Question 27019: I need help on a mixture problem, I have tried to set the problem up, but these (percents) mess me up. How many oz. of an 8% alcohol solution must be mixed with a 12% alcohol sloution to obtain 100 oz. of a 12.2 alcohol solution?
I think it is 12.2(100)-.08x = .12(100=x), but i can't get it to work out into an answer I have to chose from.
Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website! .08(S) + .12(100 - S) = .122 (100) = 12.2
where S is the amount of 8% alcohol solution
.08S + 12 - .12S = 12.2
-.04S = .2
S = -50
I don't think the problem, as stated, has a solution
when S = 0, then the left side is 12 (less than 12.2)
when S = 100, the left side is 8 (also less than 12.2)
|
|
|
