154^ The Cornell Reading-Courses 



Problem 20: 



A farmer producing 225 pounds of milk daily, testing 3.5 per cent fat, 

 lives an equal distance from two cheese factories. Cheese-maker A is 

 very careless in his methods and the average loss of fat in his whey is .37 

 per cent; while cheese-maker B takes more pains and the average loss of 

 fat in his whey is .29 per cent. When cheese is selling at 13 cents per 

 pound, now much more will the farmer receive in 30 days if he delivers 

 his milk to B rather than to A? 



225 lbs. milk X 30 = 6,750 lbs. of milk delivered 



•37% — -29% = -08% more fat retained in cheese made by B than by A 



6,750 (lbs. of milk) X .85 == 5,737.5, approxim.ate weight in pounds of 

 whey 



5.737-$ X .0008 = 4.59, nimiber of pounds more fat retained in cheese 

 by B than by A 



If each pound of fat yields 2.68 lbs. of cheese, 



4.59 X 2.68 = 12.3, number of pounds more cheese made by B than 

 by A 



$.13 X 12.3 = $1.60. Answer. 



Problem 21: 



Two farm.ers each deliver daily 385 pounds of milk to the same cheese 

 factory. A's milk tests 3.5 per cent fat and B's tests 4.5 per cent fat. 

 If they are paid for their milk on the basis of the yield of cheese, how 

 much more will B receive than A in one month, cheese selling for 13 cents 

 per pound? 



385 lbs. milk X 30 = 11,550 lbs. milk delivered by each in 30 days 

 11,550 X .035 = 404.25, number of pounds fat delivered by A 

 Each pound of fat in 3.5% milk yields 2.68 lbs. of cheese (see table, 



P- 159)- 



404.25 X 2.68 = 1,083.39, number of pounds of cheese from A's milk 

 $.13 X 1,083.39 = $140.84, amount that A should receive for his milk 

 11,550 X .045 = 519.75, number of pounds fat delivered by B 

 Each pound of fat in 4.5% milk yields 2.61 lbs. of cheese 

 519.75 X 2.61 = 1,356.55, number of pounds of cheese from B's milk 

 $.13 X 1,356.55 = $176.35, amount that B should receive for his milk 

 $176.35 — $140.84 = $35-51, amount that B should receive more than 



A for his milk. Answer. 



COMPUTING OVERRUN IN BUTTER 



Overrun is the gain in butter over the fat, or it is the sum of 

 the moisture, salt, and casein of the butter minus the losses in 

 manufacture. 



