:
1
1
ce
.
Horatius St Paul,Sacri Romani
Imperii.Comes De St Paul.
Roger Sc.2750.
:
UNIVERSITY OF MICHIGAN
LUANVSV
SI QUERIS PENINSULAM AMUIYAM
CIRCUMSPICE
181:7
ARTES
SCIENTIA
VERITAS
LIBRARY
OF THE
Inition
TE
TIENIOR
.ل. د . د بل ' اي
laminton KLHLUTOILUN
Salt
:וננווונו
uut diutamu
Minianiuoni NOW
umaupung
James Adinston
TREATISE
OF
Practical Arithmetic,
BOTH
Integral and Fractional.
WITH
The MENSURATION of all sorts of Bo-
dies, both Superficially and Solidly.
'The Whole after a New Method.
Accommodated to the Capacity of Beginners.
By ALEXANDER WRIGHT, M. A,
Writing-Maſter and Accomptant at Aberdeen.
miunu
Si
@
1 ONDOV
Printed for J. OSWALD, at the Rofe and Crown, near the
Manſion Houſe in the Poultry. 1740.
Where may be hade
A TREAT 15E of FRACTIONS, by the ſame Author,
Pluce this near the Title
To the Right Honourables
Baillies.
WILLIAM CHALMERS EsởL? Provoſt.
Mexander Robertson
Alex"Minhel
of Colpna
Tames Moorijon Irun,
MWilliamFordyce
WÁMouat Tun.Dean of Gild.
Andren: Logie, Theaſurer.
And to the hemanent Members of
Town Counalof ABERDEEN
The following Treatise
is most humbly
Inscribed
by their much Obliged &
most Obedient ervant
Alexander Wright.
Felicifsimæ Indotis
Optimaque per
Adolescentri
D. Andrea Skene
r
de Lethenty
Hunce trithmeties Präclicæ
Tractatum
1
imoris et Obrerinntir er
rg
D.D.6.2.
Alexander Wright
.
WA
ون (1، s1 . : : :
$۱ .
1.1?
THE
37661
P R E F
E F A C E.
or A
HE Books already publiſhed 092 Arithmetic,
are lo very rimerous, that the Public will
poſſibly be ſurprized to ſee a New One 02
that Subject ; eſpecially when it is confider'd
that by the late Improvements this Science
has received, there can little be ſaid on it, but what has
been advanced by ſome one Author or other. It would
ſeem therefore that all that can be expected now, is to
Jee the moſt eaſy, plain, ſhort, and familiar Method of
inſtručting Youth in this excellent Art, to which I flatter
myſelf I have (at leaſt) paved the way in the follow-
ing Sheets; a Specimen whereof I ſhall bere briefly lay
before the Reader. The Method is very much new .
122
the Integral Part I have retrench'd the Number of what
they call the Rules, and have reduced them all (after
the firſt five general ones) to three, viz. the Rule of
Three Numbers, Rule of Five Numbers, and Rules of
Practice, comprehending in the firſt of theſe, thoſe that
are commonly called the Rules of Intereſt, Diſcount,
Tare and Trect, Fellowſhip, Gain and Loſs, Alligation,
Barter and Exchange. The Examples adduced are choice,
very practical, and contain a vaſt Variety, each of which
I have illuſtrated ; so that there can remain 720 Diffi-
culty in underſtanding the Method and Reaſon of ope-
rating. In the Fractional Part, which is ſo superfici-
ally handled by moſt Authors, I have been very copious
and plain, that the Learner may not be diſcouraged in pro-
ſecuting ſo uſeful and neceſſary a Part of the Science, and
have placed them immediately after Integral Diviſion,
As judging it proper they ſhould be learned before one
A 2
com
PRE FACE.
commenceth the Rule of Three ; becauſe their Uſe
and Application moſtly appears there, and in what fól-
lows afterwards. To the End of each Rule I have ſub-
joined some uſeful Practical Queſtions, ſbewing its im-
mediate Application to Buſineſs. The Appendix contains
a further Explication of Intereſt, both Simple and Com-
pound, with all the Tables that are neceſſary for that
purpoſe; as alſo a fuller and plainer Account of the
Menſuration of all Šorts of Bodies, than is commonly to
be met with in Treatiſes of Arithmetic. So that a
Learner may, without the Aliktance of a Teacher, ſooner
attain to a competent Knowledge of Numbers, by studying
them in the Method here laid down, than any other I
have had the opportunity of ſeeing. In the Decimal
Part, I have purpoſely omitted Infinites, Circulates, and
Approximates, which the Reader may fee fully explain'd,
if he pleaſes to conſult a Treatiſe of Fractions lately
publiſhed by me, to which I refer him.
Si quid noviſti rectius iſtis,
Candidus imperti: Si non, his utere mecum.
HORAT:
From my School in the Upper
Kirk-ſtreet, Aberdeen.
ERRAT A.
Ag. 5. for 11, 111,111, lege 12, 101, 111. P.45, in
Exa. 3. the Diviſor 8 is wancing. P. 48. l. 19. for
lege 1. P. 58.1. 15. for 7 d. read 17 d. p.60.1. 11. for
15 d. lege 15 1: -l.is. lege 11 s. 6 d. -- lege 13. s.
INTRODUCTION.
I
HAVE thought it proper in this place, to pre-
miſe the Definitions of ſome Terms, as alſo the
Significations of ſome Marks or Symbols, made
Uſe of in the following Sheets for Brevity's ſake.
1. A Number is either Unity or Multitude.
2. An Integer, or whole Number, is I, or any
Number of Units.
3.' A Fraction is a Part or Parts of Unity, accor-
ding as the ſame is divided.
4. An Aliquot Pare is a leffer Number, which is
contained a certain Number of Times in a greater,
without a Remainder; as 4 is an aliquot Part of 8, 12,
16, 20, ec.
5. An Aliquant is a leſſer Number, which is con-
tained in a greater a certain Number of Times, with
ſomething remaining: Thus 4 is an aliquant Part of
loI418, pc.
6. An Abstract Number is a Number conſidered
abſolutely in itſelf, without any Name or Denomina-
tion applied to it.
7. An Applicate Number is ſuch as hath ſome
Name applied to it: Thus 6, taken abſolutely, and
ſignifying ſimply ſix, is Abſtract; but if we annex to
it the word Men, Pounds, Gallons, &c. it is Ap-
plicate.
8. The Common Meaſure to any two or more
Numbers, is a Number which can divide thoſe Nuni-
bers, without a Remainder: Thus 6 is a Common
Meaſure to 18, 24, 36, C.
9. The Greateſt Common Meaſure to any two or
more Numbers, is the greatest Number which can
2
INTRODUCTION.
divide the propoſed Numbers, without a Remainder :
Thus 6 is the greateſt Common Meaſure to 18 and
24; likewiſe 4 is the greateſt Common Meaſure to 8,
22, 20.
10. A Prime Number is that which has no Mez-
ſure bur itſelf and Unity: Thus 3, 5, 7, II, 13, 234
29, br. are Prime Numbers.
11. Thc Multiplier and Multiplicand go frequeni-
ly by the name of Factors, becauſe by being multi-
plied together they make the Product.
12. When a Sum of Money is lent,, or lies out for
any Time, it is called a Principal, and the Money
paid for the Uſe or Forbearance of t'he Principal, it
commonly called Intereſt, and is always at ſome cer-
tain Rate per cent. per Anunt, fuch as 4, 5, 6, &C.
13. Simple Intereſt is that which is produced by
the Principal lent or forborn for any Time. But,
14. Compound Intereſt is that which is produced
by the Principal and the Simple Intereſt lying out un-
paid for any Time, both in one Sum, For Example,
L. 100 lent at 5 per cent. makes at the End of The
Year L.105; and if this L.105 be let lie in the Bor:
rower's hands, and become a new Principal to bear
Intereſt the next Year, then it it is called Compound
Intereſt: And ſo on for any Number of Years, by
ſtill adding the preceding Year's Intereſt to its Prin-
cipal.
15. The Amount of any Principal for any Time is
the Sum of that Principal and all the Intereſts due
upon the ſame.
26. When a Sum of Money, payable at any Time
hence, and not bearing Intereſt till after it is due, is
to be paid preſently, the Creditor muſt allow the Deb-
tor ſo much for advancing it before the Time, as,
being put out to Intereſt from this Time to that of
the Payment of the Debt, would amount to the Intea
reſt of the Debt for the ſame time, at any Rate per
Cent, agreed on: and the Money fo allowed is called
Dif-
INTRODUCTION. 3
Diſcount or Rebate; which Diſcount being ſubtract-
ed from the Debt firſt due, the Remainder is the pre-
ſene Worth
17. An Annuity is an annual Payment, or a Sum of
Money payable every Year for a certain Time or for
ever,, whether it be due on Lands, Houſes, or Mo.
ney in Stock or Bank, dr. And when an Annuiry
lies unpaid for any Time, it is ſaid to be in Arrears.
18. Bartering is the exchanging of one Commodity
for another.
19. Tare is an Allowance for the Weight of the
Hogſhead, Cask, Cheſt or Bag, c. which contain
the Goods, and may either be known ſeparately, from
the Weight of the Goods, or is accounted to much
per Cent, or 112 lib.
Trete is an Allowince on ſome Sort of Goods for
Walte, Duft, Refuſe or Inlack, &c.
Cloff is an Allowance of 2 lib. on every Draught
above
3
C. to Freemen of London,
When the Tare is fubtracted, the Remainder is cal-
led Suctle Weight; and when all Allowances are de-
ducted, what remains is called Net Weight or Weight
payable.
EXPLANATION of the MARKS
or SYMBOLS.
The Sign + (plus or more) is the Sign of Addition
The Sign -- (minus or leſs) deno:es Subtraction.
The Sign + (divided by) is the Sign of Diviſion.
The Sign x (multiply'd by), is the Sign of Multiplica-
tion.
The Sign = (equal to) is the Sign of Equality.
B 2
ARITH
4
ARITHMETIC
I
Sihe Art of Numbering.
The Characters whereby Numbers are expref-
ſed, are,
2. 3. 4. 5. 6. 7. 8. 9. O.
1.
NO TATION
TE
Eacheth to expreſs any Sum or Number by thefe
Figures, according to the following
TABLE,
13 Billions.
12 C Th. Mills,
II X Th. Mils.
10 M Millions.
9 C Millions.
8 X Millions.
y Millions.
6 C Thouſands
5 Ten Thour.
4 Thouſand.
3 Hundred
2 Tens.
I Unirs.
SOWEN
•senet
Or thus ::
Th. Bills. Bills. Th. Mills. Mills. Thouf. Units,
cxUn. CẦUn.cXVn.CX Un.CXVn.CxUn.
&c.
Read thus : Units, Tens. Hundreds of Units; U-
nirs, Tons, Hundreds of Thouſands, &c. ſo that
when a Sum is propoſed to be read or expreſſed in
Words, divide the ſame into Periods, each conſiſting
ot
NO I Am To'Ñ.
5
of 3 Figures, and read each Period as if it was an en-
tire Number by itlelf, applying always its general
Name.
For Examp. Let the Number 14356485 be propo-
ſed to be read or expreſſed in Words; firſt divide
into Periods thus, 14,356,485, and you'll find it con-
ſiſt of 3 Periods, that is, Units, Thouſands Millions,
and therefore is read 14 Mill. 3 hund. 56 Thouſand,
4 Hund. eighty five. In like manner,
2758,7630747 758 Mill. 7763 Thouſ. 747
12,469,721,766 12Th 49 M. 72 Th 766
4,317,467,005,701
4 4. B. 317th. 467 M. 5th.701
2400,073,000,465
+
th: 73 Mill. 465.
1,000,000,000,000 i Billion.
read
1S
4008
Set down in Figures Four choufind and eight
Set down Fifty four th. Millions 54,000,000,000
Set down Eleven hundred and eleven
I,III
Set down Elev, hund, and elev. th, and elev. 1,111,011
Ser down Eleven hundred and eleven thou- ?
fand, one hundred and eleven $
1,111,111
Set down Eleven Mill. Eleven hund.
thouf. eleven hund. and eleven
}
II,II III
11
1,
Of the Roman Notation.
The Romans exprcſs'd Numbers by the Capital Let-
ters of their Alphaber, and their ſimple Characters.
were there,
1.V.X.L.C. D. M equal to
1.5 610.50 - 100
50 - 100 . 500
And the intermediate Numbers betwixt theſe, they ex-
preſled by a Repcriiion of the fam“, ſeriing them
a fier one anoiher in a Line, the Characters of the
greateſt Value being placed to the left: Thus Il=2,
Vimmer, Vi=8, LX=60, DC=600, DCC=700.
But ſometimes they fer the Character of the leffer
Value
1000
B 3
6
NOTATION.
Value to the left of the greater, and this fignified the
Difference betwixt the two, which they uſed for a ſhor-
ter Expreſſion: Thus IV=4,IIV=3,IX=9,XL=40,
CD400, CM=900; and for D=500 they wrote
I], and by adding another , it expreſs'd ten Times
as much: Thus 15D=5000, 10=50000. Allo for
M they wrore CD, and by placing Cand I on each
hand, it expreſſed ten Times as much, for CCID =
10000. They expreſſed their Thouſands ſometimes
by a Line drawn over the Top of any Number lels
than 109, thus, V=5000, X=10000.
70 LXX
2 II.
3 II.
10o C.
Io X.
1 l.
.
89 LXXXIX.
4 III or IV.
200 CC.
SV.
500 D or I”.
6 VI.
600 DC or 1°C.
9 VII.
700 DCC or ICC.
8 VIII or IIX.
9 VIII or IX.
1000 M or CI° or 1.
2000 CIO CIO or T.
II XI.
3000 CIO CIO CIO or TT.
I2 XII:
4000 CIO CIO CIO CIO or TV.
13 XIII.
5000 15 or
XVIII
19 XVIIII or XIX.
20 XX.
50000 1000 or i
100000 CCCII. or
40 XXXX or XL.
Sooooo I9000.
49 XLIX.
1000000 CCCCIOor M.
2000000 MM.
CIO1OCCXXXVIII or
OF
60 LX.
1738
69 LXIX.
CHAP
y
1 xviil or ]IXX. 10000 CCJ or X.
T.
30 XXX.
45 XLV.
900 CM.
50 L.
59 LIX.
CHAP. II. Addition
Finds the sum of two or more Numbers propoſed.
1 SHALL begin with the Addition of Simple Abſtract
or Simple Applicate Numbers, for which obſerve
the following
RULE. Set down the ſeveral Mambers under one
another, so as Units may ſtand under Units, Tens urf-
der Tens, doc. and, beginning at the right Hand, add
rogerher che Figures in each Column ſeverally, and ſet
down the Sum, if it does not exceed 9; or, if it is 10
or above, ſet down the Exceſs, carrying i for every
10 to the next Column: And thus proceed 'till you
come to the laſt Column, ferting down the Sum of it,
becauſe you have no more Columns to add. The
Figures thus found are called the Amount, Total, Ag-
gregate or Sum of the feveral Particulars given to be
added.
Exa. I.
4265
1532
Exa. 3.
Yds. 1187
Exa. 2.
L. 3658
783+
Sum 11492
356
Sum 5797
Sum 1543
Exa. 4.
Exa. 5.
26370
285
16325
176
1483
3.7
218
105
3
186
400
EX8. 6.
4000
2000
600
300
490
1000
700
100
76
42
27
II2
17
Sum 18364
Sum 27498
Sum 9100
1. Of
8 Addition of Mixt Applicate Numbers.
1. Of. Engliſh Money.
The leaſt Piece of Money now uſed in England is
a Farthing, whereof 4 make i Penny, 12 Pence Shil-
ling, and 20 Shillings i Pound : In there Accounts
are kept, and are marked with the Characters L. so
d. g. or f.
Beſides thoſe they have other Coins, ſome real and
others imaginary.
4. Farth. ) ri Penny. The real Coins now
4
Pence
i Groat. current, and com-
6 Pence
i Tefter monly known, are,
12 Pence
i Shilling Of copper,
i Crown. A Farth, and a Half-
6 Sh. 8 d.
i Noble.
penny
ìo Shill.
i Angel. Of Silver,
I Mark. A Penny, Tivo
I Pound. pence, Three. ponce,
Four-pence, Six-pence, a Shilling, Half a Crown, á
Crown.
And of Gold, Half a Guinea=10 s. 6 d. and a
Guinea=21 Shill. They have alſo ſeveral other Pieces
both of Gold and Silver, but they are not ſo com-
mon.
In Scor!and we uſe L. D. d. and L. 12 Scotch are e-
qual to 1 L. Sterling. But llerchants, and People of
Fashion, generally account in Englijh Money.
5 Shill.
smake
13 S. 4 d.
20 Shil.
Exa.
Addition of Mixt Applicate Numbers. O
Exa. 2.
S.
d.grs.
Exa; I.
(20)(12) (4)
L.
145: 19: II: 3
26: 8: 6: I
35:11: 10: 2
16: 8
43: 17: 10:3
54:13:
18: 6: 8: 2
20: 13:
9: 1
(20) (12) (+)
b. d.qrs.
13: 4: 3
7: 9:3
4: 10: 2
9:0
18: 5:3
0:
7:2
4:0
I:
2: 1
7:
8: 3
16: 10:1
Sum 362: 3 : J0:0
Sum 3: 10:11: 2
As I have ſer down, on the Top of each Denomina-
cion, the Numbers the Learner muſt carry at, I ſhall
not further illuſtrate the above Examples; but, in the
Column of Shillings, let him rather add the Unics
Place (without pointing) ferting down the Exceſs o-
ver the ro's, and carrying one for each 10 to the Tenis
Place, and thence I for every 2 to the Place of L;
And thus I have added the ad Example.
The Farth. and Halfpence are frequently marked
thus, ) , which fignify a Fourth of a Penny or i
Farth. one Half of a Penny or 2 Farth. 3 Fourths of
a Penny or 3 Farth. as in the following Example.
L.
S.
d.
120: 14:6
74: 8:72
12: 17:4
6: 4
Here for every d. I rec-
kon 2 Farth. and add as in
Exa. I.
17:
38: 15:5
319: 16:43
2: O
Sum 589: 19:11
To Addition of Mixt Applicate Numbers,
2. Of Troy Weight.
The Original of all Weights uſed in Ergland (we are
told) was a Grain of Wheat taken out of the middle
of the Ear, and being well dried, 32 of them were to
make i Penny-weight, 20 Penny-weight i Ounce,
and 12 Ounces 1 Pound; but the Penny-weighe was
afterwards divided into 24 equal Parts called Grains,
which is the leaſt Weight now commonly uſed. By
this Weight are weighed Jewels, Gold, Silver, Liquors
and Bread.
TABLE.
Gr.
24 11 d. wr.
That is, 24 Grains make
i d. wt. 20 d. wt. or 480
410 20 Oz.
Gr. make 1 Oz. 12 OZ.
5700 240 12 1 lib.
or 240 d. wt. or 5760 Gr.
make i lib. Therefore you muſt carry or point at
every 24, 20, 12, as in the following Examplc.
(12)(20)(24)
lib. oz. d.wr.gr,
148: II: 19:23
25: 7: 6: 18
10: 10: 14: 6
44: 312:
8
31: 11: 8: 15
18:
9: 10; 17
Sum 280: 6: 12:15
1
3. Of
Addition of Mixt Applicate Numbers. 11
3. Of Apothecaries Weight.
This is the ſame with Troy Weight, unleſs that they
divide the Lib. after a different manner, viz. into
Grains, Scruples, Drams and Ounces. By this Weight
the Apothecaries compound their Medicines, but buy
and ſell their Drugs by Aver-du-poiſe Weight.
TABLE for Apothecaries Weight.
Gr.
That is, 20 Gr. make
2011 Scr.
1 Scruple, 3 Scr, i Dr.
6c
11 Dr.
8 Dr. í Oz and 12 OZ.
i Lib. I uſe theſe Con-
480 24 811 Oz.
tractions, Gr. Scr. Den
57601288196 121 Lib.
Oz. Lib.
3
Example.
(12)(8) (3) (20)
Lib. oz. dr. ſcr. gr.
36: 11: 7: 2:19
8: 2:0 : 14
7: 10: 3:2; 6
7:Q:I: 12
3: 6: 5: 1: 16
1: 10: 6: 2: 15
0:
2:
Sum 53: 7: 3:0:
2
4. Of Aver-du-poiſe Weight.
By this Weight are weighed all coarſe Commodities,
and ſuch as are ſubject to Waſte, as Iron, Lead, Salt,
Hemp, Flax, Butter, Cheeſe, &c. It is divided into
two Denominations, viz. the Greater and the Leffer:
'The Greater confits of Tuns, Hundred Weights,
Quar-
12 Addition of Mixt Applicate Numbers.
Quarters and Pounds, which are thus mark'd, T. C.
rs. lib.The Leffer conſiſts of Stones, Pounds, Ounces
and Drams, thus marked, St. lib. oz. dr.
TABLE for the Greater. TABLE for the Leſſer.
lib.
dr.
28][ qr.
16I OZ.
112 411 C.
256 16. lib.
224013020
30 20 1 Tun.
4096 259116) 1 Stone.
That is, 28 lib. make iqr. That is, 16 dr. make i
4 qrs. i C. and 20 C.'i
oz. 16 oz. I lib. and 16.
Tun.
lib. 1 St..
Examples.
(20)(4) (28)
T. C. grs. lib.
34: 19: 3: 27
12: 10: 2: I
41: 5: 1:
8
6: II: 3:14
3: 9:0:20
5:13: 1: 16
(16) (10)(16)
St. lib. oz. dr.
70: 15: 15:15
18: 6:12: 8
45: 13: 6:
4
5
6: 15: 8: 10
17: 9: II:
3
20:
8:
3:
Sum 104: 10: 1: 19
Sum 180: 5: 9:13
Wool Weight in England hath theſe Denomina-
tions, aiz Lib. Cloves, Stones, Tod, Wey, Sack,
Laſt, according to the following Table.
Lib.
Addition of Mixt Applicate Numbers. 13
Lib.
711 Clove.
I
21 Stone.
28
4
2/1 Tod.
182 26
26
13/ 61 Wey.
· 3641 52
26 13 21 Sack.
43681624) 312 156 2412 1 Laft.
Mr. Ward ſays, that, by a very nice Experiment,
he found, that I lib. Aver-du-poiſe is equal to 14 OZ.
I d.wt. 15 gr. Troy Weighr.
5. Of Liquid Meaſures.
Liquid Meaſure is founded on Troy Weight; for
S lib. Troy Weight, taken out of the middle of the Ear
and well dried, was to make a Gallon of Wine, Ale
or Corn, than which no other Meaſure was at firſt al,
lowed, tho' others have been introduc'd ofterwards.
TABLE of Wine Meaſure.
Pints..
2/1 Quart.
8 4
[ Gall.
168 42 [ Tierce.
501 252 63.11 Hhd.
672 3301 84 2 2,1;fi Puncheon:
1008
But or Pipe
2
2016110081252 ol
336
بابا
504126
3 2/1 Tun.
C
Accounts
· 14. Addition of Mixt Applicate Numbers.
Accounts are kept in theſe Denominations, Tuns,
Pipes. hhds. gall. pints. Thus,
T. P.hhds.gal. pts.
54: 1: 1: 62: 7
3:0: 1:15: 4
I: I:0: 12:
ng
2:0:0:18:
4
3: 1: 1: 25: 3
2: 0: I: 19:
Note, 18 Gall. make i Rund-
let, and 31 i Gall. make a
Wine or Vinegar Barrel.
In the Exciſe the Wine Gall. is ſuppoſed to contain
231 Cubic Inches, conſequently the Pint muſt con-
tain 28 } Cubic Inches; but the Beer or Ale Gallon is
ſuppoſed to contain 282 Cubic Inches, and therefore
the Pint muſt contain 35 g ſuch Inches.
TABLE of Ale and Beer Meaſure.
Pints.
211 Quart.
A Firkin of
8
411 Gall.
Soap or Her-
rings is the ſame
32
i Firkin
with that ofAle.
128
6410 21 Kilderkje. 9 Gallons make
a Firkin of Beer,
25112832 4.
384 | 1924816
I
64
2 1 Barrel.
31Hhd.
EXA.
Addition of Mixt Applicate. Numbers. 15
Exa, in Ale Meaſure.
In Beer Meaſure.
(48) (+)(2)
(54)(4) (2)
bhd. gall. qr.pr.
hhd. gall.qc.pr.
3:47: 3: I
13:53: 3:1
I: 18: 2:0
8:42:1:0
O: 28: 1:0
3:16: 2: 1
O: 13: 3:1
I: 2.4: 2:1
O: 26: 2: I
2:15: 3:1
2: 15:1: I
I: 48:2:0
Sum 9:
9: 6:2:0
Sum 31: 39: 3:0
In Scotland Accounts of Liquid Meaſure are kept
in theſe Denominacions, hhds. gall. pints, murchkins,
gills, according to the following Table
4
Gills.
I Murch.
Note, 2 Murchkins make
1 Chopin, and 2 Pints i
4 [ Pine.
Quirt
The Engliſh Pint
33
81 Gill.
is fornewhat larger than the
2048 512li28 116) 1 Hhd. Scotch Murchikin.
1o)
1:2,1
Example.
(16) (8) (4) (+)
hhd.gall.pr.mut gills.
I: 15: 7:3:3
0: 8: 4: 1:0
3: 12: 4:2:2
4: 6:3:1
I: 0:5 : 2:2
4: 10: 3: I: 3
2:
Sum 14: 5:0:2:3
C2
6.0f
16 Addition of Mixt Applicate Numbers.
6. Of Engliſh Dry Cor Corn) Meaſure.
Dry Meaſure is different both from Wine and Ale
Meaſure; for the Gall. here contains only 268 Cu-
bic Inches, and the Buſhel 2150.42. The Denomina-
tions in this Meaſure are according to the following
Table.
Note, 5 Decks make a
Bulli, Water Meaſure.
Pints.
Iard ſays, that 10
Quarters make a Wey,
i Quart.
and 12 Weys a Laft, but
I Pottle.
all oilier Authors have
[ Gall.
4
it as in the annex'dTable.
8
i Peck.
2
2
1
2
Ö
197*2011
IO
2
ÓL
32
IÓ
8
}
1
SI 2
256
12,8
22
2
[ Bull.
250! 138
64 32 16 I Coomb.
64
1 Quarter.
2018 1024
2:6 128;2174Chald.
2500 1280
04.0320160
to lir
Tål Tun or W.
$120 2560 1280 1640132611012
i Laſt.
10
.
Accounts of this Meaſure are kept in Chald. grs.
buſh. pecks, gall. pts.
(4) (8) (4) (2)(8)
Ch.qrs.buſh.p.gal.pe.
13: 3:7; 3: 1:7
2: O: 4:2:0:4
I: 2:6: 2: 1:0
O: 2.5 : I: 0:3
5: 3: 1: 2: 1:6
O: 1: 2: I: 0:4
Sum 24: 2; 4: 2:0.: 0
IA
Addition of Mixt Applicaté Numbers. inay
In Scotland the Dinominations of Dry Mealurc are,
Chalders, Bolls, Firlocs, Pecks, Lipies, and in theſe
Accounts are kept.
The Engli Buſhel is
Lipies.
leſs than the Scotch Firlot,
I Peck.
for 5 Buthels' make Boll
4
Scotch Meaſure.
16
4 1 Firlot.
4
I Boll..
10211256157 coli Chald.
0.
:
Example
(16)/4),(4)(4)
Ch. bolls.for.peck.lip
2: 15: 3:3 : 3
I: 10:.2: 0 : 2
4: 4:1:1
3
2: 12; 2: 2:: i
3: Ş: 2 : 3:3
DI:
5: I: 2:0
Sum 26: 6:2: 2:0
2. Of Engliſh Meaſures of Length:
The Original of Meaſures of Lengths (as of Weights)
is from a Corn of Wheat taken out of the middle
of the Ear and well dried, 3 of theſe in Length were
to make i Inch, and thence all orber Meaſures are
computed, as in the following Table
C 3
Bar.
18 Addition of Mixt Applicate Numbers.
Bar. Corns.
31 Inch.
36
1211 Foot.
30 311 Yard.
594
198
5i Pole or Perch.
237601 7920
01 Furlong
190080.163260152801 770013201 81 1 Mile.
108
162
نے
7920,60
Accounts are kept in Miles, Cloth Meaſure con-
furlongs, yards, feet and Gifts of theſe Deno-
inches.
minations, yds, qrs.
Dails.
Exa.
Exa.
(8)(220)(3)(12)
(4) (4)
M. fur. yds. feet in.
yds. grs nai.
1:7: 219: 2: 11
43: 3:.3
0:4: 140: 1: 8
5
7:0:2
3: 3: 200: 2:
4: 1: 3
2; 4: 175: 1: 8
5: 2:
1: 6: 134:2: 7
33: 1:2
6:2:1
0: 2:
17:0:
1
2: 1
Sum 10: 6:
8:2: 4
Sum 81:0:0
The eaſieſt method of Note, That 1 yd.
adding the above yds, is or 5 quarters, or 45
to begin with the Units inches, make an Eil
Place, carrying at ro's, Engliſh, alſo 2 yds
and then adding the other make a Fathom.
2 Columns together, car-
rying at 22.
The
Addition of Mixt Applicate Numbers. 19
The Scotch Ell contains 37 Scotch Inches, and is di-
vided into 4 Quarters, and a Quarter into 4 Nails.
The Foor Meaſure contains 12 of thoſe Inches, where-
of the Ell contains 37. The Scotch Ell is į Part of an
Inch Engliſh
, more than 37 Inches Engliſh; ſo that the
Scotch Foot is to the Engliſh as 186 to 185.
The Scotch Mile contains 8 Fu:longs, á Furlong 40
Falls, and a Fall 18 Feet Scotch, according to the fol,
lowing Table.
Inches.
I2
I Foot.
37 37
I Elj.
184
61 1 Falt.
8880
740 · 40 401 Furlong.
71040! 5920 | 1yat 32:18 11 Mile.
222
Thus the Scotch Mile is 5920 Scotch Feet; ſo that
if the Difference betwixt the Engliſh and Scotch Feet is
not regarded, the Scotoh Mile will be to the Engliſh as
37 or 33, that is 37 Engliſh Miles are equal to 33
Scotch Miles,
Of Land Meaſure.
The Denominations uſed in meaſuring of Land, are,
Poles, Roods, Acres, Miles, and are always underſtood
to be ſquare.
40 ſquare Poles make 1 Rood, 4 Roods an Acre,
and 640 Acres a Mile. See the following Table.
19.
20 Addition of Mixt Applicate Numbers,
fg. Inches.
144] : [q. Foot.
1296
9) 1 ſq. Yd.
39204 272 3011 ſq. Pole or Perch.
1568160108901210401 [q. Furlong or Rood.
62726401435601484011601 4.
al
life. Mile.
144
TABLE for Scotch Superficial Meaſures.
ſquare Inches.
i fq. Foor.
1269 914 ſq. Ell.
49284 3424
3611 fq Fall:
1975260 I?' On | 141.0
4011 Rood.
7885.47154.60157601501-1 1 Acre.
A Scotch Acrc is to an Engliſh Acre as 55353.6 10
43;60.
8. Of Time.
The moſt common Diviſion of Time is into Years,
Months, Days, Hours, Minutes, Seconds, Thirds,
&c. but Seconds being the leaſt Part of Time which
can be meafured, with them I begin the following
Table.
Seconds.
601 Minute. .
3( O
6011 Hour.
864.00 1,149
2+1, 1 Day.
31556937'5-594,8705365 D. 5 Ho. 48 Min.
157 Sec.
I
1 Year.
1
But
Addition of Mixt Applicate Numbers. 21
But for eaſe in Calculation, we make the Year to
conſiſt of 365 Days, 6 Ho. and in Leap (or every 4th)
Year, of 366 Days. Therefore 60 Mio. make í Ho.
24 Ho. 1 Day, 28 Days 1 Mon. and 13 Months and
1 Day make one Year; this is called the Cominon Year,
and is divided into 1:2 unequal Monchs, viz.
January 3:1 Days
Fuly
3:1
February 28 and in Leap Year 29 Auguſt 31
March.
Septeinber. 30
April
30
Otober.
31
May
31
30
31
November. 30
December. 31
Fune
Some Things are accounted by the Dozen, the
higheft Denomination whereof is a Groſs. Thus, 12
Particulars make i doz, and 12 doz. i Groſs.
Example.
Groſs. doz. part.
4: II: II
1:
8:
7
0: 2: 4.
2:
5: 5
Note, 12 Bara
I: IO: 8 rels make it
3: 6:3
Laſt.
Sum 14: 9:
2
Of
II2
22
Of the Proof of Addition.
1. In order to prove the work of Addition, you
may diſtribute the Numbers added into ſeveral Par-
cels; and after having added theſe Parcels together, if
their Sum is the ſame with chat firſt found, the Work
has been right. To illuſtrate this I ſhall repeat-Exa.5.
2. Addition may be proved by
26370
caſting out the o's, thus: Take each
285 of the given Numbers ſeparately, and
26655 add their Figures together as ſimple
186
Units, and when the Sum is equal to
400
or greater than 9, but leſs than 18,
76
neglect the 9, adding wüat is over to
662 the next Figure, and ſo proceed 'cill
42
you have gone thro' them all, mark-
27
ing whar is over or under 9 at the
last Figure: Do the fame wiíh all the
181 Numbers given to be added, ſetting
down the Exceſſes of 9 together in
Total 27498 one Line, and ſum them up the ſame
way, marking the Excifs of 9 as be-
fore, or what the Sum is leſs than 9. Laſtly, do the
fame with the Toral Suin, and what is under 9, or o-
ver any Number of y's in this, muſt be equal to the
Exceſs or Number leſs chan 9 lait marked, elle the
Work is falſe.
Here I ſay 4+1=11, which is 20-
Ex&.
ver, then 2-+6+8=16, which is
4768 1.c 7 ver, and therefore I ſet ir down over a-
1378 i gainſt this firſt Line; and thus proceed-
793
ing, I find the Exceſſes over 9's in the
574 7 other Lines to be 1, I, 7; the Sum of
all which is 16, from which caſting a-
7513 7 way 9 there remains 7, which l ſer
down under; and caſting out the o's
the ſame way from the total Sum 7513, I find 7 over
70-
I
1
.
alſos
Of the Proof of Addition.
23
allo; wherefore the work is right. But this being te
dious, eſpecially where there are many Particulars,
3. The beſt Method, in my opinion, is to begin at
the Top and add downwards, whereas it was formerly
done by beginning below.
Queſtions to exerciſe the Learner in this Rule.
Qu. 1. I bought a Parcel of Goods, the firſt Coſt
whereof was L. 103: 15. I paid for Packing 8 h.6 d,
for Carriage L. 1:2: 4. and at Bargaining 4 16. 8 d.
What did the Goods ſtand me in, in all ?
L. fo.
. d.
103: 15: o firſt Coſt.
8: 6 Packing.
4 Carriage.
0:. 4, 8 Bargaining.
105: 10: 6Tot.Charge.
0:
2:
7
Qu. 2. I lent my Friend at one Time L. 48: 10.
at another L. 100. at another 18 M. 10 d. at another
L. 31: 14. at another L. 40:8 d. How much comes
it to in all ?
L. 48: 10:
IQO:
: 18:10
31: 14:
40:
: 8
L. 221:
3: 6 Antwering
Qu. z. A Father was 22 Years 6 Mon, old (reca
koning 28 Days to a Mon, and 13 Mon. to a Year)
when his eldeſt Son was born ; betwixt the eldeſt and
ſecond were 11 Mon. 12 Days; betwixt the ſecond
and
24 7
Queſtions in Addition.
and third were 2 Years 8 Mon. when the third is 14
Years 2 Mon. how old is the Father?
Y. M. D.
22: 6:00
O: II: 12
2: 8:00
14:
2: 00
40:
1: 12 Anſwer.
Qui 4: A has ſuch a Sum of Money, that if
L. 13: 15:8 be taken from it, what remains is e-
qual to L. GI; 10:7.
L. 13: 15: 8
61: 10:7
75: 6: 3 Anſwer.
2
CHAP. III. Of SUBTRACTION.
BY
Y this Rule we find the Difference betwixt any two
Numbers or Sums propoſed.
The Number from which the Subtraction is to be
made is called the Minuend, the Number to be ſub-
tracted is called the Subirabend, and what remains is
called the Remainder, Difference or Exceſs.
Exa
Subtraction of Mixt Applicate Numbers. 25
Exa.
Minuend 6538
Subtrahend 4125
From 3867
Take 1645
Rem. 2412 Difference 2222
When the Figure to be ſubtracted happens to be
greater than that which ſtands over it, add 10 to the
upper Figure, and then ſubtract or take the lower Fi-
gure from the Sum of both, paying 1 to the next low
Figure for the 10 thus borrowed.
Exe.
4632
What is the Difference betwixt 4632 3654
and 365+?
978 Ani
From 179864
Take 57698
420
I So
3000
2816
2713
1080
Remi 122166
240
: 184
1633
I proceed next to give Examples in Mixt Applicate
Numbers, where you'll mind to borrow (in caſe you
want) as you carried in Addition.
:
Exa, in Money
L.
d. f.
From L. 147: 13:6:3 From 1245: 8:4:1
Take 89: 10: 4:2
Take 486: 17:8: 3
Rem. 58: 3: 2: I
Rem. 758: 10: 7: 2
"
D
Troy
26 Subtraction of Mixt Applicate Numbers
Troy Weight.
lib. oz. d.wt.gr.
From 1+2: 7:
2: 15
Take 128: 4: 14: 20
Aver-du p. Weight Leffer.
It. lib. oz. dr.
From 18: 12: 12: 10
Take 6: 10: 13:14
Rem. 14:2:
7:19
Rem.' 12:
I: 14: 12
Apothecaries Weight. Engliſh Liquid Meaſure.
lib. oz. dr. ſcr.gr.
T. P. bhd.gal.pt.
From 36: 4:5: 1: 12
From 20: 1: 0: 20:6
Take 15:8:6:0: 14
Take 15:0: 1:40:4
Rem. 20: 7:7: 0: 18. Rem. 5:0:0: 43 : 2
Aver-du-poiſe Weight Gr. Scots Liquid Meaſure.
T. C. qrs. lib. bhd. gal.pt.mut.gil.
From 40: 15: 2: 20 From 150: 12: 2: 1:0
Take 18: 16: 1: 25 Take 86: 10: 6: 3: 2
Rem. 21 : 19:0:23
Rem. 64: 1: 3: 1: 2
Engliſh Dry Meaſure.
Ch.qrs.buſh.p.gal.pt.
From 4: 2:6: 2; 0:4
Take 1: 3: 4: 3:1:6
Scots Dry Meaſure.
Ch. B. fir. p. lip.
From 5: 8: 2:1: 2
Take 4: 12: 3: 2:3
Rem. 2: 33 1: 2: 0:6
Rem. 0:11: 2: 2: 3
Long
Subtraction of Mixt Applicate Numbers. 27
Long Meaſure.
Në, fur. yds, feet in.
From 12. 4: 124: 1:
Take 8:7: 215: 1: TI
Cloth Meaſure.
yds. qrs. n.
From 45: 2:1
Take 13: 3:3
IO
Rem. 3: 4: 128: 2: 11
Rem. 26: 2:2
1738
Queſtions to exerciſe the Learner in this Rile.
Qu. 1. A Man was born in the Year
of our LORD 1686, how old is he this
1686
preſent Year 1738?
Anfi. 52
Qu. 2. Suppoſe the fame Perſon ſhould 1760
live to the Year 1760, how old will he 1686
be then
74. Ans.
Qui 3. I lent my Friend L.375:14:6. whereof
he has paid me L. 198: 16:8. how much does he
yet owe mc
L. 375: 14: 6'Lent.
198: 16: 8 Received.
176: 17: 10 Anfr.
u. 4. I bought 50 Tuns 5 C. 1 r. 20 lib. 10 oz.
of Tron, whereof I have ſold 24 Tuns 8 C. 17 lib.
12 02. How much remains unſold?
T. C. gr. lib. OZ.
Bought 50:
5: 1:20: 10
8
: 0: 17 : 12
Sold 24:
Unſold 25: 17: 1:
D 2
2: 14
Gu.
28 Queſtions in Subtraétion.
Qu. 5. A Church is 215 Years ſtanding this pre-
ſent Year, what Year was it built in?
1738
215
1523 Anſwer.
04. 6. What Sum of Money added to L. 130: 17
will make it L.413: 12 i 7?
413; 12:7
130: 17:
282: 15: 7 Anſwer.
This Queſtion is the ſame, as if a Man was owing
Ł. 413: 12: 7 to be paid preſently, and had only
L. 130; 17, and it ſhould be demanded, how much
he must borrow preciſely to pay the ſame.
56:
Ox. 7. Lene my Friend at one Time L. 172: 13:
at another Time L. 20:12, at another Time L.
16, whereof he has paid me L. 299. What does he
yet owe me?
L. 172: 13:4
20:12:
56: 16:0
250:
209:
I: 4 Lent in all,
2
Paid me..
Anfi. 41:
1: 4 Sum yet due to me,
Queſtions in Subtraction:
29
Qu. 8. There are two Numbers, the greater of
which is 128, and their Difference is 57. What is the
leffer?
128
57
71 Anſwer.
Qu. 9. The Sum of the Ages of two Men is 1407
Years, and the Difference is 47 Years 9 Mon: What
is the Age of each? (accounting 12 Mon. to the Year.)
Y. M.
! 0 70 3
Half the Sum is 70: 3 23: ics
40 }Add
47: 9
Half the Difference 23 : 101
94: '1] Elder's Age.
70: 3
94: I
1
23: 10; } Sub
46: 43
46: 43 Younger's Age.
140: 6 Proof.
Qu. 10. A Father was 21 Years 10 Months and 6
Days old when his eldeſt Son was bord, and is now
68 Years 1 Month 20 Days; how old is the Son ? (ac-
counting 30 Days to, the Mon, and 12 Mon. to the
Year.)
68: I: 20
21: 10: 6
10: 3:14 Answer,
D.2
Proof
30
Proof of the Work of Subtraction.
This may be made two Ways, viz. 1. By adding
the Subtrahend to the Remainder, for the Sum ought
to be the ſame with the Millend. Or, 2. By ſub-
tracting the Remainder from the Minuend, for this
Difference ought to be the ſame with the Subtrahend.
BASADOREE
CHAP. IV. Of MULTIPLICATION.
MY Ultiplication is the taking or repeating any Num.
bet a certain number of Times.
TABLE.
I
2
6
3
2
8
4.
7
9
I2
6
2
8110
16
4
12 14
18
24
9 121.15 18 21 24
27
36
-
lal
lalalalu
36
48
8
4
8 12 16 20 24 28 32
5/10/15/20/25/30/35/40
6 12 18 24 30135 42 48
45
60
1
- 54
54
72
1
7114212835424956
63
84
1
1
|
8 16 24 32 40 485664
72
96
1
918 27136451 54 63 72
81
108
1
12/24136/48160172 84196
108/
144
This
Multiplication of Abſtract Numbers. 31
This Table the Learner ought to get by Heart per-
rectly before he proceed.
The Number to be multiplied is called the Multi-
plicand, that Number which multiplies is called the
Multiplier, and the Number reſulting from the multi-
plying the one by the other is called the Product. The
firſt two are otherwiſe called Factors.
I, To Multiply Abſtract Numbers.
RULE. Firſt ſet down the Multiplicand, and un-
der it orderly the Multiplier, and beginning with the
Figure which ſtands in the Units. Place of the Multi-
plier, by it multiply the Figure ſtanding above it in
the Multiplicand; and if the Product is leſs than 10,
ſet it down in its proper Place (and proceed to mul-
tiply the next Figure of the Multiplicand ;) but if the
Product is above 10 or ro's, ſet down only the Exceſs,
and carry the 10 or 10's to the Product of the next
Figure when multiplied, ſetting down the Exceſs as be-
fore: And thus go on to multiply all the Figures of the
Multiplicand by this firſt Figure of the Multiplier ;
then multiply the Multiplicand by the ſecond Figure
of the Multiplier (if this laſt con Gifts of more Places
than one) after the ſame manner, ſetting down the
Product as before, and proceed to multiply by each
Figure of the Multiplier till you have done; only mind
to ſet down the firſt Figure of the Product of each
particular Multiplication under the multiplying Figure;
then adding all the partial Products together, the
Sum is the Anſwer. The Rule will be betrer under-
ſtood by attending to the Operation of the following
Examples
Exm
32 Multiplication of Abſtract Numbers.
Exa. I.
Multiplicand 4687
Multiplier 4
Exa. 2.
17468
7
Product
18748
122276 Product.
Exa. 3.
Exa, 4
18694
I2
Multiply 3576
by
24
Product 42912 74776 Product by 4.
37388 Product by 2.
1
448656 Total Product:
Exa. 5.
7365
35
Exa. 6.
