Preliminary mathematics

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Austin. F„ E. 1873-
F" r e 1 iminary mathematics

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PRELIMINARY

MATHEMATICS

BY

PROF. F. E. AUSTIN, B.S., E.E.

HANOVER, N. H.

Author of;

“How to Make Low-Pressure Transformers”

“Hozv to Make High-Pressure Transformers”
“Examples in Alternating-Currents”

“Examples in Magnetism”

“Examples in Battery Engineering”

“Generator and Motor Examples”

COPYRIGHT 1917

PREFACE

The ultimate aims in the study of mathematics may be to obtain a
practical knowledge of numbers or to discipline the mental powers.
The realization of either aim demands calculation, and calculation is
an art.

That book which best assists in the realization of knowledge of
numbers or in mental training must be comprehensive in principles and
extensive in details. The author of this book does not expect that this
book will displace fiction or relieve the student of occasion for mental
effort.

For many generations Algebra has followed Arithmetic in educa¬
tional and scientific courses, and the change from a study of numbers
to a system of reasoning solely conducted by symbols and letters,
gives the average student such an educational jolt as to induce dis¬
couragement, and largely contributes to the cessation of school edu¬
cation at or before the high school stage.

If this small book should prove in a limited degree a connecting link
between the study of Arithmetic and the study of Algebra for the
boys and girls of any country, the writer would consider the labor
of preparation well repaid.

While this book was designed originally for the use of those whose
educational training had been limited, it has been so remodelled as to
adapt it for the use of pupils attending the eighth grade and the high
school, or the “Junior High School.”

The subject matter up to page 77 is suitable for pupils in the eighth
grade and below, while the remaining portion of the text will prove
of assistance to pupils in the high schools.

As this book is to be used as an auxiliary, in conjunction with other
text books, many points are explained herein that are passed over in
ordinary text books. The chief object of this book is to show how
to solve problems.

Many of the examples and problems presented in this book are
original, having been evolved in the process of many years of teaching;
others have been taken from standard text books, and some have been
received for solution from students studying to pass college entrance
examinations.

One aim in the preparation of this book was to show practical
applications of the theory discussed.

The value of the book as a reference book will be found to be
greatly increased by the insertion of a comprehensive index, referring
directly to pages.

Useful tables pertaining to interest, and to weights and measures
have been added to the subject matter.

HOW TO STUDY

With mental, as with physical development, success depends upon
regular, systematic training. The athlete to become successful, is
obliged to undergo a certain period of physical training, consisting
of certain forms of muscular exercise, repeated regularly every day;
of a regular diet consisting of carefully selected foods, and an absti¬
nence from certain actions, drugs and intoxicating liquors that have
been proved, by scientific investigation to be detrimental to the human
system.

It seems logical that those features adopted in the training for
physical development, should also be adopted in training for mental
development.

When training for physical development, it is usual not to neglect
the mental condition; as is evidenced in the training of college athletes
and professionals.

On the other hand it is not advisable for those who are undergoing
mental training, to neglect their physical welfare.

Health is wealth,—not to be evaluated in the term of dollars. Active
healthy brains are not likely to continue in efficient operation in
unhealthy bodies.

The fundamental principle underlying the college education is con¬
stant systematic application. It is this required attention to daily
lessons that makes a college course a success in the development of an
educated person.

Any educational system to be in any degree a successful one, must
be logically arranged. Some college courses are more successful and
beneficial than are others, simply because of the logical sequence of
their subjects and studies.

SUCCESS is not a talent but an ACQUISITION.

All great achievements and notable professional careers, have been
possible only as the result of continuous application, in the building up
or assembling, day by day, detail upon detail, lesson upon lesson,
experiment after experiment.

The great author simply assembles a multitude of individual ideas
in the form of a story or a book.

The noted artist assembles his separate ideas on the canvas in the
shape of a painting.

The inventor has all the characteristics of the author and painter in
the assembling of his ideas of the application of the laws of Physics,
Mechanics, Chemistry, and Electricity, to make a complete, useful,
properly operating, and valuable device.

Prelitninary Mathematics

5

The greatness of all of these, results from bringing together, for
harmonious comprehension, a number of single ideas, or facts, or
detached portions of scientific knowledge.

In studying any scientific subject one should begin in a systematic
and logical manner; should observe extreme care to study a certain
definite amount each day; even if no longer than a half-hour; should
each day learn one new fact pertaining to the subject, and should
constantly think of various applications of each new fact; either as
applied to some invention, or as applied in the economic operations
of daily life.

Above all other considerations of importance, one should never
allow any event, except perchance one’s own death, to interrupt the
continuity of this daily routine.

By proper daily routine it lies within the power of any individ¬
ual to make whatever one pleases of one’s self.

The incentive to become a useful citizen or a noted professional
man must come from within the individual. Nobody can kick goodness
or greatness into you. The most noted teacher may teach you each
day during four years, but you may not learn anything as a result.
A teacher cannot make you learn a subject if you decide not to. Each
individual usually has preference as to the kind of work to be engaged
in during life. After one decides what particular branch to pursue,
then should be formulated the resolution to apply one’s energies to
become proficient in this special line.

The method adopted in the presentation of these courses is such as
to arrive at results in the very shortest time possible, without sacrific¬
ing accuracy or clearness. This may perforce mean that some students,
because of limited educational advantages in', early life, will find it
necessary to consult other books dealing with certain fundamental
processes in Arithmetic, Algebra, Geometry, and Physics. Wherever
such consultation is considered timely or valuable throughout the
courses, the names of desirable books will be mentioned.

The purchase of these books will result in the possession of a
valuable reference library.

The predominating idea in all engineering progress is utility and this
idea is prominent throughout this course.

6

Preliminary Mat he ma tics

GENERAL INSTRUCTIONS

All general instructions should be read carefully several times before
engaging in any work that is to be submitted, and should be carefully
consulted when in doubt as to the proper method to pursue.

The calculations of all problems should be plainly copied, with ink,
into the proper blank spaces reserved for them throughout the book.

The arrangements of the given “data” and the successive steps in the
mathematical process may be observed in Example 7, page 32, and in
Example 15, page 68. These Examples should serve as guides in
solving other problems.

Always read the directions and rules applying to mathematical
processes several times very carefully before proceeding to per¬
form calculations.

A problem may be defined as a question proposed, to which an intell¬
igent answer is desired. A problem is to be solved or to be worked
out. The statement of the problem usually lays down certain condi¬
tions, or states certain relations that exist between certain known
quantities. The problem is solved by obtaining the numerical value
of (or the relation of) one or more unknown quantities, from the
known quantities and their relation with the unknown.

A problem is very different from an example.

An example is a problem so solved as to show the process of solu¬
tion, step by step, serving as a guide in the solution of problems of a
similar nature.

Every individual is a problem; both a physical and a psychological
one, but not every individual can be said to be either a physical or a
psychological example: for others to copy.

DIRECTIONS APPLYING TO WORK IN PRELIMINARY

MATHEMATICS

PRELIMINARY MATHEMATICS is divided into lessons, each of
proper length to be studied and learned in about two hours by a
student of ordinary ability. Concentrated thought and continuous
application will accomplish more than quickness and flashy brilliancy.

Complete all of the problems, some marked with numerals; some
with numerals and letters, before beginning the study of a succeed¬
ing lesson.

The lessons are of increasing difficulty as they continue; but the
last lessons will require no more time than the first, if the regular
sequence is adhered to.

Preliminary Mathematics

7

It will be better to solve the problems on scrap paper, using a pencil,
and then carefully copy the complete solution step by step with ink, in
the spaces left blank for them throughout the book.

THE GREEK ALPHABET

Symbol

Name in English

English Equivalent

A a

Alpha

A a

B /3

Beta

B b

r 7

Gamma

G g

A 5

Delta

D d

E e

Epsilon

E e (short)

Z S'

Zeta

Z z

H 7]

Eta

E e (long)

0 d

Theta

Th

I t

Iota

I i

Iv K

Kappa

K k (or hard c)

A X

Lambda

L 1

M /x

Mu

M m

N v

Nu

N n

3 £

Xi

X x

0 0

Omicron

0 o (short)

IT 7T

Pi

P p

P P

Rho

R r

2 a

Sigma

S s

T r

Tau

T t

T u

Upsilon

(u)y

<p

Phi

Ph

X x

Chi

Kh

\p ^

Psi

Ps

0 to

Omega

0 o (long)

8

Preliminary Mathematics

INTRODUCTORY

Quantity; Measurement; Number

QUANTITY is that which is capable of being increased, diminished,
and measured; such as time, distance, and weight: the three most
important quantities which enter into our common experience. From
these fundamental quantities are obtained other important quantities,
as area, volume, energy, work and power.

Any quantity may be considered as made up of a number of smaller
amounts of the same quantity. These smaller amounts are called units.

A quantity is measured by the process of finding how many times the
quantity contains the smaller quantity or unit.

In measuring any quantity, the unit used in the process, must be of
the same kind as the quantity itself.

The expression denoting a quantity usually consists of two parts or
symbols; one part or symbol denotes the amount of the quantity; while
the other part denotes the kind.

The symbol denoting the amount is sometimes called the numeric or
the coefficient.

Examples: 10 seconds ; 5 miles ; 2000 pounds ; 10 apples ; 5 amperes ;
110 volts; all of these quantities named are measurable quantities and
are called concrete quantities.

It may be said then that the relation between a quantity and its unit
is an abstract number. In the preceding examples, the coefficients 10,
5 and 110 are abstract numbers. An abstract number denotes some
operation or process of measuring, and is sometimes alluded to as an
operator.

ALL UNITS ARE ARBITRARILY ASSUMED. Some units have
been chosen with some degree of reason while others have been
assumed without much reason.

The so-called “ English” system of units contains units that have been
wisely chosen as regards convenience, being in this respect far more
desirable than the so-called “Metric” system. The only point worthy
of consideration that has been urged in favor of the Metric system is
that it is a decimal system ; that is, the units increase by ten or multi¬
ples of ten, and decrease by “ tenths” or sub-multiples of ten.

1 he fundamental unit in the English system is the foot, which is
a very convenient length for all common transactions in our daily
experience.

1 he foot being divided into twelve equal parts gives the inch which
is a very convenient length for purposes for which the foot is too long.

The yard, which is three feet in length is also a very convenient unit.

Preliminary Mathematics

9

The fundamental unit in the Metric system is the meter, which is
equal to very nearly (ninety-one and four-tenths), 91.4 inches.

The one-hundredth part of a meter is a small unit called the centi¬
meter, which is altogether too small for ordinary uses. The decimeter
would be one-tenth of a meter, or ten centimeters, which is a very in¬
convenient length for ordinary use.

The argument that 10 is a more convenient number for common use
than 12, is questionable. For example the number 10 is divisible by
1, 2, 5 and 10, an even number of times without a remainder: The
number 12 is divisible by 1, 2, 3, 4, 6 and 12, without a remainder: two
more divisors than 10 has. Take the number 36 (36 inches in a yard).
This number is divisible by 1, 2, 3, 4, 6, 9, 12, 18, and 36, without a
remainder. The number 100 is divisible by 1, 2, 4, 5, 10, 20, 25, 50 and
100. Here is a number nearly three times as large as 36, which only
has the same number of divisors. The number 72, which is twice 36,
has twelve divisors.

As regards performing mathematical operations, if operations should
be performed, employing common fractions, much labor would be
saved, and more satisfactory results obtained than when the decimal
system is employed, with its repeating decimal fraction.

Examples of this will be given from time to time, which should be
carefully noted.

As a matter of fact each country fixes by law the standard units for
the use of its inhabitants.

The standard units are kept very carefully either at the capitol or at
an important city of each country.

For example, the yard is fixed by United States law as the distance
between two marks on a certain bar of metal, when at a certain tem¬
perature. One-third of this standard distance is called one foot.

The yard is the standard of length, but the FOOT is the common
unit of length.

Standards of volume and weight are fixed in a similar manner.

While the use of the metric system of units is legal in the United
States the system is not the commonly adopted system.

The standard length, adopted by the French, is called the meter;
which in reality is the distance between two marks on a certain bar of
platinum-iridium, carefully kept in the International Metric Bureau
at Sevres, near Paris.

The commonly adopted unit of length is T f )0 of the meter, and is
called a centimeter, tcmtt) of a meter is called a millimeter. T -„Vo of
a meter is A of a centimeter.

10

Preliminary Mathematics

LESSON I

SYMBOLS OF NUMBER AND NOTATION

The symbols usually employed to denote number are the Arabic
numerals, 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0, and the letters of the English
alphabet. The Arabic numerals or characters are alluded to as figures;
sometimes called digits. The tenth character as printed above, is called
“cipher” or “zero”, and denotes a quantity too small to be measured.
In direct contrast to 0 another symbol, 00 is used to denote a quantity
that is too large to be measured; called infinity. The sign or symbol
00 denotes a quantity greater than any that can be denoted by any
arrangement of the Arabic numerals.

The numerals or figures denote known or definite numbers; and
mathematical operations or calculations, employing the numerals, con¬
stitutes the science of Arithmetic..

The letters of the alphabet may denote any numbers or values what¬
ever ; either known or unknown, and mathematical operations, or
processes, employing letters, or a combination of numerals and letters,
constitutes the science of Algebra.

The method of calculation is identical for both Arithmetic and
Algebra, and many of the symbols adopted are common to both. In
both these subjects the various symbols employed, represent numbers,
their relation with one another, and the operations performed with
them.

The use of symbols lessens mental labor and requires much less
space in printing or writing.

Everyone is familiar with many symbols met with in daily experi¬
ence.

Any quantity may be designated by a symbol, such as a letter of the
English alphabet or a letter of the Greek* alphabet, or by any other
conventional sign. The sign $ placed before the number 10 means a
certain amount of money; $10, denotes ten dollars.

1 he flexibility of the system employing the Arabic numerals is won¬
derful. By properly arranging these 10 symbols, the largest and the
smallest conceivable number may be expressed.

I he following arrangements of the figure 1 and the symbol 0 are
important.

* See page 7 for Greek alphabet.

Preliminary Mathematics

11

10 denotes ten.

100

1,000

10,000

100,000

1,000,000

10,000,000

100,000,000

1,000,000,000

u

((

u

u

ii

u

u

one hundred.

one thousand.

ten thousand.

one hundred thousand.

one million.

ten million.

one hundred million.

one billion; or one thousand million.

The use of the decimal point (see page 26) with this system still
further adds to simplicity of expression.

Example 1: Suppose a collection of Arabic numerals denotes a
number as : 137896 ;

which translated into common language is, one hundred and thirty-
seven thousand eight hundred and ninety-six. The figures present a
brief expression; yet everyone who reads can understand its meaning.
Note the time and the ink, as well as the physical and mental exertion
saved by employing the symbols or the figures.

The symbols as arranged, should properly be put into English as
follows :

One hundred thousand, and thirty thousand, and seven thousand, and
eight hundred, and ninety, and six.

As an arithmetical process the arrangement would indicate:

100000

30000

7000

800

90

6

137896

This arrangement will be referred to later under exponents, page 32.

To denote a number less than 1, the Arabic numerals are arranged
one over the other and the arrangement is called a fraction.

i denotes one-half of one; which is the reciprocal of two.

| denotes one-half of three; or three halves of one; or simply three
halves.

:V denotes one-third of one.

To denotes one-tenth of one.

h°- denotes one-third of ten; or ten-thirds of one; or simply ten-thirds.

12

Prelimina ry Mathcma tics

TiryVo« therefore denotes a comparatively small number; as com¬
pared with 1.

Suppose a decimal point is placed between the 7 and the 8 as:
137.896.

Then the meaning is: one hundred and thirty-seven, and eight hun¬
dred and ninety-six one thousandths; which may be expressed in sym¬
bols as :

137

8 9_6 •
1 0 0 0 ?

or as:

1 3 789 6 — n 3 7 0 0 0 I 8 9J3 \

1 0 0 0 V 1 0 0 0 I 1 0 0 0 )•

Other examples are:

i ooo. oo i=i ooo 1 0 \ T o — 1 ‘W’o Q o” 1 ;

100.01 =ioo f * 0 =-W;

10.1 =io*=w.

PROBLEM 1: Express the following in the blank spaces using
Arabic numerals or characters:

Seven thousand and twenty-one.

Answer.

PROBLEM la: One million, twenty-seven thousand, and forty-
two.

Answer.

PROBLEM lb: Sixty-six million, sixty thousand, and sixty.

Answer.

PROBLEM lc: Seven hundred ninety-six, and twelve one hun¬
dredths. (decimal.)

Answer.

PROBLEM Id: Seven hundred ninety-six, and twelve thou¬
sandths. (decimal.)

Answer.

PROBLEM le:

Arabic numerals:

Three-fourths.

Answer

Express the following in the blank spaces, using

Eight-ninths. Two-tenths.

Answer Answer

PROBLEM If: Fourteen
thousandths.

Answer.

one-hundredths. Twenty-seven

Answer

one-

Preliminary Mathematics

13

LESSON II

USE OF LETTERS IN MATHEMATICS

The use of the letters of the English alphabet in mathematical pro¬
cesses will be explained more fully later on, and will be used very
extensively in all the various subjects of the different courses. Suffice
it to say if the following numerical values be assigned to the given
letters.

a=l
6=2
c=3
d —4
c=5
f=6
9=7
h —8
i— 9

then 137896 could be expressed by aeghif; but unfortunately when
letters are used to denote numbers their arrangement has a very
different meaning attached to it, than has the corresponding arrange¬
ment of numerals. According to the adopted method of expressing a
number by the use of letters in algebraical operations,

aeghif with the numerical value assigned to each letter as above,
would indicate:

282,480; (flX C X gX hXiXf)

being the product of the numerals denoted by the individual letters.

That is, in Algebra, be means b multiplied by c, and substituting the
numerical equivalents assumed above, be would equal 2 times 3, or 6.

Compare with the arrangement of numerals as explained on page 20.

The Roman Numerals I, II, III, IV, V, VI, VII, VIII, IX, X,
corresponding to the Arabic numerals on page 10, cannot be used con¬
veniently in mathematical processes.

Expressing 1918 in Roman numerals as

MDCCCCXVIII

will serve to show how cumbersome is the mere expression of a num¬
ber, using this system, not to mention the complexity of multiplying
the date as expressed in the Roman system, by IV. In the Roman sys¬
tem in addition to the equivalents already given, M denotes 1000; D
denotes 500; C denotes 100; and L denotes 50. To express 600 it will
be necessary to write DC.

14

Preliminary Mathematics

The following is a statement recently issued:

Washington, Oct. 9. Secretary Mc-
Adoo today instructed the supervising
architect of the treasury to use Arabic
instead of Roman numerals on all public
buildings.

The order was issued because of the
difficulties the average citizen finds in
quickly interpreting Roman numerals.

Example 2. Using the Roman numerals as characters, the number
2241 would be expressed by

MMCCXLI

Attention may be directed to the position of the characters relative
to each other as determining the value of a number.

For example: IV means 4, while VI means 6.

IX means 9, while XI means 11.

XL means 40, while LX means 60.

A table showing the comparative arrangement of some of the Arabic
and Roman numerals is given below:

1,

I.

21,

XXI.

41, XLI.

61,

LXI.

2,

II.

22,

XXII.

42, XLII.

62,

LXII.

3,

III.

23,

XXIII.

43, XLIII.

63,

LXIII.

4,

IIII or IV. 24,

XXIV.

44, XLIV.

64,

LXIV.

5,

V.

25,

XXV.

45, XLV.

65,

LXV.

6,

VI.

26,

XXVI.

46, XLVI.

66,

LXVI.

7,

VII.

27,

XXVII.

47. XLVII.

67,

LXVII.

8,

VIII,

28,

XXVIII.

48, XLVIII.

68,

LXVIII.

9,

IX.

29,

XXIX.

49, XLIX.

69,

LXIX.

10,

X.

30,

XXX.

50, L.

70,

LXX.

11,

XI.

31,

XXXI.

51, LI.

71,

LXXI.

12,

XII.

32,

XXXII.

52, LII.

72,

LXXII.

13,

XIII.

33,

XXXIII.

53, LIII.

73,

LXXIII.

14,

XIV.

34,

XXXIV.

54, LIV.

74,

LXXIV.

15,

XV.

35,

XXXV.

55, LV.

75,

LXXV.

16,

XVI.

36,

XXXVI.

56, LVI.

76,

LXXVI.

17,

XVII.

37,

XXXVII.

57, LVII.

77,

LXXVII.

18,

XVIII.

38,

XXXVIII.

58, LVIII.

78,

LXXVIII.

19,

XIX.

39,

XXXIX.

59, LIX.

79,

LXXIX.

20,

XX.

40,

XL.

60, LX.

80,

LXXX.

90, XC.

100, C.

200, CC. 300, CCC. 400,

cccc.

500, D.

1000, M.

2000, MM.

Prel im in a ry M at hematics

15

PROBLEM 2: Express the following numbers in the blank
spaces, using Roman numerals.

450. Two hundred and twenty.

Answer. Answer.

PROBLEM 2a:

800 .

Answer.

PROBLEM 2b:

2582.

Answer.

PROBLEM 2c: Using Roman Numerals, express

1885. 1915.

Nine thousand nine hundred and ninety-nine,

Answer.

Answer.

Answer.

16

Preliminary Mathematics

LESSON III

algebraic symbols

The symbols employed in Arithmetic and in Algebra may be sepa¬
rated into four classes as follows:

1. Symbols denoting Quantity.

2. Symbols denoting Relation.

3. Symbols denoting Operation.

4. Symbols denoting Abbreviation.

The symbols denoting quantity and operation have already been
discussed to some extent. The other symbols will be considered as

follows:

Any arrangement of letters with algebraic symbols is called an
Algebraic Expression.

Thus a+b; 5a— 2b; az+bx-cy are all algebraic expressions.

Numerical expressions might be; 5 + 17 = 22; l + s-T- and
86—5—19=62, in which no letters are used.

THE SIGN OF EQUALITY, = is expressed in words as “equals’
or “is equal to”, a = c denotes that a number denoted by a is equal
to another number denoted by c. The expression denoting the equal¬
ity of two or more numbers or two or more quantities is ca e an
“equation”. Thus x — y = 36, is an equation ; as is also a - c, an

xy=4p.

The SIGNS OF INEQUALITY, > and < are sometimes em¬
ployed. > means “is grea ter than” and < means is less than . I he
symbol of inequality always points towards the smaller quantity.

For example: a > b means, a is greater than b.
while ci b means, ci is less them b.

In a given problem, it often happens that numbers occupy corre¬
sponding relations, while differing in value. In such cases, a may
denote one value, a' another, and a” another. These are read:

a'; a prime.

a"; a two prime ; or a double prime,
a"'; a three prime.

Sometimes letters may be used with subscripts, as.

a t ; read “a sub one”.
a 2 ; read “a sub two”.
a a ; read “a sub three’.

Preliminary Mathematics

17

PROBLEM 3: Express the following in the form of a numerical
expression, observing the proper consecutive order of arrangement.
An individual received 250 dollars; spent 75 dollars; earned 15 dol¬
lars; lost 17 dollars. How much money has he to his credit?

Answer.

PROBLEM 3a: Express the following as an algebraic expres¬
sion; To a is added c, and subtracted b; giving as a result, b, less a,
less c.

Answer.

PROBLEM 3b: Express the relation, using the proper symbols;
that 16 is equal to a.

Answer.

PROBLEM 3c: That 16 is greater than b.

Answer.

PROBLEM 3d: That 4 is less than 5.

Answer.

PROBLEM 3e: That x is less than 25.

Answer.

PROBLEM 3f: That 481 is less than y.

Answ r er.

18

Preliminary Mathematics

LESSON IV

SYMBOLS OF OPERATION

In almost all mathematical processes, whether employing figures or
letters, time may be saved by adopting certain symbols denoting
“operation”.

Some of the common symbols denoting mathematical processes or
operations are as follows :

THE SIGN OF ADDITION:

Read “plus”.

25—{-7 means that to twenty-five is added seven, making their sum
thirty-two.

a-{-b means that a number denoted by a, is to be added to another
number denoted by b. If a denotes, or stands for 25, while b denotes
7, then a-\-b=32. a-^-b-^-c means that a number denoted by a is to be

added to a number denoted by b and their sum is to be added to a
third number denoted by c.

When numbers are added together the resulting number is called
their sum.

The order in which numbers are added, will make no difference in
the numerical value of their sum.

THE SIGN OF SUBTRACTION:

Read “minus”.

25—7 means that from the number twenty-five is taken the number
7, making their difference eighteen.

a—b means that from a number denoted by a is taken or subtracted
a number denoted by b. If a denotes 25 and b denotes 7 then a —fr=18.

The number from which another number is taken is called the
minuend; while the number taken away is called the subtrahend.

I J re limiua ry M at lie m atics

19

THE SIGN OF MULTIPLICATION:

Read “times” or “multiplied by”.

25 X 7 means that 25 is taken seven times; or is multiplied by
seven.

Multiplication is simply a short method of performing addition.
If the number 25 is written down seven times, and all seven of die
twenty-fives added together, the result is :

25

25

25

25

25

25

25

175

The same result would have been obtained had seven been added
twenty-five times.

The addition of the seven twenty-fives is further illustrated by:

25

25

25

25

25

25

25

35

140

175

Seven fives are first added together, giving 35; next seven twenties
are added together, giving 140, and the two sums added, giving 175.
Expressing the operation as one of multiplication gives :

25

7

35

140

175

20

P retimina ry Mat he ma tics

The result obtained by multiplying two or more numbers together
is called the product, while the numbers which are multiplied to¬
gether are called “factors” of the product. aXb means that a num¬
ber denoted by a is to be multiplied by a number denoted by b. If a
denotes 25; while b denotes 7, aXb=\75.

aXbXc indicates that the product of a multiplied by b is to be
multiplied by c. a, b and c are called the factors of the result.

Suppose a denotes 25, b denotes 5, and c denotes 2; then aXbXc—

25x7x2=350.

In the operation of multiplying two numbers, the number to be
multiplied is called the “multiplicand”, while the number by which the
multiplicand is multiplied is called the “multiplier”.

As indicated above, 25 is the multiplicand and 7 is the multiplier.
It is true that 7 might be the multiplicand and 25 the multiplier; but
it is common practice to call the larger number the multiplicand, and
the smaller number the multiplier.

When letters are used to denote numbers, as in Algebra, the sign
of multiplication is usually omitted; thus abc means the same as
aXbXc.

When expressing a number by using the Arabic numerals, it is the
sign of addition that is omitted; the sign of multiplication is never
omitted in arithmetical expression.

A point is used by some writers to denote multiplication instead
of the sign X. In such cases 4*5*3 means 4x5x3. This is not a
practice to be encouraged. The word “of” is sometimes used instead
of the multiplication sign X, to denote multiplication. This word
“of” is 1 used very often when multiplying fractions. For example;
instead of writing §X90, the expression f of 90 is used.

The saving effected by the common method of multiplication is
apparent when dealing with large numbers. Suppose the problem is
given to multiply 5280 by 123. By addition it would be necessary
to write down 5280 one hundred and twenty-three times, forming a
column, and then adding. By multiplication:

5280

123

15840

10560

5280

649440

Preliminary Mathematics

21

The partial products are shown as follows:

5280

123

240=80X3
600=200X3
15000=5000X3
1600=80X20
4000=200X20
100000=5000X20
8000=80 x 100
20000=200X100
500000=5000X100

649440 649440

The Italian Short Proof Method of Checking Multiplication. It is
sometimes of value to be able to quickly “check” the result of multi¬
plying two large numbers by each other.

One method is called the “ Italian Short Proof Method” and is
explained as follows :

Suppose 172856 is to be multiplied by 375, expressed as;

A.172856 Add digits giving 29. Divide 29 by 9 giving 3 and remainder of 2.

B.375 Add digits giving 15. Divide 15 by 9 giving 1 and a remainder of 6.

864280

1209992

518568

C.64821000 Add digits giving 21. Divide 21 by 9 giving 2 and a remainder of 3.

Proof. In line A, add together all the digits, divide the sum by 9
and place the remainder in the square designated by 1, figure 1.

} 3X5280= 15840

J

| 20X5280=105600
^-100 X 5280=528000

In line B, add together all the digits, divide the
sum by 9, and place the remainder in the square
designated as 2 in figure 1.

Multiply the number in square 1 by the number in square 2, divide
the product by 9 and place the remainder in square designated as 3.
In this case 2X6=12 and 12 divided by 9=1; with a remainder of 3.

22

Preliminary Mathematics

In line C, add together all the digits, divide the sum by 9, and if the
multiplication has been properly performed, the remainder will be the
same as the number placed in square designated by 3.

Place the remainder in square designated by 4. It happened in this
particular case that the remainder number in square 3, was 3.

American Short Proof Method of Checking Multiplication.
Another method of proving the result or product of multiplying two
numbers together, which might be called the “ American Short Proof
Method,” is as follows:

Suppose 24532 is to be multiplied by 13685.

A.24532 Adding digits gives 16; adding again; 1+6=7

B.13685 Adding digits gives 23; adding again; 2+3=5

122660 Multiplying 7 by 5=35

196256 Adding 3 and 5=8

147192
73596

24532 D-E

C.335720420 Adding digits gives 26; adding again gives 8.

Proof. Add the digits in the multiplicand; designated by A. In
this case giving the number 16. Again add digits giving 7.

Add the digits in the multiplier B, giving in this case 23. Again
add digits, giving 5. Multiply 7 by 5 giving 35 and adding digits 3+5
giving 8.

Add all the digits in the product C, giving in this case 26. Again add
digits 2+6 giving 8.

Always add digits until a number containing only one figure results.

If the result of the operations indicated above the line D E is the
same as the result of the operations indicated below the line, the multi¬
plication has been performed correctly.

PROBLEM 4: Express the following, using Arabic numerals and
symbols. Fourteen, plus ten, minus eight, multiplied by sixty.

Answer.

PROBLEM 4a: Fivd times five, minus twenty, times two, plus
ten, minus twenty.

Answer.

Prel itn inary Mathematics

23

PROBLEM 4b: Write the numerical result of the following:

Ten times four, minus fifteen, plus five, times five, minus one hundred
and fifty.

Answer.

PROBLEM 4c: Minus twenty-five, plus one hundred and fifty,
times two, times twenty-two, plus twenty-five.

Answer.

PROBLEM 4d: Multiply 155 by 5, and check the product by ad¬
dition. That is, prove by addition that the result is correct.

Answer.

PROBLEM 4e: Multiply 15 by itself and check the result by the
process of addition.

Answer.

PROBLEM 4f: Multiply 999999 by 888 and check the result by the
Italian short proof method.

Answer.

PROBLEM 4g: Multiply 123456789 by 4321 and check the result
by both the Italian and the American short proof methods.

Answer.

24

PreIiminary Ma th cmatics

LESSON V

THE SIGN OF DIVISION:

Read “divided by.”

