What is wrong ratio from a percentage?

A 12\% decay is R = 0.88, not 0.12. Growth of 40\% is R = 1.4, not 0.4.

§-Syllabus dot point

WAMathematics ApplicationsSyllabus dot point

How do we model quantities that grow or shrink by a constant ratio?

Recognise geometric growth and decay, use recurrence relations and the explicit rule for geometric sequences, and model compound and reducing situations.

How to model constant-ratio change with geometric sequences, switch between recurrence and explicit rules, and apply them to compound interest, depreciation and population change.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Constant-ratio change
Growth versus decay
Finding the ratio or starting value
Doubling time and half-life
Distinguishing the model in worded problems

What this dot point is asking

You must recognise constant-ratio change, use both forms of the rule, and apply them to compound, decaying and population situations.

Constant-ratio change

A quantity changes exponentially when it gains or loses a fixed percentage each period, so it multiplies by the same factor every step. This is exactly a geometric sequence.

The difference from arithmetic change is the multiplier: a fixed percentage means multiply, while a fixed amount means add (arithmetic). Spotting "percentage" versus "amount" in the wording selects the model.

Growth versus decay

Growth ( $R > 1$ ): the quantity rises ever faster, an upward curve. Compound interest, growing populations, spreading information.
Decay ( $0 < R < 1$ ): the quantity falls ever more slowly towards zero, never reaching it. Reducing-balance depreciation, cooling, radioactive decay, drug concentration.

Worked example

Population growth and a doubling time

A town of $15\,000$ grows by $2.5\%$ each year. Find the population after $12$ years and estimate how long it takes to grow to $20\,000$ .

Step 1: Identify the common ratio and write the explicit rule.

A growth rate of $2.5\%$ per year means the population is multiplied by the same factor every year, making this a geometric sequence. Adding $2.5\%$ keeps $100\% + 2.5\% = 102.5\%$ of the previous value, so the common ratio is $R = 1.025$ . The explicit rule is $P_n = 15000 \times 1.025^n$ , where $n$ is the number of years elapsed.

Step 2: Calculate the population after 12 years by substituting $n = 12$ .

Substituting directly into the explicit rule gives $P_{12} = 15000 \times 1.025^{12} = 15000 \times 1.34489 = 20173$ , so the model predicts about $20\,173$ people after $12$ years.

Step 3: Find how long it takes to reach $20\,000$ by testing successive values.

We need to solve $15000 \times 1.025^n = 20000$ , which rearranges to $1.025^n = 1.3333$ . Rather than using logarithms, we test successive whole-number values of $n$ : $1.025^{11} = 1.3121$ (below the target) and $1.025^{12} = 1.3449$ (above the target), so the population first reaches $20\,000$ during year $12$ .

Step 4: Interpret the result in context.

The model predicts the town reaches $20\,000$ residents in about $12$ years from now. Notice that exponential growth accelerates over time: the town adds more people in each successive year than in the previous one, even though the percentage rate stays constant at $2.5\%$ , because the base on which the percentage is calculated keeps growing.

Finding the ratio or starting value

When given two values, divide to find $R$ . If $A_0 = 500$ and $A_4 = 800$ , then $500 R^4 = 800$ , so $R^4 = 1.6$ and $R = \sqrt[4]{1.6} = 1.1247$ , an annual growth of about $12.5\%$ . When given a later value and the ratio, divide to recover the start.

Doubling time and half-life

Two standard questions ask how long a model takes to double (growth) or to halve (decay, the half-life). Solve $R^n = 2$ or $R^n = 0.5$ by trial. For $6\%$ annual growth, $1.06^n = 2$ gives $1.06^{11} = 1.898$ and $1.06^{12} = 2.012$ , so doubling takes about $12$ years. For a substance decaying at $9\%$ per year, $0.91^n = 0.5$ gives $0.91^7 = 0.517$ and $0.91^8 = 0.470$ , so the half-life is during year $8$ . Report the first whole period that crosses the target.

Distinguishing the model in worded problems

The decisive clue in a worded question is whether the change is a fixed amount or a fixed percentage. "Increases by $200$ each year" is a fixed amount, so it is arithmetic (linear, a straight line). "Increases by $4\%$ each year" is a fixed percentage, so it is geometric (exponential, a curve). Many SCSA questions deliberately offer both kinds in one scenario and ask you to compare the two models at a particular time, where the geometric model eventually overtakes the arithmetic one because it compounds on a growing base.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20216 marksA bacterial culture starts at

200

cells and increases by

40\%

each hour. (a) Write the explicit rule for the number

N_n

after

n

hours. (b) Find the number after

6

hours. (c) In how many whole hours does the culture first exceed

5000

cells?

Show worked answer →

A $40\%$ hourly increase multiplies by $1.4$ each hour.

(a) Common ratio $R = 1.40$ , so $N_n = 200 \times 1.4^n$ . (2 marks)

(b) $N_6 = 200 \times 1.4^6 = 200 \times 7.5295 = 1505.9$ , so about $1506$ cells. (2 marks)

(c) Solve $200 \times 1.4^n > 5000$ , so $1.4^n > 25$ . Testing, $1.4^9 = 20.66$ and $1.4^{10} = 28.93$ , so the culture first exceeds $5000$ in hour $10$ . (2 marks)

Markers reward the ratio $1.4$ , the explicit substitution, and the first whole hour above the threshold.

WACE 20235 marksA radioactive sample of

80

mg decays by

12\%

each year. (a) State the common ratio and write the recurrence relation. (b) Find the mass remaining after

10

years. (c) Explain why the mass never reaches exactly zero.

Show worked answer →

A $12\%$ loss keeps $88\%$ each year.

(a) Common ratio $R = 1 - 0.12 = 0.88$ , so $M_{n+1} = 0.88 M_n$ , with $M_0 = 80$ . (2 marks)

(b) $M_n = 80 \times 0.88^n$ , so $M_{10} = 80 \times 0.88^{10} = 80 \times 0.27851 = 22.28$ mg. (2 marks)

(c) Because the mass is multiplied by $0.88$ (a positive ratio less than $1$ ) each year, it always remains positive and shrinks towards zero without reaching it; geometric decay approaches zero as a limit. (1 mark)

Markers reward the decay ratio $0.88$ , the substitution, and the asymptotic-to-zero reasoning.