NumberAQA GCSE Mathematics (8300)

Measures and Accuracy

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

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Worked examples4 Key formulae4 Definitions6 Common mistakes4

Step-by-step worked examples

01.Write the error interval for a length rounded to the nearest centimetre
Foundation
Question
A plank of wood is measured as 85 cm, correct to the nearest centimetre. Write the error interval for the true length $l$ .
Solution
1. 1
  Identify the unit of rounding.
  $u = 1 \text{ cm}$
2. 2
  Calculate the half-unit.
  $\frac{u}{2} = 0.5 \text{ cm}$
3. 3
  Find the lower bound.
  $\text{LB} = 85 - 0.5 = 84.5 \text{ cm}$
4. 4
  Find the upper bound.
  $\text{UB} = 85 + 0.5 = 85.5 \text{ cm}$
5. 5
  Write the error interval. Lower bound is included; upper bound is excluded.
  $84.5 \leq l < 85.5$
Answer
$84.5 \leq l < 85.5$
Examiner note
AQA awards 1 mark for both bounds correct and 1 mark for the correct inequality notation ( $\leq$ and $<$ ). A common error is writing $\leq$ at both ends.
02.Find the maximum value of a speed calculation using bounds
Higher
Question
A cyclist travels a distance of $18$ km, measured to the nearest kilometre, in a time of $45$ minutes, measured to the nearest minute. Find the upper bound for the cyclist's average speed in km/h. Give your answer to 3 significant figures.
Solution
1. 1
  Convert time to hours and find bounds. $45$ minutes to the nearest minute: unit = 1 min = $\frac{1}{60}$ h.
  $\text{LB(time)} = \frac{44.5}{60} \text{ h}, \quad \text{UB(time)} = \frac{45.5}{60} \text{ h}$
2. 2
  Find bounds for distance. $18$ km to the nearest km: unit = 1 km.
  $\text{LB(dist)} = 17.5 \text{ km}, \quad \text{UB(dist)} = 18.5 \text{ km}$
3. 3
  Upper bound of speed = UB(distance) ÷ LB(time). To maximise a fraction, maximise the numerator and minimise the denominator.
  $\text{Max speed} = \frac{18.5}{44.5/60} = \frac{18.5 \times 60}{44.5}$
4. 4
  Evaluate.
  $= \frac{1110}{44.5} = 24.943\ldots \approx 24.9 \text{ km/h (3 s.f.)}$
Answer
$24.9$ km/h (3 s.f.)
Examiner note
Full marks require: correct bounds for both quantities (B1B1), correct formula UB(dist) ÷ LB(time) (M1), correct answer to stated accuracy (A1). The most common error is dividing UB(dist) by UB(time).
03.Estimate the value of a calculation
Foundation
Question
Estimate the value of $\dfrac{\sqrt{98.6} \times 4.93}{0.198}$ . Show your working.
Solution
1. 1
  Round each value to 1 significant figure.
  $\sqrt{98.6} \approx \sqrt{100} = 10, \quad 4.93 \approx 5, \quad 0.198 \approx 0.2$
2. 2
  Substitute the rounded values.
  $\frac{10 \times 5}{0.2}$
3. 3
  Evaluate the numerator.
  $10 \times 5 = 50$
4. 4
  Divide by 0.2 (equivalent to multiplying by 5).
  $\frac{50}{0.2} = 50 \times 5 = 250$
Answer
$250$ (exact answer: $\approx 249.6$ )
Examiner note
AQA mark schemes award marks for (i) rounding all values to 1 s.f. and (ii) a correct calculation with those rounded values. Show your rounded values explicitly — the method mark is awarded even if the final arithmetic is incorrect.
04.Find the minimum of a difference using bounds
Higher
Question
Two lengths are measured as $p = 12.6$ cm and $q = 9.8$ cm, each correct to 1 decimal place. Find the minimum value of $p - q$ .
Solution
1. 1
  Find bounds for $p$ . Unit = 0.1 cm, half-unit = 0.05 cm.
  $\text{LB}(p) = 12.55 \text{ cm}, \quad \text{UB}(p) = 12.65 \text{ cm}$
2. 2
  Find bounds for $q$ . Unit = 0.1 cm, half-unit = 0.05 cm.
  $\text{LB}(q) = 9.75 \text{ cm}, \quad \text{UB}(q) = 9.85 \text{ cm}$
3. 3
  Minimum of $p - q$ : use LB( $p$ ) − UB( $q$ ). Make $p$ as small as possible and $q$ as large as possible.
  $\text{Min}(p - q) = 12.55 - 9.85$
4. 4
  Evaluate.
  $= 2.70 \text{ cm}$
Answer
$2.70$ cm
Examiner note
The mark scheme awards B1 for correct LB( $p$ ) and UB( $q$ ), M1 for subtracting LB( $p$ ) − UB( $q$ ), A1 for $2.70$ (or $2.7$ ). A very common error: computing LB( $p$ ) − LB( $q$ ) = 2.80, which is the wrong combination.

