Compound Growth and Decay Revision Notes for AQA GCSE Maths

Compound growth and decay

Understanding the basics

Compound growth and decay is different from simple interest because the percentage change is calculated on the new amount each time, rather than always using the original amount. This creates an exponential pattern where changes become more dramatic over time.

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The key to mastering this topic is learning and understanding the main formula. Once you know this formula, compound growth and decay problems become much more manageable.

The essential formula

The formula for compound growth and decay is:

$N = N\_0 \times (\text{multiplier})^n$

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Variable definitions:

N = the final amount after n time periods
N₀ = the initial amount (starting value)
multiplier = the percentage change converted to a multiplier
n = the number of time periods (days, months, years, etc.)

Working out the multiplier

The multiplier is calculated differently depending on whether you have growth or decay:

For increases: multiplier = 1 + percentage change (as a decimal)
For decreases: multiplier = 1 - percentage change (as a decimal)

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Worked Example: Calculating Multipliers

For a 5% increase:

multiplier = 1 + 0.05 = 1.05

For a 26% decrease:

multiplier = 1 - 0.26 = 0.74

Compound interest problems

Compound interest is a common application where money grows because interest is added to the account each time period, and then the next interest calculation is based on this new higher amount.

When you see "per annum" in a question, this simply means "each year".

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Worked Example: Compound Interest Calculation

Daniel invests £1000 at 8% compound interest per annum for 6 years:

Step 1: Identify the values

Initial amount (N₀) = £1000
Percentage increase = 8% = 0.08
Multiplier = 1 + 0.08 = 1.08
Time period (n) = 6 years

Step 2: Apply the formula $\text{Amount} = 1000 \times (1.08)^6 = £1586.87$

Depreciation problems

Depreciation questions involve items that decrease in value over time, such as cars, machinery, or technology. The same formula applies, but now you're dealing with a multiplier less than 1.

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Worked Example: Car Depreciation

Susan's car depreciates by 9% each year. If the car cost £6500 originally, what's its value after 3 years?

Step 1: Understand the depreciation

The car loses 9% of its value each year
This means it retains 91% of its value each year
Multiplier = 1 - 0.09 = 0.91

Step 2: Apply the formula $\text{Value} = 6500 \times (0.91)^3 = £4898.21$

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Sometimes you might need to work backwards or use trial and error to find how many years it takes for something to reach a certain value.

Population and disease applications

The compound growth and decay formula can also describe population changes, bacterial growth, or disease spread.

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Worked Example: Bacterial Growth (Working Backwards)

Bacteria in a sample increases by 30% each day. After 6 days there are 7500 bacteria. Find the original number.

Step 1: Identify the known values

Final amount = 7500 bacteria
Growth rate = 30% per day
Multiplier = 1 + 0.30 = 1.3
Time period = 6 days

Step 2: Set up the equation $7500 = N\_0 \times (1.3)^6$

Step 3: Solve for N₀ $N\_0 = 7500 \div (1.3)^6 = 1554 \text{ bacteria originally}$

Problem-solving strategies

When tackling compound growth and decay problems, follow these systematic steps:

Identify the type: Is it growth (multiplier > 1) or decay (multiplier < 1)?
Find the multiplier: Add or subtract the percentage from 1
Identify all the variables: What's the initial amount, time period, and what are you solving for?
Apply the formula: Substitute your values into $N = N\_0 \times (\text{multiplier})^n$
Check your answer: Does it make sense given whether it's growth or decay?

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For questions where you need to find the time period, you might need to use trial and error, testing different values of n until you get close to the target value.

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Key Points to Remember:

The compound growth and decay formula is $N = N\_0 \times (\text{multiplier})^n$ - learn this formula thoroughly
For increases: multiplier = 1 + percentage (as decimal), for decreases: multiplier = 1 - percentage (as decimal)
"Per annum" means "each year" - a common phrase in compound interest problems
The same formula works for money (compound interest), depreciation, and population problems
When working backwards or finding time periods, trial and error with different values of n can be helpful

Compound Growth and Decay (AQA GCSE Maths): Revision Notes