Mathematics Standard • Year 11 • Module 3 • Lesson 13

Depreciation

Build fluency with both depreciation methods — straight-line S = V₀ − Dn and declining balance S = V₀(1 − r)ⁿ — and the language cues that signal which to use.

Build · Skill Drill

1. Quick recall

Answer each part in the space provided. 1 mark each

Q1.1 Write each depreciation formula and name the variables:

Straight-line: S = ____________. V₀ = ______, D = ______, n = ______.

Declining balance: S = ____________. r = ______.

Q1.2 Match each phrase to the correct method (write SL or DB):

"depreciates by $1,800 per year" → ____ "depreciates at 12% per annum" → ____

"same dollar amount each year" → ____ "same percentage each year" → ____

Q1.3 An asset depreciates at 16% per annum declining balance. Write the yearly multiplier (1 − r) = ____________

Stuck? Revisit lesson § Choosing the Correct Depreciation Model — language cues table.

2. Worked example — declining balance depreciation

Follow each line. Every step has a reason on the right.

Problem. A car is purchased new for $38,500 and depreciates at 22% per annum declining balance. Find (a) its value after 3 years, and (b) the total depreciation.

Step 1 — Form the multiplier.

r = 0.22, so (1 − r) = 1 − 0.22 = 0.78

Reason: declining balance multiplies by (1 − r) each year — write this multiplier explicitly to avoid sign errors.

Step 2 — Substitute into S = V₀(1 − r)ⁿ.

S = $38,500 × (0.78)³

(0.78)³ = 0.474552 (calculator)

Reason: showing the value of (1 − r)ⁿ on its own line earns a method mark even if the arithmetic slips.

Step 3 — Multiply by V₀ for part (a).

S = $38,500 × 0.474552 = $18,270.25

Reason: this is the salvage value after 3 years.

Step 4 — Total depreciation = V₀ − S for part (b).

Total depreciation = $38,500 − $18,270.25 = $20,229.75

Reason: do NOT try to multiply the yearly dollar amount by n — the dollar amount changes each year under declining balance. Always use V₀ − S.

Conclusion. (a) S = $18,270.25; (b) total depreciation = $20,229.75.

3. Faded example — fill in the missing steps

A piece of industrial equipment was purchased for $85,000 and depreciates by $7,400 per year using straight-line. Find its value after 6 years. Fill in each blank. 3 marks

Step 1 — Identify the model. "depreciates by $7,400 per year" → ____________ depreciation (SL or DB?)

Step 2 — List V₀, D, n: V₀ = $ __________, D = $ __________, n = ____

Step 3 — Apply S = V₀ − Dn:

S = $85,000 − ($ __________ × ____) = $85,000 − $ __________ = $ __________

Conclusion. Value after 6 years = $ ____________

Stuck? Revisit lesson § Worked Example 1 — Straight-Line Depreciation.

4. Graduated practice — depreciation calculations

Identify the model first, then substitute. Show one line of substitution and clearly label your final answer.

Foundation — single-step substitution (4 questions)

Q	Problem	Answer
4.1 1	A computer worth $2,800 depreciates by $420 per year straight-line. Find its value after 4 years.
4.2 1	An asset is bought for $50,000 with salvage value $18,000 after 8 years straight-line. Find the annual depreciation D.
4.3 1	An asset depreciates at 18% per annum declining balance. State the yearly multiplier (1 − r).
4.4 1	A vehicle worth $45,000 depreciates at 25% per annum declining balance. Find its value after 2 years.

Standard — typical HSC difficulty (6 questions)

Show one line of substitution; for declining balance, show (1 − r)ⁿ as a separate line.

4.5 A printer costing $6,200 depreciates by $540 per year straight-line. Find its value after 5 years. 2 marks

4.6 A boat purchased for $64,000 depreciates at 12% per annum declining balance. Find its value after 3 years. 2 marks

4.7 A machine bought for $24,000 depreciates at 10% per annum declining balance. Find the total depreciation over 4 years. 2 marks

4.8 A van purchased for $42,000 depreciates by $3,800 per year straight-line. After how many complete years does its value first fall below $20,000? 2 marks

4.9 Office furniture is purchased for $16,000. Method A: straight-line at $1,800/year. Method B: declining balance at 14% per annum. Find the value after 5 years under each method. 2 marks

4.10 A laptop worth $2,400 new depreciates at 28% per annum declining balance. Find its salvage value after 3 years, then state the total depreciation. 2 marks

Extension — comparison and unknown rate (2 questions)

4.11 Using 4.9 above, state which method gives the higher salvage value after 5 years, and by how much. Comment on why the answer makes sense given the percentage cue. 3 marks

4.12 A machine was purchased for $28,000 and is worth $15,950 after 4 years under declining balance. Find the annual depreciation rate r, correct to 2 decimal places. 3 marks

Stuck on 4.12? Rearrange S = V₀(1 − r)ⁿ: divide both sides by V₀, take the 4th root (raise to the power 1/4 = 0.25), then subtract from 1.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 (straight-line) I subtracted $420 four times — I did NOT use (1 − 0.42) or any percentage method.

