Ex. 14. 1457924651÷1204........ 15. 28101418481÷1107. 19. 347382600435-727...... ........... 34. 31461738478379387.......... =3351628687 :623768798478871 :965711266756148 =8403898585064 =982108215759- 35. 555777999444777÷891 47. 1000000000000000÷111........ =900900900900911 900090009000??? 48. 1000000000000000÷÷÷-1111...... 50. 555555555555÷123456 ......... 51. 555555555555÷÷654321 54. 3784926474826-384365........ .......... 59. 167817938473÷87700........... 98787 -4500028123456 84905684821 8264517040 13878059l f 9847219113! =2227704461 3 3 =1007720388 30572824223 -17535833786 8 19135454 -60. 267817938115-1360000........= 1969241388 61. 367817938473-87000............ 422779284133 62. Devis, the highest mountain in the neighbourhood of Belfast, is 1542 feet high; and Mont Blanc 15,680 feet: how many mountains, each as high as the former, must be piled one above another to equal the height of the latter? Answ. 10. 63. In 1821, the population of Belfast and the suburbs, was about 40,000, and that of London was 1,274,800: how many towns, each containing as many inhabitants as Belfast, must be combined, to form one containing the same population as London? Answ. 313888. 64. The linen exported from Ireland in 1809, was 43,904,382 yards of how many pieces, each containing 25 yards, did this quantity consist? Answ. 1,756,175,7. 65. The entire quantity of tea sold by the East India Company in 1799, was 24,853,503 pounds: how many chests, each containing 87 pounds, would this quantity fill ? Answ. 285,67284. 66. If it be supposed, as in common circumstances is found to be nearly true, that as many persons die in 33 years as are equal to the entire population, it is required to find how many persons die each year, at an average, in the British dominions in Europe, the population (in 1821,) being 21,481,152. Answ. 650,944. 67. How many lessons of ninety-five lines each, are contained in Virgil's Eneid, the number of lines contained in that poem being nine thousand, eight hundred, and ninety-two? Answ. 10413. 68. The following is the quantity, in hundreds weight, of the butter exported from Ireland, from 1800 till 1811 inclusive: required the quantity exported each year, at an average: in 1800, 263,288, in 1801, 178,496; in 1802, 304,666; in 1803, 396,353; in 1804, 334,251; in 1805, 320,155; in 1806, 294,415; in 1807, 338,508; in 1808, 333,998; in 1809, 346,856; in 1810, 385,953; in 1811, 390,833. Answ. 323,981. 69. The following is the quantity, in pounds, (avoirdupois,) of cotton-wool imported into England, from 1805 till 1810 inclusive: required the quantity imported each year, at an average: in 1805, 59,862,406; in 1806, 58,176,283; in 1807, 74,925,306; in 1808, 43,605,982; in 1809, 92,812,282; in 1810, 136,570,103. Answ. 77,658,727, ABBREVIATIONS IN THE FUNDAMENTAL RULES. THE rules already given for performing the fundamental operations in Arithmetic, are of a nature completely general, and are fully adequate to the performance of all operations that can occur. Hence the pupil should be made substantially acquainted with them, before he proceeds to any thing else. As the operations, however, are often of a tedious and laborious nature, it is desirable to be able to employ easy and expeditious methods of performing them, when such can be obtained Some of the most usefu! of these will be found in what follows. Others might have been added, but not so easy or useful, or so frequently applicable. With respect to those which are given, it will be well to proportion to the capacity and proficiency of the pupil, the number of them which he will be required to learn. 41257 63783 56558 38416 The principal abbreviation admissible in Addition, is the adding of two or more figures at once. This is peculiarly convenient, when the sum of two or more figures is exactly ten. Thus, in the annexed example, we may say 10 and 14 are 24, and 10 are 34. Then, carrying 3, we may proceed, 10 and 10 are 20, and 10 (5+5) are 30, and 8 are 38. 10 (3+6+1) and 16 (4+5+7) are 26, and 2 are 28. 14 (2+5+7) and 14 (8+6) are 28, and 4 are 32. 14 (3+4+7) and 14 (3+5+6) are 28, and 4 are 32. When, by attention and practice, facility in this mode of proceeding is acquired, it will be found to be of more value than might at first be imagined. 77198 45672 322884 384729 X 15 5770935 To multiply by 5: Add a cipher, or rather conceive it to be added, to the multiplicand, and take half the result. This is evidently the same as multiplying by 10, and taking half the product. To multiply by 15: Conceive a cipher to be added to the multiplicand, and to the result add half of itself. To multiply by 25: Conceive two ciphers to be added to the multiplicand, and take one-fourth of the result. This is the same as multiplying by 100, and taking one-fourth of the product, which is done because 25 is one-fourth of 100. In like manner, because 125 is one-eighth of 1000; to multiply by 125: Conceive three ciphers to be added to the multiplicand, and take oneeighth of the result. To multiply by 75: Conceive two ciphers to be added to the multiplicand, and from the result take one-fourth of itself. To multiply by 35: Conceive two ciphers to be added, take one-fourth, and set it so that it may be added to the multiplicand, with one cipher annexed. In a similar manner, the product by 225 may be found from the product by 125. To multiply by 175: Divide 700 times the multiplicand by 4. To multiply by 275: Divide 1100 times the multiplicand by 4. 