Cambridge IGCSE Mathematics 0580 — 2025 May/June Paper 2 · Variant 1

1 Simplify. 7 c - 5 d + c + 3d ................................................. [2]

Mark scheme: Question Answer Marks Partial Marks 1 8c – 2d final answer 2 B1 for answer 8c – kd or kc – 2d or for correct answer seen and spoilt

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2 x° NOT TO SCALE w° y° 158° 76° The diagram shows two parallel lines intersecting two straight lines. Find the values of w, x and y. w = ................................................ x = ................................................ y = ................................................ [4]

Mark scheme: 2 [w =] 158 4 B1 for w correct [x =] 76 B1 for x correct [y =] 82 B2FT for y = 158 – their x correctly evaluated or B1 for 22 (identified) or 82 in position vertically opposite to y or for y = 158 – their x

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3 Sally invests $1500 at 3% per year simple interest. Work out the total value of her investment at the end of 6 years. $ ................................................ [3]

Mark scheme: 3 1770 3 B2 for 270 1500 3 6 or M2 for 1500 + 100 1500  3  6  or M1 for 100

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4 Work out. 5 2 3 - # 6 3 8 ................................................. [3]

Mark scheme: 4 7 3 6 oe M1 for oe 12 24 M1 for correct use of common denominator in 5 6 20 6 subtraction – their , e.g. and oe 6 24 24 24 1 If 0 scored, SC1 for answer oe 16

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5 The interior angle of a regular polygon is 150°. Find the number of sides of this polygon. ................................................. [2]

Mark scheme: 5 12 2 360 180 ( n − 2 ) M1 for or for = 150 180 − 150 n

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6 y 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 x The line x + y = 7 is drawn on the grid. (a) On the grid, draw the line y = 2x + 1. [2] (b) Use your graph to solve these simultaneous equations. x + y = 7 y = 2x + 1 x = ................................................ y = ................................................ [1]

Mark scheme: 6(a) Correct line drawn 2 M1 for a line with gradient 2 or for a line with positive gradient and intercept at y = 1 6(b) [x =] 2, [y =] 5 1 FT intersection of their (a) with the given line

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7 Write the recurring decimal .026o as a fraction. Give your answer in its simplest form. ................................................. [3]

Mark scheme: 7 4 3 24 B2 for oe 15 90 or M1 for 26.66… – 2.66… oe or for 90x = 24 oe 2 6 or for + oe 10 90

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11 8 8 m = e o n = e o 5 - 3 (a) Find 2m - n. f p [2] 5 (b) The vector e o has a magnitude of 7. y Find the value of y. y = ................................................ [2]

Mark scheme: 8(a)  14  2  14   k   22    B1 for   or for   or for    13   k   13   10  8(b) 24 2 M1 for y + 52 = 72 or better

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9 The table shows some information about the marks scored by a group of students in a test. Test mark 4 5 8 Frequency 2 4 n The mean mark is 6. Work out the value of n. n = ................................................ [3]

Mark scheme: 9 4 nfww 3 M2 for 8 + 20 + 8n = 6(2 + 4 + n) or better 2 +4 4 +5 n  8 or M1 for [= 6] oe 2 + 4 + n or 8 + 20 + 8n or 6(2 + 4 + n)

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10 A D E NOT TO SCALE O 35° 40° C B A, B and C are three points on a circle, centre O. DE is a tangent to the circle at A. Angle ACO = 35° and angle BCO = 40° . Find (a) angle AOC Angle AOC = ................................................ [1] (b) angle ABC Angle ABC = ................................................ [1] (c) angle DAC Angle DAC = ................................................ [1] (d) angle OAB. Angle OAB = ................................................ [1]

Mark scheme: 10(a) 110 1 10(b) 55 1 FT their 110  2 10(c) 55 1 FT their (b) Provided their (b) < 90 10(d) 15 1 FT 70 – their (b) Provided their (b) < 70

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11 The diagram shows the graph of y = f ( x) and the point P ( - 2 , 11) . y 15 P 10 5 – 2 – 1 0 1 2 3 x – 5 – 10 The tangent from P touches the graph of y = f ( x) at the point (a, b). The values of a and b are integers. (a) By drawing this tangent, find the value of a and the value of b. a = ................... , b = ................... [2] (b) Find the equation of the tangent. Give your answer in the form y = mx + c . y = ................................................ [3]

