A mixture of milk and water contains 66 (2/3)% water and 100 litres milk. '2p' litres of mixture is now removed and 20 litres milk is added. Find the quantity of water in resultant mixture if quantity of water is 50% more than that of milk in resultant mixture.

Total quantity of initial mixture = (100/0.3333) = 300 litres Initial quantity of water = 200 litres Ratio of initial quantity of milk and water respectively = 100:200 = 1:2 Quantity of water removed = (2/3) X 2p = (4p/3) litres Quantity of milk removed = (1/3) X 2p = (2p/3) litres Quantity of water in resultant mixture = 200 - (4p/3) = [(600 - 4p) /3] litres Quantity of milk in resultant mixture = 100 - (2p/3) + 20 = [(360 - 2p) /3] litres So, [(600 - 4p) /3]: [(360 - 2p) /3] = 3:2 Or, 2 X (600 - 4p) = 3 X (360 - 2p) Or, 1,200 - 8p = 1,080 - 6p Or, 2p = 120 Or, 'p' = 60 Therefore, required quantity of water = (600 - 4 X 60) ÷ 3 = (360/3) = 120 litres