Quantitative Aptitude
MIXTURES AND ALLEGATIONS MCQs
Alligations And Mixtures
Percentage of milk in the mixture formed = [80/100 (180) + 36] / (180 + 36) * 100% = (144 + 36)/216 * 100% = 5/6 * 100% = 83.33%.
Let the capacity of the can be T litres.Quantity of milk in the mixture before adding milk = 4/9 (T - 8)After adding milk, quantity of milk in the mixture = 6/11 T.6T/11 - 8 = 4/9(T - 8)10T = 792 - 352 => T = 44.
Quantity of milk in the mixture = 90/100 (70) = 63 litres.After adding water, milk would form 87 1/2% of the mixture.Hence, if quantity of mixture after adding x liters of water, (87 1/2) / 100 x = 63 => x = 72Hence 72 - 70 = 2 litres of water must be added.
Let us say the ratio of the quantities of cheaper and dearer varieties = x : yBy the rule of allegation, x/y = (87.5 - 7.50) / (7.50 - 6) = 5/6
B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k.
Quantity of water in B = 8k - 5k = 3k.Quantity of water in container C = 8k - 3k = 5kContainer: A B CQuantity of water: 8k 3k 5kIt is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water.5k - 148 = 3k + 148 => 2k = 296 => k = 148The initial quantity of water in A = 8k = 8 * 148 = 1184 liters.
Quantity of alcohol in vessel P = 62.5/100 * 2 = 5/4 litresQuantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litresQuantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litresAs 6 litres of mixture is formed, ratio of alcohol and water in the mixture formed = 4.75 : 1.25 = 19 : 5.
Let the initial quantity of milk in vessel be T litres.Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.Quantity of milk finally in the vessel is then given by [(T - y)/T]n * T For the given problem, T = 90, y = 9 and n = 2.Hence, quantity of milk finally in the vessel = [(90 - 9)/90]2 (90) = 72.9 litres.