Key formulae

Lower bound (rounded value)
$\text{LB} = x - \frac{u}{2}$
- $x$
  — The stated (rounded) value
- $u$
  — The unit of rounding (e.g. 0.1, 1, 10)
When to use
Any time a measurement has been rounded and you need the smallest possible true value. Identify the unit of rounding first.
Example
Length stated as 7.4 cm to the nearest 0.1 cm: $\text{LB} = 7.4 - 0.05 = 7.35$ cm.
Upper bound (rounded value)
$\text{UB} = x + \frac{u}{2}$
- $x$
  — The stated (rounded) value
- $u$
  — The unit of rounding
When to use
Finding the largest value the true measurement could take (exclusive — the true value is strictly less than UB).
Example
Length stated as 7.4 cm to the nearest 0.1 cm: $\text{UB} = 7.4 + 0.05 = 7.45$ cm.
Maximum of a fraction (compound measure)
$\text{Max}\!\left(\frac{a}{b}\right) = \frac{\text{UB}(a)}{\text{LB}(b)}$
- $a$
  — Numerator quantity (e.g. distance, mass)
- $b$
  — Denominator quantity (e.g. time, volume)
When to use
Finding maximum speed, density or pressure when both quantities are rounded. Use UB of the top and LB of the bottom.
Example
Max speed $= \dfrac{\text{UB(distance)}}{\text{LB(time)}}$ .
Minimum of a fraction (compound measure)
$\text{Min}\!\left(\frac{a}{b}\right) = \frac{\text{LB}(a)}{\text{UB}(b)}$
- $a$
  — Numerator quantity
- $b$
  — Denominator quantity
When to use
Finding minimum speed, density or pressure. Use LB of the top and UB of the bottom.
Example
Min density $= \dfrac{\text{LB(mass)}}{\text{UB(volume)}}$ .

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Significant figureExaminer keyword: A meaningful digit in a number. The first significant figure is the first non-zero digit from the left. Trailing zeros after a decimal point are significant; leading zeros are not.
Related: decimal place, rounding
Error intervalExaminer keyword: The range of values that a rounded or truncated measurement could take. Written as $\text{LB} \leq x < \text{UB}$ for a rounded value; the lower bound is included, the upper bound is excluded.
Related: lower bound, upper bound, inequality notation
Lower boundExaminer keyword: The smallest value that a rounded measurement could have. For a value rounded to unit $u$ : LB = stated value $- \frac{u}{2}$ . The lower bound is included in the error interval ( $\leq$ ).
Related: upper bound, error interval
Upper boundExaminer keyword: The smallest value that a rounded measurement could NOT have without rounding to a higher stated value. For a value rounded to unit $u$ : UB = stated value $+ \frac{u}{2}$ . The upper bound is excluded from the error interval ( $<$ ).
Related: lower bound, error interval
Truncation: Cutting off digits after a given position without rounding up. A truncated value always rounds down. Error interval for truncated value $t$ with unit $u$ : $t \leq x < t + u$ .
Related: rounding, error interval
EstimationExaminer keyword: An approximate answer obtained by rounding all values to 1 significant figure before calculating. Used to check the order of magnitude of a calculated result.
Related: significant figure, approximation

