SOLUTION: A Chemist has one solution that is 60% chlorinated and another that is 40% chlorinated. How much of the first (60%) solution is needed to make a 100L solution that is 50% chlorinat

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A Chemist has one solution that is 60% chlorinated and another that is 40% chlorinated. How much of the first (60%) solution is needed to make a 100L solution that is 50% chlorinat      Log On


   

Question 41034: A Chemist has one solution that is 60% chlorinated and another that is 40% chlorinated. How much of the first (60%) solution is needed to make a 100L solution that is 50% chlorinated?
--so i put it like this right?
--x(.60) + (100-x)(.50) = 100(.50)
.6x + 50 - .5x = 50
.1x = 100
x = 1000
100-x = -100
Is this right???
Thank you
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
.60X+(100-X).4=100*.5 OR .6X+40-.4X=50 OR .2X+40=50 OR .2X=10 OR X=10/.2 OR
X=50L OF 60% & 50L OF 40%