**Compound Growth and Decay** *GCSE Revision and Worksheets*

## What you need to know

The most typical example of **compound growth **is **compound interest**. Interest is a sum of money that gets added on to a bigger sum of money (e.g., the balance of your bank account), typically once a year. The amount of interest added is determined by a percentage value of the bigger sum.

Suppose you have a bank account containing £100. Compound interest works like this: each year, the bank calculates the interest for you based on a percentage interest rate on your account, say 3%, and adds it your total balance. In this case, 3% of 100 is 3, so your £100 would become £103. Then, at the end of next year, it adds another 3%, but now that 3% amounts to more, because we’re taking 3% of £103 – in other words, we’re getting interest on last year’s interest, which is precisely what compound interest is. So, at the end of the two years your money increases by £3.09 rather than just £3, meaning you’d have £106.09 in total.

There is an alternative called **simple interest**. You don’t really see this so much in real-life scenarios, but you might be asked a question about it regardless and fortunately for you, as the name states, it’s not so difficult. Simple interest works like this: you don’t get any interest on your interest, only interest on the original balance. In the above example, if it were a 3% *simple* interest rate (as opposed to a *compound* interest rate), then the interest you would earn each year would remain at £3 and not increase. So, after 2 years you would have £106 rather than £106.09. Hardly a tragic loss, but when numbers get bigger and many years pass, these things can make a big difference.

You may have noticed by now that this is going to require a good understanding of how to use percentages, so now is a good a time as any to brush up on them here (https://mathsmadeeasy.co.uk/gcse-maths-revision/percentages-gcse-revision-and-worksheets/). We’ll see in this next example how using percentage multipliers can save us a lot of time.

**Example: **James deposits £24,000 into a savings account. Work out the difference between his bank balance after 6 years with an annual simple interest rate of 4% and an annual compound interest rate of 4%.

Firstly, we’ll sort out the simple interest case.

4\%\text{ of } 24,000 = 0.04 \times 24,000 = \pounds 960

This is an annual rate, and we’re looking for the total after 6 years, so we need to add six lots of this interest onto his total to get our answer (in the simple interest case).

\text{Total after 6 years } = 24,000 + (6 \times 960) = \pounds 29,760

Next up, the compound interest case. In this case, because the interest is included in the calculation of new interest each year, and we’re looking at 6 years in total, there could be a lot of working out involved. Fortunately, there’s a better way! Roll on percentage multipliers.

The percentage multiplier for a 4% increase is 1 + \dfrac{4}{100} = 1.04. What this means is that in order to increase a number by 4%, all you need to do is multiply that number by 1.04. That reduces the amount of work we have to do down to multiplying by 1.04 six times, which can be reduced even further by considering that

24,000 \times 1.04 \times 1.04 \times 1.04 \times 1.04 \times 1.04 \times 1.04 = 24,000 \times 1.04^6

Now, all we need to do to work out the result of 6 years of 4% compound interest on a balance of £24,000 is calculate the value of 24,000 \times 1.04^6. We get:

\text{Total after 6 years } = 24,000 \times 1.04^6 = \pounds 30,367.66

That means, the difference between the two totals is

30,367.66 - 29,760 = \pounds 607.66

I told you it could make a big difference.

In practice, all you need to do is times your total by the appropriate multiplier (here, 1.04) raised to a power equal to the number of times the interest is being added (here, 6 times). It’s almost always annually, but there’s no reason they might try and trick you by asking a question about interested being added every 6 months.

**Compound decay **is very much the same thing, only the numbers go down. This means that your multiplier should be a number below 1 rather than above 1. Other than that, we figure out our answer in exactly the same way.

**Example: **Aza buys a car for £17,000. The value of this car will experience compound decay at a rate of 25% per year. Work out the value of the car after 8 years.

The percentage multiplier for a 25% decrease is 1 - \dfrac{25}{100} = 0.75.

We will multiply the starting value by 0.75 raised to the power of 8, since the car will be undergoing a 25% decrease 8 times.

\text{Value after 8 years } = 17,000 \times 0.75^8 = \pounds 1,701.92

Questions on this topic are worth a lot of marks but if you’re smart about answering them, they can be sorted out in no time at all, so get practising.

## Example Questions

1) Sun wins the lottery and chooses to deposit $1,400,000 into a savings account which offers 2.4% annual compound interest. How much money will she have in this account after 4 years?

The percentage multiplier for a 2.4% increase is 1 + \dfrac{2.4}{100} = 1.024.

So, we will multiply the starting amount by 1.024 raised to the power of 4 to account for the 4 years of compound interest gained. This will give us our final answer.

1,400,000 \times 1.024^4 = $ 1,539,316.28

2) The population of an endangered species of tiger is currently 234 and is predicted to decrease at a rate of 18% per year. Riley thinks there will be less than 100 tigers of this species left in 5-years-time. Assuming the predicted decrease will happen, is she correct?

The percentage multiplier for an 18% decrease is 1 - \dfrac{18}{100} = 0.82.

So, we will multiply 234 by 0.82 to the power of 5, to check if the population will fall below 100 after 5 years.

234 \times 0.82^5 = 87 \text{ tigers, to the nearest whole number}

This is less than 100, therefore Riley is correct.

3) (TRICKY) Hernando buys a house worth £268,000, and its value rises at a compound rate of x\% per year. After 2 years, it is worth £292,662.70. Work out the value of x.

This is a tricky one. As we know the outcome but not the percentage we’re going to have to set up an equation which we can then solve to find the value of our percentage, x. To do this, let’s work through the steps of the normal process, putting x where the percentage value would typically be.

So, the percentage multiplier for an x\% increase is 1 + \dfrac{x}{100}. We will leave it in this form (we don’t really have much choice). We know the result of squaring this and multiplying it by the starting value, so let’s express that as an equation.

268,000 \times \left(1 + \dfrac{x}{100}\right)^2 = 292,662.70

Now, we must solve for x. First, we divide both sides by 268,000.

\left(1 + \dfrac{x}{100}\right)^2 = \dfrac{292,662.70}{268,000}

Then, take the square root of both sides.

1 + \dfrac{x}{100} = \sqrt{\dfrac{292,662.70}{268,000}}

Then subtract 1 from both sides and multiply by 100.

\dfrac{x}{100} = \sqrt{\dfrac{292,662.70}{268,000}} - 1

x = 100\left(\sqrt{\dfrac{292,662.70}{268,000}} - 1\right)

Putting it into the calculator, we get

x = 4.5

Hernando’s house increased in value by 4.5% per year.

## Compound Growth and Decay Revision and Worksheets

## Compound Growth and Decay Teaching Resources

Compound depreciation resources can be accessed via this dedicated page. If you are also looking for decay and compound depreciation questions you will also find them as part of this Maths Made Easy resource page. Whether you teach GCSE Maths in Leeds or you are a Maths tutor in London, you will find these compound interest questions and resources useful.