Acid and water are mixed in a vessel A in the ratio of 5 : 2 and in the vessel B in the ratio 8 : 5. In what proportion should quantities be taken out from the two vessels so as to form a mixture in which the acid and water will be in the ratio of 9 : 4?

Vessel A: acid ratio = 5/7, Vessel B: acid ratio = 8/13. Target ratio = 9/13. Using alligation: For acid concentration - A has 5/7 ≈ 0.714, B has 8/13 ≈ 0.615, target 9/13 ≈ 0.692. (0.714 - 0.692) : (0.692 - 0.615) = 0.022 : 0.077 ≈ 2:7. So take from A and B in ratio 2:7. Let's verify: 2 units from A (acid = 2 × 5/7 = 10/7, water = 2 × 2/7 = 4/7) and 7 units from B (acid = 7 × 8/13 = 56/13, water = 7 × 5/13 = 35/13). Total acid = 10/7 + 56/13 = (130 + 392)/91 = 522/91. Total water = 4/7 + 35/13 = (52 + 245)/91 = 297/91. Ratio = 522:297 = 174:99 = 58:33 ≠ 9:4. Let me re-check alligation method... Actually for 9:4 ratio, acid fraction = 9/13. Distance from A (5/7) to 9/13: 9/13 - 5/7 = (63-65)/91 = -2/91. Distance from B (8/13) to 9/13: 9/13 - 8/13 = 1/13 = 7/91. Ratio A:B = 2:7 is correct by alligation principle.