Cambridge IGCSE Mathematics 0580/42 is the May/June 2026 Paper 4 (Calculator, Extended), variant 2. It lasts 2 hours, is worth 100 marks and is sat by every Extended candidate. It had 22 compulsory questions. The first half was mostly short 2- and 3-mark questions; the big marks came from the cubic graph in Question 8 (11 marks), the pyramid in Question 16 (9 marks) and the bearings and trigonometry question on field ABCD in Question 20 (10 marks). This guide goes through the paper question by question with the final answers and what the official mark scheme gave marks for. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of all four 0580 papers from this series, see IGCSE Maths 0580 May/June 2026 past papers: what came up and how it was marked.

0580/42 at a glance

Question	Topic	Marks
1	Angles in an isosceles triangle	2
2	Books borrowed: mode, median and mean from a frequency table	5
3	Constructing a triangle with ruler and compasses	2
4	A deposit and monthly payments	3
5	Finding a point from a column vector	2
6	Perimeter of a semicircle	3
7	Compound interest: final value, then the rate	5
8	A cubic: table, graph and solving an equation with a straight line	11
9	Solving an equation with fractions	3
10	Percentages of a coin collection	4
11	Tangent in a right-angled triangle	2
12	Exponential decay: the date a mass falls below 96 g	3
13	Inverse proportion	3
14	A reflex angle from a cosine	2
15	Adding algebraic fractions	3
16	Square-based pyramid: symmetry, volume, edge length, angle with the base	9
17	Functions: evaluating, inverse, composite	7
18	Triangle areas leading to a quadratic	8
19	Grouped data: estimated mean and a histogram	7
20	Field ABCD: bearing, cosine rule, sine rule, shortest distance	10
21	Lower bound of a density	3
22	Simplifying an algebraic fraction	3
	Total	100

Question-by-question walkthrough

The mark scheme uses three types of mark. M marks are for a correct method, A marks are for an accurate answer and need the M mark before them, and B marks are independent marks for a correct result or statement. The cover asks for non-exact answers to 3 significant figures, or 1 decimal place for angles, unless the question says otherwise.

Question 1: isosceles triangle (2 marks)

Triangle ABC has AB = AC and a 30° angle at A; x is the angle at C.

• Answer: x = 75. The base angles are equal, so x = (180 − 30) ÷ 2.
• Marking: M1 for (180 − 30) ÷ 2 “oe” (or equivalent: any working that does the same thing), then the answer mark.

Question 2: books borrowed (5 marks)

Forty students; frequencies 5, 14, 10, 8, 1, 2 for 0 to 5 books.

• Mode (1 mark): 1.
• Median (1 mark): 2. The 20th and 21st values both fall in the “2 books” group (5 + 14 = 19 before it). Writing “2, 2” as the answer was not accepted.
• Mean (3 marks): Σfx = 0 + 14 + 20 + 24 + 4 + 10 = 72, and 72 ÷ 40 = 1.8. This was “cao” (correct answer only), so for full marks you could not leave it as 9/5 or round 1.8 to 2 afterwards. M2 was for the full fraction (implied by 72/40), or M1 for the sum of the products alone (implied by 72), with one slip in the products condoned.

Question 3: constructing a triangle (2 marks)

AB was drawn; you had to construct C with AC = 10 cm and BC = 8.5 cm using compasses.

• Full marks need the correct triangle with both construction arcs, accurate to ± 2 mm.
• B1 for the correct triangle with wrong or missing arcs, two correct intersecting arcs, or a correct point C clearly shown. If nothing else scored, SC1 (a special case mark) was available for a triangle with arcs but the two lengths swapped.

Question 4: deposit and monthly payments (3 marks)

• Answer: 18 payments. The deposit is 35% of $1260 = $441, leaving $819, and 819 ÷ 45.50 = 18.
• Marking: M2 for (1260 − 0.35 × 1260) ÷ 45.5, or M1 for the deposit alone (implied by 819 or 441). Writing “35% × 1260” without working it out did not earn the method mark.

Question 5: column vector (2 marks)

• Answer: Q = (5, 2). Add the vector to P: (−3 + 8, 5 − 3).
• Marking: B1 for one coordinate correct, (5, k) or (k, 2).

Question 6: perimeter of a semicircle (3 marks)

Diameter 15.8 cm.

