In this article, you will learn about the uniform rate of growth and depreciation. Depreciation refers to the decrease in the value of the fixed assets, so it becomes important to measure this decrease in the value of an asset and also account for it. The relative increase in a quantity is called growth and a growth per unit of time is called the rate of growth.
There are various methods of providing depreciation through the depreciation formula. We can classify the methods into Uniform Charge Methods, Declining Charge Methods, and Other Methods. On this page, you will practice the problems based on the Uniform Rate of Growth and Depreciation from here and learn about the Concept of Uniform Rate of Growth and Depreciation better by going through this entire article, learn How to Apply Principal of Compound Interest to the Combination of a Uniform Rate of Growth and Depreciation.
Also learn important Formulas, Solved Examples on the concept of Uniform Rate of Increase or Decrease from this article. You have Step-by-Step Solutions for all the Problems provided and get a good knowledge of them.
Also, Read:
- Uniform Rate of Growth
- Uniform Rate of Depreciation
- Factors Affecting Interest
Uniform Rate of Growth and Depreciation – Definitions
Uniform Rate of Growth: The uniform rate of growth is defined as, we call the Increase in Quantity as Growth, and the Growth Per Unit of Time is known as the Rate of Growth. If a growth rate occurs at the same Rate then we call it a Uniform Increase or Uniform Growth Rate.
Uniform Rate of Depreciation: Depreciation will be 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.
How do you find the Uniform Rate of Increase or Decrease?
In the below modules, we will discuss how to find the Uniform Rate of Growth or Depreciation in detail in the below modules.
If a quantity P grows at the rate of r1% in the first year and depreciates at r2% in the second year and grows at r3% in the third year, the quantity becomes Q after 3 years.
Take r/100 with a positive sign for each growth or appreciation of r % and r/100 with a negative sign for depreciation of r%.
Also, Refer:
- To find Principal when Time Interest and Rate are Given
- In Simple Interest when the Time is given in Months and Days
How to apply the Principle of Compound on Uniform Rate of Growth and Depreciation? | Uniform Rate of Growth and Depreciation Formula
Let us discuss, how to find the uniform rate of growth using the principle of compound interest in detail in the below modules. The following are a few important formulas:
- When the growth of organization or production in any particular industry is taken into consideration and the formula Q = P(1 + R/100)^n is used as:
Production after n years = Initial (original) production (1 + r/100)^n, where the rate of growth in production is r%. - The growth of plants, growth of inflation, the formula Q = P (1 + R/100)^n is used.
- If the present value P of a quantity increases 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 growth = Q – PÂ i.e., P{(1 + r/100)^n – 1}.
(i) If the present population of a town = P, rate of growth of population = r % p.a. so the population of the town after n years is Q, where
Q = P(1 + r/100)^n and the growth of population = Q – P that is P{(1 + r/100)^n – 1}.
(ii) If the present price of a house = P, rate of appreciation in the price of the house = r % p.a. then the house price after n years is Q, where
Q = P(1 + r/100)^n and appreciation in price = Q – P that is P{(1 + r/100)^n – 1}. - If the present value 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 uniform rate of decrease in 1st year be r1%, in 2nd year be r2%, in 3rd year be r3%and ….. at last in the nth year be rn%, then the total decrease after n years = P(1+r1/100)(1+r2/100)(1+r3/100)….(1+rn/100)
Students of 10th Grade Math will find the concept extremely beneficial as we have covered step by step.
Uniform Rate of Growth and Depreciation Examples
Problem 1:
A man starts a business with a capital of $1000000. He incurs a loss of 4% during the first year, he makes a profit of 5% during the second year on his remaining investment. During the third year, he makes a profit of 10% on his new capital. What is his total profit at the end of three years?
Solution:
As given in the question,
The initial capital P = 1000000,
The loss for the first year = r1% is 4%,
The gain for the second year = r2% is 5%
The gain for the third year = r3% = 10%
Now, we will find the total profit at the end of three years.
