SOLUTION: Ooops. I just asked one question and was looking at the other problem. Sorry. Here is the correct question and the way I set it up. 60 fluid ounces of a 15% muriatic acid s

Question 25921: Ooops. I just asked one question and was looking at the other problem. Sorry.
Here is the correct question and the way I set it up.
60 fluid ounces of a 15% muriatic acid solution is needed to kill algae in a pool. If the technician has a 5% solution and a pure solution on hand, how many ounces of each must be combined to create the 15% solution?
I don't even know how I would set this one up, I know I am supposed to be seeing a pattern on how to set mixture problems up, but I keep putting the wrong numbers in the wrong places.
Thanks for your help.
Donna
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
OK, first define what you need.

Let x = volume of 5% acid.

Now for the leap of faith part...how much of the pure acid is needed? Answer is (60-x)! This is the hardest part done now.

Now for the solution.
BEFORE = AFTER

0.05x + 1.00(60-x) = 0.15*60
0.05x + 60 - x = 9
-0.95x = -51
--> x = -51/-0.95
x = 53.68 fl.oz (to 2 dp)

so we need 53.68 fl.oz of the 5% acid and we will need 6.32 fl.oz of the pure acid too.

jon.

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