gal. } Note. This weight is used in compounding medicines., Drugs and medicines, however, are bought and sold by Avoirdupois weight. The pound and ounce of this weight are the same as those of Troy Weight. The subdirisions or smaller divisions are different. 37. How many grains in 8 Z. 33. 23. ? Ans. 4,060. 38. How many 3. in 1 lb. ? Ans. 96. 39. How many 3. in 17. ? Ans. 24. 40. How many grs. in 7 ib. ? Ans. 40,320. 41. How many grs. in 57 lb. 93.73.23. 17 grs. ? Ans. 333,117. 42. How many grs. in 275 lb. 113. 53. 13. 8 grs. ? 43. How many grs. in 2,134 kb. 93. 63. 20. 19 grs. ? The following are the denominations of DRY MEASURE. 1 pottle pot. 2 pottles 1 gallon 8 quarts 1 peck pk. 4 pecks 1 bushel bu. 4 bushels 1 coomb coomb. 2 coombs or 1 quarter 8 bushels qr. 36 bushels 1 chaldron “ ch. 5 quarters 1 load load. Note. This measure is used to measure grain, salt, seeds, oysters, coal. roots, lead ore, and all dry goods of his kind. 44. How many pints in a peck ? Ans. 16. 45. How many qts. in 13 bu. ? Ans. 416. 46. How many bu, in 58 ch.? Ans. 2,088. 47. How many pts. in 36 bu. 3 pks. 5 qts. 1 pt. ? Ans. 2,363. 48. How many pks. in 6,254 bu. ? How many qts. ? How many pts.? 49. How many pts. in 21,857 bu. 3 pks. 7 qts. I pt. ? 50. How many pts. in 578,656 bu. 2 pks. 1 qt. 1 pt. ? Ans. 37,034,019. 51. How many bu. in 629 ch.? 52 How many bu. in 235,079 ch. ? Aris. 8,462,8.14. 53. How many pks. in 21,953,205 qrs.? The following are the denominations of ALE OR BEER MEASURE. 1 gallon 1 firkin of ale fir. 9 gallons 1 firkin of beer fir. 2 firkins 1 kilderkin kil. 2 kilderkins 1 barrel bar. 3 kilderkins 1 hogshead lihd. 3 barrels 1 butt (of ale) buti. gal. 100 years Note. This measure is used in measuring ale, beer and milk. In London the Ale firkin contains only 8 gallons, while the Beer firkin contains 9. 08 course the higher measures differ in the same proportion. 54. How many gallons in the ale hogshead ? Ans. 48. 55. How many in the beer hogshead ? Ans. 54. 56. How many kils. in 15 butts ? Ans. 90. 57. How many qts. in: 13 Ale hogsheads ? Ans. 2,496. 58. How many pts. in 16 butts, 7 gal. 3 qts. of Ale ? Ans. 12,350. 59. How many pts. in 527 hhds. 6 gal. 2 qts. 1 pt. of Beer ? 60. How many pts. in 325 bar. 1 kil. 1 fir. 5 gal. 3 qts. 1 pt. of Beer ? The following are the denominations of TIME. 1 hour h. 24 hours 1 day d. 7 days 1 week 4 weeks 1 month mo. i { common or Julian year yr. 1 century. Note. The year is called in the table the Julian ycar, because the calendar was regulated by Julius Caesar, Emperor of Rome. The true solar year contains 365 days, 5 hours, 48 minutes, and 50 seconds, instead of 365 days, 6 hours, according to the table. The calendar months, that is, the months, by which wa reckon the year in the Almanac, or calendar, are inequal in length, but the number of days in each may easily be remembered by the following stanza. "Thirty days hath September, February's days are twenty nine.” But, besides the even days, we have seen that the Julian calendar allovs 6 hours to the year. 6 hours is the quarter of 24 hours, or of one day, and therefore, in four years, this excess will amount to another day. An allowance is made for this, by giving February 29 days, instead of 28, every fourth year, called Bissextilc, or leap year. But we have seen, also, that the true excess is 5 honrs, 48 minutes, 50 seconds, which is 11 minutes, 10 seconds, less than 6 hours. The Julian calendar, therefore, made the year 11 minutes, 10 seconds, too long. This would make 1 day in about 130 yeaze, or nearly 3 days in 400 years. Pope Gregory XIII observed this error and rectified it, by ordering three leap years in every 400 years to be considered as common years, and no additional day to be annexed to February. As every centurial year, (that is, year on which a century is completed, as 1700, 1800, &c ,) is a leap year, it was ordered, that three successive centurial year should be reckoned as common years, and the fourtha centurial year as a leap year. There is a slight error still , but it will hardly amount to a day in 4,000 years. Pope Gregory found, likewise, that, in consc quence of the error in the Julian Calendar, the months liad fallen back from their true places in the year. That is, since the year was made longer than it ought to have been, January, and of course, the other months, began later than they