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Let x be the volume of the pure antifreeze under the question, in quarts.
8 quarts of the 50% antifreeze solution contain 0.5*8 quarts of the pure antifreeze.
When we add x quarts of the pure antifreeze, the amount of pure antifreeze becomes x + 0.5*8 quarts.
The total volume of liquid solution becomes (x+8) quarts.
Therefore, the antifreeze concentration after adding becomes .
According to the condition, it must be 60% = 0.6. It gives you an equation
= 0.6.
To solve it, first multiply both sides by (x+8). You will get
x + 4 = 0.6*(x+8), or
x + 4 = 0.6x + 4.8 ====> x - 0.6x = 4.8 - 4 ====> 0.4x = 0.8 ====> x = = 2.
Answer. 2 quarts of the pure antifreeze must be added.
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
in this site.
Among them, find the lesson on antifreeze marked by (*).
Read these lessons and become an expert in solving 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".