8065
68
36825 Product by 5.
22095 Product by 3.
64520
48390
257775 Total Product.
548420 Product
Exa.. 7.
25968
769
Exa. 8.
39745
4673
233712
155808
181776
119235
278215
238470
158980
19969392
185728385
1. When
Multiplication of Abſtract Numbers. -33
1. When the Multiplier or Multiplicand, or both,
end in a o or o's, write the ſignificant Figures under
one another, and multiply, neglecting the o’s ’till you :
have done; but then you muſt add as many o's co the
·Product when found as are contained in both Factors.
Exa. 2.
Exa. 3.
Exa. I.
Mult.
3260
by 24
325
240
38000
2400
1364
652
1300
650
152
76
78240
78000
91200000
2. When å o' or o's happen to ſtand between ſigni-
ficant Figures in the Multiplier, neglect them alto-
gecher.
Ixa. 1.
Exa. 2.
4376
608
13706
Gook
35008
20256
109648
82336
82345648
2660608
3: When the Mukiplier is I with any Number of
o's annexed, viz. iQ, 100, 1000, cc. you have only
to add to the Multiplicand as many o's as are contains
ed in the Multiplier, and the Work is done
Thus, 486X10 = 4860
486X100 =43600
486X 1000=486000 do.
4. It
34 Multiplication of Abſtract Numbers.
4. It is convenient to make the leaſt of the two
given Numbers Multiplier, tho' it may be propoſed as
Multiplicand, ſeeing the Product will be the ſame.
Exa. Ler ic. be propoſed to multiply 375 by 483.5.
- Operation, The Multiplier may ftand
4836 uppermoft, thus,
375
46
4837
24180
33852
29022
1450.8
19348
1813500
222502
5. When the Multiplier is a Number that has 9 ir
all its Places, as 9,99,999, co ſet as many o's on the
right of the Multiplicand as there are y's in the Mul-
tiplier, and ſubtract it from itſelf to increaſed. Thus.
if 346 was to be multiplied by 9, 99, 999 or 9999,
the Products would be (as by the following Opera-
tions)
3460
346
34600
346
346000
346
3460000
376
3114
34254
345654
3459654
6. When either Factor conſiſts of 2 Figures, which
are che Product of any 2 Digits, as' 2424 times 6,
multiply by one of theſe Digits, and the Product by
the other, and the laſt Product is the Anſwer: Thus
to multiply 8475 by 240 the Operation will ſtand as
below.
8475
Multiplication of Numbers.
35
8475
33900
6
203400 Product.
Exa. 2. Multiply 346 by 54, which is 6 times g.
, 9
6
2076
9
18684 Product.
And we may uſe the ſameMethod when the Multiplier
is'the Product of any 3 or more Digets : For Exa.
to multiply 346 by 168, which is 4 times 6 times 7.
Thus found,
41
+2
4
41168
346
1 2 3
1384
8304
7
:58128 Product.
You'll ſee the Uſe of ſuch Multiplication further ond
II. To
36 Multiplication of Mixt Applic. Numbers.
II. To Multiply Mixt Applicate Numbers.
When the Multiplicand is Mixt and Applicate, and
the Multiplier a ſingle Abſtract Number, or the Pro-
duct of two or more Figures, the Operation may be
performed without reducing the Multiplicand to the
loweſt Species mentioned, as you'll ſee by the follow-
ing Examples.
Multiply L. 14: 12: 8: 3
Here I carry 1 for every
by
4, for every 12, for eve.
ry 26, in the Place of far.
102: 9:1:1 d. and the reſpectively.
t
St. lib. oz. dr.
Exa. 2. Multiply 10: 4: 13: 5 by 8.
8
Here I carry i for
82:6; 11:0 every 16, 16, 16.
Vide Table 2. p. 12.
Exa. 3. Where the Multiplier is the Product of two
Digits;
Multiply L. 17: 10: 7:2 45 is 5 times 9.
by 45
5
:
87: 13: 1: 2
9
788: 18: 1:2 Anſwer,
Exa.
Múltiplication of Mixt Applic. Numbers. 37
St. lib. Oz dr.
Exa. 4. Multiply 25: II: 6: 10 2.8 is
4 times 7.
by 28.
4
102: 13:10: 8
7
719: 15: 9: 8 Anſwer.
Ch. b. fir. pilip.
Exa. 5. Multiply
5: 12: 2: 1 : 3
by 22.
4 224 is 4x7x8
23:
22+
2: I: 3:0
412
so
162; 1: 0: 1:0
8
1296: 8: 2:0:0 Anſwer.
By duly attending to the Method of operating theſe
5 Eximples, you may eaſily perform others of the fame
Nature. Wherefore I proceed to
20
R E DUCTION
From an higher to a lower Denomination (called
Reduction Deſierding) performed by Multiplication.
Money.
In L. 374: 14: 10: 1 how many Farthings?
Here I multiply the L, by 20, and
the Product is ll. this I multiply
by 12, and the Product is de which
laſt I multiply by 4, and the Pro-
89938 d.
duct is fer. But note, that I take
4 in the odd ſb. d. and far, in their
proper Places as I go along.
7494 JB.
I 2.
359753 far.
E
Troy
38
Reduction by Multiplication.
Troy Weight.
In 148:4: 10: 15 how many Grains ?
148:4: 10:15
Here I multiply
the lib. by 12 to
bring them to oz.
1780 oz.
theſe by 20 to re-
duce them to d. wt.
and theſe laſt by 24
35610 d. wt.
24
12
20
for gr.
14.2445
71221
854655 gr. Anſwer.
Aver-du-poiſe Weight Greater.
T. C. grs. lib.
Ir 120: 12: 2:17 how many lib.
Anfr. 270217 lib.
Aver-du.poiſe Weight Leller.
St. lib. oz. dr.
In 40: 11: 12: 7 how many Drams?
Anfr. 166855.
Engliſh Liquid Meaſure,
T. P. hhd. gal. pes.
In 17: 0: 1:44: 5 how many pinis ?
Anſi. 35133
Scots Liquid Meaſure.
hhd.ga'.pt.mut.gil.
In 2: 13: 5: 2: 1 how many Gills ?
Anfr.5849
Engliſh
Reduction by Multiplication.
39
Engliſh Dry Meafire.
Ch. qrs.buſh.p. gal. pt.
In 15: 2: 7: 2:1:4 how many Pines ?
Anfr. 32236.
Scots Dry Meaſure.
Ch. bol. fir. p. lip.
In 18: 15: 2: 3 : 3 how many Lipies?
Anſt. 19439.
I have left the Operation of the preceding Queſtions
to the Learner's Exerciſe.
Queſtions in Multiplication.
Qu. 1. If any one Thing coſt 4 m. what is the
Value of 42 ſuch 'Things at the fame Rate?
4.2
܇
163. Anfr, or L. 8: 3
*. 2. If I ſpend 71). a Day, how much is it a
Year?
305
7
L. 1,
2555 fin. Anfi. 127; 15
Qu. 3IF 63 gall
. make a hhd. how many gall.
are there in 40 hhds.
40
2520 Anfr.
E 2
49 Queſtions in Multiplication.
Qu. 4. IF I C. of any thing coſt L. 1: 16:4 what
is the value of 8 C. at: che fame Rate?
: :
14: 10:8 Anfr.
Qu. 5. If
any one Thing coſt L. 3: 15: 10 how
much will 28 ſuch Things coſt at the ſame Rate?
L. 3: 15: 10
4
Here I multiply by
4 and 7, inſtead of 28.
15: 3: 4
7
L.106: 3: 4 Anfr.
Qu. 6. A Father gave his Daughter for her Dow-
ry 24 Boxes, and in each Box were 16 lefſer Boxes, in
each leſſer Box were 12 Purſes, and in cach Purſe were
16 leſler Purſes, in each of which laſt were 4 d. Scots,
What was her Portion?
24
16
384 Number of lefſer Boxes,
12
4608 Number of bigger Purſes.
16
2764.8
4008
73728 Number of leffer Purſes.
4
24912 d. Anfr. or L. 1228: 16
found by dividing by 12 and 20.
Qu
Queſtions in Multiplication. 4
41
Qu.. I have 20 Casks of Raiſins, the net Weight
of each is 7 Stone, lib. 10 OZ. what is the net
Weight of the whole?
St. 7:9:10
4
Here I multiply by
4 and 5 inſtead of 20.
5
30: 6: 8
152: o: 8 Anfr.
Qu. 8. How many N. d. and Halfpence are con-
tained in L. 10000 Sterl.?
10000 L,
20
200000 l.
12
2400030 d.
2
4800003 Halfpence.
:
24 Jo What Number divided by & gives 21 for
the Quote?
.8
168 Arifi
QH. 10. . What Sum of Money muſt be divided e-
qually amongſt 18 Men, ſo that cach may receive
1. 14: 6:181
E
by
42
Queſtions in Multiplication.
L. 14:6:81
18 is 3x6.
3
43:0:1
6
258: 0:9 Axfwer.
Qu. 11. In 50 Groſs 8 doz. of Pairs of Stockings
bow many ſinglé Pairs ?
Gr. doz.
50:8
12
608
12
7296 Pairs, Anſwer.
Qu. 12. What is the Total Weight of 26 Packs of
Wool, each weighing 17 Stone 12 lib. 6 oz. net ?
, .
17: 12:6
4
71: I: 8
6
426: 9:0 for 24 times.
35:
8: 12 for 2 times.
462: 1:12 Anſwer.
Qu. 13. What Number divided by 3, 5, 7, 17,
ill have no Remainder?
Anfr. 1260.
Proof of Multiplication.
Multiplication is beſt proved by Diviſion; thus, in
yide the Product by the Multiplier, the Quote is the
Mui.
Proof of Multiplication.
43
Multiplicand; or, divide the Product by the Multipli-
cand, and the Quote is the Multiplier.
I ſhall finish this Rule, when I have obſerved, that
toth Factors cannot be Applicate Numbers, but that
the one of them muft neceffarily be Abſtract. Where-
fore it is abſurd to propoſe (for Exa.) L. 4: 6:8 to
be multiplied by L. 3: 2: 6. &c. If the Propoſers
of ſuch Queſtions would be ſo good as to tell us, how
oft they would have ſuch a Sum taken or repeated
(which muſt be the meaning of Multiplication, elſe it
has no meaning) I ſhould do my beſt to give them a
fatisfactory Anſwer; but 'till they explain themſelves,
I think they deſerve none. My deſigned Brevity does
not allow me to profecute this Subject, and therefore
I refer the Reader to Malcolm's Arith. pag. 85. Edit,
Lond. 1730.
ano
20200026
CHAP. V. Of Division.
DI
Iviſion finds how oft one Number is contained in
another, or what Part it is of the ſame.
The Number to be divided is called the Dividend,
the Number which divides is called the Divifor, and
the Number reſulting by dividing the one by the o-
ther, is called the Quotient or Quote.
RULE.
Set down the Diviſor and
Dividend as in the annexed Diviſor Dividend Quote
Exa. and if the Divifor is 4) 7376 (
contained in an equal Num-
ber of Places of the Dividend, fet a Point under or 1-
bove that Figure of the Dividend; but if the Diviſor
is not contained in an equal Number of Places of the
Dividend, ſet the Point under or above that Figure
which ſtands one place further to the righe; then
conſider how often the Diviſor is contained in that
Num-
44
Diviſion
Number of Places of the Dividend, and let down the
Anſwer on the right of) within the crooked Line,
and multiplying the Divifor by the faid Figure; place
the Product orderly under the Dividual, and ſub-
tract it, writing the Remainder below: Then take
down the next Figure of the Dividend, and annexing it
to the right of the Remainder, conſider how off the Bi-
viſor is contained in that Sum, and fet the Anſwer in the
Quote, on the right of the Figure which ſtands there
already, multiplying and ſubtracting as before; and to
the Remainder take down the next Figure; and thus
proceed till all the Figures are taken down, and the
Quote completed.
Note 1. "The Figure ſet in the Quote muſt expreſs
the greateſt Number of Times the Diviſor is con-
tained in that part of the Dividend, and conſequent-
ly the Remainder muſt always be leſs than the Divi-
for.
2. The Product of the Quote Figure multiplied
into the Diviſor muſt never exceed the Dividual.
Exa. I.
Ex. 2.
4) 7376 (1844 6) 87493.(14582
4.
6.
1
33
32
27
24
1
17
16
34
30
42.
16
16
48
mukaan
1-3
12
Rem. I
Diviſion.
45.
Exam. 3:
Exa.- 4.
201634 (25204
16:
5204
57) 38426:(674
342
41:
40
422
399
IQ
16
236
228.
Rem. 8
3+
32
Rem.2
1
Exa: 5.
457) 1246874 (2728
914":
Exa. 6.
1579) 1087647 (688
947-4
3328
14024
12672
3-199
12.97
914
13927
12032
Rem. 1295
3834
3556.
Rem. 178
E**.
46
Diviſion.
Exa. 7.
Exa. 8.
63087)41090746846513 7469)170196846(22787
378522.4
14938
323854
315435
20816
14938
84196
63087
58788
52283
211098
189261
6505+
59752
Rem. 21837
53026
$2283
Rem. 743
More Examples for Praltice.
817467.9) 417690850638 (51095 5029133
674100746) 19203764274674 (28487.6503 2 3 3 72
341968257644) 46287346747467 (135 -21031901522
96100700673846)4173867474628468(43415373 45653 ago
Contractions in Divifion.
1. When the Diviſor is a ſingle Figure, or even 11
or 12, the Diviſion may be perform'd by ſubtracting
mentally, and the Quote found by one Line,
Exa. I.
Exa. 2,
4) 7384
1846 Quore.
6) 327968
54661 ; Quote.
ESA,
Contractions in Diviſiorr.
47
Exa. 3.
Exa. 4.
8) 279685
34960; Quote.
9) 1326754
147417 5 Quote
6168357
24}
11) 428764
12) 391467
389787i Quote. 32622 Quote.
2. When the Diviſor is the Product of one or
more ſingle Figures (or Digits) divide by any one
of theſe, and the Quote by any other, and ſo on, and
the laſt Quote is the Anſwer: Thus, if I was to di-
vide 68357 by 24, I would work as in the Margin.
When any thing remains after
6 17089 + the firſt Diviſion, it muſt be ſet
284.8 down with a Line drawn through
Or thus
it, as you ſee in the annexed Exa.
and if any thing remains over the
4111392 § next Divilion, it muſt not be ſet
28487. down, but you are to multiply it
(mentally) into the firſt Diviſor,
addiug the firſt Remainder to the Product for the true
(total) Remainder; ſo I remaining after I divided by
6, I lay once 4 is 4, which with the I remaining for-
merly makes s for the true Remainder, as you may
prove by dividing the long Way.
Exa. 2.
Let it be required to find the 63d Part of
39485.
63 is =7X9
5640
626 43 Quote.
Exa. 3. Required the 27th Part of 16346
27=3x9.
5448 %
605 14 Quote.
Exa. 4. Required the 64th Part of 16736.
781 16736
s &
2092 ♡
261 or 1 Quote
63}}/39485
27}}|
}}/16346
64
48
Contractions in Diviſion.
.
Here the firſt Remainder being o, I don't multiply
the 2d Remainder 4 into the firſt Diviſor, becauſe
there is nothing to add to it, and therefore the Re-
mainder is the ſame as if ic had been only a fimple Di-
viſion by 8. However, the Remainder would have
been equivalent, if I had multiplied it into the firſt
Diviſor, according to the Kule, only it would have
come come out in a higher Term, for 24 is when re-
duced or
If you want to know the Reaſon of thus managing
the Remainders, let us reſume the 3d Exa. where, after
dividing by 3, there remains 2, which is }; then there
is 54481 to be divided by 9, and
|
afrer dividing the whole Part there
27
5448 remains 32 which reduced to a
60521 Fraction, having for its Denomi-
nator 3, (vis. the Denominator of
the Fraction to be divided) inakes ; to which adding
the j remaining over the firſt Divilion, the Sun is Š
this divided by 9 quotes as is plain from 3d Café,
Diviſ. Com. Fract. But this laſt is not ſo eaſy as the
firſt Method.
Here the firſt Remainder is 1,
which I let down; the 2d Re-
3163805 .mainder is 3, which I multiply in-
75
to the firſt Diviſor 32. and adding
4253 10 the firſt Remainder, the Sum is
850
10, which I mark; the 3d Ren
mainder is alſo 3, which I multi-
ply into the Product of the 2 firſt Diviſors, viz. 3
and 5, and it makes 45; to which adding the Remain-
der 10, the Sum is 55 for the total Remainder.
Exa 6 Required to divide 3867395 by 756 after
this contracted Method.
E%0. 5.
1
55
750
Contractions in Diviſion. 49
756
Here in order to find the
31 3867395 Diviſors, I divide 756 by 3,
612891312 the Quote by 6, this by 6,
214855 5 and chis by 7; and fo of o-
35809 23 thers: Thus,
5115457
3 756
252
6
42
7 7
I
So I find that 3x6x6x7=756.
3.
Wlien the Diviſor ends in o or o's, cut off
theſe o's and as many Places of the Dividend, and ſo
the Operation will be ſhorter.
Exa. 2.
4100) 864120 741100) 4630145 (62 Quote,
216 Quote
444
· Exa. I.
190
18
Rom. 42846
Exa. 3.
E:ca. 4
2413 } 3940115
45100}}|
}; 1179860120
9850
| ୨
359720
1641 78 Quote.
3996. Quote.
4.
When both Diviſor and Dividend can be re-
duced lower, reduce, and the Work will be eaſier :
Thus, 468) 36810 (
4 468) + 36849
Here firſt I divide both by 42
117) 9210
and theſe Quotes by 3, and
39). 3070 (7.8 the Diviſor becomes 39, and
the Dividend 3070.
273
3
340
312
Rem.
28
F
Some
50 Contractions in Diviſion.
Some:imes by this Method the Diviſor becomes 1 :
Thus, 10 divide 11232 by 432.
+32) 61.1232 So the Quote is 26 for 1.) 26 (25
.952) 91872 times.
88) 8 209
1)
26
6
5. Diviſion may be contracted, by omitting to write
down the ſeveral Products of the Quote Figures into
the Diviſor, but fubtracting mentally, and ſetting down
only the Remainders, as in Exa. 4, and 5, repeated.
Exa. 4
Exa. 5.
-57) 38426 (674
422'.
457) 1246874 (2728
3328.
236
Rem. 8
1297
3834
Rem.
178
4238
Some daſh a Line through each Figure of the Divi
dual as they ſubtract ; thus, in Exa. 4 repeated.
57) 38426 (6 57) 38426 (67 53842.5 (674
42
423.
Ź
Exa. 5. repeated.
457) +246874 (2728
Note, the Figures 178 want-
ing the Dith, are the Remain-
der.
+297
3834
178
Divifox in Mixt Applicate Numbers.
CASEI. When the Dividend is a simple or Mixt
Applicate Number, and the Diviſor Abstract: See
the following Examples.
3328
L.
Diviſion in Mixt Appl Nicumbers. 57
S.
L.
3+) 2629: 1775 L.
238
L. d. f
76) 2747 : 15: 19:36;6 L.
228
· 407
249
238
456
II
II Remi.
20
20
) 220 (6 6.
204
70) 236 (37.
228
16 Remi
8
12
12
) 192 (5.di
170
70) 105 (11.
76
22 Remi
4
301
4
).88(2 f.
764123 (1 g
76
1
20-
47
So that the 34th Part of L. 2629 is 773 6:52
279. And the 76th Part of L. 2747 16: 10: 3 is
L. 36: 3:1:,14%.
F 2
In
52 Diviſion in Mixt Applicate Numbers,
.
I
In Troy Weight.
lib. oz. d.wtogr.
37) 253: 8: 10:14 (14. lib,
17
17) 30 (1 d.wt.
17
68
20
.83
13
35
2+
12
17) 188 (11 oz.
17
56
27
17) 326 (19 gr.
17
18
17
156
153
3
But when the Diviſor is a ſingle Figure, or the Prom
duct of two or more Figures, work as is directed,
[ 47,
inus, What is the 6th Part of L. 1076: 18: 4?
6) 1076: 18:4
179: 9:8: 2 Anſwer.
Exa. 2. What is the 8th Part of L. 3419 ?
8) 3419
427: 7:6 Anfuer.
Where the Remainders are reduced, and the Sub-
traction made mentally. See alſo the following Ex-
amples.
L. d. f.
L. S. d. f.
24 1755: 10: 11: I
197: 18:6:3
so
35
438: 17:
39:11:8:2 +
73:
2.: II: le
5:13: 1:035
S.
2+} 4
8:35
Avere
Diviſion in Mixt Applicate Numbers: 53
Aver-du-p. Weight . Leffer. Scots Liquid Meaſure.
St. lib.oz. dr. hhds.gal: pts.mur.gill.
}}|
25: 8:5: 7
4. 12:2:2: I
8:8:1:13
:,
42} |
0:12: 5:3:0 H
1: 3:7: I ģ
1: 6:2:0
And fo of others.
21
CASE II. When Diviſor and Dividend are both
ap:
plicate to the fame kind of Things.
RULE. Reduce both to ſimple Numbers of the
loweſt Denomination mentioned in either (if neceſſary)
and then divide: The Quote is abſtract, and ſhews
how oft the Diviſor is contained in the Dividend, or
what Part it is of the ſame.
Exa. I.
Ex. 2.
L. L
f..
di do
4) 28 (7. 4) 28 (7 4) 28 (7
28
28
28
That is, 4 L. is the 7th Part of L 28, 4/b. the 7th
Part of 28 h. and 4 d. the ſame Part of 28 d.
L. d.
38: 12) 169:16: 8:
Exa. 3.
L.
S.
S.
20
20
772
12
3396
12
So that L. 38: 12 is the 41h
Part of L. 169: 16: S, and
ſomewhat more.
9264
14076014
37056
3704
F3
W bat
54 Diviſion in Mixt Applicate Numbers.
ft. OZ.
lib. oz. dr.
What Part of 1400: 13: 10 is 12: 8?
ft. lib. oz.
lib. oz. dr.
12: O: 8) 1400: 13: IO
16
16
dr. dr.
49280):358618 (9
22413 oz.
344950
16
13658
1160
35.8618 dri
193
192 lib.
16
3080 oz:
16
Anſwer, the 7th Párt, and ſomewhat
more. And ſo of others.
49.280 dr.
RÉDUCTION
12
9211 d. if
From a lower to an higher Species (commonly called
Reduction Aſcending) perform'd by Diviſion
Exa. in Money.
In 36845 far. how many L.lb. and d?
41 36845 f.
Here I divide the far.
by 4, and the Quote is
250) 7617 5. 7 d: if.. di the d I divide by. 12,
38 L. 76.7d. If and the Quote is . this
haft I divide by 20, and the Quote is L.
Troy Weight.
Apothecary Weight
In 13865 gr. how many lib. In 13867 gr: how many lib.
210) 138015
3:466 +
3) 693: 5
210) 5767:17
8). 231:0:5
12) 28:17:17
12) 28: 7:0:5
2: 4:17:17 Anfr. 2:4: 7:0:5
243.4.1 1386
Aver
Reduction by Diviſion. 555
ny. Tuns?
ny Tuns?
28141 635163
28
Aver du-poiſe Weight.
Wine Meaſure.
in 635163 lib.' how mía- In 76385 Pints; how.ma-
8) 76385
7 158790 3
271 9518:1
4) 22684:11 91 1364 Ø
0
210) 56711:0:11 2) 151:35:1
283: 11:0:11 2) 75: 1:35:11
37: I: 1:35:1
And fo of others.
63}}
Queſtions to exerciſe. Diviſion.
Qu. 1. What is the Value of one Yard of Cloth
when 26 yds. of the ſame coſt L:2: 18:6?
L. d.
26) 2: 18:6 (OL.
S.
20
) 58 (232
52
d.
Anfr. 2: 3
is alt
2
978 (d.
78
qu. 2. There is L. 1372: 1416 to be divided
equally amongſt ſix Men, what is each's Share ?
L. B. d.
6) 1372: 14:6
228: 15:9 Anſwer,
இழ
56
Queſtions in Diviſion
Qu. 3. If I ſpend 186 L. a Year, how much is it-
a Week
?
52) i 86(3 La
-156
30
20
52) 600 (ins.
52
1
80
52
28
12
52) 336(6 d.
312
L. S. d. f,
Anfr.: 3:11: 6:11
24
4:
52)96(1.f.
52
44
Qu. 4. How many poor People may L. 20° be di.
vided amongſt, ſo as to give Half a Crown to each?
s. d. L.
2:6) 20
.
I 2
20.
30
400
12
310) 48010
160 Anſwer.
9.
Queſtions in Diviſion. 57
Du 5. How many Tuns Burthen is that Ship of,
which can carry L. 11000 Sterling, when converted
into Scots Halfpence, each of which being of an
Ounce Aver-du-poiſe?
I1000
240
d. in ii
2640000
2
4) 5280000 Halfpence or 4ths of an Ounce,
16}4 / 1320000 OZ.
4
330000 $
28 84 | 82500 lib.
71 20625 %
4) 2946: 12. lib.
216) 7316 C. 2 grs. 12 lib.
36 T. OC. 2 qrs. 12 lib. Anſwer:
Qu. 1:
Follow Sorre practical Queſtions for further Ex-
erciſe in the foregoing Rules; to the Solution
of which there is nothing further neceſſary than
what has been already delivered.
I went to a Market with L. 150 Sterlings
and a Horſe which coft me 5 Guineas and a Half; I
fold the Horſe for n Guineas 10 hi I bought Linen
Cloth to the value of L. 15: 12, whereof the Seller
diſcounted me a Crown. I bought a Horſe for L. 4:
15: 6; and ſpent of Charges before I returned 12 ſh.
4 d. As I was coming home, I had the Misfortune to
drop a Purſe of 50 Guineas. How much Money re-
mained?
Anfr. L. 84: 12: 2
Qu. 2. I owe L. 137: 10: 8 to be paid preſently,
whereof I have lying by me only L. 86: 6 d.. how:
much muſt I borrow preciſely to pay this Debt ?
Anfri L. 51: 10: 2
qu
58
Mixt Practical Queſtions
Qu. 3. I lent my Friend at one time L. 50 Ster--
ling, at another 4 Guineas, at another 35 Merks and
a half Scots; whereof he paid me at one time 2 Twen-
ty Shillings Notes, at another he gave me 16 Yards
of fine Linen ar 2 ſb. per yd. and at another Time in
Calh L. 41 Scots. I drew a Bill on him for L. 39:
17: 4 Scots payable to C. D. How much does he yer
owe me?
Anfr.
Qu. 4. A Butcher ſent his Servant to a Fair with
300: 12 Scoisy to buy Sheep at L. 2: 5 cach, Cows
at L. 25 each, and Oxen at L. 30 each, of each a like
Number, allowing him Half a Crown for Charges :
How many of each did the foreſaid Sum purchaſe,
and how much Money was returned to the Maſter?
Anfr. 5 of each, and L. I: 7 d. Sterling returned,
Qu. 5. Suppofe a Privateer makes a Prize to che
Value of L. 3578: 14: 6, how much is due to each
of the Crew (being 47 in all) allowing the Captain
and the other Officers (who are 4 in Number) each
Ti Part of the Remainder, and what remains over and
above being to be equally diftributed amongſt the pri-
vate Men
Captain's Share, L. 223: 13:: 4:35
Anfr. Each Officer's Share, 104: 16: 10:34
Each private Man's Share, 69: 17: 11: 133
24. 6. Suppoſe I have ſpent of Cafhi theſe 8 Years
laft paft L. 8381 1), and have run in debt to the Sum
of. L. 75: 1:9 over and above, how much comes it to
per Year, per: Month, per Week and per Day?
L. 114: 6:3. per Ann.
Anfr.
8: 15:44per Mon.
3: 1013 per Week.
6: 343 per Day.
qu. 7. Suppoſe l'ſpent laſt Year L. 114: 6: 3,
bow much will ſerve me 20 Years at the ſame Rate?
Anfr. L. 2343; -8: 13
16)
2:
in Addition, Subtraction, &c.
59
Q:. 8. Suppoſe I have iš of a Ship, and her Freight
for 3 Voyages is L. 345 Sterling, how much of the
ſame falls to me, there being L.13: 11 to be paid by
the Partners for Charges on the ſaid Voyages?
Anfr. L. 20: 14: 33
Qu. 9.
I went to a Market with L. 60: 52 where
I bought 400 yds. of Linen at 10 d. per yd. alſo
250 yds. of Plaiden at 5 d. per yd. allo 4 i Groſs
coarſe Stockings at 13.1. 4 d. per Doz. I ſpent of
Charges 7 p. 6 d. When I returned home I had pre-
ciſely L. I: 2: 4. whether oughe I to have had more
or leſs?
Anfr. I ſhould have had a Shilling more.
Qu. 10. I bought 3 Laſt of Salmons at L. 2: 5
per Bar. which I ſent abroad. I ſpent of Charges ac
Shipping for Freight, &c. L. 2: 10. My Factor advi-
ſed me that he had ſold the ſame for ready Money, the
neat Proceeds whereof (all Charges deducted) amount-
ed to L. 103: 10, for which he ſent me in return
4:14. lib. of Tea. How much was gained by the Sal-
mon, and what did the Tea coſt me per lib. as alſo
how may I fell the ſame to gain L. 32: 1:6 on the
Salmons ?
L. 20 Gain on the Salmons.
Anfr.5 lb. primc Coſt of the Tea per lih.
5 ]]. 7 d. Price of the Tea at the propoſed Gain,
Qu. 11. Of what Quantity is 39 Bolls 12 Pecks
the 17th Part?
Auf. 40: 1:12
Qu. 12. A Merchant has 4 hḥds. of Wine, con-
taiding each (ſuppoſe) 130 Pines Scots, which he de-
ſigns to draw off in Chopin Bottles; how many Groſs
will he need for this purpoſe?
Gr. doz. bottles.
Anfr. 7:
8
21.13. I lent my Friend L.56: 18 Sterl. in return of
which he gave me 45 Piſtoles, valued each at 17. Jh.
Ch. B. pecks.
2:
6
రం Mixt Practical Queſtions
:
6 d. and 15 Carolus's of Gold at 23 Jh. each ; was I
fully paid? Anfr. He ſtill owed me 5. fb. 6 d.
Qu. 14. How many ſquare Stones of 10 Inches,
and ii Inch thick, will pave a Floor which is 5
yds. long, and half as broad?
Anfr. 162.
Qu. 15. I have to fell 4 doz. fine. Stockings, a
Merchant has offer'd me 5ilb. 6 d. per Pair, and to
pay for the marking of chem: Another tells me he'll
give me L.3: 6 per doz. without being at the Charge
of the Stamp; and a. Third offers me 15 Ştone of
Wool worch 15 d. per lib. I being at the Charge of
the Stamping. Which of all theſe is the beſt Bargaini,
and by how much does it exceed the other two?
Anfr. The laſt is the beſt Bargain, viz. better than
the firſt by. Is. 6 d. and better than the other by 3.fb.
(I paying the Stamp.)
Nóte. The Stamp is 2 d. Scots, or ; d. Sterling pemu
Pair.
Qu. 16. How much Sterling Money is contained
in L. 1874: 10: 6 Scots?
Anfr. L. 156: 4:21, found by dividing by 12.
Qu. 17. In L. 156: 4:: 2. Sterling, how much
Scots Money?.
Anfr. 1874: 10: 6, found by multiplying by 12.
Qu. 18. In 4176 L, how many Merks ?
Anfr. 6264 Merks, found by multiplying by 3,
and dividing the Product by 2; or by adding of
4176 to itſelf.
Qu. 19. In 3126 Merks how many L.?
Anfr. L.2084, found by multiplying by 2, and divi-
ding the Product by 3; or by ſubtracting í of 3126 :
from itſelf.
Qu. 20. In 20600 Merks Scots how much Sterling
Money?
Anfr. L: 1144: 8: 10 Ž, found by dividing by 18.
2
214.
Mixt Practical Queſtions, &c. 6
1. 21. In L. 145: 14: 10 Sterl. how many Scots
Merks?
Anfr. 2623 Merks, and 4 1 d. Sterl. over and above.
24. 22. In 7000 Guilders at 23 per Guilder, how
much Sterling and how much Scots Money?
A11f7. L.685:8: 4 Sterling, and
L. 8225:0:0 Scors.
Qu. 23. In 436 Guineas at 21 m. how many L?
Anfr. L. 457: 16, found by multiplying by 21, and
dividing the Product by 20; or by adding ză of 436
to itſelf.
CHAP. VI. Of Arithmeticel Progreffion.
WHEN a Rank of Numbers do either increaſe or
decreaſe by a common Difference, thoſe Num-
bers are ſaid to be in Arithmetical Progreſſion: Thus,
s 1. 2. 3. 4. 5. 6. 7. c. S Here the com. Diffe-
27. 6. 5. 4. 3. 2. I.
{
rence is I.
$1. 3. 5. 7. 9. II. 13. &c. Here the com. Diffe-
{
13. II. 9. 7. 5. 3. I.
rence is 2.
And ſo of any other Series whoſe Difference is 3, 4,
5. &c.
In every Arith. Progreſſion there are 5 Things con-
ſiderable, viz.
The two Extremes, i.e. the firſt and laſt Term of
the Series.
The Common Difference.
The Number of Terms. And
The Sum of all the Series.
And the moſt uſeful Caſes are theſe, wherein are 3
given, to find the ocher 2.
G
CASE
62 Arithmetical Progrefron.
CASE :I. Having the Extremes and Number of
Terms, to find the common Difference aud Sum of all
the Series.
RULE. Divide the Difference of the Extremes by
the Number of Terms leſs. I, and the Quote is the
Difference: And multiply the Sum of the Extremes by
the Number of Terms, and half chis Product is the
Sum of the Series.
Exa. A Gentleman laid by for a younger Son's Por.
tion on the firſt Day of January L. 12, and continued
to lay by a certain Sum in Arich. Progreſſion on the
firſt Day of each Month to the End of the Year, the
Jaſt Sum being L. 45. How much did each exceed the
former, and what did his Portion amount to in all?
45-12=33 Difference of Extremes.
12-1=Number of Terms leis 1.
Then II) 33 (3 common Difference or Excels.
And 12+45357 Sum of the Extremes.
12 Number of Terms.
2) 684 (342 his Portion, or Sum of the
Terms.
Exa: 2. Suppoſe 100 Eggs were placed in a right
Line, a Yard diftant from each other, and the firſt Egg
a Yard diſtant from a Basket; how many Yards will
one walk (beginning at the Basket) before he can ga-
ther up thoſe 100 Eggs fingly one after another, retur-
ning with each to the Basket, and putting them in
without breaking ?
2+200=202 Sum of the Extremes.
100 Number of Terms.
2;
-200
1. 10 Number of Yards Anſwer.
And 1760) 10100 (5 Miles and 3 Quarters ferè,
CASE
Arithmetical Progreſſion. 63
CASE 2. Having the Extremes and common Dif-
forence, to find the Number of Terms and Sum of the
Series.
RULE. Divide the Difference of che Estreines
by tlic common Difference, and to the Quote adding
I, the Sum is the Number of Terms: then find the
Sum of the Series by Caſe it.
Exa. A Gentleman laid by for a younger Son's
Portion on the firſt Day of January L. 12, and con-
tinued to lay by a Sum on the firft Day of each Monch
for a certain Number, increaſing each by L. 3, the .
laſt wherof was 45 L. How many Months did he
continue ſo to do, and what did his Porrion amount
to in all?
45-12=33 Difference of Extremes.
3)33(11, and uti=12 Number of Terms.
Thén 127:15357 Sum of the Extremes.
12
2) 6-3,4 (342 L. Sum of the Series,
.
CASE 3. Having one of the Extremes, the Num.
ber of Terms and the common Difference, to find the
other Extreme and the Sum of all the Terins.
RULE. If it is the leſſer Extreme that's given,
multiply the common Difference by the Number of
Terms leſs 1, and to the Product add the given Ex-
treme; but if it is the greater Extreme that's given,
ſubtract it, and the Sum or Remainder is the other
Extreme, and fiud the Sum of all the Terms by Cafe
I or 2.
Exa.' A Gentleman laid by for a younger Son's
Portion on the firſt of January L. 12, on the firſt of
February L. 15, and ſo on to the End of the Year,
increaſing each by L. 3. How much did he lay by on
the first of December, and what was his Portion?
G2
3
64
Arithmetical Progreſſion.
3 common Difference.
31 Number of Terms leſs 1.
33+12=45 laid by on the firſt of December,
Then 12+ 45357 Sum of the Extremes.
12 Number of Terms.
2) 684 (342 Sum of all the Terms.
CASE 4. Having the Extremes and Sum of all
the Series, to find the Number of Terms and com
mon Difference.
RULE. Divide double the Sum by the Sum of
the Excremes, the Quote is the Number of Terms:
then find the common Difference by one of the pre-
ceding Caſes
Exa. A Gentleman laid by for a younger Son's
Portion on the firſt of January L. 12, and continued
ſo to do on each firſt Day of the ſucceeding Months
for a certain Number, increaſing each by Aritämeri-
cal Progreſſion, ſo that the laſt Sum laid by was L. 45.
How many Months did he continue to lay by, and by
how much did each Sum exceed another, the Sum of
all he laid by being L. 342 ?
342 Sum of all the Series.
12+45557 Sum of Extr.
57)684 (12 Number of Terms.
Then by Caſe iſt to find the common Difference.
45–12=33 Diff. of Extremes.
12-1=I1 Num-
ber of Terms leſs i. Therefore 11) 33 (3 common
Difference,
2
CASE 5. Having one Extreme, Sum of the Se-
ries and Number of Terms, to find the other Ex-
treme and the common Difference.
RULE. Divide double the Sum of the Series by
the Number of Terms, and from the Quote ſubtract
the
Arithmetical Progreſión: 65
Exa.
the given" Extreme, the Remainder is the Extreme
fought; then find the Difference by one of the pre-
ceding Caſes.
A Gentleman laid by for a younger Son's
Portion on the firſt Day of January L. 12, and on the
firſt Day of the ſucceeding Months of the Year a cer-
tain Sum, each increaſing in Arithmetical Progreſſion,
ſo that he laid by in all L. 342. What was the Sum he
laid by on the firſt of December, and by how much did
each Sum exceed another?
342.
2
12) 684 (57 Sum of the Extremes.
12
45 laid by on the firſt of December..
Then by laſt Cafe to find the common Difference.
45-12=33; and 12-=II. Then 11)33(3 com. Diff.
CASE 6. Having the Sum of the Series, com-
mon Difference and Number of Terms, to find the
Extremes.
RULE. By the Sum and Number of Terms find
che Sum of the Extremes, as in laſt Examp. then, by
means of the cominon Difference and Number of
Terms, find the Difference of the Extremes chus,
Multiply the common Difference by the Number of
Terms leſs 1, and the Product is the Difference of
the Extremes. Laſtly, having the Sum and Diffe-
rence of the Extremes, find the Extremes thus, half
their Sum + half their Difference is the greater, and
half their Difference ſubtracted from half their Sum
is the leſſer.
Exai A Gentleman laid by for a younger Son's
Portion on the firſt Day of January a certain Suin,
and fo continued to do on the firſt Day of each liv.
ceeding Month to the End of the Year, increauit
G3
Ć
66 Arithmetical Progreſſion.
each by L. 3. ſo that the Amount of all came to
L. 342: How much did he lay by on the firſt Day of
Jan. and alſo on the firſt Day of Desember?
342
2
12) 684(57 Sum of the Extremes.
3 comm. Diff. 12-ISI Number of Terms leſs I.
II
33 Difference of Extremes.
2) 57 (28 half the Sum of the Extremes.
2) 33 (16 3 half the Diff. of the Extremes.
Sum 45 greater Extreme, and 57—45=12 for the
Jeffer Extreme.
CASE 7. Having one Extreme, Sum of the Series
and common Difference, to find the other Extreme
and Number of Terms.
RULE. If the Jeffer Extreme is required, ſquare
the greater Extreme, to which add its Product by the
common Difference, the 4th Part of the Square of
the common Difference, from which Toral ſubtract
the Product of twice che Difference by the Sum of
the Series; out of which Remainder extracting the
ſquare Root, to ic add the common Difference, and
the Sum is the leſſer Extreme. But, if the greater
Extreme is required, To the Square of the leffer add
the Product of the Sum of the Series into twice the
common Difference, and to this Sum add the 4th
Part of the Square of the common Difference, from
which Sum ſubtracting the Product of the given Ex-
treme by the common Difference, take the ſquare
Root of the Remainder, and from this Remainder ſub-
tract half of the common Difference, and the Re-
mainder is the greater Extreme.
СНАР.
I
Geometrical Progreſion.
67
CHAP. VII. Of Geometrical Progreffion.
2
WHEN a Rank of Numbers do either increaſe by
a common Multiplicator, or decreaſe by a
common Diviſor, theſe Numbers are ſaid to be in
Geometrical Progreſſion: As
{
2. 4. 8. 16. 32. &c. where 2 is the com. Multiplier.
S2
Or
2. 6. 18. 54. 162, óc. { where is the com.
Multiplier
162. 54. 18. 6. 2. where 3 is the com. Diviſor.
.
In Geometrical Progreſſion (as in Arithmetical) there
are 5 Things conſiderable, viz.
The 2 Extremes:
The Ratio, i.e. the Quote of the greater Term di-
vided by the leſſer, or the common Multiplier.
The Number of Terms.
The Sum of the whole Series.
And the moſt uſeful Caſes are thoſe, in which are
given any of theſe 3 Things, to find the other 2.
CASE 1. Having the Extremes and Ratio, to
find the Sum and Number of Terms.
RULE. For the sum of the Series, divide the
Difference of the Extremes by the Ratio leſs I, the
Quote is the Sum leſs the greater Extreme. And to
find the Number of Terms, divide the greater. Ex-
treme by the lefier, and raiſe the Racio to a Power e-
qual to that Quote, and the Index or Number of
Multiplications is the Number of Terms leſs 1.
.
CASE 2. Having the Extremes and Number of
Terms, to find the Sum and Ratio,
RULE
68
Geometrical Progreſſion:
RULE
1. For the Ratio, divide the greater Extreme by
the leffer, extract ſuch a Root of the Quote whoſe In-
dex is the Number of Termis leſs , and it is the Ra-
rio: And find the Sum by. Caſe 1.
CASE 3. Having the Extremes and Sum, to find
the Ratio and Number of Terms.
RULE
I. For the Ratio, divide the Difference of the Sum
and leſſer Extreme, by the Difference of the Sum and
greater Extreme, and the Quote is the Ratio. And for
the Number of Terms, find it by Cafe 1.
CASE 4. Having either of the Extremes, the Ra-
tio and Number of Terms, to find the other Extreme
and the Sum of the Series.
RULE. If it is the leſſer Extreme that's given,
multiply it by the Ratio multiplied into itſelf as oft as
is the Number of Terms leſs i, and the Product is the
greater Extreme.
But if the greater Extreme is given, divide it by the
Ratio multiplied into itſelf as oft as is the Number of
Terms lefs 1, and the Quote is the leſſer Extreme:
And for the Sum of the Series, find it by Caſe I.
CASE 5. Having the Sum of the Series, Num-
ber of Terms and Ratio, to find the Extremes.
RULE. To find the leffer Extreme, multiply the
Ratio as oft into itfelf as is the Number of Terms;
then multiply the Sum of the Series by the Ratio leſs
1, and divide the Product by the foreſaid Power of
the Ratio leſs 1, the Quote is the leffer Extreme; and
for the other Extreme, find it by laſt Caſe.
CASE 6. Having either Extreme, the Sum and Ra-
tio, to find the other Extreme and Number of Terms.
RULE.
Geometrical Progreſſion. 69
RULE. If the greater Extreme is given, to find
ihe leſſer, multiply the Difference of the Sum and
greater Extreme by the Ratio leſs 1, ſubtract the Pro-
duct from the greater Extreme, and the Remainder is
the leſſer : And find the Number of Terms by Caſe v.
But, if the leffer Extreme is given, to find the great-
er, multiply the Difference of the Sum and lefſer Ex-
creme by the Rario leſs 1, and divide the Product by
the Ratio, and to the Quote adding ihe leffer Extreme,
the Sum is the greater; finding the Number of Terms
as before.
Now to exemplify ſome of the preceding Cafes
take the following Queſtion.
A Servanr, skill'd in Numbers, agreed with a Gentle-
man to ſerve him for 12 Months, provided he would
give him a Farthing for bis firſt Month's Service, a
Penny for the ſecond, 4 Pence for the third, and ſo
Of, quadrupling, each preceding Month's Wages, to
the end of the Year: What did his Hire amount to?
By Caſe 4. given the leſſer Extreme i far.
And 4 rais'd to the auth Power is #194304. for the
greater Extreme, or laſt Month's Wayes.
Then 4194304-1=4194303 Differ. of Extremes.
