OCR GCSE Mathematics: Rounding, estimation, error intervals and bounds (J560) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow section 4.01 of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026): rounding (4.01a), estimation (4.01b) and error intervals and bounds (4.01c). Questions 1 to 6 and 12 are Foundation and Higher content. Questions 7 to 11 are (Higher tier only). Learn the methods first from the study guide and the revision notes. The OCR GCSE Mathematics hub, the checklist and the 10-minute diagnostics are linked for planning.

Questions

1. Round each number as stated.

(a) 5482 to the nearest 100. [1] (b) 0.04068 to 2 significant figures. [1] (c) 12.996 to 2 decimal places. [1]

2. A ribbon 9.4 m long is cut into 7 equal pieces. Work out the length of one piece. Give your answer to a suitable degree of accuracy. [2]

3. Do not use a calculator.

(a) Estimate the cost of 4.1 kg of grapes at £5.95 per kg. [2] (b) Estimate the value of (5.82 × 39.6) ÷ √80.9. Show the rounded values you use. [3]

4. Write down an error interval for each value.

(a) a = 8.4, rounded to 1 decimal place. [1] (b) b = 7600, rounded to 2 significant figures. [1] (c) c = 0.9, truncated to 1 decimal place. [1] (d) d = 15 000, rounded to 2 significant figures. [1]

5. A speaker is 60 cm wide, to the nearest centimetre.

(a) A shelf gap is 61 cm wide, to the nearest centimetre. Can you be sure the speaker fits in the gap? Show your reasoning. [3] (b) A second gap is 60 cm wide, to the nearest centimetre. Can you be sure the speaker fits? Explain. [2]

6. A timer shows an athlete’s time as 4.7 seconds, truncated to 1 decimal place.

(a) Write an error interval for her time t. [1] (b) A second athlete’s time is shown as 4.8 seconds, truncated to 1 decimal place. Can you be sure the second athlete was slower? Explain. [2]

7. (Higher tier only) A floor is 8.4 m by 5.7 m, with each length given to 1 decimal place.

(a) Calculate the upper bound of the area. [2] (b) Calculate the lower bound of the area. [2] (c) Write the area to a suitable degree of accuracy, justifying your answer. [2]

8. (Higher tier only) A = 15.8 and B = 6.4, each to 1 decimal place.

(a) Find the upper bound of A − B. [2] (b) Find the lower bound of A ÷ B, to 3 significant figures. [3]

9. (Higher tier only)

(a) A fair has 1200 visitors to the nearest hundred. Write the possible range for the number of visitors n. [2] (b) A parcel has mass 1200 g to the nearest 100 g. Write the possible range for the mass m. [2] (c) Each visitor paid exactly £8. Find the greatest and least possible total takings. [2]

10. (Higher tier only) A cyclist rides 840 m, to the nearest 10 m, in 105 seconds, to the nearest second. Find the upper and lower bounds of her average speed in m/s, then give the speed to a suitable degree of accuracy. [6]

11. (Higher tier only) A metal block measures 4.2 cm by 3.5 cm by 2.0 cm, each to 1 decimal place. Its mass is 118 g to the nearest gram.

(a) Find the upper and lower bounds of the volume. [3] (b) Find the upper and lower bounds of the density (mass ÷ volume). [3] (c) Give the density to a suitable degree of accuracy, with a reason. [2]

12. Mia buys 19 tiles at £3.85 each and 2.4 kg of adhesive at £6.15 per kg.

(a) Estimate her total bill by rounding each number to 1 significant figure. [3] (b) Work out the exact bill. Was your estimate an overestimate or an underestimate? [2] (c) Her bank shows the bill as £88 to the nearest pound. Write an error interval for the bill B, and say whether £87.91 is consistent with it. [2]

Answers

1. (a) 5500 [1]. (b) The first significant figure is 4. Two significant figures gives 0.040, then the next digit 6 rounds up: 0.041 [1]. (c) 12.99|6 rounds up through the 9s: 13.00 [1]. [3] Examiner insight: the final zeros in 13.00 show the stated accuracy, so keep them.

2. 9.4 ÷ 7 = 1.342857… [1]. 1.34 m (3 significant figures, matching the data) [1]. [2] Examiner insight: quoting every digit of the display claims more accuracy than the measurements support.

3. (a) 4 × 6 [1] = £24 [1]. [2] (b) 5.82 → 6, 39.6 → 40, √80.9 → √81 = 9 [1]. 6 × 40 = 240 [1]. 240 ÷ 9 ≈ 27 [1]. [3] Examiner insight: write the rounded values down. A bare number gives the reader nothing to follow, so show each rounding.

4. (a) 8.35 ≤ a < 8.45 [1]. (b) The unit is 100, half is 50: 7550 ≤ b < 7650 [1]. (c) Truncated, so the unit is a whole 0.1: 0.9 ≤ c < 1.0 [1]. (d) The unit is 1000, half is 500: 14 500 ≤ d < 15 500 [1]. [4] Examiner insight: the lower end uses ≤ and the upper end uses <. Writing ≤ at both ends is a common slip.

