P1 fill a tank in 30hr where as P2 empties the tank in 60 hr. If P1 is opened in the first hour and P2 in the second hour and so on. What is the earliest time in which the tank is filled??

• In 116 hrs, 58/60 tank is full.
  In next hr, 1/30 or 2/60 part of tank is filled and tank is full now.
  so earliest time in which the tank is filled is
  58*2+1 = 117 hrs
• 13 years agoHelpfull: Yes(15) No(4)
• in two hrs, 1/30-1/60=1/60
  so total times= 120 hrs Ans...
• 13 years agoHelpfull: Yes(7) No(4)
• since p1 is the first and p2 fills after p1 ,therefore p2 will be 1 less than p1
  x/30-(x-1)/60=1.....gives x=59hrs.
• 13 years agoHelpfull: Yes(3) No(2)
• in first hour only P1 is opend so 1/30 tank is filled..from second hour P1 is filling 1/30 simultaneously p2 is emptying 1/60 so net is (1/30-1/60) which equals 1/60...so (1/30+x*1/60)=1 2+x=60 x=58 but 1/30 is included..implies 58+1 = 59 hrs is solution
• 13 years agoHelpfull: Yes(2) No(2)
• 120hr...
  if both work alternate hrs then tank is full in 2 hr is
  (1/30)-(1/60)=(1/60)
  so for full the tank require time is 60*2= 120 hr.
• 13 years agoHelpfull: Yes(1) No(7)
• dipin is ryt bcoz dey are asking for earliest tym ...so ans is 117 but if they ask 4 actual tym den ansis 120
• 13 years agoHelpfull: Yes(1) No(0)