Allo 4-1=3=Ratio leſs 1, and 3/4194303(1398101
Laſtly, 1398101 +419430455592405 far.=
L. 5825 : 8:5: 1 Anſwer.
Which Queſtion you may vary like that in Aritha
Progreſſion, and ſo apply the other Cafes.
CHAP .
70
Of Fractions.
CHAP. VIII. Of FRACTIONS.
THERE are only ewo Kinds of Fractions now ge-
nerally uſed, viz. Vulgar or Common, and Decimal,
A Common Fraction conſiſts of two Members,
viz, a Numerator and a Denominator, which are
written as in the following Example or 3/4, the Fi-
gure 3 being the Numerator, and 4 the Denominator.
The firſt Form ſeems moſt convenient for Calculati-
ons, and the other for keeping Accompts.
The Denominator fhews into how many Parts the
Integer or whole Thing is divided, or ſuppoled to be
divided.
The Numérator ſhews how many of theſe parts are
contained in the given Fraction: Thus in the adduced
Example the Denominator ſhews, that the whole
Thing (whatever it is) is divided, or ſuppoſed to be
divided into 4 equal Parts; and the Numerator thews,
that 3 of theſe equal Parts are contained in the Frac-
tion. In like manner,
One half.
Two third Parts, or 2 Thirds.
ſignifies
4 fifth Parts, or 4 Fifths.
9 fixtecnth Parts, or 9 Sixteenths.
,
Or.we may read them thus: i divided by 2; 2 di-
vided by 3; 4 divided by 5; 9 divided by io; and ſo
of others : For the Numerator is always the Dividend,
and the Denominator the Diviſor.
A Common Fraction is Proper or Improper, Simple
or Compound.
A Proper Common Fraction is ſuch whoſe Nume-
rátor is leſs than its Denominator, and confequently the
Fraction is leſs than the Integer, to which it refers:
Exa. dc.
An
!}
Of Fractions.
71
An Improper Common Fraction is ſuch whoſe Nu-
merator is equal to or greater than its Denominator,
and therefore the Fraction is equal to or greater than
is relative Integer. Exa. $% %2% 41%, &c.
And here note, thar if the Numerator and Denomi-
nator are both the ſame, the Fraction is equal to the
Integer, as $ % 1, 4, Sc.
A Simple Common Fraction has only one Nume-
rator and one Denominator, whether it be Proper or
Improper. Exa. , SM Br.
But a Compound Fraction is a Fraction of a Frac-
tion, and hath ſeveral Numerators and ſeveral Deno-
Ininators; that is, it conGifts of ſeveral Simple Fractions,
the Particle of ſtanding betwixt each cwo of them.
Exa. of by which is meant, firſt, that the whole
Thing or Integer is divided into 3 equal Parts, two of
which Parts make then, that this Fraction is di-
vided into 5 equal Parts, and 4 of theſe Parts taken.
In like manner of of 'şof of of ză, are
Compound Fractions.
A Mixt Number (in Fractions) is a whole Number
with a Fraction annex’d, as 4 , 201, which are read
four and one half, 20 and 3 fourth Parts, where the
Fractions are ſuppoſed to refer to the fame Integers
with the 4 and 20 reſpectively, ſo that if 43 be 4 L.
its Value is L.4: 10, and if the 20%be 20. yds, the
Value is 20 yds and 3 Quarters.
CHAP. IX. Of Reduction of Common
Fractions.
HERE
ERE I would adviſe yan ? (arne: to make himſelf
Maſter of the following X I. Cales, becauſe upon
the
072 Reduction of Common Fractions.
the right underſtanding of them depends the moſt of
what is to be delivered in Addition, Subtraction, &c.
and they are as it were the very Hinge upon which
(almoſt) all the Operations, where Fractions are con.
cern'd, do turn.
CASE I.
To reduce a Fraction to its loweſt Term, that is, to
find another equivalent Fraction to a given one, but
expreſs'd in lower Terms (when poſſible.)
RULE.
'Divide the greater Term by the leſſer, that is, the
Numerator or Denominator, the one by the other;
and if there is a Remainder, by it divide the firſt Di-
viſor ; and if any thing yet remains, let it be a Diviſor
to the laſt Diviſor, and thus go on, by ſtill dividing
the preceding Diviſor by the laſt Remainder, till o re-
main: the ſeveral Quotes are to be neglected, and
the laſt Diviſor is to be aſſumed as the greateſt com-
mon Meaſure to the Terms of the propoſed Fraction,
by which common Meaſure dividing the Numerator
and Denominator ſeverally, you'll have the Fraction
expreſs'd in its loweſt equivalent Terms.
Exa. 1. Reduce to its loweft Terms.
Operation.
16) 5613
Or thus
3
48
8) 16 (2
16
16/2
8
O
Their greateſt common Meaſure is 8, wherefore by
dividing both Terms by ir, (thus, 8) (*) the Frac-
tion is reduced to 4, its loweſt equivalent Expref-
fon.
Exei
Reduktion of Common Fractions. 73
Exa. 2.
Reduce 1% to its leaſt Terms.
Operation.
Or thus,
1284814
2832)12848 (4
283 11
11328
152011
1
1520) 2832(1
1520
131216
1312) 1520 (1
1312
2083
6414
10
208) 1312 (6
1248
64) 208 (3
192
16) 64 (4
64
Their greateſt common Meaſure (thus found) is 16,
wherefore 2832—16=177 Numerator, and 12848
16=803 Denominator, and ſo 281 in its lowed
Terms is 73.
Exa. 3. Reduce their to its loweſt Expreſſion.
28) 384 (13
28
28
10+
104
84
Or thus,
384113
[
2012
20) 28(1
8
812
20
8) 2012
16
1970
4) 8 (2
8
H
Thes
74
Reduction of Common Fractions.
Then 4) (99 Anſwer in its loweſt Terms.
1. Obſerve, If Unity is the laſt Diviſor, the Frac-
tion is irreducible, and is already in its loweſt Terms;
as in the Exa. following, 154.
Operarion.
54) 197.(3
Or ſhorter thus,
162
1973
54/1
35) 54 (I
35
351
191
19) 35 (1
19
1615
313
-16) 18 (T
I
3) 16 (5
15
1) 3 (3
3
2. Obſerve, That the general Rule is not ſo conve.
nient for Practice as the taking $ $ $ 6c. (or divi-
ding by 2, 3, 4, &c.) of both Numerator and Deno-
minator (as you ſee practicable) to divide them with-
cut a Remainder. Such Diviſors, after a little Practice
and Experience, will diſcover themſelves almoſt at firſt
view, without any previous Trial.
Exa. *** in its loweſt Terms is so found thus:
129417981214
3
.
1330 112 16
3. Obſerve, If Numerator and Denominator end in
an even Number, they may both be divided by 2,
without any Remainder; and if they are both even af-
ter the firſt Diviſion, you may divide again the Quote
by
Reduction of Common Fraštions. 75
2
2)
by 2, and ſo continue to divide by 2 as oft as poflible,
and after that by any Figure that will do it, till you
come to its lowelt Terms.
Exia. 7 is $; 183 is thus found,
192
261*48)*241*12/2012
288 (144) 721 361 181 913
Here I divide s times by 2, and once by 3. Now
tho' 2 will certainly divide every even Number, yer
you'll frequently at firſt Sight obſerve, that ſome greater
Number will do it, which you ought to take, becauſe
the greater the Diviſors are, the fewer there muſt be
of them, and the Operation is the ſooner ended. So
in the laſt Exa. to have divided by 8 and 12 would
have produced the fame Effect.
4. Obfèrve, If both Numerator and Denominator
end in o or o's, cut off an equal Number from borbi,
and you have the Fraction lower, which when praci-
cable you may reduce, till you come to the loweſt.
Esca. is ; o Co is firſt 76, and by dividing
by 9, it becomes loweit. This Abbreviation is the
fame with dividing both by 10, 100, 1000, doc.
5. Obferve, IF Numerator and Denominator, the
one of them end in o and the other in 5, they are both
diviſible by 5, without a Remainder. Exa. Tô=;;
2=* Alſo if both end in 5, thus } =}, and
-
Use. This Reduction is neceſſary for facilitating
any Operation with them: Allo a Fraction muſt be
reduced to its loweſt Terms, before its Roor can be
extracted.
CASE II.
To reduce an Improper Fraction to its equivalent
whole or mixt Number.
RULE. Divide the Numerator by the Denomina-
tor, the Quote is che whole Number; and if there is
a Remainder, it muſt be annexed to the Integral Part
in form of a Fraction, and ſo the Anſwer will be a
mixt
75
T85
IS
37.
H 2
76 Reduction of Common Fractions.
25
mixt Number Exa. =1;=4; "=3 $;
+37; 2=4.
USE. 'This Reduction is of uſe for the better uns
derſtanding their Value.
CASE III.
To reduce a whole Number to an equivalent (Im-
proper) Fraction, having any aſſigned Denominator.
RULE. Multiply the whole Number by the af-
[igned Denominator, the Product is the Numerator,
which fet over the aſſigned Denominator gives the
Anſwer. Exa. Reduce 6 to an Improper Fraction,
whoſe Denominator is 8. Thus, 6x8=48; then is
the Thing fought: This is the Converſe of the laſt
Care.
Use. This is of uſe when a Fraction is to be ad-
ded to or ſubtracted from a whole Number, viz. thao
the Sum or Remainder may be in form of a Frac-
tion.
CASE IV.
To reduce a whole Number to an equivalent (Im-
proper) Fraction, when no Denominacor is given.
RULE. Set the whole Number for a Numerator,
and i for a Denominator, and it is done. Exa. is
4 ; 12 is 3.
Use. Such Reduction is neceſſary when a whole
Number is to be multiplied or divided by a Fraction.
CASE.V.
To reduce a mixt Number to an equivalent Impro-
per (ſimple) Fraction.
RULE. Multiply the whole Number by the De-
nominator of the annexed Fraction), adding to the Pro-
duct its Numerator, the Sum is the Numerator, and
the Denominacor given is the Denominator of the An-
firer. Exa. Reduce 3 } to a ſimple Fraction; thus,
3*5=15, and 15+2=17; ſo the Anſwer is .
Exa. 2. Reduce 16- to a ſimple Fraction; thus,
16x=112, and 112+3=115; fo the Anſwer is 113.
USE
Reduction of Common Fractions. 27
Use. This Reduction is neceſſary, when a mixt
Number is to be added to, multiplied or divided by
a Fraction; or when a Fraction of a different Deno-
minator is to be ſubtracted from it, or it from another
mixt Number of a different Denominator.
CASE VI.
To reduce a Compound Fraction to a Simple equia
valent one.
RULE. Multiply all the Numerators continually
into one another for the Numerator, and all the De-
nominators together for the Denominator of the Ana
(wer, which you may reduce to its loweft, if it is not
ſo after the Multiplication. Exa. Reduce of } to a
Simple Fraction; thus, 4X2=8 Numeracor, and
5*3=15 Denominator; therefore iş is the Anſwer.
Alſo of g of <=38 for the Simple equivalent Frac-
tion, which in its loweſt Terms is zs
1. If one of the Members of a Compound Frac-
tion is a whole Number greater than 1, then this
whole Number being turned into an Improper Frac-
tion (by Caſe 4th), let the Reduction be inade as be-
fore. Exa. 'of 4 is of 4=; of 8 is of
==6; j of 175 of 1=3=U*
2. When one of the Members is a mixt Number,
reduce it by Caſe 5th to a Simple Fraction, and work
as before. Exa. of 43 is } of =i=27.
3. When all or any one of the given Fractions can:
be reduced to lower Terms, let that be done firſt, and
then apply the general Rule; for by this Means the
Anſwer will come out in lower Terms, tho' not al-
ways in its loweſt. Exa. 4 of 1 is of 1=iori;
of is of 1=1
4.
It matters nor in what order the Members of a:
Compound Fraction are taken, for j. of ! is the ſame
with 1 of 3
USE. This Reduction is of uſe for the more ready
comprehending their Value, as alſo in order to prepare
them for Addition, Subtraction, Multiplication and
Diviſion,
CASE
8
H 3
78 Redu&tion of Common Fra&tions,
CASE VII.
To reduce Fractions of different Denominar ors to
other equivalent Fractions having the ſame Denomina-
tor
Tator.
16
15
RULE. Multiply all the Denominators continual-
ly into one another, the Product is the common De-
nominator fought; then multiply the Numerator of
cach Fraction into the Denominators of all the other
continually, and the Product of each is a new Nume-
Exa. Reduceſ and to the ſame Denomina-
tor: Thus 3x8=24 their common Denominator;
then 2x8=16 firſt Numerator, and 5*3=15 the
other Numerator; ſo the equivalent Fractions found
are . and is; for } is zh, and is
21: The
Denominator may be ſet down only once, thus,
Reduce this and ž to a common Denominator ;
thus, 7x18 126, and 126*5=620 for the Deno-
minator; then 4 x 18X5=360 firſt Numerator; next
13 X 7X5=455 the ſecond Numerator; laſtly 2 X 18
x7=252 the laſt Numerator. So the equivalent Frac-
tions found are %10
- is, and=}, and
may ſtand thus, 360
i. After the common Denominator is found, you
may find the new Numerator thus: Divide the com-
mon Denominator by the Denominator of each of
che Fractions, and multiply the Quotes by the Nume-
Tators reſpectively, and fo you have the new Numera-
tors fought.
2. When there are two Fractions to be reduced,
and one of the Denominators is a Multiple of the o-
ther, divide; and by the Quote multiply the Numera-
tor and Denominator of that Fraction whoſe Denomi-
nator you divide by, and thus they'll be both reduced
to the fame Denominator. Exa. To reduce and
to the ſame Denominator, I divide the Denominator
8 by 4 the Denominator of the other Fraction, and
the Quote is 2;. by which multiplying the Frac-.
tion, it is reduced to $; having the ſame Denomina-
4 455
7) 639
360 455 253
630
tor.
Reduction of Common Fractions. 79
.
tor with the other Fraction. The Advantage of this
Method is, that it gives the Fractions frequently, tho'
not always, in their loweſt Terms.
USE. This Reduction is neceſſary before they can
be added or ſubtracted, as alſo to know which of two
Fractions propoſed is the greater (if they do not dif-
cover themſelves at firſt ſight.) It is further uſeful for
finding two Integers in the ſame Proportion to one a-
nother with the Fractions given: For inſtance, to find
two Integers in proportion to one another, as į is to
* ;, thus, ģ:::: i that is; as 8 to 9; ſo that
8 and 9 are the Integers ſought.
CASE VIIT.
To reduce a Fraction of an Unit of an higher
Denomination or Name, to an equivalent Fraction of
any Unit of a lower Species (of the ſame Kind with
the higher.)
RULE: Multiply the Numerator of the given Frac-
tion by the Number of Units in the next inferior
Species that makes an Unit of the Denomination of
your Fraction, and the Product multiply by the Num.
ber of Units in the next inferior Denomination that
make an Unit of the laſt Denominacion, and thus
proceed 'till you come to the lowest you deſign, then
make the laſt Product a Numerator to the Denomis
nátor of the Fraction given. Exa. Reduce L. 10
an equivalent Fraction in the Denomination of id.
3 Numerator of the given Fract.
20 Number of . in 1 L.
So that 2 L. is
=?20 d. or 144
60
d lowelt.
12 Number of d. in 1 fl.
720
EXA.
80 Reduction of Common Fractions.
Exa. 2. Reduce { lb. to the Fraction of 1 farth,
I the Numerator given.
12 Number of d. in i fi
So that th.
is = 45 far.
I-2
4 Number of far. in 1 d.
1
48
Exa. 3. Reduce C. to the Fraction of 1 oz.
2
4
So that įC. is equal
to 35 8 OZ.
8 qrs.
28
224 lib.
16
3584 oz.
Or you may expreſs them in form of Compound
Fractions, and then reduce them to Simple ones; ſo
in the firſt Exa. L. is } of 20 s. of 12 d. = d of ².
of 729, as before. In Exa. 3. C. is of 4
grs. of 28. lib. of 16 oz. Ź of 4 of 28 of 9
U oZs The Use of this Reduction will appear in
that of the next following Caſe.
CASE IX.
To reduce a Fraction of an Unit of a lower Deno-
mination, to an equivalent Fraction in the Denomina-
sion of an higher.
RULE. Multiply the Denominator of the propc-
fed Fraction by the Number of Units of the ſaid Frac-
tion that is equal to an Unit of the next ſuperiour De-
nomination, and the Product by ſuch a Number of
Units of its Denomination as is equal to an Unit of
the next above it; and thus go on till you come to the
higheſt
Reduction of Common Fractions. 8.1
higheſt Species required, and the laſt Product is a De-
nominator to the Numerator of the Fraction giveni.
Exa. Reduce 3 d. to the Fraction of iL.
5
Denominator given.
I2
1
So that} d. is = 11aL.
or om loweſt.
60
20
1200 = Denominator required.
Exa. 2.
Reduce ozi: to the Fraction of 1 C.
y Denominator given,
16
So thar - oz.
I I2
28
7774 C.
Inté loweit
.896..
224
3136
4
12544 = Denominator required.
But it is (perhaps) as convenient to expreſs them in
Compound Fractions, and reduce ihe fame to Simple
ones for the Anſwer. Thus in Exa. I. } d. is of it
of 1 of 1 L=1263 L. In Exa. 2. 4 oz. is 1 of 1
of of of 1 C.=73 C.
Use. This Reduction is abſolutely neceſſary be-
fore we can add, ſubtract or divide Fractions that be-
long to different Unics."
CASE X.
To expreſs a whole or mixt Number of an inferiour
Denomination, by a Fraction of ſome ſuperiour Unit
(of the ſame Kind.)
RULE..
1
82 Reduction of Common Fractions.
Rule. If it is a simple Number, let it be the
Numerator of the Fraction, and for its Denominator
take the Number of that inferiour Species, which makes
I of the ſuperiour you would have it expref'd in, and
you have the Fraction required.
Exa. 1. Expreſs 4 d. in the Fraction of 1 L. In
i L. are 240 d. wherefore 4 d. is 396, or că L. loweſt:
Exa. 2. Expreſs 3. qrs. in the Fraction of ini.
In 1ſ. are 48 far. wherefore 3 far. is 7 . or 15
Jh. loweſt.
Exa. 3. Expreſs 2 lib. in the Fraction of 1 C.
In 1 C. are 112' lib. wherefore 2 lib. is 17ž C. or 55
loweſt.
2. If it is a mixt Number, let it be firſt reduced
to a fimple Number of the loweſt Species mentioned,
and that Number is the Numerator; and for the De-
nominator take the Number of that Species you re«.
duce your mixt Number to, that makes i of the ſupe-
riour. Exa. 1. Expreſs 3 1. 8 d. in the Fraction of
1L. ' 3.fb. 8 d. is 44 d. wherefore there being 240 d.
in i L. 31. 8 d. is 4* L. or loweſt.
Exa. 2. Expreſs 66. 2 d. 3 far. in the Fraction
of iL. The mixt Number reduced to a ſimple one
makes 299 f. for the Numerator, and the Denomina-
cor is 960, there being that Number of far, in 1 L.
ſo the Anfwer is ! L.
Exa. 3. Expreſs 4 lib. 7 oz. in the Fraction of
IC.
4
lib. 7 oz. is 71 oz. then becauſe in a C. are
1792 oz. the Anſwer is 1771 C.
Use. This ſeems to be of Uſe principally when
a fimple or mixt Number of an inferiour Species is
to be reduced to an equivalent Decimal of an higher.
CASE XI.
To find the Value of a ſimple Fraction in Integers
of a lower Species, when the Fraction given is not of.
the loweſt Denomination.
RULE.
Reduction of Common Fractions. 83
RULE. Multiply the Numerator of the given
Fraction by the Number of Units of the next inferi-
our Species that makes I of the Denomination of
your Fraction, and divide the Product by its Denomi-
nator, the Quote is ſo many Integers of that lower
Species, and if there is a Remainder, reduce it to the
next inferiour Species, and divide again; and thus go
on by reducing and dividing, till you come to the
loweſt Species,
and the ſeveral Quotes, with the Re-
mainder, if any be, which is always a Fraction of the
loweſt Species, are the Anſwer. Exa. 1. What is į L.?
2
20
20
3) 40 (131
3
IO
9
I
f. d..
12
Anfr. 13:4
3) 12 (4.d.
12
Exa. 2.
What is the Value of * .?
4
12
5) 48 (2 15.
45
nal
5):12 (2 d.
d. f.
Anfr. 9:23
IO
Exa.
84 Reduction of Common Fractions.
Exa. 3. What is the Value of 33% C.?
34
4
356) 136 (o qrs.
28
1088
272
356) 3808 (10 lib.
356
24.8
16
1488
248
356) 3968 (11 OZ.
356
408
356
52
16
lib. oz. dr.
Anfr. 10: 11:23
356) 832 (2 dr.
712
120
Ex4.
Reduktion of Common Fractions.
85
8
Exa. 4. What is the Value of 4 Boll?
4
- 16
7) 64 (9 pecks.
63
p: lip
Anfi. 9:01
I
4
7) 410 24 lip.
USE. 'The Uſe of ſuch Reduction is evident.
If the given Fraction is Improper, you muſt firſt re:
duce it, and the Quote is an Integer of the fame
Species with the Fraction; then reduce the Remain-
der as before. Eva. What is the Value of 73 lb.?
8) 77 jb.
M. d. f.
9:7:2
Anfr. 9:7: 2
WX XXXX XXXX%2:22
CHAP. X. Of Addition of Commor
Fractions.
GENERAL Rule.
ENERAL RULE. Add together the Numera-
tors for the Numerator of the Sum, and ſetting
iť over the common Denominator, you have the An-
fwer; which, if it is an Improper Fraction, you may
reduce it to a whole or mixt Number by Caſe II.
Chap. IX. or if it is not in its loweſt Expreſlion, de
preſs it by Cafe I.
CASE L.
When the Fractions belong all to the ſame Unit or
Integer, and alſo have the fame Denominator ?
I
RULE.
86 Addition of Cominon Fractions.
IS
and 7
I 32
LI
27
283
RULE. Follow the General Rule, and you have
the Anſwer. Exa. What is the Sum of and of
any Thing? += Exa. 2. What is the Sum
of', and of any thing? +4=19=1} Aufr.
Exa. 3. What is the Sum of $ and ? +=;
or Anfr.
CASE II.
When all the Fractions belong to the ſame Integer,
but have different Denominators.
Reduce them to Fractions having one common De-
nominator by Cafe VII. Chap. IX. and perform the
reſt of the Work as in the preceding Caſe. Exa.
Add to & -tý
21 or Sum.
Exa. 2. Add together and The Fractions re-
108-192-4 120
duced are
==
I za š or I
loweſt, Anfr.
CASE MI
When the Fractions have the ſame or different De-
nominarors, hui belong nor to the ſame Integer.
RULE. Reduce them first to Fractions of the
ſame Integer by Caſe VIII. or IX. Chap IX. and
then to one Denominator, if neceſſary, and apply the
general Rule. Exa. Add į L. to } lb. The Fraction
jf. being reduced to the equivalent Fraction of i L.
is co? L. wherefore ;L + To L. when reduced to
the ſame Denominator are
200 to 21
ŽE L. for
700
the Sum, = 6b. 3 d. 378 f. or 37} loweſt.
You may either reduce the Fraction of the lower
Species to an equivalent of the higher, or contrariwiſe,
for the final Reſult will be the ſame in both, as you
may try by the precedirg Exa. If three or more ſuch
Fractions are given, let all the inferior ones be reduced
to equivalent Fractions of the higheſt mentioned, and
the reſt of the Work is the ſame as before.
CASE
Addition of Common Fractions.
87
CASE IV.
To add mixt Numbers.
1. Add the Fractions by themſelves, according to
the preceding Cafes, and join their Amount to that of
the Integers. Or,
2. Reduce the mixt Numbers to Iniproper Frac-
tions, and add as above; reducing the Sum to a whole
or mixt Number, and the fractional Part (if any be)
so its loweſt Ternis. Exa. Add 2005. By the
firſt Method I take the two Fractions and redu:
cing them to the ſame Denominaror, they become
23 and to, whoſe Sum is 4 or I ; to this I add the
Sum of the Integers 2 and 5 (viz. 7) and the Total
Aggregate is 8 for the Anſwer.
By che ſecond Method I reduce the mixt Numbers
10 Improper Fractions, and they become and 43;
which in the fame Denominator are 13 and 243
- whole Sum is 37 87, as before.
Of theſe two. Methods I prefer the firſt, as being
eaſieſt and moſt expeditious, eſpecially when the mixt
Numbers to be added are three or more.
CASE V.
When all or any are Compound Fractions.
RULE. Reduce them firſt to Simple Fractions,
and add according to the preceding Caſes. Exa. Add
of stoof of iſ of, alſo of of
i= til or 3%, then 7 + za=
13
840
Sum is : for the Anſwer.
45
.
Take the following Queſtions for your further
Exerciſe in this Rule.
1. What is the Sum of 1 tý tits? Anfr. 176.
2. What is the Sum of 1 L. +1h. tid. Anſi.
1997L. = 60. 7 d. 2139 far.
I 2
3.
88 Subtraction of Common Fractions.
3. What is the Sum of Ti 11. te of 18 d. tio
far. Anfr. 437.= 30.2, far.
4. Add together L. 148 + L 115 + 1911 15. to
4 d. Anfr. L. 26: 9: 7: o ſ} far.
0000000000000000000000
Thus 19 - 25
5 27
CHAP. XI. Of Subtraction.
GENERAL RULE. Subtract the leffer Nume-
rator from the greater, and the Remainder fer
over the common Denominator gives the Anſwer.
CASE 1.
When the Fractions belong to the fame Unit and
have the ſame Denominator, apply the general Rule,
and you have the Anſwer.
Exa. 1. From } L. take } L. Thus, 3-1.52
Numerator, therefore į L. is the Anſwer. Exa. 2.
From '1 of any thing take ţ of the ſame thing:
2 17wherefore che Anſwer is looking
.
CASE II.
When both refer to the ſame Integer, but have dif-
ferent Denominators.
RULE. Reduce them to the fame Denominator,
and then apply the Rule. Exa. I. From $take.
The Fractions reduced to the ſame Denominator are
18 and Ins; the Difference of their Numerators is 142
which ſer over the common Denominator gives or
Iż-for the Anſwer. Exa. 2. From & take ž. Thus
&=15-12 = iš or loweſt, for the Diffe-
18
rence.
CASE III.
When the Fractions have the ſame or different De-
nominators, but belong not to the fame Integer.
RULE.
I
Subtraction of "Common Fractions. 89
200
21
L. 5
RULE. Reduce them firſt to Fractions of the
fame Integer, next to Fractions having the fame Deno-
minator (if need be) and apply the general:Rule. Exa.
From L. take } lb. f. is 7oL. wherefore L.
153 L.
77. L.
5 fb. 1 d. it's
700
Anſwer.
If you want the Difference betwixt any two ſuch
Fractions in Integers of known Species, you need not
reduce them after the above Method, but reduce them
at firſt to their known Values, and then ſubtract. So in
the laſt Example, L. is (by laſt Cafe
of Reduction) found to be '. 8 d.
23 far. and lh. is, by the ſame Cale, 5:8: 2
7 d. of far. which fet down, and ſub-
: 7:0
tract, as in the Margin.
5:1:1 35
CASE. IV.
When one or both are mixt Numbers. .
1. If one or both are mixed, and the Fractions
have the fame Denominator, there is no need of any
Reduction, but the Work may be performed as in
the following Examples.
Ex4. I.
Exa. 2.
Exa.
4.
From 6-3 6 6}
63
7
43 43
47
d. f.
17
Exa. 3.
Exa. 5.
Take 4
1
2 orang
Rem. 23 15
in or 1
675
2. But if they are both mixt, and th: Fracticns
have different Denominators, you muſt reduce the
mixt Numbers into Improper Fractions, and there to
the ſame Denominator, and apply the geijeral Rule.
Exa. From 6 take 44. When reduced they become
27 and 3;; then in the fame Denominator they are
if and 3$, ſo that their Difference is for 278.
Or you may reduce the 2 Fractions to the ſame De-
nominator, and annexing them to their proper Inte-
gers,
I 3
90 Subtraction of Common Fra&tions.
3
I
4 2
gers,
fubtract as in Part I. of this Caſe. Thus
in the Exa. 6 4}, and in the fame 6
Denominator are it and which I annex 10
their proper Integers, and ſubtract, as in the
Margin, beginning with the Fractions.
215
CASE V.
When one or both are Compound Fractions:
RULE. Reduce them to Simple ones, and to one
Denominator, and then apply the general Rule.
Exa. From 4 of fake of. Thus of
1 , or in their loweſt Terms
ſo that their Difference is 21%. And ſo
210
of others.
of
6
I 2
55
Wil launin
8
3
45 - 28
2
More Examples for further Exerciſe in this
Rule may be ſuch as follow.
1. What is the Difference betwixt Tooő and T7.
Anfr. Too
2. What is the Difference betwixt 7 Jh. and I of
1 far. Anfr. 6 d. ii's far.
3. From 20 L. 139 h. 11 d. take 19 iş L. 18
*? fb. 10 d. 2} far. Anfr. L. 1471 lb. I ja d.
Iffar. = 19. 6 d. 373 far.
4. From i C. take ţ qrs. 19 } lib. 4'} oz. 10 dr.
Anfr. 2 qrs. 7 } lib. 10} oz. 6 dr. = 2 qrs. 27 lib.
3 oz. 10 dr.
5. A has tő of a Ship, and B has į of the ſame,
What is the Difference of their Shares ? Anfr. zo.
СНАР.
Multiplication of Common Fractions. 91
CHAP. XI. Multiplication in Com-
mon Fractions.
THE multiplying any
Number, whether Integral or
Fractional by a Fraction, is the taking ſuch a
Part or Parts of that Integer or Fraction, as the mul-
tiplying Fraction exprefferh.
GENERAL RULE. Multiply the two Numera-
tors the one by the other, as alſo the two Denomina-
tors into one another, for the Numerator and Deno-
minator of the Product reſpectively, which, if it
happen to be in high Terms, let it be depreſs'd to its
loweſt;" or, if it is an Improper Fraction, reduce it to
its equivalent whole or mixt Number. You may alſo
abbreviate the Fractions to be multiplied (when pof-
fible) before you begin, and ſo they become more
manageable.
CASE I.
When they are both Simple Fractions referred to the
fame Integers.
RULE. Obſerve the general Rule. Exa. Mul-
tiply by }. Thus, *x}= or i, Product. Exa.
2. {x= Anfr. Exa. 3. * * = or 13
Product.
CASE JI.
When the one is a whole Number and the other
a Fraction:
RULE. Reduce the whole Number to a Fracti-
on, and then apply the general Rule. Exa. Multiply
6 by 3. Thus X=i=1 loweſt, Anfr. Exa.
2. Multiply 56 by. Thus sex = -2= 32 Anfr.
Or multiply tbe whole Number by the Numerator,
and divide the Product by the Denominator.
CASE
92 Multiplication of Common Fractions.
CASE III.
When one or both are mixt Numbers.
RULE. Reduce the mixt Numbers to Simple
Fractions, and multiply as before. Exa. Multiply 16
by 21 Thus x = 8 = 40 Anfr. Exa. 2.
34x83"}=104=26 Anfr. Exa. 3. Multiply
58 by 1; } Thus 31 x4=2881368 Anfr.
1. But if the Multiplier is a whole Number, and the
Multiplicand mixt (or you may make either of them
Multiplier, ſeeing the Product will be the fame) you
need not reduce, but beginning with the Fraction,
multiply it, (viz. its Numerator) and if the Product
is Improper, reduce it, ſerting down the Remainder,
and carrying the Integral part or Quote of the Diviſi-
on, to the next Product. Exa. Multiply 81 by 6.
Here I ſay 6 times is ?), which is 4 to
82
carry, and į or 1 to be ſet down; then 6
6
Times 8 is 48 +4 Carried is 52. See the
Margin.
52ą or
Exa. 2. Multiply 5 by 19 ý: Here I take
19} the Integral Number for the Multiplier, be-
5 cauſe the Work will be eaſier.
+
3
528
981
Alſo if one of them is the Product of
17
2 or more Digits, the Anſwer may be
3 found without Reduction as before. Exa.
1. Multiply 17 by 18. Here I multi-
ply by 3 and 6 inſtead of 18. See the
Margin.
Exa 2. Multiply 135 9 by 112. Here
317 ß or I multiply by 4, 4, 7, becauſe 4 x 4x7
=112, and the Anſwer is 15216, as you
may try at your Leiſure.
2. Alſo when both are mixt, the Work may ſome-
times be performed without Reduction too, and ſo
muclt
Multiplication of Common Fractions. 93
نہ اب نما+
48
36
much compendized, as in the following Example to
multiply 48 í by.151
Here I multiply 48 by 15, ſetting down
48 the partial Products as in common Mul-
15
tiplication, then I multiply 48 by the
Fraction, and the Product 36 I place
210
under the other two Products, viz. in
Units and ro's Place; next I mulciply 15
by the other Fraction ſ, and the Product
10 or
10 I ſet under the reſt, according to the
lait Form; laſtly, I multiply the two Frac-
7661 Anfr. sions together, and their Product or !
I annex to the Integers, and then add all
together. But this Method can be conveniently uſed
only when the Integers are diviſible by the Denomi-
nators of the alternate Fractions without a Remain-
der, that is, when theſe Denominators are aliquot
Parts of their oppoſite lategers.
CASE IV.
When one or both are Compound Fractions.
RULE. Reduce to Simple ones, and proceed as
before. Exa. Multiply of by of *; when re-
duced they become išX=18 or I's Anfr.
1. When the Multiplicand is a mixt applicate Num-
ber, with a Fraction annexed to its loweſt Species,
and the Multiplier a Sinple Fraction or reduced to
fuch, then reduce the mixt Number to its loweſt
Species mentioned, next to a Fraction of the Denomina-
tor of the Fraction annex'd, and then multiply accord-
ing to the general Rule, the Product is the Anſwer in
that loweſt Species, which you may reduce as you ſee
neceſſary. Exa. Multiply L. 10: 8:4'} by 2}; be-
ing reduced they become 2500_ d. x="2.0 d. X
237548 d. = 5937 28d. =L. 27: 14: 9 18 for
the Anſwer.
But if the Multiplier is any of the 9 Digics (or if it
is the Product of any 2 or more of them) there is no
need
94 Multiplication of Common Fractions.
need of a previous Reduction of the Multiplicand:
Thus to multiply L. 38: 8:46 by 7, I ſer thein down
thus, and multiply, Carrying at the ſeveral Denomina-
iions as in Addition,
L. 38: 8:45
7
268: 18:5Product.
Exa. 2. Multiply L. 20: 16: 10 } by 24. Here
I multiply by 4 and 6, inſtead of 24.
L. 20: 16: 10%
4
1
83: 7:58
7: 57 or Prod. by 4.
6
Anfr. 500: 4: 9 Product by 6.
3. If the Multiplicand is mixt (as before) and the
Múltiplier a pure Fraction, and that a ſmall one, mula
tiply the mixt applicate Number by the Numerator
of the Fraction, and divide the Product by its Deno-
minator. Exa. What is * of L. 64: 16: 8.7
L. 64: 16: 8
3. Numerator..
Denominator +) 194: 10:0
48: 12:6 Anſwer.
Or you may firſt divide by the Denominator, and
then multiply the Quote by the Numerator. Thus,
4) 64: 16: 8
16: 4:2
3
48: 12: 6 Anfr. as before:
Exa.
Multiplication of Common Fractions. 95
1
Exa. 2.
What is & of 187 L. 13./6. 4$ d.?
187: 13:45 By the other Method.
5
6) 187: 13:45
31: 5:62
6)938: 6: 83
5
156: 7: 972
156: 7: 917
4. When a Fraction, whether Proper or Improper,
is to be multiplied by a whole Number, and the De-
nominator of the Fraction is the ſame with that whole
Number, take the Numerator of the Fraction, and it
is the Anſwer without any Multiplication. Exa. ?X
8 :7; for i=s=7. Exa. 2. X 5
X = 12,
for x==12.
5. If you are to multiply a Fraction by 2, you may
either multiply the Numerator for the Numerator of
the Product, retaining the Denominator given; or
you may take 1 of the Denominator, retaining the
given Numerator; but this laſt Method is only prac-
cicable when the Denominator is evera. Exa. } X 2
is g by multiplying the Numerator, or it is by hal,
ving the Denominator, which are equivalent.
6. It may ſeem ſtrange, that Multiplication by a
Proper Fraction ſhould produce ſomething leſs than
the Multiplicand, whereas in whole Numbers the Pro-
duct is always greater than either Multiplier or Multi-
plicand, (excepting in one pariicular Caſe, where the
Product is juſt the ſame with either Factor;) but this
Difficulty is eaſily removed, by confidering, that if
the multiplying a whole Number by i produce the
Multiplicand itſelf, to multiply any Nurber by a Num-
ber leſs than 1, (which is a proper Fraction) the Pro-
duct muſt be leſs than the Multiplicand, in the ſame
Proportion as the multiplying Fraction is remov'd from
Unity; and this I take to be as clear and intelligible
a Reaſon as can be aſſigned for it. Exa. To multi-
ply 12 by I, the Product is preciſely 12, but the Pro-
duct
96 Multiplication of Common Fractions.
duct of 12 multiplied by { muſt be only the half of
12, viz. 6, becauſe the latt Multiplier 1 is only the
half of the firſt Multiplier i. So gives only :
for the Product, which is leſs than the Multiplicand or
Multiplier. Now you are to obſerve, that this Effect
is only produced when one of the Factors is a proper
Fraction; for if it is Improper, the whole Number,
or the Fraction multiplied, is really increaſed, and con
ſequently the Product is greater than either the Multi-
plicand or Multiplier.
Now follow ſome Examples for further Exer-
ciſe in this Rule.
1. If any one Thing coft far. what is the Value
of 25 { ſuch things at the fame Rate. Anfi. 3 d.
38 far.
2. I bought 7's Bags of Hops, each weighing I C.
9}
2 qrs. 8. lib. how many lib. in all, and what is the
toal Coſt, at 7 d. 1 far. per lib.? Anfr. 1340 lib.
in all, and L.41: 4:0:3ž total Coſt.
3. What is zo of L. 12: 10: 8:3}? Anfr. L.
19: 2:05
4. What is the Product of 38 C. 2} qrs. 1411
lib. 6 oz. multiply'd by 4$? Anfr. 189C. 7 lib.
141 oz.
5. What Number divided by 14$, gives 173 18 for
the Quote? Anfr. 407
55
407043
160
CHAP. XII. Of Diviſion of Common
Fractions.
GENERAL RULE. Multiply the Numerator of
the Dividend into the Denominator of the Di-
viſor,
Diviſion of Common Fraktions.
97
+
4
viłor, and the Product is the Numerator of the Quote;
next multiply the. Denomina:or of the Dividend into
the Numerator of the Diviſor for the Denominator
of the Quote.
CASE T.
When they are both Simple Fractions belonging to
the ſame Unit.
RULE. Apply the general Rule, and you have
the Anſwer. Exa. Divide by $: Thus -, or
thus $ ) (46 Quote. “Exa. 2. Divide şi by īſ: Thus
22
3; 1==
ili or its loweſt.
CASE II.
When they are both Simple Fractions, but belong
not to the ſame Unit.
Rule. Reduce one of them to an equivalent Fraction
of the Denomination of the other, and then apply the
general Rule.
Exa. Divide ž L. by $. Thus žL,
$. is L. 10*L=28=7 loweſt.
CASE III.
When the one is a whole Number, and the other a
FraErion.
RULE. Reduce the whole Number to an Im-
proper Fraction, and work as before. Exa. }) 8 is
1) (24 or 24 Quote. Exa. 2. Divide by 8: Thus
== Quote. Alſo 40} divided by 6, quotes
612
CASE IV.
When one or both are mixt Numbers.
RULE. Reduce the mixt Numbers to Simple Frac-
tions, and then apply the general Rule. Exa. I. 71
+21="%) 1 (734 Quote. Exa. 2. 51-6=
5%= Quote.
CASE V.
When one or both are Compound Fractions.
RULE. Reduce the Compound Fractions to
Simple ones, and then work by the general Rule
I
3
2
K
Exa
98 Diviſion of Common Fractions.
4
5
Exa. z of 4 = of when reduced become -
=}{ or 2 11 Quote.
1. You may either write the Dividend and Diviſor
thus, 3 =1where the Dividend ſtands to the
left, or you may place them as in Integral Arithme-
tic thus $ ) (15; only obſerve, that in whatever Form
you place them, you make the Dividend always your
Sandard, by beginning the Multiplication with its Nu-
merator for the Numerator of the Quote, and then
multiplying contrariwiſe for the Denominator.
2. When the Quote is found, you may depreſs it to
its loweſt Terms.
3. If thiey have both the ſame Denominator, the
Quote is ſooneſt found by dividing the one Numera-
tor by the other. Exa. ) $ (2 Quote. Exa, 2. 1)
1 (1 or Quote.
4. If the Numerator and Denominator of the Di-
vidend can be divided without a Remainder by the
Numerator and Denominator of the Diviſor reſpec-
tively, divide; and the two Quotes are the Aniwer
ſought. Exa. )T$ (& or loweſt. Exå. 2. 3) 28 (
Quote. Or if you can find any Number which will
divid: the Terms of the Diviſor or Dividend without
a Remainder, divide, and take theſe Numbers inſtead
of the given ones, and ſo the Quote will come out in
lower Terms than if no ſuch Diviſion had been made.
Exa. 11)=)(=Il Quote.
Exa. 2. 21)
2. =) Quote. And fo of others.
5. When the Numerator of the Dividend can be
divided without a Remainder by the Numerator of
the Diviſor, divide, and by the Quote multiply the
Denominator of the Diviſor for the Numcrator of
the Anſwer, retaining the Denominaror of the Divi-
dend for the Denominator of the Qucte. Exa. *) **
(Quote. Exa. 2. }) ({} or iti Quote.
6. When the Denominator of the Dividend is divi-
fible without a Reinainder by the Denominator of the
Diviſor,
Diviſion of Common Fractioris.
99
Diviſor, divide, and multiply the Quote by the Nu-
merator of the Diviſor for the Denominator of the
Anſwer, retaining the Numerator of the Dividend for
the Numerator. Exa. 4) zi (1 Exa. 2. 3) 27 (
Quote.
7. When the Numerator of the Dividend can be
divided without a Remainder by the Numerator of
the Diviſor, and alſo the Denominator of the Divifor
by the Denominator of the Dividend, then divide the
one Numerator by the other, as alſo the one Deno-
minator by the orher, and the Product of theſe two
Quotes is the Anſwer fought. Exa. 1'}) $ (6 Quote.
Exa. 2. zi) " (9 Quote.
8. When the Divifor is an abſtract Fraction, and
the Dividend a ſimple applicate Number, work by the
general Rule, and the Quote is applicate to the ſame
Integer with the Dividend. Exa: 4) L. or of what
Sum of Money is L. the Part: Thus 4) (L.
= 18.b. 8 d. ſo that L. is ģ of i L. or of 18 ).
8 d. But if both are applicate, the Quore is abſtract.
Exa. À L.) L. (19; which denotes that i L. is con-
cained in L. If times, which is very near once, and
conſequently \ L. 'is nearly equal to L. that is “ L.
exceeds L. by } L. Another Exa. What Part or
Parts of 2 C. 3 qrs. 12} lib. is 6 C. 2 qrs.? When
reduced they become 242 lib.) 60 lib. (3916 Anfir.
9. When the Diviſor is an abſtract mixt Number,
and the Dividend a mixt applicare, reduce the Divi-
for to an Improper Fraction, and the Dividend to the
loweſt Species mentioned, and then apply the general
Rule. Exa. 3 §) 16 L. 10 b. 4. d. (by Reduction)
11893 d.
1081 Tiid. L.4: 10: 111
Anſwer. But this is ſooner done by dividing the mixt
applicate Number L. 16: 10: 4. by the Numerator of
the Improper Fraction, and then multiplying the
Quote by its Denominator. Thus,
TI
2204.d.
I
K. 2.
100 Diviſion of Common Fractions.
II) 16: 10:4
I: 10:01
3 Denominator.
II
4: 10:11} Anfr. as before.
10. A Fraction is halved or divided by 2, by halv.
ing its Numerator if even, or by doubling the Deno-
minator if the Numerator is odd. Thus + 2 is ž,
and 2 is I.
