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ARITHMETIC
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ARITHMETIC
FOR
SCHOOLS
BY
CHARLES SMITH, M.A.
it
MASTER OF SIDNEY SUSSEX COLLEGE, CAMBRIDGE
REWRITTEN AND REVISED BY
CHARLES L. HARRINGTON
HEAD MASTER OF DR. J. SACHS'S SCHOOL
FOR BOYS, NEW YORK
MACMILLAN AND CO.
AND LONDON
1895
All rights reserved
Copyright, 1895,
By MACMILLAN AND CO.
NortoaoD -press :
J. S. dishing & Co. — Berwick & Smith.
Norwood, Mass., U.SA
PREFACE.
In the following work it has been the endeavor to put
the science of Arithmetic on a sound basis, and to give
clear and complete explanations of all the fundamental
principles und processes. It has not been the aim to
introduce novelties, but to promote accuracy and clearness
of conception, so as to make the study of Arithmetic not
only of practical utility, but also of great educational value.
I am indebted to many friends for their kindness in
looking over the proof sheets, for help in the verification
of the answers, and for valuable criticisms and sugges-
tions. My special thanks are due to Mr. J. Barnard, M.A.,
Head Mathematical Master at Christ's Hospital.
CHARLES SMITH.
184003
CONTENTS.
Chapter
I. Numeration — Notation
II. Addition — Subtraction
ION ....
Multiplication — Drvrs-
III. Factors and Multiples — Square Boot — Highest
Common Factor — Least Common Multiple —
Page
1
13
Parenthesis — Canc
ELLATION
. 59
IV. Fractions ....
87
V. Decimal Measures .
125
VI. Non-Decimal Measures
139
VII. Approximation .
176
/"III. Areas — Volumes
185
Carpeting, Papering, Plastering
189
Dimensions of Circles
193
Specific Gravity
201
IX. Ratio and Proportio
v
208
Partnership.
217
Mixtures .
221
Work and Time
223
Races and Games
226
X. Percentages
231
Profit and Loss .
234
Trade Discount
.
237
Commission and Brokerage
239
Taxes and Duties
241
ix
X
CONTENTS.
Chapter Page
XL Interest 244
Promissory Notes 244
Interest for Days at 6 % 249
Interest for Days at Other Rates than 6 % . . 252
6 % for Long Times 254
Annual Interest 256
Commercial Discount . . - . . . . 258
Exact Interest 268
Partial Payments 269
XII. Exchange — Domestic and Foreign . . . 275
XIII. Stocks and Bonds 282
XIV. Progressions 291
XV. Cube Root 297
OF THE
UNIVERSITY
OF
ARITHMETIC.
CHAPTER I.
NUMEKATION — NOTATION.
1. The idea of number is first acquired from the obser-
vation of groups of distinct objects, actions, sounds, etc. ;
we thus learn to speak of two boys, three balls, four
strokes of a clock, etc.
A single object of any kind, or any group of objects
considered as a whole, is called a unit.
Thus, one ball, one inch, one dozen, one, one ten, are units.
2. Arithmetic is the science which treats of numbers and
of the different operations to which they are subject.
3. The first few numbers in order are, one, two, three,
four, Jive, six, seven, eight, nine, and ten.
4. It will be observed that the names of the first ten
numbers are in no way connected with one another.
Now it is obvious that the knowledge of numbers and
of their relations to one another must always have re-
mained very limited if every successive number had had
a special name given to it independent of the names of
the preceding numbers ; for it would be almost impossi-
ble to remember, in their order, many such names.
b 1
2 NUMERATION — NOTATION. [Chap. I.
5. Successive numbers have therefore been named
according to a systematic plan which requires the use of
as few independent names as possible.
The method by which numbers are expressed in words
according to some systematic plan is called Numeration.
Numeration.
6. To show how all numbers can be named by means
of a few special words, imagine a collection of objects of
the same kind, for example, a heap of apples ; and sup-
pose that we wish to know how many apples there are,
and to give a name to this number.
If there are not more than ten apples altogether, we
find the number at once by counting them, that is by say-
ing in order the names one, two, three, etc., each time
separating one of the apples from the original heap ; and
the name which is said with the last of the heap gives
the number of the apples.
If there are more than ten apples in the heap, count
off ten and put them apart, and go on making groups of
ten until there are fewer than ten apples left. Suppose
there are seven groups of ten each and five apples over,
then we could call the number seven tens and five.
By separating the whole heap into groups of ten in
this way we at once find, and can give a name to, the
number of the apples, provided there are not more than
ten of the groups. Thus, our original ten names suffice
to name all numbers up to that which is made up of ten
groups, each containing ten apples, and we have a new
name, namely one hundred, for the number which con-
sists of ten tens.
If there are more than ten of the groups each of which
contains ten apples, the groups can be arranged in sets of
Arts. 5-7.] NUMERATION. 3
ten, so that there will be one hundred apples in each of
these sets. Suppose that there are five of these sets and
six groups over and four single apples besides, then the
number is made up of five hundreds, six tens, and four.
Thus no new name is necessary until we come to the
number which consists of ten hundreds, and this number
is called a thousand.
It will be seen at once that the names in actual use are only
slightly modified forms of the names which naturally arise from
the above method of division into groups of ten. Instead of
saying two tens, three tens, four tens, etc., we say twenty, thirty,
forty, etc., and we say seventy -five instead of seventy and five.
Also instead of the names ten and one, ten and two, ten and
three, ten and four, etc., we use the names eleven [Gothic ainlif,
ain one and lif ten], twelve [Gothic twalif, twa two and lif ten],
thirteen, fourteen, etc.
If an apple be cut into ten equal parts, any number of
these parts may be put with some apples already counted.
Each part is but one out of ten parts and may be counted
as one tenth.
If one of these tenths be cut into ten equal parts, each
new part is but one out of ten parts of one tenth, and
may be counted as one one-hundredth. Thus by separat-
ing into ten equal parts, etc., we do not really require new
names. If, now, we have five sets, six groups, four single
apples, and three tenths and seven hundredths, then the
number is five hundred sixty-four and thirty-seven hun-
dredths (for three tenths is equal to thirty hundredths).
The principle of the ordinary system of numeration
will now be apparent.
7. The English names which are employed in the
system of Numeration which is universally used are the
following: one, two, three, four, Jive, six, seven, eight, nine,
ten, eleven, twelve, thirteen, fourteen, fifteen, sixteen, seven-
4 NUMERATION — NOTATION. [Chap. I.
teen, eighteen, nineteen, twenty, thirty, forty, fifty, sixty,
seventy, eighty, ninety,
a hundred
which
is ten tens,
a thousand
u
" ten hundreds,
a million
a
" a thousand thousands,
a billion
a
" a thousand millions,
a trillion
u
" a thousand billions,
a quadrillion
a
" a thousand trillions,
and so on.
The names billion, trillion, etc., are very rarely used.
8. The numbers one, ten, a hundred, a thousand, ten
thousand, a hundred thousand, a million, etc., are often
called units of the first order, of the second order, of the
third order, etc. ; and ten units of any order are required
to make one unit of the next higher order.
To give a name to any number it is sufficient to state
the number of units of each different order that the
number contains.
Thus the number which is made up of three millions two hun-
dreds-of-thousands three tens-of-thousands four thousands five hun-
dreds seven tens six units two tenths and four hundredths is called
three million two hundred thirty-four thousand five hundred
seventy-six and twenty-four hundredths ; also the number which is
made up of two hundreds-of-millions three tens-of-millions two
tens-of-thousands three thousands and five hundreds is called two
hundred thirty million twenty-three thousand five hundred.
It should be remarked that the parts of a number are
mentioned in the order of their magnitude, the largest
being given first; in English, however, this order is
reversed for the numbers between ten and twenty. It
should also be noticed that all the thousands, all the mil-
lions, all the billions, etc., are grouped together, as in the
above two cases.
Arts. 8-11.] NOTATION. 5
9. The system of numeration described above is called
the Decimal system, since ten units of any order are
required to make up one unit of the next higher order.
The decimal system of numeration is employed by all
people who have any names at all for numbers greater
than ten, and the origin of the system was doubtless the
natural habit of counting on the fingers.
Notation.
10. We have now to show how numbers can be repre-
sented in a simple manner by means of a few symbols
called figures or digits.
The method by which numbers are expressed by means
of symbols according to some systematic plan is called
Notation.
11. The first nine numbers in order are represented
by the symbols
1, 2, 3, 4, 5, 6, 7, 8, 9.
The same figures are also employed to represent the
first nine collections of tens, of hundreds, of tJwusands,
etc., but on the understanding that the figures are to be
written in a row, and that the figure which represents
the units of the highest order named in a number is to
be written as the left-hand figure of the row; while
the figure which represents the units of the lowest order
named is to be written as the right-hand figure. Thus,
any figure placed just to the right of another represents
units of the order next below that represented by the
other.
To distinguish the figures which represent units, tens,
hundreds, etc., from those which represent tenths, hun-
dredths, etc., a dot, called the Decimal Point, must be
6 NUMERATION — NOTATION. [Chap. I.
written between the figures which represent units and
tenths.
Forty- five is written 45.
Four hundred seventy-two is written 472.
Three and two-tenths is written 8.2
Fifty-one and twenty-seven hundredths is written 51.27
Thirty -five hundredths is written .35
12. The decimal point serves to separate the figures
which represent the wholes from those which represent
the tenths, hundredths, etc. That part of the number
to the left of the decimal point is called the Integral part ;
that part to the right is called the Decimal part.
In 2.5, the 2 is integral, and the .5 is decimal ; in 45.627, the 45
is integral, and the .627 is decimal.
The decimal point is omitted, if the number contains
no decimal.
13. Besides the nine symbols already specified, it is
necessary to have an additional symbol to meet the case
when units of one or more of the different orders are
absent. This symbol is ; its name is naught or cipher.
It has no value by itself, and is used to indicate that
there are no units of the particular order corresponding
to the place in which it occurs.
The other figures are sometimes distinguished from
the naught by being called significant figures.
Thus, 20 represents two tens and no ones; that is, the num-
ber twenty. Again, 2005 represents two thousands, no hundreds,
no tens, and five ones ; that is, the number two thousand five, the
naughts serving to bring the significant figures into the places
intended for them.
It should be noticed that a naught placed at the beginning of
an integral number, or at the end of a decimal number, does not
affect the value of the number • 056, .7020, 0102,60 are the same
as 56, .702, 102.6.
Arts. 12-15.] . NOTATION. 7
14. Any figure, say 5, has two values ; namely (1) its
digit value, which is indicated by its shape, and in virtue
of which it always represents Jive units of some kind,
and (2) its local value, which depends only on its place,
and in virtue of which it represents units of the order
which corresponds to its position in the horizontal row ;
so that if 5 be in the first place to the left of the decimal
point, it represents five units ; if in the second place, it
represents five tens, and so on ; if it be in the first place
to the right of the decimal point, it represents 5 tenths,
and so on.
Thus, in 45 the 5 represents five ones; in 567, it represents five
hundreds; in 2.35, it represents five hundredths. Again, 534.62
represents five hundreds three tens four ones six tenths and two
hundredths; therefore the number represented is Jive hundred
thirty-four and sixty-two hundredths.
15. The names of the units of the orders in common
use, and the positions which correspond to them, are
shown below :
I s I J s § § 1 £ I J I
■5 E
eo E
3d 2d
Period Period
INTEGRAL PERIODS DECIMAL PERIODS
AND ORDERS AND ORDERS
It will be seen that the names are repeated in groups of three ;
and that the decimal periods and orders correspond in name to the
integral periods and orders.
8 NUMERATION — NOTATION. . [Chap. I.
The first twelve integral periods are as follows :
First, Units. Seventh, Quintillions.
Second, Thousands. Eighth, Sextillions.
Third, Millions. Ninth, Septillions.
Fourth, Billions. Tenth, Octillions.
Fifth, Trillions. Eleventh, Nonillions.
Sixth, Quadrillions. Twelfth, Decillions.
16. To write in figures any number expressed in words,
it is necessary only to write the figures which represent
the number of the units of the different orders in their
proper places as shown above, filling up the vacant places,
if any, with naughts.
Thus, to write in figures the number two hundred forty-three,
we must have 2 in the place for hundreds, 4 in the place for tens,
and 3 in the place for units. To write in figures the number five
hundred twenty-four thousand six hundred seven, we must have 5
in the place for hundreds of thousands, 2 in the place for tens of
thousands, 4 in the place for units of thousands, 6 in the place for
hundreds, in the place for tens, and 7 in the place for units, as
follows : 524,607. Sixteen million is written 16,000,000. To write
in figures one thousand thirty and seven thousandths, we must
have 1 in thousands' 1 place, in hundreds' place, 3 in tens' 1 place,
in units' place, in tenths' 1 place, in hundredths" place, and. 7 in
thousandths' place, as follows : 1030.007.
17. To express in words any number given in figures,
first divide the integral and the decimal parts separately
into groups of three, beginning at the right in each case
(the left-hand groups will often be incomplete) ; begin-
ning at the left, read each group of the integral part as if
it were alone and give it the name of the period to which
it belongs, then read the decimal part as if it were inte-
gral and give it the name of the order on the right.
For example, to express in words the number represented by
24160523, we can separate off two complete groups of three figures,
and 24,160,523 is then read twenty-four million one hundred
sixty thousand five hundred twenty-three.
Arts. 16, 17.] EXAMPLES. 9
To express in words the number represented by 36405.4916,
we can separate off one complete group in the integral part and
one in the decimal part, and 36,405.4,916 is then read thirty-six
thousand four hundred five and four thousand nine hundred six-
teen ten-thousandths.
To 'read off' decimals, it is, however, the common
practice merely to name the digits in order.
Thus .615 is read ' decimal six, one, five ' ; 15.0524 is read ' fifteen,
decimal naught, five, two, four'; and 1567.0082 is read 'one
thousand five hundred sixty-seven, decimal naught, naught, eight,
two.'
In reading numbers the word ' and ' should be used only when
we reach a decimal point.
EXAMPLES I.
1. For what does 5 stand in the numbers 15, 1.57, 514,
352167, and 3561234, respectively ?
2. For what does 7 stand in the numbers 70, 37123,
125.479, 274126315, and 370001002003, respectively ?
3. Name all the figures which represent their digit
value of hundred's in 314, 2167, 50412, and 31024.
4. Name all the figures which represent their digit
value of thousands in 2314, 56123, 60417, and 3005167.
5. Express in words the separate value of every figure
in 3.5, 15.7, 125.34, 12.53, 800.17, 1200.63, .875, 50.037,
5.00107, 560002.19007.
6. Express in words the numbers 27, 349, 560, 3.06,
1204, and 5020.
7. Express in words the numbers 200.9, 6050, 12345,
10.305, 40050, and 1.20463.
8. Express in words the numbers 518618, 602010,
100010, 504075, 420040, and 107.005.
10 NUMERATION — NOTATION. [Chap. I.
9. Express in words 111111111, 1203405, 2314100,
504.0314, 20050060, and 30300074.
10. Express in words the numbers 3012004, 1101.11011,
201201201, 1000040305101, and 604102000300004.
11. "Write in figures the numbers fifty-eight, eighty-
five, two hundred eleven, three thousand twelve, six
thousand forty, and nine thousand three hundred.
Write in figures the following numbers :
12. Twelve and three-tenths.
13. Three hundred four and nine-tenths.
14. Twenty-five, three-tenths, and four hundredths.
15. Eour, six-tenths, and seven-thousandths.
16. One million, four-tenths, and three-millionths.
17. Write in figures the numbers eleven hundred
eleven, fourteen hundred sixty, twelve hundred thou-
sand sixteen hundred, six million twelve hundred six-
teen, and eleven billion eleven hundred eleven.
18. Write in figures twenty million twenty thousand,
seventeen million fifty thousand nineteen, one hundred
four million six hundred two thousand eleven, and six
thousand three hundred seven million two thousand fifty
six.
18. The ordinary system of notation was introduced
into Europe by the Arabians, and is still called the Ara-
bic system of Notation although it is now known that the
Arabians derived their knowledge from the Hindoos.
Eoman Numerals.
19. Besides the Arabic system of notation some use
is still made of the cumbrous system employed by the
Romans.
Arts. 18-22.] ROMAN NUMERALS. H
The symbols which were used by the Romans, and
which are called Roman Numerals, are the following :
I for I, V for 5, X for 10, L for 50, C for 100, D for 500,
M for 1000.
A horizontal line over any numeral increases its value
one thousand fold : thus V stands for 5000, X for 10000,
etc.
Roman numerals are arranged in order of magnitude
from left to right, and are repeated as often as may be
necessary.
Thus, 2 is represented by II, 30 by XXX, 233 by CCXXXIII, and
1887 by MDCCCLXXXVII.
20. To avoid some of the troublesome repetitions
which are common to the Roman system of notation, a
numeral is in certain cases placed before another of
greater value to denote that the value of the larger is to
be diminished by the amount of the smaller.
Thus, IV denotes one less than five, that is, 4 ; IX denotes one
less than ten, that is, 9 ; XL denotes ten less than fifty, that is, 40 ;
and XC denotes ten less than one hundred, that is, 90 ; so also,
CCXC denotes 290.
21. The symbols CIO, CCIOO, CCCIOOO, etc., were
anciently employed to denote respectively 1000, 10,000,
100,000 etc. ; also 10, IOO, 1000, etc., to denote respect-
ively 500, 5000, 50,000, etc. In fact, M and D are only
modified forms of CIO and 10 respectively.
22. Roman numerals were used only to register num-
bers, and were never employed in making numerical
calculations. The Romans made their calculations by
means of counters and a mechanical apparatus called an
Abacus. The counters used were often pebbles (Latin,
calculus), whence our word calculation.
12 NUMERATION — NOTATION. [Chaps. L, II.
EXAMPLES II.
1. Express all the numbers from 1 to 20 by means of
Eoman numerals.
2. Express by means of Eoman numerals the num-
bers 20, 30, 40, 50, 60, 70, 80, 90, 200, 400, 600, 800,
and 900.
3. Express by means of Eoman numerals the numbers
39, 49, 59, 69, 79, 89, 99, 96, 444, 1294, and 1889.
4. Write the numbers LVIII, XXXIX, XLIV, XCIV,
XCIX, CXCIX, and MMDCCXCIX, in the Arabic
notation.
Arts. 23, 24.] ADDITION. 13
CHAPTER II.
ADDITION— SUBTRACTION — MULTIPLICATION —
DIVISION.
Addition.
23. The process of finding a single number which
contains as many units as there are in two or more given
numbers taken together is called Addition ; and this single
number is called the Sum.
The sum of the numbers of the units in two or more
groups would therefore be found by forming a single
group containing them all, and then counting the number
of the units in this single group.
24. The following fundamental truth is evident :
TJie number of the things in any group will always be
found to be the same in whatever order they may be counted.
From this it follows that the sum of the numbers of the
things in any two groups will be found by first counting
all the things in the first group and then proceeding to
the second ; that is, by increasing the number in the first
group by as many units as there are in the second. The
same sum will also be found by increasing the number in
the second group by as many units as there are in the first.
Thus, the sum, of 3 and 5 is found by counting five
onwards from three, namely four, five, six, seven, eight;
or by counting three onwards from five, namely six, seven,
14 ADDITION. [Chap. II.
eight In the first case we are said to add 5 to 3, and in
the second case we are said to add 3 to 5 ; but the results
must be the same.
25. Addition is indicated by the sign -f, which is read
' plus.*
Thus, 5 + 4 is read five plus four, and denotes that 5 is to be
increased by 4, that is, that 4 is to be added to 5 ; also, 5 + 4 + 3
denotes that 4 is to be added to 5, and then 3 added to the result.
26. The sign = , which is read 'equals' or 'is equal to,'
is used to denote the equality of two numbers.
Thus, 5 + 4 = 9 denotes that the sum of 5 and 4 is 9.
27. When children first begin to add they make use
of their fingers, but all counting on the fingers, or with
any other real objects, should be discontinued as soon as
possible, and the results of adding numbers not greater
than nine should be given instantaneously.
Tables of the results of the addition of any two numbers
each not greater than 10 might at first be made by the
pupil, arranged in lines ; as for example, 8 and 1 are 9,
8 and 2 are 10, 8 and 3 are 11, etc.
EXAMPLES III.
Oral Exercises.
These examples should be practised until great rapid-
ity is attained.
1. Add 1 and 9, 3 and 8, 2 and 6, 4 and 7, 6 and 3,
4 and 4.
2. Add 7 and 8, 7 and 6, 3 and 9, 5 and 4, 3 and 5,
9 and 8.
3. Add 4 and 3, 9 and 9, 8 and 8, 6 and 9, 7 and 2,
3 and 3.
Arts. 25-28.] EXAMPLES. 15
4. Add 5 and 9, 9 and 4, 6 and 8, 5 and 7, 2 and 9,
8 and 5.
5. Add 7 and 7, 5 and 5, 6 and 6, 8 and 4, 6 and 4,
9 and 7.
6. Add 8 to 15, to 25, to 35, to 45, to 65, and to 95.
7. Add 13 and 7, 23 and 7, 43 and 7, 63 and 7, 83
and 7, 93 and 7.
8. Add 9 to 17, to 27, to 57, to 67, to 87, and to 97.
9. Begin with 7 and add 2 again and again up to 27.
Do not say 7 and 2 are 9 and 2 are 11 and 2 are 13, etc., but
state results; thus, 7, 9, 11, 13, etc.
10. Begin with 2 and add 3 again and again up to 35.
11. Begin with 85 and add 4 again and again up to
101.
12. Begin with 50 and keep on adding sevens until
the sum exceeds 100.
13. Begin with 15 and keep on adding nines until the
sum exceeds 100.
14. Add the following numbers in order, first begin-
ning at the right and then at the left :
(1) 2, 7, 4, 0, 6, 9, 5, 2, 6, 5, 9, 3, 4, 8.
State results only ; thus, 2, 9, 13, 13, 19, 28, etc.
(2) 7, 9, 5, 4, 0, 8, 6, 7, 3, 5, 9, 8, 2, 6.
(3) 3, 5, 6, 9, 0, 7, 8, 4, 3, 6, 2, 5, 7, 9.
(4) 9, 6, 7, 4, 2, 8, 1, 3, 7, 5, 4, 6, 5, 8.
28. The sum of any two numbers may be found by
counting onwards from the first as many units as there
are in the second, but this method would obviously be
very troublesome except when the second number is very
small.
16 ADDITION. [Chap. II.
Now numbers are divided, as we have already learned,
into groups of units, tens, hundreds, tenths, hundredths,
etc. ; and when numbers are to be added, the parts into
which they are divided may be added in any order we
please, provided they are all counted; hence we may first
add the units of one order, then the units of another order,
and so on.
29. In order to add numbers, they should first be
arranged so that their decimal points are in a vertical
column. This will ensure that all the tenths shall be in
the same vertical column, and so for the hundredths,
etc. ; and so also for the units, tens, hundreds, etc. This
arrangement is for convenience only.
The following examples will show how this principle
enables us readily to find the sum of any given numbers.
Ex. 1. Add 235.7 and 524.2.
Since we wish to add the tenths by themselves, the units by
themselves, etc. , we write the numbers so that the decimal points
are in a vertical column ; thus,
235.7
524.2
Now 2 tenths and 7 tenths make 9 tenths, 4 units and 5 units
make 9 units, 2 tens and 3 tens make 5 tens, and 5 hundreds and 2
hundreds make 7 hundreds. The required sum is generally placed
just under the numbers to be added and separated from them by a
horizontal line ; thus,
235.7
524.2
759.9
Ex. 2. Add 548.6, 789, and 197.8.
Write the numbers as in Ex. 1 ; thus,
548.6
789.
197.8
1535.4
Arts. 29, 30.] EXAMPLES. 17
Now 8 tenths and 6 tenths make 14 tenths, that is, 1 unit and 4
tenths. The 4 tenths can be put in the column for tenths, but the
1 unit must be counted with the other units. We then have 1 unit,
7 units, 9 units, and 8 units, which make 25 units ; that is, 2 tens
and 5 units. The 5 is put in the column for units, but the 2 tens
are ' carried ' (as it is called) and added with the other tens. So we
proceed until all the columns are added.
Note. Since ten units of any order make one unit of the next
higher order, the figures in any column may be added without
specifying the kind of units they represent ; that is, without call-
ing them tens, or hundreds, or thousands, etc., as the case may be.
Also, we should never use as many words as in the above explana-
tions, but should say (see ex. 2) only 8, 14 ; 1 (carried), 8, 17, 25 ;
2 (carried), 11, 19, 23 ; 2 (carried), 3, 10, 15. Of course the 4, 5, 3,
and 15 are the figures to be written. In all cases the sums of num-
bers should be more prominent than the numbers themselves.
30. To detect mistakes in addition it is well to add
each line of figures twice, once from bottom to top and
once from top to bottom. An error is much more likely
to be detected in this way than by simply repeating the
addition in the same order, for the same mistake is very
likely to be made again.
Pupils should not be allowed to add more than one
column at a time.
EXAMPLES IV.
Written Exercises.
1. Add 3104, 297, 5649, and 989.
Find the sum of
2. 21.63, 5.24, 170.63, 27.59, 17.
3. 301.7, 30.17, 3.017, .3017, .03017.
4. 319, 562, 1230, 857, 4908, and 9087.
5. 235, 796, 804, 987, 359, and 856.
6. 170.2, 3.605, 17.35, 15.609, .0086.
18
ADDITION.
[Chap. II.
7. .0037, 21.85, 169.4, 17.9375, .90087.
8. 4.1372, 41.372, 4137.2, .41372, 41372.
Add the numbers in each column and in each row of
the squares. Do not change the positions of the numbers.
9.
1
15
14
4
10.
9
16
23
12
5
12
6
7
9
13
2
10
19
21
8
10
11
5
20
24
11
3
7
13
3
2
16
1
8
17
25
14
22
15
4
6
18
11.
1
101
80
59
38
17
117
96
75
54
33
21
121
89
68
47
26
5
105
84
63
42
30
9
109
88
56
35
14
114
93
72
51
39
18
118
97
76
55
23
2
102
81
60
48
27
6
106
85
64
43
22
111
90
69
57
36
15
115
94
73
52
31
10
110
78
77
45
24
3
103
82
61
40
19
119
98
86
65
44
12
112
91
70
49
28
7
107
95
104
74
83
53
62~
32
41
11
20
100
120
79
99
58
67~
37
^6~
16
25
116
4
113
92
71
50
29
8
108
87
66
34
13
Perform the additions indicated below :
12.
3157
13. 589.761
14. 412.64506
294
35.71
39.17412
16903
840.693
246.82441
8057
392.75
49.1733
62934
1569.4242
387.198207
998
359.177
129.38946
JT. 31.]
EXAMPLES.
1!
15. 50971
16.
314569
17. 842713
8265
73985
9185
13926
387648
38977
78912
930807
796359
34056
186794
246824
19389
389548
135791
8747
153875
924678
In the next three examples do not change the positions
of the numbers.
18. Find 30.1 + 297 + 35.16 + 1079 + 8.017 + 10.053.
19. Find 93084 + 15614 + 3801.76 + 536174 + 123456
+ 40.404.
20. Find 218904 + 37.215 + .199 + 582163 + 397157
+ 81.429 + 7.9163.
21 . Add six hundred ninety-five, one thousand seventy-
four, eleven thousand four hundred eighty-nine, and fifty-
four thousand three hundred seventy.
22. Add three million four hundred seventeen thou-
sand thirty-five, nine hundred forty-six thousand seven
hundred, fifteen million fifteen thousand fifteen, and sixty
million sixteen hundred twenty-four.
23. Add six million five hundred nine thousand seven
hundred six and twelve thousand four hundred thirty-
two hundred-thousandths, three hundred ninety thousand
and four hundred twelve thousandths, eighteen million
forty and six ten-thousandths.
31. Thus far we have studied numbers without refer-
ence to objects.
When numbers are used without reference to any
particular units, they are called Abstract Numbers.
Two and five are abstract numbers.
20 ADDITION. [Chap. II.
When numbers are associated with particular units,
they are called Concrete Numbers.
Two feet and five tons are concrete numbers.
32. Concrete numbers can be added only when the unit
is the same. For example, 3 horses and 4 cows do not
make 7 horses nor 7 cows ; they do, however, make 7
animals; because regarding them as animals the unit is
the same. Also the sum of 3 feet and 4 inches is not 7
feet nor 7 inches.
EXAMPLES V.
Written Exercises.
1 . In 1890 the population of each of the New England
States was as follows : Maine, 661000 ; New Hampshire,
377000; Vermont, 332000; Massachusetts, 2239000;
Ehode Island, 346000 ; Connecticut, 746000. What was
the total ?
2. In a town, noted for the number of its schools,
there were 225 boys in a military school, 175 girls in a
school for girls, 126 young men in a theological school,
163 boys in a training school, 23 children in a kinder-
garten, and 1500 pupils in the public schools. How many
pupils in all ?
3. A man paid 527.37 dollars for 14 cows, 1463.80
dollars for twelve horses, and 918.36 dollars for 153 pigs.
How many animals were there, and how much was paid
for them all ?
4. The population of each of the six northern counties
of England is as follows : Cumberland, 250647 ; Durham,
867258 ; Lancashire, 3454441 ; Northumberland, 434086 ;
Westmoreland, 64191; and Yorkshire, 2886564. What
is the total population ?
Arts. 32-37.] SUBTRACTION. 21
Subtraction.
33. The process of finding how many units are left
when a number is taken aivay from a larger number is
called Subtraction. The result is called the Remainder,
or the Difference.
Any two numbers can be added ; it is, however, impossible to
subtract one number from a smaller number.
34. The larger of the two numbers is called the
Minuend.
The smaller of the two numbers is called the Sub-
trahend.
Illustration. 8 Minuend.
5 Subtrahend.
3 Remainder.
35. It is clear that the remainder is that number ivhich,
when added to the subtrahend, will give the minuend.
Thus, to subtract 5 from 12 is to find the number which, when
added to 5, will make 12.
The question involved in subtraction may be put in different
ways. Thus, it may be asked :
(1) What is the remainder when 5 is taken from 12 ?
(2) What must be added to 5 to make 12 ?
(3) By how many is 12 greater than 5 ?
(4) By how many is 5 less than 12 ?
36. Subtraction is indicated by the sign — , which is
read 'minus.'
Thus, 9 — 4 is read nine minus four, and denotes that 9 is to be
diminished by 4, that is, that 4 is to be subtracted from 9 ; also,
5 — 4 + 3 denotes that 4 is to be taken from 5, and then 3 added
to the result.
37. The knowledge of the results of the addition of
numbers not greater than ten will furnish us with the
22 SUBTRACTION. [Chap. II.
results of the subtraction of small numbers. Examples
of subtractions of this kind should be practised until
great rapidity is attained.
EXAMPLES VI.
Oral Exercises.
1. How many are left when we take 7 from 14,
5 from 10, 6 from 12, 8 from 12, 4 from 10, and 7 from
16, respectively ?
2. How many are left when we take 5 from 14,
4 from 13, 8 from 14, 7 from 12, 9 from 11, and 5 from
13, respectively ?
Find the difference between the numbers in each of the
following pairs :
3. 5 and 12, 7 and 16, 9 and 18, 3 and 11, 6 and 14,
8 and 15.
4. 3 and 8, 5 and 11, 6 and 13, 8 and 14, 7 and 15,
9 and 16.
5. Begin with 50 and go on diminishing by fours as
many times as possible.
6. Begin with 53 and go on diminishing by fives as
many times as possible.
7. Begin with 70 and go on diminishing by sixes as
many times as possible.
8. What must be added to 5 to make 8, to make 13,
to make 10, to make 12 ?
9. What must be added to 7 to make 9, to make 12,
to make 10, to make 15 ?
10. What must be added to 8 to make 10, to make 12,
to make 14, to make 16 ?
Art. 38.]
SUBTRACTION.
23
11. 9 and
4 and
Fill up the blanks below.
make 10, 3 and
make 11, 6 and
12. 7 and make 15, 6 and
4 and make 9, 3 and
13. 3 and make 7, 9 and
6 and make 15, 7 and
make 11, 2 and
make 9, 4 and
make 13, 9 and
make 8, 8 and
make 18, 8 and
make 9, 3 and
make 8
make 8
make 12
make 17
make 16
make 6
38. The consideration of the following examples will
show how the difference between any two numbers can
be found.
Ex. 1. Subtract 524.63 from 759.85.
The smaller number should be placed just under the greater, so
that one decimal point is vertically over the other. (See Art. 29.)
759.85
524.63
Beginning with the lowest order, we find the remainder when
3 hundredths are taken from 5 hundredths, 6 tenths from 8 tenths,
4 units from 9 units, 2 tens from 5 tens, and 5 hundreds from
7 hundreds ; thus,
759.85 Minuend.
524.63 Subtrahend.
235.22 Remainder.
Ex. 2. Subtract 35.7 from 78.3.
78.3 Minuend.
35.7 Subtrahend.
42.6 Remainder
Now 7 tenths are more than 3 tenths, therefore we cannot sub-
tract : if, however, we take 1 unit from the 8 units and change
that unit to 10 tenths, we shall have 13 tenths in all. Now 7 tenths
from 13 tenths leave 6 tenths, 5 units from 7 units leave 2 units,
and 3 tens from 7 tens leave 4 tens. Remainder = 42.6.
24 SUBTRACTION. [Chap. II.
Mental Work Illustrated. We may omit names of orders. (See
note, Art. 29.)
Ex. 3.
468.27
186.49
281.78
9 from 17,
4 from 11,
6 from 7,
8.
7.
1.
8 from 16,
8.
1 from 3,
2.
20.07
12.6
7.47
from 7,
6 from 10,
2 from 9,
7.
4.
7.
1 from 1,
0.
Ex. 4.
In this example 1 ten is taken from 2 tens and changed to 10
units ; one of these units is changed to ten tenths. The operation
may be represented thus :
20.07 =19. 10 7
12.6 =12.6
Remainder = 7. 4 7
39. One concrete number cannot be subtracted from another
unless both are expressed in terms of the same unit. For example,
we cannot subtract 5 tons from 7 miles ; nor can we subtract 3
feet from 60 inches, unless either 3 feet is expressed in inches or
60 inches expressed in feet.
40. It is easily seen that if from a given number several num-
bers be taken in succession the result will be the same as if the
sum of those numbers were subtracted from the given number.
Ex. Subtract the sum of 366, 648, and 759 from 2314.
~^* 9, 8, and 6 make 23 ; subtract the 3 from the 4 and carry
j?^j the 2 ; 2, 5, 4, and 6 make 17 ; subtract the 7 from 11
759 and carry the 1 ; 1, 7, 6, and 3 make 17, which is to be sub-
541 tracted from 22.
Mental Work.
9, 17, 23, 3 from 4 = 1.
2,7,11,17, 7 " 11=4.
1, 8, 14, 17, 17 " 22 = 5.
Arts. 39-42.] EXAMPLES. 25
41. When several operations of addition and subtraction have
to be performed in succession the result is the same in whatever
order the operations are performed.
Hence, to find 28 - 15 + 26 - 17 - 14 + 12, first find the sum of
28, 26, and 12, the numbers to be added ; then the sum of 15, 17,
and 14, the numbers to be subtracted ; and finally taking the
difference of these two sums ; thus,
28 15
26 17
12 14
66 - 46 = 20.
42. To detect mistakes in subtraction, add the remain-
der to the subtrahend, and the sum should equal the
minuend ; or subtract the remainder from the minuend,
and the new remainder should equal the subtrahend.
EXAMPLES VII.
Written Exercises.
1. Subtract 129.6 from 3145, 81.7 from 3002, and
123.4 from 432.1.
2. Subtract 15.97 from 79.15, 18235 from 1000000,
and 135.79 from 24680.6.
3. Find the values of 645-378, 307-149, 294-
208, 2179 - 1984, 3206 - 1679, and 120573 - 98765.
Fi
nd the difference between
4.
3.726 and 5.949.
8. 3.008 and 3.08.
5.
14.753 and 6.876.
9. .217 and .271.
6.
1 and .888.
10. 20 and .675.
7.
.00013 and .00175.
11. .8017 and .00693.
12. Find the values of
(1) 31 + 97 - 23 + 175 - 184.
(2) 151 - 77 + 94 - 111.
26 SUBTRACTION. [Chap. II.
(3) 315 - 127 - 172 + 358 - 265.
(4) 742 - 329 - 197 + 215.
13. Find 3.17 + 4.216 - 5.8004 + 2.0097 - .99873.
14. Find 21.09 - 3.985 - 7.0095 + .09372 - 4.38009 +
2.60009.
15. Subtract from 11.214 the sum of 2.301, 1.7293,
2.0507, and 3.62743.
16. Subtract from 20 the sum of 3.416, 2.6008, 5.73124,
and 1.5063.
17. Subtract from 121097 the sum of 7916, 1214, 1397,
and 34162.
18. Subtract from 1000000 the sum of 421654, 127,
31562, 1795, and 123456.
19. Subtract 27 from 80, and then 27 from the re-
mainder, and so on as many times as possible ; and find
the final remainder.
20. What number must be taken from 81 to leave 37
as remainder ?
21. By how much does the sum of 3.5612 and 4.71305
exceed the sum of 1.70862 and 5.91927?
22. What number must be taken from one hundred
thousand to leave five thousand four hundred eighty-
seven as remainder ?
23. The difference between two numbers is 145, and
the greater is 597 ; what is the smaller ?
24. The sum of two numbers is 1000, and one of them
is 594 ; what is the other ?
25. On a man's birthday in 1891 he was 63 years old.
In what year was he born ?
26. In 1891 a man of 65 was on his birthday just 37
years older than his son. In what year was the son born ?
Arts. 43, 44.] MULTIPLICATION. 27
27. Add the sum of 516 and 784 to the difference be-
tween 314 and 176.
28. Add the difference between 1925 and 1789 to the
difference between 3421 and 1679.
29. In an orchard there are 1572 fruit trees; of these
352 are apple trees, 275 are pear trees, and 187 are plum
trees. How many other trees are there ?
30. The population of each of five towns is as fol-
lows: ^4,3789; 5,7861; 0,2893; D, 756; #,847. If
B and D were united, the new town would be how much
larger than A, C, and E together ?
Multiplication.
43. A short process of adding two or more equal num-
bers is called Multiplication.
Ex. 1. 5 + 5 + 5 + 5 = 20 ; i.e., 4 fives = 20.
Ex. 2. 3 + 3 + 3 + 3 + 3 = 15 ; i.e., 5 threes = 15.
If we say (Ex. 1) 5, 10, 15, 20, or (Ex. 2) 3, 6, 9, 12,
15, we are adding by a long process.
If we say 4 fives = 20, or 5 threes = 15, we are adding
by .a short process called multiplication.
44. The number which is to be thus increased is called
the Multiplicand.
The number which indicates how many equal numbers
are to be added is called the Multiplier.
The result of multiplication is called the Product.
The multiplicand and multiplier are called Factors of
the product.
Ex. 1. Multiply 5 by 4. Ex. 2. Multiply 3 by 5.
5 Multiplicand. 3
\ Multiplier. _5
20 Product. 15
Factors of 20 { \ ^r* ! !* Factors of 15.
I 4 Multiplier. 5
28
MULTIPLICATION.
[Chap. II.
45. The multiplication of any two numbers not greater
than nine is easily found by actual addition. It will be
shown that every case of multiplication can be reduced
to a series of cases of multiplications of numbers not
greater than ten; it is therefore essential to learn by
heart all the products of such numbers. These products
are given in the following table, called the Multiplication
Table.
1
2
3
4
5
6
7
8
9
10
11
12
2
4
6
8
10
12
14
16
18
20
22
24
3
6
9
12
15
18
21
24
27
30
33
36
4
8
12
16
20
24
28
32
36
40
44
48
5
10
15
20
25
30
35
40
45
50
55
60
6
12
18
24
30
36
42
48
54
60
6Q
72
7
14
21
28
35
42
49
56
63
70
77
84
96
8
16
24
32
40
48
56
64
72
80
88
9
18
27
36
45
54
63
70
72
80
81
90
90
100
99
110
108
120
10
20
30
40
50
60
11
22
33
44
55
66
77
88
99
110
121
132
144
12
24
36
48
60
27
84
96
108
120
132
Any horizontal line in the table gives the products of
the number which begins the line by the first twelve
numbers in order. Thus the fourth line can be read 1
four is 4, 2 fours are 8, 3 fours are 12, 4 fours are 16, etc.
It is usual and desirable, though not absolutely necessary, to
learn the Multiplication Table as far as 12 times 12. This table
Arts. 45-47.] THEOREMS. 29
should be made again and again by the pupil himself by actual
addition.
46. Multiplication is indicated by the sign x , which
is read 'multiplied by' or 'times.'
Thus, 5 x? 4 is read 5 multiplied by 4, which means 4 times 5 ;
also, 5x4x3 denotes that 5 is to be multiplied by 4, and this
product multiplied by 3.
When one number is multiplied by two or more other
numbers in succession, the result is called the Continued
Product.
47. Before considering how to find the product of any
two numbers, certain general truths, which hold good for
all numbers whatever, must be investigated. For this
purpose consider the following arrangement of dots :
The total number of the dots is independent of the way
in which they are counted.
Now there are 10 dots in each row and 5 rows ; the
whole number of the dots can therefore be counted as 10
repeated 5 times, or 6 repeated 10 times ; i.e., 10 x 5
= 5 x 10. It is clear that this result would hold good
however many rows and columns there might be ; thus
we are led to
Theorem I. The product of any number by any second
number is the same as the product of the second by the
first.
30 MULTIPLICATION. [Chap. II.
Again, if we consider separately the two parts divided
by the vertical line, we see that the whole number of
dots is the sum of 6 repeated 5 times and 4 repeated 5
times, the 6 and 4 together making 10, so that 10 x 5 is
the same as6x5-f4x5; thus we are led to
Theorem II. The product of any two numbers is the same
as the sum of the products of the multiplier and any two or
more numbers ivhich together make up the multiplicand.
Now consider the following arrangement :
5 5 5 5 5 5
5 5 5 5 5 5
5 5 5 5 5 5
Here the sum of all the numbers consists of 5 repeated
6x3 times. But in each row 5 is repeated 6 times and
the rows are repeated 3 times ; hence the sum of all the
numbers is 5x6x3. Again, in each column 5 is re-
peated 3 times and the columns are repeated 6 times ;
hence the whole sum is 5 x 3 x 6. Thus, to multiply by
6 and 3 in succession, in any order, gives the same result
as to multiply at once by 6 x 3, that. is by 18. It is clear
that the same would be true for any other numbers what-
ever ; thus we are led to
Theorem III. To multiply by two or more numbers in
succession gives the same result as to multiply at once by
their product.
48. To multiply by 10, 100, etc. Numbers are sepa-
rated into groups of units, tens, hundreds, etc. ; tenths,
hundredths, etc. ; and a number is multiplied by 10 when
each of its parts is multiplied by 10 [Theorem II] ; i.e.,
when each of its parts is raised to the next higher order.
For example, 62.34 is multiplied by 10 when its 4 hundredths
are made 4 tenths, its 3 tenths are made 3 units, its 2 units are
Arts. 48, 49.] TO MULTIPLY BY 10, ETC. gl
made 2 tens, and its 6 tens are made 6 hundreds ; this is accom-
plished by moving the decimal point one place to the right ; i.e.,
62.34 x 10 = 623.4.
Also, 623.4 x 10 = 6234 ; 6234 x 10 = 62340.
Multiplying by 10 and then by 10 again is the same
as multiplying by 100, and it will be noticed that in mul-
tiplying 62.34 by 100, the decimal point is moved two
places to the right.
Hence, to multiply by 10, 100, 1000, etc., move the decimal
point as many places to the right as there are naughts in the
multiplier.
Examples. 1. 164.2789 x 10 = 1642.789.
2. 164.2789 x 1000 = 164278.9.
3. 340 x 100 = 34000.
49. The following examples will show how to find the
product of any two numbers.
Ex. 1. Multiply 52.34 by 7.
In multiplication, the multiplier is placed under the multiplicand,
so that the right-hand figures shall be in the same vertical column.
This is only for convenience.
52.34 Multiplicand.
7 Multiplier.
366.38 Product.
By Theorem II, we multiply the units of the different orders
separately by 7, and add the results. Now, 4 hundredths x 7 = 28
hundredths, or 2 tenths and 8 hundredths ; the 8 is put in hun-
dredths' column, and the 2 must be counted with the tenths, or
'carried.' Next, 3 tenths x 7 = 21 tenths, which with the 2 tenths
carried = 23 tenths, or 2 units and 3 tenths ; write the 3 in tenths'
column, and carry the 2 to units' column. Next, 2 units x 7 = 14
units, which with the 2 units carried = 16 units, or 1 ten and 6
units ; write the 6 in units' column, and cany the one to tens'
column. Finally, 5 tens x 7 — 35 tens, which with the 1 ten carried
= 36 tens ; write the 5 in tens' column, and the 3 in hundreds'
column. The product is 366.38.
32 MULTIPLICATION. [Chap. II.
Ex. 2. Multiply .3 by .2.
.3 Multiplicand.
.2_ Multiplier.
.06 Product.
Now, 3 tenths x 2 = 6 tenths. Since 2 is 10 times 2 tenths,
the product obtained by multiplying by 2 is 10 times the product
obtained by multiplying by 2 tenths. If, therefore, we separate
the 6 tenths, which equals 60 hundredths, into 10 equal parts,
one of those parts must be the true product. We find, from the
multiplication table, that 6 hundredths is one of the 10 equal parts
of 60 hundredths. Therefore .06 is the true product.
Ex. 3. .08 x 3 = .24 ; .08 x .3 = .024.
.18 x 4 = .72 ; .18 x .4 = .072.
.53 x 2 = 1.06 ; .63 x .2 = .106.
Observe that the number of decimal places in the product is
equal to the number of decimal places in both multiplicand and
multiplier.
Ex. 4. Multiply 5.64 x 302.6.
5.64 Multiplicand.
202.6 Multiplier.
1st partial product 3.384 = 5.64 x 6
2d partial product 11.28 = 5.64 x 2
3d partial product 1692. = 5.64 x 2 x 100
Sum of partial products 1706.664 = Product.
We multiply 5.64 separately by .6, by 2, and by 200, writing the
partial products in form for addition; i.e., so that the decimal
points are in column.
It is not necessary to specify, as we have done above, the kinds of
units which are being multiplied at any stage. (See note, Art. 29.")
Ex. 5. Multiply 321 x 218.
321 Multiplicand.
218 Multiplier
2568 = 321 x 8
3210 = 321 x 10
64200 = 331 x 200
69978 = Product.
The naughts are omitted in practice, because they count for
nothing in addition.
Arts. 50-52.] INVOLUTION. 33
Ex. 6. 31642 Ex. 7. 27.006 Ex. 8. 27.68
100506 2001.908 36000
189852 21 6048 16608
158210 24 30 54 8304
31642 27 00 6 996480.00
3180210852 54012 -
. 54063.527448
Naughts at the right of multiplicand or multiplier are omitted
in the partial products and annexed to the significant figures of
the answer before writing the decimal point.
50. In all cases of multiplication the multiplier must
be an abstract number, for to repeat anything 5 shillings
times or 3 tons times is absurd; the multiplicand, how-
ever, may be either a concrete number or an abstract
number.
Thus, we can multiply 3 feet by 5, but not 3 feet by 5 feet, nor
5 by 3 feet.
In case a person is reckoning the cost of 5 pounds of tea at 60
cents a pound, he is not multiplying 60 cents by 5 pounds, but is
multiplying 60 cents by 5, since he is finding the sum of as many
60 cents as there are pounds.
51. To test the answer, multiply the multiplier by the
multiplicand. This should give the same result as mul-
tiplying the multiplicand by the multiplier [Theorem I,
p. 26].
52. The continued product of a number by itself is
called a Power of that number.
Thus, 5 x 5 is called the second power or the square of
5; 5 x 5 x 5 is called the third power or the cube of 5;
5x5x5x5 is called the fourth power of 5 ; and so on.
The squares of the first nine numbers are 1, 4, 9, 16, 25,
36, 49, 64, 81. The cubes are 1, 8, 27, 64, 125, 216,
343, 512, 729.
A small figure placed above and to the right of a
number is used to denote a power of that number, and
34 MULTIPLICATION. [Chap. II.
is called an Exponent, or Index. For example, 5 4 denotes
5x5x5x5. Also, 10 2 = 10 x 10 = 100, 10 3 = 10 x
10 x 10 = 1000, and 10 6 = 1000000.
It should be noticed that the second power of 10 contains two
naughts, the third power three naughts, and so on.
When a number is raised to a power, the process is
called Involution.
Involution, then, is but a name for a special case of
multiplication.
EXAMPLES VIII.
Oral Exercises.
The pupils should read the answers while looking at
the examples.
Multiply each of the following by 10 :
1. 23, 2.34, 51.67, 21.08, 700, 314.5.
2. 769, 7.123, .562, 1000.007, .0034.
3. 2.006, .00006, 150.27, 5000000.
Perform the multiplications indicated below :
4. 2.157 x 100. 10. .07 x 1000.
5. 1.2308 x 1000. 11. 3.501 x 1000.
6. 1.27 x 100. 12. .00039 x 10000.
7. 6573 x 100. 13. 98.764 x 1000.
8. .0067 x 100. 14. .00001 x 1000.
9. 2.1345 x 1000. 15. 16.02 x 100000.
16. Find the squares of 4, 7, 2, .3, 5.
17. Find 9 2 , .9 2 , 12 2 , 1.2 2 , .12 2 .
18. Find 5 2 , .5 2 , .6 2 , .7 2 , .08 2 .
19. Find the cubes of 2, 3, .2.
20. Find .2 3 , .02 3 , .03 3 , .003 3 .
RT. 52.]
EXAMPLES
3,
EXAMPLES :
EX.
Multiply
Written Exercises.
I. 37 by 3.
7.
1083 by
11.
13. 5.7 x 8.
2. 65 by 5.
8.
3408 by
12.
14. 31.09 x 23.
3. 253 by 9.
9.
597 by :
LI.
15. 1.25 x 7.
4. 197 by 8.
10.
1.6 x 4.
16. 3.72 x 9.
5. 384 by 7.
11.
12.56 x
17.
17. 92.74 x 37.
6. 909 by 6.
12.
142857 by 7.
18. 7.314 x 84.
19. 12345679 by {
d.
39.
125 by 47.
20. 25 by 20.
40.
384 by 65.
21. 27 by 30.
41.
908 by 73.
22. 36 by 70.
42.
18 by 12345679.
23. 318 by 50.
43,
63 by 12345679.
24. 527 by 60.
44.
12.34 x 2.4.
25. 894 by 80.
45.
38.24 x 3.9.
26. 125 by 700.
46.
.1729 x .24.
27. 389 by 600.
47.
.3462 x .75.
28. 239 by 900.
48.
3.4165 x 3.57.
29. 21670 by 4000.
49.
.2675 x 3.85.
30. 5790 by 8000.
50.
697 by 123.
31. 6175 by 8000.
51.
587 by 358.
32. 821400 by 5000.
52.
399 by 586.
33. 25 by 25.
53.
2809 by 702.
34. 27 by 39.
54.
1973 by 904.
35. 79 by 97.
55.
3097 by 807.
36. 38 by 56.
56.
.0827 x .2413.
37. 79 by 87.
57.
.0237 x .5214.
38. 98 by 39.
g
58.
3156 by 2065.
36
MULTIPLICATION.
[Chap. II.
59. .13579 x .0246.
60. 7802 by 2005.
61. 2.31575 x 4.0824.
62. 325.1 by 35.79.
63. .021628 x .002828.
64. 13579 by 21695.
65. .01 x .01 x .01.
66. .5 x .05 x .005.
67. 6 x .6 x .06.
68. 2.5 x .25 x .025.
69. 3109.72 by 90.706.
70. 823156 by 753698.
71. 826075 by 1509607.
72. 8257.314 by 78167094.
73. 2178 by 5506.
74. 3008 by 2345.
75. 687.4 by .2468.
76. 12837 by 56294.
77. 305.009 by 72809.
78. 123.456 by 65.4321.
79. 5009826 by 7090.068.
80. 21840376 by 9287915.
81. 3.9017 x .215.
82. .00167 x .0589.
83. 2.1046 x 4.0035.
84. .21089 x .003904.
85. .1 x .1 X .1.
86. .31 x .41 x .51.
87. 20 by 125.
88. 50 by 350.
89. 800 by 125.
90. 20 by 315.
91. 400 by 125.
92. 600 by 8745.
93. 8000 by 1250.
94. 12000 by 28971.
Find the continued products of
95. 12, 18, and 15. 98. 10, 11, 12, 13, and 14.
96. 17, 18, and 19. 99. 2, .6, 73, and 5.
97. 3, 4, 5, 6, 7, 8, and 9. 100. .7, .3, .006, and 1000.
101. Find 3 x 7 x 9 x 11 X 13 X 37.
Find the squares of
102. 170. 103. 220. 104
360.
Find
106. 125 2 .
107. 5.37 2 .
108. 4690 2 .
109. 64700 2 .
105. 430.
110. 17495 2 .
111. 215.729 2 .
Arts. 53, 54.] NUMBERS IN PARTS. 37
112. 80 3 . 116. 425 3 . 120. .6 3 .
113. 160 3 . 117. 1608 3 . 121. .76 3 .
114. 800 3 . 118. 3507 3 . 122. .006 3 .
115. 1600 3 . 119. 16730 3 .
123. There are 2240 pounds in one long ton. How
many pounds are there in 517 long tons ?
124. There are 168 hours in one week. How many
hours are there in 506 weeks ?
125. There are 24 sheets in a quire of paper, and 20
quires in a ream. How many sheets are there in 524
reams ?
126. There are 86400 seconds in a day. How many
are there in 365 days ?
127. In an orchard there are 57 rows of gooseberry
bushes, and there are 256 bushes in each row. How
many bushes are there altogether ?
128. A book has 312 pages, on each page there are 32
lines, and in each line there are 42 letters. How many
letters are there altogether ?
53. It is often convenient to express a number in parts,
connected by the sign -f ; thus, we may write 8 -f 4, or
11 + 1, or 7 + 3 + 2, instead of 12.
54. A number expressed in parts may be multiplied just
as if expressed as a whole ; thus,
~ l ~ m Here we have 6 units plus 5 units to be multiplied by
4 2 _|_ 35 7 J the answer has the same value as 11 x 7.
Note. A parenthesis may be used to indicate that the several
parts compose one number.
94- 6
38 MULTIPLICATION. [Chap. II.
Again,
9x4+3x8 + 10 Here we have (9 x 4) units plus (3 x 8)
4 units plus 10 units, all to be multiplied by 4 ;
144 + 96 + 40 the answer has the same value as 70 x 4.
3 + 2 Again, to multiply 5 by 5, or to find 5' 2 , we may
3 + 2 multiply 3 units plus 2 units first by 2 and then by 3
6 + 4 and add the partial products ; the answer has the same
value as 5 x 5. It is evident that we may easily square
9 + 12 + 4 an y num ber between 10 and 100 after separating it
into two parts, — its tens and its units. For example, 32 2 = the
square of 32 after being expressed as the sum of its tens and its
units.
30 + 2
30 + 2
60 + 4
900+ 60
900 + 120 + 4
EXAMPLES X.
Written Exercises.
1. Multiply (9 + 8) by 12 ; (6 + 11) by 8.
2. Multiply' (4 x 3 + 6) by 7 ; (9 x 2 +• 8 x 3) by 2.
3. Multiply (6 + 4 + 2) by 3; (80 + 4 x 8) by 4.
4. Find 13* ; 21 2 ; 47 2 ; 94 2 ; 69 2 .
55. The following methods are practical and are of
great value in saving time. Pupils should become pro-
ficient in performing examples by these methods, and
should use them constantly.
I. Multiplication table for numbers between 12 and 20.
Multiply the units of multiplicand and multiplier, and
write the unit figure of the product ; then add the tens
figure (if any) of the product, the multiplier, and the
units of the multiplicand. ,
Art. 55.] SHORT PROCESSES. 39
Ex. 1.
16 Mental Work.
6 x 3 = 18 ; write 8, then 1 + 13 + 6 = 20 ; write 20.
13
208
Ex.2.
13 Mental Work.
-JL 3 . 3x3 = 9: write 9, then 13 + 3 = 16 ; write 16.
16.9
II. To multiply by 11.
Write the unit figure of the multiplicand; then add
units and tens, tens and hundreds, etc., separately, writ-
ing the right-hand figures of the several sums and carry-
ing the left-hand figures ; finally, write the last figure of
the multiplicand after adding what was carried.
7
Ex.1. 469047 7 + 4 = 11
11 1 (carried) +4= 5
5159517 9 + lZll
1 + 6 + 4 = 11
1 + 4= 5
The last figures in. the column are the ones to be written in the
product.
Ex. 2. 89006.037 Ex. 3. 49.8769
n .11
979066.407 5.486459
Mental Work. Mental Work.
