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ADAmS'
NEW ARITHMETIC,
SUITED TO HALIFAX CURRENCY;
':"\„'^ '•'■!;*-."" IN WHICH THt -'
PRINCIPLES OF OPERATING BY NUMBERS
v«,
■' , . - ► •-it*-
■■/P.
ANALITICALLY EXPLAINED,
AND
SYNTHETICALLY APPLIED
')
THUS. COMBINING THE ADVANTAGES TO BE DERIVED BOTH FROM
'THE INDUCTIVE AND SYNTHETIC MODE OF INSTRUCTING :
rHE WHOLE MADE FAMILIAR BY A GREAT VARIETY OF USEFUL AND
INTERESTING EXAMPLES, CALCULATED AT ONCE TO ENGAGE THE
PUPIL IN THE STUDY, AND TO GIVE HIM A FULL KNOWLEDGK
OF FIGURES IN THEIR APPLICATION TO ALL THE
> ' PRACTICAL PURPOSES OF LIFE. ^
^!'- :• .1' -^
'■•'>■•
^ . - V ■ » '•
1 .
-^ DESIGNED FOR THE USE OF
SCHOOLS & ACADEMIES IN THE BRITISH PROVINCES.
.f', •'■
*:^-^-«
'<
BY DANIEL, ADAMS, M. D.
■ ■ -^r
:. ..■^■■
SHERBROOKE,C..E.
PUBLISHED BY WILLIAM BROOKS.
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PRINTED BY J. S, WALTON, SHERlQnoOKE, CANADA EAST»
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Therm afe two xnehods of teach ing-: the synthetic, and th6
^nalytici In the synthetic methud,the pupil is tirst presented
urith a general view of the science he is siudyins. and after-^
wards with the particulars of vrhich it consists. The analytic
method reverses this order t the pupil is first presented with the
particulars, from which he is lea, by certain natural and easy
gradations, to those views \^'hieh are more general and conv;;^
prehensive.
The Scholar's Arithmetic pilblished in 1801, is synthetic. If
that is a faolt of the work, it is a fault of the times in which it
appeared. The analytic or inducftive ' method of teaching, as
how applied to elementary instruction^ is among the improve*'
ments of later years^. Its introduction is ascribed to Pcstaloz"
zi, a distinguished teacher in Switzerland. It has been applied v
to arithmetic, with gre^t ingenuity, by^ Mh CotBOmi, in our
own country.
The analytic is unquestionably the be^ method of acquiring
J^nowledge ; the Synthetic is the belt method of recapitulating
or reviewing it. In a treatise designed for school education,
both methoila are useful. Such is the plan of the present un-
dertaking which the author, occupied as he Is wim other ob-
jects and pursuits, would \villingly have forborne, but that, the
demand fofr the Scholar's Arithmetic still continuing,* an obli"
gation, incurred by long-couthiued and extended patronage, did
not allow him to tiecline the labor of a revisal, which should
adapt itto the present lAore enlightened views of teaching this
science in our schools. In doing this, however, it has been
necessary to make it a ne«^ work* ' »
In the exeCutW of this design, an analysis of each rule is
iii-st given, containing a familiar explanation of its various
principles ; after which follows a synthesis of these principles,
with questions in form of a supplement. Nothing is taught
dogmatically i no technical term is used till it h<is ^st been de«>
fined, nor any principle inculcated without a previous develope*
ment of its trdth ; and the pupil ismade to understand the xe^
6on of each process as he proceedst
The examples under each rule are mostly of a practical na-
ture, beginning with those that are Very easy, apd gradually
advancing to those more difficult, till one is introduced cob-
taining larger numbers, and which is not easily solved in the
•knind>; then in a plain, familiat manner, the pupil is showa
4 ■" PREFACE.
how thB solution may be facilitated bv figures. In this way
he iff made to see at onoe their use anu their application.
At the close of the fundamental rules, it has been thought
advisable to collect into one clear, view the distinguishing prop-
erties of those rules, and tomve a number of examples involv-
ing one or more of them. These exercises will prepare the
pupil more readily to understand the application of these to the
succeeding rules ; and besides, will serve to interest him in the
science, smoe he will find himself able, by the application of I
a very few principles, to solve many curious^uestions.
The arrangement of the subjects is that, which to the author
has appeared most natural. Fractions, haye received all that
consideration which their importance demands. The^ princi-
ples of a rule ci^Ued Practice are exhibited, but its detail of
cases omitted, as unnecessarV, since the adoption and general
use of federal money. The Kule of Three^ or Proportion, is re-
tained «nd the solution of questions invdlving th^ principles of
proportion, by analysis, is distinctly shown.
The articles Alligation, Arithmetical and Geometrical Pro-
gression^ Atmuities and Permutation, were prepared by Mr. Ira
YouNO, a member of Dartmouth College, from whose knowledge
of the subject, and experience in teaching, I have derived im-
portant aid in other parts o{^ the work.' f
The numerical paragraphs are chiefly for the purpose of ref-
erence ; these references the pupil should not be allowed to
neglect. His attention also ought to be particularly directed,
bv his instructor, to the illustration of each particular prinoi-
ple, from which general rules are deduced j for this purpose,
recitations by classes ought to be instituted in every school
where arithmetic is taught. '
11 le supplements to the rules, and the geometrical demon-
strations of the extraction of the square and cube roots, are the
only traitSnOf the old work preserved in the new.
DANIEL ADAMS.
',. ?u ;.iT
y-"
; J1 •■11' J ,. '•
PUBLISHER'S PREFACE.
The author of the following practical treatise upon
[Arithmetic, has made himself favourably known in the
United States, and to a considerable extent in the Canadas,
I for a great number of years, by his works, designed for the
use of Academies and primary schools. The " Scholars'
Arithmetic," published in the year l801, continued in al-
most universal use, until within a very short time past. —
JButjuster views beginning to prevail, and sounder princi-
Iples becoming established in the public mind, apon the sub-
iject of elementary education, a revision of the work seem-
led necessary. At this time, " Adams' New Arithmetic,"
I was published. This seems evidently to have been pre-
Ipared with much care. The author has recognised in it
Ithroughout, this .important kw in relation to the mind, that
Jit must first be made acquainted with particular facts, or
[there will be no ability to arrive at correct general conclu-*
Isioas. Particular examples are therefore given upon eaeh
jsubject, and from them, in a manner obvious to the ypung
lind, all the general rules, are deduced. In other wQrds,
the author has care^lly and prudently pursued, in his book,
rhat is called the antUytic method. ' 'The care used in cle-
ining necessary terms, which might not be quite clear, the
)rdctical character of the examples given under each rule,
the methodical disposition of the different parts of each
subject, and of the different subjects, the general per-
spicuity, simplicity and accuracy of the work, render it '.n^
|valuable to the pupil.
It is due the author to observe, that ^Adams' New
arithmetic," for its adaptation to the capacities of young
ind ordinary minds, is justly considered the best practical
[reatise which has been offered to the public. *
In the present edition, the main purpose in view was to
idapt Adams' work to the clirrency of the British Provin-
ces. No separate article, as in the original, has been allott-
d to Federal Money ; for this the pupil has been referred
^o Decimal Fractions, in which also almost all tbe exam-
)les will be found in the money of the United States. Ad-
litional examples in the compound rules have been given,
A2
^
l^tiBiisltit^A ^Util^ilbK:
and the old ones r^ained, under the titl^ of tialifrft elih
Irency ; and generally throUffhout the book, where denom-'
initiotiit of tttoney oceurj IlalifaX currency has been sub<
«tituted fof Federal money.
The tules and eitamples in R,eduction of CutteMiea
have bUen es^ntially changed j tod in Reduction, after the
Table of £ngli9h Mon^y. which is called the Table of
Halifax (JUrreney, A list of the Gold and Silver Coins cur-
rent in the I^rovinc^, has been inserted. This may be dc
][)ended upon as entii'ely Accurate. The tables of f^rench,
and Dry, Loilg, Square, ftnd Solid Measure, have been giv<<
en-^and wh&t a^e the weights and measures establisheaby
Uw in this Province i? also stated.
Th6 most novel feature in the book will be found in the
(ui^df Interest, (jertainly an innovation, but it is believed,
an intprovement, has be6n made. The pounds in any giv^
en sum upon which interest is to be cast, are left to stand
Bi the nnits^ i^nd the shillings and pence are reduced to
decimal paHs ot a pounds The interest is then obtained
the same as in. I^^deral Money, and the decimal parts in
the tesult teducdd to shillings and penca It is considered
that this method is ttior^ simple and concise, and will b^
found in practice to be more conveilient than any othet.^~
ijut setting asi^e considerations of temporary eonvenience^
if thi^ change and attempted amelioration, shall assist in
Mttit ^ery jsHght degree in turning men's minds toward the
Decinial AatiPi ana ihducing them to look forward to a
period when all the denominations of money, weights and
measures, throughout the world, shall be expressed in dec-
hiAT.s, it cannot be affirmed that no benefit has been obtained.
The importance of the principal and essential alteration
\A we book, viz ; the adaptation of it to the ('utrency o'f I
the cdunt!fy» will not fail to be observed by et^ery one* It I
is indeed singular, that hitherto, no Canadian Arithmetic
in the tenglish language, has been published* Mercantilr,
HgricuitUfal, and gerterallvthe business men of the country,
\yill be aWaTe of a benefit to Ije realized, and it is consider-
ed that something also bearing a relation to political advan:|
Jtage, maybe in the result ^ ;. ; ,
^hcrkraakc, L, C, June 6^ 1849,
I -I
.<,.,.-5f
i^^ ♦-..'•v
1,
<•:.'« '*s
I,
^
index;
■•» iini • »7r
I*' .
.yjitioh, i. - *
Alliffation, - ■*
Arithmetical Projrression, '■
Compound Numbert, •
Addition of,
I/'
..-f*.
-Subtraction ofj * *
-Multiplication audi Division of,
> 194
'* • 58
80
85
89
154
59
213
'^ 39
• * 199
* 173
204
- 64
* 98
- 100
C'Om mission,
Coins, Table of, - *
Cube Root, Extractten of,
Division, * ^ -»
Duodecimals, * *
-^-^ *- — - — Multiplication of,
Kquation of PuymentSj -
Evolution, ".''''
Federal iVloncy, - »
^'ructionfi, - « a
Proper and Improper, " . " •* •*
!f(i\ i' To change an Improper Fraction to a whole or
'<»t"- TM '^■'■^ mixed number, - - - 100
To reduce a whole or mixed numbel' to an Im-
proper Fraction, i. - ^ 101
TiJ redude a Fraction to its lowest terms, - 102
To divide a Fraction bv a Whole humbet) '^-^ *• 105
107
109
no
112
114
116
121
128
(
,'•(
■■PI
To multiply " «' "
ii J.^ whole niimber by a Fraction,
- — ^*-^ ^one Fractioti by another, -
T'j divide a whole number bv a Friuclioti,
^ — ■■ — ^♦-^one fraction by anptiier.
Addition and Subtj-aclion of Fractions,
){eduction of
Decimal Fractions,
a
-*• Addition and Subtraction of, 132
—Multiplication of, -
^Division of. « *
* 135
137
■ -^*^* 143
llrfduciion of Vulgar Fractions to Decimals, 140
1*eno\vship, - - - - - - 190
ilalifax. Currency, * ^ * * - $3
-Reduction of.
8
INOBX.
Insurance, - > "^ '
Interest, - - .
l-Compound, -
Involution, - • ,j(t,r>
Multiplication, Simple, *
Numeration, - - -
Proportion, or the Rule of Three,
Compound,
- IM
151
- 166
203
- 28
9
- 176
184
r 234
58
Permutation,
Reduction,
Tables of Money, Weights, Measures, &c. 59—78
——of Currencies, - - - - 149
Ratio of the Relation of Numbers, - - - 174
Subtraction, Simple, - ^
^uaie Root, Extraction of,
■ iv^
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44
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I
: MISCELLANEOUS EXAMPLES.
.;'^■ •■1' 1 ,
Baiter,ex. 20— 31. '^ | Position, ex. 88— 107.
I'o dnd the area of a Square or Parallelogram, ex. 147 — 153.
— a triangle, ex. 154 — 158.
Having the Diameter of a Circle, to find the Circumference ; or
having the Circumference, to find the Diameter, ex. 170 — 174.
To find the Area of a Circle, ex. 175 — 178. ..
^ a Globe, ex. 179, 180.
'i'o rind the Solid contents of a Globe, ex. 181, 183.
-— * Cylinder, ex. 184 — 186.
. Pyramid, or Cone, ex. 187, 188.
. any Irregular Brtly, ex. 201.
Guaging, ex. 189, 190. j Mechanical Powers, ex. 191—200.
;•:• 1-
<■ ^I't^/'t, '■■-. • r - ■ -■■■ .-.
Forms of Notes, Receipts, Orders an J Bills of Parcels, page 260.
Book Keepinij, - - ^- - - 261.
I I
NUMERATION. . ^
f 1. A BiNOLK or individual thinj; is called a unit, unity
or one ; one and one more are called two ; two and one
more are called three , three and one more are called four ;
four and one more are called five ; five and one more are
called six ; six and one more are called seven ; seven and one
more are called eight ; eight and one more are called nine ;
nine and one more are called ten, 6lc.
These terms, which are expressions for quantities, are
called numbers. There are two methods of expressing
numbers shorter than writing them out in words ; one called
the Roman method by letters,* and the other the Arabic
method by figures. The latter is that in gjeneral use.
^n the Arabic method, the nine first numbers have each
an appropriate char/icter to represent them. Thus,
*In the Roman method by letiert, I repreientii one, \ Jive, X ten,
L fifty, C one hundred, D ftbe hundr^, and M one thoueand.
Aa onen aa any letter ia repeated, ao many timea ia ita value repented,
unleaa H be a letter represenlini; a lesa number, placed before one rep-
reaenting a greuter , then, the ieaa number is taken from the greater,
thus| IV ropreaenU four, IX nine, Acfia will be aeen in the follow-
ing TABLE :—
Ninety
One hundred
Two hundred
Three hundred
Four hundred
Five hundred
Six hundred
Snven hundred
Eight hundred
Nine hundred
One thousand
Five Thousand
XXXX.or XL Ten thousond
L Fifty thousand
LX Hundred thousand
LXX One million
LXXX Two millions
*IC> is used instead of D to represent five hundred, and for every ad-
ditional Q annexed at the righ hand, the number is increased ten times.
fClQ ia used to represent one thousand, and for every C and q put
at each endj the number is increased ten timea.
I A line drawn over any number increases its value t thousand ti*es.
One
Two
Three
Four
Five
Six
Seven
Eighi
Nine
Ten
Twenty
Thirty
Forty
Fifty
Sixty
Seventy
Eighty
I
II
III
nil, or IV
V .,„.,.., .
VI
VII
VIII
Vim, or IX
X
XX
XXX
LXXXX, or XC
C
CO
>i;'.J'/'vv
CCC
;''.'i'' .;V:
cccc
1
D, orlD*
f
DC
iy >.
DOC
iA
DCCC
■*''
DCCCC
.1
M, or Clot
> t '
lOD. or Vt_
CCIo3,orX
1 000
■%'• ■
CCCIooa, or
C '
M
M M
^S
10
NtJifEitAtloKi
fli
A tiAii^ unity f or ontt is tepteaenttA by this chatactei^j 1
Fiv4
Six
■iieoeA
Eight
Nine \
■%>>
B
1
•''^=U.
u '"fa Ik •»; u ■ b ■*'■'' 4
Ten has "ho a{iipro^iate chai^abter ^o i^^^t ft t ^*^ is
consider fid as forming a u&it of a second of higher
order, consisting of tens, represented by the same
character (I) as a unit of the first br lowe^ order,
but is written in the second place ffom the tight
hand, that is, on the left hand side of Units ; and
as, in this caLe^ there are no units to be written
with it, we write in the place of units, a cipher, 0,
^ff^'Whichof itself sjgnifies 'nothing; thus, Ten
One ten and one iXnit a^e called ^r^^x^Mi^i ^Eleven
two " '•.,- ,/M>y'im0i'
three " :#»-#'?.^<?^-'^|
four •' ' **■' 'TM^^^ ^^■'"^*^-'
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7.
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ft
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I'wo tens are
Three "
Four •• '
Five
«ix
•Seven
Eight
Nine
ti
t'l
It;
II
hve
six
seven '*
eight
nine
'.M
rt ,,
*l.
*VttoeUt
Thirteeti
Fourteen
Fifteen
Sitteen
Seventeen
^Eighteen
'"^i Nineteen
'Tioenty
^Thirty
Forty
.Pifty '
nfv
10.
11;
12.
13.
14.
16
17.
la.
19.
20i
30.
40.
60.
60.
70.
80
90.
^^Sixty
i' 7^^'%;li>0.i^'p\rl^ Seventy
:v,:^vi;^»'**||^. . ■■■■ ^ Fighti/
^* Ninety
Ten tens aife called a hundred. Which, fofmS a Unit of a still
higher order, consisting of hundileds, represented by the same
thaifacter (1) as a unit of ea«;hof the ibriftgoing orders, but is
written one place further toward the left hand, that is on the
left hand side of tens ; thus, - - ^ one hundred • 100.
One hundred, one ten and one unit, arei called
One hundred and eleven 111.
■<*•■•.
^■■■•■■■:S
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NUMBHATiOIf,
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6.
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8.
a
3utis
igher
same
irder,
tight
«
; and
Titten
ler, 0,
fi "'
io.
n
IL
at
12.
teeii
13.
'teen
14.
'trt
15^^
.en
16;
aeeh
17.
iteen
la.
t
tecri
19>
nty
20;
t!f
30.
40.
60.
; '
60.
ntt/
70.
it^
80.
Bty
90>
it of
a still
y the same
iers»
but is
■it is on the
1
M
.100
yen
^ 9l ' There are three hiindre4 sixty-Ave days in a year,
1(1 this number are contained all the orders now described^
viz. units, tens, and hundreds. Let it be recollected, units
occupy the first place on the right hand ; tens^ the second
place from the right hand ; hundreds^ the third place, ^ This
number may now be decomposed^ that is, separated into parts,
exhibiting each order by itself, as fbllows i-^The highest or.>
der, or hundreds, are tkreCy represented by this character,
3 ; but, that it may be made to occupy the third place, counti.
ing from the right hand, it mustbe followed by Jtwo ciphers^
thus, 300, (three hundred.) The next lower order, or tens^
are six, (six tens are sixty,) represented by this character,
6; but, that it .may occupy the second place, which is the
place of tens, it must be followed by one cipher, thus 60,
(sixty.) The lowest order, or MM2^s, are five, re{^resented
by a single character, thus, 5, (five.)
We may now combine all these p^irts together, first writ-t
ing down the five units for the right hand figure, thus, 5 ;
then the six tens (60) on the left hand of the units, thus 65 ^
then the three hundreds (300)' on the left hand of the sij;
tens, thus, 365, which number, so written, may be read
three hundred, six tens, and five units ; or, as is more usual,
three hundred and sixty-five. i. 4 :rf. ^ « f
tl 3. Hence it appears, that figures have % different
value according to the place they occupy, counting fron^
the right hand towards the left,
V
M^vi->
''\]U3' A\v:}^
i^^l-*!^
m-i-
s'!^'^
Ill
Take for example the number 3 3 3, made by the same
figure three times repeated. The 3 on the right hi^nd, or in
i\vei first place, signifies 3 units ; the same figure, in the sec-
ond place, signifi^ 3 tens^ or thirty ; its value is now in->
creased ten times. Again, the same figure in the ^Airc? pi ace,
signifies neither 3 units^ nor 3 tens, but 3 hundreds, which
is ten times the value of the same figure in the place immen
diately preceding, that is, in the place of tens ; ^nd this ia
a fundamental law in notation, that a removal of one plcu^e
towards the left increases the value of a figure ten times.
Ten hundred make a thousand, or a unit of the fourth
or<ier, ^hen follow tens and hundreds of thousands, in the
1
12
IflHltitATMnf
T 3.
ssnie manner as tensjuid littiidreds of units. To tfaotsands
siieoeed miilims, biUhnt, &^. , to each of which, as to units
and to tboosands, are appropriated -Mree jdaces,* as exhib-
ited in the followinf^ exaropies :
Example Ist.
JSXAMPUB 3d.
3 174592887463512
3, 1 7 4,5 9 2,8 3 7,4 6 3,6 1 2
^o
ji'i li si's ^Jt i'iit'ii
iiiiii!i|Hiii|rM
(S scr ^ seh^ sSS ^k'S ^H.2 ^S
To facilitate the reading of large numbers, it is frequently
practised to point them off into periods of Mree.yf^res each,
as in the 2d example. The names and the order of the pe-
riods being known, this division enables us to read num-
bers consisting of mtmy figures as easily as we can read
three l^gures onlyt Thus, the above examines are read 3
(three) Cluadrillions, 174 (one hundred setenty-four) Tril-
lions, 592 (five hundred •ninety-two) Billions, 837 (eight
hundred thirty-seven) Millions, 463 (four hundred sixty-
three) Thousands, 512 (five hundred and twelve.)
After the same manner are read the numbers contained in
the following
«ThM U according to the Frencfa method o^ couoting. The English,
•Itwr httndr«dt of aullions, initead of prpeeediiM to billiont,. reckon
tbooeands, tens and handreds of tbousiuMls of aiUlionS} appropriating
«ii places, instead of threei to millions, billions, &c.
' • ■ 'Ti'-^;--- '•■■■■■ "* ■■•" ■•" ■
.•,vV'-b:V.;-V^.".\-'^, V
,?'V
113
NUMERATION.
NUMERATION TABLE.
13
I'hose words at the head of the
table are applicable to any sum or
number, and must be committed
perfectly to memory, so as to be
readily applied on any occasion.
Of these characters, 1 , 2, 3, 4, 5, B H ^ ffl H H W h ^3
6, 7, 8, 9, 0, the ninejirst are some-
times called significant figures, or
digits, in distinction from the lasty
which, of itself, is of no value, yet,
placed at the right hand of another
figure, it increases the value of
that figure in the same ten fold pro-
portion as if it had been followed by
any one of the significant figures.
8
5
6
4
2
3
4
7
8 6
3 3
5 4
7 1
9
Note. Should the pupil find any difficulty in reading the
following numbers, let him first transcribe them, and point,
them off into periods.
5769 52831209 286297314013
34120 175264013 .6203845761204
701602 3456720834 13478120673019
6539285 25037026531 341246801734526
The expressing of numbers, (as now shown,) by figures,
is called Notation. The reading of any number set down
in figures, is called Numeration.
After being able to read correctly all the numbers in the
foregoing table, the pupil may proceed to express the fol-
lowing numbers by figures :
1. Seventy-sibc.
2. Eight hundred and seven.
3. Twelve hundred, (that is, one thousand and two hun-
dred.)
4. Eighteen hundred. /
B
14 ADDITION OF SINGLE NUMBERSTV^ ^l 3, 4.
5. Twenty-seven hundred and nineteen. ''*>mn^-
6. Forty-nine hundred and sixty. •:,---
7. Ninety-two thousand and forty-five, "^^ryi^ ts^oirt:
'; 8. One hundred thousand, vw^-^fc^^' .'^•;J'te^'V!.'?.•
•''^ 9. Two millions, eighty thousands, and seven hundreds.
10. One hundred millions, one hundred thousand, one
hundred and one.
11. Fifty-two millions, six thousand, and twenty.
13. Six billions, seven millions, eight thousand, and nine
hundred.
13. Ninety-four billions, eighteen thousand, one hundred
and seventeen.
■■ 14. One hundred thirty-two billions, two hundred mill-
ions, and nine. . > .■■
15. Five trillions, sixty billions, twelve millions, and ten
thousand.
16. Seven hundred trillions, eighty-six billions, and seven
millions.
Addition of Siiii^ci UTtiinbers.
51 4. 1. James had five peaches, his mother gave him
3 peaches more ; how many peaches had he then 1
2. John bought *one book for 9 pence, and another for 6
pence ; how many pence did he give for both ?
3. Peter bought a wagon for lO shillings, and sold it so
as to gain 4 shillings ; how many shillings did he get for it ?
4. Frank gave 15 walnuts to one boy, 8 to another, and
had 7 left ; how many walnuts had he at first ?
5. A man bought a carriage for 54 pounds ; he expended
8 pounds in repairs, and then sold it so as to gain 5 pounds ;
how many pounds did he get for the carriage ?
: 6. A man bought 3 yoke of oxen ; for the first he gave
16 pounds, for the second he gave 18 pounds, and for the
third he gave 20 pounds ; how many pounds did he give for
the three?
7. Samuel bought an orange for four pence, and some
walnuts for three pence ; then he bought a knife for 1 shil-
ling, and a book for 4 shilling ; how many shillings did he
spend, and how many pence ?
If 4.
ADDITION OF SIMPLK NUMBERS).
15
8. A man had 3 c^ves worth 10 shillings each, 4 calves
worth 15 shillings each, and 7 calves worjlh 2 pounds each;
how may calves had het ■i'^>^ f (?* tv^r ,.i-^' ? . ., m, -^ l
9. A man sold a cow for 4 pounds, some corn for 5 poands,
wheat for 7 pounds, and butter for 2 pounds; how many
pounds must he receive t
The putting together two or more numbers, (as W the
foregoing examples,) so as to make one wJiole number, is
called Addition, and the whole number is called the sum, or
amount. ^- ' ' ' - — •■ - •- •-' ■■■■■• ; ^.'i. --.>;: r<^ ,
10. One man owes me 5 pounds, another 6 pounds, anoth-
er 14 pounds, and another 3 pounds ; what is the amount
due to me ?
11. What is the amount of 3, 7, 2, 4, 8, and 9 pounds?
12. In a certain school 9 study grammar, 15 study arith-
metic, 20 attend to writing, and 12 study geography ; what
is the whole number. of scholars? . , ,
SiG^s. A cross, +, one line horizontal and the other per-
pendicular, is the sign of addition. It shows that numbers,
with this sign between them, are to be added together. It
is sometimes resLdj^lus; which is a Latin word, signifying
more. : .... ;. „-.^ ■■. /.
Two parallell, horizontal lines, =, are the sign of equality.
It signifies that the number before it is equal to the number
after it. Thus, 5-f-3=8 is read 5 and 3 are 8 ; or, 5 plug
(that is, more) 3 is equal to 8. -../// •^^ n
In this mani
following
2-f0=
2
2+1=
3
2+2=
4
2+3=
5
2+4=
6
2+5—
7
2+6=
8
2+7=
9
2+8=
10
2+9=
11
ICC I.11C ^u^ix uc mail uuic
ADDITION TABLE.
3+0= 3
4+0= 4
5+0= 6 ,,
' '"t
3+1= 4
4+1= 5
5+1= a ;
r
3+2= 5
4+2— 6
5+2— 7 -
.'.y
3+3= 6
4+3= 7
5+3= 8 .,,
,|.
3+4= 7
4+4= 8
5+4= 9 ..
3+5— 8
4+5= 9
5+5= 10
^.'.ift
3+6= 9
4+6=10
5+6= 11
3+7= 10
4+7= 11
5+7= 19 ,
■y-
3+8— 11
4+8— 12
5+8=13
'-•/,
3+9= 12
4+9= 13
5+9b=14
■■■'-'■.<'
II 5.
ADDITION OF SIMPLE NUMBERS.
16
^ 6-f-0= 6
6+1= 7
64-2= 8
. 6-1-3= 9
6+4=10
, 6H-6= 11
, 6+6= 12
, 6+7=13
6+8=14
6-1-9= 15
7+0= 7
7+1= 8
7-f-2= 9
7+3=10
7+4=11
7+5= 12
7+6= 13
7+7= 14
7+8= 15
7-f-9=16
8-
8-
8-
8-
8-
8-
8-
8-
8-
8-
-0:
-1:
-2:
-3-
-4:
■5:
■6:
-7=
•8=
9=
8
9
10
11
12
13
14
15
16
17
9+0= 9
9+1= 10
9+2= 11
9+3= 12
9+4= 13
9+6= 14
94-6= 15
9+7= 16
9+8= 17
9-j-9=18
6+9=: how many ?
8-1-7= how many ?
ii_i_aj_Q — '
.m^-f'
4 ^ .
. 8-1-7= how many ?
J 4-j-3+2= how many?
6+4+5=: how many ?
2-f-0-j-4+6= how many?
7-f-8+0+8= how many?
9+3+3+4=: how many?
• ' 8-j-2-f-8-f-3+5= how many?
;" 54.7+6+1+8= how many?
*'^- 8-f_9_[.7+0+5+6= how many
' 4+1+0+4+4+5= how many
2-|-5-|-2-f-3-f-7-[-3= how many
A ■ I ..
i. ' ' * .
s- • >
" <' ■
51 S. When the numbers to be added are small, the ad-
dition is readily performed in the mind ; but it will frequent-
ly be more convenient, and even necessary, to write the
numbers down before adding them.
13. Harry had 43 bopks in his little library, his father
gave him 25 volumes more; how many volumes had he
then ?
One of these numbers contains 4 tens and 3 units. The
other number contains 2 tens and 5 units. To unite these
two numbers together into one, write them down one under
the other, placing the units of one number directly under
units of the other, and the tens of one number directly un-
der tens of the other, thus :
' 43 volumes. Having written the numbers in this
25 volumes, manner, draw a line underneath.
■^-'■■■■■^'f- ■
ADDITION OF SIMPLE NUMBERS.
ilM
43 volumes,
25 volumcSy
8 it''*.;
4;3 volumes,
We thon begin at the right hand, and
add the 5 units of the lower number to
the 3 units of the upper number, making ;
8 units, which we set down in unit's
place.
We then proceed to the next column,
and add the 'Z tens of the lower number
25 volumes, to the 4 tens of the upper number, mak-
— ing 6 tens, or 60, which we set down in
[j'is.68 volumes, ten's place, and the work is done.
It now appears that Harry's whole number of volumes is
|0 tens and 8 units, or 68 volumes ; that is, 43-4-25=68.. ■^f? .
14. A gentleman bought a carriage for 214 pounds, a
Ihorse for 30 pounds, and a saddle for 4 pounds ; what was
It lie whole amount?
Write the numbers as before directed, with units uncler
units, tens under tens. &/C. . ; ; »<i '» iuw i" '
OPERATION.
\Carr iage, '^I'i pomids, Add as before. The units will
Horse, 30 pounds, be 8, the tens 4, and the hundreds
iSaddle, 4 pounds, 2, that is, 2144-30+4=248.
ii-i(-
Answer, 248 pounds.
After the same manner are performed the following ex-
unples :
15. A man had 15 sheep in one pasture, 20 in another
)asture, and 143 in another ; how many sheep had he in
[he three pastures ? 15-j-20-|-143= how many ?
16. A man has three farms, one containing 500 acres,
linother 213 acres, and another 76 acres ; how many acres
n the three farms ? 500-f 213-f 76= how many ?
17. Bought a farm for 625 pounds, and afterward sold it
as to gain 150 pounds; what did 1 sell the farm for?
f25-|-150=:how many ?
Hitherto the amount of any one column, when added up,
las not exceeded 9 ; consequently has been expressed by a
tingle figure. But it will frequently happen tiiat the amount
y a single cotumn will exceed 9, requiring two or more fig-
ures to express it.
18. There are three bags of money. The first contains
B 2
'?»f;
18
ADDITION OF SIMPLE NUMBERS.
IT 6.
876 pounds, the second 653 pounds, the third 524 pounds ;
what is the amount contained in all the bags 1
OPERATION.
I\rst bag,
Second bag.
Third bag.
Amount.
Writing down the numbers as aU
876 ready directed, we begin with the
653 right hand, or unit column, and find
524 the amount to be 13, that is, 3 units
and 1 ten. Setting down the 3 1
2053 units, or right hand figure, in unit's
place, directly under the column,
we reserve the 1 ten, or lefl hand figure, to be added with
the other tens, in the next column, saying, 1, which we re-
served, to 2''makes 3, and 5 are 8, and 7 are 15, which is 5
units of its own order, and 1 unit of the next higher order,
that is, 5 tens and \ hundred. Setting down the 5 tens, or
right hand figure, directly under the column of tens, we re-
serve i\ie left hand figure, or 1 hundred, to be added in the!
column of hundreds, saying 1 to 5 is 6, and 6 are 12, andl
8 are 20, which, being the last column, we set down the!
whole number, writing the 0, or right hand figure, directlyl
under the column, and carrying forward the 2, or left haudl
figure, to the next place, or place of thousands. Where-f
fore we find the whole amount of money contained in the|
three bags to be 2053 pounds — the answer.
Proof. We may reverse the order, and beginning at thd
top, add the figures downward. If the two results are alike]
the work is supposed to be right.
From the examples and illustrations now given, we de
rive the following RULE.
I. Write the numbers to be added, one under anotherj
placing units under units, tens under tens, &c. and draw
line underneath.
II. Begin at the right hand or unit column, and add to
gether all the figures contained in that column ; if th|
amount does not exceed 9, write it under the colunm ; bi)|
if the amount exceed 9, so that it shall require two or morj
figures to express it, write down the unit figure only unde
the column ; the figure or figures to the left hand of unitJ
being tens, are so many units of the next higher ordef
which, being reserved, must be^carried forward, and addej
to the first figure in the next column.
III. Add each succeeding column in the same manneij
and set down the whole amount at the last column.
-<►■--
115
ADDITION OF SIMPLE NUMBERS.
19
EXAMPLES FOR PRACTICE.
19. A man bought four loads of hay ; one load weighed
1817 pounds, another weighed 1950 pounds, another 2156
pounds, and another 2210 pounds ; what was the amount of
hay purchased ?
20. A person owes A 100 pounds, B 522 pounds, C 785
pounds, D 92 pounds ; what js the. amount of his debts?
21. A farmer raised in one year 1200 bushels of wheat,
850 bushels of IndiRU corn, 1000 bushels of oats, 1086 bush-
els of barley, and 74 bushels of peas ; what was the whole
amount? Ans. 4210.
22. St. Paul's Cathedral, in London, cost 800,000 pounds
sterling; the Royal Exchange 80,000 pounds; the Man-
sion-House 40,0<)^ounds ; Black Friars Bridge 152,840
pounds; Westminster Bridge 389,000 pounds, and the
Monument 13,000 pounds; what is the amount of these
sums? -4ws. 1,474,840 pounds.
' 23. If at the rensXis in 1831, the population of the fol-
lowing counties \v a.-« as follows : — Lower Canada : Gaspe,
4,171, Dorche.<ter, 11,946; Nicolet, 12,504; Sherbrooke,
6,814; St.instpad, 8,272: Upper Canada: Gore, 23,552;
Home, 32,871; Niagara, 21,974; London, 26180; Otta-
wa, 4,456 ; what was the whole number of inhabitants in
these counties at that time? Ans. 152,740.
24. From the creation to the departure of the Israelites
from Egypt was 2513 years ; to the siege of Troy, 307 years
more ; to th^ building of Solomon's Temple, 180 years ; to
the building of Rome, 251 years ; to the expulsion of the
kings from Rome, 244 years; to the destruction of Car-
thage, 363 years; to the death of Julius Cesar, 102 years;
to the Christian era 44 years ; required the time from the
creation to the Christian era.
25.
2863705421061
3107429315638
625 30 34 792
2 4 7 13 5
867 3
Ans. 4004 years.
26.
43658302146 3 4
175 2 349713620
608 12753062 1 7
5653 174630128
87032634720 1 3
so SUPPLEMENT TO NUMERATION AND ADDITION. 51 5.
'27.
1 2 9 5 G 2 8 9 3 3 1 2 2
4 16439303468 1
7459601245786
123568934 2 1 5 5
9 7 3 2 1 5 4 6 7 1 U 9 8
28
2 8 9 5 4 3 6 10 8 3 2
3 4 6 2 I 8 5 (i I 3 2 5
5 7 8 3 2 1 4 5 (5 7 9 3 2
8 4 3 2 14 5 7 9 3 1
1346793 2 4578 2
.' 29. What is the amount of 5674,3335, and 9S6 pounds ?
■ ' 30. A man has three orchards; in the iirst there are 140
trees that bear apples, and 64 trees that bear peaches ; iii
the second, 234 trees bear apples, and 73 bear cherries ; in
the third, 47 trees bear plunus, 36 bear^ears, and 25 bear
fcherries ; how many trees in all the orchards ?
'.r'^
SUPPLEMENT
TO NUMERATION AND ADDITION.
QUESTIONS.
1. What \fA »'mf\e or indidividual thing called ? 2. What is nota-
tloii 1 3. Whlil are the methods of notation now in use ? 4. How ma-
ny are the Arabic characters or figures ? 5. VVhat is numer>ition ? G.
What is a fundamental law in notation 1 7. What is addition ? 8.
What is the rule for addition 1 9. What is the result, or number kought,
called ? 10. VVhat is the sign of addition ? 11. ——of equality 7 12.
How is addition proved '?
.*«*>-^f' "" ' EXERCISES.
t. Washington was born in the year of our Lord 1 /32 ;
he was 67 years old when he died ; in what year of our Lord
did he die ?
2. The invasion of Greece by Xerxes, took place 481 years
before Christ ; how long ago is that this current year 1849?
3. There are two numbers, the less number is 8671, the
difference between the numlsers is 597 ; what is tlie greater
number?
4. A man borrowed a sum of money, and paid in part
684 pounds ; the sum left unpaid was 876 pounds, what was
the sum borrowed ? .
6, 6.
SUBTRACTION OF ilMPLE NUMBES.
1 8
:j'i
(5 I 3
2 r.
() 7 9
3 2
670
3 1
457
8-i
^G pounds?
3re are
Hi)
Baches
; in
lerries
: in
nd^^o
bear
6. There are four numbers, the first 317, the second 812,
ihe third 1350, and the fourth as much as the other three;
^hat is the sum of them all ?
6. A gentleman lefl his daughter 16 thousand, 16 hun*
Ired and 16 pounds ; he left his son 1800 more than his
laughter ; what was his son's portion, and what was the
lount of the whole estate ?
Ans.
i Son's portion,19,416.
) Whole estate,37,032.
7. A man, at his death, left his estate to his four chil<
iren, who, after paying debts to the amount of 1476 pounds,
Received 4768 pounds each ; how much was the whole es>
ite? ^«5. 20548.
8. A man bought four hogs, each weighing 375 pounds ;
low much did they all weigh? Ans. 1500.
9. The fore quarters of an ox weigh one hundred and
^ight pounds each, the hind quarters weigh one hundred
pd twenty-four pounds each, the hide seventy-six pounds,
id the tallow sixty pounds ; what is the whole weight of
le ox? Ans. 600.
10. A man, being asked his age, said he was thirty-four
llears old when his eldest son was born, who was then fif-
3n years of age ; what was the age of the father ? ^
11. A man sold two cows for five pounds each, twenty ^^*'
^ushels of corn for three pounds, and one hundred pounds
' tallow for two pounds; what was his due? . .:. .
■ . i
>A.:)
V
:i\
'%.
ISubtraction or Simple IViimbei's*
T[ 6. 1. Charles, having 11 pence, bought a book, for
[hich he gave 5 pence ; how many pence had he left ?
2. John had 12 apples; he gave 5 of them to his brother ;
)w many had he left?
3. Peter played at marbles; he had 23 when he began,
It when he had done he had only 12; how many did be
se?
<'2'
%l.
SUBTRACTION OF MMPLI HUMBBES.
tl«.
4. A man bought an article fi^r 17 shillings and sold it
again for 33 shillings ; how many shillings did he gain ?
5. Charles is 9 years old, and Andrew is 13; what is the
difference in their agcs?{
6. A man borrowed 50 pounds, and paid all but 18; how
many pounds did he pay? that is, take 18 from 50, and
how many would there be leil?
7. John bought several articles for 10 shillings ; he gave
for 4 books 6 shillings; what did the other articles coet
him? ■>. ' ' 'i.» ■:;' , «■■ ; .;,.
8. Peter bought a trunk for 17 shillings, and sold it for
23 shillings ; how many shillings did he gain by the bar*
gain?
«0. Peter sold a wagon for 32 shillings, which was 5 shilU
ings more than he gave for it ; how many shillings did be
give for the wagon ?
10. A boy, being asked how old lie was, said that he was
S5 years younger than his father, whose age was 33 years;
how old was the boy ?
•i.U A'- 1.
The taking of a less number from a greater (as in the
foregoing examples) is called Subtraction. The greater
number is called the minuend, the less number the suhtror
hmdy and what is leil after subtraction, is called the differ'
enu QX remainder. . o > , .'. ; r v >
11. If the minuend be 8, and the subtrahend be 3, what j
is the remainder ?
13. If the subtrahend be 4, and the minuend 16, what is|
the remi^inder ?
, 13. Samuel bought a book for 11 pence; he paid downj
4 pence ; how many pence more must he pay ?
Sign. A short horizontal Ijine, — , is the sign of sub-
traction. It is usually read minus, which is a Latin word I
signifying less. It shows that the number after it is to be I
taken from the number fi^ore it. Thus, 8—3=5, is read!
8 minus or less 3 is equal to 5 ; or 3 from 8 leaves 5. The!
latter expression is to be used by the pupil in committing!
the following
:.^^^iM'i^-
HV.
SUBTRACTION OP SIMPLE NUMBERS.
that he was
s 33ycai8;
r (as in the
^he greater
the suhtra-
i the differ-
2—2:
3—2:
4—2:
6—2:
(5—2:
7—2:
8—2:
0—2:
10^2:
SUBTRACTION TABLE.
^t'
:0
:1
:2
:3
:4
:i>
:()
:7
:8
3—3:
4—3:
6— iJ:
7—3=
8—5=
9—4=
12—3=
13—4=
-J)
-A
6—3=3
7—3=4
8—3=5
9—3=6
10—3=7
4—4=0
5—4=1
6—4=2
7—4=3
8—4=4
9—4=5
10—4=6
5—5=0
(;-5=i
7—5 2
8—5=3
10—5=5
7—7=0
8—7=1
9 7=2
10^7=3
8—8=0
9—8=1
6—6=0
7— ()=1
8 6-2
9 6—3
10— <J 4
10—8=2
9—9=0
10—9=1
■J,
t.
how many ?
how many ?
how many ?
how many ?
how many 1
18— 7
28— 7
22—13
33— 5
41—15
how manyt
how many 1
how many ?
how many ?
how many?
sign of sub"
Latin word
it is to be
s=5, is read
ives5. The
committing
^ T. When the numbers are small, as in the foregoing
examples, the taking of a less number from a greater, is
readily done in the mind; but when the numbers are large^ "
I the operation is most easily performed part at a 'time, and
therefore it is necessary to write the numbers down before
[performing the operation.
14. A farmer having a flock of 237 sheen, lost 114 of
I them by disease; how many had he left?
Here we have 4 units to be taken from 7 units, 1 ten to
I be taken from 3 tens, and 1 hundred to be taken from 2
jhundreds. It will therefore be most convenient to write the
jless number under the greater, observing, as in addition, to
[place units under units, tens under tens, &c., thus:
We now begin with the
OPKRATION.
^rom 237, the minuend,
\Take 114, the subtrahend,
123
units, saying, 4(units) from
7 (units,) and there rem^a
3 (units,) which we set
down directly under the
column in unit's place. —
I V '
: I
i iit:^u,
S>
&4
SUPPEEMENT TO SUBTRACTION.
!I7.
15 shillings.
7 shillings.
Then, proceeding to the next column we sny 1 ten from 3
(tens,) and there remain 2 (tens,) which we set down in
ten's place. Proceeding to tlic next column, we say, 1
(hundred) from 2 (hundreds,) and there remains 1, (hun-
dred,) which we set down in hundred's place, and the work
is done. It now appears, that the numher of sheep left was
123: that is, 237— 114r=123.
After the same manner are performed the following ex-
amples :
15. There are two farms ; one is valued at 073 pounds, and
the other at 421 pounds; what is the difference in the value
of the two farms I
16. A man's property is worth 2170 pounds, but he has
debts to the amount of 1110 pounds ; what will remain after
paying his debts ?
17. James having 15 shillings bought a book for which
hfe gave 7 shillings ; how many shillings had he left?
OPERATION.
A difficulty presents itself here; for we
cannot take 7 from 5; but we can take 7
from 15, and there will remain 8.
8 shillings left.
18 A man bought several articles for 85 pounds, and oth-
er articles for 27 pounds ; what did the former cost him more
than the latter?
OPERATION. The same difficulty meets us here as inl
Pirst articles, 85 the last example; we cannot take 7 fronij
Other articles, 27 5 ; but in the last example the larger num
— ber consisted of 1 ten and 5 units, which!
Difference, 58 together make 15; we therefore took 'ij
from 15. Here we have 8 tens and 5 units. We can nowj
in the mind, suppose 1 ten taken from the 8 tens, whic
would leave 7 tens, and this 1 ten we can suppose joined to
the 5 units, making 15. We can now take 7 from 15, a^
before, and there will remain 8, which we set down. Tli
taking of 1 ten out of 8 tens, and joining it with the 5 units,
is called borrowing ten. Proceeding to the next higher or
der, or ticns, we must consider the upper figure 8, from whici
we borrowed, 1 less, calling it seven ; then, taking 2 (tens
from 7 (tens) there will remain five (tens,) which we sej
down, making the difference 58. Or, instead of making tin
j>i
7-
To
subtrah
csqual to
Top
the amo
duce it,
Thus
and 7—
From
(be folio
I. Wj
pJacing
lino und(
II. Be
the loiDer
mainder
III. V§
wt over
'"■'■ ■■'"' "'T. '-'•
■1 ' ' ■ f 7, 8. BHBTRACTION OF BIMPLB NUMBEBB.
26
en from 3
: down in
we say, 1
s 1, (him-
1 the work
op left was
lowing ex-
jounds, and
n the value I
, but he has I
cmain after
i for which!
;left?
lere; for wel
e can take 7|
n 8.
ds, and oth-j
lost him morel
IS here as inl
take 7 froinl
klarger numT
units, whici
;efore took 1
^Ve can now,
tens, whicli
ose joined tij
' from 15, aj
down. TM
h the 5 unitsJ
jxt higher orj
;, from whiclj
,ving 2 (tensi
[vliich we sej
if making tlij
ingthe upper figure, IJess, calling it 7, we^^Qr make the
lower figure 1 more, cdling it 3, and the result will be tbtt
same ; for 3 from 8 leaves 5, the same as 2 from 7.
19. A man borrowed 713 pounds, and paid 471 pounds ;
how many pounds did he then owet 713 — 471= how
many? Ans. 242 pounds.
20. 1612—465=howmany? Ans. 1147.
21. 43751— 6782=how many? Ans, 36969.
^ 8. The pupil will readily perceive, that subtraction is
the reverse of addition.
22. A man bought 40 sheep, and sold 18 of them ; how
many had he left ? 40 — IS^how many ? Ans. 22 sheep.
23. A man jold 18 sheep, and had 22 left; how many
had he at first ? 18 -|-22 =:how many ?
24. A man bought some articles for 75 pounds, and otk"
ers for 16 pounds ; what was the difference of costs ?
75 — 16=how many ? Reversed, 59-|-16=how many ?
25. 114— 103 = how many? Reversed, ll-f,103= how
Dany?
27. 143—76 = how many ? Reversed, 67+76 = how
many ?
Hence, subtraction may be proved by addition, as in tht
Cbregoing examples, and addition by subtraction.
To prove subtraction, we may add the remainder to the
subtrahend, and, if the work is correct, the amount will be
equal to the minuend.
To prove addition^ we may subtract, successively, from
the amount, the several numbers which were added to pro-
duce it, and if the work is right, there will be no remainder.
Thus 7+8+6 = 21; i>roo/, 21—6 = 15, and 15—8=7,
and 7—7=0.
From the remarks and illustrations now given, we deduee
fbe following
RULE.
I. Write down the numbers, the less under the greater,
placing units under units, tens under tens, &c., and draw a
line under them.
II. Beginning with units, take successively each figure in
the lower number from the figure over it, and write the re-
mainder directly below.
III. When the figure in the loweilr number exceeds the fi|^
I ve over it, suppose 10 to be added to the upper figure ; but
C
m
HK
m
■Mil
SUPPLEMENT TO SUBTRACTION-
US.
in this case we must add 1 to the lower figure in the next
ciyiumn bej^ Subtracting. This J^^alled borrowing 10.
EXAMPLES FOTl PRACTlCEv
27. If a farm and the buildings on it, be valued at 3000
pounds, and the buildings alone be valvcd at loOO pounds,
what is the value of the land ?
28. The population of Lower Canada, at the last census,
was 690782, at the census previous the census was 511917 ;
what was the difFerence in the two enumerations ?
29. What is the difference between 7,748,203 and
928,671 ?
30. How much must you add to 353,642 to rriako
1,487,945? "
31. A man bought an estate for 3798 pounds, and sold it
for 4137 pounsd ; did he gain or lose by it? arid how much !
:32. From 354,931,347,543 take 27,412,507,543.
33. From 824,264,213,909 take 631,245,653,3;j6.
34 . From 127,245,775,075,635 take 978,567,076,250.
SUPPLEMENT TO SUBTRACTION
, /■
QUESTIONS.
1. What Is subtraction? 1. What is the greater number call-
ed ? 3.
the less number? 4. What is the resvJt or answer
called? 5. What is the sif^n of subtraction? 6. What is iIih
rule? 7. What is understood by borrowing ten 1 8. Of what is
subtraction the reverse? 9. How is subtraction proved ? 10.
How is addition proved by subtraction ?
» FXERCISES.
1. How long from the discovery of" America by Colum-
];us, in 1492, to the period of the cession by France of all
li'.^r possessi(>ns in North America toGreat Britain in 1763?
2. Supposing a nmn to have been born in the year 1773, j
h.jw old was he in 1H48?
3. Su])posing a man to have been 105 years old in the!
TP.^r io4iS, in v;hat year was he born?
4. T'.ori? arc two numbers, whose difference" is 8 O'i; tiioj
^raat.cr number is 156&7; I demand the less?
tl 8-
SUPPLEMENT TO SUBTRACTION.
27
5. What number is that which taken from 3794, ileaves
865?
6. What number is that to which if you add 789, it will
become 63o0 1
7. In a certain city, there were 123707 inhabitants ; in
another 43,940 ; how many more inhabitants were there \v.
one than in the other?
8. A man possessing an estate of twelve thousand pounds,
jjave two thousand five hundred pounds to each of his two
daughters, and the remainder to his son; what was his son's
share?
0. From seventeen million take fifty-six thousand, and
what will remain ?
10. What number, together with these three, viz, 1301,
2501, and 3120, will make ten thousand?
11. A man bought a horse for 35 pounds, and a chaise
for 47 pounds ; how much more did he give for the chait-fe
than for the horse ?
12. A man borrows 7 ten dollar bills, and three one dol-
lar bills, and pays at one time 4 ten dollar bills and 5 one
dollar bills ; how many ten dollar bills and one dollar bills
must he afterwards pay to cancel the debt?
Ans. 2 ten doll, bills and 8 one do!
13. The greater of two numbers is 24, and the less is 16 ;
what is the difterence ?
14. The greater of two numbers is 24, and their differ-
ence 8 ; what is the less number ?
15. The sum of two numbers is 40, the less is 16; what
is tiie greater ?
10. A tree 08 feet high, was broken off by the wind ; the
top part which fell was 49 feet long ; how high was the
stump which was left ?
17. Elizabeth became Queen of England in 1558; how
many years since?
18. A man carried his produce to market; he sold his
pork for 14 pounds, his cheese for 11 pounds, and his but-
ter for 9 pounds ; he received, in pay, salt to the value of
pounds, 3 pounds worth of sugar, two pounds worth cf
molasses, and the rest in money; how much money did he
receive ? Ans. 23 pounds.
19. A boy bought several sleds for 13 shillings, and gave
shillings to have them repaired ; he sold them for 18 shili-
^■J
IIULTIPLICATION OF SIMPLB NUMBERS. 51 ^i 9l
II i^
I 't-
I :.,
p
I Iv
ings ; did he gain or lose by the bargain ? and how much t
. 20. One man travels 67 miles in a day, another man fol-
lows at the *rate of 42 miles in a day ; if they both start
from the same place at the same time, how far will they be
apart at the close of the first day ? of the second ?
of the third? of the fourth?
21. One man starts from Toronto Monday morning, and
travels at the rate of 40 miles a day ; another starts from
the same place Tuesday morning, and follows at the rate of
70 miles a day; how far are they apart Tuesday night?
Ans. .10 miles.
22. A man owing 379 pounds, paid at one time 47 pounds.
At another time, 84 pounds, at another time, 27 pounds, and
at another 143 pounds ; how much did he then owe ?
Ans. 82 pounds.
23> A man has property to the amount of 34764 pounds,
but there are demands against him to the amount of 14297
pounds ; huw many pounds will be left after the payment of
his debts ?
24. Four men bought a lot of land for 482 pounds ; the
first man paid 274 pound, the second man 194 pounds less
than the first, and the third man 20 pounds less than the
second ; how much did the second, third, and the fourth
man pay ? { The second paid 80.
Ans. < The third paid 60.
, t The fourth paid 68.
25. A man, having 10,000 pounds, gave away 9 pounds;
how many had he left ? Ans. 9991.
JHultiplication of Sfiniple Jluinber§.
ff 9. 1. If one orange cost 2 pence, how many pence
must I give for 2 oranges ? how many pence for 3 or-
anges ? for 4 oranges ?
2. One bushel of apples cost 3 shillings ; how many
•hillings must I give for 2 bushels ? for 3 bushels ?
3. One gallon contains 4' (juarts ; how many quarts in
8 gallons ? in 3 gallons ? in 4 gallons ?
4. Three men bought a horse; each man paid 6 pounds
V
y^\
V6.
IIUPTIPLICATION OF SIMPLE NUMBERS
for his share; how many pounds did the horse cost them?
5. A man has 4 farms worth 95 pounds each ; how ma>
ny pounds are they all worth?
6. In one pound there are 20 shillings ; how many shil*
lings in 5 pounds ?
7. How much will 4 pair of shoes cost at 9 shillings ft
pair ?
8. How much will 3 pounds of tea cost at 5 shillings a
pound ?
9. There are 24 hours in 1 day ; how many hours in
3
days?
10.
m
3 days'
m
4 days'
io 7 days ?
Six boys met a beggar and gave him 9 pence each ;
how many pence did the beggar receive ?
When questions occur, (as in the above examples,) wher«
the same number is to be added to itself several times, tb9
operation may be , facilitated by a rule, called Multipli-
cation, in which the number to be repeated is called ths
multiplicand, aad the number which shows how many times
the multiplicand is to be repeated is called the multiplier.
The multiplicand and multiplier, when spoken of collectively
are called the factors, (producers,) and the answer is called
the product.
11. There is an orchard in which there are 5 rows of treei
and 27 trees in each row ; how many trees in the orchard f
In this example, it is
In the first row, 27 trees,
second " 27
<(
a
*t
<(
third
((
27
fourth
((
27
fifth
<(
27
((
((
((
evident that the whoid
number of trees will ba
equal to the amount of
five27's added together.
In adding, we find
that 7 taken five times
In the orchard 135 trees. amounts to 35. We write
doi^n the five units, and reserve the 3 tens ; the amount of
2 taken five times is 10, and the 3, which we reserved,
makes 13, which, written to the left of units, makes tb«
whole number of trees 135.
If we have learned that 7 taken 5 times amounts to 35,
and that 2 taken 5 times amounts to 10, it is plain we need
write the number 27 but once, and then, setting the multi-
plier under it, we may say, 5 times 7 are 35, writing down
e 2
,4V-
30
UULTIPLICATION OF SIMPLE NUMBERS. ^ 9, 10.
« '■
B'H
i
the 5, and reserving the 3 (tens) as in addition. Again .'j
times 2 (tens) are
27 trees in each row.
5 rows.
Multipticand,
Multiplier, ^
Product, 135 trees, Ans.
10 (tens,) and 3,
(tens,) which w«i
reserved, make 13,
(tens,) as before.
*
«
iy 10, 12, There are on a board 3 rows of spots, and 4
spots in each row ; how many spots on the board ?
A slight inspection of the figure will
show that the number of spots may be
found either by taking 4 three times, (3
times 4 are 12,) or by taking 'i four tijties,
(4 times 3 are 12;) for we may say therf
are throe rows of 4 spots each, or 4 rows of 3 spots each :
therefore, we may use either of the given numbers for a
multiplier, as best suits our convenience. We generally
write thp numbers as in subtraction, the larger number up-
permost, with units under units, tens under tens, &.c. Thus,
Multiplicand, 4 spots. Note. 4 and 3are the /ac/cri-,
3Tultiplicr, 3 rows. which produce the product 12
Product, 12 4«^'
it ■ ■ •
Hence, — Multiplication is a short tfo/y of pctformin^
mam/ additions ; in other words — It is the method of repea-
ting any number any given number of times.
Signs. Two short lines crossing each other in the form
of the letter X, are the sign of multiplication. Thus, 3X4
=12, signifies that 3 times 4 are equal to 12, or 4 times 3
are 12.
Note. Before any progress can be made in this rule, th*?
following table must be committed perfectly to memory.
'rnW'^
..i..i^,;.»-iil . ;.
51 9, 10, ■ 5[ 10. MULITPLICATION Of SIMPLE NUMBGRJ^.
31
Again .'S
(tens) are
s,) and 5{,
which we
[,niakc IJJ,
IS before.
)ots, and 4
fiaure will
ots may he
'c times, {'i
four tijiiff,
y say tlierf
spots each ;
nbors for a
I generally
lumber uj>-
&.C. Thus,
ihe far furs,
roduct 12
pcrfnrminsi
d of rrpta-
in the form
rhus, 3X4
r 4 times U
lis rule, th*?
memory.
MULTICATION TABLE.
3X0
1
o
4
5
t) X
1 X
'IX
2 X
ii X
1 X «
■J X 7
^2 X 8
!2X 9
2X10
I 2 Xll
2 X12
:
: 2
: 4
: f)
= 8
= 10
= 12
= 14
= 16
— 18
= 20
— 22
1=24
4 X 10
4 X 11
12
^
40
44
48
\X
3 X 1
J X 2
^ X 3
:3 X 4
;j X r,
3X6
3 X 7
3X8
i X i)
:i xio
Xll
Xl2
3
6
9
12
15
18
21
24
27
30
33
36
o X
5 X
5 X
5 X
o X
r> X
."> X
5 X
•'S X
5 X
1
2
3
4
^
6
7
8
9
=
10
15
20
X 10
X II
X 12
25
30
35
40
45
50
X
X
X
X
X
X
7=
8:
0=
10:
1[=
12:
4X0
14 X 1
4X2
4X3
|4 X 4
4X5
4X6
|4 X 7
4X8
|4 X 9
4
8
12
16
20
24
28
32
36
i
2
3
4
6 X
6 X
OX
OX
OX
OX
OX
OX
OX
OX
OX 10
OX 11
ox 12
8X
8X
8X
8X
8X
8X
8X
8X
55 8 X
00,8 X
-^8X
8X
8X
5
6
7
8
9
6
12
18
24
30
30
42
'48
54
00
60
0z=
\-
o —
3::=
4=
5=
0=
1-
8=
9=
10=
11 =
12=
49
50
03
70
77
^
8
H)
24
32
40
48
50
04
72
80
88
9i
X
X
X
X
1
2
3
4
5
9X
9 X
9 X
9 "
9 "
9 "
9 "
9 "
9 "
9 "
, 7
14
21
28
35
42
0=
1=
o-
3=
A
5=
0=
7=
8r
9=
10=
11=
12=
^
10"
10" 1
10" 2=
10" 3
= 9
= 18
= 27
^- 30
= 45
-- 54
= 03
= 72
= 81
= 90
-- 99
= 108
1
=
= 10
= 20
-• 30
X 4=
X 5=
X 6=
X 7::=
X 8=
X 9=
xio=
Xll=
X12=:
40
55
00
70
80
90
100
110
1'^
X 0=
X 1 =
X 2zr:
3=
4=
5=
6=
X
X
X
X
X 7=
X 8=
X 9=
XIO-
Xll=
12=
<(
0=:
1 =
2=
3—
5=:
6=
7=
8^
9=
10=
11=
12=
f
11
2y
33
44
55
66
/ 4
88
99
no
121
ija
12
24
36
48
60
72
84
96
108
120
133
144
MVLTIPLICATION OF SIMPLE NUIIBERI. <D ' ^J lO.
,
%
9 X 2t=how many ?
4 X'6 =how many ?
8X9 =:how many T
8 X 7*=how many ?
i X 5^=how many ?
4X3X2.^:24.
3X2X5 L~how many ?
7X1X2 =liow many T
8X3X2 =:liow many ?
3X2X4X5=how manyf
arisATioiv.
253
dns. 3036
13. What will 84 barrels of flour cost at 2 pounds a bai^
fel ? Ans. 168 pounds.
14. A merchant bought 12 dozen hats at the rate of 12
pounds per dozen ; what did they cost 1 Ans. 144 pounda.
How many inches are there in 253 feet, every foot being
12 inches ?
The product of 12, with each of the signify
cant figures or digits, having been committed
to memory from the multiplication table, ii
is just as easy to multiply by 12 as by a singla
figure. Thus, 12 times 3 are 36, &/C.
16. What will 476 barrels of fish cost at 3 pounds a baiu
lel ? • Ans. 1428 pounds.
17. A piece of very valuable land, containing 33 acres,
was sold for 246 pounds an acre ; what did the whole come to t
As 12 is the largest number, the product of which, with
the nine digits, is found in the multiplication table, ther^
fore, when the multiplier exceeds 12, we multiply by each
figure in the multiplier separately. Thus :
OPERATION.
^ ^ 246 pounds, the price of one acre. *' The multipR-
33 number of acres. ~^ er consists of 3
■ ' tens and 3 units.
7*SS pounds, the price of three acres, First, multiply-
738 pounds,thepriceof thirty acres, ing by the 3
units gives ns
Ans. QWS pounds, the price of 33 acres. 738 pounds thfi
price of 3 acres. We then multiply by the 3 tens, writing
the first figure of the product (8) in ten''s place, that is, di-
rectly under the figure by which we multiply. It now ajv
paars that the product by the 3 tens consists of the same figures
as the product by the 3 units ; but there is this difference--
tlie figures in the product by the 3 tens are all removed ons
place further to the left hand, by which their value is in-
creased tenfold, which is as it should be, because the priee
^ ' ■ H 1^' MULTIPLICATION OP SIMPLE NUMBERS.
T
?
lanyf
jnds a baN
68 pounds.
rate of 13
44 pounds.
foot being
the signifi-
committed
n table, it
by a singia
mds a bai^
2S pounds,
g 33 acres,
le come to f
rhich, with
ible, ther&>
)ly by each
le multipR.
msists of 3
ind 3 units.
multiply-
by the 3
gives in
)ounds ths
, writing
:hat is, d>
now ap-
ime figures
fferenc^—
noved oma
alue is in-
the pri««
of 30 acres is evidently ten times as much as the price of
3 acres, that is, 70380 pounds ; and it is plain that theas
two products, added together, give the price of 33 acres.
These examples will be sufficient to establish the follow-
ing
* RULE. »
I. Write down the multiplicand, under which write thi
multiplier, placing units under units, tens under tens, &c.,
and draw a line underneath.
II. When the multiplier does not excd^d 12, begin at
the right hand of the multiplicand, and multiply each figurt
contained in it by the multiplier, setting down and carrying
the same as in addition.
III. When the multiplier exceeds 12, multiply by each
figure separately, first by the units, then by the tens, &c.,
remembering jdways to place the first figure of each pro-
duct directly under the figure by which you multiply.
Having gone through in this manner with each figure io
the multiplier, add their several products together, and tha
•um of them will be the product required.
EXAMPLES FOR PRACTICE.
18. There are 320 rods in a mile ; how many rods aa
there in 57 miles ?
19. Suppose it to be 706 miles from Halifax to Quebec-;
how many rods is it ?
20. What will 784 chests of tea cost, at 17 pounds a
chest?
21. If 1851 men receive 758 pounds apiece; how many
pounds will they all receive ? Ans. 1403058 pounds.
22. There are 24 hours in a day ; if a ship sail 7 mile»
ill an hour, how many niiles will she sail in 1 day, at thai
rate ? how many miles in 36 days ? how many miles in 1
year, or 365 days ? Ans. 61320 miles in 1 year.
23. A merchant bought 13 pieces of cloth, each pieca
containing 28 yards, at 2 pounds a yard ; how many yards
were there, and what was the whole cost 1
Ans. The whole cost was 728 pounds.
24. Multiply 37864 by 235. Product, 889804a
25. « 29831 " 952. " 28399M3.
26. ♦* 93956 **8704. " 817793024.
34
CONTRAC^riOXS I.V MULTIPLICATION.
IT 11.
-^r' CONTRACTIONS IN MTTLTIPLICATJON.
1. When the multiplier is a composite number.
^ 11. Any number, which may hi prochiccd by the
multiplication of two or more numbers, is called a rompos-
ite number. Thus, 15, which arises from the multiplica-
tion of 5 and 3, (;»X'i-^i«'5,) is a composite number, and
the numbers 5 and W, which, multiplied together, produce
it, are called component parts, or factors, of that number.
So, also, 24 is a composite number ; its component parts, or
factors riKiy be '2 and 1*2, ('iX VH^^'IA ;) or they niay be 4
and 0, (4X0=24;) or they may be 2, 3, and 4. (2X3
X4-z:24.)
What will 15 pieces of cloth cost, at 4 pounds a piece ?
15 pieces are equal to 5X3 pieces. The cost
of 5 pieces would be 5X4=20 pounds; and be-
cause 15 pieces contains 3 times 5 pieces, so the
cost of 15 pieces will evidently be 3 times the
cost of 5 pieces, that is, 20 pounds X3=-(iO pounds.
Ans. 60 pounds.
1.
4
6
%)
60
Wherefore, If the multiplier he a composite number, we
may, if we please, multiply the nr'OiplicandJirst by one of
the component parts ; that product by the other, and so on, \{
tJje component parts be more than two; and, having in this
fe way multiplied by each of the component parts, the last pro-
' duct will be the product required.
2. What will 130 tons of potashes come to, at 24 pounds
^ per ton ?
6X4=24. It follows, therefore, that 6 and 4 are cora-
p<inent parts or factors of 24. Hence,
I 1 36 tons.
Gone of the component pans, or factors.
- 816 pounds, the price of 6 tons.
4 the other component part, or factor.
An?,. 3264 pounds, the price of 136 tons.
3. Supposing 342 men to be employed in a certain piece
of work, for which they are to receive 28 pounds, each,
how much will they all receive ?
7x4-=;28 Ans. 9576 pound.^.
^ 12, 13. CONTRACTIONS IN MULTiruCXTlON.
4. Multiply J3G7 by 48.
ti..
853
108G
^ if
n
7-4.
Product, : 701(1.
477()H.
11. When the multiplier is 10, 100, 1000, &-c..
^ 12. It will be recollected, (if 3.) tbjit any figure, on
being removed one place towards the Irft hand, has its vai-
vio increased tmjhld ; hence, to multiply any number by 10
it is only necessary to write a cipher on the ri^ht hand of it.
Thus, 10 times 25 arc 250; for the 5, which was units be-
fore, is now made tcn.'i, and the 2 which was tens before, is
now made hundreds. So, also, if any figure be removed
/j/'f/ places towards the left hand, its value is increased 100
times, &C.
Hence,
\Vhen the multijdicr is 10, 100, 1000, or 1 with any num-
htr of ciphers amAxed, annex as many ciphers to the niulti-
plicand as there are ciphers in the multiplier, and the nuil-
liplicand, so increased, will be the propuct required. Thus,
Multiply 40 by 10, the product is 400 ,
83 j)y 100, '' 8300 .
" 95 by 1000, •* - ' 95000
EXAMPLES FOR PRACTICE.
are com-
1. What will 70 loads of corn cost at 10 pounds a load ?
2. If 100 men receive 32 pounds each, how many pounds
will they all receive?
3. What will 1000 pieces of broadcloth cost, estimating
loach piece at 78 pounds?
4. Multiply 5082 by 10000.
5.
82134 " 100000.
^ 13. On the principle suggested in the last ^, it fol-
Jows,
When there are ciphers on the right hand of the multipli-
cand, multiplier, either or both, we may at first neglect
these ciphers, multiplying by the significant figures only ;
ifter which we must annex as many ciphers to the product
IS there Jire ciphers on the right hand of the multiplicand
id multiplier, counted together.
CONTRACTIONS IN MVLTIPLICATION.
*
EXAMPLES FOR PRACTICE.
Via
1. If 1300 men receive 460 pounds apiece, how tnanj
pounds will they all receive ?
OPERATION. The ciphers in the multiplicand
460 and multiplier, counted together,
1300 are three. Disregarding these, w«
■ write the significant figures of ths
138 multiplier under the signijicant fig-
40 ures of the multiplicand, and mt^
tiply; after which we annex thres
Ans. 598000 pounds, ciphers to the right hand of ths
product, which gives the true answer.
2. The number of distinct buildings in New England,
appropriated to the spinning, weaving, and printing of coU
ton goods, was estimated, in 1826, at 400, running, on an
average, 700 spindles each ; what was the whole numbsr of
^)indles ?
3. Multiply 257 by 63000.
4. " 8600 " 17.
6. " 9340 " 460.
6, ♦• 5200 " 410.
T. " 378 " 204.
OPERATION.
378
204
1512
000
756
77112
In the operation it will be f een, that mnV
tiply ing by ciphers produces nothing. Th«r»*
fore,
ni. When there are ciphers between the significant jif^
mres of the multiplier^ we may omit the ciphers, multiply-
ing by the signijicant figures only, placing the first figurs
of each product directly under the iigtire by which we mul- 1
EXAMPLES FOR PRACTICE.
& Multiply 154326 by 3007.
I liV
V If
li », I-. .Ikii i
li.K '■■li' . ' I
•;: .V ,'•
.»:of/'»» f'ltv vn<f rr T/*»f.i.»''" ^?
hi;
svrrvcMCNT tu multu'jlication. . -v | 37
I ^/
> ' . '
, , , , 1080283
,if
'(/if rt 'i'
9. Multiply 543 by 206. ,, ,.. ,'..,,, i- ' ^ '
10. " 1620 •• 2103.
11. J, i" 36243 " 32004.
:Un 11. "^ •' / n* ' 11'
^UP^lfeAlENf h^b MtfLtlPLte'ATtbN.
'Xt.
QUESTIONS.
1. What is Tnultiplicsition ? 2. What is the number /o &«
mvltiplied ici^Wtid 1 3. to multiply by called? 4. .What
is the result or answer called ? .5. Taken iiM^divcly, what arc
the multiplicand and multiplier called ? 7. What ia the sign of
I multiplication? 7. What does it show? 8. In what orcfcr m^sl
[the given numbers be placotl for multiplication ? 6. How do
I you proceed when the multiplier is less than 12 ? 10. When it
\vxeeeds 12, what is the method of procedure? 11. What is u
\com})Osite nurahet V 12* What is to be understood by the corn -
lf)on«n^ partSf or /actors, of any number? 13. How may yoti
Iproceed when the multiplier iasi composite number f 14. To
Imultiply by 10, 100, 1000, &c., what suffices? 15.. Why?
