Simple and compound interest

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

• Differentiate between simple and compound interest and extrapolate the advantages and disadvantages of each in specific situations.
• Calculate simple and compound interest over different periods at specific rates.
• Do calculations using computational tools efficiently and correctly and verify solutions in terms of the context.
• Use solutions to calculations effectively to define the changes that occur over a period.

What you should know

Before you start this unit, make sure you can:

• Explain what is meant by income and expenditure. If you need help with this, review Unit 1 of Subject outcome 5.1.
• Solve linear equations. If you need help with this, review Unit 1 of Subject outcome 2.3.
• Solve literal equations. If you need help with this, review Unit 2 of Subject outcome 2.3.
• Describe what a straight line graph or linear function is. Refer back to Unit 1 in Subject outcome 2.1 if you need help with this.
• Describe what an exponential function is. Refer back to Unit 4 in Subject outcome 2.1 if you need help with this.

Introduction

In the previous subject outcome, we learnt about the importance of personal budgets and how these help us to follow two of the basic principles of good money management and wealth creation; namely, save at least [latex]\scriptsize 10\%[/latex] of your income, and control your expenses so that you do not spend more than you earn.

In that subject outcome, we mentioned that while saving is important, keeping the money under your bed is of little value. You need to make sure that your savings are growing and making more money for you. In other words, you need to invest your savings. It is important, however, that you invest wisely. Being able to analyse and calculate the expected rate of return of your investments is an important skill and to do this you need to understand what simple and compound interest are, and how they differ.

In the context of savings and investments, interest is the money that you earn (your fee) for lending your money to someone else. In the context of loans, interest is the money you have to pay to someone else (their fee) for them lending their money to you.

Simple vs compound interest

An excellent way to discover what simple and compound interest really are is to see a practical example of each type of interest.

Activity 1.1: Simple vs compound interest

Time required: 30 minutes

What you need:

• a pen or pencil
• a piece of paper
• a calculator

What to do:

1. Imagine you invest [latex]\scriptsize \text{R}1\ 000[/latex] once-off in a savings account. The account pays [latex]\scriptsize 15\%[/latex] simple interest on your money each year. Simple interest means that you earn interest only on the money that you initially invest. How much money will be in your savings account after one year?
2. Calculate how much money will be in your savings account at the end of the second year at [latex]\scriptsize 15\%[/latex] simple interest per year. Remember that simple interest means that you only earn interest on your initial deposit.
3. Now copy the following table onto a blank piece of paper. Complete the simple interest column for [latex]\scriptsize 10[/latex] years calculating the total value of your investment at the end of each year.
   
4. Now suppose that the same savings account paid compound interest rather than simple interest. Compound interest means that you earn interest on the money that you initially invest as well as on any previous interest you have earned. How much money will be in your savings account after one year at [latex]\scriptsize 15\%[/latex] compound interest?
5. Calculate how much money will be in your savings account at the end of the second year at [latex]\scriptsize 15\%[/latex] compound interest per year. Remember that compound interest means that you earn interest on your initial deposit as well as on any interest you have already earned.
6. Complete the compound interest column for [latex]\scriptsize 10[/latex] years calculating the total value of your investment at the end of each year.
7. Now compare the values at the end of each year for simple and compound interest. Which kind of interest is better for you as an investor? How long does it take for you to double your investment at both interest rates?
8. What is happening to the difference in the balance in the savings account under simple and compound interest? Why do you think this is the case?
9. On the same set of axes, draw graphs of the growth of each balance. What kind of function do you think each graph represents?
10. Calculate the difference in these balances after [latex]\scriptsize 20[/latex] years.

What did you find?

1. After one year, you would earn [latex]\scriptsize 15\%[/latex] of the initial deposit in interest. . So, at the end of year one, your balance would be [latex]\scriptsize \text{R}1\ 000+\text{R}150=\text{R}1\ 150[/latex].
2. At the end of the second year, you will also earn [latex]\scriptsize 15\%[/latex] but this will be [latex]\scriptsize 15\%[/latex] of your initial deposit only. Therefore, you will earn [latex]\scriptsize \text{R}1\ 000\times 0.15=\text{R}150[/latex] in interest. At the end of the second year, your balance would be [latex]\scriptsize \text{R}1\ 150+\text{R}150=\text{R}1\ 300[/latex].
3. Here is the table with the completed simple interest column. Each year you earn [latex]\scriptsize \text{R}150[/latex] in interest and so each year your balance increases by [latex]\scriptsize \text{R}150[/latex].
   
