SOLUTION: mix a solution that is 50% acid with a solution that is 100% water to make 4 liters of a solution that is 10% acid. How much of each solution is used? This is what I have (4-x)*

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Question 317379: mix a solution that is 50% acid with a solution that is 100% water to make 4 liters of a solution that is 10% acid. How much of each solution is used?
This is what I have (4-x)*(50/100) = 4x*10/100
50(4-x) = 40
200-50x = 40
200(-200)-50X = 40-200
-50x = -160
-50x = -160
50 50
x = -3.2
So -3.2 liters of water with 0.8 liters of 50% solution to get 4 liters with a 10% percent solution of acid.
I'm not sure if this is right. I have been working on it for so long I keep making stupid mistakes.
Found 2 solutions by nyc_function, solver91311:
Answer by nyc_function(2741) About Me  (Show Source):
You can put this solution on YOUR website!
Start by finding the amount of acid that would be in the 4 liter solution you end with. So that would be 10% of 4 liters, or .4 liters. So that .4 liters of acid has to all come from the solution of half acid and half water. So .4L of acid = 50% of the total amount of mixed solution, which means that there will be .8 solutions of that solution. If .8 liters of the final 4 liter solution is made up of the half acid half water solution, then the final 4 - .8 = 3.2 liters is pure water.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


You almost have it. It is just that your answer comes up with the wrong sign. Doesn't make much sense to add a negative amount of water to an acid solution, now does it?

You actually set the problem up correctly, but you made a small error when you divided both sides by 50.





So



and therefore



Meaning 3.2 liters of water and 0.8 liters of the 50% acid solution.

There is a much less complex way to do this one. First off, let represent the amount of acid solution needed. Then, realizing that 50% is the same as one-half, and 10 percent of 4 liters is 0.4 liters, you can write:





And we get the same answer: 0.8 liters of acid solution and 3.2 liters of water.

John