Teaching unit · UK Year 10 (Key Stage 4 / GCSE Foundation, ages 14 to 15)

Growth and decay: compound interest

The multiplier method, compound interest and depreciation, and why compound interest overtakes simple interest

About four lessons of 45 to 60 minutes

Start here · hook

Would you rather have simple interest, or compound interest?

Two banks both offer 5% interest on a £500 deposit. Bank A pays simple interest: exactly £25 every single year, forever. Bank B pays compound interest: 5% of whatever is currently in the account, so the amount you earn each year keeps growing. After 1 year they pay exactly the same, £25. Most people would say it barely matters which bank you choose.

After 10 years, Bank A has paid £250 in interest, for a total of £750. Bank B has paid £314.45 in interest, for a total of £814.45, over £64 more, without you adding a single extra penny. The gap only gets bigger the longer the money sits there, because compound interest earns interest on its own interest. This unit is about calculating exactly that: using a single decimal multiplier to grow (or shrink) an amount by a fixed percentage, again and again.

£500 saved at 5% for 10 yearssimple interest earns £250; compound interest earns £314.45, because year 2's interest is calculated on year 1's total, not the original £500
A new car losing 20% of its value every yearcompound DECAY: an £8,000 car is worth £4,096, not £3,200, after 3 years
A town's population growing by 4% a yearthe same multiplier method as money, just with people instead of pounds
A phone's resale value dropping year after yeareach year's drop is a percentage of a SMALLER starting value than the year before

Learning objective

What students will be able to do

Students will convert a percentage increase or decrease into a single decimal multiplier, apply that multiplier repeatedly to model compound growth and compound decay (depreciation) over several years, use the formula Total accrued = P $(1 + r/100)^{n}$ , and compare compound interest with simple interest on the same investment to explain why they diverge.

Success criteria

I can write the decimal multiplier for a given percentage increase (e.g. +8% is x 1.08) or decrease (e.g. -8% is x 0.92).
I can apply a multiplier repeatedly, once per year, to find a value after several years of compound growth or decay.
I can use the formula Total accrued = P $(1 + r/100)^{n}$ to find the total value of an investment.
I can find the interest earned, or the value lost, by subtracting the original amount from the final amount.
I can calculate both simple interest and compound interest on the same investment and explain why compound interest grows faster.

Curriculum anchor

Standards this unit teaches

GCSE Ratio, Proportion & Rates of Change #16UK GCSE Mathematics (DfE, England)
Growth and decay problems, including compound interest
Subject content statement (Department for Education, "GCSE mathematics: subject content and assessment objectives", published 1 November 2013, reference DFE-00233-2013, "Ratio, proportion and rates of change" section, item 16, https://www.gov.uk/government/publications/gcse-mathematics-subject-content-and-assessment-objectives): students should "set up, solve and interpret the answers in growth and decay problems, including compound interest and work with general iterative processes". This is standard-type content, so ALL GCSE students (Foundation and Higher tier) are taught and assessed on it, not only the highest-attaining. The compound interest formula, "Total accrued = P $(1 + r/100)^{n}$ ", is given in the same document's Appendix: Mathematical formulae, section 2 (formulae candidates should derive or informally understand; not given in the examination).
GCSE Ratio, Proportion & Rates of Change #9UK GCSE Mathematics (DfE, England)
Percentage change and simple interest
Subject content statement (same document, item 9): students should "define percentage as 'number of parts per hundred', interpret percentages and percentage changes as a fraction or a decimal, interpret these multiplicatively, express 1 quantity as a percentage of another, compare 2 quantities using percentages, and work with percentages greater than 100%" and "solve problems involving percentage change, including percentage increase/decrease and original value problems, and simple interest in financial mathematics". This unit uses simple interest only as the point of comparison against compound interest (item 16); single-step simple interest itself is already covered by this site's Grade 8 "Financial Mathematics" worksheets.

Before you start

Prior knowledge

This unit builds on skills students should already have met. Revisit any that are shaky first.

