SOLUTION: An automobile radiator contains 18 quarts of a solution which is 20% alcohol and 80% water. How much of the solution must be drained off and replace by pure alcohol to give a sol

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Question 1172281: An automobile radiator contains 18 quarts of a solution which is 20% alcohol and 80% water. How much of the solution
must be drained off and replace by pure alcohol to give a solution which is 30 alcohol?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
An automobile radiator contains 18 quarts of a solution which is 20% alcohol and 80% water. How much of the solution
must be drained off and replace by pure alcohol to give a solution which is 30 alcohol?
~~~~~~~~~~~~~~~~~~~ 


Let V be the volume of the original solution to drain off from 18 quarts and to replace by pure alcohol.


Step 1:  Draining.  After draining,  you have 18-V quarts of the 20% solution.

                    It contains 0.20*(18-V) of pure alcohol.


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

                     After the replacing,  you have the same total liquid volume of 18 quarts.

                     It contains 0.20(18-V) + V quarts of the pure alcohol.



So, the alcohol concentration after replacement is  %280.20%2A%2818-V%29%2BV%29%2F18. 

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



Therefore, your "concentration equation" is

    %280.20%2A%2818-V%29%2BV%29%2F18 = 0.3.    (1)    


The setup is done and completed.


To solve the equation (1), multiply both sides by 18. You will get

    0.20*(18-V) + V= 0.3*18,

    3.6 - 0.20V + V = 5.4,

    0.8V = 5.4 - 3.6 = 1.8  ====>  V = 1.8%2F0.8 = 2.25 quarts.


Answer.  2.25 quarts of the 20% antifreeze must be drained and replaced by 2.25 quarts of the pure alcohol.


Check.   %280.20%2A%2818-2.25%29%2B2.25%29%2F18 = 0.3.    ! 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 "drained off..." and "replaced by..." language is only used to confuse you.

Essentially the problem is just mixing 20% alcohol and 100% alcohol to get 30% alcohol.

For a quick and easy non-algebraic method for solving the problem, note that you are starting with a 20% alcohol solution and adding 100% alcohol, stopping when you get to 30% alcohol.

With that way of thinking of the problem, observe that 30% is 1/8 of the way from 20% to 100% (20 to 30 is 10; 20 to 100 is 80; 10/80 = 1/8); that means 1/8 of the final mixture is the added 100% alcohol.

Since the capacity of the radiator is 18 quarts, the amount of 100% alcohol in the mixture is 18/8 = 2.25 quarts.

ANSWER: 2.25 quarts of the original solution must be drained off and replaced by 100% alcohol to obtain 18 quarts of 30% alcohol.