Cambridge IGCSE Mathematics 0580 — 2025 Oct/Nov Paper 4 · Variant 2

1 Write the ratio 60 grams : 3 kilograms in the form 1| n . 1 | ................................................ [2]

Mark scheme: Question Answer Marks Partial Marks 1 50 2 M1 for 60 : 3000 oe

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2 Solve. 8x - 17 = 27 x = ................................................ [2]

Mark scheme: 2 1 11 2 17 27 5.5 or 5 or M1 for 8x = 27 + 17 or x – = oe 2 2 8 8

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3 Write down the order of rotational symmetry of a regular decagon. ................................................. [1]

Mark scheme: 3 10 1

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4 Pedro makes cards. (a) He makes cards at a rate of 9 cards every 20 minutes. Work out the number of cards he makes in 8 hours. ................................................. [2] (b) Each card costs 12 cents to make. Pedro sells each card for 50 cents. Work out his percentage profit on each card. ..............................................% [2]

Mark scheme: 4(a) 216 2 M1 for 8 × 60 ÷ 20 oe or 9 × 60 ÷ 20 oe 4(b) 317 or 316.6 to 316.7 2 50 − 12 M1 for  100  oe 12 50 or for  100 [–100] oe 12

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5 Nuwa is buying a phone. One website sells the phone for 953 Yuan. A different website sells the same phone for $141. The exchange rate is 1 Yuan = $0.152 . Calculate the difference between these phone prices. Give your answer in dollars, correct to the nearest cent. $ ................................................ [2]

Mark scheme: 5 3.86 cao 2 M1 for 953 × 0.152 oe isw If 0 scored, SC1 for answer 25.4 or 25.36 to 25.37

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6 One morning, a dentist has appointments for 10 patients. The stem-and-leaf diagram shows the waiting time for 8 of these patients. 0 1 4 1 0 2 9 2 1 5 5 Key: 1 | 0 represents 10 minutes The times for the two other patients, P and Q, are not shown in the stem-and-leaf diagram. The mean waiting time of all 10 patients that morning is 16 minutes. The range of waiting times is 26 minutes. Patient P waits longer than patient Q. Find the waiting time for each of patient P and patient Q. Patient P .........................................min Patient Q .........................................min [4]

Mark scheme: 6 [Patient P =] 27 4 B1 for answer P = 27 [Patient Q =] 16 M2 for 10 × 16 – (1 + 4 + 10 + 12 + 19 + 21 + 25 + 25 + their 27) oe isw or M1 for 10 × 16 oe 1 + 4 + 10 + 12 + 19 + 21 + 25 + 25 + P + Q or for = 16 oe 10

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7 The diagram shows a parallelogram. NOT TO SCALE x m 6.51 m The parallelogram has the same perimeter as a circle with radius 4 m. (a) Show that x = 6.06 m , correct to 2 decimal places. [4] (b) NOT TO SCALE 6.06 m 36° 6.51 m The floor of a room is in the shape of this parallelogram. It costs $18 per square metre to tile the floor. Calculate the total cost of tiling the floor. $ ................................................ [4]

Mark scheme: 7(a) 2 × π × 4 = 6.51 × 2 + 2x M3 M1 for 2 × π × 4 oe or better M1 for 6.51 × 2 + 2x soi 6.055 to 6.058 [= 6.06] A1 7(b) 574 or 575 4 M2 for 6.51 × 6.06 × cos 36 oe or 574.1 to 574.5… or M1 for [height = ] 6.06 × cos 36 oe seen M1 for their area × 18 leading to answer Alternative method: 1 M2 for 2 × × 6.51 × 6.06 × sin(90 + 36) oe 2 1 or M1 for × 6.51 × 6.06 × sin(90 + 36) oe seen 2 M1 for their area × 18 leading to answer

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8 y 7 6 5 A B 4 3 2 1 x – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 – 1 – 2 – 3 – 4 – 5 (a) On the diagram, draw the image of 1 (i) shape A after a translation by the vector e o [2] - 7 (ii) shape A after a reflection in the line y = x + 1. [3] (b) Describe fully the single transformation that maps shape A onto shape B. ..................................................................................................................................................... ..................................................................................................................................................... [3]

Mark scheme: 8(a)(i) Image drawn at (–2, –3), 2 1  k  B1 for translation by  or by   (–3, –2), (–4, –2) and k  −7  (–3, –4) 8(a)(ii) Image drawn at (2, –3), 3 B2 for correct size and orientation but wrong position (3, –2), (4, –3) and (4, –4) or for quadrilateral with 3 correct vertices or B1 for y = x + 1 drawn soi long enough to be recognised 8(b) Rotation 3 B1 for each 90° anticlockwise oe (–2, 6)

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9 The diagram shows the speed–time graph for part of a car journey. 80 NOT TO Speed SCALE 60 (km / h) 0 0 15 25 Time (minutes) Find the total distance travelled in the 25 minutes. ........................................... km [3]

