Cambridge IGCSE Mathematics 0580 — 2025 May/June Paper 4 · Variant 3

1 Find the median of these numbers. 5 11 11 13 17 21 ................................................. [1]

Mark scheme: Question Answer Marks Partial Marks 1 12 1

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2 In triangle ABC, AC = 21 m and BC = 15.9 m. Using a ruler and compasses only, complete the scale drawing of the triangle. Use a scale of 1 cm to represent 3 m. The side AB is drawn for you. A B Scale : 1 cm to 3 m [3]

Mark scheme: 2 Correct triangle constructed with 3 B2 for correct triangle with no incorrect arcs or AC = 7 cm and BC = 5.3 cm for a triangle with reversed sides with correct and intersecting arcs arcs or B1 for any triangle with a side of 7 cm or a side of 5.3 cm or for 7[cm] or 5.3[cm] soi

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3 2 15 23 144 - 2 0.8 5 From this list, write down (a) a natural number ................................................. [1] (b) an irrational number. ................................................. [1]

Mark scheme: 3(a) 23 or 144 1 Accept 12 for 144 3(b) 1 15

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4 The diagram shows a pentagon made from two congruent trapeziums. 11.4 cm NOT TO x cm SCALE 14 cm The area of the pentagon is 130.2 cm2. Calculate the value of x. x = ................................................ [3]

Mark scheme: 4 1 3 1 14 130.2 7.2[0] or 7 M2 for × × ( x + 11.4 ) = oe or 5 2 2 2 better 1 14 or M1 for × × ( x + 11.4 ) 2 2 ALTERNATIVE METHOD 1 1 M2 for 14x + × (11.4 – x) × 14 = 130.2 oe 2 1 or M1 for 14x + × (11.4 – x) × 14 2 ALTERNATIVE METHOD 2 M2 for 1 14 14 × 11.4 – 2 × × × (11.4 – x) = 130.2 oe 2 2 or M1 for 1 14 14 × 11.4 – 2 × × × (11.4 – x) 2 2 If 0 scored, SC1 for height of obtuse-angled triangle = 4.2 (condone their choice of variables)

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5 (a) Convert 780 cm2 into m2. ............................................ m2 [1] (b) Convert 0.037 m3 into cm3. .......................................... cm3 [1]

Mark scheme: 5(a) 0.078[0] oe 1 5(b) 37 000 1

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6 One titanium atom has a mass of .795 # 10 -23 grams. Calculate the number of titanium atoms in 1 kg of titanium. Give your answer in standard form. ................................................. [2]

Mark scheme: 6 1.26 × 1025 2 1000 M1 for oe or 1.257 × 1025 to 1.258 × 1025 23 7.95 × 10− or B1 for 1.26 × 10k or 1.257 × 10k to 1.258 × 10k

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7 % = {x: x is an integer and 1 G x G 12 } E = {even numbers} M = {multiples of 3} (a) Find n(M ). ................................................. [1] (b) Write down the set E + M . { ............................................... } [1] (c) y ! ( E , M ) l Write down a possible value of y. ................................................. [1]

Mark scheme: 7(a) 4 1 7(b) 6, 12 1 7(c) 1, 5, 7 or 11 1

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8 Factorise. (a) 28xy - 12x ................................................. [2] (b) y - 6x + 2xy - 3 ................................................. [2]

Mark scheme: 8(a) 4x(7y – 3) final answer 2 B1 for 4x(7y – 3) seen then spoilt or final answers 4(7xy – 3x) or x(28y – 12) or 2x(14y – 6) 8(b) (2x + 1)(y – 3) final answer 2 B1 for 2x(y – 3) + y – 3 or 2x(y – 3) + 1(y – 3) or y(2x + 1) – 3(2x + 1) or correct answer seen then spoilt

More questions on Algebraic manipulation

9 (a) Simplify. 5x 5 y # 7x 3 y 2 ................................................. [2] (b) 7 n = 5 7 Find the value of n. n = ................................................ [1]

Mark scheme: 9(a) 35x8y3 final answer 2 B1 for two correct parts from: 35, x8 and y3 or for correct answer seen then spoilt 9(b) 1 1 oe 5

