many pounds of each should be mixed together in order to get 240-lb mixture that is 7% portein how many pounds of the cornmeal should be in the mixture? How many pounds of the soybean meal sould be in the mixture?
You omitted some vital information.
To determine the number of pounds of cornmeal and soybean meal needed for the mixture, we need to set up a system of equations based on the given information and solve for the values.
Let's denote the pounds of cornmeal as C and the pounds of soybean meal as S.
Based on the problem, we have two pieces of information:
1. The total weight of the mixture is 240 pounds.
Therefore, we can write the first equation as:
C + S = 240 (Equation 1)
2. The mixture is 7% protein.
This means that the protein in the mixture is 7% of its total weight.
The amount of protein in the cornmeal is 0%, while the amount of protein in the soybean meal is 100%.
To calculate the protein content of the mixture, we can use the following equation:
(0% of C) + (100% of S) = 7% of (C + S) (Equation 2)
Now, let's solve this system of equations to find the values of C and S.
Rewriting Equation 2:
0.00C + 1.00S = 0.07(C + S)
0.00C + 1.00S = 0.07C + 0.07S
Moving terms around:
1.00S - 0.07S = 0.07C - 0.00C
0.93S = 0.07C
Now, let's rearrange Equation 1 to express C in terms of S:
C = 240 - S
Substituting this into the rearranged Equation 2:
0.93S = 0.07(240 - S)
Expanding:
0.93S = 16.80 - 0.07S
Combining like terms:
0.93S + 0.07S = 16.80
1.00S = 16.80
Dividing both sides by 1:
S = 16.80
Now, we can substitute the value of S back into Equation 1 to find C:
C + 16.80 = 240
C = 240 - 16.80
C = 223.20
Therefore, to obtain a 240-lb mixture that is 7% protein, you would need approximately 223.20 pounds of cornmeal and 16.80 pounds of soybean meal.