A mixture of milk and water contains 70% water and 90 litres milk. '2p' litres of mixture is now removed and 40 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 = (90/0.30) = 300 litres Initial quantity of water = 0.70 X 300 = 210 litres Ratio of initial quantity of milk and water respectively = 90:210 = 3:7 Quantity of water removed = (7/10) X 2p = (7p/5) litres Quantity of milk removed = (3/10) X 2p = (3p/5) litres Quantity of water in resultant mixture = 210 - (7p/5) = [(1050 - 7p) /5] litres Quantity of milk in resultant mixture = 90 - (3p/5) + 40 = [(650 - 3p) /5] litres So, [(1050 - 7p) /5]: [(650 - 3p) /5] = 3:2 Or, 2 X (1,050 - 7p) = 3 X (650 - 3p) Or, 2,100 - 14p = 1,950 - 9p Or, 5p = 150 Or, 'p' = 30 Therefore, required quantity of water = (1,050 - 7 X 30) ÷ 5 = (840/5) = 168 litres