7, 10, 4, 6, 6, 0, 9, 17, 9. 9, 15, 14, 16, 18, 14, 5.
III. To square a number of two figures and ending in 5.
Square the units and write the whole product; then
square the tens, add the tens to this square, and write the
sum.
Ex. 1. 35 Ex. 2. 6.5 2 = 42.25.
35 For 5 2 = 25
1225 an d 62+6=42.
Here 5 x 5 = 25
and 3 2 + 3 = 12,
40 MULTIPLICATION. [Chap. II.
EXAMPLES XI.
Multiply
1. 14 by 18. 6. 19 by 19.
2. 19 by 13. 7. 1.8 by 18.
3. 1.6 by 1.3. 8. .16 by .16.
4. .18 by 1.5. 9. 6 by 3 by 14.
5. 1.3 by .13. 10. 2 by 8 by 19.
Multiply the following by 11 :
11. 26751. 15. 64.9786. 19. .463.
12. 498.67. 16. 800960. 20. 590001.
13. 94600. 17. 493.006. 21. 9000095.
14. 888. 18. 2.76398. 22. 399678.
Find the following :
23. 55 2 ; 5.5 2 ; .55 2 . 26. 65 2 ; 7.5 2 ; 85 2 .
24. 15 2 ; .15 2 ; 150 2 . 27. 25 2 ; .25 2 ; 250 2 .
25. 45 2 ; 4.5 2 ; .045 2 . 28. 35 2 ; 350 2 ; 3500 2 .
This method may be used in finding the squares of 105,
115, and 125.
After a little practice, much of the above work may be done
without writing anything but the results.
Division.
56. A short process of finding out how many equal
numbers may be together subtracted from another number
is called Division.
For example, to divide 12 by 4 is to find out how many fours
may be together subtracted from 12, — to find out how many fours
there are in 12. The simplest method of finding the required
number of fours is to subtract 4 from 12, and then 4 from the
remainder, and so on, as many times as possible. It will be found
that there is no remainder after subtracting 3 fours. Hence there
are 3 fours in 12.
Arts. 56-59.] DIVISION. 41
57. The number which is to be thus diminished is
called the Dividend.
One of the equal numbers to be subtracted is called
the Divisor.
The result of division is called the Quotient.
Ex. 1. Divide 24 by 8. Ex. 2. Divide 30 by 10.
24 is the dividend. 30 is the dividend.
8 is the divisor. 10 is the divisor.
3 is the quotient. 3 is the quotient.
58. Since the dividend contains the divisor as many
times as there are units in the quotient, the dividend is
equal to the product of the divisor and the quotient.
We may say then that division is the process by which
one factor may be found when the product and the other
factor (or factors) are given.
Thus, division is the inverse or undoing of multiplica-
tion, just as subtraction is the undoing of addition.
59. Division can be looked upon from two different
points of view, the distinction between which is best
seen by taking as an example the division of a concrete
number.
We have 5 feet x 7 = 35 feet ; and in connection with
the undoing of this multiplication, there are the two
distinct questions:
(1) How many times is 5 feet contained in 35 feet ?
The answer to which is 7 times.
(2) If 35 feet be divided into 7 equal parts, what will
will each part be ? Or, what length is contained 7 times
in 35 feet ?
The answer to which is 5 feet.
Thus, in division, either the divisor is an abstract
number and the quotient a quantity of the same kind as
42 DIVISION. [Chap. II.
the dividend, or else the divisor is a quantity of the same
nature as the dividend, and the quotient is an abstract
number.
60. Division is indicated by the sign ^-, which is read,
' divided by,' or, 'by.'
Thus, 24 h- 4 is read 24 divided by 4, and denotes that 24 is to be
divided by 4 ; also, 24 -+■ 4 -4- 3 denotes that 24 is to be divided by
4 and the result divided by 3, and 24 + 4 x 3 denotes that 24 is to
be divided by 4 and the result multiplied by 3.
61. Inexact Division. If we try to divide 14 by 4, we
find that after subtracting 3 fours there are 2 units left.
The number left over is called the Remainder.
One number is said to be exactly divisible by another
when it is divisible without remainder.
62. It follows from the definition of division that the
product of the divisor and the quotient plus the remainder
is equal to the dividend; that is,
Divisor x Quotient + Remainder = Dividend.
Hence, if any three of these four numbers be given, the
remaining one can be found.
Ex. 1. The divisor is 5, the quotient is 20, and the remainder
is 2. What is the dividend?
The dividend must exceed the product of the divisor and quotient
by 2. Hence,
Dividend = 5 x 20 + 2 = 102.
*
Ex. 2. The dividend is 59, the quotient 7, and the remainder 3.
What is the divisor?
The dividend must exceed the product of the quotient and divisor
by 3. Hence, the product of the quotient and divisor is 59 - 3 = 56,
and the divisor = 56 -*- 7 = 8.
63. Division could always be performed by successive
subtractions of the divisor, as in Art. 56 ; but, except in
Arts. 60-64.] DIVISION BY 10, ETC. 43
the case of very small numbers, the process would be
extremely tedious, and the necessity for these successive
subtractions is obviated by a knowledge of the results of
multiplication.
For example, to divide 75 by 9.
Since we know that 8 nines are 72, and that 9 nines are 81, we
see that 75 -f- 9 gives 8 for quotient and 3 for remainder.
EXAMPLES XII.
Oral Exercises.
Give the quotient in each of the following cases, and
the remainder whenever the division is not exact :
1. 12-4. 11. 64-8. 21. 80-9.
2. 18-9. 12. 45-9. 22. 55-9.
3. 35-7. 13. 15-4. 23. 53-7.
4. 56-1-8. 14. 17^5. 24. 48-5.
5. 60-10. 15. 18-7. 25. 92-9.
6. 49-7. 16. 17-3. 26. 87-8.
7. 81-9. 17. 37 -h 9. 27. 80-7.
8. 72-8. 18. 43 -=-5. 28. 63-5.
9. 56-^-7. 19. 68-7. 29. 70-6.
10. 36-^-6. 20. 70-8. 30. 100-9.
64. Division by 10, 100, etc. To divide any number
by 10, it is necessary only to move the decimal point one
place to the left. For this divides each of the parts of
the number by 10.
For example, 623.4 (see ex., Art. 48) is divided by 10 when its
6 hundreds are made 6 tens, its 2 tens are made 2 units, its 3 units
are made 3 tenths, and its 4 tenths are made 4 hundredths ; i.e.,
623.4 - 10 = 62.34.
Also, 62.34 ■*■ 10 = 6.234 ; 6.234 - 10 = .6234.
44 DIVISION. [Chap. II.
Dividing by 10 and by 10 again is the same as divid-
ing by 100, and it will be noticed that in dividing 623.4
by 100 the decimal point is moved two places to the left.
Hence, to divide by 10, 100, 1000, etc., move the decimal
point as many places to the left as there are naughts in the
divisor.
Examples. 1. 268706 -=-10 = 26870.6.
2. 46000 * 100 = 460.
3. 26783 - 1000 = 26.783.
65. Short Division. — When the divisor is not greater
than 12, the process of division can be written in a
very compact form. The method will be seen from the
following example :
Ex. Divide 43251 by 8.
The operation is set down in the following form :
8 ) 43251
5406, remainder 3.
Explanation. First, 43 -=- 8 gives quotient 5 and remainder 3 ;
we pat 5 under the 3 of the dividend, as the 5 represents units of
the same order as the 3 (namely, thousands', in the present case).
Then, the remainder 3 is equal to 30 units of the next lower order,
and taking into account the next figure of the dividend, namely 2,
we have 32 which when divided by 8 gives quotient 4 and
remainder ; we put down 4 next to 5, and have nothing to ' carry.'
Then, 5 -=- 8 gives quotient and remainder 5 ; we put down next
to 4 and 'carry' 5. The 5 carried and 1, the next figure of the
dividend, make 51 which when divided by 8 gives quotient 6 and
remainder 3. Thus, the complete quotient is 5406 with remainder 3.
EXAMPLES XIII.
Written Exercises.
Divide
1. 92 by 4. 3. 75 by 5. 5. 7.85 by 5.
2. 87 by 3. 4. 234 by 6. 6. 91.8 by 9.
Arts. 65, 66.] LONG DIVISION. 45
7. 72.15 by 5. 11. 7568 by 11. 15. 823507 by 8.
8. 6.402 by 6. 12. 35.628 by 12. 16. 2104316 by 6.
9. .3564 by 9. 13. 72156 by 9. 17. 123456 by 7.
10. 6822 by 12. 14. 346089 by 7. 18. 987654 by 9.
19. 563753696 by 11. 20. 1374819756 by 12.
Divide without uniting the terms of the dividend
21. (8 + 14 + 6) by 2.
22. (6 x 2 + 15x5) by 3.
23. (14 x 3 + 21x5) by 7.
24. (18 x 4 + 33 x 12) by 3, and the result by 2.
66. Long Division. — When the divisor is greater than
12 the process of division is written in a long form so
that the mind will not become confused.
Ex. 1.
Divide 1026 by 18.
18)1026(57
The full operation may be
thus expressed :
90
126
126
18)1026(50 + 7
900
126
126
First, beginning at the left, we use the smallest part of the number
that can be divided by 18. Now, neither 1 nor 10 can be divided
by 18, but 102 can be. 102 -t- 18 = 6, with a remainder of 12. The
quotient 5 is of the same order as the last figure of the dividend
used in the first division (just as in short division). The remainder
12 we reduce to units of the next lower order and add the 6 of that
order, and we have 126 to be divided by 18. Now 126 -*- 18 = 7.
Ex. 2. Divide 102739 by 29.
29)102739(3542
87
157 The last figure of the dividend used in
145 the fi rs t division is 2, and in thousands'
J^3 place. Therefore the first quotient figure
obtained is thousands'.
79
58
21 remainder.
46 DIVISION. [Chap. II.
Ex. 3. Divide 44393 by 145.
In this example, the first remainder (8
i l hundreds) reduced to tens and the 9 tens
aqz. added makes 89 tens, which does not con-
g^ tain 145. Therefore there are no tens in
870 the answer and we write a naught, and
~23 remainder, proceed by reducing the 89 tens to units,
adding 3 units.
The same reasoning applies for a deci-
' ' 1 — — mal dividend as for an integral dividend.
The first figure obtained in the quotient is
of the same order as the last figure used in the first division.
This fact determines the position of the decimal point.
Ex. 5. Ex. 6. Divide .019 by 125.
9). 288
~032 125).019000(.000152
125
650
625
250
250
Ex. 7.
48)5.220(.108
— - The remainder is the same in name as the last
3 81 figure of the dividend. In this case it is .036.
"36
Ex. 8.
15)474000(31600 In this example, it is unnecessary to extend
45 the written work beyond dividing 90 hun-
^^ dreds by 15. Since, however, every order of
the dividend must have a corresponding figure
in the quotient, we write naughts in tens' and
units' places.
90
90
67. To divide, when the divisor is partly or wholly a
decimal.
Art. 67.] DIVISION BY A DECIMAL. 47
Here we make use of the following principle :
Multiplying both dividend and divisor by the same num-
ber does not change the quotient.
Thus, 24 -=- 4 = 6, and if both 24 and 4 be multiplied by 2, we
shall have 48 + 8 = 6 ; also, 3.6 + .6 = 6, and if both 3.6 and .6
be multiplied by 10, we shall have 36 -s- 6 = 6.
Hence, to divide any number by a decimal, we first
multiply both dividend and divisor by that power of 10
which will make the divisor a whole number, and then pro-
ceed as in the case of division by a whole number. We
perform these multiplications by moving the decimal
points.
Ex. 1. Divide 11.68 by 1.6
Move the decimal points one place to the right. Then
1.6J11.68(7.3
112
48
48
Do not forget that the first significant figure of the quo-
tient is of the same order as the last of those figures of the
dividend which are used in the first division. This will
indicate the position of the decimal point in the quotient.
Ex. 2. Divide .21 by .0125 Ex. 3. Divide .0697585 by 1.33.
.0125 ).2lJ^B.8 1.33J.06 97585C.05245
125 '
850
750
100
100
6J35
325
266
598
532
665
665
Note. Always let the old decimal point remain, and indicate
the new one by a mark similar to those in Ex. 1.
48 DIVISION. [Chap. II.
68. Note. It should be noticed that, although the quotient is
unchanged by multiplying both dividend and divisor by the same
number, the remainder, if any, is not unchanged, but is equal to
the original remainder multiplied by the number by which the
original divisor and dividend were multiplied.
For example, 26 -=- 6 = 4, with a remainder of 2 ; and 8 times 26
divided by 8 times 6 equals 4, with a remainder of 8 times 2.
Therefore, we must divide the remainder by the multiplier, if we
wish the remainder obtained by using the original numbers, as the
remainder is the part of dividend not used.
Ex. 1. Divide 17.8 by 1.4.
1.4J17.8 v (12
14 The remainder would have been 1
3 8 unit if we had not multiplied by 10.
28
Remainder 1 units.
Ex. 2. How many pieces each 1.02 inches long can be cut from
a rod ichose length is 18 inches f
We can find the quotient by dividing 1800 by 102. Thus
102)1800(17
102
780
714
66
Hence there are 17 pieces ; and since the original divisor and
dividend were multiplied by 100, the remainder left over is
(6Q ■*■ 100) inches = .66 inches.
69. Division by Factors. — We have seen that to multi-
ply by two or more numbers in succession gives the same
result as to multiply at once by their product. It there-
fore follows, conversely, that to divide by two or more
numbers in succession gives the same result as to divide
at once by the product of the numbers.
Arts. 68-70.] BY FACTORS. 49
Ex. 1. Divide 11445 by 35,
Since 35 = 7 x 5, we may divide by 7 and 5 in succession.
7 )11445
5 )1635
327
Ex. 2. To divide 315637 by 20.
20 )31563J
15781, remainder 17.
Dividing both dividend and divisor by 10, as indicated, we have
31563 to be divided by 2. The quotient is 15731 and the remainder
1, which must be multiplied by 10 and the figure cut off by the
decimal point annexed, making 17 as the true remainder.
70. When one number is divided by several others in
succession, the method of finding the remainder will be
seen from the following example :
Ex. 1. Divide 11467 by 35.
7 )11467
5) 1638 sevens and 1 unit over.
327 thirty-fives and 3 sevens over.
The whole remainder is therefore 3 sevens and 1 unit, that is, 22.
From the above it will be seen that the whole remain-
der is found by multiplying the remainder after the second
division by the first divisor and then adding the remainder
after the first division.
Ex. 2. Divide 251633 by 8 x 6 X 7.
3 )251633
5 )83877 groups of 3 each and 2 units over.
7)16775 groups of 3 x 5 each and 2 groups of 3 each over.
2396 groups of 3 x 5 x 7 each and 3 groups of 3 x 5 each over.
The whole remainder is therefore 3 groups of 3 x 5 each + 2
groups of 3 each +2 = 3x3x5 + 2x3 + 2 = 45 + 6 + 2 = 53.
Thus, if there are more than two successive divisions the whole
remainder is found by multiplying each remainder by all the di-
visors preceding that from which the remainder arises, and then
adding these results to the first remainder.
50 DIVISION. [Chap. II.
71. The work of finding some products may be short-
ened by making use of multiplication and division at the
same time.
Ex. 1. Multiply 6174 by 25.
Since 25 = 100 h- 4, we shall multiply by 25 if we first multiply
by 100 and then divide by 4. For by multiplying by 100 we get 4
times too much, which is put right when we divide by 4. To mul-
tiply by 25 we may therefore affix two naughts and divide by 4 ;
thus,
4 )617400
154350
Ex. 2. Multiply 6174 by 125.
Since 125 = 1000 + 8, we multiply by 1000 and then divide by 8,
that is, we affix three naughts and divide by 8 ; thus,
8 )6174000
771750
The methods adopted in the following examples are also worth
notice.
Ex. 3. Multiply 7964 by 9998.
Since 9998 = 10000 - 2, we can multiply by 10000 and by 2, and
take the difference of these products.
7964
9998
79640000
15928
79624072
Ex. 4. Multiply 7.964 by 9998.
7.964
79640.
15.928
79624.072
72. To test the answer in division, multiply the quo-
tient by the divisor (not divisor by quotient), and to the
product add the remainder (if any) ; the result should
equal the dividend. [Art. 58.]
Arts. 71-73.]
EXAMPLES.
51
73. Some saving of time in division will be effected by
performing the multiplication of the divisor and the sub-
traction from the dividend simultaneously ; this method
should, however, be attempted only by those who show
some aptitude for numerical calculations, for the slight
gain in speed by no means makes up for the increased
liability to error.
The method will be understood from the following
example :
Divide 102739 by 29.
29)102739(3542
157
123
79
21 rem.
Explanation. Instead of multiplying 29 by 3 and subtracting
the whole product from 102, we subtract the several figures of the
product as we go along. Thus, 3 times 9 are 27, and 7 from 12
leaves 5 ; we write 5, and carry 3 (2 from the 27, and 1 from the
12). Then, 3 times 2 are 6, and 3 (carried) are 9, and 9 from 10
leaves 1. The remainder is 15, which with the 7 of the dividend
makes 157 for the next partial dividend. And so on to the end.
EXAMPLES XIV.
Written Exercises.
Divide
1. 182 by 13. 4. 399 by 19. 7. 702 by 26.
2. 204 by 17. 5. 575 by 23. 8. 1054 by 34.
3. 221 by 17. 6. 899 by 29. 9. 4185 by 31.
10. 1591 by 37. 14. 430686 by 71.
11. 6016 by 94. 15. 415242 by 59.
12. 710007 by 87. 16. 426713 by 47.
13. 435435 by 65. 17. 562171 by 53.
52 DIVISION. [Chap. II.
18. 850902 by 78. 26. 21112 by 104.
19. 1173021 by 97. 27. 185745 by 305.
20. 1034550 by 95. 28. 801738 by 567.
21. 2706420 by 86. 29. 8035370 by 2674.
22. 11336 by 109. 30. 9570744 by 1593.
23. 22563 by 207. 31. 407514744 by 6724.
24. 160335 by 315. 32. 31587678 by 5067.
25. -39483 by 123. 33. 266 + 126 + 210 by 14.
34. 6164 + 5226 by 67.
35. The trees in an orchard are arranged in 153 rows,
with the same number of trees in each row, and there
are 16371 trees altogether. How .many trees are there
in each row ?
36. There are 86400 seconds in a day; in how many
days are there 13564800 seconds ?
b.
In division of decimals, the quotient should be continued until
there is no remainder, unless otherwise directed. This can be
accomplished by annexing naughts to the dividend, as in Ex. 6,
Art. 66. In general practice three or four decimal places in the
quotient are considered sufficient.
Divide
1. 16.4 by 2. 3. 17.2 by 4. 5. .288 by 9.
2. 32.7 by 3. 4. .156 by 6. 6. .135 by 9.
7. 125.6 by 20. 13. 5.22 by 48.
8. 31.83 by 30. 14. .171 by 72.
9. 11.7215 by 50. 15. .012 by 1600.
10. 215.4 by 80. 16. .027 by 45.
11. .0321 by 60. 17. 2.355 by 75.
12. .174 by 120. 18. 2.715 by 48.
Art. 73.]
EXAMPLES.
53
19. 52.7 by 17.
20. 43.7 by 23.
21. 166.6 by 119.
22. 3751.5 by 123.
23. 3.7515 by 1230.
24. 375.15 by 125.
25. 37.515 by 1250.
Find, to 4 places of decimals,
32. 12.15 -r- 148. 35.
33. 2.374 -=-156. 36.
34. 41.75-89. 37. 121 -f- 170.
38. Simplify .026 x .0493 -r- 221.
26. 67.77 by 135.
27. .006777 by 1350.
28. 1.036656 by 207.
29. .1036656 by 5008.
30. .001036656 by 2070.
31. .651714 by 3156.
135 -=- 17.
17 -=- 135.
Divide
1. 6.2 by .01.
2. .347 by .001.
3. 12.3 by .0001.
4. 3.5 by .5.
5. .75 by .05.
6. 1.25 by .005.
7. 62.5 by 2.5.
8. .625 by .025.
9. 625 by .0025.
10. 1.1 by .125.
11. .019 by 1.25.
12. 170 by .00125.
25. Find, to 4 places
.0167 -*- 3.17.
13. 1.5 by 2.4.
14. 5.76 by 4.8.
15. 8.1 by .36.
16. 159.1 by 3.7.
17. 6.016 by .94.
18. 70.992 by 8.7.
19 .435435 by .0065.
20. 430.686 by .0071.
21. 415.242 by .0059.
22. .185745 by 3.05.
23. 4.07514744 by .006724.
24. .9570744 by 159.3.
of decimals, 43.21 -f- 123.4, and
54 DIVISION. [Chap. II.
26. Simplify 360 -- 7.2 -=- .16.
27. Simplify .0441 -- .21 -h .56.
28. Simplify 1.953 -- 8.68 x .035.
29. How many lengths eacli 2.56 inches are there in a
rod 120 inches long ; and how much is left over ?
30. How many packets of tea, each containing 1.85
ounces, can be made up out of a chest containing 2400
ounces ; and how much is left over ?
d.
Divide, using factors not greater than 12,
1. 396 by 18. 3. 625 by 25. 5. 8820 by 36.
2. 816 by 24. 4. 3753 by 27 6. 15750 by 42.
7. 1958528 by 64. 18. 21574 by 20, 40, and 60.
8. 59081805 by 81. 19. 123456 by 20, 30, and 40.
9. 13339728 by 108. 20. 158937 by 20, 50, and 70.
10. 10654069140 by 132. 21. 2167 by 30, and 50.
11. 316794 by 45. 22. 16819 by 30, and 80.
12. 7196243 by 35. 23, 17943 by 40, and 60.
13. 2106935 by 36. 24. 21985 by 50, and 90.
14. 9172143 by 72. 25. 217943 by 500.
15. 22222222 by 99. 26. 712415 by 700.
16. 123456789 by 132. 27. 217643 by 216.
17. 32163 by 20, 30, and 40. 28. 1234567 by 242.
e.
Multiply, using the short process,
1. 74562 by 25. 4. 387.4 by 125.
2. 4.162 by 25. 5. 79.624 by 99.
3. 12678 by 125. 6. 1897 by 999.
Art. 73.] MISCELLANEOUS EXAMPLES. 55
7. 29075 by 998. 10. .6003 by 12.5.
8. .79184 by 9999. 11. 786 by 250.
9. 6729 by 12.5. 12. 34.65 by .0125.
EXAMPLES XV.
Miscellaneous Examples, Chapters I and II.
1. Express in words 3015602, and in figures eleven
million five hundred thousand two hundred fourteen.
2. Find the sum of 30157, 12.468, 31947, and 3.6539.
3. By how many is 13018 greater than 12997?
4. Multiply 8000 by 1250, and 3200 by 12345.
5. How many times can 317 be subtracted from 1389,
and what is the remainder ?
6. Express MDCCCLXXIX in the Arabic notation,
and 1449 by means of Roman numerals.
7. Find 1325 + 3016 +79 + 90167.
8. Find 316 - 179 + 257 - 89 - 185 + 398 - 485.
9. Multiply 1234 by 4321 and 9009 by 31562.
10. How many nineteens are there in five thousand,
and how many are over ?
11. By how much does the sum of 3.72 and 10.015 fall
short of the sum of 7.216 and 6.52 ?
12. Express in words 1632057 and 300416720J 500.
13. Subtract the sum of 3158, 2016, and 5143 from
mil.
14. Multiply the difference between seventy-six mil-
lion seventy-six and four hundred forty thousand four
hundred forty, by eleven hundred fourteen.
56 MISCELLANEOUS EXAMPLES. [Chap. II.
15. A farmer has 197 sheep and three times as many
lambs. How many sheep and lambs has he altogether ?
16. Find by short divisions how many thirty-fives
there are in 31578, and how many are over.
17. Add 31.057, 156.0083, 2.61759, and .008347.
18. Subtract the difference between 3.14 and 1.0625
from the sum of 1.00172 and 2.127.
19. By how many is one million eight thousand nine
hundred seventy-four less than two million eleven hun-
dred twelve ?
20. Find 3142 - 1250 - 989 + 6217 - 3587 - 1924.
21. A farmer had 2000 bags of wheat. He sold 527
bags to one man and 255 bags to each of three others.
How many bags were left unsold ?
22. How many letters are there in a book of 375
pages, each page of which contains 32 lines, and each line
45 letters?
23. Multiply 31.025 by .032, and .0625 by .00125.
24. By what number must 59755 be divided in order
that the quotient may be 19 ?
25. Divide 7.0175 by 17.5, and 7.5 by .00625.
26. In one school there are one hundred seventy-six
boys and one hundred and twelve girls ; and in another
school there are half as many boys and twice as many
girls. How many scholars altogether are there in the
two schools?
27. The sum of two numbers is 317205 and one of
them is 185964 ; what is the other ?
Art. 73.] MISCELLANEOUS EXAMPLES. 57
28. A farmer sold 75 cattle at 24 dollars a head and
bought with the money sheep at 2 dollars each. How
many sheep did he buy ?
29. Divide .04312 by .0044, and 9.0225 by .225.
30. Divide 358 by 15 by short divisions.
31. What is the least number which must be added to
57914 in order that the sum may be exactly divisible by
315?
32. Divide the product of 37.5 and .1248 by .005625.
33. Express MDCCCXCIY in the Arabic notation, and
2875 by means of Roman numerals.
34. In a school of four hundred and ninety children
there are two hundred and seventy-six girls. How many
more girls than boys are there ?
35. In a train there are 37 cars each having seats for
36 people, and there are 375 passengers in the train ; how
many seats are empty?
36. Simplify 1.702 x 2.9015 -h .0005803.
37. Divide the product of .0374 and .0075 by the dif-
ference between .675 and .6375.
38. Show that the sum of the squares of three thousand
nine, and four thousand twelve, is equal to the square of
five thousand fifteen.
39. What is the least number which must be subtracted
from 2146537 in order that the remainder may be exactly
divisible by 4275 ?
40. Subtract nine hundred five million eight thousand
nine hundred sixty-five from eleven hundred million two
thousand three hundred, and express the result in words.
58 MISCELLANEOUS EXAMPLES. [Chaps. II. , III.
41. At an election, the successful candidate, who ob-
tained 12597 votes, had a majority of 1479 over the un-
successful candidate. How many votes were given
altogether ?
42. Find 2197-1982 + 374 + 10085-8216 + 11597
- 7986.
43. Find the squares of 2.15 and .0324.
44. Multiply 16777216 by 131072, also divide 16777216
by 131072, and express the results in words.
45. Find the least number of repetitions of 3745 whose
sum is greater than a million.
46. Divide .378 by 262.5, and 37.8 by .02625.
47. Express the numbers 29, 47, 158, 679, 1464, and
10385 by means of Roman numerals.
48. How many figures are there in all the numbers
from 1 to 100? How many in the numbers from 1 to
1000?
49. A certain number when divided by 3008 gives a
quotient 3875 and a remainder 2794. What is the-
number ?
50. Divide 999999 by the continued product of 3, 7,
11, and 13.
51. The sum of two numbers is 315642, and one of the
numbers is twice the other : find them.
52. Divide 2722.05 by .345, and .0272205 by 3.45.
53. Divide (144.4 + 152x4.6) by 19; prove your
answer by dividing after uniting the terms of the
dividend.
54. Divide, by factors, (6.3 x 6 + 4.9 x 18) by 21.
See Art. 44 for definition of factors.
Arts. 74, 75.] FACTORS. 59
CHAPTER III.
FACTORS AND MULTIPLES — SQUARE ROOT — HIGHEST
COMMON FACTOR — LEAST COMMON MULTIPLE.
Factors.
74. An exact divisor of a number is called a Factor
of that number ; thus,
2, 3, 4, 6, and 12 are factors of 24.- [Art. 44.]
A factor is also called a Measure.
A number that is exactly divisible by another number
is called a Multiple of that number ; thus,
12, 30, 54, 72, and 90 are multiples of 6.
It will be seen at once that a number has a limited number of
factors, but an unlimited number of multiples.
75. A number which is not divisible by any number
except itself and 1 is called a Prime Number, or a Prime.
Thus, 2, 3, 5, 7, etc., are primes.
Every number which has other factors beside itself
and unity is called a Composite number.
Thus, 4, 6, 8, 9, etc., are composite numbers.
Two numbers, both of which cannot be divided by the
same number (except unity), are said to be prime to one
another.
Thus, 4 and 9 are prime to one another; both, however, are
composite numbers.
60 FACTORS AND MULTIPLES. [Chap. III.
76. Numbers divisible by 2 are called Even numbers.
Numbers not divisible by 2 are called Odd numbers.
2, 14, 30, and 74 are even numbers.
3, 7, 27, and 51 are odd numbers.
The following simple conditions of divisibility will be
found to be useful :
(i) A number whose last digit expresses an even num-
ber is divisible by 2.
248 and 100694 are divisible by 2.
(ii) A number whose last digit is 5 or is divisible
by 5.
25, 55, and 600 are divisible by 5.
(iii) A number whose last two digits express a num-
ber divisible by 4 or by 25 is divisible by 4 or by 25,
respectively.
67215736 is divisible by 4.
23798675 is divisible by 25.
(iv) A number the sum of whose digits is divisible by
3 or by 9 is divisible by 3 or by 9, respectively.
The sum of the digits of the number 56174154, namely,
5 + 6 + 1 + 7 + 4 + 1+5 + 4, is 33;
and 33 is divisible by 3, but is not divisible by 9. Thus, the
number 56174154 is divisible by 3, but not by 9.
(v) A number is divisible by 11 when the difference
between the sum of the first, third, fifth, etc., digits and
the sum of the second, fourth, sixth, etc., digits is zero
or a multiple of 11, and not otherwise.
Thus, 3572129 is seen to be divisible by 11, since 9 + 1 + 7 + 3
differs from 2 + 2 + 5 by 11.
Arts. 76, 77.] EXAMPLES. 61
EXAMPLES XVI.
Oral Exercises.
Which of the numbers, 2, 4, 8, 3, 9, 5, 25, 125, 11, can
be seen by inspection to be factors of
1. 964. 4. 7326. 7. 94680. 10. 49125.
2. 225. 5. 6975. 8. 29304. 11. 307890.
3. 1925. 6. 4125. 9. 76164. 12. 264792.
77. The following are important general theorems :
I. Every divisor or factor of each of several numbers is
a divisor of their sum.
If, for example, each of several numbers is divisible by 12, then
each can be arranged in groups of twelve, and therefore their sum
consists of a certain number of twelves. Similarly for any other
divisor.
II. Every divisor of a number is a divisor of any mul-
tiple of that number.
If, for example, any number is divisible by 12, it can be arranged
in groups of twelves, and so also can any number of repetitions of
the number.
III. Every divisor of two numbers is a divisor of the
sum, or of the difference, of any multiples of the numbers.
If, for example, two numbers are both divisible by 12, they can
both be arranged in groups of twelves, and so also can any multiples
of either. These multiples can then be added, or one can be taken
from the other, without taking to pieces any of the groups.
To make the above theorems quite clear to a beginner,
it would be well to have actual counters to deal with,
which could be tied up by twelves in bags or bundles.
The pupil would then see that the different additions
and subtractions could be performed without undoing any
62 FACTOKS AND MULTIPLES. [Chap. III.
of the bags or bundles, and therefore the final resnlt must
be a certain number of twelves.
78. The Sieve of Eratosthenes. — The different prime
numbers can be found in order by the following method,
called the Sieve of Eratosthenes.
Write in their natural order the numbers from 1 to any extent
that may be required ; thus,
1, 2, 3, 4, 5, 6, 7, 8, 9, i6,
11, 12, 13, i4, i5, 16, 17, i8, 19, 20,
2i, 22, 23, 24, 25, 26, 27, 28, 29, 30,
31, 32, 33, 34, 35, 36, 37, 38, 39, 40, etc.
Now take the first prime number, 2, and over every second number
from 2 place a dot : we thus mark all the multiples of 2. Then,
leaving 3 unmarked, place a dot over every third number from 3 : we
thus mark all multiples of 3. The number next to 3 left unmarked
is 5 ; and, leaving 5 unmarked, place a dot over every fifth number
from 5 : we thus mark all multiples of 5. And so for multiples
of 7, etc.
By proceeding in this way all multiples of the prime numbers,
2, 3, 5, 7, etc., are struck out ; also multiples of all composite numbers
are necessarily struck out at the same time : for example, all multi-
ples of 6 are struck out as being multiples of either of its prime
factors 2 or 3. Hence all the numbers which are left unmarked are
primes, for no one of them is divisible by any number (except unity)
which is smaller than itself.
We can thus find in order as many prime numbers as we please.
The primes less than 100 will be found to be
1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.
79. To find whether a given number is or is not a prime,
we have only to see whether it is divisible by any one of
the prime numbers, 2, 3, 5, 7, etc.
Arts. 78-80.]
PRIME FACTORS.
63
Ex. 1. Is 233 a prime number f
By trial it will be found that 233 is not divisible by 2, nor by 3,
nor by 5, nor by 7, nor by 11, nor by 13, nor by 17. Now it is not
necessary to try any other primes, for 233 -^-17 gives a quotient less
than 17 ; if, therefore, 233 were divisible by a prime greater than 17,
the quotient would be less than 17, and 233 would be divisible by
this quotient, that is by a number less than 17, which we know is
not the case. Hence 233 is a prime number.
80. Resolution into Prime Factors.
The following examples will suffice to show how to
express any number whatever as the product of factors
each of which is a prime.
The method is applicable to all numbers however large, provided
we find as many prime numbers as may be necessary by means
of the ■ sieve ' ; the method would, however, be extremely tedious
in the case of a very large number.
Ex. 1. Express 28028 as the product of prime factors.
28028 = 2 x 14014
= 2 x 2 x 7007
= 2 x 2 x 7 x 1001
= 2x2x7x7x143
= 2x2x7x7x11x13.
These continuous divisions may
be thus expressed :
28028
14014
7007
1001
143
18
Ex. 2. Find the prime factors of 3978.
2
3978
3
1989
3
663
13
221
17
The answer is 2, 3 2 , 13, and 17.
Ex. 3. Obtain two factors of 14 + 22.
14 + 22 = 2 multiplied by (7 + 11).
64 FACTORS AND MULTIPLES. [Chap. III.
EXAMPLES XVII.
Express the following numbers as products of prime
factors :
Oral Exercises.
1. 6, 9, 10, 15, 24, 30, 36, 39, 45, 48.
2. .6, .9, 1.5, 2.4, 3.6, 3.9, 4.5, 4.8.
3. .09, .15, .24, .36, .39, .45, .48.
4. 49, 50, 54, 60, 5.4, 75, 81.
5. 3.2, 100, 120, 130.
"Written Exercises.
6. 184, 196, 275, 273, 391, 525.
7. 350, 459, 715, 728, 792, 999.
8. 1092,3885.
9. 51051, 74613, 462462.
10. Obtain two factors of (6 + 15).
11. Obtain three factors of (30 + 70).
12. Obtain two factors of (2x6 + 4x5 + 2 x 17).
Square Root.
81. Obtain the two equal factors of 4 ; of 9 ; of 25 ;
of 0.4.
Obtain the three equal factors of 8 ; of 27 ; of .008.
Obtain the four equal factors of 16 ; of 81.
One of the equal factors of a number is called a Root
of the number ; thus, 3 is a root of 9 ; 5 is a root of 25 ;
3 is a root of 27 ; .2 is a root of .04 ; .2 is a root of .008.
If a number is the product of tivo equal factors, its root
is called a second root, or Square Root.
If a number is the product of three equal factors, its
root is called a third root, or Cube Root.
Likewise we have fourth smdjifth roots, etc.
Arts. 81-85.] SQUARE ROOT. 6$
82. It was shown in Art. 52 that a square is obtained
when the multiplicand equals the multiplier.
Here it is seen that a square root is obtained when the
quotient equals the divisor.
83. The squares of the first 12 whole numbers should
be known : they are
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144.
It will be seen at once that the square root of an integer is by
no means always an integer ; in fact the only numbers between 1
and 100 which have an integral square root are 4, 9, 16, 25, 36, 49,
64, and 81.
It will be seen later on that the square root of an integer which
is not the square of a whole number can be found approximately*
only.
An integer (or a decimal) which is the square of
another integer (or decimal) is called a Perfect Square.
Thus, 16 and .09 are perfect squares ; namely, the squares of 4
and .3, respectively.
84. The sign ^/ is used to indicate a root, and is called
the Radical Sign.
If any other root than the second is to be indicated,
a small figure called an Index is placed just above the
radical sign ; thus,
-y/9 indicates the square root of 9 ;
■y/8 indicates the cube root of 8 ;
.J/243 indicates the fifth root of 243.
85. In simple cases, the square root of a given number
can be found by separating it into factors which are
squares, and making use of the principle that the product
of the squares of two or more quantities is equal to the
square of the product of those quantities.
66 FACTORS AND MULTIPLES. [Chap. III.
For example, to find V 324 -
324 = 4 x 81 = 22 x 9 2 = (2 x 9)2 ;
hence, V 324 = V( 2 x 9) 2 = 2 x 9 =18.
Also, V 1 - 44 = V( 22 x - 62 ) = V( 2 x .6)2 = 2 x .6 = 1.2.
EXAMPLES XVIII.
Written Exercises.
Find the square roots of the following numbers :
1. 196; 1.96. 6. 576; 5.76. 11. 2601.
2. 225; 2.25. 7. 676; 6.76. 12. 3969; .003969.
3. 324. 8. 1089;. 1089. 13. 4225; 42.25.
4. 400; 4.84. 9. 1225; 12.25. 14. 7056.
5. 441; 4.41. 10. 2025. 15. 11025.
In each of the following numbers, what is the least
multiplier that will produce a perfect square ?
16. 12. 18. 24. 20. 126. 22. 1176.
17. 20. 19. 52. 21. 140. 23. 1344.
24. State a number which has a second and a fourth
root ; a second, third, and sixth root.
86. The above method cannot be easily used in all
cases, but the method which can be used will be under-
stood from the following explanation. [Arts. 86, 87, 88.]
Let it be required to find 63 2 . This may be done in
the usual way, and the square is found to be 3969.
Now 63 2 may be written (60 + 3) 2 , which equals the square of
60 + twice the product of 60 by 3 + the square of 3.
60 + 3
60 + 3
60 x 3 + 3 2
60 2 + 60 x 3
60 2 + 2(60x3)+3 2
Arts. 86-88.] SQUARE ROOT. 67
The square of the sum of any other pair of numbers
can be expressed in a similar form.
Hence, the square of the sum of any two numbers is equal
to the sum of their squares plus twice their product.
87. Since
.01 2 = .0001, 10 2 = 100,
.l 2 = .01, 100 2 = 10000,
l 2 . = 1, 1000 2 = 1000000,
and so on, it follows that if a number has one digit, its
square has either one or two digits ; if a number has two
digits, its square has either three or four digits; if a
number has three digits, its square has either five or six
digits ; and so on.
Hence, if we mark off the digits of a given number,
beginning at the units' digit, into periods of two, the last
of the periods on the left containing either one or two
digits ; then the number of these periods will be equal to
the number of digits in the square root of the given number.
For example, by pointing off the numbers, 961, 54.76, 36.8449,
1522756, thus, 9'61, 54'.76, 36'.84'49, 1'52'27'56, we see that the square
roots of these numbers contain 2, 2, 3, and 4 figures, respectively.
88. To find the Square Root of Any Number.
The method will be seen from the following examples :
Ex. 1. To find the square root of 3969.
39'69' f 60 + 3 By pointing off the digits into P eriods
36 00 °^ two ' we see tliat tliere are two digits
2 x60+3 — 123Y369" * n tne re( l u i re d root ; and, since 60 2 =
3 69 3600 and 70 2 = 4900, we see that the
root lies between 60 and 70. The tens'
digit must therefore be 6, and we have now to find the units' digit.
If we subtract 60 2 from the given number, the remainder is 369 ;
and, by Art. 86, this remainder is equal to (2 x 60) times units'
digit + (units' digit) 2 , or units' digit times (2 x60 -f units' digit) ;
68 FACTORS AND MULTIPLES. [Chap. III.
i. e. , 369 is the product of the unknown digit by (2 x 60 + the
unknown digit).
Hence, if we use 2 x 60 as a trial divisor, we obtain a quotient,
namely 3, which is either equal to or greater than the required digit.
Put this quotient for the unknown digit, and we have (2 x 60 + 3),
or 123, as a true, or complete, divisor. Now dividing 369 by 123,
we find that 3 is the correct digit for units' place.
39'69'f63 The P rocess is shortened, as in ordinary division,
3g by the omission of zeros ; the periods, of two figures
123)369 each, are brought down one at a time, one figure of
369 the root corresponding to each period.
Ex. 2. Find the square root of 114244.
11'42'44'(300 +30 + 8 11 '42 '44 (338
9 00 00 _9
600 + 30)2 42 44 63)2 42
1 89 00 1 89
660 + 8) 53 44 668) 53 44
53 44 53 44
There are here three periods and therefore three digits in the root,
the first of which is 3, since 114244 is between 300 2 and 400 2 . Using
300 x 2 as a trial divisor in order to find the second figure in the root,
we obtain the quotient 40 ; this, however, is too great, for (600 + 40),
the complete divisor, is not contained 40 times in the dividend ; we
therefore try 30, which proves to be correct.
The process is usually indicated in the shortened form, any
trial divisor being the product of the quotient already found by
2 and 10 continuously, while the corresponding complete divisor is
the trial divisor with its naught displaced by the quotient figure
obtained in using the- trial divisor : thus, in Ex. 2, the first trial
divisor is 3 x 2 x 10 = 60, while the complete divisor is 63 ; also
the second trial divisor is 33 x 2 x 10 = 660, while the complete
divisor is 668.
Ex. 3. Find the square root of 50126400.
Here there are four periods and therefore
50'12'64'00'(7080 f 0U r figures in the root. A figure of the
root corresponds to each period brought
1408)1 12 ^ down in the shortened process ; and in the
^ present case two figures of the root are
naughts.
Arts. 88, 89.] SQUARE ROOT. 69
Ex. 4. Find y/U. 44.
In the case of a decimal, the pointing must be begun at the
decimal point, and carried to the left for the integral part, and to
the right for the decimal part.
14.'44'(3.8
9
68)5 44
5 44
Ex. 5. Find y/SVo.
Having used both periods of the given
3'15.'00'00'(17.74 + number, there is a remainder of 26. We
_J place a decimal point after the units' figure
27)2 15 f both dividend and quotient, and then
m continue the periods by using naughts.
' 94 2Q ^ke process would never terminate, hence
or.A± \ i 7i on 315 is not a perfect square. We obtain,
X 41 76 however, an approximate answer by stop-
29 24 ping after the second or third decimal
place.
89. Since the square of a number cannot end with a naught
unless the number itself ends with a naught, it follows that, if the
process of finding a square root does not terminate when the
last significant figure is brought down, the process will never
terminate.
Expressions such as -^/S, y/2.5, which cannot be found exactly are
called Surds.
Although no definite number can be found whose square is exactly
equal to 3, the process of Art. 88, Ex. 5, if continued far enough,
will enable us to find a decimal whose square differs from 3 by as
small a quantity as we please.
EXAMPLES XIX.
Written Exercises.
Find the square roots of
1. 729. 4. .1849. 7. 16.81.
2. 3481. 5. 2209. 8. 56169.
3. 11.56. 6. 6084. 9. 4157521.
70 FACTORS AND MULTIPLES. [Chap. III.
10. 49126081. 13. 9345249. 16. 13.69.
11. 26625600. 14. 934.5249. 17. 136.9.
12. 182.493081. 15. 1369. 18. 1.369.
19. .00022201. 20. 2.2201.
Find, to three decimal places,
21. V 5 - 24 « V 125 - 4 - 27- V-081.
22. V 19 - 25 - V 31 - 046 - 28 - V- 01735 -
23. V 21 - 5 - 26 - V- 4 - 29 - V- 0002 -
Highest Common Factor.
90. A number which exactly divides two or more num-
bers is called their Common Factor.
For example, 2, 3, and 6 are common factors of 18 and 24.
The largest number which exactly divides two or more
numbers is called their Highest Common Factor (H.C.F.) ;
called, also, the Greatest Common Measure (G.C.M.), and
the Greatest Common Divisor (G.C.D.).
Thus, 6 is the H.C.F. of 18 and 24,
or the G.C.M. of 18 and 24,
or the G.C.D. of 18 and 24.
91. After numbers have been resolved into their prime
factors, their H.C.F. can be found by inspection.
Consider, for example, the numbers 30 and 42.
30 = 2 x 3 x 5
42 — 2 x 3 x 7 Here we see that 2 an( * ' are tlie on ^ y
primes that are divisors of both 30 and 42.
H.C.F. =2x3 Therefore the H.C.F. =2x3 = 6.
= 6.
Again,
720 = 2* x 3 2 x 5
1080 = 2 3 x 3 8 x 5 Here 2 is a common factor three times,
H.C.F. = 2 3 x 3 2 x 5 3 is common twice, and 5 is common once.
= 360.
Arts. 90-92.] HIGHEST COMMON FACTOR. 71
The H.C.F. of two or more numbers must be the
continued product of all the common prime factors of the
numbers.
Further Illustrations.
Ex. 1. Ex. 2.
792 = 2 3 x 3 2 x 11 2730 = 2 x3x5x7xl3
4368 = 2* x 3 x 7 x 13 5304 = 23x3 x 13 x 17
H.C.F. = 2 3 x3 780 = 2 2 x 3 x 5 x 13
= 24. H.C.F.= 2 x3 xl3
= 78.
EXAMPLES XX.
Oral Exercises.
Find the H.C.F. of '
1. 12 and 18. 3. 30 and 42. 5. 60 and 84.
2. 20 and 25. 4. 18 and 30. 6. 54 and 90.
"Written Exercises.
7. 45 and 105. 10. 189 and 273. 13. 693 and 819.
8. 72 and 90. 11. 132 and 252. 14. 792 and 924.
9. 126 and 315. 12. 315 and 357. 15. 891 and 1221.
16. 48, 60, and 72. 18. 264, 360, and 600.
17. 72, 108, and 180. 19. 630, 756, and 1155.
92. We must now show how to find the H.C.F. of two
numbers without going through the troublesome process
of expressing the numbers as the product of prime
factors.
The method depends on the following theorem, proved
in Art. 77 :
Any common factor of two numbers is also a factor of
the sum, or of the difference, of any multiples of the
numbers.
72 FACTORS AND MULTIPLES. [Chap. III.
Suppose that we have two numbers whose H.C.F. is
required.
If we divide the greater number by the smaller, then,
by the nature of division,
(i) the remainder is equal to the difference between
the greater number and some multiple of the smaller ;
(ii) the greater number is equal to the sum of the
remainder and some multiple of the smaller.
From (i) it follows that any common factor of the
original numbers is a factor of the remainder, and there-
fore is a common factor of the remainder and the smaller
number.
From (ii) it follows that any common factor of the
remainder and the smaller number is a factor of the
greater number also, and therefore is a common factor of
the two original numbers.
The H.C.F. of the two original numbers must therefore
be the same as the H.C.F. of the smaller number and the
remainder.
Thus the problem of finding the H.C.F. of the two
original numbers is reduced to that of finding the H.C.F.
of the smaller number and the remainder.
Ex. 1. Find the H.C.F. o/3663 and 5439.
Divide the greater by the less.
3663)5439(1
3663
1776
Hence the H.C.F. required is the same as the H.C.F. of 1776
and 3663. Divide the greater of these by the less.
1776)3663(2
3552
111
Art. 02.] HIGHEST COMMON FACTOR. 73
The problem is now reduced to finding the H.C.F. of 111 and
1776. Again divide.
111)1776(16
111
666
666
Thus, 111 is a factor of 1776, and therefore 111 is the H.C.F.
of 111 and 1776.
But the H.C.F. of 111 and 1776 is the H.C.F. required.
The successive divisions are usually written in a more compact
form, as follows :
3663)5439(1
3663
1776)3663(2
3552
111)1776(16
111
666
666
Ex. 2. Find the H. C. F. of 311 and 331.
311)331(1
311
20)311(15
20_
111
100
11)20(1
11
9)11(1
9
2)9(4
8
1)2(2
2
Here the H.C.F. of 311 and 331 is the same as the H.C.F. of
1 and 2, so that the numbers are prime to one another.
In this example, it would be a great waste of time to proceed to
the end; for the H.C.F. required is the H.C.F. of any divisor and
the corresponding dividend, and as soon as it is obvious that one
such pair have no common factors it is not necessary to proceed
further. Now the only prime factors of 20 are 2 and 5, and by
inspection neither of these is a factor of 311.
1.080 =
:2* X
33 x
5 x
.001
.072 =
23 X
3 2
X
.001
H.C.F. =
23 X
3 2
X
.001
.072,
Find the H.C.F.
of .108 and .072.
.108 =
2 2 x
33 x
.001
.072 =
2 3 x
32 x
.001
H.C.F. =
22 x
32 x
.001
—
.036.
74 FACTORS AND MULTIPLES. [Chap. III.
93. The H.C.F. of numbers containing decimals is not
often needed. The process, however, for finding such is
as follows :
Arrange all the numbers, by annexing naughts, so that
all shall have the same number of decimal places ; then
proceed as before.
Ex. 1. Find the H. C.F. of 1.08 and .072.
Ex. 2.
Ex. 3. Find the H. C.F. of 366.3 and 54.39.
366.30 = 2 x 32 x 5 x 11 x 37 x .01
54.39 3 x 72 x 37 x .01
H.C.F. = 3 x 37 x .01
= 1.11.
EXAMPLES XXI.
Written Exercises.
Find the H.C.F. (or G.C.M.) of
1. 221 and 247. 8. 4899 and 5893.
2. 357 and 391. 9. 9709 and 22849.
3. 899 and 1073. 10. 11663 and 12091.
4. 663 and 923. 11. 17947 and 29737.
5. 1517 and 1927. 12. 11453 and 12961.
6. 1785 and 2485. 13. 3834038 and 4169594.
7. 3499 and 3953. 14. 132038 and 369792.
15. 5411728 and 10902416.
Arts. 93-95.] LEAST COMMON MULTIPLE. 75
To find the H.C.F. of three or more numbers, we have
only to find the H.C.F. of the first two numbers ; then the
H.C.F. of this result and the third number ; and so on.
Ex. Find the H. C.F. of 286, 338, and 585.
The H.C.F. of 286 and 338 is 26. Then the H.C.F. of 26 and
585 is 13.
Find the H.C.F. of
16. 165, 198, 242. 19. 2387, 2821, 4433.
17. 312,429,572. 20. 3157,3321,4059.
18. 222,370,550. 21. 4732,5824,6643.
22. What would be the answers to 7, 9, and 18, if
they were as follows ?
(7) 3.499 and 395.3.
(9) 97.09 and 2.2849.
(18) 2.22, 370, and 550.
Least Common Multiple.
94. A number which is exactly divisible by two or
more numbers is called a Common Multiple of those
numbers.
For example, 200 is a CM. of 20 and 25.
The smallest number which is exactly divisible by two
or more numbers is called the Least Common Multiple
(L.C.M.) of those numbers.
For example, 100 is the L.C.M. of 20 and 25.
95. When numbers are resolved into their prime fac-
tors, their L.C.M. can be found by inspection:
Consider, for example, the numbers 120, 252, and 3575.
120 = 23 x 3 x 5
252 = 22 x 3* x 7
4125 = _ 3 x 5 3 x 11
L.C.M. = 23 x 32 x 5 3 x 7 x 11
= 693000.
76 FACTOKS AND MULTIPLES. [Chap. III.
Here we see that a common multiple must contain the prime
factors 2, 3, 5, 7, and 11. Also a multiple of the first number must
contain the third power at least of the factor 2 ; a multiple of the
second must contain the second power at least of the factor 3 ; a
multiple of the third number must contain the third power at least
of the factor 5 ; and we must also have the first power at least of
the factors 7 and 11.
The least common multiple must therefore be
2x2x2x3x3x5x5 x5x7x 11.
The L.C.M. of two or more numbers must be the con-
tinued product of the highest powers of all the different
prime factors of the numbers.
105 = 3 x 5 x 7
Illustration. 126 = 2 x 3 2 x 7
196 = 22 x 7 2
L.C.M. =22 X 3 2 x 5 x 7 2
= 8820.
96. The form for finding the L.C.M. given in Art. 95 is ex-
cellent, because of the ease with which it may be applied to any
example in which the H.C.F. or the L.C.M. is to be found, and
because it involves more or less mental work, which is stimulating.
In obtaining the prime factors, the highest powers of the lowest
primes should always be taken out first; thus, 2, 2 2 , 2 3 , etc., then
3, 3 2 , 3 3 , etc., should be the order in which the factors should be
taken out.
The usual method by which, the L.C.M. of numbers is
found is as follows :
Having written the numbers in a row, first strike out any num-
bers which are factors of any of the others {for every multiple of
a number is a multiple of any factor of that number). Then
divide by any prime which divides at least two of the numbers, and
put the quotients below those numbers which can be divided, and
bring down all those numbers which are not divisible by that prime.
Operate on the second row in the same maimer as on the first, and
go on until a row is arrived at in which no two numbers have any
[ UNIVERSITY j
Art. 96.] EXAMPLES. 77
common factor. Then the L.C.M. is the continued product of all
the divisors and all the numbers left in the last row.
Thus, to find the L.C.M. of 2, 5, 15, 24, 25, 30, 36.
The process is written as follows :
2 )2, ft m 24, 25, 30, 36
2)12, 25, 15, 18
3) 6, 25, 15, 9
2, 25, ft 3
Hence, the L.C.M. is2x2x3x2x25x3 = 1800.
The reason the L.C.M. is given by the above process is that the
L.C.M. of the numbers in the first row = 2 x L.C.M. of the num-
bers in the second row ; and so on to the end.
EXAMPLES XXII.
Oral Exercises.
Find the L.M.C. of
1. 4,10. 5. 9,12. 9. 30,50.
2. 6, 8. 6. 3, 7, 10. 10. 10, 35.
3. 5, 10, 12. 7. 12, 20. 11. 3, 4, 5, 6.
4. 7, 6, 2. 8. 8, 12, 3. 12. 2, 8, 12, 20.
Written Exercises.
Find the L.C.M. of
13. 12,16. 16. 8,12,20. 19. 30,45,54.
14. 20, 25. 17. 6, 8, 12. 20. 10, 14, 35.
15. 25, 30. 18. 15, 20, 30. 21. 6, 12, 18, 63.
22. 9, 36, 45, 81. 25. 16, 20, 22, 33, 36.
23. 3, 11, 18, 33, 36. 26. 3, 4, 10, 12, 14, 16, 18.
24. 25, 27, 33, 55. 27. 3, 14, 12, 56, and 28.
78 FACTORS AND MULTIPLES. [Chap. III.
28. 16,18,20,24,30,36.
29. 12, 15, 18, 21, 25, 35, 210.
30. 12,42,49,54.
31. 30,35,42,60,72.
32. 18, 54, 90, 102, 120, 144.
97. To find the L.C.M. of numbers which cannot be
readily resolved into factors, we must first find the H.C.F.
For example, to find the L.C.M. of 4592 and 5371.
The H.C.F. is 41. And by division
4592 = 41 x 112.
5371 = 41 x 131.
Hence, as 112 and 131 are prime to one another, the L.C.M. is
41 xll2 x 131.
Hence the L.C.M. of two numbers equals one of the
numbers multiplied by the quotient obtained by dividing
the other number by their H.C.F.
It should be noticed that the L.C.M. x G.C.M. = 41 xll2 x 41
X 131 = product of the numbers.
To find the L.C.M of more than two numbers we can
find the L.C.M. of two of the numbers, and then the
L.C.M. of this result and of the third number, and so
on to the end.
EXAMPLES XXIII.
Written Exercises.
Find the L.C.M. of
1. 357,391. 3. 3497,4035. 5. 165,198,242.
2. 851,943. 4. 4899,5893. 6. 312,429,572.
7. 360, 1350, 1500. 9. 420, 630, 1050, and 1470.
8. 195, 546, 286. 10. 1365, 2288, 2.640.
Arts. 97-101.] PARENTHESIS. 79
98. By using the form of work as given in Art. 95, we
may find the H.C.F. and the L.C.M. at the same time.
Ex. Find the H.C.F. and the L.C.M. of 24, 60, and 72.
24 = 2 3 x 3
60 = 2 2 x 3 x 5 '
72 = 2 3 x 3 2
H.C.F. = 22 x 3 =12.
L.C.M. = 2 3 x 3 2 x 5 = 360.
It seems best to make a digression here to explain the
use of the sign of parenthesis, and the use of cancella-
tion.
Parenthesis.
99. The sign of parenthesis is made thus, ( ) ; in
mathematics this sign is often called parenthesis.
100. Several numbers are included in parenthesis when
they are to be treated as a tvhole, some sign being put outside
the parenthesis to show hoiv this whole is to be treated.
