Financial mathematics: Use mathematics to plan and control financial instruments

# Unit 1: Work with simple and compound growth formulae

Natashia Bearam-Edmunds

### Unit outcomes

By the end of this unit you will be able to:

- Understand and apply the simple interest formula.
- Understand and apply the compound growth formulae.
- Understand and apply the compound growth formulae with interest compounded more than once a year.
- Calculate the values of [latex]\scriptsize A,~\text{ }P,\text{ }~i~[/latex] and [latex]\scriptsize n[/latex].

## What you should know

Before you start this unit, make sure you can:

- do basic compound interest calculations. You can revise financial maths in level 3 subject outcome 5.1 and level 3 subject outcome 5.2.

## Introduction

You have seen that financial mathematics has many practical applications. Some real-life examples include taking a bank loan, buying furniture on hire purchase and making monthly instalments on a car. You may have also heard of someone who lends money to others (referred to colloquially as ‘Mashonisa’).

What do all these situations have in common? Interest! Interest must be paid, or interest is being charged. Interest is earned by the lender and interest is paid by the borrower.

When a person borrows money they must pay an amount back, which includes interest added over time. Therefore, the cost of borrowing money is the interest you must pay back on the loan amount.

The benefit of investing money is the interest earned on the amount invested. When someone saves money in a bank account or invests money in an investment account, they earn interest at a given interest rate over the time that the money is saved or invested.

Let’s have quick recap of the financial concepts you learnt in level 3. Remember that there are two different ways in which interest is calculated; simple interest or compound interest.

## Simple interest

Simple interest is calculated on only the initial or principal amount invested or borrowed. The interest received or charged for each period will always be the same.

The formula for calculating simple interest is:

[latex]\scriptsize A=P(1+i\cdot n)[/latex]

[latex]\scriptsize \displaystyle A~=\text{accumulated(final) amount }\!\!~\!\!\text{ }[/latex]

[latex]\scriptsize P=\text{initial }\!\!~\!\!\text{ or }\!\!~\!\!\text{ principal }\!\!~\!\!\text{ amount}[/latex]

[latex]\scriptsize i=\text{interest }\!\!~\!\!\text{ rate }\!\!~\!\!\text{ }\left( {\text{written }\!\!~\!\!\text{ as }\!\!~\!\!\text{ a }\!\!~\!\!\text{ decimal}} \right)[/latex]

[latex]\scriptsize n=\text{period }\!\!~\!\!\text{ }\left( {\text{time }\!\!~\!\!\text{ in }\!\!~\!\!\text{ years }\!\!~\!\!\text{ for }\!\!~\!\!\text{ simple }\!\!~\!\!\text{ interest}} \right)[/latex]

A hire purchase (HP) agreement, also known as an instalment plan, is an application of simple interest. In HP agreements a person agrees to buy an item at a certain interest rate over a stated period and will usually pay a deposit in order to secure the item.

### Example 1.1

Ayesha buys a fridge on hire purchase for [latex]\scriptsize \displaystyle \text{R}15\text{ }000[/latex]. She pays a [latex]\scriptsize 10\%[/latex] deposit. The store charges [latex]\scriptsize \displaystyle 20\%[/latex] simple interest per annum for two years. What are Ayesha’s monthly instalments?

*Solution*

Write down the key information you have been given.

This is an HP agreement so we will use [latex]\scriptsize A=P(1+i\cdot n)[/latex].

To work out Ayesha’s monthly instalments we need to first calculate the accumulated amount after interest is added.

Deposit is [latex]\scriptsize 10\%[/latex] of [latex]\scriptsize \displaystyle \text{R}15\text{ }000[/latex].

[latex]\scriptsize 0.1\times \text{R}15\text{ }000=\text{R}1\text{ }500[/latex]

So the amount she must pay off is: [latex]\scriptsize \displaystyle \text{R}15\text{ }000-\text{R}1\text{ 5}00=\text{R}13\text{ 5}00[/latex].

