Mathematics Standard • Year 11 • Module 3 • Lesson 14
Interest and Depreciation — Exam Practice
Apply the full Interest and Depreciation toolkit to mixed real-world scenarios — loans, investments, asset purchase decisions and capstone comparisons.
Problem 1 — Two loan offers for the same renovation
Reece needs $15,000 for a kitchen renovation over 4 years.
Lender A: Simple interest at 8.4% per annum.
Lender B: Compound interest at 7.5% per annum compounding annually.
Set up: What are we solving for?
(i) Calculate the total amount Reece would repay under Lender A. 1 mark
(ii) Calculate the total amount Reece would repay under Lender B. Show (1 + r)4 as a separate line. 2 marks
(iii) State which lender is cheaper, by how much, and explain in one sentence why the lower rate did or did not win. 2 marks
Stuck? Revisit lesson § Investment and Loan Comparison Questions — for a BORROWER, lower total repayment wins.Problem 2 — "Buy a ute or invest the cash?" (small-business decision)
Naomi has $40,000 in cash and is deciding between:
Option A: Buy a ute for $40,000 (depreciates at 20% per annum declining balance).
Option B: Invest the $40,000 at 4.8% p.a. compounding annually.
Set up: What are we solving for?
(i) Calculate the ute's book value after 5 years. 2 marks
(ii) Calculate the investment's value after 5 years. 2 marks
(iii) State the difference between the two final values. Then, in one sentence, explain why this is NOT the only thing Naomi should consider when choosing — name one factor the maths does not capture. 2 marks
Stuck? Revisit lesson § Worked Example 2 — find each independently, then compare. The qualitative factor could be earning use of the ute, fuel costs, business need, etc.Problem 3 — Reverse-engineering the depreciation rate (vehicle resale)
Sam bought a motorbike new for $19,500. After 4 years he is offered $9,250 as a trade-in. The dealer states this is consistent with declining balance depreciation at a constant annual rate.
Set up: What are we solving for?
(i) Set up the equation $9,250 = $19,500 × (1 − r)4. Divide both sides by $19,500 to isolate the bracket. 1 mark
(ii) Take the 4th root and solve for r as a percentage, correct to 2 decimal places. 2 marks
(iii) Using your rate from (ii), predict the trade-in value the dealer would offer Sam at the end of year 7. 2 marks
Stuck? Revisit lesson § Worked Example 3 — keep the unrounded (1 − r) on the calculator when projecting forward to year 7.Problem 4 — Comparing two compound interest products
Yuki has $12,000 to invest for 4 years.
Product A: 5.1% per annum compounding annually.
Product B: 4.9% per annum compounding monthly.
Set up: What are we solving for?
(i) Calculate the final amount under Product A. 1 mark
(ii) Calculate the final amount under Product B. State the adjusted r and n explicitly. 2 marks
(iii) State which product gives more and by how much. In one sentence, explain why the more-frequent compounding option did NOT overtake the higher-rate option in this case. 2 marks
Stuck? Revisit lesson § Worked Example 4 — Two Compound Investment Products.Problem 5 — Capstone: combined loan + depreciating asset
A café owner borrows $24,000 from the bank at 6.6% per annum compounding annually to buy a new espresso machine that costs $24,000. She agrees to repay the loan as a lump sum after 4 years. The espresso machine depreciates at 15% per annum declining balance.
Set up: What are we solving for?
(i) Calculate the lump sum the café owner must repay the bank after 4 years. 2 marks
(ii) Calculate the book value of the espresso machine after 4 years. 2 marks
(iii) The café owner's "net asset effect" of this borrow-and-buy decision = (book value of machine at year 4) − (lump sum repaid). Calculate this net effect and state whether it is a gain or a loss. State by how much. 3 marks
Stuck? Revisit lesson § A Reliable Sequence for Mixed Financial Mathematics Problems — calculate each side fully before comparing.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Reece's renovation loan
Set up. Calculate total repaid under both lenders; pick the smaller.
(i) Lender A: I = 15,000 × 0.084 × 4 = $5,040.00; total = $15,000 + $5,040.00 = $20,040.00.
(ii) Lender B: (1.075)4 = 1.335469…; A = 15,000 × 1.335469 = $20,032.03.
(iii) Difference = $20,040.00 − $20,032.03 = $7.97. Lender B (compound) is cheaper by $7.97 over 4 years. The lower nominal rate (7.5% vs 8.4%) overcomes the effect of compounding because the rate gap is large enough for 4 years of compounding to leave Lender B slightly ahead — but barely.
Problem 2 — Naomi: ute vs invest $40,000 over 5 years
Set up. Find the year-5 value under each option, then subtract.
(i) Ute (20% DB): (0.80)5 = 0.32768; S = 40,000 × 0.32768 = $13,107.20.
(ii) Investment (4.8% CI annual): (1.048)5 = 1.264118…; A = 40,000 × 1.264118 = $50,564.73.
(iii) Difference = $50,564.73 − $13,107.20 = $37,457.53. The investment ends $37,457.53 ahead of the ute's book value, but the maths ignores qualitative business factors — e.g. Naomi may NEED the ute to operate her business and earn revenue, which is not captured by the depreciation calculation alone.
Problem 3 — Sam's motorbike: find r and project to year 7
Set up. Rearrange S = V₀(1 − r)n, find r, then project forward.
(i) (1 − r)4 = 9,250 ÷ 19,500 = 0.474359 (to 6 d.p.).
(ii) 1 − r = (0.474359)0.25 ≈ 0.829800. r ≈ 1 − 0.829800 = 0.170200 = 17.02% per annum.
(iii) Year-7 trade-in: keep (1 − r) = 0.829800 (full precision). S = 19,500 × (0.829800)7 = 19,500 × 0.271193 ≈ $5,288.27.
Problem 4 — Yuki: Product A vs Product B on $12,000, 4 years
Set up. Calculate both final amounts; pick the larger and explain.
(i) Product A: (1.051)4 = 1.220143…; A = 12,000 × 1.220143 = $14,641.72.
(ii) Product B: r = 0.049 ÷ 12 = 0.004083333…; n = 48. (1.004083333)48 = 1.216042…; A = 12,000 × 1.216042 = $14,592.50.
(iii) Product A − Product B = $14,641.72 − $14,592.50 = $49.22. Product A is larger by $49.22. The boost from monthly compounding under Product B (effective rate ≈ 5.01%) was not enough to close the 0.2 percentage-point gap to Product A's nominal 5.1% annual rate.
Problem 5 — Café owner: bank loan + machine depreciation, 4 years
Set up. Calculate the lump-sum loan repayment (CI annual) and the machine's book value (DB), then compare.
(i) Lump sum: (1.066)4 = 1.290881…; A = 24,000 × 1.290881 = $30,981.13 owed to the bank.
(ii) Machine: (0.85)4 = 0.522006; S = 24,000 × 0.522006 = $12,528.15.
(iii) Net asset effect = $12,528.15 − $30,981.13 = −$18,452.98. This is a net LOSS of $18,452.98 — the loan has grown to almost $31k while the machine has lost more than half its book value. (In reality the café would also have earned trading revenue from the machine, but on the pure borrow-and-buy maths alone, the position is significantly negative.)