ON PERMUTATIONS AND COMBINATIONS.
204. If a number of counters, distinguished by different letters, be
placed on the table, and any number of them, say four, be taken away,
the question is, to determine in how many different ways this can be
done. Each way of doing it gives what is called a _combination_ of
four, but which might with more propriety be called a _selection_
of four. Two combinations or selections are called different, which
differ in any way whatever; thus, _abcd_ and _abce_ are different,
_d_ being in one and _e_ in the other, the remaining parts being the
same. Let there be six counters, _a_, _b_, _c_, _d_, _e_, and _f_;
the combinations of three which can be made out of them are twenty in
number, as follow:
_abc_ _ace_ _bcd_ _bef_
_abd_ _acf_ _bce_ _cde_
_abe_ _ade_ _bcf_ _cdf_
_abf_ _adf_ _bde_ _cef_
_acd_ _aef_ _bdf_ _def_
The combinations of four are fifteen in number, namely,
_abcd_ _abde_ _acde_ _adef_ _bcef_
_abce_ _abdf_ _acdf_ _bcde_ _bdcf_
_abcf_ _abef_ _acef_ _bcdf_ _cdef_
and so on.
205. Each of these combinations may be written in several different
orders; thus, _abcd_ may be disposed in any of the following ways:
_abcd_ _acbd_ _acdb_ _abdc_ _adbc_ _adcb_
_bacd_ _cabd_ _cadb_ _badc_ _dabc_ _dacb_
_bcad_ _cbad_ _cdab_ _bdac_ _dbac_ _dcab_
_bcda_ _cbda_ _cdba_ _bdca_ _dbca_ _dcba_
of which no two are entirely in the same order. Each of these is
said to be a distinct _permutation_ of _abcd_. Considered as a
_combination_, they are all the same, as each contains _a_, _b_, _c_,
and _d_.
206. We now proceed to find how many _permutations_, each containing
one given number, can be made from the counters in another given
number, six, for example. If we knew how to find all the permutations
containing four counters, we might make those which contain five
thus: Take any one which contains four, for example, _abcf_ in which
_d_ and _e_ are omitted; write _d_ and _e_ successively at the end,
which gives _abcfd_, _abcfe_, and repeat the same process with every
other permutation of four; thus, _dabc_ gives _dabce_ and _dabcf_.
No permutation of five can escape us if we proceed in this manner,
provided only we know those of four; for any given permutation of five,
as _dbfea_, will arise in the course of the process from _dbfe_, which,
according to our rule, furnishes _dbfea_. Neither will any permutation
be repeated twice, for _dbfea_, if the rule be followed, can only
arise from the permutation _dbfe_. If we begin in this way to find the
permutations of two out of the six,
_a_ _b_ _c_ _d_ _e_ _f_
each of these gives five; thus,
_a_ gives _ab_ _ac_ _ad_ _ae_ _af_
_b_ ... _ba_ _bc_ _bd_ _be_ _bf_
and the whole number is 6 × 5, or 30.
Again, _ab_ gives _abc_ _abd_ _abe_ _abf_
_ac_ ... _acb_ _acd_ _ace_ _acf_
and here are 30, or 6 × 5 permutations of 2, each of which gives 4
permutations of 3; the whole number of the last is therefore 6 × 5 × 4,
or 120.
Again, _abc_ gives _abcd_ _abce_ _abcf_
_abd_ ... _abdc_ _abde_ _abdf_
and here are 120, or 6 × 5 × 4, permutations of three, each of which
gives 3 permutations of four; the whole number of the last is therefore
6 × 5 × 4 × 3, or 360.
In the same way, the number of permutations of 5 is 6 × 5 × 4 × 3 ×
2, and the number of permutations of six, or the number of different
ways in which the whole six can be arranged, is 6 × 5 × 4 × 3 × 2
× 1. The last two results are the same, which must be; for since a
permutation of five only omits one, it can only furnish one permutation
of six. If instead of six we choose any other number, _x_, the number
of permutations of two will be _x_(_x_-1), that of three will be
_x_(_x_-1)(_x_-2), that of four _x_(_x_ -1)(_x_-2)(_x_-3), the rule
being: Multiply the whole number of counters by the next less number,
and the result by the next less, and so on, until as many numbers
have been multiplied together as there are to be counters in each
permutation: the product will be the whole number of permutations of
the sort required. Thus, out of 12 counters, permutations of four may
be made to the number of 12 × 11 × 10 × 9, or 11880.
EXERCISES.
207. In how many different ways can eight persons be arranged on eight
seats?
_Answer_, 40320.
In how many ways can eight persons be seated at a round table, so that
all shall not have the same neighbours in any two arrangements?[30]
_Answer_, 5040.
[30] The difference between this problem and the last is left to the
ingenuity of the pupil.
If the hundredth part of a farthing be given for every different
arrangement which can be made of fifteen persons, to how much will the
whole amount?
_Answer_, £13621608.
Out of seventeen consonants and five vowels, how many words can be
made, having two consonants and one vowel in each?
_Answer_, 4080.
208. If two or more of the counters have the same letter upon them, the
number of distinct permutations is less than that given by the last
rule. Let there be _a_, _a_, _a_, _b_, _c_, _d_, and, for a moment,
let us distinguish between the three as thus, _a_, _a′_, _a″_. Then,
_abca′a″d_, and _a″bcaa′d_ are reckoned as distinct permutations in
the rule, whereas they would not have been so, had it not been for the
accents. To compute the number of distinct permutations, let us make
one with _b_, _c_, and _d_, leaving places for the _a_s, thus, ( ) _bc_
( ) ( ) _d_. If the _a_s had been distinguished as _a_, _a′_, _a″_,
we might have made 3 × 2 × 1 distinct permutations, by filling up the
vacant places in the above, all which six are the same when the _a_s
are not distinguished. Hence, to deduce the number of permutations of
_a_, _a_, _a_, _b_, _c_, _d_, from that of _aa′a″bcd_, we must divide
the latter by 3 × 2 × 1, or 6, which gives
6 × 5 × 4 × 3 × 2 × 1
--------------------- or 120.
3 × 2 × 1
Similarly, the number of permutations of _aaaabbbcc_ is
9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
---------------------------------
4 × 3 × 2 × 1 × 3 × 2 × 1 × 2 × 1.
EXERCISE.
How many variations can be made of the order of the letters in the word
antitrinitarian?
_Answer_, 126126000.
209. From the number of permutations we can easily deduce the number of
combinations. But, in order to form these combinations independently,
we will shew a method similar to that in (206). If we know the
combinations of two which can be made out of _a_, _b_, _c_, _d_, _e_,
we can find the combinations of three, by writing successively at the
end of each combination of two, the letters which come after the last
contained in it. Thus, _ab_ gives _abc_, _abd_, _abe_; _ad_ gives _ade_
only. No combination of three can escape us if we proceed in this
manner, provided only we know the combinations of two; for any given
combination of three, as _acd_, will arise in the course of the process
from _ac_, which, according to our rule, furnishes _acd_. Neither will
any combination be repeated twice, for _acd_, if the rule be followed,
can only arise from _ac_, since neither _ad_ nor _cd_ furnishes it. If
we begin in this way to find the combinations of the five,
_a_ _b_ _c_ _d_ _e_
_a_ gives _ab_ _ac_ _ad_ _ae_
_b_ ···· _bc_ _bd_ _be_
_c_ ···· _cd_ _ce_
_d_ ···· _de_
Of these, _ab_ gives _abc_ _abd_ _abe_
_ac_ ···· _acd_ _ace_
_ad_ ···· _ade_
_bc_ ···· _bcd_ _bce_
_bd_ ···· _bde_
_cd_ ···· _cde_
_ae_ _be_ _ce_ and _de_ give none.
Of these, _abc_ gives _abcd_ _abce_
_abd_ ···· _abde_
_acd_ ···· _acde_
_bcd_ ···· _bcde_
Those which contain _e_ give none, as before.
Of the last, _abcd_ gives _abcde_, and the others none, which is
evidently true, since only one selection of five can be made out of
five things.
210. The rule for calculating the number of combinations is derived
directly from that for the number of permutations. Take 7 counters;
then, since the number of permutations of two is 7 × 6, and since two
permutations, _ba_ and _ab_, are in any combination _ab_, the number of
combinations is half that of the permutations, or (7 × 6)/2. Since the
number of permutations of three is 7 × 6 × 5, and as each combination
_abc_ has 3 × 2 × 1 permutations, the number of combinations of three is
7 × 6 × 5
----------.
1 × 2 × 3
Also, since any combination of four, _abcd_, contains 4 × 3 × 2 × 1
permutations, the number of combinations of four is
7 × 6 × 5 × 4
-------------,
1 × 2 × 3 × 4
and so on. The rule is: To find the number of combinations, each
containing _n_ counters, divide the corresponding number of
permutations by the product of 1, 2, 3, &c. up to _n_. If _x_ be the
whole number, the number of combinations of two is
_x_(_x_ - 1)
-------------;
1 × 2
that of three is
_x_(_x_ - 1)(_x_ - 2)
---------------------;
1 × 2 × 3
that of four is
_x_(_x_ - 1)(_x_ - 2)(_x_ - 3)
------------------------------; and so on.
1 × 2 × 3 × 4
211. The rule may in half the cases be simplified, as follows. Out of
ten counters, for every distinct selection of seven which is taken, a
distinct combination of 3 is left. Hence, the number of combinations
of seven is as many as that of three. We may, therefore, find the
combinations of three instead of those of seven; and we must moreover
expect, and may even assert, that the two formulæ for finding these two
numbers of combinations are the same in result, though different in
form. And so it proves; for the number of combinations of seven out of
ten is
10 × 9 × 8 × 7 × 6 × 5 × 4
--------------------------,
1 × 2 × 3 × 4 × 5 × 6 × 7
in which the product 7 × 6 × 5 × 4 occurs in both terms, and therefore
may be removed from both (108), leaving
10 × 9 × 8
----------,
1 × 2 × 3
which is the number of combinations of three out of ten. The same may
be shewn in other cases.
EXERCISES.
How many combinations of four can be made out of twelve things?
_Answer_, 495.
What number { 6 } { 8 } { 28
of combinations { 4 } out of { 11 } _Answer_, { 330
can be made of { 26 } { 28 } { 378
{ 6 } { 15 } { 5005
How many combinations can be made of 13 out of 52; or how many
different hands may a person hold at the game of whist?
_Answer_, 635013559600.
BOOK II.
COMMERCIAL ARITHMETIC.
SECTION I.
WEIGHTS, MEASURES, &C.
212. In making the calculations which are necessary in commercial
affairs, no more processes are required than those which have been
explained in the preceding book. But there is still one thing
wanted--not to insure the accuracy of our calculations, but to enable
us to compare and judge of their results. We have hitherto made use of
a single unit (15), and have treated of other quantities which are made
up of a number of units, in Sections II., III., and IV., and of those
which contain parts of that unit in Sections V. and VI. Thus, if we are
talking of distances, and take a mile as the unit, any other length may
be represented,[31] either by a certain number of miles, or a certain
number of parts of a mile, and (1 meaning one mile) may be expressed
either by a whole number or a fraction. But we can easily see that in
many cases inconveniences would arise. Suppose, for example, I say,
that the length of one room is ¹/₁₈₀ of a mile, and of another ¹/₁₇₄
of a mile, what idea can we form as to how much the second is longer
than the first? It is necessary to have some smaller measure; and if
we divide a mile into 1760 equal parts, and call each of these parts a
yard, we shall find that the length of the first room is 9 yards and
⁷/₉ of a yard, and that of the second 10 yards and ¹⁰/₈₇ of a yard.
From this we form a much better notion of these different lengths,
but still not a very perfect one, on account of the fractions ⁷/₉ and
¹⁰/₈₇. To get a clearer idea of these, suppose the yard to be divided
into three equal parts, and each of these parts to be called a foot;
then ⁷/₉ of a yard contains 2⅓ feet, and ¹⁰/₈₇ of a yard contains ³⁰/₈₇
of a foot, or a little more than ⅓ of a foot. Therefore the length of
the first room is now 9 yards, 2 feet, and ⅓ of a foot; that of the
second is 10 yards and a little more than ⅓ of a foot. We see, then,
the convenience of having large measures for large quantities, and
smaller measures for small ones; but this is done for convenience only,
for it is _possible_ to perform calculations upon any sort of quantity,
with one measure alone, as certainly as with more than one; and not
only possible, but more convenient, as far as the mere calculation is
concerned.
[31] It is not true, that if we choose any quantity as a unit, _any_
other quantity of the same kind can be exactly represented either by
a certain number of units, or of parts of a unit. To understand how
this is proved, the pupil would require more knowledge than he can be
supposed to have; but we can shew him that, for any thing he knows
to the contrary, there may be quantities which are neither units nor
parts of the unit. Take a mathematical line of one foot in length,
divide it into ten parts, each of those parts into ten parts, and so
on continually. If a point A be taken at hazard in the line, it does
not appear self-evident that if the decimal division be continued
ever so far, one of the points of division must at last fall exactly
on A: neither would the same appear necessarily true if the division
were made into sevenths, or elevenths, or in any other way. There may
then possibly be a part of a foot which is no exact numerical fraction
whatever of the foot; and this, in a higher branch of mathematics, is
found to be the case times without number. What is meant in the words
on which this note is written, is, that any part of a foot can be
represented as nearly as we please by a numerical fraction of it; and
this is sufficient for practical purposes.
