a=14% protein in soybean b=7% protein in cornmeal
how many pounds mixed together to get 280lb mixture that is 9% protein
a= ____lbs soybean
b= ____lbs cornmeal
Thank you from the bottom of my heart,
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Looks like a classical algebra problem to me.
280lbs=S+C weight equation
280*.09=.14S+.07C protein equation
How to solve?
I would multiply the top equation by .07, then subtract one equation from the other.
To solve this problem, we'll need to set up an equation based on the given information.
Let's assume the amount of soybean required is "x" pounds.
Therefore, the amount of cornmeal required would be 280 - x pounds, as the total mixture is 280 pounds.
Now, let's calculate the amount of protein from the soybean and cornmeal separately:
Protein from soybean = (14/100) * x pounds
Protein from cornmeal = (7/100) * (280 - x) pounds
Since the resulting mixture is 9% protein, we can set up the equation:
Protein from soybean + Protein from cornmeal = 9% of 280 pounds
(14/100) * x + (7/100) * (280 - x) = (9/100) * 280
Now, let's solve this equation step by step:
(14/100) * x + (7/100) * (280 - x) = (9/100) * 280
Multiply all terms to eliminate the fractions:
0.14x + 0.07(280 - x) = 0.09 * 280
0.14x + 19.6 - 0.07x = 25.2
Combine like terms:
0.07x + 19.6 = 25.2
Subtract 19.6 from both sides:
0.07x = 25.2 - 19.6
0.07x = 5.6
Divide both sides by 0.07:
x = 5.6 / 0.07
x ≈ 80
Therefore, you would need approximately 80 pounds of soybean and 280 - 80 = 200 pounds of cornmeal to get a 280-pound mixture that is 9% protein.
So, the answer is:
a = 80 pounds of soybean
b = 200 pounds of cornmeal