Question

Two pipes can fill a tank in 10 hours and 12 hours respectively while a third pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be filled?

• A. 7 hours 30 mins
• B. 8 hours 30 minutes
• C. 15 hours and 30 minutes
• D. 9 hours and 22.7 minutes

Hint:

To avoid fractions, assume the capacity of the tank is the LCM of the times (60 units).
- Rate of Pipe 1 = $+6$ units/hr
- Rate of Pipe 2 = $+5$ units/hr
- Rate of Pipe 3 = $-3$ units/hr
- Net Rate = $6 + 5 - 3 = 8$ units/hr.
- Time = $\frac{60}{8} = 7.5$ hours.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a standard work-and-time problem involving inlets (positive work) and outlets (negative work).
Key Formula or Approach:
Let the net rate of the three pipes working together be $R$: \[ R = \frac{1}{T_1} + \frac{1}{T_2} - \frac{1}{T_3} \]

Step 2: Detailed Explanation:
The given times are:
- Pipe 1 (inlet): $T_1 = 10\text{ hours}$
- Pipe 2 (inlet): $T_2 = 12\text{ hours}$
- Pipe 3 (outlet): $T_3 = 20\text{ hours}$
Calculate the net rate: \[ R = \frac{1}{10} + \frac{1}{12} - \frac{1}{20} \] Find the LCM of 10, 12, and 20, which is 60: \[ R = \frac{6 + 5 - 3}{60} = \frac{8}{60} = \frac{2}{15} \text{ part per hour} \] Therefore, the time required to fill the tank is the reciprocal of the rate: \[ T = \frac{1}{R} = \frac{15}{2} = 7.5\text{ hours} \] Convert 0.5 hours to minutes: \[ 0.5 \text{ hours} = 0.5 \times 60 = 30 \text{ minutes} \] The tank will be filled in 7 hours and 30 minutes.

Step 3: Final Answer:
The time matches Option (A).