.
Let x = the amount of the 50% alcohol wine needed (in liters).
Then the amount of the 75% alcohol wine needed is (4-x) liters.
The amount of the "pure" alcohol in these solutions is 0.5*x + 0.75*(4-x).
It must be exactly 60%, or 0.6 of 4 liters, i.e. 0.6*4 liters.
So, your "pure alcohol" equation is
0.5*x + 0.75*(4-x) = 0.6*4.
Simplify and solve it for x:
0.5x + 3 - 0.75x = 2.4,
-0.25x = 2.4 - 3,
-0.25x = -0.6,
x = = 2.4.
Answer. 2.4 liters of 50% alcohol wine and 4-2.4 = 1.6 liters of 75% alcohol wine are needed.
Check. 2.4*0.5 + 1.6*0.75 = 2.4 = 4*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.
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.