Question 53

A sum becomes 8 times of itself in 7 years at the rate of compound interest (interest is compounded annually). In how many years will the sum becomes 4096 times of itself?

Solution

Let Principal be Rs.P
Then, Amount after 7 years = Rs.8P
Let rate of interest = R%
$$P(1+\dfrac{R}{100})^7 = 8P$$

$$(1+\dfrac{R}{100})^7 = 8$$

$$(1+\dfrac{R}{100}) = 8^{\frac{1}{7}}$$ -- (1)

Let Amount after T years be Rs.4096P
$$P(1+\dfrac{R}{100})^T = 4096P$$

=> $$(1+\dfrac{R}{100})^T = 4096$$

Substituting (1) in above equation
=> $$(8^\frac{1}{7})^T = 8^4$$
=> $$8^\frac{T}{7} = 8^4$$
=> $$\dfrac{T}{7} = 4$$
=> $$T = 28$$
Therefore, In 28 years, the given principal will become 4096 times of itself.


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