Overview
What you’ll learn
- Round numbers to a given number of decimal places or significant figures, keeping trailing zeros that are significant.
- Estimate calculations by rounding each value to 1 significant figure and computing the simplified expression.
- Choose a sensible degree of accuracy for a final answer in context.
- Write the upper and lower bounds of a measurement rounded to a stated accuracy, as an interval or inequality.
- Find bounds of calculated results — perimeters, areas, speeds and densities — by combining the bounds correctly (Extended).
Syllabus checklist (0580)
Statements are paraphrased and grouped by tier. Core is studied by everyone; Extended content is examined on Papers 2 and 4 only.
Core — every candidate
- 1.9Round values to a specified degree of accuracy, including decimal places and significant figures.
- 1.9Make estimates for calculations involving numbers, quantities and measurements, e.g. by rounding each number to 1 significant figure.
- 1.9Round answers to a reasonable degree of accuracy in the context of a given problem.
- 1.10Give upper and lower bounds for data rounded to a specified accuracy.
Extended — Extended candidates only
- 1.10Find upper and lower bounds of the results of calculations which have used data rounded to a specified accuracy, e.g. bounds of a perimeter, area or speed. (Extended)
In 30 seconds
- 1Decimal places count after the point; significant figures count from the first non-zero digit — 0.0006491 to 2 s.f. is 0.00065.
- 2A rounding digit of 5 or more rounds up; kept zeros matter: 84.976 to 3 s.f. is 85.0, not 85.
- 3Estimate by rounding every number to 1 s.f., then working exactly with the easy values: (5.98 × 19.7) ÷ 0.492 ≈ (6 × 20) ÷ 0.5 = 240.
- 4A value m to the nearest unit lives in the interval m − ½ ⩽ m < m + ½: the lower bound counts, the upper bound does not.
- 5Extended — bounds of calculations: sums use all lower (or all upper) bounds; products of positive quantities maximise with UB × UB; quotients like speed maximise with UB distance ÷ LB time.
- 6Round the final bounds sensibly (often to the accuracy of the data or just beyond) and show the unrounded values first.
Notes · 1 of 4
Rounding to decimal places and significant figures
The two counting systems, and the trailing-zero trap.
Round to a number of decimal places (d.p.) by looking at the next digit: 27.483 to 1 d.p. is 27.5 (the 8 rounds the 4 up). Round to significant figures (s.f.) by counting from the first non-zero digit: 0.005392 to 2 s.f. is 0.0054 — the leading zeros never count.
| Instruction | 84.976 | 0.004826 |
|---|---|---|
| 2 d.p. | 84.98 | 0.00 |
| 1 d.p. | 85.0 | 0.0 |
| 3 s.f. | 85.0 | 0.00483 |
| 1 s.f. | 80 | 0.005 |
Notice 84.976 to 3 s.f.: the third significant figure is the 9, which rounds the 8 up to 9, which carries again — 85.0. The final zero is doing real work: it says “this digit was checked”. Writing 85 (2 s.f.) loses the accuracy the question asked for. In context, round answers to a sensible accuracy: money to 2 d.p., measurements to the accuracy of the instrument or a matching number of s.f.
Notes · 2 of 4
Estimating calculations
Round everything to 1 s.f., compute the friendly version, keep the ≈ honest.
To estimate a messy calculation, round every number to 1 significant figure and evaluate the clean version. For (5.98 × 19.7) ÷ 0.492: round to (6 × 20) ÷ 0.5, which is 120 ÷ 0.5 = 240. The estimate tells you the size of the true answer — if your calculator gives 239.4…, the estimate just saved you from a slip.
Use the ≈ symbol for estimates, and show the rounded values you used — the marks are for the method. Estimates also catch impossible answers: a fraction near 20 ÷ 5 can never be 2000, no matter what the calculator says. Division by a number rounded down (like 0.492 → 0.5 is rounded *up*) shifts the estimate slightly; that is fine — estimates are for scale, not for digits.
Notes · 3 of 4
Upper and lower bounds
Where a rounded measurement really lives.
