Question 11 mark
Question
A quadrilateral has • rotational symmetry of order 2 • two diagonals that are the only lines of symmetry. Write down the geometrical name of this quadrilateral. …… [1]
Mark scheme
1 rhombus 1
Mathematics 0580/41 October/November 2025: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), October/November 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
0580_w25_qp_410580_w25_ms_410580/41/O/N/25
- Paper
- Paper 4 (Extended)
- Total marks
- 100
- Time
- 2 hours
- Code on the paper
- 0580/41/O/N/25
0580/41 mark scheme, question by question
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question 22 marks
Question
Solve. 11 - 2x = 4 x = …… [2]
Mark scheme
2 1 7 2 M1 for 11 – 4 = 2x or –2x = 4 – 11 3.5 or 3 or 11 2 x 4 2 2 or − = or better 2 2 2
Question 32 marks
Question
x° NOT TO SCALE 55° The diagram shows two parallel lines and a straight line. Find the value of x. x = …… [2]
Mark scheme
3 125 2 M1 for 125 or 55 correctly placed on diagram
Question 41 mark
Question
A train journey starts at 22 16. The journey takes 5 hours 52 minutes. Find the time the train journey finishes. …… [1] * 0000800000004 * DFD 4
Mark scheme
4 [0]4 08 1
Question 55 marks
Question
A B (a) In triangle ABC, AB = 8 cm, AC = 7 cm and BC = 5 cm. Using a ruler and compasses only, construct triangle ABC. The side AB has been drawn for you. [2] (b) Measure angle ACB. …… [1] (c) Triangle ABC is a scale drawing of a field. (i) The scale is 1 : 10 000. Find the actual distance from A to B. Give your answer in kilometres. …… km [1] (ii) B is due east of A. Find the bearing of A from B. …… [1]
Mark scheme
5(a) Triangle accurately completed 2 B1 for accurate triangle with no/incorrect arcs or with arcs at C. for accurate triangle with arcs but with AC and BC reversed 5(b) 80 to 85 1 FT their triangle if 0 scored in (a) 5(c)(i) 0.8 1 5(c)(ii) 270 1
Question 61 mark
Question
Expand. g (3 - 2g) …… [1] * 0000800000005 * DFD 5
Mark scheme
6 3 g − 2 g 2 final answer 1
Question 74 marks
Question
(a) x –5 –4 –3 –2 –1 0 1 2 3 Write down the inequality represented in the diagram. …… [2] (b) Write down the integer values of x that satisfy the inequality - 4 1 2x G 8 . …… [2]
Mark scheme
7(a) –3 < x ⩽ 2 2 B1 for –3 < x or x ⩽ 2 7(b) –1, 0, 1, 2, 3, 4 2 B1 for 5 correct (and no extras) or 6 correct with one extra or B1 for –2 < x ⩽ 4
Question 81 mark
Question
Calculate. 1 (3.5 2 - 2.2 3 ) 4 …… [1]
Mark scheme
8 1.13 or 1.125… 1
Question 92 marks
Question
An athlete runs at a speed of 9.5 m/s. Convert this speed into km/h. …… km/h [2] * 0000800000006 * DFD 6
Mark scheme
9 34.2 2 M1 for figs 342 or for k × 60 × 60 or for k ÷ 1000
Question 102 marks
Question
A is the point (2, 1). 2 AB = e o 4 Find the coordinates of B. ( …… , …… ) [2]
Mark scheme
10 (4, 5) 2 B1 for each
Question 114 marks
Question
In a sale, the prices of coats are reduced by 15%. (a) The original price of a coat is $60. Calculate the sale price of the coat. $ …… [2] (b) The sale price of a different coat is $58.14 . Calculate the original price of this coat . $ …… [2] * 0000800000007 * DFD 7
Mark scheme
11(a) 51 2 100 − 15 M1 for 60 × oe 100 or B1 for 9 11(b) 68.40 2 100 − 15 M1 for [ ] × = 58.14 oe 100
