Cambridge IGCSE International Mathematics 0607/22 is the May/June 2026 Paper 2 (Extended Non-calculator), variant 2. It lasts 1 hour 30 minutes, is worth 75 marks and is sat by every Extended candidate. It had eighteen compulsory questions. Most were worth 1 to 6 marks, but Question 16 on functions carried 10, and the paper ended with a 6-mark “show that” and factorising question on a car journey. 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 six 0607 papers from this series, see IGCSE International Maths 0607 May/June 2026 past papers: what came up and how it was marked.

0607/22 at a glance

Question	Topic	Marks
1	Measuring a reflex angle, angles with parallel lines	3
2	Probability a train is not on time, expected number	2
3	Time interval running into the next day	1
4	Translation, describing a rotation	5
5	Simple interest: finding the rate	3
6	Vectors: column vector, magnitude, finding a point	6
7	Adding mixed numbers	3
8	Formula with a cube root: substituting, rearranging	3
9	Surface area of a cylinder and hemisphere in terms of π	4
10	Circle with an isosceles triangle: finding angle OAB	3
11	Linear equation with fractions	3
12	Venn diagram: counting, probability, two picks without replacement	6
13	Indices, including a fractional power	4
14	Area of a triangle from a line’s intercepts, perpendicular line	6
15	Exact values of cos 30° and cos 330°	2
16	Functions: evaluating, inverse, composite, logarithmic function and its inverse	10
17	Surds: simplifying, rationalising the denominator	5
18	Two-stage journey: forming and solving a quadratic by factorising	6
	Total	75

Question-by-question walkthrough

The mark scheme uses a few letters throughout. M marks are method marks for a valid method; A marks are accuracy marks for a correct answer, and they need the M mark before them; B marks are independent marks for a correct result or statement. M2 or B2 means a mark worth 2 for a fuller step. FT (follow through) means a later mark can still be earned using your own earlier wrong answer, and oe means “or equivalent”. Other abbreviations are explained where they first appear.

Question 1: angles (3 marks)

• Measure the reflex angle r (1 mark): 320°, with anything from 318° to 322° accepted. Measure the acute angle and subtract it from 360° if your protractor only goes to 180°.
• Angle m between two parallel lines (2 marks): two lines met on the upper parallel line, making angles of 35° and m° there, and one of them crossed the lower line at 70°. The parallel lines carry the 70° angle up to the top line, and angles on a straight line then give m = 180 − 70 − 35 = 75. B1 for 35°, 70° or 110° marked in the correct position, or M1 for 180 − 70 − 35 or 110 − 35.

Question 2: probability (2 marks)

The probability that a train arrives on time was 5/6.

• Not on time (1 mark): 1 − 5/6 = 1/6. 0.167 or 16.7% (or better) were accepted.
• Expected number on time out of 72 (1 mark): 72 × 5/6 = 60.

Question 3: a time interval (1 mark)

• From 18 35 to 09 40 the next day (1 mark): 18 35 to midnight is 5 h 25 min, and midnight to 09 40 is 9 h 40 min, so the total is 15 h 5 min.

Question 4: transformations (5 marks)

Triangle A had vertices (1, 1), (1, 2) and (3, 1); triangle C had vertices (3, 1), (3, 3) and (4, 3).

• Translate A by the vector (1, 3) (2 marks): the image has vertices (2, 4), (2, 5) and (4, 4). B1 for a translation with one component right, or for one using (3, 1), the vector with its components swapped.
• Describe the single transformation from A to C (3 marks): rotation, 90° clockwise, centre (3, 1). B1 for each fact. Naming more than one transformation scored 0, and the centre had to be a point: “3, 1” was condoned, but a column vector was not.

Question 5: simple interest (3 marks)

• Find the rate (3 marks): the interest over 2 years is 212 − 200 = $12, so $6 a year, and 6 ÷ 200 × 100 = 3%. B2 for (212 − 200) ÷ 2, or for stating that one year’s interest is $6; B1 for 212 − 200 alone. The formula route (200 × r × 2 ÷ 100 = 12) earned the same marks.

Question 6: vectors (6 marks)

P was (3, 7) and Q was (7, 4).