11. As in Multiplication, if the Multiplier is a: Pro-
per Fraction, the Product is leſs than the Multiplicand;
ſo in Diviſion, if the Diviſor is a Proper Fraction,
the Quote is greater than the Dividend: Both which
are contrary to Multiplication and Diviſion in whole
Numbers. I have already briefly accounted for the for-
mer, and now it remains that I explain .che larter,
which I ſhall do after this manner. Diviſion finds how
of one Number is contained in another, conſequent-
ly the Quote declares, how oft the Divifor is contain-
ed in the Dividend. Now to divide by a Proper Frac-
tion is only finding how oft that Proper Fraction is
contained in the Dividend, or wbac Part it is of the
Dividend. Hence it is plain, that if the Dividend is a
whole Number, the Quote muſt be greateſt; becauſe
if when the Diviſor is 1, the Quote is preciſely the
fame with the Dividend, it follows, that when the Di-
viſor is leſs than 1, the Quote muſt be greater than
the Dividend, (for the greater the Diviſor the leſs is
the Quote, and vice verſa ;) ſo if 24 is to be divided
by $it is evident that part is contained in 24 more
than 24 times; for 1 is contained in 24, 24 times,
therefore, which is leſs than 1, muſt be contained oft-
ner, and by conſequence the Quote muſt be greateſt.
Moreover, tho? the Dividend is a Proper Fraction
leſs than the Diviſor, yet the Quote is a Fraction
greater alſo than the Dividend; for Exa. ſ) $ (12. For
may not one ask how oft ž is contain'd in , or what
Part it is of $? 'Tis evident cannot be contained -
3
any
Diviſion of Common Fractions. TOI
any Integral Number of Times in $, becauſe the Di-
viſor here is greater than the Dividend; therefore ic
muſt be contained ſome Fractional Number of Tines,
and the Quote 7% denores that is iš Paris of the
Dividend , or that it is contained in , I Parts of
1; that is, if the Dividend was divided into 16 equal
Parts, then į is preciſely it of theſe Parts.
Or it may be proved thus : The Quote muſt bave
the ſame Proportion to the Dividend as Unity has to
the Diviſor; lo }) }(%, the Quote :: 1 : Ž. Af-
ter all, if the Diviſor and Dividend are reduced to
the fame Denominator, the Difficulty is quite remov-
ed, for ihen you divide the one Numerator by the
other as in Integers, neglecting the common Denomi-
nator, as you may prove at your Leiſure'; laſtly, you
are to obſerve that the common Rule for dividing
Fractions does not ſo much divide as it produces new
Terms, viz. a Numerator to be divided by a Deno-
minator; thus ) (where you ſee the Diviſion
produces 4, that is 4 to be divided by 2; but if the
Fractions are reduced to one Denominator before you
begin, there is no more to do but to make one ſimple
Diviſion, and ſo the Quote is found; for Exa. to di-
vide I by I, when reduced they become ) $, and ſo
the Quote is 2 found by (neglecting the common De-
nominator 8, and) dividing the one Numerator 4 by
the other Numerator 2.
More Examples for Practice.
1. What is the Price of a lib. of any thing, when
12 lib. of the fame coſt L. 5: 17: 10? Anfr.
9.b. 5 d. 2 136 far.
2. What Part of L. 345: 10: 3: IŠ is L. 771?
Anfr. 483662.
3. Of what is 11 gall. 3 pints 2 mutchkins 2 'š gills
(Scots Meaſure) the zi Part? Anfr. Of 20 gall. 2 p:
3 mut. 31 gills.
K 3
4.
102 Of Decimal Fractions.
4. What Number multiplied by si produces 112 *?
Anfr. 203.
2
CHAP. XIII. Of Decimal Fractions.
IN
N Decimal Fractions the Unit, Integer, or whole
Thing is divided, or ſuppoſed to be divided into
10 equal Parts, and each of theſe 10 into 10, (ſo that
the Unit is tereby divided into 100 Parts) and each of
theſe laſt into 10, (whereby the Unit is divided into
1000 Parts, and ſo on infinitely: which Parts are called
Decimal or Tench Parts, and any Number of them is
called a Decima! Fraction. So that a Common Frac-
tion may have any Number: for iis Denominator, that
is, an Unit tliere may be divided, or ſuppoſed to be
divided into any Number of Paris: But the Denoni-
nator of a Decimal is always 10, 100, 1000 or 10000,
& c. that is, it conſiſts of I with any Number of o's
annexed; but ſuch Denominators being never expreſs’d,
all Operations with them are render'd almoſt as eaſy
as wich Integers, and there is thereby ſaved a vaſt
Trouble, which would be inevitable by uang the com
mon Expreſſion.
Notation of Decimal Frations.
RULE. Firſt fer down the Numerator of the
Fraction, whether Proper or Improper, then conſider
how inany o's there are in the Denominator, and bea
ginning at the Place of Units, or right hand of the
Nurnerator, reckon towards the left as many Places
as the Denominaror contains o's; and if there are 110t
to many, ſupply the Defect with o's ſet on the left of
tie ſignificant Figures of the Numerator, and for the
laſt
Of Decimal Fractions. 103
laſt Place of the Denominator (in which always ſtands
I) make a Point thus (.). If the Fraction is Impro-
per, then an equal Number of Figures with that of
the o's in the Denominator being pointed off, the Fi-
gures remaining towards the left are the Integral Part,
and the other the Decimal: All which ſhall be exem-
plify'd by Fractions tranſlated from the Common to the
Deciinal Form.
To is expreſs'd in the Decimal Form 5.
Ext. 2. 765 is .54. Exa. 3. 1ooo is .026.
Tool is .0035.
Exa. 5. 1& is 7.5. Exc. 6. 3.784 is
37.84 Exa. 7. 431867 is 431.267. Exa. 8. Todoo
is .000001.
Exa. 40
II. How to read a Decimal Freilion.
RULE. Take the whole Rank of Figures in the
Decimal (neglecting the o's that are prefixed to the
left of the lignificant Figures) for the Numerator, and
for the Denominator take I with as many o's as the De-
cimal contains Figures both ſignificant and Cypheis.-
Exa. 5. is read ; .026 is To8; .0035 is 100;
000001 is 7500506, C. So that you ſee they are
hereby reduced back again from the Decimal to the
common Form. But there are ſeveral other Ways of
reading or expreſſing (in Words) a Decimal Frac-
tion
1. By calling the Figure next to the Point ſo many
Primes, the ſecond Figure ſo many Seconds, the third
Figure ſo many Thirds, 6c. Thus the Decimal .5873
read after this Way is 5 Primes, 8 Seconds,.7 Thirds
and 3 Fourths.
2. Others would expreſs the ſame Deeimal thus, 5
thouſand 8 hund. feventy-three Fourths.
3. The other Method, as practiſed by ſome, is no-
thing elſe than calling it ſo much of a Decimal as the
Numerator expreſſes ; ſo the laſt propoſed Decimal
read this Way is 5 thouſ, 8 hundred ſeventy-chree of
104 Of Decimal Fractions.
a Decimal, or it is five, eight, ſeven, three of a Dei
cimal; which, tho' not the moſt natural, I think the
ſhorteſt and eaſieſt Method of expreſſing them, and
as intelligible as any other, eſpecially if their Nature
and Genelis be well underſtood.
And now for further Explication of what I have al-
ready ſaid, the Learner is carefully to conſider che fol-
lowing Remarks
1. A Decimal Fraction is fo many roths of the
whole, or ic is ſo many r'o ths of a roth, or ſo many
loths of 1 of 16, &c. the two laſt of which are
Compound Fractions, and being reduced you have
their equivalent Simple Expreſſion; for Exa. 1} of 15
of
1ő is made in the common Form, and .003. in the
Decimal. So this Decimal Fraction .6852 (ſuppoſe
of 1 L.) is 6 tenth Parts, 8 hund. Paris, 5 thouſ.
Parts, and 2 ten-thouſ. Parts; or it is 6 tenths, to-
gether with 8 tenihs of 1 of the laſt cenths, with 5
tenths of 1 of the laſt tenths, with 2 tenths of 1 of
the laſt tenths.
2: A Decimal Fraction being written always without
its Denominator, is diftinguiſh'd from a whole Num-
ber by a Point, called by ſome the Separatrix, prefixt :
but ſome Authors uſe a Comma (,), others a perpen-
dicular Line :0), and others this Mark (L), and ſome
uſe other Symbols; but I prefer the Point, as being
the ſimpleſt, and alſo becauſe it is moſt common.
3. As the Value of a whole Number is increaſed
in a decuple or tenfold Proportion, by annexing a fig-
nificant Figure or Cypher to the Place of Units or
right hand; fo by prefixing any Figure to the left of
a Decimal (and right of the Poin) its Value is de-
creaſed in a ſub-decuple Proportion. Thus 6, which
is ſo many Units or ones when ſtanding by itſelf, if
you annex to it a ſignificant Figure (ſuppoſe 3:) or a
Cypher (o), then its value is changed from 6 to 60,
and is read 63 and 60: ſo if 6 is a Decimal. it is .6.
or
Of Decimal Fractions.
IOS
or o, but if you prefix 3 to it, then it is only too
(and with the 3 prefixt makes 736) its Value being
changed from 60.06 or Too
If inſtead of 3 you
prefix o, it is preciſely ob; if another o is prefixt, it
is only 1ooo or .006; and if you prefix another Og it
it becomes thor or .0006; and by ſtill prefixing o's
or ſignificant Figures to its left, its value is decreaſed
infinitely
4. As Cyphers ſet on the left of Integers neither
increaſe nor diminiſh their Value, ſo Cyphers ſet on
the right of Decimals neither increaſe nor diminiſh
their Value. Thus if you ſhould prefix ever ſo many
o's to 6 being an Integer, for Exa. 0006, it is only 6.
In like manner, ſuppoling 6 ro be a Decimal (thus .o)
if you annex to its rigl: any Number of o's, it wilí
be but .6 or to; becaule whatever Number of o's you
annex, the ſame Number is ſuppoſed to be annexed
to its Denominator, and ſo by cutting off an equal
Number of o's from Numerator and Denominator
(as was taught in C: ſe I. of Reduct. of Com. Fract.) it
is reduced to its firſt State.
5. A Proper Fraction in Decimals hatb all its Fi-
gures ſtanding on the right of the Point, becauſe the
Denominator muſt have more Places than the Nume-
rator, and therefore the Point mult fall without the
Figures of the Numerator; but an Improper Decimal
Fraction has Figures on both ſides of the Point, viz.
the Integral part on its left, and the Fractional on the
right; thus 7% is .37, but 375 is 3.78.
6. A Proper Deciinal Fraction may be divided into
as many leſſer ones as it contains fignificant Figures,
their Numerators being the ſeveral Figures of the Nu-
merator given, and their Denominators having as
many o's as chere are places after the point to theſe
ſeverally. Thus .536 is .5+.037.000; alío .0486
is .04 7:008+.0001.
CHAP ..
106 Addition of Decimal Frattiqrzs.
CHAP. XIV. Of Addition of Des
cimal Fractions.
Ddition of Decimals is the ſame with that of
.whole Numbers, reſpect being had to the true
and orderly placing of the Numbers to be added; for
which obſerve the following
RULE. Whether they are pure Decimals or mixt
(that is, Integers and Decimals) diſpoſe them ſo the
one under the other, that all the Points may ſtand in
one Column, and the Figures in ſeveral Columns,
each according to its Place and Degree; then begin-
ning at the right hand, add together the Figures in
each Column, and for every 10 in the Sum carry I
to the next Column (as in Addition of ſimple abſtract
Numbers in Integral Arithmetic) ſetting down the
Exceſs, and minding to place the point in the Sum un,
der thoſe in the Numbers given.
Exa. I.
Exa. 2.
yds.
L.
.17
.5
.05
.125
.13
.05
137
.8
.25
.384
.076
.025
.605
Sum 2.614
Exa. 6.
137.354
48.057 24-4183
.27
565
5.
573.673
15.138
·3527
.6
.62
46.975
.87904
.033
.1826
21.0.74.104 75.6
650.007
Theſe
Exa. 3.
:56
.908
02
.258
•312
Sum .996
Exa. 4
Sum 1.75
Exa. 5.
3.16
5.0056
18.73
Subtraction in Decimal Fractions. 107
Theſe Examples need (I ſuppoſe) no Illuſtration,
and therefore I ſhall finiſh this Rule, after I have ob-
ſerved, I. That if the Sum of the Decimals in the
firſt Column, viz. to the right hand, is a preciſe Num-
ber of 10's, the o needs not be ſet down, but you may
proceed, carrying your Number of 10's to the next
Column; and you may do, the fame, if there ſhall
happen a o in the next place. (See Exa. 2, 5, 6,
where the 'o's Places are left vacant.) But after a lig-
nificant Figure theſe Cyphers muſt not be neglected,
See Exa. 4. and 6.'
చటం లేటలోలోతులో లేటులో
:
CHAP. XV. Subtraction in Decią
mal Fractions.
RULE. Set down the Minuend and Subtrahend
the ſame way as was directed in Addition; viz.
by placing the two Points the one under the other,
and each Figure of the Subtrahend orderly under thoſe
of the Minuend, and begin at the right hand and ſub-
tract as in Integers, paying. I to the next Place, when
you have Occaſion to borrow 10.
E%0. I. Exa. 2. Exa. 3. Exa. 4.
Minuend
•405
.27348
Subtrahend .462 .032
.006
.198
•738
.0083
..276 373 .0023 .07548
1. If one or both is a mixt Number, ſet down the
Integral Part towards the left hand, and the Fractio-
nal on the right, obſerving the above Direction.
Exa. 6.
From 16.2805
60-437
Take .135
7.9 59.262
Exa. 5.
Exa. 7.
143.81
16.1455
135.91
1.075
27
108 Subtraction in Decimal Fra£tions.
2. If the Decimals in the Minuend are fewer than
thoſe in the Subtrahend, then as many o's muſt be
placed, or ſuppoſed to be placed over the Subtrahend's
Figures as will fill up the Defect.
Exa. 8.
Exa. 10
Exa. II.
58.2
12.75
7.25
.6831
•4593 99.999
Exa. 9.
IO,
I.
IOO.
.001
50.95 12.0669
-5407
3. As in Addition, ſo likewiſe here, if a Cypher
ſhall happen on the right of the firſt ſignificant figure
of the Remainder, the fame may be neglectedy as be.
ing of no Uſe.
Exa. 12.
14.137
20.0084
.165
5.237
:7084
Ex. 13.
Exa. 14
-485
-32
8.9
19.3
WASABABBE
CHAP. XVI. Multiplication in De-
cimal Fractions.
ENERAL RULE. Set down the Multiplicand
and Multiplier as in whole Numbers, and mul-
tiply them together as ſuch; then point off ſo many
Figures to the right of the Product as there are Deci-
mal Places in both Factors, and if there happen not to
be ſo many in the Product, prefix as many o's on the
left of the Decimals found, as is the Difference; which
Product, according to the Quality of the Factors,
will be either a pure Fraction, (as in Exa. 1, 2, 3.)
or a mixt Number, (as in Exa. 4. 5.) or an Integer
(as in Exa. 6.)
Exa,
Multiplication in Decimal Fractions, 109
Exa. 3.
Exa. I.
Mult. .63
by 45
Exa. 2.
.046
.005
4
.23
.00023
.92
315
252
62835
Exa. 4.
Ex4. 5
26.185
3.54
Exa. 6.
6.25
.372
68
6.4
2976 10474
2500
2232 130925 3750
78555
25.296
40.
92.6949
1. When a Decimal Fraction or a mixt Number is
to be multiplied by 10, 100, or 1000, &c. there is no-
thing to do but to remove the Point as many Places
towards the right hand as there are o's in the Multi-
plier, (ſuch o or o's on the right of the Product being
to be neglected.) Thus,
If the Decimal Fraction .5736 was to be multi-
plied by
5.736
the Product would be
1000
10000
5736.
For .5736 x 100 = 57.3600, where the two o's are
inſignificant.
In like manner, if the mixt Number 38.754 was to
be multiplied by
387.54
the Product would be
38754
10000, &c.]
1387540, &c.
L
IO?
100
57.36
573.6
IO
3875.4
100
1000
110 Multiplication in Decimal Fractions.
I
.01
produces
.00I
2. When a Decimal Fraction is to be multiplied by
.I, .01, .001, .0001, there is no more to be done,
but to prefix to the Multiplicand as many o's (with
the Decimal Point on the left of all) as there are Fi-
gures in the Multiplier. So
.05736
:5736 multiplied by
.005736
.0005736
dc. .0001J & L.00005736, &c.
3. Becauſe the Work of Multiplication of Deci-
mals, when there are many Places in the Multiplicand
or Multiplier, or both, is very tedious; and ſeveral of
the Decimal Places of the Product when found, being
to be rejected as of no uſe, ſome Authors propoſe a
compendious Method of performing the Work true,
or nearly ſo, to as many Decimal Places in the Product
as you incline, viz. to 2, 3, 4, &c. Places (after the
Point) as you ſhall think you may have uſe for; for
which I ſhall ſet down the Rule, and make an Obſer-
vation or two upon the Advantage that is gain'd by
this Method.
RULE. Having written down the Multiplicand,
conſider how many Decimal Places you incline to
have in the Product, and ſet the Units Place (viz. of
Integers) of the Multiplier under that Decimal Place
of the Mulciplicand which ſtands as far from the
Point as you delign the Product ſhould have Decimal
Places, that is, under the firſt Figure after the Point,
f you would have only one Decimal Place in the Pro-
duct, under the ſecond if you would have 2, &c. then
set the remaining Figures of the Multiplier in the re-
verſe Order, and multiply as uſual; only ift, you muſt
begin with that Figure in the Multiplicand which ſtands
over the multiplying Figure, neglecting (as it were)
the Figures ſtanding to its right. But, 2dly, conſider
what would have been carried from the Product of
theſe right-hand Figures if you had actually multiplied
them, (which may for the moſt part be found by ta-
cirely
Multiplication in Decimal Fractions. I } I
citely multiplying the 2 right-hand Figures next that
you begin with, and carrying the so's of the nearest
Figure thereto.) 3dly, Set down the partial Products,
ſo as they may all ſtand even, or in the ſame Column
on the right, which is contrary to the common Me-
thod, placing the reſt of the Figures in diſtinct Co.
lumns to the left, and then add then together. But
yet, 4thly, in ſo doing you muſt gueſs as near as you
can, what would have been carried from the sum of
the preceding Columns, if none of the Figures in the
Multiplicand had been neglected in the ſeveral particu-
lar Multiplications, ſo as the ſame may be added to
the Sum of the firſt Column. Exa. I. Let it be pro-
poſed to multiply 35.423786 hy 3.628, fo as four De-
cimals in the Product may be true.
Operation at large. Operation contracted.
35.423786
35.423786
826.3
3.628
283390288
70847572
212542716
106271358
1062713
212542
7084
2833
128.5174
128.517495608
If the Multiplier is a pure Decimal, ſet a o in the
Unirs Place (or imagine it fu) and the Fraction in
the reverſe Order, as above. Thus,
Exa. 2. Let 824.6537 bemulriplicd by.4657, fo as
there may be three Decimal Places true in the Produkt,
Operation at large. Operation contracted.
824.65 37
824.6537
.4057
756.4.0
57725759
329861
41232685
4.9479
49479222
4.123
32986148
577
384.04.122809
384.04.1
Exa.
112 Multiplication in Decimal Fractions.
Exa. 3.
Let 59.48576 be multiplied by .0473, ſo
as to have 3 Decimal Places exact in the Product.
Operation at large. Operation contracted.
59.48576
59.48576
04.73
3740.0
17845728
41640032
23794304
2379
416
17
2.81367644.8
2.813
When there are not ſo many Decimal Places in the
Multiplicand as you would have in the Product, an-
nex as many o's to the Multiplicand as you want; and
it ir is an treger, you muſt reckon the Cyphers added
mbe Decimals, and therefore a Point is to be ſet be-
[wixt them and the Integral Part.
And now you may obſerve (which is very obvious)
that after we have multiplied by this contracted Me-
thod, it is altogether a Gueſs wbat Allowance to make
when we come to add, for the Increaſe of the ne-
glected Figures of the Multiplicand, if they had been
actually inultiplied, and their ſeveral Products ſer down
and added according to common Form; and conte-
quently we may err ſometimes in Excefs, ſometimes in
Defect, (for it is impoſſible to preſcribe a Rule which
will be exact but in a few Cales) tho' the Difficulty
in making Allowance when multiplying is not ſo great.
Therefore it is not every one that can promiſe upon
(perhaps) tolerable Exactneſs in uſing this Method; but
only ſuch as have had much Practice therein, and can
readily judge what Allowance to make in every com-
mon Caſe.
2. That when we incline to be pretty exact, the
beſt Way is to multiply a Figure or two more chan
we intend the Product ſhould have of Decimal Places,
and after Addition to caſt away chefe as uſeleſs.
3:
Multiplication in Decimal Fractions. It 3
3. That in fome Calculations, where much Accu-
racy is required, this Method ought not to be uſed at
all. And even in moſt caſes, when we uſe the Deci-
mal Way, I would take the common Method of mul-
tiplying, I mean by ſetting the Multiplier in the uſual
Form; and if the Multiplicand has many Decimal
Places, I'd begin to multiply at its third, fourth, &c.
Place after the Point, as I ſhould judge convenient,
rejecting all the Figures to the right of theſe as fuper-
fluous, and in the mean time make ſome Allowance
for the Increaſe, as nearly true as I could gueſs.
You are further to take norice, that as in Multipli-
carion of Common Fractions, when both Factors are
proper Fractions, the Product is leſs than either; fo.
in Multiplication of Decimals, when the factors are
borh Proper, that is, when they have no Figures on
the left of the Point, the fame Effect is produced, for
the Reaſon there alligned. Vide Exa. 1, 2,
કછ છછ છછછછછછછછ છછછછો
CHAP. XVII. Diviſion in Decimal
Fractions.
D Tviſion of Decimals being reckoned the hardeſt
Task in Commori Arithmetick, I ſhall endea-
vour to be very particular in it, and to render it as
plain and intelligible as Words will do it, and my de-
ſigned Brevity will permit.
The Difficulty then lies not in the Diviſion itſelf,
(for that is the ſame in every reſpect wih Integers)
but only in qualifying the Quote after the Diviſion is
over; that is, in knowing certainly, whether it be all
Integral or all Fractional, or a mixt Nuniber ; and if
the laſt, what the Integral and what the Fractional
Parr is.
L3
RULE,
114 Diviſion in Decimal Fractions.
RULE.
Conſider the Dividend and Diviſor as
whole Numbers, without regarding the Points, and
divide in every reſpect as if they were ſuch: But be-
fore you begin, if the Diviſor contains a greater Num-
ber of Figures than the Dividend, you muſt annex as
many o's to the Dividend as will at leaſt make their
Number equal, (or you make the Dividend greateſt,
which if it is a whole Number, the o's added are to
be reckoned its Decimals, and muſt have a Point pla-
ced before them;) then proceed in your Diviſion, by
adding o's to the Remainders, 'cill you have carried
the Quote as far as ſhall be neceſſary; and v hen the
Diviſion is finiſh'd, the Quote muſt be qualify'd after
this manner, viz. Conſider how many Decimal Places
there are in the Diviſor, and alſo how many in the
Dividend, including in this laſt all the o's you added
and made uſe of; then, becauſe the Number of De-
cimal Places in both muſt either be equal, or the Di-
viſor has moſt, or the Dividend exceeds the Diviſor,
(for there is not a fourth Caſe ;) therefore,
1. If the Number of Decimal Places in both are
equal, the Quote is a whole Number. Vide Exa. I,
2, 3, 17.
2. If the Diviſor has moſt Decimal Places, annex
as many o's as is the Difference, to the right of the
Quote, and it is a whole Number. Vide Exa. 42 52
6, 10, 13, 15, 16.
3. If the Dividend has the greateſt Number of De-
cimal Places, take the Difference betwixt them and
thoſe in the Diviſor, and point off as many to the
right of the Quote, ſupplying the Defect (if any be)
with a o or o's, and the Quore will either be a Frac-
tion or a mixt Number. Vide Exa. 70 8, 9, 11,
12, 14
Exa,
Diviſion in Decimal Fractions. 115
Exa. I.
Exa. 2.
2167).835 (5 .2007).2996 (428
835
28: :
1
14
19:
14
5
56.
Exa. 4.
Exa. 3.
14.0032) 1036.2368 (74
980224
(253) 88.55 (350
759
560128
560128
1265
1265
Exa. 5.
10872) 17.44 (200.
17 44
Exa. 6..
.000035).140 (4000
140
Exa. 73
7:
36) .20808 (578
18o
Exa. 8.
3.8) 100.85.2 (26.54
76
.
280
252
248
228
288
288
205
190
152
152
Exa?
116 Diviſion in Decimal Fractions,
Exa 9.
Exa. 10.
:06).804 1.13.4
6
0125).500. (40
500
+
20
18
medad
2 ito
24
EX. II.
:5).0125 (.025
Exa. 12.
100) .100 .001
IOO
IO
25
25
A
Ex4. 13:
Exa. 140
Exa. 15
1) 100 (1000
4) .20 1.05
2) 4 (20
I
20
4
00
Exq. 16.
.000001) 1 (1000000
Exa. 17.
,05).10 (2
ΙΟ
The Rule itſelf is ſo very plain, that I ſhall not uſe
any Iluſtration for the above Examples, but proceed
to lay down another eaſy Method of knowing the
true Place and Degree of the Quotient Figures; and
it is this:
Conſider under what Figure of the Dividend the
Units Place of the Diviſor would ſtand at the firſt
Demand, the firſt Figure of the Quote is always of
that ſame place and Value, by which are eaſily known
the Places of che other Figures of the Quote.
This will be beſt underſtood by Examples; and to
let you ſee the Conſonancy of this Method with the
preceding; I ſhall repeat ſome of thoſe already addu-
ced
Diviſion in Decimal Fractions.
117
ced, and begin with Exa. I. viz. .167).835 ( where
becauſe they are pure Fractions, I prefix a o to each
for their Place of Units, thus 0.167) 0.835 (
and find that the Place of Units of the Divi: 0.835
for ſtands under the Place of Units of the Din 0.167
vidend, as in the Margin; wherefore I conclude
according to the Rule, that the Quote 5 is in the firſt
Place of Integers alſo.
In Exa. 6. the Units Place of the Diviſor ſtands
under the fourth Place (or Place of Thouſands of lu-
tegers) of the Dividend, con-
fequently the firſt Figure of 0.000035) 0000.140
the Quote muſt be four thou-
0.000035
fand; and becauſe no other
Figures follow it, I annex three o's to make it ſo.
The Effect of theſe two Rules is one and the ſame;
you may therefore uſe either of them you incliné,
only you need not ſer down the Diviſor under the Di-
vidend after the above Form, but you may imagine
it fo: I have done it for no other Reaſon than to lec
you see the Method.
1. When pure or mixt Decimal is to be divided
by 10, 100, 1000, or 10000, Quc. remove the Point
as many Places towards the left as there are o's in the
Diviſor, and it is done. Exa. Let the mixt Num-
ber 3875.4 be divided by
IO
387 54
38.754
and the Quote is 3.8754
3875+
.038754
Which you may prove by actually dividing and quali-
fying the Quote according to the Rule.
2. When a Decimal Fraction or a mixt Number is
to be divided by .I, .01, .001, .0001, doc. remove
the Decimal Point of the Dividend as many Places
towards the right as there are Figures in the Diviſor:
Thus if .38754 was divided by
100
IOOO
10000
100000
I
118 Diviſion in Decimal Fractions.
I
.ΟΙ
.00 I
I 지
​3.8754
the Quote would be
38.754
387.54
.0001
3875.4
3. Hitherto I have ſuppoſed no Remainder. I ſhall
next thew how to value the Quote exactly, when any
thing happens to remain after the Diviſion.
RULE. Proceed in your Diviſion as is before
taught, 'till all the Figures in the Dividend are taken
down; and if the Quote is not then carried to a ſuf-
ficient Number of Places, ſet a o to the Remainder,
and divide, placing another Figure in the Quote;
and if this is not ſufficient, add another o to the laſt
Remainder, and ſo go on by adding o's to the Re-
mainders, and dividing, 'till your Quote is carried to
as many. Places as ſhall be judged neceſſary (qualifying
the ſame as is already taught,) and value your laſt Re-
mainder thus;
Make it the Nuinerator of a Fraction, whoſe De-
nominator is the Denominator of the Dividend, taking
in all the Cyphers added and made uſe of in the Ope-
ration, and you have the true Remainder; it matters
not whether you expreſs it in the Common or Deci-
mal Form, which Remainder if you would add to the
Quote already found, to make the ſame complete, you
muſt
Take the Remainder in its true Value, as above,
and divide it by the Diviſor, taken alſo in its true Va-
lue, and the Quote is the additional Part ſought. Or
you may do it after this Manner; ſet the Remainder
for the Numerator of a Common Fraction, whoſe
Denominator is the Diviſor, if the Number of Deci-
mal Places in both is equal. Vide Exa. 24. But if the
Dividend has moſt, take the Difference and annex as
many o's to the Denominator. Vide Exa. 18, 19, 20.
Laſtly, if there are more Decimal Places in the Divi-
for, take the Difference, and annex as many o's to the
Numeratory, and you have the Remainder exact in the
3.
com-
Diviſion in Decimal Fractions. 119
common Expreſſion, to be added to the Quote already
found, to make the ſame complete. Vide Exa. 21, 22.
Exa. 18.
-46).942(2.04787760061.358)487-3(1361.17+3557
92
Exa. 19.
I 14
35800
358
220
184
1293
1074
360
2190
322
2148
420
380
368
358
12
620
Exa. 20.
358
80.279)3876.1375(48.28333877790663
321116
2620
2506
664977
642232
14
227455
160558
668970
642232
267380
240837
In Exa. 18, after I have
carried the Quote to 5 Places
and qualified it, there re-
mains 12, which is .000012
or in a Vulgar Fraction
105076, and being to add
it to the Quote 2.0478 to
make the ſame complete,
divide it by the Diviſor 46
(taken in its true Value, viz.
.46 or 48, and the Quote
is 430 공중 ​for the additional
Member. Or I find it thus ;
I make the Remainder 12 a
Numerator to the Diviſor
46,
265430
240837
245930
240837
5093
120 Diviſion in Decimal Fractions.
46, thus ž, and becauſe the Decimal Places of the
Dividend exceed in Number thoſe of the Diviſor by
4, I annex 4 o's to the Denominator, and it becomes
Fodobé as before.
Take alſo theſe following Examples.
Exa. 21.
1.005)1748.28(349650+3=349656 preciſely.
15
Exa. 22.
..0006)25384.31423070001100=42307166%
24 - 24
20
48
13
12
45
.
18
18
32
30
43
28
25
42
1
Rem. 1=500
3
Ex. 23
Exa. 242
..001) 4387.6(4387600
4
8.37) 74896.8 (8948 +354
6696
3
3
7936
7533
8
8
4038
3348
olaalver cool we
7
7
6900
6696
6
6
Rem. 2045494
06
Diviſion in Decimal Fractions. 121
Obferve 1. When the Number of Decimal Places
made uſe of in the Dividend is equal to, or greater
than, that in the Diviſor, the Remainder will always
be a Proper Fraction, as in Exa. 18, 19, 20; ſee alſo
Exa. 24, where the Number of Decimal Places in
both being equal, the Remainder with the Diviſor is
a Proper Fraction, the reſt of the Quote being Inte-
gral.
2 When the Dividend has feweſt Decimal Places,
the Remainder will be an Improper Fraction, as in
Exa. 21, by which you ſee, that if the Diviſion was
carried further on, the Integral part of the Quote
would become greater; ſo that by reducing this Im-
proper Fraction to a whole Number, and adding the
fame to the Quote (already found) in its proper Place,
the Quote would have all its Integral Figures, and the
Remainder, if any were, would be a Proper Fraction,
as in Exa. 22.
Moreover, if the additional Member is an Improper
Fraction equal to a whole Number, it ſhews, that af-
ter ſo many Steps as the whole Number has Figures,
the Diviſion would have been complete without a
Remainder, as in Exa. 21.
3. That therefore the firſt part of the Quote may
not want 1 of the Truth, and conſequently the Re-
mainder may be a Proper Fraction, the Diviſion is to ·
be continued, 'till the Number of Decimal Places in
the Dividend is equal to, or excecd, that in the Divi-
for, unleſs the Divition is finiſhed without a Remain-
der before you conie to that; for then the Quotc qua-
lified according to the Rule will be complere. See
4. When the Quote is carried to a ſufficient Num-
ber of Decimal Places, according to the Nature and
Circumſtances of the Queſtion, and in common Caſes,
when the Quote is not to be multiplied, nor very
much Accuracy required, five or fix Places after the
Point will be ſufficient; tho' the farther the Diviſion
Exa. 23
M
is
122 Reduction in Decimal Fractions.
is continued, the Quote will ſtill be the more exact,
and want leſs of its true Value) the Remainder may be
fafely neglected as of no uſe. But how far the Divi-
fion is to proceed muft be left to every one's owo
Diſcretion, fince no Rule can be preſcribed for that
Purpoſe. However, you'll find, when we come to
Reduction of Decimals, how to manage the Quote,
so as it ſhall not want an Unit of the loweſt Denomi-
nation, nor any aſſignable Fraction of iis true Value.
00:00 00:00
CHA P. XVIII. Reduction of Decimals.
REduction of Decimal Fractions we ſhall conſider
under the following Caſes, viz.
1. To reduce a Common (Proper) Fraction to a De-
cimal one equivalent, or nearly fo.
2. To reduce a Decimal of an higher Species to
an equivalent one of a lower.
3. To reduce a Decimal of an inferior Species to
an equivalent one of an higher.
4. To reduce Integers of a lower Species or Deno-
mination to equivalent Decimals of higher Species.
5. To find the value of a Decimal Fraction in In-
tegers of known Species.
6. To reduce a Decimal Fraction to a Common
Ona.
CASE I.
To reduce a Common Fraction to a Decimal.
RULE. Say, by the Rule of Three Numbers, as
the Denominator of the Common Fraction is to its
Numerator, fo is 10, 100, 1000, @c. (as you intend
the Decimal thould have Places) to the Anſwer.
Or rather thus:
Add a competent Number of o's to the Numera-
cor of the Cominon Fraction (viz. 'till it be equal to,
Or
Reduction in Decimal Fractions. 123
or greater than its Denominator,) and dividing by the
Denominator, the Quote is the Decimal fought. Ob-
ſerve, that if the Diviſion is not finiſhed, after you
have made uſe of all the o's you added to the Nunc-
rator, you may annex more to the Remainders (as was
hinted at already) and continue your Diviſion as far as
you pleaſe, qualifying the Quote according to the Rule.
Allo ſet a Point betwixt the o’s you added to the Nu-
merator and its own proper Figures, that you may not
miſtake when qualifying the Quote, which you do (if
you pleaſe) after this Method, viz. Let the Quore
Have as many Decimal Places as you annexed o's,
wherein if it comes ſhort, let the Defect be ſupplied
with as many o's as is the Difference prefix's.' Or
thus, if one o being added, the Numerator is leſs
than the Denominator, ſer a o after the Decimal Point
in the Quote; and if, another o being added, it is
Rill leſs, fet another o in the Quote, and thus proceed
by placing o's in the Quote 'till you come to the firſt
ſignificanc Figure, and carry on the Diviſion to what
Degree of Exactneſs you pleaſe. Vide Exa. 4 and 5
following
Exa. 1. Reduce to a Decimal.
Say 8: 1 :: 1000: .125 Answer,
Or rather thus,
8) 1.000 (-125 Antwer.
8
Perman
20
16
ndan
40
40
Exa. 2.
Reduce to a Decimal.
4) 3.00(-75 Anſwer.
28
20
20
M. 3.
E.co.
124
Reduction in Decimal Fracti077s,
Exa. 3. Reduce g to a Decimal.
7) 6.0000000
.8571428, &c. the Remainder being &
Exa. 4. Reduce zto a Decimal. .
25) 2.00 .08 Anfr.
200
Exa: 5. Reduce 575 to a Decimal.
476)1.00coc( 0021008
952
480
476
4000
3808
192
In Exa. 3. there remains 4, fo the Quote wants
ghôātost of its complete Value, that is , =.8571428
+ 700077*. In Exa. 5. there remains 192, which,
according to the Rule for valuing the Remainder, is
1050, ſo that the Quote wants #7777övbl of its
complete Value. And fo of others.
Becauſe all Common Fractions cannot be reduced
to their equivalent Decimals without a Remainder
(for every Number is not an aliquot Part of another,
elſe there could be no Remainder) it is neceſſary to
ſhew, how far the Rednction is to be carried, to that
its correſponding Decimal ſhall not want an Unit of
the loweſt Species, nor any aſſignable Fraction of its
true Value.
RULE. Conſider to what Integer or Unit refers ;
and if it has as niany Places (after the Point) as that
Integer has figures when reduced to its loweſt Species,
then the Decirdal found does not want an Unit of the
loweſt Species. For Exa. If the Decimal of 1 L. has
bree Places after the Point, it does not want too,
and conſequently does not want i far, which is greater,
viza, TL, of i L. In like manner, if the Decimal
of
Reduction in Decimal Fractions. 125
of a C. is carried to fix Places after the Point, it can-
not want 1000 of 1 C. confequently it cannot
want i gr. of a Dram, which is greater, viz. 114683
of i C.
Now if a Decimal is to be multiplied by any Nun-
ber (ſuppoſe of 6 Places) that the Product may not
want an Unit of the loweſt Species, let the Decimal
be carried to as many places after the Point, as is the
Sum of the Figures of the Multiplier to the Number
of Figures of the loweſt Denomination that make an
Unit of the Denomination of the Decimal. So if
the Decimal of 1 L. was to be multiplied by a Num-
ber of 6 Places, let the propoſed Decimal be carried
Places (before you begin to multiply) and the
Product ſhall not want i far. of the Truth. And ſo
of others. Now that the Quote may not want any af-
fignable Fraction of its true Value, carry the ſame to
as many places as is the Number of o's in the Deci-
mal Denominator of the aſſigned Fraction. Thus,
let be reduced to a Decimal, which ſhall not want
IoJoDo of the Truth.
7) 3.00000( 42857 Here the Remainder is 1,which
28
according to the Rule
is toouoo conſequently
the Quote wants Jā Too
14
which is leſs than 15050.
to
1
y
20
1
60
56
40
35
50
49
In like manner, if the Decirnal of 1 L. muſt not
want Touhoć of a far. let the Quote be carried to
8 Places, and is thall not want the propoſed Fraction.
M3
The
126 Reduction in Decimal Fractions.
The Reaſon is; if it be carried to 3 Places it cannot
want a far. therefore if it is carried s Places further,
then it is carried to as many places as the Denominator
of the propoſed Fraction has o's, and conſequently can-
not want that Fraction.
In dividing a whole Number by another, if there
happens to be a Remainder, it is oftentimes conve-
nient (eſpecially if the Quote is to be multiplied) to an-
nex o's, ſetting a Point immediately after the Integral
Part of the Quote, and to continue the Diviſion as far
as is neceſſary, the Figures found in the Quote, by
this Addition of o's to the Remainders, being all De-
cimals, and the more there are of them, the Quote is
the nearer to be complete.
Exauples.
58)3682 (63-4827 24) 7827(326.125
348
72
202
62
43
174
280
232
147
144
480
464
30
24
160
60
48
116
4.40
120
406
I20
Rem. 34
In Exa. 1. there remains 343 which is .00342 ſo
that the Quote wants 3750** of its complete Value.
In Exa. 2. after the Integral Diviſion is over, there
remains 3, to which annexing a o, and dividing, I place
1 (after a Point) in the Quote, and there remains 6;
to
Reduktion in Decimal Fractions. 127
to which I annex another o, and ſet 2 in the Quote;
laſtly there remains 12, wherefore I annex another on
and ſetting 5 in the Quote, the fame is compleat with
out any Remainder.
In like manner, if the Dividend is a ſimple appli-
cate Number, and the Diviſor abſtract, inſtead of re-
ducing the Remainder to the next known Denomina-
tion, and dividing, you may carry on the Diviſion de-
cimally, and the Decimal Part of the Quote is equiva-
lent (or nearly ſo) to theſe inferiour Species, which
would have been found by reducing the Remainders,
and carrying on the Diviſion vulgarly: Take an Exa.
done both Ways. What is the 7th Part of L.248 ?
Vulgarly.
Decimally.
7)248 (35 L. 7) 248 (35-42857
21
21
38
38
35
35
1
3
20
30
28
20
7)60(8 .
56
14
60
4
12
50
7)48 (6 d.
42
40
35
6
4
50
49
1
7) 24 (3 far.
1
21
3
The
12:8 Reduction in Decinial Fractions.
The Value of the Decimal .42857 (as ſhall be taught
preſently) is 8.1). 6 d. 3 far. and ſomewhat more, as
by the ocher Method.
Noie, This Queſtion may be ſhorter wrought, thus,
Vulgarly.
Decirnally
7)248
7)248
35: 8: 6:33
35-42857, c.
2. In moſt caſes, tho' you ſhould continue the Di-
viſion to ever ſo many Places in the Quote, there will
ſtill be a Remainder, and the Quote will never be
complete, becauſe there will happen a Remainder the
ſame with the former; ſo that all che intermediate Fi-
gures in the Quote betwixt theſe two Remainders,
will conſtantly be repeated, the Diviſion going on for
ever. Such Decimals we call circulating or repeating
Decimals, becauſe of the continual Return or Repe-
tition of the fame Figure or Figures; as alſo infinite
or indeterminate Decimals, becauſe the Diviſion can
never come to an end, and conſequently the Deci-
mal wants ſtill ſomething of the true Value of the
Common Fraction. For Exa. reduced to a Dcci-
mal is -333, &c. infinitely, for
3) 1.000
.
333, &c.
Where the Dividend being always 10, the Quote
muſt ſtill be 3. In like manner, f in a Decimal Ex-
preſlion is .666 infinitely, becauſe the Dividend is al-
ways the ſame, for
3)2.000
.666, &c.
Likewiſe in a Decimal is .428571428571, bir
the Figures 428571 being continually repeated.
7)
Reduction in Decimal Fractions. 129
7) 3.000(-4285714, here the Circulation begins,
28
20
14
60
56
40
35
50
49
IO
7 영
​Amen
30. the ſame with be firſt Dividual,
28
2
So that whenever the Quote begins to circulate, you
need proceed no further in the Diviſion; but obſery-
ing what Figure or Figures would ſtill be repeated,
fet down as many in the Quote as you pleaſe.
3. The Diviſion needs only to be concinued ſo far,
that the Defect may be inconliderable, and the Re-
mainder neglected as of no Value, the Method of
which has been already taught, p. 124.
CASE II.
To reduce a Decimal of an higher Species to an e-
quivalent Decimal of a lower.
RULE. Multiply the given Decimal by ſuch a
Number of Units as makes 1 of your Decimal De-
nomination, and that Product by ſuch a Number of
Units as makes i of the laſt Denomination, and ſo on
till you come to the loweſt required, minding to point
off
130 Reduction in Decimal Fractions.
off as many for Decimals in the laſt Product of all, as
there were Places in the given Decimal.
Reduce .005 L. to the Decimal of id.
.005
Anfr. 1: 2 d.
Exa. 1.
20
OI00
Here the Anſwer is a mixt Num-.
ber, or an Improper Decimal.
12
1.200
Exa. 2. Reduce .05 L. to the Decimal of 1 d.
.05
20
Anfr. 12d. Here the Anſwer
is Integral.
100
I2
EXA. 3:
12.00
Reduce .153 C. to the Fraction of 1 L.
.153
4
_28
612
28
Anfr. 17.136 C.
4896
1224
17.136
CASE III.
To reduce a Decimal of an inferiour Species to an
equivalent of an higher.
Rule. This caſe being the Converſe of the laſt,
muſt require a contrary Operation, and therefore di-
vide here as you multiplied there.
Exa. I.
Reduce 125 far. to the Decimal of i L.
4)
Reduction in Decimal Fractions. 131
.
28}
4).125 f.
12).03125 d.
20).00 2604, &c. .
.0001302, br. Anfr.
Exa. 2. Reduce .847 d. to the Decimal of 1 L.
12.847 d.
20.0705833, doc. a.
.0035291, bc. Anfr.
Exc. 3. Reduce :16 oz. to the Decimal of 1 gr.
of I C.
161.16 oz.
4
01 lib.
7.0025
.00035714@c. Anſi.
CASE IV.
To reduce an Integer of a lower Species or Deno-
mination to an equivalent Decimal of an higher
Species.
RULEI. If it is a simple Number, turn it firſt
into a Common Fraction, and then reduce that Com-
mon Fraction to a Decimal by this Chap.
Exa. I.
Reduce 7. Jh. to the Decimal of 1 L.
7. is zā L. therefore, 20) 7.0
:35 L. Anfr.
Exa. 2. Reduce 3 far. to the Decimal of 1 L.
3 far. is gol, therefore,
960) 3.0000 (.003125 L. Anfr.
2880
1200
960
2400
1920
4800
4800
Exa
132 Reduction in Decimal Fractions.
Reduce 5 oz. to the Decimal of 1 L,
5 oz. is lib. therefore,
16) 5.00(.3125 lib. Anfr.
Exa. 3.
48
20
16
40
32
80
Exa.
.
4.
Reduce 1 oz. to the Decimal of 1 C.
