SOLUTION: How many liters of a 40%-alcohol solution must be mixed with 10 liters of a solution that is 80% alcohol to get a solution that is 60% alcohol? I need to translate this problem int

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Question 334816: How many liters of a 40%-alcohol solution must be mixed with 10 liters of a solution that is 80% alcohol to get a solution that is 60% alcohol? I need to translate this problem into a pair of linear equations in two variables. Can anyone help?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let A be the number of liters of 40%-alcohol solution.
Look at the concentration equation.
%2840%2F100%29A%2B%2880%2F100%29%2810%29=%2860%2F100%29%28A%2B10%29
40A%2B800=60%28A%2B10%29
40A%2B800=60A%2B600
-20A=-200
A=10
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10 liters of 40% solution plus 10 liters of 80% will provide 20 liters of 60% solution.