Imágenes de páginas
[ocr errors]

Also, it is at Figueira past noon..

4 25 36 Lemnos...

6 42 4 Pekin.......

.12 46 54 Cape Mendocino.....

.20 44 36 Or, in other words, if it is noon on the first of June at Philadelphia, it is, at Figueira, 25 m. 36 sec. past 4 in the afternoon, or 4 h. 25 m. 36 sec., P.M. ;* at Lemnos, 6h. 42 m. 4 sec., P.M.; at Pekin, it is 46 m. 54 sec. past midnight, or, June 2nd, 46 m. 54 sec., A.M.; at Cape Mendecino, it is 8 h. 44 m. 36 sec., A.M.

3. Suppose that the above places are all situated on the pa. rallel of 40° 4' N.,-a degree on this parallel being 45,916 nautical or geographical miles—what are the respective distances of those places in such miles, proceeding eastward, as in the preceding example ?

From Philadelphia to Figueira... ..3048,87
« Figueira to Lemnos.

1566,52 +
16. Lemnos to Pekin..

.4187,75 " Pekin to Cape Mendocino.......... 5483,60

“ Cape Mendocino to Philadelphia.. 2243,03 4. Reduce the distances in the preceding example to common miles.

English miles. Answer: From Philadelphia to Figueira. ..3514,67 « Figueira to Lemnos...

.1805,85 Lemnos to Pekin........ ..4827,55

Pekin to Cape Mendocino.... ..6321,37 " Cape Mendocino to Philadelphia. 2585,72

Nautical miles.




311. This is the method of finding the equivalent of any number of units of a higher grade in units of a lower, or of any number of units of a lower grade in terms of those of a

* P.M., Post Meridiem, Latin, afternoon. forenoon.

A. M., Ante Meridiem,

higher. The first of these two is called reduction descending, and the last, reduction ascending.

312. By reduction descending, which is performed by multiplication, we find the equivalent of a compound number in units of its lowest order, thus : descending in regular succession, from the highest order to the lowest, we reduce the units of each higher order to those of the next lower, by multiplying them by the number which shows the ratio of the unit of the higher to that of the lower, taking care to add the digits of the lower order to those of the product, which are of the same kind. For example, to reduce £5 16 s. 4 d. to pence, we proceed thus :

£ d.
5 16 4



1396 In multiplying by 20, it is easy to see that the unit figure of the shillings to be added, will always be the unit figure of the product; we therefore say 0, but 6 is 6; then, twice 5 is 10, and 1 is 11, which gives 116 s. for the value of £5 16 s.

In multiplying by 12, we say, 12 times 6 is 72, and 4 is 76; six and go 7: then 12 times 11 is 132, and 7 is 139. Thus we have 1396 for the number of pence, equivalent to £5 16 s. 4 d.

313. By reduction ascending, which is performed by division, we find the different denominations of a compound number, from their equivalent expressed in units of its lowest order.

For example, to have the value of 1396 pence in pounds, shillings, and pence: as 12 d. make 1 s., it is evident that as often as 12 is contained in 1396, there will be so many shillings; we therefore divide by 12, and have 116 s. 4 d. for the quotient.

Again, to bring 116 s. to pounds, we divide by 20—because 20 s. make £1—and we have £5 16 s. for the quotient; which, together with the 4 d., gives £5 16 s. 4 d. for the value of 1396 d., as was required. The operation stands thus :

12) 1396
2,0) 11,6 s. 4 d.

£5 16 s. 4 d.

To divide by 20, we divide first by 10, and then by 2. Now, to divide by 10, as we separate, by a comma, one figure to the right, the figure separated is tenths ; but, in dividing these tenths by 2, they will (165) become twentieths—that is, twentieths of a pound, or shillings: wherefore, we consider the figure separated as signifying a number of shillings. In dividing the figures on the left of the comma by 2, if 1 remains, as this, divided by 2, is £j 10 s., we always write this remainder 1, as ten, on the left of the figure separated. In reducing cwts. to tons, we divide by 20 in the same manner.

Hence, we see that, to reduce units of a higher order to those of a lower, we multiply by the number which shows how many of the lower make one of the higher; and, to reduce units of a lower order to those of a higher, we divide by that same number.

Again, to reduce 4 t. 13 cwt. 3 qrs. 14 lbs. to pounds, the operation is as follows:

T. cwt. qrs. Ibs.
4 13 3 14
93 value in ewts. of 4 t. 13 cwt.

375 value in qrs. of 4 t. 13 cwt. 3 qrs.


10514 value in lbs. of 4 t. 13 cwt. 3 qrs. 14 lbs. To reduce 10514 lbs. to tons, we proceed thus:

28 S 7) 10514


4) 1502

4) 375 — 2 X7= 14 lbs. 2,0) 9,3 - 3 qrs.