I oz. is 1772 of 1 C. therefore,
1792) 1.00000(.000558, Qc. C. Anfr.
8960
10400
8960
14400
14330
64
Exa. Reduce 1 d. to the Decimal of 1 L.
Tá d. is te of iż of zo=787L. therefore,
3840) 9.00000.00234375 L. Anfr.
7680
13200
11520
16800
15360
14400
11520
28800
26880
1920
1920
But
Reduction in Decimal Fractions. 133
But ſuch Reduction is ſooner made by dividing gra-
dually from one Species to another, as ſhall be illuſtra-
ted by repeating the preceding Examples.
Exa. 2.
41 3.000
16
415.000
• 750
S41.250
.0625
.003125 L.
Exa. 3.
Exa, I.
20) 7.00
:35 L.
I2
20
•3125 L.
Exa. 4.
Exa.5.
16
16541 2.25000
28}
4
}
1.000000
4
9.00000
4 .250000
4 .062500
121.56250
7 .015625
.20) 0468875
41 .0022321
.002344375
.0005580 &c. C.
2. But if the Number to be reduced is mixt, that is,
if it conſiſts of ſeveral Denominations, then firſt reduce
each of them to the Decimal of the Denomination re-
quired, and add theſe together for the Anſwer. Or,
ſecondly, reduce the mixt Number to a ſimple one of
the loweſt Species mentioned, and thence into a De-
cimal.
Or, thirdly, reduce the Number of the loweſt Spe-
cies to the Decimal of the next ſuperiour (whether
there is any Number of that ſuperiour Species in the
Queſtion or not) and to it add the Number of that
Species in the Queſtion (if there is any) and reduce
the Sum to the next higher Species, adding to the De-
cimal laſt found, the Number of that Species given in
the Queſtion, and ſo go on till you come to the de-
ſigned Integer
Exa. I. Reduce 7 ß. 11 d. 1 f. to the Decimal of
I L. By the firſt Method I work thus; 7 s. is z L.
=:35 L.next 11 d. is 24: L. = .04583333&c. L. and
laſtly i far. is ooL.=.001041666 &c. Their Sum is
-39687499 (or.396875, becauſe the two laſt Figures
are o's) for the Anſwer.
N
Ву
134 Reduction in Decimal Fractions.
*
12
20
By the ſecond Method I reduce the 7. lh. II d. if
to far. and they make 381, which is 200 L.=.396875
L. as before.
By the laſt Method I work thus; 1 far. is 25 d. to
which I add 11 d. making the Sum 11. 25 d. This I
divide by 12 and find .9375 M. to which adding the
4 1.00000 71. I divide the Sum by 20, and
II.25000 the Quote is .396875 L. as before.
7.93750
And ro of others.
-396875
Note 1. The Numbers taken in, are marked with an
Aſterisk. 2. The eaſieſt and exacteſt of theſe Me-
thods I take to be the laſt, becauſe of the great Trou-
ble that ariſes from the Reduction and Diviſion in the
two firſt Methods, eſpecially the firſt, where there
are ſo many Diviſions.
In like manner, I gr. 19 lib. 5 oz. 10 dr. reduced to
the Decimal of 1 C. is .422781808, thus
10.000000 dr.
16
2.500000
16
4
5.625000 oz.
4
1.406250
19.35156250 lib.
7 4.837890625
4
1.691127232 grs.
.422781808 C. Anfr.
But there are ſeveral other Methods for ſuch Re-
ductions, as iſt in general, by Decimal Tables of Mo-
ney, Weight and Meaſure, &c. framed and conſtrued
for the purpoſe, ſhewing the Decimal of any inferiour
Species in the Denomination of any ſuperiour (of the
ſame kind) which Tables, tho' commonly met with
in Books on this Subject, I ſhall not infert'; ſave that
of Money, becauſe they are ſeldom uſed.
*
}
*
284
28}
Decimal
Reduction in Decimal Fractions. 135
I
2. I
61.3
Decimal Table of Money.
Integer 1 L. Integer 1 L.
I .0010416-11.11.05
2 .0020833–
3.003125
3.15
Penny. 1.0041666 4.2
2-0083-
5.25
3 .0125
41:016-
71:35
51.22083–
6
·025
9.45
7.02916–
101.5
8
03333
II.55
91.0375
10.04166
13.65
II .04583
--- 14.7
15 1.75
161.8
17.85
8.4
12.6
18!.9
19.95
Note. The Sign annexed to ſome of the Deci-
mals, fignifies that they are too licule, and not exactly
equivalent to their correſponding Integers.
2. More particularly, by finding the Decimal of an
Unit of any inferiour Species in the Denomination of
any ſuperiour (when it can be found without a Re-
mainder, which more feldom happens) and uſing the
ſame as a Standard for finding the Decimals of the reſt
of that inferiour Species. Thus the Decimal of 1 fb.
in the Denomination of 1 L. being found to be .05;
any Number of Shillings multiplied by it gives their
Decimal Value; for Example, 9 h. reduced to the
Decimal of iL this way is 45 L. for 9x.05.45.
And ſo of others.
N2
But
1:36 Reduction in Decimal Fractions.
But there is another eaſier Method of reducing
Shillings to Decimals of 1 L. thus: If the Number of
Shillings is even, take its Half, and ſet a Point before
it if odd, ſuppoſe a o annexed, and then take its Half.
Thus 6 . is 3 L. and 13 fb. is 65 L.
In like manner, the Decimal of 1 Penny in the De-
nomination of 1 L. being .004166, &c. any Number
of Pence multiplied hereby gives their Decimal Value
in the Denomination of i L. but becauſe this Standard
is incomplete, we muſt make ſome Allowance when
multiplying, according to the Figures multiplied.
Alſo the Decimal of 1 far. in the Denomination of
i L. being .00104166, &c. this multiplied by 2 gives
.00208333, &c. for the Decimal of 2 far. And ſo on.
After the ſame manner you may make up Tables
for all our common Weights and Meaſures, having ſtill
a due Regard to the Circulation of the Root (if the
fame happens to be indeterminate) yet a ſeparate Re-
duction, tho' more tedious, is the ſafeſt way.
CASE V.
To find the Value of a Decimal Fraction in Inte-
gers of known Species.
RULE. Multiply the given Decimal by the Num-
ber of Units of the next inferiour Denomination, as
make an Unit of your Decimal Denomination; and
then cut off ſo many Figures towards the Right of the
Product, as there are places in the Decimal; the Fi-
gures ſtanding to the Left of thoſe which are cut off,
being ſo many Integers of that inferiour Denomination;
and thoſe on the Right are the Remainder, which you
muft multiply by ſuch a Number of Units of the next
inferiour Denomination, as make an Unit of the laſt
Denomination, and point off as before, and thus go
thro' all the Denominations, till you come to the
loweſt.
Еха, .
Reduction in Decimal Fractions. 137
Exa. 1. What is the Value, Exa. 2. What is the Value
of
of
.65 1.
•546 L.
20
I2
. 10.920
d. 7.80
4
I2
d. 11.040
f. 3.20
Anſw. 10 ſh. 11 d. 16. Anfr. 7 d. 3 to far.
Exa. 3. What is the Value Exa. 4. What is the Value
of
of
48 d.
4
4
•384 C.
f. 1.92
97.
1.536
28
Anſw. 2 f. ferè.
4288
1072
lib.
15.008
16
oz.
0. 128
16
768
128
dr. 2.048
gr. lib. oz; dr.
Anfr. 1-15-0--27466
N3
Exa.
138 Reduction in Decimal Fractions.
Exa.5. What is the Value Exa. 6. What is the Value
of .4106 of a Scotch Gal.
of .736 of a Year.
13
8
4106
9.568
28
Pints. 3.2848
4
4 544
I136
Mut.
1.1392
4
15.904
24
Gills.
0.5568
2
3616
1808
Halfgills. 1.1136
pts.mut.gills. halfgills.
Anfr. 3-1-0-1702.
21.696
60
41.760
Mon. D. be, min.
Anſ". 9-15--21-41766
The Value of any Decimal of a Pound may be
found, either by the Table or Inſpection ; but on this
I ſhall not infift. See my Treatiſe of Fractions, Page
119, 120, 121.
CASE VI.
To reduce a Decimal Fraction to a common one.
RULE. Set the Decimal for a Numerator of a com-
mon Fraction; whoſe Denominator will be I with as
many o's as there are Figures in the Decimal: which
common Fraction thus found, you may reduce to its
loweſt Terms.
Exa. 1. 56 is 786 or loweſt. Exa. 2. .038 is
or
Exa. 3. .0648 is 16644
Tigo
loweſt: And ſo-of o hers.
38
TO
81
I 62
2500
or
CHAP
Rule of Three Numbers.
139
000:000000000000000000
CHA P. XIX. The Rule of Three
IS S ſo called, becauſe therein are 3 Numbers given to
find a fourth Proportional : tho' ſometimes there
are more Numbers in the Queſtion, which are all fu-
perfluous, as not entering into the Stating, but ſer-
ving only to complete the Senſe. Theſe are always
known by being repeated or mentioned oftener than
once, either in the very fame Words, or in others of
the ſame Import and Significarion.
This Rule being of moſt general Uſein Arithmetical
Calculations, I ſhall endeavour, by a Variety of choice
Examples, to make it very plain and intelligible to the
meaneſt Capacity; and ſhall be the more prolix and
particular in it, as intending to caſt together here ſuch
Queſtions as all other Authors on this Subject treat of,
as belonging to particular Rules: Whereas it is plain
they are nothing but a further Application of the Rule
of Three. By theſe I underſtand what are commonly
called the Rules of Intereſt, Diſcount, Barter, Ex-
change, Fellowſhip, Gain and Loſs, Tare and Trett,
Alligation, &c. As for the Rules of Practice, and
the Rule of Five Numbers, I Thall treat of them by
themſelves.
Now, as a great part of the Difficulty of this Rule
lies in the right ſtating of the Terms, you are carefully
to obſerve the following
DIRECTIONS
Conſider what Name the thing fought is of, viz. if
it is required to find a Sum of Money, a Quantity of
Meaſure, Weight or Time, br. and ſet that Number
ar Term which is like it in the Queſtion middleniost:
then take that Number on which the Demand lies,
and
140 Rule of Three Numbers.
and place it on the Right, and the other (which be-
longs to the Suppoſition, and is always of the fame
Name, either general or particular, with the third or
Riglit-hand Number) ſet on the Left. Thus the ſu-
perfluous Terms, if there are any, will preſently appear:
The Terms being thus ſtated, you muſt reduce them
all to ſimple Numbers, if they are mixt, and the two
Extremes, that is, the firſt and third Numbers, to the
ſame particular Denomination, if they are not ſo al-
ready. Then you are to conſider, whether the fourth
Number ſought (or Anſwer to the Queſtion) ought to
be greater or leſs than the middle Number. If it
ought to be greater, you muſt multiply the middle
Number by the greater Extreme, and divide the Pro-
duct by the leffer ; but if the fourth Number, or thing
ſought, muſt (according to the Senſe of the Queſtion)
be leſs than the middle Number, this laſt is to be mul-
tiplied by the leſſer Extreme, and the Product divided
by the greater. The Quote reſulting from this Divi-
fion is the fourth Term or Anſwer to the Queſtion in
the fame Name or Denomination with the middle
Number juſt when you multiplied, which if in a low
one muſt be brought higher, to the more ready Com-
prehenſion of the Mind.
Note 1. When the firſt Term divides, the Queſtion
is ſaid to be in direct Proportion; and when the third
Term is Diviſor, it is called Indirect or Inverſe.
2. The middle Number is never uſed for Diviſor,
but is always either multiplied by, or multiplies one of
the Extremes, according to the Nature of the Que-
ſtion.
3. When the firſt and ſecond Numbers are of, or
can be reduced to the ſame Denomination, it is not
neceſſary to reduce the third to the Denomination of
the first, but the Anſwer will come out in the ſame
Name with the third. Vid. 24. 35.
4. After you have diſcovered which Term is Multi-
plier to the middle one, it is eaſieſt and ſhorteſt to
multiply by the lefſer of the two.
5. When
Rule of Three Numbers.
141
5. When after Reduction the third Term is 1, the
Queſtion is ſolved by Diviſion alone, without any
Multiplication ; and when the firſt Term is I, it is fol-
ved by ſimple Multiplication without Diviſion. Vide
Queft. 2, 6, 9.
6. Many Queſtions in this Rule are complex, and
require ſeveral previous Operations independent of one
another, before you can come to a Proportion in or-
der to a Solution. Vide Queſt. 20, 27, c.
I have here purpoſely avoided that common Diſtinc-
tion of the Rule of Three into Direct and Inverſe, ſo
much uſed by the Generality of the Authors in this
Science, there being not the leaſt Neceflity for ſuch
Diſtinction to any Perſon who can take up the Mean-
ing and Demand of a Queſtion propoſed. For is it not
eaſy for one of common Senſe to know whether the
Anſwer to a Queſtion ought to be greater or leſs than
the middle Term? If it ought to be greater, it is plain
the lefſer Extreme muſt be Diviſor to the Product of
the ocher two; and if the fourth Term (according to
the Senſe of the Queſtion ought to be leſs, then it is
as clear, that the Product of the other two muſt be
divided by the greater Extreme. The Reaſon of both
which is, that to find a greater Quote, or to make the
Anſwer greater than the middle Term, we muſt take
a leſs Diviſor than the Number we mulciplied by; and
to find a leſs Quotient, or to make the Anſwer leſs
than the middle Number, we muſt take a greater
Number for the Diviſor than is the multiplying Num-
ber; for the greater the Diviſor, the leſs is the Quote,
and contrariwiſe.
Qu. 1. If 4 Yards of Cloth coſt 23 h. what is
the Value of 15 Yards at the ſame rate?
Srate
142 Rule of Three Numbers.
yds. ſh. yds.
ILLUSTRATION.
State 4:23 : : 15
Here I mulciply the mid-
15
dle Number by the greater
Extreme 15, becauſe the
115
Anſwer to the Queſtion
23
muſt be greater, and di-
viding the Product by the
41345
firſt, there reſulteth 86 .
215)816: 3
(the middle Number being
L.4:6:3
jb.) and I over which I re-
duce (mentally) to d. and it makes 12, which divided
by 4 quotes 3 d. So the Anſwer is 86 f. 3 d. or
L. 4:6:3
Qu.2. What is the Value of an Ounce of any thing,
when 8 lib. of the fame coſt 19 f. 7 d. Sterling?
lib. fo. d. oz. Here I reduce the 8 lib.
State 8 : 1947::I
to oz. to correſpond with
16
the third Number, alſo I
reduce the middle Num-
128 '235 (1 d. ber to d. and after divid-
128
ing by the firſt Term,
(for it would have been
107
unneceſſary to have multi-
4
plied by the third Term
the Quote is 1 d. and
128) 428 (3 f.
107 d. remaining, which
384
I reduce to f. and dividing
as before, the Quote is 3f.
44
and 44 remaining; ſo the
Anſwer is i d. 3111 f. or fi loweſt.
2. 3. If I buy 5 Yards of Cloth for L. I--4,
how many
Yards at the ſame Rate may I have for
L. 29-10-3?
I 2
Srate
Rule of Three Numbers.
143
20
20
L. s. yds. L. S. d. Here I reduce the
State 1-4 : 5::29-10-3
Remainder 279 to
qrs. and divide again
by 288, and there
24
590
remains 252, which
I reduce to Nails,
and dividing again, the
288
7083
total Anſwer is 122
5
yds. 3 qrs. 3 nails, and
288) 35415 (122 yds. nail for 4 =
288
12
1 2
661
576
855
576
279
4
) 1116 ( 3 qrs.
864
252
4
1
) 1108 ( 3 nails.
064
144
24. 4. If 7 Men ſpend L. 37--15-6 in 12 Days,
how long will L. 119-5 ſerve them at the ſame rate
of ſpending?
State
· 144 Rule of Three Numbers.
days.
State L. 37–15-6: 12 :: L. 119—5
20
20
755
12
2385
12
9066
28620
12 mid.numb.
9066) 343440 (37 Days.
27198
71460
63462
7998
24
9066) 191952 (21 ho.
18132
10632
9066
Here the 7 Men is the ſuperfluous Term, for the
Anſwer would have been the fame, whatever Number
of Men had been mentioned, ſince the Suppoſition and
Demand are equally affected by it.
Qu. 5. Of what Weight ought the Penny Loaf of
Bread to be, when the Peck of Wheat is ſold for iß.6 d.
if I get 8 oz. for a Penny, the Wheat being at 2 ſl.
So M. d. Here the ſuperfluous terms
State 2: 8
:
1-6. are í Loaf and I Peck:
alſo I conſider that I'll have
18 d. more Weight for my Mo-
8
ney when the Wheat is
192
cheaper, therefore I mul-
64
tiply the middle Number
10 » 138 by the greater Extreme,
. oz.
8 :
12
I2
24d.
18}}|
or
Rule of Three Numbers.
145
12.
I212
or which is the ſame, this by that; ſo the Anſwer is,
10 02. 13 d. wt. 8 gr. (Troy Weight.)
Qus. 6. If a Nail of Velvet coſt 16 d. Sterling, what
is the Value of 75 Yards 3 qrs. at the ſame rate ?
d. yds. grs.
Flere the firſt Term
State 1 : 16:: 753
being 1, dividing by
4
it is unneceſſary, and
therefore the An-
303
ſwer is found in
Pence by Multipli-
4
Cation only; which
Pence I reduce to
16
L. by dividing by
12 and 20.
12) 19392
) 15115
L. 80~16 Anfr.
Qu. 7. If I travel 600 Miles in 24 Days, when the
Day is 12 Hours long, in how many Days, at the ſame
rate of Travelling, ought I to finish the ſame Journey,
when the Day is 16 Hours long (ſuppoſing in each char
I travel from Sun-riſing to Sun-ſeccing?)
ho. long. Days. ho. long. Here the fuperfluous
State
24
16 Term is 600 Miles,
and I multiply che
middle Numbor by
288
the leffer Extreine,
72
becauſe the Anſwer
IS
muſt be leſs than
the ad Term, fo the Quote is 18 Days.
Qu. 8. What time will 10 Men cake to do a Piece
of Work, when 4 ocher ſuch could do the ſame in
15 Hours, 12 Minutes ?
I2
:
:
I2
IÓ
}
Bu / 288
o
State
146 Rule of Three Numbers.
20
Men. bo. min. Men. Here I multiply the
State 4 : 15-12 : : 10 middle Number by
4
the leffer Extreme,
becauſe 10 Men will
10)60—48
take leſs time. The
6-
ſuperfluous Term is
1 Piece of Work. Obſerve alſo, that I don't reduce
the middle Number, becauſe the Multiplier is ſmall.
Qu. 9. How much will L. 1-6-3 per Day amount
to in a Year?
D. L. S. d. D.
In a Year are 365
State 1 : 1-6-3 : : 365 Days. The iſt Term
315. being 1, the Queſti-
on is performed by
26
1825 pure Multiplication,
365 and the Anfr. comes
1095
out in d. which re-
315
duced makes L. 479
12) 114975
I-3
210)95811-3
479-I-3
24.10. How many C. of Sugar, at 64 d. per Pound,
may I purchaſe for L. 43-18-4?
d. lib. L.
d. Here I reduce the
State 6: 1 :: 43-18-4
Extremes to Half-
pence, and the Anfr.
is 162172 lib. which
13
878
reduced makes 14C.
I gr. 2573 lib.
12
$.
2
20
I2
10540
2
13) 21080
28
4 1 162113
$ 7405+
4157-25
Anfr. 14--1-2513
u.
Rule of Three Numbers: 147
12
is L. 5:
Qu. II. What principal Sum will gain L. 5 in 7
Months, when L. 100 gains the ſame in a Year, or 12
Months ?
Mon. L. Mon,
Here 7 Months will
State
12 : 100 : 7
require a greater
Principal, becauſe
the time is ſhorter.
7)1200
The ſuperflu. Term
L. 171-8--69
5.
Qu. 12. If 474 Yds. of 6 gr. broad Cloth hang a
Room, Bc. how many Yds. of Yard broad Cloth will
do it? Or rather thus, How many Yds. of Yard broad
Cloth are equal preciſely in Content to 474 Yds. of
6 qrs. broad
?
grs. yds.
qrs.
Here I conſider that
State 6: 473 :: 4
there muſt be taken
more in length of
the narrow Cloth,
4)286—2
to equalize that of
71-22
the broader; there-
fore I multiply by the greater Extreme, and the Anfr.
is 71 Yds. 2 qrs: 2 nails. This Queſtion may be
wrought ſhorter, by adding 3 of 47–3 to itſelf; thus
2)4743
23--3-2
71-2-2 Anfr. as before.
Q11. 13. If 560 Men, being in Garriſon, have Pro-
vilion only for 3 Months, how many of the ſaid Num.
ber muſt be turned out, that the Proviſion may ſerve
the remaining Men other 5 Months
Mon. Men. Mon.
Here 3 Mon. and
State 3: 560 :: 8
other 5 make 8 Mon.
3
and conſidering that
fewer can be main-
8)1680
tained 8 Mon. on
the fame Quantity,
than
210
02
148 Rule of Three Numbers.
than can be done for 3 Mon. I multiply by the lef-
fer Extreme.
560
210 to be kept in.
Anfr. 350 to be turned out.
Qu. 14. If when Wheat is 6h. 5 d. the Buihel, the
Penny-Loaf weighs 12 oz. what ought the Wheat to be
fold at, when the Loaf weighs 17 oz. 10 d. weight?
oz. n. d. d. wt. Here I conſider that
State 12 : 6-5: : 17-10
the Wheat muſt be
cheaper when the
Loaf weighs more,
240 77
350 therefore the Pro-
240
portion is indirect.
02.
20
I2
20
308
154
I
51184810
369-2
12152--37 or
ſh. 4-4-35 Anfr.
Qu. If 8 Men do a piece of Work in 6 Days, in
what time will they do 16 times as much ?
W. D.
W
State I: 6: : 161/ 를
​6
99 Days Anfr.
9. 16. What is the Intereſt of L. 37-13-4. for a
Year, when L. 7 is the Intereſt of L. 100 for the
ſame time?
Statc
Rule of Three Numbers. 149
L. f.d. Here inſtead of di-
Scate L. 110: 7 :: 37–13--4 viding by 100 the
Ź long way, I cut off
the two o's, and two
2163--13--4 Figures for them,
reducing the Re-
mainder, and the
12173
Anſr.is
L. 2—12-8-35
20
I2
8180
4
The fame may be
wrought thus
3120
100 : 7 :
: 37-13 man ito
266-13-4
IOO
0}1:
26-
2--12-8-3 as before
Qu. 17. If a Servants daily Wages be 6d. 1 far.
how much will it amount to being forborn 3 Years,
8 Months, and 10 Days ?
D. d.
D.
In a Year are 13
State 1 :61::3
38-10 Months and 1 Day,
13
therefore I add 3
Days for the 3 Years.
47
28
y. m.
376
95
1326 Days.
3 Days add.
1329 Days in all.
67
O 3
3
7974
150 Rule of Three Numbers,
7974
· 332
12183061
210)6912-21
L. 34.-12-21 Anfr.
Qu. 18. If I receive 30 Stone 12 oz. of Soap for a
Debt of L. 8-7-4, how much doth it ſtand me
per lib?
St. lib, oz. L. s. d. lib.
State 30-0-12 : 8-74 : : I
16
16
20
480
16
167
16 0%.
I2
7692 oz.
2008
IÓ
7692)32128(4 di
30768
1360
4
Anfr. 4per lib.
Ansp. $S95
5400(0 far.
Qu. 19. A owes B L. 117.-10-2 and compounds
the Debt by paying 17.1.4 d. per L. how much ought
B to receive according to this Diſcount, and how
much doch he loſe?
L. S. d.
d. Here I ſay, if I L.
State I: 17-4
: : 117
fall to 176. 4.d. how
much will the Sum
due fall to ?
2350
Lui
S,
20
I2
20
20 208
12
I2
24.0
28202
Rule of Three Numbers. 15,1
28202 From 117-10-2
208 Take 101–16-gis receives.
240
225616
15-13-415 loſes.
56404
415 8660116
61466503
12/24441176
210)20315-9
L. 101-16-91731
qu. 20. What will 3 Pieces of Holland Cloth come
to, the firſt Piece containing 154 Yds. the ſecond 1933
and the third 104 at 17 1.4 d. for 3 Yards ?
yds.
State
3: 17-4:: 451
19
10
halfyds. 6 208 91 half yds.
45 yds. in all.
91
yds. f. d.
15
2
12-
2
208
1872
6) 18928
12) 3154-23
210) 2612-10-25
L. 13-2--10-23 Aufr.
Or without reducing the Extremes to half yards, thus
yds.
3 : 208 : : 451
457
yds. d.
1040
2
104 of 208
319464
3154–2Ž=L.13:2:10:23 as before
இய.
152
Rule of Three Numbers.
Qu. 21. A Gentleman has an Eſtate of L. 1460.-
10—4 per Annum, how much does he lay by yearly,
ſpending daily at the rate of L. I-18-6?
D. L. M. D.
Here I firſt find
State 1:1-181:: 365 how much he ſpends
381
yearly, which fub-
tracted from his
381 2920 yearly Income gives
1095
the Anſi.
1821
20
HIN
2l5) 140512
L. 702--12--6
From 1460-10-4 yearly Income.
Take 702-12-6 yearly Expences.
757–17-10 lays by. Anfr.
Qu. 22. What is the Intereſt of L. 313-11--G for
a Year and 74 Days at L. 5-10 per cent. per anz-
nun ?
L., L. L. fb. d. Here I firſt find its
State ilbo : 51:313-11-6
Int. for a Year, nexo
si for 74 Days, and the
Sun of both is the
1567-17-6
Anfr.
156-15-9
17124-13-3
20
4193.
I 2
11119
Again,
Rule of Three Numbers.
153
D. L. f. d.
Again, 365: 17-4-11 :: 74
D.
20
344
12
L. f. d.
Int. for a Year
I7-4-II
Int. for 74 Days 3-D II
4139
74
Anfr. 20mm 14--10
16556
28973
121
365) 306286 ( 839
2920 210) 613--11
1428 3.-II
1095
3336
3285
51
But ſuch Queſtions are more exactly wrought by
one Operation, as you'll ſee further on.
Qu. 23. How many Ducats at 7 f. 4 d. per Piece,
are equal in Value to 300 Dollars, at 4. fl. 8 d. per
Dollar?
Dol. fb. d.
Dol.
Here I firſt find the
Value of the Dol-
lars, and from, that
I have the Value of
the Ducats, by
300
I:41
8:: 300
I2
56
16800 Value of the Doll.
ſaying
154 Rule of Three Number's.
M. 4 Duc. d.
faying 7-4:1:: 16800
--
12
88) 16800 ( 190
88
Anfr. 190 Duc. and
So d. over and a-
bove.
800
792
J. d.
12
d. taken to equa-
Rem. 80 d.
But all Queſtions of this nature are better folved by
one Operation, thus
m. d.
Here I conſider,
4-68: : 300 :7-4
that there muſt be
12
fewer Pieces at 85.
4
88
lize a Number of
300
other Coin at 4fb.
8 d. cherefore mul-
88) 16800 (190
tiply by the leffer
88
Extreme. Take an-
other Exa. wrought
800
after this method.
792
56
Rem. 80
Qu. 24: How many lib. of Tea, at 9 m. 6 d. per
lib. muft' B give A for 22 Reams of Paper, at 5 ſh.
10 d. per Ream ?
b. d. f. d. So that B muſt give
State 5-10:22 ::9-6 A 13 lib. of I ea.
and 58 d. more for
his 22 Ream, at 5.
70
114
10 d. per Ream.
1:14). 1540 (13 lib
114
400
400
12
I 2
22
Rule of Three Numbers.
1.55
400
342
Rem. 58 d.
Qu. 25. Lent upon Bond L. 546 on the 18th Fan.
1737, what will be due at Whitſunday 1739, the Rate
of Intereſt being at 6 per Cent. per annum ?
Betwixt 18th Jan. 1737 and 18th Jan. 1739 are 2
Years =730 Days; and betwixt 18th Jan. and 15th
May are 117 Days, thus found,
L. L. L.
State 10:6::546
Jan. 13
6
Feb.
28
March 31
L. 32176
20
April 30
May 15
ſh. 15120
I2
117
730
d. 2170
847
4
f. 1160
Days. L. 1. d. f. Days
. .
Again 365 : 32—15--2-1::117
20
655
12
7862
4
31448
117
220135
220136
156
Rule of Three Numbers.
220136
31448
31448
365) 3679416 ( 10080
365 12)2520
210)2110
2941 10-10
2920
216
Anfr. L. 622:0^-41
L. 32~15~2~I Intereſt for 1 Year.
2
65-10-4-2 Intereſt for 2 Years.
10--10Oo Intereit for 117 Days.
76-0
---2 Toral Intereſt
546-o-o-o Principal lent.
But this, like Qu. 22, can be wrought by one Ope-
ration.
Qu. 26. In L. 4176--11-- Scotch, how much
Sterling?
L. Sc. L.St.
Here I have no
State 12 : 1:
12 : 1 :: 4178-1-6
4178-1-6 more to do but to
12 4178-11-6 divide by 12, be-
Anfr. 348-4-1 cauſe 12 L. Scots
are equal to 1 L. Sterling.
Qu. 27. In 42 Bags of a certain Commodity, weigh-
ing Groſs 146 C. 2 qrs. 24. lib. Tare ? lib. per 112 lib.
and Trett 4 lib. per 1047 how many lib. neat?
.I 12
Rule of Three Numbers.
157
TI2
1.04
7 fubtract
4. fub.
112 : 105:: 146--2--24 Again 104: 100 :: 15405
4
15405
104 1540500 (14812
104
5.86
28
1
832
00
4692
416
1174
845
16432
105
130
82160
104
16432
260
4/1725360
208
X12
41 431340
107835
52
15407 lb.Suttle
Anfr. 148127 lib. Nett.
This Method is ſomewhat tedious. Vide Page 169,
where you'll find a ſhorter.
Qu. 28. A and B make a joint Stock. A puts in
L. 127, and B L. 31: In Trading they gain or loſe
L. 16: What is each proportional Share of the ſame?
A 127
B
31
158 Sum of the Stocks
State
158
48
22
.
Rule of Three Numbers.
1. 1. 1.
1. I. 1.
State 158:16:: 127 Again 158:16:: 31
16
31
762
16
127
158) 2032 ( 12 1.
) 496 ( 31.
158
474
452
) 440 ( 2 s.
316
) 2720 (175.
124
158
12
1140
) 1488 (9 d.
1106
1422
34
66
12
4
) 408 (2d.
158
316
20
136
20
I
316
106
92
4
368 ( 25.
316
A 12-17--2--2758
B 3-2-list
52 Proof 16-OOO
Here I add the Stocks together for the firſt Term,
then I ſay, If the whole Stock gain or loſe ſo much,
what will each particular Stock gain or loſe? But
Queſtions of this nature are capable of a more com-
pendious Solution.
Vide Page 187
இய.
Rule of Three Numbers.
159
Qu. 29. A, B, and C bought a Parcel of Cloch
containing in all 748 Yards, for l. 212, whereof A
paid I. 187, B 110, and C the reſt. How much of
the Cloth belongs to each?
L. Yds. L.
Yds. Qrs. Nails.
State 412 : 748:: 187
$ 339--2--021
412 : 748::110 to the Anfrs. 199-2-3119
412 : 748 :: 115
208-3--02
Proof 748-0-0
Vide Page 187
Qu. 30. If a Pound of any thing is bought for
16.4 d. and fold for 17 ſh. 6d. how much is gained
on 1. 187?
By Subtraction I find 16
16-4
sh. 4 d. gains 1 j. 2 d.
therefore I ſay how much
164: 1-2: : 100 will l. 100 gain? where I
don't reduce the l. 100.
Vid. Note 3. Page 149.
196 14
17-6
12
12
100
196) 1400 (7 km
1372
28
20
196) 560 ( 216.
392
168
12
196) 2016 (10 do
196
5.6
P2
qui
60 Rule of Three Numbers.
Qu. 31. What will l. 54 amount to, being forborn
15 Years, at 6 per cent. per annum, fimple Intereſt
State 100 ; 6: : 54
Here I find the Intereſt of
6 of l. 54 for a Year, which
multiplied by 15 gives the
3124 Intereſt for 15 Years,
which added to the Princi-
pal makes l. 102-12 for
4180
the Amount.
20
12
SLSO
4
2140
1. S. d. f.
3-4 g27
15
48-12-0 Int.
54-O
o Principal.
102-12-Oo Amount.
Qu. 32. A, B, and C purchaſe a Ship for l, 3000,
whereof A paid l. 1400, B 1000, and C'che reft. Her
Freight for the firſt Voyage was l. 374, what muſt each
Partner receive of the ſame proportionally to his Share
in the Purchaſe ?
A 1400 3cfoo: 374 :: 14100
31300:374::11300
B 1000
14
C 600
31374
1496
124-13-4
374
3515236
30130: 374 :: 6100
3000
30 8610474
4
174-10-8
3036448—16
A
74-16
Rule of Three Number's, 161
A 400
B 500
C 700
A 174-10-8
B 124-13-4
C 74-16-
Proof 374-00-
Qu. 33. A Man is indebted to A, B, and C the fol.
lowing Sums, viz. to A L. 400, to BL. 500, and to
C L. 700, but is only worth L. 979. How much of
the ſame muſt each receive, and what Loſs do they
ſuſtain ſeverally?
Or ſhorter thus, by finding the
Proportional Share of L. I.
1600)9761.61 Then
16100:976::4100 16100:976:: 5100 7.61X400—244
4
5
.61X500=305
.61X9003427
413904
976
411220
244
305
16100: 776 :: 7100
Then by ſubtracting what
7
each receives, from what he
ought to have received, you'll
find A loſes L. 156, BL. 1957
41708
427
Qu. 34. A has 40 Yards of Cloth at 8.11.4 d. per
Yard, ready Money, but in Barter he will have 10 i
2 d. B hath Wool at 20 d. per lib. ready Money; How
much Wool muſt be delivered to A for his Cloth, and
how much is the Price of the Wool to be raiſed in
Barter?
1621
4/4880
102 443
162416833
and C L. 273.
I
Id.
162 Rule of Three Numbers,
Id. h. d. rds.
d. lib. de
1:8-4::40
Then 20:1::4000
12
I
100
40
217)40002
200 lib. of Wool.
4000 d. Value of the Cloth in ready Money.
fh. d. f. d. d.
Laſtly, 8-4: 10-2:: 20
12
12
1
100
122
20
d. f.
24-118 Value of the Wool
per lib. in Barter.
24140
4
Il50
After finding the Value of the Cloth at the ready
Money Price, I ſay, If 20 d. buy i lib. of Wool, how
much will the Value of the Cloth buy? Anf. 200 lib.
Then to find how much the Price of the Wool is to be
saiſed in Barter, I ſay, If 8-4 ready Money riſe to
10.2 d. in Barter, what will 20 d. ready Money riſe
to? And the Anſwer is, 24. d. It or far.
Qu. 35. What is the Diſcount of L. 542 : 14: 8 for
3 Months at 5 per cent. per annum ?
M. L. M. 1. M. L. M. L. M. d.
State 12:5:3 10–5: Imam 5 542-14 -8
3
::
20
20
20
2025
25
10854
32)15
15 Int. of
L. 100 for 3 Months.
12
130256
130256
Rule of Three Numbers. .
163
Brought over
130256
25
651280
260512
5
3256700
651280
130256
14472
2025
9
1608 2015
200
1211608
Here, and in all Que-
215)1311O
ſtions of this Nature, I
6--14--0722.5 or muſt firſt find the Intereft
of L. 100 for the Time,
and at the Rate propoſed, which adding to L. 100,
I ſay, if chat Aggregate loſe the Intereſt of L. 100 al
ready found, what will the Sum propoſed loſe and
the Quote is the Anſwer required; which taken from
the Sum propoſed, leaves the Sum to be paid preſent-
ly, and to find this at once (without ſubtracting) let
Li 100 be the middle Term in the former Stating.
L. f. L. L. f. d.
Thus 101–5:100 :: 542-148
20
512025
2025 2000 10854
51405
9181
919
130256
So that 2025 is
5X5X9X9.
260512000
52102400
2025
10420480
1157832
12/128647 1825 or 2}
Brought
20
20
12
I
2000
164 Rule of Three Numbers.
Brought over
12/128647 18kt os}}
210)10721073
L.5360731 Ans. as by the
firſt Method : For,
L. fb. d.
From 542-14-8
Take 6--14-09
Rem.536--00—781
Vide a ſhorter Method; Appendix, of ſimple Intereſt;
Tab. I.
Qu. 36. If I buy a Yard of Cloth at 3 lb. ready Mo-
ney, and fell it immediately at 3 h. 4 d. with eight
Months Credit, what ſhould I gain if I ſhould fell
L. 100 worth at the ſame Rate, with twelve Months
Credit ?
M. d. L.
M, L. 1. d. M.
State 3: 4:
Then 8:11-02-2
4
8
:: 100
::12
12
30
36{91408
2
3
12138-17-95
66-13-4
L. 7-08-iyor if
L. II-02-2
By the firſt Operation I find, that if 3.ſh. gain 4 d.
in any Time, L. 100 in the ſame Time will gain
L. 11-2-23; then I ſay, If 8 Months gain ſo much,
12 Months will gain leſs, (in this Cafe) becauſe the
longer Goods are credited, the Seller has the leſs Pro-
fit. But this may be wrought by one Operation.
Qu. 37. Bought 40 doz. Pair of Stockings for L.42,
how may I fell them per doz. to gain L. 15 per cent.
Rule of Three Numbers.
165
L. L. L. Doz. L. 1. Doz.
State 1100:115::42 Then 40:48–06::1
42
I
230
460
48-06
819-13_2 %
L. I-04-12 or 3}f.
L. 4.8130
20
M6100
Qu. 38. A Farmer mixes 8 Pecks of Wheat at 16 d.
per Peck, with 9 Pecks at 18 d. with 12 Pecks at 17 d.
what is a Peck of the Mixture worth?
d.
P. d P
8 ac 162128
Then zy: 494:1 Ol
9 ar 187162
12 at 17204
-29)494(170's d. Ans:
494
29
I
29
204
203
1
If it was required what 2, 3 or 4, dvo, Pecks would
coſt, make, 2, 3, or 4, &c. the laſt Term (inſtead of i
here, and you'll have the Anſwer.
Qu. 39. What is the Compound Intereſt of L.
387-10-forborn 3 Years at 6 per Cent. per annum,
and what is the Amount?
State
166 Rule of Three Numbers,
State l. 100:6::387--10 387-10~o ift Principal.
6 23-0500 iſt Year's Int.
23:25_o 410-15--02d Principal.
24---12-10 2d Year's Int.
20
5100
Again 100:6:: 410m15
6
435-07-10 3d Principal.
26-02-05 3d Year's İnt.
461-10-03 Amount Ans,
24164-10
20
12190
12
10180
L. L. L. fo. d.
Laſtly, 100:6::435710
.6
L. 23-05-00
24-12-10
26-02-05
26112-0
20
Sum of Int. 74-00-03
214.7 Prin. lent 387-10.00
12
Amount 461-10-03 as before.
5864
So you ſee I always add the Intereſt to the Princi-
pal, which Sum becomes a new Principal, and have as
many Operations, as is the Number of years the Mo-
ney lies out. See Chap. 2. Tab. IV. of compound Inte-
reſt, where you'll find a ſhorter Method.
Qu. 40. A and B entered into Company, A put in
L. 55, which continued in Stock 4 Months, and B
put in L. 131 for 7 Months; they gained or loft L. 48,
what is each Man's Share in the ſame?
55
Rule of Three Numbers.
167
55x4=220
131X7=917
1137
L. L. L.
1137:48::220
48
L. L. L.
Again 1137:48 :: 917
917
1
1760
880
7336
3668
1:137)10560196.
10233
1137)44016(38 l.
3411
327
20
9906
9096
810
1137)654015.M.
5685
20
855
)16200(14.11.
1137
12
1137) 10260(9 d.
10233
4830
4548
27
282
12
1137)3384 (2 d.
2274
IIIO
27
A
B
L. 9-05-9117
38-14-21137
Proof 48-00-
Here
168 Rule of Three Numbers.
Here I multiply each's Money by the Time it con-
tinues in Scock, the Sum of both which is the firſt
Term, the Gain or Loſs the ſecond Term, and the
Product of each's Stock by his Time the third Term,
in two diftinct Operations.
Vide Pag. 187, where you'll find a more compen-
dious Method,
Qu. 41. A hath 36 lib. of Galls at ind. per lib. rea-
dy Money, in Barter 12 d. 81 Yards of Cloth at 16 d.
per Yard ready Money, in Barcer 17 d. 54 Reams of
Paper at sſb.10 d. per Ream ready Money, and in
Barter 6 .
B has Wine at 13. M. per Gallon,
Silk at 15 m. per Yard, and ready Money;
Brandy at 29 l. per Cask,
How many Gallons of Wine, Yards of Silk and Casks
of Brandy, of each a like Number, muſt B give A for.
his Goods, advancing his own proportionally?
36 lib. at 11 d. 396
81 Yards at 16 d. R1296
54. Reams at 70 d. =3780
5472 d. Total Value of A's
Goods at the ready Money Price.
13 lb.
15 lb.
29 m.
57 M. or 684 d. Value of 1 of each of B's Things
at the ready Money Price. Then I ſay, If 684 d. buy
I of each of B's, how many of B's Things will the
total Value of A's Things buy?
d. of each.
d.
684: I
:: 5472
I
684)5472 (8 of each, Anfr.
5472
Rule of Three Numbers. 169
1
that of 24. 27
2011
Qu. 42. In 36 C. I gr. 12 lib. groſs, how much
nett Weight, Tare at 8 lib. per 112 lib. and Trect 4.lib.
per 104? C. gr. lib.
52 | 36:1:12
14{
Here I uſe a fhor-
7 | 18:0:20
ter Method than
2:2:10:13
for I conſider thar
261 33: 3:01:02 Suttle
8 lib. is a of 112
1:1:05:0631
and 4. lib. is of
104.
Anf. 32:1:11:11 Nett.
Therefore, if the Tare is at 2 lib. per 112 lib. ſub-
tract , if at 4 lib. ſubtract is, if at 6 lib. ſubtract
75 + 1 of that is ; if at 14. lib. ſubtract , &c.
qu. 43. My Friend lent me L. 46:15, for 5 Months;
and he having occaſion afterwards to borrow from me
L. 100, I demand how long he ought to uſe it in or-
der to a full Requital of his former Kindneſs?
L. M. M. L.
State 46-15:5:: 100
Here I conlider that
L. 100 being a greater
Sum, muſt not lie out
935
ſo long as L. 46-15,
5
therefore I multiply the
middle Number by the
21300)41575(2 M.
lefſer Extreme and di-
41
vide by the greater.
20
20
2000
។
675
30
)20250(10 Days.
'M. Days.
Anf.2 : 10
Qu. 4.4. A Merchant is to receive L. 500 Sterling,
either in Dollars at 4M. 4 d. per Piece, which are
worth but 4.1.3 d. or in French Crowns at 6m. Id.
a
per
170 Rule of Three Numbers.
I2
12
I
2
2
43
+
per Piece, which are worth but 6 . which of theſe
ought he to take, in order to ſuſtain the least Loſs?
D. d. d. M. d.
4:4
4-4:1::6-11
So that in receiving
4:3
the Dollars, each 6-11
loſeth only 1432;
52
73
whereas, if he ſhould
take the Crowns, eve-
ry 6 loſeth 1 d.
104 )147(124 which is greater than
104
I 43., and therefore he
ought to take the Dol-
43
lars.
Qu. 45. If: 35 Ells at Vienna make 24 at Lyons, and
3 Ells ar Lyons make 5 at Antwerp, and 100 at Ant-
werp make 125 at Frankfort, how many Ells at Frank-
fort make 42 at Vienna?
Ant. Frank. Ant. Frank.
1. If 100 : 125
GA
Ly.
Ly.
3 :
Vien.
Vien.
3. If 35 : 50 :: 42
: 60 Anfr.
Or thus :
35 V. = 24
L. Then 24X5X125X42630000
3 L. =
for a Dividend; and
100 A=125 F.
35X3X100 =10500, for a
42 V. =?
Diviſor.
Wherefore 105100)6300100(60 Anſwer, as before.
630
5.
2. If
64 :: 2+
: 50
5A.
O
Qu. 46. If 35 Ells at Vienna make 24 at Lyons, and
3 Liis at Lyons make 5 at Antwerp, and 100 Ells at
- Antwerp make 125 at Frankfort, how inany at Vienna
make bo at Frankfort ?
L,
Rule of Three Numbers. 171
V.
L.
L. V.
1. If 24: 35 :: 3 : 41
A.
A.
2. If 5: 4
5: 45:: 100: 87
F.