A measurement rounded to the nearest unit could be anything in a half-unit window either side. A length of 7 m to the nearest metre was measured as anything from 6.5 m up to (but not including) 7.5 m: the lower bound is 6.5 and the upper bound is 7.5. As an inequality: 6.5 ⩽ length < 7.5 — the length could equal 6.5 (it rounds up to 7) but could never equal 7.5 (that would round to 8).
| Stated value | Rounded to | Lower bound | Upper bound |
|---|---|---|---|
| 45 m | nearest metre | 44.5 m | 45.5 m |
| 1.2 m | nearest cm | 1.195 m | 1.205 m |
| 12.5 s | nearest 0.1 s | 12.45 s | 12.55 s |
| 3.4 kg | nearest 0.1 kg | 3.35 kg | 3.45 kg |
The half-step is always half of the rounding unit: nearest centimetre gives ±0.5 cm, nearest 0.1 s gives ±0.05 s. Write bounds with the units attached, and remember the upper bound is approached but not reached — which is why the interval notation uses a strict inequality at the top.
Notes · 4 of 4
Bounds of calculated results
Pushing bounds through sums, products and quotients. (Extended)
Extended. When a calculation uses rounded data, the answer also lives in an interval. To maximise a result, push every input in the helpful direction; to minimise, push the other way. For a rectangle with sides 8.4 cm and 3.6 cm to the nearest centimetre: the upper-bound area is UB × UB = 8.45 × 3.65 = 30.8425 ≈ 30.8 cm², and the lower-bound area is LB × LB = 8.35 × 3.55 = 29.6425 ≈ 29.6 cm². Bounds of a perimeter add all upper (or all lower) bounds: 2 × (8.45 + 3.65) = 24.2 and 2 × (8.35 + 3.55) = 23.8 cm.
For quotients, the directions split: speed = distance ÷ time is largest with the UB of distance over the LB of time, and smallest with LB over UB. A distance of 250 m (nearest metre) covered in 34 s (nearest second) gives speed bounds 249.5 ÷ 34.5 = 7.23… up to 250.5 ÷ 33.5 = 7.48…, so the speed is 7.4 m/s to 2 s.f., somewhere between 7.2 and 7.5. Densities behave the same way: mass ÷ volume maximises with UB mass over LB volume.
Revise
Key terms
The words examiners expect you to use precisely.
- Decimal place (d.p.)
- A position after the decimal point used for rounding.27.483 to 1 d.p. is 27.5.
- Significant figure (s.f.)
- A digit counted from the first non-zero digit of a number.0.005392 to 2 s.f. is 0.0054.
- Estimate
- A quick approximation found by rounding inputs, marked with ≈.(6 × 20) ÷ 0.5 = 240.
- Lower bound
- The smallest value a rounded measurement could have taken.7 m to the nearest metre has LB 6.5 m.
- Upper bound
- The value just above the largest a rounded measurement could have taken.7 m to the nearest metre has UB 7.5 m (not included).
- Error interval
- The inequality showing where a rounded value lies.6.5 ⩽ length < 7.5.
Show 2 more key termsShow fewer key terms
- Rounding unit
- The step the stated accuracy rounds to.Nearest cm → unit 0.01 m → bounds ±0.005 m.
- Bounds of a calculation (Extended)
- The interval containing the true result of a calculation on rounded data.Area between LB × LB and UB × UB.
Revise
Common mistakes
Where students most often lose marks in this topic — and what to do instead.
Mistake: Dropping a significant trailing zero when rounding.
85 is 2 s.f.; the question asked for 3 s.f.
Instead: 84.976 to 3 s.f. is 85.0 — the zero records the checked digit.
Mistake: Counting leading zeros as significant figures.
In 0.004826 the zeros only locate the decimal point.
Instead: The first non-zero digit starts the count: 4, then 8.
Show 10 more mistakesShow fewer mistakes
Mistake: Rounding mid-calculation.
Early rounding feeds wrong digits through every later step.
Instead: Keep full values in the calculator; round only the final answer.
Mistake: Estimating without showing the rounded values.
The method — which numbers you used — carries the marks.
