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.)