A vessel is full of a mixture of milk and water, comprising 60% milk. 6 litres of the mixture are withdrawn and replaced with pure water. If the milk now comprises 36% of the resultant mixture, how much does the vessel hold?

• 15 litres
  
  Ratio Milk : Water = 60 : 40
  Let the vessel is of 'x' capacity, then Milk= 0.6x, Water=0.4x
  Quantity of milk remained, when 6 litres of mixture is taken out = 0.6x - 6*60/100 = (0.6x - 3.6) litres.
  As the withdrawn quantity is replaced with water, so total quantity remains same i.e. x litres.
  Now % of milk in new mixture = (0.6x - 3.6) / x = 36/100
  => (0.6x - 3.6) / x = 9/25
  => 25(0.6x - 3.6) = 9x
  => 15x - 90 = 9x
  => 6x = 90 or x=15
• 6 years agoHelpfull: Yes(0) No(0)