Growth/decay models | Year 11

01 / LEARN

Understand the idea.

Choose the correct factor

Turn the percentage rate into a decimal before constructing a factor. Growth at rate $r$ uses $1 + r$ ; a decrease of rate $r$ , with $0 < r < 1$ , uses $1 - r$ . Raise that factor to the number of equal periods.

A compound model assumes the rate continues and no other changes intervene. Deposits, withdrawals, changing conditions or a capacity limit can invalidate the prediction. State the time unit so that the exponent matches the percentage period.

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

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

Recover a starting value or time

To work backwards, divide by the whole compound factor. A later amount $P$ after $n$ periods gives starting amount $A = \frac{P}{b^{n}}$ . Subtracting the percentage of the later amount does not reverse growth.

For an unknown number of periods, use a table, graph or logarithms, then interpret the time restriction. Keep calculator precision until the final result; money usually needs cents, while a modelled population may need an approximate whole-number report.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Repeated growth

A culture starts at 500 units and increases 15% per hour. Predict its size after 3 h.

Worked solution

1
Build the hourly factor.
$b = 1 + 0.15 = 1.15$
Fifteen-percent growth retains the whole current amount and adds fifteen percent.
2
Write the time model.
$P (t) = 500 {(1.15)}^{t}$
The culture starts at five hundred units and compounds once for each hour.
3
Evaluate three hours.
$P (3) = 500 {(1.15)}^{3} = 760.4375$
Three hours requires three successive multiplications.
4
Report a sensible estimate.
$P (3) \approx 760 units$
The model prediction is rounded to a whole unit only at the end.

Answer

About 760 units.

EXAMPLE 02

Asset depreciation

An asset worth 18000 dollars loses 12% per year. Find its value after two years.

Worked solution

1
Find the fraction retained each year.
$b = 1 - 0.12 = 0.88$
A twelve-percent loss leaves eighty-eight percent of the previous value.
2
Compound for two years.
$V (2) = 18000 {(0.88)}^{2}$
Both annual losses apply to the current depreciated value.
3
Evaluate and express cents.
$V (2) = 13939.20 dollars$
The computed value is thirteen thousand nine hundred and thirty-nine dollars and twenty cents.

Answer

13939.20 dollars.

03 / YOUR TURN

Try it yourself.

Check your understanding

An amount becomes 1250 after one year of 25% growth. Find its original value.

Show answer & reasoning

1000.

One year multiplies the original value by $1.25$ . Reverse that multiplication: $A = \frac{1250}{1.25} = 1000$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Write a model for 600 increasing by 8% per year.
2 marks
A machine worth 20000 dollars depreciates 10% annually. Find its value after 3 years.
2 marks
A quantity reaches 864 after two steps of 20% growth. Find its start.
2 marks
A quantity starts at 40 and doubles every 3 h. Find it after 9 h.
2 marks
A quantity changes from 500 to 320 over two equal steps. Find its positive step factor and percentage decrease.
3 marks
Use a scientific calculator to find when $100 {(1.1)}^{t}$ reaches 150. Give continuous time and first whole interval.
3 marks

Read the worked solutions

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

Write a model for 600 increasing by 8% per year.
2 marks
Convert 8% to 0.08; use factor 1.08.
$P = 600 {(1.08)}^{n}$ .
A machine worth 20000 dollars depreciates 10% annually. Find its value after 3 years.
2 marks
$20000 {(0.9)}^{3} = 14580$ .
14580 dollars.
A quantity reaches 864 after two steps of 20% growth. Find its start.
2 marks
$A = \frac{864}{{(1.2)}^{2}} = \frac{864}{1.44} = 600$ .
600.
A quantity starts at 40 and doubles every 3 h. Find it after 9 h.
2 marks
9 h contains three doubling intervals.
$40 {(2)}^{3} = 320$ .
320.
A quantity changes from 500 to 320 over two equal steps. Find its positive step factor and percentage decrease.
3 marks
$b^{2} = \frac{320}{500} = 0.64$ .
Thus $b = 0.8$ .
Factor 0.8; decrease 20%.
Use a scientific calculator to find when $100 {(1.1)}^{t}$ reaches 150. Give continuous time and first whole interval.
3 marks
$t = \frac{log (1.5)}{log (1.1)} \approx 4.2542$ .
The first integer at or above this time is 5.
About 4.25 intervals; first whole interval 5.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: algebraic and graphical modelling of linear/exponential change, with interpretation and evaluation.

Official curriculum source ↗

In this lesson

Understand the idea.

Choose the correct factor

Recover a starting value or time

See the method in action.

Repeated growth

Build the hourly factor.

Write the time model.

Evaluate three hours.

Report a sensible estimate.

Asset depreciation

Find the fraction retained each year.

Compound for two years.

Evaluate and express cents.

Try it yourself.

Check your understanding

Keep practising