Lesson Draining-replacing mixture problems

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Draining-replacing mixture problems

Problem 1

The car radiator holds 20 liters. It contains a 25% antifreeze solution. Since the weather is getting colder, the mechanic should make
the solution 40% antifreeze. How much of the 25% solution should be drained and replaced with pure antifreeze to give a 40% solution.

Solution

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 25% antifreeze.

                    It contains 0.25*(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 0.25(20-V) + V liters of pure antifreeze.



So, the antifreeze concentration after replacement is  . 

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



Therefore, your "concentration equation" is

 = 0.4.    (1)    


The setup is done and completed.


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

0.25*(20-V) + V= 0.4*20,

5 - 0.25V + V= 8,

0.75V = 8 - 5 = 3  ====>  V =  = 4 liters.


Answer.  4 liters of the 25% antifreeze must be drained and replaced by 4 liters of pure antifreeze.


Check.    = 0.4.    ! Correct !

Problem 2

A tank holds 80 liters of a chemical solution. Currently, the solution has a strength of 30%.
How much of this solution must be drained and replaced with a 70% solution to get a strength of 40%?

Solution

Let W be the volume to drain off from 80 liters of solution.


Step 1:  Draining.  After draining,  you have (80-W) liters of the 30% acid solution.



Step 2:  Replacing.  Then you add W liters of the 70% solution (the replacing step).

                     After the replacing,  you have the same total liquid volume of 80 liters.


It contains  0.3*(80-W) + 0.7*W   of pure solvent.



So, your "concentration equation" is


     = 0.4.    (1)



At this point, the setup is done and completed.



Now you need to solve your basic equation  (1).

As the first step, multiply both sides by 80, and then simplify


    0.30*(80-W) + 0.7*W = 0.4*80

    24 - 0.3W + 0.7W = 32

    0.4W = 32 - 24 = 8

    W =  = 20.


ANSWER.  20 liters of the original 30% solution should be drained and replaced by 20 liters of the 70% solutions.


CHECK.    = 0.4.    ! Precisely correct !

Problem 3

A 2.5 liters container has a mixture of 25% alcohol. How many liters of the mixture must be drained out and replaced with pure alcohol
in order to obtain a mixture containing 40% alcohol?

Solution

Let W be the volume to drain off from 2.5 liters of the original mixture.


Step 1:  Draining.  After draining,  you have 5-W liters of the 25% mixture.


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

                     After the replacing,  you have the same total liquid volume of 2.5 liters.


It contains  0.25(2.5-W) + W  of the pure alcohol.


So, your "concentration equation" is


 = 0.4.    (1)


The setup is done and completed.


Now you need to solve your basic equation  (1).


0.25*(2.5-W) + W = 0.4*2.5

0.25*2.5 - 0.25W + W = 0.4*2.5

0.75W = 1 - 0.625

W =  = 0.5.


Answer.  0.5  of a liter of the original mixture should be drained and replaced with the pure alcohol.

My other lessons on word problems for mixtures in this site are
    - 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 dry substances like coffee beans, nuts, cashew and peanuts
    - Word problems on mixtures for dry substances like candies, dried fruits
    - Word problems on mixtures for dry substances like soil and sand
    - Word problems on mixtures for alloys
    - Typical word problems on mixtures from the archive
    - Advanced mixture problems
    - Advanced mixture problem for three alloys
    - Unusual word problem on mixtures
    - Check if you know the basics of mixtures from Science

    - Special type mixture problems on DILUTION adding water
    - Increasing concentration of an acid solution by adding pure acid
    - Increasing concentration of alcohol solution by adding pure alcohol
    - How many kilograms of sand must be added to a mixture of sand and cement
    - How much water must be evaporated
    - Advanced problems on draining and replacing
    - Using effective methodology to solve many-steps dilution problems

    - Entertainment problems on mixtures

    - OVERVIEW of lessons on word problems for mixtures

Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.

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