Alligation or Mixture problems
In a container, there is 960 ltr of pure milk from which 48 ltr of milk is replaced with 48 ltr of water, again 48 ltr milk is replaced by same amount of water, as this process is done once more. Now, what is the amount of pure milk?
Number of liters of water in 150 liters of the mixture = 20% of 150 = 20/100 x 150 = 30 liters.P liters of water added to the mixture to make water 25% of the new mixture.Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).(30 + P) = 25/100 x (150 + P)120 + 4P = 150 + P => P = 10 liters.
Water in 60 gm mixture=60 x 75/100 = 45 gm. and Milk = 15 gm. After adding 15 gm. of water in mixture, total water = 45 + 15 = 60 gm and weight of a mixture = 60 + 15 = 75 gm. So % of water = 100 x 60/75 = 80%.
Milk in 1-litre mixture of A = 4/7 litre. Milk in 1-litre mixture of B = 2/5 litre. Milk in 1-litre mixture of C = 1/2 litre. By rule of alligation we have required ratio X:Y X : Y 4/7 2/5 \ / (Mean ratio) (1/2) / \ (1/2 ? 2/5) : (4/7 ? 1/2) 1/10 1/1 4 So Required ratio = X : Y = 1/10 : 1/14 = 7:5