soybean meal is 18% protein; cornmeal is 9% protein. How many pounds of each should be mixed together in order to get 360-lb mixture tha is 13% protein?
How many LBS of the cornmeal should be in the mixture and how many lbs of the soybean meal should be in the mixture?
let the amount of cornmeal be x lbs
then the amount of soybeans is 360-x lbs
.09x + .18(360-x) = .13(360)
multiply by 100
9x + 18(360-x) = 13(360)
I am sure you can finish it from there.
To solve this problem, we can set up a system of equations. Let's say we need x pounds of soybean meal and y pounds of cornmeal in the mixture.
We are given the following information:
- The soybean meal is 18% protein.
- The cornmeal is 9% protein.
- The total mixture should weigh 360 pounds.
- The mixture should be 13% protein.
From the given information, we can set up the following equations:
Equation 1: x + y = 360 (total weight equation)
Equation 2: (0.18x + 0.09y) / 360 = 0.13 (protein content equation)
Let's solve this system of equations to find the values of x and y.
Step 1: Simplify equation 2 by eliminating the denominator:
0.18x + 0.09y = 360 * 0.13
0.18x + 0.09y = 46.8
Step 2: Multiply equation 1 by 0.09 to eliminate y:
0.09x + 0.09y = 32.4
Step 3: Subtract equation 2 from equation 1:
0.09x + 0.09y - (0.18x + 0.09y) = 32.4 - 46.8
-0.09x = -14.4
Step 4: Divide both sides of the equation by -0.09:
x = 14.4 / 0.09
x = 160
Now we have the value of x, which represents the pounds of soybean meal in the mixture. To find y, the pounds of cornmeal in the mixture, we can substitute the value of x into equation 1:
160 + y = 360
y = 360 - 160
y = 200
Therefore, 160 pounds of soybean meal and 200 pounds of cornmeal should be mixed together to obtain a 360-pound mixture that is 13% protein.