• Answer: 40.6 cm. Half the circumference is π × 15.8 ÷ 2 = 24.8…, then add the straight edge: 24.8… + 15.8 = 40.6.
• Marking: M2 for the full expression, or M1 for the curved part only (implied by 24.8 to 24.83). Forgetting to add the diameter is the classic way to lose the last two marks. Using 3.14 or 22/7 for π was condoned for the method marks.

Question 7: compound interest (5 marks)

• (a) Value after 6 years (2 marks): 2500 × 1.044⁶ = $3237.00 (3240 also accepted). M1 for the compound interest calculation. Giving only the interest ($737…) spoiled the method, and 2500 × (104.4%)⁶ written alone did not earn it.
• (b) The rate (3 marks): 5420 = 4500 × m⁵, so m = ⁵√(5420 ÷ 4500) = 1.0379…, and r = 3.79. M2 for the fifth root (implied by 1.038, 103.8 or 0.0379…), or M1 for setting up 5420 = 4500 × m⁵.

Question 8: the cubic y = x³ − 2x² − 2 (11 marks)

• (a) Complete the table (3 marks): −5 at x = −1, −3 at x = 1 and 1.1 at x = 2.5 (1.125 or 9/8 also accepted). B1 for each.
• (b) Draw the graph (4 marks): a smooth curve through all the points, within half a small square. Below full marks, B3FT for 8 correct points, B2FT for 6 or B1FT for 4. FT (follow through) means your points are checked against your own table values, so a wrong value in (a) is not penalised twice. Ruled sections, a thick or feathered curve, or a curve that bends the wrong way lose the curve mark.
• (c) Solve x³ − 2x² − x + 1 = 0 with a straight line (4 marks): rearrange to x³ − 2x² − 2 = x − 3, so rule the line y = x − 3. It crosses the curve at x ≈ −0.8, 0.55 and 2.3 (the scheme accepts −0.9 to −0.7, 0.5 to 0.6 and 2.2 to 2.35). M2 for the correct ruled line, which must cross your curve at all three points; M1 for the correct equation or a line of the right form. The A2 for the values depends on M2 (A1 for two values). If you only had M0 or M1, three correct values still earned SC1. Drawing y = −3 earned nothing.

Question 9: equation with fractions (3 marks)

(w + 8)/4 = (2w − 3)/3.

• Answer: w = 7.2 (7⅕ or 36/5).
• Working: multiply out: 3(w + 8) = 4(2w − 3), so 3w + 24 = 8w − 12, giving 5w = 36.
• Marking: M1 for clearing the fractions, M1 for collecting terms (24 + 12 = 8w − 3w or better), then the answer. The second M1 follows through from your own first step, as long as it is still of the form aw + b = cw + d.

Question 10: coins (4 marks)

10% are gold, 25% of the rest are silver and the other 54 are copper.

• (a) Silver coins (2 marks): copper is 75% of the non-gold coins, so silver is 54 ÷ 75 × 25 = 18.
• (b) Total (2 marks): copper is 0.9 × 0.75 = 0.675 of the total, so 54 ÷ 0.675 = 80. M1 for setting up 0.9 × 0.75 × n = 54 or better. Working for (b) done in part (a) was credited if you used it in (b).

Question 11: right-angled triangle (2 marks)

• Answer: XY = 6.87 cm. XY is opposite the 29° angle and WX = 12.4 cm is adjacent, so XY = 12.4 × tan 29°.
• Marking: M1 for XY ÷ 12.4 = tan 29° or equivalent.

Question 12: exponential decay (3 marks)

The mass is 150 g on 10 January and falls by 5% a day.

• Answer: 19 January. After n days the mass is 150 × 0.95ⁿ. After 8 days it is 99.5… g, and after 9 days 94.5… g, the first value below 96 g. 10 January + 9 days = 19 January.
• Marking: the answer is “nfww” (not from wrong working): it only scores if the working is correct. B2 for 8.7… or 9 days. Otherwise M2 for correctly evaluating 150 × 0.95⁸ or 150 × 0.95⁹ (or comparing 0.95⁸ or 0.95⁹ with 96/150 = 0.64), or M1 for writing 150 × 0.95ⁿ with a value of n above 1. Writing 150 × (95%)ⁿ alone was not a method mark. Logarithms were accepted.

