NSWMaths Standard 2Syllabus dot point

How are reducing-balance loan repayments calculated, and how does each repayment split between interest and principal?

Use recurrence relations and amortisation tables to calculate loan repayments, outstanding balances and the total interest paid on a reducing-balance loan

A focused answer to the HSC Maths Standard 2 dot point on reducing-balance loans. Recurrence model for the outstanding balance, building an amortisation table, the interest vs principal split of each payment, and worked Australian mortgage examples at current RBA cash rate levels.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to model a reducing-balance loan with a recurrence relation, build an amortisation table by hand for a few periods, compute the outstanding balance after $n$ payments using the closed-form formula, and split a payment into its interest and principal components.

The answer

How a reducing-balance loan works

A loan of $P$ dollars is repaid by equal payments $M$ at the end of each compounding period. Each period:

Interest is added to the opening balance at the per-period rate $r$ .
The payment is subtracted.

The closing balance is what is still owed. Let $B_n$ be the balance just after the $n$ th payment.

B_n = B_{n-1}(1 + r) - M.

This is the recurrence model. $B_0 = P$ .

Amortisation table

A standard format for the first few periods:

Period	Opening	Interest	Payment	Principal repaid	Closing
IMATH_12	IMATH_13	IMATH_14	IMATH_15	IMATH_16	IMATH_17
IMATH_18	IMATH_19	IMATH_20	IMATH_21	IMATH_22	IMATH_23
...	...	...	...	...	...

Each row: interest is the opening balance times the per-period rate; principal repaid is payment minus interest; closing balance is opening plus interest minus payment.

Closed-form for the balance

Iterating the recurrence using a geometric series gives

B_n = P(1 + r)^n - M \cdot \frac{(1 + r)^n - 1}{r}.

The first term is what the loan would grow to without payments. The second term is the future value of payments made so far. The difference is what is still owed.

Interest vs principal split

For the $k$ th payment:

Interest portion: $I_k = r \cdot B_{k-1}$ (interest on the previous balance)
Principal portion: IMATH_26

Early in the loan, most of each payment goes to interest. Near the end, almost all goes to principal. This is why making extra repayments early saves a lot more interest than the same dollar amount later.

Total interest paid

If the loan is repaid by $n$ payments of $M$ :

\text{total interest} = n M - P.

That is, total cash paid out minus the original loan amount.

Australian mortgage context

A typical Sydney mortgage in 2025: $\$ 700000 $borrowed at around$ 6.2% $per annum (typical major-bank variable rate, RBA cash rate$ \sim 4.35% $plus bank margin). Over$ 25$ years monthly:

$r = 0.062/12 \approx 0.005167$ , $n = 300$ .

The repayment formula (also on the reference sheet) gives $M \approx \$ 4576 $per month. Total paid over$ 25 $years$ \approx \ $1.37$ million. Total interest $\approx \$ 673000$, almost as much as the original principal.

Worked example

Amortisation by hand

$\$ 50000 $at$ 0.5% $per month, payment$ \ $1100$ .

Month $1$ : opening $50000$ , interest $50000 \times 0.005 = 250$ , payment $1100$ , principal $850$ , closing $49150$ .

Month $2$ : opening $49150$ , interest $49150 \times 0.005 = 245.75$ , payment $1100$ , principal $854.25$ , closing $48295.75$ .

Month $3$ : opening $48295.75$ , interest $48295.75 \times 0.005 \approx 241.48$ , payment $1100$ , principal $\approx 858.52$ , closing $\approx 47437.23$ .

Notice the principal portion grows each month (from $850$ to $854.25$ to $858.52$ ) and the interest portion shrinks.

Balance after many periods (closed form)

For the same loan after $5$ years ( $n = 60$ ):

$(1.005)^{60} \approx 1.34885$ .

$B_{60} = 50000 \times 1.34885 - 1100 \times \frac{0.34885}{0.005}$

$= 67442.50 - 1100 \times 69.770 = 67442.50 - 76747.00 = -9304.50$ .

A negative balance means the loan would be paid off before $60$ months. To find the actual term, solve $B_n = 0$ :

$50000(1.005)^n = 1100 \times \frac{(1.005)^n - 1}{0.005}$

$50000(1.005)^n = 220000((1.005)^n - 1)$ .

$(1.005)^n (50000 - 220000) = -220000$

$(1.005)^n = \frac{220000}{170000} = 1.2941$ .

$n = \frac{\log 1.2941}{\log 1.005} = \frac{0.1119}{0.00217} \approx 51.6$ months.

So the loan is paid in $52$ months (the final payment is slightly less than $\$ 1100$).

Australian mortgage (2025 rates)

$\$ 650000 $at$ 6.0% $per annum compounded monthly over$ 30$ years.

$r = 0.005$ , $n = 360$ . Using the repayment formula $M = \frac{P r}{1 - (1 + r)^{-n}}$ :

$(1.005)^{-360} \approx 0.16604$ , so denominator is $0.83396$ .

$M = \frac{650000 \times 0.005}{0.83396} = \frac{3250}{0.83396} \approx \$ 3896.79$ per month.

Total paid: $360 \times 3896.79 \approx \$ 1402845 $. Total interest:$ \ $752845$ .

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q265 marksA car loan of

\

25000

at

7.2\%

per annum compounded monthly is repaid by monthly instalments of

498.21

. Find the balance owing immediately after the 12th payment. Round to the nearest cent.

Show worked answer →

Per-period rate: $r = \frac{0.072}{12} = 0.006$ . Monthly payment $M = 498.21$ .

The balance after $n$ payments is

$B_n = P(1 + r)^n - M \cdot \frac{(1 + r)^n - 1}{r}$ .

$(1.006)^{12} \approx 1.07442$ .

$B_{12} = 25000 \times 1.07442 - 498.21 \times \frac{0.07442}{0.006}$

$= 26860.50 - 498.21 \times 12.4033$

$= 26860.50 - 6179.91 \approx \$ 20680.59$.

Markers reward the outstanding-balance formula, accurate intermediate values, and the answer rounded to cents.

2021 HSC Q244 marksUse an amortisation table to find the balance after

3

monthly repayments of

\

1200

on a

60000

loan at

6\%

per annum compounded monthly.

Show worked answer →

Per-period rate: $r = 0.005$ .

Month $1$ : opening $60000$ , interest $60000 \times 0.005 = 300$ , payment $1200$ , principal repaid $900$ , closing $59100$ .

Month $2$ : opening $59100$ , interest $59100 \times 0.005 = 295.50$ , payment $1200$ , principal repaid $904.50$ , closing $58195.50$ .

Month $3$ : opening $58195.50$ , interest $58195.50 \times 0.005 \approx 290.98$ , payment $1200$ , principal repaid $909.02$ , closing $\approx \$ 57286.48$.

Markers reward a clean amortisation table layout, interest computed on the opening balance each period, principal repaid as payment minus interest, and the closing balance.