A mixture contains milk & water in the ratio of 7:4 & another mixture contain them in the ratio of 3:5. How many li t r e s of the f o rmer m u st be mixed with 12 litr e s of the other so that the result ant mixture may contain equal quantities of milk & water?

Let the quantity of former mixture mixed be x litres. In former mixture, Milk : Water = 7 : 4 So, quantity of milk in x litres of former mixture = (7/11)x Quantity of water in x litres of former mixture = (4/11)x In the other mixture, Milk : Water = 3 : 5 Total quantity of other mixture = 12 litres So, quantity of milk in 12 litres of other mixture = 12 × (3/8) = 4.5 litres Quantity of water in 12 litres of other mixture = 12 × (5/8) = 7.5 litres According to the question, Resultant mixture contains equal quantities of milk and water. So, (7/11)x + 4.5 = (4/11)x + 7.5 Or, (7/11)x - (4/11)x = 7.5 - 4.5 Or, (3/11)x = 3 Or, x = 3 × 11/3 = 11 Therefore, 11 litres of the former mixture must be mixed.