Uniform Rate of Depreciation | How to find Rate of Depreciation?

# Uniform Rate of Depreciation | How to find Rate of Depreciation?

## How to Calculate the Rate of Uniform Decrease or Depreciation?

Let us discuss in detail how to find the Uniform Rate of Decrease or Depreciation in the coming sections.

If the Present Value P of a Quantity Decreases at the rate of r % per unit of time then Value Q of the Quantity after n units of time is given by

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Q = P(1-r/100)n

Depreciation in Value = P – Q

= P – P(1-r/100)n

= P{1-(1-r/100)n}

Efficiency of a Machine after regular use, Decrease in the Value of Furniture, Buildings, Decrease in the Number of Diseases, etc. come under Uniform Rate of Decrease or Depreciation.

### Solved Examples on Uniform Rate of Depreciation

1. The value of a residential flat constructed at a cost of Rs.1,20,000 is depreciating at the rate of 5% per annum. What will be its value 4 years after construction?

Solution:

From the Given Data P = Rs. 1,20,000

r = 5%

n = 4 Years

We know the formula to find Q = P(1-r/100)n

Substitute the input values in the above formula and we have the equation as such

Q = 1,20,000(1-5/100)4

= 1,20,000(95/100)4

= 1,20,000(0.8145)

= Rs. 97,740

Therefore, the Value of a Residential Flat after 4 Years would be Rs. 97,740.

2. The price of a motor vehicle depreciates by 10% every year. By what percent will the price of the car reduce after 2 years?

Solution:

From the given data

Consider the Price be P

r = 10%

n = 2 Years

We know the formula to find Q = P(1-r/100)n

Substitute the input values in the above formula and we have the equation as such

Q = P(1-10/100)2

=P(90/100)2

= P(9/10)(9/10)

= 81P/100

Reduction in Price = P -81P/100

= 19P/100

Percent Reduction in Price = (19P/100)/P*100%

= 19%

3. The cost of a school bus depreciates by 8 % every year. If its present worth is $27,000. What will be its value after three years? Solution: From the given data P =$27,000

r = 8%

n = 3 years

Formula to Calculate the Price of Depreciated Value Q = P(1-r/100)n

= 27,000(1-8/100)3

= 27,000(92/100)3

= 27,000(0.92)3

= 27,000(0.778)

= $21, 024 The Cost of a School Bus after the Depreciation is$ 21,024.

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