Question 122 marks
Question
Some students are asked if they like football (F) or rugby (R). The Venn diagram shows the results. % R F 33 7 6 4 (a) Find the number of students who do not like rugby. …… [1] (b) Use set notation to describe the region containing students who like rugby but not football. …… [1]
Mark scheme
12(a) 37 1 12(b) F ′ R oe 1
Question 133 marks
Question
T NOT TO 12.6 m SCALE E 1.5 m A P 33 m The diagram shows two vertical poles, AE and PT, standing on horizontal ground, AP. Calculate the angle of elevation of the point T from the point E. …… [3] * 0000800000008 * DFD 8
Mark scheme
13 18.6 or 18.59… 3 12.6 − 1.5 M2 for tan = oe 33 or M1 identifying the correct angle of elevation or for 18.6 or 18.59… seen and spoiled After 0 scored SC1 for [angle ETP =] 71.4 or 71.4[0…] or 71.41
Question 1410 marks
Question
A cube contains a solid metal sphere. The sphere touches all the faces of the cube. The side length of the cube is 8 cm. 256 (a) Show that the volume of the sphere is r cm 3 . 3 [1] (b) Calculate the percentage of the cube that is not occupied by the sphere. …… % [3] (c) The density of the metal of the sphere is 7.86 g/cm 3 . Calculate the mass of the sphere. Give your answer in kilograms. [Density = mass ' volume] …… kg [2] * 0000800000009 * DFD 9 (d) The sphere is melted down and made into a solid cylinder with radius 3.1 cm. Calculate the total surface area of the cylinder. …… cm 2 [4]
Mark scheme
14(a) 4 256 1 × π × 43 [= π] 3 3 14(b) 47.6 nfww or 3 50 47.63 to 47.64… nfww B2 for 52.4 or 52.35 to 52.37 or π nfww 3 OR 256 83 − π M2 for 3 × 100 oe 83 256 256 83 − π π or M1 for 3 [×100] oe or 3 × 100 83 83 oe 14(c) 2.11 or 2.107… 2 256 M1 for π ×7.86 3 14(d) 233 or 233.3 to 233.4 4 256 M1 for π × 3.12 × h = π 3 2 M2dep for 2 × π × 3.1 + 2 × π × 3.1 × their h or M1dep for 2 × π × 3.1 × their h or M1 for 2π × 3.12
Question 153 marks
Question
B T NOT TO O SCALE 67° C A A, B and C lie on a circle, centre O. TA and TB are tangents to the circle at A and B. Calculate angle ATB. Angle ATB = …… [3] * 0000800000010 * DFD 10
Mark scheme
15 46 3 M2 for [angle ATB] = 180 – 2 × 67 oe OR M1 for [obtuse] angle AOB = 2 × 67 M1 for angle OAT or angle OBT = 90 OR M1 for angle TBA = 67 or angle TAB = 67 M1 for angle TDA or TDB = 90 where D is the intersection of OT and AB OR M1 for angle BOT = 67 or angle AOT = 67 M1 for angle OTB = 180 – 67 – 90 or angle OTA = 180 – 67 – 90
Question 164 marks
Question
The table shows some information about the heights of 50 plants. Height (h cm) 5 1 h G 10 10 1 h G 12 12 1 h G 20 Frequency 3 24 23 Calculate an estimate of the mean height. …… cm [4]
Mark scheme
16 13.09 4 M1 for mid-values 7.5, 11 and 16 soi M1 for Σfx where x-values in correct interval (including boundaries) Σfx M1dep on second M1 for 50
Question 173 marks
Question
Find the equation of the straight line that passes through the points (2, 0) and (0, 4). Give your answer in the form y = mx + c . y = …… [3] * 0000800000011 * DFD 11
Mark scheme
17 [y =] –2x + 4 3 4−0 M1 for gradient = oe 0−2 M1 for answer mx + 4
Question 184 marks
Question
