Alligation or Mixture problems
A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is:
By the rule of alligation, we have: Strength of first jar Strength of 2nd jar 40% MeanStrength 26% 19% 7 14 So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2 ∴ Required quantity replaced = 2 3
Using Alligation rule, (Quantity of cheaper tea) / (Quantity of dearer tea) = (d - m) / (m - c) = 7/3Therefore, they must be mixed in the ratio of 7 : 3.
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.
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
EXAMIANS