ARITHMETIC OF MILK AND MILK PRODUCTS I99 



In this case we add nothing, so that C equals o and 

 B — C=5 — 0=5. The results mean that for 5 pounds 

 of the milk, we should remove i.i pounds of skim- 

 milk, thus reducing 5 pounds of milk containing 3.9 

 per cent, of fat to 3.9 pounds of modified milk contain- 

 ing 5 per cent, of fat. 



Applying these results to a specific case, how much 

 skim-milk should be removed from 980 pounds of 

 3.9 per cent, milk to increase the fat to 5 per cent? 

 Divide 980 by 5 (B — C), which gives 196, and mul- 

 tiply this by I.I (B — A) which gives 215.6 pounds of 

 milk-serum or skim-milk to be removed, leaving 764.4 

 pounds of modified 5 per cent. milk. 



(3) JVhen the per cent, of fat is to be decreased by 

 adding skim-milk. Rule — From the per cent, of fat 

 in the milk to be modified subtract the per cent, of fat 

 desired in the modified milk, and the result is the num- 

 ber of pounds of skim-milk to be used. From the per 

 cent, of fat desired in the modified milk, subtract the 

 per cent, of fat in the skim-milk, and the result is the 

 number of pounds to use of the milk to be modified. 

 Example: How much skim-milk containing .1 per 

 cent of fat should be added to milk containing 5 per 

 cent, of fat to reduce the fat to 3.9 per cent.? 



^Iilk B— C=3.8 (pounds of 5 per cent. milk). 



B=3-9 



Skim-niilk 



'a— B=i.i (pounds of skim-milk). 



(a) How much skim-milk should be added to i,ooo 

 pounds of 5 per cent, milk to produce 3.9 per cent, 

 milk? Divide 1,000 by 3.8, giving 263, and multiply 

 the result by i.i, which gives 289, the number of 



