IGCSE Maths Past Paper Solutions: 0580/03 Specimen Paper 2025 (Paper 3 Core Calculator)
Full worked solutions and answers to the Cambridge IGCSE Mathematics (0580) Paper 3 Core Specimen Paper for examination from 2025 (0580/03/SP/25). All 22 questions are solved step by step, every diagram is redrawn so you can follow along, and every answer has been checked against the question paper. Try each question first, then open “Show solution” to compare your working.
Question 6 — Temperature change with negative numbers
The lowest temperature recorded at Scott Base in Antarctica is −57.0°C. The highest temperature recorded at Scott Base is 63.8°C more than this. Calculate the highest temperature recorded at Scott Base.
Show solution
“More than” means add: −57.0 + 63.8 = 6.8°C
71 markCoreExchange ratesSpecimen 2025 · Paper 3 (0580/03)
Question 7 — Converting dollars to euros
Lee changes $450 into euros. The exchange rate is $1 = 0.8476 euros. Calculate the amount in euros that Lee receives.
Show solution
Each dollar is worth 0.8476 euros, so multiply: 450 × 0.8476 = 381.42 euros
82 marksCoreSubstitutionSpecimen 2025 · Paper 3 (0580/03)
92 marksCoreConstructionsSpecimen 2025 · Paper 3 (0580/03)
Question 9 — Constructing a 6 cm, 7 cm, 8 cm triangle with ruler and compasses
A triangle has sides 6 cm, 7 cm and 8 cm. Using a ruler and compasses only, construct the triangle. The 6 cm line has been drawn for you. Show all your construction arcs.
Show solution
Leave both arcs visible — they are the evidence of construction.
Label the ends of the 6 cm line A (left) and B (right).
Set the compasses to 7 cm. With the point on A, draw an arc above the line.
Set the compasses to 8 cm. With the point on B, draw an arc that crosses the first arc. Label the crossing point C.
Use a ruler to join AC and BC. Leave both arcs showing — the arcs earn the method mark.
Either way round is accepted (7 cm from A and 8 cm from B, or the reverse).
Exam tip: work out the top and bottom separately (or use the fraction key). Typing it in one line without brackets divides only by −0.31.
117 marksCoreCirclesSpecimen 2025 · Paper 3 (0580/03)
Question 11 — Tangent, chord, radius from area, and circumference in millimetres
(a) The line XY touches the circle at the point R. Write down the mathematical name for the line XY.
(b) Points P and Q lie on the circle. Write down the mathematical name for the line PQ.
(c) The area of the circle is 43.5 cm². Calculate the radius of the circle.
(d) The diameter of a different circle is 6.4 cm. Calculate the circumference of this circle. Give your answer in millimetres.
Show solution
(a) A straight line that touches a circle at exactly one point is a tangent
(b) A straight line joining two points on the circumference is a chord
(c) Area = πr², so r² = 43.5 ÷ π = 13.846… r = √13.846… = 3.7211… = 3.72 cm (3 s.f.)
(d) Circumference = π × diameter = π × 6.4 = 20.106… cm 1 cm = 10 mm, so 20.106… × 10 = 201.06… = 201 mm (3 s.f.)
Common mistake: leaving (d) in centimetres (20.1) loses the final mark. Read the units on the answer line. More circle vocabulary in our geometrical terms notes.
123 marksCoreRangeSpecimen 2025 · Paper 3 (0580/03)
Question 12 — Range from a stem-and-leaf diagram
The stem-and-leaf diagram shows the scores of each of 27 students in a test.
Stem
Leaf
2
8
8
9
3
2
5
6
6
7
8
8
4
0
1
1
2
3
4
6
7
9
5
1
3
4
5
5
7
8
6
2
Key: 2 | 8 represents a score of 28
(a) Find the range of the scores.
(b) When the score for another student is included in the diagram the new range is 38. Find the two possible scores for this student.
(b) To make the range 38, the new score must become either the new lowest or the new highest. New lowest: 62 − 38 = 24 New highest: 28 + 38 = 66 2466
1311 marksCoreTravel graphsSpecimen 2025 · Paper 3 (0580/03)
Question 13 — Travel graph: average speed, stopping time and where two cyclists meet
Jason leaves Town A at 09 00 and cycles to Town C. The travel graph shows Jason’s journey.
Each small square is 6 minutes across and 1 km up.
(a) Find Jason’s average speed, in kilometres per hour, from Town A to Town B.
