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3 parts of the 40% and 1 part of the 80%, i.e. 9 liters of the 40% and 3 liters of the 80%, IF TO SOLVE THE PROBLEM MENTALLY
Let's check: = = = 0.5 = 50%.
Yes, it works.
Now I'll show you how to solve it using equation.
Let x be the amount (the volume in liters) of the 40% solution.
Then the amount of the 80% solution is (12-x) liters.
The "alcohol contents/concentration equation" is
= 0.5.
Multiply by 12 both parts:
0.4x + 0.8*(12-x) = 6 ====>
0.4x + 9.6 - 0.8x = 6 ====>
9.6 - 6 = 0.8x - 0.4x ====> 0.4x = 3.6 ===> x = = 9.
So, we get the same answer: 9 liters of the 40% solution and (12-9) = 3 liters of the 80% solution.
Solved.
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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.
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.