SOLUTION: Nicolas mixes three solutions of acid with concentrations of 10%, 15%, and 5%. He wants to make 30 L of a mixture that is 12% acid and he uses four times as much of the 15% solutio

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Nicolas mixes three solutions of acid with concentrations of 10%, 15%, and 5%. He wants to make 30 L of a mixture that is 12% acid and he uses four times as much of the 15% solutio      Log On


   

Question 1055238: Nicolas mixes three solutions of acid with concentrations of 10%, 15%, and 5%. He wants to make 30 L of a mixture that is 12% acid and he uses four times as much of the 15% solution as the 5% solution. How much each of the three solutions must he use?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Three-part mixture but you can turn this into something that looks like a two-part mixture problem. 
VOLUME     CONCENTRATION
x            5%
y           10%
z           15%

z%2Fx=4

Already an equation is identified.

system%28%285x%2B10y%2B15z%29%2F30=12%2Cx%2By%2Bz=30%29
and again, you know one already found equation, useful as z=4x.

system%28%285x%2B10y%2B15%284x%29%29%2F30=12%2Cx%2By%2B4x=30%29

system%2865x%2B10y=360%2C5x%2By=30%29

system%2812x%2B2y=72%2C5x%2By=30%29
If you think carefully for a few seconds, you can continue this partly through Elimination Method.

( x=6 )

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

Nicolas mixes three solutions of acid with concentrations of 10%, 15%, and 5%. He wants to make 30 L of a mixture that is 12% acid and he uses four times as much of the 15% solution as the 5% solution. How much each of the three solutions must he use?
Correct answers: 

Again, IGNORE THE OTHER PERSON'S USUAL ERRONEOUS solution, if it can be called a solution!