'X', 'Y', and 'Z' are three mixtures of milk and water.

Solution

Let the quantities of mixtures 'X', 'Y', and 'Z' in the final mixture be 3a, 5a, and 2a, respectively. Now, calculate the amount of milk in each mixture: Milk in 'X' = (4/9) × 3a = (12a/9) = (4a/3) Milk in 'Y' = (2/5) × 5a = (10a/5) = 2a Milk in 'Z' = (3/7) × 2a = (6a/7) Total amount of milk = (4a/3) + 2a + (6a/7) LCM of 3 and 7 = 21, so: (4a/3) = (28a/21), (2a) = (42a/21), (6a/7) = (18a/21) Total amount of milk = (28a + 42a + 18a) / 21 = (88a/21) Now, calculate the total quantity of the final mixture: Total mixture = 3a + 5a + 2a = 10a The concentration of milk = (Total amount of milk / Total mixture) × 100 = (88a/21) / 10a × 100 = (88/210) × 100 = 41.90% ≈ 42% Correct option: b