116. When there are cipher's on the righi hand of the mullipii-
cand, multiplier, either or both, how may we proceed? 17.
When there are ciphers between the significant figures of the
tiulti plier, how are they to be treated ?
EXIJRCISES.
1. Ar army of 10700 men having pliindercJ a city, took
o much money, that, when it waa shared among them, each
an received 46 pounds ; what was the sum of money tak^n ?
2. Supposing the number of houses in a certain town to
145, each house, on an average, containing two famiHc»>
nd each family 6 members, what vi ould be the number of
nhabitants in that town? ;', \ Am. 1740
3. If 46 men can do a piece of work in CO days, how
any men will it take to do it in one day ?
D
• ■:!..', 7
•( ,! .
»f
^
SUPPLEMENT TO MtLTIPLICATIOIf.
II 13.
> iX
4 Two men depart from the same place, and travef in
opposite directions, one at the rate of 27 miles a day, the
other 31 miles a day ; how far apart will they be at the end
of 6 days ? Ans. 348 miles,
5 Whet number is that, the factors of which are 4, 7,6, 1
and 20? iln*. 3360.
6. If 18 men can do a piece of work in 90 days, how |
long will it take one man to do the same? ' ^
7. What sum of money must be dividtsd be«ween 27 men, |
so that each man may receive 1 15 pounds ?
8. There is a certain number, the factors of which are|
89 and 265 ; what is that number ?
9. What Is that number, of which 9,^12, and 14 are fac«
tors?
IP. If a carriage wheel turn round 346 times in running!
1 mile, how many times will it turn round in the distance]
from Quebec to Montreal it being 180 miles.
'■/':■::,_ ".\. ''"''' .i^-^.-''i--'*>i- ' -^n*. 62280.1
II. In one mihute ate 60 seconds; how many secondsj
in 4 minutes? -^^i— in 5 minutes?—^ — in 20 minutes i|
* in 40 minutes ?
12 In one hour are 60 minutes; how many seconds inl
,«n hour ? — — in two hours ? How many seconds firoin|
nine o'clock in the morning till noon?
13. In one pound are 4 dollars ; how many dollars in 3|
pmmds ? - — " in 300 pounds ? in 467 pounds ?
14. Two men, A and B, start from the same place at thel
same time, and travel the same way ; A travels 52 miles al
day, and B 44 miles a day ; how far apart will they be at|
the end of 10 days?
15. If the interest of 100 pounds, for one year^ be sixl
pounds, how many pounds will be the interest for 2 years!!
for 4 years? for 10 years ? for 35 years ?|
for 84 years ?
16. If the interest of one hundred pounds, for one year,!
|>e six pounds, what is the interest for two hundred pound/
the same time ? 7 hundred pounds? 8 hundred!
pounds ? 5 hundred pounds ?
17. A farmer sold 468 pounds of pork, at 3 pence ij
pound, and 48 pounds of cheese, at 4 pence a pound ; ho>v|
hiany pence must he receive in pay ?
* 18. A boy bought 10 oranges ; lie kept 7 of them, and sold
^1'- Y ■- " ''['^-V
the otheril for 5 peiice a piece; how muiy pence did he riv
ceiveT '•■■.;■•/'■ -a-''
10. The ccunponent parts pjf a certain huqaber ieure 4| 5, 7,
6,9, 8, and 3; what is the number?
SO. In 1 hogshead are^3 gallons.; how inany gallons in
6 hogsheads? In 1 gallon are 4 quarts; how many quarts
in 8 hogsheads.' In 1, quart are 2 pints; how manj piats
in8h<^ea48^ .no ^^.. hm .qu'xi:- ni; bu. ;m.;,A Ml
.-{A;?
»'!. '^\-
' ' . j^^^^ of Siiuple numbers,
5| 14. 1. James divided 12 apples undng four boys; :
how many did he. give each boy ^
^. 'James would divide 12 apples fmppg three bo^rsi ^w
many >must he giye each boy .' ';'!m ^oif h^''^ m v» ' •< i' ^>^rfi
3. John had 15 apples, and gave them to his playmates,
who received 3 appl^each; ^ow many boys did he g^ve
thew to/
4. If you had ^ pence, how piany cakes could you buy
at i2 pence a piec^? "..:...;;.=,. ..;.. ..■>,>..,. n \^.v.^■^y \>jv-->>,
5. How many yards of cloth could you buy for ^ pounds,
at 2 pounds a yard .'
6. If you pay 250 shillings for 10 yards of cloth, what
is one yard wprth 7
7. A man works 6 days for 42 shillings ; how many shil-
lings is that for one day ? '•
8. How many quarts in 4 pints? - ■ in 6 pints,' — .—
in 10 pints?
9. Ilow many times is 8 pooLtaioed in 88.' - < :' < h, ■-;
10. If a man can travel 4 miles an hour, how many hours
would it talo him to travel 24 miles?
11. In an orchard ther^ ajre 28 trees standing in rows, ;,
and there are 3 trees in a row ; how many roivs are there ?
Remarh. When any one thing is divided into two equal ,
I parts, one of those parts is called a half; if into 3 equal,
parts, one of those parts is <^s^pd a third t if into four e-
qual parts, oiie part is called a quarter of n. fourth; if into .
Uivie, one part is called, aj(/ii(ft, and so on.
12. A boy had two apples, and gave one hi^f an Apple to
Neacb of his compapioi^; how man^ were his companioopf
DIVISION OF SIMPLE NUMBERS. ^J 14, 15.
Ti. A boy divided f£itiV appres amcHig*liiis companions^ 1}j '
' ffiving ihem one third of ai) appje each : amoa^ bow msLny
iid'hlaiViiie^M^lpiJMv •'•'• /^'^"- '- ^^^ ■ -"^^^ ;
14.. H-.V many quarters in three oranses ? ^ ,.,. ' ".'
15. How many oranges would it take to give J^ boyv
one
19. A man had 30 sheep, and sold one finn oif ttiem ;
how. many of them did he sell/
20. A man purchased shipep fo^* 13 shillings api^tc^, and
paid for theitt siR lift slnirings;' what Was theif rtiiinber ?
21. How many oranges, at 3 pep.ce each, mav be bought
for i^'^##c^ ?;.'■>'■■""'' -^hl" -f fc^;'^-''t-' • ; -J-'- ^ -' i '
It is plain, that as many tlm^^ '9 p^c^''c'^*B^'t|(k^^'
frdfei 13 p^ce, S^ many "oranges mky be bbti^ht if tfiti bbiect
therefore, is to find how many tiiioe^ 3 is ' cbntain^ in ISl
,»otn/i!7t;[<[ -■ ]^>^ene^; ''^''' - ' ^'^ t^'-'^i^V *-' '*^'' "*^'^*' •*■
Ftm'of^^^l' ^.•ptn(^/''- ' "^e'sfee m'«fiisfe^aih^fe;^^
. — we may take 3 from 12 /our
vf,d ffor l,fi(»-9? >•> 7t tlme^, «ftfer'wto there 'is lio
^ec£m(/ 0ra»^e, 3 pence. remainder; consieiqiiently,^ sti^'
,H>»arjoq ih. iui.vjM' mrjAiji^ctidn alone is sufficient for i
. 6 ^ the operation Voiit«^eiflky|c;bme
Tkird oi^titt^e, ^ 3 pfettcie.^ ^ id* the same' result by a" jirofces^,
~^ — in most pases ' ]6iiudi shorted,
-luh vrrnm v;r.il : ^nJlal. galled DmHon:''-. '■' -; '■: ■ ^
Jfowrth orange,J^ pence. v y^|,i> ,,j;<» mt buli j^i ^^,
■^ eii.iu tjQr:^ \ otfUT I- a; • •(; fi) (iijau T'^H '."'
•fy 15, It is plain, that the cost of one orange, (3 jjjfeticie^)
multiplied by the number of oranges; ^4,) is'equaf to the
cdW of al! 'tlie orangfes,' (12 pence j) 12 is, th^ttefore, ijiro-
duct, and 3 one of its factcfl^s*; 4nd' t6 ftnd^hoir nr&ny iiAies
3 i^'Cfc^ntainfed iii 12 is to 'find fhtt iyihef factorVlvhichJ p\rl-
ti^HM'Aito 3, will produce 12. .This factor W0 find by Wi^,
to be 4,* |[4><8==12;) consequently, ii^contiitied in 12, 4
iiihHl' '•'^''"' ■' '■■■'■^ .i:i '■.;':> ^» ^^-'^'V'^ns. 4 oranges.
i82,jA^niari^w6uld divide' liibrdng^'equaily artiohr 3
childteiir'hd^ inany orah^te wcj'uld e^ch chrl4 have? i
Here the object is to dime the 12 orange^ jnt'o 3 eqpal
paMSjr ami' id iisbertdin the '^mberof pranj[to i^ eacii o/
/
fi 14, 15. H ^ J5. DIVISION OF SIMPLE NUMBERS.
41
those parts. The operation is evidently as in the last ex-
ample, and consists in finding a number, which, multiplied
by 3, will produce 12. This number we have already found
to be 4. Ans. 4 oranges apiece.
As, therefore, multiplication is a short way of performing
many additions of the same number; so division is a short
way of performing many subtractions of the same number ;
and may be defined. The method of finding hoto many times
one number is contained in another; and also o{ dividing a
number into any number of equal parts. In all cases, the
process of division consists in finding one of the factors
of a given product, when the other factor is known.
The number given to be divided, is called the dividend,
and answers to the product in multiplication The number
given to divide by is called the divisor, and answers to one
of the factors in multiplication. The result, or answer
sought, is called the quotient, (from the Latin word quoties,
how many ?) and answers to the other factor.
Sign. The sign for division is a short horizontal line
between two dots, -r. It shows thjit the number before it is
to be divided by the number after it. Thus, 27-r9 =3, is
read, 27 divided by 9 is equal to 3 ; or to shorten the expre*-'
sion, 27 by 9 is 3; or 9 in 27 3 times. In place of the dots,
the dividend is often writteii over the line, and the divisor
under it, to express division ; thus, y =3, read as before.
The reading used by the pupil in committing the follow-
ing table may be 2 by 2 is 1, 4 by 2, &c., or 2 in 2 one
time, 2 in 4 two times, &c. ,
' DIVISION TABLE.
J=l
* =1
t -1
i=l
1=1
f = 1
1-2
f =^2
1=2
V»=2
V=2
V*_2
^r =3
t =3
^2 = 3
¥-=3
¥=3
V - 3
f -4
^=4
^^=4
¥==4
¥=4
^8 = 4
y>=5
¥=5
^0=5
¥=5
¥-5
V = 5
^ = 6
V«=6
V— 6
¥=6
¥ «
v=«
V_7
¥=7
2^8—7
V~7
V 7
V = 7
V«=8
¥=8
^2 -—8
¥ -8
V-8
V — H
V« =9
V =9
3i« = 9
V ~9
¥-9
V = 9
D2
m
V
V
DIVISION OF SIMPLE NUMBERS. f| 15, 16 ■ i[| 16.
DIVISION TABLE— CONTINUED.
if=l
11=3
tf=4
if =6
ff=8
W=9
=1
1=1
i«=l.
H-1
==a
1^=2
fe— 2
ff=2
=3
V=3
i^=3
f*=3
=4
3^=4
H-4
H=4
=5
V=5
f^=5
H-5
=6
V-6
H=6
ff=6
=7
V=7
H='7
H-7
==8
V-8
f^-8
f!-8
=9
V-9
H-9
ff-9
28-
42-
54-
32-
33
-7, or ^^= how iriany ? 49-^-7, or 4y9=how many?
-6, or ^= how many? 32^-4, or 3^=how many?
-9, or ^^= how many ? 99-^11, or ff=how many?
-8, or ^— how many? 84-^-12, or f|=how many ?
-11, or If =how many ? 108-^12, or V^^^how many?
H 10. 23. How many yards of cloth, at 4 shillings a
yard, can be bought for 856 shillings?
Here the number to be divided is 856, which therefore
is the dividend; 4 is the number to divide by, and therefore
the divisor. It is not evident how many times 4 is contain-
ed in so large a number as 856. This difficulty will be
readily overcome, if we decompose this number, thus :
856=800-1-40+16.
Beginning with the hundreds, we readily perceive that 4 is
contained in 8 2 times ; consequently, in 800 it is contain-
'"•GO times. Proceeding to the tens, 4 is contained in 4 1
time, and consequently in 40 it is contained 10 times.
Lastly, in 16 it is contained 4 times. We now have
200-f-10-j-4=:214 for the quotient^vor the number of times
4 is contained in 856. Ans. 214 yards,
We may arrive to the same result without decomposing
the dividend, except as it is done in the mind, taking it b)
parts, in the following manner :
Dividend. For the sake of convenience, we|
Divisor, 4 ) 856 write down the dividend with the di
visor on the left, and draw a line be
Quotient, 214 tween them ; we also draw a linei
underneath." Then, beginning on
the left hand, we seek how often the divisor (4) is contain^
U 15, 16 I m 16.
DIVISION OF SIMI>LE NUMBERS.
43
ed in 8, (hundreds,) the left hand figure ; finding it to be 2
times we write 2 directly under the 8, which falling in the
place of hundreds, is in reality 200. Proceeding to tens,
4 is contained in 5 (tens) 1 time, which we set down in
ten's place, directly under the 5 (tens.) But aft;er taking 4
times ten ouLof the 5 tens, there is 1 ten left. This 1 ten
we join to tR 6 units, making 16. Then, 4 into 16 goes 4
times, which we set down and the work is done.
This manner of performing the operation is called Short
Division. The computation it may be perceived, is car-
ried on partly in the mind, which is always easy to do when
the divisor does not exceed 12.
RULE.
From the illustration of this example, we derive this gen-
eral rule for dividing, when the divisor does not exceed 12:
I. Find how many times the divisor is contained in the
first figure, or figures, of the dividend, and, setting it di-
rectly under the dividend, carry the remainder, if any, to
the next figure as so many tens.
II. Find how many times the divisor is contained in this
dividend, and set it down as before, continuing so to do till
all the figures in the dividend are divided.
Proof. We have seen, (^ 15,) that the divisor and quo-
tient are factors, whose product is the dividend, and we
I have also seen, that dividing the dividend by one factor is
merely a process for finding the other.
Hence division and multiplication mutually prove each
i other.
To prove division, we may multiply the divisor by the quo-
I tient, and, if the work be right, the product will be the same
as the dividend ; or we may divide the dividend by the quo-
Uient^ and, if the work is right, the result will be the same
las tne divisor.
To prove Multiplication, we may divide the product by
lone factor, and if the work be right, the quotient will be the
other factor.
EXAMPLES FOR PRACTICE.
24. A man would divide 13,462,725 pounds among 5
[men; how many ponnds would each receive?
44
DIVISION OF SIMPLE NUMBERS.
II 16, 17. ■ fl 17.
OPERATION.
Dividend.
Divisor, 5)13,462,725
Quotient, 2,692,545
Proof.
Quotient.
2,692,515
5 divisor.
In this example, as we cannot
have 5 in the first figure, (1) we
take two figures, and say 5 in 13
will go 2 times, and there are 3
over, which, joined to 4, the next
figure, makes 34 ; and 5 in 34
will go 6 times, &-c.
In proof of this example, we
multiply the quotient by the di-
visor, and, as the product is the
same as the dividend, we con-
elude that the work is right. —
13,462,7515 From a bare inspection of the
above example and its proof, it is plain, as before stated,
that>division is the reverse of multiplication, and that the
two rules mutually prove each other.
25. How many yards of cloth can be bought for 4,354,560
shillings, at 2 shillings a yard? at 3 shillings? at
4 shillings ? at 5 shillings ? at 6 shillings ? at
7? at8?— T-at9?atl0?
Note. Let the pupil be required to prove the foregoing,
and all of the following examples.
26. Divide 1005903360 by 2, 3,4, 5, 6, 7, 8, 9, 10. 11,
and 12.
27. If 2 pints make a quart, how many quarts in 8 pints?
in 12 pints ? in 20 pints ? in 24 pints ?
in 248 pints ? in 3674 pints? in 47632 pints?
28. Four quarts jnake a gallon ; how many gallons in 8
quarts ? in 12 quarts ? in 20 quarts ? in 36
quarts ? in 368 quarts ? in 4896 quarts ? in
5436144 quarts?
29. A man gave 86 apples to 5 boys ; how many apples
would each boy receive ?
Dividend. Here, dividing the
Divisor, 5)86 number of the apples
— (86) by the number of j
Quotient^ 17 — 1 Remainder. boys, (5,) we find, that
each boy's share would be 17 apples j but there is 1 apple
left.
1] 1 7. 5)86 In order to divide all the apples equal-
ly among the boys, it is plain, we must
17^ divide this one remaining apple into 5
DIVIBION OF SIliPLE NUMBERS.
45
tqMl parts, and give one of these parts to each of the boys.
Tl}«n ,^aph boy'ssljare vWQujd.be 17,ppples, and o^^ fifth
part pf WPthpr applq ; ;,^hiph is jyritten thus.^ ,lf7ff(appJle*:A
,,i, ; ,;.,;; 1 -4i»^. 17^ appleaeach.
Tfjet^ 17, expressing pohof^ , ;app|es„ . ^ue caUi^d . int^^rs^, ;>
(that i^j vH<! n««»^f^),i'ith^ i,(p9e,,6Al?)of an^iippW^-'ii
expressing part of a broken or divided apple, is c^lledia, .|
\ fraction, (that is a broken number.)
Fra^tipip^ as^e l^ere see, arjS VKrjtteiii wit^ two, iimpbers^o,^
one ^fre^tly j([^,eji;, tUe, Qther, ; wit^ a shor,t line; betwee»f thf^ttif. ,[j
I showing that the upp^r^ nuQij^er is. tp b|e divided by tiic^ |
/ou'er. , Tl^f |Upper , nmi^ber, or dividei^, 19, in fracti.(W(8»;^
cmedti^f(^e,nfipl^na^or,[ , :*, .<:f . >! tu .'> (Tsr H
iV(^0. A nuniber like 17-^, composed of integers (17)
I and a fraction^ {-^y) is cadled a mixe^ number, i> ^^< i< {'« ^
lo the pTfecedin^ example, the dn^ apple, whi6h W^s len
I after eawyirtgitlife divisioh ai^ 'far as ooiild be by whole ndiii-''^
bers, i6 cdied the f*(^^iniEfer,'and is evidiently ja part of the ^
/fiw'ffiiitf yet lindivided. ; In order td c6rtii)Jete the divisiprj^^"
this remain i6r, a» w^ before remai'ked, must be divide!^ into *
5 equal purta; biit thtt' divisor itself ex]f)resses the numbef ,
of parts.' If, now, we examine the fraction, we shall se6;' '
that it consists of the remaihdier (l)'f6r its nvmeraior. and
the divisor <(&) for ltd a^ndmina^er' ..
Therefore, if there' be i remainder, mi ii down,' at file
right )iand of the quotient for iiie numerator of a fraction,
under^ which write the divisoir for it^ ■denominator. .,'
In proving this example, we fiud
it necessary to, multiply our fi-action
by 5 ; but this is^ ^^}^}(\ <lone, if yv.e
• — consider, tl^att]i^/raciion I express*
86 es owe psy-t pf"aq, apple divided into
equal parts ; hencQ, _5 tinies \ is 4:^5:1, |tha|: is,, oiie j^hol^
[apple, which, we reserv* to, oe .^^ed tp (he if,ni/s, saying,^ p
times 7 are 35,, aiiii.one we reserved makes 36, &c. j,,. ;,: , ,
30. 'Eighi nl^ri drew a'.'bduniy of >^53 pounds from goV-j,
jrnment, pow ihany pciuiids 'did each, irecg;v.e?;j,\ a,-, , oi^cn ^
\ Proof of last example.
. ... m '
-■•' •'■.3-
46 DIVIfllON OF SIMP1.E NUMBKRB. . 51 l^i l^'.
> " ' Dividend. ' Here, after earrying- the division aii' '
I>im)>M', 8)463 far aa possible hy whdle numhetB, we
' have a remainder of 5 pounds, which,
Qutftient^' 66| written as above directed, giVes for the
ans#er 66 pounds and f (five eighths) of another potind, fo
«aeh man, , , ' »
5T 18. Here we may notice, that the eighth part of 5
pounds is the same as 5 times the eighth part of I pound,
that iSf the eighfth part of 5 pouhds is | Of a pound. Henee,
f ei^reaees the quotient of 5 divided by 8.
Proof. f is 5 parts, and 8 times 6 is 40, thait \k, V=^«
S6| wtiieh, resei^ed and added to the prbdnc^t oif 8
8 times 6, makes 53, &c. Hence, to muHipty a'
—— fraction^ we may mattiply by the numerator,
4«'>3 and divide the product by the denominator.
Qr, in proving division, we may multiply the whole num*
ber in the quotient only, and to the product adx) the remain-
der | asd this, till the pt^pil shall be more particularly taught
in fractions, will be more easy in practice. Thus, 56X6;==
448, and 4484-^> the remainder, =453, as before.
31. T'lere ^re 7 ,day» i|»>a weftt; hpw many weeks in
365 days? ..(,... ,. ilras. 52^ weeks.
32. When flour is worth 2 pounds a barrel, how many
barrels may be bought for 25 pounds? how many for 51
pounds t for 487 pounds ? for 7631 pounds ?
93. Divide ^40 pounds among 4 men, ''
640-7-4, or «f 0=160 pounds, ilna.
/ ',.r Am. 113.
■*\
liH!
'W \
Ana. 3d4f,
34. 678^6 or 6|8=howmany?
35. *'^«>=how many?
36. Ty*=^howmany1
37. »^*=how many?
38. «|f *=how many!
39. 4o|oi— howmany? ' ' '"
40. »o ij4jt> 1 2^how many f
il 19. 4L Divide 4<370 pounds equity among 21 men.
When, as in this example, the dfi visor exceeds 12, it is
evident that the computation cannot be readily carried on
in the mind, as in the foregoing examples. Wherefore, it
is more convenient to write down the computation at length
Vi^ the following manner :
1119.
DIVISION OF SIMPLE NVMBCRS.
47
" OPERATION. We may write the divisor
Divisor, Dividend, Quotient, and dividend a» in short di-
,. 21 )4370(208/r. vision, but instead of writing
42 ., , the quotient undtr the divi-
^, " ' dend, it will be found more
170 ( convenient to set it to the
168 ,! right hand, 'A
2 Taking the dividend by
parts, we seek how oflen we can have21 in 43 (hundreds;)
finding it to be 2 times, we set down 2 on the right hand
of the dividend for the highest figure in the quotient. The
43 being hundreds, it follows, that the 2 must be hundreds^
This, however, we need not regard, for it is to be followed
by tens and units, obtained from the tens and units of the
dividend, and will therefore, at the end of the operation,
be in the place of hundreds, as it should be.
It is plain that 2 (hundred) times 21 pounds ought now
to be taken out of the dividend ; therefore, we multiply the
divisor (21) by the quotient figure 2 (hundred) now found,
making 42 (hundred,) which, written under the 43 in the
dividend, we subtract, and to the remainder, 1, (hundred,)
bring down the 7, (tens,) making 17 tens.
We then seek how often the divisor is contained in 17,
(tens;) finding that it will not go, we write a cipher in tho
quotient, and bring down the next figure, making the whole
170. We then seek how often 21 can be contained in 170,
and, finding it to be 8 times, we write 8 in the quotient, and
multiplying the divisor by this number, we set the product,
168, under the 170 ; then subtracting, we find the remain-
der to be 2, which, written as a fraction on the right hand
of the quotient, as already explained, gives 208^ pounds,
for the answer.
This manne'- of performing the operation is called Long-
Division. It jnsists in writing down the tcAo/ie computation.
From the above example, we derive the following
, ^ :.>:■■*;. ■ RULE. ,;/' ■
I. Place the divisor on the left of the dividend, separate
them by a line, and draw another line on the right of the
dividend to separate it from the quotient.
II. Take as piany figures^ on the left of the dividend, a»
d
I..
<\i
S.'f.',
0(1 1
DIVISION OF, SIMPLE NUMBERS. ^ 19,
'f'cbntttiritliediv'iaoi' oti<iebr rnorc; seek Kow ftia'tiy (Jmesthey
J contain ii, and place 'the answer on the right hand of the
- dividend fbr the 'first fixture in the quotient.
•>■' 'III. Multii^ty the t'ivii^r by this quotient figure, and
'- 'Write the prodiict under that part of the dividend taifen.
IV. Subtract the product from the figures above, and to I
' tlie remainder brin^ down the next figure in the dividend,
V anddiVid^ th^tiumber it makes up, as before. So continue
^tod<>;till all the figures in the dividend shall have been
''■ b!*ought doWn and dividedl'^ ;« •^•"» ' ^^ ;'''*,' ^\ >";';" 1
i'* ' i Note 1 . Having bi*6ught AowA a figibtb t(Nhe remainder,
'if thenurriber it makes up be less than the divisor,' write a|
'cipher in the quotient^ and bringdown the next figure.
' '■ 'iVo*i?2. ' If the product of thfe divisor, by any quotient I
.' fi^ure^ be ^rgft^fer than the part of the dividend taken, it is
an evidence that the quotient figure is too ?ar^^, knd must
^' be diittini^^di it the remainder tit any time be gf eaten
■ than, the divisor, or equal to it, the quotient figure is too\
'' ii/ia/f^ a^d mtist be increased. '
:>:i! .U bl fJji) V.'l'l EXAMPLES 'TOR fAACnCE/'); ''^ ,-''^"; ■!',|;*
(, 1) j^l , 'll^W' maliy hrtgshet ds of molasses, at 7 poun^^' a hogs-
head, may be bought for 6318 pounds ?
."^' i • Ans. 903| hogsheads.
;•>'*' ;8. If ti Wah's tTiciome be 124B poiinds a year, how much |
h that pet' week^ there beirig ^2 weeks in a year f ' ^•'; V^/
j.'ki rn !.,.;■. u\ .- '•■.; ir: i 1 - i.-.'M. ' j^j^g. 24 fiounds perAveek,
i>nf- .». What Will be the '(^utrtient of 153598, divided by 29 ?l
.'•"'*•■>' ■■■'■■ ' '• i4»5.'5296f|,
- ii!' '4. MtiW mariy tim^s is 6? coritairied in 30131 ?
^"'-Urtsi478|f times; tha| is, 478 time^, and^J of anbtherl
5. What will be the several qitotientsof 7652, divided hv
\ 16"; SJ3, 34, 86, and 92?
'"6. If a farm, containing 256 acres, be worth 1850 Jjpunds J
what Is that per acre ?''^' '^ , , aiz: i.-<^ , i; . .
7. What will be the quotient of 97403^, divided by 365 f
■^^^•^- {- .' ' '' ' ^»s. 2671^V6
' '^- 8. Divide 3228242 potinds equally among 563 men ; howl
many pounds must each man i-eceive 1 Ans. 5734pQtinds,[
*f ■'^. tf 67624 be divided into 216' 586, arid 976 equal!
parts, what will be the magnitude of one of each of thesej
equal parts ?
fl 20, 21.
C0?iT9ACT10MB IN DIVMION,
times they
md of the
igure, and
I taken.
Bve, and to I
B dividend,
o continue
have been
rema,inderJ
ibr,' write a|
figure,
ny quotient I
taken, it is I
^,knd must
be gfeaten
igure is too\
Ans. The magnitude of one of the last of these equal
parts will be 59^^'.
10. How many times does 1030603615 contain 3215 ?
Ans. 320561 times.
11. The earth' in its annual revolution round the sun, is
said to travel 596088000 miles ; what is that per hour,
there being 8766 hours in a year? - » / ' ■'• •'
12. »2 34^f|8 90;_ how many? i^V'** ;» ; vA,nv.Hm\f,
13. 40|§-fJ20:s= how many? ■/} M"^''^'^'-. '^M >'*>*/** "^
14. 98^^^8»=how many? -^i^. 4^^ ^' -^^ 11^ J«a, :;-^%
,_v iti'T. ''j^''\vv5{ :.'"•*',:! ■.je:ih-: .■^''^■^ ^ - '•^'''.'j^ ' -'-''■ '•'}**'i'. '**
J, divided!))
.ill . I
[SSO founds J
led by 365 1
IS. 2671:jVi
l3 men ; howl
I734 pounds!
1 976 equall
Ich of these!
'i CONTRACTIONS IN DIVISION, -i* .^'
1. Tf^en fAe DIVISOR 15 a composite number. v«t,j|.f.|
f[ 30. 1. Bought 15 yards of cloth for 30 pounds ; how
much was that per yard?
15 yards are 3X5 yards. If there had been but 5 yards,
the cost of one yard would be ^B=Q|t pounds ; but as there
are 3 times 5 yards, the cost of one yard will evidently *- e
but one third p irt of 6 pounds ; that is, ^«b:2 pounds, Ans,
Hence, when the divisor is a composite number, we may,
if we please, divide the dividend by one of the component
I parts, and the quotient, arising uom that division, by the
I other ; the last quotient will be the answer. V'*j» -v "*< '•■' ^m
2. If a man can travel 24 miles in a day, how many days
I will it take him to travel 264 miles?
It will evidently take him as many days as 264 contains 24.
OPERATION.
|24=6X4. 6)264 -w^ ^
or,
■m^^ jH^x ,'.: 4)44
1 i'<.
,H.',
.'\::
24)264(11 days, An5.
24 ...... U.„.,...^
24- .---'^'••■:
•4ms
'^■i'^:' '11 days.' /
3. Ditide 576 by 48=(8x6.)
4. Divide 1260 by63=:(7X9.)
5. Divide 2430 by 56.
IT. To divide by 10, 100, 1.000, fcc. 't.'fr ;^^ ,
^31. 1. A note of 2478 pounds is owned by 10 me|^
hat is each mah's share ? **^' •-
-M
>; Jl
■J y^i
M)'
CONTBACTIONS IM DIVISION.
1I2f,22.
I
rv
i I
II'
Each man's share will be equal to the number of tens con-
tained in the whole sum, and, if one of the figures be cut
off at the right hand, all the figures to the left may be con-
sderedso many tens; therefore' each man's share will be
^147^^0 pounds.
It is evident, also, that if 2 figures had been cut off from
the right, all the remaining figures would have been so ma-
ny hundreds ; if 3 figures, so many thousands, &c. Hence,
we, derive this general Rule for dividing hy 10, 100, 1 000,
6lc. : Cut off from the right of the dividend so many figures
as there are ciphers in the divisor ^ the figures to the leftof
the point will express the quoiitnt^ and those to the right,
the remainder. ' .
2. How many 100 in 424001 .4it5. 424.
4<24I00 Here the divisor is 100; we therefore cut off 2
' figures on the right hand, and all the figures to the i
hft (424) express the number of Ijundreds.
a. How many 100 in 34567 7 Ans. 345^Vir- 1
4. How many huiMlreds in 4567640 hundreds ? >7 ...
... 5. How many hundreds in 345600 hundreds?
6. How many 100 in 42604 hundreds ? Ans, 426^^7.
, 7. How many thousands in 4Q00? in 25000?
8. How many thousands in 6487 thousands ? Ans. 6^j^.
9. How m^ny thousands in 42863 thousands? in
368456 thousands? in 96842378 thousands?
10. How many tens in 40? in 400? in 20? in 468? in
487? in 34640?
III. When there are ciphers on the right hand of the divisor. |
^ tt^. 1. Divide 480 pounds among 40 men ?
OPERATION.
4[0)48|0 In this example, our divisor,!
f^>) is a composite number,
12 pounds, tIw.?. (10x4=:40;) we may there^
fore, divide by one component part, (10,) and that quotient
by the other, (4 ;) but to divide by 10 we have seen, is but
to cut off the right hand figure, leaving the figures to the
left of the point for the quotient, which we divide by 4, and
the work is done. It is evident, that, if our divisor hadj
been 400, we should have cut off 2 figures, and have divi-
ded in the same manner; if 4000, 3 figures, &c. Hence,!
this general Rule : ^Vhen there are ciphers at the rightl
"'.i . -. i:' Wt _' .:' '.t'uJ. :. .^ . ='*..»,''- . . V.
5122.
■UPPLBMENT TO DtVItlOIf.
51
ftht divisor.
hand of the divitor, cut them oflf, and also as many places in
the dividend ; divide the remaining figures in the dividend,
by the remaining figures in the divisor ; then annex the fig-
urea out oflT firom the dividend, to the remainder.
2. Divide 748346 by 8000. » ^^ «'
Dividend. • * > «
Di»i««r,8|000)748)346 \ '
1. 1 ' I .
5;^
Quotient, 93.-4346 Remainder. Ans. 9^Uh
3. Divide 46720367 by 4200000.
Dividend. -i'-rr.t. ; \. J ...
' . 42|00000)467|20367(U^,«y|yQMO<t«i<. ' "•*
^lir.A,^. . - 42 .. .- ;■.'■ (.'I'i.
■ -. »■,,■■
it
'< ..V t
47
4«
•»
( ' I
520367 Remainder.
4. How many pieces of cloth can be bought for 346500
pounds, at 20 pounds per piece ?
5. Divide 76428400 by 900000.
6. Divide 345006000 by 84000.
7. Divide 4680000 by 20, 200, 2000, 20000, 3000, 4000,
50,600, 70000, and 80.
SUPPLEMENT TO DIVISION. '
QUESTIONS.
1. What ia division ? 2. In what does the procaia of diviaion con-
sist *? 3. Division is the revtrte of whati 4. What is the number to
be divided called; and to what does it answer in multiplication ? 5.
What is the number to divide by called, and to what does it answer,
&c.t 6. What is the result or answer called, &c. 1 7. What is the
tign of dirision, and what does it show ? 8. What is the other way of
expressing division 1 9. What is ehort division, and how is it per-
formed 1 10. How is division proved ? 11. How ia multiplication
proved? 12. W hit ire integers, ur whole numbers? 13. What are
/ractUma, or broken numbers 1 14. What is a mixed number 1 15.
When there is anf thing; left afler division, what is it called! hnd
how is it to be written ? 16. How are fractions written 1 17. What
is the upper number called? 18. the lotver number? 19.
How do you multiply a fraction ? 20. To what do the numerator and
the denominator of a fraction answer in division ? 21. What is long
division? 22. Rule? 23. When the divisor ia a compbsite number^
how may we proceed ? 24r When the divisor is 10, 100, or 1000, &.c.
liow may the operation be contracted? 25. When there are ciphers
at the ri^bt hand of the divisor bow may we proceed ?
'')
«l
SUPPLBHBNT TO DIVISION.
til
Vm
II It
,^^.-. ,/""■■ EXERCISES. :^ .-v .:,:, . \
1.; An tamy of 1500 men, having plundered a cicjr, took
3625000 pounds ; what was each man's share.?
2. A certain number of men were concerned in th^pay*
mentof 18950 pounds, and each man ^aid 25 pounds ; what
was the number of men ?
3. If 7412 eggs be packed in 34 baskets, how many in a
basket?
4. What number must I muitiply by 185 that the product
may be 505710. ;.K«;;
5. Light moves with sucii amazing rapidity,' as to pass
from the sun to the earth in about the space of 8 minutes. —
Admitting the distance, as usually computed to be 95000000
miles, at what rate per minute does it travel ?
6. If the product of two numbers be 704, and the multi>
plier be 11, what is the multiplicand ? Ans. 64.
7. If the product be 704, ai|d the multiidicand 64, what
is the multiplier ? '^^ ^^^-'* '»^^'''>. ^'- ^'^--i^^'^''-'^ Ms. ■ 11.
8. The divisor is 18, aiid the dividend 144; what is the
quotient?
9. The quotient of two numbers is S, and the dividend
144 ; what is the divisor ?
10. A man wishes to travel 585 miles in 13 days ; how
many miles must he travel each day ?
11. If a man travels 45 miles a day, in how many days
will he travel 585 miles?
12. A man sold 140 cows for 560 pounds; how much
wasthat for each cow ? ^ *; ' V ; , '
13.. A man, selling his cows for 4 pounds each, received
for all 560 pounds ; how many cows did he sell ?
14. If 12 inches make a foot, how many feet are there in
364812 inches?
15. If 364812 inches are 30401 feet, how many inches
make 1 foot ?
Id If you would divide 48750 pounds among 50 men,
how many pounds would you give to each one ?
17. If you distribute 48750 pounds among a number of |
men, in such a 'manner as to give to each one 975 pounds,
'how maiiy men receive a share?
18. A man has 17484 pounds of tea in 186 chests ; ho?
many pounds in each chest ?
'..'^i.^,: .'^i».,*i..':.j.K'/iilk.,fcSif4:;iX ,l,..ji'«i';.;,^j
-';.^". , , ^ ,J .M%&':
1120.'
MISCELLANEOUS QUESTIONS.
53
19. A man would put up 17484 pounds of tea into chests
containing 94 pounds each ; how many chests must he have?
20. In a certain town there are 1740 inhabitants, and 12
persons in each house ; how many houses are there ? in
each house are 2 families, how many persons in each fam-
ily?
21. If 2760 men can dig a certain canal in one day, how
many days would it take 46 men to do the same ? How
many men would it take to do the work in 15 days ? in
5 days? in 20 days? 40 days? in 120 days?
22. If a carriage wheel turns round 62280 times in run-
ning from duebec to Montreal, a distance of 180 miles, how
many times does it turn in running 1 mile ? Ans. 346.
23. Sixty seconds make 1 minute ; how many minutes in
3600 seconds? in 86400 seconds ? in 604800 sec-
onds ? in 2419200 seconds ?
24. Sixty minutes make one hour ; how many hours in
1440 minutes ? in 10080 minutes? ., in 40320
minutes ? in 525960 minutes ?
25. Twenty-four hours make a day ; how many days in
168 hours ? in 672 hours ? in 3766 hours ?
26. How many times can I subtract forty-eight from four
hundred and eighty ?
27. How many times 3478 is equal to 47854 ?
28. A bushel of grain is 32 quarts ; how many quarts
must I dip out of a chest of grain to make one half (^) of a
bushel ? for one fourth (^) of a bushel ? for
one eighth (^ ) of a bushel ?
29. How many is ^of 20?
247? ^ of 847?
Ans. to the last, 4 quarts.
^of 48? ^ of
4 of 345878.^ 1 of
204030648 ? Ans. to the last, 1 02015324.
30. How many walnuts are one third part (^) of 3 wal-
nuts 7 i of 6 walnuts 7 ^ of 12 walnuts ?
^ of 30.^ 1-^ of 45 .' ^ of 300 7-
of 478.^
31.
What is I
4 of 3456320?
of 4.^
i of 7843 7
Ans.
iof
to the last,
^0?
1152l06f.
i of 320.?
Ans. to the last, 196Gf .
MISCELLANEOUS QUESTIONS,
Involving the principles of the preceding rttles.
Note. The preceding rules, viz. Numeration, Addition,
E 2
4
M
MISCELLANEOUS QtfeSTIONS.
IT 51
^Subtraction, Multiplication, and Division, are called the
JFundcunmtal Joules of AritkmetiCf because they are the
^<fouQdati(Hi of all other rules. ' tS^fc**^ iijf ^Hi^
1. A man bought a chaise for 57 pounds, and a horse for
:94 pounds; what did they both cost?
J2. If a horse and chaise cost 91 pounds, and the chaise
.^eost 57 pounds, what is the cost of the horse ? If the horse
jGost 24 pounds, what is the cost of the chaise? Ui- 'iim^
3. If the suni of 2 numbers be 487, and the greater nunr-
het be 348, what is the less number? If the less number
,ibe 139, what is the greater number?
4. If the minuend be 7842, and the. subtrahend 3481,
ivhat is the remainder? If the remainder be 4361, and
ithe minuend be 7842, what is the subtrahend ?
^ ^3. ,When the minuend and the subtVahend are
given, how do you find the Remainder ?
When the minuend and remainder are given, how do you
find the subtral^end? .i < » ^..-..,.^... \ ; ,: -< . .! »n
When the subtrahend and the remainder are given, how
sdo you find the minuend ?
When" you have the sum of two numbers, and one of them
given, how do you find the other?
When you have the greater of two numbers, and their
tiifference given, how do you find the less number ?
When you have the less of two numbers, and their differ-
jsnce given, how do you find the greater num\)er ?
5. The sum of two numbers is 48, and one of the num-
' bers is 19 ; what is the other ?
6. The greater of two numbers is 29, and their difer-
,encc 10 ; what is the less number ?
7. The less of two numbers is 19, and their difference is
10 '; what is the greaief 1
8. A man bought 5 pieces of cloth at 44 pounds a piece ;
974 dozen of shoes, at 3 pounds a dozen ; 60t) pieces of |
calico, at 6 pounds a piece ; what is the amount ?
9. A man sold six cows at 5 pounds each, and a yoke of |
oxen, for 19 pounds; in pay, he received a chaise, worth 31
pounds^ and the rest in money ; how much money did he
receive 1
10. What will be the cost of 15 pounds of butter^ at 7
pence per po^nd ?
. ..Kiy'-Jj !i*>*'
% ^^ I IT ^^» ^^' MISCELLANEOUS aUESTIt)Nli.
56
sailed the
f are the
L horse for
;he chaise
the horse
eater nunr-
ss number
«nd 3481,
4361, and
ii^v^^ ':":'
ahend are
low do you
given, how
(me of them
, and their
;r?
their rf{^cr-
the num-
leir difer-
difference is
ids a piece ;
ft pieces of
t?
id a yoke of I
le, worth 31
>ney did he
jutter. at 7
11. How many bushels cif wtat 6an you buy for If70
shillings, at 8 shillings per bushel?
t[ ^4« When the price of on« pound, otie bushel, &/C.
of any commodity -is given, how do you find the cost of any
number of pounds, or bushels, &/C. of that commodity ? If
the price of the 1 pound, &-c. be in shillings, in what will
the whole cost be? If in pence, what?
When the cost of ani/ given number of pounc^, or bushels,
fee. is given, how do you find the price of one pound or
bushel, &c. In what kind of money will the answer be ?
When the cost of a number of pounds, &.c. is given, and
also the price of one pound, &c. how do you find the num-
ber of pounds, &-C.
12. When rye is 4 shillings per bushel, what will be the
cost of 948 bushels ?
13. If G48 pounds of tea cost 173 pounds, (that is 4152©
pence) what is the price of one pound ?
When rh'^ 'motors are givfen,howdoyoufindthe product?
When tri . Juct and one factor are given, how do you
find the oti .aCtor ?
When the divisor and quotient are given, how do you
find the dividend ?
When the dividend and quotient are given, how do you
find the divisor ? '- > ^
14. What is the product of 754 and 25? ,
15. What number, multiplied by 25, will produce 18850 ?
16. What number, multiplied by 754, will produce 18850 ?
17. If a man save 5 pence a day, how many pence would
he save in a year, (365 days,)? how iiiany in 45 years ?
How many cows could he buy with the money, ^t 742
pence each?
18. A boy bought a number of apples ; he** gave away
ten of them to his companions, and afterwards bought thir-
ty-four more, and divided half of what he then had among
four companions, who received 8 apples each ; how many
apples did the boy first buy ?
Let the pupil take the' last number of apples, 8, and re-'
verse the process. • Ans. 40 apples.
19. There is a certain number, to which, if 4 be added^
I and 7 be substracted, arid the difference be multiplied by 8,
and the product divided by 3, the quotient will be 64 ; what
is that number ? ' Ans, ^7.
i
«$
MISGBLLANEOtTSr QUSSTIONS«
I
IT 25.
4te
B
20. A board has 8 rows of 8 squaifes.each; how many
squares on the board ? ' * l> jg 4?^ . ; ^
.^ U 3«S. 21. There is a spot of ground 5 rods long, and 3
rods wide ; how many square rods does .it contain 1
Note. A square rod is a
square (like one of those in
in the annexed figure) meas-
uring a rod on each side.
By an inspection of the fig-
ure, it will be «een, that
there are as many squares in
a row as rods on one side, and
that the number of rows is
equal to the number of rods, on the other side ; therefore,
5X3=15, the number of squares. Ans. 15 square rods.
A figure, like A, B, C, D, having its opposite sides equal
and parallel, is called a parallelogram or oblong.
22. There is an oblong field, 40 rods long, and 24 rods
wide ; how many square rods does it contain ?
j 23. How many square inches in a board 12 inches long,
and 12 inches broad ? Ans. 144.
24. A certain township is six miles square ; how many
square miles does itcontain ? Ans. 36.
25. A man bought a lot of land for 2246 pounds ; he
sold one half of it for 1175 pounds at the rate of 3 pounds
per acre ; how many acres did he buy 7 and what did it cost
him per acre ? ■
26. A boy bought a sled for 56 pence, and sold it again
for 8 quarts of walnuts ; he sold one half of the nuts at 8
pence a quJTrt, and gave the rest for a penknife, which he
sold for 18 pence ; how many pence did he lose by his bar-
gains.^
27. In a certain school-house, there are 5 rows of desks;
on each row are six seats, and each seat will accommodate
2 pupils ; there are also two rows, of 3 seats each, of the
same size as the others, and ^ne long seat where 8 pupils ,
may sit; how many scholars will this house accommodate?
Ans. 80,
,28. How many square feet of boards will it take for the
% ?5^.;>
>m9^h^mm^iSv^fim,
67
floor of a room 16 feet long ti^ 15 feet wide, if we allow
12 square feet for waste .' ;
39. There is a room 6 yards long ai»d 5 yards wide ; how
many yards of carpeting, a yard wide, will be sufficient to
cover the floors, if the hearth and fireplace occupy 3 square
yards? ,, '.^^khr^^n- n h: -r^iM' .I;:-
30. A board 14 feet long, contains 28 square feet ; what
is its breadth ?
|l^31. How many pounds of pork, ijirorUi 4 pence a, pQund,
can be bought for 144 pence ? , ,i|,,{,.,f .v<t* !»i »m11^ '"Mi
32. How many pounds of butter, at 9 pence per pound,
must be paid for 25 pounds of tea, at 38 pence per pound ?
33. 4-f-5-f-6-|-l-|-8= how many? „ ^, .... ,
34. 4+34-10— 2— 4-1-6— 7= how many. ^
35. A man divides 30 bushels of potatoes among 3 poor
men; how many bushels does each man receive.' What
is ^ of thirty ? how many af^ % (^tro«thirds) of 30 ?
36. How many are one-third (J) of 3 .' of 6 f
of 9 ? of 282 of 45674312 ?
37. How many are two thirds (§) of 3? of 6?
of 9 ? of 282 ? — of 45674312 ?
38. How many are ^^ of 40 ? — ^ — ^^of ^;of 40 ? i of
160? -I of 60?— .^i of 80]-^r-^of IWi -.rn? of
|24d876?| of 94687C(? ' . .
39. jEtow many is ^ of 80 ? — ^^ of 80 ? f of lOOt
40. An inch is one twelfth part {^^) of a foot how many
jfeet in 1? inches? — — in 24 inches ? in 36 inches ?
— in 12243648 inches?
41. if 4 pounds of tea cost 128 pence, what does 1 poi^nd
jost? .- 2 pounds.? 3 pounds? : 5 pounds?
— 100 pounds ?
42. When oranges are worth 4 pence apiece, how many
[can be bought for 1464 pence ?
43. The earth in moving round the sun, travels at the rate
68000 miles an hour; how many miles does it travel
in one day, (24 houris ?) how many miles in one year, (365
lays?) and how many days would it take a man to travel
^his last distance, at the jrate of 40 miles a day ? how many
^ears ? Ans. to the last, 40800.
44. How many pence can a man earn in 20 M^eeks, at 35
)ence per day, Sundays excepted ?
^9. A man married at the age of 23 ; he livedo with his
n
,1
i.i
58
COMPOUND NUMBERS — ^BBDUCTION. ^ 26, 27.
'!»»■;.
wife 14 years; she then died, leavinff him a daughter, 12
years of age; 8 years after the daughter was married to a
itian 5 years older than herself, who was 40 years of age
when ^he father died; bpw old was the father at his death?
- 46. There is a field 20 rods long, and 8 rods ufidk ; how
inany square rods does it contain t Ans. 160 rods.
47. What is the width of a field, which is 20 rods long,
and contains 160 square rods.
48. What is the length of a field, 8 rods wide, and con-
taining 160 square rods? i'"^; ".^ ' ' "^ U*" '.^i '
59. What is the width of a piece oif Iand,'25'rbd8 long,
and containing 400 square rods?
;i-ijF>u <• ;;;^nMPTi; '.'>:>;;. /ixj
i
^?:
i
••ni i(/i
■■••/
i^iri.! '^f
■ > .- (
4-
f,-
''^l> COMPOUND NUMBERS. ^ < f^' >
IT 36* A number expressing things of the same kind is
called a simple number ; thus, 100 men, 56 years, 75 cents, |
are each of them simple numbers; but when a number ex*
presses things of different kinds, it is called a compound\
number ; thus, 46 pounds 7 shillings and 6 pence, is a com*
pound number ; so 4 years 6 months and 3 days, 4^ dQllarsI
525 cents and 3 mills, are compound numbers. ^ '[^ .' *
Note. Different kinds, or names, are usa ally called differ-
ent denominations.
..•..,i.,
Reduction.
tr 27. In this Province as in England, money is reckoned
in pounds, shillings pence and farthings. In the United!
States, money is reckoned in dollar<s, cents and mills. These!
are called denominations of money. Time is reckoned in I
years, months, weeks, days, hours, minutes, and seconds,
called denomination if time. Distance is reckoned in!
miles, rods, feet, and i 'les, called denominations of mea-!
sure, &c.
The relative valr j of these denominations is exhibited!
in tables, which the pupil must commit to memory.
.-tj:,.';*; ■ ;..''/;,^.iv-''.i'i.^i.iK^ J
:J.i.-ilkl.'>:
1127.
HALIFAX CURRENCY.
HALIFAX CURRENCY.
59
M-tf
The present currency of Lower Canada, is called Halifax
currency, having been introduced into this Province, after
its cession to Great Britain, by France, in 1763, from Nova
Scotia. The denominations are the same in name as the
denominations of English money, i. e. pounds, shillings,
pence, and farthings ; and the ratios of the different denom-
inations to each other are the same as in English mon^v. i. e.,
the shilling is one twentieth of the pound, the peh - ,.ne
twelfth of the shilling, and the farthing one fourth of the
penny. In value Uiey are different, as will be seen in
the II upon reduction of curjrencies; where the ratio of
each to the other, ,and of both to Federal Money is exhi-
bited, with the method of ascertaining them in practice, for
particular sums. .
Ik ■ ■ \
2 farthings (qrs.) make
4 «
12 pence
20 shillings
«
«
«
.'I.'".?
half-penny, marked ^d. ;
penny, " d, ,
shilling, " 8. '?
1 pound, J
«
j€.
Note. Farthings are often written as the fraction of a
penny ; thus, 1 farthing is written ^d.^ 3 farthings, ^d., 3
farthings, |d. , , . ...■■^.... ^ . ;.
It will be proper here to insert an abstract from the Pro-
vincial statute passed in 1842, fixing, the value at which the
gold and silver coins of other countries shall pass current in
this Province. ,,..,,,. ;
The values assigned to the several coins by law in Cana-
da, are not arbitrary, but are proportioned (except in the
I case of British silver) to the quantity of pure gold or silver
in each. The £ currency was and is equal to 4 dollars of
account ; and a note for $100 either in Upper or Lower
Canada, is now, as it has always been payable by ^^25 cy.,
in any ccins equivalent by law to that sum. By the curren-
cy Act the Provincial dollar of account is made equal in
value to that of the United States. . ; i..
/
^>.
60
HAUPAX CVERINCY.
The coins, current by lav, are : ^/ ^'f
British gold coins at the rate of ;C1 4s 4d cy. to £\ stg.
American Eagles coined before 1st July 1834, at ;^ 13s
4d cy — Do. coined between 1st July, 1834, and 1st Janua-
ry, 1841, at J^ 10s, — aftd at the same r"tes for half Ea-
gles, &,c.
The above ar^ a legal tender by tale if within two grains
of full weight, deducting ^ cy. for each 4^ of a grain want-
British gold and Am^rlc^n gold <Soined before 1834, at
94s lOd cy. per oz. troy, —
* American gold coined between 1811^ and ?d41, at 93s cy.
per oz. troy,— '■>rf''/ '■^'ni^i'?t;H'9.-io .'f??fiiwr*r-»'p -iuiK^'-y^.^. j
Coined f Gold coiii of Prance, at OSs Id cy. per oz. troy.|
before
Apr. 26
1841.
((
<(
((
cur-
Do. of Laplata & Columbia, at 89s 5d
Do. of Portugal & Brazil, at 94s 6d '■'
Sp. Mex. &j Chilion Doubloons at 89s 7d
— if offered respectively in sums of not less than df '
rency at one time. «
British silver as above stated. •.
The dollars of Spain, United States, Peru, Chili, Central I
America, States of South America and of Mexico, coined I
before 1841, at 5s Id currency, and half dollars at 2s 6^dl
currency. ClUarters.at Is 3d. Eights at 7^ and sixteenths!
at 3^d, if legal weight. The parts less than halves being al
tender at the said rates by tale to the amount of <£2 10s inl
one payment, until they have lost one twenty-fifth of theirj
weight, and not aftewards. ,'**t*i *ii ^'
f French 5 franc silver pieces, coined^ before 26th April 1842,|
at 4s 8d each.
) Gold and silver coins of the same nations of later dates!
may be made current by proclamation to be issued as afore{
said.-rJ'^i }'>''- ■',<• t*?(iiv)n'i-.»".{>i!i,^ ■»»,;; >*>^i, ,V;trl1«iMi. -'^ ^
• Ciopper coiiiiof the United Kiiigdom, (or aiiy tobecoinj
ed by Her Majesty of not less than five-sixths the weight o^
fuch coin) at th^ir nominal rates.
< The least legal weight of a Sovereign is, 5dwts. 2^ grsj
—of an Eagle coined before 1834, 11 dwts. 6 grs., aftei[
1834, 10 dwts. 18 grs.— of a Dollar, 17 dwts. 4 grs.— of
5 franc piece 16 dwts.
REDUCTION.
61
••t:(;
• / w
'\ ,.Jj
The £ sterling, in any act or contract made after the pas-
sing of the Currency Act, [proclaimed 26 April, 1842] is
to be understood as equivalent to £1 is Id cy., but in any
act or contract made before that time, the word sterling is
to be construed according to the intention of the Legislature
or of the parties. .... .,;,,., i
How many farthings in one
penny? in 2 pence ?
in 3 pence? in 6 pence ?
in 8 pence? in 9 pence?
in 12 pence ? in
1 shilling ? in 2 shil-
lings? ,! i
How many pence in 2 shil-
lings? in 3 s.?
in 4s. ? in 6s. ? in
8s. ? — - in 10s. ? — - in 2
Shillings and 2 pence ?
in 2». 3d. ? in 2s. 4d. ?
in 4s. 3d. ?
How many shillings in 1
pound? in 2 ]f
in 3^ ? in 4 £ in
1 4^ 6s. ? in 6£ 8s. ?
in 3^ 10s.? -
ore 1834, at I
U,at93scy.
>er 01. troy,
d
98 7d "
m£' cur-
Chili, Central
sxico, coined!
ars at 28 e^dj
nd sixteenths!
lalves being al
of .£2 10s inl
-fifth of theirL
The changing of one kind, or denomination, into another
th April 1842,1 kind, or denomination, without altering their value, is call-
ed Reduction. (^ fS7.) Thus, when we change shillings
into pounds, or pounds into shillings, we are said to reduce
them. From the foregoing examples, it is evident, that,
when we reduce a denomination of greater value into a de-*
nomination of less value, the reduction is performed by mul-
tiplication ; and it is then called Reduction Descending. —
But when we reduce a denomination of less value into one
of greater value, the reduction is performed 6y divsion ; it
is then called Reduction Ascending. Thus, to reduce pounds
-in 2£ 15s. ?
How many pence in 4 far-
things? in 8 farthings?
in 12 farthings ? in
124 farthings ? in 32 far-
things? in 36 farthings?
— in 48 qrs. ? How many
shillings in 48 qrs? — — in
96 qrs?
How many shillings in 24
pence ? -= — in 36d. ? in
48d.? m72d.? in
96d. ? in 120d. ? ■
in 26d. ? in 27d.?
in 28d. ? in 30d. ?
in42d.? inSld.?
How many pounds in 20
shillings ? in 40s. ?
in 60s. ? — — in 80s. ?
in 86s. ? ^ in 128s; ?
in 70s.? in 55s.?
of later dates!
sued as aforej
iny tobecoinj
the weight oi
Sdwts. 2i grsj
6 grs., afterj
I. 4 grs. — of '
REDUCTION.
51558.
to shillings, it i» plain we must multiply by 20. And again,
to reduce shillingH to pound?, we must di' ide by 20; It
follows, therefore, that reduction , descending and ascending
reciprocally prove each other.
2. In 1697Marthings, how
many pounds?
OPERATION.
Farlhings in a penny 4)16971 3q»".
Pence in a shilling, 12)4242 <)<!.
Shillings in a ppunU 2|0)'35|3 I3«
• . .. ,; . — : 17^, •
AnsAl£ 13s 6|d.
I. In 17^. 13s. 6^d. how
many farthings.' -
OPERATION.
A s. d. qrs.
,i>Vj 17 13 6 3
'■•*v, ■ 20s.
jm't
i!t-
:•?
• IgI 353s in 17^. I3s.
;tf - 4342d..' *^
4q.
i?.
-1,
i'< ).' <
Farthings will be reduced
to pence i if we divide them
by 4, because every 4 far-
things make 1 penny. There-
fore, 16971 farthings, divided
by 4, the .quotient is 4242
10971 5r5. the Ans.
In the above example, be-
cause 20 shillings make
pound, therefore we multiply
17^. by 20, increasing the' p;^;;";;^ 7^ V;;;;f„"^^^^^^^^
product by the addition of the J^ ^^^^^ -^ forthings, of the
given shillings (13,) which,
it is evident, must always be
done in lik^ cases; then, be-
cause 12 pence make 1 shil
ling, we multiply the shillings
(ti53) by 12, adding in the
given pence, (6.) Lastly,
because 4 farthings make 1
penny, we"^ multiply the pence
(4242) by 4, adding in the
given farthings, (3.) We
then find, that in 17<£. 13s.
6fd., are contained 16971
same name as the dividend.
We then divide the pence
(4242) by 12, reducing them
to shillings ; and the shillings
(353) by 20, reducing them
to pounds. The last quotient
17 J?., with the several re-
mainders, I3s. 6d. 3qr8. cqi>-
stitute the answer.
Note. In dividing 353s. by
20, cut off the cipher, &.C.,
as taught^ 22.
furthings. ..
51 2?^. The process' in the foregoing examples, if care-
fully examined, will will be found to be as follows, viz.
To reduce high denominations To reduce low dimmtinotions
to higher. — Divide the lowest
denomination given by that
to lower, — Multiply the high-
est denomination by that num-
r y,
5128.
3d again,
y 20: It
isccndittg
ings,ho>«r
16971 3qr.
5)4242 Od.
0)351313*
€ 13s 6|<1.
be reduced
Lvide them
fexy 4 far-
ny. There-
ngs, divided
mt is 4242
tmainder . of
ings, of the
dividend,
the pence
iucing them
he shillings
cing them
ast quotient
several re-
3qr8, CQi>-
'" . >■'
ig 353s. by
)her, &.C.,
[es, if caie-
/s, viz.
linotions
the lowest
in by that
REDUCTION. 63
number which it takes of th^
same to make I )0f the next
higher. Proceed in the «ame
manner with eagh^ucceeding
denomination^ until you have
brought it to the denomina-
tion required,. . - . ,
•.( , : kH '(■< a "•''>* yi>HA ill '
5128.
her wliich it takes of t1ie next
less to make 1 of this higher,
(increasing the product by the
number given if any of that
less denomination.) Pi'O'ceed
in the same manner with each
succeeding denomination, un-
til you have brought it to the
denomination required.
In the two examples, from which the above general rules
arc deduced, the denominations are pounds, shillings, pence
and farthings, considered as in Halifax Currency ; but it is
obvious that these rules can be applied to all currencies
where the den minations are the same ; or to currencies in
which the denominations are different; and in general to
all compound numbers.
EXAMPLES FOR PRACTICE.
3. Reduce 20<£- i^&- '^d- to pence.
4.
5.
7.
8.
.!>i.
9.
<(
<t
((
it
.•A
Ans. 4970;
Ans.'UmiO^
Ans. 15912.
Ans. 151680.
10. R,educe 32^. los. 8d,
to farthings.
12. In 29 guineas, at \£
3s. 4d. each, how many qrs. ?
14. Reduce $163, at 6s.
each, to pence ?
16^ In ISguineas, how ma-
ny poinds?
24<£. to farthings.
66^. 68. 6d. to pence*
158*^. to farthings.
1234^. 15s, 7d. to farthings. Ans. 1185388.
337587 farthings to pounds, &,c.
Ans. 351je. 13s. Od, 3q.
118538B farthings to pounds, &c.
' ' ' Ans. USi£. 15s. 7d.
11. Reduce 31472 farthings^
to pounds.
13. tn 38976 farthings, how;
many guineas ?
15. Reduce 1 1736 pence to
dollars. -^ r!';^
17. Reduce 21^. to ^ih-
eas. '
Note. We cannot reduce guineas directly to pounds, but
we may reduce the guineas to shillings, and then the^ shil-
lings to pounds. •■ -■','.; .v. •. . .. ■'-.1 H> > i'--! •"
.ys':A"'
I !JI , l>
•''■$■■"
; >!f ff ;'t
64 REDUCTION.
y . OLD CURRENCY.
<*; .! 12 deniers make ' i- -«*j.'« '¥ i sou.
'^ - 20 sous " vmW.:^^ i Uy^e, or franc.
' The livrd^ lOd Halifax currency.
In 32 livres lO sous how many sous? , . ;, . ; :'
In 97 livres 11 sous, how many sous? -;^.,. ,; ,/>;
In 650 sous, how many livres? , - > ^ n -!
In 1951 sous, how many livres?
In 10 livres 6 sous 9 deniers, how many deniers ?
' How many pounds currency in 96 livres ?
!I20.
.1'.
If \\ ■ :>,
'7 •
#«, ft-.
<u.
ii'
,„ v,t.-.,r..riif') <|i 'FEDERAL MONEY. '' ' : •'
^ ^B9. Federal money is the coin of the Imited States.
The kinds or denominations, are eagles, dollars, dimes,
cents, and mi'^s. tn* ; i ,•;».'; <?!
TABLE.
• ■ • ■ . ■ ' . I ' '' ■•■'
10 mills ' - - - areeqiladto' '-'' "Icent.
10 cents, (t=100 mills,) - - . - =1 dime.
10 dimes, \=100 cents=1000 mills,) ' ii* =1 dollar.
10 doll's., («l00dimes=l000cent8=l600dm's)=l eagle*
Sign. • This character, $, placed before a' number, shows
it to express;/crfera/ ikoney.
As 10 mills mak6 a cent, 10 cents a dime, 10 dimes a
dollar, &c. it is plain, that the relative value of mills, cents,
dime^, dollar^ and eagles corresponds to the orders of units,
tens, hundreds, &,c. in simple numbers. Hence, they may
be read either in the Iqwest denpmination, or partly in a
higher y and pattly in the lowest denomination. Thus :
5 S s js" «-
^=? S =5
,>■'.,•; .)■
3 4 6 5 2 may be read, 34652 mills ; or 3465 cents ahd 2
mills ; or, reckoning the eagles tens of dollars, and the
•The eagle is a gold coin, the dollar and dime are silvey coins
the centis a copper coin. The mill is only tmagmary, there being
no Coin of that denomination. There are half eiagles, half dol-
lars, half dimes, and half cents, real coins.
1120.
. ■;!'
.;, -.M
rr ;
V ; i:./i
tic.
1
1. H'
!i • .>
rst
•;;T'kr.
' til
f ■
• t ■ '
•.1".ll .
ited States,
urs, dimes,
1. • )i
=1 dime.
=1 dollar.
=1 eagle*
iber, shows
dimes a
nills, cents,
rs of units,
3, they may
artly in a
hu?:
. '! I •
-.15 .1- ..^
sents aiid 2
rs, and the
I I ~
[silvey coins
I there being
Ss,halfdol-
^ 29. ' REDUCTION. ' ' * (15
dimes tens of cents, which is the usual practice, the whole
may be read, 34 dollars 65 cents and 2 mills.
For ease in calculating, a point, (') called a scparatrh,'^
is placed between the dollars and cents, showing that all the
tigitres at the left hand express dollars, while the two jirst
fffures at the right hand express cents, and the third, mills.
Thus, the above example is written $34'G52; that is, 34
dollars 65 cents 2 mills, as above. As 100 cents make a
dv/llar, the cents may be any number from 1 to 99, often re-
quiring two figures to express them; for this reason, two
places are appropriated to cents, at the right hand of the
point, and if the number of cents be less than ten, requiring
but one figure to express them, the ten's place must be filled
with a cipher. Thus, 2 dollars and 6 cents are written 2'Gt}.
10 mills make a cent, and consequently the mills never ex-
ceed 9, and are always expressed by a single figure. Only
one place, therefore, is appropriated to mills, that is, the
place immediately following cents, or the third place from
the point. When there are no cents to be M'riiten, it is ev-
ident that we must write two ciphers to fill up the places of
cents. Thus, 2 dollars and 7 mills are written 2'007'. Six
cents are written, 06, and 7 mills are written *007..
Note. Sometimes 5 mills =3^ a cent is expressed frac-
tionally: thus, *125 (twelve cents and five mills) is ex-
pressed 12^ (twelve and a half cents.) ^ .
17 dollars and 8 mills are written, 17*008 , \
4 dollars 5 cents, - - - - 4*05 ,
' 75 cents, , - '75
24 dollars, 24*
^^■"- 9 cents, '09
4 mills, '004
6 dollars 1 cent and 3 mills, - 6'Or?
Write down 470 dollars 2 cents ; 342 doll,?r;- 40 cents-
and 2 mills ; 100 dollars, 1 cent and 4 mills ; 1 mill ; 2
mills ; 3 mills ; 4 mills ; ^ cent, or 5 mills ; 1 cent and 1
I mill ; 2 cents and 3 mills ; six cent^' and one mill ; sixty
cents and one mill ; four dollars and one cent ; three cents ;
I five cents ; nine cents.
*The character used for the scparatrix, in the " Scholars' A-
jrithmetic," was the comma, the comma inverted is here adopted^
(to distinguish it from the comma used in punctuation.
F 2
6a
REDUCTION OF FEDERAL MONEY.
^30.
REDUCTION OF FEDERAL MONEY.
tf SIO. How many mills in one cent? — in 2 cents?
— in 3 cents ? — in 4 cents ? — in 6 cents ? — in 9
cents? — in 10 cents? — in 30 cents? — in 78 cents?
— in 100 cents, (=1 dollar)? — in 2 dollars? — in 3
dollars? — in 4 dollars? — in 484 cents? — in 563
cents ? — ^in 1 cent and 2 mills ? — in 4 cents and 5 mills ?
How many cents in 2 dollars ? — in 4 dollars ? — in
8 dollars? — in 3 dollars and 15 cents? — in 5 dollars
and 20 cents ? — in 8 dollars and 20 cents ? — in 4
dollars and 6 cents?
How many dollars in 400 cents ? — in 600 cents ? -r-
in 380 cents ? — in 40765 cents ? Hmv many cents in
1000 mills? How many dollars in 1000 mills ? — in 3000
mills ? — in 8000 mills ? — in 4378 mills ? — in
^46732 mills ?
As there are 10 mills in one cent, it is plain that cents are
changed or reduced to mills by multiplying them by 10, that
is, by merely annexing a cipher, (^ 12.) 100 cents make a
dollar ; therefore dollars are changed to cents by annexing 2
ciphers, and to mills by annexing 3 ciphers. Thus, 16 dol-
lars =1600 cents =16000 mills. Again, to change mills
b ick to dollars, we have only to cut off the three right hand
figures, (IT 21 ;) and to change cents to dollars, cut off the
two right hand figures, when all the figures to the left will
Jae dollars, and the figures to the right, cents and mills.
Reduce 34 dollars to cents. Ans. 3400.
Reduce 240 dollars and 14 cents to cents.
Ans. 24014 cents.
Reduce $748'143 to mills.
Reduce 748143 mills to dollars.
Reduce 3467489 mills to dollars.
Reduce 48742 cents to dollarrs.
ileduce 1234678 mills to dollars.
Reduce 3469876 cents to dollars.
Reduce $4867/467 to mills.
Reduce 984 mills to dollars.
Reduce 7 mills to dollars.
Reduce $ *014 to mills.
Reduce 17846 cents to dollars.
Ans, 748143 mills.
J.ns. $748443.
Ans. 3467*489.
Ans. $487*42.
Ans. $ *984.
A^v.z. $ 007.
!13L
RfiDrCTION.
67
Reduce 984321 cents to mills. */
Reduce 9617^ cents to dollars. Ans. $9G'17^.,
Reduce 2064^1^ cents, 503 cents, 106 cents, 921^ cents,
500 cents, 726 J^ cents to dollars.
Reduce 86753 mills, 96000 mills, 6042 mills, to dollars.
TROY WEIGHT.
11 31. It is established by law, that the pound Troy, with
its parts, multiples, and proportions, shall be the standard
weight for weighing gold* and silver in coin or bullion,
drugs, and precious stones. The denominaticms of Troy
weight are pounds, ounces, pennyweights and grains.
TABLE.
I pennyweight, marked pwt.
1 ounce, - - - - oz.
1 pound, - - - - lb.
24 grains (grs.) make
20 pennyweights - -
12 ounces - - -
1. How many grains m a
silver tankard weighing 3 lb.
5oz. ?
3. Reduce 210 lb. 8 oz. J2
pwts. to pennyweights.
5. In 7 lb. 11 oz. 3 pwt.
9 grs. of silver, how many
grains?
o
In 19680 grains
many pounds, &c.
how
4. In 50572 pwt. how ma-
ny pounds ?
6. Reduce 45681 grains to
pounds.
• APOTHECARIES' WEIGHT.
Apothecaries' weightt js used by apothecaries and phy-
sicians, in compounding medicines. The denominations are
poundts, ounces, drams, scruples, and grains.
TABLE. '
20 grains, (grs^) make 1
3 scruples - - - 1
8 drams - - - - 1
12 ounces - - - - 1
scruple, marked g.
dram, - - - 3.
ounce, - - - §.
pound, - - - lb.
•The fineness of gold is tried by fire, and is rerkoned in carats,
by which is understood the 24th part of any quantity ; if it lose noth-
ing by the trial, it is said to be 24 carats fine ; if it lose 2 carats, it
is then 22 carats fine; which is the standard for gold.
Silver which abides the tire without loss is said to be 12 ounces fine.
The standard for silver coin is 11 oz. 2 pwts. of fine silver, and 18
pwts. of copper melted together.
