# Uniform Rate of Growth and Depreciation

We will discuss here about the principle of compound interest in the combination of uniform rate of growth and depreciation.

If a quantity P grows at the rate of r$$_{1}$$% in the first year, depreciates at the rate of r$$_{2}$$% in the second year and grows at the rate of r$$_{3}$$% in the third year then the quantity becomes Q after 3 years, where

Take $$\frac{r}{100}$$ with positive sign for each growth or appreciation of r% and $$\frac{r}{100}$$ with negative sign for each depreciation of r%.

Solved examples on the principle of compound interest in the uniform rate of depreciation:

1. The present population of a town is 75,000. The population increases by 10 percent is the first year and decreases by 10% in the second year. Find the population after 2 years.

Solution:

Here, initial population P = 75,000, population increase for the first year = r$$_{1}$$% = 10% and decrease for the second year = r$$_{2}$$% = 10%.

Population after 2 years:

Q = P(1 + $$\frac{r_{1}}{100}$$)(1 - $$\frac{r_{2}}{100}$$)

⟹ Q = Present population(1 + $$\frac{r_{1}}{100}$$)(1 - $$\frac{r_{2}}{100}$$)

Q = 75,000(1 + $$\frac{10}{100}$$)(1 - $$\frac{10}{100}$$)

Q = 75,000(1 + $$\frac{1}{10}$$)(1 - $$\frac{1}{10}$$)

Q = 75,000($$\frac{11}{10}$$)($$\frac{9}{10}$$)

⟹ Q = 74,250

Therefore, the population after 2 years = 74,250

2. A man starts a business with a capital of $1000000. He incurs a loss of 4% during the first year. But he makes a profit of 5% during the second year on his remaining investment. Finally, he makes a profit of 10% on his new capital during the third year. Find his total profit at the end of three years. Solution: Here, initial capital P = 1000000, loss for the first year = r$$_{1}$$% = 4%, gain for the second year = r$$_{2}$$% = 5% and gain for the third year = r$$_{3}$$% = 10% Q = P(1 - $$\frac{r_{1}}{100}$$)(1 + $$\frac{r_{2}}{100}$$)(1 + $$\frac{r_{3}}{100}$$) ⟹ Q =$1000000(1 - $$\frac{4}{100}$$)(1 + $$\frac{5}{100}$$)(1 + $$\frac{10}{100}$$)

Therefore, Q = $1000000 × $$\frac{24}{25}$$ × $$\frac{21}{20}$$ × $$\frac{11}{10}$$ ⟹ Q =$200 × 24 × 21 × 11

⟹ Q = $1108800 Therefore, profit at the end of three years =$1108800 - $1000000 =$108800

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