Algebra
posted by Anonymous .
Suppose a 2% acid solution is mixed with a 3% acid solution. Find the percent of acid in each mixture.
a mixture that contains an equal amount of 2% acid solution and 3% acid solution
a mixture that contains 3 times more 2% acid solution than 3% acid solution

equal amounts: 2.5%
if 3% solution is x times the 2% solution, then if the resulting concentration is y%,
.02 + .03x = (1+x)*y
You can see that if x=1, y=.05/2 = .025 = 2.5%
If x = 3, .11 = 4y, so y = .11/4 = .0275 = 2.75%
Makes sense, since the new concentration is 3/4 of the way from 2% to 3%. 
Good thing you reposted it. I misread the problem, and solved it for having 3 times as much 3% as 2%. See bobpursley's solution for the correct answer.
Having read my algebra carefully, though, you should have been able to do it right.
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