Let V be the volume to drain off from 26 quarts of antifreeze.
Step 1: Draining. After draining, you have 26-V quarts of the 50% antifreeze.
It contains 0.50*(26-V) of pure antifreeze.
Step 2: Replacing. Then you add V quarts of the pure antifreeze (the replacing step).
After the replacing, you have the same total liquid volume of 26 quarts.
It contains 0.50*(26-V) + V quarts of pure antifreeze.
So, the antifreeze concentration after replacement is .
It is the ratio of the pure antifreeze volume to the total volume.
Therefore, your "concentration equation" is
= 0.6. (1)
The setup is done and completed.
To solve the equation (1), multiply both sides by 26. You will get
0.50*(26-V) + V= 0.6*26,
13 - 0.5V + V= 15.6,
0.5V = 15.6 - 13 = 2.6 ====> V = = 5.2 liters.
Answer. 5.2 quarts of the 50% antifreeze must be drained and replaced by 5.2 quarts of pure antifreeze.
Check. = 0.6. ! Correct !
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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.
Notice that among these lessons there is one on antifreeze solutions 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.