end, add to the minuend as many units as equal one of the next higher denomination, and subtract; then consider the units of that higher denomination of the minuend as one less, and subtract as before. (1) Problems. (3) 90 A. 7 sq. rd. 10 sq. yd. 40 bu. 1 pk. 6 qt. 75 100 30 7 mi. 75 rd. 3 yd. 3 100 5 (4) (5) 37 3 7 (6) 30 sq.rd. 20 sq. yd. 6 sq. ft. 25 8 8. 150 T. 15 cwt. 25 lb. 5 da. 59 min. 57 sec. 100 T. 18 cwt. 50 lb. 9. 75 hhd. 50 gal. 2 qt. 3 gi. — 56 hhd. 60 gal. 1 pt. 10. 60° 10' 30" 55° 15′ 45"; £100 - £50 9 s. 10 d. 11. 100 mi. 180 rd.-90 mi. 250 rd. 5 yd. 2 ft. 12. Boston is in latitude 42° 21′ 30′′ N., and New York in 40° 42′ 43′′ N. What is the difference in their latitude? 13. Washington is in longitude 77° 0′ 15′′ W., and San Francisco in 122° 46′ 48′′ W. What is the difference in their longitude? Take the difference from 180°. 14. The sum of two compound numbers is 400 mi., and one of them is 176 mi. 200 rd. 2 ft. What is the other? rd. 6 sq. ft., and the Find the subtrahend. 15. The minuend is 47 A. 30 sq. remainder is 29 A. 30 sq. yd. 8 sq. ft. 16. Of 16 mi. 5 rd. 2 yd. 1 ft. of a telegraph line, 13 mi. 5 yd. 2 ft. are finished. What is yet to be put up? 17. From a farm containing 100 A. 30 sq. rd., there were sold 50 A. 50 sq. rd. 30 sq. yd. What remained? 18. How many years, months, and days from December 14, 1799, to July 4, 1876? SOLUTION. 1876 yr. 7 mo. 4 da. 1799 12 14 EXPLANATION. - Since the later of two dates expresses a greater period of time than the earlier, write it as the minuend, and the earlier as the subtrahend, placing to the right of each year the number of the month, and the number of the day. Consider 30 days as a month, and 12 months as a year, and subtract (425). 76 yr. 6 mo. 20 da. 19. How long from Apr. 13, 1899, to Jan. 1, 1901? How long from May 10, 1900, to Dec. 25, 1901 ? 20. A note given July 10, 1900, was paid Mar. 1, 1901. How long had it run? If paid Sept. 9, 1901? 21. The sum of three numbers is 200 A. 10 sq. rd., and two of them are 98 A. 60 sq. rd. 8 sq. ft., and 89 A. 30 sq. yd. 72 sq. in. What is the other? 22. A contractor engaged to build 75 mi. of telegraph; after building 25 mi. 250 rd., a storm blew down 2 mi. 10 rd. 2 ft. What remained to be done? 23. From a piece of land containing 100 A. 150 sq. rd., there were sold at one time 25 A. 30 sq. rd., and at another time 50 A. 10 sq. rd. 5 sq. ft. How much remained? SECTION XXIV. MULTIPLICATION OF COMPOUND NUMBERS. Written Exercises. 427. Example. Find the product of 4 miles 100 rods 4 yards multiplied by 6. Six times 4 yd. are 24 yd., or 4 rd. 2 yd.; write 2 yd. under yards, and carry 4 rd. to add to the product of the rods. Six times 100 rd. are 600 rd.; 600 rd. and 4 rd. are 604 rd., or 1 mi. 284 rd. Write 284 rd., etc. Six times 4 mi. are 24 mi.; 24 mi. and 1 mi. are 25 mi., which write under miles, giving the entire product 25 mi. 284 rd. 2 yd. 428. Rule for Multiplication of Compound Num bers. I. Write the multiplier under the lowest denomination of the multiplicand. II. Begin with the lowest denomination, and multiply the units of each denomination separately. III. If the product is less than one of the next higher denomination, write it under the denomination multiplied. If the product is equal to, or greater than, one of the next higher denomination, change it to that higher denomination, write the remainder under the denomination multiplied, and add the ones of the higher denomination to the product of that denomination. 275 4.80 sq.rd. 8 sq. ft. 70 sq.in. 345 mi. 40 rd.5 yd.2 ft. 10 in. 75 'Find the product of 98 6. 