Depreciation & Appreciation (WJEC GCSE Maths & Numeracy (Double Award)): Revision Note

Exam code: 3320

Depreciation & Appreciation

What does depreciation mean?

  • Depreciation is where an item loses value over time

    • E.g. cars, mobile phones, etc

  • Depreciation is usually calculated as a percentage decrease at the end of each year

    • This works the same as compound interest, but with a percentage decrease

What does appreciation mean?

  • Appreciation is the opposite of depreciation

  • It is where an item gains value over time

    • E.g. Houses or antiques

  • Appreciation is usually calculated as a percentage increase at the end of each year

    • This works the same as compound interest

How do I calculate depreciation?

  • A similar method to compound interest can be used

  • Change the multiplier to one which represents a percentage decrease

    • e.g. a decrease of 15% would be a multiplier of 0.85

  • If a car worth £ 16 000 depreciates by 15% each year for 6 years

    • Its value will be 16 000 × 0.856, which is £ 6034.39

  • If you are asked to find the amount the value has depreciated by:

    • Find the difference between the starting value and the new value

How do I calculate appreciation?

  • A similar method to compound interest can be used

  • Use a multiplier which represents a percentage increase

    • e.g. an increase of 7% would be a multiplier of 1.07

  • If an antique worth £ 2 000 appreciates by 7% each year for 5 years

    • Its value will be 2 000 × 1.075, which is £ 2 805.10

  • If you are asked to find the amount the value has appreciated by:

    • Find the difference between the starting value and the new value

Depreciation formula

  • An alternate method is to use the following formula to calculate the final value

    • Final value = P open parentheses 1 minus r over 100 close parentheses to the power of n space end exponent where

      • P is the original amount,

      • r is the % decrease

      • and n is the number of years

    • Note that all of 1 minus r over 100 is the multiplier

      • e.g. 0.75 for a 25% depreciation using r = 25

  • This formula is not given in the exam

Appreciation formula

  • An alternate method is to use the following formula to calculate the final value

    • Final value = P open parentheses 1 plus r over 100 close parentheses to the power of n space end exponent where

      • P is the original amount,

      • r is the % increase

      • and n is the number of years

    • Note that all of 1 plus r over 100 is the multiplier

      • e.g. 1.25 for a 25% appreciation using r = 25

  • This formula is not given in the exam

Worked Example

Mercy buys a car for £20 000. Each year its value depreciates by 15%.

Find the value of the car after 3 full years.

Answer:

Method 1

Identify the multiplier

100% - 15% = 85%

1 - 0.15 = 0.85

Raise to the power of number of years

0.853

Multiply by the starting value

£20 000 × 0.853

£12 282.50

Method 2
Learn and use the formula for the final value: P open parentheses 1 minus r over 100 close parentheses to the power of n space end exponent
Substitute P = 20 000, r = 15 and n = 3 into the formula 

20 space 000 open parentheses 1 minus 15 over 100 close parentheses cubed

£12 282.50

Worked Example

Becky buys a house for £260 000. Its value appreciates by 2% each year for 5 years.

Find the new value of the house to the nearest thousand pounds.

Answer:

Method 1

Identify the multiplier

100% + 2% = 102%

1.02

Raise to the power of number of years

1.025

Multiply by the starting value

£260 000 × 1.025 = £287 061

Round to the nearest thousand pounds

£287 000

Method 2
Learn and use the formula for the final value: P open parentheses 1 plus r over 100 close parentheses to the power of n space end exponent
Substitute P = 260 000, r = 2 and n = 5 into the formula 

260 space 000 open parentheses 1 plus 2 over 100 close parentheses to the power of 5 equals £ 287 space 061

Round to the nearest thousand pounds

£287 000

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