25-^5 indicates that 25 is to be divided into 5 equal parts. This
operation may be also expressed by ^

Applying to Algebra, a-i-b indicates that a number denoted by a is
to be divided by a number denoted by b. This may also be expressed

by - which does away with the symbol of division.
b

Division is a short method of performing subtraction. (Note the
relation between multiplication and division.)

For example: suppose 1492 is divided by 2, giving 746; 1492 minus
746=746. Again suppose from the number 25 is subtracted the number
5, giving 20; from which is subtracted 5 giving 15; from which is sub¬
tracted 5 giving 10, and from which is subtracted 5 giving as a result
5. This operation proves that 25 divided by 5 is equal to 5.

It is thus evident that division is an exact reverse operation from
that of multiplication.

When a given number is divided by another number, the given num¬
ber is called the dividend, the other number is called the divisor, while
the result is called the quotient. Proof of the correctness of the re¬
sult of division: Multiply the quotient by the divisor, which should give
as a product the same numerical value as that of the dividend.

Example. Divide 175 by 7.

25

7)175

14

35

35

To prove that the quotient is correct, multiply 25 by 7 which gives
as a product, 175.

Preliminary Mat he mat i c s

25

PROBLEM 5: Divide 649,440 by 123, and prove that the result is
correct.

Answer.

PROBLEM 5a: Divide 11190 by 746, and prove the result.

Answer.

PROBLEM 5b: If the divisor is 746 and the quotient is 30, find
the dividend. Prove result.

Answer.

26

Preliminary Mathematics

LESSON VI

DECIMALS

The word decimal is derived from the Latin word decimus, mean¬
ing tenth.

A decimal system would be a system based upon tenths. The Metric
system is such a system.

To facilitate expression, a point or period is employed to denote the
decimal relation of numbers.

For example:
jV) may be expressed 0.1.
ih by .01.
too it by .032.

20 /u by 20.08.

The decimal point, to a certain extent replaces the common frac¬
tion. Some fractions cannot be expressed in absolute value, as a
decimal, as may be demonstrated by such a fraction as l, which can

only be approximately expressed as a decimal as 0.33333 .(See

page 38 for explanation of the sign of continuation .) This may be
illustrated as follows:

-l (one-third) = 1 divided by 3.

.3333 etc.

3JT0000

9

10
9

10

9

10

9

The table on the next page gives the decimal equivalents for a few
of the common fractions.

Preliminary Mathematics

27

TABLE OF DECIMAL EQUIVALENTS OF
COMMON FRACTIONS

1

l> 4

—

.015625

5

1 6

or

1 0
3 2

or

2 0
6 4

—

.3125

JL

32

or

2

(ft

mm

.03125

8

8

or

12
3 2

or

2 4

(5 4

—

.375

3

0 4

mm

.046875

7

1 8

or

1 4

3 2

or

2 8
6 3

—

.4375

JL

1 l>

or

o

3 2

—

.0625

1

•>

or

1 6
3 2

or

3 2
6 4

mm

.5

3

3 2

—

.09375

<1

1 0

or

1 8
3 2

or

3 (i
6 4

—

.5625

1

8

or

4

3 2

or A

=

.125

5

8

or

1 0

1 (!

or

2 0
3 2

mm

.625

9

i) 4

mm

.140625

1 1

1 6

mm

.6875

5

3 2

or

1 0
0 4

mm

.15625

8

4

—

.75

l 1
114

mm

.171875

1 8
l

—

.8125

3

1 6

or

6

3 2

or a

—

.1875

7

8

—

.875

■h

or

1 4
0 4

—

.21875

1 5

1 <»

mm

.9375

i

4

or

8

3 2

or It

—

.25

The application and use of the decimal point may be comprehended
by examples as follows:

0.1= T V “one-tenth” (equal to ten “one hundredths”)

.01= T lo “one hundredth” (equal to ten “one thousandths”)

.001= , don “one thousandth” (equal to ten “ten thousandths”)

.0001= , oo 0 o “one ten thousandth” (equal to ten “hundred thousandths”
.00001= iooVou “one hundred thousandth” (equal to ten “millionths”)
.000001= 10 0 0 0 00 “one millionth”

Applying the Arabic numeral system, it is evident that:
0.2=two-tenths ; 0.3=three tenths; 0.4=four-tenths; etc.

.02=two hundredths; .03=three hundredths; etc.

.002=two thousandths ; .003=three thousandths; etc.

Next combining tenths and hundredths:

0.23=twenty-three hundredths; that is two-tenths, being equal to
twenty “one hundredths” added to three hundredths, gives twenty-
three “hundredths”.

Likewise; 0.235 means two hundred and thirty-five, “thousandths”;
since “two-tenths” equals two hundred “one thousandths” , and “three
one hundredths” equals thirty “thousandths ”.

It is evident from the foregoing explanation that decimals have to
do with tenths and submultiples of tenths.

28

Prelim inary M a t hematics

The multiplication of decimals by numbers that are not fractions,
called “integers” or “whole numbers”, is important and several ex¬
amples will be given.

Example 6: Multiply six hundred and twenty-five ten thousandths,
by eight.

.0625

8

.5000

Example 6a: Multiply fifteen thousand six hundred and twenty-
five millionths by thirty-six.

.015625

36

93750

46875

.562500

Compare the decimal values of ^ and T 9 6 on page 27. The follow¬
ing example shows how to multiply one decimal by another decimal.

Example 6b: Multiply one hundred and twenty-five thousandths,
by twenty-five hundredths.

.125

.25

625

250

.03125

Rule for “pointing off” or for locating the position of the decimal
point in the product of two decimals.

Beginning at the right of the product “point off” as many figures
(digits) as there are decimal places in both multiplicand and multi¬
plier. In the last example there are three decimal places in the mul¬
tiplicand, and two places in the multiplier: a total of five places. So
beginning at the right hand figure (which is 5) of the product, count

Preliminary Mathematics

29

to the left five decimal places, and place the point to the left of the
fifth figure. In this case there were only four figures, so a cipher
must he added as indicated. Had there been only three figures, then
two ciphers would have been added.

In this case compare the multiplication of the decimal equivalent of
% and J4, page 27.

.03125

lX}=h- ;l V= 32)1.00000

96

40

32

80

64

160

160

The multiplication of decimals may be proved, the same as the
multiplication of other numbers, by dividing the product, by the mul ¬
tiplier, which should give the multiplicand.

The division of a decimal number by an integer has already been
shown, on pages 26 and 29. The proper location of the decimal point
is evident from these examples.

The division of one decimal number by another decimal number is
illustrated by:

Example 6c : Divide three thousand one hundred and twenty-five,
hundred thousandths, by one thousand eight hundred and seventy-five,
ten thousandths.

1166....

.1875).0312 500
187 5

12500

11250

12500

11250

30

Prel im in ary M at hematics

The quotient in this case is an example of a so-called “repeating
decimal” a number that cannot be expressed by a proper or a com¬
mon fraction.

The further the process of division is carried out, the more sixes
will appear in the quotient.

It is interesting to “ prove” such an example:

.1875

.166

11250

11250

1875

.0311250

The product differs from the dividend after the first two figures.

Had the division been carried to .16666 and the proof applied, the
product would have been more nearly like the dividend. It would re¬
quire an “infinite” number of sixes to prove exactly.

PROBLEM 6: Express as a decimal number, fifteen thousand six
hundred and twenty-five, hundred thousandths.

Answer.

PROBLEM 6a: Express as a decimal, I„, and express the an¬
swer in words.

Answer.

PROBLEM 6b: Multiply .004 by 1000.

Answer.

Preliminary Mathematics

31

PROBLEM 6c: Divide 0.625 by .015625.

Answer.

PROBLEM 6d: Express 0.328125 as a common fraction.

Answer.

Note.

As a result of extended experience, it may be stated that col¬
lege students make more errors in placing the decimal point than
in any other operation. Students should exercise the utmost care
in all mathematical processes involving decimals.

PROBLEM 6e: Express 18as a fraction.

Answer.

32

Prelim in ary Ma t hematics

LESSON VII

MULTIPLICATION OF POSITIVE AND NEGATIVE

NUMBERS.

The rule for ascertaining the “sign” to prefix to the product of two
numbers is very simple, but should be firmly fixed in mind, since it
applies in both Arithmetic and in Algebra.

The product obtained by multiplying two numbers with unlike signs
before them should be preceded by a minus sign. If the two numbers
are preceded by like signs, the sign of the product is plus.

Example 7: Multiply —J—55 by —25.

+55

—25

275

110

— 1375

Multiply —55 by —25.

-55

—25

275

110

+ 1375

The same rule applies to algebraical multiplications ; a multiplied by
—b = —ab ; — a multiplied by — b = ab.

EXPONENTS.

If a number is multiplied by itself any number of times, the result
or product is called a “power” of that number.

If 25 be multiplied by itself, the result, 625, is said to be the second
power of 25. If 25 be multiplied by itself, and the product again by
25, the result, (625 X 25)=15625, is called the third power of 25.

The third power of 3 is 27.

The sixth power of 2 is 64.

The third power of 10 is 1000.

P reliminary Mat he matics

33

In order to indicate to what power any number is raised, a small
figure is placed to the right and slightly above the number, to indicate
the power ; or to indicate how many times the number is to be multi¬
plied.

This small figure is called an “exponent” or index, and may be a
negative as well as a positive number, and may also be a fraction.

For example:

5 2 =5 x 5=25.

5 3 =5 x 5 x 5=125.

4 2 =4 X 4=16.

4 3 =4 X 4 X 4=64.

3 2 =3 X 3=9.

3 3 =3 X 3 x 3=27.

The powers of 10 are of considerable importance in engineering
calculations and some explanations will be made regarding them.

10 1 =10*, read simply 10; or ten to the first power.

10 2 =10 X 10=100, read “ten squared”.

10 3 =10x 10 X 10=1000, read “ten cubed”, or ten to the “third power”.

10 4 =10x 10 X 10 X 10=10000, read ten to the “fourth power”.

10 6 = 1,000,000 (one million), read ten to the “sixth power”.

10 8 = 100,000,000 (one hundred million), read ten to the “eighth

pozver”.

10 9 = 1,000,000,000 (one thousand million or one billion), read ten to
the “ninth power”.

Suppose a given number is 42,000,000; by adopting the “power of
ten” method of expressing the number, it would be 4.2 X10 7 .

It saves space and inlo to write 10 with its exponent instead of
printing or writing all the figures denoting the larger numbers, hence
the following rule regarding the expressions for powers of ten.

Rule. In order to zvrite out the number that is expressed by 10
zvith any exponent, add as many zeros to 1 as indicated by figure of
the exponent.

This rule is illustrated by the powers of 10 indicated above; 10 9 is
expressed by 1 with nine ciphers after it.

*\Yhen no exponent is written, 1 is always understood.

34

Preliminary Mathematics

In Algebra the exponent has the same meaning as when used in
Arithmetic.

For example:

a 2 means “a squared” or "a to the second power”, same as a X a.

a 3 means “a cubed”, or “a to the third power”, same as a'Xa'Xa.

a 1 is the same as a, “a to the first power”.

Rule. When multiplying two or more similar numbers by each
other, each having exponents, add their exponents together.

For example:

10 2 X 10 4 X lO^lO 1 ^ 100,000,000,000 : or one hundred billions.

Applying the rule to literal factors:

a 2 X a 5 X a 1 =a s .

Suppose it is given that 4.2 X 10 7 is to be multiplied by 5 X 10 4 ; then
5X4.2X 10 7 X 10 4 is the arrangement, and the result will be 21.0 X10 11
or 2.1 X 10 12 . Considerable time is saved in the operation of multiply¬
ing, by adopting the power of 10 expression.

Rule. When dividing one number by another similar number, but
each having different exponents, subtract exponents.

Example 7. Divide 25 3 by 25 -2 .

Subtracting exponents gives -f~ 3 — ( —2) — 5.

Therefore 25 3 X 25 +2 =25 s =9,765,625.

FRACTIONAL EXPONENTS. Exponents may be fractional as
well as integers.

j.

4‘ means four raised to the one-half power, and is often expressed
by use of the so-called "radical sign” V.

Using this sign, 4" is the same as V 4. The one-half power of a
number is also called the “square root” of the number.

The cube root of a number is expressed by the fractional exponent

or by the sign V.

A few numerical examples are given as follows:

25-= V 25 = ±5; V~625 = ±25; ^l25~= 5 ; ^15625 = 25.

Preliminary Mathematics

35

PROBLEM 7:

Multiply —25 by 25 and the result by —25.

Answer.

PROBLEM 7a:

Multiply 5 3 by 4 2 .

Answer.

PROBLEM 7b: Divide 5 5 by 5 3 .

*

Answer.

PROBLEM 7c: Using the “pozver of ten” method, express the
product of 1.5X10 6 by 9.2 X10 2 .

Answer.

36

Preliminary Mathematics

LESSON VIII

NEGATIVE EXPONENTS

RECIPROCAL. The reciprocal of a number, is 1 divided by that
number (Refer to page 10).

The reciprocal of 10 is of 25 is ^; of x is

The reciprocal of 64, 32, 16, 8, 4 and 2 are shown as fractions in
the table on page 27.

An application of reciprocals to powers may be alluded to here;
since it becomes necessary at times to deal with numbers having nega¬
tive exponents. Suppose a -1 is given in some mathematical process.
The question naturally is asked how can a number be raised to a
negative power?

Any number having a negative exponent is the same as the recipro¬
cal of that number with a positive exponent having the same numerical
value as the original negative exponent. That is:

1

; or a

a

In proof of this, apply the rule for finding the exponent of the
product of two or more powers of any number: (See page 34.)

Then a" 1 = a -2 X a 1 .

Dividing both members of this equation by a" 2 gives:

-1

-1

a

a

^a 1 which is the same as —— ^a 1 ; obtained

or

-2

-2

a

a

a

1

by dividing both numerator and denominator by a 1 ; or — = a -1

a

This proof may be better understood after studying Lesson XII,
page 53.

SYMBOLS DENOTING ABBREVIATION.

In mathematical processes or operations much time and labor is
saved by employing certain signs or symbols denoting a long or
complicated process or numerous short processes.

THE SIGNS OF AGGREGATION OR COLLECTION are,

the parentheses, ( ) ;

the braces, ^

the brackets, [ ] ;

and the vinculum,

1

Preliminary Mathematics

37

all of which indicate that, the numbers included by them are to be
considered collectively, as a unit.

For example:

(a-fb)c;

j a + b | c;

[ a-|-b] c;
a+b X c,

all mean that the sum resulting from adding a to b is to be multiplied
by c.

Each of the above four expressions may of course be written
ac-f-bc.

Further examples of aggregation are:

(a+b) 2

[a+b] 2

a+b*

each of which denotes that the sum of a and b is to be multiplied by
itself or is to be squared.

The following example shows the use of the signs of aggregation.

Example 8.

2b [8 (a+b) 2 —5a] =2b [8 (a 2 +2ab+b 2 ) —5a.]

2b [8a 2 +16ab+8b 2 —5a] =16a 2 b+32ab 2 +16b 3 —lOab.

The student should be very careful indeed in all calculations where
these signs of aggregation are employed. Very many mistakes are
made by not observing the proper relations of numbers with which
these symbols are used.

THE SIGN OF DEDUCTION:

Example:

ab = 10;

1 o

Read “ therefore” or “hence”.

38

Preliminary Mathematics

THE SIGN OF CONTINUATION;

read “and so on”.

Example:

*> x + y, x-\-2y, x-\-3y .

read x, x -f- y, x + 2y, x + 3y “and so on”.

Another example:

1 1 1

l+l H-1-1—— .

1x2 1x2x3 1X2X3X4

read “one plus one, plus one divided by 1 times 2, plus one divided by
1 times 2 times 3, plus 1 divided by 1 times 2 times 3 times 4, and so
on”.

See pages 26 and 29 for application of the sign of continuation.

PROBLEMS.

PROBLEM 8: Find the numerical value of 5+ 25~ 2 ; 4+
625+; 1000000-”.

Answer.

PROBLEM 8a: Find the numerical value of (2—J—6)8.

Answer.

PROBLEM 8b: Find the numerical value of 7[9(3+7)—125].

Answer.

Preliminary Mathematics

39

PROBLEM 8c:

Find the numerical value of 5 [ 10 (5—j-5) 2 —1250]

Answer.

PROBLEM 8d:

--3-3

Find the numerical value of 5—f-5' -f-5—2 —-49.

Answer.

.

PROBLEM 8e.

Find the numerical value of [2(5-f-3) ] 2 .

Answer.

40

Preliminary Mathematics

LESSON IX

RELATING TO ALGEBRAIC EXPRESSIONS

AN ALGEBRAIC EXPRESSION, may be defined as the represen¬
tation of a quantity by algebraic symbols; which may include both
letters and figures. All of the following are algebraic expressions:

a; 2a-|-3b-|-x; x—~; F=|C-)-32; x 2 —y 2 —36=0.

(The last two expressions are equations; explained on page 53.)

A “TERM” is that part of an algebraic expression separated from
the rest of the expression by the signs -j- or —. The terms in the
expression x 2 — y 2 — 36 = 0 are x, — y, and — 36.

The terms in a 2 -|-2ab-j-b 2 =625 are a 2 , 2ab, b 2 and 625.

A POSITIVE TERM is one that is preceded by a plus sign, as
a 2 , x 2 , 2ab, and 625. This is why the sign is often called the positive
sign.

Whenever no sign precedes a term, -)- is always understood.

A NEGATIVE TERM is one that is preceded by a minus sign,
as —y, —36. For this reason the minus sign —, is often called the
negative sign.

Great care should be observed never'to omit the minus sign before
a negative term. If no sign appears before a term, plus is understood.

A “MONOMIAL” is an algebraic expression consisting of only
one term;, as 4ax; —§,x; —a; 90mx.

A “BINOMIAL” is an algebraic expression consisting of two
terms; as 2x-f-2y; 3x—a; 2ax—3by, and —az+l.

A “TRINOMIAL” is an algebraic expression consisting of three
terms; as 2x-\-2y-\-z ; —az-|-l—2ax, and a 2 +2ab-f-b\

A “POLYNOMIAL” is an algebraic expression consisting of
more than one term, as x 2 -(-y 2 ; a 2 -|-2ab-|-b 2 ; a 2 -|-2ab-|-b 2 =625.

COEFFICIENTS. Any numerical or literal symbol, placed as a
multiplier, before another numerical or literal symbol or combination
of numerical and literal symbols, is called a coefficient. Greek letters
are often used as literal coefficients to distinguish them readily from
other symbols. As an example the expression for the area of a circle
may be given; A=7rR 2 . The Greek letters more commonly used as
coefficients in engineering literature and calculations, are /3, 7 , p, p. and

Preliminary Mathematics

41

it. See page 7 for the Greek alphabet. The following letters of the
English alphabet are also used ’considerably as coefficients: a, b, c, d,
k, l, in, n and p.

Any figure or number may be employed as a coefficient, and may be
either positive or negative in sign.

In the expression 4x 2 -\-2xy-\-y 2 —0, the figures 4 and 2 may be con¬
sidered as “ numerical” coefficients; while in the expression y 2 =px, p
is the " literal” coefficient.

In the expression for the area of a circle in terms of the radius
of the circle A=7rR 2 , A denotes the area in square units, while R de¬
notes the radius in linear units of the same denomination. If R is
expressed in inches then A is in square inches; while if R denotes
feet, then A denotes square feet. The Greek letter ir (pi) is a so-
called constant; having the same numerical value regardless of the
size of the circle.

ALGEBRAIC ADDITION AND SUBTRACTION.

A few examples will be given to explain more fully algebraical
addition and subtraction.

Example 9: Add 2a, 4 b, 3c, 8 b and 9a.

Place like terms directly under each other and add, with proper
regard for the signs of the terms, as follows :

2a 4 b 3c

9 a 8b

Answer is 11a 12a 3c This illustrates the addition of “monomials

Example 9a: Add 2x-\-2y and 2.r—2v.

2x-\-2y
2x —2y

Answer is 4.r This illustrates the addition of '‘binomials".

Example 9b: Add 2xy-\-3ax —20v+10a.r and 20y—lOa.r— 3ac —2.
2xy —20y+ 3ax
20v-f-10a.r

— lOa.a*— 3ac —2

Answer is: 2xy -f-3a.r— 3ac —2

This illustrates the addition of “polynomials

42

Preliminary Mathematics

An example illustrating algebraic subtraction will be given as fol¬
lows :

Example 9c. Subtract 5 xy — 2x-\-y from 6 xy — 2x-\-y.

The rule for algebraic subtraction is; change all the signs of the
subtrahend and proceed as in algebraic addition. Hence the arrange¬
ment will be:

bxy — 2x-\-y
-5xy-\-2 x—y

and adding gives: „ry+0+0 or simply xy.

EVALUATING EXPRESSIONS.

The numerical value of any algebraic expression is the number ob¬
tained after substituting the numerical value of each symbol that has
been ARBITRARILY assigned to it, and then performing all the
operations indicated by the symbols of operation.

A few examples of finding the numerical value of algebraic expres¬
sions will be given as follows:

Let it be arbitrarily assumed that

a denotes 1.
b denotes 2.
c denotes 3.
x denotes 0.
y denotes 10.

from which assumptions find the numerical value of

(a+fr) X (a-\-b). Making the proper substitutions,
gives : (l-)-2) X (l-j-2)

=3x3=9 Answer.

Evaluate:

ax-\-by-{-c
(lx0) + (2xl0)+3
0+20+3

=23 Answer.

Evaluate:

x 2 -\~ a xy-\-y~

o+(ixoxio)+io 2

= 100 Answer.

Any number multiplied by 0 (zero) is zero.

Preliminary Mathematics

43

Assuming that:

a=l
b=2
c=3

z—0

x—S

y— 10

Find the numerical value of the following algebraical expressions

PROBLEM 9:

y 2 =2ax.

Answer.

PROBLEM 9a:

x 2 +axy+y\

Answer.

PROBLEM 9b:

x 2 +y 2 =a\

Answer.

PROBLEM 9c:

o •>

+ = 1 .

a l 7 2

a b

Answer.

PROBLEM 9d:

3y 2 =25x.

Answer.

PROBLEM 9e:

5x 2 =9y.

Answer.

44

Preliminary Mathematics

PROBLEM 9f:

If x — 1 . Find the numerical value of C

X X

(C a +C n )

•

Answer.

PROBLEM 9g:

3' = x~*.

Answer.

The following problems illustrate algebraical addition:

PROBLEM 9h:

Add 2x; 3xy; — 5z; 8yz; —10a-; 3 z.

Answer.

PROBLEM 9i:

Add 2cy —3 as; 2yc-\-2az, and — 02 :+1.

Answer.

PROBLEM 9j:

and 1—5 b-\-a.

Add 2x-\~3y —5a-f-5fr; 5a— 5b —3 y; 5b — 2x

Answer.

Preliminary Mathematics

45

PROBLEM 9k: Subtract 3x-\-2 from 2x-\-2.

Answer.

PROBLEM 91: Subtract 2ax-\-Scx 2 — 2by from ax — by-\-lcx~.

Answer.

46

Preliminary Mathematics

LESSON X

ROOTS

SPECIAL CASES OF ALGEBRAICAL AND
ARITHMETICAL MULTIPLICATION

For the present consideration, suppose a denotes 20 and b denotes 5.
Then a-\-b= 25. The square of 25=25 2 =25 X 25=625. The squar¬
ing of 25 is performed as follows:

25
25

or -

125
500

625 625

The squaring of (a-\-b) may be performed as follows:

a-\-b

a+b

ab+b 2
a 2 -\-ab

a 2 +2ab+b 2

The letters being multiplied by one another, as were the figures in
the process of squaring 25.

Substituting the assumed numerical values of the letters, in the last
algebraical product the following is true:

a 2 =400

2ab=2W

b 2 = 25

25

25

125

50

or a 2 -f2ab-\-b 2 = 625

The preceding operations illustrate a simple application of “Alge¬
bra”, to a mathematical process; showing the relation between an
arithmetical process and an algebraical process. The arithmetical case
is a specific one; while the algebraical case is a general one, when no
numerical values are assigned to the letters. That is, the expression
a 2 -\-2ab-\-b 2 is a typical expression, which may denote a variety of
numerical conditions.

For example, if a=5 and b— 5, then a 2 -\-2ab-{-b 2 =l00. Likewise
a 2 -\-2ab-\-b 2 might equal 144.

Ft clitninary Mathematics

47

FACTORS AND ROOTS.

Considering the number 625, it has been shown to be obtained by
multiplying 25 by 25 ; it might also have been obtained by multiply¬
ing 125 by 5; or by 5 X 5 X 5 X 5 which shows that 625 is made up of
the product of several numbers.

Those numbers which when multiplied together give a certain num¬
ber, are called “factors” of the given number.

, Two or more equal factors of any given number are called the
“roots” of that number. (Refer to page 32 under exponents.) Each
of two equal factors is called the square * root.

The number 125 is obtained from 5x5x5; hence 5 is said to be
the “cube root” of 125. On the other hand, 125 is the “cube” of 5.

By following out the foregoing theory, the “fourth root” of 625 is 5.

It is interesting to note how rapidly “cubes” increase.

Take the case of 4, which is twice 2:—

The cube of 2 is 2X2X2=8.

The cube of 4 is 4X4X4=64.

The cube of 4 is eight times as great as the cube of 2.

In the algebraical case just considered, a 2 -f-2ab J r b 2 is the square of
(a-{-b), and (a-)-fr) is the square root of a 2 -\-2ab-\-b 2 .

The cube of (a-f-b) may be obtained as follows:

(a-f-fr) X (a-\-b)=a 2 -\-2ab-\-b~

a+fr

a 2 &+2a^ 2 +6 3
cf:-\-2a 2 b-\- ah 2

a 3 +3a 2 6+3a& 2 +6 3 .

Then (a-j-fr) is the “cube root” of a s -{-3a 2 b-\-3ab 2 -{-b 3 .

When multiplying algebraical expressions by one another, always
set like terms directly under each other for convenience in addition.

Note that in multiplying literal factors by each other their expo¬
nents are added, (algebraical addition.)

*The reason for the statement “ square ” root may be evident by considering a square.
Multiplying together the numerical values of two sides of any square will give as a
product the area of the square. If the side of any square is 4 feet, the area of the square
is 4x4 - 4 2 =16 square feet.

48

Preliminary Mathematics

PROBLEM 10:

Find the square of 2; 4; 6; 8 and 9.

Answer.

PROBLEM 10a:

Find the cube of 2; 4; 6; 8 and 9.

Answer.

•

PROBLEM 10b:

0 + 2 y).

Calculate the square of (+ + 3>); (2,ir —)— y) ;

Answer.

PROBLEM 10c:

Calculate the cube of (2 x-\-y) ; (4+3?) ; (++8).

Answer.

PROBLEM lOd:

a-\-b.

Find the algebraical expression for the cube of

Answer.

Prelimin ary Mat he matics

49

PROBLEM lOe: Find the algebraical expression for the cube of
a — b.

Answer.

PROBLEM lOf: Write the algebraical expression for the result
of O—:y) J +0— x)*.

Answer.

50

Preliminary Mathematics

LESSON XI

RATIO AND PROPORTION

A ratio expresses the relative magnitude of two quantities of the
same kind.

A ratio is the relation of two values as expressed by division.

The ratio of 4 to 5 is expressed by This is simply a fraction

whose numerical value is 4-f-5; which expressed as a decimal, is 0.8.

Therefore the ratio of f=0.8. (f Xf=A); multiplying both numer¬
ator and denominator of the fraction f by 2.)

Since it is not logical to compare quantities of different kinds, the
result of dividing one quantity by another quantity of the same kind,
but possibly different in amount, can be only a number, not denoting
any particular kind; hence a ratio is sometimes alluded to as a “pure
number”. It might be called an abstract number.

It would not be logical or sensible to compare 3 horses with 7
houses; nor 3 apples with 7 automobiles, but it would be more con¬
sistent to compare 3 horses with 7 horses; 5 pounds with 10 pounds.

If 5 pounds be compared with 10 pounds and expressed as a ratio
it will be:

5 pounds.

10 pounds.

The result is evidently not T 5 o pounds; but simply or or 0.5; an
abstract or “pure number”.

The physical meaning in this case being that 10 pounds is twice as
great as 5 pounds.

This consideration has a very important application in efficiency
engineering, in comparing the useful output of any machine or device
with the total input to the same machine or device.

The efficiency of any machine is the ratio of its useful output to
the total input to the machine.

There are many classes of efficiency. One important efficiency met
with in practice is power efficiency.

The power efficiency of any machine or device is the ratio of
the useful power output from the machine or device, to the total
power input to the same machine or device.

While there are many different kinds of efficiency, efficiency in
general is always a ratio whose numerical value is always less than 1.

Percent efficiency is the fractional value multiplied by 100.

A ratio, then, is really the number of times one quantity is contained
in another quantity of the same kind.

Prelim inary Mathematics

51

In Algebra, the ratio of a to b is expressed as a : b or as -, and a

b

is called the first “term”, and b is called the second “term” of the
ratio.

A proportion is the expression of the equality of ratios.

The following is an expression of a numerical proportion:

3 : 7 : : 9 : 21, and means that 3 is to 7 as 9 is to 21. The same
relation may be expressed by:

3 : 7=9 : 21; or by f= 2 9 T .

In any proportion the two middle figures are called the “means”
while the two outer figures are called the “extremes”.

The symbol : means “is to”, while : : means “as”, or equals.

It is well to note that the numerical value of a ratio may be 1,
more than 1, or less than 1.

PROBLEM 11: Find the numerical value of the ratio of 3,730
to 746.

Answer.

PROBLEM 11a: Find the ratio of the power output (power
efficiency) to the power input for a motor whose output is 7460 watts
(10 horse-power) and whose power input is 8288.8 watts. (11.1
horse-power.)

Answer.

PROBLEM lib: Find the numerical value of the ratio of the
area of a square to the area of a circle if the area of the square is 4
square inches and the area of the circle is 3.14 square inches.

Answer.

52

Preliminary Mathematics

PROBLEM 11c: Given a : 10= 15 : 150, find the numerical
value of a.

Answer.

PROBLEM lid: Given find the numerical value of a.

Answer.

Preliminary Mathematics

53

LESSON XII

SOLUTION OF EQUATIONS

As the word " equation” implies, an equation is an expression of
equality between two quantities. The expression may be one involv¬
ing either numbers or letters, or both. That is, there are arithmeti¬
cal equations and algebraic equations; as 2X4 = 9 — 1, and 2x = 5y.

The study of the subject of engineering involving as it does, the
consideration of equations of all kinds, it will be well to carefully
consider a few important characteristics of equations and in particu¬
lar the solutions of a variety of equations.

The following will be taken as an illustration of an equation con¬
taining both letters and figures; or numbers.

F=fC+32.

The sign of equality is considered to separate any equation into two
parts, the left-hand side and the right hand side as onq faces the
equation. The sides of an equation are sometimes alluded to as the
left-hand “ member” or the first member, and the right-hand “mem¬
ber”, or the second member.