• Show that PQ = (4, −3) (1 mark): subtract the position vectors: (7 − 3, 4 − 7) = (4, −3). The working had to be in vector form.
• Magnitude of PQ (2 marks): √(4² + (−3)²) = √25 = 5. M1 for 4² + (−3)². Writing 4² + −3² (which reads as 16 − 9) did not get the M1 unless you recovered from it.
• R where PR = 2PQ (3 marks): 2PQ = (8, −6), so R = (3 + 8, 7 − 6) = (11, 1). B2 for one coordinate correct, B1 for (8, −6), or M1 for a sketch with QR the same as PQ.

Question 7: mixed numbers (3 marks)

• 1 5/6 + 3 1/4 as a mixed number in simplest form (3 marks): 22/12 + 39/12 = 61/12 = 5 1/12. This was cao (correct answer only), so 61/12 earned B2, not 3. M2 for both fractions over a common denominator, or M1 for converting one correctly.

Question 8: a formula with a cube root (3 marks)

The formula was y = ∛t ÷ w.

• y when t = 27 and w = 9 (1 mark): ∛27 = 3, so y = 3/9 = 1/3. isw (ignore subsequent working) applied if you then tried to convert it to a decimal.
• Make t the subject (2 marks): multiply by w to get wy = ∛t, then cube both sides: t = (wy)³, or w³y³. M1 for wy = ∛t or y³ = t ÷ w³; B1 for an answer with only one of the letters cubed.

Question 9: surface area of a cylinder and hemisphere (4 marks)

The cylinder was 10 cm long with radius 3 cm, and a hemisphere of radius 3 cm closed one end. The total surface area was kπ.

• Find k (4 marks): the curved hemisphere is ½ × 4π × 3² = 18π; the curved cylinder is 2π × 3 × 10 = 60π; the flat circular end is π × 3² = 9π. The total is 87π, so k = 87. Each of the three areas was a separate M1. 87π was condoned for full marks, but the flat end is the easy one to forget.

Question 10: angles in a circle (3 marks)

A, B and C were on a circle with centre O, with AB = BC and angle ABC = 130°.

• Angle OAB (3 marks): angle ABC stands on the major arc, so the reflex angle AOC is 2 × 130 = 260°, and the other angle AOC is 100°. In the quadrilateral OABC, angles OAB and OCB together make 360 − 130 − 100 = 130°, and the shape is symmetrical, so angle OAB = 65°. M2 for (360 − 130 − 100) ÷ 2. Otherwise B2 for two, or B1 for one, of 260°, 100°, 25° or 40° in the correct position (or angles of 50° at the centre, AOB or COB).

Question 11: a linear equation (3 marks)

• Solve (t + 1)/7 = (4 − t)/2 (3 marks): cross-multiply to get 2(t + 1) = 7(4 − t), so 2t + 2 = 28 − 7t, 9t = 26 and t = 26/9 (2 8/9). M1 for 2t + 2 = 28 − 7t, then M1 for collecting to kt = c from your four-term equation. 2.89 was allowed.

Question 12: a Venn diagram (6 marks)

Of 24 students, 6 studied German only, 9 both, 8 Spanish only and 1 neither.

• Neither language (1 mark): 1. Writing {1} was not accepted.
• n(G ∩ S′) (1 mark): German but not Spanish, 6.
• German or Spanish but not both (1 mark): (6 + 8) ÷ 24 = 14/24 (7/12). Decimals to 3 significant figures were fine; ratios or words were not.
• Two German students: one studies Spanish, one does not (3 marks): 15 study German, 9 of them Spanish and 6 not. The two orders are 9/15 × 6/14 and 6/15 × 9/14, so the probability is 2 × 54/210 = 18/35. M2 for 9/15 × 6/14 × 2 (or the other order × 2), M1 for any one of 9/15, 9/14, 6/15 or 6/14. Remember both orders: 9/15 × 6/14 on its own is only M1.

Question 13: indices (4 marks)

• (3y³)⁴ (2 marks): 81y¹². B1 for ky¹² or 81y^k. Leaving it as 3⁴y¹² scored only B1.
• (16x¹⁶)^¼ (2 marks): the fourth root of 16 is 2 and 16 × ¼ = 4, so 2x⁴. B1 for kx⁴ or 2x^k.

Question 14: a straight line (6 marks)

Line L was y = 3x − 4.

