GCSE · Maths · AQA · Spec 8300 · Foundation

Growth and decay problems including compound interest

£100 earns 10% interest a year for 10 years. Ten lots of 10% makes £200… right? You’d actually have £259.37 — so where did the extra £59.37 come from?

Grab the end of the line and lift it

Change each period, r 10 %

Change each period, r: 10 %. Value after 10 periods: 259.37

Every corner on the line is the value after one more period, starting from 100. Lift the end point to change r, the percentage change each period. Positive r is an increase; negative r is a decrease, so r = −10 means “down 10% each time”. Try r = 10 first. Ten lots of 10% sounds like it should reach 200 — but the end point reads 259.37, and the steps get taller as you go. Each 10% is taken of a bigger number than the one before.

Watch out: Now pull the end point down until r = −20 and read the value: 10.74. The line keeps falling, but more and more gently, and it never touches zero — each period takes 20% of what is left, so something is always left.

Where the multiplier comes from

r% of A = (r/100) × A

“Per cent” means “out of 100”, so r% of an amount is r/100 of it. For example, 12% of 250 is 12/100 × 250 = 30.

1 / 6

Which of these do you believe?

Be honest — which is closest to what you think about repeated percentage changes?

4% interest for 3 years is 12% altogether, so £500 becomes 500 × 1.12 = £560.If a price goes up 10% and then down 10%, it ends up back where it started.If something loses 10% every year, after 10 years there’s nothing left.Each change is taken of the latest value, so I multiply by the same multiplier once for every period.

How sure are you?

I'm sureI think soI'm guessing

4% interest for 3 years is 12% altogether, so £500 becomes 500 × 1.12 = £560.

This is a common misconception

That treats every 4% as 4% of the original £500. But year 2’s interest is 4% of £520, not £500, and year 3’s is 4% of an even bigger amount. The correct total is 500 × 1.04³ = £562.43. The extra £2.43 is interest earned on earlier interest — the same reason the steps on the graph got taller.

If a price goes up 10% and then down 10%, it ends up back where it started.

This is a common misconception

It feels balanced, but the 10% drop is taken of the NEW, bigger price. £40 goes up to £44, then 10% of £44 is £4.40, leaving £39.60. In multipliers: 1.1 × 0.9 = 0.99, so the price ends 1% lower than it started.

If something loses 10% every year, after 10 years there’s nothing left.

This is a common misconception

10 × 10% = 100% is the adding mistake again. Each year keeps 0.9 of what is left, so after 10 years you have 0.9¹⁰ ≈ 0.349 of it — about 35%. Set the graph to r = −10 and the end point reads 34.87 out of 100. Decay never quite reaches zero.

Each change is taken of the latest value, so I multiply by the same multiplier once for every period.

This matches the evidence

Exactly right. £500 at 4% for 3 years is 500 × 1.04 × 1.04 × 1.04 = 500 × 1.04³ = £562.43 to the nearest penny. One multiplier, used once per period — that’s why a power appears.

Never add the percentages up. Find the multiplier, then use it once for every period.

Your turn: fill in the missing steps

Sam puts £2400 into a savings account that pays 3.5% compound interest per year. (a) How much is in the account after 6 years? (b) How much interest has Sam earned?

P = 2400, r = 3.5 (% per year), n = 6 yearsStart by pulling the three numbers out of the words.
missing step

Which line is step 2?

Multiplier = 1 + 3.5/100 = 1.035Multiplier = 3.5/100 = 0.035Multiplier = 1 + 3.5 = 4.5

Spot where this answer goes wrong

A car is bought for £18 500. Its value depreciates (goes down) by 15% each year. Work out its value after 4 years, to the nearest pound.

A student’s answer — which line goes wrong?

Depreciation means a decrease of 15% each year, for 4 years.
Multiplier = 1 + 15/100 = 1.15
Value = 18 500 × 1.15⁴
= 18 500 × 1.74900625 = 32 356.62
Value after 4 years ≈ £32 357

Higher

Iteration: feed the answer back in

Problem

A sequence follows the rule xₙ₊₁ = 0.8xₙ + 30, starting from x₀ = 200. Find x₁, x₂ and x₃.