The first rule to be observed in solving any equation is; when a
TERM* is changed from one side of the equation to the other side
the algebraic SIGN of the term must also be changed. If the sign
of the term is -f- before the term is changed over, then the term must
be preceded by a — sign after being changed. If—before change, it
must be -f- after change. The principle underlying this rule may be
illustrated by the following: given: x — 5 = 2 to find the numerical
value of x.

Add 5 to both members of the equation giving:

■r — 5 + 5 = 2 +5, (-5 + 5 = 0)

whence x = 7. The same result is obtained by placing — 5 on the right
hand side of the equation and changing the sign to -{-.

Given y-f-20 = 44; to find the value of y.

Subtract 20 from both sides, giving:

y + 20 — 20 = 44— 20;
whence y = 24.

Another rule is that multiplying or dividing both sides of any equa¬
tion by the same number, does not alter the value of the equation.

* See page 40.

54

Preliminary Mathematics

Also adding to or subtracting from both sides of any equation the
same number does not affect the value of the equation,f and changing
the sign of all the members of an equation does not change the value
of the equation.

Suppose it is desired to solve the equation F—|C -|- 32 for C; the
term containing C must be placed on the left-hand side of the equality
sign, and F must be placed on the other side. Then the equation
would be expressed:

—£C= —F-j-32, which after changing all the signs, becomes:

|C=F—32. As the equation now appears, |C is the first or left-
hand member, and (F—32) is the second or right-hand member.
When solving any equation for a particular quantity, it is usual to
divide both members of the equation by a number that shall make
the coefficient of the particular quantity designated, equal to unity.
In the above case both members of the equation must be divided by
, giving:

C = f (F-32).

Note the use of the parenthesis as explained on page 37. Perhaps
the simplest form of an equation is when both sides are alike; as
y x — x y ) 2 —|- 4 = 4 2. Such equations are called “identical”

equations.

Another class of equations is very common, in which the two sides
are equal, numerically, only upon some condition being imposed on
some factor or term. An example is 2v=10.

2,r=10 only upon condition that x = 5; then 2 X 5 = 10. If any
other value is assumed for x, the two members would not be numer¬
ically equal to each other.

In such a case, x is called the unknown quantity.

That value of the unknown quantity in any equation, which makes
the two members equal to each other, is called a “root” of the equa¬
tion.

When the root of the equation is properly substituted in the equa¬
tion it is said to " satisfy” the equation.

The process of finding the root or roots of an equation is called
solving the equation.

Conditional equations are divided into classes or orders, accord¬
ing to their degree, or according the power of the unknown quantity.

The “degree” of an equation is the same as the highest degree or
power of its unknown quantity.

t Equal quantities, equally affected, remain equal.

Preliminary Mathematics

55

As examples:

2x — a; cx — lOy, and x -f- y — 5, are equations of the first degree.
y 2 — 2b, and y 2 -\-2x -\-b =z A, are of the second degree.

Equations of the second degree are called “quadratics”.

5a- 3 =10, and y s -\-3y 2 b-\-5yb 2 -\-b 9 , are of the third degree.

Equations of the third degree are called cubical equations, or cubics.
If the.side of a square is multiplied by itself three times, the product
will give the volume of a cube.

The exponents of the unknown quantities might be either posi¬
tive, negative, fractional or integral, and each equation might have
more than one unknown quantity.

PROBLEM 12: If F in the equation F=-C-|-32, is equal to

5

—40 find the numerical value of C.

Answer.

PROBLEM 12a: Given the equation R = pj, find the numerical
value of R if p = 10.8, l = 1 and A = 4108.8.

Answer.

PROBLEM 12b: Using the assumed numerical values; .r=5
and y=10, find the effect of multiplying the equation x 2 -\-x=3y, by 3.

Answer.

PROBLEM 12c: Given §—10 = 20 to find the value of 2 .

Answer.

56

Preliminary Mathematics

PROBLEM 12d: Given 8-t' -J- 4 r= 10 to find the value of x.

Answer.

PROBLEM 12e: Given y = nur + b; find the numerical value of
m, when y — 2, x = 4 and b = 5.

Answer.

Preliminary Mathematics

57

LESSON XIII

CONSTANTS AND VARIABLES

Any quantity . or any value may continue to be constant as time
continues, or it may change or vary.

Any varying quantity is called a variable; while any value that
continually remains the same is called a constant.

Of course a quantity may be constant for a certain interval of time
and then may vary for a certain time. It is customary to state the
conditions of variability when considering any particular problem, or
during any mathematical discussion. A few examples of physical
constants may be cited; as the force of gravity, denoted by g, which
is constant for any given locality on the earth’s surface; it has, of
course, different values in different localities.

The diameter of the earth may be considered to be a constant.

The ratio of the circumference of any circle to its diameter is a
constant; denoted by 7r and is taken as 3.1416 for ordinary calcula¬
tions.

It is customary to denote constants or constant values, by using the
first few letters of the alphabet, while the last few letters of the alpha¬
bet are used to denote variables.

Constants are denoted by ; a, b, c, d, g, h, k, m, and n. Variables
are denoted by p, q, s, t, x, y, z.

Some of the letters of the Greek alphabet, as k, 7 r, and w are used to
denote constants and such letters as 0 , 6 , 77 are used to denote varia¬
bles.

FUNCTIONS.

Whenever a quantity, denoted by y, depends upon another quantity,
denoted by x, in such a manner that no change can be made in x

without producing a corresponding change in y, then y is said to be a

function of x.

The symbol f ( x ) is used to denote a function of x, and is read
the “f function of x”.

In a like manner 0 ( x, y) denotes a function of x and y, and is
read “the <p function (phi function) of x and y”.

Further; the symbol

y—f (x) is often used to express the fact that y is a function of x.

The fact that a change may be made in a quantity denoted by x or

y, indicates at once that x and y denote variables. For brevity, x and
y are therefore themselves called variables.

58

Preliminary Mathematics

In any equation involving x and y, y is a function of x, if x is
taken as the independent variable; while x would be a function of y
if y be taken as the independent variable.

Example: x 2 — y 2 —36=0, is an equation involving the two variables
x and y. One of these variables as x may be made the independent
variable; then y will be the dependent variable. Suppose the above
equation is solved for y, as follows:

—y 2 =:—jr 2 +36;
or y 2 = x 2 —36
and y =± V x 2 —36

Now the function is expressed directly in terms of the independent
variable x, and is said to be an explicit function, or y is said to be
an explicit function of x.

If any numerical values are put in place of x in this last expres¬
sion, corresponding numerical values may be obtained for y. One
value will be assumed for x for the present. Let ,r=10; then .r 2 =100,
and

y = V100—36 = V64~ = ± 8.

A large number of different values might be substituted for x and
corresponding values for y found by the process of Arithmetic. This
question will be considered more at length, later on.

Perhaps the meaning of the word function and the relation of
functions may best be understood by considering a few concrete illus¬
trations. It is well known that the area of a square is found by mul¬
tiplying the values of two of its sides together. If a denotes one side
of a square, then its area = a X a = a 2 . The area of a square is a
function of one of its sides. It makes no difference what the size
of a square is, the above relation always holds. Suppose a side of a
square varies; then its area varies as the square of its variable side.
If y denotes the area of any square and x a side, then:

y = * 2

If x is doubled, y is increased four times.

PROBLEM 13: Given the equation x 2 -\-y 2 =37, find the numeri¬
cal value of x, when y=l.

Answer.

Preliminary Mathematics

59

PROBLEM 13a: Given the equation jv 2 —(- 10-ir——8y-f-41=36,
compute and properly tabulate the values of y corresponding with the
following values of x; 0, 1, —1, 2, —2, 3, 4 and 5.

Answer.

PROBLEM 13b: Given the equation F:=^C-|-32, compute and

tabulate values of F corresponding with the following values of C ;
—40; 0; 32; 100; 1000.

Answer.

60

Preliminary Mathematics

LESSON XIV

LOGARITHMS

The word logarithm is made up of the two Greek words (\S 70 p)
logos, meaning " ratio”, and ( apiOp ) arithmos, meaning “number”.

The strict meaning of the word is, a “ratio number”.

Logarithms enable us to perform in a short time certain mathemati¬
cal calculations that otherwise require considerable time and labor.

For example; multiplication and division, which are sometimes long
and difficult operations, may be quickly and easily performed by
methods involving logarithms and “logarithmic tables”.

Logarithms may also be employed to easily and quickly check the
results of long and complicated processes of multiplication or divi¬
sion.

The important application of logarithms is in finding the numerical
values of numbers that are raised to fractional powers; such for
example as 2 . 8 1,6 ; two and eight-tenths raised to the one and six-
tenths power. It would be difficult to perform such an operation by
ordinary arithmetical processes, but it is a very simple process by use
of logarithms.

As used in ordinary practice the logarithm of any number is the
EXPONENT denoting the power to which a number, called the
BASE of the system must be raised, to produce the given number.

Instead of writing the word logarithm, the abbreviation log is used.

For example if

y =a x , and y' = a x ' then x and x '

are the logarithms of y and y' in that particular system whose base
is a; then

x = log y, and x’ = log y'

To consider a numerical example, let 2 be taken as the base, and
suppose 64 be a given number.

Then 64=2 6 and log of 64=6.

With a base 2, a table of logarithms may be composed as follows :

2 °= 1 and log 1=0
2 1 — 2 and log 2=1
2 2 = 4 and log 4=2
2 8 = 8 and log 8=3
2 4 =16 and log 16=4
2 5 =32 and log 32=5

Preliminary Mathematics

61

The logarithm of 3 in this system is evidently between 1 and 2,
and it is 1.584; that is, 2 must be raised to the 1.584 power to equal
the number 3.

The raising of 2 to the 1.584 power may be expressed as follows:

2 ^ 1-5 84 )_ 2^ 1 + l6 °^ ff ^ _

In this system the logarithm of 5 is between 2 and 3; as is also the
logarithm of 6 , and of 7.

If 4 be taken as a base then:

4° = 1, and log 1=0;

4g= 4, and log 4=1; instead of 2 as in the system with base 2.

4“ =16, and log 16=2; instead of 4 as in the system with base 2.

4 d =64, and log 64=3; instead of 6 as in the system with base 2.

A large number of systems of logarithms might be worked out,
using as bases any positive number (whole number or a fraction or
a whole number and a fraction) except 1.

In any system of logarithms the fractional part of the logarithm
of any number is called the “mantissa”, (plural mantissae) and the
integral part of the logarithm is called the “characteristic”.

In the system with 2 as its base, the logarithm of 3 being 1.584, the
fraction 0.584 is the mantissa of the logarithm of 3; while 1 is the
characteristic.

Only two systems of logarithms have come into general use,—the
one first published by John Spidwell, London, in 1619, called the ‘‘nat¬
ural” system, using as a base 2.718 281 828; a number obtained by
continuing the process indicated by

0 00 _

W»15 8 4

1 +} + £ + 2^-3 + +

2 X 3 X 4X 5

the other, and the one most commonly adopted and used, worked out
and published by Henry Briggs in 1624, using 10 as the base, and
called the common system. From what has been shown regarding the
forming of a table of logarithms, it is evident that the logarithm of
the same number will be different in the two systems mentioned.

For example: the log of 180 is 2.25527 in the common system, base
10; while the log of 180 in the so-called natural system is 5.1930.

In the common system, 10 is raised to the 2.25527 power to amount
to 180, while in the natural system 2.718 281... .must be raised to
the 5.1930 power to equal 180.

To designate which system is employed, the following is adopted;
logic 276; meaning the logarithm of 276 using the base 10; or log e 450;
meaning the logarithm of 450 using the base 2.718281 .... The base of
the natural system is usually denoted by e; (epsilon).

62

Preliminary Mathematics

In general, loga-r means the logarithm of x, with the base a.

If 10 is used as the base of a system of logarithms the following
table of the logarithms of 10 and multiples of 10 could be extended as
desired:

Number.

Logarithm.

1

0

10

1

100

2

1000

3

10000

4

The numbers in the second column denotes the power to which 10
must be raised to equal the corresponding number in the first column.

For example:

10 = 10 "; 100 = 10 2 ; 1000 = 10 3 .

With 10 as a base it is necessary to find the exponent, or the power
to which 10 must be raised to give the numbers between 1 and 10 ,
between 10 and 100 , between 100 and 1000 and so on.

Such a table has been arranged and printed on page 65. Only the
mantissae are printed in this table, as is customary in tables of log¬
arithms employing the base of 10 ; the characteristic being readily seen
by inspection of the given number.

The mantissa of the logarithm of 10 as given in the table on page
65, is 000 0000 ; but since the characteristic is 1 , (10 raised to the 1st
power=10) the logarithm of 10 is 1.000 0000. The logarithm of 25 is
1.397 9400; the logarithm of 250 is 2.397 9400, and the logarithm of
2500 would be 3.397 9400.

In the table the characteristic of the logarithms of the numbers
from 1 to 10 is 0 ; from 10 to 100 is 1 , and from 100 to 1000 is 2.

The logarithm of 24100 could be found by use of the table, by pre¬
fixing the proper characteristic to the logarithm of 241.

The log 24100 is 4.3820170. The log 244000 is 5.387 3898.

Another table of logarithms, using the base of 10 is arranged some¬
what differently on page 64, showing the common logarithms of a few

Prelim inary M athematics

63

numbers from 9950 to 100009. As usual only the fractional part
(mantissae) of the logarithm is printed; the integral part (character¬
istic) being supplied by inspection.

This is a “ seven-place” table; the first three figures being omitted in
all the columns but the second, headed 0, to save space and repetitions.

Also the first two figures of the number are omitted in many cases,
in the first column; headed “number”. The figures are in different
type to facilitate location. The mark o indicates that the first three
figures of the mantissa following the part of the mantissa containing
the sign, should be prefixed, instead of the three figures preceding.
For example the log 99542 is 4.998 0064; while the log 99540 is
4.9979976. From this table the log 100009 is 5.000 0391.

64

Preliminary Mathematics

Number

0

1

2

3

4

5

6

7

8

9

9950

997

8231

827418318

8362

8405

8449

8493

8536

8580

8624

51

8667

8711

8755

8798

8842

8885

8929

8973

9016

9060

52

9104

9147

9191

9235

9278

9322

9365

9409

9453

9496

53

9540

9584

9627

9671

9715

9758

9802

9845

9889

9933

54

9976

5020

0064! o!07

6151

6195

6238

0282

0325

o369

55

998

0413

0456

0500

0544

0587

0631

0674

0718

0762

0805

56

0849

0893

0936

0980

1023

1067

mi

1154

1198

1241

57

1285

1329

1372

1416

1460

1503

1547

1590

1634

1678

58

. 1721

1765

1808

1852

1896

1939

1893

2026

2070

2114

59

2157

2201

2245

2288

2332

2375

2419

2463

2506

2550

9960

998

2593

2637

2681

2724

2768

2811

2855

2899

[2942

2986

61

3029

3073

3117

3160

3204

3247

3291

3335

3378

3422

62

3465

3509

3553

3596

3640

3683

3727

3771

3814

3858

63

3901

3945

3988

4032

4076

4119

4163

4206

4250

4294

64

4337

4381

4424

4468

4512

4555

4599

4642

4686

4729

65

4773

4817

4860

4904

4947

4991

5035

5078

5122

5165

66

5209

5252

5296

5340

5383

5427

5470

5514

5557

5601

67

5645

5688

5732

5775

5819

5862

5906

5950

5993

6037

68

6080

6124

6167

6211

6255

6298

6342

6385

6429

6472

69

6516

6560

6603

6647

6690

6734

6777

6821

6864

6908

9970

998

6952

6995

7039

7082

7126

7169

7213

7256

7300

7344

71

7387

7431

7474

7518

7561

7605

7648

7692

7736

7779

72

7823

7866

7910

7953

7997

8040

8084

8128

8171

8215

73

8258

8302

8345

8389

8432

8476

8519

8563

8607

8650

74

8694

8737

8781

8824

8868

8911

8955

8998

9042

9086

75

9129

9173

9216

9260

9303

9347

9390

9434

9477

9521

76

9564

9608

9651

9695

9739

9782

9826

9869

9913

9956

77

999

0000

0043

0087

0130

0174 0217

0261

0304

0348

0391

78

0435

0479

0522

0566

0609

0653

0696

0740

0783

0827

79

0870

0914

0957

1001

1044

1088

1131

1175

1218

1262

9980

999

1305

1349

1392

1436

1479

1523

1567

1610

1654

1697

81

1741

1784

1828

1871

1915

1958

2002

2045

2089

2132

82

2176

2219

2263

2306

2350

2393

2437

2480

2524

2567

83

2611

2654

2698

2741

2785

2828

2872

2915

2959

3002

84

3046

3089

3133

3176

3220

3263

3307

3350

3394

3437

85

3481

3524

3568

3611

3655

3698

3742

3785

3829

3872

86

3916

3959

4003

4046

4090

4133

4177

4220

4264

4307

87

4350

4394

4437

4481

4524

4568

4611

4655

4698

4742

88

4785

4829

4872

4916

4959

5003

5046

5090

5133

5177

89

5220

5264J

5307

5351

5394

5438

5481

5524

5568

5611

9990

999

5655

5698

5742

5785

5829

5872

5916

5959

6003

6046

91

6090

6133

6177

6220

6263

6307

6350

6394

6437

6481

92

6524

6568

6611

6655

6698

6742

6785

6828

6872

6915

93

6959

7002

7046

7089

7133

7176

7220

7263

7307

7350

94

7393

7437

7480

7524

7567

7611

7654

7698

7741

7785

95

7828

7871

7915

7958

8002

8045

8089

8132

8176

8210

96

8262

8306

8349

8393

8436

8480

8523

8567

8610

8653

97

8697

8740

8784

8827

8871

8914

8958

9001

9044

9088

98

9131

9175

9218

9262

9305

9349

9392

94351 9479

9523

99

9566

9609

9653

9696

9739

9783

9826

98701 9913

9957

10000

000

0000

0043

0087

0130

0174

0217

0261

0304(0347

0391

Pre l imin ary Mat he matics

65

No.

Logarithm

No.

Logarithm

No.

Logarithm

No.

Logarithm

No.

Logarithm

0

50

698

9700

100

000

0000

150

176

0913

200

301

0300

1

000

0000

51

707

5702

101

994

3214

151

178

9769

201

303

1961

2

301

0300

52

716

0033

102

008

6002

152

181

8436

202

305

3514

3

477

1213

53

724

2759

103

012

8372

153

184

6914

203

307

4960

4

602

0600

54

732

3938

104

017

0333

154

187

5207

204

309

6302

5

698

9700

55

740

3627

105

021

1893

155

190

3317

205

311

7539

6

778

1513

56

748

1880

106

025

3059

156

193

1246

206

313

8672

7

845

0980

57

755

8749

107

029

3838

157

195

8997

207

315

9703

8

903

0900

58

763

4280

108

033

4238

158

198

6571

208

318

0633

9

954

2425

59

770

8520

109

037

4265

159

201

3971

209

320

1463

10

000

0000

60

778

1513

110

041

3927

160

204

1200

210

322

2193

11

041

3927

61

785

3298

111

045

3230

161

206

8259

211

324

2825

12

079

1812

62

792

3917

112

049

2180

162

209

5150

212

326

3359

13

113

9434

63

799

3405

113

053

0784

163

212

1876

213

328

3796

14

146

1280

64

806

1800

114

056

9049

164

214

8438

214

330

4138

15

176

0913

65

812

9134

115

060

6978

165

217

4839

215

332

4385

16

204

1200

66

819

5439

116

064

4580

166

220

1081

216

334

4538

17

230

4489

67

826

0748

117

068

1859

167

222

7165

217

336

4597

18

255

2725

68

832

5089

118

071

8820

168

225

3093

218

338

4565

19| 278

7536

69

838

8491

119

075

5470

169

227

8867

219

340

4441

201301

0300

70

845

0980

120|079

1812

170

230

4489

220

342

4227

21

322

2193

71

851

2583

121

082

7854

171

232

9961

221

344

3923

22

342

4227

72

857

3325

122

086

3598

172

235

5284

222

346

3530

23

361

7278

73

863

3229

123

089

9051

173

238

0461

223

348

3049

24

380

2112

74

869

2317

124

093

4217

174

240

5492

224

350

2480

25

397

9400

75

875

0613

125

096

9100

175

243

0380

225

352

1825

26

414

9733

76

880

8136

126

100

3705

176

245

5127

226

354

1084

27

431

3638

77

886

4907

127

103

8037

177

247

9733

227

356

0259

28

447

1580

78

892

0946

128

107

2100

178

250

4200

228

357

9343

29

462

3980

79

897

6271

129

110

5897

179

252

8530

229

359

8355

30

477

1213

80

903

0900

130

113

9434

180

255

2725

230

361

7278

31

491

3617

81

908

4850

131

117

2713

181

257

6786

231

363

6120

32

505

1500

82

913

8139

132

120

5739

182

260

0174

232

365

4880

33

518

5139

83

919

0781

133

123

8516

183

262

4511

233

367

3559

34

531

4789

84

924

2793

134

127

1048

184

264

8178

234

369

2159

35

544

0680

85

929

4189

135

130

3338

185

267

1717

235

371

0679

36

556

3025

86

934

4985

136

133

5389

186

269

5129

236

372

9120

37

568

2017

87

939

5193

137

136

7206

187

271

8416

237

374

7483

38

579

7836

88

944

4827

138

139

8791

188

274

1578

238

376

5770

39

591

0646

89

949

3900

139

143

0148

189

276

4618

239

378

3979

40

602

0600

90

954

2425

140

146

1280

190

278

7536

240

380

2112

41

612

7839

91

959

0414

141

149

2191

191

281

0334

241

382

0170

42

623

2493

92

963

7878

142

152

2883

192

283

3012

242

383

8154

43

633

4685! 93

968

4829

143

1 155

3360

193

285

5573

243

385

6063

44

643

4527

94

973

1279

144

158

3625

194

287

8017

244

387

3898

45

653

2125

95

977

7236

145

161

3680

195

1290

0346

245

389

1661

46

662

7578

96

982

2712

146

164

3529

196

292

2561

246

390

9351

47

672

0979

97

986

7717

147

167

3173

197

294

4662

247

392

6970

48

681

2412

98

1991

2261

148

170

2617

198

1296

6652

248

394

4517

49

690

1961

99

1995

6352

149

| 173

1863

199

1298

8531

249

396

1993

50]698

9700

1001000

0000

150

|176

0913

200

1 301

0300

| 250

397

9400

66

Preliminary Mathematics

PROBLEM 14: If 3 is given as the base of a system of logarithms
find the logarithms of the following numbers: 9; 27; 243; 2187 and
59049.

Answer.

PROBLEM 14a: From the table on page 65 find the logarithm
of the following numbers: 9; 27; 82; 180 and 243. Do not fail to
supply the proper characteristic in each case.

Answer.

PROBLEM 14b: Using the table on page 65 and prefixing the
proper characteristics, write the logarithms of 2700; 181000 ; 200 ; 2000,
and 2000000.

Answer.

Preliminary Mathematics

67

LESSON XV.

GENERAL PROPERTIES OF LOGARITHMS.

Theorem I. In any system of logarithms the logarithm of unity is
zero.

Let a denote any base, then:

a m Xa°=a <m+0) = a m ; from which;

a. m

a° = TUT = 1 ; or a° = 1

d

• lOgal = 0

It is often useful in solving equations to know the above fact.

Theorem II. In any system of logarithms the logarithm of the
base itself is unity.

If a denotes any base, then;

a 1 = a ; or log a a = 1

If the base is 10 then 10^=10; or log , 0 10=1 (see page 62).

Theorem III. In any system of logarithms having a base greater
than unity, the logarithm of zero is minus infinity.

If base a > 1, then a _00 -f- ^oo = & — 0 (see page 15 for >).

A number raised to a greater and greater power gives a constantly
increasing result, provided the given number is greater than 1 .

This is seen on page 33 where 10 is raised to different powers.

Any number divided by a number that is too great to be measured
must give as a quotient a number too small to be measured.

• *. loga 0 = — oo (see page 9 for oo).

Theorem IV. In any system of logarithms whose base is less than
unity, the logarithm of zero is infinity.

For if a < 1, a 00 = 0, log,, 0 =oo.

To show that a °°=0 when a is less than 1 ; suppose a= $; then
a 2 =£, a 3 =|, a 4 = T V, a 5 = 3 L and the higher the power to which a = l is
raised, the less the numerical value of the result of dividing 1 by
the increasing number becomes.

Theorem V. In any system of logarithms, having a positive num¬
ber for its base, the logarithm of a negative number is imaginary.

68

Preliminary Mathematics

Since the base is positive, no power of the base can ever become a
negative number. Many mathematical operations may however be
performed with negative numbers, using logarithms, by proceeding as
if the numbers were positive and prefixing the proper sign at the end
of the process.

The arithmetical complement of a logarithm is the remainder ob¬
tained by subtracting the logarithm from 10.

The practical utility of any system of logarithms is really based
upon the two following theorems.

Theorem VI. In any system of logarithms, the logarithm of the
product of two or more factors is equal to the sum of the logarithms
of the factors. If a denotes the base of the given system then :

let a« = m ;) f which S
* y = n; S (

S loga m — X,

l and loga n = y.

then multiplying member by member, a®Xa^ = m X n, or a (x *v ) = mn,
and logamn = x-\-y. Also adding member to member, logam+logan=
x + y. And logamn = log a m -f- log a n. (Two quantities equal to the
same quantity are equal to each other.)

The same theorem may likewise be proved for the product of three
or more factors.

Example 15: Given the log 10 of 2 = .30103 and the log 10 of
3 = .47712, find the log 10 of 288.

288=2X2X2X2X2X3X3

But Iog 10 288=log 1 o2+log 10 2-flog 1 o2+ log 10 2-flog 10 2+log 10 3+log 10 3
.Mog 10 288 = 5 X (.30103)+ 2 X (.47712) + 2.45939.

Example 15a: Suppose it is desired to find the logarithm of
15625. This number is equal to 25x25x25; therefore the log 15625
=log 25+log 25-f-log 25;

=1.397 9400+1.397 9400+1.397 9400;

=4.193 8200. (The characteristic 4, is checked by inspection ).

This example may be checked by the following:

15625=5 x5x5x5x5x5.

log 15625=log 5+log 5+log 5+log 5+log 5+ log 5.

=0.698 9700X6;

=4.1938200.

It may be noted that 15625=5 6 (The sixth power of 5=15625).

Preliminary Mathematics

69

This theorem leads to another important principle that may be
treated as a separate theorem. This has to do with evaluating the
powers or the roots of any number.

While it is a simple process to find the 6th or any other power of a
number, it may require considerable time; the use of logarithms in
such a case greatly lessens the time and the labor. The same is true
as to finding any root, (as the 27th root) of a number.

Theorem VII. To find the logarithm of the power of any num¬
ber, multiply the logarithm of the number by the index of the power ,
and the product will be the logarithm of the power of the given num¬
ber.

This is true whether the index is a whole number or a fraction.

Example 15b: Find the logarithm of 5 3 from the table on page 65.
log 5=0.698 9700; 3 X 0.6989700=2.0969100.

This may be checked by looking up the log 125 in the same table.

Example 15c: Find the logarithm of 243 The logarithm of
243=2.385 6063; multiplying this logarithm by }, gives 0.5964016;

which is the log 243 ^

Theorem VIII. In ANY system of logarithms the logarithm of
the quotient of two numbers is equal to the logarithm of the dividend,
minus the logarithm of the divisor.

This theorem may also be expressed as follows:

In ANY system of logarithms the logarithm of a fraction is equal
to the logarithm of the numerator minus the logarithm of the denom¬
inator.

Let the base of a given system be denoted by a,

and let a*= m, ) . h J log a m = x*
and a^= n, ) { log a n = y

Dividing the above two equations, member by member gives:

— = “or a ™ And loga ™ =x — y. But subtracting the above

a# n n -n

two equations, member by member, gives log ft m — log a n = x — y.

.'. log a ™= log a m— og a n. (Quantities equal to the same quantity are equal to
each other.)

*This is true because x log a = log m and the log, a ( logarithm of the base of the
system), is = 1. .*. * = logm.

70

Preliminary Mathematics

Example 15d: Find the logarithm of 248 divided by 62 or the log- 8 fi V.

log 248=2.395 4517 and log 62=1.792 3917.

log W=log 248—log 62.

=2.394 4517—1.792 3917.

= .6020600.

The number corresponding with this log is found from the table on
page 65, to be 4, which is the quotient.

Letting x and y denote any numbers whatever, and a the base of
the system, the preceding theorems may be recapitulated for handy
reference as follows:

log 1=0

log a=l

log a>1 0= —oo

log, <,0= oo

log— x is imaginary

log .ry=log .v+log y

logf=log x —log y.
y

log x n —Ti log x

n_

log V x— * log x

log i = —log *

Having explained the theory of logarithms, and how they may be
obtained, a few words will be devoted to the practical application or
the use of logarithms. While the logarithm of many numbers could
be obtained by applying the principle already mentioned, the use of the
table is limited, in finding the logarithm of a number. If a logarithm
is given to find the number corresponding to it, the limits of the table
are evident. The larger the tables the more useful, and this is why
they are usually published in a separate volume.

The more figures there are in the logarithms the more accurate the
answer obtained by their use. For all ordinary calculations the
“seven-place” logarithmic table, as illustrated on page 65, is sufficient.

Preliminary M at hematics

71

Suppose it is desired to find the numerical value of 20‘ (twenty
raised to the fifth power.)

According to Theorem VI, page 68, look up the log 20, multiply it
by 5, and from a table of logarithms look up the number correspond¬
ing to the logarithm found by multiplying the log 20 by 5. The log
20=1.301 0300 times 5=6.505 1500. The number corresponding with
.505 1500 is 32 (see table page 65). Since the characteristic is 5, there
must be six digits* in the answer. The answer is therefore 3,200,000.
This operation is just the reverse of Theorem VI.

PROBLEM 15: Given the log 10 2=0.30103; log 10 3=0.4771213;
log 10 4 = 0.60206, and logi 0 7 = 0.845098, find the logarithm of 32.

Answer.

PROBLEM 15a: Given the logs as in problem 15, find the log 10 ,
196.

Answer.

PROBLEM 15b: Given the logs as in problem 15, find the log™.
336.

Answer.

PROBLEM 15c: Given the logs as in problem 15, find the log™

886 .

Answer.

* See page 61 .

72

Preliminary Mathematics

PROBLEM 15d: Given the logs as in problem 15, find the log l0
1323.

Answer.

PROBLEM 15e: With the values of logs given in problem 15,
find log 10 3929.

Answer.

PROBLEM 15f: With the values of logs given in problem 15,
find logm 9261.

Answer.

PROBLEM 15g: With the values of logs given in problem 15,
find log 10 37044.

Answer.

PROBLEM 15h: With the values of logs given in problem 15,
find log 10 54.

Answer.

Preliminary Mathematics

73

PROBLEM 15i: Using the method of logarithms, multiply 8 by 9.

Answer.

PROBLEM 15j: By method of logarithms find the quotient of
225 divided by 15.

Answer.