IThe pound and ounce apothecaries' weight and the pound and
ounce Troy, are tho sainc» only differently divivideiJ, and subdivided.
68
REDUCTION.
t[31.
7. In 9 fb. 8 §. 1 3. 2 9| 8. Reduce 55799 grs. to
19 grs., how many grains. (pounds.
I .,-t'.
AVOIRDUPOIS WEIGHT.*
It is established by law that the pound Avoirdupois
with Jts parts &c. shall be considered as the standard for
weighing every thing commonly sold by weight, except
those articles, in weighing which, Troy weight is used.
The denominations are tons, hundreds, quarters, pounds,
ounces, and drams.
TABLE.
16 drams, (drs.) malce
16 ounces - - - -
28 pounds - - - -
4 quarters
* 20 hundred weight
ounpe,
pound,
- marked
oz.
lb.
I quarter, ----- qr.
1 hundred weight - - - cvvt.
1 ton, T.
Note 1. In this kind of weight, the wordls gross and net
are used. Gross is the weight of the goods, together with
the box, bal«, bag, cask, &c, which contains them. Net
weight is the weight of the goods only, after deducting the
weight of the box, bale, bag, or cask, &/C., and all other al-
lowances.
Note 2. A hundred weight, it will be perceived is 1*12 lb.
Merchants at the present time, in the principal sea ports of
the United States, buy and sell by the 100 pounds.
9. A merchant would put
109 cwt. qrs. 121b. of rais-
ins into boxes, containing 26
ib. each ; how many boxes
will it require?
11. In 12 tons, 15 cwt.
1 qr. 19ib. 6 oz. 12 dr. how
many drams?
13. In 28Ib. avoirdupois,
how many pounds Tro) ?
10, In 470 boxes of raisins,
containing 26 lb. each, how
many cwt. ?
12. In 7323500 drams, how
many tons?
14. In 34 Ib. oz. 6 pwt.
16 grs. Troy, how many
pounds avoiadupois?
•175 oz. Troy-192 oz. ayuirdupois, and 1751b. troy=144lb avoin
pois, lib. troy=5760 grains, and 1 ib. avoirdupoi8=70U0 grains troy
-,-■•^, ■ v^p-^.
yr-Tj^-iT.
1131.
REDUCTION'
CLpTiI MEASURE.
69
"• ■■.•'•(-.';• 'fir-'. "^V^-. ' f\^i
Cloth medifiure is \ised in selling cloths and other goods
sold by the yard, or eJI. It is established by law that the
English yard with its parts &:-c. shall be the standard for
measuring all kinds of cloth or stuffs made of wool, flax &.c.
the English ell, when there is a special contract for it may
be used with its parts. The denominations are ells, yards,
quarters and nails.
''TABLE.
4 nails, (na.) or 9 inches make 1 quarter, marked qr,
4 qiiaftlers or 36 inches, - - 1 yard, - - - yd.
3 quarters •--.--- 1 ell Flemish, - - E. Fl.
5 quarters " - -' i - - - 1 ell Engli'sh, - - E. E.
6 quarters '•- - i /-•'•-'•- 1 ell French, - - E. Fr.
17. In 9173 nails, how-
many yards I
19. * In 188f yards, how
many ells English ?
?.'
16. In 573^ yds.! c^: t^.
how many nail^.i^^ ,'^ ' '
18. ih 151 ells "Eng. ho^V
many yards.? '•"/'
Note. €6lis\ilt^ 2S ex. 16.
i . , . \ . »
..^^;\),;:tir.^'U-i;!LONG MEASURE.
:'■>;;!',' .^ ;-\ '^^ •■■ ' ■
Lon^ measure is us^ in measuring distances, or other
things, where length is considered without regard to breadth.
The denominations ate degrees, leagues, miles, fur]dngs,
rods, ysirds^ feet, inches, and barley-corns.
•.■, :.i>:*'r /It' f' •■•/, -^ — "
•\ '.■'■. \ 1 •■ ,-. ■ ■ s> iv •. ] '■ ■ TABLE.
'-* .,:. ..::.„•. .
marked
d barly-coins, (bair.) make 1 inch.
12 inches
3 feet
5^ yards, or 16^ f^et, -
40 rods, or 220 yards, -
8 furlongs, or 320 rods, -
' 3miles, .,-,..j; V - -
60geographicki„or G9|J^ )
st,atute miles, . - i
360 degrees, - - - |
1 foot,
1 yard, - - 1 - -
1 rod, perch, or pole,
1 furlong, - - -
1 mile, - - -• -
1 leauge, - - - .
m.
ft.
yd.
r. p.
fur.
M.
L.
1 degree.
deg. or °
a great circle, or circumfer-
ence of the earth.
1131
RBpUCTION.
.Ul
70
It is established by law, tliAtthe JrM*5 foot with its parts,
&c. shall be the standard meaiiure of length, for measuring
land, wood, timber, stone, masons', car'^enters', and joiners'
work. The English foot may be used wlien there is a spe-
cial contract for it. , ./ ," , •/ "^
TABLE.
12 lines make 1 inch.
3 toises make 1 rod.
12 inches - 1 foot.
' 10 rods - . - 1 arpent.
6 feet - 1 toise.
84 arpents - 1 leauge
1 French foot :^i-^-^fj English feet.
20. How many barley-corns
will reach round the globe, it
being 360 degrees ?
Note. To multiply by 2 is
to take the multiplicand 2
times ; to multiply by 1 is to
take the multiplicand 1 time ;
to multiply by ^ is to take
the multiplicand half a time,
that ia, the half of it. There-
fore, to reduce 300 degrees to
statute miles, we multiply
first by the whole number,
09, and to the product add
half , the multiplicand. Thus :
4)360 .
eoi
3240
2160
180 half the multiplicand.
25020 satute miles in 360°.
22. How many inches iVom
Quebec to Three Rivers, sup
posing it to Be 90 miles ? >
24. How many tim^s will
a wheel 10 feet and inches
in circumference, turn round
in the distance from Quebec
to St. Annes, supposing it to
be 60 miles ?
21. In 4755801600 barley,
corns, how many degrjses?
-i'i
Note. The barley-corns be-
ing divided by 3, and ihat
quotient by 12r, we have
13210a500 feet which are tof
be redu,qe|^ to rods. We c?in-
not easily divide by 16^ on
account of the fraction ^ ; but
16^ feet = 33 kalffeet, in 1
rod; and 132105600 /ccf =
264211200 half feet, which
divided by 33)-giveB 8006400
rods. . . '^^.v,.' .> i»f »;'■ : .■■■''}.ii I
Hence, when the divisor Is
encumbered with a fraction,
^ or ^, &c., we may reduce
the divisor to halves or fourths
&,c., and reduce the dividend
to the sarft^; then the' qho-
tient will be the true answer.
23. In ' 30539520 inches,
how many miles ?
25. If a wheel 16 feet in.
in circumference, turh round
19200 times in gbiftjg from
Quebec to St. Aanes, what ii
the distaijce !
A^-
.-70
th its parts,
measuring
ind joinej-s'
•e is a spe-
1 rod.
1 arpent.
1 leauge.
IGOObarley-
legr^esT
ley-corns be-
3, and ihat
, we have
vhich are to|
IS. Wec^n-
! by 16^ on
ictjon^; but
ilffeet, in 1
)5600/ece =
feet, which
ves 8006400
the divisor is
a fraction,
reduce
ox fourths
|the dividend
en the'qtio-
Irue answer.
|520 inches,
1132.
REDUCTION.
71
may
16 feet ^ in.
turib round
rbitfg from
mes, what it
26. In 28 leagues, 43 ar-
pents, how many feet ? how
many toises ? how many rods ?
27. In 7000 feet how many
rods? how^many.arpents?
t LAND OR SQUARE MEASURE.
Square measure is used in measuring land, and any other
thing, where length and breadth are considered. The de-
nominations are miles, acres, roods, perches, yards, feet and
inches.
fl «IS. 3 feet in length make a yard in long measure ; but
it requires 3 feet in length, and 3 feet in breadth, to make
a yard in square measure ; 3 feet in length and 1 foot wide,
make 3 square feet; 3 feet in length, and 2 feet wide,
make 2 times 3, that is, 6 square feet; 3 feet in length and
3 feet wi(^e make 3 times 3, that is 9 square feet. This
will clearly appear from the annexed figure. jj; „
3 feet =-1 yard. ... ... - , . 1 .
It is plain, also that a square foot,
that is, a^^quare 12 inches in length
and 12 iriches in breadth, must con-
tain 12X12=B»144 square inches.
1-
TABLE.
144 square inches=:12Xl2; that is, \
12 inches in length and 12 inch- > make 1 square foot,
es in breadth, ----- j
9 quare feet==3X3; that is, 3 feet )
in length and 3 feet in breadth }
1 30^ square y ai:ds=5^ X 5 1, or 272| V
square feet=3l 64X16^ - - f
40 square rods, --------
4 roods, or 160 square rods, - •
i()40 acres, --------
Note. Gunter's chain, used in measuring land is 4 rods
lin length. It consists of 100 linkf^, each link being '7-f^xs
linches in length; 25 links make 1 rod long measure and
|625 square links make 1 scpiare rod.
1 squa»re yard.
( 1 square rod.
\ perch or pole
1 rood.
1 acre.
1 .square mile.
n
RXiDUCTION.
FRENCH SaUARE MEASURE.
1131.
. 144 squafe inches make 1 square foot.
36 - feet - - 1 toise.
9 - toises - - 1 rod.
100 - rods - - - 1 arpent.
7056 - arpents - - 1 league.
62500 French feet ^71289 English feet.
Reduce 16 leagues to feet, to toises, to rods.
Reduce 98764321 feet to toises, to rods, — —to ar-
pents, — —to leagues.
29. In 776457 square feet,
how many acres?
Note. Here we have 776457
square feet to be divided by
272^. Reduce the divisor to
take a fourth part of the mu\-\fourths, that is to the low-
28. In 17 acres 3 roods 12
rods, how many square feet ?
Note. . In reducing rods to
feet, the multiplier will be
272^. To multiply by |, is to
tiplicand. The principle is
the same as shown in fl 28,
ex. 20.
30. Reduce 64 square miles
to S4uare feet. ?
32. There is a town 6 miles
square ; how many square
miles in that town? how
many acres ?
est denomination contained in
it ; then reduce the dividend
to fourths, that is, to the same
denomination, as shown H 31,
ex. 21. .
31. In 1,784,217,600 sq.
feet, how many square miles?
33. Reduce 23040 acres to
square miles.
'jiri
II
SOLID OR CUBIC MEASURE.
< ■ < i f I
Solid or cubic measure is used in measuring things that
have length, breadth, and thickness ; such as timber, wood,
stone, bales of goods, &/C. The denominations are cords, |
tons, yards, feet and inches.
U 33. It has been shown, that a square yard contains I
3x3=9 square feet. A cubic yard is 3 feet long, 3 feet
wide, and 3 feet thick. Were it 3 feet long, 3 feet wide
and one foot thick, it would contain 9 cubic feet ; if 2 feet
thick, it would contain 2X9=: 18 cubic feet; and, as it is
1I3i; 1 1183. ,
RBBUOTION.
im;
3 feet thick, it does contain 3Xda37 cubic feet. This
will clearly appear fir<Mn the an-
. lyi=3ftlong. nexed figure.
■iiiiiiiiiiiiiiiiiiin mil iMiiii * "
It is plain, also, that a cabie
foot, that is, a solid 12 inches ia
length, 12 inches in breadth, and
12 inches in thickness, will con-
tain 12X12X12:^1728 solid or
cubic inches.
TABLE
1728 solid inches,=12X 12X12,)
that is, 12 inches in length, > make one solid foot.
12 in breadth, 12 in thickness, }
27 solid feet,=3x3X3
40 feet of round timber, or 50 \
feet of hewn timber, |
128 solid feet,=8x4X4, that is, i
8 feet in length, '4 feet in >
width, and 4 feet in height, j
Note. What is called a cord foot, in measuring wood, is
16 solid feet ; that is, 4 feet in length, 4 feet in width, and
1 foot in height, and 8 such feet, that is 8 cord feet make 1
cord.
FRENCH SOLID MEASURE.
1728 solid inches make 1 solid foot.
216 - - feet make 1 toise.
1000 French feet =1218,186432 English feet
1 solid yard.
1 ton or load.
1 cord of wood.
32. Reduce 9 tons of round
timber to cubic inches.
34. In 37 cord feet of wood
how many solid feet?
36. Reduce 64 cord feet of
wood to cords.
38. In 16 cords of wood,
how many cord feet? how
many solid feet ?
40. In 12 toises how many
inches?
G
33. In 622080 cubic inch-
es how many tons of round
timber.'
35. In 592 solid feet of
wood, how many cord feet ?
37. In 8 cords of wood,
how many cord feet ?
39. In 2048 solid feet of
wood, how many cord feet ;
how many cords ?
4L In 834692773 inches
how many feet; how many
, toises T
/ 'I.
\
■.,■■. ^ -l.:; iitjA.i'
:a
y
'',1
i
f
^i')i
f<v«^. ' ^!i4r|b-'>
a-
REDUCTION. -"
WINE MEASURE.
' ' 1133,1
< It is established by law that the wine gallon with its
{)arts, &c. shall be the standard liquid measure, for measurJ
ing wine, cider, beer^ and all other liquids commonly sold
by gauge,' or measure of capacity. The denominations are
tuns, pipes, hogsheads, barrels, gallons, quarts, pints, ai^dj
gills.
?■> J^;tv> -A^'
tf-i
4 gills (gi.)
2 pints *
4 quarts
31^ gallons
03 gallons
2 hogsheads
'■•■l:.ui TABLE. ^ -■
make * 1 pint, marked
- 1 quart,
* 1 gallon,
1 barrel,
*• 1 hogshead,
4 pipe.
'- i^!ip» .
2 pipes, or four hc^sheads 1 tun.
pt.
qt-
gal,
bar.
hhd.
P.
T.
Note. A gallon wine measure, contains 231 cubic inches.
42. Reduce 12 pipes of wine
to pints.
44. In 9 P. 1 hhd. 22 gals.
3 qts. how many gills ?
46. In a tun of cider, how
many gallons ? ^
43. In 12096 pints of winej
how many pipes 1
45. Reduce 39032 gills to|
pipes.
47. Reduce 252 ctallons to|
tuns.
-.i.
ALE OR BEER MEASURE.
Ale or beer measure is used in measuring aie, beer, and]
milk. The denominations are hogsheads, barrels, gallensj
quart!4> and pints< m
V TABLE.
2 pints (pts.) make 1 quart, marked
4 quarts - ; * i- >/ 1 gallon, - *
36 gallons - -• - 1 barrel,
r»4 gallons - - - 1 hogshead, -
qt.
gal,
bar,
hh(i,|
ntfir. A gallon beer measure, contains 282 cubic indie?
V Reduce 47 bar. 18 gal
of ale to pints.
50, In 29 hhds. of beer,
how manyi pints ?
49. In 13680 pints of ale]
how many barrels?
51. Reduce 12528 pints ttj
hogsheads.
252 £rallons tol
i -ivf.'ioKn.' V .
a 'J.^ "I
bu. ^^ '
fl33. ' REDUCTION.* a ' 7i^'
DRY MEASURE. ^ t
Dry measure is used in measuring all dry goods, such as
grain, fruit, roots, salt, coal, &.c. The denominations are
chaldron^, bushels, pecks, quarts, and pints,
- TABLE,
2 pints (pts.) make - 1 quart, •• marked
8 quarts - »• - 1 peck, -
4 pecks - - - 1 bushel, •> r
36 bushels - - - 1 chaldron,
Note. A gallon dry measure, contains 268f cubic inches^
A Winchester bushel is IS^ inches in diameter, 8 inches
deep, and contains 2l50f cubic inches.
It is established bj law that the Canada Minot, with its
parts, multiples, and proportions, shall be the standard io
Dry Measure,
1 pot==116'94569 English 'cubic feet
20 pots malyB one minot,
OLD MEASURE.
16 litrons - make - 1 . . * * . boisseau.
3 boisseaux --- 1 ^-.-, minot. -..
2 minots ---- 1 ,--,, mine.'
2 mines --■-* 1 ...,,.• setier.
12 setiers ---- 1 muid.
40 French cubic inches=::l litron.
The mandard measure for the sale and purchase of coal,
for this Province, is the chaldron of 36 minots, each minot
to be heaped up.
52. In 75 bushels of wheat
how many pints ?
54. Reduce 42 chaldrons of
coal to pecks.
53. In 4800 pints, how mar
ny bushels ?ii
65. In 6048 pecks, how ,
many chaldrons .' ' '.i. . '■ .
.■v;
- TIME.
The denominations of time are years, months, weeks,
days, hours, minutes, and seconds. ' ' ^
TABLE.
60 seconds (s.) - make - 1 minute, marked m.
60 roiuutes , - - 1 hour, - - h.
*A
i
fi
:,1
hi
I
76
SM'Hiours
RBDUCTION.
/
V
day,
week, - -
month, -
common, or )
Julian year, )
!I34.
d.
mo.
yr-
February, 2d,
March, 3d,
- 28
- 31
April, 4th,
May, 5th,
June, 6th,
. 30
- 31
- 30
July, 7th,
August, 8th,
September 9th,
October 10th,
- 31
- 31
- 30
- 31
November 11th,
. 30
December 12th,
- ^ - 31
7 dtys
'4 weeks
13 months, 1 day and 6 hours,
or 365 days and 6 hours,
51 S4» The year is also divided into 12 calendar months,
which in the order of their succession are numbered as foU
lows, viz.
January, 1st month, has 31 days.
Note. When any year
can be divided by 4 with-
out a remainder, it is cal-
led leap year, in which
February has 29 days.
The number of days in each month may be easily fixed
in the mind by committing to memory the following lines :
' "• Thirty days hath September,
April,, June and November,
. February twenty-eight alone ; ' /
All the rest have thirty-one.
The first seVen letters of the alphabet. A, B, C, D, E, F, G,
are used to mark the several days of the week, and they are
disposed in such a manner, for every year, that the letter A
shall stand for the 1st day of January, B for the 2d, &c. In
pursuance of this order, the letter which shall stand for Sun-
day, in any year, is called the 2>omtntca/ letter for that year.
The Dominical letter being known, the day of the week on
which each month comes in may be readily calculated
from the following couplet:
At Dover Dwells George Brown Esquire,
Good Carlos Finch And David Fryer.
These words correspond to the 12 months of the year, and
the first Utter in each word marks the day of the week on
■; ;• J
1?31
I
"/I -If
mEOUCTIOlV.
k»i.i>
77
which each corresponding month conies in; whence any other
day may be easily found. For example, let it be required
to find on what day of the week the 4th ai July falls, in the
year 1827, the Dominical letter for which year is G. Good
answers to July ; consequently, July comes in on a Sunday;
wherefore the 4th of July falls on Wednesday.
Nott. There are two Dominical letters in hap years,
ont for January and February, and another for the rest of
the year
56. Supposing your age to
be 15y. 19d. J lli. 37m. 45s.,
how many seconds old are
you, allowing 365 days 6
hours to the year /
58. How many minutes from
the 1st day of January to the
I4th day of August, inclu
sively ?
60. How many minutes from
the commencement of the war
between America and Eng'
land, April 19th, 1775, to the
settlement of a general peace
which took place Jan. 20th,
1783/
57. Reduce 475047465 se-
conds to years.
59. Reduce 335440 minutes
to days.
61. In 4079160
how many years ?
mmutcs,
\
CIRCULAR MEASURE, OR MOTION,
Circular measure is used in reckoning latitude and lonifi-
tude; also in computing the revolution of the earth and
other planets rouna the sun. The denominations are cir>
cles, signs, degrees, minutes and seconds.
TABLE.
60 seconds (") make 1
60 minutes - - - 1
30 degrees - - - 1
12 signs, or 360 degrees, - 1
Note. Every* circle whether
into 360 equal parts, called degrees.
minute, marked ''
degree, .. - t»
sign, - - s
circle of the zodiac,
great or sni ill, is divisible
62. Reduce 9s.
second.s.
G2
130 'Zry to
63. In 1020300 , how many
degreca ?
itPPLEMKNT to RBBUCTION.
^M.
II
u
1 dozen.
1 gross.
1 great gross.
1 score.
.;>: t
"^ The following are denominations of things not includcU in
the tables: — ^' .>
12 particular things make
)' • 13 dozen ....
' 12 gross, or 144 dozen,
Also,
20 particular things make
6 points make 1 line, ) used in measuring the length of
12 lines - 1 inch ) the rods of cluck pendulums.
4 * hA 1 h H I ^^^^ '" measuring the- height of
*" ' an ^ liorses.
6 feet - 1 fathom used in measuring depths nt sea.
112 pounds make - 1 quintal offish.
24 sheets of paper make 1 quire.
20 quires - - - 1 ream.
SUPPLEMENT TO REDUCTION.
QUESTIONS.
1. What is reduction 'i 2. Of how many vnri^lies is reduction ? 3.
what is underatood by different denominations, as of money, weight,
mensure, &cc. ^ 4. How are high denominations brouj^ht into lower i
5. How are low denominations brought into higher I 6. What are
the denominations of Halifax currency ? 7. What name is given to
the currency of this Province l 8. And why? 9. Are the ratios of
the diffi^rent denominations to each oilier the same as in English mo-
ney ? 10. Will the rule for reduction of one denominetion to another
in Halifax currrency; apply to all currencies in which the denominn-
lions arc of the same name? 11. What is the usd of Troy weight
and, what are the denominations'? 12. ^avoirdupois weight '{
> the denominations ? 13. What distinction do you make between
gross and net weight 1 14. What distinction do you make between
long, square, and cubic measure ? 15. What nre the denominations
in long mensure 1 16. — ■— square measure? 17. — —in cubic
measure '{ 18. How do you multiply by 1-2? 19. When the divisor
contains a fraction how do you proceed ? 20. How is the superficial
'contents of a square figure found? 21. How is the solid contents of
any body found in cubic measure? 22. How many solid or cubic feri
'of wood maiiC a cord 1 23. What is underslocd by & cnrd foot ? 24.
I^owrminy such feet make a cord ? 23. What are the dennmin'itionH
■of dry measure 1 26. of wine measure ? 27. of time ? 28.
■■ of circular measure 1 29. For what is circular measure used /
30. How many rods in length is Gunter's chain? of how many links
does it consist 1 how many)inkf> make a rod? 31. How many rods
in a mile 1 '32, How many square rods in an acre 1 33. How many
pounds make 1 cwt, ?
1IS4.'.
•' I
SUVPLVMENT TO REDUCTION.
at gross.
re.
the length of
eiitlulums.
16" height of
lepths at sea.
of fish.
N.
IS reduction ? 3.
money, weight,
Hht into lower 1
1 6. What ore
me is given to
re llie ratios of
in Enftllsh mo-
lion to another
the denoniinu-
j( Troy weight
upois weight l
make between
make between
denominations
— ._ in cubic
jen the divisor
the superticial
ilid contents of
id or cubic ferl
cord foot ? 24.
dennminxtiuhii
of time? 28.
[measure used /
ow many linU
ow many rods
. How many
EXERCISES.
1. In 154 dollars, at 6s. each, how many pounds, 6lc.
Ans. d8£. lOs.
2. In 36 guineas, at l£. 3s. 4d each, how many crowns,
at 5s. 6d. ? Ana. 131 crowns and Ss. lOd. over.
3. How many rings, each weighing 5pwt. 7grs., mfey be
made of 3Ib. 5oz. IGpwt. Sgrs. of gold ? Ans. 158.
4. Suppose a bridge to be 212 rods in length, how many
times will a chaise wheel, 18 feetGinches in circumference,
turn round in passing over it 1 Ans ISOj^ times.
5. In 470 boxes sugar, each 261b., how many cwt.?
6. In 101b. of silver, how many spoons, each weighing
loz. lOpwt. ?
7. How many shingles, each covering a space 4. inches
one way and (5 inches the other, would it take to cover 1
square foot? How many to cover a roof 40 feet long, and
24 wide? {See ^ 25.) Ans. to the last, 5760 shingles.
8. How many cords of wood in a pile 26 feet long 4 feet
wide, and 6 feet high ? Ans. 4 cords, and 7 cord feet.
i). There is a room 18 feet in length, 16 feet in width,
and 8 feet in height ; how many rolls of paper, 2 feet wide,
and containing 1 1 yards in each roll, will it take to cover
the walls? ylw^-. 8^g.
10. How many cord feet in a load of wood 6-^ feet long,
2 feet wide, and 5 feet high? Ans. i-^^r cord feet.
11. If a ship sail 7 miles an hour, how fur will she sail,
at that rate, in 3vv. 4d. 16h?
12. A merchant sold ]2 hhds. of brandy, at 83 a gallon;
how much did each hogshead'come to, and to how much in
currency did the whole amount ?
13. How much clot^ at 7s. a yard, may be bought ior
29.£. Is?
14. A goldsmith sold a tankard for 10,£8s. at the rate
of 5s. 4d. per ounce ; how much Hid it weigh ?
15. An ingot of gold weighs 21bs. 8oz. 16pwt. ; how
much is it worth at 3d. per pwt. ?
16. At 11 pence a pound, what will 1 T. 2cwt. 3qr.s.
161b. "of lead come to ?
17. Reduce 14445 ells Flemish to ells English.
18. There is a house, the roof of which is 44^ feet in
length, and 20 feet in width, on each of the two sides ; if
3 shingles in width cover one foot in length, how many
V,
so
ADDITION OF COMPOUND NUMBERS. t[ 34, 35.
fihinglea will it take to lay one course on this roof? if 3
courses make one foot, how many courses will there be on
0ne side of the' roof? how many shingles will it take to
cover one side ? to cover both sides ?
Ans. 16020 shingles.
19. How many steps, of 30 inches each, must a man
take in travelling 54^- miles ?
20. How many seconds of time would a person redeem
in 40 years, by rising each morning ^ hour earlier than he
now does?
21. If a man lay up ^ a dollar each day Sundays except-
ed, how many pounds would he lay up in 45 years ?
22. If 9 candles are made from 1 pound of tallow, how
many dozen can be made from 24 pounds and 10 ounces ?
23. If one pound of wool make 60 knots of yarn, how
many skeins, of ten knots each, may be spun from 4 pounds
6 ounces of wool ?
Addition of Compound J¥unibers.
^ 93, 1. A boy bought a knife for 9 pence, and a comb
for 3 pence j how much did he give for both ? A7is. 1 shil-
ling.
2. A boy gave 2s. 6d. for a slate, and 4s. 6d. for a book ;
how much did he give for both ?
3. Bought one book for Is. 6d., another for 2s. 3d., an-
otlier for 7d, ; how much did they all cost ? Ans. 4s. 4d.
4. How many gallons are 2qts.-j-3qts.-f-lqt. ?
5. How many gallons are 3 qts. -f- 2 qts. -f- 1 qt. + 3
qts. -j- 2qts. ?
6. How many shillings are 2d.-f 3d.-i-5d.-f 6d.+7d ?
7. How many pence are Iqr. -J- 2 qrs. -\- 3 qrs.-j-2 qrs.
-flqr.?
8. How many pounds are 4s. -|- 10s. -|- 15s. -{- Is. ?
9. How many minutes are 30sec. -f- 45sec. -f- 20sec ?
10. How many hours are 40 min. -f- 25 niin. -{- (Jmin. ?
1 1. How many days are 4h. 4-8h. -f lOh. -f 20h. ?
12. How many yar<.\. in length are If. -j" 2f. t|- 1^-
I ill
■•»■"-
,,,,.-, , ^,-., _.J.^,,,J
5135.
> •- ■ ^ ■ *
ADDITION or COMPOUND NUMBERS. 'tt^"
£.
s.
d
15
14
6
20
2
8
5
6
4
13. How many feeet are 4 in. -f- 8 in. -f" ^^ >n. -)- 2in.4*>
1 inch? ■: |. • ', ': • • ,,,- •-.■(i' ■ i ■• • .-
14. How much \» the amount of 1yd 2ft. 6in' -{-^ yda.
Ifl. 8 inches?
15. What is the amount of 2s. 6d.-|.48. 3d.-|-7s. 8d. ?
16. A man has 2 bottles, which he wishes to fill with
wine ; one will contain 2 gal^ 3 qts. 1 pt. and the other 3
qts. ; 4iow much wine can be put in them ?
17. A man bought ahorse for 15c£ 14s. 6d., a pair of
oxen for 20£. 2s. 8d., and a cow for 5^. 6s. 4d. ; what did
he pay for all ?
When the numbers are large it will be most convenient
to write them down, placing those of the same kind, or de-
nomination, directly under each other, and, beginning with
those of the least value, to add up each kind separately. ,
OPERATION.
In this example, adding up the
column of pence, we find the amount
to be 18 pence, which being = Is.
6d., it is plain that we may write
Ans.^ 3 6 down the 6d. under the column of
1 — pence, and reserve the Is. to be add-
ed in with the other shillings.
Next, adding up the column of shillings, together with
the Is. which we reserved we find the amount to be 23s.
=1^. 3s. Setting the 3s under its own column, we add
the 1£. with the other pounds, and, finding the amount to
be 41^. we write it down, and the work is done.
Ans. 4l£. 3s. 6d.
Note. It will be recollected, that, to reduce a lower into
a higher denomination, we divide by the number which it
takes of the lower to make one of the higher denomination.
In addition, this is usually called carrying for that number :
thus, between pence and shillings, we carry for 12, and be-
tween shillings andl pounds, for 20, &.c.
The above process may .be given in the form of a general
KvLK for the Addition of Compound Numbers.
I. Write the numbers to be added so that those of the
same denominatiun may stand directly under each other.
II. Add together the numbers in the column of the lowest
denomination, and carry for that number which it lakes of
er
t'. *i
ADDITION OF COMOPUND NUMBERS. ''
tl 86.
the (fuhe.to tttiike 1 of the next higher denominiation. Pro-
ceed in this manner with all the denominations, till you
come to: the last, whose amount is written as in simple
numbers. : -ii '
Proof, The same as in addition'iof simple numbers, '
^ V '
HALIFAX CURRENCY.
£
s.
d, qr. £ s. d.
£ s. d.
46
11
3 2' 72 9 U
183 19 4
16
7
4 4 18 10^
8 17 10
538
19
7 1 36 16 6|
15 4
' £
8.
d. ' > £ s. d.
.£ s. d. "
14
7+ 37 15 8
61 3 2^
8
15
3 14 12 9f
7 16 8
63
4
7 17 14 9
29 13 10^
4
17
8 23 10 9i *
12 16 2
23
4f 8 6
7 5f
6
6
7 14 51
24 13
91
10^ 54 2 7J^
5 lOf 1
No examples in Federal Money are here introduced, al-
though the general rule for the addition of all compound
numbers is precisely applicable to the addition of Federal
Money, since that consists of different denominations. In
Federal Money the denominations increase and decrease in
a dedmal ratio. The pupil is therefore referred to the rules
for the Addition, Subtraction Multiplication and Division of
Decimals, which are the same absolutely with the rules for
the addition, subtraction, multiplication and division of
Federal Money. . '
TROY WEIGHT.
tb.
oz. pwt.
^^•
oz. pwt.
jn.
9z. pwt. gr.
3(3
7 10
11
6 14
9
18
42
6 9
13
9 G
16
13 16
81
7 16
15.
3 11
10
3 7 4
1135.
-3Wf
ADDITION QF gOMJPOtND NUMBEAS.
83
Bought a silver tankard, weighing 21b. 3 oz., a silver
cup, weighing 3 oz. 10 pwt. and a silver thimble, weighing
2 pwts. 13 grs. ; what was the weight of the whole *? ,
AVOIRDUPOIS WEIGHT,
2r,
T. cwt. qr. lb. oz. dr.
14 11 1 16 5 10
2 11 8 15
7 18 25 11 9
cwt. qr. lb. oz. dr,
16 3 18 6 14
3 16 8 12
22 11 10
/
A pnan bought 5 loads of hay, weighing as follows, viz,
23 cwt (=: 1 T. 3 cwt.) 2 qrs. 17 lb. ; 21 cwt. 1 qr. 19 lb. ;
19 cwt. qr. 24 lb. ; 24 cwt. 3 qr.; 11 cwt. qr. 1 lb. ;
how many tons in the whole ?
CLOTH MEASURE. ^
yds. qr. ??.
36 1 2
41 2 3
(}5 7
E. F. qr. na.
41 1 2
57 5 8
57. 3
EE qr. nn:
75 4 2
35 7 C
28 3 1
There are four pieces of cloth, which measure -s follows,
viz., 37 yds. JJ-qrs. 1 na. ; 18 yds. 1 qr. 2na. ; 40yt's. 3qrs.
3 na. ; 12 yds. qr. 3 na. ; how many yards in the whole ?
LONG MEASURE.
deg. mi. fur. r. ft. in. bar.
;>9 46 6 29 15 10 2
246 39 1 36 14 6 I
678 53 7 24 9 7 1
mi. fur. pol.
3 7
8 6 27
iia;
m
ADDITION OF COMPOUND NUMBERS.
Tras.
LAND OR saUARE MEASURE.
Pol. ft. in. A. rood. pol. ft. in.
36 179 137 56 3 37 245 228
19 248 119 29 1 28 93 25
12 96 75 416 2 31 128 119
There are 3 fields which measure as follows, viz. 17 A.
ar. 16p. : 28 A. 5r. 18p. ; UA. Or. 25p.; how much land
4in the three fields ? ^^
SOLID OR CUfiC MEASURE.
Ton. ft. in. yds. ft. in. cords, ft.
29 36 1229 75 22 1412 37 119
12 19 64 9 26 195 4 110
8 11 917 3 19 1091 48 127
WINE MEASURE.
hhds. gal. qts. pts,
50 53 1 7
27 39 3
9 13 1
tun. hhd. gal. qts,
37 3 44 5
19 1 50 1
28 2
Sll
![3
for Is.
2. A
how m
3. A
how m
4. bJ
A merchant bought two casks of brandy, containing as
follows, viz. 70 gal. 3 qts. ; 67 gal. Iqt. j how many hogs-
heads of 63 gal. each in the whole?
DRY MEASURE.
Bush. p. qt. pt.
36 2 5 1
19 3 7
Ch. bus. p. qt.
48 27 3 5
6 29 1 7
r ■ \:-} 3i
1135.
^ S6. SUBTRACTION OP COMPOUND NUMBERS.
TIME.
Y. mo. IP. d. Ji. m. s.
75 11 3 6 23 55 11
84 9 2 16 42 18
32 6 5 5 18 5
27
8S
Y. mo. w. d.
40 3 1 5
16 7 4
2
nz. 17 A.
nuch land
cords, ft.
37 119
4 no
48 127
5
1
Intaining as
lany hogs-
qt.
5
7
Siibtraetion of Compound JVnmbcris
^ 36. 1. A boy bought a knife for 9 pence, and sold it
for Is. 4d. ; how much did he gain l-y the bargain?
2. A boy bought a slate for 2s. 6d., and a book for 3s. 6d,;
how much more was the cost of the book than of the slate?
3. A boy owed his playmate 2s. ; he paid him Is. 6d. ;
how much did he then owe him?
4. Bought two books; the price of one was 4?;, 6d.. the
price of the other 3s. 9d, ; what was the difference of their
costs?
5. A boy lent 5s. 3d.; he received in payment 2s. 6d. ;
how much was then due ?
6. A man has a bottle of wine containing 2 gallons and 3
quarts ; after tumiug out 3 quarts how much remained •
7. How much is 4 gal, less 3 gal.? 4 ga^. — (less) 2qt. ?
4 gal.— Iqt. ? 4 gal. — 1 gal. Iqt. ? 4 gal. — 1 gal. 2qts ?
1 4 gal, — 1 gal. 3qts ? 4 gal. — 2 gal. 3qts? 4 gal. 1 qt. —
1 gal. 3 qts.?
8. How much is 1ft. — (less) 6in? 1ft. — 8 in 7 6ft, 3
[inches, — 1 ft. 6 in. 7ft. Sin. — 4ft. 2in? 7ft. Sin. — 5ft,
lOin?.
9. What is the difference between A£ Qi. and 1^ ^'s. ?
10. How much is 3oe~(less) Js.? 3i:— 2s. 3i:— 3s. ?
\U —15s ? Z£ 4s.-^2^ 6s ? 10^ 4s. -^£ 8s .'
11. A man bought a horse for 30=^4'^. 8d., and a cow
IforS.^ 14s. 6d. ; what is the difference of their costs?
H
^ ' ■^ti
m
m
86
STTBTRACTION OF COMPOUND NUMBERS.
1136.
js ■
,\iiU%
OPERATION.
£. s. d.
Minuend, 30 4 8
Subtrahend y 5 14 G
As the two numbers are large,
it will be conveniertt to write
them down, the less under the
greater, pence under pence, shil-
lings under shillings, &c. We
Ans. 24 10 2 \ may now take 6d. from 8d., and
there will remain 2d. Proceeding to the shillings, we can-
not take 148 from 48., but we may borrow as in simple num-
bers, one from the pounds,=20s., which joined to the 4s,
makes 24s. from which taking 14s. leaves 10s, which we
iset down. We must now carry 1 to the 5^ making Q£
which taken from 30=£ leaves 24<£ and the work is done.
Note. The most convenient way in borrowing is, to t»ub-
tract the subtrahend from the figure borrowed, and add the
difference to the minuend. Thus, in the above example, 3 4
from 20 Jeaves 6, and 4 is 10,
The process in the foregoing example may lie presented
in the form of a Rule for the Subtraction of Compound
Numbers.
I. Write down the sums or quantities, the less under the
greater, placing those numbers which are of the same de-
nomirlation directly under each other.
II. Beginning with the least denomination, take succes-
isively the lower number in each denomination from theup-
))er, and write the remainder underneath, as in subtraction
of simple numbers.
III. If- the lower number of any denomination be greater
than the upper, borrow as many units as make one of the
next hiijher denomination, subtract the lower number there-
from, and to the remainder add the upper number, remem-
bering always to add one to the next higher denomination for
tl;.at which you boirv^wef^.
Proof Add the vemainder and the subtrahend together,
as in subtraction of simple numbers; if the work be right,
the amount will be equal to the minuend.
;examples for practice.
HALIFAX CURRENCY.
£. s. d. £. s. d.
79 17 8 103 3 2
35 12 4 71 12 5
fl36.
SUBTRACTION OP COMPOUND NUMBERS.
87
£ s. d.
81 10 11^
29 13 3
tB s. d.
245 12
27 9 4^
520 U 3
109 17 4
631 14 7
6 19 9
MISCELLANEOUS EXAMPLES.
1. A merchant sold goods to the amount of 136.£ 7s. 6|d ,
and received in payment 50<£ 10s. 4f d ; how much remain-
ed due ? Ans. 85<£ 17s. If d.
2. A man bought a farm for 1256<£ 10s, and, in selling
it, lost 87.£ 10s. 6d ; how much did he sell it for ,"
^ws. 1168^ 19s. 6d.
3. A man bought a horse for 27^ and a pair of oxen for
19c£ 12& 8^d J how much was the horse valued more than
the oxen 1
4. A merchant drew from a hogshead of molasses, at one
time, 13gal. 3qts; at another time, 5gal. 2qts' Ipt; what
<juantity was there left ? Ans, 43gaJ. 2qts. Ipt.
5. A pipe of brandy, containing 118 gal. sprang aleak,
when it was found only 97gal. 3qts. Ipt. remained in the
cask ; how much was the leakage ?
6. There was a silvei* tankard which weighed 31b. 4c»z. ;
the lid alone weighed 5oz. 7pwt. 13grs ; how much did the
tankard weigh without the lid ?
7. From 151b, 2oz. 5pwt. take9oz. 9pwt. lOgrs.
8. Bought a hogshead of sugar, weighing 9cwt. 2qrs.
I71b ; sold at three several times as follows, viz. 2cwt. Iqr,
llib. 5oz; 2qrs. 181b. lOoz ; 251b. 6oz ; what was t^e
.V3ight of sui 11 which remained unsold ?
iins. Gcwt. Iqr. 171b. lloz.
9. Bought a piece of black broadcloth, containing 36yds,
2qrs ; two piec s of blue, one containing 10 yds. 3qrs. 2na.
the other 18 yds. 3qrs. 3na ; how much more was there of
the black than of the blue.?
10. From 28 miles, 5 fur. 16r. take 15m 6 far. 26r 12ft
11. A farmer has two mowing fields; one containing 13
i
I
;>v
<i8 SUBTRACTION OF COMPOUND NUMBERS. ^ 36, 37.
m
acres 6 roods ; the other, 14 acres 3 roods : he has two
pastures also ; one containmg 26 A. 2r. 27p; the other,
45 A. 5r. 33p : how much more has he of pasture ''han of
mowing f
12. From 64A.'2r. lip. 29ft. take 26A. 5r. 34p. 132ft.
13. From a pile of wood, containing 21 cords, was sold,
at one time, 8 cords 76 cubic feet ; at another time, 5 cords
7 cord feet ; what was the quantity of wood left ?
14. How many days, hours and minutes of any year will
be future time on the ^h day of July, 20 minuses past 3
o'clock, P. M ? Ans. 180 days, 8 hours, 40 minutes.
15. On the same day, hour and minute of July, given in
the above example, what will be the difference between the
past and future time of that month ?
16. A note, bearing date Dec. 28th 1826, was paid Jan.
2d, 1827 ; how long was it at interest ?
The distance of time from one date to that of another may
be found by subtracting the first date from the last, observ-
ing to number the months according to their order. (1T34.)
OPERATION.
NotCu In casting in-
terest, each month is
reckoned 30 days.
Ans, 4 days.
17. A note, bearing date Oct. 20th, 1823, was paid April
25th, 1825 ; how long was the note at interest ?
18. What is the difference of time from Sept. 29, 1816,
^o April 2d, 1819 ? Ans, 2y. 6m. 3d.
19. London is Sl^ 32', and Montreal 45o 30', N. lati*
tude ; what is the difference of latitude between the two
places ? . Ans. 6® 2.'
20. Montreal is 73*^ 20', and the city of "Washington is
770 43/ \y_ longitude ; what is the difference of longitude
between the two places? Ans. 4° 23'.
21. The island of Cuba lies between 74o and 85o W.
longitude; how many degrees in longitude does it extend?
IT 37, 1. When it is 12 o'clock at the most easterly ex-
tremity of the island of Cuba, what will be the hour at
the most westerly extremity, the difference in longitude be-
ing 110 1
Note. The circumference of the earth being 360<*, and
the ^arth performing one entire revolution -in 24 hours^ it
A Y. { 1827. Istm. 2d day.
, • "' \ 1826. 12 ^28
-f?'*' jr^T-^
H 36, 37.
le has two
the other,
ire ^han of
). 132ft.
, was sold,
ae, 5 cords
y year will
i^es past 3
minutes,
y, given in
letween the
s paid Jan.
tnother may
ast, observ-
ier. (1134.)
a casting in-
|i month is
days.
\ paid April
. 29, 1816,
2y. 6m. 3d.
W, N. lati«
en the two
Ans. 6° 2.'
ishington ia
longitude
ns. 40 23'.
d 850 W.
it extend 1
asterly ex-
|he hour at
gitude be-
3600, and
|4 hours^ it
fl 37, 38. MULTIPLICATION AND DIVISION, &C.
89
follows, that the motion of the earth on its surface, from
west to east, is
150 of motion in 1 hour of time; consequently,
10 of motion in 4 minutes of time, and
1' of motion in 4 seconds of time.
From these premises it follows, that, when there is a dif-
ference in longitude between two places, there will be a
corresponding difference in the hour, or time of the day,
The difference in longitude being 15o, the difference iii time
will be one hour, the place easterly having the time of the
day 1 hour earlier than the place westerly, which must be
particularly regarded.
If the difference in longitude be 1®, the -difference in
time will be 4 minutes, &c.
Hence, — If the difference in longitude, in degrees and
minutes, between two places, be multiplied by 4, the pro-
duct will be the difference in time, in minutes and seconds,
which may be reduced to hours.
We are now prepared to answer the above question. '
110 Hence, when it is 12 o'clock at the
4 most easterly extremity of the island,
— it will be 16 minutes past 11 o'clock
44 minutes. at the most western extremity.
2. Montreal being 730 20' W. longitude and Washington,
77043'; when it is 3 o'clock at the city of Washington,
what is the hour at Montreal ?
Ans. 17 minutes 32 seconds past 3 o'clock.
3. Lower Canada being about 73o, and the Sandwich
Islands about 155o W. longitude, when it is 28 minutes past
^ o'clock, A. M. at the Sandwich Islands, what will be the
hour in Lower Canada?
Ans. 12 o'clock at noon, lacking 4 minutes.
Hultiplicafion it Diyisionof Compound
JVnmber^.
tf 38. 1. A man bought 2 yards of cloth, at Is. 6d. per
[yard; what was the cost ?
H2
y
V«j-... - - •. y,;c^
00
MULTIPLICATION AND DIVISIO.V.
. m
11 j».
2. If 2 yards of cloth cost 3 shillings, what is that per
yard?
3. A man has three pieces of cloth, each iiienKuring 10
yds. 3qr3. ; how inn.ny y.irds in the whole?
4. If 3 equal pieces of cloth contain 32yu3. 1 qr., how
much does each piece contain ?
5. A man has five hottles, each containing 2 gal. 1 qt.
1 pt. ; how much wine do they all contaii^ ?
6. A man has 11 gal. 3qts. Ipt. of wine, which he would
divide equally. into 5 bottles; how much must he put into
each bottle ?
7. How many shillings are 3 times 8d? — 3X0d ''
— 3Xl(W? — 4X7d? -7X0d? lOXDd?
2X3qrs? 5X2qrs?
8. How much is one third of 2 shillings? — J of 2s 3d^
— ^ of 3s. 6d ? — 1^ of 2s. 4d ? — ^ of 3s. Gd ?— J
'of 7s. 6(1 ? — ^ of l^d ?— J of 2|d?
10
9. At \£!Ss. 8fd. per yard,
A\'hat will 6 yards of cloth
cost ?
10. If G yards of cloth cost
7£ 14s. 4^(1, what is the price
per yard ?
Here, as the numbers are large, it will be most convenient
to write thera down before multiplying and dividing.
OPER>riON. OPERATION,
£ s. d. qr.
G )7 14 4 'Z cost of (S yards.
£ s. d. qr.
15 8 fS price of \ yard
6 number of yds.
Ans. 7 14 4 2 cost of 6 yards
6 times 3 qrs. are ISqrs.rr
4d. and 2qrs. over; we set
down the two qrs; then, G times
8d. are 48d, and 4 to carry
makes 52d. = 4s. and 4d.
over, which we write down ;
a^ain 6 times 5s. are 30s.
and 4 to carry makes 34s. =
1<^ and 14s. over; G times
1^ are 6£, and one to carry
makes 1£, which we write
down, and it is plain, that
15 8 'ii price of 1 yarc\
Proceeding after the man-
ner of short division, 6 is con-
tained in 1£ 1 time, and l£\
over; we write down. the |
quotient, and reduce the re-
mainder {]£) to shillings,
(20s,) which, with the given I
shillings, (I4s,) make 34s:
G in 343. goes 5 times, and
4s. over ; 4s. reduced to pence]
:48d, which with the giv-
en pence, (4d,) make 52d ; d
in 52d. goes 8 times, and 4d,(
the united products arising|over ; 4d, = 16 qrs. which,
ION.
^^-
n^-
/,.. OP COMPOUND NUAIBSRS.
91
Ts, what is that per
each measuring 10
e?
I 32yus. 1 qr., how
itaining 2 gal. 1 qt.
aii^?
ine, which he would
ch must he put into
les 8d?— '3X9(1?
lOXDd?
ings? — ^ of 29 3(11
. ^of3s. 0(1?— T^o
' G yards of cloth cost
4^d, what is the price
I?
ill be most convenient
and dividing.
TION.
d. qr.
4 2 cost of 6 yards.
8 ^price of 1 yar(\
eding after the man-
lort division, G is con-
7^ 1 time, and li
! write down, the
and reduce the re-
(Ji:) to shillings,
hicb, with the given
(I4s,) make 34s;
goes 5 times, and
4s. reduced to pence
[which with the giv
;, (4d,) make 52d ; '
roes 8 times, and 4(1.
'td. = 16 qrs. which,
from the several cknomina-
tions is the real product aris-
ing from the whole compound
number. . ,; , .. .
U. Multiply ^£ 4s. Gd.
by 7.
13. What will be the cost
of 5 pairs of shoes at lOs. Gd.
a pair ?
15. In 5 barrels of wheat,
, each coritaining 3 bus. 3 pks.
Gqts, how many bushels ?
17. IIoW many yards of
cloth will be required for 9
coats, allowing 4 yards Iqr.
3na. to each ?
19. In 7 bottles of wine,
each containing 2(|ts. Ipt. 3
gills, how many gallons ?
21. What will be the
weight of 8 silver cups, each
weighing 5oz. 12pvvt 17grs?
23. How much sugar in 12
hogsheads, each containing
9cwt. 3qrs.211b?
25. -In 15 loads of hay,
each weighing IT. 3cwt. 2qrs.
how many tons ?
with the given qrn. (2) =s= 18
([rs; G ^n IBqrs. gies 3 times
and it is plain, that the unit-
ed (quotients arising from the
several denominations, is the
real quotient arising from the
whole compound number.
12. Divide 22je lis. 6d.
by 7.
14, At2£ 12h 6d. for 5
pairs of shoes, wli • is that a
pair :
IG. If 14bu8.'^^ -,. Gqts.of
wheat be e(iually divided into
5 barrels, how many bushels
will each contain ?
18. If 9 coats contain 39
yds. 3qrs. 3na, what does 1
coat contain ?
20. If 5 gal. 1 gill of wine
be divided equally into 7 bot-
tles, how much will each con-
tain ?
22. If 8 silver cups weigh
3lb. 9oz. Ipwt. IGgrs., what
is the weight of each I
24. If 119cjvt. Iqr. of su-
gar be divided into 12 hogs-
heads, how much will each
hogshead contain?
26. If 15 teams be loaded
with 17T. 12cwt. 2qrs. of hay,
how much is that to each team?
When the multiplier or divisor, exceeds \2, the operations
of multiplying and dividing are not so easy, unless they be
composite nnmbers ; in that case, we may make use of the
component parts, or factors, as was done in simple numbers.
Thus 15, in the example 15 being a composite num-
above is a composite number, ber and 3 and 5 its compo-
ptoduced by the multiplica- nent parts, or factors, we miy
I
IMAGE EVALUATION
TEST TARGET (MT-3)
1.0
1.1
U£i28 |2.5
•^ IM III 2.2
2.0
U!
■ 4.0
1.8
1.25 1 ,.4 , ,.6
^ 6"
►
vl
/:
^ei
^J
■/A
^5^V#
'/
Photographic
Sciences
Corporation
23 WEST MAIN STREET
WEBSTER, N.Y. 14580
(716)872-4503
'r.^^r
■J
MULTIPLICATION AND DIVIBION
1138.
w
tkm of 3 and 5, ( 3X 5 =
16.) We may therefore,
multiply IT. 3cwt. 2qrs. by
one of those component parts,
or factors, and that product by
the other, which will give the
true answer, as has been al
ready taught, (H 11.)
OPERATION.
T. evDt. qK ' /
1 3 2
3 one of tkef actors.
3 10 2
5 the other factor.
divide 17T. 12cwt. 2qrs. by
one of these component parts
or factors, and the quotient
thence arising by the other,
which will give the true an-
swer, as already taught,
(1120.)
17 12 2 the answer.
27, What will 24 barrels
of flour cost, at 2jS, 12s. 4d.
a barrel ?
29. What will 1121b. of su-
gar cost at 7:^d. per lb?
Note. 8, 7, and 2, are fac-
tors of 112.
31. How much brandy in
Omfactor,
OPERATION.
T. ciot. jr
3)17 12 2
The other factor ^S)& 17 2
MSf 1 3 2
28. Bought 24 barrels of
flour for6S^ 16s; how much
was that per barrel?
30. If Icwt. of sugar cost .
3jB, 7s. 8d., what is that per '
lb?
82.
\
Bought
84 pipes of
84 pipes, each containing 112jbrandy, containing 9468 gal.
gal. 2qts. lpt^3g?
33. What wUl ISQyds. of
cloth cost, ait 3£, 66. 5d. per
yard?
139 is not a composite num<
ber. We may, however, de-
compose this number thus,
139^100+304-9.
We may now multiply the
price of 1 yard by 10, which
will give the price of 10 yards,
and this product again by 10,
which will give the price of
100 yards.
Iqt. Ipt ; how much in a pipe ?
34. Bought 139 yards of
cloth for 461£ lis. lid;
what was that per yard ?
When the divisor is such a
number as cannot be produced
by the multiplication of small
numbers, the better way is to
divide after the manner of
long division, setting down
the work of dividing and re*
ducing in manner as follows :
-.i4t'ii-"-i-i. ;
^k'yiiijj-"-. .,.Ji'.s.i. ■; ,,jj(i*.,:i.j»^^-.j-.,'.'4- i^'.w.j
1138.
OP COMPOUND NUMBERS.
d3
We may then fnultiply the
price of 10 yards by 3, which
will give the price of 30 yards
and the price of 1 yard by 9,
which will give the price of
9 yards, and these three pro-
ducts, added together, will
evidently give the price of
139 yards ; thus :
£ s. d.
3 6 5 price of J t/ard,
10
33 4 2 price of 10 yards.
10
332 1 8 price of \00yds.
99 12 6 price of QOyds.
29 17 9 price of 9 yds.
461 11 U price of \2» yds.
30 yards, and in multiplying
the price of 1 yard {3£ 6s.
9 yards, the multipliers, 3
and 9, need not be written
down, but may be carried in
the mind.
£ 5. d.
139)461 11 11(3.^
417
lit
30 ,
■•;)
,b9\(Q
834
57
695 {5d.
695
i . • r?.
Note. In multiplying the
price of 10 yards (33^ 4s
2d.) by 3, to get the price of]6 times, (6s,) and a remainder
The divisor, 139, is contain-
ed in 461^ 3 times (3^,) and
a remainder of 44^, which
must now be reduced to shil*
lings, multiplying it by 20,
and bringing in the given ^hil*
lings, (lis,) making 891s, in
which the~ divisor is contained
of 57s, which must be reduc-
ed to pence, multiplying it by
5d.) byfi, to get the price of 12, and bringing in the given
pence, (lid,) together mak-
ing 695d, in which -the divi-
sor is contained 5 times, (5d,)
and no remainder.
The several quotients, 3£
I. 5d. evidently make the
answer.
The processes in the foregoing exaropWs mayjaow be pre-
sented in the form of a '
KvLEfor the Multiplication ofKvhEf/tr the Division of Com'
Compound Numbers.
1. When the multiplier does
not exceed 12, multiply suc-
cessively the numbers of each
denomination, beginning with
pound Numbers.
1. When the divisor does
not exceed 12, in the manner
of short division, find how
many times it is contained in
;■■+'
04
MULTIPLICATION AND DIVISION
1138.
the least, as in multiplication
of simple numbers, and carry
as in addition of compound
numbers, setting down the
whole product of the highest
denomination.
II. If the multiplier exceed
12, and be a composite num-
ber, we may multiply first by
one of the component parts,
that product by another, and
s5 on, if the component parts
be mere than two; the last
product will be the product re-
quired.
III. When the multiplier
exceeds 12, and is no^ a com-
posite, multiply first by 10,
and this product by 10, whic|i
will give the product for 100 ;
and if the hundreds in the mul-
tiplier be more than one, mul-
tiply the product of 100 by the
number of hundreds ; for the
tens, multiply the product of
10 by the number of tens ; for
the units, multiply the multi-
plicand ; and these several pro-
ducts will be the product re-
quired.
the highest denomination, un-
der which write the quotient,
and if there be a remainder,
reduce it to the next less de-
nomination, adding thereto the
number given, if any, of that
denomination, and divide as
before; so continue to do
through all the denominations
and the several quotients will
be the answer.
II. If the divisor exceed 12,
and be a composite, we may di-
vide first by one of the com-
ponent parts, that quotient by
another, and so on, if the com-
ponent parts be more than
two, the kst quotient will be
the quotient required.
III. When the divisor ex-
ceeds 12, and is not a compos-
ite number, divide after the
manner of long division, set-
ting down the Work of divid-
ing and reducing.
I
'examples for PRACTICE,
\ «
E^ALIFAX
CURRENCY.
'
£
,S.
d.
£
5,
d.
Multiply
81
6
5
93
4
11
by
f
17
»
) '
48
Vi*^--.^:.-:
1138. 1 ^38 39.
OF COMPOUND NPKBERS.
95
)mination, un-
5 the quotient,
a remainder,
next less de-
ling thereto the
if any, of that
and divide as
tntinue to do
denominations
i quotients will
jrisor exceed 12,
site, we may di-
le of the com-
tiat quotient by
on, if the com-
be more than
iiotient will be
quired.
the divisor ex-
s not a compos- 1
iivide after the
ig division, set-
Work of divid-
ing.
5.
d.
4
11
48
L- ,
£ s. d.
Mult.p]y98 3 10
by^ 78
£ s. d.
64 11 2
986 11 4
73
93
93
' ' ,
892
5 3
145
Divide 77 11 9 by 18.
« 140 2 3 " 21.
" 360 5 2" 133.
" 7856 8 9" 197.
£ s. d.
143 2 3 by 21.
1950 7 4" 98.
47 9 6 " 11.
562 8 3" 20.
MISCELLANEOUS EXAMPLES.
at 4s.
7^d.
per
1. What will 359 yards of| 2. Bought 359yds. of cloth
for83.£0s /•J-d; what was
that a yard ?
4. If 441r.wt. 131b. of flour
be contained in 241 barrels^
how much in a barrel ?
6. If 37lbu. Ipk. of wheat
be divided equally into 135
bags, how much will each
bag contain ?
8. At 759^ 10s. for 35cwt.
of tobacco, what is that per
lb?
10. If 14 men build 92 rods
12 feet of stone wall in 74
days, how much is that per
day?
cloth cost,
yard? ,
3 In 241 barrels of flour,
each containing Icwt 3qr.
91b; how many cwt?
5. How many bushels of
wheat in 135 bags, each con
taining 2 bu. 3 pks ?
3X9X5=135.
7. What will 35cwt. of to-
bacco cost, at 38 lO^d. per
lb?
9. If 14 men build 12 rods
6 feet of wall in one day, how
many rods will they build in
7^ days ?
II 39. 1. At 10s. per yard, what will 17849 yards of
cloth cost ?
Note. Operations in multiplication of pounds, shillings,
pence, or of any compound numbers, may be facilitated by
I
96
MULTIPLICATION AND DIVISON, &,C.
TI39.
N
taking aliquot parts of a higher denomination. Thus, in this
last example, if the price had been 20s. i. e. \£ per yard,
it is clear, the price of the whole would havq^been equal t»
the whole number of' yards in pounds, 17849; but the price
is 10s. i. e. ^£ per yard, and so the price of the whole
will be equal to ^ the number of yards, ^'^f*^ in pounds;
8924i£, or 8924^: 10s.
When one quantity is contained in another exactly 2, 3,
4, 5, &;C. times, it is called an aliquot or even part of that
quantity ; thus 6d. is an aliquot part of a shilling, because
6d.X2=l shilling; so 3d. is an aliquot part of a shilling;
3d. X4=ls. ■ So 5s. is an aliquot part of a pound, for 5s.
X4=lc£: and 3s. 4d. is an aliquot part of a pound, for
3s. 4d.XG=loe, &c. '
From the illustration of the last example it appears, that,
when the price per yard, pound, &,c. is one of these aliquot
parts of a shilling, or a pound, the cost may be found by
dividing the given number of yards, pounds, i^c. by that
number which it takes of the price to make Is. or 1£. If
the price be Gd. we divide by 2 ; if 5s. we divide by 4 ; if
3s. 4d. by 6, fee. &:-c. This manner of calculating by ali-
quot parts, is called Practice.
2. What cost 34648 yards ^f cloth, at 10s. or ^i^ per
yard I at 5s.=r^^ per yard ? at 4s.=^,£ per
yard? at 3s. 4d.=^^ per yard? at 2s.:=j^^£
per yard ? Ans. to last, 3464c£ 1 6s.
3. What cost 7430 pounds of sugar, at 6d.=: J^s. per lb ?
at 4d.==^s.per lb? at 3d.=;^s per lb?
at 2d.=^s. per lb ?
at l^d.=|s. per \h?
Ans. to the last, 7 4^0s.=928s. 9d.=46^8s. 9d.
4. At 3i? 16s. per cwt, what will 2qrs.=J^cwt. cost?
what will lqr.:fc^cwt. cost? what will ]61b.=
|cwt. coat ? what will 141b.=^cwt. cost ? what
will 81b.=^\cwt. cost? Ans. ^o the last, 5s. 5|d.
5. What cost 340 yards of cloth, at 12s. 6d. per yard ?
12s. 6d.=10s. {=^£) and 2s. 6d. (=i£) ; therefore,
^)l)340
170<.€ := cost at 10s. per yard.
42^ 10s.=at 2s. 6d. per yard.
Ans. 2\2£ M)s.=at I2s. 6d. per yard.
1139.
'hus, in this
£ per yard,
len equal t»
•ut the price
■ the whole
in pounds ;
exactly 2, 3,
part of that
ng, because
f a shilling ;
»und, for 5s.
a pound, for
ippears, that,
these aliquot
be found by
&>€. by that
is.orl^. If
vide by 4 ; if
lating by ali-
)s. or ^^per
4s.=|^ per
at ^s.=j^xj£
,3464.£ 16s.
zr J^s. per lb 1
lb?
-A6£ 8s. 9d.
^cwt. cost?
will 161b.=
;? what
J last, 5s. 5|d.
|6d. per yard ?
^erefore,
rd.
rd.
ird.
1139.
8VPPLBMENT TO COMPOUND NtJMBESfl,
t
Or, ,, , ...V.-..V
97
iOB.=^£)MO
2g. 6d=i of I0f,)170j6' at 10s. per yard.
42^ lOg. at2s. 6d. perybd.
Ans. 212^. 10s. at 12s. 6d. per yard.
SUPPLEMENT TO ^COMPOUND NUMBERS. .
QUESTIONS.
1. What dbtinetioti do you make between slmpltf nnd eompwM
numbers 1 (P 26.) 2. What is the«ule for addition of coDspouiid
numbers ? 3. -—— for subtraction of, &,e. 4. There are three roii'*
ditiona in the rule giten for tRuItipticaliiiii of cc^pdund htimlfer« i
what are they, and the methods of procedure under each? 6. The
same questions in respect ttf the dttision of (Sompound numbers f 0.
Whan the multiplijpr or divisor is encumbered with a fraction, how
d& you proceed t 7. How is the distance of time from one date ttf
another found ? 8. How many degrees dbes the earth re^^olva frotn
Wfest to cast in 1 hour t 9. In what tiine does it revolte ]e 1 Where
Is the time or hour of the day earlier-^at the place most easterly of
most westerly ? 10. The difference in longitude between two placet)
b^ingl known, how ii the difference in time calculated? 11. How
nfay operations^ in the fhultiplieation of compound numbers be fa«
eilitated 1 12. What are some of the aliquot parts of £l f «•— of
Is. 1 " of Icwtl 13. What is this manner of operating usually
called 1
EXERCISES.
1. A gentlennan 13 possessed of l^dciz«tl of silver, spodtts,
each weighing 3oz. 5pwt; 2 doz. of tea spoons^ eaeh weigh*
ing ]5pwt. 14gr; 3 silver cans, «ach 9oz. 7pwt ; 2 «il?#r
tankards, each 21oz. I5pwt ; and 6 silver .porrmgefrs^each
Uoz. ISpwt; what is the weight of the whol^?
>4rt5; I8Ib. 4o«. 3pwt.
Note. Let the pupil he required to reverse and prove the
following examples :
2. An English guinea should #eigh ^pwf. Cgr,' a p\eee'
of gold weighs 3pwt. 17gr ; how buch is that short of the
tveight of aguinea?
3. What is the weight of 6 chests of iei, each wtfighiitf
dcwt. 2€(rs. 91b ?
4. In 35 pieces df cloth, eaek ffleafufing S7 j^rdSfhcw
many yards ?
9d SUPPLEMENT TO COMPOUND NUMBERS. H ^f ^0.
5. How much brandy in 9 casks, each containing 45 gal.
3qts. Ipt?
6. If 31cwt. 2qrs. 201b. of sugar be distributed equally
into 4 casks, how much will each contain .'
7. At 4^. per lb. what cost Icwt. of rice ? 2cwt ;
3cwt?
Note. The pupil will recollect that 8, 7, and 2 are fac-
tors of 112, and may be used in place of that number.
8. If 800cwt. of cocoa cost 18i£ 13s. 4d. what is that
per cwt 1 what is it per lb. ?
9. What will 9:|cwt. of copper cost at 5s. 9d. per lb ?
10. If 6j^cwt. of chocolate cost 72<f . 16s, what is that
per lb? . . #
11. What cost 456 bushels of potatoes, at 2s. 6d. per
bushel ?
Note. 2s. 6d. is i of 1^ (See H 39.)
12. What cost 86 yards of broadcloth, at 15s. per yard ?
Note. Consult ^ 39, ex. 5.
13. What cost 7846 pounds of tea, at 7s. 6d. per lb. ?
at 14s. j^er lb 1 13s. 4d ?
14. At $94*25 per cwt. what will be the cost of 2qrs. of
tea? of 3 qrs? of 141bs? of 21 lbs?
-cfl61b8? of241b8.^
Note. Consult ^ 39, ex. 4 and 5.
15. What will be the cost of 2 pks. and 4qts. of wheat,
at 8s. 6d. per bushel ?
16. Supposing a meteor to appear so high in the heavens
as to be visible at Montreal, 73° 20', at the city of Wash-
ington, 77° 43', and at the Sandwich Islands, 155o W. lon-
gitude and that its appearance at the city of Washington
be at 7 minutes past 9 o'clock in the evening; what will be
the hour and minute of its appearance at Montreal and
at the Sandwich Islands ?
Fractions^
ff 40. We have seen, (t| 17,) that numbers expressing
tvhole things are called integers or whole numbers ; but that
in division, it is often necessary to divide or break a whole
thing into parts, and that these parts are called fractions,
or broken numbers.
V,,,.,.:,
39, 40.
g 45 gal.
1 equally
— 2cwt ;
I are fac-
nber.
lat is that
erlb? '
hat is that
Is. 6d. per
. per yard?
)d. per lb. ?
tofSqrs. of
lbs?
. of wheat,
Ithe heavens
■y of Wash-
iS® W. lon-
ashington
hat will be
>ntreal and
1140.
FRACTIONS.
99
expressing
rs; bat that
jak a whole
td fractions,
It will be recollected, (f[ 14, ex. 11,) that when a thing
or unit is divided into 3 parts, the parts or fractions are call-
ed thirds ; when into four parts, fourths ; when into six parts,
sixths ; that is, the fraction takes its name or denomination
from the number of parts into which the unit is divided. Thus
if the unit be divided into 16 parts, the parts are called six-
teenths, and 5 of these parts would be 5 sixteenths, expressed
thus, ^. The number below the short line,. (16,) as before
taught, (1[ 17,) is called the denominator, because it gives
the name or denomination to the parts ; the number above
the line is called the numerator, because it numbers the parts.
The denominator shows how many parts it takes to make
a unit or whole thing; the numerator shows how many of
these parts are expressed by the fraction.
1. if an orange be cut into 5 equal parts, by what frac-
tion is 1 part expressed ? 2 parts ? — — 3 parts ?
— — 4 parts? ^ 6 parts ? how many parte make unity
or a whole orange ?
2. If a pie be cut into 8 equal pieces^ and 2 of these
pieces be given to Harry, what will be his fraction of the
pie? iif 5 pieces be given to John, what will be his fr^tion t
what fraction or part of the pie will be left ?
It is important to bear in mind, that fractions arise from
division f (If 17,) and that the numerator maybe considf^eda
dividend, and the denominator a divisor, and the ralue of the
fraction is the quotient ; thus, ^ is the^. quotient of 1 (^.
numerator) divided by 2 (the denonainator ;) ^ is^tke quo-
tient arising from 1 divided by 4, and f is 3 times as much,
that is, 3 divided by 4 ; thus, one fourth part of ^ is the
same as 3 fourths of 1. ,;
Hence, in all cases a frftotjon is dtways expressed by the
sign of division.
f expresses the quotient, ( 3 ittlte4iyidend, omumerator,
of which ( 4 is the divisor or denomiiiator.
3. If 4 oranges be equally divided among 6 boys, what
part of an orange is each boy'-s share ?
A sixth part of an orange is ^, and a sixth part of 4 oranges
is 4 such pieces,=s5;|. Ans. | of an orange.
4. If 3 apples be equally divided among 5 boys, what part
of an apple is each boys share 'f if 4 apples, what ? if 2t
apples, what ? if 5 apples, what ?
100
FRACTIONS.
H 40, 41.
5. What is the quotient of 1 divided by 3 ?-^— of 2 by 3 ?
^— K)f 1 by 4? ;of 2 by 4? of 3 by 4? of 5
by7? -of 6 by8? of4by5?— of2 by 14?
6. What part of an orange is a third part of 2 oranges ?
one fourth of 2 oranges ? ^ of 3 oranges ?
^of threeoranges? j of 4? Jof 2? -- — f of 5?
-^|of3? iof2?
A proper fraction. Since the denominator sho\(rs the num-
ber of parts necessary to make a whole thing, or 1, it is plain
that when the numerator is less than the denominator, the
fraction is less than a unit, or whole thing ; At is then called
^proper fraction. Thus, 1, |/&'C. are proper fractions.
An improper fraction. When the numerator equals or ex-
ceeds the denominator, the fraction eaualsor exceeds unity,
or 1 ,' and is then called an imprmer fraction. 'J^hns, f , ^^
f , V^, are improper fractions.
A milled number ^ as already shown, is one composed of a
whole number and a fraction. Thus, 14j, 13|, d&c. ate
^ixed numbers.
* 7. A father bought 4 oranges, and cut each orange into
6 equal parts ; he gave to Samuel 3 pieces, to James 5
pieces, to Mary 7 pieces, and to Nancy 9 pieces ; what was
each one's inaction ?
Was James' fraction proper or improper ? WJiy? ''^^
Was Nancy's fraction proper, or imprq>er? Why?
To change an improper fiaction to a whole or mixed number.
IT 41. It is evident that every improper fraction must
contain one or more whole ones, or integers.
1. How many whole aisles are there in 4 halves (f) of
^ apple? inf?-: — Uif.^ y? ? in ^
in V? in »f«? in»f*?
— in I of a yard ?
in V? in\^
2. How many yards in f of a yard ?
— -inf? inf» in^? —
^ — in ^r I in 9^ ? in V ?
3. How many bushels in 8 pecks? that is, in f of a bush*
el f in V> ? — in V ? in V* *? in \* ?
int^o? inV?
