Question 62

A sum of money becomes 3 times in 10 years at the rate of compoundinterest (compounded annually). In how many years will it become 81 times?

Solution

amount with compound interest = $$P(1+\frac{R}{100})^n$$

3P= $$P(1+\frac{R}{100})^{10} $$

3= $$(1+\frac{R}{100})^{10}$$ ->( 1)

amount with compound interest = $$P(1+\frac{R}{100})^n$$

81P= $$P(1+\frac{R}{100})^T$$

$$3^4$$ =$$(1+\frac{R}{100})^T$$ ->( 2)

Equating both the equations

$$((1+\frac{R}{100})^ {10})^4$$ =$$(1+\frac{R}{100})^T$$
T = 40 years

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