A 42 litres of alcohol and water mixture placed in a can in which alcohol and water ratio is 4 : 3. Then 7 litres of mixture is removed from the can and replaced with pure alcohol. This procedure is done a total for 2 times. After this two times removal, the difference between the quantity of water in the original mixture and the resultant mixture is – Correct Answer 5.5 litres Given: Quantity of alcohol in mixture = 42 × 4 / 7 = 24 litres Quantity of water in the mixture = 42 – 24 = 18 litres Calculation: For 7 litres mixtures, Quantity of alcohol = 7 × 4 / 7 = 4 litres Quantity of water = 7 × 3 / 7 = 3 litres After adding, 7 litres of alcohol, Quantity of alcohol in mixture = 24 – 4 + 7 = 27 litres Quantity of water = 18 – 3 = 15 litres Ratio of Alcohol and water in mixture = 27 : 15 = 9 : 5 For another replacement, Quantity of alcohol in 7 litres mixture = Quantity of alcohol = 9 / 14 × 7 = 4.5 litres Quantity of water = 5 / 14 × 7 = 2.5 litres On adding 7 litres of pure alcohol, Quantity of alcohol = 27 – 4.5 + 7 = 29.5 litres Quantity of water = 15 – 2.5 = 12.5 litres Required difference = 18 – 12.5 = 5.5 litres