Common mistakes and misconceptions

Mistake
Using both upper bounds when finding the maximum of $a - b$ or $a \div b$ .
Why it happens
Students associate 'maximum' with 'use upper bounds everywhere' — a rule that works for addition and multiplication but not for subtraction or division.
How to avoid it
Ask yourself: to make $a - b$ as large as possible, $b$ should be as small as possible. So use LB( $b$ ). Same logic for $a \div b$ : a smaller denominator makes the fraction larger, so use LB( $b$ ).
Mistake
Writing $\text{LB} \leq x \leq \text{UB}$ (both inclusive) for a rounded error interval.
Why it happens
Students apply the familiar 'between two values' inequality pattern without remembering that the upper bound rounds to the next stated value.
How to avoid it
The upper bound is the first value that would round up — a true value equal to UB would give a different stated value. So always write $< \text{UB}$ , never $\leq \text{UB}$ .
Mistake
Counting leading zeros as significant figures.
Why it happens
Students start counting significant figures from the decimal point rather than from the first non-zero digit.
How to avoid it
Scan left to right and find the first non-zero digit — that is s.f. 1. In $0.00472$ , the first significant figure is 4. Rounding to 2 s.f. gives $0.0047$ , not $0.00$ .
Mistake
Applying the rounding lower bound formula (stated $- \frac{u}{2}$ ) to a truncated value.
Why it happens
Truncation and rounding have different error intervals; students apply the rounding rule by default.
How to avoid it
Check the question: 'truncated' means the LB equals the stated value itself. If $x$ is truncated to 5 (unit 1), then $5 \leq x < 6$ — not $4.5 \leq x < 5.5$ .

NumberAQA GCSE Mathematics (8300)