This finding how many integers, or whole things, are
contained in any improper fraction is called reducing an
improper fraction to a wMe or mixed number.
S!
'<
y
SdT-^ri^-.
1141,42.
FRACTIONS.
101
4. If I give 27 children j- of an orange each, how many
oranges will it take ? It will take ^/ ; and it is evident,
OPERATION. that dividing the numerator 27, (=: the
4)27 number of parts contained in the frac-
tion,) by the denominator 4, (= the
Ans. OJ oranges, number of parts in 1 orange,) will give
the number of whole oranges.
Hence, To reduce an improper fraction to a whole or mixed
number, — Rule : Divide the numerator by the dei|ominar
tor; the quotient will be the whole or mixed nnmber.
EXAMPLES FOR PRACTICE.
5. A man, spending ^ of a pound a day, in 83 days would
spend ^ of a pound ; how many pounds would that be ?
Ans. l^£.
6. In *|^^ of an hour, how many whole hours?
TheGOthpartof an hour is a minute; therefore the ques-
tion is evidently the same as if it had been, in 1417 min-
utes, how many hours? Ans. 23g^ hours.
7. In ^{^^ of a shilling, how many units or shillings.^ •
Ans. 730^ shillings.
8. Reduce *m® to a whole (w mixed number.
1). Reduce f^, '^,Uh iiU, \W, to whole or mixed
numbers.
To redtkcc a whole or mixed number to an improper fraction.
IT 4:^8. We have seen, that an improper fraction may be
changed to a whole or mixed number ; and it is evident
that by reversing the operation, a wholc^ or mixed number
may be changed to the form of an improper fraction.
1. In 2 whole apples, how many halves of an apple ? Ans. 4
halves ; that is f . In 3 apples how many halves ? in 4
apples? in 6 apples? in 10 apples? in 34? in 60? iu
170? in 492?
2. Reduce 2 yards to thirds. Ans. f , Reduce 2f yards
to thirds. Ans. f . Reduce 3 yards to thirds -
3§ yards. 5 yards. 5f yards.-
Reduce 2 bushels to fourths. — ^ — 2# bu. — G bush-
3
els.
4
-3^ yards.
-G^
yards.
•6^ bushels. —
-7f bushels. 25f bushels.
In IGy^ pounds, how many y'^ of a pound ?
make 1 pound : if therefore, we multiply 16 by 12,
that is, multiply the whole number by the deaominatcr, the
J.2
12
lOS
PRACTIONt.
IT 42, 43
product will be the number of ISths in IQ£ : 16X 12=1U3
and this, increased by the numerator of the fVaction, (5,)
evidently gives the whole number of 12ths ; that is, t^ut'
a pound, Ans.
OPBRATION.
16 Vb pounds
12
192=12th8 in 16 pounds, or the whole number.
5aal2th8 contained in the fraction.
\97=z\^., the answer. <
Hence,^ Te reduce a mied number to an improper frac-
tion, — Rule : Multiply the whole number by the denomin-
ator of the fraction, to the product add the numerator, and
write the result over the denominator.
i E^XAMPLES FOR PRACTICE.
5. What is the improper fraction equivalent to 23^^
hours? ilni. »|^7 of an honr.
, 6. Reduce 730^^ shillings to 12ths.
As ^ of a shilling is equal to 1 penny, the question id
evidently the same as, in 730s. 3d., how many pence ?
Ans. ^ff ^ of a shillintr ; that is 8763 pence.
7. Reduce l^f , 17|f , 8/^, 41^,^^^, and K^ to impro-
per fractions. ' ^
8. In 1562j^|^ days, how many 24ths of a day ?
Ans. 3Jfi =3701 hours.
9. In 342f gallons, how many 4ths of a gallon ?
Ans. ^^^ o{ a gallon==1371 quarts.
To reduce a fraction to its lowest or most simple terms.
^ 4«l. The numerator and the denominator, taken to-
gether, are called the terms of the fraction. . % .
If ^ of an apple be divided into 2 equal parts, it becomes j .
The effect on the fraction is evidently the same as if we had
multiplied both of its terms by 2. In either case, theparta
are made two times as many as they were before ; but thy
are only half as large ; for it will take 2 tiflies as many
fourths to make a whole one as it will take halves ; and
hence it is that f is the same in value or quantity as 4.
1^ is 2 parts-; and if each of these parts be again divided
into 2 equal parts, that is, if both terms of the fraction be
: t a-'^is-'.^A-iiA'- f ^i-^
1143.
nUCTlONB.
103
multiplied by 3, it becomes | . Hence, Jf=faB=|, and the
reverse of this is evidently true, that |=>|=^.
It follows therefore, by n^ultiplying or dividing both terms
of the fraction by the same number , toe change its terms
without alttringits value.
Thus, if we reverse the above operation, and divide both
terms of the fraction | by*3, we obtain its equul, ^ ; divid-
\Vi\t again by 2, we obtain ^ , which is the most simple form
of the fraction, because the terms are the least possible by
which the fraction can be expressed.
The process of changing | into its equal y, is called redu-
cinjr the fraction to its lowest terms. It consists in dividing
both terms of the fraction by any number which will divide
them both without a remainder , and the quotient thence aris'
ing in the same manner, and so on, till it appears that no
number greater than I will again divide them.
A number which will divide two or more numbers with-
out a remainder, is called a common divisor, or common meas-
ure of those numbers. The greatest number that will do
this is called the greatest common divisor. • .
J. What part of an acre are 128 rods ?
One rod is t^tt <>** an acre and 128 rods afc -f jJg of an
acre. Let us reduce this fraction to its lowest terms. We
find, by trial, that 4 will exactly measure both 128 and 160
and, dividing, we change the fraction to its equal ^^. Again
we find that 8 is a divisor common to both terms, and, di-
viding, we reduce the fraction to its equaj ^, which is now
in its lowest terms, for no greater number than 1 will again
measure them. The operation may be presented thus :
■• ■' " S) ■
vl28 32 4 ^
*) Too = 40=^5 ''^ ^" ^'''y '""''•
2. Reduce 1^^, /^V, j|g, and j^fff to theii^ lowest terms.
^«3. h i, }, ii"d l
Note. If any number ends with a cypher, it is evidently
divisible by 10. If the two right hand figures are divisible
by 4, the whole number is also. If it ends with an even
number, it is divisible by 2 ; if with a 5 or 0, it is divisible
by 5. •
3. Reduce |^^, ^^^, ^f ^, and §^ to their lowest terms.
hi
104
PftACTlONS.
!I44.
IT 4-i • Any fraction may evidently be reduced to its low-
est terms by a single division, if we use the greatest common
divisor of the two terms. ^ The greatest common measure of
any two numbers may be found by a sort of trial easily made.
Let the numbers be the two terms of the fracticii ^ff. The
common divisor cannot exceed the less number, for it must
measure it. We will try, therefore, if the less number, 126,
which measures itself, will-also measure or divide 160.
128)160(1
128
'■'flVr'^
Ci*
128 in 160 goes 1 time, and 32 re-
main; 128, therefore, is not a divisor of
— 160. We will now try whether this re-
32) 128(4 mainder be not the divisor sought ; for if
128 32 be a divisor of 128, the former divi-
sor, it must also be a divisor of 160, which consists of 128
-|^32. 32 in 129 goes 4 times, without any remainder.
Consequently, 32 is a divisor of 128 and 160. And it is
evidently the greatest common divisor of these numbers;
for it must be contained at least once more in 160 than in
J 28, and no number greater than their difference, that is,
greater than 32, can do it. > '• -^^ v .^;#
Hence, the rule for finding the greatest common divisor of
tido numbers^ — Divide the greater number by the less, and
that divisor by the remainder, and so on, always dividing
the last divisor by the last remainder, till nothing remain.
The last divisor will be the greatest common divisor required.
Note. It is evident, that,^when we would find the great-
est common divisor of more than two numbers, we may first
find the greatest common divisor of two numbers, and then
of that common divisor and one of the other numbers, and
so on to the last number. Then will the greatest common
divisor last found be the answer.
4. Find the greatest common divisor of the terms of the
fraction f |, and, by it, reduce the .fraction to its lowest terms.
OPERATION.
^^ 21)35(1 ^ . ^
14)21(1
14
Greatest divis. 7)14(2.
14
Then, 7)H=|in5.
-^.i;,
f|44. ■ fl44,46.
PKACTIOIffl.
lOS
r4.
/*'
••».J'
4>w. ^.
. Ans. J
6. Reduce ^ to its lowest terms. *^^»^'^*^^« :4„5. j^.
iVb^e. Let these examples be wrought by both methoas;
bj several divisprs^ and also by finding the greatest common
divisor. ' ' V ■ V I s-
6. Reduce -^^ to its lowest terms.,
7. Reduce ||f to its lowest ^rms. !
8. Reduce ^^ to its loweil terms. '
9. Reduce j^ff | to its lowest terms.
Tq divide afrtution by a vfkoh number. />^^ •.
5| 45. 1. If 2 yards of cloth cost f of & pounds what
does 1 yard cost? how much is :§ divided by 2i ' 1'^'^''^
2. If a cow consume f of a bushel of meal in 3 days^
1)0W much is that per day .' f H- 3= how much ?
3. ff a boy divide | of an braqge among 2 boys, how
much ^ill^he give each one ? f-?- Se^how mueh /
4. A boy iKrtight 5t cakes for f} of a shilling; ivhat did
I cake cost? •fjJ-J-Srshow much? -^ ■ ,..;?' ij *
5. If 2 bushels of apples cost -^ of apouhd, what latitat
per bushel ?
1 bushel is the half of 2 bushels; the half ^ is ^, »>
^^mff «;»;«'.,H:.-^mn«^-l»,;. ^.*■UH* -mi ^ b.v. ■ j^^ -ji^ pouiid.
6. If 3 horses consume |f of a ton of hay in a months
vAiti will 1 horse consume in tl^^aan^e tJm^}
•ff are 13 parts ; if 3 horses consume 12 such parts hi a
month, as many times as 3 are contained in 12, so many
parts 1 horse wUl consume. '^- ''« *« r ^ Ans. -^ of ato»n.
7. If f § of a barrel of flour be divided equdly among 5
families, how much will eaph family receive ?
|| is 25 parts ; 5 into 2£»^goes 5 times. Ans. ^ of a barrel
The process in the foregoing examples is evidently di<
viding a' fraction by a whole number; and consists, as may
be seen, in dividing the numerator ^ (when it can be done
without' a remainder,) and under the quotient writing the
denominator. But it not unfirequently happens,^ that the
numerator will not contain the whole number without a
remainder. • /'■''•' '•"'■ '^j--^ i--> ■ t _ ^ r'^H
8. A maq divided ^of a pound equally amppg 2 persons;
what part of a pound did he give to each '} fr ',
^ of a pound divided into 2 equal parts Vjl be 4tNi
Ans. He gave I of a pound to each.
'''■'^'''^T'
106
FRACTIONS.
tl 45, 46. 1 fl 46, 47
9. A mother divided ^ a pie among 4 children ; what part
of the pie ''id she give to each ? ^ -r 4 = how much ?
. id. A boy divided ^ of an orange equally among 3 of his
companions ; what was- each one's share ? ^ -7- 3 = how
much?
11. A man divided f of an apple equally between 2 chil.
dren ; what part did he gi||e to each ? f -7- by 2 = what
part of a whole one ? > . . ..
f is 3 parts : if each of these piu'ts he divided into 2 equal
parts, they will make 6 parts. He may now give 3 parts to
one, and 3 to the other : but 4ths divided into 8 equal parts
become 8ths. The parts are now twice so mantff but they
are only half so large; consequently, § is only half so much
as f . Ans. f of an apple.
In these last exainples, the fraction has beqn divided by
'multiplying the denominator, without changing the numera*
tor. The reason is obvious; for, by multiplying ]the denom-
iinator by any number, the parts ate made so many tinies I
4nia//«r, since it will take so many mpte of them to m;^a
whole one ; and if no more of these smaller parts be taken |
than were before taken of the larger, that is, if the aumer-
ator be not changed, the value of the fraction is evidently |
made so many times less. c ,..;.,;..% a. ,.:>•,( t?
If 4L6. Hence, we have two ways to divide a fraction \
hy a whole number.
I. J^ivide the numerator by the whole number, (if it will I
contain it widjout a remainder,) and under Ui« quotient |
"write the denomiaator. Otherwise, , ,, ^ .,. - -u
II. Multiply the denominator by the whole number, and|
over the product write the numeratcu*. , , ,. ; ^
..^ EXAMPLES FOR PRACTICE,
> ;, 1. If 7 pounds of tobacco cost f^ of a pound, what is |
that per pound ? f|-f-7=how much 1 Ans. ^o(vi lb.
,,jt 2* If ^ of an acre produce 24 bushels, what part of ao|
acre will produce 1 bushel ? ^$-7-24=:how much ?
3. If 12 yards of silk cost ^^ of a pound, what is that a |
yard? |f-4-I2=howmuch?
,..,4» Divide f by 16.
Note. When the divisor is a composite number, the in*
telligent pupil will perceive, that he can first divide by on«|
component part, and the quotient thence arising by the oth-
■ ■ ..^i.
U 45, 46. ■ fl 46, 47.
FRACTIONS.
107
^»
er; thus he may frequently shorten the operation. In the
last example, 16=8X2 and |-;-8=s^, and ^-r2— i ^. .
Ans, 1^.
5. Divide ^^ by 12. Divide ^ by 21. Divide f |>y 24.
6. If 6 bushels of wheat cost £1% what is it per bushel ?
Note. The mixed number may evidently be reduced to an
improper fraction, and divided as before.
Ans. it=2T ^^ ^ pound, expressing the fraction in its
lowest terms. (H 43.)
7. Divide ^4|4 by 9. Cluot. ^^^ of a pound,
a Divide 12f by 5. Q«o^ V^==2f.
9. Divide 14f by 8. Qmo^ If^.
10. Divide 184^ by 7. Quot. 26^.
Ab^c. When the mixed number is large, it will be most
convenient, first to divide the whole number, and then re-
duce the remainder to an improper fraction ; and, after di-
viding, annex the quotient of the fraction to the quotient of
the whole number ; thus, in the last example, dividing 184^
by 7, as in whole numbers, we obtain 26 integers, with 2^
[ s=^ remainder, which divided by 7, gives -^ and 26-|-f^
=26/j,i4Ms.
11. Divide 2786^ by 6. Ans. 464f .
12. How many times is 24 contained in 7646^^ ?
Ans. 318ff^.
13. How many times is 3 contained in 462^ 1
Ans. 154^.
To multiply a fraction by a whole number.
t[ 47. 1. If 1 yard of cloth cost ^ of a pound, what will
1 2 yards cost? -}X2=:how muctt?
2. If a cow consume ^ of a bushel of meal in 1 day, how
I much will she consume in 3 days .^ ^X3=:how much?
3. A boy bought 5 cakes, at f of a shilling each; what
: did he give for the whole ? ^^X5=how much?
4. How much is 2 times ^ ? ■ > 3 times ^ ? _ 2
times f?
5. Multiply ^ by 3.
tby2.
iby7.
6. If a man spend f of a shilling per day, how much will
he spend in 7 days ?
f is 3 parts, u he spend 3 such parts in 1 day, he will
[evidently • spend 7 times 3, that is, \* = 2| in 7 days.
^. "^
108
fl&ACTIONS.
U 47, 4g,
Hence, we perceive, a fraction is multiplied hy multipli/ing
the numerator, without changing the denominator.
But it has been made evident, (tl 46,) that multiplying
the denominator produces the same effect on the value of the
fraction, as dividing the numerator : hence, also, dividing
the denominator virill produce the samd etTect on the value of
the fraction, as m.uttiplying the numerator. In all cases,
therefore, vrhere one of the terms of (he fraction is to be
multiplied tH^ same result will be effected by dividing the
other ; and where one term is to be divided, the same result
may be effected by multiplying the other.
This principle, borne distinctly in mind, will frequently
etiable the pupil to shorten the operations of fractions. Thus,
in the following example : ,
At ^ of a pbund, for 1 pound of sugar, what will 11
pounds cost ?
Multiplying the numerator by 11, we obtain for the pro*
duct f f =i of a pound for the answer.
tf 4^. But by applying the above principle, and dlvid'
ing the denominator, instead of multiplying the numerator we
at once come to an answer, f in much lower terms, ll^nce,
there are two ways to multiply a fraction by a whole number,
I. Divide the denominator by the whole number, (when
it can be done without a remainder,) and over the quotient
write the numei'atof'. Otherwise,
II. Multiply the numerator by the whole number, and an*
der the product write the. denominator. If then it be an
improper fraction, it may be reduced to a whole or mixeji
li umber.
EXAMPLES FOft PAACTlCE.
1. tf 6ne man consume ^of a barrel'of flour in a months
how much will 18 men consume in the same time 1 — — 6
ftien ? — 9 men 1 Ans, to the last, 1^ barrels.
2. What is the product of ^ multiplied by 40? ^^X
40=equal how much ? Ans. 23f .
3. Multiply T^^T by 10. by 20. ■ by 18. -— by
36. by 48. — by 60.
^ote. When the multipllei' is a composite number, ik
pupil will recollect (IT 11)« that he may multiply first by
one component part, and tjiat product by the other. Thtts^
U 47, 48. I ^ ^^ 49.
FHACTIONI.
109
in the last example, the multiplier 60 is equal to 12X5;
therefore, TVVXl2=if, and jfX5= fj«»A , Ans.
4. Multiply 5i by 7. Ans. 40^.
Note. It is evident that the mixed number may be re>-
duced to an improper fraction, and multi[died, as in the pre-
ceding examples; 'but the operation will usually be shorter,
to miutifriy the fraction and whol^ number separately^ and
add the results together. Thus, in the last example, 7 times
5 are 85; and 7 times f are ^ss^^ which added to 35,
make 40^^ Ans.
Or, we may multiply the fraction first, and, writing down
the fraction^ reserve the integers, to be carried to the pro*
duct of the whole number.
5. What will 9^ tons of hay come to at ^ per ton ?
Ans. 2a£ 198.
6. If a man travel 2 ^ miles in one hour, how far will he
travel in 5 hours ? in 8 hours ? — — in 12 hours ?
■ in 3 days, suppose he travel 12 hours each day ?
Ans. to the last, 77f miles.
Note. The fraction is here reduced to its lowest iermsy
the same will be done in all the following examples.
To multiply a whole number by a fraction.
tl 49. 1. If 36 pounds be paid for apiece of cloth, what
cost f of it.' 36Xf^how much?
f of the quantity will cost f of the price; | of a time 36
pounds, that is, f of 36 pounds, implies that 36 be first di-
vided into 4 equal parts, and then that one of these parts be
taken 3 times ; 4 into 36 goes 9 times, and*^3 times 9 is 27.
Ans. 27 pounds.
From the above example it plainly appears that the object
in multiplying by a fraction, whatever may be the multipli'
cand, is to take of the multiplicand a part, d^oted by the
multiplying fraction] and that this operation iff composed of
two others^ viz. a division by the denominator of the maIti*-~
plying fraction, and a multiplication of the quotient by the
numerator. It is a matter of indifference, as it respects the
result, which of these operations precedes the other, for 36
X3-^4=37, the same as 36-M X 3=27.
Hence, — To multiply by a fraction, whether the multi-
plicand be a whole number or a jTrcicfion,— Rule :
Dividt the multiplicand by the denominator of the multi-
N
m
FRACTIONS.
V V
N ^
use.
fAy'mg frtotion, andjpiukiply the quotient by the numerator;
or, (which will oftenbe found more convenient in practice,)
first muitiply by the numerator, and diyide tl)e;prp4uc( by
.thjC denominator. .<,^ ',..r.?F* ,..,■! i- ».:;^ ^i.ti.. -. ■» .,^.j'
Muitiplipatioo, therefore, when implied to fractions, does
■ot always imply aogmeiitatMm, or incr-ease, as in whole
Viumbers; for, v^aesi the muh^>Iier is less than vnitfft it will
Always require the |M;oduct to be less than the multiplicaa4i
to'which it would 'he oi^yequaljif the multiplier wei« ).
We have seen, (IJIO,) that, when two numbers are mul>
Aiplied together, either c^ them maiy be made the multiplier
without i^e<^ii|g the iresult. In the last example, therefore,
Instead of multiplying 16 by ^ we m%y milMipIy ^ iby 161,
(^ 47,) and the result w,iU be the same, ^^t i j <>
^%1iMfi.9» FOR PRApTlOE.
% What will 40 bmweis of meal come to at | of a poun4
per^bMcel? 4(lX|*^PW mficht
9. What will Ayaxde of cloth cost at f of a pound per
jvti'i ^X|7=h0wmuQh<?
4. How fiiuoh is^^^idO ? --^-H- § of 360 ? ^ — ^ of 45/
6. Multiply ^S^byy;,. Mi^ltiply 30 by i.
(To tuuH^^ > cWe ^ fraction by lOnoiAen. i -> . '
IT «1^0. 1. A man owning ]|- of a farm, sold ^ of his share;
-what part 'of the H^hote »f^m lUd he 'Sell t i of ^ is how
muchl
We have just seen, (^ 40,') that to tmultipjl^y byajftaictian,
is to divide the muUij^Uediitd^hy ihe^enaii^iHator, and to mul'
t^ly the quotient by the numerator. ^ divided by 3, the de-
nominator of the mult^yui^ £raction, (YT 46,) is ;^, which,
multiplied by 2, the numerator, (^ 4^,) is :^, Ans.
The process, if carefiiUy considered; will befound to con-
sist in multiplying togMhefthettpomumertdw^.fer a new m'
meratoTy nndike two denwniilteiiorsjor a new dtnomindtitr.
BXAMPL£8 FOR PRACTieE.
% A man, having f of a pound, gave ^V of it for a din-
ner what did the dinner cost him? Ans. ^ pound.
3. Multiply i by f Multiply i^ by f FrodMct.^y
4. How much is f of § of |- of |^? un* v« •, > v^t.j, .
Note. Fracticms like the above, connected by the word
H 5». ■ ^ 50, 6i;
FRACTIONS*
111
of^ are sometimes ealled compound ftnctitms. The word of
implies their continual multiplietUion into each other.
When there are several fractions to be imritiplied eonthi-
ually together, as the severai numefators tae factors of the
ticw ntnneratov, and the stverai (knominator^ wt^ factors of
the new deacnHinator, the operation ma^ be shortened k^
droppiiig those factors which are the same in hath terms j on
the principle eiqdained in f[ 4^ Thus, in the \dst exMnpIe,
I, f , {, ^f we And a 4 and a 3 both among the numerators
afed uiKHig theidteiloRiinfrtors ; tJi^Peibre we drop them multi*
plying tog)etlM» oiAf the vemainuiif tivtmemUnt^f 9XVsrl4,
for ft BOW nuBMrator, and theremaininf d«Donmnter», SXo
3540, for a new denemhiater, making jfsE^, Ans, ae before.
5. I of I of f of I of ^^ of ^of ^howmuch? Am. ^.
6. WlMt is the ooMiiioal prt^ec of T, f , f of | and 3| ?
Nine. The mt^r 7 may be reduoed to the form of an
impropeip Jhietion, ij writing a nnit nnder it ibr a den<«ii-
nator, thus, |. Ans.H^.
7. At ^ of a pound a yard, what wiR ( of a j^ard of
8. At If pounds per barrel for floor, what wilt -^ of a
barrel eostf
lj2sV then VX^=^!M' ^^' ' •' '
d. At f of a pound, per yard, what eeet 7| yards? - ^
'^ •■'"■'•■ • . Ans. \
10. At i9| per yard, what ooit d| yardis.' Ana. f ll
11. What li the continued prodnet of 0, |« ^ of j, ^,
and U of f of I f -4n». Mf .
5T ol« The Rous for the ninltiplieation of frat^tlens
may now be presented at one view : ''' "^ " •
I. 7*0 mk^iph ttf inaction by m lekole numher, at a whole
numhet fry a fiiutum.'^Divide the denonnnator by fh^
whole number, when it can be done without a remainder ;
otherwise, multiply the numerator by it, and under the pr»-
duet write the denominator, which may then be reduced to
a whde or mixed number.
II. To muU^ly it mixed ntcmfrer frj^ a whok xtimfrer;—
Multiply the fraction and integers, separately, and add their
products together.
\Wk Tq multiply one fraction by another^ — ^Multiply to*
119
PRACTI0N8.
U Hh «2.
ftther the numeraiors for a new numerator, and the deno-
minators for a new denominatcM'.
Note. If either or both are mixed numbers^ they may first
be reduced to improper fractions.
EXAMPLES FOR PRACTICE.
1. At fj( per yard, what cost 4 yafds of cIotH ? 5
yds? 6 yds? . u 8 yds? -r— «0 yds?
2. Multiply 148 by ^ -, by |
Product, f
Product, i&i.
Product, 2|.
Product, I.
Product 3.
i4n5. to the last, 15^.
^byA bytlr.
Ztost product, 44 •^.
3. If 2 1^ tons of hay keep 1 hoyse through the winter,
how much will it take to keep 3 horses the same time ?
7 horses? .13 horses ? Ans. tq the last, 37-^ tons.
4. What will 8^ barrels of cider come ^^ ai 7 shillings
per barrel? ,m,? ^ii^i^.r.'^ 'ut ] 'i , ^/-'iM >(,-i' } .
5. At 14|;f per cwt. what will be the cost of 147 cwt ?
0. A owned f of a note ; B owned -^ of the same ,* the
note amounted to.lOQQ^; what was each 9n9> share of the
mcNiey f
7. Multiply ^ off by ^ off,. ; ^ ni:
8. Multiply 7i by 2^.
9. Multiply ^ by 2J. i ;,. hi vIh
10. Multiply! of 6 by f
11. MuKiply J of 2 by ^ of 4. . m ,.
12. Multiply continually together ^ of 6, f of 7, | of 9.
and I of 10. Product,^.
13. Multiply 1000000 by |. Product, 55555d|.
7*0 divide awkah number hy a fraction.
^ «I3. We have already shown (5T 46,) how to ^xt\6fi a
fraction by a whole number ', we now proceed to shpyr how
to divide a whole number- by a fraction.
1. A man divided Q£ among some poor people, giying
them I of a pound each ; how many were the pejrs(»\s who
received the money ? 9~£g=4iow many ?vS--*.>4^ }«»??. ^4.
1 pound is 1^, and 9 poi^nds is 9 times as many, that is,
^/ ; then f is contained in ^ as many times as 3 is con-
tained in 3G. Ans. 12 persons,
That is, — Multiply the dividend by the denominator of the
dividing fraction, (thereby reducing the dividel;^l to parts
of the same magnitude as the divisor) and divide ^Ae pro\
duct by the numerator. ,1. m --.s.*! . ,
tF52,63.
FRAvrONP..
m
Il3i
2. How many time« 19 f contained in $ 7r i-H^^^w
.;. V OEBRATION. _.. . ■ y,) 1; s,f,i . jU .1;
,f .,,,;;■.. . .8 Dividend. - '.- :..,r,!U> n^rft ;; ii .i
6 Denominator ati i-i,' itoi' i^
•wp« S'li./i 'i > »5f<:jit..ii J lit ii
V'
(V-
IfflftJJl' Y«
nr.
Numeifatar, 3 ) 4j0 vrr^^J^-
A-
^.'l»f.
.^!
Quotient, 13^ times the answer. < - '1> ••
To mnlti|>ly by a fraction, wie 1)av$ Sjsen, {^ 4d>) implies
two operations— a division and a multiplicaiim i 90 {^fK>, to
divide by a fraction inipiiestwo opera|ion4-a ffiM/%A'ca<toit
md a diviiion, ...
(if
51 63. Division is the reverse of multiplication.
»n .r.
To multiply hj a, i^action^
whether tlie multii^icand be
a whole number or a fraction
ad has already been 9hown,
(fl 49,) we divide by the de-
nominator of the muHyplying
fraction, dndmuitiply th€ quo-
To divide by a fraction,,
whether the dividend be a
whole number or ^ fraction ^
we mvUi^fy by the denomir
nator of the dividing, fractiofi
and divide the product by the
numerator* , v >/^ .> *.■ i'
tient by the numerator.
Note, In either case, it is matter of indiflierence, as it
respects the result, which of these operations precedes the
other ; but in practice it will frequently be more conveni>
.ent, that the multipdiplication precede the division.
12 multiplied by |, Uie pro-
duct is 0.
In multiplication,, the mul-
tiplier being ies5 than unity,
or 1, will require the product
to be less than the multipli
cand, (51 49,) to ,^hich it is
only equal when the multipli-
er is 1, and greater when the
multiplier is more than 1.
12dividle(iby f, the quo-
tient is 16.
In division, the divisor be-
inipless than unity, or 1, will
be contained a greater num-
ber of times; consequently
will require the quotient to be
greater than the dividend, to
which it will be equal when
the divisor is 1, and less when
the divisor is more than 1.
EXAMPLES -FOR PRACTICE. ,
1. How many times is ^ contained in 7 1 7-j-^=How
many? • ,• ,:i. . . .^^^ , ■ '■ ■ \ ^:..-.
K2
114
FRACnONf.
1F«3,54.
2. How many timeB eui I draw |^ of a gallon of wine
out of a cask containing 26 gallons ?
3. Divide 3 by f 8 by f . 10 by ^
4. If a man drink -^ of a quairt of rum a day, how long
will 3 gallons last him I / ^>.*i»v* ,*
5. If 2f bushels of oats sow an aere, how many acres
will 22 bushels sow? 22-r-2f=:how many times? v
Note, Reduce the mixed number to an improper frac*
tion, 2|=y. .Vi.v.ii:! '■>"* '''^"li/ t^'t v^ Ans. 8 acres.
6. At IpS a yard, how many yards of cloth may be
bought for S7£ ? Ans. 26f yards.
7. How many times -^/^ contained in 84 ?
Ans. 90-^ time».
8. How many times is ^ contained in 6 ?
Ans. ^ of I time.
9 How many timeftia 8f contained in 53 ?
Ans. 6}^ time;^.
10. At f of a pound for building 1 rod of stone wall,
how many Fodi may be built for 67 jf ? 87-rfeBhow ma-^
ny times? " ••
Tq divide one Jr action hy another .
^54. 1. At § of a pound per parrel^ how much rye n^iay
be bought for f of a pound X f is contained in ^ how ma*
ny times? -'^
Had the rye been 2 whcik pounds per barrel, instead of
f of a pound, it is evident, that f of a pound must have been
divided by 2, and the quotient wouki have been ^ ; but
the divisor is 3ds, and 3ds will b» contained 3 times where
a like number of whole ones are contained 1 time ; conse-
quently the quotient ^ is 3 times too smally and must
therefore in order to give tlie true answer, be multiplied by
3, that is, by the denominator of the divisor ; 3 times ^%=
-^fl- barrel, answer.
The process is that already described/ ff 52 and 53. If
oarefully considered;, it will be perceived,' that the numerator
of the divisor is multiplied into the denominator of the di-
viidcnd, and the denominator of the divisor into the numer^
ator of 1}he dividend ; wherefore in practice, it will be more
convenient to invert the divisor ; thus, §^inverted becomes f;
(hen multiply together the two upper terms for a numerator
uifKd tM ttp(k lower terms for a denominator ^^hs in the mul.ti»
\U-:J:.
^»M. ■ ^54,55/
PItACTlONff.
115
e : conse-
plication of one firttetion by wiother. Thus, in the %bofe
example, 3X8 9 ^ -ku .-u.um .. . s - . w ..
...;.*. - .-=:—, M before. '' •':'-•<■.•>•■;■■'-•>'-;
'" *'* KXAMPLCS FOR PRACTICE. ' ' '
2. At I of a pound per bushel for wheat, how many
bushels may be bought for | of a pound? How many
times is ^ contained in 1 1 Ans. 3^ bushels.
3. If I of a yard of cloth cost j^ of a pound, what is that
per yard ? It will be recollected (tf 24) that when the cost
of any quantity is given to find thepme of a unit, we diftrtde
the cost by the quantity. Thus, f (the cost) divided by {
(the quantity) will give the price of 1 yard.
Ans. f ^ of a pound per yard.
Proof. If the work be right, (^ 16, " Proof,") the pro-
duct of the quotient into the divisor will be equal to the
dividend ; thus, MXi=i. This, it will be perceived, is
multiplying the price of one yard (ff) by the quantity (l).
to find the cost (^ ;) and is, in fact, reversing the question ;
thus, if the price of one yard be f | of a pound, what will |
ol a yard cost ? Ans. f of a pound.
Note. Let the pupil be required to reverse and prove
the succeeding examples in the same manner.
4. How many bushels of wheat at ^ of a pound per
bushel, may be bought for f of a pound ? Ans. 4f bushels.
5. If 4^ pounds of butter serve a family 1 week, how
many weeks will 36^ pounds serve them ?
The mixed numbers, it ^dl be recollected, may be re-
duced to improper fractions. Ans. Syf ^ weeks.
6. Divide ^ by i, Quot. 1 Divide ^ by ^ Quot. 2.
7. Divide | by }, Quot. 3 Divide | by ^^ Quot ff.
8 Divide 2^ by 1^, Quot. If Divide lOf by 2^
Quot. 4if .
9. How many times is ^ contained in f ? Ans. 4 times.
10. How many times is ^ contained (n 4^ ?
^ns. 11 f times.
11 Divide f off by J off Quot. 4.
«
t[ titl. The RuLEyor division of fractions may now be
presented at one view : —
I. To divide a fraction hif a whole number ^-r-DivideHlaSi
1W
FRACTIONS.
1f«6,W
numerator by the whole number, wh«n it otn be done with-
out ti remainder, and under the quotient write the denoinin^
tor ; otherwise, multiply the d^oHfinaKar by it, and qver the
product write the numerator. , , *
II. To divide a yikole number hy afractioHf — Multiply
the dividend by the denominator of th§ fraction, and divide
the product by the numerator,
III. To divide one fraction by another, ^-Invert the divisor
and multiply together the two upper terms for a numerator^
and the twdlower^rms for ^ denominator.
Note. If either or both are inixed numbers, they may be
reduced to improper fractions.
EXAMPLES FOB FHACTICE.
1. If 7 tb of tobacco cost -^ of a pound, what is it per
pound ? ^^-7-7=how much '^ | of -^ is how mudi t
2. At|A for I of a parrel of cider, what is that per bar-
rel?
3. If 4 pounds of sugar cost ^ofi pound, what does 1
pound cost? , ^.;[ ,-
4. If I of a yard cost 13s. what is the price per yard?
^ If 14 1 yards cost 43<£, what is the price per yard ?
Ans, 2Hi.
6. At 4^ pounds for 10^ bafrels of cider,, what is that
per barrel ? Ans. ^£.
7. How many times is f contained in 746 T Ans. 1969^.
8. Divide j^ off by ^.
Quot. |.
9. Divide i of ^ by f off.
10. Divide ^ of 4 by ^.
11. Divide 4f by f of 4.
12. Divide ^ of 4 by 4^.
Divide I by ^ off.
Quot. ^^.
Quot. ^.
Quot. 3.
» ' Quotil^.
<i Quot.^.
«n i»
ADDITION AND SUBTRACTION OF FRACTIONS.
fI56. 1. A boy gave to one of his companions | of an
orange, to another f , to another ^; what part of on orange
did he give to all ? f-|.|-f^=:how much ? Ans. }.
2. A cow consumes in one month i^ of a ton of hay ; a
horse, in the same time, consumes ^ of a ton ; and a pair
of oxen t\ ; how much do they all consume ? how much
more d.Q^s the horse consume than the cow I rr-^ the oxen
1iaa,67,
fBACTiOVf.
UZ
A=show maehf -^ — ^^^skow
niuoh t , - ,' Mr^-
^=bowmach? |-r4«vhow much ? ' ''
"h ^ 'f ^> -hrf^=i6^w . nucb I II — f^— ^»
6. A boy htring | of an tpple, giT6 ^ of it to hit tiater ;
what part of the apple had he left ? f--^how much t
When ike denominators of two or more firactiona are
alike f (aa in the foregoing examples) they are said to hare
a common denominator. The parts are then in the same
denomination, and, consequently, of the same magnitude or
value. It is etldent, thereibre, that they may be added or
subtracted, by adding or subtracting their uumeraion, that
is, the number of then: parts, oare Uiing taken to write un-
der the' result their piopf r denominatof , ThM, ^^ff^^sz
TT» 8 l"*i' *'
6. A boy having an orange, gave f of it to hls^iste^
and ^ to his brother ; what part ^ the orange did he gire
away f
4ths and 8ths being parts of diferent Magnitudes, or
value, cannot be add^ together. We must therefore firsi
reduce them to parts of the same magnitude, that 'is, to a
common denominator, f are three parts. If each of these
parts be divided into 2 equal parts, that is, if we multiply
both terms of tl^ fr-ac^ion ^ by 2, (tl 43) it will be changed
to f ; then f and ^ are |. Ans. |^ of an <Hrange.
7, A man had § of a hogshead of molasses in one cask,
andf of a hogshead in another; how much more in one
cask than in the ott\er } • ^ • V4 < •
Here, 3ds cannot be so divided as to become 5ths, nor
can 5ths be so divided aa to become Sds ; but if ^e 3ds be
each divided into 5 equal parts, and the 5ths ^ach into 3
equal parts, they will aU biecome 15ths. The f will be-
come i%. and the f will become -^ ; then ^ taken from |^
leaves -j^ A^s» • :...-jt,ui «; ^K>; ^h iU-' ^'U?
•«iI7*
I -mi vf.I
U Si7. From the very process of dividing each of the
parts, that is, of increasing the denominators by multiplying
them, it (bllows that ettch denominator must be k factor of
the common denominator ; now, multiplying all the denomii
payors together will evidently produce such, a numb^r^ >;
118
MACTIONt.
1167.
' Henee,-^To ^eihtte fractions ofdiftnnt demnmnnt^rs tn
equivalent fractions, having a common denominator ^-^KvLf: :
Multiply togetbeir all the denomfaiators for a common deno-
minator ; and as by this process each denominator is mul-
tij^lied by all the others, so, to retain the value of each flrae-
tion, maltiply each numerator by all the denominators, ex-
cept its own, fer a new numerator^ and under it, write the
common denominator.
m,':
fiXAMPLBt FOR PRAC^IClft.
1. Reduce ^, | and ^ to fraetioii» of equal value, having
ft common dCTiominator.
3X4X5£=:60, the common denomittator. <' '- ' «
2x4X^=^» tJie new numerator for the first fVaetiom.
3X3X^=45, the new numerator for the second fraction.
3X4x4aBs48, the new numerator for the third fraction.
The new fractions, therefore, are |^, |4, and ff . By an
inspection of the operation, the pupil will perceive tliaf the
numerator and denominator of eaeh fVaction have been mul-
tiplied by the same numbers : consequently, (51 43) that
their value has not been altered.
3. Reduce to equivalent fractions of a common denomi"
nator, and add together ^^ | and|, '
^ni. U-¥U+ U ^ U * ^ Hh amount.
4. Add together f and f. Amount, l^f.
6. What is the amount ofj-j- j -f j-f-^ ? Ans. ft^l^.
6. What are the fractions of a common denominator
equivalent to f and | ? Ans. ^ and ^, or ^ and ^.
We have already se^n (^ 56, ex. 7,) that the common de-
nominator may be any number, of which each given deno-
minator is a factor, that is, Imy humbei which may be divi-
ded by each of them without a remainder. Such a number
is called a common multiple of all its common divisors, and
the least number that will do this is called their leaet com-
mon multiple ; therefore, the least common denominator of any
fractions is the least common multiple of all their denom-
inators. Though the rule idready given will always find a
commim multiple of the given denominators, yet it will not
always find their /eojt common multiple. In the last ex-
ample, 24 is evidently a common multiple of 4 and 6, for it
fl «7, 58.
niACTIONt.
il9
minai&rs tft
rr,— Rule :
Bmon deno-
ator 18 mui-
if each fVae-
linCitors, ex-
it, write the
alue, having
Vaetioii.
ad inaction,
fraotioni'
1 1*. By Ml
leive that the
ive been mul-
(^ 43) that
mon denomi'
will exactly measutie both of them ; but 12 will do the saitie,
and as 12 is the least number that will do this, it i* thft
least oommoB muhi|^ of 4 and 6. It will theref(n<e be
coavevient to have a rule for finding this least common mui*
tiple. Let the nunnbera he 4 and 6.
It is evident that one number is ji multiple of another,
when the former contains all the factors of the latter. The
factors of 4 are 2 and 2 (2x2=4). The factors (^ 6 are
2 and 3, (2X3=sa) consequently, 2x2X3=)b12 contaii^
the factors of 4, that '}», 2X2; and also contains the fae^
tors of 6, that k, 2x3. 12 then, is a common mul-
tiple of 4 and 6, and it is the ktoit commcm multiple,
because it does not contain any factor ^ except those which
make up the numbers 4 and 6 ; nor either of those repeated
more than is necessary to produced and 6. Hence it fol-
lows, that when any two nwnhers have a fai^tor comm^m to
both, it may be once omitted ; thus, 2 is a factor common
both to 4 and 6, and is consequently onoe (xnitted.
U «S8. On this principle is founded the Ksht for J^'
ing the least common multiple of two or more numbers.
Write down the numbers in a line, and divide them by any
number that will measure two or more of them ; and write
the quotients and undivided numbers in a line beneath.
Divide this line as before, and soon, until there are no two
numbers that can be measured by the same divisor ; then
the continual product of all tlie divisors and numbers in the
last line will be the least ccmimon multiple required.
Let us Itpply the rule to find the least common multiple
of 4 and 6.
4 and 6 may both be measured by 2 ; the
2) 4 - 6 quotients are 2 and 3. There is no number
1 greater than 1, which will measure 2 and 3.
2 - 3 Therefor|, 2X2X3=12 is the least commou
multiple' <of 4 and 6.
If the pupil examine the process, he will see that the di-
visor 2 is a factor common to 4 and 6, and that dividing 4
by this factor gives for a quotient its other factor, 2. In the
same manner, dividing 6 gives its other factor, 3. There^
fore the divisor and quotients make up all the fact<»s of the
two numbers, which, multiplied together, must give the
QOBimon multiple.
IM
PAACTIONI.
If 68.
t 7. Reduce f, ^^ f and ^ to eqiiifaleflit firactions of the
leaiC conimoa denominator.
Then^ 3X3X%s:12, least common
denominator. It is erident we need
not mttltipljr by the Is, as this would
not altor the number;
OmAVOK.
f ) 4 . 3 - 8 i^ 6
3)2-1-3-3
2.1-1*1
To find the new numerators, that is^ how many ISths
each fraction is, we may take }, j^, f , and | of 12, thus
|ofl2s=9
ofl2«t«
ofl2-a8
f ofl2-d
New numerators, which, i -^*^
^ of 12«b6 r written bter the common T .^c=s^
}ofl2aa8( denominators, give * J -i4=l
vnm* V ;4 ;« ^n AttS. -fji ^^ -^^ and ■^.
ft Reduce }^ f , and | to fractions having the least com^
mon dentrnitnatoT) and add them together.
Ams. ^ -t A 'f jg — Hg^% amount.
9. Reduce ^ and |- to fractions orUie least common de-
nomiuattHr) uid subtract cme irMn the othe#.
:-K^^mt<\^^ v\*m -Kj *^m_ \*--^^:;Ans. ^ — A*=tV» difference.
10. What is the least number that 3> 5| 8 and 10 will
measure? Ans. 120.
11. There are 3 pieces of cloth, <Mie containing 7| yards,
another 13^ yards, and the other 15^ yards; how many
yards in the 3 piecesi^.v/- h' futvyfu wa s^;.-' > ;. '.•;(.■■.;'.
Before adding;, reduce the fractional parts to their least
con^mon denominator ; this being done, we shall have,
* Ackling together all the 24ths, viz. 18-|-20
7f= 7^1 ) H-21, we obtain 69, that is, ^1=2^. We
13|=1^^ > write down the fraction ^ under the other
15|3sl5|^ ) fractions, and reserve the 2 integers to be
— carried to the amount of the other integers,
Ans. 97^ making in the whde 37^ An84
12. There was a piece of 'doth ooniaihing 34f yards,
from which were taken 12§ yards ; how much was there left ?
We cannot take 16 twenty-fourths
(]j^) from, 9 twenty-fourths, (^) we
must theriefiH'e borrow 1 integer7:24
twenty-fourths, (f^) which, with ^,
makes ^ ; we can now take ^ from
f|, and there will remain ^ ; but as
121=12^1 '
4ns. 21^ yds.
fl 59, 60.
REDUCTION OF FRACTIONS.
121
ions of the
i0t common
»Bt we need
I this would
many IStbs
' 12, thus :
■i T^, and -^.
be least com^
1^, amount,
oommon de-
ji
^i difference.
3 and 10 will
*y Ans. 120.
ling T| yards,
j; how many
to their least
dl have,
vi2. 18+20
L=:2ii. '^^
ier the other
ntegers to be
ther integers,
ig 34f yards,
uras there left?
wenty-fourths
irths, M I]
1 integer5c24
iich, with A,
take ^ from
iin^;buta8
we borrowed, so also we must carry 1 to the 12, which makes
it 13, and 13 from 34 teaves 21. Ans. 21^|.
13. What is the amount of ^ of f of a yard, f of a yard,
and I of 2 yards ?
Note. The compound fraction may be reduced to a sm-
pk fraction ; thus, ^ of f =| ; and I of 2=f ; then, f-|-§
-f-f=Tfo=lT¥(T yds., anstocr.
ff 59. From the foregoing examples we derive the fol-
lowing Rule : — To add or subtract fractions, add or sub-
tract their numerators, when they have a common denomina-
tor ; otherwise, they must first be reduced to a common de-
nominator.
Note. Compound fractions must be reduced to simple
fractions before adding or subtracting.
EXAMPLES FOR PRACTICE.
1. What is the amount of f, 4§ and 12? Ans. 17^|.
2. A man bought a farm, and sold f of ^ of it ; what
part of the farm had he left ? Ans. |.
3. Add together ^, |, I, -^u, } and |^? Amount. 2f|.
4. What is the difierence between 14y\ & ^^liu ' ^ns.
From 1^ take f .
From 3 take ^.
From 147^ take iSi
1 16
Remainder, f
Remainder, 2f
Rem. 981
Retn.
5.
6.
7.
8. From i of j\ take J- of ^j.
9. Add together 112.^, 31 1§, and lOOOf.
10. Add together 14, 11, 4f, ^V and
11. From I take ^. From ^ take f.
What is the difference between ^ and ^ ? -j and I ?
37
h
12.
and
13.
V
and f ? f and f ? f and
3 ?
4
How much is 1—^ ? 1— J ? 1— f
2—4? 2^—1? 3f— J^? 1000-
. 1 ?
1 5
REDUCTION OF FRACTIONS.
^ 60, We have seen (1127,) that integers of one de-
nomination may be reduced to integers of another denomi-
nation. It is evident that fractious of one dcnotnimition,
after the same manner, and by the same rules, may be re-
duced io fractions of another denomination; that is, frac-
tions, like integers, may be brought into lower denomin;;-
tious by multiplication, and into liigher denonlinati<•ll^^ l»\-
(livision.
]2-;3
REDUCTION OF FRACTIONS.
IT 60.
To reduce higher into LOWEp
denominations.
(Rule. See H 28.)
1 . Reduce ^^(^ of a pound
to pence, or the fraction of a
penny.
Note. Let it be recollect-
ed that a fraction is multipli-
ed either by dividing its de-
nominator, or by multiplying
its numerator.
jlTy^.X20=,JjS. X 12 =
^d. Ans.
Or thus: -.jU of "i^ of '/=
a|o=|. of a penny, Ans.
;}. Reduce xaVir of a pound
to the fraction of a farthing ?
TaW*X20=:j§^^:iy s. X 1'*
=i¥^dx4=f¥8%=fq.
Or thus :
Num. 1
20 s. in I^.
20
12 d. in 1 s.
240
4 q. in 1 d.
' 960
Then Wxi^—H ^«>'-
5. Reduce 2^^^ of a guin-
ea to a fraction of a penny.
7. Reduce f of a'guinea to
the fraction of a pound.
Consult 11 28, ex. 12.
1>. Reduce f of a moidore,
at i£. 10s. to the fraction of
a guinea.
II. Reduce 2T of a pound,
Troy, to the fraction of an
ounce.
To reduce lower into higher
^Icnominations.
(Rule. See H 28.)
2. Reduce f of a penny to
the fraction of a pound.
Note. 'Division is perform-
ed either by dividing the nu-
merator, or by multiplying the
denominator.
f d. -i- 12 = tVs- -^ 20==:
^1^^. Ans,
Or thus : f of -^^ of
s^-~
4. Reduce f of a farthing
to the fraction of a pound.
-|-20 — ^^xs — r^ViF'^-
Or thus :
Dcnom. 4
4 q. in Id.
16
12d. in Is.
192
208. in ^l.
3840
Then ■s^\js='£t^- Ans*
6. Reduce f of a penny to
the fraction of a guinea.
8. Reduce | of a pound to
the fraction of a guinea.
10. Reduce f J of a guinea
to the fraction of a moidore.
12. Reduce | of -an ounce
to the fraction of a pound
Troy.
-/ «
1160.
fl 60, 61.
REDUCTION OP FRACTIONS.
123
HIGHER
'S.
128.)
1 penny to
tund.
5 perform-
ig the nu-
plyingiiiG
of 2^ =
' a farthing
L pound.
.12=Tt^s.
71* •
a penny to
ruinea.
a pound to
uinea.
of a guinea
|a moidore.
>f "an ounce
»f a pound
V3^ Reduce ^'g of a pound
avoirdupois, to the fraction
of an ounce.
15. A man has j}w of ^
hogshead of wine ; what part
iri that of a pint ?
17. A cucumber grew to
the length of ijiyVti "^ ^ ^^^^^ 5
what part is that of a foot ?
19. Reduce f of l of a
pound to the fraction of Is.
21. Reduce i of ^j of 3
pounds to the fraction of a
penny.
!I 61. It will frequently
be required io find the value
of a fraction^ that is io re-
duce a fraction to integers of
less denominations.
1. What is the value ©f f
of a pound ? In other words,
reduce f of a pound to shil-
lings and pence.
|ofa.£ is^o--i3i shil-
lings; it is evident from ^ of
a shilling may be obtained
some pence ; ^ of a shilling is
i-2=:4d. — that is, multiply
the numerator by that num-
ber which will reduce it to
the next less denomination,
and divide the product by the
denominator ; if there be a
remainder, tnuitiply and di-
vide as before, and so on ; the
several quotients, placed one
after another in their order,
will be the answer.
14. Reduce f of an ounce
to the fraction of a pound
avoirdupois.
16, A man has {^^ of a pint
of wine ; what part is that of
a hogshead ?
18. A cucumber grew to-
the length of 1 foot 4 inches
=ifi.=| of a foot ; what part
is that of a mile?
20. f ^ of a shilling is f of
what fraction of a pound '?
22. ^f of a penny is ^ of
what fraction of 3 pounds?
W ^^ ^ penny, is ^ of what
part of 3 pounds ? ^^ of a
penny is ^ of -^y of how many
pounds ?
It will frequently be re-
quired to reduce integers to
the fraction of a greater de-
nomination. •
2. Reduce 13s. 4d. to the
fraction of a pound.
13s. 4d. is 160 pence ; there
are 240 pence in a pound?
therefore, 13s. 4d. is ]J^f8=|
of a pound, ^hat is, reduce
the given sum or quantity to
the least denomination men-
tioned in it, for a numerator ;
then reduce an integer of
that greater denomination (to
a fraction of which it is re-
quired to reduce the given
sum or quantity) to the same
denomination, for a denomi-
nator, and they will form the
fraction required.
124
REDUCTION OF FRACTIONS.
^01.
EXAMPLES FOR PRACTICE.
3. What is the value of f
of a shilling ?
OPERATION.
Numer. 3
12
«Denom. 8)36(4d. 2q. Ans,
32
4
4
16(2q.
16
5. What is the value of f
of a pound Troy?
7. What is the value of f
of a pound avoirdupois 1
9. I of a month is how ma-
ny days, hours and minutes ?
11, Reduce f^ of a mile to
Its proper quantity.
13. Reduce -^^ of an acre
to its proper quantity.
15. What is the value of
■{% of a dollar in shillings,
pence, &:.c. ?
17. What is the value of
^g of a yard?
19. What is the value of
fr^ of a toil.
4. Reduce 4d. Scj. to the
fraction of a shillitio-.
OPERATION.
4d. 2q. Is.
4 12
12
4
48 Denom.
18 Numer.
il=l Ans.
6. Reduce 7 oz. 4 pwt. to
the fraction of a pound Troy.
8. Reduce S oz. 14f dr.
to the fraction of a pound
avoirdupois.
Note. — Both the numerator
and the denominator must be
reduced to 9ths of a dr.
10. 3 weeks Id. 9h. 36m.
is what fraction of a month ?
12. Reduce 4 fur. 125 yds.
2 ft. 1 in. 2| bar. to tne frac-
tion of a mile.
14. Reduce 1 rood 30 poles
to the fraction of an acre.
16. Reduce 4s. 8;|^d. to the
fraction of a dollar.
18.
Reduce 2 ft. 8 in. l^b.
to the fraction of a yard.
20. Reduce 4 cvvt. 2 qr.
12 lb, 14 oz. 12/^ dr. to the
fraction of a ton.
Note. Let the pupil be required to reverse and prove the
following examples :
^ 61.
SUPPLEMENT TO FRACTIONS.
125
'21. What is the value of y\ of a guinea?
22. Reduce 3 roods, 17^ poles to the fraction of an acre.
23. A man bought 27 gal. 3 qts. 1 pt. of molasses; what
pirt is that of a hogshead ?
24. A man purchased fV of 7 cwt. of sugar ; how much
sugar did he purchase ?
25. 13h. 42m. 51 fs. is what part or fraction of a day ?
SUPPLEMENT TO FRACTIONS.
1. What are fractions f 2. Whence is it that the parts into which
iuiy Ihinn cr any number niay be divided, t;ika their name'? 3. How
arc fractions represented by figures ? 4. Wh»l is the aumher above the
line called ?— Why is it so culled i 5. What is th i numbor below the
line called ?— Why is it so called ?— What does it show 1 T.. What is
it which determines the mu^nilude of tliB purls ? — Why ? 7. What is
a simple or proper fraction ? — — an improper fraction a mixed
number ? S. How is an improper fraction reduc«d to a wiiole or mixed
number? 9. How is a mixed number reduced to an improper frac-
tion t a whole number ? 10. What- ib understood by the terms of
thefra.'tion ? 11. Haw is a fraction reduced to its most smp/c or
l)west ierms? 12. What is understood by a coniman divisor?
by the greatest common divisor? 13. How is it found ? 14. How
miay ways are there to multiply a fra;-tio.i by a \vhole number ? 15,
How does it appear, that dividing the denominator multiplies the frac-
tion? 16. How is a mixed number multiplied ■? 17. What is implied
ill niulliplying by a fraction'? 18. Of how many operations docs it
consist?— What arf! they? 19. When the multiplier is k'is than a
unit, what is the product compered wiih the muiliplicaud ? 20. How
(io you multiply a whole number by a fraction "? 21. How do you
multiply one fraction by another ? 22. How do you multiply a mixed
number by a mixed number ? 23. How does it api>ear, that in nf^ulti-
plying both terms of the fraction by the same number the valuj of the
fraction is not altered 1 24. How many ways are thereto divide a
ftaclion by a whole nnmlier 1 What are they ? 25. How does it appear
lh:il a fraction is divided by multiplying its 'Jcnominalor ? 20. How
does dividing by a fraction differ from multiplying by a fraction (" 27.
When the rftuisor is /.!sa* thin a uiiil, what is ihj qtjoiienl compared
with the dividend ? 118 What is understood by a fo/;i,7io;i dunoiuiua-
lor ? the least comimm donomiuator .'' 2\i, Huw dur-s il appear
llial each g-jrendtiiioniinalor must be a factor of the co;!ur<'jii diinomin-
alor '{ 3D. How is the common denoniitntor to two or more fraeiions
found? 31. What i.s understood by a multiple.^ by a comman
niull>;,'ii:? by the least common muliipti^ ? Uiiit is tiie pro-
Cfist. of iMidingil'? '.]2. How are frin-Lions adJed and .subtracted.'* 3'i.
How is a faction cf a gieiUiir dtinomiuaiion reduced to oiie of a less ?
of a less to n gieater? 34. How are fractions o( a ujreatrr Ac-
nomination rjduced to integers of a Icbd ? io'c^or-^ <)\ a les*
iic!ii);niua'iori to Itie fraction of agrcu'er?
L2
126
SUPPLEMENT TO FRACTIONS.
!IQ1.62.
EXERCISES.
1. What is the amount of ^ and f 7 of ^ and f ?
of 12^, 3f and 4f ? Ans. to the last, 20^.
2. To 1^ of a pound add f of a shilling. Amount, 18|s.
Note. First reduce both to the same denomination.
3. I of a day added to f of an hour, make how many
hours ? what part of a day ? Ans. to the last, f f d.
4. Add ^ lb Troy to /g- of an ounce.
Amount, 6 oz. 11 pwt. 16 gr.
5. How much is | less ^ ? i^^i of f of |^ ?
Ans. to the last, ^|3.
6. From ^ shilling take f of a penny. Rem. 5^d.
7. From ^ of an ounce take |- of a pwt.
Retn. 1 1 pwt. 3 gr.
8. From 4 days 7^ hours, take 1 day 9^g hours.
Rem. 2 days, 22 hours, 20 min.
9. At £^ per yard, what costs f of a yard of cloth ?
1] 69. The^me of unity, or 1, being given to find the
cost of any quantity, either less or more than unity, multi-
ply the price by the quantity. On the other hand, the cost
of any quantity, either less or more than unity, being given,
to find the price of unity, or 1, divide the cost by the quantity.
Ans.
£±ii
1, If 1^ ib of sugar cost ^5 of a shilling, what will ff of
a pound cost ?
This example will require two operations : first, as above,
to find the price of 1 lb ; secondly, having found the price
of 1 lb, to find the cost of |f of a pound. /-jS.-^l^ {\f of
t7^s. H 54)=t9^Vs. the price of 1 ib. Then, tV^s.XU (f!
of TV5S. 11 50)z=:f 9-Hs-=4d. m-m- the answer.
Or we may reason thus : first to find the price of 1 lb ;
J-^ ib costs -/-s. li we knew what y'vj lb would cost, we
might repeat this 13 times, and the result would be the
price of 1 lb, -\^ is 11 parts. If y^^ lb costs -j\s. it is evi-
dent y^^ lb will cost j\ of T\=Ti5S- ^"^^ it ^ ^^il^ cost 15?
times as much, tliat is, ^^g'^s.=the price of 1 lb. Then, f 4
o^^^-=nU^- the cost of -H of a pound, f ^H^.^^cl.
'^f ff5&4' ^^ before. This process is called solvfinr the ques-
tion by analysis.
After the same manner, let the pupil solve the folUmin?
questions :
1IQ1,62.
fl62»
SUPPLEMENT TO FRACTIONS.
127
,d f ?
t, 20i^.
mnt, I8^s.
tion.
how many
St, If d.
pwt. 16 gr.
le last, ^f ^.
Rem. 5^d.
1 pwt. 3 gr.
irs.
irs, 20 min,
cloth ?
1 to find the
inity, multi-
mi, the cost
being given,
he quantity.
' ns. £^^.
X will f f of
st, as above,
id the price
■H (if of
>s.xe(e
■
:e of 1 tb;
[Id cost, we
)uld be the
js. it is evi-
Ivill cost V^
Then.e
];if tke qifcs-
io folUm'in?
2. If 7 lb of tobacco cost J of a pound, what is that a
pound ? I of f =how much ? What is it for 4 tb ? | of
|=how much ? What for 12 lb ? Vof f =how much ?
Ans. to the last, £lf.
3. If G^ yards of cloth cost £{i, what cost 9^ yards ?
Ans. £4. 5s. 4^d.
4. If 2 oz. of silver cost lis. 3d. what costs ^ of an oz 1
Ans. 4s. 2d. 2^q.
5. If f oz. costs 4s. Id. what costs 1 oz ? Atis. 5s. 8|d.
6. If + ib less by } costs 13^d. what costs 14 lb less by
1 of 2 tb ? Ans. £4. 9s. 9^\d.
7. If J yard costs £l, what will 40.i yards cost.
Ans. £50. Ig. 2fd.
8. If 1^5^ of a ship costs c£25l, what is ^^ of her worth ?
Ans. £5^. 15s. 8^d.
9. At ^£3| per cwt. what will 9f tb cost ?
10. A merchant owning | of a vessel, sold f of his share
tor c£39. 5s. what was the vessel worth? Ans. .£448 lis. 104d.
11. Iff yards cost £^, wliat will f^ of an ell Eng. cost.
Ans. 1 7s. Id. 2fq.
12. A merchant bought a number of bales of cloth, each
containing 129^|- yards, at the rate of £7 for 5 yards, and
sold them out at the rate of <£ll for 7 yards, and gained
of200 by the bargain ; how many bales were there ?
First find for what he sold 5 yards ; then what he gained
on 5 yards — what he gained on 1 yard. ^Then, as many
times as the sum gained on 1 yard is contained in <£200, so
many yards there must have been. Having found the num-
ber of yards, reduce them to bales. Ans. 9 bales.
13. If a staffs^ feet in length, cast a shado.-/ of G feet,
how high is that steeple whose shadow measures 153 feet ?
Ans. 144i feet.
14. If IC men finish a piece of work in 28^ days, how
Icmg will it take 12 men to do the same work ?
First find how long it would take 1 man to do it ; then
12 men will do it in j\ of that time. Ans. 37^ days.
15. How many pieces of merchandise, at 20^s. apiece,
mast be given fjr 240 pieces, at 12^s. apiece? Ans. 149^^-^.
16. How many yards of booking that is Ijyd. Wide will
1)0 sutficient to line 20 vds. of camlet that is } of a vard wide ?
123
DECIMAL FilACTIONS.
!I 62, 63.
First find the contents of the camlet in square measure ;
then it will be easy to find how many yarda in length of
booking that is 1^ yd. wide it will take to make the same
quantity. Ans. 12 yardti of camlet.
17. If 1] yd. in breadth require 20^ yds. in length to
make a cloak, what in length that is ^ yd wide will be re-
(juired to make the same? Ans. {Ml yds.
18. If 7 horses consume 2| tons of hay in ti weeks, liow
many tons will 12 horses consume in 8 weeks ?
If we knew how much 1 horse consumed in I week, it
would be easy to find how much 12 horses would consume
in 8 weeks.
2^='j-' tons. If 7 horses consume y tons in 6 weeks;
one horse will consune } of iji=J^^ of a ton in weeks;
and if a horse consume ^^ of a ton in 6 weeks, he will con-
sume } of ^i=iVV of a ton in I week. 12 horses will con-
sume 12 times t'\5'a=f^| in 1 week, and in 8 M'eeks they
will consumes times |§|='^\2—-(j^ tons, answer.
19. A man with his family, which in all were 5 persons,
did usually drink 7f gallons of cider in 1 week ; how much
will they drink in 22^ weeks when 3 persons more are added
to the fiimily ? " yl«5. 280| gallons.
20. If 9 students spend oflO^ in 18 days, how mucli will
20 students spend in 30 days ? Ans ct'39. 18s. 4|5^(l
fle'ciitial Fractions.
ff 63* We have seen, that an individual thing or num-
ber may be divided into any aumber of equal parts, and that
these parts will be called halves, thirds, fourths, fifths, sixtlis,
&.C., according to the number of parts into which the thinor
or number may be divided; andthateachof these parts may
be again divided into any other number of equal parts, and so
on. Such are called common or vulgar fractions. Their
denominators are not uniform, but vary with every varying
division of a unit. It is this circumstance v/hich occasions
the chief difficulty in the operations to be performed on
them; for when numbers are divided into difterent kinds
or parts, they cannot be so easily compared. This difli-
culty led to the invention o{ decimal fractions, in which an
individual thing or number is supposed to be dividfMl first
into ten equal parts, which will be tenths, and each of tlic-ic
<1 fi4.
DECIMAL FRACTIONS.
129
parts to be again divided into ten other equal parts, which
will be hundredths ; and each of these parts to be still fur-
ther divided into ten other equal parts, which will be thoU'
sandths ; and so on. Such are called decimal fractions,
(from the Latin word decern, which signifies ten,) because
they increase and decrease in a tenfold proportion, in the
same manner as whole numbers.
^ 61. In this way of dividing a unit, it is evident,
that the denominator to a decimal fraction will always be
10, 100, 1000, Of 1 with a number of ciphers annexed ;
consequently, the denominator to a decimal fraction need
not be expressed, for the numerator only, written with a
point before it, (') called the separatrix, is sufficient of it-
self to express tlie true value. Thus,
6
TJ
685
TOdff
are written '6. .
• • • • /^ • •
'685.
The denominator to a decimal fraction, although not ex-
pressed, is always understood, and is 1 with as many ci-
phers annexed as there are places in the nuinCia'or. Thus,
'3705 is a dacimal consisting of four places ; consequently,
I with four ciphers annexed, (10000) is its proper denomi'
nator. Any decimal may be expressed in the form of a
common fraction by writing under it its proper denomina-
tor. Thus, '3705 expressed in the form of a common frac-
tion, is TV.iVb-
When the whole numbers and decimals are expressed to-
gether, in the same immber, it is called a mixed number.
Thus, 25'63 is a mixed number, 25', or all the figures on
the left hand of the decimal point, being whole numbers,
and *63, or all the figures on the right hand of the decimal
point, being decimals.
The names of the places to ten-millionths, and, generally,
how to read or write decimal fractions, may be seen from
tiie following
130
DECIMAL FRACTIONS.
TABLE.
!I 64.
t5
ill -^f
orca oZzir
C(«r*r
55
3 ^
cr =p
n o
^'^ ^O *■« iS
p a
w
^1
(«l (V
CO
o
B
•a
ad
2(1
Ist
place.>o» II II II II II
place. M 4*
place. )^;jf -^ o
II Hundreds.
Tens.
Units.
Ist place. © C5 o QL Ci © o
2d place. © M o c: gc oi ii
;Jd place, o o o 01 o
^ 4th place, o oom
5tli place. ^ cjr,
Oth place. ^ ^
7tli place. ^
J,, Tenths.
Hundredths.
Thousandtlis.
Ten-Thousandths.
Hundred-Thousandths
Millionths.
Teu-Millionths.
c: is ^f gp
. p ss
s a.
■ji
From the table it appears, that the first figure on the right
hand of the decimal point signifies so many tenth parts of a
unit ; the second figure, so many hundredth parts of a unit ;
the third figure, so many thousandth parts of a unit, 6lc.
It takes 10 thousandths to make 1 hundreth, 10 hundredths
to make 1 tenth, and 10 tenths to make I unit, in the same
manner as it takes 10 units to make 1 ten, 10 tens to make
1 hundred, &c. Consequently, we may regard unity as a
starting point, from whence whole numbers proceed, con-
tinually increasing in a tenfold proportion towards the left
fl(>5, 00.
DECIMAL FUACTIONB.
131
hand, and decimuls conUnuaWy derrr as in ^ in the same pro-
portion, towards tlic right hund. But iia decimHKs docrcnHO
towards the right hand, it foUows of course, that they in-
crease tovvardH the left hand, in the same manner as whole
mnnhcrs.
^ 0*1. The value of every figure is determined by its
place from unifn. Consequently, ciphers placed at the ri'^'/i^
li;uid of decimals do not alter their value, since every sig-
nificant figure contiinies to possess the same place from
unity. Thus, '5, '50, '500, are all of the same value, each
being ecjual to -^f*^ or ^.
But every cipher placed at the frft hand of decimal frac-
tions diminishes them tenfold, by removing the significant
figures further from unity, and consequently making each
part ten tin»es as small. Thus, '5, *05, '005, are of diflTer-
etit value, *5 being eqal to y^, or }, '05 being equal to ^^g,
or v},f, and '005 being eqal to jr/ij!T> <>r -j^xj'
Decimal fractions, having different denominators, arc rea-
dily reduced to a common denominator, by annexing ciphers
until they are equal in number of places. Thus, '5, *00,
mi may be reduced to '500, *0G0, '234, each of which has
1000 for a common denominator.
11 00. Decimals are read in the same manner as whole
numbers, giving the name of the lowest denomination, or
right hand figure, to the whole. Thus, '0853 (the lowest
denomination, or right hand figure, being ten-thousandths)
is read 0853 ten-thousandths.
Any whole number may evidently be reduced to decimal
parts, that is, to tenths, hundreths, thousandths, &c,, by
annexing ciphers. Thus, 25, is 250 tenths, 2500 hun-
dredths, 25000 thousandths, &c Consequently, any mixed
number may be reAd together giving it the name of the low-
est denomination or right hand figure. Thus, 2ral'03 may
be read 2503 hundredths, and the whole may be expressed
in the form of a common fraction, thus, ^'^V-
The denominations in federal money are made to cor-
respond to the decimal divisions of a unit now described, dol-
lars being units, or whole numbers, dimes tenths, cents hun-
dredths, and mills thousandths of a dollar; consequently,
132
DECIMAL FRACTIONS.
Tf t)6, 67.
the expression of amj sum in dollars, cents and mills, is sim-
ply the expression of a mixed number in decimal fractions.
Forty-six and seven tenths=4e/y==46'7.
Write the following numbers in the same manner :
Eighteen and thirty-four hundredths.
Fiily-two and six hundreths.
Nineteen and four hundred eighty-seven thousandths.
Twenty and forty-two thousandths.
One and five thousandths.
135 and 3784 ten thousandths.
9000 and 342 tep thousandths.
10000 and 15 ten-thousandths.
974 and 102 millionths.
320 and 3 tenths, 4 liundredths and 2 thousandths.
500 and 5 hundred thousandths.
47 millionths.
Four hundred and twenty-three thousandths.
ADDITION AND SUBTRACTION OF DECIMAL
FRACTIONS.
If 07. As the value of the parts in decimal fractions in-
creases in the same proportion as units, tens, hundreds, &:,c.,
and may be read together, in the same manner as whole
numbers, so, it is evident that all the operations on decimal
fractions may he performed in the same manner as on lohole
numbers. The only difficulty, if any, that can arise, must
be in finding lohere to place the decimal point, in the result.
This, in addition and subtraction, is determined by the
same rule ; consequently, they may be exhibited together.
I. A man bought a barrel of flour for $8, a firkin of but-
ter for $Jd*50, 7 pounds of sugar for 83^ cents, an ounce ol
pepper for G cents; what did he give for the whole 1
Note. See the table of Federal Money, ^ 27. Let the
pupil go back now and read carefully all that is said respect-
ing Federal Money in Reduction. From wtiat is there stated
it is plain, that we may readily reduce any sums in federal
money to the same denominations, as to cents or mills, and
'*•"■•'■<>''.■ '"^•■'v'-'-y''^ ''V '
;>■'■'
f*f.
DECIMAL FRACTIONS.
