GCSE Maths Practice: percentages

Question 2 of 10

A car is bought new for £25,000. It loses 12% of its value each year. Work out its value after 5 years. Give your answer to the nearest pound.

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\( \begin{array}{1}\text{A car is bought new for £325,000.}\\ \text{It depreciates by 12% each year.}\\ \text{What will it be worth after 5 years?}\\ \text{Give your answer to the nearest pound.}\end{array} \)

Choose one option:

Form the multiplier (1 − 0.12) = 0.88 and raise to the power of the number of years. Keep full precision until the final rounding.

Depreciation as Compound Percentage Decrease

In GCSE Higher Tier, depreciation problems model repeated percentage loss each year. Unlike a one-off reduction, compound decrease means each year’s drop is taken from the current value, not the original. This produces an exponential decay curve rather than a straight line.

The General Model

The standard formula is

\[ N = P(1 - r)^n, \]

where \(P\) is the initial value, \(r\) the annual depreciation rate (as a decimal), and \(n\) the number of years. Because \(0 < 1 - r < 1\), powers of \(1 - r\) get smaller as \(n\) increases — capturing the compounding loss.

Why It’s Not Simple Subtraction

A common misconception is to subtract \(nr\) from 1 (simple decrease). For example, with a 9% loss over 3 years, simple decrease would suggest \(1 - 0.27\), but the correct compound factor is \((0.91)^3\). These are not equal. Compound depreciation always multiplies repeatedly by \(1 - r\).

Two Reliable Methods (Avoiding Our Question's Numbers)

  • Multiplier method: For a laptop worth £1800 that depreciates \(15\%\) per year for 2 years, the multiplier is \((1 - 0.15)^2 = 0.85^2\). Final value: \(1800\times0.85^2\).
  • Year-by-year table: A bike at £900 loses \(8\%\) yearly.
    Year 1: \(900\times0.92=828\).
    Year 2: \(828\times0.92=762.\overline{--}\) (continue without rounding until the end for accuracy).

Accuracy and Rounding

In exam settings, keep full calculator precision until the final step, then round to the required accuracy (e.g. nearest pound). Rounding early each year can drift the final result by tens of pounds across several years.

Checking: Estimation & Bounds

Use a quick bound to sense-check. With an annual loss around 10%, five years would lose a bit under half the value. So a starting amount near £20k might end near £10–£12k; a starting £8k might end near £4–£5k, etc. For other rates, compare with simple subtraction as a rough lower bound and note that compound will be a little higher than the simple bound for small \(n\).

Common Higher-Tier Variations

  • Mixed change: Decrease \(6\%\) for 2 years, then increase \(5\%\) the year after: \(P\times0.94^2\times1.05\).
  • Reverse depreciation: Final value \(N\) and \(r, n\) are known; find \(P\): \(P=N/(1-r)^n\).
  • Different time bases: Monthly depreciation \(r_m\) applied for \(12t\) months: \(P(1-r_m)^{12t}\).

Typical Pitfalls

  • Using \(1-rn\) instead of \((1-r)^n\).
  • Entering \(1-12\) instead of \(1-0.12\) (percentage–decimal confusion).
  • Rounding the multiplier prematurely (e.g. using 0.88^5 ≈ 0.53 too early).

Exam Tips

  • Write the multiplier first (e.g. \(0.91^3\)), then substitute numbers.
  • Show at least one exact power or an interim value to secure method marks.
  • State rounding clearly: “to the nearest pound”.

Further Practice (Different Numbers)

  • Phone value £950 depreciates \(18\%\) per year for 3 years: \(950\times0.82^3\).
  • Machine £12,000 loses \(7\%\) annually for 6 years: \(12000\times0.93^6\).
  • Camera £2,400 decreases \(25\%\) for 1 year, then \(10\%\) for 2 years: \(2400\times0.75\times0.9^2\).

Summary

Depreciation is compound percentage decrease. Model with \(N=P(1-r)^n\), avoid early rounding, and sense-check using estimation. Mastering this unlocks reverse problems and mixed growth/decay questions at Higher Tier.