Year 10 Maths Advanced Earning Money

Compound Interest

Q: What is the compound interest formula?

A equals P times (1 plus r) to the power n, where P is the principal, r is the interest rate per compounding period as a decimal and n is the number of periods. The interest earned is A minus P.

Q: How do you calculate compound interest compounded monthly?

Divide the annual rate by 12 and multiply the number of years by 12. For example, 3.6 percent p.a. for 3 years becomes 0.3 percent, or 0.003, for 36 months.

Q: How do you find the interest rate in compound interest?

Divide the final value by the principal, take the nth root, and subtract 1. That gives the rate per period. If the periods are months, multiply by 12 for the annual rate.

Q: How long does it take money to double with compound interest?

Solve (1 plus r) to the power n equals 2 by testing whole numbers of periods. At 8 percent p.a. compounded annually it takes 10 years, because 1.08 to the power 9 is just under 2 and 1.08 to the power 10 is just over.

20 practice questions 1 video lesson Theory + worked examples

Compound interest is interest earned on the interest as well as on the original amount, so an investment grows faster each period. It uses $FV=PV(1+r)^n$, with $r$ the rate per compounding period and $n$ the number of periods. In NSW Year 10 Advanced maths interest is compounded yearly, six-monthly, quarterly, monthly or daily (NESA outcome MA5‑FIN‑C‑02).

You will find future values and interest, work backwards to the present value, compare compounding periods, and compare compound with simple interest. You will find the rate that doubles money in a given time, the least number of periods needed to pass a target, and the effective yearly rate of two bank offers. Problems include saving for a car while prices rise, adding money part-way and a change of rate. It is part of Year 10 Advanced Earning Money.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 1 video(s)

Compound Interest - Video - Compound Interest Watch

Theory

Compound interest is interest paid on the principal and on the interest already earned, so the balance grows by a percentage each period. This Year 10 Advanced maths page covers the formula $A = P(1+r)^n$, monthly, quarterly and daily compounding, working backwards to the principal or the rate, and comparing compound with simple interest.

The principal $P$ is the amount first invested or borrowed. With simple interest the interest is the same every year, $I = Prn$, because it is always worked out on the principal.

With compound interest the interest is added to the balance at the end of each compounding period, and the next lot of interest is worked out on the new, larger balance. Interest earns interest.

The final value (or future value) $A$ is the balance at the end. The formula is also written as $FV = PV(1+r)^n$: $FV$ is the same as $A$, and the present value $PV$ is the principal $P$. The interest earned is the final value minus the principal.

Rates are quoted per annum (p.a.). If interest is compounded more often than once a year, divide the annual rate by the number of periods in a year, and count the time in those periods.

Simple interest adds the same amount each year. Compound interest also earns interest on past interest, so the gap keeps widening.

Convert the annual rate and the time into compounding periods before you use the formula.

For a principal $P$, a rate $r$ per period (as a decimal) and $n$ periods:

\[A = P(1+r)^n \qquad I = A - P\]

A = P {(1 + r)}^{n}, I = A - P

Rearranged to find the principal or the rate:

\[P = \dfrac{A}{(1+r)^n} \qquad r = \sqrt[n]{\dfrac{A}{P}} - 1\]

P = \frac{A}{{(1 + r)}^{n}}, r = \sqrt[n]{\frac{A}{P}} - 1

Converting the rate and the time:

Compounded	Periods per year	Rate per period	$n$ for $t$ years
annually	$1$	annual rate	$t$
six-monthly	$2$	annual rate $\div 2$	$2t$
quarterly	$4$	annual rate $\div 4$	$4t$
monthly	$12$	annual rate $\div 12$	$12t$
daily	$365$	annual rate $\div 365$	$365t$

Simple interest for comparison. \(I = Prn\) with \(r\) per year and \(n\) in years, so the final value is \(P(1 + rn)\). Set this equal to \(P(1+r)^n\) and the \(P\) cancels, so an equivalent rate does not depend on the amount invested.

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.

