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anonymous

  • 5 years ago

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?

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  1. anonymous
    • 5 years ago
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    x+y=20 .4x+.7y=(20)(.5)

  2. anonymous
    • 5 years ago
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    then use the substition method to solve

  3. anonymous
    • 5 years ago
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    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)

  4. anonymous
    • 5 years ago
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    zbay, I don't think you set that up properly.

  5. anonymous
    • 5 years ago
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    Or rather I think you misread the question.

  6. anonymous
    • 5 years ago
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    .4(20) + .7(x) = .5(20+x) Find x.

  7. anonymous
    • 5 years ago
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    i think if you do the system of equations it sould work out, but let me work it out

  8. anonymous
    • 5 years ago
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    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.

  9. anonymous
    • 5 years ago
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    Alright i think you got me, it's not working out

  10. anonymous
    • 5 years ago
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    I think the answer is 10 liters, can you verify this for me?

  11. anonymous
    • 5 years ago
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    Yes. It is 10 liters.

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spraguer (Moderator)
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