.
Let "x" the amount of the 65% antifreeze to be mixed, in gallons.
Then the amount of the 90% antifreeze is (90-x) gallons.
Your equation is
= 0.75.
The numerator of this formula is the amount of the pure antifreeze in the mixture,
that came from each of the two input solutions.
Divided by the total volume of 90 gallons, it is the concentration of the antifreeze in the mixture,
which must be 0.75 = 75%, according to the problem requirement.
To solve the equation, multiply both sides by 90. You will get
0.65x + 0.9*90 - 0.9x = 0.75*90
-0.25x = 0.75*90 - 0.9*90 = -13.5 ====> x = = 54.
Answer. 54 gallons of the 65% and (90-54) = 36 gallons of the 90% antifreeze should be mixed to get 90 gallons of the 75% antifreeze.
Check. = 0.75 = 75% ! Correct !
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
- Advanced mixture problems
- Advanced mixture problem for three alloys
in this site.
You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization, from very detailed to very short.
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.