AQA GCSE Mathematics Paper 2 (Higher), June 2025: Question 5

Error Intervals & Bounds: Fence Panel Problem

📋 Revision Summary

What this question tests

Calculating Bounds: Finding the upper and lower bounds of a measurement rounded to the nearest 10 cm.
Inequality Notation: Writing error intervals using standard notation ( Lower Bound ≤ value < Upper Bound ).
Unit Consistency: Converting freely between metres and centimetres ( 1 m = 100 cm ).
Mathematical Proof / Justification: Demonstrating maximum possible combined values to confirm or refute a statement.

Question 5 (a) — Finding an Error Interval [2 marks]

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

✅ Correct Answer

155 cm ≤ length < 165 cm

Mark Scheme:
• B2 for fully correct inequality: 155 ≤ length < 165
• Partial credit (B1): Seeing 155 or 165 in the correct position.
• Special Case (SC1): Reversing the numbers ( 165 ≤ length < 155 ).

💡 Key Knowledge: The Half-Unit Rule

Find the degree of accuracy: here, it is 10 cm.
Divide this accuracy by 2: 10 ÷ 2 = 5 cm .
Lower Bound (LB): 160 - 5 = 155 cm
Upper Bound (UB): 160 + 5 = 165 cm
Remember: The upper bound uses a strict inequality ( < ) because 165 rounds up to 170.

📐 Step-by-Step Calculation

Identify rounding unit: Nearest 10 cm .
Calculate half-step: 10 ÷ 2 = 5 cm .
Subtract for lower limit: 160 - 5 = 155 cm .
Add for upper limit: 160 + 5 = 165 cm .
Fill into blanks: 155 cm ≤ length < 165 cm.

❌ Common Errors to Avoid

Writing 164.9: You must write 165. The strictly less-than sign ( < ) already takes care of excluding 165 itself.
Dividing by 10 instead of 2: Adding/subtracting 10 cm instead of 5 cm (giving 150 to 170).
Swapping the positions: Placing the larger number on the left.

Question 5 (b) — Applying Bounds to a Real-World Problem [2 marks]

"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."

✅ Model Solution

Method 1 (Total Upper Bound):

Upper bound of 1 panel = 2.05 m (or 205 cm )
Maximum total length = 3 × 2.05 = 6.15 m (or 615 cm )
Since 6.15 m < 6.2 m (or 615 cm < 620 cm ), Kim is correct.

Mark Scheme:
• M1: Identifying the upper bound as 2.05 (or 205 ) OR calculating 6.2 ÷ 3 = 2.066...
• A1: Showing both 2.05 × 3 and 6.15 (or matching cm values 615 and 620 ) with a concluding comparison.

🧠 Exam Technique: "Show that..."

To prove a total is always less than 6.2 m, you must test the worst-case scenario (maximum possible length).
Always state which bound you are using. Here, you need the Upper Bound.
Work in consistent units! Either convert everything to metres or everything to centimetres before calculating.

📐 Step-by-Step Working

Convert units:
2 m = 200 cm and 6.2 m = 620 cm
Find Upper Bound of 1 panel:
Rounded to nearest 10 cm → half unit = 5 cm = 0.05 m
Upper Bound = 2 m + 0.05 m = 2.05 m (or 205 cm )
Calculate maximum length of 3 panels:
3 × 2.05 m = 6.15 m (or 3 × 205 cm = 615 cm )
Compare and conclude:
6.15 m < 6.2 m , therefore the total length is definitely less than 6.2 m. Kim is correct.

❌ Common Examiner Traps

Unit Mixing: Adding 10 cm directly to 2 m to get "2.10 m" without halving, or mixing metres and centimetres ( 2 + 5 = 7 ).
Using Lower Bound: Finding 3 × 1.95 = 5.85 m . Showing the minimum is less than 6.2 m does not prove the total is always less than 6.2 m!
Missing the comparison: Stopping at 6.15 m without explicitly comparing it to 6.2 m or stating Kim is correct.