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 =
where
P is the original amount,
r is the % decrease
and n is the number of years
Note that all of
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 =
where
P is the original amount,
r is the % increase
and n is the number of years
Note that all of
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:
Substitute P = 20 000, r = 15 and n = 3 into the formula
£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:
Substitute P = 260 000, r = 2 and n = 5 into the formula
Round to the nearest thousand pounds
£287 000
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