• Area of triangle OAB (4 marks): A, the y-intercept, is (0, −4); B, the x-intercept, comes from 0 = 3x − 4, so x = 4/3. The area is ½ × 4 × 4/3 = 8/3 (2 2/3). B3 for ½ × 4 × 4/3; otherwise B1 for the x-coordinate of B, B1 for the y-coordinate of A, and M1 for ½ × your two lengths. Negative lengths were allowed in the method, but the final area had to be positive. 2.67 was accepted, but not 2.66.
• Perpendicular line through (0, 5) (2 marks): the gradient of L is 3, so the perpendicular gradient is −1/3, giving y = −⅓x + 5. B1 for the right gradient with the wrong intercept, or the right intercept with the wrong gradient.

Question 15: exact cosines (2 marks)

• cos 30° (1 mark): √3/2.
• cos 330° (1 mark): cosine is positive in the fourth quadrant and 330° = 360° − 30°, so cos 330° = √3/2. This was FT from your answer to (a).

Question 16: functions (10 marks)

• f(2) for f(x) = 2x + 1 (1 mark): 2 × 2 + 1 = 5.
• f⁻¹(x) (2 marks): write x = 2y + 1 and rearrange: f⁻¹(x) = (x − 1)/2. M1 for x = 2y + 1 (or the equivalent first step). (1 − x)/−2 was also accepted.
• gg(x) for g(x) = 1/(x + 2), as a fraction in its simplest form (3 marks): gg(x) = 1 ÷ (1/(x + 2) + 2). Writing the denominator as a single fraction gives (1 + 2(x + 2))/(x + 2), so gg(x) = (x + 2)/(2x + 5). M1 for substituting g into itself; M2 for reaching the combined form.
• Solve h(x) = 3 for h(x) = 1 + log₃ x (2 marks): log₃ x = 2, so x = 3² = 9. M1 for log₃ x = 2.
• h⁻¹(x) (2 marks): swap x and y to get x = 1 + log₃ y, so log₃ y = x − 1 and h⁻¹(x) = 3^(x − 1). M1 for the swapped equation or x − 1 = log₃ y.

Question 17: surds (5 marks)

• √200 + √98 (2 marks): √200 = 10√2 and √98 = 7√2, so the answer is 17√2. B1 for either surd simplified.
• Rationalise 5/(5 + √5) and simplify (3 marks): multiply top and bottom by 5 − √5. The denominator becomes 25 − 5 = 20, so the fraction is 5(5 − √5)/20 = (5 − √5)/4. M1 for multiplying by (5 − √5)/(5 − √5), and M2 for the fully expanded fraction. The scheme’s answer is the cancelled form; 5(5 − √5)/20 appears in its guidance as working for the M2.

Question 18: a two-stage car journey (6 marks)

Stage 1 was 80 km at x km/h, stage 2 was 450 km at (x + 50) km/h, and the whole journey took 7 hours.

• Show that 7x² − 180x − 4000 = 0 (3 marks): time = distance ÷ speed, so 80/x + 450/(x + 50) = 7 (M1). Multiply through by x(x + 50): 80(x + 50) + 450x = 7x(x + 50) (M1). Expand and collect: 80x + 4000 + 450x = 7x² + 350x, which leads to 7x² − 180x − 4000 = 0 (A1). As a “show that”, the A1 needed at least one line of working after clearing the fractions and no errors or omissions; writing the given equation at the end is not enough.
• Solve by factorising (3 marks): (7x + 100)(x − 40) = 0, so x = 40. M2 for the correct factors, or M1 for brackets (7x + a)(x + b) with ab = −4000 or a + 7b = −180, or for a correct grouping step. B1 for x = 40. The other root, −100/7, is negative and cannot be a speed: giving 40 and −100/7 together did not earn the B1.

Where the marks are on 0607/22

• Algebra carried 16 of the 75 marks: the cube-root formula (Question 8), the linear equation (Question 11), indices (Question 13) and the journey quadratic (Question 18).
• Number carried 12 marks: time (Question 3), simple interest (Question 5), mixed numbers (Question 7) and surds (Question 17).
• Vectors and transformations carried 11 marks: Questions 4 and 6.
• Functions carried 10 marks, all in Question 16, the biggest question on the paper.
• Sets and probability carried 8 marks: Questions 2 and 12.
• Geometry carried 6 marks (angles in Question 1 and the circle in Question 10) and coordinate geometry 6 marks (Question 14).
• Mensuration carried 4 marks (Question 9) and trigonometry 2 marks (Question 15).

