A chemist has two containers A and B of equal volumes of 35 litres each. Container A initially contains a mixture of X and water in the ratio 2 : 3, while container B initially contains a mixture of Y and water in the ratio 3 : 4. The chemist decides to prepare a mixture in which the concentration of liquids X and Y should be 33 1 3 % of the total solution. However, while transferring the liquids, the chemist accidentally pours 20% of the mixture from container A and 30% of the mixture from container B. Now how much water should be mixed in the resultant mixture to get the targeted result?

Pouring 20% of mixture from container A means transferring 7 liters, containing 2.8 liters of X and 4.2 liters of water. Pouring 30% of mixture from container B means transferring 10.5 liters, containing 4.5 liters of Y and 6 liters of water. The total resulting mixture is 17.5 liters, and to achieve a 33.33% concentration of X and Y, the required amount of these liquids must be 17.5 / 3 = 5.833 liters; since the available amount is 2.8 + 4.5 = 7.3 liters, the required amount of water to be mixed is 17.5 - 5.833 - 7.3 = 4.367 liters.

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