Question 12 marks
Question
The nth term of a sequence is 5 - 2n . (a) Find the 6th term of this sequence. …… [1] (b) Find the greatest number in this sequence. …… [1]
Mark scheme
1(a) –7 1 1(b) 3 1
Mathematics 0580/43 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_430580_w25_ms_430580/43/O/N/25
- Paper
- Paper 4 (Extended)
- Total marks
- 100
- Time
- 2 hours
- Code on the paper
- 0580/43/O/N/25
0580/43 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 24 marks
Question
The stem-and-leaf diagram shows the age of each of 16 adults. 3 2 3 3 5 6 7 4 0 1 5 5 6 8 9 5 1 1 1 Key: 3 | 2 represents age 32 years (a) Find the mode. …… years [1] (b) Find the median. …… years [1] (c) Find the percentage of the 16 adults with an age of less than 38 years. …… % [2] * 0000800000004 * DFD 4 4
Mark scheme
2(a) 51 1 2(b) 43 1 2(c) 37.5 2 6 M1 for oe 16
Question 31 mark
Question
G = m2n 5 Find the value of G when m = 6 and n = 15. G = …… [1]
Mark scheme
3 432 1
Question 44 marks
Question
(a) The scale diagram shows the position of town A on a map. Town B is 12 km from town A on a bearing of 080°. Using a scale of 1 cm represents 2 km, mark the position of town B on the diagram. North A Scale: 1 cm to 2 km [2] (b) The bearing of C from D is 130°. Work out the bearing of D from C. …… [2] * 0000800000005 * DFD 5
Mark scheme
4(a) B marked 6 cm from A on a bearing of 2 M1 for a correct bearing of 080° but 080°. incorrect length or for a 6 cm line from A 4(b) 310 2 M1 for 180 + 130 oe or indicates required bearing on a sketch
Question 54 marks
Question
(a) The diagram shows a regular decagon. AB is a line of symmetry of the decagon. A NOT TO d° SCALE B Work out the value of d. d = …… [3] (b) The exterior angle of a regular polygon with n sides is 45°. Work out the value of n. n = …… [1]
Mark scheme
5(a) 72 3 180 × (10 − 2 ) 360 M2 for or 180 – 10 10 360 or M1 for 180 × (10 – 2) or 10 5(b) 8 1
Question 63 marks
Question
Simplify. y5 (a) y2 …… [1] (b) 3x 3 # 5x 5 …… [2] * 0000800000006 * DFD 6
Mark scheme
6(a) y3 1 6(b) 15x8 final answer 2 B1 for kx8 or 15xk as final answer or for correct answer spoilt
Question 75 marks
Question
y 12 11 10 9 8 D 7 6 5 4 F 3 2 1 x –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 –1 E –2 –3 –4 –5 –6 –7 –8 (a) Describe fully the single transformation that maps triangle D onto triangle E. …… [2] (b) Describe fully the single transformation that maps triangle D onto triangle F. …… [3] * 0000800000007 * DFD 7
Mark scheme
7(a) Translation 2 B1 for translation 2 2 B1 for −8 −8 7(b) Enlargement 3 B1 for each 1 [scale factor] − 3 [centre] (–3, 4)
Question 82 marks
Question
m is a positive integer. Write these values in order of size, starting with the smallest. 1 m m 33% of m of m 320% of 3 10 …… , …… , …… , …… [2] smallest 3
Mark scheme
8 m 1 2 B1 for three in the correct order 320% of 33% of m of m m 10 3
Question 91 mark
Question
Draw a ring around the calculation that is equivalent to n ' 2 . 5 5 5 13 1 13 1 5 n#2 n# n# # # 3 13 5 n 5 n 13 [1]
Mark scheme
9 5 1 n× identified 13
Question 102 marks
Question
Solve the simultaneous equations. You must show all your working. 3x + 5y = 5 2x - 5y = 45 x = …… y = …… [2] * 0000800000008 * DFD 8
Mark scheme
10 x = 10 2 B1 for each y = –5
Question 112 marks
Question
2.89 # 10 -1 1.3 # 10 12 8.3 # 10 1 9 # 10 11 2.03 # 10 -5 0.3 # 10 -2 Use a number from the box to complete each statement. The number that is not written in standard form is …… . The largest number is …… . The smallest number is …… . [2]
Mark scheme
11 0.3 × 10–2 2 B1 for two correct 1.3 × 1012 2.03 × 10–5
Question 122 marks
Question
A vase contains flowers that are red or pink or white. Ruth picks a flower at random from the vase. The probability that the flower is not red is 0.9 . The probability that the flower is not pink is 0.65 . Find the probability that the flower is white. …… [2]
Mark scheme