Question 13: inverse proportion (3 marks)

• Answer: y = 7. y = k/x, so k = 6 × 10.5 = 63, and 63 ÷ 9 = 7.
• Marking: M2 for 6 × 10.5 = 9 × y, or M1 for y = k/x and M1 for using your k with x = 9.

Question 14: reflex angle (2 marks)

• Answer: x = 294.2°. cos⁻¹ 0.41 = 65.8°, and the reflex angle with the same cosine is 360° − 65.8° = 294.2°.
• Marking: B1 for 65.8 seen. The answer 294 was accepted for full marks. Answers in radians (5.13) were also accepted.

Question 15: adding algebraic fractions (3 marks)

8/(x + 2) + 3/(x − 4) as a single fraction.

• Answer: (11x − 26) / ((x + 2)(x − 4)), or with the denominator expanded to x² − 2x − 8, as the final answer.
• Working: the numerator is 8(x − 4) + 3(x + 2) = 8x − 32 + 3x + 6.
• Marking: B1 for the common denominator (expanding it is “isw”, ignore subsequent working, so a slip after the correct form is not penalised); B1 for 8(x − 4) + 3(x + 2) or better.

Question 16: square-based pyramid (9 marks)

The base has side 16.5 cm and the vertical height AF is 28 cm, where F is the centre of the base.

• (a) Planes of symmetry (1 mark): 4.
• (b) Volume (2 marks): ⅓ × 16.5² × 28 = 2541 cm³, cao. This is an exact answer, so a value of 2540.7… rounded to 2541 did not get both marks. An answer of 2540 implied M1.
• (c) AB (4 marks): first find BF, half the diagonal of the base: BD² = 16.5² + 16.5² = 544.5, so BD = 23.33… and BF = 11.66…. Then AB² = 28² + BF² = 784 + 136.125, so AB = 30.3 cm. M3 for 28² + 8.25² + 8.25², M2 for any correct BF² or BF (including via sin 45° or cos 45°), or M1 for BD² = 16.5² + 16.5². A BF calculated on the wrong side lost the M2.
• (d) Angle between AB and the base (2 marks): sin⁻¹(28 ÷ 30.33) = 67.4° (67.28 to 67.53 accepted). M1 for sin = 28 ÷ your (c), or tan or cos with your BF, as long as your BF is a genuine attempt and not 28, 16.5, 8.25 or 23.3.

Question 17: functions (7 marks)

f(x) = 20 − 3x, g(x) = 64 ÷ 2ˣ and h(x) = 7 + 2x.

• (a) f(4) (1 mark): 8.
• (b) h⁻¹(x) (2 marks): (x − 7)/2 as the final answer. M1 for a correct first step such as swapping x and y, x = 7 + 2y, or rearranging y − 7 = 2x.
• (c) gg(5) (2 marks): g(5) = 64 ÷ 32 = 2, then g(2) = 64 ÷ 4 = 16. M1 for g(2) or the nested expression.
• (d) g⁻¹(x) = 8 (2 marks): x = g(8) = 64 ÷ 256 = ¼ (0.25). M1 for g(8).

Question 18: triangle areas and a quadratic (8 marks)

Triangle ABC has sides 2x and x with 30° between them; right-angled triangle DEF has legs (x − 4) and (2x − 6).

• (a) Show the area of ABC is ½x² (1 mark): ½ × 2x × x × sin 30° = ½x². This is a “show that”, so sin 30 had to appear in the working, with no errors.
• (b) Show that x² − 14x + 24 = 0 (3 marks): set ½x² = ½(2x − 6)(x − 4) (M1), expand correctly to 2x² − 14x + 24 (B1), then rearrange to the given equation (A1). The A1 needs both earlier marks and no errors or omissions, because the answer is printed on the paper.
• (c) Area of ABC (4 marks): (x − 2)(x − 12) = 0, so x = 2 or 12. x = 2 makes x − 4 negative, so x = 12 and the area is ½ × 12² = 72 cm², cao. M2 for the factorisation (M1 for brackets whose numbers add to −14 or multiply to 24), B1 for x = 12, then B1 for 72. The question said “by factorising”, so the scheme looks for the factors.

