Mathematics Advanced • Year 12 • Module 7 • Lesson 14
Calculating Loan Repayments
Practise HSC-style writing on the loan-repayment formula and an extended response on term/rate trade-offs.
1. Short-answer questions
1.1 A home loan of $300,000 is taken at 6.0% p.a. compounded monthly over 20 years.
(a) State r per month and n.
(b) Find the minimum monthly repayment M to the nearest cent. 3 marks Band 3
1.2 A borrower can afford $1,500 per month at 4.8% p.a. compounded monthly over 15 years.
(a) State r per month and n.
(b) Find the maximum loan P the borrower can take to the nearest dollar. 3 marks Band 3-4
1.3 A $450,000 home loan is taken at 5.0% p.a. compounded monthly over 30 years.
(a) Find the minimum monthly repayment.
(b) Find the total amount repaid over the loan's life.
(c) Find the total interest paid. 4 marks Band 4
2. Extended response
2.1 A buyer is comparing three options for a $500,000 home loan at 5.0% p.a. compounded monthly. All three options have the same rate; only the term differs.
Option A: 20-year term.
Option B: 25-year term.
Option C: 30-year term.
(a) Compute the minimum monthly repayment M for each option using M = Pr ÷ [1 − (1+r)−n], showing r and n explicitly.
(b) Compute the total amount repaid and the total interest paid over the life of each loan.
(c) State and explain in 2-3 sentences why the longer-term loan has a smaller monthly payment but a larger total interest cost. Reference the structure of the formula. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — correct r = 0.05 ÷ 12 ≈ 0.004167 stated.
• 1 mark — at least two of three monthly repayments correct to within $5.
• 1 mark — all three monthly repayments correct to within $1.
Part (b) — 3 marks
• 1 mark — at least one option's total interest correct to within $200.
• 1 mark — two options correct to within $500.
• 1 mark — all three totals (repaid and interest) correct to within $500.
Part (c) — 2 marks
• 1 mark — explicitly notes that increasing n shrinks Pr ÷ [1 − (1+r)−n] because the denominator approaches 1, so M falls.
• 1 mark — explicitly notes that the longer term means more periods over which interest is paid, so total interest = M × n − P rises despite M being lower.
Your response:
Stuck on (c)? Reference the lesson's "Misconceptions" callout: doubling the term does not double the repayment, but it does massively increase total interest.How did this worksheet feel?
What I'll revisit before next class:
1.1 — $300,000 at 6%, 20 years (3 marks)
Sample response. (a) r = 0.06 ÷ 12 = 0.005; n = 20 × 12 = 240. (b) Pr = 300,000 × 0.005 = $1,500. (1.005)−240 ≈ 0.30212. Denom = 0.69788. M = 1,500 ÷ 0.69788 ≈ $2,149.29/month.
Marking notes. 1 mark — correct r and n. 1 mark — correct substitution into the formula with (1.005)−240 shown. 1 mark — correct M to nearest cent.
1.2 — M = $1,500 at 4.8%, 15 years, find max P (3 marks)
Sample response. (a) r = 0.048 ÷ 12 = 0.004; n = 15 × 12 = 180. (b) (1.004)−180 ≈ 0.48774. 1 − 0.48774 = 0.51226. Annuity factor = 0.51226 ÷ 0.004 = 128.065. P = 1,500 × 128.065 ≈ $192,098.
Marking notes. 1 mark — correct r and n. 1 mark — correct annuity factor. 1 mark — correct P to the nearest dollar.
1.3 — $450,000 at 5%, 30 years (4 marks)
Sample response. r = 0.05 ÷ 12 ≈ 0.004167; n = 360. (a) Pr = 450,000 × 0.004167 = $1,875. (1.004167)−360 ≈ 0.22384. Denom = 0.77616. M = 1,875 ÷ 0.77616 ≈ $2,415.70/month. (b) Total repaid = 2,415.70 × 360 ≈ $869,652. (c) Total interest = 869,652 − 450,000 ≈ $419,652.
Marking notes. 1 mark — correct M (within $1). 1 mark — correct multiplication M × 360. 1 mark — correct total repaid. 1 mark — correct total interest (subtract P).
2.1 — $500,000 at 5%, three terms (8 marks): sample Band-6 response
Sample Band-6 response.
(a) Repayments. r = 0.05 ÷ 12 ≈ 0.004167 throughout; Pr = $2,083.33.
Option A (n = 240): (1.004167)−240 ≈ 0.36879; denom = 0.63121; M = 2,083.33 ÷ 0.63121 ≈ $3,300.46.
Option B (n = 300): (1.004167)−300 ≈ 0.28722; denom = 0.71278; M = 2,083.33 ÷ 0.71278 ≈ $2,922.95.
Option C (n = 360): (1.004167)−360 ≈ 0.22384; denom = 0.77616; M = 2,083.33 ÷ 0.77616 ≈ $2,684.11.
[3 marks]
(b) Total repaid and total interest.
Option A: Total = 3,300.46 × 240 ≈ $792,110. Interest ≈ $292,110.
Option B: Total = 2,922.95 × 300 ≈ $876,885. Interest ≈ $376,885.
Option C: Total = 2,684.11 × 360 ≈ $966,280. Interest ≈ $466,280.
[3 marks]
(c) Why M falls but total interest rises with longer terms. The repayment formula has the denominator 1 − (1+r)−n, which grows as n increases but is bounded above by 1; doubling n therefore shrinks M less than proportionally — the curve flattens. So a 30-year loan only pays about 18% less per month than a 20-year loan despite spreading repayment over 50% more time. However, total interest = M × n − P scales nearly with the number of years, because the borrower still owes interest on the slowly-shrinking balance for each extra year. Option C pays $174,000 more in interest than Option A — buying a smaller monthly commitment at a six-figure long-term cost. [2 marks]
Total: 8/8.
Band descriptors for marker.
Band 3: One repayment correct, total interest attempted but inconsistent. Part (c) restates "longer = more interest". ≈ 3-4 marks.
Band 4: All three repayments correct, two of three total-interest values correct, Part (c) notes lower M and higher interest without referencing the formula. ≈ 5-6 marks.
Band 5: All calculations within ±$500, Part (c) identifies the denominator's role in shrinking M. ≈ 6-7 marks.
Band 6: Full calculations to within $5/$200, and Part (c) explicitly references the asymptotic behaviour of 1 − (1+r)−n and the M × n − P identity for total interest. 8/8.