12 0.55 oe 2 B1 for P(red) = 0.1 or P(pink) = 0.35 or M1 for 0.9 + 0.65 – 1 or 1 – (0.1 + 0.35)
Question 133 marks
Question
The point (5, 1024) lies on the curve y = c x , where c is a whole number. Find the y-coordinate of the point on the curve with x-coordinate -2. …… [3] * 0000800000009 * DFD 9
Mark scheme
13 1 3 B2 for c = 4 soi 0.0625 or or M1 for 1024 = c5 16
Question 145 marks
Question
These expressions are all equal in value. 5x - 2 10 - x y + 11 3 Find the value of y. y = …… [5]
Mark scheme
14 –5 5 B3 for x = 4 or M2 for 5x + 3x = 30 + 2 or better or M1 for 5 x − 2 = 3 (10 − x ) or better M1 10 – their x = y + 11 or better 5 × their x − 2 or = y + 11 or better 3 Alternative method: B4 for 3y + 5y = –5 – 2 – 33 or B3 for –5 – 5y – 2 = 3y + 33 5 ( −1 − y ) − 2 or B2 for = y + 11 3 or B1 for x = –1 – y
Question 155 marks
Question
The population of a town is 54 000. The population is decreasing exponentially at a rate of 2% per year. (a) Calculate the decrease in the population at the end of 4 years. …… [3] (b) Find the number of complete years it takes for the population of 54 000 to first fall below 44 000. …… years [2] * 0000800000010 * DFD 10
Mark scheme
15(a) 4192 or 4193 3 B2 for 49 807 to 49 808 4 2 or M2 for 54 000 − 54 000 × 1 − oe 100 4 2 or M1 for 54 000× 1 − oe 100 15(b) 11 nfww 2 2 n M1 for 54 000× 1 − evaluated 100 with n = 10 or n = 11 or B1 for 10.1 or 10.13 to 10.14
Question 164 marks
Question
Expand and simplify. (a) 7(x + 2) + 4 (3x - 5) …… [2] (b) (3x - y) (5x + 2y) …… [2]
Mark scheme
16(a) 19x – 6 final answer 2 B1 for 7x + 14 or 12x – 20 or for 19x – 6 seen then spoilt 16(b) 15x2 + xy – 2y2 final answer 2 B1 for 15x2 + 6xy – 5xy – 2y2 with at least three terms correct
Question 173 marks
Question
Make t the subject of the formula. 7t x= 5-t t = …… [3] * 0000800000011 * DFD 11
Mark scheme
17 5x 3 M1 for correctly clearing the oe final answer x+7 denominator and expanding bracket M1FT for correctly collecting terms in t on one side and terms not in t on the other M1FT for correct factorising and for correct division Maximum 2 marks for an incorrect answer
Question 186 marks
Question
102° 41.3 cm x° y° 64.5 cm NOT TO SCALE 52.1 cm 70.2 cm (a) Calculate the value of x. x = …… [3] (b) Calculate the value of y. y = …… [3] * 0000800000012 * DFD 12
Mark scheme
18(a) 38.8 or 38.77 to 38.78 3 41.3sin102 M2 for [sin x =] oe or better 64.5 64.5 41.3 or M1 for = oe sin102 sin x 18(b) 73.2 or 73.16… 3 64.52 + 52.12 − 70.22 M2 for 2 × 64.5 × 52.1 or M1 for 70.22 = 64.52 + 52.12 –2 × 64.5 × 52.1 × cos y
Question 198 marks
Question
f (x) = 5 x g (x) = 3x - 2 h (x) = x 2 + 1 (a) Find f (5) . …… [1] (b) Find g (8x) . …… [1] (c) Find g -1(x) . g -1(x) = …… [2] (d) Find the positive solution of gh (x) = 364 . x = …… [3] (e) Find ff -1(12). …… [1] * 0000800000013 * DFD 13
Mark scheme
19(a) 3125 1 19(b) 24x – 2 or 2(12x – 1) final answer 1 19(c) x+2 x 2 2 M1 for correct first step or + final answer 3 3 3 y 2 e.g. x = 3y – 2 or y + 2 = 3x or = x− 3 3 oe 19(d) 11 3 M2 for 3x2 + 1 = 364 or better or M1 for 3(x2 + 1) – 2 Alternative method: M2 for x2 + 1 = (364 + 2) ÷ 3 or M1 for 3x – 2 = 364 19(e) 12 1
Question 204 marks
Question
y = x 3 + 3x 2 - 13x dy (a) Find . dx …… [2] (b) Find the gradient of the curve y = x 3 + 3x 2 - 13x at the point where x = 3. …… [2]
Mark scheme
20(a) 3x2 + 6x – 13 final answer 2 B1 for two terms correct or correct answer seen then spoilt 20(b) 32 2 dy FT their dep on B1 earned in (a) dx M1 for correct substitution of x = 3 into dy their dx
Question 214 marks
Question
A dressmaker takes 75 hours to make 31 dresses. In week 1, she takes a total of 12 hours 30 minutes to make the first 4 dresses. In week 2, she makes the remaining 27 dresses at a constant hourly rate. Work out the percentage increase in her hourly rate of making dresses from week 1 to week 2. …… % [4] * 0000800000014 * DFD 14
