Banks and other financial institutions allow us to borrow money in the form of a loan, but in return they charge interest on the balance of the loan. While being charged compound interest on the balance of the loan, you also make regular repayments to reduce this balance and eventually repay the loan entirely. So we call these reducing balance loans, and as the amount owing on the loan gets smaller after repayments are made, so does the amount of interest that is charged each period.
Exploration 1
Consider the example of using a small loan to purchase a new pair of shoes for $\$300$$300. Interest of $10%$10% per week is charged on the balance of the loan, and we are able to pay back $\$100$$100 each week.
The interest charged in the first week is $300\times0.1=30$300×0.1=30, so if $\$100$$100 is repaid, the amount due at the end of the first week is $300+30-100=\$230$300+30−100=$230.
The interest added in the second week is then $230\times0.1=23$230×0.1=23, so if another $\$100$$100 is repaid, the amount due at the end of the second week will be $230+23-100=\$153$230+23−100=$153.
The reducing balance for the first $4$4 weeks of the loan is shown in the table.
| Period (n) | Value at beginning of period | Interest added this period | Repayment | Value at end of period |
|---|---|---|---|---|
| 1 | $\$300$$300 | $\$30$$30 | $\$100$$100 | $\$230$$230 |
| 2 | $\$230$$230 | $\$23$$23 | $\$100$$100 | $\$153$$153 |
| 3 | $\$153$$153 | $\$15.30$$15.30 | $\$100$$100 | $\$68.30$$68.30 |
| 4 | $\$68.30$$68.30 | $\$6.83$$6.83 | $\$75.13$$75.13 | $\$0$$0 |
We can see that even though we took out a loan of $\$300$$300, and made repayments of $\$100$$100, it took us $4$4 weeks (rather than $3$3) to repay the loan. This is due to the interest that was added to the loan each week.
If we want to find the total loan amount (i.e. the total amount we end up repaying), we can find the sum of all the repayments we made.
$100+100+100+75.13=\$375.13$100+100+100+75.13=$375.13
Note that as the size of the outstanding balance on the loan decreased, so too did the amount of interest added each week. This means that less and less of each repayment is taken up by interest, and more and more of each repayment goes towards paying off the initial loan.
When considering borrowing money, it can be beneficial to compare interest rates and play with different repayments and loan terms. Even small changes in the interest rate can make a big difference in the total amount of interest charged over the term of the loan.
Exploration 2
A student takes out a $\$3000$$3000 loan for their first car and makes yearly repayments of $\$700$$700. The interest on the loan is $12%$12% p.a. compounded monthly. We want to calculate how much is owed after the second repayment.
The loan accrues interest throughout the first year, which is compounded monthly. We can calculate this in the same way we calculate compound interest for an investment. There are $12$12 periods of interest before the first repayment.
First year
| $r$r | $=$= | $\frac{12}{12\times100}$1212×100 | (Converting into monthly interest) |
| $FV$FV | $=$= | $PV\left(1+r\right)^n$PV(1+r)n | (The compound interest formula) |
| First year loan with interest | $=$= | $3000\times\left(1+\frac{12}{1200}\right)^{12}$3000×(1+121200)12 | (Substituting values into formula) |
| $=$= | $\$3380.48$$3380.48 | (Evaluating to two decimal places) |
We can now subtract the first loan repayment from the amount owing on the loan.
| First year: Balance of loan amount at end of period | $=$= | Loan with interest - Repayment |
| $=$= | $3380.48-700$3380.48−700 | |
| $=$= | $\$2680.48$$2680.48 |
Second year
We can then use the end balance of the first year as the starting balance of the second year, and perform the same calculations.
| $FV$FV | $=$= | $PV\left(1+r\right)^n$PV(1+r)n | (The compound interest formula) |
| Second year loan with interest | $=$= | $2680.48\times\left(1+\frac{12}{1200}\right)^{12}$2680.48×(1+121200)12 | (Substituting values into formula) |
| $=$= | $\$3020.43$$3020.43 | (Evaluating to two decimal places) |
| Second year: Balance of loan amount at end of period | $=$= | Loan with interest - Repayment |
| $=$= | $3020.43-700$3020.43−700 | |
| $=$= | $\$2320.43$$2320.43 |
So the amount owing after the second repayment is $\$2320.43$$2320.43.
Question 1
Gwen takes out a loan to purchase a surround sound system. She makes $11$11 equal loan repayments. The total loan amount paid is $\$6600$$6600.
What is the value of each repayment?
Question 2
Laura takes out a loan of $\$86000$$86000 to renovate her home. The loan accrues interest at $8%$8% p.a. compounded monthly. Repayments of $\$11008$$11008 are made annually.
How much money does Laura's loan increase by in the first year? Give your answer to the nearest cent.
How much money does Laura owe after the first repayment?
How much money does Laura's loan increase by in the second year?
How much money does Laura owe after the second repayment?
Question 3
A car loan of $\$6000$$6000 accrues interest at $8%$8% p.a. compounded annually. Repayments of $\$720$$720 are made annually. The table below tracks the loan amount over three years.
Fill in the table below.
Time Period (n) Value at beginning of time period Interest at end of time period Repayment this period Amount at end of time period 1 $\$6000$$6000 $\$$$$\editable{}$ $\$$$$\editable{}$ $\$$$$\editable{}$ 2 $\$$$$\editable{}$ $\$460.80$$460.80 $\$720$$720 $\$$$$\editable{}$ 3 $\$5500.80$$5500.80 $\$$$$\editable{}$ $\$720$$720 $\$$$$\editable{}$
Question 4
Each of the following graphs 1, 2, 3 and 4 represents a reducing balance loan in which the repayments are made annually. Match each of the graphs to one of the following scenarios.
a. Interest is added semi-annually and the loan is repaid in more than $3$3 years.
b. Interest is added quarterly and the loan is repaid in more than $3$3 years.
c. Interest is added every $6$6 months and the loan will never be repaid.
d. Interest is added annually and the loan is repaid in $3$3 years.
Graph 1
Graph 2
Graph 3
Graph 4