What is the formula for compound interest?

A = P(1 + r)^n, where P is the amount invested or borrowed (the principal), r is the interest rate per compounding period as a decimal, n is the number of compounding periods, and A is the final amount. The interest earned is I = A − P. For $5000 at 4% p.a. compounded yearly for 3 years, A = 5000 × 1.04³ = $5624.32, so the interest is $624.32.

What is the difference between simple and compound interest?

Simple interest is worked out on the original amount only, so it is the same every year (I = Prt). Compound interest is worked out on the balance, which already includes earlier interest, so it grows each year. $8000 at 5% p.a. for 10 years earns $4000 simple interest but $5031.16 compound interest (compounded yearly).

How do I calculate compound interest compounded monthly?

Divide the yearly rate by 12 and multiply the number of years by 12. At 6% p.a. compounded monthly for 5 years, r = 0.06 ÷ 12 = 0.005 per month and n = 5 × 12 = 60 months, so A = P × 1.005⁶⁰.

How do I find the principal in a compound interest question?

Rearrange A = P(1 + r)^n to P = A ÷ (1 + r)^n. To have $10 000 in 4 years at 3.5% p.a. compounded quarterly, P = 10 000 ÷ 1.00875¹⁶ ≈ $8698.88.

Is depreciation the same as compound interest?

It uses the same idea, but the value goes down. Use A = P(1 − r)^n. A $30 000 car that loses 15% of its value each year is worth 30 000 × 0.85⁴ ≈ $15 660.19 after 4 years.

What year level learns compound interest?

In the Victorian Curriculum 2.0, Year 9 covers simple interest, and Year 10 uses mathematical modelling for growth and decay, including compound interest. Year 10A and senior maths (General and Methods) build on it with exponential functions and logarithms.

How to Calculate Compound Interest: Formula, Examples and Calculator (Year 10)

The compound interest formula

A = P(1 + r)^n

$P$ is the principal: the amount you start with (invested or borrowed).
$r$ is the interest rate per compounding period, written as a decimal. $4\%$ is $0.04$ .
$n$ is the number of compounding periods: years, months or quarters, depending on how often interest is added.
$A$ is the final amount (principal plus all the interest).

The interest itself is the difference:

I = A - P

Why a power? Adding $4\%$ is the same as multiplying by $1.04$ . Do that once a year for $3$ years and you've multiplied by $1.04 \times 1.04 \times 1.04 = 1.04^3$ .

Compound interest compounded yearly

Worked example

Maya invests $\text{\textdollar}5000$ at $4\%$ p.a. compounded yearly. Find the value of the investment after $3$ years and the interest earned.

A = P(1 + r)^n

A = 5000 \times (1 + 0.04)^3

A = 5000 \times 1.04^3

A = \text{\textdollar}5624.32

Interest earned: $I = 5624.32 - 5000 = \text{\textdollar}624.32$

Here's the same thing one year at a time, so you can see the interest grow:

Year	Start	Interest ( $4\%$ )	End
1	$\text{\textdollar}5000.00$	$\text{\textdollar}200.00$	$\text{\textdollar}5200.00$
2	$\text{\textdollar}5200.00$	$\text{\textdollar}208.00$	$\text{\textdollar}5408.00$
3	$\text{\textdollar}5408.00$	$\text{\textdollar}216.32$	$\text{\textdollar}5624.32$

Simple interest would pay $\text{\textdollar}200$ every year. With compounding, year 2 pays $\text{\textdollar}208$ because the first year's $\text{\textdollar}200$ is now earning interest too.

Compound interest calculator (with a yearly table)

Enter the amount, the yearly interest rate, the time and how often it’s compounded. You’ll get the final amount, the interest, the working, a comparison with simple interest and the balance at the end of every year.

Amount invested or borrowed ($)Interest rate (% p.a.)Time (years)Compounded

$5,624.32 after 3 years ($624.32 interest)

Compound interest formula: $A = P(1 + r)^n$ , where $r$ is the interest rate per compounding period (as a decimal) and $n$ is the number of periods.
Interest is added once a year, so $r = 4\% = 0.04$ and $n = 3$ .
Substitute: $A = 5000 \times (1 + 0.04)^{3} = 5000 \times 1.04^{3}$ .
Work it out in one go on your calculator: $A \approx \text{\textdollar}5{,}624.32$ (rounded to the nearest cent).
Interest earned: $I = A - P = \text{\textdollar}5{,}624.32 - \text{\textdollar}5{,}000.00 = \text{\textdollar}624.32$ .
Simple interest at the same rate would be $I = Prt = 5000 \times 0.04 \times 3 = \text{\textdollar}600.00$ , so compounding earns $\text{\textdollar}24.32$ more.