Thus, 17 + (4+8) denotes that 8 is to be added to 4 and that
this result is then to be added to 17.
Again, 117 — 6 x (4 + 8) denotes that 6 is to be multiplied by the
sum of 4 and 8, and that the product is to be subtracted from 117.
Also, 36 -=- (24 -f- 4) denotes that 24 is to be divided by 4, and that
36 is to be divided by this result.
When two parentheses come together without any sign
between them, the sign of multiplication must be under-
stood.
Thus, (5 + 7) (9 - 3) is put for (5* + 7) x (9 - 3).
Also, 6 (5 + 7) is put for 6 x (5 + 7).
101. Instead of enclosing numbers in parenthesis a
line, called a vinculum, is sometimes drawn over them.
Thus, 17 + 4+8 may be used instead of 17 + (4 + 8).
80 FACTORS AND MULTIPLES. [Chap. III.
102. Sometimes a parenthesis is put within a paren-
thesis : to avoid confusion the parentheses are made of
different shapes and are named as follows :
( ), sign of parenthesis,
j j, braces,
[ ], brackets.
For example, 150 - 3 {13 - (9 - 3)}.
In order to simplify when there is more than one bracket, it will
be found convenient to clear away the innermost bracket first.
Thus, 150 - 3 {13 - (9 - 3)} =150-3 {13 - 6} = 150 - 3 x 7
= 150 - 21 = 129.
Operations of multiplication and division are to be
performed in order from left to right, and each sign is a
direction to multiply or divide by the number that follows
next after it.
For example, 60 + 6x3= 10 x 3 = 30 ;
and 60 x 6 - 3 = 360 - 3 = 120 ;
and 60-10-2= 6-2= 3.
EXAMPLES XXIV.
1. 8+ (7+3).
2 . 15 -(9 + 3).
3. 27 -(11 -4).
4. 15-2(8-5).
5. 3 7-2-4(8-
6. 7(3 + 9) -5 (12
13. 325-(17-2)(24-5).
14. 27 - 3(20 - 11) + 4(8 - 3).
15. 18+S17-2-(15-4)J.
16. 23 -[41- {2 + lj -27^6].
Written Exercises.
7.
(.9 + .7)(.5 + l.l).
8.
(13-
.5)(1.5-
-.7).
9.
(23-
12) (28 -
-12).
10.
3(13-
-4)(15-
-7).
6).
11.
8(1.7
-1.1)(1.
3 - .06),
-4).-
12.
5(11 + 5)(11-
-5).
Arts. 102-104.] CANCELLATION. 81
17. (3.8016 -2.794) (1.8093 -. 078).
18. 12-6 x 2.
19. 12 + (6x2). 22. 28+- 7-3 + 2. •
20. 18x6 +-2. 23. 28 +-(7 -3) +2.
21. 18x(6+-2). 24. 28 +- [7 - (3 + 2)].
Cancellation.
103. Here we make use of the following principle :
Dividing both dividend and divisor by the same number
does not change the quotient.
Thus, 24 -7- 4 = 6, and if both 24 and 4 be divided by 2, we shall
have 12 -4- 2 = 6 ; also, if both 24 and 4 be divided by 10, we shall
have 2.4 -f- .4 = 6.
104. Note. It should be noticed that although the quotient
is unchanged by dividing both dividend and divisor by the same
number, the remainder, if any, is not unchanged, but is equal to
the original remainder divided by the number by which the original
divisor and dividend were multiplied.
Therefore we must multiply the remainder, if any, by the number
used in dividing, if we wish the remainder obtained by using the
original numbers.
Ex. 1. 44 -4- 8 = 5, with a remainder of 4 ;
while ll--2 = 5, " " " " 1.
It is evident that the remainder 1 compared with the divisor 2 is
just as large as the remainder 4 compared with the divisor 8.
Ex. 2. How many pieces, each 24 inches long, can be cut from a
string 231 inches long, and what will be the length of the part left
over ?
231 -*- 24 = 77 + 8 = 9, with a remainder of 5.
This remainder must be multiplied by 3, if we wish to know how
many inches of string remain. Ans. = 9 pieces, with 15 inches
remaining.
82 FACTORS AND MULTIPLES. [Chap. III.
105. The principle of Art. 103 may be used in shortening the
process of division, especially when the dividend and divisor can be
factored at sight. Thus,
Ex. 1. 99 )3663 gives the same quotient as 11)407
37 = Ans.
Ex. 2. 36 )4884 = 6 )814
135, rem. 4.
4884 -f- 36 = 135, rem. 24. [Compare Art. 68.]
106. Division has thus far been indicated by the sign
-h ; but division is often indicated by writing the dividend
just above the divisor with a line between ; thus,
36 -7- 8 is sometimes written ^-,
and what is true of the first expression is true of the
second. In either case dividing both 36 and 8 by the
same number does not change the quotient ; thus,
36-j-8 = 9-j-2 = 4, with a remainder of 1. ) . , _, A .
Art. 104.
¥ =
= f =4,
of 1. ")
" I- >
107. The process of dividing both dividend and divisor
(or their factors) by the same factor is called Cancellation.
A thin line drawn through a dividend or divisor indicates that a
factor has been cancelled ; thus,
20
^ = — = 2, with a remainder of 6.
n 7
7
Here 6 is cancelled from dividend and divisor, and the quotients
are written, one above and one below.
Ex. (5 x 48 x 28) + (10 x 14 x 9) may be written
5 x 48 x 28
10 x 14 x 9*
The latter is the better form when we wish to cancel ; thus,
1 16 i
$ x S. x 8 - 1? = 5, with a remainder of 1.
Xfixlix 9 3
% % 3
Arts. 105-107.] MISCELLANEOUS EXAMPLES. 83
Here 5 is cancelled from 5 and 10, 3 from 48 and 9, 7 from 28
and 14, and the two 2's thus obtained in the divisor cancel with
the 4 in the dividend. The result is the same as if there had been
no cancellation.
EXAMPLES XXV.
Written Exercises.
\m
lplify
by cancellation :
l.
»
4- *fc
7- t 7 A-
10.
444
T6T'
2.
»
5- iff
8. m-
11.
tu-
3.
if-
13.
6. iH-
7x22
11 x 63"
9- Iff
12.
rn-
14.
(8 x 38 x 41) -
(19 x 4).
15.
16.
6(42 - 3)
13x8 '
1.26 x 3.5
.6x7
17. Why can we not cancel in " .
EXAMPLES XXVI.
Miscellaneous Examples, Chap. IU.
1. Divide the product of 8978 and 55112 by 5561.
2. Divide 210 dollars between A and B, so that A
may have 5 times as much as B.
3. Find the H.C.F. of 3465 and 3696.
4. Multiply 606.78 by 11.
5. Find35 2 ; 105 2 ; 7.5 2 ; V 24 ™09.
6. What number is the same multiple of 7 that 21560
is of 55?
84 MISCELLANEOUS EXAMPLES. [Chap. III.
7. What is the price of a silver bowl weighing 50
ounces, at 1.25 dollars an ounce ?
8. Two equal sums were respectively divided among
12 men and a certain number of boys. Each man received
5 dollars, and each boy 1 dollar. How much was divided
altogether, and how many boys were there ?
9. Exactly 20 years ago, a man was four times as old
as his son, whose present age is 28. What is the present
age of the father ?
10. Eindl9 x 16; 656x125.
11. A certain chapter of a book begins at the top of
the 357th page and ends at the bottom of the 435th page.
How many pages are there in the chapter ?
12. After multiplying 375 by 29, and 131 by some
other number, the results when added amounted to 13888.
What was the other number ?
13. Find the H.C.F. of 5610, 11781, and 1309.
14. Find the least number by which 222 must be
multiplied in order that the product may be a multiple
of 1295.
15. Four bells toll at intervals of 3, 4, 5, and 6 seconds,
respectively. If they all begin to toll at the same instant,
how long will it be before they again all toll together ?
16. Add fourteen hundred seventeen, four thousand
eleven hundred nine, six million fifteen thousand, and
eighteen million twelve hundred nineteen.
17. A certain number was divided by 35 by ' short '
divisions ; the quotient was 72, the first remainder was
2, and the second remainder was 6. What was the
dividend ?
Art. 107.] MISCELLANEOUS EXAMPLES. 85
18. Multiply 700630.0003 by 1006.07, and prove by
dividing the product by the multiplier.
19. Find the continued product of 18, 13, and 11 ;
obtain the square root of the product to two decimal
places.
20. Divide 126819 by 21, using factors.
21. What is the least number of times that 315 must
be added to 1594 that the sum may exceed a million ?
22. Multiply 67412 hf 9997 as shortly as you can.
23. Divide 789 by .10063 to 3 decimal places.
24. Find the H.C.F. of 10481 and 17617.
25. Four men can walk 30, 35, 40, and 45 miles a day,
respectively ; what is the least distance they can all walk
in an exact number of days ?
26. Find the L.C.M. of 12, 64, 80, 96, 120, 160.
27. Find the prime factors of 1176 and 19404, and
hence write down their G.C.M. and L.C.M.
28. The quotient is twice the divisor, and the remainder
which is 50 is one-fifth part of the quotient. Find the
dividend. .
29. Simplify ; obtain the answer in two
forms. 125X219'
30. Find the least number which can be divided by
7, 20, 28, and 35, and leave 3 as remainder in each case.
31. What number is that which after being subtracted
19 times from 1000 leaves a remainder of 12 ?
32. Multiply three thousand eighty-seven by seventy-
two thousand nine hundred thirty. What numbers less
than 12 will exactly divide the product ?
86 MISCELLANEOUS EXAMPLES. [Chaps. III., IV.
33. (a) Simplify 650 x 1.25 - .5.
(b) The answer is a multiple of which of the
following numbers : 5, 15, 25, 65, 105, 125 ?
Obtain (6) by first obtaining primes of the answer.
34. Find 19 x 17 x 11 X 2.5 x 1.25.
35. Find65 2 x .11.
36. Simplify (a) \ 2 - x 4 - 2 + 6 (18 - 14).
(P) ¥-X (4-2) + 6(18-14).
(c) 2(¥x4) -.J2 + 6(18 -14)j.
37. If a number when divided by 35 give a remainder
27, what remainder will it give when divided by 7 ?
38. What is the greatest and what is the least number
of four digits which is exactly divisible by 73 ?
39. Find the H.C.F. and the L.C.M. of 21, 22, 24, 28,
32, 33; also of 16, 18, 20, 24, 30, 36.
40. Find the number nearest to 1000 and exactly
divisible by 39.
41. Multiply 7863 by 999, and see if the product is
divisible by 3.
42. Find V4912.888464.
43. Find V^ 6 -
(a) Divide the following numbers by 2.
(b) Prove your answers by first simplifying the
numbers, and then dividing by 2.
44. 3(6 + 8); 3(6x8).
45. 4(6 + 8); 4(6x8).
46. 4(18 + 6); 4(18-6).
47. (6x2) (8 + 10).
48. 6(12-3) + 8 6 + 4 + 2.
49. 28 -f- [7 - (3 + 2)].
Arts. 108-110.] FRACTIONS. 87
CHAPTER IV.
FRACTIONS.
108. If a unit be divided into 2, 3, 4, 5, etc., equal parts,
these parts are called halves, third-parts, fourth-parts,
fifth-parts, etc., or more shortly and more generally, halves,
thirds, fourths, fifths, etc.
If the unit quantity be divided into any number of
equal parts, one or more of these parts is called a Fraction
of the unit.
For example, if a unit quantity, as one apple, be divided into
sevenths, three of these parts constitute three sevenths, and the
three sevenths is a fraction of seven sevenths, the unit quantity.
109. The number which indicates how many parts of
the unit quantity are to be used is called the Numerator.
The number which indicates into how many parts the
unit quantity is divided is called the Denominator.
110. The expression formed by writing a numerator
just above a denominator with a line between is called a
Common Fraction.
Thus, f, y 8 ^ (eight-thirteenths), ^ (one twenty-third), are com-
mon fractions (called briefly fractions).
Common fractions are also called vulgar fractions.
Note. A fraction is an expression of division, the numerator
and denominator corresponding to the dividend and divisor respec-
tively. What is true of dividend and divisor is true of numerator
and denominator. When the indicated division is performed, the
quotient is generally a decimal.
88 FRACTIONS. [Chap. IV.
Ex. & = 3-s-24.
24)3.000(.125
24
60
48
120
120
111. If we have 3 units, and divide each of them into
5 equal parts, and then take one of the parts from each
divided unit, we shall take one part out of every five, that
is, one-fifth of the whole three units j but each of the parts
is one-fifth of a single unit and we take 3 of them : we
therefore take 3 fifths of one unit.
Thus, 3 fifths of 1 unit is the same as 1 fifth of 3 units.
Hence, f, which by definition denotes 3 fifths of 1 unit,
may also be considered to stand for 1 fifth of 3 units.
The same holds good for all other fractions ; for example,
f of 1 dollar = £ of 3 dollars ;
and | of 1 foot = } of 7 feet.
EXAMPLES XXVII.
1. Write in figures the following fractions: five-
ninths, six-elevenths, eleven twenty-thirds, sixteen twenty-
sevenths, seventeen ninety-firsts, ninety-five one hundred
fourths.
2. Write in words : f, f, ^ A> A> Vb &> tVd J*&
112. The numerator and denominator of a fraction are
called its Terms.
When the numerator is less than the denominator, the
fraction is called a Proper Fraction ; and when the numera-
tor is equal to or greater than the denominator, the fraction
is called an Improper Fraction.
Arts. 111-114.] MIXED NUMBERS. 89
A number made up of an integer and a fraction is called
a Mixed Number.
Thus, 2\ (2 and \), which means 2 + £, is a mixed number.
Changing the form of an expression, or changing the
units in terms of which any quantity is expressed, is
called Reduction.
113. Reducing a mixed number to an improper fraction.
Consider, for example, 3|.
Each unit contains 7 sevenths, therefore 3 units contain 3 times
7 sevenths.
21-4-2 23
Hence, 31 = 3 times 7 sevenths + 2 sevenths = — ±— = —
7 7 7
a™^ 72 7x9 + 2 65
Again, 7f = = -•
From the above it will be seen that a mixed number is equivalent
to an improper fraction whose denominator is the denominator of
the fractional part, and whose numerator is obtained by multiplying
the integral part by the denominator of the fraction and adding its
numerator.
It should be noticed that a whole number can be expressed as
a fraction with any given denominator. For example,
6 = 6x7 sevenths = - 4 7 2 - ; also, 6 = 6 x 13 thirteenths = f f .
114. Conversely, reducing an improper fraction to a
whole or mixed number.
Consider, for example, - 2 ^.
Since 7 sevenths make 1 unit,
- 2 f = 3x? sevenths + 2 sevenths = 3 + 2 sevenths = 3f .
Again, - 2 ¥ * = 6x4 fourths = 6, since 4 fourths = 1.
From the above it will be seen that an improper fraction is
reduced to a mixed number by dividing its numerator by its denom-
inator ; the quotient will form the integral part, while the remainder,
if any, will form the numerator of the fractional part, whose
denominator must be the denominator of the improper fraction.
90 FRACTIONS. [Chap. IV.
EXAMPLES XXVIII.
Oral Exercises.
Express as improper fractions :
1. 11 If, 2|, 3*. 2. 7^, 6 T % 5f 3f
3. 4^, 9f, 12fc 11^.
4. Express 3, 5, and 9 as fractions with a denominator 7.
Express as whole or mixed numbers :
s. ¥> ¥>¥>¥• e. ft ?> ft if 7. ¥>¥>¥>«•
Written Exercises.
Express as improper fractions :
8. 4^,5^,18^. [Art. 55,1.]
10. Reduce 13 to fifteenths, and 41 to twenty-fifths.
11. Express 427 as a fraction with a denominator 99.
Express as whole or mixed numbers :
12- ft ft If 13. tflifoW 14 - ftW,W-
See Art. 70 for examples in 14.
115. Reducing a fraction to its lowest terms.
A fraction is said to be in its Lowest Terms when the
numerator and denominator have no common factor.
Thus, the fractions, §, f , |§, are in their lowest terms ; but the
fractions, f , |£, § £ , are not in their lowest terms, for in each case
the numerator and denominator have 2 as common factor.
The following is a very important truth :
The value of a fraction is not changed by dividing the
numerator and denominator by the same number.
This truth is but a repetition of the principle stated in Art. 103.
Art. 115.] REDUCTION TO LOWEST TERMS. 91
Ex. B educe T %\ 5 u to its lowest terms.
To reduce to the lowest terms we must divide by the H.C.F. of
the numerator and denominator ; for we thus obtain an equivalent
fraction whose numerator and denominator have no common factors.
In the present case the H.C.F. will be found to be 55.
825 825 -55 15
Thus,
1540 1540 ■*■ 55 28
Instead of reducing a fraction to its lowest terms by dividing
the numerator and denominator by their H.C.F., we may divide
by any common factor, and repeat the process until the fraction is
reduced to its lowest terms. Thus,
We see at once that 5 is a common factor ; we therefore divide
the numerator and denominator by 5, and obtain the equivalent
fraction iff. We now see that 11 is a common factor, and having
divided the numerator and denominator by 11, we have the equiva-
lent fraction £f , which is at once seen to be in its lowest terms.
EXAMPLES XXIX.
Oral Exercises.
Reduce to their lowest terms :
*• ¥> "6"? T5"? T6"> T5- 3 * "2T> "5T> JTi H'
2- «,«,*, if, it 4. ft ft ft ft, ft
Written Exercises.
Reduce to their lowest terms :
»• fit s- m ii- mi- i4. tut
e- m- 9- m i2- \m- is- Hft
»• m io. AVt is- A%- is- A- 2 A-
92 FRACTIONS. [Chap. IV.
116. Reducing fractions to equivalent fractions having
the lowest common denominator.
The following is a very important truth :
The value of a fraction is not changed by multiplying the
numerator and denominator by the same number.
This truth is but a repetition of the principle stated in Art. 67.
Consider the fractions, f , f , and f .
The L.C.M. of the denominators 4, 6, and 9 is easily seen to
be 36. Since 36 is a multiple of each denominator, all the fractions
can be reduced to equivalent fractions which have 36 for denom-
inator, provided the numerator and denominator of each of the
fractions be multiplied by a suitable number, namely, by the num-
bers 36 -=- 4, 36 -r- 6, and 36 -4- 9, respectively ; that is, by 9, 6, and
4, respectively.
Thus,
and
3
3
x
9
27
4
4
X
9
36
5
5
X
r>
30
6
6
X
6
36
8
8
X
4
32
9
9
X
4
36
Again, reduce £}, ^r, ^ ? , to equivalent fractions having the lowest
common denominator.
Full Work Illustrated.
18 = 2 x3 2
30 = 2 x 3 x 5
24 = 23 x 3
L.C.M =2 3 x 3 2 x 5
11 11 x 2* x 5
220
18 18 x 22 x 5
360*
7 7 x 22 x 3
84
30 30 x 2 2 x 3
360'
5 5x3x6
24 24 x 3 x 5
_ 75
360
117. Comparison of Fractions.
Of two fractions which have the same denominator, the greater
is that which has the greater numerator ; for, the parts being the
same, the greater fraction is that which has the most of them.
Arts. 116, 117.] EXAMPLES. 93
Again, of two fractions which have the same numerator, the
greater is that which has the smaller denominator ; for, the number
of parts being the same in both, the greater is that in which the
parts are the greater ; that is, in which the unit has been divided
into the smaller number of equal parts.
We can therefore see at once which of a number of fractions is
the greatest, and which is the least, provided the fractions are first
of all reduced to equivalent fractions with the same denominator.
For this particular purpose it would do equally well to reduce the
fractions to equivalent fractions with the same numerator, but it
is for other purposes much less convenient to reduce fractions to
equivalent fractions with the same numerator.
Ex. Which is the greatest and which is the least of the fractions,
As in the preceding article, the fractions are equivalent to f|, § £ ,
and ||, respectively; they are therefore in ascending order of
magnitude.
EXAMPLES XXX.
Written Exercises.
Eeduce to equivalent fractions with the lowest common
denominator, and arrange in ascending order of magni-
tude:
*• h |> T 5 - t> A> A*
*• "3~> 9? TT m 6. y^, -Jy, -g-g-.
3 - h A? A- 7 - J? "fi p
!3- Afiff 16.
I 4 - hh iA>*< 17.
is- A> «>**>» is- f>A>«,H,ffr
Eeduce to equivalent fractions which have the lowest
common numerator :
19 5.10.15 Oft 121618 oi 15 18 20
xy ' 8' IT? 2T' *"• TT> T9> ft' /J1 ' ST> ft? if-
9.
2 3 5
3"? T> T
10.
i, h A-
11.
H> if. H-
12.
**» it, It-
4
A,
A. it-
3 5 7 11
8? TT> TS> TO"'
94
FRACTIONS.
[Chap. IV.
118. Addition of Fractions.
Fractions which have the same denominator are called
Similar Fractions.
If fractions are dissimilar they must be made similar
[Art. 116] ; then their numerators may be added, and the
sum written as a numerator for the common denominator.
[Compare Art. 32.]
Ex. 1. Find f + T V + f •
5 5x6 30
6 6x6 36
Or,
5_
6
30
7 7x3 21
12 12 x 3 36
7 _
12
21
4 _ 4 x 4 _ 16
9 9x4 36
4_
9
16
Sum=^ =
36
If*.
Sum =
~67~
36
m
After a little practice the middle column might be omitted.
Ex. 2. Find 24 + 3A.
3& = »&
Sum = 6&
Here the 12ths are added, and 1 is carried to units. The pro-
cess is similar to that represented in Ex. 2, Art. 29.
The result should in all cases be reduced to its lowest
terms, and an improper fraction should be expressed as a
mixed number.
EXAMPLES XXXI.
Oral Exercises.
Find the sum of the following fractions :
1. £andf. 4. JLand^. 7.
2. | and \. 5. f and f. 8.
3. ^and T %. 6. H and ff- 9 -
i and f .
| and £
f and f
Art. 118.] EXAMPLES. 95
10. f andf 14. 20^andlO T V
11. f andf 15. 3£and4£.
12. 2Jand3f. 16. 8f and 6 A
13. 4 1 ^and6 1 ^. 17. 12^- and 6^.
18. 4f and6A-
Written Exercises.
Find the sum of
19. A and fa 22. 2| and 1-^. 25. 2f and 3|.
20. A and fa 23. 5| and 2^. 26. 3f and 1£.
21. A and A- 24. 7f and f 27. Iff and 7£f.
28. *, ♦, A, and «. 30. A> A> A> A> tt> **& H ■
29. f,f,f,andf. 31 . A A> "B> tt «> and ff .
32. Findf + A + A + A + A-
33. Find # + A + * + & + &
34. Find3f + A + 5A + 7 T V
35. Find 3J + A + 5^ + 3^ + 7ff
36. Find 3J + 5f + tt + 3&.
37. Fmdff + 2A + UA + 5ff
38. Find3f + 7A + A + 2if.
39. Find fttfr + BJfa
40. Find3AV + 5AV + llMt-
41. Find 10jfj + ll^ + 7fff
96 FRACTIONS. [Chap. IV.
119. Subtraction of Fractions.
Subtraction can be performed with fractions only when
they are similiar. [Compare Art. 39.]
Ex. 1. Subtract f from ft. Ex. 2. Find 6ft - 3f.
Difference = f T . Remainder = 2^ T .
Ex. 8. Subtract 5\\from 5f.
Here f| cannot be subtracted from £f,
5f = Hi therefore we take 1 unit from the 5 units and
3 H = 3|f add it (changed to 24ths) to ft, making §£ ;
Remainder = lft. now II from f I equals ft, and 3 from
4 equals 1.
The operation is similar to that represented in Ex. 2, Art. 38.
Ex. 4. Simplify 3 J
- 2f + 8f -
-S&.-SA + & [See Art. 41.]
3*= %
8|= 8ff
2f= 2H
*& = 2ft
12H
- io|f = iff.
EXAMPLES XXXII.
Oral Exercises.
Simplify (give lowest terms in your answers)
"*• A— A-
5- f-f
9.
5 _ 2
2- f-f
6. f-f.
10.
3A-V-
3- A- A
7- l-f
11.
6^-4|.
4# 3"2 ~~ A*
8- A-f
12.
6*-2A
Simplify ;
Written Exercises.
13. H-A
15. ft-»
17.
*-A
"• A- A
is. A-A-
18.
T2~ — 7*
Arts. 119, 120.] MULTIPLICATION. 97
19. A -A- 23 - 3 i~ 2 i- 27 - 7 *- 5 A-
20. A -A 24- m-5U- 28. 19JT-12A-
21- A~» 25. 2f|-2||. 29. r^-flA-
22. A~to 26. 3f-2J y 30. 3^-2^.
31. 6J-2||. 33. 16ft -5^
32- 5ft- Sft 34. 9ft-4ftU.
Find the difference between
35. 3^- and 5^. 39. 8JJ- and 12 Jf.
36. 72 8 T and82 7 ¥ . 40. 6fJ- and 12JJ.
37. 6^andl5|f 41. 143^ and 127^.
38. 7iiand5||. 42. 85^r and 72/^.
Simplify :
43.
2i + 3i-4i.
44.
GH-Si + IA-
45.
5|-3A + ^-2^.
46.
15^-13A + 16A-9M-
47.
12^-10 + 7^-t-5f£.
48.
4f-2A + 2H-3^.
49.
6A-|"2A + 6rlr + TTfW
120. Multiplication by a Whole Number.
Fractions may be treated as concrete numbers ; therefore,
as 3 times 5 tons = 15 tons,
so 3 " 5 sevenths = 15 sevenths ;
i.e., 3 " .: f =¥•
Again, 3 " ■& = |f = f (by cancellation)
5 5
i.e., 3
18-5-3 6
Hence, to multiply a fraction by a whole number, we must
multiply the numerator, or (when possible) divide the denomina-
tor, by that whole number.
98 FRACTIONS. [Chap. IV.
The product should always be reduced to its lowest terms, and
an improper fraction should be expressed as a mixed number.
Ex. Multiply T 5 g by 15.
±xid=*kH=I5 ==?*=* 4*.
18 18 18 6 *
EXAMPLES XXXIII.
Oral Exercises.
Multiply and reduce to their simplest forms :
1. * by 2. 3. 2 4 iby3. 5. f by 4.
2. ^- by 3. 4. T 3 T by4 6. & by 4.
7. | by 3. 9. ^by6.
8. |f by 17. 10. ^by8.
Written Exercises.
Perform the following examples (see Art. 107) :
11. tVxlO. 15. 7 T \x 25. 19. -V/X22.
12. T % x 8. 16. |f x 15. 20. 44^ x 26.
13. 2|x6. 17. 9fix25. 21. 99 x ^.
14. 5f x 10. 18. || x 16. [Art. 55.] 22. ^ffg x 9.
121. Multiplication by a Fraction.
We understand multiplication to be the taking one
number as many times as there are units in another.
Thus, to multiply 5 by 4, we take as many fives as there
are units in 4. Now 4 is 1 + 1 + 1+1, and 5 x 4 is
5 + 5 + 5 + 5.
Thus, to multiply one number by a second is to do to the
first what is done to the unit to obtain the second.
For example, to multiply f by f , we must do to f what is done
to the unit to obtain | ; that is, we must divide f into 4 equal parts
Art. 121.] MULTIPLICATION. 99
and take 3 of those parts. Each of the 4 parts into which $ is
divided will he — — , and hy taking 3 of these parts we get x •
7x4 7x4
Thus, ^x 3 5x8
7 4 7x4
Hence, the product of any two fractions is another fraction
whose numerator is the product of their numerators and whose
denominator is the product of their denominators.
The continued product of any number of fractions is
obtained by continued application of the above rule.
Thus, to find the continued product of §, f , and |.
2 x i x - = 2 x4 x - = 2 x4x8 = i*£.
3 5 9 3x5 9 3x5x9 135*
Hence, the product of any number of fractions is another frac-
tion whose numerator is the product of their numerators and
whose denominator is the product of their denominators.
It should be noticed that the product of one fraction by
a second is equal to the product of the second by the first.
It should be noticed also that an integer x a fraction
equals (the integer x the numerator) h- the denominator.
Ex. 1. Multiply j% by fy.
2 1
i x l = M_ = i. [Art. 107.]
35 27 3^x2/ 45
5 9
Ex. 2. Simplify f X f X f.
1 1 1
3 X ^ X 5~5
1 %
1
Ex. 3. Multiply 2\ by 3|.
The mixed numbers must first be reduced to improper fractions.
Thus, 2}x3! = ?x 2 -9 = 9A-?9 = 261
* *4 8 4x8 32
100 FRACTIONS. [Chap. IV.
EXAMPLES XXXIV.
Simplify: ° ral Exercises.
MX}- 7. |X|. 13. |xf{.
2. }xf 8. fx*f 14. ||x|f
3. fxf. 9. fx||. 15. «XA-
4. fxf 10. fx& !6- J|XA-
5. Hx21 ii. _Z_ X 22. 17. (|) 2 .
6. 2J X 3|. 12. f X ft. 18. (f) 2 . 19. (|)3.
Written Exercises.
20. 5^X3^. 27. frx5&x{i.
21. | X | X f 28. 2 T V X 3f X 6£.
22. fxfxf 29. 1^X21X1^.
23. fxAxA- 30. 6 T Vx||xl|f.
24. «xl|x5f 31. ^ x 2^ x 5J r x T W
25. 2^X3^X4^. 32. 1^ X Iff X f$ X l^V
26. 5ix71x^. 33. tt xl|x5^x2|f.
34. (|)«. 35. (A) 2 . 36. (#)«.
122. Division by a Whole Number.
Just as 15 tons h- 3 = 5 ions,
so also 15 sevenths -*■ 3 = 5 sevenths,
that is, - 1 / h- 3 = f
Again, to divide f by 3.
5 5x3
Here 5 is not a multiple of 3. But, since - = — - — ,
6 6x3
5^ o _ 5x3 . g _ 5 _ 5
6 ' ~ 6 x 3 ' ~ 6 x 3 ~ 18*
Art. 122.]
EXAMPLES.
101
We see at once, that a fraction is divided by a whole number
by multiplying its denominator by the whole number. For
example, in
x3
there are the same number of parts as in
|, namely five, but the unit in the former case is divided into 3
times as many parts as in the latter, and therefore each of the
parts in is one-third of each of the parts in 4.
F 6x3 *
Hence, to divide a fraction by a whole number, we must divide
the numerator, or multiply the denominator (only when neces-
sary), by that whole number.
Ex. 1. Divide 31 by 7.
34- - 7 = ^ -r- 7
5 5
16
5x7
16
35'
Ex. 2. Divide 215f by 9.
23 + 8f -=- 9
23
When the integral part of the dividend
= 23M.
is large, we first divide the integer by the
» "*" divisor ; then the remainder + the frac-
tional part is to be divided by the divisor.
Simplify :
1. f + 2.
2. i+3.
3. A + 4-
4. ^ + 3.
EXAMPLES XXXV.
Oral Exercises.
5. f-r-8.
6. ^^4.
7. M-s-5.
ft 42 _l_ 7
9. ^ + 25.
10. H
11. a + 12.
12. H
17. [Art. 55.]
Written Exercises.
13.
tt+6.
17.
2**11
14.
11 + 16.
18.
71-6.
15.
If + 30.
19.
9|-s-8.
16.
M + 7.
20.
« + !&
21. 5f-r-5.
22. 7I--6.
23. 8 T \ + 15.
24. 12f-f.ll
102 FRACTIONS. [Chap. IV.
25. 85$-*- 9. 26. 214 T V-r-7. 27. 174f£ - 18.
28. 7|i-r-15. 29. 254^-25.
123. Division by a Fraction.
If the fraction -J be divided by 1 (unity), the quotient
is -J; but, if the unit be separated into thirds, and one
of these thirds be used (instead of unity) as the divisor,
the quotient is 3 times as large as before.
Thus, | + l=fj but | + * = J x 3 = JyL.
Now, if the second divisor (i) be multiplied by 2, the quotient
(y) must be divided by 2 ; thus,
| 4- f = f x 3 -^ 2
The same reasoning will apply to all cases.
Hence, to divide by a fraction, we must multiply by the fraction
inverted.
Note. Sometimes a short method of dividing a fraction by a
fraction is to divide the numerator and denominator of the dividend
by the numerator and denominator of the divisor, respectively ;
thus, | + f '= f .
Ex. 1. Divide § by |f.
1 16
6^15 = ^ >< ^ = 16 =17
6 ' 32 W 9 *
3 3
Ex. 2. Dmde 2\ by ljf.
The mixed numbers must first be expressed as improper fractions.
1 9
Arts. 123-125.] COMPOUND FRACTIONS.
103
Simplify :
EXAMPLES XXXVI.
Written Exercises.
1.
*-*-*
12.
4 _i_ 2
6" * "3'
23.
IT'S" : "5T"
2.
f+i
13.
2J+2J.
24.
121 _:_ 143
tt¥ • rfv
3.
2 _t_ 3
9 ' 4"*
14.
6A + 1A-
25.
9 6 -11
4.
t + f
15.
6* -If.
26.
4_4 _s_ 129
5.
i + f
16.
*H + iA-
27.
Ifl | 1 7
6.
A + tt
17.
6|-11.
28.
1 23 _^_ 1 37
-'-TIT * J^TT^*
7.
*+*
18.
6f-=-9.
29.
143 _:_ 959
8.
T ? 6^tV
19.
2 2 • TT
30.
K7 • 913
9.
i-i-t
20.
2H^7.
31.
11^-12,%
10.
i + *
21.
«+*•
32.
2W+«Ht-
11.
t + i
22.
tt-*tt
124. When unity is divided by any number, the quo-
tient is called the Reciprocal of the number ; thus,
l is the reciprocal of 5 ; f is the reciprocal of f ; 5 is the re-
ciprocal of \.
Any number x its reciprocal = 1.
125. A fraction of a fraction is called a Compound
Fraction.
Thus, § of f is a compound fraction.
To take f of f , we must divide f into 3 equal parts and take 2 of
those parts.
Hence, § of f is the same as \ x f .
Ex. 1. Multiply f of 2i by f of If.
f of 2* = f x ¥» sod f of If =* f x i j
3 11 S T 11
hence the required product = £x — x " x - = —
- M M 12
104 FRACTIONS. [Chap. IV.
EXAMPLES XXXVII.
Written Exercises.
1. State the reciprocals of 12, f, if-, f and f^-.
Simplify :
2. | Off 9. f of ^ of ^.
3. | off. 10. 21 of 31 off
4. fof^. 11. 6* Of 2ft Of If
5. lfof2f 12. | off X^of2J.
6. 3iof6i. 13. $ of 2f x 1 A of 21
7. 7fof2f 14. lfofajxiixof.
8. 3£ of 3£. 15. If of 3| x 5J of 7f .
126. A fraction whose numerator, or denominator, or
both, are fractional is called a Complex Fraction.
3 2. 1 I 1
Thus, JL, JL, and 2 Z I are complex fractions.
Complex fractions are simplified by dividing the numer-
ator (simplified) by the denominator (simplified).
Ex. 1. Simplify ±-
j
| 3 " 7 3 5 15
Ex. 2. Simplify ttk.
l+i
? 3
3
Caution. Dividing by the sum of two fractions is not equivalent
to multiplying by the sum of the reciprocals of those fractions.
Art. 126.] COMPLEX FRACTIONS. 105
In solviDg the above example the following would be wrong :
|^|=(m)x(t + f)-
A complex fraction is unchanged in value by multi-
plying its numerator and denominator by the same
number.
5
g 9 = 45
7 H 28
4
For
example,
5
7 _
1"
-fx
11
11
For
f"
►*«
fx
1=1
and
IxiiH
4x
11 =
Ex. 1,
Multiply the numerator and denominator by 24, the L.C.M. of
3, 4, 6, 8. Then we have
(f-f)24 _18- 1 26_2_ 9
(f -|) 24 21-20 1
Ex. 2. Simplify —
5 + — A
7 2-
4 + i
* = *•
5 + =^- 5 +
2 5 + ^f
4 + i
First, multiply the numerator and denominator of the lowest
q
complex fraction, namely — ■ — , by 2, and we get f . Next, multiply
4 + i o
the numerator and denominator of the fraction by 9, and we
7-1 o
get if. Then multiply the numerator and denominator of — - —
5 + if
by 57, and we get i£i, which is then reduced to its lowest terms.
A fraction of this type is called a Continued Fraction.
106 FRACTIONS. [Chap. IV.
EXAMPLES XXXVIII.
Simplify :
1. _. 4. Z5. 7. H. 10
I lo i H
Written Exercises.
1 2
10'
7.
*
4'
8.
3
7
A
I
1
9.
«i
1 5. 1. 8. 1. 11. 9.
I * A i*
I. 6. 1 9. ?i. 12. It.
HI
3 l~lj 21 3 £ of 4 ^
2|-| • 2iof6^
. 7 A-4A 22 . § + 2i
5A-2f 1A + 2*
2 _i_ 1
i + i
tt-t
tt + *
i + A
i + A
A- A
15. U.H. 19. 15 A-1°^ 23. 3 * + 4 ^
18f-16A 6j + l T v
20. fLzll. 24. i + *+£
I °i I A+t**+A
A ~"TT
14 1 18
-A-l
1 4- 1 —
TO" tf
■A-A
29.
1
2-
22 3 +
26. — *V* 2_i
1 + 8Tf 30. *
1 +
27. —^—.
5-
7-f
31.
3 + 4
7
28 -2^S-
5 3i + 2J 3-f
Art. 127.] COMPLEX FRACTIONS. 107
32. 3 33. ?& .
Q I jj Ql jj
127. We now proceed to give examples of a more
complicated nature; it will be well, however, for the
student to consider carefully the following cases in which
mistakes are frequently made in the meaning of the signs
employed.
I. Operations of multiplication and division are to be
performed in order from left to right, and each sign is a
direction to multiply or divide what precedes the sign by
the number that follows next after it.
For example, 36 x 6 + 3 = 216 + 3 = 72,
36 -6-3 = 6-3 = 2,
and
36 -=- 6 x 3 = 6 x 3 = 18.
So also,
2 X 3. 8_2x3 . 8 _ 2 x 3 16 .
3 5 * 15 3 x 5 15 3x5 8 "
2.5. 4_2 6 . 4_2 v 6 5 ,
3 6 5~35 * 5~3 X 5 4 '
_3
"4'
and
2 . 6 4_2 6 4 16
3 " 6 5 3 5 5 25
II. Numbers connected by the sign ' of ' must be con-
sidered as a single number, just as if they were enclosed
in brackets.
Thus, M^ of 7 = U^2_^ = U x 5 2 ^ = 8 >
15 5 8 15 5x8 15 2x7 3
Again,?of^§of^ = ^^^§^ = 3 -^ >< §JL6 = §.
8 4 8 8 6 4x8 8x6 4x8 3x5 2
III. Before performing any operations of addition or
subtraction, all multiplications and divisions must be
performed, and complex and compound fractions must
be reduced to simple fractions.
108 FRACTIONS. [Chap. IV.
Thus,MofM = 2 + ^ + §
'3 4 6^4 3^4x6^4
= ?, 6 ■ 3_ 16+15 + 18 = 49_ Q1
3" 1 " 8 + 4 24 24 * ¥ '
It is a very common mistake to work a question of this kind as
ifitmeaat + of +
EXAMPLES XXXIX.
Written Exercises.
12
Simplify :
1. fxfxf 15. 284f of }f -h 17J
2. i + *Xf 16. ^ + l|«f4J«rflf
3- l-f^-f 17- lof|^13Joff
4. | + A+f 18- 4foflf + 4£of2f
5. H+i*2i> 19. ty+fafft + q.
6. 1J + JX2J. 20. li + SJoff x6J.
7- }+|xA+f 21. 2j-lfof 1^ + 4
s- A+fxA+q. 22 - * + i«fi-f
9. 6Jx4t-f.q-r.2f. 23. |of| + |-|.
10. | + 4of^. 24. ^-Jofl-J.
11. frff-l-A- 25. |ofi-|off
12. f«rfq-»-f 26. 2| + 1\ of 2£ - 3J .
13. ^of3f-=-2f 27. 2£ of q + 21, of 3f
i4. q-t-ifofq 28. 3|-Aof2j-if
29. fof3f-2|of ^of2^-.
so. liofi+q-A-
31. f of 2f - 4| of 5£ -=- 2| of 3}.
Art. 128.] " EXAMPLES. 109
32. f + U<**t-i-**ifr-
33. 3lofl T V + 7i-l|-|of^.
35. 2 i-|° fl J
iof3i + if
' » + *« 2 ^i
37 * + » + & . T 5 I Qf2 i
* i- T V ' * at 4*
21-jofll + f
39.
128. To express one number or quantity as a fraction of
another, we proceed as follows :
Ex. 1. Express 174 as a fraction o/188.
Now 1 = T is of 188 ;
... 174 - i%± of 188
sfg of 188.
Ex. 2. Express 2\ dollars as a fraction 0/8 dollars.
Now 1 dollar = $ of 8 dollars ;
.-. 2 1 dollars = ?f of 8 dollars
= T \ of 8 dollars.
That number or quantity which is the part must be the numerator,
while the other number must be the denominator, of the required
fraction.
HO FRACTIONS. [Chap. IV.
EXAMPLES XL.
Oral Exercises.
1. Express 27 as a fraction of 81.
2. Express 140 pounds as a fraction of 280 pounds.
What fraction of
3. 9 is 3? 6. 49 is 7? 9. 9 is2j?
4. 11 is 7? 7. 56 is 49? 10. 16 is 2$ ?
5. 20 is 5? 8. 88 is 4? 11. 2iis T \?
Written Exercises.
12. How many times does 8 J feet contain 2 J feet ?
13. Express -£$ of 4 dollars as a part of 7 dollars.
14. Eeduce 2\ of 11 cents to the fraction of 5f of 15
cents.
15. What would be the measure of |- of 23 tons, if \ of
4 tons were used as the unit ?
16. If the income of A is -f- of f of 1260 dollars, and the
income of B is T 8 T of g 1 ^- of 5440 dollars, how large is A's
income compared with B's ? How large is B's income
compared with A's ?
What fraction of
17. (8-2T3)(6+ 7-3 2 )is2 3 ?
18. 16 X 17 is 4[6 - jll -(3 + U) j + 2] ?
K20|-6|-2) 2 " J
1Q 488x11—1 of 7500 . fttl + MS 2 ,,, . 55- -=
3*x5+l is (6+14) -| of J/xi 22-5.
129. Reduction of Decimals to Common Fractions.
Decimals may be considered as fractions with powers
of 10 for denominators.
Thus, .6 = ^; M = J&; .002007 = T &°&W
Arts. 129, 130.] REDUCTION TO DECIMALS. HI
Ex. 1. Beduce .76 to a common fraction.
7« _ 7 6 —19
Ex. 2. Beduce 4.012 to a mixed number.
4.012 = 4 -f T Ho = M o-
130. Reduction of Common Fractions to Decimals.
Ex. 1. Express / 5 as a decimal.
Since / s may be considered as the quotient obtained by dividing
4 by 25, we have only to perform this division. Thus,
25)4.00(.16
2 5
1 50
1 50
Ex. 2. Beduce, to 3 places of decimals, the common fractions,
|, £}, <md f ; and thus show that the fractions are in ascending
order of magnitude.
The decimals required are .75, .854..., and .857....
EXAMPLES XLI.
Oral Exercises.
leduce
i to decimals :
i. i.
4. f.
7. f.
10.
f
2. J,
5. f.
8. »?>
11.
t-
3. |.
6. 1
9. f.
12.
H-
Written Exercises.
13- ft.
16.
**&•
19. £V 22- 7 T VA-
I*- m
17.
T2"T'
20. ^. 23. W^yy^-
15- tIt
18.
21
l2~0"'
21- 3»Hr
112 FRACTIONS. [Chap. IV.
Circulating Decimals.
131. We have hitherto considered examples of division
of decimals in which by proceeding far enough an exact
quotient is found with no remainder. This, however, is
by no means always the case ; in fact, it is very rarely
the case.
Consider, for example, the division of 5 by 3.
3 1 5.0000
1.6666...
We may here continue the process of division to any extent, but
each figure of the quotient will be_6, and the remainder will always
be 2.
Again, divide 2 by 7.
7 1 2.000000000
.285714285...
Here the six digits, 2, 8, 5, 7, 1, 4, come over and over again in
the same order, and we shall never arrive at a stage at which there
is no remainder.
When a decimal ends with digits which are repeated
over and over again without end in the same order, the
decimal is called a Recurring or Circulating decimal, and
the digit, or set of digits, which is repeated, is called the
Circulating Period, called also the Repetend.
Thus, 2.45555..., .014141414..., and 5.1246246246... are circu-
lating decimals with circulating periods of one, two, and three
figures, respectively.
A circulating period is denoted by placing dots over
the first and last of the figures which recur.
Thus, 2.45 denotes 2.45555..., .014 denotes .014141414..., and
5.1246 denotes 5.1246246246...
A circulating decimal is said to be Pure or Mixed,
according as all the figures after the decimal point do or
do not recur.
Art. 131.] EXAMPLES. 113
Thus, 5.6, 31.24, and 14.i35 sue pure circulating decimals; and
.56, 3.124, and .14135 are mixed circulating decimals.
A decimal which contains a definite number of figures
is called a Terminating decimal, to distinguish it from a
circulating decimal, which contains an unlimited number
of figures.
Note. Although it is not possible to reduce any common frac-
tion to a terminating decimal, it is always possible to find a decimal
which is equal to the common fraction to any degree of accuracy
that may be required.
For example, $ lies between .333 and .334, so that the difference
between £ and .333 is less than one one-thousandth, so also the
difference between ^ and .333333 is less than one one-millionth;
and so on.
Now there is no species of magnitude which can be measured
with perfect accuracy. It would, for instance, be difficult to deter-
mine the length or the weight of a body without a possible error as
great as one one-thousandth of the whole. Hence the measure of
any quantity can be expressed as accurately by means of decimals
as by means of fractions.
EXAMPLES XLII.
Written Exercises.
Express the following quotients as circulating decimals :
1. 1.5-2.7. 4. .035 -.072. 7. 3.1-7.
10 - .03.
5.
.316 -r- 2.4.
8. 15.6 -.07.
1.7 - .09.
6.
.312 - 8.8.
9. 1.25-13.2.
0. 5.193 -r-
.0168.
13.
.3157 - .259.
1. .0235 -^
.00616.
14.
27.31 - 6.475.
2. 16.72-
.0143.
15.
693.11 - .011396.
Reduce the following common fractions to circulating
decimals :
16. |. 18. f. 20. T V 22. &.
17. f 19. ^ 21. ft. 23. iV
114 FRACTIONS. [Chap. IV.
24. 2&. 26. 5Jf 28. 11^. 30. 2fif.
25. 3if 27. 7^. 29. 13 T V 9 ,. 31. 5^V
132. Reduction of a Circulating Decimal to an Equivalent
Common Fraction.
We have seen (Art. 129) that a terminating decimal can
be expressed as a common fraction. We have now to show
that a circulating decimal may be expressed as a common
fraction.
Consider the decimals, .31, .5216, and .15607.
In each case multiply the decimal by that power of 10 which will
move the decimal point to the end of the first recurring period ; also
(if necessary) multiply the decimal by that power of 10 which will
move the decimal point to the beginning of the first recurring period.
Subtract the second product from the first, and notice the result.
0)
.31
X
100
= 31.31
.31
X
1
= .3i
.3i
X
99
= 31.
.-.
.31
_31
99*
No advantage will be gained by repeating the .31 in the minuend
or subtrahend ; we obtain only an integer in the remainder.
(ii) .5216 x 10000 = 5216.5216
.5216 x 1 = .5216*
.5216 x 9999 = 5216.
.-. .5216 = ffi!-
(hi) .15607 x 100000 = 15607.607
.15607 x 100= 15.607
.15607 x 99900 = 15592.
.-. .15607 = iUW
Art. 132.] CIRCULATING DECIMALS. H5
Three facts concerning the fraction equivalent to a cir-
culating decimal are easily noted :
1. The numerator is the whole decimal minus the number
expressed by the non-recurring digits.
2. The number of 9's in the denominator equals the
number of recurring digits.
3. The number of naughts in the denominator equals the
number of non-recurring digits.
Ex.1. i = f, Ex.3. .156= #* = *■
Ex. 2. .07 = ft. Ex. 4. 3.3L2 = 3|ff = 8||f.
It should be noticed that by the above rule .9 = f = 1. This
result can be seen independently ; for the differences between 1 and
the decimals, .9, .99, .999, etc., are respectively .1, .01, .001, etc.,
each difference being one-tenth of the preceding, and therefore
when a large number of nines is taken, the difference between 1 and
.99999... becomes inconceivably small.
Since .9 = 1, .09 = .1, .009 = .01, and so on, a recurring 9
can always be replaced by 1 in the next place to the left ; for
example, .79 = .8 and .249 = .25.
EXAMPLES XLIII.
■Written Exercises.
Find common fractions in their lowest terms equivalent
to the following circulating decimals :
1.
.3.
7.
.185.
13.
.04878.
2.
.09.
8.
.396.
14.
.07317.
3.
17.27.
9.
.142857.
15.
9.23.
4.
.15.
10.
.285714.
16.
.79.
5.
1.027.
11.
.428571.
17.
6.36.
6.
.037.
12.
.012987.
18.
.315.
116
19.
.116.
20.
.0254.
21.
.016.
22.
.749.
FRACTIONS.
[Chap. IV.
23. .2027.
27.
11.3021976.
24. .19324.
28.
.542857L
25. .402439.
29.
.012345679.
26. .304878.
30.
.135802469.
It should be noticed that if a common fraction in its
loicest terms be equivalent to a terminating decimal, the
denominator of the fraction can contain only the prime
factors 2 and 5.
133. Addition, Subtraction, Multiplication, and Division
of circulating decimals are performed after first reducing
to common fractions. The answer in each case should
be reduced to a circulating decimal.
134. An exact divisor of a number is sometimes called
an Aliquot Part of the number.
2* is an aliquot part of 10 ; 16| is an aliquot part of 100.
This enables us to use a short process of multiplication
(or division) in cases where the multiplier (or divisor) is
an aliquot part of some power of 10.
Tox 31,
x 10 and -
-3.
To-
8J,h
- 10 and x 3.
To x 12i,
x 100 and -
=-8.
To-f
- 12i, -
r- 100 and x 8.
To x 16|,
x 100 and -
-6.
To-
- 16f , -
- 100 and x 6.
To x 25 ,
x 100 and -
-4.
To-
25 , -
- 100 and x 4.
To x 33|,
X 100 and -
-3.
To-
83*,-
- 100 and x 3.
To x 125,
x 1000 and -
-8.
To-
125, -
- 1000 and x 8.
Kead the signs ' multiply by ' and ' divide by '.
3 2
Square Roots of Fractions.
/3V = 3 x3 ^
\±J 4x4 4 2
it follows conversely that
135.
Since
4
9 = 3 = y9
16 4 V 16
Arts. 133-135.] SQUARE ROOT. H7
Thus, the square root of a common fraction is equal to a
fraction whose numerator and denominator are respectively
the square roots of the numerator and denominator of the
given fraction.
Ex. 1. Find the square roots o/|||, l T 9 g, .4, and 2.086419753.
;144 _ y!44 _ 12 . V ^JX = /25 _ y25 _ 5 ,
169 V 169 13 ' "Vm V 16 4 '
1144
\169
■V s
V 9 3'
and V2.086419753 = y/2fMtftffo = y/2%
/ 169 _ yi69 = 13 = i 4
81 V 81 9 ~ " '
Ex. 2. Find, to four places of decimals,
(i) ^| (") ^ (iii) V-3, and (iv) JL
(i) ,»/- = -^-- = - V 5 * which can be fonnd as in Art. 88.
(ii) In examples in which the denominator is not a perfect
square, the fraction should be expressed as a decimal. In the
present case */- = ^.8 = ..., etc.
Or thus : J| -J5 .Jflnlv»..-, etc*.
\5 \25 V25 6
(iii) .3 = .33'33'33'33' ... Then proceed as in Art. 88.
(iv) JL = _A*^L = 6 v3 = ..., etc.
The change of form from — to - V 3 will save labor.
V3 3
EXAMPLES XLIV.
"Written Exercises.
Find the square roots of
1 -Mr
3. iff.
5. 39^. 7.
.004.
o 1 2 5
4- lift
6. .1. 8.
.134.
9.
1.361.
10. 4.38204.
118 FRACTIONS. [Chap. IV.
Find, to four places of decimals, the square roots of
11. tV. 13. 3£. 15. 2.4. 17. .083.
18. 3.5i62.
12. T V
14.
8f
16. .041.
136.
The H.C.F.
and L.C.M
. of Fractions.
By the H.C.F. of two or more fractions we mean a frac-
tional H.C.F. The quotients, however, are integral.
A fraction -f- a fraction = an integer only when the numerator
and denominator of the dividend divided by the numerator and
denominator of the divisor respectively, produce an integer and
the reciprocal of an integer ; thus,
27 81 }
Here, 14-4-2 is an integer, and 27 -r- 81 is the reciprocal of an
integer ; i. e. , the numerator of the divisor is a factor and the
denominator of the divisor is a multiple ; also, the numerator of
the dividend is a multiple, and the denominator of the dividend is
a factor.
Hence, the H. C.F. of two or more fractions must have for its
numerator the H. C.F. of the given numerators, and for its denom-
inator the L.C.M. of the given denominators.
Also, the L.C.M. of two or more fractions must have for its
numerator the L. C. M. of the given numerators, and for its denom-
inator the H.C.F. of the given denominators.
Note. Before obtaining the H.C.F. or the L.C.M., the given
fractions must be in their lowest terms, and mixed numbers
must be reduced to improper fractions. The L.C.M. may be
integral.
Ex. 1.
Ex. 2.
f The H.C.F. of f and i| = & ;
\ The L.C.M. of f and J| = - 5 /.
The H.C.F. of | and § = ft ;
The L.C.M. of f and T % = 5,
Art. 136.] EXAMPLES. 119
EXAMPLES XLV.
Written Exercises.
Find H.C.F. and L.C.M. of
1. |,2V, and -y, 4. Jf ||, and ||,
2. A>6 6 5>andLf. 5. fftandftf
3. ]»-&{> and ft 6. ff and Iff
7. A 8 5. t 6 A> and 39 &.
EXAMPLES XLVI.
Miscellaneous Examples. Chap. IV.
1. Reduce 5-f, 8y 3 T , and 25^- to improper fractions.
2. Simplify |_3 + _5___5_ + _7__ _7_
3. What must' be added to 5J that the sum may
bel2f?
4. Multiply 2| of 5f by 3| - 6f
3 1 1 n f S 5 '
T4 -h 3 0I ¥ — 2T
31 _ 5 v 61 3
5. Simplify ?* f T4
6. Arrange, in ascending order of magnitude, the
fractions, T \, f, T 8 5, ft
7. From the sum of % and i take the difference be-
tween \ and |.
8. Simplify 2f of 4§ of 5f
9. Simplify 3 ^ + 2 ^ 7 4 » x A.
12 ~f
10. What fraction of 350 equals f of 168 ?
11. Reduce ft$$f and ||}|| to their lowest terms.
12. Eeduce to a common denominator ^ ¥ , g- 2 -^-, T ^- S}
and T |^.
120 FRACTIONS. [Chap. IV.
13. Simplify 3| + 2f of 1$ - 4$.
14. Simplify t'ttt" 1 ? -
15. A and B started on a tour with 192 and 156 dollars
respectively, and they had equal sums left at the end.
A spent J of his money ; what fraction did B spend of his ?
16. Add T V, &> A> lWr> A 7 o> and ^V
17. Subtract 5Jf from 7^ ; also, 6f + 2f from 12 J.
18. Divide2 T ^4-2||-3/ T by2 i V + 3i-4f
19 simplify l* + 2jaf5t-12 t >
20. What is the value of f of a property, if -f of it is
worth 750 dollars ?
21. Eeduce ffjf, £ft£M, and i^ff* to their lowest
terms.
22. Show that 1 + - t + ^— 5 — ; — =—7: is less
2 2x3x4 2x3x4x5x6
than ^-, but greater than T \.
23. Simplify fff X f f X &
2 j + 1 of lj - 3 4-
3 _|_ 1 _;_ 2 _ 1_
1" I 4 * 3" 8
25. Find the G.C.M. of 5J and 4|; express the answer
as a circulating decimal and obtain the square root.
26. Simplify 4 + | + f + f - ¥ ~ 2V
27. Subtract 2|f from 5£f, and 8if from 12^.
28. Multiply 3£ of 5| by 4-J- of f , and divide the result
by 41 of 1J.
24. Simplify ^ 3 ~Vi ZT l
4" 1 4 "*" 3" ~~ ¥
Art. 136.] MISCELLANEOUS EXAMPLES. 121
29. Sunphfr f^^ Vi^f of jL_ .
30. Find the L.C.M. of J and §.
31. By how much does the sum of 1^, f, and ^ fall
short of the sum of -J, |-, J, and ii ?
