A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

• Let Original Volume of Mixture = x litre and it's to be replaced by y litre of water.
  
  Ratio of Acid & Water in mixture = 80% : 20% = 4:1 =5
  
  volume of the Acid in x - y litre of Mixture = 4/5 (x - y) litre;
  
  and volume of water in x - y litre of Mixture = 1/5 (x - y) Litre
  
  But volume of water will be increased by y litre = 1/5 (x - y) litre + y Litre. ==> (x + 4y)/5
  
  Now new ratio of Acid to Water is given 4 : 3
  
  Thus 4/5 (x - y) : (x + 4y)/5 = 4:3
  
  Solving we get (4x - 4y) / (x + 4y) = 4 : 3
  
  ==> 16x - 12y = 4x + 16y
  
  ==> 12x = 28y
  
  ==> y/x = 12/28 = 2/7
• 10 years agoHelpfull: Yes(2) No(1)