What is wrong split of a payment?

The interest part is balance times rate; the principal part is payment minus interest, not the other way round.

What is annual rate in a monthly model?

Divide the nominal rate by the periods per year first.

What is total interest as principal times rate?

Total interest is total repaid minus the amount borrowed; it is not a single percentage of the loan.

§-Syllabus dot point

WAMathematics ApplicationsSyllabus dot point

How does a loan reduce to zero when interest is charged and regular repayments are made?

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

How to model a reducing-balance loan with a recurrence relation, read and build an amortisation table, and find the balance, repayment and total interest paid.

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

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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.

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.

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.

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.

Exam-style practice questions

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

WACE 20227 marksA

\

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 20245 marksA loan of

\

15\,000

at

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.

Show worked answer →

Solve for the payment, then compare total paid with the principal.

(a) Finance solver: $PV = 15000$ , $I\% = 9$ compounding monthly, $N = 24$ , $FV = 0$ ; solve for $PMT$ . The monthly repayment is about $\$685.27$ . (3 marks)

(b) Total paid $= 685.27 \times 24 = 16446.48$ ; total interest $= 16446.48 - 15000 = 1446.48$ , about $\$1446$ . (2 marks)

Markers reward the solver setup for $PMT$ and total interest as total repaid minus principal.