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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
Let the time taken by A to fill the tank alone be x hours, which implies the time taken by B to empty the tank alone is (x-1) hours (B is the drainage pipe), and the time taken by C to fill the tank is y hours.
It is given that when pipes A, B, and C are turned on together, the empty tank is filled in two hours.
Hence, $$\frac{1}{x}-\frac{1}{x-1}+\frac{1}{y}=\frac{1}{2}$$ .... Eq(1)
It is given that 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.
Hence, B worked for 1 hour, and C worked for 2 hours 15 minutes, which is equal to $$\frac{9}{4}$$ hours.
In 1 hour, B worked $$-\frac{1}{x-1}$$ units, and in $$\frac{9}{4}$$ hours, C worked $$\frac{9}{4y}$$ units.
Hence, $$\frac{9}{4y}-\frac{1}{x-1}=1$$ .... Eq(2)
Solving both equations, we get $$y=\frac{3}{2}$$, and $$x=3$$
Hence, the time taken by C is $$\frac{3}{2}$$ hours, which is equal to $$90$$ minutes.
The correct option is A
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