How to solve a compound interest problem

Convert the annual rate to a rate per period, as a decimal, and the time to a number of periods.
Write $A = P(1+r)^n$ and substitute what you know.
Solve for the unknown: multiply out for $A$, divide for $P$, take the $n$th root for $r$, or test whole numbers for $n$.
Answer the question asked: final value, interest ($A - P$), or an annual rate (multiply a monthly rate by $12$).
Round money to the nearest cent at the end, and check the size makes sense.

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).

Example 1 — Final value and interest

$\$9500$ is invested at $3.6\%$ p.a., compounded monthly, for $3$ years. Find, to the nearest cent:
(i) the final value
(ii) the interest earned.

Solution

Change the rate and the time into months first.

$r$	$=$	$\dfrac{3.6\%}{12} = 0.3\% = 0.003$
$n$	$=$	$3 \times 12 = 36$

(i)

$A$	$=$	$P(1+r)^n$
	$=$	$9500(1.003)^{36}$
	$=$	$10\,581.7426\ldots$
	$\approx$	$\mathbf{\$10\,581.74}$

A = 9500 {(1.003)}^{36} \approx 10581.74

(ii)

$I$	$=$	$A - P$
	$=$	$\$10\,581.74 - \$9500$
	$=$	$\mathbf{\$1081.74}$

I = A - P = 1081.74

The formula gives the final value, not the interest. Subtract the principal to get the interest.

Example 2 — Working backwards

Priya wants $\$20\,000$ in $4$ years for a car.
(i) How much must she invest now at $4.8\%$ p.a., compounded quarterly?
(ii) She has only $\$16\,000$. What annual interest rate, compounded annually, would grow this to $\$20\,000$ in $4$ years? Give the rate to $2$ decimal places.

Solution

(i) Here $A$ is known and $P$ is not. Use $r = 4.8\% \div 4 = 0.012$ and $n = 16$ quarters, then divide.

$P(1.012)^{16}$	$=$	$20\,000$
$P$	$=$	$\dfrac{20\,000}{1.012^{16}}$
	$=$	$16\,525.0124\ldots$
	$\approx$	$\mathbf{\$16\,525.01}$

P = \frac{20000}{{1.012}^{16}} \approx 16525.01

(ii) Now the rate is unknown. Divide both sides by $16\,000$, then take the fourth root.

$16\,000(1+r)^4$	$=$	$20\,000$
$(1+r)^4$	$=$	$1.25$
$1+r$	$=$	$\sqrt[4]{1.25}$
	$=$	$1.057371\ldots$
$r$	$\approx$	$0.0574 = \mathbf{5.74\%}\text{ p.a.}$

1 + r = \sqrt[4]{1.25}, r \approx 5.74 %

On the calculator, $\sqrt[4]{1.25}$ is $1.25^{(1 \div 4)}$. Keep the brackets round $1 \div 4$.

Example 3 — How long until it passes a target?

$\$7000$ is invested at $4.4\%$ p.a., compounded six-monthly. Find the least number of whole half-years until the investment is worth more than $\$10\,000$. Give the time in years and months.

Solution

The rate is $4.4\% \div 2 = 2.2\%$ per half-year. The time $n$ is unknown, so test whole numbers of periods.

$7000(1.022)^n$	$\gt$	$10\,000$
$n = 16:\ \ 7000(1.022)^{16}$	$\approx$	$\$9915.45$
$n = 17:\ \ 7000(1.022)^{17}$	$\approx$	$\$10\,133.59$

7000 {(1.022)}^{17} > 10000

After $16$ half-years it is still under $\$10\,000$; after $17$ it is over. So the answer is $\mathbf{17}$ half-years, which is $\mathbf{8}$ years $\mathbf{6}$ months.

A good first guess saves time: $10\,000 \div 7000 \approx 1.43$, so find the power of $1.022$ that first passes $1.43$.

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. At $4.8\%$ p.a. compounded quarterly, use $0.012$ for $16$ quarters in $4$ years. Using $0.048$ with $16$, or $0.012$ with $4$, are both common mistakes.

The formula gives the final value. If the question asks for the interest, subtract the principal. If it gives the interest, add the principal back or use $P\big((1+r)^n - 1\big) = I$.

