Pipe A and B fill a tank in 24 hours. Pipe A and C fill same tank in 16 hours. Pipe A can empty a tank in 20 hours. Find the time taken by pipe B and C together to fill a tank.

The correct answer is $\frac{240}{49}\text{hours}$.

Step 1: Determine the rate of Pipe A.
Let the capacity of the tank be 1 unit.
Since Pipe A empties the tank in 20 hours, its rate is negative (emptying rate):

$\text{Rate of A}=-\frac{1}{20}\text{tank/hour}$

Step 2: Find the rate of Pipe B.
Pipes A and B together fill the tank in 24 hours:

$\text{Rate of (A + B)}=\frac{1}{24}\text{tank/hour}$

Substitute the rate of Pipe A into the combined rate equation:

$-\frac{1}{20}+\text{Rate of B}=\frac{1}{24}$

$\text{Rate of B}=\frac{1}{24}+\frac{1}{20}$

Taking the LCM of 24 and 20, which is 120:

$\text{Rate of B}=\frac{5+6}{120}=\frac{11}{120}\text{tank/hour}$

Step 3: Find the rate of Pipe C.
Pipes A and C together fill the tank in 16 hours:

$\text{Rate of (A + C)}=\frac{1}{16}\text{tank/hour}$

Substitute the rate of Pipe A into the combined rate equation:

$-\frac{1}{20}+\text{Rate of C}=\frac{1}{16}$

$\text{Rate of C}=\frac{1}{16}+\frac{1}{20}$

Taking the LCM of 16 and 20, which is 80:

$\text{Rate of C}=\frac{5+4}{80}=\frac{9}{80}\text{tank/hour}$

Step 4: Calculate the combined rate of Pipe B and Pipe C.
Add the filling rates of Pipe B and Pipe C:

$\text{Rate of (B + C)}=\frac{11}{120}+\frac{9}{80}$

Taking the LCM of 120 and 80, which is 240:

$\text{Rate of (B + C)}=\frac{22+27}{240}=\frac{49}{240}\text{tank/hour}$

Step 5: Find the total time taken by Pipes B and C together.
The time required to fill the tank is the reciprocal of their combined rate:

$\text{Time taken}=\frac{1}{\left(\frac{49}{240}\right)}=\frac{240}{49}\text{hours}$

Hence, the time taken by Pipe B and Pipe C together to fill the tank is $\frac{240}{49}\text{hours}$.