Given that A takes x hours to fill the tank alone, it follows that B (the drainage pipe) needs (x-1) hours to empty the tank alone, and C needs y hours to replenish the tank.
Given that pipelines A, B, and C are activated simultaneously, the empty tank will fill in two hours.
So \(\frac{1}{x}-\frac{1}{x-1}+\frac{1}{y}=\frac{1}{2}....(1)\)
Pipe C will take an extra hour and fifteen minutes to fill the remaining tank if pipes B and C are turned on simultaneously when the tank is empty and Pipe B is turned off after an hour.
Thus, B worked for one hour while C worked for two hours and fifteen minutes, or nine times four hours.
B completed \(-\frac{1}{x-1}\) units in an hour, whereas C completed \(\frac{9}{4y}\) units in \(\frac{9}{4}\) hours.
So, \(\frac{9}{4y}-\frac{1}{x-1}=1.....(2)\)
After solving both equations, we have \(x=3\) and \(y= \frac{3}{2}. \)
As a result, C took \(3 \frac{1}{2} \)hours, or 90 minutes, to complete.
The correct option is (D): 90.