SOLUTION: How many liters of a blue dye that costs S1.60 per liter must be mixed with
18 L of anil that costs S2.50 per liter to make a mixture that costs S1.90 per
liter?
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-> SOLUTION: How many liters of a blue dye that costs S1.60 per liter must be mixed with
18 L of anil that costs S2.50 per liter to make a mixture that costs S1.90 per
liter?
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Question 1198449: How many liters of a blue dye that costs S1.60 per liter must be mixed with
18 L of anil that costs S2.50 per liter to make a mixture that costs S1.90 per
liter? Found 2 solutions by MathLover1, greenestamps:Answer by MathLover1(20850) (Show Source):
The solution from the other tutor shows a typical formal algebraic method for solving the problem.
If formal algebra is not required, here are two variations of a method that can be used to solve any 2-part mixture problem like this quickly and easily.
(1) The cost of the mixture, $1.90 per liter, is "twice as close" to the $1.60 per liter cost of the blue dye as it is to the $2.50 per liter cost of the anil ($1.90-$1.60 = $0.30; $2.50-$1.90 = $0.60).
That means the amount of blue dye must be twice the amount of anil.
Since there are 18 L of anil, the required amount of blue dye is 2*18 = 36 L.
ANSWER: 36 L
(2) Look at the three costs per liter -- $1.60, $1.90, and $2.50 -- on a number line and observe/calculate that $1.90 is 30/90 = 1/3 of the way from $1.60 to $2.50. That means 1/3 of the mixture is the more expensive anil.
So 2/3 of the mixture is the blue dye -- twice as much as the anil. And since there are 18 L of anil, there must be 36 L of blue dye.
