Cambridge IGCSE Mathematics 0580 — 2025 Feb/March Paper 4 · Variant 2

1 Write down a factor of 28 that is a prime number. ................................................. [1]

Mark scheme: Question Answer Marks Partial Marks 1 2 or 7 1

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2 Simplify. 4y 2 + 3 y - y 2 + 2y ................................................. [2]

Mark scheme: 2 3y2 + 5y or y(3y + 5) final answer 2 B1 for 3y2 or 5y correct in final answer or for correct answer seen then spoilt

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3 Shade two more small squares to make a pattern with two lines of symmetry. [1] 3

Mark scheme: 3 1

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20.24 - 3 30 4 Calculate . 6.5 Give your answer correct to 1 decimal place. ................................................. [2]

Mark scheme: 4 2.6 cao 2 B1 for 2.64 or 2.635 to 2.636 If 0 scored, SC1 for their more accurate value seen rounded correctly to 1 decimal place

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5 y D A (2, 5) NOT TO SCALE C B (4, 1.5) 0 x A pattern is formed by 3 congruent rectangles. Each rectangle is a rotation of 90° around one vertex of the rectangle next to it. The point A has coordinates (2, 5). The point B has coordinates (4, 1.5). Work out the coordinates of point C and point D. C ( ...................... , ...................... ) D ( ...................... , ...................... ) [3]

Mark scheme: 5 [C = ] (7.5, 3) 3 B2 for [C = ] (7.5, 3) or [D = ] (9.5, 6.5) [D = ] (9.5, 6.5) or for [C = ] (7.5, k) and [D = ] (9.5, m) or for [C = ] (q, 3), q ≠ 8 and [D = ] (s, 6.5) or B1 for [C = ] (7.5, p) or (q, 3), q ≠ 8 or [D = ] (9.5, r) or (s, 6.5)

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6 Each week Nisha is paid $12 per hour for the first 40 hours that she works. She is paid 30% more per hour for any extra hours that she works. One week Nisha works for 45.5 hours. Calculate how much she is paid that week. $ ................................................ [3]

Mark scheme: 6 565.8[0] 3  30  M2 for 12 × 40 + 5.5 × 12 ×  1 +  oe  100  or M1 for 12 × 40 oe or 12 × 45.5  30  or [5.5 ×] 12 ×  1 +  oe  100   30  or [12 ×] 5.5 ×  1 +  oe  100  or 12 × 0.3 × 5.5

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7 The table shows the age and mass of each of 10 babies. Age (weeks) 9 12 15 2 5 6 9 7 1 11 Mass (kg) 4.3 5.8 6.2 3.0 3.9 4.0 4.6 4.5 2.5 5.3 (a) 7 6 5 Mass (kg) 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Age (weeks) Complete the scatter diagram. The first six points have been plotted for you. [2] (b) What type of correlation is shown in the scatter diagram? ................................................. [1] (c) On the scatter diagram, draw a line of best fit. [1] (d) Use your line of best fit to find an estimate of the mass of a 14-week old baby. ............................................ kg [1]

Mark scheme: 7(a) 4 points correctly plotted 2 B1 for 2 points correctly placed 7(b) Positive 1 7(c) Ruled line of best fit 1 7(d) FT reading at 14 weeks from their 1 straight line of best fit with positive gradient

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8 The scale drawing shows the positions of boat A and boat B. The scale is 1 cm represents 0.5 km. North A B Scale: 1 cm to 0.5 km (a) Find the actual distance between boat A and boat B. ........................................... km [2] (b) Lighthouse L lies to the east of boats A and B. L is 4.4 km from boat A and 3.3 km from boat B. Using a ruler and compasses only, construct and label the position of L. [3]

Mark scheme: 8(a) 2.1 2 B1 for 4.2 [cm] or M1 for answer their written measurement × 0.5 correctly evaluated 8(b) Correct construction for position of L 3 B2 for correct position of L indicated with with intersecting arcs 8.8 cm from A no/incorrect arcs and 6.6 cm from B or for completely correct with intersecting arcs but L to the west of A and B or B1 for 8.8 cm from A or 6.6 cm from B soi If 0 scored, SC1 for fully correct with arcs but distances reversed 6.6 cm from A and 8.8 cm from B or for position of L to the east with intersecting arcs 4.4 cm from A and 3.3 cm from B

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9 (a) Write 0.007 09 in standard form. ................................................. [1] 2 (b) Work out `4 # 104j . Give your answer in standard form. ................................................. [2]

Mark scheme: 9(a) 7.09 × 10–3 1 9(b) 1.6 ×109 2 B1 for 16 × 108 oe

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10 y 7 6 5 4 A 3 2 1 x -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 -1 -2 B -3 -4 -5 (a) Draw the image of (i) triangle A after a reflection in the line y = 1 [2] - 6 (ii) triangle A after a translation by the vector b l . [2] 1 (b) Describe fully the single transformation that maps triangle A onto triangle B. ..................................................................................................................................................... ..................................................................................................................................................... [3]