Key vocabulary

Words to teach and display

Multiplier: the single decimal number you multiply by to increase or decrease an amount by a given percentage in one step
Compound interest: interest calculated on the current total (original amount plus any interest already earned), so it grows faster each year
Simple interest: interest calculated only on the original amount, so it adds the same fixed amount every year
Principal: the original amount of money invested or borrowed, before any interest is added
Depreciation: a decrease in the value of an item (such as a car or a phone) over time, often modelled as compound decay
Iterative process: repeating the same step, using the previous result as the input to the next step, such as multiplying by the same multiplier every year

Teaching sequence

Teach it: concrete, pictorial, abstract

The lesson moves from things students can hold, to pictures and diagrams, to the written maths. The diagrams below are drawn from data, so they are accurate and print cleanly. Teach straight from them.

1. Turning a percentage change into a multiplier

Concrete

Every percentage increase or decrease can be written as ONE decimal multiplier. An increase of 8% means the new amount is 100% + 8% = 108% of the original, so the multiplier is 108/100 = 1.08. A decrease of 8% means 100% - 8% = 92% of the original, so the multiplier is 0.92.

This matters because it turns a two-step calculation (find 8% of the amount, then add or subtract it) into a single multiplication. £500 increased by 8% is £500 x 1.08 = £540, not £500 + (£500 x 0.08) worked out separately, even though both give the same answer.

Worked example

Write the multiplier for a decrease of 15%, then use it to find 15% off a £60 jacket.

A decrease of 15% leaves 100% - 15% = 85% of the original.
The multiplier is 85/100 = 0.85.
£60 x 0.85 = £51.00.

Answer: Multiplier = 0.85. The jacket costs £51.00 after the decrease.

Check for understanding, ask

What is the multiplier for an increase of 20%?
What is the multiplier for a decrease of 3%?
Why is 100% always the starting point?

2. Compound growth: applying the multiplier again and again

Pictorial

Compound growth means applying the SAME multiplier once per period, to the NEW total each time, not to the original amount every time. £500 invested at 5% compound interest does not earn £25 every year forever; it earns 5% of a slightly bigger number each year.

£500 at 5% compound interest: Year 1: £500 x 1.05 = £525.00. Year 2: £525.00 x 1.05 = £551.25 (not £500 x 1.05 again). Year 3: £551.25 x 1.05 = £578.81 (rounded to the nearest penny). Each year's answer becomes the START of the next year's calculation, which is exactly what 'iterative process' means.

£500 growing at 5% compound interest: the curve bends upward, because each year's growth is calculated on a bigger amount than the year before.

Worked example

£500 is invested at 5% compound interest. Find its value after 3 years, and the total interest earned.

Multiplier for +5% is 1.05.
Year 1: £500.00 x 1.05 = £525.00.
Year 2: £525.00 x 1.05 = £551.25.
Year 3: £551.25 x 1.05 = £578.8125, which rounds to £578.81.
Interest earned = £578.81 - £500.00 = £78.81.

Answer: Value after 3 years = £578.81. Interest earned = £78.81.

Check for understanding, ask

Why is Year 2's interest calculated on £525.00, not on the original £500.00?
Using the formula Total = P $(1 + r/100)^{n}$ , what are P, r and n in the worked example?

3. Compound decay: the same idea, going down

Pictorial

Depreciation works exactly like compound interest, but the multiplier is below 1 because the amount is shrinking. A car losing 20% of its value every year is multiplied by 0.8 each year, not by 0.2 (that would only find the amount LOST, not what remains).

An £8,000 car depreciating by 20% a year: Year 1: £8,000 x 0.8 = £6,400.00. Year 2: £6,400.00 x 0.8 = £5,120.00. Year 3: £5,120.00 x 0.8 = £4,096.00. Notice the value drops by a SMALLER amount in pounds every year (£1,600, then £1,280, then £1,024), even though the percentage rate (20%) never changes, because 20% of a smaller number is a smaller number.