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

Every repeated percentage change is one multiplier, used once for every period.

What you need to know

An increase of r% multiplies a quantity by (1 + r/100); a decrease of r% multiplies it by (1 − r/100).
After n equal percentage changes: value = original × (1 + r/100)ⁿ for growth, or original × (1 − r/100)ⁿ for decay.
Compound interest: Total accrued = P(1 + r/100)ⁿ, where P is the principal, r is the interest rate per period and n is the number of times the interest is compounded.
Interpreting the answer: the interest earned is the total accrued minus the principal, and money is rounded to the nearest penny.
An iterative process applies the same rule to the result of the previous step: x₁ = f(x₀), x₂ = f(x₁), and in general xₙ₊₁ = f(xₙ).

The big picture

When a quantity changes by the same percentage again and again, each change is taken of the latest value, not the original. So an increase of r% is multiplying by (1 + r/100), a decrease of r% is multiplying by (1 − r/100), and after n changes the original has been multiplied by that multiplier n times: value = original × (1 ± r/100)ⁿ. Compound interest is the same idea with money: Total accrued = P(1 + r/100)ⁿ. Growth curves get steeper; decay flattens out but never reaches zero.

Key points

1Each percentage change is taken of the LATEST value, so equal percentage changes give unequal steps.

2A multiplier above 1 means growth; a multiplier below 1 means decay.

3n equal changes means the multiplier is used n times — so it goes in the power, never “× n”.

4Percentages don’t add up over time: 4% for 3 years is not 12%, and +10% then −10% leaves you 1% down.

5Decay keeps shrinking but never reaches zero, because each step removes a fraction of what is left.

6Always check what the question asks for: the total, or only the part that grew (the interest).

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Worked example

Problem

A town has a population of 12 000. The population increases by 2.5% each year. (a) Work out the population after 8 years. (b) After how many whole years does the population first go above 14 000?

⚠ Watch out

Adding the percentages instead of multiplying: treating 5% a year for 4 years as 20%, or thinking +10% then −10% gets you back to the start. Each change is taken of the latest value, so use the multiplier once per period.

🧠

Memory hook

Find the multiplier, then power it up: “ONE number, n TIMES.” Above 1 it climbs, below 1 it sinks — and it never adds.

✓

Check yourself

A £3000 antique gains 6% in value each year. What single calculation gives its value after 5 years? (Answer: 3000 × 1.06⁵ = £4014.68, to the nearest penny.)

Flashcards

(12)

The multipliers for an r% increase and an r% decrease

Increase: 1 + r/100. Decrease: 1 − r/100. For 4%, that’s 1.04 and 0.96.

Why does increasing by r% mean multiplying by (1 + r/100)?

You keep all of the amount (the 1) and add r/100 of it on top.

Value after n equal percentage changes?

original × (multiplier)ⁿ — the multiplier is used once for every period.

Compound interest formula, and what P, r and n stand for

Total accrued = P(1 + r/100)ⁿ. P = principal, r = interest rate per period, n = number of times the interest is compounded.

How can you tell growth from decay by looking at the multiplier?

Above 1 → growth. Below 1 → decay.

Why isn’t 5% a year for 3 years the same as 15%?

Each 5% is taken of the latest (bigger) value, so the total change is 1.05³ = 1.157625, a 15.7625% increase.

Does a 10% increase followed by a 10% decrease get you back to the start?

No. 1.1 × 0.9 = 0.99, so you end up 1% lower.

Does something that decays by a fixed percentage ever reach zero?

No. Each step removes a fraction of what is left, so there is always something left.

Total accrued vs interest earned

Total accrued is everything in the account. Interest earned = total accrued − principal.

What happens to the size of each step when a quantity grows by the same percentage each period?

The steps get bigger each time, because each percentage is taken of a bigger amount — the graph curves upwards.

What is an iterative process?

Applying the same rule again and again, each time to the previous result: xₙ₊₁ = f(xₙ), starting from a given x₀.

Write growth of r% per period as an iteration

xₙ₊₁ = (1 + r/100)xₙ — each new value is the old value times the multiplier.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

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Grab the end of the line and lift it

Which of these do you believe?

Your turn: fill in the missing steps

Spot where this answer goes wrong

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