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 2 years ago
i need help setting this up.
one solution containing 20 percent alcohol is mixed with another solution containing 60 percent alcohol to make 10 gallons of a solution that is 32 percent alcohol. How much of each solution is used?
 2 years ago
i need help setting this up. one solution containing 20 percent alcohol is mixed with another solution containing 60 percent alcohol to make 10 gallons of a solution that is 32 percent alcohol. How much of each solution is used?

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amcdowns
 2 years ago
Best ResponseYou've already chosen the best response.1So this is basically what you want to do: 0.32(10)= 0.2(x)+0.6(10x) Set it up and find what x equals, which is the amount of the 20 percent solution needed. Once you know that, you can find the 60% solution by subtracting the amount of the 20 percent solution by 10, or the total solution. I hope this makes sense, do you understand??

krissywatts
 2 years ago
Best ResponseYou've already chosen the best response.0yes it makes more sense then my set up. Thanks

amcdowns
 2 years ago
Best ResponseYou've already chosen the best response.1What did you get for your answer? (I am studying this too and want to compare)

krissywatts
 2 years ago
Best ResponseYou've already chosen the best response.0x= 71/5 and y= 21/5

amcdowns
 2 years ago
Best ResponseYou've already chosen the best response.1Let's think, does it make sense that one would be negative? I don't think so. So lets work through the equation and do what we can: 3.2=0.2x+60.6x We can combine like terms and we get −2.8=0.4x, or 2.8+0.4x We know that 1/5 of x is 1.4, so x=7 x represents the number of gallons used in the 20% solution, so we can subtract 7 from 10 to get the number used in the 60% solution, to get 3. I hope you understand! Anymore questions?

krissywatts
 2 years ago
Best ResponseYou've already chosen the best response.0oh okay, i put a ten by the six on accident
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