Cambridge IGCSE Mathematics 0580/22 is the May/June 2026 Paper 2 Non-calculator (Extended), variant 2. It lasts 2 hours, is worth 100 marks and is sat by every Extended candidate, alongside Paper 4. It had 25 compulsory questions and no question was worth more than 7 marks; the biggest were transformations (Question 5), a region from four inequalities (Question 12) and a tangent and perpendicular line found by differentiation (Question 23). This guide goes through the paper question by question, giving the final answer and showing how the official mark scheme split the marks. 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.

0580/22 at a glance

Question	Topic	Marks
1	Scatter diagram: plotting, correlation, line of best fit, estimate	4
2	Net of an open box: total area and volume	5
3	Reciprocal of a mixed number	2
4	Dividing fractions	2
5	Translation, reflection in y = x, describing a rotation	7
6	Substituting into a formula and making c the subject	5
7	Factorising	2
8	Exact values of cos 45° and tan 60°	2
9	Prime factors, LCM and an algebraic HCF	6
10	Recurring decimal to a fraction	3
11	Sides of a regular polygon from its interior angle	2
12	Drawing four lines and shading a region of inequalities	7
13	Arc length in terms of π	2
14	Sequences: terms from an nth term, nth term of a geometric sequence	4
15	Circle theorems with reasons	5
16	Radius of a sphere using π ≈ 3	2
17	Surds: expanding brackets and rationalising a denominator	5
18	Probability with and without replacement	6
19	Expanding three brackets	3
20	Similar bottles: volume and height	3
21	Completing the square and the turning point	3
22	Vectors in a parallelogram and collinearity	5
23	Gradient of a tangent by differentiation, equation of a perpendicular line	7
24	Solving an equation with indices	3
25	Where a straight line meets a curve	5
	Total	100

Question-by-question walkthrough

The mark scheme uses a few letters throughout. M marks are for a correct method, even if the answer is then wrong. A marks are for an accurate answer and need the linked M mark. B marks are independent marks for a correct value or statement. FT (follow through) means you can still score using your own earlier wrong answer. oe means “or equivalent”, cao means “correct answer only” (no other form or follow through), isw means “ignore subsequent working”, and SC is a special case: a mark the scheme gives for a specific partly right answer.

Question 1: scatter diagram of fuel consumption and CO₂ emissions (4 marks)

• Plot a point at (6.2, 162) (1 mark): it had to lie strictly between 6.15 and 6.25 across and between 160 and 164 up.
• Type of correlation (1 mark): positive. Words like “strong” were ignored, “+ve” was condoned, but ”+” on its own was not.
• Line of best fit (1 mark): a ruled line passing within set windows at fuel consumption 4.35 and 6.5. Very good freehand was condoned.
• Estimate the fuel consumption at 140 g/km (1 mark): read from your own line (strict FT), as long as it had a positive gradient. The tolerance was strictly less than one small square, so misreading the scale cost the mark.

Question 2: net of an open box (5 marks)

The net was drawn accurately but unlabelled, so you had to measure it: the box was 3 cm by 5 cm with a height of 2 cm.

• Total area of the net (3 marks): base 3 × 5 = 15, two faces 2 × 3 = 6 and two faces 2 × 5 = 10, so 15 + 12 + 20 = 47 cm². B1 for finding the lengths 2, 3 and 5 (or 2, 5 and 11), and M1 for the correct sum of areas, follow through on your own whole-number lengths written on the diagram. An answer of 62 cm² (the surface area of a closed box, adding the missing lid) was a special case worth 2 marks.
• Volume (2 marks): 2 × 3 × 5 = 30 cm³, with full follow through from your own whole-number measurements, or M1 for the product.

Question 3: reciprocal of 3⅓ (2 marks)

3⅓ = 10/3, so the reciprocal is 3/10 (cao, as a fraction in its simplest form). B1 for 10/3 or for the answer 0.3. A negative answer, −3/10, did not earn the method mark.

