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hscscience Maths Std · Y12
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Module 7 · L7 of 12 ~30 min MS12-5 ⚡ +50 XP available

Loans and Amortisation

You borrow $400,000 to buy a house. Over 30 years, you will pay back nearly $800,000. Where does the extra $400,000 go? It goes to interest — and understanding exactly how it flows is the key to smarter financial decisions. Amortisation is the process of gradually paying off a loan through regular repayments, where each payment splits between interest and principal. Early payments are mostly interest; later payments are mostly principal.

Today's hook — A $300,000 mortgage at 5% over 25 years costs about $1750/month. In the first year, roughly how much of your $21,000 in payments goes to interest vs principal? Predict before reading.
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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

01
Recall — your gut answer first
+5 XP warm-up

A $300,000 mortgage at 5% over 25 years costs about $1750/month. In the first year, roughly how much of your $21,000 in payments goes to interest vs principal? Predict before reading.

Before reading on — write your gut feeling. We will revisit this at the end of the lesson.

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02
Key ideas for this lesson
reference

Every amortisation problem uses the same four equations applied period by period.

Monthly repayment: $M = PV \times \dfrac{r}{1 - (1+r)^{-n}}$

Interest for period: $I = \text{Balance} \times r$

Principal for period: $P = M - I$

New balance: $\text{Balance}_{new} = \text{Balance}_{old} - P$

REPAYMENT FORMULA M = PV x r / [1-(1+r)^-n] PV = loan, r = periodic rate n = total periods Early payments = mostly interest
Key insight: Early in a loan, most of each payment is interest. Extra repayments early save the most because they reduce the balance future interest is calculated on.
r is the periodic rate
6% p.a. monthly compounding: r = 0.06 ÷ 12 = 0.005. Never use the annual rate directly.
n = total periods
25 years monthly = 300 periods. Match r and n to the same time unit.
Total interest
Total interest = (M × n) − PV. Subtract the original loan from total amount repaid.
03
What you will master
Know

Key facts

  • Amortisation formula
  • Interest vs principal split
  • Reducing balance concept
Understand

Concepts

  • Why early payments are mostly interest
  • How extra repayments save money
  • The true cost of long-term loans
Can do

Skills

  • Build amortisation schedules
  • Calculate total interest
  • Compare loan scenarios
04
Key terms
AmortisationThe process of gradually paying off a loan through regular repayments split between interest and principal.
Principal componentThe portion of each repayment that reduces the outstanding loan balance.
Amortisation scheduleA table showing the interest, principal, and closing balance for every repayment period.
Reducing balanceInterest is charged only on the outstanding balance — not the original loan amount.
Comparison rateA rate that includes most fees in addition to the interest rate, allowing fair comparison between loan products.
Extra repaymentAny payment above the minimum, which reduces principal faster and saves significant interest.
05
Amortisation schedules — tracking every dollar
core concept

An amortisation schedule shows how each repayment is split between interest and principal, and how the balance reduces over time.

Example: $20,000 car loan at 7.2% p.a. compounded monthly over 3 years.

$r = 0.072 \div 12 = 0.006$, $n = 36$. Using the repayment formula:

$$M = 20000 \times \frac{0.006}{1 - (1.006)^{-36}} = \$618.96 \text{ per month}$$
Month Opening Balance Interest Principal Closing Balance
1$20,000.00$120.00$498.96$19,501.04
2$19,501.04$117.01$501.95$18,999.09
3$18,999.09$113.99$504.97$18,494.12
36$615.27$3.69$615.27$0.00
Pattern to notice: Interest decreases each month because the balance decreases. The principal component increases each month because more of the fixed repayment is left after paying interest. The loan reaches exactly $0 at the final payment.
What to write in your book
  • Amortisation schedule = table of opening balance, interest, principal, closing balance for every period.
  • Interest for period = Balance × r (periodic rate).
  • Principal for period = M − I. New balance = old balance − principal.
  • Early payments are mostly interest; later payments are mostly principal.

Quick check: In Month 1 of a $20,000 loan at 7.2% p.a. compounded monthly (r = 0.006), the interest component of the repayment is:

06
Extra repayments — the power of paying more
core concept

Adding even small extra amounts to your regular repayment dramatically reduces total interest and loan term.

Example: $400,000 mortgage at 4.8% p.a. compounded monthly over 30 years.

