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Rule.-Reduce both terms to like denominations, and divide as in simple numbers. 2. Divide 1 oz. by 60 pwt.

Ans. 3. 3. Divide .625 cwt. by 500 oz.

Ans. 2. 4. Divide 9.75 b. by 24 da.

Ans. 71. 5. Divide 170 circ. by 1224.90.

Ans. 5. 6. Divide 4.81 cd. by 3014 cd. ft.

Ans. 11. 7. Divide yr. by 9.81 mo.

Ans. 1. 8. Divide .236 pwt. by 6.109 gr.

Ans. 18. 9. If a farmer raise 50 bu. of wheat on 440 sq. rd. of land, how much can be raise on 205 acres ? Ans. 375 bu.

10. If Weston should walk 4 miles in an hour, how long would it take him to walk 425 miles, walking 10 b. per day?

Ans. 9 da. 1 h. 4 min. 174 sec.

THE GREATEST COMMON DIVISOR OF

DENOMINATE NUMBERS. 464. The Greatest Common Divisor of two or more denominate numbers is the greatest denominate number which will exactly divide them.

1. Find the greatest common divisor of 2 bu. 3 pk. 6 qt. and 4 bu. 1 pk. 5 qt. SOLUTION.—2 bu. 3 pk. 6 qt. equals

OPERATION. 94 qt., and 4 bu. 1 pk. 5 qt. equals

2 bu. 3 pk. 6 qt.=94qt. 141 qt.; the greatest common divisor

4 bu. 1 pk. 5 qt.=141 qt. of 94 qt. and 141 qt. we find, by

G. C. D.=47 qt. Art. 179, is 47 qt. which, by reduc

47 qt.=1 bu. 1 pk. 7qt., Ans. tion, we find equals 1 bu. 1 pk. 7 qt. Hence the following rule:

Rule.Reduce the numbers to like denominations; find the greatest common divisor of the results as in simple numbers, and then reduce it to any convenient denomination.

NOTE.-The greatest common divisor of denominate numbers, though new, will be found interesting and practical.

Find the greatest common divisor 2. Of 1646|qt. and .93ļ hhd. Ans. 58 gal. 3} qt. 3. Of 337.0 C. and 52000°. Ans. 48 C. 1 S. 23° 20'. 4. Of 1 rd. 4 yd. 1 ft. 31 in. and 1.527 rd. Ans. 2 ft. 3} in.

5. Of 150 gal. 2 qt. 1 pt. 3.36 gi. and 4220.44 pt.

Ans. 75 gal. 1 qt. 3.68 gi. 6. A farmer has 3 lots of hay weighing respectively of a ton, 444 cwt. and 5.75 cwt. 65 lb. ; what is the heaviest bale into which he can divide them? Ans. 3 cwt. 20 lb.

7. A man has a triangular field whose sides are respectively 3 rd. 3 yd. 2 ft. 4 in., 6 rd. lyd. 1 ft. 5 in., 7 rd. 2 yd. 2 in. long; wbat is the greatest length of boards that he can use in fencing it, without cutting the boards? Ans. 6 ft. lin.

8. What are the largest sized house lots of equal extent into which 3 fields, containing respectively 2 A. 100 P., 3 A. 177 P., and 2 A. 115. P. may be divided ?

Ans. 15 P. 16 sq. yd. 7 sq. ft. 36 sq. in. THE LEAST COMMON MULTIPLE OF DENOMI

NATE NUMBERS. 465. The Least Common Multiple of two or more denominate numbers is the smallest denominate number that is a whole number of times each of them.

1. Find the least common multiple of 2 bu. 3 pk. 6 qt. and 4 bu. 1 pk. 5 qt. SOLUTION.—2 bu. 3 pk. 6 qt.

OPERATION. equals 94 qt., and 4 bu. 1 pk.

2 bu. 3 pk. 6 qt.= 94 qt. 5qt. equals 141 qt.; the L. C.

4 bu. 1 pk. 5 qt.=141 qt. M. of 94 qt. and 141 qt. is (Art.

L. C. M.= 282 qt. 187) 282 qt., which by reduction, we find equals 8 bu. 3 pk. 2 qt.

282 qt. = 8 bu. 3 pk. 2 qt., Ans. Hence the following rule:

Rule.- Reduce the numbers to the same denomination, find the least common multiple of the results, as in simple numbers, and reduce it to any convenient denomination.

NOTE.—The least common multiple of denominate numbers, though new, will be found interesting and practical.

Find the least common multiple
2. Of 40 lb. ; 30 lb.; and 26 lb. 10% oz.

Ans. 2 cwt. 40 lb. 3. Of 30 sq. yd. 6 sq. ft. ; 690 sq. ft. ; 57 sq. yd.

Ans. 8 A. 22 P. 21 sq. yd. 1 sq. ft. 72 sq. in, 4. Of 2 bu.pk. ; 44 pk. ; 19.96875 bu.

Ans. 99 bu. 3 pk. 3 qt.

5. Of.391633;.342413; 117.5 gr. Ans. 13 53 22 21 gr. 6. Of 10 min. .i sec.; 1 h. 10 min. . I sec.; 1.5 h. 1 sec.

Ans. 10 h. 30 min. 7 sec. 7. What are the contents of the least possible keg that will hold an exact number of times the contents of each of three kegs, holding respectively it of a gallon, 2 gal. 3 qt., and 4 gal. 34 qt. ?

Ans. 24 gal. 3 qt. 8. What is the area of the smallest square lot that can be enclosed by boards 5 ft. 3 in., 10 ft. 6 in., or 15 ft. 9 in. in length without cutting the boards ?

