Mathematics Advanced • Year 12 • Module 7 • Lesson 4

Depreciation

Build fluency with flat-rate (linear) and reducing-balance (exponential) depreciation formulas and their differences.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the formulas:

Flat rate: S = ____________________________________________

Reducing balance: S = ____________________________________________

Q1.2 Which of the two depreciation methods can produce a negative book value if n is large enough?

Q1.3 In one sentence, describe the relationship between reducing-balance depreciation and compound interest.

Stuck? Revisit lesson § Formula Reference and § Misconceptions.

2. Worked example — find the reducing-balance rate

Follow each line. Each step has a short reason.

Problem. Industrial machinery purchased for $120,000 has a book value of $28,000 after 6 years under reducing-balance depreciation. Find the annual rate r.

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

28,000 = 120,000(1 − r)⁶

Reason: this is the reducing-balance formula with all values plugged in.

Step 2 — Isolate (1 − r)⁶.

(1 − r)⁶ = 28,000 / 120,000 = 0.23333

Step 3 — Take the 6th root.

1 − r = (0.23333)^1/6 = 0.79263

Reason: undo the exponent before solving for r.

Step 4 — Solve for r.

r = 1 − 0.79263 = 0.20737 ≈ 20.74% p.a.

Conclusion. The machinery depreciates at 20.74% per annum on a reducing-balance basis.

3. Faded example — laptop depreciation (flat rate vs reducing balance)

A laptop purchased for $2,400 depreciates at 25% p.a. Fill in each blank. 4 marks

Step 1 — Flat-rate annual depreciation:

D = r × V₀ = ______ × ______ = $______ per year

Step 2 — Flat-rate book value after 3 years:

S = 2,400 − ______ × 3 = $______

Step 3 — Flat-rate book value after 5 years:

S = 2,400 − ______ × 5 = $______ (this is unrealistic because ___________________)

Step 4 — Reducing-balance book value after 5 years:

S = 2,400 × (1 − ______)⁵ = 2,400 × ______ = $______

Conclusion. Flat rate gives a non-physical answer at year 5; reducing balance preserves a positive value of about $______.

Stuck? Revisit lesson § Flat Rate Depreciation and § Reducing Balance.

4. Graduated practice — book values and rates

For each, identify the method, show the substitution, and state the final value.

Foundation — single-step substitution (4 questions)

Q	Scenario	Working & book value
4.1 1	V₀ = $10,000, flat rate 15%, n = 3 years
4.2 1	V₀ = $10,000, reducing balance 15%, n = 3 years
4.3 1	V₀ = $25,000, reducing balance 10%, n = 5 years
4.4 1	V₀ = $8,000, flat rate 20%, n = 4 years

Standard — typical HSC difficulty (6 questions)

4.5 A delivery van is purchased for $45,000 and depreciates at 15% p.a. flat rate. Find its book value after 4 years and after 7 years. Comment on the year-7 figure. 2 marks

4.6 The same $45,000 van depreciates at 15% p.a. reducing balance. Find its book value after 4 years and compare with 4.5. 2 marks

4.7 A car worth $35,000 depreciates at 18% p.a. flat rate. Calculate the value after 4 years and the total depreciation. 2 marks

4.8 Equipment purchased for $65,000 has a book value of $18,000 after 5 years under reducing balance. Find r to 2 dp. 2 marks

4.9 An asset is purchased for $80,000. Compare its book value after 3 years under (a) flat rate 20% p.a. and (b) reducing balance 20% p.a. 2 marks

4.10 An asset depreciates at 12% p.a. reducing balance. After how many full years does it drop below 50% of its original value? Use logarithms. 2 marks

Extension — combine concepts (2 questions)

4.11 An asset of initial value V₀ is depreciated at r p.a. flat rate. Show that it reaches zero book value at n = 1/r years (with r as a decimal). Hence find the rate at which an asset reaches $0 at exactly 5 years. 3 marks

4.12 Reducing-balance depreciation can be written as S = V₀(1 + i)ⁿ with i = −r. Use this to explain why an asset depreciated at 20% reducing balance for 5 years retains 0.8⁵ ≈ 32.77% of its value, while at 20% flat rate it would reach $0 at exactly 5 years. 3 marks

Stuck on 4.11? Set V₀ − rV₀ · n = 0 and solve for n.

5. Self-check the easy 3

Tick once you have checked your method.

For 4.1 I subtracted (rate × V₀ × n), not (rate × current value × n).

For 4.2 I used (1 − r)ⁿ, not (1 + r)ⁿ — the sign matters.