(b) Jason leaves Town C at 12 00. Jason continues to Town D at a constant speed of 15 kilometres per hour.
Calculate the time Jason takes to travel from Town C to Town D. Give your answer in hours and minutes.
On the travel graph, complete Jason’s journey.
(c) Find the total time, in minutes, that Jason stopped between Town A and Town D.
(d) Calculate Jason’s overall average speed, in kilometres per hour, from Town A to Town D.
(e) Lisa leaves Town C at 11 00 and arrives at Town A at 13 42. Lisa cycles at a constant speed on the same road as Jason, without stopping.
Draw a line on the travel graph to show Lisa’s journey.
Find the distance from Town A when Lisa and Jason pass each other.
Show solution
Reading the graph: A to B from 09 00 to 10 00; stopped at B from 10 00 to 10 42; B to C from 10 42 to 11 30; stopped at C from 11 30 to 12 00.
Blue: Jason C → D. Gold: Lisa C → A. Red: where they pass.
(a) 18 km in 1 hour (09 00 to 10 00): speed = 18 ÷ 1 = 18 km/h
(b)(i) Distance C to D = 50 − 30 = 20 km. Time = 20 ÷ 15 = 1.333… hours = 1 hour + 0.333… × 60 minutes = 1 h 20 min
(b)(ii) Draw a straight line from (12 00, 30 km) to (13 20, 50 km). It passes through (13 00, 45 km).
(c) The horizontal sections are the stops: 10 00 to 10 42 = 42 minutes, and 11 30 to 12 00 = 30 minutes. Total = 42 + 30 = 72 min
(d) Jason arrives at Town D at 12 00 + 1 h 20 min = 13 20. Total time from 09 00 = 4 h 20 min = 4.333… hours. Average speed = total distance ÷ total time = 50 ÷ 4.333… = 11.538… = 11.5 km/h (3 s.f.)
(e)(i) Draw a straight line from (11 00, 30 km) down to (13 42, 0 km).
(e)(ii) Lisa’s line crosses Jason’s line while he is cycling from B to C. Read across from the crossing point: 27 km Check by calculation: Lisa covers 30 km in 162 min; Jason covers 12 km in 48 min from 10 42. Setting their distances equal, they meet about 35 minutes after 10 42, at about 26.8 km ≈ 27 km.
Common mistake in (d): overall average speed uses the total time including stops. Using only moving time (3 h 8 min) gives the wrong answer. Writing 4 h 20 min as 4.2 hours is another frequent slip — 20 minutes is 0.333… of an hour. Practise more in our graphs practice questions.
149 marksCoreNets & cuboidsSpecimen 2025 · Paper 3 (0580/03)
Question 14 — Net of a cuboid: volume, surface area, edge length and perimeter in terms of x and y
(a) The diagram shows the net of a cuboid.
(i) Find the volume of the cuboid.
(ii) Show that the total surface area of the cuboid is 148 cm².
(iii) Calculate the total length of the edges of the cuboid.
(b) In this part, all measurements are in centimetres.
This is the net of a cuboid with edges of length x, y and (x − 1). Find an expression, in terms of x and y, for the perimeter of the net. Give your answer in its simplest form.
Show solution
(a)(i) The cuboid measures 6 cm by 4 cm by 5 cm. Volume = 6 × 4 × 5 = 120 cm³
(a)(ii) A cuboid has three pairs of identical faces: 6 × 4 = 24, 4 × 5 = 20 and 6 × 5 = 30. Surface area = 2 × (24 + 20 + 30) = 2 × 74 = 148 cm² ✓
(a)(iii) A cuboid has 12 edges: 4 of each length. Total = 4 × 6 + 4 × 5 + 4 × 4 = 24 + 20 + 16 = 60 cm
(b) Go round the outside of the net and label every edge (gold outline below).
x-edges: (x − 1) + x + x + x + x + (x − 1) = 6x − 2
y-edges: 8 edges of length y = 8y Perimeter = 6x + 8y − 2
Exam tip: tick each outside edge as you add it. There are 14 outside edges here; missing one is the usual way to lose a mark. Revise nets in our surface area and volume notes.
154 marksCoreSpheresSpecimen 2025 · Paper 3 (0580/03)
Question 15 — Radius and volume of a sphere from its surface area
A sphere has a surface area of 177 cm².
(a) Calculate the radius of the sphere.
(b) Calculate the volume of the sphere.
