SOLUTION: Pure acid is to be added to a 10% acid solution to obtain 54L of a 20% acid solution. What amount of each should be used?

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Pure acid is to be added to a 10% acid solution to obtain 54L of a 20% acid solution. What amount of each should be used?      Log On


   

Question 507422: Pure acid is to be added to a 10% acid solution to obtain 54L of a 20% acid solution. What amount of each should be used?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Pure acid is to be added to a 10% acid solution to obtain 54L of a 20% acid solution.
What amount of each should be used?
:
Let x = amt of pure acid required
Since the resulting amt is to be 54L:
(54-x) = amt of 10% acid required
:
A typical mixture equation
:
x + .10(54-x) = .20(54)
x + 5.4 - .10x = 10.8
x - .1x = 10.8 - 5.4
.9x = 5.4
x = 5.4%2F.9
x = 6 L of pure acid to be combined with: 54-6 = 48 L of 10% solution