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# 9.2 Simple interest

## 9.2 Simple interest (EMA6M)

Simple interest

Simple interest is interest calculated only on the initial amount that you invested.

As an easy example of simple interest, consider how much we will get by investing $$\text{R}\,\text{1 000}$$ for $$\text{1}$$ year with a bank that pays $$\text{5}\%$$ p.a. simple interest.

At the end of the year we have

\begin{align*} \text{Interest} & = \text{R}\,\text{1 000}\times \text{5}\% \\ & = \text{R}\,\text{1 000}\times \frac{5}{100} \\ & = \text{R}\,\text{1 000} \times \text{0,05} \\ & = \text{R}\,\text{50} \end{align*}

With an opening balance of $$\text{R}\,\text{1 000}$$ at the start of the year, the closing balance at the end of the year will therefore be

\begin{align*} \text{Closing balance}& = \text{Opening balance } + \text{ Interest} \\ & = \text{R}\,\text{1 000} + \text{R}\,\text{50} \\ & = \text{R}\,\text{1 050} \end{align*}

The opening balance in financial calculations is often called the principal, denoted as $$P$$ ($$\text{R}\,\text{1 000}$$ in the example). The interest rate is usually labelled $$i$$ ($$\text{5}\%$$ p.a. in the example and “p.a.” means per annum or per year). The interest amount is labelled $$I$$ ($$\text{R}\,\text{50}$$ in the example).

So we can see that

$I = P \times i$

and

\begin{align*} \text{Closing balance } & = \text{ Opening balance } + \text{ Interest} \\ & = P + I \\ & = P + P \times i \\ & = P\left(1 + i\right) \end{align*}

The above calculations give a good idea of what the simple interest formula looks like. However, the example shows an investment that lasts for only one year. If the investment or loan is over a longer period, we need to take this into account. We use the symbol $$n$$ to indicate time period, which must be given in years.

The general formula for calculating simple interest is

\begin{align*} A & = P\left(1 + in\right) \\ \text{Where: } & \\ A & = \text{ accumulated amount (final)} \\ P & = \text{ principal amount (initial)} \\ i & = \text{interest written as decimal} \\ n & = \text{number of years} \end{align*}

## Worked example 1: Calculating interest on a deposit

Carine deposits $$\text{R}\,\text{1 000}$$ into a special bank account which pays a simple interest rate of $$\text{7}\%$$ p.a. for $$\text{3}$$ years. How much will be in her account at the end of the investment term?

### Write down known values

\begin{align*} P & = \text{1 000} \\ i & = \text{0,07} \\ n & = 3 \end{align*}

### Write down the formula

$A = P\left(1 + in\right)$

### Substitute the values

\begin{align*} A & = \text{1 000}\left(1 + \text{0,07}\times 3\right) \\ & = \text{1 210} \end{align*}

At the end of $$\text{3}$$ years, Carine will have $$\text{R}\,\text{1 210}$$ in her bank account.

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## Worked example 2: Calculating interest on a loan

Sarah borrows $$\text{R}\,\text{5 000}$$ from her neighbour at an agreed simple interest rate of $$\text{12,5}\%$$ p.a. She will pay back the loan in one lump sum at the end of $$\text{2}$$ years. How much will she have to pay her neighbour?

### Write down the known variables

\begin{align*} P & = \text{5 000} \\ i & = \text{0,125} \\ n & = 2 \end{align*}

### Write down the formula

$A = P\left(1+in\right)$

### Substitute the values

\begin{align*} A & = \text{5 000}\left(1 + \text{0,125}\times 2\right) \\ & = \text{6 250} \end{align*}

At the end of $$\text{2}$$ years, Sarah will pay her neighbour $$\text{R}\,\text{6 250}$$.

We can use the simple interest formula to find pieces of missing information. For example, if we have an amount of money that we want to invest for a set amount of time to achieve a goal amount, we can rearrange the variables to solve for the required interest rate. The same principles apply to finding the length of time we would need to invest the money, if we knew the principal and accumulated amounts and the interest rate.

Important: to get a more accurate answer, try to do all your calculations on the calculator in one go. This will prevent rounding off errors from influencing your final answer.

## Worked example 3: Determining the investment period to achieve a goal amount

Prashant deposits $$\text{R}\,\text{30 000}$$ into a bank account that pays a simple interest rate of $$\text{7,5}\%$$ p.a.. How many years must he invest for to generate $$\text{R}\,\text{45 000}$$?

