SOLUTION: A 25% SOLUTION OF ALCOHOL IS TO BE MIXED WITH A 40% SOLUTION TO GET 50L OF A FINAL MIXTURE THAT IS 30% alcohol. How much of each of the original solution should be used?
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: A 25% SOLUTION OF ALCOHOL IS TO BE MIXED WITH A 40% SOLUTION TO GET 50L OF A FINAL MIXTURE THAT IS 30% alcohol. How much of each of the original solution should be used?
Log On
Question 54769: A 25% SOLUTION OF ALCOHOL IS TO BE MIXED WITH A 40% SOLUTION TO GET 50L OF A FINAL MIXTURE THAT IS 30% alcohol. How much of each of the original solution should be used? Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! .25X+.40(50-X)=50*.30
.25X+20-.40X=15
-.15X=15-20
-.15X=-5
X=-5/-.15 [CORRECTION I USED .25 INSTEAD OF .15 - SORRY]
X=33.33 L OF 25% ALCOHOL &
50-33.33=16.67 L OF 40% ALCOHOL ARE USED TO CREATE A MIX OF 50 L WITH 30% ALCOHOL.