Compound Growth and Decay Worksheets | Questions and Revision

# Compound Growth and Decay Worksheets, Questions and Revision

Level 6-7

## Compound Growth and Decay

Compound growth and decay are an extension on percentages and are used to model real world applications such as interest, disease and population.

Make sure you are happy with the following topics before continuing.

Level 1-3

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## Simple Interest

Simple interest is where we increase the original value by the same percentage, over and over. You don’t get interest on your interest, only on the original amount. For more simple interest questions visit our dedicated page.

Example: A bank account containing $\textcolor{blue}{£100}$ gets $\textcolor{red}{3\%}$ percent simple interest each year.

If we had a $\textcolor{red}{3\%}$ simple interest rate, then the interest you would earn each year would remain at $£3$ and not increase. So, after $1$ year you would have $\textcolor{limegreen}{£103}$ and after $2$ years you would have $\textcolor{maroon}{£106}$.

Level 1-3

## Compound Growth and Decay Formula

The following formula for compound growth and decay enables you to substitute in values to calculate the growth or decay. Knowing and understanding this formula is essential.

$\textcolor{purple}{N} = \textcolor{blue}{N_0} \, \times$ $\bigg( 1 \textcolor{red}{\pm \dfrac{\text{Percentage}}{\text{100}}} \bigg) ^{\textcolor{orange}{n}}$

$\textcolor{purple}{N} = \text{\textcolor{purple}{Amount after the period of time}}$

$\textcolor{blue}{N_0} = \text{\textcolor{blue}{The original amount}}$

$\textcolor{red}{+} \, \text{\textcolor{red}{when it is growth}}$ ; $\textcolor{red}{-} \, \text{\textcolor{red}{when it is decay}}$

$\textcolor{orange}{n} = \text{\textcolor{orange}{Number of periods}}$ $\text{\textcolor{orange}{(Days/hours/minutes etc.)}}$

Level 4-5
Level 4-5

## Compound Interest

Compound interest is where we take an original value and increase it by a percentage. In the next time period we then take this new value (unlike simple interest) and increase it by the same percentage, and so on. You get interest on your interest.

Example: A bank account containing $\textcolor{blue}{£100}$ gets $\textcolor{red}{3\%}$ compound interest. How much will be in the account after $4$ years?

Year 1:

$\textcolor{red}{3\%}$ on top of $100$ = $\textcolor{limegreen}{£103}$.

Year 2:

$\textcolor{red}{3\%}$ on top of $103$ = $\textcolor{limegreen}{£106.09}$.

And so on …

To find compound growth after $\textcolor{orange}{n}$ years, we can substitute all the values into the compound growth and decay formula

Here $\textcolor{Orange}{n = 4}$

$\textcolor{purple}{N} = \textcolor{blue}{£100} \times \bigg(1 \textcolor{red}{+ \dfrac{3}{100}} \bigg) ^{\textcolor{orange}{4}} = \textcolor{purple}{£112.55}$ ($2$ dp)

Level 4-5

## Compound Decay

Compound decay (or depreciation) works in the same way as compound interest but you deduct the percentage each time period instead of adding it on.

Example: A car bought for $\textcolor{blue}{£15000}$ depreciates at $\textcolor{red}{5\%}$ per year. Calculate the value of the car after $\textcolor{Orange}{5}$ years.

This time we need to use $\textcolor{red}{-}$ instead of $\textcolor{red}{+}$.

Using $\textcolor{Orange}{n} = \textcolor{orange}{5}$ we can put our values into the formula:

$\textcolor{purple}{N} = \textcolor{blue}{£15000} \times \bigg(1 \textcolor{red}{- \dfrac{5}{100}} \bigg) ^{\textcolor{orange}{5}} = \textcolor{purple}{£11606.71}$ ($2$ dp)

Level 4-5
Level 4-5

## Finding the Number of Periods ($\textcolor{Orange}{n}$)

This is a type of problem we face where the number of periods is unknown, and we have to use trial and improvement to find this value.

