A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to 8 times?
A. 9 years B. 8 years C. 27 years D. 12 years Answer: Option DShow Answer
Solution(By Apex Team)
$\begin{array}{l}\text{Let,}\\
\text{P}=\text{Rs.}\ 100\\
\text{A}=\text{Rs.}\ 200\\
\text{Rate}=r\%\\
\text{Time}\ \left(t\right)=4\ \text{years}\\
\text{Now,}\\
\begin{array}{l}A=P\times\left[1+\left(\frac{r}{100}\right)\right]^n\\
200=100\times\left[1+\left(\frac{r}{100}\right)\right]^4\\
2=\left[1+\left(\frac{r}{100}\right)\right]^4——(i)\\
\end{array}\end{array}$
If sum become 8 times in time 8 years.
Then,
$\begin{array}{l}8=\left(1+\left(\frac{r}{100}\right)\right)^n\\
2^3=\left(1+\left(\frac{r}{100}\right)\right)^n——(ii)\\
\text{Using equetion (i) put in (ii)}\\
\text{we get;}\\
\begin{array}{l}\left(\left[1+\left(\frac{r}{100}\right)\right]^4\right)^3=\left(1+\left(\frac{r}{100}\right)\right)^n\\
\left[1+\left(\frac{r}{100}\right)\right]^{12}=\left(1+\left(\frac{r}{100}\right)\right)^n\\
\text{Thus, n}=12\ \text{years.}\end{array}\end{array}$
HI 
EN 