Cambridge IGCSE Mathematics 0580 — 2025 May/June Paper 2 · Variant 3

1 The probability of picking a green pen from a box is 0.17 . Find the probability of not picking a green pen from the box. ................................................. [1]

Mark scheme: Question Answer Marks Partial Marks 1 0.83 oe 1

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2 A x° E 30° y° F 104° NOT TO SCALE D 130° B 70° C ABCD is a quadrilateral. CDE is a straight line. AFB is an isosceles triangle. Find the value of x and the value of y. x = ................................................ y = ................................................ [4]

Mark scheme: 2 x = 84 4 B2 for x = 84 y = 75 or M1 for CDA = 76 or 360 – 70 – 130 – (180 – 104) oe B2 for y = 75 or M1 for (180 – 30) ÷ 2

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3 E NOT TO SCALE D C M A B The diagram shows a pyramid ABCDE with a square base. M is the centre of the square base. E is vertically above M. (a) Write down the number of planes of symmetry of this pyramid. ................................................. [1] (b) Using two of the letters from A, B, C, D, E and M, complete the statement about the pyramid. The axis of rotational symmetry passes through the points ............... and ............... [1]

Mark scheme: 3(a) 4 1 3(b) E and M 1

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4 The number of ice creams sold increases as the temperature rises. What type of correlation does this statement describe? ................................................. [1]

Mark scheme: 4 Positive 1

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5 y 9 8 7 6 B 5 4 3 A 2 1 – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 7 8 9 10 11 x – 1 – 2 – 3 – 4 – 5 – 6 – 7 (a) Describe fully the single transformation that maps triangle A onto triangle B. ..................................................................................................................................................... ..................................................................................................................................................... [3] (b) On the grid, draw the image of (i) triangle A after a reflection in the line x =-1 [2] (ii) triangle A after a rotation 90° clockwise, centre (1, -2). [2]

Mark scheme: 5(a) Enlargement 3 B1 for each [Scale factor] 3 (1, 2) 5(b)(i) Triangle at (–6, 1), (–3, 3), (–4, 4) 2 B1 for reflection in x = k or in y = –1 5(b)(ii) Triangle at (4, –5), (6, –2), (7, –3) 2 B1 for correct 90° anticlockwise rotation about (1, –2) or correct orientation, wrong centre

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6 There are 15 giraffes in a group. The table gives information about the heights of the 15 giraffes. One giraffe has a height of 2.6 m No giraffe is shorter than 2.5 m The range of heights for the 15 giraffes is 2.3 m More than 3 giraffes have the same height The modal height for the giraffes is 3.9 m The stem-and-leaf diagram shows information about the height of 9 of these giraffes. 2 5 3 2 7 7 4 1 1 4 5 7 Key: 4|1 represents a giraffe height of 4.1 m Use the information in the table to complete the stem-and-leaf diagram for the group of 15 giraffes. [3]

Mark scheme: 6 3 B1 for each correct row 2 5 6 3 2 7 7 9 9 9 9 4 1 1 4 5 7 8

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7 Work out. 5 7 (a) - 6 12 Give your answer as a fraction in its simplest form. ................................................. [2] 1 8 (b) 1 ' 3 15 Give your answer as a mixed number in its simplest form. ................................................. [3]

Mark scheme: 7(a) 1 2 M1 for correct method to find common cao 4 10 7 denominator e.g. and , 12 12 60 42 and 72 72 7(b) 1 3 4 15 2 cao M2 for  2 3 8 20 8 or  oe with common 15 15 denominator 4 their k 15 or M1 for or for  , where 3 3 8 k > 3

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8 (a) Write 42 as a product of its prime factors. ................................................. [2] (b) Find the highest common factor (HCF) of 84 and 70. ................................................. [2]

Mark scheme: 8(a) 2 × 3 × 7 2 B1 for 2, 3, 7 or M1 for correct factor tree/diagram/table 8(b) 14 2 B1 for answer 2 or 7 or M1 for 2 × 7 as final answer or 84 = 2 × 2 × 3 × 7 and 70 = 2 × 5 × 7 or 2 correct factor trees or tables

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9 (a) Solve. 5x 2 = 12 - 17 x x = .................. or x = .................. [4] (b) ax 2 + a = b where a and b are integers. One solution of this equation is x = 6 . Write down the other solution. x = ................................................ [1]

Mark scheme: 9(a) 3 4 M1 for 5x2 + 17x – 12 [= 0] or 0.6 5 M2 for correct method to solve their and –4 ax2 + bx – c [= 0] e.g. factorising (5x – 3)(x + 4) [= 0], completing the square or using the formula or M1 for (5x + a)(x + b) where ab = –12 or 5b + a = 17 or correct partial factorisation e.g. 5x(x + 4) – 3(x + 4) or for 17 2 − 4 ( 5 )( −12 ) or better −17 + d −17 − d or or 2 ( 5 ) 2 ( 5 ) 9(b) –6 1