Example: A train ticket from Leeds to Liverpool costs $\textcolor{blue}{£50}$. The ticket increases in price at a rate of $\textcolor{red}{10\%}$ per year. How many years will it take for the train ticket to have a future cost of $\textcolor{purple}{£80}$? Give your answer to the nearest year.

We substitute our known values into the compound growth and decay formula

$\textcolor{blue}{£50} \times \bigg( 1 \textcolor{red}{+ \dfrac{10}{100}} \bigg) ^{\textcolor{orange}{n}} = \textcolor{purple}{£80}$

We now substitute various values of $\textcolor{orange}{n}$ into this equation, until the right-hand side is equal to $\textcolor{purple}{£80}$:

$n=2$ gives $£60.5$

$n=4$ gives $£73.21$

$n=6$ gives $£88.58$

$n=5$ gives $£80.5$

$\textcolor{orange}{n = 5}$ gives the closest answer to $\textcolor{purple}{£80}$, so it takes approximately $\textcolor{orange}{5}$ years for the train ticket to increase from $\textcolor{blue}{£50}$ to $\textcolor{purple}{£80}$.

Level 4-5
Level 4-5

## Example 1: Compound Growth

James deposits $\textcolor{blue}{£29,760}$ into a savings account with annual compound interest rate of $\textcolor{red}{4\%}$.

Calculate how much money he will have after $\textcolor{orange}{6}$ years assuming he takes no money out.

[2 marks]

Using the compound growth and decay formula, we can substitute in the numbers given in the question as shown below:

$\textcolor{blue}{£29760}\times \bigg(1 \textcolor{red}{+ \dfrac{4}{100}} \bigg)^{\textcolor{orange}{6}}= \textcolor{purple}{£37655.89}$

So after $\textcolor{orange}{6}$ years James would have $\textcolor{purple}{£37655.89}$

Level 4-5

## Example 2: Compound Decay

Aza buys a car for $\textcolor{blue}{£17,000}$. The value of this car will experience compound decay at a rate of $\textcolor{red}{25\%}$ per year. Work out the value of the car after $\textcolor{orange}{8}$ years.

[2 marks]

This time we use the minus symbol in the same equation as shown:

$\textcolor{blue}{£17000}\times \bigg(1 \textcolor{red}{- \dfrac{25}{100}} \bigg)^{\textcolor{orange}{8}}= \textcolor{purple}{£1,701.92}$

So Aza’s car will be worth $\textcolor{purple}{£1701.92}$ after $\textcolor{orange}{8}$ years.

Level 4-5

## GCSE Maths Revision Cards

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### Example Questions

Sun’s investment will increase by $2.4\%$ each year.  To calculate the amount of money in her account after each year, we will need to multiply her balance by $1.024$.

Since Sun will receive $2.4\%$ interest each year, then we can calculate her balance after $4$ years by multiplying the starting balance by $1.024$ four times (or $1.024$ to the power of $4$):

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

If something decreases by $18\%$, this means that it is worth $82\%$ of its original value.  Therefore, the multiplier which we need to use for an $18\%$ decrease is $0.82$.

Since we are being asked to multiply the tiger population after five years, we need to multiply the tiger population by $0.82$ five times.

Therefore, we need to multiply $234$ by $0.82$ to the power of $5$, to see what the tiger population is after $5$ years.

$234 \times 0.82^5 = 87$ tigers, to the nearest whole number

This is less than $100$, therefore Riley is correct.

This is a challenging question, but if we tackle it slowly and logically, we should be fine!

In most questions of this type, we have a starting value which we multiply by a percentage multiplier in order to arrive at the final value.  In this question, we have the final value and not the percentage multiplier, so we will need to do some rearranging.