Balance at the end of each year
Year	Start	Interest	End
Year 1	$5,000.00	$200.00	$5,200.00
Year 2	$5,200.00	$208.00	$5,408.00
Year 3	$5,408.00	$216.32	$5,624.32

Each row's interest is bigger than the last because interest is earned on the interest already added. Amounts are rounded to the nearest cent for display only.

Compounding monthly, quarterly or half-yearly

Rates are quoted per annum (p.a.), but interest is often added more often than once a year. Change both the rate and the number of periods to match:

Compounded	Rate per period	Periods in $t$ years
Yearly	$r$	$t$
Half-yearly	$r \div 2$	$2t$
Quarterly	$r \div 4$	$4t$
Monthly	$r \div 12$	$12t$

Worked example: monthly

Josh puts $\text{\textdollar}12\,000$ in a term deposit paying $6\%$ p.a. compounded monthly. How much is it worth after $5$ years?

Rate per month: $r = 0.06 \div 12 = 0.005$

Number of months: $n = 5 \times 12 = 60$

A = 12\,000 \times 1.005^{60}

A \approx \text{\textdollar}16\,186.20

So the interest is about $\text{\textdollar}4186.20$ .

The more often interest is added, the more you end up with, but the differences get small. $\text{\textdollar}10\,000$ at $6\%$ p.a. for $5$ years grows to $\text{\textdollar}13\,382.26$ compounded yearly, $\text{\textdollar}13\,468.55$ quarterly and $\text{\textdollar}13\,488.50$ monthly.

Compound interest vs simple interest

Worked example: comparing

$\text{\textdollar}8000$ is invested for $10$ years at $5\%$ p.a. Compare simple interest with interest compounded yearly.

Simple interest: $I = Prt = 8000 \times 0.05 \times 10 = \text{\textdollar}4000$ , so the final amount is $\text{\textdollar}12\,000$ .

Compound interest:

A = 8000 \times 1.05^{10} \approx \text{\textdollar}13\,031.16

Compounding gives $\text{\textdollar}13\,031.16 - \text{\textdollar}12\,000 = \text{\textdollar}1031.16$ more.

Over a short time the two are close. Over a long time compound interest pulls further and further ahead, because simple interest grows in a straight line and compound interest grows exponentially. That's why it matters for savings, and for credit card and loan debt.

Finding the principal (working backwards)

If you know the final amount and need the starting amount, divide instead of multiply:

P = \frac{A}{(1 + r)^n}

Worked example

How much should Priya invest now at $3.5\%$ p.a. compounded quarterly so that she has $\text{\textdollar}10\,000$ in $4$ years?

Rate per quarter: $r = 0.035 \div 4 = 0.00875$

Number of quarters: $n = 4 \times 4 = 16$

P = \frac{10\,000}{1.00875^{16}}

P \approx \text{\textdollar}8698.88

Check by going forwards: $8698.88 \times 1.00875^{16} \approx 10\,000$ . Type the whole calculation into your calculator in one go, using brackets, rather than rounding $1.00875^{16}$ first.

Depreciation: the same idea going down

When something loses a percentage of its value each year, subtract the rate instead of adding it:

A = P(1 - r)^n

Worked example

A car bought for $\text{\textdollar}30\,000$ depreciates by $15\%$ each year. What is it worth after $4$ years?

A = 30\,000 \times (1 - 0.15)^4

A = 30\,000 \times 0.85^4

A \approx \text{\textdollar}15\,660.19

Losing $15\%$ a year for $4$ years is not losing $60\%$ . Each year's $15\%$ comes off a smaller value, so the car keeps a little over half its value, not $40\%$ .

Common mistakes

Using the percentage instead of the decimal. $5\%$ is $0.05$ , so the multiplier is $1.05$ , not $1.5$ or $6$ .
Forgetting to change the rate and the periods together. $6\%$ p.a. compounded monthly for $5$ years is $1.005^{60}$ . Using $1.06^{60}$ or $1.005^{5}$ gives answers that are far too big or too small.
Giving $A$ when the question asks for the interest. "How much interest" means $I = A - P$ . "What is it worth" or "how much is owed" means $A$ .
Multiplying before the power. $5000 \times 1.04^3$ means cube $1.04$ first, then multiply. Working out $(5000 \times 1.04)^3$ gives a huge wrong answer.
Rounding in the middle. Rounding $1.00875^{16}$ to $1.15$ changes the answer by several dollars. Keep everything in the calculator and round to the nearest cent at the end.

