A person deposited some amount of money in the bank at simple interest. After 20 years, the amount became nine times the amount deposited. In how many years will the amount become 13 times the amount if the interest rate remains the same?

The correct answer/option is 30.

Let the principal amount deposited by the person be $P$ and the rate of simple interest be $r$% per annum.

According to the formula for simple interest, the simple interest ($SI$) earned over $t$ years is given by:
$SI=\frac{P\times r\times t}{100}$
The total amount ($A$) after $t$ years is the sum of the principal and the interest:
$A=P+SI=P+\frac{P\times r\times t}{100}=P\left(1+\frac{r\times t}{100}\right)$

Step 1: Find the rate of interest using the first condition.
We are given that after $t=20$ years, the amount becomes $9$ times the principal amount ($A=9P$).
Substituting these values into the amount formula:
$9P=P+SI$
This means the simple interest earned in 20 years is:
$SI=9P-P=8P$

Now, substitute $SI=8P$ and $t=20$ into the simple interest formula:
$8P=\frac{P\times r\times 20}{100}$
We can divide both sides by $P$ (since $P\ne 0$):
$8=\frac{20r}{100}$
Simplify the fraction:
$8=\frac{r}{5}$
Solving for $r$:
$r=8\times 5=40$% per annum.

Step 2: Find the time required for the amount to become 13 times the principal.
Let the required time be $t_2$ years. Here, the new amount ($A_2$) is $13$ times the principal ($A_2=13P$).
The new simple interest earned is:
$S{I}_2=13P-P=12P$

Using the simple interest formula with the rate $r=40$%:
$12P=\frac{P\times 40\times t_2}{100}$
Dividing both sides by $P$:
$12=\frac{40t_2}{100}$
Simplify the fraction:
$12=\frac{2t_2}{5}$
Solving for $t_2$:
$2t_2=12\times 5$
$2t_2=60$
$t_2=30$ years.

Thus, the amount will become 13 times the principal in 30 years.