According to the problem we start with 6000 gal of E10 which means there is 10% of ethanol in this 6000, with ths information we can calculate the amount of ethanol already in the mixture:
[tex]6000gal*0.1=600gal[/tex]
The relationship between ethanol and total fluid can be represented like this:
[tex]\frac{600}{6000} =0.1[/tex]
Based on this relationship we can build the one that we are looking for:
[tex]E9=9Ethanol91Gasoline=\frac{Ethanol}{Total}=0.09[/tex]
[tex]\frac{TotalEthanol}{TotalSolution} =0.09[/tex]
the total solution will be:
[tex]TotalSolution=6000+X[/tex]
where X is the E5 solution volume
and the TotalEthanol would be:
[tex]TotalEthanol=600+E5Ethanol[/tex]
where E5ethanol (5% Ethanol 95% Gasoline) would be:
[tex]\frac{E5Ethanol}{X}=0.05\\E5Ethanol=0.05X[/tex]
Now replacing in the relationship we found before:
[tex]\frac{600+0.05X}{6000+X}=0.09[/tex]
Solving for X:
[tex]600+0.05X=0.09(6000+X)\\600+0.05X=540+0.09X\\0.09X-0.05=600-540\\0.04X=60\\X=\frac{60}{0.04} \\X=1500[/tex]
We should mix 1500 gal of E5 to make an E9 mixture
