A simple interest investment is when the investment will increase by the same amount each year. It is another example of linear growth. The interest charged is always based on the initial amount, or principal, so the added percentage, or interest, remains constant.

When an investment is made by putting money into a bank account, the bank is borrowing money from you, and so pays you interest.

When money is borrowed from a bank (or another lending agency), you have to pay the bank interest for taking out a loan with them.

The amount of simple interest earned over a period of time can be calculated using the formula:

I=PRT where P is the principal (the initial amount borrowed or invested), R is the interest rate per time period, expressed as a decimal or fraction and T is the number of time periods (the duration of the loan).

If we are given an interest rate of r\% per annum, then to convert it to the R used in the formula above, we need to divide it by 100. So: R=\dfrac{r}{100}

Examples

Example 1

Calculate the simple interest earned on a loan \$8580 at 2\% p.a. 10 months.

Worked Solution

Create a strategy

Convert the given T to years and use the simple interest formula.

Apply the idea

There are 12 months in a year, so to find T in years we divide the given months by 12. So we have the given values: P=\$8580, \, R=2\%, and T=\dfrac{10}{12}.

\displaystyle I	\displaystyle =	\displaystyle PRT	Write the formula.
	\displaystyle =	\displaystyle 8580 \times 2\% \times \dfrac{10}{12}	Substitute the given values
	\displaystyle =	\displaystyle 5000\times 0.02 \times \dfrac{10}{12}	Convert the percentage to a decimal
	\displaystyle =	\displaystyle \$ 143.00	Evaluate

Example 2

Elizabeth has \$1300 to put into a savings account, which earns simple interest at a rate of 0.75\% p.a.

If she wants to earn \$20 in interest, how long will she have to wait?

Worked Solution

Create a strategy

Use the formula: I = PRT

Apply the idea

We are given: I=\$20, { } P=\$1300 and R=0.75\%=0.0075

\displaystyle I	\displaystyle =	\displaystyle PRT	Write the formula.
\displaystyle 20	\displaystyle =	\displaystyle 1300 \times 0.0075 \times T	Substitute the given values
\displaystyle 20	\displaystyle =	\displaystyle 9.75 \times T	Evaluate the multiplication
\displaystyle T	\displaystyle =	\displaystyle 2.05 \text{ years}	Divide both sides by 9.75

Idea summary

The amount of simple interest earned over a period of time can be calculated using the formula:

\displaystyle I = PRT

\bm{P}

is the principal (the initial amount borrowed or invested)

\bm{R}

is the interest rate per time period expressed as a decimal or fraction

\bm{T}

is the number of time periods (the duration of the loan)

If we are given an interest rate of r\% per annum, then to convert it to the R used in the formula above, we need to divide it by 100. So: R=\dfrac{r}{100}

Simple interest loans and investments

Modelling a simple interest investment or loan with a recurrence relation is the same as that of the flat rate and unit cost depreciation models except instead of subtracting an amount each period, an amount is added each period.

Interest on loans and investments can be modelled using the following recurrence relation:V_{n+1}=V_n +d, V_0=kWhere V_{n+1} is the value of the investment or loan after (n+1) time periods, d is the amount added per time period, calculated as a percentage of the principal, k is the initial value of the investment or loan; the principal value.

Note that a simple interest recurrence model is another example of an arithmetic sequence.

Examples

Example 3

An investment of \$6000 pays simple interest at a rate of 4.2\% per annum and is modelled by the recurrence relation V_{n+1}=V_n +252, \, V_0=6000 where V_{n+1} is the value of the investment after n years.

Use the sequence facility on your calculator to answer the following questions.

Calculate the value of the investment after 5 years.

Worked Solution

Create a strategy

Enter the given recurrence relation along with n=5 into your calculator.

Apply the idea

We get: V_6= 12\,100.

The value of the investment after 5 years will be \$7260.

How much interest has been earned in 5 years?

Worked Solution

Create a strategy

Subtract the initial investment from the value of the investment after n years.

Apply the idea

In part (a) the value of the investment after 5 years is \$7260 and the initial investment, V_0 is \$6000.

\displaystyle \text{interest earned}	\displaystyle =	\displaystyle 7260-6000	Substitute the values
	\displaystyle =	\displaystyle \$1260	Evaluate

After how many years will the investment double?

Worked Solution

Create a strategy

Input the recurrence relation given in the question into your calculator. Find the first value of n that is greater than or equal to 12\,000.

Apply the idea

By looking through the list generated by your calculator, you should find that the first value greater than or equal to \$12\,000 is \$12\,048 which occurs when n=24.

So the investment will double after 24 years.

Idea summary

Interest on loans and investments can be modelled using the following recurrence relation:

\displaystyle V_{n+1}=V_n+d, \, V_0=k

\bm{V_{n+1}}

is the value of the investment or loan after n+1 time periods

\bm{d}

is the amount of added per time period, calculated as a percentage of the principal

\bm{k}

is the initial value of the investment or loan; the principal value

Note that a simple interest recurrence model is another example of an arithmetic sequence.

Outcomes

U3.AoS2.6

demonstrate the use of a recurrence relation to determine the depreciating value of an asset or the future value of an investment or a loan after 𝑛 time periods for the initial sequence

U3.AoS2.3

the concepts of financial mathematics including simple and compound interest, nominal and effective interest rates, the present and future value of an investment, loan or asset, amortisation of a reducing balance loan or annuity and amortisation tables

U3.AoS2.4

the use of first-order linear recurrence relations to model compound interest investments and loans, and the flat rate, unit cost and reducing balance methods for depreciating assets, reducing balance loans, annuities, perpetuities and annuity investments

5.04 Simple interest investments and loans

Ideas

Simple interest

Examples

Example 1

Example 2

Simple interest loans and investments

Examples

Example 3

Outcomes

U3.AoS2.6

U3.AoS2.3

U3.AoS2.4

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