Ans. 3 P. 19 sq. yd. 41 sq. ft. MISCELLANEOUS PROBLEMS. 1. A shipper sold 16 bales of hay, each weighing 4 cwt. 96 lb., at $2.371 per cwt.; what did it amount to?

Ans. $188.48. 2. A can dig 21 rd. 4 yd. 2 ft. of ditch in 3 days, and B can dig 35 rd. 2 yd. 1ị ft. in 4 days; how much can they together dig in a week ? Ans. 96 rd. 4 yd. 2 ft. 6 in.

3. Mr. James was born Feb. 29th, 1832, and died May 10th, 1864; how many birthdays did he see and what was his exact age? Ans. 9 birthdays; age, 32 yr. 2 mo. 11 da.

4. The Knickerbocker Ice Company have an ice-house 50 ft. long, 40 ft. wide, and 24 ft. high ; how many tons of ice will it hold, a cubic foot weighing 58} lb. ? Ans. 1395.

5. A druggist bought 3 lb. 8 oz. Av. of drugs for $52.50, and made them up into pills of 5 grains each, which he retailed at 36¢ a dozen ; what was his gain ? Ans. $94.50.

6. The longitude of Cairo is 108° 17' 6'' east of Washington, and that of St. Joseph 17° 40' 44'' west; what is the time at St. Joseph when it is midnight at Cairo?

Ans. 3 h. 36 min. 8ş sec. P. M. 7. A lady wished to make tucks of an inch wide, leaving of an inch between the edge of one tuck and the stitching of the next; how many can she make in yard-wide muslin?

Ans. 41 tucks and finch remaining. 8. I bought 400 tons of coal in England for 3 s. 6 d. a cwt., paid $1.50 a ton for transportation, and sold them in New

York at 62 cents a cwt.; did I gain or lose, and how much? .

Ans. Lost, $1813.10. 9. A merchant traded 16 gross and 4 dozen of buttons worth $5 a gross, for hats worth $3 apiece, and 3 dozen caps at $7 a dozen ; how many hats did he receive?

Ans. 20. 10. The circumference of the fore wheels of a wagon is 10 ft. 4 in., that of the hind wheels 15 ft. 3 in.; how far must the wagon move that the wheels may hold the same relative position to each other as when it started ? Ans. 630 yd. 1 ft.

11. What are the dimensions of the least possible pile that can be made either out of scantling 3; ft. long, 21 in. wide, and 1.9 in. deep; 155 ft. long, 12. in. wide, and 94 in. deep; or 9} ft. long, 74 in. wide and 517 in. deep?

Ans. 463 ft. ; 37} in.; 28.1 in. 12. A merchant had a cask of vinegar from which there leaked away 26 gal. 3 qt. 1 pt. ; he then put in 19 gal. 2 qt., and sold 24 gal. 3 qt., and found there lacked 39 gal. 1 qt. 1 pt. of being 60 gal. ; how much was in the cask at first ?

Ans. 52 gal. 3 qt. 13. A, B, and C started on the morning of the same day to travel round a lake 121 miles in circumference; A traveled 3 mi. 106; rd., B, 10 miles, and C,16 mi. 213} rd. a day; how many days must they travel before they will meet again at the place where they started ? Ans. 3 days.

14. G. W. Whitney & Co. bought of Johnson & Co., London, July 15, 1877, 151 reams Bath post paper @ £1 2 s. 6 d.; 5000 envelopes at 17 s. 9 d. q M. ; 20 gross Gillott's steel pens @ 4 s. 3 d. ; 5 gross Faber's lead pencils @ 18 s. 9 d.; and 4 dozen diaries at 10 s. a dozen. Make out and receipt Whitney & Co.'s bill.

Ans. £32 16 s. 3 d. 15. What are the dimensions of the largest possible sticks of timber of equal size that can be used to make 3 piles, respectively 28 ft. long, 17 ft. 6 in. wide and 14 ft. high; 46 ft. 8 in. long, 24 ft. 6 in. wide, and 18 ft. high; and 65 ft. 4 in. long, 31 ft. 6 in. wide, and 22 ft. high ?

Ans. 9 ft. 4 in. ; 3 ft. 6 in. ; 2 ft.

16. An American traveling in a coach going about 10 English miles an hour, inquired the distance to Berlin and. was told it was 15 miles; at the end of an hour and a half, seeing no signs of the city, he again inquired how long it would be before they arrived, and was told that it would take about 5 hours. What was the distance in English miles at the time of the first inquiry? Ans. 70-77.

17. Two men, A and B, on opposite sides of a pond, which is 97 rd. 2 yd. 1 ft. 6 in. in circumference, start simultaneously to go around it in the same direction. A walks 16 rods in one minute, and B 22} yards in 15 seconds; how often will B circumambulate the pond before they arrive together at the place from which B started? Ans. 17 times.

18. A vessel sailed from Philadelphia, and after being out 30 days, the captain took an observation and found the solar time to be 2 h. 16 min. 24 sec. P. M., the chronometer at the same time marking 11 h. 36 min. 40 sec. A. M.; required the longitude of the vessel, supposing the chronometer to have lost 41 sec. per day. Ans. 35° 44' 35" W.

19. Two pedestrians are on a straight road on opposite sides of a gate, and distant from it 2 mi. 120 rd. and 4 mi. 80 rd. respectively, and travel each towards the original station of the other. The first travels 120 rods in 10 minutes, and the second 133} rods in 10 minutes; how long must they travel before they are equally distant from the gate ?

Ans. 1st time 831 min.; 2d time 450 min.

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