[latex]\scriptsize \displaystyle \therefore P=\text{R}13\text{ }500[/latex]

[latex]\scriptsize \begin{align*}i&=\displaystyle \frac{{20}}{{100}}\\&=0.2\end{align*}[/latex]

[latex]\scriptsize n=2[/latex]

Use your calculator and round off only at the very end of the question.

[latex]\scriptsize \begin{align*}A&=13\text{ }500(1+(0.2)\cdot 2)\\&=\text{R}18\text{ 9}00\end{align*}[/latex]

To work out the monthly payments we must divide the final amount by the number of months in two years.

Monthly payments:

[latex]\scriptsize \displaystyle \frac{{\text{R}18\text{ 9}00}}{{24}}=\text{R787}\text{.50}[/latex]

### Exercise 1.1

- Determine the value of an investment of [latex]\scriptsize \displaystyle \text{R}13\text{ }000[/latex] at [latex]\scriptsize \displaystyle 12\%[/latex] p.a. simple interest for three years.
- The value of an investment grows from [latex]\scriptsize \displaystyle \text{R}2\text{ }500[/latex] to [latex]\scriptsize \displaystyle \text{R4 5}50[/latex] in eight years. Determine the simple interest rate at which it was invested.
- Jamie buys a sofa on an HP agreement to be paid off over three years. If the sofa costs [latex]\scriptsize \text{R}6\text{ }000[/latex] and he pays a deposit of [latex]\scriptsize 8\%[/latex], and is charged [latex]\scriptsize 22\%[/latex] simple interest p.a., what are his monthly instalments? (Round off to the nearest rand).

The full solutions are at the end of the unit.

## Compound growth

Compound growth allows interest to be earned on interest. The interest is calculated on the sum of the initial amount and the accumulated interest of an investment or loan. Compound interest makes the value of an investment or loan grow at a faster rate than simple interest does.

The formula for calculating compound growth is:

[latex]\scriptsize A=P{{(1+i)}^{n}}[/latex]

[latex]\scriptsize \displaystyle A~=\text{accumulated(final) amount }\!\!~\!\!\text{ }[/latex]

[latex]\scriptsize P=\text{initial }\!\!~\!\!\text{ or }\!\!~\!\!\text{ principal }\!\!~\!\!\text{ amount}[/latex]

[latex]\scriptsize i=\text{interest }\!\!~\!\!\text{ rate }\!\!~\!\!\text{ }\left( {\text{written }\!\!~\!\!\text{ as }\!\!~\!\!\text{ a }\!\!~\!\!\text{ decimal}} \right)[/latex]

[latex]\scriptsize n=\text{period}[/latex]

Let’s look at some examples.

### Example 1.2

Sandi wants to invest [latex]\scriptsize \displaystyle \text{R2 500}[/latex] for eight years. Safe Bank offers a savings account which pays simple interest at a rate of [latex]\scriptsize \displaystyle 15\%[/latex] per annum, and Buck Bank offers a savings account paying compound interest at a rate of [latex]\scriptsize \displaystyle 12\%[/latex] per annum. Which savings account would give Sandi the better bank balance at the end of the eight year period?

*Solution*

Final amount using the simple interest formula at Safe Bank:

[latex]\scriptsize A=P(1+i\cdot n)[/latex]

[latex]\scriptsize \begin{align*}A&=2\text{ }500(1+(0.15)(8))\\&=\text{R}5\text{ }500\end{align*}[/latex]

Final amount using the compound interest formula at Buck Bank:

[latex]\scriptsize A=P{{(1+i)}^{n}}[/latex]

[latex]\scriptsize \begin{align*}A&=2\text{ }500{{(1+(0.12))}^{8}}\\&=\text{R6 190}\end{align*}[/latex]

The Buck Bank savings account would give Sandi a better bank balance at the end of the [latex]\scriptsize \displaystyle 8[/latex] year period.

