Growth & Decay

SPEC R9, R16 • RATIO, PROPORTION & RATES OF CHANGE

Compound interest, depreciation, repeated percentage change — and iterative processes (Higher).

Why This Topic Matters

Compound interest, depreciation and repeated percentage change appear on every paper, typically as 3–4 mark calculator questions. The same formula covers savings accounts, car values, populations, bacteria counts and half-life problems.

The Big Idea

When a quantity changes by the same percentage each period, the growth (or decay) is exponential, not linear. You apply the multiplier once per period — which means raising it to the power n for n periods.

One Formula Covers It All

New amount = Original × (multiplier)ⁿ

Where multiplier = 1 + r/100 for growth, and 1 − r/100 for decay. Memorise this single idea — every exam question uses it.

Simple vs Compound Interest

Two very different types of interest — get them mixed up and every mark is gone. The key distinction: does the base grow, or stay fixed?

Simple Interest — Linear Growth

Interest is calculated each year on the original principal, never on the interest earned. A fixed amount added each year.Interest = P × r × t ÷ 100where P = principal, r = rate %, t = time (years). Total amount after t years = P + Interest.

Compound Interest — Exponential Growth

Interest is calculated each year on the current balance (principal + any interest already earned). The interest itself earns interest.Amount = P × (1 + r/100)ⁿThis formula is given in the exam on the formulae sheet.

The Gap Widens

For short time periods and small rates, simple and compound are close. Over long periods (say 10+ years) compound pulls far ahead. The default for UK savings accounts is compound — read the question carefully to check which is asked.

Seeing the Difference — £1 000 at 10% for 5 Years

Exactly the same starting amount and rate, but simple and compound diverge year by year.

End of yearSimple (linear)Compound (multiplier 1.1)Gap

0£1 000£1 000.00—

1£1 100£1 100.00£0

2£1 200£1 210.00£10

3£1 300£1 331.00£31

4£1 400£1 464.10£64.10

5£1 500£1 610.51£110.51

Reading the Table

Simple adds a flat £100 every year (10% of the original £1 000).

Compound multiplies by 1.1 every year — so the interest itself earns interest.

After 5 years, compound is £110.51 ahead. Over decades, the gap becomes enormous.

Growth vs Decay — The Graphs

Compound growth and decay produce exponential curves: growing ones curve upwards faster over time; decaying ones drop towards (but never reach) zero.

Key Shape Features

Both curves start at the same initial amount P.

Growth (multiplier > 1) rises more and more steeply.

Decay (multiplier < 1) drops quickly at first, then flattens out — approaching zero but never reaching it.

The Multiplier — Your Best Friend

Every percentage change can be expressed as a single multiplier. Multipliers make compound problems easy — just raise to the power of the number of periods.

ChangeHow the Multiplier is FoundMultiplier

Increase of 4%1 + 0.041.04

Increase of 25%1 + 0.251.25

Decrease of 15%1 − 0.150.85

Decrease of 8%1 − 0.080.92

Halving (50% decrease)1 − 0.50.5

Why Multipliers Are So Powerful

For repeated percentage change, you don't need to work out the interest each year. Just raise the multiplier to the power n:

Final = Initial × (multiplier)ⁿ

Don't Forget the "1 +"

A 4% increase is ×1.04, not ×0.04. Multiplying by 0.04 would reduce the amount to 4% of its value — a catastrophic decrease. The "1 +" keeps the original amount; the 0.04 is the addition on top.

The MCE Procedure for Compound Problems

Every compound interest or depreciation question uses the same three-step pattern. Using MCE keeps the method visible for method marks.

Method

Identify principal P, rate r, and periods n. Decide: growth (×) or decay (÷ from original)?

Calculate

Write the multiplier. Raise to power n on the calculator. Multiply by P. Do not round intermediate values.

Evaluate

Round the final answer to the required accuracy (often nearest penny for money). Include units.

Rounding Rule

Compound calculations carry many decimal places through. Only round on the very last line. Rounding in the middle (e.g. truncating 1.26531 to 1.27) introduces errors that grow with each step — sometimes tens of pounds wrong.

Worked Example — Compound Interest01

Sarah invests £2500 in an account paying 4% compound interest per year. How much is the investment worth after 6 years? Give your answer to the nearest penny. [3 marks]

M

Use the compound interest formula: Amount = P × (1 + r/100)ⁿ. Here P = 2500, r = 4, n = 6.

C

Multiplier = 1 + 0.04 = 1.04.

Amount = 2500 × 1.04⁶ = 2500 × 1.2653190185… = 3163.297546…

E

Round to the nearest penny: £3 163.30.

