A DAIRY LABORATORY GUIDE 3! 



The following example will illustrate how a 

 knowledge of the per cent of S. N. F. of a sample 

 of milk is utilized in figuring adulteration per cents : 



Before adulteration a sample of milk had 4.2 per 

 cent of fat and 9.0 per cent of S. N. F. After adul- 

 teration the sample contained 3.1 per cent fat and 

 7.5 per cent of S. N. F. The following questions 

 can be answered from calculations with the above 

 data: 



1. What per cent of water was added to the 



milk? 



2. At what rate per cent was the water added? 



3. What per cent of the fat was adulterated 



by skimming? 



4. What per cent of the fat was adulterated 



by watering? 



(1) Since there were 9 parts of S. N. F. before 

 adulteration and 7.5 parts of S. N. F. after adultera- 

 tion, 1.5 must have been removed by adulteration. 



97-5=1.5 



1.5-^-9=16.66% of S. N. F. removed by adding 



the water. 



Since 16.66 per cent of S. N. F. was removed, 

 water is the only thing that could have taken its 

 place; therefore 16.66 per cent of water was added 

 to the milk. 



(2) For every 7.5 parts of S. N. F. in the milk 

 1.5 parts were removed, or in other words water 

 was added at the rate of 1.5 parts of water to every 

 7.5 of S. N. F. 



The rate per cent at which water was added 

 would be i. $-7-7.5=20%. 



Before adulteration the sample had 4.2 parts fat 

 and after adulteration 3.1 parts fat, i.i parts of fat 

 were removed, which equals 26.19 per cent. 

 1.1-7-4.2=26.19%. 



