Cambridge IGCSE International Mathematics 0607/42 is the May/June 2026 Paper 4 (Calculator, Extended), variant 2. It lasts 1 hour 30 minutes, is worth 75 marks and is sat by every Extended candidate. It had nineteen compulsory questions. Many were short 1- to 3-mark skills questions, but three long ones, the quadrilateral in Question 15 (11 marks), the graph sketch in Question 5 (10 marks) and the grouped data in Question 17 (10 marks), carried 31 marks between them. This guide goes through the paper question by question and shows 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.

0607/42 at a glance

Question	Topic	Marks
1	Naming a quadrilateral from its symmetry	1
2	Collecting like terms	2
3	Multiplying algebraic fractions	1
4	Factorising	2
5	Sketching a rational function: zero, local maximum, range of k for one solution, a translation	10
6	HCF of two numbers, LCM of two powers	3
7	Reverse percentage: the price before a 20% reduction	2
8	Number of sides of a regular polygon	2
9	Quadratic formula, answers to 2 decimal places	3
10	Inverse proportion with a cube	2
11	Sequence with alternating signs: next term and nth term	3
12	Depreciation at 10% a year, forwards and backwards	6
13	Standard form calculations	4
14	Similar solids where one volume is twice the other	2
15	Quadrilateral: area with ½ab sin C, perimeter with the cosine rule, shortest distance to a line	11
16	Radius of a sector from its area	3
17	Grouped data: mean, cumulative frequency, median, quartile, probability of two from one interval	10
18	Square-based pyramid: face area, surface area, angle with the base	4
19	Simplifying an algebraic fraction by factorising	4
	Total	75

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 method mark first, and B marks are independent marks for a correct result. Accuracy follows the front cover: 3 significant figures for non-exact answers, or 1 decimal place for angles, unless the question says otherwise.

Question 1: a quadrilateral from its symmetry (1 mark)

One line of symmetry and no rotational symmetry: kite or isosceles trapezium. Either was accepted.

Question 2: collecting like terms (2 marks)

3c − d + 4c − d = 7c − 2d, as a final answer. B1 for getting one of the two terms right (7c − kd or jc − 2d), and the scheme allowed B1 for 7c on its own.

Question 3: multiplying algebraic fractions (1 mark)

5x/y × 3/p = 15x/(yp), as a final answer.

Question 4: factorising (2 marks)

2p² − 8pt = 2p(p − 4t). A partial factorisation, p(2p − 8t) or 2(p² − 4pt), earned B1, as did a correct answer that you then spoiled.

Question 5: sketching a rational function (10 marks)

The function was f(x) = 5/x² + 1/x³, with x ≠ 0, to be sketched for x between −2 and 2 on axes running from −5 to 25.

• Sketch (3 marks): the left branch rises from near the x-axis to a local maximum, then falls steeply, crossing the x-axis and heading down beside the y-axis; the right branch comes down from the top of the grid next to the y-axis and flattens towards the x-axis. B2 for the left branch, B1 for the right branch. A slight error in the left branch (such as curling back) dropped it to B1, and drawing both branches correctly but joined still scored B2. The branches could touch the axes but not cross them.
• Zero of f(x) (1 mark): x = −0.2 (or −1/5). Writing it as the point (−0.2, 0) was not allowed, because a zero is an x-value.
• Local maximum (2 marks): (−0.3, 18.5), one B mark for each coordinate. The exact y-value, 500/27, was also accepted.
• Range of k for exactly one solution of f(x) = k (2 marks): k ≤ 0 or k > 18.5. Above the local maximum, only the right branch reaches the line y = k; at or below zero, only the left branch does. B1 for each inequality. Writing k < 0 was condoned, but k ≥ 18.5 was not. Combining them wrongly as 0 ≥ k > 18.5 earned only a special case mark (SC1).
• Transformation from f(x) to h(x) = 5/(x + 1)² + 1/(x + 1)³ (2 marks): a translation (B1) by the vector (−1, 0), written as a column vector (B1). Describing it in words (“1 to the left”) was allowed, but “shift to the left by −1” and coordinate form were not, and naming more than one transformation scored 0.

Question 6: HCF and LCM (3 marks)

• HCF of 180 and 96 (2 marks): 180 = 2² × 3² × 5 and 96 = 2⁵ × 3, so the HCF is 2² × 3 = 12. A common factor that is not the highest (2, 3, 4 or 6) earned B1. M1 was also available for both prime factorisations, or two correct factor trees.
• LCM of x¹² and x¹⁴ (1 mark): x¹⁴.

Question 7: reverse percentage (2 marks)

The sale price is 80% of the original, so the original price is 284.80 ÷ 0.8 = $356. M1 for (original) × (1 − 20/100) = 284.80 or better. Writing 1 − 20% in your working was not accepted unless you recovered it.