### Write down the known variables

\begin{align*} A & = \text{45 000} \\ P & = \text{30 000} \\ i & = \text{0,075} \end{align*}

### Write down the formula

$A = P\left(1 + in\right)$

### Substitute the values and solve for $$n$$

\begin{align*} \text{45 000} & = \text{30 000}\left(1 + \text{0,075}\times n\right) \\ \frac{\text{45 000}}{\text{30 000}} & = 1 + \text{0,075}\times n \\ \frac{\text{45 000}}{\text{30 000}} - 1 & = \text{0,075}\times n \\ \frac{\left(\frac{\text{45 000}}{\text{30 000}}\right)-1}{\text{0,075}}& = n \\ n & = 6\frac{2}{3} \end{align*}

It will take $$\text{6}$$ years and $$\text{8}$$ months to make $$\text{R}\,\text{45 000}$$ from $$\text{R}\,\text{30 000}$$ at a simple interest rate of $$\text{7,5}\%$$ p.a.

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## Worked example 4: Calculating the simple interest rate to achieve the desired growth

At what simple interest rate should Fritha invest if she wants to grow $$\text{R}\,\text{2 500}$$ to $$\text{R}\,\text{4 000}$$ in $$\text{5}$$ years?

### Write down the known variables

\begin{align*} A & = \text{4 000} \\ P & = \text{2 500} \\ n & = 5 \end{align*}

### Write down the formula

$A = P\left(1 + in\right)$

### Substitute the values and solve for $$i$$

\begin{align*} \text{4 000} & = \text{2 500}\left(1 + i\times 5\right) \\ \frac{\text{4 000}}{\text{2 500}} & = 1 + i\times 5 \\ \frac{\text{4 000}}{\text{2 500}} - 1 & = i\times 5 \\ \frac{\left(\frac{\text{4 000}}{\text{2 500}}\right) - 1}{5}& = i \\ i & = \text{0,12} \end{align*}

A simple interest rate of $$\text{12}\%$$ p.a. will be needed when investing $$\text{R}\,\text{2 500}$$ for $$\text{5}$$ years to become $$\text{R}\,\text{4 000}$$.

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Textbook Exercise 9.1

An amount of $$\text{R}\,\text{3 500}$$ is invested in a savings account which pays simple interest at a rate of $$\text{7,5}\%$$ per annum. Calculate the balance accumulated by the end of $$\text{2}$$ years.

\begin{align*} P & = \text{3 500}\\ i & = \text{0,075}\\ n & = 2\\ A & = ?\\\\ A & = P(1 + in) \\ A &= \text{3 500}(1 + (\text{0,075})(2))\\ A &= \text{3 500}(\text{1,15})\\ A &= \text{R}\,\text{4 025} \end{align*}

An amount of $$\text{R}\,\text{4 090}$$ is invested in a savings account which pays simple interest at a rate of $$\text{8}\%$$ per annum. Calculate the balance accumulated by the end of $$\text{4}$$ years.

Read the question carefully and write down the given information: \begin{align*} A & = ? \\ P & = \text{4 090}\\ n & = 4 \\ i & = \frac{8}{100} = \text{0,08} \end{align*}

\begin{align*} A &= P(1+in)\\ &= \text{R}\,\text{4 090} \left(\text{1} + \left(\text{0,08}\right) \times \text{4}\right) \\ &= \text{R}\,\text{5 398,80} \end{align*}

An amount of $$\text{R}\,\text{1 250}$$ is invested in a savings account which pays simple interest at a rate of $$\text{6}\%$$ per annum. Calculate the balance accumulated by the end of $$\text{6}$$ years.

Read the question carefully and write down the given information: \begin{align*} A & = ? \\ P & = \text{1 250}\\ n & = 6 \\ i & = \frac{6}{100} = \text{0,06} \end{align*}

Simple interest formula: \begin{align*} A &= P(1+in)\\ &= \text{R}\,\text{1 250} \left(\text{1} + \left(\text{0,06}\right) \times \text{6}\right) \\ &= \text{R}\,\text{1 700,00} \end{align*}

An amount of $$\text{R}\,\text{5 670}$$ is invested in a savings account which pays simple interest at a rate of $$\text{8}\%$$ per annum. Calculate the balance accumulated by the end of $$\text{3}$$ years.

