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Cambridge 0580 IGCSE  ·  Topic C1.10

C1.10 Limits of Accuracy

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01 — Explanation

What does "correct to the nearest X" actually mean?

Any rounded measurement is hiding a small range of possible true values. For example, a length given as 6 m6\text{ m} to the nearest metre could really be anywhere from 5.5 m5.5\text{ m} up to (but not including) 6.5 m6.5\text{ m}.

Core tier note: you only need the bounds of a single rounded measurement — finding bounds of a calculation built from several rounded values is Extended-only.

02 — Explanation

How do I find the upper and lower bounds of a measurement?

Halve the unit of accuracy (not the measurement itself), then add and subtract:

Lower Bound=Recorded Value−u2,Upper Bound=Recorded Value+u2\text{Lower Bound} = \text{Recorded Value} - \frac{u}{2}, \qquad \text{Upper Bound} = \text{Recorded Value} + \frac{u}{2}

Example: a bag of rice has a mass of 2.4 kg2.4\text{ kg}, correct to the nearest 0.1 kg0.1\text{ kg}.

  1. Unit of accuracy =0.1 kg= 0.1\text{ kg}, so half-unit =0.05 kg= 0.05\text{ kg}.
  2. Lower bound =2.4−0.05=2.35 kg= 2.4 - 0.05 = 2.35\text{ kg}.
  3. Upper bound =2.4+0.05=2.45 kg= 2.4 + 0.05 = 2.45\text{ kg}.

Common Slip

halving the measurement itself instead of the rounding unit. Always identify what unit the value was rounded to first.

03 — Explanation

How do I write this as an error interval?

State it as a double inequality, with the lower bound included and the upper bound excluded:

Lower Bound⩽x<Upper Bound\text{Lower Bound} \leqslant x < \text{Upper Bound}

For the rice bag: 2.35⩽m<2.452.35 \leqslant m < 2.45.

The upper bound is strict (<<, not ⩽\leqslant) because reaching exactly 2.45 kg2.45\text{ kg} would actually round up to 2.5 kg2.5\text{ kg}, not 2.4 kg2.4\text{ kg}.

04 — Worked example

Worked exam-style question

The length of a room is given as 6 m6\text{ m}, correct to the nearest metre. Write down the error interval for the true length LL.

  1. Unit of accuracy =1 m= 1\text{ m}, so half-unit =0.5 m= 0.5\text{ m}.
  2. Lower bound =6−0.5=5.5 m= 6 - 0.5 = 5.5\text{ m}.
  3. Upper bound =6+0.5=6.5 m= 6 + 0.5 = 6.5\text{ m}.
  4. Error interval: 5.5⩽L<6.55.5 \leqslant L < 6.5.

05 — Explanation

Does the method change if the measurement is rounded to a decimal place instead of a whole unit?

No — the method is identical, you just need to correctly identify the unit of accuracy first. Rounding "to the nearest 0.10.1" or "to 1 decimal place" both mean the unit of accuracy is 0.10.1.

Example: a plank of wood has a length of 3.8 m3.8\text{ m}, correct to 1 decimal place. Find the upper and lower bounds.

  1. Unit of accuracy =0.1 m=0.1\text{ m}, so half-unit =0.05 m=0.05\text{ m}.
  2. Lower bound =3.8−0.05=3.75 m=3.8-0.05=3.75\text{ m}.
  3. Upper bound =3.8+0.05=3.85 m=3.8+0.05=3.85\text{ m}.
  4. Error interval: 3.75⩽x<3.853.75 \leqslant x < 3.85.

Exam Tip

whatever the unit of rounding — whole numbers, tenths, hundredths, or even "nearest 50" — the process is always the same: find the unit of accuracy, halve it, then add and subtract from the recorded value.

06 — Common mistakes

Common mistakes to avoid

  • Writing 6.496.49 as the upper bound instead of 6.56.5 — the strict inequality already excludes 6.56.5 from rounding back down, so you should still state the value as 6.56.5.
  • Halving the wrong quantity — always halve the unit of accuracy, not the recorded measurement.
  • Using ⩽\leqslant on both sides. The convention is inclusive lower bound, strict upper bound.