Mark scheme: 10(a)(i) Triangle drawn at (2, 0), (1, –1) 2 B1 for reflection in y = k and (3, –3) or for reflection in x = 1 10(a)(ii) Triangle drawn at (–4, 3), (–5, 4) and 2  −6   k  B1 for translation by or by (–3, 6)      k   1  10(b) Rotation 3 B1 for each 90° anticlockwise oe [centre] (3, –1)

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11 (a) Midhil invests $1500 at a rate of 4.2% per year compound interest. Calculate the value of the investment at the end of 5 years. $ ................................................. [2] (b) Hitanshi invests some money at a rate of x% per year compound interest. At the end of 11 years the value of the investment has doubled. Calculate the value of x. x = ................................................ [3]

Mark scheme: 11(a) 1842.59 2 5  4.2  M1 for 1500 ×  1 +  oe  100  11(b) 6.5[0…] 3 1 M2 for 11 2 or 2 11 or M1 for [k×]y11 = 2[×k] oe

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12 A cone has sloping edge 12 cm and base radius 5 cm. Calculate the total surface area of the cone. ......................................... cm2 [2]

Mark scheme: 12 267 or 267.0 to 267.1 2 M1 for π  12 +5 π  52

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13 The table shows the first 5 terms of sequences A, B and C. 1st term 2nd term 3rd term 4th term 5th term nth term Sequence A 5 12 31 68 129 10 9 8 7 6 Sequence B 3 4 5 6 7 Sequence C 4 8 16 32 64 Complete the table to show the nth term of each sequence. [6]

Mark scheme: 13 n3 + 4 oe final answer 6 B2 for n3 + 4 oe final answer or B1 for any cubic expression and or for 3rd diff = 6 with at least two shown or for correct answer seen then spoilt 11 − n oe final answer 11 − n n + 2 B2 for oe final answer n + 2 and or B1 for fraction with numerator or denominator correct or for correct answer 2n + 1 oe final answer seen then spoilt B2 for 2n + 1 oe final answer or B1 for answer of form [a ×]2n [+ b] b − n  1  or [a ×]    2  or for correct answer seen then spoilt

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14 f ( x) = 5 - 4x (a) Find f `– 3j. ................................................. [1] (b) Find f `3 - 2 xj. Give your answer in its simplest form. ................................................. [2] (c) Find f – 1 `xj. f – 1 `xj = ................................................ [2]

Mark scheme: 14(a) 17 1 14(b) 8x – 7 final answer nfww 2 M1 for 5 – 4(3 – 2x) oe or better 14(c) 5 −x 2 y 5 oe final answer M1 for x = 5 – 4y or y – 5 = –4x or = − x 4 4 4 or better

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15 Virat records the height of each of 80 sunflowers. The results are shown in the table. Height (h m) 1.2 1 h # 1 .5 1.5 1 h # 1 .6 1 .6 1 h # 1 .7 1.7 1 h # 1.9 Frequency 12 20 34 14 (a) Calculate an estimate of the mean height. ............................................. m [4] (b) Draw a histogram to show the information in the table. 400 300 Frequency density 200 100 0 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 h Height (m) [3]

Mark scheme: 15(a) 1.606 25 4 M1 for mid-points soi (1.35, 1.55, 1.65, 1.8 ) M1 for use of fm with m in correct interval including both boundaries M1 dep (dep on 2nd M1) for fm  80 15(b) Correct histogram 3 B2 for 3 correct blocks or B1 for 2 correct blocks If 0 scored, SC1 for 3 correct freq densities soi (40, 200, 340, 70)

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16 The graph shows the speed of a cyclist during a journey of 30 seconds. 8 6 Speed (m/s) 4 2 0 0 5 10 15 20 25 30 Time (seconds) (a) Write down the acceleration of the cyclist between 15 seconds and 25 seconds. ........................................ m/s2 [1] (b) By drawing a tangent, find an estimate for the acceleration of the cyclist at 7.5 seconds. ........................................ m/s2 [2] (c) Work out the average speed of the cyclist between 15 seconds and 30 seconds. ........................................... m/s [3]

Mark scheme: 16(a) 0 1 16(b) Ruled tangent to curve at time = 7.5 B1 seconds 0.25 to 0.4 B1 Dep on tangent correct or close attempt 16(c) 6.666 to 6.733… 3 (4.75 to 5)  8 M2 for (10 to 10.25) × 8 + oe 2 or M1 for attempt at one relevant area under graph for time between 15 and 30 seconds

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17 North NOT TO A SCALE 65 m B 95 m 38° C The diagram shows three points A, B and C on horizontal ground. The bearing of C from A is 145° and angle ACB = 38°. AC = 65 m and BC = 95 m. (a) Find the bearing of B from C. ................................................. [2] (b) Show that AB = 59.3 m, correct to 1 decimal place. [3] (c) Angle BAC is obtuse. Work out the bearing of B from A. ................................................. [4]