587 X 75 14675 44025 463 X 35 11575 16205 To multiply by 9: Conceive a cipher to be added to the multiplicand, and from the result subtract the multiplicand; or, simply, conceive a cipher to be added, and from each figure of the result take the one before it. Thus, 5 from 10 and 5 remain: and 7 are 8; 8 from 15, and 7 remain: 1 and 4 are 5; 5 from 7, and 2 remain: 6 from 14, and 8 remain: carry 1; 1 from 6, 6475 X 9 58275 6483 X 11 71313 1438 X 99 142362 and 5 remain. To multiply by 11: Conceive a cipher to be annexed to the multiplicand, and to the result add the multiplicand; or, simply, comceive a cipher to be annexed to the multiplicand, and to each figure of the result add the one which immediately precedes it. Thus, 3 and 0 are 3: 8 and 3 are 11: carry 1; 1 and 4 are 5, and 8 are 13: 1 and 6 are 7, and 4 are 11: 1 and 6 are 7. To multiply by 99: Conceive two ciphers to be annexed to the multiplicund, and from the result subtract the multiplicand; or, simply, conceive two ciphers to be annexed to the multiplicand, and from each figure in the result sub tract the figure nearest it on the left side, except one. Thus, 8 from 10, and 2 remain: 1 and 3 are 4; 4 from 10, and 6 remain: 1 and 4 are 5; 5 from 8, and 3 remain: 1 from 3, and 2 remain: 0 from 4, and 4 remain: 0 from 1, and 1 remains. To multiply by 101: Conceive two ciphers to be annexed to the multiplicand, and to the result add the multiplicand; or, simply, conceive two ciphers to be annexed to the multiplicand, and to each figure of the result add the figure nearest it on the left side, except one. Thus, 8 and 0 are 8; 3 and 0 are 3; 9 and 8 are 17; 1 and 5 are 6, and 3 are 9; 0 and 59 are 59. In general, to multiply by 9, 99, 999, &c.: To the multiplicand annex as many ciphers as there are 9's, and from the result subtract the multiplicand: and to multiply by 11, 101, 1001, &c: To the multiplicand annex as many ciphers as there are digits in the multiplier, wanting one, and to the result add the multiplicand. 5938 X 101 599738 The principle on which these last rules depend, may be generalized in its application in the following manner: If the multiplier be a little less than 100, 1000, or any other number expressed by a unit with ciphers annexed, which number may be called the approximate multiplier; or if it be a little greater than one of these numbers, take the difference between it and whichever of them is nearest to it; in the former case, call this difference the complement, in the latter, the excess; then, to the multiplicand annex as many ciphers as there are in the approximate multiplier; and in the one case subtract from the result the product of the multiplicand and the complement; in the other, add to the result the product of the multiplicand and the excess.* It is obvious that this rule will be principally useful, when the complement or excess does not exceed 12. It may indeed be employed with considerable facility, when the complement or excess is between 12 and 20, if the pupil have committed to memory the additional part of the Multiplication Table given in the note in page 17, or by means of the method of multiplying by 13, 14, &c. illustrated in the following example: Here, 7 times 9 are 63; carry 6; 7 times 2 are 14, and 6 are 20, and 9 the following figure,) are 29; carry 2: 7 times 3 are 21, and 2 are 23, and 2, (the following figure,) are 25; carry 2: 7 times 7 are 49, and 2 are 51, and 3 are 54; carry 5: 7 times 0 is 0, but 5 and 7 are 12. It is obvious that we may multiply by any number ve 7329×17 124593 673849 168428 2695396 18867772 113206632 Another abbreviation, which may perhaps be more frequently useful than any of the above, consists in deriving a product from others already found; and in doing this, and indeed in all cases, it should be considered, that it is a matter of indifference by which figure of the multiplier we multiply first, by which secondly, &c. provided we assign to the partial products their proper places. In the annexed example, where the multiplier is 168428, the product by 4 is first found; this product is then multiplied by 7, to find the product by 28, because 28=4X7. The product thus found for 28 is then multiplied by 6, to find the product by 168, because 168 28 X 6. The sum of these partial products, properly arranged, is the product required. Even when no farther abbreviation can be employed, it sometimes renders the work less laborious to derive the product for a single digit from one found already: thus, the product by 6 will be found by doubling the product by 3; the product by 8 is 4 times the product by 2, &c. It is scarcely necessary to say, that when the same digit occurs twice, or oftener, in the multiplier, it is sufficient to multiply once by it, as the product thus found may be copied, when that digit recurs in the multiplier. 113495039372 To divide by 5: Double the dividend, and cut off the last figure of the result, the half of which figure will be the remainder. This is obviously nothing else than doubling the dividend, and dividing by 10. To divide by 15, 35, 45, or 55: Double the dividend, divide the result by 30, 70, 90, or 110, respectively, and for the true remainder take half the remainder thus found. To divide by 25: Multiply by 4, cut off two figures from the result, and take one-fourth of the number expressed by them for the remainder. To divide by 125: Multiply the dividend by 8, cut off three figures from the result, and take one-eighth tween 20 and 30, by multiplying by the units' figure, and adding each time, along with what is carried, twice the following figure of the multiplicand; and the principle ma be readily extended to numbers above 30, but the operation becomes more complicated and difficult. C3 |