Ratio, Proportion & Rates of Change
Grade 7-8Growth & Decay
Growth and decay problems apply the same percentage change repeatedly over a number of time periods, using a single multiplier raised to a power rather than repeating the calculation step by step. This lesson covers using a growth or decay multiplier, finding the percentage rate from a before-and-after value, and solving problems where a value needs to cross a target threshold.
Written by Asad, Co-Founder of Teachably
Video walkthrough coming soon
The written lesson below covers everything you need in the meantime.
Using a growth or decay multiplier
Write a percentage increase as a multiplier above 1, and a percentage decrease as a multiplier below 1, then raise it to the power of the number of time periods and multiply by the original value.
A population of bacteria starts at 200 and grows by 8% every hour. After 5 hours, the population is 200 × 1.08⁵ ≈ 294.
Applying decay in the same way
A percentage decrease of r% is written as a multiplier of (1 − r/100), and is used in exactly the same way as a growth multiplier, raised to the power of the number of time periods.
A radioactive substance has a mass of 80 g and decays by 12% per year. After 4 years, its mass is 80 × 0.88⁴ ≈ 48.0 g.
Finding the percentage rate from a before-and-after value
Divide the final value by the original value to find the overall multiplier, then take the appropriate root to find the multiplier per time period, and convert this to a percentage.
A population grows from 500 to 605 over 2 years. The overall multiplier is 605 ÷ 500 = 1.21, and the square root of 1.21 is 1.1, so the annual growth rate is 10%.
Solving threshold problems
To find when a value first exceeds or falls below a target, test successive whole-number powers of the multiplier until the value crosses the threshold, then state the number of complete time periods needed.
A population of 1000 bacteria grows at 10% per hour. After 7 hours it is 1000 × 1.1⁷ ≈ 1948.72, which is still below 2000, but after 8 hours it is 1000 × 1.1⁸ ≈ 2143.59, which exceeds 2000, so it takes 8 hours.
Worked Examples
Three exam-style questions, fully solved.
A population of bacteria starts at 200 and grows by 8% every hour. Find the population after 5 hours, giving your answer to the nearest whole number.
Easy- 1.Write the growth multiplier: 1.08
- 2.Raise it to the power of the number of hours and multiply by the starting value: 200 × 1.08⁵
Answer: 294 bacteria
A population grows from 500 to 605 over 2 years at a constant annual percentage growth rate. Find the annual growth rate.
Medium- 1.Find the overall multiplier: 605 ÷ 500 = 1.21
- 2.Find the annual multiplier by taking the square root: the square root of 1.21 is 1.1
Answer: 10%
A population of 500 grows at a rate of 20% per year. Find the number of complete years until the population first exceeds 1000.
Hard- 1.Test 3 years: 500 × 1.2³ = 864.00, which is too small
- 2.Test 4 years: 500 × 1.2⁴ = 1036.80, which exceeds 1000
Answer: 4 years
Avoid These
The most common mistakes students make.
Using the wrong multiplier, such as using 1.08 for an 8% decrease instead of 0.92, or the reverse for a growth problem.
Multiplying the original value by n lots of the percentage instead of raising the multiplier to the power of n, which confuses repeated percentage change with a fixed one-off increase.
When finding the growth or decay rate from a before-and-after value, forgetting to take the correct root for the number of time periods involved.
In "first exceeds" or "first falls below" threshold problems, testing one power too many or too few, or misreading which result is too small and which has crossed the target.
When a problem combines more than one growth or decay phase, applying the multipliers in the wrong order or over the wrong number of time periods.
FAQ
Questions parents and students ask.
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