Pipe A can fill a tank in 12 hours and pipe B can empty a tank in 18 hours. Both pipes are opened at 6:00 am, after some time pipe B is closed, and the tank is full at 8:00 pm on the same day.
Answers
Step-by-step explanation:
Let V denotes the total capacity of the given tank.
It is mentioned that
Pipe A & pipe B separately can fill the tank’s full capacity V in respectively 8 hrs & 12 hrs.
So in 1 hr, pipe A & pipe B separately can fill the fractional amounts respectively V/8 & V/12 of the given tank.
It is mentioned that
Pipe A is opened at 7:00 AM and pipe B is opened at 9:00 AM.
In 2 hrs time (from 7:00 AM to 9:00 AM) pipe A will fill the fractional amount (2*V/8 =) V/4 of the given tank.
Since 9:00 AM both the pipes A & B will continue to remain open. Let T denotes the time (in hrs) in which the open pipes A & B together will fill the remaining empty portion (V - V/4 =) 3*V/4 of the given tank.
So, we get the following relation,
T*(V/8 + V/12) = 3*V/4 or T*5/24 = 3/4 or T = 18/5 (hrs)
Hence, total time required to fill the tank = 2 + 18/5 = 28/5 = 5 (3/5) (hrs) = 5 hrs 36 min
Therefore the tank will be full at (7:00 AM + 5 hrs 36 min =) 12:36 PM