$r = 0.004$, $n = 360$.

$$M = 400000 \times \frac{0.004}{1 - (1.004)^{-360}} = \$2098.64 \text{ per month}$$

Total payments = $2098.64 × 360 = $755,510. Total interest = $355,510.

If you pay $2,200/month instead (only $101.36 extra per month):

  • New term ≈ 25 years — saves 5 years of repayments
  • Total interest ≈ $260,000 — saves approximately $95,000
Why it works: Every extra dollar directly reduces the principal. This means next month's interest is calculated on a smaller balance, so more of the minimum repayment goes to principal — a compounding snowball effect over the life of the loan.
What to write in your book
  • Extra repayments reduce principal early, saving interest on every future payment.
  • Small regular extras (e.g. $100/month) can save tens of thousands of dollars over 25–30 years.
  • Fortnightly payments (half the monthly amount, 26×/year) = 13 full payments per year instead of 12.

True or false: Paying $100 extra per month on a 30-year mortgage saves more money if you start in Year 1 than if you start in Year 20.

PROBLEM 1 · AMORTISATION SCHEDULE

$250,000 mortgage at 5.4% p.a. compounded monthly over 20 years. Find the monthly repayment, total interest, and remaining balance after 5 years.

1
$r = 0.054 \div 12 = 0.0045,\quad n = 20 \times 12 = 240$
Convert annual rate to monthly; convert years to number of periods
PROBLEM 2 · EXTRA REPAYMENTS

$15,000 personal loan at 9.6% p.a. compounded monthly over 5 years. Find monthly repayment, total interest, and savings from paying $50 extra per month.

1
$r = 0.096 \div 12 = 0.008,\quad n = 60$
Convert to monthly rate and count periods
08
Comparing loans — which is cheaper?
core concept

When comparing loans, always consider:

  • Interest rate: Lower is better, but watch for fees.
  • Fees: Establishment fees, monthly account fees, early repayment penalties.
  • Features: Offset account, redraw facility, repayment flexibility.
  • Comparison rate: Includes most fees — the fairest basis for comparison.

Example: Loan A: 5.0% with $0 fees. Loan B: 4.8% with $10/month fee on $300,000 over 25 years.

Lower rate on Loan B saves ≈ $150/year on interest, but pays $120/year in fees. Net saving = only $30/year. Loan A is very slightly better despite the higher headline rate.

Always check the comparison rate. Australian lenders are legally required to disclose the comparison rate alongside the headline rate. A lower headline rate with high fees can cost more than a slightly higher rate with no fees.
What to write in your book
  • Comparison rate = interest rate + most fees expressed as a single % — use this for fair comparisons.
  • Lower headline rate + high fees can be more expensive than higher rate + no fees.
  • Offset account: money held reduces the balance that interest is calculated on.

Fill the gap: For a $20,000 car loan at 7.2% p.a. compounded monthly over 3 years, the monthly repayment is $618.96. The total interest paid over the 36 months is $ .

Trap 01
Using annual rate instead of periodic rate
Always divide the annual rate by the number of periods per year. For 6% p.a. monthly: r = 0.06/12 = 0.005 — never use 0.06 directly in the repayment formula.
Trap 02
Forgetting to match n and r
n must be in the same unit as r. If r is monthly, n = years × 12. If r is fortnightly, n = years × 26. Mixing units produces wildly wrong answers.
Trap 03
Calculating total interest incorrectly
Total interest = (M × n) − PV. Do NOT multiply the annual rate by the original loan — this ignores the reducing balance and will always be wrong.
What to write in your book
  • Periodic rate r = annual rate ÷ periods per year.
  • n = term in years × periods per year. Keep r and n in the same units.
  • Total interest = M × n − PV. Not P × r × t.

Match each term to its meaning:

1

Create the first 4 months of an amortisation schedule for a $30,000 loan at 6% p.a. compounded monthly over 4 years. For each month, show opening balance, interest, principal, and closing balance.

2

Compare total interest on a $300,000 mortgage at 5% p.a. compounded monthly for 25 years vs 30 years. How much extra does the longer term cost?

Top 3 list: Name THREE strategies a borrower can use to reduce the total interest paid on a mortgage. For each, explain the mathematical reason why it works.

10
Revisit your thinking

Monthly interest in Month 1 = $300,000 × 0.05/12 = $1,250. Monthly payment ≈ $1,750. So in Month 1, interest = $1,250 and principal = only $500. Over the first year, interest totals about $14,500 and principal about $6,500. This means roughly 69% of your first year's payments go to interest. This is why making extra repayments early is so powerful — every extra dollar reduces the balance that next month's $1,250 interest is calculated on.

