SOLUTION: A landscaping company needs 120 gallons of 13​% fertilizer to fertilize the shrubs in an office park. They have in stock 20​% fertilizer and 10​% fertilizer. How much of each

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Question 1135355: A landscaping company needs 120 gallons of 13​% fertilizer to fertilize the shrubs in an office park. They have in stock 20​% fertilizer and 10​% fertilizer. How much of each type should they mix​ together?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52797)   (Show Source): You can put this solution on YOUR website!
.
Let x = the amount of the 20% fertilizer (in gallons) and y = the amount of the 10% fertilizer.


Then you have two equations


         x +     y = 120       gallons  of the mixture                (1)
and
     0.20x + 0.10y = 0.13*120  gallons  of the pure  fertilizer       (2)


So, your system of equations is THIS


         x +     y = 120         (1)

     0.20x + 0.10y = 0.13*120    (2)
         

It can be solved by Substitution or Elimination method, on your choice.

If you solve by Elimination, multiply eq(1) by 0.2 (both sides) and keep eq(2) as is, with no change. You will get

      
     0.20x + 0.20y = 0.20*120    (1')

     0.20x + 0.10y = 0.13*120    (2)

----------------------------------------- Now subtract eq(2) from eq(1')

                                          The terms " 0.20x " will cancel each other, and you will get


             0.20y - 0.10y = 0.20*120 - 0.13*120,   

which gives you

             y =  = one click in my Excel = 84.


ANSWER.  84 gallons of the 10% fertilizer and  the rest (120-84) = 36 gallons of the 20% fertilizer.


CHECK.  0.1*84 + 0.2*36 = 15.6 gallons of the pure fertilizer = 0.13*120.   ! Correct !

Solved.

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

It is a standard and typical mixture problem.

For introductory lessons covering various types of mixture word problems see
    - Mixture problems
    - More Mixture problems
    - Solving typical word problems on mixtures for solutions
    - 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 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)   (Show Source): You can put this solution on YOUR website!


Here is a very short and quick alternative to the traditional algebraic solution method shown by the other tutor.

(1) The target percentage of 13% is 3/10 of the way from 10% to 20% -- the percentages of the two ingredients.
(2) Therefore 3/10 of the mixture should be the higher percentage ingredient.

ANSWER: 3/10 of 120 = 36 gallons of 20% fertilizer; 7/10 of 120 = 84 gallons of 10% fertilizer.
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