SOLUTION: How many liters of a blue dye that costs $1.90 per liter must be mixed with 22 L of anil that costs $2.50 per liter to make a mixture that costs $2.10 per liter? _____L

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Question 1196971: How many liters of a blue dye that costs $1.90 per liter must be mixed with 22 L of anil that costs $2.50 per liter to make a mixture that costs $2.10 per liter?
_____L
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
"anil"? ??

x liters of $1.90 per liter
22 liters of $2.50 per liter
MIXTURE x+22 liters resulting in $2.10 per liter

1.9x%2B2.5%2A22=2.1%28x%2B22%29
The algebra is very simple , and then computation.
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Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


First a typical formal algebraic solution....

x liters at $1.90 per liter, plus 22 liters at $2.50 per liter, yields (x+22) liters at $2.10 per liter:

x%281.90%29%2B22%282.50%29=%28x%2B22%29%282.10%29
1.90x%2B55=2.10x%2B46.20
8.80=0.20x
x=8.80%2F0.20=44

ANSWER: 44 liters

And then a quick and easy informal solution, if formal algebra is not required....

Look at the three prices per liter -- $1.90, $2.10, and $2.50 -- (on a number line, if it helps) and observe/calculate that $2.10 is 1/3 of the way from $1.90 to $2.50.

That means 1/3 of the mixture needs to be the more expensive ingredient -- i.e., there needs to be twice as much of the less expensive ingredient.

Since there are 22 liters of the more expensive ingredient, there need to be 2*22 = 44 liters of the less expensive.

ANSWER (again, of course): 44 liters