Uniform Rate of Growth and Depreciation | How to find Uniform Rate of Increase or Decrease?

# Uniform Rate of Growth and Depreciation | How to find Uniform Rate of Increase or Decrease?

Table of Contents

Practice the Questions based on the Uniform Rate of Growth and Depreciation from here. Learn about the Concept of Uniform Rate of Growth and Depreciation better by going through this entire article. You will find How to Apply Principal of on Combination of Uniform Rate of Growth and Depreciation. Check out Formula, Solved Examples on the concept of Uniform Rate of Increase or Decrease from this article. Get Step by Step Solutions for all the Problems provided and get a good hold on the concept.

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## How to find the Uniform Rate of Increase or Decrease?

Let us 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 then 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%.

### Solved Examples on Uniform Rate of Growth or Depreciation

1. The current population of a 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:

Initial Population = 60,000

r1 =10%, r2 = 5%

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

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

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

= 60,000(1.1)(0.95)

= 62,700

Therefore, the Population after 2 Years is 62,700.

2. The count of a certain breed of bacteria was found to increase at the rate of 4% per hour and then decrease by 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 2,00,000.

Solution:

Since the Population of bacteria increases and decreases we use the formula

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

= 2,00,000(1+4/100)(1-2/100)

= 2,00,000(1+0.04)(1-0.02)

= 2,00,000(1.04)(0.98)

= 2,03,840

Bacteria at the end of 2 hours is 2,03,840

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

r1 = 10% r2 = 15%

Since the Price of a Car depreciates we use the formula

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

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

= 2,50,000(90/100)(85/100)

= 2,50,000(0.9)(0.85)

= $1,91,250 Value of Car after 2 Years is$1,91,250.

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