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in the decimal, and, as 15s. contains 2s. 7 times, we shall have .7. There is a shilling over. This is a half of i or a half of %==.05. For 15s. then, we have the decimal.75. Thus far, the method is perfectly accurate.

:

1 farthing is of a £., because 1£. contains 20×12 X4 960 farthings. Now 960+ of itself=1000.Therefore, any number of farthings, or 960ths. of £1., increased by of itself, will express the same value in 1,000ths of £1. If the farthings be 12, a 24th. part of them will be if more than 12, more than . In this case, then, if we add 1 to the farthings, they will be 1,000ths., with an error, less than a 1,000th. If they are 36, a 24th. part will be 1; if more, more than 1. In this case, then, if we add 2 to the farthings, they will be 1,000ths. with an error, less than a thousandth. If they are 24 or 48, a 24th. is 1, or 2, which added, gives 1,000ths exactly.

39

=

Now in the example above, we have 93d.=39 qrs.= of a pound, of a pound =.041, the decimal for the pence and farthings. Hence the whole decimal, for shillings, pence and farthings, is .75+.041=.791, corresponding to one of the answers above.

Hence, to express shillings, pence and farthings in the decimal of a pound,

CALL EVERY 2 shillings, 1 tanTH, EVERY ODD SHILLING 5 HUN

DREDTHS, AND THE FARTHINGS IN THE PENCE AND FARTHINGS, SO MANY THOUSANDTHS, adding 1 if tHE NUMBER IS BETWEEN 12 AND 36, AND 2, IF ABOVE 36.

NOTE. As 48 farthings make a shilling, we never have occasion to go above this number. The above mode of finding the decimal of a pound, is called find. ing the decimal BY INSPECTION; that is, by sight, or without the trouble of a written calculation. Take the following examples.

10. Reduce 17s. 6d. 3qrs. to the decimal of a pound?

Ans. £0.878.

11. Reduce £19; 8; 7; 2, to a decimal expression.

Ans. £19.431.

12. Reduce £18; 16; 9, to a decimal expression.

Ans. £18.840.

13. Reduce 13s. 3d. 1qr. to the decimal of a pound. Ans. £0.664.

14. Reduce 14s. 8d. 3qrs.; 19s. 104d.; 16s. 74d. and 18s. 91d. to decimals.

15. Reduce £16; 17; 3; £11; 19; 54; £13; 9; 1, and £15; 8; 2, to decimals.

ADDITION.

¿LXI. 1. A man paid debts as follows; to A, $37.3, to B, $94. 05, to C, $127.063, and to D, $1,843.375. How much did he pay in all ?

We have given this first example, in Federal Money, because the pupil will see at once that dimes should be added to dimes, cents to cents, &c. Of course he will place the numbers so that these denominations shall stand under each other.

$ 37.3

$

94.05 $ 127.003 $1,843.375

Ans. $2,101.728

2. Add 37.3; 94.05; 127.003, and 1,843.375. These numbers are the same as those in the last example, except that they are not Federal Money. They must therefore evidently be written down and added as before; for as dimes must be added to dimes, cents to cents, &c., so tenths must be added to tenths, hundredths to hundredths, &c. This, it will be observed, brings all the decimal points in the given numbers under each other: and the point in the result, of course, immediately under them. Hence the rule,

WRITE THE NUMBERS SO THAT THE DECIMAL POINTS MAY BE UNDER EACH OTHER; ADD AS IN WHOLE NUMBERS, AND PLACE THE POINT IN THE ANSWER DIRECTLY UNDER THE OTHER POINTS.

NOTE. Repetends should be extended, at least as far, and usually one place farther, than the longest finite decimal, before adding. When numbers are given in vulgar Fractions, they must be reduced to decimals, before addition takes place; and when the numbers are compound, the lower denominations should be made decimals of the higher.

EXAMPLES FOR PRACTICE.

3. A man purchased at one time, 7 cwt. of sugar; at another, 17% cwt.; at another, 15 cwt.; at another, 20 cwt. How many cwt. did he purchase in all? A. 60.744+

4. In 3 pieces of cloth, are contained as follows: in the first, 25 yds. 3 qrs. 2 na.; in the second, 19 yds. 1 qr. 3 na,; and in the third, 17 yds. 1 qr. 1 na. What is the whole amount in yards and A. 62.625.

decimals.

5. In one cask are contained 28 gals. 2 qts. 1 pt.; in another, 36 gals. 3 qts.; in another, 27 gals, 3 qts.; and in another, 16 gals. 1 pt. How much in all ? A. 109.25 gals.

