Question

Soybean meal is 16% protein; cornmeal 8% protein. How many pounds of each should be mixed together in order to get 320-b mixture that is 14% protein. How many pounds of the cornmeal should be in the mixture?

Answer 1

Hint: To find the mixture of cornmeal, we need to consider the given data that, there is total cornmeal 8% protein in which Soybean meal is 16% protein, mixed together in order to get 320-b mixture that is 14% protein; hence considering this we need to solve for the quantity of cornmeal to be in the mixture.

Complete step by step solution:
Soybean protein content = 16%
Cornmeal protein content = 8%
Target protein content =14%
Total weight of mixture to be 320lb.
Let the amount to find Soybean be s.
Let the amount to find of Cornmeal be c.
Then, we have:
 \[s + c = 320lb\]
 \[ \Rightarrow s = 320 - c\] ……………….. 1
Target protein content is 14%, then we have:
 \[ \to \dfrac{{14}}{{100}} \times 320 = 44.8lb\]
Equating the protein content with respect to soybean and cornmeal, we have:
 \[ \Rightarrow \dfrac{{16}}{{100}}s + \dfrac{8}{{100}}c = 44.8lb\] …………………… 2
As, we need to determine how many pounds of the cornmeal should be in the mixture, i.e., here we need to determine the value of c, so substitute for s in equation 2 using equation 1 as:
 \[\dfrac{{16}}{{100}}s + \dfrac{8}{{100}}c = 44.8lb\]
Substituting value of s as \[\left( {320 - c} \right)\] from equation 1 we get:
 \[ \Rightarrow \dfrac{{16}}{{100}}\left( {320 - c} \right) + \dfrac{8}{{100}}c = 44.8\]
Now, simplify the terms, as:
 \[ \Rightarrow \dfrac{{16}}{{100}}\left( {320} \right) - \dfrac{{16}}{{100}}c + \dfrac{8}{{100}}c = 44.8\]
 \[ \Rightarrow \dfrac{{5120}}{{100}} - \dfrac{{16}}{{100}}c + \dfrac{8}{{100}}c = 44.8\]
Simplify the terms, and combine the like terms, to evaluate, we get:
 \[ \Rightarrow \dfrac{{256}}{5} - \dfrac{8}{{100}}c = 44.8\]
 \[ \Rightarrow 51.2 - \dfrac{8}{{100}}c = 44.8\]
Rearranging the terms to solve for c, hence we have:
 \[ \Rightarrow \dfrac{8}{{100}}c = 51.2 - 44.8\]
 \[ \Rightarrow \dfrac{8}{{100}}c = 6.4\]
 \[ \Rightarrow c = \dfrac{{100}}{8} \times 6.4\]
Therefore, cornmeal in the mixture is:
 \[ \Rightarrow c = 80lb\]
Now, let us determine the value of s from equation 1 as:
 \[s = 320 - c\]
 \[ \Rightarrow s = 320 - 80\]
Therefore, soybean in the mixture is:
 \[ \Rightarrow s = 240lb\]
So, the correct answer is “s = 240lb”.

Note: There is a direct relationship in the mixture between the two foods. If you have some weight of soybean the amount of cornmeal is 320 - weight of soybean. Hence, by considering the weight of just one of the foods you are by default implying the weight of the other. If you have no soybean that is all cornmeal at 8% protein and if you have all soybean at 16% protein then there is no cornmeal. Hence, this is the key point we need to note while solving.