The measures which are used in this country are not those which would
have been chosen had they been made all at one time, and by a people
well acquainted with arithmetic and natural philosophy. We proceed
to shew how the results of the latter science are made useful in
our system of measures. Whether the circumstances introduced are
sufficiently well known to render the following methods exact enough
for the recovery of _astronomical_ standards, may be matter of opinion;
but no doubt can be entertained of their being amply correct for
commercial purposes.
It is evidently desirable that weights and measures should always
continue the same, and that posterity should be able to replace any
one of them when the original measure is lost. It is true that a yard,
which is now exact, is kept by the public authorities; but if this were
burnt by accident,[32] how are those who shall live 500 years hence to
know what was the length which their ancestors called a yard? To ensure
them this knowledge, the measure must be derived from something which
cannot be altered by man, either from design or accident. We find such
a quantity in the time of the daily revolution of the earth, and also
in the length of the year, both of which, as is shewn in astronomy,
will remain the same, at least for an enormous number of centuries,
unless some great and totally unknown change take place in the solar
system. So long as astronomy is cultivated, it is impossible to suppose
that either of these will be lost, and it is known that the latter is
365·24224 mean solar days, or about 365¼ of the average interval which
elapses between noon and noon, that is, between the times when the sun
is highest in the heavens. Our year is made to consist of 365 days,
and the odd quarter is allowed for by adding one day to every fourth
year, which gives what we call leap-year. This is the same as adding ¼
of a day to each year, and is rather too much, since the excess of the
year above 365 days is not ·25 but ·24224 of a day. The difference is
·00776 of a day, which is the quantity by which our average year is too
long. This amounts to a day in about 128 years, or to about 3 days in 4
centuries. The error is corrected by allowing only one out of four of
the years which close the centuries to be leap-years. Thus, A.D. 1800
and 1900 are not leap-years, but 2000 is so.
[32] Since this was first written, the accident has happened. The
_standard yard_ was so injured as to be rendered useless by the fire at
the Houses of Parliament.
213. The day is therefore the first measure obtained, and is divided
into 24 parts or hours, each of which is divided into 60 parts or
minutes, and each of these again into 60 parts or seconds. One second,
marked thus, 1″,[33] is therefore the 86400ᵗʰ part of a day, and the
following is the
MEASURE OF TIME.[34]
60 _seconds_ are 1 _minute_ 1 m.
60 _minutes_ ” 1 _hour_ 1 h.
24 _hours_ ” 1 _day_ 1 d.
7 _days_ ” 1 _week_ 1 wk.
365 _days_ ” 1 _year_ 1 yr.
214. The _second_ having been obtained, a pendulum can be constructed
which shall, when put in motion, perform one vibration in exactly
one second, in the latitude of Greenwich.[35] If we were inventing
measures, it would be convenient to call the length of this pendulum a
yard, and make it the standard of all our measures of length. But as
there is a yard already established, it will do equally well to tell
the length of the pendulum in yards. It was found by commissioners
appointed for the purpose, that this pendulum in London was 39·1393
inches, or about one yard, three inches, and ⁵/₃₆ of an inch. The
following is the division of the yard.
MEASURES OF LENGTH.
The lowest measure is a barleycorn.[36]
3 _barleycorns_ are 1 _inch_ 1 in.
12 _inches_ 1 _foot_ 1 ft.
3 _feet_ 1 _yard_ 1 yd.
5½ _yards_ 1 _pole_ 1 po.
40 _poles_ or 220 _yards_ 1 _furlong_ 1 fur.
8 _furlongs_ or 1760 _yards_ 1 _mile_ 1 mi.
Also 6 _feet_ 1 _fathom_ 1 fth.
69⅓ _miles_ 1 _degree_ 1 deg. or 1°.
[33] The minute and second are often marked thus, 1′, 1″: but this
notation is now almost entirely appropriated to the minute and second
of _angular_ measure.
[34] The measures in italics are those which it is most necessary that
the student should learn by heart.
[35] The lengths of the pendulums which will vibrate in one second are
slightly different in different latitudes. Greenwich is chosen as the
station of the Royal Observatory. We may add, that much doubt is now
entertained as to the system of standards derived from nature being
capable of that extreme accuracy which was once attributed to it.
[36] The inch is said to have been originally obtained by putting
together three grains of barley.
A geographical mile is ¹/₆₀th of a degree, and three such miles are one
nautical league.
In the measurement of cloth or linen the following are also used:
2¼ inches are 1 nail 1 nl.
4 nails 1 quarter (of a yard) 1 qr.
3 quarters 1 Flemish ell 1 Fl. e.
5 quarters 1 English ell 1 E. e.
6 quarters 1 French ell 1 Fr. e.
215. MEASURES OF SURFACE, OR SUPERFICIES.
All surfaces are measured by square inches, square feet, &c.; the
square inch being a square whose side is an inch in length, and so on.
The following measures may be deduced from the last, as will afterwards
appear.
144 square inches are 1 square foot 1 sq. ft.
9 square feet 1 square yard 1 sq. yd.
30¼ square yards 1 square pole 1 sq. p.
40 square poles 1 rood 1 rd.
4 roods 1 acre 1 ac.
Thus, the acre contains 4840 square yards, which is ten times a square
of 22 yards in length and breadth. This 22 yards is the length which
land-surveyors’ chains are made to have, and the chain is divided into
100 links, each ·22 of a yard or 7·92 inches. An acre is then 10 square
chains. It may also be noticed that a square whose side is 69⁴/₇ yards
is nearly an acre, not exceeding it by ⅕ of a square foot.
216. MEASURES OF SOLIDITY OR CAPACITY.[37]
Cubes are solids having the figure of dice. A cubic inch is a cube each
of whose sides is an inch, and so on.
1728 cubic inches are 1 cubic foot 1 c. ft.
27 cubic feet 1 cubic yard 1 c. yd.
[37] ‘Capacity’ is a term which cannot be better explained than by its
use. When one measure holds more than another, it is said to be more
capacious, or to have a greater capacity.
This measure is not much used, except in purely mathematical questions.
In the measurements of different commodities various measures were
used, which are now reduced, by act of parliament, to one. This is
commonly called the imperial measure, and is as follows:
MEASURE OF LIQUIDS AND OF ALL DRY GOODS.
4 _gills_ are 1 _pint_ 1 pt.
2 _pints_ 1 _quart_ 1 qt.
4 _quarts_ 1 _gallon_ 1 gall.
2 _gallons_ 1 _peck_[38] 1 pk.
4 _pecks_ 1 _bushel_ 1 bu.
8 _bushels_ 1 _quarter_ 1 qr.
5 _quarters_ 1 _load_ 1 ld.
The gallon in this measure is about 277·274 cubic inches; that is, very
nearly 277¼ cubic inches.[39]
217. The smallest weight in use is the grain, which is thus determined.
A vessel whose interior is a cubic inch, when filled with water,[40]
has its weight increased by 252·458 grains. Of the grains so
determined, 7000 are a pound _averdupois_, and 5760 a pound _troy_.
The first pound is always used, except in weighing precious metals and
stones, and also medicines. It is divided as follows:
[38] This measure, and those which follow, are used for dry goods only.
[39] Since the publication of the third edition, the _heaped_ measure,
which was part of the new system, has been abolished. The following
paragraph from the third edition will serve for reference to it:
“The other imperial measure is applied to goods which it is customary
to sell by _heaped measure_, and is as follows:
2 gallons 1 peck
4 pecks 1 bushel
3 bushels 1 sack
12 sacks 1 chaldron.
The gallon and bushel in this measure hold the same when only just
filled, as in the last. The bushel, however, heaped up as directed by
the act of parliament, is a little more than one-fourth greater than
before.”
[40] Pure water, cleared from foreign substances by distillation, at a
temperature of 62° Fahr.
AVERDUPOIS WEIGHT.
27¹¹/₃₂ _grains_ are 1 _dram_ 1 dr.
6 _drams_, or _drachms_ 1 _ounce_[41] 1 oz.
16 _ounces_ 1 _pound_ 1 lb.
28 _pounds_ 1 _quarter_ 1 qr.
4 _quarters_ 1 _hundred-weight_ 1 cwt.
20 _hundred-weight_ 1 _ton_ 1 ton.
[41] It is more common to divide the ounce into four quarters than into
sixteen drams.
The pound averdupois contains 7000 grains. A cubic foot of water weighs
62·3210606 pounds averdupois, or 997·1369691 ounces.
For the precious metals and for medicines, the pound troy, containing
5760 grains, is used, but is differently divided in the two cases. The
measures are as follow:
TROY WEIGHT.
24 _grains_ are 1 _pennyweight_ 1 dwt.
20 _pennyweights_ 1 _ounce_ 1 oz.
12 _ounces_ 1 _pound_ 1 lb.
The pound troy contains 5760 grains. A cubic foot of water weighs
75·7374 pounds troy, or 908·8488 ounces.
APOTHECARIES’ WEIGHT.
20 _grains_ are 1 _scruple_ ℈
3 _scruples_ 1 _dram_ ʒ
8 _drams_ 1 _ounce_ ℥
12 _ounces_ 1 _pound_ lb
218. The standard coins of copper, silver, and gold, are,--the penny,
which is 10⅔ drams of copper; the shilling, which weighs 3 pennyweights
15 grains, of which 3 parts out of 40 are alloy, and the rest pure
silver; and the sovereign, weighing 5 pennyweights and 3¼ grains, of
which 1 part out of 12 is copper, and the rest pure gold.
MEASURES OF MONEY.
The lowest coin is a farthing, which is marked thus, ¼, being one
fourth of a penny.
2 _farthings_ are 1 _halfpenny_ ½_d_.
2 _halfpence_ 1 _penny_ 1_d_.
12 _pence_ 1 _shilling_ 1_s_.
20 _shillings_ 1 _pound_[42] or _sovereign_ £1
21 _shillings_ 1 _guinea_.[43]
219. When any quantity is made up of several others, expressed in
different units, such as £1. 14. 6, or 2cwt. 1qr. 3lbs., it is called a
_compound quantity_. From these tables it is evident that any compound
quantity of any substance can be measured in several different ways.
For example, the sum of money which we call five pounds four shillings
is also 104 shillings, or 1248 pence, or 4992 farthings. It is easy to
reduce any quantity from one of these measurements to another; and the
following examples will be sufficient to shew how to apply the same
process, usually called REDUCTION, to all sorts of quantities.
I. How many farthings are there in £18. 12. 6¾?[44]
[42] The English pound is generally called a _pound sterling_, which
distinguishes it from the weight called a pound, and also from foreign
coins.
[43] The coin called a guinea is now no longer in use, but the name is
still given, from custom, to 21 shillings. The pound, which was not a
coin, but a note promising to pay 20 shillings to the bearer, is also
disused for the present, and the sovereign supplies its place; but the
name pound is still given to 20 shillings.
[44] Farthings are never written but as parts of a penny. Thus, three
farthings being 3/4 of a penny, is written 3/4, or ¾. One halfpenny may
be written either as 2/4 or ½; the latter is most common.
Since there are 20 shillings in a pound, there are, in £18, 18 × 20, or
360 shillings; therefore, £18. 12 is 360 + 12, or 372 shillings. Since
there are 12 pence in a shilling, in 372 shillings there are 372 × 12,
or 4464 pence; and, therefore, in £18. 12. 6 there are 4464 + 6, or
4470 pence.
Since there are 4 farthings in a penny, in 4470 pence there are 4470 ×
4, or 17880 farthings; and, therefore, in £18. 12. 6¾ there are 17880
+ 3, or 17883 farthings. The whole of this process may be written as
follows:
£18 . 12 . 6¾
20
--------
360 + 12 = 372
12
-----
4464 + 6 = 4470
4
-----
17880 + 3 = 17883
II. In 17883 farthings, how many pounds, shillings, pence, and
farthings are there?
Since 17883, divided by 4, gives the quotient 4470, and the remainder
3, 17883 farthings are 4470 pence and 3 farthings (218).
Since 4470, divided by 12, gives the quotient 372, and the remainder 6,
4470 pence is 372 shillings and 6 pence.
Since 372, divided by 20, gives the quotient 18, and the remainder 12,
372 shillings is 18 pounds and 12 shillings.
Therefore, 17883 farthings is 4470¾_d_., which is 372s. 6¾_d_., which
is £18. 12. 6¾.
The process may be written as follows:
4)17883
-----
12)4470 ... 3
----
20)372 ... 6
£18 . 12 . 6¾
EXERCISES.
A has £100. 4. 11½, and B has 64392 farthings. If A receive 1492
farthings, and B £1. 2. 3½, which will then have the most, and by how
much?--_Answer_, A will have £33. 12. 3 more than B.
In the following table the quantities written opposite to each other
are the same: each line furnishes two exercises.
£15 . 18 . 9½ | 15302 farthings.
115ˡᵇˢ 1ᵒᶻ 8ᵈᵚᵗ | 663072 grains.
3ˡᵇˢ 14ᵒᶻ 9ᵈʳ | 1001 drams.
3ᵐ 149 yds 2ᶠᵗ 9 in | 195477 inches.
19ᵇᵘ 2ᵖᵏˢ 1 gall 2 qᵗˢ | 1260 pints.
16 ʰ 23ᵐ 47ˢ | 59027 seconds.
220. The same may be done where the number first expressed is
fractional. For example, how many shillings and pence are there in ⁴/₁₅
of a pound? Now, ⁴/₁₅ of a pound is ⁴/₁₅ of 20 shillings; ⁴/₁₅ of 20 is
4 × 20 4 × 4 16
------, or ----- (110), or ---,
15 3 3
or (105) 5⅓ of a shilling. Again, ⅓ of a shilling is ⅓ of 12 pence, or
4 pence. Therefore, £⁴/₁₅ = 5_s._ 4_d._
Also, ·23 of a day is ·23 × 24 in hours, or 5ʰ·52; and ·52 of an hour
is ·52 × 60 in minutes, or 3ᵐ·2; and ·2 of a minute is ·2 × 60 in
seconds, or 12ˢ; whence ·23 of a day is 5ʰ 31ᵐ 12ˢ.