PROBLEM 15k: Find the numerical value of 22‘, using loga¬
rithms.

Answer.

PROBLEM 151: Find the numerical value of 50 2 , using loga¬
rithms.

Answer.

74

Preliminary Mathematics

PROBLEM 15m: Find the logarithm of the square root of 20.

Answer.

PROBLEM 15n: Find the logarithm of the V15.

Answer.

PROBLEM 15o: Find the log 10 216s.
Answer.

PROBLEM 15p: Find the numerical value of 128?.

Answer.

PROBLEM 15q: Compute the numerical value of the square
root of 3.14159. Find the same by logarithms to check result.

Answer.

Preliminary Mathematics

75

PROBLEM 1
Answer.

PROBLEM 2

“American Short

Answer.

PROBLEM 3
Answer.

PROBLEM 4
Answer.

PROBLEM 5

REVIEW PROBLEMS.

Compute the decimal equivalent of

: Multiply 654321 by 1234, and prove result by the
Proof Method”.

: Multiply 50 2 by 25" 2 .

Divide 25 2 by 50 a .

Express algebraically, the cube of a — c.

Answer.

76

Preliminary Mathematics

PROBLEM 6: Express algebraically, the cube of c—a.
Answer.

PROBLEM 7: Find the numerical value of y, in the equation

f = 36.

Answer.

PROBLEM 8: Find the numerical value of x, in the equation
x 2 -\-y 2 =:61 if y = 5.

Answer.

PROBLEM 9: If Rj =5, and R 2 = 4; find the numerical value
of R in the equation R — x —-

Ro + Ri

Answer.

PROBLEM 10: If *>=10.79; /=1000, and A=4,107, find the
value of R in the equation R—p^.

Answer.

Preliminary Mathematics

77

PART II.

The following examples and explanations have been assembled and
added to this book for the purpose of assisting high school pupils
who desire an opportunity to receive help outside of school instruc¬
tion, and also to aid any who have had limited opportunity for
school instruction but who desire to pursue the study of Algebra
by themselves. The following examples and explanations will be of
considerable aid to those who propose to take college entrance exam¬
inations. Several college entrance examinations have been added to
the text, and the problems properly solved.

While many of the problems presented for solution in ordinary text
books dealing with Algebra are absolutely without practical applica¬
tion, one should not ignore the importance of the mental training one
receives from their logical and careful solution.

One unconsciously employs mental training at many times in one’s
career, and many years of careful and continuous practice are needed
to realize the attainment of a notable success that is effected in a
few moments. In other words, while an individual’s reputation is
apparently made in a few short moments, it has required years of
patient and continuous labor to enable the individual to successfully
meet brief demands at a critical time.

Students who study the examples and problems presented through¬
out this book should constantly keep in mind that these are typical,
serving as a guide in the process of solving similar problems, and
that many problems presenting different numerical values, may be
readily solved by a proper substitution of values, in the type forms
here presented.

78

Pro li mi nary Ma them a tics

LESSON XVI

DEFINITIONS.

Examples of different kinds of equations were cited on page 53
and several rules given pertaining to the solution of equations. There
arq a variety of algebraical equations, and a few of these will be
given particular consideration in the following portion of the text.

Symmetrical Equations, are those which are not affected by any
interchange of the unknown quantities.

x 2 -f- 2xy -f- y 2 = 25 is a symmetrical equation.

a — b —-4 is not a symmetrical equation.

Homogeneous Equations, are equations having their terms all of
the same degree, as regards the unknown quantities.

x 2 -f- 2xy -)- y 2 = 25, x 2 — 4 xy — 16y 2 = 10, and

* 2 + / = xy are homogenous equations.

Literal Equations, are equations whose terms consist entirely of
letters, x 2 -f- y 2 = xy, ax bx — mn, and

a _ b

r = c are all examples of literal equations.

Absolute Term: The so-called absolute term of any equation is
the term which does not contain any unknown quantity.

In the equation x 2 -f- 2 xy -(- y 2 = 25, the absolute term is 25.

Every equation may be considered to be the expression, in alge¬
braic language, of a particular question.

Thus the equation x -f- x = 24 is the algebraic expression of the fol¬
lowing question:

What number added to itself will give 24?

If the answer to this question be required it might be arrived at by
the following process:

By adding x to itself

2x = 24, and dividing both members of the equation by 2;
■arm 12, results.

Pre limin ary Ma thema t ics

79

The solution of a question by algebraic methods consists of two
distinct parts:

1st. To make the STATEMENT: that is to express the condi¬
tion of the question algebraically.

2nd. To solve the equation: or in other words to find a numerical
value from the unknown quantities; or to separate the known from
the unknown quantities.

Unfortunately no specific rules can be given to guide in the alge¬
braical statement of questions or problems. It is necessary to care¬
fully study the given problem to see if the algebraical equation is not
immediately evident, or if it is not possible to discover new condi¬
tions from which an equation may be formed.

The best method to pursue in order to become expert in stating
problems, is to state as many as possible.

The solution of problems, after the statement has been made, is
comparatively easy.

LINEAR EQUATIONS: EQUATIONS OF THE

FIRST DEGREE

EXAMPLE 16: A cistern is filled by a pipe, and the amount
let into the cistern during 8 minutes is 45 gallons more than the
quantity let in during five minutes. Find the number of gallons
let in per minute.

Let x denote the number of gallons let in during 5 minutes.

Then x -f- 45 will denote the number of gallons let in during 8 min¬
utes.

The average input, or the number of gallons let in per minute will
be expressed by;

x . ; from which is obtained,

5x -j- 225 = 8^-
x = 75

The number of gallons let into the tank per minute, is therefore 15.
In 8 minutes there would be 8 X 15= 120 gallons let in, and in 5 min¬
utes there would be 5 X 15 = 75 gallons. 120 — 75 = 45.

80

Preliminary Mathematics

PROBLEM 16: A cistern being filled by a pipe, the amount let
into the cistern during the first 8 minutes is 60 gallons more than the
quantity let in during the next 4 minutes. Find the number of gallons

let in per minute. 15 gallons. Answer.

%

Problem 16a: While a cistern is being filled, the amount let in
during the first 10 minutes is 50 gallons more than the quantity let
in during the next 4 minutes. Find the number of gallons let in
per minute.

EXAMPLE 17: A man is 33 years old, and his son is 12 years
old. How many years ago was the father four times as old
as the son?

Let x denote the number of years ago when the father was 4
times as old as his son.

Then; 33 — x= 4(12 — x )

and 33 — x = 48 — 4x

3x = 15
x — 5

That is, 5 years ago the father was four times as old as his son.

Proof: 33 — 5 = 28.

12 — 5= 7.

4 X 7 = 28.

PROBLEM 17: A man is 60 years old, and his son is 20 years old.
How many years ago zvas the father five times as old as the son;

10 years ago. Answer.

Problem 17a: A man is 50 years old, and his son is 15 years old.
How mnny years ago was the father six times as old as the son?

EXAMPLE 18: A man is 36 years old, and his son is 11 years
old. In how many years will the father be twice as old as his
son.

Let x denote the number of years hence, when the father will be
twice as old as his son.

36 -j- x = 2 (11 -f- x)

36 -f- x = 22 -f- 2x

x = 14

Then;

Preliminary Mathematics

81

In 14 years the father will be twice as old as his son.

Proof: 36 + 14 = 50

11 + 14 = 25
50 = 2 X 25

PROBLEM 18: A man is 50 years old and his son is 20 years
old. In how many years zvill the father be twice as old as the son?

In 10 years. Answer.

Problem 18a: A man is 49 years old and his son is 18 years old.
In how many years will the father be twice as old as the son?

EXAMPLE 19: A has $20, and B has $30. How many dollars
must B give to A in order that A shall have four times as much
as B has?

Let x denote the number of dollars that B must give to A.

Then; 20 + ;r = 4(30 — x)

20 + x = 120 — 4x
5x = 100
x = 20

Or B must give $20 to A.

Proof: 20 + 20 = 40

30 — 20 = 10

40 = 4 X 10

PROBLEM 19: A has $100, and B has $1000. How many dollars
must B give to A in order that A shall have 5 times as much money
as B has? $816 f. Answer.

Problem 19a: A has $25, and B has $35. How many dollars must
B give to A in order that A shall have three times as many dollars
as B has?

EXAMPLE 20: Find two consecutive numbers whose sum is
243.

Let x denote one number, and x + 1 denote the other.

Then ; x + (x + 1) = 243

2x = 242
x = 121

One number is 121, the other is 121 + 1 = 122.

Proof: 121 + 122 = 243.

82

Pre l im in ary Mathema tics

PROBLEM 20: Find the numerical value of two consecutive num¬
bers whose sum is 267. 133 and 134. Answer.

Problem 20a: Find the numerical value of two consecutive num¬
bers whose sum is 365.

Example 21: The difference of two numbers is 8, and their
sum is five times the smaller. Find the two numbers.

Let x denote the smaller number.

Then x -f- 8 denotes the larger number, and according to the given
conditions;

x + (■** + 8 ) = 5x
2x -(- 8 — 5.r
3x — 8

X =r J- —2f .

The smaller number is therefore 2| and the larger number is
2f-f-8 = lOf.

PROBLEM 21: The difference of two numbers being 20, and their
sum being 4 times the smaller, find the two numbers.

Smaller — 10, and larger = 30. Answer.

Problem 21a: The difference of two numbers being 35 and their
sum being 9 times the smaller, find the value of the two numbers.

EXAMPLE 22: The night at Petrograd (St. Petersburg) on
December 21, lasts 13 hours longer than the day. What is the
duration of the day, in hours?

Let x denote the length, in hours, of the day.

Then 24 — x denotes the length of the night, in hours.

By condition ; 24 — x — x -f- 13

2x=U

■ .slj

x = V =5£ hours.

The day is therefore only 5£ hours in length.

PROBLEM 22: The night at Chicago on December 21, lasts 4-
hours longer than the day. What is the duration of the day, in hours?

x

9 7 hours. Answer.

4

Preliminary Mathematics

83

Problem 22a: The night at New Orleans, on December 21, lasts
5 hours longer than the day. What is the duration of the day, on the
same date, in hours? •

EXAMPLE 23: A line that is 45 inches long is divided into
two parts. Two times the longer part exceeds three times the
shorter part by 30 inches. How many inches are there in each
part?

Let * denote the longer part.

Then 45 — x denotes the shorter part.

By condition ; 2* — 30 = 3 (45 — x)

2x — 30 =135 — 3x
5x = 165
x = 33

The longer part is 33 inches long.

The shorter part is therefore 45 — 33 = 12 inches long.

Proof: 2 X 33 — 30 = 3 X 12

66 — 30 = 36

PROBLEM 23: How may a line 77 inches long be divided into
two parts one of which is 2i times as long as the othert

Shorter is 22 inches long, and longer is 55 inches long.

Answer.

Problem 23a: How may a line 55 inches long be divided in two
parts such that two times the longer part exceeds three times the
shorter part by 20?

EXAMPLE 24: The difference of the squares of two numbers
is 221; if their sum is 17, what are the numbers?

Let x denote the larger number.

Then 17— x will denote the smaller number.

By the conditions imposed in the example;

x 1 —(17 — *) 2 = 221 (See page 46).

Then; x 2 —(289 — 34* a" 2 ) = 221, after changing signs of

all terms in parentheses,

x 2 — * 2 + 34* — 289 = 221

From which; 34* = 221 -{- 289 = 510

*= Vi° =15

The larger number being 15, the smaller is 17 —15 = 2.

Proof: 15 2 — 2 s = 225 - 4 = 221.

84

Prelim in ary M athematics

PROBLEM 24: The difference of the squares of tzvo numbers be¬
ing 400, and the sum of the two numbers being 40, find each of the
two numbers. One is 15; the other 25. Answer.

Problem 24a: The difference in the squares of two numbers is 7
and the sum of the two numbers is 7; find the value of each of the
numbers.

EXAMPLE 25: A has $16 less than B. If B gives $20 to A,
then A will have five times as many dollars as B. How many
dollars has each?

Let x denote the number of dollars that A has.

Then x +16 will denote the number of dollars B has.

And by the given conditions ; ;r + 20 = 5 (x + 16 — 20) = 5 (x — 4)

x -f- 20 = 5x — 20
4x = 40
x = 10

Therefore A has $10, and B has 10 -f- 16 = $26.

Proof: 26 —20 = 6; 10 + 20 = 30. 30 = 6 x 5.

EXAMPLE 26: A began in business with three times as
much capital as B. During the first year A lost $600, B gained
$200, and A had then only twice as much as B. What amount
of capital did each start with?

Let 3x denote A’s capital; in dollars

Then x will denote B’s capital at the start; in dollars.

By condition ; 3x — 600 = 2 (x + 200)

3x — 600 = 2x + 400

x = $1000. B started with $1000 and

A started with $3000.

EXAMPLE 27: A man has $6.25 in half dollars and quarters.
He has three times as many quarters as he has half dollars.
How many half dollars, and how many quarters has he?

Let x denote the number of half dollars.

Then 3x denotes the number of quarters.

Then by condition ; 50.r + 25 (3.r) = 625

(dividing by 25) 2 x3x = 25

5.r = 25
x = 5

Preliminary Mathematics

85

That is the man has 5 half dollars and 3X5=15 quarters.

Proof: 50X 5=250

25X15=375

625 cents or $6.25

EXAMPLE 28: The amount of $3.50 is made up of half dol¬
lars and dimes. If there are 19 coins altogether, how many are
there of each coin?

Let x denote the number of half dollars.

Then will 19 —x denote the number of dimes.

By condition; 50.r-f-10 (19— x) =350 cents

(dividing by 10) 5x-\-19 —;r=35

4.r=16
x= 4

There are 4 half dollars; therefore there are 19—4=15 dimes.

Proof: 4 X 50=200

15X10=150

350 cents or $3.50

86

Preliminary Mathematics

LESSON XVII

EXAMPLE 29: A merchant adds to his capital one-fourth of
it each year. At the end of each year he deducts $1200 for ex¬
penses. At the end of the third year he has, after the deduction
of the last $1200, one and a half times his original capital, minus
$950. What was his original capital?

Let x denote his original capital.

Then will x-\-1 x denote the amount at the end of the first year,
and (*+£*)+ i* denote the amount at the end of the second year.

By the condition stated in the example;

*+f.r—3 (1200) =l$x —950

From which is obtained; ar-f-fx—§# = 3600—950

{v++—f_r= 2650
Jv—f *= 2650
\x= 2650
*=10600

His original capital was therefore $10,600.

To verify this result proceed as follows;

3 X $1200 = $3600 expenses for three years.

$1^00X 3~ $7,950 total additions to capital.

$7,950— $3,600= $4,350 net addition to capital.

$10,600+ $4,350=$14,950

$14,950+ $950=$15,900 which is 1| of original capital.

EQUATIONS INVOLVING TWO UNKNOWN

QUANTITIES.

Whenever there are two unknown quantities in a given equation,
one of them may be eliminated by substituting its value in terms of
the other unknown quantity.

EXAMPLE 30: Given a room of such dimensions, that the
difference of the sides multiplied by the lesser side is equal to
36, and the product of the sides is equal to 360. Find the
numerical value of each side.

P relim in ary M at he ma tics

87

Let x denote the lesser side.

And y denote the greater side.

Then by the first condition ; (y — x) *= 36,
and by the second condition; .ry=360

From the first equation is obtained;

x y — x 2 — 36

360

Solving the second equation for y gives; y= -

x

Substituting this value of y, in the first equation gives;

x —x 2 =36. From which is obtained; * 2 =324

x

The value of x is therefore ± V 324=±18.

If *=±18, then by substituting this value in *y=360 is obtained;

± 183 )=360, y=^=±20.

18

The same result could also have been obtained by subtracting
xy —* 2 =36 from *y=360 as follows;

xy— 360
-j-* 2 — xy— — 36

* 2 = 324

x=± 18

The shorter side is 18, and the longer is 20 feet long.

EXAMPLE 31: What two numbers have a product equal to
30 and quotient equal to 3i?

Let * denote one number.

And y denote the other number.

Then by the first condition;

*y=30

and by the second condition, ^=3£=u>

30

From the first equation x=~ and from the second equation x=^y.
It is therefore evident that

I 0 _ 3 0

y‘ =?*=9

Hence; y=3.

Substituting in the first equation the value 3 just obtained for y
gives

*3=30

*=-V=10.

Prelim h i a ry Mat he mat ics

Verification of the foregoing;

3X10=30, and ^=3*.

It seems to be common experience that equations and problems
involving literal factors and terms are most difficult of solution,
therefore a number of typical cases are presented.

EXAMPLE 32: If the product of two numbers is a and their
quotient b, find the numbers.

Let x denote one number.

And y denote the other number.
Then by the first condition,

xy=a, and by the second condition,

-=b.

From the first equation and from the second equation x=hy.

Therefore,

Hence,

by= a

y

3

2 a

V?

V a

VF

And by properly substituting- in the first equation the value of y
j ust obtained;

V a

x —— =a
Vb

x=y/b

a

V a

= V uV b

EXAMPLE 33: If the sum of the squares of two numbers is
a, and the difference of their squares is b, what are the numbers?

Let x denote one number.

And y denote the other number.

Then by the first condition;

x^-\-y 2 —a, and by the second condition;
x 2 — y 2 —b.

Prelimina ry Ma th cm a tics

89

By adding the two equations ;

2x 2 =a+b
2 _(/—(— b

- r= V

a+b

By subtracting the two equations ;
2 v 2 — a — b

y

EXAMPLE 34: Find two numbers which are to each other
as m to n, and the sum of whose squares is equal to a 2 .

Let x denote one number.

And y denote the other number.

Then ~ ~ ~ and x 2 -f- v 2 = a 2
y n 1 -

From the first equation; x= —y and from the second equation

n'

x=± V a 2 — y r .

Therefore, — y=± V a 2 — y 2 .

n

Squaring both members gives;

Jtl 2 2 „,2

y —a — y
n 2

And solving for y gives;

90

Preliminary M at hematics

Substituting in the first equation the value obtained for y gives;

x _ m
a n n

V m 2 +n 2

a n m

From which is obtained; x= ,—> y —

V m +n n

a m

V wi 3 +« 2

EXAMPLE 35: What are the two numbers that are to each
other as m to n, and the difference of whose squares is equal
to b 2 ?

Let .i' denote one number.

And y denote the other number;

X

m

and

2 2 7.2

.i' —y —b .

.v

n

m

2 nf 2

X

y or

x = _ 0 y

n

n

Substituting the value for x 2 , just found, in the second equation
will give;

2

ff'l 2 2 I 2

2 y —y —b

n

, 2 2
o / m —n

y

y

/m —n ^

n

b 2 n 2

2 2

m —n

_ , I b 2 n _ _

3'-A/ —2-; ,

\ m — n ±

By proper substitution ; .r—

b n

1 / 5 2

V m — n
m b

± V m 2

n

If numerical values are assigned to b, m, and n, it may be observed
that m 2 must be greater than n 2 , in order that their difference shall be
a positive number. Otherwise the V m 2 — n 2 will mean the square root
of a negative number.

Preliminary Mathematics

91

EXAMPLE 36: A person distributes a sum of money among
a number of boys and girls. The number of boys is to the num¬
ber of girls as 3 to 4. The girls receive one-half as many dollars
as there are individuals, and the boys receive twice as many
dollars as there are girls. Altogether they receive $138. How
many boys and how many girls were there?

Let x denote the number of boys.

And y denote the number of girls.

Then by condition; _

v 4

x +y. 2y = 138
2

From the first equation; x — %y

The second equation may be reduced to;

x + y + 4 y = 276
' x —f- 5y — 276

Substituting in the last equation the value x—%y, gives;

fy+5y
‘ 5fy
x~y
y

Hence there are 48 girls. Since there are f as many boys as girls,
there are |X 48=36 boys.

276

276

276

1104

23

= 48

92

Prelimina ry Mat he n uiti c s

LINEAR EQUATIONS WITH THREE UNKNOWN

QUANTITIES.

EXAMPLE 37: Find three numbers such that the second is
three times the first, the third being four times the first, while
the difference between the second and the third is five.

Let x denote the first number.

Then 3x denotes the second number.

And 4x denotes the third number.

By condition ; 4x —3a'=5

x—5

The first number being 5, the second is 15, and the third is 4X5=20.

Proof: 20—15=5.

EXAMPLE 38: Find three numbers such that the sum of the
first two is 14, the third being twice the first, while the third
is greater than the second by 4.

Let x denote the first number.

Then 14— x denotes the second number.

And 2 x denotes the third number.

By condition; 2x —4=14— x

3.i*= 18
x— 6

The first number is 6; the second is 14—6=8, and the third is
2X6=12.

PROBLEM 3S: Find three numbers such that the second number
exceeds the first number by 3, the third exceeds the first by 10, while
the third is twice the first.

First =10; second =13; third = 20. Answer.

PROBLEM 39: Find each of three numbers, such that the sum
of the first and second is 5, the sum of the first and third is 6, while
the third is twice the first.

First = 2, second = 3, third = 4. Answer.

PROBLEM 40: If the difference between tzvo numbers is 4, while
a third number is 5 less than the sum of the first and second, and
the sum of the first and third numbers is 14, zvhat arc the numbers?

First number is 5, second is 9, and third is 9. Answer.

Prcli u i in a ry Mathematics

93

EXAMPLE 41: A man has 5 sons each of which is three
years older than the next younger. If the age of the oldest three
years hence will be three times the present age of the youngest,
what is the age of each son?

Let x denote the present age of the oldest son.

Then ; x —3 denotes the age of the second.

By conditions ;

2.r=39

*=194

The oldest son is 194 years old.

The second son is 16i

The third son is 134

The fourth son is 10 £ “ “

The fifth son is 7\ “

Proof: 194-f-3=3 X 74.

PROBLEM 41: A man has four sons each of which is four years
older than the next younger. If the age of the oldest three years
hence will be three times the present age of the youngest, what is the
present age of eachf

Oldest is 194; next is 15i; next is 114, and youngest is 74.

Answer.

Problem 41a: A man has three sons each of which is four years
older than the next younger. If the age of the youngest three years
hence is twice the present age of the oldest, what is the present age of
each son;

EXAMPLE 42: The sum of the three angles of any triangle
is 180 degrees (180°). If the second angle of a triangle is 4
degrees (4°) larger than the first, and the third is twice the sum
of the first and second, find the value, in degrees, of each angle.

Let x denote the first angle.

Then x-j-4 will denote the second angle.

And 2 [x-\-(x-\-4) ] will denote the third angle.

By the stated condition; x -f- (.r-j-4) -f-2 [x -f- (x + 4) ] =180

2x + 4 -f 4x + 8 = 180

6 .r = 168
x = 28

94

Preli m inary Mat he matics

The first angle is 28°; the second angle is therefore 28 + 4=32°,
and the third angle is 2 (28 + 32) = 2 X 60 = 120. The sum of the
three angles should be 180°.

28° + 32° + 120° = 180°

PROBLEM 42: If the second angle of a triangle is 5° larger than
the first, and the third angle is three times the sum of the first and
second, find the value of each angle of the triangle.

x + (jt + 5) + 3 [x + (x +5) ] = 180

First angle = 20°; second angle = 25°; third angle = 135°.

Answer.

Problem 42a: If one angle of a triangle is 6° larger than another,
while a third angle is 4 times the sum of the other two angles, find
the value of each angle of the triangle.

EXAMPLE 43: Find the value of each of three consecutive
numbers whose sum is 216.

Let x denote one number.

Then (.r+1) will denote the next consecutive number.

And (x + 2) will denote the next number.

By condition ; x + (x + 1) + (x + 2) = 216

3x = 213
x = 71

One number is 71; the next is 72, and the next is 73.

Proof: 71 + 72+ 73 = 216.

PROBLEM 43: Find the value of four consecutive numbers whose

sum is 26. 5; 6; 7, and 8. Answer.

Problem 43a: Find the value of each of four consecutive numbers
whose sum is 10.

EXAMPLE 44: The sum of three numbers is 100. If the first
and second are respectively 24 and 11 greater than the third,
what are the numbers?

Preli m ina ry M a thematic s

95

Let x denote the value of the third number.

Then will x -\-\\ denote the value of the second.

And x + 24 will denote the value of the third.

By condition ; x -\- x \ \ x -\-24 — 100

3x — 100 — 35 = 65
v = 21*

The numbers are therefore 45f, 32 f, and 2If. Answer.

PROBLEM 44: The sum of three numbers is 100. If the first and
second are 40 and 30 respectively greater than the third, find the
numerical value of the third number.

Third number = 10. Answer.

SOLUTION OF EQUATIONS BY FACTORING

EXAMPLE 45: Solve by factoring 12+— 7x 2 — lOx = 0.

The factors are; ( 4x 2 — 5.r) (3.r + 2) = 0

3x + 2 = 0
x=—i

Also; 4+ — 5.r = 0

x 2 —f .v = 0

+ =fjr. Dividing both members of this equation by x gives; x =f.

Another arrangement might be as follows;

x (12+ — 7x — 10) = 0 from which x = 0.
Also; 12+ — 7 x =10

x —

12 ■

111
1 2

Completing the square; + — ^jc+^j) 2 ={|+(o\) 2

r — i-— +V4P- + £3

_ x 24 — ~ 5 V 6 — 1 14

V — _j 2 1}—3 0 /-».

A — 2 4 - 1 - 2 4 — 2 4

r _16

-5 2 0

— 5 7 ti

—I or — f

The three correct values of x are therefore — f, f and 0. Each of
these values substituted in the original equation will satisfy the equa¬
tion.

It is evident that the method of solving equations by factoring,
involves much less time and mental effort than the method of “com¬
pleting the square”.

96

Preliminary Mathematics

SIMULTANEOUS EQUATIONS

EXAMPLE 46: Solve for x and y the simultaneous equations;

x

2 y

-x

23

x

= 20 +

2x — 59

2

y

— y ~ 3 = 30 — 73 ~~3y

x

18

The first equation may be reduced to;

23,r — x 2 — 2y + x _40 + 2x — 59 J2x — 19
23 — x 2 2

— 2x 2 — 4y + 48.r = 46a* + 19^ — 2x 2 — 437

— 4y + 48a* = 46^r + 19.r — 437

— 1 7 x — 4y —437

^ - 4y + 437

The second equation may be reduced to;

xy — 18y — y + 3 _ 90 — 73 + 3y _ 17 + 3y
x —18 ~ 3 ~ 3

3xy — 54y — 3y + 9 = 17x — 306 + 3xy — 54y
From which; \7x =9

1 7

x —

Preli m i nary M a the mat ics

97

LESSON XVIII

EXTRACTION OF THE SQUARE ROOT OF NUMBERS

The mention of square roots and squares of numbers was made on
pages 46 and 47, where the square or second power of a number was
stated to be the product resulting from multiplying any number by
itself once and the square root of a number to be that number which
multiplied by itself once, will produce the given number.

Evolution and Involution: The process of finding the square
root of any number is called evolution, while the reverse process:
finding the square of any number, is called involution.

Although there is no number of which any given power may not
be found exactly, there are many numbers of which exact roots can
not be obtained; but, by the use of decimals, may be approximately
evaluated to any assigned degree of exactness.

Rational Roots and Surds : Approximate roots are called surd
roots; while perfect, or accurate roots are called rational roots.

While the square of any number whether a whole number or a
fraction is always easily found by multiplying the number (either whole
or fractional) by itself once, the extraction or obtaining the square
root of a number is sometimes attended with difficulty and requires
particular explanation.

The first ten numbers are:

123456789 10

and their squares are:

1 4 9 16 25 36 49 64 81 100

From which it may be seen that the numbers in the second line are
the squares of corresponding numbers in the first line; while the
numbers in the first line are exact square roots of corresponding
numbers in the second line.

It may be noted that the square root of a number whose value
is between two numbers in the second line, will be between two cor¬
responding square roots appearing in the first line.

For example: the square root of 42, that occurs between 36 and 49,
will be between 6 and 7. Likewise the square root of 22, which occurs
between 16 and 25, will be a number between 4 and 5.

98

Preliminary Mathematics

The square of a number consisting of a single figure, will consist of
two figures; or the square of a number consisting of but one figure,
will consist of a number containing no figure of a higher denomina¬
tion than tens. Also the square of a number expressed by two figures,
will be a number containing no higher denomination than thousands.

The largest number expressed by a single figure is 9, and the
square of 9 is 81. The largest number expressed by two figures is 99,
and the square of 99 is 9801.

The square of 999 is 998,001.

The square of 9999 is 99,980,001.

For every figure in the given number there are two figures in its
square.

Every number of two figures, may be considered as made up of a
certain number of tens and a certain number of units. Thus 81 is
made up of 8 tens and 1 unit, and may be expressed in the form of
80+1.

If now the tens be represented by a and the units by b, the follow¬
ing will be true:

a -f- b = 81
(a + by — (81) 2

or a 2 + 2 ab -f- If = 6561 ; which proves the square of a number com¬
posed of tens and units, contain the square of the tens, plus twice the

product of the tens by the units, plus the square of the units.

The preceding example may be expressed as follows:

If a = 80 and b = 1

a 2 = 6400
lab = 160
b*= 1

6561

Suppose a cipher is added to each of the figures in the first line on
page 97 giving:

10 20 30 40 50 60 70 80 90 100

and the squares are:

100 400 900 1600 2500 3600 4900 6400 8100 10000

From which it is seen that the square of one ten is 100, the square
of two tens is 400; and in general, that the square of tens will contain
no figure of a less denomination than hundreds, nor of a higher de¬
nomination than thousands.

Preliminary Mathematics

99

To find the square root of 6724:

8 2

67 24
64 00

160

2

•

324

162

324

It is to be noted that 324 consists of twice the product of the tens
by the units plus the square of the units.

The operation that has in reality been performed is subtracting
from the number 6724 the square of 8 tens or 80; (80' = 6400) twice
the product of the tens by the units; (2 X 80 X 2 = 320) and finally
the square of the units; (2 2 = 4) that is the three components that
enter into the composition of the square of 80 + 2, and since the
result of the subtraction is 0, it follows that the square root of
6724. is 82.

To find the square root of 675684:

8 2 2

67 56 84

64 00 00

160+2= 162! 356 00

324 00

1640 + 2= 1642 ; 32 84

32 84

For the extraction of the square root of any number, the following
general rule applies :

RULE.

I. Separate the number into periods of two figures each, starting
at the right hand (the extreme period on the left may often contain
but one figure).

II. Find the greatest square contained in the first period on the
left and place its root figure over the period, similar to the figure of
a quotient in division. Subtract the square of the root from the first
period, and to the remainder add the second period for a dividend.

100

Prelim inary M a thematics

III. Double the root number already found, and place it on the left
for a divisor. Ascertain how many times the divisor is contained in
the dividend, excluding the right hand figure, and place the figure,
denoting the number of times, with the root number already found,
and also add it to the trial divisor.

IV. Multiply the divisor thus constituted, by the last figure of the
root number, and subtract the product from the dividend, arid to the
remainder add the figures of the next period for a new dividend. In
case any of the products should be greater than the dividend, diminish
the last figure of the root number.