5. (a) Speaker: 59.5 ≤ w < 60.5 [1]. Gap: 60.5 ≤ g < 61.5 [1]. The speaker is always narrower than 60.5 and the gap is at least 60.5, so yes, it fits [1]. [3] (b) The speaker could be up to just under 60.5 while the gap could be as small as 59.5 [1]. No, you cannot be sure [1]. [2] Examiner insight: compare the largest possible object with the smallest possible space, not the stated sizes.

6. (a) 4.7 ≤ t < 4.8 [1]. (b) The second time is at least 4.8 and the first is below 4.8 [1], so yes, the second was slower [1]. [3] Examiner insight: name both intervals before you decide. A truncated interval starts at the stated value.

7. (a) Upper limits 8.45 and 5.75 [1]. 8.45 × 5.75 = 48.5875 m² [1]. [2] (b) Lower limits 8.35 and 5.65 [1]. 8.35 × 5.65 = 47.1775 m² [1]. [2] (c) To 2 significant figures the bounds are 49 and 47, which differ, but to 1 significant figure both are 50 [1]. Area = 50 m² (1 significant figure) [1]. [2] Examiner insight: justify the accuracy by comparing the rounded bounds. Quoting 48 m² with no reason leaves the choice unsupported.

8. (a) Upper limit of A is 15.85 and lower limit of B is 6.35 [1]. 15.85 − 6.35 = 9.5 [1]. [2] (b) Lower limit of A is 15.75 [1]. Upper limit of B is 6.45 [1]. 15.75 ÷ 6.45 = 2.4418… = 2.44 [1]. [3] Examiner insight: for a difference or a quotient, one value takes its upper limit and the other its lower. State which you use.

9. (a) Counted, so whole numbers only: 1150 ≤ n ≤ 1249 [2]. [2] (b) Measured, so any value is possible: 1150 ≤ m < 1250 [2]. [2] (c) Least: 1150 × 8 = £9200 [1]. Greatest: 1249 × 8 = £9992 [1]. [2] Examiner insight: for a count, the greatest possible value is 1249. Using 1250 or 1249.9 does not describe a number of people.

10. Distance: 835 ≤ d < 845. Time: 104.5 ≤ t < 105.5 [1]. Upper bound of speed = 845 ÷ 104.5 [1] = 8.086… m/s. Lower bound of speed = 835 ÷ 105.5 [1] = 7.914… m/s [1]. Both round to 8 to 1 significant figure [1], so speed = 8 m/s (1 significant figure) [1]. [6] Examiner insight: the largest speed combines the largest distance with the smallest time. Pairing the wrong limits gives the wrong bound.

11. (a) Limits: 4.15 to 4.25, 3.45 to 3.55, 1.95 to 2.05 [1]. Upper bound = 4.25 × 3.55 × 2.05 = 30.93 cm³ [1]. Lower bound = 4.15 × 3.45 × 1.95 = 27.92 cm³ [1]. [3] (b) Mass limits are 117.5 and 118.5 [1]. Upper bound = 118.5 ÷ 27.919 = 4.24 g/cm³ [1]. Lower bound = 117.5 ÷ 30.929 = 3.80 g/cm³ [1]. [3] (c) Both bounds round to 4 to 1 significant figure, but not to 2 [1]. Density = 4 g/cm³ [1]. [2] Examiner insight: keep the unrounded bounds for the next step, and round only when you decide the final accuracy.

12. (a) 19 → 20, 3.85 → 4, 2.4 → 2, 6.15 → 6 [1]. 20 × 4 = 80 and 2 × 6 = 12 [1]. ≈ £92 [1]. [3] (b) 19 × 3.85 = 73.15, 2.4 × 6.15 = 14.76, so the bill is £87.91 [1]. The estimate is bigger, so it is an overestimate [1]. [2] (c) 87.5 ≤ B < 88.5 [1]. £87.91 lies inside this interval, so it is consistent [1]. [2] Examiner insight: give the direction of the error with a reason. Saying “close” is not enough.

Where marks are usually lost

• Dropping a final zero when rounding to a stated number of decimal places or significant figures.
• Writing ≤ at both ends of a rounded interval for a measured quantity.
• Adding half a unit to a truncated value instead of a whole unit.
• Comparing the stated sizes in a fit question instead of the worst cases.
• Using the lower bound of the denominator when you need the upper bound of a quotient.
• Rounding bounds before using them in the next stage of a calculation.
• Quoting an answer to more figures than both bounds share, with no justification.
• Giving 250 or 249.9 as the greatest value of a counted quantity.
• Showing an estimate with no rounded values written down.

Next steps

• Review the revision notes and the study guide.
• Open the OCR GCSE Mathematics hub and the printable checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017, section 4.01 Approximation and estimation (4.01a to 4.01c).