SOLUTION: Crazy Chris has a 1% alcohol solution mix with a 12.5% alcohol solution. He wants 500 mL of 2.5% alcohol. How much of each does he need to mix? LOst.

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: Crazy Chris has a 1% alcohol solution mix with a 12.5% alcohol solution. He wants 500 mL of 2.5% alcohol. How much of each does he need to mix? LOst.      Log On

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Question 26088: Crazy Chris has a 1% alcohol solution mix with a 12.5% alcohol solution. He wants 500 mL of 2.5% alcohol. How much of each does he need to mix?
LOst.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Let "x" be the amount of 1% solution.
Then "500-x" is the amount of 12.5% solution.
EQUATION:
Follow the acid:
1%x + 12.5%(500-x)=2.5%(500)
Multiply by 1000 to get rid of the % and decimals:
10x+125(500-x)=25(500)
10x+62500-125x=12500
-115x=-50000
x=434.7 mL (You need to have 434.7 mL of 1% solution)
500-x=65.3 mL (You need 65.3 mL of 12.5% solution)