Appreciation and Depreciation

Most things you buy change in value after you buy them. A new phone or car loses value every year: that’s depreciation. A house or a rare hockey card might gain value: that’s appreciation. Knowing how values change helps you make smart choices, like whether to buy something new or used.

Key ideas

Appreciation and depreciation

• Appreciation is an increase in value over time. Things that often appreciate: homes and land, some collectibles (rare coins, trading cards, limited-edition sneakers), and art.
• Depreciation is a decrease in value over time. Things that usually depreciate: cars, phones, laptops, game consoles, and furniture. They wear out, and newer models come out.

Nothing is guaranteed. A collectible can lose value if people stop wanting it, and home prices can fall as well as rise. Looking at how similar items changed in value in the past is one of the best ways to estimate what might happen next.

Straight-line depreciation

With straight-line depreciation, an item loses the same dollar amount every year.

A laptop that costs $1200 and loses $240 per year:

Year	$0$	$1$	$2$	$3$	$4$	$5$
Value ($)	$1200$	$960$	$720$	$480$	$240$	$0$

The value goes down by the same amount each year, so this is a linear relation. The initial value is $1200$ and the rate of change is $-240$ dollars per year:

$$V = 1200 - 240n$$

where $n$ is the number of years. Its graph is a straight line.

Percent depreciation

With percent depreciation, an item loses the same percent of its current value every year. Since the value keeps shrinking, the dollar amount lost each year shrinks too.

If a car loses $20\%$ per year, it keeps $100\% - 20\% = 80\%$ of its value each year. So multiply by $0.8$ each year:

Year	Value ($)	Lost that year ($)
$0$	$30\,000$	—
$1$	$30\,000 \times 0.8 = 24\,000$	$6000$
$2$	$24\,000 \times 0.8 = 19\,200$	$4800$
$3$	$19\,200 \times 0.8 = 15\,360$	$3840$

The graph curves: it drops steeply at first, then levels off. The value never quite reaches $0$, because you always keep $80\%$ of something.

Percent appreciation

With percent appreciation, an item gains the same percent of its current value each year. If it gains $5\%$ per year, multiply by $1 + 0.05 = 1.05$ each year.

Change each year	Multiply by
loses $20\%$	$1 - 0.20 = 0.80$
loses $15\%$	$0.85$
gains $5\%$	$1 + 0.05 = 1.05$
gains $8\%$	$1.08$

Spotting the pattern

Multiplying by the same number every year is repeated multiplication, so you can use an exponent. The $30\,000$ dollar car above is worth $30\,000(0.8)^n$ after $n$ years. In general, an item with starting value $P$ that changes by a rate $r$ each year is worth

$$V = P(1 - r)^n \quad\text{(depreciation)} \qquad V = P(1 + r)^n \quad\text{(appreciation)}$$

This is the same pattern as compound interest, which you’ll study more later. In Grade 9, a year-by-year table is a perfectly good way to work these out.

Reading value graphs

The graph below compares the two kinds of depreciation for a $30\,000$ dollar car.

A $30 000 car: losing $3000 each year (straight-line) vs. losing $20\%$ each year (percent).

To read a value, go up from the year on the horizontal axis to the graph, then across to the value axis. For example, after $5$ years the straight-line value is $15 000 and the percent value is just under $10 000. Notice that the percent model loses much more in the first few years. That’s why a car loses so much value as soon as it’s driven off the lot.

Worked examples

Example 1: Straight-line depreciation

A gaming console costs $600 and depreciates by $75 per year. Make a table for the first $4$ years, write an equation for its value, and find when it will be worth $0.

Solution. Subtract $75$ each year:

Year	$0$	$1$	$2$	$3$	$4$
Value ($)	$600$	$525$	$450$	$375$	$300$

The initial value is $600$ and the rate of change is $-75$ per year:

$$V = 600 - 75n$$

Set $V = 0$:

$$600 - 75n = 0 \quad\Rightarrow\quad 75n = 600 \quad\Rightarrow\quad n = 8$$

It will be worth $0 after $8$ years.

Example 2: Percent depreciation

A new e-bike costs $2000 and loses $25\%$ of its value each year. Find its value after $3$ years and the total amount it lost.