Check

Sanity check: £2500 at 4% simple for 6 years would earn only 2500 × 0.04 × 6 = £600, giving £3 100. Compound gives a bit more (£3 163.30) — that's the compounding effect.

Worked Example — Depreciation02

A new car is worth £18 000. Each year it depreciates by 15% of its value at the start of that year. Work out the value of the car after 4 years. [3 marks]

M

Depreciation is exponential decay. Value = Initial × (1 − r/100)ⁿ. Here Initial = 18 000, r = 15, n = 4.

C

Multiplier = 1 − 0.15 = 0.85.

Value = 18 000 × 0.85⁴ = 18 000 × 0.52200625 = 9 396.1125.

E

Value after 4 years = £9 396.11.

Note

Each year, the car loses 15% of its current value — not 15% of the original. That's why a flat £18 000 × 0.15 × 4 = £10 800 loss would be wrong.

Iterative Processes Higher

An iterative process is a rule that produces the next term from the previous one. The rule is applied repeatedly to move step by step. Compound interest is one special type of iteration, but iteration is more general — it can include additions, subtractions, or any combination.

Iterative Formula — Notation

Written in the form:Aₙ₊₁ = (rule involving Aₙ)with a given starting value A₀. To find A₂, first find A₁ from A₀, then A₂ from A₁.

Worked Example — Loan Repayment

Rick borrows £1 500. Each month, 2% interest is added and he pays off £100. The amount he still owes at the end of month n+1 is given by:

Aₙ₊₁ = 1.02 × Aₙ − 100, A₀ = 1 500

Find what he owes at the end of month 2.

A₁ = 1.02 × 1 500 − 100 = 1 530 − 100 = 1 430.

A₂ = 1.02 × 1 430 − 100 = 1 458.60 − 100 = £1 358.60.

Don't Shortcut It

You cannot use the compound interest formula here because £100 is being subtracted each month. Every iteration must be done step by step. Write out A₁, A₂, … separately.

Beyond Money — Other Applications

The same formula covers any quantity that changes by a fixed percentage per period. Recognise the pattern and apply the multiplier the same way.

ContextFormulaExample

Population growthNₙ = N₀ × (1 + r/100)ⁿTown of 5 000 grows at 3%/year

Bacteria doublingNₙ = N₀ × 2ⁿEach hour the count doubles

Radioactive decay (half-life)Nₙ = N₀ × 0.5^t/hAmount halves every half-life h

Cooling (news spread, virus models)Same formula, with a decay multiplierProportion remaining falls by r% each period

Half-Life Formula in Detail

If the half-life is h, and time elapsed is t, then the number of halvings is t/h. So the remaining amount is:

N = N₀ × (0.5)^t/h

Example: 800 g with half-life 5 years, after 15 years: N = 800 × 0.5³ = 800 × 0.125 = 100 g.

Common Errors to Avoid

These five errors account for most lost marks on growth-and-decay questions. Master them so you don't repeat them.

Students often writeCorrect versionWhy it matters

P × r × n/100 for compound interestP × (1 + r/100)ⁿThat's the simple interest formula — wrong tool

4% increase → multiplier 0.041.04 (you must add the 1)×0.04 would shrink the amount to 4% of its value

15% depreciation/year × 4 years = 60% offUse 0.85⁴ = 0.522… so 47.8% remainsPercentages don't add across years — they compound

Round 1.04⁶ to 1.27 then multiplyKeep full precision, round only at the endEarly rounding causes errors of £ or more

Answer given with no unitsInclude £, or pence, or appropriate unitsMissing unit can cost a mark

Examiner Alerts

Examiner Alert — Simple vs Compound

The question wording is the key. Look for "compound interest", "per year" applied to a changing balance, or phrases like "this is repeated for n years" — these all mean compound. Look for "simple interest" for the linear version. If the question doesn't state explicitly, the default for savings/loans is compound.

Examiner Alert — Rounding too Early

Students who round 1.04⁶ to 2 dp (1.27) and multiply by £2500 get £3175 — wrong by £12. Keep the full calculator output until the final line, then round once.

Reverse Problems — Finding the Rate

If asked to find the rate needed for a certain final amount, solve by rearranging:

(1 + r/100)ⁿ = Final ÷ Initial ⟹ 1 + r/100 = (Final ÷ Initial)^1/n

Take the n-th root on the calculator, then subtract 1 and multiply by 100.

Practice Question 1

£5 000 is invested in an account paying 3% compound interest per year. Work out the amount in the account after 2 years. [2 marks]

M

Amount = P × (1 + r/100)ⁿ, P = 5000, r = 3, n = 2.

C

Amount = 5000 × 1.03² = 5000 × 1.0609 = 5304.5.