The longest question was Question 16 (10 marks over five parts), followed by four 6-mark questions: vectors (6), the Venn diagram (12), the straight line (14) and the journey (18). The single biggest parts were 9 and 14(a), each worth 4.

How to self-mark 0607/22 with the mark scheme

1. Check the answer line first. A correct final answer usually scores full marks, but not on “show that” questions: in 6(a) and 18(a) the working is what is marked.
2. Watch for cao and final-form demands. Question 7 wanted a mixed number in its simplest form, so 61/12 lost a mark. 13(a) left as 3⁴y¹² scored only 1 of the 2 marks, and 17(b) expects the cancelled answer (5 − √5)/4.
3. Count your method marks when the answer is wrong. Question 9 gave M1 for each of the three areas, 14(a) gave B1 for each intercept, and 12(d) gave M1 for a single correct fraction. Those marks survive an arithmetic slip.
4. Apply follow through only where the scheme says FT. 15(b) followed through from 15(a); most other parts did not.
5. Check the rejected forms. {1} in 12(a), a column vector for a centre of rotation in 4(b), 4² + −3² in 6(b), and two roots in 18(b) all cost marks even with the right idea.
6. Log each lost mark as missing knowledge, an algebra or arithmetic slip, an unsimplified answer or a misread question, and revise in that order.

Sit 0607/22 as a timed mock

The paper allows 90 minutes for 75 marks, a little over one minute per mark. A workable plan:

Block	Marks	Time
Questions 1–3	6	6 minutes
Questions 4–5	8	9 minutes
Questions 6–7	9	10 minutes
Questions 8–9	7	8 minutes
Questions 10–11	6	7 minutes
Question 12	6	7 minutes
Questions 13–14	10	11 minutes
Questions 15–16	12	14 minutes
Question 17	5	6 minutes
Question 18	6	8 minutes
Check, especially simplest forms and signs	–	4 minutes

Question 18 comes last and needs careful algebra, so do not let the early short questions eat into its time.

Other May/June 2026 0607 papers

• 0607 Paper 12 (Core Non-calculator)
• 0607 Paper 32 (Core Calculator)
• 0607 Paper 42 (Extended Calculator)
• 0607 Paper 52 (Core Investigation)
• 0607 Paper 62 (Extended Investigation and Modelling)

Practise and download

• Download 0607/22 and its mark scheme on the Learning Portal →
• All 0607 May/June 2026 papers: overview and marking patterns
• Try a free IGCSE International Mathematics quiz

Need help with a 0607 topic? Book a free trial lesson with an IGCSE International Mathematics tutor →

Frequently asked questions

What topics were on 0607 Paper 22 May/June 2026? +

Eighteen questions: angles, probability, time, transformations, simple interest, vectors, mixed numbers, a cube-root formula, the surface area of a cylinder and hemisphere, a circle theorem, a linear equation, a Venn diagram, indices, a straight line, exact cosines, functions, surds and a two-stage journey leading to a quadratic.

How many marks is 0607/22 and how long is it? +

It is out of 75 marks and lasts 1 hour 30 minutes, with no calculator. Question 16, on functions, is the biggest at 10 marks.

What was the answer to the car journey question in 0607/22? +

x = 40. Part (a) asked you to show that 80/x + 450/(x + 50) = 7 leads to 7x² − 180x − 4000 = 0. Part (b) factorised it as (7x + 100)(x − 40) = 0. The negative root, −100/7, is not a speed, and giving both values lost the final mark.

What was the inverse of h(x) = 1 + log₃ x in 0607/22? +

h⁻¹(x) = 3^(x − 1). Swap x and y to get x = 1 + log₃ y, so log₃ y = x − 1 and y = 3^(x − 1). In the part before, h(x) = 3 gave x = 9.

Is 0607/22 the Core or Extended paper? +

Extended. Paper 2 is the Extended non-calculator paper, sat alongside Paper 4 (Extended calculator) and Paper 6 (Investigation and Modelling). Core candidates sit Paper 1 instead.

Where can I download 0607/22 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge IGCSE International Mathematics 0607 past papers page.

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