Question

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?

Answer 1

So this is basically what you want to do:
0.32(10)= 0.2(x)+0.6(10-x)

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??

Answer 2

yes it makes more sense then my set up. Thanks

Answer 3

What did you get for your answer? (I am studying this too and want to compare)

Answer 4

x= 71/5 and y= -21/5

Answer 5

am i correct?

Answer 6

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+6-0.6x
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?

Answer 7

oh okay, i put a ten by the six on accident

Answer 8

Haha totally okay!