Simple Interest (HSC SSCE Mathematics Standard): Revision Notes

Worked Example: Calculating Simple Interest on a Loan

Abbey applied for a flat-rate loan of $40,000 at 9% per annum simple interest. She plans to repay the loan after two years and six months.

a) How much interest will be paid?

Write the simple interest formula:

$I = Prn$

Identify the values:

• $P = 40000$
• $r = 0.09$ (9% expressed as a decimal)
• $n = 2.5$ (2 years and 6 months = 2.5 years)

Substitute into the formula:

$I = 40000 \times 0.09 \times 2.5$

Evaluate:

$I = \$9000$

Therefore, the simple interest owed is $9,000.

b) What is the total owing at the end of two years and six months?

Write the amount owed formula:

$A = P + I$

Substitute $P = 40000$ and $I = 9000$:

$A = 40000 + 9000$

$A = 49000$

Therefore, the amount owed is $49,000.

Worked Example: Interest on Savings Account

The table below shows the entries in Shane's bank account. If the bank pays interest at a rate of 3% per annum on the minimum monthly balance, find the interest payable for the month of May correct to the nearest cent.

Solution:

Determine the minimum monthly balance for May:

Looking at the table, the minimum balance during May was $440.00 (after the cash withdrawal on 4 May).

Write the simple interest formula:

$I = Prn$

Substitute the values:

• $P = 440.00$
• $r = 0.03$
• $n = \frac{1}{12}$ (one month is one-twelfth of a year)

$I = 440 \times 0.03 \times \frac{1}{12}$

Evaluate:

$I = \$1.10$

Therefore, the interest payable for May is $1.10.

Worked Example: Finding the Principal

Noah applied for a simple interest car loan with an interest rate of 9% p.a. He was told the total simple interest would be $6,300 for 3½ years. What was the principal?

Solution:

Write the simple interest formula:

$I = Prn$

Substitute the known values:

$6300 = P \times 0.09 \times 3.5$

Make $P$ the subject by dividing both sides by $(0.09 \times 3.5)$:

$P = \frac{6300}{0.09 \times 3.5}$

Evaluate:

$P = \$20000$

Therefore, the principal is $20,000.

Worked Example: Calculating Fortnightly Loan Repayments

Jessica wishes to buy a lounge suite priced at $2,750. She chooses to buy it on terms by paying a 10% deposit and borrowing the balance. Interest is charged at 11.5% p.a. on the amount borrowed. Jessica makes fortnightly repayments over 3 years. Calculate her fortnightly repayments.

Solution:

Calculate the deposit:

$\text{Deposit} = 10\% \text{ of } \$2750$

$= 0.10 \times 2750$

$= \$275$

Calculate the balance (amount to be borrowed):

$\text{Balance} = 2750 - 275$

$= \$2475$

Write the simple interest formula:

$I = Prn$

Substitute $P = 2475$, $r = 0.115$, and $n = 3$:

$I = 2475 \times 0.115 \times 3$

$= \$853.875$

Calculate the total to be paid:

$\text{Total} = \text{Principal + Interest}$

$= 2475 + 853.875$

$= \$3328.875$

Calculate the number of repayments:

There are 26 fortnights in a year, so over 3 years:

$\text{Number of repayments} = 3 \times 26 = 78$

Calculate the fortnightly repayment:

$\text{Repayment} = \frac{3328.875}{78}$

$= 42.67788462...$

$\approx \$42.68$

Therefore, Jessica's fortnightly repayments are $42.68.

Worked Example: Drawing a Simple Interest Graph

Draw a graph showing the amount of simple interest earned over a period of 4 years if $1,000 is invested at 6% p.a. Use the graph to estimate the interest earned after 8 years.

Solution:

Write the simple interest formula:

$I = Prn$

Substitute $P = 1000$ and $r = 0.06$:

$I = 1000 \times 0.06 \times n$

$I = 60n$

Create a table of values by substituting different values of $n$:

Draw the graph with $n$ on the horizontal axis and $I$ on the vertical axis. Plot the points $(0, 0)$, $(1, 60)$, $(2, 120)$, $(3, 180)$, and $(4, 240)$:

To estimate the interest after 8 years, extend the line to $n = 8$ and read the corresponding value of $I$ from the graph.

From the graph, when $n = 8$, $I \approx `$480$`.

Therefore, the interest after 8 years is approximately $480.