SOLUTION: pure acid is to be added to a 10% acid solution to obtain 90L of 32% soulution. what amounts of each should be used? how many liters of 100% pure acid should be used to make the so

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Question 1134842: pure acid is to be added to a 10% acid solution to obtain 90L of 32% soulution. what amounts of each should be used? how many liters of 100% pure acid should be used to make the solution?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let x be the pure acid volume (in liters), and

let y be the volume of the 10% acid solution to be mixed.


Then you have these two equations:


    for the total liquid volume

        x + y = 90             (1)


    for the pure acid volume

        x + 0.1y = 0.32*90     (2)


Subtract equation (2) from equation (1). You will get


            0.9y = 90 - 0.32*90.

Next divide both sides by 0.9. You will get


               y = %2890-0.32%2A90%29%2F0.9 = 68.


Thus you need 68 liters of the 10% acid solution.


To find x, use equation (1)


    x + 68 = 90


and get from it


    x = 90 - 68 = 22.


The problem is just solved.  You need  22 liters of the pure acid and  68 liters of the 10% acid solution.    ANSWER

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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.


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


There is an alternative to the standard algebraic method for solving mixture problems like this, as shown in the solution from tutor @ikleyn.

The alternative method is based on where the mixture percentage lies between the percentages of the two ingredients.

The given numbers in this problem make the solution almost trivial. Here is what the method looks like for this problem.

(1) The 32% target percentage is 22/90 of the way from the 10% of the first ingredient to the 100% of the second ingredient. (100-10 = 90; 32-10 = 22; so 22/90)
(2) That means 22/90 of the mixture has to be the second ingredient.
(3) 22/90 of 90L is 22L

So 22L of the 100% pure acid, 90-22 = 68L of the original 10% acid.