Mark scheme: 9 24.2 or 24.16 to 24.17 3 60 + 80 15 80  10 M2 for  + oe 2 60 2  60 or M1 for a correct partial area under line 60 + 80 15 80  10 e.g.  or oe 2 60 2  60 OR If 0 scored, SC2 for final answer 1450 60 + 80 80  10 or SC1 for  15 + oe seen 2 2

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10 Find the nth term of each sequence. (a) 17, 9, 1, – 7, -15, f ................................................. [2] (b) 3, 12, 27, 48, 75, f ................................................. [2]

Mark scheme: 10(a) 25 – 8n oe final answer 2 B1 for answer k – 8n (any k) oe or 25 – jn (j ≠ 0) or for correct answer seen then spoilt 10(b) 3n2 oe final answer 2 B1 for quadratic or for second difference = 6 (at least two, with none incorrect) or for correct answer seen then spoilt

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11 The table shows some values for y = x 3 - 2x + 3 . Where appropriate, values of y are given correct to 2 decimal places. x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 y -1 4 3.88 3 2.13 2 7 (a) Complete the table. [2] (b) Draw the graph of y = x 3 -2x + 3 for - 2 G x G 2 . y 7 6 5 4 3 2 1 x 0 – 2 – 1 1 2 – 1 [4] (c) By drawing a suitable straight line on the grid, solve the equation x 3 - 2.5 x + 1 = 0 . x = ....................... or x = ....................... or x = ....................... [4]

Mark scheme: 11(a) 2.63 and 3.38 2 B1 for each 11(b) Correct graph 4 B3FT for 8 correct plots or B2FT for 6 correct plots or B1FT for 4 correct plots 11(c) y = 2 + 0.5x ruled M2 M1 for [y =] 2 + 0.5x oe soi e.g. x3 – 2x + 3 = 2 + 0.5x or y = k + 0.5x ruled or y = kx + 2 ruled, but not y = 2 –1.85 to –1.7 , 0.4 to 0.5, A2 A1 for two correct values dep on M2 1.25 to 1.4 If M0 or M1 scored, SC1 for three correct values

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12 NOT TO B SCALE 52° A y° O x° 65° E C D A, B and C lie on a circle centre O. DE is a tangent to the circle at C. Angle ABC = 52° and angle BCE = 65°. (a) Find the value of x. Give a geometrical reason for your answer. x = ........................... because .................................................................................................... ..................................................................................................................................................... [2] (b) Find the value of y. y = ................................................ [2]

Mark scheme: 12(a) 104 2 B1 for each and angle at the centre is twice the angle at the circumference 12(b) 27 2 B1 for angle BAC = 65 or angle ACD = 52 or M1 for OCB = 25 and reflex AOC = 360 – their 180 − their 104 104 or for OAC or OCA = 38 FT 2

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13 Simplify. 7 3 + 2m 8 m ................................................. [2]

Mark scheme: 13 31 2 28 3 final answer M1 for + oe 8m 8m 8m

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14 (a) Carlos invests $24 000 at a rate of 3.2% per year compound interest. Calculate the value of his investment at the end of 4 years. $ ................................................ [2] (b) Carlos buys a painting for $x. He sells the painting for $40 870. He makes a profit of 34%. Calculate the value of his profit. $ ................................................. [3] (c) Carlos also buys a car with a value of $32 500. The value of the car decreases exponentially by 23% each year. Find a formula for the value, $V, of the car at the end of n years. ................................................. [3]

Mark scheme: 14(a) 27 223 2 4  3.2  M1 for 24 000 ×  1 +   100  14(b) 10 370 3 B2 for 30 500 40 870 or M2 for  34 oe 134  34  or M1 for  1 +  x = 40 870 oe  100  14(c) V = 32 500 × 0.77n final 3 B2 for answer 32 500 × 0.77n answer or for correct explicit formula seen in working, may be unsimplified or M1 for 32 500 × (1 – 0.23)n oe seen or for correct implicit formula seen, may be unsimplified or answer of form V = 32 500 × kn or answer of form V = p × (0.77 oe)n

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15 A NOT TO D SCALE 140° 112 m 180 m 300 m C B The diagram shows a field, ABCD, in the shape of a quadrilateral. BD is a straight path across the field. (a) Calculate BC. BC = ............................................ m [3] (b) Calculate angle DBC. Angle DBC = ................................................ [3] (c) The total area of the field, ABCD, is 35 900 m2. Work out the length of the shortest distance from D to AB. ............................................. m [4]