Again, suppose it required to find what part of a pound 6_s_. 8_d_. is.
Since 6_s._ 8_d._ is 80 pence, and since the whole pound contains 20
× 12 or 240 pence, 6_s._ 8_d._ is made by dividing the pound into 240
parts, and taking 80 of them. It is therefore £⁸⁰/₂₄₀ (107), but ⁸⁰/₂₄₀
= ⅓ (108); therefore, 6_s._ 8_d._ = £⅓.
EXERCISES.
⅖ of a day is 9ʰ 36ᵐ
·12841 of a day 3ʰ 4ᵐ 54ᔆ·624[45]
·257 of a cwt. 28ˡᵇˢ 12ᵒᶻ 8ᵈʳ·704
£·14936 2ˢ 11ᵈ 3ᶠ·3856
[45] When a decimal follows a whole number, the decimal is always of
the same unit as the whole number. Thus, 5ᔆ·5 is five _seconds_ and
five-tenths of a _second_. Thus, 0ᔆ·5 means five-tenths of a second;
0ʰ·3, three-tenths of an hour.
221, 222. I have thought it best to refer the mode of converting
shillings, pence, and farthings into decimals of a pound to the
Appendix (See Appendix _On Decimal Money_). I should strongly recommend
the reader to make himself perfectly familiar with the modes given in
that Appendix. To prevent the subsequent sections from being altered in
their numbering, I have numbered this paragraph as above.
223. The rule of addition[46] of two compound quantities of the same
sort will be evident from the following example. Suppose it required to
add £192. 14. 2½ to £64. 13. 11¾. The sum of these two is the whole of
that which arises from adding their several parts. Now
¾_d._ + ½_d._ = ⁵/₄_d._ = £0 . 0 . 1¼ (219)
11_d._ + 2_d._ = 13_d._ = 0 . 1 . 1
13_s._ + 14_s._ = 27_s._ = 1 . 7 . 0
£64 + £192 = 256 . 0 . 0
-----------
The sum of all of which is £257. 8 . 2¼
This may be done at once, and written as follows:
£192 . 14 . 2½
64 . 13 . 11¾
----------------
£257 . 8 . 2¼
[46] Before reading this article and the next, articles (29) and (42)
should be read again carefully.
Begin by adding together the farthings, and reduce the result to pence
and farthings. Set down the last only, carry the first to the line
of pence, and add the pence in both lines to it. Reduce the sum to
shillings and pence; set down the last only, and carry the first to the
line of shillings, and so on. The same method must be followed when the
quantities are of any other sort; and if the tables be kept in memory,
the process will be easy.
224. SUBTRACTION is performed on the same principle as in (40), namely,
that the difference of two quantities is not altered by adding the same
quantity to both. Suppose it required to subtract £19 . 13. 10¾ from
£24. 5. 7½. Write these quantities under one another thus:
£24. 5. 7½
19. 13. 10¾
Since ¾ cannot be taken from ½ or ²/₄, add 1_d._ to both quantities,
which will not alter their difference; or, which is the same thing,
add 4 farthings to the first, and 1_d._ to the second. The pence and
farthings in the two lines then stand thus: 7⁶/₄_d._ and 11¾_d._ Now
subtract ¾ from ⁶/₄, and the difference is ¾ which must be written
under the farthings. Again, since 11_d._ cannot be subtracted from
7_d._, add 1_s._ to both quantities by adding 12_d._ to the first, and
1_s._ to the second. The pence in the first line are then 19, and in
the second 11, and the difference is 8, which write under the pence.
Since the shillings in the lower line were increased by 1, there are
now 14_s._ in the lower, and 5_s._ in the upper one. Add 20_s._ to the
upper and £1 to the lower line, and the subtraction of the shillings in
the second from those in the first leaves 11_s._ Again, there are now
£20 in the lower, and £24 in the upper line, the difference of which is
£4; therefore the whole difference of the two sums is £4. 11. 8¾. If we
write down the two sums with all the additions which have been made,
the process will stand thus:
£24 . 25 . 19⁶/₄
20 . 14 . 11¾
------------------
Difference £4 . 11 . 8¾
225. The same method may be applied to any of the quantities in the
tables. The following is another example:
From 7 cwt. 2 qrs. 21 lbs. 14 oz.
Subtract 2 cwt. 3 qrs. 27 lbs. 12 oz.
After alterations have been made similar to those in the last article,
the question becomes:
From 7 cwt. 6 qrs. 49 lbs. 14 oz.
Subtract 3 cwt. 4 qrs. 27 lbs. 12 oz.
----------------------------
The difference is 4 cwt. 2 qrs. 22 lbs. 2 oz.
In this example, and almost every other, the process may be a little
shortened in the following way. Here we do not subtract 27 lbs. from 21
lbs., which is impossible, but we increase 21 lbs. by 1 qr. or 28 lbs.
and then subtract 27 lbs. from the sum. It would be shorter, and lead
to the same result, first to subtract 27 lbs. from 1 qr. or 28 lbs. and
add the difference to 21 lbs.
226. EXERCISES.
A man has the following sums to receive: £193. 14. 11¼, £22. 0. 6¾,
£6473. 0. 0, and £49. 14. 4½; and the following debts to pay: £200 .
19. 6¼, £305. 16. 11, £22, and £19. 6. 0½. How much will remain after
paying the debts?
_Answer_, £6190. 7. 4¾.
There are four towns, in the order A, B, C, and D. If a man can go from
A to B in 5ʰ 20ᵐ 33ˢ, from B to C in 6ʰ 49ᵐ 2ˢ and from A to D in 19ʰ
0ᵐ 17ˢ, how long will he be in going from B to D, and from C to D?
_Answer_, 13ʰ 39ᵐ 44ˢ, and 6ʰ 50ᵐ 42ˢ.
227. In order to perform the process of MULTIPLICATION, it must be
recollected that, as in (52), if a quantity be divided into several
parts, and each of these parts be multiplied by a number, and the
products be added, the result is the same as would arise from
multiplying the whole quantity by that number.
It is required to multiply £7. 13. 6¼ by 13. The first quantity is made
up of 7 pounds, 13 shillings, 6 pence, and 1 farthing. And
1 farth. × 13 is 13 farth. or £0 . 0 . 3¼ (219)
6 pence × 13 is 78 pence, or 0 . 6 . 6
13 shill. × 13 is 169 shill. or 8 . 9 . 0
7 pounds × 13 is 91 pounds, or 91 . 0 . 0
--------------
The sum of all these is £99 . 15 . 9¼
which is therefore £7. 13. 6¼ × 13.
This process is usually written as follows:
£7 . 13 . 6¼
13
-------------
£99 . 15 . 9¼
228. DIVISION is performed upon the same principle as in (74), viz.
that if a quantity be divided into any number of parts, and each part
be divided by any number, the different quotients added together will
make up the quotient of the whole quantity divided by that number.
Suppose it required to divide £99. 15. 9¼ by 13. Since 99 divided by 13
gives the quotient 7, and the remainder 8, the quantity is made up of
£13 × 7, or £91, and £8. 15. 9¼. The quotient of the first, 13 being
the divisor, is £7: it remains to find that of the second. Since £8 is
160_s._, £8. 15. 9¼ is 175_s._ 9¼_d._, and 175 divided by 13 gives the
quotient 13, and the remainder 6; that is, 175_s._ 9¼_d._ is made up of
169_s._ and 6_s._ 9¼_d._, the quotient of the first of which is 13_s._,
and it remains to find that of the second. Since 6_s._ is 72_d._,
6_s._ 9¼_d._ is 81¼_d._, and 81 divided by 13 gives the quotient 6
and remainder 3; that is, 81¼_d._ is 78_d._ and 3¼_d._, of the first
of which the quotient is 6_d._ Again, since 3_d._ is ¹²/₄, or 12
farthings, 3¼_d._ is 13 farthings, the quotient of which is 1 farthing,
or ¼, without remainder. We have then divided £99. 15. 9¼ into four
parts, each of which is divisible by 13, viz. £91, 169_s._, 78_d._, and
13 farthings; so that the thirteenth part of this quantity is £7. 13.
6¼. The whole process may be written down as follows; and the same sort
of process may be applied to the exercises which follow:
£ _s._ _d._ £ _s._ _d._
13)99 15 9¼( 7 13 6¼
91
--
8
20
---
160 + 15 = 175
13
---
45
39
--
6
12
--
72 + 9 = 81
78
--
3
4
--
12 + 1 = 13
13
--
0
Here, each of the numbers 99, 175, 81, and 13, is divided by 13 in the
usual way, though the divisor is only written before the first of them.
EXERCISES.
2 cwt. 1 qr. 21 lbs. 7 oz. × 53 = 129 cwt. 1 qr. 16 lbs. 3 oz.
2ᵈ 4ʰ 3ᵐ 27ˢ × 109 = 236ᵈ 10ʰ 16ᵐ 3ˢ
£27 . 10 . 8 × 569 = £15666 . 9 . 4
£7 . 4 . 8 × 123 = £889 . 14
£166 × ₈/₃₃ = £40 . 4 . 10⁶/₃₃
£187 . 6 . 7 × ³/₁₀₀ = £5 . 12 . 4¾ ²/₂₅
4_s._ 6½_d._ × 1121 = £254 . 11 . 2½
4_s._ 4_d._ × 4260 = 6_s._ 6_d._ × 2840
229. Suppose it required to find how many times 1s. 4¼_d._ is contained
in £3. 19. 10¾. The way to do this is to find the number of farthings
in each. By 219, in the first there are 65, and in the second 3835
farthings. Now, 3835 contains 65 59 times; and therefore the second
quantity is 59 times as great as the first. In the case, however, of
pounds, shillings, and pence, it would be best to use decimals of a
pound, which will give a sufficiently exact answer. Thus 1s. 4¼_d._ is
£·067, and £3. 19. 10¾ is £3·994, and 3·994 divided by ·067 is 3994 by
67, or 59⁴¹/₆₇. This is an extreme case, for the smaller the divisor,
the greater the effect of an error in a given place of decimals.
EXERCISES.
How many times does 6 cwt. 2 qrs. contain 1 qr. 14 lbs. 1 oz.? and 1ᵈ
2ʰ 0ᵐ 47ˢ contain 3ᵐ 46ˢ?
_Answer_, 17·30758 and 414·367257.
If 2 cwt. 3 qrs. 1 lb. cost £150. 13. 10, how much does 1 lb. cost?
_Answer_, 9_s._ 9_d._ ¹³/₃₀₉.
A grocer mixes 2 cwt. 15 lbs. of sugar at 11_d._ per pound with 14
cwt. 3 lbs. at 5_d._ per pound. At how much per pound must he sell the
mixture so as not to lose by mixing them?
_Answer_, 5_d._ ¾ ¹⁵³/₉₀₅.
230. There is a convenient method of multiplication called PRACTICE.
Suppose I ask, How much do 153 tons cost if each ton cost £2. 15. 7½?
It is plain that if this sum be multiplied by 153, the product is the
price of the whole. But this is also evident, that, if I buy 153 tons
at £2. 15. 7½ each ton, payment may be made by first putting down £2
for each ton, then 10s. for each, then 5_s._, then 6_d._, and then
1½_d._ These sums together make up £2. 15. 7½, and the reason for this
separation of £2. 15 . 7½ into different parts will be soon apparent.
The process may be carried on as follows:
1. 153 tons, at £2 each ton, will cost £306 0 0
2. Since 10s. is £½, 153 tons, at 10_s._ each,
will cost £15³/₂, which is 76 10 0
3. Since 5_s._ is ½ of 10_s._, 153 tons, at 5_s._,
will cost half as much as the same number at
10_s._ each, that is, ½ of £76 . 10, which is 38 5 0
4. Since 6_d._ is ⅒ of 5_s._, 153 tons, at 6_d._
each, will cost ⅒ of what the same number
costs at 5_s._each, that is, ⅒ of £38 . 5,
which is 3 16 6
5. Since 1½ or 3 halfpence is ¼ of 6_d._ or 12
halfpence, 153 tons, at 1½_d._ each, will cost
¼ of what the same number costs at 6_d._ each,
that is, ¼ of£3 . 16 . 6, which is 0 19 1½
-----------
The sum of all these quantities is 425 10 7½
which is, therefore, £2 . 15 . 7½ × 153.
The whole process may be written down as follows:
or what
153 tons
would
| £153 0 0 cost at £1 per ton.
| ----------- -----------
£2 is 2 × £1 | 306 0 0 2 0 0
10_s._ is ½ of £1 | 76 10 0 0 10 0
5_s._ is ½ of 10_s._ | 38 5 0 0 5 0
6_d._ is ⅒ of 5_s._ | 3 16 6 0 0 6
1½_d._ is ¼ of 6_d._ | 0 19 1½ 0 0 1½
| ------------ ----------
Sum | £425 10 7½ £2 15 7½
ANOTHER EXAMPLE.
