anonymous
  • anonymous
brad has to mix 20 liters of 40% acid with a 70% acid solution to get a mixture that is 50% acid. how many liters of the 70% acid does he need?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
x+y=20 .4x+.7y=(20)(.5)
anonymous
  • anonymous
then use the substition method to solve
anonymous
  • anonymous
The amount of acid is a constant. The volume of the final solution is the sum of the other two volumes. 20Liters * (.4 Acid/Liter) + XLiters * (.7Acid/Liter) = (20 + X)Liters * (.5 Acid/Liter)

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anonymous
  • anonymous
zbay, I don't think you set that up properly.
anonymous
  • anonymous
Or rather I think you misread the question.
anonymous
  • anonymous
.4(20) + .7(x) = .5(20+x) Find x.
anonymous
  • anonymous
i think if you do the system of equations it sould work out, but let me work it out
anonymous
  • anonymous
I don't think so because your system is claiming that the sum of their volumes is 20 liters. I'm claiming that the volume of the final solution will be more than 20 liters. Very different problems.
anonymous
  • anonymous
Alright i think you got me, it's not working out
anonymous
  • anonymous
I think the answer is 10 liters, can you verify this for me?
anonymous
  • anonymous
Yes. It is 10 liters.

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