### Example 1.3

James decides to open an investment account with [latex]\scriptsize \text{R}25\text{ }000[/latex]. What compound interest rate must the investment account achieve for him to double his money in [latex]\scriptsize 10[/latex] years? Give your answer correct to one decimal place.

*Solution*

**Step 1**: Write down the known variables and the compound interest formula

[latex]\scriptsize \begin{align*}A&=25\text{ }000\times 2\\&=\text{R}50\text{ }000\\P&=\text{R}25\text{ }000\\n&=10\end{align*}[/latex]

[latex]\scriptsize A=P{{(1+i)}^{n}}[/latex]

**Step 2**: Substitute the values and solve for [latex]\scriptsize i[/latex]

[latex]\scriptsize \begin{align*}50\text{ }000&=25\text{ }000{{(1+i)}^{{10}}}\\{{(1+i)}^{{10}}}&=2\\i&=\sqrt[{10}]{2}-1\\&=0.0717...\end{align*}[/latex]

We must round up to a rate of [latex]\scriptsize 7.2\%[/latex] p.a. so that James doubles his investment in the period.

### Example 1.4

Calculate how much interest John will earn if he invests [latex]\scriptsize \displaystyle \text{R}2\text{ }000[/latex] for [latex]\scriptsize \displaystyle 4[/latex] years at [latex]\scriptsize \displaystyle 5.4\%[/latex] p.a. compound interest.

Solution

**Step 1**: Write down the known variables and the compound interest formula

[latex]\scriptsize \begin{align*}P&=\text{R}2\text{ }000\\n&=4\\i&=0.054\\A&=?\end{align*}[/latex]

[latex]\scriptsize A=P{{(1+i)}^{n}}[/latex]

**Step 2**: Substitute the values and solve for [latex]\scriptsize A[/latex]

[latex]\scriptsize \begin{align*}A&=2\text{ }000{{(1+0.054)}^{4}}\\&=\text{R}2\text{ }468.27\end{align*}[/latex]

**Step 3**: Calculate the interest earned

[latex]\scriptsize \begin{align*}\text{Interest}&=A-P\\&=\text{R}2\text{ }468.27-2000\\&=\text{R}468.27\end{align*}[/latex]

John earns [latex]\scriptsize \text{R}468.27[/latex] interest over the four years.

### Exercise 1.2

- Calculate the value of [latex]\scriptsize \displaystyle \text{R6 5}00[/latex] invested at[latex]\scriptsize \displaystyle ~8.6\%[/latex] p.a. compound interest for five years.
- Bongani invested [latex]\scriptsize \displaystyle \text{R}10\text{ }000[/latex] for six years. If the value of his investment is [latex]\scriptsize \text{R18 }500[/latex], what compound interest rate did it earn?
- If an investment is worth [latex]\scriptsize \text{R22 0}00[/latex] at the end of five years at a compound interest rate of [latex]\scriptsize 11.5\%[/latex], what was the initial amount invested?

The full solutions are at the end of the unit.

### Compounding period

So far, we have looked at compound growth using annual interest rates only. However, interest is often compounded more than once a year. The compounding period tells us the number of times interest is charged or earned in a year.

Interest can be compounded daily, weekly, monthly, quarterly, half-yearly (semi-annually or biannually), annually, biennially (every 2 years) or even continuously. Generally, regardless of the compounding period, the interest rate is stated as an annual rate also called the **nominal rate**.

When the compounding period is not annual we must adjust the compound interest formula to reflect the different compounding period. We make the following adjustments to the compound interest formula:

[latex]\scriptsize A=P{{(1+\displaystyle \frac{i}{m})}^{{t\times m}}}[/latex]

To take into account compounding that occurs more than once per annum, let’s say [latex]\scriptsize \displaystyle m[/latex] times a year, we multiply the number of years [latex]\scriptsize \displaystyle t[/latex] in the compound interest formula by [latex]\scriptsize \displaystyle m[/latex] and we also divide the interest rate [latex]\scriptsize i[/latex] by [latex]\scriptsize \displaystyle m[/latex].