What do 1735 lbs. cost at 9_s._ 10¾_d._ per lb.? The price 9_s._
10¾_d_. is made up of 5_s._, 4_s._, 10_d._, ½_d._, and ¼_d._; of which
5_s._ is ¼ of £1, 4_s._ is ⅕ of £1, 10_d._ is ⅙ of 5_s._, ½_d._ is ¹/₂₀
of 10_d._, and ¼_d._ is ½ of ½_d._ Follow the same method as in the
last example, which gives the following:
or what
1735 tons
would
| £1735 0 0 cost at £1 per lb.
| ------------ -----------
5_s._ is ¼ of £1 | 433 15 0 0 5 0
|
4_s._ is ⅕ of £1 | 347 0 0 0 4 0
|
10_d._ is ⅙ of 5_s._ | 72 5 10 0 0 10
|
½_d._ is ¹/₂₀ of 10_d._ | 3 12 3½ 0 0 0½
|
¼_d._ is ½ of ½_d._ | 1 16 1¾ 0 0 0¼
| ------------- ------------
by addition ... | £858 9 3¼ £0 9 10¾
In all cases, the price must first be divided into a number of parts,
each of which is a simple fraction[47] of some one which goes before.
No rule can be given for doing this, but practice will enable the
student immediately to find out the best method for each case. When
that is done, he must find how much the whole quantity would cost if
each of these parts were the price, and then add the results together.
[47] Any fraction of a unit, whose numerator is unity, is generally
called an _aliquot part_ of that unit. Thus, 2_s._ and 10_s._ are both
aliquot parts of a pound, being £⅒ and £½.
EXERCISES.
What is the cost of
243 cwt. at £14 . 18 . 8¼ per cwt.?--_Answer_, £3629 . 1 . 0¾.
169 bushels at £2 . 1 . 3¼ per bushel?--_Answer_, £348 . 14 . 9¼.
273 qrs. at 19_s._ 2_d._ per quarter?--_Answer_, £261 . 12. 6.
2627 sacks at 7_s._ 8½_d._ per sack?--_Answer_, £1012 . 9 . 9½.
231. Throughout this section it must be observed, that the rules can be
applied to cases where the quantities given are expressed in common or
decimal fractions, instead of the measures in the tables. The following
are examples:
What is the price of 272·3479 cwt. at £2. 1. 3½ per cwt.?
_Answer_, £562·2849, or £562. 5. 8¼.
66½lbs. at 1_s._ 4½_d._ per lb. cost £4. 11. 5¼.
How many pounds, shillings, and pence, will 279·301 acres let for if
each acre lets for £3·1076?--_Answer_, £867·9558, or £867. 19. 1¼.
What does ¼ of ³/₁₃ of 17 bush. cost at ⅙ of ⅔ of £17. 14 per bushel?
_Answer_, £2·3146, or £2. 6. 3½.
What is the cost of 19lbs. 8oz. 12dwt. 8gr. at £4. 4. 6 per
ounce?--_Answer_, £999. 14. 1¼ ⅙.
232. It is often required to find to how much a certain sum per day
will amount in a year. This may be shortly done, since it happens that
the number of days in a year is 240 + 120 + 5; so that a penny per day
is a pound, half a pound, and 5 pence per year. Hence the following
rule: To find how much any sum per day amounts to in a year, turn it
into pence and fractions of a penny; to this add the half of itself,
and let the pence be pounds, and each farthing five shillings; then
add five times the daily sum, and the total is the yearly amount. For
example, what does 12_s._ 3¾_d._ amount to in a year? This is 147¾_d._,
and its half is 73⅞_d._, which added to 147¾_d._ gives 221⅝_d._, which
turned into pounds is £221. 12. 6. Also, 12_s._ 3¾_d._ × 5 is £3. 1.
6¾, which added to the former sum gives £224. 14. 0¾ for the yearly
amount. In the same way the yearly amount of 2_s._ 3½_d._ is £41. 16.
5½; that of 6¾_d._ is £10. 5. 3¾; and that of 11_d._ is £16. 14. 7.
233. An inverse rule may be formed, sufficiently correct for every
purpose, in the following way: If the year consisted of 360 days, or
³/₂ of 240, the subtraction of one-third from any sum per year would
give the proportion which belongs to 240 days; and every pound so
obtained would be one penny per day. But as the year is not 360, but
365 days, if we divide each day’s share into 365 parts, and take 5
away, the whole of the subtracted sum, or 360 × 5 such parts, will
give 360 parts for each of the 5 days which we neglected at first.
But 360 such parts are left behind for each of the 360 first days;
therefore, this additional process divides the whole annual amount
equally among the 365 days. Now, 5 parts out of 365 is one out of
73, or the 73d part of the first result must be subtracted from it
to produce the true result. Unless the daily sum be very large, the
72d part will do equally well, which, as 72 farthings are 18 pence,
is equivalent to subtracting at the rate of one farthing for 18_d._,
or ½_d._ for 3_s._, or 10_d._ for £3. The rule, then, is as follows:
To find how much per day will produce a given sum per year, turn
the shillings, &c. in the given sum into decimals of a pound (221);
subtract one-third; consider the result as pence; and diminish it by
one farthing for every eighteen pence, or ten pence for every £3. For
example, how much per day will give £224. 14. 0¾ per year? This is
224·703, and its third is 74·901, which subtracted from 224·703, gives
149·802, which, if they be pence, amounts to 12_s._ 5·802_d._, in which
1_s._ 6_d._ is contained 8 times. Subtract 8 farthings, or 2_d._, and
we have 12_s._ 3·802_d._, which differs from the truth only about ¹/₂₀
of a farthing. In the same way, £100 per year is 5_s._ 5¾_d._ per day.
234. The following connexion between the measures of length and the
measures of surface is the foundation of the application of arithmetic
to geometry.
[Illustration]
Suppose an oblong figure, A, B, C, D, as here drawn (which is called
a _rectangle_ in geometry), with the side A B 6 inches, and the side
A C 4 inches. Divide A B and C D (which are equal) each into 6 inches
by the points _a, b, c, l, m_, &c.; and A C and B D (which are also
equal) into 4 inches by the points _f, g, h, x, y_, and _z_. Join _a_
and l, _b_ and _m_, &c., and _f_ and _x_, &c. Then, the figure A B C D
is divided into a number of squares; for a square is a rectangle whose
sides are equal, and therefore A _a f_ E is square, since A _a_ is of
the same length as A _f_, both being 1 inch. There are also four rows
of these squares, with six squares in each row; that is, there are 6
× 4, or 24 squares altogether. Each of these squares has its sides 1
inch in length, and is what was called in (215) _a square inch_. By the
same reasoning, if one side had contained 6 _yards_, and the other 4
_yards_, the surface would have contained 6 × 4 _square yards_; and so
on.
[Illustration]
235. Let us now suppose that the sides of A B C D, instead of being
a whole number of inches, contain some inches and a fraction. For
example, let A B be 3½ inches, or (114) ⁷/₂ of an inch, and let A C
contain 2½ inches, or ⁹/₄ of an inch. Draw A E twice as long as A B,
and A F four times as long as A C, and complete the rectangle A E F G.
The rest of the figure needs no description. Then, since A E is twice
A B, or twice ⁷/₂ inches, it is 7 inches. And since A F is four times
A C, or four times ⁹/₄ inches, it is 9 inches. Therefore, the whole
rectangle A E F G contains, by (234), 7 × 9 or 63 square inches. But
the rectangle A E F G contains 8 rectangles, all of the same figure
as A B C D; and therefore A B C D is one-eighth part of A E F G, and
contains ⁶³/₈ square inches. But ⁶³/₈ is made by multiplying ⁹/₄ and
⁷/₂ together (118). From this and the last article it appears, that,
whether the sides of a rectangle be a whole or a fractional number of
inches, the number of square inches in its surface is the product of
the numbers of inches in its sides. The square itself is a rectangle
whose sides are all equal, and therefore the number of square inches
which a square contains is found by multiplying the number of inches in
its side by itself. For example, a square whose side is 13 inches in
length contains 13 × 13 or 169 square inches.
236. EXERCISES.
What is the content, in square feet and inches, of a room whose sides
are 42 ft. 5 inch. and 31 ft. 9 inch.? and supposing the piece from
which its carpet is taken to be three quarters of a yard in breadth,
what length of it must be cut off?--_Answer_, The content is 1346
square feet 105 square inches, and the length of carpet required is 598
feet 6⁵/₉ inches.
The sides of a rectangular field are 253 yards and a quarter of a mile;
how many acres does it contain?--_Answer_, 23.
What is the difference between 18 _square miles_, and a square of 18
miles long, or 18 _miles square_?--_Answer_, 306 square miles.
237. It is by this rule that the measure in (215) is deduced from
that in (214); for it is evident that twelve inches being a foot, the
square foot is 12 × 12 or 144 square inches, and so on. In a similar
way it may be shewn that the content in cubic inches of a cube, or
parallelepiped,[48] may be found by multiplying together the number of
inches in those three sides which meet in a point. Thus, a cube of 6
inches contains 6 × 6 × 6, or 216 cubic inches; a chest whose sides are
6, 8, and 5 feet, contains 6 × 8 × 5, or 240 cubic feet. By this rule
the measure in (216) was deduced from that in (214).
[48] A parallelepiped, or more properly, a _rectangular_
parallelepiped, is a figure of the form of a brick; its sides, however,
may be of any length; thus, the figure of a plank has the same name. A
cube is a parallelepiped with equal sides, such as is a die.
SECTION II.
RULE OF THREE.
238. Suppose it required to find what 156 yards will cost, if 22 yards
cost 17_s._ 4_d._ This quantity, reduced to pence, is 208_d._; and if
22 yards cost 208_d._, each yard costs ²⁰⁸/₂₂_d_. But 156 yards cost
156 times the price of one yard, and therefore cost
208 208 × 156
---- × 156 pence, or --------- pence (117).
22 22
Again, if 25½ French francs be 20 shillings sterling, how many francs
are in £20. 15? Since 25½ francs are 20 shillings, twice the number of
francs must be twice the number of shillings; that is, 51 francs are
40 shillings, and one shilling is the fortieth part of 51 francs, or
⁵¹/₄₀ francs. But £20 15_s._ contain 415 shillings (219); and since 1
shilling is ⁵¹/₄₀ francs, 415 shillings is
51 × 415
⁵¹/₄₀ × 415 francs, or (117) -------- francs.
40
239. Such questions as the last two belong to the most extensive rule
in Commercial Arithmetic, which is called the RULE OF THREE, because in
it three quantities are given, and a fourth is required to be found.
From both the preceding examples the following rule may be deduced,
which the same reasoning will shew to apply to all similar cases.
It must be observed, that in these questions there are two quantities
which are of the same sort, and a third of another sort, of which last
the answer must be. Thus, in the first question there are 22 and 156
yards and 208 pence, and the thing required to be found is a number
of pence. In the second question there are 20 and 415 shillings and
25½ francs, and what is to be found is a number of francs. Write the
three quantities in a line, putting that one last which is the only one
of its kind, and that one first which is connected with the last in
the question.[49] Put the third quantity in the middle. In the first
question the quantities will be placed thus:
22 yds. 156 yds. 17_s._ 4_d._
In the second question they will be placed thus:
20_s._ £20 15_s._ 25½ francs.
[49] This generally comes in the same member of the sentence. In some
cases the ingenuity of the student must be employed in detecting it.
The reasoning of (238) is the best guide. The following may be very
often applied. If it be evident that the answer must be less than the
given quantity of its kind, multiply that given quantity by the less of
the other two; if greater, by the greater. Thus, in the first question,
156 yards must cost more than 22; multiply, therefore, by 156.
Reduce the first and second quantities, if necessary, to quantities of
the same denomination. Thus, in the second question, £20 15_s._ must
be reduced to shillings (219). The third quantity may also be reduced
to any other denomination, if convenient; or the first and third may
be multiplied by any quantity we please, as was done in the second
question; and, on looking at the answer in (238), and at (108), it
will be seen that no change is made by that multiplication. Multiply
the second and third quantities together, and divide by the first. The
result is a quantity of the same sort as the third in the line, and is
the answer required. Thus, to the first question the answer is (238)
208 × 156 17_s._ 4_d_. × 156
----------pence, or, which is the same thing, -------------------.
22 22
240. The whole process in the first question is as follows:[50]
yds. yds. _s._ _d._
22 : 156 ∷ 17 . 4
12
---
208 pence.
156
----
1248
1040
208
-----
22)32448(1474¾_d._ and ¹⁴/₂₂, or ⁷/₁₁ of a farthing,
22 or (219) £6 . 2 . 10¾-⁷/₁₁.
---
104
88
----
164
154
----
108
88
--
20
(228) 4
--
80
66
--
14
[50] It is usual to place points, in the manner here shewn, between the
quantities. Those who have read Section VIII. will see that the Rule
of Three is no more than the process for finding the fourth term of a
proportion from the other three.
The question might have been solved without reducing 17_s._ 4_d._ to
pence, thus:
yds. yds. _s._ _d._
22 : 156 ∷ 17 . 4
156 (227)
----------
22)£135 . 4 . 0(£6 . 2 . 10¾-⁷/₁₁ (228)
132
---
3 × 20 + 4 = 64
44
--
20 × 12 = 240
220
---
20 × 4 = 80
66
--
14
The student must learn by practice which is the most convenient method
for any particular case, as no rule can be given.
241. It may happen that the three given quantities are all of one
denomination; nevertheless it will be found that two of them are of
one, and the third of another sort. For example: What must an income
of £400 pay towards an income-tax of 4_s._ 6_d._ in the pound? Here
the three given quantities are, £400, 4_s._ 6_d._, and £1, which are
all of the same species, viz. money. Nevertheless, the first and third
are income; the second is a tax, and the answer is also a tax; and
therefore, by (152), the quantities must be placed thus:
£1 : £400 ∷ 4_s._ 6_d._
242. The following exercises either depend directly upon this rule,
or can be shewn to do so by a little consideration. There are many
questions of the sort, which will require some exercise of ingenuity
before the method of applying the rule can be found.
