Year 10 Maths Advanced Earning Money

Depreciation

Q: What is the depreciation formula?

S equals V0 times (1 minus r) to the power n, where V0 is the original value, r is the rate of depreciation per year as a decimal, n is the number of years and S is the value after n years.

Q: How do you calculate the depreciation of a car?

Multiply the price by (1 minus the rate) once for every year. A 30 000 dollar car depreciating at 15 percent p.a. is worth 30 000 times 0.85 cubed, about 18 423.75 dollars, after 3 years. The depreciation is the price minus that value.

Q: How do you find the rate of depreciation?

Divide the later value by the original value, take the nth root, and subtract the result from 1. Write the answer as a percentage.

20 practice questions 1 video lesson Theory + worked examples

Depreciation is the loss in value of an item such as a car, a phone or a machine as it ages. When it falls by the same percentage each year, the value after $n$ years is $S=V_0(1-r)^n$, the compound interest formula with a minus sign. In NSW Year 10 Advanced maths you find values, losses, rates and original prices (NESA outcome MA5‑FIN‑C‑02).

You will find the value of an asset after several years, the total depreciation, and the percentage of the original value lost. You will work backwards to an original price or an annual rate, find when an item first drops below a given share of its price, and compare buying a car with investing the money. Harder questions combine two rates, match the values of two cars, and explain why a rise and a fall of the same percentage do not cancel. It is part of Year 10 Advanced Earning Money.

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Depreciation - Video - Depreciation of Car Word Problem Solution Watch

Theory

Depreciation is the fall in the value of an item, such as a car or equipment, as it ages. With reducing balance depreciation the value falls by the same percentage each year, so $S = V_0(1 - r)^n$. This Year 10 Advanced maths page covers the value after $n$ years, total depreciation, and working back to the rate or the original price.

Depreciation is the loss in value of an asset over time. Cars, phones, machines and computers all depreciate.

In reducing balance (declining balance) depreciation the asset loses a fixed percentage $r$ of its value at the start of each year. It is compound interest in reverse: the value is multiplied by $(1 - r)$ every year.

$V_0$ is the original value (purchase price) and $S$ is the salvage value, the value after $n$ years. The total depreciation is $V_0 - S$.

Because the percentage is taken from a smaller value each year, the dollar loss shrinks every year, and the value never quite reaches zero.

Reducing balance depreciation takes the same percentage each year, so the amount lost gets smaller as the value falls.

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

For an original value $V_0$, a depreciation rate $r$ per year (as a decimal) and $n$ years:

\[S = V_0(1 - r)^n \qquad \text{total depreciation} = V_0 - S\]

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

Rearranged:

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

V_{0} = \frac{S}{{(1 - r)}^{n}}, r = 1 - \sqrt[n]{\frac{S}{V_{0}}}

Loss during year $k$ is the rate times the value at the start of that year:

\[\text{loss in year } k = r \times V_0(1 - r)^{k-1}\]

loss = r V_{0} {(1 - r)}^{k - 1}

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 multiplier $1 - r$. For $13\%$ p.a. it is $0.87$.
Substitute into $S = V_0(1 - r)^n$ and solve for the unknown: multiply for $S$, divide for $V_0$, take the $n$th root for $r$, or test whole numbers for $n$.
Answer the question asked: the value, the total depreciation $V_0 - S$, or a percentage of the original.
Check the value has fallen and the dollar losses shrink each year.

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 4).
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 — Value and total depreciation

A tractor is bought for $\$58\,000$ and depreciates at $13\%$ p.a. Find, to the nearest cent:
(i) its value after $6$ years
(ii) the total depreciation over the $6$ years.

Solution

Each year the tractor keeps $100\% - 13\% = 87\%$ of its value, so the multiplier is $0.87$.

(i)

$S$	$=$	$V_0(1 - r)^n$
	$=$	$58\,000(1 - 0.13)^6$
	$=$	$58\,000 \times 0.87^6$
	$=$	$25\,150.3196\ldots$
	$\approx$	$\mathbf{\$25\,150.32}$

S = 58000 \times {0.87}^{6} \approx 25150.32

(ii) The total depreciation is the value lost.

$D$	$=$	$V_0 - S$
	$=$	$58\,000 - 25\,150.32$
	$=$	$\mathbf{\$32\,849.68}$

D = 58000 - 25150.32 = 32849.68

Check: after $6$ years the tractor is worth $0.87^6 \approx 43.4\%$ of its price, so it has lost about $56.6\%$.

Example 2 — Finding the rate

A ute bought for $\$36\,000$ is worth $\$21\,000$ after $5$ years. Find the annual rate of depreciation, correct to $1$ decimal place.

Solution

Substitute into $S = V_0(1 - r)^n$. Divide to get the multiplier on its own, then take the fifth root.

