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1 TV i re CXI v of* PROF, F. B. AUSTIN, i:. 1 lcpJ: i'I-ID 1 ! U; aA 39 „A85 Austin. F„ E. 1873- F" r e 1 iminary mathematics » \ ' ■ I \. c \ - 4 ' * • - •' • • . * .* , 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¬ duce electrical pressure from 110 volts to about 8 volts as a minimum, for experimental purposes, such as operating low voltage tungsten lamps, ringing door bells, etc. The directions should enable an amateur to make a successful transformer having good efficiency. The author states that the efficiency of transformers made in accordance with his instruc¬ tions has been found to be over 90 per cent, in many cases, and never below 85 per cent. Review in “Electrical World,” July 17, 1915. This booklet should prove useful to amateurs wishing to construct small transformers. The directions are clear, and satisfactory results should be obtainable by any reader who will carefully follow the in¬ structions. From “Electrical Engineering,” July, 1915. How to Make Transformers for Low-Pressures is the title of a small book designed for “Young America” by Professor F. E. Austin, Hanover, N. H. The book gives specific directions for procedure in constructing transformers, and the many transformers built according to the specifications have shown wonderfully high efficiencies for small devices. The small transformers, when connected with the ordinary alternating current house circuits, may be used to operate small elec¬ tric lights, doorbells, small arc lights and direct current toy railways, operating five or six loaded trains at one time. The text is sufficiently suggestive throughout to invite initiative on the part of the reader to deviate from the 19 clauses of the specifications and effect many varia¬ tions. The price of the book is 40 cents. From Mining and Scientific Press, San Francisco. This little book is a companion to the author’s work on the design, etc., of high pressure transformers, and from it amateurs and others will be able to construct their own apparatus. Transformers appear to be complicated to the layman, and here is everything easily explained. If your book dealer cannot supply, address PROF. F. E. AUSTIN, Hanover, N. H., Box 441 BOYS CAN MAKE THEIR OWN TRANSFORMERS by following directions in this book. From “Power,” New York. How to Make Transformers for Low-Pressures. By Prof. F. E. Aus-| tin. Published by the author at Hanover, N. H. It frequently happens that the operating man, student or amateur electrician desires to construct a small transformer to operate on a 110- volt lighting circuit and produce a low secondary voltage for the opera¬ tion of bells, signals or alarm circuits. This little booklet contains full directions for making such a transformer, with suggestions for differ- i ent coil connections so as to produce various secondary voltages. It is excellent in so far as telling “how to do the work.” From “Canadian Engineer,” Canada’s leading Technical Weekly. How to Make a Transformer for Low-Pressures. By Prof. F. E. Aus¬ tin, Professor of Electrical Engineering, Thayer Scliooi of Civil Engineering, Hanover, N. H. Published by the author. Illus¬ trated, 5x7 ins., cloth. Price, 40c. net. Those interested in transformer construction will find this little book exceedingly interesting and useful. It answers a number of ques-1 ^ un ^ amen f a ^ principles and solves numerous problems f which the amateur transformer maker is likely to meet. The book de¬ scribes the process of construction step by step. From “Lighting Journal,” New York. Hozv to Make a Transformer for Low Pressures , edited and published by F. E. Austin. Illustrated. Price, 40 cents. Ihis contains step by step directions for making step-down trans- I formers, to work on voltages of about 110. A list of the materials neces- 1 sary is included as well as the detailed instructions for assembling. This second edition also contains some additional information on transformer I operation and methods of connection for securing various voltages. This book is intended to aid the amateur in constructing low volt¬ age transformers for the lighting of small lamps, ringing bells, etc. Remit amount with order to PROF. F. E. AUSTIN, Box 441 Hanover, N. H. Distributors for England, E. & F. N. SPON, Ltd., 57 Hay- market, London. * DIRECTIONS FOR DESIGNING, MAKING AND OPERATING HIGH-PRESSURE TRANSFORMERS The Book for those who want to make their own Wireless Apparatus Bound in Cloth, Postpaid . 65 cents Review from “Electrical Age”, New York. A new book entitled “DIRECTIONS FOR DESIGNING, MAK¬ ING AND OPERATING HIGH PRESSURE TRANSFORMERS,” written for those experimenters who desire to construct their own ap¬ paratus, is also published by Prof. F. E. Austin. The book is a com¬ panion volume of “How to Make a Transformer for Low-Pressures,” but containing more working directions and useful talks such as loss due to Hysteresis, per cubic inch of iron core for various flux densities and frequencies; and data applying to copper magnet wire. The book is well illustrated with half-tone and line cuts showing special methods of procedure, fundamental theories, and finished apparatus. It is written in simple English, is full of technical information and new ideas relating to methods of design and construction, and will prove of great assistance to those who are pursuing correspondence courses or regular college courses. The price of the book is 65 cents. Review from “The Electrical Experimenter”, New York. This volume is a brief but valuable treatise for those interested in the construction of high-tension transformers. The author tells in plain language how to calculate and obtain the various dimensions for different sizes of closed core high voltage trans¬ formers for use on any ordinary low-tension circuits. The copper and iron losses and their usual values are explained; also the method of calculating them. A table of the loss in watts at 15, 25, 60 and 100 cycles frequency for a cubic inch of transformer iron is given. An example is given for the calculation of a 20,000-volt, 1 kilowatt, closed core trans¬ former, for use on a 110-volt, 60-cycle circuit. Suggestions are offered on the manner of assembling the iron core laminations, and the sectional secondary method of construction is illustrated in detail. The possibili¬ ties of a transformer being used as a frequency changer are mentioned, as well as the method of connecting primary coils to produce different secondary potentials. IMPORTANT NEW BOOKS IN PRESS Examples in Battery Engineering, cloth, $1.25. This book treats of both primary and storage cells and batteries from the engineering standpoint. Every operator of an automobile, flying machine, or submarine should own a copy. Generator and Motor Examples, $2.00. A very useful and valuable book for those desiring a clear explana¬ tion of the fundamental principle of direct-current generator and motor design and operation. Remit amount to the author or to the follow¬ ing: E. & F. SPON, Ltd. Distributors for England, 57 Haymarket, London, England Review from “The Wireless World,” June, 1916, London, Eng. Directions for Designing, Making, and Operating High-Pressure Trans¬ formers, by Professor F. E. Austin, Hanover, N. H. 3s. net. This is an interesting and clearly written little book, particularly valuable to the serious student of wireless and to the operator who is 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 ■ ■ I I T* , . \ ' ' • I c V