SOLUTION: How many liters of a 14% alcohol solution must be mixed with 20 L of a 50% solution to get a 30% solution? Show you process. Thank you!!

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Question 423888: How many liters of a 14% alcohol solution must be mixed with 20 L of a 50% solution to get a 30% solution? Show you process.

Thank you!!
Answer by dnanos(83) About Me  (Show Source):
You can put this solution on YOUR website!
let x litters of the 14% solution.
Then we obtain x+20 litters of the 30%.
The pure alcohol litters before and after mixing must be equal.
So we obtain an equation
%2814%2F100%29%2Ax%2B%2850%2F100%29%2A20=%2830%2F100%29%2A%28x%2B20%29
14%2F100%2Ax%2B50%2F100%2A20=30%2F100%2A%28x%2B20%29
Multiply by 100
14x%2B50%2A20=30%28x%2B20%29
14x%2B1000=30x%2B600
1000-600=30x-14
400=16x
16x=400
Divide by 16
x=400%2F16
x=25
So 25 litters 14% needed.