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Mixture and Alligation #24

24. 3 litre of water is added to 11 litre of a solution containing 42% of alcohol in the water. The percentage of alcohol in the new mixture is
A. 20%B. 33%
C. 30%D. 25%

Answer: Option B

Explanation:

Solution 1

We have a 11 litre solution containing 42% of alcohol in the water.
=> Quantity of alcohol in the solution $=\dfrac{11 × 42}{100}$

Now 3 litre of water is added to the solution.
=> Total quantity of the new solution = 11 + 3 = 14

Percentage of alcohol in the new solution $=\dfrac{\dfrac{11 × 42}{100}}{14}× 100$
$=\dfrac{11 × 3}{100} = 33\%$


Solution 2

%Concentration of alcohol in pure water = 0
%Concentration of alcohol in mixture = 42
Quantity of water : Quantity of mixture = 3 : 11
Let the %concentration of alcohol in the new mixture $=x$

By rule of alligation,

%Concentration of alcohol
in pure water (0)
%Concentration of alcohol
in mixture(42)
Mean %concentration $(x)$
$42-x$$x-0=x$

But $(42 - x) : x = 3 : 11$
$\Rightarrow 11(42-x)=3x\\\Rightarrow 42 × 11 - 11x = 3x\\\Rightarrow 14x = 42 × 11 \\\Rightarrow x = 3× 11 = 33$

i.e., Percentage of alcohol in the new mixture is 33%

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