F. V.
3. If 125: 871: : 60 : 42 Anfr.
Or thus :
35 V. = 24 L. Then 35x3x100x60=630000 for
a Dividend; and
100 A. =125 F. 24X5X125=15000 for a Diviſor.
60 F. ?
Wherefore 15l000)6301000(42 Anſwer as before.
60
And ſo of others, whe-
ther they confift of more
30
or fewer Terms.
30
3 L..
5 A.
Qu. 47. A and B made a joint Stock, A put in
L.
300,
and at the End of 3 Months took out L. 752
and at the End of 7 Months after put in L. 50.
B put in at firſt L. 180 for 5 Months, and then puc
in L. 48 more, and two Months after shat took oue
At the End of 12 Monchs they find L. 100 of
Gain. How mucli of the ſame is due to each?
L. 300 for 3 Months is 900
.
L. 40.
75 Subtr.
225 for 7 Months is 1575
50
225 for 2 Months is 550
L. 3025 Sum of A's Stocks
into his Times.
#
Qa
L. 180
172 Rule of Three Numbers.
L. 180 for 5 Months is 900
48 Add
228 for 2 Months is 456
40 Subtr.
188 for 5 Months is 940
2296 Sum of B's Stocks in-
to his Times.
A 3025
B 2296
5321 Sum of both.
Then 5321)100.000.0187934, &*c.
Wherefore 3025X.0187934–56.850035=A's Share.
And 2296X.0187934343.149646~B's Share.
Proof 99.999681=100 L. fere.
The Reaſon why it makes not L. 100, is becauſe the
Decimal.o187934 is incomplete.
Qu. 48. If I pay for 45 Pints of Brandy L. 4:10,
how muſt one Pine be ſold to gain L. 10 per cent?
100: 108::4.5:4.86 Price of the whole 45 Pints
at the propoſed Gain. Then to find the Price of 1
Pint,
45
91.972
108 L. 2 ſ. id. 31;far. Anfr.
Qu.49. How ought L. 350 to be divided among 3
Men, A, B, and C, ſo as A may have , B, and CT?
= Here the Sum of the Parts
24
is greater than the whole: And therefore 'tis impoſſible
to give them the Parts propoſed; but the Meaning of
the Queſtion muſt be to divide L. 350 in ſuch Propor-
tion as į į and bear to one another. Wherefore,
{tite=12+16+6
ner.
Rule of Three Numbers. 173
IO
14), 55+
Anfrs.
Anfrs.
1871
]
IO
1:4
neglecting the Denominator 4, I find a common Mul-
tiplier thus: 34)350(10.29412+
Then 10.29412X12=123.52944=123: 10:7
And 10.29412X16=164.70592=164:14:11 ferè,
Allo 10.294 12x 6= 61.76472= 61:15:33
L. 350: 0:0 Proof.
Qu. 50. A Merchant mingles Tobacco thus: He
takes ſome at 10 per lib. ſome at 15
d.
per
lib. ſome at
18 d. per lib. and ſome ar 13 d. per lib. how many lib.
muſt he take of each Sort to Tella lib. of the Mixture at
14, ſo as neither to gain nor loſe ?
1071
15
or thus : 14)
18
1344
13
Or thus :
I:4 15
4
15 4:1
or thus: 14)
5
4:1
18
4
4
13
4
4
13
I
Or thus :
1:45
15
4:1 5
14),
18
5
13
1:415
So that there are 5 different Anſwers, and all true.
Here, and in all Queſtions of this Nature, you muſt
ſet the ſimple Prices under one another, and connect
them fo together, as that one leſſer and one greater
than the mean Price may be always joined together,
and then take the Difference betwixt the mean Price
and that of each ſimple, ſetting the ſame over againſt
the ſimple wherewith it ſtands connected, and the fe-
veral Differences are the Anſwers fought.
Qu. 51. How many Crowns at 56% d. are equal to
L. 600:12:8 Sterling?
State
14) IT
18
' 'salny
3
IO
4:1
Q3
174
Rule of Three Numbers.
d. Cr. L. f. d.
State 56%:1::600--1208
6
20
337
12012
12
144.152
6
337)864912(2566 CX.
674
1909
1685
2241
2022
2192
2022
170
3
25.10(1 Livre.
337
173
20
)3460(10 fols.
337
90
Cr. Liv. Sols.
Anfr: 2566 : : 10
I
С НА Р.
175
CHAP. XX. Contrations in the Rule of
Three Numbers, otherwiſe called Rules of
Practice,
CASE I.
WHEN the Value of an Unit is an Aliquot Part
of any ſuperior Species, or equal to the Sum of
two or more Aliquot Parts; alſo if it is a Part or Parts
of ſuch Aliquot Parts, take that Part, or Parts of the
Number given, whoſe Value is ſought, and you have
the Anſwer in the fame Name with that ſuperior Spe-
cies, reducing the Remainder, if any be, and finding
its Value.
A. d.
Imm8
TABLE I. Of the Aliquot Parts of 1 L. Sterl.
r
200h
12th
2-O
Ioth
2-6
8th
> is a
3-4
Oth
5th
59
028
3d
100
2d
of į L. Sterling
4o
4th
10
30?
of the Aliquant or uneven Parts of 1 L
riotis
十​a or £5.6
7
8-
is
9
10
“ fin met alin mo
2
t
1
Ilmö
376
Contractions in the
II
o
12_o
itzo
333 좋
​13-Oist
14 34
15
용
​16-
tao
I
10
17-
I'mo
190
2 / 420
tš
I-O
TABLE II. Of the Aliquot or even Parts of I ſh.
d. f.
I2th
8th
is a
Oth
of i Shill:
4th
4-
3d
6-
2d
I-2
1
3-
ہستی
Of the Aliquant or uneven Parts of i ſh.
d. f.
5-O
ti
I
12
4+5
is a
8-
9
IOO
IIO
»m alt alm mit mit mit
of i Shill.
st
tos
TABLE III. Of the Aliquot Parts of id.
I
is
of 1 Penny.
3
Examples of this firſt CASE.
Exa. 1. 480 Yards at 1 lb.per Yardim. is zo l.
Therefore 210)4810
24. L. Anfr.
Here
Rule of Three Numbers. 177
šof laſt for i llb. 15-15 Add
Here I conſider that 480 Yards at L. 1.per Yard
would make L. 480, therefore at 1 lb. per Yard, it
muſt be only the 20th Pare of what it was formerly,
becauſe 1 . is che 20ch Part of i L. And this accounts
for the Method of operating this and the ſubſequent
Examples
Exa. 2. 376 lib. at 4 m. per lib. 4.1. is } l. and
5)376
the I over, after
L. 75-4, Anfr. Diviſion is l. or
4.
Exa. 3. 538 oz. at 2 d. per. oz. 2 d. is b. there-
6)538
fore I divide by 6,
2p)819-8
and the Anſwer is
L. 4-9--8 Anfr.
89.J. and 4 over,
which is door
g d. The ſh. I reduce to l. by dividing by 20.
Exa. 4. 95 Stone Weight at 6fb. 8 d. per Stone.
3)95
66. 8 d. is į l. After Diviſion
L. 31-13-4
there reinains 2, which is l. or
13.4d.
Exa. 5. 120 lib. Weight at 1 d. per lib.
8)120
Iish
15 Mh. Anfr.
Exa. 6. 83 Yards at 3. f. 4 d. per Yard.
6)83
3. 4 d. is I I. after divi .
L. 13-16—8 Anfr. ding there remains 5, which
is I. or 5 times 3.fb.4.d.
FI61. 8 d.
Exa. 7. 315 Yards at 6 p. per Yard.
315 6. is 1.4.2 l. or it is
$1.7 of that
for 5 h.
L. 94-10 Anfra
Exa.
178 Contractions in the
Exa. 8. 413 0%. at 9 d. per oz.
413
9 d. is 21. wherefore I multiply
3
by 3, and divide the Quote by 4.
4)1239
212)30192
L. 15-9-9 Anfri
Or I might have divided by 4 and then multiplied
the Quote by 3, thus:
4)413
1033
3
2103019-9
L. 1999
Exa. 9. 210 Ells at 15%.
151. is 10. +51).
210
for 10. 105
of last for 5 fl. 52--10
L. 157-10
EX8. 10. 470 lib. at II d. 11 ist & fl.
3
Or the Anſwer may be
4)1410
found in /b. by ſubtract-
ing A, thus :
352-6=
470
78--=%
i's=39--2
310)4310---10
L. 21-10-10 Anfr. 1.430-10 as before.
EMA.
Rule of Three Numbers.
179
ܪ݂
Ex&. II. 111 Ells at 11ſ. It d. 3 far. per Eli.
III
for roll.
to of laſt for ilh.
t'i of 10 m. for 10 d.
its of laſt for 1 d.
of laſt for 2 far.
į of laſt for i far.
15510
5-IL
4-12-0
4- 2
0-2-3-3
L. 66-9---I Afr.
The Work is plain.
Exa. 12. 60 Stone Weight at 11 o. per Stone.
60
II. is it sol. or it
is į l. tis ott
for lo b.
30
to of laſt for i ſh. 3
2
oj
L. 33 Anfr.
Exa. 13. 419 Yards at 19 b. per Yard.
20)419
19/h. is l. +.
20-19
Or rather ſubtract no.
1
L.398- I Anfr.
When the Price of an Unit is an Aliquot Part, ſuch
Diviſion is certainly the moſt expedicious Way of fol-
ving the Queſtion: But when they are Aliquant or
Compound Parrs, the common Metliod of Reduction
is preferable, as being ſhorter and leſs perplexed.
CASE II.
When the Price of an Unit is an uneven Number
of Jh. multiply the Number whoſe Value is ſought, by
the Half of the Shillings, doubling the firſt Figure or
Unit's Place of the Product for th. the reſt are L.
Exa.
180 Contractions in the
Exa. 1. What is the Value of 347 Yards at 8 m. per
Yard ?
347
4
L. 138-16 Anfr.
Exa. 2. What is the Value of 316 lib. at L. 1.-12
per lib?
316
L. 1--12 is 32 b. of
which is 16.
16
189---12
316
L. 505-I 2 Anfr.
The Reaſon of this Method may be thus accounted
for: To mulciply by an even Number of lh. and divide
the Product by 20, is the ſame as to multiply by one half
of the Shillings and divide the Product by 10: But to
divide by 10 is the ſame as cutting off the Unit's Figure;
thoſe that remain to the left being entire L. and the Fi-
gure
fo cut off being ſo many roth Parts of a L. each
of wbich is equal to 2 ſb. wherefore che Figure ſo cut
off, muſt be doubled or multiplied by 2.
CASE III.
When the Price of an Unit conſiſts of L. J. and do
or L. and ſb. firſt multiply the Number given by theſe
L. and for the lh. and d. take one of the preceding
Caſes. The Sum is the Anſwer.
Exa. I. Wluat is the Value of 16 C. at L.4----8
16
Here I firſt mulci-
4
ply by 4, and for
66. 8 d. I take
64
of the Number gi-
of 16 5-6-8
ven, and their Sum
is the Anſwer.
Lögma'y--3 A1:
Exe.
per C?
3
Rule of Three Numbers. 181
Exa. 2. 413 lib.at L. I---16-8 per lib.
at il.
413
I for 10 m. 206-10
of laſt for 5 . 103—5
of laſt for iſ.8 d. 348-4
757-3-4 Anfr.
Exa. 3. 149 C. at l. 2--8-6 per C.
2
298
of laſt for 8
59-12
its of laſt for 6 d. 3-14-6
5
L. 361–66 Anfi.
CASE IV.
When that Extreme, which is the Multiplier, is an
Aliquot Part or Aliquor Parts of the Unit, which is
Diviſor, take ſuch Part or Parts of the middle Number,
and their Sum is the Anſwer.
Exa. 1. What is the Value of 7 lib. when i Quarter
of a C. coſts l. 43-10?
4)43-IO
q lib. is * Quarter.
L. 10--17--6 Anfr.
Exa. 2. What is the Value of 1 Quarter 2 Nails at
-38 p. per Yard.
18
Here I take for i
Quarter, and for 2
for 1 Quarter,
Nails, and the Sum
for 2 Nails 2-3
6jb.9 d. is the Anfr.
4-6
3
per Stone
1. 6-9 Anfo.
Exa. 3. What is the Value of 4. lib. at l. 1-15-6
I-I56 4 lib. is of 1 Stone.
4
R
CASE
#
182
Contractions in the
CASE V.
When after a Queſtion in the Rule of Three is fta-
ted, and reduced according to the Directions already
given, that Extreme which is Diviſor, and any of the
other two are diviſible by the ſame Number, divide
both by that Number, and work with the Quotes in
their ſtead. By this means the Diviſor and that other
Term will be reduced lower, and conſequently the
Work eaſier; and ſometimes one of them will be re-
duced to 1, ſo that the Operation will be performed
by a ſimple Multiplication or Diviſion. This Rule
holds good whether the Proportion be direct or inverſe.
Exa. 1. What is the Value of 20 Yards of Cloth,
wben 8 Yards coft L. 6-12-8?
State 4)8:6--12--8:: 4)20 Here I divide
Quote 2:
5 5 Quote the Diviſor 8,
and che other
2133-03-4
Extreme 20 by
16-11-8 Anfr.
4; ſo that the
ftating after this
Reduction is the ſame as to ſay, If 2 Yards coft
1.6--12-8, what will s Yards coft ?
Exa. 2. If 1-2 . buy 8 Yards of Ribbon, how ma.
Yards
may Ibave for L. 4-4 at the ſame Rate ?
jb: yds. l. ll.
State 4)12:+)8::4-+ Here, after reducing the
20
third Termn top. I di-
vide the Diviſor 12 and
3: 2::84 the middle Number 8,
each by 4, and ſo the
Stating is, If 3 fb. buy
2 Yards, how many
56 Anfr. Yards will 848h. buy?
ny
2
3)168
Or,
Rule of Three Numbers. 183
Or, being it is the ſame, which of the other 2 Terms
I divide by that Number which meaſures the Diviſor,
I divide the Diviſor 12 and the other Extremne 842
each by 12, and the Operation is ſhorter yet.
Thus : 12) 12:8::12) 84
1:8::
7
7
M.
I
56 Anfr. as before.
Exa. 3. How long will 45 Men take to do a Piece
of Work, when 9 other ſuch could do it in 18 Hours?
M. Ho.
Here I divide
State 99:18::945
the Diviſor 45
1:18::
5 and the other
Extreme, each
by 9, the Quotes
5)18
are 5 and 1 ; ſo
69. 3--36 Anfr. that the Stating
is reduced to
this Form, If i Man take 18 Days, how long will 5
Men take?
Or I might have divided 45 and 18 by 9, and the re-
duced Stating would have been
M. H. M.
9 23:5
9
A
5) 18
3-36 Anfr. as before.
Exa. 4. If I buy 400 Yards for 120 L. how much
Will 1200 Yards coft at the ſame Rate?
K
Ids
184
Contrations in the
I
Ids. L Ids.
State 4130:120:: 12700 Here I divide the Divi-
4)4 : 120::4)12 for and the cther Ex-
:120::
3 treme 1200, each by
3
100, (or which is the
fame Thing I cut off
360
two o's from each, and
they are reduced 0 4
and 12, then I divide each of theſe by 4; ſo that the
Queſtion becomes the ſame as if I should ſay, If i Yard
coft 120 L. what will 3 Yards colt? The ſame Rc-
ſult would have been, it the Diviſor 4 and the middle
Term had been abbreviated, by dividing boi h by 40.
Thus: 40)400:40)120 :: 1200
Second State
3
Then by dividing the Divifor 10 and the other Ex-
treme both by 10, the State would be 1:3:: 120, the
Reſult of which is a ſimple Multiplication, as before.
IO
:: 1200
CAS E VI.
When a Queſtion is to be ſolved by ſeveral Opera-
tions of the Rule of Three Numbers, and the two
firſt Terms in each Operation are the fame, divide
the fecond Number by the firſt, and the Quote will
be a common Multiplier, by which you muſt multiply
all
your
third Numbers, and theſe Products are the
Anſwer
Exa. Let it be propoſed to divide 500 L. amongſt
Men, A, B, C and D, ſo that as oft as A has L B
may have 3; and as oft as B has 3 L. C may have 5 I.
and as oft as C has 5 L. D may have 6 L. how much
muſt each receive ?
Ву
Rule of Three Numbers.
18:5
By the coromon Method, the Proportions are,
16 : 500 :: 1:
31-57
3
16 : 500 : :3
93--15
5.
5 1565
7
16: 500 :: 7
218--15
I
16 : 500:
-500 Proof.
16
But by the contracted Method I work thus :
16)500(31.25
Then 31.25*I=31.25 Alfo
31.25x5=156.25
And 3.1.25*3=93.75 Laſtly 31.25x7=218.75
I ſhall not further inſiſt on theſe Contractions, leſt I
ſhould ra:her perplex than inſtruct the Learner, and in-
deed I'm of Opinion that thoſe lix Caſes already men-
tioned, comprehend the moſt, if not all that can be
pertinently ſaid on the Subject. I own, that if I ſhould
follow the Method of ſome Authors, by beginning
with the Price of an Unit at i far, and thus aſcending
till I came to L. 1, I might make their Number much
greater. Beſides, many of the Examples I have addu-
ced and innumerable others of the fame Nature) can
be ſooner wrought by the common Method of Re-
duction, and as there may be ſeveral Ways of work-
ing one and the ſame Queſtion, the chuſing the short-
eſt and ealieſt depends entirely on Practice and Expe-
rience. 'Tis true, when the Price of an Unit is an
Aliquot Part of the next ſuperior Species, (Cafe I.)
there can be no ſhorter nor eaſier Method than what I
have propoſed; but when it is an Aliquant, viz. a
Compound Part, or Paits of Parts, the Trouble, by
this Method, is conſiderably greater, and that by Re-
duction is preferable. So that when the Price is an
Aliquant Part of L. 1, ſuch as 3, 6, 7, 8, 9, 11, &c.J!.
or of 1. ſuch as 5, 7, 8, 9, 10, 11 d. I would in moſt
Examples find the Anſwer by multiplying, and then
reducing it to L. I now proceed to
R3
СНАР.
186
Rule of Three
CHAP. XXI. Rule of Three in common
Fractions.
1.STATE the three Terms as in Integral A.
rithmetick; and perform the Work by multiply-
ing and dividing, according to the Nature and Senſe of
the Queſtion, obſerving in the Operation the Rules of
Fractions already delivered.
2. After you have ſtated the three Terms as they
are given in the Queſtion; if any of them is a whole
or mix'd Number, or a compound Fraction, let it be
reduced to a ſimple one.
3. If the Terms can be reduced lower, let that be
done before you muliiply, and ſo the Operation will be
much facilitated by having them in lower Expreſſions.
4. If the Extremes are not Fracions of the ſame
Integer, reduce the one of them to the Denomination
of the other, (and it ſeems moſt convenient that the
Fraction of the lower Unit be reduced to an Equivalent
Fraction of the higher.)
5. If you cannot readily know which of the Extremes
is greateſt (and conſequently at a Loſs how to mulri-
ply and divide) reduce them to one Denominator, and
take the Numerators inſtead of the given Fractions,
and work with them as with Integers.
6. If the Anſwer comes out in high Terms, or if it
is an improper Fraction, let it be reduced; alſo, if
you want the Value of it in known Species, find it by
The laſt Caſe of Reduction of common Fractions.
Exa. I. If of a Yard coſt L. what is the Price
of of a Yard at the faine Rate?
Id. L. Id.
4
Here Yard coſts leſs than
x = 1, then fai(ii L. , therefore I multiply the
4. Jib. 9 d. Anfr. middle Number by the
Jeffer Extreme, and divide
the
1
in common Fractions.
187
the Product by 4, and the Quote (in its loweſt Terms
found by 5th Note of Cafe V. Divif.Common Fra&tions)
is ali L. whoſe Value is found by Caſe Xl. of Reduction.
Exa. 2. If z d. buy 2 0%. How much will
4
L.
buy ?
d. oz. L.
State:23:4
Here I reduce the
L. Oz.
L.
d. to the Fraction
When reduced 716::: 4
of i L. the mixt
X=? then zio)? (6840 Oż. Number 2 to a
4273 lib, Anfr.
ſimple Fraction and
the 4 L. to a Frac-
tion; and after multiplying, I divide the ſhorteſt Way
by 7th Note, Cafe V. of Diviſion, and the anſwer is
oz. which I reduce to lib. by 16.
Exa. 3. If of Cloth that is 1 Yard broad, 2 Yards
will make a Coat, how much in length of another
Cloth which is Yards in Breadth will make another
Coat of the ſame Dimenſions? Or rather thus, How
much of Yard broad Cloth is equal to 2 Yards of
broad
:
Here becauſe the Extremes
5x435 then 3)54(have the fame Denominator,
-4-2 Yd. 07 4 Yds. 2 Ors. I neglect it, and work with
I
the Numerators.
Exa. 4. If 4 }.coft 9 d. how much may be bought
for b. at the ſame Rate ?
å
State 1:14
• 3
4
Nail.
d. lib. fo.
28
2
3
State
93:45
By reducing the 2 firſt to Fractions,
Jo.
By reducing the firſt to the Fract. of 1/b. 38:25
Then by contracting the firſt Term, 1:
And by neglecting the Denominators 9, 7:
Therefore 21x2=4 and 43 == finfr.(by 4th Note
Caſe V. of Diviſion) =ij lib.
IAM Searlo
2
21.
5
2
Exa.
188
Rule of Three
Ex4.5. What is the Value of Yard at 17 Mh. 9 d..
per Yard?
8
:
Yd. b. d. rd.
State 1 : 17-9 ::
Here I reduce the firſt
:19
Term to the Denomi-
8:17--9
nator of the third, and
5
neglect the Denomina-
tors becauſe they are the
8)88-9
fame in both Extremes.
11-ij Anfr. But in all Queſtions of
this Nature, the Anſiver
is ſooner found by multiplying the middle Number by
the Number of the third Term, and dividing the Pro-
duct by its Denominator, as you ſee it happens to be
here, by means of the Method I have uſed in redu-
cing.
Exa. 6. If lib. be worth 12 5. 7 d. what is the
Price of I lib. at the ſame Rate ?
State
12--7::1
By reducing the 3d Term to a Fract.
having the ſame Denominator :: 12—7::
with the firſt Term,
By neglecting the Denominators; 3:127::4
3)50-4
1649 Ž. Anfr.
But in all Queſtions of this Nature, where the first
Term is a Fraction and the laſt an Integer both of the
fame Name, the Anſwer is ſooner found by multiply.
ing the middle Number by the Denominator of the
Fraction, and dividing the Product by its Numerator.
Here follow fome more Queſtions with their Anſwers, for
the Learner's further Exerciſe in this Rule:
Qu. 1. In what Time will B perform a Piece of
Work, when A alone can do it in 17 Days, and A
and B together in 15 Days? Anfr. 123 Days.
பெ.
in common Fractions.
189
Qu. 2. Suppoſing 3 to be of 12, what would be
of 20
Anfi.34.
Qu. 3. What is the Value of 50%. at 34 d. per lib.
Anfr. 3 far.
Qu. 4. I bought at one time 194 lib. of a certain
Commodity at 18š d. per lib. at another time I bought
of the ſame Commodity 58. lib. at the Rate of i į d.
per oz. what came each to at their reſpective Prices ;
and which of them was the betier Bargain, and by how
much per lib?
Anfr. The Value of 19 } lib. at 18 } d. per lib. is
L. IM9-4-0}}; the Value of 58 lib. at id. per
oz. is L. 4-2-0--285, and the laſt is the cheapeſt
Birgain by is d per lib.
To extract the Roots of common Fractions,
RULE. Reduce them to their loweſt Expreſlions,
and extract the Rout of Numerator and Denominator
ſeverally. Exa. 1. What is the Square Root of ?
This Fraction reduced to its loweſt Terms is m, the
Root of which is . Exa. 2. What is the Cube Root
of ? The Cube Root of 64-is 4, and that of 343
is 7; ſo the Anſwer is 2.
తడితడి
అంతలు తంతు తడి తడి తడి తడి తడికి
CHAP. XXII. Rule of Three in Decimal
Fractions.
STATE the Queſtion in the ſame Terms where-
in it is propoſed.
2. Reduce the common Fractions to Decimals; and
if any of the Terms are mix'd, reduce the inferior Spe-
cies to Decimals of the higheſt, annexing theſe Deci-
mals, inſtead of the common Fractions or inferior
Species
190
Rule of Three
Species to their reſpective Integers (if there are any)
with the Decimal Point betwixt them.
3. If the Extremes are not Fractions of the ſame U-
nit, ler the one of them be reduced to the Denomina-
tion of the other.
4. The Preparatory Work being made, multiply and
divide according to the Rules already delivered.
5: When the Quote is found, qualify it, that ſo you
may know how much of it is Integral, and how much
Fractional, and then you may reduce the Fractional
Part to inferior Species of the Integer, and thus you
will have the Anſwer complete, or nearly ſo, in a whole,
mix'd, or fractional Number, according to the Na-
ture of the Queſtion.
Exa. I. If d. buy oz. what will buy at the ſame
Rate?
State 1:1::
Reduced to Dec..75:.125::.8
123
?
:75)1.000(.1333&c. 0%. Anfr.
75
By Com. Fract.
fb, then
250
102. 1's Anfr. 225
250
225
250
225
25
The Anſwer to this Queſtion wrought Decimally is
.1333 (the 3 ſtill repeating) which wants of the com-
plete Value found by the Rule for valuing the Remain-
der, 7 godz. oz. or 10000
Ex4.
in Decimal Fractions.
191
Exa. 2. If 24 lib. coft 3& L. what will 40 & lib.
coſt at the ſame Rate?
State 24:33:: 40%
By reducing the 2.571428 : 3.55555::40.83333
Fract. co Decimals.
40.83333
1066665
1066665
1066665
1066665
2844440
1422220
L.
2.571428)145.1849464815(56.460825
12857140
20
16613546
15428568
9.216500
12
11849784
10285712
2.598000
4
15670728
15428568
2.392000
21216015
20571424
6445910
5142856
13030540
12857140
173400
After reducing the common Fractions to Decimals,
and multiplying and dividing according to the Rules, i
continue the Quote to 2 Places further, by adding two
o's to the Remainders, and after the Quote is quali-
fied,
192
Rule of Three
fied, I find the Value of the Fractional Part: And here
you are to obſerve, that all the Decimals being Circu-
lates or incomplete ; the Anſwer is not allogether ex-
act, becauſe I have made no Allowance for the Increaſe
that would have come out in the Product, if I had car-
ried the Decimals of the laſt Terms to further Places
before multiplying: However, the Defect is very in-
conſiderable, which will be manifeſt by comparing the
Decimal with the common Way.
Exa. 3. If 2. fo. buy 4 oz. of any thing, how much
will 5 L. buy? Jb. oz. l.
State 22:4:5
When reduced to 2.75:4 5
the Fract. of 1 L.
1375 : 4:15
5
OZ.
.1375)20.000(145.4545
1375
16
6250
5500
7.2720
oz. dr.
Anſi. 145-, and ſomewhat
more,
7500
6875
6250
5500
7500
6875
6250
5500
7500
6875
625
Here
in Decimal Fractions.
193
Here I reduce 2. to the Decimal of a L. (be-
Cauſe the third Term is L) and after multiplying and
dividing, I qualify the Quote, and it makes 145.45452
and there is till a Remainder of 45, which would hap-
pen though I ſhould have continued the Diviſion in
infinitum : However the Defect is ſo very inconſider-
able, that the Quote wants only 1375 of a Quarter of
2 Dram, as you may prove by common Fractions.
Exa. 4. A, B, C and D enter into Company: A
put in 1375 L. B 180 L. C 975 L. and D 547 L.
By trading they gain 463 L. How much of the ſame
ought each Partner to receive in Proportion to his
Stock ?
A 1375
B
180
C.975
D 5+7
3077 Sum of the Stocks.
Then 1375X. 15047123=206.89794125
IL. 2006-17-II2 for A's Sharc.
And 180X. 15047123==27.0848214
L. 27-1-8-I for B’s Share.
Allo 975X.15047123=146.70944925
= L. 146mm 14--2-1 for C's Share.
Laſtly, 5477.15047123=82.30776281
SL. 82-6-1-3 for D's Share.
L.
L. L.
3077:463 :: 1
I
3077) 463.001.15047123, &c. Proportional
3977
Share of I Los
15530
15385
14500
S
14500
194
Rule of Three
Brought over 14500
12308
21920
21539
3810
3077
7339
6154
11760
9231
2529
A 206–17-II-2
B 27.-i-8
Add
Ç 146-14- 2-I
82-6-I
I-3
Ladd
Share of
LD 82
Here the Loſs is 1 far. 462-19-11-3 Proof.
Here I take the Sum of the Stocks, and with it and
the Gain I find the Proportion Gain of 1 L. lo be
.15047123, etc. which multiplied by each's Stock gives
their ſeveral Gains. The Reaſon why the Proof wants
of the toral Gain is, becauſe the Gain of 1 L. was ini-
complete. And ſo of others.
Exa. 5. What is the Value of 17. C. 3 qrs. 16 lib. of
any Commodity, when 2 C. I gr. 17 lib. of the ſame
coſt L. 6-19-6?
74 17.000000
2.
7 4.250000
4
*1.607143 ferè
.4017857 Decimal of 1 gr. 17 lib.
28
in Decimal Fractions.
195
28
7
4
16.00000
4 00000
4 3-57143 ferè
.892857 Decimal of 3 qrs. r6. lib.
6.000
19.500
.975 Decimal of 19.1. 6 d.
C. L. C.
Siate 2.4017:6.975 :: 17.892
I2
e 1
20
6.975
89400
125244
161028
107352
In
2.4017).124796700(51.961821.51-10-2-3
Here I reduce the mix'd Numbers of cach Term)
to Decimals of the higheſt, by the expeditious ive-
thod mentioned before.
L.
100
.
Exa. 6. If 6 do gain 1 far. how much per cent. is
gained at the fame Rate?
1b. B.
State.5 :.025 :: 100
Here I reduce the
2 firſt Terms, each to
the Decimal of im.
:5)2.500 (5 h Anfr.
and the Anſwer comes
25
our in L. becauſe the
Proportion is, as ſb.are
to M. ſo are L. 10.L.
This laſt Queſtion wrought by Reduction the com-
mon Way, will ſtand as below. -
S. 2
196
Rule of Three
d. f.
L,
6:14: : 100
5
6
2000
12
24000
6 the middle Number.
Firſt Term 6)144000
5)24000
4)4900 far,
12)1200 d.
213) 1015 m.
5 L. Anfi.
Or thus :
Or thus:
d. f. L.
d. f. L.
6:1}:: 100
6:1}:: 100
20
20
2000
12
2000
12
24000
i 13 /
24000
When reduced,
d. f.
24000
4800
6:11: : 24000
1:15:: 4000 by dividing by 6.
It
6) 28800
4)4800
12)1200
2p) 1012
4000
800
5 L.
4)4800
12) 1200
2lb)rols
SL.
Rule of Five Numbers. 197
Or thus:
L. L. f. L.
2015: :: 199
@x
102=60=120
Then I'm)"29(4800 far.= L. 5, as by the other four
Methods
There are yet ſeveral other ways of working this
ſame Question; but I ſhall forbear inſerting them, ha-
ving dwelt too long upon this Rule already. Take
only one Queſtion more with the Anſwer, the Opera-
tion being left for your Exerciſe.
Exa. 7. A Gentleman has L. 100 Sterling, which
with the Intereſt thereof (ſuppoſe at 5 per cent.) he is
to ſpend in 20 Years, ſo as to ſpend equally each Year,
and lo exhauſt the whole Sum and Intereſt at the End
of 20 Years: Quær. How much he muſt ſpend yearly?
Anfr. L. 80--4-10-I ferè per annum.
CHAP. XXIII. Rule of Five Numbers.
INI
N this Rule are given five Numbers to find a fixth,
of which three contain a Suppofition, and the other
two a Demand.
All Queſtions in this Rule may be ſolved, either by
two Operations of the Rule of Three Numbers; or,
by one Diviſion. I ſhall begin with the firſt Method;
for which take the following Directions.
1. Of the three Terms of the Suppoſition, for firſt
down that one which is like the Thing fought, the o-
ther two place on the left, (ic matters not in what Or-
der) and thoſe that belong to the Demand, ſet on the
right. ſo as the fourch may be of the ſame Name with
the firſt, and the fifth of the fame Name with the fe-
cond.
2. Take the firſt, third and fourth Terms, and irih
S.
chele
1
198 Rule of Three, &c.
theſe make a Queſtion in the Rule of Three, and
find an Anſwer.
3. Take the ſecond Term, the Anſwer to this firſt
Queſtion, and the fifth Term, and make another Que-
ſtion of the Rule of Three, the Anſwer to this laſt is
the Anſwer to the general Queſtion, or the Thing re-
quired.
Exa. 1. If 8 Men are boarded 6 Months for L.
202
how much will ſerve 32 Men for 4 Months?
Men Mon. L. Men Mon. By the firſt Opera-
State 8 : 6:20.:: 32 : 4
tion I find, that if 8
Men L. Men.
Men ſpend L. 20
iſt, 8-: 20:: 32
(in any Time) 32
20
Men (in the ſame
Time) will ſpend
8)640
L. 80.
80
)
2dly, Mon. L. Mon.
6: 80:: :4
4
By the ſecond Operation, I
find that if 8 Men (or any
Number of Men) be inain-
tained 6 Months for L. 80
the ſame Number will be
maintained 4 Months for
L. 53-6-8. Both theſe
Proportions are direct.
6),320
L.53-6-8
Exa. 2. If 12 C. of any thing carried 100 Miles
coft L. 5--12, how many Ć. may be carried 150 Miles
for L. 12-12?
L
Rule of Five Numbers. 199
20
20
L. fb. M. C. L. fhM.
State 5--12:100: 12:: 1:
2-12 : 150
Iſt, 5--12:12::12--12 In the ſecond Opera-
tion I conſider, that
if 27 C. be carried 100
252 Miles (for any Sum)
leſs Weight muſt be
carried 1-50 Miles for
112)3024(27
the ſame Sum :. So that
224
the firſt Proportion is
direct and the other
784 inverſe.
784
LI
I2
M.
C.
M.
2dly, 100:27: : 150
1:00
1510)27010
18 c. Anfr.
Exa. 3. What Principal Sum will gain L. 3-7–6
in 9 Months, when L. 100 gains L. 6 in 12 Months ?
L. M. L. L. b. 4. M.
State 6: 12: 100 :: 3-7--06:9
Iff, 6:100::3-7-6 Firſt, I find that if L. 6
is gained by L. 100 (in
any time) L. 3-7--6
67
will be gained by L.
56ms in the ſame time,
20
I 2
810
Io0
6)31000
12) 13500
21011215
56-5
2dly,
200 Rule of Three Numbers,
12
2dly, M. L. M. M. In the ſecond Operation
12 : 56-5::9 I conſider, that if 12
Months require L. 56-5
for any other Principal)
9) 675-00
to gain any Sum, 9
L. 75-00 Anfr. Months will require a
greater Principal to gain
the fame Sum, becauſe the time is ſhorter, therefore
the Proportion is inverſe.
Exa. 4. If the Produce of 6 Pecks of Corn be 4
Bolls 14 Pecks, i Lippy in a Year, what would 19
Bolls 8 Pecks yield in 7 Years at the ſame Race (ſup.
poling that 19 Bolls 8 Pecks be ſown each of theſe 7
Years?)
P. Y. B. P. lip. B. P. F.
State 6:1:4-14-1::19-8:7
r. lip r.
ift, 6:4-14-1::19-8 2dly,, 1: 16276::7
16
16
7
122
78
4
19
16
4)113932
4128483 Pecks.
241 7123 3:
1780-3
312
313
312
626
313
939
.
6)97656
B. Pecks.
16276 lip. Anfr. 1780--3
Exa. 5. If 6 L. is the Intereſt of 100 L. for 12
Months, what is the Intereſt of 75. L. for 9 Monibs ?
L. M. L. L. 11.
State 100 : 12:6 :: 75: 9
11.
Rule of Five Numbers.
201
M. L. 1. M.
iſt, 100:6::75 2dly, 12:410 :: 9
6
9
,
4150
12) 40--10
20
Iolo
L. 3-7--6 Anfr.
Exa. 6. If 75 L in 9 Months gain L. 3--7--6, what
is the Intereſt of L. 100 for 12 Months ?
State 75:9:3-7er 6::100: 12
L. d. L.
M. d. Mon.
117, 75:810::100 2dly, 9: 1080:: 12
IOO
12
75)81000(1080 d. 9) 12960
12)1440
210) 1210
6 L. Anfi.
Exa. . What Principal Sum will raiſe L. 6 in 12
Months, when 75 L. raiſes L. 3amla in 9 Months ?
L. b. d. M. L. L. M.
State 3--7-6:9: 75
: 75 :: 6 : 12
M. L. M.
iſt, 3--6: 75 :: 6 2dly, 9:1331:12
20
20
9
67
120
12)1200
100 L. Anff.
12
12
810
1440
75
7200
10080
8.1103/1080ch
L. 133
ExA.
3
202 Rule of Five Numbers.
Exa. 8. What Time will L. 75 take to gain L.
3-7--6, when L. 100 gains L. 6 in 12 Months ?
State 100: 6.:: 12:: 75 : 37-6
Ift, 100: 125 : 75 . 2dly, 6:.16:: 376
100
20
20
15)1200016
75
67.
120
67
I 2
12
1440
450
450
810
16
4.860
810
1441231211788
୨
Mon. Anfi.
Exa. 9. If 75 2. in 9 Months gain l. 3--7-6, what
Time will l. 100 take to gain 1.6?
State 75: 3----7~6:9:: 100:6
L. p. d. M. L:
ift, 75:9::100 2dly, 3---7-6:63:: 6
9
20
20
Iloo)6175(62
67
I20
12
12
8:10
1440
6314
8640
360
720
83109197219
1108
12 Mon. Anfr.
Anda
Rule of Five Numbers.
203
And thus I bave varied the chird Example, and turn'd
it into all its Shapes, by continually altering the De-
mand.
Exa. 10. How many Buſhels of Corn will 150
Horſes eat up in 40 Days at the Rate of 14. Buthels
for three Horſes in 5 Days ?
H. D. B. H. D. By the firſt Operation
State 3:5:14:: 150:40 I find, that if Horſes
H. B. H.
up 14 Buthels
1.ft, 3:1f::150
in any Times. 150
14
Horſes in the ſame
Time will eat up 700
3)2100,
Buthels.
700
D. B. D. Then, if 5 Days ſpend 700
2dly, 5 : 700:: 40 Buſhels, 40 Days will ſpend
40
5600 Busheis.
eat
5)28030
5600
In the Operations of theſe 10 Queſtions, I have uſed
none of the contracted Methods mentioned in. Con-
tractions in the Rule of Thrée, particularly in Cafe V.
that the Learner may the better underſtand the Rule;
though I might have conſiderably abbreviated the most
of them.
When there is any Remainder in the firſt Operation,
let it be reduced to the loweit Denomination, and if
any Thing yet remains, annex it fractionwiſe to the
Number of the loweſt Species, and make it a part of
the middle Number in the ſecond Operation: For if
you ſhould neglect the firſt Remainder, or even the
laſt, the fival Anſwer would be incomplete, and in ma-.
ny Caſes the Defect very conſiderable.
Now I proceed to thew how all Queſtions in this
Rule may be ſolved by one Operation, which Method
is much eaſier, more exact, and more compendious
than the preceding.
RULE.
204
Rule of Five Numbers,
RULE.
State the five Numbers as before, and reduce the
correſponding Terms, viz. Iſt and 4th, alſo ad and 5th
to the ſame Denomination ſeverally (if neceſſary) and
the middle Number (if mix'd) to a ſimple one.
2. Compare the firſt, third and fourth Terms, and
find what Extreme would be Diviſor, if working by
two Operations, which mark : Then compare the
ſecond, third and fifth, marking the Extreme that
would be Diviſor there.
3. Which two Terms thus marked, multiply toge-
ther for a Diviſor, and the other three for a Dividend,
the Quote reſulting from this Diviſion is the Anſwer
to the Queſtion, in the ſame Name with the middle
Number.
For Examples, we ſhall repeat thoſe adduced for
exemplifying the Rule of Five Numbers by two Oce-
rations, to let you fee the Conſonancy of boch Me
thods: But be ſure to have a clear Underſtanding of
the firſt Method before you commence this.
Exa. I. repeated.
Men Mon. L. Men Mon.
X8: X6 : 20 :: 32 : 4
6
20
48
640
4
1
48
6)2560
:
28) 426-13-
L. 53- Omad
Here I ſay, If 8 Men ſpend L. 20 (im any Time)
32 Men will ſpend more in the ſame Time; there-
fore I divide by the lefler Extreme, viz. 8, which I
mark thus X. Then I ſay, IF 6 Months ſpend
If 20 l.
(or
by one Operation.
205
(or any Sum) 4 Months will ſpend leſs; therefore, I
divide by the greater Extreme, viz. 6; which two ſo
marked, I multiply together for a Diviſor, and the o-
ther three for a Dividend, and the Anſwer is L.53-6-8,
as before.
Exa. 2. repeated.
L. 1. M. C. L. Ñ. M.
5-12 : 100: 12:: 12-12:X 150
20
20
XII2
150
252
100
560
25200
12
II2
168120
)3024p0(18 C.
168
1344
1344
Here I reduce the firſt and fourth Terms to b. and
finding that if 112 h. carry 12 C. (any length) 252 fb.
will carry more weight the ſame length, mark the
leíſer Extreme 112 for one of the Diviſors ; chen if
12 C. be carried 100 Miles for any Sum of Money, I
conſider that leſs Weight will be carried 150 Miles
for the fame Sum; therefore I mark the greater Ex-
treme, viz. 150 for the other Diviſor, and after mul-
tiplying and dividing according to the Rule, the Quote
is 18 C. for the Anſwer.
T
206
Rule of Five Numbers
Exa. 3. repeated
6:12 : 100:: 3—7-6:X9
20
120
120
67
12
12
810
X 1440
9
100
12960
81000
12
1296ł0)9720010(75 L. Anfr. as before.
9072
6480
6480
Exa. 4 repeated.
P. r. B. p. lip. B. p. 7.
x6:X1 : 4-14-1:: 19-8: 7.
16
16
78
4
122
19
*
313
312
313
936
312
936
97050
7
61683592
Brought
by.one Operation. 207
Brought over 61683592
41113932
28483
16
71203
1780mm3 Anfr. as before.
Exa. 5. repeated.
M. L. L. M.
X 100: X12:6: : 75: 9
9
L.
573
In
I 200
6756
10 4051
12 40-IO
L. 3—7~6 Anfo. as before.
Exa. 6. repeated.
L. M. L. h. d. L. M.
X 75 : X9 : 37mob:: 100:-1.2
9
20
675
67
I 2
810
IOO
81000
I2 Ile
675)972000 (1410 d.
675...210)12121.
6 L. Anfr.as before.
2970
2700
2700
T 2
Exo.
2700
208
Rule of Five Numbers
Exa. 7. repeated:
3-7-6:9:75 :: 6:12 X
20
20
67
120
12
12
x 810
1440
75
12
9720
7200
1008
108000
9
1
9720)972000(100 l. Anfr. as before.
Exa. 8. repeated.
I. L. M. L. L. M. d.
100: 6:12:: X 75: 3-7-6
20
20
120
67
12
12
X 1440
810
75
12
7200
10080
9720
Ιοο
1081000
9721300(9 M. Anfr. as before.
972
Exa.
by one Operation.
209
Exa. 9. repeated.
75 : 3-7-6:9:: X 100:6
20
20
67
120
12
12
X 810
100
1440
9
81lood
12960
75
64800
90720
}/
81
9721000
108
12 Mon. Anfr. as before.
Exa. 10. repeated.
H. D. B. H. D.
X 3:X5 : 14 ::150:40
5
40
*
15
6000
14
15}
}}|
31 84000
28000
5600 Buſh. Anfr. as before.
T 3
Exs.
210 Rule of Five Numbers.
Exa. 11. If 12 Men build a Wall 30 Foot long; 6
Foot high, and 3 Foot thick, in 15 Days, in how ma-
ny Days will 60 Men, build 300 Foot long, 8 Foot
high and 6 Foot thick (according to the fame Rate ?)
30x6x3=540 folid Feet, and 300x8x614400 ſolid
Feet.
M. Feet D. · M. Feet.
12:X 540 : 15:: X 60: 14400
60
15
Diviſor 324100
216000
I 2
324)25920700(80 Days, Ans.
2592
0
APPENDIX.
2II
APPEN D I X.
CH A P I.
1. Of Simple Intereſt.
ALL Queſtions relating to limple Intereſt may be
folved by the Rule of Three Numbers, or that
of Five Numbers, after the Method of Exa. II, 16,
22, 25, 31, Pages, 147, 148, 152, 155, and 160; or
of Exa. 3, 5, 6, 7, 8, 9, Pages, 199, 200, 201 and
202.