For 4.3 (18% DB), the multiplier is 0.82 (not 0.18, not 0.92) — I subtracted 0.18 from 1, not 1 from 18.

For 4.4 (25% DB, 2 years), I raised 0.75 to the power 2 before multiplying by $45,000.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Formulas and variables

Straight-line: S = V₀ − Dn. V₀ = initial value; D = depreciation per period (dollars); n = number of periods.
Declining balance: S = V₀(1 − r)ⁿ. r = depreciation rate per period (decimal).

Q1.2 — Method match

"$1,800 per year" → SL; "12% per annum" → DB; "same dollar amount" → SL; "same percentage" → DB.

Q1.3 — Multiplier for 16% DB

(1 − r) = 1 − 0.16 = 0.84.

Q3 — Faded example (straight-line, $85,000, $7,400/yr, 6 yrs)

Step 1: SL (straight-line) — the cue is "$7,400 per year".
Step 2: V₀ = $85,000, D = $7,400, n = 6.
Step 3: S = $85,000 − ($7,400 × 6) = $85,000 − $44,400 = $40,600.
Conclusion: Value after 6 years = $40,600.00.

Q4.1 — Computer $2,800, $420/yr SL, 4 years

S = 2,800 − (420 × 4) = 2,800 − 1,680 = $1,120.00.

Q4.2 — D from $50,000 → $18,000 over 8 years SL

D = (50,000 − 18,000) ÷ 8 = 32,000 ÷ 8 = $4,000.00 per year.

Q4.3 — Multiplier for 18% DB

(1 − r) = 1 − 0.18 = 0.82.

Q4.4 — Vehicle $45,000 at 25% DB for 2 years

(0.75)² = 0.5625. S = 45,000 × 0.5625 = $25,312.50.

Q4.5 — Printer $6,200, $540/yr SL, 5 years

S = 6,200 − (540 × 5) = 6,200 − 2,700 = $3,500.00.

Q4.6 — Boat $64,000 at 12% DB for 3 years

(1 − 0.12) = 0.88; (0.88)³ = 0.681472. S = 64,000 × 0.681472 = $43,614.21.

Q4.7 — Machine $24,000 at 10% DB, total dep. over 4 years

(0.90)⁴ = 0.6561. S = 24,000 × 0.6561 = $15,746.40. Total depreciation = 24,000 − 15,746.40 = $8,253.60.

Q4.8 — Van $42,000, $3,800/yr SL, when does value first fall below $20,000?

Set 42,000 − 3,800n < 20,000 ⇒ 3,800n > 22,000 ⇒ n > 5.789… So n = 6 complete years. Check: after 5 yrs, S = 42,000 − 19,000 = $23,000 (still above); after 6 yrs, S = 42,000 − 22,800 = $19,200 (first below $20,000). ✓

Q4.9 — Furniture $16,000 over 5 years, both methods

Method A (SL, $1,800/yr): S = 16,000 − (1,800 × 5) = 16,000 − 9,000 = $7,000.00.
Method B (DB, 14%): (0.86)⁵ = 0.47043…; S = 16,000 × 0.47043 = $7,526.83.

Q4.10 — Laptop $2,400 at 28% DB, 3 years

(0.72)³ = 0.373248. S = 2,400 × 0.373248 = $895.80. Total depreciation = 2,400 − 895.80 = $1,504.20.

Q4.11 — Compare 4.9

Method B ($7,526.83) > Method A ($7,000.00) by $526.83. Comment: declining balance at 14% takes 14% of a shrinking balance each year, so total dollars lost is less than 5 × $1,800 = $9,000 over 5 years. The percentage-on-shrinking-balance model retains more value here.

Q4.12 — Find r when V₀ = $28,000, S = $15,950, n = 4

15,950 = 28,000 × (1 − r)⁴ ⇒ (1 − r)⁴ = 15,950 ÷ 28,000 = 0.569642857…
1 − r = (0.569642857…)^1/4 ≈ 0.868762
r = 1 − 0.868762 ≈ 0.131238 = 13.12% per annum (to 2 d.p.).