Compound Growth and Decay Worksheets | Questions and Revision

# Compound Growth and Decay Worksheets, Questions and Revision

Level 6 Level 7

## What you need to know

### Compound Growth and Decay

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 – meaning you’d have £106.09 in total. Fortunately there is a compound interest formula for GCSE that can and should be used in all questions as shown below.

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

There is an alternative to compound interest called simple interest. 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. For more simple interest questions visit our dedicated page.

Topics that will help with covering compound interest include:

### Compound Growth and Decay Formula

The following formula for compound interest and decay enables you to substitute in values and calculate the growth or decay without too much difficulty. The manual method highlighted above becomes to much effort and doesn’t work for all different question types, so knowing and understanding this formula is essential.

### Example 1: Compound Growth

James deposits £29,760 into a savings account with annual compound interest rate of 4%. Calculate how much money he will have after 6 years assuming he takes no money out.

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

$29760\times(1 + \dfrac{4}{100})^6= 37655.89$

So after 6 years James would have £37655.89. The compound interest formula makes this easy but you do have to memorise it!

### Example 2: Compound Decay

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.

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

$29760\times(1 - \dfrac{25}{100})^8=\pounds 1,701.92$

Questions on compound growth and decay are often 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

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 1024 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.

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

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 divide by 100:

$(1.045 – 1) \div 100 = 4.5\% \text{ increase}$

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 0.94³):

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

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 0.96²):

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

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:

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

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:

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

Level 4-5

Level 4-5

Level 4-5

Level 4-5

GCSE MATHS

GCSE MATHS

GCSE MATHS

### Learning resources you may be interested in

We have a range of learning resources to compliment our website content perfectly. Check them out below.