For 4.5 year-7 I obtained a negative book value and noted that real assets stop at $0 salvage.

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

Flat rate: S = V₀ − Dn (where D = rV₀); equivalently S = V₀(1 − rn). Reducing balance: S = V₀(1 − r)ⁿ.

Q1.2 — Which can go negative?

Flat rate. With S = V₀ − Dn, once n exceeds 1/r the value is negative — a mathematical artefact corrected in practice by a salvage cap of $0.

Q1.3 — Reducing balance vs compound interest

Reducing balance is mathematically compound interest with a negative rate: writing S = V₀(1 + (−r))ⁿ shows it is the compound-interest formula where each year's "growth" is a loss of r% of the current value.

Q3 — Faded example: $2,400 laptop at 25% p.a.

D = 0.25 × 2,400 = $600/year. Year 3 (flat): S = 2,400 − 600 × 3 = $600. Year 5 (flat): S = 2,400 − 3,000 = −$600 (unrealistic; real assets can't have negative value). Year 5 (reducing balance): S = 2,400(0.75)⁵ = 2,400 × 0.23730 = $569.53. Reducing balance preserves ≈ $570 even after the flat-rate calculation has gone negative.

Q4.1 — V₀ = 10,000, flat 15%, n = 3

S = 10,000 − 0.15 × 10,000 × 3 = 10,000 − 4,500 = $5,500.

Q4.2 — V₀ = 10,000, RB 15%, n = 3

S = 10,000(0.85)³ = 10,000 × 0.614125 = $6,141.25 — about $641 more than flat rate at the same year.

Q4.3 — V₀ = 25,000, RB 10%, n = 5

S = 25,000(0.90)⁵ = 25,000 × 0.59049 = $14,762.25.

Q4.4 — V₀ = 8,000, flat 20%, n = 4

S = 8,000 − 0.20 × 8,000 × 4 = 8,000 − 6,400 = $1,600.

Q4.5 — Van at 15% flat rate

D = 0.15 × 45,000 = $6,750/year. Year 4: S = 45,000 − 6,750 × 4 = $18,000. Year 7: S = 45,000 − 6,750 × 7 = −$2,250 (impossible — flat-rate without salvage cap). In practice the value drops to $0 between year 6 and year 7.

Q4.6 — Van at 15% reducing balance

S = 45,000(0.85)⁴ = 45,000 × 0.52200625 = $23,490.28 — about $5,490 higher than the flat-rate Q4.5 figure. Reducing balance preserves more value because the annual amount of depreciation shrinks as the book value shrinks.

Q4.7 — $35,000 car at 18% flat rate, n = 4

D = 0.18 × 35,000 = $6,300/year. S = 35,000 − 6,300 × 4 = $9,800. Total depreciation = $25,200.

Q4.8 — Equipment $65,000 → $18,000 in 5 yr (RB)

(1 − r)⁵ = 18,000/65,000 = 0.27692; 1 − r = 0.27692^0.2 = 0.76851; r = 1 − 0.76851 = 0.2315 = 23.15% p.a.

Q4.9 — $80,000 at 20% for 3 yr

(a) Flat: S = 80,000 − 16,000 × 3 = $32,000. (b) RB: S = 80,000(0.80)³ = 80,000 × 0.512 = $40,960. RB retains $8,960 more.

Q4.10 — Time to drop below 50% at 12% RB

Solve (0.88)ⁿ < 0.5 ⇒ n > ln 0.5 / ln 0.88 = −0.6931 / −0.12783 = 5.42 years. The first whole year below 50% is year 6.

Q4.11 — Flat-rate zero at n = 1/r

Set S = 0: V₀ − rV₀ · n = 0 ⇒ V₀(1 − rn) = 0 ⇒ rn = 1 ⇒ n = 1/r. For zero at exactly 5 years: 5 = 1/r ⇒ r = 1/5 = 0.20 = 20% p.a.

Q4.12 — RB at 20% vs flat at 20% over 5 yr

Writing S = V₀(1 + i)ⁿ with i = −0.20: S = V₀(0.8)⁵ = V₀ × 0.32768, i.e. ≈ 32.77% retained. Flat rate at the same 20% for 5 years gives S = V₀(1 − 0.20 × 5) = V₀ × 0 = $0. The reason for the difference is structural: flat rate subtracts the same fixed amount (rV₀) each year regardless of how much value remains, while reducing balance shrinks the amount of depreciation each year because it is calculated on the current book value — exponential decay never reaches zero.