32. Simplify 1\ - \ of 4J + 2| + 3* X 2J - ^.
33. How many pieces each f of 1 inch can be cut from
a wire whose length is 5] inches ; and what will be the
length of the piece left over ?
34. Simplify ^ - 3j + 5 tV _ llf - 5 T y
35. Find L.C.M. and H.C.F. of ^fo |fj, and T |§ v
36. Take the sum of f and \ from the sum of £ and f .
37. After taking away \ and f of a certain quantity,
what fraction of the whole will be left ?
38. Multiply \\ + 3| by 3J + 2f, and divide the result
by 51 of 5|.
39 Simplify 2 H ~ A of 4 T y + 3^ + 7f
39 ' SimpM > 3 A oflO T V-3i-^4 t *
40. Find the value of 3 T 3 T of 4f of If of Jj,
41. Add |, f|-, -J^-, TT0 -, and 28 3 o6 .
5
42. Simplify —
91 _2 ft f15
43. Simplify ? '/V,' , and
T OI ° 3" T "8
i L
2 + -!-
122 FRACTIONS. [Chap. IV.
44. There are three partners in a certain business, one
of whom provided -f of the whole capital, and another
provided f. What fraction of the whole was supplied
by the third partner ?
45. A man gives \ of his money to his wife, \ of the
remainder to his son, and \ of what then remains to his
daughter ; and has still left a sum of 1350 dollars. How
much was there at first ?
1
1-
46.
Simplify 9i 9 ^" 6 8 9 , 7 7 t and r^
2 + i
47. Divide 1£ of 5| by 2f of 7^-.
»-^£»-»
49. Find the value of f of f of 5 dollars - 1 of £ of 2
dollars, and express the difference as a fraction of 11.25
dollars.
50. Reduce to its simplest form
(3t + 5|-A-)(4i-3i)
lA + 2*-(2A-i-A)
51. After spending J- of his money, a man found that
f- of the remainder was 63 cents ; how many cents had he
at first?
52. I purchased some square tiles for a room 483
inches long and 266 inches broad ; the manufacturer sent
me the largest tiles I could use ; how long was each tile ?
53. A man travelled f of a certain distance by railway,
-^ of the whole distance by coach, and walked the rest
of the way, which was 15 miles. What was the length of
the whole journey ?
Art. 136.] MISCELLANEOUS EXAMPLES. 123
54. By what must 7§ of 3| be multiplied that the prod-
uct may equal 4f of 2f ?
55. Simplify g*I^M^M±lli ; ^^ ^
answer into two fractions so that one factor shall be a
perfect square.
56. Find the H.C.F. and the L.C.M. of ££ and f f
57. Subtract 10ft from 23|, and 16|f from 20| J.
58. Simplify (3* + 21 + 4J) + (J + f) of (| - *)•
59. Simplify
Q 2
3 _ 2 v I — 1 °T¥
U (4 i -2|) + (3i-li)' <■ ; 2J-J-4
4— 4
60. A man gives away f of his money and afterwards
-J of the remainder. What fraction of the whole had he
then left?
61. Reduce to a common denominator, and arrange in
order of magnitude the fractions, -j^-, T 7 ^, -g 9 ^, if? if •
62. Multiply the difference between 3£ of 1 T V + 7|
and 2\ -+- f of -^ by the sum of _ and —
63. Simplify I of 6 9~ 2 f^ K¥ + 19 t).
64. After spending -| of his money, a boy found that $
of the remainder was 2| dollars. What had he at first ?
65. Reduce to their lowest terms f^J, y^ff, and
66. A gave i of his marbles to B, i to C, i to D, Jg-
to E, and then had 105 left. How many did each
receive ?
124 FRACTIONS. [Chaps. IV., V.
67. Find 2||of {2J| -4<rf (3i of 2f - 5|) of 1J - Af.
68. 64 feet of brass rods cost 121 cents a foot; what
was the cost of the rods ?
69. Nashville, Tenn., Jan. 1, 1895.
J. S. Cushing & Co. To H. A. Armstrong, Dr.
For 2 lb. Sugar @ 7 cents
" 5 " Tea "50 "
" 11 " Coffee "34 "
" 17 " Starch "14 "
Find the amount of the above bill ; answer in dollars
and cents, letting 100 cents equal one dollar.
What would be the answer in dollars, and the decimal
of a dollar ?
^ 70. Find the H.C.F. and L.C.M. of 1.485 and 12.6.
71. Simplify m*<
81x32x34
72. Add, without changing positions: 67.04, 12, 5^,
9 T \, 4.17, 243^, 14, S T \.
73. A certain lake is .327 of a mile long; what is its
length compared with the length of a second lake 2| miles
long?
(Answer must be reduced to a circulating decimal.)
74. Three tanks contain 924, 1500, and 2520 gallons of
water respectively ; what is the largest number of gallons
that can run from each of the tanks per minute and allow
all to be emptied in a whole number of minutes, the rate
of flow from each tank being the same? How many
minutes are required to empty each tank ?
• I
Arts. 137, 138.] MAGNITUDE — QUANTITY. 125
CHAPTER V.
DECIMAL MEASURES.
137. Anything which, can be increased or diminished
is called a Magnitude.
Lengths, areas, weights, etc., are magnitudes.
To measure a magnitude is to compare it with some
known magnitude of the same kind, which is taken as
a unit, and to say how many times the unit must be
repeated in order to make up the magnitude in question.
For example, to measure any given length of string, is to find
how many times some known length, say a foot, must be repeated
to make up the given length ; and this number of times is called
the measure of the length.
A measured magnitude is called a Quantity.
Thus, any quantity is expressed by a number and a
unit of the same kind as itself.
138. Numbers are first used in connection with distinct
objects, and are afterwards used in measuring continuous
magnitudes of any kind. If the continuous magnitude
cannot be measured by one unit, a series of units smaller
and smaller in value may be used.
For example, to measure a string, some definite length, say a
yard, is fixed on as a unit. Suppose the given string contains
Q\ yards. We may use a second unit, say a foot, to measure the
126 DECIMAL MEASURES. [Chap. V.
^ yard. If there are 3 feet in one yard, the \ yard will be \\ feet,
and the string will measure 6 yards \\ feet. This \ foot may be
expressed in a smaller unit still, say an inch ; if there are 12 inches
in a foot, the \ foot will be 6 inches, and the string will measure
6 yards 1 foot 6 inches.
139. Quantities expressed in terms of a single unit
are called Simple Quantities, and quantities which are
expressed in terms of more than one unit are called
Compound Quantities.
To measure every different kind of quantity, some
standard unit is employed, and also other units which
are obtained by subdivisions and repetitions of the
standard unit.
Units which require 10 of one to make one of the next
higher are the simplest to use. Such units are called
Decimal Units.
In numeration of quantities, units of different kinds
are called units of different denominations.
Tables of Decimal Units.
140* Table of United States Money.
Money is a measure of values.
10 mills (m.) = 1 cent (ct.).
10 cts. = 1 dime (d.).
10 d. = 1 dollar ($).
10 $ = 1 eagle (e.).
The eagle is usually called ten dollars, and the dime is
usually called ten cents; so that the only names generally
used are dollars and cents.
* It is advisable to study numeration and notation of decimal
measures at the same time.
Arts. 139-143.] U. S. COINS. 127
Thus, $25.35 is read, '25 dollars 35 cents,' and not, '2 eagles
5 dollars 3 dimes 5 cents ' ; also, $.20 is read, ' 20 cents.'
The notation is as follows :
The figure representing eagles is put in tens' place,
u " dollars " " units' place,
" " dimes " " tenths' place,
" u cents " " hundredths' place,
" " mills " " thousandths' place.
141. A sum of money represented in any denomina-
tion may be represented in higher denominations by
moving the decimal point to the left, one place for each
denomination. A reduction is made to lower denomina-
tions by moving the decimal point to the right ; thus,
6742 mills = 67.42 dimes = 6.742 dollars ;
4671 dollars = 46.71 cents = 467.1 mills.
142. We have already shown how to perform the
operations of addition, subtraction, multiplication, and
division of decimals ; and the application of these rules
to sums of money will require no further explanation,
except to state that in cases of addition and subtraction
care must be used in writing units of the same denomina-
tion in the same vertical column. This is not a necessity
— only a convenience.
143. The coins in use are as follows :
Gold coins : the dollar, the quarter-eagle, the half-eagle,
the eagle, and the double-eagle.
Silver coins: the dollar, the half-dollar, the quarter-
dollar, and the dime.
Nickel coin : the five-cent piece.
Bronze coin: the cent.
The mill is used only in computation.
128 DECIMAL MEASURES. [Chap. V.
EXAMPLES XLVII.
Written Exercises.
1. Write the following in figures : two dollars thirteen
cents, sixty dollars forty cents, three hundred dollars two
cents, sixteen cents, six cents, three cents five mills.
2. Add $14.15, $37.24, $156.50, $.75, and $1204.06.
3. Add $2.04, 26.7 ct., 49.62 m., and 4.338 ct.
4. By how much is $1507.45 greater than $1429.78?
5. After spending $145.45 a man had $13.55 left;
how much had he at first?
6. A man had originally $1345.40. How much had
he left after paying away $135.25, $416.67, and $575.48 ?
7. What will 250 barrels of apples cost at $2.75 per
barrel ?
8. A man bought 150 horses at $125 each. He sold
50 at $145 each and the rest at $137.50 each. How
much did he gain ?
9. What is the value of 1400 bushels of wheat at 67
ct. a bushel ?
10. A man bought 1250 bushels of oats at 381 c t. a
bushel, and 1500 bushels of wheat at 65|- ct. a bushel.
What was the whole cost ?
11. A man bought 15 pounds of cheese at $.14 a
pound, 9 pounds of coffee at $.25 a pound, and 13
pounds of butter at 18 ct. a pound. How much did the
whole cost?
12. Two men had between them $1595, and one had
$155 more than the other. How much had each ?
Art. 144.] THE METRIC SYSTEM. 129
13. Multiply $684.93 by 6.75.
(Give answer to two decimal places, remembering that 5 or
more mills increase the number of cents by one ; anything under
5 mills is not considered.)
14. Multiply $71.41 by .23.
15. Divide $5687.98 by 27.3.
(Be sure in the answer to find out whether or not the mills will
be as many as five.)
16. A man spent $4.86 in buying beef at $.09 per
pound. How many pounds did he buy ?
17. A man bought wheat at 64 ct. a bushel, and spent
$736 altogether. How many bushels did he buy ?
18. How many d. in $560.1 ?
19. How many e. in $270 ?
20. How many m. in $41.90 ?
21. How many ct. in 86420 m. ?
22. How many $ in 86420 m. ?
23. Divide 784 d. by 2.75, and write the answer as e.
24. Multiply $76 by .0025; of what denomination
is the answer ?
The Metric System.
144. In almost all civilized countries, the United
States and England being unfortunately exceptions, the
different weights and measures have been arranged on
the decimal system.
In France, Belgium, and Switzerland all sums of
money are expressed in terms of the Franc, with its sub-
unit the Centime (y^- of a franc). In Italy, Spain,
and Greece the standard unit of money is of exactly
the same value as the franc, but is called by different
names.
130 DECIMAL MEASURES. [Chap. V.
In Germany the standard nnit is the Mark, with its
sub-unit the Pfennig (y^ of a mark).
In Austria the standard unit is the Gulden, with its
sub-unit the Kreutzer ( T ^-g- of a gulden).
In the United States the Metric System is used in
scientific investigations and is authorized to be used in
the Mint and Post Ofhce.
145. In the Metric System of weights and measures,
the fundamental unit is called a Meter. The meter is
approximately the one ten-millionth part of the distance
from the equator to the north pole. (A slight error was
made in obtaining the meter, but its length remains as at
first calculated.)
The standard units of area, volume, capacity, and weight
are derived from the meter.
Decimal divisions of a standard unit are distinguished
by the Latin prefixes deci-, centi-, milli-.
Decimal multiples of a standard unit are distinguished
by the Greek prefixes deka-, liekto-, kilo-, myria-.
Tables of Decimal Units. — Continued.
146. Table of Linear Measures.
Length is distance in a straight line between two
points.
The unit of linear measure is a meter.
10 millimeters (mm) = 1 centimeter (cm).
10 cm =1 decimeter (dm).
10 dra =1 meter (m).
10 m =1 dekameter (Dm).
10 Dm =1 hektometer (Hm).
10 Hm = 1 kilometer (Km).
10 Km =1 myriameter (Mm).
Arts. 145-148.]
LINEAR MEASURES.
131
If the figure representing meters is put in units' place,
then " " dm " " tenths' place,
" " cm u *« hundredths' place,
" " Dm " " tens' place,
" " Hm " " hundreds' place, etc.
147. Length, represented in any denomination may-
be represented in higher denominations by moving the
decimal point to the left, one place for each denomina-
tion; a reduction is made to lower denomi-
nations by moving the decimal point to the
right; thus,
14.45 m =144.5 d ™ = 14450 "«■ ;
1256.4 «n - 12.564 ™ = 1.2564 »m ;
2 Km _ 200 Dm = 20000 dm ;
24.6 Hm =2.46 Km = .246 Mm .
The same methods are used in operations here as
in decimals [Arts. 29 and 49]. Units of the same
denomination should be in the same vertical column.
The teacher should have a meter stick, properly-
graduated, and keep it constantly before the class.
Not a word should be said about any other linear
measure, known or unknown to the class.
148. Figures representing decimal measures
of any kind are read just as figures represent-
ing integral and decimal numbers are read,
and then the name of the denomination repre-
sented is read ; thus,
14.45 m is read 'fourteen and forty-five hundredths
meters, ' (which means the same as if it were read ' one
dekameter four meters four decimeters and five
centimeters').
.246 Km is read 'two hundred forty -six thousandths
kilometers.'
1
% :
S
s*
g
o
o
i— i
U
II
E
o
O
t-H
*
II
U
a>
I
6 :
o
<x>
t-H
7 ;
8 :
9 '-
10 ;
132 DECIMAL MEASURES. [Chap. V.
EXAMPLES XL VIII.
"Written Exercises.
1 . Cut from cardboard a narrow strip, l dm long, and
mark it accurately into tenths and hundredths.
2. Obtain the measure of the length of a book, and
state the answer in dm and mm.
3. Mark your height on the wall, and obtain its
measure in m ; also in dm.
4. Measure a room in m; obtain length, breadth, and
height.
5. Express 25 m as Dm; as Mm; as cm; as mm.
6. Write 126.73 Dra as m; as Km; as dm; as mm.
7. Add 14™, 6 dm , 5027 mm , and 6.5 Bm . Answer in m.
8. Find the number of Dm in 12.62 Mm + 4267 m -f
845 cm .
9. How much longer is a room 12.65 m than a room
106 dm long?
10. Find 8469 ra + 46892 mm - 468 Dm + 12 dm - 186 cm in m.
11. Multiply 78.6 dm by 125. Answer in m; also in
Hm.
12. Four m of ribbon cost 16| ct. per m; find total
cost [Theorem I, Art. 47 ; also Art. 50]. The answer is
what fractional part of $1 ?
13. Divide 7469 mm by 11. Answer in three denomi-
nations.
149. Table of Surface Measures (Square Measures).
That which has length and breadth, but no thickness,
is called a Surface; thus,
The surface of a book has length and breadth.
Arts. 149, 150.] SQUARE MEASURES.
133
A portion of a surface bounded by lines is called a
Figure.
A plane figure bounded by four equal sides,
and whose four angles are equal, is called a
Square.
Any square may be used as a unit of surface
for instance, a square centimeter, or a square
Square
Centimeter.
measure
meter.
100 square millimeters (qmm) = 1 sq. centimeter (qcm).
100 qcm = 1 " decimeter (qdm).
100 qdm = 1 " meter (qm).
100 qm = 1 " dekameter (qDm).
100 qDm =1 " hektometer(qHm).
100* Hm =1 " kilometer (qKm).
For Land Surveying.
l qm is called a centar (ca).
lqD» U U an ^ ^
lq Hm U « a l ie k tar ( Ha ) #
Sq. cm, etc., are often used instead of qcm, etc.
150. It is evident from the figure that, if one square
10 units
lo
ngi
I unit long.
□
is 10 times as long as
another, its surface is 100
times as large j there-
fore,
A surface represented
in any denomination may
be represented in higher
denominations by moving
the decimal point to the
left, two places for each
denomination; a reduction is made to lower denomina-
134 DECIMAL MEASURES. [Chap. V.
tions by removing the decimal point to the right, two
decimal places for each denomination; thus,
15.6<i m = .156<i Dm ; 10625<i mm = 1.0625 i dm ; 12 a = .12 Ha .
1.49qdm = i49qcm . i.0625<i Hm = 10625 i m ; 1.6 a = 160 ca .
15.6 <J m is read ' fifteen and six-tenths square meters.'
EXAMPLES XLIX.
"Written Exercises.
1. Cut from cardboard a piece to represent one qdm,
and mark it accurately into qcm.
2. How many qcm in \ a qdm ? How many in the
square of ^ a dm ?
3. Mark out on the floor a qm. What would that be
called if it were marked on the ground ?
4. Write 30 qm 15 qdm 21 qcm as qm ; as qHm ; as qmm.
5. Express 31 qm as qcm; 14.1 qm as qdm ; .5 qm as qcm;
120.7 qKm as qm.
6. Read 15.14 qDm as qm ; as qcm. Read l.l qHm as
qDm; as qmm. Read 121 a as ca; as Ha.
7. Multiply 78.141 qm by 16, and answer in qmm.
8. Represent 15.6789 qm as qmm ; 140 (icm as qHm.
9. In 1.49 qdm what might the .49 be called ?
10. Read 15.6 qm as qDm and qdm.
11. Represent l qKm , 12 qHm , l qm , 4 qmm as qm.
12. Read, stating the number of units of each denomi-
nation represented, 167.08193 qHm .
13. Divide 78965 qDm by 5, and answer in qm.
14. Add 167 qm , 200 qKm , 18.67 qDm , and 160003 qmm .
Arts. 151, 152.]
CUBIC MEASURES.
135
15. From 12.6 a subtract 4 ca .
16. A piece of ground containing 400 a is 2000 dm long;
what is its breadth ?
151. Table of Volume Measures (Cubic Measures).
A solid bounded by six equal square sur-
faces is called a Cube.
Any cube may be used as a unit of cubic
measure ; for instance, a cubic centimeter or
a cubic meter.
1000 cubic millimeters (cmm) = 1 cubic centimeter (ccm),
1000 ccm = 1 cubic decimeter (cdm),
1000 cdra = 1 cubic meter (cum).
For measuring wood, 1 cubic meter is called a ster (st).
152. It is evident from the figure that, if one ciibe is
10 times as long as an-
A
/
/—
_j
Cubic
Centimeter.
Si
/ / / /
-/-
p^g
m
other, its volume is 1000
times as large; there-
fore,
A volume represented in
any denomination may be
represented in higher de-
nominations by moving
the decimal point to the
left, three places for each
denomination; a reduction
is made to lower denomi-
nations by moving the
decimal point to the right, three places for every denomi-
nation.
Thus, 4678 cdm = 4.678 cum ; 8.67 cum = 8670000 ccm .
4.678 cum i s rea d i f our an( i s i x hundred seventy-eight thousandths
cubic meters,' which is equivalent to saying four cubic meters and
six hundred seventy-eight cubic decimeters.
10 units long.
s
unit long.
136
DECIMAL MEASUEES.
[Chap. V.
EXAMPLES L.
Written Exercises.
1 . Cut from wood (or rubber, or cork) a piece to rep-
resent l ccm .
(Each student should carry in his
pocket such a piece of wood, so that
he can constantly refer to it.)
2. Cut from bristol board a
piece shaped like the figure,
having each of its six squares
1*» long. Cut nearly through
the cardboard in places repre-
sented by dotted lines, and
make small flaps as shown.
Such pieces can be made into cubes ; a little mucilage on
the flaps will keep the cubes in shape. Flaps should be
out of sight in the finished cubes.
3. Mark the sides of the cube into qcm and learn
how many ccm, like the one in your pocket, would be
required to make a block as large as the cardboard cube.
4. How many ccm in J- a cdm? How many ccm in
the cube of i a dm; i.e., in a cube 5 cm on an edge?
5. Write 105 cum 215 cdm as cum; 27 cdm ascum.
6. Eead 10.516 cum as cdm; as cDm.
7. Eead 10067 cdm as st.
8. Kead 100601.41 cum , stating the number of units of
each denomination represented.
9. Add 14.1 cmm , 14.1 cdm , and 14.1 cDm .
10. Divide 14.4 ccm by 12, and write the answer as ccm,
and as cmm.
Arts. 153, 154.] LIQUID — WEIGHT. 137
153. Table of Volume Measures (Liquid Measures).
The cubic decimeter is used as the unit of measure,
and is called a Liter.
10 milliliters (ml) = 1 centiliter (cl).
10 cl = 1 deciliter (dl).
lO 41 = 1 liter (1).
10 1 = 1 dekaliter (Dl).
10 D1 = lhektoliter (HI).
10 m =lkiloliter (Kl).
Comparing the above table with the one in Art. 151,
we find that
•J^cdm __ -J^l
1 ccm __ 1 ml
-Jcum__ ^Kl
EXAMPLES LI.
"Written Exercises.
1. Add 4.5 ", 2 dl , 47 cl , and 673 ml .
2. Express the answer to Ex. 1 in Kl, HI, and cl.
3. How many liters of water in 4 cum ?
4. Change 46.0949 Kl to 1; to ml; to dl; to ccm.
5. Multiply .678 cl by 2693; express the answer in
cdm, and in cum.
154. Table of Measures of Weight.
The attraction which the earth and any other body (on
or off the earth) have for each other is called Gravity.
The amount of this attraction is called the Weight of
the body.
The weight of l ccra of water is the unit of weight, and
is called a Gram.
138 DECIMAL MEASURES. [Chaps. V., VI.
10 milligrams (mg)= 1 centigram (eg).
10 cg =1 decigram (dg).
10 dg =lgram(g).
10* = 1 dekagram (Dg).
10 Dg = 1 hektogram (Hg).
10 Hg = 1 kilogram (Kg).
10 Kg =1 myriagram (Mg).
10 Mg = 1 quintal (Q).
10 Q = 1 tonneau (T).
Observe, in the case of water, that
l ml (=l ccm ) weighs l g ;
l 1 (=l cdm ) weighs l Kg ;
1 K1 (= l cum ) weighs 1 T .
155. Kilogram is called Kilo. Quintal is not often used.
The cubic centimeter of water, which, is used as the standard
unit, must be distilled, must be at a temperature of 39.2° F.
(4° C), and must be weighed in a vacuum at the level of the sea.
EXAMPLES LII.
Written Exercises.
1. Eead 64.95 g as dg, cg, mg, and Mg.
2. Eead 1256 ng as Kg, Q, T, and g.
3. What is the weight of l ml of standard water? Of
10 ml ? Of l cl ? Of 10 cl ? Of 3 dl ? Of 3 1 ? Of 1000 ccm ?
4. Iron is 7.8 times as heavy as water ; what is the
volume (in cdm) of 29.25 Kg ? What is the weight of 2 cdm ?
Of 55 ccm ? Of 7.2 ccm ? Of 1.67 ccm ? Of 125 cum ?
5. Find the value (in grams) of 4 Kg - 18 dg + 18 g
+ 67.896 mg - 126.73 cg + 4 T - 11.6 Mg .
6. Gold is 19.5 times as heavy as water; what is the
weight of l ccm ? Of one cubic meter ?
Arts. 155-157.] NON-DECIMAL MEASURES. 139
CHAPTER VI.
NON-DECIMAL MEASURES.
156. The simplicity of calculations when using decimal
measures is due to the facts that changes can be easily
made from one denomination to another by moving the
decimal point, and that several denominations can be
expressed together in one set of figures.
In Non-Decimal measures, called also Denominate num-
bers and Compound Quantities, a variety of divisors is
used in the different tables in order to change from low
denominations to higher ones ; also, it is unusual to ex-
press several denominations together in one set of figures.
For example, consider the case of the string mentioned in Art.
138. There, 12 inches equal 1 foot, and 3 feet equal 1 yard ; and
the length of the string must he expressed, not with the denom-
inations together in one set of figures, hut each denomination
separately, — 6 yards, 1 foot, 6 inches.
To express a compound (Art. 139) quantity, express the number
of units of each denomination separately, indicating the denomi-
nations, as in the above illustration.
To read compound quantities, read them exactly as expressed.
157. Table of Measures of Time.
The Standard Unit of Time is the Mean Solar Day;
that is, the mean interval between two successive pas-
sages of the sun across the meridian of any place. A
day is supposed to begin at midnight.
140 NON-DECIMAL MEASURES. [Chap. VI.
60 seconds (sec.) = 1 minute (min.).
60 min. = 1 hour (hr.).
24 hr. = 1 day (da.).
7 da. =1 week (wk.).
365 da. = 1 common year (yr.)
366 da. = 1 leap year.
The year is divided into 12 months, called Calendar
Months, which contain an unequal number of days,
namely: January 31, February 28, March 31, April
30, May 31, June 30, July 31, August 31, September 30,
October 31, November 30, and December 31.
Every fourth year contains 366 days, and is called Leap
Year, and in these years February has 29 days. It is a
Leap Year when the number of the year is exactly di-
visible by 4 ; thus, 1896 will be a Leap Year.
The Solar Year contains 365 da. 5hr. 48 min. 46 sec, very
nearly. Now it would clearly be very inconvenient to reckon by
years which did not contain an exact number of days ; hence, as
the Solar Year contains very nearly 365^ days, we have 3 years
(called Civil Years) of 365 days each, and then one year of 366 days.
The Solar Year is, however, somewhat less than 365^ days, and the
necessary correction is made by omitting three Leap Years in every
400 years, the years which are not counted as Leap Years (although
divisible by 4) are the years which end the Centuries, and are such
that the number of the Century is not divisible by 4. Thus, 1800
was not a Leap Year, and 1900 will not be a Leap Year ; the year
2000 will, however, be a Leap Year.
158.* Reduction of Compound Quantities.
The method by which a compound quantity can be
expressed as a simple quantity will be seen from the
following example.
* The methods of reductions of compound quantities, also ad-
dition, etc., will be illustrated by the use of the above table because
the different units are familiar to all.
Arts. 158-160.] COMPOUND ADDITION.
141
Ex. Reduce 7 da. 3 hr. 12 min. 26 sec. to seconds.
7 da. 3 hr. 12 min. 26 sec.
24
168
3
171 hr.
60
10260
12
10272 min.
60
616320
26
616346 sec.
7 da. = 168 hr.
Adding the 3 hr., 7 da. 3 hr. = 171 hr.
171 hr. = 10260 mm.
Adding 12 min., 171 hr. 12 min. =a 10272 min.
10272 min. = 616320 sec.
Adding 26 sec, 10272 min. 26 sec. = 616346 sec.
6 t )1467 t 8 sec.
6 v )24 t 4 min. 38 sec.
4hr. 4 min. 38 sec.
159. To reduce a Simple Quantity to a Compound Quantity.
Ex. Reduce 14678 sec. to hr., min., and sec.
Since 60 sec. make lmin., if we
divide the number of sec. by 60,
we shall obtain the number of min.
equivalent to 14678 sec, i.e., 244
min. , but shall have 38 sec. over. We then divide the number of
min. by 60 and obtain the number of hours with 4 min. over.
160. Addition, Subtraction, Multiplication, and Division
of Compound Quantities.
It will be seen that no new principle is involved.
Care, however, must always be taken in regard to the
number of units of one denomination required to make
one unit of the next higher.
(a) Compound Addition [see Art. 142].
Ex. Find the sum of 14 da. 41 min. 11 sec, 121 da. 18 hr. 16 min.
29 sec, 201 da. 13 hr. 4 sec, and 11 hr. 23 min. 30 sec.
da.
hr.
14 41 11
121 18 16 29
201 13 4
11 23 30
337 19 21 14
142 NON-DECIMAL MEASURES. [Chap. VI.
Here the sum of the seconds equals 74 = 1 min. 14 sec. ; write
the 14 and carry the 1. The number of min. = 81 = 1 hr. 21 min. ;
write the 21 and carry the 1. The number of hr. equals 43 = 1 da.
19 hr. ; write the 19 and carry the 1. The number of days = 337.
(6) Compound Subtraction.
Ex. From 16 da. 12 min. and 50 sec. subtract 4 da. 12 hr. 13 min.
and 54 sec.
da.
hr.
min.
sec.
16
12
50
4
12
13
54
11 11 58 56
Here 54 cannot be subtracted from 50 ; therefore we take 1 min.
from the 12 min., change it to sec, and we have with the 50 sec.
110 sec. in all ; subtract 54 sec. from 110 sec, and we have 56 sec
remainder. Now 13 from 11 we cannot take, therefore we take
1 hr. from the next column and proceed as before.
(c) Compound Multiplication.
Case I. When the multiplier is not greater than 12.
Ex. Multiply 9 da. 10 hr. 31 min. 14 sec. by 7.
da.
hr.
min.
sec.
9
10
31
14
7
66 1 38 38
Here 14 sec. x 7 = 98 sec. = 1 min. 38 sec. ; write the 38 and
carry 1. 31 min. x 7 = 217 min. ; 217 min. + 1 min. = 218 min. =
3 hr. 38 min. ; write the 38 and carry the 8. 10 hr. x 7 = 70 hr. ;
70 hr. + 3 hr. = 73 hr. = 3 da. 1 hr. ; write the 1 and carry the 3.
Finally, 9 da. x 7 = 63 da. ; 63 da. -f 3 da. = 66 da. Ans. = 66 da.
1 hr. 38 min. 38 sec.
Case II. When the multiplier can be seen to be the
product of factors each not greater than 12.
Ex. Multiply 9 da. 10 hr. 31 min. 14 sec. by 35.
9
10
31
14
7
66
1
38
38
5
330 8 13 10
Art. 160.] COMPOUND DIVISION. 143
Case III. When the multiplier cannot be seen to be
the product of factors each not greater than 12.
The following example will explain the method to be adopted,
which will he seen to differ very little from the method adopted
in the multiplication of simple quantities, the only apparent dif-
ference arising from the fact that we cannot at once write down
the result of multiplying by 10, 100, etc.
Ex. Multiply 9 da. 10 hr. 31 min. 14 sec. by 257.
da. hr. min. sec.
9
10
31
14
10
94
9
12
20 =
10
multiplicand
x 10
943
20
3
20 =
2
u
x 100
1887
2d line x 5 462
1st " X 7 66
16
22
1
6
1
38
40 =
40 =
38 =
M
X2001
x 50 [
x 7j
2416 15 46 58 = " x 257
(d) Compound Division.
In division there are two cases to consider, according
as the divisor is an abstract number or a concrete
quantity of the same kind as the dividend [Art. 59].
Case I. To divide a compound quantity by an abstract
number.
Ex. 1. Divide 22 da. 1 hr. 13 min. 1 sec. by 6.
da. hr. min. sec.
6 )22 1 13 1
3 16 12 10^
Here, dividing 22 da. by 6, we have 3 da. with an undivided re-
mainder of 4 da., which must be reduced to hr. ; then we have 97
hr. in all to be divided by 6 ; the quotient equals 16 hr. with 1 hr.
over. One hour and 13 min. = 73 min. ; 73 min. -^ 6 = 12 min. with
1 min. over. Finally, 1 min. = 60 sec. ; 61 sec. + 6 = 10^ sec.
Ex. 2. Divide 9 wk. 6 da. 21 hr. 13 sec. by 33.
wk. da. hr. min. sec.
3 )9 6 21 13
11 )3 2 7 41
2 2 49 5|f
144 NON-DECIMAL MEASURES. [Chap. VI
Case II. When the divisor is a concrete quantity of
the same nature as the dividend.
Ex. Divide 37 da. 20 hr. 6 min. 48 sec. by 12 da. 14 hr. 42 min.
16 sec. [Compare Art. 50.]
37 da. 20 hr. 6 min. 48 sec. = 3269208 sec.
12 da. 14 hr. 42 min. 16 sec. = 1089736 sec.
3269208 sec. -=- 1089736 sec. = 3 (an abstract number).
(e) To multiply or divide by a fraction.
[Arts. 121 and 123.]
Ex. 1. Multiply 14 da. 2 hr. 12 sec. by f .
da.
hr.
min.
sec.
14
2
12
5
)70
10
1
10 1 25 61 f-
Ex. 2. Divide 14 da. 2 hr. 12 sec. by %.
da.
hr.
min.
sec.
►)14
2
12
2
19
36
f
19 17 12 16f
Both operations are understood because the nature of a fraction
has been explained.
EXAMPLES LIII.
Written Exercises.
1. Add 17 da. 14 hr. 22 min. 12 sec, 13 da. 11 hr. 24
min. 18 sec, and 15 da. 33 min. 40 sec.
2. From 6 da. 12 sec. subtract 2 da. 4 hr. 12 min. 59
sec.
3. Multiply 7 da. 12 hr. 14 min. 25 sec. by 5.
4. Multiply 7 da. 12 hr. 14 min. 25 sec. by 18.
5. Multiply 7 da. 12 hr. 14 min. 25 sec. by 347.
Art. 161'.] AVOIRDUPOIS WEIGHT. 145
6. A steamer makes a trip of 800 miles in 4 da. 8 hr.
12 min. 20 sec. ; how many such trips could she make in
18 da. 10 hr. 52 min. 25 sec. ?
7. Multiply 8 da. 5hr. 8 min. 48 sec. by f .
Measures of Weight.
The units of measure for all weights are derived from
the weight of a kernel of wheat taken from the middle of
a ripe ear.
The name of such a weight is one Grain (gr.).
161. Table of Avoirdupois Weight.
The unit is a pound consisting of 7000 grains.
16 drams (dr.) = 1 ounce (oz.).
16 oz. = 1 pound (lb.).
100 lb. = 1 hundred-weight (cwt.).
20 cwt. = 1 ton (t.).
112 lb. = 1 long hundred-weight.
2240 lb. = 1 " ton (l.t.).
Avoirdupois weight is used in weighing all ordinary substances.
The long ton is used in the Custom House, and in certain whole-
sale transactions.
The English Standard unit of weight is the Imperial Pound
(Avoirdupois), and is the weight of a certain piece of platinum
kept in the Exchequer Office.
English i
EXAMPLES LIV.
"Written Exercises.
1. Keduce 2 cwt. 151b. 12 oz. to dr.
2. How many gr. in an oz. avoirdupois ?
3. Add 4 cwt. 721b. 14 oz. 11 dr., 341b. 12 oz. 2 dr.,
8 cwt. 14 dr., 14 cwt. 56 lb. 3 oz., and 8 lb. 2 oz. 6 dr.
146 NON-DECIMAL MEASURES. [Chap. VI.
4. Eeduce 1687649 dr. to units of higher denomina-
tions.
5 . Keduce 16000 oz. to long tons, etc.
6. One boy weighs 125 lb. 10.5 oz. ; how many such
boys together weigh 22 cwt. 61 lb. 13 oz. ?
162. Table of Troy Weight.
The unit is a pound consisting of 5760 grains.
24 grains (gr.) = 1 pennyweight (pwt. or dwt.).
20 pwt. = 1 ounce (oz.).
12 oz. = 1 pound (lb.).
Troy weight is used in weighing gold and silver.
Diamonds and other jewels are spoken of as weighing so many
carats. The carat is a little more than 3£ grains.
The United States Standard unit of Weight is the Troy
Pound (same as the English Pound Troy), and is the
weight of a certain piece of brass in the custody of the
Director of the U. S. Mint.
EXAMPLES LV.
Oral Exercises.
1. How many gr. in 3 pwt. ? In 1\ dwt. ?
2. How many gr. in 2 oz. ?
3. How many oz. in SJ lb. ? In 51 lb. ?
4. How many oz. in 70 pwt. ? In 45 dwt. ?
5. How many oz. in 480 gr. ?
6. How many lb. in 78 oz. ? In 43 oz. ? In 400 pwt. ?
Written Exercises.
7. Eeduce 7563 dwt. to lb., etc.
8. Reduce 6 lb. 14 gr. to gr.
Arts. 162, 163.] APOTHECARIES' WEIGHT. 147
9. How many bronze cents weigh. 1 lb., the 1 ct. piece
weighing 48 gr. ?
10. One hundred gold dollar pieces weigh 5 oz. 7 pwt.
12 gr. ; what is the weight of one piece ?
163. Table of Apothecaries' Weight.
The unit is a pound consisting of 5760 grains.
gr. 20 = 1 scruple (3).
33 = 1 dram (3).
38 = 1 ounce (5).
5 12 = 1 pound (lb.).
Apothecaries' weight is used by physicians when writing pre-
scriptions and by druggists when selling drugs in small quantities.
Avoirdupois weight is used by them when dealing in large quantities.
The symbols are always written at the left of the figures.
EXAMPLES LVI.
Oral Exercises.
1. How many gr. in 3 3 ? In % 1 ? In3l£?
2. How many % in lb. 2J ? In lb. 3f ?
3. How many 3 in gr. 480 ? In $ 48 ?
4. How many $ in 5 2 ? In gr. 60 ? In lb. 1 ?
5. How many lb. in 5 42? In396? InS103l6?
Written Exercises.
6. Eeduce $ 7563 to lb.
7. Eeduce lb. 4 52 to gr.
8. Multiply lb.6 %1 3l3 2 gr. 15 by 16.
9. Divide ft). 12 S3 3 7 3 2 gr.4by4.
148 NON-DECIMAL MEASURES. [Chap. VI.
Measures of Length, Surface, and Volume.
164. Table of Linear Measures.
The English Standard unit of Length is the Imperial
Yard fixed by Act of Parliament to be the distance between
two marks on a bar of metal kept in the Exchequer Office.
The U. S. Standard unit of Length is the same as that
of England.
12 inches (in.) = l foot (ft).
3 ft. =1 yard (yd.).
5* yd.) =lrod(rd.).
16} ft. ) V '
320 rd. =lmile (mi.).
For Land Surveying.
7.92 inches = 1 link (li.).
100 li. | = 1 chain (ch.).
4 rd. )
80 ch. = 1 mi.
1 mi. = 1760 yd. = 5280 ft.= l statute mile.
Kods are sometimes called poles and perches. A furlong (fur.)
= 40 rods = \ mi. Civil engineers use a chain 100 feet in length.
EXAMPLES LVII.
Oral Exercises.
1. Express 4 yd. as in. ; 7 ft. as yd.
2. Reduce 18 ft. to rd., ft., and in.
3 . How many in. in 2 yd. 1 ft. ?
4. How many ft. in a surveyor's chain?
5 . How many li. in 1 rd. ?
6. How many yd. in 7 rd. ?
7. Practise frequently the drawing (freehand) of a
straight line 1 ft. long.
Arts. 164, 165.] SQUARE MEASURES. 149
"Written Exercises.
8. Reduce 40 rd. 6 ft. 7 in. to in.
9. Reduce 1 mi. to ft.
10. 803 in. to rd. and in.
165. Table of Square Measures.
The unit is any square, usually a square which is 1 ft.
long.
144 square inches (sq. in.) = l square foot (sq. ft.).
9 sq. ft. = 1 square yard (sq. yd.)
m\TS] =1 square rod (sq.rd.)
160 sq. rd. = 1 acre (A.).
640 A. =1 square mile (sq. mi.).
1A. = 160 sq. rd. = 4840 sq. yd. = 43560 sq. ft.
For Land Surveying.
16 sq. rd. = 1 square chain (sq. ch.).
10 sq. ch. = 1 A.
Square measure is used for measuring land, flooring, and in fact,
everything in which length and breadth have to be taken into
account.
EXAMPLES LVIII.
"Written Exercises.
1. Draw on the board (freehand) a figure representing
a square foot, marking it accurately into square inches.
2. How many sq. in. in i a sq.ft.? How many in
the square of i a ft. ?
3. Having in mind a square 1 yd. long, how many
sq. ft. in such a square ?
4. Find by a figure the number of sq. yd. in 1 sq. rd.
150 NON-DECIMAL MEASURES. [Chap. VI.
5. Find by multiplication the number of sq. yd. in
1 sq. rd.
6. How many sq. in. in a square 2 ft. long ? In a
square 3 in. long?
7. Kepresent 3 A. 4 sq. rd. 50 sq. ft. as sq. ft.
8. Divide 58 A. 84 sq. rd. 3 sq. yd. 4 sq. ft. by 8.
166. Table of Cubic Measures.
The unit is any cube, generally a cube 1 in. long, or a
cube 1 ft. long.
1728 cubic inches (cu. in.) = 1 cubic foot (cu. ft.).
27 cu. ft. = 1 cubic yard (cu. yd.).
For Measuring Wood.
16 cu. ft. = 1 cord foot (cd. ft,).
8 cd. ft. = 1 cord (cd.).
Cubic measure is used for measuring solid bodies in which
length, breadth, and thickness have to be taken into account.
EXAMPLES LIX.
Written Exercises.
1 . Make two cubes similar to the one in Ex. 2, p. 136,
one cube an in. long, the other 4 in. long.
2. Mark the sides of the large cube into sq. in., and
calculate how many cubes equal to the small cube might
be cut from a block equal to the large cube.
3. Reduce 2 cu. yd. 1201 cu. in. to cu. in.
4. How many cu. in. in a cube 2 in. long ? 5 in. long ?
5. How many cords of wood in a pile containing 1541
cu. ft. ? State answer to two decimal places.
6. Multiply 18cu.yd. 9cu.ft. 1063 cu. in. by 4.
Arts. 166-169.] LIQUID — DRY — FLUID. 151
167. Table of Liquid Measures.
The unit is a Gallon of 231 cu. in. (the old English
wine gallon).
4 gills (gi.)= 1 pint (pt.).
2 pt. =1 quart (qt.).
4 qt. =1 gallon (gal.).
311 gal. = 1 barrel (bbl.).
2 bbl. = 1 hogshead (hhd.).
A gallon of water weighs 8.33 lb.
The quart is a volume of 57 £ cu. in.
The English Imperial Gallon contains 277.274 cu. in.
168. Table of Dry Measures.
The unit is a Bushel of 2150.42 cu. in. (the old English
Winchester bushel).
2 pints (pt.)= J quart (qt.).
8 qt. =1 peck (pk.).
4 pk. = 1 bushel (bu.).
The quart is a volume of 67 £ cu. in.
The English Imperial bushel is 8 Imperial gallons = 2218.192
cu. in.
169. Table of Apothecaries' Fluid Measures.
60 minims, or drops (i*L)=l fluid dram (f3).
f38 =1 fluid ounce (fS).
fS16 =lpint(0).
08 =1 gallon (Cong.).
EXAMPLES LX.
Oral Exercises.
1. How many pt. in 5 qt.? In 3 gal.?
2. How many pt. in 3 pk.? In 1 bu.?
3. Is the pt. in Ex. 1 equal to the pt. in Ex. 2 ?
4. Eeduce 3 qt. 1 pt. to gi. j 1 bu. to qt.
152 NON-DECIMAL MEASURES. [Chap. VI.
5. How many «t in f§l?
6. How many f§ in ^960?
Written Exercises.
7. Eednce Cong.l to K.
8. Divide 14bu. 3 pk. 5 qt. 1 pt. by 5.
9. Divide 7 fglO f36 "l59§ by f.
10. Keduce 750 dry qt. to liquid qt.
11. How many bushels of potatoes in a bin 4 ft. x 3 ft.
X 2 ft. ?
12. How many piles of 3 bu. each could be made out
of a pile containing 8 cu. ft. ?
13. Reduce 1 hhd. to cu. ft.
Foreign Monies.
170. Table of English Money.
The unit is the Pound (£).
4 farthings (/cw\) = l penny (&).
12d = 1 shilling (*.).
20s. = 1 pound (£).
21s. =1 guinea (ga.).
One farthing is written \d. ; two farthings, or one half -penny,
is written \d. ; and three farthings is written \d. Thus, eightpence
farthing is written S\d.
The coins in use in England are as follows :
Gold coins : the sovereign (20s.) and half-sovereign (10s.).
Silver coins: crown (5s.), half-crown (2s. 6d.), florin (2s.),
double florin (4s.), shilling, sixpence, and threepence.
Copper coins : penny, half-penny, and farthing.
171. German Money.
100 pfennigs (j)/.) = l mark (m.).
Arts. 170-174.] FRENCH MONEY. 153
172. French Money.
100 centimes (c.) = 1 franc (/.).
Note. See Art. 262 for equivalents in American money.
173. The value of a given fraction of a given concrete
quantity is found as follows :
For example, to find T 5 f of 3 lbs. 4 oz. Troy, we must divide
3 lbs. 4 oz. into 18 equal parts, and then take 5 of those parts ;
that is, we must divide by 18, and then multiply by 5. We may,
however, first multiply by 5 and then divide by 18. Thus,
18
174. The value of a given decimal of a given concrete
quantity is found as follows :
Ex. Find .54375 of 1 lb. Troy in lower denominations.
.54375
lb. oz. pwt. gr.
13_4
|l 8
lb.
3
oz. pwt. gr.
4
5
2 4 10}
b
«}S
16
8
8
4
11 2 5£
11 2 5}
Here .54375 of 1 lb. = .54375 of 12 oz.,
which = 6.525 oz. ;
.525 of loz. = .525 of 20 pwt.,
which = 10.5 dwt. ;
.5 of ldwt. =.5 of 24 gr.,
which = 12 gr.
12
20
dwt. =10.500
24
gr. = 12.0
. \ 6 oz. 10 dwt. 12 gr. = Ans.
EXAMPLES LXI.
Written Exercises.
1. Find 2| of 3 da. 12 hr.
2. Find 4f of 3 cwt. 36 lb.
3. Add | of 2s. 6d, 2£ of Is. Sd., and l^ of 6s. 5d
154 NON-DECIMAL MEASURES. [Chap. VI.
4. Find 1.625 of 1 da. - .02 of 1 wk.
5. Express 4.3125 lb. Troy in lb., oz., etc.
6. Find .436 of 1 mi.
7. By how much does f of 1 mi. exceed ^ of 310 rd.
lyd.?
175. To express one quantity as a fraction of another,
we proceed as follows :
Ex. 1. Express 14s. 6d. as a fraction of lbs. Sd.
Us. 6d. = lUd.
lbs. Sd. = 188c?.
Now Id = %h of 188 <*-
... i74tf. = i|4 of issd.
Ex. 2. Express 2£ of Is. 7\d. as a fraction of £1.
2£ of Is. l\d. = -U of ¥* = - 4 tW-
£1 = 240<J.
Now Id. = j£* of £1.
. •. Wd. = W x & of £1 = £ ^\\
176. To express one quantity as a decimal of another,
we proceed as follows :
The method here is the reverse of that in Art. 174.
Ex. Express Id oz. II dwt. 12 gr. as the decimal of lib. Troy.
12 gr. Divide the grains by 24 to reduce to pwt.; add
11.6 the 11 pwt. and divide 11.5 pwt. by 20 to reduce
10.575 to oz ; add the 10 oz. and divide 10.675 oz. by 12
.88125 to reduce to lb.
An excellent method is to express one quantity as a
fraction of the other [Art. 175], and then reduce this
common fraction to a decimal [Art. 130] ; thus,
Arts. 175, 176.] EXAMPLES. 155
10 oz. lldwt. 12 gr. = 5076 gr.
1 lb. = 5760 gr.
Now, *W =' 88125 '
EXAMPLES LXII.
Written Exercises.
1. Express 251b., 601b., 121b. 8 oz., and 61b. 4oz. as
fractions of 1 cwt.
2. What would be the measure of 4 yd. 2 ft. 8 in. if
1 yd. 1 ft. 7 in. were taken as the unit ?
3. Express 1 oz. 6 dwt. 6 gr. as a decimal of 1 lb. Troy.
4. What decimal of an acre is 20 sq. rd. 5 sq. ft. 72
sq. in.?
5. Express £5. 12s. 6& as a decimal of £10.
6. Express 2 mo. 7 da. as a decimal of 1 yr.
7. Express 7 mo. 12 da. as a decimal of 1 yr.
8. Express 10 mo. 15 da. as a decimal of 2 yr.
EXAMPLES LXIII.
Simple Examples in Reduction for Written Work.
Reduce :
1 . 1 1. 3 cwt. 10 lb. to pounds.
2. 3t. 12 cwt. 161b. to pounds.
3. 6hr. 12min. 10 sec. to seconds.
4. 12 hr. 5 min. 24 sec. to seconds.
5. 13 yd. 2 ft. 11 in. to inches.
6. 17 yd. 2 ft. 7 in. to inches.
7. 12 mi. 3 fur. 10 rd. to rods.
8. 13 mi. 5 fur. 26 rd. to rods.
9. 8 bu. 3 pk. 4 qt. to quarts.
156 NON-DECIMAL MEASURES. [Chap. VI.
10. 5 gal. 3 qt. 1 pt. to pints.
11. 5 A. 27 sq. rd. to square rods.
12. 17 A. 135 sq. rd. to square rods.
13. 13 sq. yd. 6 sq. ft. 100 sq. in. to inches.
14. 8 sq. yd. 7 sq. ft. 90 sq. in. to inches.
15. 61b. 7 oz. 10 dwt. 15 gr. to grains.
16. 18 lb. 9 oz. 15 dwt. 20 gr. to grains.
17. 3 wk. 5 da. 12 hr. to hours.
18. 16 da. 22 hr. 40 min. 35 sec. to seconds.
19. 12 1. 13cwt. 751b. 7 oz. to ounces.
20. 5 1. 17 cwt. 68 lb. 14 oz. to ounces.
21. 2 mi. 3 fur. 80 yd. 2 ft. to feet.
22. 12 mi. 1200 yd. 1ft. 7 in. to inches.
Reduce to tons, cwt., etc. :
23. 14621b. 25. 115971b. 27. 57812 oz.
24. 135741b. 26. 56214 oz. 28. 81974 dr.
Reduce to acres and square rods :
29. 315 sq. rd. 31. 1574 sq. rd.
30. 5142sq.rd. 32. 3725 sq. rd.
Reduce to yards, feet, etc. :
33. 156 in. 34. 342 in. 35. 417 in. 36. 1179 in.
Reduce to lb., oz., dwt., gr. :
37. 517 dwt. 41. 13407 gr.
38. 574 dwt. 42. 24709 gr.
39. 3156 gr. 43. 35937 gr.
40. 4215 gr. 44. 51940 gr.
Reduce to bushels, pecks, etc. :
45. 156 pt. 46. 1472 pt. 47. 416 qt. 48. 1875 pt.
Art. 177.] DIFFICULT REDUCTIONS. 157
Keduce to square yards, etc. :
49. 1462 sq. in. 50. 2156sq.in. 51. 3564 sq. in.
Keduce to days, hours, etc. :
52. 31572 sec. 53. 257672 sec. 54. 7142169 sec.
Calculate the number of
55. Sq. rd. in 1 sq. mi. 56. Sq. rd. in 1 A.
57. A. in 1 sq. mi.
58. Change 3.12 rd. to the decimal of a mi.
59. Change .2 sq. rd. to the fraction of an A.
60. Change lb. .00694 to the fraction of a 3.
61. How many cd. of wood might be packed into a
shed the size of your school-room ?
177. The following cases are somewhat more difficult
than those previously considered because one rod does
not equal an exact number of yards.
Ex. 1. Beduce Bird. 3yd. 2 ft. Win. to inches.
31rd. 3 yd. 2 ft. 11 in.
5.5
15 5
155 Here we have 31 rods to be multi-
173.5 yd., the 3 yd. included, plied by 5^ = 5.5. We might have
§ multiplied by JUk
522.5 ft., "2 ft.
12
6281.0 in., " 11 in.
Caution. In adding the yards or feet of the example while
multiplying, care must be used in regard to the decimal point.
Ex. 2. Beduce 1885 sq. rd. 16 sq. yd. 6 sq.ft. to sq.ft.
1885 sq. rd.
30.25
57037.25 sq. yd., including the 16sq.yd.
9
513341.25 sq. ft., " " 6 sq.ft.
158 NON-DECIMAL MEASURES. [Chap. VI.
Ex. 3. Reduce 6281 in. to units of higher denominations.
First Method.
12 )6281 in. 5.5,)174.0 % (31
3)523 ft. 5 in. 165
5.5)174 yd. 1ft. ?0
31 rd. 3.6 yd.
55
35 [See Art. 68.]
rd. yd. ft. in.
Integral part of answer =31 3 1 5
Decimal " " M =.5 yd. = 1 6
Sum = 31 3 2 11 Ans.
Second Method.
12)6281 in.
3)523 ft.
5 in.
174 yd.
2
11)348
31 rd.
1 ft. It is shorter to multiply by T 2 T
than to divide by 5.5.
. . . 1 yd. [Art. 68.]
rd. yd. ft. in.
Integral part = 31 1 5
Fractional part = f yd. = 3 1 6
Sum = 31 3 2 11 Ans.
Ex. 4. Beduce 513341.25 sq.ft. to sq. rd., sq. yd., and sq.ft.
9)513341.25
30.25 )57037 sq. yd. 8.25 sq. ft.
1885 sq.rd. 15.75 sq. yd.
sq. rd. sq. yd. sq. ft. sq. in.
Integral part = 1885 15 8
_ . . . r .25 sq. ft. = 36
Decimal part i _- , ~
* I .75sq.yd. = 6 108
Sum = 1885 16 6 Ans.
Here it is shorter and easier to divide by 30.25 than to divide by
ifJ- (i.e., to multiply by T | T ).
It will be noticed that only the integral part of any dividend is
to be divided ; the decimal part, if any, is to be regarded as a
decimal part of the remainder.
Art. 178.] MISCELLANEOUS MEASURES. 159
178. In some cases of Reduction we cannot pass
directly from one denomination to the other.
Ex. How many lb. Troy are there in 144 lb. Avoir, f
Since 1 lb. Avoir. = 7000 gr.,
144 lb. Avoir. = 7000 gr. x 144.
These grains are now reduced to lb. Troy in the usual manner.
Stationery.
Miscellaneous Measures. Numbers.
3 barleycorns = 1 in. 12 units = 1 dozen.
4 in. =1 hand. 12 dozen = 1 gross.
40 rd. =1 furlong. 12 gross = 1 great gross.
i ~~~~.~i.t -> 20 units = 1 score.
1 geographi-) lknot#
cal mi. =6080 ft. J
3 knots = 1 league.
6 ft. =1 fathom. 24 sheets = 1 quire.
20 quires = 1 ream.
1 cu. ft. of pure water weighs 2 reams = 1 bundle.
1000 oz. = 62 1 lb. 5 bundles = 1 bale.
EXAMPLES LXIV.
Reduce : Written Exercises.
1. 10 rd. 2 yd. 1ft. to feet.
2. 5rd. 3 yd. 2 ft. to inches.
3. 1 mi. 3 fur. 20 rd. 1 yd. to yards.
4. 6 mi. 5 fur. 30 rd. 3 yd. to yards.
5. 18 mi. 11 rd. 3 yd. 1ft. 6 in. to inches.
6. 27 mi. 273 rd. 2 yd. 2 ft. 7 in. to inches.
7. 6 mi. 52 yd. to yards.
8. 18 mi. 5 rd. 160 yd. 2 ft. 11 in. to inches.
9. 3 A. 16 sq. rd. to square yards.
10. 15 A. 24 sq. rd. to square yards.
*The knot recognized by the U. S. Coast and Geodetic Survey
equals 6080.20 ft.
160 NON-DECIMAL MEASURES. [Chap. VI.
11. 3 A. 85 sq. rd. 16 sq. yd. 6 sq. ft. to square inches.
12. 16 sq. rd. 18 sq. yd. 5 sq. ft. 100 sq. in. to square
inches.
Reduce to miles, etc. :
13. 6974 yards.
16.
6315 feet.
14. 21571 yards.
17.
51621 inches.
15. 15737 yards.
18.
158743 inches.
Reduce to acres, sq. rd.,
etc.
19. 20812 sq. yd.
21.
5172400 sq. in.
20. 38599 sq. yd.
22.
8156179 sq. in.
Reduce :
23. 36 lb. Avoir, to lb. Troy.
24. 720 lb. Avoir, to lb. Troy.
25. 1 cwt. Avoir, to Troy weight.
26. 11 lb. 8 oz. Avoir, to Troy.
27. 350 oz. Troy to oz. Avoir.
28. 4 lb. 3 oz. 20 gr. to lb. and oz. Avoir.
29. lcwt. 91b. to lb., 3, etc.
30. lb. 9 56 3 6 3 2 gr.5 to lb., etc., Avoir.
EXAMPLES LXV.
Written Exercises.
Add:
da. hr. mini. da. hr. min. sec.
1. 5 17 42 3. 5 17 27 45
3 11 53 6 11 39 56
7 19 37 17 21 49 40
11 7 21 6 11 11 31
2.
hr. min. sec.
cwt. lb.
oz.
