CASE II. 321. To find the difference of longitude of two places when their difference of time is given. 1. The difference of time between two places is 26 min. utes; what is their difference of longitude ? SOLUTION.-Since 1 h. of time corresponds to 15° OPERATION. of longitude, and 1 min. of time to 15' of longitude, h. min. 15 times the number of hours and minutes difference 0 26 in time will equal the number of degrees and minutes 15 Cixerence in longitude. Multiplying by 15 we have 6° 30'. Hence the following 6° 30' Rule.-Multiply the difference of time expressed in H. MIN. SEC. by 15; the result will be the difference of longitude in o 11. WRITTEN EXERCISES. 2. The difference of time between Philadelphia and Cin. cinnati is about 37 min. 20 sec. ; what is the difference of longitude ? Ans. 9° 20'. 3. The time at St. Louis is about 53 minutes earlier than the time at Washington; what is the difference in longitude? Ans. 13° 15'. 4. When it is noon at London it is about 7 o'clock, A. M., in Philadelphia; required the difference of longitude. Ans. About 75°. 5. In traveling from New York to Cincinnati I find my watch is 41 min. 32 sec. too fast; required the difference of longitude. Ans. 10° 23'. SUPPLEMENTARY PROBLEMS, To be omitted unless otherwise directed. 6. In coming from San Francisco to Philadelphia I find my watch is 3 h. 9 min. 8f sec. too slow; what is the longitude of San Francisco, that of Phila. being 750 9' 5'/? Ans. 1220 26' 15'. 7. The longitude of Cambridge, England, is 5' 21" east, and the difference of time between it and Cambridge, Mass., is 4 h. 44 min. 504 sec.; required the longitude of the latter place. Ans. 710 77 21" west. 8. The longitude of New York is 74° 3' west, and of Jerusalem is 350 32' east; when it is 41 o'clock A. M. at New York, what is the time at Jerusalem ? Ans. 48 min. 20 sec. past 11 A. M DENOMINATE FRACTIONS. 322. A Denominate Fraction is one in which the unit of the fraction is denominate; as, į of a pound. 323. Denominate Fractions may be expressed either as common fractions or as decimals. REDUCTION OF DENOMINATE FRACTIONS. 324. Reduction of Denominate Fractions is the process of changing them from one denomination to another without altering their value. 325. There are two general cases, reduction ascending and descending, which, for convenience of operation, are subdivided into several other cases. REDUCTION DESCENDING. OPERATION. CASE I. 326. To reduce a common denominate fraction to a fraction of a lower denomination. 1. Reduce of a shilling to farthings. SOLUTION.Since there are 12 pence in one shilling, 12 times the number of shillings off x*xt=i far. equals the number of pence; and since there are 4 farthings in 1 penny, 4 times the number of pence equals the number of farthings; hence go of a shilling equals Xaxi farthings, which by cancelling and multiplying becomes i of a farthing. Therefore, etc. Rule.-Express the multiplication by the required multipliers, and reduce by cancellation. WRITTEN EXERCISES. Reduce Ans. Ans. 3. 1440 of an oz. to the fraction of a grain. 4. Todo of a day to the fraction of a minute. Ans. 24. Ans. 4. 5. zág of a gal. to the fraction of a gill. 6. of a rod to the fraction of an inch. Ans. 7. 22:00 of a ton to the fraction of an ounce. Ans. 44. 8. 8157 of a sq. rd. ta the fraction of a sq. in. Ans. 315. 8. Toto of a mile to the fraction of an inch. Ans. 3. 1386 CASE 11. 327. To reduce a common denominate fraction to integers of lower denomination. 1. What is the value of of a pound Troy? SOLUTION.- There are 12 oz. in one pound, OPERATION. hence 12 times the number of pounds equals fx12=956 oz. the number of ounces ; 12 times & equals se #x20 ==137 put. or 6} oz.: there are 20 pwt. in one oz., there X 24=8 gr. fore 20 times the number of oz. equals the Qumber of pwt.; 20 times equals e, or 13} pwt., etc. OPERATION. SOLUTION 2d. — of a pound equals $ of 5 lb.; lb. oz. pwt. gr. and $ of 5 lb. we find by dividing is 6 oz. 13 pwt. 9)5 000 &gr. 6 13 8 Rule I.— Reduce the fraction until we reach an integer and a fraction of a lower denomination, set aside the integer and reduce the fraction as before, and thus continue as far as necessary. Rule II.