FERRELL'S ADVAW^OED ARITHMETIC UC-NRLF mi m ?P3 « P r Crane & Company, Topeka^ Kansas Ji LIBRARY University of California. GIFT OK EdijA «4H Accession .9.3.45^ ^^^^ Digitized by the Internet Archive in 2007 with funding from IVIicrosoft Corporation http://www.archive.org/details/ferrellsadvancedOOferrrich FERRELL'S ADVANCED AEITHMETIC. BY J. A. FEERELL, B. S., O. E., Superintendent of Sedan City Schools, Sedan, Kansas, author of " Teachers' and Students' Manual of Arithmetic," and 'Terrell's Elementary Arithmetic." Cbane & Company, Publishers ToPEKA, Kansas 1901 TERRELL'S ARITHMETICS. ^kS\ 1 Ferrell's Teachers' and Students' Manual of Arithmetic. 165 Pages. — : ^'^a ft.,^ Ferrell's Elementary Arithmetic. 217 + xiii Pages. Ferrell's Advanced Arithmetic. 408 + xvi Pages. CRANE & COMPANY, PUBLISHERS, TOPEKA, KAN. (^^^£A^-e^- Copyrighted, 1901, by Crane & Co. PREFACE. This book was prepared especially as an advanced arithmetic for the public schools. It is the companion book of FerrelPs Elementary Arithmetic, which contains the mathematics for the third, fourth, and fifth school years. Briefly, the plan of the book is as follows : Numbers Part I. Part II. Problems Number t Part III. Arabic notation ( pp. 1-15). Processes (15-78). {Equation (79-92). Ratio (92-99). Proportion (99-105). Nature and classes (105-111). Methods of solution | ^-'j^lfi^ll-g^l^^, r Denominate numbers (148-179). j Involution and evolution (179-197). I Literal notation (197-209). ' Formulas and their use (210-213). Mensuration (213-246). [ Problems J Percentage (247-308). Mechanics (308-320). i Miscellaneous (321-332). Positive and negative (333-335). Processes (335-357). ' One base (357-365). Two bases (365-386). Three bases (386-393). Quadratics (393-399). , Progressions (399-408). Numbers Problems The author claims for this book the following special features of excellence : I. It is Complete. — It comprehends every phase of the study of numbers and problems, properly belonging to public-school (iii) 93457 IV PREFACE. arithmetic. It includes the fundamental and derived processes of numbers expressed by figures and letters — integral and fractional, positive and negative. It gives all the operations of simple equations, and a limited discussion of quadratics. It classifies and discusses the nature of all kinds of arith- metical problems, and illustrates and explains the various methods of solution. II. It is Scientific. — If any one thought more than another has inspired the author in the preparation of this book, it has been this : There should he system in studying and teaching arithmetic. Part I gives the pupil (1) a thorough and practical discus- sion of the fundamental and derived processes (except involu- tion and evolution) of numbers expressed by figures, and (2) one complete method by equations and one by proportion for solving all the simpler arithmetical problems. There is not a new plan of reasoning and a new method of solution for each particular problem. Solutions are given complete. This part contains the philosophical study of problems. Part II (1) gives the pupil such a knowledge of expressing numbers by letters as will enable him to understand and inter- pret a formula; (2) it teaches him how to classify problems into general types ; and (3) how to develop and use the formu- las for these types. This part applies the science of arithmetic to the problems of practical Ivi^ivess life. Part III gives the pupil (1) a knowledge of positive and nega- tive numbers, (2) the method of solving more difficult prob- lems — problems of one, two, three or more unknown numbers, and (3) a limited knowledge of quadratic equations and \)T0- PREFACE. gressions. This part prepares the pupil for the study of higher mathematics. III. It IS Teachable. — The teacher and the pupil are not left to guess at the author's methods of presenting the various subjects. Every part is explained and illustrated, and numer- ous examples are given. The pupil always has before him a model to study and follow, and there is no excuse for his efforts being haphazard and aimless. The plan of the book is adapted to the developing mind both in arrangement and explanations. For suggestions on teaching this book, see Introduction. J. A. F. Sedan, Kansas, August, 1901. TABLE OF CONTENTS. PART I. Page. , Study of Numbers 1-78 A. Introduction 1-4 Definitions 1 B. Notation and numeration op integers 4-15 Definitions 4 Methods 5 Figures 6 Reading and writing numbers to one thousand 6 Notation and numeration of integers in general 9 Reading and writing dollars and cents 14 0. Addition op integers 15-18 Definitions, signs, and principles 15 Process of addition 16 D. Subtraction of integers 18-20 Definitions, signs, and principles 18 Process of subtraction 18 E. Multiplication op integers 21-25 Definitions, sign, and principles 21 Process of multiplication 23 F. Division op integers 25-35 Definitions, signs, and principles 25 Process of short division 26 Three applications of division 27 Different ways of expressing division 28 Disposing of the remainder 29 Long division 30 General review 32 G. SiMPLIPYING numerical EXPRESSIONS 35-36 A term 35 Process 36 H. Derived operations 37-50 1. Factoring 37-44 Definitions and principles 37 (vii) Vlll CONTENTS. Page. Factoring by inspection 40 Factoring by division 41 Factoring expressions composed of two or more terms 42 Principles of multiplication and division relating to factor 43 2. Cancellation 45-46 Definition and principle 45 Process 45 3. Great common divisor 46-48 Definition and principles 46 Process 47 4. Least common multiple 48-50 Definition and principles 48 Process 49 I. Common fractions 51-64 Definitions and principles 51 Reducing fractions to higher terms 52 Reducing fractions to lowest terms 53 Reducing mixed numbers to improper fractions 54 Reducing improper fractions to integers or mixed numbers 55 Reducing fractions to L. C. D 55 Addition of fractions 56 Subtraction of fractions 57 Multiplication of fractions 58 Division of fractions 60 Greatest common divisor of common fractions 62 Least common multiple of common fractions 64 J. Decimal fractions 65-78 Definitions 65 Reducing decimals to higher terms 68 Reducing decimals to lower terms 68 Addition of decimals 69 Subtraction of decimals 70 Multiplication of decimals 71 Division of decimals 72 Reduction of common fractions to decimals 74 Reducing decimals to common fractions 77 II. Study of Problems 79-147 A. Means of expressing solutions 79-105 1. Equations 79-92 Definitions 79 CONTENTS. , IX Classification as to source v 79 Transformation 80 To turn an equation around 80 Transposition of terms 81 Simplifying members 83 To multiply an equation 83 To divide an equation 89 2. Ratio 92-99 Definitions 92 Writing the corresponding abstract terms of a ratio 93 Process of finding the numerical ratio 97 3. Proportion 99-105 Definitions 99 Denominations of terms 100 Solving a proportion with abstract terms 100 Solving a proportion with concrete terms 102 B. Problems op one base — nature and classification 105-111 Nature 105 Stating the parts of a problem 106 Classification 110 C. Problems op one basis — solution 112-147 Two methods of solution 112 1. Equation method 112-138 Arranging the parts of a problem 112 Process of solving problems 116 Integral solutions 116 Fractional solutions 122 Reciprocals 125 Solutions shortened 127 Percentage solutions 131 2. Proportion method 138-147 Process 138 PART II. I. Study of Numbers 148-209 A. Denominate numbers 148-168 Definitions '. 148 1. English system 149-168 Linear measures 149 CONTENTS. Page. Surface measures 154 Solid measures 156 Measures of capacity 157 Measures of mass 159 Measures of time 162 Measures of value 164 Review and rapid drill work 164 2. French system ^69-174 Linear measures 169 Surface measures 171 Solid measures 172 Measures of capacity 172 Measures of mass 174 Measures of value 174 3. Compound denominate numbers 174-179 Addition 174 Subtraction 176 Multiplication 177 Division 178 B. Involution 179-183 Definitions 179 Process 180 Another method 181 C. Evolution 183-197 Definition 183 1. Square root 184-190 First process 184 Second process 184 Second process shortened 187 2. Cube root 190-196 First process 190 Second process 190 Second process shortened 194 3. Roots of higher degree 196-197 Process 196 D. Numbers expressed by letters 197-209 Explanations 197 Addition ." 198 Subtraction 200 Multiplication 201 CONTENTS. , XX Page. Division 202 Involution , 203 Evolution 205 Equations containing letters 207 II. Study of Problems 210-332 A. Development and use op formulas 210-213 General numbers and general problems 210 Solving the general problem or type 211 B. Mensuration 213-246 Definitions 213 1. Plane surfaces 216-232 Parallelograms 216 Triangles 221 Other quadrilaterals 225 Regular polygons and circles 227 2. Solids 232-241 Polyhedrons 232 Solids having curved surfaces 237 Similar figures 241 Review 244 B. Percentage 247-309 1. Percentage formula 247-251 Elements . , 247 Developing the percentage formula 248 2. Percentage without time 251-279 Profit and loss 251 Trade discount 257 Commission 261 Stocks and bonds 264 Taxes 275 Duties 276 Insurance 278 3. Percentage with time 281-308 Terms used in interest 281 Time 281 Simple interest 284 Notes 290 Partial payments 292 Annual interest 294 Xll CONTENTS. Page. Compound interest 295 Bank discount 297 True discount 300 Exchange 302 0. Mechanics 308-320 Force 308 Work 313 Activity or power 315 Simple machines 317 D. Miscellaneous 321-332 Partitive proportion 321 Mixture problems 322 Equation of payments 324 Partnership 326 Longitude and time 327 PART III. I. Study of Numbers 333-408 A. Integral numbers 333-351 Positive and negative numbers 333 Numbers expressed by means of letters 334 Addition 335 Subtraction 339 Multiplication 343 Division 346 Factoring 348 G. C. D 350 L. C. M 350 B. Fractional numbers 351-357 Reduction 351 Addition 353 Subtraction 354 Multiplication 356 Division 357 II. Study of Numbers 357-408 A. Problems op one basis 357-365 Solving equations of one unknown number 357 Problems 360 CONTENTS. . Xlll Page. B. Problems op two bases 365-386 Solving equations of two unknown numbers 365 Problems , 373 C. Problems of three or more bases 386-393 Solving equations of three or more unknown numbers 386 Problems 388 D. Problems containing quadratic equations 393-399 Quadratic equations 393 Problems 396 E. Progressions 399-408 Arithmetical progressions 399 Falling bodies 403 Geometrical progressions 405 INTRODUCTION. To the Teacher: I. Plan. — Our best geometries first solve and give full ex- planations of all the fundamental problems and propositions of a book (or part) of geometry for the pupil's study ; and then give at appropriate places or at the close of the part a list of problems and propositions for the pupil to solve. The logic of this plan appears in this : That the pupil is taught the nature of problems and correct methods of solving them before he is required to solve them for himself. This is in general the author's plan of teaching not only ge- ometry, but all the branches of mathematics, especially all of the lower branches. Note.