SOLUTION: How many ml of a 20% solution should be added to 50ml of a 40% solution to obtain a 25% solution?

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Question 1137340: How many ml of a 20% solution should be added to 50ml of a 40% solution to obtain a 25% solution?
Answer by ikleyn(52884) About Me  (Show Source):
You can put this solution on YOUR website!
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Let x be the required amount (volume) of the 20% acid solution (in milliliters).


Then the equation for the pure acid volume is 


    0.2x + 0.40*50 = 0.25*(x + 50).


It is your basic equation for this problem. 

Simplify it


    0.2x - 0.25x = 0.25*50 - 0.4*50


Express x from the equation and calculate the answer


    x = %280.25%2A50-+0.4%2A50%29%2F%280.2+-+0.25%29 = 150.


Answer. 150 milliliters.


CHECK.  (0.20*150 + 0.40*50)/(50 + 150)}}} = 0.25,  or 25%.    ! Correct !

Solved.

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It is a standard and typical mixture problem.

In this site, there is a huge collection of lessons covering various types of mixture word problems.
Start from these introductory lessons

    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Typical word problems on mixtures from the archive

You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at different levels of detalization,  from very detailed to very short.

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".


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.