Practice Maths

L13 — Depreciation and Appreciation

Key Terms

Depreciation
A reduction in an asset’s value over time.
Straight-line (SL) depreciation
The asset loses the same fixed dollar amount each year; V = P(1 − rn). Value eventually reaches zero at n = 1/r years.
Reducing balance (RB) depreciation
The asset loses the same percentage of its current value each year; V = P(1 − r)n. Value never reaches zero.
Appreciation
An increase in an asset’s value over time; V = P(1 + r)n — identical structure to compound interest.
Book value
The asset’s recorded value at any point in time.
Scrap value
The estimated residual value at the end of an asset’s useful life.

Three Models

ModelFormulaGrowthReaches zero?
Straight-line depreciation V = P − Dn  or  V = P(1 − rn) Linear decrease Yes, at n = 1/r years
Reducing balance depreciation V = P(1 − r)n Exponential decay Never (approaches 0)
Appreciation V = P(1 + r)n Exponential growth —
  • P = initial value (purchase price), r = rate per year as a decimal, n = years, D = fixed annual dollar depreciation
  • Straight-line: the asset loses the same dollar amount each year.
  • Reducing balance: the asset loses the same percentage each year — it depreciates faster at first, then slows down.
SL RB $0 $2500 $5000 $7500 $10000 0 2 4 6 8 10 Years (n) Value (V) P = $10 000, r = 10% p.a.
Straight-line (blue) reaches $0; reducing balance (orange) stays above it and never reaches $0.
Hot Tip: Both depreciation methods give the same value after exactly year 1. From year 2 onwards, reducing balance always retains more value than straight-line at the same rate — the exponential curve lies above the straight line.

Straight-Line Depreciation

Each year, the asset loses the same fixed dollar amount. The value decreases in a straight line.

V = P − Dn   or   V = P(1 − rn)

where D = Pr is the annual depreciation in dollars. The value reaches zero when n = 1/r (or n = P/D).

Worked Example 1 — Straight-Line Depreciation

A machine costs $24000 and depreciates by 8% p.a. straight-line. Find its value after 5 years.

V = P(1 − rn) = 24000(1 − 0.08 × 5) = 24000 × 0.60 = $14400

Annual depreciation: D = 24000 × 0.08 = $1920 per year. At year 5: 24000 − 5 × 1920 = $14400 ✓

Reducing Balance Depreciation

Each year, the asset loses the same percentage of its current value. The value decreases exponentially — fast at first, then slower. The value never reaches exactly zero.

V = P(1 − r)n

Worked Example 2 — Reducing Balance Depreciation

A car costs $24000 and depreciates at 15% p.a. reducing balance. Find its value after 5 years.

V = 24000 × (1 − 0.15)5 = 24000 × (0.85)5 = 24000 × 0.44371 ≈ $10649

Appreciation

Appreciation means the value increases over time (e.g. property, collectibles). This follows the same exponential formula as compound interest.

V = P(1 + r)n

Worked Example 3 — Appreciation

A house purchased for $420000 appreciates at 5% p.a. Find its value after 8 years.

V = 420000 × (1.05)8 = 420000 × 1.47746 ≈ $620531

Finding Unknowns

To find the number of years, use trial and error or rearrange the formula. For example:

  • SL: V = P(1−rn) ⇒ n = (P−V)/(Pr)
  • RB: V = P(1−r)n ⇒ divide both sides by P, then use trial and error to find n.

  1. Straight-line depreciation. Fluency

    • (a) P = $8000, r = 10% p.a. Find the value V after 3 years.
    • (b) P = $15000, annual depreciation D = $1500. Find V after 4 years.
    • (c) P = $6000, r = 8% p.a. Find V after 5 years.
    • (d) P = $20000, D = $2000 per year. After how many years does V = $12000?

  2. Reducing balance depreciation. Fluency

    Use V = P(1 − r)n. Give answers to the nearest dollar.

