SOLUTION: How man milliliters of 100% acid must be added to 50 milliliters of a 40% acid solution to obtain a 50% acid solution? I had set up and figured the problem like this: 1(x)+.

Question 701981: How man milliliters of 100% acid must be added to 50 milliliters of a 40% acid solution to obtain a 50% acid solution?
I had set up and figured the problem like this:
1(x)+.4(50)=.5(50+x)
20x=25+.5x
19.25x=25
x=50/39 mL
But I have a gut feeling that x does not equal 50/39.
What did I do wrong?

Found 2 solutions by jim_thompson5910, josgarithmetic:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
1(x)+.4(50)=.5(50+x)

x+20=25+0.5x

x - 0.5x = 25 - 20

0.5x = 5

x = 5/(0.5)

x = 10

So you need to add 10 mL of 100% acid
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
How much pure acid will be in the solution? Letting x equal how much of the 100% material to add,
Pure acid in the resulting solution is 1*x+50*(0.40).
How much solution must result when the two parts are added and mixed?
The amount of resulting solution is x+50 milliliters.
You want a result of 50% acid, or as fraction, 0.5 as acid.

Solve for x.
Skipping a step,...

Continue....Finish...
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