First of all, we need to consider what we would normally to do the percentage multiplier for a compound percentage change over a two-year period.  If a value is being increased by $x\%$ each year, then for two years we would multiply the starting value by $x^2 \%$.  Let’s write this as an equation using the values presented to us in this question:

$\pounds268,000 \times x^2 = \pounds292,662.70$

If

$\pounds268,000 \times x^2 = \pounds292,662.70$

then

$\pounds292,662.70 \div \pounds268,000 = x^2$

Therefore

$x^2 = 1.092025$

To find $x$, we simply need to take the square root of $1.092025$:

$\sqrt{1.092025} = 1.045$

So, the multiplier in our equation is $1.045$.  What does this means in terms of a percentage increase or decrease?  Since the value is more than $1$, we know it is an increase and not a decrease (we know this anyway since the question states that the house rises in value).  If it is not obvious to you how to convert the multiplier to the percentage amount, subtract $1$ and then multiply by $100$:

$(1.045 – 1) \times 100 = 4.5\%$ increase

In order to solve this question, we need to work out the value of the property after $3$ years, and then after $5$ years.

Since in the first $3$ years, the property is decreasing in value at a rate of $6\%$ every year, this means that at the end of each year, the property is worth $94\%$ of what is was worth at the start of the year.  Therefore, we need to multiply the original value of the house by $0.94$ three times (which is the same as multiplying the original value by $0.94^3$):

$\pounds850,000 \times 0.94^3 = \pounds705,996.40$

For the subsequent $2$ years, the property is continuing to decrease in value, but at a rate of $4\%$ every year, which means that at the end of each year, the property is worth $96\%$ of what is was worth at the start of the year.  Therefore, we need to multiply the value of the house after three years by $0.96$ twice (which is the same as multiplying the original value by $0.96^2):$

$\pounds705,996.40\times 0.96^2 = \pounds650,646.2822$

This means that after $5$ years, the property is worth $£650,646.2822$.  If the original value of the house was $£850,000$, then to work out the overall percentage decrease in value, we simply need to work out the difference between the original value and the new value after $5$ years, and work this out as a percentage of the original value, which can be calculated as follows:

$\dfrac{\pounds850,000 - \pounds650,646.2822}{\pounds850,000} \times 100 = 23\%$ to the nearest whole number

This is a challenging question where we will need to find the value of $x$ before we can calculate the value of the speedboat after $5$ years.

In most questions of this type, we have a starting value which we multiply by a percentage multiplier in order to arrive at the final value.  In this question, we have the final value and not the percentage multiplier, so we will need to do some rearranging.

First of all, we need to consider what we would normally to do the percentage multiplier for a compound percentage change over a three-year period.  If a value is being decreased by $x\%$ each year, then for three year we would multiply the starting value by $x^3 \%$.  Let’s write this as an equation using the values presented to us in this question:

$\pounds36,000 \times x^3 = \pounds15,187.50$

If

$\pounds36,000 \times x^3 = \pounds15,187.50$

then

$\pounds15,187.50\div \pounds36,000 = x^3$

Therefore

$x^3 = 0.421874$

To find $x$, we simply need to take the cube root of $0.421874$:

$\sqrt[3]{0.421875} = 0.75$

So, the multiplier in our equation is $0.75$.  What does this means in terms of a percentage increase or decrease?  Since the value is less than $1$, we know it is a decrease and not an increase decrease (we know this anyway since the question states that the speedboat decreases in value).

$0.75$ is the multiplier for $75\%$.  If we are working out $75\%$ of an amount, then this means that the amount is reducing in value by $25\%$, so this speedboat is depreciating at a rate of $25\%$ per year.

Now that we know the depreciation rate, we can work out the value of the speedboat after $5$ years.

The value of the speedboat after $3$ years is $£15,187.50$, so if it depreciates by $25\%$ for another two years, its value can be calculated as follows:

$\pounds15,187.50 \times 0.75^2 = \pounds8,543$ to the nearest pound

### Worksheets and Exam Questions

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