What is the difference between simple interest and compound interest in Mathematics?

In Mathematics, particularly in the Edexcel IGCSE curriculum, simple interest is calculated on the original principal amount throughout the investment period. In contrast, compound interest is calculated on the initial principal and also on the accumulated interest from previous periods. This means compound interest can grow faster than simple interest over time.

How do you calculate simple interest for an Edexcel IGCSE Mathematics exam?

To calculate simple interest, use the formula: Simple Interest = Principal × Rate × Time. Here, 'Principal' is the initial amount of money, 'Rate' is the annual interest rate (expressed as a decimal), and 'Time' is the period the money is invested or borrowed for, in years.

What is the formula for compound interest in the context of the Edexcel IGCSE Mathematics syllabus?

The formula for compound interest is: Compound Interest = Principal × (1 + Rate)^Time - Principal. In this formula, 'Principal' is the initial sum of money, 'Rate' is the annual interest rate (as a decimal), and 'Time' is the number of years the money is invested or borrowed. This formula calculates the total amount including interest, so subtracting the principal gives the compound interest earned.

Why is compound interest considered more beneficial than simple interest in long-term investments?

Compound interest is often more beneficial for long-term investments because it allows the interest to be calculated on both the initial principal and the accumulated interest from previous periods. This results in exponential growth of the investment over time, as opposed to simple interest, which grows linearly.

Can you provide an example of calculating compound interest for an Edexcel IGCSE Mathematics problem?

Certainly! Suppose you invest £1,000 at an annual interest rate of 5% compounded annually for 3 years. Using the compound interest formula: Amount = £1,000 × (1 + 0.05)^3 = £1,000 × 1.157625 = £1,157.63. The compound interest earned is £1,157.63 - £1,000 = £157.63.

Why is "Using the simple interest formula for a compound problem (or vice versa)" a common mistake in Simple and Compound Interest?

Why it happens: Reading rushed. How to avoid it: Read the question for keywords. 'Compound' or 'compounded annually' = compound formula. Otherwise simple. When in doubt, compound is more common at Higher Tier.

Why is "Using $1 + r/100$ for depreciation" a common mistake in Simple and Compound Interest?

Why it happens: Same formula structure. How to avoid it: Depreciation = decrease. Multiplier = 1 - r/100. £18000 × 0.85 for 15\% off, not £18000 × 1.15.

Why is "Using $r$ as a decimal in $I = Prt/100$" a common mistake in Simple and Compound Interest?

Why it happens: Confusing with multiplier method. How to avoid it: In the Edexcel formula, r is the percentage (e.g. 5 for 5\%, NOT 0.05). The /100 is built in.

Why is "Rounding intermediate values, getting wrong final amount" a common mistake in Simple and Compound Interest?

Why it happens: Habit. How to avoid it: Keep extra decimals. Round only the FINAL answer. £5000 × 1.04^3 → keep 1.124864, multiply, round to nearest pence at the end.

Numbers and the Number SystemPearson Edexcel IGCSE Mathematics 4MA1 Higher

Simple and Compound Interest

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Simple and Compound Interest Study Notes — Edexcel IGCSE 4MA1 Higher Tier (2026 onwards)

How money grows over time. Simple = linear. Compound = exponential. Master both for Paper 2H, plus depreciation as the 'compound decrease' twin.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

1.13 — Calculate simple interest.
1.13 — Calculate compound interest.
1.13 — Calculate compound depreciation.
1.13 — Use the multiplier method consistently.

Simple interest

Interest on the original principal only. Linear growth.

Definition. Simple interest is calculated on the ORIGINAL principal, regardless of accumulated interest. Same amount each year.

Formula:

$I = \dfrac{P \cdot r \cdot t}{100}$

where:

$P$ = principal (initial amount).
$r$ = annual interest rate as a NUMBER (5 for 5%).
$t$ = time in years.
$I$ = interest earned.

Total amount in account: $A = P + I$ .