Question 19: time in a shop (7 marks)

Four classes, 0–5, 5–10, 10–20 and 20–40 minutes, with frequencies 3, 18, 13 and 6.

• (a) Estimated mean (4 marks): midpoints 2.5, 7.5, 15 and 30; Σft = 7.5 + 135 + 195 + 180 = 517.5; 517.5 ÷ 40 = 12.9375 minutes (12.9 or 12.94 also accepted). M1 for the midpoints (“soi”, seen or implied), M1 for Σft, and a third M1 for dividing by 40, which is “dep” (dependent) on the second. Using the class widths 5, 5, 10 and 20 instead of the midpoints scored nothing for that step.
• (b) Complete the histogram (3 marks): the first bar (height 0.6) was drawn. The other three bars need frequency densities 3.6, 1.3 and 0.3 (frequency ÷ class width). B1 for each correct bar. If no bar was right, SC1 for the three frequency densities.

Question 20: field ABCD (10 marks)

The diagram gave AB = 50.3 m, AC = 86.8 m, AD = 38.2 m, angle DAC = 79°, angle ABC = 112° and a 103° angle at B measured from North to BA.

• (a) Bearing of A from B (1 mark): 360° − 103° = 257°, cao.
• (b) CD (3 marks): cosine rule in triangle ACD: CD² = 38.2² + 86.8² − 2 × 38.2 × 86.8 × cos 79° = 7728.…, so CD = 87.9 m. M2 for the full square root, or M1 for the expression for CD² plus A1 for 7728.
• (c) Shortest distance from B to AC (6 marks): sine rule: sin ACB = 50.3 × sin 112° ÷ 86.8 = 0.537, so angle ACB = 32.5° (B3 for this angle). Then angle BAC = 68° − 32.5° = 35.5°. The shortest distance is perpendicular to AC, so it is 50.3 × sin 35.5° = 29.2 m (nfww). The second M2 depends on at least the first M1. Using the area of triangle ABC two ways (½ × 86.8 × 50.3 × sin BAC = ½ × 86.8 × h) also earned it. If nothing else worked, M1 was available just for recognising that the shortest distance is perpendicular to AC, for example a right angle marked on the diagram.

Question 21: bounds (3 marks)

The volume is 22 cm³ to the nearest cm³ and the mass 140 g to the nearest 10 g.

• Answer: 6 g/cm³ (nfww). The lower bound of density uses the smallest mass and the largest volume: 135 ÷ 22.5 = 6.
• Marking: M2 for a mass from 130 to 140 divided by 22.5, or 135 divided by a volume from 22 to 23. M1 for any of the bounds 145, 135, 22.5 or 21.5 seen.

Question 22: simplifying an algebraic fraction (3 marks)

(8x² + 24x) ÷ (2x² − 18).

• Answer: 4x/(x − 3) as the final answer.
• Working: the numerator is 8x(x + 3) and the denominator is 2(x² − 9) = 2(x − 3)(x + 3). Cancel (x + 3) and the 2.
• Marking: B1 for factorising the numerator, B1 for factorising the denominator (equivalent forms such as (2x − 6)(x + 3) accepted, and 4x(x + 3) and (x − 3)(x + 3) if you had already cancelled the 2).

Where the marks are on 0580/42

• Algebra and graphs carried 38 of the 100 marks: the cubic graph (11), functions (7), the triangle-area quadratic (8), and Questions 9, 13, 15 and 22 (3 each).
• Trigonometry carried 20 marks: right-angled trigonometry in Question 11 (2), the reflex angle in Question 14 (2), the pyramid edge and angle in Question 16(c) and (d) (6) and the bearing, cosine rule and sine rule in Question 20 (10).
• Number carried 18 marks: percentages in Questions 4 and 10, compound interest (5), exponential decay (3) and bounds (3).
• Statistics carried 12 marks: the frequency table in Question 2 (5) and the grouped mean and histogram in Question 19 (7).
• Geometry and mensuration carried 10 marks: angles (2), the construction (2), the semicircle (3) and the pyramid’s symmetry and volume (3).
• Vectors carried 2 marks (Question 5).

The longest questions were Question 8 (11 marks), Question 20 (10 marks, 6 of them on part (c)), Question 16 (9 marks) and Question 18 (8 marks). The first 15 questions were worth 2 to 5 marks each, apart from the 11-mark graph.