Some of the common compounding periods are listed below. For example, if compounding is half- yearly, interest is added two times in the year.

COMPOUNDING PERIOD |
VALUE OF [latex]\scriptsize \displaystyle m[/latex] |

Monthly | [latex]\scriptsize 12[/latex] |

Half-yearly/ biannually |
[latex]\scriptsize 2[/latex] |

Quarterly | [latex]\scriptsize 4[/latex] |

Weekly | [latex]\scriptsize 52[/latex] |

Daily | [latex]\scriptsize 365[/latex] |

Biennially | [latex]\scriptsize \displaystyle \frac{1}{2}[/latex] |

### Example 1.5

You invest [latex]\scriptsize \displaystyle \text{R}6\text{ }000[/latex] at [latex]\scriptsize \displaystyle 9\%[/latex] p.a. compounded monthly. After seven years you withdraw the full amount. How much will you be able to withdraw in total?

*Solution*

Since compounding occurs monthly, we must multiply [latex]\scriptsize \displaystyle t[/latex] by [latex]\scriptsize 12[/latex], as there are [latex]\scriptsize 12[/latex] months in a year, and divide [latex]\scriptsize \displaystyle i[/latex] by [latex]\scriptsize \displaystyle 12[/latex] in the compound interest formula.

[latex]\scriptsize \displaystyle \begin{align*}A&=P{{(1+\displaystyle \frac{i}{m})}^{{t\times m}}}\\&=6\text{ }000{{(1+\displaystyle \frac{{0.09}}{{12}})}^{{7\times 12}}}\\&=6\text{ }000{{(1.0075)}^{{84}}}\\&=\text{R}11\text{ }239\end{align*}[/latex]

You will be able to withdraw [latex]\scriptsize \displaystyle \text{R}11\text{ }239[/latex] at the end of seven years.

The amount of compound interest earned on an investment or paid for a loan depends on the frequency of compounding; the higher the number of compounding periods, the greater the compound interest.

### Example 1.6

Nimrod and Mike each invest [latex]\scriptsize \text{R10 }000[/latex]. Nimrod invests his money for five years at [latex]\scriptsize \displaystyle 8\%[/latex] p.a. compounded monthly. Mike invests his [latex]\scriptsize \text{R10 }000[/latex] for five years at [latex]\scriptsize \displaystyle 8\%[/latex] p.a. compounded half-yearly.

What is the difference in their accrued amounts at the end of the five-year period?

*Solution*

Nimrod:

[latex]\scriptsize \displaystyle \begin{align*}A&=P{{(1+\displaystyle \frac{i}{m})}^{{t\times m}}}\\&=10\text{ }000{{(1+\displaystyle \frac{{0.08}}{{12}})}^{{5\times 12}}}\\&=10\text{ }000{{(1+\displaystyle \frac{{0.08}}{{12}})}^{{60}}}\\&=\text{R14 898}\end{align*}[/latex]

Mike:

[latex]\scriptsize \displaystyle \begin{align*}A&=P{{(1+\displaystyle \frac{i}{m})}^{{t\times m}}}\\&=10\text{ }000{{(1+\displaystyle \frac{{0.08}}{2})}^{{5\times 2}}}\\&=10\text{ }000{{(1.04)}^{{10}}}\\&=\text{R14 802}\end{align*}[/latex]

Nimrod earns [latex]\scriptsize \displaystyle \text{R14 898}-\text{R14 802}=\text{R}96[/latex] more than Mike. This makes sense since his investment has more frequent compounding than Mike’s.

### Example 1.7

Sarah invests [latex]\scriptsize \text{R5 }000[/latex] at [latex]\scriptsize 6.5\%[/latex] p.a. compounded monthly. Two years later she adds [latex]\scriptsize \text{R3 }000[/latex] to the savings account. Calculate the amount in her account six years after she invested the first amount.