4 t. 13 cwt. 3 qrs. 14 lbs. We here first divide by 28, because 28 lbs. make 1 qr. ; then by 4, because 4 qrs. make 1 cwt. ; and, lastly, by 20, bebecause 20 ewt. makes 1 t.; observing that the 3 which we separate in dividing by 10, being 3 tenths, will, (165,) in dividing by 2, become 3 twentieths; we therefore consider it

[ocr errors]

cwts. also, the unit which remains in dividing by 2, being { t. is 10 cwt., and is therefore placed on the left of 3, making 13 cwt.

Examples. 1. In 7s. 111 d., how many farthings? Answer, 382. 2. In 19 s. 11 d., how many farthings ? Answer, 959.

3. How many barleycorns will reach round the earth on the equator? (305.)

Answer, 4732231680. 4. How many seconds are there in a solar year? (297.)

Answer, 31556931. 5. How many cubic inches in a cubic mile?

Answer, 254358061056000. 6. In 29 t. 17 cwt. 3 qrs. 27 lbs. 15 oz. 13 dr., how many drachms?

Answer, 17145853. 7. In 382 farthings, how many shillings ?

Answer, 7 s. 113 d. 8. In 959 farthings, how many shillings?

Answer, 19 s. 11 d. 9. In 4732231680 barleycorns, how many miles ?

Answer, 24896. 10. Reduce 31556931 seconds to days.

Answer, 365 d. 5 h. 48 m. 51 sec. 11. Reduce 254358061056000 cubic inches to cubic miles.

Answer 1. 12. Reduce 17145853 drachms, Avoirs., to tons.

Answer, 29 t. 17 cwt. 3 qrs. 27 lbs. 15 oz. 13 dr. 314. If, under a given number of units of any order, we write, in the form of a fraction, the number which shows how many of those units are required to make a unit of some higher order; as this expresses the division of the upper number by the lower, it is plain (313) that this fraction, whether proper or improper, is a fraction of the higher order; because, as often as the given number contains this divisor, so often does it contain a unit of the higher order. Thus, to reduce 7 d. to the fraction of a shilling, as 12 d. make 1 s., we say 7d.=is. To reduce 7 d. to the fraction of a pound: as 240 d. make £1, we say,

7 d.= ão £. To reduce 440 yds. to the fraction of a mile, as the mile is 1760 yds., we place this number under

1760 this is nothing else than to divide 440 yds. by 1760. But, (313,) in dividing yards by 1760, we reduce them to miles the fraction 146, or ļ, is therefore the fraction of a mile.

[ocr errors]

315. From what has been said, we can easily reduce all the inferior parts of a compound number to a fraction of the principal unit. For example, to reduce 17 s. 6 d. to the fraction of a pound, we first reduce 6 d. to the fraction of a shilling, which (314) gives i=s. We then have 173 s., or 35- S. Now to reduce - 35 s. to the fraction of a pound; we divide by 20—that is, (165,) we multiply the denominator by 20—which gives , or , for the fraction required.

Or thus : we reduce 17 s. 6 d. to pence, which gives 210 d., under which we place 240, the number of pence in a pound, and have ž48=j, as before. Also, 210 d. and 240 d. are homogeneous numbers; and, (188,) by placing 210 over 240, we show what part 210 is of 240. But 210 d. is the value of 17 s. 6 d., and 240 d. that of £l: therefore, žiế, or , is the part or fraction that 17 s. 6 d. is of £1.

Again, to reduce 2 fur. 26 p. 3 yds. 2 ft. to the fraction of a mile, we may proceed thus : 2 ft. =- i yd.; 3; yds. = 1-yd.; '11:5}=1:1

11 2


3 11 26; p. = 80 p.; 80 = 40=fur.; 2; fur. = $fur.;

lastly, +8=jm. Wherefore, the compound number 2 fur. 26 p. 3 yds. 2 ft. is

of a mile; or thus: 2 fur. 26 p. 3 yds. 2 ft. 1760 feet, and 1 m. = 5280 ft. :

then 388=of a mile, as before. Hence we see that, to reduce the lower denominations of a compound number to a fraction of the principal unit, we reduce its lowest denomination to a fraction of the next higher, to which we annex it. We then first reduce this mixed number to an improper fraction, and then to a proper fraction of the next higher order, to which we annex it, and thus we proceed till we arrive at the fraction required.

Or, we reduce the whole to units of its lowest order, and place the result over the number of units of this order contained in the principal unit, reducing the fraction, if necessary, to its lowest terms.

The scholar may operate by both methods, which will serve as proof.

« AnteriorContinuar »