1 41 15
4 5 16
10
6 17 39
3 39
6
7 35 42
7 -47
14
5 16 13
1 25
9
Art. 178.] EXAMPLES. 161
lb. oz. dr. lb. oz. dwt. gr.
5. 5 12 8 12. 5 11 16 18
4 13 12 2 9 11 13
7 9 15 7 10 15 21
3 11 14 3 7 9 16
t. cwt. lb. oz. yd. ft. in.
5 15 17 3 13. 5 2 9
1 12 67 12 11 1
15 17 20 11 13 2 7
3 9 21 7 6 11
lb. oz. dr. yd. ft. in.
7. 5 9 13 14. 16 1 7
7 14 12 9 2 10
18 6 9 20 8
3 11 11 11 2 11
t. cwt. lb. oz. yd. ft. in.
8. 16 17 19 14 15. 15 9
119 16 47 3 2 7
72 12 37 13 18 1 11
65 15 24 8 8 9
mi. rd. yd.
9. 6 24 10 16. 6 100 2
cwt.
lb.
oz.
6
24
10
17
78
12
14
7
14
11
41
2
lb.
oz.
dwt.
6
4
19
13
9
7
2
11
17
7
10
13
oz.
dwt.
gr.
1
17
23
2
8
11
5
15
7
7
4
21
3 140 4
18 97 3
2 15 2
mi. rd. yd. ft. in.
10. 17. 1 190 2 1 4
3 3 11
2 84 4 2 7
3 180 3 1 9
mi. rd. yd. ft. in.
1. 17 23 18. 5 300 2 2 1
15 3 1 9
1 187 4 2 11
2 74 5 9
162 NON-DECIMAL MEASURES. [Chap. VI.
A.
sq. yd.
bu.
P k.
qt.
pt.
19.
5
12
23.
3
2
5
1
17
25
1
3
3
3
18
10
6
1
4
30
2
3
4
1
A.
sq. yd.
m.
5
3
9
gr.
20.
1
27
24.
4
10
6
2
5
16
19
3
8
5
2
15
8
22
1
1
1
6
19
7
2
7
19
gal.
qt. pt.
Cong.
0.
*3
f3
21.
5
2 1
25.
1
6
12
4
6
3 1
2
5
13
7
4
1
1
2
3
5
1
gal.
2 1
qt. pt.
gi-
2
1
5
3
lb.
5
3
3
22.
18
3 1
2
26.
1
11
7
2
4
1
3
2
9
1
1
6
2 1
1
3
6
6
1
1 .1
1
cu. yd.
cu.ft.
5
cu. in.
4
5
1
27.
5.2
22.1
16.4
1.3
19.2
126.9
3.3
3.
14.3
5.4
8.2
9.2
Answer in exact units.
EXAMPLES LXVI.
"Written Exercises.
Subtract :
1. 5 da. 16 hr. 22 min. from 11 da. 18 hr. lOmin.
2. 15 da. 17 hr. 13 min. 42 sec. from 31 da. 9hr. 11
min. 40 sec.
3. 5 cwt. 73 lb. 11 oz. from 7 cwt. 11 lb. 9 oz.
Art. 178.] EXAMPLES. 163
4. 61b. 10 oz. 11 dr. from 161b. 9 oz. 5 dr.
5. 7t. 13cwt. 151b. 12 oz. from 10 1. 11 cwt. 10 oz.
6. 3 lb. 4 oz. 10 dwt. from 9 lb. 1 oz. 5 dwt.
Find:
7 731b.4oz.l0pwt. _ 2(51b 10oz 18pwt 1Q5gr)
o
8. 10 yd. -5 yd. 1ft. 10 in.
9. 29 yd. 1ft. 4 in. -17 yd. 2 ft. 11 in.
10. 17 mi. lfur. 150 yd. -6 mi. 3 fur. 164 yd.
11. From lb. 4 56 gr.17 subtract lb. 2 %1 3 3 gr. 15.
12. From 18 sq. yd. 3 sq. ft. 17 sq. in. take 6 sq.yd.
7 sq.ft. lOOsq.in.
13. From 215 sq. yd. 3 sq. ft. 84 sq. in. take 118 sq.yd.
6 sq. ft. 112 sq. in.
14. From 25 A. take 15 A. 120 sq. rd. 10 sq. yd.
15. From 23 A. 40 sq. rd. 10 sq.yd. take 6 A. 125
sq. rd. 25 sq. yd.
16. Find 6 cu. yd. 24 cu. ft. 1200 cu. in. - 3 cu. yd. 25
cu. ft. 8 cu. in.
bu. pk. qt. pt. gi.
17. 7 1.2 1 3.6
3 2.4 1.3 2.4
18.
gal.
10
5
qt. pt.
1
2 1
Answer in exact units.
0.
19. 7
*3
10
f3
5
50
3
14
6
51
EXAMPLES LXVII.
Written Exercises.
Multiply :
1. 5 hr. 10 min. 33 sec, (i) by 5, (ii) by 7, (iii) by 9.
2. 5 cwt. 39 lb., (i) by 7, (ii) by 8, (iii) by 9.
164 NON-DECIMAL MEASURES. [Chap. VI.
3. 6t. 17cwt. 641b. 6oz. 5 dr., (i) by 4, (ii) by 6,
(iii) by 9.
4. 8 lb. 10 oz. 15 dwt. 20 gr., (i) by 5, (ii) by 7, (iii)
by 12.
5. tt>6 34.1 32.3 31.2 gr.ll by 5.
6. 10 yd. 1 ft. 7 in., (i) by 8, (ii) by 11, (iii) by 12.
7. 8 mi. 215 yd., (i) by 5, (ii) by 8, (iii) by 12.
8. 1 mi. 20 rd. 4 yd., (i) by 7, (ii) by 56.
9. 15 sq. yd. 7 sq. ft. 100 sq. in., (i) by 6, (ii) by 11.
10. 4 en. ft. 163 cu. in., (i) by 8, (ii) by 11.
11. 3 bu. 2 pk., (i) by 5, (ii) by 11.
12. 3 gal. 2 qt. 1 pt., (i) by 5, (ii) by 7.
13. 3 da. 17 hr. 10 min. 15 sec, (i) by 35, (ii) by 45.
14. 15 1. 12 cwt. 16 lb., (i) by 42, (ii) by 72.
15. 8 lb. 11 oz. 15 dwt. 18 gr., (i) by 49, (ii) by 84.
16. 3 yd. 2 ft. 10 in., (i) by 44, (ii) by 132.
17. 3 yd. 1ft. 7 in. by 350.
18. 5bn. 2pk. by 420.
19. 12 da. 13 hr. 14 min. 12 sec. by 65.
20. 5t. 7 cwt. 151b. by 94.
21. 31b. 4oz. 12 dwt. 12 gr. by 124.
22. 3 cwt. 75 lb. 5 oz. by 257.
23. 15sq.yd. 7 sq.ft. 82sq.in. by 1212.
24. 6t. 15 cwt. 71b. 3 oz. by 2341.
25. 21b. 4oz. 16 dwt. 18 gr. by 3124.
26. 1 mi. 2 fur. 15 rd. 4 yd., (i) by 5, (ii) by 9.
Art. 178.] EXAMPLES. 165
EXAMPLES LXVIII.
Written Exercises.
Divide :
1. 22 da. lhr. 12min. by 6.
2. 37 cwt. 3 lb. by 7.
3. 441b. 2oz. 8 dr. by 8.
4. 52 lb. 10 oz. 13 dwt. by 9.
5. 153 yd. 2 ft. lin. by 11.
6. 95 A. 64sq.rd. by 12.
7. 1851b. 8oz. 17 dwt. by 54.
8. 123 da. 10 hr. 45 min. by 50.
9. 1052 yd. 1ft. by 132.
10. 251 A. 133sq.rd. by 121.
11. 19 1. 14 cwt. 81b. 3oz. 4 dr. by 500.
12. 214 1. 10 cwt. 441b. by 196.
13. 12 1. 3 cwt. 9 lb. by 37.
14. 309 1. 12 cwt. 141b. by 47.
15. lOt. 6 cwt. 701b. loz. by 57.
16. 37 yd. 2 ft. 3 in. by 151.
17. 35 1. 2 cwt. 631b. 2 oz. by 289.
18. 2237 bu. 1 pk. 7 qt. by 253.
19. 61 1. 1 cwt. 75 lb. by 2896.
20. 24 mi. 58 yd. 2 ft. 4 in. by 1234.
21. 36 mi. 4 fur. 23 rd. 3 yd. 1ft. 6 in. by 10.
• 22. 55 mi. 7 fur. 26 rd. 1yd. 1ft. by 43.
23. 298 A. 39 sq. rd. 18 sq. yd. 2 sq. ft. 108 sq. in. by 73.
166 NON-DECIMAL MEASURES. [Chap. VI.
EXAMPLES LXIX.
Written Exercises.
1. Divide 2 tons 5 cwt. by 9 cwt.
2. Divide 6 oz. 10 dwt. by 13 dwt.
3. Divide 3 A. 50 sq. rd. by 19 sq. rd.
4. Divide 20 bu. 1 pk. by 2 bu. 1 pk.
5. How many pieces each 3 yd. 1 ft. long can be cut
from a rope whose length is 180 yd.?
6. A wheel revolves once every 2 m. 15 sec. ; how
many times does it revolve in 1 hr. 48 m. ?
7. The circumference of a tricycle wheel is 12 feet;
how many times does the wheel turn round in a journey
of 10 miles ?
8. A field of 13 A. 80 sq. rd. is divided into allot-
ments, each containing 1 A. 20 sq. rd. ; how many allot-
ments are there ?
9. A man's average step is 2 ft. 11 in. ; how many
steps does he take in walking 3 J miles ?
10. How many jars, each containing 2 gal. 3 qt. 1 pt.,
can be filled out of a cask containing 46 gal. ?
11. How many rails, each weighing 4 cwt. 37 lb., can
be made out of 58 1. 19 cwt. 90 lb. of iron ? What will
each rail cost at 3 ct. a lb. ?
12. How many times does 2 miles 76 yd. contain 14
yd. 1ft. 6 in. ?
13. Each of a certain number of articles weighs 14 lb.
1 oz., and the total weight is 3 t. 75 lb. ; how many are
there ?
14. How many times is 361b. 3oz. 3 dwt. contained in
543 lb. 11 oz. 5 dwt. ?
Art. 179.]
CIRCULAR MEASURES.
167
15. How many bullets, each, weighing 2 \ oz., can be
made from a quantity of lead weighing 7 cwt. 35 lb. ?
16. A sovereign weighs 123 grains ; how many can be
made out of 3 lb. 5 oz. of standard gold ?
179. Table of Circular Measures.
The plane figure whose bound-
ing line is a curve everywhere
equally distant from the centre
is called a Circle.
The bounding line of a circle
is called its Circumference.
Any part of a circumference
is called an Arc.
If the circumference be di-
vided into 360 equal parts, one of these parts is called an
arc of one Degree (1°).
The unit is an arc of 1°.
60 seconds (") = 1 minute (*).
60' = 1 degree (°).
360° = 1 circumference (C).
EXAMPLES LXX.
Written Exercises.
. Ac
Id 5°
21'
15"
27°
41'
23"
196°
12'
39"
150°
2'
10"
2. From 182° 1' 49"
Subtract 12° 50' 50"
3. How many seconds in 90° ?
4. How many degrees in 5678" ?
5. How many circumferences in 1800° ?
6. Eeduce 100000" to units of higher denominations.
168
NON-DECIMAL MEASURES.
[Chap. VI.
180. Longitude and Time.
EXAMPLES LXX. — Continued.
Oral Exercises.
7. Let the figure represent a globe rotating on its
axis ; how many degrees does c move towards the present
position of g while the globe is making I of a rotation
•L of a rotation 5
8. The earth is a rotating globe, and a point, as c
or r, moves once around its circle in 24 hr. ; how long
does it take c to move to the present position of e, the arc
ce being 30° ? To the present position of g ? Of d ?
9. How long does it take r to move to the present
position of s ?
10. How long does it take the arc ac to reach the
present position of the arc (meridian) ae ?
11. How many degrees does the earth rotate in 1 hr. ?
In 1 min. ?
Arts. 180, 181.] LONGITUDE AND TIME.
169
12. How many arc minutes does the earth rotate in 1
min. ? In 1 sec. ?
13. How many arc seconds does the earth rotate in
1 sec. ?
Since 15° rotation require 1 hr.
and 15' rotation require 1 min.
and 15" rotation require 1 sec,
we may change time measure to circular measure by mul-
tiplying hr., min., and sec. by 15 ;
we may change circular measure to time measure by dividing
°, ', and " by 15.
The meridian distance (the difference in longitude)
between two places is measured in units of circular meas-
ure, or in units of time measure.
181. Difference in Longitude, and in Time.
Longitude is reckoned either east or west from the merid-
ian passing through Greenwich. It is evident that if two
170 NON-DECIMAL MEASURES. [Chap. VI.
places are either east of, or west from, Greenwich, the dif-
ference in longitude is found by subtraction ; if one place is
east and the other west, the difference is found by addition.
Ex. Find difference in time between Cleveland, 81° 40' 30" W.,
and St. Paul, 93° 4' 55" W.
93° 4' 55"
81° 40' 30"
15 )11° 24' 25"
45 min. 37 sec. Ans.
EXAMPLES LXXI.
"Written Exercises.
Find the difference in time between
1. Portland (Me.), 70° 15' 40" W., and Detroit, 82°
58' W.
2. New York, 74° 0' 3" W., and Chicago, 87° 37' 30" W.
3. New York and Washington, 77° 2' 48" W.
4. Berlin, 13° 23' 53" E., and Paris, 2° 20' 22."5 E.
5. Berlin and New York.
6. Boston, 71° 3' 30" W., and San Francisco, 122° 24'
15" W.
7. Greenwich and Washington.
8. What is the longitude of St. Louis, the difference
in time between New York and St. Louis being 1 hr. 5
min. 1 sec. ?
9. The difference in time between Philadelphia and
Chicago is 49 min. 50 sec. ; what is the difference in lon-
gitude ? What is the longitude of Philadelphia ?
10. When it is 4 o'clock (p.m.) at Greenwich, what
time is it at Washington ?
11. When it is 1 o'clock (a.m.) at New York, what time
is it at Berlin ?
Art. 181.] EXAMPLES. 171
EXAMPLES LXXII.
Reduction of Metric Numbers to Non-Metric Numbers; also, of
Non-Metric Numbers to Metric Numbers.
1 . How many cm in 1 in. ?
2. How many yd. in 17.6 m ?
3. How many t. in l Mg of water ?
4. How many sq. ft. in l qm ?
5. How many cu. in. in l 1 ?
6. How many lb. in l cum of water ?
7. How many Mg in 1 l.t. ?
8. How many ml in 1 qt. (liquid) ?
9. How many g in lb 5 ?
10. How many gr. in 15 cg ?
11. How many gr. in 500 ccm of water ?
12. How many HI in 5 pk. ?
13. How many bu. in 3 K1 ?
14. If either a qt. or a liter of milk cost 6 ct., which
would you prefer to purchase ?
15. Which would you prefer to buy, 1 A. or 2.5 Ha for
the same money ?
16. Find the value of 3 13 in g.
17. Find the value of l Kg in lb.
18. Express 2 gal. lpt. 3gi. as liters.
19. How many sters in 100 cu. ft. ?
20. How many A. in 7 Ha ?
21. Express 1-^-mi. as m and as Hm.
22. What cost 4 kilos of sugar at 5±- ct. per lb. ?
23. What costs £ a kilo of gold at $1 a pwt. ?
172
NON-DECIMAL MEASURES.
[Chap. VI.
Tables for Convenient Reference.
Time.
Square Measures.
60 sec. = 1 min.
144 sq. in. = 1 sq. ft.
60 min. = 1 hr.
9 sq. ft. = 1 sq. yd.
24 hr. = 1 da.
30£ sq. yd. = 1 sq. rd.
365 da. = 1 yr.
160 sq. rd. = 1A.
366 da. = 1 leap yr.
640 A. =lsq.mi.
Troy Weight.
16 sq. rd. = 1 sq. ch.
10 sq. ch. = 1 A.
24 gr. = 1 pwt.
20 pwt. = 1 oz.
Cubic Measures.
12 oz. = 1 lb.
1728 cu. in. = 1 cu. ft.
Avoirdupois Weight.
27 cu. ft. = 1 cu. yd.
16 dr. = 1 oz.
16 cu. ft. = 1 cd. ft.
16 oz. = 1 lb.
128 cu. ft. = 1 cd.
100 lb. as 1 cwt.
20 cwt. = 1 1.
Liquid Measures.
112 lb. =11. cwt.
4 gi. =1 pt.
22401b. =ll.t.
2 pt. =1 qt.
4 qt. =1 gal.
Apothecaries' Weight.
31£ gal. = 1 bbl.
gr. 20 = 3 1
2 bbl. = 1 hhd.
3 s= 51.
1 qt. = 57| cu. in.
3 8= 51.
512 = ft). 1.
Dry Measures.
2 pt. =1 qt.
Linear Measures.
8 qt. =1 pk.
12 in. = 1 ft.
4 pk. = 1 bu.
3 ft. =1 yd.
1 qt. = 67| cu. in.
5 ^ d '} = lrd.
16^ ft. i
320 rd. = 1 mi.
Apothecaries' Fluid
Measures.
ni60 = f31.
f38 = f51.
7.92 in. = 1 li.
100 li. = 1 ch.
f316 = 01.
80 ch. = 1 mi.
08 = Cong. 1.
Art. 181.]
SYNOPTIC CONVERSION.
173
Synoptic Conversion of English and Metric Units.*
English to Metric.
lin.
lyd.
1 mi.
a 2.54 cm .
= .9144 m .
= 1.60935 ki
Metric to English.
1™ =39.37 in.
lKm = 1093.61 yd.
8 Kra =5mi. nearly.
1 sq.
yd.
.83613 q«".
Iqm
= lca =
10.7639 sq
ft.
1A.
—
.404687 Ha .
1» =
lHa =
119.599 sq.
2.471 A.
yd.
leu. in. = 16.3872"™
leu. yd. ss. 76456 c » m .
lqt. (U. S.)=. 946361.
leu m_ 61023.4 cu. in.
= 35.3145 cu. ft.
= 1.30794 cu. yd.
Icdm 1
J, | = 61.023 cu. in.
==.26417 gal. (U.S.)
== 1.05668 qt. (U. S.).
lgr. =64. 7989 m *.
1 lb. avoir. = .45359*2.
It. (20001b.) =907. 18 K g.
1 l.t. (2240 lb.) = 1.01605 t
is = 15.4324 gr.
l K s = 2.20462 lb. avoir.
IT = 2204.62 lb. avoir.
IT =.98421 l.t. (2240 1b.).
Weights.
1 bu. wheat
= 60 lbs.
1 stone = 14 lbs
1 " potatoes
= 60 "
1 bbl. pork =200 "
1 " beans
= 60 "
1 " flour = 196 "
1 ** corn
= 56 "
1 cental of grain = 100 "
1 " barley
= 48 «
1 quintal of fish =100 "
1 " oats
= 32 "
* Arranged from the Smithsonian Tables,
black type should be memorized.
Figures printed in
174
NON-DECIMAL MEASURES. [Chap. VI.
Inches.
1 1
,l,\l,
il.'.l.
.i.f.i,
MM Mil IMI Mill
1
! }
i
i i
1 1
3
1
i
Centimeters.
lin. = 2.54 cm .
D
l** 1 . 1 sq. in.
1 sq. in. = 6.45 i cm .
A
/
>
y
l ccm - 1 cu. in.
1 cu. in. = 16.387 ccm -
Art. 181.]
1T5
Diameter and height of a cylindrical liter
measure and of a cylindrical quart measure.
li = 61.063 cu. in. = 1000 ccm .
1 qt. = 57.75 "
176 APPROXIMATION. [Chap. VII.
CHAPTER VII.
APPROXIMATION.
182. No continuous magnitude can be measured with
perfect accuracy. When, for example, we endeavor to
make two pieces of wire equally long, all that we can
ensure is, that they shall be of the same length so far as
the eye, or other instrument, can judge; however, they may,
and probably will, differ by some thousandths or even
hundredths of an inch.
In all questions involving continuous magnitude, such
as length, weight, etc., we must, therefore, be content
with approximations (more or less accurate) to the true
measure. It follows that calculations dependent upon
measurement can give only approximately accurate
results.
For example, if we are told that a slab of stone is 17.6 inches
long, and 12.4 inches wide, we are not to conclude that these are
perfectly accurate measurements, but only that the measurements
are near enough for practical purposes, the real length and breadth
being at any rate less than 17.7 and 12.5 respectively.
If the given measurements were accurate, the area of the slab
would be 17.6 x 12.4 square inches. The actual area may, how-
ever, have any value between 17.6 x 12.4 square inches and
17.7 x 12.5 square inches ; that is, between 218.24 square inches
and 221.25 square inches.
183. When the measure of any quantity is given, for
example, as 3.628, it generally means that the measure is
Arts. 182-184.] APPROXIMATION. 177
not less than 3.628, and not greater than 3.629, the possible
error made by stopping at the third decimal place being
an error in defect less than one one-thousandth of the
unit. Now, if the above measure had to be given as far
only as hundredths of the unit, 3.63 would be more accu-
rate than 3.62. This principle is often employed when
approximate measures are given. Thus the quantity
whose measure is 6.57684 would be most accurately given
by 6.5768, 6.577, or 6.58 to four, three, or two decimal
places respectively, the possible error in excess or defect
being now not greater than half the unit represented by
the last decimal place retained.
184. To find the sum of any numbers to any given num-
ber of decimal places, it would be necessary to consider
the figures two places beyond, in order to see what had
to be ' carried.'
Ex. 1. Find, to 3 places of decimals, the sum of 14.61825,
3.17924, .518479, and 154.017235.
14.618
3.179
.518
154.017
172.333
25
24
479
235
Ex. 2. Find, to within one one-thousandth of the whole, the sum
of 5.3184, 27.5162, 18.4196, and 23.0135.
5.31
27.51
18.41
23.01
74.27
Here we have to find the sum correct to the first four figures.
The sum of the numbers in the fifth column is 27, which is nearer
to 30 than to 20. Hence, the most accurate sum to four figures
will be 74.27.
178
APPROXIMATION.
[Chap. VII.
185. The method of finding a product or a quotient to
any required degree of accuracy will be seen from the
following examples.
Ex. 1. Find, to two places of decimals, the product of 4.163 and
5.784.
4.10
3
5.7
84
20.81
5
2.91
41
.33
30
.01
66
24.08
Arrange with the decimal point of the multiplier as ahove, and
begin the multiplication from the left of the multiplier. The verti-
cal line on the left gives the figures which are to be finally retained ;
it is, however, necessary to go two places beyond to see what
should be ' carried ' to the last column retained.
Multiply as usual so long as all the figures are to be retained.
In the present case all the figures in the first two rows are to be
retained.
Before multiplying by 8, cross out the last figure of the multipli-
cand, namely 3 ; then multiply 416 by 8, putting down the first
figure of the product (adding in mentally what would be carried
from the multiplication of the figure crossed out) in the last
column. Now cross out another figure of the multiplicand, and
multiply what remains by 4, again putting down the first figure of
the product (with what must be carried from the multiplication of
the last figure crossed out) in the last column. Proceed in this
way to the end.
Since the sum of the figures in the fifth column is 18, the most
accurate product we can give to two places of decimals is 24.08.
Ex. 2. Find, to within one one-millionth
of the whole, the product of 51.6243 and
112.4167.
Here we have to find the product, correct
to the first 7 figures. •
61.622
11
5162.43
516.243
130.248
20.649
.516
.309
.036
5803.433
2.4
8
72
24
74
13
167
Art. 185.]
EXAMPLES.
179
Ex. 3. Find, to within one one-millionth, the quotient
516.24175 -*- 123.456.
122.4^)516.24175(4.181585
493 824
22 4177
12 3456
10 07215
9 87648
19567
12345
7222
6172
1050
987
63
61
We have here to find the first seven figures of the quotient. Having
found the first three figures in the ordinary way, the remaining four
figures, being less by two than the number of figures in the divisor,
can be found by a shortened process ; namely, instead of annex-
ing a naught at every stage on the right of the remainder as usual,
we strike out the last figure on the right of the divisor instead,
taking care, however, to use the last figure struck out to see what
should be 'carried'.
Ex. 4. Find, to the nearest penny, the value of
£51.3125 x 17.1874.
Since \d. = £ .001 nearly, it will be unnecessary to retain more
than four decimal places in the product.
Thus, &b\J8ffl
17 .1874
513.125
359.1875
5.1312
4.1050
.3591
205
5
8
2
'£881.9285
20
s. 18.5700
12
d. 6.84 Ans. £881. 18s. Id.
180 APPROXIMATION. [Chap. VII.
EXAMPLES LXXIII.
Written Exercises.
Find the following to the nearest thousandth of the
whole :
1. 14.625x31.857. 4. 138.714x89.47.
2. 15.816 x 19.714. • 5. 314.2108 -j- 18.306.
3. 156.423x175.45. 6. 81.4623 -s- 129.54.
7. 15.8193 x 6.7149 -j- 1.3425.
8. 115.416 x 123.518 - 119.417.
Find, to within a millionth of the whole :
9. 198.4653x5.194238. 10. 8.10976429-5-15.623.
Find, to within one one-thousandth of the whole, the
areas of the rectangles whose dimensions are :
11. 17.215 in. by 34.827 in.
12. 184.27 yd. by 112.53 yd.
13. Find, to 4 places of decimals :
(i ) 1 + i+ i + ^-L- n +
1 1x2 ' 1x2x3 1x2x3x4
1 1x2 1x2x31x2x3x4
Find the value, to the nearest farthing, of
14. £31.625x12.8743. 15. £ 119.48125 x .46127.
Find, to the nearest cent, the value of
16. $15.23x18.24. 18. $315.80x175.297.
17. $17.32x112.428. 19. $30.47x2180.3079.
Art. 185.] MISCELLANEOUS EXAMPLES. 181
EXAMPLES LXXIV.
Miscellaneous Examples, Chapters V, VI, VII.
Written Exercises.
1. Find 18 x 19 x 25 x 16f.
2. Express .035, .625, .12288 as common fractions in
their lowest terms.
3. How many times is 14 yd. 1 ft. 6 in. contained in
244 yd. 3 in. ?
4. Reduce 3 lb. 5 oz. 16 dwt. to gr., and express 1 oz.
16 dwt. 11 gr. in avoirdupois weight.
5. Find H.C.F. and L.C.M. of 936 and 2925.
6. Arrange -^, yl^, and -fe in order of magnitude.
7. Find the cost of 25cwt. 25 1b. 12 oz. of a sub-
stance at $ 16 per cwt.
8. Find the value of 51 things, any four of which are
worth £ 19. 3s. Id.
9. Simplify |f (1 - ff) + fx|(| + A)-
10. What is the least number which must be added to
1000000 that the sum may be exactly divisible by 573 ?
11. Multiply 4 mi. 31 rd. 4|yd. by 3, and divide the
result by 37.
12. The circumferences of the large and the small
wheels of a bicycle are 143 in. and 40 in. respectively ;
how many more turns will the latter have made than
the former in a distance of 13 mi. ?
13. A man spends 7.75 francs a day ; how much does
he save in a year (of 365 days) out of a yearly income
of 3000 francs ?
14. A man spends 9.35 marks a day; how much in
English money does he spend in a year (of 365 days),
taking a mark to be worth ll|d ?
OF THE "^
I'NfVFDQlTV
182 MISCELLANEOUS EXAMPLES. [Chap. VII.
15. Afield is 192 m long and 57.75 m wide; how many
Ha does it contain, and what wonld it cost at 7500
francs per Ha ?
16. Eeduce 772642 sq.yd. to A., sq. rd., and sq. yd.
17. Find, in hr., min., and sec, .6575 of a day.
18. What fraction of 8 lb. 11 oz. 2 dwt. 17 gr. is 10
lb. 9oz. 16 dwt. 11 gr.?
19. Reduce ||, -£fe, and fl yg ft to decimals.
20. A certain number was divided by 105, by 'short'
divisions ; the quotient was 192, the first remainder was 1,
the second was 4, and the third was 6. What was the
dividend ?
21. Find by factors the square root of 23716.
22. What is the greatest sum of money of which both
$ 11.05 and $ 188.50 are multiples ?
23. How much would it cost to put gravel to a depth
of a dm all over a court-yard 7.5 m by 5.75 m , the gravel
and labor costing 8 francs per ster ?
24. A grocer buys 15 cwt. of goods for $24.50; at
what rate per lb. must he sell to gain $ 5.50 ?
25. A druggist buys 50 1b. of a certain drug; how
many weeks will it last if he uses lb 1 5 6 3 1 3 2 gr. 10
per week in putting up prescriptions ?
26. Find If of 8 bu. 1 pk.
27. How many numbers, each 567, must be added that
the sum may be greater than a million ?
28. What is the greatest number of Sundays there can
be in a year ? On what day of the week will the first of
February fall when the number of Sundays in a year of
365 days is greatest ?
Art. 185.] MISCELLANEOUS EXAMPLES. 183
29. How many times can 3 yd. 1 ft. 7 in. be sub-
tracted in succession from 115 yd. 2 ft. 11 in., and what
will be the last remainder ?
30. A bar of metal weighing 100 oz. 16 dwt. is made
into coins, each weighing 1 oz. 8 dwt. ; how many coins
are made from the bar ?
81. Simplify If of |L=^J2f-! of -U.
32. A surveyor measured some ground and found it to
be 10 ch. long and 4 ch. broad ; how many A. were there ?
33. What is the smallest number of exact acres that
can have the form of a square ?
34. What decimal of 1 mi. is 119yd. 2ft. 4 in.?
35. Find the value of 21b. 6 oz. 10 dwt. 12 gr. of
gold at $ 216 per lb.
36. Find 105 2 ; 48 x 33£; 850 - 16|.
37. Express lb. 1 as the decimal of 1 lb. Avoir.
38. Having given that a meter is 39.37 in., prove that
the difference between 5 mi. and 8 Km is nearly 51 yd.
39. Add
i ° f feJ ° f • 4 - 65 to ft- ° f 3*3 ° f • Li5 -
°3 % + * ±J -TT ^ — l S
40. Find Vlwt an( ^ re( iuce the answer to lowest terms.
41. Express .88125 cwt. in lb. and oz.
42. Express 15 yd. 2 ft. 8 in. as the decimal of a mi.
43. Reduce 11.2765625 lb. to lb., oz.. pwt., and gr.
44. Find to the nearest cent $ 48.96 x 72.8967.
45. Reduce 1000 sq. yd. to qm.
46. Reduce 1000 l to pt.
184 MISCELLANEOUS EXAMPLES. [Chaps. VII., VIII.
47. Express .136 x 7.3 -s- .43 as a decimal.
48. Find the value of 43 sq. rd. 24^ sq. yd. of building
land at $ 1815 per acre.
49. Find the greatest length of which both 1 mi. 4 fur.
16 rd. 2 yd. and 1 mi. 1 fur. 10 rd. 2 yd. are multiples.
50. Subtract 161 x ** from 5 f of 7 \ .
51. Find y.004 to 4 decimal places.
52. Reduce 4 Hg to pounds Troy.
7
53. Simplify
5 +
3-f
54. Find the annual cost of repairing a road 9 mi.
120 rd. 177 yd. long at $ 88 per mi.
55. A vessel steams 18 knots an hour; to how many
statute miles is this equivalent ?
56. If a ccm of iron weighs 7.788 g , what will be the
weight of a cu. ft. ?
57. How many pieces each .17 in. long can be cut from
a wire 21.09 in. long ; and how long will be the piece
left over?
58. Add .5125 of a yd., .62734 of a rd., and .018325 of
a fur. ; subtract the result from .0049 of a mi., and ex-
press the answer in yd., also in dm.
59. Find V4900546043.21156004.
60. What is the least number which when
divided by 15 leaves a remainder 3, when
it
« 18
u
u
a
6,
((
" 24
a
a
a
12?
Arts. 186, 187.]
AREAS.
185
CHAPTER VIII.
AREAS — VOLUMES.
186. A plane figure [Art. 149] bounded by four straight
lines, and whose four angles are equal, is called a
Rectangle.
An equilateral rectangle is a Square.
Rectangles.
Square.
The amount of surface included within the bounding
lines of a figure is called its Area, and the area is measured
by some square unit, — one sq. in., one sq. yd., or one
qm, etc.
B
C
A
G
D
F
B
187. To find the Area of a Rectangle. — Let ABCD be
the rectangle whose area is required.
186 AEE AS — VOLUMES. [Chap. VIII.
Suppose, for example, that AB is 4 in. and that AD is 3 in.
Divide AB into four equal parts and AD into three equal parts,
and draw lines parallel to the sides as in the figure on the left.
Then the rectangle is divided into squares each of which is a
sq. in. ; and the number of these squares is clearly the product
of the number of in. in AB by the number of in. in AD.
The above reasoning applies to all cases, both the length and
the breadth of the rectangle being an integral number of in.
Now suppose, for example, that in the figure on the right AB is
§ in., and that AD is £ in.
Let AEFG be one sq. in. Divide AE into two equal parts,
and AG into five equal parts, and GD into two equal parts.
Then the subdivisions of AB will be all equal, as also those of AD.
Hence, if lines be drawn as in the figure, ABCD will be divided
into 3x7 equal rectangles, such that the square inch AEFG will
3x7
contain 2 x 5 of these rectangles. Hence AB will contain
square inches ; that is, (§ x |) square inches.
From the above it follows that the number of square
inches (or square feet, etc.) in a rectangle is equal to the
product of the number of inches (or feet, etc.) in the length
by the number of inches (or feet, etc.) in the breadth.
It should be noticed that the length and breadth must
both be expressed in terms of the same unit.
For example, the area of a rectangle whose length is 2 ft. and
breadth 6 in. is (2 x T 6 2) sq.ft , or (24 x 6) sq. in.
The above rule for finding the area of a rectangle is often ex-
pressed shortly by the statement that area = length x breadth.
188. Now that we find the area of a rectangle, we can
see that the relations between the different units given in
the Table for Square Measure, on page 149, follow at once
from the relations between the corresponding units in
linear measure.
For, since 12 in. make 1 ft., (12 x 12) sq.in. make 1 sq.ft.
Since 3 ft. make 1yd., (3 x 3) sq.ft. make 1 sq. yd.
Arts. 188, 189.] EXAMPLES. 187
Since 5|yd. make 1 rd., (5£ x 5|) sq. yd. make 1 sq. rd.
Again, 22 yd. make 1 ch., therefore (22 x 22) sq.yd. = 484 sq.
yd. make 1 sq. ch.
Thus, 4840 sq. yd. = 10 sq. ch. = 1 A.
Also, 1 sq. mi. = (1760 x 1760) sq. yd. = 1760 x 1760 «■ 4840 A.
= 640 A.
Ex. Find the acreage of a rectangular field whose length is 132
yd. and whose breadth is 38 1 yd.
The area = (132 x 38£) sq. yd.
= 5082 sq.yd. = ffff A.
= |iA. = 1A. 8sq. rd.
189. If the area of a rectangle be known, and also the
length, the breadth can be at once found.
For example, to find the breadth of a rectangle whose length is
15 ft. and whose area is 200 sq. ft.
Since the product of the number of ft. in the breadth by the
number of ft. in the length is equal to the number of sq. ft. in
the area, we have
breadth = (200 + 15) ft. = 13£ ft. = 13 ft. 4 in.
EXAMPLES LXXV.
Written Exercises.
Find the areas of the rectangles whose lengths and
breadths are as follows :
1. 14 ft., 12 ft. 6. 10 yd., 23 ft.
2. 22 ft, 17it. 7. 5 yd. lft., 3 yd. 2 ft.
3. 25 yd., 17 yd. 8. 21 yd. 2 ft., 18 yd.
4. 122 in., 114 in. 9. 13 ft. 4 in., 9 ft. 2 in.
5. 5 ft., 17 in. 10. 11 ft. 9 in., 8 ft. 7 in.
188 AREAS— VOLUMES. [Chap. VIII.
Find the acreage of the rectangular fields whose lengths
and breadths are as follows :
11. 319 yd., 275 yd. 15. 550 yd., 400 yd.
12. 363 yd., 240 yd. 16. 125 yd., 49£ yd.
13. 400 yd., 214|-yd. 17. Length x breadth = ?
14. 178|yd., 162fyd. 18. Area -i- length = ?
19. Area -r- breadth = ?
20. Find the area of a rectangular field whose length
is 119.5 m and whose breadth is 96.2 m .
21. How many stones having rectangular tops 2 dm
X 1.2 dm will be required to pave a street 5 Hm long and
16.8 m wide, provided no spaces are left between the
stones ?
22. Find the area of a rectangular field whose length
is 9 ch. 12 li. and whose breadth is 6 ch. 25 li.
23. Find the area of a rectangular field whose length
is 9 ch. 25 li. and whose breadth is 7 ch. 75 li.
24. The area of a rectangle is 925 sq. in., and its
breadth is 25 in. ; what is its length ?
25. What is the length of a rectangular table the area
of whose top is 71 sq. ft. 16 sq. in., and the breadth 6 ft.
8 in.?
26. The area of a rectangular court-yard is 52 sq. yd.
2 sq. ft. 36 sq. in., and its length is 14 yd. 9 in. ; what is
its breadth ?
27. What will it cost to paint the ceiling of a room
whose length is 24 ft. 6 in. and breadth 16 ft. 6 in. at
$.60 per sq. yd. ?
28. What is the area of a square floor 7 m long ?
29. What is the length of a square room whose area is
4225 qdm ?
Art. 190.] CARPETING, ETC. 189
Carpeting, Papering, Plastering.
190. Examples like the following are of frequent
occurrence :
Ex.1. How much will be the cost of a cm-pet for a room 16 ft.
x 20 ft. 3 in. with carpet 27 in. wide at $.75 a yd., the strips running
lengthwise ?
Number of strips = 20 ft. 3 in. +- 27 in. = 9.
Total length of carpet = 16 ft. x 9 = 144 ft. = 48 yd.
Cost = 48 yd. x $.75 = $36. [Art. 50.]
Ex. 2. How much will be the cost of paper for the walls of a
room 19/£. 3 in. long, 15/£. 9 in. wide, and 12 ft. high, the paper
being 21 in. wide and costing 5 ct. per yard ?
Area of a wall = its length x its height.
. •. Area of 4 walls = distance around the room x height
= 70 ft. x 12 ft.
= 840 sq. ft.
Length of paper = 840 sq. ft. -r- f \ ft.
= 480 ft = 160 yd.
Cost of paper = 160 yd. x 5 ct.
= $8.00.
Note. In the preceding questions we have found the quantity
of carpet (or wall paper) which would be required if it were of
one uniform color throughout. When, as is almost invariably the
case, there is a pattern on the carpet or paper, there must be a
certain amount of waste, if the different lengths are properly
fitted together. Moreover, wall papers are sold in lengths of 8
yards, called rolls ; if, therefore, as in Ex. 2, 160 yards of paper
were required, 20 rolls would have to be bought. American wall
papers are generally 18 inches wide.
Ex. 3. A room 21 ft. by 19/£. has a Turkey carpet in it, a border
3ft. wide all round being left uncovered by the carpet. The
border was stained at a cost of $ .45 a square yard, and the carpet
cost $4.50 a square yard; what was the total cost ?
Since the border is 3 ft. wide all round the room, the length of
the carpet must be 21 ft. - 3 ft. x 2 = 15 ft., and the breadth must
be 19 ft. - 3 ft. x 2 = 13 ft.
190 AREAS — VOLUMES. [Chap. Vlll.
Area of carpet = 15 ft. x 13 ft. = - 6 / sq. yd.
Price of carpet = $4.50 x 6 ^- = $97.50.
Border = area of room — area of carpet
a= (21 x 19) sq. ft. - (15 x 13) sq.ft.
as 204 sq. ft. ss - 6 3 8 - sq. yd.
Cost of staining = $.45 x -^ = $10.20.
Total cost = $ 97. 50 + $ 10.20
= $107.70.
EXAMPLES LXXVI.
Written Exercises.
1. How much carpet 27 in. wide will cover a room
22 ft. 6 in. long and 15 ft. 9 in. wide, carpet running
lengthwise ? What will be the cost at $1.20 per yd. ?
2. A room is 8.3 m long and 5 m wide; how many
meters of carpet must be purchased for such a room, the
strips being 7 dm wide and running crosswise ? How much
in width must be turned, under ? In surface ?
3. If you were carpeting a room 9 m x6 m , which way
would you have the strips run if they were 6.8 dm wide?
How many less qm would be used than by running the
strips the other way ?
4. A room is 10 yd. 2 ft. long and 7 yd. 1ft. 6 in.
wide ; find the cost of covering it with Turkey carpet at
$1.25 a sq. yd.
5. Find the cost of carpeting a room 8^ yd. long by
6 yd. 2 ft. broad with carpet 2\ ft. wide at 84 ct. a yd.
6. What would be the expense of carpeting a room
24 ft. 6 in. by 18 ft. with carpet 27 in. wide, and which
costs $1.20 a yd. ?
Art. 190.] CARPETING, ETC. 191
7. How much carpet 27 in. wide would be required
for a room 32 ft. by 23 ft., a margin 4 ft. wide being
left uncovered ?
8. Find the area of the four walls of a room 15 ft.
long, 14 ft. wide, and 10 ft. high.
9. Find the area of the four walls of a room 16 ft.
4 in. long, 13 ft. 8 in. wide, and 11 ft. 4 in. high.
10. Find the area of the four walls of a room 10.5 ■
long, 5 ra wide, and 4.9 m high.
11. Find the qm of the four walls of a room 7 m x 4 m
X 3.2 ra , leaving out 3 windows, each 2 m x 1.1 m , and one
door 2.4 m x 1.3 m .
12. Find the area of the four walls of a room 14 ft.
6 in. long, 13 ft. 10 in. wide, and 10 ft. 8 in. high.
13. A room is 18 ft. long, 13 ft. 6 in. wide, and 12 ft.
high ; how much paper 21 in. wide will be required to
cover the walls, and what will be the cost at $ .75 per
piece of 12 yd. ?
14. How much will it cost to paper a room 17 ft. 6 in.
square, and 14 ft. 3 in. high, with paper 1 ft. 9 in. wide
at 12 ct. a yd. ?
15. A room is 6.1 ra x 5 m x 4.2 m ; find the cost of plas-
tering at 62^- ct. per qm, allowing 7 Qm for windows,
door, and base-board. Do not forget the ceiling.
16. A room 8 m x 5.6 m x 4.2 m has 4 windows, each
2.1 m x l m , 2 doors, each 2.8 m x 1.4 m , and a base-board
24 dm high; fi n d (i) the cost of plastering at 50 ct. per
qm, (ii) the cost of paper 5 dm wide at $3.50 per roll of
10 m , (iii) the cost of a carpet 6.2 dm wide at $2 per m,
all for this room. Find the total cost, allowing $ 15 for
labor in putting on the paper and laying the carpet.
192 AREAS — VOLUMES. [Chap. VIII.
Board Measure.
191. A board which is one foot square and one inch or
less in thickness has a measurement called one Board
Foot.
Boards and squared timber are sold by the Board Foot.
192. The number of board feet in a board one inch or
less in thickness is the same as the number of square feet
in the surface.
The number of board feet in a stick of timber more
than one inch thick is the number of square feet in the
surface multiplied by the number of inches in the thick-
ness.
Ex. 1. How many board feet in a board 20/£. x2ft. x f of
an inch f
20 ft. x 2 f t. = 40 board feet.
Ex. 2. How many board feet in a board 15 ft. long, IS in. wide
at one end and 14 in. wide at the other end, and \ an inch thick ?
15 ft. x 1 J ft. = 20 board ft.
In this case the average width is used.
Ex. 3. How many board feet in a stick of timber 21.6 ft. long,
14 in. wide, and 3§ in. thick ?
21.6ft. x lift. = 25.2 board ft.,
if the stick were 1 in. or less in thickness. But we must multiply
this result by 3|, since the timber is 3| in. thick. Thus,
21.6 ft. x | ft. x- : f = 92.4 board ft.
EXAMPLES LXXVII.
Written Exercises.
Find the number of board feet in the following :
1. A board 20 ft. x 2 ft. x 1£ in.
2. A board 19 ft. 8 in. x 1 ft. 9 in. x £ in.
Arts. 191-193.] DIMENSIONS OF CIRCLES. 19S
3. A timber 13 ft. x 1.1 ft. x 4J in.
4. A joist 11 ft. x 5 in. x 2 in.
5. A joist 16 ft. x 6 in. x 2\ in.
6. Find the cost of each of the above five pieces at
$20 per M; i.e. f by the thousand (board feet).
7. A hall 75 ft. x 50 ft. has two layers of boards for
its floor, one kind costing $ 12 per M, and the other cost-
ing $21 per M; the floor timbers, 70 in number, are
placed crosswise, and cost $ 16 per M. How much is the
cost of material, the boards being fin. thick, and the
timbers 8 in. x 3 in.
Dimensions of Circles.
193. Cut from cardboard a circular piece having a
known radius, as 3 cm , or 3 in. Roll the circle (held
upright) along a straight line and measure the line trav-
ersed in one complete rotation of the circle. This line
will be found to be about 3^ times the diameter.
We have no means of finding the exact measure of the
circumference in terms of the diameter, but by means of
geometry we learn that the measure is 3,1416 (nearly)
times the diameter. This is more exact than 3^.
Hence, Diameter x 3.1416 = Circumference,
and C -f- 3.1416 = D.*
Using the results obtained in geometry for areas of
circles, we have,
Area = R 2 x 3.1416,
R 2 = Area -- 3.1416,
R = VArea -h 3.1416.
* Z), B, and C stand for diameter, radius, and circumference,
respectively.
194 AREAS — VOLUMES. [Chap. VIII.
Ex. 1. The diameter of a circle is 10 dOT ; find the circumference
and area.
C=Dx 3.1416 Area = i? 2 x 3.1416
= 10 dm x 3.1416 = 25^ x 3.1416
= 31.416 dm . = 78.54 «*".
Ex. 2. The area of a circle is 50.2656 sq, in. ; find the radius
and the circumference.
JR= VArea-s-3.1416 C = D x 3.1416
= V50.2656 *■ 3.1416 = 8 x 3.1416
s Vl6 =25.1328 in.
= 4 in.
EXAMPLES LXXVIII.
Written Exercises.
Find the circumference when
1. D = 14 in. 3. E = 15 cm . 5. Z) = 56yd.
2. 2) = 75 ft. 4. B = l$ m . 6. iJ = -J-mi.
Find
7. R when = 314.16 cm . 8. D when C = 1 mi.
9. D when O = 153.9384 yd.
10. iZwhen C = 47.124 m .
Find area when
11. 5 = 14ft 13. C=37.6992 dm .
12. Z) = 20 m . 14. C = 251.328 rd.
15. How many" sq. ft. in the floor of a circular room
whose diameter is 28 ft. ?
16. The bottom of a round liter measure has a surface
of 500 qcm ; find the approximate radius.
17. Find the cost of concreting a circular fountain
basin whose diameter is 20 ft., the work and material
costing f 3.27 per sq. yd.
Arts. 194, 195.] RECTANGULAR SOLIDS.
195
Rectangular Solids.
194. That which has length, breadth, and thickness is
called a Solid.
A solid bounded by six rectangular [Art. 186] faces is
called a Rectangular Solid.
A cube [Art. 151] is one form of a rectangular solid.
Any substance (water, air, wood, etc.) may be a rectan-
gular solid in form.
The space included between the bounding surfaces of
a solid is called its Capacity (or Volume), and the capacity
of a solid is measured by some cubic unit — :Oiie cu. in.,
one cu. ft., l ccm , or l cum , etc.
195. To find the Capacity of a Rectangular Solid.'
Suppose, for example, that the dimensions of the solid
are 5 in. by 4 in. by 3 in. We can divide the edges re-
spectively into 5, 4, and 3 parts, each being one inch ; and
if planes be drawn through the points of division parallel
to the outer faces of the solid, as in the figure, the whole
solid will be divided into equal cubes each of which is a
cubic inch.
There will be as many layers of cubes as there are inches
in the height of the solid, and the number of cubes in each
layer will be the product of the number of inches in the
length by the number of inches in the breadth.
196 AREAS — VOLUMES. [Chap. VIII.
Thus, the number of cubic inches (or cubic feet, etc.) in a
rectangular solid is equal to the continued product of the
number of inches (or feet, etc.) in its length, breadth, and
thickness.
196. Now that we can find the capacity of a rectangular
solid, we can find the relations between the cubic yard,
the cubic foot, and the cubic inch.
For 1 cu. yd. = (3 x 3 x 3) cu. ft.,
and 1 cu. ft. = (12 x 12 x 12) cu. in.
Ex. 1. Find the volume of a rectangular block of stone 12 ft.
long, 1ft. wide, and lft. 6 in. high.
Volume = (12 x 7 x If) cu. ft. = 126 cu. ft.
Ex. 2. A beam lft. 6 in. wide and lft. Sin. high contains 46\
cubic feet of timber ; what is its length ?
Since volume = length x breadth x thickness,
lengths volume -•
breadth x thickness
4g i
Hence, length required = 1 — 241 ft.
\\ x l£ft. 3
Ex. 3. How many gallons of water will a cistern hold if it is
6 ft. long, 4 ft. 6 in. wide, and 3 ft. 6 in. high ? [A gallon con-
tains 231 cu. in.~]
The cistern will hold
(72 x 54 x 42) cu. in. = 163296 cu. in.
Hence, the number of gallons required = 163296 -4- 231 = 706.90.
Ex. 4. The external dimensions of a rectangular stone tank are :
length 12 ft. 6 in., breadth 8 ft., and height 4 ft. The interior is also
rectangular, and the sides and bottom are 3 in. thick. Find the
number of cu.ft. of stone in the tank.
The internal length = 12 ft. 6 in. - 3 in. x 2 = 12 ft.,
the internal breadth = 8 ft. - 3 in. x 2 = 7 ft. 6 in.,
and the internal height «= 4 f t. — 3 in. = 3 ft. 9 in.
Art. 196.] EXAMPLES. 197
Now the volume of the stone is the difference between the
volumes given by the external and internal dimensions.
Hence, volume required
= (12$ x 8 x 4 - 12 x 7£ x 3|) cu. ft.
= (400 - 337i) cu. ft. = 62£ cu.ft.
EXAMPLES LXXIX.
Written Exercises.
Find the volumes of the rectangular solids whose
dimensions are
1. 5 ft. by 4 ft. by 2 ft.
2. 12 ft. by 6 ft. by 4 ft.
3. 3 yd. by l^yd. by 2 ft.
4. 5 yd. by 21 yd. by 4 ft.
5. 6 ft. 4 in. by 4 ft. 3 in. by 2 ft. 6 in.
6. 7 ft. 9 in. by 5 ft. 3 in. by 3 ft. 6 in.
7. 5 yd. 1ft. by 3 yd. 2 ft. by 2 ft. 9 in.
8. 6 yd. 9 in. by 2 yd. 1ft. by 2 ft. 7 in.
9. A rectangular block of stone 4 ft. long and 2 ft.
6 in. broad contains 17-J cu. ft. of stone ; what is its
height?
10. Find the length of a rectangular beam which con-
tains 98 cu. ft. of timber and whose cross-section is 2 ft.
square.
11. How many loads (cu. yd.) of gravel would be
required to cover a path 150 yd. long and 4 ft. wide to
a depth of 2 in. ?
12. A school-room whose floor is 60 ft. by 40 ft. has
accommodation for 360 children, allowing 100 cu. ft. of
air for each child; what must be the height of the room ?
198 AREAS — VOLUMES. [Chap. VIII.
13. If 1 gal. = 231 cu. in. and 1 gal. of water weighs
8.355 lb. Avoir., find the number of gal. and the weight
of the water which would fall on an area of an A. during
a rainfall of one in.
14. A tank is 21ft. 4 in. long, 3 ft. wide, and 2 ft.
deep ; it is filled with water to within 3 in. of the top.
What is the volume of the water, and what is its weight ?
[A cu. ft. of water weighs 1000 oz.]
15. What weight of water will fall on a road \ a mi.
long and 30 ft. wide during a rainfall of an in. ?
16. A level tract of land 20 mi. long and f of a mi.
broad is flooded to a depth of 4 ft. Given that a cu. ft.
of water weighs 62.5 lb., find in t. the weight of the
water on the land.
17. What is the capacity of a tank 20 m x 8 m x 2 m ?
How many T of water will it hold ? Eeduce the T to
t. (tonneaux to tons).
18. Find the total surface of the stone in Ex. 9.
19. Find the inner surface of the tank in Ex. 17.
20. A square room is 5 m long and 3 m high; how many
cu. in. of air will the room contain ?
21. A square room 9 ft. 3 in. high has a capacity of
1563 \ cu. ft. j what is the length of the room ?
Cylinders.
197. A solid whose ends are circles and whose curved
surface is perpendicular to the ends is called a Right
Circular Cylinder.
The ends are called Bases,
For example, a common lead pencil is a right circular cylinder ;
and some tin measures used for liquids are right circular cylinders.
Note. When cylinders are mentioned in this book, right circu-
lar cylinders are meant.
Arts. 197, 198.] CYLINDEKS. 199
198. The total surface of a cylinder consists of two
flat surfaces (circles)
and a curved surface
called the Lateral Sur-
face.
If a piece of paper be
fitted to a cylinder so as
to cover all its lateral sur-
face and then unrolled, it
will be a rectangle whose
length is the circumference
of the cylinder and whose
breadth is the height of the
cylinder.
Hence, lateral surface = Cx height ;
which (by Art. 193) = D x 3.1416 x H;
whence H= lateral surface -r- (D x 3.1416),
and D = lateral surface -r- (3.1416 x H).
Ex. 1. Find the total surface of a cylinder 8 in. high and the
radius of whose base is 2 in.
Total surface = 2 bases + lateral surface
= 2 iF 2 x 3.1416 + D x 3.1416 x H
= 2B x 3.1416 (22 + #)
= 4 x 3.1416 x 10
= 125.664 sq. in.
Note. H = height ; D = diameter ; C = circumference.
Ex. 2. The lateral surface of a cylinder is 188.496 i dm ; find the
height when D = 6.
H= 138.496 - (D X 3.1416)
= 188.496 -4- 18.8496
= 10 ^
200
AREAS— VOLUMES.
[Chap. VIII.
199. If a cylinder is 10 in. high, it is evident that it
will contain 10 times as many cubic inches las if it were 1
in. high; since the number of cubic
inches in a cylinder 1 in. high is the
same as the number of square inches in
the base, the
Volume of a cylinder = base x H,
or (Art. 193), = R 2 x 3.1416 x H,
Volume
and
and
and
H=
R 2 =
R 2 x 3.1416'
V
5.1416 x H'
*=v
3.1416 x H
Ex. 1. Find the volume of a cylinder whose height is 10 in. and
the radius of whose bast is 6 in.
V=B*x 3.1416 xH
= 36 x 3.1416 x 10
= 1130. 976 cu. in.
Ex. 2. Find the radius of the base of a cylinder whose volume
is 125.664 cdm and whose height is l m .
JB
-V
= ^V 3.1416 x 10
b 2 dm .
3.1416 x H
125.664
EXAMPLES LXXX.
Written Exercises.
Find, in a cylinder, the
1. Lateral surface when R = 3 dm and H= 14 dm .
3. Lateral surface when R = 1 in. and H = 5 in.
Arts. 199, 200.] SPECIFIC GRAVITY. 201
3. Total surface when D = 10 m and H= 8 m .
4 . H when lateral surface = 1413.72 i dm and D = 15 dm .
5. Fwheni2 = 3 dm and i7=14 dm .
6. V when R == 1 in. and H= 5 in.
7. FwhenD = 10 m and J fiT=8 m .
8. IT when V= 125.664 cu. ft. and B = 2 ft.
9. jRwhen F=1570.8 1 and 17 = 20 dm .
10. Measure in centimeters the height and diameter of
some cylinder and calculate how many cubic centimeters
of liquid it would hold if hollow. How many grams of
water would it hold ?
Specific Gravity.
200. Weigh accurately a stone. Then place it in a jar
brimful of water and weigh the water which runs over.
Now divide the weight of the stone by the weight of the
water which ran over, and you will know how many times
the weight of the stone is greater than the weight of
the same volume of water.
The number of times that the weight of a substance is
greater than the weight of the same volume of water is
called the Specific Gravity (S.G.) of the substance.
A body floating in water displaces a weight of water
equal to its own weight.
EXAMPLES LXXXI.
Written Exercises.
1. The S.G. of iron is 7.8 ; how much does a cu. ft. of
iron weigh ? A cu. in. ? A ccm ?
2. What is the weight of a cdm of silver, its S.G. being
10.5?