Examples:

£ $1000$ at $5\%$ for $4$ years:

$I = \dfrac{1000 \times 5 \times 4}{100} = £200$ .
Total: £ $1200$ .

£ $3000$ at $4\%$ for $5$ years:

$I = \dfrac{3000 \times 4 \times 5}{100} = £600$ .
Total: £ $3600$ .

When is simple interest used?

Some short-term loans.
Some bonds.
Less common in modern banking — most savings accounts and mortgages are compound.

Worked qualitative. A bond pays $6\%$ simple interest per year on £ $1000$ . After how many years has the interest equalled the principal?

Set $I = P = 1000$ .
$1000 = 1000 \times 6 \times t / 100$ .
$1000 = 60t$ , so $t = 16.67$ years.
About $16$ years $8$ months.

Edexcel tip. Always check if 'simple' or 'compound' is specified. Simple is typically Foundation Tier; Higher Tier focuses on compound.

$I = Prt/100$ .
$r$ is the percentage NUMBER (5, not 0.05).
Same interest each year.
Used for some bonds and short-term loans.

Compound interest

Interest added to principal, then earns its own interest. Exponential growth.

Definition. Compound interest is calculated on the principal PLUS all previously earned interest. Money grows exponentially.

Formula:

$A = P \times \left(1 + \dfrac{r}{100}\right)^{n}$

where:

$A$ = final amount (principal + interest).
$P$ = principal.
$r$ = annual rate.
$n$ = number of years.

The multiplier: $1 + r/100$ tells you how much money grows each year.

Examples:

£ $5000$ at $4\%$ for $3$ years:

Multiplier: $1.04$ .
$A = 5000 \times 1.04^{3} = 5000 \times 1.124864 = £5624.32$ .

£ $10{,}000$ at $7\%$ for $5$ years:

$A = 10000 \times 1.07^{5} = 10000 \times 1.402552 = £14{,}025.52$ .

The 'rule of 72' (rough mental check):

Money roughly DOUBLES in $72/r$ years.
$5\%$ : doubles in ~ $14.4$ years.
$10\%$ : doubles in ~ $7.2$ years.

Compound vs simple — the gap.

£ $1000$ at $5\%$ for $10$ years:

Simple: $I = 1000 \times 5 \times 10 / 100 = £500$ . Total £ $1500$ .
Compound: $A = 1000 \times 1.05^{10} = £1628.89$ . Interest £ $628.89$ .
Compound exceeds simple by £ $128.89$ .

For longer periods or higher rates, the gap is dramatic.

Simple interest adds a fixed amount each year; compound interest curves upward as interest earns interest.

Worked qualitative. Why is compound so much better than simple over time?

Simple grows by $r\%$ of $P$ EACH year — constant.
Compound grows by $r\%$ of the current balance — which itself grows.
After $1$ year: same. After $2$ years: compound has earned interest on year-1 interest. Gap widens each year.
Famous: 'Compound interest is the eighth wonder of the world' — Albert Einstein (probably misattributed but memorable).

Edexcel tip. Always state the formula. $A = P(1 + r/100)^{n}$ . Show the substitution. Compute step by step. Round only at the end.

$A = P(1 + r/100)^{n}$ .
Multiplier each year.
Exceeds simple after year 1.
Doubles in roughly $72/r$ years.

Compound depreciation

Same formula, with $1 - r/100$ as multiplier.

Definition. Compound depreciation: value decreases by a percentage each year. Same exponential structure as compound interest, but downward.

Formula:

$A = P \times \left(1 - \dfrac{r}{100}\right)^{n}$

Common contexts:

Cars losing value.
Equipment writing off.
Antiques (sometimes).
Buildings (sometimes).

Each year the value is multiplied by 0.85, so the curve falls steeply then flattens towards zero.

Example. Car £ $18000$ , $15\%$ depreciation/year, after $4$ years:

Multiplier: $1 - 0.15 = 0.85$ .
$A = 18000 \times 0.85^{4} = 18000 \times 0.522006 = £9396.10$ .

