SOLUTION: A chemist mixes a 5% salt solution with an 11% salt solution. How many milliliters of each should be used to make 600 milliliters of a 7% salt solution?

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Question 1138413: A chemist mixes a 5% salt solution with an 11% salt
solution. How many milliliters of each should be used to make
600 milliliters of a 7% salt solution?

Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39621)   (Show Source): You can put this solution on YOUR website!
H = 11
L = 5
T = 7
M =600
x = unknown amount of "H" material
M-x = unknown amount of "L" material


.
.
the steps will yield .
Substitute and evaluate.

Answer by ikleyn(52818)   (Show Source): You can put this solution on YOUR website!
.

In this problem, concentrations are the ratios of a salt mass (in grams) to the volume of a mixture (in milliliters).



Let x be the amount of the 5%  solution needed (in milliliters), and

let y be the amount of the 11%  solution needed.



The mass of the salt in the 5% mixture  is 0.05x grams.

The mass of the salt in the 11% mixture is 0.11y grams.

The resulting mixture contains  0.05x + 0.11y grams of the dissolved salt and has the volume of 600 mL.


Thus you have these two equations


    x + y = 600    milliliters           (1)    (the total volume)

     = 0.07.                 (2)    (the resulting mixture concentration)



From equation  (1), express  x = 600 - y.  Substitute it into equation (2) and multiply both sides of this equation by 600. 
You will get


    0.05*(600-y) + 0.11y = 0.07*600.


From the last equation express y and calculate


    y =  = 200 mL of the 11% mixture are needed.


Then from equation (1),  x = 600 - 200 = 400 mL of the 5% mixture are needed.


Answer. 200 mL of the 11% mixture  and  400 mL of the 5% mixture are  needed.


Check.   = 0.07 = 7%.   ! Correct concentration !

The problem is just solved.
I used 2-equation setup and the Substitution method.

There are other solution methods, too.

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

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
    - Advanced mixture problems
    - Advanced mixture problem for three alloys
in this site.

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(13203)   (Show Source): You can put this solution on YOUR website!


Here is a solution using an alternative to the usual algebraic methods. If you understand this method, it will get you to the answer to mixture problems like this much faster and with far less work than any of the algebraic methods.

(1) The target percentage (7) is 1/3 of the way from the lower percentage (5) to the higher percentage (11).

(It might help you see what I'm saying there by plotting the 5, 7, and 11 on a number line; 7 is 1/3 of the way from 5 to 11.)

(2) That means 1/3 of the mixture has to be the higher percentage.

ANSWER: 1/3 of 600ml, or 200ml of 11%; so 400ml of 5%.

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