FARM DAIRYING 445 



Computing percentage of fat, pounds of product, or pounds of fat, 

 having any two of these quantities given. 



Problem 5 



How many pounds of milk testing 4.5 per cent fat would be required 

 to furnish 157.5 lb. of fat? 



? lb. milk X .045 = 157.5 



Therefore,^ 157.5 -H .045 = 3500, number of pounds of 4.5% milk. 

 Answer. 



Problem 6 



250 lb. of cream contained 75 lb. of fat. What did the cream test? 



250 X ? = 75 lb. fat 



Therefore, 75 -r 250 = .30 



.30 X 100 = 30%, fat test. Answer. 



Problem 7 



How many pounds of 22 per cent cream can be obtained from 3500 

 lb. of 4 per cent milk? 



In 3500 lb. of 4% milk there are 140 lb. of fat (3500 X .04 = 140). 



This amount of fat will be contained in the 22 % cream. The prob- 

 lem is to find the number of pounds of cream testing 22%. 



? lb. cream X .22 = 140 



Therefore, 140 -5- .22 = 636.36, number of pounds of cream. 

 Answer. 



Standardizing milk and cream. 



Standardizing milk or cream consists in raising or lowering the fat- 

 content to a fixed standard. This is done by adding to the material to 

 be standardized, milk or cream of a higher or lower percentage of fat. 

 In standardization there are two classes of problems involved : first, 

 one in which a certain fixed amount of milk is to be made up or a certain 

 amount of standardized milk is desired ; and second, one in which a 

 certain amount of milk or cream is to be used and enough of another 

 product added to make the mixture test a certain percentage of fat. 

 In the latter case the amount of the mixture is indefinite. 



The original method of computing problems in standardization is 

 long and difficult, but a scheme has been devised that is comparatively 

 simple. 1 The method is as follows : 



Draw a rectangle and place in the center of it the percentage of fat 



1 By R. A. Pearson, at that time Professor of Dairy Industry, 

 Cornell University. 



