Three taps A,B and C can fill a tank in 20,30 and 40
minutes respectively. All the taps are opened
simultaneously and after 5 minutes tap A was closed and
then after 6 minutes tab B was closed .At the moment a
leak developed which can empty the full tank in 60
minutes. What is the total time taken for the completely
full?

• 24 minutes
  
  Filling upto 5 min.=5*[(1/20)+(1/30)+(1/40)]=13/24
  For next 6 min. only taps 'B' & 'C' are opened, so filling=6*[(1/30)+(1/40)]=7/20
  Total filling upto first 11 min.=(13/24)+(7/20)=107/120 so remaining=13/120
  As after this only tap 'C' and leakage @1/60 per min. are opened,
  so its 1 min. filling rate= (1/40)-(1/60)=1/120
  Time taken to fill remaining 13/120 tank= (13/120)/(1/120)= 13 min.
  So total time taken to completely fill the tank= 5+6+13=24 min.
• 8 years agoHelpfull: Yes(3) No(0)
• (5/20)+((5+6)/30)+((5+6+x)/40)-(x/60)=1
  x=13
  total time= 13+5+6=24
• 8 years agoHelpfull: Yes(1) No(0)