Alligation or Mixture problems
In a 40 litre mixture of alcohol & water, the ratio of alcohol and water is 5 : 3. If 20% of this mixture is taken out and the same amount of water is added then what will be the ratio of alcohol and water in final mixture?
Quantity of alohol in the mixture = 40 x 5/8 = 25 lit Quantity of water = 40 - 25 = 15 lit According to question, Required ratio = 20 - 40 x 20 100 x 5 8 15 - 40 x 20 100 x 3 8 + 40 x 20 100 = 20 15 - 3 + 8 = 1 : 1
% of milk in first bottle = 64% % of milk in second bottle = 100 - 26 = 74% Now, ATQ 64% 74% 68% 6 4 Hence, by using allegation method, Required ratio = 3 : 2
General Formula: Final or reduced concentration = initial concentration x 1 - a m o u n t b e i n g r e p l a c e d i n e a c h o p e r a t i o n t o t a l a m o u n t n where n is the number of times the same operation is being repeated. The "amount being replaced" could be pure or mixture as per the case. similarly ,"total amount" could also be either pure or mixture. Here amount being replaced denotes the quantity which is to be withdrawn in each time. Therefore, 50 × 1 - 5 50 3 = 36.45 L
Let cost price of spirit be Re. 1 per liter. Then SP of mixture = Re. 1 per liter Gain = 25% So, CP of mixture = 1 × (100 / 125) = Re. 4 / 5We assume that CP of water is zero. Using allegation rule on cost price, Water should be mixed to spirit in the ratio (1 / 5) : (4 / 5) or 1 : 4
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.
By the rule of alligation: C.P. of 1 kg sugar of 1st kind C.P. of 1 kg sugar of 2nd kind Therefore, Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3. Let x kg of sugar of 1st kind be mixed with 27 kg of 2nd kind. Then, 7 : 3 = x : 27 or x = (7 x 27 / 3) = 63 kg.
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