7 da. 5 hr. 50 min. × 5; by 7; by 9; by 11; by 25. 7. 10° 15′ 35′′ × 6; by 8; by 10; by 12; by 20; by 75. 8. £10 10 s. 8 d. × 3; by 6; by 9; by 12; by 110. 9. 5 T. 3 cwt. 90 lb. x 4; by 9; by 14; by 35; by 100. 10. 10 lb. 4 oz. 15 pwt. 18 gr. × 7; by 12; by 17; by 125. 11. 85 A. 25 sq. rd. 8 sq. ft. × 25; by 46; by 127; by 176. 12. 98 mi. 80 rd. 2 ft. × 154; by 234; by 345; by 567. 13. If an engine is run at the average rate of 25 mi. 10 rd. 5 yd. per hour, how far does it run in 12 hours? 14. If a team can haul 1 cord 20 cu. ft. of wood in one load, how much can it haul in 50 loads? In 75? 15. How much time is there in 100 solar years, each 365 da. 5 hr. 48 min. 49.7 sec.? In 150? In 175? In 234? 16. If the average weight of a bushel of wheat is 59 lb. 81 oz., what is the weight of 50 bushels? Of 65? 78? 17. A piece of land was divided into 25 lots, each 75 sq. rd. 30 sq. yd. How much land was in the piece? 18. From 16 mi. 8 rd. 4 yd. 1 ft. take 10 mi. 20 rd. 5 yd., and multiply the remainder by 9; by 10; by 15. 19. A farmer sold 40 A. 100 sq. rd. from a farm containing 60 A. 15 sq. rd. 3 sq. yd., and bought 4 times as much as the remainder. How much land had he then? 20. Of 125 mi. 80 rd. 5 yd. of railroad, 4 sections, each 25 mi. 100 rd. 4 yd., have been finished. How much remains to be done? SECTION XXV. DIVISION OF COMPOUND NUMBERS. Written Exercises. 429. Example 1.- Divide 41 mi. 20 rd. 5 yd. by 8. SOLUTION. 8)41 mi. 20 rd. 5 yd. 3 8 5 mi. 42 rd. 3 yd. EXPLANATION.-Write the divisor, etc. Begin with the highest denomination, and divide the units of each denomination separately. One eighth of 41 mi. is 5 mi. with a remainder of 1 mi. Write 5 mi. as the miles of the quotient, and change the 1 mi. to rods. 1 mi. is 320 rd., and 320 rods plus 20 rd. are 340 rd. One eighth of 340 rd. is 42 rd. with a remainder of 4 rd. Write 42 rd. as the rods of the quotient, and change the 4 rd. to yards, etc. One eighth of 27 yd. is 3 yd. with a remainder of 3 yd., or 3§ yd., which write, etc.; giving the entire quotient 5 mi. 42 rd. 3 yd. 3 8 430. Example 2.-138 gal. 1 pt. ÷ 27 gal. 2 qt. 1 pt. SOLUTION. 138 gal. 1 pt. 27 gal. 2 qt. 1105 pt. = 1105 pt. 221 pt. 5 EXPLANATION. Since the dividend and the divisor are similar numbers, the quotient may be most easily found by changing both compound numbers to the lowest denomi nation mentioned in either, and then dividing as in simple numbers. 138 gal. 1 pt. equal 1105 pt., and 27 gal. 2 qt. 1 pt. equal 221 pt. 221 pt. are contained in 1105 pt. 5 times. 431. Rules for Division of Compound Numbers. When the divisor is a simple number, I. Write the divisor at the left of the dividend. II. Begin with the highest denomination, and divide the units of each denomination separately. III. If after dividing the number of any denomination there is a remainder, change it to the next lower denomination, and to the result add the given number of that denomination; then divide as before. The several results will be the quotient required. When the divisor is a compound number, Change the dividend and the divisor to the lowest denomination mentioned in either, and then divide as in simple numbers. (1) Problems. (2) (3) 7)10 mi. 8 rd. 3 yd. 8)20 A. 25 sq. rd. 9)50 bu. 2 pk. 6 qt. 10)25 cd. 10 cu. ft. 100 cu. in. 12)7 mi. 100 rd. 5 yd. 2 ft. 10 in. Find the quotient of 6. 12 gr. gro. 7 gro. 10 doz. ÷ 6; 15 rm. 10 sheets ÷ 7. 7. 15 yr. 5 hr. 10; 365 da. 5 hr. 48 min. 49 sec. 11. 8. 90° 30" 12; £15 10 s. 81 d. ÷ 13; £100 6 d. ÷ 14. |