Mark scheme: 11(a) For correct ruled tangent and 2 B1 for correct ruled tangent or both values [a =] 1, [b =] 2 correct without a correct tangent 11(b) [y =] 5 – 3x 3 3FT their (a) provided m < 0, c ≠ 0 B1 for (their –3)x + c rise or M1 for correct for their line run B1 for mx + c where c is the correct intercept for their graph, m ≠ 0

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12 The time spent on the internet by each of 120 adults is recorded for one day. The cumulative frequency diagram shows this information. 120 100 80 Cumulative 60 frequency 40 20 0 0 2 4 6 8 10 Time (hours) (a) Use the cumulative frequency diagram to find an estimate of the interquartile range. .............................................. h [2] (b) 70% of the adults spent less than k hours on the internet. Use the cumulative frequency diagram to find an estimate of the value of k. k = ................................................ [2]

Mark scheme: 12(a) 1.4 2 B1 for [UQ =] 5.6 or for [LQ =] 4.2 or SC1 for 0.7 12(b) 5.4 2 B1 for 84 seen

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13 4 cm 10 cm NOT TO h cm SCALE 6 cm Solid A Solid B The diagram shows solid A and solid B. Solid A is made from a hemisphere and a cone each with radius 6 cm. The cone has sloping edge 10 cm. Solid B is a cylinder with radius 4 cm and height h cm. The total surface area of solid A is equal to the total surface area of solid B. (a) Work out the value of h. h = ................................................ [5] (b) Work out the height of solid A. ............................................ cm [3]

Mark scheme: 13(a) 12.5 oe 5 M4 for 8[]h = 100[] OR M3 for π 6 10 + 2 π 6 2 = 2 π 4 2 + 2 π 4 h oe OR M1 for π 6 10 + 2  π  6 2 oe M1 2 π 4 2 + 2 π 4 h oe OR SC2 for answer 16.5 13(b) 14 3 2 2 M2 for 10 − 6 or M1 for 62 + x2 = 102

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14 f ( x) = 3x - 4 g ( x) = 4x + 1 (a) Find f ( - 2 ) . ................................................. [1] (b) Find f -1 ( x) . f -1 ( x) = .................................................. [2] (c) fg ( x) = ax + b Find the value of a, and the value of b. a = .................... b = ................... [2] (d) Simplify. 2 5 - f ( x) g ( x) Give your answer as a single fraction in terms of x. ................................................. [3]

Mark scheme: 14(a) –10 1 14(b) x + 4 2 M1 for correct first step oe y + 4 = 3x or x = 3y – 4 3 y 4 or = x − 3 3 14(c) a = 12, b = –1 2 B1 for either a or b correct or M1 for 3(4x + 1) – 4 14(d) 22 − 7 x 22 − 7 x 3 B1 for 2(4x + 1) – 5(3x – 4) oe or better isw or ( 3 x − 4 )( 4 x + 1) 12 x 2 − 13 x − 4 B1 for common denominator (3x – 4)(4x + 1) final answer oe isw

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15 (a) Expand and simplify. ( 2 - 5 )( 1 - 3 5 ) ................................................. [2] (b) Rationalise the denominator. Give your answer in its simplest form. 6 10 ................................................. [2]

Mark scheme: 15(a) 2 B1 for 3 correct terms from 17 – 7 5 final answer 2 – 6 5 – 5 + 3  5  5 oe 15(b) 3 10 2 6 10 cao M1 for  oe 5 10 10

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16 Expand and simplify. ( x + 4)( x - 3)( 3x + 2 ) ..................................................................... [3]

Mark scheme: 16 3x3 + 5x2 – 34x – 24 final answer 3 B2 for correct expansion unsimplified or simplified four-term expression of correct form with 3 terms correct or B1 for one pair of brackets expanded with at least 3 terms out of 4 correct

More questions on Algebraic manipulation

17 (a) A bag contains 6 red marbles, 3 green marbles and 1 blue marble. Two marbles are picked at random from the bag with replacement. Find the probability that both marbles are green. ................................................. [2] (b) Another bag contains 4 red counters and 2 yellow counters. Two counters are picked at random from this bag without replacement. (i) Complete the tree diagram. First Second counter counter Red 3 5 Red 4 6 Yellow ............ Red ............ 2 6 Yellow Yellow ............ [2] (ii) Find the probability that one of the two counters is yellow. ................................................. [3]

