Asked 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
-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
Answers
Answered by
Steve
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%.
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%.
Answered by
Steve
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.
Having read my algebra carefully, though, you should have been able to do it right.
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