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

• A. 94%
• B. 92%
• C. 90%
• D. 89%

Answer 1

Answer: (B)

92%

Answer 2

The correct answer is (B): $92\%$

(Let the quantity of solutions $A$ and $B$ mixed initially be $p$ and $3p$ respectively.

After an additional $4p$ of solution $A$ is added $60\%$ of $(1p+4p)+x\%$ of $3p = 72\%$ of $(1p+4p+3p)⇒x=92$