6. One cask of sugar weighs 1 cwt. 2 qrs. 14 lb.; another, 3 cwt. 14 lb.; another, 2 cwt. 1 qr.; and another, 1 cwt. 14 lb. How much do all weigh? A. 8.125 cwt.

7. If I travel 35 mls. 3 fur. 20 rds., one day; 41 mls. 2 iur. 10 rds., another; 17 mis. 4 fur., another; and 26 mls. 3 fur. 20 rds., anoth. er; how far do I travel in all ? A. 109.65625 mls.

8. If I pay $21 for one hhd.. of molasses; $19 for another; $17 for another; 23 for another; and 18 for another; how much do I give for all? A. $101.00 9. Add $12.34565; $7.891; $2.34; $14 and $0.0011. A. $36.57775.

10. Add .014; .9816; 1.32; 2.15914; .72913 and 3.0047.

A. 8.20857.

11. Add 27.148; 918.73; 14,016; 294,304.001; .7,138 and 221.701. A. 309,488.2938.

12. Add 312.984; 21.3918; 2,700.42; 3.173; 27.2 and 581.07. A. 3,646.2388.

MULTIPLICATION.

§ LXII. 1. At $1.25 per yd. what cost 3 yds. of cloth? This may be performed, either by Addition or by Multiplication. BY MULTIPLICATION.

BY ADDITION.
$1.25
1.25

1.25.

$1.25

3

$3.75 Ans.

$3.75 Ans.

We have exhibited both modes, to show where the point ought to be placed, as this is the only difficulty in decimal calculations. When the multiplier is a whole number, then, there must be as many decimals in the product, as in the multiplicand.

2. At $125 a pipe, what cost .3 of a pipe of wine? $125. On the left we have found what 3 whole

$125

.3

3 pipes cost, by multiplying by 3. Now .3 is the tenth part as much as 3 whole ones. $375 Of course, we must take the tenth part of $37.5 Ans. the price of 3 whole pipes, for the price of .3 of a pipe. But (§ xxx.) cutting off one figure divides by 10. Of course, we have one decimal place in the product by .3 This is shown on the right. When the multiplicand is a whole number, then, there must be as many decimals in the product, as in the multiplier.

3. At $1.25 per lb. what cost .3 pound of tea?

If it were 3 whole pounds, we should obtain $3.75, as in Ex. 1. But it is the tenth part of 3 lb., being .3 lb., and therefore, the price will be a tenth part as much=$0.375 Ans.

When neither factor is a whole number, then, there must be as many decimals in the product as in both factors. Observation of the following, may, perhaps, render the principle more clear. We take the same numbers, making them at first whole, and moving the decimal point one place at each step.

125

12.5

1.25

.125

.125

.125

.0125

[blocks in formation]

MULTIPLY AS IN WHOLE NUMBERS, AND POINT OFF AS MANY DECIMALS

IN THE PRODUCT, AS THERE ARE IN BOTH FACTORS.

NOTE.

From observing the three latter illustrations, the pupil will see that when there are not as many figures in the product as the rule requires, cyphers must be prefixed to make out the number.

EXAMPLES FOR PRACTICE.

4. At $3 pr. lb., what cost 16 lb. of tea? A. $55.20. 5. At $2 pr. gal. what cost & gal. of wine? A. $1.20. 6. At $5.27 pr. yd., what cost 7 yds. of cloth?

A. $36.9954.

7. At 3 cts. pr. lb., what cost 7,853 lbs. of rice ?

A. $274.855.

8. At $133 a bl., what cost 464 bls. of sugar?

9. Multiply .004 by .04.
10. Multiply 24.61 by .0529.
11. Multiply .0007853 by .035.
12. Multiply .051 into .091.
13. Multiply .0217 into .00431.

A. $627.12.
A. .00016.
A. 1.301869.
A. .0000274855.

A. .004641.
A. 000093527.

NOTE. The truth of the rule will further appear, by writing the decimals in the form of vulgar Fractions, and multiplying by the rule § XLIV. Thus, multiply .02 by .03.02 and .03 180 X 80 = 10000 =.0006, where the decimal places of the product are equal to those in both factors.

