3. If a floor be 79 ft. 8' by 38 ft. 11', how many square yards does it contain? Ans. 344 yd. 4 ft. 4′ 4′′. 4. If a block of marble be 7 ft. 6′ long, 3 ft. 3′ wide, and 1 ft. 10' thick, what are the solid contents? Ans. 44 ft. 8' 3". 5. How many solid feet in 7 sticks of timber, each 56 ft. long, 11 inches wide, and 10 inches thick? Ans. 299 ft. 5' 4". 6. How many feet of boards will it require to inclose a building 60 ft. 6' long, 40 ft. 3′ wide, 22 ft. high, and each side of the roof 24 ft. 2', allowing 523 ft. 3' for the gables, and making no deduction for doors and windows? Ans. 7880 ft. 5'. CONTRACTED METHOD. 391. The method of contracting the multiplication of decimals may be applied to duodecimals, the only modification being in carrying according to the duodecimal, instead of the decimal, scale. 1. Multiply 7 ft. 3' 5" 8" by 2 ft. 4' 7" 9"", rejecting all denominations below seconds in the product. OPERATION. 7 ft. 3' 5" 8" 9" 7" 4' 2 ft 14 ft. 6' 11" 2 ft. 5' 2" 4' 3" 5" ANALYSIS. We write 2 ft., the units of the multiplier, under the lowest order to be reserved in the product, and the other terms at the left, with their order reversed. Then it is obvious that the product of each term by the one above it is seconds. Hence we multiply each term of the multiplier into the terms above and to the left of it in the multiplicand, carrying from the rejected terms, thus; in multiplying by 2 ft., we have 8′′ × 2 ft. 16′′ 1′′ 4'"', which being nearer 1' than 2", gives 1" to be carried to the first contracted product. In multiplying by 4', we have 5' x 4' 20′ = 1′′ 8′′, which being nearer 2′′ than 1", gives 2" to be carried to the second contracted product, and so on. = 17 ft. 4' 9", Ans. = EXAMPLES FOR PRACTICE. 1. Multiply 7 ft. 3' 4" 5"" by 5 ft. 8' 6", extending the product only to primes. Ans. 41 ft. 7'±. 2. How many yards of carpeting will cover a floor 36 ft. 9′ 4′′ long, and 26 ft. 6' 9" wide? 3. How many cu. ft. in a block of marble measuring 6 ft. 2′ 7′′ in length, 3 ft. 3' 4" wide, and 2 ft. 8' 6" thick? 4. Find the product of 7 ft. 6′ 8′′, 3 ft. 2′ 11′′, and 3 ft. 8′ 4′′, correct to within 1'. Ans. 90 ft. 6'±. DIVISION. 392. 1. Divide 41 ft. 8' 7" 6" by 7 ft. 5'. down the next term of the dividend, we have 4 ft. 7′ 7′′ for a new dividend. Reducing the first two terms to primes, we have 55′ 7′′, whence by trial division we obtain 7′ for the second term of the quotient, and 3′ 8′′ for a remainder. Completing the division in like manner, we have 5 ft. 7′ 6′′ for the entire quotient Hence the following RULE. I. Write the divisor on the left hand of the dividend, as in simple numbers. II. Find the first term of the quotient either by dividing the first term of the dividend by the first term of the divisor, or by dividing the first two terms of the dividend by the first two terms of the divisor; multiply the divisor by this term of the quotient, subtract the product from the corresponding terms of the dividend, and to the remainder bring down another term of the dividend. III. Proceed in like manner till there is no remainder, or till a quotient has been obtained sufficiently exact. EXAMPLES FOR PRACTICE. 1. Divide 287 ft. 7' by 17 ft. 2. Divide 29 ft. 5' 4" by 6 ft. 8'. Ans. 16 ft. 11'. Ans. 4 ft. 5'. 3. A floor whose length is 48 ft. 6′ has an area of 1176 ft. 1' 6"; what is its width? Ans. 24 ft. 3'. 4. From a cellar 38 ft. 10' long and 9 ft. 4' deep, were excavated 275 cu. yd. 5 cu. ft. 1' 4" of earth; how wide was the cellar? Ans. 20 ft. 6'. CONTRACTED METHOD. 393. Division of Duodecimals may be abbreviated after the manner of contracted division of decimals. 