Alligation or Mixture problems
The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?
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
Let he mixes the oils in the ratio = x : y Then, the cost price of the oils = 60x + 65y Given selling price = Rs. 68.20 => Selling price = 68.20(x+y) Given profit = 10% = SP - CP => 10/100 (60x + 65y) = 68.20(x+y)-(60x + 65y) => 6x + 6.5y = 8.20x + 3.20y =>2.2x = 3.3y => x : y = 3 : 2
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
According to figure we can find that the ration would be 1 : 7.Quantity sold at 10% profit = 1 / (1 + 7)× 160 = 20 kgs. Quantity sold at 6% loss = (160 ? 20) = 140 kgs.
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
EXAMIANS