IGCSE Maths Past Paper Solutions: 0580/03 Specimen Paper 2025 (Paper 3 Core Calculator)
Full worked solutions to the Cambridge IGCSE Mathematics (0580) Paper 3 Core Specimen Paper for examination from 2025 (0580/03/SP/25) — 22 questions, 80 marks, calculator allowed. Attempt each question yourself first, then click “Show solution” to check your working step by step.
The pictogram shows the number of text messages sent by five students in one day.
- (a) Kim sent 15 text messages. Complete the key.
- (b) Find the number of text messages sent by Hana.
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- (a) Kira’s row shows 3 full circles plus a ¾ circle. If 3¾ symbols represent 15 messages, then one full symbol represents 15 ÷ 3.75 = 4 text messages
- (b) Hana’s row shows 2 full circles plus a small quarter-circle piece. Using the key (1 symbol = 4 messages): 2 × 4 = 8, plus ¼ × 4 = 1, giving 8 + 1 = 9 text messages
Write down all the factors of 68.
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Check each whole number up to √68 (≈8.2) to find factor pairs: 1×68, 2×34, 4×17. No other whole numbers up to 8 divide exactly into 68.
1, 2, 4, 17, 34, 68
Insert one pair of brackets to make this statement correct: 4 × 6 − 2 + 1 = 17
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Testing 4 × (6−2) + 1: brackets first gives 6−2=4, then 4×4=16, then 16+1=17 ✓
4 × (6 − 2) + 1 = 17
Write down the reciprocal of 4.
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The reciprocal of a number is 1 divided by that number: 1 ÷ 4 = 0.25 (or ¼)
Find the value of:
- (a) 24²
- (b) ∛2197 (cube root of 2197)
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- (a) 24 × 24 = 576
- (b) Since 13³ = 13×13×13 = 2197: 13
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.
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−57.0 + 63.8 = 6.8°C
Lee changes $450 into euros. The exchange rate is $1 = 0.8476 euros. Calculate the amount in euros that Lee receives.
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450 × 0.8476 = 381.42 euros
W = t⁄2 (7t − 4). Find the value of W when t = 18.
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7t − 4 = 7(18) − 4 = 126 − 4 = 122. Then t⁄2 = 18÷2 = 9.
W = 9 × 122 = 1098
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.
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Method: Label the ends of the given 6 cm line A and B. Open the compasses to 7 cm and draw an arc centred at A. Open the compasses to 8 cm and draw an arc centred at B, crossing the first arc. Label this crossing point C, then join AC and BC with straight lines. Leave all construction arcs visible — do not rub them out.
Calculate 13.7 + 14.02⁄−0.31 + ∛15.625. Give your answer correct to 2 decimal places.
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Numerator: 13.7 + 14.02 = 27.72
Cube root of 15.625: since 2.5³ = 15.625, ∛15.625 = 2.5. Denominator: −0.31 + 2.5 = 2.19
27.72 ÷ 2.19 = 12.6575… → 12.66
The diagram shows a circle. The line XY touches the circle at point R. Points P and Q lie on the circle.
- (a) Write down the mathematical name for the line XY.
- (b) 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.
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- (a) A line that touches a circle at exactly one point is called a tangent
- (b) A straight line joining two points on the circumference is called a chord
- (c) Area = πr², so r² = 43.5 ÷ π = 13.846. Taking the square root: r = √13.846 = 3.72 cm (3 s.f.)
- (d) Circumference = π × diameter = π × 6.4 = 20.106 cm. Converting to millimetres (×10): 201 mm (3 s.f.)
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.
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- (a) Highest score = 62, lowest score = 28. Range = 62 − 28 = 34
- (b) A new range of 38 must come from either a new minimum or a new maximum:
If the new score is lower than 28: 62 − new score = 38 → new score = 24
If the new score is higher than 62: new score − 28 = 38 → new score = 66
Answer: 24 and 66
Jason leaves Town A at 09 00 and cycles to Town C. The travel graph shows Jason’s journey (Town B at 18 km, Town C at 30 km, Town D at 50 km from Town A).
- (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.
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- (a) Town A to Town B: 18 km covered from 09 00 to 10 00 (1 hour). Speed = 18 ÷ 1 = 18 km/h
- (b)(i) Distance Town C to Town D = 50 − 30 = 20 km. Time = distance ÷ speed = 20 ÷ 15 = 1⅓ hours = 1 h 20 min
- (b)(ii) Leaving Town C (30 km) at 12 00, draw a straight line rising to 50 km, arriving at 13 20.
- (c) Reading the flat (stationary) sections of the graph: Jason is stopped from 10 00 to 10 45 (45 minutes) at Town B, and from 11 30 to 12 00 (30 minutes) at Town C. Total stopped time = 45 + 30 = 75 minutes
- (d) Jason arrives at Town D at 13 20 (from part (b)(i), 1 h 20 min after leaving Town C at 12 00). Total journey time from 09 00 to 13 20 = 4 hours 20 minutes = 4⅓ hours. Total distance = 50 km.
