Reducing-balance loans and amortisation in WACE Mathematics Applications Unit 4

Syllabus dot point

“Model a reducing-balance loan with a recurrence relation, build an amortisation table, and find balances, repayments and total interest.”

WACEMathematics ApplicationsUnit 48 min readUpdated 2 Jun 2026

Quick answer

A reducing-balance loan charges interest on the balance still owing, then a fixed repayment is made each period. Model it with $A_{n+1} = R A_n - d$ , where $R = 1 + i$ and $d$ is the repayment. An amortisation table splits each payment into interest (on the current balance) and principal reduction. Early payments are mostly interest; later payments are mostly principal. Total interest is total repaid minus the amount borrowed.

Jump to a section

What this dot point is asking
The loan recurrence
The amortisation table
Finding the repayment and total interest
The effect of changing the repayment
Reading an amortisation table question
Exam-style questions

What this dot point is asking

You must model the loan with a recurrence, build an amortisation table, and find balances, repayments and total interest.

The loan recurrence

Each period, interest is added to the balance, then the repayment is subtracted.

Reducing-balance loan

$A_{n+1} = R A_n - d$ , with $R = 1 + i$ , $i$ the per-period rate, $d$ the regular repayment and $A_0$ the amount borrowed. The loan is repaid when the balance first reaches zero.

Because interest is charged only on what is still owed, the interest shrinks each period as the balance falls, so each fixed payment chips away more principal than the last.

The amortisation table

An amortisation table breaks each payment into its interest and principal parts.

Amortisation table columns

For each period:

Interest $=$ (current balance) $\times i$ .
Principal reduction $=$ payment $-$ interest.
New balance $=$ current balance $-$ principal reduction.

The pattern is the heart of the topic: the interest column falls and the principal column rises across the life of the loan, even though the total payment is constant.

Early payments are mostly interest

At the start, the balance is large, so most of each payment covers interest and little reduces the principal. As the balance falls, less interest is charged and more of each payment reduces the principal. This is why paying extra early saves the most interest.

Worked example

Building an amortisation table

A $\$20\,000$ car loan at $12\%$ per annum compounding monthly is repaid by $\$900$ per month. Build the first three rows.

Step 1: Convert the annual rate and write the recurrence relation.

Because interest compounds monthly, we divide the annual rate by $12$ to get the per-period rate: $i = \dfrac{0.12}{12} = 0.01$ . The growth factor is $R = 1.01$ , giving the recurrence relation $A_{n+1} = 1.01 A_n - 900$ , with $A_0 = 20000$ . This relation says: multiply the current balance by $1.01$ (add one month's interest), then subtract the $\$900$ repayment.

Step 2: Complete the first row of the amortisation table (Month 1).

We split the $\$900$ payment into its interest and principal parts. Interest is charged on the current balance: $20000 \times 0.01 = 200$ . The portion of the payment that reduces the debt is the remainder: principal reduction $= 900 - 200 = 700$ . The new balance is $20000 - 700 = 19300$ .

Step 3: Complete the second row (Month 2).

The balance is now $\$19300$ , so the interest charged is smaller: $19300 \times 0.01 = 193$ . A smaller interest charge means more of the fixed $\$900$ goes to principal: principal reduction $= 900 - 193 = 707$ . New balance $= 19300 - 707 = 18593$ .

Step 4: Complete the third row (Month 3).

Interest $= 18593 \times 0.01 = 185.93$ . Principal reduction $= 900 - 185.93 = 714.07$ . Balance $= 18593 - 714.07 = 17878.93$ .

Step 5: Identify the amortisation pattern.

Comparing the three rows reveals the key pattern: the interest portion falls ( $\$200 \to \$193 \to \$185.93$ ) while the principal portion rises ( $\$700 \to \$707 \to \$714.07$ ). The total payment stays fixed at $\$900$ . This shift occurs because each month's interest is charged on a smaller balance, leaving a growing share of the fixed payment free to attack the principal.

Finding the repayment and total interest

Use the finance solver to find an unknown repayment ( $PV$ the loan, $FV = 0$ , $N$ periods, $I\%$ rate; solve for $PMT$ ) or the number of periods. The total interest paid over the loan is the total of all repayments minus the amount borrowed: total interest $=$ (payment $\times$ number of payments) $-$ principal. The final payment is usually smaller than the regular one, to clear the small remaining balance exactly.

The effect of changing the repayment

A higher monthly repayment clears the loan faster and cuts the total interest sharply, because more of each payment attacks the principal and less is lost to interest on the shrinking balance. Conversely, a lower repayment extends the term and raises total interest; if the repayment is set below the first period's interest, the balance actually grows and the loan never reduces. A useful check before any amortisation is to compare the first repayment with the first interest charge: the repayment must exceed it for the loan to make progress.

Reading an amortisation table question

SCSA often supplies a partly completed amortisation table and asks you to fill gaps. Work along a row using the three relations: interest $=$ balance $\times i$ , principal reduction $=$ payment $-$ interest, new balance $=$ old balance $-$ principal reduction. Given any one missing entry, the others follow. A common task is to find the interest rate from a row by rearranging interest $=$ balance $\times i$ to $i = \dfrac{\text{interest}}{\text{balance}}$ , then multiplying by the number of periods per year to recover the nominal annual rate.

Common errors

Charging interest on the original principal each period: Interest is charged on the current balance, which falls each period; this is what makes it reducing-balance.
Wrong split of a payment: The interest part is balance times rate; the principal part is payment minus interest, not the other way round.
Annual rate in a monthly model: Divide the nominal rate by the periods per year first.
Total interest as principal times rate: Total interest is total repaid minus the amount borrowed; it is not a single percentage of the loan.

Exam-style questions

Questions in the style of SCSA 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.

WACE 2022-style7 marks

\$320\,000

home loan is taken at

5.4\%

per annum compounding monthly, repaid by

\$2100

each month. (a) Write a recurrence relation for the balance. (b) Construct the first two rows of an amortisation table (interest, principal reduction, balance). (c) State what happens to the interest portion of each payment over time.

Show worked answer →

Each month, interest is added then the repayment is subtracted.

(a) Monthly rate $= \dfrac{0.054}{12} = 0.0045$ , so $A_{n+1} = 1.0045 A_n - 2100$ , with $A_0 = 320000$ . (2 marks)

(b) Month 1: interest $= 320000 \times 0.0045 = 1440$ ; principal reduction $= 2100 - 1440 = 660$ ; balance $= 320000 - 660 = 319340$ . Month 2: interest $= 319340 \times 0.0045 = 1437.03$ ; principal $= 2100 - 1437.03 = 662.97$ ; balance $= 319340 - 662.97 = 318677.03$ . (4 marks)

(c) The interest portion decreases each month (charged on a smaller balance) while the principal portion increases, even though the total payment stays at $\$2100$ . (1 mark)

Markers reward the recurrence, an amortisation row splitting interest and principal, and the shifting interest-to-principal pattern.

WACE 2024-style5 marks

A loan of

\$15\,000

9\%

per annum compounding monthly is repaid over exactly

24

months. (a) Use the finance solver to find the monthly repayment. (b) Find the total interest paid over the loan.