In this article, you will learn about the uniform rate of growth. The relative increase in a quantity is called growth and a growth per unit of time is called the rate of growth. Practice the Questions based on the Uniform Rate of Growth from here. Learn about the Concept of a Uniform Rate of Growth better by going through this entire article.

You will learn How to Apply Principal of Compound Interest to the Combination of a Uniform Rate of Growth, Formula, Solved Examples of the concept of a Uniform Rate of Increase or Decrease from this article. On this page, you have step-by-step solutions for all the problems provided and get a good hold of the concept.

Also, Read:

### Uniform Rate of Growth – Definition & Meaning

It is defined as, when 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. An increase in population, an increase in the number of students in academic institutions, increase in production in the fields of agriculture and industry are examples of uniform Growth or increase.

**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. This will be the main difference between compound interest and simple interest.

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

Where, Q =Â P (1 + r/n)Â ^{(nt)} – P.

**Simple Interest:** The interest is not added to the principal while calculating the interest during the next period in simple interest while in the compound interest the interest is added to the principal to calculate the interest.

### How to apply the Principle of Compound on Uniform Rate of Growth? | Uniform Rate of Growth Formula

Let us discuss, how to find the uniform rate of growth using the principle of compound interest in detail in the below modules. In general, the word growth can be used in several ways, like

(i) In a Country, the growth of industries.

(ii) The growth of plants or inflation, etc are rapidly growing.

- The rate of growth occurs at the same rate, we call it uniform growth or uniform increase. 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%. - In the same way, the growth of plants, growth of inflation, the formula Q = P (1 + R/100)^nÂ 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. then 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}. - Take r/100 with a positive sign for each growth or appreciation of r % and takeÂ r/100 with a negative sign for each depreciation of r%.

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

### Uniform Rate of Growth Examples

**Problem 1:
**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 two years, if the present population is 16000?

**Solution:
**Given that,

P =Initial population = 16000

R =Rate of growth of population =5% per annum

n is Number of years =2

Now, we will find the basic population of the town after two 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 2 years is 16000 x ( 1+ 5/100)^2.

i. e., P = 16000 x (1 +1/20)^2

P= 16000 x (21/20)^2

So, the population after 2 years is 16000 x (21/20) x (21/20) = 40 x 21 x 21 = 17640.

Therefore, the population after 2 years is 17640.

**Problem 2:**

The present population of Berlin is 2000000. If the rate of the population of Berlin is increased at the end of a year is 2% of the population at the beginning of the year, find the population of Berlin after 3 years?

**Solution:
**As given in the question, the present population of berlin is 2000000

**Now, we need to find the Population of Berlin after 3 years.**

As we know that, Q = P(1 + r/100)^n.

The given values are substituted in the above formula, we get

Q = 2000000(1 + 2/100)^3

Q= 2000000(1 + 1/50)^3

i.e., Q= 2000000(51/50)^3

So, the population Q is 2000000(51/50) Ã— (51/50) Ã— (51/50)

Q = 2122416

Hence, the population of Berlin after 3 years is 2122416.

**Problem 3:**

If an amount of $5,000 is deposited into a savings account at an annual interest rate of 5%, compounded monthly. Find the value of the investment after 10 years?

**Solution:**

Given in the question, the data is

P = 5000.

r = 5/100 = 0.05 (decimal).

n = 12.

t = 10.

The formula is, Q = P(1 + r/100)^n.

Put the given values into the formula, we get as

A or Q = 5000 (1 + 0.05 / 12)^(12 * 10).

So the Q value is 8235.05

Hence, the investment balance amount after 10 years is $8,235.05.

**Problem 4:**

The count of a certain breed of bacteria was founded to increase at the rate of 2% per hour. What is the value of bacteria at the end of 2 hours if the count was initially 600000?

**Solution:
**Given that, the population of bacteria increases at the rate of 2% per hour.

Now, we need to find out the bacteria count at the end of 2 hours.

we use the formula is,Â A = P(1 + R/100)^n

After substituting the value is,

A = 600000 (1 + 2/100)^2.

=600000(1 + 0.02)^2

=600000(1.02)^2 = 624240.

Hence, at the end of 2 hours, the population is 624240.

**Problem 5:**

The current population of the town is 60,000. The population increases by 10 percent in the first year and decreases by 5% in the second year. Find the population after 2 years?

**Solution:
**Given that, the initial Population is 60,000

Increased by r1 is 10%,

Decreased by r2 is 5%

The formula is Population after 2 Years Q = P(1+r1/100)(1-r2/100).

Substitute the values in a formula, we get

Q= 60,000(1+10/100)(1-5/100)

Q= 60,000(1+0.1)(1-0.05)

Q= 60,000(1.1)(0.95)

Q= 62,700

So, the Population after 2 Years will be 62,700.

### FAQ’s on Uniform Rate of Growth

**1. How do you calculate the uniform rate?**

Applying the principle of compound interest, we can find the values, when subjected to a uniform rate of growth by the formula Q = P(1 + r/100)^n-1. The population of a town grows uniformly at the rate of 4% every year.

**2. What are the applications of Compound Interest?**

Some of the applications of compound interest are listed below:

1. Increase in population or decrease in population.

2. Growth of bacteria.

3. Rise the value of an item.

4. The value of an item will be depreciation.

**3. How are simple interest and compound interest different?**

The interest is typically expressed as a percentage, it can be either simple interest or compound interest. Based on the principal amount of a loan or deposit will be calculated interest is called Simple interest. Whereas the compound interest is based on the principal amount and the interest that accumulates on it in every period.