Pipe A can fill 50% of the tank in 6 hours and Pipe B can fill 100% capacity of tank in 18 hours. If Pipe C can empty the full tank in 9 hours, then find the time taken by the all the pipes to fill the tank together (in hours).

To find the time taken by all three pipes working together to fill the tank, we can determine the individual work rates of each pipe.

Let the total capacity of the tank be represented as 1 unit (or a volume of V).

Step 1: Find the rate of Pipe A
Pipe A can fill 50% (which is $\frac{1}{2}$) of the tank in 6 hours.
Therefore, the time taken by Pipe A to fill 100% (the full tank) is:
$6\times 2=12\text{hours}$
So, the rate of Pipe A (part filled per hour) is:
$R_A=\frac{1}{12}$

Step 2: Find the rate of Pipe B
Pipe B can fill the full tank in 18 hours.
So, the rate of Pipe B (part filled per hour) is:
$R_B=\frac{1}{18}$

Step 3: Find the rate of Pipe C
Pipe C is an emptying pipe that can empty the full tank in 9 hours.
Since it empties the tank, we represent its rate as negative.
The rate of Pipe C (part emptied per hour) is:
$R_C=-\frac{1}{9}$

Step 4: Find the net rate when all three pipes are open together
The net rate of filling per hour is:
$R_{\text{net}}=R_A+R_B+R_C$
$R_{\text{net}}=\frac{1}{12}+\frac{1}{18}-\frac{1}{9}$

To simplify this expression, find the Least Common Multiple (LCM) of the denominators 12, 18, and 9. The LCM is 36.
Now, rewrite each fraction with a denominator of 36:
$\frac{1}{12}=\frac{3}{36}$
$\frac{1}{18}=\frac{2}{36}$
$\frac{1}{9}=\frac{4}{36}$

Substitute these back into the rate equation:
$R_{\text{net}}=\frac{3}{36}+\frac{2}{36}-\frac{4}{36}$
$R_{\text{net}}=\frac{3+2-4}{36}$
$R_{\text{net}}=\frac{1}{36}$

Step 5: Calculate the time taken to fill the tank
The time taken to fill the entire tank is the reciprocal of the net rate:
$\text{Time}=\frac{1}{R_{\text{net}}}=36\text{hours}$

Therefore, the correct answer option is 36.