Measures and Accuracy

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

PreviousNext

Master the topic

Worked examples4 Key formulae4 Definitions6 Common mistakes4

Step-by-step worked examples

01.Write the error interval for a length rounded to the nearest centimetre
Foundation
Question
A plank of wood is measured as 85 cm, correct to the nearest centimetre. Write the error interval for the true length $l$ .
Solution
1. 1
  Identify the unit of rounding.
  $u = 1 \text{ cm}$
2. 2
  Calculate the half-unit.
  $\frac{u}{2} = 0.5 \text{ cm}$
3. 3
  Find the lower bound.
  $\text{LB} = 85 - 0.5 = 84.5 \text{ cm}$
4. 4
  Find the upper bound.
  $\text{UB} = 85 + 0.5 = 85.5 \text{ cm}$
5. 5
  Write the error interval. Lower bound is included; upper bound is excluded.
  $84.5 \leq l < 85.5$
Answer
$84.5 \leq l < 85.5$
Examiner note
AQA awards 1 mark for both bounds correct and 1 mark for the correct inequality notation ( $\leq$ and $<$ ). A common error is writing $\leq$ at both ends.
02.Find the maximum value of a speed calculation using bounds
Higher
Question
A cyclist travels a distance of $18$ km, measured to the nearest kilometre, in a time of $45$ minutes, measured to the nearest minute. Find the upper bound for the cyclist's average speed in km/h. Give your answer to 3 significant figures.
Solution
1. 1
  Convert time to hours and find bounds. $45$ minutes to the nearest minute: unit = 1 min = $\frac{1}{60}$ h.
  $\text{LB(time)} = \frac{44.5}{60} \text{ h}, \quad \text{UB(time)} = \frac{45.5}{60} \text{ h}$
2. 2
  Find bounds for distance. $18$ km to the nearest km: unit = 1 km.
  $\text{LB(dist)} = 17.5 \text{ km}, \quad \text{UB(dist)} = 18.5 \text{ km}$
3. 3
  Upper bound of speed = UB(distance) ÷ LB(time). To maximise a fraction, maximise the numerator and minimise the denominator.
  $\text{Max speed} = \frac{18.5}{44.5/60} = \frac{18.5 \times 60}{44.5}$
4. 4
  Evaluate.
  $= \frac{1110}{44.5} = 24.943\ldots \approx 24.9 \text{ km/h (3 s.f.)}$
Answer
$24.9$ km/h (3 s.f.)
Examiner note
Full marks require: correct bounds for both quantities (B1B1), correct formula UB(dist) ÷ LB(time) (M1), correct answer to stated accuracy (A1). The most common error is dividing UB(dist) by UB(time).
03.Estimate the value of a calculation
Foundation
Question
Estimate the value of $\dfrac{\sqrt{98.6} \times 4.93}{0.198}$ . Show your working.
Solution
1. 1
  Round each value to 1 significant figure.
  $\sqrt{98.6} \approx \sqrt{100} = 10, \quad 4.93 \approx 5, \quad 0.198 \approx 0.2$
2. 2
  Substitute the rounded values.
  $\frac{10 \times 5}{0.2}$
3. 3
  Evaluate the numerator.
  $10 \times 5 = 50$
4. 4
  Divide by 0.2 (equivalent to multiplying by 5).
  $\frac{50}{0.2} = 50 \times 5 = 250$
Answer
$250$ (exact answer: $\approx 249.6$ )
Examiner note
AQA mark schemes award marks for (i) rounding all values to 1 s.f. and (ii) a correct calculation with those rounded values. Show your rounded values explicitly — the method mark is awarded even if the final arithmetic is incorrect.
04.Find the minimum of a difference using bounds
Higher
Question
Two lengths are measured as $p = 12.6$ cm and $q = 9.8$ cm, each correct to 1 decimal place. Find the minimum value of $p - q$ .
Solution
1. 1
  Find bounds for $p$ . Unit = 0.1 cm, half-unit = 0.05 cm.
  $\text{LB}(p) = 12.55 \text{ cm}, \quad \text{UB}(p) = 12.65 \text{ cm}$
2. 2
  Find bounds for $q$ . Unit = 0.1 cm, half-unit = 0.05 cm.
  $\text{LB}(q) = 9.75 \text{ cm}, \quad \text{UB}(q) = 9.85 \text{ cm}$
3. 3
  Minimum of $p - q$ : use LB( $p$ ) − UB( $q$ ). Make $p$ as small as possible and $q$ as large as possible.
  $\text{Min}(p - q) = 12.55 - 9.85$
4. 4
  Evaluate.
  $= 2.70 \text{ cm}$
Answer
$2.70$ cm
Examiner note
The mark scheme awards B1 for correct LB( $p$ ) and UB( $q$ ), M1 for subtracting LB( $p$ ) − UB( $q$ ), A1 for $2.70$ (or $2.7$ ). A very common error: computing LB( $p$ ) − LB( $q$ ) = 2.80, which is the wrong combination.

Key formulae

Lower bound (rounded value)
$\text{LB} = x - \frac{u}{2}$
- $x$
  — The stated (rounded) value
- $u$
  — The unit of rounding (e.g. 0.1, 1, 10)
When to use
Any time a measurement has been rounded and you need the smallest possible true value. Identify the unit of rounding first.
Example
Length stated as 7.4 cm to the nearest 0.1 cm: $\text{LB} = 7.4 - 0.05 = 7.35$ cm.
Upper bound (rounded value)
$\text{UB} = x + \frac{u}{2}$
- $x$
  — The stated (rounded) value
- $u$
  — The unit of rounding
When to use
Finding the largest value the true measurement could take (exclusive — the true value is strictly less than UB).
Example
Length stated as 7.4 cm to the nearest 0.1 cm: $\text{UB} = 7.4 + 0.05 = 7.45$ cm.
Maximum of a fraction (compound measure)
$\text{Max}\!\left(\frac{a}{b}\right) = \frac{\text{UB}(a)}{\text{LB}(b)}$
- $a$
  — Numerator quantity (e.g. distance, mass)
- $b$
  — Denominator quantity (e.g. time, volume)
When to use
Finding maximum speed, density or pressure when both quantities are rounded. Use UB of the top and LB of the bottom.
Example
Max speed $= \dfrac{\text{UB(distance)}}{\text{LB(time)}}$ .
Minimum of a fraction (compound measure)
$\text{Min}\!\left(\frac{a}{b}\right) = \frac{\text{LB}(a)}{\text{UB}(b)}$
- $a$
  — Numerator quantity
- $b$
  — Denominator quantity
When to use
Finding minimum speed, density or pressure. Use LB of the top and UB of the bottom.
Example
Min density $= \dfrac{\text{LB(mass)}}{\text{UB(volume)}}$ .