4. After one year, you would earn [latex]\scriptsize 15\%[/latex] of the initial deposit in interest. . So, at the end of year one, your balance would be [latex]\scriptsize \text{R}1\ 000+\text{R}150=\text{R}1\ 150[/latex].
5. At the end of the second year, you will earn interest not only on your initial deposit but also on the interest you have already earned. Therefore, you will earn [latex]\scriptsize \text{R}1\ 150\times 0.15=\text{R172}\text{.5}[/latex]. So, your balance at the end of the second year will be [latex]\scriptsize \text{R}1\ 150+\text{R}172.5=\text{R}1\ 322.5[/latex].
6. Here is the table with the completed compound interest column. Each year you earn [latex]\scriptsize 15\%[/latex] in interest on your balance from the previous year.
   
7. Clearly, as an investor, compound interest is much better. You earn far more interest over time. Under simple interest, the investment doubles in value by the end of the seventh year. Under compound interest, it doubles in value after five years.
8. As time progresses, the difference between the simple interest and compound interest balances gets bigger and bigger. This is because the amount of interest earned under simple interest is fixed so the balance goes up by the same fixed amount each year. However, under compound interest, the amount of interest earned is always increasing and so the balance increases each year by an ever-increasing rate.
9. .
   
   The simple interest graph is a straight line or linear function. The compound interest graph is an exponential function. Refer back to Unit 1 and Unit 4 in Subject outcome 2.1 if you don’t understand what these terms mean.
10. To calculate the simple interest balance after [latex]\scriptsize 20[/latex] years is reasonably simple. You will earn [latex]\scriptsize \text{R}150[/latex] in interest each year. Therefore, in [latex]\scriptsize 20[/latex] years you will earn a total of [latex]\scriptsize \text{R}150\times 20=\text{R}3\ 000[/latex] in interest. Your total balance after [latex]\scriptsize 20[/latex] years will be [latex]\scriptsize \text{R}1\ 000+\text{R}3\ 000=\text{R}4\ 000[/latex].
    .
    Calculating the compound interest balance is a bit more difficult without a nice formula but it can be done by calculating the balance manually at the end of each year. If you do this, the balance after [latex]\scriptsize 20[/latex] years is [latex]\scriptsize \text{R}16\ 366.54[/latex]. Therefore, the difference in the balances will be [latex]\scriptsize \text{R}16\ 366.54-\text{R}4\ 000=\text{R}12\ 366.54[/latex]

After completing Activity 1.1 we can clearly see the differences between simple and compound interest and, for an investor, compound interest is significantly more beneficial, especially over the long term.

As a borrower, however, the opposite is true. If you borrow with a compound interest rate, you are not only charged interest on the initial loan but also on the interest that that loan attracts.

In most cases in the real world, both investments and loans are almost always calculated using compound interest.

Take note!

Simple interest: You earn interest only on the initial investment. You are charged interest only on the amount of the loan.

Compound interest: You earn interest on the initial investment as well as on the interest you have earned. You are charged interest on the initial loan as well as on the interest that the loan attracts.

Simple interest

Now that we know what simple interest is, let’s explore it in a little more detail. There is a formula that we use to work out the value of an accumulated amount after an initial amount has been growing due to simple interest. That formula is below.

Example 1.1

Jeff deposits [latex]\scriptsize \text{R}1\ 500[/latex] into a savings account which pays a simple interest rate of [latex]\scriptsize 6.5\%[/latex] per annum (p.a.) for three years. How much will be in his account at the end of the investment term?

Solution:

The first thing we need to take note of is that we are dealing with simple interest. In this case we are given that [latex]\scriptsize P=\text{R}1\ 500[/latex], [latex]\scriptsize i=\displaystyle \frac{{6.5\%}}{{100}}=0.065[/latex] per year (p.a. is an abbreviation for ‘per annum’ and ‘annum’ is another word for year) and [latex]\scriptsize n=3[/latex] years. We need to find the value of [latex]\scriptsize A[/latex].

