Let’s translate your full solution into English: Let the principal amount be \(x\) rupees, which becomes \(4x\) in \(n\) years under compound interest. So, \(4x = x(1 + \cfrac{r}{100})^n\) ⇒ \(4 = (1 + \cfrac{r}{100})^n\) ⇒ \(1 + \cfrac{r}{100} = 2^{\cfrac{2}{n}}\) — (i) Now, suppose the same principal \(x\) becomes \(8x\) in \(y\) years. So, \(8x = x(1 + \cfrac{r}{100})^y\) ⇒ \(2^3 = (2^{\cfrac{2}{n}})^y\) ⇒ \(2^3 = 2^{\cfrac{2y}{n}}\) ⇒ \(\cfrac{2y}{n} = 3\) ⇒ \(y = \cfrac{3n}{2}\) ∴ At a fixed rate of compound interest, if a sum becomes 4 times in \(n\) years, it will become 8 times in \(\cfrac{3n}{2}\) years.