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10 Solve the simultaneous equations. 4x - 5y = 13 3x - 2y = 8 x = ................................................ y = ................................................ [4]

Mark scheme: 10 [x =] 2 4 M1 for correctly equating one set of [y =] –1 coefficients or for making x or y the subject of one equation M1 for correct method to eliminate one variable A1 for x = 2 A1 for y = –1 If M0 scored, SC1 for 2 values satisfying one of the original equations

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11 Angela picks a number at random from the numbers 1, 2 and 3. She then picks a number at random from the numbers 4, 5 and 6. She adds the two numbers to find the total. (a) Complete the table to show the possible outcomes. First number + 1 2 3 4 5 6 7 Second 5 number 6 [2] (b) Given that the total is odd, find the probability that one of the numbers Angela picks is 3. ................................................. [2]

Mark scheme: 11(a) 2 B1 for 4 correct First number + 1 2 3 4 5 6 7 Second number 5 6 7 8 6 7 8 9 11(b) 2 2 c oe B1 for answer where c < 5 5 5 2 or seen in working 5

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2 12 (a) v = e o - 3 Find 5v. f p [1] (b) H is the point (-3, 8) and K is the point (-4, 0). 7 HJ = e o - 2 Find JK . ................................................. [4]

Mark scheme: 12(a)  10  1    −15  12(b) 10 4 B3 for answer 10 OR M3 for (–4 – 4)2 + (0 – 6)2 oe or better OR  −4 4 M2 for  − oe  0 6  −1  or B1 for [J =] (4, 6) soi or    −8  M1 for (their –8)2 + (their –6)2 oe or better

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13 F A 60° 35° NOT TO SCALE E D B C A, B, C and D are points on a circle. EF is a tangent to the circle at A. AB is parallel to DC. (a) Find angle DCB, giving a geometrical reason. Angle DCB = ........................ because ................................................................................... .................................................................................................................................................. [2] (b) Find angle DBC. Angle DBC = ................................................ [2]

Mark scheme: 13(a) 120 2 B1 for 120 or a fully correct reason and opposite angles of a cyclic quadrilateral sum to 180° 13(b) 25 2 M1 for (180 – 120) – 35 or B1 for ABC = 60 or ABD = 35

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14 Find the lowest common multiple (LCM) of 15xy 3 and 18x 4 y . ................................................. [2]

Mark scheme: 14 90x4y3 final answer 2 B1 for two correct parts from: 90, x4 and y3 or for 3×5×x×y×y×y and 2×3×3×x×x×x×x×y or for correct answer seen then spoilt

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15 (a) Simplify. 27 + 12 ................................................. [2] 40 8 (b) = k , where k is an integer. 5 2 Find the value of k. k = ................................................ [2] (c) Rationalise the denominator. 1 3 - 5 ................................................. [2]

Mark scheme: 15(a) 5 3 2 B1 for 3 3 or 2 3 15(b) 16 2 8 4 2 M1 for 40  2 2 , 8  2 2 , or 2 40 8 2 or better 5 2 2 or B1 for 8 4 or 256 15(c) 3 + 5 2 1 3 + 5 M1 for  oe 4 3 − 5 3 + 5

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o o16 Write .0328 as a fraction in its simplest form. ................................................. [3]

Mark scheme: 16 65 3 65 cao B2 for a fraction equivalent to e.g. 198 198 325 990 3 28 or M1 for 328.28 − 3.28 oe or + 10 990 k oe or 990x = 325 or 990

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17 Solid A is mathematically similar to solid B. H cm NOT TO SCALE 7 cm Solid A Solid B The height of solid A is 7 cm and its surface area is 60 cm2. The surface area of solid B is 540 cm2. Calculate the height of solid B. ............................................ cm [3]

Mark scheme: 17 21 3 540 60 M2 for 7 × oe or 7 ÷ oe 60 540 540 60 or M1 for oe or oe or 60 540  7 2 60 oe  =  H  540

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18 Make t the subject of the formula. m ( 1 - t) 2 = pt t = ................................................ [4]

Mark scheme: 18 m 4 M1 for correctly clearing their fraction  t = final answer M1 for correct expansion 2 p + m M1 for correctly collecting their terms in t on one side and other terms on the other side, not dividing by 1 – t M1 for correct factorisation and division of their two-term expression in t To a maximum of 3 marks for an incorrect answer

