Mathematics Advanced • Year 12 • Module 7 • Lesson 14
Calculating Loan Repayments
Apply the loan-repayment formula and its transpositions to affordability, comparison and loan-design problems.
Problem 1 — A $600,000 / 5.5% / 30-year mortgage (Think First scenario)
A couple borrows $600,000 at 5.5% p.a. compounded monthly over 30 years.
Set up: What are we solving for?
(i) State r per month and n. 1 mark
(ii) Use M = Pr ÷ [1 − (1+r)−n] to compute the monthly repayment to the nearest cent. 3 marks
(iii) Recompute M if they switch to a 20-year term, same rate. Express the increase as a percentage of the 30-year repayment, and comment on whether the increase is closer to 20%, 30% or 50%. 3 marks
Stuck? Revisit lesson § Think First and § Worked Example.Problem 2 — The 28% affordability rule
A household earns $9,000/month gross. The lender will not allow loan repayments to exceed 28% of gross monthly income.
Set up: What are we solving for?
(i) Compute the maximum allowable M. 1 mark
(ii) At 5.2% p.a. compounded monthly over 25 years, what is the maximum loan P this household can afford? 3 marks
(iii) If the median house price in their suburb is $850,000 and they have a $150,000 deposit, can they afford the median house? Show the gap (or surplus) in dollars. 2 marks
Problem 3 — Reverse lookup (find the term)
A car loan of $20,000 is taken at 7.2% p.a. compounded monthly. The borrower can afford M = $500/month.
Set up: What are we solving for?
(i) Verify that M = $500 exceeds the breakeven interest charge in Month 1 (otherwise the loan can never clear). 1 mark
(ii) Use n = −ln(1 − Pr ÷ M) ÷ ln(1 + r) to find n (months) and convert to years and months. 3 marks
(iii) Compute the total interest paid over the life of the loan. 2 marks
Problem 4 — Two products at competing rates
A $500,000 home loan is offered by two banks. Bank A: 5.0% p.a. compounded monthly, 30-year term. Bank B: 4.6% p.a. compounded monthly, 30-year term.
Set up: What are we solving for?
(i) Compute M and total interest under each bank's offer. 3 marks
(ii) State the monthly saving and the lifetime interest saving by choosing Bank B. 2 marks
(iii) Explain in 1 sentence why a 0.4 percentage-point rate cut is worth more than a 0.4 percentage-point fee cut on a super fund of the same starting balance and term. 2 marks
Problem 5 — Budget-first loan design
A buyer can comfortably pay $2,000/month. They want to choose between three rate/term combinations, all at the same lender.
Option 1: 5.4% p.a. compounded monthly, 25-year term.
Option 2: 5.4% p.a. compounded monthly, 30-year term.
Option 3: 4.8% p.a. compounded monthly, 30-year term.
Set up: What are we solving for?
(i) For each option, find the maximum P the buyer can borrow at M = $2,000/month. 3 marks
(ii) Rank the options from highest to lowest borrowing capacity. 1 mark
(iii) Comment in 1-2 sentences on the trade-off: which factor (rate vs term) buys more borrowing capacity at the same monthly outlay? 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — $600,000 / 5.5% / 30 yr
Set up. Compute the minimum repayment for two terms and compare.
(i) r = 0.055 ÷ 12 ≈ 0.004583; n = 30 × 12 = 360.
(ii) Pr = 600,000 × 0.004583 = $2,750. (1.004583)−360 ≈ 0.19447. Denom = 0.80553. M = 2,750 ÷ 0.80553 ≈ $3,414.02/month.
(iii) 20-year: n = 240. (1.004583)−240 ≈ 0.33277. Denom = 0.66723. M = 2,750 ÷ 0.66723 ≈ $4,121.55/month. Increase = (4,121.55 − 3,414.02) ÷ 3,414.02 ≈ 20.7%. Closest to 20% — much less than the "50%" some students predict, because the denominator 1 − (1+r)−n grows non-linearly with n.
