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

The Correct Option is D

Assuming A fills the tank in \( x \) hours, B (the drain) empties it in \( (x-1) \) hours, and C fills it in \( y \) hours.

Step 1: First Equation Formulation

With pipes A, B, and C operating concurrently, the tank fills in 2 hours. This scenario is represented by the equation:

\[ \frac{1}{x} - \frac{1}{x-1} + \frac{1}{y} = \frac{1}{2} \tag{1} \]

Step 2: Second Scenario Formulation

In the second scenario, pipe B operates for 1 hour, and pipe C operates for 2 hours and 15 minutes. The work done by each pipe is:

• Pipe B: \( -\frac{1}{x-1} \) units in 1 hour.
• Pipe C: \( \frac{9}{4y} \) units in \( \frac{9}{4} \) hours.

The equation for this scenario is:

\[ \frac{9}{4y} - \frac{1}{x-1} = 1 \tag{2} \]

Step 3: System of Equations Resolution

The two equations are:

• Equation (1): \( \frac{1}{x} - \frac{1}{x-1} + \frac{1}{y} = \frac{1}{2} \)
• Equation (2): \( \frac{9}{4y} - \frac{1}{x-1} = 1 \)

Solving these equations yields:

• \( x = 3 \)
• \( y = \frac{3}{2} \)

Step 4: Final Computation

Pipe C requires \( 3 \frac{1}{2} \) hours (equivalent to 90 minutes) to complete its task. Therefore, the correct selection is:

The correct answer is (D): 90 minutes.