138^
Is, is sim-
ctions.
ler :
indths.
dthe
ECIMAL
fractions in-
kdreds,&'C.,
Ir as whole
on decimal
\as on whok
larise, must
Let the
laid respect-
Ithere stated
in federal
mills, and
and add or subtract them as simple numbers. Or, what is
the same thing, we may set down the sums, taking care to
write dollars under dollars, cents under cents, and mills un-
der mills, in such order that the sepaifating points of the
several numbers shall fall directly under each other, and
add them as simple numbers, placing the separatrix in the
amount directly under the other points.
OPERATION.
$g' = 8000 mills, or lOOOths of a dollar.
3<50 = 3500 mills, or lOOOths.
«835=: 635 mills, or lOOOths. .
^06 = , 60 mills, or lOOOths. ^
An$. $12<395=12395 mills, or lOOOths.
As the denominations of fe/ieral money correspond with
the parts of decimal fractions, so the rules for adding add
subtracting decimals are exactly the same as for the s^me
operations; in federal money.
2. A man owing $375, paid $17575; how much did be
then owe 1
OPERATION. ^
$375' == 37500 cents, or lOOths of a dollar.
175'75= 17575 cents, or lOOths of a dollar.
$199'25= 19925 cents, or lOOths.
Wherefore, — In addition and subtraction of decimal
fractions, — Rule : Write the numbers undqr each other,
tenths under tenths, hundredths under hundredths, according
to the value of their places, and point off in the result as
many places for decimals as are equal to the greatest num-
ber of decimdl places in any of the given numbers.
EXAMPLES FOR PRACTICE.
3. Bought 1 barrel of flour for 6 dollars and 75 cents, 10
lb. of coffee for 2 dollars 30 cents, 71b. of sugar for 92 cents,
1 lb. of raisins for 12^ cents, and 2 oranges for 6 cents ;
what was the whole amount? Ans. $10455.
4. A man is indebted to A, $237'62 ; to B, $360 ; to C,
$8642^; to D, $9'62Jf ; and to E, $0*834; what is the
amount of his debts? vlns. $684*204.
5. A man has three notes specifying the following suiiiii^
viz. three hundred dollars, fifty dollars sixty cents, and nina
M
i\
i^-'
1&4 ADDITION AND SUBTRACTION OF DECIMALS. ff 67.
Sip
dollars eight eents; what ia^the amount of the three notes?
CV-MV.i;-^;tM.i?'i .*-<*;*'-. n> . atfrii.!-;-. ;, ... - Jns. $359*68,
6. A mail gave 4 dollars 75 cents for a pair of boots, and
2 dollars 12^ cents for a pair of shoes ; how much did the
boots cost more than the shoes?
OPERATION. OPERATION. '
;.,N
4750 mills.
2125 mills.
w,
75
2425
/
2625 mills= $2*625 Arts. 82'625 Ans.
7. A man bought a cow for eighteen dollars, and sold her
again for twenty-one dollars thirty-seven and a half cents;
how much did he gain ? Ans. 3'37o.
S. A man bought a horse for 82 dollars, and sold hiin
•ngftin for seventy-nine dollars seventy-five cents ; did he
!gainorlo8e? and how much ?"' * : m ,
9. A man sold wheat at several times as follows, viz,
13'25 bushels ; 8'4 bushels ; 23*«5l bushels ; 6 bushels,
and *75 of a bushel; how much did he sell in the whole?
Ans. 51*451 bushels,
10. What is the amount of 429, 21^^^^^, 355, j^kxiy hh,
and 1 /^ ? ;' ^- ; ' Ans. 808^*<y3^ , or 808' 143.
11. What is the amount of 2 tenths, 80 hundredths, 89
thousandths, 6 thousandths, 9 tenths, and 5 thousandths ?
Ans. 2,
' 12. ' What is the emount of three hundred and twenty-nine I
and seven tenths ; thirty-seven and one hundred sixty-two j
thousandths, and sixteen hundredths?
13. A man, owing $4316, paid $376*86.5 ; how much I
did he then owe ? Ans. $3939*135 (
14. From thirty-five thousand thake thirty-five thou-
Ans. 34999*965,1
Ans. 1*5507
Ans. 234*9925.
1793*13 and 8171
Ans. 976*07307
sandths.
15. From 5*83 take 4*2793. ■
. 16. From 480 take 245*0075.
17. What is the difference between
05693 ?
. 18. From ij^jj take 2 j\y. Tlcmaindcr, 1 j^^»^ or 1*98,1
:, 19. What is the amount of 29 j%, 374 ittt) giruir, ^^M
• 315 ToV(T. 27, and lOO^^j, ? Ans. 942*95700Hl
i
IA% or r98.
fl 68. MULTIPLICATION OP DECIMALS. 135
MULTIPLICATION OF DECIMAL FRACTIONS.
^ OS. 1. How muc^i h^y 'm, 7 loa/ds^ each cojit^ning
23,571 cwt? ■< Y--'^ ; ■ ••' - -•; ■■^- ■■^■■:- ■
, i OPERATION.
23*571 cwt.= 23S71 lOOOths of a ewt.
7 7
■ST*
; ^;.
vif ./
Ams. 164'997 cwt.= 164997 lOOOths of a cwt.
We may here, (^ 66,) consider the multiplicand so, many
thousandths of a cwt., and then tl^e product will evidently
be thousandths, and will be reduced to a mixed or whole
number by pointing off 3 figures, that is, the same number
as are in the multiplicand ; and as either factor may be made
the multiplier, so, if the decimals had been in the multiplier",
the same number of places must have been pointed off for
decimals. Hence it follows, we must always point off in the
product as many places for decf/nah qs there ar», decimal
places in toth factors.
2. Multiply *75 by '25.
OPERATION. In this example, we have 4 deci-
*75 mal places in both factors; we
*25 must therefore point off 4 places
. for decimals in the product. , The
375 reason of pointing off this num-
150 ; ^ , ' ber may appear still flaore plain, if
we consider the two factors as
'ISlfS Product, common or vulgar fractions. Thus,
«75 is f^, and '26 is -^^ : now, i^%X^xk=^^^^^J='l^''^,
ilns. same as before. / r ,
3. Multiply 425 by '03.
Here, as the number of significant
figures in the product is not equal to
the number of decimals in both fac-
tors, the deficiency must be supplied
'00375 ' by prefixing ciphers, that is, placing
them at the lefl hand. The ccwrectness of the rule may
appear from the following process : '125 is -jV^j ^^^
'03 is T§^ : now, TV<jVXT§Ty=T<jV(jW=*00375, the same
as before. ,!.,..
These examples will be sufficient to establish tjie fpllow-
OPERATION.
'125
'03
>/-":■. :.h
Vi-
136
MULTIPLICATION OF DECIMALS.
H 68.
M,
kuLE.
la the multiplication of decimal fractions, multiply as in
yivhxM numbers, and from the product point off so many fig-
ures for decimaJis as there are decimal places in the multi-
plicand and multiplier counted together, and, if there are
,not so many figures in the product, supply the deficiency by
prefixing ciphers.
As the denominations of federal money correspond with
the parts of decimal fractions ; the rules for the multiplica-
tion and division of both are the same.
EXAMPLES FOR PRACTICE. «
4. At $5'47 per yard, what cost 8*3 yards of cloth ?
Ans. 45'401.
• 5. At $'07 per pound, what cost26'5 pounds of rice ?
•v Ans. $V855 cwt.
6. If a barrel contain 1*75 cwt. of flour, what will be the
weight of '6^ of a barrel ? Ans. 1*1025.
7. If a melon be worth $0*9 what is *7 of a melon worth ?
^ Ans. 6^ cents.
8. Multiply five hundredths by seven thousandths
Product, *000a>.
9. What is *3 of 116.' , >4n5. 34'8.
10. What is *85 of 3672! . Ana. 3l2r2.
* 11. What is *37 of *0563t Ans. *020831.
: 12. Multiply 572 by *58.
13. Multiply eighty-six by four hundredths.
Product, 3'44.
14. Multiply *2062 b, *0008.
15. Multiply forty-seven tenths by one thousand eighty-
six hundredths.
16. Multiply two hundredths by eleven thousandths.
17. What will be the cost of thirteen hundredths of a
ton of hay, at $11 a ton?
18. What will be the cost of three hundred seventy-five
thousandths of a cord of wood at $2 a cord ?
19. If a man's wages be seventy-five hundreths of a dol-
lar a day, how much will he earn in four weeks, Sundays
excepted?
20. What will 250 bushels of rye come to at $0*88^ per
bushel f ilns. $22r25.
24. What is the value of 86 barrels of flour, a^ *'*'^'^ '
barrel?
>u!*/^..,<t/'''i
^69.
DIVISION OP DECIMALS.
137
22 What will be the cost of a hogshead cf molasses
containing 63 gallons, at28^cents agnilcn? Ans. $I7'955.
23. If a man spend 12} cents a day, what will that a-
mount to in a year of 365 days? what will it amount to in
five years ? Ans. $22842^ in 5 years.
DIVISION OF DECIMAL FRACTIONS.
51 09. Multiplication is proved by division. We have
seen, in multiplication, that the decimal places in the pro-
duct must always be equal to the number of decimal places
in the multiplicand and multiplier counted together. The
multiplicand and multiplier, in proving multiplication, be-
come the divisor and quotient in division. It follows of
course, in division, that the number of decimal places in the
divisor and quotient counted together, must always he equal
to the number of decimal places in the dividend. This will
still further appear from the examples and illustrations which
follow :
1. If 6 barrels of flour cost $44'718, what is that a bar-
rel?
By taking away the decimal point, $44*718=44718 mills,
or lOOOths, which, divided by 6, the quotient is 7453 mills,
=$7*453, the answer.
Or, retaining the decimal point, divide a§ in v.'hole num-
bers :
OPERATION. As the decimal places in the di-
6)44'718 visor and quotient, counted togeth-
er, must be equal to the number of
Ans. 7*453 decimal places in, the dividend,
there being »o decimals in the divisor, — therefore point off
three figures for decimals in tlic quatiini, equal to tlie ruuii-
ber of decimals in tlie dividend, which brings us to the same
result as before.
2. At 64'75 a barrel fur cider, how many barrels may be
bo>ight for 631 /
In this example, there are decimals in the divisor, and
none in the dividend. $4*75:^^475 cents, and $31, by an-
nexinrr two ciohers r=:3100 cents ; that iy, reduce the divi-
dead to parts of the same denomination as the divisor. —
M2
138
DIVISION OF DECIMALS.
U 69.
itik'
Then, it is plain, as many times 475 cents are contained in
3100 cents, so many barrels may be bought.
475)3100(6fi^^ -barrels, the answer; that is, 6 barrels
2850 and f ^§ of another barrel.
But the remainder, 250, instead of bp.
250 ing expressed in the form of a common
fraction, may be reduced to lOths by annexing a cipher,
which, in effect, is multiplying it by 10, and the divisor
continued, placing the decimal point after the 6, or wlioio
ones already obtained, to distinguish it from the decimals
which are to follow. The points may be withdrawn or not
from the divisor and dividend.
OPERATION.
4'75)3r00(6'526+barrels, the answer, that is t> barrels
2850 and 526 thousandths of anotlier bar-
rel.
2500 By annexing a cipher to the first
2375 remainder, thereby reducing it to
lOths, and continuing the division, wo
1250 obtain from it '5, and a still further
950 remainder of 125, which, by anuex-
V ing another cipher, is reduced to
3000 lOOths, and so on.
2850 The last remainder, 150, is |-^ft ot
a thousandth part of a barrel, wiiicli
150 is of so trifling a value, as not to merit
notice.
If now we count the decimals of the dividend, (for overv
cipher annexed to the remainder is evidently to be counted
a decimal of the dividend,) we shall find them to he Jim,
which corresponds with the number of decimal 'plnce.s in
the divisor and quotient coui^ted together. •
3. Under II 68, ex. 3, it was required to multiply 'l'J5 In
*03 ; the product was '00375. Taking thi« product for a
dividend, let it be required to divide *00375 by *125. One
operation will prove the other. Knowing that tlic nu'inbor
of decimals in the quotient and divisor, counted together,
will be equal to the decimal places in the dividelid, we may
divide as in whole numbers, being careful to retain tiie de-
cimal points in their proper places. Thus :
«1 69, 70.
DIVISION OF DECIM.4LS.
139
reduced to
OPERATION.
'l-25)*00375('03 The divisor, 125, in 375 goes 3
375 times and no remainder. We have
only to place the decimal point in
000 the quotient and the work is done.
There are five decimal places in the dividend ; conse-
quently there must be five in the divisor and quotient count-
ed together ; and, as there are three in the divisor, there
must be two in the quotient ; and since we have but one
figure in the quotient, the deficiency must be supplied ^by
prefixiniT a cypher.
The operation by vulgar fractions will bring us to the
*=!ime result. Thus, '{'^5 is ^0^)%, and '00375 is xaWrra ••
"ow, ^y3^7^^^-i-^ij3y5jy=-j..^i-Sc)oo^ =^3 ^='03 the same as
betbre.
*j 71>. The foregoing examples and remarks are .suffi-
cient to establish the following
. RULE.
In the division of decimal fractions, divide as in whole
niuribers, and from the right hand of the quotient point off
as many figures for decimals as the decimal figures in the
dividend, exceed those in t) j divisor, and if there are not
.so many figures in the quotient, supply the detficiency by
prefixing ciphers.
If at any time there is a remainder, or if the decimal
iigurcs in the divisor exceed those in the dividend cyphers
may be annexed to the dividend or the remainder, and the
quotient carried to any necessary degree of exactness ; but
the ciphers annexed must be counted so many decimals of
the dividend.
EXAMPLES FOR PRACTICE.
4. If $472,875 be divided equally between 13 men, how
much will each one receive .^ Ans. 83(),375.
5. At $'75 per bushel, how many bushels of rye can be
bought for $141 ? Ans. 188 bushels.
; 6. At Ql cents apiece, liow many oranges may be bought
for $8 ? Ans.. 128 oranges,
7. If '6 of a barrel of flour cost $5, what is that per bar-
rel ? Ans. 8'333+
Divide 2 by 53'1. Qmt. '037-]-
140 REDUCTION OP VULGAR FRACTIONS. &-C. ^ 70, 71-.
9. Divide *012 by '005.
10. Divide three thousandths by fl^ur hundredths.
Quot. '075.
11. IIow many times h *I7 contiined in 8.^
12. If I pa" $40S'75 for 75 ) pounds of wool, what is the
value of I pound ? Am^. $0'625; or thus aO'624
13. If a piece of cloth, mer.saring I25yards^, costf I8l"i5
Wh:it is that a yard ?, Ans. 81*45.
14. If 536 quintals of fish cost $1913,52, how much is
that a quintal ? An:;. $3*57.
15. Bought a farm, containing}^;! acres, for $^3213 ; what
did it cost me,per acre ? ^Ins. $38*25.
16. At $954 for 3816, yards of flannel, what is that per
yard.' Ans. $0*25.
REDUCTION OF COMMON OR VULGAR FRAG-
TIONS TO DECIMALS.
•»"
fj 71, 1. A man hns '^ of a barrel of flcur ; what is
that expressed in decimal parts ?
As many times as the denominator of a fraction is con-
tained in the numerator, so many whole ones are contained
in the fraction. We can obt lin no whole ones in f , be-
cause the dc.-iominator is not contained in the numerator.
We may, however, reduce the numerator to tenths, (1]69,
ex. 2,) by annoxing a cipher to it, which, in effect, is mul-
tiplying it by 10, making 40 tenths, or 4*0. Then, as ma-
ny times as the denominator, 5 is contained in 40, so ma-
ny tenths, are contained in the fi'action. 5 into 4-0 goes 8
times and no remainder. »,^ns. *8 of a bush.
2. Express ^ of a dollar in decimal parts.
The numerator, 3, reduced to tenths, is f ^, 3*0, which,
divided by the denominator, 4, Tlie quotient is 7 tenths, and
a remainder of 2. This remi^inder must new be reduced to
hundredths by annexing another cipher, making 20 hun-
dredths. Tilth, as many times as the denominator 4, i.s
contained in S), so many hiindredihs also may be obtainec'.
4 into 29 goes 5 times, and no remainder. ^ of a dolljf,
therefore, reduced to decimals, is 7 tenths and 5 hundredth?,
that is, *75 of a dollar.
f -.^-i
V »
If 71. RBOUCTIOliI^ or VULGAR FRACTIONS, 6i^C.
Ul
The operation may be presented in form as follows :-
Num.
Dtnom. 4)3*0('75 of a dollar, the answer,
28
20
*J0
3. Reduce ^^ to a decimal fraction.
The numerator must be reduced to hundreths by annex-
ing two ciphers, before the division can begin.
66)4'00('0606-f-, the answer.
396
400
396
As there can be no tenths, a cipher must
be placed in the quotient, in tenths place.
4
Not^. -^g cannot be reduced exactly ; for, however lonj;
the division be continued, there v.-ili stxJ be a remainder.*
it is sufficiently exact for most putpoies, if the decimal be
ex tended to three or four places.
• Decimal figures which continually repp.at, like *06, in this ex-
ample, are called Repdends, or Circu'.aivr.g Decimals. If only
CM figure repeats, as '3333 or '7777, &c, it is called a single re-
petend: If two or more figures circulate alternately, as *060606,
'234234234, &c. it is called a compound repetend. If olhor fi*;-
ures arise before those which circulate as '743833, '1430I010I,
kc. the decimal is a mixed repetend.
A single repetend is denoted by writing: only the circulating fig-
ure, with a point over it thus : '3, signi es that the 3 is to hn
I'ontinually repeated, forming an infinite or never ending series
dVa.
A compound repetend is denoted b' i point over ihefirnt ami
laat repeating figure : thus, 234 sigi les that 234 is to be contin-
ually repeated.
It may not be amiss, here to show how the value of any repe-
tend may be found, or in other words, how it may be reduced lo
its equivalent vulgar fraction.
If we attempt to reduce |^ to a decimal, we obtain a con-
tinual repetition of the figure 1 : thus, *lllll, that is, the
repetend *i The value of the repetend *i then is ^; the val-
ue of '22,2, &,c. the repetend '2 will httwice as much; that
142
REDUCTION OF VULGAR F^ATIONS.
II Tl.
From the foregoing examples we may deduce the follow-
ing general Rule : To reduce a common to a dccimnl frac-
tion : — Annex one or more ciphers, as may he necessary, to
the numerator, and divide it by the denomiaator. If then
there be a remainder, annex another cipher, and divide as
before, and so continue to do so long as there shall continue
is, f. In the same manner, *3=f, and '4^|, and *5=^,
and so on to *9, which=f=:l.
1. Whatis the value of *^? Ans. |.
2. What is the value of 'Gt^w-s. |=f . What is the val-
ue of *3 of *7 ? of 4 ? of 5 ? of '9 ? ' 1?
If g'g be reduced to a decimal, it produces '010101, or
■ • • •
the repetend'Ol. The repetend '02, being 2 times as much,
• • • •
must b*e /^ and *03=:^\, and *4 8, being 48 times as much,
must be |f , and '74=Jf , &c.
If g^y be reduced to a decimal, it produces '001 ; conse"
quently, *002==5f^, and '037 = ^/^, and 425==|f f , &c.
As this principle will apply to any number of places, we
have this general Rule, for reducing a circuhting decimal
to a vulgar fraction. — Make the given repetend the numer-
ator, and the denominator will be as many 9s as there are
repeating ^figures.
3. What is the vulgar fraction equvialent to '704 ?
Ans. ffl.
4. What is the value of '003? 014? '324?
' 01021 ? —'2*463 ; —'002103 ? Ans. to the last,r^^^^\^.
i. What is the value of '43 ?
In this fraction, the repetend begins in the second place,
or place of hundredths. The first figure, 4, is -^xf, and the
repetend, 3, is ^ of ^, that is, /(, ; these two parts must be
added together. T*(T-|-^iy=M=^^, cms. Hence, to find
the value of a mixed repetend, — Find the value of the two
parts separately, and add them together.
6. What is the value of '153? ^■i'^%Tf^U=N^xj^ns.
7. What is the value of • 138 ? ' 16 ? ' 4123 ?
It is plain, that circulates may be added, subtracted,
multiplied, and divided, by first reducing them to their
equivalent vulgar fractions.
fl 71,72. REDUCTION OP DECfMAL FRACTIONS.
148
to be a remainder, or until the fraction shall be reduced to
any necessary degree of exactness. The quotient will be
the decimal required, which must consist of as many deci-
mal places as there are ciphers annexed to the numerator ;
and if there arc not so many figures in the quotient, the de-
ficiency must be supplied by prefixing ciphers
- • , : ■ • ■ • f
EXAMPLES FOR PRACTICE.
4. Reduce ^, l, 5^, and tt^^tj to decimals.
Ans. *5; '25; '025; '00797-f
5. Reduce f ^, j-^^xf, TT^a, and ^ttV\)f *« decimals.
Ans. '602-1- ; '003 ; '0028-}- ; 'OOOlSS-f-
6. Reduce' ^^1, -^^ft b^oo to decimals.
7. Reduce |, g^, -^h* i. f» tV. 'h> ^h to decimals.
8. Reduce i, g, |, ^, f, f , ^, ^, ^l, VV to decimals.
REDUCTION OF DECIMAL FRACTIONS.
1] 73. Fractions, we have seen, (^ 60) like integers,
are reduced from iow to higher denominations by division,
and from high to lower denominations by multiplication.
To reduce a compound num-
ber to a decimal of the high-
est denomination.
I. Reduce 7s. 6d. to the
decimal of a pound.
6d. reduced to the decimal
of a shilling, that is, divided
by 12, is '5s, which annexed
to the 7s, making 7'5s, and
divided by 20, is '375<£, the
answer.
The process may be pre-
sented in form of a rule, thus :
Divide the lowest denomina-
tion given, annexing to it one
or more ciphers, as may be
necessary, by that number
which it takes of the same to
make ojic of the next higher
denomination, and annex the
To reduce the decimal of(j^
higher denomination to inte-
gers of lower denomination.^.
2. Reduce '375<£ to inte-
gers of lower denominations.
ft '375<£redacedto shillings,
that is, multiplied by 20, is
7'50s. ; then the fractional
part, *50s, reduced to pence,
that is, multiplied by 12, is
6d. Ans. 7s. 6d.
That is, multiply the given
decimal by that number which
it takes of the next lower de-
nomination to make owe of this
higher, and from the right
hand of the product point ofiT
as many figures for decimals
as there are figures in the
given decimal, and so con-
144
REDUCTION OF DECIMAL FRACTIONS.
u n.
■4
•y
^•j
quotient, as a decimal to that
higher <leaoinination ; so con-
tinue to do, until the whole
■hall be reduced to the deci-
mal required.
EXAMPLES FOR PRACTICE.
3. Reduce 1 oz. 10 pwt.
to the fraction of a pound.
OPERATION.
20)10'0 pwt.
Hfe lb. Ans.
■ 1
5. Re<luce 4 cwt. 2f qrs.
to' the decimal of a ton.
Note. 2^=2*6.
7. Reduce 38 gals 3*52
qts. of beer to the decimal of
a hhd.
9. Reduce 1 qr. 2n. to ihe
decimal of a yard.
11. Reduce 17h. 6m. 43s.
to the decimal of a day.
13. Reduce 2l8. lOjd. to
the decimal of a guinea.
15. Reduce 3cwt. Oqr. 71b.
8oz. to the decimal of a ton.
tinue to do through all the de-
nominations; the several num-
bers at the lefl hand of the
decimal points will be the
value of the fraction in the
proper denominations.
EXAMPLES FOR PRACTICE.
4. Reduce *125lbTroy to
integers of lower denomina-
tions.
lb.
OPERATION.
'125
12
oz.
I'oOO
20
pwt. lO'OOO. Ans loz. lOpwt.
6. What is the value of
'2325 of a ton ? •
a What is the value of *72
hogshead of beer ?
10. What is the value of
'375 of a yard ?
12. What is the value of
"713 of a day?
14. What is the value of
•78125 of a guinea?
16. What is the value of
•15334821 of a ton?
Let the pupil be required to reverse and prove the fol-
lowing examples;
17. Reduce 4 rods to the decimal of an acre.
18. What is the value of '7 of a lb of silver ?
19. Reduce 18 hours, 15m. 50*4 sec. to the decimal of a
day. , .
20. What is the value of *67 of a league ?
Reduce 10s. 9^d. to the fraction of a pound.
^ 73. There is a method of reducing shillings, pence
fl 73, 74. REDUCTION OP DECIMAL FRACTIONS.
145
all the de-
eral num-
id of the
1 be the
on in the
ns.
ACTICE.
lb Troy to
lenomina-
oz. lOpwt.
value of
alue of '72
value of
value of
value of
value of
e the fol-
cimal of a
g9, pence
and farthings to the decimal of a pound, by inspection, more
simple and concise than the foregoing. The reasoning in
relation to it is ns follows :
•f\y of 20s. is 2s. ; therefore every 2s. is-j^g^, or '!,£. Ev-
ery shilling is T^Jj = yg^, or *0.5c£. Pence are readily re-
duced to farthings. Every farthing is ^^,7y<£. Had it so
Iiappened, that 1000 farthings, instead of 9(j0, had made a
pound, then every farthing would have been jj'^tt, or 'OOli^.
960-1-40=1000 ; that is, 3^ of 900 added to 960 is 1000.
Taking J^ of a numhor, and adding to that number, is the
same as mnltiphjiu^ the number by unity and the fraction
2^, Ij^lj-. Suppose you have the fractidn ^. If you mul-
tiply both the numerator, and denominator by 1^, you do
not change the value of the fraction. Do this, and you ob-
tain jUir- {\% then is equal to ^g^^r- ^^j is 24 fartli-
ings ; of course it follows that 24 farthings is equal to
^xMiT- Wherefore, if the number of farthings, in the
given pence and farthings, be »/orc than 12, ^^Ij-part will be
more than 4 ; therefore, add 1 to them ; if they be more than
36, 2V P"t will be more than l^J ; therefore add 2 to them ;
then call them so many thousandths, and the result will be
correct within less than ^ oC-^TiViT of a pound. Thus, 1 7s.
5fd is reduced to the decimal of a pound as follows r 16s=
ii«8and ls=^«05. Then 5£d=23 farthings, which, in-
creased by 1, (the number being more than 12, but not ex-
ceeding 36) is c£*024, and the whole is j£^*874, the answer.
Wherefore, to reduce shillings, pence and farthing. -i to the
decimal of a pound hy inspection, — Call every two shilliufrs
one tenth of a pound; every odd shilling, five hundredtli^;
and the number of farthings, in the given pence and far-
things, so many thousandths, adding one, if the number be
more than twelve and not exceeding thirtyrsix, and two if
the number be more than thirty-six.
fl TI. Reasoning as above, the result, or the three first
figures in any decimal of a pound, may readily be reduced
back to shillings, pence and fi^things, by inspection. Dou-
ble the Jfr.sf figure, or tenths, for shillings, and, if tlie second
figure, or hundredths, be Jive or fuore than five, reckon
another shilling ; then, after the five is deducted, call the fio-
ures in the second and third place so many farthino-s, abat-
N °
146
SUPPLEMENT TO DECIMAL FRACTIONS.
1174.
iBg one when they are above twelvei, and two when above
thirty-six, and the result will be the answer, sufficiently ex-
act for all practical purposes. Thus, to find the value of
*876£ by inspection : —
*S tenths of a pound - • - = 16 shillings.
*05 hundredths of a pound. - . =s 1 shilling.
*036 thousandths, abating 1, =35 farthings =3 s. 6j d.
<876 of a pound
= 17 8.
6id.
Am.
EXAMPLES FOR PRACTICE.
1. Find, by inspection, the decimal expressions of 9s. 7d.
and 12s. Of d. Ans. '479«£, and '603^^.
2. Find, by inspection, the value of '523 <£, and <694<£.
"Ans. 10s. 5^d., and 138. lOi^d.
3l Reduce to decimals, by inspection, the following sums,
and find their amount, viz : 15s. 3d. ; Ss. 11^. ; lOs. 6j^d. ;
Is. 8^d. ; ^d. and 2^d. ^ Amount, £V8dS.
4. Find the value of '47.^.
Note. When the decimal has but two figures, after tak*
ing out the shillings, the remainder, to be reduced to tkou"
sandths will require a cipher to be annexed to the right
heind, or supposed to be so. Ans. 98.. 4fd.
5. Value the following decimals, by inspection, and find
their amount, viz. '786^. ; '357^. ; '916£. ; '7i£. ; *&£. ;
♦25£. ; '09i:.; and '008^. Ans. 3^. 12s. lid.
SUPPLEMENT TO DECIMAL FRACTIONS.
QUESTIONS.
1. What are decimal rractions ? 2. Whence is the term derived?
3. How do decimals differ from common fractions 7 4. How are deci-
mal fractions written 1 5. How can the proper denominator to a deci-
mal fraction be known, if it be not expressed'? 6. How is the value
ol every figure determined 1 7. What does the firist figure on the right
b&nd of the decimal point signify 1 the second figure 7 — — the
third figure 1 — — (ourlh figure 1 8. Huw do ciphers, placed at the
rif^ht hand of decimals affect their value t 9. Placed at the left hand
how do ihey sffect their value 1 10. How are decimals read ? 11. How
are decimal fractions, having different denominators, reduced to a com*
mun denominator '/ 1'2. What is a mixed number ? 13. Howmayanf
whole numle be reduced to decimal paits 7 14. How can any mixed
number be read together, and the whole expressed in the form of a
common traction 7 15. What is observed respectins; the denomina*
1174.
8UPPLEMBNT TO DECIMAL FRACTIONS.
147
tions in federal money f 16. What is the rule for addition and sub*
traction of dacioials, particularly as respects placing the decimal point
in the results 1 inultii)lication 7 division ? 17. How is a com*
mnn or vulgar fraction reduced to a decimxl ? 18. What is the rule
for reducing a compound number to a decimal ofJ^e highest denomina-
tion contained in it ? 19. What is the rule fot finding the value of
any given decimal of a higher denomination in tdrms of a lower 1 20-
What is iho rule for reducing shillings, pence and farthings to the deci*
mal of a pound, by inspection? 21. VVhat is the reasoning in rela*
tion to this rule 1 32. How may the three first figures oi i^ny decimal
of a pound be reduced to shillings, pence and farthings, by inspection ?
EXERCISES. I
1. A merchant had several remnants of cloth, measuring
as follows :
7 1 yds. ^
U
3iV
«
«
u
tt
((
How many yards in the whole, and what WQuld
the whole come to at ^3*67 per yard ?
Note. Reduce the common fractions to deci-
mals. Do the same wherever they occur in the
examples which follow.
Ans. 36'475 yards. $133'863-t-,'cost.
2. From a piece of cloth containing 36| yds. a merchant ,
sold, at one time, 7^ yds. and at another time, 12 § yards ;
how much of the cloth had he left f Ans. 16*7 yds>
3. A farmer bought 7 yards of broadcloth for ^^8^, a
barrel of flour, for £^^^, a cask of lime for ^^If, and 7 lbs.
of rice for £^ ; he paid 1 ton of hay at £S^^, I cow &i
^6f , and the balance in pork at £-^ per tb ; how many
were the pounds of pork ?
Note. In reducing the common fractions in this example,
it will be sufficiently exact if the decimal be extended to
three places. Ans. 108f lb.
4. At 12^ cents per lb, what will 37f lbs of butter cost ?
Ans. $4'718^.
5. At $17'37 per ton for hay, what will U^ tons cost ?
Ans. $201'92|.
6. The, above example reversed. At $20l*92f for 11 §■
tons of hay, what is that per ton ? Ans. il7'37.
7. If '45 of a ton of hay cost $9, what is that per ton?
Consult ^ 62. Ans. $20.
6. At '4 of a dollar a gallon, what will '25 of a gallon of
molasses cost? Ans. 4 of a dollar.
t
148 SUPPLEMENT TO DECIMAL FRACTIONS. ^ 74.
9. At 9 dollars per cwt. what will 7 cwt. 3 qrs. 16 lbs. of
sugar cost 1
. Note. Reduce the 3 qrs. IC lbs. to the decimal of a cwt.
extending the defl^al in this, and the examples which fol*
low to /owr places. ',^- - v • -"^ Ans. 7r035-|-
10. At $69*875 for 5 "cwt. i qr. 14 lbs. of raisins, what
is that per cwt. Ans. $13.
U. What ^ill 2300 Ib^ of hay come to at 7 mills per lb?
12. What will 765^ lbs. of coffee come to at 18 cents
per lb? ^ ^w.s. $137'79.
13. What will 12 gals. 3 qts. 1 pt. of gin cost, at 28 cents
^ quart ?
Note. Reduce the whole. quantity to. quarts and the de-
cimal of a quart. ; ' .^ i4«5. $14*42.
14. Bought 16yds. 2qrs. 3na. of broadcloth for $100*125.
what was that per yard ? Ans. $6,
15. At $1*92 per bushel, how much wheat may be pur-
chased for $*72 ? Ans. 1 peck 4 qts.
: 16. At $92*72 per ton, how much iron may be purchased
■for $60*268/,:.,,; ,...,^ ,.;:"^^;«.^,P IV-;,.. Ans. 13cwt.
17. Bought a load of hay for $9*17, paying at the rate of
$16 per ton ; what was the weight of the hay ?
Ans. 1 1 cwt. 1 qr. 23 lbs.
. 18. At $302*4 per tun, what will 1 hhd. 15 gals. 3 qts.
of wine cost? . v • . .<:, . wi ^»5. $94*50.
19. The above reversed. At $94*50 for I hhd. 15 gals.
3 qts. of wine, what is that per tun ? Ans. $302*4.
Note. The following examples reciprocally prove each
other, excepting when there are some fractional losses, as
explained above, and even then the results will be sufficiently
exact for all practical purposes. If, however, ^rcafcr exact-
ness be required, the decimals must be extended to a greater
number of places.
20. At $1*80 for 3^ qts. of
wine, what is that per gallon ?
22. If f of a ton of potash
cost $60*45, whai; is that per
ton?
21. At $2*215 per gallon,
what cost 3^ qts ?
23. At $96*72 per ton for
potash, what will |^ of a ton
cost ?
1174.
REDUCTION OF CURRENCIES.
t4d
Reduction of Currencies.
In the United States, since the act of Conf ress in 1786, establishing
Federal money, calculations in aioney hare generfllr been niAde ill
dollars, cents and mills. In England, the denomtoations, though the
same in name as the currency of this Province, are diSbrent in value.
In the United States, previous to the act of Congress, it was the cus-
tom to reckon in pounds, shillings &c. ; and now, though all accounts
are kept in federal money, snnall' sums are mentioned frequently ia
these denominations. There are different currencies of the same nunc
in different parts of the United States. It may be necessary often in
commercial dealings, and in the course of ordinary business, to change
Tttues in foreign currencies into the currency of the Procinces.
Supposing there is a sum in federal money— $21 '604* We find by
the table of coins, IT 27, that 1 dollar is equal to 5 shillings, and of
course 4 dollars are equal to 1 pound, there being 4 times 5 snillings in
20 shillings. The faiue of pounds then, it is clear, is 4 times that of
dollars, and of course dollars are reduced to pounds by dividing tht
given sum, by 4.
4) 24 dollars. <
h ii Is.
There remain, however, $^604, 6 " . w and 4 mills to be changed to
Halifax currency. By reterrin; to Decimal Fractions, V 66, we 8e»
that dollars are the units in federal money, and cents and mills decimal
parts : cents hundredths, and mills thousandths. We have then simply
to divide these decimals of a dollar by 4, and the quotient will be In
decimal parts of a pound, thus :
4) '604 of a dollar.
*151 of a pound.
This an be reduced to shillings, pence and farthidgs by inspection,
(see IT 73) as follows : £451 equal to 3j. Od. Iqr. We find that $24<604
is equal io £6 3s. Od. Iqr. jins.
The following then, is the general rule to reduce federal money to
Halifax currency— (fit;t(2e the given sum by 4^ and the quotient will be
in pounds and decimal parts of a pound, which can be reduced to shil-
lingSf pence and farthings by inspection.
EXAMPLES FOR PRACTICE. %
Reduce $500 to Halifax currency, Jtns £125
do 27'304 do do ' 6 lis 6d !qr.
do I18'25 do do 29 lis 'M ■
do 236'50 do do 59 23 i'A
Reduce $490 to Halifax currency $56'03 $93 81 1 $836:^
i^l977'642 respectively to Halifax currency.
To reduce Halifax currency to federal money, we must reverse i!ie
process in the above examples. The rule is as follows ; Reduce thf;
siiillings, pence kc. if any, to the decimal of a pound, by inspection,
N2
150
REOUCTIOif Of CURRENCIES.
11 74.
and multiply the given turn by 4, the product will be in the denomina-
tiont of federal money.
EXAMPLES FOR PRACTICE.
Reduce £125 Halifax curreney to federal money. Jins. $500
do 69 9i 6d do do do 236'50
In order to change English sterling money, and the currencies in
some degree in use in the different parts of the United States, into
Halifax currency : since the denominations are the same in namci it will
be necessary te take some other currency, the denominations of which
are different, as a common object of comparison for these currencies,
and for HaliAui currency. By this means we shall be able to ascertain
the ▼ftlues of the former relaUvely to those of the latter. We will take
federal money as this common object of comparison, and will compare
with a unit of one of ite denominations, the dollar ^ one or more units
of a denomination of Halifax currency, and the before mentioned cur*
rencies, the thiUing.
In Halifax currency - 5s. =$1./
In English, or sterling money* '(4s. 6d.)4^s.=$l.
In New England currency - • - - 6s. =$1.
In New York currency 8s. =$1.
Iiv Pennsylvania currency (7s. 6d. ) - - 7^s.= $ 1 .
In Halifax currency 5s. are equal to $1, and in English
sterling money, 4^s. are equal to $1. Sterling money, then,
is to Halifax currency as 5 to 4J-, or to avoid the fraction »
as 10 to 9, since 2X5=10, and 2x4^=9. Therefore, to
change sterling money into Halifax currency, multiply by
*^, or, take once the given sum, and add ^, thus —
9) <^48 13s 9d sterling money.
5 8 1
54 10 Halifax currency.
Reduce J£56 17s. 6d. sterling to Halifax currency. Ms. £C3 3s. lOd.
do 92 4s. 6d. do do do
In New England currency, 6s. are reckoned to the dollar. New
England currency, then, is to Halifax currency as 5 to 6. There foie,
to reduce New England currency to Halifax currency, take live'sixths
of the given sum, thus —
6) J£14 5 4 New England currency.
2 7
5f
5
<£M 17 4^ Halifax currency.
* Without Premium, which varies from 5 to 8 per cent.
-'•an
=;J>T,-,^jip^W.
n 74, 75.
INTEREST.
151
Reduce J^60 4s. lOd. New England currency to Halifax
currency. Ans. £50 4s O^d.
To reduce Halifax currency to sterling money, or to re*
duce Halifax to New England, it is only necessary to re-
verse the process in the foregoing operations.
To reduce sterling to Halifax, we multiply by ^, there-
fore, to reduce Halifox currency to sterling tmoney,---divide
the given sum by \p, or, what is the same, multiply by -^^y
that is, take -^ of the given sum ; e. g.
10) ^54 10 Halifax currency.
5 8 1, • ' ■ :■
9
jB48 12 9 sterling money.
In the same manner, to reduce Halifax currency to New
England, — take f of the given sum, or, add \ to the given
sum.
From the foregoing rules and illustrations the pupil him-
self will be able, by pursuing a similar course, to reduce,
with facility, any currency, the denominations of which are
pounds, shillings, &c. to any other in which the denomina-
tions are the same.
The following is the general rule for finding a multiplier
to reduce any currency to the par of anotlier : Make the
number of shillings that are equal to a dollar in the cur-
rency to he reduced^ the denominator of a fraction ; and over
this, for enumerator, write the number of shillings that are
equal to a dollar in the currency to which the given sum is
to be reduced.
Let the pupil find multipliers to reduce New York and
Pennsylvania curreficies to Halifax, and then Halifax cur-
rency to those.
lil[TER£8T.
U TiS. Interest is an allowance made by a debtor to a
creditor for the use of money. It is computed at a certain
number of pounds for the use of each hundred pounds, or
so many dollars, for each hundred dollars, &/C. one year,
and in the same proportion for a greater or less sum, or fgr
a longer or shorter time.
15a
INTEREST.
1175
'01.
'005.
The number of pounds so paid for the use of a hundred
pounds, one year, is called the rate per cetU or per centum ;
the Yfotda per cent, or per centum signifying 6^ the hundred.
The highest rate allowed by law m the Canadas is 6 per
cent,* that is, 6 pounds for 100 pounds, 6 shillings for 100
shillings ; in other words, j%ij of the sum lent or due is paid
for the use of it one year. This is called legal interest, and
will here be understood when no other rate is mentioned.
Let us suppose the sum lent or due to be one pound,
The hundredth part of one pound or ^jj of a pound is, de-
cimally expressed, thus, '01, and yf^ of a pound, the legal
interest, written as a decimal fraction, is '06. So of any
rate per cent.
1 per cent expressed as a common fraction, is
t^jy ; decimally,
1 per cent is a half of one per cent, that is,
i per cent is a fourth of one per cent, that is, - '0025.
^ per cent is three times a quarter per cent, that is, '0075.
Note. The rate per cent is a decimal carried to two places
that is, th hundredths ; all decimal expressions lower than
hundredths are parts of one per cent. |- per cent, for
instance, is '625 of I per cent, that is, '00625.
Write 2^ per cent as a decimal fraction.
2 per cent is '02, and ^ per cent is '005. Ans. '025.
Write 4 per cent asLa decimal fraction. 4^ per cent
— 4f per cent. 5 per cent. 7^ per cent.
8 per cent. 8f per cent. 9 per cent. 9| per
cent. 10 per cent. (10 per cent is ^V^; decimally '10)
10^ per cent. 11 per cent. 12 J per cent.
15 per cent. /
1. If the interest of one pound for a year be '06 of a
pound, what will be the interest on ^5 for the same time?
It will be 25 times 6 or 6 times 25, which is the same
thing : —
25
'06
1*50 answer; that is, ^1 and 5 tenths The 5 tenths
must be reduced to shillings, pence and fart i. as by th e rule
• In the New England States the legal ra' ; is the same as in
the Canadas. In England it is 5 per cent.
1175.
INTEREST.
153
'01.
•005.
•0025.
, •OOTS.
places
oer than
ent, for
for the reduction of decimals; or with sufficient exactness
by inspection. See U 73. '50, or 'o of a pound equal 10
shillings. The interest of <£25 for a year is then £1 10s.
To find the interest on any sum for one year, it is evi-
dent we need only multiply it by the rate per cent written
as a decimal fraction. The product, observing to place the
point as directed in multiplication of decimal fractions, will
be the interest required.
Note, Principal is the money due, for which interest is
paid. Amount is the principal and interest added together.
2. What will be the interest of of 32 3s. for one year, at
4^ per cent ?
Vie are to multiply the principal by the rate per cent, 4 j-,
expressed '\h the form o^a decimal '045 ; we must therefore
reduce the 3s. in the principal, to decimal's by inspection.
We find 3s. equal to '15. There being five decimal places
£2li^\^ principal. in the multiplicand and mul-
•045 rate per cent. , tiplier, 5 figures must be point-
ed off for decimals from the
• 16075 product, which gives the ans-
12860 ^ wer 1 pound and 44675 hun-
' dred thousandths. Anything
<£1'44675 less than thousandths need not
be regarded ; hence, .£1*446 is sufficiently exact for the
answer. The *446 must be reduced to shillings, pence and
farthings by inspection. Double the '4 for shillings, equals
8s; call the *046 so many farthings, deducting 2, because
one 36 equals 44 farthings. In 44 qrs. there are lid.
^*446=8s. lid. The interest, then, of ^32 3s. for one
year, at 4^ per cent, is .£1 8s. lid. answer.
Always, then, if there are shillings, pence and farthings,
or either denomination, in the given principal, reduce them
to the decimal of a pound by inspection, before multiplying
hy the rule. After obtaining the answer in decimals, reduce
the tenths, hundredths and thousandths to shillings, pence
and farthing A, by inspection. The method of effecting each
reduction, is exhibited in U 73 and 74, and must be made
perfectly familiar to the pupil's mind. <
3. What will be the interest of £\ 1 3s. 4d. for one year,
at 3 per cent ? at 5^ per cent ? -' at 6 per cent ?— ■ —
at 1\ per cent ? at 8^ per cent 1 at 9f per cent ?
,>
"M^r-
154
INTEREST.
1175,76. I fl76.
at 10 per cent? at 10^ per cent' at 11 per
cent? at 11^ per cent? at 12 per cent? at
12^ per cent ? '
4. A tax on a certain town is £406 ISs. 102d. on which
tKe collector is to receive 2^ per cent for collecting ; what
will he receive for collecting the whole tax at th -' rate ?
In this example, the shillings, &lg. reduced to the decimal
of a pound equal '795. Multiply therefore, <£406'795 by
the rate 2^, thkt is '025. The answer, in decimals, is
<£10'169; the tenths, d&c. reduced to shillings, &c. equal
3s. 4^ The answer then, is i^lO 3s. 4d.
Note. In the same way are calculated commission, insu>
ranee, buying and selling stocks, loss and gain, or anything
else rated at so much per cent without respect to time.
5. What must a man, paying 37^ per cent on- his debts,
pay on a debt of ^£132 5s. ? Ans. <£49 Us. iO^d.
6. A merchant having purchased goods to the amount
of <£5d0, sold them so as to gain 12^^ per cent, and in the
same proportion for a greater or less sum ; what was his
whole gain, and what was the whole amount for which he
sold the goods.' Ans. His whole gain was £12 10s. ; whole
amount, £652 10s.
7. A merchant bought a quantity of goods fojr £173 15s.
how much must he sell (hem for to gain 15 per cent ?
- Ans. £199 16s. 3d.
51 76* Commission is an allowance of so much per
cent to a person called a correspondent , factor^ or hroher^
for assisting merchants and others in purchasing and selling
goods.
8. My correspondent sends me word that he has pur-
chased goods to the amount of £1286 on my account ; what I
will his commission come. to at 2^ per cent ? Ans. £32 3s.
9. What must I allow my correspondent for selling goods
to the amount of £2317 9s. 2^. at a commission of 3^ per |
cent? -4ns. £75 6s. 4d.
Insurance vis an exemption from hazard, obtained by
the payment of a certain sum, which is generally so much [
per cent on the estimated value of the property insured.
Premium is the sum paid by the insured for the insurance.
i '.^ 'isc- -N^,'. -A; ' ■
H 75, 76. I II 76.
INTEREST.
155
Policy is the name given to the instrument or writing,
by which the contract of indemnity is effected between the
insurer and insured. ^
10. What will be the premium for insuring a ship from
Montreal to Liverpool, valued at 9450jP, at 4^ per cent ?
« Ans. <£425 5s.
11. What will be the annual premium for insurance on
a house against loss by fire, valued at 875^ at f per cent '?
By removing the separatrix 3 figures towards the left, it
is evident, th6 sum itself may be made to express* the pre-
mium at 1 per cent, of which the given rate parts may be
taken; thus, one per cent on 875x is 8*75 and f of 375^
a&5m£. Ans. 6£ lis. 3d.
12. What will be the premium for insurance on a ship
and cargo valued at 6310<£ at ^ per cent ? at f-per
cent 1 at ^ per cent ? at f per cent .^ at |
per cent? Ans. at | per cent the premium is d9£ 7s. 8^d.
Stock is a general name for the capital of any trading
company or corporation, >or of a fund established by gov-
ernment.
The value of stock is variable. When 100 pounds of
stock sells for 100 pounds in moneys the stock is said to be
atjpar, which is a Latin word signifying equal; when for
more^ it is said to be above ^^ar; when- for less, it is said to
be below par.
13. What is the value of 756^ of stock, at 12^ per
cent ? that is, when 1 pound of stock sells for 1 pound 12^
hundredths in money, which is 12j- per cent above par, or
12^ per cent advance, as it is sometimes called.
Ans. 850ir lis.
14. What is the value of 3700^ of bank atock, at 95^
per cent 1 that is 4^ per cent below par ? Ans. 3533£ lOe.
15. What is the value of I20je of stock, at 92^ per cent ?
at 8QI per cent? at 67| per cent ? at 104^
per cent? at 108^ per cent? at 115 per cent?
at 37^ per cent advance ?
Loss AND Gain. 16. Bought a hogshead of molasses
for 15<£; for how much must I sell it to gain 20 per cent ?
Ans. 18je.
17. Bought broadcloath. at 12s. 6d. per yard; but, it be-
r ill
156
INTEREST.
TT 76, 77.
I
lag damaged, I am willing to sell it so is to lose 12 per
cent; how much will it be per yard? Ans. Us.
18. Bought calico at Is. per yard ; how must I sell it to
gain 5 per cent ? 10 per cent? 15 per cent ? .
to lose 20 per cent? Ans. to the Iast,9J^(l.
U 77m We have seen how interest is cast on any sum
of money when the time is one year ; but it is frequently
necessary to cast interest for months and days.
Now, the interest on \£ for 1 year, at 6 per cent, )eiiig
*06, is
*01, one hundredth for 2 months,
*005 five thousandth (or^ a hundredth) for 1 month of 30
*. ) days, (for so we reckon a month in casting inter-
-liff" -; 'f est,) and ' ; ' ■ ^vul >■ :^;' »<■ . >
*001 one thousandth for every 6 days ; 6 being contained
5 times in 30.
Hence, it is very easy to cast in the mind, the interest
on 1.^, at G per cent for any given time. The hundredth^
it is evident, will be equal to half the greatest even num-
ber of months ; the thousandth will be 5 for the odd month,
if there be one, and 1 for every time 6 is contained in the
given number of the days.
Suppose the interest of 1=£, at 6 per cent, be required
for 9 months and 18 days. The greatest even number of
the months is 8, half of which will be the hundredths *04;
the thousandths, reckoning 5 for the odd month, and 3 for
the 18 (3X6=18) days, will be *008, which, united with
the hundredths (*048) give 4 hundredths and 8 thousandths;
4 hundredths, and 8 thousandths, or, *048j6 redticedz=:lld.
Ans. lid.
1. What will be the interest on \£ for5 monthsGdays?
G months 12 days? 7 months 1 8 months
34 days ?
11 months
6
9 months 12 days?
days ? > 12 months
months 6 days ? — ;— 16 months?
10 months?
18 days? —
15
Odd Days. — 2. What is the interest of £1 for 13 months
16 days .5' ' ''■ ■
The hundredths will be 6, and the thousandths 5, for the
odd month, and 2 for 2 times 6 === 12 days, and there is a
remainder of 4 days, the interest for which will be such
;. i^.^i
^ 76, 77. I fl 77.
INTERKST.
w
)se 12 per
Ans. lis.
I sell it to
cent ?
e last, 9^(1.
Ml any sum
1 frequently
cent, )eing
nonth of 30
isting inter-
r contained
the interest
hundredth,
t even num-
odd month,
ained in the
be required
number of
[redths'04;
|h, and 3 for
united with
;housandths;
ced=lld.
Arts. lid.
|nths6days?
8 months
.nths? ■
s? 15
13 months
IS 5, for the
Id there is a
IviH be such
part of 1 thousandth as 4 days is part of 6 daya, that is, |
ss $ of a thousandth. Ans. '067f .
3. What will be the interest of £1 for 1 month 8 days ?
2 months 7 days ? 3 months 15 days ? ~<— 4
months 22 days ? 5 months 1 1 days ? 6 months
17 days ? 7 months 3 days? 8 months U days ?
9 months 2 days? 10 months 15 days ? 11
months 4 days ? 12 months 3 days ?
Note. If there is no odd months and the number of days
be less than 6, so that there are no thousandths, it is evident,
a cipher must be put in the place of thousandths ; thus, in
the last example, — i2 months 3 days, — the hundreths will
be '06, the thousandths 0, the 3 days } a thousandth.
Ans. Is. ^d.
4. What will be the interest of £\ for 2 months 1 day ?
. 4 months 2 days ? 6 months 3 days ? 8 months
4 days? 10 months 5 days ? for 3 days?
for 1 day ? for 2 days ? for 4 days ? for 5 days ?
5. What is the interest of £56 2s. 7f d. for 8 months 5
days.' The interest of £1, for the given time, is '040^;
therefore,
i) and ^)jf5643 principal.
'040|
interest of £1 for the given time.
224520 interest for 8 months.
> 2806 interest for 3 days. v ,
1871 interest for 2 days. *
^'29197 =.£2 5s. 9fd. .
5 days=3 days -f-2 days. As the multiplicand is taken
once for every six days, for 3 days take ^, for 2 days take ^,
of the multiplicand. ^-|- ^= |^. So also, if the odd days
be 4 = 2 days -f-2 days, take ^ of the multiplicand twice ;
for 1 day, teike |.
Frcwn the illustrations now given, it is evident, — Tojin^d
the interest of any sum in Halifax currently, or any other cur-
rency of which-the denominations are pounds, shillings, &c.
at 6 per cent, it is only necessary to multiply the given prin-
cipal, after having reduced the shillings and pence in it to
the decimal of a pound by inspection, by the interest of \£
for the given time, found as above directed and written as
15H
INTEREST.
■ II 77.
a decimal fraction ; after pointing off as many places for
decimals in the product hh there are decimal places in both
the factors counted together, these can be reduced back
ttgain to shillings and pence by inspection^
^ iout '— • EXAMl'IiES FOR I'RACITICE. liUI".,
6. What is the interest of .^'87 3s. Of d. for 1 year 3
months? Ans. £6 lOs. 9^d.
7. Interest of .€116 Is. 7f for 11 mo. 19 days?
Ans. cf6 158. 0|d.
Interest of .£200 for 8 mo. 4 days ? £8 2s. Tfd.
of 17s. for 19 mo. ? Is. 7|d.
of £8 lOtf. for 1 year 9 mo. 12 days ?
18s. 2fd.
of ^675 for 1 mo. 21 days ? £5 14s. 8|d.
9.
10.
.«(
11.
12.
13.
H.
15.
16.
17.
of .£8673 for 10 days?
of 14s. 7id. for 10 mo. ?
of i:96 for 3 days ?
of .£73 10s. for 2 days ?
of ^180 158. f rSdays?
of jeiSOOO for 1 day /
Handth, the pounds themselves express the interest in thou-
sandth for six days, of which we may take part"
Thus, 6)15000 thousandths,
<(
(t
t(
((
£U 9s. l^d.
8jd.
1 Note. The
I interest oi'£\
J for 6 days be-
J ing 1 thou.
2'500, that is, £'2 10s. Ans. to the last.
When the interest is required for a large number of years,
it will be more convenient to find the interest for one year,
and multiply it by the number of years ; after which find
the interest for the months and days, if any, as usual.
18. What is the interest of £0)00 for 120 years?
Ans. .£7200.
19. What is the interest of ^520 Os. Of d. for 30 years
and 6 months? Ans. .£951 13s. 5fd.
20. What is the interest on .^400 for 10 years 3 months
and 6 days? ^W5. .£246 8s.
21. What is the interest of .£220 for 5 years ? for
J 2 year^ ? 50 years ? Ans. to the last, .£660.
i^2. What is the amount of £"86, at interest 7 years ?
Ans. £1^2 2s. 4|d.
23. What is the interest of $48*30 for 1 year?
It must be clear to the pupil's mind, that to obtain the
II 77. I H 78, 79.
INTEREST.
IfiO
interest upon any sum in federal money, fur any time, we
proceed just as wo do in Il.ilifax currency ; only' we are not
coiupclled to reduce any part of the given sum to decimals,
since all the denominations of federal money arc in a deci-
mal ratio. T.ie answer to the last example is $!2,899.
What is the interest of IG4 for 2 years ? Ans. $VGS.
Whit is the interest of $i)8'50 for 7 years, months and
10 days ? Ans. >I44'489.
5T «&i. 1. What is the interest of 36 pounds for 8
months, at 4j- per sent ? _■,. . • • i i . ^ m
Note. When the rate is any other than six per cent, first
find the interest at six per cent, then divide the interest so
found by such part as the interest, at the rate required, ex-
ceeds or falls short of the interest, at six per cent, and the
quotient added to or subtracted from the interest at six per
cent, as the case may be, will give the interest required.
£m
'04 i^ per cent is f of six per cent ; therefore
from the interest at six per cent subtract ^ ;
^)144 the remainder will be the interest at 4^ per
'36 cent. . , ,, ,
£ViM £i Is. 7id. answer.
3. Interest of <^54 16s. 2|d. for eighteen months, at tive
per cent ? Ans. £4 2s. 2J^d.
3. Interest of ^500 for nine months and nine days, at
ciorht per cent? Ans. £Sl.
\ Interest of ^62 2s. 4Jd. for one month and twenty
days, at four per cent ? Ans. 6s. lO^d.
5. Interest of .£85 for ten months and fifteen days, at
12} per cent ? Ans. £9 Ss. lOfd.
6. What is the amount of .£53 at ten per cent for seven
months? Ans. £5& Is. 9^d.
The time^ rate per cent and amount given, to find the
principal.
ff 7d» 1. What sum of money, put at interest at 6 per
cent, will amount to £61 Q3. 4f d. in I year 4 months ?
The amount of ^l at the given rate and time i& <^r08;
hence ^6l*02-i-£l'08=.56'50, the principal required; that
is, find the amount of ^1 at the given rate and time, by
which divide the given amount ; the quotient will be the
principal required. Ans. <£56 lOs.
160
INTEREST
1IT9.
2. What principal, at 8 per cent, in 1 year 6 months,
will amount to .£85 2s. 4f d. 1 Ans. .£76.
3. What principal, at 6 per cent, in 11 months 9 days,
will amount to .£99 6s. 2f d. ? An&. ^94.
4. A factor receives .£988 to lay out after deducting his
commission of 4 per cent ; how much will remain to be
laid out 1
It is evident he ought not to receive commission on his
own money. This question, therefore, in principle, does
not differ from the preceding.
NoU. In questions like this, where no respect is had to
time, add the r.att to i^l. Ans. .£950.
5. A factor receives J61008 to lay out after deducting his
commission of 5 per cent ; what does his commission
amount to .' Ans. .£48.
jj^
llii'i
Discount. — 6. Suppose I owe a man .£397 10s. to be
paid in 1 year, without interest, and I wish to pay him now,
how much ought I to pay him when the usual rate is 6 per
cent ? I ought to pay him such a sum as, if put at interest,
would, in one year, amount to jf397 10s. The question,
therefore, does not differ from the preceding. Ans. if 375.
Nott. An allowance made for the payment of any sum of
money before it comes due, as in the last example, is called
discmtnt^
The sum which, put at interest, would, in the time and
at the rate per cent for which discount is to be made, amount
to the given sum, or debt, is called the present worth.
7. What is the present worth of .£834 payable in 1 year,
7 months and 6 ()ays, discounting at the rate of 7 per cent ?
Ms. .£750.
8. What is the discount on .£321 12s. 7^d. due 4 years
hence, discounting at the rate of 6 per cent ?
Ans. ^62 5s. 2f d.
9. How much ready money must be paid for a note of
£'18, due fifteen months hence, discounting at the rate of 6
per cent.' Ans. £16 14s. lO^d.
10. Sold goods for .£650, payable one half in 4 months,
and the other half in 8 months ; what mqst be discounted
for present payment? 4w5, £\S.
The
fl 80, 81, 82/
INTEKEST.
•161
11. What is the present worth of jf56 4s. payable in 6ne
year eight months, discounting at 6 per cent? at 4^ per
cent ? "at 5 per cent ?- — at 7 per cent 7 at 1^ per ^
cent ? at 9 per cent ? Ans. to the last £A% 178. 4|d.
The time, rate per cent, and interest being given to find the
, principal
^ 80. i. What sum of money put at interest sixteen
months, will gain ^10 10s. at 6 per cent ?
£{ at the given rate and time, will gain *08 ; hence,
£lO'50-^Je'03=^l3^25, the principal required; that is—
find the interest of £\. at the given rate and time, by which
divide the given gain or interest ; the quotient will be the
•principal required. Ans. <£131 5s.
2. A man paid £i 10s. 4fd. interest at the rate of 6 per
cent at the end of 1 year 4 months ; what was the principal ?
Ans. £5& 10s.
3. A man received for interest on a certain note at the
end of one year ,£20; what was the principal, allowing the
rate to have been 3 per cent t Ans. .£333 6s. 8d.
The principal, interest and time being given, to find the
rate per cent./
^81. 1. If I pay ^3 15s. 7^^. interest for the use of
£36 for 1 year 6 months, what is that per cent ?
The interest on £36 at one per cent, the given time, is
£'54 ; hence £3*78-^-£'54='07, the rate required ; that is,
find the interest on the given sum, at one per cent, for. the
given time, by which divide the given interest ; the quotient
will be the rate at which interest was paid. Ans. 7 per ct.
2. At £2 6s. 9^d. for the use of £468 for a month, what
is the rate per cent ? Ans. 6 per cent.
3. At £46 16s. for the use of £520 for two years, what
is that per cent ? Ans. 4] per cent.
The prices at which goods are bought and sold, being given^
to find the rate per cent o/'gain or loss.
^ 89. 1. If I purchase cloth at £1 2s. a yard, and sell
ii at £1 7s. 6d. per yard ; what do I gain per cent ?
This question does not differ essentially from those in the
foregoing paragraph. Subtracting the cost from the price
02
16^
INTEREST.
■'^' !F82,83.
w
'io
\
at sale» it is evident I gain ^'275 on a yard ; that is f ,{&
of the first cost, f '^^='25 per cent, the answer. That
is,— make a common fraction, writing the gain or loss for
the numerator, and the price at which the article was bought
for the denominator , then reduce it to a decimal.
2. A merchant purchases goods to the amount of JS550 ;
what per cent profit must he make ^o gain <£66 ?
• * • ' Ans. 12 per cent.
3. What p^r cent profit must he make on the same
purchase to gain ^38 10s. ? to gain j£24 15s. ?
to gain <£2 15s. *
Note. The last gain gives for a quotient *005, which is
^ per cent. The rate per cent, it will be recollected, (IF 75,
note,) is a decimal carried to two places, or hundredths ; all
decimal expressions lower than hundredths are parts of one
per cent.
4. Bought a hogshead of liquor, containing 114 gallons,
at .£'96 per gallon, and sold it at £\ Os. Od. 3|qrs. per gal.
what was the whole gain, and what was the gain per cent ?
Ans. £4c 18s. 5f d. whole gain. — 4^ gain per cent.
5. A merchant bought a quantity of tea for £365, which,
proving to have beeij, damaged, he so!d for .^332 3s. ; what
did he lose per cent ? Ans. 9 per cent,
6. If I buy cloth at £'2 per yard, and sell it for <£2 10s.
per yard, what should I gain in laying out<i^lOO. Ans. £'26,
7. Bought indigo at 6s. per lb. and sold the same at 4s.
6d. per lb. ; what was the loss per cent ? Ans. 25 per cent.
8. Bought 30 hogshead of liquors at c£600 ; paid in duties
.£20 13s. 2f d. ; for freight £40 15s. 7|d. ; for porterage
£Q Is. ; and for insurance ^30 16s. 9|d, ; if I sell them at
£26 per hogshead, how much shall I gain per cent ?
Ans. 11*695 per cent.
The principal, rate per cent, and interest being given, to
Jind the time.
tl 83, 1. The interest on a note of <£36, at 7 per cent,
was ^3 15s. 7fd. ; what was the time?
The interest on .£36 for a year, at 7 per ct, is <£2 10s. 4f d.
£3'78-7-£2'52=:r5 years, the time required ; that is — find
the interest for one year on the principal given, at the given
rate by which divide the given interest j the quotient will
fl 82,83. I 1183.
INTEREST.
163
be the time required in years and decimc^' parts of a year ;
the latter may then be reduced to months and days.
Ans. 1 year 6 months.
2. If ^31 14s. 2fd. interest be paid on a note of £226
lOs. what was the time, the rate being 6 per cent ?
Ans. 2*33^=2 years 4 months.
3. A note of .£600, paid interest £20, at 8 per cent ;
what was the time ^
Ans '416-{-=:5 months so nearly as to be called 5, and
would be exactly 5, but for the fraction lost.
4. The interest on a note of £217 5s. at 4 per cent was
£28 48. lOd. ; what was the time ? Ans. 3 yrs. 3 moi^.
Note. When the rate is 6 per cent, we may divide the
interest by half the principal, removing the separatrix ^i^o
places to the left, and the quotient will be the answer in
months.
The method given above, of finding the mterest upon any
sum in Halifax currency, for any time, and at any rate,
will be found sufficiently exact in practice, and as simpTe
and concise, perhaps, as any that could be proposed.
The teacher will do well to see that the scholar under-
stands perfectly the process by which the reciprocal re-
ductions are effected by inspection, and the reason of this
process.
If greater exactness be required, the reductions can be
effected by the ordinary rules for the reduction of decimal
fractions.
The following is a method of casting interest by vulgar
fractions.
To obtain the interest upon any sum for any time, at
any rate : — Multiply the lowest terms of a fraction, the nu-
merator of which is the given rate, and the denominator
100, by the given number of years ; multiply the lowest terms
of a fraction, the numerator of which is the given rate, and
the denominator 1200, by the given number of months ;
multiply the lowest terms of a fraction the numerator of
which is the given rate, and the denominator 3600, by the
given number of days ; then reduce these several fraction.*)
to one common denominator ; add them together, and by
the resulting fraction multiply the given principal.
161
INTEREST.
II 8a
^1
tP(
or»
Find the interest of jf 100 for 2 yrs. 6 mo. 10 dy. at 6 per ct.
■^ (or tJjt) X 2 yewB^xr -'-''^^ ^'» '•
iritr (or xAiy) X 6 raonths=jf i^
¥I^I^ (or ss^ov) X 10 day8=ff J^^rr^ J,y
2§Tr> if^Tf V^ to be reduced to one common dene-
minator. , Neglect the ciphers in the denominators —
6 X 2 X 6=60; 1 4-2 -|- 2=5, the number of ciphers.
The common denominator is then 60 and 5 ciphers.
6 X 2 X 6=72 ; this with 4 ciphers is first numerator.
5 X 6 X 6=180 ; this with 3 ciphers is 2d numerator.
5 X 2 X 1=10 ; this with 3 ciphers is 3d numerator.
Each numerator has as many as 3 ciphers; cut offthtee
from each, and three from the common denominator ; ^^^^
'\'hU-^M^^x^u\=^i^' Then ^100, the given princi-
pal, multiplied by ^j^=£^^=:£\5 3s. 4d.
The reasons of the different steps in the foregoing pro-
cess will appear : when the rate, as in the above example,
is 6 per cent, it is obvious that the interest of ^ny given
principal for one year is yf^y or -^^ of that principal. For
any number of years, the interest must be as many times
^^y of the principal as there are units in the given number
of years. In the example, 2 is the given number of years ;
multiply then ^^ by 2 ; or multiply the lowest terms of a
fraction, tht numerator of which is the given rate, and the
denominator 100, by the given number of years, -^jj of the
given principal then is the interest for 2 years, y^jyof the
given principal is the interest for 1 month; for there are 12
months in a jear, and y^^ X iV = t^ott or g-^^. ^^^^^ of
the given principal is the interest for 1 day ; for there are
30 days in I month, and y/j^ X i^if = ^^xTu =FTrVir- We
have then -^^ of given principal, as the interest for 1 year ;
2"^ J of sam3, for 1 month, and ^a^jyjy for 1 day. For 2 years,
we have ^^y X 2 = ^(y: for 6 months ^^^ X f»=^ishl for
ten days, ^^Vt X 10=^^^^^=^^^- ^^, j^^ and ^^^ then
of the given principal are the interest of ,£100 for 2 years,
6 months and' 10 days. It is clear now, why we reduce
these several fractions to one common denominator, add
them together, and by the resulting fraction multiply the
given principal.
Find the interest upon £78 4s for 3 years, 9 months and
6i.diys, by this method, at 6 per cent and also at 5 per cent.
■, -i-x-—--.-^'
1184. •
•: ' INTEREST.
m
To find the intereet due on Notes, S^c. when partial pay-
ments have been made.
IT 84. There is no statute in this Province, prescribing
any particular form or method of casting interest upon
notes or other obligations. It is believbd the following
method is f^enerally allowed before the courts of the country,
and also is that which has obtained to the greatest extent
in mercantile transactions.
Rule. — Compute the interest upon the value for which
the note or other instrument was given, tp the time of pay-
ment, which add to the principal ; find the amount also of
each endorsement to the time of payment, which several
amounts add together, and the sum subtract from the
amount of the value upon the face of the note, or other
instrument. , : .
1. For value received, I promise to pay Louis Rousseau,
or order, one hundred pounds fifteen shillings, with interest.
iflOO 15s. . John Burton.
May 1, 1822.
On this note were the following endorsements . •
Dec. 25, 1822, received ^10
July 19, 1823, "
Sept. 1, 1824, "
June 14,1825, "
1 4s.
3 6s.
21 15s.
April 15, 1826, "
What was due Aug. 3, 1827.1'
54 9s.
Ans. £U 3s. Id.
The whole time is, from May 1st, 1822, to Aug. 3, 1827,
which is 5 years, 3 months, 2 days. The interest of .£100
153. for this time is ^31 15s. 4f d. This added to the value
for which the note h as given i8 £100 15s.-f-<£31 15s. 4|d.=
£iS2 10s. 4fd. which is equal to the amount of the value
for which the note was given. The first endorsement is
£10; the date of this endorsement is Dec. 25, 1822; the
time of payment is Aug. 3, 1827. The time, therefore, for
which interest is to be cast upon this endorsement, is 4 yrs,
7 mo. 8 ds. The interest for this time is £'2 15s. 3d.
which, added to the endorsement, makes its amount £\^
15s. 3d. In the same way find the amount of each other
endorsement, by casting the interest upon it from the day
of its date to the day of the payment of the note, and add
this interest to the principal, that is, the endorsement. .
,, ■■ (
16^
COMPOUND INTEREST.
The 2d endorsement is
3d
• 4th " :
5th
The time for which interest is to be cast upon the
2d endorsement is - -4 years, months, 23 days
3d " - - - 2 " 11 " 2 "
4th 'f - - - 2 " 1 " 19 "
^84,85. I ^^^'
£ 1 4s.
3 68.
21 15s.
54 9s.
5th
- 1
((
((
t(
it
((
((
The interest upon the 2d endorsement is
" '• 3d
4th
5th
The amount of the 2d endorsement is
»3d
" 4th
'" 5th
The amount of 1st endorsement we found to be 12
The sum of the ^mounts of all the endorsements 101
The value upon the face of the note is - 100
The amount of this value is - - - 132
Subtract the sum of amounts of endorsements 101
(<
((
((
£
2
4
1
3
24
58
18 "
s. d.
