SOLUTION: How many liters of a 30% alcohol solution must be mixed with 80 liters of a 80% solution to get a 50% solution? I have made a table and tried to work many algebraic solutions bu

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: How many liters of a 30% alcohol solution must be mixed with 80 liters of a 80% solution to get a 50% solution? I have made a table and tried to work many algebraic solutions bu      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 336175: How many liters of a 30% alcohol solution must be mixed with 80 liters of a 80% solution to get a 50% solution?
I have made a table and tried to work many algebraic solutions but cannot come up with one.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How many liters of a 30% alcohol solution must be mixed with 80 liters of a 80% solution to get a 50% solution?
-------------------
Equation:
alcohol+alcohol = alcohol
0.30x + 0.80*80 = 0.50(x+80)
Multiply thru by 100 to get:
30x + 80*80 = 50x + 50*80
-20x = -30*80
x = (3/2)80
x = 120 liters (amt of 30% solution needed in the mixture)
=================
Cheers,
Stan H.