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SAGeneral MathematicsSyllabus dot point

How does a loan balance fall as we make regular repayments against compounding interest?

Model a reducing-balance loan with a recurrence relation, track the balance after each repayment, and analyse the effect of changing the repayment.

How to model a reducing-balance loan with a recurrence, build an amortisation schedule splitting each payment into interest and principal, and see how repayment size changes the loan term.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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What this dot point is asking
How a reducing-balance loan works
The amortisation schedule
Analysing changes to the repayment
Offset accounts and extra repayments
The total cost of a loan
The final repayment

What this dot point is asking

You must model the loan with a recurrence, build an amortisation schedule, and analyse how changing the repayment affects the term and total interest.

How a reducing-balance loan works

Interest is charged only on the amount still owed. Each period, interest is added to the balance, then the repayment is taken off. The recurrence is:

B_{n+1} = B_n(1 + i) - R

where $B_n$ is the balance after $n$ repayments, $i$ is the interest rate per period (decimal) and $R$ is the regular repayment.

Worked example

A $\$10\,000$ loan is charged 12% per annum compounded monthly, with repayments of $\$900$ per month. Find the outstanding balance after two repayments and identify how much of the first repayment goes to interest.

Step 1: Find the monthly interest rate

The annual rate of 12% applies each month at one-twelfth of the annual rate. Dividing by 12 converts it to the period rate the recurrence uses:

i = \frac{0.12}{12} = 0.01.

Step 2: Apply the recurrence to find the balance after month 1

The recurrence $B_{n+1} = B_n(1 + i) - R$ says: first grow the balance by one month's interest, then subtract the repayment. The order matters, because the lender charges interest on the balance before the repayment arrives.

B_1 = 10\,000(1.01) - 900 = 10\,100 - 900 = 9200.

Of the $\$900$ repayment, $0.01 \times 10\,000 = \$100$ covers interest and the remaining $\$800$ reduces the principal. The balance falls from $\$10\,000$ to $\$9200$ .

Step 3: Apply the recurrence again for month 2

With the new starting balance of $\$9200$ , interest is now charged on a smaller amount, so slightly less of the second payment goes to interest and slightly more reduces the debt:

B_2 = 9200(1.01) - 900 = 9292 - 900 = 8392.

Interest for month 2 is $0.01 \times 9200 = \$92$ , and $\$808$ reduces the principal.

Final answer: After two repayments, $\$8392$ is still owing. The interest portions were $\$100$ and $\$92$ respectively, with the principal reduction increasing each month.

The amortisation schedule

An amortisation schedule splits each repayment into the interest charged that period and the principal repaid (the rest of the payment).

Interest for the period $= i \times B_n$ (the balance at the start of the period).
Principal repaid $=$ repayment $R -$ interest.
New balance $= B_n -$ principal repaid.

Worked example

For the $\$10\,000$ loan at 12% per annum compounded monthly with monthly repayments of $\$900$ , build a two-line amortisation schedule that splits each of the first two repayments into their interest and principal components.

Step 1: Calculate the interest charged in month 1

Interest is always charged on the balance outstanding at the start of that period, before the repayment is made. The monthly rate is $i = 0.01$ , and the balance at the start of month 1 is the full $\$10\,000$ :

\text{interest}_1 = 0.01 \times 10\,000 = 100.

Step 2: Find the principal repaid and the new balance in month 1

The repayment of $\$900$ first covers the $\$100$ interest charge; the remainder goes directly to reducing what is owed. This is why the balance falls by more than just the interest but less than the full repayment:

\text{principal repaid}_1 = 900 - 100 = 800, \qquad B_1 = 10\,000 - 800 = 9200.

Step 3: Repeat for month 2

The interest for month 2 is charged on the reduced balance of $\$9200$ , so it is slightly less than in month 1. More of the fixed $\$900$ payment now goes to reducing the debt:

\text{interest}_2 = 0.01 \times 9200 = 92, \qquad \text{principal repaid}_2 = 900 - 92 = 808, \qquad B_2 = 9200 - 808 = 8392.

This confirms the recurrence result $B_2 = 8392$ from the earlier calculation.

Final answer: Month 1: $\$100$ interest, $\$800$ principal, balance $\$9200$ . Month 2: $\$92$ interest, $\$808$ principal, balance $\$8392$ .

Analysing changes to the repayment

The repayment $R$ must exceed the first period's interest, or the balance will grow instead of shrink. Increasing the repayment clears the loan in fewer periods and reduces the total interest paid; decreasing it lengthens the term and raises total interest.

Offset accounts and extra repayments

Two features change how quickly a loan clears. An offset account holds savings that are subtracted from the loan balance for the purpose of calculating interest, without reducing the principal itself, so interest is charged on a smaller amount and more of each repayment attacks the principal. Making extra repayments, or repaying more frequently, also reduces the balance faster and cuts the total interest substantially, because interest is always charged on the outstanding balance. SACE questions often ask you to compare the total interest paid with and without such a change, which you compute by running the recurrence under each scenario.

The total cost of a loan

The total amount repaid is the regular repayment multiplied by the number of repayments (adjusting the final, smaller payment). Subtracting the original principal from this total gives the total interest paid over the life of the loan. For long mortgages this interest can approach or exceed the amount borrowed, which is why even a small reduction in the rate or term, or a modest extra repayment, can save a large sum. Being able to compute and interpret this total cost is a standard examined outcome.

The final repayment

The last repayment is usually smaller than the rest, because only the remaining balance plus its final period's interest is owed. To find it, take the second-last balance, add one period's interest, and that total is the final payment.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20222 marksCalculator-assumed. Keisha took out a home loan of 650000 dollars over 20 years at 2.88% per annum, compounded monthly. Show that the minimum monthly repayment was approximately 3570 dollars.

Show worked answer →

Use the amortisation model. Monthly rate $i = \dfrac{0.0288}{12} = 0.0024$ ; payments $n = 20 \times 12 = 240$ ; loan $650000$ , future value $0$ .

R = \frac{650000 \times 0.0024 \times (1.0024)^{240}}{(1.0024)^{240} - 1} \approx 3570 \text{ dollars per month}.

Marks: one for the monthly rate and number of payments, one for arriving at about 3570 dollars. A financial solver gives the same result.

SACE 20232 marksCalculator-assumed. Grace took out a home loan of 356000 dollars over 25 years at 5.11% per annum, compounded monthly. Show the monthly repayment is approximately 2100 dollars.

Show worked answer →

Monthly rate $i = \dfrac{0.0511}{12} \approx 0.0042583$ ; payments $n = 25 \times 12 = 300$ ; loan $356000$ .

R = \frac{356000 \times 0.0042583 \times (1.0042583)^{300}}{(1.0042583)^{300} - 1} \approx 2100 \text{ dollars per month}.

Marks: one for the rate and 300 payments, one for the repayment of about 2100 dollars.

SACE 20211 marksCalculator-assumed. Dakota's 25-year home loan of 350000 dollars has a minimum monthly repayment of approximately 1880 dollars. Calculate the total cost of the loan.

Show worked answer →

Total cost is the repayment times the number of repayments. Payments $n = 25 \times 12 = 300$ , so total $= 1880 \times 300 = 564\,000$ dollars.

So Dakota repays about 564000 dollars; compared with the 350000 dollars borrowed, the extra roughly 214000 dollars is interest. The mark is for the multiplication.