Question 8: a regular polygon (2 marks)

The diagram showed the interior angle at B as 4x° and the exterior angle as x°, on the straight line ABD. So 5x = 180, x = 36, and the number of sides is 360 ÷ 36 = 10. B1 for x = 36, seen or implied (soi), for example by 4x = 144. “Decagon” was accepted if 10 appeared in the working.

Question 9: the quadratic formula (3 marks)

The equation was 3p² − p − 3 = 0, to be solved with the formula, showing all working.

• M2 for the correct substitution: p = (−(−1) ± √((−1)² − 4(3)(−3))) ÷ 2(3). M1 for just the square root part, or for the rest of the formula with any number under the root.
• B1 for p = −0.85 and p = 1.18, to 2 decimal places as the question asked.
• The scheme allowed 1 for −(−1), 1 for (−1)², 6 for 2(3) and √(1 + 36), but not √37 in place of the substitution: the substituted values have to be seen. The square root sign had to reach past −4ac and the division line had to cover the ± sign. Leaving the answers as surds scored 0 for the final mark.

Question 10: inverse proportion (2 marks)

y was inversely proportional to x³, with y = 1/4 when x = 2. So y = k/x³, 1/4 = k/8 and k = 2, giving y = 2/x³. M1 for 1/4 = k/2³.

Question 11: an alternating sequence (3 marks)

The first five terms were 80, −40, 20, −10, 5.

• Next term (1 mark): −2.5.
• nth term (2 marks): each term is the one before multiplied by −1/2, so the nth term is 80 × (−1/2)ⁿ⁻¹ (oe, or equivalent; for example 160 × (1/2)ⁿ × (−1)ⁿ⁻¹ was full marks). M1 for a multiplier of (−1/2) to some power or division by (−2) to a power. If nothing else scored, 80 × (1/2)ⁿ⁻¹, which loses the alternating sign, earned SC1.

Question 12: depreciation (6 marks)

A car loses 10% of its value each year and is now worth $22 680.

• Value 2 years ago (2 marks): 22 680 ÷ 0.9² = $28 000. M1 for (value) × (1 − 10/100)² = 22 680 or better.
• Complete years until it is first below $10 000 (4 marks): 8 years, marked nfww (not from wrong working). After 7 years it is worth about $10 848, and after 8 years about $9763.
  • Solving 22 680 × 0.9ⁿ = 10 000 with logarithms gives n = 7.77; that answer alone earned B3, because the question asked for complete years.
  • Without the final answer: M3 for n log 0.9 = log(10 000/22 680), a good sketch showing the value between 7 and 8, or correct trials reaching both 7 and 8 years; M2 for 0.9ⁿ = 10 000/22 680, a suitable sketch or at least three correct trials; M1 for the equation, or two correct trials. Values to 3 significant figures were fine for trials.
  • Starting from $28 000 by mistake and answering 10 earned a 2-mark special case only.

Question 13: standard form (4 marks)

• (6.2 × 10⁸⁷) × (3 × 10⁸⁷) in standard form (2 marks): 18.6 × 10¹⁷⁴ = 1.86 × 10¹⁷⁵, cao (correct answer only). B1 for 18.6 × 10¹⁷⁴ or for any number × 10¹⁷⁵.
• (4 × 10ʸ) + (8 × 10ʸ⁻¹) in standard form (2 marks): 40 × 10ʸ⁻¹ + 8 × 10ʸ⁻¹ = 48 × 10ʸ⁻¹ = 4.8 × 10ʸ, cao. B1 for the figures 48 × 10 to any power.

Question 14: similar solids (2 marks)

Solid B had twice the volume of solid A, and A was 8 cm tall. The volume scale factor is 2, so the length scale factor is ∛2 and h = 8 × ∛2 = 10.1 (10.07 to 10.08 also accepted). M1 for (h/8)³ = 2 or ∛2 seen. Using 8³ = 512 and 512 × 2 = 1024 also earned the M1.

Question 15: a quadrilateral with trigonometry (11 marks)

ABCD was made from a right-angled triangle ABC (right angle at B, BC = 4.5 cm, AC = 9 cm, angle BCA = 60°) and a triangle ACD (CD = 13 cm, angle ACD = 35°).