— It is bad pedagogy, contrary to business judgment, and a great waste of time, to put a pupil at work trying to solve problems before he has a knowledge of the nature of such problems as he is expected to solve, or a knowledge of the plan by which such problems are to be solved. Throughout this book the matter is arranged in two divisions or parts: explanatory part^ and exercises. The explanatory part consists of definitions, principles, rules, explanations, and ex- amples of processes and solutions. This part is to be learned by the pupil. The exercises are to be worked out by the pupil. II. How TO Use the Explanatory Part. — Assign explanatory matter for study and recitation. (1) At the time of assigning a lesson, go over it with the pupil, and explain such (and only such) things as, in your judgment, your pupil will be unable to understand for himself. (2) Require pupils to commit definitions and principles to memory. Note. — Do not tolerate " parrot" memorizing. Before the pupil at- tempts to commit a definition, he should find out what it means. Half (XV) XVI INTRODUCTION. of the work of memorizing is done when the pupil understands the mean- ing of the language used. The author is an advocate of memory work; but let it be intelligent memory work, and not ''words, words, words." (3) After reciting on principles, definitions, and the like, the pupil should be sent to the blackboard without his book ; and, after reaching the board, examples should be assigned and he be required from his own recollection and reasoning to write out and thoroughly explain the processes or solutions. Note. — It is the least of the author's intentions to permit the pupil to pass over examples as something to w^hich he may refer when in trouble — something to be studied or neglected at his pleasure. A thorough un- derstanding of the example comes first in time and importance in study- ing and teaching this book. III. The Exercises. — (1) In the number work, the exercises are given mainly for practice. Accuracy and speed should be the watchwords in number work. (2) In the problem work, the exercises are given to test the pupil's ability in applying the principles, solutions, and formulas which he has already studied; also as a drill, better to fix principles, models, and formulas in the mind. IV. Additional Exercises. — Exercise LVI has only 26 prob- lems. Nearly all the problems in Exercises CVII to CXIV are percentage problems, and the pupil should be able to solve any of them by the method given in Article 80. It is recom- mended that the pupil be given a thorough knowledge of the solution of percentage problems by the equation method. V. Answers. — Answers are not put in this book. The author believes that the presence of answers, either with the problems or at the close of the book, fosters dependency and lack of self- confidence in the pupil. For teachers, who may need answers to save time and labor in correcting and grading school work, the answers are printed in a pamphlet, and may be obtained from Crane & Co. ADVANCED ARITHMETIC. PART I. I. STUDY OF NUMBERS. A. INTRODUCTION. 1. Definitions. — Arithmetic is the introductory branch of the science of numbers and problems. Science is classified knowledge. A Number is one or more units, considered as forming one quantity, or amount. An expression of one or more units is also called a number. The word number is a term signifying one or more units, or an expression of one or more units. Note. — A distinction should be made between ^^ number" as used in the sentence, A number of apples were eaten, where the word number means quantity or amount; and *' number" as used in the sentence, The divisor is the number on the left, where the word number means the expres- sion on paper. A Unit is a single thing, or one. Units are classed as con- crete or abstract, integral ov fractional. A Concrete Unit is one thing. As, 1 book, 1 man, 1 box. An Abstract Unit is simply one, a ratio, and not a thing. As, J XVI INTRODUCTION. of the work of memorizing is done when the pupil understands the mean- ing of the language used. The author is an advocate of memory work; but let it be intelligent memory work, and not ^'words, words, words." (8) After reciting on principles, definitions, and the like, the pupil should be sent to the blackboard without his book ; and, after reaching the board, examples should be assigned and he be required from his own recollection and reasoning to write out and thoroughly explain the processes or solutions. Note. — It is the least of the author's intentions to permit the pupil to pass over examples as something to which he may refer when in trouble — something to be studied or neglected at his pleasure. A thorough un- derstanding of the example comes first in time and importance in study- ing and teaching this book. III. The Exercises. — (1) In the number work, the exercises are given mainly for practice. Accuracy and speed should be the watchwords in number work. (2) In the problem work, the exercises are given to test the pupil's ability in applying the principles, solutions, and formulas which he has already studied; also as a drill, better to fix principles, models, and formulas in the mind. IV. Additional Exercises. — Exercise LVI has only 26 prob- lems. Nearly all the problems in Exercises CVII to CXIV are percentage problems, and the pupil should be able to solve any of them by the method given in Article 80. It is recom- mended that the pupil be given a thorough knowledge of the solution of percentage problems by the equation method. V. Answers. — Answers are not put in this book. The author believes that the presence of answers, either with the problems or at the close of the book, fosters dependency and lack of self- confidence in the pupil. For teachers, who may need answers to save time and labor in correcting and grading school work, the answers are printed in a pamphlet, and may be obtained from Crane & Co. ADVANCED ARITHMETIC PART I. I. STUDY OF NUMBERS. A. INTRODUCTION. 1. Definitions. — Arithmetic is the introductory branch of the science of numbers and problems. Science is classified knowledge. A Number is one or more units, considered as forming one quantity, or amount. An expression of one or more units is also called a number. The word number is a term signifying one or more units, or an expression of one or more units. Note. — A distinction should be made between '^ number" as used in the sentence, A number of apples were eaten, where the word number means quantity or amount; and *^ number" as used in the sentence, The divisor is the number on the left, where the word number means the expres- sion on paper. A Unit is a single thing, or one. Units are classed as con- crete or abstract, integral or fractional. A Concrete Unit is one thing. As, 1 book, 1 man, 1 box. An Abstract Unit is simply one, a ratio, and not a thing. As, J 2 ADVANCED ARITHMETIC. A unit, not considered as a part of any other unit, is an Integral Unit. As, 1, 1 book, 1 foot. A unit, considered as a part of some other unit, is a Frac- tional Unit. As, ^ foot, i dollar, -J pint. Note.— The Hs 1 fifth of 1 foot. Then, the unit, ^ is a part of the unit, 1 foot ; so, the i is a part of $1 and the i is a part of 1 pint. The unit of which the fractional unit is a part is called the Unit of the Frac- tion. A number composed of concrete units is a Concrete Number. As, 6 men, 15 dollars, 20 miles. A number composed of abstract units is an Abstract Number. As, 6, 15, 20. When a concrete number and an abstract number have each the same number of units, they are called Corresponding num- bers. Every concrete number has its corresponding abstract num- ber. Note. — This fact will be important, when you come to solve problems by proportion and by formulas. A Simple Number is a number whose integral units are of the same kind and size. As, 250 bushels, 75 men. A Compound Number is a number whose integral units are of the same kind but of two or more sizes. As, 5 yards 2 feet; 4 bushels 2 pecks 5 quarts. ADVANCED ARITHMETIC. 8 A number composed of integral units is an Integral Number, or an Integer. As, 12 men, 25. Note. — An integer may be either concrete or abstract. A number composed of fractional units is a Fractional Number, or a Fraction. As, f dollar, t5_. Note. — A fractional number may be either concrete or abstract. A number composed partly of integral units and partly of fractional units is a Mixed Number. As, 5i dollars, Of. Note. — A mixed number may be either concrete or abstract. A Problem is a question proposed for solution. Note. — A problem is not always stated in the form of a question. For example, Find the cost of 10 books, at SOf each. But such problems have all the essentials of questions, and may be easily so stated. Thus, What is the cost of 10 hooks, at SOf each f EXERCISE I. 1. What is arithmetic ? 2. Are there other branches of the science of number ? 3. What is meant by " introductory branch " ? 4.. What is science f 5. What then is the science of number ? 6. What is a number ? 7. What is quantity f 8. What is a unit f 4 ADVANCED ARITHMETIC. 9. In the sentence, The number on the right is the quotient, does the word number signify one or more units or does it sig- nify the expression that is written ? 10. In the sentence, I do not know the number of marbles that were lost, is the word number and the real number of mar- bles the same thing ? What is the difference, if any ? 11. In the sentence, There are 7 books on the table, is the ex- pression, "7 books," and the real number of books the same thing ? What is the difference, if any ? 12. Number is sometimes defined as a collection of units. Do the units have to be collected, or may they be considered together as forming one quantity without being in reality col- lected ? Illustrate your answer with an example. 13. Give the classification of units. H. What is a concrete unit ? Give an example. 15. What is an abstract unit ? Give an example. 16. What is an integral unit ? Give an example. 17. What is 2i fractional unit f Give an example. 18. What is a concrete number ? Give an example. 19. What is an abstract number ? Give an example. 20. What are corresponding numbers ? Give an example. 21. What is an integer? Give an example. 22. What is a fraction f Give an example. 23. What is the unit of a fraction ? Give an example. 2J/.. What is a mixed number? Give an example. 25. What is a problem ? 26. Is a problem always in the form of a question ? Illus- trate your answer by examples. 27. Are all questions problems ? Illustrate your answer by examples. B. NOTATION AND NUMERATION OF INTEGERS. 2. Definitions. — ]N"otation is the process of express- ing numbers by means of words, letters, or figures. ADVANCED ARITHMETIC. 5 Numeration is the process of reading numbers, expressed in words, letters, or figures. Note. — A pupil that can correctly write numbers can read them, for the same knowledge of the method, of use of terms and characters, is re- quired in both. Hence, both processes are treated together in this book. 3. Metliods. — Five methods of expressing numbers are in common use : (1) By means of words that express the number of units considered. As, three, nine, twenty-five. (2) By means of letters that express the number of units considered. As, III, IX, XXV. Note. — Neither of the above methods is used in arithmetical compu- tations. Their study belongs to the reader rather than to the arithmetic. (3) By means of words that do not express the number of units considered. As, cost price, John's age, B's money. Note. — This method of expressing number is used in the solution of arithmetical problems. It is self-explanatory, and does not require spe- cial study at this time. (4) By means of letters that do not express the number of units considered. As, X, 2 ay, m*. Note. — This method is usually called algebraic notation. It is studied and used in Parts II and III. (5) By means of figures that express the number of units considered. As, o o ok Note.— This method as herein used is called the Arabic method. It is the common method of expressing known numbers in arithmetical computations. 6 ADVANCED ARITHMETIC. 4. Figures. — The Arabic method of notation employs ten characters, called Figures or Digits. Their forms and names are as follows : 0, 1, 2, s, 4, 5, 6, 7, 8, 9. zero, one, two. three, four. five, six, seven, eight, nine. Note. — The is also called naught. Some authors claim that there are only nine digits, and that 0, representing no value, is not a digit. 