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

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.

Step 1: First Equation Setup 

When all pipes (A, B, and C) are activated simultaneously, the empty tank will fill in 2 hours. The equation representing this scenario is:

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

Step 2: Second Scenario Setup

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

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

So, the equation for this scenario is:

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

Step 3: Solve the System of Equations

Now, we solve the two equations:

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

By solving these equations, we find:

• \( x = 3 \)
• \( y = \frac{3}{2} \)

Step 4: Final Calculation

Now, we know that pipe C takes \( 3 \frac{1}{2} \) hours (or 90 minutes) to complete the task. Therefore, the correct option is:

The correct answer is (D): 90 minutes.