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.