    • (a) P = $10000, r = 15% p.a., n = 5 years.
    • (b) P = $5000, r = 20% p.a., n = 3 years.
    • (c) P = $12000, r = 10% p.a., n = 6 years.
    • (d) P = $8000, r = 25% p.a., n = 4 years.

  3. Appreciation. Fluency

    Use V = P(1 + r)n. Give answers to the nearest dollar.

    • (a) P = $300000, r = 5% p.a., n = 3 years.
    • (b) P = $50000, r = 8% p.a., n = 10 years.
    • (c) P = $1000, r = 3% p.a., n = 5 years.
    • (d) P = $20000, r = 4% p.a., n = 20 years.

  4. Find the unknown. Fluency

    • (a) V = $6000, P = $10000, r = 10% p.a. (straight-line). Find n.
    • (b) V = $4000, P = $10000, n = 5 years (straight-line). Find the annual depreciation amount D.
    • (c) P = $10000, r = 10% (reducing balance). By checking V at n = 5, 6, 7, find the first year V falls below $5000.
    • (d) A property worth $100000 appreciates to $133100 after 3 years. Find the annual appreciation rate r.

  5. Read from the graph: straight-line vs reducing balance. Understanding

    The graph below shows two assets, each initially worth $10000, depreciating at 10% p.a. Blue (SL) = straight-line; orange (RB) = reducing balance.

    SL RB $0 $2500 $5000 $7500 $10000 0 2 4 6 8 10 Years (n) Value (V)
    • (a) Read the approximate value of each asset after 5 years.
    • (b) At year 1, both assets have the same value. From which year onwards does reducing balance always retain more value?
    • (c) When does the straight-line asset reach $0?
    • (d) A 15-year warranty is offered. Estimate the reducing balance value at n = 15 using the formula V = 10000 × (0.9)15.

  6. Useful life of a machine. Understanding

    A machine is purchased for $50000 and depreciates by straight-line at 12.5% p.a.

    • (a) Write the formula for its value after n years.
    • (b) After how many years is the machine worth $25000?
    • (c) In what year does the machine reach a value of $0 (scrap value)?
    • (d) If the same machine used reducing balance at 12.5% p.a. instead, find its value at the same year it reaches $0 under straight-line.

  7. Comparing two depreciation methods. Understanding

    Two identical cars are both purchased new for $25000. Car A uses straight-line at 15% p.a.; Car B uses reducing balance at 15% p.a.

    • (a) Find the value of each car after 2 years.
    • (b) Find the value of each car after 6 years.
    • (c) Which car has a higher resale value after 6 years? Explain why.
    • (d) After how many complete years does straight-line depreciation reach $0?

  8. Property appreciation. Understanding

    A house is purchased for $450000 and appreciates at 4.5% p.a.

    • (a) Write the formula for its value after n years.
    • (b) Find its value after 5 years.
    • (c) Find its value after 10 years.
    • (d) By how much did the house value grow over 10 years?

  9. Car depreciation. Problem Solving

    A car is purchased new for $32000 and depreciates by reducing balance at 18% p.a.

    • (a) Write the formula and find the value after 3 years.
    • (b) By testing n = 5 and n = 6, find the first complete year in which the car is worth less than $10000.
    • (c) The owner decides to sell when the car's value is between $20000 and $21000. By testing n = 2 and n = 3, find in which year this occurs.
    • (d) If the car instead depreciated by straight-line at 15% p.a., in what year would it first be worth less than $20000? (Solve V = P(1−rn) < 20000.)

  10. Investment property — variable appreciation. Problem Solving

    Mia buys an investment property for $500000. It appreciates at 5% p.a. for the first 5 years, then at 3% p.a. for the next 5 years.

    • (a) Find its value after the first 5 years.
    • (b) Using the answer from (a) as the new principal, find its value after the full 10 years.
    • (c) What single constant annual rate r would give the same final value over 10 years? Set up the equation 500000(1 + r)10 = your answer from (b) and solve by trial and error.
    • (d) By how much has the property grown in value over the 10 years?
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