The car's value after each year of 20% compound depreciation. Each bar is 80% of the one before it, so the bars shrink by a smaller AMOUNT each year even though the RATE stays at 20%.

Worked example

A laptop bought for £600 depreciates by 10% per year. Find its value after 2 years, and the value lost.

Multiplier for -10% is 0.90.
Year 1: £600.00 x 0.90 = £540.00.
Year 2: £540.00 x 0.90 = £486.00.
Value lost = £600.00 - £486.00 = £114.00.

Answer: Value after 2 years = £486.00. Value lost = £114.00.

Check for understanding, ask

Why is the multiplier 0.8 rather than 0.2 for a 20% decrease?
Does a fixed depreciation rate lose the same number of pounds every year? Explain.

4. Simple vs compound interest: why they diverge

Abstract

Back to the hook: simple interest adds a fixed amount every year (the same percentage of the ORIGINAL amount, every time); compound interest adds a growing amount (the same percentage of the CURRENT total, which keeps increasing). In year 1 they are identical. After that, compound interest is always at least as large, and the gap grows every year.

Simple interest formula: interest = P x r x n / 100 (a straight line if you plotted it against n). Compound interest formula: Total = P $(1 + r/100)^{n}$ (a curve that bends upward). For £500 at 5% over 3 years: simple interest total is £500 + (£500 x 5 x 3 / 100) = £500 + £75 = £575.00. Compound interest total (from section 2) is £578.81, which is £3.81 more, even over just 3 years.

Check for understanding, ask

For the same principal and rate, is compound interest ever LESS than simple interest after year 1? Why or why not?
£500 at 5% simple interest earns £75 over 3 years. Compound interest earns £78.81. Which formula grows faster as n gets larger, and why?

Watch for

Common misconceptions and how to address them

MisconceptionA 20% decrease means multiplying by 0.2.

Why it happens: Students confuse the percentage LOST with the percentage REMAINING.

How to address it: Always start from 100%. A 20% decrease leaves 100% - 20% = 80% remaining, so the multiplier is 0.8, not 0.2 (which would leave only a fifth of the original by ONE multiplication, an 80% decrease).

MisconceptionCompound interest just means multiplying the original amount by the multiplier, n times, added together.

Why it happens: Students try to reuse the simple-interest 'same amount every year' pattern instead of compounding.

How to address it: Compound interest multiplies the CURRENT total (already including previous interest) by the multiplier each year, one multiplier applied n times in a row: P x multiplier x multiplier x ... (n times), which is P x multiplierⁿ, not P x multiplier x n.

MisconceptionA 10% decrease followed by a 10% increase brings you back to the original amount.

Why it happens: Students assume opposite percentages cancel out, treating percentage change like a simple addition/subtraction rather than a multiplication.

How to address it: £100 decreased by 10% is £100 x 0.9 = £90. £90 increased by 10% is £90 x 1.1 = £99, not £100, because the second 10% is calculated on the smaller £90, not the original £100.

Do it together

Guided practice (with answers)

1. Write the multiplier for an increase of 12%.
Answer: 1.12, because 100% + 12% = 112%, and 112/100 = 1.12.
2. Write the multiplier for a decrease of 25%.
Answer: 0.75, because 100% - 25% = 75%, and 75/100 = 0.75.
3. £400 grows at 10% compound interest for 2 years. Find its value.
Answer: £484.00, because £400 x 1.10 = £440.00 after year 1, then £440.00 x 1.10 = £484.00 after year 2.
4. A £1,000 computer depreciates by 25% per year. Find its value after 2 years.
Answer: £562.50, because £1,000 x 0.75 = £750.00 after year 1, then £750.00 x 0.75 = £562.50 after year 2.
5. £400 at 10% SIMPLE interest for 2 years earns how much interest in total?
Answer: £80, because 10% of £400 = £40 per year, and £40 x 2 years = £80 (compare this to the £84 that 10% COMPOUND interest would earn on the same £400 over 2 years: £484.00 - £400.00 = £84.00).

