Mixtures and Replacements 1/1

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1. A cask contains alcohol solution with concentration 18%. If 6 Liters of this mixture is taken out and replaced with water, then concentration drops to 15%. What is the original volume in the cask.
a. 30 litres
b. 36 litres
c. 40 litres
d. 54 litres

Answer: B

Explanation:

You can observe from the diagram, concentration remains the same after 6 liters were removed from the cask.  Concentration changes inversely proportional to the volume of water added to the mixture.  So change in the concentration happened only from second stage to third change.

Method 1:
We know that total alcohol component in the mixture is equal to alcohol component in the mixture that has been taken out plus remaining alcohol in the cask. so assume initial or final volume is V liters.
18% (V) = 18% (6) + 15% (V)
3% (V) = 18% (6)
V = 36

Method 2:
Volume and concentration are inversely proportional to each other.
$IV$ = initial volume; $IC$ = initial concentration; $FV$ = Final volume; $FC$ = Final concentration
$IV \times IC = FV \times FC$
$(V-6) \times 18% = V \times 15%$
$\dfrac{{V - 6}}{v} = \dfrac{6}{5}$
So $V = 36$

Method 3:
We can also solve this problem by using alligation rule. It states that in what ration two components are mixed to get a targeted concentration. We can apply this rule to this problem for the second stage. We added water which is at 0% concentration to a mixture of 18% concentration to get a solution with 15% concentration.

So we understand from the above diagram Mixture and water should be mixed in the ration 15:3 to get desired concentration 15%. But we know that 3 units of water is equal to 6 liters so 15 units of mixture is equal to 30 liters. Total volume is equal to 30 +6 = 36. Please note that in the second stage the volume is equal to (V - 6)

Method 4:
Initial Condition 18 % = ( A : W) = 18 : 82
Final Condition 15 % = (A : W) = 15 : 85

We know that there is no change in the Pure alcohol component from second stage to third stage. so we can equate alcohol component in the above two equation by multiplying with appropriate numbers. Now we observe a change in the water components from 41 to 51. This is due to the water we added to the mixture. We added 6 liters of water which is equal to 10 units change in the mixture. so
10 Units = 6 Liters
60 Units = 36 Liters

Method 5:
We can use this formula
$ \Rightarrow FC = IC{\left( {1 - \displaystyle\frac{x}{v}} \right)^n}$
$ \Rightarrow 15\% = 18\% {\left( {1 - \displaystyle\frac{x}{v}} \right)^n}$
Here n = 1 because we made this substitution only once.
$ \Rightarrow \displaystyle\frac{{15}}{{18}} = \left( {1 - \frac{6}{v}} \right)$
$ \Rightarrow \displaystyle\frac{5}{6} = 1 - \frac{6}{v}$
$ \Rightarrow \displaystyle\frac{6}{v} = 1 - \frac{5}{6}$
$ \Rightarrow v = 36$

2. 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 ?
a. 3 : 5
b. 4 : 7
c. 5 : 7
d. 3 : 2

Answer: D

Explanation:
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,
$\displaystyle\frac{1}{3}{\rm{ = }}\displaystyle\frac{3}{7} \times {\left( {{\rm{1 - }}\displaystyle\frac{{10}}{V\strut}} \right)^1}$
$ \Rightarrow \displaystyle\frac{7}{9}{\rm{ = 1 - }}\frac{{10}}{V\strut}$
$ \Rightarrow \displaystyle\frac{2}{9}{\rm{ = }}\frac{{10}}{V\strut}$
$ \Rightarrow V = 45$

3. 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.
a. 4 : 9
b. 4 : 5
c. 5 : 4
d. 9 : 4

Answer: B

Explanation:
Applying formula:
$ \Rightarrow \displaystyle\frac{2}{5}{\rm{ = K \times }}\left( {{\rm{1 - }}\displaystyle\frac{{10}}{{100}}} \right)$
Here 2/5 is the milk concentration.
$ \Rightarrow \displaystyle\frac{2}{5}{\rm{ = K \times }}\displaystyle\frac{9}{{10}} \Rightarrow K = \displaystyle\frac{4}{9}$
Milk : Water = 4 : 5

4. A beaker had 20 L of 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.
a. 15%
b. 18%
c. 24%
d. 76%

Answer: D

Explanation:
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%
Final alcohol concentration $ = 80\left( {1 - \dfrac{4}{{20}}} \right)\left( {1 - \dfrac{5}{{20}}} \right)\left( {1 - \dfrac{{10}}{{20}}} \right)$
$ = 80\left( {\dfrac{{16}}{{20}}} \right)\left( {\dfrac{{15}}{{20}}} \right)\left( {\dfrac{{10}}{{20}}} \right)$
$\require{cancel} = 80\left( {\dfrac{\cancel{16}^4}{\cancel{20}^5}} \right)\left( {\dfrac{\cancel{15}^3}{\cancel{20}^4}} \right)\left( {\dfrac{\cancel{10}}{\cancel{20}_2}} \right)$
$= 24%$
Glycerol concentration = $100 - 24 = 76%$

5. 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.
a. 10 : 7
b. 12 : 7
c. 23 : 7
d. 18 : 7

Answer: C

Explanation:
Here, we are adding pure milk. So we have to put water concentration in the formula.
Final volume of the mixture = 80 - 24 + 16 = 72
Replacement quantity = 16
Applying formula,
$ \Rightarrow FC = \displaystyle\frac{3}{{10}} \times \left( {1 - \displaystyle\frac{{16}}{{72}}} \right)$
$ \Rightarrow FC = \dfrac{3}{{10}} \times \displaystyle\frac{7}{9} \Rightarrow \displaystyle\frac{7}{{30}}$
Water : Milk = 7 : 23
So Milk and Water after replacement = 23 : 7

Alligation and Mixtures: Formulas Alligation Rule: Exercise Mixtures and Replacements: Exercise