A can fill in 20 mins and B can empty in 30 mins . If A and B be kept open alternately for one minute each, how soon will the cistern be filled?

• A ll fill 1/20 part of the tank in 1 min
  B ll empty 1/30 part of the tank in 1 min
  Assuming if A is opened at the 1st min nd B at the next min
  1/20 - 1/30 = 1/60 part filled in 2 mins
  
  1/60 part ----> 2
  1 part ----> x
  x = 60*2 = 120 mins
  
  Ans : 120 mins
• 9 years agoHelpfull: Yes(7) No(3)
• 119 min.
  
  First A fills for 1 min.= 1/20 & then B empty for 1 min.=1/30
  2 min. filling= (1/20)-(1/30)= 1/60
  Filling upto 118 min. =(118/2)*(1/60)=59/60
  So at 119 min. A will again fill= (59/60)+(1/20)=62/60 (cistren is filled)
• 9 years agoHelpfull: Yes(6) No(0)
• Ans is 119 min
  
  on last min there is already tank is full
• 9 years agoHelpfull: Yes(2) No(1)
• 115 minutes is the answer.
  1/20-1/30 = 1/60 part in 2minutes ..
  so 1/120 part in 1 minute ..
  now A fills 1/20 part in a minute.
  last time a fills ..then remaining will be 19/20 parts..
  time required to fill 19/20 parts will be (19/20) / 1/120 = 114 minutes
  total time = 114 + 1 minute = 115 minutes.
• 9 years agoHelpfull: Yes(1) No(2)
• 115mins
  
  assume capacity of tank= l.c.m of 20,30=60 units
  tap A fills 3 units/min
  Tap B empty 2 units/min
  so in 2 min i.e.1st min(tap A)+2nd min(tap B)=(3-2) = 1 units/min (say as) 1 cycle
  1 cycle fills 1 units
  for 57 cycles i.e in 114 mins fills 57 units
  in next 115th min A pipe fills 3 units.....So total units (57+3)=60 completed....
  
  Ans is 115min
• 9 years agoHelpfull: Yes(1) No(2)
• 24 min
  
  assume capacity of tank= l.c.m of 20,30=60 units
  tap A allows 3 units/min
  Tap B allows 2 units/min
  so in 2 min i.e.1st min(tap A)+2nd min(tap B)=5 units/min (say as) 1 cycle
  1 cycle fills 5 units
  x cycles fills 60 units
  x=60/5
  x=12cycles
  x=12*2 min=24 min
  
• 9 years agoHelpfull: Yes(0) No(3)
• 120+130=I12
  ANS IS 12
• 9 years agoHelpfull: Yes(0) No(5)