A vessel contains a mixture of milk and water in the respective ratio 3 : 1. 32 litre of mixture was taken out and replaced with the same quantity of milk so that the resultant ratio between the quantities of milk and water in the mixture was 4 : 1 respectively. If 10 litre of mixture is again taken out from the vessel, what is the resultant quantiy of water in the mixture ?

Initial ratio 3:1 means 4 parts total. Let 4x = total mixture. Milk = 3x, water = x. After removing 32L and adding 32L milk, new ratio is 4:1, so milk = 4y, water = y. Water unchanged: y = x - 32 × (1/4) = x - 8. Total quantity same: 4x = 4x - 32 + 32 = 4x. From ratio: 3x - 32 × (3/4) + 32 = 4(x - 8). Solving gives x = 40. Initial water = 40L, final water = 40L. After removing 10L mixture (removes 2.5L water), water = 37.5L. Wait - let me recalculate. Initial: 4x total, 3x milk, x water. After replacement: (3x - 24 + 32) : (x - 8) = 4:1. So (3x + 8)/(x - 8) = 4/1, giving 3x + 8 = 4x - 32, so x = 40. Final water = 40 - 8 = 32L. After removing 10L mixture (which has 1/5 water = 2L), water = 32 - 2 = 30L.