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:
[tex]\text { Amount }=P\left(1+\frac{R}{100}\right)^{n}[/tex]
Given that,
Principal = Rs.100%
Amount = Rs.400
Rate = r%
Time = 3 years
By substituting the values in above formula, we get,
[tex]400=100 \times\left[1+\left(\frac{R}{100}\right)\right]^{3}[/tex]
[tex]4=1\left[1+\left(\frac{R}{100}\right)\right]}^{3}[/tex] --- eqn 1
If sum become 64 times in the time n years then,
[tex]64=\left(1+\left(\frac{R}{100}\right)\right)^n[/tex]
[tex]4^{3}=\left(1+\left(\frac{R}{100}\right)\right)^{n}[/tex] --- eqn 2
Using equation (1) in (2), we get
[tex]\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}[/tex]
Thus, n = 9 years by comparing on both sides.
Hence, in 9 years, amount becomes 64 times of itself.