Question 4: dividing fractions (2 marks)

1/6 ÷ 2/7 = 1/6 × 7/2 = 7/12 (oe; 0.583 or better was also accepted). M1 for writing it as a multiplication by 7/2, or for converting both to 42nds.

Question 5: transformations (7 marks)

• Translation by the vector (−2, 5) (2 marks): image at (−6, 7), (−4, 7), (−4, 5), (−5, 5). B1 for a translation that got only one of the two components right.
• Reflection in y = x (2 marks): image at (0, −2), (2, −2), (2, −4), (0, −3). B1 for three correct vertices. If neither mark was earned, reflecting in y = −x instead was a special case worth 1 mark.
• Describe the single transformation from A to B (3 marks): rotation, 90° anticlockwise, centre (3, 3), one mark each. 270° clockwise was also accepted. Words such as “turn” or “spin” were not, and naming more than one transformation scored 0 for the whole part.

Question 6: the formula E = 5c² − d² (5 marks)

• Find E when c = √18 and d = 2 (2 marks): c² = 18, so E = 5 × 18 − 4 = 90 − 4 = 86. M1 for the substitution, or B1 for seeing 18, 90 or −4.
• Make c the subject (3 marks): add d², divide by 5, then square root: c = √((E + d²) ÷ 5). One M1 for each step, in any order, but the most you could score with a wrong final answer was 2. The square root sign had to cover the whole fraction.

Question 7: factorising 6wx − 4xy (2 marks)

2x(3w − 2y). A partial factorisation, such as 2(3wx − 2xy) or x(6w − 4y), earned B1.

Question 8: exact trig values (2 marks)

From the list given, cos 45° = 1/√2 and tan 60° = √3, one B1 each. Equivalent exact forms such as √2/2 were condoned, but decimals alone scored 0.

Question 9: prime factors, LCM and HCF (6 marks)

• 63 as a product of primes (2 marks): 3² × 7 (or 3 × 3 × 7). B1 for listing 3, 3, 7 not written as a product, or M1 for a correct factor tree.
• LCM of 63 and 42 (2 marks): 42 = 2 × 3 × 7, so the LCM is 2 × 3² × 7 = 126. B1 for another common multiple such as 252, or M1 for the prime factors of 42.
• HCF of 63a⁵c² and 42a²c⁵ (2 marks): 21a²c². B1 for getting either the number or the powers right (for example 21 alone, or a number times a²c²).

Question 10: recurring decimal to a fraction (3 marks)

0.7333… : 100x − 10x = 73.3… − 7.3… = 66, so 90x = 66 and x = 66/90 = 11/15 (cao). B2 for an unsimplified fraction equal to 66/90, as long as it contains no decimals; M1 for the subtraction step.

Question 11: regular polygon with interior angle 140° (2 marks)

The exterior angle is 180 − 140 = 40°, and 360 ÷ 40 = 9 sides. M1 for 360 ÷ (180 − 140), or for 180(n − 2) = 140n.

Question 12: lines and a region of inequalities (7 marks)

• Draw x = 8, y = 2, x + y = 8 and 2y − x = 4 (5 marks): B1 for each of the first three lines and B2 for 2y − x = 4 (B1 for a diagonal line through (0, 2) or (8, 6), or with gradient ½). Lines had to be ruled and solid. The scheme withheld 1 mark the first time a dashed line appeared anywhere in Question 12.
• Shade the unwanted regions and label R (2 marks): R is to the left of x = 8, above y = 2, to the right of x + y = 8 and below 2y − x = 4, a quadrilateral with corners at (4, 4), (6, 2), (8, 2) and (8, 6). B1 for a region that satisfies three of the four inequalities.

Question 13: arc length (2 marks)

90/360 × 2 × π × 12 = 6π cm (cao). M1 for the method; a decimal such as 18.8 showed the method but did not earn the answer mark, because the question asked for the answer in terms of π.

