Three pipes P, Q, and R are attached to a tank. P and Q individually can fill the tank in 3 hours and 4 hours...

Pipe P is faster so it must be kept open as much as possible

11 am => Pipe P filled the tank for 1 hour and pipe R drained it for 1 hour
The volume of the tank filled = 1/3 - 1/5 = 2/15

11:15 am =>Pipe Q filled the tank for 15 minutes (1/4 hour) and pipe R drained it for 1/4 hour

$$Total\ emulative\ volume\ of\ \tan k\ filled\ =\ \frac{2}{15}+\left(\frac{1}{4}-\frac{1}{5}\right)\cdot\frac{1}{4}=\frac{2}{15}+\left(\frac{1}{20}\cdot\frac{1}{4}\right)$$
$$=\frac{2}{15}+\frac{1}{80}=\frac{160+15}{1200}=\frac{175}{1200}=\frac{7}{48}$$
Therefore, for every 1 hour 15 minutes (5/4 hours) 7/48 volume of the tank is filled 5/4 hours then 48/48 volume of the tank will be filled in

$$\frac{5}{4}\div\frac{7}{48}\ hours$$
$$\frac{5}{4}\cdot\frac{48}{7}\ =\frac{240}{28}=8.57\ hours$$
Since P opens at 10am, time taken to fill the tank = 10:00am + 8.57 hours
=10:00am + 8.6 hours = 18.6 hours
= 6:36pm

The closest option = option C