V. Double the whole root number already found, for a nezv
divisor } and continue the operation as before, until all the periods have
been employed.

It should be noted as a general principle, that the number of figures
in the square root number, will always be equal to the number of
periods into which the given number is separated.

As the student will receive greater benefit from the careful study
of examples showing the principle pertaining to the extraction of the
square root of numbers, than by memorizing the “rules” given, a num¬
ber of typical examples and problems will be given.

EXAMPLE 47: Find the square root of 36729.

19 1.6 4

3 67 29.00 00
1

2 91 267
261

38 I) 629
381

382 6j 24800
22956

3832 4 1 184400

153296

31104 Remainder.

Preli minary Mat hem a t i cs

101

In this example there are two periods of decimals, which gives two
figures in the decimal portion of the square root number.

As there is a remainder, the root number is only approximate in
value.

Problem 47a: Find the square root of 2268741.

1506.23+. Answer.

Problem 47b: Find the square root of 7596796.

2756.22 +. Answer.

Problem 47c: Find the square root of 101.

10.04987 +. Answer.

V2 = 1.41421 VT0= 3.16228

VT= 1.73205 V~l2 = 3.4661

VeT= V2 V 3 = 2.44949

Problem 47d: What is the square root of 96? Find result to
three decimal figures.

SQUARE ROOT OF FRACTIONS.

The second power or the square root of a fraction is obtained by
squaring the numerator and denominator separately, and writing the
result as a fraction. That is, the square root of a fraction is equal
to the square root of the numerator divided by the square root of
the denominator.

*>

For example, the square root of +2 ^ anc ^ square root of

x~ + 2xy -f- y~ _ x -f- y
x 2 — 2,ry -f- y 2 x — y

EXAMPLE 48: Find the square root of f.

VT= 1 V"4 = 2

vl=i

PROBLEM 48: Find the square root of , 9 ti .

Problem 48a: Find the square root of £f-

3

4 ’

Answer.

102

Preliminary Mathematics

If neither the numerator nor the denominator of a fraction is a per¬
fect square, multiply both numerator and denominator by the denom¬
inator, and divide the square root of the new numerator by the origi¬
nal denominator; giving the approximate root.

EXAMPLE 49: What is the square root of ?

4 X f=V V 28 = 5.2914

, 5.2914

anc * ^ = 1.3228. Answer.

PROBLEM 49: What is the square root of—?

1.2472. Answer.

Problem 49a: What is the square root of § ?

Another method of finding the square root of a fraction is to
change the fraction into a decimal, and extract the square root.

EXAMPLE 50: What is the square root of-?

] reduced to a decimal, by dividing the numerator by the denomi¬
nator is 1.75, and the square root of 1.75 is 1.3228. Compare with
Example 49.

PROBLEM 50: What is the square root of - ?

0.9354. Answer.

Problem 50a: What is the square root of £ ?

A few examples and problems in literal expressions will be given
as follows:

EXAMPLE 51: Find the square root of 25a'x 4 .

Sax'. Answer.

The proof of the above is to multiply the result by itself.
Sax 2 X Sax 2 — 25 a 2 x*. Observe rule on page 34, regarding exponents.

PROBLEM 51: Find the square root of 225a 6 b*x 2 .

15 a*b'x. Answer.

Problem 51a: Find the square root of 81a 8 y 4 A* 2 .

Preliminary Mathematics

103

After having found a value for the square root of a number,
always multiply the value found, by itself, in order to verify the
correctness of the result obtained.

It may be observed that when a monomial* is a perfect square, its
coefficient is a perfect square and all the exponents of the literal fac¬
tors are even numbers.

If a monomial is not a perfect square its approximate square root
may be found as follows:

What is the square root of 98a/; 4 ? Since the square root of the
product of two or more factors is equal to the product of the square
roots of the factors, the given expression may be as follows;

V98 ah* = V49 X V2 X Va X VT 4
= 7x V"2 Xfl- xb 2
= 7 X 1.414 X a-/; 2
= 9.898 J b 2

In a similar manner;

V45 a*b*c 2 d= V 9a 2 b 2 c 2 X 5 bd = 3abc VJbd

RULE FOR EXTRACTING THE SQUARE ROOT OF

MONOMIALS.

I. Find the square root of the coefficient .f
II. Divide the exponent of each letter by 2.

To determine if a given number has any factor that is a perfect
square, divide the given number by each of the perfect squares,

4, 9, 16, 25, 36, 49, 64, 81, etc.,

and if it is not divisible without a remainder, it does not contain a
factor that is a perfect square.

EXAMPLE 52. A certain general has in command an army
of 141376 men. Find the necessary number in “rank and file*’
to form them into a square.

The necessary number in “rank and file” must be the square root
of 141376, or 376 men.

* See page 40.

+ See page 40 for definition of coefficient.

104

Preliminary Mathematics

PROBLEM 52: Find the greatest possible number of hills of corn
that can be planted on a plot of land comprising a square acre, if the
centers of the hills are to be no nearer each other than three and
one-half feet.

4165 hills. Answer.

EXAMPLE 53: Divide 100 into two numbers, such that the
sum of their square roots will be 14.

Let * denote one number, and (100 — .r) denote the other number.

Then by condition;

V x —j— V 100 — x = 14
and squaring both members gives;

* + 2 V * V100 — * + (100 — *) = 196

Which may be reduced as follows:

2 V* V 100 — * = 96

V* V 100 — x = 48 (dividing both members by 2)
x (100 — *) = 2304 (squaring both members)

100* — * 2 = 2304

Rearranging and changing signs;

* 2 - 100* = — 2304

Completing the square, by adding to both members the square of
one-half the coefficient of *, gives;

* 2 — 100* + 2500 = 2500 — 2304

* — 50 = ± VT96 = ± 14

* = 50 ± 14 = 64 or 36

If * = 64 then 100 — * = 36.
x = 36 then 100 — * = 64.

The square root of 36 = 6, and the square root of 64 = 8.

6 + 8=14.

The preceding Example 53 may be better comprehended after a
study of page 109.

PROBLEM 53: Divide 106 into two numbers, such that the sum
of their square roots may be 14.

25 and 81. Answer.

Prelim in ary Mathemat i cs

105

Problem 53a: Divide 10000 into two numbers, such that the sum
of their square roots may be 140.

EXAMPLE 54: The sum of a number plus its square root
is 90, what is the number?

Let x denote the number, then;
x + V~x = 90

x VF+G) 2 = 90 + (i) 2 = 90+± = 36i
V~r + i = ± V 1 ! 1 = ± V 9
V x — — \ ± = 9 or — 10

Therefore, x — 81 or 100.

81 is the number required.

106

Preliminary Mathematics

LESSON XIX

EXPLAINING EQUATIONS OF THE SECOND DEGREE.

QUADRATIC EQUATIONS

Mention was made on page 54 as to the meaning of the degree
of an equation. It may be said further that if an equation contains
two unknown quantities, it is of the second degree when the greatest
sum of the exponents of the unknown quantities in any term, is equal
to 2.

Thus; x'~a, ax 2 bx — c, and xy -J- x = e 2 are equations of the
second degree.

Equations of the second degree may be separated into two classes:

1st. Equations involving only the square of the unknown quantity,
and known terms; called Incomplete Equations.

2nd. Equations which involve the first and second powers of the
unknown quantity, and known terms; called Complete Equations.

Examples of incomplete equations are;

x 2 + 3x 2 -4 = 9
3x 2 — 5x 2 5 — a

Examples of complete equations are;

3x 2 — 5x -3x 2 -\-b — c

2x 2 — 8x 2 — x — a = b

Every incomplete equation can be reduced to an equation consisting
of two terms, of the general form;

x 2 = m, which is sometimes called a pure quadratic.

By extracting the square root of both members, the equation be¬
comes ;

x = V m; or x = ± V m

An example illustrating the foregoing will be given.

EXAMPLE 55: What number multiplied by itself will give
6561?

Let x denote the number.

Then x X x or ,r* = 6561.

x= V 6561 =81. Answer.

Preliminary Mathematics

107

PROBLEM 55:,
49a*b 2 x 6 ?

What number multiplied by itself will give

7a~bx 2 . Answer.

Problem 55a: What number, multiplied by itself will produce
1521 aWy 4 ?

Problem 55b:

625c* d*y*?

What number multiplied by itself will give

25 c 2 d'y*. Answer.

A root* of an equation may be defined as any expression which sub¬
stituted for the unknown quantity, will satisfy the conditions of the
equation; that is, will render the two members numerically equal to
each other.

Thus in the equation x 2 = 25,

there are two roots, -j- 5 and — 5 ; since either of these numbers sub¬
stituted for x will satisfy the condition.

Hence it may be stated that;

\

1st. Every complete equation of the second degree has two roots.
2nd. That these roots are numerically equal, but have unlike signs.

EXAMPLE 56: What are the roots of the equation 3x 2 -f 6 =
4x 2 — 10?

Rearranging the terms gives:

— x 2 = — 16

And changing all signs ; x 2 = 16
Therefore; x — — 4

The roots are -f- 4 and — 4. Answer.

2

PROBLEM 56: Find the roots of the equation ^x 2 — 8 = ^ -j- 10.

x — + 9 and x = — 9. Answer.

Problem 56a: Find the roots of the equation 4x l 4~ 13 — 2x 2 = 45.

* See page 47.

108

Prelim in ary M a t hematics

EXAMPLE 57: Find the numerical value of a number such
that if one-third of it is multiplied by one-fourth of it the
product shall be 108.

Let x denote the number.

Then by conditions imposed ;

x , x

3 X 4

108

108

x
12

x 2 = 1296
x — ± 36

PROBLEM 57: Find the number whose square, plus 18, shall be
equal to one-half its square plus 301.

Number — 5. Answer.

EXAMPLE 58: Find the value of two numbers which are
to each other as 5 to 6, and the difference of whose squares is 44.

Let x denote the greater number.

Then £,r will denote the lesser number.

Bv the conditions ; x 2 — x 2 = 44

36.r 2 — 25.r 2 = 1584
ll.r 2 = 1584
x 2 = 144
*=12

If * — 12, then f of 12:= 10. Therefore one number is 10 and the
other is 12.

PROBLEM 58: Find the numerical value of two numbers which
are to each other as 3 to 4, and the difference of whose squares is 28.

The two numbers are 6 and 8. Answer.

Problem 58a: Find the numerical value of two numbers which
are to each other as 5 to 3, and the difference of whose squares is
400.

Problem 58b: Find the numerical value of two numbers which
are to each other as 5 to 11, and the sum of whose square is 584.

The two numbers are 10 and 22. Answer.

Preliminary Mathematics

109

Those who intend to enter the engineering profession should
very carefully train themselves to solve quadratic equations. It is
necessary for Mechanical, Steam, and Electrical engineers, on many
occasions, to be able to readily solve equations of the second degree,
in connection with the computing of all kinds of efficiency, and ques¬
tions of power. Especial consideration is therefore given to the prin¬
ciples of second degree equations in this book and students are
strongly advised to carefully work out as many equations as possible
from other text books; employing as guides the Examples in this
book, and employing as many methods of proving or verifying results,
as is possible.

A few general principles relating to quadratics will now be con¬
sidered.

Suppose a given equation is;

x 2 + 4x — 12 = 0

Transposing the —12 gives; jr 2 -f- 4^r = 12. If now 4 be added to
both members of the equation, the left hand member becomes a per¬
fect square,* giving;

.r 2 + 4x + 4 = 12 + 4 = 16

Taking the square root of each member gives;

x -f 2 = ± V 16 = ± 4

From which, x — — 2 ± 4 = -)- 2 or — 6

It should be noted that 4, the number added to both members of the
equation, is equal to the square of | the coefficient of the term con¬
taining x. That is £ of 4=2, and 2 2 = 4.

Again suppose the following equation is given ;

4x 2 + 32x — 132 = 0

Transposing the absolute term to the right hand side gives:

4x s + 32x = 132

Dividing the coefficient of each term by 4, gives;

x 2 -f 8x = 33

Adding to both members the square of one-half the coefficient of
x gives;

,r 2 + 8x 4 -16 = 33 + 16 = 49

Taking the square root of each member gives;

x 4- 4 = ± 7

x — — 4 ± 7 = + 3 or — 11

*lt also happens that the second member is a perfect square in this case.

110

Prclimi nary M a them at i cs

Substituting the value —11 in the given equation there results;

4 X 121 + 32 X (—11) = 132
484 — 352 = 132
132= 132

Substituting the value x = + 3 gives ;

4x9 + 32x3 = 132
36 + 96= 132
132=132

A form denoting a perfectly general case may be expressed as fol¬
lows ;

ax 2 + bx + c = 0

Dividing each member by a gives;

; . b c

* + a x = - 5

If to both members is added the square of one-half the coefficient
of x there results;

*’+‘- h +/ = (£) 2

f _ Av _ f _ i _
a 4a ~ a a Ma

Expressing the square root of each member gives :

c )

c)

This
lows;

a = 4
b = 32

may be applied to the equation

] Substituting these equivalent
132 j sion for x gives;

4.r 2 + 32.r—132 = 0 as fol-

values in the above expres-

.r = — + ± V £(1014+132;

= — 4± V £(64+132)

= — 4 ± VW= ~ 4± V49

= — 4 ± 7 = + 3 or —11.

While any equation may be solved by substituting the proper values
in the typical or general form, it is far better to treat each equation
by itself, applying the following general rule.

Preliminary Mathematics

111

1st. Place the absolute term on the right hand side of the equation.

2nd. Divide the coefficient of each term by such a number as to
reduce to unity, the coefficient of the term containing the second
power of the unknown quantity.

3rd. After arranging the terms according to the descending powers
of the unknown quantity, change all the signs so as to cause the sign
of the term containing the second power to be positive.

4th. Add to both members of the equation, the square of one-
half the coefficient of the term containing the first pozver of the un¬
known quantity.

5th. Extract the square root of both members of the equation.

6th. Solve for x, by placing x on the left hand side of the equa¬
tion and all the known or absolute quantities on the right hand side.

These rules may be expressed differently as follows :

To solve a complete quadratic;

Reduce the equation to the general form x 2 -px — q, and com¬
plete the square by adding to both members the square of one-half
the coefficient of x. Extract the square root and solve for x the
equation of the first degree thus formed.

The general form in this case will be;

.r 2 + px + (;) = ( 2 ) +1

x + l = ± }tY±S

x — — l (p — V p 2 + 4<7)

Applying this to the equation, 4x 2 -\-32x —132 = 0, the following
is obtained;

x 2 + Sx — 33

p = 8 and q = 33.

Then ; x = — *(8 ± V 64+ 132)

= — \ (8 ± VT96) = — i (8 ± 14)

= — |X 22 or — i (— 6 ) = + 3
= — 11 or + 3 ; as found on page 109.

In every case thus far considered, the sign of the term involving
the first power of x has been positive. It will be well to now con*
sider the case when the sign of this term is negative.

112

Preliminary Mathematics

Suppose the given equation is of the form;

x 2 — px -\- q = 0

Then ; x 2 — px — — q.

Completing the square gives;

x

. , P±_

P* + 4 — + 4

Q-

And extracting the square roots;

It should be noted that when the sign of the term containing the first power of x was
positive the root of the first member was expressed by j? + ^but when this term is nega¬
tive, the root is expressed by x — ^

Solving for x ;

■ r =2 ± yj-

Z 1 ' 4<7

=i (P ± V> 2 “

4 q)

A careful study of the following examples and problems will enable
anyone to readily solve any quadratic found in ordinary practice.

After one has found the squares of a few algebraical expressions,
many square root values may be found by a simple inspection of the
given expressions. Consult pages 46 and 47.

A few typical expressions may be stated as follows;

(a+fc) 2 = a 2 + 2ai> + & 2 ;

(a — b) 2 = a 2 -2ab + b 2 ;

(a ~f- 2b) 2 = a 2 4ab -f- 4£> 2 ;

(5a — 2 b) 2 = 25a 2 — 20 ab + 4 b 2 ;

(5a + 2f>) 2 = 25a 2 + 20 ab + 4 b 2 ;
(jr —(— 5) 2 = -i* 2 -f- lO.r -f- 25 ;

(x — 5) 2 = .r - 10.r + 25 :

(* + y) 2 =r* 2 + 2*y + y 2 ;

(x — y) 2 = x 2 — 2xy -j- y 2 ;

(o + 4) 2 =a 2 +a+i; '

(a— IY = a 2 — i-a-fiV;

(a b) = square root of d'2abb 2

(a — b) = square root of a 2 — 2 ab-\-b 2

(a -f- 2b) = square root of a 2 -f- 4ab -j- 4b 2

(5a — 2b) = square root of 25a 2 — 20 ab -f- 4 b

(5a-{-26) — square root of 25a 2 -f~ 20ab -j- 4&

{x + 5) = square root of x 2 -(- 10.r -f- 25

(x — 5) = square root of x 2 —lO.r-j-25

(^r + y) = square root of x 2 -j~ 2xy - f-y 2

(x — y) = square root ot x 2 — 2xy -j- y 2

(a-j-i) — square root oia 2 -\~ aJ r^

(a —4) = square root of a 2 —4a-f-yV

See Problem 10b: page 48.

EXAMPLE 59: Solve for x; 3ax 2 -f- 2bx -f- c = 0.

Rearranging gives;

3ax 2 -j- 2bx = — c
Dividing each term by 3a gives;

Pre lim inary M a the ma tics

113

Completing the square by adding the square of 4 the coefficient
of x to both members of the equation gives;

&)*=-£+<,-&)•

3/7 1 9 a 2

— 3ac -f- h 2

Ir — 3a c

9 a 2

Extracting the square root gives;

b

v-f

3 a

V

b 2 — 3ac

x -~3a ±

9 a 2
Vb 2 — 3ac

±

9cC

V5~ — 3a c
3a

3a

b 4-

3ac

or

b — V b — 3ac

3a 3a

This example may also be solved by substituting the proper values
in the general equation given on page 111.

In this case p — f-, and q = — /

c

3/7

SIMPLE QUADRATICS

EXAMPLE 60: Solve the following equation to obtain the
value of x.

V 4x -f- 2 4 — V x

4 -f- V x V x

The above is the same as;

2Vx + 2 4— V x

4+ y/~x y/x

Multiplying both members by V x gives;
2x -f- 2 V x

4 —f- V x
4 — V x gives ;

=4 — V x, and multiplying both members by

2x -f- 2 V .r — 16 — x, which combines to give;

3„r -f- 2 V x — 16. Dividing each term by 3, gives;

jr-hlrV^—W 5 Completing the square by adding to both members the
square of $ the coefficient of V x gives;

114

Preliminary Mathemcitics

x + * v a- + ay= v+( i ) 2 =y+i=v-+i-=-v

Extracting the square root of both members gives;

V x +i =— v v == — I

y/x = — h±i=%or —|

From which .r = (f) 2 or (— f) 2 = or V=4 or °,, 4

EXAMPLE 61. Solve the following equation to obtain an
expression for x. In short, solve for x.

1

11 1

a r b~^~ x

a -J- b + x a b
Reduce the second member to a common denominator, giving;

1 _ bx -|- ax -|- ab

a -f- b -f- x ab. r

Multiplying both members by abx gives;

abx , | , ,

a -f- b -f- x

Multiplying both members by a -f- b + x, gives ;

abx — (a -f- b -|- x) (bx ax -f - ab)

— a 2 b -f- a 2 x -f- 3abx -f- ab' + b'x -f- ax 2 -J- bx 2

From which is obtained;

ax 2 -f- bx 2 -f- c?x -f- Ifx -j- 2 abx arb -j- ab 2 — 0

(a -)- b)x‘ -f (a 2 -j- 2 ab + b 2 )x — — a 2 b — ab 2

Dividing both members by ( a-\-b ) gives;

.r 2 + (a + b)x= - T~P~= - ,'T = — ab

a + b a -f- b

Adding the square of 4the coefficient of x to both members gives;

(a -)- b ) 2 _ (q-t-fr ) 2 (a + b ) 2 — lab

a* 2 -f (q + b) x

4 4 4

Extracting the square root of both members, gives;

■ q + b • q — b
x 4- 1 -= ±-

2 2

From which x — — a — ±

2

q

q

b 4- q

b or — a

EXAMPLE 62: A square picture is surrounded by a frame.
The side of the picture exceeds by an inch the width of the
frame, and the number of square inches in the frame exceeds by
124 the number of inches in the perimeter of the picture. Find
the area of the picture, and the width of the frame.

Preli in i n ary Mat he in a tics

Let x denote the area of the picture.

Let one side of the picture be denoted by V x.
Let V x — 1 denote the width of the frame.

Fig. 2

[2(V.r—1) -f- V.r] 2 — x — area of frame.
Then;

[2( y/~x — 1) + V x] 2 — x — 124 = 4 V~x
(3y~x — 2) 2 — x — 124 =4 yjx
9x — 12 \fx + 4 — * — 124 = 4V~r
8.r — 16 V x = 120
x — 2 y/ x — 15

x-2Vx+ (l) 2 = 15 -f- (1) 2 = 16

V x — 1 = ± 4

V x = 5 or — 3

,r = 25 or nine. Answer.

116

Preliminary Mathematics

EXAMPLE 63: The circumference of the front-wheel of a
carriage is less by 4 feet than that of the back-wheel. In trav¬
elling 1200 feet, the front-wheel makes 25 revolutions more
than the back-wheel. Find the circumference of each wheel.

Let * denote the circumference of the back-wheel.

Let x — 4 denote the circumference of the front-wheel.
1200

Then

-denotes the number of revolutions of the back-wheel

x

and

1200
x — 4
1200

Then;

denotes the number of revolutions of the front-wheel.

1200

x x — 4
1200* — 4800 =

— 25

= 1200* -f 25* z

100 *

25* 2 — 100* = 4800
* z — 4* = 192
* 2 — 4* + 4 = 196
* — 2 = ± 14

* — 16 or — 12

Circumference of back-wheel = 16 feet,

Circumference of front-wheel = 16 — 4 = 12 feet. Answer,

The diameter of either wheel may be found by dividing each cir¬
cumference by 3.1415.

Preliminary Ma the m a t ics

117

LESSON XX

SIMPLE QUADRATIC EQUATIONS

EXAMPLE 64: A and B gained $2100 in trade. A’s money
was in the firm during 15 months, and he received in principal
and gain, $3900. B’s money, which was $5000, was in the firm
12 months. How much money did A put into the firm?

Let x denote A’s principal put into the firm.

Then 3900 — x denotes A’s gain; while 2100—(3900 — x) denotes
B’s gain.

3900 — x _5 x x

2100 — 3900-f -x ~4 X 5000~4000

15,600.000 — 4000* = 2100.r — 3900* + * 2
* 2 + 2200.r = 15,600,000
.I* 2 + 2200.1* + lTOO 2 = 15,600,000 + 1,210,000
* + 1100 =±4100

x = ± 4100 — 1 100 — $3000. Answer.

A’s gain — $900. B’s gain = $1200. While B’s capital was in the
firm during a shorter time, he received $300 more than A did, be¬
cause of the larger amount invested.

EXAMPLE 65: The sum of $120 was divided between a cer¬
tain number of persons. If each person had received $7 less,
he would have received as many dollars as there were persons.
Find the number of persons.

Let x denote the number of persons.

Then

120

x

denotes number of dollars each received.

By the condition imposed;

120 _ 7 — r

x

120 — 7.i* = .i* 2

* 2 + 7x = 120

x 2 +7x+ (I) 2 =120+ G) 2 = •-+’
x + | = ± % :!

x = ± 3 .f— 1=8 (or — 15). Answer.

118

Preliminary Mathematics

EXAMPLE 66: A person’s income is $5000. After deducting
a percentage for income tax, and then a percentage, less by one
than that of the income tax, from the remainder, his income is
reduced to $4656. What is the rate per cent of the income tax?

Let x denote the rate of the income tax.

Then; 5000 X —— denotes the amount of income tax.

100

5000 — ^r^r~' r denotes the remainder after deducting the income tax.

100

(5000 — 50.r)

x —1

100

- denotes extra tax on remainder.

Then by conditions of the statement;
5000 — 50.r — (5000 — 50*) - — 4656

100

5000 — 50.r

— (50* — 50 — -= 4656

5000 — 50* — 50* -j- 50 -j- --- = 4656

9 ?

*

2

201

x = — 394

x 2 — 201.r* = — 788
/ 201 \

* 2 —201* + (+r'

201 __ j __
2 ' 2

x

) =- 7 88 + (f)

193

= +

17149

201 193 1QV ,

x — — ± — — 197 or 4.

2 2

Answer.

The rate could not be over 100, therefore the value 4 is correct.

EXAMPLE 67: If $2000 amounts to $2205 when put at com¬
pound interest for 2 years, the interest being compounded an¬
nually, what is the rate per cent per annum.

Let x denote the rate of interest, in per cent.

Then the amount at the end of the first year is expressed by;

2000-f ( - X 2000 ^

Vioo /

The amount at the end of two years is expressed by;

[ 2<X)0 + (rs x 2000 )]+1 2000 +(4 x 2000 ) 41 = 2205

This reduces to;

2000 +20x + 20x + - = 2205

Preliminary Mathematics

119

4- 40.r = 205
5

.r + 200.r = 1025

x 2 -f 200.r + 100 * = 1025 + 10000 = 11025
x + 100 = ± V 11025 = ± 105

x — — 100 -f- 105 = 5 per cent. Answer.

EXAMPLE 68. A man travelled 105 miles. Had he gone 4
miles more in an hour, he would have completed the journey in
9 : ‘, hours less than he did. How many miles an hour did he
travel?

Let x denote his rate of travel in miles per hour.
Then according to the condition ;

105 _K)5_

x x + 4

From which is obtained;

105 105

x x -f- 4

This reduces to ;

105 315 +28.r + 112

x 3x -f 12

315.1' -f 1260 = 315.1* -f- 28.r 2 + 112x
28.F+ 112jt = 1260
x 2 + 4x = 45

x 2 + 4x + (2) 2 = 45 + 4 = 49
x + 2 =±7

x — — 2 ± 7 = 5 miles per hour.

Answer.

EXAMPLE 69: A crew can row a boat 8 miles down stream
and back again in 4^ hours; if the rate of the stream is 4 miles
an hour, find the rate at which the crew can row the boat in
still water.

The solution is based on the fact that in general a rate =

distance

time

That is; that the rate of any moving body is equal to the distance
passed over by the body, divided by the time required in passing;
assuming uniform motion. This is expressed in symbols by;

d
t

rate

120

Preliminary Mathematics

Let x denote the rate in still water.

Then the rate going down streams (>-f-4), and the rate going up
stream is (x — 4). The distance going down is 8 miles, and is the
same going up.

Let t denote the time in going down and t' the time in going up.
Then;

t — —--and t' — —-- But the total time is

x + 4 x — 4

hours.

Then t+t' = —^ —+—— = M

x -\- 4 ’ x — 4 5

Or; 8.r — 32 -f 8x + 32 = 2 / (x 2 — 16) x*-l6='x+4) (x-4)

80x = 24 (x 2 — 16)

Rearranging gives; x 2 — V° A' = 16

Completing the square; x 2 — 1 3 Q v4-( J 6°) 2 =16+

Whence x — — V 5 or :1 , ; 6 = 6 miles per hour. Answer.

EXAMPLE 70: A rectangular garden is surrounded by a
walk 7 feet wide; the area of the garden is 15,000 square feet,
and of the walk 3696 square feet. Find the length and breadth
of the garden.

Figure 3 shows the arrangement.

Let x denote the length of the outer
rectangle. Then 15000 + 3696 or

X X

denotes the breadth of the outer rectangle.

The length of the garden is therefore
denoted by (x — 14), and the breadth of

the garden by

x — 14

The area of the garden is equal to its
length times its breadth, and is 15000 sq. feet.

Therefore;- 1 ? 000 = Mg<L_ 14
x — 14 x

From which;

15000.1- — 18696.1- — 14 X 18696 = 14.r 2 +

Area of walk — 3696 sq. ft.

-7—»

Area m
15000 sq. ft.

Fig. 3

196 x

Preliminary Mathematics

121

Which reduces to;

14 a- 2 — 3892a* = — 14 x 18696
.r 2 - 278a- = — 18696

x“ — 2 78x + (139) 2 3= — 18696 + (139) 2 = 625
x — 139= ± V625 = ±25
.r = 139 ± 25
— 164 or 114

The length of the garden = 164 — 14 = 150 feet.

The breadth of the garden = 100 feet.

150

EXAMPLE 71: A merchant sold a quantity of wheat for
$56, and gained as many per cent as the wheat cost dollars.
What was the cost of the wheat?

Let x denote the cost of the wheat, in dollars.

One per cent is therefore denoted by if the gain was x per

cent, the gain would be denoted by y^fo (— x X T * Tf ). The gain
added to the cost will give the selling price.

The condition is therefore expressed by;

A- + /o 8 o = 56

-V "T IOOat — 5600 multiplying each term by 100

Completing the square;

a- 2 + 100a* + (50) 2 = 5600 + (50) 2 = 8100
Extracting roots ; x + 50 — ± 90

Cost of the wheat was; x = — 50 ±90 =40 dollars. Answer.
This may be verified as follows;

The wheat cost $40. 40% of $40 = $16. $ 40-f-$16 = $56.

EXAMPLE 72: A farmer bought a number of sheep for $378.
Having lost 6, he sold the remainder for $10 a head more than
they cost him, and gained $42. How many did he buy?

Let x denote the number of sheep purchased.

178

Then — = cost in dollars for each sheep.

1 £

(^+10 ) (x - 6) = $378 + $42

122

Preliminary Mathematics

.178 -f 10a- —-60 = $420. Multiplying by x, gives ;

318.V + 10.r — 2268 = 420a*

or 10.r 2 — 102 a* = 2268. Dividing by 10 gives ;

Completing the square gives;

21 sheep. Answer.

It is obvious that the converse of the preceding Example would be
the actual practical reality. The farmer would know how many he
purchased at first, and the number lost, and selling the remainder at
a certain price, would realize a certain sum; either gaining or losing.

EXAMPLE 73: Find a number such that the sum of its cube,
twice its square, and the number itself, is twenty times the next
higher number.

Let x denote the number, then the condition stated is expressed by;

x s -f- 2x 2 -f- x = 20 (x —j- 1) (The next higher number is the

x 3 -j-2x 2 -j-x = 20x-j-20 given number plus one.)

From which;
a- 3 + 2x~ — 19a- — 20 = 0

One method of solving such an equation will be to substitute differ¬
ent numerical values for x until one is found that satisfies the equa¬
tion. In applying this method it will be logical to begin with l and
continue with the positive integers in order, until one is found that
satisfies the equation.

Substituting 1; gives l —(— 2 — 19 — 20 = 0

or 3 — 39 = 0; which is evidently not correct

Substituting 2; gives 8 -f- 8 — 38 — 20 = 0

or 16 — 58 = 0; also not correct.

Substituting 3 ; gives 27+18 — 57 — 20 = 0

or 45 — 77 = 0; not correct.

