SOLUTION: How much of​ a 60% orange juice drink must be mixed with 9 gallons of a 40% orange juice drink to obtain a mixture that is 50% orange​ juice? The amount of 60% orange juice

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Question 1184118: How much of​ a 60% orange juice drink must be mixed with 9 gallons of a 40% orange juice drink to obtain a mixture that is 50% orange​ juice?
The amount of 60% orange juice drink is x gallons.
​(Type an integer or a​ decimal.)
Please help me find "x" thank you
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
How much of​ a(n) 60% orange juice drink must be mixed with 9 gallons of a 40% orange juice drink
to obtain a mixture that is 50% orange​ juice?
The amount of 60% orange juice drink is x gallons.
​(Type an integer or a​ decimal.)
Please help me find "x" thank you
~~~~~~~~~~~~~~~~~~ 

If the amount of the 60% juice is x gallons, the the total volume of the final mixture is (x+9) gallons.


x gallons of the 60% juica contain  0.6x   gallons of the pure juice.

9 gallons of the 40% juica contain  0.4*9  gallons of the pure juice.


The final mixture should contain  0.5(x+9)  gallons of the pure juice.


Now you write equation


    0.6x + 0.4*9 = 0.5(x+9),


saying that the total of pure juice in ingredients is the same as it is in the final mixture.


    +--------------------------------------------------------------+
    |    As soon as you get this equation, the setup is complete,  |
    |            and now your task is to solve it.                 |
    +--------------------------------------------------------------+


From the equation


    0.6x - 0.5x = 0.5*9 - 0.4*9

        0.1x    =     0.9

           x    =     0.9/0.1 = 9.


ANSWER.  9 gallons of the 60% juice must be mixed.

Solved.

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

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.



Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
This is mixing a weaker solution of OJ with a stronger solution of OJ and
ending up with a medium strength solution of OJ.

You can do this one in your head.  Since 60% is just as much above 50% as
40% is below 50%, you would mix the same amount of each.  So the answer
should be 9 gallons.

But that's not really working it the way your teacher expects.  Most mixture
problems can't be done in your head.  This one was special.  Let's do it as
your teacher expects.

Most teachers suggest that you should make a chart with such word problems,
at least until you get a little more proficient.  One column for the amounts
of liquid, one for the percents, and for the third we multiply what's in the
other two columns, to get the amount of pure substance contained in the two
liquids.

------------------------------------------------------------
                 |        | percent as |   gallons of      |
                 | gallons|  a decimal | pure OJ contained |
============================================================
weaker solution  |   x    |     0.6    |      0.6x         |  
------------------------------------------------------------
stronger solution|   9    |     0.4    |    0.4(9)         |
------------------------------------------------------------
medium-strength  |  x+9   |     0.5    |   0.5(x+9)        |
       solution  |        |            |                   |
------------------------------------------------------------

The equation comes from the last column



Remove the decimals by multiplying through by 10:



Distribute:




Just as we figured, 9 gallons of each.

Edwin
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