5 10
11 6f
15 8|
4 11^
9 10
17 GJ
10 8|
13 l\i
15 3
7 3f
15
10 4f
7 Si-
Balance due Aug. 3d 1827, ^ 31 3 1
2. For value received, I promise to pay Thomas Wilson,
or order, two hundred thirty-eight pounds eighteen shillings,
with interest.
£238 18s. Charles Stewart.
Jan. 6, 1820.
On this note were the following endorsements, viz :
April 16, 1823, received
April 16, 1825,
Jan. 1, 1826, "
• What was due July 11, 1827 ?
£
s.
rf
23
10
19
4
87
19
COMPOUND INTEREST.
Tf S5» A. promises to pay B. ^256 in three years, with
interest annually ; but at the end of one year, not finding it
convenient to pay the interest, he consents to pay interest
IF 84,85. I ^^^'
TEWART.
COMPOUND INTEREST.
Ki7
on the interest from that time, the same as on the principal.
Note. — Simple Interest is that which is allowed for the
principal only ; compound interest is that which is allowed
for both principal and interest, when the latter is not paid
at the time it becomes due.
Compound Interest is calculated by adding the interest
to the principal at the end o£ each year, and making the
amount the principal for the next succeeding year.
1. What is the compound interest of £256 for three
years, at 6 per cent ? i , '. '
• * £256 given sum or first principal.
'06
15 .^6 interest, > . u jj j * ^l
n-atnix • • i ? to be added together.
2oo*00 prnicipal, ) °
27r36 amount or principal for second year.
*06
16'2816 compound interest 2d year, ) added
271*36 principal, , do | together
2S7'6416 amount or principal for 3d year.
♦06
17*258496 compound interest 3d year, ) added
287*641 principal, . do « j together
304*899 amount.
256 first principal subtracted.
^'48*899 compound interest for three years.
Ans. £48 17s. llfd.
2. At 6 per cent, what will be the compound interest,
and what the amount of £1 for two years 7 what the
amount for 3 years ? for 4 years ? for 5 years ?
for 6 years ? for 7 years? for 8 years ?
Ans. to the last, £1 lis. lO^d.
It is plain that the amount of £2, for any given time,
will be two times as much as the amount of ill ; the amount
of £3 will be three times as much, &c.
168
COMPOUND INTEREST.
1185.
Hence, we may form the amounts of one pound, for seve-
ral years, into a table of multipliers for finding the amount
of any sum, for the same time.
TABLE,
Showing the amount of One Pound or One Dollar ^c.for
any number of years not exceeding 24, at the rates of 6
and 6 per cent Compound Interest.
Years
1
2
3
4
W0
o
6
7
8
9
10
11
12
5 per cent
1<05
1*1025
1*15702+
r215504-
l'27628-l-
l'34009-[-
1*40710-1-
1*47745-1-
1*55132-1-
1*62889-1-
1*71033-1-
1*79585-1-
6 per cent
1*06
V1236
1*19101+
1*26247-1-
1*33822-}-
1*41851-1-
1*50363-1-
1*693844-
1*68947-1-
1*79084-1-
l*89829-f-
2'01219-|-
Years
13
14
15
16
17
18
19
20
21
22
23
24
5 per cent
1*88564+
1*97993+
2*07892-1-
2*18287-1-
2*29201+
2*40661-1-
2*52695
2*65329+
2*78596-f-
2*92526-1-
3*07152+
3*22509-1-
6 per cent
2*13292+
2*26090-}-
2*39655-1-
2*54035-f.
2*692774-
2*85433-}-
3*02559-}-
3*20713-}-
3*39956+
3*60353-}-
3*81974+
4*04893-}-
Note 1. Four decimals in the above numbers will be
sufficiently accurate for most operations.
Note 2. When there are months and days, you may first
find the amount for the years, and on that amount cast the
interest for the months and days ; this added to the amount,
will give the answer.
3. What is the amount of X600 10s. for 20 years at 5
per cent, compound interest ? j- at 6 per cent ?
£1 at 5 per cent by the table is ^£2*65329 ; therefore,
2*65329X600*50=£l593*30+is £1593 6s. Ans. at 5 per
cent; and 3*20713X600*50= £l925*881+is £1925 17s.
7^d. ans. at 6 per cent.
4. What is the amount of £40 4s. at 6 per cent com-
pound interest, for 4 years .^ for 10 years? for 18
years ? for 12 years ? • for 3 years and 4 months?
for 24 years, 6 months and 18 days ?
Ans. to the last £168 2s. 8|d.
Note. Any sum at compound interest will, double itself
in 11 years, 10 months and 22 days.
i <
H 85. 1 ff 85.
COMPOOND INTBRE8T.
160
ibers will be
From what has now been advanced, we deduce the fi^
lowing general ....;..,;
RULE.
I. To find the interest when the time is one year, or, to
find the rate per cent on any sum of money, without respect
to time, as the premium for insurance, commission, &c. —
Multiply the principal or given sum, after having reduced
the shillings and pence in it to the decimal of a pound, by
the rate per cent, written as a decimal fraction ; afler point-
ing off as many places for decimals in the product as there
are decimals in both the factors, and reducing these deci-
mals back to shillings and pence, we shall obtain the inter-
est required.
II. When there are months and days in the given time,
to find the interest on any sum of money at 6 per cent, —
Multiply the principal, reducing the shillings and pence by
inspection, by the interest on one pound for the given time
found by inspection, and the product, as before, will be the
interest required, taking care ^o reduce the decimal parts
to shillings and pence by inspection.
III. To find the interest on one pound at 6 per cent,
for any given time by inspection, — It is only to consider
that half the greatest even number of months will denote
hundredths of a pound, and that there will be five thou-
sandths of a pound fo. he odd month, (if there be one) and
one thousandth for every six days.
IV. If the sum given be in federal mone^, — The deno-
minations bef ng in a decimal ratio, we are saved from the
necessity of effecting the reciprocal reductions, at the be-
ginning and end of the process, otherwise proceed precisely
as in Halifax currency.
V. If the interest required be at any other rate than six
per cent, (if there be months, or months and days in the
given time,) — First find the interest at six per cent ; then
divide the interest so found by such part or parts, as the in-
terest, at the rate required, exceeds, or fsdls short of the
interest at six per cent, and the quotient, or quotients, ad-
ded to or subtracted from the interest at six per cent, as the
case may require, will give the interest at the rate required.
Note. The interest on any number of pounds, for 6 days
at 6 per cent, is readily found by cutting off the unjit or
P
170
.1
INTIREST. %i:or)
!185.
3^!
right hand figure ; those at the left hand will «how the in-
terest in hundredths for 6 days.
'>')
EXAMPLES FOR PRACTICE.
1. What is the interest of J^1600 for 1 yoar 3 months?
^ Ans. £120.
^S.JW'hat is the interest of i£5 16s. for 1 year IL months?
Ans. 13s. 4d.
3. What is the interest of <£2 5s. 9^d. for 1 month 19
days, at 3 per cent f ■ Ans. 2^d.
4. What is the interest of <£18 for 2 years 14 days at 7
per cent ? Ans. £'2 1 Is. 4^d.
5. What is the interest of i^l7 13s. 7^d. for 11 months
28 days? Ans. £1 Is. Id.
6. What is the interest of .£200 for 1 day? 2 days?
3 days ? 4 days ? 5 days ?
Ans. for 5 days, 3s. 3f d.
7. What is the interest of half £'001 for 567 years?
. , i. ; , Ans. 4d.
8. What is the interest of £81 for 2 years 14 days, at }
per cent ? | per cent ? | per cent ? 2 per
cent ? ■ 3 per cent ? I j- per cent ? 5 per cent ?
per cent ?
9 per
7 per cent ?
cent ? 10 per
7^ per cent ? 8 per
cent ? 12 per
12^ per cent? Ans. to last, £20 12s. lO^d.
What is the interest of £*09 for 45 years, 7 months,
11 days? ^»5. 48. 10|d.
10. A.'s note of £175 was given Dec. 6, 1798, on which
was endorsed a year's interest; what was due 1st Jan. 1803?
Note. Consult Ex. 16, Supplement to Subtraction of
Compound Numbers. Ans. £207 4s. 4fd.
11. B.'s note of £56 15s. was given June 6, 1801, on
interest after 90 days; what was there due 9th Feb. 1802?
Ans. £58 3s. d^d.
12. C.'s note of £365 was given Dec. 3, 1797 ; June 7,
1800, he paid £97 3s. 2id. ; what was there due 11th Sept.
1800? " ^ns. £327 Os. 7id.
13. Supposing a note of £422, dated July 5, 1797, on
which were endorsed the following payments, viz. Sept. 13,
1799, £208 4s. ; March 10, 1800, £96; what was there due
Ist Jan. 1801 ?
1185/
,;r, 7«.^'T
tom.iMCli¥'Tb iilYBiietT.
Snpplement to Interest.
QUESTIONS.
171
1 . Whdt ii interest f 2. How ia it computed 1 3. What is uodor-
(tood by rate per cent 7 4. by principal t 5. — — > by amount f
6. by legal Interest ? 7. by commiisionl^ 8.— —Insur-
ance f 9. premium t 10. policy ? 11. Stock t 12.
What is understood by stock bein); at par ? 13. ahovo par T 14.
below par? 15. The rate per cent is a decimal carried to bow
many places I 16. What are decimal expressions lower than hun-
dredths 1 17. How is interest (when the time is one year) eommis-
sian, insurance, or anything else rated at so much per cent without
respect to time, found? 18. When the rate is one per cent, or less,
how majf the operation be contracted t 19. How is the interest on one
pound at 6 per cent, t>r any givsn time, found by inspection ? 20.
How is interest cast at 6 per cent, when there are months and days
in the given time 1 21. When the given time is less than 6 days, how
is the interest most readily found ? 22. If the sum given be in federal
money, how is interest cast 1 23. When the rate is any other than 6
per cent, if there be months and days in the given time, how is the in-
terest found ? 24. What is the rule for casting interest on notes, Ac.
ffhen partial payments have been made ?.. 25. How may the principal
be found, the time, rate per cent and amount being given ? 26. What
is understood by discount ? 27, — — by present worth 1 28, How
is the principal found, the time, rate per cent and interest being given?
29. How is the rate per cent of gain or loss found, the prices at which
^oods are bought and sold bein^ given ? /30. How is the rate per cent
found, the principal, interest and time being given ? 31. How ia the
time found, the principal, rate per cent and interest being given 1 32.
How may interest be cast by vulgar fractions ? 33. What is the
reasoning in regard to this rule f 34. What is simple interest ? 35.
compound interest 1 36. How is compound interest computed ?
EXERCISES. '
1. What is the interest of £279 10s. ^d, for 1 year 10
days, at 7 per pent? v s ; ' " 't Ans. i£19 13s. 6^.
2. What is the interest of J?486 for 1 year 3 months 19
days, at 8 per cent ? Ans. £50 138. 4Jd.
3. D.'s note of ^6203 was given Oct. 5, 1808, on interest
after 3 months; Jan. 5, 1809, he paid ^50; what was there
due 2d May, 1811? Ans. .£175 7s. 2d.
4. E.'s note of ^870 was given Nov. 17, 1800, on inter-
est after 90 days ; Feb. II, 1805, he paid ^186 ; what was
there due 23d Dec. 1807 ? Ans. ^1009 lis. 6fd,
5. What will be the annual insurance, at f per cent, on
a house valued at ^1600 ? Ans. £\9,
m
ii< '>:f
m
uiinum$s:^ to i»TiREfi'»
1186.
6. What will be the insurance of a thip and cargo, valued
at £5643 at 1^ per cent ? at ^ per cent ? at fg
per cent ? at -f^ per cent ? at ^ per cent ?
Note. Consult II 76, ex. 11. Arts, at ^ per cent i^42 Gs. ^d.
7. A man having compromised with his creditors at (i'2^
per cent, what must he pay on a debt of <;fl37 9s. 2jd. ?
Ans. ;£85 188. 3d.
8. What is the value of £800 Montreal Bank stock, at
1 12 j- per cent / Ans. .£900.
9; What is the value of <£560 15s. of stock, at 93 per
cent? iltt5. Je521 9s. ll^d.
10. What principal, at 7 per cent, will, in 9 months 18
days, amount to <£422 8s. ? Ans. .£400.
11. What is the present worth of ^426, payable in 4
years 12 days, discounting at the rate of 5 per cent ?
In large sums, to bring out hundredths and thousandths
correctly, it will sometimes be necessary to extend the de<
cimal in the divisor to five places. Ans. £354 10s. l^d.
12. A merchant purc|iased goods for £250, ready money,
and sold them again for £300, payable in 9 months ; what
did he gain, discounting at 6 per cent ? Ans. £37 1 s. 7^d.
13. Sold goods for £3120, to be paid one half in three
months, and the otheV half in six months ; what must be
discounted for present payment ? Ans. £6S 9s. lOd.
14. The interest on a certain note for 1 year 9 months
was £49 17s- 6d. ; what was the principal ? Ans. £475.
15. What principal, at 5 per cent, in 16 months 24 days,
will gain £35? ilns. £500.
16. If I pay £15 lOs. interest for the use of £500, nine
months and nine days, what is the rate per cent ?
17. If I buy candies at $'167 per Id, and sell them at
20 cents, what shall I gain in laying out $100 ?
^115. $19*76.
18t Bought hats at 4b. a-piece, and sold them again at
48. 9d. ; what i9> the profit in laying out £100 ?
Ans. £18 15s.
19. Bought 37 gallons of brandy at $1'10 per gallon,
and sold it for $40 ; what was gained or lost per cent ?
20. At 4s. 6d. profit on one pound, how much ib gained
in laying out £100, that is, how much per cent ?
j/l^ - ilns. ^2108.
W;
I4VATI0N OF rATMBNTt. .^'«'
173
go, valued
at ^,
It?
2 68. 5{d.
^rs at (J2i
. 2id. ?
L88. ad.
stock, at
f. ^900.
at 93 per
s. U^d.
[nonths 18
. ^400.
irable in 4
mi?
lousandths
nd the de-
Os. l^d.
Ldy money,
iths ; what
17 Is. 7id.
If in three
it must be
9s. lOd.
9 months
s. ^^476.
IS 24 days,
^'500.
'500, nine
;11 them at
$19'76.
again at
18 153.
er gallon,
cent.'
ib gained
|22 lOs.
21. Bought cloth at $4*48 per yard ; how muit I idl it
to gain 13^ per cent? -f v Ans. $(i*Oi.
22.' Bought a barrel of powder for £4t; for how much
must it be sold to loae 10 per cent ? Ans. £B 128.
23. Bought cloth at 16i. per yard, which, not proving
so good as I expected, I am content to lose '17^ per cent ;
how must I sell it per yard? Ans 128. 4jd.
24. Bought 50 gallons of brandy at 92 cents per gallon,
but by accident, ten gallons leaked out ; at what rate must
I sell the remainder per gallon, to gain upon the whole
cost at the rate of ten per cent ? i4ns. $1*265 per gal.
25. A merchant bought ten tons of iron for I960 ; the
freight and duties came to $145, and his own charges to
$25 ; how must he sell it per lb, to gain twenty per cent by
it? Ans. 6 cents per lb.
Equation of Payment!.
^ HS* Equation of Payments is the method of finding
the mean time for the payment of several debts due at
different times.
1. In how many months will one pound gain as much as
five pounds will gain in six months?
2. In how many months will one pound gain as much as
forty pounds will gain in fifteen months ? " Ans. 600.
3. In how many months will the use of five pounds be
worth as much as the use of one pound for forty months ?
4. Borrowed of a friend one pound for twenty months ;
aflerwards lent my friend four pounds ; how long ought he to
keep it to become indemnified for the r ,e of the one pound ?
5. I have three notes against a man ; one of <£12, due in
three months ; one of £9, due in five months ; and the other
of £6, due in ten months; the man wishes to pay the whole
at once ; in what time ought he to pay it ?
£12 for 3 months is the same as £1 for 36 months,
9 5"" 1 45
6 10 " " 1 60
<(
27
141
P2
174 RATIO, OE TnB RBLiVnOIf OF NVVBER8. 1186,87: I ^87^
He miglil thtf efore have one ptmnd 141 inotith0;!and'he
may fc«^ twenty-seven pounds ^V part es k»^g; that U,
yy* 3= five months- 6-f-day»,,i4»i. IrnrW <t jii^woJ* «;,■
Henoe, — Te find the mean time for eeveral paymenta,—
RvLB ; Multiply each eum by its time of payment, and di^
vide the sum' of the j»r*oJuc^5 by the sum of the payments,
and the quotient will be the answer.
Note. This rule is founded on the supposition that what
is gained by keeping a debt a certain time after it is due,
is the same as what is lost by paying it an equal time before
it is due ; but in the first case the gain ia evidently equal
to the interest on the debt for the given time, while in the
second case the loss is only equal to the discount of the
debt for th^t time, which is always less than the interest ;
therefore, the rule is not exactly true. The error, however,
is so trifling, in most questions that occur in business, as
. scarce to merit notice.
6. A merchant has ov/ing to him .£300, to be paid as
follows : £5Q in two months, 6^100 in five months, and the
rest in eight months ; and it is agreed to make one payment
of the whole ; in what time ought that payment to be ?
Ans. 6 months.
7. A. owes B. <£136, to be paid in ten months; ^96 to
be paid in seven months ; and <£260 to be paid in 4 months ;
what is the equated time for the payment of the whole ?
Ans. 6 months 7 days-}-.
' 8. A. owes B. $600, of which 200 is to be paid at the
present time, 200 in four months, and 200 in eight n^onths ;
wliat is the equated time for the payment of the whole ?
Ans. 4 months.
9. A. owes B. $300, to be paid as follows : ^ in three
months, ^ in four mouths, and the rest in six months ; what
is the equated time ? . • • ^»s. 4^ months.
>'it-,::
:t
. ^,:f ,
...1
Ratio : or Kclation of TVumbers.
U 87. 1. What part of a gallon is three quarts.' oi
gallon is four quarts, and three qu, :ts is f of four quarts.
Ans. f of a gallon.
,::..« ' ,.;.
....,, ,.p,„.
tl 86, ST. I If 87; ratio; or tbb ublavioh of numbers.
175
tha^^and'he
gr that w,
r
aymenla,—
int, and dh
3 payments,
n that ivhat
er it is due,
time before
lently equal
tvhile in the
ount c^ the
he interest ;
)r, however,
business, as
) be paid as
ths, and the
►ne payment
to be?
1 months,
ths ; £m to
n 4 months ;
whole ?
' days+.
paid at the
[ht n]ionths ;
whole?
months.
^ in three
)nths ; what
months,
crs.
[juarts ? one
)ur quarts,
a gallon.
2. Whatt part of 3 quarts ia ofie gallon? 1 gallon being 4
quarto, is ^ of 3 quarts ; that is, 4 quarts is 1 time 3 quarts
and I of another time. ^ns. ^=s\^.
3. What part of five bushels is twelve bushels ?
Finding what part one number is of another, is the same
as finding what is called the ratio or relation of one number
to another ; thus, the question, What part of five bushels is
twelve bushels ? is the same as What is the ratio of five
bushels to twelve bushels ? The answer is ^=2f .
RatiOf therefore, may be defined the number of -times
one number is contained in another ; or, the number of
times one quantity is contained in another quantity of the
same kind.
4. What part of eight yards is thirteen yards? or, What
is the ratio of 8 yards to 13 yards.'
13 yards is -^ of 8 yards, expressing the division /rrtc-
tionally. If now we perform the division, we have for the
ratio 1|; that is, 13 yards is one time 8 yards, and f of
another time. ; .
We have seen (51 15, sign,) that division may be expres-
sed /rac<8ona%. So also the ratio of one number to another,
or the part one number is of another, may br expressed
fi-actionally ; to do which, make the number which is called
the jpar/, whether it be the larger or the smaller number,
i\iQ numerator of a fraction, under. which write the other
number for a denominator. When the question is. What
is the ratio, &-c.? the number last named ts the /)ar< ; con-
sequently it must be made the numerator of the fraction,
and the number j^rs^ named the denominator.
5. What part of 12 pounds is 11 pounds? or, 11 pounds
is what part of 12 pounds? 11 is the number which ex-
presses the part. To put this question in the other form,
viz. What is the ratio, &:.c., let that number which expresses
the part, be the number last named ; thus, What is the ratio
ol 12 pounds to 1 1 pounds ? Ans. \^.
6. What part of i6'l is 2s. 6d. ? or, What is the ratio of
£\ to 2s. 6d. ?
£1:=240 pence, and 2s. 6d.s=30 pence; hence, ^W^gj
is the answer.
7. What part of 13p. 61 is ^1 lOs. ? or, What is the
ratio of 13s. Od. to £i lOs. ? Ans.^.
176
.^?;m
SULK or thrbb.
* i
1188.
fl88, 89
il'
8. What is the ratio of 3 to 5 ? of 5 to 3 ? of
7 to 19? of 19 to 7? of 15 to 90? of 90 to
15? — - of 84 to 160 ? of 160 to 84 ? of 615
to 1107 ? of 1107 to 616? Ans. to the last f.
PROPORTlOnr:
t
OR
THE SINGLE RULE OF THREE.
^ 88, 1. If a piece of cloth 4 yards long, cost ^12,
what will be the cost of apiece of the same cloth seven yds.
long ?
Had this piece contained twice the number of yards of
the first piece, it is evident the price would have been twice
as much ; had it contained three times the number of yards,
the price would have been three times as much ; or had it
contained only half the number of yards, the price would
have been only half as much; that is, the cost of seven yds.
will be such part of -^12 as seven yards is part of four yards.
Seren yards is | of 4 yards ; consequently, the price of 7
yards must be J of the price of 4 yards, or I of £12 : I of
^12, that is, 12X|=V=^21, answer.
2. If a horse travel 30 miles in 6 hours, how many miles
will he travel in 1 1 hours at that rate ?
11 hours is y of 6 hours, that is, 11 hours is one time 6
hours, and | of another time ; consequently, he will travel,
in 1 1 hours, Ltime 30 miles, and | of another time ; that
is, the ratio between the distances will be equal to the ratio
between the times.
V of 30 miles, thatis,30X V=^F=55 miles,
no error has been committed, 55 miles must be
miles. This is actually the case ; for ^^=: V
Ans. .;5 miles.
Q,uantilies which have the same ratio between them are
said to ht proportional. Thus, these four quantities —
HOURS HOURS. MILES. MILES.
If, Uien.
V of 30
the first i
is, the ra
ratio bet\
proportio
bination
numbers
Tode
bers 6, 1
which is
same par
as many
or relatio
the antec
portion t
viz. the h
the consie
the propc
the cons(
Thecc
numerate
fraction,
first ratio
tios are e
Thetv
by reduci
of the on
and, cons
same pro
case, for
if four n
and last,
the secor
Hence
given, to
ing that
times or
tion thus
6.
11,
30.
55,
written in this order, being such, that the second contains
!I88. 11188,89.
RULE OF THRES.
177
? Of
-of 90 to
— of 615
last |.
EE.
;ost ^12,
even yds.
■ yards of
een twice
of yards,
or had it
ce would
seven yds.
pur yards,
jricc of 7
f 12 ; I of
any miles
ne time 6
ill travel,
me ; that
the ratio
If, Uien.
V of 30
miles,
them are
lies —
contams
the first as many times as the fourth contains the third; that
is, the ratio between the third and fourth being equal to the
ratio between the first and second, form what is called a
proportion. It follows, therefore, that proportion is a com-
bination of two equal ratios. Ratio exists between iwn
numbers ; but proportion requires at least three.
To denote that there is a proportion between the num-
bers 6, 11, 39, 55, they are written thus —
6 11 : : 30 : 35
which is read, 6 is to 11 as 30 is to 55 ; that is, is the
same part of 11 that 30 is of 55 ; or, 6 is contained in 1 1
as many times as 30 is contained in 55 ; or, lastly, the ratio
or relation of 11 to 6 is the same as that of 55 to 31).
fl 89. The first term of a ratio, or relation, is^( tiled
the antecedent, arid the second the consequent. In a pro-
portion th§re are two antecedents, and two consequents,
viz. the antecedent of the first ratio, and that of the second ;
the consequent of the first ratio and that of the second. In
the proportion 6:11 :: 30 : 55, the antecedents are 6, 30 ;
the consequents 11, 55.
The consequent, as we have already seen, is taken for the
numerator, and the antecedent for the denominator of the
fraction, which expresses the ratio or relation. Thus, the
iirst ratio is y, the second ^^=y ; and that these two ra-
tios are equal, we know, because the fractions are equal.
The two fractions ^ ^"d ^^ being equal, it follows that
by reducing them to a common denominator,^ the numerator
of the one will become equal to the numerator of the other,
and, consequently, that 11 multiplied by 30 will give the
same product as 55 multiplied by 6. This is actually the
case, for 1 1 X 30=330, and 55X6=330. Hence it follows
if four numbers be in proportion, the product of the first
and last, or of the two extremes, is equal to the product of
the second and third, or of the two means.
Hence it will be easy, having three terms in a proportion
given, to find the fourth. Take the last example. Know-
ing that the distances travelled are in proportion to the
times or hours occupied in travelling, we write the pi'opor-
tion thus —
JIOURS.
6
HOURS,
11
MILES.
30
MILKS.
Win'
178
RULE OF THREE.
Now, since the product of the extremes is equal to the
product of the means, we multiply together the two mean8,
Jl and 30, which makes .330, and, dividing this product by
the known extreme, 6, we obtain for the result 55, that is,
55 miles, which is the other extreme or term sought,
3. At ^54 for 36 barrels of flour, how many barrels mat
be purchased for £1861
In this question, the unknown quantity is the number of
barrels bought for .£186, which ought to contain the 36
barrels as many times as ^186 contains £54 ; we tlius get
the following proportion :
Pounds, Pounds. Barrels. Barrels. . '
54 : 186 :: 36 : :
36
1116
558
64)6696(124 barrels, answer.
54
129
108
"2T6
216
The product 6696
®* the t\Vo means,
divided by 54, the
known extreme, gives
124 barrels for the
other extreme, which
is the term sought,
or answer.
Any three terms of a proportion being given, the opera-
tion by which we find the fouTth, is called the Rule of
Three. A just solution of the question will some times re-
quire that the order of the terms of proportion be changed.
This may be done, provided the terms be so placed, that the
product of the extremes shall be equal to that of the means.
4. If 3 men perform a certain piece of work in ten days,
how long will it take 6 men to do the same ?
The number of days in which six men will do the work,
being the term sought, the known term of the same kind,
viz. ten days, is made the third term. The two remaining
terms are 3 men and 6 men, the ratio of which is t. But the
more* men there are employed in the work, the less time will
* The rule of three lias sometimes been divided into direct and in'
vtrse, a distinction which ia lolally useless. It may not hotvever be
aniitts to explain, in this place, in what this distinction consists.
The Rule of Three Direct is when more requires more, or /ess re-
RULE OF THREE.
179
be required to do it ; consequently the days will be less in
proportion as the number of men is greater. There is still
a proportion in this case, but the order of the terms is in-
verted ; for the number of men in the second set being two
times that in the first, will require only one half the time.
The firjtt number of days, therefore, ought to contain the
second as many times as the second number of men con-
tains the first. This order of the terms being the reverse
of that assigned to them in announcing the question, we say
that the number of men is in the inverse ratio of the number
of days. With a view, therefore, to a just solution of the
question, we reverse the order of the two first terms, (in do-
ing which, we invert the ratio,) and instead of writing the
proportion 3 men : 6 men (f ) we write it ii men : 3 men, (g)
lluit is, men. men. days. days. ■
6:3 :: 10
Note. We invert the ratio when we reverse the order ol
the terms in the proportion, because then the antecedent
takes the place of the consequent, aud the consequent that
of the antecedent ; consequently, the terms of the fraction
which express the ratio are inverted ; hence the ratio is
inverted. Thus, the ratio expressed by f=2, being inver-
ted, is ^=^:
Having stated the proportion as above, we divide the
product of the means, (10x3=30,) by the known extreme
6, which gives 5, that is, 5 days, for the other extreme or
term sought. w4ns. 5 days.
From the examples and illustr.ii; ions now given, we de-
duce the following general
quires lesSy as in this example .— if 3 men dig a irench 48 feet long in
a certain lime^ how many feel will 12 men dig in the same time ? Here
it is obvious that the more men there are employed, the more work
will be done ,* and therefore, in this instance, more requires mor;;.
Again — if 6 men dig 48 feet in a given time, how much will 3 men dig
in the same time ? Here less requires less, for the less men there are
employed, the less work will be dune.
The Rule of Three Inverse is when more requires less, or less requires
more, as in this example : — If 6 men dig a certain quantity of trench
in 14 hours, how many hours will it require 12 men to dig the same
quantity ? Here more requires loss ; that is, 12 men being more than
ti, will require less time. Again — if 6 men perform a piece of work in
seven days, how long will three men be in performing the same work/
Here less requiros more j for the number of men being less, will require
more time.
''-m
■'■.m
■m
m
180
RVLE OF THREE.
RULE.
Of the three given numbers, make that the third term
which is of the same kind with the answer sought. Then
consider, from the nature of the question, whether the
answer will be greater or less than this term. If the answer
is to be greater J, place the greater of the two remaining num-
bers for the second term, and the less number for the first
term ; but if it is to be less, place the less of the two re-
maining numbers for the second term, and the greater for
the first ; and, in either case, multiply the second and third
terms together, and divide the product by the first for the
answer, wlJch will always be of the same denomination as
the thi a* term.
Ni*te, L If the first and second terms contain different
d!"n(»ur, :'a. jns, they must both be reduced to the same de-
noiwia:!.*^.
If 8 ; 'itth of cloth cost £\ 4s. what will li64 qrs. cost ?
yds. qrs.
8 : 364 : : £1 43.
Reduce 8 yards and 364 quarters to the same denomina-
tion, by dividing the 364 quarters by 4, which will bring it
into yards. 3|*=91.
yds. yds,
8 : 91 :: ^1 4s.
Note 2. If the third term be a compound nur.iber, it must
either be reduced to integers of the lowest denomination,
or the low denominations must be reduced to a fraction of
the highest denomination contained in it.
ydi. yds.
8 ' : 91 : : ^1 4s.
20
24s. ♦^
Now multiply the 24s. by 91, and divide the product by
8 ; the answer will be shillings, which can be reduced to
pounds ; or, the 4s. can be reduced to the fraction of a pound,
4S.-7-20, that is, ^Tr=i of a pound ; so ^1 4s.=^l|. Or,
we can reduce the 4s. to the decimal of a pound ; 20)40
which, annexed to the £\y is equal to £V2. —
'2
.■* ,.
«ULB OP THEfcB,
m
The first method is most usually practised. ' ' 'i' ' ' .
Note 3, The same rule is applicable, whether the giv.en
KjuaDtities he iIH^gra], fractjional, or decimal,
EXAMPLES FOR PRACTICE,
5. If 6 horses consume 21 bushels of oats in three v/efii(.ii,
fiov inany bushels will serve 20 horses the same time ?
Ans, 70 bushels,
6. The above question reversed. If 20 horses consume
70 bushels of oats in 3 weeks, how fn^jiy bushels will servj^
6 horses the same time ? 4w«. Ql bushels,
7. If 365 men consume 75 barrels of provisions in niyi^
rrioriths, how piuch will 500 me^ consume in the sam^
time ? Ans, J02^^ barrels,
8. Tf 500 men consume 102^^ barrels of" provisions iij j)
lYionths, how much will 365 mei) consume in the saniit^
time 1 "^ Ans, 75 barrels,
9. A goljlsmith sold a tankard for ,£10 12s. at the rait«
of 5s, 4d, per ounce ; I demand the weight of it,
Ans. 39 oz, 15 pw^t,
10. If the moon move 13 ^ 10' 35" in a day^ iij wi)ajt
time does it perform one revolution ? Ans, 27d, 7h. 43ui,
11. If a person whose rent is £33, pay £8 28. parivslj
taxes, how much should a person pay whose rent is ^£"97 ?
Aiifi. £9 2s. 2|f d,
12. If I buy 7 fcs, of sugar for 3s, Od, how many poundi^
ean I buy for £1 10s. ? , Ans, 56 tha,
13. If2 lbs. of sugar cost Is, 3d., what will 100 lbs, ot
coffee cost, if 8 }bs of sugajr are worth 5 Jbs, of coffee ?
Ans, ot',5/
14. If I give £6 for the use of £100 for 12 months, whai
jDust J give for the use of £983 the same time ?
15. There is a cistern which has 4 pipes ; the first will
fill it in ten minutes, the second in twenty minutes, the vT4
in forty minutes, the fourth in eighty minutes ■ in wha.t Uff^
yvill all four, running together, fill it ?
iV + 5V +5V + ffV ^ ih ^^istern in ! minute,
Ans,5\- m'iTiiitf«^
16. If a family of 10 persons spend 3 bushela of mBlf. i(j
a month, how many bushejs will serve thgnji when thera ttf^
30 in the famWy ? An.s, 9 hu«t)f 5«;
,..■,),.
1
m
182
RULE OF THREE.
TI89.
i
I
Note. The rule of Proportion, although of frequent use,
is not of indispensable necessity ; for all questions under it
may b9 solved on general principles, without the formality
of a proportion ; that is, by analysis, as already shown, jj 62
ex. 1. Thus, in the above example, — If 10 persons spend
3 bushels, 1 person, in the same time, would spend -j^y of
3 bushels, that is, -^ of a bushel ; and 30 persons would
spend 30 times as much, that is f$=9 bushels, as before.
17. If a staffs feet 8 inches in length, cast a shadow of
6 feet, how high is that steeple whose shadow measures
153 feet ? , Ans. U4h feet.
18. The same by analysis. If 6 feet shadow require a
staff of 5 feet 8 inches=68 inches, one foot shadow will re-
quire a staff of '^ of ^ inches, or ^^ inch; then 153 feet
shadow will require 153 times as much; that is, ^^ X 153
-_io|.o4_i734 inches=144^ feet as before.
19. If i£3 sterling be equal to JG3^ Halifax, how much
Halifax is equal to .£1000 sterling ? Ans. .£1 1 11 2s. 2§d.
20. If £1111 23. 2§d. Halifax be equal to jeiOOO sterling,
how much sterling is equal to £3^ Halifax? Ans. £3.
21. If £1000 sterling be equal* to £1111 2s. 2§d. Hali-
fax, how much Halifax is equal to £3 sterling ? Ans. £3^.
' 22. If j£3 sterling be equal to £3;^ Halifax, how much
sterling is equal to £1111 2s. 2fd. Halifax ? Ans. £1000.
23. Suppose 2000 soldiers had been supplied with bread
sufficient to last them 12 weeks,' allowing each man 14 oz.
a day ; but, on examination, they find 105 barrels, contain-
ing 200 lbs. each, wholly spoiled ; what must the allowance
be to each man, that the remainder may Ir.^t them the same
time 1 Ans. 12 ounces a day.
24. Suppose 2000 soldiers were put to an allowance of
12 oz. of bread per day for 12 weeks, having a seventh part
of their bread spoiled, what was the whole weight of their
bread, good and bad, and how much was spoiled?
. i The whole weight, 147000 lbs.
'^"*' ( Spoiled, ^i"«" "
21000
25.
— 2000 soldiers, having lost 105 barrels of bread,
weighing 200 lbs. each, were obliged to subsist on 12 oz. a
day for J 2 wetk?^ : had none been lost, they might have had
long ?
31.
days a
51 80. ■ ^ 89.^^
RULE OF THREE.
Ids
Ans.
14 oz. a day ; what was the whole weight, including what
was lost, and how much had they to subsist on I
i Whole weight, 147000 lbs.
( Left to subsist on, 126000 "
26. 2000 soldiers, after losing one seventh part of
their bread, had each 12 oz. a day for 12 weeks; what was
the whole weight of their bread, including that lost, and
how much might they have had per day, each man, if none
had been lost ? ) Whole weight, 147000 lbs.
Ans. SLoss, 21000 "
j 14 oz, per day, had none been lost.
27. There was a certain building raised in 8 months by
120 workmen ; but, the same being demolished, it is re-
quired to be built in 2 months ; I demand how many men
must be employed about it. Ans. 480 men.
28. There is a cistern having a pipe which will empty it
in ten hours; how many pipes of the same capacity will
empty it in 24 minutes ? JJns. 25 pipes.
29. A garrison of 1200 men has provisions for 9 months,
at the rate of 14 oz. per day ; how long will the provisions
last, aJt the same allowance, if the garrison be reinforced by
four hundred men ? Ans. 6f months.
30. If a piece of land, 40 rods in length and 4 in breadth,
make an acre, how wide must it be when it is but 25 rods
Ans. 6% rods.
31. If a man perform a journey in 15*
(lays are 12 hours long, in how many will
the days are but 10 hours long ?
32. If a field will feed 6 cows 91 days, how long will it
feed 21 cows.-* Ans. 26 days.
33. Lent a friend £292 for 6 months ; sonle time after,
he lent me .^806 ; how long may I keep it to balance the
favor ? Ans, 2 months 5-f-days.
34. If 30 men can perform a piece of work in 1 1 days,
how many men will accomplish another piece of work, four
times as big, in a fifth part of the time ? Ans. 600 men.
35. If -f^ lb. of sugar cost ^^ of a shilling, what will |§
of a pound cost ?, Ans. 4d. 3f g^4^ q.
Note. See tl 62, ex. 1 , where the above question is solved
by analysis. The eleven following are the next succeeding
examples in the same paragraph.
long ?
days when the
he do it when
Ans. 18 days.
A
•% ('
m
ktht OP tHIttii
flsd.oo. I ^
v>>*'.
-4'
&6i If 7 lbs. of sugar cost f of 5s. what tiost 12 lbs.
Ans. ()^s;
87. if 6^ yards of oloth eost i^3, what cost 9^ yards ?
.?:?'/.. . : ' Ans. ^4 5s. 4^(1.
88. If 3 oz. of silver cost lis 9fd- what costf oz. ?
A715. 4s. Sfd.
80. If ^ osjj cost 4-/2S., what costs 1 Oz. 1 Ans. Gs. 5d.
40. If ^ tbi less by I ib eost 13|d., what cost 14 lbs. less
l)y I of 2 Ibs; Ans. £i 9s. 9ir\d.
41. If f of a yafd cost £^i what will 40^ yards cost ?
Ans. £59 Is. 2^d.
42. if f^p of a ship cost £'iiolf what is -^ of her worth ?
ilns. ^^53 15s. 8^d.
43. At <^3| per cwt., what will Of Ibs; cost ?
Ans. 6sv 3^d
44. A merchant owning 4 of a vessel^ sold § of his share
for .£957; what was the vessel worth ? Ans, <j€1794 7s. Cd.
45. If i of a yard cost £^, what will -^^ of an ell English
tost ? Ans. 17s. Id. 2f q.
40, A merchant bought a number of bales of velvet, each
Containing 129^f yards, at the rate of £7 for 5 yards, and
Sold them out at the ral^^ jf .£11 for 7 yards, and gained
£200 by the bargain ; how mrjiy bales were there ?
Ansi 9 bales.
47. At £9 for 6 barrcjls of floui', Vvhat ftiust be paid for
t78 barrels ? Ans. £267.
48. At 9s. 6d. for 3 cwt. of hay, how much is that pet
ton ? Ans. £3 3s. 4d.
49. if 2*5 tbsi of tobacco cost 75 cents^ how miiCh will
185, lbs. cost 1 Ans. $5'55.
50. What is the value of *15 Of a hogshead of lime, at
lis. 11 |d. per hhd.? Ans. Is. 9^d.
51. If '15 of a hhdw of lime cost ls< 94^d., what is it per
hhd. T n, .,,. . y ,1,. ^ns< lis. ll^d.
, -~' " — '— ■^
COMPOUND PROI^ORTION.
1[ 00 < It frequently happens that the relation of the
quantity i'e<)dired^ to the given quantity of the same kindy
depends upon several circumstances combined together ; it
is then caQled Compound Proportion, or Double Mule o/
Three.
1|8d,90. I ^®^*
COMPOUND PROPORTION.
185
t 12 lbs.
Ans. 6^si
9^ yards?
. £\ 6s. 4 jd.
3t f OZ. ?
An5. 4s. 2fdj
ilws. Gs. 5d.
)st 14 lbs. less
£\ 9s. 9jf\d.
ards cost ?
.£59 Is. 2fd.
f her worth ?
£53 15s. 8^d.
4ws. 6s. 3^^d.
§ of his share
^1794 7s. Cd.
an ell English
. 17s. id. 2fq.
of velvet, each
•r 5 yards, and
Is, and gained
there 1
AnSi 9 balesi
Lst be paid fof
AnSi .£267.
ch is that pet
nsi £'^ 3s. 4d.
low much will
AnSi $5*55.
ad of lime, at
Ans. Is. 9^d.
what is it per
ns, lis. ll^d.
elatiori 6f the
le same kind<
d togeither ; it
ouhle Rule of
1. If a man ttavel 273 miles in 13 days, travelling only
seven hours in a day, how many miles will he travel in 12
d'iys, if he travel 10 hours in a day?
This question may be solved several ways. First, by
analysis —
If we knew how many miles the man travelled in one
hour, it is plain we might take this number 10 times, which
would be the number of miles he would travel in ten hours
or in one of these long days ; and this again taken 12 times,
would be the number of miles he would travel in 12 days,
travelling 10 hours each day.
If he travel 273 miles in 13 days, he will travel -^ of 273
miles ; that is, ^^ miles, in 1 day of 7 hours ; and f of Yi^
miles is ^j^ miles, the distance he travels in 1 hour ; then,
10 times y^ = ^^^^o niiles, the distance he travels in ten
hours; and 12 times 27 3o_3y^6o_=360 miles, the distance
he travels in 12 days, travelling ten hours each day.
Ans. 360 miles.
But the object is to show how the question may be solved
by proportion —
First, it is to be regarded that the number of miles tra-
velled over depends upon two circumstances, viz. the num-
ber of dai/s the man travels, and the number of hours he
travels each day.
We will not at first consider this latter circumstance, but
suppose the number of hours to be the same in each case ;
the question then will be — If a man travel 273 miles in 13
days, how many miles will he travel in 12 days ? This will
furnish the following proportion : —
13 days : 12 days : : 273 miles : miles,
which gives for the fourth term or answer, 252 miles.
Now, taking into consideration the other circumstance,
or that of the hours, we must say — If a man travelling seven
hours a day for a certain number of days, travels 252 miles,
how far will he travel in the same time, if he travel. ten
hours in a day ? This will lead to the following proportion :
7 hours : 10 hours : : 252 miles : miles.
This gives for the fourth term or answer, 360 miles.
We see, then, that 273 miles has to the fourth term, or
answer, the same proportion that 13 days has to 12 days,
Q2
>.^. .^0-::
IMAGE EVALUATION
TEST TARGET (MT-S)
1.0
I.I
1.25
y^l^S 12.5
2.2
1^ 1^
It? i^
2.0
1.4
m
I
1.6
V
vl
Photographic
Sciences
Corporation
23 WEST MAIN STREET
WEBSTER, N.Y. 14580
(716) 873-4503
• .
I
189
COMPOUND PROPORXION.
and that seven hours has to ten hours. Stating this in the
form of a proportion, we have
13 days :.12days ) ^^^ •, .,
7 hours : 10 hours \ ' ' ^^^ "^^^"' * "^'^^^
by which it appears that 273 is to be multiplied by both 12
and 10 ; that is, 273 is to be multiplied by the product of
12X10, and divided by the product of 13X7, which, being
done, gives 360 miles for the fourth term, or answer, as
before. - >
In the same manner, any question relating u compound
proportion, however complicated, may be stated and solved.
2. If 248 men, in 5 days of 11 hours each, can dig a
trench 230 yards long, 3 wide, and 2 deep, in how many
days of 9 hours each, will 24 men dig a trench 420 yards
long, 5 wide and 3 deep ?
Here the number of days, in which the proposed work
can be done, depends on five circumstances, viz. the num-
ber of men employed, the number of hours they work each
day, the length, breadth and depth of the trench. We will
consider the question in relation to each of th^se circum-
stances, in the order in which they have been named^ —
L2.0 5.
Ji'i
Bit::
1st. The number of men employed. Were all the circum-
stances in the two cases alike, except the number of men
and the number of days, the question would consist only in
finding in how many days 24 men would perform the work
which 248 men had done in 5 days ; we should then Jiave
24 men : 248 men : : 5 days : days.
2d. Hours in a day. But the first laborers worked 11
hours in a day, whereas the others worked only 9 ; less
hours will require more days, which will give
9 hours : 1 1 hours : : 5 4ays : days.
3d. Length of the ditches. The ditclies being of unequal
length, as many more days will be necessary as the second
is longer than the first ; hence we shall have
230 length : 420 length : : 5 days : days.
4th. Widths. Taking into consideration the widths,
which are different, we have
3 wide : 5 wide : : 5 days days.
5th. Depths. Lastly, the depths being different, we have
2 deep : 3 deep
5 days
days.
•«iiK'„ii,.!
fWI
COMPOUND PROPORTIOK.
187
It would seem, therefore, that 5 days has to the fourth
term, or answer, the same proportion
that 24 men has to 248 men, whose ratio is ^^,
9 hours " 1 1 hours, the ratio of which is V,
230 length " 420 length
3 width " 5 width
2 depth " 3 depth
ail of which, stated in form of a proportion, we have
((
((
((.
Men, 24
. 248}
Hours, 9
11
Length, 230
420 y
Width, 3 .
5
Depth, 3
: 3J
common term.
: : 5 days : ■
davs.
Tf 91. The continued product of all the second terms
•348X11X420X5X3, multiplied by the third term, 5 days,
and this product divided by the continued product of the
first terms, 24X9X230X3X2, gives 288^%*^^^^% days for
the fourth term, or answer. 288^jj^,
But the first and second terms are the fractions y-^, y,
If g, ^ and f , which express the ratios of the men and of
the hours, of the lengths, widths and depths of the two
ditches. Hence it follows, that the ratio of the numbe^of
days given to the number of days sought, is equal to the
product of all the ratios, which result from a comparison of
the terms relating to each circumstance of the question.
The product of all the ratios is found by multiplying to-
gether the fractions which express them, thus —
248X11X420X5X3 17186400 1718G400
= and this frac.
24 X 9 X230X3X2 298080 .298080
represents the ratio of the quantity required to the given
quantity of the same kind. A ratio resulting in this manner
from the multiplication of several ratios, is called a com-
pound ratio.
From the examples and illustrations now given, we de-
duce the following general
RULE
for solving questions in compound proportion, or Double
Rule of Three, viz. — Make that number which is of the
same kind with the required answer, the third term ; and of
...%;>; .•J«\:rfeAVWi'i:i.';. i ■'^^/.'\ii^i
.■y.'Vf .^^V*^' *■■"
188
COMPOUND PROPORTION.
1I9I.
the remaining numbers, take away two that are of the same
kind, and arrange them according to the directions givea
in simple proportion : then any other ,two of the same kind,
and so on till all are used.
Lastly, multiply the third term by the continued product
of the second terms, and divide the result by the continued
product of the first terms, and the quotient will be the 4th
term, or answer required.
EXAMPLES FOR PRACTICE.
1. If 6 men build a wall 20 feet long, 6 feet high, and 4
feet thick in 16 days, in what time will 24 men build one
200 feet long, 8 feet high and 6 feet thick? Ans. 80 days.
2. If the freight of 9 hhds. of sugar, each weighing 12
cwt. 20 leagues, cost ^16, what must be paid for the freight
of 50 tierces, each weighing 2^ cwt 100 leagues?
^ns.^92 lis. lOfd.
3. If 56 ibs. of bread be sufficient for 7 men 14 days,
how much bread will serve 21 men 3 days? Ans. 36 Ids.
T/ie same by analysis. If 7 men consume 56 lbs. of bread,
1 man, in the same time, would consume f of 56 lbs.=
V^ lbs. ; and if he consume ^ lbs. in 14 days, he would
consume ^^ of ^6=|| lbs. in one day. 21 men would con-
sume 21 times so much as 1 man ; that is, 21 times |f=
*il^ lb.s. in i day, and in 3 days they would consume 3
, times as much; that is, ^^^^z=ZQ lbs. as before.
Ans. 36 lbs.
Note. Having wrought the following examples by the
rule of proportion, let the pupil be required to do the same
by analysis.
4. If 4 reapers receive £2 15s. 2^. for 3 days' work,
how many men may be hired 16 days for ^25 15s. 2^d. ?
Ans. 7 men.
5. If 7 oz. 5 pvvt. of bread be bought for 4fd. when corn
is 4s. 2d. per bushel, what weight of it may be bought for
Is. 2d. when the price per bushel is 5s. 6d. 1
Ans. I lb. 4 oz. 3|^f pwts.
6. If <^100 gain £Q in 1 year, what will ^400 gain in 9
months ?
Note. This and the three following examples reciprocally
prove each other.
^■jt -.
of the same
itions given
same kind,
ed product
i continued
i be the 4th
high, and 4
n build one
«s. 80 days,
i^eighing 12
r the freight
i?
5 lis. lOfd.
en 14 days,
ins. 36 lbs.
bs. of bread,
>f 56 lbs.=
, he would
would con-
times 11=
consume 3
ins. 36 lbs.
)les by the
o the same
days' work,
5s. 2^d. ?
ins. 7 men.
when corn
bought for
3|Jf pwts.
gain in 9
eciprocally
If 01. eL'ppLiJMErit tcr Single ri/Le of 'tHiUEji, . 180
7. If .^100 gain £Q in I'yeary \ti what time will ^400
gain .£18?
8. If £400 gain .£18 in 9 mottths, what is the rate per
sent per annum ?
9. What principaly at C per cent per annum will gain
jf 18 in 9 tnonths ?
10. A Usurer put out $75 at interest, and at the end of
8 months, received, for principal and interest, $79; I de-
mand at what fate per cent he received interest.
Ans. 8 per cent.
IL If 3 men receive £Q^^ for 19;i^ days' work, hoW
Much must 30 men receive for 100^ days 7
Ans. £305 Os. 8d.
"?»|;i(
(^iipplrment to Sliig'lc Rule of Three.
QUESTIONS.
1. What is propdrlioiA ? 2^ ^a^v many nuniibeiCs are required io
form a ratio f 3< HoMi many to form a proportion / 4. \Vh;il ii> tlie
iirst term of a ratio called / d. ^ ttte second term 1 6< Which is
taken for the numerator, and which f6r the denominator of the fraction
elpressing the ratio ? 7. How may il be known when 4 numbera are
In proportion 7 8. Hating three terms m the prdportion given, how
(iiay the fiurlh term be found 7 9. \Vhiil is the opurdtion, by which
the fourth ternd is found; calleid 1 lO. How does a ratio beco:ne in'
terted % 11. What is the ruld in proportion % \%, In what denomi-
httion will the 4ih term or anstter be found 1 19. If the fust and sc
cohd terms contii.i different denominations, what is to b« done 1 II.
What is compound proportion^ or double rule of three 1 15< Hale %
m
EXfiRCISE^
1. if I buy 76 yards of cloth fcr £28 6s.' lOdr f^ qrs/
What does it cost per ell English 1 Jins. 9s. 3^d/
2. Bought 4 pieces of Holland, each containing -24 ells
English for .£24 ; how much was that per yard? Ans 4s.
2. A garrison had provisions for 8 months, at the rate of
15 ounces to each person per day ; how much must be al-
lowed per day in order that the provisions may last 9^
months? Ans. 12|f oz*
4. HovV milch land at 12s. 6d. per acre, must be given
in exchange for 360 acres, at 188. 9d. per acre ?
Ans, 540 acree.
f
i90
'.J,
TELLOWSHIP.
! t/
!I 01,92. It! 92.
6. Borrowed 185 quarters of corn when the price was
19s. ; how much must I pay when the price is 17s. 4d. 1
Ans. 202f|
6. A person owning f of a coal mine, sells f of his share
for iifflTl ; what is the whole mine worth? Ans, £380.
7. If I ot a gallon cost |^ of a pound, what cost ^ of a
tun? _ Ans. £UQ.
8. At ^l^per cwt. what cost 3^ lbs.? Ans. lO^d.
9. If4J-cwt. can be carried 36 miles for 35 shillings,
how many pbunds can be carried 20 miles for the same
money ? Ans. 907^ lbs.
10. If the sun appears to move from east to west 360
degrees in 24 hours, how much is that in each hour ?
in each minute ? in each second ?
Ans. to the last, 15" of a deg.
11. If a family of 9 persons spend c£ 1 12 lOs.. in 5 months,
how much would be sufficient to maintain them 8 montjis
if 5 persons more were added to the family ? Ans. .£280.
Note. Exercises 14th, I5th, 16th, 17th, 18th, 19th and
20th, •' Supplement to Fractions" afford additional exam-
ples in single and double proportion, should more examples
be thought necessary.
FEIiLOWl^HIP.
^ 9^. 1.. Two men own a farm ; the first owns ^, and
i\\9 second owns f of it ; the farm is sold for ,^40 ; what is
each man's share of the money .^
2. Two men purchase a horse for 20 pounds, of which
one pays 5 pounds, and the other 15 pounds ; the horse is
sold for 40 pounds ; what is each man's share of the money ?
3. A- and B. bought a quantity of cotton ; A. paid 100
pounds, and B. 200 pounds : they sold it so as to gain 30
pounds ; what were their respective shares of the gain ?
The process of ascertaining the respective gains or losses
of individuals engaged in joint trade, is called the rule of
Fellowship.
The money, or value of the articles employed in trade, is
called the capital or stock ; the gain or loss to be shared is
called the dividend.
^91,92. In92.
FELLOWSHIP.
191
It is plain that each man's gain or loss ought to have the
same relation to the whole gain or loss, as his share of the
stock does to the whole stock.
Hence we have this Rule : — As the whole stock : to
each man's share of the stock : : the whole gain or loss : his
share of the gain or loss.
4. Two persons have a joint stock in trade ; A. put in
i:-250, and B. .£350 ; they gain £400 ; what is each man's
share of the profit ?
OPERATION.
A.'s stock, £250 ) Then,
B.'s " 350 C 600 : 250 :: 400 : £166 13s. 4d. A.'s
Whole stock, £600 ^ 600 : 350 :: 400 : 233 6s. 7d. B.'s
The pupil will perceive that the process may be con-
tracted by cutting off an equal number of ciphers from the
first and second, or first and third terms ; thus, 6 : 250 :: 4 :
^166 13s. 4d. &c.
It is obvious, the correctness of the work may be ascer-
tained by finding whether the sums of the shares of the
gains are equal to the whole gain ; thus, £166 13s. 4d.-|-
£233 6s. 7d.=£400, the whole gain.
5. A. B. and C. trade in company ; A.'s capital was £175,
B.'s £200, and C.'s £500 ; by misfortune they lose £250 ;
what loss must each sustain? r£ 50 A.'s loss.
Ans. I 57 2s. lO^d. B.'s "
C.'s
i
57 2s. lO^^d.
142 17s. lid.
/"
6. Divide $600 among 3 men, so that their shares may
be to each other as 1, 2, 3, respectively.
^ws. $100, $200 and $300.
7. Two merchants, A. and B. loaded a ship with 500
hhds. of rum ; A. loaded 350 hhds. and B. the rest ; in a
storm, the seamen were obliged to throw overboard 100
hhds. ; how much must each sustain of the loss ?
ylHs. A. 70, and B. 30 hhds.
J?. A. and B. companied; A. put in £45, and took out f
of the gain; how much did B. put in ? Ans. £30.
Note. They took out in the same proportion as they put
in ; if f of the stock is £45, how much is f of it?
9. A. and B. companied, and trade with a joint capital of
If
ilri«'
m m
yl i
!SIi'<:H
m
PELLOWSHIV.
1IW,W.|f93.
ti-
i^400; A;, receiv.ftf for his sh^re of the gain, } as muchM
^ ; what was the stock of ea«h ?
., )rfJ33 6ii, 74. A's stock,
^»«- f £266. 13s. 4d. B's stock,
10. A banlfrupt is indebted to B $780, to C $4|80, and to
.P $760 ; bis e»tate is worth only $($00 ; bow must it be di»
vided? , : .
Note. The question evidently involves the priociples of
fellowship, ana may be wrought by it,
Ans. B $234, C $138, and P $228,
11. B and C venture e(}ual stocks in trade, and clear
<^164 ; by agreement, B was to have 5 per cent of the prof*
its, because be managi^ the concerQs; C was to have but
2 per cent 4 what was <}ach one's gain? and bow muob did
B receive for his trouble ?
Ans. B.'s gain was <£U7 2e, lO^d. and C.'a£40 li7s, l^d,
and B. received ^7Q 5s. 8^d, for his trouble,
12. A cotton factory, viUued at ,£12000, is 4iviHed into
100 shares ; if the promts amount to 15 per c;ent yearly,
what will be the profit accruing to J share ,^-'-.— to 2 shares?
— ^to 25 shares ? Ans, to the last £450,
13. In the above-mentioned factory, repair^ are to be
made which will cost ^340 ; what will be the tax on each
^are, necessary to raise the wm ? ^" ' on 2 shares ? «• — '
on 3 shares ? on 10 shares? Ans, to the last, ^£34,
14. If a town raise a tax of <£1850, and the whole town
he vajued at ^37000, what will that be on ^1 ? What will
be the tax of a man whose property is valued at ^1780?
Ans, Is. on a pound, and £8^ on 4^1780,
51 93. In assessing taxes, it is necessary to have au
inventory of the property, bcih real and personal, of the
^vhole town, and also of the whole number of the polls; an4
;as the poIJs are rated at so much each, we must jirst take
out from the whole tax what the polls amount to, and the
remainder is to be assessed on the property. We may then
.find the tax upon one pound, and make a table pontainipg
the taxes on one, two, thr^ep, &.c, to ten pounds j then on
twenty, thirty, &-c, to a hundred ; then on 100, 200, &.c, to
1000 pounds. Then knowing the inventory of any indivi*
4u^.l, l\ is easy to find the tax upon his property*
15. A
^2259 1
what is
real ests
^874, ai
It will
taxes to
process
exactnes
will be r
will be n
final ans
540X
2259*90
property,
tax on o]
Tax. on
Now,!
by the ta
The tax
In like m
Two pol]
i67'62=
1FW,l».|ff93.
FELLOVrSUIP.
198
. A's stock,
I. B'9 stock,
MjSO.andto
lusjt it be di»
riocipleji of
nd P 1228,
, and clear
of the proff
10 have but
V iQuoh did
*6 173. l^d,
divided into
pent yearly,
-to 2 shares?
e Jast ^450,
s are to be
taK on eajch
liares ? "• — '
le last, ^34,
)vhole town
What will
^1780?
on 4PJ780,
to have an
onal, of the
e polls ; an4
ist ^rst take
to, and the
^e may then
i containing
Js^hen on
200, &e, to
any indJvi'
15. A certain town, valued at jf 64530, raises. A tax of
i2259 18s. ; there are 540 polls, which are taxed 3s. each ;
what is the tax on a pound, and what will be B.'s tax, whose
real estate is valued at jf 1340, his personal property at
f 874, aad who .pays for two polls. ^ w
It will be better in questions relating to the assessment of
taxes to use decimals, as we have done in interest. The
process will be shorter, and the result will be obtained with
exactness. The shillings, therefore, in the given values,
will be reduced to the decimal of a pound, and the table
will be made out decimally, and the decimal parts in the
final answer can be reduced to shillings and pence,
540X'60"(3s.)=v£324, amount of the boll taxes, and
2259'90 (.£2259 I8s.)— je324c=1935'90, to be assessed on
property. .£64530 : 1935'90 ;; ^l'03i or »ff|t^='08
tax on one pound, . . i
TABLE.
Tax, on
£ £
£
£
1 is '03
Tax on 10 is
I '30
2 '06
SO
'60
3 '09
30
«90
4 '12
id
1*20
5 '15
50
1'50
6 '18
50
1*80
7 '21
. 70
2'10
8 '24
80
2*40
9 '27
90
2'70
Tax on
jg.,:'
i)
100 iE
( 3'
200
6*
300
9'
400
11'
500
15'
600
18'
700
21'
800
24'
900
27'
[000
30'
Now, to find B.'s tax, his real estate being £1340, 1 find
by the table that
£ £ *
The tax on - - 1000 is - - 30'
300 9'
40 1'20
The tax on his real estate - . - 40'20
In like manner I find the tax on his personal
property to be - • - - - ^ 26*22
Two polls at '60 each, are . , - 1*20
.£67'62r=£67 12s. 4^d. ansmr. Amount, 67'63
VH
tmtLOWMtt.
16. What will C/b tax amoam to whoso inrentary is 874
dollars retd^ and 210 doUars personcU property, nnd who
pagra for three polU ? ^ ^fi5. $34'%i.
17. What will be the tax of a man paying for one polU
whose property is valued at $34*8* ? at $768 ? — -|
at $940 T ' at $4657 ? ^4^.9. to last, 1 140*91 .
18. Two men paid $W for the use of a pasture 1 motith;!
A. kept in 24 cows, and B^ 10 cows; how much shouldj
each pay /
19. Two men hired a pasture for $10; A. put in 8 cowsl
9 months, and B. put in 4 cows 4 months ; how much|
should each pay ?
• ^ 04* The pasturage of 8 cows for 3 months is thel
same as S^ cotws for 1 month ; and the pasturage of 4 com'sI
for 4 months is the same as of 16 cows for one month. Thel
shares of A. and B. therdfore^ are 34 to 10, as in, the formerl
question. Hence, when time is regarded in fellowship,— I
Multiply ;each one's stock by the time he continues it inl
trade, and use the product for his share. This is calledl
l^ouble Fellowship. Ans. A. $16, and B. $4
20. A. and B. enter into partnership ; A. puts in £100 sixl
months, itnd then puts itt <£50 more; B. puts in £200 fourl
months, and then takes out £80 ; at the close of the year,!
they find that they hiame gained £95 ; what is the profit of]
each? J ( £43 14s. 2^d. A.'s share.
21. A. with a capital of $500, began trade Jan. 1, 1826,
and meeting with success, took in B. as a partner, with a
capital of $600, on the 1st March following ; four months
after, they admit C. as a partner, who brought $800 stock
at the close of the year, they fmd'the gain to be $700; how
must it be divided among the partners ?
r $250 A.'s share,
Ans. I 250B.'s *<
( 200C.'s "
3.
^ QUESTIONS.
■/'■•' '.'■ *# ' ■
1. What is fellowship 1 2. What is the rule for operating'?
When time is regarded in fieiiowship, what is it called? 4. What is
the method of operating in double fellowship ? 5, How are taxes
assessed ? 6. Flow is fsllowship proved 1
iOXlOATION.
Its
ALLI«AT10]¥.
a I!! rr.
^ IKS. Alligation is the method of mixing two or morr
himples, of different qualities, so that the compoeitioa may*
I be of a mean or middle quality. . .
Whep the quantities and prices of the simples are gives
I to find the mean price of the mixture compounded of them,
I the process is called ^Alligation Medial^
1. A farmer mixed together 4 bushels of wheat, worth 66
I pence per bushel, 3 bushela of rye, worth 32 pence per
bushel, and 2 bushels of corn, worth 28 pence per bushel ;
' what is a bushel of the mixture worth ?
It is plain that the cost of the whole, divided by the nui»>
I bar of bushels, will give the price of one bushel. ; .:C' < •
4 bushels, at ^ pence, cost 264 pence.
3 " .: . 88 "96
2 " .1 m " 56
«<
d bushels cost
u
.'(-'
.11
416 pence. ?
*^6=i46f pence, Am.
2. A grocer mixed 5 lbs. c^ sugar, worth lOd. per lb. 8
IS. worth 12d. 20 lbs. worth 14d. ; what is a poand of the
mixture worth? i4n«. 12|^d.
3. A goldsmith melted together 3 ounces of gold 20 ca>*
rats fine, and 5 ounces 22 carats fine ; what is the fioeness
of the mixture ? Ans. 21^.
4. A grocer puts 6 gallons of water into a cask contain-
ing 40 gallons of rum, w«rth 2s. 7d. per gallon ; what is a
gallon of the mixture worth?) ! i ; v < Ans. 2s. 2||d.
5. On a certain day the inercilry was observed t6 9tand
in the thermometer as follows :—^ hours of the day it stood
k 64 degrees ; 4 hour^at 70 degrees ; 2 hours at 75 degrees,
aad 3 hours at 73 degrees ; what was the nuaa, tempNature
forthal'day? .-vl ,.■';'■><)■ '\m\: -f <, t(. 'i'--^- .■ ;„
It is plain this question does not dif&i', in the mode of its
operation .from the former. Ans. 69-j^ degrees.
il HO* When the mean price or rate, and the prices or
riitOs of the several simples ,are given, to find the propor-
tions or quantities of each simple, the process is caiUed alU->
gation alternate i alligation iQternate is, therefore, the re-
verse of alligation mediq^l, and may be proved by^k. .
196
ALLIGATION.
HOe
1. A man has corn worth 40d. per bushel, which he
winhes to mix with rye worth 60d. per bushel, so that the
mixture may be worth 42d. per bushel ; what proportions or
quantities of each must he take 1
Had the price of the mixture required erceeded the price
of the corn, by just as much as it fell short of the price of
the rye, it is plain he must have taken equal quantities of
corn and rye ; had the price of the mixture exceeded the
price of the corn by only half as much as it fell short of the
price of the rye, the compound would have required twice
as much corn aa rye ; and in all cases the lexs the difierence
between the price of the mixture and that of one of the sim^
pies, the greater roust be the quantity of that simple, in pro-
portion to the other ; that is, the quantities of the simples
must be inversely as the difF:irences of theif prices from the
price of the mixture ; therefore, if these differences be mu»
tually exchanged, they will directly express the relative
quantities of each simple necessary to form the compound
required. In the above example, the price of the mixture
is 4^d. and the pricie of the corn is 40d. ; consequently the
'diflfbrence of their prices is 2d. ; the price of the rye is 50d.
Mrhich differs from the price of the mixture by'8d. There'
fore, by exchanging these differences, we have 8 bushels of
corn to 2 bushels of rye for the proportion required.
Ahs. 8 bushels of corn to 2 bushels of rye^ or in that pro-
portion.
The correctness of this result may now be ascertained by
the last rule; thus, the cost of 8 bushels of com at 40 pence
is 320 pence ; and 2 bushels of rye at 50 pence is 100 pence ;
then, 320r|-100^^20, and 420 divided by the number of
bushels, (84*2):zrl0, gives 42 pence for the price of the
inixturei< : • I ;■' '" ■' • ' '' '• -.'■' " " ■'
2w A merchant has several kinds of tea • some at 8s. 'some'
at 9s. some at lis. and some at 12s. per lb. ; what propor-
tions of each must he mix, that he may sell the compound at
lOs. pet lb.
Here we have 4 simples ; but it is plain that what has just
been proved of ^«>o will apply to any number ofpaifs, if in
each pair the price of <me simple is greater, and that of the
other iessythjoi the price of the mixture required. Hence
we have ;Uii»
lOs.
«I9fl.
ALLIGATION.
197
roportions or
nces be mu«
RULE.
The mean rate and the several prices being reduced to
the same denomination, — connect with a ctmtinued lino
each price that is loss than the mean rate with one or more
tliat is greater, and each price greater than the mean rate
witii one or more that is less.
Write the difference between the mean rate, or price, and
the price of each simple opposite the price with which it is
connected ; (thus the difference of the two prices in each
pair will be mutually exchanged) then the sum of the differ-
ences, standing against any price, will express the relative
quantity to be taken of that price.
By attentively considering the rule, the pupil will perceive
that there may be as many different ways of mixing the sim-
ples, and consequently as many different answers, as there
are different ways of linking the several prices.
We will now apply the rule to solve the last question : —
OPERATIONS.
1 =1
— l-f2=3
-2 =2
Here we set down the prices of the simples, one directly
under another, in order, from least to greatest, as this is
most convenient, and write the mean rate (10s.) at the left
hand. In the first way of linking, we find that we may take
in the proportion of 2 pounds of the teas at 8 and 12s. to I
pound at 9 and lis. l\\ the second way, we find for the
answer 3 pounds at 8 and lis. to 1 pound at 9 and 12s.
3. What proportion of sugar, at bd. lOd. and 14d/per lb.
will compose a mixture worth 12d. per lb.
Ans. In the proportion of 2 lbs. at 8 and 10 pence to six
pounds at 14 pence.
Note. As these quantities only express the proportions of
each kind, it is plain that a compound of the same mean
price will be formed by taking 3 times, 4 times, one half, or
any proportion ,of each quantity. Hence,
When the quantity of one simple is given, after finding
the proportional quantities by the above rule, we may say —
As the proportional quantity : is to the given quantity : : so
R2
Iv
198
ALLIGATION.
1196,
is each of the other proportional quantities : to the required
quantities of each.
4. If a man wishes to mh i gallon of brandy worth 16s,
with rum at 9s. per gallon, so that the mixture may be worth
lis. per gallon, how much rum must he use ?
Taking the differences ns above, we find the proportion:
to be 2 of brandy to 5 of rum ; consequently, one gallon ol
brandy will require 2J- gallons of rum. Ans. 2^ gals
5. A grocer has sugars worth 7d. 9d. and 12d. per pound
which he would mix so as to form a compound worth lOd.
per lb. ; what must be the proportions of each kind ?
Ans. 2 lbs. of the 1st and 2nd to 4 lbs. of the 3rd kind
lb. of the 1st kind, how much must he take
—if 4 lbs. what ? if 6 lbs. what ?
if 20 lbs. what?
6. If he use I
of the others ?—
if 10 lbs. what?-
Ans. to the last, 20 lbs. of the 2nd and 40 of the 3rd,
7. A merchant has spices at I6d. 20d. and 32d per lb. :
he would mix 5 lbs. of the first sort with the others, so as
to form a compound worth 24d. per lb ; how much of eacli
sort must he use ?
Ans. 5 lbs. of the 2nd and 7J- lbs. of the 3rd,
8. How many gallons of water of no value must be mixed
with 60 gallons of rum, worth 48d. per gallon, to reduce its
value to 42d. per gallon ? Ans. 8f gallons,
9. A man would mix 4 bushels of wheat at 90d. per bushel,
rye at 70d. corn at 70d. and barley at 30d. so as to sell the
mixture at 48d. per bushel ; how much of each may he use?
10. A goldsmith would mix gold 17 caiats fine with some
19, 21 and 24 carats fine, so that the compound may be 22
carats fine ; what proportions of each must he use ?
Jim. 2 of the 3 first sorts to 9 of the last.
1 1 . If he use one ounce of .the first kind, how much must
he use of the others ? What would be the quantity of the
compound? Ans. to the last, Ih ounces,
12. If he would have the whole compound consist of 15
ounces, how much must he use of each kind? if of 30
ounces, how much of each kind? if of 37i ounces how
much ? Ans. to last, 5 oz. of the 3 first, 22^ oz. of the last,
Hence, when the quantity of the compound is given, we
may say — As the sum of the proportional quantities found
by the above rule, is to the quantity required, so is each
• fl96,
) the required
dy worth 16s,
may be worth
le proportions
one galJon of
Ans. 2^ gals,
ttl. per pound,
id worth lOd.
kind?
the 3rd kind,
must he take
3. what ? —
10 of the 3rd,
32d per lb. •
others, so as
nuch of each
)s. of the 3rd,
ust be mixed
to reduce its
s. 8f gallons,
d. per bushel,
as to sell the
1 may he use ?
ne with some
id may be 22
jse ?
