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

The Correct Option is B

To solve the problem, let us denote the volume of solution A by \( V_A \) and solution B by \( V_B \). According to the problem, solutions A and B are mixed in the ratio 1:3. Hence, we can write:

\( V_B = 3V_A \) 

The volume of the initial mixture becomes:
\( V_{\text{mixture}} = V_A + V_B = V_A + 3V_A = 4V_A \)

This initial mixture is then doubled by adding solution A, making the new volume:
\( V_{\text{new}} = 4V_A + 4V_A = 8V_A \)

We know that the resulting mixture has 72% alcohol. Therefore:
\( \text{Alcohol in new mixture} = 0.72 \times 8V_A = 5.76V_A \)

Since solution A contains 60% alcohol, the alcohol content from solution A in the new mixture is:
\( \text{Alcohol from A} = 0.60 \times 5V_A = 3V_A \)

The remaining alcohol must come from solution B:
\( \text{Alcohol from B} = 5.76V_A - 3V_A = 2.76V_A \)

The fraction of alcohol in solution B is:
\( \text{Alcohol fraction in B} = \frac{2.76V_A}{3V_A} = 0.92 \)

Therefore, the percentage of alcohol in solution B is:
92%

Answer 2

Let the quantity of alcohol in solution A be \(x\) and in solution B be \(3x\). 

Let the concentration of alcohol in solution B be \(X\%\).

Total initial mixture = \(x + 3x = 4x\)

An equal amount of solution A (i.e., \(4x\)) is added.

New total quantity = \(4x + 4x = 8x\)

New amount of solution A = \(x + 4x = 5x\), amount of solution B = \(3x\)

Given that the final alcohol concentration is 72%, we write:

\[\frac{5x \times 60 + 3x \times X}{8x} = 72\]

Multiply both sides by \(8x\): 
 

\[300x + 3xX = 576x\]

Subtracting \(300x\): 
 

\[3xX = 276x \Rightarrow X = \frac{276x}{3x} = 92\]

Therefore, the concentration of alcohol in solution B is 92%.