Quantitative aptitude for competitive exams MIXTURE AND ALLIGATION Part 2

Quantitative aptitude for competitive exams MIXTURE AND ALLIGATION Part 2

SUCCESSIVE REPLACEMENT:

Consider an example to understand the concept of successive replacement.

If a vessel has x litres of milk and y litre of milk is removed from it and replaced with y litres of water the milk left in the vessel after the replacement is (x-y) litres and the total milk water solution available in the vessel is again x litres (x – y +y). so, after the replacement the proportion of milk in the milk water solution will be equal to:

Some examples:-

1. From a solution containing milk and water in the ratio 3 : 4, 10 L is removed and replaced by water. If the resultant solution contains milk and water in the ratio 1 : 2 then what was the amount of the original solution?
   Solution:
   Here also we are replacing with water.  So FC and IC must be milk concentrations.
   Initial concentration of the milk = 3/7
   Final concentration of the milk = 1/3
   Applying formula
   1/3=3/7×(1−10/V)¹
   ⇒7/9=1−10/V
   ⇒2/9=10V
   ⇒V=45
2. 10% of a solution of milk and water is removed and then replaced with the same amount of water. If the resulting ratio of milk and water is 2 : 3, find the ratio of milk and water in the original solution.
   Solution:
   Applying formula:
   ⇒2/5=K×(1−10/100)
   Here 2/5 is the milk concentration.
   ⇒2/5=K×9/10

⇒K=4/9

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2. A beaker had 20 L of the alcohol-glycerol mixture in the ratio 4 : 1 by volume. In the first round, 4 L of the mixture is removed and replaced with glycerol. In the second round, 5 L of the resultant solution is removed and replaced with glycerol. Finally, 10 L of the resultant mixture is removed and replaced with glycerol. What is the final quantity of glycerol in the mixture
   Solution:
   Here we are replacing the mixture with glycerol.  So we have to take Alcohol concentrations for IC and FC.
   Initial concentration of alcohol is 4/5 = 80%
   Applying the formula for the first replacement:
   ⇒FC¹=80%×(1−4/20)
   Here FC¹ is the concentration after first replacement.
   Second replacement:
   ⇒FC²=FC¹×(1−5/20)
   Third Replacement:
   ⇒FC³=FC²×(1−10/20)
   Now substituting the FC¹ and FC² in FC³ we get

⇒FC³=80%×(1−4/20)×(1−5/20)×(1−10/20)
⇒FC3=80%×4/5×3/4×1/2

⇒FC³=24%

3. In a mixture of 80 L, milk and water are in the ratio 7:3. If 24 L of this mixture is replaced by 16 L if milk, find the final ratio of milk and water.
   Solution:
   Final volume of the mixture = 80 – 24 + 16 = 72
   Replacement quantity = 16
   Applying formula,
   ⇒FC=3/10×(1−16/72)
   ⇒FC=3/10×7/9⇒7/30
   So Milk and Water after replacement = 23 : 7

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