Two pipes A and B fill up a half tank in 1.2 hours. The tank was initially empty. Pipe B was kept open for half the time required by pipe A to fill the tank by itself . Then pipe A was kept open for as much time as was required by the pipe B to fill up 1/3 of the tank by itself. It was found that the tank was 5/6 full. The least time in which any of the pipes can the fill the tank fully is ?
Work done per minute by pipe A = Wa
Time taken by pipe A to fill up a tank for x minutes = Wa * x = Volume ( V )
x = V/ Wa
Work done per minute by pipe B = Wb
pipe B kept open for x/2 min = Wb * x/2
Work done by pipe B to fill up in one-third of tank in y min = Wb *y min = V / 3
y = V / 3Wb
=> Wb = V/ 3y = (Wa x) / 3y
=> Wa * y min = kept open
( Wa + Wb ) * 72 min = V/2 ---(i)
=> (Wb * x/2 ) + ( Wa *y ) + 5/6 = 1
Substituting y = V / 3Wb .
( ( Wa x ) /3y * V/ 2*Wa ) + ( Wa * V/ 3Wb ) + 5/6 = 1
Then?
Work done per minute by pipe A = Wa
Time taken by pipe A to fill up a tank for x minutes = Wa * x = Volume ( V )
x = V/ Wa
Work done per minute by pipe B = Wb
pipe B kept open for x/2 min = Wb * x/2
Work done by pipe B to fill up in one-third of tank in y min = Wb *y min = V / 3
y = V / 3Wb
=> Wb = V/ 3y = (Wa x) / 3y
=> Wa * y min = kept open
( Wa + Wb ) * 72 min = V/2 ---(i)
=> (Wb * x/2 ) + ( Wa *y ) + 5/6 = 1
Substituting y = V / 3Wb .
( ( Wa x ) /3y * V/ 2*Wa ) + ( Wa * V/ 3Wb ) + 5/6 = 1
Then?