AQA GCSE Mathematics Paper 1 (Higher), June 2025: Question 12

3 marks · Medium difficulty · Multi-step Problem

Calculate the total amount of money shared between Oscar and Nikita in the ratio 8 : 5 given that Oscar receives £27 more than Nikita.

Practise this question

Question

Question 12 asks: 'Oscar and Nikita share some money in the ratio 8 : 5. Oscar has £27 more than Nikita. How much do they have altogether?' followed by working space and an answer line with 'Answer £' worth 3 marks.
Question text

12 Oscar and Nikita share some money in the ratio 8 : 5

Oscar has £27 more than Nikita.

How much do they have altogether?

[3 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 12 shows three alternative methods. Method 1: M1 for 27 ÷ (8 - 5) or 9, M1dep for 9 × (8 + 5) or 72 + 45, A1 for 117. Method 2 uses algebraic equations (e.g. 8/5 = (x + 27)/x), M1dep for total, A1 for 117. Method 3 uses listing multiples to find 72 and 45 (M1), adding them (M1dep), and 117 (A1). Special credit SC1 of £43.81 to £43.94 for misinterpreting Oscar having £27 total.

Q Answer Mark Comments

Alternative method 1: works out the value of one share

27 ÷ (8 – 5) or 27 ÷ 3 or 9 oe

M1

implied by 72 or 45 identified or used

their 9 × 8 + their 9 × 5 oe

or 72 + 45

M1dep

or

their 9 × (8 + 5) or their 9 × 13

117 A1 SC1 43.81 or 43.87 or 43.88 or 43.94

Alternative method 2: sets up and solves an equation

8 x + 27 oe equation

= or x = 45

5 x any letter

or

8 y

= or y = 72 M1

5 y − 27

or

(8 − 5)w

= 27

8 + 5

their 45 + their 45 + 27 oe

or

their 72 + their 72 – 27 M1dep

or

27 × (8 + 5) ÷ (8 – 5)

117 A1 SC1 43.81 or 43.87 or 43.88 or 43.94

The mark scheme for question 12 continues on the next page

Alternative method 3: trial and improvement with values in the correct ratio

8, 16, 24, 32, 40, 48, 56, 64, 72… repeated multiples of 8 and 5 until the

difference is 27

and M1

if the lists continue, 72 or 45 must be

5, 10, 15, 20, 25, 30, 35, 40, 45…

identified or used

cont 72 + 45 M1dep oe

117 A1 SC1 43.81 or 43.87 or 43.88 or 43.94

Additional Guidance

SC1 is for thinking Oscar has £27, with rounding at different stages

How to answer it

Solving Ratio Problems Involving Differences

📌 What this question tests

This question assesses your ability to solve unstructured ratio problems where a difference between quantities is given rather than a total sum. Specifically, it tests:

  • Recognising that the difference in value corresponds directly to the difference in ratio parts.
  • Finding the unitary value (the worth of exactly 1 part/share).
  • Scaling up to find individual shares or the total amount shared.

Question 12 (3 Marks)

Oscar and Nikita share some money in the ratio 8 : 5. Oscar has £27 more than Nikita. How much do they have altogether?

✅ Correct Answer

£117

📐 Step-by-Step Calculation (Unitary Method)

Step 1: Find the difference in ratio parts

Oscar has 8 parts and Nikita has 5 parts.
Difference in parts = 8 - 5 = 3 parts

Step 2: Calculate the value of 1 part (Method Mark [M1])

The difference of 3 parts is equal to £27.
Value of 1 part = £27 ÷ 3 = £9

Step 3: Calculate the total amount (Dependent Method Mark [M1dep])

Approach A (Total parts first):
Total parts = 8 + 5 = 13 parts
Total money = 13 × £9 = £117

Approach B (Individual amounts first):
Oscar = 8 × £9 = £72
Nikita = 5 × £9 = £45
Total money = £72 + £45 = £117

Step 4: Final statement (Accuracy Mark [A1])

Altogether they have £117 .

📊 Visualising with a Bar Model:
Oscar: [ £9 ][ £9 ][ £9 ][ £9 ][ £9 ] [ £9 ][ £9 ][ £9 ] (8 blocks)
Nikita: [ £9 ][ £9 ][ £9 ][ £9 ][ £9 ] (5 blocks)
The 3 extra blocks represent the £27 difference. Thus, each block is £27 ÷ 3 = £9.

💡 Key Knowledge

  • The "Difference" Rule: When a question states someone has "more than" or "less than" someone else, divide the amount by the difference in ratio parts, NOT the sum of parts.
  • Ratio terminology: A ratio a : b means for every a parts one person receives, the other receives b identical parts.
  • Check your work: £72 - £45 = £27 (Oscar has £27 more) and £72 : £45 simplifies to 8 : 5 by dividing by 9.

🧠 Exam Technique & Mark Scheme Notes

  • Show working clearly: Writing 27 ÷ 3 = 9 secures the first method mark immediately, even if an arithmetic error happens later.
  • Implied marks: Stating £72 or £45 automatically awards M1 because it proves you found the value of one part.
  • Alternative methods accepted:
    • Algebra: Solving 8/5 = (x + 27)/x gives Nikita's share x = 45 .
    • Listing multiples: Listing 8s (8, 16... 72) and 5s (5, 10... 45) until the difference is 27.

❌ Common Errors & Examiner Pitfalls

  • Dividing by the sum of parts: Many students automatically add the parts ( 8 + 5 = 13 ) and calculate 27 ÷ 13 . This is incorrect because £27 is NOT the total amount.
  • Assuming £27 belongs to Oscar: Thinking Oscar has £27 results in 27 ÷ 8 = 3.375 per part, leading to answers like £43.88. The mark scheme only awards a consolation Special Case mark ( SC1 ) for this blunder.
  • Stopping too early: Finding £9 (the value of 1 share) or £72 / £45 and failing to sum them together to answer "altogether". Always re-read the final line of the question!
Mark Breakdown:
• M1: For 27 ÷ (8 - 5) or 27 ÷ 3 or finding 9 (also implied by 72 or 45 seen).
• M1dep: Dependent on the first M1. For finding total: their 9 × 13 or their 72 + their 45 .
• A1: Fully correct final answer of 117 .

Topics

Ratio, proportion and rates of change · 3.3 Ratio, proportion and rates of change

Question and mark scheme from the AQA GCSE Mathematics examination, Paper 1 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.