In this article, You will learn about Depreciation. The word Depreciation refers to the decrease in the value of the fixed assets due to normal wear and tear, efflux of time, or obsolescence due to technology. Thus, it becomes important to measure this decrease in the value of an asset and also account for it. There are various methods of providing depreciation through the depreciation formula. However, we can classify the methods into Uniform Charge Methods, Declining Charge Methods, and Other Methods.

On this page, you will about the concept of a uniform rate of depreciation, the formula, and solved examples of the concept of a uniform rate of depreciation. You will find How to Apply the Principal of Compound Interest on the Uniform Rate of Depreciation. Practice the Questions based on the Uniform Rate of Depreciation from here. you have a Step-by-Step Solution for all the Problems provided and get a good knowledge of this concept.

Read More:

- Uniform Rate of Growth and Depreciation
- To find Time when Principal Interest and Rate are Given
- Factors Affecting Interest

### Uniform Rate of Depreciation – Definition

Depreciation is defined as the reduction of the recorded cost of a fixed asset in a systematic manner until the value of the asset becomes zero or negligible. The total amount of depreciation each year, which is represented as a percentage, is called the depreciation rate. If the rate of decrease is uniform, we called it a uniform decrease or depreciation. The r/100 with a negative sign is depreciation of r%.

### How to Apply the Principle of Compound on Uniform Rate of Depreciation? | Uniform Rate of Depreciation Formula

Let us discuss, how to find the uniform rate of depreciation using the principle of compound interest in detail in the below modules.

**Compound Interest:** Compound interest means the interest imposed on a loan or deposit amount. It is the most commonly used concept in our daily life. The compound interest for an amount depends on the Principal and interest gained over periods.

The compound interest formula is, CI= Q-P.

Where, Q = P (1 + r/n)^(nt) – P.

- If the present value P of a quantity decreases at the rate of r% per unit of time then the value Q of the quantity after n units of time is given by Q = P(1 – r/100)^n and the depreciation in value is P{1 – (1 – r/100)^n}.
- If the present population of a car = P, the rate of depreciation is r% per annum then the price of the car after n years is Q, where Q = P(1 – r100)^n and the depreciation in value is P{1 – (1 – r/100)^n}.

Also, Refer: What is Similar Interest

### How is the Uniform Rate of Depreciation Calculated?

In this section, we will discuss how to calculate a Uniform Rate of Decrease or Depreciation in detail:

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

Q = P(1-r/100)^n

The Depreciation in Value is P – Q.

So, the depreciation value is P – P(1-r/100)^n = P{1-(1-r/100)^n}

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

Students of 10th Grade Math will find the concept extremely beneficial as we have covered step by step.

### Uniform Rate of Depreciation Examples with Answers

**Problem 1:**

The price of a car is $ 2,50,000. The value of the car depreciates by 10% at the end of the first year and after that, it depreciates by 15%. What will be the value of the car after 2 years?

**Solution:
**Given that,

The initial Price of the Car is $ 2,50,000.

The value of the car depreciates by (r1) 10%.

The value of the car depreciates by (r2 ) 15%.

Now, we need to find the value of the car after 2 years.

Since the Price of a Car depreciates, we use the formula is

Q = P(1-r1/100)(1-r2/100)

Substitute the given values in the above formula, we get

Q = 2,50,000(1-10/100)(1-15/100)

Q= 2,50,000(90/100)(85/100) = 2,50,000(0.9)(0.85)

Q= $1,91,250

Hence, the value of a car after 2 Years is $1,91,250.

**Problem 2:**

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 the value of 4years after construction?

**Solution:
**As given in the question, the data is

The constructed flat cost is (P) = Rs. 1,20,000

The depreciating rate is r = 5% per annum.

The time period is n = 4 Years.

Now, we will find the value after 4years of construction.

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

Substitute the input values in the above formula. we get,

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

Q = 1,20,000(95/100)^4 = 1,20,000(0.8145)

Q = Rs. 97,740.

Hence, the value of a Residential Flat after 4 Years would be Rs. 97,740.

**Problem 3:**

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

**Solution:
**The given data is,

The Price will be P

The depreciation rate is r = 10% every year.

Time is n = 2 Years

We will find the reduced percentage of the price of the car after 2years.

We know the formula Q = P(1-r/100)^n.

Substitute the values in the above formula and we have the equation,

Q = P(1-10/100)^2

Q = P(90/100)^2 = P(9/10)(9/10)

Q = 81P/100

So, the Reduction in Price = P – 81P/100 = 19P/100

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

Thus, the percent reduction in price after 2years is 19%.

**Problem 4:**

The cost of a school bus depreciates by 8 % every year. If its present worth is $ 27,000. After three years, What is the value?

**Solution:
**The given data in the question is,

P = $27,000

r = 8%

n = 3 years

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

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

Q = 27,000(92/100)^3 = 27,000(0.92)^3

Q = 27,000(0.778) = $ 21, 024

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

**Problem 5:**

The cost of a school bus depreciates by 10% every year. If its present worth is $ 18,000. Find its value after three years?

**Solution:
**In the question, the given data is

The present population P = 18,000,

The depreciation Rate is (r) = 10.

The Unit of time being year (n) is 3.

Now, we need to find the value after 3 years.

Applying the formula of depreciation we get:

Q = P(1 – r100)^n

Substituting the values, we get

Q = $18,000(1 – 10/100)^3

Q = $18,000(1 – 1/10)^3 = $18,000(9/10)^3

Q = $18,000 × (9/10) × (9/10) × (9/10)

i.e., Q = $18,000 × (9×9×9/10×10×10)

The value Q is $18 × 81 × 9 = $13,122.

Therefore, the value of the school bus after 3years will be $ 13,122.

### FAQs on Uniform Rate of Depreciation

**1. What are the features of Depreciation?**

The following are the 3 principal features of Depreciation:

1. Depreciation is decreasing the book value of fixed assets.

2. Depreciation involves the loss of value of assets due to the passage of time and obsolescence.

3. Depreciation is an ongoing process until the end of the life of assets.

**2. How do determine the depreciation rate?**

The annual depreciation rate is calculated using the formula:(100 x Number of Periods In Year)/Number of periods in expected life. Each period of depreciation amount is calculated by using the formula is annual depreciation rate/ number of periods in the year.

**3. What are the 3 depreciation methods?**

Depreciation accounts for decreases in the value of a company’s assets over time. The four depreciation methods include straight-line, declining balance, the sum of the years’ digits, and units of production. The most commonly used method is the straight-line method, it is the simplest method to use.

**4. What is depreciation explain with an example?**

In accounting terms, depreciation is defined as the reduction of the recorded cost of a fixed asset in a systematic manner until the value of the asset becomes zero or negligible. An example of fixed assets are buildings, furniture, office equipment, machinery, etc.,