How Much Alcohol :P

Replies

• 

  durga ch

  initial volumn 1st vesel=20lts-filled
  initial volumn 2nd vessel=20lts-empty
  
  volumn transferred=xlts
  
  volumn in first vessel=20-xlts
  and volumn in seoncd vessel=xlts
  
  amount of volumn of water added =20-x in second vessel
  
  now what ever amount of seoncd vessels liquid is transferred, the volumn solution in first vessel will be = 20
  
  but since content to be transfered is x... now the total volumn in seocnd vessel is 20-x where as volumn in vessel 1 is 20
  
  transfering 20/3 lts to second vessel makes the volumnequal.
  
  20-20/3=20-x
  
  x=20/3lts
  
  
  I have not considered the concentrations though
• 

  electron1212

  sorry thats not the correct answer.
• 

  durga ch

  oh okie!!😔
  will give it a second go!!
• 

  electron1212

  44+ views but still no correct answer/explanation and i thought this one was easy.:sshhh:
• 

  mech_guy

  Lets say, Alcohol transferred from first vessel to second is "V"
  
  So, alcohol in first vessel = 20 - V
  alcohol in 2nd vessel = V
  
  Now, we add (20-V) lt of water in second vessel and mix it well.
  To completely fill first vessel V lt. solution should be taken from vessel two.
  Alcohol in this V lt of solution will be = V/(20/V) = V*V/20
  
  Amount of alcohol in first vessel now = (20 - V + V*V/20) = (20 - X) in 20 lt solution
  Amount of alcohol in second vessel now = (V - V*V/20) = X
  
  where, V - (V*V/20) = X
  
  Now (20/3) lt of solution is taken from first vessel.
  Alcohol in this (20/3) lt of solution will be = (20 - X)/3
  
  So, now alcohol left in first vessel = (20 - x) - [(20 - x)/3} = 2*(20 - X)/3
  and alcohol in second vessel = X + (20 - X)/3 = (20 + 2*x)/3
  
  equating both will give X = 5
  i.e. V - (V*V/20) = 5
  which gives V = 10 lts
  
  So originally 10 lts of alcohol was transferred from first vessel to second.
  
  Regards
• 

  pradeep_agrawal

  Solution looks good to me, good work mech_guy.
  
  -Pradeep
• 

  Anil Jain

  mech_guy

  Lets say, Alcohol transferred from first vessel to second is "V"
  
  So, alcohol in first vessel = 20 - V
  alcohol in 2nd vessel = V
  
  Now, we add (20-V) lt of water in second vessel and mix it well.
  To completely fill first vessel V lt. solution should be taken from vessel two.
  Alcohol in this V lt of solution will be = V/(20/V) = V*V/20
  
  Amount of alcohol in first vessel now = (20 - V + V*V/20) = (20 - X) in 20 lt solution
  Amount of alcohol in second vessel now = (V - V*V/20) = X
  
  where, V - (V*V/20) = X
  
  Now (20/3) lt of solution is taken from first vessel.
  Alcohol in this (20/3) lt of solution will be = (20 - X)/3
  
  So, now alcohol left in first vessel = (20 - x) - [(20 - x)/3} = 2*(20 - X)/3
  and alcohol in second vessel = X + (20 - X)/3 = (20 + 2*x)/3
  
  equating both will give X = 5
  i.e. V - (V*V/20) = 5
  which gives V = 10 lts
  
  So originally 10 lts of alcohol was transferred from first vessel to second.
  
  Regards

  Looks fine..
  
  -CB
• 

  jhbalaji

  need clue mate...