Average speed = 50 ÷ 4⅓ = 50 ÷ 4.3333 = 11.5 km/h (3 s.f.) - (e)(i) Lisa’s journey: plot a straight line from (11 00, 30 km) to (13 42, 0 km).
- (e)(ii) Lisa’s speed = 30 km ÷ 162 minutes. Setting Lisa’s position equal to Jason’s position during the interval where Jason cycles from Town B to Town C (10:45–11:30) and solving the two straight-line equations gives a crossing point at approximately 26.7 km from Town A.
(a) The diagram shows the net of a cuboid, with measurements 6 cm, 4 cm and 5 cm marked.
- (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.
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- (a)(i) The net folds into a cuboid with edges 6 cm, 4 cm and 5 cm. Volume = 6 × 4 × 5 = 120 cm³
- (a)(ii) A cuboid has 3 pairs of identical rectangular faces: 6×4, 6×5 and 4×5.
Surface area = 2(6×4) + 2(6×5) + 2(4×5) = 2(24) + 2(30) + 2(20) = 48 + 60 + 40 = 148 cm² ✓ - (a)(iii) A cuboid has 12 edges: 4 of each of the 3 different lengths.
Total edge length = 4(6 + 4 + 5) = 4 × 15 = 60 cm - (b) This net has the same “cross” shape as part (a): a row of four rectangular faces (widths x, y, x, y in turn, each of height (x−1)), with the top and bottom faces (each x by y) attached above and below the first rectangle.
Tracing the outer boundary of the whole net and adding every outside edge gives:
Perimeter = 4x + 8y + 2(x − 1) = 4x + 8y + 2x − 2
Perimeter = 6x + 8y − 2
A sphere has a surface area of 177 cm².
- (a) Calculate the radius of the sphere.
- (b) Calculate the volume of the sphere.
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- (a) Surface area = 4πr², so r² = 177 ÷ (4π) = 14.0855. Taking the square root: r = 3.75 cm (3 s.f.)
- (b) Volume = &frac43;πr³ = &frac43; × π × (3.75306)³ = &frac43; × π × 52.876 = 221 cm³ (3 s.f., using the unrounded radius)
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.
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- (a) The difference in ratio parts is 15 − 7 = 8 parts, which equals $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 amount = 56 000 − 6720 − 14 000 = $35 280.
As a percentage: 35 280 ÷ 56 000 × 100 = 63% - (b)(iii) Compound interest: 14 000 × (1.024)⁴ = 14 000 × 1.099512 = $15 393.17
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.
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- (a) 2⁴ = 16, 3² = 9. N = 16 × 9 = 144
- (b) Prime factors: 120 = 2³ × 3 × 5, and N = 2⁴ × 3². The HCF takes the lowest power of each common prime factor: 2³ × 3¹ = 8 × 3 = 24
- (c) The LCM takes the highest power of every prime factor present: 2⁴ × 3² × 5¹ = 16 × 9 × 5 = 720
- (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.
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- (a) Going from 7 to 31 takes 4 equal steps (7→a→b→c→31), so 4d = 31−7 = 24, giving d = 6.
a = 7+6 = 13, b = 13+6 = 19, c = 19+6 = 25 - (b)(i) The common difference is 7 (11−4=7, 18−11=7, etc). The nth term has the form 7n + k. Using the 1st term: 7(1)+k=4 → k=−3.
nth term = 7n − 3 - (b)(ii) Set 7n − 3 = 361: 7n = 364, n = 52. Since 52 is a positive whole number, 361 is the 52nd term of the sequence.
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.
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Total score of adults = 12 × 43.25 = 519. Total score of children = 16 × 39.75 = 636.
Combined total = 519 + 636 = 1155, over 12+16 = 28 people.
Mean = 1155 ÷ 28 = 41.25
Luca walks at a speed of 5.4 kilometres per hour. Write this speed in metres per second.
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Convert km to m (×1000) and hours to seconds (÷3600): 5.4 × 1000 ÷ 3600 = 5400 ÷ 3600 = 1.5 m/s
The diagram shows a circle, centre O, with diameter PQ. R is a point on the circumference. Angle RPQ = 32° and PR = 6.2 cm.
- (a) Give a geometrical reason why angle PRQ is 90°.
- (b) Calculate the length of PQ.
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- (a) The angle in a semicircle (subtended by a diameter) is always 90°
- (b) Triangle PRQ is right-angled at R, with the 32° angle at P. PR is adjacent to this angle, and PQ (the diameter) is the hypotenuse.
cos(32°) = PR ÷ PQ, so PQ = PR ÷ cos(32°) = 6.2 ÷ 0.8480 = 7.31 cm (3 s.f.)
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.
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This forms a right-angled triangle with the ladder as the hypotenuse (5.6 m), and the two legs being the height up the wall and the 1.5 m base distance.
Using Pythagoras’ theorem: height² = 5.6² − 1.5² = 31.36 − 2.25 = 29.11
height = √29.11 = 5.40 m (3 s.f.)