Alligation or Mixture problems
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
Suppose the vessel initially contains 8 litres of liquid. Let x litres of this liquid be replaced with water. Quantity of water in new mixture = ❨ 3 - 3x + x ❩ litres 8 Quantity of syrup in new mixture = ❨ 5 - 5x ❩ litres 8 ∴ ❨ 3 - 3x + x ❩ = ❨ 5 - 5x ❩ 8 8 ⟹ 5x + 24 = 40 - 5x ⟹ 10x = 16 ⟹ x = 8 . 5 So, part of the mixture replaced = ❨ 8 x 1 ❩ = 1 . 5
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.
Here withdrawal of liquid A and B result into making the container empty. Hence percentage of two liquids withdrawn are two components of the percentage by which the container becomes empty. Applying the rule of alligation, we getA : B = 10 : 20 or 1 : 2 Quantity of liquid = 1 (1 + 2) × 90 = 30 liters Quantity of liquid B = 90 ? 30 = 60 liters.
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
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