Difference from simple depreciation.

Simple depreciation (rare in real world): same amount each year. £ $18000$ at £ $2700$ /year for $4$ years would lose $4 \times 2700 = £10{,}800$ , leaving £ $7200$ . Different number!

Compound is more realistic — assets typically lose proportionally less as they age.

Worked qualitative. Why does compound depreciation feel slower than expected?

After year 1: car worth £ $15300$ .
After year 2: $15300 \times 0.85 = £13005$ .
The decrease is on a smaller amount each time.
Asymptotically approaches zero but never reaches it (mathematically).

Edexcel tip. Same setup as compound interest, just $1 - r/100$ instead of $1 + r/100$ . Identify whether the problem is growth or decline.

$A = P(1 - r/100)^{n}$ .
Multiplier slightly less than 1.
Common: cars, equipment.
Asymptotically approaches 0.

Reverse problems — find rate or time

Use logs (or calculator) to back out $r$ or $n$ .

Reverse problem 1: Find $r$ given $A, P, n$ .

Set up: $A = P \times (1 + r/100)^{n}$ .

Solve:

$\dfrac{A}{P} = (1 + r/100)^{n}$ .
$\sqrt[n]{A/P} = 1 + r/100$ .
$r = 100 \times (\sqrt[n]{A/P} - 1)$ .

Example. £ $2000$ → £ $2315.25$ in $3$ years. Find rate.

$\dfrac{2315.25}{2000} = 1.157625$ .
Cube root: $1.157625^{1/3} = 1.05$ .
$r = 100 \times 0.05 = 5\%$ .

Reverse problem 2: Find $n$ given $A, P, r$ .

Set up: $A = P \times (1 + r/100)^{n}$ .

Solve using logs:

$A/P = (1 + r/100)^{n}$ .
$\log(A/P) = n \times \log(1 + r/100)$ .
$n = \log(A/P) / \log(1 + r/100)$ .

Example. £ $1000$ at $5\%$ until reaches £ $1500$ . Find $n$ .

$1500/1000 = 1.5 = 1.05^{n}$ .
$\log 1.5 = n \log 1.05$ .
$n = \log 1.5 / \log 1.05 \approx 0.176 / 0.0212 \approx 8.31$ years.
So takes about $9$ years.

Worked qualitative. Why might Edexcel give 'how many full years?' rather than exact $n$ ?

Logs give non-integer answers.
Real-world banks compound at year-end → only whole years matter.
'After how many years is value at least £ $X$ ?' rounds up.

Edexcel tip. For 'find $r$ ' use cube/ $n$ th root. For 'find $n$ ' use logs. Both are calculator skills on Paper 2H.

Find $r$ : take $n$ th root.
Find $n$ : use logs.
Round time UP to nearest year for 'reaches' problems.
Calculator paper only.

How it’s examined

Interest appears every Paper 2H (calculator). Typical: compound interest for $n$ years (3-4 marks), depreciation (3 marks), reverse problem to find rate or years (4-5 marks). Examiner reports flag (1) using simple formula for compound problem, (2) wrong sign in depreciation multiplier, (3) rounding too early.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H Jan 2024, 4MA1/2H Jun 2023. Last reviewed 2026-05-07.

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Step-by-step worked examples — Simple and Compound Interest

Step-by-step solutions to past-paper-style questions on simple and compound interest, written exactly the way a tutor would explain them at the board.

