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Mathematics 05801.5

Estimation, bounds and limits of accuracy

Rounding, estimating, and the upper and lower bounds of a measurement.

Learning objectives

What you need to be able to do

Teacher-mapped phrasing — check against the official Cambridge syllabus for exact wording.

  • 1.5.1Round values, estimate calculations, and find upper and lower bounds.
  • 1.5.2Use bounds in calculations involving sums, differences, products and quotients.Supplement

6 minute read

Estimation, bounds and limits of accuracy

Estimating

Round every number to 1 significant figure, then calculate. It is meant to be quick and done without a calculator, giving a rough check on a real answer. (4.87 × 19.6) / 0.51 ≈ (5 × 20) / 0.5 = 200

Bounds

A measurement given to the nearest unit could be anything within half a unit either side. A length of 8 cm to the nearest cm:

  • Lower bound = 7.5 cm
  • Upper bound = 8.5 cm

The upper bound is written as 8.5 even though 8.5 would itself round up — this is the convention, and 8.5 is the value the real length approaches but does not reach. For 1 decimal place, half a unit is 0.05; for 2 decimal places it is 0.005.

Calculating with bounds

This is where marks are won or lost, because you must think about which combination gives the extreme:

  • Maximum of a sum: UB + UB
  • Minimum of a sum: LB + LB
  • Maximum of a difference: UB − LB (biggest minus smallest)
  • Minimum of a difference: LB − UB
  • Maximum of a product: UB × UB
  • Maximum of a quotient: UB ÷ LB (biggest divided by smallest)
  • Minimum of a quotient: LB ÷ UB

The division rules are the counter-intuitive ones. Dividing by a smaller number gives a bigger answer, which is why the maximum uses the lower bound on the bottom.

Think of it like this

A rounded measurement is a range in disguise. "8 cm" is really a promise that the true value lives somewhere between 7.5 and 8.5 — so any calculation using it inherits that uncertainty.

Worked examples

Method, step by step

A rectangle has length 12 cm and width 7 cm, each to the nearest cm. Find the upper bound of its area.

  1. 1Length: LB = 11.5, UB = 12.5. Width: LB = 6.5, UB = 7.5.
  2. 2For the maximum area, use the largest possible values of both.
  3. 3Maximum area = 12.5 × 7.5 = 93.75 cm².

93.75 cm²

A car travels 100 m (to the nearest metre) in 8 s (to the nearest second). Find the maximum possible speed.

  1. 1Distance: UB = 100.5 m, LB = 99.5 m. Time: UB = 8.5 s, LB = 7.5 s.
  2. 2Speed = distance ÷ time, so the maximum needs the largest distance and the smallest time.
  3. 3Maximum speed = 100.5 ÷ 7.5 = 13.4 m/s.

13.4 m/s

Common misconceptions

  • Using UB ÷ UB for the maximum of a division. It is UB ÷ LB, because dividing by less gives more.
  • Writing the upper bound as 8.49 or 8.499. The convention is the exact half-unit, 8.5.
  • Adding or subtracting a whole unit instead of half. For values to the nearest cm, the bounds are ±0.5, not ±1.

In the exam

  • Write LB and UB for each quantity on a separate line before combining them — it makes choosing the right pairing obvious.
  • For a maximum, ask "what makes this as big as possible?" rather than memorising all six rules. Division then follows naturally.