Mathematics Standard • Year 11 • Module 3 • Lesson 13
Depreciation
Practise HSC Mathematics Standard 2-style writing on depreciation — short answers and a structured extended response with explicit marking criteria.
1. Short-answer questions
1.1 A laser cutter is purchased for $46,000 and depreciates at 16% per annum declining balance. Calculate its value at the end of 4 years and the total depreciation over the period. 3 marks Band 3
1.2 A delivery truck is bought for $78,000. The owner uses straight-line depreciation with D = $7,200 per year.
(a) Calculate the truck's value after 7 years.
(b) After how many complete years does the value first fall below $35,000? 3 marks Band 3-4
1.3 A boat was purchased for $74,000 and is worth $39,000 after 5 years under declining balance depreciation.
(a) Find the annual depreciation rate r, correct to 2 decimal places.
(b) Using your rate from (a), calculate the boat's projected value after 8 years (one calculation, not by re-computing each year). 4 marks Band 4
2. Extended response
2.1 A small construction firm buys a new excavator for $185,000. The firm's accountant proposes two depreciation methods to track the asset's book value over the first 6 years.
Method X: Straight-line, $19,500 per year.
Method Y: Declining balance, 14% per annum.
(a) Calculate the excavator's book value after 6 years under Method X.
(b) Calculate the excavator's book value after 6 years under Method Y. (Show (1 − r)6.)
(c) State which method gives the higher book value at the end of year 6 and by how much. Comment on which method represents a more conservative estimate of the asset's value in the EARLY years (years 1-2) and justify your answer with one example calculation. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — correct book value using S = V₀ − Dn.
Part (b) — 2 marks
• 1 mark — correct multiplier (0.86) and (0.86)6 shown on its own line.
• 1 mark — correct book value with units.
Part (c) — 4 marks
• 1 mark — correct numerical comparison at year 6 (difference).
• 1 mark — explicit conclusion sentence naming which method is higher and by how much.
• 1 mark — correctly identifies which method is more conservative in the early years (depreciates the asset faster).
• 1 mark — supports the claim with an example calculation at year 1 or year 2 under both methods.
Your response:
Stuck on the "conservative in early years" part? Calculate year-1 book values under both methods. The lower year-1 value = more conservative (asset already worth less on the books).How did this worksheet feel?
What I'll revisit before next class:
1.1 — Laser cutter $46,000 at 16% DB for 4 years (3 marks)
Sample response.
(1 − 0.16) = 0.84.
(0.84)4 = 0.497871…
S = $46,000 × 0.497871 = $22,902.06.
Total depreciation = $46,000 − $22,902.06 = $23,097.94.
Marking notes. 1 mark — correct multiplier and (0.84)4 shown. 1 mark — correct S with units. 1 mark — correct total depreciation using V₀ − S. Common error: multiplying $46,000 × 0.16 × 4 = $29,440 — this misapplies the DB rate as if it were straight-line.
1.2 — Delivery truck $78,000, SL $7,200/yr (3 marks)
(a) Sample response. S = 78,000 − (7,200 × 7) = 78,000 − 50,400 = $27,600.00.
(b) Sample response. Set 78,000 − 7,200n < 35,000 ⇒ 7,200n > 43,000 ⇒ n > 5.972…⇒ smallest n = 6 complete years. Check year 5: $42,000 (still ≥ $35,000); year 6: $34,800 (first < $35,000). ✓
Marking notes. (a) 1 mark for correct S with units. (b) 1 mark for setting up and solving the inequality; 1 mark for rounding UP correctly and verifying with check values.
1.3 — Boat $74,000 → $39,000 after 5 years DB (4 marks)
(a) Sample response.
(1 − r)5 = 39,000 ÷ 74,000 = 0.527027…
1 − r = (0.527027)1/5 = (0.527027)0.2 ≈ 0.878156.
r ≈ 1 − 0.878156 = 0.121844 = 12.18% per annum (to 2 d.p.).
(b) Sample response. Using full precision for (1 − r) = 0.878156…:
S = 74,000 × (0.878156)8 = 74,000 × 0.357025… ≈ $26,419.83 (to 2 d.p.).
Marking notes. (a) 1 mark — correctly divides S by V₀; 1 mark — correctly takes the 5th root and finds r with units (%). (b) 1 mark — uses (1 − r)8 in one calculation (not eight); 1 mark — correct value with units. Common error in (a): rounding 0.527027 to 0.53 before the 5th root introduces noticeable error.
2.1 — Excavator $185,000 over 6 years (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Method X — straight-line, $19,500 per year.
SX = 185,000 − (19,500 × 6) = 185,000 − 117,000 = $68,000.00. [1 mark — Method X correct.]
(b) Method Y — declining balance at 14%.
(1 − 0.14) = 0.86; (0.86)6 = 0.404567… [1 mark — multiplier and (0.86)6 shown.]
SY = 185,000 × 0.404567 = $74,844.94. [1 mark — Method Y correct with units.]
(c) Comparison and conservatism.
Difference at year 6: $74,844.94 − $68,000.00 = $6,844.94. [1 mark — correct comparison.]
Conclusion: Method Y (declining balance at 14%) gives the higher book value at year 6, by $6,844.94. [1 mark — explicit conclusion.]
In the EARLY years, Method Y is the more conservative method — it depreciates the asset faster initially because 14% of $185,000 is a larger dollar amount than $19,500. [1 mark — identifies Y as more conservative early.]
Year 1 example: under Method X, S = 185,000 − 19,500 = $165,500. Under Method Y, S = 185,000 × 0.86 = $159,100. Method Y's year-1 book value ($159,100) is $6,400 lower than Method X's ($165,500), confirming Y is more conservative in early years. [1 mark — example calculation supports the claim.]
Total: 7/7.
Band descriptors for marker.
Band 3: Method X correct; cannot apply declining balance correctly (e.g., uses (1.14) instead of (0.86), or multiplies by 0.14 instead of by (1 − r)n). ≈ 2-3 marks.
Band 4: Both methods calculated correctly with (1 − r)n shown; conclusion sentence present but either the "conservative early" claim is missing/wrong, or no supporting example given. ≈ 4-5 marks.
Band 5: Full calculations, correct comparison, correct identification of conservatism, but the supporting example is computed incorrectly or compares year 6 not early years. ≈ 6 marks.
Band 6: Complete, correct, conclusion sentence with units, and an EARLY-year example showing Method Y's lower year-1 value supports the conservatism claim. 7/7.