*Solution*

First, calculate the amount she has at the end of two years:

[latex]\scriptsize \begin{align*}A&=5\text{ }000{{(1+\displaystyle \frac{{0.065}}{{12}})}^{{2\times 12}}}\\&=\text{R}5\text{ 692}\text{.144665}\end{align*}[/latex]

Next, add [latex]\scriptsize \text{R3 }000[/latex] to the accumulated amount.

[latex]\scriptsize \displaystyle \text{R3 }000+\text{R}5\text{ 692}\text{.144665}=\text{R8 692}...[/latex]

The new amount will stay in her account for four more years at the same interest rate. So final amount will be:

[latex]\scriptsize \begin{align*}A&=8\text{ 692}...{{(1+\displaystyle \frac{{0.065}}{{12}})}^{{4\times 12}}}\\&=\text{R}11\text{ }265.20\end{align*}[/latex].

### Exercise 1.3

- Ntombi opens accounts at a number of clothing stores and spends freely. She gets herself into terrible debt and she cannot pay off her accounts. She owes Fashion World [latex]\scriptsize \displaystyle \text{R}5\text{ }000[/latex] and the shop agrees to let her pay the bill at a nominal interest rate of [latex]\scriptsize \displaystyle 24\%[/latex] compounded monthly.
- How much money will she owe Fashion World after two years?
- What are her monthly instalments?

- Jackson invests [latex]\scriptsize \displaystyle \text{R2}0\text{ }000[/latex] for [latex]\scriptsize \displaystyle 18[/latex] months at [latex]\scriptsize 10\%[/latex] interest compounded quarterly. Calculate how much money he will have at the end of the period.
- A financial advisor promises that she will treble the value of an investment at the end of six years. If the interest rate is fixed and compounded monthly, calculate the annual rate of interest that she offers.

The full solutions are at the end of the unit.

### The effect of inflation

The price of a dozen eggs now is not the same as the price paid for a dozen eggs ten years ago. We all know that over time the prices of things increase. The average increase in the price of goods over time is called **inflation**.

Inflation changes continuously. The rate of inflation is quoted as a percentage per annum. We can use the compound growth formula to calculate the increase in the price of goods if we know the inflation rate.

Inflation affects the rate of interest that you receive from an investment or pay on a loan. If interest rates increase, money invested will increase. Similarly, interest on loans will increase and people will have to pay back more money.

If interest rates decrease you receive less on investments and loans will become cheaper.

### Example 1.8

The average rate of inflation over the past five years was [latex]\scriptsize 4.5\%[/latex] per annum. The current price of a eggs is [latex]\scriptsize \text{R}35[/latex].

- How much did a dozen eggs cost five years ago?
- Calculate the expected price of a dozen eggs in three years’ time if the average rate of inflation stays the same.

*Solutions*

- The current price of a dozen eggs is [latex]\scriptsize \text{R}35[/latex][latex]\scriptsize \therefore A=35[/latex] if we are looking back.

[latex]\scriptsize \begin{align*}i&=0.045,\text{ }n=5\\A&=P{{(1.045)}^{5}}\\35&=P{{(1.045)}^{5}}\\P&=\text{R28}.09\end{align*}[/latex] - The current price of a dozen eggs is [latex]\scriptsize \text{R}35[/latex][latex]\scriptsize \therefore P=35[/latex] since we are looking forward in time.

[latex]\scriptsize \begin{align*}i&=0.045,\text{ }n=3\\A&=35{{(1.045)}^{3}}\\&=\text{R39}\text{.94}\end{align*}[/latex]

### Exercise 1.4

- A painting costs [latex]\scriptsize \text{R}25\text{ }000[/latex] in April 2021. If the price of the painting increases at an average rate of [latex]\scriptsize 6.5\%[/latex] p.a. How much will the painting be worth in April 2030?
- Kate’s monthly salary is [latex]\scriptsize \text{R}15\text{ }000[/latex]. Her monthly mortgage loan payment is [latex]\scriptsize \text{R}4\text{ }500[/latex]. Her salary will increase this year by [latex]\scriptsize 6\%[/latex] and the interest rate on her mortgage will increase by [latex]\scriptsize 1\%[/latex]. Calculate:
- Her new monthly salary.
- The increase in her monthly loan payment.

