Depreciation
Assets lose value over time — but does a car lose the same dollar amount each year, or the same percentage? Straight-line depreciation uses a fixed dollar amount ($S = V_0 - Dn$); declining balance uses a fixed percentage ($S = V_0(1-r)^n$). Learn both, compare them, and find unknown rates.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
A new car is purchased for $30,000. After one year it might be worth only $24,000 — a loss of $6,000 in year one. Does it lose another $6,000 in year two, or something different? Does an asset lose the same dollar amount each year, or the same percentage each year? Do you think one method is more realistic for describing how assets like cars actually lose value?
Without calculating — write your gut feeling. We'll revisit this at the end of the lesson.
Depreciation in Maths Standard uses two formulas that mirror simple and compound interest — but in reverse (value decreases instead of grows).
Straight-line: $S = V_0 - Dn$ loses the same dollar amount $D$ each year — linear, like simple interest in reverse. Declining balance: $S = V_0(1-r)^n$ loses the same percentage each year — exponential decay, like compound interest but shrinking.
Key facts
- Straight-line formula: $S = V_0 - Dn$
- Declining balance formula: $S = V_0(1-r)^n$
- How to find $D$ from initial value, salvage value, and time
- That total depreciation = $V_0 - S$ for both methods
- That declining balance never theoretically reaches zero
Concepts
- Why the declining balance multiplier is $(1-r)$, not $(1+r)$
- Why straight-line is linear and declining balance is exponential
- How to identify which method from question wording
- Why declining balance depreciates faster in early years
Skills
- Apply $S = V_0 - Dn$ including solving for $D$ or $n$
- Apply $S = V_0(1-r)^n$ for any number of periods
- Compare both methods for the same asset
- Find depreciation rate $r$ by rearranging $S = V_0(1-r)^n$
Straight-line depreciation reduces an asset's value by the same fixed dollar amount $D$ each period, producing a linear decrease in value over time — identical in structure to simple interest, but going down instead of up.
For example, a machine worth $24,000 that depreciates by $3,000 per year will be worth $24,000 − ($3,000 × 5) = $9,000 after 5 years. An asset depreciated to zero is "fully depreciated."
If the initial value, final salvage value, and time are all known, calculate $D = (V_0 - S_n) \div n$.
Declining balance depreciates faster in early years; straight-line is constant.
What to write in your book
- $S = V_0 - Dn$ — $D$ is always a dollar amount, never a percentage.
- If $D$ is unknown: $D = (V_0 - S_n) \div n$.
- For "below $X$ after $n$ years" questions: set up inequality $V_0 - Dn < X$ and solve for $n$, then round up to the next whole year.
Quick check: An asset depreciates at 18% per annum declining balance. What multiplier should be used each year?
Declining balance depreciation reduces the asset's value by a fixed percentage of its current value each period — so the dollar amount of depreciation decreases as the asset becomes worth less.
The formula $S = V_0(1-r)^n$ mirrors the compound interest formula structurally — except instead of multiplying by $(1+r)$ to grow, we multiply by $(1-r)$ to shrink. For example, a vehicle purchased for $32,000 depreciating at 18% per annum after 4 years: $S = \$32{,}000 \times (0.82)^4 = \$32{,}000 \times 0.45212 = \$14{,}468$.
To find total depreciation: always use $V_0 - S$ — do not multiply the annual percentage by $n$, as the dollar amount changes each year.
What to write in your book
- $S = V_0(1-r)^n$ — same structure as $A = P(1+r)^n$ but decreasing.
- Label the multiplier: $(1-r) = 0.82$ when $r = 18\%$.
- Total depreciation = $V_0 - S$. Never $r \times n \times V_0$.
True or false: Under declining balance depreciation, the dollar amount lost each year stays the same.
Worked examples · 4 problems, reveal step by step
A piece of industrial equipment was purchased for $85,000. It depreciates by $7,400 per year using the straight-line method. (a) What is its value after 6 years? (b) After how many complete years will its value first fall below $30,000?
$S = V_0 - Dn = \$85{,}000 - (\$7{,}400 \times 6) = \$40{,}600$
$\$85{,}000 - \$7{,}400n < \$30{,}000$
$\$7{,}400n > \$55{,}000$
$n > 7.43\ldots$
After 7 years: $\$85{,}000 - \$51{,}800 = \$33{,}200 > \$30{,}000$ ✓
After 8 years: $\$85{,}000 - \$59{,}200 = \$25{,}800 < \$30{,}000$ ✓
$\therefore$ After 8 complete years, the value first falls below $\$30{,}000$.
A car is purchased new for $38,500 and depreciates at 22% per annum declining balance. Calculate: (a) its value after 3 years, and (b) the total depreciation over this period.
