Question

A certain sum quadruples in 3 years at compound interest, interest being compounded annually. In how many years will it become 64 time itself?

Answer 1

ANSWER:  

In 9 years, amount becomes 64 times of itself.

SOLUTION:

Given, a certain sum quadruples in 3 years at compound interest, interest being compounded annually.  

We know that, When interest is compound annually:

$\text { Amount }=P\left(1+\frac{R}{100}\right)^{n}$

Given that,

Principal = Rs.100%

Amount = Rs.400

Rate = r%

Time = 3 years

By substituting the values in above formula, we get,

$400=100 \times\left[1+\left(\frac{R}{100}\right)\right]^{3}$

$4=1\left[1+\left(\frac{R}{100}\right)\right]}^{3}$ --- eqn 1

If sum become 64 times in the time n years then,

$64=\left(1+\left(\frac{R}{100}\right)\right)^n$

$4^{3}=\left(1+\left(\frac{R}{100}\right)\right)^{n}$ --- eqn 2

Using equation (1) in (2), we get

$\begin{array}{c}{\left(\left[1+\left(\frac{R}{100}\right)\right]^{3}=\left(1+\left(\frac{R}{100}\right)\right)^{2}\right.} \\ {\left[1+\left(\frac{R}{100}\right)\right]^{9}=\left(1+\left(\frac{R}{100}\right)\right)^{n}}\end{array}$

Thus, n = 9 years by comparing on both sides.

Hence, in 9 years, amount becomes 64 times of itself.