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, what is the time required to fill the tank ?

• A. 7 hours 30 minutes
• B. 6 hours
• C. 6 hours 30 minutes
• D. 8 hours

Hint:

When pipes fill or empty a tank simultaneously, consider their rates as positive (for filling) and negative (for emptying). The reciprocal of the net rate gives the total time taken.

Answer 1

The Correct Option is A

Step 1: Find the rate of filling for each filling pipe.
Rate of the first pipe = \( \frac{1}{10} \) of the tank per hour. Rate of the second pipe = \( \frac{1}{12} \) of the tank per hour. Step 2: Find the rate of emptying for the emptying pipe.
Rate of the third pipe (emptying) = \( \frac{1}{20} \) of the tank per hour. Step 3: Find the net rate of filling when all pipes operate simultaneously.
Net rate = (Rate of first pipe) + (Rate of second pipe) - (Rate of third pipe) Net rate = \( \frac{1}{10} + \frac{1}{12} - \frac{1}{20} \) Step 4: Calculate the common denominator and simplify the net rate.
The least common multiple of 10, 12, and 20 is 60. Net rate = \( \frac{6}{60} + \frac{5}{60} - \frac{3}{60} = \frac{6 + 5 - 3}{60} = \frac{8}{60} = \frac{2}{15} \) of the tank per hour. Step 5: Find the time required to fill the tank at the net rate.
Time = \( \frac{\text{Total work}}{\text{Net rate}} = \frac{1}{\frac{2}{15}} = \frac{15}{2} \) hours. Step 6: Convert the time to hours and minutes.
\( \frac{15}{2} \) hours = 7.5 hours = 7 hours and \( 0.5 \times 60 \) minutes = 7 hours and 30 minutes.