[latex]\scriptsize \begin{align*}A & =P(1+in)\\\therefore A & =1\ 500(1+0.065\times 3)\\ & =1\ 792.50\end{align*}[/latex]

[latex]\scriptsize \text{R}1\ 792.50[/latex] will be in Jeff’s account at the end of the investment term.

Note: It is important when calculating with the simple interest formula that you round off only your final answer, if necessary.

Example 1.2

Fatima borrows [latex]\scriptsize \text{R}3\ 450[/latex] from her neighbour at an agreed simple interest rate of [latex]\scriptsize 11.2\%[/latex] p.a. She will pay back the loan in one lump sum at the end of four years. How much will she have to pay her neighbour?

Solution:

This is a simple interest problem.

[latex]\scriptsize P=\text{R3}\ 450[/latex], [latex]\scriptsize i=0.112[/latex] and [latex]\scriptsize n=4[/latex] years

[latex]\scriptsize \begin{align*}A & =P(1+in)\\\therefore A & =3\ 450(1+0.112\times 4)\\ & =4\ 995.60\end{align*}[/latex]

Fatima will need to pay back [latex]\scriptsize \text{R4}\ 995.60[/latex] after four years.

Example 1.3

Big Burgers deposits [latex]\scriptsize \text{R}35\ 000[/latex] into a bank account that pays a simple interest rate of [latex]\scriptsize 4.5\%[/latex] p.a. How many years must Big Burgers invest for to generate a total of [latex]\scriptsize \text{R}50\ 000[/latex]?

Solution:

This is a simple interest problem. However, in this case we need to find the number of years ([latex]\scriptsize n[/latex]) of the investment. [latex]\scriptsize P=\text{R35}\ 000[/latex], [latex]\scriptsize i=0.045[/latex] and [latex]\scriptsize A=\text{R50}\ 000[/latex].

[latex]\scriptsize \begin{align*}A & =P(1+in)\\\therefore \displaystyle \frac{A}{P} & =1+in\\\therefore \displaystyle \frac{A}{P}-1 & =in\\\therefore n & =\displaystyle \frac{{\displaystyle \frac{A}{P}-1}}{i}\\ & =\displaystyle \frac{{\displaystyle \frac{{50\ 000}}{{35\ 000}}-1}}{{0.045}}\\ & =9.52\end{align*}[/latex]

After [latex]\scriptsize 9.52[/latex] years the investment will have grown to [latex]\scriptsize \text{R}50\ 000[/latex].

Example 1.4

Makhize is planning to buy a new car in three years’ time. He has [latex]\scriptsize \text{R}42\ 850[/latex] to save and expects he will need [latex]\scriptsize \text{R}55\ 000[/latex] for the new car. At what p.a. simple interest rate does he need to invest his money?

Solution:

This is a simple interest problem. However, in this case we need to find the interest rate p.a. ([latex]\scriptsize i[/latex]) of the investment. [latex]\scriptsize P=\text{R42}\ 850[/latex], [latex]\scriptsize n=3[/latex] and [latex]\scriptsize A=\text{R}55\ 000[/latex].

[latex]\scriptsize \begin{align*}A & =P(1+in)\\\therefore \displaystyle \frac{A}{P} & =1+in\\\therefore \displaystyle \frac{A}{P}-1 & =in\\\therefore i & =\displaystyle \frac{{\displaystyle \frac{A}{P}-1}}{n}\\ & =\displaystyle \frac{{\displaystyle \frac{{55\ 000}}{{42\ 850}}-1}}{3}\\ & =0.0945\\ & =9.45\%\end{align*}[/latex]

He will need to invest his money at a simple interest rate of [latex]\scriptsize 9.45\%[/latex] p.a.

Exercise 1.1

1. An amount of [latex]\scriptsize \text{R}7\ 850[/latex] is invested in a savings account which pays simple interest at a rate of [latex]\scriptsize 6.23\%[/latex] per annum. Calculate the balance accumulated by the end of six years.
2. Joseph deposited [latex]\scriptsize \text{R}5\ 724[/latex] into a savings account on his son’s fifth birthday. When his son turned [latex]\scriptsize 18[/latex], the balance in the account had grown to [latex]\scriptsize \text{R17}\ \text{892}\text{.31}[/latex]. If simple interest was paid, calculate the rate at which the money was invested.
3. Salim wants to invest [latex]\scriptsize \text{R}10\ 800[/latex] at a simple interest rate of [latex]\scriptsize 8.25\%[/latex] p.a. How many years will it take for the money to grow to [latex]\scriptsize \text{R}33\ 000[/latex]? Round up your answer to the nearest year.

