Arithmetic on the Productive System: Accompanied by a Key and Cubical Blocks |
From inside the book
Results 1-5 of 30
Page 22
... received ? MISCELLANEOUS QUESTIONS . VIII . 1. A man bought 12 bushels of wheat for 30 dollars , and sold 9 bushels for 25 dollars . How many bushels had he left , and what did they cost him ? 2. Suppose a stage goes 120 miles in 12 ...
... received ? MISCELLANEOUS QUESTIONS . VIII . 1. A man bought 12 bushels of wheat for 30 dollars , and sold 9 bushels for 25 dollars . How many bushels had he left , and what did they cost him ? 2. Suppose a stage goes 120 miles in 12 ...
Page 25
... receiving 4 dollars a month more than the boy ; what would the wages of each amount to in a year ? 60. A man , woman , and boy were hired a week for 21 dollars ; the woman to receive 5 dollars more than the boy , and the man 5 dollars ...
... receiving 4 dollars a month more than the boy ; what would the wages of each amount to in a year ? 60. A man , woman , and boy were hired a week for 21 dollars ; the woman to receive 5 dollars more than the boy , and the man 5 dollars ...
Page 44
... received ? A. 2,106 dollars . 9. If a man earns 3 dollars a day , how much will he earn in a year , which contains 313 working days A. 939 dollars . 10. If 8,011 men could build a bridge in 9 days , how long would it require 1 man alone ...
... received ? A. 2,106 dollars . 9. If a man earns 3 dollars a day , how much will he earn in a year , which contains 313 working days A. 939 dollars . 10. If 8,011 men could build a bridge in 9 days , how long would it require 1 man alone ...
Page 51
... receiving at the rate of 3 dollars per bushel . How much wheat did he sell ? A. 9010 bushels . A. 903 days . 22. How long would it take 3 men , working at the same rate , to do what 1 man is 2,709 days about ? 23. How many times is 6 ...
... receiving at the rate of 3 dollars per bushel . How much wheat did he sell ? A. 9010 bushels . A. 903 days . 22. How long would it take 3 men , working at the same rate , to do what 1 man is 2,709 days about ? 23. How many times is 6 ...
Page 68
... receiving each successive year 50 dollars advance ; what sum did he receive each year ? NOTE . - The 2nd year he received 50 more than for the 1st .; the 3rd year 100 more ; the 4th year 150 more ; then 50 × 100 + 150 = 300 , the excess ...
... receiving each successive year 50 dollars advance ; what sum did he receive each year ? NOTE . - The 2nd year he received 50 more than for the 1st .; the 3rd year 100 more ; the 4th year 150 more ; then 50 × 100 + 150 = 300 , the excess ...
Other editions - View all
Common terms and phrases
12 cents 50 cents acres amount angles annexing apiece barrels of flour bought breadth bushels called cent pieces ciphers circumference common divisor composite number compound interest compound number contain cows cube root cubic decimal denominator diameter difference discount Divide dividend Division dollars equal example excess factors farthings Federal money feet long Find the sum fraction frustrum gallons given number greater greatest common divisor Hence hogshead horses hundred improper fraction inches indorsed least common multiple length merchant miles million mills minuend mixed number months multiplicand Multiply ounces payment pence pints pounds present worth proportion purchase quadrillion quantity quarts quotient rate per cent ratio reckoning Reduce remainder Repeat the Table sell shillings sold solid feet square rods square root subtract subtrahend Suppose tens thousand TROY WEIGHT units vulgar fraction whole number yards of cloth
Popular passages
Page 242 - Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
Page 115 - Find a common measure, by dividing the greater term by the less, and this divisor by the remainder, and so on, always dividing the last divisor by the last remainder, till nothing remains; the last divisor is the common measure.* 2. Divide both of the terms of the fraction by the common measure, aud the quotients will make the fraction required. * To find the greatest common measure...
Page 116 - Mnltiple of two or more numbers is the least number that can be divided by each of them without a remainder ; thus 30 is the least common multiple of 10 and 15.
Page 35 - This is no denial of the mathematical proposition that the whole is equal to the sum of all its parts...
Page 207 - The first term of a ratio is called the ANTECEDENT, and the second, the CONSEQUENT, and both together form a COUPLET, as 12 : 3.
Page 121 - Multiply the numerators together for a new numerator, and the denominators together for a new denominator.
Page 254 - Multiply the last term by the ratio, from the product subtract the first term, and divide the remainder by the ratio, less 1; the quotient will be the sum of the series required.
Page 215 - Lastly, multiply the third term by the continued product of the second terms, and divide the result by the continued product of the first terms, and the quotient will be the fourth term, or answer required.
Page 186 - Compute the interest to the time of the first payment ; if that be one year or more from the time the interest commenced, add it to the principal, and deduct the payment from the sum total. If there be...
Page 277 - ... above the upper deck ; the breadth thereof at the broadest part above the main wales, half of which breadth shall be accounted the depth of such vessel, and...