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Reducing-balance loans (SACE Stage 2 General Mathematics)

Syllabus dot point

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

8 min read

Quick answer

A reducing-balance loan charges interest on the balance still owing, then a repayment is subtracted. The recurrence is Bn+1=Bn(1+i)−RB_{n+1} = B_n(1 + i) - R, where ii is the period rate and RR is the repayment. As the balance falls, the interest portion of each payment falls too.

Jump to a section
  1. What this dot point is asking
  2. How a reducing-balance loan works
  3. The amortisation schedule
  4. Analysing changes to the repayment
  5. Offset accounts and extra repayments
  6. The total cost of a loan
  7. The final repayment
  8. Exam-style questions

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:

Bn+1=Bn(1+i)−RB_{n+1} = B_n(1 + i) - R

where BnB_n is the balance after nn repayments, ii is the interest rate per period (decimal) and RR is the regular repayment.

Reducing-balance loan

A loan on which interest is charged each period on the outstanding balance only. Because the balance reduces with every repayment, the interest charged shrinks over time, so a growing share of each fixed repayment goes towards the principal.

Worked example

A $10 000\$10\,000 loan is charged 12% per annum compounded monthly, with repayments of $900\$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=0.1212=0.01.i = \frac{0.12}{12} = 0.01.

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

The recurrence Bn+1=Bn(1+i)−RB_{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.

B1=10 000(1.01)−900=10 100−900=9200.B_1 = 10\,000(1.01) - 900 = 10\,100 - 900 = 9200.

Of the $900\$900 repayment, 0.01×10 000=$1000.01 \times 10\,000 = \$100 covers interest and the remaining $800\$800 reduces the principal. The balance falls from $10 000\$10\,000 to $9200\$9200.

Step 3: Apply the recurrence again for month 2

With the new starting balance of $9200\$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:

B2=9200(1.01)−900=9292−900=8392.B_2 = 9200(1.01) - 900 = 9292 - 900 = 8392.

Interest for month 2 is 0.01×9200=$920.01 \times 9200 = \$92, and $808\$808 reduces the principal.

Final answer: After two repayments, $8392\$8392 is still owing. The interest portions were $100\$100 and $92\$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×Bn= i \times B_n (the balance at the start of the period).
  • Principal repaid == repayment R−R - interest.
  • New balance =Bn−= B_n - principal repaid.
Worked example

For the $10 000\$10\,000 loan at 12% per annum compounded monthly with monthly repayments of $900\$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=0.01i = 0.01, and the balance at the start of month 1 is the full $10 000\$10\,000:

interest1=0.01×10 000=100.\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\$900 first covers the $100\$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:

principal repaid1=900−100=800,B1=10 000−800=9200.\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\$9200, so it is slightly less than in month 1. More of the fixed $900\$900 payment now goes to reducing the debt:

interest2=0.01×9200=92,principal repaid2=900−92=808,B2=9200−808=8392.\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 B2=8392B_2 = 8392 from the earlier calculation.

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

Interest falls, principal rises

With a fixed repayment, early payments are mostly interest because the balance is large. As the balance shrinks, the interest portion falls and more of each payment reduces the principal, so the loan clears faster towards the end.

Analysing changes to the repayment

The repayment RR 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.

Worked example

Will the loan ever be repaid?

For the $10 000 loan at 12% per annum monthly, what minimum monthly repayment is needed for the balance to fall?

First-month interest is 0.01×10000=1000.01 \times 10000 = 100. The repayment must be more than $100, otherwise the balance stays the same or grows. Any R>100R > 100 reduces the balance; the larger RR is, the faster the loan clears.

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.

Track the balance to the cent

Carry full accuracy through the recurrence and round only the final answer. Rounding the balance each line can accumulate error over many months and give a final payment that is several dollars wrong.

Common mistake

Do not subtract the repayment before adding the interest. The correct order is to grow the balance by (1+i)(1 + i) first, then subtract RR. Also do not assume the same number of full repayments clears the loan exactly; the final payment is almost always a smaller, adjusted amount.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style2 marks
Calculator-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=0.028812=0.0024i = \dfrac{0.0288}{12} = 0.0024; payments n=20×12=240n = 20 \times 12 = 240; loan 650000650000, future value 00.

R=650000×0.0024×(1.0024)240(1.0024)240−1≈3570 dollars per month.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 2023-style2 marks
Calculator-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=0.051112≈0.0042583i = \dfrac{0.0511}{12} \approx 0.0042583; payments n=25×12=300n = 25 \times 12 = 300; loan 356000356000.

R=356000×0.0042583×(1.0042583)300(1.0042583)300−1≈2100 dollars per month.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 2021-style1 mark
Calculator-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=25×12=300n = 25 \times 12 = 300, so total =1880×300=564 000= 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.

Practise this

Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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