Lesson 2 of 4Financial Literacy

Unit 8 · Lesson 2 of 4

Appreciation and Depreciation

Model things that lose value (straight-line and percent depreciation) and things that gain value (appreciation), calculate values year by year or with a formula, and read the shapes of their graphs.

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Key things to remember

1Straight-line depreciation: $V = P - d t$ (same dollar loss $d$ each year; graph is a straight line).
2Yearly straight-line loss with a salvage value: $d = \frac{P - S}{n}$ .
3Percent depreciation: $V = P (1 - r)^{t}$ (same percent loss each year; curve that flattens).
4Appreciation: $V = P (1 + r)^{t}$ (same percent gain each year; curve that gets steeper).
5Losing $r$ per year means multiplying by $1 - r$ ; losing 15% means $\times 0.85$ , not $\times 0.15$ .

A new car that costs $28,000 starts losing value the day it leaves the lot. If it loses 15% of its value each year, three years later it is worth only about $17,195.50, even if it still runs perfectly. A rare trading card bought for $150 can go the other way and be worth more each year. A loss of value over time is called depreciation; a gain in value over time is called appreciation. Knowing how fast something loses or gains value helps you decide whether to buy new or used, how long to keep something, and whether a "deal" really is one. There are two common ways to model the change: by the same dollar amount each year, or by the same percent each year. They give very different numbers and very different graphs.

Straight-line depreciation: the same amount each year

In straight-line depreciation, an item loses the same number of dollars every year. If it starts at a purchase price $P$ and loses $d$ dollars per year, then after $t$ years its value is $V = P - d t$ . This is a linear relation: the initial value is $P$ and the rate of change is $- d$ dollars per year, so its graph is a straight line sloping down. The model reaches a value of zero after $P \div d$ years; after that the formula would give negative values, which make no sense for a price, so the model stops there. If an item is expected to be worth a salvage value $S$ (what you could still sell it for) after $n$ years, the yearly loss is $d = \frac{P - S}{n}$ .

V = P - d t (straight-line: lose d dollars every year)

Fig 1A 1,500 dollar e-bike that loses 250 dollars a year: its value for years 0 to 6 is 1,500, 1,250, 1,000, 750, 500, 250 and 0 dollars. The dashed boxes show the same 250 dollar loss each year, and the dashed line through the bar tops is V = 1500 − 250t.

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A $900 tablet loses $150 of value each year (straight-line). What is it worth after 2 years?

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An item's value is

V = 3000 - 400 t

dollars after

t

years. What does the 400 tell you?

Percent depreciation: the same percent each year

Many things, such as cars and electronics, are better described by percent depreciation (also called declining-balance depreciation): each year the item loses the same percent of its current value. Losing $r$ (as a decimal) each year leaves $1 - r$ of the value, so each year you multiply by $1 - r$ . A 20% loss means multiplying by $0.80$ every year. After $t$ years the value is $V = P (1 - r)^{t}$ . Because each year's loss is a percent of a smaller and smaller value, the dollar loss shrinks every year, and the graph is a curve that drops steeply at first and then flattens out. As long as the rate is between 0% and 100%, the model's value gets closer and closer to zero but never actually reaches it.

V = P (1 - r)^{t} (percent depreciation: multiply by 1 - r every year)

Year	Straight-line value (2,000 dollars a year)	Percent value (20% a year)	Percent model: loss that year
0	20,000.00	20,000.00	none
1	18,000.00	16,000.00	4,000.00
2	16,000.00	12,800.00	3,200.00
3	14,000.00	10,240.00	2,560.00
4	12,000.00	8,192.00	2,048.00
5	10,000.00	6,553.60	1,638.40
10	0.00	2,147.48	536.87

A 20,000 dollar car under two depreciation models, values in dollars rounded to the nearest cent

Fig 2Value of a 20,000 dollar item over 10 years under three models: straight-line depreciation of 2,000 dollars a year (a straight line reaching 0 at year 10), 20 percent depreciation a year (a curve that flattens out, reaching 2,147.48 dollars at year 10), and 6 percent appreciation a year (a curve rising to 35,816.95 dollars).

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An item loses 25% of its value every year. What do you multiply its value by each year?

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A $400 phone loses 25% of its value every year. What is it worth after 2 years?

Appreciation: growing by the same percent

Appreciation works the same way as percent depreciation, but the value grows. Gaining $r$ each year means multiplying by $1 + r$ , so after $t$ years $V = P (1 + r)^{t}$ . Each year's gain is a percent of a larger value than the year before, so the dollar gain grows and the graph curves upward more and more steeply. Things that can appreciate include some collectibles, land and homes. Appreciation is never guaranteed: home prices fell in many Canadian markets after peaking in early 2022, and the Canadian Real Estate Association's national home price index was still 3% lower in August 2026 than a year earlier (CREA, 2026). A collectible can lose value when it goes out of fashion. A model with a fixed rate is an estimate, not a promise.