$21\,000$	$=$	$36\,000(1 - r)^5$
$(1 - r)^5$	$=$	$\dfrac{21\,000}{36\,000}$
	$=$	$0.5833\ldots$
$1 - r$	$=$	$\sqrt[5]{0.5833\ldots}$
	$=$	$0.8978\ldots$
$r$	$=$	$0.1021\ldots$
	$\approx$	$\mathbf{10.2\%}\text{ p.a.}$

1 - r = \sqrt[5]{\frac{21000}{36000}}, r \approx 10.2 %

Keep the full value of the fraction in the calculator: $\left(\dfrac{21\,000}{36\,000}\right)^{(1 \div 5)}$. Check: $36\,000 \times 0.8978^5 \approx 21\,000$.

Example 3 — One year's loss, and a threshold

A server costs $\$12\,500$ and depreciates at $20\%$ p.a.
(i) How much value does it lose during its third year?
(ii) After how many full years is it first worth less than $25\%$ of its price? Justify your answer.

Solution

(i) The loss in a year is $20\%$ of the value at the start of that year. The third year starts after $2$ years.

$\text{Value after 2 years}$	$=$	$12\,500 \times 0.8^2 = \$8000$
$\text{Loss in year 3}$	$=$	$0.2 \times 8000 = \mathbf{\$1600}$

0.2 \times 12500 \times {0.8}^{2} = 1600

(ii) After $n$ years it is worth $0.8^n$ of its price, whatever the price. Test whole numbers:

$0.8^6$	$=$	$0.2621\ldots\ \ (26.2\%)$
$0.8^7$	$=$	$0.2097\ldots\ \ (21.0\%)$

{0.8}^{6} > 0.25, {0.8}^{7} < 0.25

After $6$ years it is still worth more than $25\%$; after $7$ it is worth less. Answer: after $\mathbf{7}$ years.

Example 4 — 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 $20\%$.

Common pitfalls

Use 1 minus the rate. At $13\%$ p.a. the multiplier is $0.87$, not $0.13$. Using $0.13^n$ makes the value almost vanish.

Depreciation is not a fixed amount. $20\%$ of $\$20\,000$ is $\$4000$ in the first year only. Later years lose less, because the percentage is taken from a smaller value.

Value or depreciation? $S$ is what the item is still worth. If the question asks how much value was lost, subtract: $V_0 - S$.

Undo with division, not addition. To go back a year, divide by $1 - r$. Adding $20\%$ back to a value that fell by $20\%$ does not return the original.

Lost or kept? After $5$ years at $10\%$ p.a., $0.9^5 \approx 59\%$ of the value is kept and about $41\%$ is lost. Read which one the question wants.

Frequently asked questions

What is the depreciation formula?

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

How do you calculate the depreciation of a car?

Multiply the price by $(1 - r)$ once for every year. A $\$30\,000$ car depreciating at $15\%$ p.a. is worth $30\,000 \times 0.85^3 \approx \$18\,423.75$ after $3$ years. The depreciation is the price minus that value.

How do you find the rate of depreciation?

Divide the later value by the original value, take the $n$th root, and subtract the result from $1$: $r = 1 - \sqrt[n]{\dfrac{S}{V_0}}$. Write the answer as a percentage.

How is depreciation different from compound interest?

They use the same idea, a percentage change each year on the current value. Compound interest multiplies by $1 + r$ so the value grows; depreciation multiplies by $1 - r$ so the value falls.

Does an asset ever depreciate to zero?

Not with reducing balance depreciation. Each year it keeps a fixed percentage of its value, so the value gets smaller and smaller but never reaches zero.

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\(S\)	\(=\)	\(V_0(1 - r)^n\)
\(\)	\(=\)	\(58\,000(1 - 0.13)^6\)
\(\)	\(=\)	\(58\,000 \times 0.87^6\)
\(\)	\(=\)	\(25\,150.3196\ldots\)
\(\)	\(\approx\)	\(\mathbf{\$25\,150.32}\)

\(D\)	\(=\)	\(V_0 - S\)
\(\)	\(=\)	\(58\,000 - 25\,150.32\)
\(\)	\(=\)	\(\mathbf{\$32\,849.68}\)

\(21\,000\)	\(=\)	\(36\,000(1 - r)^5\)
\((1 - r)^5\)	\(=\)	\(\dfrac{21\,000}{36\,000}\)
\(\)	\(=\)	\(0.5833\ldots\)
\(1 - r\)	\(=\)	\(\sqrt[5]{0.5833\ldots}\)
\(\)	\(=\)	\(0.8978\ldots\)
\(r\)	\(=\)	\(0.1021\ldots\)
\(\)	\(\approx\)	\(\mathbf{10.2\%}\text{ p.a.}\)

\(\text{Value after 2 years}\)	\(=\)	\(12\,500 \times 0.8^2 = \$8000\)
\(\text{Loss in year 3}\)	\(=\)	\(0.2 \times 8000 = \mathbf{\$1600}\)

\(0.8^6\)	\(=\)	\(0.2621\ldots\ \ (26.2\%)\)
\(0.8^7\)	\(=\)	\(0.2097\ldots\ \ (21.0\%)\)

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

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

Depreciation

📖 Theory

How to solve a depreciation problem

Harder problems

Common pitfalls

Frequently asked questions

Theory