A cistern can be filled by one of the pipes in 30 min and by the other in 36 min .both pipes are opended for a certain time but being partially clogged only 5/6 of the quantity of water flows though the former and only 9/10 through the later the obstruction however being suddenly removed the cistern is filled in 31/2 min form that moment .how long was it before the full of water began?

• In normal condition at 1 min when both the pipe opened it would fill =(1/36 +1/30)=11/180
  in 31/2 min it would fill =31/2 * 11/180= 341/360
  ATQ, remaining part i.e (1- 341/360)= 19/360 was filled at partially clogged condition of both the pipe.
  At clogged condition the former pipe flows in 1 min=(1/30 * 5/6)=1/36
  and 2nd pipe in 1min=(1/36 * 9/10)=1/40
  let x be min during which both the pipe was in clogged condition.
  so,
  x(1/36 +1/40)=19/360
  x=1min
  ans is 1min
• 9 years agoHelpfull: Yes(39) No(0)
• let x wil be the time for which 5/6 and 9/10 0f water flows.
  if 5/6 of water flows it take 36 minutes to fill the tank
  if 9/10 of water flows it take 40 minutes fill the tank.
  x(1/36+1/40)+(1/30+1/36)=1
  therefore x=1minute.
  therefore it take 1 minute for the removal of obstruction.
• 9 years agoHelpfull: Yes(2) No(7)
• 4.49 minutes
• 9 years agoHelpfull: Yes(0) No(3)
• 1.74 mnt...
• 9 years agoHelpfull: Yes(0) No(3)