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Module 7 · L14 of 20 ~40 min ⚡ +95 XP available

Calculating Loan Repayments

The single most important number in any loan is not the interest rate, it is the repayment. Too high and you cannot afford it; too low and you pay a fortune in extra interest. In this lesson you will learn the formula that banks use to calculate your minimum repayment, how to transpose it to find any missing variable, and how to evaluate whether a loan is affordable.

Today's hook, A couple borrows $\$600,000$ at 5.5% over 30 years. Is their monthly repayment closer to $\$2,000$, $3,000, or $4,000? And if they switch to 20 years, does the repayment increase by 20%, 30%, or 50%? The answer surprises most people, and the formula explains why.
0/5QUESTS

Get oriented

Recall what you already know and settle the terms for loan repayments.

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
Think first, your gut answer
+5 XP warm-up

A couple borrows $600,000 at 5.5% p.a. compounded monthly over 30 years.

Question 1: Do you think their monthly repayment is closer to $2,000, $3,000, or $4,000?

Question 2: If they switch to a 20-year term, will the monthly repayment increase by roughly 20%, 30%, or 50%?

Predict before calculating your gut instinct now, formula later.

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What you'll master
Know

Key facts

  • The loan repayment formula
  • How to transpose for $P$, $M$, $r$, or $n$
  • How banks calculate minimum repayments
Understand

Concepts

  • Why the repayment formula is the PV annuity formula solved for $M$
  • How term affects repayment and total interest
  • The affordability threshold (28–30% of income)
Can do

Skills

  • Calculate minimum repayments for any loan
  • Find maximum borrowable amount given a budget
  • Compare loans with different terms and rates
  • Assess loan affordability
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Key terms
Repayment ($M$)The fixed amount paid each period to service the loan. Covers both interest and principal reduction.
Present value annuityThe current value of a series of equal future payments; the repayment formula derives from this.
Minimum repaymentThe smallest $M$ that ensures the loan is fully repaid within the agreed term. Pay less and it never clears.
TranspositionRearranging a formula to isolate a different variable, e.g., solving the repayment formula for $P$ or $n$.
Affordability ruleA common guideline: keep monthly repayments below 28–30% of gross monthly income.
Loan term ($n$)Total number of payment periods over the life of the loan.

Meet the repayment formula

See what each symbol in the repayment formula stands for.

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The loan repayment formula
core concept

The repayment formula is derived from the present value annuity formula. If $P$ is the amount borrowed, then $P$ must equal the present value of all future repayments:

$$P = M \times \dfrac{1 - (1+r)^{-n}}{r}$$

Solving for $M$ gives the repayment formula:

$$M = \dfrac{P \times r}{1 - (1+r)^{-n}}$$

$M$ = repayment per period  |  $P$ = principal (amount borrowed)  |  $r$ = interest rate per period  |  $n$ = total number of periods

Example: $P = \$400{,}000$, $r = 0.05/12 = 0.004167$, $n = 360$ months (30 years).

$M = \dfrac{400{,}000 \times 0.004167}{1 - (1.004167)^{-360}}$

$= \dfrac{1{,}666.67}{1 - 0.2238} = \dfrac{1{,}666.67}{0.7762} \approx \$2{,}147.29$/month

Pay less than $2,147 and the loan never clears
The denominator is the key. The term $(1-(1+r)^{-n})$ increases as $n$ grows, but it approaches 1 asymptotically. This is why doubling the term does NOT halve the repayment, it only reduces it by about 25–30%. The bank knows this; most borrowers don't.

Repayment formula: $M = \dfrac{Pr}{1-(1+r)^{-n}}$; Find principal: $P = M \times \dfrac{1-(1+r)^{-n}}{r}$

Pause, copy the repayment formula $M = \dfrac{Pr}{1-(1+r)^{-n}}$ and the reverse $P = M \times \dfrac{1-(1+r)^{-n}}{r}$ (maximum borrowable amount) into your book.

Did you get this? True or false: doubling the loan term (e.g., from 15 to 30 years) roughly halves the monthly repayment.

Transpose for any variable

Make the term, the rate or the principal the subject instead.

