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A. As in Simple Addition.
More Exercises for the Slate.
3. James is 147 years old, Rufus 15106, and Thomas 16707 what is the sum of all their ages ? A. 46,5 years.
4. William expended for a chaise $25510, for a wagon $37145, for a bridle $ 70, and for a saddle $ 111000; what did these amount to ? " A. $ 304,455.
5. A merchant bought 4 hhds. of molasses; the first contained 627 gallons, the second 721037 gallons, the third 50% gallons, and the fourth 5570 gallons; how many gallons did he buy in the whole ? A. 240,6157 gallons.
6. James travelled to a certain place in 5 days; the first day he went 40% miles, the second 2813 miles, the third 421 miles, the fourth 22T0%O miles, and the fifth 29100 Jo miles ; how far did he travel in all ? A. 162,0792 miles.
7. A grocer, in one year, at different times, purchased the following quantity of articles, viz. 427,2623 cwt., 2789,00065 cwt., 42,000009 cwt., 1,3 cwt., 7567,126783 cwt., and 897,62 cwt.; how much did he purchase in the whole year? 'A. 11724,309742
8. What is the amount of ,' 245137, 610, 24510000 1τσ88σσ τουσ, 427τσσσσσ, 4ος Τσοτσσσ, and 1925 ? A. 2854,492472.
9. What is the amount of one, and five tenths; forty-five, and three hundred and forty-nine thousandths; and sixteen hundredths ? A. 47,009.
whers place de for other el for et did
SUBTRACTION OF DECIMALS. & LIV. 1. A merchant, owing $270,42, paid $192,625; how much did he then owe ? OPERATION.
$270,42 For the reasons shown in Addition, $ 192,625
we proceed to subtract, and point off,
as in Subtraction of Federal Money. Ans. $77,795
Hence we derive the following
A. As in Simple Subtraction.
More Exercises for the Slate.
2. James rode from Boston to Charlestown in 4,75 minutes, Rufus rode the same distance in 6,25 minutes ; what was the difference in the time? A. 1,5 min.
3. A merchant, having resided in Boston 6,2678 years, stated his age to be 72,625 yrs. How old was he when he emigrated to that place ? A. 66,3572 yrs.
Note.—The pupil must bear in mind, that, in order to obtain the answer, the Egures annexed to each question, are first to be pointed off, supplying ciphers, if
then added together as in Addition of Decimals. 4. From ,65 of a barrel take ,125 of a barrel-525; take 2 of a barrel 45; take ,45 of a barrel-2; take ,6 of a barrel-5.; take ,12567 of a barral. Fr, A. 2,13933 barrels.
---2433; take ,26 of a barrel-39. 5. From som
CU,9 pipes take 126,45 pipes-29445; take ,625 of 2, pipe_420275; take 20,12 pipes-40078; take 1,62 pipes11.928; take 419,89 pipes-101 ; take 419,8999 pipes-10001. A. 1536,7951 pipes.
TLV. 1. How many yards of cloth in 3 pieces, each piece containing 2008 yards? OPERATION.
In this example, since multiplication
is a short way of performing addition, 20,75
it is plain that we must point off as in 3 addition, viz. directly under the separat
ing points in the multiplicand; and, as Ans. 62,25 yds. either factor may be made the multipli
cand, had there been two decimals in the multiplier also, we must have pointed off two more places
for decimals, which, counting both, would make 4. Hence, we must always point off in the product as many places for deci. mals, as there are decimal places in both the factors.
2. Multiply ,25 by ,5. OPERATION.
In this example, there being 3 decimal
places in both the factors, we point off 3 ,5
places in the product, as before directed.
The reason of this will appear more eviAns. 125
dent by considering both the factors
common fractions, and multiplying by 1 XLI., thus ; ,25 = 100, and ,5= 6; now, = To, which, written decimally, is ,125, Ans., as before.
3. Multiply ,15 by ,05.
In this case, there not being so many ,15
figures in the product as there are deci. mal places in both the factors (viz. 4),
we place two ciphers on the left of 75, ,0075
to make as many. This will appear evi
dent by the following; ,15=1t, and ,05 = T[; then 16 16o=1000=,0075, Ans., the same as before.
