Year 10 Maths - 10A Indices and Surds

Depreciation

Q: What is the depreciation formula?

The formula is S = V0(1 - r)^n, where V0 is the original value, r is the yearly depreciation rate as a decimal, n is the number of years and S is the value after n years.

Q: How do you calculate depreciation?

Find the value after n years with S = V0(1 - r)^n, then subtract it from the original value. The difference is the total depreciation.

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Theory

Depreciation is the fall in value of an item, such as a car or a machine, as it gets older. With reducing balance depreciation it loses the same percentage of its value each year, so $S = V_0(1 - r)^{n}$. This Year 10A maths page treats depreciation as exponential decay in a financial context: value and total depreciation, reading a depreciation formula, percentages of the original value, the rate, a rise followed by a fall, and working backwards from two values. In the Year 10A course (Victorian Curriculum 10A) this is Level 10 content. Curriculum: VC2M10A15 / AC9M10A04.

Depreciation is the loss in value of an asset, such as a car, a computer or a machine, as it ages. Most depreciation questions use a fixed percentage per year of the value at the start of that year.

The initial value $V_0$ is the price when new. The salvage value (or depreciated value) $S$ is what it is worth after $n$ years. The total depreciation is $V_0 - S$.

Depreciation works like compound interest in reverse. Losing $15\%$ each year means keeping $85\%$, so the value is multiplied by $0.85$ every year.

Each year the item loses the same percentage of its value, so it loses fewer dollars each year.

Multiply the multipliers. A rise then an equal percentage fall leaves less than you started with.

\[S = V_0(1 - r)^{n}\]

\[\text{depreciation} = V_0 - S\]

S = V_{0} {(1 - r)}^{n}

$r$ is the depreciation rate per year as a decimal and $n$ is the number of years.

\[V_0 = \dfrac{S}{(1 - r)^{n}}\]

\[\text{fraction of value left} = (1 - r)^{n}\]

V_{0} = \frac{S}{{(1 - r)}^{n}}

Compare with compound interest. Growth multiplies by \(1 + r\) each period; depreciation multiplies by \(1 - r\). Using \(1 + r\) for depreciation makes the value grow, which is a quick sign of an error.

Reading a formula. In \(V = 3000 \times 0.9^n\), the number in front is the value when new (\(\$3000\)) and the base is the multiplier, so \(1 - r = 0.9\) and the rate is \(10\%\) p.a.

Percentages of the original. After \(n\) years an asset is worth \((1 - r)^n\) of its price, so it has lost \(1 - (1 - r)^n\). These do not depend on the price.

How to solve a depreciation problem

Write the rate as a decimal and work out $1 - r$, the fraction kept each year.
Substitute into $S = V_0(1 - r)^{n}$ and use the power key.
Answer the question asked: the value $S$, or the depreciation $V_0 - S$. Round money at the end.

Harder problems

Different rates in different years: multiply by each year's multiplier in turn, e.g. $25\%$ in year one then $12\%$ p.a. gives $V_0 \times 0.75 \times 0.88^{n-1}$.
A rise then a fall: multiply the multipliers (Diagrams tab). Equal percentage changes up and down do not cancel.
First-year loss given: the rate is the loss divided by the price.
Total depreciation given: $V_0 - V_0(1 - r)^n = D$, so $V_0\big(1 - (1 - r)^n\big) = D$. Factor out $V_0$, then divide.
A year's loss given: write it in terms of $V_0$, e.g. the loss in year $2$ is $r(1 - r)V_0$, then solve.
Two values a year apart: the later one divided by the earlier one is $1 - r$ (Example 3).
Two assets with the same value: find the known value first, then solve for the other rate with an $n$th root.
Buy or invest? The asset is multiplied by $1 - r$ each year; money invested is multiplied by $1 + r$. Compare the two final values.

Example 1 — Year by year, then the formula

A tablet is bought for $\$1600$ and depreciates at $25\%$ p.a.
(i) Find its value at the end of each of the first $3$ years.
(ii) Check with $S = V_0(1 - r)^{n}$.

Solution

(i) Each year it loses $25\%$ of its value at the start of that year.

Year	Value at start	Loss (25%)	Value at end
$1$	$\$1600$	$\$400$	$\$1200$
$2$	$\$1200$	$\$300$	$\$900$
$3$	$\$900$	$\$225$	$\$675$

(ii) $V_0 = 1600$, $r = 0.25$, $n = 3$.

$S$	$=$	$1600 \times (1 - 0.25)^{3}$
	$=$	$1600 \times 0.75^{3}$
	$=$	$\mathbf{\$675}$

S = 1600 \times {0.75}^{3} = 675

The losses shrink each year ($\$400$, $\$300$, $\$225$) because each is $25\%$ of a smaller value.

Example 2 — Value and total depreciation

A ute costs $\$46\,500$ new and depreciates at $18\%$ p.a. Find
(i) its value after $5$ years
(ii) the total depreciation over the $5$ years. Give answers to the nearest cent.

Solution

(i) The ute keeps $82\%$ of its value each year, so multiply by $0.82$ five times.

$S$	$=$	$46\,500 \times (1 - 0.18)^{5}$
	$=$	$46\,500 \times 0.82^{5}$
	$=$	$17\,239.402\ldots$
	$\approx$	$\mathbf{\$17\,239.40}$

(ii) Depreciation is the value lost: original value minus final value.

$\text{depreciation}$	$=$	$46\,500 - 17\,239.40$
	$=$	$\mathbf{\$29\,260.60}$

46500 - 46500 \times {0.82}^{5} \approx 29260.60

Example 3 — Working backwards

A camper trailer depreciates at a constant annual rate. It is worth $\$15\,360$ after $2$ years and $\$12\,288$ after $3$ years. Find:
(i) the annual rate of depreciation
(ii) the price when new.

