Two pipes a and b can separetly fill a cistern in 60min and 75min respectively. there is a third pipe in the bottom of the cistern to empty it. if all the three pipes are simultaneously opened, then the cistern is full in 50min. in how much time the third pipe alone can empty the cistern?

• a's 1 min work = 1/60
  b's 1 min work = 1/75
  Let 3rd pipe's 1 hr work = 1/x
  
  in 50 mins , it ll be full if all r opened
  So in 1 min 1/50 part of d tank is filled
  
  1/60 + 1/75 -1/x = 1/50
  1/x = 1/60 + 1/75 -1/50
  1/x = 5 + 4 - 6/300 = 3/300 = 1/100
  
  In 1 min , 3rd pipe ll empty 1/100 part of the tank
  So it ll empty d fll tank in 100 mins
  
  Ans : 100min
• 10 years agoHelpfull: Yes(9) No(0)
• 100 min
  
  Let x be the time taken by 3rd pipe to empty the cistern
  Together they fill the cistern in 50 min
  ===>
  1/60+1/75-1/x = 1/50
  1/x = 3/100 - 1/50
  =1/100
  ===>x = 100 min
  
  3rd pipe can empty cistern in 100min
• 10 years agoHelpfull: Yes(3) No(0)
• given a+b-c=1/50(since c is empting the tank)
  so,1/60 + 1/75 -c = 1/50
  c=1/60 + 1/75 - 1/50
  =1/100
  so,c completly empties the tank
  in 100 mins...
• 10 years agoHelpfull: Yes(1) No(0)
• a and b fills in 1 min= 1/60+1/75=3/100 portion
  let c empties the cistern in C min
  so in 1 min c empties 1/C portion
  in 1 min cistern is filled=3/100-1/C
  now, 3/100-1/C=1/50
  C=100min
• 10 years agoHelpfull: Yes(0) No(0)