AQA GCSE Mathematics Paper 2 (Higher), June 2025: Question 16

Compound Depreciation: Value Halving Problem

📋 What this question tests

This question assesses your ability to model compound percentage decrease (exponential decay) over multiple time periods. Key skills include:

Finding the correct decimal multiplier for a percentage decrease (e.g. 100% − 12.1% = 87.9% → 0.879).
Setting up an exponential model using powers: Initial Value × (multiplier)ⁿ or just (multiplier)ⁿ .
Testing integer values of n using a calculator to identify when the remaining value falls below 50% (0.5).
Providing clear, written numerical evidence for your final integer answer.

Question 16 Walkthrough

A second-hand car decreases in value by 12.1% per year. Find the number of full years until it loses more than half its value. [3 marks]

✅ Final Answer

6 years

After 5 years, the car retains ~52.4% of its value (less than half lost). After 6 years, it retains ~46.1% of its value (more than half lost).

💡 Key Knowledge

Multiplier for decrease: Subtract from 100%, then convert to decimal:
100% − 12.1% = 87.9% = 0.879
Losing "more than half": If it loses >50%, the remaining value must be < 50% (or < 0.5 of initial price).
Exponential Formula:
Remaining fraction = 0.879ⁿ

📐 Step-by-Step Calculations

Method: Using Multipliers Directly

Step 1: Calculate the single-year multiplier
Each year, the car retains 100% − 12.1% = 87.9% of its value.
Multiplier = 0.879 .

Step 2: Trial powers on your calculator
We want to find the first integer power n where 0.879ⁿ < 0.5 :

For n = 4: 0.879⁴ ≈ 0.5966 (59.7% remaining — not half yet)
For n = 5: 0.879⁵ ≈ 0.5244 (52.4% remaining — still more than half left!)
For n = 6: 0.879⁶ ≈ 0.4610 (46.1% remaining — less than 50% remains, so over 50% is lost!)

Step 3: State the conclusion clearly
At 5 years, 52.4% remains (has not lost half).
At 6 full years, 46.1% remains, which means it has lost 100% − 46.1% = 53.9% of its value. This is more than half.
Therefore, the number of full years is 6.

Alternative starting price approach: You can choose an arbitrary starting price such as £100 or £1000.
£1000 × 0.879⁵ = £524.44 (> £500)
£1000 × 0.879⁶ = £460.99 (< £500) → Answer: 6 years

🧠 Exam Technique & Mark Scheme Breakdown

M1 (Method): Awarded for finding 0.879 or 87.9% or showing 100 − 12.1 .
M1dep (Dependent Method): Awarded for evaluating 0.879⁵ (gives 0.52 to 0.525) or 0.879⁶ (gives 0.46 to 0.462), or equivalent with an assumed price.
A1 (Accuracy): Awarded for the correct answer 6 along with supporting evidence showing the evaluated value for power 5 or power 6.
Tip: You must show working. Writing just "6" without showing the calculation 0.879⁶ ≈ 0.46 or similar risks scoring 0 marks because the prompt explicitly states "You must show your working".

❌ Common Mistakes to Avoid

Treating it as simple depreciation: Calculating 50 ÷ 12.1 ≈ 4.13 years. This is compound decay, not simple subtraction!
Wrong multiplier: Using 0.121 or 1.121 instead of 1 − 0.121 = 0.879 .
Stopping at year 5: Seeing 0.524 and rounding it to 0.5, thinking it has reached half. 52.4% is still more than half the value, so the car has not yet lost half.
Off-by-one error: Calculating year 6 correctly but writing down 5 because they think of elapsed whole intervals incorrectly.

Examiner Insight: Students who picked a starting value like £100 or £1000 often found it easier to verify when the value dropped below £50 or £500. However, whichever method you use, always write down your calculator values to at least 2 or 3 significant figures to guarantee the method marks!

AQA GCSE Mathematics Paper 2 (Higher), June 2025: Question 16

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