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

1 The temperature at 4 am is -12 °C. The temperature at 4 pm is 21 °C. Find the increase in temperature from 4 am to 4 pm. ............................................. °C [1]

Mark scheme: Question Answer Marks Partial Marks 1 33 1

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2 Find all the common factors of 8 and 12. ................................................. [1]

Mark scheme: 2 1, 2, 4 1

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3 A cuboid has length 6 cm, width 5 cm and height 2.5 cm. Work out the volume of the cuboid. .......................................... cm3 [2]

Mark scheme: 3 75 2 M1 for 6 × 5 × 2.5 oe

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4 A computer costs $560. In a sale, this cost is reduced by 20%. Find the cost of the computer in the sale. $ ................................................. [2]

Mark scheme: 4 448 2 M1 for 560 × 0.8 oe or B1 for 112

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5 The pie chart shows some information about the way 600 students travel to school. Walk 60° Bicycle (a) Work out the number of students that walk to school. ................................................. [2] (b) 120 of the students travel to school by car. The remaining students travel by bus. Complete the pie chart. [3]

Mark scheme: 5(a) 150 2 90 M1 for  600  oe 360 5(b) Correctly completed pie chart including 3 B2 for 72 or 138 labels of car [72] and bus [138] 120 or M1 for  360 oe 600 or B1FT for drawing one correct sector for their angle(s)

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6 B A NOT TO 75° SCALE 80° C E D ABCD is a quadrilateral. E lies on CD and AE is parallel to BC. EA = ED. Find (a) angle ABC Angle ABC = ................................................ [1] (b) angle AED Angle AED = ................................................ [1] (c) angle DAB. Angle DAB = ................................................ [2]

Mark scheme: 6(a) 105 1 6(b) 80 1 6(c) 125 2 FT (180 – their (b)) ÷ 2 + 75 B1 for EAD or EDA = 50 or B1FT for 180 − their (b) EAD or EDA = correctly 2 evaluated or M1 for (180 – their (b)) ÷ 2 + 75

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7 Represent the inequality - 4 1 x G 3 on the number line. x – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 [2]

Mark scheme: 7 2 B1 for open circle at –4 and closed circle at 3 or for a line between –4 and 3 (any circles)

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8 Kemi buys p pens that each cost 40 cents. She pays with $20. Write an expression, in terms of p, for the change, in cents, Kemi receives from the $20. ......................................... cents [2]

Mark scheme: 8 2000 – 40p final answer 2 B1 for 40p or 0.4p in final answer or 2000 – 40p seen then spoilt

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9 Rajid has a full bottle of juice. 1 He drinks of the full bottle on Monday. 3 3 He drinks of the full bottle on Tuesday. 7 Find the fraction of the bottle of juice remaining. ................................................. [3]

Mark scheme: 9 5 3 16 B2 for oe 21 21 OR 1 3 M1 for 1 − − oe 3 7 a b M1 for converting and to two 3 7 fractions with a common denominator 1  1  3 e.g. (or their  1 −  ) and to 3  3  7 7 k 9 k and 21k 21k 14k 9 k or and where k is an integer 21k 21k

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10 b = dm + 2mk (a) d = 3.14 , m = 7.92 and k = 10.16 . By rounding each value correct to 1 significant figure, work out an estimate for b. ................................................. [3] (b) Rearrange the formula to make m the subject. m = ................................................ [2]

Mark scheme: 10(a) 184 3 B1 for 3, 8 and 10 with 3, 8 and 10 shown M1 for 3 × 8 + 2 × 8 × 10 or for correct substitution of unrounded, truncated or incorrectly rounded values 10(b) b 2 M1 for [b =] m(d + 2k) [m =] final answer d + 2k

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11 P NOT TO S T SCALE Q R In the diagram, S lies on PQ and T lies on PR. ST is parallel to QR. (a) Explain why triangle PQR is mathematically similar to triangle PST. Give a geometrical reason for each statement you make. ..................................................................................................................................................... ..................................................................................................................................................... ..................................................................................................................................................... ..................................................................................................................................................... [3] (b) ST = 3 cm, QR = 9 cm and PS = 5 cm. Work out PQ. PQ = ............................................cm [2] (c) The area of triangle PST is 2k cm2. Find, in terms of k, the area of quadrilateral QRTS. .......................................... cm2 [2]