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10 Expand and simplify. (a) 2 x - x ( 5 - x 2 ) ................................................. [2] (b) ( x + 4)( x - 5)( x + 3) ................................................................................ [3]

Mark scheme: 10(a) x3 – 3x final answer 2 B1 for x3 or – 3x in the final answer or for 2x – 5x + x3 or for x3 – 3x seen then spoilt 10(b) x3 + 2x2 – 23x – 60 final answer 3 B2 for correct expansion of three brackets unsimplified or for simplified four-term expression of correct form with three terms correct or B1 for correct expansion of two brackets with at least three terms out of four correct

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11 The table shows three sequences. 1st 2nd 3rd 4th 5th 6th nth term term term term term term term Sequence 28 22 16 10 4 A Sequence 1 2 3 4 1 B 6 7 8 9 2 Sequence 1 1 1 2 4 C 4 2 Complete the table. [8]

Mark scheme: 11 8 Mark the table first 6th term nth term B1 for –2 –2 34 – 6n B2 for 34 – 6n or B1 for answer –6n + c or 34 – kn (k ≠ 0) 6 n or for a correct answer seen and spoilt 11 n + 5 –3 6 8 2n B1 for 11 n B1 for n + 5 B1 for 8 B2 for 2n –3 or B1 for answer 2n – k (k can be 0) or for a correct answer seen and spoilt

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10 40012 = 10k 10 220 # 10 80 Find the value of k. k = ................................................ [3]

Mark scheme: 12 50 nfww 3 100 10 200 B2 for 10 or 150 or answer k = 1050 10 or for 10100 = 102k or M1 for any correct first step from: 10 220 × 1080 = 10 300 , 10 400 ÷ 10 220 = 10180 , 10 400 ÷ 1080 = 10 320 , 10 400 = 10200, 10 220 = 10110, 1080 = 1040 If 0 scored, SC1 for correct manipulation of 400, 220 and 80 to give 100 or for correctly dealing with the square root of an expression involving power(s) of 10 to find an explicit value for k 10 20 or for seen 1010 3

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13 Line L has the equation y = 2 ( x - 5) . (a) Find the coordinates of the point where line L intersects the y-axis. ( ....................... , ....................... ) [1] (b) Find the x-coordinate of the point where line L intersects the line y = 19 . x = ................................................ [2] (c) Line P is perpendicular to line L. Line P passes through the point (2, 3). Find the equation of line P. Give your answer in the form y = mx + c . y = ................................................ [3]

Mark scheme: 13(a) (0, –10) 1 13(b) 1 2 19 14.5 or 14 M1 for = x – 5 2 2 or for 19 = 2x – 10 or better 13(c) 1 3 B2 for a correct answer in the wrong form y = − x + 4 or y = –0.5x + 4 final 2 OR answer 1 1 M1 for y = − x + c or gradient = − soi 2 2 M1 for correct substitution of (2, 3) 1 into y = (their − )x + c or 2 1 into y – k = (their − )(x – h) 2

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14 Jess invests $1400 in an account. The account pays compound interest at a rate of r % per year. At the end of 4 years the value of her investment is $1523.15 . Find the value of r. r = ................................................ [3]

Mark scheme: 14 2.128 to 2.13[0…] 3 1523.15 4 oe M2 for 1400 or M1 for 1523.15 = 1400 ( k ) 4 oe for any k

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15 m is inversely proportional to the square root of ( t + 2 ) . m = 0.5 when t = 23 . Find m when t = 98 . m = ................................................ [3]

Mark scheme: 15 1 3 M2 for m 98 + 2 = 0.5 23 + 2 0.25 or 4 OR k M1 for m = t + 2 their k M1 for m = 98 + 2

More questions on Ratio and proportion

16 y 7 6 5 4 3 2 1 – 4 – 3 – 2 – 1 0 1 2 3 4 5 x – 1 – 2 By shading the unwanted regions of the grid, draw and label the region R which satisfies these three inequalities. y 1 3 x G 2 y 1 x + 3 [5]