5. Reading and Writing Integers to One Tliousand. — In reading and writing numbers by the Arabic method — ( 1 ) The integers up to ten are named and written with fig- ures as follows : one, 1 five, 5 two, 2 six, 6 tliree, 3 seven, 7 four, 4 eiglit, 8 nine, 9 (2) In integers of more than nine units, the units are grouped as far as possible into groups of ten each. Groups of tens up to ten tens are named and written with figures as follows : 1 ten, ten, 10 5 tens, fifty, 50 2 tens, twenty, 20 6 tens, sixty, 60 8 tens, tliirty, 30 7 tens, seventy, TO 4 tens, forty, 40 8 tens, eiglity, 80 9 tens, ninety, 90 (3) In integers of 10 tens or more, the groups of ten each are grouped as far as possible into larger groups of 10 tens each. One of these larger groups is called a hundred. Hun- ADVANCED ARITHMETIC. dreds up to 10 hundreds are named and written with figures as follows : one liundred, 100 five liundred, 500 two liundred, 200 six liundred, 600 three hundred, 300 seven hundred, TOO four hundred, 400 eight hundred, 800 nine hundred, 900 (4) Integers between 10 and 20 are named and written with igures as follows : 1 ten 1 unit; eleven ; 11 1 ten 2 units twelve ; 12 1 ten 3 units thirteen ; 13 1 ten 4 units fourteen ; 14 1 ten 5 units fifteen ; 15 1 ten 6 units ; sixteen ; 16 1 ten 7 units seventeen ; 17 1 ten 8 units ; eighteen ; 18 1 ten 9 units nineteen ; 19 (5) Any integer between 20 and 100 is written with figures by placing the number of tens in the second place from the right or tens^ place and the number of units in the right-hand j)lace or units^ place. If there are no units, place a in units' place. These numbers are read by naming the tens and units in succession. Thus: 21, twenty-one 47, forty-seven 78, seventy-eight 86, eighty-six 55, fifty-five 34, thirty-four 92, ninety-two 63, sixty-three 99, ninety-nine (6) Any integer between 1 hundred and 10 hundred is writ- ten by placing the number of hundreds in the third place from O ADVANCED ARITHMETIC. the right, or hundreds'' j)lace ; the number of tens in tens' place ; and the number of units in units' place. Fill vacant places with O's. These numbers are read by naming the hundreds, tens, and units in succession. Thus: 125, one liundred twenty-five 790, seven liundred ninety 467, four liundred sixty-seven 806, eiglit liundred six 512, iive liundred twelve 272, tTvo liundred seventy-two 638, six hundred thirty-eiglit 301, tliree liundred one 999, nine hundred ninety-nine Note. — Do not use " and " in reading these numbers. Since, 1 liundred = 10 tens, and 1 ten = 10 units, a figure in hundreds' place represents 10 times as many as the same figure in tens' place; a figure in tens' place represents 10 times as many as the same figure in units' place. EXERCISE II. Read rapidly : 1. 2. 3. 4. 5. 78 34 800 065* 259 19 60 139 815 734 25 71 240 725 813 82 88 327 899 004* 96 07* 444 868 117 57 56 516 992 629 43 99 607 408 999 * In integers, a standing on the left of other figures has no effect upon the number. 07 is 7. 065 is 65. 004 is 4. ADVANCED ARITHMETIC. Write : 6. 7. 8. 9. twenty-one fifty-four thirty-nine sixty-nine seventy ninety eighty-one thirty-seven eighty-nine seventy-two thirteen ninety-three thirty-eight nineteen seventy-three forty-five forty-six sixty-six ninety-eight eighty-four 10. Four hundred ninety-four Two hundred nineteen Five hundred fifty-five Seven hundred seventy-nine Eight hundred six Nine hundred sixty-nine Eight hundred sixty Seven hundred ten Six hundred sixty-six Four hundred 11, Eight hundred seventy- two Three hundred thirty-three One hundred fourteen Nine hundred three Eight hundred seventy-six Six hundred twenty-four Five hundred nineteen Seven hundred ninety-four Two hundred forty-eight Nine hundred ninety-nine 6. Notation and Numeration of Integ-ers in Greneral. — By the Arabic method, integers larger than nine are represented by figures placed side by side. Thus: 274139568 The 8 represents I's, or units of the first order ; The 6 represents lO's, or units of the second order; The 5, units of the tliird order ; The 9, units of the fourtli order ; and so on. The order of any figure or unit is the number of its place, counting from the right. >i 10 ADVANCED ARITHMETIC. This method of representing integers is based upon the fol- lowing law : Law. — Ten units of any one order make one unit of the next higher order. In reading and writing integers, the figures are considered, as far as possible, iij groups of three figures each, commencing at the right hand. These groups are called Periods. The first twelve periods, counting from the right, are named as follows : units quintillions tliousands sextillions millions septillions billions octillions trillions nonillions . quadrillions decillions Note. — If the word, period, follows the name, the possessive sign should be used. Either period of thousands or thousands' period is correct. Let it be required to read the number, 346524207985325. Separating it into periods, and naming the periods, we have — trillions^ billions, millions, thousands, units. 346, 524, 207, 985, 325. How many units ? Ans., 325. How many thousand ? Ans., 985. How many million ? Ans., 207. How many billion ? Ans., 524. How many trillion ? Ans., 346. Then read the number, beginning with the left-hand period. Thus, 346 trillion, 524 billion, 207 million, 985 thousand, 325 (units). Note. — In reading numbers, drop the word units, and say hundred, thousand, million, &c,, not hundreds, thousands, millions, &c. Never use " and " in reading integers. The figures in each period are read by methods explained in Article 5 ; then the name of the period is added. ADVANCED ARITHMETIC. 11 Let it be required to write seventy-five hillion, nine hundred forty-eight million, five hundred one thousand, three hundred sixty- two. How many billion ? Arts., 75. How many million ? Ans., 948. How many thousand ? Ans., 501. How many units ? Ans.y 362. Then write the number. Thus, 75,948,501,362. Note.— When it is desired to separate the periods of a number, com- mas are used ; as in the last number above. Since a period has three places, the place on the right hand is units of the period, the second place is tens and the third place is hundreds. To illustrate, the first place in thousands is units of thousands, or simply thousands; the second place, ten- thousands ; the third place, hundred-thousands. The first place in millions is millions; the second, ten-millions; the third, hundred-millions; and so on. The method of grouping three figures into a period, as ex- plained above, is the method used by the people of the United States and France, and is called the Frencli Method. The English people and people of many other European na- tions consider a period as made up of six figures. This method of grouping the figures is called the Englisli Metliod. The first six periods according to this method are named as follows : units trillions millions quadrillions billions cxuintillions Let it be required to read 346524207985825 by the English method. 12 ADVANCED ARITHMETIC, Separating the number into periods, we have — billions, millions, units. 846, 524207, 985325 How many units ? Ans., 985325. How many million ? Ans., 524207, How many billion ? Ans., 346. Then read the number, beginning with the left-hand period. Thus, 346 billion, 524207 million, 985325 (units). Note. — The figures in each period are read by the methods already explained, then the name of the period is added. EXAMPIiES. 1. Read 3462465075. Process : 3,462,465,075; 3 bilHon, 462 million, 465 thousand, 75. Note.— Use the French method unless the English method is called for or indicated. Commas are not commonly used to separate the peri- ods, unless there are three or more periods. In smaller numbers, the periods are located mentally. 2. Read 672000024941. Process : 672,000,024,941 ; 672 billion, 24 thousand, 941. S. Read 873002403400540 ( English method ). Process : 873,002403,400540; 873 billion, 2403 million, 400540. i. Write 894 billion, 1 thousand, 127. Written : 894,000,001,127. Note. — There are no millions ; fill the orders with O's. There are no hundreds of thousands or tens of thousands ; fill these orders with O's. 5. Write 1753 billion, 201400 million, 843004. Written : 1753,201400,843004. ADVANCED ARITHMETIC. IB 6. Write 5 units of the 6th order, 7 units of the 3d order, 1 unit of the 2d order, 8 units of the 1st order. Written: 500718. Note.— Fill the 5th and 4th orders with O's. EXERCISE III. i. What is the law upon which the Arabic method of read- ing and writing numbers is based ? 2. What is the order of a figure ? How are orders numbered ? S. How many figures in a period according to the French method ? Where do you commence to point off a number into periods ? 4-. How many figures in a period according to the English method ? 5. Name in order twelve periods of the French method. 6. Name in order six periods of the English method. 7. Do the French and English methods both conform to the law of the Arabic method ? Read by the French method : 8. 43250 13, 78604501257943 9. 678426 U, 10030040250641 10. 9407081 15. 83000034562135 11. 5675694073 i^.' 786429536742 12. 889970708432 17. 182103243576842 Read by the English method : 18. 765302456901327. 19. 1234567878785821. 20. 986485200840561256. Write : 21. 55 thousand, 421. 22. 704 million, 78 thousand, 25. 14 ADVANCED ARITHMETIC. 23. 846 billion, 721 million, 1 thousand. 24.. Twenty-five thousand seven hundred one. 25. Seventy-six billion, eight hundred one million, sixteen thousand three hundred seven. 26. Seven units of the 5th order, nine units of the 4th order, six units of the 8d order, one unit of the 2d order, 8 units of the 1st order. 27. 8 units of the 7th order, 2 units of the 5th order, 5 units of the 4th order, 9 units of the 1st order. 28. 125 quadrillion, 125 trillion, 125 million, 125. 29. Five hundred sixty-six thousand two hundred eighty-one million, eighty-nine thousand twenty-one. 30. 675421 billion, 604001 million, 845762. 7. Reading' and Writing Dollars and Cents.— The dollar-mark, $, is always placed before the number with which it is used. Thus, , $5, five dollars. $908, nine hundred eight dollars. Cents may be expressed with dollars or with a dollar-mark. When so expressed, tivo places are always used for cents. 100 cents = 1 dollar; then, any number of cents less than a dollar may be written with two figures, of which the right-hand figure is for the units of cents, and the left-hand figure is for the tens of cents (dimes). A period (.) is always placed between dollars and cents and the word and is used there in reading. Thus, $5.10, 5 dollars and 10 cents; $12.06, 12 dollars and 6 cents; $.05, 5 cents; $807.50, 807 dollars and fifty cents. ADVANCED ARITHMETIC. 15 Cents may be written by using the cent-mark, /. Thus, 3/, 3 cents ; 20/, 20 cents ; 85/, 85 cents. EXERCISE IV. Read : 1. 2. S. 4. $7 $10.10 $.40 25/ $59 $72.55 $.75 5/ $408 $246.05 $.03 70/ $3906 $109.13 $.99 99/ 5, Write with $ : Two dollars and forty cents ; seventeen dol- lars and seventy-five cents ; nine thousand five hundred sixty- three dollars and twenty-five cents ; two cents ; fifty cents. 6. Write with / ; Twenty-five cents ; nine cents ; eighteen cents; one hundred twenty-five cents ; ninety-five cents. C. ADDITION OF INTEGERS. 8. Definitions, Signs, and Principles. — Addi- tion is the process of uniting two or more numbers into one. The numbers to be added are called Addends. The result of addition is called the Sum. or Amount. The Sign of Addition, +, is read plus. The Sign of Equality, =, is read equals. 17+9 = 26 is read 17 plus 9 equals 26, and means that 9 added to 17 gives a result of 26. 17 and 9 are addends, and 26 is the sum. In the fundamental processes of mathematics, certain self- evident principles must be observed. In addition they are as follows : Principles: 1. Only similar numbers can be added, 2. The sum contains all the units of all addends. 3. The sum and the addends must be similar numbers. 16 ADVANCED ARITHMETIC. If.. The sum is the same, regardless of the order in which the num- bers are added. Note. — Similar numbers, as here used, are those whose integral units are of the same kind and size. 9. Process of Addition. — Write the addends so that figures of the same order will stand in a column. Always be- gin with units to add. Ten units of one order make one of the next higher order. Then, in adding numbers of more than one order, if the sum of all the units of any order is ten or more, the tens of that order are considered as units of the next higher order, and are so added. Add 576, 842, 75, 981, 203. Process. Explanation : (1) The sum of the units of the first col- 576 umn (the one on the right) is 17, or 1 unit of the 2d order 342 and 7 units of the first order. Write the 7 below, and po? carry the 1 to the next column. 203 (2) The sum of the units of the 2d column, including 2177 result *^^ ^ carried, is 27, or 2 units of the 3d order and 7 units of the 2d order. Write the 7 below and carry the 2 to the next column. (3) The sum of the units of the 3d column, including the 2 carried, is 21, or 2 units of the 4th order and 1 unit of the 3d order. Write the 1 and the 2 below in their proper orders. Tests of Accuracy: (1) Revieiv the work, or (2) add the col- umns in reverse order. Note. — Various other tests of accuracy in addition are suggested by different authors. Such as, casting out the 9's, or by separating the addends into two or more groups, adding each group, then adding their sums. Such methods are not resorted to once in a thousand times to test the accuracy of addition. ADVANCED AEITHMETIC, 17 EXERCISE V. Copy and find sums : 1. 