CAT 2020 Exam Shift 3 Solved Paper

Question : 72 of 76

Marks: +1, -0

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}

Solution:

Let the quantity of solutions

A

and

B

mixed initially be

p

and

3 p

respectively.
After an additional

4 p

of solution

A

is added

60 \%

(1 p+4 p)+x \%

3 p=72 \%

(1 p+4 p+3 p) ⇒ x=92

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