A cistern is provided with two pipes A and B. A can fill it in 20 minutes and B can empty it in 30 minutes. If A and B be kept open alternately for one minute each, how soon will the cistern be filled?

• ans is 2 hours
• 7 years agoHelpfull: Yes(2) No(2)
• TAP -A -20 mints -> 3 L/mint
  TAP -B - 30 mints -->2 L/mint
  Tank will be full in 60 mints
  
  so A-B+A-B+A-B..................... = TF
  => 1.5-1 = 0.5
  1 * 60 = 60 mints
  0.5 * 120 mints or 2 hrs
• 7 years agoHelpfull: Yes(1) No(4)
• Total unit of work = 60(lcm of 20 and 30)
  a's efficiency=3, b's efficiency= -2
  therefore net amount of work done in 2 min = 1 unit
  now in 57 min, i.e actually 114 min, work done is 57/60
  remaining 1/20 work will be done by a in 1 min, i.e 115 min is the answer
  It can also be done by fraction:
  1/20 - 1/30 = 1/60
  and proceeding as above
• 7 years agoHelpfull: Yes(0) No(0)
• A 1 min work=1/20
  B 1 min work= -1/30
  A/q
  =>A+B+A+B+....=1(total work)
  1/20-1/30+1/20-1/30....=1
  1/60+1/60+1/60+....=1
  means 1/60 work is done in 2 min
  1 work is done in=2*60=120min or 2hr (ans)
• 7 years agoHelpfull: Yes(0) No(0)