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How many gallons of a 70% antifreeze solution must be mixed with 100 gallons of 10% antifreeze
to get a mixture that is 60% antifreeze? Use the six- step method
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I don't know what the six-step method is;
therefore, I will solve the problem following to a standard procedure and common sense.
100 gallons of 10% antifreeze contain 0.1*100 = 10 gallons of pure antifreeze.
If we add x gallons of the 70% of the antofreeze solution, the total liquid volume will be 100+x gallons,
and the content of antifreeze in it will be 10+0.7x gallons.
Now, the concentration of the antifreeze in the new mixture is C = gallons of antifreeze
per gallon of the mixture.
It gives an equation to find x
= 0.6.
To solve it, multiply both sides by (100+x) to get
10+0.7x = 0.6*(100+x).
Simplify and find x
10 + 0.7x = 60 + 0.6x
0.7x - 0.6x = 60 - 10
0.1x = 50
x = 50/0.1 = 500.
ANSWER. 500 gallon of the 70% antifreeze solution should be added.
CHECK. = 0.6 = 60%. ! Correct concentration !
Solved.
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It is a standard and typical mixture problem.
In this site, there is a bunch of lessons, covering various types of mixture problems. See introductory lessons
- Mixture problems
- More Mixture problems
- Solving typical word problems on mixtures for solutions
- Word problems on mixtures for antifreeze solutions
- Typical word problems on mixtures from the archive
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 online textbook 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.