Paying off a Loan (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Calculating Amount Owing After Six Years

Let's say Yianni and Eleni borrow $20,000 at 12% per annum, compounded monthly. They make monthly payments of $300 starting one month after taking out the loan.

Question: How much do they still owe after six years?

Solution:

First, convert the interest rate to a monthly rate:

Monthly rate = $12\% \div 12 = 1\%$

So $1 + R = 1.01$

Calculate the loan growth over 72 months (6 years):

Initial loan after 72 months = $20000 \times 1.01^{72}$

Now calculate how each instalment grows:

• 1st instalment (invested for 71 months) = $300 \times 1.01^{71}$
• 2nd instalment (invested for 70 months) = $300 \times 1.01^{70}$
• ...continuing this pattern...
• 71st instalment (invested for 1 month) = $300 \times 1.01^1$
• 72nd instalment (invested for 0 months) = $300$

The total of all instalments forms a GP with:

• First term $a = 300$
• Common ratio $r = 1.01$
• Number of terms $= 72$

Amount still owing after 72 months:

$A_{72} = 20000 \times 1.01^{72} - (300 + 300 \times 1.01 + \ldots + 300 \times 1.01^{71})$

Using the GP sum formula $S_n = \frac{a(r^n - 1)}{r - 1}$:

$A_{72} = 20000 \times 1.01^{72} - \frac{300(1.01^{72} - 1)}{0.01}$

$A_{72} = 20000 \times 1.01^{72} - 30000(1.01^{72} - 1)$

$A_{72} \approx \$9529$

Finding the interest paid:

Total instalments paid = $300 \times 72 = 21600$

Reduction in loan = $20000 - 9529 = 10471$

Interest charged = $21600 - 10471 = 11129$

Notice this is more than half the original loan amount!

Worked Example: Determining Monthly Instalment Amount

Ali takes out a $10,000 loan to buy a car. He will repay it over 5 years with 60 equal monthly instalments, beginning one month after taking out the loan. Interest is charged at 6% per annum, compounded monthly.

Question: What should the monthly instalment be?

Solution:

Monthly interest rate = $6\% \div 12 = 0.5\%$

So $1 + R = 1.005$

Let each instalment be $M$ dollars.

Calculate the amount owing after 60 months, then set it equal to zero (since the loan is fully repaid).

Initial loan after 60 months = $10000 \times 1.005^{60}$

The instalments form a GP with:

• First term $a = M$
• Common ratio $r = 1.005$
• Number of terms $= 60$

Amount owing after 60 months:

$A_{60} = 10000 \times 1.005^{60} - \frac{M(1.005^{60} - 1)}{0.005}$

$A_{60} = 10000 \times 1.005^{60} - 200M(1.005^{60} - 1)$

Since the loan is fully paid off, set $A_{60} = 0$:

$10000 \times 1.005^{60} - 200M(1.005^{60} - 1) = 0$

$200M(1.005^{60} - 1) = 10000 \times 1.005^{60}$

$M(1.005^{60} - 1) = 50 \times 1.005^{60}$

$M = \frac{50 \times 1.005^{60}}{1.005^{60} - 1}$

$M \approx \$193.33$

Worked Example: Finding Loan Repayment Duration

Natasha and Richard take out a $200,000 loan on 1st January 2002 with monthly instalments of $2200. Interest is charged at 12% per annum, compounded monthly.

Part a: Find a formula for the amount owing after $n$ months.

Solution:

Monthly interest rate = $1\%$, so $1 + R = 1.01$

Initial loan after $n$ months = $200000 \times 1.01^n$

The instalments form a GP with:

• First term $a = 2200$
• Common ratio $r = 1.01$
• Number of terms $= n$

$A_n = 200000 \times 1.01^n - \frac{2200(1.01^n - 1)}{0.01}$

$A_n = 200000 \times 1.01^n - 220000(1.01^n - 1)$

$A_n = 220000 - 20000 \times 1.01^n$

Part b: How much is owing after 5 years?

Substitute $n = 60$:

$A_{60} = 220000 - 20000 \times 1.01^{60}$

$A_{60} \approx \$183,666$

This is still almost as much as the original loan!

Part c(i): How long to repay the full loan?

Set $A_n = 0$:

$220000 - 20000 \times 1.01^n = 0$

$20000 \times 1.01^n = 220000$

$1.01^n = 11$

Convert to a logarithmic equation:

$n = \log_{1.01} 11$

Using the change-of-base formula:

$n = \frac{\log_{10} 11}{\log_{10} 1.01}$

$n \approx 241 \text{ months}$

This equals approximately 20 years and 1 month.

Part c(ii): How long to repay half the loan?

Set $A_n = 100000$:

$220000 - 20000 \times 1.01^n = 100000$

$20000 \times 1.01^n = 120000$

$1.01^n = 6$

$n = \log_{1.01} 6 = \frac{\log_{10} 6}{\log_{10} 1.01}$

$n \approx 180 \text{ months} = 15 \text{ years}$

Notice this is about three-quarters, not half, of the total loan period!

Part d: Why would $1900 monthly instalments never repay the loan?

With a $200,000 loan at 1% monthly interest:

Initial monthly interest = $200000 \times 0.01 = 2000$

At the start of the loan, $2000 of the instalment is needed just to cover the interest. With payments of only $1900, the debt would actually increase rather than decrease.