9 of the last,
Y much must
antity of the
t, 7^ ounces,
consist of Iti
if of 30
\ ounces how
z. of the last,
is given, we
ntities found
, so is each
H 96, 97.
DUODECIMALS.
199
Then 2-f 2
8d.
lOd.
14d.
ns.
proportional quantity, found by the rule,' to the required
quantity of each.
1.3. A man would mix a hundred pounds of sugar, some at
8d. som» at lOd. and some at I4d. per lb., so that the com-
pound may be worth 12d. per lb. ; how much of each kind
must he use ?
We find the proportions to be 2, 2 and 6.
4-6=10, and ( 2 : 20 lbs.
10 : 100 : : '^ 2 : 20 "
l6:G0 «•
14. How many gallons of water of no value, must be
mixed with brandy at 120d. per gallon, so as to fill a vessel
of 75 gallons, which may be worth 92d. per gallon ?
Ans. 174 gallons of water to 57^ of brandy,
15. A grocer has currants at 4d. Gd. 9d and Md. per lb.
and he would make a mixture of 240 lbs., so that the mix-
ture may be sold at 8d. per lb. ; how many pounds of each
sort may he take ?
Am. 72, 24, 48 and 96 lbs. ; or 48, 48, 72, 72, &.c.
Note. This question may have five different answers.
1.)
QUESTIONS.
1, What is alligation.^ 2. medial?
operating? 4. What is alli<;atiun ullernute ?
3, the rule for
5. When the price of
the ir.i.\lure, and the price of the several sitnples are given, how do you
find the proportional (]Uurilitie3 of each simple f 6. When the quantity
of one simple is given, how do you And the others ! 7. VV hen the
quantity of the whole coiiipound ia givtt>> how do you liiid the quantity
of each bimplu ? .
^ Oy. Duodecimals are fractions of a foot. The word
is derived from the Latin word duodecim, which signifies
twelve. A foot, instead of being divided decimally into ten
equal parts, is divided duodecimally into twelve equal parts,
called inches, ox primes, marked thus, ('). Again, each of
these parts is conceived to be divided into twelve other equal
parts called seconds^ ("). In like manner, each second is
conceived to be divided into twelve equal parts, called thirds
('"); each third into twelve equal j/arts called /owr^As, {"")
and so on to any extent.
200
^ •
MULTIPLICATION OP DUODECIMALS.
TI97.
In this way of dividing a foot, it is obvious that
1' inch or prime is - - - - tV of a foot,
1" second is ^ of ^
— xii
((
((
I'" thirdis^Vof-i^of-jJ^ - . =^^^^
i"" fourth is -^ of ^^ of ^ of ^^ . = Yuis^
V"" fifth is tV of tV of rV of tV of tV = ^f^Viij "
Duodecimals are added and subtracted in the same man-
ner as compound numbers, 13 of a less denomination
making one of a greater, as in the following
TABLE.
12"" fourths make
12"' thirds
12" seconds
12' inches or primes
Note. The marks, ', ", "', "", &c. which distinguish the
different parts, are called the indices of the parts or deno-
minations.
V" third,
1" second,
I' inch or prime,
1 foot.
MULTIPLICATION OF DUODECIMALS.
Duodecimals are chiefly used in measuring surfaces and
solids.
1. How many square feet in a board 16 feet 7 inches
long, and 1 foot 3 inches wide ?
Note. Length Xbreadth=superficial contents, (^ 25.)
OPERATION.
7 inches or primes= j^ of a
foot and 3 irtches=f'j of a foot ;
consequently, the product of 7'
X^'=-h\ of a foot, that is, 21'
= 1' and 9", wherefore, we set
down the 9", and reserve the
1' to be carried forward to its
kns. 20 8' 9" proper place. To multiply 16
feet by 3' is to take ^^2^ of tP = A|^ that is 48'; and th. ''
which we reserved makes 49',=4 feet 1'; we therefore set
down the 1', and carry forward the four feet to its proper
place. Then, multiplying the multiplicand by the one foot
in the multiplier, and adding the two products together, we
obtain the answer, 20 feet, 8', 9".
The only difficulty that can arise in the multiplication of
duodecimals is, in finding of what denomination is the pro-
ft-
Length 16 7'
Breadth 1 3'
4 1'
16 7'
9"
197.
duct o
as abbi
the _prfl
denomi
the ab(
16
4
■■7t:r^';
T[97.
hat
of a foot,
((
e same man-
enomination
J,
►r prime,
itinguish the
Its or deno-
lALS.
surfaces and
jet 7 inches
(!I 25.)
es=y^ of a
^2 of a foot ;
oduct of 7'
that is, 21 '
ore, we set
reserve the
ward to its
multiply 16
and th( V
herefore set
its proper
he one foot
ogether, we
)lication of
is the pro-
.,.^„. ■■-...
fld7;
»i^
v; .!
MttLTlPLTCATION OT DUODECFBIAtli.
28f
duct of any two denominations. This may be ascertained
as above, and in aU cases it will be found to hold true that
the product of any two denominations will always he of the
denomination denoted by the sum of their indices. Thus, in
the above example the sum of the indices of 7'X3' is " ;
consequently, the {Product is 21" ; and thus primes multi-
plied by primes will produce seconds ; primes multiplied by
seconds produce thirds ; fourths multiplied by 5ths produce
ninths, &/C.
It is generally most Convenient, in practice, to multiply
the multiplicand first by the feet of the multiplier, then by
the inches, &c. thus : —
ft-
16 7'
1 3'
16
4
7'
I'
9'
16 feet X 1 foot = 16 feet ; and 7'X 1
foot = 7'. Then, 16 feet X 3'=48'^
4 feet, and 7'X3'=21"=1' 9'. The two
products added together, give for the
answer, 20 feet 8' 9", as before.
ii
20 8' 9" '
2. How many solid feet in a block 16 feet 8' long, 1 foot
5' wide, and 1 foot 4' thick ?
ft-
Length, 15 8'
Breadth, 1 5'
t I.
,it
15
8'
6
6'
4//' •
22
2'
4"
s 1
4'
1
22
2'
4"
7
4'
9" 4'"
The length multiplied by
the breadth, and that pro-
duct by the thickness, gives
the solid contents.
(1133.)
:!.:•
. i • • ^
Ans. 29 7' 1" 4'" ■
From these examples we detive the following Rule :—
Write down the denominations as compound numbers, and
in multiplying, remember that the product of any two de-
nomiiiations will always be of that denomination denoted by
the sum of their indices.
im
m:
MVLTtPLlCATlON OF DUODECIMALS.
H 97, 98.
-i
BXABIPL'KS for FRAOTICIi!.
3. How many square feet in a stock of 15 boards, 12 feet
S' in length, and 13' wide ? Ans. ^5 feet 10',
4. What is the product of 371 feet 2' G" multiplied by
181 feet 1' 9" t n ^ ; Ans, 67243 feet 10' 1' 4 " 6" '.
Note. Painting, plastering, paving, and some other kmda
of work, are done by the square yard. If the contents in
square feet be divided by 9, the quotient, it is evident, will
be square yards. ' A
5. A man painted the walls of a room 8 feet 2' in height,
and 72 feet 4' in compass ; that is, the measure of all its
sides ; how many square yards did he paint ?
Ans. 65 yards 5 feet 8' 8".
6. How many cord feet of wood in a load 8 feet long, 4
fee^ wide, and 3 feet 6 inches high ?
Note. It will be recollected that 16 solid feet make a
itord foot. Ans. 7 cord feet.
7. In a pile of wood 176 feet in length, 3 feet 9' wide,
and 4 feet 3' high, how many cords?
Ans. 21 cords, 7-i^ cord feet.
8. How many cord feet of wood in a load 7 feet long, 3
feet wide, and 3 feet 4' high ; and what will it come to at
2s. per cord foot ?
Ans. 4^ cord feet, and will come to 8s. 9d.
9. How much wood in a load 10 feet in length, 3 feet 9^
in width, and 4 feet 8' in height ? and what will it cost at
$1*92 per cord ?
Ans. 1 cord and 2 j| cord feet, and it will come to $2'62^.
T 98. Remark. — By some surveyors of wood, dimen-
sions are taken in feet and decimals of a foot. For this pur*
pose, make a rule or scale 4 feet long, and divide it into feet
and each foot into ten equal parts. On one end of the rule
for 1 foot, let each of these parts be divided into ten other
«qual parts. The former division will be tenths, and the
latter hundredths of a foot. Such a rule will be found very
convenient for surveyors of wood and lumber, for painters,
joiners, 6i,c. ; for the dimensions taken by it being in feet
and decimal parts of a foot, the casts will be no other than
:^o many operations, in decimal fracticms.
10. How many square feet in a hearth stone, which, by 3
^0'»^' Iff 98,99.
INTOLUTiaN.
003
\A
ards, 12 feet
205 feet 10',
luhiplied by
1" 4'" 6"".
other kinds
contents in
evident, will
2' in height,
re of all its
5 feet 8' 8".
feet long, 4
feet make a
7 cord feet,
feet 9' wide,
f^ cord feet,
feet long, 3
coine to at
ne to 8s. 9d.
gth, 3 feet 9'
ill it cost at
le to $2*GQi,
rood, diinen-
For this pur*
le it into feet
rd of the rule
nto ten other
iths, and the
>e found tery
for painters,
being in feel
other than
, which, by a
rule, as abore described, measures 4'5 feet in lehgth, and
3^6 feet in widt^ ? And what will be its cost, at 75 cents per
st^re foot T Ans. 11'7 feet ; and it will cost $8*7T5.
11. How many cords in a load of wood 7*5 feet in length,
3*6 feet in- width, and 4'8 feet in height ? Ans. 1 cord 1 ^ ft.
13. How many cord feet in a load of wood 10 feet long,
3*4 feet wide, and3'5 feet high? > - Ans.J^^,
V , QUESTIONS.
1. What are ^liotlecimats ? 2. From what is the word tier ired ?
3. Into how many parts is a Toot usually divided, and what are the
parts called? 4. What are the other denominations ? 5. What is
understood by the indices of the denominations ? 6. In what are duo-
decimals chiefly used f 7. How are the contents of a surface bounded
by straight lines found 'I 8. How are the contents of a solid found 1
9. How is it known of what denomination is the product of any two
denominations t 10. How may a scale or rule be formed for taking
divaenaions in feet and decimal parts of a foot 1 :m ...
IJVYOIilJTIOJV. ;
f[ 99. Involution, or the raising of powers, is the mul-
tiplying any given number into itself continually a certain
number of times. The products thus prodaced are called
the powers of the given number. The number itself in called
the first power or root. If the first power be multiplied by
itself, the product is called the second power or square : if
the square be multiplied by the first power, the product is
called the third power, or cube, &c: thus :
5 is the root, or first power of 6,
5X5= 25 is the 2d power, or square of 5, =5^
5X5X5=.125 3d " cube, of 5, =53
5X5X5X5=625 4th " biquadrate, of 5,=5*
The number denoting the power is called the index, or
exponent; thus, 5* denotes that 6 is raised or involved to the
4th power. ■■ ^s',. ,.
1. What is the square or 2d power of 7.^ Ans. 49.
2.
3.
4.
5.
6.
((
({
<(
of 30 ? An3 900.
« of 4000? ' Ans. 16000000.
cube or 3d power of 4 ? Ans. 64,
" of 800.? ilns. 512000000.
4th power of 60 ? Ms. 12960000.
204
EVOLUTION.
!:/
7. What is the square of 1 ?
of 4?
8. What is the cube of 1 ?-
of 4?
!I 99, 100.
of 2? of 3?
Ans. 1, 4, 9, and 16.
.of 2? of 3? .
^101
Ans, 1, 8, 27, and 64.
f|? of|?.
^ws. I, ^, If,
10. What is the cube of § ? -, — of | ? of | ?
9. What is the square of §?-
^ns. ^, /A, and ^f f .
• the 5th power of ^ ?
Ans. ^, and -j^.
— the cube ?
11. What is the square of ^ ? —
12. What is the square of 1'5?
Ans, 2'25, and 3'375.
13. What is the 6th power of r2'? Ans, 2*985984.
14. Involve 2^ vo the 4th power:
Note, A mixed number like the above may be reduced to
an improper fraction before involving : thus, 2^=J ; or it
irtay be reduced to a decimal j thus, 2i=:;:2*25.
Ans. m^=25ifi.
15. What is the value of 7*, that is, the 4th power of 7 .^
Ans. 2401,
16. How much is 9^ ? 6& ? 10* ?
^«s. 729, 7776, 10000.
17 How much is 2^ ? 3^ 1 4^ ? 53 ? — ^
6* ? 103 1 jing^ to the iggt^ 100000000.
The powers of the nine digits, from the first power to
the fifth, may be seen in the following
. TABLE.
Roots
1 1
2|
3|
4|
o\
6|
7|
8|
9
Squares
1
4|
9|
16 1
25 1
36 1
49 j
64|
81
Cubes
1
«l
27 i
64 1
125 1
216 1
343 1
512 1
729
Biquadratsl
|16|
81!
256 1
625 1 1296 1
2401 1
4096 1
6561
59049
Stursolids
1
1 32 1 243 1
1024 1
3125 1
7776 1 16807 1 32768 |
CV0LUTIO]¥,
^ 100. Evolution, or the extracting of roots, is the
method of finding the root of any power or number.
' The root, as we have seen, is that number which, by a
continual multiplication intd itself, produces the given power.
The square root is a number which, being squared, will
U 99, 100.
of 3?
,9, and 16.
f3?
27, and 64.
[)f|?
*• t» ir* 1 4"'
,\, and Iff.
iver of^?
s. ;J, and 7^.
, and 3'375.
s, 2*985984.
e reduced to
2^=1 ; or it
VV=25^fi-
lower of 7 ?
Ans. 2401.
r776, 10000.
— 53? — -
100000000.
irst power to
fllOl
EXTRACTION OF THE SQUARE ROOT.
265
8|
9
64|
81
512 1
729
4096 1
6561
{2768 1
59049
produce the given number ; and the cube, at third root, is a
number which, being cubed or involved to the third power,
will produce the given number ; thus, the square root of 144
is 12, because 12^=144 ; and the cube root of 343 is 7, be-
cause 73, that is, 7X7X7=343 ; and so of other numbers.
Although there is no number which will not produce a
perfect power bv involution, yet there are many numbers
of which precise roots can never be obtained. But by the
help of decimals, we can approximate, or approach towards
the root to any assigned degree of exactness. Numbers,
whose precise roots cannot be obtained, are called surd
numbers, and those whose roots can be exactly obtained,
are called rettional numbers*
The square root is indicated by this character \/ placed
before the number ; the other roots by the same character
with the index of the root placed over it. Thus, the square
root of 16 is expressed \/16 ; and the cube root of 27 is
3 5
expressed v27, and the 5th root of 7776, V 7776.
When the power is expressed by several numbers, with
the sign -j- or — between them, a line, or vinculum, is
drawn from the top of the sign over all the parts of it ; thus
the square root of 21 — 5 is \/21 — 5, &c.
roots, is the
iber.
which, by a
given power,
squared, will
Fxtraction of the Square Root.
IT 101. To extract the square root of any number is
to find a number, which, being multiplied into itself, shall
produce the given number. .-
1. Supposing a man has 625 yards of carpeting, a yard
wide, what is the length of one side of a square room, the
Soor of which the carpeting will cover ? that is, what is one
side of a square, which contains 625 square yards ?
We have seen (IT 32) that the contents of a square sur-
face is found by ipultiplying the length of one side into itself,
that is, by raising it to the second power ; and hence, hav-
ing the contents (625) given, we must extract its square
rout to find one side of the room.
This we liiust do by a sort of. trial, and
ist. We will endeavour to ascertain how many figures
■H..
806
EXTRACTION OF THE 8QUABE ROOT.
nioi.
OPERATION.
625(2
4
225
Fig.X,
there will be in the root. This we can easily do, by point-
ing off the number, from units, into periods of two figures
each ; for the square of any root always contains just twice
as many, or one figure k»s than twifce as many figures, as
are in the root ; of which truth the pupil may easily satisfy
himself by trial. Pointing off the number, we find that the
root will consist of two figures
— a ten and a unit.
2d. We will now seek for
the first figure, that is, for the
tens of the root, and it is plain
that we must extract it from
the left hand period 6, (hun-
dreds ) The greatest square
in 6 (hundreds) we find, by
trial, to be 4, (hundreds) the
root of which is 2, (tens==:20)
therefore, we set 2 (tens) in
the root. The rooty it will be
recollected, is one side of a
square. Let us, then, form a
square, (A. fig. 1,) each side
of which shall be supposed 2
tens, = 20 yards, expressed
by the root now obtained.
The contents of this square are 20x20=400 yards, now
disposed of, and which, consequently, are to be deducted
from the whole number of yards,'(625) leaving 225 yards.
This deduction is most readily performed by subtracting
the square number 4, (hundreds) or the square of 2, (the
figure in the root already found) from the period 6, (hun-
dreds) and bringing down the next period by the side of the
remainder making 225, as before.
3d. The square A. is now to be enlarged by the addition
of the 225 remaining yards ; and in order that the figure
may retain its square form, it is evident the addition must
be made on two sides. Now, if the 225 yards be divided by
the length of the two sides, (20-|-20=40) the quotient will
be the breadth of this new addition of 225 yards to the sides
c d and 6 c of the square A.
■a
o
a
!1 101.
do, by point-
f two figures
ins just twice
ly figures, as
easily satisfy
; find that the
of two figures
nit.
low seek for
hat is, for the
ind it is plain
[tract it from
iriod 6, (hun-
eatest square
we find, by
tundreds) the
2, (tens=20)
t 2 (tens) in
ooty it will be
me side of a
then, form a
,) each side
i supposed 2
expressed
obtained.
) yards, now
)e deducted
225 yards.
subtracting
•6 of 2, (the
iod 6, (hun-
e side of the
the addition
it the figure
ddition must
e divided by
quotient will
tothesidcft
TlOl.
EXTAACnON OF THE SQUARE ROOT.
907
But our root already found, =2 tens, is the length of one
side of the figure A ; we therefore take double this root,s=4
tens, for a divisor.
OPERATION CONTINUED.
625(25
4
45)225
225
Fig.^.
20 yds.
5 yds.
m
20
5
•a
B
5
5
o
a—
D—
100
25
1
c
A
'
c
■
/
•o
^
§
20
20
20
5
490
100
a
b
20 yds.
5 yds.
The divisor 4 (tens)
is in reality 40, and we
are to seek how many
times 40 is contained
in 225, or, which is the
same thing, we may
seek how many times 4
(tens) is contained in
22, (tens) rejecting the
oi right hand figure of the
'S, dividend, because we
* have rejected the ciph-
er in the divisor. We
find our quotient, that
is, the breadth of the
g addition, to be 5 yards;
^ but if we look at}?^. 2,
S* we shall perceive that
this addition of 5 yards
to the 2 sides does not
complete the square;
for there is still want-
ing in the comer D, a
small square, each side
of which is equal to this last quotient, 5; we must therefore
add this quotient 5, to the divisor 40, that is, place it at the
right hand of the 4 (tens) making it 45 ; and theh the whole
divisor, 45, multiplied by the quotient, 6, will give the con-
tents of the whole addition around the sides of the figure
A, which, in this case, being 225 yards, the same as our
dividend, we have no remainder, and the work is done.
Consequently, J?^. 2 represents the floor of a square room,
25 yards on a side, which 625 square yards of carpeting
will exactly cover.
The proof may be seen by adding together the several
parts of the figure, thus : —
m
EXTRACTION OF THU SQUARE ROOT.
!l 10!
, r. "
<(
((
((
(I
Or we may prove it
by involution, thus :—
25X25=:GJ3, as be-
fore.
l^be square A contains 400 yards.
* figure B " 100
" C " 100
** D « 25
"; vf,-sf' ■. , . • r- — ^
..v|,n. .» iVoo/ 625 ,..
From this example and illustration, we* derive the follow-
ing general , . _
RULE
*^ ' Fbf the JSxtraction of the Square Root.
1. Point off the given number into periods of two figures
each, by putting et, dot over the units, another over the hun-
dreds, and 80 on^ These dots show the number of figures
of which the root will consist.
• I II. Fiitd the greatest square number in the left hand pe-
riod, and write its root as a quotient in division. Subtract
the square number from the left hand period, and to the,
remainder bring down the next period for a dividend.
III. Double the root already found for a divisor ; seek
liow many times, tjie divisor is contained in the dividend,
excepting the right hand figure, and place the result in the
root, and also at the right hand of the divisor ; multiply the
divisor, thus augmented, (by the last figure of the root, and
subtract the product frorn the dividend; to the remainder
bring down the; neict period for a new dividend.
IV. Double tUe root already tbund for a new divisor, and
continue the operation as before, until all the periods arc
brought downf
Note 1. IC We double the right band figure of the last
divisor, we sh^U have the double of the root.,
Note'i. As the value of figures, whether integers or de-
cimals, is determined by their distance from the place of units
do we must always begin at unit's place to point off the givea
number, and,.if it be a mixed number, we must point it off
both ways £rom units, and if there be a deficiency in any pe-
riod of decimals, it may be supplied by a cipher. Tt is plain
the root must always consist of so many integers and decimals
3ks there are periods belonging to each in the given number.
,« ;■" EXAMPLES FOR PRACTICE.
%. What i^ the sq^uane root of 1,0342^6 ?
If 101
may prove it
tion, thus :—
:6'i5, as be-
ve the follow-
90t.
)f two figures
over the hun-
ber of figures
left hand pe-
•11. Subtract
, and to the.
lividend.
divisor ; seek
the dividend,
result in the
multiply the
the root, and
e remainder
I.
divisor, and
periods are
! of the last
egers or de-
)laceofunitB
off the given
t point it off
;y in any pe-
r. Tt is plain
ind decimals
en number.
T 101, 103. BXTR ACTION OP THE ■QVAM HOOT.
^ ; OPERATION.
' t ■ 1
l6342t)56 (3316 Am.
, • .' ' , ■
9
!
1 t
63) 134
134
(
641) 1036
t
What is
641
':
6436) 38556
38550
5.
the square root of 43364 ?
1 1
OPERATION.
What is
43264 (308, An$.
4
408) 3264
3364
4.
the square root of 993901 ?
Ans. 999.
5.
<(
234' 99?
Ans. ]5'3.
6.
<(
984*5192369241?
^«5.31'05671.
7.
tt
'001296?
Ans. '036.
8.
«' ,
•3916?
Ans. '54.
9.
«
36373961
? Ans. 6031.
10.
«
" " 164 ?
Atis. 13'8-f-
IT I OS. In this last example, as there was a remain((er
after bringing down all the figures, we continued the opera-
tion to decimals, by annexing two ciphers for a new period,
and thus we may continue the operation to any assigned
degree of exactness; but the pupil will re iidily perceive that
he can never in this manner obtain the precise root ; for the
last figure in each dividend will always be a cipher, and the
last figure in each divisor is the same as the last quotient
figure; but no one of the nine digits u>i>ltiplied into itself.
S3
9tit
•UrriiBMBNT TO Till SQUARB BOOT.
Uioa
1102
produces a number ending with a cipher ; therefore, what-
ever be the quotient figure, there will still be a remainder.
11. What is tiie square root of 51? -- Ans. 1'73-|-.
12. " •♦ •• 10 7 ylwi-. 34«4-.
13. " " " 184*2? AnitA^'57-\-.
14. " " " ^?
Note. — We have seen (IT 99, ex. 9,) that fractions are
squared by squaring both the numerator and the denomina-
tor. Hence it follows, that the square root of a fraction is
li)und by extracti.ig the root of the nuir»erator and of the
denominator. The rcx)t of 4 is 2, and the root of 9 is 3.
Ans. ^.
15. What is the square root of -J^? Ans. ^.
1«. " " •' t'd^? Ans.^^.
17. •• ♦* ♦• j\^T Ans. ^2=1
la " " " 2dir Ans. A}.
When the numerator and denominator are not exad
squares, the fraction may be reduced to a decimal, and the
approximate root found, as directed above.
1 9. What is the square root of f =75 ? Ans. 'SfiG-f.
20. " •• " JJ^? Ans. *9\2'\-.
SUPPLEMENT TO THE SQUARE ROOT.
QUESTIONS.
I. Whnt ia involution ? 2. What is understood by a power if 3.
the first, the second, Ur.; third, the fourth power 1 4. What is
the index, or exponent? 5. Hoiv do you involve a number to any le-
(|uircd power ? 6, Whiit is cvotulion ? 7. What ia a root 1 8. Can
Kie precise raot <>f nil numbera be found ? d. What is a s<ird number f
lU. — — a rational 1 11. Whnt is it to extract the square root of any
number? 12. Why is Ihs given sum pointed into periods of two
figures each ? 1^. Why do we double ihe root for a divisor? 14.
Why do we, in dividing, reject tUe right hand figure of the dividend ?
15. Why do we place the quotient figure lo the rizht hand of thtj divi-
•or! 1Gb Huw may we prove the work 1 17. Why do we point off
mixed numbers both ways fiom units ? 18. When there ia a rtimain-
der, how may wc continue the oi>eration 1 19, Why can we never ob-
tain the precise root of »uid numbera ? 20. How do we extract the
square root of vulgar fractions 'i
EXERCISES.
1> A general has 409i> men; how many must he place in
tank and tile, to form theiu into a square 1 .<4»s. 04.
2. If J
rods doe
3. H(
tainini^r
4. Th
IS 5184
of equal
5. A.
other CO
field con
how mai
6. If
how mu(
iiig 4 tin
7. Ift
of one 4
times as
large ?
8. Iti
of a part
length ai
Note.
two equa
9. Iw
that my
how mat
each rov
10. T
square r
required
11. T
is the di
Note.
to the sc
Therefo
diamete
iquare i
l± 1
IT 103
sforc, what-
remainder.
[ns. 1'73-|-.
Iwi. 3'1«4-.
u. 13'57-|-.
actions are
; denomina-
1 fraction is
and of the
of 9 is 3.
Ans. ^.
/1ms. 4.
Ans. -^ff.
Ahs. -^7=^.
Ans. 4^.
e not exact
lal, and the
Ins. '866-1-.
Ins. '9I2-I-.
lOOT.
power ? 3.
4. Whxlis
)er to any re-
wtl 8. Can
s'lrd number f
re root of any
sriods of two
divisor? 14.
the dividend f
id of tbd divi-
xie point off
i is a rtimain-
we never ob-
re extract the
he place in
.<4»s. 04.
1102
•UPPLEMBNT TO THE IQUilFV: ROOT.
311
2. If a square field contains 2025 square rods, how many
rods does it measure on each side ? Anf, 45.
3. How many trees in each row of a square orchard con-
tainin^jT 5625 trees ? Ans. 75.
4. There is a circle wliose area^ or superficial contents,
is 5184 fc>et ; what will be the len^h of the side of a square
of equal area? \/5184-=72 feet, i4n.<!.
5. A. has two fields, one containing 40 acres, and the
other containing 50 acres, for which B. offers him a square
field containing the same number of acres as both of these;
how many rods must each side of this field measure ?
Ans. 120 rods.
6. If a certain square field measure 20 rods on each side,
how much will the side of a square field measure, contain-
ing 4 times as much.' /v/20X 20X4=40 rods, Ans.
7. If the side of a square be 5 feet, what will be the side
ol'one 4 times as large? 9 times as large.' 16
times as large ? — — 25 times as large ? — ■ — 36 times as
large ? .4«s. 10ft. 15ft. 20ft. 25ft. and 30ft.
8. It is required to lay out 288 rods of land in the form
of a parellplogram, which shall be twice as many rods in
length as it is in width.
Note. If the field be divided in the middle, it will form
two equal squares. Ans. 24 rods long and 12 rods wide.
9. I would set out, at equal distances, 784 apple trees, so
that my orchard may be four times as long as it is broad •
how many rows of trees must I have, and how many trees in
each row ?^ Ans. 14 rows, and 56 in each row.
10. There is an oblong piece of land, containing 192
square rods, of which the width is ^ as much as the length ;
required, its dimensions. Ans.^16 by 12.
11. There is a circle whose diameter is 4 inches; what
is the diameter of a circle 9 times as large?
Note. The areas, or contents of circles are in proportion
to the squares of their diameters, or of their circumferences.
Therefore, to find the diameter required, square the given
diameter, multiply the square by the given ratio, and the
square root of the product will be the diameter required.
\/4X4X9=12 inches, Ans.
ViL There are two circular ponds in a gentleman's plea^*
JfR.'.; .■Jij'vi
212
BUPPLBMJBNT TO THB SQUARB ROOT. 51 lOSlff 104.
I^'l
sure ground; the diameter of the less is 100 feet, and the
greater is 3 times as large; what is its diameter ?
^/i5. 173'2+ft.
13. If the diameter of a circle be 12 inches, what is the
diameter of one l as large ? . Ans. 6 inches.
TI I03» 14. A carpenterhas a large wooden square;
one part of it is 4 feet long, and the other 3 feet long; what
is the length of a pole that will just reach from one end to
the other (
„ A ^
JVote. — A figure of three sides
is called a triangle, and if one of
the corners be a square corner,
or right angle, like the angle at
B. in the annexed figure, it is
called a right-angled triangle,
of which the square of the long-
est side A. C. (called the hypo-
tenuse) is equal to the sum of
the squares of the other 5i sides,
A. B. and B. C.
3
u
a
u
Busti.
a
4*=16, and 32=9 ; then \^9-\-lG=^^ feet, Ans.
15. If, from the corner of a square room, 6 feet be mea-
mired off one way, and 8 feet the other way, along the sides
of the room, what will be the length of a pole reaching from
point to point ? Ans. 10 feet.
16. A wall is 33 feet hi^h, and a ditch before it is 24
feet wide ; what is the length of a ladder that will reach from
the top of the wall to the opposite side of the ditch ?
Ans. 40 feet.
17. If the ladder be forty feet, and the wall SI feet, what
is the width of the ditch? Ans. 24 feet.
18. The ladder and ditch given, required the wall,
Ans. 32 feet.
19. The distance between the lower ends of two equal
rafters is 32 feet, and the height of the ridge above the beam
on which they stand is 12 feet ; required, the length of each
rafter. Ans. 21) feet.
20. There is a building 30 feet in length and 22 feet in
width, an
side ; th(
building,
is ten fee
is the dis
eaves? ai
middle oj
one end ?
Ans. in o
21. Tl
what is tl
22. Tl
many roc
rods apai
23. Tl
tance is 1
of the sic
length oi
length, b
conseque
power, g
Hence
number (
body, of
number
1. Wl
each sid
2. Ho
on each
3. He
containi:
Note.
4. Vti
feet ? —
feet?
II 108
set, and the
;. 173*2-f-ft.
what is the
s. 6 inches.
len square;
long; what
one end to
r three sides
md if one of
tare corner,
the angle at
figure, it is
jrf triangle,
of the long-
jd the hypo-
the sunt of
her 5i sides,
5 feet, Ans.
ieet be mea-
ng the sides
iching from
ins. 10 feet,
ore it is 24
I reach from
ch?
ins. 40 feet.
I feet, what
ins. 24 feet.
walJ,
ins. 32 feet,
f two equal
ve the beam
igth of each
ins. 2!) feet.
221 feet in
ffl04.
EXTRACTION OF THE CUBE ROOT.
813
width, aad the eaves project beyoni the wall a foot on every
side ; the roof terminates in a point at the centre of the
building, and is there supported by a posty the top of which
is ten feet above the beams on which the rafters rest; what
is the distance from the foot of the post to the coi'ners of the
eaves? and what is the length of a rafter reaching' to the
middle of one side ? a rafter reaching to the middle of
one endl and a rafter reaching to the corners of the eaves T
Ans. in order, 20ft. ; 15*62-j-ft. ; 18'86-|-ft. ; and 22'36-|-ft.
21. There is a field 800 rods long and 600 rods wide ;
what is the distance between two opposite corners ?
Ans. 1000 r'^ds.
22. There is a square field containing 90 acres; how
many rods in length is each side of the field T and how many
rods apart are the opposite corners ?
If/ /-'I J li Ans. 120 rods; and 169*7+ n)d3.
23. There is a square field containing 10 acres ; what dis-
tance is the centre fi-om each corner ? Ans. 28'28*f- rods.
EXTRACTION OF THE CUBE ROOT.
tl 10^. A solid body, having six equal sides, and each
of the sides an exact square, is a cube, and the measure in
length of one of its sides is the root of that cube ; for the
length, breadth, and thickness of such a body are all alike;
consequently, the length of one side, raised to the third
power, gives the solid contents. See ^ 33.
Hence it follows, that extracting the cube root of any
number of feet, is finding the length of one side, of a cubic
body, of which the whole contents will be equal to the given
number of feet.
1 . What are the solid contents of a cubic bloclj, of which
each side measyres 2 feet.^ Ans. 23=2X2X2=8 feet.
2. How many solid feet in a cubic block, measuring 5ft.
on each side.' " , .4ns. 5^^=125 feet.
3. How many feet in length is each side of a cubic block
containing 125 solid feet? Ans. v'12o=5 feet.
Note. The root may be found by trial.
4. W'hat is th6 side qf a cubic block containing 64 solid
feet ? 27 solid feet ? -^^216 solid feet ? 512 solid
feet? '. . r u ) - i4n5. 4ft. ; 3ft.; 6ft.; and 8ft.
i
S14
EXTRACTION OP THE CUBE ROOT.
moi
OPERATION.
13824 (2
8
5. Supposing a man has 13824 feet of timber, in sepa-
rate blocks^ of one cubic foot each ; he wishes to pile them
up in a cubic pile ; what will be the length of each side of
such a pile ?
It is evident, the answer is found by extracting the cube
root of 13824 ; but this number is so large, that we cannot
so easily find the root by trial as in the former examples ;
We will endeavor, however, to do it by a. sort of trial; and
1st. We will try to ascertain the number of figures, of
which the root will consist. This we may do by pointing
the number off into periods of 3 figures each (fj 101, ex. 1.)
Pointing off, we see, the
root will consist of two fif^ures,
a ten and a unit. Let us then
seek for the first figure, or
tens of the root, which must
be extracted from the left
hand period, 13 (thousands.)
The greatest cube in thirteen
(thousands) we find by trial,
or by the table of powers, to
be 8 (thousands) the root of
which is 2 (tens;) therefore,
we place 2 (tens) in the root.
The root, it will be recollect-
ed, is one side of a cube. Let
us then form a cube, (fig. 1.)
each side of which shall be
supposed 20 feet, expressed
by the root now obtained.
The contents of this cube are
20X20X20=8000 solid feet
which are now disposed of.
6824
Fig. I.
C 20
D
400
20
8000 feet, Contents.
and which, consequently, are to be deducted from the whole
number of feet, 13824. 8000 taken from 13J^24, leave
5824 feet. This deduction is most readily performed by
subtracting the 'cubic number, 8, or the cube of 2, (the
figure of the root already found) from the period 13, (thou-
sands) and bringing down the next period by the side of the
remainder, making 5824, as before.
2d. The cubic pile A D is now to be enlarged by tlw
t|104
Divis.
20
^ 51 104.
iber, in sepa<
to pile them
' each side of
ting the cube
at we cannot
er examples ;
of trial; and
of figures, of
) by pointing
n01,ex. 1.)
we see, the
)f two figures,
Let us then
St figure, or
which must
om the left
(thousands.)
)e in thirteen
find by trial,
jf powers, to
the root of
;) therefore,
) in the root.
be recoUect-
' a cube. Let
ube, (fig. 1.)
ich shall be
t, expressed
w obtained.
this cube are
00 solid feet
disposed of,
)m the whole
i?24, leave
erformed by
of 2, (the
d 13, (thou-
le side of the
rged by tli«
n04
EXTRACTION OF THE CUBE ROOT.
215
addition of 5824 solid feet, and, in order to preserve the
cubic form of the pile, the addition must be made on one
half of its sides, that is, on three sides, a, 6, and c. Now,
if the 5824 solid feet be divided by the square contents of
these three equal sides, that is, by 3 times, (20X20=400)
=1200, the quotient will be the thickness of the addition
made to each of the sides a, 6, c. But the root 2, (tens)
already found, is the length of one of these sides ; we there-
fore square the root 2, (tens)=20X 20=400, for i\ie square
contents 6f one side, and multiply the product by three, the
number of sides, 400x3=1200, or, which is the same in
effect, and more convenient in practice, we may square the
2, (tens) and multiply the product by 300, thus, 2X2=4,
and 4X300=1200, for the divisor, as before.
The divisor, 1200, is con-
OPERATIONS CONTINUED. ^^.^^^ j^ ^^^ ^.;. ^^^ j' ^ ^^^^^ .
13824 (24 Root, consequently, 4 feet is the
^ thickness of the addition made
to each of the 3 sides a, h, c.
Dims. 1200)5824 jDiridenrf. ^nd 4X1200=4800, is the
solid feet contained in these
4800
960
64
5824
0000
Fig. IL
20
additions ; but if we look at
fig. 2, we shall perceive that
this addition to the 3 sides
does not complete the cube ;
for there are deficiencies in
the three corners n, n, n. Now
the length of each qf these de-
ficiencies is the same as the
length of each side, that is, 2
(tens)=:20, and their width
and thickness are*^ each equal
to the last quotient figure (4) ;
their contents, therefore, or
20 the number of feet required to
Jill these deficiencies, will be
found by multiplying the
square of the last quotient
20 figure (42)=16, by the length
of all the deficiencies, that is,
by 3 times the length of eacA
S)6
EXTRACTION OF THE CUBE ROOT.
I*
'i I.,
If 104
side, which is expressed by the former quotient figure, 2
(teas.) a times ii (tens) are (i (tens)=60 ; or, what is the
same in effect, and more convenient in practice, we may
multiply the quotient figure 2 (tens) by 30, thus, 2X30=
(iO, as before; then, (iUX 10=900, contents of the three
Aleficiencies, «, w, ?t. , ,
Looking at fig. 3, we per-
ceive there is still a deficiency
in the corner where the last
blocks meet. This deficiency
is a cube, each side of which
is equal to the last quotient
20 fiorure, 4. The cube of 4,
therefore, (4X4x4=64) will
be the solid contents of this
corner, which, in figure 4, is
seen filled.
Now, the sum of these seve-
ral additions, viz. 4800+960
-f64=5824, will make the
subtrahend, which, subtracted
from the dividend, leaves no
remainder, and the work is
donej
Figure 4 shows the pile
which 13824 solid blocks of
one foot each would make,
when laid together, and the
root 24 shows the length of
24 re«;t one side of the pile. The
correctness of the work may be ascertained by cubing the
side now found, 243, ti^ug 24X24X24=13824, the given
number ; or it may be proved by adding together the con-
tents of all the several parts, thus—
/Ve^— SOO'Jrrrcontents of fig. 1.
4800=addition to the sides a, 6, c, fig. 1.
960= " to fill the deficiencies n, n, n, fig. 2.
64= " to fill the corner c, e, c, fig. 4.
Fig. IV.
24 (let.
13S24=contents of the whole pile, figure 4, — 21 feet
on each side.
If 104
ient figure, 2
r, what is the
;tice, we may
hus, 2X30=
I of the three
g. 3, we per*
II a deficiency
here the last
liis deficiency
side of which
last quotient
cube of 4,
X 4=64) will
iitents of this
n figure 4, is
of these seve-
5. 4800+960
11 make the
;h, subtracted
d, leaves no
the work is
ws the pile
id blocks of
i^ould make,
ler, and the
le length of
pile. The
cubing the
4, the given
ler the con-
11.
n, M, fig. 2.
Ifig. 4.
4,-21 feet
II 104.
EXTRACTION OF THE CVBB ROOT.
^^S^r
From the foregoing example and illustration, we derive
the following • ■:r;!'r t, ' - . ^^
RULE
For Extracting the Cube Root.
I. Separate the given number into periods of three figures
each, by putting a point ever the unit figure, and every
bird figure beyond the place of units.
II. Find the greatest cube in the left hand period, and
)ut its root in the quotient.
III. Subtract the cube thus found from the said period,
and to the remainder bring down the next period, and call
this the dividend.
IV. Multiply the square of the quotient by 300, calling
it the divisor.
V. Seek how many times the divisor may be had in the
dividend, and place the result in the root ; then multiply
the divisor by this quotient figure, and write the product
under the dividend.
VI. Multiply the square of this quotient figure by the
former f^ure or figures of the root, and this product by 30,
and place the product under the last ; under all, write the
cube of this quotient figure, and call their amount the sub-
trahend.
VII. Subtract the subtrahend from the dividend, and to
the remainder bring down the next period for a new divi-
dend, with which proceed as before ; and so on, till the
whole is finished.
Note 1. — If it happens that the divisor is not contained
in the dividend, a cipher must be put in the root, and the
next period brought down for a dividend.
Note 2. — The same rule must be observed for continuing
the operation, and pointing olT for decimals, as in the square
root.
Note 3. — The pupil will perceive that the number which
we call the divisor, when multiplied by the last quotient
figure, does not produce so large a number as the real sub-
trahend ; hence, the figure in the root must frequently bQ
smaller than the quotient figure.
EXAMPLES FOR PRACTICE.
6. What is the cube root of 1860867 ? . -V; "^
T
ill
m ^^
wm
r;"'»
ti'"
Sf 1
ill
i
I \
218
8UPPLXMENT TO THE CUBE AOOT/
;1li04
^tri'Trs: .>," /no-Ji^-iijijiiw operation.
1860867 (123 Ms. ^":'' '
- 1
M ...v..
12x300=300) 860 first Dividend.
600 ..«;•!.: i :
, 22 X 1x30= 120 ;;, .. ^
23 = 8 ..; .,;/;;■ .:,■• ';
.M'[i/t;i uii i ;? 728 first Subtrahend.
122^X300=43200) 132867 second Dividend.
82X12X30=
., .. .33=
129600
3240
27
the
>i .1
000000
7. What is the cube root of 373248? '
8. " " " 21024576?
132867 second Subtrahend.
9.
10.
11.
12.
. ■■>■
«
84*604519?
'000343?
8 7
Ans. 72.
ilns. 276,
Ans. 4'39
il»5. '07.
^ns. 1'25-f.
ilH5. ^
Note. See tf 99, ex. 10, and H 102, ex. 14.
13. " " " ^ff? ^4«s.f
M4. " , f, " « t5?^^? .; ilns.-jV
15. " V, " " sixf^ ^«s.'125+.
16. " ..V " T^? . . Ans.i
SUPPLEMErjfT TO THE CUBE ROOT.
QUESTIONS.
1. What is a cube ? 2, What is understood by the c^le root? 3.
What is it to extract the cube root ? 4. Why is the square of the quo
tivo.nt multiplied by 300 Tor a divisor 1 5, Why, in finding the subtra
hend, do we multiply the square of the last quotient figure by 30 times
the former figure of the root 1 6, Why do we cube the quotient figure ?
7. How do we prove the operation 1
^105.
1. Wl
eet long
2. Th
many sol
3. Ho
what woi
4. Th.
what woi
— 64
5. Th(
what is i
— 64
cube
heir conl
iroportioi
all solid f
6. If a
ill be thi
lbs.?
7. Ifa
will be th(
8. Ifa
what is thi
12
9. The
er, and tl
mailer gl
10. Ift
he diatne
ivould it U
11. Ift
earth, and
diameter c
12. Th
ftf the less
■ I
11104
s.
end.
rahend.
1 Dividend.
If 105.
SUPPLEMENT TO THE CUBE ROOT.
319
'.•t
\
Arts. 72.
ilns.276,
Atts. 4<39.
Ans. *07.
Ans. 1'25-f,
./Ins. ^,
ilns. y'j,
^ns.'125-i-
Ans. i
lOOT.
c-l e j-oot ? 3.
lare of the quo
ling the subtra-
ire by 30 times
[uotieni figure ?
5.
v: ^'"-^ ".'? • EXERCISES.!' "^'^ -- • «' • V
1. What is the side of a cubical mound, equal to one 288
feet long, 216 feet broad, and 48 feet high ? Ans. 144 ft.
2. There is a cubic box, one side of which is 2 feet ; how
many solid feet does it contain ? Ans 8 feet.
3. How many cubic feet in one 8 times as large ; and
what would be the length of one side? > y i u<i
Ans. 64 solid feet, and <Mie side is 4 feet-
4. There is a cubical box, one side of which is 5 feet ;
what would be the side of one containing 27 times as much ?
64 times as much ? 125 times as much ?
Ans. 15, 20, and 25 feet.
There is a cubical box measuring 1 foot on each side ;
what is the side of a box 8 times as large ? 27 times ?
— 64 times 1 Ans. 2, 3, and 4 feet.
^ lOtS. Hence, we see that the sides of cubes are as
he cube roBts of their solid contents^ and consequently,
heir contents are as the cubes of their sides. The same
d Subtrahen(l.ftf(>po''tioii is true of the similar sides, or of the diameters of
all solid figures of similar forms.
6. If a ball weighing 4 lbs. be 3 inches, in diameter, what
vill be the diameter of a ball of the same metal, weighing
)2 lbs. ? 4 : 32 : : 3* : 63 . Jlns. 6 inches,
7. If a ball, 6 inches in diameter,, weigh 32 pounds, what
will be the weight of a ball 3 inches in diameter ? Ans. 4 lbs.
8. If a globe of silver, one inch in diameter, be worth $6,
what is the value of a globe one foot in diameter ?
^ns. $10368.
9. There are two globes ; one of them is 1 foot in diame-
ter, and the other 40 feet in diameter ; how many of the
mailer globes would it take to make one of the laiig^r ?
Ans.^Aim.
10. If the diameter of the sun is 112 times as much as
the diameter of the earth, how many globes like the earth
would it take to make one as large as the sun ? Ans. 1404928.
11. If the planet Saturn is 1000 times as large as the
eartl^ and the earth is 7900 miles in diameter, what is the
diameter of Saturn ? Ans. 79{X)0 miles.
12. There are two planets of equal density ; the diameter
of the less is to that of the larger as 2 to 9 ; what is the ratio
t)i" th>#tolidities ? Ans. , f ^ ; or, as 8 to 729.
Ih'
:i
m ■ ■
lift,,
mi
220
ARITHMETICAL PROGRESSION. ^ 105, 106.
f|106.
Note. The roots of most powers may be found by the
square and cube root only : thus, the biquadrate or 4th root
is the square root of the square root ; the6th root is the cube
root of the square root ; Ihe 8th root is the square root of
the 4^h root ; the OtKroot is the cube root of the cube root,
d&c. Those roots, viz. the 5th, 7th, 11th, &x. whith arc
not resolvable by the square and cube roots, seldom occur;
and when they do, the work is most easily performed by
logarithms ; for if the logarithm of any number be divided
by the index of the root, the quotient will be the logarithm
ef tlie root itself. . -
ARlTHi«IETl€AL PR0«R£SS10iV.
II 106, Any rank or series of numbers more than two,
increasing- or decreasing by a constant difference, is called
an Arithmetical Series, or Prf^ression.
When the numbers are formed by a continual addition of
the common difference, they form an ascending series ; but
when they are formed by a continual 5t/6^rac^}0» of the com*
mon difference, they form a descending series,
rnt i 3, 5, 7, 9,11,13,15, &c. is an ascending series,
^""^' ) 15,13,11, 9, 7, 6, 3, &c. is a descending "
The numbers which form the series are called the tema
of the series. The first and last terms are the extremes^
and the other terms are called the means.
There are five things in arithmetical progression, any
three of which being given, the other two may be found :—
1st. The /frs* term.
2d. The last term.
3d. The number of terms.
4th. The camTnon difference.
5th. The sum of all the terms.
1. A man bought 100 yards of cloth, givilig 4d. for the
first yard, 7d. for the second, lOd. for the third, and so on
with a common difference of 3d. ; what was the cost of the
last yard ?
As the common difference, 3, is added to every yard ex-
cept the last, it is plain the last yard must be 99X3; = 297
)>ence more than i\iQ first yard. Ans. 301 |ii^nce,
5T 105, 106.
found by the
te or 4th root
tot is the cube
iquare root of
he cube root,
>c. yrhith arc
eldom occur ;
performed by
»er be divided
the logarithm
ssiorv.
lore than two,
mee, is called
ja) addition of
ng series ; but
on of the com-
ending series,
ending "
led the termi
the extremes,
gression, any
be found :—
f|106.
ARITHMETICAL PROGRESSION.
231
g 4d. for the
rd, and so on
e codt of the
[very yard ex-
|9X3;=297
U.aOl^nce.
Hence, when the first term, the common difference, and
the number of terms are given, to find the last term, — Mul-
tiply the number of terms, less one, by the common differ-
ence, and add the first term to the product for the last term.
2. If the first term be 4, the common difference 3, and
the number of terms 100, what is the last term ? Ans. 301.
3. There are ' in a certain triangular field, 41 rows of
corn ; the first row, in one corner, is a single hill ; the se-
cond contains three hills, and so on, with a common differ-
ence of 2 ; what is the number of hills in the last row ?
Ans. 81 hills.
4. A man puts out <£! at 6 per cent simple interest, which
in one year amounts to £i-^(f, in two years to iSf l-^o, and so
on, in arithmetical progression, with a common difference
of'£-^ ; what would be the amount in 40 years? /• ■
Ans. £^^.
Hence we see, that the yearly amounts of any sum, at
simple interest, form an arithmetical series, of which the
principal is the first term, the last amount is the last term,
the yearly interest is the common difference, and the rfumber
of years is one less than the number of terms.
5. A man bought 100 yards of cloth in arithmetical pro-
gression ; for the first yard he gave 4d., and for the last 301
pence; what was the common increase of the price, on each
succeeding yarfl ?
This question is the reverse of example 1 j therefore, 301
—4=297, and 297-f-99=:3, common difference.
Hence, when the extremes and number of terms are given
to find the common difference, — Divide the difference of
the ejctremes by the number of terms, less 1 , and the quo-
tient will be the common difference.
6. If the extremes be 5 and 605,, and the number of terms
151, what is the common difference ? Ans. 4.
7. If a man puts out £1 at simple interest, for 40 years,
and receives at the end of the time ^3f^, what is the rate ?
If the extremes be 1 and 3|3-, and the number of terms
41, what is the common difference ? Ans. -^.
8. A man had 8 sons whose ages differed alike; the
youngest was 10 years old, and the eldest 45; what was the
common difference of their acres ? Ans. 5 years.
9. A man bought 109 yards of cloth in arithmetical series;
T2
mfi
ARITHMETICAL PROQRESSION,
IT 106
he give 4 pence for the first yard, and 301 pence fur the
last yard; what was the average price per yard, and what
was the amount of the whole ? '
Since the price of each succeeding yard increases by a
constant excess, it is plain the average price is as much less
than the price of the last yard as it is greater than the jwice
of the first yard ; therefore, one half the sum of the first and
last price is the average price.
^One half of 4d -f 301d. =: 152Jd. = average )
price ; and the price, 152^d.Xl^^=!li>^0d,= V Ans.
jC^i 10s. lOd., whole cost. )
Hence, when the extremes and the number of terms arc
j^iven, to find the sum of all the terms,, — Multiply half the
sum of the extremes by the number of terms, and the pro-
duct will be the answer.
10, If the extremes be 5 and C05, and the nun. 2r of
terius be 151, what is the sum of the series ? An,i. 4G05o.
Jl. What is tl:9 sum of tlie first 100 numbers, in their
natural order, that is, 1,2, 3, 4, &-c. Ans 5050.
1*2. How HKiny times does a common clock strike in 12
hours ? Ans. 78,
1 3. A man rents a house for £o^ annu ally, to be paid at the
•-.lose of each year ; what will the rent am« i^int to in 20 years,
.illowing G per cent simple interest for the use of the money ?
The last year's rent will evidently be c€50 without inter^
f:.st, t!ie last but one will be the amount of .£50 for 1 year,
the liist but two the amount of .£50 for 2 years, and so on,
\n arithmetical Series, to the first, which will be the amount
of ^'51) for 19 years==£l07.
Iftlie first term be 50, the last term 107, and the number
oC terin^ 20, what is the sum of the series ? Ans. ^1570,
14. What is the amount of an annual pension of ^£100,
he'uis, in arrears, that is, remaining unpaid, for 40 years,
allowiijcf 5^ per cent simple interest ? Ans. ,£7900.
1.3. There arc, in a certain triangular field, 4.1 rows of
Cv)rn ; the first row being in one corner, is a single hill, and
the 1 i-^t row, on the side opposite, contains 81 hills ; how
many hills of corn in the field? Ans. 1631 liills.
10. If a triangular piece of land, 30 rods in length, be
2) rods wide at one end, and come to a point at the other,
v/lHt auinber of square rod.s does it contain ? Ans. 300
17.
arithi
what
the s(
••' IT 106
pence for tlic
ard, and what
increases by a
s as much less
than the jwice
of the first and
Ans.
^i
5r of terms arc
dtiply half the
, and the pro-
he nuu. 3r of
An.. 40055.
ibers, in their
Ans 5050.
-k strike in 12
Ans. 78,
be paid at the
to in 20 years,
nf the money ?
without inter-
50 for ] year,
rs, and so on,
)e the amount
d the number
Ans. ^1570.
sion of o£:iOO,
for 40 years.
Ans. ,£7900.
d^ 4,1 rows of
ngle Jiill, and
il hills ; how
16S1 hills.
in length, be
at the otlier,
Ans. 300.
^ 106, 107. GEOMETRICAL PROUReSSION.
223
17. A debt is to be discharged at 11 several payments, in
arithmetical series, the first to be £5, and the last .£75 ;
what is the whole debt .' common difference between
the several payments ?
Ans. whole debt X440 ; common difference £7.
18. What is the sum of the series 1, 3, 5, 7, 9, &lc. to
lOOU ^115.251001.
Note. By the reverse of the rule under ex. 5, the differ-
ence of the extremes 1000, divided by the common differ-
ence 2, gives a quotient, which, increased by 1, is the num-
ber of <cn«s=:501. '
19. What is the sum of the arithmetical scries 2, 2^, 3,
3|, 4, 4^, &LC. to the 50th term inclusive ? Ans. 712^.
20. What is the sum of the decreasing series 30, 29jf,
29^, 2,9, 2Si, &LC. down toO?
Note. 30-7-^-|-l=:91, number of terms. Ans. 1365.
QUESTIONS. .
1, What is nn BrUhmelical progression? 2, When is the series
called ascending'? '>i,-~— when descending? 4, What are the iiurn*
bers Ibrminj^ the progression rallmi? n, What are the first and last
term? called '? 6, What are the other terms called '{ 7, When the
first term, common diflerence, and number of tcrnis are g'ven, how do
you find the last term ? 8, Hotv may urithmelical progression be ap-
plied to simple interest ? 9, When the extremps and t.urn'or of torus
Hre given. Iiuw do you find the common diin^rcnce ? 1 0, — — how do
you lind the sum of all the terms 1
CiiEOnETRICAL PRO(>iUE88IO.\.
IT 107. Any series of numbers, continmllySncreasing
by a constant multiplier, or decreasing by a constant divi-
sor, is called a Geometrical Progression. Thus, 1,2,4,8,16,
tSic. is an increasing geometrical series, and 8, 4, 2, 1, J-, -|,
&c. is a decreasing geometrical series.
As in aritlun€tical, so also in geometrical progression,
there are five things, any three of which being given, the
other two may be found : —
1st. The first term; "Zd. The /rt.sf term ; 3d. Tht number
of terms ; 4th. The ratio ; 5th. The sum of all the terms.
234
OEOMETRICAl. PROOREStlON.
11107
i
ill
■i'
The ratio is the multiplier, or divisor, by which the series
is formed.
1. A man bought a piece of silk, measuring 17 yards,
and, by agreement, was to give what the last yard would
come to, reckoning 3 pence for the first yard, G pence for
the second, and so on, doubling the price to the last ; what
did the piece of silk cost him?
3X2X2X2X2X2X2X2X2X2X2X2X2X2X2X2
X2=196608 pence,=:;£819 4s. Ans.
In examining the process by which the last term (106608)
h-ys been obtained, we see that it is a product of wiiich the
ratio (2) is sixteen times a factor, that is, am time less than
the number of terms. The last term, then, is the sixteenth
power of the ratio, (2) multiplied by the first term, (3.)
Now, to raise 2 to the 16lh power, we need not produce
all the intermediate powers; for 2*=2X2x2X2=16, is a
product of which the ratio 2 is 4 times a factor ; how, if 16
be multiplied by 16, the product, 256, evidently contains
the same factor (2) 4 times-|-4 times,=8 times; and 256X
256=65536, a product of which the ratio (2) is 8 times -j-
8 times,=16 times, factor; it is, therefore, the 16th power
of 2, and, multiplied by 3, the first term, gives 196608, the
last term, as before. Hence,
When the first term, ratio, and number of terms, are
given, to find the last term, —
' I. Write down a few leading powers of the ratio with
their indices ,over them.
II. Add together the most convenient indices, to make
an index less by one than the number of the term sought.
III. Multiply together the powers belonging to those in-
dices, and their product, multiplied by the first term, will
be the term sought.
2. If the first term be 5, and the ratio 3, what is the 8th
term ?
Powers of the ratio with 1 1, 2, 3, -j- 4=7*
their indices over them V 3, 9, 27, X 81=2187 X 5, first
J term,= 10935, Ans.
3. A man plants 4 kernels of corn, which, at harvest,
produce 33 kernels ; these he plants the second year ; now,
fluppi
woull
nelni
4.1
interl
allo«
»>|
5.
yard]
the
■<*v
. 11107
hich the series
'> :^'-^ .,"(:;;
Ing 17 yards,
St yard uould
d, (i pence for
the last ; what
<2X2X2X2
erm (19G608)
t of which the
time less than
the sixteenth
term, (3.)
J not produce
X2=1G, is a
►r ;'how, if 16
ntly contains
(s; and 256 X
is 8 times -j-
e 16th power
s 196608, the
3f terms, are
le ratio with
ces, to make
m sought,
to those in-
st term, will
lat is the 8th
37 X 5, first
at harvest,
year; now,
11107.
GBOMRTAICAL PKOOKEmOlV.
supposing the annual increase to continue 8 fold, what
would be the produce of the 16th year, lUlowing 1000 ker-
nelntoapint? Amj). 2199023355*552 buaheli.
4. Supposing a mar had put out one penny at compound
interest in 1620, what would have been the amount in 1624/
allowing it to double once in 12 years ?
2> 7 S131072. ^115. je54G 2s. 8d.
5. A man bought 4 yards of cloth, giving 2d. for the first
yard, 6d. for the second, and so on in 3 fold ratio ; what did
the whole cost him ?
S-(-6-|- 18+54=^80 pence '' i4n5. 80 pence.
In a long series, the process of adding in this manner
would be tedious. Let us try, therefore, to devise some
shorter method of coming to the same result. If all the
terms, exceptiig the last, viz. 2-f-^H'l^> ^ multiplied by
the ratio, 3, the product will be the series 0-f-18-f-54, sal>>
tracting the former series from the latter, we have for the
remainder, 54 — 2, that is, the last term less the first term^
which is evidently as many times the first series (2-f-&-f-18)
as is expressed by the ratio, less one ; hence, if we divide
the difference of the extremes (54 — 2) by the ratio, less 1,
(3 — 1) the quotient will be the sum of all the terms, except
the last, and, adding the last term, we shall have the whole
amount. Thus, 54— 2=52, and 3—1=2 ; then 52-^-2=
26, and 54 added, makes 80. Ans. as before.
Hence, when the extremes and ratio are given to find the
sum of the series, — ^Divide the difference 6f the extremes
by the ratio less I , and the quotient, increased by the greater
term, will be the answer.
6. If the extremes be 4 and 131072, and the ratio 8,
what is the whole amount of the series ?
, 131072—4 I
l-f 31072=149796. Ans.
8—1
7. What is the sum of the descending series 3, 1, ^, ^,
]>;■, &c. extended to infinity ?
It is evident the last term must becone 0, or indefinitely
near to nothing ; therefore, the extremes are 3 and 0, and
the ratio 3. Ans. 4^.
8. What is the value of the ihfinite series 14'iH~y6"f"BV»
&.C.? Ans, 1^.
.:.-i„
\.J..Mt^,-'
!nL-\- :■•-. t.::!f.
/Ml
1 n
Jl
226
GEOMETRICAL PROGRESSION.
^ 107.
9. What is ♦he value of the infinite series, i^.-i-i^ "h
TT^<T "h TTF^vxT' ^^- > ^^t ^^^ ^^^ ^^^ sauie, the decimal
*ll 111, &c. coutinually repeated ? Ans. ^.
10. What is the value of the infinite series, T^Ty~l~Tir§TrTj>
&,c., descending by the ratio 100 ; or, which is the same,
the repeating decimal '020203, &c. Ans. ^^.
11., A gentleman whose daughter was married on a new
year's day, gave her ^1, promising to tripple it on the first
day of each month in the year ; to how much did her por-
tion amount .'
Here, before finding the amount of the series, we must
find the last term, as directed iu the rule after ex. 1.
Ans: £265720,
The 2 processes of finding the last term, and the amount,
may, however, be conveniently reduced to one, thus : —
When the first term, the ratio, and the number of terms,
are given, to find the sum or amount of the series^ — Raise
the ratio to a power whose index is equal to the number of
terras, from which subtract 1 ; divide the remainder by the
ratio, less 1, and the quotient, multiplied by the first term,
will be the answer.
Applying this rule to this last example, 3 V^ss531441 and
o31441— I
— : Xl=c£265720. ilMs. as before, i
3—1
12. A man agrees to serve a farmer fojty years, without
any other reward than 1 kernel of corn for the first year, 10
for the second year, and so on,, in 10 fold ratio, till the end
of the term ; what will be the amount of his wages, allowing
1000 kernels to a pint, and supposing he sells his corn for
30 pence per bushel ?
IQio— 1 1^ 1,111, 111,111, 111,111,111, 111, HI,
— —- XI— ^ 111,111,111,111,111, kernels.
iln-s. j^:2,170,13i3,388,888,888,8i6S,888,888,886,888,a^,
!7s. 9^d.
13. A gentleman dying, left his estate to his 5 sons, to
the youngest .£1000, to the second o£'J500, and ordered that
each son should exceed the younger by the ratio of H;
what was the amount of the estate ?
^
is
!I 107. I fl 108.
OEOMETRICAL PROGRESSION.
227
the decimal
Ans. ^.
is the same,
Ans. ^g.
ied on a new
t on the first
did her por-
ies, we must
BX. 1.
'.s\ ^65720.
the amount,
thus : —
ber of terms,
riesj — Raise
le number of
linder by the
ne first term,
=531441 and
lars, without
first year, 10
till the end
fes, allowing
his corn for
1,111,111,
kernels.
5 sons, to
ordered that
atio of 14 ;
Note. Before finding the power of the ratio 1^, it may be
reduced to an improper fraction=s^y or to a decimal, V^.
^^—1 1'5^— 1
— X 1000 =r: 13187^; or, X 1000 =
<£13187'50 = ^^13187 10s. Ans. 1*5—1
I Compound Interest hy Progression. "■ ' " '
^ 108. 1. What is the amount of <£4 for 5 years, at 6
per cent compound interest ?
We have seen (fl 86) ihtiX compound interest is that which
arises from adding the interest to the principal at the close
of each year, and, for the next year, casting the interest on
that amount, and so on. The amount of £\ for one year
is 1'06; if the principal, therefore, be multiplied by r06^
the product will be its amount for one year ; this amount
multiplied by TOO, will give the aniount (compound ir>ter-
est) for two years ; and this second amount multiplied by
r06, will give the amount for three years ; and so on.
Hence, the several amounts arising firom any sum at com-
pound interest, form a geometrical series, of which the prin-
cipal is the first term ; the amount of ^1 or $1, &;C. at the
given rate per cent, is the ratio ; the time, in years, is one
less than the number of terms ; and the last amount is the
last term.
The last question may be resolved into this : If the first
term be 4, the number of terms 6, and the ratio .1'0(), what
is the last term ?
1'065=1«338, and I*338x4=je6'362-|-. Ans. £b 7s. ^A.
Note 1. The powers of the amounts of <£1, at 5 and at 6
per cent, may be taken from the table under IT 85. Thus,
opposite 5 years under 6 per cent, you find 1*038, &.c.
Note 2. The several processes may be conveniently exhi-
bited by the use of letters, thus :-^
Let P represent the Principal.
R " Ratioortheamountof.^ Ij&c.for 1 yr.
T " Time in years.
A " Amount.
When two or more letters are joined together, like a
word, they are to be multiplied together. Thus, PR. im-
plies, that the principal is to be multiplied by the ratio.
When one letter is placed above another, like the index of
•5«3
GEOMETRICAL PR0VRE6SI0N.
1I108li|i09.
W'Am
i<i.
a power, tWJirst is to b6 raised to a power, whose index
is denoted by the second. Thus Rt* implies that the ratio
is to be raised to a power whose index shall be equal to the
itnte, that is, the number of years. ^' '
2. What is the amount ofi£40 for 11 years, at 5 per cent
'Compound interest ?
Rt. xP=A ;|therefore, I'OS* » X40==:68'4. Ans. £G8 Ss,
. 3. What is the amount of ^6 for 4 years, at 10 per cent
compound interest ? Ans, £8 15s. 8d.
4. If the amount of a certain sum for 5 years at 6 per
cent compound interest, be £5 7s. O^d., what is that sum,
or principal ?
If the number of terms be 6, the ratio 1'06, and the last
term 5*352, what is the first term 1
This question is the reverse of the last; therefore,
A 5'352
= P;or = 4. Ans.£l
Rt. 1*338
5. What principal, at 10 per cent compound interest,
will amount, in 4 years, to i^8*7846. Ans, £t
6. What is the present worth of ^68 8s., due 11 years
hence, discounting at the rate of 5 per cent compound ic>
terest ? JUns. £4Q.
7. At what rate per cent will £6 amount to <£8'7846 in
4 years ?
If the first term be 6, the last term 8'7846, and the num-
ber of terms 5, what is the ratio ?
A 8'7846
— = Rt. that is, =z 1'4641 = the 4th power of
P 6
the ratio ; and then, by extracting the 4th root, we obtaio
I'lO for the ratio. Ans. 10 per cent.
8. In what time will ^6 amount to .£8'7846, at 10 per
cent compound interest ?
A 8'7846
— =Rt. that is, z=r4641=l'10T ; therefore, if
P 6
we divide 1*4641 by 140, and then divide the quotient
thence arising by TIO, and so on, till we obtain a quotient
that will not contain 140, the number of these divisions will
be the number of years. ^ JJns. 4 years.
«1109.
GEOMETRICAL PROGRlJpSION.
229
and the num-
9. At 5 per cent compound interest, in what time Will
£40 amount to £68 8s. ?
Having found the power of the ratio 1*Q5, .is belbre,
which is 1*71, you may look for this number in thie table
under the given rate, 5 per cent, and against it you will
find the number of years. »^ns. 11 years.
10. At 6 per cent compound interest, in what time will
£A amount to £o 7s. O^d. Ans. 5 years.
Annuities at Compound Interest, , , .
IT 109* It may not be amiss, in this place, briefly to
show the application of compound interest, in computing
the amount and present worth of annuities.
An annuity is a sum payable at regular periods of one
year each, either for a certain number of years, or during
the life of the pensioner, or for ever.
When annuities, rents, &c. are not paid at the time they
become due, they are said to be in arrears.
The sum of all the annuities, rents, &c. remaining un-
paid, together with the interest on each, for the time they
have remained due, is called the amount.
1. What is the amount of an annual pension of ^100,
which has remained unpaid 4 years, allowing 6 per cent
compound interest ?
The last year's pension will be ^100, without interest ;
the last but one will be the amount of .£100 forgone year ;
the last but two the amount (compound interest) of £W{)
for two years, and so on ; and the sum of these several
amounts will be the answer. We have then a series of
amounts, that is, a geometrical series, (51 108]^ to find the
sum of all the terms.
If the first term be 100, the number of terras 4, and th<»
ratio 1*06; what is the sum of all the terms ?
Consult the rule under ^ 107, ex. 11.
1*06*— 1 .
*06
■X 100=437*45.
Ans. c€437 ys
Hence, when the annuity, the time, and rate per cent,
are given, to find the amount — Raise the ratio (the amount
of £1, &,c. for one year) to a power denoted by the num-
ber of years ; from this power subtract 1 , then divide the
U
230
GEOMETRICAL PROGRESSION.
51110
'li
m
remaindfr by the ratio less 1, and the quotient multiplied
by the annuity, will be the amount.
Not«. T\\e powers of the ainounts, at 5 and 6 per cent
up to the 24th, may be taken from the table under 11 85.
2. What is the amount oi' an annuity of .£50, it being in
arrears 20 yeajrs, allowing 5 per cent compound interest ?
Ans. of 1653 5s. 9id.
3. If the. annual rent of a hcJuse, which is .£150, be in
arrears 4 years, what is the amount, allowing ten per cent
compound interest ? A ns. <£69(> 3s.
4. To how much would a salary of .£500 per annum
amount in 14 years, the money being improved at six per
cent compound interest 1 in 10 years ? in 20 years ?
'-in 22 years ? in 24 years ?
Ans. to the last, £25407 15s.
If 110. If the annuity is paid in advance, or if it be
bought at the beginning of^the first year, the sum which
ought to be given for it is called the present worth.
5. What is the present^ worth of an annual pension of
£100, to continue for four years, allowing 6 per cent com-
pound interest ?
The present worth is evidently a sum which, at six per
cent, compound interest, would, in four years, produce an
amou t equal to the amount of the annuity in arrears the
same time.
By the last rule we find the amount=£437'45, and by
the directions under 1| 108, ex. 4, we find the present worth
=£34651. ^ Ans. £346 10s. 4^d.
Hence, to find the present worth of any annuity, — First
find its amount in arrears for the whole time ; this amount,
divided by that power of the ratio denoted by the number
of years, will givg the present worth.
6. What is the present worth of an annual salary of £100
to continue twenty years, allowing five per cent ?
^«s. .£1246 4s. 4fd.
The operations under this rule being somewhat tedious,
we subjoin a
^ 110.
GEOMETRICAL PROGRESSION.
231
TABLE
Showing the present worthorjCl or $1 annuity, at 5 and 6 per cent,
compound iulerest, for any number of years from 1 to 34.
653 5s. 9id.
£25407 15s.
Vears
1
2
3
4.
5
7
8
9
10
11
12
13
14
15
16.
17
5 per cent.
0'95238
1*85941
2*72325
3'54595
4'32948
5*07569
5'78637
6'4632l
7*10782
7*72173
8'3064l
8*86325
9*39357
9*89864
10*37966
10'83777
11*27407
6 percent.
0*94339
1*83339
2*67301
3*4651
4*21236
4*91732
5*58238
6*20979
6*80169
7*36008
7*88687
S'38384,
8*85268
9*29498
9*71225
10*10589
10*47726
Years
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
5 per cent.