Mark scheme: 17(a) 287 2 M1 for North line at C with angle 35 or 145 marked on diagram at C isw or for 360 – 35 – 38 oe 17(b) M2 M1 for 652 + 952 −2 65  95cos38 652 + 952 −2 65  95cos38 A1 for 3518[.0] to 3518.1 59.31… A1 17(c) 244.4 to 244.6 4 B3 for [BAC =] 99.5 or 99.6 or 99.49 to 99.58… isw or for answer 225.4 to 225.5… OR 95sin38 M2 for [sin A = ] oe 59.3 59.32 + 65 2 − 95 2 or [cos A = ] 2  59.3  65 59.3 95 or M1 for = oe sin38 sin A or 952 = 59.32 + 652 −2 59.3  65cosA oe M1dep for 145 + their BAC leading to answer

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18 (a) Factorise. 18a 2 - 98 ................................................. [2] (b) Expand and simplify. `x + 4j`2x - 1j`x - 2j ................................................. [3]

Mark scheme: 18(a) 2(3a – 7)(3a + 7) final answer 2 M1 for 2(9a2– 49) or (6a – 14)(3a + 7) or (6a + 14)(3a – 7) isw or for correct answer seen and spoilt 18(b) 2x3 + 3x2 –18x + 8 final answer 3 B2 for correct expansion unsimplified or for answer simplified 4 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

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19 Solve the equation 2 + 5 cos x = 0 for 0° # x # 360°. x = .................or x = ................... [3]

Mark scheme: 19 113.6 and 246.4 3 B2 for one correct angle as answer or M1 for cos x = –0.4 oe If M1 or 0 scored, SC1 for 2 angles that add and round to 360.0

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20 A piece of metal has volume 1240 cm3, correct to the nearest 20 cm3. The mass of the piece of metal is 7800 g, correct to the nearest 100 g. Calculate the lower bound of the density of the metal. [Density = mass ÷ volume.] ...................................... /gcm3 [3]

Mark scheme: 20 6.2 nfww 3 7800 − 50 7700 to 7800 M2 for or 1240 to 1260 1240 + 10 or M1 for 7850 or 7750 or 1250 or 1230 seen

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21 B E NOT TO n A SCALE C m O OAB is a triangle. C is the midpoint of OA. OC = m and CB = n . E lies on AB and AE : EB = 4 : 5 . Find, in terms of m and n, the position vector of E. Give your answer in its simplest form. ................................................. [4]

Mark scheme: 21 14 4 4 4 m + noe simplified final answer B3 for 2m + ( −m + n ) oe 9 9 9 or M2 for 4 4 AE = ( −m + n ) or EA = (m − n ) 9 9 5 5 or BE = (m − n ) or EB = (n − m ) 9 9 M1 for OB = m + n or BO = –m – n or AB = n – m or BA = m – n or for a correct vector route along the lines on the diagram

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22 The line y = 4x + 12 intersects the curve y = 2x 2 - x - 3 at point P and point Q. Find the coordinates of P and Q. You must show all your working and give your answers correct to 2 decimal places. ( ......................., .......................) ( ......................., .......................) [6] Question 23 is printed on the next page.

Mark scheme: 22 2x2 – 5x – 15 [= 0] M2 M1 for 4x + 12 = 2x2 – x – 3 or better or or y2 – 34y + 144 [= 0] oe 2  y − 12   y − 12  y = 2 − − 3 or better      4   4  2 M2 FT their 3-term quadratic in x or y −−( 5)  ( −5 ) − 4(2)( −15) oe or M1 for ( −5) 2 − 4(2)( −15) or better 2(2) −−( 5) + q −−( 5) − q or for oe or oe 2 2  2 2  2 5 15  5  2 or for  +   oe  5  4 2  4  or for  x −   4  (4.26, 29.04) B2 B1 for one correct pair or both x-values and correct or both y – values correct (– 1.76, 4.96)

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23 NOT TO 3 m SCALE O 2.5 m The diagram shows the major segment of a circle, centre O, radius 2.5 m. The segment is the cross section of a tunnel with height 3 m. The length of the tunnel is 800 m and it has the same cross section throughout its length. Calculate the volume of the tunnel. ........................................... m3 [7]

Mark scheme: 23 9835 to 9844 nfww 7 B5 for [area of segment =] 12.29 to 12.31 nfww M1 for 12.29 to 12.31 × 800 OR B2 for angle at centre = 157 or 156.9 or 156.92 to 156.93 or for [reflex angle =] 203 to 203.1  3 − 2.5  or M2 for 2 × cos–1   oe  2.5  3 − 2.5 or M1 for cos x = oe 2.5 360 − their 2 M2dep for π 2.5 360 [360 −]their 2 or M1dep for π 2.5 seen 360 1 M1dep for  2.5  2.5  sin(their) oe 2 or 2 × 0.5 × 2.52 − 0.52 × 0.5 oe M1 for their area × 800 leading to answer

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