8 NOT TO SCALE Speed (m/s) 0 0 2 10 Time (seconds) The diagram shows part of the speed–time graph for an athlete in a race. (a) Calculate the distance the athlete runs in the first 10 seconds. …… m [2] (b) The length of the race is 100 m. After 10 seconds, the athlete continues to run at a speed of 8 m/s until the end of the race. Calculate the total time the athlete takes to complete the 100 m race. …… s [2] * 0000800000012 * DFD 12
Mark scheme
18(a) 72 2 1 M1 for 8 × 8 or for × 2×8 2 1 or (10 + 8) × 8 2 18(b) 1 2 B1 for 3.5 oe 13.5 or 13 2 100 − their 72 or M1 for [+ 10] 8
Question 193 marks
Question
The mass of a radioactive substance decays exponentially at a rate of 10% per day. The initial mass of the substance is 20 g. Find the number of whole days it takes for the mass of the substance to first be less than 1 g. …… days [3]
Mark scheme
19 29 3 B2 for 28.4[3…] OR M2 for 20 × 0.928 or 20× 0.929 evaluated to at least 1 dp or for 0.928 or 0.929 evaluated to at least 2 dp or M1 for at least two trials of [20 ×] 0.9 n soi n 10 or for 1 > 20 1 − oe 100
Question 203 marks
Question
These are the first five terms of a sequence. 48 24 12 6 3 (a) Find the next term. …… [1] (b) Find the nth term. …… [2]
Mark scheme
20(a) 1 3 1 1.5 or 1 or 2 2 20(b) 1 n 2 1 n+ k 96 × oe final answer M1 for answer 96 × oe 2 2 n 1 or for 96 × oe seen 2
Question 212 marks
Question
In triangle STU, ST = 8 cm, SU = 9 cm and angle TSU = 50°. Calculate the area of triangle STU. …… cm 2 [2] * 0000800000013 * DFD 13
Mark scheme
21 27.6 or 27.57 to 27.58 2 1 M1 for × 9 × 8 × sin 50 oe 2
Question 223 marks
Question
Alex invests $200 at a rate of r% per year compound interest. At the end of 25 years the value of this investment is $301.10 . Find the value of r. r = …… [3] 1
Mark scheme
22 1.65 or 1.649[9…] 3 301.10 M2 for 25 oe 200 or M1 for 200 × [ ]25 = 301.10
Question 232 marks
Question
y\ x+1 When x = 8, y = 3. Find y in terms of x. y = …… [2] * 0000800000014 * DFD 14
Mark scheme
23 9 2 k 9 [y =] oe final answer M1 for 3 = oe or for seen and x +1 8 +1 x +1 spoiled
Question 2411 marks
Question
Martha walks a distance of 10 km at a speed of x km/h. She then runs a distance of 5 km at a speed of (x + 4) km/h. The total time taken for the whole journey is 3.5 hours. (a) Write down an expression in terms of x for the time Martha is walking. …… h [1] (b) Show that 7x 2 - 2x - 80 = 0 . [4] (c) Solve 7x 2 - 2x - 80 = 0 , giving your answers correct to 2 decimal places. You must show all your working. x = …… or x = …… [3] (d) Calculate the difference between the time Martha is walking and the time she is running. Give your answer in hours and minutes correct to the nearest minute. …… h …… min [3] * 0000800000015 * DFD 15
Mark scheme
24(a) 10 1 x 24(b) 10 5 7 M1 their + = oe x x+4 2 20 x + 80 + 10 x = 7 x 2 + 28 x oe M2 Strict FT for correctly clearing fractions from their three-term equation with two algebraic denominators in x and x + 4 and expanding all brackets Strict M1FT for correctly expressing their two algebraic fractions with two denominators in x and x + 4 as a single fraction or with a common denominator within a correct equation or for correctly clearing fractions from their three-term equation with two algebraic denominators in x and x + 4 but