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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? 

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?  Correct Answer 1 hr. 24 min.

GIVEN :

Pipes A & B can together fill a tank in 4 hours.

C & D can empty the tank in 10 hours and 12 hours respectively. 

 

CALCULATION :

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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