IGCSE Maths · Number · Core & Extended

Estimation and Limits of Accuracy: Rounding, Sig Figs and Bounds

Rounding and estimating are identical for both tiers — every measurement in real life is an approximation, and knowing how to round sensibly, and sanity-check an answer, is a skill you’ll use in every topic from here on. Limits of accuracy splits a little: Core stops at writing bounds for a single measurement, while Extended goes further and works out bounds for quantities calculated from rounded measurements — tagged below.

Why this matters No ruler, scale, or stopwatch is perfectly precise — “the plank is 2 m long” almost always means “2 m, to the nearest something.” Builders, engineers, and scientists all need to know exactly how much error that rounding could be hiding, especially when several rounded measurements get combined into one calculation.

1. Rounding to Decimal Places

Look at the digit after the place you’re rounding to. If it’s 5 or more, round up; if it’s less than 5, round down (leave it as is).

Worked examples Round 3.14159 to 2 decimal places.
The digit after the 2nd decimal place is 1 (from …4159), so round down: 3.14

Round 0.0876 to 2 decimal places.
The digit after is 7, so round up: 0.09
Try it — Decimal Places
  1. Round 7.3821 to 2 decimal places.
  2. Round 15.996 to 1 decimal place.
  3. Round 0.4058 to 3 decimal places.
Show answers
  1. 7.38
  2. 16.0
  3. 0.406

2. Rounding to Significant Figures

The first significant figure is the first non-zero digit, counting from the left. Leading zeros are never significant; zeros between or after other digits usually are, depending on context.

Worked examples Round 5764 to the nearest thousand (1 significant figure).
5764 is between 5000 and 6000, closer to 6000 → 6000

Round 5764 to 3 significant figures.
The first 3 significant digits are 5, 7, 6 — look at the next digit (4) to decide rounding: round down → 5760

Round 0.03482 to 2 significant figures.
The first significant figure is 3 (leading zeros don’t count) — the first two are 3 and 4, look at the next digit (8) to round up → 0.035
Try it — Significant Figures
  1. Round 48,392 to 1 significant figure.
  2. Round 48,392 to 3 significant figures.
  3. Round 0.006758 to 2 significant figures.
  4. Round 7.0962 to 3 significant figures.
Show answers
  1. 50,000
  2. 48,400
  3. 0.0068
  4. 7.10

3. Estimating Calculations

To sanity-check an answer without a calculator, round every value in the calculation to 1 significant figure first, then work it out. This won’t give the exact answer, but it will catch major errors — like a misplaced decimal point.

Worked example Estimate the value of (19.6 × 0.48) ÷ 3.9, by rounding each number to 1 significant figure first.
19.6 ≈ 20,   0.48 ≈ 0.5,   3.9 ≈ 4
(20 × 0.5) ÷ 4 = 10 ÷ 4 = 2.5
(The actual value is about 2.41 — close enough to confirm an answer like “2.41” is sensible, and to flag something like “24.1” or “0.241” as an error.)
Try it — Estimating
  1. Estimate (39 × 0.21) ÷ 5.8, rounding each value to 1 significant figure.
  2. Estimate (212 × 4.9) ÷ 9.7, rounding each value to 1 significant figure.
  3. A calculation gives an answer of 340. Estimate first to check whether this is sensible for (58 × 6.1).
Show answers
  1. 40 × 0.2 ÷ 6 ≈ 8 ÷ 6 ≈ 1.3
  2. 200 × 5 ÷ 10 = 100
  3. 60 × 6 = 360, close to 340 — so 340 is sensible.

4. Upper and Lower Bounds for a Measurement

When a measurement is rounded, the true value could be anything within half a unit either side of the rounded figure. These limits are called the upper bound and lower bound.

Lower bound = rounded value − (half the rounding unit)
Upper bound = rounded value + (half the rounding unit)
Worked example A length is recorded as 8 cm, to the nearest cm. Find the lower and upper bounds.
Half of 1 cm is 0.5 cm.
Lower bound = 8 − 0.5 = 7.5 cm,   Upper bound = 8 + 0.5 = 8.5 cm
So the true length, L, satisfies 7.5 ⩽ L < 8.5.
Worked example — a different rounding unit A mass is recorded as 240 g, to the nearest 10 g. Find the lower and upper bounds.
Half of 10 g is 5 g.
Lower bound = 240 − 5 = 235 g,   Upper bound = 240 + 5 = 245 g
Try it — Bounds of a Single Measurement
  1. A time is recorded as 15 minutes, to the nearest minute. Find the bounds.
  2. A distance is recorded as 60 km, to the nearest km. Find the bounds.
  3. A mass is recorded as 3.5 kg, to the nearest 0.1 kg. Find the bounds.
  4. A crowd size is recorded as 4,000 people, to the nearest 100. Find the bounds.
Show answers
  1. 14.5 ⩽ t < 15.5 (minutes)
  2. 59.5 ⩽ d < 60.5 (km)
  3. 3.45 ⩽ m < 3.55 (kg)
  4. 3,950 ⩽ n < 4,050

