IGCSE Mathematics 0580 · Topic 1.9

IGCSE Mathematics: Estimation and Limits of Accuracy Practice Questions

A rounded measurement has an upper bound and a lower bound half a unit either side of it. When combining bounds, use the largest values for a maximum and the smallest for a minimum, except in division, where the divisor works the opposite way.

Cambridge IGCSE Mathematics (0580) · Topic 1.9: Estimation and Limits of Accuracy

Topic 1.9 of Cambridge IGCSE Mathematics 0580 is one of the highest scoring Extended topics because the method is fixed. The one part that catches students out is division, where the upper bound of the answer needs the lower bound of the divisor. The questions below build up to that.

What you need to know for Estimation and Limits of Accuracy

IGCSE Mathematics Estimation and Limits of Accuracy questions and answers

4 exam-style questions written to the 0580 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.

Question 1[3 marks]

The length of a rod is 24 cm, correct to the nearest centimetre. Write down an inequality for the possible length L of the rod.

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Answer: 23.5 is less than or equal to L, which is less than 24.5.
  1. The rod is measured to the nearest centimetre, so the rounding unit is 1 cm.
  2. Half of that unit is 0.5 cm, and the true value lies within 0.5 cm of the stated value.
  3. Lower bound = 24 minus 0.5 = 23.5 cm. Upper bound = 24 plus 0.5 = 24.5 cm.
  4. Written as an inequality: 23.5 is less than or equal to L, which is less than 24.5.
How the marks are awarded. 1 mark for a lower bound of 23.5. 1 mark for an upper bound of 24.5. 1 mark for the correct inequality symbols, inclusive at the lower end and strict at the upper end.
Where students lose the mark. Giving 23.5 to 24.4 as the range. The upper bound is exactly 24.5, excluded by a strict inequality, not 24.4.
Question 2[4 marks]

A rectangle has length 12.4 cm and width 8.6 cm, each measured correct to 1 decimal place. Calculate the upper bound of its area.

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Answer: 107.6925 cm2
  1. The rounding unit is 0.1 cm, so half of it is 0.05 cm.
  2. Upper bound of the length = 12.4 plus 0.05 = 12.45 cm.
  3. Upper bound of the width = 8.6 plus 0.05 = 8.65 cm.
  4. For a maximum area, multiply both upper bounds: 12.45 x 8.65 = 107.6925 cm2.
How the marks are awarded. 1 mark for using 0.05 as the half unit. 1 mark for an upper bound length of 12.45. 1 mark for an upper bound width of 8.65. 1 mark for 107.6925 cm2.
Where students lose the mark. Using 0.5 rather than 0.05 as the half unit. The half unit depends on the accuracy stated, and 1 decimal place means a unit of 0.1.
Question 3[4 marks]

A runner covers 100 m, measured to the nearest metre, in 12.5 seconds, measured to 1 decimal place. Calculate the lower bound of her average speed, giving your answer to 3 significant figures.

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Answer: 7.93 m/s
  1. Bounds for the distance: 99.5 m to 100.5 m, since the half unit is 0.5 m.
  2. Bounds for the time: 12.45 s to 12.55 s, since the half unit is 0.05 s.
  3. For the lowest possible speed, use the smallest distance and the largest time.
  4. Lower bound of speed = 99.5 divided by 12.55 = 7.9282 m/s, which is 7.93 m/s to 3 significant figures.
How the marks are awarded. 1 mark for correct distance bounds. 1 mark for correct time bounds. 1 mark for choosing the lower distance with the upper time. 1 mark for 7.93 m/s.
Where students lose the mark. Using the lower bound of both values. In a division, the smallest answer needs the smallest numerator and the largest denominator.
Question 4[3 marks]

Two masses are recorded as 15 g and 8 g, each to the nearest gram. Calculate the upper bound of the difference between them.

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Answer: 8 g
  1. Bounds for the first mass: 14.5 g to 15.5 g. Bounds for the second: 7.5 g to 8.5 g.
  2. The difference is the first mass minus the second mass.
  3. To make the difference as large as possible, take the largest possible first mass and the smallest possible second mass.
  4. Upper bound of the difference = 15.5 minus 7.5 = 8 g.
How the marks are awarded. 1 mark for both sets of bounds. 1 mark for combining the upper bound of the first with the lower bound of the second. 1 mark for 8 g.
Where students lose the mark. Subtracting the two upper bounds, giving 7 g. For a maximum difference you need the largest first value and the smallest second value.

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Estimation and Limits of Accuracy FAQs

How do I find upper and lower bounds?

Take half of the unit the measurement was rounded to, then add it for the upper bound and subtract it for the lower bound. A value given to the nearest whole number uses 0.5, one given to 1 decimal place uses 0.05, and one to 2 decimal places uses 0.005.

Which bounds do I use for division?

For the maximum of a divided by b, use the upper bound of a and the lower bound of b, because a smaller divisor produces a larger answer. For the minimum, use the lower bound of a and the upper bound of b. This reversal is the most common source of errors.

Why is the upper bound written with a strict inequality?

A value of exactly 24.5 would round up to 25 rather than to 24, so it cannot be part of the range. The lower bound of 23.5 does round to 24, so it is included. This is why the notation uses an inclusive symbol at the lower end and a strict one at the upper end.

How do I find the maximum area from rounded measurements?

Find the upper bound of each dimension by adding half the rounding unit, then multiply those upper bounds together. For a minimum area, multiply the lower bounds. Multiplication is straightforward because both bounds move the answer in the same direction.

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Written to the published Cambridge IGCSE Mathematics (0580) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.