Mathematics Standard • Year 12 • Module 7 • Lesson 1
Simple Interest — Past-Paper Style
Practise HSC Mathematics Standard 2-style writing on simple interest — three short-answer questions and one longer scenario with marking criteria.
1. Short-answer questions
1.1 Calculate the simple interest payable on a $7,200 loan at 6.5% p.a. simple interest for 4 years. 2 marks Band 3
1.2 An investment of $4,500 earns $945 in simple interest after 3.5 years.
(a) Calculate the annual simple-interest rate, as a percentage.
(b) State the total amount in the account at the end of the 3.5 years. 3 marks Band 3-4
1.3 A car loan of $18,000 is taken at 7.5% p.a. simple interest. The borrower makes equal monthly repayments of $450 (this is principal plus interest combined).
(a) Calculate the total interest payable on the loan if it runs for 5 years.
(b) Calculate the total amount the borrower will pay over 5 years if they pay $450 per month for 60 months.
(c) Briefly compare your answers in (a) and (b) and state what the difference tells you about whether the loan is set at exactly 7.5% simple interest. 4 marks Band 4
2. Extended response
2.1 Jaylan has $20,000 to invest for 4 years and is comparing three simple-interest offers from different institutions.
Offer A: 5.4% p.a. simple interest for the full 4 years.
Offer B: 6.2% p.a. simple interest, but a one-off $300 set-up fee deducted from the interest at the end of year 4.
Offer C: A flat $4,800 in total interest payable at the end of year 4, no fees.
(a) Calculate the total interest Jaylan would earn from Offer A after 4 years.
(b) Calculate the net interest Jaylan would earn from Offer B after 4 years (after the $300 fee).
(c) Express Offer C's $4,800 total interest as an equivalent simple-interest rate per annum.
(d) Recommend the best offer for Jaylan and justify your recommendation in one to two sentences. 6 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — correct I for Offer A using I = 20,000 × 0.054 × 4.
Part (b) — 2 marks
• 1 mark — correct gross interest for Offer B using I = 20,000 × 0.062 × 4.
• 1 mark — correct subtraction of the $300 fee to give net interest.
Part (c) — 1 mark
• 1 mark — correct equivalent rate using r = I / (P n) = 4,800 / (20,000 × 4).
Part (d) — 2 marks
• 1 mark — explicit comparison of the three net interest figures (or equivalently the three rates) on the same basis.
• 1 mark — a clear recommendation sentence naming the chosen offer and the reason (e.g. "highest net interest after 4 years").
Your response:
Stuck on (d)? After computing the three net interest amounts, the winner is simply the largest figure — but the justification sentence is essential for the final mark.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Simple interest on $7,200 (2 marks)
Sample response.
I = P r n = 7,200 × 0.065 × 4 = $1,872.00.
Marking notes. 1 mark — correct substitution with r as 0.065 and n as 4. 1 mark — correct final answer with $ sign. A bare "$1,872" with no working scores 1/2; using r = 6.5 (not 0.065) typically gives $187,200 and scores 0/2.
1.2 — Find rate, find total (3 marks)
(a) Sample response. r = I / (P n) = 945 / (4,500 × 3.5) = 945 / 15,750 = 0.06 = 6% p.a.
(b) Sample response. Total = $4,500 + $945 = $5,445.00.
Marking notes. (a) 1 mark — correct rearrangement r = I/(Pn). 1 mark — correct percentage with units. (b) 1 mark — correct total (must be P + I, not just I).
1.3 — Car loan interpretation (4 marks)
(a) Sample response. I = 18,000 × 0.075 × 5 = $6,750.00.
(b) Sample response. Total paid = $450 × 60 = $27,000.00.
(c) Sample response. If interest were exactly $6,750 then total = $18,000 + $6,750 = $24,750 — but the borrower actually pays $27,000, which is $2,250 more. So the actual rate the lender is charging is higher than 7.5% p.a. simple interest (the extra $2,250 implies an effective rate closer to 10% p.a. simple interest).
Marking notes. (a) 1 mark — correct I. (b) 1 mark — correct total = M × n. (c) 2 marks — 1 for noting the gap ($2,250 more paid than the stated rate would imply), 1 for the conclusion that the actual rate is higher than 7.5%.
2.1 — Comparing three simple-interest offers (6 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Offer A interest.
I_A = 20,000 × 0.054 × 4 = $4,320.00. [1 mark — correct simple interest on Offer A.]
(b) Offer B net interest.
Gross interest = 20,000 × 0.062 × 4 = $4,960.00. [1 mark — correct gross interest.]
Net interest = $4,960 − $300 = $4,660.00. [1 mark — correctly subtracts the $300 fee.]
(c) Offer C equivalent rate.
r = I / (P n) = 4,800 / (20,000 × 4) = 4,800 / 80,000 = 0.06 = 6.0% p.a. [1 mark — correct equivalent rate.]
(d) Comparison and recommendation.
Net interest: A = $4,320, B = $4,660, C = $4,800. [1 mark — all three figures compared on the same "net interest after 4 years" basis.]
Conclusion: Jaylan should take Offer C, which yields the highest net interest of $4,800 over 4 years — $140 more than Offer B and $480 more than Offer A, with no fees. [1 mark — clear recommendation sentence naming the chosen offer and the dollar reason.]
Total: 6/6.
Band descriptors for marker.
Band 3: Calculates Offer A correctly, calculates Offer B gross interest but forgets the $300 fee, no equivalent rate for C. ≈ 2-3 marks.
Band 4: Three offers calculated correctly (including the fee on B and equivalent rate on C), but no comparison sentence — answer ends with three separate numbers. ≈ 4 marks.
Band 5: Full numerical solution with comparison of the three figures, but the conclusion is bare numbers without naming the chosen offer or the dollar gap. ≈ 5 marks.
Band 6: Complete: three correct calculations, explicit comparison on the same basis, AND a clear conclusion sentence that names Offer C and the size of the gain. 6/6.