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.