Mark scheme: 15(a) 392 or 392.4 to 392.5 3 2 2 M2 for 300 + 112 −2 300 112  cos140 OR M1 for 3002 + 1122 − 2  300  112  cos140 A1 for 154 022[…] 15(b) 10.6 or 10.7 or 10.55 to 3 112sin140 M2 for oe 10.69 their (a) 300 2 + ( their (a) ) 2 − 112 2 or cos[ DBC ] = 2  300  their (a) 112 their (a) or M1 for = oe sin DBC sin140 or 112 2 = 300 2 + ( their (a) ) 2 − 2  300  their (a)  cos DBC oe 15(c) 279 or 278.9… 4 M3 for 1  1  (35 900 –  112  300  sin140 ) ÷   180  oe 2  2  OR 1 M1 for  112  300  sin140 oe 2 M1 for recognition that the shortest distance from D is perpendicular to AB

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16 The histogram shows information about the masses of some coconuts. The masses are classified into four categories A, B, C and D. 200 150 Frequency density 100 C 50 B D A 0 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 Mass (kg) (a) Show that there are 10 coconuts in category D. [1] (b) Two of the coconuts from those in category C and category D are chosen at random. Find the probability that both are from category D. ................................................. [3] (c) Calculate an estimate of the mean mass of the coconuts. ............................................ kg [4]

Mark scheme: 16(a) (1.6 – 1.35) oe × 40 [= 10] 1 16(b) 5 3 10 9 oe M2 for  oe 39 10 + 170  0.1 9 + 170  0.1 10 9 9 10 or M1 for or or or oe 10 + 17 10 + 16 10 + 17 10 + 16 seen 10 9 or  oe with k > 10 and an integer k k − 1 m m − 1 or  oe with 0 < m < 27 and an integer 10 + 17 10 + 16 16(c) 1.26 or 1.259… nfww 4 M1 for frequencies 6, 15, 17 soi M1 for midpoints soi (1, 1.175, 1.3, 1.475) M1 for use of their fm  their f with m in correct interval including both boundaries

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

Mark scheme: 17 2x3 + 7x2 –10x –24 final 3 B2 for correct expansion unsimplified answer or for simplified four-term expression of correct form with three terms correct or B1 for one pair of brackets expanded with at least three terms out of four correct

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18 C B NOT TO SCALE b O a A In the diagram, OA is parallel to CB. OA | CB = 4 | 3 OA = a and OB = b . (a) Find AB in terms of a and b. AB = ................................................ [1] (b) M is the midpoint of OC. Find AM in terms of a and b. Give your answer in its simplest form. AM = ................................................ [3]

Mark scheme: 18(a) b – a 1  18(b) 11 1 3 B2 for a correct unsimplified vector for AM seen − a + b oe simplified 8 2  1 3 or for OM = b − a oe soi final answer 2 8  3 or B1 for OC = b − a oe soi 4  or for a correct vector route for AM along lines on the diagram

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19 Solve the simultaneous equations. You must show all your working and give your answers correct to 2 decimal places. y = 5 - 2 x y = 3 x 2 - 7x - 6 x = ………………… y = ………………… x = ………………… y = ………………… [6]

Mark scheme: 19 3x2 – 5x – 11 [= 0] M2 M1 for 3x2 – 7x – 6 = 5 – 2x  5 − y  2  5 − y  or 3y2 – 20y – 19 [= 0] or for y = 3   − 7   − 6  2   2  [ −−( )]5  ([ −]5) 2 −−4 3 11 M2 [ −−( )]5 + k [ −−( )]5 − k M1 for oe or oe 2  3 2  3 2  3 or for ([ −]5) 2 −−4 3 11 oe OR OR 2 5 11  5  2 x − =  + oe  5    M1 for  x −  6 3  6   6  x = 2.92, y = –0.84 B2 B1 for one correct solution for x and y and or for x = 2.92 and –1.25 x = –1.25, y = 7.51 or for y = –0.84 and 7.51 If B0 scored and at least two method marks scored, SC1 for correct substitution shown of both of their x-values or their y-values into y = 3x2 – 7x – 6 or y = 5 – 2x

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20 A solid metal prism has a mass of 4810 g, correct to the nearest 10 g. The density of the metal is 7.7 g/cm3, correct to 1 decimal place. Calculate the lower bound for the volume of the prism. [Density = mass ' volume] ......................................... cm3 [3]

Mark scheme: 20 620 nfww 3 4810 − 5 4800 to 4810 M2 for or oe 7.7 to 7.8 7.7 + 0.05 or M1 for 4815 or 4805 or 7.75 or 7.65 oe seen

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21 y is inversely proportional to ( x + 2 ) 2 . w is proportional to x. When y = 8, x = 3. 5 . When w = 15, x = 90 . Find y in terms of w. y = ................................................ [4]

Mark scheme: 21 242 4 8(3.5 + 2) 2 y = oe final answer M3 for may be done in stages 2 2 (6 w + 2)  90   w + 2   15  or for correct answer seen then spoiled OR 242 M2 for y = ( x + 2) 2 k or M1 for y = oe or better ( x + 2) 2 x B1 for w = oe 6

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