Two inlet pipes A and B and one outlet pipe O are attached to a cistern.Pipe A fills the cistern in 8 hours, pipe B in 12 hours, and the outlet O can empty a full cistern in 24 hours. If all three pipes are opened simultaneously, how long will it take to fill the cistern?

The correct answer is Option A: 6 hours.

Let's solve the problem step-by-step by determining the work done by each pipe per hour.
First, we find the rate of work for each pipe:

Pipe A fills the cistern in 8 hours. Therefore, the rate of work of Pipe A is:
Rate of Pipe A = $\frac{1}{8}$ of the cistern per hour.

Pipe B fills the cistern in 12 hours. Therefore, the rate of work of Pipe B is:
Rate of Pipe B = $\frac{1}{12}$ of the cistern per hour.

Outlet pipe O empties the full cistern in 24 hours. Because it empties the cistern, its work is negative:
Rate of Pipe O = $-\frac{1}{24}$ of the cistern per hour.

Now, if all three pipes are opened simultaneously, their combined rate of work per hour is:
Combined Rate = Rate of A + Rate of B - Rate of O

$\text{Combined Rate}=\frac{1}{8}+\frac{1}{12}-\frac{1}{24}$

To add these fractions, find the least common multiple (LCM) of 8, 12, and 24, which is 24. We rewrite the fractions with the common denominator:
$\frac{1}{8}=\frac{3}{24}$
$\frac{1}{12}=\frac{2}{24}$

Substitute these back into the equation:
$\text{Combined Rate}=\frac{3}{24}+\frac{2}{24}-\frac{1}{24}$

$\text{Combined Rate}=\frac{3+2-1}{24}=\frac{4}{24}=\frac{1}{6}$

This means that together, the three pipes fill $\frac{1}{6}$ of the cistern in one hour.

Therefore, the total time required to fill the entire cistern is the reciprocal of the combined hourly rate:
Total Time = $\frac{1}{1/6}=6$ hours.