Solution. It keeps $75\%$ each year, so multiply by $0.75$:

Year	Value ($)
$0$	$2000$
$1$	$2000 \times 0.75 = 1500$
$2$	$1500 \times 0.75 = 1125$
$3$	$1125 \times 0.75 = 843.75$

After $3$ years it’s worth $843.75. It lost $2000 - 843.75 = 1156.25$ dollars.

Check with the pattern: $2000(0.75)^3 = 2000 \times 0.421875 = 843.75$. ✓

Example 3: Appreciation

Jaden buys a rare hockey card for $150. Similar cards have gained about $8\%$ per year. If that continues, in which year will the card first be worth more than $200?

Solution. Multiply by $1.08$ each year. Keep the full calculator value and round only when you write it down:

Year	Value ($)
$0$	$150$
$1$	$150 \times 1.08 = 162$
$2$	$162 \times 1.08 = 174.96$
$3$	$174.96 \times 1.08 \approx 188.96$
$4$	$188.9568 \times 1.08 \approx 204.07$

The card first passes $200 in year $4$, when it’s worth about $204.07.

Remember this is a model based on the past: the card’s value depends on what collectors want, so it could grow faster, slower, or even fall.

Example 4: New or used?

A new car costs $30 000. Cars like it lose about $20\%$ of their value per year, so a $3$-year-old one sells for $15 360. Compare how much value each car loses over the next $3$ years.

Solution. Use the multiplier $0.8$ three times for each car.

New car: from the table in Key ideas, after $3$ years it’s worth $15 360.

$$\text{value lost} = 30\,000 - 15\,360 = 14\,640$$

Used car:

$$15\,360 \times 0.8^3 = 15\,360 \times 0.512 = 7864.32$$ $$\text{value lost} = 15\,360 - 7864.32 = 7495.68$$

Over the same $3$ years, the new car loses $14 640 of value and the used car loses about $7495.68, roughly half as much. Depreciation is a real cost of owning something, so buying used can save a lot. Of course, the used car may need more repairs, so a smart buyer weighs both.

Common mistakes

Taking the percent of the original price every year. If a $30 000 car loses $20\%$ per year, it loses $6000 in year $1$ but only $4800 in year $2$ ($20\%$ of $24 000). Taking $6000 off every year is straight-line depreciation, not percent depreciation.

Multiplying by the percent lost instead of the percent kept. For a $20\%$ loss, multiply by $0.8$, not $0.2$. Multiplying by $0.2$ gives what was lost, not what is left.

Mixing up the multipliers for gains and losses. A $6\%$ gain means multiplying by $1.06$. A $6\%$ loss means multiplying by $0.94$. Ask yourself: should the value go up or down?

Rounding too early. In Example 3, rounding each year to the nearest dollar can make the later values drift off. Keep the full calculator value in each step and round only at the end.

Thinking a loss and an equal gain cancel out. A $100 item that loses $20\%$ is worth $80. If it then gains $20\%$, it’s worth $80 \times 1.2 = 96$ dollars, not $100. The second percent is taken of a smaller number.

Expecting percent depreciation to reach zero. Under straight-line depreciation the value hits $0. Under percent depreciation, you always keep a fraction of the value, so it gets small but never reaches $0.

Practice

1. (Warm-up) Does each item usually appreciate or depreciate?

• (a) a new smartphone
• (b) a house in a growing city
• (c) a new pickup truck
• (d) a rare coin

Solution

(a) Depreciates. (b) Usually appreciates. (c) Depreciates. (d) Usually appreciates (if collectors still want it).

2. (Warm-up) A $900 phone depreciates by $150 per year (straight-line). What is it worth after $4$ years? When is it worth $0?

Solution

After $4$ years: $900 - 4 \times 150 = 900 - 600 = 300$, so $300.

Worth $0 when $900 - 150n = 0$, so $n = 900 \div 150 = 6$ years.

3. (Warm-up) What do you multiply by each year if an item:

• (a) loses $15\%$ per year?
• (b) gains $4\%$ per year?
• (c) loses $30\%$ per year?

