Mathematics Standard • Year 11 • Module 3 • Lesson 14
Interest and Depreciation — Exam Practice
Practise HSC Mathematics Standard 2-style writing across the full Interest and Depreciation toolkit — short answers and a structured capstone extended response with explicit marking criteria.
1. Short-answer questions
1.1 $25,000 is invested for 5 years.
(a) Calculate the interest earned at 6% per annum simple interest.
(b) Calculate the interest earned at 5.6% per annum compounding annually.
(c) State which option earns more interest and by how much. 4 marks Band 3-4
1.2 A machine costs $60,000 and depreciates by $5,500 per year using straight-line depreciation. After how many complete years does its value first fall below $20,000? Justify your answer with a check at the year before and the year of the answer. 3 marks Band 3-4
1.3 A boat was purchased for $54,000 and is worth $29,160 after 4 years under declining balance depreciation. Find the annual depreciation rate, correct to 2 decimal places, and verify your answer numerically. 3 marks Band 4
Stuck on 1.3? Rearrange S = V₀(1 − r)n; divide both sides by V₀, take the 4th root (power 0.25), subtract from 1. Verify by computing $54,000 × (1 − r)4 ≈ $29,160.2. Extended response — capstone
2.1 A small business has $50,000 in capital and is choosing between two strategies for the next 5 years.
Strategy 1 — Buy equipment. Spend the full $50,000 on a piece of equipment that depreciates at 16% per annum declining balance.
Strategy 2 — Invest in a term deposit. Invest the full $50,000 at 5.4% per annum compounding semi-annually.
(a) Calculate the book value of the equipment at the end of year 5 under Strategy 1.
(b) Calculate the term deposit's value at the end of year 5 under Strategy 2. (Show adjusted r and n explicitly.)
(c) State the difference between the year-5 values of the two strategies. Identify which strategy ends in a stronger numerical position, by how much, and explain in one sentence why a real-world business might still prefer the other strategy. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct multiplier (0.84) and (0.84)5 shown.
• 1 mark — correct book value with units.
Part (b) — 2 marks
• 1 mark — correctly adjusts r = 0.054 ÷ 2 = 0.027 and n = 5 × 2 = 10.
• 1 mark — correct final amount with (1.027)10 shown.
Part (c) — 3 marks
• 1 mark — correct numerical difference.
• 1 mark — explicit conclusion sentence naming the stronger numerical strategy and the dollar amount.
• 1 mark — sensible qualitative reason (e.g. the equipment generates revenue, or is necessary for trading) that the bare maths does not capture.
Your response:
Stuck on (c)? After subtracting, name the higher strategy with the dollar amount. Then add one sentence — "But a business might still prefer Strategy 1 because…" naming a real reason (revenue, operational need, etc.).How did this worksheet feel?
What I'll revisit before next class:
1.1 — Simple vs compound on $25,000 for 5 years (4 marks)
(a) Sample response. ISI = 25,000 × 0.06 × 5 = $7,500.00.
(b) Sample response. (1.056)5 = 1.313174…; A = 25,000 × 1.313174 = $32,829.36; ICI = $32,829.36 − $25,000 = $7,829.36.
(c) Sample response. Compound interest earns more by $7,829.36 − $7,500.00 = $329.36 over 5 years, despite the lower nominal rate.
Marking notes. (a) 1 mark — correct SI calculation. (b) 1 mark — correct A; 1 mark — correct I via A − P. (c) 1 mark — correct difference with a conclusion sentence naming the winning option. A bare "$329.36" without naming compound interest loses the conclusion mark.
1.2 — Machine $60,000, $5,500/yr SL, when S < $20,000? (3 marks)
Sample response.
60,000 − 5,500n < 20,000
5,500n > 40,000
n > 40,000 ÷ 5,500 = 7.272…
So the smallest whole number of complete years is n = 8 years.
Check year 7: S = 60,000 − 38,500 = $21,500 (still ≥ $20,000). ✓
Check year 8: S = 60,000 − 44,000 = $16,000 (first time < $20,000). ✓
Marking notes. 1 mark — sets up and solves the inequality (or equivalent). 1 mark — rounds UP correctly to 8. 1 mark — verifies with both year-7 and year-8 check values. Stopping at "n > 7.27, so n = 7" loses 2 marks (wrong rounding direction).
1.3 — Boat $54,000 → $29,160 after 4 years DB; find r (3 marks)
Sample response.
(1 − r)4 = 29,160 ÷ 54,000 = 0.54.
1 − r = (0.54)0.25 ≈ 0.857310.
r ≈ 1 − 0.857310 = 0.142690 = 14.27% per annum (to 2 d.p.).
Verify: 54,000 × (0.857310)4 = 54,000 × 0.540000 = $29,160. ✓
Marking notes. 1 mark — correctly divides S by V₀. 1 mark — correctly takes the 4th root (power 1/4 = 0.25) and computes 1 − (S/V₀)1/4. 1 mark — correct r in percentage form to 2 d.p. AND verification step. Common error: rounding 0.54 → 0.5 before taking the root, which gives r ≈ 16% (wrong).
2.1 — Equipment vs term deposit on $50,000 for 5 years (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Strategy 1 — equipment at 16% DB for 5 years.
(1 − 0.16) = 0.84; (0.84)5 = 0.418212. [1 mark — multiplier and (0.84)5 shown.]
S = 50,000 × 0.418212 = $20,910.59. [1 mark — book value with units.]
(b) Strategy 2 — term deposit at 5.4% p.a. semi-annual for 5 years.
r = 0.054 ÷ 2 = 0.027 per half-year; n = 5 × 2 = 10 half-years. [1 mark — adjusted r and n.]
(1.027)10 = 1.305453…
A = 50,000 × 1.305453 = $65,272.66. [1 mark — final amount with (1.027)10 shown.]
(c) Compare and conclude.
Difference = $65,272.66 − $20,910.59 = $44,362.07. [1 mark — correct numerical difference.]
Conclusion: Strategy 2 (term deposit) ends in a stronger numerical position, by $44,362.07 after 5 years. [1 mark — explicit conclusion with dollar amount.]
However, a business might still rationally prefer Strategy 1 because the equipment generates revenue during those 5 years (e.g. the equipment is needed to make and sell the business's products), and the depreciation calculation captures only the falling book value, not the income the equipment produces. [1 mark — sensible qualitative reason the maths does not capture.]
Total: 7/7.
Band descriptors for marker.
Band 3: One side calculated correctly (usually the depreciation). Cannot adjust r and n for semi-annual compounding, or chooses the wrong formula for the investment. ≈ 2-3 marks.
Band 4: Both sides calculated correctly with (1 ± r)n shown; numerical difference correct; concluding sentence missing OR no qualitative reason given for (c). ≈ 4-5 marks.
Band 5: Full numerical work, correct adjustment of r and n, explicit conclusion sentence, but qualitative reason is generic ("it's a business decision") rather than naming something the maths actually misses. ≈ 6 marks.
Band 6: Complete, correct, conclusion sentence with units, AND a specific qualitative reason (revenue generation, operational necessity, customer-facing value, etc.) that the depreciation/compound interest maths does not capture. 7/7.