There are three pipes connected with a tank. The first pipe can fill \( \Large \frac{1}{2} \) part of the tank in 1 h, second pipe can fill \( \Large \frac{1}{3} \) part of the tank in 1 h. Third pipe is connected to empty the tank. After opening all the three pipes, \( \Large \frac{7}{12} \) part of the tank can be filled in 1 h, then how long will third pipe take to empty the full tank?
Correct Answer: Description for Correct answer:
1st pipe takes 1 h to fill \( \Large \frac{1}{2} \) part of the tank.
So, time taken to fill the whole tank (m) = 2 h
2nd pipe takes 1 h to \( \Large \frac{1}{3} \) part of the tank
So. time taken to fill the whole tank (n) = 3 h
Let 3rd pipe takes P h to empty the tank = x
Therefore, \( \Large \frac{1}{m} \)+ \( \Large \frac{1}{n} \)- \( \Large \frac{1}{x} \)= \( \Large \frac{7}{12} \)
= \( \Large \frac{1}{2} \)+ \( \Large \frac{1}{3} \)- \( \Large \frac{1}{x} \) = \( \Large \frac{7}{12} \)
= \( \Large \frac{1}{x}=\frac{6+4-7}{12}=\frac{3}{12}=\frac{1}{4} \)
Therefore, x = 4 h
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