Question 14: sequences (4 marks)

• First three terms of n² − 3 (2 marks): −2, 1, 6, in that order. B1 for one correct value in the right position. Starting from n = 0 (−3, −2, 1) was a special case worth 1 mark.
• nth term of 4, 12, 36, 108 (2 marks): each term is 3 times the last, so the nth term is 4 × 3ⁿ⁻¹ (oe, such as (4/3) × 3ⁿ). B1 for an expression built on a power of 3.

Question 15: circle theorems with reasons (5 marks)

ABCD is a cyclic quadrilateral with a tangent EF at A, angle ADB = 20° and angle BCD = 75°.

• Angle BAD = 180 − 75 = 105°, because opposite angles of a cyclic quadrilateral add up to 180°.
• Angle ABD = 180 − 105 − 20 = 55° (angles in triangle ABD).
• Angle EAD = angle ABD = 55°, by the alternate segment theorem.

The marks split 3 for the working and 2 for the reasons. B3 for 55°; B2 for angle ABD = 55°; B1 for angle BAD = 105° or angle BAF = 20°. Then B1 for each of the two reasons above, as long as no reason was linked to wrong working. The scheme was strict about wording: “alternate segment” alone, “AST”, “alternate angles” or “opposite segment theorem” did not count.

Question 16: radius of a sphere (2 marks)

Using π ≈ 3, 4/3 × 3 × r³ = 500, so 4r³ = 500, r³ = 125 and r = 5 cm. M1 for reaching r³ = 125 or an equivalent expression; ±5 did not get full marks.

Question 17: surds (5 marks)

• Expand (4 − √7)(3 + 2√7) (2 marks): 12 + 8√7 − 3√7 − 14 = −2 + 5√7 (cao). B1 for the four-term expansion with at least three terms correct.
• Rationalise 2/(3 − √5) (3 marks): multiply top and bottom by 3 + √5. The denominator is 9 − 5 = 4, so the result is (6 + 2√5)/4 = (3 + √5)/2 (cao, though simplified equivalents such as 1.5 + 0.5√5 were allowed). B2 for (6 + 2√5)/4; otherwise M1 for multiplying by (3 + √5)/(3 + √5) and M1 for expanding the denominator correctly.

Question 18: probability with numbered tiles (6 marks)

The ten tiles were 1, 2, 2, 3, 3, 4, 4, 4, 4, 4.

• Same even number, with replacement (3 marks): P(2, 2) + P(4, 4) = (2/10)² + (5/10)² = 4/100 + 25/100 = 29/100. M2 for the sum of the two products, or M1 for one product. Without replacement (11/45) was a special case worth 1 mark.
• Three tiles summing to 11, without replacement (3 marks): the only way is 4, 4 and 3, in any of three orders. 3 × (5/10 × 4/9 × 2/8) = 120/720 = 1/6. M2 for one order multiplied by 1, 2 or 3; M1 for a product of three fractions over 10 × 9 × 8. With replacement (3/20) was a special case worth 1 mark.

In both parts the scheme did not accept a ratio, such as 29 : 100, and withheld 1 mark the first time an answer was written in words, such as “29 out of 100”.

Question 19: expanding three brackets (3 marks)

(3x + 1)(x + 4)(x − 2) = 3x³ + 7x² − 22x − 8. B2 for a correct unsimplified expansion, or a four-term answer of the right form with three terms correct; B1 for one pair of brackets expanded with at least three of the four terms correct.

Question 20: similar bottles (3 marks)

The volume scale factor is 27/216 = 1/8, so the length scale factor is ∛(1/8) = 1/2, and h = 10 × 1/2 = 5 cm. M2 for h = ∛(27/216) × 10, or M1 for taking a cube root of the volumes (or a length scale factor of ½ or 2).

