Year 10 Maths Core Earning Money

Compound Interest

Q: What is the compound interest formula?

The formula is FV = PV(1 + r)^n, where PV is the amount invested, r is the rate per compounding period as a decimal, n is the number of periods and FV is the final value.

Q: How do you calculate interest compounded monthly?

Divide the yearly rate by 12 to get the monthly rate and multiply the number of years by 12 to get the number of months. Then use FV = PV(1 + r)^n with those values.

Q: How do you find the present value?

Rearrange the formula: PV equals FV divided by (1 + r)^n. This tells you how much to invest now to reach a target amount later.

20 practice questions 1 video lesson Theory + worked examples

Compound Interest in NSW Year 10 Core maths is interest earned on interest (NESA outcome MA5‑FIN‑C‑02). Working year by year shows the balance growing faster each time, and $FV = PV(1 + r)^n$ gives the future value in one step, where $r$ is the rate per period and $n$ the number of periods.

You will build compound interest tables, use the formula with annual, quarterly and monthly compounding, and find the total interest earned. You will also compare simple and compound interest on the same investment, work backwards to find the amount to invest now, the rate or the number of years, and find how long it takes an investment to double. It is part of Year 10 Core Earning Money.

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Compound Interest - Video - Compound Interest Watch

Theory

With compound interest, each period's interest is added to the balance, so the next period earns interest on it too. This Year 10 maths page shows how to work it out year by year and with $FV = PV(1 + r)^{n}$, how to handle monthly and quarterly compounding, how to find the present value, and how it compares with simple interest.

With compound interest, the interest for each period is added to the balance. The next period's interest is worked out on the new, bigger balance, so the money earns interest on interest.

The present value ($PV$) is the amount invested now. The future value ($FV$) is what it grows to. The interest earned is $FV - PV$.

The compounding period is how often interest is added: annually (once a year), quarterly ($4$ times), monthly ($12$ times). The rate is quoted per annum, so divide it by the number of periods in a year, and count the total number of periods.

Compound interest earns interest on interest, so its graph curves upward and pulls away from simple interest.

Adding $10\%$ each year is the same as multiplying by $1.1$. Over $n$ years, multiply by $1.1^{n}$.

\[FV = PV(1 + r)^{n} \qquad \text{interest} = FV - PV\]

F V = P V {(1 + r)}^{n}

$r$ is the rate per compounding period as a decimal and $n$ is the number of compounding periods.

Compounded	$r$	$n$ for $t$ years
annually	yearly rate	$t$
quarterly	yearly rate $\div 4$	$4t$
monthly	yearly rate $\div 12$	$12t$

\[PV = \dfrac{FV}{(1 + r)^{n}}\]

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

Why the formula works. Adding \(r\) of the balance is the same as multiplying by \(1 + r\). Doing that \(n\) times multiplies by \((1 + r)^{n}\).

How to solve a compound interest problem

Find $r$ per period (as a decimal) and $n$, the total number of periods.
Substitute into $FV = PV(1 + r)^{n}$, using brackets or the power key on the calculator.
Round money to the nearest cent at the end, and subtract $PV$ if the interest is asked for.

Other questions you will meet

Year by year or a table: interest $=$ rate $\times$ value at the start of that year; the end value starts the next row.
Interest in one year only: value at the end of that year minus value at the end of the year before.
Comparing accounts: work out each final value with its own $r$ and $n$, then compare. More frequent compounding helps, but a higher rate can matter more.
Reaching a target: find $FV$ and compare with the target, or find the $PV$ needed (Example 3).
Finding the rate: $(1 + r)^{n} = \dfrac{FV}{PV}$, so take the $n$th root: $1 + r = \sqrt[n]{FV \div PV}$.
Finding the time: try whole numbers of periods until the value passes the amount wanted (Example 4).

Example 1 — Year by year

Aria invests $\$2500$ at $4\%$ p.a. interest, compounded annually.
(i) Work out the interest each year to find the value after $3$ years.
(ii) Check with $FV = PV(1 + r)^{n}$.

Solution

(i) Each year the interest is $4\%$ of the value at the start of that year, so it grows.

Year	Value at start	Interest (4%)	Value at end
$1$	$\$2500.00$	$\$100.00$	$\$2600.00$
$2$	$\$2600.00$	$\$104.00$	$\$2704.00$
$3$	$\$2704.00$	$\$108.16$	$\$2812.16$

(ii) $PV = 2500$, $r = 0.04$, $n = 3$.

$FV$	$=$	$2500 \times 1.04^{3}$
	$=$	$\mathbf{\$2812.16}$

F V = 2500 \times {1.04}^{3} = 2812.16

Simple interest would give only $\$2800$. The extra $\$12.16$ is interest earned on interest.

Example 2 — Compounding monthly

Ben invests $\$8000$ at $4.8\%$ p.a., compounded monthly, for $3$ years. Find
(i) the final value
(ii) the interest earned. Give answers to the nearest cent.

Solution

Change the rate and the time to months.

$r$	$=$	$0.048 \div 12 = 0.004 \text{ per month}$
$n$	$=$	$3 \times 12 = 36 \text{ months}$

$FV$	$=$	$8000 \times 1.004^{36}$
	$=$	$9236.419\ldots$
	$\approx$	$\mathbf{\$9236.42}$
$I$	$=$	$9236.42 - 8000 = \mathbf{\$1236.42}$

F V = 8000 {(1.004)}^{36} \approx 9236.42

Using $r = 0.048$ with $n = 36$ would treat $4.8\%$ as a monthly rate and give a far too big answer.

Example 3 — How much to invest now

Chloe wants $\$15\,000$ in $5$ years. Her account pays $3.2\%$ p.a., compounded quarterly. How much must she invest now? Give your answer to the nearest cent.

