1538 The Cornell Reading-Courses 



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 (lbs. fat) -7- 140 (lbs. milk) = .0367 

 .0367 X 100 = 3.67%, fattest. Answer. 



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



4.2% + 3.0% + 4-5% + 5-o% + 3-5% = 20.2%, sum of tests 

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



Note: Had all the cows given the same amount of milk, the latter method would give 

 the correct answer. 



COMPUTING THE PERCENTAGE OF FAT IN A VAT OF CREAM AFTER STARTER 



IS ADDED 



If there is added to a vat of cream some skimmed-milk starter, the 

 number of pounds of fat in the cream is not changed but the percentage 

 of fat is decreased. 



Problem ij: 



A vat contained 300 pounds of 35-per cent cream. To this amount 

 of cream was added 2 5 per cent of skimmed-milk starter. What percentage 

 of fat was there in the mixture? 



Note: There is not enough fat in skimmed milk to be considered. 



The 300 lbs. of cream was increased by 25% of its own weight. 



300 X -25 = 75, number of pounds of starter added 



300 -f- 75 = 375, number of pounds of cream and starter 



Another way to find the number of pounds of the mixture is to regard 

 the weight of the cream as 100%. Since 25% was added, the weight of 

 the mixture was 125% of what it was before adding the starter, 

 300 X 1.25 = 375, number of pounds of cream and starter. 



In the original amount of cream there was 105 pounds of fat 

 (300 X .35 = 105). There was the same number of pounds of fat after 

 the starter was added. We have given, therefore, the number of pounds 

 of the mixture and the number of pounds of fat in the mixture, to find 

 the percentage of fat. This is done by dividing the number of pounds of 

 fat by the total weight of the mixture. 



105 -^ 375 = -28 

 .28 X 100 = 28%, fat in mixture. Answer. 



