An empty tank be filled with an inlet pipe A in 42 minutes. After 12 minutes an outlet pipe B is opened which can empty the tank in 30 minutes. After 6 minutes another inlet pipe C opened into the same tank, which can fill the tank in 35 minutes and the tank is filled. Find the time taken to fill the tank?

• For first 12 mins tank is 2/7 filled(because only pipe A is opened). For first 18 mins tank is 8/35 filled. i.e pipe A is opened for 18 mins and pipe B is opened for 6 mins. (3/7)-(1/5)=8/35. that means after 18 mins tank is 8/35 filled.
  so, after 18 mins all the 3 pipes work simultaneously.
  now we calculate the time required to fill the tank by 3 pipes,
  (1/42)+(1/35)-(1/30)=2/105. It means 105/2 mins.
  But the tank is already 8/35 filled. so,we have to find the time required to fill 27/35 of the tank.
  time taken to fill 27/35 of the tank=(105/2)*(27/35)= 40.5 mins.
  time taken to fill the tank= 18 mins+40.5 mins = 58.5 mins.
  Ans:58.5 mins
• 10 years agoHelpfull: Yes(20) No(5)
• pipe A is open for 12 mins, so (12/42)=(2/7) of the tank is filled. pipe A and B are open together for the next 6 mins. So 6((1/42)-(1/30))+(2/7)= (8/35) is filled in 18 mins. then all three pipes are open together. they can fill (1/42)-(1/30)+(1/35)=(2/105) of the tank. now only 1-(8/35)=(27/35) needs to be filled. So time taken to fill the remaining part is (27/35)/(2/105)=40.5 mins.
  total time = 18+40.5 = 58.5 mins
• 10 years agoHelpfull: Yes(9) No(0)
• 58.5 minute
• 10 years agoHelpfull: Yes(4) No(2)
• let total capicity of tank =100 and total time=t
  so t(10042)-(t-12)(1030)+(t-18)(10035)=100
  on solving this equation we get t=58.52
  so t=58.52
  
  
• 10 years agoHelpfull: Yes(4) No(1)
• Pipe A in one minute fills 1/42 of the tank. Pipe A is totally kept open for 12 + 6 -> 18 minutes before Pipe C is opened. Thus in 18 minutes Pipe A fills 18/42 -> 3/7 capacity of the tank.
  
  Pipe B can drain the tank in 30 minutes and in one minute can drain 1/30 of the tank.
  
  Pipe B is kept open for 6 minutes before Pipe C is opened. In this 6 minutes it would have drained 6/30 -> 1/5 capacity of the tank.
  
  Hence, after 18 minutes the tank is filled up to – > 3/7 – 1/5 -> 8/35 capacity.
  
  The remaining capacity to be filled is -> 27/35
  
  All the three pipes are now running to fill this quantity.
  
  Pipe C in one minute can fill 1/35 capacity.
  
  Now in one minute both Pipes A and C fills and C drains. The quantum filled in one
  
  Minute is -> 1/42 + 1/35 – 1/30 -> 2/70.
  
  Thus to fill the remaining capacity of 27/35 the time taken will be 27/35 * 70/2 -> 27 minutes.
  
  Hence, the total time taken to fill the tank is 12 + 6 + 27 = 45 minutes.
• 9 years agoHelpfull: Yes(1) No(3)
• first take the lcm of 42,30and 35 we will get an 210
  take this has units i.e 210 units
  a=210/42 = 5
  B = 210/ 30 = 7
  c= 210/35 = 6
  till 18 min A is opened and only 6min b is opened then
  5*18=90 and 7*6=42
  90-42=48
  210-48=162
  if a=5,b=-7 c=6
  then a+b+c = 4
  162/4 = 40.5
  18+40.5 = 58.5
  
• 9 years agoHelpfull: Yes(1) No(0)
• could any one explain...i am getting 35.5 mins
• 10 years agoHelpfull: Yes(0) No(4)
• please expalin these question clearly and step by step process
• 10 years agoHelpfull: Yes(0) No(0)
• 45mins is the correct 200% sure
• 9 years agoHelpfull: Yes(0) No(2)