Show solution
(a) Surface area = 4πr², so r² = 177⁄4π = 14.085… r = √14.085… = 3.7530… = 3.75 cm (3 s.f.)
Question 16 — Ratio difference, percentage of profit and compound interest
Jo and Mira buy a shop.
(a) They pay for the shop in the ratio Jo : Mira = 7 : 15. Mira pays $84 000 more than Jo. Work out how much they each pay.
(b) The shop makes a profit of $56 000. Jo receives 12% of the profit. Mira receives $14 000 of the profit. The rest of the profit is put into a bank account.
Calculate how much money Jo receives.
Calculate the amount put into the bank account as a percentage of the profit.
Mira invests $14 000 at a rate of 2.4% per year compound interest. Calculate the value of this investment at the end of 4 years.
Show solution
(a) The difference is 15 − 7 = 8 parts = $84 000, so 1 part = 84 000 ÷ 8 = $10 500. Jo = 7 × 10 500 = $73 500 Mira = 15 × 10 500 = $157 500
(b)(i) 12% of 56 000 = 0.12 × 56 000 = $6720
(b)(ii) Bank account = 56 000 − 6720 − 14 000 = $35 280 As a percentage: 35 280⁄56 000 × 100 = 63%
Common mistake in (a): sharing $84 000 in the ratio 7 : 15. The $84 000 is the difference (8 parts), not the total. More in our percentages notes (compound interest) and ratio practice questions.
174 marksCoreHCF & LCMSpecimen 2025 · Paper 3 (0580/03)
Question 17 — Prime factors, HCF and LCM of 120 and N = 2⁴ × 3²
The number, N, is written as a product of its prime factors. N = 2⁴ × 3²
(a) Work out the value of N.
(b) Find the highest common factor (HCF) of 120 and N.
(c) Find the lowest common multiple (LCM) of 120 and N.
Show solution
(a)N = 16 × 9 = 144
(b) 120 = 2³ × 3 × 5 and N = 2⁴ × 3². Put the prime factors in a Venn diagram:
HCF = product of the overlap = 2³ × 3 = 24
(c) LCM = product of everything in the diagram = 2⁴ × 3² × 5 = 16 × 9 × 5 = 720
186 marksCoreSequencesSpecimen 2025 · Paper 3 (0580/03)
Question 18 — Missing terms, nth term 7n − 3, and showing 361 is a term
(a) These are the first five terms of a sequence: 7, a, b, c, 31. In the sequence, the same number is added each time to obtain the next term. Find the value of each of the terms a, b and c.
(b) These are the first five terms of another sequence: 4, 11, 18, 25, 32.
Find the nth term of the sequence.
Show that 361 is a term in the sequence.
Show solution
(a) There are 4 equal steps from 7 to 31: 31 − 7 = 24, so each step is 24 ÷ 4 = 6. a = 13b = 19c = 25
(b)(i) The common difference is 7, so the nth term starts 7n. When n = 1, 7n = 7, but the first term is 4, so subtract 3. 7n − 3
(b)(ii) Set 7n − 3 = 361 → 7n = 364 → n = 52. 52 is a whole number, so 361 is the 52nd term
Exam tip: in “show that” questions you must state the conclusion — “n = 52 is a whole number, so 361 is in the sequence”. Practise with our sequences questions.
193 marksCoreMeanSpecimen 2025 · Paper 3 (0580/03)
Question 19 — Combined mean of two groups of different sizes
In a quiz, the mean score of each of 12 adults is 43.25. In the same quiz, the mean score of each of 16 children is 39.75. Calculate the mean score of the 28 people.
Show solution
Adults’ total = 12 × 43.25 = 519. Children’s total = 16 × 39.75 = 636.
Mean = (519 + 636) ÷ 28 = 1155 ÷ 28 = 41.25
Common mistake: averaging 43.25 and 39.75 gives 41.5, which is wrong because the groups are different sizes. Always go back to the totals.
202 marksCoreSpeed unitsSpecimen 2025 · Paper 3 (0580/03)
Question 20 — Converting 5.4 km/h to metres per second
Luca walks at a speed of 5.4 kilometres per hour. Write this speed in metres per second.
Show solution
5.4 km = 5400 m and 1 hour = 60 × 60 = 3600 s.
5400 ÷ 3600 = 1.5 m/s
Shortcut: to convert km/h to m/s, divide by 3.6. To go from m/s to km/h, multiply by 3.6.
