Question 366493
Let  {{{c}}} = pounds of cornmeal needed
Let {{{s}}}= pounds of soybean meal needed
given:
Pounds of protein in {{{c}}} pounds of cornmeal:
{{{.07c}}}
Pounds of protein in {{{s}}} pounds of soybean meal
{{{.14s}}}
(1) {{{c + s = 280}}}
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In words:
(pounds of protein in final mixture)/(total pounds of final mixture) = 11%
{{{(.07c + .14s)/280 = .11}}}
{{{.07c + .14s = 30.8}}}
{{{7c + 14s = 3080}}}
divide both sides by {{{7}}}
(2) {{{c + 2s = 440}}}
Subtract (1) from ()
(2) {{{c + 2s = 440}}}
(1) {{{-c - s = -280}}}
{{{s = 160}}}
and, since
{{{c + s = 280}}}
{{{c = 120}}}
120 pounds of cornmeal and 160 pounds of soybean meal are needed
check answer:
{{{(.07c + .14s)/280 = .11}}}
{{{(.07*120 + .14*160)/280 = .11}}}
{{{(8.4 + 22.4)/280 = .11}}}
{{{8.4 + 22.4 = 30.8}}}
{{{30.8 = 30.8}}}
OK