It costs money to borrow money. The extra money that banks and other lenders charge us to borrow money is called interest. Interest may also refer to the additional money that is earned from investing money, such as into a savings account. There are different types of interest and in this lesson we are going to look at simple interest.
What is simple interest?
Simple interest, or flat rate interest, describes a method of calculating interest where the interest amount is fixed (i.e. it doesn't change). The interest charge is always based on the original amount borrowed (or invested), and does not take into account any interest earned along the way (that is, interest on interest is not included).
Many financial institutions express their interest rates as a percentage "per annum", often abbreviated to "p.a.", which means "per year". For example, an interest rate might be given as $3%$3% p.a.
Simple interest is calculated as:
$I=Pin$I=Pin
- $P$P is the principal amount invested (or borrowed)
- $i$i is the interest rate per time period expressed as a decimal or fraction
- $n$n is the number of time periods (the duration of the loan or investment)
The total total value of an investment or loan $A$A is then calculated as the principal plus interest:
$A=P+I$A=P+I
Caution: To use the formula, ensure that the time periods for the rate and $n$n match. For example and annual rate and the duration given in years.
Worked example
example 1
An investment of $\$8580$$8580 is deposited at $2%$2% p.a. flat rate for $4$4 years.
a) Find the interest earned in $1$1 year.
Think: $2%$2% interest is earned each year. So we need to find $2%$2% of $\$8580$$8580. Can you see this in the same as using the formula with $P=\$8580$P=$8580, $i=2%$i=2% and $n=1$n=1 year? For the formula express $i$i as a decimal or fraction, thus $i=\frac{2}{100}$i=2100 or $0.02$0.02.
Do:
| $I$I | $=$= | $Pin$Pin | |
| $=$= | $\$8580\times0.02\times1$$8580×0.02×1 | ||
| $=$= | $\$171.60$$171.60 |
For financial questions give answers to $2$2 decimal places |
Thus, $\$171.60$$171.60 in interest would be earned in $1$1 year for this investment.
b) Find the interest earned in $4$4 years.
Think: For simple interest the same amount is earned each year, so we could multiply the figure found in part (a) by $4$4. Can you see this is the same as using the formula with the values $P=\$8580$P=$8580, $i=0.02$i=0.02, $n=4$n=4 years?
Do:
| $I$I | $=$= | $Pin$Pin |
| $=$= | $8580\times0.02\times4$8580×0.02×4 | |
| $=$= | $\$686.60$$686.60 |
c) What is the total value of the investment after $4$4 years?
Think: The total value is the original investment plus the interest that has accrued over $4$4 years.
Do:
| $A$A | $=$= | $P+I$P+I |
| $=$= | $\$8580+\$686.60$$8580+$686.60 | |
| $=$= | $\$9266.40$$9266.40 |
Practice questions
Question 1
Calculate the simple interest earned on an investment of $\$2540$$2540 at $9%$9% p.a. for $2$2 years.
Give your answer to the nearest cent.
Question 2
Buzz takes out $6$6-year $\$50000$$50000 loan at $8%$8% p.a. flat interest.
What is the total interest he will have to pay?
What is the total amount he will have to pay?
Different rates and times periods
In the previous examples the rate given was an annual rate and the time was given in years. We could also use the formula for a monthly rate and the duration given in months. However, if the time period for the rate and $n$n don't match we will need to convert one or both. Converting rates to annual and the duration to years is a common approach.
For example, if we are given a rate which is not annual we can convert it to annual: a rate of $1.5%$1.5% per quarter is the same as an annual rate of $1.5%\times4=6%$1.5%×4=6% p.a., since there are $4$4 quarters in a year.
Or when calculating simple interest for time periods that are not years, such as months, weeks or days, we can express the the time given as a fraction of a year. For example, for an investment of $10$10 months $n=\frac{10}{12}$n=1012 or for a loan for $30$30 days $n=\frac{30}{365}$n=30365.
| $1$1 year | $=$= | $12$12 months |
| $=$= | $52$52 weeks | |
| $=$= | $4$4 quarters (of $3$3 months each) | |
| $=$= | $365$365 days |
Worked example
example 2
A loan of $\$2000$$2000 is taken out at $5%$5% p.a. flat rate for $18$18 months. Find the interest due on the loan after $18$18 months.
Think: We can substitute the values into the formula using $P=\$2000$P=$2000, $i=5%=0.05$i=5%=0.05 and $n=\frac{18}{12}$n=1812 years.
Do:
| $I$I | $=$= | $Pin$Pin |
| $=$= | $\$2000\times0.05\times\frac{18}{12}$$2000×0.05×1812 | |
| $=$= | $\$150$$150 |
Practice questions
Question 3
Question 4
Calculate the simple interest earned on an investment of $\$7000$$7000 at $1.8%$1.8% per quarter for $9$9 years.
Give your answer to the nearest cent.
Question 5
Calculate the simple interest earned on an investment of $\$5530$$5530 at $3%$3% p.a. for $5$5 weeks.
Assume that a year has $52$52 weeks.
Give your answer to the nearest cent.
Finding rate, principal or time
The formula for finding interest can also be rearranged to find an unknown rate, principal amount or time.
Rearranging $I=Pin$I=Pin for each variable we obtain:
| $P=\frac{I}{i\times n}$P=Ii×n | $n=\frac{I}{P\times i}$n=IP×i | $i=\frac{I}{P\times n}$i=IP×n |
| Unknown principal | Unknown time | Unknown rate |
Note:
- The formula is the similar for each with the interest earned on top of the fraction and the two known variables on the bottom
- $n$n is the is the number of time periods (the duration of the loan or investment)
- $i$i is the interest rate per time period expressed as a decimal or fraction, multiply by $100%$100% to convert to a percentage
Worked example
EXAMPLE 3
An investment of $\$2800$$2800 increases to a total value of $\$3052$$3052 in $2$2 years. What simple interest rate was the investment earning?
Think: For the formula to find a rate we need the interest earned by the investment, the principal and the time in years. In this case, we have not been given the interest earned directly but we can find it. The interest earned on the investment is the difference between the principal and final value. Hence,
| $I$I | $=$= | $\$3052-\$2800$$3052−$2800 |
| $=$= | $\$252$$252 |
Then we have the values necessary for our formula, with $I=\$252$I=$252, $P=\$2800$P=$2800 and $n=2$n=2 years.
Do:
| $i$i | $=$= | $\frac{I}{P\times n}$IP×n | |
| $=$= | $\frac{\$252}{\$2800\times2}$$252$2800×2 | ||
| $=$= | $0.045$0.045 | ||
| $\therefore i$∴i | $=$= | $0.045\times100%$0.045×100% |
Multiply by $100%$100% to convert the rate to a percentage |
| $=$= | $4.5%$4.5% p.a. |
The p.a. indicates the rate quoted is an annual rate. |
Practice questions
Question 6
For a simple interest rate of $6%$6% p.a. , calculate the number of years $T$T needed for an interest of $\$1174.20$$1174.20 to be earned on the investment $\$1957$$1957.
Give your answer as a whole number of years.
Enter each line of working as an equation.
Question 7
For his investment into government bonds , Buzz was paid simple interest of $9%$9% p.a.
Calculate the size of Buzz's initial investment $\$P$$P if he earned $\$294.12$$294.12 interest after $2$2 years.
Write each step as an equation.