Solving Applied Problems with Two Equations

Question

Soybean meal is 18% protein; cornmeal is 9% protein. How many pounds of each should be mixed in order to get 360-lb mixture that is 10% protein?

How many lbs of cornmeal should be in the mixture?

How many lbs of soybean meal should be in the mixture?

Answer 1

S = Soybean (lbs)

C = Corn (lbs)

 

multiply each amount of soybeans and corn times the ratio of protein

that will then equal the total amount times the desired ratio of protein (=0.10)

 

1)  S*0.18 + C*0.09 = 360*0.10                (18% = .18,  9% = 0.09,   10% = 0.10 )

 

2)  S + C  = 360    (total lbs)

 

we'll use substitution from eq #2   C = 360 - S    

in eq #1

 

           S*.18 + (360-S)*.09 = 360*0.10 = 36

 

           S*.18  +  360*.09  - S*.09 = 36      

           S*(.18-.09) + 32.4            = 36

 

           0.09*S = 36 - 32.4 = 3.6

 

           S = 3.6/.09 = 40     (Soybeans = 40 lbs,  so Corn = 320 lbs)

 

as a check, use eqn #1.

 

             S*.18 +    C*.09    = 360*0.10

           40*.18  +  320*.09  =  36   ???

              7.2    +   28.8       =  36          

 

                                 36 = 36  (Checks!)