202 AREAS — VOLUMES. [Chap. VIII.
3. A rectangular iron tank weighs 25 kilos, and it
floats on water; what is the weight of the water dis-
placed ? What is the volume of water displaced ? What
is the volume of iron in the tank ?
4. A cubical liter measure weighs 150 grams ; if put
in water, what pressure must be added to its own weight
to make it sink ?
5. The S.G. of gold is 19.5; if a person can lift 125
lb., how many cu. in. of gold can he lift at one time ?
6. What is the weight of a cdm of gold in pounds
Avoir. ? In pounds Troy ? What is its value at $1 per
pwt. ?
If the stone mentioned in Art. 200 be weighed in air
and then in water, the loss of weight will be found equal
to the weight of the water which ran over. Therefore
if we divide the weight of a substance by its loss of weight
in water, we shall obtain its S.G.
7. A substance weighs 2501b. in air and 1251b. in
water ; what is its S.G. ? How much water would run over
if the substance were put into a jar brimful of water?
What is the volume of the substance ?
8. A piece of wood, S.G. .25, floats on water and dis-
places 40 ccm of water ; what is the volume of the wood ?
How much iron must be attached to the wood to make it
float under water ?
9. How much weight will a ccm of iron lose when
weighed in air and then in water ?
Note. S.G. = weight in air -f- loss in water.
10. A person weighing 146^ lb. has a S.G. of 1.0417 ;
how much does he weigh in water ?
Art. 200.] MISCELLANEOUS EXAMPLES. 203
EXAMPLES LXXXII.
Miscellaneous Examples, Chapter VIII.
1. Find the cost of graveling a carriage drive 69 ft.
9 in. long and 16 ft. wide, at 30 ct. per sq. yd.
2. The outer and inner boundaries of a gravel path
are squares, and the path is 4 ft. wide. The side of the
square enclosed by the path is 50 yd. How much would
it cost to gravel the path at 37^ ct. a superficial yd. ?
3. Find the cost of turfing a lawn tennis court which
is 78 ft. long and 39 ft. wide, making a margin of grass
12 ft. wide at each end and 6 ft. wide at each side ; the
turf costing 4d. a sq. yd.
4. Find the prime numbers from 100 to 125 by using
the sieve. [Art. 78.]
5. What will be the lowest cost of carpeting a room
33 ft. long and 24 ft. wide with carpet 27 in. wide and
costing 85 ct. a yd., a border one yd. wide being left
uncovered ? How broad a strip must be cut off, or turned
under ?
6. Multiply 688.4 by 99; 460.01237 by 11.
7. Find the number of cu. ft. in a school-room by using
its length, breadth, and height ; find also the area of its
six surfaces, including windows, etc.
8. Find the cost of covering the floor of a hall 39 ft.
41 in. long by 20 ft. llf in. wide with tiles each 5J in.
by 4|in., and costing (including the labor of laying
them) $4.95 a hundred.
9. What is tile weight of an iron girder 20 ft. long,
and having 54 sq. in. sectional area, the weight of iron
being 480 lbs. per cu. ft. ?
204 MISCELLANEOUS EXAMPLES. [Chap. VIII.
10. A hall is 103.23 ft. long and 83.25 ft. broad, and
it is to be paved with equal square tiles ; what is the size
of the largest tile which will exactly fit, and how many
of them will be required ?
11 . A room 22 ft. 3 in. long, 17 ft. 9 in. wide, and 12 ft.
6 in. high has two windows each 5 ft. 3 in. by 3 ft. 4 in.,
a door 7 ft. by 3 ft. 9 in., and a fireplace 5 ft. 3 in. by
4 ft. 4 in. How many pieces (each 12 yd. long) of
paper 21 in. wide would have to be bought to paper the
room?
12. How many sq. ft. of boards are required for the
floor of a circular hall 100 ft. in diameter ?
13. What would be the weight of a beam of oak 5 dem
square and 7 m long, on the supposition that the S.G-. of
oak is .895 ?
14. Divide 8921045 by 385 by the method of Art. 69.
15. Supposing a postage stamp to be 1 in. long and
| of an in. broad, how many stamps would be required
to cover a wall which is 15 ft. 6 in. long and 10 ft. 8 in.
high?
16. What is the cost of paper for the hall of Ex. 12,
the hall being 25 ft. high, at $.57 a roll, allowing 500
sq. ft. for windows, etc. ?
17 . A cubical cistern, open at the top, costs £16. 13s. 4d.
to line with lead at 2d. per qdm; how many cum of
water will it hold ?
18. I have a rectangular box cover \ of a meter long
and A of a meter broad to be painted in squares ; what
is the largest square I can use ?
19. Find v to three decimal places.
Art. 200.J MISCELLANEOUS EXAMPLES. 205
20. Find the number of board feet in a stick of timber
8 \ in. square at one end, 8% in. by 5} in. at the other,
and 21 ft. long.
21. Find, by dividing by factors, 40579 -r- 72.
22. How many cdm of water are required to fill a
cylindrical tank whose radius is 10 dm and whose height
is2 ra ?
23. A cylindrical tank holds 1884.96 \ and its height
is 16f dcm ; what is its radius?
24. How many cu. in. of wood are there in a wooden
box whose external dimensions are 4 ft. 4 in. by 3 ft.
10 in. by 3 ft. 6 in., the wood being everywhere 1 in.
thick ?
25. An iron safe is everywhere l^in. thick, and its
external dimensions are 6 ft. by 4 ft. 6 in. by 3 ft. 6 in.
How much does the iron weigh? [The S.G. of iron has
been given several times.]
26. It cost 81.75/. to gravel a rectangular court-yard
8.25 m x 4.8 m with gravel costing 7.5/. per st. What was
the thickness of the layer of gravel ?
27. An importer received 427 T of goods at 125/ per
T; he paid the custom house $1602.92; his expenses
of cartage, etc., were $212 ; for how much must he sell
the goods per cwt. in order to make a profit of $2000 ?
28. The velocity of flow of water through a pipe 6 cm in
diameter is 7.6 ** per sec; how many 1 flow through
in 11 sec. ?
29. A room is 18 ft. 1 in. long, lift. 8 in. wide, and
11 ft. 3 in. high ; how many rolls of paper would be used
in papering the walls, supposing that windows, etc., make
up i of the whole surface of the walls ?
206 MISCELLANEOUS EXAMPLES. [Chap. VIII.
30. How much would it cost to carpet the room of Ex.
29, the carpet being 1 yd. wide, at $1.66§ per yd. ?
31. The S.G-. of a piece of wood is .5; the wood being
8 m x 8 dm x 8 cm , how many kilos would be required to sink
the wood if placed on water ?
32. Find J\ 2 |600 - (4 150 - 25.5 - 16) + 3}
- (4.1 12-2) - (*i x 90 + 50
33. The S.G. of gold is 19.5; find the weight of l cdm
and the volume of 1 kilo.
34. A pile of wood is 8 ft. long, 4 ft. high, and 4 ft.
broad ; what is its volume in cu. ft. ?
35. How many such volumes (Ex. 34) in a pile 49.6
ft. long, 8ft. high, and 4ft. thick?
36. A quadrangle 120 ft. x 100 ft. has, in the center, a
grass-plot 80 ft. x 60 ft. ; find the cost of graveling the
rest of it to a depth of 6 in. at 54 ct. a cu. yd.
37. At a certain place the annual rainfall was 24.15
in. ; find the number of gal. which fell on each sq. mi.
38. A slate cistern open at the top is everywhere 1
in. thick, and the external dimensions are length 6 ft.
4 in., breadth 3 ft. 2 in., and height 4 ft. 8 in. Find the
weight of the slate employed, assuming that 1 cu. ft. of
slate weighs 2880 oz.
39. Find the number of gal. the cistern in the previous
question will hold.
40. What is the height of a cylindrical liter measure
if the radius of its base is 5 cm ?
Answer to the nearest tenth of a mm.
Art. 200.] MISCELLANEOUS EXAMPLES. 207
41. State the squares of 23, 34, 38, 47, 78, and 96,
using the method indicated in Arts. 54 and 86.
42. Find by factors the H.C.F. and L.C.M. of 168,
2772, 4368, and 12474.
43. Find the weight of 12 m of alcohol, its S.G. being
.81.
44. A room is 24 ft. 2 in. x 18 ft. 11 in. ; which way
would it be the cheaper to run the carpet strips, each
strip being 27 in. wide ?
45. Find the dividend when the divisor is yf and the
quotient is J4.
208 RATIO — PROPORTION. [Chap. IX.
CHAPTER IX.
RATIO — PROPORTION.
201. The quotient of one number divided by another
of the same kind may be called the Ratio of the first to
the second.
Thus, the ratio of 6 ft. to 2 ft. is 3, and the ratio of 62 cwt. to
16 cwt. is ||, or 3|.
A ratio is expressed by the sign : placed between the
two quantities ; this sign means the same as the sign -j-.
Thus, 52 m : 26 m means 52 m -=- 26 »
The first is read ' the ratio of 52 m to 26 m ' ; the second is read
'the quotient obtained by dividing 52 m by 26 m .' The answer is
the same in both cases. A ratio may be also expressed in the form
52 m
of a fraction ; as,
' 26 m
202. The two quantities compared in a ratio are called
the terms of the ratio.
The old names, antecedent and consequent, for the first and
second terms respectively of a ratio, are still sometimes used.
The terms of a ratio must be of the same kind of
magnitude; for we cannot compare, for example, tons
with weeks, or acres with gallons.
When a ratio and one of its terms are given, the other
term can at once be found.
Akts. 201, 202.] EXAMPLES. 209
Ex. 1. A ratio is 3, and its first term is 6 ; find its second term.
Here a dividend 6 and a quotient 3 are given ; hence, the divisor
= 6-7-3 = 2 = the second term of the ratio.
Ex. 2. A ratio is 4.5, and its second term is 10 ; find its first
term.
Here a quotient 4.5 and a divisor 10 are given ; hence, the
dividend = 4.5 x 10 = 45 = the first term of the ratio.
Ex. 3. What sum of money has to $30 the ratio of 5:8?
Here the ratio is §, and its second term is $30 ; hence, $30 x $
= $18.75 = the first term.
EXAMPLES LXXXIII.
Oral Exercises.
Find the indicated ratios, and in lowest terms :
1. 9:3; 16:4; 50:5; 20:9; 12:16.
2. 18:30; 75:100; 90:30; .5:5; 5 : .5.
3. 121b. :51b.; 20 gr. : 33; $ 5 : $ 55 ; $ .50 : $5.50.
4. 24 g :36 g ; 50 cum :20 cum .
5. A square 2 ft. long : a square 1 ft. long.
6. A cube 2 in. long : a cube 1 in. long.
7. A circle 6 m in diameter : a circle 2 m in diameter.
8. What is the ratio of a square to another square
half as long? Twice as long? One-third as long?
Three times as long?
9. What is the ratio of a cube to another cube half
as long ? Twice as long ?
10. What is the ratio of a circle to another circle
having twice the diameter ? Three times the diameter ?
Five times ?
p
210 RATIO — PROPORTION. [Chap. Dt.
Written Exercises.
Find, in lowest terms, the indicated ratios :
11. $5 : $7.50 ; $3.25 : $12.50 ; $5.44 : $39.10.
12. 3t.5 cwt. : It. 15 cwt.; 3cwt. 641b. : 4cwt. 761b.
13. 7 mi. 208 rd. : 4 mi. 277 rd. ; 2 oz. 11 pwt. 18 gr. :
51 32 &1 gr.l.
14. Find what has to $1.20 the ratio of 2 : 3.
15. Find what has to 1 da. 4hr. 20 min. the ratio of
3:4.
16. Find what has to 11 cwt. 55 lb. the ratio of 7 : 11.
17. Find what has to £ 11. 14s. 9d. the ratio of 2.
18. Find the second term, the first term being 4 cd.
24 cu. ft., and the ratio being f, or 8 : 9.
Proportion.
203. When fonr quantities are such that the ratio of
the first to the second is equal to the ratio of the third
to the fourth, the four quantities are said to be Propor-
tionals.
For example, the ratio $5: $15 = the ratio 3t. : 9t. Hence,
the four quantities, $5, $15, 3 1., and 9t. , are proportionals.
The notation is as follows :
$5 : $15 = 3 1. : 9 1. ;
or, $5 : $15 : : 3 1. : 9 t.
This is read, '$5 is to $15 as 3t. is to 9t.' ; meaning that the
ratio between $5 and $15 is the same as that between 3 1. and 9 1.
A proportion may be expressed in the form
$5 = 3 t.
$15 9t.'
Arts. 203-205.] PROPORTION. 211
It follows that, when two fractions are equal, the terms
of the fractions, taken in the order in which they are
written, are proportionals.
Since the two terms of any ratio must be of the same
kind of magnitude, the first and second terms of a propor-
tion must be of the same kind, and the third and fourth
terms must be of the same kind.
204 .* The first and fourth, of four quantities in pro-
portion, are called the Extremes, and the second and
third are called the Means.
In the above case, the ratios are T \ and | respectively ; and
since T 5 7 = §, it follows that 5 x 9 = 15 x 3.
Thus, in this, and similarly in other cases, the product of the
extremes is equal to the product of the means.
205. When any three terms of a proportion are known,
the remaining term can be found.
Ex. 1. What quantity : 18 lb. : : $ 4 : $12 ?
For convenience, let x stand for the quantity to be found ; then,
x lb. : 18 lb. : : $4 : $12,
whence [Art. 204] 12 x x = 18 x 4.
. •. once x = 18 x T % = 6 lb.
This is equivalent to saying that T 4 ^ is a ratio, and 18 lb. is its
second term ; the first term (or dividend) must be 18 x T V
[Art. 58; also 202, Ex.2.]
Ex. 2. 27 oz. : 15 oz. : : what quantity : 25 in. f
27 oz. : 15 oz ; : x in. : 25 in.
15 x x = 27 x 25 ;
once x = f| x 25
= 45 in. = Ans.
* It will be noticed that, in any proportion, when the first term
is larger than the second, the third is larger than the fourth ; that,
when the first is smaller than the second, the third is smaller than
the fourth.
212 RATIO — PROPORTION. [Chap. IX.
We observe from the above that
either extreme = P^ct of means .
the other extreme
Likewise, either mean = P^ct of extremes,
the other mean
206. When the second term equals the third term, we
have but three different quantities in the proportion ; the
second is then called a Mean Proportional between the first
and third, and the third is called a Third Proportional
to the first and second.
Thus, in 4 : 6 : : 6 : 9, 6 is a mean proportional between 4 and 9,
and 9 is a third proportional to 4 and 6.
Here 9 _ square of the second and fl = ./ product of the extremes.
the first
EXAMPLES LXXXIV.
1 . Find the fourth proportional to 4, 7, and 12.
2. Find the fourth proportional to 9, 8, and 3.
3. Find the fourth proportional to -1-, ^, and f.
4. Find the mean proportional between 4 and 9.
5. Find the mean proportional between 8 and 18.
6. Find the mean proportional between | and f.
7. What quantity has to $1.32 the same ratio that
4 ft. 3 in. has to 2 ft. 9 in. ?
8. To what sum has Is. 9d the same ratio as 6 days
10 hr. has to 7 days 8 hr. ?
9. Fill up the blank in the proportion
£ 1. 12s. 6d. : £ 2. 2s. 6d : : 1 cwt. 18 lb. : .
Questions in which a missing term of a proportion has to
be found, and other questions of a similar nature, are best
treated by a method which we now proceed to consider.
Arts. 206, 207.] THE UNITARY METHOD. 213
The Unitary Method.
207. The method will be seen from the following
examples :
Ex. 1. If 5 lbs. of tea cost $2.75, how much will 8 lbs. cost at the
same rate ?
Since cost of 5 lbs. = $2.75,
« " 1 lb. = $2.75 - 5.
.-. " " 8 lbs. = $2.75 -5x8 = $4.40.
Ex. 2. If lcwt. 24lb. of sugar cost $8.06, how much will 2cwt.
46 lb. cost at the same rate?
Since cost of lcwt. 241b. (1241b.) =$8.06,
the " " lib. =$8.06-124
= $.065,
and " "2 cwt. 46 lb. (246 lb.) = $.065 x 246
= $15.99.
Ex. 3. How long would 24 horses take to consume the same
quantity of food that 45 horses eat in 16 days ?
Since 45 horses eat the food in 16 days,
45 x 16 horses would eat the food in 1 day ;
.-. 45 x 16-24 " " " " 24 days.
Thus, the number of days required = 45 x 16 — 24 = 30.
EXAMPLES LXXXV.
Oral Exercises.
1. If 5t. cost $35, what will 20 1. cost at the same
rate?
2. A man walked 12 mi. in 3hr. ; how far would he
walk in 1\ hr. ?
3. A certain quantity of food would be consumed by
18 persons in 15 da. ; how long would it last 90 persons ?
214 RATIO — PROPORTION. [Chap. IX.
4. If 18 yd. are bought for $3.60, how much will 45
yd. cost ?
5. How far should 15 1. be carried for the money
charged for carrying 12 t. 5 mi. ?
Written Exercises.
6. If 27 men can mow a field in 8 hr., how long
will 36 men take to mow the same field ?
7. If 18 yd. are bought for $16.50, find the price
of 111 yd.
8. How far should 100 t, be carried for the money
charged for carrying 75 t. a distance of 120 mi. ?
9. If 19 men do a certain piece of work in 117 da.,
how long will it take 13 men to do the same work ?
10. If 19 horses can be bought for $475, how many
can be bought for $700 at the same rate ?
11. . If 25 cows cost $1387.50, how much will 6 cost at
the same rate ?
12. If a train runs 704 yd. in 12 sec, how long will
it take to go half a mi. ?
13. How many men would do in 20 da. the same
amount of work as 15 men can do in 16 da. ?
14. How long would 75 horses take to consume the
same quantity of food that 40 horses eat in 15 da. ?
15. If I lend a man $100 for 14 weeks, how long ought
he to lend me $175 in return ?
16. If 7cwt. 41b. of steel cost $133.76, what will
3 cwt. 4 lb. cost at the same rate ?
17. If 1 cwt. 19 lb. of coffee cost $41.65, how much
will 5 cwt. cost at the same rate ?
Art. 208.] EXAMPLES. 215
18. If 123 yd. of silk cost $165.05, how much can
be bought for $58.05 at the same rate?
19. A man walks 9 mi. in 2 hr. ; how long will he take
to walk 12 mi. at the same rate ?
20. If I lend a man $350 for 34 weeks, how long ought
he to lend me $170 in return ?
21. If gold is worth $18.60 per oz., what is the
value of a cup weighing 7 oz. 5 dwt. 12 gr. ?
22. Find the value of 12 things any 7 of which are
worth $26.46.
23. If 3^- lb. can be bought for $5.46, how much can
be bought for $26.52 ?
24. If a t. of sugar cost $110, how much will 8 cwt.
26 lb. cost at the same rate ?
208. Each of the examples in the last exercise may be
solved by the method of Art. 205.
For instance, in 7, the price of 18 yd. holds the same ratio to the
price of 111 yd. that 18 holds to 111 ; hence,
18 : 111 : : $16.50 : x ;
x = $16.50 x -W-
= $101.75.
Here T X T 8 T is the ratio and $16.50 is the first term ; therefore, we
must divide $16.50 by ^ to find the second term.
Again, in 6, the time required for 36 men is f| (ratio) of the
time required for 27 men ; therefore,
27 : 36 : : x : 8 ;
x = ft X 8
= 6 hours.
EXAMPLES LXXXVI.
Written Exercises.
Perform examples 8-15 in the last exercise, using the
method of Art. 208.
216 RATIO — PROPORTION. [Chap. IX.
Similar Figures — Similar Solids.
209. Figures or Solids which have the same shape are
called Similar Figures or Similar Solids.
In the cases of rectangular figures, and rectangular
solids, and cylinders [note, Art. 197], sameness in shape
is determined by the ratios which exist between lines
having the same relative positions. If these ratios are
equal, the figures or solids have the same shape.
For instance, two rectangles, 12 ft. and 8 ft. in length, and 3 ft.
and 2 ft. in height, are similar, because 12 ft. : 8 ft. : : 3 f t. : 2 ft.
Likewise, two cylinders, 15 dm and 9 dm in height, and 10 cm and
6 cm in diameter, are similar, because 15 dm : 9 dm : : 10 cm : 6 m .
From preceding examples we have learned that heights
of squares are proportional to their lengths; that circum-
ferences of circles are proportional to their diameters; that
surfaces of squares or circles are proportional to the squares
of lengths or diameters; and that volumes of cubes are pro-
portional to cubes of lengths or heights or breadths.
What is true of squares, circles, and cubes, is true of
all similar figures ; viz.,
(1) Lines are proportional to lines (height to length, etc.) ;
(2) Surfaces are proportional to squares of corresponding
lines ;
(3) Volumes are proportional to cubes of corresponding
lines.
EXAMPLES LXXXVII.
Written Exercises.
1 . The circumference of a circle is 12 in. ; what is the
circumference of a circle whose diameter is 3 times as
great ?
Arts. 209, 210.] PROPORTIONAL PARTS. 217
2. A rectangle 16 m x 10 m is similar to another rec-
tangle whose length is 4 m ; what is the height of the
second rectangle ?
3. What is the area of a rectangle 5 in. long when a
similar rectangle 9 in. long has an area of 32.4 sq. in. ?
4. What are the comparative areas of two similar
figures whose lengths are 8 cm and 17 cm ?
5. Two cubes are 13 m and l m long; how large is the
first in terms of the second ?
6. Two similar cylinders have diameters of 5 dm and 3 dm
respectively : compare their lateral surfaces ; their bases ;
their volumes.
7. A cylindrical bin will hold 300 bu. of wheat; a
similar one 3 times as high will hold how many bu.?
8. A cylinder 2 m high and 9 dm in diameter will hold
how many kilos of water ? What will be the diameter
of a similar cylinder which will hold 10178784 g ?
Proportional Parts.
210. Partnership.
When the ratio between the parts of a given quantity
are known, the parts themselves can be at once found.
Ex. 1. Divide $100 between A and B so that A may have $3
for every $2 that B has.
For every $3 that A receives, B will receive $ 2, and the two
together will receive $3 + $2 = $5.
Hence, A receives $3 out of every $5 of the whole ;
. •. A " | of the whole = § of $100 = $60.
Also, B " | of the whole = f of $100 = $40.
Ex. 2. The profits of a business are to be divided between the
partners A, B, and C, so that A may have 4 parts, B 3 parts, and
C 2 parts. How much does each get out of a profit of $4500 ?
218 RATIO — PROPORTION. [Chap. IX.
If A has 4 parts to B's 3 parts and C's 2 parts, A ivill have 4
parts out of (4 + 3 + 2) parts divided between them all.
Hence A will have of the whole ;
4 + 3 + 2
.-. A will have $ of $4500 = $2000.
B « « f of $ 4500 = $ 1500,
and C " «| of $4500 = $ 1000.
Ex. 3. Divide $23.50 between A, B, and C, so that A's share
may be to B's share as 4 : 5, and B's share to C's share as 3 : 4.
Here A's share = f of B's share, and B's share = f of C's share ;
.-.A's " = f of f of C's share = f of C's share.
Hence A, B, and C have together (f + f + 1) of C's share ;
i.e.,
$23.50 =
.12 + 15 + 20
20
C's
$10.
share =
= t?
of
$23.50 =
:$10
; A'S = | Of I
Or thus :
A
's share
:B's
= 4:5,
B's:
C's= 3:4.
Now multiply the terms of the two ratios by such numbers that
the numbers corresponding to B's share may be the same in both.
In the present case, multiply by 3 and 5 respectively. Then
A's share : B's : C's = 12 : 15 : 20.
Thus, A gets 12 parts out of (12 + 15 + 20) parts altogether, etc.
Ex. 4. Divide £ 11. 12s. between 12 men, 8 women, and 20 chil-
dren, giving to each man twice as much as to each woman, and
to each woman three times as much as to each child.
A man's share = a woman's share x 2 = a child's share x 6.
Hence 12 men, 8 women, and 20 children will have
(12 x 6 + 8 x 3 + 20) shares of a child.
Hence a child's share x (72 + 24 + 20) = £ 11. 12s. = 232s. ;
. •. a child's share = f f |s. = 2s.
Whence it follows that each man has 12s., and each woman 6s.
Art. 210.] EXAMPLES. 219
Ex. 5. Divide 532 into three parts proportional to f , f , and $.
Since f , f , and f are respectively ff , ££, and £§, we have merely
to divide into parts proportional to 40, 45, and 48.
40
Hence, as in Ex. 3, the parts are - of the whole, etc.
40 + 45 + 48
EXAMPLES LXXXVIII.
Written Exercises.
1. Divide $245 into parts in the ratio 3 : 4.
2. Divide $165 into parts in the ratio 2\ : 3.
3. Divide $33.15 into parts in the ratio § : f .
4. Divide $54 into three parts proportional to the
numbers, 1, 2, and 3.
5. Divide $90.19 into parts proportional to the num-
bers, 7, 9, and 13.
6. Divide £17. lis. into three parts proportional to 5,
51, and 7f
7. A sum is divided into parts proportional to the
fractions, f , f , J ; what fractional part of the whole is the
first part ?
8. The profits of a business are to be divided between
the three partners in proportion to the numbers, 5, 3, and
2 ; how much does each receive out of a total profit of
$6237 ?
9. In a certain business A has 7 shares, B 5, C 3, and
D 1 share. The profits are $2410. Find each partner's
share.
10. A provides $5000, B $3000, and C $1250 to
carry on a business. How much should each get out of a
profit of $555 ?
220 RATIO — PROPORTION. [Chap. IX.
11. A, B, and are partners in a business and have
shares in proportion to the numbers, 4, 3, and 2, respec-
tively, after ^ per annum has been paid on the capital.
The capital is $20000, of which sum A provided $12000,
and B the remainder. How much does each receive out
of a total yearly profit of $3400 ?
12. A, B, and C are partners in a business ; C as mana-
ger receives y 1 ^ of the net profits, the remainder being di-
vided between A, B, and C in proportion to the numbers,
5, 4, and 3, respectively. In a certain year A's share of
the profits amounted to $1520 ; what were the shares of
B and C ?
13.* The shares of A, B, and C of the capital in a
business are as 4 to 3 to 2. After 4 months A withdraws
half his capital, and the profits at the end of the year are
$1518. How should this be divided between A, B,
andC?
Hint. A has { 4 f or 4 mos ' } = 32 for 1 mo.
( 2 for 8 mos. >
14. Divide $157.50 between A, B, C, and D, so that
A may have as much as C and D together, B as much as
A and C together, and D twice as much as C.
15. Divide $80 among 22 men, 26 women, and 82
boys, so that 2 men may have as much as 3 women, and
1 woman as much as 2 boys.
16. If 8 men can do as much as 14 women, and 5
women as much as 9 boys, divide $270 among 4 men, 6
women, and 9 boys in proportion to the work they do.
* When partners put capital into a business for the same length
of time, the case is one of Simple Partnership.
When capital is put into a business for different lengths of time,
the case is one of Compound Partnership.
Arts. 211,212.] MIXTURES. 221
17. Divide $1519.10 among three persons, A, B, and
C, so that A may get one-fourth as much as B receives,
and C may get one-tenth as much as A and B together.
18. Three partners, A, B, and C, had shares in a busi-
ness proportional to the numbers, 4, 5, and 6, respectively.
C retired and received as his share of the business #15000.
How much of this money should be paid by A and B
respectively in order that after C's retirement their shares
might be equal ?
19. A and B, whose capitals were as 3 to 4, joined in
business, and at the end of 4 months they withdrew f
and f respectively of their capitals from the business.
How should a gain of $624 be divided between them at
the end of the year ?
20. The volumes of three substances contained in a
certain mixture are proportional to the numbers, 2, 1, and
4, respectively ; also the weights of equal volumes of the
substances are as the numbers, 1, 32, and 16, respectively.
Find the weight of the first substance contained in 3 lb.
1 oz. of the mixture.
Mixtures.
211. The cost of a mixture of given quantities of two
different ingredients is at once found when the prices of
the separate ingredients are known.
Ex. 8 lb. of tea costing 30 ct. per pound is mixed with 3 lb. of
tea costing 55 ct. per pound; ichat is the cost of the mixture?
The mixture cost 30 ct. x 8 + 55 ct. x 3 = 405 ct.
Hence, each pound of the mixture cost 405 ct. -f- 11 = 36 T 9 T ct.
212. The ratio in which two different ingredients must
be taken in order to make a mixture whose cost is any
given sum intermediate between the costs of the separate
ingredients, will be seen from the following examples.
222 RATIO — PROPORTION. [Chap. IX.
Ex. 1. In what ratio must tea costing 30 ct. per lb. be mixed with
tea costing 55 ct. per lb. that the mixture may cost 45 ct. per lb. ?
The loss on the better quality is 10 ct. per lb.
The gain on the poorer quality is 15 ct. per lb.
The ratio between the loss and gain being |, we equalize loss
and gain by making
the number of lb. of the better quality _ 3
the number of lb. of the poorer quality 2
Ex. 2. In what way must 3 kinds of tea worth 30 ct., 35 ct., and
50 ct. per lb. respectively, be mixed that the mixture may be worth
38 ct. per lb.?
When there are 3 (or more) kinds of commodity, and only the
price of the mixture fixed, there is an indefinite number of ways
of satisfying the condition.
In the present case the gain on the lower two grades of tea,
namely, 11 ct. on 2 lb. (1 lb. of each grade) must just balance
the loss on the best grade, namely, 12 ct. per lb. The ratio be-
tween gain and loss = \\. Hence, we must have 12 lb. of each of
the lower grades and 11 lb. of the best grade.
Or, we may say that the gain on 2 lb. of 30 ct. tea with the gain
on 1 lb. of 35 ct. tea (19 ct. in all) must just balance the loss (24 ct.)
on a certain number of 2 lb. packages of 50 ct. tea. Here the ratio
of gain to loss is |$, Hence, we must have twenty-four 3 lb.
packages (each package consisting of 2 lb. of 30 ct. tea and 1 lb. of
35 ct. tea) and nineteen 2 lb. packages of the 50 ct. tea.
Ans. =48 lb., 241b., and 381b.
Or, gainon{ llb - 30ct - tea = 8c H = 20ct.;
1 4 lb. 35 ct. tea = 12 ct. J
loss on 4 lb. 50 ct. tea = 48 ct. ;
.-. gain : loss : : 5 : 12.
Hence, we must have twelve 5 lb. packages (1 lb. of first kind
with 4 lb. of second kind) and five 4 lb. packages of third kind.
Ans. = 12 lb., 48 lb., and 20 lb.
Art. 213.] WORK AND TIME. 223
EXAMPLES LXXXIX.
Written Exercises.
1. What would be the cost per lb. of a mixture of
4 lb. of tea at 30 ct., and 6 lb. at 40 ct. ?
2. What will be the cost of a mixture of 3 gal. of spirit
at $2.80 per gal. and 5 gal. at $3.50 a gal. ?
3. If 180 lb. of sugar which cost 4 ct. per lb. be mixed
with 120 lb. which cost 5 ¥ ct. per lb., at what price must
the mixture be sold so as to gain let. per lb.
4. A milkman buys milk at 20 ct. per gal. He adds
£ as much water as he buys milk, and sells the mixture
at 28 ct. per gal. What is his gain per gal. ?
5. In what ratio must two kinds of tea, which cost
respectively Is. 3d. and Is. 9& per pound, be mixed in
order that the mixture may cost Is. 5d. per pound ?
6. In what ratio must biscuits worth respectively
11 ct. per lb. and 15 ct. per lb. be mixed that the mixture
may be worth 12 ct. per lb. ?
7. How much sugar worth 7Jct. per lb. must be
mixed with 112 lb. of sugar worth 4£ ct. per lb. in
order that the mixture may be worth 7 ct. per lb. ?
8. Tea at 66 ct. a lb. is mixed with tea at 78 ct. a
lb. In what proportion must they be mixed, so that
by selling the mixture at 77 ct. a lb. a profit of y 1 ^ of
the cost may be made ?
Work and Time.
213. We now consider problems with reference to work
done in various times. These can all be solved by con-
sidering the fractional parts of the whole work which are
done in a definite time.
224 RATIO — PROPORTION. [Chap. IX.
Ex. 1. One man can mow afield in 30 hr., and another man
can mow the field in 60 hr. ; how long would it take them working
together to do it f
The first man mows fa of the whole in 1 hr.,
the second man mows -X of the whole in 1 hr. ;
And, as the two together would mow (fa + fa) = fa of the
whole in 1 hr., they would mow the whole in 20 hr.
Ex. 2. A cistern could be filled in 20 min. by its supply pipe
and emptied in 35 min. by its waste pipe. If the cistern be empty
and both pipes be opened, how long would it take to fill it ?
The supply pipe fills fa of the cistern in 1 min.,
the waste pipe empties fa of the cistern in 1 min. ;
hence, together they fill (fa — fa) = T | ¥ of the cistern in 1 min.
And, as T f ^ of the whole is filled in 1 min., the whole will be
filled in 1 min. -?- T f 7 = 46 1 min.
EXAMPLES XC.
Written Exercises.
1. A can mow a field in 3 da., and B can mow the
same field in 6 da. ; in how many da. will they do it
working together ?
2. A bath could be filled by its cold water pipe in 15
min. and by its hot water pipe in 30 min. ; in what time
will it be filled when both are opened ?
3. A can do a piece of work in 12 da., and B can do
the same in 20 da. A works at it for 3 da. How long
would it take B to finish it ?
4. A can mow a field in 15 hr., and B can mow the
same field in 25 hr. They work together for 1\ hr.,
when A goes away. How long will it take B to finish
the work ?
Art. 213.] WORK AND TIME. 225
5. Two men together can do in 20 days a piece of
work which one of them alone could do in 30 days ; how
long would it take the other man to do the work alone ?
6. When the hot and cold water pipes are both
opened a bath is filled in 6 minutes ; and when only the
cold water is turned on, the bath is filled in 10 minutes.
In how long would the bath be filled if the hot water
pipe only were opened ?
7. A and B could together finish a piece of work in 25
days. They work together for 15 days, and then A
finished it by himself in 20 days. How long would it
take them to do the whole, working separately ?
8. A and B could together do a piece in 22^- days.
A worked at it alone for 10 days, and then B finished it
alone in 60 days. How long would it take them sepa-
rately to do the whole work ?
9. A can do a piece of work in 2\ days, B can do it
in 3 days, and C can do it in 3f days ; how long would it
take them to do it, all working together ?
10. A cistern is filled by one pipe in 48 minutes, by
another in an hour, and by a third in half an hour ; in
what time would it be filled if all three pipes were open
together ?
11. A cistern can be filled by one pipe in 3 hours, by
another in 3 hr. and 40 min., and it can be emptied by a
third pipe in 2 hr. 20 min. ; if it be empty, and they are
all opened together, in what time will the cistern be
filled ?
12. C does half as much in a day as A and B can do
together, and B does half as much again as A ; if all
three working together can mow 20 acres of barley in 16
days, how long would each, working by himself, take to
mow 5 acres ?
226 RATIO — PROPORTION. [Chap. IX.
13, A can do a piece of work in 6 days, B in 8 days,
and C in 12 days. B and C work together for 2 days, and
then C is replaced by A. Find when the work will be
finished.
14. A and B together can perform a piece of work in
24 hr., A and C in 30 hr., and B and C in 40 hr. ; in
what time would each be able to perform it when work-
ing separately ?
Races and Games.
214. The following are examples of questions of this
nature.
Ex. 1. In a 100 yards race A can give B 5 yards start and just
win ; also, B can give C 5 yards start ; how much could A give C ?
A runs 100 yards while B runs 95, and B runs 100 yards while
C runs 95.
Hence, C's distance in any time = T 9 ^ of B's = fifo x T 9 o 5 ^ of A's.
Hence, while A runs 100 yards, C will run T 9 ^ x T V<j- of 100 = 90£
yards.
Thus, A can give 9| yards to C.
Ex. 2. In a certain game A can give B 1 point in 5, B can give
C 1 point in 5, and G can give D 1 point in 8 ; how many points
in 100 can A give D t
A : B =5:4 = 25 : 20
B : C = 5:4 = 20 : 16
C : D = 8:7= 16 : 14.
Hence, as in Ex. 3, Art. 210,
A : B : C : D : : 25 : 20 : 16 : 14
: : 100 : 80 : 64 : 56.
Thus, A can give (100 - 56 =) 44 points in 100 to D ; i.e., A can
make 100 points while D makes 56 points.
Art. 214.] EXAMPLES. 227
EXAMPLES XCI.
Written Exercises.
1. A can give B 10 yards start in a race of 100 yards,
and B can give C 10 yards start over the same distance.
How many yards start can A give C ?
2. A can give B 20 yards and C 51 yards start in a
quarter of a mile race. How many yards could B give
C in a quarter of a mile ?
3. A can beat B by 5 yards in a 100 yards race, and
B can beat C by 10 yards in a 200 yards race ; by how
much could A beat C in a 400 yards race, supposing that
they always run at the same pace ?
4. A wins a race of 100 yards, beating B by 19 yards
and C by 10 yards ; how many yards start ought C to
give B in 200 yards that they may run a dead heat ?
5. In a certain game A can give B 1 point in 10, B
can give C 1 point in 6 ; how many can A give C in 100 ?
6. At a certain game A scores 100 points while B
scores 85, and B scores 100 while C scores 80; how
many will C score in the time that it takes A to score 500 ?
7. A can make 9 articles while B makes 14, and B
can make 7 while C makes 6 ; how many can C make in
the time that A makes 30 ?
8. In a certain game A can give B 1 point in 10, B
can give C 1 point in 12, and C can give D 1 point in 15 ;
how many can A give D in 1000 ?
9. A can give B 20 yards and can give C 41 yards
start in a race of a quarter of a mile, and B can give C a
start of 3 seconds over the same distance; how long
does each take to run a quarter of a mile ?
228 RATIO — PROPORTION. [Chap. IX.
10. In a certain game A can give B 1 point in 5, B can
give C 1 point in 8, and C can give D 3 points in 10 ;
how many can A give D in 100 ?
215. The following examples are worth notice.
Ex. 1. A starts at 10 o'clock to walk along a road at the rate of
4 miles an hour ; B starts on a tricycle at 45 minutes past 10 and
rides after A at the rate of 9 miles an hour. When will B over-
take A ?
When B starts, A has already traveled £$ of 4 miles ; that is, 3
miles.
B gains on A at the rate of (9 — 4 = ) 5 miles an hour.
B will overtake A when he has gained 3 miles, which he will do
in (3 -r- 5) hours = 36 minutes.
Ex. 2. At what time between 4 and 5 o'clock will the hands of the
clock be together ?
At 4 o'clock the minute-hand is 20 minute-spaces behind the
hour-hand. In one hour the minute-hand passes over 60 minute-
spaces, and the hour-hand passes over 5 minute-spaces.
Thus, the minute-hand gains 55 minute-spaces in an hour.
Now, when the two hands are together, the minute-hand must
have gained on the hour-hand 20 minute-spaces, and the time
required for this = §£ of an hour = 2l T 9 T minutes.
Thus, the time required is 21 T 9 T minutes past 4.
Ex. 3. A train traveling at the rate of 45 miles an hour is
observed to completely pass a certain telegraph post in 5 seconds ;
it also completely passed in 4 seconds a second train which was
traveling along a parallel line of rails in the opposite direction at
the rate of 30 miles an hour. How long were the trains ?
The time the first train takes to completely pass a post is the
time it takes to travel a distance equal to the length of the train ;
and, since the train goes at the rate of 45 miles an hour, it goes in
5 seconds a distance = jf^ of 45 miles = 110 yards. Thus, the
first train is 110 yards long.
Again, in the time the trains take to completely pass one another
the distance traveled by the two trains together must be the sum
Art. 215.] EXAMPLES. 229
of the lengths of the trains; and in 4 seconds the trains will
together travel ^% 7 of 45 miles + ¥ ^ of 30 miles = 146 yd. 2 ft.
Hence the length of the second train
= 146 yd. 2 ft. - 110 yd. = 36 yd. 2 ft.
Ex. 4. Seven fowls are worth 6 ducks, 7 ducks are icorth 2 geese,
10 geese are worth 7 turkeys, and a turkey is icorth 17s. 6d. ; how
much is a fowl worth ?
" " f of f of the worth of a goose,
" " T \- of f of f of the worth of a turkey,
" " ^of f of f of 17s. 6c?.
= T V x f x f x ^ shillings = 3s.
EXAMPLES XCII.
"Written Exercises.
1. One boy runs at the rate of 100 yards in 15 sec-
onds, and has a start of 40 yards in front of another boy
who runs at the rate of 100 yards in 12 seconds ; when
will the first boy be overtaken ?
2. One cyclist rides at the rate of 15 miles an hour
and starts half-an-hour after another who rides along the
same road at the rate of 12 miles an hour; when will
the first rider be overtaken ?
3. At what time between 5 and 6 o'clock will the
hands of a clock be together ?
4. At what time between 2 and 3 o'clock will the
hands of a clock be at right angles ?
5. A train traveling at the rate of 45 miles an hour
is observed to completely pass a certain point in 9 seconds ;
find the length of the train.
230 RATJO — PROPORTION. [Chaps. IX., X.
6. A man on the platform of a station observed that
a train passed him in 10 seconds, and passed completely
through the station, which is 308 yards long, in 24 seconds ;
how long was the train, and how fast was it going ?
7. A passenger train, moving at the rate of 45 miles
an hour, overtook a mineral train twice as long as itself
and which was going along a parallel line of rails in the
same direction at the rate of 23 miles an hour ; and the
passenger train completely passed the mineral train in
22^- seconds. How long was each train ?
8. A person lights two candles, 12 and 10 inches long
respectively, at 6 p.m. The former diminishes 5 inches
in length in 4 hours, and the latter 1 inch in 2 hours.
If kept alight, at what time will the former be two inches
shorter than the latter ?
9. If 3 pears are worth as much as 4 apples, 5 apples
as much as 3 plums, 8 plums as much as 3 peaches, and
if pears cost 36 ct. a dozen, what is the price of a peach ?
10. Twelve fowls are worth as much as 11 ducks, 5
ducks are worth as much as 4 pheasants, 10 pheasants as
much as 3 turkeys, and 7 turkeys as much as 10 geese ;
also a fowl and a pheasant are together worth 6s. 6d.
Find the cost of a goose and a turkey together.
Arts. 216,217.] PERCENTAGE A PATIO. 231
CHAPTER X.
PERCENTAGES.
Percentage a Ratio.
216. In many cases the ratio of one number to another,
or of one quantity to another of the same kind, is ex-
pressed by the number of hundredths the first is of the
second, and this is called the per cent the first is of the
second.
For example, 2 apples = T 2 ^ of 8 apples, or 25 per cent of 8
apples.
This means that the ratio of 2 apples : 8 apples is .25. The first
term is sometimes called the percentage, the second is called the
base, and the quotient is, as formerly, the ratio.
Per cent is expressed by the sign %, or by writing the numerator
as a decimal ; thus, T 2 ^ = 25%, or .25 ; and we write ' 5 is 25% of
20,' or '5 is .25 of 20.'
217. The following examples will show how to ex-
press any given quantity as a per cent of any other given
quantity of the same kind.
Ex. 1. Five dollars is what % o/$40 ?
$5 = & of $40 = i of $40 = i?i of $40 =12 \ % of $40.
Ex. 2. In a town whose population was 243200 there were 15504
children born in a year. Find the per cent the number of births was
of the population.
The ratio of births to population is 15504 : 243200 ;
15504 + 243200 = .06| ; .-. Ans. = 6j %.
232 PERCENTAGES. [Chap. X.
218. The following examples will show how to find a
given per cent of a given quantity.
Ex. 1. Find 12£% of $18.
12|% of $18 = .12| of #18 = $2.25.
Ex. 2. In a town whose population was 243200 the birth rate in a
year was 6|% of the population ; how many children were born in
the year.
6f % of 243200 = .06| times 243200 = 15504.
Ex. 3. Of what is 9 ct. 22 \% ?
Since 9 ct. is 22 \%, 100%, or the whole, must be ~5 of 9ct.
= -V- of 9 ct. = 40 ct. 22 *
219. Frequently in finding percentage it is best to
multiply by the common fraction which is equivalent to
the per cent expressed decimally ; thus,
6\% of 96 = T \ of 96 ; 12 l% of 432 = | of 432 ;
16f % of $36.85 = i of $36.85 = $6.14^. [Art. 134.]
EXAMPLES XCIII.
Oral Exercises.
What fractions are denoted by the following per cents ?
1. 50. 4. 10. 7. 12J. 10. 3^.
2. 25. 5. 5. 8. 16}. 11. 6J,
3. 20. 6. 2\. 9. 33|. 12. 5|.
What per cents are equivalent to the following fractions?
13. \. 16. \. 19. A. 22. T V
14. f. 17. T V 20. \\. 23. &.
15. f. 18. fr 21. ^. 24. if
Written Exercises.
Find the per cent the first is of the second in
25. $10, $25; 45 ct, $2.70; $1,121 $3.
26. 7.s. 6d, £2. 27. 7 lb., 1 cwt.
Arts. 218,219.]
EXAMPLES.
233
28. 1 hr. 12 min., 1 da.
29. 3oz. 15dwt., lib. lOdwfc
30. 3216 % 1.608 *
Find
31. 5% of £7. 10s.
32. 10% of $85.63,
33. 12|% of $492.64.
34. The population of a certain town increased 50 %
in the 10 years from 1881 to 1891, and the population
in 1891 was 34617 ; what was the population in 188.1 ?
35. Find the % of error in the statement that 1 oz.
Troy is equal to 1.1 oz. Avoir.
36. Fill the blanks in the following table by giving
the per cents of the 1889 amounts to the nearest tenth.
Receipts from —
1890.
16S9.
Increase.
Amount.
Per
Cent.
Ordinary passengers . .
Season-ticket holders . .
Excess baggage, mails, etc.
Total
26983000
2316000
5029000
25678000
2196000
4757000
1305000
120000
272000
34328000
32631000
1697000
37. Fill the blanks in the following table by giving
the per cents of the 1889 amounts to the nearest tenth.
Receipts from —
1S90.
18S9.
Increase.
Amount.
Per
Cent.
Mineral traffic
General mdse traffic . .
Live stock
Total
17543000
23300000
1377000
$
17052000
22694000
1340000
1
491000
606000
37000
42220000
41086000
1134000
234 PERCENTAGES. [Chap. X.
Profit and Loss.
220. When anything is sold for more than it cost, it is
said to be sold at a profit, and when it is sold for less
than it cost, it is said to be sold at a loss. Profit or loss
is often expressed as a, percentage, and this percentage is
always to be reckoned on the cost price.
Thus, if goods which cost $50 are sold for $60, the percentage
gain, or profit, is $10, and the per cent gain is 10 : 50 = 20 % on
the original outlay.
221. The following examples will show how to treat
questions involving profit or loss.
Ex. 1. A house was bought for $400 and sold for $480 ; what
was the profit per cent ?
The total profit = $480 - $400 = $80.
And the ratio of $80 : $400 = 20%.
Ex. 2. An article cost $10.40 and was sold at a loss of 15% ; for
what was it sold t
Selling price = cost — 15% of cost ;
.-. " " = 85% of cost = .85 of $10.40 = $8.84.
Ex. 3. What was the cost of goods which were sold for $56, at a
gain of 12%?
Selling price = cost + 12% of cost = 112% of cost ;
.♦. cost = i£f of selling price = $50.
Ex. 4. By selling tea at 50 ct. a pound a grocer would gain 5%
more than by selling it at 48 ct. a pound ; what was the cost of the
tea ?
50 ct. - 48 ct. is 5% of the cost ; hence 2 ct. is 5% of the cost ;
.-. 40 ct. = the cost.
Arts. 220, 221.] PROFIT AND LOSS. . 235
Ex. 5. A manufacturer sells at a profit of 20% to a wholesale
dealer, who sells at a profit of 15% to a retail dealer, and the retail
dealer sells for $2.76 and makes a profit of 25%. Find the cost
of manufacture.
It cost the retail dealer {%$ of $2.76
" " wholesale dealer fff of }f§ of $2.76
" " manufacturer ff § of if § of iff of $2.76
Thus required cost = {§-<>- x |fo x i go of $ 2 .76 = $1.60.
EXAMPLES XCIV.
Written Exercises.
What was the gain or loss % in the following cases ?
1. Cost price $20, selling price $ 24.
2. Cost price $2.00, selling price $2.28.
3. Cost price 40 ct., selling price 44 ct.
4. Cost price $3, selling price $3.60.
5. Cost price $140, selling price $130.
6. Cost price $1.20, selling price $1.62.
7. Cost price 84 ct., selling price 98 ct.
8. Cost price $7.80, selling price $8.97.
9. Cost price $74, selling price $70.30.
10. Cost price $15.20, selling price $20.52.
11. Cost price $12.40, selling price $10.23.
12. Cost price $147, selling price $122.01.
13. If an article be bought for $4.20 and sold for
$6.60, what is the gain % ?
14. What was the cost price of tea which is sold for
80 ct. a pound and at a gain of 25% ?
15. If a grocer buys 60 lb. of tea for $21.00, at what
price per lb. must he sell it so as to make 20% profit?
236 PERCENTAGES. [Chap. X.
16. An article was sold for 56 ct., at a gain of 12% ;
what did it cost ?
17. The profit on an article if sold for $3.00 is 25% ;
what would be the profit if it were sold for $2.88 ?
18. By selling a house for $759 a builder gained 10% ;
what would he have lost % if he had sold for $621 ?
19. If a profit of 22-1-% is made by selling an article
for $2.94, what would be the selling price if the profit
were only 5% ?
20. A person bought a carriage and sold it for $37.80
more than he gave for it, thereby clearing 7% ; what
did he give for it ?
21. A house is sold for $4000, and 25% profit is made ;
how much % profit would be made by selling for $3360 ?
22. A tradesman by selling an article for $1.62 gains
35% ; what would he have gained % if he had sold it
for $1.98 ?
23. A man bought apples at the rate of 6 for 2 ct.,
and an equal number at the rate of 10 for 2 ct. ; and he
sold the whole at the rate of 5 for 2 ct. What profit %
did he make ?
24. If 5% more be gained by selling an article for 24
ct. than by selling it for 23 ct., what was the original
price ?
25. If 3% more be gained by selling a horse for
$399.60 than by selling for $388.80, what must have
been the original cost ?
26. If a woman gains 12% by selling 5 herrings for
14 ct., what % would she gain by selling them at 6 for
18 ct.?
27. If a woman buys eggs at 20 ct. a dozen, how many
ought she to sell for 18 ct. in order to gain 8% ?
Art. 222.] TRADE DISCOUNT. 237
28. A man who had been paying $25.20 for 4t. of
coal changed his coal merchant and then got 5 1. for
$20.16 ; how much did he save % ?
29. A draper bought 240 yd. of silk. He sold J at a
gain of 25%, J at a gain of 20%, and the remainder at
a loss of 15%, and received $800 in all. What was the
cost price per yd. ?
30. A draper bought a piece of silk 35 yd. long ; and,
after cutting off 2 yd. which were damaged, he sold the
remainder so as to clear 10% on his outlay. How much
% was the selling price of a yd. higher than the cost
price ?
31. A manufacturer sold at a profit of 25% to a
wholesale dealer, who sold at a profit of 12% to a retail
dealer, and the retail dealer sold for $3.22 and made a
profit of 15% ; what was the cost of manufacture ?
32. A quantity of wheat was sold in succession by
three dealers, each of whom made a profit of 5 % . The
last of the three sold for $3087 ; how much did it cost
the first?
33. A house was sold by the builder at a profit of 30%,
and the purchaser sold it again at an advance of $117 in
the price, and gained 20% on his outlay; how much
did the house cost the builder ?
Trade Discount.
222. Merchants often sell goods at a certain price with
a certain % discount ; thus,
Macmillan & Co. may sell books at $1.60 per copy less
15% ; this means that they sell for $1.60 - 15% of $1.60,
or for $1.60 - $.24 = $1.36.
238 PERCENTAGES. [Chap. X.
223. Sometimes after a given % discount is allowed, a
second allowance of another % is made, and even a third
allowance is made.
Ex. 1. Goods sold for $2500 with a discount of 20%, 5%, and
1 1 % bring what price f
$2500 - 20 % of $2500 = $2000 ;
$2000 - 5 % of $2000 = $1900 ;
$1900 - I* % of $1900 = $1871.50 = Ans.
Ex. 2. Which is cheaper, to buy goods at a discount of 30% and
5%, or with $&\%off?
The marked price less 30% = 70% of marked price ; 70% -5%
of 70 % = m\ %. It is cheaper to buy at a discount of 30 % and 5 %
than at a discount of 33^ %.
EXAMPLES XCV.
Oral Exercises.
What is paid for goods marked
1. $50 with a discount of 10% ?
2. $50 with a discount of 10% and 10% ?
3. $50 with a discount of 20% and 5% ?
4. $600 with a discount of 33|% ?
5. $900 with a discount of 16|% ?
6. $1000 with a discount of 27% and 10% ?
7. $1000 with a discount of 20%, 10%, and 1% ?
What is the marked price of goods sold for
8. $90 after a discount of 25% ?
9. $63 after a discount of 30% and 10% ?
10. $49 after a discount of 121%, and 12 J % ?
11. $45 after a discount of 16} % and 10% ?
Arts. 223, 224.] COMMISSION — BROKERAGE. 239
Written Exercises.
12. Find what was received for goods marked $1200
if a discount of \ and 15% is allowed.
13. For what % of the marking price are goods sold
if an allowance of \, 10%, and 6|% is made ?
14. Goods are marked $170 and sold for $144.50;
what % discount was allowed ?
15. Goods marked $16 were sold at 6J% discount and
5% off for cash ; what was the selling price ?
16. Goods cost a merchant $1600; he wishes to make
a profit of 25% after making a discount of 20% and 16|% ;
what was the marked price '?
17. At what % above the cost must goods be listed
that a merchant may allow a discount of 20% and realize
a profit of 12% ?
18. A merchant allows on $2000 worth of goods (list
price) a discount of 15%, 9%, and 5% for cash, then |%
to clinch the bargain; how much, cash did he receive and
what profit did he make, his % of profit being 8 ?
Commission and Brokerage.
224. An agent employed to buy, or sell, goods, or to
collect rents, is usually paid a percentage on the price of
the goods, or on the amount of rent. This percentage is
called Commission.
To insure against loss of life, or damage by fire, some
persons pay money to an Insurance Company. In return
for this money, the Company undertake to compensate
the person insured for any loss caused by fire, or to pay
a specified sum to relatives of the deceased. The money
240 PERCENTAGES. [Chap. X.
paid to the Company is a percentage on the value of the
property insured, or on the specified sum, and is called a
Premium.
Kx. 1. The total rental of an estate is $8474.40, and the agent
is paid a commission of 5%; how much is the commission/
$8474.40 x .05 = $423.72.
Ex. 2. What is the annual premium for insurance on a building
worth $7500 at the rate of 24 ct. for $250 ?
— x $7500 = $7.20.
250
EXAMPLES XCVI.
Written Exercises.
1. After paying 5% to his agent, a man received
$ 1436.40; what was the agent's commission?
2. What is the amount of annual premium for the in-
surance of a building for $8520 at -fa%?
3. A landlord allowed his tenants 20% reduction from
their rents; what was the nominal rent of a tenant whose
reduced rent was $1800 ?
4. A commission merchant sells goods for $2864 and
sends to his principal $2824.62 after deducting com-
mission ; what was the % commission ?
5. A commission merchant is asked to purchase $6800
worth of goods at 2|% commission; how much money
was paid by his principal ?
6. A commission merchant received $6953 with which
to purchase goods after deducting 2\ % commission ; what
was paid for the goods ?
7. An agent sold goods for $5672; his bill for ex-
penses was $56.72, and his commission was 11% ; what
f of the selling price did the principal receive ?
Art. 225.] TAXES AND DUTIES. 241
8. A. man insured his life for $5000 at an annual
premium of 2|-% ; how much had he paid at the end of
13 years ?
9. A cargo is insured for $254500, its full value, at
2% ; the ship is insured for $120000 at 2|% ; the owner
of the cargo pays all insurance and sells his goods"at the
end of the voyage at an advance of 9% over total cost,
allowing $2000 for freight; what was the selling price?
10. The premium for insuring a building at 2J% is
$1136.25; find the insurance.
11. A company insured a building and the goods it
contained for $117944, the goods being worth 15% of the
value of the building. The merchant paid 2% premium
on the building and \\% premium on the goods; what
was the total premium ?
12. A man sold through an agent some merchandise,
paying the agent 5% commission. The agent invested
the proceeds in two parts after taking out commissions
of $325 at 5%, and $260 at 4%, respectively; what was
the value of the merchandise ?
13. A man had two houses, each costing $5000; he
insured one for $4000 at \\% and the other for $6000
at \\°fo : find the difference between the loss on one and
the gain on the other, both houses having been burned
on the day after insurance.
