The reason of this rule, which is of frequent use in Arithmetic, and other branches of the Mathematics, is obvious from the principles delivered in page 82. Exam. A farm of 97 a. 3 r. 5 p. is to be divided into two parts, such that the one may be three-fourths of the other. What are the parts? Here, the parts are evidently in the ratio of 4 and 3, the sum of which is 7. Therefore, as 7: 4 :: 97 a. 3 r. 5 p. : 55 a. 3 r. 20 p., the greater part; and as 7:3:: 97 a. 3 r. 5 p. : 41 a. 3 r. 25 p., the less part. The sum of these parts is 97 a. 3 r. 5 p., which proves the operation. Ex. 1. Suppose a traveller to proceed from Belfast for Killarney at the rate of 6 miles hour, and another at the same time from Killarney for Belfast at the rate of 5 miles hour: where will they meet, the distance between the two places being 224 miles? Answ. 122 miles from Belfast. 2. Divide 398 into three parts which will be to one another as the numbers 5, 7, and 11. Answ. 8613, 121, and 190,. 3. Divide 80 miles into four parts in the ratio of 10, 9, 8, and 7. Answ. 23, 21, 1814, and 16. 4. Divide £5000 among three persons in such a manner, that the share of the second may be one-half greater than that of the first, and the share of the third, one half greater than that of the second. Answ. £1052 12 71, £1578 18 11, and £2368 8 5. In this exercise it is easy to see, that the parts will be as the numbers 1, 1, and 24, or as 4, 6, and 9. 5. Pure water is composed of two gases, or kinds of air, called oxygen and hydrogen, in such proportions, that the weight of the former is to that of the latter as 15 to 2 Required the weight of each contained in a cubic foot, or 1000 ounces, avoirdupois weight, of water. Answ. 882 oz., and 1171 oz. 6. How much copper and how much tin will be required to make a cannon weighing 16 c. 1 q. 20 lbs., gun-metal being composed of 100 parts of copper, and 11 of tin? Answ. 14 c. 3 q. 5, and 1 c. 2 q. 143 tbs. bs., 7. The British standard gold for coinage consists of 11 parts of pure gold, and 1 part of alloy; (usually a mixture of silver and copper:) how much pure gold and how much alloy are contained in a guinea? (See exercise 77, page 80.) Answ. 4 dwts. 221 gr. & 10 grs. 8. The British silver coin consists of 37 parts of silver and 3 of copper; how much of each does the half crown (2/6) contain, each pound, troy weight, being coined into 66 shillings? Answ. 8 dwts. 9 grs., and 16 grs. 9. How much tin and copper are contained in a bell weighing 150 lbs, bell-metal being composed of three parts of copper and 1 of tin? Answ. 112 its., and 371⁄2 lbs. 10. Pewter is composed of 112 parts of tin, 15 of lead, and 6 of brass; how much of each ingredient is requisite to make a ton of pewter? Answ. 16 c. 3 q. 10 lbs., 2 c. 1 q. 013 lb. and 3 q. 17, fbs. 11. Proof spirits are composed of 48 parts of alcohol, or pure spirit, and 52 parts of water. How much of each of these is contained in 84 gallons of proof spirits? Answ. 40 gal. and 431} g. 12. 76 parts of nitre, 14 of charcoal, and 10 of sulphur, compose gun-powder: how much of these ingredients will be requisite to form a hundred weight of powder? Answ. 3 q. 125 lb. 1513 tbs., and 113 lbs. The following questions are illustrative of some applications of this rule which are often very useful to the Mathematical student. They are not useful however, nor even intelligible, to the mere Arithmetical pupil. By him, therefore, they should be omitted. Exam. 2. Required the natural sine of 54° 35′ 43′′, those of 51° 35′ and 54° 36′ being 8149593 and 8151278. Here the difference of the two sines is 0001685; then, as 1' or 60", (the difference of the arcs whose sines are given):43" (the excess of the intermediate arc above the less,) :: 0001685 0001208, the part of the difference corresponding to 43′′; and the sum of this and the less sine is 8150301, the sine required. The ciphers might have been rejected, and 1685 used as a whole number, and the result 1208 added to the smaller sine, as if it had been a whole number.