AQA GCSE Mathematics Paper 3 (Foundation), June 2025: Question 10
4 marks · Medium difficulty · Multi-step Problem
Find the modal coin value when paying £4.56 with the minimum number of coins, and determine the missing number that makes the median of a list equal to 7.
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AQA GCSE Maths: Mode, Median & Money Problem Solving
This two-part problem assesses fundamental data handling and numerical reasoning under exam conditions:
- Greedy coin selection: Decomposing a money amount (£4.56) into UK currency denominations using the minimum number of coins.
- Mode (Modal value): Identifying the most frequently occurring value in a real-world set, including required money units.
- Median from an unordered list: Ordering a data set and reverse-engineering the position and value of an added number when given the median of an even set of values.
Question 10 (a) — Minimum Coins & Modal Value
2 Marks · Mathematical Reasoning & Statistics
📐 Step-by-Step Solution
Step 1: Find the smallest number of UK coins to make £4.56
Always subtract the largest possible legal UK coin denomination at each stage:
- £4.56 − £2 = £2.56
- £2.56 − £2 = £0.56 (56p)
- 56p − 50p = 6p
- 6p − 5p = 1p
- 1p − 1p = 0p
The unique set of minimum coins is:
£2, £2, 50p, 5p, 1p (or written entirely in pounds: £2, £2, £0.50, £0.05, £0.01).
Step 2: Find the modal value
The mode is the denomination that appears most often:
• £2 appears 2 times
• 50p, 5p, and 1p each appear only 1 time.
Modal value = £2
✅ Correct Answer & Mark Scheme
Answer: £2 (or 200p )
Working shown: £2, £2, 50p, 5p, 1p
• B2: Correct modal answer of £2 AND the full correct list of coins ( £2, £2, 50p, 5p, 1p ) with correct units throughout.
• B1: Awarded for just £2 alone without working, OR for an otherwise complete list of coins totaling £4.56, OR for correctly finding the mode of an incorrect coin list.
• Special Case (SC1): Correct values given but missing units entirely (e.g., writing 2 with list 2, 2, 50, 5, 1 ).
❌ Common Mistakes to Avoid
- Omitting units: Writing just 2 instead of £2 loses a mark immediately. Units must be visible.
- Invalid decimal notation: Writing 0.50p instead of 50p or £0.50 . Note: 0.50p means half of one penny!
- Using non-existent coins: Using a fictitious coin (e.g., 6p or £4) drops you to a maximum of 1 mark.
- Not using the smallest count: Using four £1 coins instead of two £2 coins will fail the condition "smallest possible number of coins".
🧠 Exam Technique
- Notice "You must show your working": You cannot get both marks without writing out the individual coins you chose.
- Keep units consistent: Either use standard pound signs throughout (£2, £2, £0.50, £0.05, £0.01) or clear mixed notation (£2, £2, 50p, 5p, 1p).
Question 10 (b) — Working Backwards from the Median
2 Marks · Data Analysis & Reverse Problem Solving
📐 Step-by-Step Solution
1 Order the original list:
Unordered: 5, 9, 4, 16, 8
Ascending order: 4, 5, 8, 9, 16
2 Understand the new list length:
With an extra number added, there are now 6 numbers (an even amount). The median of 6 numbers is the mean of the 3rd and 4th numbers.
3 Set up the median relationship:
Let the middle two numbers be a and b:
(a + b) ÷ 2 = 7 ⇒ a + b = 14
4 Locate where the extra number fits:
Looking at our sorted values ( 4, 5, 8, 9, 16 ), one of the middle numbers must be 8.
If b = 8, then a = 14 − 8 = 6.
Check the full ordered list with 6 inserted:
4, 5, 6, 8, 9, 16
Middle numbers: 6 and 8. Median = (6 + 8) ÷ 2 = 7. It matches!
✅ Correct Answer & Mark Scheme
Answer: 6
• M1 (Method Mark): Awarded for writing out the list in order ( 4, 5, 8, 9, 16 ), OR writing an ordered list with any extra number inserted, OR explicitly identifying that the two middle values must be 6 and 8.
• A1 (Accuracy Mark): Correct final answer of 6.
💡 Key Knowledge: Median Rules
- Always sort first: You cannot find or manipulate a median without ordering the list in ascending or descending order.
- Odd total of items (n): Middle value is at position (n + 1) ÷ 2.
- Even total of items (n): The median is the halfway point (mean) between the two central numbers at positions (n ÷ 2) and (n ÷ 2 + 1). For 6 numbers, this is the 3rd and 4th values.
❌ Common Mistakes to Avoid
- Assuming the extra number is 7: Many students see "median is 7" and assume the missing number must be 7. If you add 7, the list is 4, 5, 7, 8, 9, 16 , making the median (7 + 8) ÷ 2 = 7.5, not 7!
- Confusing Mean and Median: Trying to add all values and set the total to 6 × 7 = 42. That is the method for the mean, not the median.
- Giving the full list as the answer: The question asks to work out the extra number. If you write the list 4, 5, 6, 8, 9, 16 but leave the answer line blank or write 7, you lose the A1 mark.
Topics
Statistics · Number · 3.6 Statistics · 3.1.1 Structure and calculation · 3.1.3 Measures and accuracy
Question and mark scheme from the AQA GCSE Mathematics examination, Paper 3 (Foundation), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.