CAT 2020 Question Paper (Slot 3) Question 51

Question 51

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


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%

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6 months, 1 week ago

Nice explanation


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