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ExamPLE III.-Reduce 40,480 yards to miles. We first reduce the yards to perches, by diving by 51,- or by multiplying by 2 and dividing by II ; next, the perches to fürlongs, by dividing by 40; and then the furlongs to miles, by dividing by 8. In 40480 yds. 40,480 yards there are(40,480X2=)80,960 half-yards, and in a perch there are(51 X2=) II half-yards. Dividing 80,960 by 11, 11)80960 half-yds. therefore, we find that in 80,960 half-yards, or 40,480 yards, there are 7,360 lengths of 40) 7360 per.

half-yards or 51 yards each--that is, 7,360 perches. Dividing 7,360 by 40, we

8)184 fur. next find that in 7,360 perches there are 184 lengths of 40 perches each—that is, 184

23 m. furlongs. Lastly, dividing 184 by 8, we find that in 184 furlongs there are 23 lengths of 8 furlongs each—that is, 23 miles. So that 40,480 yards are equal in length to 7,360 perches, or to 184 furlongs, or to 23 miles.

EXAMPLE IV.–Reduce 31,398 yards to miles. Reducing the yards to half-yards, and dividing by 11 (the number of half-yards in a perch), we obtain 5,708 for dividend, and 8 for re- 31398 yds. mainder—that is, 5,708 perches and 8 half-yards. Instead, however, of the 8 half-yards, we set down (8=2=) 4 yards. 11)62796 half-yds. Dividing by 40, we next find that in 5,708 perches there are 142 furlongs and 28 40)5708 per. 4 yds. perches. Lastly, dividing by 8, we find that in 142 furlongs there are 17 miles 8) 142f. 28p. 4 yds. and 6 furlongs. So that 31,398 yards are equal in length to 17 m. 6 fur. 28 per. 17m.6.28p. 4y. 4 yds.

NOTE.—In dividing by 40, we employ the factors 10 and 4. Thus, dividing 5,708 [Ex. IV.] by 10, we obtain 570for quotient, and 8 for remainder. Next, dividing 570 by 4, we obtain 142 for quotient, and 2 for remainder. This second quotient we set down as the "true" one. To obtain the “ true” remainder, we multiply 10, the first divisor, by 2, the second remainder, and add 8, the first remainder, to the product: 10X2=20; 20+8=28.

74. Rule for Ascending Reduction : First, reduce the given number to the next higher denomination, by dividing by the number which indicates how many units of the given denomination are contained

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in one of the next higher; then, reduce this new denomination-if it be not the required one-to the next higher, and continue the process until the required denomination is obtained.

THE COMPOUND RULES.

COMPOUND ADDITION. 75. Addition is called COMPOUND when the addends are compound numbers (of the same kind).

76. Compound addition is performed by means of Simple Addition and Ascending Reduction.

EXAMPLE I.-A merchant received £16 12s. 7d. on Monday, £23 178. 8£d. on Tuesday, £39 188. 4ļd. on Wednesday, £65 IIS. 3}d. on Thursday, £46 198. 5 d. on Friday, and £50 168. 23d. on Saturday: how much did he receive during the week ?

Arranging the addends as in the margin, we find, by means of Simple Addition, that, altogether, the merchant received 239 pounds, 93 shillings, 29

£ s.

d. pence, and 13 farthings (each halfpenny being 16 12 73 reckoned 2 farthings). In practice, however, 23 17 8 a sum of money is not written under this form 39 18 41 -as much as possible of the amount being ex- 65 II 33 pressed in pounds, as much as possible of the 46 19 58 remainder in shillings, and as much as possible 50 16 23 of what then remains in pence: so that the number of shillings can never exceed 19, nor 239 93 294 the number of pence ii, nor the number of farthings 3.

We therefore reduce the farthings to pence, which we “carry” to the pence column; we next reduce the pence to shillings, which we carry to the shillings

£ $. d. column; and we then reduce the shillings to 16 12 7 pounds, which we carry to the pounds column. 23 17 8.1 The required sum is thus found to be

39 18 41 £243 155. 8ļd. The work proceeds in this 65 II 31 way :-2+3+2+1+2+3=13; 13 farthings 46 19 5 3}d.; [set down] £d., and carry 3 ;--3 (the

50 16 2 carried figure) +2+5+3+4+8+7=32; 32 pence 2s. 8d.; [set down] 8d., and carry 243 15 81

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2 ; &c.

EXAMPLE II.--A farmer has four farms, containing, respectively, 734. 2R. 30P., 654. 3R. 24P., 494. IR. 17P., and 36A. 2R. 35P. : how much land does he hold altogether?