Read the question carefully and write down the given information: \begin{align*} A & = ? \\ P & = \text{5 670}\\ n & = 3 \\ i & = \frac{8}{100} = \text{0,08} \end{align*}

Simple interest formula: \begin{align*} A &= P(1+in)\\ &= \text{R}\,\text{5 670,00} \left(\text{1} + \left(\text{0,08}\right) \times \text{3}\right) \\ &= \text{R}\,\text{7 030,80} \end{align*}

Calculate the accumulated amount in the following situations:

A loan of $$\text{R}\,\text{300}$$ at a rate of $$\text{8}\%$$ for $$\text{1}$$ year.

\begin{align*} P & = 300 \\ i & = \text{0,08} \\ n & = 1 \\ A & = ? \\ \\ A & = P(1 + in) \\ A & = 300(1+ (\text{0,08})(1))\\ A & = 300(\text{1,08})\\ A & = \text{R}\,\text{324} \end{align*}

An investment of $$\text{R}\,\text{2 250}$$ at a rate of $$\text{12,5}\%$$ p.a. for $$\text{6}$$ years.

\begin{align*} P & = \text{2 250} \\ i & = \text{0,125} \\ n & = 6 \\ A & = ? \\ \\ A & = P(1 + in) \\ A & = \text{2 250}(1+ (\text{0,125})(6))\\ A & = \text{2 250}(\text{1,75})\\ A & = \text{R}\,\text{3 937,50} \end{align*}

A bank offers a savings account which pays simple interest at a rate of $$\text{6}\%$$ per annum. If you want to accumulate $$\text{R}\,\text{15 000}$$ in $$\text{5}$$ years, how much should you invest now?

Read the question carefully and write down the given information: \begin{align*} A & = \text{R}\,\text{15 000} \\ P & = ? \\ i & = \frac{6}{100} = \text{0,06} \\ n & = 5 \end{align*}

Simple interest formula: \begin{align*} A &= P(1 + in)\\ \text{R}\,\text{15 000} &= P \left(\text{1} + \left(\text{0,06}\right) \times \text{5}\right) \\ P & = \frac{\text{R}\,\text{15 000}}{\text{1,3}} \\ &= \text{R}\,\text{11 538,46} \end{align*}

Sally wanted to calculate the number of years she needed to invest $$\text{R}\,\text{1 000}$$ for in order to accumulate $$\text{R}\,\text{2 500}$$. She has been offered a simple interest rate of $$\text{8,2}\%$$ p.a. How many years will it take for the money to grow to $$\text{R}\,\text{2 500}$$?

\begin{align*} A & = \text{2 500} \\ P & = \text{1 000} \\ i & = \text{0,082} \\ n & = ? \\ \\ A & = P(1 + in) \\ \text{2 500} &= \text{1 000}(1+ (\text{0,082})(n))\\ \frac{\text{2 500}}{\text{1 000}} &= 1 + \text{0,082}n\\ \frac{\text{2 500}}{\text{1 000}} - 1 &= \text{0,082}n \\ \left(\frac{\text{2 500}}{\text{1 000}} - 1\right) \div \text{0,082} & = n\\ n & = \text{18,3} \end{align*}

It would take $$\text{19}$$ years for $$\text{R}\,\text{1 000}$$ to become $$\text{R}\,\text{2 500}$$ at $$\text{8,2}\%$$ p.a.

Joseph deposited $$\text{R}\,\text{5 000}$$ into a savings account on his son's fifth birthday. When his son turned $$\text{21}$$, the balance in the account had grown to $$\text{R}\,\text{18 000}$$. If simple interest was used, calculate the rate at which the money was invested.

\begin{align*} A & = \text{18 000} \\ P & = \text{5 000} \\ i & = ? \\ n & = 21 - 5 = 16 \\ \\ A & = P(1 + in) \\ \text{18 000} = \text{5 000}(1+ (i)(16)) \\ \frac{\text{18 000}}{\text{5 000}} & = 1 + 16i\\ \frac{\text{18 000}}{\text{5 000}} - 1 & = 16i \\ \left(\frac{\text{18 000}}{\text{5 000}} - 1\right) \div 16 & = i\\ i & = \text{0,1625} \end{align*}

The interest rate at which the money was invested was $$\text{16,25}\%$$.

When his son was $$\text{6}$$ years old, Methuli made a deposit of $$\text{R}\,\text{6 610}$$ in the bank. The investment grew at a simple interest rate and when Methuli's son was $$\text{18}$$ years old, the value of the investment was $$\text{R}\,\text{11 131,24}$$.

At what rate was the money invested? Give the answer correct to one decimal place.

Read the question carefully and write down the given information: \begin{align*} A & = \text{R}\,\text{11 131,24} \\ P & = \text{R}\,\text{6 610} \\ i & = ? \\ n & = 18 - 6 = 12 \end{align*}

The question says that the investment “grew at a simple interest rate”, so we must use the simple interest formula. To calculate the interest rate, we need to make $$i$$ the subject of the formula: \begin{align*} A &= P(1+in)\\ \frac{A}{P} &= 1+in\\ \frac{A}{P}-1 &= in\\ \frac{\frac{A}{P}-1}{n} &= i\\ \text{Therefore } i &= \frac{\left(\frac{\text{}\, \text{11 131,24}}{\text{6 610}}\right) - 1}{\text{12}}\\ & = \text{0,057}\\ & = \text{5,7}\% \text{ per annum} \end{align*}

When his son was $$\text{6}$$ years old, Phillip made a deposit of $$\text{R}\,\text{5 040}$$ in the bank. The investment grew at a simple interest rate and when Phillip's son was $$\text{18}$$ years old, the value of the investment was $$\text{R}\,\text{7 338,24}$$.