Mark scheme: 11(a) Three correct pairs with correct reasons 3 B2 for two correct pairs with correct reasons PQR = PST and corresponding angles or B1 for one correct pair with correct PRQ = PTS and corresponding angles reason QPR = TPS and common angle OR Two correct pairs with correct reasons and correct similarity condition e.g. AAA 11(b) 15 2 3 5 M1 for = oe or better 9 PQ 11(c) 16k 2 2 1 M1 for 32 or  oe 3

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12 A fitness club has 100 members. 60 swim (S ). 70 cycle (C ). 25 do not swim or cycle. % S C (a) Complete the Venn diagram. [3] (b) One member of the fitness club is chosen at random. For this member, find (i) P(C ) ................................................. [1] (ii) P(S + C ) ................................................. [1] (iii) P(S , C l). ................................................. [1]

Mark scheme: 12(a) 3 If values placed: S C B2 for 55 correctly placed OR 5 55 15 B1 for 25 correctly placed M1 for total S = 60 and total C = 70 with intersection not equal zero/not blank 25 OR or Alt method if probabilities placed: B2 for 0.55 oe correctly placed S C OR B1 for 0.25 oe correctly placed 0.05 0.55 0.15 M1 for total S = 0.6 and total C = 0.7 with intersection not equal zero/not blank 0.25 12(b)(i) 70 1 oe 100 12(b)(ii) 55 1 their 55 oe FT from their Venn diagram 100 100 12(b)(iii) 85 1 their 25 + their 5 + their 55 oe FT or 100 100 their 15 1 − from their Venn diagram 100

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13 M = 2 7 # 3 3 # 5 2 (a) Write 14M as a product of its prime factors. Give your answer in index form. ................................................. [2] (b) R is an integer. M is a cube number. R Find the smallest possible value of R. R = ................................................ [2]

Mark scheme: 13(a) 28 × 33 × 52 × 7[1] cao 2 M1 for [14 =] 2 × 7 soi 13(b) 50 2 M1 for 23 × 33 [× 50] or 26 × 33 [× 50] oe or or 24 × 52 or 2 × 52 oe –86 400 or SC1 for answers –50, [–]400, [–]1350, [–]3200, [–]10 800, 86 400 oe

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14 Find the value of (a) 3 2 ' 3 -2 ................................................. [2] - 23 (b) 16 . ................................................. [2]

Mark scheme: 14(a) 81 2 B1 for 34 or 32 × 9 or 9 × 9 14(b) 1 2 1 1 [  ]64 M1 for 4 3 or 64–1 or 4096

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15 Factorise. (a) x 2 - 64 ................................................. [1] (b) 5 x ( x - 2y) + 6 ( x - 2y) 2 ................................................. [2]

Mark scheme: 15(a) (x + 8)(x – 8) final answer 1 15(b) (x – 2y)(11x – 12y) oe final answer 2 M1 for (x – 2y)(5x + 6(x – 2y)) or for (x + ay)(11x + by) where ab = 24 or 11a + b = – 34 or for correct answer seen and spoilt. If 0 scored, SC1 for 11x2 – 34xy + 24y2

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16 D NOT TO C SCALE 14° A B AB and BD are two sides of a regular 15-sided polygon. AB and BC are two sides of a regular n-sided polygon. Angle DBC = 14°. Work out the value of n. n = ................................................ [4]

Mark scheme: 16 36 4  360  M3 for 360   − 14  oe  15  or B2 for 24 or 156 (15 − 2 ) 180 or M1 for 360 ÷ 15 or oe 15

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17 B is the point (-3, 1) and D is the point (-5, 9). BD is a diagonal of the kite ABCD. (a) The ratio of the lengths of the diagonals BD : AC = 2 : 3. Work out the length of AC. Give your answer as a surd in its simplest form. ................................................. [5] (b) Find the coordinates of the midpoint of BD. ( ...................... , ...................... ) [2] (c) The diagonal AC of the kite passes through the midpoint of BD. Find an equation of AC. Give your answer in the form y = mx + c . y = ................................................ [4]

Mark scheme: 17(a) 3 17 cao 5 B4 for answer equivalent to 3 17 but 3 68 not in the correct form e.g. , 2 6 17 , 153 2 OR B3 for 68 or 2 17 or M2 for (9 – 1)2 + (–5 – – 3)2 oe or M1 for (9 – 1) or (–5 – – 3) oe and 3 M1 for their 68  oe 2 17(b) (– 4, 5) 2 B1 for each coordinate 17(c) 1 4 − 1 y = x + 6 final answer M1 for  grad BD = 9 oe 4 −−−5 3 −1 M1 for  grad AC =  their grad BD M1 for their (– 4, 5) substituted into y = their mx + c oe

