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How to Calculate Compound Interest: A Step-by-Step GCSE Maths Guide

How to Calculate Compound Interest: A Step-by-Step GCSE Maths Guide

Compound interest is a core topic in GCSE Maths, appearing on all major exam boards (AQA, Edexcel, OCR). It's a real-world concept used in savings, loans, and investments. Many students find it tricky at first, but once you understand the multiplier method, it becomes straightforward. This guide will walk you through it from first principles to the efficient formula.

What is Compound Interest?

Simple interest is calculated only on the original amount. Compound interest is different: interest is calculated on the original amount and on any interest already added. This means your money grows faster (or a debt increases faster) because you earn "interest on interest".

The Long Way: Understanding the Process

Let's start with an example to see what's actually happening. This is crucial for understanding, but you wouldn't want to do it this way in an exam!

Example 1: £800 is invested in a savings account paying 3.5% compound interest per annum (per year). Find the value after 5 years.

Step-by-step (the long method):

Start: £800

Year 1: Interest = 3.5% of £800 = 0.035 × £800 = £28.

New total = £800 + £28 = £828.

Year 2: We start with £828.

Interest = 3.5% of £828 = 0.035 × £828 = £28.98.
New total = £828 + £28.98 = £856.98.

Year 3: Start with £856.98.

Interest = 3.5% of £856.98 ≈ £29.9943 ≈ £30.00 (to nearest penny).
New total = £856.98 + £30.00 = £886.98.

Year 4: Start with £886.98.

Interest = 3.5% of £886.98 ≈ £31.0443 ≈ £31.04.
New total = £886.98 + £31.04 = £918.02.

Year 5: Start with £918.02.

Interest = 3.5% of £918.02 ≈ £32.1307 ≈ £32.13.
New total = £918.02 + £32.13 = £950.15.

After 5 years, the investment is worth £950.15.

See how tedious that was? Just five years involved lots of steps and rounding. For 10 or 20 years, it would be a nightmare! This shows why we need a better method.

The Efficient Method: The Multiplier

The key is to spot a pattern. Each year, we are finding 103.5% of the previous year's amount.

Finding 103.5% is the same as multiplying by 1.035 (since 100% + 3.5% = 103.5%, and 103.5/100 = 1.035).

This number (1.035) is called the multiplier.

Let's retry the same problem using this powerful idea.

Original Amount: £800
Interest Rate: 3.5%
Multiplier: 1 + (3.5/100) = 1.035
Time: 5 years

Instead of doing 5 separate steps, we can do one calculation:

Final Amount = Original Amount × (Multiplier)^{Number of Years}

Or, using the common formula:
A = P(1 + r/100)^n
Where:

A = final amount

P = principal (original amount)

r = percentage rate

n = number of time periods (years)

Calculation:
Final Amount = £800 × (1.035)^5

We now use our calculator carefully:

Calculate 1.035^5.

Type: 1.035, then the x^y or ^ button, then 5, then =.

You should get: 1.187686 (keep this on your calculator screen).

Now multiply by 800.

With the previous answer still on screen, press × 800 =.

Answer: 950.1488...

Round to the nearest penny (two decimal places for currency).

£950.15

We got the same answer with one calculation! This method is essential for the exam.

flowchart TD
A["Start: Principal P, Rate r%, Time n"] --> B["Calculate multiplier: 1 + r/100"]
B --> C["Calculate multiplier^n (use x^y button)"]
C --> D["Multiply by P"]
D --> E["Round to 2 d.p. for currency"]
E --> F["Final Amount A"]

What About Depreciation?

Depreciation (like a car losing value) works exactly the same way, but the multiplier is less than 1.

For interest (increase): Multiplier = 1 + (percentage/100)

For depreciation (decrease): Multiplier = 1 – (percentage/100)

Example 2 (Depreciation): A new car costs £15,000. It depreciates in value by 12% per year. Find its value after 6 years.

Method:

P = £15,000

Rate of decrease = 12%

Multiplier = 1 – (12/100) = 1 – 0.12 = 0.88

n = 6 years

Calculation:
Final Value = £15,000 × (0.88)^6

Calculate 0.88^6 = 0.464404

Multiply by 15,000 = 6966.06

After 6 years, the car is worth £6,966.06.

K

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The diagram shows a cube and a square-based pyramid.
Diagram for the exam question
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