Mathematics Standard • Year 12 • Module 7 • Lesson 7

Loans and Amortisation — Skill Drill

Build fluency in the amortisation routine: use M = PV × r / [1 − (1+r)⁻ⁿ], split each repayment into interest and principal, and walk a balance down month by month.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Convert each annual nominal rate to the equivalent monthly rate r used in M = PV × r / [1 − (1+r)⁻ⁿ].

6% p.a. → r = ____________ 4.8% p.a. → r = ____________ 7.2% p.a. → r = ____________

Q1.2 Complete the amortisation row formulas.

Interest_this month = ____________ × r Principal_this month = M − ____________ New balance = Old balance − ____________

Q1.3 A 5-year monthly loan has n = ____________ payments. A 25-year monthly mortgage has n = ____________.

Stuck? Revisit lesson § Key Ideas — Monthly repayment formula and Amortisation Schedule.

2. Worked example — first month of an amortisation schedule

Follow each line of working. Every step has a reason on the right.

Problem. A $20,000 car loan at 7.2% p.a. compounded monthly over 3 years has a monthly repayment of $618.96. Find the interest, principal repaid and closing balance for Month 1.

Step 1 — Convert the annual rate to a monthly rate.

r = 0.072 / 12 = 0.006 per month

Reason: the period in M, n and the balance update must all be in the same time unit (months).

Step 2 — Interest charged in Month 1 = Opening balance × r.

I₁ = $20,000 × 0.006 = $120.00

Reason: in a reducing-balance loan, interest each period is charged only on the current balance.

Step 3 — Principal repaid = Repayment − Interest.

P₁ = $618.96 − $120.00 = $498.96

Reason: the repayment first pays the interest charged, and only the leftover reduces the debt.

Step 4 — Closing balance = Opening balance − Principal repaid.

B₁ = $20,000 − $498.96 = $19,501.04

Reason: Month 2 will then charge interest on this smaller balance — that is the "reducing" part.

Conclusion. Month 1: interest $120.00, principal $498.96, closing balance $19,501.04.

3. Faded example — fill in Month 2 of the same loan

The Month 1 closing balance was $19,501.04. The monthly repayment is still $618.96 and r = 0.006. Fill in each blank. 4 marks

Step 1 — Interest for Month 2:

I₂ = $ ____________ × 0.006 = $ ____________

Step 2 — Principal repaid in Month 2:

P₂ = $618.96 − $ ____________ = $ ____________

Step 3 — Closing balance for Month 2:

B₂ = $19,501.04 − $ ____________ = $ ____________

Compare. Is the interest in Month 2 larger or smaller than Month 1? ____________ Is the principal larger or smaller? ____________

Stuck? Revisit lesson § Amortisation Schedule — the worked table for $20,000 at 7.2%.

4. Graduated practice — amortisation calculations

Show your working below each part. Round dollars to 2 dp and rates to 4 sf where needed.

Foundation — convert rates and read one period (4 questions)

Q	Problem	Answer
4.1 1	Convert 6% p.a. to a monthly rate r.
4.2 1	How many monthly payments are in a 4-year loan?
4.3 1	A loan balance is $15,000 and the monthly rate is 0.005. Find the interest charged this month.
4.4 1	A repayment is $850. Interest this month is $250. Find the principal repaid this month.

Standard — typical HSC difficulty (6 questions)

Show the formula step before substituting, and label final answers with units.

4.5 Calculate the monthly repayment on a $30,000 personal loan at 6% p.a. compounded monthly over 4 years. 2 marks

4.6 A $30,000 loan has the monthly repayment from 4.5. For Month 1, calculate (a) interest, (b) principal repaid, (c) closing balance. 3 marks

4.7 Calculate the monthly repayment on a $350,000 mortgage at 5.4% p.a. compounded monthly over 25 years. 2 marks

4.8 For the mortgage in 4.7, calculate the total amount repaid and the total interest paid over the full term. 2 marks

4.9 A loan has opening balance $349,442.86 in Month 2 at monthly rate r = 0.0045 with repayment M = $2,132.14. Find (a) Month 2 interest, (b) principal, (c) closing balance. 3 marks