14. Multiply .02 by 10. 15. Multiply .003 by 100.

6

A. .2

A. .3

It will be seen that multiplying a decimal by 10 does not alter the significant figures, but only moves the separatrix one place towards the right. This makes the multiplicand 10 times greater, because it causes every figure to stand one place higher. So, multiplying by 100 moves the separatrix two places towards the right; by 1,000, three places, &c. Hence, to multiply by a unit with cyphers annexed,

REMOVE THE SEPARATRIX AS MANY PLACES TO THE RIGHT AS THERE ARE CYPHERS IN THE MULTIPLIER, ANNEXING CYPHERS IF NECESSARY. NOTE. If the multiplier be any number with cyphers annexed, it may be considered as a composite number, one of whose factors is 10, 100, 1000, &c., and we may first employ this factor, according to the above rule, and afterwards, the other, as usual.

16. Multiply .38179 by 10; by 100; by 1,000; by 10,000.

17. Multiply 1.876 by 1,000; by 10,000; by 1,000,000. 18. Multiply 17.9 by 1,00; by 1,000; by 100,000,000.

§ LXIII. It often happens that, in practice, we have no occasion for as many decimal places as the rule will give us. The lower decimals are so small that they may be rejected as of no importance. We will now propose a mode, by which we may obtain decimals, accurately as far as we please without the trouble of finding the others.

1. Multiply 13.1346 by 12.2876.
COMMON METHOD.
13.1346
12.2876

788076 91 9422 1050 768 2626 92 26269 2 131346

161.392 71096

I wish to perform this multipli-
plication, and retain only three
places of decimals. Now I know
(SLXII.) that when the multiplier
is whole numbers I have just
as many decimals in the product
as in the multiplicand. If then
I multiply only three decimal
figures of the multiplicand by
the 2 in the units' place of the
multiplier, I shall have only
three decimal places in the pro-

CONTRACTION. 13.1346

6782.21

131.346

26.269

2.627

1.050

.092

.008

161.392

luct, thus obtained. If I multiply four decimal figures of the multiplicand by the 1 in the tens' place of the multiplier,=10, I shall have four decimal places in the product, but the last one, (being a cypher) may be neglected, leaving three places, as before. If I multiply two figures of the multiplicand by the 2 in the tenths' place in the multiplier, I shall have three decimal places in the product. (§ LXII.) If I multiply, in like manner, one decimal place by the 8 in the hundredths' place, I shall have again three decimal places in the product. So, if I go on, multiplying one place less in the multiplicand, for every place lower in the multiplier, shall still have three decimal places in the product. I may therefore arrange all these products, thus obtained, with their right hand figures under each other, and put the separatrix, in each, three places from the right. They may then, it is obvious, be added as they stand, and the sum will be the total product, with only three decimal places.

It is a convenient way of determining where to commence multiplying with each figure, to put the units' place of the multiplier under that decimal of the multiplicand, which you wish should be the last in the product; and then write all the other figures before and after it, in an inverted order, as in the contraction above. Then in employing each figure of the multiplier, commence with the figure of the multiplicand immediately above it, neglecting all the figures to the right, and place the right hand figures of the several products exactly under each other. They will then stand in proper order to be added. Thus, in the last example, units stand under thousandths; and, of course, the first figure of the product is thousandths; tenths stand under hundredths, and the last figure of the product is in like manner thousandths; hundredths under tenths, which produce thousandths, as before, &c.

One thing more. In order to ensure sufficient accuracy, when we begin to multiply by any figure, we must first multiply the preceding figure of the mul tiplicand by the figure we are using as a multiplier, and carry from that product, to the first figure which we set down; always carrying for the nearest number of tens, whether above or below. Thus when multiplying by the 2 (units) in the above example, the first figure in the product is 2X3=6. But to this we carry 1, making 7, because 2X4 (the preceding figure) -8, which is nearer ( 10 than it is to 0. In multiplying by 8, (tens,) the first figure obtained is 8X18, to which we carrry 2, making 10, because 8X3=24, which is nearer to 3 tens than to 3 tens, &c. Hence, the rule.

I. WRITE THE UNIT FIGURE OF THE MULTIPLIER UNDER THE DECIMAL PLACE WHICH IS TO BE LAST IN THE PRODUCT, AND THE OTHER FIGURES IN AN INVERTED ORDER.

II. IN MULTIPLYING, REJECT ÅLL THE FIGURES IN THE MULTIPLICAND,

TO THE RIGHT OF THE FIGURE YOU ARE MULTIPLYING BY; CARRY TO THE FIRST FIGURE OF THE PRODUCT FOR THE NEAREST NUMBER OF TENS CON TAINED IN THE PRODUCT OF THE PRECEDING FIGURE IN THE MULTIPLICAND, AND SET DOWN THE RIGHT HAND FIGURES OF THE SEVERAL PRODUCTS UNDER EACH OTHER.

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