2, 3. ^. 5, 6. 346 76d 555 276 821 482 425 675 - 462 549 874 888 879 578 888 695 999 476 276 989 951 598 846 827 942 876 877 846 594 848 379 768 894 589 666 555 566 898 856 444 389 776 7. 8. 9. 10. 11, 12. 942 872 584 829 938 456 224 884 882 559 986 472 837 724 486 629 729 829 246 654 785 478 259 464 461 542 682 545 454 928 729 421 428 882 793 846 288 899 987 765 455 999 13. U^ 15. 16. 17. 12467 99999 95487 57684 88579 86492 86897 58987 66666 54865 94876 58987 89987 57868 47986 87594 68859 66666 47984 55555 59187 45958 97869 88946 84484 67895 78986 77777 98979 98897 78957 54852 12345 78787 48864 98798 77958 99999 57978 87758 78978 68879 98765 86868 69589 67594 59987 88888 69699 57868 98979 68827 48767 87879 48573 64888 77989 56789 87654 48452 18 ADVANCED ARITHMETIC. 18. 123456789 547695487 865796543 759876987 666666666 989898989 465798788 543543543 999988888 789789789 19. 876543219 34768492 99999 7586493 59876538 789437598 888888888 958764356 579867958 576576576 24678 579642 8978687 59875879 4769876543 989898989 65943756 987689 8976576438 9578495789 D. SUBTRACTION OF INTEGERS. 10. Definitions, Sign, and Princii)les. — Sub- traction is the process of taking one number from another. The number taken is called the Subtraliend. The number from which the subtrahend is taken is called the Minuend, The result of subtraction is called the Remainder or Dif- ference. The Sign of Subtraction, — , is read minus. Thus, 17-9 = 8 is read 17 minus 9 equals 8, and means that 9 taken from 17 leaves 8. 17 is the minuend ; 9, the subtrahend ; and 8j the remainder. Principles: 1. The minuend, subtrahend, and remainder must be similar numbers. 2. The remainder equals the minuend minus the subtrahend. 3. The subtrahend equals the minuend minus the remainder. }f. The minuend equals the sum of the subtrahend and remainder. 11. Process of Subtraction. — Place the subtrahend under the minuend, units under units, tens under tens, etc. Always begin with units to subtract. One of any order (except the first order) makes ten of the next ADVANCED ARITHMETIC » 19 lower order. When a figure of the minuend represents fewer units than the corresponding figure of the subtrahend, one unit is taken ("borrowed"-) from the next higher order of the minuend and considered as ten of the lower order. EXA1CPI.ES. 1. From 43 take 26. Process. Explanation complete : ( 1 ) "We cannot take 6 units from 3 43 units ; but we can take one ten of the 4 tens, and add it to the 26 3 units. This will make 13 units. Then, 13 units -6 units = 7 ]7 units. Write the 7 below. (2) One ten has been taken from the 4 tens, which leaves but 3 tens in the minuend. 3 tens -2 tens leaves 1 ten. Write the 1 below. 2. From 704 take 875. Process. Explanation complete: (1) We cannot take 5 units from 4 704 units, and there are no tens in the minuend ; then we must 375 take 1 hundred from the 7 hundred : 1 hundred = 10 tens. 329 Now, we can take 1 of the 10 tens and add it to the 4 units. 1 ten +4 units = 14 units. 14 units -5 units = 9 units. Write the D below. (2) Of tho 10 tens we took from hundreds' place, we have used 1 ten and have 9 tens left. 9 tens - 7 tens = 2 tens. Write the 2 below. (3) We have used 1 hundred of the 7 hundred ; then we have only 6 hundred left. 6 hundred - 3 hundred = 3 hundred. Write the 3 below. 3. From 8801 take 5674. Process. Explanation shortened : (1) Looking upon 1, think 11 (why ? ). 8801 11 - 4 = 7. Write the 7 below. 5674 (2) Looking upon the 0, think 10, then 9 (why?). 9-7 = 2. 3127 Write the 2 below. (3) Looking upon the ^, think 7 (why?). 7-6 = 1. Write the 1 below. (4) 8-5 = 3. Write the .? below. Tests of Accuracy: (1) Review the work; or (2) the minu- end minus the remainder should equal the subtrahend; or (3) the sum of the subtrahend and remainder should equal the minuend. 20 ADVANCED ARITHMETIC. EXERCISE VI. Copy and find remainders : 1. ^. 3. -^. 5. 764 591 932 846 821 178 486 763 487 295 6, 7. 8. 9. 10. 887 700 500 800 400 639 583 479 865 128 11, 12. IS. U^ 15. 8426 7363 8342 1264 1848 7544 5486 ' 486 947 796 16. i7. 18. 19. 20. 1920 1980 2004 2245 6472 749 806 1759 1892 2808 21. 22. 23. 2J^. 25. 7038 8388 1675 2008 3004 5984 2845 597 1009 2106 Find remainders : 26. 94675- -67897 37. 345678-95876 27. 58406- -57643 38. 193846-87909 28. 70698- -67899 39. 150000-74567 29. 54868- -48794 1,0. 183456-89999 30. 60054- -54876 Ul. 100000-65482 31. 92845- -56789 1,2. 240000-186426 ,^^. 80125- -48056 U3. 170984-145096 33. 72010- -35476 U^ 333888-278747 ,5^. 82904- -68926 J,5. 811221-456789 55. 92500- -47882 U6. 830405-640500 36. 76543- -34567 J^7. 900002-712345 ADVANCED ARITHMETIC. 21 E. MULTIPLICATION OF INTEGERS. 12. Definition, Sign, and Principles.— Multi- plication is a process, shorter than addition, for finding the sum when one number is to be used as an addend several times. The number to be used as the addend is called the Multipli- cand. The number showing how many times the multipli- cand is to be used is called the Multiplier.' The result of multiplication is called the Product. The expression, ^^ multiplied by,''^ used to indicate multij^lication, means that the number placed before it is to be used as the multiplicand and the number placed after it is to be used as the multiplier. The expression, ^^ multiplied together,^ ^ is u^ed to indicate the multiplication of one number by another without designating which number is to be used as the multiplier or multiplicand. The Sign of Multiplication, X, is read times* *'iX"is read one time, or once; ";^X," two times, or twice. 5x7 = 35 is read 5 times 7 equals 35, and means that the sum obtained by using 7 as an addend 5 times is 35. The 5 is the multiplier ; the 7, the multipli- cand ; and the 35, the product. Principles : 1. The multiplicand may be either concrete or ab- stract. 2. The multiplier must be abstract. 8. The product must be similar to the multiplicand. If. When the two numbers to be multiplied together are both ab- stract, the product is the same, whichever number is taken as the multiplier. Note. — Before proceeding with multiplication, the pupil should have a thorough knowledge of the multiplication table. *Many authors also read the sign, X, multiplied by. In that reading, the multiplicand must come before and the multiplier after the sign. For reasons which It would not be profitable to discuss here, the author prefers to read the sign times^ and adheres to that reading throughout this book. 22 ADVANCED ARITHMETIC. MULTIPLICATION TABLE. IX 0= 2X 0= 8X 0= 4X 0= ix 1= 1 2X 1= 2 8X 1= 8 4X 1= 4 IX 2= 2 2X 2= 4 8X 2= 6 4X 2= 8 IX 8= 3 2X 8= 6 8X 8= 9 4X 8 = 12 IX 4= 4 2X 4= 8 8X 4=12 4X 4=16 IX 5= 5 2X 5 = 10 8X 5 = 15 4X 5 = 20 IX 6= 6 2X 6=12 8X 6 = 18 4X 6 = 24 IX 7= 7 2X 7=14 8X 7 = 21 4X 7 = 28 IX 8= 8 2X 8 = 16 8X 8 = 24 4X 8 = 82 IX 9= 9 2X 9=18 8X 9 = 27 4X 9 = 86 IX 10=10 2X10 = 20 8X10 = 80 4x10 = 40 1X11 = 11 2X11=22 8X11=88 4x11=44 1X12=12 5X 0= 2 X 12 = 24 8x12 = 86 4X12 = 48 8X 0= 6X 0= 7X 0= 5X 1- 5 6X 1= 6 7x 1= 7 8X 1= 8 5X 2 = 10 6X 2 = 12 7x 2 = 14 8X 2=16 5X 8 = 15 6X 8 = 18 7x 8 = 21 8X 8 = 24 5X 4 = 20 6X 4 = 24 7X 4 = 28 8X 4 = 82 5X 5 = 25 6X 5 = 80 7X 5 = 85 8X 5 = 40 5X 6 = 80 6X 6 = 86 7X 6 = 42 8X 6 = 48 5x 7 = 85 6X 7 = 42 7x 7 = 49 8X 7 = 56 5x 8 = 40 6X 8 = 48 7X 8 = 56 8X 8 = 64 5X 9.= 45 6X 9 = 54 7x 9 = 68 8X 9=72 5x10 = 50 6X10 = 60 7x10 = 70 8X10 = 80 5X11=55 6X11=66 7x11 = 77 8X11=88 5X12 = 60 9X 0= 6X12 = 72 7 X 12 = 84 8X12 = 96 10 X 0= 11 X 0= 12X 0= 9X 1= 9 10 X 1= 10 11 X 1= 11 12X 1= 12 9X 2= 18 10 X 2= 20 11 X 2= 22 12 X 2= 24 9X 8= 27 10 X 8= 80 11 X 8= 88 12 X 8= 86 9X 4= 86 10 X 4= 40 11 X 4= 44 12 X 4= 48 9X 5= 45 10 X 5= 50 11 X 5= 55 12 X 5= 60 9X 6= 54 10 X 6= 60 11 X 6= 66 12 X 6= 72 9X 7= 68 10 X 7= 70 11 X 7= 77 12 X 7= 84 9X 8= 72 10 X 8= 80 11 X 8= 88 12 X 8= 96 9X 9= 81 10 X 9= 90 11 X 9= 99 12 X 9 = 108 9x10= 90 10x10=100 11X10=110 12x10=120 9X11= 99 10X11 = 110 11X11 = 121 12X11 = 182 9X12 = 108 10X12 = 120 11X12=182 12X12=144 ADVANCED ARITHMETIC. 28 13. Process of Multiplication. — Place the multi- plier under the multiplicand ; multiply by each figure of the multiplier, beginning with units. Always place the right-hand figure of each product under that figure of the multiplier used in obtaining it. Adxi these products for the complete product. EXAMPLES. 1. Multiply 379 by 67. Process. 379 67 2653 2274 25393, result. Explanation: (1) 7x9 = 63. Write the 3 below and carry the 6. 7x7 = 49. 49 + 6 (carried) =55. Write the 5 (units) below, and carry the 5 (tens). 7x3 = 21. 21 + 5 (carried ) = 26. Write the 26 below. (2) Multiply by the 6 in the same way. (3) Add these products. The complete product is 25393. Note. — The 6 of the multiplier is 6 tens, and the first figure of its product, which is 4, is also 4 tens. Therefore the 4 is placed under the 6, or in the second order. Note.— The ''Long Way*' given in the following examples is for ex- planation, and not to be practiced by the pupil. 2. Multiply 57 by 40. Long Way. 4Q Note.— In the Short Way, we place the 4 under the 7, and the to the right of the 4. Bring down the and multiply by 4 only. 00 228 Short Way, 57 40 2280 2280 3. Multiply 870 by 500. Long Way. Note. — In the Short Way, we place the 5 under the 7, letting the O's fall to the right. Multiply by the 5, and bring down three O's — one for the in the multiplicand, and two for the O's in the multiplier. 185000 Short Way. 370 500 185000 24 ADVANCED ARITHMETIC. U. Multiply 246 by 502. Long Way. Short Way. 246 gQ2 Note. — We need not multiply by the in ^aq — —- the multiplier. But we must be careful to 502 QQQ put the of the partial product obtained by 7^ 1230 multiplying by the 5 under the 5. 1230 123i92 123492 5. Multiply 678 by 1200. 678 1200 Note.— Multiply by 12, not by 2 and 1. 813600 Tests of Accuracy: (1) Review the work, or (2) multiply the multiplier by the multiplicand, considering the multiplicand abstract. EXERCISE VII. Find products : 1. 400x583* 16. 724x6743 2. 530x475 17. 594X3948 3. 640X873 18. 475x9873 i. 370x582 19. 649X8479 5. 420x897 20. 709x4377 6. 730X948 21. 806x9835 7. 890x345 22. 639x3829 8. 900x760 23. 738x5893 9. 730x800 2J^. 1263x85643 10. 1200X548 25. 2095x56789 11. 1100X675 26. 4075X83064 12. 246X375 27. 8764x93725 13. 856x948 28. 4609x85274 U, 987x987 29. 5007x95683 15. 876X955 30. 4199x87878 *The multiplier Is placed before and the multiplicand after the sign " X-" ADVANCED ARITHMETIC. 25 31. Multiply 504768 by 943000 J^l. 75x94x237 32. Multiply 724256 by 23462 J^2. 184x276x329 S3. Multiply 958740 by 37542 Jf.S. 101x279x834 3J^. Multiply 818475 by 30724 U- 642x200x730 35. Multiply 798648 by 48056 J^5. 820x540x422 36. Multiply 829874 by 58876 J^6. 25 X 84 X 68 X 75 37. Multiply 941072 by 86754 ^7. 38 X 240 X 820 X 97 38. Multiply 480762 by 10000 J^8. 39x800x600x240 39. Multiply 708840 by 25000 J!^9. 42 X 560 X 720 X 980 1^0. Multiply 984682 by 92845 50. 125x836x457x789 F. DIVISION OF INTEGERS. 14. Definitions, Signs, and Principles. — Divi- sion is the process of finding how many times one number contains another. The number contained is called the Divi- sor. The number containing the divisor is called the Divi- dend. The result of division is called the Quotient. The Sign of Division, -^, is read divided by. The Sign of Ratio, : , which is also a sign of division, is read the ratio of. ...to Principles: 1. The dividend, divisor and remainder must be similar numbers. 2. The quotient must be abstract. 3. The dividend equals the 'product of the divisor and quotient, plus the remainder. Note.— Many authors hold that the divisor may be abstract and the quotient similar to the dividend. Such a principle is inconsistent with the above definition of division. The author believes that the above definition is broad enough to cover every application of division. (See Art. 16.) 35^5 = 7 is read 35 divided by 5 equals 7. 26 ADVANCED ARITHMETIC. 35:5 = 7 is read the ratio of 35 to 5 equals 7. Each means that 35 things contain 5 of those things 7 times. ( See further explanation, Art. 17.) The 35 is the dividend ; the 5, the divisor ; and the 7, the quotient. 