SOLUTION: If a 7% saline solution and a 4% saline solution are mixed to make 500 milliliters of a 5% saline solution, how much of each solution, to the nearest milliliter, is needed?

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Question 710342: If a 7% saline solution and a 4% saline solution are mixed to make 500 milliliters of a 5% saline solution, how much of each solution, to the nearest milliliter, is needed?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
CHOOSING VARIABLES FOR ALL NUMBERS
Concentration of Low percent saline, L
Concentration of High percent saline, H
Target concentration of saline wanted, T
Amount of Low concentration saline to use, u, unknown
Amount of High concentration saline to use, v, unknown
Amount of saline solution made, M, or 500 ml. in the example

From problem description:
L=4%, H=7%, T=5%, M=500 ml.

Create Equations to Represent Situation:
%28uL%2BvH%29%2FM=T
u%2Bv=M
Solve that system for u and v.
uL%2BvH=MT
Use the volume sum equation for known mixture volume amount to substitute for v.
Using v=M-u,
uL%2B%28M-u%29H=MT
uL-uH=-MH%2BMT
u%28L-H%29=M%28T-H%29
highlight%28u=%28M%28T-H%29%29%2F%28L-H%29%29

You could substitute the given values into this formula for u, and then compute u. Then, use the u+v=M relationship to compute v.