SOLUTION: 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 s

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: 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 s      Log On


   



Question 1195943: 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 4000gal of E10to make an E9 mixture?

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The typical algebraic setup for solving the problem would look something like this:

4000 gallons of 10% ethanol, plus x gallons of 5% ethanol, yields (4000+x) gallons of 9% ethanol

.10%284000%29%2B.05%28x%29=.09%284000%2Bx%29

I'll leave it to you to find the answer by that method.

Here is another very different way to solve any 2-part mixture problem like this which, if the numbers are "nice", gets you to the answer faster and with less work.

You are mixing 5% and 10% ethanol to get a mixture that is 9% ethanol.
Consider the three percentages on a number line: 5, 9, 10.
Observe/calculate that 9% is 4/5 of the way from 5% to 10%.
That means 4/5 of the mixture must be the 10% ethanol, so 1/5 of the mixture is the 5% ethanol; that means the amount of 5% ethanol must be 1/4 the amount of 10% ethanol.

There are 4000 gallons of the 10% ethanol, so the number of gallons of the 5% ethanol must be (1/4)(4000) = 1000.

ANSWER: 1000 gallons

CHECK:
.10(4000)+.05(1000)=400+50 = 450
.09(5000) = 450