It may be now noted that as the numbers substituted, increase in
value, the more nearly the required condition is fulfilled.

Preliminary Mathematics

123

Substituting 4 ; gives 64 -f- 32 — 76 — 20 = 0

or 96 — 96 = 0; 96 = 96 which does fulfill the
condition, x is therefore = 4.

If x =■ 4 then x — 4—0, and (a*—4) is one of the factors of
at 3 + 2x 2 — 19 x — 20 = 0

Dividing a- 3 -(- 2a* 2 — 19a* — 20 = 0 by (x — 4) gives a* 2 -f- 6x -j- 5, the
factors of which may be seen by inspection to be (A'-f-l) and
(a* -f- 5).

It is evident therefore that;

(x — 4) ( x -f- 1) (a- -f 5) = 0

Dividing both members by (a* — 4) (a*-|-1) gives;

x -j— 5 — 0

from which; x= — 5; another value that fulfills the condition.

Also ; x —1 = 0

x = — 1; a third correct value.

It is obvious that negative numbers could have been substituted in
the original equation as well as the positive integers. Then the
value — 1 could have been substituted and then; — 1 —j— 2 — 1=0, or
2 = 2 would have resulted at once.

It is further evident that fractional values may always be substi¬
tuted as well as integers; and negative fractions as well as positive
ones.

EXAMPLE 74: A man bought a horse for a certain sum.
and after a time sold the horse for $24. By the' sale the man
loses as much per cent upon the purchase price as the horse
cost him. What did the man pay for the horse?

Let x denote the purchase price in dollars.

Then x — 24 expresses the loss in dollars.

Since the loss was x per cent, the loss was — on each dollar; upon

100

o

I*

x dollars the loss would be x X —— — ——

100 100

The following will then express the condition ;

— = a* — 24

124

Preliminary Mathematics

x 2 =100* — 2400
x 2 — 100 a' = — 2400
x 2 — 100.r + (50) 2 = 2500 — 2400
x — 50—± VTOO = ± 10
x = $60 or $40

Both values satisfy the conditions.

EXAMPLE 75: Find four consecutive numbers such that if
the first two be taken as the digits of a number, that number is
equal to the product of the other two.

Let the four numbers be denoted by x

x+1
x -f- 2
x -f- 3

Then according to the conditions ;

lO.r -f- (x -{- 1) = (x -)- 2) (x -J- 3)

10.r —)- x -)- 1 — .r 2 -f- 5.r -)- 6
— x 2 + 6.r — 5
x 2 — 6x = — 5
*>_6* + 9 = -5 + 9 = 4

Extracting roots ; x — 3 = ±V4=±2

x = 3 ±: 2 =5 or 1. Answer.

The first number is therefore 5 or 1.

If the first number is 5 the others are 6, 7, and 8. Answer.

If the first two are taken as the digits of a number the number
is 56, and 56 = 7 X 8.

If the first number is 1 the others are 2, 3, and 4.

The first two as digits give 12 and 12 = 3 X 4.

This example shows the advisability of retaining the negative value
of the root numbers as well as the positive values. Using both values
in this case gives two correct answers.

EXAMPLE 76: A man in mowing his lawn, starts on the
outside of the lawn and goes round and round the lawn, which
is a rectangle 100 feet by 120 feet. What width will he have
to cut to be | done? To be f done? To be 4 done?

Area of the lawn = 120 X 100 = 12000 square feet.

j of the area = 4000 sq. feet.

Let x denote the width cut over; consult figure 4.

Prc 1 iminary Mathematics

125

t

w

8

T

Then ; (100 — 2x) (120 — 2x) =8000

12000 —440.r + 4.r 2 = 8000= f of 12000
4.r 2 — 440x = — 4000
* 2 — 110* = — 1000

x 2 — 110* + (55) 2 = (55) 2 — 1000 = 3025 — 1000 = 2025

x - 55 = 55 ± V 2025

x — 55 ± 45 = 100 or 10 feet. Answer.

Adopting the value 100 will mean the complete lawn is mown,
which is contrary to the hypothesis. The value of 10 feet is the
correct value to be selected in the given case.

The correctness of the result may be checked as follows;

120 x 10 = 1200; 1200 X 2 = 2400 (100 — 20) 10 = 800

800 X 2 = 1600 ; 2400+ 1600 = 4000

4000 square feet is l of total area of 12000 square feet.

Fig. 4

Second Part

To find the width when f of the lawn has been mown; let x t
denote the width cut over in this case.

Then by condition;

(100 — 2x\) (120 — 2xt) — 4000 = of total area.

12000 — 440.r, + 4.r, 2 = 4000

4x 2 — 440.rj = — 8000
x 2 — 1 10a*! = — 2000

x 2 — llO^q + (55) 2 = (55) 2 -2000= 3025 -2000 = 1025
x x — 55 = ± V1025
x\ = 55 ± 32.015

= 87.015 or 22.985. Answer.

126

Prcliminary Mathematics

It is evident that the value 87.015 cannot logically fulfill the con¬
ditions; since 2 X 87 = 174, which is greater even than the length (120
feet) of the lawn.

Employing 22.985 as the proper value,^the following must be true;

100 — 2 X 22.985 = 100 — 45.970 = 54.03

120 — 2 X 22.985 = 120 — 45.970 = 74.03

54.03 X 74.03 = 3999.84 square feet. This is the approximate value,
which may be increased to a value nearer to 4000, the true value, by
carrying out the results to more decimal figures.

When the lawn is 4 cut over, the condition is expressed by;

(100 — 2.r) (120 — 2.t') =6000; From which .r = 16 feet.

EXAMPLE 77: The electric light poles along a certain road
are set at equal intervals. If there were two more in each mile,
the interval between the poles would be decreased by 20 feet.
Find the number of poles per mile.

Let x denote number of poles per mile.

Then --- denotes the distance, in feet, between poles.

x

By the given conditions the following must be true;

5280 _ 5280 _

x + 2 x

5280.r = 5280;r + 10560 — 20+ — 40.r

20+ + 40x — 10560
x 2 + 2x = 528

+ + 2.r + ( 1) 2 = 528+(l ) 2

.i- + 1 = ± V +29 = ± 23
x — 22 or — 24

There are 22 poles per mile. In this case the distance between
poles is 5 f|° = 240 feet.

If there were 24 poles per mile, the distance between poles would
be '-|| a =220 feet; which is 20 feet less than 240 feet.

Such a problem is of no particular value except insofar as it fur¬
nishes a certain amount of mental training.

The practical condition would be to find the saving effected by
using 22 poles in place of the 24, if the cost of the poles set in posi¬
tion were $7.00 per pole.

Decreasing the number of poles per miles will effect a saving in
poles, insulators and cross arms, but allows danger of breaking of
wire due to wind and sleet acting on the longer spans of wire.

Preliminary Mathematics

127

SIMULTANEOUS QUADRATIC EQUATIONS

EXAMPLE 78: Solve the following for x and y.

8.r 2 — 3xy — y 2 = 40
9x 2 -f- %y + 2 y 2 — 60

Multiply the first equation by 3 and the second by 2, then;

24.f" — 9xy — 3y 2 = 120
18.r 2 -f- 2xy -f- 4y 2 — 120

24.t* 2 — 9 xy — 3y~ — 18.r 2 -f- 2 xy -f- 4y 2 . Which properly combined re¬
duces to; 6x 2 — Wxy — 7y 2 = 0. The factors of the last equation
are; (3.r — 7y ) (2.r-f-y) = 0

Then;

2x — — y and 3.r = 7y
x— — | x~\y

Substituting these values in either one of the original equations
gives;

8f -f-3fy — y 2 = 40
2y- -I- f y 2 — y~ — 40
(1 +f)y =40
|y 2 . z= 40
y~ = ¥ = 16

y = ± 4 ; therefore x = T 2
Also y — ± f and x = T t 6 ,

EXAMPLE 79: Solve the following equations for x and y.

2x 2 + 3.ry + 12 = 3y 2
3x -(- 5y 1 = 0

The value of x from the second equation is;

25y~‘ -4- lOy -f- 1
9

Substituting these values in the first equation gives;

Sy + 1

x = —from which .r 2
3

2 / 25 - v " rtr 1Q y -h . 1 ) + 3y -f 12 = 3y 2

128

Prcliininary Mathematics

From which is obtained;

50y~ -f- 2 O 3 ' -j~ 2 3y -)~ 15y~ ^ 2 j 2

9 ~ 3 ~

50y 2 + 20y + 2 — 9y — 45 y 2 — 2 7y 2 —

— 22 y 2 +lly= —110
y 2 — \ y — 5

Completing the square gives;

/-irhVj=5 + A

,,_1 - \/Sl1— 4 - »

3 4-V 1 6- x 4

From which ; y —f or — 2
Therefore ; x — — I or 3

Preliminary Mathematics

129

LESSON XXI

RATIO AND PROPORTION

The fundamental definitions pertaining to ratio and proportion are
given in Lesson XI, page 50, but since “ratio and proportion” really
play a very important role in our daily existence, it will be well to
consider the matter more at length.

It is well to keep in mind that a ratio is a quotient arising from
dividing one quantity by another quantity of the same kind.

A ratio is not a proportion.

Proportion: A proportion is an expression of the equality of two
ratios.

A ratio is commonly expressed as 4:10; or 3:9; in each of the
two ratios expressed, the first figure is called the antecedent, and
the second figure is called the consequent. The antecedent and the
consequent together are called the terms of a ratio.

Rule: Multiplying or dividing both terms of a ratio by the same
number does not change the value of the ratio.

O V 3 6

For example; f -=-— = |— 1 — t

2X4 |

The following are equivalent expressions;

_ 4 _dividend numerator

divisor denominator
These might be arranged differently as follows :
dividend

4 10 =

4 -

1 O'

4: 10

divisor

numerator

denominator

antecedent

consequent

antecedent

consequent

One very important relation pertaining to ratios should be kept
clearly in mind; the relation of one number to another of the same
kind expressed by the quotient of the first divided by the second, is
called the ratio of the first to the second.

For example; the ratio of 2 to 4 is f (= |) or 2:4, not 4:2 or i
(=2). It is evident that the ratio of 2 to 4 is very different from
the ratio of 4 to 2.

130

Preliminary Mathematics

Rule: In any proportion the product of the means is equal to the
product of the extremes.

This rule is important, and its application should be carefully un¬
derstood.

The following will illustrate a few applications of the rule. Given
the proportions;

4: 10 = 20: 50, then 4 X 50 = 10 X 20.

3:9 = 6: 18, then 3 X 18 = 9 X 6.

x : y — a: b, then bx = ay.

2a : 3b = 4c : 6d, then 2x6XnXrf = 3X4x£Xc.

The fundamental theory of the above rule may be explained as
follows:

The proportion 4:10 = 20:50 may be written 1 4 0 =fo

Suppose both members are first multiplied by 10;

, 4 X 10 200 „ 200

then —--— ==r- or 4

10

50

50

Next suppose both members of the last equation be multiplied bv

50;

then 4 X 50 = 20050 or 200 = 200

50

This proves that the product of the means is equal to the product of
the extremes.

EXAMPLE 80: What is the ratio of 3 days to 4 weeks?

Since a ratio is the quotient of similar quantities, weeks must be
reduced to days. 4 weeks = 4 X 7 = 28 days. Therefore the ratio
of 3 days to 4 weeks may be expressed by ^ or by 3 : 28. Answer.

PROBLEM 80: What is the ratio of one inch to one foot?

One inch is j 1 ., of one foot. or 1:12. Answer.

EXAMPLE 81: Find two numbers in the ratio of 16 to 9
such that, if each number be diminished by 8, they will be in
the ratio of 12 to 5.

Preliminary Mathematics

131

Let x denote one number, and 3 ; denote the other number.

Then ~ = or 9x = 16y. From which x=-\ e y or y = X %x
From the second condition the following is true;

x—8

y-8

V = 5.r — 40 = 12 y — 96 = 5x — 12 y

56

By substituting in the last equation, the value of y = ,VL the fol¬
lowing results ; 5x— \% s -x = — 56

8 0 v _ 1 0 8

1 *'»

1 6

X

= — 56

_ 2 8

1 6 x

— 56

• #=11 x 56—i 56= 2 f 4 =32

Therefore one number is 32. Answer. Substituting this value

32

in the first equation gives; — = \ r 6 -from which y = T 9 T .32= 2 T 2 ^= 18.

Answer.

PROBLEM 81: Find two numbers in the ratio of 5 to 4 such that,
if each number be diminished by 4, they will be the ratio of 2 to 3.

x — 2 ~° and y=-V. Answer.

Problem 81a: Find two numbers in the ratio of 4 to 5 such that,
if each number be diminished by 4, they shall be in the ratio of 2 to 3.

EXAMPLE 82: Divide 36 into two parts such that the
greater diminished by 4 shall be to the lesser increased by 3, as 3
is to 2.

Let x denote the greater part.

Then 36 — x denotes the lesser part.

By condition;

x — 4 _..

(36 — x) -{- 3

Which may be reduced as follows;

2x — 8 = 108 — 3x -f 9 = 117 — 3x
5x=125

x = 25 = greater part. Answer.

36 — 25= 11 = lesser part. Answer.

PROBLEM 82: Divided 575 into two parts such that the greater
diminished by 25 shall be to the lesser increased by 50, as 11 is to 13.

Greater = 300; lesser = 275. Answer.

Problem 82a: Divide 575 into two parts such that the greater
diminished by 25 shall be to the lesser increased by 50, as 13 is to 9.

132

Preliminary Mathematics

LESSON XXII

SERIES; ARITHMETICAL PROGRESSIONS

Series: An arrangement of numbers, succeeding each other

according to some fixed law, is called a series.

Terms of a Series: The successive numbers form the terms of a
series.

A series may be denoted by symbols and letters as well as by num¬
bers.

Two quantities of the same kind may be compared with each other
in two ways :—

1st. By considering how much greater or less one is than the
other; indicated by their difference.

2nd. By considering how many tunes one is greater or less than
the other, which is shown by their quotient.

By comparing the numbers 4 and 16 with respect to their differ¬
ence, it is seen that 16 exceeds 4 by 12, and in comparing them with
reference to their quotient, it is seen that 16 contains 4, four times;
or that 16 is four times as great as 4.

The first of these methods is called Arithmetical Progression, and
the second is called Geometrical Progression.

Hence, Arithmetical Progression considers the relation of quanti¬
ties with respect to their difference, and Geometrical Progression the
relation of quantities zvith respect to their quotient.

Arithmetical Progression: An arithmetical progression may be
defined as a series, each term of which ( except the first), is formed
from the preceding term by the addition or subtraction of a constant
number.

The constant number which is added, is called the common differ¬
ence.

The first term of an arithmetical progression is called an antece¬
dent, and the second term a consequent. Any number of an arith¬
metical progression may be considered as an antecedent, and its
succeeding number or term, a consequent.

Given the five numbers;

2, 4, 6, 8, and 10.

Preli in in a ry M at hematics

133

The difference between the first and second, and between the second
and third, etc., is 2. That is, the common difference is 2. By adding
2 to the last term given, another term can be derived and the process
continued.

Two other progressions are given as follows;

1, 4, 7, 10, 13, 16, 19, 22

60, 56, 52, 48, 44, 40, 36, 32.

The first of these may be called an increasing progression or an
ascending progression, while the second is called a decreasing or
descending progression.

In the first the common difference is 3, while in the second the
common difference is — 4.

The natural series of numbers 1, 2, 3, 4, 5, 6 , 7, 8 , 9 .... constitute
an arithmetical progression. The common difference is unity.

An example of an ascending progression expressed algebraically
would be;

a, a -f- d, a - f- 2d, a -f- 3d, a -J- Ad,
and of a descending one;

а, a — d, a — 2d, a — 3d, a — Ad.

See also, series at top of page 38.

The consideration of the principles of arithmetical progression may
be made general by considering the following;

Let a, a -f- d, a i -f- 2 d, a -{- 3d, a + Ad denote an ascending pro¬
gression ; a denoting the first term and d denoting the common dif¬
ference.

It may be observed that any term of this series is equal to the first
term plus as many times the common difference as there are preced¬
ing terms.

Suppose the letter 1 is used to denote any term, and the letter n
used to denote the place or position of this term ; then the expres¬
sion for any general term will be;

/ = a + (n —1) d.

For example the 10th term of the series;

б , 9, 12, 15, 18 would be found

from 7 = 6 -f- (10 — 1)3 = 6 -f- 9 X 3 = 33.

The letter 1 is used to denote the desired term, since 1 is the first
letter of the word last; the desired term, denoted by 1 really becomes
the last term in a given series.

134

Preliminary Mathematics

EXAMPLE^ 83: If the first term of a series is 5, and the
common difference is 4, what is the 6th term?

The general equation is; DATA.

l = a-\-(n — 1 )d a — 5

Making the proper substitutions gives ; d — 4

7 = 5 + (6 —1)4 21 = 6

= 5-(-20 = 25. Answer.

PROBLEM S3: If the first term of a series is 8, and the common
difference is 5, what is the 10th term?

Tenth term is 53. Answer.

Problem 83a: If the first term of a series is 40, and the common
difference is 20, what is the 50th term ?

EXAMPLE 84: Find the last term of a decreasing progres¬
sion if the first term is 60, the number of terms is 20, and the
common difference is 3.

For a decreasing arithmetical progression the general equation is;

7 = a _ („—\)d

DATA.

Making the proper substitutions in a = 60

the general equation gives the fol- n — 20

lowing: <7 = 3

7 = 60—(20— 1)3
= 60 — 19 x 3 = 60 — 57

= 3. The last term is therefore 3. Answer.

PROBLEM 84: The first term of a decreasing series is 100, the
number of terms is 40, and the common difference is 2, what is the
last term?

The last term is 22. Answer.

Problem 84a: The first term of a decreasing series is 800, the
number of terms is 200. and the common difference is 2, what is the
last term?

Preliminary Mathematics

135

It is sometimes desirable to find the numerical value of the sum
of the first few, or of the first n terms of a progression or series.

If the letter S is used to denote the value of the desired sum, the
condition may be expressed by;

A simple way to remember this formula is that S = (the average of
the first and last terms) times (the number of terms).

EXAMPLE 85: Find the sum of the first fifty terms of the
arithmetical progression, 2, 9, 16, 23 ....

GIVEN DATA.

The value of the 50th term is to
be found from;

a — 2
d— 7
n = 50

/ = a + (n -J- 1) d
= 2 + 61 X 7
= 429

COMPUTED DATA.

7 = 429

The desired sum is found by making the proper substitutions in;

S = ” (« + /)

=A>°- (2 + 429) =25 X 431

= 10775. Answer.

PROBLEM 85: Find the sum of the natural numbers 1, 2, 2,

4, 5 from 1 to 25 inclusive.

S = V (25 + 1) = 25 X 13 = 325. Answer.

Problem 85a: Find the sum of 100 terms of the series 1, 3, 5,

7, 9 ....

EXAMPLE 86: The sum of three numbers in arithmetical
progression is 15; the square of the second exceeds the product
of the other two by 4. Find the number.

The following expression denotes the first mentioned condition;

a +(a + d) + (a + 2d)= 15
By the second condition; (a+d) 2 — 4 = a(a + 2d)

a 2 + 2ad + d 2 — 4 = a 2 + 2ad
Rearranging gives ; d 2 = 4

d — 2

The common difference is therefore found to be 2.

136

Preliminary Mathematics

Then from the first equation;

a+(a + 2) + (a + 4) = 15
3a+ 6 =15
3a = 9
a — 3

The numbers are therefore a = 3,

(a + d)=3 + 2 = 5,
and (a -f- 2d) = 3 -}- 4 = 7

The square of 5 is 5 X 5 = 25

Product of 1st and 3rd is — 3 X 7 = 21

**

and difference is — 4

EXAMPLE 87: Find four integers in arithmetical progres¬
sion such that their sum shall be 24, and their product 945.

GIVEN DATA.

S = f (<* + /) and /=a + (n — l)d. S = 24

n = 4

From the given conditions, the sum of the four integers may be
expressed by;

a -(- (a -f- d) + (a -f- 2d) -f- (a -f- 3d) = 24
4a -j- 6d = 24

24 —6d ___ 12 — 3d
4 ~ 2

From the first two expressions, by substituting the value of / in
the first expression;

(»= 4 )

Which verifies the first method

above for finding an expression for a.

According to the second condition given in the example;

a X (a -(- d) X ( a + 2d) X (a -f- 3d) = 945

12 — 2a . .

Substituting in the last equation the value d — o obtained
from one of the previous equations gives;

a (a+- L2 ' ~~ 2a )\a+ 1(12 — 2a) ] (a + 12 - 2a) = 945

S=- (2a -\-(n — 1)d)
2

24 = 2 (2a + 3d).

24 = 4a + 6d

24 —6d _ 12 — 3d
4 ~ 2

Prelim in a ry Mathematics

137

Which may be reduced to the following;

o(12 +a) (24 — o) (12 -a) =8505

If in this expression the values 1, 2 and 3 are successively substi¬
tuted for a, it will be found that 3 will fulfill the condition, so that
o = 3. That is, the first number of the arithmetical progression is 3.
Substituting the proper values in;

S = ”(o + /) gives 24 = £(3 + 7)

From which 7 is found as follows;

24 = 2(3 + /)= 6+ 27

7 — V* = 9 (This serves as a check on the result)

The last of the four numbers is therefore 9.

Substituting the value of a = 3 in;

a +(o + d) + (o + 2d) + (a + 3d) = 24

gives the common difference d = 2.

If o = 3, then the second number is 3 + 2 = 5; the third number is
5 + 2=7, and the fourth number is 7 + 2 = 9.

EXAMPLE 88: Find the sum of all positive integers con¬
sisting of three digits which are multiples of 11.

The last term of the progression cannot be greater than 999; the
nearest multiple of 11 is 990. The last term is therefore 990.

The first term is 110.

The difference is 11.

7 = a + (n — 1) = a + nrf —d

/ — a + d

from which n = ^

_ 990— 110+ 11 _891_

11 “ 11 - 81

S= f (a+ 7)

S = V (110 + 990) = 81 1 \p° = 81 X 550 = 44550. Answer.

LESSON XXIII

GEOMETRICAL PROGRESSION

A geometrical progression is a series of numbers so arranged that
the quotient of any term by the preceding term is always the same.

The letters G. P. are sometimes used to designate geometrical pro¬
gression.

The constant quotient mentioned above is also called the ratio,
and is designated by r.

A geometrical progression is a progression by quotients.

The following is an example of a decreasing geometrical progres¬
sion ;

64:32: 16:8:4:2: 1

while the following illustrates an increasing geometrical progression

1:3:9:27:81:243 ....

From the last example it is evident that, = 3, or the ratio r=3.
If the first term of any geometrical progression be denoted by a,
then the second term may be denoted by a r, the third term by a r~,
the fourth by a r 3 , etc.

Any term that has n — 1 terms preceding it may be expressed by;

1 ~ a r (n_1)

The sum of any number, (as n) of terms of a geometrical pro¬
gression may be designated by;

-- in which a denotes the first term and

r — 1

1 denotes the nth term.

EXAMPLE 89: Find the 5th term of the progression;

2, 4, 8, 16 ... .

DATA.

a — 2

Substituting the proper values in T — 2

the equation; n = 5

1 — a r n_1

gives / —2 X 2 * _1 = 2 X 2 4 = 32. Answer.

Preliminary Mathematics

139

PROBLEM 89: Find the 8th term of the progression;

2, 6, 18, 54.

8th term = 2 X 3 7 = 2 X 2187 = 4374. Answer.

Problem 89a: Find the 6th term of the progression;

4, 12, 36, 108.

EXAMPLE 90: Find the sum of eight terms of the progres¬
sion;

2, 6, 18, 54, 162.

DATA.

a — 2

First find the value of the 8th term. n~8

1 — 2 X 3 7 = 4374 r — 3

Then the sum of the first eight terms is found from ;

It —a 4374 X 3
r — 1 - 3 — 1

— 6560. Answer.

PROBLEM 90: Find the sum of ten terms of the progression;

2, 6, 18, 54, 162, ....

Sum of ten terms = 59048. Answer.

Problem 90a: Find the sum of five terms of the progression;

2, 4, 8, 16, 32.

The application of geometrical progression may be appreciated by
the following consideration;

Suppose it is desired to find the absolute value of the recurring
or repeating decimal .333333.

This decimal may be written in form,

33 , 33 j 33 +

100 (100) 2 (100) 3
In this case a = yVo T — too and the value to infinity of the series

is;

5 oo ‘

:i:{
l 0 0

.33

1 — r 1-4,, -99

To find the absolute value of the recurring decimal .5555..

~ _ a _.55_ 5

00 “ 1 — r .99 “

To find the absolute value of the recurring decimal .66666.

a .66

S'co —

1 — r .99

See Lesson VI, page 26.

-li-2

' <)

140

Preliminary Mathematics

EXAMPLE 91: A man purchased 10 bushels of wheat on
the condition that he should pay 1 cent for the first bushel, 3
cents for the second, 9 for the third, and so on to the last.
What did he pay for the last bushel, and for all ten bushels?

The last or the tenth bushel cost;

1— 1 X 3 9 = 19683 cents or $196.83.

The ten bushels cost;

S = 19683 X - 3 -— - — 29524 cents or $295.24.

3 — 1

PROBLEM 91: What debt can be paid in 12 months, by paying $1
the first month, $2 the second month, $4 the third month, and so on,
each payment being double the preceding one? What will be the last
payment?

Debt is $4095, and last payment is $2048. Answer.

Problem 91a: A farmer planted 4 bushels of corn, and the har¬
vest yielded 32 bushels; these were planted the next year, and yielded
256 bushels; these were planted the following year and again yielded
8 fold; this process was continued during 16 years. How many
bushels were the result of the last harvest, and what was the total
number of bushels of corn handled during the 16 years?

Result of last harvest =140737488355328 bushels.

Preli w ina ry M a the unities

141

LESSON XXIV

CLOCK PROBLEMS

As the many questions that may be asked regarding the various
time relations of the hands of a clock offers a very attractive field for
examiners, a few of the numerous and impractical problems, which
offer great possibilities for mental training will be considered.

The fundamental relation of the various positions of the hands of
a clock or of a watch are of considerable importance and should be
carefully considered.

For instance at 12 o’clock the two hands coincide. This time and
position of the two hands might be taken as the datum from which
all other relations could be computed; however other positions may
offer easier solutions for some problems.

It may be advisable to consider a circle and circular measure, in
connection with clock problems.

A circle may be defined as a curve, every point of which is equally
distant from a point within the curve, called its center. The term
circle is assumed to mean the area included by the curved line;
while the curved line itself is referred to as the circumference of the
circle. Portions of the circumference are referred to as arcs. Every
circumference or circle is supposed to be made up of 360 units
called degrees (expressed by ° ; thus 45° means forty-five degrees).

A semicircle =^f a =180°; one fourth of a circumference = : T'
= 90°, and is called a quadrant. One-twelfth of a circumference will
equal 30°.

The minute hand of a watch passes over 360° while the hour hand
passes over 30°. The minute hand moves 12 times as fast as the
hour hand.

EXAMPLE 92: When will the hands of a watch next coin¬
cide, after 12 o’clock?

When the hour hand has passed over an angle of x degrees, the
minute hand has passed over an angle of x360°.

Since the minute hand moves 12 times as fast as the hour hand, the
following relation must be true;

12.r — x -\- 360°

\\x = 360
x = : V\° degrees.

142

Preliminary Mathematics

Since the hour hand moves over 30° in one hour; or during 60
minutes; one degree must correspond with two minutes, for this
hand.

The actual time, in minutes, corresponding with \ 6 , 0 degrees, must
be

X 2 = 65 vY minutes.

The two hands next coincide at 65 T 5 T minutes after 12 o’clock; or

at 1-5 T \-o’clock. That is at 5y\ minutes past one o’clock.

Answer.

The next coincidence will be l-5/ T -j-l-5i 5 T or at 2-10T?, and so on.

PROBLEM 92: Where is the hour hand of a zvatch, when the
minute hand is at 12-15 o’clockt

At li minutes past 12. Answer.

Problem 92a: Starting at 12 o’clock what is the value of the
angles, in degrees, each of the two hands have passed over at the
end of 25 minutes?

EXAMPLE 93: At what time between one and two o’clock
do the hands of a watch coincide?

The angle between thq hands, at one o’clock is 30°. After one
o’clock the hands begin to move and the minute hand moves 12 times
as rapidly as the hour hand. In one hour the hour hand passes over
30° of angle.

Let x denote the speed of the hour hand; then 12.r* denotes the
speed of the minute hand.

The condition stated in the example may therefore be expressed by;

12.r — 30° = x

From which ll.r=:30 o

and x = 2 A°=f?° \\

For the minute hand a movement of 1 degree =-^- 0 = }. of a minute,
and for the hour hand 1 degree = 12 X = 2 minutes. During one
complete rotation of either hand it sweeps over 360 degrees of angle
or of arc.

If „r =f ^degrees and one degree for the hour hand means 2 min¬
utes, in the given case the time interval must be f^X2= ii = 5j 3 T min¬
utes.

At 5 minutes past one o’clock the two hands coincide.

Answer.

Preli /// in ary M a them a tics

143

PROBLEM 93: At what time between 2 and 3 o’clock arc the
hands of a clock together?

12 .tr — 60° = .r
ll.r = 60°

x=^° At 10minutes after 2 o’clock. Answer.

The minute hand passes over %% Q - = 6° every minute.

The hour hand passes over ! \¥° = 30° every hour or 4 ° every min¬
ute ; or 1 degree every 2 minutes.

The converse of the last problem would be; when the hands of a
clock coincide between 2 and 3 o’clock, what time is it?

EXAMPLE 94: What time is it when the hands of a watch
coincide, between 3 and 4 o’clock?

At 3 o’clock the angle between the hands is 90°. The condition
for the coincidence of the two hands after this time is; 12„r — 90° — x
where x denotes the speed of the hour hand.

Then 1 l.r = 90° or x = ?f °

The hour hand has moved over ° = 8 degrees, and since the min¬
ute hand moves 12 times as rapidly, it must have moved 12 X 8 T 2 T °
= 98 / j °.

Since for the minute hand 6° — 1 minute, the minute hand must

indicate 1 f| 9 =l& T 4 x minutes after three o’clock.

For the hour hand|°— 1 minute. Therefore if this hand has
passed over 8 °, it also must indicate W=16r 4 i minutes after 3 o’clock.

EXAMPLE 95: At what time between 2 and 3 o’clock are
the hands of a watch at right angles with each other?

The condition is expressed by;

12.r — 60° — x + 90°

From which 1 \x = 150°

and x=*ff 0 =13 1 \ °

13 , T , X 2— 27 minutes past 2 o’clock. Answer.

PROBLEM 95: At what time betzveen 1 and 2 o’clock are the :
hands of a watch at right angles with each other?

At 21 minutes past 2 o’clock. Answer.

