166. To reduce fractional compound numbers to whole numbers. First reduce the given numerator to the next lower denomination ; (Art. 161, 1 ;) then divide the product by the denominator, and the quotient will be an integer of the next lower denomination. Proceed in like manner with the remainder, and the several quotients will be the whole numbers required. 2. Reduce of £l to shillings. Ans. 12s. 3. How many shillings and pence in £f? 4. How many shillings, &c., in £? 5. In of 1 week, how many days, &c. ? 6. In of 1 day, how many minutes, &c.? 7. Change of 1 league to miles, &c. 8. Change of 1 mile to furlongs, &c. 9. Reduce of 1 hundred weight to quarts, &c. 10. In of 1 ton, how many hundred weight, &c. ? 11. In şof 1 bushel, how inany quarts, &c. ? 12. In of 1 peck, how many quarts, &c. ? 13. Reduce of £1 to shillings. Suggestion. Since the numerator, when reduced to the denomination required, cannot be divided by the denominator, the division must be represented. Ans. 69 s. Note.- This, in effect, is reducing &z of £1 to the fraction of a shilling 14. Reduce 365 of £l to pence. Ans. 365d. 167. From the last two examples it is manifest, that a fraction of a higher denomination may be changed to a fraction of a lower denomination, by reducing the given numerator to the denomination of the required fraction, as in the preceding article, and placing the result over the given denominator. Quest.-166. How are fractional compound numbers reduced to whole ones? 167. How is a fraction of a higher denomination changed to a fraction of a lower denomination ? 15. Reduce tả7 of £1 to the fraction of a shilling. Ans. 127 16. Reduce io of 1 week to the fraction of a day. 17. Change 14 of 1 mile to the fraction of a rod. 18. Change são of 1 rod to the fraction of a foot. 19. Changes of 1 yard to the fraction of a nail. 20. Change 100doo of 1 ton to the fraction of a pound. ADDITION OF COMPOUND NUMBERS. £ S. d. far. 11 un 1. What is the sum of £4, 8s. 60. 2 far. ; £3, 12s. 8d. 3 far.; and £8, 6s. 9d. 1 far. ? Operation. Having placed the farthings under farthings, pence under 4 9 6 2 pence, &c., we add the column 3 12 8 of farthings together, as in siin8 6 9 1 ple addition, and find the sum is 6, which is equal to 1d. and 16 9 2 Ans. 2 far. over. Set the 2 far. der the column of farthings, and carry the ld. to the column of pence. The sum of the pence is 24, which is equal to 2s. and nothing over. Place a cipher under the column of pence, the 2s, to the column of shillings. The sum of the shillings is 29, which is equal to £l and 9s, over. Write the 9s. under the column of shillings, and carry the £l to the column of pounds. The sum of the pounds is 16, the whole of which is set down in the same manner as the left hand column is in siniple addition. (Art. 25.) The answer is £16, 9s. Od. 2 far. 168. Hence we derive the following general RULE FOR ADDING COMPOUND NUMBERS. I. Write the numbers so that the same denominations shall stand under each other. Quest.–168. How are compound numbers written for addition ? Which denomination is added first ? When the sum of any column is found, what is to be done with it? II. Beginning with the lowest denomination, find the sum of each column separately, and divide it by that number which it requires of the column added to make one of the next higher denomination. Set the remainder under the column, and carry the quotient to the next column. III. Proceed in this manner with all the other denominations except the highest, whose entire sum is set down as ir simple addition. (Art. 29.) Proof:— The proof is the same as in Simple Addition. (Art. 28.) Obs. The process of adding numbers of different denominations is called Compound Addition. It is the same as Simple Addition, except the method of carrying from one denomination to another. 2. What is the sum of £10, 6s. 7d. ; £18, 12s. 10d., £5, 3s. 4d.? Ans. £34, 2s. 9d. 9. Add 7 lbs. 9 oz. 16 pwts. 10 grs.; 3 lbs. 10 oz. 8 pwts. 9 grs. ; 8 lbs. 3 oz. 1 pwt. 4 grs. 10. An Englishman-bought a carriage for £35, 12s.; a horse for £27, 8s. 10d.; a harness for £7, 16s. 11d. : how much did he give for the whole ? Quest.