A vessel contains a mixture of milk and water in the respective ratio of 10: 3. Twenty six litre of this mixture was taken out and replaced with 8 litre of water. If the resultant respective ratio of milk and water in the mixture was 5: 2, what was the initial quantity of mixture in the vessel? (In litre)

Let the initial quantity be 13x liters (since milk:water = 10:3). Initial milk = 10x, water = 3x. After removing 26 liters (which maintains 10:3 ratio): milk removed = 20L, water removed = 6L. Remaining milk = 10x-20, remaining water = 3x-6. After adding 8L water: new milk = 10x-20, new water = 3x-6+8 = 3x+2. The new ratio is 5:2, so (10x-20)/(3x+2) = 5/2. Cross-multiplying: 20x-40 = 15x+10. Solving: 5x = 50, x = 10. Initial quantity = 13×10 = 130L. Verifying: 10(10)-20 = 80L milk, 3(10)+2 = 32L water, ratio = 80:32 = 5:2 ✓