SOLUTION: One 1700-pound load of feed mixture contains 25% corn, 40% bran, and roughage. How many pounds of corn should be added to a load of feed mixture to make the feed 40% corn?

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Question 1198048: One 1700-pound load of feed mixture contains 25% corn, 40% bran, and roughage. How many pounds of corn should be added to a load of feed mixture to make the feed 40% corn?
Found 3 solutions by ikleyn, josgarithmetic, greenestamps:
Answer by ikleyn(52797)   (Show Source): You can put this solution on YOUR website!
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One 1700-pound load of feed mixture contains 25% corn, 40% bran, and roughage.
How many pounds of corn should be added to a load of feed mixture to make the feed 40% corn?
~~~~~~~~~~~~~~~ 

Let x be the mass of the corn to add, in pounds.


Then you have this equation

    0.25*1700 + x = 0.4*(1700+x)


saying that the final mixture is 40% of corn.


From the equation

    0.25*1700 + x = 0.4*1700 + 0.4x

      x - 0.4x    = 0.4*1700 - 0.25*1700

        0.6x      =     255

           x      =     255/0.6 = 425.


ANSWER.  425 pounds of the corn should be added.

Solved.



Answer by josgarithmetic(39618)   (Show Source): You can put this solution on YOUR website!
               percent    pounds            wanted %        pounds result

CORN            25       (0.25)(1700)        40             c+(0.25)(1700)

BRAN           40

ROUGHAGE        35

TOTAL         100          1700                              c+1700

c, how many pounds of corn to add.


.
.

Compute this.

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


The original mixture is 1700 pounds, of which 25%, or 425 pounds, is corn.

Add x pounds of corn to the mixture; there are now 425+x pounds of corn, with a total of 1700+x pounds of mixture.

When that has been done, the corn is 40% (=2/5) of the mixture:







ANSWER: 425 pounds
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