Two taps A and B can fill a vessel in 12 and 15 minutes, respectively, but a third pipe C can empty the full tank in 60 minutes. A and B is kept open for 5 minutes in the beginning and then C is also opened. In what time will the vessel be emptied?
Answer:
45 minutes
Attempted solution:
Portion filled in 5 min by A and B = (1/12 + 1/15)*5 = 9/12.
When all the three pipes are open, the vessel has to be emptied. So the rate of emptying the vessel should be higher than the rate of filling the vessel. But the rate of emptying the vessel (1/60 portion per minute) is slower than the rate of filling the vessel by taps A and B individually (1/12 th portion per minute and 1/15th portion per minute ). So the tap should fill up instead of getting emptied!
Nevertheless,
Since the tank has to be emptied in lets say time t, we have the following equation:
(1/60 - 1/12 -1/15)*t = 9/12
=>(1-5-4)t/60 = 9/12
which results in a negative t = -45/8 minutes.
Please help me with this. Please correct me if my concept is wrong.
Portion filled in 5 min by A and B = $\left(\frac{1}{12} + \frac{1}{15}\right) * 5 = \frac{9}{12} = \frac{3}{4}$
After 5 minute,
C - A - B = $\left(\frac{1}{60} - \frac{1}{12} - \frac{1}{15}\right)$
= $\frac{-8}{60}$ or $\frac{-2}{15}$
Answer is negative because speed of empty pipe is so slower.
So tank will be not emptying it is filling.
Edit - The question asked by OP has an answer 45 minutes but he is getting answer -45/8 minutes. It means vessel is not emptying. But the above question has some mistake C empty in 6 minutes not 60. So if we do above question using 6 minutes we have correct result 45 minutes (Given below).
Original question -
Solution -