We know the formula, Q = P(1 – r1/100)(1 + r2/100)(1 + r3/100)
Substitute the values in the formula, we get
Q = $1000000(1 – 4/100)(1 + 5/100)(1 + 10/100)
Therefore, Q = $1000000 × 24/25 × 21/20 × 11/10
Q = $200×24×21×11 = $1108800
Next, the profit at the end of three years is $1108800 – $1000000 = $108800.
Hence, at the end of three years, the profit is $108800.
Problem 2:
The price of a bike is $1,50,000. The value of the bike depreciates by 10% at the end of the first year and after that, it depreciates by 15%. What is the value of the bike after 3 years?
Solution:
In the given question,
The Initial Price of the bike = $ 1,50,000
The value of the bike depreciates by r1 is 10%
After that it will depreciatted by r2 is15%
For finding the Price of a bike depreciates, we use the formula is
Q = P(1-r1/100)(1-r2/100)
Substituting the value in a formula, we get
Q = 1,50,000(1-10/100)(1-15/100)
Q = 1,50,000(90/100)(85/100)
Q = 1,50,000(0.9)(0.85)
Q = $1,14,250
So, the value of a bike after 2 Years is $1,14,250.
Problem 3:
The present population of the town is 65,000. The population increases by 10 percent in the first year and decreases by 10% in the second year. Find the population after 2 years.
Solution:
The data given in the question is,
The initial population P = 65,000.
The population increase for the first year is r1% = 20%
It will be decrease for the second year is r2% = 10%.
Now, we need to find the population after 2 years.
The formula for finding the Population after 2 years is
Q = P(1 + r1/100)(1 – r2/100)
After substituting the values, the formula is
Q = 65,000(1 + 20/100)(1 – 10/100)
Q = 65,000(1 + 1/10)(1 – 1/10)
Next, the value of Q is 65,000(11/10)(9/10)
Q = 64,250
Therefore, the population after 2 years is 64,250.
Problem 4:
The population of a town is increasing at the rate of 5% per annum. What is the population of the town on the basis after three years, if the present population is 18000?
Solution:
Given that,
The Initial population (P) is 18000
R = The Rate of growth of the population is 5% per annum
n is the Number of years =3
Now, we will find the basic population of the town after three years.
We know the formulae,
The Population after ‘n’ years is P(1+R/100)^n.
Substitute the given values in the above formula. We get,
The Population after 3 years is 18000 x ( 1+ 5/100)^3.
i. e., P = 18000 x (1 +1/20)^3
P= 18000 x (21/20)^3
So, the population after 3 years is 18000 x (21/20) x (21/20) x (21/20) = 20837.
Therefore, the population after 3 years is 20837
Problem 5:
The price of a motor vehicle depreciates by 10% every year. What is the percent value if the price of the car reduces 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%.
FAQs on Uniform Rate of Growth and Depreciation
1. What is depreciation explain with an example?
Depreciation is defined as the reduction or decrease of the recorded cost of a fixed asset in a systematic manner until the value of the asset becomes zero or negligible. Examples of fixed assets are buildings, furniture, office equipment, machinery, etc.,
2. What is a uniform rate of growth?
We call the Increase in Quantity as Growth and the Growth Per Unit Time is known as the Rate of Growth. The Growth Rate will occur at the same Rate then we call it as Uniform Increase or Uniform Growth Rate.
3. Explain Compound Interest?
Compound interest means the interest imposed on a loan or deposit amount. This 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. This will be the main difference between compound interest and simple interest.
4. What is the formula of compound interest with an example?
The compound interest formula is given below:
Compound Interest = Amount – Principal
Where the amount is given by:
A = P(1 + r/n)^{nt} P = Principal
r is an annual interest rate as a decimal
n is the number of compounding periods
t is Time (in years)
For example, If john deposits Rs. 4000 into an account by paying 6% annual interest compounded, then the money will be in his account after five years will be calculated as:
First, Substituting the values in the formula, P = 4000, r = 0.06, n = 4, and t = 5 in A. A = P(1 + r/n)^{nt}, we get A = Rs. 5387.42
5. Which depreciation method is best?
Straight-Line Method is one of the methods in the depreciation method. This method is the most commonly used method for calculating depreciation.