214 marksCoreCircle theoremsSpecimen 2025 · Paper 3 (0580/03)
Question 21 — Angle in a semicircle and finding a diameter with cosine
The diagram shows a circle, centre O, with diameter PQ. R is a point on the circumference.
(a) Give a geometrical reason why angle PRQ is 90°.
(b) Calculate the length of PQ.
Show solution
(a)The angle in a semicircle is 90°
(b) Triangle PRQ is right-angled at R. From the 32° angle at P, PR is the adjacent side and PQ is the hypotenuse, so use cosine.
cos 32° = 6.2⁄PQ → PQ = 6.2⁄cos 32° = 7.3109… PQ = 7.31 cm (3 s.f.)
Check: the hypotenuse must be the longest side — 7.31 cm > 6.2 cm ✓. Revise SOH CAH TOA in our trigonometry notes.
223 marksCorePythagorasSpecimen 2025 · Paper 3 (0580/03)
Question 22 — Ladder against a wall using Pythagoras’ theorem
A ladder of length 5.6 m rests against a vertical wall. The bottom of the ladder is 1.5 m from the bottom of the wall, on horizontal ground. Calculate the distance from the top of the ladder to the base of the wall.
Show solution
The ladder is the hypotenuse of a right-angled triangle; the unknown height h is a shorter side, so subtract the squares:
h² = 5.6² − 1.5² = 31.36 − 2.25 = 29.11
h = √29.11 = 5.395… = 5.40 m (3 s.f.)
Common mistake: adding the squares (5.6² + 1.5²) gives 5.80 m, which is longer than the ladder — impossible. Finding a shorter side means subtract.
0580/03 Specimen 2025 Answer Key
Final answers only, for quick marking. Open each question above for the full method.
Q
Answer
Q
Answer
1
(a) 4 (b) 9
13
(a) 18 km/h (b)(i) 1 h 20 min (c) 72 min (d) 11.5 km/h (e)(ii) 27 km
2
1, 2, 4, 17, 34, 68
14
(a)(i) 120 cm³ (ii) 148 cm² (iii) 60 cm (b) 6x + 8y − 2
3
4 × (6 − 2) + 1
15
(a) 3.75 cm (b) 221 cm³
4
¼ or 0.25
16
(a) Jo $73 500, Mira $157 500 (b)(i) $6720 (ii) 63% (iii) $15 393.16
5
(a) 576 (b) 13
17
(a) 144 (b) 24 (c) 720
6
6.8°C
18
(a) 13, 19, 25 (b)(i) 7n − 3 (ii) n = 52
7
381.42 euros
19
41.25
8
1098
20
1.5 m/s
9
Construction with arcs
21
(a) Angle in a semicircle (b) 7.31 cm
10
12.66
22
5.40 m
11
(a) tangent (b) chord (c) 3.72 cm (d) 201 mm
12
(a) 34 (b) 24, 66
How to score well on IGCSE Maths 0580 Paper 3 (Core)
Use the marks as a clock. 80 marks in 90 minutes is just over a minute per mark, leaving about 10 minutes to check.
Keep full calculator values. Round only the final answer to 3 s.f. (Q11, Q15, Q21 and Q22 all depend on this).
Read the units on the answer line. Q11(d) wants millimetres; Q20 wants m/s.
Show working on multi-mark questions. Method marks are available even if the final answer slips.
Leave construction arcs. In Q9 the arcs are what earn the marks.
Yes. From 2025, Paper 3 is the Core calculator paper: 1 hour 30 minutes, 80 marks, worth 50% of the Core qualification. Paper 1 is the Core non-calculator paper.
What grades can I get on IGCSE Maths Core?
Core candidates (Papers 1 and 3) can achieve grades C to G. To aim for A* to B you need the Extended papers, Papers 2 and 4.
What accuracy should I give my answers to?
Non-exact answers to 3 significant figures and angles to 1 decimal place, unless the question says otherwise. Use your calculator’s π or 3.142.
What are the answers to Question 13 (travel graph)?
(a) 18 km/h, (b)(i) 1 h 20 min, (c) 72 minutes, (d) 11.5 km/h, (e)(ii) about 27 km from Town A. See the full working and completed graph.
Which topics come up in this paper?
Number work, constructions, circles, statistics, travel graphs, nets and mensuration, ratio and compound interest, HCF/LCM, sequences, trigonometry and Pythagoras. See the topic map for every question.
