The quantity of water in vessel A is how much percent more than the quantity of P1 in vessel B?

          Milk       Water       P1 A         5x            2x             - B          2y               -             y                C           -              2x Vessel A, => 5x + 2x = 28 => x = 4 Quantity of milk in vessel A = 5x = 20 Quantity of water in vessel A = 2x = 8 Quantity of milk in vessel B, => 2y + 20 = 40 => y = 10 litres Quantity of milk in vessel B = 20 litres Quantity of P1 in vessel C = (1500/60) = 25 litres Now, ratio of quantities of mixture in B and C is 6:13. => 30/C = 6/13 => C = 65 Quantity of mixture in C = 65 litres So, 2x + 25 = 65 => 2x = 40 litres Quantity of milk in vessel A = 40 litres Quantity of water in vessel A = 100 litres Required % = [(40 – 10)/10] x 100 = 300%