From the solution of this and similar problems we derive the following rules for the addition of compound numbers: RULES. 1. Write the numbers so that units of the same denomination shall stand in the same column. 2. Beginning with the lowest denomination, add as in simple numbers, and reduce the sum to the next higher denomination, writing the remainder, if any, under the column added, and adding the quotient, obtained by the reduction, to the next column. 3. Proceed in the same manner with all the columns to the last. NOTE.-If any places are wanting, supply ciphers. 5 T. 8 cwt. 50 lb.; 8 cwt. 15 10 T. 27 lb. 12 oz. Ans. 38 T. 9 cwt. 19 lb. 8 oz. 13. Add 16 bu. 2 pk. 6 qt.; 2 pt.; 7 bu. 3 pk. 4 qt. 1 pt.; 12 bu. 3 pk. 4 qt. 14. What is the sum of 17 lb. 8 12 lb. 2 oz. 9 pwt. 15 gr.; 1 lb. 11 12 lb. 3 oz. 18 pwt. 10 gr.? Ans. 56 bu. 3 pk. 2 qt. 15. Add 13 mi. 3 fur. 39 rd. 1 yd. 7 in.; 7 mi. 7 fur. 32 rd. 2 ft. 8 in.; 15 mi. 6 fur. 23 rd. 1 yd. 1 ft. 6 in.; 23 mi. 7 fur. 28 rd. 1 ft. 9 in.; 30 mi. 5 fur. 27 rd. 2 ft. 4 in. Ans. 91 mi. 7 fur. 29 rd. 4 yd. 2 ft. 10 in. 16. Add 71 T. 13 cwt. 1 qr. 16 lb. 11 oz.; 14 T. 13 cwt. 3 qr. 13 lb. 14 oz.; 14 T. 17 cwt. 2 qr. 15 oz.; 13 T. 17 cwt. 1 qr. 15 lb. 13 oz.; 1 T. 6 cwt. 2 qr. 13 oz. Ans. 116 T. 8 cwt. 2 qr. 23 lb. 2 oz. 17. Add 12 mi. 5 fur. 39 rd. 12 ft. 7 in.; 16 mi. 6 fur. 18 rd. 5 ft. 7 in.; 10 mi. 5 fur. 32 rd. 4 ft. 6 in. Ans. 40 mi. 2 fur. 10 rd. 2 yd. 2 in. 18. A farmer raised in four fields the following quantities of wheat: 75 bu. 2 pk. 3 qt.; 108 bu. 1 pk. 1 pt.; 600 bu. 1 pt.; and 106 bu. 3 pk. 7 qt.: how much did he raise in the four fields? Ans. 890 bu. 3 pk. 3 qt. 19. How far is it from A, which is east of a given point 3 mi. 3 fur. 9 rd. 3 yd., to B, which is west of the same point 6 mi. 1 fur. 2 yd. 6 in.? Ans. 9 mi. 4 fur. 9 rd. 5 yd. 6 in. 20. How long is it from 3 hr. 16 min. and 6 sec. before 12 o'clock to 15 min. past 9 o'clock in the evening? Ans. 12 hr. 31 min. 6 sec. 21. The four sides of a field measure as follows: 1st, 16 rd. 3 yd. 2 ft. 9 in.; 2d, 18 rd. 3 in. ; 3d, 12 rd. 4 yd. 1 ft. ; 4th, 15 rd. 2 yd. 2 ft. 11 in.: how far is it around the field? Ans. 63 rd. 11 in. 2. SUBTRACTION OF COMPOUND NUMBERS. 133. Subtraction of Compound Numbers is the process of finding the difference between two similar compound numbers. EXAMPLE. From 120 lb. 6 oz. 13 dwt. 20 gr. take 75 lb. 9 oz. 8 dwt. 22 gr. SOLUTION. Explanation.-We write the numbers as in addition of compound numbers, and 120 lb. 6 oz. 13 dwt. 20 gr. begin to subtract at the lowest denomi 75" 9 " 66 8 22" 44 lb. 9 oz. 4 dwt. 22 gr. nation. We cannot take 22 gr. from 20 gr.; we therefore take 1 dwt., equal to 24 gr., which added to 20 gr. equals 44 gr.; 22 gr. from 44 gr. leave 22 gr.; 8 dwt. from 12 dwt. leave 4 dwt. We cannot take 9 oz. from 6 oz.; we therefore convert 1 lb. to ounces; 12 oz. and 6 oz. are 18 oz.; 9 oz. from 18 oz. leave 9 oz. ; 75 lb. from 119 lb. leave 44 lb. NOTE.