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

Compound Interest when Interest is Compounded Yearly

Compound Interest when Interest is Compounded Half-Yearly

Compound Interest when Interest is Compounded Quarterly

Problems on Compound Interest

Variable Rate of Compound Interest

Difference of Compound Interest and Simple Interest

Practice Test on Compound Interest

Uniform Rate of Growth

Uniform Rate of Depreciation

● Compound Interest - Worksheet

Worksheet on Compound Interest

Worksheet on Compound Interest when Interest is Compounded Half-Yearly

Worksheet on Compound Interest with Growing Principal

Worksheet on Compound Interest with Periodic Deductions

Worksheet on Variable Rate of Compound Interest

8th Grade Math Practice 

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