1. Divide 35 ft. 11' 11" by 4 ft. 3' 7" 3"", and find a quotient correct to seconds. OPERATION. 4 ft. 3′ 7′′ 3′′ ) 35 ft. 11′ 11′′ (8 ft. 4′ 5′′ 34 ft. 4' 10" ANALYSIS. Having obtained by trial, 8 ft. for the first term of the quotient, we multiply three terms of the divisor, 4 ft. 3′ 7′′, carrying from the rejected term, 3′′ × 8 = 24′′′ = 2′, making 34 ft. 4' 10", which subtracted from the dividend leaves 1 ft. 7' 1" for a new dividend. In the next division, we reject 2 terms from the right of the divisor, and at the last division, 3 terms, and obtain for the required quotient, 8 ft. 4′ 5′′. EXAMPLES FOR PRACTICE. 1. Divide 7 ft. 7' 3" by 2 ft. 10' 7", extending the quotient to seconds. Ans. 2 ft. 7' 8"±. 2. Separate 64 ft. 9′ 8′′ into three factors, the first and second of which shall be 7 ft. 2' 4" and 4 ft. 7′ 9′′ 8"" respectively, and obtain the third factor correct to within 1 second. Ans. 1 ft. 11' 3"±. 3. What is the width of a room whose area is 36 ft. 4' 8" and whose length 7 ft. 2' 11"? SHORT METHODS. 394. Under the heads of Contractions in Multiplication and Contractions in Division, are presented only such short methods as are of the most extensive application. The short methods which follow, although limited in their application, are of much value in computations. FOR SUBTRACTION. 395. When the minuend consists of one or more digits of any order higher than the highest order in the subtrahend. The difference between any number and a unit of the next higher order is called an Arithmetical Complement. Thus, 4 is the arithmetical complement of 6, 31 of 69, 2792 of 7208, etc. 1. Subtract 29876 from 400000. OPERATION. 400000 29876 370124 ANALYSIS. To, subtract 29876 from 400000 is the same as to subtract a number one less than 29876, or 29875, from 399999 (Ax. 2). We therefore diminish the 4 of the minuend by 1, and then take each figure of the subtrahend from 9, except the last or right hand digit, which we subtract from 10. Hence the RULE. I. Subtract 1 from the significant part of the minuend and write the remainder, if any, as a part of the result. II. Proceeding to the right, subtract each figure in the subtrahend from 9, except the last significant figure, which subtract from 10. EXAMPLES FOR PRACTICE. 1. Subtract 756 from 1000. 6. Subtract 599948 from 1000000. Ans. 244. Ans. 3991424. 7. What is the arithmetical complement of 271? Of 18365? Of 3401250? FOR MULTIPLICATION. CASE I. 396. When the multiplier is 9, 99, or any number of 9's. Annexing 1 cipher to a number multiplies it by 10, two ciphers by 100, three ciphers by 1000, etc. Since 9 is 10 1, any number may be multiplied by 9 by annexing 1 cipher to it and subtracting the number from the result. For similar reasons, 100 times a number 1 time the number 99 times the number, etc. Hence, RULE. Annex to the multiplicand as many ciphers as the multiplier contains 9's, and subtract the multiplicand from the result. 397. When the multiplier is a number a few units less than the next higher unit. Were we required to multiply by 97, which is 100 - 3, we could evidently annex 2 ciphers to the multiplicand, and subtract 3 times the multiplicand from the result. Were our multiplier 991, which is 1000 - 9, we could subtract 9 times the multiplicand from 1000 times the multiplicand. Hence, RULE. ciphers. I. Multiply by the next higher unit by annexing II. From this result subtract as many times the multiplicand as there are units in the difference between the multiplier and the next higher unit. EXAMPLES FOR PRACTICE. 1. Multiply 786 by 98. Ans. 77028. Ans. 415392. |