a pipe can fill the tank in x hours and other can empty in y hours.if the tank is 1/3 rd full then the number of hours n which they will together fill it in is

• 2xy/[3(y-x)]
  
  Part filled in 1 hour by both the pipes = 1/x - 1/y = (y-x)/xy
  As tank is 1/3 full, so remaining to be filled=1 - 1/3 = 2/3
  Hence required time to fill 2/3 tank = (2/3) / (y-x)/xy = 2xy/[3(y-x)] hours.
• 7 years agoHelpfull: Yes(0) No(1)