SOLUTION: how many ounces of a 25 % alcohol solution must be mixed with 11% of a 30% alcohol solution to make a 26 % alcohol solution?

Question 1070925: how many ounces of a 25 % alcohol solution must be mixed with 11% of a 30% alcohol solution to make a 26 % alcohol solution?
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let x = the number of ounces of 25% alcohol solution.

let y = the number of ounces of 11% of the 30% solution.

your formula becomes:

.25 * x + .30y = .26 * (x + y)

solve for x to get x = 4y

this means that the number of ounces of 25% alcohol solution must be 4 times the number of ounces of 11% of the 30% solution.

for example:

assume you have 100 ounces of 30% solution.

11% of that would be 11 ounces of 30% solution.

this means that y is equal to 11, because y represents 11% of the 30% solution.

4 times that would be 44 ounces of 25% solution.

if you mix 44 ounces of 25% solution with 11 ounces of 30% solution, you should get a solution that is 26% alcohol.

44 * .25 + 11 * .30 = 14.3 ounces of alcohol.

44 + 11 = 55 ounces of solution.

14.3 / 55 = .26 = 26%

this means that the 55 ounces of combined solution is 26% alcohol.

Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.
The condition in your post is written incorrectly.

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