5 per cent. is
Or, by finding the Intereſt of 1 L. for 1 Day, at
the Rate per cent. propoſed, and thereby multiplying
the Sum whoſe Intereſt is required, and the Product
by the Number of Days for the Anſwer.
Now, the Int. of 1 L. for 1 Day at
.00013698, &c.
Alſo, the Int. of 1 L. for 1 Day at
6 per cent. is
.00016438,&c.
Found thus:
L. L. L.
D. L. D.
100:5:: 1:.05. Then, 365:05::1:.00013698, do.
Alſo,
L. L. L.
D. L. D.
100:6:: 1:.06. Then, 365:.06::1:.00016438, 8,
And ſo for any other Rate per cent:
Exa
212
APPEN D I X.
Exa. 25, Page 175. repeated and wrought this Way.
.00016438
546
L. P. d.
00098628 Anfr. 76-0-41
00065752
00082190
00.08975148
847
62826036
35900592
71801184
76.01950356 Intereft.
Another Exainple. What is the Intereſt of L. 4.20
for 160 Days at L. 5 per cent. per annum ?
.00013698
420
27396
54792
.05753160
160
L. 9.205056 Anfr. or L. 9-4-14 ferè by this
Method.
The ſame Example wrought by the Rule of Five.
100: 365:5:: 420 : 160
160
67200
5
365100)336oloo(9 L.
3285
Brought
Of Simple Intereſt.
213
Brought over 3285
75
20
)1500(4 1. By this Method the An-
1460
ſwer is more complete.
40
I 2
)480(1 d.
365
115
4
46011 f.
365
95
II. Of Rebate or Diſcount at Simple Intereſt.
This you have in Exa. 35. Page 162, which it is
unneceſſary to repeat. I ſhall therefore propoſe an-
other Method; namely, Find the preſent Worth of
I L. due at the end of any Number of Years to come,
diſcounting at a propoſed Rate per cent, by which mul
tiplying the Sun propoſed, the Product is the preſent
Worth required.
Now, to find the preſent Worth of 1 L. due at the
end of any Number of Years to come, diſcounting
ſimple Intereſt at any Rate per cent. uſe the following
Method.
If
214
A P P E N D I X.
1
2
IIO : 100
If the Intereſt is at 5 per cent. ſay,
Years. L. L. L. L.
105 : 100 :: 1 :.9523817
:I:.9090917
3 115 : 100 :: 1 :
I:.869565-
4 120 : 100 :: 1 :.833333-
And ſo on to as many Years as you pleaſe.
But if the Diſcount be computed at L. 6 per cent.
per annum, the Proportions will be,
Years. L. L. L. L.
106 : 100 :: 1.943396
: 1 .892857
3
118 : 100 : ; 1.847457
4 124 : 100 : : 1.806451
And ſo for any other Rate per cent. Thus are cal-
culated the Numbers in the following Table I. ex-
tending to 30 Years, which may be continued to any
Number of Years you pleaſe. And tho' I have only
made one ar 5 per cent. you may by the fame Me.
thod (mutatis mutandis) compoſe other Tables at other
Rates,
1
2
I I2 : 100
TABLE
Of Simple Intereſt.
215
T A BL E I.
Shewing in Decimal Parts of i L. the preſent worth
of 1 L. due at the End of any Number of Years to
come under 31, diſcounting at the Rate of 5 per cent.
per annum, Simple Intereſt.
rears,
Years.
11-952381-41167-555555
2.9090914 17.540540-
3.869565+ 181.526;16-7-
4 .833333— 191:512820–
51.8
2015
0.7692317 21.487804-
- 740 740—22 +761917
81.714286+ 23 -465114
9.639655= 24.454545
10666666-251.4
11.645161-26 -134783
121.625
27-445532+
13.606060–281.415666-
141.588235-29 -408163
151.571428-1301.4
Such as are marked with the Sign + are ſome-
what more than the Truth, and thole marked with the
Sign -- are ſomewhat leſs.
Exa. What is the preſent Worth of L. 150 due at
the end of 5 Years, Rebate being computed at 5 L.
per cent. per annum, Simple Intereſt ?
Tabular Number againit 5 Years is
-.8
Which multiplied by
150
Gives L. 120 for the Apfwer
120.0
Operation
216
A P P E N D I X.
Operation after the common Way.
125 : 100:: 150
150
125)15000(120
125.
250
250
o And fo of others.
III. Of Annuities in Arrears at Simple Intereſt.
When an Annuity is in Arrear for any Number of
Years,
and you want to know the Amount of it,
Simple Intereſt being computed for each particular
Payment from the time it became due to the End of
the Number of Years propoſed, work as in the fol-
lowing Example.
There is an Annuity of L. 150 forborn to the End
of 5 Years, what is then due, Simple Intereſt being
computed at 5 per cent ?
Here you muſt find the Intereſt of L. 150 for
+
Years, 3 Years, 2 Years and 1 Year: All which be-
ing added to the Sum of the yearly Annuity, the Ag-
gregate is the Anſwer or Amount of L. 150 Annuity
foreborn for 5 Years at Simple Intereſt.
7:10
Now, the Intereſt of L. 150
at 5 per cent. for
3
4
30:
71
Years is 15:
22: 10
con
Sum of the Intereſts,
75:
5 times 150, viz. the Sum of the Annuity, 750:
Amount, L. 825:
But
Of Simple Intereſt, 217
But the Intereſt of any Sum for 4 Years + 3 Years
+ 2 Years + 1 Year, is equal to the Intereſt of that
Sum for 10 Years; wherefore, to find the Sum of the
Intereſt at once, take the natural Series of Numbers,
1, 2, 3, &c. to the Number of Years lefs 1, and
find the Intereſt for the Sum hereof, which added to
the Sum of the yearly Annuity gives the Amount as
before.
Thus in the laſt Exa. It2+3+4=10 Years.
L. r. L. L. r.
And 100: 1:5:: 150 : 10
IO
1500
5
Intereſt, L. 75 100
Sum of the Annuity, 750
L. 825 Anfr.
Hereupon is grounded the Conſtruction of the fol-
lowing
T A B LE II.
Shewing in L. and Decimal Parts of a L. 'the A-
mount of i L. Annuity being forbore to the end of any
Number of Years under 31, Sinple Intereſt being
computed at 5 per cent. per annum.
I.
2
2.0511.
Il
13.75121 31.5
15.3
22
33.55
3 3.15 316.9 23 35.65
4 4.3 18 55/24 37.8
5 5.5 59120.25250.
6 6.75 16/22.
7 8.05 17123.8 2744.55
8
94
1825.651 28 +6.9
9 10.8 119127.5529149.3
101 12.25, 121 29.5 3.
ir og 5
U
261 42.25
Eiga.
218 A P P E N D I X.
Exa. There is an Annuity of L. 150 forborn to the
End of 5 Years, what is then due, Simple Intereſt be-
ing computed at 5 per cent. per annum?
Tabular Number againſt 5 Years, 5.5
The Annuity,
150
Anfr. L. 825.0
IV. Of the preſent Worth of Annuities at Simple
Intereſt.
When an Annuity is to be ſold for ready Money at
Simple Intereſt, its Value will be found by the Method
of operating the following Queſtion.
There is an Annuity of L. 150 to continue 5 Years
to come, what is it worth in ready Money, Rebate be-
ing allowed at 5 per cent. Simple Intereſt ?
Here you muſt find the preſent Worth of L. 150
due at the End of 1 Year, the preſent Worth of L. 150
due at the End of 2 Years, its preſent Worth due at
the End of the 3d Year, likewiſe its preſent Worth
due at the End of the 4th and sch Years, the Sum of
all which is the preſent Worth of the Annuity.
There ſeveral Worths are thus found :
Year.
105 :
150 : 142.85714285
150 : 136.36363636
3
100 :: 150 : 130.43478261
150 : 125
125 :
IGO : 120.
I
IIO:
II:
I20 :
1
vt vu to
L. 654.65556182 Anſa.
or L. 654:13:1 and a Fraction more.
Whence
Of Simple Intereſt. 119
Whence is conſtrued the following TABLE III.
Newing the preſent Worth of L, I Annuity to con-
tinue any Number of Years under 31, Simple Intereſt
being computed at 5 per cent. Thus:
Preſ. Worth of I l. due
105 : 100 : :
:
Preſ. Worth of il. due
at 2 years end.
1.861471 { Pre W, of 1 2. Ann
Pref. W. of Il. Ann.
due at 2 years end.
Preſ. Worth of 1 l. due
115 : 100 :: 1 : .869565
869565 {
at 3 years end.
Pref. W. of 1. I. Ann.
2.731036 due at 3 years end .
Pref. Worth of il. due
120 : 100 :: 1 :
:I: .833333
at 4 years end.
3.564369
Pref. W.of il. Ann.
due at 4 years end.
U2
TABLE
220
Α Ρ Ρ Ε Ν DIX.
TABLE III.
I
2
Shewing the preſent worth of 1 L. Annuity to con-
tinue any Number of Years under 31, Simple late-
reſt being computed at 5 per cent.
Tears.
Years.
952381|16|11.536386
1.861471 17 12.076926
3 2.731036 18 12.603242
4 3.56+36919 13.116062
54.364369 20 13.616062
6 5.133600 21 14.103866
7 5.874340 22 14.580057
8 6.588626 23 15.045171
୨ 7.278281 24 15.499716
IO 7.944947 25 15.944160
8.590108 26 16.378943
9.215108/27 16.804475
139.821168 28 17.221 141
14 10.40940329 17.629304
15 / 10.980831/30/18.029304
Exa. What preſent Money will ſatisfy for an An-
nuity of L. 150 to continue 5 Years, Rebate being
made at 5 per cent. per annun, Simple Intereſt?
Tabular Number for 5 Years, 4:364369
150
II
12
21821845
4364369
Anfr. L. 654.65535
And now, for the Solution of all Queſtions con-
cerning Annuities in Arrears at Simple Intereſt, take
the following Caſes.
CASE
Of Simple Intereſt.
12I
CASE I.
Having the Annuity, Time and Rate (viz. the Inte.
reſt of i L. for a Year) to find the Amount.
RULE. "Take the natural Series of Numbers, 1,
2, 3, es to the Number of Years leſs 1, the Sum of
this Series multiçly by one Year's Intereſt of the An-
nuity, and the Product is the whole Intereſt due upon
the è unuity, vi, which adding the Total of the An-
nuities, the Sun is the Amount required.
Ex41. What is the Amount of L. 150 Annuity for
5 Years, allowing Simple Intereſt for each Year after
it falls due at 5 per cent?
1+2+3+4=10. Then .05X150=37.5 and 7.5X10
75 and 150x5=750. Laſtly, 750+75=825the
Amount iequired.
CASE II.
Having the Amount, Rate and Time, to find the
Annuity.
RUL E. Take the Sum of the natural Series of
Numbers, Y, 2, 32 &c. as before, to the Number of
Years leſs 1, which multiply by the Rate, and to the
Product adding the Years by this Sum; divide the A-
mount, and the Quote is the Annuity.
Exa. What Annuity will in 5 Years amount to L.
825 at 5 per cent. Simple Intereſt?
Thus it2+3+4=10. Then 1oX.053.5 and.5.45
55.5. Laſtly, 5.5)825(150 Anfr.
CASE. III.
Having the Annuity, Amount and Time, to find the
Rate.
RUL E. Take the Difference betwixt the Amount
and the Produ£t of the Annuity and Time, which di-
vide
U 3
722 Α Ρ Ρ Ε Ν DI X.
vide by the Product of the Annuity, multiplied into
the Sum of this Series, 1, 2, 3, Qur. to the Number
of Years leſs 1, and the Quote is the Rate.
Exa. At what Rate of Intereſt will an Annuity of
L. 150 amount to L. 825 in 5 Years, Simple Inte-
reſt?
150X55750; then 825_-750575 for a Dividend,
and 150x10=1500 for a Diviſor.
1500)75.000.05 Anfr.
CASE IV.
Having the Annuity, Amount and Rate, to find the
Time.
RULE. Divide twice the Amount by the Pro-
duct of the Rate into the Annuity, then ſubtracting
the Product of the Rate into ihe Annuity, from double
the Annuity, divide the Difference by the Product of
the Rate into the Annuity, and ſquare the Quote, di-
viding che Product by 4; which laſt Quote add to the
firſt, and from the Square Root of the Sum fubtract
I of the Number you ſquared, and the Remainder is
the Anſwer or Time ſought.
Exa. What time will L. 150 Annuity take to a-
mount to L. 825 at 5 per cent. per annüm, Simple
Intereſt?
825
150
.05
2
1650 twice the Amount. 7.50 Product of the Rate
into the Annuity.
7.5)1650(220
150
Of Simple Intereſt.
123
Then 39x3951521 and
4)1521
150
2
300
7.5 Subtr.
380,25 } Add
7-5)292.5(32
600.25(245
4
44)200
176
485)2425
2425
Laſtly, = 195 and 24.5–19.5=5 Anfr.
Of the Purchaſe of Annuities at Simple Intereſt,
CASE V.
Having the Annuity, Rate and Time, to find the
preſent Worth.
For the Solution of this caſe, work as in Article
4th, Page 217;
CAS E VI.
Having the preſent Worth, Rate and Time, to find
the Annuity.
RUL E. Take any Annuity at Pleaſure, and find
its preſent Worth as before, then fay, As that preſent
Worth is to its Annuity, ſo is the given preſent Worth
to its Annuity.
Exa. Whiat Annuity to continue 5 Years is worth
654.65535 preſent Money, allowing Simple Intereſt ac
5 per cent?
The
I 24
A P P E N D I X.
The preſent Worth of i L. Annuity for 5 Years
at 5 per cent. is 4.364369; therefore,
4.364369:1 :: 654.65535: 150 Anfr.
CASE VII.
Having the Annuity, preſent Worth and Rate, to
find the Time
RULE. Divide the given Annuity by the Series
of the Amounts of i L. Annuity for 1, 2, 3, &c.
Years, taking the Sum of the Quotes at every Steps
and thus proceed till you find a Sum equal to the given
preſent Worth; and the Number of Diviſions is the
Number of Years required.
Exa. What Time muft an Annuity of 150 L. con-
tinde, to be worth 654.6555617, ready Money, at the
Rate of 5 per cent. Simple Intereſt ?
1.05)150(142.857142851 Year.
1.1 )150(136.3636363=2 Years.
279.2207791
1.15)150(130.4347826=3 Years.
409.6555617
I. 2)150(125
=4 Years.
534.6555617
1.25)150(120.
55 Years.
654.6555617 Anfr. 5 Years.
СНАР.
Of Compound Intereſt.
125
CH A P. II.
Of Compound Intereſt.
WHEN the Amount of any Sum at Compound
Intereſt is required, work as in Queft. 39. Page
165, only it is eaſier to do it by Decimal Fractions.
But all ſuch Queſtions are more expeditiouſly ſolved
by finding the Amount of 1 L. for any Number of
Years at any Rate per cent. by which multiplying the
Sum propoſed, the Product is the Anſwer.
Thus, Intereſt being at 5 per cent.
Tears.
100 : 105
: 105 :: 1 : 1.05
100, : 105
105 :.: 1. 05.: .1.1025
3
100': 105 : : 1.1025 :: 1.157625
4 100: 105 : : 1.157625 :: 1.21550625
And thus is conſtrued the following
i
I
2
TABLE
1 26
A P P E N D I X.
T ABLE IV.
Shewing the Amount of 1 L. forborn to the End
of any Number of Years under 31, Compound Inte-
reſt being computed at 5 per cent.
Years,
Years.
Il1.05
16,2.1828746
211.1025
17 2.2920183
3.1.157625 18 2.4066192
4 1.2155063 192.5269502
5 1.2762816 202.6532977
1.3400956 21 2.7859626
711.4071006 22 2.9252607
8 1.47745542313.0715238
91.5513282 24 3:225099
10 1.6288940 25 3:3863549
II 1.7103393 26 3.5556727
12 1.795856327 3.7334563
131 1.8856491 28 3.9201289
14 1.9799316 29.4.1161356
1512.07892823014.3219424
Exa. What will L. 315:10 amount to, being for-
born 3 Years, Compound Intereſt being computed ac
5 per cenzt. per annum ?
Tabular Number for three Years, 1.157625
315.5
Ariſt. L. 365:4:7
5788125
5788125
1157625
3472875
365.2306875
Il, of
Of Compound Intereft.
(127
II. Of Rebate or Diſcount at Compound Intereſt.
Exa. There is 365.2306875 due at the End of
3
Years to come, I demand how much preſent Money
will ſatisfy for the ſaid Debt, diſcounting at the Rate
of 5 per cent. per annum, Compound Intereſt ?
L. L. L.
L.
105 : 100 :: 365.2306875 : 3+7.83875
105 : 100 :: 347.83875 : 331.275 .
105 : 100 ; : 331.275 : 315.5 Anfr. or L.315:10
And upon this is founded the Calculation of the fol-
lowing Table: For,
L. L. L, L.
105 : 100 : : 1.952381 fere .
105 : 100 : : .952381 :.907029, &c.
105 : 100 : : .907029 : .863837, &c.
1
TAB L E V.
Shewing the preſent Worth of 1 L. due at the end
of any
Number of Years to come under 31, diſcount-
ing at 5 per cent. per annum, Compound Intereſt.
2
I! .952381161 .4581115
:9070294 171 4362967
3.863837618
4155207
.822702519 .3957339
-7835262 20 :3763895
-7462154 21 :3589424
7 710681322 ·3418499
6768394 23 3255713
9.6446089 24 .3100679
.6139133 25 2953028
: 584679 26.2812407
12 5568374/27 .2678483
13
-5303213 28
.2550936
14 505067929 .2429403
.48101711301 .2313774
IO
15
Exa
228 APPENDI X.
Exa. What preſent Money will ſatisfy for a Debt of
365.2306875 due at the end of 3 Years to come, dil-
counting at 5 per cent. Compound Intereſt?
365.2306875
25566148125
10956920625
2.9218455000
10956920625
21913841250
29218455000
315.4997813979375 A 12fr. L. 315 : 10 ferè,
becauſe the Tabular Number is ſomewhat too ſmall.
III. Of the Amount of Annuities at Compound
Intereſt.
There is an Annuity of L. 150 forborn to the end
of 5 Years, what is then due, Compound Intereſt be-
ing computed at 5 per cent. per aninam ?
L. L. L. L.
100 : 105 : : 150: 157.5
100 : 105 : : 157.5 : 165.375
100 : 105 : : 165.375 : 173.64375
100 : 105 : : 173.64375 : 182.3269375
150.
Here you muſt work as if it was a Principal lying
out for any Number of Years at Compound Intereſt.
For,
150
Of Compound Intereſt.
229
165.375
150 L. due at the end of the 5th Year is 150.
And 150 L. due at the end of the 4th Year,
will at the end of the 5th Year amount to 157.5
And 150 L. due at the end of the 3d Year,
will at the 5th Year's end amount to
And 150 L. due at the end of the 2d Year,
will at the 5th Year's end amount to 173.64375
And 150 L. due at the Iſt Year's end, will
at the 5th Year's end amount to
182.326937
The Sum of all which is what is due at
the 5th Year's end.
But ſuch Queſtions are more eaſily ſolved by the
following
} 828.845687
T A BL E VI.
I.
I
1
2
18 28.1323799
Shewing the Amount of 1 L. Annuity to continue
any Number of Years under 31, Compound Intereſt
being computed at 5 per cent. per annum.
Tears,
Tears.
16, 23.6574918
2.05 17 25.8403664
3.1 3.1525
4 4.310125 1930.5389989
51 5.525631220133.0659189
6.8019128 21 35.7192518
71 8.1420079 22 38.5052144
81 9.5491079 23 41.4304679
9 11.0265639 24 44.501999
10 12.5778925 25 +7.7270899
I14.2067871 2651.1134537
12 15.9171265 27 54.6691264
13 17.7129799 : 28 58.4025827
14 19.5986289 29 62.3226999
15121.5785599 : 30166.4388474
X
For
230
Α Ρ Ρ Ε Ν DI X.
For 1 L. Annuity due at the end of 1 Year is 1.000000
And I L. Ann. due at the end of 2 Years is 2.05
Thus found;
100: 105 :: 1:1.05, to which adding the
I L. Annuity, the Sum is 2.05.
Alſo, I L. Ann. due at the end of 3 Years is 3.1525
Thus found;
100: 105:: 2.05: 2.1525, to which adding the
I L. Annuity the Sum is 3.1525.
And i L. Annuity due at the end of 4 Years is 4.310125
Thus found;
100:105 :: 3.1525: 3.310125, to which ad-
ding the i. L. Ann. the Sum is 4.310125.
And after the ſame Manner is the preceding Table
calculated, where the Numbers are ſomewhat deficient
of their true Value; ſo that the Anſwer found thereby,
may perhaps want i Farthing, and in ſome caſes
I Halfpenny.
Exa. What is the Amount of L. 150 Annuity for-
born 5 Years, at 5 per cent. per annum, Compound
Intereſti
5.5256312
150
2762815600
55256312
Anfr. L. 828.8446800
IV. Of the preſent Worlb of Annuities at Com-
pound Interest.
The preſent Worth of Annuities at Compound In-
tereft, is found after the Meihod of the following Ex-
ample.
Exa. What is the preſent Worth of an Annuity of
150 L. to continue 5 Years, rebating at 5 per cent.
Compound Interest?
L.
Of Compound Intereſt. 232
L.
L.
150.
: 142.8571428531. year.
L. L. 142.85714285:136.05613605=2. year.
105:100 :: 136.05613605:129.57727243=3. year.
129.57727243: 123.4069261254. year.
123-40692612:117.53040583=5. year.
Sum of the ſaid preſ. Worths 649.42788328 Anſi.
Hence is caculared the following
T ABLE VII.
1
2
Shewing the preſent worth of 1 L. Annuity payable
by yearly Payments, and to continue any Number of
Years not exceeding 30, diſcounting at 5 per cent.
Compound Intereſt
Tears.
Years.
-952381 16 10.8377695
1.8594103 17 11.2740662
32.7232479 18 11.6895869
4 3.5459499 19 12.0853208
4.3294759 20 12.4622103
6 5.0756999 21 12.8211527
5.7863734 22 13.1630026
8 6.64632122313-4885739
9 7.1078217 24 13.7986418
7.72 1734925 14.0939445
8.3064142 26 14.3751853
8.8632516 27 14.6430336
139.3935689 28 14.8981272
141 9.89864082915.1410735
15 / 10.37965811 30115.3724509
5
7
IO
II
12
X 2
For
232
A P P E N D I X.
L. L. L. L.
For 105 : 100 :: 1:.9523815ift Year.
105 : 100::.952381 .907029
1.859410=20 Year.
105 : 100::.907029: .863837
2.723247=3d Year.
105 : 100::.863837 : .822702
3.54594954th Year, &c.
Exa. What is the preſent Worth of an Annuity of
L. 150 to continue 5 Years, Rebate being at 5 per
cent. Compound Intereſt.
Tabular Number for five Years, 4.329475
150
21647375
4329475
649.421250 41f1.
Here follow the Solutions of the moſt common
Caſes coi.cerning Annuities in Arrear at Compound
Intereſt.
CASE 1,
Having the Annuity, Rate and Time, to find the
Amount.
RULE. Multiply the Annuity by the Amount of
I L. for the Time, and at the Rate propoſed, and the
Product is the Anſwer.
CAS E II.
Having the Amount, Rate and Time, to find the
Annuity.
RULE. Divide the Amount given by that of 1 L
An-
Of Compound Intereſt.
233
Annuity for the Time, and at the Rate propoſed, and
the Quote is the Anſwer.
Exa. What Annuity will amount to L. 828.84465
in 5 Years at 5 per cent. per annum, Compound In-
tereft?
5.525631)828.84465(156 Anfr.,
CASE. III.
Having the Annuity, Rate and Amount, to find the
Time.
RULE. Find a Principal of which 1 Year's Inte-
reft is equal to the given Annuity, the Sum of this and
the given Amount is the Amount of that Principal for
the given Rate and Time fought : Then divide that
Amount by its Principal, and multiply the Rate conti-
nually by itſelf till the Product be equal to the former
Quote, and the Number of Multiplications is the An-
ſwer or Time fought.
Exa. In what Time will L. 150 Annuity amount to
828.84465 at 5 per cent. Compound Intereſt?
.05:1:: 150: 3000,and 3000+828,60=3828.84465
Then '3000)3828.84465(1.2762815531.05 raiſed to
the 5th Power, or multiplied 5 times into itſelf. So
the Anſwer is 5 Years.
.
Of the Purchaſe of Annuities at Compound Intereſt.
CAS E IV.
Having the Annuicy, Rate and Time, to find the
preſent Worth.
RULE. Find the preſent Worth of each Year's
Annuiry by itſelf, the Sum of all which is the preſen,
Worth fought. Or,
2. Multiply the Annuity by the preſent Worth of
I L. Annuity for the Time, ani at the Rate propoſed,
and the Product is the Anſwer.
X 3
CASE
234
APPEND I X.
5 per cent?
CASE V.
Having the preſent Worth, Rate and Time, to find
the Annuity
RULE. Divide the preſent Worth given by that
of 1 L. Annuity, and the Quote-es che Anſwer.
Exa. What Annuity, to continue 5 Years, will
649.42125 purchaſe, allowing Compound Intereſt at
4:329475)649.42125(150 Anfr.
CAS E VI.
Having the Annuity, preſent Worth and Rate, to
find the Time.
RULE. Find a Principal whereof 1 Year's Inte-
reſt is the Annuity given, from which ſubtract the pre-
fent Worth, and the Remainder is the preſent Worth
of chat Principal, conſidered as a Sum due at the End
ofthe Annuity ; then find what Time this preſent Worth
will take, to amount to the Principal found.
Exa. What time muſt an Annuity of L. 150 con-
tinue to be worth 649.42125 in ready Money?
:05:1:: 150 : 3000; then 3000-649.42 125=
2356, 7875 preſent Worth of L. 3000, due at the
end of the Time ſought.
Then 2350.57875)3000.00000(1.276281,6r. which
being the 5th Power of 1.05,5; is the Number of Years
fought.
V. To find the preſent Value of a Free-hold Eſtate
or Annuity to continue for ever.
RULE. Find a Principal of which 1 Year's Inte-
reſt is the Rent or Annuity given, and this is the Price
fought.
Exa. What is the Price or preſent Value of an An-
nuity of 150 L. to continue for ever, diſcounting at 5
der cent. Compound Intereſt?
05:1:: 150: 3000 Anfr.
OF
Place this beforPay. 2355
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9
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Fig. 10.
Fig.11.
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b
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Fig. 13.
D
IT
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Fig. 112.
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Fig 15
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235
OF
MENSURATION
OF
SUPERFICIES and SOLIDS.
SECT. I. Of Superficies.
PRO P. I.
HAVING the Diameter of a Circle, to find the
.
RULE. The Diameter of a Circle, being 1, the
Circumference is 3.1416 fere, and all Circles being
to one another as their Diameters; therefore, as 1 :
3.1416 :: ſo is the Diameter of any Circle to its Cir-
cumference.
Exa. What is the Periphery or Circumference of a
Circle, whoſe Diameter is 15 (Inches, Feet, Yards, &c.)
I: 3.1416::15:47.124. Circumference.
PROP. JI.
Having the Circumference of a Circle, to find the
Diameter.
RULE. This being the Converſe of the laſt Prop.
fay, As 3.1416 is to I, ſo is the Circumference given,
to the Diameter required.
Exa. What is the Diameter of a Circle whoſe Cir-
cumference is 47.124?
3.1416:1:: 47.124: 15 Diameter.
Or, becauſe I divided by 3.1416 quotes :3183
therefore the Circumference of any Circle multiplied
by : 3183 gives the Diameter, for 47.124X.3183315
ferè, for it is =14.999, &c.
PROP.
23.6 MENSURATION OF
PROP. II.
Having the Diameter of a Circle, to find the Area.
RULE. Multiply of the Circumference by the
Diameter, and the Product is the Anſwer.
Or thus : The Diameter of a Circle being 1, the
Area is .7854 ferè; wherefore, multiplying the Square
of the Diameter of any Circle by the foreſaid Num-
ber, you have the Area.
Exa. What is the Area of a Circle whoſe Diameter
is 15?
By Prop. I. the Circumference is 47.124, where-
47.124
fore, XI5=11.781X15=176.715 Area.
4
By the ſecond Method 15X153225, and 225X.78.54
=176.7153 as before.
PRO P. IV.
Having the Area of a Circle, to find the Diane'er.
RULE. Divide the given Area by .7854, and he
Quote is the Square of the Diameter, whole Root is
the Thing fought.
Exa. What is the Diameter of a Circle, whoſe Arca
is 176.-15?
-7854) 176.715(225 Square of the Diameter, whoſe
Root is 15
PRO P. V.
Having the Circumference of a Circle, to find the
Area.
RULE. Say, As I is to .079577, ſo is the Square
of the Circumference to the Area.
But this being too tedious, find the Diameter by
Prop. 2d, and thence the Area by Prop. 3d.
Exa. What is the Area of a Circle whoſe Circum-
ference is 47 134:
By the firſt Method 47.124 ſquared, is 222.671376,
which
Superficies and Solids.
237
which multiplied by .079577 produces 176.698.82,66.
for the Area.
By Prop. 2. the Diameter is found to be 15, and
by Prop. 3. the Area is 176.715.
PRO P. VI.
Having the Area, to find the Circumference.
RULE. Divide the given Area by .07957.7, and
the Quote is the Square of the Circumference, whoſe
Root is the Thing fought.
Exa. What is the Circumference of a Circle whoſe
Area is 176.715?
176.715
=2220.68, whoſe Root is 47.124.
.079577
PRO P. VII.
To find the Area of a Semicircle, as abdemc, Fig. I.
RULE, Multiply the Square of the Diameter by
-3927 (viz. į of.7854) and the Product is the An-
fwer.
PROP. VIII.
To find the Area of a Quadrant, as acdb, Fig. I.
RULE. Multiply the Square of twice the Radius
by .19635(viz. 1 of.7854) and the Product is the
Anſwer.
PRO P. IX.
To find the Area of a Sector and Segment of a
Circle, as the Sector bced, and the Segment bed, Fig. 1.
In order to find the Area of the Segment of a Circle,
there muſt always be given the Circle's Diameter, or
elſe its Circumference or Area, to find the Diameter ;
and 2dly, the Segment's Baſe, otherwiſe called the
Chord, as be, or the vers’d Sine dh, which is the Seg-
ment's Height: Then 'tis plain, that if the Area of the
Triangle bce, be taken from the Area of the Sector
bred,
238 MENSURATION of
borit, ihe Remainder will be the Area of the Segment
bhed; and if the Area of the Segment bhed be taken
from that of the whole Circle, the Remainder will be
that of the other Segment bakme:
Let the Diameter am be 15
Then cd is
7.5
Let db be
3
Then bc is
4.5
nd bh is found by fubtracting the Square of he, viz.
20.25 from the Square of bc, viz. 56.25 ; and taking
the Square Root of the Remainder, viz. 6=bh. Now,
the Arch bd, or the Quantity of the Angle bed, muſt
be found by Trigonometry, thus, bc: bh:: Radius :
Sine of c; that is, 7.5:6:: 10.0000: 53° 7'; Ther, as
the Circumference of the whole Circle in Degrees, is
to the Circumference in equal Parts, ſo is any Arch in
Degrees (here bd 53°7') to the ſame Arch in e-
qual Parrs. The Circumference of a Circle whoſe
Diameter is 15 is (by Prop. I.) 47.124: therefore,
360° : 47.124::5307: 6.948=bd, conſequently 6.948
x7.5552.u (found by Prop. 18. Sect. 1.) Area of the
Sector bced; and bh being 6, be muſt be 12 ; where-
fore, by the ſame Prop. 18. Sect. 1. 12X2.25=27, Area
of the Triangle bce; and their Difference is 25.11, for
the Area of the Segment bed.
The Area of the Segment of a Circle may be other-
wiſe found, viz.
RULE. Take 2 times the Square of the Semi-
diameter of the Circle, from which ſubtract 13 times
the Product of the Semidiameter into the Difference
betwixt che Segment's Height and the Circle's Semidi-
ameter, as alſo the Square of the ſaid Difference; the
Remainder divide by il times the Semidiameter +
the Difference betwixt the Semidiameter and the Seg.
mene's Height, which Quote multiply by half the Seg.
ment, and the Product is the Segment's Area.
7.5x7.5X2}=131.25 and 7.5X4.5X11=45.
4.5X4.5=20.25. The Sum of thele two laſt is 65.25;
therefore
Allo,
Superficies and Solids.
239
therefore 131.25-65.25=66, which divided by 15.75,
viz. 11 times the Semidiameter, br. quores 4.19,
and this multiplied by half the Segment, viz. 6, gives
25.14 for the Segment's Area.
PRO P. X.
To find the Area of the Lune, or ſuch a part of a
Circle as armca.
RULE. Find the Area of the Semicircle abdeme,
from which ſubtracting that of the Segment armia,
the Remainder is the Area of the Lune armia.
PROP. XI.
To find the Area or ſuperficial Content of a Square.
A Square is a Figure conſiſting of 4 equal Sides, and
as many Right Angles, as ABC”, Fig. 2.
RUL E. Multiply the length of a Side by itſelf,
and the Product is the Anſwer.
Exa. How many Square Feet are contained in a
Table, each of whoſe Sides is 3 Feet 6 Inches, or 42
Inches?
3.5X3.5=12.25=12 Sq. Feet.
Or, 31x3=121 as before : For
Or thus: 42
31
42
3:6
10:6
84
168
I : 9
12 : 3
144)1764(121 as before.
(1
1728
36
Exa. 2. There is a ſquare Piece of Ground, the
length of cach of whoſe Sides is 136.25 Yards, what
is the Content in Acres?
136
240 MENSURATION of
136.25X136.25=18564.0625, which divided by 4840,
A. R. P. Tds. Feet. Ilicha.
quotes 3.83555 Acres=3.: 3:13 : 20 : 7 : 44
P RO P. XII.
To find the Area or ſuperficial Content of a Rec-
tangle, Parallelogram or Oblong, Fig. 3.
This Figure conſiſts of t Right Angles, having its
oppoſite Sides equal.
RULE. Multiply the Length by the Breadth, and
the Product is the Area ſought.
Exa. There is a Table in form of a long Square or
Oblong, whoſe Length is 5 Feet 4 Inches, and Breadth
3 Feet 8 Inches: How many ſquare Feet doch it con-
tain ?
Feet. Inch,
4:5–53 Inchi
3 : 8544 Inch. Then, 53X4432332 fq. Inches,
which divided by 144 (the ſq. Inches in a ſq. Foot)=
16 fq. Feet, and 28 ſq. Inches, Anfr.
Or thus; 4:5
3:8
for 3
13 : 3
2 : 2.5 for 6 Inches.
0 : 8.833 for 2 Inches.
16 : 2.333, &c.
that is, 16 ſq. Feet tiç or of a fq. Foot +_333;&c.
1000,&c.
Parts of i'a of a ſquare Foot: So that the Anſwer is
16} fq. Feet fere.
Exa. 2. There is a Room in form of a long Square,
whoſe Length is 20 Feet and Breadth 15: How many
Yards of 3 Quarter Broad Cloth will be ſufficient to
hang the ſame, its Height being 7 Feet ?
20
Szuperficies and Solids.
241
1
204-15535 and 35x2570 and 70x7=490= the ſq.
Feet contained in the 4 Walls=70560 fg. Inches, and
27X36==972 the fq. Inches contained in i Yard ;
rds. Ors. Nails.
wherefore 972)70560172--2-I Anfr.
PROP. XIII.
To find the Area or ſuperficial Content of a Rhom.
bus or Rhomboides, Fig. 4 and 5.
A Rhombus has 4 cqual Sides, whereof the two op-
polite Sides are equal, and the Sides parallel, and may
be repreſented by a Diamond, or a Square out of its
true Poſition, as Fig. 4.
A Rhomboides has 4 equal Sides, whereof the two
oppoſite are equal and parallel, and the oppoſite Angles
equal, being a Parallelogram out of its crue Poſition,
as Fig. 5.
RUL E. In either of theſe Figures, multiply the
Baſe into the perpendicular height, and the Product is
the Antwer.
Exa. There is a piece of Wainſcot in form of a
Rhombus, whoſe Baſe or Lengch is 2 1 Feet, and the
Height or Perpendicular CH, 2 Feet 4 Inches; What
is the Area in ſquare Feer?
2 Feet
30 fuch.
21 Feet = 28 inch. Then 30x28=3.10, and this
divided by 144 quotes sis fq. Feet = 5 q. Feet and
120 19. Inches.
Or chus;
Or thius;
2 : 4
2 : 4
2 : 6
21
4:8
I : 2
4:8
I : 2
5:10
5:10
Y
PROP
242
MENSURATION of
PRO P. XIV.
To find the Area or ſuperficial Content of a Trape-
zium.
This Figure conſiſts of 4 Sides and 4 Angles, which
are generally neither parallel nor equal, as Fig. 6.
RULE. Divide it into two Triangles, by drawing
a Diagonal Line from one of its acute Angles to the
oppoſite one, and thereon letting fall 2 Perpendiculars
from the oppoſite Angles, multiply half the length of
the Diagonal into the Sum of the Perpendiculars (or
the Length of the Diagonal into half the Sum of the
Perpendiculars) and the Product is the Area.
Exa. Suppoſe the Diagonal AC 40 Yards, the Per-
pendicular Pp 19 Yards, and the other Perpendicular
PP 12 Yards; What is the Area of the Figure ?
=20; and 19+12=31. Then 20x31=620 ſq.
Yards for the Anſwer.
14
2
PROP. XV.
To find the Area of any regular Polygon, as Fig. 7.
A Regular Polygon is a Figure conſiſting of more
than 4 equal Sides.
RULE. Circumſcribe it with a Circle touching
the angular Points; and from the Centre det fall a Per-
pendicular (on any Gde) half of which multiplied into
the Sum of the sides gives the Area.
Exa. Required, the Area of the Pentagon ABHDE,
each of whoſe Sides is 75 Yards, and the Perpendicu-
·lar CG 51 Yards ? 75x5=375Sum of the Sides,
and 255 half the Perpendicular, wherefore, 3754.25.5
=9562.5 ſquare Yards, which divided by 4840, the
ſquare Yards in an Acre, quotes 1.976 ferè Acres for
the Anſwer.
PROP.
Superficies and Solids.
243
PRO.P. XVI.
To find the Area or ſuperficial Content of an irre-
gular Polygon, as Fig. 8.
An irregular Polygon is a Figure cooſiſting of more
than 4 Sides, all of which, as alſo the Angles, are ge-
nerally unequal.
RULE. Reduce it into Triangles by drawing Dia-
gonals, and the Sum of their Areas is the Area requi-
red.
Exn. Required the Area of the irregular Polygon,
HIKLM, the Length of the Side HI=5, IK=51,
KL= 81 LM=9, and MH-7, and the Perpendicu-
lar li=4.5, Mm=6, and Kk-6.5.
3.5X4,5=15.755 Area of the Triang. HIM.
2.75X6=16.5 Area of the Triang: IMK.
4.5X6.5=29.25 Area of the Triang: KML.
61.5 Area of the whole Fig. HIKLM.
PROP. XVII.
To find the Area of a Parallelopleuron, as Fig. 9.
This Figure conſiſts of 4 Sides, having two oppoſite
ones parallel, and the 2 Angles at each end equal to
one another.
RULE. Draw a Diagonal, on which letting fall
2 Perpendiculars, by of their Sum multiply the Dia,
gonal, and the Product is the Area.
Exa. Suppoſe the Diagonal AB 36, the Perpendicu-
lar Pp 21, and the other Perpendicular pp 13.21 +13
=34 and 342=17; and 36X17=618 for the Area
of the Figure ApBP.
PROP. XVII.
To find the Area or ſuperficial Content of a Trip
angle, as Fig. 1o.
RULE. Of what kind foever it be, multiply the
Lungth of the Baſe by half the Length of the Perpen-
dicular
Y 2
244
MENSURATION of
dicular, or Height: Or, multiply the Length of the
Perpendicular by half the Length of the Baſe, and you
have the Square Content ſought; for every Triangle
is half its circumſcribing Parallelogram.
Exa. In the Triangle ABC, ſuppoſe the Baſe BC 43>
and the Perpendicular AD 27 ; What is the Area ?
43
=21.5 and 21.5X27=580.5 Area ſought.
PRO P. XIX.
To find the Area or ſuperficial Content of an Ellipſe
or Oval, as Fig. II.
An Oval or Ellipſe is formed by cutcing a Cone
through flaunt-wiſe, or not perpendicular to the Baſe.
RULE. Multiply the greateſt Diameter by the
leaſt, and the Product by .7854, and this laſt Product
is the Area required.
Exa. Suppoſe in the Ellipfis ABCD, the greateſt
Diameter AC is 41, and the leaſt Diameter BD 28;
What is the Area?
41X2851148 and 1148x.78543901.6392 for the A-
rea or ſquare Content.
PROP. XX.
To find tbe Area of a Parabola, as Fig. 12.
This Figure is form’d by cutting a Cone parallel to
irs oppoſite Side.
RULE. Multiply the Height by the greateſt
Breadth, and of the Product is the Anſwer.
Exa. Suppoſe the Height AB 46, and the greateſt
Breadth CD 32, what is the Area
46x32=1472, which multiplied by 2, and the pro-
duct divided by 3, quotes 931 } for the Anſwer.
PROP. XXI.
To find the Area or ſuperficial Content of a Cute.
A Cube is a Solid, bounded by 6 equal Squares in
form of a Die.
RULE.
Superficies and Solids. 245
RULE. Find the Area of any of the Sides 'by
Prop. II, and multiply it by 6 for the Anſwer.
Exa. Suppoſe a Cube having its Side 16, what is the
fuperficial Content?
16x16=256 and 256X6=1536 Anfr.
PRO P. XXII.
To find the Area or ſuperficial Content of a Globe
or Sphere.
A Globe or Sphere is a Solid, bounded or included
within one regular Superficies; and is formed by the
Rotation of a Semicircle about its Diameter.
RULE. Multiply the Diameter or Axis into the
Circumference of a great Circle upon it: Or, multi-
ply the Area of a great Circle upon it by 4; and the
Product is the Anſwer.
Exa. What is the Arca of a Sphere or Globe whoſe
Diameter or Axis is 16 Inches, Feet, &c?
By Prop. 1. the Circumference of a great Circle
upon it is 50.2656, which multiplied by the Diameter
16, gives 804.2796, for the fuperficial Content re-
quired
By the other Method, I find the Area of a great
Circle upon it (Prop. 3.) to be 201.0624, which mul-
tiplied by 4, gives 804 2496, as before.
PRO P. XXIII.
To find the Area of a Fruitum of a Globe or Sphere,
as the Fruítun ACF, Fig. 13.
RULE. First, find the Altitude of theother Fruſtum,
by dividing the Square of the Semidiameter of the
given Fruſtum's Bale by its Altitude; which add to
the given Fruitum's Altitude; then fay, As the Axis of
the whole Sphere is to its Superficies, ſo is the Height
or Axis of the Segment or Fruftum given, to the Curve
Superficies thereof; to which adding the Area of the
Segmene's Baſe, the Sum is the ſuperficial Content of
che whole Segment.
Y3
Exa:
246 MENSURATION of
Exa. What is the Superficial Content of a Fruftum
of a Sphere, whoſe Altitude is 6 Inches, and the Dia-
meter of the Fruſtum’s Baſe 24 Inches?
The Square of the Semidiameter of the Fruſtum's
Bale (12X12) is 144, which divided by 6 (the Fru-
ſtum's Alcicude) makes 24 Inches for the Altitude of
the other Fruftum; which added to 6, makes 30 for
the Axis or Diameter of the whole Sphere, whoſe Su-
perficial Content is found to be (by laſt Prop.) 2827-44;
therefore, 30: 2827.44 ::6:565.488, the Curve Su-
perficies of the Fruſtum, to which adding the Area of
its Baſe 452.3904 (found by Prop. 3.) gives 1017.8784
for the Anſwer or Area of the Fruſtum.
PRO P. XXIV.
To find the Area or Superficial Content of a Priſm.
A Priſm is a Solid, contained under ſeveral Planes,
two of which being oppoſite, viz. the two Ends, are
called the Baſes, and theſe are parallel and equal, and
the other Planes are Parallelograms, in which a Right
Line may be every where applied from Baſe to Bale.
Priſms are either triangular, multangular, circular or
elliptical, dr. according to the Figure of the Baſe;
thus a Cube is a Priſm, bounded by 6 equal Square
Planes; a Parallelopipedon is a Priſm having its Sides
bounded by 4 equal Parallelograms, and 2 Square Baſes
or Ends; a Cylinder (or Solid, like a Rolling Stone in
a Garden) is a circular or round Priſm.
