Every limits of accuracy question from 12 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 1.10. 12 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/11/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Number past paper questions.
0580 Limits of Accuracy past paper questions: Core (Papers 1 and 3)
May/June 2026 · Paper 1 · 0580/11/M/J/26 · Q22
The time, t seconds, for the winner of a race is 12.7 seconds, correct to 3 significant figures. Complete the statement … ≤ t < …
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- 12.7 to 3 significant figures is correct to the nearest 0.1 second.
- Half of 0.1 is 0.05. Subtract and add this to 12.7.
- Lower bound = 12.7 − 0.05 = 12.65. Upper bound = 12.7 + 0.05 = 12.75.
- 12.65 ≤ t < 12.75
May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q23
The mass, m grams, of some butter is 650 grams, correct to the nearest 50 grams. Complete this statement: … ≤ m < …
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- To the nearest 50 g means the true mass is within half of 50 g, which is 25 g, of 650 g.
- Lower bound = 650 − 25 = 625. Upper bound = 650 + 25 = 675.
- 625 ≤ m < 675
Feb/Mar 2026 · Paper 1 · 0580/12/F/M/26 · Q23
The height, h metres, of a tree is 37.4 m, correct to the nearest 10 centimetres. Complete the statement … ≤ h < …
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- 10 centimetres = 0.1 metres, so the height is correct to the nearest 0.1 m.
- Half of 0.1 is 0.05. Subtract and add this to 37.4.
- Lower bound = 37.4 − 0.05 = 37.35. Upper bound = 37.4 + 0.05 = 37.45.
- 37.35 ≤ h < 37.45
Oct/Nov 2025 · Paper 1 · 0580/12/O/N/25 · Q26
The length, l metres, of a rope is 16.32 metres, correct to 2 decimal places. Complete the statement … ≤ l < …
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- Correct to 2 decimal places means correct to the nearest 0.01.
- Half of 0.01 is 0.005. Subtract and add this to 16.32.
- Lower bound = 16.315. Upper bound = 16.325.
- 16.315 ≤ l < 16.325
Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q20
The height, h cm, of a door is 180 cm, correct to the nearest centimetre. Complete the statement … ≤ h < …
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- Correct to the nearest centimetre means the true value is within half a centimetre.
- Lower bound = 180 − 0.5 = 179.5. Upper bound = 180 + 0.5 = 180.5.
- 179.5 ≤ h < 180.5
May/June 2025 · Paper 1 · 0580/12/M/J/25 · Q22
The mass, m kg, of a stone is 3.2 kg, correct to the nearest 100 g. Complete this statement about the value of m. ….. ≤ m < …..
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- 100 g = 0.1 kg, so the mass is correct to the nearest 0.1 kg.
- Half of 0.1 is 0.05. Subtract and add this to 3.2.
- Lower bound = 3.2 − 0.05 = 3.15. Upper bound = 3.2 + 0.05 = 3.25.
- 3.15 ≤ m < 3.25
0580 Limits of Accuracy past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q21
The volume of a solid metal cuboid is 22 cm³, correct to the nearest cm³. The mass of the metal cuboid is 140 grams, correct to the nearest 10 grams. Calculate the lower bound of the density of the metal. (Density = mass ÷ volume)
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- Find the bounds. Volume: 21.5 ≤ V < 22.5. Mass to the nearest 10 g: 135 ≤ m < 145.
- Density = mass ÷ volume. To make it as small as possible, use the smallest mass and divide by the largest volume.
- Lower bound of density = 135 ÷ 22.5 = 6 g/cm³
May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q16
A rectangle measures 4.2 cm by 6.8 cm. Each measurement is given correct to the nearest millimetre. Calculate the lower bound of the area of the rectangle.
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- To the nearest millimetre means to the nearest 0.1 cm, so each length can be up to 0.05 cm less.
- Lower bounds: 4.2 − 0.05 = 4.15 cm and 6.8 − 0.05 = 6.75 cm.
- Lower bound of area = 4.15 × 6.75 = 28.0125 cm²
Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q21
The length of a cuboid is 2 cm and the width is 4 cm, each correct to the nearest integer. The volume of the cuboid is 123.8 cm³, correct to 1 decimal place. Find the lower bound of the height of the cuboid.
Show solution
- Find the bounds first. Length: 1.5 ≤ l < 2.5. Width: 3.5 ≤ w < 4.5. Volume: 123.75 ≤ V < 123.85.
- Height = volume ÷ (length × width). To make the height as small as possible, use the smallest volume and divide by the largest length and width.
- Lower bound = 123.75⁄2.5 × 4.5 = 123.75⁄11.25 = 11 cm
Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q9
Nina walks at an average speed of 5 km/h, correct to the nearest km/h. She walks for exactly 2 hours. Work out the lower bound for the distance Nina walks.
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- 5 km/h correct to the nearest km/h means the speed could be as low as 5 − 0.5 = 4.5 km/h.
- The time is exact, so the smallest distance uses the smallest speed.
- Lower bound of distance = 4.5 × 2 = 9 km
Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q20
A solid metal prism has a mass of 4810 g, correct to the nearest 10 g. The density of the metal is 7.7 g/cm³, correct to 1 decimal place. Calculate the lower bound for the volume of the prism. (Density = mass ÷ volume)
Show solution
- Find the bounds. Mass to the nearest 10 g: 4805 ≤ m < 4815. Density to 1 d.p.: 7.65 ≤ d < 7.75.
- Volume = mass ÷ density. For the smallest volume, use the smallest mass and divide by the largest density.
- Lower bound = 4805 ÷ 7.75 = 620 cm³
Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q25
Jenna has a length of wire measuring 68 cm, correct to the nearest cm. From this wire she cuts off a piece of length 4.7 cm and a piece of length 10.0 cm, each correct to the nearest mm. Work out the lower bound and the upper bound for the length of wire remaining.
Show solution
- Find the bounds of each length.
- Wire to the nearest cm: 67.5 ≤ w < 68.5. Pieces to the nearest mm: 4.65 ≤ a < 4.75 and 9.95 ≤ b < 10.05.
- Remaining = wire − pieces.
- Lower bound: smallest wire minus largest pieces = 67.5 − 4.75 − 10.05 = 52.7 cm
- Upper bound: largest wire minus smallest pieces = 68.5 − 4.65 − 9.95 = 53.9 cm
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