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Quantitative aptitude for competitive exams –  MIXTURE AND ALLIGATION Part 2

Quantitative aptitude for competitive exams -  MIXTURE AND ALLIGATION Part 2:SUCCESSIVE REPLACEMENT

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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  1. 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%

 

  1. 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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March 19, 2018

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