SOLUTION: A chemist has three different acid solutions. The first acid solution contains 20 % acid, the second contains 35 % and the third contains 55 % . He wants to use all t

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Question 1142928: A chemist has three different acid solutions. The first acid solution contains
20
%
acid, the second contains
35
%
and the third contains
55
%
. He wants to use all three solutions to obtain a mixture of
90
liters containing
40
%
acid, using
3
times as much of the
55
%
solution as the
35
%
solution. How many liters of each solution should be used?
Found 3 solutions by ikleyn, josgarithmetic, greenestamps:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

Let x be the amount of the 35% solution, in liters.


Then the amount of the 55% solution is 3x liters,

and the amount of the 20% solution is  (90-x-3x) = (90-4x) liters.


The balance equation for the pure acid is


    0.2*(90-4x) + 0.35x + 0.55*(3x) = 0.4*90.


At this point the setup is completed, giving you one single equation for one unknown x.


Simplify and solve for x.


    18 - 0.8x + 0.35x + 1.65x = 36

    1.2x = 36 - 18 = 18.

    x             = 18%2F1.2 = 15 liters.


Thus the amount of the 35% solution is 15 liters;

     the amount of the 55% solution is 3*15 = 45 liters;

     the amount of the 20% solution is the rest  90 - 15 - 45 = 30 liters.


CHECK.  0.2*30 + 0.35*15 + 0.55*45 = 36 liters of the pure acid;

        0.4*90 = 36 liters of the pure acid.   ! Two values coincide - the solution is CORRECT !


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

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

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.


Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
               VARIABLE     CONCENTRATION%    PURE ACID

First           x            20                0.2x

Second          y            35               0.35y

Third         3y             55               0.55*3y

TOTAL         90             40              0.4*90

system%28x%2B4y=90%2C0.2x%2B2y=36%29

x=90-4yandsubstitute:

0.2%2890-4y%29%2B2y=36
--
18-0.8y%2B2y=36
1.2y%2B18=36
1.2y=18
y=15 
SUMMARY

         CONCENTRATION  VOLUME(Liters)
First     20%            30
Second    35% Acid       15
Third     55% Acid       45

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


The solution by tutor @josgarithmetic is not correct; the initial equations do not match the given information. Use the solution from tutor @ikleyn.

While the formal algebraic solution method shown by the other tutors is fine (and should be used if a formal algebraic solution is required), I find it much faster to solve mixture problems like this using logical reasoning rather than formal algebra. This method determines the proper portions of the ingredients by looking at where the target percentage lies between the percentages of two given ingredients.

In this problem, with three ingredients, we can work with two ingredients at a time. Specifically, here is how I would work the problem.

We are using 3 times as much of the 55% solution as the 35% solution. The amounts are then 3x and x, so 3/4 of this part of the mixture must be the 55% solution. And 3/4 of the way from 35% to 55% is 50%; that means these two ingredients together will be 50% acid.

So now we have two ingredients, with percentages of 20% and 50%; and our target percentage is 40%. Do the same kind of calculations as in the preceding paragraph. 40% is 2/3 of the way from 20% to 50%; so 2/3 of the mixture must be the mixture of 50% acid solution that we got from the 55% and 35% ingredients.

So 2/3 of the 90 L of the final mixture, or 60 L, is our 50% solution; since that solution contains 3 times as much of the 55% acid ingredient as it does of the 35% acid ingredient, that means 45 L of the 55% ingredient and 15 L of the 35% ingredient.

And then the other 1/3 of the final mixture, or 30 L, is the 20% acid ingredient.

ANSWER: 30L of 20% acid; 15 L of 35% acid; 45 L of 55% acid.

CHECK:

.2(30) + .35(15) + .55(45) = 6 + 5.25 + 24.75 = 36;
.40(90) = 36