not all brackets expanded Leading to 7 x2 − 2 x − 80 = 0 A1 No errors or omissions 24(c) − ( −2 ) ± ([ −]2)2 − 4 ( 7 )( −80 ) B2 or B1 for ([−]2)2 − 4 ( 7 )( −80 ) oe or for oe 2 (7) −(−2) − p −(−2) + p oe or for oe 2(7) 2(7) 2 2 or x − 14 –3.24 and 3.53 B1 24(d) 2h 10min 3 B2 for 2.168 to 2.18 [h] or for 130.08 to 130.8 [min] or for 2hours 10.08 min to 2 hours 10.8 min OR 10 5 M2 for − their positive x their positive x + 4 or 10 5 M1 for or their positive x their positive x + 4
Question 253 marks
Question
Kai sorts parcels into two types, light and heavy. Type of parcel Mass (m kg) Light 01mG2 Heavy 21mG5 The histogram shows some information about the number of parcels Kai sorts in one day. 40 30 Frequency 20 density 10 0 m 0 1 2 3 4 5 Mass (kg) (a) Find the number of light parcels. …… [1] (b) There are 102 heavy parcels. Complete the histogram. [2] * 0000800000016 * DFD 16
Mark scheme
25(a) 40 1 25(b) Correct column, from 2 to 5 and 2 102 height 34 M1 for 5−2
Question 264 marks
Question
Solve the simultaneous equations. You must show all your working. y = 2x 2 - 3x - 7 y = 2x - 7 x = …… , y = …… x = …… , y = …… [4] * 0000800000017 * DFD 17
Mark scheme
26 2x2 − 5x[= 0] or better or M2 M1 for 2 x 2 − 3x − 7 = 2 x − 7 oe or for 2 2 y2 + 18 y + 28[= 0] or better y+7 y+7 y = 2 − 3 −7 2 2 x = 0, y = –7 B2 B1 for x = 0, y = –7, or for x = 0 and x = 2.5 or x = 2.5, y = –2 for x = 2.5, y = –2 or for y = –2 and y = –7 If M1B0 or M2B0 scored then SC1 for correct substitution seen of both of their x-values or their y-values into y = 2 x 2 − 3 x − 7 or y = 2 x − 7
Question 274 marks
Question
H G D C NOT TO 7 cm SCALE 5 cm E F 15 cm A 10 cm B The diagram shows a prism of length 15 cm. The cross-section of the prism is a trapezium. Angle DAB = 90° and angle ADC = 90°. AB = 10 cm, AD = 5 cm and DC = 7 cm. Calculate the angle the diagonal AG makes with the base ABFE. …… [4] * 0000800000018 * DFD 18
Mark scheme
27 16.8 or 16.80 to 16.81 4 5 M3 for tan = oe or 15 + 7 2 2 5 sin = oe or 15 + 72 + 52 2 152 + 72 cos = oe 152 + 72 + 52 or M2 for 152 + 7 2 or 152 + 7 2 + 52 or M1 for indication of correct angle
Question 283 marks
Question
f (x) = 7 x - 4 Find the value of x when (a) f (x) = 1 x = …… [1] (b) f - 1 (x) = 1. x = …… [2] * 0000800000019 * DFD 19
Mark scheme
28(a) 4 1 28(b) 1 2 M1 for [x =] f(1) or better 343
Question 298 marks
Question
B 13 cm 10 cm NOT TO SCALE y° 14 cm A C 38° 97° D (a) Calculate the value of y. y = …… [3] (b) Calculate BD. BD = …… cm [5] * 0000800000020 * DFD 20
Mark scheme
29(a) 63.0 or 63.02 to 63.03 3 102 + 142 − 132 M2 for [cos y =] oe 2 × 10 × 14 or M1 for 132 = 102 + 142 – 2 × 10 × 14 × cos y oe 29(b) 15.1 or 15.13 to 15.14 5 14sin38 M2 for [AD] = sin97 AD 14 or M1 for = oe sin 38 sin 97 M2 for 102 + (their AD)2 – 2 × 10 × their AD × cos(their y + 180 – 97 – 38) or M1 for angle BAD = their y + 180 – 97 – 38 soi
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Where can I find the 0580/41 October/November 2025 mark scheme?
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Mathematics 0580/41 for October/November 2025 is out of 100 marks, with 2 hours to answer it.
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