144

Preliminary Mathematics

Problem 95a: At what; times between 3 and 4 o’clock are the
hands of a watch at right angles with each other?

A rough check on such problems as the preceding, should always
be applied by the student, in some such manner as follows; From
the conditions mentioned in Problem 95a, it is evident that the hour
hand must be located below the figure 3 and if the minute hand is
90° in advance of the position of the hour hand it must be beyond
the figure 6 , since there are only 90° between the figure 3 and
the figure 6 . This means that it must be more than 3-30. Another
position will be when the minute hand is between 12 and 1. A com¬
prehensive problem of no great practical importance would be asking
at what times during the day are the hands of a clock at right
angles? Obvious times are 3 o’clock and 9 o’clock.

EXAMPLE 96: What is the time interval, in minutes, be¬
tween the two perpendicular positions of the hands of a watch
between the hours of 4 and 5 o’clock?

At some time between 4 o’clock and 4-10 the two hands are per-
pendicular with each other. As the two hands move around they will
later coincide, and still later will again be perpendicular with each
other. The time interval is required between the two perpendicular
positions.

Let x denote the angle passed over by the hour hand.

Then \2x will express the angle passed over in the same time by
the minute hand.

At 4 o’clock the angle between the two hands is 120°.

After the hour hand has moved x degrees, the minute hand has
moved 12 ;r degrees.

Starting with 12 o’clock* as datum, the following relation is true
for the first perpendicular position;

Angle passed over by minute hand x 4(360°) i

Angle passed over by hour hand x 1 ’

from which 12.r = x + 1440°

i-liiO °

— ii

For the second perpendicular position the following is true;

.v 1 -}- 4(360°)+ 180° __ i

x 1 '

From which x 1 =

11

The difference between the two positions is J fp— -^degrees

The time interval is therefore 2 X W Q ~ — 32 minutes.

Answer.

* Four o’clock could be assumed as datum if desired.

Prelim inary Mathematics

145

2 pints.
8 quarts

4 gills.

2 pints.

4 quarts...
M l / 2 gallon
2 barrels..

TABLE OF MEASURES

Dry Measure

. 1 quart—qt. 4 pecks.1 bushel—bu.

..1 peck—pk. 36 bushels.1 chaldron

Liquid or Wine Measure

.... 1 pint—pt.
... 1 quart—qt.
. 1 gallon—gal.
,. 1 barrel—bbl.
hogshead—hhd.

U. S. Standard Gallon

231 cubic inches

Beer gallon.282 cubic inches

36 beer gallons.1 barrel

Time Measure

60 seconds. 1 minute

60 minutes. 1 hour

24 hours.1 day

7 days.1 week

4 weeks.1 lunar month

28, 29, 30, or 31 days.1 calendar month

30 days (in computing interest) .1 month

52 weeks and 1 day or 12 calendar months.1 year

365 days, 5 hours,

48 minutes, and 49 seconds.

Circular

Measure

60 seconds.

.1 minute

90 degrees.

.1 quadrant

60 minutes.

4 quadrants or

360 degrees

30 degrees.

1 circle

Long Measure

12 inches.

40 rods.

.1 furlong

3 feet.

.1 yard

8 furlongs.

... 1 statute mile

5*4 vards.

.1 rod

3 miles.

.1 league

Square Measure

144 square inches.

.1 square foot

40 square rods

.1 rood

9 square feet.

.1 square yard

4 roods.

30^4 square yards

..1 square rod

640 acres.

. . . . 1 square mile

Cubic Measure

1728 cubic inches.

..1 cubic foot

128 cubic feet .

..1 cord (wood)

27 cubic feet.

40 cubic feet. .

1 ton (shipping)

2,150.42 cubic inches.

.1

standard bushel

268.8 cubic inches. 1 standard gallon

1 cubic foot.about four-fifths of a bushel

Surveyors' Measure

7.92 inches.1 link

25 links. 1 rod

4

10 square chains or 160 square rods

640 acres.

36 square miles (6 miles square)

rods

. 1 chain

. 1 acre

1 square mile
.... 1 township

146

Preliminary Mathematics

Cloth Measure

2y A inches.

4 quarters..

.1 yard

4 nails.

.1 quarter

Mariners’

Measure

6 feet.

.1 fathom

5,280 feet..

.1 statute mile

120 fathoms.

1 cable length

6,085 feet..

.1 nautical mile

7]/ 2 cable lengths

Miscellaneous

3 inches.

18 inches..

.1 cubit

4 inches.

21.8 inches.

.1 Bib 1 ^ cubit

6 inches.

2y 2 feet....

TABLE OF

WEIGHTS

Troy Weight

24 grains (gr) . .

1 pennyweight—pwt.

20 pennyweights.

3.2 grains.

By this weight gold, silver and jewels only are

weighed. The ounce

and pound in this

are the same as

in apothecaries’ weight.

Apothecaries’ Weight

20 grains.

8 drachms.

3 scruples.

12 ounces..

Avoirdupois Weight

16 drachms.

4 quarters..

-100 weight—cwt.

16 ounces.

20 hundred

weight.1 ton

25 pounds.

. 1 quarter—qd.

5,760 grains apothecaries’ or troy weight.1 pound

7,000 grains avoirdupois weight.1 pound

Therefore, 144 pounds avoirdupois equal 175 pounds apothecaries’
or troy.

1 gallon oil weighs.

1 gallon distilled water....

I gallon sea water.

1 gallon proof spirits.

1 grain — 0.00229 ounce.

= 0.064799 gram.

= 0.00014 pound.

1 ounce = 437.5 grains.

= 28.3495 grams.

= 0.0625 pound.

1 pound = 7,000 grains.

= 453.6 grams.

= 0.4536 kilogram.

Liquids

.7.32 pounds avoirdupois

.8.33 pounds

.8.55 pounds

.7.68 pounds

1 Long Ton = 2,240 pounds.

1 Short Ton — 2,000 pounds.

1 Metric Ton = 1,000 kilograms.

1 Acre = 43,560 square feet.

= 4840 square yards.

= .0015625 square mile.

1 Square Mile = 27,878,400 square feet
= 3,097,600 square yards.

1 Square Yard = 1296 square inches.

Preliminary Mathematics

147

METRIC EQUIVALENTS IN LINEAR MEASURE

Linear

1 millimeter.0.03937 inch

1 centimeter.0.3937 in.

1 decimeter. .3.937 in. = 0.328 ft.

1 meter_39.37 in. r= 1.0936 yds.

1 dekameter.1.9884 rods

1 hectometer.328.091 ft.

1 kilometer.0.62137 mile

1 kilometer.1093.63890 yards

1 myriameter... 10936.3890 yards

1 sq. centimeter
1 sq. decimeter.

1 sq. meter.

1 are.

1 hektar.

1 sq. kilometer.,

Square

0.1550 sq. in.
.0.1076 sq. ft.
.1.196 sq. yd.
.3.954 sq. rd.
...2.47 acres
.. 0.386 sq. m.

Measure

1 cu. centimeter.0.061 cu. in.

1 cu. decimeter.0.0353 cu. ft.

1 cu. meter.1.308 cu. yd.

1 stere.0.2759 cord

1 liter.

1 dekaliter....
1 hektoliter...

S 0.908 qt. dry
} 1.0567 qt. liq.

1 2.6417 gal.

{ .135 pks.
....2.8375 bushels

Measure

1 inch.2.54 centimeters

1 foot.3.048 decimeters

1 yard.0.9144 meter

1 rod.0.5029 dekameter

1 mile.1.6093 kilometers

Measure

1

sq. inch.

. .6.452 sq. centimeters

1

sq. foot.

.9.2903 sq. decimeters

1

sq. yd.. .

.0.8361 sq. meters

1

sq. rod..

1

acre.. ..

.0.4047 hektar

1

sq. m... .

...2.59 sq. kilometers

of Volume

1

cu. in...

. 16.39 cu. centimeters

1

cu. ft...

.28.317 cu. decimeters

1

cu. yd... .

.0.7646 cu. meters

1

cord....

1

qt. dry..

1

qt. liq...

.0.9463 liter

1

gal.

.0.3785 dekaliter

1

peck....

.0.881 dekaliter

1

bushel. . .

.'.0.3524 hektoliter

Weights

1 milligram.0.0154 grains

1 centigram.0.1543 grains

1 decigram.1.5434 grains

1 gram.....15.4340 grains

1 gram.0.03527 ounce

1 decagram.154.3402 grains

1 hectogram 3.527 oz. Avoirdupois

1 kilogram.2.2046 lbs.

1 metric ton.. 1.1023 English ton

1 ounce.28.85 grams

1 lb.0.4536 kilogram

1 English ton..0.9072 metric ton

APPROXIMATE METRIC EQUIVALENTS

1 decimeter.4 inches

1 meter.1.1 yards

1 kilometer. % of mile

1 hektar.2*4 acres

1 stere or cu. meter % of a cord

1 liter 1.06 qt. liq. = 0.9 qt. dry

1 hektoliter.2-^ bushels

1 kilogram.21/5 pounds

1 metric ton.2,200 pounds

1 quart

0.94636 liter 1 gallon

3.78543 liters

148

Preli mina ry M a them a t ics

RULES FOR COMPUTING INTEREST

The following will be found to be excellent rules for finding the
interest on any principal for any number of days. When the prin¬
cipal contains cents, point off four places from the right of the result
to express the interest in dollars and cents. When the principal
contains dollars only, point off two places.

Two per Cent.—Multiply the principal by the number of days to
run, and divide by 180.

Two and one-half per Cent.—Multiply by number of days, and
divide by 144.

Three per Cent.—Multiply by number of days, and divide by 120.

Three and one-half per Cent.—Multiply by number of days, and
divide by 102.86.

Four per Cent.—Multiply by number of days, and divide by 90.

Five per Cent.—Multiply by number of days, and divide by 72.

Six per Cent.—Multiply by number of days, and divide by 60.

Seven per Cent.—Multiply by number of days, and divide by 51.43.

Eight per Cent.—Multiply by number of days, and divide by 45.

Nine per Cent.—Multiply by number of days, and divide by 40.

Ten per Cent.—Multiply by number of days, and divide by 36.

Twelve per Cent.—Multiply by number of days, and divide by 30.

Fifteen per Cent.—Multiply by number of days, and divide by 24.

The 4th root is the square root of the square root.

The 6th root is the square root of the cube root.

The 9th root is the cube root of the cube root.

Preliminary Mathematics

149

EXAMINATION IN ALGEBRA

1. Express the following, using positive exponents;

x* y~ s ; lx 3 y 3 ; m -2 n 2 p 3

2. Express the following, using radical signs;

« , 7jr :i y , Py 6 ; # :! ; 5v 4 y 5

3. Write with fractional exponents the following;

VP; VP;

V * 7

4. Perform the operations indicated, and simplify;

1

7T

4 k. < 4 \ 4 w / n i \ 4 ^ A"

* « 3 ; (sr X (34V; —

Vo

5. Simplify;

<7_ b

3_3

a\b ’
2 1 3

2v + —

l r—

x —2

2v-

X-£

6. Separate into factors;

3o 2 — 6ab + 9a 2 P;

o 4 + o 2 + 1

7. Find three numbers whose sum is 20, and the first plus twice
the second plus three times the third is 44, and twice the sum of the
first two minus four times the third is — 14.

8. Two passengers together have 500 pounds of baggage. One
pays $1.25 and the other $1.75 for excess weight. If the baggage had
belonged to one person he would have paid $4.00 for excess. How
much baggage is allowed free to each person.

9. Solve;

x + y = 11

(1)

y + z — 13

(2)

z + x — 12

(3)

2.r -J- 4y — 3z = 3

(1)

3x — 8y -j- 6z — 1

(2)

8.r — 2y — 9z — 4

(3)

10. Solve;

150

Preliminary Mathematics

Solution of the Foregoing Examination

1.

x

2. Va 2 ;

,5 y

N'T?- 7 —

i

xs

7 - ;

ys

orff ^ x
1 yi =7 '

n

m p

/- 6

V-rVy; V a n ; 5Vx s,-,

3. A'o ■ vf- i/ £ = ( 2Vi _ 2

p., o, y«3 Va3/ J flio

, V4 V id

4. a« (adding exponents) ; ~zzX zr

v 5 \/ 3

2 V 10 2 V 15 V 10

2 x V 5 x V 3 x V 2 x V 5
15

V15 15

2 X 5 x V2X V3 _

15

I V 6;

xV a

V a

3a — 2fr 2.v 2 — 3 .v
3 + 2& ; 2.r 2 — 5x

6. 3a (a — 26 —{— 3ai> 2 ).

7. Let x, y, and z denote the three numbers, then, according to
the first mentioned condition,

x -f- y z — 20 (1)

By second condition; x -J- 2y -f- 3z = 44 (2)

By third condition; 2{x-\-y) —4 z— — 14

or 2x-\-2y — 4 z — — 14 (3)

The three equations may be designated as (1), (2), and (3).
Multiplying the first equation by 2 and subtracting the result from
(3) gives;

2x -j-2y — 4 z — — 14

2x 2y -j- 2z — 40

— 6z — — 54

From which z — 9

Substituting this value for z in (2) and (3) gives;

x -J- 2y -f- 27 = 44 = x -\-2y = \7

2x + 2y — 36 = — 14 = 2x -f 2 y = 22
Subtracting the last two equations gives x = 5.

Substituting in (1) the values x — 5 and z — 9, gives;

5 + y + 9 = 20
y = 20-14= 6

Preliminary Mathematics

151

Verification; 5 —6 —9 = 20

5-f2x6+3X9= 44

2(5 + 6) —4 X 9 = — 14

Another method is as follows;

x + y + z = 20

x + 2y + 2>z = 44

2x + 2 y — 4z = — 14

Subtracting the first two equations gives ; y + 2z = 24

Subtracting the last two equations gives ; 2 y + \0z = 102 after

multiplying the second equation by 2.

These two resulting equations may be arranged as follows and sub¬
tracted.

2y + 4s = 48

— 2y — 10z= — 102

— 6j=— 54 from which z = 9.
Substituting this value for z in first two of the original equations
gives;

x + y + 9 =20 x + y = 11
x + 2y + 27 = 44 or .r + 2y = 17
Subtracting gives ; — y = — 6 or y = 6.

Substituting the values found for y and z in the first of the original
equations gives;

x + 6 + 9 = 20 or x = 5.

The three numbers desired are therefore, 5, 6, and 9.

8. Let x denote the pounds of baggage allowed free to each per¬
son.

Let y denote the pounds of baggage belonging to one passenger;
then 500 — y will denote the pounds of baggage belonging to the
other passenger.

125

Then

and

Then

Also

Then

— = cost per single pound of excess baggage.
y — x

175

——- = cost per single pound of excess baggage.

500 — y — x

125 175

y — x 500 — y — x
400

500 — x

400 125

(a)

500 — x y — x

= cost per single pound of excess baggage.

(b)

152

Preliininary Mathematics

9

(a) reduces to; 5&r — 300y =— 62500

or to x — 6 y = — 1250
Multiplying by 11 gives — 66y \\x =— 13750

(b) reduces to 400y— 275x = 62500

or to 16y — \\x = 2500

— 66y + ll.v = — 13750

Adding gives; — 50J? = —11250

y — 225 pounds.

If one passenger has 225 pounds, the other must have 500 — 225 =
275 pounds.

Also x — 100 pounds; amount allowed free to each person.

Charge is one cent per pound of excess baggage.

9. Subtracting equation (2) from (1) gives;

x — 2 — — 2

Subtracting this from equation (3) gives;

— 2z = — 14 or 2 — 7.

Substituting this value in (2) gives y — 6. Substituting this value
in (1) gives x = 5.

10. Multiplying equation (1) by 2 gives;

4x -f- 8y — 62 = 6 Add equation (2)
to last equation; 3x — 8 y -f- 62 = 1

Adding gives; 7x — 7 or x =1

Multiply equation (3) by 4 giving;

32:r — 8 y — 362 = 16
Substitute value 1 for x, giving;

— 8y — 362 = — 16

Substitute value 1 for x in equation (2) giving,

— 8 y -\-62 — — 2

Subtracting last two equations gives;

— 8y — 36z = — 16

— 8y + 62 = — 2

— 42z = — 14

^ JLi — 1

* 4 2 3

Multiplying equation (2) by 3 and substituting x = 1 gives;

— 24y-|-18.s= — 6. Mult. (3) by 2 and substituting
x = 1 gives ; — 4y — 18r = — 8

Adding gives ; — 28y= —14

y = i

x = 1 and y — 4 in equation (1) gives ; 2 = 1.

Substituting

Preliminary Mathematics

153

EXAMINATION FOR ADMISSION, BROWN UNIVERSITY

Algebra, Wednesday, September 20, 1912

_ 6 9 2 3

1. (a) Divide x 3 f a 2 by x 3 -\-a 2

(b) Determine the least common multiple of
x 2 — a ’; x -\- a; x s — a 3

2. Find the square root of 347, correct to two decimal places.

3. Solve for x, y and z,

3x — 4y -f- 2s = —3
2x + y — 3z = 84
x 2z — — 3

4. Solve for x ; 3ax 2 -(- 2 bx -f- c = 0.

5. A man buys 8 lbs. of tea and 5 lbs. of sugar for $2.39; and at

another time 5 lbs. of tea and 8 lbs. of sugar for $1.64, the price being
the same as before. What were the prices?

6. A man had one dollar in silver and copper coins; each copper
coin was worth as many, cents as there were silver coins; there were
in all 27 coins. How many of each were there?

7. The sum of three numbers in arithmetical progression is 15;

the square of the second exceeds the product of the other two by 4.

Find the numbers.

Answers to the foregoing examination;

1. (a)

3 2 | 6

a ~x 3 I a 2

x a
x

| 5— L

—a ~x °\ a

3 a

—a 2 x 3 a

1)

n

9

4 X

Answer

6 2 i !>

»> —— o

a ~x °i a -

r, 2 , s»

a 2 x 3 i a 2

(b) Factoring the expression gives;

O -f-o) (x — a), (x a), (x—a) (x 2 + ax + a')

The least common multiple is therefore;

(x + a) (x — a) (x 2 + ax + a 2 ) = x 4 - ax 3 + 2aV — a 3 x + a\

Answer.

154

Preliminary Mathematics

2. 18.260. Answer.

3. x — 1; y = L and z = — 2. Answer.

4. Dividing each term by 3a gives;

b 2 — 3ac

x ~ + *'«*+(*«) ~~ 9a 2

Extracting the square root of each member of the equation gives;

I b 2 — 3ac
\ 9a*

V b 2 — 3ac
3a

b ^ V b' 2 — 3a c
3a 3a

= -±-(—b± V £> 2 — 3ac)

3a

5. Let x denote the price of tea.

Let y denote the price of sugar.

Then by condition; 8 a* -f- 5y = 239 (1)

Also; 5x + 8 y = 164 (2)

Multiplying (1) by 8 , and (2) by 5 gives;

6 Ax -f 40 y = 1912

23 x -f- 40y = 820

Subtracting gives; 39.r = 1092

And x — 28

Tea cost 28 cents per pound. Answer.

Substituting value x = 28 in equation (2) gives;

140 —(- 83 ; = 164

From which y = 3

Sugar cost 3 cents per pound. Answer.

Verification ; 8 lbs. X 28 cts. -f- 5 lbs. X 3 cts. = 239 cents $2.39.

5 lbs. X 28 cts. 8 lbs. X 3 cts. = 164 cents = $1.64.

6 . Let x denote the number of silver coins.)

Let y denote the number of copper coins.

Each copper coin is worth x cents.

Each silver coin is worth y cents.

It is evident that the number of copper coins times the value in
cents of each coin, added to the product of the number of silver coins
times the value in cents of each silver coin must total 100 cents.

All the silver coins are worth xy cents and all the copper coins are
worth yx cents.

Preliminary Mathcmatics

155

Then by the conditions;

yx + yx = 100 or 2 xy = 100
50

From which y =

x

Also by condition ; x -f- y = 27.

50 •

Substituting the value y =_in the last equation gives;

x

, 50
x +

21

x

x 2 + 50 = 27X
x 2 — 2 lx = — 50

x 2 — 21 x + (V-) 2 = (V-) 2

50= 7 f n — 50 = 72!, 7 20 ° :

- 5 2 9
' 4

Extracting square root; x — V =± V 5 |

2 9

2 3

= 5 2° or f = 25 or 2

Substituting the value x = 25 in x + y = 27, gives y = 2. Sub¬
stituting the value .r = 2 then y = 25.

If there are 25 silver coins there are 2 copper coins. The
silver coins being worth 2 cents each, and the copper coins 25
cents each.

2 x 25 + 25 x 2 = 100

If there are 2 silver coins then there are 25 copper coins.

Then 25 X 2 = 2 X 25 = 100

7. By first condition;

x + (x + d) + (++ 2 d) =15. (See page 133 on progression.)

By second condition;

(x + d) 2 — 4 = x(x + 2d)
x 2 + 2 dx + d 2 — 4 = a- 2 + 2 dx
From which d 2 = 4

Or d = ± 2

Substituting the value d = 2 in the first equation gives;

x + x + 2 + x + 4 = 15 or 3.r = 9 x = 3
The first number is therefore 3, the second is 3 + 2 = 5 and the
third is 3 + 4 = 7.

If the value d = — 2 is used then from
x + (x + d)-\-(x + 2d)~ 15

x = 7. Then the second number is 7 — 2 = 5, and the third
is 7 — 4 = 3.

Verification ; 3 + 5 + 7 = 15

25 — 4 = 21
7 + 5 + 3 = 15
25 — 4 = 21

Second case;

156

Preliminary Mathematics

DARTMOUTH COLLEGE

Entrance Examination—September, 1915

(Students taking examination in Elementary Algebra complete
(14 units credit) will take questions 2 and 4 of mathematics A1 and
all the questions of mathematics A2. Such students will be allowed
2 hours and 20 minutes.)

Mathematics A1 (Algebra to Quadratics)

1. Solve each of the following equations for x:

(a)

( b )

2,r + 7 3,r -f- 8 _ 4 a- + 3

4 5x -f- 3 8

x — b x — a 2 (a — b)

x — a x — b x — a — b

2 . (a) Factor:

75xy s — 130*y — 9x"y 5

3 3 l 2 2

x y — xy x y — xy
4 (x — 3) —(- x (x — 3) {4x — 4)

( b ) Simplify:

x

i+: i+:

i i

X ' 3’

3. (a) Solve:

2 x — y -}- z — 1
3x -f y + z — 2
x + y + z = 0
(b) Simplify the fol¬
lowing fraction by ra¬
tionalizing its denomi¬
nator ;

1

1 + V2+ V3

4. (a) Multiply x%y by x*y + — 1-f-jr *y4-

(b) Find the square root of:

& n .3 11 | 0 i 1 | -6 14 0 1 7 , B

Xo—La ~>x •> -j-La->x->-f~a ax •> — La^xs— )— a->

5. A garrison of 1000 men having provisions for 60 days was re¬
inforced after 10 days, and from that time the provisions lasted only
20 days. Find the number in the reinforcement.

6. If the numerator of a certain fraction be increased by one and
its denominator diminished by one, its value will be one. If the
numerator be increased by the denominator and the denominator be
diminished by the numerator, its value will be four. Find the frac¬
tion.

Preliminary Mathematics

157

Mathematics A2 (Quadratics and Beyond)

1. (a) Find to two decimal places the values of * which satisfy
the following equation:

x ■

( b ) Solve for x:

1 111

-= - + “4--

x —|— a —j— b x 1 a 1 b

2. (a) Find the values of x which satisfy the following equation:

V3* + 10 — V* — 1= V2J^ 1

(6) For what relation in r, s, t, will the equation

xP -|- 2 sx + t — 0

have equal roots? Prove your answer to be correct.

3. Solve the following pair of simultaneous equations :

* 2 + y 2 — 25 = 0
7x + y — 25 — 0

Draw graph illustrating each equation and their solution.

4. (a) Expand by binomial theorem and express each term of ex¬
pansion in simplest form:

( b ) The sum of the first and fourth terms of an arithmetical pro¬
gression is 19 and the third and sixth terms is 31. What is the first
term ?

5. A broker sells certain railroad shares for $3240. A few days
later, the price having fallen $9 per share, he buys for the same
sum five more shares than he had sold. Find the price and the num¬
ber of shares transferred on each day.

Solution of the Foregoing Examination

1. (a) (4*+14) (5*+ 3) — 24* — 64= (4*+ 3) (5*+ 3)

20+ + 82.r + 42 — 24* — 64 = 20.+ + 27 x + 9
82* — 51* = 9 + 64 — 42
31* = 31

*=1. Answer.

158

Preliminary Mathematics

( b ) (x‘ — 6) 2 (x — a — b ) — (x — a ) 2 (;r— a — fr)=2(a— b)
(x — a) (x — b)

•»3

3* 2 fc + 34 t6 2 — + 2 xab — ab 2 — 6 3 — + — 3+a — 3.ra 2 + a 3 +

b.x * 2 — 2;rab + = 2+a — 2 xa 2 + 2a 2 & — 2+fr — 2^*6 2 — 2 ab~

From which; x = ^ (5 2 ~ --Answer.

5b — a

2. (a)

( b )

•ry

vy 3 (15 + .ry) (5 — 9.ry). Answer.

xy (x +• y -f-1) (x — y). Answer.

4 (x — 3) (x 2 — # + 1) • Answer.
xy 2 xy

y + x r x + y y -j- x

3. (a) 3x + y + -s’ = 2

,r 4~ y ^ — 9

Subtracting; 2x = 2

x — 1.
y = 0.

= 0. Answer.

Answer.

Answer.

2 x — y + 2
3.r + y + s
Sx + 2 z
5 + 2,

1

2

3
3

= — 1. Answer.

(b ) Multiply both numerator and denominator by 1 — V2 — V3

giving; - ——Multiply both numerator and denominator

— 4 —2V6

of the last fraction by—4 + 2V6 giving;

(1 _ V2— V3) (—4 + 2 V6) 2- V2V3+- V2

T=r 8~ 4

4. (a) vy" 1 —a’ 4^ t+j;iy ^+2v^y i+2v i>^+vv'^ 3 »^+v^y—2.

vy 1 +2viy _ i+x ^yi—jr^y ^—2—v ^y^+v^y ^'+2v'4y4+x 1 y Answer.

(b) x» — a Answer.

5. Let x denote portion of provisions one man receives.

Let y denote number of men in the reinforcement.

Then 60 (1000.r) = 1 or x = —-—

60000

and 10 (lOOO.r) + 20.r (1000 + y) = 1

Substituting the value x ——--in the last equation gives;

60000

y = 1500 men. Answer.

6. Let x denote the numerator and y the denominator.

Then x = 1 and A - =4
9 — 1 y — x

x — 3 and y — 5. Therefore the fraction is ?. Answer.

Preliminary Mathematics

159

Answers to Mathematics A2 (Quadratics and Beyond)

1. (a) The given equation may be reduced to;

x 2 — 3 +- 2x -f- 2,r = 0. From which x = .65 or — 4.65. Answer.

Checks ; .42 -f- 2.66 — 3 = 0 and 21.6 — 18.6 — 3 = 0 ( See page 30)

(b) The given equation may be reduced to;

a'b + air -j- ax 2 -f- bx 2 + a 2 x -f- b 2 x -f- 2abx =0
From which a (x + b) + x (x -j- b) = 0

(a -f- x) (b + x) =0

x = — a or — b. Answer.

The answer may be proved by substituting each of these values for
x in the given equation.

2. Squaring both members and reducing gives;

3*+10 — 2 V 3x + 10 VF^1 + j — 1 = 2*— 1

x + 5 = V 3x 2 + 7x — 10. Again squaring both members and
reducing gives; (2„r + 7) (x — 5)=0.

x = — I or 5. Answer.

o

( b ) Both roots are equal when s 2 = rt; then t = —

r

*>

s *

Proof: substitute — for t in the given equation;

r

o

r»“ t £

then; rx -\-2sx — - from which x =--

r r

3. x = 4 or 3 and y = — 3 or 4.

x 2 + y 2 = 25 is the equation of a circle and lx + y — 25 = 0 is
the equation of a straight line intersecting the circle in one point
whose coordinates are x = 3 and y = 4, and in another point whose
coordinates are x = 4 and y — — 3.

4. (a) According to the binomial theorem. (See Higher Algebra
by Fisher and Schwatt, Philadelphia, Pa., University of Pennsyl¬
vania.)

_1_0
a~ 3

fA#' 4 '<++8 § 0tf+ " a — ~ tf-l + A# %a~ —

• >

ay

32

Which reduces to;

10
a~ 3

_u

\a

t+i a

- *3L

3

fflo + iVn

ay

32'

Answer.

(b) Let .v denote the first term, and y the common difference.

Then x + (x + 3y) = 19 or 2x + 3y = 19

and (x + 2y) + (x + 5y) = 31 or 2x -j- 7y = 31

From which; x—S. Answer. Subtracting; — Ay = — 12

y= 3

160

Preliminary Mathematics

5. Let x denote price in dollars per share at first sale.

Let y denote number of shares.

Then ' xy — 3240 1 _ \ xy — 3240

(x — 9) (y -f- 5) = 3240 \ \ xy — 9y -f- 5x = 3825

3240

Substituting the value y = - in the last equation gives;

x

- 29160 + Sx‘ = 3285.V
x

From which is obtained; x 2 — 9x — 5832. (See page 111)

x = $81. y = 40 shares. Answer.

If 5 more shares were purchased the second day the number must
have been 45, and the price was — $72.

Proof ; 81 X 40 = 3240 = 72 X 45.

WORCESTER POLYTECHNIC INSTITUTE
Entrance Examination

1. Algebra I. Wednesday, June 16, 1915. 2 to 3.35 P. M.

1. (a) Divide;
by fa 2 — \ab—kb~.

( b ) Multiply; x^— 4^—(——4.y _1 3’t-}-v"\v 2

by x^y

2. Simplify:

2 a -j- b
a -f- h

and ^

a-\- b

l + c +

2c 2

1 — c

3. Solve the following equation for x:

2x + 1 _8_ 2x — 1

2x — 1 ~4x 2 —1~2x+ 1

4. A merchant adds to his capital one-fourth of it each year. At
the end of each year lie deducts $1200 for expenses. At the end of
the third year he has, after the deduction of the last $1200, one and
a half times his original capital, minus $950. What was his original
capital ?

Preliminary Mathematics

161

5. Solve for x and y the simultaneous equations

x —

2 y — ,r

23 — x

v — 3

20 +

x — 18

— 30

2x — 59
2

73 — 3y

Answers to the foregoing examination

1. i a 3 + $- a?b — kab'-\-ib z . Answer.

2. (a) 2 a + b_ j 2o + 6
a + b _

a

2 a — a

a

1. Answer.

a + b

a + b

a

a

(b)

1 +

1 + f 2
1 — c

Answer.

