SOLUTION: Pure acid is to be added to a 10% acid solution to obtain 21L of a 40% acid solution. what amount of each should be used?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: Pure acid is to be added to a 10% acid solution to obtain 21L of a 40% acid solution. what amount of each should be used?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 157315: Pure acid is to be added to a 10% acid solution to obtain 21L of a 40% acid solution. what amount of each should be used?
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
x+.1(21-x)=.4*21
x+2.1-.1x=8.4
.9x=8.4-2.1
.9x=6.3
x=6.3/.9
x=7 gallons of pure acid is needed.
21-7=14 gallons of 10% acid is needed.
Proof:
7+.1*14=8.4
7+1.4=8.4
8.4=8.4