At what rate was the money invested? Give your answer correct to one decimal place.

Read the question carefully and write down the given information: \begin{align*} A & = \text{R}\,\text{7 338,24} \\ P & = \text{R}\,\text{5 040} \\ i & = ? \\ n & = 18 - 6 = 12 \end{align*}

The question says that the investment “grew at a simple interest rate”, so we must use the simple interest formula. To calculate the interest rate, we need to make $$i$$ the subject of the formula: \begin{align*} A &= P(1+in)\\ \frac{A}{P} &= 1+in\\ \frac{A}{P}-1 &= in\\ \frac{\frac{A}{P}-1}{n} &= i\\ \text{Therefore } i &= \frac{\left(\frac{\text{}\, \text{7 338,24}}{\text{5 040}}\right) - 1}{\text{12}}\\ & = \text{0,038}\\ & = \text{3,8}\% \text{ per annum} \end{align*}

When his son was $$\text{10}$$ years old, Lefu made a deposit of $$\text{R}\,\text{2 580}$$ in the bank. The investment grew at a simple interest rate and when Lefu's son was $$\text{20}$$ years old, the value of the investment was $$\text{R}\,\text{3 689,40}$$.

At what rate was the money invested? Give your answer correct to one decimal place.

Read the question carefully and write down the given information: \begin{align*} A & = \text{R}\,\text{3 689,40} \\ P & = \text{R}\,\text{2 580} \\ i & = ? \\ n & = 20 - 10 = 10 \end{align*}

The question says that the investment “grew at a simple interest rate”, so we must use the simple interest formula. To calculate the interest rate, we need to make $$i$$ the subject of the formula: \begin{align*} A &= P(1+in)\\ \frac{A}{P} &= 1+in\\ \frac{A}{P}-1 &= in\\ \frac{\frac{A}{P}-1}{n} &= i\\ \text{Therefore } i &= \frac{\left(\frac{\text{}\, \text{3 689,40}}{\text{2 580}}\right) - 1}{\text{10}}\\ & = \text{0,043}\\ & = \text{4,3}\% \text{ per annum} \end{align*}

Abdoul wants to invest $$\text{R}\,\text{1 080}$$ at a simple interest rate of $$\text{10,9}\%$$ p.a.

How many years will it take for the money to grow to $$\text{R}\,\text{3 348}$$? Round up your answer to the nearest year.

Read the question carefully and write down the given information: \begin{align*} A & = \text{R}\,\text{3 348} \\ P & = \text{R}\,\text{1 080} \\ i & = \frac{\text{10,9}}{100} = \text{0,109} \\ n & = ? \end{align*}

To calculate the number of years, we need to make $$n$$ the subject of the formula: \begin{align*} A &= P(1+in)\\ \frac{A}{P}&= 1+in\\ \frac{A}{P}-1 & = in\\ \frac{\frac{A}{P}-1}{i} & = n\\ \text{Therefore } n & = \frac{\left(\frac{\text{3 348}}{\text{1 080}}\right) - \text{1}}{\text{0,109}} \\ & = \text{19,3}\\ & = \text{20} \text{ years} \quad \Leftarrow \text{round UP to the nearest integer} \end{align*}

Andrew wants to invest $$\text{R}\,\text{3 010}$$ at a simple interest rate of $$\text{11,9}\%$$ p.a.

How many years will it take for the money to grow to $$\text{R}\,\text{14 448}$$? Round up your answer to the nearest year.

Read the question carefully and write down the given information: \begin{align*} A & = \text{R}\,\text{14 448} \\ P & = \text{R}\,\text{3 010} \\ i & = \frac{\text{11,9}}{100} = \text{0,119} \\ n & = ? \end{align*}

To calculate the number of years, we need to make $$n$$ the subject of the formula: \begin{align*} A &= P(1+in)\\ \frac{A}{P}&= 1+in\\ \frac{A}{P}-1 & = in\\ \frac{\frac{A}{P}-1}{i} & = n\\ \text{Therefore } n & = \frac{\left(\frac{\text{14 448}}{\text{3 010}}\right) - \text{1}}{\text{0,119}} \\ & = \text{31,9}\\ & = \text{32} \text{ years} \quad \Leftarrow \text{round UP to the nearest integer} \end{align*} Rounding up to the nearest year, it will take $$\text{32}$$ years to reach the goal of saving $$\text{R}\,\text{14 448}$$.