SOLUTION: suppose that 50 mL of 25% acid are added to x mL of a 70% acid solution. How much 70% acid should be added to the 25% solution to produce a new solution that is 45% acid? Ive t

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: suppose that 50 mL of 25% acid are added to x mL of a 70% acid solution. How much 70% acid should be added to the 25% solution to produce a new solution that is 45% acid? Ive t      Log On

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Question 904339: suppose that 50 mL of 25% acid are added to x mL of a 70% acid solution. How much 70% acid should be added to the 25% solution to produce a new solution that is 45% acid?
Ive tried this so far
(50)(0.25)+(x)(0.70)=(x+50)(0.45) but I end up getting that x is 0.025 mL which doesn't seem right.
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
For thorough help, study this - http://www.algebra.com/my/Two-Part-Mixture-with-one-material-quantity-unknown.lesson?content_action=show_dev

To substitute the given values early,
highlight_green%28%2850%2A25%2Bx%2A70%29%2F%28x%2B50%29=45%29
Just be careful with the arithmetic and the decimal usage, and you should have no real trouble.

%2850%2A25%2Bx%2A70%29%2F%28x%2B50%29=45

%2810%2A25%2B14%2Ax%29%2F%28x%2B50%29=9

14x%2B250=9%28x%2B50%29

14x%2B250=9x%2B450

5x=200

x=200%2F5

highlight%28x=40%29, this number of mL


Does that work?
%2850%2A25%2B40%2A70%29%2F%2840%2B50%29=45
Yes.