1 + c

2^ 2

1+1 + C 2

1 — C

1 — c

(2x+ l) 2 — 8:
4.1' 2 + 4x - 7 :

8at

X

(2x — l) 2
4 a* 2 — 4x + 1
8

1. Answer.

4. See Example 29, page 86.

5. 2x(23 — x)-(2y — x)2 = 20(23 - x)2 -j-(2x — 59) (23 — x)
Reduces to; — 17a* — 4y = 437

3y (x — 18) — 3 (y — 3) = 90 (x — 18) — (73 — 3y) (x — 18)
Reduces to ; — 17a* — 3y — — 1629 + 1314 = — 315

17.r + 4 y= 437

— 17.v — 3y — — 315

y = 122, and x — — 3. Answer.

Let x denote his original capital.

Then x-\~ix — $1200 will denote capital at the end of first year;

(a* +4a* — 1200) + ' * 3+ 1-^QQ —$1200 denotes capital at end of

second year; and ^ .r — 1875 — 1500 — 1200 denotes capital at end of
third year.

By condition stated in problem; x H-x — 4575 — §x — 950.

From which; at = $8000; the original capital. Answer.

162

Prelim inary Mathematics

WORCESTER POLYTECHNIC INSTITUTE
Entrance Examination

2. Algebra II. Wednesday, June 16, 1915; 3.45 to 5.15 P. M.

1. A man has a rectangular house which occupies 1200 sq. feet
in the middle of a rectangular lot 8000 sq. feet in area. The lot ex¬
tends 30 feet beyond the house at each end and 25 feet at each side.
What are the dimensions of the lot?

2. Show that the equation 3.i* 2 -f- 5 jt —(- 1 — k 2 — 0 has two real
roots for any real value that may be given to k.

3. According to Boyle’s Law, the volume of a gas is inversely
proportional to the pressure on it. A tank contains 4 cu. ft. of air
under a pressure of 60 lbs. per sq. in. How much space will be occu¬
pied by the air when expanded so as to be under atmospheric pres¬
sure, 15 lbs. per sq. in?

4. (a) Given that G is the Geometric Mean of two numbers a and
b, that H is the Harmonic Mean of a and b, and that M is the Arith¬
metic Mean of a and b. Prove that G is also the Geometric Mean
of H and M.

( b ) The first term of an Arithmetic Progression is 1, the last
term 7, and the sum of all the terms is 13. Find the Progression.

5. (a) Compute (0.99) 5 by expanding (1 — ,01)‘ by the Binomial
Theorem, and then simplifying.

/ _ i v i o

( b ) Find the fifth term of ^ x 4" ^ j Simplify it.

The foregoing examination is worked out as follows:

1. Let x denote the width, in feet, of the lot.

Then will denote, in feet, the length of the lot.

X

By the given conditions; (x — 50) (^02 — 60) = 1200

Which reduces to; (x — 50) (8000 — 60.r) = 1200,r

And performing the multiplication, to ; 8000.V — 60.r 2 — 400000 -f-

3000,r = 1200.r.

Giving; .r 2 — 4 f°.r = — 2 -°§ 00

Completing the square by adding to both members of the equa¬
tion the square of 4 the coefficient of x, gives;

Preliminary Mathematics

163

r 2 =ifvr + m*-y=-**t**+ m»y

J- _ 4 9 0 - -+- y ^ 4 9 0 y- 20000

r - 490 -4- \/ 2 4 010 0 _ 240000

.1 - 9 — v jfg "36

- 4JL0 -t- a/ 1J)0 490 4. 10

- ' 6 ' — v 3 6 6 111 6

— 419 01* -’l l 0
6 '- u 6

= 80 or 83^

If the value of x = 80 feet as the width of the field is accepted,
then the length will be 8 g$°- = 100 feet. Answer.

If the width is taken as 83 i feet,, the length will be 96 feet.

Answer.

2. If the equation is solved for .t* the result is;

__ 5± V25 — 12 + 12/r _ — 5 ± V 13 + 12ft 2

6 ~~ 6

If the quantity expressed by the radical is positive the roots are
real; if the value of the radical is negative the roots are unreal or
imaginary. If k has any value whatever assigned to it the value of
the radical is always positive and the equation will have two real
roots.

3. According to Boyle’s law pv — k.

Let z\ denote the volume under 15 pounds pressure.

Then according to Boyle’s law;

= from which v x — —— =16 cubic feet. Answer.

4. (a) Given the Geometric Progression
Given the Harmonic Progression
Given the Arithmetic Progression

a -f- G -f- b
a -f- H -f- b
a -f- A -(- b

To show that G 2 = A H
From (3) A —a — b — A

From (2) a - — a and «(H — b)=b(a — H )
b H -f - b

from which H =

a-\-b

J G 2 = ab

p 2 _ A T-r<! 2A — aJ r b

G — A HI 2 ab

A =

a -f- b
(b) See page 132.

1,

11 19 27 35
3 3 3 3

( 1 )

( 2 )

(3)

Answer.

164

Preliminary Mathematics

5. (a) (1-.01) 3 = (l) 6 -5 (1) 4 +10 (l) 3 (.Ol) 2 — 10 (l) 2
(.01) 3 -f- 5 (1) (.Ol) 4 - (.Ol) 5

— 1 _ .05 + .001 — .00001 + .00000005 -f .0000000001

— 1.00100005 — . .0500100001

=: 0.9509899499. Answer.

( b ) The general expression for the r th term of (a-{-£)"
n (n — 1) (n —2) ... (n— r + 2) («- r +l) (r-n

1 X 2 X 3 ... (r —1) a b

In the given case

n 3= 10
r 33: 5

a — V x

1

b 33

3 V x

Substituting these values in the general expression gives;

10 (9) (8) (7)
1 x 2 x 3 x 4

(V*)

210.v

8Lr

.3

J2,

= \%x.

Answer.

YALE ENTRANCE EXAMINATION, JUNE, 1912
Algebra A. [Time allowed, one hour.]

[Omit one question in Group II and one question in Group III.]

Group /.

1. Resolve into prime factors;

(a) 6.r 2 — 7x — 20

(b) (x 2 — 5x) 2 — 2 (x 2 — 5x) — 24

(c) a* -f- 4a 2 -J- 16

2. Simplify ( 5 - ^-19V )^( 3 _a-|£.)

' a — 4.i* / ' a — 2.t' /

3. Solve

2 (.v —7) ■ 2 —jr a ■+3

x 2 + 3.1- -28 4 —x x + 7

=0

Group II.

V2”-4- 2 VT

4 Simolifv ~~~ - -— an d compute the value of the fraction to

V 2- V12

two decimal places.

5. Solve the simultaneous equations

tf-£+23r£=I
2x~i- jr*=|-

Preliminary M at hematics

165

Group III.

6. Two numbers are in the ratio of c : d. If a should be added to
the first and subtracted from the second, the results will be in the
ratio of 3:2. Find the numbers.

7. A dealer has two kinds ot coffee, worth 30 and 40 cents per
pound respectively. How many pounds of each must be taken to
make a mixture of 70 pounds, worth 36 cents per pound?

8. A, B, and C can do a piece of work in 30 hours. A can do
half as much again as B, and B two-thirds as much again as C. How
long would each require to do the work alone?

Answers to the foregoing examination:

1. (a) (3x -f- 4) (2x — 5).

(b) [ (x s - 5x) - 6] [ o' - 54-) + 4] = (x -6) (x + D
X (x — 4) (x — 1 ).

(c) a 4 -f- 4a 2 -f- 16. * Answer.

2 5a 2 — 20,r' — a 2 -f- 19.r 2 3a — 6.v — a -f- 5.v 4a 2

a 2 — 4x 2 ' a — 2.r a 2 -

(2a—.r) (2ax) (a —2,r) 2a -j- x

4x

X

a — 2x
2 a — x

(a — 2x ) (a-(-2.t') (2a — x) a 2x

Answer

(2,r — 14) + U+ 7) (2-x) - (4-.r) (.r + 3)

— x 2 — 3.r + 28

2x — 14 — x 2 — 5x + 14 + .v 2 — x — 12
— x 2 — 3x + 28

= 8.r — 16 = 0; from which x = 2. Answer.

4.

V2 + 2V3 V2 — 2 V3 1+V2V3

V2”— V~3 V"4 V2— VTV2V2

1 4 - 1.414 x 1,732 -3.449

-1 _ 1.414 x 1.732 - 1.449 — 2 - 38 '

1 _ V 2 V 3

Answer.

5. Multiply the first equation by 2, then subtract second from
first.

2*-H-4r i =¥=i

y-i= -j

5 v -s=f Vy = 3 y = 9. Answer.

x — 4 . Answer.

166

Preliminary Mathematics

6. Let x denote one number and y denote the other; then;

a c.

- =“,and x=y

yd

d

x 4- a

y

c

5 ad

an< ^ y — a = f from which 2x-\-2 a = 3y — 3a, and y = — — _^

Sad

x — - v_

d x 2c — 3d

Sac

2 c—3d

Answer.

7. Let x denote the number of pounds at 30 cents.

Let y denote the number of pounds at 40 cents.

Then 30x + 40 y = 70 X 36 = 2520

and x -f- y = 70. From which x — 28 lbs. and y = 42 lbs.

Proof : 42 X 40 -j- 28 X 30 = 2520 cents.

8. Let x denote number of hours required by A to do the work
alone.

Let y denote number of hours required by B to do the work alone.

Let z denote number of hours required by C to do the work alone.

Then, =Zo> also ^r=fy and y=f^. y— 93 hrs. x — 62 hrs.

x y Z

£=155 hrs. Answer.

Algebra B. [Time allowed, one hour.]

(Omit one question in Group I and one in Group II. Credit will
be given for five questions only.)

Group /.

1. Solve

x+a x-Gb
x -(- b ' x -j- a

5

2

( r V

2. Solve the simultaneous equations j '

Arrange the roots in corresponding pairs.

3. Solve ; 3.t"f -f- 20,r* = 32.

-j- 28 xy — 480 = 0
Zv + y = 11

Group II.

4. In going 7500 yards a front wheel of a wagon makes 1000
more revolutions than a rear one. If the wheels were each a yard
greater in circumference, a front wheel would make 625 more rev¬
olutions than a rear one. What is the circumference of each?

5. Two trolley cars of equal speed leave A and B, which are 20
miles apart, at different times. Just as the cars pass each other an

Prelim in ary Ma t hematics

167

accident reduces the power and their speed is decreased 10 miles per
hour. One car makes the journey from A to B in 56 minutes and
the other from B to A in 72 minutes. What is their common speed?

Group III.

6 . Write in the simplest form the last three terms of the expan¬
sion of (4at—

7. (a) Derive the formula for the sum of an arithmetical pro¬
gression.

(b) Find the sum to infinity of the series 1, — 4, p,—i.

Also find the sum of the positive terms.

Solution of the foregoing examination;

1 . 2 (r -f a) (x -f- a) -j- 2 (x -j- b) (x -j- b ) = 5 (x -f- b) (x -(- a)

2 ( x 2 -j- 2ax -j- a 2 ) -f- 2 (x 2 -f- 2 bx -f- b 2 ) = 5 (x 2 -f- ax -j- bx -f - ab)
x 2 -f- a.t* — 2 a 2 -j- bx + 5ab — 2b 2 = 0

.r 2 -f (a + b)x = 2a 2 — 5 ab + 2b 2 *

/a-f-b \ 2 a 2 -f- 2 ab -\- b 2 ,92 e 1 19 / 2

,r ! -+(a + fc).r+ ) = 4 - 2a ~ Sab + 2h

a 2 -j- 2 ab -j - b 2 -\- 8a 2 — 20 ab -f- 8 b 2

= 4

9a 2 — l8ab + 9b 2
~ 4

a + b , V 9a' — 18aft + % 2 _ — a _ ft ± 3a — 3ft

•V ■ _ 3Z ————————————————————————————— -

2 2 2

_ 2a —- _4 ^ . — a — 2 ft. Answer.

2

or . lh? A — 2 a -f b. Answer.

2

2. The first equation may be expressed; (xy — 12) (xy -f- 40) = 0,

from which x — — or— —. By proper substitution in the second

y y

equation; 2x— =11; from which x — f or 4. Therefore y —

X

8 or 3 and x — - or 4. Answer.

2

3. The equation may be written as follows; (consult page 36)

3 20 3 . 20

_ s + 7 = 32 which is the same as = 32.

xi v* V.r V x

168

Preliminary Mathematics

The last equation may be reduced to;

V A' 3 — | V .r 3 — 3 3 2 = 0, which is equal to V .r 3 — f V .r 3 -f- 2 Xc =
The roots of which are; V x* = T 5 ff ± pg=$ or = —i
From which „r 3 = (— |) 4 or (|) 4 . From which x — T V, or

4 9
•2 r, n

4. Let x denote circumference of front wheels, in yards.
Let y denote circumference of rear wheels, in yards.

7500 7500

Then

y

7500

x

7500 _

x + 1 ~ y + 1
7500y

1000

=625

x= 7^00 _i_ 1000y ^ rom ec l uat ’ on above.

By substituting the last value of x in the second equation gives;

7y 2 — 32y — 15 = 0
(7j> + 3) (y-5)=0

y — — for 5 yards; circumference of rear wheels.

^■—3 yards; circumference of front wheels. Answer.

(Consult Example 63, page 116)

5. Let x denote common speed, and let y denote the distance
from A that the accident happened. Then 20 — a - will denote the
distance from B that the accident happened.

_ y . 20 — y 14 .20 — y t y 6

Then; +“ A: = -, r and i “rx =c

x ' x — 10 15 x x — 10 5

From which x — 25 miles per hour. Answer.

I

6. The last three terms will be the 7th, 8th, and 9th. See page 164.
solution of question 5 ( b ).

n = 8; a — 4a* ; b = — a ir«

7th = 28 (16) a 6 .r 2 = 448aV* = Answer.

8th = — 32a 5 a* 3 . Answer.

9th = aVt. Answer.

7. (a) See page 135.

For finding the sum of decreasing series.

Rule. Multiply the first term by the ratio, and divide the product
by the ratio less 1. The resulting quotient is the sum of an infinite
decreasing series.

Prc lim i nary M a the mat ics

169

( b ) In the given case the ratio is
Applying the above rule; ^ X ^ ——

1-r- —* = —2.

= 4--. Answer.
3

Applying the same rule to the positive terms gives the ratio

4X1 4 A

= 1 — i = 4, and sum = ^ ■ Answer.

The sum of the negative terms may also be found by the same

t ^ X 4_ o

rule. The ratio is 4, and sum = ^ f

The sum of the negative terms may also be found by the same
f—£=f- which confirms the first answer given under (6).

INDEX

All references are to pages

A.

Abbreviation, symbol for 16

Abstract number 50

Absolute Term 78, 109

Addition, of binomials 41

of monomials 41

of polynomials 41

sign of 18

Ages, problems on 80, 81

Algebra, defined 10

Algebraic addition 41

Algebraic subtraction 41, 42

Algebraic expression 40

Algebraic symbols 16

American Short Proof Method

22

Antecedent 129, 132

Application of Logarithms 70

Arabic Numerals 10

Area of Circle 41

Arithmetic, defined 10

Arithmetical Progression

132, 153, 157, 162

B.

Base of Logarithms 60, 61

Binomial 40, 41

Binomial Theorem 157, 159, 162

Braces 36

Brackets 36

Brown Entrance Examination

153

C.

Capital, examples on 84, 86

Centimeter, defined 9

Characteristic, of Logarithms 61
Cipher 10

Circle, area of 41

defined 141

degrees in a 141

Circumference, defined 141

Cistern, filling of 79, 80

Clock Problems 141

Coefficient, defined 8, 40

Coefficients, literal 40, 41

numerical 40, 41

Coins, examples on 84, 85

College Entrance Examinations

153

Common Fractions 11

use of 9

Common System of Logarithms

61

Conditional Equations 54

Constants 57

Consecutive Numbers 81

Consequent 129, 132

Continuation, sign of 38

Cube 47

of ten 33

Cubical Equation, defined 55

Cube Root 47

D.

Dartmouth Entrance Exam¬
ination 156

Decimal, table of equivalents 26, 27
Decimal point 11, 26

Decimals, division of 26, 27

multiplication of 29

repeating 30

Decimeter, defined 9

Decreasing Series 133, 168

Deduction, sign of 37

Degree of an equation 54

Dependent variable 58

Digits 10

Difference of two numbers 82

of squares of numbers 83, 84
Dimensions, examples on 86

Dividend 24

Division, sign of 24

E.

Efficiency, classes of 50

of a machine 50

power 50

Entrance Examinations 149 to 170

Equation, algebraical 53

arithmetical 53

cubical 54

conditional 54

complete 106

11

All references are to pages

Equation

degree of an 106

defined 53

identical 54

incomplete 106

involving two unknowns 86
members of an 53

of first degree 55, 79

of second degree 55, 106

of third degree 55

powers and roots of 69

with three unknown 92

simultaneous quadratics

12 7,166

Equality, sign of 16

Equal Roots 47, 157

Equivalent, decimal 26

Equivalents, Metric 147

Evaluating expressions 42

Evolution 97

Examinations in Algebra 149

Example, defined 6

Example on baggage 149

Example of triangles 93

Example on ages 93

on army 103, 156, 158

on Boyle’s Lave 162

on brokerage 157, 160

of capital 117, 160

on division of money 117

on coins 153

on electric poles 126

on fractions 156, 158

on garden 120

of Geometric Progression 138
on income tax 118

on interest 118

on mixtures 165, 166

on mowing a lawn 124

of numbers 108, 122, 124, 149

of proportion 130

on Progression 157

of picture 114

of series 134

on trade 121, 123

Explicit functions 58

Exponents, defined 32, 60

fractional 34

negative 36

Expression, algebraic 40

typical 46

Extreme of a proportion 51

F.

Factors 20

and roots 47

equal 47

Filling of cisterns 79, 80

Flexibility of Arabic Numerals

10

Foot, defined 8, 9

Force of Gravity 57

Fraction 11, 26

defined 11

square root of 101, 102

Fractional exponents 34

Functions, defined 57, 58

explicit 58

G.

Geometric Mean 162

Geometrical Progression 132, 138
Greek Alphabet 7

H.

Harmonic Series 163

Homogeneous Equations 78

I.

Identical Equations 54

Inch, defined 8

Index, of a number 33

Infinity, defined 10

sign or symbol for 10

Integers 28

Involution 97

Irrational roots 148

Italian Short Proof Method 21

L.

Lengths of nights 82, 83

Legal Units 9

Linear Equations 79

with three unknowns 92

Line, divisions of a 83

Literal Equations 78, 88, 89, 90

Logarithms 60

of a fraction 69

of the square root of 20 74

of the square root of 15 74

of the power of a number 69

of a quotient 69

Logarithm of unity 67

Logarithmic Tables 64, 65

Logarithms, common 61

natural 61

properties of 61

A ll references ore to pages

in

M.

Mantissa

61

Means of a proportion

51

Members of an equation

53

Meter, defined, legal

8, 9

Metric System

8, 9

Metric Equivalents

147

Millimeter, defined

9

Minuend

18

Minus Sign

18

Monomial

40

Multiplicand

20

Multiplication

Algebraical and Arithmeti-

cal

46

of positive and negative

numbers

32

special case of

46

sign of

19, 20

Multiplier

20

N.

Natural Logarithms

60, 61

Negative Exponents

36

Number, abstract

8

pure and abstract

50

ratio

60

Numerals

10

Numeric, defined

8

O.

Operation, symbols of

16, 18

P.

Parentheses

36

Per cent efficiency

50

Plus sign

18

Pointing off, rule for

28

Polynomial

40

Positive term

40

Power efficiency, defined

50

Power of a number

32

of ten

33

Prime, symbol

16

Problem, defined

6

Progression

132

ascending

133

descending

133

Proof of Multiplication

21, 22

Proportion, defined

51, 129

Properties of Logarithms

67

Pure Quadratics

106

Q.

Quadrant, defined

141

Quadratic equation

55, 109

complete

111

pure

106

rule for solving-

111

typical forms

110, 111

Quadratics, defined

54, 55

simple

113

solution of

109

Quantity, concrete

8

defined

8

how denoted

10

how measured

8

Quotient, defined

24

R.

Radical sign

34

Ratio 50,

60, 129

of circumference to diam-

cter

57

Ratio and Proportion

129

Rational Roots

97

Reciprocal

11, 36

Relation, symbols of

16

Repeating decimal

30, 139

Roman Numerals

13, 14

Roots

46, 47

cube root of 125

47

equal

47, 157

fourth root of 625

47

of an equation

107

Rowing, example

119

Rule, for checking square

root

99, 103

for computing interest 148
for exponents 34

for extracting sq. root 99, 103
for pointing off decimals 28
of proportion 130

for signs 32

for solving equations 53
for sum of a decreasing
series 168

IV

All references are to pages

s.

T.

Second Degree Equations

106

Table of decimal equivalents

27

Semicircle, defined

141

Table of Logarithms 64,

65

Series, defined

132

of interest

148

Arithmetical

132

of Metric Equivalents

147

Geometrical

133

of Roman numerals

14

Harmonic

163

of weights and measures

145

natural of numbers

132

Term, defined

40

sum of

135

negative

40

Signs of addition

18

positive

40

of continuation

38

Terms of a Ratio 51,

129

of deduction

37

Trinomial

40

of division

24

Typical, expressions

46

of equality

16

Quadratics

112

of infinity 10

of inequality 16

of multiplication 19

of ‘.subtraction 18

Simultaneous Equation, three
unknown 149, 156, 158

Simultaneous Quadratic Equa¬
tions 127, 166

Sixth Roots 148

Solution of Quadratics

110 , 111 , 112

Square Root 34, 47, 97

of fractions 101

rule for extracting 99

U.

Units, fundamental 8

legal 9

Use of letters in Mathematics 13

Value of 7 r
Variables, defined
dependent
independent
Vinculum

57

57

58
58
36

Squares of numbers

98

W.

Subscripts

16

Subtrahend

18

Weights and

Measures,

tables

Sum of at Geometric Progres

of

145, 146

sion

139

Worcester Polytechnic

Exam-

Sum of a series

135

ination

160, 162

Sum of positive integers

137

Surds

97

Y.

Symbols, algebraic

16

Yale Entrance

Examination 164

of continuation

38

Yard, defined

8 , 9

denoting aggregation

36

of operation

18

Z.

Symmetrical Equations

78

Zero

10

FOR AMATEURS

How To Make Low-Pressure Transformers

THIRD EDITION, 1916

IN CLOTH, Postpaid . 40 cents

Review in “Machinery,” May, 1915.

This pamphlet describes the construction of a transformer to re¬
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DIRECTIONS FOR DESIGNING, MAKING AND OPERATING

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anxious to understand thoroughly the principles and construction of the
component parts of his installation.

The author introduces the subject by referring to the commercial
demand and necessity for electric power at high pressure, and the reasons
why alternating current is the most useful for this purpose. A simple
but very practical explanation of the construction of the transformer
then follows, after which we find an explanation of symbols and anno¬
tation, the various losses in a transformer, power factor, and other mat¬
ters. The author next treats of the design of a 20,000 volt transformer,
entering very carefully into practical details of calculation. Following
this, we have a chapter entitled “Directions and Data for Constructing a
3-KW. 20,000 volt Transformer,” the approximate cost of materials not
being overlooked. A further chapter deals with data applying to a 4,000-
volt transformer.

We do not remember having previously seen any small book dealing
so thoroughly and practically with the construction of high pressure
transformers, nor one in which the diagrams and photographic illustra¬
tions were so happily chosen. The impression we have gained after
reading the book is that the author knows exactly what he is talking
about and how to express himself.

From “Journal United States Artillery.”

Directions for Designing, Making, and Operating High-Pressure Trans¬
formers. By Professor F. E. Austin, Box 441, Hanover, N. H.
5" x 7y ". 46 pp. 21 il. Tab. Cloth. Price, 65 cents.

This seems to be a very practical little volume. For one who already
has a knowledge of the fundamental principles of transformers and who
has any reason to construct experimental apparatus of this character,
the book will give all the practical and theoretical information desired
and that, too, in a very small compass and in a very readable and clear
form.

The illustrations are numerous and make the explanation very clear.

All the details of the mechanical process of winding, making up the
core, etc., are simply explained and clearly illustrated.

All the necessary theoretical calculations are made evident by type
examples worked out. All theoretical principles are simply and clearly
explained. Specifications, drawings, and bills of material with approx¬
imate costs are given to guide the investigator.

The book should be a great help to an instructor, to a student work¬
ing in an alternating current laboratory, or to an investigator making up
transformers for experimental purposes.

Remit amount to:

PROF. F. E. AUSTIN, Box 441, Hanover, N. H.

Distributors for England, E. & F. N. SPON, Ltd., 57 Hay-
market, London.

SECOND EDITION 1916

EXAMPLES IN MAGNETISM

Fully Illustrated with Plates and Diagrams

For students of

Physics and Eledtrical Engineering

Not a book of “problems,” but of carefully solved examples.
The fundamental equations are carefully stated and data systematically
arranged.

Price post-paid in the United States, $1.10

Remit amount with order to

PROF. F. E. AUSTIN, HANOVER, N. H. BOX 441

BOOK REVIEW

A handy and practical little work for all those who are interested
in electricity and magnetism is- Professor F. E. Austin’s “Examples in
Magnetism.” It starts with a couple of chapters devoted to a simple
explanation of trigonometric functions, formulas and problems, takes
up the metric and C. G. S. systems of measurements and passes thence
to definitions and discussions of magnetic quantities.

Practically all problems involving magnetic poles and pole strength,
fields of force and the changing of magnetic to mechanical force are
presented in a way that is easily understood by anyone with a knowl¬
edge of arithmetic and elementary algebra. The whole subject is
handled in twelve lessons. At the end are several useful tables and a
comprehensive index. It is neatly bound in flexible leather, and worker,
engineer, and student will find it well worth the price. Size 4 by 6
inches. 90 pages. $1.10 net. Technical Journal Company, Inc., 233
Broadway, New York.

EXAMPLES IN MAGNETISM, second edition, $1.10.

From “The Wireless World”, London, Eng., June, 1916.

“Examples in Magnetism for Students of Physics and Engi¬
neering.” By F. E. Austin, B.S., E.E. Published by the Author at
Hanover, N. H. 5s. net.

This is a book similar in style to “Examples in Alternating Cur¬
rents,” by the same author, reviewed in our March issue. The plates are
particularly interesting and helpful, as they show the lines of force
surrounding magnets by means of actual photographs of iron filings.
This is a great improvement on the old method of drawing an imaginary
field with a few dotted lines, and should be much appreciated by the
student.

The problems and examples seem carefully chosen and well worked
out, and should furnish a guide to students who are beginning to study
electrical engineering, and enables them to develop the process of correct
and logical thinking.

The book is well produced, and will prove valuable to both students
and instructors.

E. & F. N. SPON, Ltd., distributors for England and Australasia.
Address: 57 Haymarket, London.

Your Attention is Called to a

NEW DEPARTURE

in the Book Production, entitled

Examples in Alternating-Currents

By PROF. F. E. AUSTIN, E. E.

Second Edition With Additions

A valuable book for students, teachers and engineers.

FOR STUDENTS: The application of fundamental principles
to practice is aptly illustrated by completely worked out problems; the
process of solution being clearly outlined step by step. Class room prob¬
lems and engineering problems are fully discussed.

The collection of useful trigonometrical formulae, type integral

forms and tabulated values of 2-rrf, , ( 2irf) 2 and ^ - y -- r

2 ttj ( 27 r /) for

frequencies from 1 to 150 cycles is alone worth many times the price

of the book.

FOR TEACHERS: One important feature of the book affecting
those who teach the important theories of alternating-currents to be¬
ginners, is that of so clearly and definitely fixing important mathematical
processes and knowledge of physical phenomena in the student’s mind,
that instruction may resolve itself at the very start into emphasizing
engineering application.

FOR ELECTRICAL ENGINEERS: Perhaps the chief value
of the book to electrical engineers lies in the carefully tabulated arrange¬
ment of mathematical and electrical data; useful in many fundamental
considerations.

FOR NON-ELECTRICAL ENGINEERS: There is a very

large class of engineers not directly engaged in electrical work, but to
whom a practical working knowledge of alternating-currents is an es¬
sential element making for success. Such will find that "Examples in
Alternating-Currents” imparts the desired information in a minimum of
time.

AS A REFERENCE BOOK: The value of the book as a refer¬
ence book for all classes desiring concise and exact information on elec¬
trical matters involving the principles of alternating-currents, is greatly
enhanced because of an extensive index, referring directly to pages.

PERMANENT FEATURE: The discussions throughout the
text, dealing as they do with fundamental principles, renders the infor¬
mation of permanent value. The book will be as useful twenty-five
years hence, as it is today.

ILLUSTRATIONS: The book contains carefully arranged dia¬
grams of electrical circuits with corresponding vector diagrams of pres¬
sure and current components. Many diagrams are inserted showing the
combination of sine-curve alternating-quantities, and the derivation of
fundamental equations.

Price, Pocket Size, $2.40
Remit amount with order to
PROF. F. E. AUSTIN, Box 441, Hanover, N. H.

From “Journal United States Artillery.”

Examples in Alternating Currents. Vol. I. (For Students and Engi¬
neers.) By F. E. Austin, Box 441, Hanover, N. .H. 5" x 7^4".
223 pp. 4 plates, 70 figures. Price: $2.40.

The purpose of the book as stated by the author is to help students
of electrical engineering gain a better knowledge of fundamental princi¬
ples by solution of well selected problems involving those principles.

The problems are numerous, well chosen, and well arranged under
proper headings. Each principle is illustrated by an example worked out
in detail. Following each illustrative example are sufficient problems of
the same general character to give the student adequate practice in test¬
ing his knowledge of the particular principles involved and the method
of solution.

The first part of the book explains briefly, and illustrates the appli¬
cation of, trigonometry and differential and integral calculus to the solu¬
tion of problems in alternating currents.

The problems which follow involve the use of trigonometry, differ¬
ential calculus, integral calculus, the vector diagram, and algebra. For
students who desire to limit their study to problems most frequently met
with in practice and who do not desire to use the calculus, those in the
latter part of the book furnish examples illustrating every principle in¬
volved in practical work, and the solutions are by vector diagrams or
are algebraic.

The tables in the back of the book contain values of the variable
quantities entering the alternating current formulas that will save much
work in solving problems.

The book is of value to a student of alternating-currents who de¬
sires to ground himself in the fundamentals of the subject. An
instructor will find it of great value in furnishing a well selected and
classified list of problems suitable for tests or illustrative use.

Review by “Mining and Scientific Press.”

This work has been carefully prepared, and is published in pocket
size. It contains examples dealing with wave-length, frequency, sine
and nonsine alternating-currents and pressures, generators, power,
power-factor, capacity, inductance, and resonance, showing the process
of solution, step by step, together with the fundamental equations apply¬
ing, and the necessary data properly arranged. The diagrams are very
clear, greatly assisting in an understanding of the fundamental princi¬
ples. Electrical engineers will find the trigonometric formulae of value,
and tables of variable quantities for frequencies from 1 to 150,000 cycles.

Distributors for England,

E. & F. N. SPON, Ltd.,,

57 Haymarket, London

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