Solution

(i) From one year to the next the value is multiplied by $1 - r$, so divide the later value by the earlier one.

$1 - r$	$=$	$\dfrac{12\,288}{15\,360}$
	$=$	$0.8$
$r$	$=$	$0.2 = \mathbf{20\%}\text{ p.a.}$

1 - r = \frac{12288}{15360} = 0.8

(ii) Two years of depreciation took the price $V_0$ down to $\$15\,360$.

$V_0 \times 0.8^2$	$=$	$15\,360$
$V_0$	$=$	$\dfrac{15\,360}{0.64}$
	$=$	$\mathbf{\$24\,000}$

V_{0} = \frac{15360}{{0.8}^{2}} = 24000

Check: $24\,000 \times 0.8^3 = 12\,288$. Dividing by the multiplier undoes a year of depreciation; dividing by $0.8$ is not the same as adding $20\%$.

Example 4 — One year's loss, and when to replace

A machine is bought for $\$62\,000$ and depreciates at $20\%$ p.a.
(i) How much value does it lose during the fourth year?
(ii) It will be replaced once its value first falls below $\$20\,000$. After how many full years is that?

Solution

(i) The fourth year runs from the end of year $3$ to the end of year $4$. Find both values and subtract.

$\text{after 3 years}$	$=$	$62\,000 \times 0.8^{3}$
	$=$	$\$31\,744.00$
$\text{after 4 years}$	$=$	$62\,000 \times 0.8^{4}$
	$=$	$\$25\,395.20$
$\text{loss in year 4}$	$=$	$\mathbf{\$6348.80}$

(ii) Keep multiplying by $0.8$ until the value drops below $\$20\,000$.

$\text{after 5 years}$	$:$	$\$20\,316.16 \ (\text{still above})$
$\text{after 6 years}$	$:$	$\$16\,252.93 \ (\text{below})$

62000 \times {0.8}^{6} \approx 16252.93 < 20000

It is replaced after 6 years.

Common pitfalls

Subtract the rate from 1. For $12\%$ depreciation use $0.88$, not $0.12$ or $1.12$. The item keeps $88\%$ of its value each year.

Value or depreciation? $S$ is what the item is still worth. The depreciation is the amount lost, $V_0 - S$. Read which one the question wants.

The loss is not the same each year. $20\%$ of a smaller value is a smaller loss. Do not multiply the first year's loss by the number of years.

Brackets before the power. Work out $0.85^{4}$ first, then multiply by $V_0$.

Frequently asked questions

What is the depreciation formula?

The formula is $S = V_0(1 - r)^{n}$, where $V_0$ is the original value, $r$ is the yearly depreciation rate as a decimal, $n$ is the number of years and $S$ is the value after $n$ years.

How do you calculate depreciation?

Find the value after $n$ years with $S = V_0(1 - r)^{n}$, then subtract it from the original value. The difference is the total depreciation.

What is salvage value?

Salvage value is what an item is worth after it has depreciated for a number of years. It is the $S$ in the depreciation formula.

How is depreciation like compound interest?

Both multiply by the same factor every year. Compound interest multiplies by $1 + r$, so the value grows. Depreciation multiplies by $1 - r$, so the value shrinks.

Does an item ever depreciate to zero?

Not with percentage depreciation. Each year it keeps a fraction of its value, so the value gets closer and closer to zero but never reaches it.

Year	Value at start	Loss (25%)	Value at end
\(1\)	\(\$1600\)	\(\$400\)	\(\$1200\)
\(2\)	\(\$1200\)	\(\$300\)	\(\$900\)
\(3\)	\(\$900\)	\(\$225\)	\(\$675\)

\(S\)	\(=\)	\(1600 \times (1 - 0.25)^{3}\)
\(\)	\(=\)	\(1600 \times 0.75^{3}\)
\(\)	\(=\)	\(\mathbf{\$675}\)

\(S\)	\(=\)	\(46\,500 \times (1 - 0.18)^{5}\)
\(\)	\(=\)	\(46\,500 \times 0.82^{5}\)
\(\)	\(=\)	\(17\,239.402\ldots\)
\(\)	\(\approx\)	\(\mathbf{\$17\,239.40}\)

\(\text{depreciation}\)	\(=\)	\(46\,500 - 17\,239.40\)
\(\)	\(=\)	\(\mathbf{\$29\,260.60}\)

\(1 - r\)	\(=\)	\(\dfrac{12\,288}{15\,360}\)
\(\)	\(=\)	\(0.8\)
\(r\)	\(=\)	\(0.2 = \mathbf{20\%}\text{ p.a.}\)

\(V_0 \times 0.8^2\)	\(=\)	\(15\,360\)
\(V_0\)	\(=\)	\(\dfrac{15\,360}{0.64}\)
\(\)	\(=\)	\(\mathbf{\$24\,000}\)

\(\text{after 3 years}\)	\(=\)	\(62\,000 \times 0.8^{3}\)
\(\)	\(=\)	\(\$31\,744.00\)
\(\text{after 4 years}\)	\(=\)	\(62\,000 \times 0.8^{4}\)
\(\)	\(=\)	\(\$25\,395.20\)
\(\text{loss in year 4}\)	\(=\)	\(\mathbf{\$6348.80}\)

\(\text{after 5 years}\)	\(:\)	\(\$20\,316.16 \ (\text{still above})\)
\(\text{after 6 years}\)	\(:\)	\(\$16\,252.93 \ (\text{below})\)

Depreciation

📖 Theory

How to solve a depreciation problem

Other questions you will meet

Harder problems

Common pitfalls

Frequently asked questions

Theory