.
Let W be the volume to drain off from 2.5 liters of the original mixture.
Step 1: Draining. After draining, you have 2.5-W liters of the 25% mixture.
Step 2: Replacing. Then you add W liters of the pure alcohol (the replacing step).
After the replacing, you have the same total liquid volume of 2.5 liters.
It contains 0.25(2.5-W) + W of the pure alcohol.
So, your "concentration equation" is
= 0.4. (1)
The setup is done and completed.
Now you need to solve your basic equation (1).
0.25*(2.5-W) + W = 0.4*2.5
0.25*2.5 - 0.25W + W = 0.4*2.5
0.75W = 1 - 0.625
W = = 0.5.
Answer. 0.5 of a liter of the original mixture should be drained and replaced with the pure alcohol.
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There is entire bunch of 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 dry substances like coffee beans, nuts, cashew and peanuts
- Word problems on mixtures for dry substances like candies, dried fruits
- Word problems on mixtures for dry substances like soil and sand
- Word problems on mixtures for alloys
- Typical word problems on mixtures from the archive
- Advanced mixture problems
- Advanced mixture problem for three alloys
- OVERVIEW of lessons on word problems for mixtures
in this site.
A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.
Read them 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.