05
Finding any variable, transposition
+5 XP to read

We just saw the repayment formula $M = \dfrac{Pr}{1-(1+r)^{-n}}$, it gives $M$ directly when $P$, $r$, and $n$ are known. That raises a question: HSC questions often give $M$ and ask for $P$ or $n$, can we rearrange? This card answers it → transposing to find maximum borrowable $P$ or loan term $n$ via logarithms.

HSC questions may ask you to find $P$, $n$, or $r$ given the other values. The key transpositions are:

Find $P$ (max borrowable)
$P = M \times \dfrac{1-(1+r)^{-n}}{r}$, use the PV annuity formula directly with the given repayment.
Find $n$ (how long to pay off)
Rearrange: $(1+r)^{-n} = 1 - Pr/M$, then $n = \dfrac{-\ln(1-Pr/M)}{\ln(1+r)}$.
Find $r$
Usually requires trial-and-error or a financial calculator. Not commonly tested in HSC, but understand the principle.

Example, Find $n$: $P = \$20{,}000$, $M = \$400$/month, $r = 0.006$ (7.2% p.a. monthly).

$$n = \dfrac{-\ln(1 - 20{,}000 \times 0.006 / 400)}{\ln(1.006)} = \dfrac{-\ln(1-0.3)}{\ln(1.006)} = \dfrac{-\ln(0.7)}{0.005978} = \dfrac{0.3567}{0.005978} \approx 59.7 \text{ months}$$

So approximately 60 months (5 years). Always round up you need whole periods, and you can't do 59.7 payments.

To find $P$: use the PV annuity formula $P = M[1-(1+r)^{-n}]/r$; To find $n$: rearrange to $(1+r)^{-n} = 1 - Pr/M$, then take logs: $n = -\ln(1-Pr/M) / \ln(1+r)$

Pause, copy the two key transpositions: $P = M \times \dfrac{1-(1+r)^{-n}}{r}$ for max borrowable, and $n = \dfrac{-\ln(1-Pr/M)}{\ln(1+r)}$ for loan term (round $n$ up), into your book.

Quick check: To find the maximum loan amount a borrower can afford given $M$, $r$, and $n$, which formula do you use?

Work the examples

Follow three worked repayment problems end to end.

Worked examples · step through 3 in a row

PROBLEM 1 · CALCULATE MINIMUM REPAYMENT

Find the minimum monthly repayment for a $400,000 loan at 5% p.a. compounded monthly over 30 years.

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$r = 0.05 \div 12 \approx 0.004167$ per month; $n = 30 \times 12 = 360$
Identify and convert all variables: monthly rate and total periods.
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PROBLEM 2 · FIND MAXIMUM BORROWABLE AMOUNT

A home buyer can afford $2,800 per month. The interest rate is 4.8% p.a. compounded monthly. They want a 25-year loan. What is the maximum they can borrow?

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$r = 0.048 \div 12 = 0.004$; $n = 25 \times 12 = 300$
Convert annual rate to monthly; find total periods.
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PROBLEM 3 · FIND REPAYMENT + TOTAL INTEREST

Find the monthly repayment for a $250,000 loan at 6% p.a. compounded monthly over 20 years. Then find the total interest paid.

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$r = 0.005$; $n = 240$; $M = \dfrac{250{,}000 \times 0.005}{1 - (1.005)^{-240}}$
Set up the repayment formula with converted variables.
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Think & link: A $\$300,000$ loan at 6% p.a. monthly over 20 years. Set up (but don't fully evaluate) the expression for $M$. What is the numerator?

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Avoid the traps, then revisit

Fix the simple-average assumption, run the calculator challenge, then name what has changed.

Common errors · 3 traps that cost marks

Trap 01
Simple average repayment
Some students calculate repayment as (principal / n) + one period's interest. This massively understates the true payment because it ignores compounding on the declining balance. You must use the formula.
Trap 02
Doubling term halves repayment
The denominator $(1-(1+r)^{-n})$ grows with $n$ but approaches 1, so the repayment falls with longer terms, but not proportionally. A 15-year loan has a repayment roughly 1.4–1.5× that of a 30-year loan, not 2×.
Trap 03
Rounding $n$ down
When finding the number of periods to clear a loan, always round $n$ UP. Rounding down means you haven't made enough payments to clear the loan, a small balance would remain.