EXERCISES.
If 15 cwt. 2 qrs. cost £198. 15. 4, what does 1 qr. 22 lbs. cost?
_Answer_, £5 . 14 . 5 ¾ ¹⁸⁵/₂₁₇.
If a horse go 14 m. 3 fur. 27 yds. in 3ʰ 26ᵐ 12ˢ, how long will he be
in going 23 miles?
_Answer_, 5ʰ 29ᵐ 34ˢ(²⁴⁶²/₂₅₃₂₇).
Two persons, A and B, are bankrupts, and owe exactly the same sum; A
can pay 15_s._ 4½_d._ in the pound, and B only 7_s._ (6¾)_d._ At the
same time A has in his possession £1304. 17 more than B; what do the
debts of each amount to?
_Answer_, £3340 . 8 . 3 ¾ ⁹/₂₅.
For every (12½) acres which one country contains, a second contains
(56¼). The second country contains 17,300 square miles. How much does
the first contain? Again, for every 3 people in the first, there are 5
in the second; and there are in the first 27 people on every 20 acres.
How many are there in each country?--_Answer_, The number of square
miles in the first is 3844⁴/₉, and its population 3,321,600; and the
population of the second is 5,536,000.
If (42½) yds. of cloth, 18 in. wide, cost £59. 14. 2, how much will
(118¼) yds. cost, if the width be 1 yd.?
_Answer_, £332. 5. (2⁴/₁₇).
If £9. 3. 6 last six weeks, how long will £100 last?
_Answer_, (65¹⁴⁵/₃₆₇) weeks.
How much sugar, worth (9¾d). a pound, must be given for 2 cwt. of tea,
worth 10_d._ an ounce?
_Answer_, 32 cwt. 3 qrs. 7 lbs. ³⁵/₃₉.
243. Suppose the following question asked: How long will it take 15 men
to do that which 45 men can finish in 10 days? It is evident that one
man would take 45 × 10, or 450 days, to do the same thing, and that 15
men would do it in one-fifteenth part of the time which it employs one
man, that is, in (450 ÷ 15) or 30 days. By this and similar reasoning
the following questions can be solved.
EXERCISES.
If 15 oxen eat an acre of grass in 12 days, how long will it take 26
oxen to eat 14 acres? _Answer_, (96¹²/₁₃) days.
If 22 masons build a wall 5 feet high in 6 days, how long will it take
43 masons to build 10 feet? _Answer_, (6⁶/₄₃) days.
244. The questions in the preceding article form part of a more general
class of questions, whose solution is called the DOUBLE RULE OF THREE,
but which might, with more correctness, be called the Rule of _Five_,
since five quantities are given, and a sixth is to be found. The
following is an example: If 5 men can make 30 yards of cloth in 3 days,
how long will it take 4 men to make 68 yards? The first thing to be
done is to find out, from the first part of the question, the time it
will take one man to make one yard. Now, since one man, in 3 days, will
do the fifth part of what 5 men can do, he will in 3 days make ³⁰/₅,
or 6 yards. He will, therefore, make one yard in ³/₆6 or (3 × 5)/30 of
a day. From this we are to find how long it will take 4 men to make 68
yards. Since one man makes a yard in
3 × 5 3 × 5
----- of a day, he will make 68 yards in ----- × 68 days,
30 30
3 × 5 × 68
or (116) in ---------- days; and 4 men will do this in one-fourth
30
3 × 5 × 68
of the time, that is (123), in ---------- days, or in 8½ days.
30 × 4
Again, suppose the question to be: If 5 men can make 30 yards in 3
days, how much can 6 men do in 12 days? Here we must first find the
quantity one man can do in one day, which appears, on reasoning similar
to that in the last example, to be 30/(3 × 5) yards. Hence, 6 men, in
one day, will make
6 × 30 12 × 6 × 30
------ yards, and in 12 days will make ----------- or 144 yards.
5 × 3 5 × 3
From these examples the following rule may be drawn. Write the given
quantities in two lines, keeping quantities of the same sort under one
another, and those which are connected with each other, in the same
line. In the two examples above given, the quantities must be written
thus:
[Illustration]
SECOND EXAMPLE.
[Illustration]
Draw a curve through the middle of each line, and the extremities of
the other. There will be three quantities on one curve and two on the
other. Divide the product of the three by the product of the two, and
the quotient is the answer to the question.
If necessary, the quantities in each line must be reduced to more
simple denominations (219), as was done in the common Rule of Three
(238).
EXERCISES.
If 6 horses can, in 2 days, plough 17 acres, how many acres will 93
horses plough in 4½ days?
_Answer_, 592⅞.
If 20 men, in 3¼ days, can dig 7 rectangular fields, the sides of each
of which are 40 and 50 yards, how long will 37 men be in digging 53
fields, the sides of each of which are 90 and 125½ yards?
2451
_Answer_, 75----- days.
20720
If the carriage of 60 cwt. through 20 miles cost £14 10_s._, what
weight ought to be carried 30 miles for £5. 8. 9?
_Answer_, 15 cwt.
If £100 gain £5 in a year, how much will £850 gain in 3 years and 8
months?
_Answer_, £155. 16. 8.
SECTION III.
INTEREST, ETC.
245. In the questions contained in this Section, almost the only
process which will be employed is the taking a fractional part of a
sum of money, which has been done before in several cases. Suppose it
required to take 7 parts out of 40 from £16, that is, to divide £16
into 40 equal parts, and take 7 of them. Each of these parts is
16 16 16 × 7
£----, and 7 of them make ---- × 7, or ------ pounds (116).
40 40 40
The process may be written as below:
£16
7
-----
40)112(£2 . 16_s._
80
--
32
20
---
640
40
---
240
240
---
0
Suppose it required to take 13 parts out of a hundred from £56. 13. 7½.
56 . 13 . 7½
13
----------------
100) 736 . 17 . 1½ ( £7 . 7 . 4 ¼ ¹/₄₁
700
---
36 × 20 + 17 = 737
700
---
37 × 12 + 1 = 445
400
---
45 × 4 × 2 = 182
100
---
82
Let it be required to take 2½ parts out of a hundred from £3 12_s._ The
result, by the same rule is
£3 12_s._ × 2½ 5
-------------------, or 123 £3 12_s._ × ---;
100 200
so that taking 2½ out of a hundred is the same as taking 5 parts out of
200.
EXERCISES.
Take 7⅓ parts out of 53 from £1 10_s._
129
_Answer_, 4_s._ 1---_d._
159
Take 5 parts out of 100 from £107 13_s._ 4¾_d._
_Answer_, £5. 7. 8 and ³/₂₀ of a farthing.
£56 3_s._ 2_d._ is equally divided among 32 persons. How much does the
share of 23 of them exceed that of the rest?
_Answer_, £24. 11. 4½ ½.
246. It is usual, in mercantile business, to mention the fraction which
one sum is of another, by saying how many parts out of a hundred must
be taken from the second in order to make the first. Thus, instead of
saying that £16 12_s._ is the half of £33 4_s._, it is said that the
first is 50 per cent of the second. Thus, £5 is 2½ per cent of £200;
because, if £200 be divided into 100 parts, 2½ of those parts are £5.
Also, £13 is 150 per cent of £8. 13. 4, since the first is the second
and half the second. Suppose it asked, How much per cent is 23 parts
out of 56 of any sum? The question amounts to this: If he who has £56
gets £100 for them, how much will he who has 23 receive? This, by 238,
is 23 × ¹⁰⁰/₅₆ or ²³⁰⁰/₅₆ or 41¹/₁₄. Hence, 23 out of 56 is 41¹/₁₄ per
cent.
Similarly 16 parts out of 18 is 16 × ¹⁰⁰/₁₈, or 88⁸/₉ per cent, and 2
parts out of 5 is 2 × ¹⁰⁰/₅, or 40 per cent.
From which the method of reducing other fractions to the rate per cent
is evident.
Suppose it asked, How much per cent is £6. 12. 2 of £12. 3? Since the
first contains 1586_d._, and the second 2916_d._, the first is 1586
out of 2916 parts of the second; that is, by the last rule, it is
¹⁵⁸⁶⁰⁰/₂₉₁₆, or 54¹¹³⁶/₂₉₁₆, or £54. 7. 9½ per cent, very nearly. The
more expeditious way of doing this is to reduce the shillings, &c.
to decimals of a pound. Three decimal places will give the rate per
cent to the nearest shilling, which is near enough for all practical
purposes. For instance, in the last example, which is to find how much
£6·608 is of £12·15, 6·608 × 100 is 660·8, which divided by 12·15 gives
£54·38, or £54. 7. Greater correctness may be had, if necessary, as in
the Appendix.
EXERCISES.
How much per cent is 198¼ out of 233 parts?--_Ans._ £85. 1. 8¾.
Goods which are bought for £193. 12, are sold for £216. 13. 4; how much
per cent has been gained by them?
_Answer_, A little less than £11. 18. 6.
A sells goods for B to the amount of £230. 12, and is allowed a
commission[51] of 3 per cent; what does that amount to?
_Answer_, £6 . 18. 4¼ ⁷/₂₅.
[51] Commission is what is allowed by one merchant to another for
buying or selling goods for him, and is usually a per-centage on the
whole sum employed. Brokerage is an allowance similar to commission,
under a different name, principally used in the buying and selling of
stock in the funds.
Insurance is a per-centage paid to those who engage to make good to the
payers any loss they may sustain by accidents from fire, or storms,
according to the agreement, up to a certain amount which is named,
and is a per-centage upon this amount. Tare, tret, and cloff, are
allowances made in selling goods by wholesale, for the weight of the
boxes or barrels which contain them, waste, &c.; and are usually either
the price of a certain number of pounds of the goods for each box or
barrel, or a certain allowance on each cwt.
A stockbroker buys £1700 stock, brokerage being at £⅛ per cent; what
does he receive?--_Answer_, £2. 2. 6.
A ship whose value is £15,423 is insured at 19⅔ per cent; what does the
insurance amount to?--_Answer_, £3033. 3. 9½ ²/₅.
247. In reckoning how much a bankrupt is able to pay his creditors, as
also to how much a tax or rate amounts, it is usual to find how many
shillings in the pound is paid. Thus, if a person who owes £100 can
only pay £50, he is said to pay 10_s._ in the pound. The rule is easily
derived from the same reasoning as in 246. For example, £50 out of £82
is
50 50×20
£---- out of £1, or ----- shillings,
82 82
or 12_s._ 2½ ¹⁵/₄₁ in the pound.
248. INTEREST is money paid for the use of other money, and is always
a per-centage upon the sum lent. It may be paid either yearly,
half-yearly, or quarterly; but when it is said that £100 is lent at 4
per cent, it must be understood to mean 4 per cent per annum; that is,
that 4 pounds are paid every year for the use of £100.
The sum lent is called the _principal_, and the interest upon it is
of two kinds. If the borrower pay the interest as soon as, from the
agreement, it becomes due, it is evident that he has to pay the same
sum every year; and that the whole of the interest which he has to pay
in any number of years is one year’s interest multiplied by the number
of years. But if he do not pay the interest at once, but keeps it in
his hands until he returns the principal, he will then have more of
his creditor’s money in his hands every year, and if it were so agreed
will have to pay interest upon each year’s interest for the time during
which he keeps it after it becomes due. In the first case, the interest
is called _simple_, and in the second _compound_. The interest and
principal together are called the _amount_.
249. What is the simple interest of £1049. 16. 6 for 6 years and
one-third, at 4½ per cent? This interest must be 6⅓ times the interest
of the same sum for one year, which (245) is found by multiplying the
sum by 4½, and dividing by 100. The process is as follows:
(230) (_a_) |£1049 . 16 . 6
+--------------
_a_ × 4 | 4199 . 6 . 0
_a_ × ½ | 524 . 18 . 3
+--------------
(82) 100) 47,24 . 4 . 3(£47 . 4 . 10¹¹/₁₀₀
20
----
(228) 4,84[52]
12
------
10,11[53]
(_b_) £47 . 4 . 10¹¹/₁₀₀ Int. for one yr.
+------------------
_b_ × 6 | 283 . 9 . 0⁶⁶/₁₀₀
_b_ × ⅓ | 15 . 14 . 11³⁷/₁₀₀
+---------------------
£299 . 4 . 0³/₁₀₀ Int. for 6⅓ yrs.
[52] Here the 4_s._ from the dividend is taken in.
[53] Here the 3_d._ from the dividend is taken in.
EXERCISES.
What is the interest of £105. 6. 2 for 19 years and 7 weeks at 3 per
cent?
_Answer_, £60. 9, very nearly.
What is the difference between the interest of £50. 19 for 7 years at 3
per cent, and for 8 years at 2½ per cent? _Answer_, 10_s._ (2½)_d._
What is the interest of £157. 17. 6 for one year at 5 per cent?
_Answer_, £7. 17. 10½.
Shew that the interest of any sum for 9 years at 4 per cent is the same
as that of the same sum for 4 years at 9 per cent.
250. In order to find the interest of any sum at compound interest,
it is necessary to find the amount of the principal and interest at
the end of every year; because in this case (248) it is the amount of
both principal and interest at the end of the first year, upon which
interest accumulates during the second year. Suppose, for example, it
is required to find the interest, for 3 years, on £100, at 5 per cent,
compound interest. The following is the process:
£100 First principal.
