Compound growth, depreciation and half-life | Year 11

01 / LEARN

The idea, step by step

Convert the rate and count the periods

An annual nominal interest rate r compounded n times per year gives a per-period multiplier $1 + r / n$ . Over t years there are nt periods. This model assumes a fixed rate, no extra deposits or withdrawals, and no fees.

The effective annual rate is the actual one-year percentage change. It is generally not the nominal rate when compounding occurs more than once per year.

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

Half-life and depreciation are multiplier models

A half-life h means half the amount remains after every h time units: $M (t) = M_{0} {(1 / 2)}^{t / h}$ . A fractional number of half-lives is allowed when the model treats time as continuous.

Depreciation by a fixed percentage uses the remaining-value factor, not a fixed dollar subtraction. State whether the question needs a real-valued time or the first completed year/month satisfying a threshold.

A common trap

Using an annual percentage as a monthly percentage.

Instead: Divide the nominal annual rate by 12 and use the number of months, or use a genuinely supplied monthly rate.

02 / WORKED EXAMPLES

See the thinking

Example 1

Quarterly compounding

8000 dollars is invested at 6% per annum, compounded quarterly. Find its value after 2 years, ignoring fees.

The quarterly rate is 0.06/4=0.015 and there are 4(2)=8 periods.
$V = 8000 {(1.015)}^{8} \approx 9011.94$ .

9011.94 dollars.

Example 2

A fractional number of half-lives

A 240 mg sample has an 8-hour half-life. Find its mass after 20 hours.

$M (t) = 240 {(1 / 2)}^{t / 8}$ .
$M (20) = 240 {(1 / 2)}^{2.5} \approx 42.43$ mg.

Approximately 42.43 mg.

Check your understanding

What fraction remains after three half-lives?

Show answer

$1 / 8$

Each half-life multiplies by 1/2, so the factor is ${(1 / 2)}^{3}$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find 5000 dollars compounded annually at 4% for 3 years.
2 marks
A 20,000 dollar asset loses 15% yearly. Find its value after 3 years.
2 marks
Write a model for a 96 g sample with a 5-day half-life.
2 marks
For the previous sample, find the mass after 15 days.
2 marks
An account earns 1% each month. Find its effective annual percentage rate.
3 marks
A saver adds 100 dollars at the end of every month. Explain why P times b to the number of months does not describe the whole account balance.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find 5000 dollars compounded annually at 4% for 3 years.
2 marks
$5000 {(1.04)}^{3} = 5624.32$ .
5624.32 dollars.
A 20,000 dollar asset loses 15% yearly. Find its value after 3 years.
2 marks
$20000 {(0.85)}^{3} = 12282.50$ .
12,282.50 dollars.
Write a model for a 96 g sample with a 5-day half-life.
2 marks
The exponent is the number of five-day half-life periods.
$M (t) = 96 {(1 / 2)}^{t / 5}$ , t in days.
For the previous sample, find the mass after 15 days.
2 marks
$96 {(1 / 2)}^{3} = 12$ .
12 g.
An account earns 1% each month. Find its effective annual percentage rate.
3 marks
One year multiplies by ${1.01}^{12}$ .
The effective rate is $100 ({1.01}^{12} - 1) \approx 12.68 %$ .
Approximately 12.68%.
A saver adds 100 dollars at the end of every month. Explain why P times b to the number of months does not describe the whole account balance.
3 marks
Each deposit earns interest for a different number of periods. The balance requires extra terms or a recurrence, not one initial lump-sum exponential.
It omits the later deposits.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: exponential and logarithmic functions, growth models and numerical/algebraic prerequisites.

Official curriculum source ↗

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