Two pipes A and B can fill a tank in 40 and 45 minutes respectively. A water pipe C can empty the tank in 50 minutes. First A and B are opened. After 7 minutes, C is also opened. In how much time, the tank is full?
Time taken by pipe A to fill the tank alone = 40 mins
So, part filled by pipe A in 1 hour = (1/40)
Similarly, time taken by pipe B to fill the tank alone = 45 mins
So, part filled by pipe B in 1 hour = (1/45)
And, time taken by pipe C to empty the tank alone = 50 mins
So, part emptied by pipe C in 1 hour = (1/50)
Now, part filled by pipe A and B together = (1/40) + (1/45) = 17/360
So, part filled by both in 7 minutes = 7 x (17/360) = 119/360
Remaining part = 1-(119/360) = 241/360
Now, part filled in 1 min when A, B and C are operating at the same time = (1/40) + (1/45) - (1/50) = 49/1800
But, time taken to fill (49/1800)th part of the tank = 1 min
Or, time taken to fill 1 part of the tank = (1/(49/1800)) min = (1800/49) mins
Therefore, time taken to fill the remaining (241/360)th part of the tank = (1800/49) x (241/360)
=(1205/49) mins = 24.59 mins
Therefore, total time taken = (24.59 + 7) mins = 31.59 mins Ans.
First approach with the amount of work done by the 2 pipes in 7 minutes and after that all the three pipes will work.