11*68958
12*08.532
12*46221
12*82115
13*163
13*48807
13*79864
14*09394
14'37518
14*64303
1489813
15*14107
15*37245
15*59281
15*80268
16*00255
16*1929
6 per cent.
10*8276'
11*15811
11*46902
11*76407
12*04158
12*303538
12*55035
12*78335
13*00316
13*21053
13*40616
13*59072
13*76483
13*92908
14*08398
14*22917
14*36613
It is evident that the present worth of £2 annuity is two
times as much as that oi£l ; the present worth of <£3 will
be three times as much, &c. Hence, to find the present
worth of any annuity at 5 or 6 per cent, — Find in this table
the present worth of £1 annuity, and multiply it by the
given annuity, and the product will be the present worth.
7. What ready money will purchase an aniiiuity of .£150,
to continue 30 years at 5 per cent compound interest ?
The present worth of .£1 annuity, by the table, for thirty
years, is 15 37245 ; therefore, 15*37245X150=^2305*867
=£2305 17s. 4d. Ans,
8. What is the present worth of a yearly pension of £40,
to continue ten years at 6 per cent compound interest?
at 5 per cent ? to continue fifteen years ? ^20 years ?
25 years ? 34 years ?
Ans. to the last, £647 14s. 3fd.
When annuities do not commence till a certain period of
time has elapsed, or till some particular event has taken
place, they are said to be in reversion.
I 4iliil
111'
•'}
¥','.'
m
GEOMETRICAL PROGRESSION.
Tf 111
. 9. What is the pfesent worth of <£100 annuity, to be con
tinued four years, but not to commence till two years hence,
allowing 6 per cent compound interest ?
The present worth is evidently a sum which, at 6 per
cent compound interest, would, in two years, produce an
amount equal to the present worth of the annuity, were it
to commence immediately. By the last rule, we find the
present worth of the annuity, to commence immediately, to
be ^346'dl, and by directions under IF lOS, ex. 4, we find
the present worth of i.'346'51 for two years to be c£308'393.
Ans. £308 7s. lO^d,
Hence, to find the present worth of any annuity taken in
reversion, at compound interest, — -First, find the present
worth, to commence immediately, and this sum, divided by
the power of the ratio, denoted by the time in reversion,
will give the answer.
JO. What ready money will (iurchase the reversion of a
lease of £G0 per annum^ to continue 6 years, but not to
commence till the end of three years, allowing 6 per cent
compound interest to the piirchaser ?
The present worth to commence immediately, we find lo
295'039
be 295'039, and — ^ =247*72 Ans. .£247 14s. 4|d.
1'063
. It is plain, the same result will be obtained by finding the
present worth of the annuity, to commence immediately,
and to continue to the end of the time, that is 34-6=9
years, and then subtracting from this sum the present worth
of the annuity, continuing for the time of the reversion, 3
years. Or, we may find the present worth of £1 for the 2
times by the table, and multiply their difference by the giveii
annuity. Thus, by the table,
The whole time, 9 years=6'80169
The time in reversion, 3 " =2'67301
Difference,
442868
60
11.
..5e247'72080
^47*72080=^247 14s. 4|d. Ans.
What is the present worth of a lease of if 100, to con»
^flll.
tinue
ailowii
■ 81
ml
which
per ceil
In t|
tate is
equal tl
tiplied]
(lividec
m
and
'.V''-\^-.;'->:jifc'-^'Lj-;i^-,
;y, to be con-
years hence,
ch, at 6 per
produce an
luity, were it
we find the
mediately, to
X. 4, we find
)ede308'393.
508 7s. lO^d.
uity taken in
the present
1, divided by
in reversion,
«
aversion of a
I, but not to
ig 6 per cent
y, we find lo
•47 14s. 4|d.
finding the
mmediately,
is 3-f-6=9
resent worth
reversion, 3
l&l for the 2
)y the giveii
169
60
|80
4^d. Ans.
[00, to con»
11 111.
GEOMETRICAL PROGRESSION.
233
tinue 20 years, but not to commence till the end of 4 years,
allowing 5 per cent ? — ^what, if it be 6 years in reversion ?
8 years? 10 years ? 14 years?
11 111. 12. What is the worth of a freehold estate of
which the yearly rent is ^€60, allowing to the purchaser 6
per cent?
In this case, the annuity continuesybr ever, and the es-
tate is evidently worth a sum of which the yearly interest is
equal to the yearly rent of the estate. The principal w/w/-
tiplied by the rate gives the interest ; therefore, the interest
divided by the rate will give the principal; 60-^*06=: 1000.
Ans. .£1000.
Hence, to find the present worth of an annuity, continu-
ing for ever, — Divide the annuity by the rate per cent, and
the quotient will be the present worth. \
Note. The worth will be tlie same, whether we reckon
simple or compound interest ; for since a year's interest of
the price is the annuity, the profits arising from that price
can neither be more nor less than the profits arising from
the annuity, whether they be employed at simple or com-
pound intere' t.
13. Wha. is the worth of .£100 annuity, to continue for
ever, allowing to the purchaser 4 per cent? allowing
5 per cent? 8 per cent? 10 per cent 7 15
per cent ? 20 per cent ? Ans. to the last, .£500.
14, Suppose a freehold estate of .£60 per annum, to com-
mence two years hence, be put on sale ; what is its value,
allowing the purchaser 6 per cent ?
Its present worth is a sum which, at 6 per cent compound
interest, would in two years produce an amount equal to the
worth of the estate if entered on immediately.
60
— =£1000=the worth, if entered on immediately,
*06
£1000
and ==i:889'996=£889 19s. lid. the present worth.
r062
The same result may be obtained by subtracting from
the worth of the estate, to commence immediately, the pre-
sent worth of the annuity 60, for two years, the time of
reversion. Thus, by the table, the present worth of £1 for
U2
"<"►,
I'M
^:. 'ji 'i
I
*J:J4
PEllMttATION'.
11 la'
two years is 1 '83339 X 00=1 1 0'0034=:Dresent worth of JGliO
Tor two years, and Jei000—n0'0034 =^ .£809*9960=^889
19:>. lid. Arts, as before.
15. Wliat is the present worth of a perpetual annuity of
•£ii)\), to commence G years hence, allowing the piwchaser
•> })er cent compound interest ? what, if 8 years in re«
version ?— -10 years ? — ^4 years ? 30 years ?
Ans. to the last, .£462 15s. l^d.
The foregoing examples in compound interest have been
confined to i/earli/ payments ; if the payments are half-yearly;
we take half the principal or annuity, half the rate per cent
and twice the number of years, and work as l)efore, and so
for any other part of a year.
QUESTIONS.
i. What is a geometrical progression or series ? 2. What is the
I'utio ? 'i. When the lirst term, the ratio and the number of terms, arc
given, how do you find the last terAi ? 4. When the extremes and ra-
tio are given, how do you find the sum of all the terms 'i 5. When the
first teriii, the ratio, and the number of terms are given, how do you
find the amount of the series i G. When the ratio is a fraction, ho\'
do you proceed I 7, What is compound interest 1 8. How does it
appear that the amounts arising by compound interest, form a geomc-
triciil scries f 9. What is the ratio in compound interest ? the
number of terms ?-^^— the first term 1 — — the last term? 10. When
the rate, the time and the principal are given, how do you find the
amount i 11. Wlien A 11 and T are given, how do you find P / 12,
VVheii A P ana T are given, how do you find R ? 13. When A P and
11 are given, how do you find T ? 14. What is an annuity ? lb.
When are annuities said to be in arrears? 16. What is the amount?
17. In u geometrical series, to what is the amount of an annuity equi-
valent? 18. How do you find the amount of an annuity, at compound
interest? 19. Wha*; is the present worth of an annuity? iiow
eotnputed at compound interest '? ■ - ■ ■ • ' how found by the table 1 20.
What is understood by the term reversio'ii '( 21. How do you find the
present worth of an annuity, taUen in reversion? by the table?
22. How do you find the present worth of a freehold estate, or a per-
petual annuity ? — ^ tlie same taken in reversion ? -by the table I
fER.TlIJTATJOJV.
^f lllS. Permutation is the method of finding how
many different ways the order of any number of things may
iie varied or changed.
1. 1
they c
how m
Hac
only ii
and b
m IX
with a
next bi
and b
that is,
o
i worth of jCOO
lal annuity of
the piir chaser
S years in re*
ears?
tG-J 15s. l^d.
est have been
•e half-yearJv;
rate per cent.
)efore, and so
2. What is the
er of terms, art
xlremcs and ra-
i 5. When the
n, how do you
!i fraction, ho»»
8. How does it
, form a geomc-
erest ? the
|n? 10. When
you find the
t)u find P ? 12,
Wiien A P and
annuity ? 1.5.
9 the amount?
!» annuity equi-
f^ at compound
ty ? iiow
he table 1 20.
do you find the
• by the table ';
Late, or a per-
y the table l
inding how
■ things may
M13
Miscellaneous examples.
J235
1. Four gentlemen agreed to dine together, so long as
they could sit every day in a different order or position ;
how many days did they dine together 7
Had there been but two of them, a ahd h, they could sit
only in 2 times 1 (1x2=2) different positions, thus, a b,
and b a. Had there been three, a b and c, they could sit
in lX2X3=(i different positions; for, beginning the order
with a, there will be two positions, viz a b c, and a c b;
next beginning with 6, there will be two positions, b a c^
and bca; lastly, beginning with c, we have c ab^ and cba,
that is, in all, 1X2X3=6 different positions. In the same
manner if tlvtre be four, the different positions will be
1X2X8X4=24. y|«5. 24.
Hence, to find the number of different changes or per-
mutations, of which any number of different things are ca*
pable, — Multiply continually together all the terms of the
natural series of numbers, from one up to the given number,
and the last product will be the answer.
2. How many variations may there be in the position of
the nine digits ? Ans. 862880.
3. A man bought 25 cows, agreeing to pay for them one
penny for every different order in which they could all be
placed ; how much did the Cows cost him ?
Aas. i:6463004184T2l2441600000.
4. A cei'tain church has 8 bells ; hoAv many changes may
be rung upon them ? Ans. 40320.
MlSCELLANE^OUS EXAMPLES.
51 113. 1; 44- ;x7— 1=^=60.
A line, or vinculum, drawn over several nun\bers, signi-
fies that the numbers under it are to be taken jointly, or as
one whole numbi
Ans. 30.
2. 9— «-|-'4 X 8-f-4— 6=how many ?
Ans. 230.
3. 7-1-4— 2 4- 3-f40 X5— how many?
3-J.6— 2X4— 2
4. 2X2 =howmany? Ans.^^.
5. T^iere are 2 numbers ; the greater is 25 times 78, and
their difference is 9 times 15 j their sum and product are
I
236
MISCELLANEOUS EXAMPLES.
'f 113
113.
t)3. B
t come t
vill it tu
24.
IIS pay
le recei
25.
required. Ans. 3765 is their sum, 3539250 their product
0. Wliat is the difference between thrice five and thirty.^ ii
and thrice thirty-five ? 35-|-3— 5 X 3-}-30=60, Ans. '^[\\ \yQ
7. Wliat is the differeiice between six dozen dozen, and
half a dozen dozen ? Ans. 792.
8. What number divided by 7 will make 6488?
1). What number multiplied by 6 will make 2058 ?
10. A gentleman went to sea at 17 years of age; 8 years
after, he liad a son born, who died at the age of 35 ; after
whom the father lived twice 20 years; how old was the
father at his death? Ans. 100 years.
11. What number is that which, being multiplied by 15,
the product will be ^ ? ^_^l5=^ig., Ans.
12. What decimal is that which, being multiplied by 15,
the product will be *75 ? '75-i-I5='05, Ans.
13. What is the decimal equivalent to t^l ? Ans. '0285714
14. What fraction is that, to which if you add f , the sum
will be I ? ylns. fg.
15. What number is that, from which if you take f , the
remainder will be ^ ? Ans. |^.
10. What number is thal^, which being divided by f , the
quotient will be 21 ? Ans. 15f,
17. What number is that, from which if you take f of
itself, the remainder will be 12 ? Ans. 20,
18. What number is that, to which if you add f of ^ of
itself, the whole will be 20 ? Ans. 12.
1 9. What number is that of which 9 is the § part ? Ans. 13 J .
20. A farmer carried a load of produce to market ; he sold
A
)U9hel,
leived
ushels
26.
nust be
H
27. H
n excha
Note.
ire giver
he same
28. H
n exchai
29. A
rave him
low man
30. A
7801t}s of pork, at 3d. per lb ; 250ft)s of cheese, at 5d. per lb ; )arterin£
1541bs of butter, at lOd. per lb. In pay he received 601bs
of sugar, at 7d. per lb; 15 gallons of molasses, at 2s. 3d. per
gallon; -^ barrel of mackerel, at 18s. 9d. ; 4 bushels of salt,
at 6s. 4d. per bushel ; and the balance in money ; how much
money did he receive ? Ans. ^15 14s. 8d,
21. A farmer carried his gram to market, and sold 75 bush-
els of wheat at 7s. 3d. per bushel ; 64 bushels of rye at 4s. 9d.
per bushel ; 142 bushels of corn, at 2s. 6d. per bushel. In
exchange, he received sundry articles : — 3 pieces cloth, each
containing 31 yds. at 8s. 9d. per yd. ; 2 quintals fish, I Is. 6d.
per quintal ; 8 hhds. salt, £l Is. 6d. per hhd. and the balance
in money ; how much money did he receive ? Ans. £9 14s
v'orth IS
he broai
30d. :
■s. lO^d
ctly alii
31. If
It 28. 6d
3xchangi
32. if
33. If
ushel ?
34. II
'[ 1I3|
their product
ve and thirty,
JO, Ans.
ill dozen, and
Ans. 792.
88?
! 2058 ?
f age ; 8 years
e of 3f> ; after
old was the
ns. 100 years.
Itiplied by lo,
15=^'^, Ans.
Itiplied by 15,
5='05, An!<.
Ins. '0285714
dd f , the sum
Ans. U-
3u take f , the
Ans. il
ded by f , the
Ans. 15f.
you take f of
Ans. 20.
add f of ^ of
Ans. 12.
irt?Ans. 13^.
irket ; he sold
at5d. per lb,
ceived 60Ibs
at 2s. 3d. per
jshels of salt,
how much
^15 14s. 8d.
113.
MIHCELLANEOUS EXAMPLES.
237
sold 75 bush-
rye at 4s. 9d.
bushel. In
Es cloth, each
Ash, lis. 6d.
d the balance
Ans. ^9 14s
A man exchanges 760 gallons of molasses, at 2s. per
allon, for 064- cwt. of cheese at £i per cwt. ; how much
trill be the bahin($e iu his favor ? Ans. £0 10s.
23. Bought 84 yds. of cloth at (5s. 3d. per yd. ; how mUch did
t come to ? how many bushels of wheat at 7s. 6d. per bushel,
vill it take to pay fufr it ? Ans. to the last, 70 bushels.
24. A man sold 342tbs of beef at 4d. per lb, and received
)is pay in molasses at 2s. per gallon ; how many gallons did
le receive 7 ^ Ans. 57 gallons.
25. A man exchanged 70 bushels of rye at 4s. 6d. per
ushel, for 40 bushels of wheat at 7s. per bushel, and re-
eived the balance in oats at 2s. per bushel ; how many
)ushels of oats did he receive? Ans. 17^.
26. How many bushels of potatoes at Is. Od. per bushel,
nust be given for 32 bushels of barley it 2s. 6d. per bushel ?
Ans. 53^ bushels.
27. How much salt, at $1*50 per bushel, must be given
n exchange tor 15 bushels of oats, at 2s. 3d. per bushel ?
Note. It will be recollected that when the price and cost
ire given to find the quantity, they must both be reduced to
he same denomination before dividing. Ans. 4^ bushels.
28. How much wine, at $2*75 per gallon must be given
n exchange for 40 yards of cloth at 7s. 6d. per yard 7
Ans. 21y^Y gallons.
29. A. had 41 cwt. of hops at 30s. per cwt. for which B.
ave him <£20 in money, and the rest in prunes at 5d. per ib. ;
low many prunes did A. receive? Ans. I7cwt. 3qrs. 4tb.
30. A. has linen cloth worth 2s. 6di per yard; but in
)artering he will have 2s. 9d. per yard ; B. has broadcloth
ivorth 18s. 9d. per yard, ready money ; at what price ought
he broadcloth to be rated,_ih bartering with A. ? *
30d. : 35d. :: 225d. : 262^d. ans. Or, ,^^ of 225d.=c£l
s. lOJ^d. ans. The two operations will be seen to Be ex-
ctly alike.
31. If cloth worth 2s. per yard, cash, be rated in barter
tSs. 6d., how should wheat, worth 8s. cash, be rated in
xchange for the cloth ? Ans. lOs.
32. If 4 bushels of corn cost $2, what is it per bushel ?'
33. If 9 bushels of wlieat cost =£3 7s. 6d. what is that per
ushel.'' 'w* -,}*■• Ans. 7s. 6d.
34. If 40 sheep cost £25, what is that per head? "•■;'>
338
MI8CELLANB0US EXAMPLES.
5I11J
M'
I ,
V
m
I: ' r-
3o. If 3 bushels of oats cost 7s.. Gd. how much are the
per bushel? Ans. 2s. CdL
36. If 23 yards of broadcloth cost <£21 Os. what is th(
price per yard ? Ans. lOs (jd
07. At 2s. 6d. per bushel, how much corn can be bough
for lOs. Ans. 4 bushels
3^. A man haying £%5, would lay it out in sheep, at 12s
Cd. a<piecc, how many can ho buy ? Ans. 40
39. If 20 cows cost £75, what is the price of one cow
of 2 cows? of 5 cows ? of 15 cows?
Ans. to the last. £56 5s
y 40. If 7 men consume 241fos of meat in one week, hov
much would one man consume in the same time ?— — 2 men
5 men ? 10 men ? Ans. to the last, 34if lbs
NoU. Let the pupil, also perform these questions by th(
113.
49. 1
br ^42
how mu
50.
£od 4s.
51.
for 17
52.
rule of proportion
41. If I pay £{ 10s. for the use of <£25, how much mus
I pay for the use of .£16 15s. ? Ans .£1 2s. Cd
42. What premium must I pay for the insurance of m;
house against loss by fire, at the rate of ^ per cent, that is
^ pound for 100 pounds, if my house be valued at .£2475
Ans. ^12 7s. 6(1
43. What will be the insurance, per annum, of a ston
an4 contents, valued at .£9870 8s. at 1^ per centum ?
Ans. .£148 2s. lid
44. What commission must I receive for "selling ^6^4?!
worth of books at 8 per cent ? Ans. .£38 4s. 9^d
45. A merchant bought a quantity of goods for £TM
and sold them so as to gain 21 per cent ; how much did hi
gain, and for how much did he sell his goods 1
Ans. to the last, ^^888 2s. 9|d
46. A merchant bought a quantity of goods at Montreal
for £500, and paid £43 for their transportation ; he sol
them so as to gain 24 per cent on the whole cost ; for ho»
much did he sell them ? Ans. .£673 6s. 4|d
47. Bought a quantity of books for .£64, but for cash
discount of 12 per cent was made ; what did the books cost
/ Ans ■;
48. Bought a book, the price of which was marked i
2s. 6d., but for cash the bookseller will sell it at 33^ pei
cent discount ; what is the cash price ?
Ans. 15i
54. \
5o
C 1
I
56.
days, at|
57.
nd3d
58. ^
years ai
Note
59.
out inte
of mone
60. ^
years ai
61. ]
for £5'
Wha
were g)
what d<
62. :
5s. per
63. .
Ions foi
gallon 1
Ion? h(
Ans,
per cei
64.
gain .£
If Hi
una
much are the
Ans. 2s.' 6(
Os. what is th
Alls. lOs G(]
[I can be bough
Ans. 4f bushels
in sheep, at 12s
Ans. 4(
ce of one cow
JWS?
he last. cf56 53
one week, hov
fie t 2 men
lie last, 34^ !bs
luestions by thi
how much mus
Ans .£1 2s. G(l
insurance of m
er cent, that is,
lued at ^2475
.ns. ^12 7s. 6(1
lum, of a ston
centum ?
.£148 2s. lid
selling £Ali
,s. .£38 4s. Old
oods for .£734
>w much (lid h
Is?
.£888 2s. 9|(i
)ds lit Montreal
rtation ; he soli
; cost ; for ho»
.£673 6s. 4|d
, but for cash
the books cost
r
IS. £56 6s. 4fd
vas marked £
11 it at 33i pel
Ans. 15s
MISCeLLANBOUS EXAMPLBf.
239
49. I bought tt cask of liquor, containing 120 gallons,
for jff42; for how much myst I sell it to gain 15 per cent?
how much per gallon 1 Ans to the last, 48. 0|d.
50. Bought a cask of sugar, containing 740 pounds, for
£59 4s. ; how -must I sell it per pound, to gain 25 per cent?
Aifs. 2s.
51. Wlmt is the interest at 6 per cent, of j£71 Os. 4^d.
for 17 months 12 days ? Ans. .£6 33. 6jd.
52. What is the interest of .£487 Os. OJd. for 18 months?
Ans. i:43 16s. l^d.
53. What is the interest of $8*50 for 7 months? -
Ans. $'297;^.
54. What is the interest of .£1000 for 5 days ? Ans. 16s. 8d.
55. What is the interest of lOs. for ten years ? Ans. 68.
66. What is the interest of $84'25 for 16 months and 7
days, at 7 per cent ? Ans. $7*486-|-
57. What is the interest of $164'01 for 2 years, 4 months
and 3 days, at'5 per cent? A^ . $18'032.
58. What sum put to interest at 6 per c( .it, will in two
years and 6 months, amount to $150 ? , Ans. 8130*4344-
Note. See H 79.
59. I owe a man .£476 10s. to be paid in 16 months with-
out interest ; what is the present worth of that debt, the use
of money being worth 6 per cent ? Ans. .£440 5s. G^d.
60. What is the present worth of JElOOO payable in four
years and 2 months, discounting at the rate of 6 per cent ?
61. Bought articles to the amount of j£500, and sold them
for £575, how much was gained ?
What per cent was gained? that is, how ,many p'ounds
were gained on each .£100 laid out ? If .£500 gain .£75,
what does j6^100 gain? Ans. 15 per cent.
62. Bought cloth at .£3 10s. per piece, and sold it at £4;
5s. per piece ; howmuch was gained per centum ? Ans. 21 ^.
63. A man bought a cask of liquor, containing 126 gal-
lons for £283 10s. and sold it out at the rate of £2 15s. per
gallon? how much was his whole- gain ? how much per gal-
lon ? how much per cent ?
Ans. His whole gain .£63; per gallon 10s. which is 22f
per centum.
64. If .£100 gain .£6 in 12 months, in what time will it
gain £il .£10 ? £U ? Ans. to the last, 28 months,
240
MISCELLANEOUS EXAMPLES.
11113
65. In what time will ^54 10s. at 6 per cent, gaiA £i
3s. 7^ d. .4ns. 8 months
66. Twenty men built a certain bridge in 66 days, but i
being carried away in a freshet, it is required how man
men can re-build it in 60 days ? '
:.; .J days, days. men.
50 : 60 : : 20 : 24 men. Ans
67. If a field will feed 7 horses 8 weeks, how long will if!
feed 28 horses ? Ans. 2 weeks.
68. If a field 20 rods in length must be 8 rods in width
to contain an acre, how much in width must be a field IG
rods in length, to contain the same 7
Ans. IQ rods,
[[113.
76.
sions su
part, th;
77. I
be wort
value o
78. I
I among
khan 3
79.
water
69. If I purchase for a cloak twelve yards of plaid | of a
yard wide, how much bocking 1^ yards wide must I buy to
line it .^ * Ans. 5 yards.
70. If a man earn £18 15s. in 6 months, how long must
he work to earn ^115 ? Ans. 30§ months.
71. B. owes C. .£540, but B. not being worth so much
money, ,C. agrees to take 15s. on a pound ; what sum must
C. receive for the debt 1 . Ans. <£405.
72. A cistern whose capacity is 400 gallons, is supplied
by a pipe which lets in 7 gallons in 5 minutes ; but there is
a leak in .the bottom of the cistern which lets out 2 gallons
in 6 minutes. Supposing the cistern empty, in what time
would it be filled ?
In one minute ^ of a gallon is admitted, but in the same
time I of a gallon leaks out. Ans. 6 hours 15 minutes.
73. A ship has a leak which will fill it so as to make it
sink in ten hours ; it has also a pump Which will clear it in
15 hours ; now if they begin to pump when it begins to leak,
in what time will it sink ?
* In one hour the ship would be ^ filled by the leak, but
in the same time it would be -^ emptied by the pump.
Ans. 30 hours.
74. A cistern is supplied by a pipe which will fill it in
40 minutes ; how many pipes of the same size will fill it in
five minutes ? Ans. 8.
75. Suppose I lend a friend .£500 for foti? months, he
promising to do me a like favour ; some time afterward, I
have need of .£300 ; how long may I keep it to balance the
former favour ? , Ans. 6§ months.
-■•-•:i
tliia
113.
MISCELLANEOUS EXAMPLES.
241
cent, gaift £i
ins. 8 months
60 days, but
red how many
li men. Ans
low long will it
Ans. 2 weeks.
rods in width
t be a field IC
Ans. ip rods
of plaid I of a
I must I buy to
AnS' 5 yards
low long mus
f. 30f months
'orth so much
i'hat sum must
Ans. ^405
IS, is supplied
3 ; but there is
out 2 gallons
in what time
t in the same
sis minutes,
as to make it
trill clear it in
legins to leak,
the leak, but
le pump.
ns. 30 hours.
will fill it in
will fill it in
Ans. b.
' months, he
afterward, I
3 balance the
6f months.
76. Suppose 800 soldiers were in a garrison with provi-
sions sufficient for 2 months ; how many soldiers mu it de-
part, that the provisions may serve them 5 months ? Ans, 480.
77. If my horse and saddle are worth <£21, and my horse
be worth six times as much as my saddle, pray what is the
value of my horse ? - Ans. £18.
78. Bought 45 barrels of beef at 17s. 6d. per barrel,
among which are 16 barrels whereof 4 are worth no more
than 3 of the others ; how much must I pay ?
Ans. £S5 178. 6d.
79. Bought 126 gallons of rum for ^27 10s. how much
water must be added to reduce the first cost to 3s. 9d. per
gallon ?
Note. If 3s. 9d, buy one gallon, how many gallons will
£^7 10s. buy ? • Ans. 20^ gallons.
80. A thief having 24 miles start of the officer, holds his
way at the rate of 6 miles an hour ; the officer pressing on
afler him at the rate of 8 miles an hour, how much does he
gain in one hour ? how long before he will overtake the
thief? Ans. 12 hours.
81. A hare starts 12 rods before a hound, but is not per-
ceived by him till she has been up 1^ minutes ; she scuds
away at the rate of 36 rods a minute, and the dog, on view.
makes after at the rate of 40 rods a minute ; how long will
the course hold, and what distance will the dog run ?
Ans. 14^ minutes, arid he will run 570 rods,
82. The hour and minute hands of a watch are exactly
together at 12 o'clock ; when are they next together ?
In 1 hour the minute hand passes over 12 spaces, and the
hour hand over one space ; that is, the minute hand gains
upon the hour hand eleven spaces in one hour ; and it must
gain twelve spaces to coincide with it. Ans. Ih. 5m. 27-,\s.
83. There is an island 20 miles in circumference, and 3
men start together to travel the same way about it ; A. goes
two miles per hour, B. four miles per hour, and C. six miles
per hour ; in what time will they come together again ?
Ans. 10 hours,
84. There is an island 20 miles in circumference, and
two men start together to travel round it ; A. travels two
miles per hour, and B. six miles per hour ; how long betoro
they will again come together ? r
W
v,
:C.
;-' *
if
242
MUCRLLANE0U8 EXAMPLES.
mil (113.
B. gams 4 miles per hour, and must gain twenty miles ti earn to
overtake A. ; A. and B. will therefore be together once ii
every five hours.
ach?
Ans.
85. In a river, supposing two boats start at the sami 12 learn
time from^ places 30U miles apart ; the one proceeding u| 93. A
stream is retarded by the current two miles per hour, whili mother,
that moving down stream is accelerated the same ; if boti reese ;"
b«» propelled by a steam engine which would move them i addition
miles per hour in still water, how far from each startiiij have, an
place will the "boats meet 7
Ans. 113^ miles from the lower place, and 187^ milei
from the upper place.
86. A man bought a pipe ( 1 26 gallons) of wine for .£275
he wishes to fill 1 bottles, 4 of which contain two quarts
and 6 of them 3 pints*each, and to sell the remainder so !u
to make 30 per cent on the first cost; at what rate per gal
Ion must he sell it ? ^ Ans. ^65*936+
87. Thomas sold 150 pine apples at Is. 3d. apiece, ar«
received as much money as Harry received for a certain
number of water-melons at 9d. apiece ; how much moni j
did each receive, and how many melons had Harry ?
Ans. £9 7s. 6d. and 250 melons,
88. The third part of an army was killed, the fourth part
taken prisoners, and 1000 fled , how many were in this army!
'This and the 18 following questions are usually wrought
by a rule called Position, but they are more easily solved
on general principles. Thus, ^-j-i=T^2 of the army ; there-
fore, 1000 is -^ of the whole number of men ; and if-j^ b«
1000, how much is 12 twelfths, or the whole ?
Ans. 24000 men
89. A farmer being asked how many sheep he had, ans-
wered that he had them in 5 fields ; in the first were { of
his flock, in the second |, in the third | in the fourth jVi
and in the fifth 450 ; how many had he ? ^ins. ISOO,
90. There is a pole, ^ of which stands in the mud, ^
the water, and the rest of it out of the water ; required the
part out of the water. ^ Ans. -f^,
91. If a pole be \ in the mud, f in the water, and 6 feet
out of the water, what is the lengtli of the pole ? Jins. 00 feet,
92. The amount of a certain school is as follows : j'^ ot
the pupils study grammar, f geography, -^ arithmetic, /j
in
many ha
100-
94. Ii
I pears,
how mai
95. Ii
\ red, a
ed ; ho\
96. a
the sam
ber.
97. \^
the sum
9S. V
the sum
84 =
99. A
and f ol
niucli 7
The
2A tilTl
I
100.
and '
101.
was on
twice a
their aj
102.
.v;*
mi! [115.
)gether once i
rt at the sami
proceeding uj
per hour, whil
e same ; if hot
d move them
1 each startiii
md IST^ mile
MISCELLANEOUS EXAMPLES.
243.
twenty miles ti earn to write, and 9 learn to read ; what is the numbix ef
ivine for cf 275
ain two quarts,
emainder so i
at rate per ga
ins. je5*936-f | red, and i >
Jd. apiece, am cd ; how man}
d for a certa
' much mon(
Harry ?
id 250 melons
the fourth par
e in this armj
sually wrought
; easily solved
B army ; there-
; and if^^ be
J. 24000 men.
p he had, aiis
rst were ^ of
the fourth ^,
^tns. 1200.
the mud, | in
; required the
Ans. T^j,
er, and 6 feet
>/his. 90 feet.
bllows : j'^ ot
trithmetic, ^'j
ach?
Ans. 5 in grammer, 30 in geography, 24 in arithmetic ;
2 learn to write, and 9 learn to read.
93. A man, driving his geese to market, was met by
nother, who said, "Good morrow, sh^ with your hundred
says he, "I have not a hundred ; but if I had, in
ddition to my present number, one half as many as I now
ave, and 2^ geese more, I should have a hundred :" how
many had he ?
100 — 2^ is what, part of his present number ?
Ans. He had 65 geese.
94. In an orchard of fruit trees, ^ of them bear apples,
{ pears, } plums, 60 of them peaches, and 40, cherries ;
how many trees does the orchard contain ? Ans. 1200.
95. In a cf ri.. i village, ^ of the houses are painted white
' 3 are painted green, and 7 are unpaint-
38 in the village ? Ans. 120.
96. Sl2ven eighths of a certain number exceed four fifths of
the same number by 6 ; required the number.
I — t==i^ ; consequently, 6 is ^xf o^ 'he required num-
ber. ^n.5. 80.
97. What number is that, to which if |of itself be added,
the sum will be 30 ? Ans. 25.
OS. What number is that to which if its ^ and I be added,
the sum will be 84?
84 = 1 -f-i "|~i=|^ times the required number. Ans. 48.
99. What number is that, which, being increased by f
and f of itself, and by 22 more, will be made 3 times as
much ?
The number, being taken 1, f , and f times, will make
2/^ times and 22 is evidently what that wants of 3 times,
Ans. 30.
100. What number is thit, which being increased by f ,
I and I of itself, the sum will be 234f ? Ans. 90.
101. B, C, and D; talking of their ages, C said his age
was once and a half the age of B, and D said his age was
twice and one tenth the age of both, and that the sum of
their ages was 93 ; what was the age of each ?
Ans. B 12 years, C 18 years, D 63 years old.
102. A schoolmaster being asked how many scholars he
i
244
MISCELLANEOUS EXAMPLES.
U 113.1 H li3.
had, said, "If I had as many more as I now have, f as ma-
ny, ^ as many, ^ and ^ as many, I should then have 435 ;"
^hat was the number of his pupils? Ans. 120.
103. B and C commenced trade with equal sums of
money ; B gained a sum equal to ^ of his whole stock, and
C lost £2Q\) ; then B.'s money was double that of C's ;
what was the stock of each ?
By the condition of this question, one half of f . that is,
f of the stock, is equal to ^ of the stock, less £200 ;
consequently, J£200is f of the stock. Ans. .£500.
104. A man was hired 50 days on these conditions, —
that for every day he worked, he should receive 3s. 9d.,
and for every day he was idle, he should forfeit Is. 3d. : at
the expiration of the time, he received £2 17s. 6d. , how
many days did he^ work, and how many was. he idle?
Had he worked every day, his wages would have been 8s.
9d.X50=£9 7s. 6d. that is £2 10s. more than he received ;
but every day he was idle lessened his wages 3s. 9d,-}-ls.
3d.=5s. ; consequently he was idle 10 days.
Ans. He wrought 40, and was idle 10 days.
105. B and C have the same Income; B saves ^ of
his ; butC, by spending <£30 per annum more than B, at
the end of 8 years tinds himself <£40 in debt ; what is their
income, aud what does each spend per annum ?
Ans. Their income, £200 per annum ; B spends £175,
and C £205 per annum.
106. A man, lying at the point of death, left his three
sons his property ; to B ^ wanting £20, to C ^; and to D
the remainder, which was £10 less than the share of B ;
what was each one's share ? A ns. £80, £50, and £70.
107. There is a fish, whose head is 4 feet long ; his taij
is as long as his head and half fine length of his body, and
his body is as long as his head ^nd tail ; what is the length
of the fish ? y
The pupil will perceive thaj i^. length of the body is ^
the length of the fish. /^ Ans. 32 feet.
108. B can do a certain ^iece of work in 4 days, and C
can do the same work in £f days ; in what time would both
working together, perform it ? Ans. l^ days.
109. Three persons can perform a certain piece of work
in the following manner : B and C can do it in 4 days, C
U 113.1 !1 l»3.
MISCELLANEOUS jBXAMPLES.
245
ive, ^ as ma-
i have 435 ;"
Ans. 120.
[ual sums of
)le stock, and
that of C's ;
of f . that is,
less £200 ;
Ans. .£500.
conditions, —
eiye 3s. 9d.,
t Is. 3d. : at
7s. 6d. , how
idle?
lave been Ss.
I he received ;
3s. 9d.-f Is.
idle 10 day.s.
J saves ^ of
3 than B, at
what is their
jends £175,
ft his three
^; and to D
share of B ;
»0, and £70.
[>ng ; his tail
is body, and
the length
he body is ^
Ans. 32 feet,
days, and C
) would both
Ins. l^^ days,
ece of work
3 4 days, C
and I^ in 6 days, and B and D in 5 days : in what time can
they all do it together ? Ans. 3^ days.
110. B and C can do apiece of work in 5 days ; B can
do it in 7 days ; in how many days can C do it ? .■4 ns. 17^.
111. A man died, leaving £1000 to be divided between
his two sons, one 14 and the other 18 years of age, in such
proportion that the shire of each, being put to interest at 6
per cent, should amount to the same sum when they should
arrive at the ago of '21 ; what did each receive ? Ans. The
elder £546 3s. 0|d.+ ; the younger £453 1 6s. lid.
112. A house being let upon a lease of five years, at £15
per annum, and the rent being in arrear for the whole time,
what is the sum due at the end of the term, simple interest
being allowed at 6 per cent ; Ans. £84.
1 13. If three dozen pair of gloves be equal in value to 40
yards of calico, and 100 yards of calico to three pieces of
satinet of 30 yards each, and the satinet be worth 2s. Gd.
per yard, how many pair of gloves can be bought for 20s. ?
Ans. 8 pair.
114. B. C. and D., would divide £100 between them, so
that C. may have £3 more than B. and D. jC4 more than C ;
how much must each man have ?
Ans. B. jC30, C. £33, and D. £37.
115. A man has pint bottles, and half-pint bottles; how
much wine will it take to fill one of each sort ? now much
to fill two of each sort ? how much to fill 6 of each sort ?
116. A man would draw off 30 gallons of wine mto one
pint and two pint bottles, of each an equal number ; how
many bottles of each kind will it take to contain the thirty
giillons ? Ans. 80 of each.
117. A merchant has canisters, some holding 5 pounds,
some 7 pounds, and some 12 pounds ; how many, of each
an equal number, can be filled out of 12 cwt. 3 qis. 12 lbs.
of tea? Ans.m.
1 18. If 18 grains of silver make a thimble, and 12 pwts.
make a tea-spoon, how many, of each an equfil number, can
be made from 15 oz. 6 pwts. of silver ? Ans. 24 of each.
119. Let sixty pence be divided among three boys in such
a manner that, as often as the first bus three, the second
shall have five, and the third seven pence ; how many pence
will each receive .' Ans. 12, 20 and 23 pence.
W
T'y
246'
MISCELLANEOUS EXAMPLES.
IT U3
120. A gentleman having fifty shillings to pay among his
labourers for a day's work, would give to every boy 6d., to
every woman 8d., and to every man 16d. ; the number of
boys, women and men was the same ; I demand the num-
ber -of each? ^«.<f. 20.
121. A gentleman had £>! 17s. 6d. to pay among his la-
borers : to every boy he gave 6d., to every woman 8d., and
to every man 16d. ;. and there were for every boy three
women, for every woman two men ; 1 demand the number
of each ? Ans. 15 boys, 45 women, and 90 men.
122. A farmer bought a sheep, a cow, and a yoke of oxen
for i;20 12s. 6d. ; he gave for the cow 8 times as much a&
for the sheep, and for the oxen three times as much as for
the cow ; how much did he give for each ? Ans. For the
sheep, 12s. 6d. the cow i^5, and the oxen -£15.
123. There was a farm of which B. owned f , and C. ^\ ;
the farm was sold for £>^A\ ; what was each one's share of
the money ? Ans. B,'s £126, and C.'s i:315.
124. Four men traded together on a capital of £3000, of
which B. put in ^, C. i, P. i,,and E, -j^; at the end of 3
years they had gained £2364 ; what was each one's share of
the gain? ^ii5. B.^si:il82,.C.'s<£591,D.'s.£394,E.'s£l97.
125. Three merchants companied ; B. furnished f of the
capital, C. §, and D. the rest ; they gain <i^l250 ; what part
of the capital did D furnish,, and what is each oncs's share
of the gain, T
Ans. D- furnished ^ij of the capital ; and B.'s share of the
pain was ^5.00, C.'s ,£468 15s., and D.'s .£281 5s.
126. B, C. and D. traded in company ; B. put in c£!125,
C. £87 10s., and D. 120 yards of cloth; they gained £83
2s. 6d., pf which D.'s share was £30; what wa&the value
f>t" D 's cloth per yard, and what was B, and C.'is share of
the gain? 600 1200 48
Nvtc. D.'s gain being 430, ig = == — of the
1662^ 332& 133
nholo gain ; hence the gain oCBt aijd C is readily found ;
also the price at w *> D.'s cloth was valued, per yard.
Ans. D.'s cloth pe ard, £1, B.'s share of tlie gain, £31
55., C.'.s share, £21 i. 6d.
127. Three gar eners, B. C. and D; having bought a
c.ecc of ground, f id the profits of it amount to £120 per.
m
ann
in
£,1
mu
of
H 113
nii3.
MISCELiANBOUtr EXAMPLE*.
247
pay among hia
ery boy 6d., to
the number of
Hand the num-
Jlns. 20.
among his la-
:oman 8d., and
^ery boy three
lid. the number
J, and 90 men.
a yoke of oxen
les as much a&
as much as for
Ans. For the
5.
If, andC. ^f ;
one's share of
md C.'s £315.
al of £3000, of
at the end of 3
I one's share of
J94,E.'s£l97.
lished f of the
150 ; what part
h onej's sliare
.'s share of the
8158.
put in c€l25,
iy gained ^'83
wa& the value
C.'s share of
bo 48
=— of the
55 133
eadily found ;
per yard.
ne gain, of 31
ing bought a
to £120 per.
annum. Now the sum of money which they laid dY)wn wa»
in such proportion, that, as often as B paid £o, C. paid
£7, and as often as C. paid £4, D. paid ^£6 ; I demand how
much each man must have per annum of the gain ?
Note. By the question, so often as B paid ^5, Di paid f
of o^7. Ans. B. ^26 13s. 4d., C. £37 68. 8d ; D. £56.
128. A gentleman divided his fortune among his son-s,
giving Bw £d as often as C. £o^ and D. £2 as ^ ^ js C.
£7 ; D.'s dividend was 1537| ; to what did the ./hole es-
tate amount? An^. .£11583 88. lOd.
129. B. and C. undertake a piece of work for .£13 lOs.,
on which. B. employed 3 hands 5 days, and C. employed 7
hands 3 days ; what part of the worjc was done by B., and,
what part by €. ? what was each one's share of the money ?
Ans. B. -^ and C ^"^ ; R.'s money £G: 12s. 6d , C.'s «£7
17s. 6d.
130. B. and C. trade in company for one year only ; on
the 1st of January B. put in £300, but C. could not put
any money into the stock until the 1st of April ; what did
he then put in to have an equal sliare with B. at the end of
the year? Ans. .£400.
131. B. C. D* and E. spent 35s. at a- reckoning, and be-
ing a little dipped, agreed that B. should pay §, C. ^, D. ^,
and E. ^ ; what did each pay in this proportion ?
Ans. B. 13s. 4d., C. 10s., 1). 6s. 8d. and E. 5s.
132. There are 3 horses belonging to 3 men, employed
to^draw a load of plaister from Montreal to Stanstead, for
£6 12s. 2d. B: and C.'s horses together are supposed to do
f of the work, R. and D.'s -^q, C. and I> 's ^ J ; they are to
be paid proportionally; what is each one's share of the
money ? > r B.'s £2 Hs. 6d. (=^5)
Ans. I C;'s I 8s. 9d. {==^)
( D.'s 2 68. Od. (s^Jj)
Proof,. £6 128. 3d.
133. A person who was possessed of f of a vessel, soldf^
of his share for £375 ; what was^ the vessel worth ?
Ans. £1500.
134. A gay felloM; soon got the better of f of his fortune ;
he then gave £1500 for a commission, and his ptofusion
eontin^ued till he had! but £450. left, whicji he fotmd to be-
248
MISCELLANEOUS EXAMPLES.
11113. I ^ 113.
just J of his money after he r ad purchased his commission ;
what was his fortune at first? Ans. X3780.
135. A younger brother received £1560, which was just
^j of his elder brother's fortune, and 5^ times the elder
brothel's fortune was J as much again as the father was
worth ; what was the value of i.is estate ?
Ans. £19165 Hs. 3fd.
136. A gentleman left, his son a fortune, -^\ of which he
spent in three months; ^ of | of the remainder lasted him
9 months longer, when he had only -C537 left ; what was the
8um beciueathed him by his f ither ? Ans. £2082 18s. 2-j2^-d.
137. A cannon ball, at the first discharge, flies about a
mile in 8 seconds ; at 4his''rate, how long would a ball be
in passing from the earth to the sun, it being 95173000
miles distant ? Ans. 24 years, 40 days, 7 h. 33 min. 20 sec.
138. A general, disposing his jirmy into a square bat-
talion, found he had 231 over and above, but inc iing
each side with one soldier, he wanted forty-four to fill up the
squ.are ; of how many men did his army consist ? Ans. 19000.
139. B. and C. cleared by an adventure at sea, 45 gui-
neas, which 'was £35 per cent upon the money advanced,
and with which they agreed to purchase a genteel horse and
carriage, whereof they were to have the use in proportion
to the sums adventured, which was found to be 11 to B. as
often as 8 to C* ; what money did each adventure?
Ans. B. £104 4s. 2f3d., C £75 !5s. 9f^d.
140. Tubes may be made of gold, weighing not more
th in at the rate of jj^j^ of a grain per foot ; what would be
the weight of such a tube which would extend across the
Atlantic from Quebec to London, estimating the distance
at 3000 miles? Ans. 1 Ife 8oz. Gpwts. 3^^^ grs.
141. A military officer drew up his soldiers in rank and
file, having the number in rank and file equal ; on being re-
inforced with three times his first number of men, he placed
them all in the same form^ and then the number in rank
and file was just double what it was at first ; he was again
reinforced with three times his whole number of men, and
after placing them all in the same form as at first, his num-
ber in rank and file was 40 men each ; how many men had
he at first ? _ Ans. 100 men.
142. Supposing a man to stand 80 feet from a steeple,
and tha
100 fee
high ab
of the
of the
143
east, at
south,
apatt w
24 hour
144.
rods ; w
145.
posite c
of each
146.
tance ol
of the fi
A
147.
of carpe
of it ?
148.)
how ma
Whei
area or
Whe
given, 1
149.
20 rods
E
■ V-
1IH3.
^113.
MISCELLANEOUS EXAMPLES.
J?40
nmission ;
IS. £3780.
^l was just
the elder
father was
I..
148. 3fd.
which he
asted him
at was the
8s. 2^jd.
s about a
a ball be
95173000
n. 20 sec.
[uare bat-
nr i'lng
fill up the
s. 19000.
a, 45 gui-
idvanced,
horse and
roportion
to B. as
5s.
9Ad.
lot more
would be
cross the
distance
3A grs.
rank and
being re-
le placed
in rank
'as again
iicn, and
lis num-
men had
00 men,
steeple,
and that a line reaching from the belfry to the man is just
100 feet in length, the top of the spire is three times as
high above thf: ground as the steeple is ; what is the height
of the spire //and the length of a line reaching fr«jm the top
of the spire to the man 1 See fl 103.
Ans. to the last, 197 feet nearly.
143. Two ships sail from the same port- one sails directly
east, at the rate of 10 miles an hour, and the other directly
south, at the rate of 7^ milea an hour ; how many miles
apart will they be at the end of 1 hour ? 2 hours?
24 hours ? 3 days ? Ans. to last, 900 miles.
144. There is a square field, each side of whi^h is 5:)
rods ; what is the distance between opposite corners 1
Ans. 7U'71-f rods.
145. What is the area of a square field, of which the op-
posite corners are 70*71 rods apart ? and what is the length
of each side ? Ans. to last, 50 rods nearly.
146. There is an oblong field, 20 rods wide, and the dis-
tance of the opposite corners is 33^ rods ; what is the length
of the field ? its area ?
Ans. Length 2Gf rods ; area 3 acres, 1 rood, 13^ rods.
147. There is a room 18 feet square ; how many yards
of carpeting, 1 yard wide, will be required to cover the floor
of it ? 182=324 feet=30 yards. Ans.
148. If the floor of a square room contain 30 hquare yds.
how many feet does it measure on each side .'
Ans. IH feet.
When one side of a square is given, how do you find its
area or superficial contents ?
When the area or superficial contents of a square is
given, how do you find one side ?
149. If an oblong piece of ground be 80 rods long and
20 rods wide, what is its area ?
Note. — A parallelogram, or ohlong,
D c has its opposite sides equal and
parallel, but the adjacent sides
unequal. Thus, A. B. C. D. is a
parallelogram, and also E. F. C. D.
and it is easy to see that the con-
tents of both are equal.
Ans. 1600 rods=10 acres.
i
2.>0
MISCRLLANKOUS EXAMPLES.
1 113."
\50. Whut is tlie length of un oblong, or parullelogram,
whose areii is ten acres, and whose broadth is 12U rods ?
Ans. 80 rods.
151. If the area be ten acres, and the lenirth 80 rods,
whiit is the other side ?
When the length and breadth arc given, how do you fuid
the area of an oblong or parallelogram ?
When th« area and one side are given, how do you find
the other side ? '
152. If a board be 18 inches wide at one end, and ten
inches wide at the other, what is the mean or average width
of the board? " Ans. 14 inches.
When the greatest and least width are given, how do you
find the mean width ?
153. How many square feet in a board 16 feet long, 1*8
fuet wide at one end, and 1*3 at the other?
r8-fi'3
Mean width, =1*55 ; and r55X 16=24'8
feet, Ans. 2
154. What is the number of square fee( in a board 20
feet long, 2 feet wide at one end, and running to a point at
the other ? * Ans. 20 feet.
How do you find the contents of a straight edged board,
when one end is wider than the other ?
If the length be in feet, and the breadth in feet, in what
denomuiation will the product be ?
If the length be feet and the breadth inches, what parts
of a foot will be the p/oduct ?
155. There is an oblong field, 40 rods long and 20 rods
wide ; if a straight line be drawn from one corner to the
opposite corner, it will be divided into two equal right-
angled triangles ; what is the area of each ?
' Afis. 400 square rodszr.2 acres 2 roods.
156. What is the area of a triangle, of which the base is
30 rods, and the perpendicular 10 rods? Ans. 150 rods.
157. If the area be 150 rods and the base 30 rods, what
is the perpendicular ? Ans. 10 rods.
158. If the perpendicular be 10 rods, and the area 150
rods, what is the base ? Ans. 30 rods.
When the legs (the base and perpendicular) of a right-
angled triangle are given, how do ydu find its area ?
f lis/ I 51 113.
•nSCELLANEOUS EXAMPLKI.
251
llelogram,
rods ?
t. 80 rods.
I 80 rods,
i> you find
you find
1, and ten
age width
14 inches.
DW do you
; long, 1'8
: 16=24*8
1 board 20
a point at
'/.5. 20 feet.
fed board,
t, in what
ivhat parts
id 20 rods
ler to the
ual right-
s 2 roods.
he base is
150 rods.
ods, what
. 10 rods.
area 150
. 30 rods.
if a right-
a?
B
When tlie area and one of the legs are given, how do you
find the other leg I
Note. Any triangle may be divided into two right-angled
triangles, by drawing a perpendicular from cue corner to
the opposite side, as may be seen by the annexed figure :
c Here, A. B.C. is a triangle, divided
into two right-angled triangles, A. d
C. and d B.C.; therefore, the whole
base A. B. multiplied by one half the
\perpendicular, d C, will give the area
A of the whole. If A. B.«==60 feet, and
d C=16 feet, what is the area ?
Ans. 480 feet.
159. There is a triangle, each side of which is 10 feet ;
what is the length of a perpendicular from one angle to its
opposite side 1 and what is the area of the triangle ?
Note. It Is plain the perpendicular will divide the oppo"
site side into two equal parts.
Ans. Perpendicular, 8*66-|-feet; area, ^3'3-f-feet.
100. What is the solid contents of a cube measuring six
feet on each side ? Ans. 216 feet.
When one side of a cube is given, how do you find its
solid contents ?
When the solid contents of a cube are given, how do you
find one side of it ?
161. How many cubic inches in a brick which is 8 inches
long, 4 inches wide, and 2 inches thick ? in 2 bricks ?
in 10 bricks ? Ans. to the last, 640 cubic inches.
16*1. How many bricks in a cubic foot ? in 40 cubic
feet ? in 1000 cubic feet.' Ans. to the last, 27000.
163. How many bricks will it take to build a wall 40 ft.
in length, 12 feet high and 2 feet thick ? Ans. 25920.
164. If a wall be 159 bricks,=100 feet in length, and 4
bricks,=16 inches in thickness, how many bricks will lay
one course ? 2 courses ? 10 courses ? If the wall
be 48'courses,=:8 feet^high, how many bricks will build it?
150X4=600, and 600X48=28800, Ans.
165. The river Po is 1000 feet broad, and 10 feet deep,
and it runs at the rate of ^ miles an hour ; in what time will
it discharge a cubic mile of water (reckoning 5000 feet to
the mile) ijito the sea ? Ans. 26 days, 1 hour.
'•• I
252
MISCELLANEOUS EXAMPLES.
una
166. If the country which nupplies the river Po with
water be 380 miles long, and 12U broad, and the whole land
upon the surface of the earth be 62,700,000 square miles,
and if the quantity of water discharged by the rivers into
the sea be everywhere proportional to the extent of land by
.which the rivers are supplied, how many times greater than
the Po will the whole amount of the rivers be ?
Ans. 1375 times.
167. Upon the same supposition, what quantity of water,
altogether, will be discharged by all the rivers into the sea
in a year, or 365 days ? Ans. 19272 cubic miles.
168. If the proportion of the sea on the surface of the earth
to that of land be as 10^ to 5, and the mean depth of the sea
be a quarter of a mile ; how many years would it take, if the
ocean were empty, to fill it by the rivers running at the pre-
sent rate? Am. 1708 years, 17 days, 12 hours.
169. If p cubic foot of water weighs 1000 oz. avoirdupois,
and the w%ight pf mercury be 13^ times greater than water,
and the height of the mercury in the barometer (the weight
of which is equal to the weight* of a column of air on the
same base, extending to the top of the atmosphere) be thirty
inches ; what will be the weight of the air upon a square
foot ? a squaremile? and what will be the whole weight
of the atmosphere, supposing the size of the eiUth as in
questions 166 and 168?
Ans. 2I09'375 lbs. weight on a square foot.
52734375000 " " " mile.
10249980468750000000 " " of whole atmosphere.
170. Ifa circle be 14 feet in diameter, what is its cir-
cumference ?
Note. It is found by calculation, that the circumference
of a circle measures about 3if times as much as its diameter,
or more accurately, in decimals, 3*14159 times.
Ans. 44 feet.
171. Ifa wheel measure 4 feet across from side to side,
how many feet around it ? Ans. 12^,
172. If the diameter of a circular pond be 147 feet, what
is its circumference ? Ans. 462 feet.
173. What is the diameter of a circle whose circumfer-
ence is 462 feet ? Ans. 147 feet.
una
^113.
MISCELLANEOtS EXAMPLES.
353
jr Po with
whole land
uare miles,
rivers into
t of land by
;reater than
1375 times,
ty of water,
into the seu
mbic miles,
of the earth
h of the sea
t take, if the
r at the pre-
s, 1*2 hours.
ivoirdupois,
than water,
(the weight
f air on the
Ire) be thirty
on a square
lole weight
eiUth as in
ire foot,
mile,
tmosphere.
is its cir-
cumference
ts diameter,
ns. 44 feet,
ide to side,
Ans. 12f
f feet, what
s. 462 feet.
circumfer-
s. 147 feet.
4ns.
dif.
174. If the distance through the centre of the earth, from
side to side, be 7911 miles, how many miles around it !
7911 X3* 14159=24853 square miles, nearly. An^
175. What is the area or contents of a circle whose
meter is 7 feet, and its circumference 22 feet ?
Note. The area of a circle may We found by multiplying
half the diameter into half the circumference.
,Ans. 38^ square feet.
176. What is the area of a circle whose circumference
is 176 rods ? Ans. 2464 rods.
177. If a circle is drawn within a square, cont^.ning
one square rod, what is the area of this circle?
Note. The diameter of the circle being one rod, the cir
cumference will be 344159,
Ans. *7854 of a square rod, nearly.
Hence, if we square the diameter of any circle, and mul-
tiply the square by '7854, the product will be the area of
the circle.
178. What is the area of a circle whose diameter is ten
rods ? 102X*7854=78'54. Ans. 78*54 rods.
179. How many square inches of leather will cover a bail
3^ inches in diameter?
Note. The area of a globe or ball is 4 times as much ns
the area of a circle of the same diameter, and may be found,
therefore, by multiplying the whole circumference into tlu^
whole diameter. Ans. 38^ square inches.
180. What is the number of square miles on the surface
of the earth, supposing its diameter 7911 miles?
7911 X24853=196,G12,3?r?, Ans.
181. How many solid inches in a ball 7 inches ii diame-
ter?
Note. The solid contents ol a globe are fov.nd by multi-
plying its area by ^ part of its diameter.
Ant;. i79| solid inches.
182. What is the number of cubic miles in the earth,
supposing its diameter as above ?
Ans. 259,223,031,435 miles.
183. What is the capacity, in cubic inches, of a hollow
globe 20 inches in diameter, and how much wine will it
contain, one gallon being 231 cubic inches ?
Ans. 4188*8-j-cubic inches, and 18*13-j-gallons.
X
2o4
MISCELLANEOUS EXAMPLES.
T 113.1^ 113
•J
; :'»
V
i
; -I
' 184. There is a round log, 'all the way of a bigness • the
areas of the circular ends of it are each 3 square feet; how
^any solid feet does one foot in length of this log contain
-^ feet.in length ? 3 feet? 10 feet ? A solid o
thii^ form is called a cyKnder.
- '■ ■' fifow do you find the solid content of a cylinder, when
the area of one end and the length are given ?
1S5, What is the solid content of \ round stick 20 feet
long and 7 inches through, that is, the ends being 7 inches
in diameter ?
Find the area of one end, as before taught, and multiply
it by the length. . ' Ans. 5'347-j-cubic it^x.
If you multiply square inches by inches in length, whar
parts of Q.foot will the product be 1 if squai:^ inches by
feet in length, what part ? /
186. A Winchester bushel is 18'5 inches in diameter,
. and 8*inches deep ; how many cubic iilches does it contain ?
^/li. 2150'4-f .
It is plain, from the above, that the solid content of all
bodies, which are of uniform bigness throughout, whatever
may be the form of the ends, is found by multiplying the
area of one end into its height or length.
Solids which decrease gradually from the base till they I ''«'^ p»ir
come to a point, are generally called Pyramids. If the base I ^f^t in
be a square, it is called a square pyrdkid ; if a triangle, a
^iangular pyramid ; if a circle, a circular pyramid or a cunc.
The point at the top' of a pyramid is called the vertex, and
a line, drawn from the vertex perpendicular to the hascy
called the perpendicular height of the pyramid.
The solid cemtent of any pyramid may be found by multi-
plying the area of the hast by ^ of the perpendicular height.
187. What is the solid content of a pyramid whose base
i.s 4 feet square, and the perpendicular height 9 feet ?
''■^ ' • 42X1 = 48. Ans. 48 feet.
There is a cone, whose height'is 27 feet, and whose
Note.
addinor i
tenths, <
tors, to 1
to a cyl
Now,
'7854, (
and that
sjolid cor
231, (no
wine gal
will give
In thi
eter will
Ions, by
only mul
by '0034
for decii
.Ions, mu
Hence
Multiply
multiply I
In the
the head
2.> in.-j-
isi
188;
base is 7 feet in diameter ; what is its content?
Ans. 346 } feet.
189. There is a cask, whose head diameter is 25 inches,
bung diameter 31 inches, and whose length is 36 inches:
how many wine gallons does it contain? how many
*beer gallons ? . .^ ^
Th
fin.
190.
mater is
inches
191.
prop, on
pounds
aiiced
the pro;
Note.
faet froi
inches,
b
.(i_.-;,.
■ ■; li
)igness ; the
re feet ; how
log contuiii ?
A solid of
Under, when
itick 20 feet
ing 7 inches
md multiply
j-cubic feet.
cngtJi, wh;if
i^ inches by
u diameter,
3 it contain ?
s. 2150'4-|-.
ntent of all
It, whatever
ijjlying the
ise till they
If the base
a triangle, a
lid or a cunc.
vertex, and
;he base, is
iid by multi-
jular height.
whose base
feet?
Ins. 48 feet.
:, and whose
346 j} feet.
5 25 inches,
36 inches ;
how many
MISCELLANEOUS EXAMPLES.
w:if
Note. The mean diameter of the cask may be found by
adding 2 thirds, or, if the staves be but a little curving, 6
tenths, of the difference between the head and bung diame-
ters, to the head diameter. The cask will then be reduced
to a cylinder.
Novv, if the square of the mean diameter be multiplied by
'7854, (ex. 177) tjie product will be the area of one end,
and that, multiplied by the length, in inches, will give the
solid content, in cubic inches, (ex. 185,) which, divided by
'231, (note to table, wine meas.) will give the content in
wine gallons, and, dividedby 282, (note to table, beer meas.)
will jjive the content in ale or beer measure.
In this process, we see that the square of the mean diam-
eter will be multiplied by *7854, and divided, for wine gal-
lons, by 231. Hence we may contract the operation by
only multiplying by their quotient, •^/;/'-*=:'0034) that it^,
by *0034 (or by 34*, pointing off 4 figures from the product
for decimals.)' For the same reason we may, for beer gal-
,lons, multiply by ('vW^'O^^S, nearly) '0028, &c.
Hence this concise Hvle for guagiKg or measuring caska :
Multiply the square of the mean diameter by the length;
multiply this product by 34,/i/r wine, or by 28 for beer,
and pointing off four decimals, the product will be the con-
trnt in gallons and decimals of a gallon.
In the above example, the bung diameter, 31 in. — 25 in.
the head diameter=5 in. difference, and % of 6=4 inches;
*2> in. -{-4 in.=i9 in. mean diameter.
Then 292=341, and 841 X36 in.=30276.
( 30276X34=1029384. Ans. 102*9384 wine gals.
\ 33726X28==S47728. Ans. 84'272S beer gals.
191). How many wine gallons in a cask whose bung dia-
«n3ter is 3t) inches, head diameter 27 inches, and length 45
inches ? Ans. 166'617.
191. There is a lever 10 feet lonp, and i\\e fulcrum, or
prop, on which it turns is 2 feet froi.i one end ; how many
pounds weight at the end, 2 feet trom the prop, will be bal-
anced by a power of 42 pounds at the other end, 8 feet from
the prop.
Note. In turning around the prop, the end of the lever 8
faet from the prop will evidently pass over a space of eight
inches, while the end 2 feet from the prop passes over a
Then.
r:>
256
MISCELLANEOUS EXAMPLES.
■f
11113,
space of 2 inches. Now, it is a fundamental principle in
mechanics, that the weight and power will exactly balance
each other, when they are inversely as the spaces they pass
•ver. Hence, in this example, 2 pounds, 8 feet from the
prop^ will balance 8 pounds 2 feet from the prop ; therefore
»if we divide the distance of the power from the prop by the
distance of the weight from the prop, the quotient will al
ways express the ratio of the weight to the power ; |=4,
that is, the weight will be four times as much as the power
42X4=168. Ans. 168 lbs.
192. Supposing the lever as above, what power would it
require to raise 1000 pounds 1 Ans. ' <^°=250 lbs
193. If the weight to be raised be 5 times as much as the
power to be applied, and the distance of the weight from
the prop be 4 feet, how far from the prop roust the power
be applied 1 Ans. 20 feet
194. If the greater distance be 40 feet, and the less half
of a foot, and the power 175 lbs., what is the weight?
Ans. 14000 pounds
195. Two men carry a kettle weighing 200 pounds ; the
kettle is suspended on a pole, the bale being 2 feet 6 inches
from the hands of one, and 3 feet 4 inches from the hands
of the other ; how many pounds does each bear.
Ans. I14f lbs. and 85f lbs
196. There is a windlass, the wheel of which is 60 inches
in diameter, and the axis, around which the rope coils, is 6
inches in diameter ; how many pounds on the axle will be
balanced by 240 pounds at the wheel ?
Note. The spaces passed over are evidently as the diam-
ofers or the circumferences ; therefore, ®^^'=10, ratio.
Ans. 2400 pounds,
197. li the diameter of the wheel be 60 inches, what
must be the diameter of the axle, that the ratio of the weight
to the power may be 10 to 1 .' Ans, 6 inches,
Note. This calculation is on the supposition that there is
no friction, for which it is usual to add 1^ to the power which
is to work the machine.
19*8. There is a screw whose threads are 1 inch asunder,
which is turned by a lever 5 feet=60 inches long ; what is
the ratio of the weight to the power ?
Note. The power applied at the end of the lever will de
tf 113.
principle in
ctly balance
I'-es they pass
et from the
»p ; therefore
prop by the
lent will al-
tower ; f =4,
s the power,
ns. 168 lbs.
wer would it
p 0=250 lbs.
much as the
weight from
St the power
Ans. 20 feet.
the less half
tveight ?
iOOO pounds.
pounds ; the
I feet 6 inches
Im the hands
ir.
, and 85i^ lbs.
h is 60 inches
)pc coils, is 6
axle will be
as the diam-\
), ratio.
3400 pounds.
inches, what
of the weight
ins. 6 inches,
I that there is
; power which
inch asunder,
long ; what is
lever will de-
!I
113.
MISCELL. NK I US EXAMPLES.
257
scribe the circumference of a circle 60X2=120 inches in
diameter, while the weight is raised I inch ; therefore, the
ratio will be found by dividing the circumference of a circle
whose diaYneter is twice the length of the lever, by the dis-
tance between the threads of the screw. 120X3|=377|
377f
circumference, and =377|, ratio. Ans.
1
199. There is a screw, whose threads are ^ of an inch
asunder ; if it be turned by a lever 10 feet long, what weight
will be balanced by 120 lbs. power? Am. 30171 fts.
200. There is a machine, in which the power moves over
10 feet, while the weight is raised 1 inch ; what is the power
of that machine, that is, what is the ratio of the weight to
the power ? Ans. 120.
201. A rough stone wis put into a vessel, whose capacity
was 14 wine quarts, which was afterwards filled with 2^
quarts of water ; what was the cubic content of the stone ?
Ans. 664^ inches.
/.
•%j
X2
258
FORMS OF NOTES AND RECEIPTiT^
/ (
t^
Forms of JVotes^ Receiptor Orders and
^r,,v.>ftjit*0»y,; Bills oi Parcels*
«.?.;;.'
*-r!
n.;.
i'*-.-'«r"f :;■*■■->■ M^'^W'M" ■™.*''^'
NOTES.
No. 1.
Montreal, Oct. 22, 1849.
For value received, 1 promise to pay to Oliver Bountiful,
or order, two pounds, ten shillings and sixpence, on demand,
V'ith interest. William Trusty.
Attest, Timothy Testimony.
No. IL
^^ Kingston, Oct. 10, 1849.
For value received of A. B. in goods, wares, and mer-
chandize, this day sold and delivered, I promise to pay him
or bearer, pounds, shillings^ and pence, in
ten davs from date, with interest. C D .
No. III.
By two Persons.
Stanstead, Oct. 1, 1849.
For value received of , in this day sold and
delivered, we jointly and seTeraJly promise to pay him, or
order, pounds, shillings and peace in
days irom date, with interest. B C .
I>-
E-
I' : ti ' «.'.
RECEIPTS.
, Montreal, Oct. 20, 1849.
Received from Mr. Durance Adley, ten pounds, in ful[
of all accounts. OrvaNd Constancy.
Receipt for Money received on a Note,
York, Nov. 1, 1849.
Received of Mr. Simon Eastly (by the hand of Mr. Titus
Trusty) sixteen pounds, ten shillings and sixpence, which
is endorsed on his note of June 3, 1831.
Samson Snow.
'^i;;,*;'- " ■■■■■ ■■^■'?^• j
ORDCIIS AND BILLS OF PARCEL?.
, 259
crs and
22, 1849.
r Bountiful,
on demand.
Trusty.
10, 1849.
s, and mer-
e to pay him
— pence, in
- D .
1, 1849.
lay sold and
pay him, or
nice in
C .
E .
20, 1849.
ands, in fuH
■NSTAN'CY.
e,
. 1, 1849.
)f Mr. Titus
encc, which
)x Snow.
« 'ks' Receipt for Money received on Account. ' ' >
Stanstead, June 2, 1849;
Received of Thomas Dubois, twenty pounds, on account,
Orlando Prompt.
Receipt for Money received for another Person.
Sherbfooke, June 4, 1849.
Received from P. D. twenty-five pounds for account of
J. T. Eli Trueman.
Receipt for Interest due on a Note.
s Quebec Dec. 18, 1849.
Received of I. S. fifteen pounds, in full of one year's in-
terest of <£250, due to me on the day of last, on
note from the said I. S. Solomon Gray.
•Receipt for Money paid before it becomes Due.
Prescot, May 3, 1849.
Received of T. Z. fifteen pounds, advanced in full for
one year's rent of my farm, leased to the saiJ T. Z. ending
the first day of April next, 1850.
John Honorus.
ORDERS.
Belville, Nov 3, 1848.
Mr. Stephen Girard. For value received, pay to A. B.,
or order, five pounds and six shillings, and place the same
to my account. - .. Saul Mann.
Montreal, Sept. 1, 1848.
Mr. Timothy Titus. Please to deliver to Mr. L. D.
such goods as he may call Tor, to the amount of seven pounds,,
and place the same to the account of your obedient servant,
NicANOR Linus.
BILLS OF PARCELS,
It is usual, when goods are sold, for the seller to deliver
to the bjyer, with the goods, a bill of the articles, and their
"y: '
■J
1260
BILLS OF PARCELH. ^i^^MV
prices, with the amount cast up. Such bills arc sometimes
called Bills of Parcels. ^
Montreal, 6th May, 1849.
Mr. Abel Atlas,
Bought of Benjamin Buck,
£ . s. d.
12:i yards tii^ured Satin, at I2s. 6d. per yard, 7 10 il
8" " Sprigged Tabby, at (Js. 3d. " t> 10
Received Payment,
jeio y
«
Benj. Buck.
'tr--
Montreal, 14th May, 1849.
Mr. John Burton,
'if , ' Bought of Geo. Williams,
!5 hhds. new Rum, IIS gallons each, at Is. 6d. per gallon,
2 pipes French B.andy, 126 & 132 gal. 5s. 7d.
1 hhd. brown Sugar, 9f cwt. at £2 lis. 9d. per cwt,
3 casks Rice, 269n> each, at 3d. *' tb.
5 bags Coffee, 751b each, at Is. 2d. " "
1 chest hyson Tea, S61b, at 43. 8d. " "
Received Payment,
For George Williams,
: ] Thomas Rousseau.
Ol
Wilderness, 8th Feb. 1849.
Mr. Simon Johnson,
Bought of Asa Fullum,
r)(>S2 feet Boards, at ^1 lOs. per M. . '
2 Is. Hd; "
3 3s: 2d. " /
1 Os. Od. "
1 lOs. Od. "
12s. 6d. "
13s 9d. "
2000
830 " ^Stuff,
15!>0 " I.ithin;:.
6;'>0 " Plank,'
879 " Timber,
23G "
OJ
00
CO
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