how many oz of 25% alcohol solution must be mixed with 14oz of 30oz alcohol to make 26% alcohol
To find out how many ounces of a 25% alcohol solution are needed to mix with 14 ounces of a 30% alcohol solution to make a 26% alcohol solution, we can use a basic equation based on the principle of the mixture.
Let's assume that x represents the number of ounces of the 25% alcohol solution we need. Here's how we can set up the equation:
0.25x + 0.30(14) = 0.26(x + 14)
Now, let's break down the equation and solve for x:
0.25x + 4.2 = 0.26x + 3.64
Next, we can subtract 0.25x from both sides:
0.25x - 0.25x + 4.2 = 0.26x - 0.25x + 3.64
This simplifies to:
4.2 = 0.01x + 3.64
Now, subtract 3.64 from both sides:
4.2 - 3.64 = 0.01x + 3.64 - 3.64
Which gives us:
0.56 = 0.01x
Finally, divide both sides by 0.01 to isolate x:
0.56 / 0.01 = x
Therefore, x = 56.
So, 56 ounces of the 25% alcohol solution must be mixed with 14 ounces of the 30% alcohol solution to make a 26% alcohol solution.