Mark scheme: 17(a) 9 2 3 3 oe M1 for  100 10 10 17(b)(i) 2 2 2 B1 for 5 5 4 5 1 5 17(b)(ii) 16 3 3FT their tree diagram dep on probabilities < 1 oe 30 4 2 2 4 M2FT for  +  6 5 6 5 4 2 2 4 or M1FT for  or for  6 5 6 5

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18 One day, Anya runs 12 km at a speed of x km/h. The next day she walks 10 km at a speed of ( x - 4 ) km/h. (a) Write down an expression, in terms of x, for the time she spends running. .............................................. h [1] (b) Write down an expression, in terms of x, for the time she spends walking. .............................................. h [1] (c) The time Anya spends walking is 1 hour more than the time she spends running. Write an equation in terms of x and show that it simplifies to x 2 - 2 x - 48 = 0 . [4] (d) Use factorisation to solve the equation x 2 - 2 x - 48 = 0 . x = .................. or x = .................. [3] (e) Find the time Anya spends running. .............................................. h [1]

Mark scheme: 18(a) 12 1 x 18(b) 10 1 x − 4 18(c) 10 12 M1 their – their = 1 oe x − 4 x 10x – 12x + 48 = x2 – 4x M2 Correctly multiplying their brackets and clearing algebraic fractions  12  e.g. ( x − 4 )  + 1  = 10  x  48 leading to 12 − + x − 4 = 10 and then x 12 x − 48 + x 2 − 4 x = 10 x or M1 for correctly clearing, or correctly collecting into a single fraction, two fractions both with different algebraic denominators e.g. 10x – 12(x – 4) = x(x – 4) or 10 x − 12 ( x − 4 ) [= 1] x ( x − 4 ) Leading to 0 = x2 – 2x – 48 A1 With no errors or omissions seen, dep on M3 18(d) (x + 6)(x – 8) M2 M1 for x(x – 8) + 6(x – 8) or x(x + 6) – 8(x + 6) or (x + a)(x + b) where ab = –48 or a + b = –2 –6, 8 B1 18(e) 1 1 12 1.5 or 1 FT 2 their 8

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- 2319 Find the value of 27 . ................................................. [2]

Mark scheme: 19 1 2 1 1 M1 for 9–1 or for or 9 3 2 3 729

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20 NOT TO 6 cm SCALE 30° x cm Find the exact value of x. x = ................................................ [4]

Mark scheme: 20 4 1 3 6 3 oe B1 for tan30 = or 3 3 3 1 or for sin60 = and sin30 = 2 2 6 6sin ( 60 ) M2 for or tan30 sin ( 30 ) 6 sin60 sin30 or M1 for = tan30 or = x x 6

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21 A B n NOT TO P 2m SCALE O C OABC is a rhombus and O is the origin. The diagonals of the rhombus intersect at P. OP = 2 m and AP = n . (a) Find, in terms of m and n, in its simplest form (i) OA OA = ................................................ [1] (ii) OC. OC = ................................................ [1] (b) D is the point such that AD = 10m - 3n . Show that OADC is a trapezium. [3]

Mark scheme: 21(a)(i) 2m – n 1 21(a)(ii) 2m + n 1 21(b) CD = 10m – 5n M2 Allow M2 for equivalents CD = –2n + 10m – 3n or CD = − ( 2m + n ) + 2m − n + 10m − 3n For M2, FT their (a) e.g. CD = their CO + their OA + 10m − 3n or M1 for correct route for CD using the lines of the diagram with AD e.g. CD = CA + AD oe CD = 5 OA A1 Dependent on M2 leading to CD is parallel to OA [ OACD is a trapezium]

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22 A curve has equation y = x n + qx 2 + 9x . dy 2 = 3 x - 12 x + 9 dx (a) Find the value of n, and the value of q. n = .................... q = ................... [2] (b) Work out the coordinates of the turning points of the curve. (............. , .............) and (............. , .............) [4]

Mark scheme: 22(a) [n =] 3, [q =] – 6 2 B1 for each correct value 22(b) (1, 4) and (3, 0) 4 B3 for (1, 4) or (3, 0) or for two correct values of x or M2 for [3](x – 1)(x – 3) [ = 0] oe −−( 12 )  ( −12 ) 2 −4 3 9 or x = oe 2  3 d y or M1 for writing = 0 d x or for 3 x 2 − 12 x + 9 = 0

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23 Simplify. 2x 2 + 10 x x 2 - 25 ................................................. [3]

Mark scheme: 23 2 x 3 B1 for 2x(x + 5) final answer x − 5 B1 for (x – 5)(x + 5)

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