4.10 Complete the first three rows of the amortisation schedule for a $30,000 loan at 6% p.a. compounded monthly over 4 years, repayment $704.55. Use the headings Opening, Interest, Principal, Closing. 3 marks

Extension — compare scenarios (2 questions)

4.11 A $500,000 mortgage at 4.5% p.a. compounded monthly has monthly repayment $2,533.43. Calculate (a) total amount repaid over 30 years and (b) total interest. State as a percentage how much of the total paid is interest. 3 marks

4.12 For a $300,000 mortgage at 5% p.a. compounded monthly, the monthly repayment is $1,610.46 over 30 years (total interest $279,766) and $1,753.77 over 25 years (total interest $226,131). State (a) the extra monthly repayment to switch to 25 years, and (b) the interest saved by doing so. 3 marks

Stuck on 4.12? Difference in M × 1 month, and difference in total interest over the full term. The extra per month is small; the lifetime saving is large.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I divided by 12 (months in a year), not by 100.

For 4.3 I used I = Balance × r (the monthly rate), not Balance × annual rate.

For 4.4 I computed Principal = M − I, not I − M.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Monthly rates

6% p.a. → r = 0.005. 4.8% p.a. → r = 0.004. 7.2% p.a. → r = 0.006.

Q1.2 — Amortisation formulas

Interest = Opening balance × r. Principal = M − Interest. New balance = Old balance − Principal.

Q1.3 — Number of payments

5-year monthly: n = 60. 25-year monthly: n = 300.

Q3 — Faded example (Month 2 of $20,000 at 7.2% over 3 yr)

I₂ = $19,501.04 × 0.006 = $117.01.
P₂ = $618.96 − $117.01 = $501.95.
B₂ = $19,501.04 − $501.95 = $18,999.09.
Compare: interest is smaller than Month 1; principal is larger.

Q4.1 — Monthly rate

r = 0.06 / 12 = 0.005.

Q4.2 — Number of payments

n = 4 × 12 = 48 months.

Q4.3 — Monthly interest

I = $15,000 × 0.005 = $75.00.

Q4.4 — Principal repaid

P = $850 − $250 = $600.00.

Q4.5 — $30,000 at 6% over 4 years

r = 0.005, n = 48. M = 30,000 × 0.005 / [1 − (1.005)⁻⁴⁸] = 150 / 0.21285 = $704.55/month.

Q4.6 — Month 1 split of the $30,000 loan

(a) I₁ = 30,000 × 0.005 = $150.00. (b) P₁ = 704.55 − 150 = $554.55. (c) B₁ = 30,000 − 554.55 = $29,445.45.

Q4.7 — $350,000 mortgage at 5.4% over 25 years

r = 0.0045, n = 300. M = 350,000 × 0.0045 / [1 − (1.0045)⁻³⁰⁰] = 1,575 / 0.73876 = $2,132.14/month.

Q4.8 — Total paid and total interest for the mortgage

Total = 2,132.14 × 300 = $639,642 (to nearest dollar). Interest = 639,642 − 350,000 = $289,642.

Q4.9 — Month 2 of the $350,000 mortgage

(a) I₂ = 349,442.86 × 0.0045 = $1,572.49. (b) P₂ = 2,132.14 − 1,572.49 = $559.65. (c) B₂ = 349,442.86 − 559.65 = $348,883.21.

Q4.10 — First three rows of the $30,000 loan schedule

Month 1: Opening $30,000.00; Interest $150.00; Principal $554.55; Closing $29,445.45.
Month 2: Opening $29,445.45; Interest $147.23; Principal $557.32; Closing $28,888.13.
Month 3: Opening $28,888.13; Interest $144.44; Principal $560.11; Closing $28,328.02.

Q4.11 — $500,000 mortgage at 4.5% over 30 years

(a) Total = 2,533.43 × 360 = $912,035 (to nearest dollar). (b) Interest = 912,035 − 500,000 = $412,035. As a fraction of the total paid: 412,035 / 912,035 ≈ 45.2% — almost half of every dollar paid is interest.

Q4.12 — Switching to a 25-year term

(a) Extra monthly = $1,753.77 − $1,610.46 = $143.31 more per month. (b) Interest saved = $279,766 − $226,131 = $53,635 over the life of the loan. Small extra each month, big lifetime saving — that is the "early principal" effect.