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

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: How many liters of a 10% alcohol solution must be mixed with 80 liters of a 80% solution to get a 50% solution?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

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) About Me  (Show Source):
You can put this solution on YOUR website!
Let A be the amount of 10% solution.
10A%2B80%2880%29=50%28A%2B80%29
10A%2B6400=50A%2B4000
40A=2400
A=60
.
.
.
60 liters of 10% solution are needed.

Answer by edjones(8007) About Me  (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.
.
64+.1x=40+.5x
24=.4x
x=60 L of the 10% alcohol must be added.
.
Ed