V = P (1 + r)^{t} (appreciation: multiply by 1 + r every year)

Fig 3A 1,000 dollar item gaining 10 percent a year: 1,000, 1,100, 1,210, 1,331, 1,464.10 and 1,610.51 dollars for years 0 to 5. The shaded tops show each year's gain, 100, 110, 121, 133.10 and 146.41 dollars, growing because each is 10 percent of a bigger value.

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A $500 coin appreciates 4% per year. What is it worth after 2 years? Round to the nearest cent.

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A collectible is modelled by

V = 80 (1.07)^{t}

. Which statement is true?

VideoAn Ontario teacher works through appreciation and depreciation problems from this course.Watch on YouTube

Reading the graphs

A straight line sloping down: the same dollar loss each year (straight-line depreciation). The slope is the yearly loss.
A curve that falls quickly and then flattens: the same percent loss each year (percent depreciation). The yearly dollar loss shrinks.
A curve that rises more and more steeply: the same percent gain each year (appreciation). The yearly dollar gain grows.
Where two value graphs cross, the two items (or two models) are worth the same amount at that time. In the figure, the straight-line value drops below the percent value between years 8 and 9.
The point where a graph meets the vertical axis (time 0) is the purchase price P.

Five years at 20 percent is not 100 percent

It is tempting to say that losing 20% a year for 5 years loses

5 \times 20% = 100%

, so the item is worth nothing. But each 20% is taken from the current, smaller value. The item keeps

0. 8^{5} \approx 0.328

, about 32.8% of its price, after 5 years. Multiply by the multiplier; do not add the percents.

Rounding money

Keep the full calculator value through every step (do not round

0.8 5^{3}

0.61

) and round to the nearest cent only at the end. Rounding early can make your answer off by several dollars on a large price.

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A value graph is a straight line from (0, 8000) down to (4, 0), with time in years. How much value is lost each year?

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An item loses 50% of its value each year. After 2 years, what percent of its original value is left?

VideoEddie Woo compares straight-line depreciation, which loses the same amount each year, with declining balance, which loses the same percent each year.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

A laptop costs $1,200 and depreciates by $240 each year (straight-line). What is it worth after 3 years? After how many years does the model say it is worth $0?

Example 2

A car costs $28,000 and loses 15% of its value each year. What is it worth after 3 years? How much value has it lost?

Example 3

A trading card bought for $150 appreciates 8% per year. What is it worth after 5 years, to the nearest cent?

Example 4

A phone costs $900 and loses 30% of its value each year. Make a table of its value for years 0 to 3, and show that the dollar loss shrinks each year.

Where marks get lost

Adding the percents: treating 5 years of 20% depreciation as a 100% loss. Each year's 20% is taken from a smaller value, so multiply: 0.8 to the power 5 is about 0.328.
Multiplying by the rate instead of the multiplier: 28000 × 0.15 gives the first year's loss (4200 dollars), not the value. The value after one year is 28000 × 0.85.
Using 1 + r for depreciation or 1 − r for appreciation. Depreciation shrinks the value (multiplier below 1); appreciation grows it (multiplier above 1).
Rounding the multiplier or a power too early (using 0.61 for 0.85 cubed), which can throw the final value off by many dollars. Round to the cent only at the end.
Assuming something will keep appreciating at the same rate forever. Real prices can level off or fall, so a fixed-rate model is only an estimate.

Flashcards 8

Depreciation

Appreciation

Straight-line depreciation formula

Percent (declining-balance) depreciation formula

+ 4 more

Practice

A bike costs $600 and depreciates by $75 each year (straight-line). What is it worth after 4 years?

A car costs $24,000 and loses 18% of its value each year. What is it worth after 2 years? Round to the nearest cent.

A graph of an item's value over time is a curve that drops steeply at first and then flattens out. Which model fits it?

Which formula gives the value of an item bought for P dollars that appreciates 6% per year, after t years?

A painting is bought for $2,000 and appreciates 5% per year. What is it worth after 3 years? Round to the nearest cent.

An item loses 20% of its value every year. After 5 years, what fraction of its original value is it worth?

What is the name of the depreciation model in which an item loses the same dollar amount every year?

A machine costs $5,000 and is expected to have a salvage value of $1,000 after 4 years. Using straight-line depreciation, how much value does it lose each year?

Match each formula with the situation it models (P is the starting value).

V = P - 500 t

V = P (0.95)^{t}

V = P (1.05)^{t}

V = P + 500 t

Four items each cost $10,000. Put them in order from lowest to highest value after 3 years.

1Loses 25% of its value a year
2Gains 4% of its value a year
3Loses 10% of its value a year
4Loses 2,000 dollars a year (straight-line)

A $20,000 car loses 20% of its value each year. After how many whole years is it first worth less than $10,000?

years

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