Question 21: completing the square (3 marks)

• x² + 8x − 7 in the form (x + a)² + b (2 marks): (x + 4)² − 23. B1 for (x + 4)².
• Turning point of y = x² + 8x − 7 (1 mark): (−4, −23), follow through from your answer to part (a) if it was in the correct form.

Question 22: vectors and collinearity (5 marks)

OABC is a parallelogram with OA = a and OC = c, and P divides AC in the ratio 1 : 4.

• AP (2 marks): AC = c − a, so AP = ⅕(c − a). M1 for AC = c − a, or for a correct route using ⅕ of AC.
• OP (1 mark): OP = a + ⅕(c − a) = ⅘a + ⅕c, FT from part (i).
• Show that O, P and Q are collinear, where AQ = ¼c (2 marks): OQ = a + ¼c = ¼(4a + c) and OP = ⅕(4a + c). B1 for showing that two of OP, OQ and PQ share a common factor (or, if not factorised, for two correct vectors followed by a correct relationship such as OQ = 5/4 OP). B1 for the conclusion: OQ is a multiple of OP, and they share the point O, so the points are collinear. This is a “show that”, so the conclusion has to be stated; the common point is part of the mark.

Question 23: tangent and perpendicular line (7 marks)

The curve was y = 2x² − 11x + 12.

• Gradient of the tangent at x = 2 (3 marks): dy/dx = 4x − 11, so at x = 2 the gradient is −3. B2 for 4x − 11, or B1 for one of its two terms right. Writing “y = −3” as the answer did not get all 3 marks.
• Equation of line L, perpendicular to the tangent at x = 2 (4 marks): at x = 2, y = 8 − 22 + 12 = −2. The perpendicular gradient is 1/3. So y + 2 = ⅓(x − 2), giving y = ⅓x − 8/3 (or 3y − x + 8 = 0). B1 for y = −2; M1 for the gradient 1/3 (or −1 ÷ your part (a)); M1 for substituting (2, −2) into y = mx + c. Like terms had to be collected for 4 marks, and rounded decimals such as −2.67 were not accepted for −8/3.

Question 24: indices equation (3 marks)

Write 49 as 7², so 7²ˣ⁺¹ = 7²⁽²⁻ˣ⁾. Then 2x + 1 = 4 − 2x, 4x = 3 and x = 3/4. B2 for 2x + 1 = 2(2 − x); M1 for rewriting either side with a common base and equating the powers.

Question 25: a line meeting a curve (5 marks)

Set 7 − 4x = 4 + 3x − 2x², which rearranges to 2x² − 7x + 3 = 0, then (2x − 1)(x − 3) = 0. So x = 0.5 or x = 3, and substituting into y = 7 − 4x gives (0.5, 5) and (3, −5).

The partial marks were M2 for reaching 2x² − 7x + 3 = 0 (or M1 for equating the two expressions), M1 FT for a correct method to solve your own three-term quadratic (factors, formula or completing the square), and B1 for one correct pair of coordinates, two correct x-values or two correct y-values. Eliminating x instead, which leads to y² − 25 = 0, was marked in the same way.

Where the marks are on 0580/22

Topic area	Questions	Marks
Number: fractions, reciprocals, prime factors, LCM and HCF, recurring decimals, surds	3, 4, 9, 10, 17	18
Algebraic manipulation and equations: formulae, factorising, expanding, completing the square, indices, simultaneous equations	6, 7, 19, 21, 24, 25	21
Sequences	14	4
Graphs and calculus: lines and inequalities, differentiation	12, 23	14
Mensuration: net and volume, arc length, sphere, similar solids	2, 13, 16, 20	12
Transformations and vectors	5, 22	12
Angles, polygons, trigonometry and circle theorems	8, 11, 15	9
Statistics and probability	1, 18	10
Total		100

• Algebra in all its forms carried 39 marks, the largest share. Questions 23 and 25 alone carried 12 marks between them, near the end of the paper.
• Seven-mark questions: 5 (transformations), 12 (four lines and a region) and 23 (tangent and perpendicular line).
• Reasons and conclusions carried marks in 15 (two reasons) and 22(b) (a stated conclusion with a common point).
• Exact answers were required throughout. With no calculator, answers came as fractions, surds or multiples of π (3/10, 11/15, 6π, (3 + √5)/2, −8/3), and decimals often lost the accuracy mark.

