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GCSE/Mathematics/AQA· Higher tier

R16Growth and decay; compound interest; iterative processes

Notes

Growth and decay; compound interest; iteration

Many real-world quantities don't grow or shrink linearly — they change by a fixed percentage each period. This is exponential growth or decay.

Compound interest

Final amount after n periods at compound rate r% per period:

A = P × (1 + r/100)ⁿ

where P = principal (initial amount), r = % rate, n = number of periods.

Worked example: £2000 invested at 4% compound interest per year for 5 years.

  • A = 2000 × 1.04⁵ = 2000 × 1.21665… = £2433.31.

Depreciation (compound decrease)

A = P × (1 − r/100)ⁿ.

Worked example: a car bought at £18 000 depreciates 20% per year. Value after 3 years?

  • A = 18 000 × 0.80³ = 18 000 × 0.512 = £9216.

Comparing simple and compound interest

Simple interest: I = P × r × n / 100 (interest only on original). Compound interest reinvests interest earned each period — grows faster than simple over time.

Exponential growth — populations, bacteria

Same formula. A = P × multiplierⁿ.

Worked example: a bacteria culture doubles every hour. Starting at 500 cells, find population after 8 hours.

  • 500 × 2⁸ = 500 × 256 = 128 000.

Iterative processes

An iterative process repeatedly applies a formula. xₙ₊₁ = f(xₙ).

Worked example: x₁ = 3, xₙ₊₁ = (xₙ + 6/xₙ)/2 (Heron's method for √6).

  • x₁ = 3.
  • x₂ = (3 + 2)/2 = 2.5.
  • x₃ = (2.5 + 2.4)/2 = 2.45.
  • x₄ = (2.45 + 6/2.45)/2 ≈ 2.4495.
  • Converging to √6 ≈ 2.4495.

Find unknowns from compound formulas

If A, P, r are known, find n by trial or logs (often trial in GCSE).

Worked example: £500 invested at 6% per year. After how many full years does it exceed £700?

  • 500 × 1.06ⁿ ≥ 700.
  • Trial: n = 5: 500 × 1.338 = 669.11 (no). n = 6: 500 × 1.418 = 709.26 (yes). Answer: 6 years.

Common mistakes

  1. Using simple interest formula instead of compound — re-read the question.
  2. Multiplying r by n in the exponent — exponent is just n.
  3. Decay sign — multiplier is (1 − r/100) for decrease, not (r/100).
  4. Forgetting to round at end — keep exact during calculation.
  5. Iteration counter — x₃ uses x₂, not x₁.

Try thisQuick check

£3500 deposited at 2.5% compound per year for 4 years.

  • A = 3500 × 1.025⁴ = 3500 × 1.10381… = £3863.34.

AI-generated · claude-opus-4-7 · v3-deep-ratio

Practice questions

Try each before peeking at the worked solution.

  1. Question 13 marks

    Compound interest

    (F/H1) £4000 is invested at 3.5% compound interest per year for 6 years. Find the amount in the account.

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  2. Question 23 marks

    Depreciation

    (F/H2) A motorbike costs £6500 new and depreciates by 18% each year. Find its value after 3 years.

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  3. Question 34 marks

    Find n by trial

    (H3) £8000 is invested at 4% compound interest per year. After how many full years does the investment first exceed £10 000?

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  4. Question 44 marks

    Population growth

    (H4) A bacteria population grows by 15% each hour. Initial population is 200. (a) Find the population after 4 hours. (b) Find the population after 24 hours, to 3 s.f.

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  5. Question 55 marks

    Iteration

    (H5) Starting with x₁ = 2, the iteration xₙ₊₁ = (xₙ + 5/xₙ)/2 is used. Find x₂, x₃, x₄ correct to 4 d.p.

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  6. Question 64 marks

    Compare simple vs compound

    (H6) £5000 is invested for 3 years. Option A: 5% simple interest. Option B: 4.5% compound interest. Which gives more, and by how much (to nearest £1)?

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  7. Question 74 marks

    Iteration to solve equation

    (H7) The equation x³ − 4x − 2 = 0 has a root near x = 2. Use the iteration xₙ₊₁ = (4xₙ + 2)^(1/3) starting at x₁ = 2 to find x₃ correct to 3 d.p.

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Flashcards

R16 — Growth and decay; compound interest; iterative processes

12-card SR deck for AQA GCSE Maths topic R16

12 cards · spaced repetition (SM-2)