Mark scheme: 16 Correct lines and region indicated 5 B1 for each correct line B2 for correct region 0 0 If R labelled, take the lines surrounding R to be the boundaries of their region and ignore 0 shading. 1 or B1dep for region identified satisfying 2 of 1 these inequalities R 1 If y = 3 and x = 2 interchanged, SC1 for a region satisfying y < x + 3, y < 2 and x < 3

More questions on Inequalities

17 In the Venn diagram, shade the region ( A , B ) + C l. % A B C [1]

Mark scheme: 17 1 A B C

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18 8 cm NOT TO SCALE 4.5 cm 13.2 cm The diagram shows a solid cuboid with sides of length 4.5 cm, 8 cm and 13.2 cm. (a) Calculate the volume of the cuboid. .......................................... cm3 [1] (b) Calculate the total surface area of the cuboid. .......................................... cm2 [3] (c) B 8 cm NOT TO SCALE 4.5 cm A 13.2 cm Calculate the angle between AB and the horizontal base of the cuboid. ................................................. [4]

Mark scheme: 18(a) 475.2 1 18(b) 402 3 M2 for [2 ×] (13.2 × 4.5 + 13.2 × 8 + 4.5 × 8) or M1 for [2 ×] 13.2 × 4.5 or [2 ×] 13.2 × 8 or [2 ×] 4.5 × 8 18(c) 29.8 or 29.83 to 29.84… 4 8 M3 for tan = oe 13.2 2 + 4.5 2 or M2 for 13.22 + 4.52 or for 13.22 + 4.52 + 82 or M1 for identifying correct angle

More questions on Surface area and volume

19 The table shows some information about the heights of 200 plants. Cumulative Height (h cm) frequency h G 30 6 h G 40 42 h G 50 76 h G 60 170 h G 70 190 h G 80 200 (a) Draw a cumulative frequency diagram to show this information. 200 160 120 Cumulative frequency 80 40 0 20 30 40 50 60 70 80 h Height (cm) [3] (b) Find an estimate of the 75th percentile. .............................................cm [1]

Mark scheme: 19(a) A correct cumulative frequency 3 B1 for correct horizontal placement for 6 plots curve through [(20, 0)], (30, 6) (40, B1 for correct vertical placement for 6 plots 42), (50, 76), (60, 170), (70, 190) B1FT dep on at least B1 for reasonable and (80, 200) increasing curve or polygon through their 6 points If 0 scored, SC1FT for 5 out of 6 points correctly plotted 19(b) Reading at 150 FT their increasing 1 curve

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20 Stephan has two bags, A and B, each containing red sweets and yellow sweets only. He picks two sweets at random, one from bag A and the other from bag B. The probability that Stephan picks a red sweet from bag A is 3. 5 2 The probability that Stephan picks a red sweet from bag A and a red sweet from bag B is . 15 (a) Find the probability that Stephan picks a red sweet from bag B. ................................................. [2] (b) Find the probability that Stephan picks a yellow sweet from bag A and a yellow sweet from bag B. ................................................. [3]

Mark scheme: 20(a) 2 2 3 2 oe M1 for × p = oe or better 9 5 15 20(b) 14 3 2 2 oe FT (1 – their ) × correctly evaluated 45 9 5 2 3 M2 for (1 – their ) × (1 – ) oe 9 5 2 3 or M1 for 1 – their or 1 – oe 9 5

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21 A NOT TO SCALE N P C B 12 cm The diagram shows an equilateral triangle ABC with all sides of length 12 cm. ANP is a sector of a circle, centre A. N lies on AC such that AN : NC = 2 : 1. P lies on AB such that AP : PB = 2 : 1. Calculate the area of the shaded region. .......................................... cm2 [4]

Mark scheme: 21 28.8 or 28.83 to 28.84… 4 M3 for 1 60  2  2 × 12 × 12 × sin60 – × π ×  × 12  2 360  3  OR k  2  2 M1 for × π ×  × 12  oe 360  3  k 1 where k < 90 or < 360 4 1 M1 for × 12 × 12 × sink oe where k < 90 2 If 0 scored, SC1 for [radius =] 8 and [∠A =] 60 seen

