SOLUTION: Michael has 1 liter of a mixture containing 69% of boric acid. How much water must be added to make the mixture 50% boric acid? (Represent answer as a fraction)

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Question 163193: Michael has 1 liter of a mixture containing 69% of boric acid. How much water must be added to make the mixture 50% boric acid? (Represent answer as a fraction)
Found 2 solutions by ptaylor, ankor@dixie-net.com:
Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website!

Let x=amount of water needed
Now we know that the amount of pure boric acid in the 69% mixture (0.69*1) plus the amount of pure boric acid in the water that is added (0) has to equal the amount of pure boric acid in the final mixture(0.50(1+x)). So our equation to solve is:
0.69*1+0=0.50(1+x) get rid of parens (distributive)
0.69=0.50+0.50x subtract 0.50 from each side
0.69-0.50=0.50-0.50+0.50x collect like terms
0.19=0.50x divide each side by 0.50
x=0.19/0.50=19/50 liter------------amount of water needed
CK
0.69*1=0.50(50/50 + 19/50)
0.69=0.50(69/50)=69/100
0.69=69/100
hope this helps----ptaylor

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
Michael has 1 liter of a mixture containing 69% of boric acid. How much water must be added to make the mixture 50% boric acid?
:
Let x = amt of water required
then
(x+1) = the amt of the resulting mixture
:
Amt of boric acid equation:
;
.69(1) = .50(x+1)
;
.69 = .5x + .50
;
.69-.50 = .5x
:
.5x = .19
x =
x = .38 liters of water required
In a fraction: = liter
:
:
Check solution
.69(1) = .50(1.38)
.69 = .69