a shape made from a triangle JKL and a semicircle with diameter JL. JKL is an isosceles right-angled triangle with JK = JL = 12.8 cm. (a) Calculate the area of this shape. ........................................... cm2 [3] (b) Calculate the perimeter of this shape. ............................................ cm [4]

Mark scheme: 7(a) 146 or 146.2 to 146.3 3 1 M1 for  12.8  12.8 2  1   12.8  2 M1 for       2   2  7(b) 51[.0] or 51.00 to 51.01… 4 1 M1 for  12.8 2 2 2 12.8 M2 for 12.8  12.8 or oe sin 45 or M1 for 12.8 2  12.8 2 oe 12.8 or sin 45  oe KL

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8 These are the first five terms of a sequence. 11 18 25 32 39 Find an expression for the nth term of the sequence. ................................................. [2]

Mark scheme: 8 7n + 4 oe final answer 2 B1 for 7n + j or kn + 4 k ≠ 0 , or 7n + 4 seen then spoilt

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9 The value of a car is $8000. Each year the value of the car decreases exponentially by 25%. Calculate the value of this car after 3 years. $ ................................................. [2]

Mark scheme: 9 3375 2 3  25  M1 for 8000  1  oe    100 

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10 Amir invests $1500 in an account. The account pays compound interest at a rate of r % per year. At the end of 8 years the value of his investment is $1656.73 . Find the value of r. r = ................................................ [3]

Mark scheme: 10 1.25 or 1.250… 3 1656.73 M2 for 8 oe 1500 or M1 for 1656.73  1500 k 8 oe for any k

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11 y 7 6 5 4 3 R 2 1 x – 1 0 1 2 3 4 5 6 7 – 1 Find the inequalities that define the unshaded region, R. .............................................................................................................................................................. [4]

Mark scheme: 11 y < x 4 B1 for y < x x < 6 B1 for x < 6 1 ⩽ y ⩽ 5 oe B2 for 1 ⩽ y ⩽ 5 or B1 for y ⩾ 1 or y ⩽ 5 If B0 scored, SC2 for y ⩽ x, x ⩽ 6 and 1 < y < 5 oe or SC1 for three correct from y = x, x = 6, y = 1 and y = 5

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12 Solve the simultaneous equations. You must show all your working. 3 x + 5y = 5 2 4x - 3y = 46 x = ................................................ y = ................................................ [4]

Mark scheme: 12 Correctly equating one set of M1 coefficients Correct method to eliminate one M1 variable x = 10, y = –2 A2 A1 for x = 10 A1 for y = –2 If M0 scored SC1 for 2 values satisfying one of the original equations.

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13 p° NOT TO SCALE 4m° 5m° (4m + 38)° The diagram shows a cyclic quadrilateral. Find the value of p. p = ................................................ [3]

Mark scheme: 13 62 3 B2 for m = 20 or M1 for 5m + 4m = 180 soi or p + 4m + 38 = 180 soi

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14 NOT TO 48° SCALE 9 cm The diagram shows a circle with radius 9 cm. Calculate the area of the shaded major sector. ........................................... cm2 [3]

Mark scheme: 14 221 or 220.5 to 220.6 3 360  48 2 M2 for   9 360 k 2 or M1 for   9 where k < 360 360 or B1 for 312

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15 Write .0146oo as a fraction in its simplest form. You must show all your working. ................................................. [3]

Mark scheme: 15 146.46  1.46 oe M1 29 A2 145 cao Al for oe 198 990 k If M0 scored SC1 for with insufficient 990 working.

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16 (a) In the Venn diagram, shade the region M l + N l. M N [1] (b) Find n ( B + ( Al , C )) . A B 33 16 3 10 18 4 20 9 C ................................................. [1]

Mark scheme: 16(a) 1 M N 16(b) 17 1

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17 B 6.7 cm 81° A NOT TO 5.9 cm SCALE C Calculate the area of triangle ABC. ........................................... cm2 [2]

Mark scheme: 17 19.5 or 19.52… 2 1 M1 for × 6.7 × 5.9 × sin 81 oe 2

More questions on Pythagoras’ theorem and trigonometry

18 y 8 7 6 5 4 3 2 1 x – 3 – 2.5 – 2 – 1.5 – 1 – 0.5 0 0.5 1 1.5 – 1 – 2 – 3 The diagram shows the graph of y = x 3 + 4x 2 - 2 for - 3 G x G 1.5 . By drawing a suitable straight line, solve the equation x 3 + 4x 2 - 2 = 2x for - 3 G x G 1.5 . x = ....................... or x = ....................... [3]

Mark scheme: 18 y = 2x ruled B1 x = –0.5 to –0.55 B2 B1 for –0.5 to –0.55 x = 0.85 to 0.9 B1 for 0.85 to 0.9

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19 Factorise completely. (a) 12m 2 - 75t 2 ................................................. [3] (b) xy + 15 + 3y + 5x ................................................. [2]

Mark scheme: 19(a) 3(2m + 5t)(2m – 5t) final answer 3 B2 for (6m + 15t)(2m – 5t) or (2m + 5t)(6m – 15t) or B1 for 3(4m2 – 25t2) or (2m + 5t)(2m – 5t) 19(b) (x + 3)(y + 5) final answer 2 B1 for x(y + 5) + 3(y + 5) or y(x + 3) + 5(x + 3)

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20 Solve the equation 8 sinx + 6 = 1 for 0c G x G 360 c . x = ....................... or x = ....................... [3]

Mark scheme: 20 218.7, 321.3 3 B2 for one correct 5 or M1 for sin x = – oe 8 If M1 or 0 scored, SC1 for two reflex angles with a sum of 540 or two non-reflex angles with a sum of 180

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21 9.1 cm E F 6.5 cm A B NOT TO SCALE H G 4 cm D C The diagram shows a cuboid. HD = 4 cm, EH = 6.5 cm and EF = 9.1 cm. Calculate the angle between CE and the base CDHG. ................................................. [4]

Mark scheme: 21 33.2 or 33.18… 4 6.5 M3 for tan  oe 4 2  9.12 or M2 for 4 2  9.12 oe or 4 2  9.12  6.5 2 oe or M1 for recognising the angle ECH

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22 Bag A and bag B each contain red counters and blue counters only. Stephan picks a counter at random from bag A and Jen picks a counter at random from bag B. The probability that Stephan picks a red counter is 0.4 . The probability that Stephan and Jen both pick a red counter is 0.25 . Find the probability that Stephan and Jen both pick a blue counter. ................................................. [4]

Mark scheme: 22 0.225 oe 4  0.25 oe M3 for  1 –  1 – 0.4   0.4  OR 0.25 M2 for 0.4 or M1 for 0.4 × p = 0.25 oe M1 for (1 –theirP(Jen red))×(1 – 0.4) oe

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