5. Bounds of Calculated Quantities Extended only

When a calculation combines two or more rounded measurements, the error can compound. The rule depends on the operation:

OperationTo find the upper bound…To find the lower bound…
Adding (e.g. perimeter)use the upper bound of each valueuse the lower bound of each value
Multiplying (e.g. area)use the upper bound of each valueuse the lower bound of each value
Dividing (e.g. speed = distance ÷ time)upper bound ÷ lower boundlower bound ÷ upper bound

Division is the one that catches people out — to make a fraction as big as possible, you want the biggest possible top and the smallest possible bottom, and vice versa for the smallest fraction.

Worked example — perimeter and area A rectangle has a length of 12 cm and a width of 7 cm, both measured to the nearest cm. Find the upper and lower bounds of the perimeter and the area.
Length: 11.5 ⩽ L < 12.5    Width: 6.5 ⩽ W < 7.5

Perimeter = 2(L + W)
Upper bound: 2(12.5 + 7.5) = 40 cm
Lower bound: 2(11.5 + 6.5) = 36 cm

Area = L × W
Upper bound: 12.5 × 7.5 = 93.75 cm²
Lower bound: 11.5 × 6.5 = 74.75 cm²
Worked example — speed (division) A distance of 100 m is measured to the nearest metre, and a time of 20 s is measured to the nearest second. Find the upper and lower bounds of the average speed.
Distance: 99.5 ⩽ d < 100.5    Time: 19.5 ⩽ t < 20.5

Upper bound of speed = (upper bound of distance) ÷ (lower bound of time) = 100.5 ÷ 19.5 ≈ 5.15 m/s
Lower bound of speed = (lower bound of distance) ÷ (upper bound of time) = 99.5 ÷ 20.5 ≈ 4.85 m/s
Try it — Bounds of Calculations Extended
  1. A rectangle has length 15 cm and width 9 cm, both to the nearest cm. Find the upper and lower bounds of the perimeter.
  2. Using the same rectangle, find the upper and lower bounds of the area.
  3. A car travels 150 km (to the nearest km) in 3 hours (to the nearest 0.1 hour, so bounds of 2.95 and 3.05 hours). Find the upper and lower bounds of its average speed.
  4. Two lengths, 8 cm and 5 cm (both to the nearest cm), are added together. Find the upper and lower bounds of the total.
Show answers
  1. Length: 14.5–15.5, Width: 8.5–9.5. Perimeter upper = 2(15.5+9.5) = 50 cm. Perimeter lower = 2(14.5+8.5) = 46 cm.
  2. Area upper = 15.5 × 9.5 = 147.25 cm². Area lower = 14.5 × 8.5 = 123.25 cm².
  3. Distance: 149.5–150.5. Speed upper = 150.5 ÷ 2.95 ≈ 51.0 km/h. Speed lower = 149.5 ÷ 3.05 ≈ 49.0 km/h.
  4. Each length: 7.5–8.5 and 4.5–5.5. Upper bound total = 8.5 + 5.5 = 14 cm. Lower bound total = 7.5 + 4.5 = 12 cm.
Common mistakes to watch for
  • Using half of the wrong unit — if a value is rounded to the nearest 10, the bounds are ± 5, not ± 0.5.
  • Mixing up which bound to use for division — for the upper bound of a quotient, you need the smallest possible denominator, not the largest.
  • Using the rounded value itself in a bounds calculation instead of the actual upper/lower bound.
  • Estimating by rounding only some of the numbers in a calculation to 1 significant figure, and leaving others as they are.
  • Forgetting that the upper bound is technically never reached (it’s a strict inequality, <, not ⩽) — worth knowing conceptually, even though in practice you calculate with the boundary value itself.

