The principal units of the French decimal system of weights and measures, with the British equivalents, are The unit of length or metre=39.37079 inches, or 3-2808992 feet, or 1·093633 yards. surface or are=119.6033 square yards=0247114 acres. volume or stere=1.308022 cubic yards. capacity or litre=61.027052 cubic inches=22009687 gallons. weight or gramme=15·434 grains troy. money or franc-9.69 pence. The kilogramme=1000 grammes=2.679514 lbs. troy or 2-204857 lbs. avoirdupois. The toise, a measure of length,=76-73511 inches, or 6·3945925 feet, or 2.1315308 yards. From these relations it is found that 1 inch 02539954 metres. 1 foot=3047945 metres. 1 cubic foot=028315301 stere. 1 shilling=1.238390093 franc. 1£. 24.76780186 francs. To show the application of these relative numbers, let it be required, 2d. To reduce 1 cwt. to kilogrammes. 1 cwt.=112 lb. (112 × 45354414) kilogr.=50·79694368 kilogr. 3d. To express 1 mile, in metres. 1 mile 1760 yds.=(1760 × 91438353) metres=1609-3150128 metres. 4th. Reduce 1000 ares to acres, &c. ?.....Ans. 24 ac. 2r. 33 p. 24 y. 8-334 ft. 317. By the rules of reduction a compound number can be expressed in terms of any single denomination contained in the weight or measure to which that compound number belongs. Therefore compound numbers, upon which any of the elementary operations of arithmetic are to be performed, may be reduced to the same denomination; and the calculation being made by the rules for whole or fractional numbers, the form of expression of the result, as a compound number, may be restored. But these reductions are tedious, especially when the numbers are large. Besides, any calculation can, in general, be more easily made with the unreduced compound numbers. Every denomination of a compound quantity, individually considered, is referred to the decimal system; hence any calculation, made with numbers of one denomination, is made as with whole numbers. The several denominations also are so arranged, that the lowest being written first, the others are made to occupy successive places on the left of the lowest and of each other, according to their relative values. The relation of these values being fixed by an arbitrary weight or measure, which involves the principle (common to it with other numbers) that a certain number of units of the lowest denomination make one unit of the second, which denomination is written immediately on the left of the lowest; a certain number of units of the second make one unit of the third, which is written on the left of the second, &c. &c. Whence, if each denomination is considered as forming a distinct order of units, it is evident that the only difference between an arithmetical operation performed upon whole and upon compound numbers lies in the different manner of passing from one order of units to another. In whole numbers the transitions are by tens and tenths; in compound numbers by multiples and parts, which vary from one denomination to another, and which are fixed for each denomination by the table of the weight or measure to which a given compound number belongs. The elementary operations of arithmetic, with whole numbers and fractions, are Addition, Subtraction, Multiplication, and Division. The same may be performed with compound numbers. ADDITION OF COMPOUND NUMBERS. 318. Since any arithmetical operation with numbers composed of different denominations differs from the same operation with numbers composed of different orders of units in respect only that 1 is carried to a higher denomination, not for 10 of the next lower, but for some other number fixed by the table of an arbitrary weight or measure, it follows that, in all other respects, the addition of compound is made like that of whole numbers. Examples of the addition of compound numbers. 1st. What is the sum of 3 cwt. 3 qrs. 12 lbs.; 5 cwt. 1 qr. 24 lbs; 2 cwt. 3 qrs. 18 lbs.; and 6 cwt. 2 qrs. 19 lbs. ? In forming the sum of compound numbers it is convenient, as in the case of whole numbers, to begin with the lowest denomination; in this instance lbs. avoirdupois. (19+18+24+12) lbs.=73 lbs. As 28 lbs. 1 qr. 27 is the greatest number which can be properly written in the place of lbs., the sum of the lbs., 73, must therefore be reduced to qrs. and lbs. Dividing by 28 (the number of lbs. which make 1 qr.), it is found that 73 lbs. are equal to 2 qrs. and 17 lbs. The 17 lbs. are reserved, as expressing the lbs. of the result, and the qrs. are carried to the denomination of qrs. Next, 2 qrs. carried+(2+3+1+3) qrs.