Rule.-I. Write the numbers to be added so that terms of the same order stand in the same column, and draw a line beneath. II. Begin at the units, add the terms of each column separately and write the sum underneath, if less than ten. III. When the sum of any column is ten or more than ten, write the units figure only, and add the tens to the next column. IV. Write the entire sum of the last column. Proof.—Begin at the top and add the columns downward, and if the work is correct the two sums will be equal. SECOND METHOD. Separate the number into two or more parts, add these parts, and then add the sums of these parts ; if the work is correct the two results will be equal. NOTES.–1. We write the figures of the same order in the same column for convenience of adding, since only units of the same order can be directly added. 2. We begin at the right to add for convenience, so that when the sum of any column exceeds 9, we may add the left-hand term of such sum to the next column. 3. Beginning at the bottom of a column to add is mostly a matter of custom ; cases may arise, however, in which it would be more convenient. 4. The proof by excess of 9's will be given hereafter. 2. What is the sum of 35246+234+7891+50673+75214 +82349? Ans. 251607. 3. What is the sum of 5462+37185+989+64732+ 34785 +6495? Ans. 149648. 4. What is the sum of 247923+568172+4136+21975+ 729186+235697 ? Ans. 1807089. 5. What is the sum of 3724679+42531+297346+4965287 +914535+6724913 ? Ans. 16669291. 6. What is the sum of 4273561+391845+72233+99+ 25673981+ 7253648 ? Ans. 37665367. Find the sum of the numbers, 7. From 223 to 232 inclusive. Ans. 2275. 8. From 3459 to 3475 inclusive. Ans. 58939. 9. From 4375 to 4400 inclusive. Ans. 114075. 10. From 78437 to 78450 inclusive. Ans. 1098209. 11. From 87692 to 87700 inclusive. Ans. 789264. PRACTICAL PROBLEMS. 1. A publisher, issuing a new work, paid $650 for the plates, $250 for the paper, $95.75 for the press-work, and $275.50 for binding; what did the first edition cost ? OPERATION. SOLUTION.—If for the plates he paid $650, for the $650.00 paper $250, for the press-work $95.75, and for the 250.00 binding, $275.50, for all he paid the sum of $650, 95.75 $250, $95.75, and $275.50, which, by addition, we find 275.50 to be $1271.25; hence the first edition cost $1271.25. $1271.25 2. A merchant laying in his spring goods, expended for calico, $765.87}; for percales, $1075.37}; for French chintz, $564.75; for Victoria lawn, $375.16; and for book muslin, $256.56; wbat was his bill ? Ans. $3037.72. 3. My grocer sends me the following bill: 25 lb. of sugar, $3.50; 7 lb. of tea, $8.75; 5 barrels of apples, $18.75; 12 lb. of coffee, $3.50; 2 barrels of flour, $19.50; 100 lb. of buckwheat flour, $3.25, and 25 lb. of corn meal, $0.75 ; what was the amount ? Ans. $58. 4. A bought a house and lot for $3000, paid $565 for building a barn, $156.35 for putting the grounds in order, $105.67 for introducing gas and water, and $75 for a new range ; for what must be sell this property to make $450.75 on his investment? Ans. $4352.77. 5. A bill of goods contains the following items: 1 piece Irish linen, $9.92 ; 3 pieces Russia crash, $4.81 ; 3 pieces plaid jaconets, $22.50; 1 piece linen drill, $13.68 ; 2 pieces Marseilles vesting, $36.30; 2 pieces linen duck, $124.37; 2 pieces brown corduroy, $46.41; required the amount of the bill. Ans. $257.99. 6. Find the sum of 8 trillion 3 billion 1 million 495 thousand, and 6 quadrillion 74 trillion 15 million 4 hundred written by the French method, and also by the English method. Ans. 6000082006085003032990800. 7. Find the sum of 950053, 420000, five hundred and one thousand one hundred, MDCCCLXXVI, MDCXCVIII, DCCCCXLIX, and DCCCLI, and express it in the Roman method. Ans. MDCCCLXXVIDXXVII. 8. Messrs. Watson & Co., rendered the following bill : 1 dozen Jones's L. H. Shovel, $4.75; } dozen auger bits, $6.30 ; dozen Hald socket chisels, $2.75; 32 lbs. axe stone, $0.42; ì rm. sand paper, ass’d, $2; 2 quires emery paper, $3.12; Ž gross table spoons, $3.75; 1 keg nails, $15.75; 1 keg horseshoes, $6.75; 3 pairs brass candlesticks, $1.35; required the amount of the bill. Ans. $46.94. 