- David plans to replace his motorbike in five years’ time with a similar make and type of motorbike. If his motorbike is currently worth [latex]\scriptsize \text{R}109\text{ 000}[/latex] and prices of motorbikes increase by [latex]\scriptsize 7.5\%[/latex] a year, how much will a new bike cost in five years’ time?

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

- How to calculate simple interest.
- How to calculate compound growth.
- How to calculate compound growth with different compounding periods.

# Unit 1: Assessment

#### Suggested time to complete: 20 minutes

- Greg enters into a five -year hire-purchase agreement to buy a computer for [latex]\scriptsize \displaystyle \text{R}8\text{ }900[/latex]. The interest rate is quoted as [latex]\scriptsize \displaystyle 11\%[/latex] per annum based on simple interest. Calculate the required monthly payment for this contract.
- Mrs. Brown retired and received a lump sum of [latex]\scriptsize \displaystyle \text{R}200\text{ }000[/latex]. She deposited the money in a fixed deposit savings account for six years. At the end of the six years the value of the investment was [latex]\scriptsize \displaystyle \text{R}265\text{ }000[/latex]. If the interest on her investment was compounded monthly, determine the nominal interest rate.
- [latex]\scriptsize \displaystyle \text{R}5\text{ }500[/latex] is invested for a period of four years in a savings account. For the first year, the investment grows at a simple interest rate of [latex]\scriptsize \displaystyle 11\%[/latex]p.a. and then at a rate of [latex]\scriptsize \displaystyle 12.5\%[/latex]p.a. compounded quarterly for the rest of the period. Determine the value of the investment at the end of the four years.
- It costs [latex]\scriptsize \text{R}32\text{ }000[/latex] per year to attend a private college. Determine the expected cost to study at this college in [latex]\scriptsize 10[/latex] years’ time if the fees increase with inflation at a rate of [latex]\scriptsize 5\%[/latex] p.a.

The full solutions are at the end of the unit.

# Unit 1: Solutions

### Exercise 1.1

- .

[latex]\scriptsize \begin{align*}A&=13\text{ }000(1+0.12\times 3)\\&=\text{R}17\text{ }680\end{align*}[/latex] - .

[latex]\scriptsize \begin{align*}4\text{ }550&=2\text{ 5}00(1+i\times 8)\\(1+i\times 8)&=\displaystyle \frac{{4\text{ }550}}{{2\text{ 5}00}}\\i&=\left( {\displaystyle \frac{{4\text{ }550}}{{2\text{ 5}00}}-1} \right)\div 8\\&=0.1025\\\text{Interest rate}&=0.1025\times 100\\&=10.25\%\end{align*}[/latex] - .

[latex]\scriptsize \begin{align*}\text{Deposit}&=0.08\times 6\text{ }000\\&=\text{R}480\\P&=6\text{ }000-480\\&=\text{R}5\text{ }520\\A&=\text{R}5\text{ }520(1+0.22(3))\\&=\text{R}9\text{ }163.20\\\text{monthly pmts}&=\text{R}9\text{ }163.20\div 36\\&=\text{R}255\end{align*}[/latex]

### Exercise 1.2

- .

[latex]\scriptsize \begin{align*}A&=6\text{ }500{{(1+0.086)}^{5}}\\&=\text{R}9\text{ }818.89\end{align*}[/latex] - .

[latex]\scriptsize \begin{align*}18\text{ }500&=10\text{ }000{{(1+i)}^{6}}\\i&=\sqrt[6]{{\displaystyle \frac{{18\text{ }500}}{{10\text{ }000}}}}-1\\&=0.10797...\end{align*}[/latex]

Compound interest rate of [latex]\scriptsize 10.8\%[/latex] - .