15. The Process of Short Division. — In Short Division, the process is performed mentally, and the dividend, divisor, and quotient only are written. Short division should be employed when the divisor does not exceed 12. Put the divisor on the left of the dividend, and begin at the left of the dividend to divide. Write the quotient below. It would be impossible to divide the whole of a large divi- dend at a glance ; but the number represented by the first one or two figures on the left can be divided ; the remainder, if any from this division, can be reduced to the next lower order and the division continued until all figures of the dividend are used. The following examples explain : EXAMPLES. i. Divide 76 by 4. Process. Explanation: (1) 7 + 4 = 1, and 3 remaining. Write the i 4 )76 below the 7. 19 (2) The remainder 3 is tens. 3 tens + 6 units = 36 units. 36 + 4 = 9. Write the 9 below. Quotient 19. 2. Divide 169 by 8. Process. Explanation : (1) 1 of the dividend is smaller than 8, 8 )169 then we use 16. 21 (2) 16-^-8 = 2. Write the 2 below the 6. i-gui 1 (3) 9 + 8 = 1, rem. 1. Write the quotient i to the right of the 2, and the rem. 1 below. Quotient 21, remainder 1. 3. Divide 7220 by 6. Process. Explanation : (1) 7-*- 6 = 1, rem. 1. AVrite the quotient 1 6 )7220 below the 7. 1203 (2) Remainder 1 is 1 thousand. 1 thousand +2 hundred rem. 2 =12 hundred. 12 + 6 = 2. Write the 2 below. (3) 2 + 6 = 0, rem. 2. Write below. (4) The remainder 2 tens = 20 units. 20 + 6 = 3, rem. 2. Write the 3 at the right of the 0, and the rem. 2 below. Quotient 1203, remainder 2. ADVANCED ARITHMETIC. 27 Tests of Accuracy: (1) Review the work, or (2) apply prin- ciple 3. EXERCISE YIIL Copy and find quotients and remainders : 1. 2. 3. 4. 5, 8)828 6)372 9)j405 7)595 5)895 6, 4 )4567 11. 10)4892 7. 6 )8946 12, 11)5948 8. 7 )5938 13. 12)8736 9. 3 )9874 U. 11)9878 10. 9 )8064 15. 10)5306 Find quotients and remainders : 16. 84695 -T- 11 17. 59060-^10 18. 73421-^12 19. 12345^12 20. 23456^9 21. 73024-T-8 22. 16295-^ll 23. 18020^6 21^. 16200-^12 25. 90000^7 26. 235061-^12 27. 456789 --11 28. 760890^9 29. 888888 -^ 12 30. 987654-f-8 16. Tliree Applications of Division. 9 )36 4 The fact of this division is, that 36 things contain 9 of those things 4 times. Note. — We must be careful not to fall into the error of thinking that 36 contains 9 as a bucket contains water. The word ^^contains,'' as here used, means ^Hs made up of," or *H*s composed of." First Application. — Problem : There are 36 apples in a lot. How many times can we take 9 apples from the lot ? Answer : From division we know that 36 apples contain 9 apples 4 times. Then, we can take 9 apples from the lot 4 times. This is the most direct application of division, and the only one which we have presented so far in this part. 28 ADVANCED ARITHMETIC. Second Application. — Problem: There are 86 apples in one lot and 9 apples in another. The first lot is how many times as large as the second lot ? ' Answer : From division, we know that 36 apples contain 9 apples 4 times. Then, 36 apples must be 4 times as large as 9 apples of the same size. This is the application nearest like the direct application, and may be called the Hatio Idea of Division. Third Application. — Problem : There are 36 apples in a lot. What is i of the lot ? Answer: From division, we know that 36 apples contain 9 apples 4 times. Suppose we take 9 of the 36 apples, and with them begin 9 new lots ; for every time we can take 9 apples from the 36 apples, we can put 1 apple in each new lot. But 36 apples contain 9 apples 4 times. Therefore we can put 4 apples in each new lot. Thus, we have separated 36 apples into 9 equal new lots, and one lot contains ^ of 36 apples, or 4 apples. Note. — If necessary, the teacher should illustrate this with objects. This may be called the Fraction Idea of Division. 17. Different Ways of Expressing Division. There are many ways or methods of expressing division. The following are in common use : (1) 86-^9 This expression is usually read, ^^36 divided by 9." It means (1) 36 contains 9, (2) the ratio of 36 to 9, or (3) i of 36. It is a general method of expressing division. ADVANCED ARITHMETIC. (2) Y This expression may be read, (1) ''S6 divided by 9," (2) ''The ratio of 36 to 9," or (3) ''36 ninths.'' It, like the first expression, is a general method of expressing division. ■^ (3) 36:9 Read, "The ratio of 36 to 9." This expresses only the Ratio Idea of Division. (4) i of 86 Read, "One-ninth of 36." This expresses only the Frac- tion Idea of Division. ' 18. Disposing of the Remainder. EXAMFIiE. 1. Divide 134 by 5. 5)134 26t Explanation : After we had obtained the quotient 26, we had a remainder of 4. This 4 is to be divided by 5. But we learned in Expression (2), Article 17, that 4 divided by 5 may be put in the form, |, and called "four-fifths." This we put on the right of the 26 and call the expression, 26|, "twenty-six and four-fifths." 26 1 is the exact quotient. When the exact quotient is an integer, the dividend is said to be divisible by the divisor. Thus, 36--9=4. 36 is divisible by 9. EXERCISE IX. Find exact quotients : 7. 347-f-7 6. 834:7 11. } of 965 16. H^ 2. 545 --9 7. 209:6 12. i of 677 17. -w 3. 806--11 8. 707:11 13. yV oi 509 18. H^ 4. 793-r-12 9. 651:5 U. tV of B27 19. iOi 5. 820-^9 10. 939: 10 15. ^ij of 436 20. -w ADVANCED ARITHMETIC. 19. Ijong Division. — In Long Division, the process is the same as in short division, but the work is written. Write the quotient above or to the right of the dividend. The author believes that placing the quotient above has some advantages. EXAMPIiBS. 1, Divide 464 by 17o Process. 27 17)464 Explanation: (1) 46 + 17 = 2+.* 2x17 = 34. 34 Bring down 4. 124 + 17 = 7 + . 7x17 = 119. 124 Quotient 27, remainder 5. 119 46-34 = 12. 124-119 = 5. 5 rem. Since we must now deal with large divisors, it is not always easy to tell just what the quotient figure should be without trying. The following examples will show how to find the quotient figure by trial. Ig 23)387 Explanation : Suppose we do not know how many times 23 23 is contained in 157, and we try 5 times. 5x23 = 115, 1^ 157-115 = 42. But 42 is larger than 23; therefore, 157 115 will contain 23 more than 5 times. Remember : When the remainder is larger than the divisor the quotient figure is too small. — - Explanation : If we do not know how many times 157 23)387 contains 23, suppose we try 7 times. 7 x 23 = 161. But 161 is larger than 157; therefore, 157 will not contain 23 23 157 7 times. 161 Remember : When the number to be subtracted is larger than the number from which you are to subtract, the quotient figure is too large. *Bead, 2+, "ficoplus." It here means 2 and a remainder. ADVANCED ARITHMETIC. 31 . 2. Divide 101655 by 251. Process. ^Q5 Explanation : After bringing down the first 5 of the div- 25l) 101655 idend, the 125 thus formed is too small to contain the IQQ^ divisor, 251. Put a in the quotient, bring down the 1255 next figure, and proceed as before. Quotient, 405. 1255 S. Divide 3400 by 200. Explanation : Out off all O's found on the right of the ^^^^^^' divisor, and an equal number of figures from the right -li — of the dividend. Divide the remaining part of the div- 2x00) 34x00 idend by the remaining part of the divisor. Quotient, 17. 4. Divide 3405 by 200. Process. 17^^ Explanation : The 5 cut off is remainder. Exact 2yS)0)3i^b quotient, mh. 5 rem. 5. Divide 3505 by 1700. Process. 2^7^,^ Explanation : If there be a remainder from the di- 17 00)35 05 viding, the part of the dividend cut off is annexed to 34 form the complete remainder. Exact quotient, 2^.?^js. 105 rem. 6. Divide 84050 by 1000. Process. Explanation : When the divisor is 10, 100, 1000, etc., 34rt^iy the part cut off of the dividend is remainder, and the 1x000)34x050 P*^* ^ot cut off is the integral quotient. Exact quo- 50 rem. tient, 34j^^j5. EXERCISE X. Find quotients : 1. 8547-^37 7. 45368--106 2. 15170-^74 8. 111366^207 3. 388564-92 9. 176868--306 4. 48418-^86 10. 258210--342 5. 373520--580 11. 150181---179 6. 597760--640 12. 177156-^259 32 ADVANCED ARITHMETIC. 13. 510816-^554 19. 429586-^347 U. 324264^-458 20. 355074-^249 15. 481866--539 21. 715464-^456 16. 363636-^468 22. 800197-f-673 17. 514201-f-579 23. 11304202^1729 18. 316386^378 2Jf.. 100812054-^5649 Find integral quotients and remainders : 25. 8581-^231 30. 600181-^937 26. 15317-^206 31. 45979^433 27. 39077-^370 32. 112559-^538 28. 49092-7-563 33. 176868-^578 29. 374813-r-644 3J^. 324264-^709 Find exact quotients : ^5. 348657-^1000 J^O. 98764379 -^ 137000 36. 4897531 -- 10000 J^l. 13951475^958 37. 7306477^45000 1^2. 32657727 -^ 75632 38. 8907601 ---10800 I^3. 7391555011^8604 39. 325093 -=-72000 U- 26083952680^36745 20. General Review. EXERCISE XI. 1. Define addition, addend, sum. 2. Give the principles governing addition. 3. Using the following addends, explain in full the process of addition : 24750, 84723, 203079. 4. Mr. Jones has real estate worth $7560; personal property, such as household goods and stock, worth $1739; a stock of merchandise worth $8420; and a bank deposit of $3124. How much is Mr. Jones worth altogether ? ADVANCED ARITHMETIC. ^6 5. The sum of 7084, 537, 9468, and one other number is 28221. Find the other number. 6. Explain fully the process of subtraction, using the follow- ing example: 3408- 1786= ( )? 7. What is subtraction ? Subtrahend ? Minuend ? Differ- ence ? 8. Give the principles governing subtraction. 9. A man worth $8750 lost by fire a house worth $1280. What was he then worth ? 10. The remainder is 756, and the subtrahend, 3146. Find the minuend. 11. The minuend and remainder are 14255 and 6728 respect- ively. Find the subtrahend. 12. Explain fully the process of multiplication, using the following example: 72x8407=( )? IS. Define multiplication, multiplicand, multiplier, product. 14.. Give the principles governing multiplication. 15. 346 X( ) = 195144? 16. ( )X 730 =88330? 17. A has 246 cattle ; but B has 26 times as many as A. How many cattle has B ? 18. Two numbers multiplied together give a product of 463686 ; one of the numbers is 654. Find the other number. 19. Define division, divisor, dividend, quotient, remainder, short division, long division. 20. Give the principles governing division. 21. Explain fully the process of short division, using the following example: 753646^9= ( )? 22. Explain fully the process of long division, using the fol- lowing example: igr,i4^^^^(^ )y 34 ADVANCED ARITHMETIC. 23. By what rule do you know when the quotient figure is too large ? 24. By what rule do you know when the quotient figure is too small ? 25. How many $1000-shares in a capital stock of $2000000 ? 26. 194788-^ ( ) = 281? 27. ( )-f- 1248= 821? 28. 289761 ^ ( ) = 875, remainder 186 ? 29. 74475-^824=229, remainder ( )? 30. How do you shorten the process of multiplication when there are O's on the right of the multiplier or multiplicand ? 31. How do you shorten the process of division when there are O's on the right of the divisor ? 32. A and B start from the same place: A goes east 560 miles; B goes west 489 miles. Find the distance between them. 33. A and B start from the same place and travel in the same direction — A, 1846 rods; B, 859 rods. Find the distance between them. 34. What is the number, from which, if 784 be taken, the remainder is 591 ? 35. What is the number, to which, if 7428 be added, the sum will be 24178? 36. Find the number, from which, if 480 be subtracted, the remainder divided by 79, the quotient will be 24. 37. What number is that, from which, if 765 be subtracted, the remainder multiplied by 184, and 1795 be added to the product, the sum will be 288347 ? 38. If 875 be subtracted from the difference between two numbers, 126 will remain. 895 is the smaller number; find the larger. 39. Four men have $7500: the 1st has $1500; the 2d, $1284; the 3d, $2179. How much has the 4th ? ADVANCED ARITHMETIC. 35 Jfi. Find the sum of four numbers, if the 1st is 177, the 2d is 316 more than the 1st, the 3d is 741 more than the 2d, and the 4th is 658 less than the 3d. G. SIMPLIFYING NUMERICAL EXPRESSIONS. 