Question 1133071: How much pure water must be mixed with 10 oz of a 60% alcohol solution to obtain 25% alcohol solution?
Answer by ikleyn(52772)   (Show Source):
You can put this solution on YOUR website! .
How much  water must be mixed with 10 oz of a 60% alcohol solution to obtain 25% alcohol solution?
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I took off the word "pure" which was in combination with water,
because in such processes "dirty" water is never used.
We have 6 oz of the pure alcohol in the 10 oz solution, making it 60% solution.
We want these 6 oz of the pure alcohol would be 25%, or  of the volume of the new solution.
Hence, the volume of the new solution must be 6*4 = 24 oz.
It means that 24 - 10 = 14 oz of water must be added. ANSWER
CHECK.  =  = = 25%. ! Correct !
Solved.
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By the way, the fact that you use the combination of words "pure water" in your post, shows that the problem formulation
was done by the person UNFAMILIAR with terminology in this class of problems.
So, be careful: your writing tells a lot about you . . ..
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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.
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