# Finance

## Finance Revision

**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

**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.

**Exam****ple:**

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:

**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:

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**:

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

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

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

**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}.

**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:

And then per hour:

£461.54\div40=£11.54

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

**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)

## Finance Example Questions

**Question 1:** Alice is running a bake sale. She buys ingredients for 140 cakes, and the cost for one is 56\text{ p}.

Alice manages to sell 75 cakes for 99\text{ p} each.

Calculate whether Alice made a profit or a loss, and the amount of the profit or loss.

**[3 marks]**

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**Question 2:** Alana buys a t-shirt and a skirt. They are both on sale. The t-shirt was originally £12 and was discounted by 25\%, and the skirt was on sale for 8\% less than its original price.

In total, Alana spent £29.24 on the t-shirt and skirt.

Calculate the original price of the skirt before it went on sale.

**[4 marks]**

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

**Question 3:** A house purchased for £105,000 increases in value by 2\% each year.

Find the value of the house after 5 years to the nearest pound.

**[4 marks]**

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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