Question 436332: Soybean is 18% protein
Cornmeal is 9% protein
How many lbs of each should be mixed together to get 360-lb mixture that is 12% protein? Found 3 solutions by stanbon, htmentor, josmiceli:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Soybean is 18% protein
Cornmeal is 9% protein
How many lbs of each should be mixed together to get 360-lb mixture that is 12% protein?
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Equations:
Quantity Eq: s + c = 360 lb
Protein Eq:0.18s+0.09c = 0.12(360)
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Multiply thru the 1st by 18
Multiply thru the 2nd by 100
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18s + 18c = 18*360
18x + 9c = 12*360
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Subtract and solve for "c":
9c = 6*360
c = (2/3)360
c = 240 lb. (amt. of cornmeal needed)
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Solve for "s":
s + c = 360
s + 240 = 360
s = 120 lb (amt of soybean needed)
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Cheers,
Stan H.
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You can put this solution on YOUR website! Let x = the number of pounds of soybeans
Then 360-x = the number of pounds of cornmeal
A 360 lb. 12% mixture contains 0.12*360 = 43.2 lb
So we can write the following equation:
0.18x + 0.09(360-x) = 43.2
Solve for x:
0.09x = 43.2 - 32.4 = 10.8 -> x = 120
So the number of lbs of soybeans = 120
And the number of lbs of cornmeal = 360-120 = 240
You can put this solution on YOUR website! Let = pounds of soybean needed
Let = pounds of cornmeal needed
given: = protein in soybeans = protein in cornmeal
(1)
(2)
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(2)
(2)
(2)
(2)
Multiply both sides of (1) by and
subtract (1) from (2)
(2)
(1)
and
(1)
120 pounds of soybeans are needed
240 pounds of cornmeal are needed
check answer:
(2)
(2)
(2)
OK
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