SOLUTION: How much of the solutions do I need to use when one container has 40% and another has 100% and 50% is needed for one gallon?I came up with a huge number.

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: How much of the solutions do I need to use when one container has 40% and another has 100% and 50% is needed for one gallon?I came up with a huge number.      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   


Question 234293: How much of the solutions do I need to use when one container has 40% and another has 100% and 50% is needed for one gallon?I came up with a huge number.
Answer by stanbon(48510) About Me  (Show Source):
You can put this solution on YOUR website!
How much of the solutions do I need to use when one container has 40% and another has 100% and 50% is needed for one gallon?
-------------
Equation:
0.40x + 1(1-x) = 0.50(1)
Multiply thru by 100 to get:
40x + 100(1-x) = 50
40x + 100 -100x = 50
-60x = -50
x = 5/6 gal. (amount of 40% solution needed in the mixture)
---
1-x = 1-(5/6) = 1/6 gal. (amount of 100% solution needed in the mixture)
===========================================================================
Cheers,
Stan H.