[latex]\scriptsize \begin{align*}22\text{ }000&=P{{(1+0.115)}^{5}}\\P&=\text{R}12\text{ }765.81\end{align*}[/latex]

### Exercise 1.3

- .
- .

[latex]\scriptsize \begin{align*}A&=5\text{ }000{{(1+\displaystyle \frac{{0.24}}{{12}})}^{{2\times 12}}}\\&=\text{R}8\text{ }042.19\end{align*}[/latex] - Monthly instalments: [latex]\scriptsize \text{R}8\text{ }042.19\div 24=\text{R}335.09[/latex]

- .
- Change [latex]\scriptsize \displaystyle 18[/latex] months to [latex]\scriptsize 1.5[/latex] years first, and then apply the compound interest formula taking into account the compounding period.

[latex]\scriptsize \begin{align*}A&=20\text{ }000{{(1+\displaystyle \frac{{0.1}}{4})}^{{1.5\times 4}}}\\&=\text{R23 }193.87\end{align*}[/latex] - .

[latex]\scriptsize \begin{align*}A&=3P\\3P&=P{{(1+\displaystyle \frac{i}{{12}})}^{{6\times 12}}}\\{{(1+\displaystyle \frac{i}{{12}})}^{{72}}}&=3\\i&=12(\sqrt[{72}]{3}-1)\\&=0.1845...\end{align*}[/latex]

Annual rate of interest that she offers is [latex]\scriptsize 18.45\%[/latex].

### Exercise 1.4

- .

[latex]\scriptsize \begin{align*}A&=25\text{ }000{{(1+0.065)}^{9}}\\&=\text{R}44\text{ }064\end{align*}[/latex] - .
- .

Her new monthly salary:

[latex]\scriptsize 15\text{ }000+0.06(15\text{ }000)=\text{R}15\text{ }900[/latex]

OR

[latex]\scriptsize \text{R}15\text{ }000\times 1.06=\text{R}15\text{ 9}00[/latex] - The increase in her monthly loan payment:

[latex]\scriptsize \text{R}4\text{ }500\times 0.01=\text{R}45[/latex]

- .
- Cost of new bike:

[latex]\scriptsize \begin{align*}\text{A}&=109\text{ 000(1+0}\text{.075}{{\text{)}}^{5}}\\&=\text{R}156\text{ 484}\end{align*}[/latex]

### Unit 1: Assessment

- .

[latex]\scriptsize \displaystyle \begin{align*}A&=8\text{ }900(1+0.11\times 5)\\&=\text{R}13\text{ }795\\\text{monthly payments:}&\\\displaystyle \frac{{\text{R}13\text{ }795}}{{60}}&=\text{R}229.92\end{align*}[/latex] - .

[latex]\scriptsize \displaystyle \begin{align*}265\text{ }000&=200\text{ 0}00{{(1+\displaystyle \frac{i}{{12}})}^{{72}}}\\{{(1+\displaystyle \frac{i}{{12}})}^{{72}}}&=\displaystyle \frac{{53}}{{40}}\\i&=12\left( {\sqrt[{72}]{{\displaystyle \frac{{53}}{{40}}}}-1} \right)\\&=0.0469...\end{align*}[/latex]

Nominal interest rate is [latex]\scriptsize 4.7\%[/latex] - .

[latex]\scriptsize \displaystyle \begin{align*}A&=5\text{ 5}00(1+0.11){{(1+\displaystyle \frac{{0.125}}{4})}^{{3\times 4}}}\\&=\text{R}8\text{ 831}.88\end{align*}[/latex] - .

[latex]\scriptsize \displaystyle \begin{align*}A&=32\text{ 0}00{{(1+0.05)}^{{10}}}\\&=\text{R}52\text{ }124.63\end{align*}[/latex]

a dozen is 12