• Area (3 marks): ½ × 4.5 × 9 × sin 60° = 17.54 and ½ × 9 × 13 × sin 35° = 33.55, giving 51.1 cm². M1 for each triangle’s area.
• Perimeter (5 marks): AB = 9 sin 60° = 7.79 cm (M1; Pythagoras, √(9² − 4.5²), also worked). By the cosine rule, AD² = 9² + 13² − 2 × 9 × 13 × cos 35° = 58.3, so AD = 7.64 cm (M2, or M1 for AD² without the square root). Perimeter = 4.5 + 7.79 + 13 + 7.64 = 32.9 cm (M1 dep, a dependent mark that needs AB and AD to have come from Pythagoras or trigonometry).
• CP, where P is the nearest point on CD to A (3 marks): the nearest point is where a perpendicular from A meets CD, so triangle ACP has a right angle at P. CP = 9 cos 35° = 7.37 cm. M2 for cos 35° = CP/9 or for CP² = 9² − (9 sin 35°)². M1 for recognising that the shortest distance is perpendicular, for example by finding the height 9 sin 35° = 5.16 or using ½ × 13 × height = area of ADC.

Question 16: the radius of a sector (3 marks)

A sector with an 80° angle had an area of 62.6 cm². From 80/360 × πr² = 62.6, r² = 89.7, so r = 9.47 (9.468 to 9.469 also accepted). M2 for 62.6 ÷ (80/360 × π) or for r² = 89.65 to 89.67; M1 for setting up the equation. Using 3.14 or 22/7 for π was condoned for the method marks.

Question 17: grouped data and cumulative frequency (10 marks)

100 students recorded their homework time: 7 in 0 < t ≤ 60, 31 in 60 < t ≤ 90, 45 in 90 < t ≤ 120 and 17 in 120 < t ≤ 180.

• Estimate of the mean (2 marks): use the mid-values 30, 75, 105 and 150: (210 + 2325 + 4725 + 2550) ÷ 100 = 9810 ÷ 100 = 98.1 minutes. M1 for at least three correct mid-values.
• Cumulative frequency table (1 mark): 38 and 83 (the table already gave 7 and 100).
• Cumulative frequency curve (2 marks): plot at the upper class boundaries, (60, 7), (90, 38), (120, 83) and (180, 100), and join them with a smooth increasing curve. B1 FT for 3 or 4 points correctly plotted, or a reasonable increasing curve or polygon through your points. Bars (blocks) alone scored 0.
• Median, lower quartile and the number spending 70 minutes or less (1 mark each): read across at 50 and 25 for the median and lower quartile, and up from 70 minutes for the last part. The scheme gave no fixed values: each mark followed through (FT) from your own curve, to within half a small square. But all three marks depended on an increasing curve or polygon, and a straight line got no follow through.
• Which interval were the two students chosen from? (2 marks): the probability of picking two from an interval of n students is n/100 × (n − 1)/99. Setting this equal to 1/5 gives n² − n − 1980 = 0, so n = 45 and the answer is 90 < t ≤ 120. M1 for trying n/100 × (n − 1)/99 with n = 7, 31, 45 or 17, or for the quadratic; A1 for the interval. The question said “show how you decide”, so the working earns the method mark.

Question 18: a square-based pyramid (4 marks)

The base had side 10 cm (BC = 10 cm) and PM, from the apex to the midpoint of BC, was 12 cm.

• Area of triangle PBC (1 mark): ½ × 10 × 12 = 60 cm².
• Total surface area (1 mark): 100 + 4 × 60 = 340 cm². This followed through from your triangle area.
• Angle between PM and the base (2 marks): the foot of the pyramid’s height is 5 cm from M, so cos θ = 5/12 and θ = 65.4° (65.37 to 65.38 also accepted). M1 for cos(angle) = 5/12 or for identifying the correct angle. Finding the height first, √(12² − 5²) = 10.9 cm, and using sine or tan also worked.

Question 19: an algebraic fraction (4 marks)

The fraction was (1 − w²) ÷ (1 + wx − w − x).

• Numerator: difference of two squares, (1 − w)(1 + w). B1.
• Denominator: group as (1 − w) − x(1 − w) = (1 − w)(1 − x). B2, or B1 for the partial step (1 − w) − x(1 − w) or (1 − x) − w(1 − x).
• Cancel (1 − w): (1 + w)/(1 − x), as a final answer, for the last mark. −(w + 1)/(x − 1) also scored full marks. Leaving the denominator half-factorised, such as 1 − w − x(1 − w), scored B0 for that step.

Where the marks are on 0607/42

• Algebra carried 17 of the 75 marks: Questions 2, 3, 4, 9, 10, 11 and 19, with the biggest single piece the 4-mark algebraic fraction.
• Number carried 15 marks: HCF and LCM (Q6), reverse percentage (Q7), depreciation (Q12) and standard form (Q13).
• Trigonometry carried 13 marks: the quadrilateral in Question 15 and the pyramid angle in 18(c).
• Functions and graphs carried 10 marks, all in Question 5.
• Statistics and probability carried 10 marks, all in Question 17.
• Geometry carried 5 marks: the quadrilateral’s name (Q1), the regular polygon (Q8) and similar solids (Q14).
• Mensuration carried 5 marks: the sector (Q16) and the pyramid’s areas in 18(a) and 18(b).

