Question 25920: Here is the question. I know that there is a way to set these up, and the instructor said I would start to see a pattern, but she does not explain clearly how to do these. I am confused.
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 tried to set it up like this:
.05X+.10(.15-X)=24(.15)
I am getting my numbers put in the wrong place. Please help me.
Thank you. Donna
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=number of ounces of the pure muriatic acid to be combined
Then 60-x=number of ounces of the 5% muriatic acid solution to be combined
Now we know that the amt of pure muriatic acid to be combined (x)(1) plus the amt of pure muriatic acid in the 5% solution (60-x)(.05) has to equal the amt of pure muriatic acid in the final mixture (60)(.15).
Thus, our equation to solve is:
(x)(1)+(60-x)(.05)=(60)(.15) simplifying, we get:
x+3-.05x=9 or
.95x=6
x=6.31 ounces of pure muriatic acid, and
60-x=60-6.31=53.69 ounces of 5% solution
Lets solve this problem again, focusing on the "otherstuff" in the muriatic acid solution
We will still let x=number of ounces of the pure muriatic acid to be combined
Then 60-x=number of ounces of the 5% muriatic acid solution to be combined
Now we know that the amt of "otherstuff" in the pure muriatic acid to be combined (x)(0) plus the amt of "otherstuff" in the 5% solution (60-x)(.95) has to equal the amt of "otherstuff" in the final mixture (60)(.85).
Thus, our equation to solve is:
(60-x)(.95)=(60)(.85) simplifying, we get:
57-.95x=51 or
-.95x=-6
x=6.31 ounces of pure muriatic acid, and
60-x=60-6.31=53.69 ounces of 5% solution
Hope this helps ------- ptaylor
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