how many ounces of a 26% alcohol solution and a 44% alcohol solution must be combined to obtain 56 ounces of a 29% solution?
If there are x ounces of 26%, then the rest (56-x) is 44%.
So, just calculating the amount of alcohol in each part, and in the total, you have
0.26x + 0.44(56-x) = 0.29*56
Now find x, and 56-x
140/3
To find out how many ounces of a 26% alcohol solution and a 44% alcohol solution must be combined to obtain 56 ounces of a 29% solution, you can use a method called the mixture equation.
Let's assume x represents the ounces of the 26% solution needed, and (56 - x) represents the ounces of the 44% solution needed. Since the total amount of the two solutions combined is 56 ounces, the sum of these two quantities must equal 56.
Now, let's set up the equation for the alcohol content in the mixture. The total amount of alcohol in the mixture is the sum of the amounts of alcohol in the two solutions combined.
For the 26% solution, the amount of alcohol is (0.26 * x) ounces.
For the 44% solution, the amount of alcohol is (0.44 * (56 - x)) ounces.
Since we want the resulting mixture to be a 29% solution, the amount of alcohol in the mixture should equal (0.29 * 56) ounces.
Putting it all together, we can set up the equation:
0.26x + 0.44(56 - x) = 0.29 * 56
Next, we can solve this equation to find the value of x, which represents the ounces of the 26% solution needed.
0.26x + 24.64 - 0.44x = 16.24
-0.18x = -8.4
x = -8.4 / -0.18
x ≈ 46.67
Therefore, approximately 46.67 ounces of the 26% alcohol solution should be mixed with (56 - 46.67) ≈ 9.33 ounces of the 44% alcohol solution to obtain 56 ounces of a 29% solution.