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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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amcdownsBest ResponseYou've already chosen the best response.1
So 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??
 one year ago

krissywattsBest ResponseYou've already chosen the best response.0
yes it makes more sense then my set up. Thanks
 one year ago

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

krissywattsBest ResponseYou've already chosen the best response.0
x= 71/5 and y= 21/5
 one year ago

amcdownsBest ResponseYou've already chosen the best response.1
Let'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?
 one year ago

krissywattsBest ResponseYou've already chosen the best response.0
oh okay, i put a ten by the six on accident
 one year ago
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