Soybean meal is 12% protein, cornmeal is 6% protein. How many pounds of each should be mixed together in order to get a 240-lb mixture that is 10% protein?
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let the amount of soybean be x lbs
then the amount of cornmeal is 240-x lbs
solve:
.12x + .06(240-x) = .1(240)
No
To determine the amounts of soybean meal and cornmeal needed to create a mixture with a specific protein content, we can use the concept of variables and equations.
Let's assume:
- x = pounds of soybean meal needed
- y = pounds of cornmeal needed
We know the following information:
- The total weight of the mixture is 240 pounds, so x + y = 240.
- The soybean meal is 12% protein, meaning that 0.12x pounds of protein come from soybean meal.
- The cornmeal is 6% protein, meaning that 0.06y pounds of protein come from cornmeal.
- The mixture is required to be 10% protein, meaning that 0.10(240) pounds of protein are needed in total.
To solve the problem, we need to set up an equation based on the protein content. The equation is:
0.12x + 0.06y = 0.10(240)
Let's solve this equation to find the values of x and y:
0.12x + 0.06y = 24
Now we can use the substitution method or elimination method to find the solution. I will use the elimination method:
Multiply the entire equation by 100 to eliminate decimals:
12x + 6y = 2400
Now, rearrange the first equation x + y = 240 to isolate x:
x = 240 - y
Substitute the modified value of x into the second equation:
12(240 - y) + 6y = 2400
Distribute 12 into the parentheses:
2880 - 12y + 6y = 2400
Combine like terms:
-6y = 2400 - 2880
Simplify:
-6y = -480
Divide both sides by -6 to isolate y:
y = -480 / -6
y = 80
Now substitute the value of y back into the equation x = 240 - y:
x = 240 - 80
x = 160
Therefore, you would need 160 pounds of soybean meal and 80 pounds of cornmeal to create a 240-pound mixture that is 10% protein.