Practice questions with answers

Try each one before you open the answer. Money is rounded to the nearest cent.

Find the final amount when $\text{\textdollar}2000$ is invested at $5\%$ p.a. compounded yearly for $2$ years.
Show the answer
$A = 2000 \times 1.05^2 = \text{\textdollar}2205$
$\text{\textdollar}6500$ is invested at $3\%$ p.a. compounded yearly for $4$ years. How much interest is earned?
Show the answer
$A = 6500 \times 1.03^4 \approx \text{\textdollar}7315.81$ , so $I = 7315.81 - 6500 = \text{\textdollar}815.81$
Find the final amount when $\text{\textdollar}4000$ is invested at $6\%$ p.a. compounded quarterly for $3$ years.
Show the answer
$r = 0.06 \div 4 = 0.015$ and $n = 3 \times 4 = 12$ . $A = 4000 \times 1.015^{12} \approx \text{\textdollar}4782.47$
A loan of $\text{\textdollar}15\,000$ is charged $4.8\%$ p.a. interest compounded monthly. If nothing is repaid, how much is owed after $2$ years?
Show the answer
$r = 0.048 \div 12 = 0.004$ and $n = 24$ . $A = 15\,000 \times 1.004^{24} \approx \text{\textdollar}16\,508.22$
How much interest does $\text{\textdollar}9000$ earn in $5$ years at $7\%$ p.a. compounded half-yearly?
Show the answer
$r = 0.035$ and $n = 10$ . $A = 9000 \times 1.035^{10} \approx \text{\textdollar}12\,695.39$ , so $I \approx \text{\textdollar}3695.39$
How much needs to be invested now at $5\%$ p.a. compounded yearly to have $\text{\textdollar}20\,000$ in $6$ years?
Show the answer
$P = \dfrac{20\,000}{1.05^6} \approx \text{\textdollar}14\,924.31$
A laptop bought for $\text{\textdollar}2400$ loses $20\%$ of its value each year. What is it worth after $3$ years?
Show the answer
$A = 2400 \times 0.8^3 = \text{\textdollar}1228.80$
Which earns more on $\text{\textdollar}10\,000$ over $3$ years: $5\%$ p.a. simple interest, or $4.8\%$ p.a. compounded monthly?
Show the answer
Simple: $I = 10\,000 \times 0.05 \times 3 = \text{\textdollar}1500$ . Compound: $A = 10\,000 \times 1.004^{36} \approx \text{\textdollar}11\,545.52$ , so $I \approx \text{\textdollar}1545.52$ . The compound option earns $\text{\textdollar}45.52$ more, even at the lower rate.
$\text{\textdollar}5000$ is invested at $6\%$ p.a. compounded yearly. After how many whole years will it first be worth more than $\text{\textdollar}10\,000$ ?
Show the answer
Try values of $n$ : $5000 \times 1.06^{11} \approx \text{\textdollar}9491.49$ (not yet) and $5000 \times 1.06^{12} \approx \text{\textdollar}10\,060.98$ . So it takes $12$ years.

Interest and growth by year level

Following the Victorian Curriculum 2.0. Schools sequence topics differently, so check your course.

Year 9

Simple interest

Simple interest with I = Prt, finding the rate or time, and simple interest loans.

Worksheets and topic test · Free PDF with answers

Year 10

Compound interest

Compound interest, different compounding periods, finding the principal, and depreciation.

Worksheets and topic test · Free PDF with answers

Year 10

Exponential equations

Growth and decay written as exponential equations, solved by trial and error or graphs.

Worksheets and topic test · Free PDF with answers

Year 10A

Logarithms and log laws

Logarithms, which let you solve for the time n exactly instead of by trial and error.

Worksheets and topic test · Free PDF with answers

How to calculate compound interest

The compound interest formula

Compound interest compounded yearly

Compound interest calculator (with a yearly table)

Compounding monthly, quarterly or half-yearly

Compound interest vs simple interest

Finding the principal (working backwards)

Depreciation: the same idea going down

Common mistakes

Practice questions with answers

Interest and growth by year level

Compound interest: common questions

More maths help, Years 7 to 10

Got an interest question wrong in homework?