Fill in the blanks. To find the term $n$ of a $\$15,000$ loan at 9.6% p.a. monthly ($r = 0.008$, $M = 350$), complete the formula below:

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Activity, loan repayment calculator challenge
+10 XP activity
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Part 1, Minimum repayment. Set $P = \$500{,}000$, $r = 4.8\%$ p.a. monthly, term = 30 years. Calculate the minimum monthly repayment, total repaid, and total interest. What percentage of total repaid is interest?

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Part 2, Term comparison. Keep $P = \$500{,}000$, $r = 4.8\%$. Change to a 20-year term. How much extra per month? How much interest saved? Calculate the "ROI": interest saved ÷ (extra paid × 240). Is the extra payment worth it?

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Part 3, Affordability. A household earns $9,000/month gross. Using the 28% rule, what is their maximum affordable monthly repayment? At 5.2% p.a. monthly over 25 years, what is the maximum loan they can afford? Research: can this household afford the median Sydney house price?

Odd one out: Three of these statements are true. Which one is FALSE?

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Revisit your thinking

Earlier you predicted whether $\$600,000$ at 5.5% over 30 years is closer to $\$2,000$, $3,000, or $4,000 per month, and how much repayments change when the term shortens to 20 years.

The answer: $M = \dfrac{600{,}000 \times 0.004583}{1 - (1.004583)^{-360}} = \dfrac{2{,}750}{0.8055} \approx \$3{,}414$/month. Your prediction of $3,000 was close.

Over 20 years: $M = \dfrac{600{,}000 \times 0.004583}{1 - (1.004583)^{-240}} \approx \$4{,}121$/month. The increase is about 21%, not 50%. Many people overestimate the impact of shortening the term because they forget the denominator in the formula grows non-linearly, as the term shrinks, the denominator falls, but only moderately.

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Practise loan repayments

Write full short-answer responses, then check them against the model answers.

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.

02
Short answer
ApplyBand 43 marks

Q1. Find the minimum monthly repayment for a $300,000 loan at 6% p.a. compounded monthly over 20 years. (3 marks)

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ApplyBand 43 marks

Q2. A borrower can afford $1,500 per month. The interest rate is 4.8% p.a. compounded monthly. They want a 15-year loan. What is the maximum amount they can borrow? (3 marks)

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AnalyseBand 54 marks

Q3. A home loan of $450,000 at 5% p.a. compounded monthly over 30 years. (a) Find the minimum monthly repayment. (b) Find the total amount repaid. (c) Find the total interest paid. (4 marks)

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Comprehensive answers (click to reveal)

Activity 1: $M \approx \$2{,}623$/month. Total repaid $= 2{,}623 \times 360 = \$944{,}280$. Total interest $= \$444{,}280$. Interest percentage $= 444{,}280 / 944{,}280 \approx 47\%$, nearly half is interest.

Activity 2: 20-year repayment $\approx \$3{,}245$. Total interest $\approx \$278{,}800$. Extra/month $= \$622$. Interest saved $= \$165{,}480$. ROI $= 165{,}480 / (622 \times 240) = 111\%$, every extra dollar paid saves more than a dollar in interest.

Activity 3: Max affordable $= 9{,}000 \times 0.28 = \$2{,}520$/month. $P = 2{,}520 \times [1-(1.004333)^{-300}] / 0.004333 \approx \$425{,}000$. Sydney median ≈ $1.1–1.3 million (2024); this household cannot afford it without a large deposit.

Q1 (3 marks): $r = 0.005$, $n = 240$. $M = \dfrac{300{,}000 \times 0.005}{1-(1.005)^{-240}} = \dfrac{1{,}500}{0.6979} \approx \$2{,}149.29$ [2]. This is the minimum monthly repayment [1].

Q2 (3 marks): $r = 0.004$, $n = 180$. $P = 1{,}500 \times [1-(1.004)^{-180}]/0.004 = 1{,}500 \times 118.50 \approx \$177{,}750$ [2]. This is the maximum affordable loan [1].

Q3 (4 marks): (a) $r = 0.004167$, $n = 360$. $M = \dfrac{450{,}000 \times 0.004167}{1-(1.004167)^{-360}} \approx \$2{,}416.24$ [2]. (b) Total repaid $= 2{,}416.24 \times 360 = \$869{,}846$ [1]. (c) Total interest $= 869{,}846 - 450{,}000 = \$419{,}846$ [1].