SOLUTION: I need help setting up the equations. I almost have it I think, but I want to make sure. Solution A is 20% acid. Find the amount of Solution A and the amount of pure water nee

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: I need help setting up the equations. I almost have it I think, but I want to make sure. Solution A is 20% acid. Find the amount of Solution A and the amount of pure water nee      Log On


   

Question 477899: I need help setting up the equations. I almost have it I think, but I want to make sure.
Solution A is 20% acid. Find the amount of Solution A and the amount of pure water needed to make 5 liters of a new solution that is 18% acid.
Thanks!!
Found 2 solutions by nerdybill, ewatrrr:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Solution A is 20% acid. Find the amount of Solution A and the amount of pure water needed to make 5 liters of a new solution that is 18% acid.
.
Let x = amount (liters) of 20% acid
then
5-x = amount (liters) of water
.
.20x +0(5-x) = .18(5)
.20x = .18(5)
.20x = .9
x = 4.5 liters of 20% acid
.
Amount of pure water:
5-x = 5-4.5 = .5 liters of pure water

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Solution A is 20% acid ...to new solution 5L that is 18% acid 

Will set question up as to the % of water

Let x represent the amount of pure water 

and (5L-x) represent the amount of 80% water solution (20% acid)

Mixing to form a 82% water solution

x + .80(5-x) = .82*5

    .20x =  .02*5

       x = .02*5/.20

       x = .5L of water