Taxes and Duties.
225. Persons owning property or importing goods, pay
to the government (for its support) a certain per cent of
their property or of the foreign value of the goods im-
ported.
The percentages paid on property are called Taxes.
242 PERCENTAGES. [Chap. X.
The percentages paid on imported goods are called
Duties.
Duties levied on articles regardless of their value are
called Specific Duties.
Duties levied at a certain per cent on the foreign values
of goods are called Ad Valorem Duties.
In some States voters pay annually a small fixed sum
of money ($1.50 or $2) before they can vote. Such
money is called a Poll Tax
EXAMPLES XCVII.
Written Exercises.
1. The expenses of a certain town are $39512.32
annually ; the tax is 16 mills on the dollar ; what is the
value of the town as fixed by the assessors ? (The asses-
sors' valuation is much smaller than the real valuation.)
2. The valuation of a certain town is $6495860, while
the assessed valuation is 25 % of that ; the polls number
1112, and the taxes are $16.25 on each thousand of as-
sessed valuation ; what are the expenses of the town ?
3. The expenses of a city are $339000, and the assessed
valuation is $16950000 ; what is the tax rate expressed
as per cent ? Expressed as dollars on a thousand ?
4. What is the duty on 5000 bbl. of hydraulic cement
at 8 ct. per bbl. ?
5. What is the duty on 125 plates of polished un-
silvered glass 24 x 30 in., at 8 ct. per sq. ft. ?
6. What is the duty on 100 doz. penknives valued at
30 ct. per doz., at 25 % ad valorem ?
7. What is the duty on 3 1. of No. 23 steel wire at
2ct. per lb.?
Art. 225.] EXAMPLES. 243
8. A merchant imported 1550 yd. of tapestry carpet
valued at 80 ct. a yd. ; what was the duty at 42\ % ad
valorem ?
9. An invoice of 150 doz. linen collars valued at $ 1.30
per doz., calls for how much duty at 30 ct. and 30 % ?
10. What does the government receive on an impor-
tation of 1000 gross of steel pens at 8 ct. per gross ?
244 INTEREST. [Chap. XL
CHAPTER XI.
INTEREST.
Promissory Notes.
226. When one person borrows money from another
person, he gives to the lender a written promise to repay
the money and to pay also a percentage on the money at
a given rate °/ per year. This percentage is called
Simple Interest, or Interest.
The form of note given in the following pages is the form in use
by the best business men in the United States. Students are
strongly advised to adhere closely to the form while practising the
making of notes.
The written promise is called a Promissory Note. For
example :
fU-67^. of(MJiA>iUe,, &MH,., fan. /, 18 W-
3AaaLu debus, alt&v otat& <J promise to pay to
the order of^^^~^^fcvnve& tZH&rv^^^^^^.^^
^^^^onA, IvuM/cL'i&cL QsVxJsu-Qs&v-&yv^^„ ^ Dollar s
Value Received, w-vttv vnt&v&oZ.
No. 82. Due fam,. d//&6>. S, 'f*.
Arts. 226-229.] PROMISSORY NOTES. 245
f250®*-. zfjvLVnc^UU, TMclm,., feu*,, f , 18 <?5.
fan cL&vyvcwicL J^^^^promise to pay to
the order of^^^^^^./fcyua^& B&l&ksA, „
at tA& (^ka/fiAyyv c/tatLcyyial fdank,^
Value received, waXA Cnt&^&oZ at 5%.
No. 763. Dae. j. @. ftoumwiondL.
227. In the case of a promissory note, it is to be noticed
that the heading indicates the names of the Town and
State in which the note is written, also the month, day, and
year. At the left is written in figures the sum of money
for which the note is given.
228* The face of the note indicates the names of the
parties (Maker and Payee) to the note, the words 'value
received, the sum of money for which the note is given
(written in full), and the time for which the note is to
run. If the note is interest-bearing, it must have the
words i with interest ' written in the face.
). Notes written as above may be transferred (sold)
by the Holder to another person (who in turn becomes the
holder), and are therefore called Negotiable notes. When
the note contains the words 'or order,' the holder must
Endorse his name on the back of the note (thus becoming
* The Maker of a note is the person who signs the note. The
Payee is the person to whom the note is made payable. The Holder
of a note is the person who owns the note (the payee or some
person to whom the payee has sold the note).
246 INTEREST. [Chap. XI.
responsible for its payment) when he sells the note.
When the note contains the words ' or bearer? no endorse-
ment is legally necessary. If the note does not contain
the words * or order? or l or bearer? it is not negotiable.
230. A note is payable at the business office of the
Maker unless otherwise specified in the note. Nearly all
notes specify the place of payment.
231. When the rate of interest is not written in the
note, as in the first of the above notes, the law of the
State in which the note is to be paid fixes the rate. If the
parties interested wish a rate different from the rate in
the State in which the note is to be paid, such rate must
be specified in the body of the note, as in the second of
the above notes, but may not be more than the maximum
allowed by such State.
232. When the words ' with interest ' are omitted from
a note, no interest is payable on that note except for the
time it may over-run. [Art. 239.]
233. The first of the above notes is called a Time note ;
the second, a Demand note.
A time note is nominally due at the date indicated in
the note, but Matures (becomes legally due) three days
later. The three days are called Days of Grace. In some
States no days of grace are allowed.
234. When a note matures on a Sunday or on a legal
holiday, it is payable in some States on the business day
next preceding, and in other States on the business day
next succeeding, such Sunday or legal holiday.
235. The dates on which a note is nominally and
legally due are indicated thus : Feb. 8/11, 1880.
Arts. 230-238.] TABLE OF KATES. 247
If the time of payment is indicated in ' days after date, 1
those days, together with three days of grace (if such be
allowed by the State), are counted forward from (not
including) the date of the note in finding the date of
maturity; thus, the first of the above notes matures
Jan - 31 /Feb.3, 1894.
236. If the time of payment is indicated in 'months
after date] calendar months, together with three days of
grace, are counted forward from the date of the note in
finding the date of maturity ; thus, the maturity of a note
for 2 months, dated Jan. 31, 1892, was Mar - 31 / A p r . 3, 1892 ;
for 3 months, it was A P r - 30 /May 3, 1892 ; for one month, it
was Fob. 29/ Mch>8> 1892.
237. If payment of a note is not made on the day of
maturity, the holder must engage a Notary Public to send
to the endorser (or endorsers) a written notice of such fact.
This notice is called a Protest. The protest must be sent
on the day of maturity, otherwise the endorser cannot be
held to the payment of the note.
Table of Rates of Interest.
238. The following table gives, for each of the States
and Territories, the Legal Bate when no rate is mentioned
in a note, the Maximum Kate allowed, the Time of pay-
ment when the day of maturity falls on a holiday (the
day before by B, and the day after by A), and indicates
by the letter G those States in which days of grace are
legal.
Notes made on or after Jan. 1, '95, and payable in New
York, bear no grace.
Notes made after July 4, '95, and payable in New Jer-
sey, bear no grace.
248
INTEREST.
[Chap. XI.
State.
"S
i
•
a
e
6
1
State.
c3
M
03
2
M
%
H
8
M
a
H
o
Alabama . . .
8
8
A.
<;.
Montana . . .
10
Any.
B.
G.
Arizona . . .
7
Any.
A.
G.
Nebraska . . .
7
10
A.
G.
Arkansas . . .
6
10
B.
G.
Nevada . . .
7
Any.
B.
G.
California . . .
7
Any.
A.
New Hampshire
6
6
B.
G.
Colorado . . .
8
Any.
B.
G.
New Jersey . .
6
6
A.
Connecticut . .
6
6
B.
G.
New Mexico
6
12
A.
G.
Delaware . . .
6
6
B.
G.
New York . .
6
6
A.
Dist. of Columbia
6
10
A.
G.
No. Carolina
6
8
B.
G.
Florida ....
8
10
B.
G.
No. Dakota . .
7
12
A.
G.
Georgia . . .
7
8
A.
G.
Ohio . . . .
6
8
B.
G.
Idaho ....
10
18
A.
Oklahoma , .
7
12
B.
G.
Illinois. . . .
5
7
B.
G.
Oregon . . .
8
10
A.
Indiana . . . .
6
8
B.
G.
Pennsylvania .
6
6
A.
G.
Indian Territory-
6
10
B.
G.
Rhode Island .
6
Any.
A.
G.
Iowa ....
6
8
B.
G.
So. Carolina . .
7
8
A.
G.
Kansas ....
6
10
B.
G.
So. Dakota.
7
12
A.
G.
Kentucky . . .
6
6
B.
G.
Tennessee
6
6
B.
G.
Louisiana . . .
5
8
A.
G.
Texas . .
6
10
B.
G.
Maine ....
6
Any.
B.orA.
G.
Utah . .
8
Any.
B.orA.
Maryland . . .
6
6
B.
G.
Vermont .
6
6
A.
Massachusetts .
6
Any.
A.
G.
Virginia .
6
6
B.
G.
Michigan . . .
6
8
A.
G.
Washington
8
Any.
B.
G.
Minnesota . .
7
10
A.
G.
W. Virginia
6
6
B.
G.
Mississippi . .
6
10
B.
G.
Wisconsin
6
10
A.
Missouri . . .
6
8
A.
G.
Wyoming
12
Any.
A.
G.
EXAMPLES XCVIII.
1. Write a time note for $ 250.67 with interest at 16 %.
2. Write a time note for $ 76 with interest at 20 %.
3. Write a time note for $468.92 for 20 da. without
interest.
4. Write a time note for $20 for 4 mo. with interest
at 13%.
5. Write a time note for $560, headed Cincinnati,
Ohio, Jan. 13th, 1892, to mature in 63 da., with interest.
6. Write a demand note for $528 with interest.
7. Write a demand note for $460 with the maximum
interest allowed by the State in which you live.
Arts.
239, 240.] SIMPLE INTEREST.
249
Find the date of maturity of each of the following
indies
ited notes :
Date.
Where Payable.
Time.
8.
Dec. 18,1895
New York
30 days
9.
Jan. 18, 1895
New York
60 days
10.
Jan. 27,1896
New York
45 days
11.
July 31, 1895
New Jersey
60 days
12.
June 6, 1895
New York
3 months
13.
Jan. 30,1895
New York
1 month
14.
Jan. 30, 1896
New York
1 month
15.
Mar. 31, 1895
New York
3 months
16.
Mar. 31, 1895
Conn.
1 month
17.
Mar. 30, 1896
Mass.
1 month
18.
Sept. 3, 1890
Nebraska
60 days
19.
Jan. 29, 1896
Cal.
30 days
20.
Jan. 29, 1896
Cal.
1 month
Simple Interest.
239. Interest on the Principal (money borrowed) is
called Simple Interest, and is computed at the given rate
per cent (per year understood) for the time elapsing
between the date and maturity of the note.
If a note is not interest-bearing and is not paid at maturity,
interest is payable after maturity and until the note is paid.
[Art. 232.]
240. The majority of notes are given for short periods
of time — say 30, 60, or 90 days, or 1, 2, or 3 months.
Now it is customary in interest computations, to regard
one year as 360 days. Therefore, by a short operation,
we may find the interest on any principal for any time
and at any rate per cent.
250
INTEREST.
Ex. 1.
[Chap. XL
$350^-. tfJUxwuif, o/t.y., fan,- /6, 18<?6.
€Lakt&eA / L cLa^uoy alt&v cCat& of promise to pay to
the order of jo-el cPuC-,
,ncwyi>^
.qI^iv&& hnA/yvcLi&cL
.^Dollars
Value received, w-ttk ImZ&v&oZ.
JVo. 68. Due &e&. S. d. &. g&nMe,
350
.06
360 ) $21.00
.051
18
$1.05
In this note, the rate is 6% and the time is 18
da. ; the interest for 1 yr. will be .06 of the princi-
pal, — $21 ; the interest for one day is found by
dividing $21 by 360, and the interest for 18 days
is found by multiplying the quotient thus found
by 18. Hence, .06 and 18 are multipliers, while
360 is a divisor.
This may be expressed as follows
350 x 6 x 18
100 x 360
.350 x 18
which becomes
by cancellation.
We observe that interest for any number of days may be found
by dividing the principal by 1000, multiplying by the number of
days, and dividing by 6.
The following is a better form for practical work :
350.
$1,050
o Here, cancelling 6 from the dividend and divisor,
we have .350 to be multiplied by 3.
Art. 241.] SAMPLE INTEREST. 251
Ex. 1. In the above note let the principal be $367.91, the rate
6%, and the time 21 da. ; find the interest.
9fffl.ffX x 18395 + We must cancel the divisor complete-
rs 1 ly ; only 5 figures will be needed for
$1.28765 the multiplicand ; keep the multiplier as
$1.29 = Ans. small as possible by cancellation.
Ex.2.
$650™-. dufuda,, 7H&., fan. 7, 18<?3.
3w4- wumMa/ att&h dat& c/ promise to pay to
the order of^^^^^./lf&nvy> Butted,
^_^__c/t^ kwruLv&cl (t^tif jjg Dollars
at^^^~jLh& &iaMAt& cAatvcyyval fdanfo
Value received, with, (mt&v&oZ.
/if&wvif JS. Reld.
No. 152 Due TnavsA 7//0, '$3.
Find the interest.
01$ JJj30. 325 When the time is ' months after
?1 @* 21 date,' calendar months are counted
005
fi ZZr in obtaining the date of maturity
1 6 825 [Art. 236], and the interest is com-
6.83 = interest, puted for the stated number of
months counting 30 da. as one month.
This note payable in Maine has three days of grace ; therefore the
interest is computed for 63 da.
241. The value of a note at its date of maturity is
called its Maturity Value, and consists of the sum for
which, the note is given plus the interest (if any). In
finding maturity value, observe whether or not the note
bears int. and ' days of grace. 9
252
INTEREST.
[Chap. XI.
EXAMPLES XCIX.
Written Examples.
What interest is due at
maturity on
each of the follow-
g indicated notes ?
Date.
Principal.
Kate.
Time.
1. Arkansas,
$763
6%
12 da.
2. New Jersey,
$1467
15 da.
3. Ohio,
$1626.75
18 da.
4. Texas,
$6000
21 da.
5. New York,
$5267.50
27 da.
6. New York,
$2675
27 da.
7. California,
$376
2 mo.
8. Kentucky,
$498
1 mo.
9. Connecticut,
$75000
108 da.
10. New Hampshire,
$704.25
201 da.
11. Illinois,
$84.75
361 da.
12. Utah,
$846
51 da.
242. For rates other than 6%, find the interest at 6%
and take such a part of that interest as the given rate is
of 6%.
Ex. 1. P = $ 4673, B = 5%, time = 33 da. ; find interest.
M
2 3365
11
6)25.7015
4.2835
$21.42 =Ans.
Here the interest at 6% is $25.7015,
and | of this is $21.42. We obtain this
result rapidly by subtracting the interest
at 1 % from the interest at 6 %.
Art. 242.] SIMPLE INTEREST. 253
Ex. 2. P = $ 26.48, B = 1\ %, time = 90 da. ; find int.
„ 1 026.48
ff0 15 Here the interest at 7^ % = f of the interest at
4) .39720 6%. The answer is obtained by adding to the
• 0993 interest at 6 % one-fourth of the interest at 6 %.*
$.50
In the j'inaZ work do not waste time writing anything but the
answer.
EXAMPLES C.
)S
fYOO -^. RaUicjk, <A.&., 0Lucf.3r, 18 <?5.
&hi&& rnxyyitko/ alt&v dat& <J promise to pay to
the order of^^^^^^^^ko-Y/ia^ $v&&n>&.
s^^^^^.$/&v-&n kuncCt&cC
at tk& @Ctiq&H&> c/tatvcyyial Bank,
Value received, w\tk vnt&v&oA, at 7 °lo.
No. 16 Due cAov-. SO / '£&& 3, /8<?5.
g Dollars
1. Find the interest on the above note.
2. Find what would be the amount of the above note
if the time were 90 da.
3. Find interest on a New York note for $ 2670, dated
Dec. 31, 1895, payable 47 da. after date, with interest.
* Eor interest at 8 % add } of interest at 6 % to itself.
710/ U 1
'2/0 f
The
70/ 11 1
5|-%subt. T V
5 % " i
H% " i
4 % " I
3 % = i
from
254
INTEREST.
[Chap. XI.
What interest is due at
lowing indicated notes ?
; maturity
on each of the fol-
4.
Date.
Saratoga, N.Y.,
Principal.
$2670
Eate. Time after
Date.
Maximum. 47 da.
5.
Springfield, Mass.,
$4893
5%
60 da.
6.
Washington, D.C.,
$289.20
6%
90 da.
7.
Buffalo, KY.,
$48.93
H%
3 mo.
8.
Utica, N.Y.,
$48.93
4%
3 mo.
9.
Baltimore, Md.,
$765
Maximum. 6 mo.
10.
Hartford, Conn.,
$4893
it
90 da.
11.
Denver, Col.,
$8695
12%
24 da.
12.
St. Louis, Mo.,
$463.50
9%
60 da.
13.
Chicago, 111.,
$873
61%
30 da.
14.
St. Paul, Minn.,
$ 487.20
m
2 mo.
15.
NewYork,N.Y.,
$286.37
4f%
3 mo.
16.
Boston, Mass.,
$ 499.99
4%
60 da.
17.
Find the amount
in each
of the
last three
examples.
The above method for computing interest is in general
use when the time is less than one year ; but if the time
is in yr., mo., and da., the 6% method is the more
frequently used.
Six Per Cent Method.
243. A demand note for $ 268.50, dated Nov. 26, >87,
was paid 4 yr. 7 mo. 18 da. after date ; what was the
interest at 6% ?
Arts. 243, 244.] SIX % METHOD. 255
At 6%
the interest on $1 = $.06 (six cents) for 1 yr.,
« " $1 = $.005 (5 mills) for 1 mo.,
" " $1 = $.0001 (I of a mill) for 1 da. ;
u
hence
u
« « $1 = $.24 for 4 yr.,
" « $1 = $.035 for 7 mo.,
" " « $1 = $.003 for 18 da.,
" « " $1 = $.278 for 4 yr. 7 mo. 18 da.
$ 268.50 Having found the interest on $ 1 for the given rate
.278 and time, we multiply this interest by the principal.
$ 74.64 [Art. 47, Theorem I.]
Note. It is evident that the interest for 2 mo. at 6 % may be
computed by moving the decimal point 2 places to the left. Thus,
the interest on $784.70 for 2 mo. is $7.85. Similarly, the interest
for 6 da. is $.78. Also the interest for 12 da. is $1.57. The 6 %
method is sometimes used when the times are less than 1 yr.
244. There is great diversity in the methods of finding
the time in a case like this. Some prominent banks and
business houses in the United States use the method of
counting the time in years and days, instead of the method
just described.
Thus, the above note was paid July 14, '92; the 4yr. were
counted forward from Nov. 26, '87, the 7 mo. were counted for-
ward as calendar months from Nov. 26, '91, and the 18 da. were
counted forward from June 26, '92. (This is not compound
addition.)
In obtaining the int. the years were reckoned as wholes, but the
months and days were reckoned in the exact number of days
found in that 7 mo. and 18 da. which began with Nov. 26, '91. Thus,
the time was 4 yr. 231 da. — 4 in Nov., 31 in Dec, 31 in Jan., 29 in
Feb., 31 in Mar., 30 in Apr., 31 in May, 30 in June, 14 in July.
The int. on $1 = $.24 for 4 yr.,
" " $1 = $.0385 for 231 da.,
" $1 = $.2785 for 4 yr. 231 da.
$268.50 x .2785 = $74.77 = Ans.
256 INTEREST. [Chap. XI.
EXAMPLES 01.
Written Exercises.
What was the amount at maturity of each of the
following indicated notes ?
Obtain answers by each method [Arts. 243, 244].
Date.
Principal.
Rate
Time.
1.
Philadelphia
July 31,
'84
$7680.95
6%
3yr.
6 mo. 12 da.
2.
Richmond,
Aug. 6,
'87
$683.42
4yr.
7 mo.
3.
Boston,
Jan. 9,
'88
$1492.88
4yr.
11 mo. 18 da.
4.
Cleveland,
Oct. 1,
'90
$2689.42
2yr.
10 mo. 18 da.
5.
Jersey City,
Aug. 1,
'95
$487.50
lyr.
7 mo. 13 da.
6.
Providence,
Aug. 8,
'79
$2000
8yr.
5 mo.
245, For rates other than 6%, proceed as in Art. 242.
EXAMPLES OIL
A few demand notes are here indicated; find the
interest on each of the first four, and the amount on each
of the others.
Obtain answers by each method.
Date.
Principal.
Rate.
Paid after Date.
1.
June 8,
'81
$468.93
41 o/
*J /o
5 yr. 9 mo. 18 da.
2.
Aug. 2,
'87
$1680.50
4 »
3yr. 6 mo. 27 da.
3.
Feb. 29,
'88
$2500
5i"
3yr. 11 mo. 6 da.
4.
Mch. 9,
'91
$155
B|"
lyr. 3 mo. 3 da.
5.
May 13,
89
$450.50
*h"
3 yr. 2 mo. 7 da.
6.
Oct. 23,
'85
$896.88
4£"
4yr. 11 mo. 29 da
7.
Sept. 15
'82
$15875
2|"
9yr. 6 mo. 15 da.
Annual Interest.
246. Some States allow interest to be collected on each
annual instalment of interest, if such instalment is not
paid when due.
Art. 245, 246.] ANNUAL INTEREST. 257
Ex.
$673^. Bwoktyn,, ofi.y., CtfriM 6, 18 <?5.
€n oLeAYvancL J ^^^^^^jpromise to pay to
the order of ftewuy ofyyvltk
^^^cfwc AsWvuOi&ct Qs&v-eMXAf-tkv&z-^-^^DoTlcLrs
Value received, with iAtt&i&oZ a/yuriMaZVu at 3 % .
No. /7> Due Jd&nja-vyvLrv jcyuLan.
If this note be paid in 3 yr. 6 mo. after date, and no interest has
"been paid meanwhile, there will be paid the principal, the simple
interest on the principal, and simple interest on each annual instal-
ment of interest from the time it is due until the note is paid.
.21
6)141.33
23.555
$117,775 = interest for 3 yr. 6 mo.
7.571 = interest on interest.
673. = principal.
$798.35 = amount to be paid.
The 1st instalment bears interest for 2 yr. 6 mo.
The 2d instalment bears interest for 1 yr. 6 mo.
The 3d instalment bears interest for 6 mo.
Interest on annual instalment
is computed for 4 yr. 6 mo.
$33.65 = annual instalment at 5 %.
.27
6)9.0855
1.5142
$7,571 = interest on interest for 4 yr. 6 mo.
258 INTEREST. [Chap. XL
EXAMPLES CIII.
Written Exercises.
1.
$/86d^.. Jt&w- Tfovk, o/t.lf, fan, /a, 18 <?&.
€.n oLeAn&nci J-^^^^jpromise to pay to
€vcfkt&&vi hwyicOvect &v?ctu-thi,&& cwici ^Dollars
Value received, w\Zk Cnt&i>&aZ a/yt/yt/waMu at^-%.
No. V-/7. Due. /if&nvif
If this note be paid 5 yr. 8 mo. and 20 da. after date,
no interest being paid meanwhile, how much will the
holder receive ?
2. Cast the interest on a note similar to the above,
when P= $897.75, R = h\%, and r=4yr. 9 mo. and
15 da., no interest being paid meanwhile.
3. How much does a man owe at the maturity of a
note similar to the above, when P= $437.25, i2 = 4%,
and T = 7 yr. 27 da., no interest having been paid ?
Commercial Discount.
247. We have been considering, in the last few pages,
cases in which money is borrowed from persons ; we have
learned that the interest is payable at the maturity of the
note.
Arts. 247-249.] COMMERCIAL DISCOUNT. 259
When money is borrowed from a bank, the interest
(simple) is paid on the day on which the money is borrowed.
The simple interest which a bank takes in advance is y
called Commercial Discount, or Bank Discount.
The borrower does not receive the principal (as when
borrowing from a person), but receives the principal minus
the simple interest on the principal; this remainder is
called the Proceeds of the note.
The following example will show the methods of calcu-
lation of discounts and proceeds.
Ex. 1.
n, TWouM,., c/tav-. 16, 18 qy-.
^kwiUf clwyo, ajt&v dat& <J promise to pay to
the order of wjm*IL~~~
<^W Aundv&cl Q,v?cty-Q,&v-&n — ~~ ^ Dollars
at^^^^^£h& S^OlqZ <Aa,tu>naZ JSc^rik^^^^^^
Value received.
No. /?. Due be*. /6//f, '<?£.
248. When a person borrows money from a bank, he makes his
note payable to himself, and at the bank which makes the loan.
The note must be endorsed.
249. The discount is computed on the maturity value
of the note ; in the above note it is computed on $267,
since the note is not interest-bearing. So, also, when a
person sells a note to a bank (Ex. 2, following), the bank
discounts its maturity value.
Find the discount and proceeds of the note in Art. 247.
260 INTEKEST. [Chap. XI.
| J20J. . 1335 It will be observed that in case money
H is borrowed from a person the borrower
$1.4685 h as the use of a larger sum of money
$267 00 than when he borrows from a bank, yet
1.47 = discount. ne P a Y s tne same interest. The bank
$265.53 = proceeds. nas tlie use of tne discount while the
borrower has the use of the proceeds.
At the maturity of a note given to a person the borrower pays
principal and interest ; at the maturity of a note given to a bank
the borrower pays the principal only, having already paid the
interest.
Ex. 2.
#700- . c/tew tfcuv-sM,, @syvwi., fcvn. 3/, 18 <?5.
fai& wuynXA a^C&v ciat& J promise to pay to
the order of fowu&o, (^kiAUf~~~~~~
^v^vvvvvvv^ofeiwt Au^c^i^d^^^^^^^YQQ Dollars
at tA& ofv^t/i (Zv-&nu& Bam^k, cA&w llo-vk, o/Jf.ll.
Value received, with, vyit&v&oZ at ¥i%.
No. 27. Due Ss6. 28/ q 5. ftenvif 3iU&.
Discounted Feb. 5, at 6%.
In this case the holder took a promissory note to some bank,
and the bank discounted the note ; i.e., the teller gave the holder
the proceeds of the note calculated on its maturity value. The
time for which a bank computes discount is the exact number of
days from the day of discount to the day of maturity, although
the time of the note may be written in months. The holder must
endorse the note. [Art. 229.]
flflJOO.
4)3.50
.875
$2.63 = interest at 4i%.
$702.63 s maturity value.
1 23
.1171
23
$2.69 = discount.
$702.63
2.69
$699.94
= proceeds.
Art. 249.] COMMERCIAL DISCOUNT. 261
The note matures without grace, because payable in N.Y. The
term of discount is the number of days from Feb. 5 to Feb. 28.
Ex. 3.
f<?78 — , BuAl\#ujL<yn,, Vt., (let. /2, 18 <?#■.
&ia>-& wvonZAa, a^t&v clat& I promise to pay to
the order of 3%&c£&vi<&fc /i-o&wv&'b
^^jcAvyi& kwncOi&ci &&v-&nty-&tf A t-^^YQo Dollars
at^ J ^^^JJv& H-ow-oaxL cftatLo-yicil Bavifc^^^s^
Value received, w-iXA int&v&at.
No. / 7. Due TftaA&k 12, ' <?5. ft&wuu Shovta*.
Discounted Dec. 31, '94, at 5%.
01978.25 a\ 1002.71 .16712
I m 25 P | 71 71
$24,456 6)11.86552
$978.25 1.97758
$1002. 71 = maturity value. $9.89 = discount.
$1002.71 - $9.89 = $992.82 = proceeds.
Ex. 4. A merchant wishes $750 for immediate use for 60 days.
What must be the principal of his note given to a New York bank 9
Here we have the proceeds and rate given, to find the principal.
Find the proceeds of $ 1 for 60 days ; this will be $ .99. Then,
proceeds of $1 : given proceeds : : $1 : principal.
.-. Required principal = g iven P roceeds
proceeds of $1
= 11^0 = $757.58.
$.99
262 INTEREST. [Chap. XL
EXAMPLES CIV.
1.
s
f8700™-. o/tcuAvUU,, &wn,., &e&. 27, 18 ft.
^
cALnetif cLaAf& a/^t&v cLat& J promise to pay to
<jj$
the order of^^^^^^^^^m^etl
8
€Lykt tkow&cwicL &&v-&ri hawicOb&cL ^Dollars
d|
at tk& BunJc a^ (7LcmvmsEA&&
m
Value received, ivaXsJv imt&vejoZ.
3
No. /J? 7. Due flfemAAf £fvm&.
Find maturity, discount, and proceeds.
The State in which a note is to be paid determines the
question of ' Days of Grace.'
$3000^-. Jb&MAWU, &ol., fan. 6, 18 <?2.
^vxjju cLoAfo, a^t&i dat& J promise to pay to
the order of^^^r~~*~^wiAftoe>lfy
c^iifi& tAowxMui^^^^^^.^ Dollars
at tk& c/tavtk £&nv-&v BamJo.
Value received.
No. 70. Due ft&vnux/H,
Find maturity, discount, and proceeds,
Art. 249.]
EXAMPLES.
263
fV-530^-.
3.
Sfcvwta, Si, Jt. m. , &e&. 28, 1 8 <?3.
3kv&& rnxmtko, oL^t&v clat& J -promise to pay to
the order of^^^^^^^^wAf&&tj s >~~~~^
c^W tAow&a'yieL ^tv-5 kwruiv&cC tkuity jjjjj Dollars
Value received.
JVo. 86. Due d/moo, Zfu&fceA,.
Find maturity, discount, and proceeds.
In all notes Date of Maturity and Rate are the first
things to consider.
3w& WLO-ntha, a^t&v clat& J promise to pay to
the order of^^~^~~—^>
at^^^^^£k& cAatian^l Ji/yrv fdasvifc.
Value received.
No. /<?- Due Ww,. fC. 7\W^mW.
^Dollars
Find maturity, discount, and proceeds.
264 INTEREST. [Chai-. XI.
5.
$685™-. JW 3>aw, TnUk., CUuf. 3/, 18 <?¥.
&tv-&nXAf elaAfb oulteAj ciaZ& <J promise to pay to
the order of^^^~~~~~^. / mjf&M
^v^v^-.c/u^ kwyioUs&cl &vyA£M-^vv-& ^Dollars
at y ^^^^ ysy ^^XA& gPCuiZ c/tatvcyyiaZ fSasvik-^^^^^.
Value received.
No. //. Due.... &. &. $a,vk^.
Find maturity, discount, and proceeds.
6.
/ aft. CHAom*, Vt., fvutif Sf, 18 87.
^ev&n mcmtho.- cvft&v date, J promise to pay to
the order of^
m Dollars
at^^^^Xh& W^&ldeAv cAatio-nai Ba/wk~~~~~~
Value received.
No. 22. Due... £exyvia<u£ j?&wm&.
Discounted at 6%.
Proceeds = $6269.25 ; find maturity and discount.
Art. 249.] EXAMPLES. 265
7.
/ cA&w* Ifcyvk., cA.lf., fan. /, 18 <?6.
£fw~o- mo-ntko, a^t&v claZ& <J promise to pay to
the order of^ wvif&el^
_ m Dollars
at^^~^~~Xk& &<ym €vc&kaM,(f& JSank
Value Received.
No. /. Due @Aa&. €. fdvown,.
Discounted at 4%.
Proceeds = $248600 ; find maturity and discount.
8. What would have been the discount in 7, if the
note had been discounted by a bank in Ohio at the same
/ Boston-, maw-., CtfU. 6, 18 W.
otvfte&n day* aft&v dat& J promise to pay to
the order of mA^&tj ~~~~
m Dollars
at r ^^^^£>A,& 3wloL&uq/ cAatvonal Bcun^^^j^^^^.
Value received.
No. /8. Due JCe&na/uL Tyioundtt.
Discounted at 4^ <J .
Proceeds = $100000 ; find maturity and principal.
266 INTEREST. [Chap. XL
10.
$3560™. cAew Ifovk, Jt.y., tfefiL. /<?, 18 <?5.
<&iv-& mo-ntA^ ait&v ctat& cj promise to pay to
the order of &ied&vU& <$ > vvne>&
£'/vv&& tAowMMicL jZv& hwvicLv&cl vLocLif ^ Dollars
at £h& ^A&wC^clI cAaLl(yvia,L Ba/nfc
Value received, with, vjit&v&oZ-
No. 56. Due__„. d. f. 0LLv(yic£.
Discounted Dec. 31, at 4J %.
Find maturity and discount.
11. A note dated N.Y., July 7, 1891, payable in Ohio
in 3 yr. after date, was discounted Jan. 16, 1892, at 4£ % ;
the principal was $5000. Proceeds =?
12.
f780!ZL. RUhwumd, Va», fum&7, 18<?2.
ofoutu da,y& a^t&v dat& J promise to pay to
the order of^^^^^^/ifcyiatio £ate&
^j^^^efgAkML fiwndv&cC eZcfAfy ^^^^ Dollars
at. Xk& ffla/nteM/ ' c^atianal BamJo^
Value received, wiZA Lnt&i&oZ.
No. 27. Due jamv&O' ffa/nutton*
Discounted June 13, at 4J%,
Find maturity and proceeds.
Art. 249.] EXAMPLES. 267
13.
f67/^-. Jt&w- yovk, c/t. If., fan. 20, 18 <?5.
cA'vn&Vu daifo, cl^I&v dat& w& promise to pay to
the order o/U^^ ^<xyyww&t &volv-&&
^^^^.c/Cpc huM<cUi&cL &&v-&ntu-cyyi& —Dollars
at the* &Oia£ cA'atio-nal ISc^nk, oft. (ZuyuoX>uyi&, &la,.
Value received, with, Imt&v&oZ.
No./2. Due ftvusA, V> C*.
Discounted Feb. 28, at 6 %.
Find maturity and proceeds.
14.
/ cJ'aoAua,, c/t./f., £&&. 3/, 18<7¥.
t^W wio-ntha.' aft&v dat& w-& promise to pay to
the order of f3&rv[a/vnAAv &vaA>&&^^^^.
.___ m Dollars
at^^^^~^XA& cfe^u^biXy- 3\mq£> &>&
Value received, iv~iXA Imt&v&aZ.
No. /02. Due Jbanyid lA>cdl&.
Discounted Feb. 3, at 5 %.
Find maturity, maturity value, and principal, when the
proceeds = $790.
268 INTEKEST. [Chap. XI.
15.
/ ?Hilw-oiiAJo&&, Wio,., TnoAf 25, 18 <?3.
&ewb wo-nt/ifr ujt&'u dat& J promise to pay to
the order of^^^^@Aa>o,. <P. fdiosko-fi,
- m Dollars
at^^^ — tk& 'i/'&wyLcvyL CLryveAA^a^n fdaswh
Value received, w~\£h, vyvteA.&oZ.
No. f/S. Due /lfo-va,&& W4vit&.
Discounted July 1, '93, at 5^%.
Proceeds = $8000 ; principals?
Ans j $7920.84 = face.
' 1 $8110.28 = maturity value.
Exact Interest.
250. Thus far interest has been computed on the basis
of a year of 360 da. Such interest is evidently \\ of the
interest computed on the basis of a year of 365 days.
The interest computed on the basis of a year of 365 da.
is called Exact Interest, and is computed for only fractions
of a year.
Exact interest is computed in interest transactions with
General Governments and in many interest transactions
of ordinary business.
Ex. 1. Find exact interest at 4i% on a note for $892, dated Feb.
16, '93, and maturing Apr. 2/5, '93.
Here we find the interest at 4-*-% for the exact
4^)7 136
i 7«d numDer of days and on the 360 days basis, and sub
1.784
73)5.352
073 answer on the 365 days basis.
$5.28
Arts. 250, 251.] PAKTIAL PAYMENTS.
269
Ex. 2. Find the exact interest at 4 % on a note for $781.20, dated
June 5, '89, and maturing Oct. 4/7, '92.
$781.20
.12
Here simple interest is com-
puted for 3 yr. , and the exact
$ 93.74 Interest for 3 yr.
9,00 Exact interest for 124 da. mterest for 124da - ( lea P y r -) 1S
$102.74 = Ans.
added.
EXAMPLES CV.
Written Exercises.
What is the amount at maturity of each of the following
indicated notes, exact interest ?
Date. Face.
1. Texas, Jan. 4/93, $890
2. K. Y., Jan. 8/96, $400
3. Mass., June 11, '91, $250
Rate. Maturity.
41% Apr. 4/7/93.
4% Apr. 7, '96.
6% Mch.10/13,'92.
4. Oregon, Apr. 19, '93, $1250 Legal July 18/21, '93.
5. Iowa, Aug. 6, '94, $46849 Legal Nov. 4/7, '94.
6. Vt, Feb. 4, '95, $2685 5% May 5, '95.
Partial Payments.
251. It often occurs that part of a note is paid at one
time, another part at another time, and so on, until all
the note is paid. Such payments are called Partial
Payments.
In case of interest-bearing notes, it becomes necessary
to compute simple interest on the different principals which
appear during the life of the note.
The sums of money paid and the times of payment are
Endorsed on the back of the note.
270
INTEREST.
[Chap. XI.
Art. 251.]
PARTIAL PAYMENTS.
271
Dates 1
the
Jan.
Mch.
May
July
July
Sept.
$540.
4.59
$544.59
75.
found
note.
16,'
8,'
13,
9,
21,
19,
1st
int,
am
1st
2d
int
am
2d
3d
int
am
3d
4th
int,
am
4th
5th
int,
on Times between
successive dates.
E 5ida -
,94 66 "
•*::::::2 -
£ co "
prin.
, for 51 da.
1't of 1st prin.
payment.
$469.59
5.17
prin.
, for 66 da.
$474.76
80.
't of 2d prin.
payment. {
$394.76
3.75
i
prin. (
for 57 da. j
$398.51
100.
't of 3d prin. ]
payment. T
$298.51
.60
<
prin.
, for 12 da.
$299.11
150.
t of 4th prin.
payment.
$149.11
1.49
. prin.
, for 60 da.
When a note is wholly within
a year, the exact number of days
between dates is found, and the
days' method is used in compu-
tation.
It will be observed that the
amount of the principal is found
for the time elapsing between the
date of the note and the date of the
first payment. The first payment
is subtracted, and the remainder is
used as a new principal. And so
on to the end.
$151.60 am't paid Sept. 19.
272 INTEREST. [Chap. XI.
Ex.2.
&-. 0(yviria,jiw-LU, Tttcl., f^}t& /<?, 1888.
Hn cL&vwcwuL J promise to pay to
the order of^^^^^/if&nvu fo-kn^o-ru^^^^^^,
wvwwwvww^ tkawQsa>nd ^Dollars
at tk& dfiaAsWv&Ws' cfiaXAs(yyial> Bank^^^^.
Value received, w-vth (mt&ve^C at ¥-°lo.
^awv'l £a>6-{>ttt.
This note carried the following endorsements :
Dec. 1, '88, $150; Mch. 1, '92, $1000;
Apr. 7, '89, $250; Mch. 1, '93, $2000.
Oct. 25, '90, $275;
Find the balance which was paid on Sept. 19, '94.
Here we find the times in years, months, and days.
Dates found on Times between Interest on $1 at 6%
the note. successive dates. for the times,
yr. mo. da.
' 88 6 19 5 12 $.027
' 88 12 X 4 6 $.021
' 89 4 7 16 18 $.093) _„,
; 90 10 25 1 4 6 $.08ll =$ ' 174
' 92 3 l 1 $.06
' 93 3 1 1 6 18 |.093
'94 9 19
Art. 252.] U. S. RULE. 273
$5000 1st principal.
90 int. for 5 mo. 12 d., at 4%.
$5090 am't of 1st prin.
150 1st payment.
$4940. 2d prin.
69.16 int. for 4 mo. 6 d., at 4%.
$5009.16 am't of 2d prin.
250. 2d payment.
$4759.16 3d prin.
552.06 int. for 2yr. 10 mo. 24 d., at 4%.
$5311.22 am't of 3d prin.
1275. 3d aDd 4th payments.
$4036.22 4th prin.
161.45 int. for 1 yr., at 4%.
$4197.67 am't of 4th prin.
2000. 5th payment.
$2197.67 5th prin.
136.26 int. for 1 yr. 6 mo. 18 da., at 4%.
$2333.93 am't paid Sept. 19, '94.
In case any payment is less than the interest due at the time of
such payment (as in the 3d payment of this note) a portion of
the interest would become a part of the new principal and would
draw interest, if we should proceed as with the 1st and 2d pay-
ments. Here compound interest is forbidden by law, and we must
find the interest on the same principal until the time when the sum
of the payments equals or exceeds the interest.
The United States Rule.*
252. Compute the amount of the principal to the time
when a payment, or the sum of two or more payments,
equals or exceeds the interest due.
Subtract from this amount the payment, or the sum of
the payments, and proceed with the remainder as a new
principal. And so on to the time of settlement.
* Vermont, New Hampshire, and Connecticut have methods of
their own for computation in partial payments, but it is not advis-
abje to consider those methods in our present study.
274 INTEREST. [Chaps. XL, XII.
EXAMPLES CVI.
1. A Kentucky note for $3500, with interest, dated
Mch. 1, '90, had the following endorsements :
Apr. 6, '90, $500. May 15, '90, $800.
" 30, '90, $300. July 11, '90, $600.
What was paid in settlement on Aug. 22, '90 ?
2. An Arizona note for $8600, with interest, dated
July 1, '87, had the following endorsements :
Oct. 2, '87, $150. Feb. 21, '88, $4000.
Nov. 7, '87, $1500.
What was due May 4, 1888 ?
3. A Louisiana note for $876, with interest, dated
Feb. 6, '86, was endorsed as follows :
Apr. 11, '86, $50. June 2, '87, $300.
Dec. 1, '86, $150. July 5, '87, $75.
What was paid in settlement on Jan. 1, '88 ?
4. A Massachusetts note for $3000, with interest at
4|%, dated Jan. 1, '91, was endorsed as follows:
Mch. 7, '91, $175. Sept. 20, '93,
May 9, '91, $300. Nov. 30, '94,
Aug. 17, '93, $400.
What was paid in settlement on Dec. 5, '94 ?
5. An Indiana note for $2500, dated Jan. 6, '94, was
endorsed as follows:
Feb. 7, '94, $250. Oct. 6, '94,
Apr. 20, '94, $180. Feb. 7, '95, $350.
July 7, '94, $75.
What was paid in settlement on Feb. 20, '95 ?
Ans. $1145.
Art. 253.] DRAFTS. 275
CHAPTER XII.
EXCHANGE.
Drafts.
253. Suppose that Wilson & Co. of Baltimore buy of
Morton & Co. of St. Paul $2500 worth of goods on 60
da. credit. When the bill is due, Morton & Co. may
make a formal request for its payment. Such a request
is called a Draft ; Morton & Co. are said to draw on
Wilson & Co. Por example :
#2500™. £ft JW, TnUn., fidy 25, 18 <?f.
^^^^^^^^^^^^JZt Qyiakt^^^JPay to the
Order of^^^^^^^@wi&&lv-&Qs w^
£Tw-&riLu-lLv-& AwyioOb&ci^ ^ Dollars
WITH EXCHANGE.
Value received and charge the same to
account of
to fc>-&, | m<nto»*eo
No. 2<f. £altvnvov&, Tfld. J
The draft is sent to Wilson & Co. through a St. Paul
bank which transmits it to a Baltimore bank. The latter
presents the draft to Wilson & Co. for payment, and the
cash is sent to Morton & Co. through the St. Paul bank.
The banks charge a small fee for their services, and the words
' with exchange ' in the draft signify that the debtor must pay
the fee.
276 EXCHANGE. [Chap. XII.
The above draft is called a Sight Draft.
A sight draft is payable on presentation (most States
not allowing grace on sight drafts), and, from its nature,
is not subject to discount.
254. Instead of waiting for the expiration of the 60
da. and then drawing 'at sight/ Morton & Co. might
make a Time Draft, payable after date. For example :
$2500™ . it. JW, Mm , fwUf 25, 1 8 <?¥.
&cybty-^iA)~& cLa,if& ap^&v dat&^^^Pay to the
Order o/_____ w ____fc^^^^&^^^^^^^^^^^
^^^^&w~emty-lAj$} k^yicLv&cL .^ Dollars
Value reeeima q$m charge the same to
^ account of
To TMUonV®*., ) motion V &o-
No. <?/6. BMvywi*, THcL- J tfefoL. 8/f/, 'w.
If Wilson & Co. accede to the request, they make a
formal acceptance of the draft by writing across its face
the word * accepted,' together with their signature. Their
acceptance is equivalent to their making & promissory note,
and the draft is regarded as such by all concerned. After
acceptance, the draft is returned (through the banks) to
Morton & Co., who now have a written promise from
Wilson & Co., whereas before they had only a verbal
promise. Morton & Co. now have the draft discounted,
exactly as if it were a promissory note, and thus obtain
the cash needed.
255. In case the time draft is made payable ' after sight'
instead of ' after date' Wilson & Co. affix to their accept-
ance the date of acceptance so that maturity may be found.
Arts. 254, 255.] DRAFTS. 277
The payee is the owner of the draft. [See also Art.
228, Note.] For example :
f2500°-°-. oft. gcuul, Mi™*,., fulif 25, 18 w.
<^aiCu-tvv-& cLaAffr a^t&i> QAsCfkt^^^JPay to the
Order o/^^^^^^^^^^^^^(5Z^^^^^^^^^^^^^
^^^^^w-e^tMh^y^Sf havn^eds^^^^—lJollars
Value receid^a arm charge the same to
account of
To TMUo-n, V €*., 1 mavtaru V &*.
,mci\
No. <?/6. 8oJXJLwlo>u,, md.) tfefit. W//3, 'W.
Ex. This draft was discounted at 6 % on July 29th ; find ma-
turity and proceeds.
M t m. .8333
fl I ' jjg 23 From day of discount to maturity
$ 19.17 = discount. was 46 da.
$ 2480.83 = proceeds.
Time drafts are rarely used, while sight drafts are very
common.
EXAMPLES CVII.
1. E. A. Winslow of Brattleboro, Vt., drew on F. B.
Crane of St. Louis, Mo., for the payment of a $650 debt
contracted Apr. 13, '92, and due in 90 da. The draft was
dated May 13, '92, and made payable ' after date.'
Write the draft, indicating acceptance, and write its
date of maturity in the lower right-hand corner.
2 . Eewrite the draft, making it payable 'after sight' and
find its maturity, it having been accepted on May 16, '92.
3. Winslow had the first draft discounted May 20;
find the proceeds.
278 EXCHANGE. [Chap. XII.
4. What would have been the proceeds of the second
draft, had it been discounted May 18, '92 ?
5. On Jan. 1, '92, S. B. Titus of Austin, Texas, drew
on Ward & Co. of Macon, Ga., for the payment of a
$1765 debt, contracted Dec. 7, '91, and due in 90 da.
Write the draft, payable ' after sight,' indicate accept-
ance on Jan. 3, '92, and write its date of maturity.
6. Draft in Ex. 5 was discounted Jan. 6, '92; pro-
ceeds = ?
256. It is evident that all the drafts thus far shown
have been requests made by a creditor to his debtor.
Now drafts may be used for paying debts as well as for
collecting debts. In this case the debtor (through his
bank) makes a draft on some bank in the city where his
creditor lives and payable to such creditor.
Domestic Exchange.
257. The main object of drafts is the payment of
debts without sending the actual money, thus avoiding
expense, and risk of loss.
The draft method of making payments between cities
in the same country is called Domestic Exchange,
Foreign Exchange.
258. The draft method of making payments between
cities in different countries is called Foreign Exchange.
259. Foreign drafts are made more extended in form
than domestic drafts, and are called Bills of Exchange.
A Bill of Exchange consists of a set of two bills, both alike,
Arts. 256-262.] FOREIGN EXCHANGE. 279
except that they are numbered. These two bills are sent
by different steamers, and as soon as one of the bills has
been paid the other becomes void.
260. The drawing of Bills of Exchange is done by
brokers, and no commission is charged for transacting
the business.
261. The actual amount paid for Bills of Exchange,
for example paid in New York for bills on London, varies
from time to time ; the current price paid for Bills, called
the 'Rate of Exchange/ cannot, however, ordinarily be
much above or below par ; for if it would cost more to
discharge a debt by means of a bill than by the actual
transmission of bullion, the latter method would naturally
be adopted.
It should be noticed that even if all countries had
exactly the same coinage, there would still be fluctuations
in the rate of exchange between two countries, as the
balance of indebtedness between those two countries
varied.
262. The following table gives the value of some
foreign coins in terms of U. S. Money as proclaimed by
the Secretary of the Treasury on Jan. 1, '95 :
Austria 1 Crown = $ .20, 3
Belgium 1 Franc = .19, 3
Brazil 1 Milreis = .54, 6
Chili 1 Peso = .91, 2
, m , r Shanghai = .67, 3
China 1 TaeW TT .. & '
I Haikwan = .74, 9
Cuba 1 Peso = .92, 6
France 1 Franc = .19, 3
Germany 1 Mark = .23, 8
Great Britain .... 1 Pound Sterling = 4.86, Q\
Holland 1 Guilder = .40, 2
Italy 1 Lira = .19, 3
280
EXCHANGE.
"hap.
XII.
.99,
7
.98,
8
.26,
8
■77,
2
.19,
3
.26,
8
.19,
3
Japan 1 Yen (gold)
Mexico 1 Dollar (gold)
Norway 1 Crown
Russia 1 Rouble (gold)
Spain 1 Peseta
Sweden 1 Crown
Switzerland 1 Franc
These values are subject to change.
263. Exchange on Great Britain is quoted at the value
of one pound sterling (£ 1) in dollars; exchange on France
is quoted at the number of francs to the dollar; exchange
on Germany is quoted at the value of four reichsmarks.
The following is copied from a daily journal :
The foreign exchange market was steady, but very quiet in
tone. Posted rates were unchanged at $4.88| for sixty-day bills
and $4.90 for demand. Actual sales were $4.87§ @ .$4.88 for
sixty-day bills, $4.89^ for demand, $4.89£ for cables, and $4.87 @
$4.87^ for commercial.
In Continental, francs 5. 17| for long and 5. 16 J for short ; reichs-
marks 95^ and 95| ; guilders at 40$ and 40f .
The following example shows the form of a Bill of
Exchange and how to find its cost.
£1200^-. c/W Ifcyyfo, cA. If., fot. 2f, 18 <?$.
ojf th& &wm>& dat& avid t&ruyu, wn/foaAxL
^^^^^^^^v^^^^^ to the
Order of ^cwyvw&l Llttl&jahyi^^^^^^
Value received and charge the same to
To I c/. mo*****, V Go., )
JUyncucyyv, (on^Canci ) '
Art. 263.] EXAMPLES. 281
On Oct. 29th sight drafts on London were quoted at $4.88|.
£1200
4.88£
$5862.00 = cost of exchange.
Ex. 2. How large a sight draft on London can be purchased
for $3890, exchange at 4.86£?
4.86^ )3890.
£800.=^ns.
EXAMPLES CVIII.
Find the cost in New York of a Bill of Exchange for
1. £500 on London at 4.86^.
2. £1750 on Glasgow at 4.85.
3. 50000 francs on Paris at 5.18f .
4. 1250 marks on Berlin at 95 \.
5. 2000 milreis on Rio Janeiro at 54.9 [cents per
milreis].
6. 3000 crowns on Vienna at par.
7. Calculate the cost at market ( a. £650.
prices (as found in some daily < b. 2400 francs,
journal) of ( c. 2000 marks.
8. What will be the face of a N. Y. draft on Bremen
costing $297.96, exchange being at 95 \ ? (Omit decimals
of the answer.)
9. How large a draft on London can be purchased for
$8554.14, exchange being quoted at 4.88J ?
10. How large a draft on Paris can be purchased for
$1920, exchange being quoted at 5.18|?
282 STOCKS AND BONDS. [Chap. XIII.
CHAPTER XIII.
STOCKS AND BONDS.
Stocks.
264. There are many business undertakings, such as
railways, banks, gas works, etc., which are on so large a
scale that many persons must combine to provide the
money necessary to carry on the business. This is gen-
erally done by dividing up the whole sum required into
1 Shares ' of definite amount, say of $10, or $50, or $100
each.
The whole body of partners is called a Company, and
the individual partners are called Stockholders.
The total amount of money raised to carry on the busi-
ness of the company is called its Capital.
The affairs of a company are managed by a small
number of elected stockholders called Directors.
The profits made by the company are called Dividends,
and are periodically divided among the stockholders ; the
dividend is declared as a percentage on the capital.
265. A stockholder in a company cannot demand the
return of the money he paid for his shares ; he can, how-
ever, sell the shares.
If the dividends of the company are high, and are
likely to continue to be high, the shares will sell for
more than they originally cost; if, however, the com-
Arts. 264-268.] PREMIUM — DISCOUNT. 288
pany is not prosperous, the shares would have to be sold
for less than they originally cost.
Thus, the stockholders in a company are continually
changing, and different stockholders may have bought
their shares at very different prices.
266. The most important point to notice is that the
amount of dividend paid to a stockholder does not depend
on the price at which his shares were bought, but simply on
their nominal value.
Thus, two men who had the same number of $100 shares in a
company would be entitled to the same amount of dividend,
although one may have bought, for example, $100 shares for $180
and the other for $50 each.
267. Shares are said to be above or below 'par' ac-
cording as they are sold for more or for less than their
nominal value. The nominal value is $ 100 per share,
unless otherwise stated.
Thus, if $100 shares sell for $110 each, since $110 is \%% of
$100, the shares are 10 per cent above par.
When the price of shares is more than their nominal
value they are said to be * at a premium, ■ and when the
price is less than their nominal value the shares are ■ at
a discount.'
268. The following are examples of the different ques-
tions which may have to be considered.
Ex. 1. $100 shares in a gas company sell for $240 each; how
much will 70 shares cost ?
Each $100 share costs $240 cash ;
. •. 70 shares cost $240 x 70 = $16800.
Ex. 2. A man bought $100 shares in a gas company for $16800,
giving $240 for each $100 share ; how many shares did he buy f
Since each share cost $240,
the number of shares = $16800 -r- $240 = 70.
284 STOCKS AND BONDS. [Chap. XIII.
Ex. 3. A gas company pays a dividend of 8% per annum; how
much does a man receive who holds 70 $100 shares ?
His share of the capital is $100 x 70 = $7000, and he receives
8% on this, or $560.
Ex. 4. A man invests money in the stock of a company, each
$100 share costing $240 ; what % does he receive on his investment
when the company pays an 8% dividend ?
He receives $8 on each share, and having paid $240 for a share,
he receives $8 on each $240 invested ; ^| ¥ = 3£%.
269. Sometimes a company does not need its full
capital to carry on its business ; and in that case only a
certain fraction of the nominal amount of the shares is
' paid up ' ; the stockholders are, however, bound to pay
the rest if it should become necessary. When a dividend
is declared at so much per cent, this percentage is paid
only on the amount paid up on the shares, and not on
their full nominal value.
Ex. What income will be obtained by investing £1008 in the
purchase of £20 bank shares, on each of which £5 is paid up, at
£24 each share, the bank paying a dividend of 18 per cent ?
Since £24 buys one share, £1008 will buy £1008 -- £24 = 42
shares. These 42 shares, on each of which £5 is paid, make up a
capital of £5 x 42 = £210. On this capital of £210 a dividend of
18% is paid ; hence, income required = £210 x T Vo = £37. 16s.
EXAMPLES CIX.
Written Exercises.
1. If $10 shares sell for $3.50, how many shares can
be bought for $9271.50 ? What is the nominal value of
shares purchased ?
2. Mining shares of $10 each are sold at $2.50 dis-
count ; what is the price of 80 shares ?
Art. 269.] EXAMPLES. • 285
3. The shares of a certain company are sold at 10%
above par ; how much must be paid for 1060 $50 shares ?
4. A company pays a dividend of 8%; how much
does A receive if he holds 50 $50 shares ?
5. A man holds 350 shares of $50 each, and the
company pays 7% dividend; how much does he receive?
6. A man sells 63 $100 shares for $180 each, and
buys with the proceeds $50 shares at $35 each; how
many shares does he buy ?
7. What is the difference between a $100 stock and
$100 worth of stock?
8. A man sold 75 $50 shares for $65 each, and in-
vested the money in $100 shares at $125 each; how
many shares did he buy ?
9. What income would be obtained by investing
$3850 in the purchase of $100 shares in a company at
$175 each, the company paying a dividend of 6% per
annum ?
10. $100 shares in a certain bank sell at $350, and
the bank pays a semi-annual dividend of 7% ; what
annual income would be obtained by investing $9450 ?
11. A company pays a dividend of 4j-%, and its $100
shares sell for 50% above par; what per cent does an
investor receive ?
12. A man buys $50 shares at $62.50, and the
company pays a 5% dividend; what percentage does he
receive, and what % on his investment ?
