SOLUTION: A mixture contains 24 quarts of water and acid solution, of which 60% is acid. How much of this solution should be drained and replaced with water for the solution to be 40% acid?

Question 1091692: A mixture contains 24 quarts of water and acid solution, of which 60% is acid. How much of this solution should be drained and replaced with water for the solution to be 40% acid?

Answer by ikleyn(52784) (Show Source): You can put this solution on YOUR website!
.

Let V be the volume of the mixture to be drained and replaced with water.


Then the pure mixture contents in the final solution is  0.6*24 - 0.6*V quarts,
while the total volume after replacement remains the same, 24 quarts.


So your concentration equation is 

 = 0.4.


To solve it, multiply both sides by 24:

0.6*24 - 0.6*V = 0.4*24,  ====>

0.6*V = 0.6*24 - 0.4*24 = 0.2*24  ====>  V =  =  = 2*4 = 8.


Answer.  8 quarts of the mixture must be drained and replaced.

Solved.

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.

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".

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