Question
In a liquid solution, there is 300 ml of alcohol, which
makes up 60% of the total volume. After removing 60 ml of the mixture, 48 ml of water is added. Then 40 ml of the new mixture is removed, and 20 ml alcohol is added. Find the difference between the quantity of alcohol and water in the final solution.Solution
ATQ,
Total quantity of mixture = (300 / 0.60) = 500 ml
Quantity of water initially = 500 - 300 = 200 ml
Ratio of alcohol to water = 300 : 200 = 3 : 2
After removing 60 ml mixture:
Alcohol = 300 - (3/5) × 60 = 264 ml
Water = 200 - (2/5) × 60 + 48 = 224 ml
After removing 40 ml mixture and adding 20 ml alcohol:
Alcohol = 264 - (3/5) × 40 + 20 = 260 ml
Water = 224 - (2/5) × 40 = 208 ml
Required difference = 260 - 208 = 52 ml
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