Study the two cases and answer in terms of relation between the two cases Case 1: Two cans are filled with the mixture of milk and water. The first can contain 72% of water and second can contains 60% of water. A new mixture is prepared by taking 9 part of first can and 6 part of second can. The percentage of milk in the new mixture is Case 2: For testing, 80 ml of milk and water mixture is taken from a container which has 62% water by weight. If in the test tube, additional 20% water of the total water is added, then the % of milk in the test tube is?

Option 2 : Case 2 > Case 1

Case 1:

Ratio of milk and water in the first can = 28% : (100 -28%) = 28% : 72% = 7 : 18

Ratio of milk and water in the second can = 40% : (100 - 40%) = 40% : 60% = 2 : 3

Amount of milk taken from the first can = 9 × 28% = 252%

Amount of milk taken from the second can = 6 × 40% = 240%

Total part of milk in new solution = 15 part

% of milk per part = (252% + 240%)/15 = 32.8%

Case 2:

Ratio of milk to water in the container = (100 - 62)% : 62% = 38% : 62% = 38 : 62

Amount of water in 80 ml mixture = 62% of 80 = (62 × 80)/100 = 49.6 ml

Amount of milk = 80 ml - 49.6 ml = 30.4 ml

20% of additional water is further added to the mixture,20% of 49.6 ml

(20 × 49.6)/100 = 9.92 ml

Total amount of water in the mixture = (49.6 + 9.92) = 59.52 ml

Amount of milk in the mixture remains same 

Total quantity = 59.52 ml + 30.6 ml = 89.92 ml

% of milk = (30.4/89.92) × 100% = 33.80%

∴ From both the solutions, it is clear that the value of case 2 > case 1