Question

Ethanol fuel mixtures have "E" numbers that indicate the percentage of ethanol in the mixture by volume. For example, E10 is a mixture of 10% ethanol and 90% gasoline. How much E5 should be mixed with 6000 gal of E10 to make an E9 mixture?

Answer 1

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:

$6000gal*0.1=600gal$

The relationship between ethanol and total fluid can be represented like this:

$\frac{600}{6000} =0.1$

Based on this relationship we can build the one that we are looking for:

$E9=9Ethanol91Gasoline=\frac{Ethanol}{Total}=0.09$

$\frac{TotalEthanol}{TotalSolution} =0.09$

the total solution will be:

$TotalSolution=6000+X$

where X is the E5 solution volume

and the TotalEthanol would be:

$TotalEthanol=600+E5Ethanol$

where E5ethanol (5% Ethanol 95% Gasoline) would be:

$\frac{E5Ethanol}{X}=0.05\\E5Ethanol=0.05X$

Now replacing in the relationship we found before:

$\frac{600+0.05X}{6000+X}=0.09$

Solving for X:

$600+0.05X=0.09(6000+X)\\600+0.05X=540+0.09X\\0.09X-0.05=600-540\\0.04X=60\\X=\frac{60}{0.04} \\X=1500$

We should mix 1500 gal of E5 to make an E9 mixture