Pipes A and B can fill a rectangular tank in 40 minutes and 120 minutes, respectively. Pipe C can empty the completely filled same tank in 240 minutes. If pipes A, B and C are opened at the same time, then how many minutes will it take to fill the tank?
34 2/7
Let the capacity of the rectangular tank be \(C\).
Pipe A fills the tank in 40 minutes, so its filling rate is \(\frac{C}{40}\) per minute.
Pipe B fills the tank in 120 minutes, so its filling rate is \(\frac{C}{120}\) per minute.
Pipe C empties the tank in 240 minutes, so its emptying rate is \(\frac{C}{240}\) per minute.
When pipes A, B, and C are opened simultaneously, the combined filling rate is:
\(\frac{C}{40} + \frac{C}{120} - \frac{C}{240} = C \left( \frac{1}{40} + \frac{1}{120} - \frac{1}{240} \right)\)
To find a common denominator, we can use 240:
\(C \left( \frac{6}{240} + \frac{2}{240} - \frac{1}{240} \right) = C \left( \frac{7}{240} \right)\)
Let \(t\) be the time in minutes it takes to fill the tank when all three pipes are open. Then:
\(C \left( \frac{7}{240} \right) t = C\)
Dividing both sides by \(C\):
\(\frac{7}{240} t = 1\)
Solving for \(t\):
\(t = \frac{240}{7}\)
Now, we convert this improper fraction to a mixed number:
\(t = 34 \frac{2}{7}\)
Therefore, it will take \(34 \frac{2}{7}\) minutes to fill the tank when pipes A, B, and C are opened at the same time.
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