SOLUTION: Pure acid is to be added to a 10% acid solution to obtain 99L of a 20% acid solution What amounts should be used ?

Question 1161215: Pure acid is to be added to a 10% acid solution to obtain 99L of a 20% acid solution
What amounts should be used ?
Answer by ikleyn(52786) (Show Source): You can put this solution on YOUR website!
.


Let x be the amount of the pure acid to add in order for to get 99L of a 20% solution.


It means that the amount of the 10% original acid was  99-x liters.


The equation to find x is THIS


    0.1*(99-x) + x = 0.2*99.    (1)


It says that the amount of the pure acid in ingredients (left side) is the same as in the mixture (right side).


Simplify and solve the equation


    0.1*99 - 0.1x +x = 19.8

    0.9x = 19.8 - 0.1*99

    0.9x = 9.9

       x =  = 11.


ANSWER.  11 liters of the pure acid to add.


CHECK.   I will check equation (1).

         Ingredients:  0.1*(99-11) + 11 = 0.1*88 + 11 = 8.8+11 = 19.8  liters of the pure acid.

         Mixture:  0.2*99 = 19.8 liters of the pure acid.

         Both sides of the equation are equal.  The check is completed.

Solved.

------------------

It is a standard and typical mixture word problem.

There is entire bunch of lessons in this site 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 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
    - OVERVIEW of lessons on word problems for mixtures

A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.

Read them and become an expert in solution the 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 online textbook 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.

RELATED QUESTIONS
Pure acid is to be added to 10% acid solution to obtain 99L of 20% solution. What amounts (answered by josgarithmetic)

Pure acid is to be added to a 10% acid solution to obtain 54L of a 20% acid solution.... (answered by longjonsilver)

Pure Acid is to be added to a 10% acid solution to obtain 27 litters of 20% acid... (answered by robertb)

Pure acid is to be added to a 10% acid solution to obtain 90 L of a 20% acid solution.... (answered by josgarithmetic)

Pure acid is to be added to a 10% acid solution to obtain 90 L of a 20% acid solution.... (answered by Alan3354)

Pure acid is to be added to a 10% acid solution to obtain 81 L of a 20% acid solutions.... (answered by ankor@dixie-net.com)

pure acid is to be added to a 20% acid solution to obtain 40L of a 40% acid solution.... (answered by josgarithmetic)

pure acid is to be added to a 10% acid solution to obtain 42 L of a 40% acid solution.... (answered by oberobic)

pure acid is to be added to a 20% solution to obtain 60 liters of a 30% acid solution.... (answered by robertb)