.
Let W be the volume of water to add, in ounces.
Then the volume of liquid is 980+W with the amount of the pure alcohol of 420 ounces.
The concentration equation is then
= 0.42,
Multiply both sides by the denominator to get
420 = 0.42*(980+W)
Simplify and find W
420 = 0.42*980 + 0.42W
W = = 20.
ANSWER. 20 ounces of water to add.
Now, when the answer is just obtained, it seems OBVIOUS to you: when you add 20 ounces of water to 980 ounces
of the original mixture, you will have 1000 ounces of the diluted mixture with 420 ounces of the pure alcohol,
which makes the final mixture 42% alcohol.
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It is a standard and typical mixture problem.
In this site, there is a bunch of lessons, covering various types of mixture problems. See introductory lessons
- 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
For dilution problems, similar to the given one, there is a special lesson
- Special type mixture problems on DILUTION adding water
Read the lessons and become an expert in solution the 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 online textbook 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.