Solution

(a) $1 - 0.15 = 0.85$

(b) $1 + 0.04 = 1.04$

(c) $1 - 0.30 = 0.70$

4. (Core) A $1500 laptop loses $25\%$ of its value each year. Make a table of its value for years $0$ to $3$, to the nearest cent.

Solution

Multiply by $0.75$ each year:

Year	Value ($)
$0$	$1500$
$1$	$1500 \times 0.75 = 1125$
$2$	$1125 \times 0.75 = 843.75$
$3$	$843.75 \times 0.75 = 632.8125 \approx 632.81$

5. (Core) A house in Kitchener is worth $600 000 and appreciates by $5\%$ per year. Find its value after $1$, $2$, and $3$ years, and the total increase.

Solution

Multiply by $1.05$ each year:

$$\begin{aligned} \text{Year 1: } & 600\,000 \times 1.05 = 630\,000 \\ \text{Year 2: } & 630\,000 \times 1.05 = 661\,500 \\ \text{Year 3: } & 661\,500 \times 1.05 = 694\,575 \end{aligned}$$

After $3$ years it’s worth $694 575, an increase of $694\,575 - 600\,000 = 94\,575$ dollars.

6. (Core) Use the depreciation graph in Key ideas (a $30 000 car).

• (a) Estimate the value of the car after $4$ years under each model. Then check by calculating.
• (b) Under the percent model, about when is the car worth half its price?
• (c) About when do the two models give the same value?

Solution

(a) Straight-line: $30\,000 - 4 \times 3000 = 18\,000$, so $18 000. Percent: $30\,000 \times 0.8^4 = 30\,000 \times 0.4096 = 12\,288$, so $12 288 (the graph shows about $12 000).

(b) Half the price is $15 000. The percent curve is at $15 360 at year $3$ and $12 288 at year $4$, so it reaches $15 000 a little after $3$ years.

(c) The graphs cross between year $8$ and year $9$. Check: at year $8$, straight-line is $6000 and percent is about $5033; at year $9$, straight-line is $3000 and percent is about $4027. The order switches, so they’re equal somewhere in between.

7. (Core) Two laptops each cost $1000. Laptop A loses $200 per year (straight-line). Laptop B loses $30\%$ per year.

• (a) Which is worth more after $2$ years?
• (b) Which is worth more after $4$ years?

Solution

(a) A: $1000 - 2 \times 200 = 600$. B: $1000 \times 0.7^2 = 1000 \times 0.49 = 490$. Laptop A is worth more ($600 vs. $490).

(b) A: $1000 - 4 \times 200 = 200$. B: $1000 \times 0.7^4 = 1000 \times 0.2401 = 240.10$. Now Laptop B is worth more ($240.10 vs. $200). Percent depreciation loses a lot early, but less later.

8. (Challenge) A $500 bike loses $20\%$ of its value, and then the next year its value rises by $20\%$ (it becomes popular).

• (a) What is it worth now?
• (b) What percent increase would have been needed to get back to $500?

Solution

(a) $500 \times 0.8 = 400$, then $400 \times 1.2 = 480$. It’s worth $480, not $500.

(b) It needs to grow from $400 to $500, an increase of $100:

$$\frac{100}{400} \times 100\% = 25\%$$

A $25\%$ increase was needed.

9. (Challenge) Phones like Rana’s lose about $35\%$ of their value per year. She can buy a new phone for $1100 or a one-year-old one for $700.

• (a) Use the model to find what the new phone would be worth after one year. Is $700 a fair price for the used phone?
• (b) Rana plans to keep whichever phone she buys for $2$ years. How much value would each phone lose in that time?

Solution

(a) $1100 \times 0.65 = 715$. The model says a one-year-old phone is worth about $715, so $700 is a fair price.

(b) New phone: $1100 \times 0.65^2 = 1100 \times 0.4225 = 464.75$. It loses $1100 - 464.75 = 635.25$ dollars.

Used phone: $700 \times 0.65^2 = 700 \times 0.4225 = 295.75$. It loses $700 - 295.75 = 404.25$ dollars.

The used phone loses about $231 less in value over the $2$ years ($635.25 - 404.25 = 231$).