Mark scheme
21 35 4 27 ÷ ( 75 − 12.5 ) M3 for × 100 [ −100] 4 ÷ 12.5 oe 27 ÷ ( 75 − 12.5) or − 1 [×100] oe 4 ÷ 12.5 27 ÷ ( 75 − 12.5 ) − ( 4 ÷ 12.5 ) or [×100] 4 ÷ 12.5 oe 12.5 62.5 or ÷ × 100[– 100] 4 27 OR M1 for 4 ÷ 12.5 oe M1 for 27 ÷ (75 –12.5) oe
Question 225 marks
Question
(a) 14 mm 10 mm x mm NOT TO 45 mm SCALE The diagram shows two mathematically similar triangles. Find the value of x. x = …… [2] (b) The surface areas of two mathematically similar containers are 124 cm 2 and 279 cm 2 . The capacity of the smaller container is 56 ml. Find the capacity of the larger container. …… ml [3] * 0000800000015 * DFD 15
Mark scheme
22(a) 63 2 14 10 M1 for = oe or better x 45 22(b) 189 3 3 3 279 2 124 2 M2 for [× 56] or [56 ÷] 124 279 oe or better 1 1 279 2 124 2 or M1 for or or better 124 279 3 2 279 x or = oe 124 56
Question 237 marks
Question
The table shows some information about the mass of each of 200 oranges. Mass (m grams) 180 1 m G 200 200 1 m G 210 210 1 m G 215 215 1 m G 230 Frequency 32 64 74 30 (a) Calculate an estimate of the mean mass of an orange. …… g [4] (b) Sarah draws a histogram to show this information. The table shows the height of one of the bars for this histogram. Complete the table. Mass (m grams) 180 1 m G 200 200 1 m G 210 210 1 m G 215 215 1 m G 230 Height of bar (cm) 7.4 [3] * 0000800000016 * DFD 16
Mark scheme
23(a) 208 4 M1 for midpoints soi 190, 205, 212.5, 222.5 M1 for use of Σfx where x is in the correct interval including boundaries 190 × 32 + 205 × 64 + 212.5 × 74 + 222.5 × 30 M1 (dep on second M1) for Σfx ÷ 200 23(b) 0.8 3.2 1 3 B2 for two correct or B1 for one correct or M1 for three of 1.6, 6.4, 14.8 and 2 seen
Question 246 marks
Question
A B 6.4 cm NOT TO D C 13 cm SCALE E F 5.1 cm H G The diagram shows a cuboid ABCDEFGH. AE = 6.4 cm, EH = 5.1 cm and AG = 13 cm. (a) Calculate EF. EF = …… cm [3] (b) Calculate the angle between the line AG and the base EFGH of the cuboid. …… [3] * 0000800000017 * DFD 17
Mark scheme
24(a) 10.1 or 10.10… 3 M2 for EF 2 + 6.42 + 5.12 = 132 or better or M1 for 6.42 + 5.12 or 132 – 6.42 or 132 – 5.12 24(b) 29.5 or 29.49… to 29.54… 3 6.4 M2 for sin [… =] oe 13 or M1 for identifying angle AGE
Question 253 marks
Question
Jenna has a length of wire measuring 68 cm, correct to the nearest cm. From this wire she cuts off two smaller pieces • a piece of length 4.7 cm, correct to the nearest mm • a piece of length 10.0 cm, correct to the nearest mm. Work out the lower bound and the upper bound for the length of the wire remaining. Lower bound = …… cm Upper bound = …… cm [3] * 0000800000018 * DFD 18
Mark scheme
25 [LB =] 52.7 3 B2 for answer [LB=] 52.7 or [UB=] 53.9 [UB =] 53.9 or M1 for 68 + 0.5 or 68 – 0.5 or 4.7 + 0.05 or 4.7 – 0.05 or 10[.0] + 0.05 or 10[.0] – 0.05. oe seen
Question 265 marks
Question
B F C m NOT TO SCALE A E D p ABCD is a parallelogram. AB = m and AD = p . F is a point on BC and BF = 4FC. E is a point on AD and AE : ED = 1 : 2. (a) Find EF , in terms of m and p, in its simplest form. …… [3] (b) EF and DC are extended to meet at the point G. Find CG, in terms of m and/or p, in its simplest form. …… [2] DFD 19 * 0000800000020 * DFD 20
Mark scheme
26(a) 7 3 M2 for correct unsimplified expression in m+ p 15 terms of m and p or M1 for a correct route 26(b) 3 2 2 1 m M1 for using the ratio [p] : [p ] 7 3 5
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Where can I find the 0580/43 October/November 2025 mark scheme?
On this page, free: the mark scheme for IGCSE Mathematics 0580/43, October/November 2025 (0580_w25_ms_43), and the question paper (0580_w25_qp_43), to read or download, with each question's marking points written out underneath.
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Mathematics 0580/43 for October/November 2025 is out of 100 marks, with 2 hours to answer it.
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