The full solutions are at the end of the unit.

Compound interest

As we saw in Activity 1.1, compound interest is extremely powerful, especially over the long term. This is why it is so important that everyone starts saving, even a little bit, as soon as possible.

[latex]\scriptsize \text{R}1\ 000[/latex] invested at [latex]\scriptsize 10\%[/latex] compound interest per annum for [latex]\scriptsize 10[/latex] years will grow to [latex]\scriptsize \text{R2}\ \text{593}\text{.74}[/latex]. The same amount invested at the same rate over [latex]\scriptsize 40[/latex] years will grow to [latex]\scriptsize \text{R45}\ \text{259}\text{.26}[/latex]. One of the secrets to building wealth is to invest over the long term. The longer you invest, the quicker your money grows as shown in Figure 1.

Your savings will grow even faster if you keep adding, even little bits of money to your savings.

As for simple interest, we have a formula with which we can calculate the growth of an amount under compound interest. It is very similar to the simple interest formula except for one small detail.

Example 1.5

Mpho wants to invest [latex]\scriptsize \text{R}25\ 000[/latex] into an account that offers a compound interest rate of 6% p.a. How much money will be in the account at the end of five years?

Solution:

The first thing we need to take note of is that we are dealing with compound interest. In this case we are given that [latex]\scriptsize P=\text{R25}\ 000[/latex], [latex]\scriptsize i=\displaystyle \frac{{6\%}}{{100}}=0.06[/latex] per year and [latex]\scriptsize n=5[/latex] years. We need to find the value of [latex]\scriptsize A[/latex].

[latex]\scriptsize \begin{align*}A & =P{{(1+i)}^{n}}\\\therefore A & =25\ 000{{(1+0.06)}^{5}}\\ & =33\ 455.64\end{align*}[/latex]

[latex]\scriptsize \text{R33}\ 455.64[/latex] will be in Mpho’s account at the end of the investment term.

Note: It is important when calculating with the compound interest formula that you round off only at your final answer, if necessary.

Example 1.6

Thabile has been given [latex]\scriptsize \text{R}6\ 525[/latex] for her thirteenth birthday. Rather than spending it, she has decided to invest it so that she can put down a deposit of [latex]\scriptsize \text{R}13\ 000[/latex] on a car on her eighteenth birthday. What compound interest rate does she need to achieve this growth?

Solution:

This is a compound interest problem. However, in this case we need to find the interest rate p.a. ([latex]\scriptsize i[/latex]) of the investment. [latex]\scriptsize P=\text{R6}\ 525[/latex], [latex]\scriptsize n=5[/latex] and [latex]\scriptsize A=\text{R13}\ 000[/latex].

[latex]\scriptsize \begin{align*}A & =P{{(1+i)}^{n}}\\\therefore \displaystyle \frac{A}{P} & ={{(1+i)}^{n}}\\\therefore \sqrt[n]{{\displaystyle \frac{A}{P}}} & =1+i\\\therefore i & =\sqrt[n]{{\displaystyle \frac{A}{P}}}-1\\ & =\sqrt[5]{{\displaystyle \frac{{13\ 000}}{{6\ 525}}}}-1\\ & =0.1478\\ & =14.78\%\end{align*}[/latex]

Thabile will need to invest her money at a compound interest rate of [latex]\scriptsize 14.78\%[/latex] p.a.

Exercise 1.2

1. An amount of [latex]\scriptsize \text{R}4\ 378[/latex] is invested in a savings account which pays a compound interest rate of [latex]\scriptsize 5.6\%[/latex] p.a. Calculate the balance accumulated by the end of seven years. As usual with financial calculations, round your answer to two decimal places, but do not round off until you have reached the solution.
2. Thobeka wants to invest some money at a compound interest rate of [latex]\scriptsize 9.2\%[/latex] p.a. How much money should she invest if she wants to reach a sum of [latex]\scriptsize \text{R}35\ 000[/latex] in [latex]\scriptsize 10[/latex] years’ time? Round up your answer to the nearest rand.
3. Bongani invests [latex]\scriptsize \text{R}7\ 650[/latex] into an account which pays out a lump sum at the end of eight years. If he gets [latex]\scriptsize \text{R}13\ 427.92[/latex] at the end of the period, what compound interest rate did the bank offer him? Give the answer correct to one decimal place.