E

£5 304.50.

Practice Question 2 — Depreciation

A new van costs £15 000. Each year the van loses 20% of its value at the start of that year. Find the value of the van after 3 years. [3 marks]

M

Value = Initial × (1 − r/100)ⁿ. Multiplier = 0.80, n = 3.

C

Value = 15 000 × 0.80³ = 15 000 × 0.512 = 7 680.

E

£7 680.

Practice Question 3 — Finding the Rate

Magana puts £500 into an account that pays compound interest. She wants at least £560 in the account after 3 years. Work out, to 1 decimal place, the minimum annual interest rate she needs. [3 marks]

M

500 × (1 + r/100)³ ≥ 560. Rearrange: (1 + r/100)³ ≥ 1.12, so 1 + r/100 ≥ 1.12^1/3.

C

1.12^1/3 = 1.03849… → r ≥ 3.849…

Check 3.8%: 500 × 1.038³ = £559.19 < £560 — too low.

Check 3.9%: 500 × 1.039³ = £560.81 ≥ £560 — OK.

E

Minimum rate (to 1 dp): 3.9%.

Practice Question 4 Higher

Rick borrows £1 500 from a bank. He pays back £100 each month. The iterative formula below works out the amount he still owes at the end of each month, including 2% interest added before the payment:

A₀ = 1 500, Aₙ₊₁ = 1.02 × Aₙ − 100

Work out the amount Rick still owes at the end of the 2nd month. [3 marks]

M

Iterate twice — A₀ → A₁ → A₂. Cannot use the compound formula because £100 is subtracted each month.

C

A₁ = 1.02 × 1500 − 100 = 1 530 − 100 = 1 430.

A₂ = 1.02 × 1430 − 100 = 1 458.60 − 100 = 1 358.60.

E

£1 358.60.

Practice Question 5 — Finding n

£2 000 is invested at 3% compound interest per year. After how many complete years will the amount first exceed £2 500? [3 marks]

M

Need 2000 × 1.03ⁿ > 2500. Try successive values of n until the amount first exceeds £2 500.

C

n = 7: 2000 × 1.03⁷ = £2 459.75 (still less than £2 500).

n = 8: 2000 × 1.03⁸ = £2 533.54 (first exceeds £2 500).

E

8 complete years.

Practice Question 6 — Half-Life

A sample of a radioactive substance has an initial mass of 800 g and a half-life of 5 years. Work out the mass of the sample remaining after 15 years. [3 marks]

M

Half-life formula: N = N₀ × 0.5^t/h. Here N₀ = 800, t = 15, h = 5. Number of halvings: 15 ÷ 5 = 3.

C

N = 800 × 0.5³ = 800 × 0.125 = 100 g.

Alternate view

After 5 years: 400 g. After 10 years: 200 g. After 15 years: 100 g. Each half-life, the mass halves.

Practice Question 7 — Set Up From Context

A scientist starts with a colony of 8 000 bacteria. The colony grows by 25% every hour.

(a) Write down the multiplier that represents the growth each hour. [1 mark]

(b) Find the number of bacteria after 2 hours. [2 marks]

(c) Explain why it is not correct to say the population has grown by 50% after 2 hours. [1 mark]

(a)

25% increase → multiplier = 1 + 0.25 = 1.25.

(b)

After 2 hours: 8 000 × 1.25² = 8 000 × 1.5625 = 12 500 bacteria.

(c)

Percentages compound — they don't add across hours. The actual overall growth is (12 500 − 8 000) ÷ 8 000 = 56.25%, not 50%. The second hour's 25% is applied to the larger population after hour 1.

Summary

The Core Formulas

Compound growth: Amount = P × (1 + r/100)ⁿ. (Given on the formulae sheet.)

Compound decay (depreciation): Value = Initial × (1 − r/100)ⁿ.

Simple interest: Interest = P × r × t ÷ 100 (linear, not the same as compound).

Half-life: N = N₀ × 0.5^t/h where h is the half-life.

The Multiplier System

Increase of r%: multiplier = 1 + r/100 (e.g. 4% → 1.04).

Decrease of r%: multiplier = 1 − r/100 (e.g. 15% → 0.85).

Repeated: raise the multiplier to the power n.

The Three Traps

Simple vs compound: read the wording. Compound needs the formula with power n.

Multiplier format: 4% increase is 1.04, not 0.04.

Rounding early: keep full precision until the final line.

Iterative Processes Higher

Form: Aₙ₊₁ = (rule using Aₙ) with a starting A₀.

Work through step by step — find A₁, then A₂, etc.

Cannot use the compound formula shortcut when amounts are added or subtracted each period.