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22 A metal cube has mass 14.2 g, correct to 1 decimal place. (a) Find the lower bound of the mass of the cube. ............................................... g [1] (b) The side length of the cube is 1.3 cm, correct to 1 decimal place. Calculate the upper bound of the density of the metal. [Density = mass ' volume ] .......................................g /cm3 [3]

Mark scheme: 22(a) 14.15 cao 1 22(b) 7.296 cao nfww 3 14.2 to 14.3 14.3 − 0.05 M2 for 3 or 3 (1.3 − 0.05 ) (1.2 to 1.3 ) or M1 for 14.25 or 1.35 or 1.25 seen

More questions on Limits of accuracy

23 4.2 cm NOT TO 4.5 cm SCALE 7.5 cm The diagram shows a frustum made by removing a small cone from a large cone. The height of the small cone is 7.5 cm. The height of the frustum is 4.5 cm. The radius of the large cone is 4.2 cm. Work out the volume of the frustum. .......................................... cm3 [4]

Mark scheme: 23 168 or 167.5 to 167.6 4 M3 for π  2  7.5  2  ×  4.2 × ( 4.5 + 7.5 ) −  4.2 ×  × 7.5  3   4.5 + 7.5   oe or M2 for 3  7.5  1 2 × × π × 4.2 × ( 4.5 + 7.5 )    7.5 + 4.5  3 1  7.5  2 or for × π ×  4.2 ×  × 7.5 3  4.5 + 7.5  1 2 or M1 for × π × 4.2 × ( 4.5 + 7.5 ) 3 OR 1 2 M1 for × π × 4.2 × ( 4.5 + 7.5 ) 3 4.2 radius M1 for = oe or better 4.5 + 7.5 7.5 1 2 M1 for × π × ( their 2.625 ) × 7.5 3

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24 The histogram shows information about the distance some people travel to work. 4 3 Frequency density 2 1 0 0 10 20 30 40 50 d Distance (km) (a) Use the histogram to complete the table. Distance (d km) Number of people 0 1 d G 5 5 1 d G 20 20 1 d G 50 [3] (b) The people who travel a distance greater than 35 km to work have a car allowance. Calculate an estimate of the number of these people who have a car allowance. ................................................. [1]

Mark scheme: 24(a) 14 3 B1 for each correct answer 37 24 24(b) 6 1

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25 (a) Sketch the graph of y = sin x for 0° G x G 360° . y 1 0 360° x – 1 [2] (b) Solve the equation 2 + 5 sinx = 1 for 0° G x G 360° . x = ............................ and x = ............................ [3]

Mark scheme: 25(a) Correct sketch to go through 2 B1 for correct sine curve shape through the (0, 0), (180, 0) and (360, 0) origin (minimum one cycle) 25(b) 191.5, 348.5 and no others 3 B2 for one correct 1 or M1 for sin x = – oe 5 If M1 or 0 scored, SC1 for two reflex angles with a sum that rounds to 540.0 or two non- reflex angles with a sum that rounds to 180.0

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26 Y 5a - 2b P NOT TO SCALE 12b T O X 3a The diagram shows a quadrilateral OXPY with diagonals meeting at T. OX = 3a and OY = 12b . (a) Find XY in terms of a and b. ................................................. [1] (b) XT : TY = 1 : 2 and YP = 5a - 2b . Find the ratio OT : TP. ......................... : ...................... [4]

Mark scheme: 26(a) –3a + 12b oe 1 26(b) 2 : 3 oe nfww 4 B3 for any two of  • OP = 5a + 10b  • OT = 2a + 4b  • TP = 3a + 6b OR M2 for any two of  • OP = 5a + 10b  1 • OT = 3a + their (a) 3  2 • TP = their (a) + 5a – 2b 3 or better OR M1 for any one of  • OP = 12b + 5a – 2b  1 • OT = 3a + their (a) 3  2 • TP = their (a) + 5a – 2b 3 or better 1 1 or for ± (–3a + 12b) or ± their (a) 3 3 2 2 or for ± (–3a + 12b) or ± their (a) 3 3

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