Pipes A & B can together fill a tank in 4 hours, while pipes C & D can individually empty the tank in 10 hours and 12 hours respectively. When the tank was empty, pipes A & B were simultaneously opened. After an hour, pipe C was also opened. After the opening of pipe C, pipe D will be opened after how much time such that the tank is only half filled in 3 hours?
Part filled by pipes A & B in 1 hr = 1/4
Part emptied by pipe C in 1 hr = 1/10
Part emptied by pipe D in 1 hr = 1/12
Now, pipe A & B were open for = 3 hours
Pipe C was open for = 3 - 1 = 2 hours
Let pipe D was opened after ‘x’ hours of opening of pipe C,
Pipe D was open for = (2 - x) hours
Hence, part filled in 3 hours = 1/2
⇒ Part filled by A & B - Part emptied by C & D = 1/2
⇒ 3 × (1/4) - 2 × (1/10) - (2 - x) × (1/12) = 1/2
⇒ 3/4 - 1/5 - (2 - x)/12 = 1/2
⇒ 45 - 12 - 10 + 5x = 30
⇒ 5x = 7
⇒ x = 7/5 = 1.4 hrs = 1 hr. 24 min.
∴ Pipe D was opened after 1 hr. 24 min. of opening of pipe C
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