Question:medium

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

Updated On: Jan 15, 2026
  • 94%
  • 92%
  • 90%
  • 89%
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The Correct Option is B

Solution and Explanation

Let \( V_A \) be the volume of solution A and \( V_B \) be the volume of solution B. Given that solutions A and B are mixed in a 1:3 ratio, we have \( V_B = 3V_A \). The total volume of the initial mixture is \( V_{\text{mixture}} = V_A + V_B = V_A + 3V_A = 4V_A \). This mixture is then doubled by adding more solution A, resulting in a new volume of \( V_{\text{new}} = 4V_A + 4V_A = 8V_A \). The final mixture contains 72% alcohol, so the amount of alcohol in the new mixture is \( 0.72 \times 8V_A = 5.76V_A \). Solution A contains 60% alcohol. The alcohol contributed by solution A to the new mixture is \( 0.60 \times 5V_A = 3V_A \). The remaining alcohol must originate from solution B: \( \text{Alcohol from B} = 5.76V_A - 3V_A = 2.76V_A \). The volume of solution B in the new mixture is \( V_B = 3V_A \). Therefore, the alcohol fraction in solution B is \( \frac{2.76V_A}{3V_A} = 0.92 \). This indicates that solution B contains 92% alcohol.

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