CAT 2020 Slot 3 QA Question & Solution
Question
Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volume of the mixture is then doubled by adding solution A such that the resulting mixture has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in solution B is
Options
Solution
Initially let's consider A and B as one component
The volume of the mixture is doubled by 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 , $\frac{\left(72-60\right)}{x-72}=\frac{1}{1}$ => x= 84
Percentage of alcohol in A = 60% => Let's percentage of alcohol in B = x%
The resultant mixture has 84% alcohol. ratio = 1:3
Using allegations , $\frac{\left(x-84\right)}{84-60}=\frac{1}{3}$
=> x= 92%