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

● Compound Interest

Compound Interest

Compound Interest with Growing Principal

Compound Interest with Periodic Deductions

Compound Interest by Using Formula