Two mixtures A and B are prepared by mixing Alcohol and Soda in the ratio 5 : 8 and 3 : 7. Now equal quantities of these mixtures are mixed and 13 litres of water is also added in it. If the final mixture has 20% water concentration, then find the ratio of Alcohol, Soda and Water in the final mixture.

Mixture A has Alcohol:Soda = 5:8, so 5/13 alcohol and 8/13 soda. Mixture B has Alcohol:Soda = 3:7, so 3/10 alcohol and 7/10 soda. Equal quantities (let's say x units each) are mixed. Total alcohol = x(5/13 + 3/10) = x(98/130) = 49x/65. Total soda = x(8/13 + 7/10) = x(171/130) = 171x/130. Total volume = 2x. Water added = 13 liters. Final water concentration = 20%, so 13/(2x + 13) = 0.2. Solving: 13 = 0.4x + 2.6, so x = 26. Then alcohol = 49×26/65 = 19.6, soda = 171×26/130 = 34.2, water = 13. Ratio = 19.6 : 34.2 : 13 = 196 : 342 : 130 = 98 : 171 : 65 after simplification.