-Had the second of the given numbers been less than the first (as is the case in the cosines,) the result of the analogy must evidently have been subtracted. Ex. 13. The logarithm of 3.1415 being 4971371, and that of 3.1416, 4971509, required the logarithm of 3-141593. Answ. 4971499. 1. The logarithmic tangent of 23° 27′ is 9·6372646, and that of 23° 28' is 9.6376106: required that of 23 27′ 5. Answ. 9.6375760. 15. Required the sun's declination at Greenwich on the 10th of July 1818, at 35 minutes past 7 o'clock in the evening, his declination on the 10th at noon being 22° 18′ 47′′, and on the 11th at noon 22° 11' 10" Answ. 22° 16′ 23′′. 16. The moon passed the meridian of Greenwich, on the 10th of October 1818, at 36 minutes past 9 o'clock in the evening, and on the 11th at 23 minutes past 10; at what time did she pass the meridian of Mexico, in longitude 99o 5′ 15′′ W.? Answ. 49 minutes past 9 o'clock. on the 10th. 17. On the 19th of August, 1818, the declination of Venus, at noon, at Greenwich, was 2° 0' S., and on the 25th at noon 5o 5'S. Required the declination on the 23d at noon, in the island of Owyhee, in longitude 155° 58′ 45′′ W. Answ. 4 16' 41" S. FELLOWSHIP. FELLOWSHIP is the method of determining the respective gains or losses of the partners in a mercantile company. Fellowship is usually distinguished into two kinds, Simple and Compound, or Single and Double. In SIMPLE or SINGLE FELLOWSHIP the stocks or sums contributed by the several partners, all continue in trade for the same time. In COMPOUND or DOUBLE FELLOWSHIP the stocks continue in trade for different periods. SIMPLE FELLOWSHIP AND BANKRUPTCY. RULE I. As the whole stock is to the whole gain or loss, so is the stock of any partner to his gain or loss. In the same way, the estate of a bankrupt may be divided among his creditors by this analogy: As the sum of all the claims on the estate is to its value, so is the claim of any creditor, to his dividend or share of the estate. This rule is merely a particular application of that contained in the last article, and therefore requires no separate illustration. The method of proof is also the same. Exam. 1. Three merchants, A, B, and C, form a joint capital, of which A contributes £700, B £1000, and C £1600. What is the share of each in a gain of £880? Here, the sum of the stocks is £3300, the whole capital. Then, as £3300 : £880, or by contraction, as 15:4 :: £700: £186 13 4, A's share; and as 15:4:: £1000: £266 13 4, B's share; and lastly, as 15:4: £1600: £426 13 4, C's share. The sum of these shares is exactly £880, which proves the operation to be correct. Exam. 2. A bankrupt owes to A £900; to B £860; to C £640; to D £150; to È £70; and to F £30; but his whole estate amounts only to £1250. Required the share of each creditor. Here the sum of the debts is £2650. Then, as £2650: £1250, or by contraction, as 53: 25:: £900 £424 10 62, A's dividend; and as 53: 25:: £860: £405 13 2, B's dividend. In the same manner we find C's share to be £301 17 83; D's £70 15 1; E's £33 0 4; and F's £14 3 04. The sum of all these is £1249 19 111, a farthing being lost by neglecting the remainders. In the division of a bankrupt's estate, it is usual first to find how much in the pound he can pay, that is, how much the creditors will receive for each pound of their respective claims. Thus, resuming the same example, we have this analogy: as £2650: £1250, or as 53 : 25 : : £1: 9/5 nearly, the sum that each creditor is to re ceive in the pound. Then, by using the method of aliquot parts as in the margin, A's share will be found to be £423 15 0, agreeing nearly with the result already found, the difference arising from taking 9/5 in £900 at 9/5 pound. 5/£...... 