SOLUTION: How many liters of a 10% alcohol solution must be mixed with 80 liters of a 80% solution to get a 50% solution?

Question 336525: How many liters of a 10% alcohol solution must be mixed with 80 liters of a 80% solution to get a 50% solution?
Found 2 solutions by Fombitz, edjones:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Let A be the amount of 10% solution.

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.
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60 liters of 10% solution are needed.

Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
80*.8=64 total liters alcohol in 80% soln.
x *.1=.1x total liters alcohol in 10% soln.
(80+x)*.5=40+.5x total liters alcohol in 50% soln.
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64+.1x=40+.5x
24=.4x
x=60 L of the 10% alcohol must be added.
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Ed
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