Reducing-Balance Loans and Amortisation - TCE Mathematics Applications (Tasmania)

Syllabus dot point

“Model reducing-balance loans with recurrence relations and amortisation tables.”

TCEMathematics ApplicationsUnit 48 min readUpdated 2 Jun 2026

Quick answer

You must model a reducing-balance loan with the recurrence $A_{n+1} = A_n(1 + r) - d$ , build an amortisation table that splits each repayment into interest and principal, and find quantities such as the balance after a payment, the interest paid, and the total cost of the loan.

What this dot point is asking

A reducing-balance loan charges interest only on the amount still owing. Because the balance falls with every repayment, the interest charged each period also falls, so a larger slice of each fixed repayment goes toward the principal as time passes.

Reducing-balance loan

A loan where interest each period is calculated on the current outstanding balance. A fixed repayment is made each period; part covers the interest just charged and the rest reduces the principal.

The recurrence relation

The balance after each period is found by adding interest then subtracting the repayment.

Loan recurrence

A_{n+1} = A_n(1 + r) - d, \quad A_0 = \text{amount borrowed}

Here

r

is the interest rate per repayment period and

d

is the fixed repayment. Multiply by

(1 + r)

first to add interest, then subtract the repayment

d

Tracking a loan balance

A loan of $\$10\,000$ charges 1 percent per month and is repaid at $\$450$ per month.

Step 1: Identify the recurrence parameters

The annual rate would need dividing by 12 to get a monthly rate, but here we are told directly that the monthly interest rate is 1 percent, so $r = 0.01$ . The fixed monthly repayment is $d = 450$ and the initial balance is $A_0 = 10000$ . These three values fully define the recurrence $A_{n+1} = A_n(1 + r) - d$ .

Step 2: Apply the recurrence for month 1

Each period we multiply the current balance by $(1 + r)$ to add one period of interest, then subtract the repayment $d$ . Interest accrues on the outstanding balance before the repayment is applied, which is why we multiply first.

A_1 = 10000(1.01) - 450 = 10100 - 450 = 9650

Step 3: Apply the recurrence for months 2 and 3

The same rule applies in every subsequent period. Because the balance is now lower, the interest charged is slightly less, so a slightly larger fraction of the $\$450$ payment reduces the principal.

A_2 = 9650(1.01) - 450 = 9746.50 - 450 = 9296.50

A_3 = 9296.50(1.01) - 450 = 9389.47 - 450 = 8939.47

After three months the balance has fallen to about $\$8939.47$ .

The amortisation table

An amortisation table records, for each period, the interest charged, the principal repaid, and the new balance. It makes the inner workings of each repayment visible.

Splitting a repayment

For each period: interest $=$ balance $\times r$ ; principal reduction $=$ repayment $-$ interest; new balance $=$ old balance $-$ principal reduction. As the balance falls, the interest portion shrinks and the principal portion grows.

One row of an amortisation table

For the loan above, examine the first repayment.

Step 1: Calculate the interest charged for the period

Interest is charged on the balance outstanding at the start of the period, before any repayment is applied. This is the defining feature of a reducing-balance loan: the lender is entitled only to interest on what is actually owed.

\text{Interest} = 10000 \times 0.01 = \$100

Step 2: Find the principal reduction

The repayment of $\$450$ must first cover the interest just charged; whatever remains goes toward reducing the debt. Subtracting the interest from the repayment gives the principal reduction.

\text{Principal reduction} = 450 - 100 = \$350

Step 3: Confirm the new balance

Deducting the principal reduction from the opening balance gives the closing balance for the period. This should match the recurrence result exactly.

\text{New balance} = 10000 - 350 = \$9650

This confirms the table entry agrees with the recurrence. Of the first repayment, $\$100$ is interest and only $\$350$ pays down the debt. In later months the interest portion is smaller because the balance is lower, so an ever-growing share of each $\$450$ payment clears principal.

Total interest paid

The total interest over the life of the loan is the sum of all repayments minus the amount originally borrowed. This is often the figure that surprises borrowers, because over many years the interest can rival the principal.

Total cost of a loan

Suppose the loan above is fully repaid by 24 monthly payments of $\$450$ and a smaller final payment, with the repayments totalling $\$11\,180$ .

