# Mixture problem of ethanol

E10 is a mixture of 10% ethanol and 90% gasoline.

How much E5 should be mixed with 5000 gal of E10 to make an E9 mixture.

The numbers indicate the percentage of ethanol in the mixture by volume.

5000 gal of E10 contains 500 gal ethanol and 4500 gal gasoline.

0.05x+500 = 0.09(x+5000)

Is this right?

Let $x$ be the volume of E5 to be mixed, then the ethanol in the mixture is $$0.05x+0.1\cdot 5000$$ To obtain E9, this number should be equal to $0.09(x+5000)$. So you have to solve the equation $$0.05x+0.1\cdot 5000=0.09(x+5000)$$ that is $0.04x=0.01\cdot 5000$ and $x=1250$.

"E10 is a mixture of 10% ethanol and 90% gasoline." Great!

"How much E5 should be mixed with 5000 gal of E10 to make an E9 mixture." You just told us what E10 is. Don't you think it would be a good idea to tell us what "E5" and "E9" are? I might guess that you meant to say that "Ex is a mixture of x% ethanol and (100- x)% gasoline", but that is only a guess.

"The numbers indicate the percentage of ethanol in the mixture by volume." Okay- a little late!

"5000 gal of E10 contains 50 gal ethanol and 450 gal gasoline." 0.05x+50 = 0.09x Is this right?"

No, because you haven't said what "x" meant! But you also have the numbers in the wrong places.

Going back to "How much E5 should be mixed with 5000 gal of E10 to make an E9 mixture" if we let x be the amount of E5, then 5000- x would be the amount of E9 in this 5000 gal mixture. There are .05x gallons of ethanol in the E5 and .9(5000- x) gallons of ethanol in the E9. If those are equivalent to 5000 gal of E10, then those must be equal to the .10(5000)= 50 gal of ethanol: .05x+ .9(5000- x)= 50.