Question

A tank is connected with two inlet pipes A and B and one outlet pipe C. A, B independently can fill the (empty) tank in 20 hours and 30 hours respectively. If all the three pipes are opened simultaneously it takes 3 hours extra than if only A and B are opened. Then the time in hours required for the pipe C alone to empty the full tank (in hours) is

• A. 45
• B. 48
• C. 54
• D. 60

Hint:

When an outlet pipe is involved, its rate is subtracted from the inlet pipes' rates. Always define the rates clearly to avoid confusion between filling and emptying times.

Answer 1

The Correct Option is D

Concept: This is a work-rate problem involving three pipes (two inlets and one outlet). We use the combined rate to determine the efficiency of each pipe and solve for the unknown.

Step 1: Determine the work rates of pipes A, B, and C.
Let the capacity of the tank be 1 unit. \[ \text{Rate of A} = \frac{1}{20} \text{ tank/hour} \] \[ \text{Rate of B} = \frac{1}{30} \text{ tank/hour} \] Let the rate of outlet pipe C be \( \frac{1}{x} \) tank/hour.

Step 2: Calculate the combined rate of A and B.
\[ \text{Rate of (A+B)} = \frac{1}{20} + \frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12} \text{ tank/hour} \] Thus, A and B together fill the tank in 12 hours.

Step 3: Find the time taken when all three are open.
The problem states that when all three are open, it takes 3 hours extra than when only A and B are open. \[ \text{Time (A+B+C)} = 12 + 3 = 15 \text{ hours} \] The combined rate of all three is: \[ \text{Rate of (A+B+C)} = \frac{1}{15} \text{ tank/hour} \]

Step 4: Set up the equation to find the rate of C.
\[ \text{Rate of (A+B)} - \text{Rate of C} = \text{Rate of (A+B+C)} \] \[ \frac{1}{12} - \frac{1}{x} = \frac{1}{15} \] \[ \frac{1}{x} = \frac{1}{12} - \frac{1}{15} = \frac{5-4}{60} = \frac{1}{60} \] Therefore, pipe C empties the tank in 60 hours. \centerline{{60}}