9. A dry goods merchant bought silk for $240, linen for $375, and woolen goods for $450; the silk was sold at a profit of $85, the cloth at a profit of $75, and the woolen goods at a profit of $150; what was the whole amount received for the goods ? Ans. $1375. 10. A gentleman leaves to each of his three sons $3500, and to his two daughters $800 apiece more than to a son, and to his wife $400 more than to a son and daughter, and the remainder of his estate, which was $500 more than he bequeathed to his family, he left to an orphan asylum; what was the amount of his estate ? Ans. $55,100. 11. Add the following ledger columns: Cash. CR DR. 1874 ) 1874 July 3 To Sundries, 725 00 July 5 By Merchandise, 19" Merchandise, 11 " Bills Payable, " Bills Recei " " Interest, ". Interest, Louis Walton, “ James Nelson, 66 Merchandise, 13 - Merchandise, 22 66 Lehigh Valley R. R. Stock, 2806 "1" Commission, 22 " Rogers & Co., 1815 " Adams & Co., 30 “ Merchandise, | 875 40 Aug./ 11 “ Am't car'd for'a, “Am't car'd for’d, CONTRACTIONS IN ADDITION. 89. Contractions in Addition are abbreviated methods of adding: CASE I. 90. To add by omituing the names of the numbers added, merely naming results. 1. Find the sum of 367, 589, 635, and 768. OPERATION. SOLUTION.—We write the numbers so that terms of 367 the same order stand in the same column, and begin 589 at the right to add: 8, 13, 22, 29; we write the 9 and 635 add the 2 to the next column: 2, 8, 11, 19, 25; we 768 write the 5, and add the 2, etc. 2359 Find the sum of the numbers Ans. 15029. Ans. 10504. Ans. 140114. Ans. 157335. Ans. 138649. CASE II. 91. To add two or more columns at the same time. 1. Find the sum of 3486, 5267, 6845 and 7654. OPERATION. SOLUTION.—54 and 40 are 94 and 5 are 99 and 60 3486 are 159 and 7 are 166 and 80 are 246 and 6 are 252; 5267 we write the 52 and add the 2 to the next column; 76 6845 and 2 are 78 and 60 are 138 and 8 are 146 and 50 are 7654 196 and 2 are 198 and 30 are 228 and 4 are 232, which we write. 23252 NOTE.-In practice name only the results, omitting the naming of the numbers added. Find the sum of the numbers 2. From 496 to 512 inclusive. Ans. 8568. 3. From 832 to 848 inclusive. Ans. 14280. 4. From 5626 to 5640 inclusive. Ans. 84495. 5. From 6987 to 7000 inclusive. Ans. 97909. NOTES.–1. When two or more terms of a column can be easily grouped together, use their sum instead of adding each separately ; combining with especial reference to tens. 2. When a term is repeated several times in a column, multiply it by the number of times it is repeated, and use the result. SUBTRACTION. 92. Subtraction is the process of finding the difference between two numbers. 93. The Difference between two numbers is a number which, added to the less, equals the greater. 94. The Subtrahend is the number to be subtracted. 95. The Minuend is the number from which we subtract. 96. The Sign of Subtraction is –, and is read minus. It denotes that the number immediately following it is to be subtracted from the number preceding it. NOTES.-1. The Sign of Subtraction is a short line in the line of writing. 2. The symbol — was introduced by Stifelius, a German mathematician, in a work published in 1544. PRINCIPLES. 1. Similar numbers only can be subtracted. 2. Units of the same order only can be directly subtracted. 3. The difference is a number similar to the minuend and subtrahend. 4. If the minuend and subtrahend be equally increased or diminished, the difference will remain the same. PROBLEM. 97. To find the difference between two numbers. 98. There are Two Methods of explaining subtraction, called the Method by Borrowing, and the Method by Adding Ten. NOTE.-The taking one from the term of the minuend is called borrowing, and the adding one to the next term of the subtrahend is called carrying. 1. Subtract 365 from 647. SOLUTION BY BORROWING.–We write the subtra- OPERATION. hend under the minuend and begin at the right to 647 subtract. 5 units from 7 units leave 2 units, which 365 we write under the units; we cannot subtract 6 tens from 4 tens, we will therefore take 1 hundred from 282 the 6 hundreds and add it to the 4 tens; 1 hundred equals 10 tens, which added to 4 tens, equal 14 tens; 6 tens from 14 tens leave 8 tens, which we write in tens place: 3 hundreds from 5 hundreds (the number of hundreds remaining after taking away 1 hundred) leave 2 hundreds, which we write in the hundreds place. |