Let x = the amount of the 35% acid solution needed (in ounces).
The the amount of the 60% acid solution needed is (40-x) ounces.
The amount of the "pure" acid in these solutions is 0.35*x + 0.6*(40-x).
It is exactly 50%, or 0.5 of 40 ounces, i.e. 0.5*40.
So, your "pure acid amount" equation is
0.35*x + 0.6*(40-x) = 0.5*40.
Simplify and solve it for x:
0.35x + 24 - 0.6x = 20,
-0.25x = 20 - 24,
-0.25x = -4,
x = = = 16.
Answer. 16 ounces of the 35% acid solution is needed.
Check. 16*0.35 + (40-16)*0.60 = 20 = 0.5*40 ounces. ! Correct !
Solved.
--------------
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.