<PRE><B>Question 452313</B>
Soybean meal is 14% protein; cornmeal is 7% protein.
 How many pounds of each should be mixed together in order to get 280lb mixture that is 13% protein?
How many pounds of the cornmeal should be in mixture?
How many pounds of the soybean meal should be in mixture?
:
Let s = amt of soy
Let c = amt of corn
:
Write two equations, one for total amt, another for percent protein
:
s + c = 280
c = (280-s); use this form for substitution
and
.14s + .07c = .13(280)
replace c with (280-s)
.14s + .07(280-s) = 36.4
.14s + 19.6 - .07s = 36.3
.14s - .07s = 36.4 - 19.6
.07s = 16.8
s = {{{16.8/.07}}}
s = 240 lb of soy required
then
280 - 240 = 40 lb of corn required
:
:
See if that checks out:
.14(240) + .07(40) = .13(280)
33.6 + 2.8 = 36.4; confirm our solutions</PRE>