Two pipes A, B can fill a cistern in 10 hours and 12 hours respectively working alone while the leak present can empty the cistern in 15 hours. If both pipes are opened together, then find the time taken to fill the cistern.
1). 7 hours
2). \(8\frac{4}{7}\) hours
3). \(7\frac{6}{7}\) hours
4). \(9\frac{5}{6}\) hours
Given, pipe A can fill a cistern in 10 hours.
In 1 hr, pipe A can fill 1/10th of the cistern.
Pipe B can fill a cistern in 12 hours.
In 1 hr, pipe B can fill 1/12th of the cistern.
Leak present can empty the cistern in 15 hours.
In hr, leak empties 1/15th of the cistern.
Now, if both the taps are opened together.
In 1 hr,
Part of cistern filled $(= \frac{1}{{10}} + \frac{1}{{12}} - \frac{1}{{15}})$
⇒ Part of cistern filled $(= \frac{{6 + 5 - 4}}{{60}} = \frac{7}{{60}})$
Thus the cistern will be filled in number of hours $(= \frac{{60}}{7} = 8\frac{4}{7}hrs\;)$