SOLUTION: How many pounds of a chicken feed that is 35% corn must be mixed with 320 lb of a feed that is 80% corn to make a chicken feed that is 75% corn?

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Question 1102919: How many pounds of a chicken feed that is 35% corn must be mixed with 320 lb of a feed that is 80% corn to make a chicken feed that is 75% corn?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52805)   (Show Source): You can put this solution on YOUR website!
.
Let  x = How many pounds of a chicken feed that is 35% corn must be used.


0.35x + 0.8*320 = 0.75*(x+320).    (It is the balance equation for corn in the chicken feed)


0.35x + 0.8*320 = 0.75x + 0.75*320,

0.8*320 - 0.75*320 = 0.75x - 0.35x 

0.05*320 = 0.4x  ====>  x =  = 40.


Answer.  40 pounds of a chicken feed that is 35% corn must be used.


Check.  0.35*40 + 0.8*320 = 270 pounds of corn.

        0.75*(40+320) = 270 pounds of corn.    ! Correct !


-------------------
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 greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Here is another, very different, way to solve this kind of "mixture" problems.

For me, this other method gets me to the answer much faster, and with far less work, than the standard algebraic method shown by the other tutor.

(1) Determine how far the percentage of the mixture is from the percentages of each ingredient: 80%-75% = 5%; 75%-35% = 40%.
(2) The ratio of the percentage differences is 5:40, or 1:8.
(3) That means the two ingredients must be mixed in the ratio 1:8; and since the mixture percentage is much closer to 80% than to 35%, the larger portion must be the 80% feed.
(4) There are 320 lb of the 80% feed; the 8:1 ratio means there must be 40 lb of the 35% feed.

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