5 First year’s interest.
---
105 Amount at the end of the first year.
(249) 5 . 5 Interest for the second year on £105.
--------
110 . 5 Amount at the end of two years.
5 . 10 . 3 Interest due for the third year.
------------
115 . 15 . 3 Amount at the end of three years.
100 . 0 . 0 First principal.
------------
15 . 15 . 3 Interest gained in the three years.
When the number of years is great, and the sum considerable, this
process is very troublesome; on which account tables[54] are
constructed to shew the amount of one pound, for different numbers of
years, at different rates of interest. To make use of these tables in
the present example, look into the column headed “5 per cent;” and
opposite to the number 3, in the column headed “Number of years,” is
found 1·157625; meaning that £1 will become £1·157625 in 3 years. Now,
£100 must become 100 times as great; and 1·157625 × 100 is 115·7625
(141); but (221) £·7625 is 15_s._ 3_d._; therefore the whole amount of
£100 is £115. 15. 3, as before.
[54] Sufficient tables for all common purposes are contained in
the article on Interest in the Penny Cyclopædia; and ample ones in
the Treatise on Annuities and Reversions, in the Library of Useful
Knowledge.
251. Suppose that a sum of money has lain at simple interest 4 years,
at 5 per cent, and has, with its interest, amounted to £350; it is
required to find what the sum was at first. Whatever the sum was, if we
suppose it divided into 100 parts, 5 of those parts were added every
year for 4 years, as interest; that is, 20 of those parts have been
added to the first sum to make £350. If, therefore, £350 be divided
into 120 parts, 100 of those parts are the principal which we want to
find, and 20 parts are interest upon it; that is, the principal is
£(350 × 100)/150, or £291. 13. 4.
252. Suppose that A was engaged to pay B £350 at the end of four years
from this time, and that it is agreed between them that the debt shall
be paid immediately; suppose, also, that money can be employed at 5 per
cent, simple interest; it is plain that A ought not to pay the whole
sum, £350, because, if he did, he would lose 4 years’ interest of the
money, and B would gain it. It is fair, therefore, that he should only
pay to B as much as will, _with interest_, amount in four years to
£350, that is (251), £291. 13. 4. Therefore, £58. 6. 8 must be struck
off the debt in consideration of its being paid before the time. This
is called DISCOUNT;[55] and £291. 13. 4 is called the _present value_
of £350 due four years hence, discount being at 5 per cent. The rule
for finding the present value of a sum of money (251) is: Multiply the
sum by 100, and divide the product by 100 increased by the product of
the rate per cent and number of years. If the time that the debt has
yet to run be expressed in years and months, or months only, the months
must be reduced to the equivalent fraction of a year.
[55] This rule is obsolete in business. When a bill, for instance, of
£100 having a year to run, is _discounted_ (as people now say) at 5 per
cent, this means that 5 per cent of £100, or £5, is struck off.
EXERCISES.
What is the discount on a bill of £138. 14. 4, due 2 years hence,
discount being at 4½ per cent?
_Answer_, £11. 9. 1.
What is the present value of £1031. 17, due 6 months hence, interest
being at 3 per cent?
_Answer_, £1016. 12.
253. If we multiply by _a_ + _b_, or by _a_-_b_, when we should
multiply by _a_, the result is wrong by the fraction
_b_ _b_
--- + _b_, or ---------,
_a_ _a_ - _b_
of itself: being too great in the first case, and too small in the
second. Again, if we divide by _a_ + _b_, where we should have divided
by _a_, the result is too small by the fraction _b_/_a_ of itself;
while, if we divide by _a_-_b_ instead of _a_, the result is too great
by the same fraction of itself. Thus, if we divide by 20 instead of
17, the result is ³/₁₇ of itself too small; and if we divide by 360
instead of 365, the result is too great by ⁵/₃₆₅, or ¹/₇₃ of itself.
If, then, we wish to find the interest of a sum of money for a portion
of a year, and have not the assistance of tables, it will be found
convenient to suppose the year to contain only 360 days, in which case
its 73d part (the 72d part will generally do) must be subtracted from
the result, to make the alteration of 360 into 365. The number 360 has
so large a number of divisors, that the rule of Practice (230) may
always be readily applied. Thus, it is required to find the portion
which belongs to 274 days, the yearly interest being £18. 9. 10, or
18·491.
274 18·491
------
180 is ½ of 360 9·246
---
94
90 is ½ of 180 4·623
--
4 is ¹/₉₀ of 360 ·205
------
9)14·074
------
8)1·564
-----
·196
13·878 = £13 . 17 . 7 _Answer._
But if the nearest farthing be wanted, the best way is to take 2-tenths
of the number of days as a multiplier, and 73 as a divisor; since _m_ ÷
365 is 2_m_ ÷ 730, or (²/₁₀)_m_ ÷ 73. Thus, in the preceding instance,
we multiply by 54·8 and divide by 73; and 54·8 × 18·491 = 1013·3068,
which divided by 73 gives 13·881, very nearly agreeing with the former,
and giving £13. 17. 7½, which is certainly within a farthing of the
truth.
254. Suppose it required to divide £100 among three persons in such a
way that their shares may be as 6, 5, and 9; that is, so that for every
£6 which the first has, the second may have £5, and the third £9. It is
plain that if we divide the £100 into 6 + 5 + 9, or 20 parts, the first
must have 6 of those parts, the second 5, and the third 9. Therefore
(245) their shares are respectively,
100 × 6 100 × 5 100 × 9
£-------, £------- and £-------, or £30, £25, and £45.
20 20 20
EXERCISES.
Divide £394. 12 among four persons, so that their shares may be as 1,
6, 7, and 18.--_Answer_, £12. 6. 7½; £73. 19. 9; £86. 6. 4½; £221. 19.
3.
Divide £20 among 6 persons, so that the share of each may be as much
as those of all who come before put together.--_Answer_, The first two
have 12_s._ 6_d._; the third £1. 5; the fourth £2. 10; the fifth £5;
and the sixth £10.
255. When two or more persons employ their money together, and gain
or lose a certain sum, it is evidently not fair that the gain or loss
should be equally divided among them all, unless each contributed the
same sum. Suppose, for example, A contributes twice as much as B, and
they gain £15, A ought to gain twice as much as B; that is, if the
whole gain be divided into 3 parts, A ought to have two of them and B
one, or A should gain £10 and B £5. Suppose that A, B, and C engage in
an adventure, in which A embarks £250, B £130, and C £45. They gain
£1000. How much of it ought each to have? Each one ought to gain as
much for £1 as the others. Now, since there are 250 + 130 + 45, or 425
pounds embarked, which gain £1000, for each pound there is a gain of
£¹⁰⁰⁰/₄₂₄. Therefore A should gain 1000 × ²⁵⁰/₄₂₅ pounds, B should gain
1000 × ¹³⁰/₄₂₅ pounds, and C 1000 × ⁴⁵/₄₂₅ pounds. On these principles,
by the process in (245), the following questions may be answered.
A ship is to be insured, in which A has ventured £1928, and B £4963.
The expense of insurance is £474. 10. 2. How much ought each to pay of
it?
_Answer_, A must pay £132. 15. (2½).
A loss of £149 is to be made good by three persons, A, B, and C. Had
there been a gain, A would have gained 4 times as much as B, and C as
much as A and B together. How much of the loss must each bear?
_Answer_, A pays £59. 12, B £14. 18, and C £74. 10.
256. It may happen that several individuals employ several sums of
money together for different times. In such a case, unless there be
a special agreement to the contrary, it is right that the more time
a sum is employed, the more profit should be made upon it. If, for
example, A and B employ the same sum for the same purpose, but A’s
money is employed twice as long as B’s, A ought to gain twice as much
as B. The principle is, that one pound employed for one month, or one
year, ought to give the same return to each. Suppose, for example, that
A employs £3 for 6 months, B £4 for 7 months, and C £12 for 2 months,
and the gain is £100; how much ought each to have of it? Now, since
A employs £3 for six months, he must gain 6 times as much as if he
employed it one month only; that is, as much as if he employed £6 × 3,
or £18, for one month; also, B gains as much as if he had employed £4 ×
7 for one month; and C as if he had employed £12 × 2 for one month. If,
then, we divide £100 into 6 × 3 + 4 × 7 + 12 × 2, or 70 parts, A must
have 6 × 3, or 18, B must have 4 × 7, or 28, and C 12 × 2, or 24 of
those parts. The shares of the three are, therefore,
6 × 3 × 100 4 × 7 × 100
£----------------------, £----------------------,
6 × 3 + 4 × 7 + 12 × 2 6 × 3 + 4 × 7 + 12 × 2
12 × 2 × 100
and £----------------------.
6 × 3 + 4 × 7 + 12 × 2
EXERCISES.
A, B, and C embark in an undertaking; A placing £3. 6 for 2 years, B
£100 for 1 year, and C £12 for 1½ years. They gain £4276. 7 How much
must each receive of the gain?
_Answer_, A £226. 10. 4; B £3432. 1. 3; C £617. 15. 5.
A, B, and C rent a house together for 2 years, at £150 per annum. A
remains in it the whole time, B 16 months, and C 4½ months, during the
occupancy of B. How much must each pay of the rent?[56]
_Answer_, A should pay £190. 12. 6; B £90. 12. 6; C £18. 15.
[56] This question does not at first appear to fall under the rule. A
little thought will serve to shew that what probably will be the first
idea of the proper method of solution is erroneous.
257. These are the principal rules employed in the application of
arithmetic to commerce. There are others, which, as no one who
understands the principles here laid down can fail to see, are
virtually contained in those which have been given. Such is what is
commonly called the Rule of Exchange, for such questions as the
following: If 20 shillings be worth 25½ francs, in France, what is £160
worth? This may evidently be done by the Rule of Three. The rules here
given are those which are most useful in common life; and the student
who understands them need not fear that any ordinary question will be
above his reach. But no student must imagine that from this or any
other book of arithmetic he will learn precisely the modes of operation
which are best adapted to the wants of the particular kind of business
in which his future life may be passed. There is no such thing as a set
of rules which are at once most convenient for a butcher and a banker’s
clerk, a grocer and an actuary, a farmer and a bill-broker; but a
person with a good knowledge of the _principles_ laid down in this
work, will be able to examine and meet his own future wants, or, at
worst, to catch with readiness the manner in which those who have gone
before him have done so for themselves.
APPENDIX TO THE FIFTH EDITION OF
DE MORGAN’S ELEMENTS OF ARITHMETIC.
I. ON THE MODE OF COMPUTING.
The rules in the preceding work are given in the usual form, and the
examples are worked in the usual manner. But if the student really wish
to become a ready computer, he should strictly follow the methods laid
down in this Appendix; and he may depend upon it that he will thereby
save himself trouble in the end, as well as acquire habits of quick and
accurate calculation.
I. In numeration learn to connect each primary decimal number, 10,
100, 1000, &c. not with the place in which the unit falls, but with
the number of ciphers following. Call ten a _one-cipher_ number, a
hundred a _two-cipher_ number, a million a _six-cipher_ number, and so
on. If _five_ figures be cut off from a number, those that are left
are hundred-thousands; for 100,000 is a _five-cipher_ number. Learn
to connect tens, hundreds, thousands, tens of thousands, hundreds of
thousands, millions, &c. with 1, 2, 3, 4, 5, 6, &c. in the mind. What
is a _seventeen-cipher_ number? For every 6 in seventeen say _million_,
for the remaining 5 say _hundred-thousand_: the answer is a hundred
thousand millions of millions. If twelve places be cut off from the
right of a number, what does the remaining number stand for?--_Answer_,
As many millions of millions as there are units in it when standing by
itself.
II. After learning to count forwards and backwards with rapidity, as
in 1, 2, 3, 4, &c. or 30, 29, 28, 27, &c., learn to count forwards or
backwards by twos, threes, &c. up to nines at least, beginning from
any number. Thus, beginning from four and proceeding by sevens, we
have 4, 11, 18, 25, 32, &c., along which series you must learn to go
as easily as along the series 1, 2, 3, 4, &c.; that is, as quick as
you can pronounce the words. The act of addition must be made in the
mind without assistance: you must not permit yourself to say, 4 and 7
are 11, 11 and 7 are 18, &c.; but only 4, 11, 18, &c. And it would be
desirable, though not so necessary, that you should go back as readily
as forward; by sevens for instance, from sixty, as in 60, 53, 46, 39,
&c.
III. Seeing a number and another both of one figure, learn to catch
instantly the number you must add to the smaller to get the greater.
Seeing 3 and 8, learn by practice to think of 5 without the necessity
of saying 3 _from_ 8 _and there remains_ 5. And if the second number be
the less, as 8 and 3, learn also by practice how to pass _up_ from 8 to
the next number which ends with 3 (or 13), and to catch the necessary
augmentation, _five_, without the necessity of formally undertaking in
words to subtract 8 from 13. Take rows of numbers, such as
4 2 6 0 5 0 1 8 6 4
and practise this rule upon every figure and the next, not permitting
yourself in this simple case ever to name the higher one. Thus, say 4
and 8 (4 first, 2 second, 4 from the next number that ends with 2, or
12, leaves 8), 2 and 4, 6 and 4, 0 and 5, 5 and 5, 0 and 1, 1 and 7, 8
and 8, 6 and 8.
IV. Study the same exercise as the last one with two figures and one.
Thus, seeing 27 and 6, pass from 27 up to the next number that ends
with 6 (or 36), catch the 9 through which you have to pass, and allow
yourself to repeat as much as “27 and 9 are 36.” Thus, the row of
figures 17729638109 will give the following practice: 17 and 0 are 17;
77 and 5 are 82; 72 and 7 are 79; 29 and 7 are 36; 96 and 7 are 103; 63
and 5 are 68; 38 and 3 are 41; 81 and 9 are 90; 10 and 9 are 19.