=11 qrs. 4 qrs. 1 cwt...11 qrs.=2 cwt. 3 qrs. The 3 qrs. are reserved for the qrs. of the sum, and the 2 cwts. are carried to the denomination of cwts. Lastly 2 cwt. carried +(6+2+5+3) cwt.=18 cwt. With cwts. the calculation ends; and the sum of the given compound numbers is 18 cwt. 3 qrs. 17 lbs. The process of addition being explained, it remains to be noticed, that if the different denominations of any given compound numbers are regarded as different orders of units, the arrangement of these numbers for addition may be made by the rule of Article 23. The arrangement and sum of the preceding compound numbers are as follow; cwts. qrs. lbs. 3 3 12 5 1 24 2d. What is the sum of 5 lb. 3 oz. 18 dwt.; 16 lb. 11 oz. ; 283 lb. 15 dwt. 9 gr. ; 9 oz. 16 dwt. 54 gr.; and 2 lb. 8 oz. 178 gr. N 178 The most direct course is, first, to arrange the given numbers as in the It is often convenient and last example, and second, to find their sum. Of sometimes necessary to make certain changes or reductions of the given numbers to render them capable of a proper arrangement for addition. the first description is the writing of zeros to occupy the places of grains in the first and of dwts. and grs. in the second of the given numbers. Of the second, are the writing of zeros to occupy the places of the deficient ounces in the third and the deficient dwts. in the last of the given numbers, and the reduction to the same denominator of the fractions gr. and gr. in the fourth and fifth of the given numbers. The more distinctly to exhibit these reductions, and the use or cause of making them, the given and the reduced numbers are subjoined in horizontal lines, the sum being found by addition of the latter. 5 lb. 3 oz. 18 dwt. 283 lb. 15 dwt. 9 gr. lb. oz. dwt. gr. 0 9 oz. 16 dwts. 5 gr. 2 lb. 8 oz. 17 gr. Detail of the addition. (+3) gr.= gr.=14 gr; is written under the fractions, and 1 carried to the grains. (1+17+5+9) gr.=32 gr.=1 dwt. 8 gr.; 8 is written under the grains, and I carried to the dwts. (1+16+15+18) dwt.=50 dwt.=2 oz. 10 dwt.; 10 is written under the dwts., and 2 carried to the ounces. (2+8+9+11+3) oz.=33 oz.=2 lb. 9 oz.; 9 is written under the ounces, and 2 carried to the pounds. (2+2+283+16+5) lb.=308 lb. ; the number 308 is written under the pounds; and with the addition of this, the last denomination, the process is ended. 3d. What is the sum of 34 £. 6 sh. 83 d. ; 8 £. 154 sh. ; 59 £. 17 sh. 65 d. ; 128 £. 5·324 d. ; and 538 £. 19 sh. 7 d. ? Reduction and arrangement of the given numbers. £ sh. d. qrs. 8 3 8 15 2.4 59 £. 17 sh. 6 d. 59 17 128 £. 5.324 d. 128 0 5 1.296 538 £. 19 sh. 7 d. Remarks on this example. 770 0 1 The farthings in this example are written as a distinct denomination. They are so written because fractional parts of a farthing occur in some of the addenda. When this does not happen, farthings are expressed as fractional parts of a penny. In the second of the given compound numbers sh. is reduced to its value in pence and farthings. This is a necessary reduction; for fractional parts of a shilling cannot be left unreduced in any compound number which contains pence as well as shillings. of a farthing are found to form part of the value of sh. This qr. is reduced to a decimal fraction, because the fourth of the given numbers contains a decimal fraction, and it is required to find the sum of the fractional parts as well as of the several denominations of the given numbers. In the third of the given compound numbers d. is reduced to farthings and decimal parts of a farthing. In the fourth, zero is written to fill the place of the deficient denomination, shillings; and 324 d. is reduced to farthings and decimal parts of a farthing. In the last, d. is written as 2 qrs. When these reductions have been made, the compound numbers arranged as in the preceding examples, and the decimal fractions as directed Art. 226, the addition is commenced with the decimals. The sum of the decimal fractions is 1.196 qrs.; of this 196 is written under the decimal figures of the addenda, and 1 qr. carried to the farthings. The sum of the column of farthings is 10, which, with 1 qr. carried,= 11 qrs. 2 d. 3 qrs. 3 is written under the column of qrs. and 2 carried to the column of pence. The sum of the column of pence is 35, which, with 2 d. carried,=37 d.= 3 sh. 1 d. 1d. is written under the column of pence, and 3 sh. carried to the column of shillings. The sum of the column of shillings is 57, which, with 3 sh. carried,=60 sh. =3 £. 