AQA GCSE Mathematics Paper 2 (Foundation), June 2025: Question 23

Error Intervals & Upper Bounds of Rounded Measurements

📋 What this question tests

This question tests your ability to:

Find the lower bound and upper bound of numbers rounded to the nearest specified degree of accuracy.
Write an error interval using standard inequality notation ( ≤ and < ).
Convert between units (metres and centimetres) when dealing with limits of accuracy.
Select and apply the appropriate bound (upper bound) to construct a mathematical proof or justification.

Question 23 (a) • 2 Marks

Error Interval for a Rounded Length

The length of a fence panel is 160 cm to the nearest 10 cm. Complete the error interval.

📐 Step-by-Step Calculation

Step 1: Halve the degree of accuracy
Accuracy = 10 cm
Half accuracy = 10 ÷ 2 = 5 cm

Step 2: Find the lower bound
Lower Bound = 160 − 5 = 155 cm

Step 3: Find the upper bound
Upper Bound = 160 + 5 = 165 cm

Step 4: Express as an inequality
155 cm ≤ length < 165 cm

✅ Correct Answer

Answer:

155 cm ≤ length < 165 cm

Mark Scheme Breakdown:
• B2: Correct interval ( 155 and 165 in correct places).
• Partial credit (B1): Either 155 or 165 placed correctly.
• Special Case (SC1): Reversed interval: 165 ≤ length < 155 .

💡 Key Knowledge

To find limits of accuracy, always divide the unit of rounding by 2:
Limit = Stated Value ± (Accuracy ÷ 2)
Error intervals always take the form:
Lower Bound ≤ x < Upper Bound
The upper bound uses a strict inequality ( < ) because 165 cm would round up to 170 cm, but any value strictly below 165 rounds down to 160.

❌ Common Errors

Writing 164.9 or 164 instead of 165 as the upper bound. In continuous measurement, the upper bound is written as exactly 165 because of the < sign.
Dividing 160 by 10 instead of dividing the rounding unit (10) by 2.
Swapping the numbers around ( 165 ≤ length < 155 ).

Question 23 (b) • 2 Marks

Worst-Case Upper Bound Justification

A different fence panel measures 2 metres to the nearest 10 cm. Kim says that the total length of three of these fence panels must be less than 6.2 m. Show that Kim is correct.

📐 Step-by-Step Proof

Step 1: Match the units
2 metres = 200 cm (or 10 cm = 0.1 m).

Step 2: Find the Upper Bound of ONE panel
Accuracy = 10 cm = 0.1 m
Half accuracy = 10 ÷ 2 = 5 cm = 0.05 m
Upper bound = 2 m + 0.05 m = 2.05 m (or 205 cm)

Step 3: Calculate the Maximum Total Length
Total length < 3 × Upper Bound
3 × 2.05 m = 6.15 m (or 3 × 205 cm = 615 cm)

Step 4: Conclude clearly
Since the absolute maximum possible total length is strictly less than 6.15 m, and 6.15 m < 6.2 m (or 615 cm < 620 cm ), Kim is correct.

✅ Model Solution

Upper bound of one panel = 2.05 m
Maximum total length for 3 panels:
2.05 × 3 = 6.15 m

Since the maximum possible length is 6.15 m, which is less than 6.2 m, Kim is correct.

Mark Scheme Breakdown:
• M1: For identifying 2.05 (or 205 cm ) OR calculating 6.2 ÷ 3 = 2.06̇ .
• A1: Completing the argument showing 2.05 × 3 = 6.15 (and comparing to 6.2 m), or comparing 2.05 < 2.06̇ .

🧠 Exam Technique & Alternative Route

You can also solve this via division:

If the total were 6.2 m, the mean panel length would be:
6.2 ÷ 3 = 2.066... m (or 206.6̇ cm)
The maximum length of any panel is only 2.05 m .
Because 2.05 m < 2.06̇ m , the total must be less than 6.2 m!
Examiner Tip: Always show the multiplication ( 2.05 × 3 = 6.15 ) or division explicitly. Merely stating "yes" scores 0 marks without supporting calculations.

❌ Common Calculation Traps

Unit Mismatch: Adding 10 cm to 2 m to get 12 m or 2.10 m instead of converting properly to 0.05 m or working in centimetres ( 200 cm + 5 cm ).
Assuming fixed lengths: Trying specific random numbers (e.g. 2.01 + 2.02 + 2.03 = 6.06 ) does not prove it for all cases. You must use the upper bound (2.05 m) to prove the worst-case scenario.
Truncating recurring decimals: If using division, writing just 2.0 or 2.06 without indicating the recurring nature or sufficient significant figures (at least 4 s.f. unless 6.2 ÷ 3 is shown).