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Module 7 · L7 of 12 ~30 min MST-12-S2-02 ⚡ +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.
0/5QUESTS

Get oriented

Recall what you already know, meet the key ideas and settle the terms.

Worksheets

Practise this lesson

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

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
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: $\text{Balance}_{new} = \text{Balance}_{old} - P$

The repayment formula for an amortising loan.
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.

Read an amortisation schedule

Track every dollar of a repayment as it splits into interest and principal.

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.

An amortisation schedule shows each repayment split into interest (= r × opening balance) and principal reduction (= repayment − interest). Opening balance for next period = previous balance − principal paid. Final balance = 0.

Pause, copy the four amortisation columns: opening balance, interest (= r × opening balance), principal paid (= repayment − interest), and closing balance (= opening − principal paid) into your book.

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:

See what extra repayments do

Work out how paying more than the minimum changes the term and the total cost.

06
Extra repayments, the power of paying more
core concept

The amortisation schedule shows that each repayment splits into interest (= r × opening balance) and a principal reduction. When you make an extra payment, every dollar above the scheduled repayment goes directly to reducing the principal, which shrinks every future interest charge, shortening the loan and saving money that compounds over the remaining term.

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.

Extra repayments directly reduce the principal, which reduces future interest charges. Total interest saved = interest that would have accrued on the extra amount over remaining periods. Even small extra payments in early periods save significant interest.

Pause, copy the principal-reduction mechanism (every extra dollar goes directly to principal, cutting all future interest charges) and note why early extra payments save the most: the principal reduction compounds over more remaining periods into your book.

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.

Compare loans and terms

Decide which loan is cheaper, and see why the term matters more than the rate.

Worked examples · reveal each step

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
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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
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08
Comparing loans, which is cheaper?
core concept

Extra repayments save interest because they reduce the balance on which future interest is calculated. When comparing two loans, different rates, terms, or repayment structures, the definitive comparison is total interest paid: total repayments (repayment × n) minus the principal borrowed. A lower interest rate does not always mean lower total cost if the term is longer.

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.

To compare two loans: calculate total repayments for each (monthly repayment × number of repayments), then subtract the principal to find total interest. The loan with lower total interest is cheaper, regardless of the rate label.

Pause, copy total interest = (repayment × n) − principal, and the comparison rule: the cheaper loan is the one with lower total interest, not necessarily the one with the lower advertised rate into your book.

08
Buy now pay later, and why term matters more than rate
core concept

The syllabus asks you to compare three ways of paying later: buy now pay later, a short-term loan and a long-term loan. They are not three points on one scale, because BNPL is not priced like a loan at all.

Take a $\$3000$ purchase. A personal loan charges $12\%$ p.a. compounded monthly; BNPL charges nothing for $12$ months provided every instalment is paid on time.

OptionTermRepaymentTotal paidInterest
BNPL, paid on time12 months$\$250.00$$\$3000.00$$\$0.00$
Short-term loan2 years$\$141.22$$\$3389.28$$\$389.28$
Long-term loan5 years$\$66.73$$\$4003.80$$\$1003.80$

Read the last two rows together. Same debt, same interest rate, and the long-term loan costs two and a half times as much interest. The only thing that changed is how long the money is borrowed for. The smaller repayment is what makes it feel cheaper, and it is exactly what makes it dearer.

The BNPL condition is doing all the work. That $\$0$ holds only while every instalment lands on time. Miss one and a late fee applies, typically around $\$10$ per missed payment, and any balance still owing after the interest-free period starts attracting interest. On a rate of $18\%$ p.a. simple, $\$800$ still owing for a further $6$ months adds $800 \times 0.18 \times 0.5 = \$72$. BNPL is the cheapest option in the table and the easiest one to lose money on.

Compare pay-later options on TOTAL amount paid, not on the repayment. At the same interest rate, a longer term always costs more interest, because interest is charged for longer: $\$3000$ at $12\%$ costs $\$389.28$ over 2 years and $\$1003.80$ over 5. Buy now pay later charges no interest during its interest-free period, but only if every instalment is paid on time; late fees and post-period interest are what make it expensive.

Pause, copy the three-row comparison and one line saying why a longer term costs more at the same rate, into your book.

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 $\$$ .

Avoid the amortisation traps

Fix the annual-versus-periodic rate slips that cost marks.

Common errors · the 3 traps that cost marks

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.

Match each term to its meaning:

  • Amortisation
  • Principal component
  • Comparison rate
  • Total interest
  • Includes fees as well as interest for fair loan comparison
  • M times n minus the original loan amount
  • Paying off a loan through regular repayments split between interest and principal
  • The part of each repayment that reduces the loan balance

Drill it and revisit

Run the drill, then return to your opening answer and name what has changed.

Quick-fire practice · 2 activities

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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Practise amortisation

Answer the question bank, then write full short-answer responses.

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] = 1575/0.739973 = $\$2128.45$/month [0.5 mark]. (b) Month 1: OB=350000, I=1575.00, P=553.45, CB=349446.55. Month 2: OB=349446.55, I=1572.51, P=555.94, CB=348890.61. Month 3: OB=348890.61, I=1570.01, P=558.44, CB=348332.17 [1 mark]. (c) Total = 2128.45×300 ≈ $\$638{,}536$ (keep the unrounded repayment on your calculator). Interest ≈ $\$288{,}536$ [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].