The full solutions are at the end of the unit.

Summary

In this unit you have learnt the following:

• How to differentiate between simple and compound interest.
• What the advantages and disadvantages of using simple and compound interest are in specific situations.
• How to use and manipulate the simple growth formula [latex]\scriptsize A=P(1+in)[/latex] to solve problems.
• How to use and manipulate the compound growth formula [latex]\scriptsize A=P{{(1+i)}^{n}}[/latex] to solve problems subject to only annual compounding being made.

Unit 1: Assessment

Suggested time to complete: 45 minutes

Questions 1, 2 and 3 taken from NC(V) Mathematics L2 Paper 1 February 2013

1. When Thembi was born her father invested [latex]\scriptsize \text{R}5\ 000[/latex] at an annual compound interest rate. On her eighteenth birthday the bank paid her [latex]\scriptsize \text{R}26\ 292.89[/latex]. What was the bank’s rate of compound interest?
2. Zack deposits [latex]\scriptsize \text{R}30\ 000[/latex] into a bank account that pays a simple interest rate of [latex]\scriptsize 7.5\%[/latex] p.a. How many years must he invest for to generate [latex]\scriptsize \text{R}45\ 000[/latex] ?
3. Sonto invested [latex]\scriptsize \text{R1}0\ 000[/latex] at [latex]\scriptsize 15\%[/latex] simple interest per annum for four years. Nomsa invested the same amount at [latex]\scriptsize 15\%[/latex] compound interest per annum for four years.
   Calculate:
   1. The total amount Sonto will receive after four years.
   2. The interest Sonto will have earned after four years.
   3. The total amount Nomsa will receive after four years.
   4. The interest Nomsa will have earned after four years.
   5. Explain which investment is better, and give a reason for your answer.

Question taken from NC(V) Mathematics L2 Paper 1 March 2012

4. Mr Masondo wanted to put some money aside for the education costs for his three children. He had [latex]\scriptsize \text{R}50\ 000[/latex] to invest in different savings plans. Each policy matured when that child turned [latex]\scriptsize 18[/latex] years old. He invested [latex]\scriptsize \text{R}10\ 000[/latex] in a policy for Nzuzo (age [latex]\scriptsize 10[/latex]) which paid [latex]\scriptsize 13\%[/latex] compound interest p.a., [latex]\scriptsize \text{R}15\ 000[/latex] in a policy for Tando (age [latex]\scriptsize 13[/latex]) which paid [latex]\scriptsize 18\%[/latex] simple interest p.a. and [latex]\scriptsize \text{R25}\ 000[/latex] in a policy for Bamukele (age [latex]\scriptsize 15[/latex]) which paid a flat rate of [latex]\scriptsize \text{R}250[/latex] per month.
   1. How many years did Nzuzo have to wait before she collected her money?
   2. How much did Nzuzo receive when her policy matured?
   3. Calculate the interest that Tando earned on her investment.
   4. How much did Bamukele receive when her policy matured?
   5. Which ONE of the three children received the largest amount of money at the end of their investments?

The full solutions are at the end of the unit.

