7) 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?

• pipe A works for 18 min ie 12+6=18 so 18/42= 3/7
  pipe B works for 6 min so 6/30=1/5
  so, 3/7-1/5= 8/35 works complete in 18 min.
  remaining 1-8/35=27/35
  pipe A and c fill in one min and b emptied in one min ie= 1/42+1/35-1/30= 2/105
  so 27/35*105/2= 81/2= 40.5
  so total time 12+6+40.5= 58.5 min
• 9 years agoHelpfull: Yes(18) No(2)
• (1/42)*12 +[(1/42)-(1/30)]*6 +[(1/42)-(1/30)+(1/35)]*x=1
  
  x=21/2
  total time taken by 3 pipes=12+6+21/2=57/2=28.5minutes
• 9 years agoHelpfull: Yes(8) No(4)
• time taken to fill the tank=
  1/42 *12 + 6* (1/42 -1/30) + x* (1/42+1/35 -1/30)=210/50
  4.2
  total time taken =12+6+4.2=22.2
  ans
  
• 9 years agoHelpfull: Yes(2) No(1)
• Assume total tank capacity = 210 Liters
  Now capacity of pipe A = 210/42 = 5 Liters
  Capacity of B = 210 / 30 = - 7 Liters
  Capacity of C = 210 / 35 = 6 min
  Assume tank gets filled in x min after the third pipe got opened.
  So x×5+6×(−2)+4x=210
  ⇒48+4x=210⇒4x=162⇒x=40.5
  Total time taken to fill the tank = 40.5 + 12 + 6 = 51.5
  
• 9 years agoHelpfull: Yes(2) No(1)
• 42A = 30B = 35C
  
  filled by A = 12A = 10C
  emptied by B = 6B = 7C
  filled by C = xC (suppose x min.)
  so 10C - 7C + xC = 35C
  hence x = 32minutes
• 9 years agoHelpfull: Yes(1) 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(0)
• Ans:21.62 min
  
  For the first time
  One inlet pipe is opened, continued for 12 min.
  In 12 min
  =12/42 = 2/7 part filled.
  
  After 12 min two taps opened, one inlet and one outlet continued for 6 min
  
  = 1/42-1/30 = -2/210 part emptied.
  
  After 8 min three taps opened.
  = 1/42-1/30+1/35
  =4/210 part filled.
  
  Summing all we get
  
  60/210 - 2/210 + 4/210 = 58/210 = 3.62 min(add 18 min)
• 9 years agoHelpfull: Yes(0) No(5)