SOLUTION: I'm struggling with this word problem: A 40% saline solution is to be mixed with 60% saline solution to obtain 8 liters of 55% saline solution. How many liters 40% solution shoul

Question 1052222: I'm struggling with this word problem:
A 40% saline solution is to be mixed with 60% saline solution to obtain 8 liters of 55% saline solution. How many liters 40% solution should be used?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
40% or 60% or anything more than about 38 % "saline" is nonsense. You probably want more help with the Mathematics of the description than advice about the physics of it.

x and y for quantities of the starting materials.
-----Mix your x and y liters to make a mixture of 8 liters.

Account for the "pure material".

and simplify this.
-

------simplified from the percentage equation.

Now you have a system of two linear equations in two unknown variables.

Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

I'm struggling with this word problem:
A 40% saline solution is to be mixed with 60% saline solution to obtain 8 liters of 55% saline solution. How many liters 40% solution should be used?

Let the amount of 40% saline solution to mix, be F
Then amount of 60% saline solution to mix = 8 - F
We then get: .4F + .6(8 - F) = .55(8)
.4F + 4.8 - .6F = 4.4
.4F - .6F = 4.4 - 4.8
- .2F = - .4
F, or amount of 40% saline solution needed = =
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