21. A Term. — When a numerical expression is separated into parts by either of the signs, plus ( + ) or minus ( — ), these parts are called Terms ; when not so separated, the expres- sion itself is a term. Thus, 5-2+7=10 5, 2, 7, and 10 are terms. Sometimes a term is composed of two or more numbers joined together by signs, X or -^. Thus, 3x4-8-^2 3x4, and 8 + 2, are terms. Two or more terms may be inclosed by parentheses, ( ), and considered as a single term. Thus, (4+5) -(10-4+2) (4+5) and (10-4+2) are terms. When an expression in parentheses is to be multiplied by a number, that number is usually placed just before the paren- theses. Thus, tr/A , rT\ Atr ' 5(4+5) =45 5(4+5) means 5 times the sum of 4 and 5y or 5x9. A dividend or a divisor may be composed of two or more num- bers joined by signs. Thus, ^g±|)=5(4+5)-^(12-8). These two expressions are equal as indicated. The dividend is 5 times the sum of 4 and 5, and the divisor is the difference between 12 and 3. Each expression is a term. do ADVANCED ARITHMETIC. 22. Process. — In simplifying a numerical expression, (1) always simplify each term by itself and (2) add or sub- tract the terms as indicated by the sign', plus or minus. EXAlCFIiES. 1. Simplify 7-(9-5)+4x5-20-^4=( )? (1) 7-(9-5)+4x5-20-^4. (2) 7-4+20-5 = 18, result. Explanation : 1st term, 7; 2d term, 9-5 or 4; 3d term, 4x5 or 20; 4th term, 20-»-4 or 5. 2. Simplify 5(12-10)+4x6-(2+9-8). (1) 5(12-10)+4x6-(2+9-3). (2) 10+24-8 = 26, result. 3. 8(9--4) _27h-9+3(5+3-7) = ( )? (1) 8(9-4) _27h-9+3(5+3-7). (2) 4-3+3 = 4, result. EXERCISE XII. Simplify : 1. 4-8+10-5+12 «. 12-^4+7-6+14 3. 24-6x3+4x5x3 4. 74-10x7+8x3 5. 8-(4-2) + (9-3)-(8-5) 6. 9+(7+l)+2(10-4)+5x6 7. 8+7 , 21-32(7+3) 10-5"^ 4+2 9-5 8. 5x8-^i^±^+(25+5)-4-(ll-5) ^- 5^+g^^-7+8x5-7(12-8+4-2) ^^- |J|^+64-^(8+5)-(5+ll-4) ADVANCED AKITHMETIC. H. DEBIVED OFEBATIONS. 1. FACTORINO. 87 23. Definitions and Principles. — A Prime Number is an integer that cannot be formed by multiplying two or more other integers together. Below is a table of prime numbers, to be used for reference: TABLE OF PRIME NUMBERS FROM 1 TO 1000. 1 59 189 288 887 489 557 658 769 888 2 61 149 289 847 448 568 659 778 887 8 67 151 241 849 449 569 661 787 907 5 71 157 251 858 457 571 678 797 911 7 78 168 257 859 461 577 677 809 919 11 79 167 268 867 468 587 688 811 929 13 88 178 269 878 467 598 691 821 987 17 89 179 271 879 479 599 701 828 941 19 97 181 277 888 487 601 709 827 947 28 101 191 281 889 491 607 719 829 958 29 108 198 288 897 499 618 727 889 967 81 107 197 298 401 508 617 788 858 971 87 109 199 807 409 509 619 789 857 977 41 118 211 811 419 521 681 748 859 988 48 127 228 818 421 528 641 751 868 991 47 181 227 817 481 541 648 757 877 997 58 187 229 881 488 547 647 761 881 A Composite Number is an integer that can be formed by multiplying two or more other integers together. As, 4, 6, 8, 9, 10, 12, 15, 16. An Even Number ends in 0, 2, 4, 6, or 8; all others are Odd Numbers. Factors of a number are those numbers which, multiplied together, form that number. As, 8x5x7 = 105. 3, 5, and 7 are factors of 105. 38 ADVANCED ARITHMETIC. A Prime Factor is a prime number that is a factor. A Composite Factor is a composite number that is a factor. Thus, 8x4=12 3 is a prime factor of 12 ; 4 is a composite factor of 12. A Common Factor of two or more numbers is a factor of each of them. Thus, 7X2 = 14; 7X3=21. 7 is a common factor of 14 and 21. Two numbers that have no common integral factor (except one) are prime to each other. Thus, 3x7=21 5X11=55 21 and 55 are prime to each other. A number can have but one set of prime factors, but a num- ber that has three or more prime factors ( besides one) may have more than one set of factors, some of which are compos- ite. Thus, 30=3X2X5; but 80=3X10=2X15=6X5. 30 has but one set of prime factors, 2, 3, 5; but it has three sets of factors, some of which are composite, 3, 10 ; 2, 15 ; 6, 5. A Multii^le of a number is a number that contains that number an integral number of times. Thus, 24-T-12=2. 24 is a multiple of 12. A Common Multij^le of two or more numbers is a number that is a multiple of each of them. Thus, 24^6=4. 24-^8=3. 24 is a common multiple of 6 and 8. ADVANCED ARITHMETIC. 39 A Divisor of a number is a number that will divide that number and give an exact integral quotient. Thus, 4 is a divisor of 24. A Common Divisor of two or more numbers is a num- ber that is a divisor of each of them. Thus, 24-f-6=4. 42-^6^:7. 6 is a common divisor of 24 and 42. General Principles: 1. The multiplier and multiplicand are factors of the product. 2. The divisor and the exact quotient are factors of the dividend. 3. Any composite number is equal to the product of all its prime factors. Jf.. All of any set of factors of a concrete number must be abstract^ except one factor ^ which must be concrete and similar to the number itself. Note. — It is immaterial which factor is considered concrete. For ex- ample, factor the number 30 bushels. 2x3x5 bushels = 30 bushels. 2 X 5 X 3 bushels = 30 bushels. 5x3x2 bushels = 30 bushels. 2x5x3x1 bushel = 30 bushels. 5. A factor of a number is a factor of any multiple of that number. 6. A factor of any two numbers is also a factor of their sum and their difference. 7. When a number is divided by one of its prime factors^ the quo- tient is the product of all the remaining prime factors of the number. Principles to be Used in Inspecting a Number for Factors : 1. One factor of a number is 2^ if the number ends in 0, 2^ -^, 6, or 8. 40 ADVANCED ARITHMETIC. 2. One factor of a number is 3, if 3 is a factor of the sum obtained by adding the digits of the number. 3. One factor of a number is 4, if 4- is a factor of the number ex- pressed by the two right-hand figures. 4-. One factor of a number is 5, if the number ends in or 5. 5. One factor of a number is 6, if 2 and 3 are factors. ( See prin- ciples 1 and 2.) 6. One factor of a number is 7, if 7 is a factor of the sum of once the first figure^ plus 3 times the second, plus 2 times the third, plus 6 times the fourth, plus 4 times the fifth, plus 5 times the sixth, plus once the seventh, plus 3 times the eighth, and so on, following the order 1, 3, 2, 6, 4, 5 times the successive figures. Thus, 7 is a factor of 84755. For, by the principle, 1x5 + 3x5+2x7+6x4+4x3 = 70, and 7 is a factor of 70. 7. One factor of a number is 8, if 8 is a factor of the number ex- pressed by the three right-hand figures. 8. One factor of a number is 9, if 9 is a factor of the sum ob- tained by adding its digits. 9. One factor of a number is 10, if the number ends in 0. Note. — The 1st, 2d and 4th of these principles are most practical. The 3d, 8th and 9th are often used. The 5th and 7th are sometimes used. But the 6th is not used in practice ; for it takes longer to apply the prin- ciple than to divide by 7. 24. Factoring by Inspection. — When numbers are small, they are usually factored by inspection. EXAMFIiES. 1. Find the factors of 9. Think: 9 = 3x3. Call: " Factors 3, 3." 2. Factor l60. ADVANCED ARITHMETIC. 41 EXERCISE XIII. Factor by inspection : 1. 25 ^. 72 7. 112 10. 210 2. 40 5. 96 8. 128 ii. 175 3. 75 (5.\91 5>. 150 12. 240 25. Factoring by Division, — When the number to be factored is large, division is employed in finding the factors. (See General Principles 2 and 7.) The successive divisors are (1) found by inspection (see Principles to be used in Inspect- ing) or (2) by trial. EXAMPI.ES. 1. Find the prime factors of 210. Explanation : (1) By Prin. 4, 5 is a factor of 5 )210 210. 3 ) 42 (2) By Prin. 2, 3 is a factor of 42. 2) 14 (3) By Prin, 1, 2 is a factor of 14. ^ .-. 5, 3, 2, and 7 are the factors of 210. 2. Find the prime factors of 5040. Find 5 composite factors. Process. (1) 2)5040 2)2520 2)1260 2) 630 5) 315 3) 63 3) 21 7 1, 2, 2, 2, 2, 5, 3, 3, 7. (2) 2x2 = 4; 2x2x2 = 8; 2x3 = 6; 2x2x3 = 12; 2x7 = 14. Note. — Two or more prime factors of a number multipHed together always give a composite factor of that number. 42 P ADVANCED ARITHMETIC. EXERCISE XIV. Find the prime factors of — i. 2304 J!^. 17160 7. 73920 10. 282960 2. 2835 5. 20304 8. 39375 11. 146146 .^. 3000 6. 32340 P. 36288 . 12. 1067220 i^i^ic? fourteen factors for each of the following : 13. 210 15. 1155 17. 770 i^. 330 . 16. 462 i^. 510 26. Factoring Expressions Composed of Two or More Terms. — It is sometimes necessary or conven- ient to factor an expression consisting of two or more terms. This may be done by dividing by a factor common to all the terms. EXAMPIiES. 1. Take the factor 5 out of 20+25. Process: 20+25 = 5(4+5). Explanation : 5 is contained in 20 4 times, and in 25 5 times. Thus, 5 is one factor and 4+5 is the other. 5(4+5) may not only be interpreted as 5 times the sum of 4 and 5, but also 5 times 4 added to 5 times 5. 2. Take the factor 5 out of 25-20. Process: 25-20 = 5(5-4). Explanation : 5(5-4) may be interpreted 5 times the differ- ence between 5 and 4, or 5 times 5 minus 5 times 4' 3. Take the common factor out of 49 — 14+28. Process: 49-14+28 = 7(7-2+4). ADVANCED ARITHMETIC. / 43 EXERCISE XV. Take out the common factors : 1. 24-4 6. 5+20-15 2. 72+18 7. 18-9+36 3. 36-16 •. 8. 48-16+64 4. 57+38 9. 25+30-45 5. 75-50 10. 14+28-21+7 27. Principles of Multii)lication and Divi- sion Relating to Factors. — Factors are of common occurrence in the processes of multiplication and division, and the following principles will be of service: Principles: 1. Multiplying any factor of a product by a num- ber, multiplies the product by that number. Example: Indicate the multiplication of 3x5x4 by 2. (3 forms.) Forms: (1) (2x3)x5x4 = 6x5x4; or, (2) 3x(2x5)x4 = 3xl0x4; or, (3) 3x5x(2x6) = 3x5xl2. 2. Dividing any factor of a product by a number, divides the product by that number. Example : Indicate the division of 4x6x8 by 2. (3 forms.) Forms: (1) (4 + 2)x6x8 = 2x6x8; or, (2) 4x(6 + 2)x8 = 4x3x8; or, (3) 4x6x(8 + 2) = 4x6x4. 3. A quotient is multiplied by a number (1) by multiplying its dividend by that number, or (2) by dividing its divisor by that number. Example : Indicate the multiplication of 24^6 by 3. (2 forms.) Forms: (1) (3x24)-h6 = 72 + 6 ; or, (2) 24h-(6h-3) = 24h-2. Note.— The dividend or divisor may be expressed as a product of two or more factors. 44 ADVANCED ARITHMETIC. Example : Indicate the multiplication of ^^ by 3. (4 forms.) Forms : .-,. (3X12) X5 _ 36X5 , ^^' 9X6 9X6 ' °^' ,^^ 12X(3X5) _ 12X15 . ^^ (2) 9X6 9X6 » ^^= /o^ 12X5 _ 12X5 . ^^' (9-^3)X6 3X6 ' " ' , .V 12X5 _ 12X5 ^ ' 9X(6^3) 9X2* 4-' A quotient is divided by a number (!) by dividing its dividend by that number, or (2) by multiplying its divisor by that number. Example 1 : Indicate the division of ^^- by 8. (2 forms.) Forms: (l)^ = |;or, ^^^ 3X2 6* 12X18 Example 2 : Indicate the division of -^^ by 3. (4 forms.) 4X1 T^ ,-,v (12^3)X18 4X18 Forms: (D '-ixiT- = ixn » °^' .„. 12X08^ _ 12X6. ^^ ^"^^ ~4xii ixn' °^' ,„. 12X18 ^12X18. ^^^^ (3X4)X11 12X11 ' ' /4\ 12X18 ^ 12X18 ^*' 4X(3X11) 4X33 • EXERCISE XVI. Indicate as above the multiplication of — 1. 4X7 by 8. (2 forms.) ^. _^ by 7. (3 forms.) ^. 9 X 12 X 3 by 5. (3 forms.) S, ^ by 5. (3 forms.) 3. -V\0- by 25. (2 forms.) 6. ^^ by 3. (5 forms.) Indicate as above the division of — 7. 12X80 by 6. (2 forms.) jq. 5^0 ^7 25. (3 forms.) 3X1 3 Process: ^, = j- But -g = ^. Therefore, the value is not changed. 2. Divide both terms of | by 2. Process : g^ = y But i = ^. Therefore, the value is not changed. o o 35. Beclucing' Fractions to Hig-lier Terms.— A fraction is reduced to higher terms by multiplying both its terms by the same number, according to principle 8. This does not change its value. EXAMPLES. 1. Reduce | to 9ths. Process: y = ^ = 9, result. Why multiply by 3 ? Because the denominator 3 is to be changed to 9. 5x3 = 9. Note.— An integer may be treated as a fraction, having 1 for a denom- inator. Y =5 or 5= y. Y is read 5 ones. ADVANCED ARITHMETIC. 53 2. Reduce 7 to Sths. Process : ^ = ^-^ = — , result. 1 5X1 o 3. Reduce | to 63ds. Process -i (1) 63-5-7 = 9. Why divide 63 by 7 ? 4. Reduce yV ^^ lOSths. Process : 15 (1) 18)195 5^ _ 9X5 _ 45 J ^"^^ 7 - 9X7 - 63' result. \^/ -^'-'/ -^»'"-' 13 65 65 (2) ^ = ^3 = 1^.--"- . Reduce — EXERCISE XX, 1. TAjto48ths. 6. W to 324ths. 2. -/y to 63ds. 7. if to 333d8. S. 11 to 37ths. 5. 9 A to 2109th8. -^. 3t\ to 209thB. P. iJ to TOSths. 5. if to 390th8. 10. 18yVr to 8181sts 36. Reducing Fractions to Lowest Terms. — A fraction is in its lowest terms, when its numerator and denom- inator are prime to each other. Divide both numerator and denominator by their G. C. D., or reject all factors common to both numerator and denominator. This does not change its value. EXAMPIiES. 1. Reduce f to thirds. Process : j = —^ = -^ , result. Why divide by 2 ? How many 3's in 6 ? 2. Reduce |f to lowest terms. Divide by G. C. D. of both terms. The G. 0. D. of 12 and 18 is 6. Process : | = |i| = §, result; or ^3 = ^= |, result. 