Compound Interest (Growth/Decay)

\( A=P\left(1+\tfrac{r}{100}\right)^{n} \)
Percentages GCSE
Question 9 of 20
← All formulas ⟵ Prev Next ⟶

£1000 grows to £1210 in 2 years with annual compounding. Find the annual rate r\%.

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
\( (A/P)^{1/2}=1+r/100. \)

Explanation

Show / hide — toggle with X

Statement

The compound interest formula models repeated growth or decay where each step is based on the previous total. The general form is:

\[ A = P\left(1 + \frac{r}{100}\right)^n \]

Here, \(A\) is the final amount, \(P\) is the initial principal (starting value), \(r\) is the percentage rate per period, and \(n\) is the number of periods.

Why it’s true

  • In simple interest, the increase each time is fixed. In compound interest, each increase is calculated from the new total.
  • After 1 period, the amount is \(P\left(1+\tfrac{r}{100}\right)\).
  • After 2 periods, the amount is \(P\left(1+\tfrac{r}{100}\right)\left(1+\tfrac{r}{100}\right)=P\left(1+\tfrac{r}{100}\right)^2\).
  • Repeating for \(n\) periods gives the formula.

Recipe (how to use it)

  1. Identify the starting amount \(P\).
  2. Decide whether it is growth (\(r>0\)) or decay (\(r<0\)).
  3. Convert the percentage rate into a multiplier: \(1+\tfrac{r}{100}\).
  4. Raise this multiplier to the power of the number of periods \(n\).
  5. Multiply by the starting amount \(P\).

Spotting it

Look for words such as “compound interest”, “grows by x% per year”, “decays by x% each cycle”, or “repeated percentage change”. These signal the use of the formula.

Common pairings

  • Finance (bank accounts, loans, investments).
  • Population growth and decay models.
  • Depreciation of assets (cars, machinery).

Mini examples

  1. \(P=1000\), \(r=5\%\), \(n=3\): \(A=1000(1.05)^3≈1157.63\).
  2. \(P=500\), \(r=-10\%\), \(n=2\): \(A=500(0.9)^2=405\).

Pitfalls

  • Using simple interest instead: Don’t just do \(P+nr\); compound means repeated multiplication.
  • Forgetting decay: If it’s decay, use \(1-\tfrac{r}{100}\).
  • Wrong percentage format: Always convert the percentage to a decimal fraction (e.g. 5% = 0.05).

Exam strategy

  • Check carefully if the problem says simple or compound.
  • Underline the rate and number of periods in the question.
  • Write out the multiplier step (e.g. \(1.05\), \(0.92\)) to avoid mistakes.
  • Round at the end, unless instructed otherwise.

Summary

The compound interest formula models repeated percentage change. Growth uses multipliers greater than 1, while decay uses multipliers less than 1. The formula \(A=P\left(1+\tfrac{r}{100}\right)^n\) ensures that each step is applied to the current total, not just the starting amount. This makes it essential for real-world growth and decay contexts.

Worked examples

Show / hide (10) — toggle with E
  1. £500 invested at 4% compound interest for 2 years. Find the amount.
    1. \( A=500(1+0.04)^2 \)
    2. \( A=500(1.04)^2 \)
    3. \( A=500×1.0816=540.8 \)
    Answer: £540.80
  2. £1000 invested at 3% for 5 years. Find A.
    1. \( A=1000(1.03)^5 \)
    2. A≈1000×1.1593
    3. \( =1159.27 \)
    Answer: £1159.27
  3. An item worth £200 depreciates by 10% each year for 3 years. Find final value.
    1. \( A=200(0.9)^3 \)
    2. \( A=200×0.729 \)
    3. \( =145.8 \)
    Answer: £145.80
  4. £400 invested at 8% for 2 years. Find the compound amount.
    1. \( A=400(1.08)^2 \)
    2. \( A=400×1.1664=466.56 \)
    Answer: £466.56
  5. Population 5000 grows by 2% per year for 4 years. Find size after 4 years.
    1. \( A=5000(1.02)^4 \)
    2. \( A≈5000×1.0824=5412 \)
    Answer: 5412
  6. Car worth £10,000 depreciates by 20% per year for 2 years. Find value.
    1. \( A=10000(0.8)^2 \)
    2. \( A=10000×0.64=6400 \)
    Answer: £6400
  7. £2500 invested at 6% for 3 years. Find final amount.
    1. \( A=2500(1.06)^3 \)
    2. \( A=2500×1.191016=2977.54 \)
    Answer: £2977.54
  8. Medicine amount 80 mg decays by 12% per hour. Find amount after 5 hours.
    1. \( A=80(0.88)^5 \)
    2. ≈80×0.5277
    3. \( =42.22 mg \)
    Answer: 42.22 mg
  9. £1200 at 5% growth for 10 years. Find total amount.
    1. \( A=1200(1.05)^10 \)
    2. \( ≈1200×1.6289=1954.7 \)
    Answer: £1954.70
  10. Bacteria colony 3000 doubles every year. Find size after 6 years.
    1. Growth rate 100% → multiplier 2
    2. \( A=3000(2^6) \)
    3. \( A=3000×64=192000 \)
    Answer: 192000