How to self-mark 0580/22 with the mark scheme

1. Mark the final answer first, then the working. A correct answer usually scores every mark. If it is wrong, go through the M and B marks: in Question 23(b), y = −2, the gradient 1/3 and the substitution are worth 3 of the 4 marks even if the final line is wrong.
2. Check whether the answer is cao. Questions 3, 10, 13, 17(a) and 17(b) were cao, so 0.3 in place of 3/10, an unsimplified 66/90 or 18.8 in place of 6π all lose the final mark.
3. Apply follow through fairly, and only where the scheme allows it. 1(d) follows your own line of best fit, 2(b) your own measurements, 21(b) your completed square and 22(a)(ii) your AP. If you made one early slip, give yourself the later marks you earned from it.
4. Look for special cases. 62 cm² in 2(a), a reflection in y = −x in 5(a)(ii), terms from n = 0 in 14(a), and the “wrong replacement” answers 11/45 and 3/20 in Question 18 each earn a fixed number of marks.
5. Check the conventions that cost marks. A dashed line in Question 12, two transformations in 5(b), “AST” as a reason in Question 15, or a probability written as a ratio or in words in Question 18 all lose marks even when the maths is right.
6. Log each lost mark as a missing method, an arithmetic slip, a non-exact or unsimplified answer, or a missing reason or conclusion, and revise in that order.

Sit 0580/22 as a timed mock

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

Block	Marks	Time
Questions 1–4	13	13 minutes
Questions 5–8	16	17 minutes
Questions 9–11	11	12 minutes
Questions 12–14	13	14 minutes
Questions 15–17	12	14 minutes
Questions 18–20	12	14 minutes
Questions 21–22	8	10 minutes
Questions 23–25	15	18 minutes
Check, especially exact forms and simplified answers	–	8 minutes

Questions 23 to 25 are multi-step algebra worth 15 marks between them, so make sure you reach them with at least 18 minutes left.

Other May/June 2026 0580 papers

• 0580 Paper 12 (Core Non-calculator)
• 0580 Paper 32 (Core Calculator)
• 0580 Paper 42 (Extended Calculator)

Practise and download

• Download 0580/22 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 22 May/June 2026? +

Twenty-five questions covering a scatter diagram, the net of an open box, fractions and reciprocals, transformations, rearranging formulae, exact trig values, prime factors with LCM and HCF, a recurring decimal, regions from inequalities, sequences, circle theorems, surds, probability, similar volumes, completing the square, vectors and collinearity, differentiation, an indices equation and a line meeting a curve.

How many marks is 0580/22 and how long is it? +

It is out of 100 marks and lasts 2 hours, with no calculator allowed. The biggest questions were worth 7 marks each: transformations (Question 5), drawing four lines and a region (Question 12) and the tangent and perpendicular line (Question 23).

What was the answer to the circle theorem question in 0580/22? +

Angle EAD was 55°. Opposite angles of the cyclic quadrilateral give angle BAD = 105°, and the alternate segment theorem links the tangent to the angles in the circle. Three marks were for the working and two were for naming both reasons correctly; 'AST' or just 'alternate segment' did not count.

What was the answer to Question 25 on 0580/22 May/June 2026? +

The line y = 7 − 4x meets the curve y = 4 + 3x − 2x² at (0.5, 5) and (3, −5). Setting the two expressions equal gives 2x² − 7x + 3 = 0, which factorises to (2x − 1)(x − 3) = 0.

Is 0580/22 the Core or Extended paper? +

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

Where can I download 0580/22 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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