-Let the pupil derive the RULE. 9. From 16 cu. yd. 3 cu. ft. 1200 cu. in. take 3 cu. yd. 12 cu. ft. 1500 cu. in. Ans. 12 cu. yd. 17 cu. ft. 1428 cu. in. 10. From 16 bu. take 3 bu. 3 pk. 3 qt. Ans. 12 bu. 5 qt. 11. From 20 T. 17 cwt. 40 lb. 5 oz. take 5 T. 15 cwt. 7 lb. Ans. 15 T. 2 cwt. 32 lb. 11 oz. 10 oz. 12. A barrel contained 36 gal. of vinegar; from it was sold 13 gal. 2 qt. 1 pt.: how much remained? Ans. 22 gal. 1 qt. 1 pt. 13. A miller had 600 bu. 2 pk. 3 qt. of wheat; he ground into flour 276 bu. 3 pk. 1 pt.: how much had he remaining? Ans. 323 bu. 3 pk. 2 qt. 1 pt. 14. From 6 yd. take 2 ft. Ans. 5 yd. 1 ft. 15. A man had 107 bu. 3 pk. of wheat; he bought 190 bu. 1 pt., and sold 250 bu. 2 pk. 5 qt.: how much had he remaining? Ans. 47 bu. 3 qt. 1 pt. 16. A grocer bought 10 bbl. of sugar of 206 lb. each, and sold 1615 lb. 6 oz.: how much had he remaining? Ans. 444 lb. 10 oz. 17. Three piles of wood contain, respectively, 16 cd. 3 cu. ft.; 19 cd. 3 cu. ft.; 27 cd. 18 cu. ft.; there were sold from them 50 cd. 3 cu. ft.; how much remained? Ans. 12 cd. 21 cu. ft. 18. A farm contains 250 A. 16 pch., of which 121 A. 3 pch. 9 sq. yd. are cleared: how much woodland is there yet on the farm? Ans. 129 A. 12 pch. 21 sq. yd. TO FIND THE DIFFERENCE OF DATES. A man was born Feb. 8, 1825, and died Oct. 6, 1870: how old was he? solution. Remark.-Let the pupil give the explanation of this 45 7 28 NOTE.-In finding the difference of time, 30 d. equal 1 mo., and 12 mo. equal 1 yr. and Feb. 2, 1876? WRITTEN PROBLEMS. 1. What is the difference in time between Mar. 28, 1840, Ans. 35 yr. 10 mo. 4 d. 2. Franklin was born Jan. 6, 1706, and died Apr. 17, 1790: how old was he? Ans. 84 yr. 3 mo. 11 d. 3. Milton was born Dec. 9, 1608: how long was that before Franklin's birth? Ans. 97 yr. 27 d. 4. How long was it from Sept. 17, 1839, to Mar. 28, 1871 ? Ans. 31 yr. 6 mo. 11 d. 5. How long does a note run if dated Jan. 16, 1873, and made payable June 10, 1876? Ans. 3 yr. 4 mo. 24 d. 6. A gentleman went to Europe May 16, 1874, leaving the station at 9 o'clock 45 min. in the forenoon, and returned to the same station at 10 o'clock 20 min. in the evening, Feb. 18, 1875: how long was he gone? Ans. 9 mo. 2 d. 12 hr. 35 min. 7. Find the exact number of days from Oct. 16, 1875, to Jan. 12, 1876. Ans. 88 d. 8. Find the exact number of days from Oct. 16, 1875, to Mar. 3, 1876. Ans. 139 d. 9. How many years, months and days are you old? 10. What is the time from 15 min. past 2 o'clock in the afternoon to 10 min. before 12 at night? Ans. 9 hr. 35 min. 3. MULTIPLICATION OF COMPOUND NUMBERS. 134. Multiplication of Compound Numbers is the process of finding the product of two numbers, one of which is compound. Multiply 7 bu. 3 pk. 5 qt. by 6. SOLUTION. 7 bu. 3 pk. 5 qt. 6 Explanation.-6 times 5 qt. are 30 qt., or 3 pk. and 6 qt.; 6 times 3 pk. are 18 pk., which added to 3 pk. are 21 pk., or 5 bu. and 1 pk.; 6 times 7 47 bu. 1 pk. 6 qt. bu. are 42 bu., which added to 5 bu. are 47 bu. Remark. Let the pupil derive a rule from the analysis of the foregoing and similar problems. |