a vessel contain milk and water in the ratio 5:4.if 27 litres of mixture is removed the same quantity of water is added...the ratio becomes 4:5.what quantity does the vessel contain?

• Let us first assume that the vessel contains 5x volume of milk and 4x volume of water (from the ration 5:4). Also, the volume of mixture taken out (27ltrs) will have the same ratio of milk and water, so we can get the volume of milk and water in 27ltrs of the mixture
  for milk ---> [ 5 / ( 5 + 4 ) ] * 27 = 15ltrs
  for water ---> [ 4 / ( 5 + 4 ) ] * 27 = 12ltrs
  
  Now that we know the volume of milk and water taken out we can say...
  ( 5x - 15 ) : ( 4x - 12 + 27 ) = 4 : 5
  Note that 27 was added because we were told that the same volume of water was added
  Then we can solve this algebraicaly....
  [ ( 5x - 15 ) / ( 4x - 12 + 27 ) ] = [4 / 5]
  
  Then we get... 25x - 75 = 16x + 60
  9x = 135
  x = 135/9 = 15ltrs
  We can now input our value of x into 5x and 4x to get the quantity contained by the vessel
  = 5 ( 15 ) + 4 ( 15 )
  = 135 liters
• 4 years agoHelpfull: Yes(1) No(0)
• total vessel size is 135 litres
• 10 years agoHelpfull: Yes(0) No(1)
• Assume quantity of milk be x then the quantity of milk is (5/9)*x and water is (4/9)*x.
  when 27lit of mixture is removed then quantity should be x-27 at this time the ratio of milk is (x-27)*5/9 and (x-27)*4/9
  given the ratio when removal quantity is added which is milk then the ratios of milk and water are (x-27)5/9 and (x-27)*4/9+27
  (((x-27)*5/9))/((x-27)*4/9)+27=4/5
  by solving this,the quantity in a vessel should be 135lit
• 10 years agoHelpfull: Yes(0) No(0)