V. In a number of two figures, practise writing down the units at the
moment that you are keeping the attention fixed upon the tens. In the
preceding exercise, for instance, write down the results, repeating the
tens with emphasis at the instant of writing down the units.
VI. Learn the multiplication table so well as to name the product the
instant the factors are seen; that is, until 8 and 7, or 7 and 8,
suggest 56 at once, without the necessity of saying “7 times 8 are 56.”
Thus looking along a row of numbers, as 39706548, learn to name the
products of every successive pair of digits as fast as you can repeat
them, namely, 27, 63, 0, 0, 30, 20, 32.
VII. Having thoroughly mastered the last exercise, learn further, on
seeing three numbers, to augment the product of the first and second
by the third without any repetition of words. Practise until 3, 8, 4,
for instance, suggest 3 times 8 and 4, or 28, without the necessity of
saying “3 times 8 are 24, and 4 is 28.” Thus, 179236408 will suggest
the following practice, 16, 65, 21, 12, 22, 24, 8.
VIII. Now, carry the last still further, as follows: Seeing four
figures, as 2, 7, 6, 9, catch up the product of the first and second,
increased by the third, as in the last, without a helping word; name
the result, and add the next figure, name the whole result, laying
emphasis upon the tens. Thus, 2, 7, 6, 9, must immediately suggest “20
and 9 are 29.” The row of figures 773698974 will give the instances 52
and 6 are 58; 27 and 9 are 36; 27 and 8 are 35; 62 and 9 are 71; 81 and
7 are 88; 79 and 4 are 83.
IX. Having four numbers, as 2, 4, 7, 9, vary the last exercise as
follows: Catch the product of the first and second, increased by the
third; but instead of adding the fourth, go up to the next number
that ends with the fourth, as in exercise IV. Thus, 2, 4, 7, 9, are
to suggest “15 and 4 are 19.” And the row of figures 1723968929 will
afford the instances 9 and 4 are 13; 17 and 2 are 19; 15 and 1 are 16;
33 and 5 are 38; 62 and 7 are 69; 57 and 5 are 62; 74 and 5 are 79.
X. Learn to find rapidly the number of times a digit is contained
in given units and tens, with the remainder. Thus, seeing 8 and 53,
arrive at and repeat “6 and 5 over.” Common short division is the best
practice. Thus, in dividing 236410792 by 7,
7)236410792
---------
33772970, remainder 2.
All that is repeated should be 3 and 2; 3 and 5; 7 and 5; 7 and 2; 2
and 6; 9 and 4; 7 and 0; 0 and 2.
In performing the several rules, proceed as follows:
ADDITION. Not one word more than repeating the numbers written in the
following process: the accented figure is the one to be written down;
the doubly accented figure is carried (and don’t _say_ “carry 3,” but
do it).
47963
1598
26316
54792
819
6686
------
138174
6, 15, 17, 23, 31, 3″ 4′; 11, 12, 21, 22, 31, 3″7′; 9, 17, 24, 27, 32,
4″1′; 10, 14, 20, 21, 2″8′; 7, 9, 1′3′.
In verifying additions, instead of the usual way of omitting one line,
adding without it, and then adding the line omitted, verify each column
by adding it both upwards and downwards.
SUBTRACTION. The following process is enough. The carriages, being
always of _one_, need not be mentioned.
From 79436258190
Take 58645962738
-----------
20790295452
8 and 2′, 4 and 5′, 7 and 4′, 3 and 5′, 6 and 9′, 10 and 2′, 6 and 0′,
4 and 9′, 7 and 7′, 9 and 0′, 5 and 2′. It is useless to stop and say,
8 and 2 make 10; for as soon as the 2 is obtained, there is no occasion
to remember what it came from.
MULTIPLICATION. The following, put into words, is all that need be
repeated in the multiplying part; the addition is then done as usual.
The unaccented figures are carried.
670383
9876
-------
4022298 18′, 49′, 22′, 2′, 42′, 4′0′,
4692681 21′, 58′, 26′, 2′, 49′, 4′6′,
5363064 24′, 66′, 30′, 3′, 56′, 5′3′,
6033447 27′, 74′, 34′, 3′, 63′, 6′0′.
----------
6620702508
Verify each line of the multiplication and the final result by casting
out the nines. (_Appendix_ II. p. 166.)
It would be almost as easy, for a person who has well practised the 8th
exercise, to add each line to the one before in the process, thus:
670383
9876
-------
4022298
50949108
587255508
6620702508
8; 21 and 9 are 30′; 59 and 2 are 61′; 27 and 2 are 29; 2 and 2 are 4′;
49 and 0 are 49′; 46 and 4 are 5′0′.
On the right is all the process of forming the second line, which
completes the multiplication by 76, as the third line completes that by
876, and the fourth line that by 9876.
DIVISION. Make each multiplication and the following subtraction in one
step, by help of the process in the 9th exercise, as follows:
27693)441972809662(15959730
165042
265778
165410
269459
202226
83756
6772
The number of words by which 26577 is obtained from 165402 (the
multiplier being 5) is as follows: 15 and 7′ are 2″2; 47 and 7′ are
5″4; 35 and 5′ are 4″0; 39 and 6′ are 4″5; 14 and 2′ are 16.
The processes for extracting the square root, and for the solution of
equations (_Appendix_ XI.), should be abbreviated in the same manner as
the division.[57]
[57] The teacher will find further remarks on this subject in the
_Companion to the Almanac_ for 1844, and in the _Supplement to the
Penny Cyclopædia_, article _Computation_.
APPENDIX II.
ON VERIFICATION BY CASTING OUT NINES AND ELEVENS.
The process of _casting out the nines_, as it is called, is one which
the young computer should learn and practise, as a check upon his
computations. It is not a complete check, since if one figure were
made too small, and another as much too great, it would not detect
this double error; but as it is very unlikely that such a double error
should take place, the check furnishes a strong presumption of accuracy.
The proposition upon which this method depends is the following: If _a,
b, c, d_ be four numbers, such that
_a_ = _bc_ + _d_,
and if _m_ be any other number whatsoever, and if _a, b, c, d_,
severally divided by _m_, give the remainders _p, q, r, s_, then
_p_ and _qr_ + _s_
give the same remainder when divided by _m_ (and perhaps are themselves
equal).
For instance, 334 = 17 × 19 + 11;
divide these four numbers by 7, the remainders are 5, 3, 5, and 4. And
5 and 5 × 3 + 4, or 5 and 19, both leave the remainder 5 when divided
by 7.
Any number, therefore, being used as a divisor, may be made a check
upon the correctness of an operation. To provide a check which may be
most fit for use, we must take a divisor the remainder to which is most
easily found. The most convenient divisors are 3, 9, and 11, of which 9
is far the most useful.
As to the numbers 3 and 9, the remainder is always the same as that
of the sum of the digits. For instance, required the remainder of
246120377 divided by 9. The sum of the digits is 2 + 4 + 6 + 1 + 2 + 0
+ 3 + 7 + 7, or 32, which gives the remainder 5. But the easiest way
of proceeding is by throwing out nines as fast as they arise in the
sum. Thus, repeat 2, 6 (2 + 4), 12 (6 + 6), say 3 (throwing out 9),
4, 6, 9 (throw this away), 7, 14, (or throwing out the 9) 5. This is
the remainder required, as would appear by dividing 246120377 by 9. A
proof may be given thus: It is obvious that each of the numbers, 1,
10, 100, 1000, &c. divided by 9, leaves a remainder 1, since they are
1, 9 + 1, 99 + 1, &c. Consequently, 2, 20, 200, &c. leave the remainder
2; 3, 30, 300, the remainder 3; and so on. If, then, we divide, say
1764 by 9 in parcels, 1000 will be one more than an exact number of
nines, 700 will be seven more, and 60 will be six more. So, then, from
1, 7, 6, 4, put together, and the nines taken out, comes the only
remainder which can come from 1764.
To apply this process to a multiplication: It is asserted, in page 32,
that
10004569 × 3163 = 31644451747.
In casting out the nines from the first, all that is necessary to
repeat is, one, five, ten, one, _seven_; in the second, three, four,
ten, one, _four_; in the third, three, four, ten, one, five, nine,
four, nine, eight, twelve, three, ten, _one_. The remainders then are,
7, 4, 1. Now, 7 × 4 is 28, which, casting out the nines, gives 1, the
same as the product.
Again, in page 43, it is asserted that
23796484 = 130000 × 183 + 6484.
Cast out the nines from 13000, 183, 6484, and we have 4, 3, and 4. Now,
4 × 3 + 4, with the nines cast out, gives 7; and so does 23796484.
To avoid having to remember the result of one side of the equation,
or to write it down, in order to confront it with the result of the
other side, proceed as follows: Having got the remainder of the more
complicated side, into which two or more numbers enter, subtract it
from 9, and carry the remainder into the simple side, in which there is
only one number. Then the remainder of that side ought to be 0. Thus,
having got 7 from the left-hand of the preceding, take 2, the rest
of 9, forget 7, and carry in 2 as a beginning to the left-hand side,
giving 2, 4, 7, 14, 5, 11, 2, 6, 14, 5, 9, 0.
Practice will enable the student to cast out nines with great rapidity.
This process of casting out the nines does not detect any errors
in which the remainder to 9 happens to be correct. If a process be
tedious, and some additional check be desirable, the method of casting
out _elevens_ may be followed after that of casting out the nines.
Observe that 10 + 1, 100-1, 1000 + 1, 10000-1, &c. are all divisible by
eleven. From this the following rule for the remainder of division by
11 may be deduced, and readily used by those who know the algebraical
process of subtraction. For those who have not got so far, it may be
doubted whether the rule can be made easier than the actual division by
11.
Subtract the first figure from the second, the result from the third,
the result from the fourth, and so on. The final result, or the rest
of 11 if the figure be negative, is the remainder required. Thus, to
divide 1642915 by 11, and find the remainder, we have 1 from 6, 5; 5
from 4,-1;-1 from 2, 3; 3 from 9, 6; 6 from 1,-5;-5 from 5, 10; and
10 is the remainder. But 164 gives-1, and 10 is the remainder; 164291
gives-5, and 6 is the remainder. With very little practice these
remainders may be read as rapidly as the number itself. Thus, for
127619833424 need only be repeated, 1, 6, 0, 1, 8, 0, 3, 0, 4,-2, 6,
and 6 is the remainder.
When a question has been tried both by nines and elevens, there can be
no error unless it be one which makes the result wrong by a number of
times 99 exactly.
APPENDIX III.
ON SCALES OF NOTATION.
We are so well accustomed to 10, 100, &c., as standing for ten, ten
tens, &c., that we are not apt to remember that there is no reason why
10 might not stand for five, 100 for five fives, &c., or for twelve,
twelve twelves, &c. Because we invent different columns of numbers, and
let units in the different columns stand for collections of the units
in the preceding columns, we are not therefore bound to allow of no
collections except in tens.
If 10 stood for 2, that is, if every column had its unit double of the
unit in the column on the right, what we now represent by 1, 2, 3,
4, 5, 6, &c., would be represented by 1, 10, 11, 100, 101, 110, 111,
1000, 1001, 1010, 1011, 1100, &c. This is the _binary_ scale. If we
take the _ternary_ scale, in which 10 stands for 3, we have 1, 2, 10,
11, 12, 20, 21, 22, 100, 101, 102, 110, &c. In the _quinary_ scale, in
which 10 is five, 234 stands for 2 twenty-fives, 3 fives, and 4, or
sixty-nine. If we take the _duodenary_ scale, in which 10 is twelve, we
must invent new symbols for ten and eleven, because 10 and 11 now stand
for twelve and thirteen; use the letters _t_ and _e_. Then 176 means 1
twelve-twelves, 7 twelves, and 6, or two hundred and thirty-four; and
1_te_ means two hundred and seventy-five.
The number which 10 stands for is called the _radix_ of the _scale of
notation_. To change a number from one scale into another, divide the
number, written as in the first scale, by the number which is to be the
radix of the new scale; repeat this division again and again, and the
remainders are the digits required. For example, what, in the quinary
scale, is that number which, in the decimal scale, is 17036?
5)17036
-----
5)3407 Remʳ. 1
----
5)681 2
---
5)136 1
---
5)27 1
--
5)5 2
-
5)1 0
-
0 1
_Answer_ 1021121
Quinary. Decimal.
_Verification_, 1000000 means 15625
20000 1250
1000 125
100 25
20 10
1 1
------ -----
1021121 17036
The reason of this rule is easy. Our process of division is nothing but
telling off 17036 into 3407 fives and 1 over; we then find 3407 fives
to be 681 fives of fives and 2 _fives_ over. Next we form 681 fives of
fives into 136 fives of fives of fives and 1 five of fives over; and so
on.
It is a useful exercise to multiply and divide numbers represented in
other scales of notation than the common or decimal one. The rules are
in all respects the same for all systems, _the number carried being
always the radix of the system_. Thus, in the quinary system we carry
fives instead of tens. I now give an example of multiplication and
division:
Quinary. Decimal.
42143 means 2798
1234 194
------ -----
324232 11192
232034 25182
134341 2798
42143
--------- ------
114332222 542812
Duodecimal. Decimal.
