Alligation or Mixture problems
The diluted wine contains only 8 liters of wine and the rest is water. A new mixture whose concentration is 30%, is to be formed by replacing wine. How many liters of mixture shall be replaced with pure wine if there was initially 32 liters of water in the mixture?
Wine Water 8L 32L 1 : 4 20 % 80% (original ratio) 30 % 70% (required ratio) In ths case, the percentage of water being reduced when the mixture is being replaced with wine. so the ratio of left quantity to the initial quantity is 7:8 Therefore , 7 8 = 1 - K 40 => K = 5 Lit
Let C.P. of 1 liter milk be Re. 1, Gain = 16 2/3 % = 50/3 %and S.P. of 1 liter mixture = Re. 1 then C.P. of 1 liter mixture = (1 x (100 x 3) / 350) = Re. (6 / 7) By the rule of alligation,Hence, required ratio = (1/ 7) : (6 / 7) = 1 : 6
Let the amount of juice and water in original mixture '4x' litre and '3x' litre respectively.
According to given data,
4x / 3x + 6 = 8/7
28x = 24x + 48
28 x ? 24x = 48
4x = 48
x = 12
Amount of juice = 4x = 4×12 = 48 litre.