Two water supply lines A and B can fill a reservoir individually in 10 hours and 12 hours, respectively. A discharge outlet D can empty the reservoir completely in 6 hours. If all three are opened together at 7:00 A.M., at what time will one-fourth of the reservoir be filled?

The correct option is 10 P.M..

Let us solve this problem step-by-step by determining the net rate at which the reservoir is being filled when all three pipes are opened simultaneously.

Step 1: Find the work done by each pipe in 1 hour.

Line A can fill the reservoir in 10 hours.
Work done by Line A in 1 hour = $\frac{1}{10}$

Line B can fill the reservoir in 12 hours.
Work done by Line B in 1 hour = $\frac{1}{12}$

Discharge outlet D can empty the reservoir in 6 hours.
Work done by Outlet D in 1 hour = -$\frac{1}{6}$ (negative since it removes water)

Step 2: Calculate the net work done in 1 hour when all three are open.

Net work done in 1 hour = $\frac{1}{10}+\frac{1}{12}-\frac{1}{6}$

To add and subtract these fractions, find the Least Common Multiple (LCM) of the denominators (10, 12, and 6), which is 60.

Net work done in 1 hour = $\frac{6+5-10}{60}$

Net work done in 1 hour = $\frac{1}{60}$

This means that in 1 hour, $\frac{1}{60}$ of the reservoir is filled.

Step 3: Calculate the total time needed to fill one-fourth ($\frac{1}{4}$) of the reservoir.

Time required = $\frac{\text{Target fraction}}{\text{Net work done in 1 hour}}$

Time required = $\frac{1/4}{1/60}=\frac{60}{4}=15\text{hours}$

Step 4: Determine the exact time.

All three lines are opened together at 7:00 A.M.
Adding 15 hours to 7:00 A.M.:
- 7:00 A.M. + 12 hours = 7:00 P.M.
- 7:00 P.M. + 3 hours = 10:00 P.M.

Therefore, one-fourth of the reservoir will be filled at 10 P.M.