SOLUTION: A chemist has a barrel with 15% salt solution and another barrel with a 24% salt solution. If he wants to create 25 gallons of an 18% salt solution, how many gallons of each solu

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: A chemist has a barrel with 15% salt solution and another barrel with a 24% salt solution. If he wants to create 25 gallons of an 18% salt solution, how many gallons of each solu      Log On

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Question 1102693: A chemist has a barrel with 15% salt solution and another barrel with a 24% salt solution. If he wants to create 25 gallons of an 18% salt solution, how many gallons of each solution will he need?
Found 3 solutions by josgarithmetic, ikleyn, richwmiller:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
system%28x%2By=25%2C15x%2B24y=18%2A25%29
-
5x%2B8y=6%2A25=150
5%2825-y%29%2B8y=150
125-5y%2B8y=150
3y=25
y=25%2F3
highlight%28y=8%261%2F3%29---------gallons of the 24% salt
-
highlight%28x=16%262%2F3%29--------gallons of the 15% salt

Answer by ikleyn(52884) About Me  (Show Source):
You can put this solution on YOUR website!
.
    x +     y = 25            (1)   (gallons of the total mixture)
0.15x + 0.24x = 0.18*25       (2)   (mass of salt, calculated based on mass of each mixture)

Equivalently

  x +   y =    25             (3)
15x + 24y = 18*25             (4)


Solve by Elimination method. For it, multiply eq(3) by 15 (both sides).

15x + 15y = 15*25             (5)
15x + 24y = 18*25             (6)


Next subtract (5) from (6). You will get

9y = 18*25 - 15*25 = 3*25  ====>  y = %283%2A25%29%2F9 = 25%2F3 = 81%2F3.


Answer.  81%2F3 gallons of the 24% solution and 25 - 81%2F3 = 162%2F3 gallons of the 15% solution.


Check.   15*50%2F3 + 24*25%2F3 = 450.

         18*25          = 450.     ! Correct !


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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
in this site.

You will find there ALL TYPICAL mixture problems with different methods of solutions,
explained at the 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 richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
We want to know how many gal. at 15% to mix with 24%.
We need a total of 25 gal. at 18%
Using the method of alligation.
24         3 1/3*25=8 1/3 gal of 24% solution
      18     8 1/3+16 2/3= 25.0 gal of 18% solution
15         6 2/3*25=16 2/3 gal of 15% solution
           9

check
0.15 * 16 2/3 + 0.24 * 8 1/3 = 0.18*25
2.5 + 2.0 = 4.5
4.5 = 4.5
ok