225 £423 15 0 stead of 9/513. In the same way the rest of the dividends would be found. The following rule, which may be readily employed by a person who has not learned a regular course of decimal fractions, will be found to be preferable, in a considerable degree, to that given above, in the more complicated questions in Fellowship; and it is evident, that it is equally applicable in every case of dividing into parts in a given ratio. RULE II. (1.) Reduce the whole stock and gain to the same denomination, if they be not so already: (2.) To the latter annex six ciphers, or more if the stocks be large, and divide by the former; the quotient will be a decimal: (3.) Multiply this decimal by the several stocks successively, taking aliquot parts for shillings and pence; and from each product cut off as many figures as there were ciphers annexed: (4.) Value the figures cut off by the method shown in Reduction of decimals, problem II. Exam. 3. If a bankrupt, whose property amounts to £2100, owe to A £826 12, to B £1263 9 6, to C £724 15 10, to D £1000, and to E £242 16 4; how much can he pay in the pound, and what is the dividend of each creditor ? ⚫5175367 In this example, the sum of the debts is £4057 13 8; and the pence in this and in £2100 are 973844 and 504000. Annexing ciphers to the latter, and dividing by the former, we obtain 5175367, the value of which is 10/44 nearly, the sum which he can pay in the pound. We then, as in the margin, multiply 5175367 by the first debt £826 12, and obtain for A's dividend £427 15 11. By proceeding in a similar manner, we should find for B's dividend £653 17 102, for C's £375 2 1}; for D's 517 10 82; and for E's £125 13 4. The sum of al these is £2100, which proves the correctness of 10/= £ 31052202 10350734 41402936 .... 2587683 517537 £427-7958362, or £427 15 11. the operation. The learner who has studied decimal fractions, K will see that in finding the decimal, besides other contractions, figures may often be cut from the divisor, and a smaller number of ciphers annexed than are prescribed in the rule. Several of the following questions are expressed briefly, and it may be a useful exercise for the pupil to write them out at large. Ex. 1. A's and B's stocks are £375 and £425 respectively required the share of each in a gain of £240. Answ. A's £112 10, B's £127 10. 2. Three merchants, A, B, and C, enter into partnership: A puts into the joint stock £329, B £289 10, and ̊C £317.` Required the share of each in a gain of £583 12 6. Answ. A's £205 5 04, B's £180 12 2, C's £197 15 33. 3. A's stock £1750, B's £1250: whole gain £565 12. Answ. A's £329 18 8, B's £235 13 4. 4. A's stock £349 16 7, B's £520: whole gain £346 18 9. Answ. A's £139 10 72, B's £207 8 14. 5. A's stock £384 18 whole gain £396 13 7. C's £232 15 04. 6. A's stock £348 whole gain £795 18 8. C's £278 14 9. 104, B's £186 17 4, C's £811 17 63: Answ. A's £110 7 1, B's £53 11 51, 16 6, B's £804 11 4, C's £621 12 2: Answ. A's £156 8 44, B's £360 15 62 7. B's stock £595 12 8, C's £701 11 6: whole gain £588 1 9. Answ. B's £270 0 7, C's £318 1 2. 8. X's stock £448 19 3, Y's £582 13 4, Z's £261 14: whole gain £718 18. Answ. X's £249 13 7, Y's £324 0 8§ 7's £145 3 8. . 9. M's stock £475 15 8, N's £346 12 4, O's £396 17 6: whole gain £279 10. Answ. M's £108 19 3, N's £79 7 73, O's £90 17 104. 10. A's stock £178 18 8, B's £236 15 8, C's £493 18 8, D's £213 17 6: whole gain £583 10 9. Answ. A's £92 18 81, B's £122 19 7†, C's £256 10 91⁄2, D's £111 1 73. 11. A debtor, the value of whose effects is only £1075 12 6, owes to A £586 13 7, to B £348 10, to C £674 5, and to D £1000. What is the dividend of each, and how much is paid in the pound? Answ. A's £241 16 8, B's £143 13 1, C's £277 18 74. D's £412 4 1; and 8/22 in the pound. 12. A, B, C, and D enter into partnership, and A puts in 987 gallons of whiskey at 9/10 gallon; B, 789 gallons of rum at 16/1 C, 238 gallons of brandy at £1 4 9, and 183 gallons of geneva at £1 1 11; nd D 497 gallons of wine at 18/44: and they contribute besides £120 in money, in equal shares. Required the share of each in a gain of £318 10 8. Answ. A's £75 1 41, B's £96 13 24, C's £76 3 94, D's £70 12 3. 13. A, B, and C, enter into partnership with a joint stock of £7500, of which £3600 belong to A, £3000 to B, and the remainder to C. At the end of a year, the gain is found to be £1679 4. Required the share of this gain which each is to re |