Now, to find the Area or ſuperficial Content of any
triangular or mulrangular Priſm:
RULE. Take the Sum of the Areas of the qua-
drilateral Figures, which terminate or bound it, and
you have the Area required.
Exa. There is a triangular Priſm, i. e. a Priſm ha-
ving a Triangle for its Baſe, the Length of the (Baſe's)
Side 12 Inches, and the Priſm's Height 30 Inches;
what is the Area or ſuperficial Content?
12X30-360 Area of one side ; and 360x3=1080 A-
rea
Superficies and Solids.
247
rea of all the ſides. Then to find the Area of the Baſe,
ſuppoſe the Perpendicular let fall from any one Angle
upon the oppoſite Baſe to be 10.3, wherefore 10.3x6
= 61.8, which doubled, makes 123.6 for the Areas of
both Baſes, and 1080+123.651203.6 fq. Inches for
the Area or ſuperficial Content of the whole Priſm.
Exa. 2. There is a multangular Priſm, having for
its Baſe a Polygon of 8 equal Sides, each of which is
2 Fect, and the Height 2 Feet ; What is the Area or
fuperficial Content?
By Prop. 12; 24x30=720 and 720x8=5760 Area
of all the Sides. Then to find the Area of the Baſe,
ſuppoſe a Perpendicular
. let fall from the Center of its
inſcribed or circumſcribed Circle on any of the Sides
to be 29; wherefore, by Prop. 15; 14.5x8=116,
which doubled, makes 232 for the Area of both Baſes;
and 5760+232=5992 ſq. Inches for the Area or ſua
perficial Content of the whole Priſm.
PROP. XXV.
To find the Area or ſuperficial Content of a circu-
lar Priſm or Cylinder.
RULE. Multiply the Circumference of one of
the Baſes into the length of the Cylinder, and to the
Product add the Area of both Baſes.
Exa. There is a Cylinder, whoſe Length is 3 Feet,
and Diameter of the Baſe 10 Inches, what is the ſuper-
ficial Content ?
By Prop. 1. the Circumference of the Baſe is 31.416,
which multiplied into the Length, 3 Feet =36 Inches,
makes 1130.976 ſq. Inches for the Curve Superficies
of the Cylinder.
Then the Area of the Baſe is found by Prop. 3d to
be 78.54, which doubled, gives 157.08 for the Area
of both Baſes; wherefore 1130.9767157.08=1288.056
[q. Inches for the Anſwer.
PROP
248 MENSURATION of
PROP. XXVI.
To find the Area or ſuperficial Content of a Pyra-
mid.
Solids, which decreaſe gradually from the Baſe till
they come to a Point, are in general called Pyramids,
and are of different kinds according to the Figure of
the Baſe. Thus, a Pyramid having a Triangle for its
Baſe, is called a Triangular Pyramid. If the Baſe is a
Parallelogiam, it is called a Parallelogramic Pyramid;
and if a Circle, it is called a Circular Pyramid, or ſimply
a Cone. The Point in which the Pyramid ends is
cilled the Vertex, and a Line drawn from the Vertex
perpendicular to the Baſe, is called its Height.
And firſt, to find the Area of a circular Pyramid
or Cone.
RULE. Multiply the Circumference of the baſe in-
to the Length of the Side, (not the Perpendicular
Height) and the Product is the Area required, when
thereto you have added the Aica of the Baſe.
Exa. There is a Cone whole Diawerer at the Baſe
is 16 Inches, and Length of the Side 48 Inches; What
is the ſuperficial Content? Fig. 14.
3.1416x16=50.2656 Circumference of the Baſe, which
multiplied by 24, viz. haif the Length of the Side,
gives 1206.3744 for the Area of the Curve Superfi-
cies. Then to find the Area of the Baſe 16x16=256,
and 256x.7854=201.0624. Laſtly, 1206.3744+
201.0624=1407-4368 [g. Inches.
Or thus, bs the Semidiameter of the Baſe is to the
Length of the Side, ſo is the Area of the Baſe to the
Area of the Convex Superficies, to which adding the
Area of the Baſe, as by ihe firſt Rule, the Sum is the
ſuperficial Conient of the whole.
PROP
Superficies and Solids.
249
2
PROP. XXVII.
To find the Area of all other Sorts of Pyramids.
RULE. Take the Sum of the Triangular Figures
which conſtitute the Pyramid, and cherero adding the
Area of the Baſe, the Sum is the Area ſought.
Exa. There is a Pyramid, having for its Baſe a
Triangle, one Side of whoſe Baſe AC is 18 Inches,
CB 16, and AB 20, and the Length of the Pyramid's
Side 48 Inches; what is the Area or ſuperficial Content?
By Prop. 18th, 18X24=132; 16X24=;84, and
20X245480, whoſe Sum is 1296 (or 18+16+20=
54and 54X24=1296) for the Content of the three
Sides of the Pyramid. Then to find the Area of the
Baſe, ſuppoſe the Perpendicular CD let fall on the
Baſe AB to be 13.75, by the ſame Prop. 20x +3.75
or 10X 13.75=137.5; to which adding 1296, the Sum
is
1433.5 ſq. Inches for the Anſwer.
PROP. XXVIIT.
To find the Area of the Fruftum of a Cone cut by
a Plane parallel to its Baſe. Fig. 14.
RULE. Add to the ſuperficial Content of the
whole Cone cwice the Area of the Baſe of the ſmall
Cone, and from that Sum taking the fuperficial Con-
tent of the ſmall Cone, you have the Anſwer or ſuper-
ficial Content of the Frultum.
Exa. Let ABCD repreſent the Fruſtum of a Cone,
the Diameter of whoſe greater Baſe AB is 16 Inches,
and that of the lefler 6 Inches, and Length of the Side
of the Fruftum 30 Inches.
Suppoſe the Height of the side of the whole Cone 48
Inches (ivhich is eally found, either by laying down
the Pyramid by Scale and Compaffes, or by applying
Lines to the Solid or oppoſite Sides, and meaſuring
from the Vertex where they incet, to the Circumfe-
rences
250
MENSURATION of
rence of the Baſe) then by Prop. 27, the ſuperficial
Content of the whole Cone is 1407-4368 ſq. Inches.
Then to find the Area of the Bale of the ſmall Cone
(or Top of the Fruſtum) 6x6=36, and 36x:78545
28.2744, which doubled, makes 56.5488, and this ad-
ded to 1407.4368, makes 1463.9856. Next, the A-
rea of the ſmall Cone is found (by Prop. 27.) to be
197.9208; therefore 1463.9856 -- 197.920831266.
0648 for the Area of the fruſtum ACDB.
After the ſame Method is found the ſuperficial Con-
tent of the Frultum of any other kind of Pyramids.
SECT. II. Of Solids.
PROP. I.
To find the Solidity of a Priſm.
RULE. Multiply the Area of the Baſe into the
Priſm's Height, and the Product is the ſolid Content.
Exa. There is a Triangular Priſm, having the Side
of the Baſe 12 Inches, and Height 30 Inches; Whar
is the Solidity in Inches?
By Prop. 18. Sect. 1. the Area of the Baſe is 61.8,
which multiplied by the Height 30, gives 1854 folid
or cubic Inches for the Anſwer; and this divided by
1728 (the ſolid Inches in 1 Foot) produces I ſolid Foot
and 126 folid Inches.
Exa. 2. There is a circular Priſm, whoſe Length
is 5 Feet, and Diameter of the Baſe 10 Inches; What
is the Solidity ?
By Prop. 3. Sect. I. the Area of the Baſe is 78.54,
which multiply by 60 Inches = 5 Feet, makes 471.24
cubic or folid Inches for the Anſwer.
Exa.
Superficies and Solids.
251
Exa. 3. There is an Elliptical Priſm, the greatett
Diameter of whoſe Baſe is 41 inches, and the leaſt
28, the Length of the Priſm being 6 Feet; What is
the Solidity in Inches?
By Prop. 19. Sect. 1. the Area of the Baſe is 901.6392,
which multiplied into the Height 72, gives 64918.0224
ſolid Inches = 37 folid Feet, 982.0224 folid Inches.
PROP. IT
To find the Solidity of a Pyramid.
RULE. Multiply the Area of the Baſe by of
the Perpendicular, Height, and the Product is the
Anſwer: for every Pyramid is of a Priſm, having the
fame Baſe and Height.
N. B. To find the perpendicular Height, having
the Height of the Side, from the Square of the Side-
Height, ſubtract the Square of the Semidiameter of
Bale, and the Square Root of the Remainder is
thé perpendicular Height.
Exa. There is a Pyramid, having a Circle for its
Baſe, whoſe Diameter is 16 Inches, and Height of the
Side 48 Inches.
Firſt, the perpendicular Height is 47:328, then by
Prop. 3. Sect. 1. the Area of the Baſe is 201.0624,
which multiplied by 47.328=15.776, produces 3171.96
3
cubic Inches, or one folid Foot and 1444 folid
Inches ferè.
Exa. 2. There is a Pyramid having a ſquare Baſe,
each of whole Sides is 12 Inches, and the perpendi-
cular Altitude 40 Inches ; What is the Solidity ?
12X12=144 the Area of the Baſe ; and 144x40=
5760 Cubic Inches, which divided by 3 quotes 1920
for the Anſwer.
PROPO
252
MENSURATION of
PROP. III.
To find the Solidicy of the Fruſtum of a Cone,
cut by a plain Parallel to its Baſe, Fig. 14.
RULE. Find the Solidity of the whole Cone by
laſt Prop. and from thence taking the Solidity of the
Imall Cone, the Remainder is che Solidity of che Fru-
ftum.
Exa. Let ABCD repreſent the Fruftum of a Cone,
the Diameter of whoſe greater Baſe AB is 16 Inches,
and that of the lefler ČD 8 Inches, and Length of
the Side 30 Inches; What is the Fruftum's Solidiny?
Firſt, 30x 30==900 ſquare of the Side Height, and
64 (viz. 8x8) the ſquare of the Semidiameter of the
Baſe; then 900—6=836, whoſe ſq. Root is 28.914
ferè che Height of the Fruitum.
Now, to find the Height of the whole Cone, and
conſequently its Solidity, Tay, 4.8:8::28.914:48.18,
that is, as the Difference betwixt the Fruitum's greatest
and leaſt Semidiameters is to che gieareſt Semidiameter,
fo is the Fruftuni's Height to thit of the whole Cone.
By Prop. 2d, the Solidity of the whole Cone is
3229.06214, and by the fame Prop. the Solidity of
the ſmall Cone is 206.59574, wherefore 3229.06214
206.5957453022.4664 for the Solidity of the Fru-
ftum.
The Solidity of the Fruſtum of a Cone may be o-
therwiſe found, viz. Find the Area of both Baies, and
take a Geometrical Mean berwixt theſe two cireas ;
which Mean add to the Suin of the Areas, and multi-
plying the Toral by of the Fruitum's Height, the
Product is the Anſwer. Now, to find a Geoinetrical
Mean between any two Numbers, you must take the
ſquare Root of their Product.
Or chus, Square the greateſt Diameter, as alſo the
leſſer, and multiply the two Diameiers together, the
Sum of which three Numbers mulcipiy by the Height,
s and the Product divide by 3.8197 ; or that Sum mul-
ciplied
Superficies and Solids.
253
tiplied into of the Height, and the Product by .78543
gives the Solidity.
If ſuch Fruſtums are cut thro' the Extremities of the
Baſes by a Diagonal Line, they are called Hoofs, and
the Solidity is found, if it be a Cone, by ſquaring the
greater Diameter, and adding thereto į the Rectangle
of the two Diameters, more the Difference betwixo
the Diameters, and multiplying the Sum by the Height,
and laſtly dividing the Product by 3.8197Or, by
multiplying the ſaid Sum into of the Height, and
multiplying the Product by .7834 for the greater Hoof
Fig. 15
And for the Content of the lefſer Hoof, ſquare the
leffer Diameter; and take the Rectangle of the two
Diameters, from the Sum of which ſubtract the Diffe-
rence betwixt the 2 Diameters, and the Remainder
multiply and divide as before.
And if it is a ſquare Pyramid,
Take the Square of the greater Diameter, the Rec-
tangle of the two Diameters, as alſo the Difference be-
twixt the Diameters, multiply the Sum of theſe three
Numbers by { of the Height for the grea c er Hoof
And rake the Square of the leſſer Diameter, and {
the Rectangle of the two Diameters, from the Sum of
which ſubtract the Difference betwixt the two Dia-
meters, and multiply the Remainder by f of the Height
for the Content of the leffer Hoof.
PROP. IV.
To find the Solidity of a Globe or Sphere.
RULE. Multiply the Superficies by → of its Axis :
Or, 2. Multiply the Cube of the Axis by 5236: Or,
3. Becauſe every Globe is į of a Cylinder of the
fame Height and Diameter of the Baſe, with the Globe's
Axis, multiply the Area of a great Circle by the Di-
amerer, and take } of the Product for the Anſwer.
Exa. What is the Solidity of a Sphere, whoſe Dia-
meter is 16?
Z
Ву
254
MENSURATION of
By Prop. 23. the ſuperficial Content is 804.2496,
which multiplied by 2.6666, br. (viz. Ź of the Axis)
gives 2144.66, buc. for the Solidity.
2. The Cube of the Diameter is 4096, which mul-
tiplied by .5236 gives 2144.66, &c. as before.
3. The Area of a great Circle upon it is 201.0624,
which multiplied by the Diameter 16, makes 3216.
-9984, of which is 2144.66, &c. as before.
PRO P. V.
To find the Solidity of the Fruftum of a Globe, ha-
ving the Diameter of the Fruſtum's Baſe and its Alti-
tude.
RULE. Multiply the Square of half the Diameter
of the Baſe by 3 times the Altitude, and to the Pro-
duct add the Cube of the Altitude, the Sum multiply
by .5236, and the Product is the Anſwer. Fig. 13.
Exa. There is a Fruſtum of a Globe, the Diame-
ter of whoſe Baſe is 24, and its Altitude 6, what is the
Solidity ?
12X12=144 Square of half the Diameter, and 18 =
3 times the Altitude. Then 144x18=2592; and the
Cube of the Altitude is 216'=6x6x6 wherefore
2592-+-2162808 Laſtly, 2808x.5236–1470.2688
for the Anſwer or Solidity of the Fruftum.
PROP. VI.
To find the Solidity of a Fruſtum of a Sphere, ha-
ving the Sphere's Axis and Height of the Fruftum.
ŘULE. From the triple Product of the Axis into
the Square of the Fruſtum's Height, ſubtract twice the
Cube of the Height, and multiply the Remainder by
.5236 for the Anſwer.
Éxa. There is a Fruftum of a Sphere, whoſe Axis
is 30 Inches, and the Fruftum's Height 6 Inches,
what is the Solidity?
36x30x3=3240 triple Product of the Axis into the
Square of the Fruſtum's Height, and 432 = twice the
Cube
Superficies and Solids.
255
Cube of the Height; wherefore 32404+322808
and 2808x.5236=1470.2688 as before.
And the Content of the one Fruſtum taken from
that of the whole Sphere, leaves the Content of the o-
ther Fruftum.
If having the Altitude of the Fruftum with the Dia-
meter of irs Baſe, you want to know the Height of the
other Fruſtum, or the Sphere's Axis, divide the Square
of the Semidiameter of the one · Fruſtum's Baſe
by the Height of the Fruſtum, and the Quote is the
Height of the other Fruſtum, whoſe Sum is the Sphere's
Axis,
PRO P. VII.
To find the Solidity of the middle Zone of a Sphere.
Fig. 13. as CDGHEF.
RULE. To twice the Square of the Diameter or
Sphere's Axis add the Square of the Diameter of the
Baſe, and dividing the Sum by 3.8197, multiply the
Quote by the Zone's Height for the Anſwer.
Exa. Suppoſe the Diameter DE or AB 30, the Di-
ameter of the Baſe 24, and the Zone's Height 18;
What is the Solidity ?
30X30X2=1800 twice the Square of the Diameter;
and 24X245576 ſquare of the Diameter of the Bale
Then 1800+57652376, and 2376 divided by 3 8197,
quotes 622,038, which multiplied by the Zone's Height
18, the Product is 11196.684 for the Anſwer.
Or from the Solidity of the whole Spherc ſubtract
that of twice the Fruſtum CAF, and the Remainder
is the Solidity of the middle Zone.
By Prop. 4. the Solidity of a Sphere whoſe Axis is
30 Inches, will be found to be 14137.2, from which
ſubtracting twice the Fruſtum CAF, found by Prop.
5, viz. 1470.2688x2=2940.5376, the Remainder is
11196.6624, the ſame as before, ferè.
Z 2
PROP.
256 MENSURATION of
PRO P. VIII.
To find the Solidity of a Spheroid, Fig. 16.
A Spheroid is a Solid, formed by the Rotation of
the Semi-cllipſis ABC, about its tranſverſe Diameter
AC, which is called the Spheroids Axis: This Body
much reſembles the Shape of an Egg.
RULE. Every Spheroid being of its circum-
ſcribing Cylinder, whoſe Baſe is = to the greateſt Dia-
meter, and its Height that of the Spheroid ; therefore
find its Content by multiplying the Area of its greateſt
Circle by Ź of its Axis.
Exa. What is the Solidity of a Spheroid, whoſe Axis
is
40 Inches, and greateſt Diameter 22 Inches ?
By Prop. 3. Sect. I. the Area of its greateſt Circle
is 380.1336, which multiplied by 40 gives 15205.3447
two third Parts whereof is 10136.896 for the Anſwer.
PROP. IX.
To find the Solidity of the middle Zone of a Sphe-
roid, as abde. Fig. 16.
RULE. To twice the Square of the Diameter DB,
add the Square of the Diameter of the Baſe ab or cd,
and divide the Sum by 3.8197, the Quote of which
Diviſion multiplied by the Height produces the An-
ſwer.
Exa. Suppoſe the Height of the middle Zone 24
Inches, the Diame:er of the Baſe 17 Inches, and the
greateſt Diameter 22 Inches; What is the Solidity?
22X22=4847 and 484x2=968
17X17=289. Their Sum is 1257, which divided by
3.8197, quores 329.057, and this multiplied by the
Height 24, gives 7897-368 for the Anſwer.
PRO P. X.
To find the Solidity of any other Fruſtum of a Sphe-
roid, as Aab.
RULE. Find its Solidity as if it was a Sphere, and
ſay,
Superficies and Solids.
257
ſay, As the Solidity of the whole Sphere is to the Soli-
dity of the whole Spheroid, fo.is any part of the Sphere
to the like Part of the Spheroid.
Exa. Suppoſe the Height of the Fruftum 6, and the
Diameter of the Baſe 16; What is the Solidity ?
By Prop. 5th of this ſection, the like Fruítum of a
Sphere would be 716.2848, and by Prop. 4. the Soli-
dity of the whole Sphere is found to be 2423.878 ;
therefore, &c.
3.2
PRO P. XI.
To find the folid Content of a Parabolic Conoid,
Fig. 12.
This Solid is form'd by the Rotation of a Semipa-
rabola CAB, about its Axis AB.
RULE. Multiply the Area of the Bafe by the
Altitude, and the Product is the ſolid Content ; for eve-
ry parabolic Conoid is equal to its circumſcribing
Cylinder.
Exa. There is a parabolic Conoid, whoſe Baſe is
Inches, and its Height 46 Inches; What is the Solidity?
-7854X1024=804.2496 Area of the Baſe, which mul-
tiplied by 23 (viz., the Altitude) gives 18497.8408.
PROP. XII.
To find the Solidity of the lower Fruſtum of a
a para-
bolic Conoid, as EFCD. Fig. 12.
RUL E. Square the Diameter at each end, and
multiply their Sum by the Height, and divide the
Product by 3.8196.
Exa. Suppoſe ÉF 24, CD 32 and bB-21.3; What is
the Solidity?
32X32=1024, and 24x243576; their Sum is 1600,
which multiplied by 31.95 (viz. of 21.3) gives
51120, and this divided by 3.8196 quotes 133836 for
the Anſwer,
2 3
PROP.
258 MENSURATION of
PRO P. XIII.
To find the Solidity of a Parabolic Spindle or Pyra
midoid.
This Solid is formed by turning the Parabola about
its Bafe. Fig. 17.
RULE. Find the Contents as if it was a Cylinder,
and take it of that Content for the Content of the Py-
ramidoid.
Exa. Suppoſe the Length AB 66 Inches, and the
greateſt Diameter CD 27; What is the Solidity ?
By Prop. 1. Sect. 2. The Contents of a Cylinder of
the above Dimenſions: is found: to be 37788.7356,
which multiplied by 8; and the Product divided by 15%
gives the Content of the Pyramidoid, viz. 20153.9256.
PROP: XIV:
To find the Solidity of the middle Fruſtum of a Pa-
rabolic Spindle, ſuch as aCcdDba. Fig. 17.
RULE: Square the Diameter CD, which double,
and thereto add the Square of the Diameter at the Baſe
ab or cd, from the Sum of which take of the Square
of the Difference betwixt the 2 Diameters, and dividing
the Remainder; by 3.8197; the Quote multiplied into
the Height is the Solidity required.
Exa. Suppoſe the Diameter CD 27, as before, the
length mm 33, at the Diameter at either Baſe ab or cd
20; What is the Solidity ?
27x27=729 and 72.9X2=1458. Then 20X20=
400, their Sum is 1858. Alſo 27--20=7, and 7x7
x=19.6. Then 1858–19:6=1838.4, which di-
vided by 3.8197 quotes 481.294, and this multiplied
by the Length 33, gives 15882.902 for the Anſwer.
PROP. XV.
To find the Solidity of a Cylindroid.
This Figure has an elliptical or oyal Baſe, whereas
a Cylinder has a circular one.
RULE.
1
Superficies and Solids.
259
RULE. Find the Area of the Baſe by Prop. 19.
Sect. 1. and this multiplied into the Height gives the
Anſwer.
Exa. Suppoſe the greateſt Diameter of the Baſe 41,
and the lefſer Diameter at the Baſe 28, and the Height
of the Cylinder 72; What is the Solidity ?
By Prop. 19th, Sect. I. the Area of the Baſe is 901.
.6392, which multiplied by 72 gives 64918.022 for
the Anſwer:
PRO P. XVI.
To find the Solidity of an hyperbolic Conoid.
This Figure is formed by the Rotation of a Semi-
hyperbola ACD about its Axis AC. Fig. 18.
RULE. To 6 times the tranſverſe Diameter Ag,
add 6 times the intercepted Diameter AC for a Diviſor.
2. To 3 times the tranſverſe Diameter add twice the
intercepted Diameter, and by the Sum multiply the
Content of a Cylinder, whoſe Height is AC, and the
Diameter of the Baſe BD; the Product is a Dividend;
which divided by the Diviſor above found, gives the
Content of the hyperbolic Conoid ABCD
This Body is more eaſily meaſured by reducing it to
1 Cylinder by help of a mean Diameter.
PROP. XVII.
To find the Solidity of any Body by immerſing it
into Water.
RULE. Immerfe it in Water in a Parallelopiped;
whoſe Sides are exactly divided into Inches and Parts,
and the Solidity of the Water raiſed, will be equal to
that of the Body immerſed.
Or thus: Fill any Veſſel with Water, and having
immerſed the Body, carefully reſerve the Water which
flows over into another Vefſel; and find the Solidity of
this Water by meaſuring it in a Veflel of a known Ca-
pacity, for the Content or Solidity of this ſhall be that
of the Body immerſed...
SECT
260
Of GAUG ING.
SECT. III. Of Gauging:
divide by
PRO P. I.
TO
*O find the Content in Ale, Wine or Corn Gallons
Engliſh Meaſure, or in Scots Pints, of a Square
Ton or Veffel.
RULE. Multiply the Length or Breadth in Inches
by itſelf, and the Product is the Area or Content at one
Inch deep, which multiplied by the Height or Depth
gives the folid Content in Inches, which to reduce to
Ale Gall.
282.
Wine Gall.
231.
Corn Gall.
268.8
Scots Pints,
102.3
Or multiply the forefaid Solidity in Inches by.003546;
6004329; .003722; .009775 reſpectively, which Mul-
tipliers are thus found.
282) 1.00000, &c. (.003546
231)1.00000, br. 1.004329
268.8)1.00000,&C.( 003722
102.3)1.00000,&C.1.009775
Exa. Suppoſe the side of a Square Veſſel 40.2 Inches,
and Height 10.3; How many Gallons of Ale, Winé
or Corn or Scots Pints doch it contain ?
40.2X40.2=1616.04, which multiplied by the Height
10.3 gives 16645.212, and this divided by 282, or
multiplied by .003546 makes 59.024 ferè Ale Gallons,
which you may alſo reduce to Wine or Corn Gallons,
&c. by dividing or multiplying, as is before taught.
PRO P. II.
To find the Content in Gallons of a Vefſel in form
of a Right-angled Parallelogram.
RULE. Multiply the Length by the Breadth; and
that Product by the Depth for the ſolid Content in
Inches, which reduce to Gallons, as before....
Еха. .
Of GAUG ING. 261
Exa. Suppoſe the Length 60 Inches, Breadth 40,
and Depth 18; What is the Content in Gallons ?
60X4052400 and 2400X18343200, which divided
by 282, or multiplied by .003546 gives 153.1872 for
Ale Gallons.
PROP. III
To find the Content in Ale, &c. Gallons of a Vefſel
of a Triangular Form.
RULE. Find the Area of the Baſe by Prop. 18th
Sect. 1. which multiplied by the Height gives the ſolid
Content in Inches, and this laft divided or multiplied
as before, gives the Content in Gallons.
Exa. Suppoſe the Length of the Baſe of any Trian-
gular Veffel be 25 Inches, the perpendicular Breadth
15, and the Depth 12 ; What is the Content in Gal-
lons ?
25x7.5=187.5. and 187.5X12=2250 folid Inches,
which divided by 282 or múltiplied by .003546 gives
7.9785 for Ale Gallons: and ſo on for Wine, Corn
Gallons and Scots Pints.
PROP. IV.
To find the Content in Gallons of a Vefſel in form
of any other regular or irregular Figure, the Veſſel be-
ing equally wide throughout.
RULE. Divide it into Triangles, and multiply the
Sum of their Areas into the Depth, the Product is the
ſolid Content in Inches, which reduce to Galions, by
multiplying or dividing as before.
PRO P. V.
To find the Content in Gallons of a Veffel that's
circular, and equally wide throughout, i. e. in form of
a Cylinder.
RULE. Multiply the Area of the Baſe in Inches
by the Veſſel's Depth, and the Product is the Solidity
in
262 Of GAUGING.
}
48.3
in Inches, which reduce to Gallons, &c. by dividing
or multiplying as before.
Exa. Ler che. Diameter of a Cylindrical Veſtel be
Inches and the Depth 60.5; how many Gallons
doch it contain ?
By Prop. 3. Sect. 1. the Area of the Baſej is 1832
-2518, which multiplied by 60.5 produces 110851
.234263 ſolid Inches; and this divided by 282, or mul-
tiplied by .003546 makes 393.0784, for Ale Gallons,
dc.
Or thus : Divide the Square of the Diameter by
359.05 for Ale Gall.
The Quotes are the Areas
294.12 for Wine Gall. at 1 Inch deep, which there-
342.24 for Corn Gall.
fore muſt be multiplied by
130.25 for Scots Pints.
the whole Depth for the
Anſwer.
N. B. Thefe Diviſors are found by dividing 282,
231,:268.8, 102.3 ſeverally by .7854 the Area of that
Circle whoſe Diameter is I.
PROP. VI.
To find the Content in Gallons of a Veffel of ani
Elliptical Bottom or in Form of a Cylindroid equally
wide throughout.
RULE. Find its Solidity by Prop. 15. Sect. 2. and
divide or multiply as before, in order to reduce it to
Gallons, &c. Or (which is the ſame Thing) multiply
the Length of the Bottom or Top by its Breadth, and
that Product by .7854 for the Area at one Inch deep,
which multiplied by the whole Depth, and the Pro-
duct divided by 282, bo. gives the Anſwer in Gal-
lons, Quc.
Exa. Suppoſe the greateſt Diameter of the Top or
Bottom 41 Inches, and the leſſer 28, and Height or
Depth of the Vefſel 72 Inches; What is the Content
in Gallons, bouc.
By Prop. 15. Sect. 2. the Solidity in Inches is
64918.022, which divided by 282, quotes 226,66 for
Ale Gallons.
But
Of G AUGING.
263
But if the Veffel, whether circular or elliptical, be
wider at the Top than at the Bottom (as they general-
ly are) the beſt Way of finding its Content, is by ta-
king the Sum of the Areas of Top and Bottom, and
multiplying the ſaid Sum by the Height or Depth,
and dividing as before, in order to reduce it to Ġal-
lons.
Exa: What is the Content of a Tub, whoſe Dia-
meter at the Top is 37 Inches, and at the Bottom 30
Inches, the Depth being 24 Inches?
37x37=1369 and 1369X.7854=1075.2126 Area at
30x30x.78545706.86 Area at the ort om.
Their Sum is 1782.0726, of which is 891.0363,
which multiplied by 24, makes 21384.8712; and this
divided by 282, gives 75.8329 Ale Gallons.
the Top
PRO P. VII.
To find the Content in Gallons of a Priſm, having
for its Sides Parallelograms ſtanding at Right Angles
with the Baſe.
RULE. Find the Solidity in Inches by Prop. I.
Sect 2. and divide the ſame by 282, Loc. for the An-
[wer in Gallons, &c.
Exa. There is a Triangular Priſm, the Length ofone
of the Sides of whoſe Bale is 12 Inches, of another 16,
and of the other 18, and the Depth 24 Inches; What is
the Content in Gallons ?
By Prop. 1. Sect. 2. the Solidity of the Priſm, (fup-
poſing the Perpendicular let fall on the greateſt Side of
the Baſe, 10.5 Inches) is found to be 2268 folid Inches;
which divided by 282, quotes 8.04 Ale Gallons.
PRO P. VIII.
To find the Content in Gallons of a Vefſel in form
of a Pyramid, having Right Lines for its Baſe.
RULE. Find its Solidity in Inches by Prop. 2. Sect. 2.
and divide by 282, &c. for the Anſwer in Gallons, &c.
Exa:
264
OF G AUGING.
Exa. How many Scots Pints doth a Pyramidical
Veſſel consain, whoſe Baſe is a Circle of 16 Inches Di-
ameter, and the Perpendicular Height or Depth of the
Veſſel 30 Inches?
16x10x.7854X10=2010.624 Solidity in Inches, and
this divided by 102.3 or multiplied by .009775, gives
19.65519 Pints, 2 Mutchkins, 2 Gills feré.
If it is a ſquare Pyramid, its Content in Gallons
may be thus found; Square the Side of the Baſe and
multiply the ſaid Square by the perpendicular Height,
dividing the Product by 3 Times the conſtant Divi-
ſors, 282, 231, 268.8, 102.3 for the Anſwer in Ale,
&c. Gallons.
PRO P. IX.
To find the Content in Gallons of the Fruſtum of
a Pyramid cut by a Plain, parallel to its Baſe.
RULE. By Prop. 3. Sect. 2. find its Solidity in
Inches, which divide by 282, br. for the Content in
Gallons, &c.
If the Veſſel is in Form of a Fruftum of an Elliptical
Pyramid, or indeed of any other Form, find the Areas
of the Top and Bottom, then a Geometrical Mean be-
twixt them, and multiply the Sum of theſe two Areas
and the Mean by į of the Fruſtum's Height, and di-
vide the Product by 282, c. for Gallons, &c.
PROP. X.
To find the Content in Gallons, doc. of a Vefſel in
Form of a Globe.
RULE. By Prop. 4. Sect. 2. find the Solidity in
Inches, and divide or, multiply as before, to reduce it
to Gallons, dc.
Or çhus. Multiply the Cube of the Axis by.001856, or
divide by 538.57 for Ale Gallons: multiply by .002266,
or divide by 441.17 for Wine Gallons, and the Quotes
or Products are the reſpective Contents. For if the
Cube of the Axis being multiplied by .5236 gives the
Con-
Of GAUGING.
265
Content in Inches, (Prop. 4. Sect. 2.) therefore by di-
viding the faid Content by 282.231, &c. the Quote
gives the Content in Gall. but
282).5236 (.001856 Multiplier for A. Gall.
231) .5 236 0.002266 Multiplier for W. Gall.
Or,
-5236) 282. (538.57 Diviſor for A. Gall.
-5236) 282 (441.17 Diviſor for W. Gall.
PROP. XI.
To find the Content in Gall. of the Segment of a
Spherical Veſſel, having the Diameter of its Baſe and
Height.
Rule. Find the Content in Inches by Prop.5. Sect. 2.
and divide or multiply as before, for the Content in
Gall.
Or thus, To the triple Square of half the Diameter,
add the Square of the Height, which Sum multiply in-
to the Height, and divide the Product by 538 57 or
multiply by .001856 for Ale Gall. and divide or mule
tiply by 441.17 and .002266 for Wine Gallons.
PRO P. XII.
To find the Content in Ale, Wine, doc. Gall. of a
Veffel, in Form of the middle Fruſtum of a Globe.
RÚLE. By Prop. 7. Sect. 2. find its Solidity in
Inches, and multiply or divide as before, for the An-
ſwer in Gallons,
PROP XUI.
To find the Contents in Gallons, doc. of Vefſels in
Form of a Spheroid, middle Zone of a Spheroid, Pa-
rabolic Conoid, lower Fruftum of a Parabolic Conoid,
Parabolic Spindle, middle Fruftum of a Parabolic Spin-
dle, Cylindric and Hyperbolic Conoid.
RULE. Find their contents in Inches by Prop. 8.
9. II. 12. 13. 14. 15. 16. Sect. 2. and divide or multi-
ply as before, for the Content in Gallons, &c.
PROP
Аа
266
Of G AUGING.
PROP. XIV.
.
To find the Content in Gall. buc. of any clare Cask,
as Fig. 19.
The Contents of ſuch Casks cannot be found by a-
ny one general Rule; and therefore they are ſuppoſed
to be either
1. The middle Zone of a Spheroid, or
2. The middle Zone of a Parabolic Spindle, or
3. The lower Fruftums of 2 equal Parabolic Co-
noids, or
4. The lower Fruſtums of 2 equal Cones.
1. If the Staves of the Cask have a great Curve as
the outward Lines of the Figure, it is ſuppoſed to be
the middle Zone of a Spheroid, whoſe Content in Inches
is found by Prop. 9. Sect. 2. and thence in Gall. Soc.
by dividing or multiplying as before.
Exa. Suppoſe the greateſt or Bung Diameter Dd,
20 Inches, the leffer Cc, 16, and Height of the Cask
AB 31. What is the Content in Ale Gallons ?
20x20 5400 & 400x2=800
16X16=256
=
and 32736 divided by 3.81917, quotes 8570.306 for
the Sólid Content in Inches, which divided by 282, or
multiplied by .003546, gives 30.39 Ale Gallons.
2. If the Staves are not ſo much curved as before,
the Cask is ſuppoſed to repreſent the middle Fruſtum
of a Parabolic Spindle, whoſe Content in Inches is
found by Prop. 14. Sect. 2.
Exa. Suppoſe (as before, the greateſt Diameter 20
Inches, the leſſer 16, and Height of the Cask 31 loches,
what is the Content in Ale Gallons?
20x20 5 400 & 400X2
400 & 400X2=800; then 16x16=256
Their Sum is 1056; and 20 – 16=4: alfo 4.XIX =
6.4 and 1056 — 6.4 = 1049.6, which divid: ci by
3:8197, quores 274.73, and this multiplied into the
Height 3i, gives 8518.18 folid Content in Inches,
which
Of GAUG ING.
267
which divided by 282, or multiplied by .003546, makes
30.205 Ale Gallons.
3. When the Staves are very little curved, the Cask
is ſuppoſed to repreſent the lower Fruftums of 2 equal
Parabolic Conoids, joined together by their greater
Baſes at the Bung, whoſe Contents may be found by
Prop. 12. Sect. 2.
Exa. Suppoſe (as before) the greateſt Diameter 20
Inches, the lefſer 16, and the Height 31. What is the
Content in Ale Gallons ?
20x20=400, and 16x16=256; their Sum is 656,
which multiplied by 91, produces 30504, and this di-
vided by 3.8197, quotes 7986.17 for the Content in
Inches; which divided by 282, or multiplied by
.003546, gives 28.31 Gall. of Ale for the Anſwer.
4. Laſtly, if the Staves are pretty ſtreight from the
Bung to the Head, as the prick'd Lines in the Figure,
the Cask is ſuppoſed to be in Form of the lower Fruſ-
tums of 2 equal Cones joined together by their greateſt
Baſes at the Bung, and its Content may be found by
Prop. 3. Sect. 2.
20X20=400
16x16=256
976X31=30256, and this divided
20x16=320
by 3.8197, quores 7921.04 folid Inches, which di-
vided by 282, or multiplied by .003546, gives 28.088
Gallons of Ale for the Anſwer.
So the ſeveral Anſwers (according to the ſuppoſed
Forms of the Cask) are,
30.39
30.205
3.
28.31
4.
28.088
But ſuch Casks as theſe ought to be reduced to a Cy-
linder, by taking a mean Diameter, for which obſerve
the following Rule.
Multiply the Difference between the Head and Bung
Diameter, by :7;.65;.6; or -55, according as the
Staves are more or leſs arching, which Product add
A a 2
}
I.
2.
to
268 Of GAUGING.
.
3
to the Head Diameter, the Sum is the mean Diameter,
and the Cask is thereby reduced to a Cylinder, whoſe
Con crit is found by Prop. 5. Sect. 3.
Moreover, if you want to know what Quantity of
Liquor is drawn out, or remains in any ſuch Cask,
when it ſtands upon one of its Baſes ;
RULE. As the Square of half the Length of the
Cask, is to the Difference between the Head and Bung
Areis ; so is the Square of any Circle's Diſtance
from the Bung to the Difference between the Bung A-
sea and that of the Circle, or of the Area of the Li-
quor's Surface: which found, froin the Bung Area take
of the foreſaid Difference, and multiply the Remain-
der by the Liquor's Diſtance from the Bung, and the
Product will ſhew what Quantity of Liquor is either
above or under the Content of the Cask.
ExQ. Suppoſe (as before) the Bung Diameter 20
Inches, the leſſer 16, and Height 31, what is the Con-
tent in Ale Gallons, when there is' 12 Inches wet ?
Half of the Cask's Length is = 15.5, whoſe Square
is 240.25 ; The Liquor's Diſtance from the Bung, is
15.5 -
12=3.5, whoſe Square is 12.25. The Area
at the Bung is 1.11404 Gall
. at the Top.71298, and
their Difference is .40106. Therefore
240.25 : .40106 : : 12.25: .02045, the Difference
between the Bung Area and that of the Circle.
Then 1.11404-.02045 =1.09359, which divid-
ed by 3, quotes : 364532 and 1.11404 - .36453'=
.74951, and this laſt multiplied by 3.5, makes 2.62328
=to what the Cask wants of being half full; where-
fore 15.195 (viz. balf the Content of the whole Cask
found by Prop. 14.) -- 2.62328=12.57172, the Num-
ber of Ale Gallons in the Cask át 12 Inches wet.
If the Cask had wanted but 12 Inches of being full,
the Content would have been 15.195+-12.57172=27.
76672 Gallons.
PROP.
Of G AUGING.
269
PROP. XV.
To find out what Liquor is in a Cask (not full) when
it lies with its Axis parallel to the Horizon.
RULE. By Prop. 9. Sect. 1. find the Area of the
Segment in Inches, which reduce to Gallons, &c. as is
before taught. But becauſe the Area of a Segment
may be readily found in Gall. &c. by Help of a Table
of Segnients, as follows: The Quantity of Liquor
drawn out or remaining in any Cask lying with its Axis
parallel to the Horizon (being firſt reduced to a Cylin-
der) is beſt found this Way, viz. From the Bung Di-
ameter ſubtract the mean Diameter, and half the Dif.
ference; alſo from the wet Inches ſubtract the ſaid half
Difference, and ſay, As the mean Diameter is to 100
(the Diameter of the Tabular Circle) ſo is the laſt
Difference to a vers’d Sine in the Table; then if the
Segment ſtanding againſt that vers'd Sine, be multi-
plied into the Content of the Cask in Gall. or Inches,
the Product will ſhew what Quantity of Liquor (in
Gall. or Inches) either remains, or is drawn out of the
Cask.
Exa. Suppoſe the Cask to repreſent the middle Zone
of a Parabolick Spindle, whoſe Bung Diameter is 20
Inches, Head Diameter 16, Length of the Cask 31,
and 8 Inches wet. What is the Content in Ale Gall?
Firſt 20 — 16 = 4, and 4x.65 = 2.6, and 2.6+16=
20-18.6
18.6 mean Diameter. Then
=.7, half the
Difference between the Bung and mean Diameters, and
7.3 Difference betwixt the wet Inches and
the foreſaid half Difference. Therefore
* 18.6: 100:: 7.3:39 3.3611 Segment.
The Content of the whole Cask is found by Prop.
5. Sect. 3. to be 29.8685 Gail. wherefore this multi-
plied by :3611 gives 10.7855 for the Number of wet
Gall. and 20-8-12, Number of dry Inches; then
2
8.7
A a 3
I 2
270 Of G AUGING.
12-.711.3; therefore 18.6 : 100: : 11.3:60 V.
Sine = .62659
.6265, which multiplied by 29.8685, gives
18.7126, the Number of dry Gall. the Sum of both
is 29.5 = the Content of the whole Cask fere.
A TABLE of the Segments of a Circle, whoſe
Area is 1, the Diameter (1.128378) being di-
vided into 100 equal Parts.
77/0.8262
V S. I Segm. 117.S.] Segm. 11V.S. Segm. 117.5.1 Segm.
110.0017|| 26 |0.2006 510.5127 76 0.8155
2 0.004827 10.217852 0.5255
310.00872810.22921530.5382
78 0.8369
410.0134 29 10.2407115410.5509
79 0.8474
5 0.91871 30 10.252311550.5635 800.8576
6 0.0245 310.26401 56 0.5762 81 0.8677
7 0.03081 32 0.27591 57 0.5888 82 0.8776
8 0.03751 33 0.287858 0.601483 0.8873
90.0446 || 34 (0.2998159 0.6140 840.8968
100.0520|| 35 10.3119||60 0.6265 85 0.9059
Il 0.05981 36 10.3241 610.6389. 860.9142
12 0.068037 0.3364|| 62 0.6514187 0.9236
130.076438 0.34861163 0.6636
88 0.9320
14 0.0851 390.3611 640.6759 89 0.9402
15 0.094.1||40|0.3735116510.6881
90 0.9480
16 10.103241 0.3860
910.9554
17 0.1127|| 420.3986116710.7122 92 0.9625
180.1224 || 430 4112
93 0.9692
19 0.1323||44 0.42381169 1.7360
941 0.9755
20 0.1424 145 10.4365117010.7477
950.9813
210.1526|| 46 0.44911711 9.7593 96 0.9866
22 0.16311147 0.461811720.7708 97 0.9913
230.1738148 0.4745 1732.7822 98 0.9952
24 0.1845 || 49 10 4373 174 5.7934 99 0.9983
25 10.1955|150 10.5000/175 10.804511100 1.0000
6610.7002
6810.724-1
of
Of G AUGING.
271
Of M A L T-GAUG IN G.
According to an Act of an Engliſh Parliament anno
1697, every round Buſhel with a plain and even Bot-
tom, 18 Inches wide throughout, and 8 Inches deep,
ſhould be eſteem'd a legal Wincheſter Bushel: Now ſuch
a Veſſel will contain 2150.42 Cubic Inches ; for
18.5X15.5 =342.25, which multiplied by :7854, gives
268.80315, and this laſt multiplied by the Height 8,
produces 2150.42. And therefore to find the Number
of Buſhels contained in any Veſſel, firſt find its Solidi-
ty in Inches, according to the Form of the Verrel, and
divide by 2150.42 for the Anſwer.
If the Malt be lying on the Floor, in order to know
the true Depth, you muſt take the Depth in ſeveral
(ſuppoſe 6, 7, 8, or more) Places, the Sum of which di-
vided by the Number of Places you took the Depth in,
quotes the mean Depth.
Exa. Suppoſe a Quantity of Malt lying on the Floor,
in Form of a Rectangular Parallelogram, Length 160
Inches, and Breadth 100 Inches, what is the Number of
Buſhels contain'd in it ?
Suppoſe I
5.5
6.
2
3
4.8
Depth to be
4
5
6
7
5.9
4.8
6. I
5.7
38.8
the Sum is 38.8, which divided by 7, the Number of
Places, the Quore 5.543 is the mean Depth. Then
160X100 = 16000, and 16000X5.543 produces 88688
Cubic Inches, which divided by 2190.42, quotes 4.1.242
Buſhels for the Anſwer.
I N
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