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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. ates; 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 disference in longitude. Multiplying by 15 we have 6° 30'. Hence the following

6° 30' Rule.-Multiply the difference of time expressed in E. MIN. SEC. by 15; the result will be the difference of longitude in °'".

WRITTEN EXERCISES. 2. The difference of time between Philadelpbia 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. A bout 75o. 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. 8j sec. too slow; what is the longitude of San Francisco, that of Phila. being 750 9' 54' ? 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 7' 21" west. 8. The longitude of New York is 740 3! west, and of Jerusalem is 350 32' east; when it is 41 o'clock A. . at New York, what is the time at Jerusalem ?

Ans. 48 min. 20 sec. past 11 A, X

DENOMINATE FRACTIONS. 322. A Denominate Fraction is one in which the unit of the fraction is denominate; as, 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 equals the number of pence; and since there

oox4xt=far. are 4 farthings in 1 penny, 4 times the number of pence equals the number of farthings; hence go of a shilling equals

*4x t'farthings, which by cancelling and multiplying becomes of a farthing. Therefore, etc.

Rule.-Express the multiplication by the required multi pliers, and reduce by cancellation.

WRITTEN EXERCISES.

Ans. .
Ans.

Reduce 2. Tło of a bu. to the fraction of a pint. 3. dto of an oz. to the fraction of a grain. 4. Too of a day to the fraction of a minute. Ans. 27. 5. 28 of a gal. to the fraction of a gill. 6. of a rod to the fraction of an inch.

Ans. 7. zz'or of a ton to the fraction of an ounce. 8. rist of a sq. rd. to the fraction of a sq. in. Ans. 375. 8. ritoo of a mile to the fraction of an inch. Ans. 375.

Ans. f.

1335

Ans. 44.

CASE 11.

OPERATION.

OPLORATION,

327. To reduce a common denominate fraction to integers of lower denomination.

1. Wbat is the value of of a pound Troy? SOLUTION. There are 12 oz. in one pound, hence 12 times the number of pounds equals 8x12=9=6oz. the number of ounces; 12 times $ equals so

* x20==131 pot or 6] ɔz.: there are 20 pwt. in one oz., there

+x24=8 g. bore 20 times the number of oz. equals the number of pwt.; 20 times equals 1, or 131 pwt., etc. SOLUTION 2d. -$ of a pound equals $ of 5 lb.;

Ib. oz. pwt. gr. and $ of 5 lb. we find by dividing is 6 oz. 13 pwt.

9)5 0

0 & gr.

6 13 8 Rule 1.-Reduce the fraction until we reach an integer and a fraction of a lower denomination, set aside the integer 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. Off of a busbel ?

Ans. 3 pk. 1 qt. 1} pt. 4. Of 4 of a mile ? 2

Ans. 213 rd. 1 yd. 21 ft. 5. Of ß 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 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 I pt. Therefore .875 gal. equals 3 qt. 1 pt.

2 1000

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. 214 ft. 4. Of.364 of an acre ?

Ans. 58 P. 78 sq. yd. 5. Of.3275 of a hogshead ? Ans. 20 gal. 2 qt. 13 pt. 6. Of .9735 of a bushel ?

Ans. 3 pk. 7125 qt. 7. Of .3218 of a ton of iron ? Ans. 6 cwt. 43 lb. 9.6 oz. 8. Of 2. 1365 of á tun (4 hbd.) of wine ?

Ans. 2 tuns, 34 gal. 1 qt. 1 pt +.

REDUCTION ASCENDING.

OPERATION.

CASB I. 329. To reduce a common denominate fraction to a common fraction of a higher denomination.

1. Reduce of a farthing to the fraction of a shilling. SOLUTION.—There are 4 farthings in a penny, wherefore of the number of farthings equals the

ixixi=of 8. number of pence: there are 12 pence in 1 shilling, therefore 1 of the number of pence equals the number of shillings; hence } far. equals dixix=st of a shilling.

Rule.-Express the division by the required divisors, and 1 educe by cancellation.

WRITTEN EXERCISES.

Reduce

Ans. ato 2. f of a grain to the fraction of an ounce.

Ans. 17600 3. % of a foot to the fraction of a mile.

Ans, sto 4. $ of a gill to the fraction of a gallon.

Ans. Tot 6. of an inch to the fraction of a rod.

Ans. Toto 6. \ of a lb. to the fraction of a ton. 7 44 oz. to the fraction of a ton.

Ans. হয়ত 8. What part of a cord of wood is a pile containing 48

Ans. & cubic feet?

CASE II.

OPERATION.

330. To reduce a compound number to a common fraction of a higher denomination.

1. Reduce 3 8. 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 tar. far.; one farthing is oto of a pound, and

£1=960 far 170 far. equals 170 times oto=178, which reduced to its lowest terms, equals 17.

378=$7, Ans. Therefore, etc.

Rule.-Reduce the number to its lowest denomination, and urite 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. 2. Reduce 3 oz. 8 pwt. 12 gr. to pounds.

Ans. 187. 3. What part of 2 bushels is 1 bu. 4 qt. ?

Ans. Por 4. Of a bar. of beer (36 gal.) is 22 gal. 2 qt. ?

Ans. 5. Of a barrel of wine is 4 gal. 3 qt. 1 pt.?

Ans. 4. 6. Of 5 3. is 23 13 10 gr.?

Ans. tr 7. Of 3 years is 3 wk. 6 da. 20 b.?

Ans. 1870 8. What part of 4 inches square is 4 square inches ? What

Ans. di 16 part of a 24 inch cube is 24 cu. in. ?

CASE III.

OPERATION.

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 busbel. SOLUTION.—There are 8qt. in 1 pk., hence } of the number of quarts equals the number of pecks;

814 t 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 pke 4 qt. equals 1.625 bu.

Rule.-I. 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 mg duoed to the required denomination.

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