Arranging the addends as in the margin, we first find that the total number of perches is (35+17+24+30=) 106. Dividing by 40, the number of perches in a rood, we convert 106 perches into 2 roods and 26 A. R. 'P. perches. Setting down the 26 perches, and 73 2 30 carrying the 2 roods, we next find that the total 65 3 24 number of roods is (2+2+1+3+2=) 10. Divid- 49 I 17 ing by 4, the number of roods in an acre, we 36 2 35 convert ro roods into 2 acres and 2 roods. Setting down the 2 roods, and carrying the 2 acres, we finish by finding the total number of

225. So that the required sum is 225A. 2R. 26P.

225 226

acres =

77. Rule for Compound Addition : Set down the addends-one below the other--in such a way that all the numbers of the same denomination shall stand in the same vertical column. Draw a hori. zontal line to separate the addends from their sum. Then find, by Simple Addition, the total number of units of the lowest denomination, and divide by the number indicating how many such units are contained in a unit of the next higher denomination ; write the remainder, if there be one, at the bottom of the most right-hand column, and add (or “ carry") the quotient to the column of the next higher denomination. Proceed in the same way with each of the other columns, successively.

COMPOUND SUBTRACTION.

78. Subtraction is called COMPOUND when the two numbers whose difference is required (and which must be of the same kind) are compound numbers ; or when only one of the numbers is compound; or, again, when both numbers are simple, but of different denominations.

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“What is the difference between 5s. and 38. 4d. ?". how much does £10 exceed 178. ?" Both these questions belong to Compound Subtraction.

79. Compound Subtraction is performed by means of Simple Subtraction* and Descending Reduction.

ExamPLE I.-Out of a purse containing £23 118. 9]d. a sum of £4 178. 6 d. was paid : how much remained ?

Let us suppose the £4 17s. 6 d. to have been paid in four instalments of id., 6d., 178., and £4, respectively. The number of farthings in the purse being only 1, the purse-bearer must have paid the first instalment (id) by giving a penny, and getting back (id. -- {d.=) £d. After payment of the first instalment, therefore, the purse contained

£ s. d. Id. less, and Id. more than at first; so that the 23 II 9] number of farthings then was (1+1=) 2. After

4 17 68 payment of the second instalment (6d.), the num- 18 14 2 ber of pence in the purse was diminished by (1+6=), 7; so that the number of pence remaining in the purse was (9-7=) 2. The number of shillings in the purse being only 11, the purse-bearer must have paid the third instalment (178.) by giving a pound, and getting back (£1-178.=) 38. After payment of the third instalment, therefore, the purse contained £1 less, and 38. more than at first; so that the number of shillings then was (11+3=)14. After payment of the last instalment (34), the number of pounds was diminished by (1+4=) 5; so that the number of pounds remaining in the purse was (23–5=) 18. The required remainder is thus found to be £18 148. 2}d.

EXAMPLE II.—A butcher had in his stall 10 cwt. 2 qrs. 18 lbs. of beef, and of this quantity he sold 7 cwt. 3 qrs. 13 Ibs.: how much had he remaining ?

Let us suppose the 10 cwt. 2 qrs. 18 lbs. to have been cut up into 10 pieces of i cwt. each, 2 pieces of i qr. each, and 18 pieces of 1 lb. each ; and let us further suppose that the quantity sold was disposed of in this way :-13 lbs. to one customer, 3 qrs. to a second, and

cwt. qrs. lbs.

18 7 cwt. to a third. After the sale of the 13 lbs. the number of 1-lb. pieces in the stall

7 3 13 was (18 – 13=) 5. After the sale of the

3 5 3 qrs., which must have been cut off one of

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Simple Subtraction must here be understood as including Simple Addition.

the 1-cwt. pieces, the butcher had (1 cwt. -3 qr3.=). I qr. more, and I cwt. less, than at first; so that the number of quarters then was (2+1=) 3. After the sale of the 7 cwt, the butcher had (7+1=) 8 cwt. less than at first, that is, had (10-8=) 2 cwt. remaining. The required remainder is thus found to be 2 cwt. 3 qrs. 5 lbs.

8c. Rule for Compound Subtraction : Write the subtrahend under the minuend, in such a way that every two numbers of the same denomination shall stand in the same vertical column. Draw a horizontal line to separate the subtrahend from the remainder. Then, take the lowest denomination in the subtrahend from that in the minuend, and set down the result as the lowest denomination in the required remainder. Should this subtraction not be possible, find the difference between the lowest denomination in the subtrahend and a unit of the next higher denomination; to this difference add the lowest denomination in the minuend; write the result as the lowest denomination in the required remainder; and carry i to the next denomination in the subtrahend. Treat each of the other denominations in the same way—always taking the “carried” i into account, when there is í to carry.

COMPOUND MULTIPLICATION. 81. Multiplication is called COMPOUND when the multiplicand is a compound number.

82. Compound Multiplication is performed by means of Simple Multiplication and Ascending Reduction.

EXAMPLE I.--- Multiply £59 175. 81d. by II. Multiplying the £d. by 11, we obtain 11 halfpence: this amount, when divided by 2 (the number of halfpence in a

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