ought, on each successive year; so that, in his time, the first day of January came where the 10th should be. He, therefore, determined to restore the months to their places, by leaving 10 days out of the calendar, all at once. This was done in the month of October, the day after the 4th being counted the 15th, instead of the fifth. Catholic countries, of course, instantly adopted this reformation, but it was long before it was adopted in those states, which did not acknowledge the authority of the Pope. It was not adopted in England, until the error had become a day greater, being eleven days instead of ten. By act of Parliament, however, in 1752, 11 days were suppressed after the 2d of September, and, by the same act, the beginning of the year was transferred from the 25th of March to the Ist of Janury. The reckoning of dates by the old method is called Old STYLE, by the new, New STYLE. To any date reckoned by Old Style, therefore, we must add 11 days, to obtain the same date according to the New Style ; and, if the date, in Old Style, be between the 1st of January and the 25th of March, we must reckon the year, 1 greater : because, carrying back the begin. ning of the year, from the latter of those days, to the former, threw all the inter. mediate days into the next following year. Thus, the first of March, 1750, Old. Style, is the same as the 11th of March, 1751, Nero Style. Russia has not yet adopted the New Style, and the difference is now 12 days. This should be re. inembered in dates coming directly from that country, and allowance made. The pupil has, no doubt, often observed in the Almanacks, that a letter is put for Sunday, while the other days of the week are denoted by the figures 2, 3, 4, &c., or by the first two letters of their names. This letter is called the Dominical letter. The first seven letters of the alphabet. were used by the Primitive Christians, to stand for the days of the week in the calendar. They called the 1st of January, A, the second, B, the third, C, and so on, to the 7th, which was G. They then repeated the same letters again, calling the 8th, A; the 9th, B, and so on. The letter, which fell upon Sunday, was called the Dominical letter. At the present day, these letters are entirely disused in many calendars. Some, however, still retain the Dominical letter. In the following couplet, the initials. or first letters, of the several words, are the letters, which fall on the first days of the months in the year. Jan. Feb. Mar. Apr. May. June. July Aug. Sept. Oct. Nov. Dec. Good Carlos Finch And David Fryer." Of course, if I know the Dominical letter, I can easily tell on what day of the week, any dry of any month will happen. Thus, if I wish to know on what day of the week the 4th of March will happen, in 1830, when the Dominical letter is C, I observe that the word, Dwells, answers to March, which, therefore, comes in on Monday. The 4th day, then, will be Thursday. In leap year, on account of the additional day in February, two Dominical letters are used, one for Janua. ry and February, and the other for the rest of the year. The latter of these, is the letter, which precedes the former in the calendar. Thus, if the Domin. ical letter, with which leap year begins, be C, it will be changed to B, after the last day of February. In the same manner, B will be changed to A, A to G, G to F, &c. Years are reckoned, in all Christian countries, from the birth of our Saviour, which is called the Christian Era. 61. How many days in 9 (not calendar) mo. ? Ans. 252. 62. How many hours in a year of 365 d. 6 h. ? How many minutes ? How many seconds ? Ans. 8,766 h. 525,960 m. 31,557,600 sec. 63. How many seconds in a year of 365 d. 5 h. 48 m. 50 sec. ? Ans. 31,556,930. 64. How many seconds in a man's age, who is 52 yrs. old ; al. lowing 365 d. 6 h. to the year ? How many, allowing 365 d. 5 h. 48 m. 50 s.? Ans. 1,640,995,200 and 1,640,960,360. DUCTION. Note. For the answers to this example, the pupil may multiply the number of seconds in one year, (as found in ex. 62 and 63,) by 52, the number of years. 65. How many seconds from the commencement of the Christian Era, to the end of the year 1,830; allowing for the length of the year as in ex. 62 ? How many, allowing as in ex. 63 ? 