1Simple interest calculation
Foundation• simple interest
Question
£ $3{,}000$ is invested at $4\%$ simple interest per year. How much interest is earned after $5$ years?
Step-by-step solution
1. Step 1
  Simple interest formula: $I = \dfrac{Prt}{100}$ .
2. Step 2
  Substitute: $I = \dfrac{3000 \times 4 \times 5}{100} = \dfrac{60{,}000}{100} = £600$ .
Answer
£ $600$
Examiner tip
Simple interest is LINEAR — same amount each year. The total amount in the account would be principal + interest = £ $3{,}600$ .
2Compound interest calculation
Higher• Adapted from 4MA1/2H Jan 2024 Q9• compound interest
Question
£ $5{,}000$ is invested at $4\%$ compound interest per year. How much is in the account after $3$ years?
Step-by-step solution
1. Step 1
  Compound interest: $A = P \times \left(1 + \dfrac{r}{100}\right)^{n}$ .
2. Step 2
  Substitute: $A = 5000 \times 1.04^{3}$ .
3. Step 3
  Compute: $1.04^{3} = 1.124864$ . So $A = 5000 \times 1.124864 = £5{,}624.32$ .
Answer
£ $5{,}624.32$
Examiner tip
Compound interest is EXPONENTIAL — interest earns interest. Always use the multiplier $1 + r/100$ , raised to the power of years.
3Compare simple vs compound interest
Higher• compare
Question
£ $1{,}000$ is invested at $5\%$ per year. Calculate (a) the SIMPLE interest after $10$ years, (b) the value of the investment with COMPOUND interest after $10$ years. By how much does compound exceed simple?
Step-by-step solution
1. Step 1
  (a) Simple: $I = \dfrac{1000 \times 5 \times 10}{100} = £500$ . Total: £ $1{,}500$ .
2. Step 2
  (b) Compound: $A = 1000 \times 1.05^{10} = 1000 \times 1.62889 \approx £1{,}628.89$ . Interest earned: £ $628.89$ .
3. Step 3
  Compound exceeds simple by $1628.89 - 1500 = £128.89$ .
Answer
Simple: £ $500$ . Compound: £ $628.89$ interest. Difference: £ $128.89$ .
Examiner tip
Compound ALWAYS exceeds simple after the first year. The longer the investment, the bigger the gap (exponential vs linear).
4Compound depreciation
Higher• depreciation
Question
A car worth £ $18{,}000$ depreciates by $15\%$ per year. What is its value after $4$ years?
Step-by-step solution
1. Step 1
  Depreciation: $A = P \times \left(1 - \dfrac{r}{100}\right)^{n}$ .
2. Step 2
  Substitute: $A = 18000 \times 0.85^{4}$ .
3. Step 3
  Compute: $0.85^{4} = 0.522006\ldots$ . So $A = 18000 \times 0.522 = £9{,}396$ (to nearest £).
Answer
£ $9{,}396$
Examiner tip
Depreciation uses $(1 - r/100)$ instead of $(1 + r/100)$ . Same formula structure.
5Find the interest rate
Higher• Adapted from 4MA1/2H Jun 2023 Q14• find rate
Question
£ $2{,}000$ is invested for $3$ years at compound interest. The investment is worth £ $2{,}315.25$ at the end. Find the annual rate.
Step-by-step solution
1. Step 1
  Set up: $2315.25 = 2000 \times \left(1 + \dfrac{r}{100}\right)^{3}$ .
2. Step 2
  Divide: $\left(1 + \dfrac{r}{100}\right)^{3} = \dfrac{2315.25}{2000} = 1.157625$ .
3. Step 3
  Cube root: $1 + \dfrac{r}{100} = 1.157625^{1/3} = 1.05$ .
4. Step 4
  Solve: $r/100 = 0.05$ , so $r = 5\%$ .
Answer
$5\%$
Examiner tip
Reverse-engineering the rate uses the cube/ $n$ th root. On Paper 2H this is calculator-allowed.

Key Formulae — Simple and Compound Interest

The formulae you need to memorise for simple and compound interest on the Pearson Edexcel IGCSE 4MA1 paper, with every variable defined in plain English and a note on when to use it.

