SOLUTION: Would someone mind taking a look at this question and see if my answer is correct. I think I am on the right track. One needs to mix a 30% acid solution with a 50% acid solution t

Question 61028: Would someone mind taking a look at this question and see if my answer is correct. I think I am on the right track.
One needs to mix a 30% acid solution with a 50% acid solution to obtain 12 ounces of a 45% acid solution. How many ounces of each of the acid solutions must be used?
Amt. of mix is 12 oz
Amt of 1st acid = x
Amt. of 2nd acid is the rest = 12-x
Amt. of acid in the mixture = .45(12)
Amt. of 1st acid = .30x
Amt. of 2nd acid = .50(12-x)

3 oz of 30% acid
9 oz of 50% acid
Found 2 solutions by ankor@dixie-net.com, josmiceli:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
That's right.
:
Check your solution:
.3(3) + .5(9) = .45(12)
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
You are right with your answers, and I like your
approach better than mine, in fact.
I used two variables where you only used one.
Amt. of mix is 12 oz
Amt of 1st acid = x
Amt. of 2nd acid is the rest = y (you said (12-x))
Amt. of acid in the mixture = .45(12)
Amt. of 1st acid = .30x
Amt. of 2nd acid = .50y
.30x + .50y = .45*12
x + y = 12
.30x + .50(12-x) = .45*12
.3x + 6 - .5X = 5.4
.2x = 6 - 5.4
.2x = .6
x = 3
y = 12 - 3
y = 9
3 oz of 30% acid
9 oz of 50% acid
you are correct
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