Asked by jason
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 14%
How many pounds of cornmeal should be in the mixture? 200
how many lbs of soybean meal should be in the mixture? 160
can someone tell me if I have this right?
How many pounds of cornmeal should be in the mixture? 200
how many lbs of soybean meal should be in the mixture? 160
can someone tell me if I have this right?
Answers
Answered by
Marth
let x = pounds of soybean meal
let y = pounds of cornmeal
There should be a total of 360 pounds. Therefore, x+y = 360
The protein % is .14. Out of 360 pounds, that would be 360*.14 = 50.4
Therefore, .18x + .09y = 50.4
substitution:
x=360-y
.18(360-y) + .09y = 50.4
64.8 - .18y + .09y = 50.4
-.09y = -14.4
y = 160
x+y = 360
x+160 = 360
x = 200
200 pounds of soybean meal
160 pounds of cornmeal
Checking my work:
.18x + .09y = 50.4
.18*200 + .09*160 = 50.4
50.4 = 50.4
let y = pounds of cornmeal
There should be a total of 360 pounds. Therefore, x+y = 360
The protein % is .14. Out of 360 pounds, that would be 360*.14 = 50.4
Therefore, .18x + .09y = 50.4
substitution:
x=360-y
.18(360-y) + .09y = 50.4
64.8 - .18y + .09y = 50.4
-.09y = -14.4
y = 160
x+y = 360
x+160 = 360
x = 200
200 pounds of soybean meal
160 pounds of cornmeal
Checking my work:
.18x + .09y = 50.4
.18*200 + .09*160 = 50.4
50.4 = 50.4
There are no AI answers yet. The ability to request AI answers is coming soon!