Option 5 : 33

**Given:**

A sum of money becomes 3 times in 6 years.

**Formula used:**

S.I = PRT/100

Where S.I = Simple interest, P = Principal, R = Rate, T = time

**Calculation:**

Let the sum be 100x and after 6 years, it will be 300x

So, S.I will be (300x - 100x) = 200x

So, according to the formula, we can say

200x = (100x × R × 6)/100

⇒ R = (100/3)%

Now, according to the question again we can say

The sum becomes 12 times, i.e., it will be 1200x

So, the S.I will be (1200x - 100x) = 1100x

So, from the formula, we can write

1100x = (100x × 100 × T)/(100 × 3)

⇒ T = 33

**∴ The sum will be 12 times itself in 33 years.**

__Alternate Method__

**Formula used:**

(N_{1} - 1)T_{1} = (N_{2} - 1)T_{2}

Where a certain amount of money becomes N_{1} time itself in T_{1} years.

The task is to find the number of years i.e. T_{2} so that the amount becomes N_{2} times itself at the same rate of simple interest

Here, N_{1} = 3, T_{1} = 6

N_{2} = 12, T_{2} = ?

(3 - 1)/6 = (12 - 1)/T

⇒ (2/6) = (11/T)

⇒ T = 33

∴ The sum will be 12 times itself in 33 years.