What has changed in your understanding? What did you get right? What surprised you?

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01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next.

Q1. A $20,000 car loan at 7.2% p.a. compounded monthly has r = 0.006 and n = 36. Which expression gives the correct monthly repayment?

Q2. In Month 1 of a $30,000 loan at 6% p.a. compounded monthly, the interest component of the repayment is:

Q3. The total interest paid on a loan is most correctly calculated as:

Q4. Over the life of a mortgage with fixed monthly repayments, the interest component of each repayment:

Q5. A comparison rate is most useful for borrowers because it:

02
Short answer
ApplyBand 42 marks

SA 1. A $350,000 mortgage at 5.4% p.a. compounded monthly over 25 years. (a) Calculate the monthly repayment. (b) Create an amortisation schedule for the first 3 months. (c) Calculate the total interest over the life of the loan. (2 marks)

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ApplyBand 42 marks

SA 2. A couple has a $450,000 mortgage at 4.8% p.a. compounded monthly over 30 years. They can afford to pay $300 extra per month. (a) Find the original monthly repayment. (b) Find the new loan term with the extra payment. (c) Calculate the total interest saved. (2 marks)

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AnalyseBand 53 marks

SA 3. (a) Explain mathematically why paying fortnightly (half the monthly amount, 26 times per year) reduces total interest compared to monthly payments. (b) For a $500,000 mortgage at 5% p.a. compounded monthly over 30 years, calculate the approximate saving from switching to fortnightly payments. (c) A bank offers a 0.2% rate discount but charges a $395 annual fee. For a $400,000 loan over 25 years, is this deal worthwhile? Show all working. (3 marks)

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Comprehensive answers (click to reveal)

MC 1 — B: The repayment formula is M = PV × r / [1−(1+r)^−n].

MC 2 — A: Interest = 30000 × 0.005 = $150.00.

MC 3 — C: Total interest = M×n − PV. Other options use simple interest on the original balance, which ignores the reducing balance.

MC 4 — D: As the balance falls each period, less interest is charged, so more of the fixed repayment reduces principal.

MC 5 — B: The comparison rate includes fees — it is the fairest basis for comparing loan products.

SA 1 (2 marks): (a) r=0.0045, n=300. M = 350000×0.0045/[1−(1.0045)^−300] = 1575/0.7387 = $2132.14/month [0.5 mark]. (b) Month 1: OB=350000, I=1575.00, P=557.14, CB=349442.86. Month 2: OB=349442.86, I=1572.49, P=559.65, CB=348883.21. Month 3: OB=348883.21, I=1569.97, P=562.17, CB=348321.04 [1 mark]. (c) Total = 2132.14×300 = $639,642. Interest = $289,642 [0.5 mark].

SA 2 (2 marks): (a) M = 450000×0.004/[1−(1.004)^−360] = 1800/0.7624 = $2360.97/month [0.5 mark]. (b) New M = $2660.97. Solve: [1−(1.004)^−n] = 450000×0.004/2660.97 = 0.6764. (1.004)^−n = 0.3236. n = ln(0.3236)/−ln(1.004) ≈ 282 months = 23.5 years [0.5 mark]. (c) Original total = 2360.97×360 = $849,949. Interest = $399,949. New total = 2660.97×282 = $750,394. Interest = $300,394. Saved ≈ $99,555 [1 mark].

SA 3 (3 marks): (a) 26 half-payments = 13 full payments per year instead of 12. The extra annual payment goes entirely to principal, reducing the balance faster. Paying more frequently also means slightly less interest accrues between payments [1 mark]. (b) Monthly: M = $2684.11, total = $966,279, interest = $466,279. Fortnightly ≈ 22.5 years, total interest ≈ $340,000. Saved ≈ $126,000 [1 mark]. (c) At 5%: M=$2338.36, total=$701,508. At 4.8%: M=$2295.25, total=$688,575. Repayment saving = $12,933. Fees = 25×$395 = $9,875. Net saving = $3,058. Yes, worthwhile [1 mark].

01
Boss battle · The Loan Calculator
earn bronze · silver · gold

Five timed questions on amortisation, interest calculations, repayment schedules and loan comparisons. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering questions on loans and amortisation. Pool: lesson 7.

Mark lesson as complete

Tick when you've finished the practice and review.