SOLUTION: 40 liters of a 50% salt solution are reduced to a 40% salt solution. How much of salt solution must be drained off and replaced with distilled water so that the resulting solution

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Question 1092040: 40 liters of a 50% salt solution are reduced to a 40% salt solution. How much of salt solution must be drained off and replaced with distilled water so that the resulting solution contains only 40% solution?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

First a traditional algebraic solution; and then I will show you the unorthodox method I use to solve mixture problems. If you understand my method, it will get you to the answer much faster and with far less work.

You are draining x liters of the 50% salt solution and replacing it with distilled water (0% salt) to obtain 40 liters of 40% salt solution. Don't get confused with the "draining and replacing"; in the end you are simply mixing two ingredients.

let x = liters of distilled water (0% salt)
then 40-x = liters of 50% salt solution

You want the 40 liters you end up with to be 40% salt. Equating the amounts of salt in the two ingredients to the amount in the final mixture gives you the algebraic equation:






You need to drain 8 liters of the 50% salt solution and replace it with distilled water to get 40 liters of 40% salt solution.

Now my method, based on the ratios in which the two ingredients are mixed.

The difference between the percentages of salt in the original salt solution and in the desired mixture is 50-40=10; the difference between the percentages of salt in the desired mixture and in the distilled water is 40-0 = 0.

Those differences of 40 and 10 mean the two ingredients must be mixed in the ratio 40:10, or 4:1. And since 40% is closer to 50% than to 0%, the larger portion must be the 50% salt solution.

But 40 liters in the ratio 4:1 means 32 liters of one ingredient and 8 of the other.

So you need 32 liters of the original 50% salt solution and 8 liters of distilled water.

No equations to solve; just a couple of subtractions and finding a ratio....
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
Couple of notices regarding this problem. 

1.  The info from Wikipedia, https://en.wikipedia.org/wiki/Saline_water

    The saturation level is dependent on the temperature of the water. At 20 °C one milliliter of water can dissolve 
    about 0.357 grams of salt; a concentration of 26.3%. At boiling (100 °C) the amount that can be dissolved 
    in one milliliter of water increases to about 0.391 grams or 28.1% saline solution.


    So, it is practically unrealistic thing to have more than 28% solution of NaCl in water.


    I fully understand that this physical info is out of your scope now,
    but it is still useful to know.


2. 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.
In particular, you will find the problems of the type "draining-replacing" in the lesson marked by (*).

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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