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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
Statement: S ≤ M < X = H ≥ B ≥ K < V
Conclusion: X > K, K = X
Statements: T ≤ K = E ≤ Q, J = Q < H ≤ S ≤ V, A ≤ V < O = Y
Conclusions: I. T < V II. O ≥ H
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Statements: R < S < T < U; T < V < W
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