Solution

The future value is known, so solve $FV = PV(1 + r)^{n}$ for $PV$ by dividing.

$r$	$=$	$0.032 \div 4 = 0.008$
$n$	$=$	$5 \times 4 = 20$
$15\,000$	$=$	$PV \times 1.008^{20}$
$PV$	$=$	$\dfrac{15\,000}{1.008^{20}}$
	$=$	$12\,790.296\ldots$
	$\approx$	$\mathbf{\$12\,790.30}$

P V = \frac{15000}{{1.008}^{20}} \approx 12790.30

Check: $12\,790.30 \times 1.008^{20} \approx 15\,000$.

Example 4 — Simple or compound, and how long

Dev invests $\$4000$ at $6\%$ p.a.
(i) How much more interest does he earn in $5$ years if the interest is compounded annually rather than simple?
(ii) With annual compounding, after how many whole years is the investment first worth more than $\$6000$?

Solution

(i) Work out both, then subtract.

$\text{simple } I$	$=$	$4000 \times 0.06 \times 5$
	$=$	$\$1200.00$
$\text{compound } FV$	$=$	$4000 \times 1.06^{5}$
	$=$	$5352.902\ldots$
$\text{compound } I$	$=$	$5352.90 - 4000$
	$=$	$\$1352.90$
$\text{difference}$	$=$	$\mathbf{\$152.90}$

(ii) Try values of $n$ in $4000 \times 1.06^{n}$.

$n = 6$	$:$	$\$5674.08 \ (\text{under})$
$n = 7$	$:$	$\$6014.52 \ (\text{over})$

4000 \times {1.06}^{7} \approx 6014.52 > 6000

It first passes $\$6000$ after 7 years.

Common pitfalls

Match the rate to the period. For $8.4\%$ p.a. compounded monthly, $r = 0.084 \div 12 = 0.007$ and $n$ is the number of months. Using $0.084$ with months gives nonsense.

The formula gives the final value. $PV(1 + r)^{n}$ is the whole amount. Subtract $PV$ to get the interest.

Brackets before the power. Work out $(1 + r)^{n}$ first, then multiply by $PV$. $5000 \times 1.03^{4}$ is not $(5000 \times 1.03)$ raised to $4$.

Interest is not the same each year. Unlike simple interest, compound interest grows each year. Do not multiply the first year's interest by the number of years.

Frequently asked questions

What is the compound interest formula?

The formula is $FV = PV(1 + r)^{n}$, where $PV$ is the amount invested, $r$ is the rate per compounding period as a decimal, $n$ is the number of periods and $FV$ is the final value.

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. Compound interest is worked out on the balance including earlier interest, so it grows each year.

How do you calculate interest compounded monthly?

Divide the yearly rate by $12$ to get the monthly rate and multiply the number of years by $12$ to get the number of months. Then use $FV = PV(1 + r)^{n}$ with those values.

How do you find the present value?

Rearrange the formula: $PV = \dfrac{FV}{(1 + r)^{n}}$. This tells you how much to invest now to reach a target amount later.

Does compounding more often earn more interest?

At the same yearly rate, yes: compounding monthly earns a little more than compounding quarterly or yearly, because interest starts earning interest sooner. A higher yearly rate can still beat more frequent compounding.

Year	Value at start	Interest (4%)	Value at end
\(1\)	\(\$2500.00\)	\(\$100.00\)	\(\$2600.00\)
\(2\)	\(\$2600.00\)	\(\$104.00\)	\(\$2704.00\)
\(3\)	\(\$2704.00\)	\(\$108.16\)	\(\$2812.16\)

\(FV\)	\(=\)	\(2500 \times 1.04^{3}\)
\(\)	\(=\)	\(\mathbf{\$2812.16}\)

\(r\)	\(=\)	\(0.048 \div 12 = 0.004 \text{ per month}\)
\(n\)	\(=\)	\(3 \times 12 = 36 \text{ months}\)

\(FV\)	\(=\)	\(8000 \times 1.004^{36}\)
\(\)	\(=\)	\(9236.419\ldots\)
\(\)	\(\approx\)	\(\mathbf{\$9236.42}\)
\(I\)	\(=\)	\(9236.42 - 8000 = \mathbf{\$1236.42}\)

\(r\)	\(=\)	\(0.032 \div 4 = 0.008\)
\(n\)	\(=\)	\(5 \times 4 = 20\)
\(15\,000\)	\(=\)	\(PV \times 1.008^{20}\)
\(PV\)	\(=\)	\(\dfrac{15\,000}{1.008^{20}}\)
\(\)	\(=\)	\(12\,790.296\ldots\)
\(\)	\(\approx\)	\(\mathbf{\$12\,790.30}\)

Compounded	\(r\)	\(n\) for \(t\) years
annually	yearly rate	\(t\)
quarterly	yearly rate \(\div 4\)	\(4t\)
monthly	yearly rate \(\div 12\)	\(12t\)

\(\text{simple } I\)	\(=\)	\(4000 \times 0.06 \times 5\)
\(\)	\(=\)	\(\$1200.00\)
\(\text{compound } FV\)	\(=\)	\(4000 \times 1.06^{5}\)
\(\)	\(=\)	\(5352.902\ldots\)
\(\text{compound } I\)	\(=\)	\(5352.90 - 4000\)
\(\)	\(=\)	\(\$1352.90\)
\(\text{difference}\)	\(=\)	\(\mathbf{\$152.90}\)

\(n = 6\)	\(:\)	\(\$5674.08 \ (\text{under})\)
\(n = 7\)	\(:\)	\(\$6014.52 \ (\text{over})\)

Compound Interest

📖 Theory

How to solve a compound interest problem

Other questions you will meet

Common pitfalls

Frequently asked questions

Theory