Unit 1: Solutions

Exercise 1.1

1. This is a simple interest problem.
   [latex]\scriptsize P=\text{R}7\ 850[/latex], [latex]\scriptsize i=0.0623[/latex], [latex]\scriptsize n=6[/latex], [latex]\scriptsize A=?[/latex]
   [latex]\scriptsize \begin{align*}A & =P(1+in)\\ & =7\ 850(1+0.0623\times 6)\\ & =10\ 784.33\end{align*}[/latex]
   The balance after six years will be [latex]\scriptsize \text{R}10\ 413.40[/latex].
2. This is a simple interest problem.
   [latex]\scriptsize P=\text{R5}\ 724[/latex], [latex]\scriptsize A=\text{R}17\ 892.31[/latex], [latex]\scriptsize n=13[/latex], [latex]\scriptsize i=?[/latex]
   [latex]\scriptsize \begin{align*}A & =P(1+in)\\\therefore \displaystyle \frac{A}{P} & =1+in\\\therefore in & =\displaystyle \frac{A}{P}-1\\\therefore i & =\displaystyle \frac{{\displaystyle \frac{A}{P}-1}}{n}\\ & =\displaystyle \frac{{\displaystyle \frac{{17\ 892.31}}{{5\ 728}}-1}}{{13}}\\ & =0.1634\\ & =16.34\%\end{align*}[/latex]
   The interest paid was [latex]\scriptsize 16.34\%[/latex] simple interest per annum.
3. This is a simple interest problem.
   [latex]\scriptsize P=\text{R10}\ 800[/latex], [latex]\scriptsize A=\text{R33}\ 000[/latex], [latex]\scriptsize i=0.0825[/latex], [latex]\scriptsize n=?[/latex]
   [latex]\scriptsize \begin{align*}A & =P(1+in)\\\therefore \displaystyle \frac{A}{P} & =1+in\\\therefore in & =\displaystyle \frac{A}{P}-1\\\therefore n & =\displaystyle \frac{{\displaystyle \frac{A}{P}-1}}{i}\\ & =\displaystyle \frac{{\displaystyle \frac{{33\ 000}}{{10\ 800}}-1}}{{0.0825}}\\ & =24.92\end{align*}[/latex]
   Salim will need to invest his money for [latex]\scriptsize 25[/latex] years.

Back to Exercise 1.1

Exercise 1.2

1. This is a compound interest problem.
   [latex]\scriptsize P=\text{R}4\ 378[/latex], [latex]\scriptsize i=0.056[/latex], [latex]\scriptsize n=7[/latex], [latex]\scriptsize A=?[/latex]
   [latex]\scriptsize \begin{align*}A & =P{{(1+i)}^{n}}\\ & =4\ 378{{(1+0.056)}^{7}}\\ & =6\ 410.96\end{align*}[/latex]
   The balance accumulated will be [latex]\scriptsize \text{R}6\ 410.96[/latex].
2. This is a compound interest problem.
   [latex]\scriptsize A=\text{R35}\ 000[/latex], [latex]\scriptsize i=0.092[/latex], [latex]\scriptsize n=10[/latex], [latex]\scriptsize P=?[/latex]
   [latex]\scriptsize \begin{align*}A & =P{{(1+i)}^{n}}\\\therefore P & =\displaystyle \frac{A}{{{{{(1+i)}}^{n}}}}\\ & =\displaystyle \frac{{35\ 000}}{{{{{(1+0.092)}}^{{10}}}}}\\ & =14\ 515.82\end{align*}[/latex]
   Thobeka will need to invest [latex]\scriptsize \text{R}14\ 515.82[/latex] now.
3. This is a compound interest problem.
   [latex]\scriptsize P=\text{R7}\ 650[/latex], [latex]\scriptsize n=8[/latex], [latex]\scriptsize A=\text{R}13\ 427.92[/latex], [latex]\scriptsize i=?[/latex]
   [latex]\scriptsize \begin{align*}A & =P{{(1+i)}^{n}}\\\therefore \displaystyle \frac{A}{P} & ={{(1+i)}^{n}}\\\therefore \sqrt[n]{{\displaystyle \frac{A}{P}}} & =1+i\\\therefore i & =\sqrt[n]{{\displaystyle \frac{A}{P}}}-1\\ & =\sqrt[8]{{\displaystyle \frac{{13\ 427.92}}{{7\ 650}}}}-1\\ & =0.0729\\ & =7.29\%\end{align*}[/latex]
   The bank gave Bongani a compound interest rate of [latex]\scriptsize 7.3\%[/latex] p.a.