-Regard the numerator as so many units of the given denomination, and divide by the denominator. WRITTEN EXERCISES. What is the value 2. Of of an ounce ? Ans. 16 pwt. 16 gr. 3. Of of a bushel ? Ans. 3 pk. 1 qt. 13 pt. 4. Of of a mile ? Ans. 213 rd. 1 yd. 21 ft. 5. Of } of a rod ? Ans. 4 yd. 2 ft. 54 in. 6. Of jy of a mile ? Ans. 288 rd. 7. Off of a sign? Ans. 26° 15'. 8. Of of an acre ? Ans. 114 P. 8 yd. 5 ft. 1134 in. CASE III. 328. To reduce a denominate decimal to integers of lower denominations. 1. Reduce .875 gal. to integers of lower denominations. SOLUTION.—There are 4 quarts in one gallon, there- OPERATION fore 4 times the number of gallons equals the number of .875 quarts; 4 times .875 equals 3 qt. and .5 qt.; there are 4 2 pints in one quart, therefore 2 times the number of 3.500 quarts equals the number of pints: 2 times .5 equals 2 1 pt. Therefore .875 gal. equals 3 qt. 1 pt. Rule - Reduce the decimal until we reach an integer and a decimal of a lower denomination, set aside the integer, and reduce the decimal as before, and thus continue as far as necessary. WRITTEN EXERCISES. What is the value 2. Of.825 of a pound Troy? Ans. 9 oz. 18 pwt. 3. Of .675 of a rod ? Ans. 3 yd. 217 ft. 4. Of .364 of an acre ? Ans. 58 P. 738 sq. yd. 5. Of .3275 of a hogshead ? Ans. 20 gal. 2 qt. 1 pt. 6. Of .9735 of a bushel ? Ans. 3 pk. 713 qt. 7. Of.3218 of a ton of iron ? Ans. 6 cwt. 43 lb. 9.6 oz. 8. Of 2.1365 of a tun (4 hhd.) of wine? Ans. 2 tuns, 34 gal, 1 qt. 1 pt +. REDUCTION ASCENDING. OPERATION. CASE I. 329. To reduce a common denominate fraction to a common fraction of a higher denomination. 1. Reduce 1 of a farthing to the fraction of a shilling. SOLUTION.—There are 4 farthings in a penny, therefore of the number of farthings equals the ixxt=d8. number of pence: there are 12 pence in 1 shilling, therefore of the number of pence equals the number of shillings; hence | far. equals xix}=sts of a shilling. Rule.--Express the division by the required divisors, and reduce by cancellation. WRITTEN EXERCISES. Reduce Ans. to 3. % of a foot to the fraction of a mile. Ans. Too 4. f of a gill to the fraction of a gallon. Ans. 5. 5 of an inch to the fraction of a rod. 6. { of a lb. to the fraction of a ton. Ans. 167oo 7. 44 oz. to the fraction of a ton. Ans. Voo 8. What part of a cord of wood is a pile containing 48 cobic feet? Ans. Ans. Tot CASE II. OPERATION, 330. To reduce a compound number to a common fraction of a higher denomination. 1. Reduce 3 s. 6 d. 2 far. to the fraction of a pound. SOLUTION.—By reduction we find 3 s. 6 d. 2 far. equal to 170 far., and also £l=960 38. 6 d. 2 far.=170 far. far.; one farthing is gło of a pound, and £1=960 far. 170 far. equals 170 times oto=378, which reduced to its lowest terms, equals 17. 370=37, Ans. Therefore, etc. Rule.- Reduce the number to its lowest denomination, and write under it the number of units of this denomination which make a unit of the required denomination ; and then reduce the resulting fraction to its lowest terms. WRITTEN EXERCISES. Ans. Ans. 14 2. Reduce 3 oz. 8 pwt. 12 gr. to pounds. Ans. 137 3. What part of 2 bushels is 1 bu. 4 qt. ? Ans. & 4. Of a bar. of beer (36 gal.) is 22 gal. 2 qt. ? 5. Of a barrel of wine is 4 gal. 3 qt. 1 pt.? 6. Of 5 Z. is 23 19 10 gr.? Ans. 16. 7. Of 3 years is 3 wk. 6 da. 20 h.? Ans. 167 615870 8. What part of 4 inches square is 4 square inches ? What part of a 21 inch cube is 21 cu. in. ? Ans. 1; 26 OPERATION. CASE III. 331. To reduce a compound number to a decimal of a higher denomination. 1. Reduce 1 bu. 2 pk. 4 qt. to the decimal of a bushel. SOLUTION.—There are 8qt. in 1 pk., hence $ of the number of quarts equals the number of pecks; 84 $ of 4 equals .5, which, with 2 pk. equals 2.5 pk.; there are 4 pk. in a bushel, hence of the number of 4 2.5 pecks equals the number of bushels; } of 2.5 equals '1.625 Ans. 625, which, with 1 bu., equals 1.625 bu.; hence 1 bu. 2 pk. 4 qt. equals 1.625 bu. Rule.-1. Divide the lowest term by the number of units which equals one of the next higher, and annex the decimal quotient to the integer of the next higher denomination. II. Proceed in a similar manner until the whole 18 re duced to the required denomination. |