Year 10 Maths - 10A Indices and Surds

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.

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Theory

With compound interest, each period's interest is added to the balance, so the next period earns interest on it too. This Year 10A maths page uses $FV = PV(1 + r)^{n}$ with different compounding periods to find final values, interest, present values, rates and times, compares rates with different compounding, and handles investments in stages.

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 $\div 12$	$12t$
six-monthly	yearly rate $\div 2$	$2t$
daily	yearly rate $\div 365$	$365t$

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

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 \((1 + r)^{n}\).

Comparing rates with different compounding. Work out what \(\$1\) grows to in one year. At \(6\%\) p.a. compounded monthly, \(\left(1 + \dfrac{0.06}{12}\right)^{12} \approx 1.0617\), so it really earns about \(6.17\%\) a year. More frequent compounding means a lower stated rate can reach the same amount.

Other letters. Some books write the formula as \(A = P(1 + r)^n\): \(A\) is the final value \(FV\) and \(P\) is the principal \(PV\).

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.

Problems that need an equation

Interest given, principal unknown: $P(1+r)^n - P = I$, so $P\big((1+r)^n - 1\big) = I$. Factor out $P$, then divide.
Money split between two accounts: call one part $x$ and the other $\text{total} - x$. Grow each part, add, and solve the linear equation.
Doubling: set $A = 2P$. The $P$ cancels, leaving $(1+r)^n = 2$.
Prices that rise each year (inflation) grow exactly like compound interest, using the growth rate as $r$.
When does compound overtake simple? Write both final values in terms of $n$ and test whole numbers until the compound value is bigger.
Several deposits or a rate change: split the time into stages (Example 4 below).

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 — Two stages

Tom invests $\$8000$ at $5\%$ p.a., compounded annually. After $3$ years he withdraws $\$2000$. The rest is moved to an account paying $5.4\%$ p.a., compounded monthly, for $2$ more years. Find the final value, to the nearest cent.

Solution

Treat each stage as its own compound interest problem. The end of stage one, less the withdrawal, is the principal for stage two.

$\text{After 3 years}$	$=$	$8000(1.05)^3$
	$=$	$\$9261.00$
$\text{New principal}$	$=$	$\$9261.00 - \$2000$
	$=$	$\$7261.00$

Stage two: $r = 5.4\% \div 12 = 0.45\% = 0.0045$ and $n = 24$ months.

$A$	$=$	$7261.00(1.0045)^{24}$
	$=$	$8087.1411\ldots$
	$\approx$	$\mathbf{\$8087.14}$

A = 7261.00 {(1.0045)}^{24} \approx 8087.14

A deposit added part-way works the same way in reverse: grow each deposit for its own number of periods, then add the values.

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.

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Comparing Simple and Compound Interest

Compounded	\(r\)	\(n\) for \(t\) years
annually	yearly rate	\(t\)
quarterly	yearly rate \(\div 4\)	\(4t\)
monthly	yearly rate \(\div 12\)	\(12t\)
six-monthly	yearly rate \(\div 2\)	\(2t\)
daily	yearly rate \(\div 365\)	\(365t\)

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

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

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

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

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

\(\text{After 3 years}\)	\(=\)	\(8000(1.05)^3\)
\(\)	\(=\)	\(\$9261.00\)
\(\text{New principal}\)	\(=\)	\(\$9261.00 - \$2000\)
\(\)	\(=\)	\(\$7261.00\)

\(A\)	\(=\)	\(7261.00(1.0045)^{24}\)
\(\)	\(=\)	\(8087.1411\ldots\)
\(\)	\(\approx\)	\(\mathbf{\$8087.14}\)

Compound Interest

📖 Theory

How to solve a compound interest problem

Other questions you will meet

Problems that need an equation

Common pitfalls

Frequently asked questions

Theory