242 SOILS AND FERTILIZERS 



value of the materials used. Suppose two fertilizers 

 are selling at $28 and $35, respectively, each having a 

 different composition, the estimated value of each could 

 be obtained in the following way : 



COMPOSITION OF FERTILIZERS 



No. i No. 2 



SELLING PRICE $28 SELLING PRICE $35 

 PER CENT PER CENT 



Nitrogen as nitrates 1.50 2.10 



Phosphoric acid, available . . .'8.00 10.00 



Phosphoric acid, insoluble . . .2.00 0.50 



Potash (water-soluble) .... 2.00 3.50 



POUNDS PER TON 



No. i No. 2 



Nitrogen . . . 1.50 x 20 = 30 2.10 x 20 = 42 

 Phosphoric acid . 8.00 x 20 = 160 10.00 x 20 = 200 

 Potash .... 2.00 x 20 = 40 3.50 x 20 = 70 



ESTIMATED VALUE 



No. i No. 2 



Nitrogen . . . . 30 x 0.16 = $ 4.80 42 x 0.16 = $ 6.72 

 Phosphoric acid . . 160x0.07= 11.20 200x0.07= 14.00 

 Potash 40 x 0.05 = 2.00 70 x 0.05 = 3.50 



$18.00 $24.22 



Difference between estimated value and selling price: No. i, 

 $10.00; No. 2, $10.78. 



The trade value of a commercial fertilizer often varies 

 widely from the actual or crop-producing value, for in 

 assigning a trade value simply the cost of the ingredi- 

 ents is considered, and this is not necessarily identical 



