Question

Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is

• A. 75
• B. 120
• C. 60
• D. 90

Answer 1

Answer: (D)

90

Answer 2

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.