SOLUTION: Feeding cattle: A farmer wants to mix 2,400 pounds of cattle feed that is to be 14% protein. Barley (11.7% protein) will make up 25% of the mixture. The remaining 75% will be ma
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Question 154648This question is from textbook College Algebra
: Feeding cattle: A farmer wants to mix 2,400 pounds of cattle feed that is to be 14% protein. Barley (11.7% protein) will make up 25% of the mixture. The remaining 75% will be made up of oats (11.8% protein) and soybean meal (44.5% protein). How many pounds of each will he use? This question is from textbook College Algebra
You can put this solution on YOUR website! Amount of barley (11.7% protein) used =2400*0.25=600 pounds
2400-600=1800 pounds the amount remaining for oats and soy meal
Let x=amount of oats (11.8% protein) used
Then 1800-x=amount of soy meal (44.5% protein)used
Now we know that the amount of pure protein in the barley (600*0.117) plus the amount of pure protein in the oats(0.118x) plus the amount of pure protein in the soy meal (0.445*(1800-x)) has to equal the amount of pure protein in the final mixture (2400*0.14). So our equation to solve is:
600*0.117+0.118x+0.445*(1800-x)=2400*0.14 get rid of parens and simplify
70.2+0.118x+801-0.445x=336 subtract 70.2 and also 801 from each side
70.2-70.2+0.118x+801-801-0.445x=336-801-70.2 collect like terms
-0.327x=-535.2 divide each side by -0.327
x=1636.7lb--------amount of oats
1800-x=1800-1637.7=163.3 lb---------------amount of soy meal
600 lb is the amount of barley
CK
600*0.117+1636.71*0.118+163.3*0.445=2400*0.14
70.2+193.13+72.67=336
336=336
