Two pipes A and B can fill a tank in 10 hours and 15 hours respectivel...
Given:
Pipe A can fill the tank in 10 hours
Pipe B can fill the tank in 15 hours
Pipe C can empty the tank in 20 hours
All pipes are opened for 5 hours and then C is closed.
Approach:
Let the capacity of the tank be 'x'.
In 1 hour, pipe A can fill x/10 part of the tank.
In 1 hour, pipe B can fill x/15 part of the tank.
In 1 hour, pipe C can empty x/20 part of the tank.
Let us assume that the tank is filled in 't' hours.
After 5 hours, the fraction of the tank filled by pipes A and B is (5/x) * [(x/10) + (x/15)] = (1/6).
So, the remaining fraction of the tank to be filled is (1 - 1/6) = (5/6).
Now, only pipes A and B are working.
So, the fraction of the tank filled by pipes A and B in 't-5' hours is (5/6).
[(t-5)/10] * (x/10) + [(t-5)/15] * (x/15) = (5/6)
Solving the above equation, we get t = 7.5 hours.
Therefore, the tank will be full in 7.5 hours.
Answer: Option C.
Two pipes A and B can fill a tank in 10 hours and 15 hours respectivel...
(1/10 + 1/15 – 1/20)*5 + (1/10 + 1/15)*T = 1. We will get T = 2.5 hrs
so total time = 5 + 2.5 = 7.5 hrs