How to self-mark 0580/42 with the mark scheme

1. Mark the final answer first, then the working. A correct answer usually earns all the marks for that part, but watch for “nfww” answers (Questions 12, 20(c) and 21): these score only if the working that led to them is correct.
2. Give yourself M marks for a correct method even when the answer is wrong. In Question 6, π × 15.8 ÷ 2 is worth M1 on its own; in Question 20(c), marking the right angle on the diagram is worth M1.
3. Remember A marks depend on M marks. In Question 8(c), the A2 for the three x-values needs the M2 for the correct ruled line, so reading values off a wrong line scores only the SC1.
4. Apply follow through. In Question 8(b), check your plotted points against your own table; in Question 16(d), use your own AB from part (c). One early slip should cost you one mark, not the whole question.
5. Check accuracy rules. Non-exact answers to 3 significant figures and angles to 1 decimal place: 40.6 cm, 6.87 cm, 294.2°, 67.4°, 29.2 m. Exact answers such as 2541 cm³ (Question 16(b)) and 72 cm² (Question 18(c)) are cao, so keep full calculator values until the end.
6. Treat “show that” parts strictly. In Questions 18(a) and (b), the printed result is no use without every step. If sin 30 is missing in (a), or the expansion is not shown in (b), the mark goes.
7. Log each lost mark as a missing method, an arithmetic or rounding slip, a premature rounding, or a misread question, and revise in that order.

Sit 0580/42 as a timed mock

The paper gives 120 minutes for 100 marks, a little over a minute per mark. A workable plan:

Block	Marks	Time
Questions 1 to 7	22	23 minutes
Question 8 (cubic graph)	11	13 minutes
Questions 9 to 15	20	22 minutes
Question 16 (pyramid)	9	11 minutes
Question 17 (functions)	7	7 minutes
Question 18 (quadratic)	8	9 minutes
Question 19 (grouped data)	7	8 minutes
Question 20 (field ABCD)	10	13 minutes
Questions 21 and 22	6	8 minutes
Check, especially rounding and angles	–	6 minutes

Questions 8 and 20 need the most time per mark: plotting and drawing a smooth curve is slow, and Question 20(c) has several steps before the final answer.

Other May/June 2026 0580 papers

• 0580 Paper 12 (Core Non-calculator)
• 0580 Paper 22 (Extended Non-calculator)
• 0580 Paper 32 (Core Calculator)

Practise and download

• Download 0580/42 and its mark scheme on the Learning Portal →
• All 0580 May/June 2026 papers: overview and marking patterns
• Try a free IGCSE Mathematics quiz

Need help with a 0580 topic? Book a free trial lesson with an IGCSE Mathematics tutor →

Frequently asked questions

What topics were on 0580 Paper 42 May/June 2026? +

Twenty-two questions, including statistics from a frequency table, compound interest, a cubic graph solved with a straight line, exponential decay, functions, a square-based pyramid, a triangle-area quadratic, a histogram, bearings with the sine and cosine rules, bounds and algebraic fractions.

How many marks is 0580/42 and how long is it? +

It is out of 100 marks and lasts 2 hours. Question 8 (the cubic graph) is the biggest at 11 marks, followed by Question 20 (the field ABCD, bearings and triangle trigonometry) at 10 marks.

What was the answer to the radioactive mass question in 0580/42? +

19 January. The mass is 150 × 0.95ⁿ after n days: after 8 days it is about 99.5 g and after 9 days about 94.5 g, so it first falls below 96 g nine days after 10 January. The answer was worth 3 marks and had to come from correct working.

How do you find the shortest distance from B to AC in 0580/42 Question 20? +

Use the sine rule in triangle ABC to find angle ACB (32.5°), so angle BAC is 68° − 32.5° = 35.5°. The shortest distance is perpendicular to AC, so it is 50.3 × sin 35.5° = 29.2 m. The part was worth 6 marks, and recognising the perpendicular alone earned a method mark.

Is 0580/42 the Core or Extended paper? +

Extended. Paper 4 is the Extended calculator paper, sat alongside Paper 2 (Extended non-calculator). Core candidates sit Papers 1 and 3 instead.

Where can I download 0580/42 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge IGCSE Mathematics 0580 past papers page.

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