Instead: Write “using 400, 8 and 2…” before the simplified computation.
Mistake: Writing the upper bound of 7 (nearest metre) as 7.5 inclusive.
A true length of exactly 7.5 would round to 8, not 7.
Instead: LB ⩽ value < UB: 6.5 ⩽ length < 7.5.
Mistake: Using the full rounding unit instead of half of it.
The window extends half a unit each side, not a whole unit.
Instead: 1.2 m to the nearest cm has bounds 1.195 and 1.205, not 1.19 and 1.21.
Mistake: Using LB × UB for the upper bound of an area.
A product of positive quantities is maximised by maximising both factors.
Instead: Upper bound of area = UB width × UB length.
Mistake: Maximising speed with UB time.
Dividing by a smaller time makes the speed larger.
Instead: UB speed = UB distance ÷ LB time; LB speed = LB distance ÷ UB time.
Mistake: Forgetting to convert units inside bound calculations.
3.4 kg with 480 cm³ must share one unit system.
Instead: Convert kg to g (or cm³ to m³) before dividing for density.
Mistake: Rounding 299.5 to the nearest ten as 290.
The 5 after the tens place rounds up, and the carry cascades through the nines.
Instead: 299.5 → 300: check the digit after the place you keep, then carry left.
Mistake: Finding bounds of a subtraction as LB − LB.
To make a − b as small as possible, b must be as big as possible.
Instead: LB(a − b) = LB(a) − UB(b), and UB(a − b) = UB(a) − LB(b).
Mistake: Counting decimal places when a question asks for significant figures (or the reverse).
Significant figures count from the first non-zero digit; decimal places count from the point.
Instead: 0.005392 to 2 sf is 0.0054, not 0.01.
Check yourself
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Question 1 of 3
- Question 1: current question
- Question 2: not answered
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Round to a number of decimal places
Round 27.483 correct to 1 decimal place.
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Practise
Topical questions
14 exam-style questions with model answers and mark schemes.
A set of structured and calculation questions matched to this topic, in three tiers, each with a model answer, mark scheme and examiner tip.
Practise topical questionsExam practice
Practice paper
A 30-mark paper set out like a Cambridge paper, with a separate mark scheme. Give yourself 40 min.
Question paper
Topic 5 Practice Paper
30 marks · 40 min · 5 pages
10 multiple-choice questions and structured questions with answer lines. Topic 5: rounding to decimal places and significant figures, estimating calculations with 1-significant-figure values, upper and lower bounds of rounded measurements, and bounds of perimeters, areas, quotients and densities.
Mark scheme
PROTopic 5 Practice Paper — mark scheme
Answers · Marking points · 3 pages
Model answers with a mark for each creditworthy point — for marking at home or in class.
Which papers examine this topic?
Paper 1 / 3
Core: 80 marks each, 1 h 30 each
Paper 1 non-calculator, Paper 3 calculator
Paper 2 / 4
Extended: 100 marks each, 2 h each
Paper 2 non-calculator, Paper 4 calculator
Accuracy
3 s.f., angles to 1 d.p.
Unless a question states otherwise
Want a full paper? Try the Paper 1-style Core, Paper 2-style non-calculator or Paper 4-style calculator sets.
Questions
Frequently asked questions
What is the difference between 2 decimal places and 2 significant figures?
Decimal places count positions after the point, significant figures count digits from the first non-zero one. For 0.005392: 2 d.p. gives 0.01 (barely useful), while 2 s.f. gives 0.0054.
Why did my 3-significant-figure answer of 85 get marked wrong?
Because 85 shows only two significant figures. Rounding 84.976 to 3 s.f. gives 85.0 — the final zero is the checked digit and must be written.
How do I pick which bounds to use for a calculation?
Ask what pushes the answer up. Sums and products of positive quantities use all upper bounds (or all lower). Quotients split: the biggest speed is upper-bound distance over lower-bound time.
Why is the upper bound never included in the inequality?
Because a value equal to the upper bound would round to the next value up, not to the stated one. 7.5 m to the nearest metre is 8 m, so the true length of a “7 m” measurement stays strictly below 7.5.
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