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19 Simplify. 7x - x 2 49 - x 2 ................................................. [3]

Mark scheme: 19 x 3 B1 for x(7 – x) final answer B1 for (7 – x)(7 + x) 7 + x

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20 A NOT TO 40° B SCALE O The diagram shows a sector of a circle, centre O. The radius of the circle is 6 cm. Calculate the length of the major arc AB. Give your answer in its simplest form in terms of r. .............................................cm [3]

Mark scheme: 20 32 2 3 360 − 40 π or 10 π cao M2 for π 6 2 3 3 360 360 − 40 22 or  6 2 oe 360 7 40 or M1 for π 6 2 oe 360 or B1 for major sector angle = 320 soi or 12π

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21 (a) Differentiate x 3 - 3 x 2 + 1. ................................................. [2] (b) Find the coordinates of the turning points of the graph of y = x 3 - 3x 2 + 1. ( ...................... , ...................... ) ( ...................... , ...................... ) [4]

Mark scheme: 21(a) 3x2 – 6x or 3x(x – 2) 2 B1 for 3x2 – kx or kx2 – 6x or correct answer spoilt or SC1 for 3x2 – 6x + 1 21(b) (0, 1) 4 B3 for one correct coordinate or for two (2, –3) correct values of x following 3x2 – 6x oe seen in (b) or M2 for [3]x(x – 2) [= 0] or −− 6  ( −6 ) 2 −4 3 0 [= 0] oe 2  3 OR d y d y M1 for their = 0 or stating = 0 d x d x M1 for correct method to solve their quadratic

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22 f ( )x = 2x + 5 g ( )x = x - 4 h ( )x = 5 x (a) Find f ( 3 ) . ................................................. [1] (b) Find f -1 ( )x . f -1 ( )x = ................................................ [2] (c) Solve fg ( )x = 25 . x = ................................................ [3] (d) Find x when h -1 ( )x = 2 . x = ................................................ [2]

Mark scheme: 22(a) 11 1 22(b) x − 5 2 M1 for correct first step y − 5 2 x = 2y + 5 or y – 5 = 2x or = x 2 y 5 or = x + 2 2 22(c) 14 3 M2 for x – 4 = (25 – 5) ÷ 2 oe or better or 2x – 8 = 25 – 5 oe or better or M1 for 2(x – 4) + 5 = 25 22(d) 25 2 M1 for h(2) or 52

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23 (a) Write down the value of cos 90°. ................................................. [1] (b) x NOT TO SCALE n 30° 30° The diagram shows two different right-angled triangles joined by a common side. Find x in terms of n. x = ................................................ [5] Question 24 is on the next page.

Mark scheme: 23(a) 0 1 23(b) 4 n 4 n 3 5 B4 for correct explicit expression to find [x =] or oe x 3 3 2 n 2 n e.g. x = or x = oe cos30 3 2 OR M3 for correct explicit method to find x n  sin 30 with 2 trig functions e.g. x = cos30 oe or a correct implicit method with one trig 2n function e.g. cos 30 = oe x OR B2 for hypotenuse of lower triangle = 2n or M2 for any correct implicit method with 2 trig functions e.g. cos 30 = n  sin 30 oe x OR M1 for seeing any correct trig value from sin 30 or cos 60 = 0.5, cos 30 or 3 3 sin 60 = or tan 30 = oe 2 3 or for hypotenuse of lower triangle = n oe sin30

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24 (a) A is the point (a, 12) and B is the point (b, 27). (i) Find the y-coordinate of the midpoint of AB. ................................................. [1] (ii) The line AB has gradient 3. Find an expression for a in terms of b. a = ................................................ [3] (b) D is the point (22, 34) and E is the point (23, 39). D is the point on CE such that 2CE = 5DE. Find the coordinates of C. ( ...................... , ...................... ) [3]

Mark scheme: 24(a)(i) 19.5 1 24(a)(ii) [a =] b – 5 oe 3 27 − 1 2 M1 for 3 = oe b − a M1 for 3(b – a) = 27 – 12 or better 27 − 1 2 e.g. [a =] b – 3 24(b) 41 53 3 SC2 for answer (25.5, 51.5) or (20.5, 26.5) or ( , ) 2 2 51 103 ( , ) 2 2 OR 5 M2 for 23 − ( 23 − 22 ) 2 5 or 39 − ( 39 − 34 ) oe 2 or sketch 12.5 2.5  2.5   −2.5  or vector e.g.   or    12.5   −12.5  or M1 for 5 1 5 5 ( 23 − 22 ) or ( 39 − 34 ) 2 2 1  −1  or vector e.g.  or   5  −5 

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