Question 1195943
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The typical algebraic setup for solving the problem would look something like this:<br>
4000 gallons of 10% ethanol, plus x gallons of 5% ethanol, yields (4000+x) gallons of 9% ethanol<br>
{{{.10(4000)+.05(x)=.09(4000+x)}}}<br>
I'll leave it to you to find the answer by that method.<br>
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.<br>
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.<br>
There are 4000 gallons of the 10% ethanol, so the number of gallons of the 5% ethanol must be (1/4)(4000) = 1000.<br>
ANSWER: 1000 gallons<br>
CHECK:
.10(4000)+.05(1000)=400+50 = 450
.09(5000) = 450<br>