From these illustrations we derive the following
A. As many as are in both the multiplicand and multiplier.
Q. If there be not figures enough in the product for this purposes.
More Exercises for the Slate
containing 311 gallons ?-20475. In ,8 of a barrel ?-252. In ,42 of a barrel ?-1323. In ,6 of a barrel ?-189. In 1126,5 barrels ?-3548475. In 1,75 barrels ?-55125. In 125,626789 barrels ?-39572438535. A. 39574,9238535 gallons.
6. What will 8,6 pounds of flour come to, at $,04 a pound ?344. At $,03 a pound ?--258. At $,035 a pound ?-301. At $,0455 pound ?-3913.
At $,0275 à pound ?-23650. A. $1,5308.
7. At $9 a bushel, what will 6,5 bushels of rye cost ?-585. What will 7,25 bushels ?-6525. Will 262,555 bushels ?2362995. Will ,62 of a bushel ?-558. Will 76,75 bushels ?69075. Will 1000,0005 bushels ?-90000045. Will 10,00005 bushels ?-9000045. A. 1227,307995.
DIVISION OF DECIMALS.
In multiplication, we point off as many decimals in the product as there are decimal places in the multiplicand and multiplier counted together; and, as division proves multiplication, by making the multiplier and multiplicand the divisor and quotient, hence there must be as many decimal places in the divisor and quotient, counted together, as there are decimal places in the dividend.
1. A man bought 5 yards of cloth for $8,75; how much was it a yard? $8,75=875 cents, or 100ths; now, 875 -5.5175 cents, or 100ths,= $1,75, Ans.
By retaining the separatrix, and dividing as in whole numbers, thus:
As the number of decimal places in OPERATION.
the divisor and quotient, when count5) 8,75
ed together, must always be equal to
the decimal places in the dividend, Ans. $1,75
therefore, in this example, as there
are no decimals in the divisor, and two in the dividend, by, pointing off two decimals in the quotient, the number of decimals in the divisor and quotient will be equal to the dividend, which produces the same result as before.
2. At $2,50 a barrel, how many barrels of cider can I have for $11 ?" $11=1100 cents, or 100ths, and $2,50 = 250 cents, or 100ths; then, dividing 100ths by 100ths, the quotient will evis, dently be a whole number, thus :
In this example, 250) 1100 (4488 barrels, Ans.
we have for an an1000
swer 4 barrels, and
298 of another bar100
fel. But, instead
of stopping here in 250
the process, we may
bring the remain der, 100, into tenths, by annexing a cipher (that is, multiplying by 10), placing a decimal point at the right of 4, a whole number, to keep it separate from tlie 10ths, which are to follow. The separatrix may now be retained in the divisor and dividend, thus:OPERATION.
We have now for an an2,50 ) 11,00 ( 4,4 Ans. swer, 4 barrels and 4 tenths 1000
of another barrel. Now, if
we count the decimals in 1000
the divisor and quotient
(being 3), also the decimals 1000
in the dividend, reckoning
the cipher annexed as one decimal (making 3), we shall find again the decimal places in the divisor and quotient equal to the decimal places in the dividend. We learn, also, from this operation, that, when there are more decimals in the divisor than dividend, there must be ciphers annexed to the dividend, to make the decimal places equal, and then the quotient will be a whole number.
Let us next take the 3d example in Multiplication, (TLV.) and see if multiplication of decimals, as well as whole numbers, can be proved by Division.
3. In the 3d example, we were required to multiply ,15 by ,05; now we will divide the product ,0075 by ,15.
We have, in this examOPERATION.
ple (before the cipher was ,15),0075 (,05 Ans.
placed at the left of 5), four 75
decimal places in the dividend, and two in the di
visor; hence, in order to make the decimal places in the divisor and quotient equal to the dividend, we must point off two places for decimals in the quotient. But as we have only one decimal place in the quotient, the deficiency must be supplied by prefixing a cipher.
The above operation will appear more evident by common fractions, thus : ,0075=10H00, and ,15=106; now todo is divided by 70% by inverting Tb (1 XLVII.), thus, 100X1080 =1360o=to=,05, Ans., as before.