Alcohol concentration mixture

CAT 2020 Slot 3 · QA · Medium · Ratios and proportions

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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

  1. A.

    94%

  2. B.

    92%

  3. C.

    90%

  4. D.

    89%

Answer

B

Explanation

Let initial mixture have 1 unit of A and 3 units of B (total volume = 4 units). Volume is doubled by adding 4 units of solution A. Final composition: 5 units of A and 3 units of B (total volume = 8 units).

Alcohol in 5 units of A = 5×0.60=3.05 \times 0.60 = 3.0 units. Let alcohol percentage in B be b%b\%. Total alcohol in final mixture = 8×0.72=5.768 \times 0.72 = 5.76 units.

3.0+3×b100=5.76  ⟹  3×b100=2.76  ⟹  b=92%3.0 + 3 \times \frac{b}{100} = 5.76 \implies 3 \times \frac{b}{100} = 2.76 \implies b = 92\%.

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