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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%

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 alcoholin pure water (0) %Concentration of alcoholin 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%