Mathematics Standard • Year 12 • Module 7 • Lesson 7
Loans and Amortisation — Past-Paper Style
Practise HSC Mathematics Standard 2-style writing on loans and amortisation — three short-answer questions and one longer scenario with marking criteria.
1. Short-answer questions
1.1 Calculate the monthly repayment on a $24,000 personal loan at 7.2% p.a. compounded monthly over 5 years. 2 marks Band 3
1.2 A $180,000 home loan at 5.4% p.a. compounded monthly has a monthly repayment of $1,098.41 over 25 years.
(a) Calculate the total amount repaid and the total interest paid over the full term.
(b) For Month 1, calculate the interest, principal and closing balance. 3 marks Band 3-4
1.3 A $250,000 mortgage at 5.4% p.a. compounded monthly over 20 years has a monthly repayment of $1,707.65 (total interest $159,836). If the borrower instead pays $1,907.65 per month (an extra $200), the loan is projected to clear in 188 months.
(a) Calculate the total amount repaid under the extra-repayment plan.
(b) Calculate the total interest saved by paying the extra $200 per month.
(c) In one sentence, explain why the saving is so large relative to the extra $200 paid per month. 4 marks Band 4
2. Extended response
2.1 Priya is comparing three options for a $400,000 mortgage at 4.8% p.a. compounded monthly.
Option A: 30-year term, monthly repayment $2,098.64 (total interest $355,510).
Option B: 25-year term, monthly repayment $2,294.83 (total interest $288,449).
Option C: 30-year term but Priya pays $2,300 per month from day one; the loan is projected to clear in 256 months.
(a) Calculate the total amount repaid under Option A.
(b) Calculate the total amount repaid under Option B and confirm the total interest of $288,449.
(c) Calculate the total amount repaid and total interest under Option C.
(d) Recommend the best option for Priya, naming the option, the interest saved compared to Option A, and one trade-off she should consider. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — correct total = M × n for Option A.
Part (b) — 2 marks
• 1 mark — correct total = M × n for Option B.
• 1 mark — correct total interest = total − PV.
Part (c) — 2 marks
• 1 mark — correct total = $2,300 × 256.
• 1 mark — correct total interest.
Part (d) — 2 marks
• 1 mark — recommendation sentence naming the chosen option AND the dollar interest saved compared to Option A.
• 1 mark — names one realistic trade-off (e.g. higher monthly repayment, no flexibility to drop back, foregone use of the extra cash).
Your response:
Stuck on (d)? Identify the smallest total-interest figure, name that option, quote the dollar gain over Option A, and add a single realistic trade-off sentence.How did this worksheet feel?
What I'll revisit before next class:
1.1 — $24,000 personal loan at 7.2% over 5 years (2 marks)
Sample response. r = 0.072 / 12 = 0.006, n = 60. M = 24,000 × 0.006 / [1 − (1.006)⁻⁶⁰] = 144 / 0.30166 = $477.39/month.
Marking notes. 1 mark — correct r and n (0.006 and 60). 1 mark — correct M with units. A bare "$477" without working scores 1/2; using r = 0.072 (not 0.006) typically scores 0/2.
1.2 — Total cost + Month 1 split (3 marks)
(a) Sample response. Total = 1,098.41 × 300 = $329,523. Interest = $329,523 − $180,000 = $149,523.
(b) Sample response. r = 0.0045. I₁ = 180,000 × 0.0045 = $810.00. P₁ = 1,098.41 − 810.00 = $288.41. B₁ = 180,000 − 288.41 = $179,711.59.
Marking notes. (a) 1 mark — correct total and interest. (b) 1 mark — correct I₁ using monthly r. 1 mark — correct P₁ and B₁ (must follow from M and I₁).
1.3 — Extra-repayment saving (4 marks)
(a) Sample response. Total = 1,907.65 × 188 = $358,638 (to nearest dollar).
(b) Sample response. Interest under extra plan = 358,638 − 250,000 = $108,638. Interest saved = $159,836 − $108,638 = $51,198.
(c) Sample response. The extra $200 each month reduces the balance immediately, so all of the following months' interest is charged on a smaller principal — those small reductions compound over 188 months into a saving more than 25 times the extra paid.
Marking notes. (a) 1 mark — correct total = M × n. (b) 1 mark — correct interest under new plan, 1 mark — correct saving compared to the original $159,836. (c) 1 mark — explanation links "extra reduces balance → less interest each month".
2.1 — Choosing the best $400,000 mortgage option (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Option A — 30-year term.
Total = 2,098.64 × 360 = $755,510. [1 mark — correct total = M × n.]
(b) Option B — 25-year term.
Total = 2,294.83 × 300 = $688,449. [1 mark — correct total.]
Interest = 688,449 − 400,000 = $288,449 ✓ (matches given). [1 mark — correct total interest = total − PV.]
(c) Option C — extra repayments on the 30-year loan.
Total = 2,300 × 256 = $588,800. [1 mark — correct total = $2,300 × 256.]
Interest = 588,800 − 400,000 = $188,800. [1 mark — correct total interest.]
(d) Recommendation.
Total interest summary: A = $355,510, B = $288,449, C = $188,800. Option C is by far the lowest.
Conclusion: Priya should take Option C — by paying $2,300 per month against the 30-year loan she saves $355,510 − $188,800 = $166,710 in interest compared to Option A. [1 mark — recommendation names the option and states the dollar saving vs Option A.] The trade-off is the higher fixed monthly outlay (about $201 more than Option A) which reduces her cashflow flexibility — if her income dropped, she would still be locked in to a higher real burden than under Option A's scheduled $2,098.64. [1 mark — one realistic trade-off named.]
Total: 7/7.
Band descriptors for marker.
Band 3: Computes totals for Options A and B correctly, omits Option C or computes only M × n without comparing interest. ≈ 3 marks.
Band 4: Correct numerical solution for all three options but no comparison sentence or no trade-off mentioned. ≈ 5 marks.
Band 5: Full numerical solution with comparison sentence and named option, but no trade-off discussion. ≈ 6 marks.
Band 6: Complete: three correct totals/interest figures, recommendation names the option and quotes the dollar saving vs Option A, AND a realistic trade-off sentence. 7/7.