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Significant figureExaminer keyword: A meaningful digit in a number. The first significant figure is the first non-zero digit from the left. Trailing zeros after a decimal point are significant; leading zeros are not.
Related: decimal place, rounding
Error intervalExaminer keyword: The range of values that a rounded or truncated measurement could take. Written as $\text{LB} \leq x < \text{UB}$ for a rounded value; the lower bound is included, the upper bound is excluded.
Related: lower bound, upper bound, inequality notation
Lower boundExaminer keyword: The smallest value that a rounded measurement could have. For a value rounded to unit $u$ : LB = stated value $- \frac{u}{2}$ . The lower bound is included in the error interval ( $\leq$ ).
Related: upper bound, error interval
Upper boundExaminer keyword: The smallest value that a rounded measurement could NOT have without rounding to a higher stated value. For a value rounded to unit $u$ : UB = stated value $+ \frac{u}{2}$ . The upper bound is excluded from the error interval ( $<$ ).
Related: lower bound, error interval
Truncation: Cutting off digits after a given position without rounding up. A truncated value always rounds down. Error interval for truncated value $t$ with unit $u$ : $t \leq x < t + u$ .
Related: rounding, error interval
EstimationExaminer keyword: An approximate answer obtained by rounding all values to 1 significant figure before calculating. Used to check the order of magnitude of a calculated result.
Related: significant figure, approximation

Common mistakes and misconceptions

Mistake
Using both upper bounds when finding the maximum of $a - b$ or $a \div b$ .
Why it happens
Students associate 'maximum' with 'use upper bounds everywhere' — a rule that works for addition and multiplication but not for subtraction or division.
How to avoid it
Ask yourself: to make $a - b$ as large as possible, $b$ should be as small as possible. So use LB( $b$ ). Same logic for $a \div b$ : a smaller denominator makes the fraction larger, so use LB( $b$ ).
Mistake
Writing $\text{LB} \leq x \leq \text{UB}$ (both inclusive) for a rounded error interval.
Why it happens
Students apply the familiar 'between two values' inequality pattern without remembering that the upper bound rounds to the next stated value.
How to avoid it
The upper bound is the first value that would round up — a true value equal to UB would give a different stated value. So always write $< \text{UB}$ , never $\leq \text{UB}$ .
Mistake
Counting leading zeros as significant figures.
Why it happens
Students start counting significant figures from the decimal point rather than from the first non-zero digit.
How to avoid it
Scan left to right and find the first non-zero digit — that is s.f. 1. In $0.00472$ , the first significant figure is 4. Rounding to 2 s.f. gives $0.0047$ , not $0.00$ .
Mistake
Applying the rounding lower bound formula (stated $- \frac{u}{2}$ ) to a truncated value.
Why it happens
Truncation and rounding have different error intervals; students apply the rounding rule by default.
How to avoid it
Check the question: 'truncated' means the LB equals the stated value itself. If $x$ is truncated to 5 (unit 1), then $5 \leq x < 6$ — not $4.5 \leq x < 5.5$ .

Measures and Accuracy

Master the topic

Step-by-step worked examples

01.Write the error interval for a length rounded to the nearest centimetre

02.Find the maximum value of a speed calculation using bounds

03.Estimate the value of a calculation

04.Find the minimum of a difference using bounds

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions

Measures and Accuracy

Master the topic

Step-by-step worked examples

01.Write the error interval for a length rounded to the nearest centimetre

02.Find the maximum value of a speed calculation using bounds

03.Estimate the value of a calculation

04.Find the minimum of a difference using bounds

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions