Compound interest

What you will learn

distinguish simple interest from compound interest and explain why compound interest grows faster,
apply the compound interest formula $A = P\!\left(1 + \dfrac{r}{n}\right)^{nt}$ ,
calculate depreciation using $A = P(1 - r)^n$ ,
solve financial modelling problems including comparing investment options and loan repayments.

Worked example 0 Real-world example: savings account comparison

Zara invests $5000 for $3$ years. Bank A offers $6\%$ p.a. simple interest. Bank B offers $6\%$ p.a. compound interest (compounded annually). Which gives more?

Bank A (simple): $I = 5000 \times 0.06 \times 3 = 900$ . Total: $A = 5900$ .
Bank B (compound): $A = 5000 \times 1.06^3 = 5000 \times 1.191016 = 5955.08$ .
Compound interest earns $55.08 more because interest in years 2 and 3 is calculated on a growing balance.

Key idea: simple interest is linear (same amount each year); compound interest is exponential (grows faster over time).

1. Simple vs compound interest

With simple interest, the interest each period is calculated on the original principal only:

A_{\text{simple}} = P(1 + rT)

With compound interest, interest each period is calculated on the current balance (principal plus previously earned interest):

A_{\text{compound}} = P(1 + r)^T \quad \text{(compounded annually)}

After one year both give the same result. After two or more years compound interest always yields more (assuming $r > 0$ ).

Simple interest (straight line) vs compound interest (curve) on the same principal over 20 years. The gap widens as time increases.

2. The compound interest formula

Formula reference

A = P\!\left(1 + \frac{r}{n}\right)^{nt}

$A$ = final amount,
$P$ = principal (initial investment or loan),
$r$ = annual interest rate (as a decimal),
$n$ = number of compounding periods per year,
$t$ = time in years.

For annual compounding ( $n = 1$ ): $A = P(1 + r)^t$ .

Worked example 1 Annual compounding

Find the value of $3000 invested at $5\%$ p.a. compounded annually for $4$ years.

$P = 3000$ , $r = 0.05$ , $n = 1$ , $t = 4$ .
$A = 3000 \times 1.05^4 = 3000 \times 1.21551 = 3646.52$ dollars.
Interest earned: $3646.52 - 3000 = 646.52$ dollars.

Worked example 2 Monthly compounding

Find the value of $2000 invested at $6\%$ p.a. compounded monthly for $3$ years.

$P = 2000$ , $r = 0.06$ , $n = 12$ , $t = 3$ .
$A = 2000\!\left(1 + \dfrac{0.06}{12}\right)^{36} = 2000 \times 1.005^{36}$ .
$1.005^{36} \approx 1.19668$ . $A \approx 2393.36$ dollars.

3. Depreciation

Depreciation is the decrease in value of an asset over time. When the rate of depreciation is constant, the model is exponential decay.

Formula reference

A = P(1 - r)^n

$A$ = value after $n$ periods,
$P$ = original value,
$r$ = rate of depreciation per period (as a decimal),
$n$ = number of periods.

Worked example 3 Car depreciation

A car costs $35000 and depreciates at $18\%$ per year. Find its value after $5$ years.

$A = 35000 \times (1 - 0.18)^5 = 35000 \times 0.82^5$ .
$0.82^5 \approx 0.37043$ .
$A \approx 12965$ dollars.

Worked example 4 Finding the depreciation rate

A laptop worth $2400 is valued at $1200 after $3$ years. Find the annual depreciation rate.

$1200 = 2400 \times (1 - r)^3$ .
$(1 - r)^3 = 0.5$ .
$1 - r = 0.5^{1/3} \approx 0.7937$ .
$r \approx 0.2063$ , so the depreciation rate is approximately $20.6\%$ per year.

4. Financial modelling problems

Financial modelling often combines the compound interest formula with additional conditions such as regular deposits, comparison of options, or target amounts.

Worked example 5 Reaching a savings goal

Ethan wants to have $10000 in $5$ years. His account earns $4\%$ p.a. compounded annually. How much must he invest now?

$10000 = P \times 1.04^5$ .
$P = \dfrac{10000}{1.04^5} = \dfrac{10000}{1.21665} \approx 8219.27$ dollars.
He needs to invest approximately $8219.27 today.

Worked example 6 Comparing two investments

Option A: $6000 at $5\%$ p.a. compounded annually for $10$ years. Option B: $6000 at $4.8\%$ p.a. compounded monthly for $10$ years.

Option A: $A = 6000 \times 1.05^{10} = 6000 \times 1.62889 \approx 9773.37$ .
Option B: $A = 6000\!\left(1 + \dfrac{0.048}{12}\right)^{120} = 6000 \times 1.004^{120}$ .
$1.004^{120} \approx 1.61222$ . $A \approx 9673.35$ .
Option A gives about $100 more despite a lower compounding frequency, because the annual rate is higher.

Practice

Fluency

Tier 1: basic skills

Calculate simple interest on $4000 at $5\%$ p.a. for $3$ years.
Find the total amount when $4000 is invested at $5\%$ p.a. compounded annually for $3$ years.
State the difference between your answers to Q1 and Q2.
Find the value of $10000 invested at $3\%$ p.a. compounded annually for $6$ years.
A computer worth $1800 depreciates at $25\%$ per year. Find its value after $2$ years.
Find the value of $5000 at $8\%$ p.a. compounded quarterly for $2$ years. (Hint: $n = 4$ .)
A painting is bought for $3000 and appreciates at $10\%$ per year. What is it worth after $4$ years?
How much interest is earned on $7000 at $6\%$ p.a. compounded annually for $5$ years?

Reasoning

Tier 2: mixed practice

$12000 is invested at $4.5\%$ p.a. compounded monthly. Find the amount after $3$ years.
A car worth $28000 depreciates at $15\%$ per year. After how many whole years is it first worth less than $10000?
Which is better over $5$ years: $8000 at $6\%$ p.a. compounded annually, or $8000 at $5.8\%$ p.a. compounded monthly? Show both calculations.
Ava invests $P at $7\%$ p.a. compounded annually. After $10$ years she has $15000. Find $P$ .
A motorbike depreciates from $9000 to $4500 in $4$ years. Find the annual depreciation rate.
Calculate the total interest earned on $20000 at $3.2\%$ p.a. compounded quarterly over $8$ years.

Reasoning

Tier 3: explain and apply

Explain why the gap between simple interest and compound interest widens as time increases. Use the formulas to support your answer.
A credit card charges $1.5\%$ interest per month on unpaid balances. What is the effective annual interest rate? (Hint: $(1.015)^{12}$ .)
Mia has $20000 to invest for $10$ years. Bank A offers $5\%$ p.a. compounded annually. Bank B offers $4.9\%$ p.a. compounded daily (assume $365$ days). Which should she choose?
A factory machine costs $50000 and depreciates at $20\%$ per year. The company plans to replace it when its value drops below $10\%$ of the original cost. After how many whole years should they replace it?

Challenge

Reasoning

Harder reasoning

Show that for a principal $P$ invested at rate $r$ p.a. compounded annually for $2$ years, the compound interest exceeds the simple interest by exactly $Pr^2$ .
Jake borrows $15000 at $8\%$ p.a. compounded annually. He makes no repayments. After how many whole years does the debt first exceed $25000? How does this compare with the Rule of 70 estimate for doubling?
An investment of $10000 grows to $10000 \times 1.06^t $after$ t $years. A second investment of \$ 15000 grows to $15000 \times 1.03^t$ after $t$ years. After how many whole years does the first investment first exceed the second? Justify your answer.