Year 10 Maths Core Earning Money

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.

20 practice questions 1 video lesson Theory + worked examples

Depreciation in NSW Year 10 Core maths is the loss in value of items such as cars, computers and equipment over time (NESA outcome MA5‑FIN‑C‑02). It works like compound interest in reverse: the value falls by the same percentage each year, giving $S = V_0(1 - r)^n$.

You will find the value of an item after one or more years, the total depreciation and the amount lost in a single year, and choose the right expression for a value. You will also work backwards to find an original price, a rate of depreciation or the number of years, compare two cars that lose value at different rates, and decide when a cafe should replace its coffee machine. It is part of Year 10 Core Earning Money.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 1 video(s)

Depreciation - Video - Depreciation of Car Word Problem Solution Watch

Theory

Depreciation is the loss in value of an item over time, usually a fixed percentage each year. This Year 10 maths page shows how to find a salvage value year by year and with $S = V_0(1 - r)^{n}$, the total depreciation, the original price, and when an item's value falls below a set amount.

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.

Depreciation at rate $r$ multiplies the value by $1 - r$ each year: over $n$ years, by $(1 - r)^{n}$.

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

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

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

\[V_0 = \dfrac{S}{(1 - r)^{n}} \qquad \text{fraction of value left} = (1 - r)^{n}\]

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.

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.

Other questions you will meet

Choosing an expression: $V_0(1 - r)^{n}$, for example $20\,000 \times 0.9^{3}$ for $10\%$ over $3$ years.
Percentage of value left: $(1 - r)^{n} \times 100\%$. Compare with $50\%$ to say if an item is worth more or less than half its price.
Loss in one year: value at the start of that year minus value at the end (Example 4).
When does it fall below a value? Try whole numbers of years until it does (Example 4).
Comparing two items: work out each salvage value, then subtract. A dearer item that loses value faster can end up worth less.
Trade-in: the extra money needed is the new price minus the salvage value of the old item.
Finding the rate: $(1 - r)^{n} = \dfrac{S}{V_0}$, so $1 - r = \sqrt[n]{S \div V_0}$. Use a square root for $2$ years, a cube root for $3$.

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 — Finding the original value

A boat depreciates at $14\%$ p.a. After $4$ years it is worth $\$11\,200$. What was its original price, to the nearest dollar?

Solution

Substitute what you know into $S = V_0(1 - r)^{n}$ and divide to find $V_0$.

$11\,200$	$=$	$V_0 \times 0.86^{4}$
$V_0$	$=$	$\dfrac{11\,200}{0.86^{4}}$
	$=$	$20\,475.014\ldots$
	$\approx$	$\mathbf{\$20\,475}$

V_{0} = \frac{11200}{{0.86}^{4}} \approx 20475

Check that the answer is bigger than $\$11\,200$: the boat was worth more when new.

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.

← Previous subtopic

Term Payments

Next subtopic →

This is the last subtopic

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

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

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

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

\(11\,200\)	\(=\)	\(V_0 \times 0.86^{4}\)
\(V_0\)	\(=\)	\(\dfrac{11\,200}{0.86^{4}}\)
\(\)	\(=\)	\(20\,475.014\ldots\)
\(\)	\(\approx\)	\(\mathbf{\$20\,475}\)

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

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

Depreciation

📖 Theory

How to solve a depreciation problem

Other questions you will meet

Common pitfalls

Frequently asked questions

Theory