Alligation or Mixture problems
A milk man sells the milk at the cost price but he mixes the water in it and thus he gains 9.09%. The quantity of water in the mixture of 1 liter is :
Profit (%) = 9.09 % = 1/11 Since the ratio of water and milk is 1 : 11, Therefore the ratio of water is to mixture = 1:12 Thus the quantity of water in mixture of 1 liter = 1000 x (1/12) = 83.33 ml
Ratio of milk and water = 2 : 1Quantity of milk = 60 X 2/3 = 40 litreQuantity of water = 20 litreTo make ratio, 1: 2, we have to double the water that of milkSo, water should be 80 litre.That means 80 ? 20 = 60 litre water to be added.
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 C.P. of 1 litre milk be Re. 1. S.P. of 1 litre of mixture = Re.1, Gain = 50 %. 3 ∴ C.P. of 1 litre of mixture = ❨ 100 x 3 x 1 ❩ = 6 350 7 By the rule of alligation, we have: C.P. of 1 litre of water C.P. of 1 litre of milk 0 Mean PriceRe. 6 7 Re. 1 1 7 6 7 ∴ Ratio of water and milk = 1 : 6 = 1 : 6.
From the given data, let the initial quantity of the mixture = 5x Then, 2 x - 16 3 x - 24 + 40 = 1 4 8 x - 64 = 3 x + 16 5 x = 80 x = 16 lit Then the initial quantity of the mixture = 5x = 5 x 16 = 80 lit.
EXAMIANS