66. How many seconds old is the world at the end of 1,830, 4,004 years having elapsed before the Christian era ; reckoning by each mode of allowance ? In the preceding examples, you have been changing numbers from one denomination to another. Changing numbers, in this way, is commonly called reducing them, and the process, by which they are reduced, is called Re Then, REDUCTION IS THE PROCESS OF CHANGING NUMBERS FROM ONE DENOMINATION TO ANOTHER, WITHOUT ALTERING THEIR VALUE. In the examples above, it has been required to reduce larger, or higher denominations, to smaller, or lower. This is called REDUCTION DESCENDING, and is performed, as we have seen by MULTIPLICATION. REDUCTION ASCENDING is the reverse of this, and is the process of changing numbers from lower denominations to higher. It is performed, as we shall see, by DiVISION. The general rule for Reduction Descending, may be expressed as follows. I. MULTIPLY THE HIGHEST DENOMINATION GIVEN, BY THE NUMBER RE. QUIRED OF THE NEXT LOWER, TO MAKE ONE OF THE DENOMINATION MULTIPLIED, AND ADD THE UNITS OF THE NEXT LOWER, IF ANY, TO THE PRO. DUCT. II. MULTIPLY THIS SUM, BY THE NUMBER, REQUIRED OF THE NEXT LOWER DENOMINATION STILL, AND ADD IN, AS BEFORE. III. PROCEED THUS, TILL YOU ARRIVE AT THE DENOMINATION REQUI. RED. EXAMPLES FOR PRACTICE. 67. If an ingot of silver weigh 3 lb. 6 oz. 13 dwt. what is it worth at 4 cts. pr dwt. ? Ans. 34.12. 68. In 11 lb. 6 oz. 19 dwt. 9 gr. how many grains ? Ans. 66,705. 69. What is the value of a silver cup, weighing 9 oz. 4 dwt. 16 gr. at 3 mills pr. gr. ? Ans. $13,296. 70. At 9 cts. a pound, what cost 3 cwt. 2 qrs. 16 lb. of sugar ? Ans. $36.72. 71. In 9th. 83. 13. 27. 19 gr. how many gr. ? Ans. 55,799. 72. At 27 cts. pr. nail, what cost 2 yds. 1 qr. 3 nls. of cloth ? Ans. $10.53. 73. How many sec. in 4 years, allowing 13 mo. 1 d. 6 h. to the Fear ? Ans. 126,230,400. 74. How many minutes are there, from the 13th July, at 43 minutes past 9 in the morning to the 5th Nov., at 19 m. past 3 in the afternoon. Ans. 165,936. 75. How many sec. in 8 yrs. 3 mo. 2 wks. 2 d. 19 h. 43 m. 57 sec., allowing 13 mo. 1 d. 6 h. to the year ? Ans. 261,171,837. 76. How many seconds old are you? Ans. 5,592,870. 78. In 19 lb. 13 oz, 7 dr. how many dr. ? Ans. 5,079. 79. In 90£. 17s. 9d. 1 qr. how many qrs. ? Ans. 87,253. 80. In 34 lbs. 5 oz. 19 dwts. 10 gr. how many gr. ? A. 198,706. 81. In 27 th. 93. 23. 10. 16 gr. how many gr. ? 82. At 12 cts a grain, what must be paid for a gold cup, weighing 1 lb. 3 oz. 13 dwt. 17 gr. ? 83. At 6 cts. a pint, what cost 27 bu. 3 pks. 7 qt. 1 pt. of wheat ? 84. At 5 cts. a pirt, what cost 16 butts, 2 bar. I kil. 1 fir. 7 gal. 3 qts. 1 pt. of ale ? 85. In 16 yrs. 24d how many sec. allowing 365 d. to the year ? Ans. 506,649,600. 86. A merchant sold 84 yds. 3 qrs. 2 na. of broadcloth at $0.50 a nail. In return he received 25 boxes of raisins at $4.00 a box, and the rest in flour at $11.00 a barrel. How much flour did he receive ? 87. A farmer sold 6 loads of wood, at $3.00 a load ; and 3 cwt. of butter, at 9 cts. a pound; and 70 bushels of wheat, at $2.125 a bushel ; and 93 bushels of rye, at 75 cts. a bushel. In return he received 3 ploughs, at $12.00 a piece ; 1 yoke of oxen, $60.00; 1 pr. of horses, at 75 dollars apiece ; and the rest in money. How much money did he receive ? Ans. $20.74. NOTE. The last two questions belong to the rule, commonly called BARTER. OBSERVATIONS ON MULTIPLICATION, FOR ADVANCED PUPILS. $ XIX. The order of writing numbers for multiplication, requi. -red by the rule, is only adopted for the sake of convenience. We need some certain method, and this is as good as any. It is manifestly of no consequence which of the figures of the multiplier is first used, since if the partial products are set down correctly, the answer will be the same, whether we begin on the left, on the right, or somewhere in the middle. In order to multiply with ease, it is necessary that the several products of all the possible combinations of two digits, should be fixed in the memory; exactly as in Addition, it was found necessary to retain the sums of the same combinations in the mind. A little reflection will show, that it is even more necessary to be acquainted with a multiplication table, than with a table for addition. For in adding, we might add numbers unit by unit, as children do; but in multiplying, this process would become very tedious, and liable to error, The following table, generally supposed to have been in. |