A sum of money, put at simple interest doubles itself in 8 years. The same sum will triple itself in:

1. 16 years

2. 12 years

3. 24 years

4. 18 years

Given:

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,

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

When A becomes 3P, ⇒ S.I. = 2P

Let the time be $t$ years

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

Hence, option 1 is the correct option.