Mathematics Standard • Year 12 • Module 7 • Lesson 1
Simple Interest — Problem Set
Apply simple-interest reasoning to realistic Australian financial scenarios — loans, savings, "buy now pay later" traps and comparative deals.
Problem 1 — Used-car personal loan
Ben borrows $14,000 from a community lender to buy a used Toyota Corolla. The lender charges 8.5% per annum simple interest, and the loan is to be repaid in equal monthly instalments over 3 years.
Set up: What are we solving for?
(i) Calculate the total simple interest payable over the 3 years. 2 marks
(ii) Calculate the total amount Ben must repay (principal + interest). 1 mark
(iii) Calculate Ben's equal monthly instalment. 2 marks
Stuck? Revisit lesson § Simple Interest — total = P + I, then divide by the number of months.Problem 2 — Comparing two term-deposit offers
Aisha has $10,000 to invest for exactly 2 years.
Deal A: A bank pays 4.2% p.a. simple interest, paid at maturity.
Deal B: A credit union pays $850 in total interest over the 2 years (flat dollar amount).
Set up: What are we solving for?
(i) Calculate the interest payable on Deal A after 2 years. 1 mark
(ii) Express Deal B's interest as an equivalent simple-interest rate (per annum). 2 marks
(iii) Recommend which deal Aisha should take and explain in one sentence. 2 marks
Stuck? Revisit lesson § Rearranging — solve for r when I, P and n are known.Problem 3 — "Buy now, pay later" trap
An electronics store offers a $3,200 laptop on the following terms: "0% interest for 12 months, then 18% p.a. simple interest on the unpaid balance." Mia pays nothing for the first 12 months. At the end of month 12 she has still not paid anything, and the unpaid balance is the full $3,200.
Set up: What are we solving for?
(i) Calculate the interest Mia accrues in the next 12 months if she still has the full $3,200 outstanding. 2 marks
(ii) Calculate the total amount Mia would have to pay after 24 months from the original purchase (laptop + interest accrued in year 2). 1 mark
(iii) Explain in one to two sentences why "0% interest for 12 months" is not the same as "interest-free" for a customer who cannot repay during the first year. 2 marks
Stuck? Revisit lesson § Activity 2 — the "buy now, pay later" trap.Problem 4 — Reverse-engineering an investment
A community charity received a generous donation that earned exactly $1,890 in simple interest over 3 years at a fixed rate. The treasurer reports that the rate paid was 4.5% p.a. simple interest.
Set up: What are we solving for?
(i) Calculate the principal (the amount originally donated). 2 marks
(ii) Calculate the total amount in the charity's account at the end of year 3. 1 mark
(iii) If the charity had instead received 5% p.a. simple interest for the same 3 years, how much more interest would they have earned? 2 marks
Stuck? Revisit lesson § Rearranging the Formula — P = I / (rn).Problem 5 — Comparing two simple-interest deals over time
Two investors each have $10,000 to invest, but receive different offers.
Investor X: 5% p.a. simple interest for 10 years.
Investor Y: 10% p.a. simple interest for 5 years.
Set up: What are we solving for?
(i) Calculate the simple interest earned by Investor X. 1 mark
(ii) Calculate the simple interest earned by Investor Y. 1 mark
(iii) Explain mathematically why the totals are equal even though the rates and times differ — write your reasoning in one sentence using the formula I = P r n. 2 marks
Stuck? Revisit lesson § Activity 2 Q3 — same product r × n produces the same simple interest.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Used-car loan
Set up. We need the simple interest on $14,000 at 8.5% for 3 years, then the total repayment, then the equal monthly instalment over 36 months.
(i) I = 14,000 × 0.085 × 3 = $3,570.00.
(ii) Total = $14,000 + $3,570 = $17,570.00.
(iii) Monthly = $17,570 / 36 = $488.06 per month (to the nearest cent).
Problem 2 — Term-deposit comparison
Set up. We need the interest from Deal A, an equivalent annual rate for Deal B, then a recommendation.
(i) Deal A interest = 10,000 × 0.042 × 2 = $840.00.
(ii) Deal B rate: r = I / (P n) = 850 / (10,000 × 2) = 0.0425 = 4.25% p.a.
(iii) Deal B is slightly better (4.25% > 4.2%), earning $10 more total interest over 2 years on the same $10,000.
Problem 3 — BNPL laptop
Set up. Calculate year-2 interest on the unpaid $3,200, then the total owed after 24 months, then comment on the offer.
(i) Year-2 interest = 3,200 × 0.18 × 1 = $576.00.
(ii) Total owed after 24 months = $3,200 + $576 = $3,776.00.
(iii) "0% for 12 months" only means the customer is not charged interest during the first year. The moment the year passes without full repayment, the full balance starts accruing 18% p.a. — Mia has paid $576 more than the laptop's sticker price, even though the offer was advertised as "0% interest".
Problem 4 — Reverse-engineering the donation
Set up. We know I, r and n; solve I = P r n for P, then add to get the total, then redo the calculation at a higher rate.
(i) P = I / (rn) = 1,890 / (0.045 × 3) = 1,890 / 0.135 = $14,000.00.
(ii) Total = $14,000 + $1,890 = $15,890.00.
(iii) At 5%: I = 14,000 × 0.05 × 3 = $2,100. Extra interest = $2,100 − $1,890 = $210.00.
Problem 5 — Identical-product comparison
Set up. Calculate I for each investor, then explain via the structure of I = P r n.
(i) Investor X: I = 10,000 × 0.05 × 10 = $5,000.00.
(ii) Investor Y: I = 10,000 × 0.10 × 5 = $5,000.00.
(iii) Simple interest depends on the product r × n. For X: 0.05 × 10 = 0.50. For Y: 0.10 × 5 = 0.50. Same product, same P, so identical interest. (Important caveat: this would NOT be true for compound interest — Investor X would actually win because of more compounding periods. Compounding is the topic of lesson 2.)