Mathematics Standard • Year 12 • Module 7 • Lesson 7
Loans and Amortisation — Problem Set
Apply amortisation reasoning to realistic Australian borrowing scenarios — car loans, mortgages, extra repayments and refinancing.
Problem 1 — Used-car amortisation
Hugo borrows $24,000 from a credit union to buy a hatchback at 6.6% p.a. compounded monthly over 5 years. His monthly repayment is $470.85.
Set up: What are we solving for?
(i) Calculate the total interest Hugo pays over the life of the loan. 2 marks
(ii) For Month 1, find the interest, principal repaid and closing balance. 2 marks
(iii) For Month 2 (opening balance from your answer to (ii)), find the interest and principal repaid. Comment on what is happening to the split. 2 marks
Stuck? Revisit lesson § Amortisation Schedule — interest each month is charged on the current balance, not the original $24,000.Problem 2 — First-home mortgage total cost
Asha and Daniel take out a $450,000 mortgage at 4.8% p.a. compounded monthly over 30 years. The monthly repayment is $2,360.97.
Set up: What are we solving for?
(i) Calculate the total amount Asha and Daniel will repay over the full 30 years. 1 mark
(ii) Calculate the total interest paid. 1 mark
(iii) Express the total interest as a percentage of the original loan. Comment in one sentence on what this means in plain English. 2 marks
Stuck? Revisit lesson § Extra Repayments — the $400,000 at 4.8% example.Problem 3 — The power of paying $300 extra
For the same Asha and Daniel mortgage from Problem 2 ($450,000 at 4.8% over 30 years, M = $2,360.97), they decide to pay $2,660.97 per month (an extra $300). The loan is then projected to be cleared in 282 months instead of 360.
Set up: What are we solving for?
(i) Calculate the total amount paid with the extra repayments. 1 mark
(ii) Calculate the total interest under the extra-repayment plan and the interest saved compared to Problem 2(ii). 2 marks
(iii) The couple's extra contribution totals just $300 × 282 = $84,600 across the term. Explain in one or two sentences why this comparatively small extra cash leads to such a big interest saving. 2 marks
Stuck? Revisit lesson § Extra Repayments — "An extra $101/month saves nearly $100,000".Problem 4 — Choosing a loan term
Maya is comparing two terms on a $300,000 mortgage at 5% p.a. compounded monthly: 25 years (M = $1,753.77) versus 30 years (M = $1,610.46).
Set up: What are we solving for?
(i) Calculate the total interest paid under each option. 2 marks
(ii) State the difference in monthly repayment and the difference in total interest. 1 mark
(iii) Recommend which term Maya should take if she can comfortably afford either monthly repayment, and explain in one sentence. 2 marks
Stuck? Revisit lesson § Activity 2 — comparing $300,000 at 25 vs 30 years.Problem 5 — A small rate cut, a big number
Ranjit has a $400,000 mortgage at 5.4% p.a. compounded monthly over 25 years (M = $2,432.69). He has the option to refinance to 4.4% p.a. compounded monthly over the remaining 25 years (M = $2,193.61), with no fees.
Set up: What are we solving for?
(i) Calculate the total amount repaid under the original 5.4% loan and under the refinanced 4.4% loan. 2 marks
(ii) Calculate the total interest saved by refinancing. 1 mark
(iii) Express the monthly saving in dollars and explain in one sentence why a "small" 1% rate cut leads to such a large lifetime saving. 2 marks
Stuck? Revisit lesson § Activity 2 Q3 — a 1% rate reduction on a $400,000 mortgage.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Used-car amortisation
Set up. We need total interest over 60 months, then the Month-1 and Month-2 split.
(i) Total paid = 470.85 × 60 = $28,251.00. Interest = $28,251 − $24,000 = $4,251.00.
(ii) r = 0.066 / 12 = 0.0055. I₁ = 24,000 × 0.0055 = $132.00. P₁ = 470.85 − 132.00 = $338.85. B₁ = 24,000 − 338.85 = $23,661.15.
(iii) I₂ = 23,661.15 × 0.0055 = $130.14. P₂ = 470.85 − 130.14 = $340.71. Interest is slightly smaller, principal slightly larger — the split is gradually shifting towards paying off principal, exactly as expected for an amortising loan.
Problem 2 — Total cost of the mortgage
Set up. Total repaid = M × n; subtract the original loan to find interest; express as %.
(i) Total = 2,360.97 × 360 = $849,949 (to nearest dollar).
(ii) Interest = 849,949 − 450,000 = $399,949.
(iii) 399,949 / 450,000 ≈ 88.9%. They pay nearly $0.89 of interest for every $1 borrowed — almost as much interest as the original loan.
Problem 3 — Extra repayments save big
Set up. Total with extra = $2,660.97 × 282; subtract original loan and compare with Problem 2(ii).
(i) Total = 2,660.97 × 282 = $750,393 (to nearest dollar).
(ii) Interest under extra plan = 750,393 − 450,000 = $300,393. Interest saved = 399,949 − 300,393 = $99,556.
(iii) The $300 extra each month immediately reduces the principal, so every subsequent month's interest is charged on a smaller balance. Compounded over 282 months, those small reductions snowball into nearly $100,000 of avoided interest.
Problem 4 — 25-year vs 30-year term
Set up. Compute total paid and total interest for each term; compare directly.
(i) 25-yr: Total = 1,753.77 × 300 = $526,131. Interest = $226,131. 30-yr: Total = 1,610.46 × 360 = $579,766. Interest = $279,766.
(ii) Monthly difference = 1,753.77 − 1,610.46 = $143.31 more per month for the 25-yr term. Interest difference = 279,766 − 226,131 = $53,635 more interest for the 30-yr term.
(iii) Recommend the 25-year term: it costs only $143.31 extra per month yet saves $53,635 in interest over the life of the loan.
Problem 5 — Refinance saving
Set up. Compute total paid under both rates over 25 years and compare.
(i) 5.4%: Total = 2,432.69 × 300 = $729,807. 4.4%: Total = 2,193.61 × 300 = $658,083.
(ii) Interest saved = 729,807 − 658,083 = $71,724.
(iii) Monthly saving = 2,432.69 − 2,193.61 = $239.08. Because that saving runs every month for 300 months, a 1 percentage-point rate cut adds up to over $71,000 in lifetime interest — even though it "feels" like only a small change.