Back to Exercise 1.2

Unit 1: Assessment

1. This is a compound interest problem.
   [latex]\scriptsize P=\text{R5}\ 000[/latex], [latex]\scriptsize n=18[/latex], [latex]\scriptsize A=\text{R26}\ 292.89[/latex], [latex]\scriptsize i=?[/latex]
   [latex]\scriptsize \begin{align*}A & =P{{(1+i)}^{n}}\\\therefore \displaystyle \frac{A}{P} & ={{(1+i)}^{n}}\\\therefore \sqrt[n]{{\displaystyle \frac{A}{P}}} & =1+i\\\therefore i & =\sqrt[n]{{\displaystyle \frac{A}{P}}}-1\\ & =\sqrt[{18}]{{\displaystyle \frac{{26\ 292.89}}{{5\ 000}}}}-1\\ & =0.0966\\ & =9.66\%\end{align*}[/latex]
   The bank’s interest rate was [latex]\scriptsize 9.66\%[/latex] compound interest p.a.
2. This is a simple interest problem.
   [latex]\scriptsize P=\text{R30}\ 000[/latex], [latex]\scriptsize A=\text{R45}\ 000[/latex], [latex]\scriptsize i=0.075[/latex], [latex]\scriptsize n=?[/latex]
   [latex]\scriptsize \begin{align*}A & =P(1+in)\\\therefore \displaystyle \frac{A}{P} & =1+in\\\therefore in & =\displaystyle \frac{A}{P}-1\\\therefore n & =\displaystyle \frac{{\displaystyle \frac{A}{P}-1}}{i}\\ & =\displaystyle \frac{{\displaystyle \frac{{45\ 000}}{{30\ 000}}-1}}{{0.075}}\\ & =6.67\end{align*}[/latex]
   He will need to invest his money for [latex]\scriptsize 6.67[/latex] years.
3. .
   1. Sonto’s investment (simple interest):
      [latex]\scriptsize P=\text{R10}\ 000[/latex], [latex]\scriptsize i=0.15[/latex], [latex]\scriptsize n=4[/latex], [latex]\scriptsize A=?[/latex]
      [latex]\scriptsize \begin{align*}A & =P(1+in)\\ & =10\ 000(1+0.15\times 4)\\ & =16\ 000\end{align*}[/latex]
      The balance after four years will be [latex]\scriptsize \text{R}16\ 000[/latex].
   2. Sonto will have earned [latex]\scriptsize \text{R}16\ 000-\text{R}10\ 000=\text{R}6\ 000[/latex] after four years.
   3. Nomsa’s investment (compound interest):
      [latex]\scriptsize P=\text{R10}\ 000[/latex], [latex]\scriptsize i=0.15[/latex], [latex]\scriptsize n=4[/latex], [latex]\scriptsize A=?[/latex]
      [latex]\scriptsize \begin{align*}A & =P{{(1+i)}^{n}}\\ & =10\ 000{{(1+0.15)}^{4}}\\ & =17\ 490.06\end{align*}[/latex]
      The balance after four years will be [latex]\scriptsize \text{R}17\ 490.06[/latex].
   4. Sonto will have earned [latex]\scriptsize \text{R}17\ 490.06-\text{R}10\ 000=\text{R7}\ 490.06[/latex] after four years.
   5. Nomsa’s investment is better as she earns a higher return (more interest) at the same rate over the same term.
4. .
   1. Nzuzo will need to wait eight years.
   2. This is a compound interest problem.
      [latex]\scriptsize P=\text{R10}\ 000[/latex], [latex]\scriptsize i=0.13[/latex], [latex]\scriptsize n=8[/latex], [latex]\scriptsize A=?[/latex]
      [latex]\scriptsize \begin{align*}A & =P{{(1+i)}^{n}}\\ & =10\ 000{{(1+0.13)}^{8}}\\ & =26\ 584.44\end{align*}[/latex]
      Nzuzo received[latex]\scriptsize \text{R2}6\ 584.44[/latex].
   3. This is a simple interest problem.
      [latex]\scriptsize P=\text{R15}\ 000[/latex], [latex]\scriptsize i=0.18[/latex], [latex]\scriptsize n=5[/latex], [latex]\scriptsize A=?[/latex]
      [latex]\scriptsize \begin{align*}A & =P(1+in)\\ & =15\ 000(1+0.18\times 5)\\ & =28\ 500\end{align*}[/latex]
      Tando earned [latex]\scriptsize \text{R28}\ 500-\text{R}15\ 000=\text{R}13\ 500[/latex] interest.
   4. Bamukele collected on her policy after three years, which was [latex]\scriptsize 36[/latex] months. Each month, her investment earned [latex]\scriptsize \text{R}250[/latex] for a total of [latex]\scriptsize \text{R}250\times 36=\text{R}9\ 000[/latex]. Therefore, she collected a total of [latex]\scriptsize \text{R25}\ 000+\text{R9}\ 000=\text{R34}\ 000[/latex].
   5. Bamukele received the largest amount.

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