64 ADVANCED ARITHMETIC. 3. Reduce y*^^ to lowest terms. G. CD. of 49 and 161 = 7. n_ 49 49-4-7 7 ,, 49 7) 49 7 ,^ Process : jei = i6i:r7 = 23' ^^^^l* 5 o'' I6I = mei = 23' ^^^ult. EXERCISE XXI. Reduce to loivest terms — ^. A\ ^' m 7. m 10. iff ^. Ml 5. *« 5. nu 11- \m 37. Reducing Mixed Numbers to Improper Tractions. — Reduce the integral part to the denomination of the fractional part and add. The value is not changed EXAMPLES. 1. Reduce 8^ to halves. Process : (D 3 = |g = |. (2) | + | = |, result. (1) Reduce the integer to halves, (2) add the |. Shorter Form : 3J = ^ + j = |, result. 2. 7| to 4ths. Process: 7f = -j- + j = j, result. 3. Reduce 17-^^ to an improper fraction. Mechanical Work : 17 51 Explanation : 17 reduced to twenty-thirds 34 will make 23x17, or 391 twenty-thirds. Add- 391 ing the 9 twenty-thirds, we have 400 twenty- 9 thirds, or \Q^. 400 Result, Vs-. EXERCISE XXII. Reduce to improper fractions : 1. 17-i\ 4. 18^ 7. OSiJ. 10. 871A 2. 24t 5. 75t\ 8. 804^ 11. 488yViV 3. 18^^ 6. 8941 ^- 132U i^. 77^AV ADVANCED ARITHMETIC. 55 38. Reducing' Improper Fractions to Inte- gers or Mixed Numbers. — This is done by division. See Article 17. The value is not changed. EXAMPLES. 1. Reduce ^-^ to an integer. Process : 4)20 Note. — As an expression of division, **^0" 5 result. is the dividend and "^ " the divisor. 2. Reduce ^W ^^ ^ mixed number. Process : 1911, result. 18)355 18 175 162 13 Note. — If the fractional part of the result is not in its low- est terms, reduce by Article 36. EXERCISE XXIII. Reduce to an integer or mixed number : 2. W 5. -W^- 8. h^i^- 11, ^U S, ^-i^ 6. -W/ 9. -\W 12. 1^^^ 39. Reducing Fractions to Least Common Denominator. — The L. C. D. of two or more fractions is the L. C. M. of their denominators. The process of reduction is that of reducing fractions to higher terms. See Article 35. The values are not changed. EXAMPLES. 1. Reduce i, |, |- to L.C.D. The L. 0. D. is 12. Then the object is to reduce the several fractions to 12ths. '' UNIVERSITY OF . 56 ADVANCED ARITHMETIC. Process a)| 6X1 6 6X2 12 (2)| _ 3X3 _ 9 3X4 12 <3)| _ 2X5 _ 10 2X6 12 Reduce ^, -fy, ^ to L. C. D. Process: A, A, U- (S) ^^IJl; 7x6=^42, 6 ^; 5x4 = 20. (1) The L. C. D. is 72. (4) 18)72 72 ~; 11x3 = 33. (2) fi, fi, ?|, result, ^^^ ^^ Note. — The part on the right, which is the process of find- ing the numerators, may be put in such form as is conven- ient. As soon as the L. C. D. is found, the denominators may be placed in (2) thus : f^, 7^, y^. Then find the numer- ators and place them above. EXEKCISE XXIV. Reduce to L. C, D, : 1' h A, ii- 6. If, 5, A, -if. 2. ^, ^, ^f . Note.— Reduce If to an improper ^, 2.2 X7 11^ fraction before commencing to find / 6 37 13 19 L. CD. Consider 5 as f. 4' TTj ftj -2^? T¥- ^ io 1; HK ^ 1 « ^' tt J tT» -8 STJ TT2- ^- 'sVj "2^5 Aj 2V- 40. Addition of Fractions. — Reduce all fractions to L. C. D., add the numerators, and place the sum over the L. C. D. ADVANCED ARITHMETIC. 57 EXAMPIiES. 1, Add t, I, i. Process; l+i + \=l%+U + io = U = H^, result. Explanation : Mechanical work necessary to reduce the fractions to L. G. D. need not be preserved. 20 thirtieths + 24 thirtieths + 5 thirtieths = 49 thirtieths, or ff. Note. — If the result is an improper fraction, reduce it to an integer or mixed number. 2. Add f , I, H. Process: f + f +f| = li+IHtl = ft = 2A = 2f, result. Note. — The i-^ should be reduced to its lowest terms, f . In results, all fractions should be in their simplest forms. 3. Add 64, lOf , 5t\. Process: (1) J+|+t^ = fH 1^ = 11 (2)6t 101 5t^ 221 Explanation: The result of adding the fractions is 1|. Write the | below the fractions, carry the 1 and add it to the integers ; result, 22f . EXERCISE XXV. Find the value of— 2. 181+27^ 7. 18A+19tV+22A 3. 46A+A 8, 8465+17A+T¥ir ^. ii+i^+¥k 10. 7if+i7|i+9MJ+24^^f 41. Subtraction of Fractions. — Reduce the frac- tions to L. C. D. before subtracting. Subtract the numerator of the subtrahend from the numerator of the minuend and place the remainder over the L. C. D. 58 ADVANCED ARITHMETIC. EXAMPIiES. 1. From f take 3%. Process : f - A = f e - rl = ^e » result. Explanation: 20 thirty-sixths - 15 thirty-sixths = 5 thirty- sixths, or ^g. In the result, always reduce the fraction to its simplest form. 2. From 17i take 14^. (2) 17^ Process : (1) i - i = A - A = rV- 14^ Si^g, result. 3. From 17| take 5|. (2) m Process: (1) f-| = ||-f| = H- 5| 11^1, result. Explanation : Since | is larger than f , 1 must be taken from 17 and added to the f , making If or ? ; f - f = i|. This leaves only 16 in the minuend, from which to subtract the 5. EXERCISE XXVI. Find the value of — 1. if-f 6. 44i-36A 2. 7i-ii 7. 86||-f| ^. 18A-« ^. 538H-26H 4. 75-fl P. 452Jo-247Hi 5. 87^-15^ ^^. 140f|-14-|f 42. Multiplication of Fractions.— When the mul- tiplier is a fraction, it may be followed by the word *'o/" or the sign, *' X." Thus, f of 10=fXlO. Reduce mixed numbers to improper fractions before multiply- ing. EXAMPIiES. 1. Find f of 85. Process Complete : (1) ^^ of 35 = ^ = 5. ( See Art. 16, 3d Ap.) 3 X (1) = (2) * f of 35 = 3 X 5 = 15, result. Teacher, show the pupil that (2) comes from (1) by multiplying by 3. * "3X(1)=(2) " Is read 3 times eqttation one equals equation two. ADVANCED ARITHMETIC. 59 5 Process Shortened: ^ = 15, result. 2. Find | of |. p , • Explanation: § is 2 + 3. Dividing 1 f 2 _ 8 7- 3 «o ,if this by 7 is the same as dividing 2 by r of .-^-7-^, result. 3^7^ ^^21. 2 + 21 = ^. 3. Find f of |. Process Complete: (1) \- of 1 = ^. 3x(l) = (2) ? of | = 3x^\ = ^ = f, result. 3x2 Process Shortened : ? of | = = f , result. 7x3 By the short process^ we express all the numbers as fractions, then multiply all numerators together for a new numerator, and all denominators together for a new denominator. Employ cancellation. 4. Find i of I of f . Process : ^ of | of f = l^i^^ = j, result. ;2x^x^ 3 3 Process: 71x^3^ = ??^=/^, result. 2x^^ 5 6. Multiply I of 8 by J of 6. Process ; f ot 8 x ^ of 6 = ^^^^^^^ = 12, result. 2 ^x3 Note. — In the results all improper fractions should be reduced to in- tegers or mixed numbers, and all fractional parts should be reduced to their lowest terms. 1. 7xtt 7. 2. AX68 8. 3. AofK 9. If^ f of \% of 150 10. 5. 17^X^1^X280 11, 6. Aof2|Xifofy 12. 60 ADVANCED ARITHMETIC. EXEKCISE XXVII. Find the value ^f — AX-rVXlfof 18| SxTAX-iVe^xOG 126|X4iXTVTof 683 16fx80|Xlif 127AX45|X450 ffx-A^x«fx|i 43, Division of Fractions. — The easiest way to divide in fractions is to invert the divisor and proceed as in multiplication of fractions. To explain, we know that — 1 apple contains | apple 2 times. 1 apple contains ^ apple 8 times. 1 apple contains \ ai)ple 4 times. 1 apple contains f apple 2 times. 1 apple contains ^ apple 6 times. 1 apple contains f apple 8 times, etc. Now, let us arrange these expressions in an abstract form : l-^i=f,or2. l-^i=f, or8. l-^i=f, or4. 1-^1=1, or 2. l--i=f, or6. l-f=:|, or8. Look at each divisor and the first form of the corresponding quotient, and you will see that, in each case, the quotient is the divisor inverted. This is an illustration of the general truth : In- verting a divisor shows how many times that divisor is contained in one. Commit this to memory. With this truth in mind, let us study the following examples : ADVANCED ARITHMETIC. 61 EXAMPIiBS. 1. Divide 14 by J. 7 8 Process Complete : (1) 1-h — = — . (Above truth.) 8 7 7 2 8 14x(f) = (2) 14+ -=;^x- = 16, result. 8 fl 7 2 8 Process Shortened : 14-f - = ^^ x - = 16, result. 2. |-^J=( )? 7 8 Process Complete : ( 1 ) 1 + — = — . 2 3 3 7 3^6 jof(l) = (2) --.- = - of ^ = -, result. 2 3 7 3^6 Process Shortened : ~-*-~ = tX^ = ~> result. ^. lf-^8 = ( )? Process Complete : (1) l-«-8 = 4 8^77 1 3 12 12 ;;2 1 3 r7of(l) = (2) -^8 = j^of^=-, result. 12 ;;2 1 3 Process Shortened : — -5-8 = — X — = — , result. 2 Note. — The first step in the complete process gives us the divisor in- verted ; this may be done mentally. The ^'Shortened Process" is simply the last step in the Complete Process, and may be expressed thus: Invert the divisor and proceed as in multiplication of fractions. Use the shorter process in practice. 4. Divide 7i by 8^. 3 ■JR q Q Process : 7i -»- 31 = — x — = -, or 2^, result. 2/0 4 2 62 ADVANCED ARITHMETIC. Note. — tt" ^^ a complex fraction, and means, like any other fraction, that its numerator, 12^, is to be divided by its denominator, IJ. Then the process is easy : 12t 61 2 122 Process: -— = 12^+11= — x- = -—, or 8^^, result. 1^ 5 d 15 When the terms of the dividend are divisible by the respec- tive terms of the divisor, division of fractions may be i3er- formed in that way. ^. t¥2-H=( )^ 88 11 884-11 8 Process : 7— -»- — = — = — , result. 132 12 132 + 12 11' EXERCISE XXVIII. Find the value of- 1. -H-^SO 4.. 18I-M50 5. 6. 7. 8. 18A-38i 9 ^^ "^^ n 8A 44. Greatest Common Divisor of Common Fractions. — The Greatest Common Divisor of two or more fractions is the largest fraction that is contained in each of them an integral number of times. One fraction is a divisor of another, when the numerator of the first is a divisor of the numerator of the second, and the denominator of the first is a multiple of the denominator of the second. Thus, /y is a divisor of |. This is plain when, according to the process of division of fractions, the divisor is inverted. ADVANCED ARITHMETIC. 63 The numerator 4 of the divisor is a divisor of the numerator 8 of the dividend, and will disappear; the denominator 27 of the divisor is a multiple of the denominator 9 of the dividend, and the 9 will disappear. When the 4 and 9 both disappear, the quotient will be an integer, and jV is a divisor of f . Principle : The Greatest Common Divisor of two or more frac- tions is that fraction whose numerator is the G. C. D. of the nu- merators of the several fractions and whose denominator is the L. C. M. of their denominators. EXAMPLES. 1. Find the G. C. D. of J|, H, ^\, and |. (1) The G.C.D. of 15, 25, 5, 5 = 5. (2) The L. C. M. of 32, 64, 16, 8 = 64. .*. -iiis, the required G. 0. D. 2. Find the G. C. D. of 14^ and 95f . (1) Expressed as fractions, 14rV and 95f = V/ and ^^. (2) The G.O.D. of 175 and 763 = 7. (3) The L. 0. M. of 8 and 12 = 24. .'. /j is the required G. 0. D. EXERCISE XXIX. Find the G. C. D. of— 1. I, A, and ||. ^. AV, If, if, and If. S- «, if, f , A%, and T% 4. 5A, 12f , and 4^. 5. 9|, 5|, 16|, and 8J. 6. f , 3f , T^, H, and 5^. r. 8^1^, 9f , -U, 8A, and 23-^. ^. nh Ih 10|, 25i, and 16f . 9. 10/^, 17|, 12^\, 26t, and 18^. iO. 3i, 2-,\, 12i, 9l,and8J. 64 ADVANCED ARITHMETIC. 45. Least Common Multiple of Common Fractions. — The Least Common Multiple of two or more fractions is the smallest number that will contain each of them an integral number of times. One fraction is a multiple of another, when the numerator of the first is a multiple of the numerator of the second, and the denominator of the first is a divisor of the denominator of the second. Thus, f is a multiple of ^. Indicate the process of division by inverting the divisor. Thus, The 8 is a multiple of 4, and the 4 will disappear; the 9 is a divisor of 27, therefore the 9 will disappear, and the quotient will be integral. Principle : The Least Common Multiple of two or more fractions is that fraction whose numerator is the L. CM. of the numerators of the several fractions^ and whose denominator is the G. C. D. of their denominators. EXAMPIiES. 1. Find the L. C. M. of f, A, and fi. (1) The L. C. M. of 3, 9, 21 = 63. (2) The G. 0. D. of 4, 16, 32 = 4. .*. ^ or 15| is the required L. 0. M. 2. Find the L. C. M. of 14f , 9^\, 16|, and 25. (1) The numbers expressed as fractions are^ ?^, Yr", ^^-, \K (2) The L. C. M. of 100, 100, 50, 25 = 100. (3) The G. C. D. of 7, 11, 3, 1 = 1. .'. ^^ or 100 is the required L. 0. M. EXERCISE XXX. Find the L. C. M. of— 1' A, A^, and ii. 6. ,V, ^%, 4^^, 12f , and 4^V. ^. I, A, ii, and \. 7. 18f , 15|, 19|, and ^. S- h A, A, and H. 8. 87i, Ih SJ, 10|, and 18f . 4. 83J, 121, 6|, and 7|-. 9. 22^^, 88|, 621, 5f , and 166J. ^. IH, Hh m. -h. and -^S- 10. 142|, 55|, 41|, and 888^. ADVANCED ARITHMETIC. 65 J. DECIMAL FRACTIONS. 