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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?
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x+y=20 .4x+.7y=(20)(.5)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then use the substition method to solve

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0zbay, I don't think you set that up properly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Or rather I think you misread the question.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0.4(20) + .7(x) = .5(20+x) Find x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think if you do the system of equations it sould work out, but let me work it out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0Alright i think you got me, it's not working out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think the answer is 10 liters, can you verify this for me?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes. It is 10 liters.
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