Alligation or Mixture problems
A beaker contains acid and water in the ratio 1 : x. When 300 ml of the mixture and 50 ml of water are mixed, the ratio of acid and water becomes 2 : 5. What is the value of x?
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
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.
Ratio of Milk and water in a vessel A is 4 : 1 Ratio of Milk and water in a vessel B is 3 : 2 Ratio of only milk in vessel A = 4 : 5 Ratio of only milk in vessel B = 3 : 5 Let 'x' be the quantity of milk in vessel C Now as equal quantities are taken out from both vessels A & B => 4/5 : 3/5 x 3/5-x x - 4/5 => 3 5 - x x - 4 5 = 1 1 (equal quantities) => x = 7/10 Therefore, quantity of milk in vessel C = 7 => Water quantity = 10 - 7 = 3 Hence the ratio of milk & water in vessel 3 is 7 : 3
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
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