Mathematics • Year 10 • Unit 1 • Lesson 7
Depreciation — Mixed Challenge
Pull together both methods (straight-line and reducing balance), connect them back to Lesson 6's compound interest (note the structural similarity!), choose the right tool, spot a classic mistake, and tackle an open-ended challenge.
1. Mixed problems — choose the right method
Each problem requires you to choose between straight-line and reducing balance, sometimes both. Decide before you start writing. 3 marks each
1.1 A photocopier costs $14,000 and has a scrap value of $2,000 after 6 years (straight-line). Find (a) annual depreciation, (b) book value after 4 years.
1.2 A laptop costs $1,800 and depreciates at 22% p.a. reducing balance. Find (a) book value after 4 years, (b) total depreciation in dollars.
1.3 A car bought for $40,000 depreciates over 6 years. Compare: straight-line with scrap $8,000 vs reducing balance at 22% p.a. State the book value after 4 years for each, and the difference.
1.4 A coffee machine bought for $9,500 has a book value of $4,275 after 3 years using reducing balance. Find the annual depreciation rate r, to the nearest whole percent.
1.5 Connection back to Lesson 6. The reducing-balance formula Book Value = P(1 − r)ⁿ looks almost identical to compound interest A = P(1 + R)ⁿ. Explain (a) what each part of P(1 − r)ⁿ represents, (b) why we use a minus sign here instead of a plus, and (c) what would happen if you accidentally used (1 + r)ⁿ on a depreciation question.
1.6 A printer cost $4,000 and is depreciated using reducing balance at 25% p.a. After how many whole years will the book value first drop below $1,000?
2. Find the mistake
A student is asked: "A car costs $25,000 and depreciates at 16% p.a. reducing balance. Find the book value after 3 years." Their working is below. Exactly one line contains a mistake. Spot it, explain why, then re-do correctly. 3 marks
Student's working — car $25,000, 16% p.a. reducing balance, 3 years:
Line 1: P = 25,000, r = 16% = 0.16, n = 3
Line 2: Book Value = P × (1 + r)ⁿ ← formula
Line 3: Book Value = 25,000 × (1.16)³
Line 4: (1.16)³ = 1.560896
Line 5: Book Value = 25,000 × 1.560896 = $39,022.40
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong, and how you should have spotted it.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? Depreciation makes things cheaper, not more expensive. If the answer is bigger than the original price, something is wrong with the formula.3. Open-ended challenge — design two depreciation schedules
This question has more than one valid answer — there are several setups that work. 4 marks
3.1 An office buys a $10,000 asset and wants the book value to be approximately $3,000 after 5 years.
Design two different depreciation schedules that achieve this (one straight-line, one reducing balance). For each:
(i) State the method.
(ii) State the parameter you needed to find (the scrap value for straight-line, or the rate r for reducing balance).
(iii) Show the calculation that confirms it gives approximately $3,000 after 5 years.
Bonus: For your reducing-balance schedule, what is the book value after 1 year and after 10 years?
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Photocopier $14,000 → $2,000 over 6 years
(a) Annual depreciation = (14,000 − 2,000) ÷ 6 = 12,000 ÷ 6 = $2,000/year.
(b) After 4 years = 14,000 − (2,000 × 4) = 14,000 − 8,000 = $6,000.
1.2 — Laptop $1,800 at 22% reducing balance for 4 years
(a) Book Value = 1,800 × (0.78)⁴ = 1,800 × 0.370146 = $666.26.
(b) Total depreciation = 1,800 − 666.26 = $1,133.74.
1.3 — Car $40,000 over 6 years: compare
Straight-line: annual = (40,000 − 8,000) ÷ 6 = $5,333.33. After 4 years: 40,000 − 21,333.33 = $18,666.67.
Reducing balance 22%: 40,000 × (0.78)⁴ = 40,000 × 0.370146 = $14,805.85.
Reducing balance is lower by 18,666.67 − 14,805.85 = $3,860.82 after 4 years.
1.4 — Coffee machine: find r
9,500 × (1 − r)³ = 4,275 → (1 − r)³ = 4,275 / 9,500 = 0.45.
1 − r = ∛0.45 ≈ 0.7663, so r ≈ 1 − 0.7663 = 0.2337 ≈ 23% p.a. (to the nearest whole %).
1.5 — Connection back to Lesson 6
(a) P = original value; r = decimal rate of depreciation per period; n = number of periods. P(1 − r)ⁿ is the book value remaining.
(b) We use a minus because depreciation subtracts a percentage of the current value each period (compound interest adds a percentage).
(c) Using (1 + r)ⁿ on a depreciation question would compute compound growth — the asset would appear to gain value over time, which is the opposite of what depreciation models. Always check whether the answer is bigger or smaller than P, and use the formula that matches.
1.6 — Printer $4,000 at 25%, below $1,000
n = 4: 4,000 × (0.75)⁴ = 4,000 × 0.316406 = $1,265.63 — over.
n = 5: 4,000 × (0.75)⁵ = 4,000 × 0.237305 = $949.22 — under!
Answer: 5 years.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The student wrote the compound interest formula by accident: (1 + r)ⁿ adds growth, but depreciation requires (1 − r)ⁿ. You should have spotted it by checking the final answer: the book value came out higher than the original purchase price, which is impossible — depreciation only ever subtracts.
(c) Corrected working:
Book Value = P × (1 − r)ⁿ = 25,000 × (0.84)³
(0.84)³ = 0.592704
Book Value = 25,000 × 0.592704 = $14,817.60.
This is precisely the cross-up flagged in question 1.5 — compound interest uses +; depreciation uses −.
3 — Open-ended challenge (sample solution)
Schedule 1 — Straight-line. Set the scrap value to $3,000 and the useful life to 5 years.
Annual depreciation = (10,000 − 3,000) ÷ 5 = $1,400/year.
After 5 years: 10,000 − (1,400 × 5) = 10,000 − 7,000 = $3,000 ✓.
Schedule 2 — Reducing balance. Solve 10,000 × (1 − r)⁵ = 3,000.
(1 − r)⁵ = 0.30 → 1 − r = ⁵√0.30 ≈ 0.7860 → r ≈ 21.4% p.a. (or about 21%).
Check: 10,000 × (0.7860)⁵ = 10,000 × 0.3001 ≈ $3,001 ✓.
Bonus. For the reducing-balance schedule with r = 21.4%:
After 1 year: 10,000 × 0.786 = $7,860.
After 10 years: 10,000 × (0.786)¹⁰ = 10,000 × 0.0901 = $901. (Reducing balance never reaches zero but keeps shrinking.)
Marking: 2 marks per schedule (1 for the parameter, 1 for the verification). Bonus is recognition. Accept any reasonable rate (≈ 21–22%).