6. Quick-Fire Challenge Round

True or False?
  1. A value of 60, rounded to the nearest 10, has bounds of 55 and 65.
  2. To estimate a calculation, you should round every number to as many significant figures as possible.
  3. For the upper bound of a division, you use the upper bound of the numerator and the lower bound of the denominator.
  4. The upper and lower bounds of a sum are found by adding the upper bounds and adding the lower bounds respectively.
Show answers
  1. True.
  2. False — for a quick estimate, round to 1 significant figure, not many; more figures defeats the purpose of a fast sanity check.
  3. True. Ext.
  4. True. Ext.
Spot the Error
  1. “A length of 20 m, to the nearest metre, has bounds of 19 and 21.”
  2. “To find the upper bound of speed = distance ÷ time, I used the upper bound of both distance and time.”
  3. “48,392 rounded to 1 significant figure is 40,000.”
Show answers
  1. The bounds should be half a unit either side, not a full unit: 19.5 and 20.5.
  2. For the upper bound of a division, the denominator should be at its lower bound, not upper — using upper bound for time makes the speed too small, not too large.
  3. The first significant figure is 4 (in the ten-thousands place); the next digit is 8, so it rounds up to 50,000, not down to 40,000.

7. Mixed Practice — Bringing It All Together

Core & Extended
  1. Round 6.0482 to 2 decimal places.
  2. Round 372,500 to 2 significant figures.
  3. Estimate (61 × 2.9) ÷ 5.8 by rounding each value to 1 significant figure.
  4. A length is recorded as 45 cm, to the nearest cm. Find the lower and upper bounds.
Show answers
  1. 6.05
  2. 370,000
  3. 60 × 3 ÷ 6 = 30
  4. 44.5 cm and 45.5 cm
Extended-level mixed set Extended only
  1. A rectangle has length 20 cm and width 13 cm, both to the nearest cm. Find the upper bound of its area.
  2. A journey of 84 km (to the nearest km) takes 1.5 hours (to the nearest 0.1 hour). Find the lower bound of the average speed.
Show answers
  1. Length upper = 20.5, width upper = 13.5. Area upper = 20.5 × 13.5 = 276.75 cm².
  2. Distance lower = 83.5, time upper = 1.55. Speed lower = 83.5 ÷ 1.55 ≈ 53.9 km/h.

Practice at Home

A large set again — take it in stages.

Part A — Decimal places and significant figures
  1. Round 12.6753 to 3 decimal places.
  2. Round 0.09456 to 2 decimal places.
  3. Round 763,208 to 1 significant figure.
  4. Round 0.005029 to 2 significant figures.
  5. Round 19.98 to 3 significant figures.
Part B — Estimating calculations
  1. Estimate (78 × 0.31) ÷ 4.1, rounding each value to 1 significant figure.
  2. Estimate (5.9 × 198) ÷ 0.48, rounding each value to 1 significant figure.
  3. A calculation gives an answer of 5,200. Estimate (312 × 16) to check whether this is sensible.
Part C — Bounds of a single measurement
  1. A length of 32 cm is measured to the nearest cm. Find the bounds.
  2. A mass of 500 g is measured to the nearest 10 g. Find the bounds.
  3. A time of 2.5 hours is measured to the nearest 0.1 hour. Find the bounds.
  4. A population of 18,000 is given to the nearest 1,000. Find the bounds.
Part D — Bounds of calculated quantities Extended
  1. A rectangle has length 24 cm and width 11 cm, both to the nearest cm. Find the upper and lower bounds of its perimeter.
  2. Using the same rectangle, find the upper and lower bounds of its area.
  3. A cyclist travels 45 km (to the nearest km) in 2 hours (to the nearest 0.1 hour). Find the upper and lower bounds of their average speed.
Part E — Applied / challenge
  1. Two lengths, each measured as 6 cm to the nearest cm, are placed end to end. Explain why the upper bound of the total length is not simply 6.5 + 6.5, but instead consider what the actual upper bound should be — then calculate it. (Hint: it isn’t a trick — check the rule for sums carefully.)
  2. A square has a side length of 9 cm, measured to the nearest cm. Find the upper and lower bounds of its area.
  3. A given calculation is estimated as roughly 4,000 by rounding to 1 significant figure. The exact answer turns out to be 3,847. Explain why this is still a “good” estimate, even though it isn’t exact.

Key Vocabulary Recap

Significant figure — a digit that contributes to a number’s precision, starting from the first non-zero digit
Estimate — a rounded, approximate answer used to sanity-check a calculation
Lower bound — the smallest value a rounded measurement could actually be
Upper bound — the largest value a rounded measurement could actually be

Syllabus Reference

This note covers Cambridge IGCSE Mathematics (0580) sections C1.9 and C1.10 (Core subject content) and E1.9 and E1.10 (Extended subject content) — Estimation, and Limits of accuracy. Estimation (rounding, significant figures, sensible estimating) is identical across tiers. Core candidates are only expected to find bounds for a single rounded measurement; Extended candidates additionally find bounds of quantities calculated from rounded data, marked above.

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