0 sh. O is written under the column of shillings, and 3 £. carried to the column of pounds. The sum of the column of pounds is 767 £., and 767 £.+3 £. carried= 770 £., which sum is written under the column of pounds. Therefore the sum of the given compound numbers is 770 £. 0 sh. 1 d. 3.196 qrs. Rule for the addition of compound numbers. Replace fractional parts of any denomination higher than the lowest which is contained in the given numbers by the values of such fractional parts expressed in terms of these lower denominations, and fractions whose denominators are different by equivalent fractions reduced to the same denominator, and render the lowest denomination of all the compound numbers the same by writing zeros in the places of such denominations as are wanting. Then arrange the addenda in such a manner that the numbers of the lowest denomination may fall in the same vertical column; the numbers of the denomination immediately superior to the lowest in a second vertical column on the left of the former; and so on, to the highest denomination, which will be on the left of all the others. Find the sum of the numbers, whole or mixed, in the right column; reduce this sum to a compound number expressed in terms of the last and the next higher denominations; write the whole or mixed number which expresses the units of the last denomination directly under the column to which it belongs, and carry the units of the next higher denomination to the second vertical column. Find the sum of the numbers contained in the second column plus the units carried from the first; reduce this sum to a compound number expressed in terms of the second and the next higher denominations ; write the units of the second denomination in their proper place; and carry the units of the third denomination to the third vertical column. And in this manner proceed till the sum of all the columns has been found. The result is the sum of the given compound numbers. Exercises in addition of compound numbers. 1st. Required the sum of 74 £. 18 s. 113 d.; 96 £. 9 s. 101 d.; 58 £. 17 s. 8 d.; and 63 £. 11 s. 9 d.? Ans. 293 £. 18 s. 4d. 2d. Required the sum of 84 £. 18 s. 11 d.; 63 £. 16 s. 104 d. ; 95 £. 4 s.; and 78 £. 10 s. 93 d ?....Ans. 322 £. 10 s. 7 d. 3d. Required the sum of 74 tons 19 cwt. 3 qrs. 27 lb. 15 oz.; 85 tons 17 cwt. 2 qrs. 24 lbs. 14 oz. ; 68 tons 13 cwt. 1 qr. 20 lb. 12 oz. ; 52 tons 18 cwt. 3 qrs. 19 lb. 8 oz.; 50 tons 10 cwt. 2 qrs. 18 lb. 6 oz.; 48 tons 9 cwt. 3 qrs. 16 lb. 10 oz.; and 97 tons 5 cwt. 1 qr. 3 lb. 15 oz.? Ans. 478 tons 15 cwt. 3 qrs. 20 lbs. 21 grs.; 4th. Required the sum of 48 lbs. 11 oz. 18 dwts. Ans. 261 lbs. 10 oz. 14 dwts. 12 grs. 5th. Required the sum of 26 ac. 3 ro. 23 po. 204 yds.; 40 ac. 6th. Required the sum 7th. Required the sum of 24 ac. 5067 sq. yds. and 50000 sq. links?......... Ans. 25 ac. 3 ro. 20 po. 25 yds. 8th. Required the sum of 25 hhds.; 11591 gal.; 13 pipes; and 4728 qts. ?......... Ans. 88 hhds. 53 gals. 2 qts. 13 pts. 9th. Required the sum of 25 cwt. 2 qrs. 17 lbs. ; 6 cwt. 86 lbs. ; and 15 cwt. 3 qrs. 12·7 lbs. ?...Ans. 48 cwt. 1 qr. 3·7 lbs. 10th. Required the sum of 126·75 £.; 13·025 sh.; 16£. 17·75 sh.; and 11.98 d.? ......Ans. 144 £. 6 sh. 2·32 d. 11th. Required the sum of 2.15 £. ; 31·017 £.; and 19 sh. 114 d. ? Ans. 34 £. 3 sh. 3.33 d. 12th. Required the sum of £. and 2 crown (1 crown=5 sh.)? Ans. 12 sh. 1 d. 13th. Required the sum of 3 lb. troy and oz. troy? Ans. 9 oz. 3 dwts. 8 grs. 14th. Required in troy weight the sum of 0678 lb. troy and 9 lbs. avoir. ?.........Ans. 1 lb. 1 oz. 18 dwts. 18.528 grs. 15th. Required the sum of 1 £. 16 sh. 9 d. ; 8% £.; and 10.58 £.? Ans. 21 £. 5 sh. Od. 16th. Required the sum of 16 mi. 30 po.; 107 fathoms; and 509 yds.?..................... ........... Ans. 16 mi. 4 fur. 2 po. 0,7 yds. sh.; 2 £. 31 sh.; and of Ans. 9 £. 1 sh. 63 d.. 17th. Required the sum of 4 £. 13 3 £. 12 sh. 6 d. ?........ 18th. Required the sum in avoirdupois weight of 33 lbs. troy and 16 lbs. avoirdupois? Ans. 18 lbs. 15 oz. 1252 dr. avoir. SUBTRACTION OF COMPOUND NUMBERS. 319. The difference of two whole numbers is obtained by taking the units of each order in the subtrahend, beginning with the lowest, from the units of the corresponding order in the minuend, borrowing, when the partial subtrahend exceeds the minuend, from the next higher order of units of the minuend, to render the subtraction possible (Art. 39). The difference of two compound numbers is obtained in the same manner (Art. 317). |