SOLUTION: please help me with this problem! ): I don't know how to do this. An 11% salt solution was mixed with an 18% salt solution to make 20L of a 16% solution. How many liters of 11%

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: please help me with this problem! ): I don't know how to do this. An 11% salt solution was mixed with an 18% salt solution to make 20L of a 16% solution. How many liters of 11%      Log On


   

Question 760343: please help me with this problem! ): I don't know how to do this.
An 11% salt solution was mixed with an 18% salt solution to make 20L of a 16% solution. How many liters of 11% salt solution were required?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Assign variables to all quantities.
L = 11%, for 11% salt
H = 18%, for 18% salt
T = 16%, the target value
M = 20 Liters, the target amount of solution
u = unknown volume of the 11% solution
v = unknown volume of the 18% solution

Form the equations for the situation.
Resulting salt concentration is (amount of pure salt)/(amount of salt solution result).
Sum of solutions used to prepare the target solution will be 20 liters.

%28Lu%2BHv%29%2FM=T and u%2Bv=M
Solve this system for u and v.

Here is a start:
Lu%2BHv=TM
and from the volume sum equation, v=M-u;
substitute for v,
Lu%2BH%28M-u%29=TM
Lu%2BHM-Hu=TM
Lu-Hu=TM-HM
u%28L-H%29=M%28T-H%29
highlight%28u=M%28T-H%29%2F%28L-H%29%29 a formula now for u,...
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