PERCENTAGE 



4582 



PERCENTAGE 



(5) 3 pecks is what per cent of 6 bushels? 



(6) A baseball team during one season won 

 83 games and lost 57 games. What was the 

 per cent of games won? Carry to one decimal 

 place. 



Number games played = 83 -f 57 = 140 



OO AAA 



Part wor 



(7) $6 is what per cent of $12? 



(8) 1 pint is what per cent of 1 quart? 



(9) 3 pecks is what per cent of 1 bushel? 



(10) 3 inches is what per cent of 1 foot? 



(11) 2 ounces is what per cent of 1 pound? 



(12) 9 inches is what per cent of 1 yard? 



(13) 1 foot 9 inches is what per cent of 14 

 feet? 



(14) 6% miles is what per cent of 20 miles? 



(15) A farmer raised corn last year on 68 

 acres. This year he adds 17 acres to his corn 

 field. What per cent has he increased it? 



(16) The drainage tax of some farm land 

 along the Illinois River last year was $45 per 

 acre. This year it is $60 per acre. What is 

 the per cent of increase? 



First tax = $4 5 



Increases $15 



Rate of increase = i% 5 = i = 



Finding a Number When a Per Cent of It Is 

 Given. (1) 25 % of Mr. Howard's farm is 78 

 acres. How many acres in the whole farm? 



.25 of farm = 78 A. 



Number A. in farm = 4 X 78 = 312 



(2) 60% of Mr. Howard's crop of wheat last 

 year was 7488 bushels. How many bushels in 

 his entire crop? 



.60 of cropr=7488 bu. 



R, 2496 

 Bu. in crop = l*x of JJ**= 12480 



% of crop=7488 bu. 



Bu. in crop = % of 7488 = 12480 



.60Xcrop = 

 Bu. in 



or 

 8 bu. 



Note in the last solution that 7488 is the 

 product of the number of bushels in the whole 

 crop and .60. 



When the product of two numbers and one 

 of the numbers are known, the other number 

 is found by dividing the product by the num- 

 ber that is known. This principle runs through 

 percentage in every division of the subject, 

 interest, commission, profit and loss, and so 

 forth. The following problems illustrate this: 



(1) What sum of money on interest at 

 for 1 year gives $337.50 interest? 



.05Xprincipal = $337.50 

 Principal = 



(2) A commission agent working at 

 earned $1687.50 selling machines. What was 

 the amount of his sales? 



.1 3!/j xsales = $1687. 50 



(3) A baseball team lost 32 games in one 

 season and their per cent loss was 45 r j7. How- 

 many games did they play? 



. 45^-7 X number = 32 



Number = 



10 



.45% 



= 70 



A kind of problem arises in percentage which 

 seems to differ from the problems just dealt 

 with, but which is really the same kind of 

 problem ; for example : 



(1) Mr. Jackman sold flour at $11.60 per 

 barrel, which was a gain of 16% to him. What 

 did he pay for it? 



Cost + .ie of cost=$li.6o 

 1.16xcost=$11.60 



In this problem $11.60 is 116% of the cost, and 

 is therefore the product of 1.16 and the cost, 

 and the cost is found by dividing the product, 

 or $11.60, by 1.16, the known factor. 



(2) An athlete's weight is 156 pounds, which 

 shows a loss of 10% from his former weight. 

 What was his former weight? 

 . 90 X former wt. = 156 Ib. 

 Former wt. in Ib. = 15 



(3) I sold 2 cows for $132 each. On one I 

 gained 20% and on the other I lost 20%. Did 

 I gain or lose on the sale? How much? 



Cost -f .20 cost=$i32 

 1.20xcost = $132 



Cost of first cow= 1 1 3 ^ 00 =$110 



Cost .20 cost = $132 

 .80Xcost = $132 



$132 ' Q 



Cost of second cow = 



.oU 



Net loss = $275 $264 = $11 

 This may be solved in common fractions: 



% of cost of first cow=$132 

 22 

 Cost of first cow = % of 



56 of cost of second cow=$132 

 33 

 Cost of second cow = % of $132 = $165 



Net loss = $275-$264 = $ll 