Simple interest
$I = \dfrac{Prt}{100}$
$I$
interest earned (added to principal)
$P$
principal (initial amount)
$r$
annual interest rate (as a number, e.g. $5$ for $5\%$ )
$t$
time in years
When to use
When interest is calculated on the ORIGINAL principal only — not on accumulated interest.
Example
£ $1000$ at $5\%$ for $4$ years: $I = 1000 \times 5 \times 4 / 100 = £200$ .
Compound interest (compound growth)
$A = P \times \left(1 + \dfrac{r}{100}\right)^{n}$
$A$
final amount (principal + interest)
$P$
principal
$r$
annual rate
$n$
number of years
When to use
When interest is added to the principal each period and earns further interest. Standard for savings accounts, investments, mortgages.
Example
£ $5000$ at $4\%$ for $3$ years: $A = 5000 \times 1.04^{3} = £5624.32$ .
Compound depreciation
$A = P \times \left(1 - \dfrac{r}{100}\right)^{n}$
$r$
depreciation rate (positive number)
When to use
When value decreases by a percentage each period. Cars, equipment, household goods.
Example
Car £ $18000$ depreciating $15\%$ /year for $4$ years: $A = 18000 \times 0.85^{4} \approx £9396$ .

Key Definitions and Keywords — Simple and Compound Interest

Definitions to memorise and the exact keywords mark schemes credit for simple and compound interest answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4MA1 sitting.

Principal
Examiner keyword
The initial amount invested or borrowed. Sometimes called the 'capital'.
Simple interest
Examiner keyword
Interest calculated on the ORIGINAL principal only. Linear growth.
Example
£ $1000$ at $5\%$ simple = £ $50$ interest each year.
Compound interest
Examiner keyword
Interest calculated on the principal PLUS accumulated interest. Exponential growth.
Example
£ $1000$ at $5\%$ compound: £ $1050$ after year 1, £ $1102.50$ after year 2.
Depreciation
Examiner keyword
Decrease in value over time, usually compound. Common for assets like cars and equipment.

Common Mistakes and Misconceptions — Simple and Compound Interest

The traps other students keep falling into on simple and compound interest questions — taken from recent Pearson Edexcel IGCSE 4MA1 examiner reports and mark schemes — and how to avoid them.

✕Using the simple interest formula for a compound problem (or vice versa)
Why it happens
Reading rushed.
How to avoid it
Read the question for keywords. 'Compound' or 'compounded annually' = compound formula. Otherwise simple. When in doubt, compound is more common at Higher Tier.
✕Using $1 + r/100$ for depreciation
Why it happens
Same formula structure.
How to avoid it
Depreciation = decrease. Multiplier = $1 - r/100$ . £ $18000$ × $0.85$ for $15\%$ off, not £ $18000$ × $1.15$ .
✕Using $r$ as a decimal in $I = Prt/100$
Why it happens
Confusing with multiplier method.
How to avoid it
In the Edexcel formula, $r$ is the percentage (e.g. $5$ for $5\%$ , NOT $0.05$ ). The $/100$ is built in.
✕Rounding intermediate values, getting wrong final amount
Why it happens
Habit.
How to avoid it
Keep extra decimals. Round only the FINAL answer. £ $5000 \times 1.04^{3}$ → keep $1.124864$ , multiply, round to nearest pence at the end.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

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Simple and Compound Interest — frequently asked questions

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Simple and Compound Interest

Short Study Notes in the form of Slides

Short Study Notes — Simple and Compound Interest

Detailed Notes

What you’ll learn

How it’s examined

1Simple interest calculation

2Compound interest calculation

3Compare simple vs compound interest

4Compound depreciation

5Find the interest rate

Simple interest

Compound interest (compound growth)

Compound depreciation

Principal

Simple interest

Compound interest

Depreciation

✕Using the simple interest formula for a compound problem (or vice versa)

✕Using 1+r/1001 + r/1001+r/100 for depreciation

✕Using rrr as a decimal in I=Prt/100I = Prt/100I=Prt/100

✕Rounding intermediate values, getting wrong final amount

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Simple and Compound Interest — frequently asked questions

✕Using $1 + r/100$ for depreciation

✕Using $r$ as a decimal in $I = Prt/100$