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 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
Variable Rate of Compound Interest
Difference of Compound Interest and Simple Interest
Practice Test on Compound Interest
● 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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