SOLUTION: A car radiator needs a 70​% antifreeze solution. The radiator now holds 20 liters of a 60​% solution. How many liters of this should be drained and replaced with​

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Question 1106104: A car radiator needs a 70​% antifreeze solution. The radiator now holds 20 liters of a 60​% solution. How many liters of this should be drained and replaced with​ 100% antifreeze to get the desired​ strength?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
https://www.algebra.com/my/drain-and-replace-antifreeze.lesson?content_action=show_dev
https://www.algebra.com/my/drain-and-replace-antifreeze.lesson?content_action=show_dev

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
A car radiator needs a 70% antifreeze solution. The radiator now holds 20 liters of a 60% solution.
How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?
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Let V be the volume to drain off from 20 liters of antifreeze.


Step 1:  Draining.  After draining,  you have 20-V liters of the 60% antifreeze.

                    It contains 0.60*(20-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 20 liters.

                     It contains 0.60*(20-V) + V liters of pure antifreeze.



So, the antifreeze concentration after replacement is  %280.60%2A%2820-V%29%2BV%29%2F20. 

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



Therefore, the concentration equation is 

%280.60%2A%2820-V%29%2BV%29%2F20 = 0.70.    (1)   


The setup is done and completed.


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

0.6*(20-V) + V = 0.7*20,

12 - 0.6V + V = 14,

0.4V = 14 - 12 = 2  ====>  V = 2%2F0.4 = 5 liters.


Answer.  5 liters of the 60% antifreeze must be drained and replaced by 5 liters of pure antifreeze.


Check.   %280.6%2A%2820-5%29%2B5%29%2F20 = 0.7.   ! 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.