Question 1155750
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If you solve the problem using formal algebra, then certainly use a single variable -- it is MUCH less work.<br>
I solve this kind of problem informally, if I can.  The numbers in this problem make it easy; being good with mental arithmetic, I solved the problem in about 15 seconds.<br>
Work with two ingredients at a time, as follows....<br>
The mixture uses twice as much $4.50 water as $3 water.  That means the average price of those two ingredients is "twice as close to $4.50 as it is to $3"; that makes the average cost of those two ingredients $4.<br>
Now the problem is to mix $4 water and $14 water to get $8 water.<br>
$8 is 4/10=2/5 of the way from $4 to $14; that means 2/5 of the total mixture is the $14 water.<br>
So 2/5 of the 400 gallons, or 160 gallons, is the $14 water.<br>
That leaves 240 gallons of the $3 water and $4.50 water.  Since the amount of $4.50 water is twice the amount of $3 water, that means 160 gallons of the $4.50 water and 80 gallons of the $3 water.<br>
ANSWER:
160 gallons of $14 water
160 gallons of $4.50 water
80 gallons of $3 water<br>