A sum of money doubles itself in 10 years. In how many years will it be triple at the same rate of simple interest?

Correct option is A

Given:
Total required time to double a sum of money itself = 10 years
Formula used:
SI = $$\frac{P\times R \times T }{100}$$​​
Where, S.I. = Simple Interest
p = The sum of money
r = Rate of interest
t = Total time
Solution:
Suppose, the sum of money = X
The sum of the money becomes double itself in ten years. Then, the simple interest = 2X - X = X
Let, rate of interest = r
Then,

​$$X = \frac{(X × r × 10)}{100}\\\ \\=> X = \frac{(X × r)}{10} \\\ \\=> r = 10$$​​
The rate of interest = 10%
Now, the sum of money triple itself. Then, the total amount = 3X
Total simple interest = 3X - X = 2X
Suppose, the required time = T
Therefore,

​$$2X =\frac{ (X × 10 × T)}{100}\\\ \\=> 2X = \frac{(X × T)}{10}\\\ \\=> 2 = \frac{T}{10}\\\ \\=> T = 2 × 10 = 20$$​​
∴ The sum of money will triple itself in
20 years.