# Finance

GCSELevel 4-5Cambridge iGCSE

## Finance

There are lots of ways maths can help us look at finance problems:

Earnings: Calculating the earning potential of jobs.

Profit and loss: Working out increases and decreases in money.

Discounts: Using percentages and reverse percentages to work out discounts.

Interest and depreciation: Looking at simple and compound interest.

## Earnings

Earnings refer to the amount of money someone earns through working a job. This may be given as an hourly rate, or a yearly salary.

Example:

If Sally works $40$ hours a week on an hourly wage of $£9.40$, how much does she earn in a week and what is her yearly salary?

To work out how much Sally earns in a week, we multiply her hourly wage by the number of hours she works per week:

$£9.40\times 40 = £376$

To work out Sally’s yearly salary, we multiple her weekly wage by the number of weeks in a year, $52$:

$£376\times 52 = £19552$

Level 4-5GCSECambridge iGCSE

## Profit and Loss

Profit refers to an increase in money, and loss refers to a decrease in money.

It is helpful to look at profit and loss when considering cost and sale prices. For example, a shop owner buys products to sell in their shop and this is a cost, and then sells these products, which is a sale.

Example:

A shop owner buys $100$ teddy bears to sell in their shop, and they each cost $£2.80$. The shop owner sells each of these for $£6.99$ in their shop. They only managed to sell $48$ teddy bears in their shop.

Work out whether they made a profit or a loss, and the amount of profit or loss.

To work out profit and loss, we need to find the difference between costs and sales.

Costs: $100\times £2.80 = £280$

Sales: $48\times £6.99 = £335.52$

As the sales were higher than the costs, the shop owner has made a profit. The amount of profit is the sales minus the costs:

$£335.52 - £280 = £55.52$

Level 4-5GCSECambridge iGCSE

## Discounts

When a shop reduces the price of a product, the product is discounted.

Discounts are often written as percentages, for example, a shop may discount a product by $25\%$.

Example:

A dress usually costs $£25$. The dress goes on sale with a $30\%$ discount. Calculate the sale price of the dress.

We use our knowledge of percentages to calculate this. If a dress has a $30\%$ discount, it will be $70%$ of the original price:

$£25\times0.7=£17.50$

We may be given a sale price and asked to calculate the discount.

Example:

A dress usually costs $£48$ and goes on sale for $£26.40$, calculate the percentage discount.

Here, we use reverse percentages. We know that $£48$ multiplied by $1 -$the discount would equal $£26.40$, so we can do the opposite to find the discount:

$£26.40\div£48=0.55$

So the sale price is $55\%$ of the normal price, meaning the discount is $45\%$.

Level 4-5GCSECambridge iGCSE

## Interest and Depreciation – Simple

Simple interest is where we increase a value by the same percentage repeatedly. This means the interest is only on the original amount and not on the interest over time.

Example

A bank account contains $£5000$ and gets $5\%$ simple interest each year. Calculate the amount in the bank account after $3$ years.

Find $5\%$ of $£5000$:

$£5000\times0.05 = £250$

Therefore, each year this bank account will increase by $£250$:

$£5000 + £250 + £250 + £250 = £5750$ after $3$ years.

Level 4-5GCSECambridge iGCSE

## Interest and Depreciation – Compound

Unlike simple interest, compound interest means you get interest on your interest. The following formula enables  you to calculate the compound interest or depreciation:

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

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

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

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

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

Example

A bank account contains $£5000$ and gets $5\%$ compound interest each year. Calculate the amount in the bank account after $3$ years.

Year 1:

$\textcolor{red}{5\%}$ on top of $5000$ = $\textcolor{limegreen}{£5250}$.

Year 2:

$\textcolor{red}{5\%}$ on top of $5250$ = $\textcolor{limegreen}{£5512.50}$.

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 $\textcolor{Orange}{n = 3}$

$\textcolor{purple}{N} = \textcolor{blue}{£5000} \times \bigg(1 \textcolor{red}{+ \dfrac{5}{100}} \bigg) ^{\textcolor{orange}{3}} = \textcolor{purple}{£5788.13}$ ($2$ dp)

This formula works in the same way for decay, but using $1 - \dfrac{\text{Percentage}}{100}$.

Level 4-5GCSECambridge iGCSE

## Example 1: Using Data from a Table

Holly is $22$ years old and is looking for a new marketing job. She wants to earn at least $£12$ an hour. She has $2$ years marketing experience and gets an interview at ‘Marketing 4 U’. They send her information about their salaries for employees working $40$ hours per week:

Calculate whether Holly would make the amount of money she wants if she took this job.

[3 marks]

We know Holly is $22$ with $2$ years experience, so her salary would be $£24000$. We need to work out what her hourly rate is. Firstly, work out how much she would earn per week:

$£24000\div52=£461.54$

And then per hour:

$£461.54\div40=£11.54$

Therefore, Holly would earn less than $£12$ an hour.

Level 4-5GCSECambridge iGCSE

## Example 2: Compound Depreciation

A car purchased for $£4500$ depreciates by $8\%$ each year.

Calculate the value of the car after $6$ years.

[3 marks]

We can use the formula we learnt earlier to answer this question:

Here $\textcolor{Orange}{n = 6}$

$\textcolor{purple}{N} = \textcolor{blue}{£4500} \times \bigg(1 \textcolor{red}{- \dfrac{8}{100}} \bigg) ^{\textcolor{orange}{6}} = \textcolor{purple}{£2728.60}$ ($2$ dp)

Level 4-5GCSECambridge iGCSE

## Finance Example Questions

Costs: $£0.56\times 140=£78.40$

Sales: $£0.99\times 75=£74.25$

The costs were higher than the sales, so Alice made a loss.

The loss amount:

$£78.40-£74.25=£4.15$

Total spend: $£29.24$

T-shirt sale price: $£12\times 0.75 =£9$

Skirt sale price: $£29.24-£9=£20.24$

Skirt original price: $£20.24\div0.92=£22$

So, each year the price is multiplied by $1.02$.

We can now put these values into the formula to find the value of the house after $5$ years:

$£105000\times 1.02^5=£115928$ to the nearest pound.

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