SOLUTION: a solution containing 8% salt is to be mixed with a solution containing 16% salt to obtain a 50 liter mixture with 10%. How many liters of each solution should be used?

Question 1093174: a solution containing 8% salt is to be mixed with a solution containing 16% salt to obtain a 50 liter mixture with 10%. How many liters of each solution should be used?
Answer by ikleyn(52788) (Show Source): You can put this solution on YOUR website!
.

let x = volume in liters of the 8% solution,
    y = volume of the 16% solution.


Then

    x +     y = 50,         (1)    (total volume equation)
0.08x + 0.16y = 0.1*50.     (2)    (salt content equation)


Express x = 50-y from eq(1), and substitute it into eq(2). You will get

0.08*(50-y) + 0.16y = 5,

4 - 0.08y + 0.16y = 5,

0.08y = 5 - 4 = 1  ====>  y =  = 12.5.


Answer.  12.5 liters of the 16% solution and 50-12.5 = 37.5 liters of the 8% solution must be mixed.


Check.    = 0.1.   ! Correct !

Solved.

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There is entire bunch of introductory lessons covering various types of mixture problems
    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Word problems on mixtures for antifreeze solutions
    - Word problems on mixtures for alloys
    - Typical word problems on mixtures from the archive
in this site.

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.

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