A sum of money, lent out at simple interest, doubles itself in 8 years. Find:

(i) the rate of interest.

(ii) in how many years will the sum become triple (three times) of itself at the same rate percent?

(i) Let the Principal amount be ₹ P.

A = 2P

T = 8 years

Let the rate be $r$.

$\because \text{A = S.I. + P}\\[1em] \Rightarrow \text{S.I. = A - P}\\[1em] \Rightarrow \text{S.I. = 2P - P}\\[1em] \Rightarrow \text{S.I. = P}$

And we know,

$\text{S.I.} = ₹ \Big(\dfrac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{P} = ₹ \Big(\dfrac{P \times r \times 8}{100}\Big)\\[1em] \Rightarrow \cancel{P} = ₹ \dfrac{\cancel{P} \times r \times 8}{100}\\[1em] \Rightarrow 1 = ₹ \dfrac{8r}{100}\\[1em] \Rightarrow r = \dfrac{100}{8}\%\\[1em] \Rightarrow r = \dfrac{25}{2}\%\\[1em] \Rightarrow r = 12.5\%$

Hence, rate of interest = $12.5\%$.

(ii) When A becomes 3P, ⇒ S.I. = 2P

Let the time be $t$ years

$\because \text{S.I.} = ₹ \Big(\dfrac{P \times R \times T}{100}\Big)\\[1em] \Rightarrow \text{2P} = ₹ \Big(\dfrac{P \times 25 \times t}{2 \times 100}\Big)\\[1em] \Rightarrow 2\cancel{P} = ₹ \dfrac{\cancel{P} \times 25 \times t}{200}\\[1em] \Rightarrow t = \dfrac{400}{25}\\[1em] \Rightarrow t = 16\\[1em]$

Hence,the time is 16 years.