Round once, at the end. Rounding $1.003^{36}$ to $1.11$ before multiplying gives $\$10\,545$, more than $\$30$ too low. Keep the full calculator value.

Watch out for a monthly rate given as the answer. If you solve for the rate per month, multiply by $12$ to give the annual rate the question asks for.

Least number of periods means whole periods. Interest is only added at the end of a period, so round the number of periods up and show the values either side of the target.

Frequently asked questions

What is the compound interest formula?

$A = P(1+r)^n$, where $P$ is the principal, $r$ is the rate per compounding period as a decimal and $n$ is the number of periods. The interest earned is $A - P$.

How do you calculate compound interest compounded monthly?

Divide the annual rate by $12$ and multiply the number of years by $12$. For example, $3.6\%$ p.a. for $3$ years becomes $0.003$ for $36$ months.

What is the difference between simple and compound interest?

Simple interest is always worked out on the original principal, so it is the same amount every year. Compound interest is worked out on the current balance, which includes earlier interest, so it grows by more each year.

How do you find the interest rate in compound interest?

Divide the final value by the principal, take the $n$th root, and subtract $1$: $r = \sqrt[n]{\dfrac{A}{P}} - 1$. That gives the rate per period. If the periods are months, multiply by $12$.

Is it better to compound monthly or annually?

For an investment at the same stated annual rate, monthly compounding gives more, because interest is added sooner and starts earning interest itself. For a loan it costs you more.

How long does it take money to double with compound interest?

Solve $(1+r)^n = 2$ by testing whole numbers of periods. At $8\%$ p.a. compounded annually it takes $10$ years, because $1.08^9 \approx 1.999$ is just under $2$ and $1.08^{10} \approx 2.159$ is over.

Compounded	Periods per year	Rate per period	\(n\) for \(t\) years
annually	\(1\)	annual rate	\(t\)
six-monthly	\(2\)	annual rate \(\div 2\)	\(2t\)
quarterly	\(4\)	annual rate \(\div 4\)	\(4t\)
monthly	\(12\)	annual rate \(\div 12\)	\(12t\)
daily	\(365\)	annual rate \(\div 365\)	\(365t\)

\(r\)	\(=\)	\(\dfrac{3.6\%}{12} = 0.3\% = 0.003\)
\(n\)	\(=\)	\(3 \times 12 = 36\)

\(A\)	\(=\)	\(P(1+r)^n\)
\(\)	\(=\)	\(9500(1.003)^{36}\)
\(\)	\(=\)	\(10\,581.7426\ldots\)
\(\)	\(\approx\)	\(\mathbf{\$10\,581.74}\)

\(I\)	\(=\)	\(A - P\)
\(\)	\(=\)	\(\$10\,581.74 - \$9500\)
\(\)	\(=\)	\(\mathbf{\$1081.74}\)

\(P(1.012)^{16}\)	\(=\)	\(20\,000\)
\(P\)	\(=\)	\(\dfrac{20\,000}{1.012^{16}}\)
\(\)	\(=\)	\(16\,525.0124\ldots\)
\(\)	\(\approx\)	\(\mathbf{\$16\,525.01}\)

\(16\,000(1+r)^4\)	\(=\)	\(20\,000\)
\((1+r)^4\)	\(=\)	\(1.25\)
\(1+r\)	\(=\)	\(\sqrt[4]{1.25}\)
\(\)	\(=\)	\(1.057371\ldots\)
\(r\)	\(\approx\)	\(0.0574 = \mathbf{5.74\%}\text{ p.a.}\)

\(7000(1.022)^n\)	\(\gt\)	\(10\,000\)
\(n = 16:\ \ 7000(1.022)^{16}\)	\(\approx\)	\(\$9915.45\)
\(n = 17:\ \ 7000(1.022)^{17}\)	\(\approx\)	\(\$10\,133.59\)

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

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

Compound Interest

📖 Theory

How to solve a compound interest problem

Problems that need an equation

Common pitfalls

Frequently asked questions

Theory