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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.

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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.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

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

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.

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.

Finding any variable — transposition

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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?

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.

$r = 0.05 \div 12 \approx 0.004167$ per month; $n = 30 \times 12 = 360$

Identify and convert all variables: monthly rate and total periods.

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?

$r = 0.048 \div 12 = 0.004$; $n = 25 \times 12 = 300$

Convert annual rate to monthly; find total periods.

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.

$r = 0.005$; $n = 240$; $M = \dfrac{250{,}000 \times 0.005}{1 - (1.005)^{-240}}$

Set up the repayment formula with converted variables.

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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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

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?

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?

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?

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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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.

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].

Boss battle · The Mortgage Broker

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering loan repayment and transposition questions. Pool: lessons 1–14.

Mark lesson as complete

Tick when you've finished the practice and review.

← Recurrence Relations for Loans Lesson 15 · Extra Repayments →

Module overview · Maths Advanced · Checkpoint 3