Step 1: Understand what "total repaid" includes

Every dollar paid to the lender over the life of the loan is either returning borrowed principal or paying for the use of that money as interest. Adding up all repayments gives the total outflow, and comparing it to the amount originally borrowed separates the two components. The total of all payments is given as $\$11\,180$ .

Step 2: Calculate the total interest paid

Because the total repaid is the original principal plus all interest charges combined, subtracting the amount borrowed leaves only the interest. This works regardless of how the interest was distributed across the repayments.

\text{Total interest} = 11180 - 10000 = \$1180

Step 3: Interpret the result for the borrower

Borrowing $\$10\,000$ costs an additional $\$1180$ in interest, making the total cost of the loan $\$11\,180$ . This figure helps borrowers compare loan options: a lower rate or shorter term reduces the total interest paid, even if individual repayments are higher.

Interest is charged before the repayment reduces the balance

Each period, interest is added to the current balance first, then the repayment is taken off. Calculating principal as repayment minus interest, rather than repayment alone, is what makes the table balance to the recurrence.

The final repayment

The last repayment is usually smaller than the rest, because a full fixed repayment would overshoot and leave a negative balance. The final payment is exactly the outstanding balance plus its final period of interest, clearing the loan to zero.

Treating it like flat-rate interest

A reducing-balance loan does not charge interest on the original amount the whole time. Interest is recalculated on the shrinking balance each period, so applying a flat rate to the starting principal massively overstates the interest. Always charge $r$ on the current balance, not the original.

Finding the balance owing partway through

A subtle exam point, tested in the first question above, is the difference between the value of payments already made and the balance still owing. The balance owing after some payments equals the present value of the payments that remain, so you must use the number of payments left, not the number made. With the finance solver, enter the original loan as $PV$ , the repayment as $PMT$ , and the number of remaining periods as $N$ , then read $FV$ as the outstanding balance, or solve for $PV$ using the remaining payments. Confusing payments made with payments remaining is the single most common error in loan questions.

A complete answer uses $A_{n+1} = A_n(1 + r) - d$ with $r$ as the per-period rate, splits at least one repayment into interest and principal, uses the remaining payments when finding a balance partway through, and where asked finds total interest as total repayments minus the amount borrowed.

Exam-style questions

Questions in the style of TASC 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.

Practice2 marks

Alex calculates how much they still owe on a home loan after 5 years. They repay

\$900

each month, interest is

7\%

p.a. compounding monthly, and the loan lasts 20 years. Their calculation uses the present-value-of-annuity formula with

n = 5 \times 12

in the exponent. What common error have they made?

Show worked answer →

(2 marks) The formula Alex used finds the present value of only the payments made over 5 years, not the balance still owing after 5 years.

To find the amount outstanding after 5 years you must use the number of payments still left, which is the remaining 15 years: $n = 15 \times 12 = 180$ , not $5 \times 12 = 60$ . By substituting 60 they have valued the wrong set of payments. The correct balance owing is the present value of the 180 remaining repayments.

Practice6 marks

Use the interest-factor table for monthly compound interest. a) Find the amount borrowed over a 10-year loan if monthly repayments of

\$1665.31

are made at

6\%

p.a. compounded monthly. b) Find the monthly repayment on a

\$200\,000

loan over 20 years at

5\%

p.a. compounded monthly. c) How long to repay

\$5000

with monthly repayments of

\$425.75

4\%

p.a. compounded monthly?

Show worked answer →

a) (2 marks) Amount borrowed $=$ repayment $\times$ factor, where the factor is read from the table for $i = 0.005$ ( $\frac{6\%}{12}$ ) and $n = 120$ . This gives amount borrowed $= \$150\,000$ .

b) (2 marks) Rearrange to repayment $=$ loan $/$ factor, using $i = \frac{5\%}{12}$ and $n = 240$ . The factor gives a monthly repayment of about $\$1319.91$ .

c) (2 marks) Here factor $= \frac{5000}{425.75} = 11.744$ . Find this factor in the $i = \frac{4\%}{12}$ column of the table; it corresponds to $n = 12$ , so the loan is repaid in 12 months (1 year).

Practise this

Sources & how we know this

General Mathematics Level 3 (MTG315123) course document (2023 onwards) — TASC