– What is done with the last column? How is the operation proved? Obs. What is the process of adding compound numbers called ? How does it differ from simple addition? 11. A merchant bought of one dairyman 5 cwt. 11 lbs. 6 oz. of butter; of another, 3 cwt. 15 lbs. 9 oz.; of another, 7 cwt. 6 lbs. 10 oz.; how much did he buy of all ? 12. A manufacturer bought of one man 73 lbs of wool; of another, 96 lbs. 6 oz. ; of another, 135 lbs. 11 oz.; of another, 320 lbs. 9 oz. ; of another, 642 lbs. 3 oz. : how much wool did he buy? 13. A man sold to one customer 2 tons, 62 lbs. 10 oz. of hay; to another, 5 tons, 40 lbs. 12 oz.; to another, 3 tons, 75 lbs, 6 oz. : how much did he sell to all ? 14. A man wove 7 yds. 3 qrs. 2 na. of cloth in one day; the next day, 6 yds. 1 qr. 3 na.; the next, 8 yds. 3 qrs. 1 na.; the next, 5 yds. 2 qrs. 3 na. : how much did he weave in all ? 15. Bought several pieces of cotton ; one contained 26 yds. 1 qr. 2 na.; another, 30 yds. 2 qrs.; another, 294 yds. 3 na.; another, 324 yds. 1 na. : how many did they all contain ? 16. A hotel keeper bought at one time, 15 bu. 2 pks. 3 qts, of oats ; at another, 10 bu. 1 pk. 2 qts.; at another, 203 bur6 qts.; at another, 185 bu. 5 qts.: what was the amount of all his purchases ? 17. Bought 4 loads of wheat; the first containing 23 bu. 3 pks. 5 qts.; the second, 205 bu 6 qts. ; the third, 261 bu.; the fourth, 21% bu. 7 qts. : how many bushels did they all contain ? 18. What is the sum of 16 m. 3 fur. 16 r.; 26 m. 1 fur. 33 r.; 10 m. 8 fur. 22 r. ; 45 m. 7 fur. 20 r ? 19. A merchant bought 3 casks of oil ; one held 2 hhds. 30 gals. 2 qts. ; another, 3 hhds. 10 gals. ; another, 1 hhd. 13 gals. 1 qt: how much did they all hold ? 20. Sold several lots of' wine, in the following quantities: 1 pipe, 1 hhd. 21 gals. 2 qts. 1 pt.; 2 pipes, il gals. 1 pt. ; 3 hhds. 15 gals, 2 qts. ; 3 pipes, 10 gals. 2 qts. 1 pt. : how much was sold in all ? 21. A mason plastered one room containing 45 square yards, 7 ft. 6 in.; another, 25 yds. 6 ft. 95 in.; another, 38 yds. 4 st. 41 in. : what was the amount of plastering in all the rooms ? 3 qts. 22. Sold 10 A. 35 r. 10 sq. ft. of land at one time; at another, 3 A. 10 r. 15 ft. ; at another, 18 A. 16 r. 23 ft.: what was the amount of land sold ? 23. A merchant received several boxes of goods ; one contained 16 cu. ft. 61 in.; another, 25 ft. 81 in.; another, 20 ft. 13 in.; another, 38 st. 72 in.: how many cubic feet and inches did they all contain ? 24. One pile of wood contains 10 C. 38 ft. 39 in.; another, 15 C. 56 ft. 73 in. ; another, 30 C. 19 ft. 44 in.; another, 17 C. 84 ft. 21 in.: how much do they all contain ? SUBTRACTION OF COMPOUND NUMBERS. d. far. 11 // Ex.). From £15,7s.6d. 3 far., subtract £6, 4s. 8d. 2 far. Operation, Having placed the less num£ ber under the greater, with far15 7" 6 3 things under farthings, pence 6 4 8 2 under pence, &c., we subtract 2 far. from 3 far., and set the 9 2 10 1 Ans. remainder 1 far. under the column of farthings. Now 8d. cannot be taken from 6d.; we therefore borrow 1 from the next higher denomination, which is shillings; and 1s. or 12d. added to the 6d. make 18d. And 8d. from 18d. leaves 10d. Since we borrowed 1, we must carry 1 to the next column, as in simple subtraction. I added 10 4 makes 5; and 5 from 7, leaves 2. 6 from 15 leaves 9. Ans. £9, 2s. 10d. 1 far 169. Hence, we derive the following general RULE FOR SUBTRACTING COMPOUND NUMBERS. I. Write the less number under the greater, so that the same denominations may stand under each other. II. Beginning with the lowest denomination, subtract the number in each denomination of the lower line from the number above it, and set the remainder below. Quest.-169. How do you write compound numbers for subtraction? Where begin to subtract ? When the number in the lower line is larger than that above it, what is to be done? How is the operation proved ? |