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a chemist needs 25 liters of a 10% alcohol solution but has only a 25% alcohol solution.
how many liters each of the 25% solution and water should he mix to make the desired 25 liters of 10% solution?
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Solution 1. One equation in one unknown.
Let W be the volume of water in liters to mix with the 25% alcohol.
Then the volume of the 25% alcohol solution is (25-W) liters.
The "alcohol content" equation is
0.25*(25-W) = 0.1*25, (1)
or
6.25 - 0.25W = 2.5 ---> -0.25W = 2.5 - 6.25 ---> -0.25W = 15.
15 liters of water is needed.
Answer. Mix 25-15 = 10 liters of the 25% alcohol and 15 liters of water, and you will get 25 liters of the 10% alcohol.
Solution 2. Two equations in two unknown.
Let W be the volume of water in liters to mix with the 25% alcohol.
Let A be the volume of the 50% alcohol to mix.
Then the volume equation is
A + W = 25 (1)
The "alcohol content" equation is
0.25A = 0.1*25. (2)
Express A = 25-W from (1) and substitute it into (2). You will get
0.25(25-W) = 0.*25. (3)
This equation coincides with the equation (1) of the Solution 1 above and, hence, has the same solution.
So, in this way you get the same answer.
There is entire bunch of the 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".
The textbook contains many other solved word problems, as well as many other interesting and useful things.