FARM DAIRYING 447 



large as in the above proportion'. The 28 % cream and 3 % milk must 

 be kept in the proportion of 1 : 24, and since the amount of 28% 

 cream is to be increased 150 times, the 3% milk must also be in- 

 creased 150 times. This would give 150 lb. of 28% cream (1 X 150) 

 and 3600 lb. of 3% milk (150 X 24 = 3600), making in all 3750 lb. 

 (150 + 3600 = 3750) of a 4% mixture. 

 This problem may also be Worked by simple proportion : 

 24 : 1 : : x : 150 

 x = 3600, the number of pounds of 3% milk required. 



Proof 



The 3750 lb. of 4% milk will contain 150 lb. of fat (3750 X .04 = 150). 

 If the 150 lb. of 28% cream and 3600 lb. of 3% milk furnish 150 lb. 

 of fat, the problem is correct. 



3600 X .03 = 108, number of pounds of fat in milk 



150 X .28 = 42, number of pounds of fat in cream 



108 + 42 = 150, number of pounds of fat in mixture. Answer. 



Computing the average percentage of fat in the milk of a herd. 



Problem 11 



Compute the average test of fat of this herd : 

 Brownie 20 lb. milk testing 4.2 per cent fat 

 Spot 50 lb. milk testing 3 per cent fat 



Red 20 lb. milk testing 4.5 per cent fat 



Nancy 10 lb. milk testing 5 per cent fat 

 Lucy 40 lb. milk testing 3.5 per cent fat 



This problem is solved as follows : 



Brownie 20 X .042 = .84, number of pounds fat 

 Spot 50 X .03 = 1.50, number of pounds fat 



Red 20 X .045 = .90, number of pounds fat 



Nancy 10 X .05 = .50, number of pounds fat 

 Lucy 40 X .035 = 1.40, number of pounds fat 

 140 5\14 



, 5.14 (lb. fat) -r- 140 (lb. milk) = .0367 

 .0367 X 100 = 3.67%, fat test. Answer. 



The common incorrect method of solving this problem is as follows : 

 4.2% + 3.0% + 4.5% + 5.0% + 3.5% = 20.2%, sum of tests 

 20.2% 4- 5 (number of tests) = 4.04%, incorrect average. 



Had all the cows given the same amount of milk, the latter method 

 would have given the correct answer. 



