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 that is 11% protein?
How many pounds of cornmeal should be in the mixture?
How many pounds of the soybean meal should be in the mixture?
let the amount of soybean be x lb
let the amount of cornmeal be 360-x lbs
so .18x + .09(360-x) = .11(360)
I would multiply the equation by 100, then its easy
To solve this problem, we can use the method of mixture problems. Let's break down the information given, and then use a system of equations to find the solution.
Let's assume x represents the number of pounds of soybean meal and y represents the number of pounds of cornmeal in the mixture.
We know that the final mixture will weigh 360 pounds, so the first equation is:
x + y = 360
According to the given information, soybean meal is 18% protein and cornmeal is 9% protein. The final mixture should be 11% protein. We can express this information using the second equation:
(0.18x + 0.09y)/(x + y) = 0.11
To solve this system of equations, we can use substitution or elimination. Let's use substitution to solve for one variable in terms of the other.
From the first equation, we have y = 360 - x. Substituting this expression into the second equation, we get:
(0.18x + 0.09(360 - x))/360 = 0.11
Simplifying this equation gives us:
0.18x + 32.4 - 0.09x = 0.11 * 360
Combining like terms:
0.09x + 32.4 = 39.6
Subtracting 32.4 from both sides gives us:
0.09x = 7.2
Dividing both sides by 0.09 solves for x:
x = 7.2 / 0.09
x = 80
Now that we have the value of x, we can find y:
y = 360 - x
y = 360 - 80
y = 280
Therefore, 80 pounds of soybean meal and 280 pounds of cornmeal should be mixed together to get a 360-pound mixture that is 11% protein.
To answer the specific questions:
1. The mixture should contain 280 pounds of cornmeal.
2. The mixture should contain 80 pounds of soybean meal.