$r = 0.22$, so $(1-r) = 0.78$
$S = \$38{,}500 \times (0.78)^3$
$(0.78)^3 = 0.474552$
$S = \$38{,}500 \times 0.474552 = \$18{,}270.25$
$V_0 - S = \$38{,}500 - \$18{,}270.25 = \$20{,}229.75$
Office furniture is purchased for $16,000. Method A: straight-line depreciation of $1,800 per year. Method B: declining balance at 14% per annum. (a) Find the value under each method after 5 years. (b) Which method gives a higher salvage value after 5 years, and by how much?
$S_A = \$16{,}000 - (\$1{,}800 \times 5) = \$7{,}000.00$
$S_B = \$16{,}000 \times (0.86)^5$
$(0.86)^5 = 0.47043\ldots$
$S_B = \$16{,}000 \times 0.47043 = \$7{,}526.83$
$S_B = \$7{,}526.83 > S_A = \$7{,}000.00$
$\therefore$ Method B gives a higher salvage value by $\$526.83$.
A machine was purchased for $28,000 and is worth $15,950 after 4 years under declining balance depreciation. Find the annual depreciation rate, correct to two decimal places.
$15{,}950 = 28{,}000(1-r)^4$
$(1-r)^4 = \dfrac{15{,}950}{28{,}000} = 0.569642857\ldots$
$1-r = (0.569642857\ldots)^{1/4} \approx 0.868762$
$r = 1 - 0.868762 = 0.131238$
$r \approx 13.12\%$ per annum
What to write in your book
- Identify method from wording: fixed $ → straight-line; fixed % → declining balance.
- Finding $r$: rearrange to $(1-r)^n = S \div V_0$, take $n$th root, subtract from 1, convert to %.
- Don't round intermediate values when finding $r$ — keep full precision until the final step.
Fill the gap: A vehicle worth $45,000 depreciates at 25% per annum declining balance. After 2 years its value is $ (use $S = 45000 \times (0.75)^2$).
Common errors · the 3 traps that cost marks
What to write in your book
- Signal words: "per year" (dollar) → SL; "per annum" (percentage) → DB.
- DB multiplier: $(1-r)$. At 18%, multiplier = 0.82. Always label this line.
- Total depreciation for both methods: $V_0 - S$.
Match each description to the correct model:
Quick-fire practice · 5 calculations
A printer costing $6,200 depreciates by $540 per year using straight-line depreciation. Find its value after 5 years.
A boat was purchased for $64,000 and depreciates at 12% per annum declining balance. Find its value after 3 years.
An asset is purchased for $50,000 and has a salvage value of $18,000 after 8 years of straight-line depreciation. What is the annual depreciation amount?
A machine originally worth $2,800 depreciates by $420 per year (straight-line). What is its value after 4 years?
An asset was bought for $24,000. Under declining balance it depreciates at 10% per annum. Compare the salvage value after 4 years with straight-line at $2,200 per year.
Top 3 list: Name THREE things that distinguish declining balance depreciation from straight-line depreciation.
Look back at what you wrote in the Think First section. Under straight-line depreciation, the car loses the same dollar amount every year. Under declining balance, it loses a fixed percentage of its current value — so the dollar amount decreases over time. Declining balance is more realistic for assets like cars, which lose a large percentage of value early on.
What has changed? What did you get right? What surprised you?
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
SA 1. A printer costing $6,200 depreciates by $540 per year using straight-line depreciation. Find its value after 5 years. (2 marks)
SA 2. A boat was purchased for $64,000 and depreciates at 12% per annum declining balance. Find: (a) its value after 3 years, and (b) the total depreciation over this period. (3 marks)
SA 3. An asset was bought for $24,000. Under straight-line depreciation it loses $2,200 per year. Under declining balance it depreciates at 10% per annum. Compare the salvage value after 4 years. Which method gives the higher salvage value, and by how much? (3 marks)
📖 Comprehensive answers (click to reveal)
Drill 1: $S = 6200 - (540 \times 5) = \$3{,}500$ · 2: $(1-0.12)=0.88$; $S = 64000(0.88)^3 = \$43{,}614.21$ · 3: $D = (50000-18000) \div 8 = \$4{,}000$ · 4: $S = 2800 - (420 \times 4) = \$1{,}120$ · 5: SL: $S = 24000 - (2200 \times 4) = \$15{,}200$; DB: $S = 24000(0.90)^4 = \$15{,}746.40$; DB higher by $\$546.40$
SA 1 (2 marks): $S = 6200 - (540 \times 5) = 6200 - 2700$ [1] $= \$3{,}500.00$ [1].
SA 2 (3 marks): $(1-r) = 0.88$ [1]. $S = 64000(0.88)^3 = \$43{,}614.21$ [1]. Total depreciation $= 64000 - 43614.21 = \$20{,}385.79$ [1].
SA 3 (3 marks): SL: $S = 24000 - (2200 \times 4) = \$15{,}200$ [1]. DB: $S = 24000(0.90)^4 = \$15{,}746.40$ [1]. Declining balance gives the higher salvage value by $\$546.40$ [1].
Five timed questions on depreciation. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering questions on depreciation. Pool: lessons 1–13.
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