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18 Rationalise the denominator and simplify. 20 4 + 6 ................................................. [3]

Mark scheme: 18 2(4 – 6 ) or 8 – 2 6 3 20 4 − 6 ( ) M2 for oe or better 16 + 4 6 − 4 6 − 6 20 4 − 6 or M1 for  oe 4 + 6 4 − 6

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19 7 cm NOT TO SCALE 5 cm 5 cm The diagram shows a box in the shape of a cuboid. Mala has a straight rod of length 10 cm. Show that this rod does not fit completely inside the box. [3]

Mark scheme: 19 99 and states this is less than 10 oe 3 B2 for 99 or M2 for 52 + 52 + 72 oe or M1 for 52 + 52 or 52 + 72 oe

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21 1 20 f ( )x = , x ! g ( )x = 3x + 4 2x - 1 2 (a) Find (i) g(2) ................................................. [1] (ii) gf(-1) ................................................. [2] (iii) f -1 ( )x . f -1 ( )x = ................................................ [3] (b) Solve f ( x) = g ( x) . x = .................. or x = .................. [5]

Mark scheme: 20(a)(i) 10 1 20(a)(ii) –17 2 M1 for g(–7) oe  21 or 3  + 4 oe soi  2 x − 1  20(a)(iii) 21 + x 21 1 3 M2 for one step from answer or + oe final answer 21 + y 21 1 2 x 2 x 2 e.g. x = or x = + 2 y 2 y 2 21 or 2 y = + 1 x 21 + x or 2xy = 21 + x or 2 y = x or M1 for correct first step e.g. 21 x = or better 2 y − 1 or y (2x – 1) = 21 or better 20(b) 5 5 5 B3 for 6x2 + 5x – 25 [= 0] and − oe 3 2 or M1 for 21 = (2x – 1)(3x + 4) oe B1 for 6x2 – 3x + 8x – 4 or better and M1 for correct method to solve their three-term quadratic e.g. (3x – 5)(2x + 5) [= 0] −5 5 2 −−4 6 25 or oe 12

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21 (a) y 1 O 180° 360° x – 1 On the diagram, sketch the graph of y = cos x for 0° G x G 360° . [2] (b) Solve the equation 2 cosx + 3 = 0 for 0° G x G 360° . x = .................. or x = .................. [3]

Mark scheme: 21(a) Correct sketch to go through (0, 1), 2 M1 for correct cos curve shape through (180, –1) and (360, 1) (0, 1) or for almost correct sketch within 1 tolerance but with an omission at either end 0 or for almost correct sketch within 360 tolerance but with incorrect curvature in one place only –1 21(b) 150 and 210 3 B2 for one correct answer 3 or M1 for cos x = − or better 2 or B1 for 30 or –30 If M1, B1 or 0 scored, award SC1 for two answers in range with a sum of 360

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22 A graph with equation y = x 2 + bx + c has a minimum point at (-5, 12). Find the value of b and the value of c. b = ................................................ c = ................................................ [3]

Mark scheme: 22 [b =] 10 3 B2 for b = 10 or c = 37 [c =] 37 OR M1 for correct method for b e.g. • (x + 5)2 + 12 • 2x + b = 0 oe −b • = −5 2 1 M1 for correct method for c e.g. • x2 + 5x + 5x + 25 [+ 12] oe • 12 = (–5)2 + their b × –5 + c oe ( their b ) 2 • c − = 12 oe 4 1

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23 A NOT TO SCALE O 12 cm B 240° The diagram shows a major sector, AOB, of a circle. The sector angle is 240° and the radius is 12 cm. (a) Show that the length of the major arc AB is 16rcm . [1] (b) OA is joined to OB to form a cone. Work out the volume of the cone. `a bj r Give your answer in the form where a is an integer and b is a prime number. 3 .......................................... cm3 [6] Question 24 is printed on the next page.

Mark scheme: 23(a) 240 M1 2 12 oe 360 23(b) 6 256 5  256 5 256 5  ( ) ( ) ( ) cao B5 for answer or 3 3 3 seen then spoilt 256 5  ( ) or an answer equivalent to 3 but not in required form OR B2 for radius of cone = 8 oe or M1 for 2πr = 16π B2 for [height of cone =] 80 oe or M1 for 122 = their 82 + h2 or better 1 2 M1dep for  their 8  their 80 oe 3 dep on attempt at Pythagoras to find height of cone

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24 27 nx = ( 9 x) 2 Find the value of n. n = ................................................ [2]

Mark scheme: 24 4 2 B1 for 33n[x] or 34[x] [n =] 3

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