A mixture has two solutions A and B mixed in a ratio of 1: 4. The total volume of the mixture is doubled by adding solution A such that the resultant mixture has 60% alcohol. If solution B has 15% alcohol, then what is the percentage of alcohol in solution A?

Let's solve the problem step by step:

Step 1: Let's assume the initial volume of solution A in the mixture is x liters. Since the ratio of solution A to solution B is 1:4, the initial volume of solution B in the mixture would be 4x liters.

Step 2: The total volume of the mixture is doubled by adding solution A. So, the final volume of the mixture would be 2 times the initial volume, which is 2(x + 4x) = 10x liters.

Step 3: We are given that the resultant mixture has 60% alcohol. This means that the total amount of alcohol in the mixture is 60% of the total volume of the mixture.

Step 4: The amount of alcohol in solution A is x liters multiplied by the percentage of alcohol in solution A (which we need to find). Similarly, the amount of alcohol in solution B is 4x liters multiplied by the percentage of alcohol in solution B, which is 15%.

Step 5: Since the total amount of alcohol in the mixture is 60% of the total volume of the mixture, we can set up the equation:

(x * percentage of alcohol in solution A) + (4x * 15%) = 60% * 10x

Step 6: Simplifying the equation, we get:

(x * percentage of alcohol in solution A) + (0.6x) = 6x

Step 7: Combining like terms, we have:

x * (percentage of alcohol in solution A + 0.6 - 6) = 0

Step 8: Simplifying further, we get:

x * (percentage of alcohol in solution A - 5.4) = 0

Step 9: Since x cannot be zero (as it represents the initial volume of solution A), we can conclude that the expression in the parentheses must be equal to zero. Therefore:

percentage of alcohol in solution A - 5.4 = 0

Step 10: Solving for the percentage of alcohol in solution A, we find:

percentage of alcohol in solution A = 5.4

Therefore, the percentage of alcohol in solution A is 5.4%.