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A chemist wants to mix 10% alcohol solution and a 20% alcohol solution to make a 12% alcohol solution.
How many cups of each solution must be mixed to make 5 cups of the 12% solution?
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OK, let us measure the volume in cups. Why not?
Let "n" be the number of cups of the 10% solution.
Then the number of the 20% solution is 5-n.
The "alcohol content" equation is
= 0.12, or
0.1n + 0.2*5 - 0.2n = 5*0.12,
-0.1n = 0.6 - 1.0,
-0.1n = -0.4,
n = = 4.
Answer. 4 cups of the 10% solution must be mixed with 5-4 = 1 cup of the 20% solution.
There is entire bunch of 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 the 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".