Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volumeĀ of the mixture is then doubled by adding solution A such that the resulting mixtureĀ has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol inĀ solution B is
Initially let's consider A and B as one component
The volume of the mixture is doubled byĀ adding A(60% alcohol) i.e they are mixed in 1:1 ratio and the resultant mixture has 72% alcohol.
Let the percentage of alcohol in component 1 be 'x'.
Using allegations ,Ā $$\frac{\left(72-60\right)}{x-72}=\frac{1}{1}$$ => x= 84
Percentage of alcohol inĀ A = 60% => Let's percentage of alcohol in B = x%
Ā The resultant mixture has 84% alcohol. ratio = 1:3
Using allegations ,Ā $$\frac{\left(x-84\right)}{84-60}=\frac{1}{3}$$
=> x= 92%
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