# Appreciation and Depreciation Appreciation and Depreciation Appreciation refers

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Appreciation and Depreciation

Appreciation and Depreciation Appreciation refers to an amount increasing.

Appreciation and Depreciation

Appreciation and Depreciation refers to an amount decreasing.

We often refer to appreciation and depreciation when discussing percentages

For example, if a house appreciates in value by 6% each year, then its value increases by 6% each year.

On the other hand, if a car depreciates in value by 10% each year, then its value decreases by 10% each year.

Example of an appreciation question

Jojo purchases a new house for \$215000. She expects it to appreciate in value by 8% each year.

Assuming she is right, how much will the house be worth

Assuming she is right, how much will the house be worth a) 1 year from now?

Assuming she is right, how much will the house be worth a) 1 year from now? b) 2 years from now?

Assuming she is right, how much will the house be worth a) 1 year from now? b) 2 years from now? c) 3 years from now?

a) 215000 x 0. 08 = 17200

a) 215000 x 0. 08 = 17200 Since the value is increasing 8%, we add 17200 to the original value

215000 + 17200 = 232200

215000 + 17200 = 232200 Therefore, the house is worth \$232200 after 1 year.

b) 232200 x 0. 08 = 18576

b) 232200 x 0. 08 = 18576 Since the value is increasing 8%, we add 18576

232200 + 18576 = 250576

232200 + 18576 = 250576 Therefore, the house is worth \$250576 after 2 years.

c) 250576 x 0. 08 = 20046. 08

c) 250576 x 0. 08 = 20046. 08 Since the value is increasing 8%, we add 20046. 08

250576 + 20046. 08 = 270622. 08

250576 + 20046. 08 = 270622. 08 Therefore, the house is worth \$270622. 08 after 3 years.

Example of a Depreciation Question

Example: Jojo’s car is worth \$10000 today, but it will depreciate by 12% each year.

How much will the car be worth:

How much will the car be worth: a)1 year from today?

How much will the car be worth: a)1 year from today? b) 2 years from today?

How much will the car be worth: a)1 year from today? b) 2 years from today? c) 3 years from today?

a) 10000 x 0. 12 = 1200

a) 10000 x 0. 12 = 1200 Since the value is decreasing 12%, we subtract 1200 from the original value

10000 – 1200 = 8800

10000 – 1200 = 8800 Therefore, the car is worth \$8800 after 1 year.

b) 8800 x 0. 12 = 1056

b) 8800 x 0. 12 = 1056 Since the value is decreasing 12%, we subtract 1056

8800 – 1056 = 7744

8800 – 1056 = 7744 Therefore, the car is worth \$7744 after 2 years.

c) 7744 x 0. 12 = 929. 28

c) 7744 x 0. 12 = 929. 28 Since the value is decreasing 12%, we subtract 929. 28

7744 – 929. 28 = 6814. 72

7744 – 929. 28 = 6814. 72 Therefore, the car is worth \$6814. 72 after 3 years.

Is there an easier way to do this?

Yes there is, Johnny

We simply need to convert the percentage that we are appreciating or depreciating to a decimal.

Then, if it’s an appreciation question, we add that quantity to 1 to give us our multiplier (i. e. , the base of the exponent).

On the other hand, if it’s a depreciation question, we subtract from 1 to give us our multiplier (i. e. , the base of the exponent).

Then, we use a formula similar to the formula that we used for compound interest and for halflives.

For example, go back to our question about Jojo and her house. It is originally worth \$215000. This is the “prior” amount, P.

We are increasing the value by 8%. So, we convert 8% to a decimal, then add it to 1 (since it’s an appreciation question)

We are increasing the value by 8%. So, we convert 8% to a decimal, then add it to 1 1 +. 08 = 1. 08

We are increasing the value by 8%. So, we convert 8% to a decimal, then add it to 1 1 +. 08 = 1. 08 This is our multiplier

After 1 year: 1 A = P (1. 08)

After 1 year: 1 A = 215000 (1. 08)

After 1 year: A = 215000 (1. 08)

After 1 year: A = 232200

After 2 years: 2 A = P (1. 08)

After 2 years: 2 A = 215000 (1. 08)

After 2 years: A = 215000 (1. 1664)

After 2 years: A = 250776

After 3 years: 3 A = P (1. 08)

After 3 years: 3 A = 215000 (1. 08)

After 3 years: A = 215000 (1. 259712)

After 3 years: A = 270838. 08

This method allows us to cut right to the answer and not have to figure out EVERY year leading up to our answer

For example, when we were discussing the value of Jojo’s car, we calculated that after 3 years, its value would have decreased to \$6814. 72

We can calculate this a lot quicker as follows

We can calculate this a lot quicker as follows We know that it is depreciating by 12%

We convert 12% to a decimal

We convert 12% to a decimal 12% = 0. 12

Then, we subtract that amount from 1 to get our multiplier (since it’s a depreciation question)

Then, we subtract that amount from 1 to get our multiplier (since it’s a depreciation question) 1 – 0. 12 = 0. 88

After 3 years: 3 A = P (0. 88)

After 3 years: 3 A = 10000 (0. 88)

After 3 years: A = 10000 (0. 681472)

After 3 years: A = 6814. 72

After 3 years: A = 6814. 72 Therefore, after 3 years the car is worth \$6814. 72