On their own

Independent practice worksheets

Practise the multiplier method, compound interest, compound decay, and comparing simple with compound interest, with computed, never-wrong answer keys.

Growth & decay: the multiplier method (Year 10)the core skill this unit opens with Compound interest & growth over time (Year 10)including the Total = P(1 + r/100)^n formula Compound decay: depreciation & reduction (Year 10)cars, laptops and phones losing value Simple vs compound interest (Year 10)the comparison from section 4 Year 11 versions of every worksheet abovethe same 4 topics, freshly generated, for review or a Year 11 class

Reach every student

Differentiation

Support

Always write out '100% +/- r% = new %' before finding the multiplier, so the 0.8-vs-0.2 misconception has nowhere to hide.
Use a year-by-year table (Year 1, Year 2, Year 3 rows) so students see the previous year's TOTAL feeding into the next row, rather than trying to do it all in one mental step.
Start compound-interest problems with a calculator allowed, so the focus stays on WHICH number to multiply, not the arithmetic.
Use whole, round principal amounts (£100, £500) and whole-number percentages (5%, 10%) before mixing in less friendly numbers.

Extension

Ask students to derive the formula Total = P $(1 + r/100)^{n}$ themselves from the repeated-multiplication pattern, before it is given to them.
Introduce fractional years or half-yearly compounding (the interest rate is halved but applied twice as often) and compare the result to yearly compounding at the full rate.
Investigate how many years it takes an amount to double at a given compound interest rate, connecting to logarithms as a look-ahead to A-level.
Compare compound growth to compound decay algebraically: show that a p% increase followed by the same p% decrease never returns to the original amount (except p = 0).

Check it stuck

Assessment: exit ticket

A three-question exit ticket sampling the multiplier method, compound growth, and the simple-vs-compound comparison.

1. Write the multiplier for a decrease of 8%.
Answer: 0.92, because 100% - 8% = 92%, and 92/100 = 0.92.
2. £300 grows at 20% compound interest for 2 years. Find its value.
Answer: £432.00, because £300 x 1.20 = £360.00 after year 1, then £360.00 x 1.20 = £432.00 after year 2.
3. £300 at 20% SIMPLE interest for 2 years earns £120 total interest, for a total of £420.00. Is the compound total from the question above bigger or smaller, and by how much?
Answer: Bigger, by £12.00 (£432.00 - £420.00 = £12.00), because compound interest earns extra interest on year 1's interest, which simple interest never does.

For the teacher

Teacher notes and timings

Rough timing across four lessons: Lesson 1 the multiplier method (section 1), Lesson 2 compound growth (section 2), Lesson 3 compound decay (section 3), Lesson 4 simple vs compound comparison plus the exit ticket (section 4 and assessment).
This unit assumes comfort finding a percentage of an amount (Grade 7-8 percentage/financial mathematics). Revisit that first if the multiplier idea itself, not just the compounding, is still shaky.
Keep the full calculator value through repeated multiplication and round only the requested final answer to the nearest penny. The year-by-year balances shown for teaching may be rounded for display, but a rounded display value must not be fed back into the next calculation. This matches the stated P $(1 + r/100)^{n}$ formula and lib/content_ukgcsemath.ts's worksheet answers.
Language to repeat: 'multiplier' converts a percentage change into ONE number to multiply by; compound growth/decay means applying that SAME multiplier once per period, to the result of the previous period, not to the original amount each time.
Curriculum note: this unit cites the DfE 'GCSE mathematics: subject content and assessment objectives' (2013), items 9 and 16 of the 'Ratio, proportion and rates of change' section, the document that governs all English GCSE maths specifications (AQA, Edexcel, OCR). Item 16 (growth and decay, compound interest) is standard-type, so it applies to Foundation and Higher tier students alike.
Present and print both work: use the Print button for a clean handout, or project the compound-growth line graph and build the year-by-year table with the class live.

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