Problem 2 — 28% affordability rule
Set up. Convert allowable repayment into maximum borrowable loan, then compare against the house price.
(i) M = 9,000 × 0.28 = $2,520/month.
(ii) r = 0.052 ÷ 12 ≈ 0.004333; n = 25 × 12 = 300. (1.004333)−300 ≈ 0.27358. 1 − 0.27358 = 0.72642. Annuity factor = 0.72642 ÷ 0.004333 ≈ 167.628. P = 2,520 × 167.628 ≈ $422,422.
(iii) They need 850,000 − 150,000 = $700,000 of borrowing, but can only afford $422,422. Gap ≈ $277,578. They cannot afford the median house without a larger deposit, a longer term, a lower rate, or a higher income.
Problem 3 — $20,000 / 7.2% / $500-per-month
Set up. Verify the repayment is feasible, then solve for n.
(i) r = 0.006. I1 = 20,000 × 0.006 = $120. M = $500 > $120, so the balance reduces in Month 1 — the loan can be cleared.
(ii) Pr ÷ M = 20,000 × 0.006 ÷ 500 = 0.24. n = −ln(1 − 0.24) ÷ ln(1.006) = −ln(0.76) ÷ 0.005982 = 0.27444 ÷ 0.005982 ≈ 45.9 months — about 3 years 10 months.
(iii) Total repaid = 500 × 45.9 ≈ $22,950. Total interest = 22,950 − 20,000 ≈ $2,950.
Problem 4 — Bank A vs Bank B on $500,000
Set up. Compute M and total interest for each bank.
(i) Bank A: r = 0.005/4.166̅, but precisely 0.05 ÷ 12 ≈ 0.004167; n = 360; Pr = $2,083.33; (1.004167)−360 ≈ 0.22384; denom = 0.77616; M = 2,083.33 ÷ 0.77616 ≈ $2,684.11. Total interest = 2,684.11 × 360 − 500,000 ≈ $466,280.
Bank B: r = 0.046 ÷ 12 ≈ 0.003833; Pr = $1,916.67; (1.003833)−360 ≈ 0.25195; denom = 0.74805; M = 1,916.67 ÷ 0.74805 ≈ $2,562.50. Total interest = 2,562.50 × 360 − 500,000 ≈ $422,500.
(ii) Monthly saving ≈ $121.61. Lifetime interest saving ≈ $43,780 over 30 years.
(iii) A rate cut on a loan reduces both the interest charge each month and the lifetime denominator, accelerating principal reduction — the saving is on a six-figure balance over 30 years. A fee cut on super only changes the growth factor at the margin and is offset by contributions; same percentage gap, but the loan version cashflow effect compounds against a much larger initial balance.
Problem 5 — Budget-first design at M = $2,000
Set up. Use P = M × annuity factor for each option, then compare.
(i) Option 1 (5.4%/25 yr): r = 0.0045, n = 300. (1.0045)−300 ≈ 0.25997. Annuity factor = (1 − 0.25997)/0.0045 = 164.451. P ≈ 2,000 × 164.451 ≈ $328,902.
Option 2 (5.4%/30 yr): n = 360. (1.0045)−360 ≈ 0.19852. Annuity factor = 178.108. P ≈ 2,000 × 178.108 ≈ $356,216.
Option 3 (4.8%/30 yr): r = 0.004, n = 360. (1.004)−360 ≈ 0.23769. Annuity factor = 190.578. P ≈ 2,000 × 190.578 ≈ $381,156.
(ii) Ranking: Option 3 > Option 2 > Option 1.
(iii) Cutting the rate by 0.6 pp (Option 2 → Option 3) buys roughly $25,000 of extra borrowing capacity; extending the term by 5 years (Option 1 → Option 2) buys about $27,000. Rate cuts and term extensions buy similar capacity at this level, but the rate cut is far cheaper in lifetime interest cost (no extra years of repayment).