13. A man sells fifty shares of $100 gas stock, paying
8% dividend, at $180; he invests the proceeds in $50
railway stock at $35 ; find the change in his income, the
railway company paying a dividend of 3^-%.
286 STOCKS AND BONDS. [Chap. XIII.
14. A man buys $100 stock in a company which pays
an 8 % dividend, and he buys at such a price as to receive
3% on his investment; what does he pay per share?
15. A bank pays a 9% dividend, and its $600 shares,
of which $200 is paid up, sell for $750 ; what % does
an investor receive on his money ?
The price of Stock is given at so much per cent; thus,
stock is said to be at 115, when $100 stock costs $115,
and so in proportion for other amounts.
16. How much will $500 stock at 75 sell for?
How much will $150 stock at 120 sell for?
How much will $60 stock at 128 sell for ?
How much will $1200 stock at 97 sell for?
17. What income will be obtained from $500 stock
when the dividend is 4% ?
18. What income will be obtained by investing
$110175 in a stock which pays 3^-%, and can be bought
at 113?
19. What income will be obtained by investing $70380
in a 31% stock at 97f?
20. What °] will a man get on his money if he invests
in a 4% stock at 125?
21. A man receives $660 a year by investing $21450
in 4% railway stock; what was the nominal value of the
stock?
22. An income of $506.25 per year is derived by
investing $15300 in a 4$-% stock ; what was the price of
the stock per share ?
23. Stock was purchased at 97-J- and sold at 103 J, and
the profit was $661.25 ; how much stock was purchased
and what was the total cost ?
Arts. 270-272.] BONDS. 287
24. In which will a man receive. the greater % on his
investment ; in a 3% stock at 95 or in a 4% at 127 ?
25. What will be the difference in income between a
4% stock at 129 and a 4i% at 145 ?
Bonds.
270. Governments borrow money to meet exceptional
expenditure, and undertake to pay a fixed rate of interest.
The promissory notes given in return for this money
are called Bonds. The bonds differ, however, from the
ordinary promissory notes in being more formal, and in
having small certificates attached to enable the holder of
the bond to easily collect his interest. These certificates
are called Coupons. There is a coupon for each 3 mo.
of interest. Therefore a twenty-year bond has eighty
coupons attached.
Eailway and other companies generally issue bonds of
a nature similar to that of government bonds.
271. A person investing money in bonds is sure of
a specified income, while a person investing in stocks
receives only his share of the profits after all expenses,
including the interest on bonds, have been paid.
272. The public debt of the United States Apr. 1, '95.
Amount of Bonds.
Rate.
When Redeemable.
$25,364,500
2%.
Option of U.S.
559,624,850
4"
July 1, 1907.
54,710
4"
100,000,000
5"
Feb. 1, 1904.
28,807,900
4"
Feb. 1, 1925.
$713,851,960.00 Total int.-bearing debt.
381,025,096.92 Non int.-bearing debt (U.S. Notes, Nat. Bank
Notes, Fractional Currency).
1,770,250.26 Debt which has matured.
$1,096,647,307.18 Total debt, exclusive of bonds issued to Pacific
railroads.
288 STOCKS AND BONDS. [Chap. XIII.
273. Stocks and bonds, except those of small com-
panies, are bought and sold at a special market, called a
Stock Exchange. The agent who is employed to buy and
sell for the public is called a Stock Broker, and the person
who deals in stocks and bonds is called a Stock Jobber.
Stock Brokers charge for their services a commission
called Brokerage; in calculating the cost of stocks and
bonds this brokerage must be added to their market
prices; the proceeds of a sale of stocks and bonds are
the market prices minus the brokerage. In previous
examples, brokerage has been allowed for in the prices.
274. Brokerage is generally \ of 1%, reckoned on the
par value of the stock ; it is therefore £ of $1 on every
$100 share bought or sold, no matter what the market
price.
(In the following examples each share is to be considered as
§ 100 par value, and }% is to be allowed for brokerage.)
Ex. A man sold out $5000 stock of a company which paid 3|%
annual dividends at 94}, and invested the proceeds in a stock which
paid 4% at 108 \ ; what was his change in income ?
$5000 x M\ = $175 = original income.
91} — } = $91} = proceeds from one share.
$91} x 50 = $4556.25 = proceeds from 50 shares.
108} + } = $101} = cost of each new share.
$4556.25 h- 101.25 = 45 = number of new shares.
$4500 x .04 = $180 . = income from new shares.
. \ he had an increase of $5 in his income.
EXAMPLES CX.
1. What is the difference between a dollar of stock
and a dollar's worth of stock ?
2. What is the difference in the interests on a hun-
dred-dollar stock and a hundred-dollar bond ?
Arts. 273, 274.] EXAMPLES. , 289
3. What amount of bonds at 97f can be bought for
$3900 ?
4t What amount of bonds at 96$ can be bought for
$5335?
5. What number of bonds at 97J can be bought for
$7154 ?
6. What number of bonds at 97^ can be bought for
$584.25?
How much would "be realized by selling
7. $1000 bonds at 96?
8. $500 bonds at 981 ?
9. $100 bonds at 118$?
10. Bonds bought at 124| pay 5% on the investment;
what rate do they bear ?
11. Bonds bought at 92$ pay 4$f % on the investment ;
what rate do they bear ?
12. What is the price of U.S. 5 per cents when the
investment produces 4 T 6 T % ?
13. I have $10000 to invest in U.S. 4's at 118$; what
is my income, and how much money is not invested ?
14. I have $7000 to invest in U.S.2's at 107f ; what
is my income, and how much money remains uninvested ?
15. U.S. 2's are bought at 114$; what rate do they
bear?
16. The trustees of a school invested, as a teachers'
fund, $40512.50 in U.S. 5's at 115f ; the salary of the
principal was $1000 ; how much was left for his assistant ?
17. A speculator invested in a company and received
a dividend of 6%, which was 8|% on the investment; at
what price did he purchase ?
290 STOCKS AND BONDS. [Chaps. Xllt, XIV.
18. A young man receiving a legacy of $48000 invested
one half in 5% railway bonds at 95J, and the other half
in 6% stock at 119 J; what income did he secure ?
19. A owns a farm which rents for $320.40 per yr.
If he should sell the farm for $8010 and invest the pro-
ceeds in U.S. 4's at 111^, will his yearly income be
increased or diminished, and how much ?
20. A capitalist drew the quarterly interest on his
U.S. 4's, amounting to $540, and afterwards sold the
bonds at $124§; what were the proceeds of the sale ?
21. A lady invested $20948.75 as follows: $6160 in
Maryland 6's at 96 \, $8225 in manufacturing stock at
87f paying 8% annual dividends, and the remainder in
steamboat stock at 73f paying 10% annual dividends;
what was her total income ?
English government bonds are called Consols.
22. A man had £2400 in the 2|% consols; he sold
out at 99 J- and invested the proceeds in 4% railway bonds,
thereby increasing his income by £6 ayr; at what price
did he buy the bonds ?
23. A man having an income of £352 a yr. in the
2f % consols, sells out at 97 and invests the proceeds in
4% railway bonds, thereby increasing his income £48 a
yr ; at what price were the bonds purchased ?
Abts. 275-277.] ARITHMETICAL PROGRESSION. 291
CHAPTER XIV.
PROGRESSIONS.
275. A series of numbers which increases or decreases
regularly is called a Progression.
For instance, 3, 5, 7, 9, 11, or 23, 20, 17, 14, 11, 8,
or 3, 6, 12, 24, or 81, 27, 9, 3, 1, |, *,
are progressions.
It will be noticed that in the first two progressions the series are
made by successive additions or subtractions, while in the last two
the series are made by successive multiplications or divisions.
The first are called Arithmetical Progressions (increasing
or decreasing).
The second are called Geometrical Progressions (increas-
ing or decreasing).
Arithmetical Progressions.
276.
There are five things to be considered
the first term,
denoted
by
a,
the last term,
«
u
I,
the number of terms,
u
ii
n,
the common difference,
a
ft
d,
and the sum of the terms, " " s.
277. Any three of these five being given, the other two
may be found.
292 PROGRESSIONS. [Chap. XIV.
In the arithmetical progression,
7, 10, 13, 16, 19, 22, 25,
it is evident that the last term is a plus six d, or that the first term
is I minus six d.
.•. I = a + (n — 1) d,
and a = I — (n — 1) d.
It is also evident that if a and I be added and the sum -f- 2, the
result will be the middle term ; and that if each term be changed
so as to contain as many units as the middle term the sum of the
new series will be the same as the sum of the original series.
2
By these formulas all examples in arithmetical progression may
be solved.
Ex. 1. a = 3, d = 5, n = 12 ; find I and s.
Now I = a + (n — 1) d _ g -j- I
= 3 + 11x5 S ~ 2
= 58. = 3 + 58 x 12
= 30.5 x 12
Ex. 2. a = 5, 1 = 17, n = 7 ; find d.
Now
1 — a+ (n — 1) d;
.-. 17 = 5 + 6d;
whence
6d = 12,
and
d= 2.
Ex. 3.
Fmd n when a = 2, Z = 30, and d = 7,
Now
Z = a + (n — 1) d;
.-. 30 = 2 + 0-1)7;
whence
7 (» - 1) = 28,
and
w-l = 4;
i.e..
n = 5.
Art. 278.] GEOMETRICAL PROGRESSION. 293
EXAMPLES CXI.
Written Exercises.
Answer the indicated questions.
1.
2.
3. 4.
5.
6.
a= 12.
5.
1. ?
?
.24.
1 = ?
41.
4.5. 351
18.
?
d= 5.
4.
? 3|.
3.
1.2.
n= 8.
?
8. 6.
6.
7.
s =
9
? ?
?
7. Insert 3
means between 2 and 12.
8. Find the
i series of 8 terms when the 3d term
is 14
and the 7th term is 26.
9. Find the series of 9 terms when a = 10.8 and the
6th term = 4.8.
10. Find 2 + 5 + 8 + 11 + ... to 37 terms.
11 . Find 8 + 7.75 + 7.5 H to 11 terms.
Geometrical Progressions.
278. There are five things to be considered :
the first term, denoted by a,
the last term,
u
" I,
the number of terms,
u
« n,
the ratio,
a
the sum of the terms,
a
" S.
(The ratio is the relation existing between any two successive
terms. It is the constant multiplier by which any term is found
from the preceding term. )
Any three of these five being given, the other two may
be found.
294 PROGRESSIONS. [Chap. XIV.
In the geometrical progression,
2, 6, 18, 54, 162,
it is evident that the last term is a times the product of r by itself
four times, i.e., a x r 4 .
Ia l = ? * Z "M formula 1.
and a = Z ■+• r» - l . J
It is also evident that
8 = 2 + 6 + 18 + 54 + 162 ; (1)
multiplying the equation by the ratio,
3s = 6+ 18 + 54 + 162 + 486; (2)
subtracting (1) from (2), we have
3s -s =486-2,
or s(3-l) = 486-2;
whence . 486-2
3-1
Now 486 = rl, 2 = a, and 3 = r ;
= rj^-a, formula 2 .
r- 1
By means of these two formulas all examples in geometrical
progression may be solved.
Ex. 1.
a = 3, r = 2,
n =
- 5 ; find I and s.
Now
1 = ar nl
= 3x2*
= 48.
s = rZ ~ a
r- 1
2 x 48 - 3
2-1
= 93.
Ex.2.
a - 3, I = 81
i n
= 4 ; ^nd d.
Now
1 = ar 71-1 ;
whence
81 = 3 x r 3 :
whence
r8 = 27;
whence
r = 3.
Art. 279.] INFINITE SERIES. 295
Ex. 3.
Find
n
when
a = 3, I
= 375,
and r -
= 5,
Now
1 =
ar n - 1
>
whence
375 =
3x5"
-l .
whence
125 =
5*- 1 ;
whence
n-l =
3;
whence
Formu
ila 2 '
be
icomej
n =
:4.
~ rl M
(2) is i
5llb
if (2) is subtracted from (1).
1 — r
This should be used in case of a decreasing geometrical progression.
EXAMPLES CXII.
Written Exercises.
Answer the indicated questions.
l.
2.
3.
4.
5.
6.
a= 2.
11.
1
2"*
?
?
1.3.
1 = ?
352.
625
rsr-
2744
2 16 •
608.
?
r = 5.
2.
?
f
2.
1.2.
71= 5.
?
5.
4.
6.
4.
s =
9
9
7
?
7. Insert 3 geometrical means between 4 and 2500.
8. Find the series of 8 terms when the 3d term is 10.8
and the 7th is 874.8.
9. Find the series of 6 terms when a = yt and the
fourth term is J-ff|.
10. Find 2£ + 6f + 19$ + — to 10 terms.
11. Find the series of 5 terms when a = 36.015 and
the 3d term is .735.
12. Find 28.8 + 14.4 + 7.2 + ... to 7 terms.
279. When a decreasing geometrical series is extended
to a large number of terms, the last term will be so small
that it will have no appreciable value.
296 PROGRESSIONS. [Chaps. XIV., XV.
Thus, if we continue f, & T fa, 7 | T , „*„, „£„, indefinitely, the
last term will be almost zero ; .-.in the formula s — a ~ r the Ir
1 -r
of the numerator may be omitted, and the formula will become
s = a , by which we may find the sum of the terms of a decreas-
1 — r
ing infinite series.
Ex.1. JP&Kg $+ $ + ^ +£ + ..,
to infinity.
s « L
--i- a
S -l-r~l-
i — i — f •
2 2
Ex. 2. jFmd Me value of .46.
Now .46 = .4 + .06 + .006 + .0006, etc.
.*. the value must equal .4 + the geometrical progression, .06,
.006, .0006, etc.
. -a .06 _ .
.-, .46 = .4 + ^ = H + A = A-
EXAMPLES CXIII.
Written Exercises.
1. Find i + A H to infinity.
2. Find ] + f + ... to infinity.
3. Find the value of 1.416.
4. Find the value of 1.53L
5. Find the value of 3.3360.
Arts. 280,281.] CUBE ROOT. 297
CHAPTER XV.
CUBE ROOT.
280. The cubes of the first 10 whole numbers should
be known : they are
l; 8, 27, 64, 125, 216, 343, 512, 729, 1000.
An integer (or a fraction) which is the cube of another
integer (or fraction) is called a Perfect Cube.
Thus, 64 and iff are perfect cubes ; namely, the cubes of 4 and
4 respectively.
281. In simple cases the cube root of a number can
be found by separating it into factors, as in Art. 80.
For example, to find ^/9261.
9261 = 9 x 1029 = 27 x 343 = 3* x 73 = (3 x 7) 8 ;
hence, ^/9261 = f/(3 x 7) 3 = 3 x 7 = 21.
EXAMPLES OXIV.
Find the cube root of each of the following numbers :
1. 10648. 3. 35937. 5. 19683.
2. 3375. 4. 13824. 6. 42875.
Find the least number by which each of the following
numbers must be multiplied in order that the result may
be a perfect cube.
7. 108. 9. 336. 11. 4032.
8. 392. 10. 441 12. 7056.
298 CUBE ROOT. [Chap. XV.
282. Since,
10 3 = 1000, 100 3 = 1000000, 1000 3 = 1000000000,
and so on, it follows that
if a number has 1 digit, its cube has either 1, 2, or 3 digits
" " 2 digits, " « 4, 5, or 6 "
" « 3 " « " 7,8, or 9 "
Hence, if we mark off the digits of a given number, be-
ginning at the units' digit, in periods of three, the last of
the periods containing one, two, or three digits ; then the
number of these periods will be equal to the number of digits
in the cube root of the given number.
For example, by pointing off the numbers, 2744, 32.768, 3511808,
as follows, namely, 2'744, 32'.768, and 3'511'808, we see that the
cube roots of these numbers contain, respectively, 2, 2, and 3 figures.
Find (60 + 3)3.
By Art. 86, (60 + 3)2 = 60 2 + 2(60 x 3) + 32
Multiplying by 60+3
60 3 +2(60 2 x3) + 60 x 3 2
60 2 x3 +2(60x32) + 38
and (60 + 3)» = 60 3 + 3(60 2 x 3) + 3(60 x 3 2 ) + 3*.
The cube of the sum of any two other numbers can be
expressed in a similar form.
Hence, the cube of the sum of any two numbers is equal
to the cube of the first plus three times the square of the first
multiplied by the second plus three times the first multiplied
by the square of the second plus the cube of the second.
The above Theorem will enable us to find the Cube
Root of any number.
283. To find the Cube Root of any number. The method
will be seen from the following examples:
Arts. 282, 283.] CUBE ROOT. 299
Ex. 1. To find the cube root of 157464.
By pointing off the figures into periods of three [Art. 282], we
see that there are two figures in the required root.
The first figure of the root is 5, since 157000 is between 50 3 and
60 3 . Subtract 50 3 from the given number, and the remainder will
be 32464.
Now this remainder must consist of 3 x 50 2 x units' digit + 3
X 50 x sq. of units' digit + cube of units' digit, and the first of
these three terms is the largest ; therefore if we use 3 x 50 2 as a
trial divisor, we obtain a quotient, namely 4, which is equal to, or
greater than, the unknown (units') digit. If now we add to the
. 157'464(50 + 4
50* = 125 000
3 x 50 2 = 75J0 j 32 464*
3 x 50 x 4*^= 600
4*= 16
3 x 502 + 3 x 50 x 4 + 42 32 464
trial divisor the last two of the above three terms (omitting the
units' digit once as a factor), we shall have as a true divisor
3 x 50 2 + 3 x 50 x 4 + 4 2 = gll6. Multiplying this by the units'
digit and subtracting the product from 32464, we have no
remainder.
Ex. 2. Find the cube root of 13312063. „~
.. 13'312'053(2H|+ 30 + 7
, 200 3 =
8^)00" 000
3 x 2002 = 120000
3 x 200 x*30 =#18000
302 _ 900
5 312 e#e
. 138900 -
4 167 000'
3 x 2302 £. 158700.
3 x 230 x 7 = 4830 '
7 2 = 49
1 145 053
163579
1 145 053
Here there are three periods, and therefore three figures in the root ;
and, since 13000000 lies between 200 3 and 300 3 , the first figure of
the root is 2. Subtract 200 3 , and the remainder is 5312053. Now
take 3 x 200 2 , that is 120000, as a 'trial divisor' ; and 5312053 h-
300 CUBE ROOT. [Chap. XV.
120000 will give 40 for quotient. It will, however, be found on
trial that 40 is too great, for
(3 x 200 2 + 3 x 200 x 40 + 40 2 ) x 40
is greater than the remainder 5312053 ; we therefore try 30. Take
3 x 200 2 + 3 x 200 x 30 + 30 2 ,
and multiply this sum by 30 and subtract the product from 5312053 ;
we shall then have subtracted altogether (200 + 30) 3 from the
given number, and the remainder will be found to be 1145053.
To find the last figure of the root use 3 x 230 2 , that is 158700, as
a 'trial divisor,' and 1145033 + 158700 gives 7 for quotient. Take
3 x 230 2 + 3 x 330 x 7 -|- 7 2 , aid multiply this sum by 7, and sub-
tract the product from 1145053. There is now no remainder ; and,
from Art. 219, we have now sub#acted altogether (230 + 7) 3 ;
hence the given number = 237 3 , so that 237 is the required cube
root.
Ex. 3. Find the cube root of 252.435968.
The pointing must be begun from the units' figure, and carried
forwards for the integral part and backwards for the decimal part.
252'.435'968'(6 + .3 + .02
68 = 216.
I?x6 2 = 108 36.435968
3 x 6 x .3 = .5.4
(.3>, 2 = .09
113.49
34.04^
3 x (6.3) 2 = 119.07
3 x 6.3 x .02 = .378
(.02)2 = .0004
119.4484
2.388968
2.388968
The process can be somewhat shortened, as in Square
Root ; it is, however, very rarely necessary to find a cube
root, and it is therefore undesirable to attempt to shorten
the ahove process.
Art. 283.]
EXAMPLES.
301
EXAMPLES CXV.
Find the cube root of each of the following numbers :
1. 1331.
2. 3375.
3. 4913.
4. 12167.
5. 29791.
6. 68921.
7. 79507000.
8. 148877000.
9. 8869743.
10. 733870808.
11. 2352637.
12. 16974393.
13. 2.197.
14. .004913.
15. .238328.
16. 125525.735343.
17. 2io.
19. 12568-g-V ♦
18. 39^.
20. 240tfftf
Find to three significant figures :
21. ^/10. 22. ^1.5. 23. .s/3.75. 24. ^/.0675.
25. Find the side of a cube wkich has the same vol-
ume as a beam 40 ft. 6 in. long, 1 ft. 4 in. wide, and f in.
thick.
26. Find the length of one edge of a cube whose vol-
ume is 2 cu. yd. 14 cu. ft. 145 cu. in.
27. Find the area of each face of a cube whose volume
is 5 cu. yd. 2 cu. ft. 1592 cu. in.
28. Find approximately the length of one edge of a
cubical vessel which contains a gallon.
29. Find approximately the side of a cube of iron
which weighs a t, assuming that a cu. ft. of iron weighs
486 lb.
30. Find, to the nearest mm, the length of a cube of
gold which weighs as much as a cum of water, the S.G.
of gold being 19.5.
302 REVIEW. [Chap. XV.
MISCELLANEOUS EXAMPLES FOR GENERAL REVIEW.
1. Express in words 5006017, and in figures thirteen
million twenty-five thousand eleven.
2. Find the least multiple of 3157 which is greater
than a million.
3. How many articles each worth $14.45 should be
given in exchange for 60 articles each worth $49.13 ?
4. Reduce 5 1. 7 cwt. 30 lb. 11 oz. to oz.
5. Find the G.C.M. also the L.C.M. of 3432 and 3575.
6. Find the sum of -J-, f, £, j-i, and ^|.
7. Divide 43^ by 28 T 2 ^-, and express the result as a
fraction of 12.
8. Divide .221312 by 5.32.
9. Add .375 of 13s. 4c7. and .07 of £2. 10s., and sub-
tract the result from £.45.
10. Find the rent of 134 A. 145 sq. rd. at $19.50 per
acre.
11. Multiply 905741 by 518963, and express the result
in words.
12. A certain number was divided by 77 by short divis-
ions ; the quotient was 137, the first remainder was 9 and
the second remainder was 6 ; what was the dividend ?
13. Reduce 15 m. 95 rd. 3 yd. to in.
14. A grocer mixed 48 lb. of tea which cost him 64 ct.
a lb. with a certain quantity which cost 60 ct. a lb. He
then sold the whole for $76.92, and gained $7.20 by the
transaction. How much tea did he sell ?
15. Express 756, 1155, and 1176 as the products of
prime factors.
Art. 283.] EXAMPLES. 308
16. Simplify 4i + i-3| + 5 J- 6H-
17. Simplify | of £ of 4& - 1| of ^.
18. Simplify 2.9015 x .01702 x .005803.
19. Find f of £2. lis. lid. - .115625 of £1 + .75 d.
20. A farm of 500 Ha 91 a is rented at $3.25 per Ha.;
what is the whole rent ?
21. Find the difference between seventy-six million
eight, and four hundred ninety-nine thousand four hun-
dred forty ; and divide the result by ninety-nine.
22. What is the greatest number which will divide
2000 with remainder 11, and will divide 2708 with re-
mainder 17 ?
23. Multiply 190 rd. 9 in. by 144.
24. Taking the average length of a lunar month from
full moon to full moon to be 29.5306 da. and the length
of a yr. to be 365.2422 da., show that 4131 lunar months
are very nearly equal to 334 yr.
25. What is the least number of gr. which is an ex-
act number both of lb. Troy and of lb. Avoir. ? If the
number of lb. Troy in a certain weight exceed the number
of lb. Avoir, by 496, what is that weight in gr.?
26 . Simp]ify 3 -iA|AM_^^ 8|f .
27. Find the value of a property if the owner of f- of
it can sell f of his share for $492.
28. Divide .00625 by 2500, and 6.25 by .0025.
29. Express 20 lb. 8 oz. 9 dwt. 6 gr. as a decimal of
2541b. 10 oz.
304 REVIEW. [Chap. XV.
30. Find the difference between the value of 13 cwt.
74 lb. of sugar at $5 per cwt., and that of 52 lb. 12 oz. of
tobacco at $120 per cwt.
31. Write MDCCCXCIX in Arabic figures, and express
1489 by means of Roman numerals.
32. A man takes 100 steps a minute, and the average
length of his step is 30 in. ; how far will he walk in
4hr.?
33. How much coal is required to supply 12 fires for
27 weeks, each fire consuming 1 cwt. 42 lb. of coal weekly ?
34. Find the greatest number by which when 4344 and
5943 are divided the remainders will be 31 and 41 respec-
tively.
35. Simplify 267f of if x (f - f - T V).
41-34; 41of4|
36. S lm phfy n L-i+ I ^ F - 1 X f
37. If if of 4 of 29£ of a certain sum is $1692.60,
what is the sum ?
38. Reduce .63, .48324, and .01654 to common fractions
in their lowest terms.
39. What decimal of $2.25 is $5 ? Find the value
of .78125 of $4 - .0625 of $1.20 - 2.75 of $.04.
40. What is the cost of a silver cup weighing 2 lb. 5 oz.
17 dwt. 12 gr. at $1.85 per oz. ?
41 . $603.42 is to be divided equally among 226 people ;
how much will each receive ?
42. The heights of 5 boys are respectively 5 ft. 41 in.,
5 ft. 2 in., 5 ft. 11 in., 4 ft. 10 in., and 4 ft. 8£ in.; what
is the average height ?
Art. 283.] EXAMPLES. 305
43. Reduce 726314 in. to mi., rd., etc.
44. A circular running path is 902 yards round. Two
men start back to back to run round, and one runs at the
rate of 10 miles and the other at the rate of 10^ miles an
hour. When and where will they meet for the first time ?
45. Find the greatest length of which both 42 yd. 9 in.
and 55 yd. 9 in. are multiples.
46. Find the least fraction which added to the sum of
-i-J, ^, and |~§ will make the result an integer.
47. What fraction of $27 is T 9 T of $1.21 ?
48. Simplify ^f x ^55 . and divide .72 by il7936,
.018 .64
expressing the result as a recurring decimal.
49. Find the value of .05 of ioi of £74. 18s. 6cl
50. A person buys 5 cwt. 46 lb. of sugar at $3.87^ per
cwt., and sells it at 4 ct. per lb; what is the gain ?
51. Find the sum of all the numbers between 100 and
200 which are divisible by 13.
52. If a person's income be $1700 a year, find what
he will save in 4 yr. after spending on an average $25.50
a week, taking 52 weeks to a yr. '
53. Divide 69 mi. 319 rd. 2 yd. 1ft. 10 in. by 136.
54. The L.C.M. of two numbers is 11160, the Gr.C.M.
is 15, and one of the numbers 465; what is the other
number ?
55. Simplify lof l^of 4|-5-^-of 4of 3f
56. Express as a simple fraction.
3+ ?
306 BEVIEW. [Chap. XV.
57. Subtract |f of J- of $21 from | of ^ of $20; and
express the difference as a fraction of an eagle.
58. Divide .37592 by .0125, and 3759.2 by .000125.
59. Express } of 2.624 H1 - *£ of 1.375 1 as Kl. Is the
answer numerically the same as cubic meters ?
60. Find the value of 7 A. 80 sq. rd. of land at $200
per A.
61. How many mi., etc., are there in a hundred
million in. ?
62. If butter be bought at $27 per cwt. and sold at
33 ct. per lb., how much will be gained on every cwt. ?
63. Find the G.C.M. of 1035, 391, and 598.
64. Simplify 4J of 3J - 2£ ■*- 5^ + 6 T 6 T -*r 3^-.
mm a . „. 2 , 1551.65 v 21
65 . Simplify — and X
65.1 20.02
66. A square cistern is 3 m long inside and when filled
contains 47.25 T of water; what is the depth of the
cistern inside ?
67. If a bankrupt pays $23 in a hundred, how much
will a creditor, to whom he owes $7866, receive ?
68. If 8 cwt. 20 lb. cost $20.50, what would a t. cost
at the same rate ?
69. The distance between two stations is 234 mi. 160 rd.
38 yd. 2 ft. An engine wheel revolves 142878 times in
traveling from one station to the other. How many
in. does it travel in one revolution of the wheel ?
70. What is the greatest weight of which both 2 t.
4 cwt. 18 lb., and 5 t. 5 cwt. 94 lb. are multiples ?
Art. 283.] EXAMPLES. 307
71. The sum of the ages of a father and of his son is
now 88 yr., and 12 yr. ago the father was three times as
old as the son ; how old are they ?
72. A number is divided by 210 in three steps, the
factors being 5, 6, 7 in order; and the remainders are
2, 3, 4 in order ; what would have been the remainders
if the number had been divided by 7, 6, 5 in order ?
73. Reduce 216875 in. to mi., etc., also 57637 sq. yd.
to A., sq. rd., and sq. yd.
74. Find, to within a thousandth of the whole, the
square roots of 15, -^, and .081.
75. Find two numbers, one of which is double the other,
and whose product is 8192.
76. Find the following :
17 x 19, 18 x 14, 13 x 16, 19 x 15,
75 2 , 95 2 , 105 2 , 115 2 .
77. Find the least length which is a multiple of 1 ft.
6 in., 4 ft. 6 in., 7 ft. 6 in., and 15 ft. 9 in.
78. What is the acreage of a rectangular field whose
sides are respectively 201 yd. 2 ft., and 60 yd. ?
79. A rectangular field contains 2 A. 134 sq. rd., and
its length is 6.25 ch. ; what is its breadth ?
80. What is the least length of carpet 27 in. wide
that would be required to cover the floor of a room 24 ft.
loner and 21 ft. wide ?
81. Simplify 4f - (^ x ^ X 10*) and 4f - ^ x T \
xlOi
82. Simplify 7 x 16 - } of 4* - 1\ x 17 - \ 4 (18 - 6)
+ (26-3)|-7.
308 REVIEW. [Chap. XV.
83. Find by factors ^1936, V2601, Vdrffir-
84. How much will it cost to paint the ceiling of a
room 15 ft. 6 in. long and 12 ft. 6 in. wide at 16 ct. per
square foot ?
85. How many loads (cu. yd.) of gravel would be re-
quired to cover to a depth of 2 in. a path 90 yd. long
and 5 ft. wide ?
86. One side of a square field of 22 J A. abuts on a
road. This side is divided into building plots 100 ft.
deep and having a frontage along the road of 30 ft. each.
The building plots are let at £12 each, and the rest of
the field at £5. 10s. an A. What is the total rental of
the property ?
87. A dealer purchased 40 tubs of butter, each contain-
ing 35 lb., at 22 ct. per lb., and sold 35 tubs of the butter
for as much as the whole cost; for how much per lb.
must he sell the remainder in order to gain 16 % and
$3.22 ?
88. What is the acreage of a rectangular field whose
length is 117 rd. and whose breadth is 55 rd. ?
89. The number of sheep on a farm increased for 4 yr.
at the rate of 20% each year, and there were originally
625 sheep ; how many were there at the end of the 4 yr. ?
90. A man makes a profit of 20% by selling an article
for 24 ct. ; how much % would he make by selling it for
25 ct. ?
91. From a vessel containing 32.3 J of kerosene 1722.5 cl
were drawn ; how many dl remained ?
92. Find the least number which when divided by 17
leaves a remainder 12, and when divided by 29 leaves a
remainder 24.
Art. 283.] EXAMPLES. 309
93. Reduce to their simplest forms :
© T 3 T(i + f + iV-i)-ii°f^r
(") <* + *+* + €>+<*+*+*+»
94. The age of a father is three times the sum of the
ages of his three sons, and two years ago the father's age
exceeded the sum of the ages of the three sons by 36
years ; how old is the father ?
95. A body weighs 60 g in air and 42 g in water; what
is its S.G. ?
96. Find the interest on a 30 da. Mass. note for
$7895.56.
97. A man bought 13$ bu. of corn for $7.77, and
sold the same at 20% profit; what was the selling price
per bu. ?
98. A man paid $45.10, including a duty of 10%, for
a watch ; how much was the duty ?
99. The distance between two places on a map is
156 mm ; what is the distance in Km if the scale of the
map is 1 to 80000 ?
100. Find the number of Km in one mi.
101. Find the prime factors of the L.C.M. of 391
and 493.
102. If 15% be lost by selling an estate for $3400, for
what must it be sold to gain 20% ?
103. Find V-6 *° the nearest thousandth.
310 REVIEW. [Chap. XV.
104. A beam 36 ft. long, and whose section is a square,
contains 182-J- cu. ft. of timber ; what is its width ?
105. Find the length of the side of a square field
which contains 10 A.
106. Divide 570326 by 63 by 'short' divisions, ex-
plaining clearly the formation of the remainder.
107. Reduce to its simplest form
ft + l)of (f+iHfof ( i + 1 vH||^fi
108. Multiply 36.2 by .057, and divide 5752.8 by
.00376, and .0025 by 3.1.
109. Find the value of a bar of gold weighing 5 lb.
10 oz. 17 dwt. 22 gr. at $20 per oz.
110. How many gallons will a cistern 6 ft. by 4 ft. by
3ft. hold?
111. The total number of votes given for two candi-
dates at an election was 127345, and the successful
candidate had a majority over the other of 17377; how
many votes did each get ?
112. Divide $875 between three persons so that the
first may have $50 more than the second, and the second
$75 less than the third.
113. A certain number less than 1000, when divided
by 56 or by 72 leaves 13 as remainder; what is the
number ?
114. A grocer mixes 91b. of coffee at 54 ct. alb. with
6 lb. of chicory at 15 ct. a lb ; at what price per lb. must
he sell the mixture in order to get a profit of 25%?
Art. 283.] EXAMPLES. 311
115. The breadth of a room is twice its height and the
length is thrice its height ; and it cost $115.20 to paint
the walls at $.08 per sq. ft. ; what is the height ?
116. How many turfs each 3 ft. by 1 ft. would be re-
quired to turf a lawn 96 ft. by 75 ft., and how much
would they cost at $1.75 a hundred ?
117. Find the weight of a rectangular solid piece of
iron 17 cm by 5 cm by 3 cm , the S.G-. of iron being 7.8.
Answer in Kg.
118. Find the interest on $672.87 for 2 yr. 7 mo.
at 4%.
119. In a room 22 ft. by 18 ft. there is a Turkey
carpet with a border 2 ft. wide all round it. The
carpet cost 20 ga. ; how much was that a sq. yd.
120. Find the length of a square field whose area is
4 A. 89 sq. rd.
121. A wire .2346 yd. long is cut up into pieces each
.007 yd. long ; how many pieces will there be, and what
length will be left over ?
122. A room is 21 ft. long, 17 ft. wide, and 12 ft. high ;
how many pieces of paper 21 in. wide and 12 yd. long
must be bought to paper the room supposing 150 sq. ft.
of the walls are left uncovered ?
123. A class contains 19 boys; and in an examination
6 boys got 56% of the full marks each, one got 90%, and
the rest got 39% each, except one boy who got no marks
at all ; what was the average % got by the boys in the
class ?
124. A rectangular block of timber is 5 ft. long and
contains 3 cu. ft. If its section be a square, find its
thickness to the nearest tenth of an in.
312 REVIEW. [Chap. XV.
125. A square field is bordered by a path one yd.
wide, the field and path together occupying two and
one half A. ; find the cost of covering the path with
gravel at 36 ct. per sq. yd.
126. A flask holding 25 ccm of water, holds 20.25 « of
alcohol ; find the S.G. of the alcohol.
' 127. Some goods cost $25 ; how much is lost by selling
them at 20% below cost ?
128. One lb. Troy is what % of one lb. Avoir. ?
129. What is the proceeds of a N. Y. note for $2040
drawn Jan. 31, '95, at 3 mo. and discounted on Feb. 25th
at 5% ?
130. What sum is invested if the investment yields
$585 per annum at 4^% ?
131. Eeduce 563147 in. to mi., etc.
132. Find the prime factors of 58212. What is the
greatest square number of which 58212 is a multiple ?
133. Simplify 2W + *(* + *)-J (* + *)
Joff-foff
134. Find the acreage of a rectangular field whose
length is 25 ch. 80 li. and whose breadth is 8 ch. 75 li. ;
find also the rent at $12 an A.
135. A certain piece of work can be done by 8 men or
16 boys in 10 da. In how many da. can the work be
done by 8 men and 16 boys ?
136. An object weighs 10 g in air and 4 g in water ; find
its S.G.
Art. 283.] EXAMPLES. 313
137. A man, after deducting $4000 from his income,
pays $170 income tax on the remainder. If the $4000
had not been deducted, the tax would have been $250.
Find the rate of taxation and the income.
138. A demand note with interest was paid 4 yr. after
date. The interest at ±\ % was $365.04; find the
principal.
139. A demand note bearing interest was paid 4 yr.
after date. The amount at 5% was $2433.60; find the
principal.
140. A train 110 yd. long was observed to pass a
certain point in 10 sec. ; how many mi. an hr. was it
then going ?
141. Determine the number which when divided by
231 by the method of ' short ' divisions, gives a quotient
583, and 2, 6, and 10 as successive remainders.
142. Find the G.C.M. of 464321 and 683111, and hence
find all the common measures of those numbers.
143. Find the weight in Kg of the air in a room 60 ft.
long, 36 ft. wide, and 21 ft. high, assuming that one
cu. yd. = .765 cura , and that air weighs 1.29 g per liter.
144. The wages of A and B together for 45 da.
amount to the same sum as the wages of A alone for 72
da. ; for how many da. will this sum pay the wages of
B alone ?
145. A room is 20 ft. 7 in. long, 15 ft. 5 in. wide, and
11 ft. high. Find the number of pieces of paper, each
12 yd. long and 21 in. wide, which would have to be
bought to paper the walls, supposing that windows,
etc., which are not papered, make up one-sixth of the
whole surface of the walls.
314 EEVIEW. [Chap. XV.
146. Ten loads (cubic yards) of gravel are spread uni-
formly over a path 180 ft. long and 4 ft. wide ; what is
the depth of the gravel ?
147. A merchant borrowed $2000 from a Philadelphia
bank for 30 da. at 5% ; find the proceeds of the note.
148. An Ohio farmer sold some sheep for $475, and took
in payment a 3 mo. interest-bearing note dated Jan. 6, '93,
rate h\%. On Mch. 1 st the farmer had the note discounted
at 5% ; how much cash did he receive from the bank ?
149. One pound of silver is weighed in water; how
many pwt. does it lose, the S.G. being 10.5 ?
150. Find the weight in dg of a cylindrical stick of
silver 10 cm long and l cm in diameter, the S.G. being 10.5.
151. Find the least length which is a multiple of 5 yd.
1 ft. 3 in., and also of 7 yd. 2 ft. 9 in.
152. Simplify (4i-2f of ^ + 2i)-K4i-2|)off-f2^.
153. Find V 783 to tne nearest tenth.
154. (i) Multiply 17 + 19 + 16 by 18.
(ii) Find mentally (30 + 4) 2 .
(iii) Find mentally 85 2 .
155. Find
[35 2 -*- 7 2 x 4 2 - J (150 x § -s- 25) + 1260 + 35|] 11.
156. (i) A ratio is 47 ; find the second term when the
first term is 235.
(ii) A ratio is -^ ; find the first term when the second
term is -fa.
(iii) Two similar rooms are respectively 8 yd. and 9 yd.
long ; how much paper will be required to paper the first
room, compared with that which will be required for the
second room ?
Art. 283.] EXAMPLES. 315
157. Sound travels at the rate of 1090ft. a second;
how far off is a thunder-cloud when the sound follows
the flash after 5| sec. ? Answer to the nearest hun-
dredth of a mi.
158. A father, who had three children, left his second
son $500 more than he left the third son, and his eldest
son twice as much as the third. They had $8500 between
them ; how much had each ?
159. In a certain examination every candidate took
either Latin or Mathematics, also 79.4% of the candi-
dates took Latin and 89.6% took Mathematics. If there
were 1500 candidates altogether, how many took both
Latin and Mathematics ?
160. For what sum must goods worth $6370 be insured
at 2% premium so that in case of loss the owner may
recover the value both of the goods and the premium ?
161 SimDlifv ^ofl-Sjof^
161 - Smphfy ^(A-Do!^
162. A bill of $301.05 was paid with an equal number
of eagles, dollar pieces, quarters, and five-cent pieces;
how many coins of each kind were there ?
163. A and B received respectively ^ and -fa of a
certain sum of money, and C received the remainder.
A received $1173; how much did C receive?
164. What is the cost of a plot of building-land 242
ft. long and 21 ft. wide at $2000 an A. ?
165. At the beginning of a year the population of a
town was 16400. The deaths during the year were 3%
of the population at the beginning of the year, and 80%
of the births. What was the population at the end of
the year, neglecting changes caused by traveling ?
316 REVIEW. [Chap. XV.
166. A liter flask was half filled with sand, and the
weight of the sand was 1375 g ; what was the S.G-. of the
sand?
167. What is the cost of concreting the bottom of a
circular pond 70 ft. in diameter, when concreting costs
11.87 per sq. yd. ?
168. Find the exact interest on $700 for 30 da., at 6%.
169. A merchant sold goods for $5650, with 20% and
5% discount, and 10% off for cash. Cash was paid; how
much did the merchant receive for his goods ?
170. The buyer of the goods in Ex. 169 sold the goods
for $5603.38 ; what was his % profit ? What was his
percentage profit?
171. Express 1887 by means of Roman numerals.
172. Express in t. and fractions of a t. the weight of
lead required to cover a flat roof, 147 sq. yd. in extent,
with sheet lead one-eighth of an in. thick, supposing
that a cu. ft. of lead weighs 820 lb.
173. Simplify
174. Find the value of a silver cup weighing 2 lb. 7 oz.
7 dwt. 12 gr. at $1.20 an oz.
175. Find the cost of painting the sides and bottom of
a cistern 3 yd. long, 5 ft. wide, and 3^ ft. deep at 3s. 9d.
per sq. yd.
176. Two similar boxes hold 125 lb. of sand and 216 lb.
of sand respectively ; the larger box is 36 in. long ; find
the length of the smaller box.
177. Find the bank discount on a note for $1460
payable in San Francisco 30 da. after date.
Art. 283.] EXAMPLES. 317
178. Find the trade discount on a bill of goods for
$1460 with 15% and 7% off.
179. The volume of a room is 2592 cu. ft. ; what is the
length of the room when the height is 9 ft. and the
breadth is 16 ft. ?
180. For economy which way would carpet strips run
in the room of Ex. 179 ?
181. What number is the same multiple of 354 that
86445 is of 765 ?
182. Subtract f of f from 1±- of f ; and divide the
result by (f -f>x (| - f).
183. A lidless cistern 10 ft. 6 in. long, 7 ft. 4 in. broad,
and 5 ft. 4 in. high is to be painted outside ; find the
cost at 4^ ct. per sq. ft.
184. A promissory note, written for 30 da. and payable
in Ohio at 4%, amounts to $1505.50 ; find the principal.
185. A promissory note, written for 45 da. and payable
in Ohio at 4%, amounts to $1960.40; what would have
been the amount if the note had been payable in N.Y. ?
186. When railroad 4's can be bought at 101^ (broker-
age i), how many such bonds can be bought for $7317 ?
187. A man buys $5000 of Government 4's at lllj
(brokerage \) ; what % is he receiving on his investment ?
188. A traveler purchases £500 at 4.88| (commission
3^) ; how many dollars does he pay ?
189. A train moves 6 in. the 1st sec, 1 ft. the 2d sec,
and so on for 75 sec, and then moves 37^- ft. per sec. for
1 h. ; how far does the train go in the 1 h. 1 min. and 15
sec ? (Ans. to the nearest thousandth of a mile.) The
answer lacks how many in. of the exact result ?
818 REVIEW. [Chap. XV.
190. In a decreasing arithmetical progression a = 12,
d = £, n = 50 ; find I and s.
191. The nearest of the fixed stars is roughly twenty
trillion mi. distant. Show that it would take light 3J
yr. to traverse this distance at the rate of 190000 mi.
a sec.
192. Simplify
17ixl6-3-(i + i)J-17i-S6-3xa + i)J.
193. Express Ida. 4hr. 31 min. h2\ sec. as a decimal
of 3 da. 4 hr. 5 min.
194. Find the value of 11 oz. 13 dwt. 8 gr. of gold at
$1.02 per dwt.
. 195. What will it cost to carpet a room 18 ft. long and
15 ft. wide, the carpet being 27 in. wide and costing $1.05
a yd.?
196. A ship is worth $45000. For what sum must it
be insured at $5 per $100 in order that the owner in case
of loss may receive the value of the ship and the amount
of premium paid ?
197. At what rate %, simple int., will $7600 amount
to $7676 in 3 mo. ? (No grace.)
198. What is the price of a 4% stock, if a man who
invests $4301 gets an income of $136 a year on his in-
vestment ? (Brokerage ■£-.)
199. A man bought $100 bonds at 89 and sold them at
95 (brokerage \ on each transaction) and made a profit of
>.25 ; how many bonds did he buy ?
200. Find the mean proportional between
(i) 4 and 36 ;
(ii) .25 and ll 2 ;
(iii) .64 and 1.44.
Art. 283.] EXAMPLES. 319
Find a third proportional to
(iv) 2.5 and 4.5;
(v) .7 and 7 ;
(vi) 15 2 and 5 2 .
201 . A train is traveling at the rate of 35 mi. an hr. ;
how many ft. does it go in a sec. ?
202. Simplify 2.42 - .0025 x .02 -- .055.
203. Find the rent of 375.4875 A. at $12.80 an A.
204. How many loads (en m) of gravel will be required
to cover a court-yard 20 m by 15 m to a depth of 5 cm ,
and how much will the gravel cost at 84 ct. a load ?
205.
fjf.75™. Boolon, Mom,., fan. 6, 18<?5.
Svtt&eM, cLum, att&v ciat& <J promise to pay to
the order of WLy&elfi
^^.3'cyuA, hwrucLn&cL &&v-&nty-fCv-&^^ ^Dollars
at tk& &I06-& cAatuyyual Bank
Value received.
No. 205. Due 3vtu& Lw&bcyvnA.
When was this note due ? What were the proceeds ?
206. How many gallons will fall on a sq. mi. in a
rainfall of -^ of an in., and how many t. will the water
weigh ? (1 gal. of water weighs 8.33 lb.)
320 REVIEW. [Chap. XV.
207. A man has an income of £525. 5s. from 2 J per
cent consols. He sells ont at 96^, and buys 4 per cent
Kussian bonds of £100 at95-J-. What will be the change
in his income ? (The prices include brokerage.)
208. Find the exact interest on $750 for 36 da. at
209. Two rooms of the same height are respectively
15 ft. and 20 ft. square ; what is the ratio between the
numbers of rolls of paper required for the walls of the
rooms ? For the ceilings of the rooms ?
210. The first of two similar rooms requires 94 qm of
plastering for its walls and ceiling, and is 6 m long ; how
many qm of plastering are required for the second room,
which is 7 m long? How many cum of mortar are re-
quired for the first room if the thickness of the plastering
be 1 cm ?
211. Express 4 min. 12 sec. as a decimal of a week.
212. A man sold 25 articles for the same price as he
paid for 35 ; what was his profit % ?
213. The number of oz. Avoir, in a certain weight
exceeds the number of oz. Troy by 17 ; what is the weight
Avoir. ?
214. If a number when divided by 391 leaves a remain-
der of 300, what will be the remainder when the number
is divided by 17 ?
215. A grocer bought tea at 32 ct. per lb. and sold so
as to gain 25 % ; the duty on tea was reduced, and he then
bought and sold at 4ct. per lb. less than before; what
was his gain %?
216. A man invested $38400 in 2f% bonds at 95£;
how much stock at 109 J- could he have bought with his
first semi-annual interest ?
Art. 283.] EXAMPLES. 321
217. A man embarks his whole property in four suc-
cessive ventures. In the first he gained 60%, and in
each of the others he lost 20% ; what was his total
loss % ?
218. A man spent one-third of his income on lodgings,
one-fourth the remainder on food, one-fifth what was left
on clothes, one-sixth of the remainder on books, and then
had $1200 left; what was his income ?
219. If a railroad stock pays a 7% annual dividend,
at what price must the stock be bought so as to yield 4%
on the investment ? (Brokerage as usual.)
220. A rectangular field, whose area is 1A. 65 sq. rd.,
is 137 yd. 1 ft. 6 in. long ; what is its breadth ?
221. A cistern 9 ft. long, 8 ft. broad, and 6 ft. deep is
supplied with water by a pump which will send in 27
gal. a min. ; how long will it take to fill the cistern ?
(Answer to the nearest sec.)
222. A cistern 3 m long, 2.5 m broad, and 2 m deep is
supplied by a pipe through which run 150 ' of water per
minute ; how many minutes will be required to fill the
cistern ?
223. In the centre of a room 23 ft. square there is a
carpet 18 ft. square and the rest of the floor is covered
with oil-cloth which is extended 6 in. under the carpet.
The carpet cost $2.25 a yd. and the oil-cloth cost 90 ct. a
sq. yd. ; what was the whole cost ?
224. A man leaves by will $3600 to his wife, and the
remainder of his property to be equally divided between
his four children ; and it was found that the share of each
child was one-seventh of the whole property ; how much
did the man leave ?
324 REVIEW. [Chap. XV.
244. In the centre of a square court is a square of
grass covering T 9 g of the whole area of the court, and the
side of the square of grass is 60 feet; find the cost of
graveling the remainder of the court to a depth of 3
in., the gravel and labor costing $ 1.08 a cu. yd.
245. A man buys eggs at 30 ct. per dozen, and sells
them at $2.80 per hundred; what is his gain %?
246. A grocer pays 24 ct., 30 ct., and 40 ct. per lb.
respectively for three different kinds of tea. If he mixes
weights of these teas proportional to the numbers, 6, 4,
and 3, respectively, and sells the mixture at 36 ct. per
lb., what profit does he make %?
247. A piece of work can be done in 48 da. by 15
men, but after 9 da. two of the men go away; in how
many more da. will the men who remain finish the
work?
248. By selling goods for $45.60 a man lost 5% ;
what would he have gained if he had sold for $57 ?
249. A dealer bought a certain number of articles at
the rate of 40 in a lb., and twice the same number at
the rate of 50 in a lb. He sold the whole at the rate of
36 in a lb. ; how much % did he gain ?
250. At $1.12| per sq. yd., it cost $506.25 to carpet a
room whose length is double its breadth, and whose
height is § its breadth ; how high is the room ?
251. Simplify
65J/3H 40\ /4j + 6f\ |
569K25 + 4iJ n ^7 A -3|;- 32 ^|-
252. A clock is set right at noon, but when it strikes
12 that night it is 80 sec. fast ; find how many minutes
it will gain in a week.
Art. 283.] EXAMPLES. 325
253. In a race of 100 yards A can beat B by 5 yards,
B can beat C by 5 yards, and C can beat D by 5 yards ;
how should they be handicapped for a 100 yards' race,
putting A at scratch and giving him the advantage of
any odd fraction of a foot ?
254. What is the least number which when divided by
15 leaves a remainder 9, when divided by 35 leaves a
remainder 29, and when divided by 42 leaves a remain-
der 36 ?
255. The commercial discount and interest on a cer-
tain sum for the same time and rate are $254.10 and
$252 respectively ; find the sum.
256. A man invests $9875 partly in a 3% stock at
104 J and partly in a 5% stock at 152 f, and he obtained
2-f|% on his outlay; how much did he invest in each
stock ?
257. Three persons, A, B, and C, working together com-
plete a piece of work which it would have taken them
respectively 9, 10, and 12 da. to complete if working
separately. They receive in payment $25.44, which they
are to divide in proportion to the quantity of work done
by each ; find their shares.
258. Which term of the series, 9, 12, 15, etc., is 636 ?
259. Find (90 + 2) 2 ; (60 + 7) 2 ; (40 + 8) 2 .
260. A man had 4% Railroad Preferred Stock which
brought him $664 a yr. He sold out at 119-J-, and invested
in Common Stock at 145^. The Common Stock paid 6%
dividends ; what was his gain in income per yr. ?
261. A square field contains 22 A. 80 sq. rd. ; how
long will it take a boy to run round the boundary of the
field at the rate of 12 mi. an hr. ?
326 REVIEW. [Chap. XV.
262. A man bought a certain number of eggs at the
rate of one for a ct., three times the number at the rate
of three for two ct., six times the number at 11 ct. per
dozen, and ten times the number at the rate of 16 ct. per
score, and sells them at the rate of 90 ct. per hundred,
gaining by the transaction $ 3.60 ; how many eggs did he
purchase, and what did he gain %?
263. A grocer has two sorts of tea, which cost him
64 ct. and 50 ct. per lb. respectively ; in what ratio must
he mix them so that he may gain 25% by selling the
mixture at 70 ct. per lb. ?
264. A room three times as long as it is broad is
carpeted at $1.08 per sq. yd., and the walls are colored at
18 ct. per sq. yd., the respective costs being $39.69 and
$20.16; find the dimensions of the room, making no
allowance for doors, etc.
265. A borrows from B $550 at 6% ; six da. after-
wards B borrows from C a certain sum at 6% ; A pays his
debt in 36 da. ; B pays his debt in 30 da. The interest
being the same in each case, what was B's debt ?
266. A man who held $24600 of 3±-'s sold out at 93
and purchased as many 4's as possible at 130 J. He sold
the 4's at 139§- and re-invested in 3^'s at 94. What was
the change in his annual income, and how much money
was not re-invested ?
267. Find the sum of 51 consecutive odd numbers, the
greatest of which is 117.
Art. 283.] EXAMPLES. 327
268.
$i/-670™-. -gvzznli&U, Ma**., Jb&e,. £<?, 18 <?/.
&hv&& ryvo-ntAfr a>j<t&v dat& c/ promise to pay to
the order of @Aa&. /?. &C&ld
Suva, th<yu&a,rul> &i/>G huAvcUs&oL &&v-&ntu ~ Dollars
at^^^~~£k& c^varMOyi (ZLowntAf Bank .
Value received, wM, vnt&v&bt at i5|%.
JVo. /y-2. Due jkw^ T^IoaXaav.
Discounted at 5% on Jan. 12, '92. Proceeds = ?
269.
f2600^. ftcvnycyb, 7n&. ; f<wne 6, 18 W.
&w~&ntAf ciaAfQy a^t&v dat& J promise to pay to
the order of^^^^^^^o^f fon^a,
otwo- tko-ubOAul ^iv-& h>u,yulv&cl c^yui ^Dollars
at^^XA& /aX cAaZio-nal JSank^, ffio-atvyi, 7fla&a>.
Value received.
No. /<?. Due flf&nvy cfwvCLfa.
Discounted June 19 at 4|%. Proceeds = ?
270. Find the cost in New York of a Bill of Exchange
on Paris for 4130/, when exchange is quoted at 5.16^.
328 REVIEW. [Chap. XV.
271. A person standing on a railway platform noticed
that a train took 21 sec. to pass completely through the
station, which was 88 yd. long, and that it was 9 sec. in
passing him ; how long was the train, and at what rate
per hr. was it traveling ?
272. Two trains start at the same time from A and B
and proceed towards each other at the rate of 35 and 45
mi. per hr. respectively. When they meet, one train has
gone 17^- mi. further than the other. What is the dis-
tance from A to B ?
273. A man bought a house and sold it so as to gain
5 per cent. Had he given 10 per cent more for the house,
and sold it for $129.60 more than he did sell it for, he
would have lost 2\ per cent. Find what he gave for the
house.
274. An example in multiplication was worked cor-
rectly, ancj. then all the figures except those 4 —
given were erased, and the lines show the posi- 3-
tions of the missing figures; find the missing 36
figures. l ~
6 — 3 —
275. What is the side of a square field which contains
3 A. 96sq.rd.?
276. A thin rectangular lamina of metal, 3 ft. 2 in.
long and 2 ft. 9 in. wide, has cut from its four cor-
ners four squares whose sides are 3 in. long. The four
projecting portions are turned up at right angles to
the rest of the lamina, and thus form a lidless box. Find
the capacity of the box.
277. Perform the last example after substituting dm
for ft. and cm for in. and find out the weight of pure
water the box would hold; answer in lb., etc., to the
nearest gr.
Art. 283.] EXAMPLES. 329
278. An accommodation train going at the rate of 25
mi. an hr. starts on a journey an hour before an ex-
press train which goes at the rate of 40 mi. an hr. The
accommodation train arrives 15 min. before the express.
Find the length of the journey in Km.
279. On a map on the scale of 6 in. to a mi. a rec-
tangular field is represented by a space 1 in. long and
\ in. broad ; find its area in A. Also find how many yd.
less of paling would be required to enclose a square field
of the same area.
280. A cubical block of metal of 7.84 in. thick weighs
.25 lb. per cu. in. A hole of square sectional area is to be
cut completely through the metal, perpendicular to a face
of the cube, in order that the weight of metal left may
be 100 lb. Find, to three places of decimals, the side
of the square section.
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