SOLUTION: It is necessary to have a​ 40% antifreeze solution in the radiator of a certain car. The radiator now has 70 liters of​ 20% solution. How many liters of this should be

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Question 1118139: It is necessary to have a​ 40% antifreeze solution in the radiator of a certain car. The radiator now has 70 liters of​ 20% solution. How many liters of this should be drained and replaced with​ 100% antifreeze to get the desired​ strength?
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Drain and Replace: Radiator Antifreeze Mixture - LESSON
https://www.algebra.com/my/drain-and-replace-antifreeze.lesson?content_action=show_dev

Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
. 
Let V be the volume to drain off from 70 liters of antifreeze.


Step 1:  Draining.  After draining,  you have 70-V liters of the 20% antifreeze.

                    It contains 0.20*(70-V) of pure antifreeze.


Step 2:  Replacing.  Then you add V liters of the pure antifreeze (the replacing step).

                     After the replacing,  you have the same total liquid volume of 70 liters.

                     It contains 0.20(70-V) + V liters of pure antifreeze.



So, the antifreeze concentration after replacement is  %280.20%2A%2870-V%29%2BV%29%2F70. 

It is the ratio of the pure antifreeze volume to the total volume.



Therefore, your "concentration equation" is

%280.20%2A%2870-V%29%2BV%29%2F70 = 0.4.    (1)    


The setup is done and completed.


To solve the equation (1), multiply both sides by 20. You will get

0.20*(70-V) + V= 0.4*70,

14 - 0.20V + V= 28,

0.8V = 28 - 14 = 14  ====>  V = 14%2F0.8 = 17.5 liters.


Answer.  17.5 liters of the 20% antifreeze must be drained and replaced by 17.5 liters of pure antifreeze.


Check.   %280.20%2A%2870-17.5%29%2B17.5%29%2F70 = 0.4.    ! Correct !


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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.

Read them and become an expert in solution mixture word problems.
Notice that among these lessons there is one on antifreeze solutions marked by (*).


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.


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The problem is presented as a drain and replace problem just to confuse you. What you really have is just a mixture problem where you want to mix 20% antifreeze with 100% antifreeze to get 70 liters of 40% antifreeze.

There are many ways of solving problems like this where you are mixing two ingredients. The response from the other tutor shows a typical algebraic method.

In my opinion, there is a much easier and faster way to solve this kind of problem, using the ratio in which the two ingredients need to be mixed.

For this problem, we are mixing 20% and 100% antifreeze ingredients to get a 40% antifreeze mixture. The ratio in which the ingredients must be mixed is directly related to where the 40% lies between the 20% and the 100%.

To be short with the explanation, 40% is 1/4 of the way from 20% to 100%. (20% to 40% is 20%; 20% to 100% is 80%; 20% is 1/4 of 80%.) That means 1/4 of the mixture must be the 100% ingredient.

So 1/4 of the 70 liters, or 17.5 liters, is the 100% antifreeze that you added; the remaining 3/4 of the 70 liters, or 52.5 liters, is the 20% antifreeze that was already in the radiator.

To put that result in the terms in which the question was asked, 17.5 liters of the 20% antifreeze must be drained and replaced with 100% antifreeze to get 70 liters of 40% antifreeze.

So in the end here is all that is required to solve this problem:

40-20 = 20; 100-20 = 80; 20/80 = 1/4; 1/4 of 70 = 17.5

If your mental arithmetic is good, it will take you about 10 seconds to work the problem.