The longest questions were 15 (11 marks), 5 and 17 (10 marks each) and 12 (6 marks). Together they are 37 marks, almost half the paper.

How to self-mark 0607/42 with the mark scheme

1. Learn the abbreviations. M is method, A is accuracy (needs the M mark), B is an independent mark, FT is follow through after an earlier error, oe means “or equivalent”, soi means “seen or implied”, cao means “correct answer only”, nfww means “not from wrong working”, dep means the mark depends on an earlier one, SC is a special case and BOD is “benefit of the doubt”.
2. Check the final answer first. A correct answer usually scores all the marks for that part. The exceptions are questions that demand working: Question 9 needs the substitution in the formula to be seen, and 17(c) needs your trial or equation.
3. If the answer is wrong, hunt for method marks. The cosine rule substitution in 15(b) is worth M2 on its own, and recognising the perpendicular in 15(c) is M1 even if you never find CP.
4. Mark follow through on graphs carefully. In 17(b)(iii), your median and quartile are marked against your own curve, but only if it is an increasing curve or polygon. If you joined the points with a straight line, those three marks are lost.
5. Check the accuracy each question asks for. Question 9 wanted 2 decimal places, and 12(b) wanted complete years (8, not 7.77). Questions 13(a) and 13(b) were cao, so the answer had to be in proper standard form.
6. Watch the special cases. SC marks are a partial safety net, such as 80 × (1/2)ⁿ⁻¹ in 11(b) or an answer of 10 from $28 000 in 12(b). They are worth less than the method marks, so do not aim for them.
7. Log each lost mark as a missing method, a rounding or accuracy slip, a calculator error or a misread question, and revise in that order.

Sit 0607/42 as a timed mock

The paper allows 90 minutes for 75 marks, about 1.2 minutes per mark. A workable plan:

Block	Marks	Time
Questions 1–4	6	6 minutes
Question 5	10	12 minutes
Question 6	3	3 minutes
Question 7	2	2 minutes
Question 8	2	2 minutes
Question 9	3	4 minutes
Question 10	2	2 minutes
Question 11	3	3 minutes
Question 12	6	7 minutes
Question 13	4	4 minutes
Question 14	2	2 minutes
Question 15	11	13 minutes
Question 16	3	3 minutes
Question 17	10	12 minutes
Question 18	4	4 minutes
Question 19	4	5 minutes
Check, especially rounding and standard form	–	6 minutes

Question 17 needs a careful cumulative frequency curve, so give the plotting enough time: every reading after it depends on the curve.

Other May/June 2026 0607 papers

• 0607 Paper 12 (Non-calculator, Core)
• 0607 Paper 22 (Non-calculator, Extended)
• 0607 Paper 32 (Calculator, Core)
• 0607 Paper 52 (Investigation, Core)
• 0607 Paper 62 (Investigation and Modelling, Extended)

Practise and download

• Download 0607/42 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 42 May/June 2026? +

Nineteen questions, including a sketch of a rational function on a graphic display calculator, HCF and LCM, a reverse percentage, a regular polygon, the quadratic formula, inverse proportion, a geometric sequence, depreciation, standard form, similar solids, a quadrilateral needing trigonometry and the cosine rule, a sector, a cumulative frequency question, a pyramid and an algebraic fraction.

How many marks is 0607/42 and how long is it? +

It is out of 75 marks and lasts 1 hour 30 minutes. Question 15 (the quadrilateral, 11 marks), Question 5 (the graph sketch, 10 marks) and Question 17 (grouped data, 10 marks) carry 31 of the 75 marks between them.

What was the answer to the car depreciation question in 0607/42? +

Two years ago the car was worth 22 680 ÷ 0.9² = $28 000. Its value first falls below $10 000 after 8 complete years: after 7 years it is about $10 848 and after 8 years about $9763. Stopping at 7.77 from logarithms earned 3 of the 4 marks, because the question asked for complete years.

What was the probability question in 0607/42 Question 17(c)? +

Two students from the same time interval had a 1/5 chance of being chosen. Setting n/100 × (n − 1)/99 = 1/5 gives n² − n − 1980 = 0, so n = 45, which is the 90 < t ≤ 120 interval.

Is 0607/42 the Core or Extended paper? +

Extended. Paper 4 is the Extended calculator paper, sat alongside Paper 2 (Extended non-calculator) and Paper 6 (investigation and modelling). Core candidates sit Paper 3 instead.

Where can I download 0607/42 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.

Ready to Excel in Your Studies?

Get personalised help from Tutopiya's expert tutors. Whether it's IGCSE, IB, A-Levels, or any other curriculum — we match you with the perfect tutor and your first session is free.

Book Your Free Trial