4_t_9)76_t_4_e_08(16687 705)22610744(32071
4_t_9 1460
----- 5074
2814 1394
2546 689
-----
28_te_
2546
------
3650
3320
-----
3308
2_t_33
------
495
Another way of turning a number from one scale into another is as
follows: Multiply the first digit by the _old_ radix _in the new
scale_, and add the next digit; multiply the result again by the old
radix in the new scale, and take in the next digit, and so on to the
end, always using the radix of the scale you want to leave, and the
notation of the scale you want to end in.
Thus, suppose it required to turn 16687 (duodecimal) into the decimal
scale, and 16432 (septenary) into the quaternary scale:
16687 16432
Duodecimals into Decimals. Septenaries into Quaternaries.
1 × 12 + 6 = 18 1 × 7 + 6 = 31
× 12 + 6 × 7 + 4
--- ----
222 1133
× 12 + 8 × 7 + 3
---- -----
2672 22130
× 12 + 7 × 7 + 2
------ -------
_Answer_ 32071 1021012
Owing to our division of a foot into 12 equal parts, the duodecimal
scale often becomes very convenient. Let the square foot be also
divided into 12 parts, each part is 12 square inches, and the 12th of
the 12th is one square inch. Suppose, now, that the two sides of an
oblong piece of ground are 176 feet 9 inches 7-12ths of an inch, and
65 feet 11 inches 5-12ths of an inch. Using the duodecimal scale, and
_duodecimal fractions_, these numbers are 128·97 and 55·_e_5. Their
product, the number of square feet required, is thus found:
128·97
55·_e_5
---------
617_ee_
116095
617_ee_
617_ee_
------------
68_e_8144_e_
_Answer_, 68_e_8·144_e_ (duod.) square feet, or 11660 square feet 16
square inches ⁴/₁₂ and ¹¹/₁₄₄ of a square inch.
It would, however, be exact enough to allow 2-hundredths of a foot
for every quarter of an inch, an additional hundredth for every 3
inches,[58] and 1-hundredth more if there be a 12th or 2-12ths above
the quarter of an inch. Thus, 9⁷/₁₂ inches should be ·76 + ·03 + ·01,
or ·80, and 11⁵/₁₂ would be ·95; and the preceding might then be found
decimally as 176·8 × 65·95 as 11659·96 square feet, near enough for
every practical purpose.
[58] And at discretion one hundredth more for a large fraction of three
inches.
APPENDIX IV.
ON THE DEFINITION OF FRACTIONS.
The definition of a fraction given in the text shews that ⁷/₉, for
instance, is the _ninth_ part of _seven_, which is shewn to be the
same thing as _seven-ninths_ of a unit. But there are various modes of
speech under which a fraction may be signified, all of which are more
or less in use.
1. In ⁷/₉ we have the 9th part of 7.
2. 7-9ths of a unit.
3. The fraction which 7 is of 9.
4. The times and parts of a time (in this case part of a time only)
which 7 contains 9.
5. The multiplier which turns _nines_ into _sevens_.
6. The _ratio_ of 7 to 9, or the _proportion_ of 7 to 9.
7. The multiplier which alters a number in the ratio of 9 to 7.
8. The 4th proportional to 9, 1, and 7.
The first two views are in the text. The third is deduced thus: If
we divide 9 into 9 equal parts, each is 1, and 7 of the parts are 7;
consequently the fraction which 7 is of 9 is ⁷/₉. The fourth view
follows immediately: For _a time_ is only a word used to express one
of the repetitions which take place in multiplication, and we allow
ourselves, by an easy extension of language, to speak of a portion
of a number as being that number taken a _part of a time_. The fifth
view is nothing more than a change of words: A number reduced to ⁷/₉
of its amount has every 9 converted into a 7, and any fraction of a
9 which may remain over into the corresponding fraction of 7. This
is completely proved when we prove the equation ⁷/₉ of _a_ = 7 times
_a_/9. The sixth, seventh, and eighth views are illustrated in the
chapter on proportion.
When the student comes to algebra, he will find that, in all the
applications of that science, fractions such as _a_/_b_ most frequently
require that _a_ and _b_ should be themselves supposed to be fractions.
It is, therefore, of importance that he should learn to accommodate his
views of a fraction to this more complicated case.
2½
Suppose we take -----.
4³/₅
We shall find that we have, in this case, a better idea of the views
from and after the third inclusive, than of the first and second, which
are certainly the most simple ways of conceiving ⁷/₉. We have no notion
of the (4³/₅)th part of 2½,
1 ( 3 )
nor of 2 ---(4---)ths
2 ( 5 )
of a unit; indeed, we coin a new species of adjective when we talk of
the (4³/₅)th part of anything. But we can readily imagine that 2½ is
some fraction of 4³/₅; that the first is _some_ part of a time the
second; that there must be _some_ multiplier which turns every 4³/₅ in
a number into 2½; and so on. Let us now see whether we can invent a
distinct mode of applying the first and second views to such a compound
fraction as the above.
We can easily imagine a fourth part of a length, and a fifth part,
meaning the lines of which 4 and 5 make up the length in question;
and there is also in existence a length of which four lengths and
two-fifths of a length make up the original length in question. For
instance, we might say that 6, 6, 2 is a division of 14 into 2⅓ equal
parts--2 equal parts, 6, 6, and a third of a part, 2. So we might agree
to say, that the (2⅓)th, or (2⅓)rd, or (2⅓)st (the reader may coin the
adjective as he pleases) part of 14 is 6. If we divide the line A B
into eleven equal parts in C, D, E, &c., we must then say that A C is
the 11th part,
[Illustration]
A D the (5½)th, A E the (3⅔)th, A F the (2¾)th, A G the (2⅕)th, A H
the (1⅚)th, A I the (1⁴/₇)th, A K the (1⅜)th, A L the (1²/₉)th, A M
the (1⅒)th, and A B itself the 1st part of A B. The reader may refuse
the language if he likes (though it is not so much in defiance of
etymology as talking of _multiplying_ by ½); but when A B is called 1,
he must either call A F 1/(2¾), or make one definition of one class
of fractions and another of another. Whatever abbreviations they may
choose, all persons will agree that _a_/_b_ is a direction to find such
a fraction as, repeated _b_ times, will give 1, and then to take that
fraction _a_ times.
So, to get 2½/4⅗, the simplest way is to divide the whole unit into 46
parts; 10 of these parts, repeated 4⅗ times, give the whole. The
[Illustration]
4⅗th is then ¹⁰/₄₆, and 2½ such parts is ²⁵/₄₆, or A C. The student
should try several examples of this mode of interpreting complex
fractions.
But what are we to say when the denominator itself is less than unity,
as in 3¼/⅖? Are we to have a (⅖)th part of a unit? and what is it?
Had there been a 5 in the denominator, we should have taken the part
of which 5 will make a unit. As there is ⅖ in the denominator, we must
take the part of which ⅖ will be a unit. That part is larger than a
unit; it is 2½ units; 2½ is that of which ⅖ is 1. The above fraction
then directs us to repeat 2½ units 3¼ times. By extending our word
‘multiplication’ to the taking of a part of a time, all multiplications
are also divisions, and all divisions multiplications, and all the
terms connected with either are subject to be applied to the results of
the other.
If 2⅓ yards cost 3½ shillings, how much does one yard cost? In such a
case as this, the student looks at a more simple question. If 5 yards
cost 10 shillings, he sees that each yard costs ¹⁰/₅, or 2 shillings,
and, concluding that the same process will give the true result when
the data are fractional, he forms 3½/2⅓, reduces it by rules to ³/₂
or 1½, and concludes that 1 yard costs 18 pence. The answer happens
to be correct; but he is not to suppose that this rule of copying for
fractions whatever is seen to be true of integers is one which requires
no demonstration. In the above question we want money which, repeated
2⅓ times, shall give 3½ shillings. If we divide the shilling into 14
equal parts, 6 of these parts repeated 2⅓ times give the shilling. To
get 3½ times as much by the same repetition, we must take 3½ of these 6
parts at each step, or 21 parts. Hence, ²¹/₁₄, or 1½, is the number of
shillings in the price.
APPENDIX V.
ON CHARACTERISTICS.
When the student comes to use logarithms, he will find what follows
very useful. In the mean while, I give it merely as furnishing a rapid
rule for finding the place of a decimal point in the quotient before
the division is commenced.
When a bar is written over a number, thus, 7︤ let the number be called
negative, and let it be thus used: Let it be augmented by additions of
its own species, and diminished by subtractions; thus, 7︤ and 2︤ give
9︤, and let 7︤ with 2︤ subtracted give 5︤. But let the _addition_
of a number without the bar _diminish_ the negative number, and the
_subtraction increase_ it. Thus, 7︤ and 4 are 3︤, 7︤ and 12 make 5, 7︤
with 8 subtracted is 1︦5. In fact, consider 1, 2, 3, &c., as if they
were gains, and 1︤, 2︤, 3︤, as if they were losses: let the addition
of a gain or the removal of a loss be equivalent things, and also the
removal of a gain and the addition of a loss. Thus, when we say that
4︤ diminished by 1︦1 gives 7, we say that a loss of 4 incurred at the
moment when a loss of 11 is removed, is, on the whole, equivalent to a
gain of 7; and saying that 4︤ diminished by 2 is 6︤, we say that a loss
of 4, accompanied by the removal of a gain of 2, is altogether a loss
of 6.
By the _characteristic_ of a number understand as follows: When there
are places before the decimal point, it is one less than the number
of such places. Thus, 3·214, 1·0083, 8 (which is 8·00 ...) 9·999, all
have 0 for their characteristics. But 17·32, 48, 93·116, all have 1;
126·03 and 126 have 2; 11937264·666 has 7. But when there are no places
before the decimal point, look at the first decimal place which is
significant, and make the characteristic negative accordingly. Thus,
·612, ·121, ·9004, in all of which significance begins in the first
decimal place, have the characteristic 1︤; but ·018 and ·099 have 2︤;
·00017 has 4︤; ·000000001 has 9︤.
To find the characteristic of a quotient, subtract the characteristic
of the divisor from that of the dividend, carrying one before
subtraction if the first significant figures of the divisor are greater
than those of the dividend. For instance, in dividing 146·08 by ·00279.
The characteristics are 2 and 3︤; and 2 with 3︤ removed would be 5. But
on looking, we see that the first significant figures of the divisor,
27, taken by themselves, and without reference to their local value,
mean a larger number than 14, the first two figures of the dividend.
Consequently, to 3︤ we carry 1 before subtracting, and it then becomes
2︤, which, taken from 2, gives 4. And this 4 is the characteristic of
the quotient, so that the quotient has 5 places before the decimal
point. Or, if _abcdef_ be the first figures of the quotient, the
decimal point must be thus placed, _abcde·f_. But if it had been to
divide ·00279 by 146·08, no carriage would have been required; and 3︤
diminished by 2 is 5︤; that is, the first significant figure of the
quotient is in the 5th place. The quotient, then, has ·0000 before
any significant figure. A few applications of this rule will make it
easy to do it in the head, and thus to assign the meaning of the first
figure of the quotient even before it is found.
APPENDIX VI.
ON DECIMAL MONEY.
Of all the simplifications of commercial arithmetic, none is comparable
to that of expressing shillings, pence, and farthings as decimals of a
pound. The rules are thereby put almost upon as good a footing as if
the country possessed the advantage of a real decimal coinage.
Any fraction of a pound sterling may be decimalised by rules which can
be made to give the result at once.
Two shillings is £·100 |
One shilling is £·050 |
Sixpence is £·025 |
One farthing is £·001 | 04⅙
Thus, every pair of shillings is a unit in the first decimal place;
an odd shilling is a 50 in the second and third places; a farthing is
so nearly the thousandth part of a pound, that to say one farthing is
·001, two farthings is ·002, &c., is so near the truth that it makes
no error in the first three decimals till we arrive at sixpence, and
then 24 farthings is exactly ·025 or 25 thousandths. But 25 farthings
is ·026, 26 farthings is ·027, &c. Hence the rule for the _first three
places_ is
_One in the first for every pair of shillings; 50 in the second
and third for the odd shilling, if any; and 1 for every farthing
additional, with 1 extra for sixpence._
Thus, 0_s._ 3½_d._ = £·014
0_s._ 7¾_d._ = £·032
1_s._ 2½_d._ = £·060
1_s._ 11¼_d._ = £·096
2_s._ 6_d._ = £·125
2_s._ 9½_d._ = £·139
3_s._ 2¾_d._ = £·161
13_s._ 10¾_d._ = £·694
In the fourth and fifth places, and those which follow, it is obvious
that we have no produce from any farthings except those above sixpence.
For at every sixpence, ·00004⅙ is converted into ·001, and this has
been already accounted for. Consequently, to fill up the _fourth and
fifth_ places,
_Take 4 for every farthing[59] above the last sixpence, and an
additional 1 for every six farthings, or three halfpence._
[59] The student should remember all the multiples of 4 up to 4 × 25,
or 100.
The remaining places arise altogether from ·00000⅙ for every farthing
above the last three halfpence; for at every three halfpence complete,
·00000⅙ is converted into ·00001, and has been already accounted for.
Consequently, to fill up _all the places after the fifth_,
_Let the number of farthings above the last three halfpence be a
numerator, 6 a denominator, and annex the figures of the corresponding
decimal fraction._
It may be easily remembered that
The figures of ¹/₆ are 166666...
” ²/₆ ... 333333...
” ³/₆ ... 5
” ⁴/₆ ... 666666...
” ⁵/₆ ... 833333...
0_s._ 3½_d._ = ·014|58|3333...
0_s._ 7¾_d._ = ·032|29|1666...
1_s._ 2½_d._ = ·060|41|6666...
1_s._ 11¼_d._ = ·096|87|5