SOLUTION: In the lab, Jim has two solutions that contain alcohol and is mixing them with each other. Solution A is 40% alcohol and Solution B is 12% alcohol. He uses 500 milliliters of Solu

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: In the lab, Jim has two solutions that contain alcohol and is mixing them with each other. Solution A is 40% alcohol and Solution B is 12% alcohol. He uses 500 milliliters of Solu      Log On

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Question 1196214: In the lab, Jim has two solutions that contain alcohol and is mixing them with each other. Solution A is 40% alcohol and Solution B is 12% alcohol. He uses 500 milliliters of Solution A. How many milliliters of Solution B does he use, if the resulting mixture is a 32% alcohol solution?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!


                   CONC.           VOL.             PURE ALC.



solution A          40%            500 ml.          0.4*500



solution B          12%             b ml.           0.12b



MIX                                b+500            0.32




%280.4%29%28500%29%2B0.12b=0.32%28b%2B500%29



.



.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
In the lab, Jim has two solutions that contain alcohol and is mixing them with each other.
Solution A is 40% alcohol and Solution B is 12% alcohol.
He uses 500 milliliters of Solution A. How many milliliters of Solution B does he use,
if the resulting mixture is a 32% alcohol solution?
~~~~~~~~~~~~~~~~~~~ 

Let x be the volmme of solution B, in mL.

The balance equation for the pure alcohol in the final mixture is

    0.4*500 + 0.12*x = 0.32*(500+x)  milliliters.


It is your basic equation. Simplify it and find x

    200 + 0.12x = 160 + 0.32x

    200 - 160 = 0.32x - 0.12x

       40     = 0.2x

        x     = 40/0.2 = 200.


ANSWER.  200 mL of solution B.

Solved.

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In 95% of cases, typical school mixture problems can be solved by reducing
to similar simple equation for the balance of the solute mass/volume,
which should be solved thereafter.

So, in 95% cases it is the same routine.

-----------------

It is a standard and typical mixture problem.

For introductory lessons covering various types of mixture word problems see
    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Typical word problems on mixtures from the archive
in this site.

You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization,  from very detailed to very short.

Read them and become an expert in solution mixture word problems.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook in the section "Word problems" under the topic "Mixture problems".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.