46. Definitions. — A Decimal Fraction has for its expressed denominator 1 with O's annexed. Thus : 10> T0"0> TO^O^J To 0* I A decimal fraction may be written without using figures for its denominator. Thus : Tfo = .05. AW^=.2045. By this method of writing decimal fractions, (1) the numer- ator is written just as an integer; (2) a period, called a Deci- mal Point, is put on the left of as many places in the numerator as there are O's in the expressed denominator. If there are not so many places in the numerator as are needed, O's are prefixed to the numerator to make the required number of places. A decimal fraction written in this way is called a Decimal. The denominator of a decimal — of one place is 10; .7 is read y\; of two places is 100 ; .46 is read -^^ ; of three places is 1000 ; .807 is read ^^^. of four places is 10000 ; .0453 is read tUoq y ^^^ ^^ ^"• The decimal places are named from the decimal point to the right, as follows : The first place, tenths ; the second place, hundredths ; the third place, thousandtlis ; the fourth place, ten-thousandths ; the fifth place, hundred-thousandths ; and so on. 66 ADVANCED ARITHMETIC. The decimal orders are numbered from tenths to the right as the integral orders are numbered from units to the left. Thus, Integral Places. Decimal Places. ' . ~~"^ ^~ — ^ -^ ~"~~^ ^"™^ CD CD a. 03 tr (/ tr to cr {/ (» 'O 'd t: t: c -»^ ,C rC X X ^ rj fl c O) 0. ■^ •4^ -*-: -tj ■»3 od § K »- += 5 c 'C t: t: •d DO So CO 73 a a> c p c p D :: c 4J t- d « 03 o O C 'O V tf O) M ,c! ^ ^ c 3 z S 43 +3 +i 3 C c O ■i 1 X 4^ ,c 5 S ^ p 'd 'd O) (D c 4- »H p T? A 13 * ^^ • • 6 ^ it ^ ) 1 5 ^ r ^ > . 8 8 9 1 7 . . rC ^ M a 4J -t-3 X X -U ■4J t3 •73 .^ a M 'O 'U ->^ -u CD »c ^ CC i-H CVJ CO Tt lO Integral Orders. Decimal Orders. The law of position (or order) in decimals is the same as that in integers. Law : Ten units of any one order make one unit of the next higher order. The denominator of a decimal is indicated by the name of its right-hand place. Thus, in .0246, the name of the right-hand place is ten-thousandths; the de- nominator is ten thousand. When a decimal is annexed to an integer the expression is called a Mixed Decimal. Thus: 5.04; and is read, 5^^^. ADVANCED ARITHMETIC. 67 A common fraction may be annexed to a decimal ; such an expression is called a Coiiij)lex Decimal, Thus : m 100* .12^; and is read, 6.87i; and is read, 5 37i 100* In an expression containing two or more decimal points all decimal points may be omitted except the one farthest to the left. Thus : 59.6.4=59.64. Explanation : 59.6.4= 59.6y\: 59f: 59 n. 59t%%= 59.64. There is no such expression in our system of notation as a common fraction with a decimal point on the left of it. Thus, •i- Explanation : J without the point is I of one integral unit. .0^ is J of of one-tenth, which is the first decimal order. Therefore there is no place in our system for such an expression as .J. EXERCISE XXXI. Read : 1, .6 6. .049 2. .25 7. 51.846 S, .08 8. .00.01 4. .73 9. 7.0884 6, 12.5 10. .759.63 Write as decimals : 16. •A 21. tW^ 17. 3i 22. ? 10 1000 18. 10 23. 8J 1000 19. AV 21,. IST^ftr 20. 37i *100 25. ^^A 11. 100.00001 12. .04567 13. 17.874 U. .06511 15. 148.00ff 26. 59_A_ 10000 27. lOOyoljW^ 28. 11 iiiiixlr 68 ADVANCED ARITHMETIC. 47. Reducing Decimals to Higlier Terms. — If a is annexed to a decimal, the numerator is multij^lied by 10; but since another decimal place is thus added, the denom- inator is also multiplied by 10, and the value of the fraction is not changed. Thus: .5 = . 50=. 500=. 5000, etc. That is, A-t^«o=i^^=tWA, etc. EXAMFIiE. Reduce .125 to lOOOOOths. Process : .125= .12500, result. Explanation : Annex O's to the decimal till the number of decimal places is equal to the number of O's in the required denominator. EXERCISE XXXII. Reduce : 1. .7 to lOOOths. 6. 5.1 to lOOths. 2. .506 to lOOOOths. 7. 7o05 to lOOOths. 3. .94 to lOOOths. 8. 56.084 to lOOOOths.. 4. .08 to lOOOOths. 9. 15* to lOOOths. 5. .004 to lOOOOths. 10. 14.506 to lOOOOOths 48. Reducing Decimals to Lower Terms. — If there are O's at the right of a decimal, they may be dropped without changing the value of the decimal. Thus : .2300=. 28; that is, tWA^tW KXAMPIiES. 1. Reduce . .750 to lOOths. Process : .750 =.75, result. 2. Reduce .24000 to lOOOths. Process ; .24000 =.240, result. ' simply place the decimal point after the 15 and add the O's. ADVANCED ARITHMETIC, 69 EXERCISE XXXIII. Reduce • 1. .200 to lOths. 6. 5.0000 to an integer. 2, .3400 to lOOths. 7. 7.5000 to lOths. 3. .10500 to lOOOths. 8- 15.0700 to lOOths. 4. .50000 to lOOths. 9. 70.0700 to lOOths. 5. .94000 to lOOths. 10. .005000 to lOOOths. 49. Addition of Decimals. — Write the numbers to be added so that the decimal points will form a column. Add as in integers, placing the decimal point in the result beneath the column of decimal points above. EXAMPIiS. 1. Add .501, 3.45, .125, 7.0034, 18.3024. Long Way. .5010 3.4500 .1250 7.0034 18.3024 Short Way, .501 3.45 .125 7.0034 18.3024 Explanation : In the long way, all decimals are reduced to the highest denomination, lOOOOths. The de- nomination of the sum is lOOOOths. Why ? 29.3818, result. In the short way, the decimals are 29.3818, result. not reduced to higher denomina- tions. The O's on the right do not affect the result. Use the short way in practice. EXERCISE XXXIV. Add: 1. .52, .936, 1.63, .0075. 2. .428, .506, .362, .521, .736, 3. .002, 5.3043, 7.5555, 81.8008, 4. 12.342, 18.9813, 85.753, .00059. 5. 8.756, .00045, .04621, .989898, .OIOIOL 6. $25.30, $846.27, $954.63, $1205.84, $746.98, $1375.30, $3.85, $840.75, $1684.35. Note. — When an expression of $'s has only two decimal places, call the decimal part cents ; if there are more than two decimal places, read the decimal part as a fraction of a dollar. 70 ADVANCED ARITHMETIC. 7. $44.56, $752.28, $984.81, $1200.50, $845,95, $7400.20, $947.25, $75.75. 8. $56,845, $824,955, $5476.875, $1260, $95,625, $4000.50, $1862.05. 9. 586.084, 7495.6084, 845.9567, 1111.1111, 9988.7766, 1200.0001, .99864. 10. .04654, .90083, .12456, .00665, .77777, .854821, .750864, .999999. 50. Subtraction of Decimals. — Write the numbers so that the decimal point in the subtrahend will be beneath the decimal point in the minuend. Subtract as in integers, plac- ing the decimal point in the result beneath the decimal points above. EXAMPIiES. 1. From 5.47 take 2.688. Process : Explanation : Reduce the minuend to the de- 5.470 nomination of the subtrahend by annexing to 2-638 it. The denomination of the result is lOOOths, 2.832, result. Why ? 2. From .54652 take .869. Long Way. Explanation : In the long way, the ^^^^^ ^^^^^ 54652 subtrahend is reduced to the de- 54652 .36900 nomination of the minuend. The '359 J77T2, result. ^^^T^Tr.""! the result is ^752, result. lOOOOOths. Why ? In the short way, the subtrahend is not reduced to higher denominations. The O's on the right do not affect the result. Use the short way in practice. EXERCISE XXXV. ^ubt\ 1. ract : .2464 from .65. 6. 8.427 from 8.245. 2. .88 from 5.001. 7. .0847 from .5087. 3. .428 from .6. 8. .0042 from .042. 4. .5748 from .8046. 9. $1.20 from $7.05. 5. 1.5644 from 5. 10. $246.75 from $4000. ADVANCED ARITHMETIC, 71 51. Multiplication of Decimals. — Multiply as in integers, neglecting O^s that may be on the left of the multiplicand and multiplier, and place the decimal point in the result so that there will be as many decimal places in the product as there are in both multiplicand and multiplier. BXAMPIilSS. 1. Multiply .852 by .35. Process, .8 5 2 .3 5 4260 2556 .29820, result. Explanation : The multiplicand is thou- sandths ; the multiplier is hundredths. Hundredths times thousandths gives hun- dred-thousandths. Show this by use of common fractions. Hundred-thousandths is expressed by a decimal of 5 places. Af- ter placing the decimal point, cancel the 0*s on the right; for the product should be expressed in its lower terms. Multiply .0037 by 2.5. Process. .00 3 7 2.5 185 74 .00925, result. Explanation : For the purpose of the multiplication, the multiplicand is 37 (neglecting the O's on the left). The product after multiplying contains but three figures ; but there must be five decimal places. Why ? Therefore two O's are prefixed and the decimal point is placed to the left. 3. Multiply 246 by 3.07. Process. 246 3.0 7 1722 738 7 5 5.2 2 Note. — Have pupil explain. 72 ADVANCED ARITHMETIC. EXERCISE XXXVI. Multiply — 1. .75 by .23. 7. 75.003 by 5.006. 2. .428 by .501. 8. 1000.001 by 100.0001. S. 4.763 by .084. 9. 1111.22 by 5.060701. 4. .00073 by .0059. 10. $420.50 by .37. 5. .5064 by .00001. 11, $4860.05 by 246. 6. 56.043 by .059. 12, $840.56 by 10.05. 52. Division of Decimals. — Divide as in integers, neglecting O^s that may be on the left of the dividend or divisor. Since the dividend is equal to the product of the divisor and quotient (Art. 14, Prin. 3), the dividend v/ill have as many decimal places as both divisor and quotient. From this, the following conclusions are evident : (1) The dividend must not have fewer decimal places than the divisor, (Commit.) Note. — If the dividend has not as many decimal places as the divisor, enough O's to make them equal should be annexed to the dividend before dividing. EXAMPIiE. Prepare to divide .15 by .0025. Explanation: Two O's must be an- Form. nexed to the dividend, because there — are four decimal places in the divisor .0025; .1500 and only two decimal places in the div- idend. (2) 7/, after dividing^ the number of decimal places used in the dividend equals the number in the divisor y the quotient is an intC' ger, (Commit.) ADVANCED ARITHMETIC. 73 Divide .15 by .0025. Process. 60, result. ' .0025). 1500 150 £XAMPI/E. Note. — In the actual process of divi- sion, the two O's on the left of the divi- sor are not used ; the divisor is 25. ■ (8) IJy after dividing^ the number of decimal 'places used in the dividend exceeds the number of decimal places in the divisor, the number of decimal places in the quotient must equal that excess. (Commit.) EXAIfPIi^S. 1. Divide .98745 by .29. 3.4 5, result. .2 9).9 8 7 45 87 117 116 145 145 Explanation : There are Jive deci- mal places in the dividend and two in the divisor. The excess is three. Therefore, the quotient must con- tain three decimal places. K Divide .002868 by .239 Process. .012, result .239). 002868 478 478 Exjilanation : After dividing the 2868 (neglecting the O's on the left) by 239, the quotient is 12; but tjje excess of decimal places in the dividend over those in the divisor is three. There- fore, the quotient must contain three decimal places. If there is a remainder after using the last figure of the divi- dend, O^s may be annexed to the dividend and the division continued at pleasure. 74 ADVANCED ARITHMETIC. EXAMPIjE. Divide 2.09 by .017. Process. 12 2.941 17 + , result. .0 17)2.09 000 00 17 39 34 50 34 160 153 70 68 20 17 30 17 130 119 11 Explanation : The plus sign after the quotient shows that there is a re- mainder, and that the di- vision could be continued farther, if desired. After dividing as far as desired, locate the decimal point according to Rule 3. EXERCISE XXXVII. ivii ie — 1. .125 by .5. 7. .025 by 5. 2. .5 by 1.28. 8. .025 by .5. 3. .12 by .0016. 9. 1000 by .001. 4. .12 by 16. 10. .5484888 by .0067. 5, 12 by .16. 11. $4826.422 by $96.1. 6. .001 by 1000. 12. 104.8576 by .001024. 53. Reducing* Common Fractions to Deci- mals. — Since a common fraction is an expression of division, if the division be performed, the result will be an integer or a decimal. In dividing, observe the rules for division of deci- mals. ADVANCED ARITHMETIC. 75 BXAMFIjSS. 1. Reduce .7^ to a decimal. Process. .18 75 16)3.0000 ^ ^ Exjilanation : By reducing the i\ to 14 a decimal and annexing it to the .7, the ^ ^ ^ result is .7.1875. But all decimal points 12 may be omitted except the one on the left. 112 80 80^ ,7.1875 = .71875, result. S. Reduce /^ to a decimal. (Continued 4 places.) Explanation : It will be observed that rrocess. ^j^j^ division w^ill never be complete ; for .2 12 1 + , result. 7 ^as the figure in the dividend at the 3 3)7.0 beginning. It was the remainder after ^ ^ the 2d division, and also after the 4th 4 division. In such a process, the divi- 33 sion will never be complete, if the de- 7 nominator has any other factor than 2 8 6 or 5. These are factors of 10, and by 4 adding O's to the numerator we multi- 33 ply by 10. The numerator can thus be 7 made to contain any number of 2's or 5's, but no other factor. In such a decimal as is obtained in example 2 above, one or more figures continue to recur in a certain order. Such deci- mals are called Recurring' or Circiilating" Decimals, or simply Circulates. The part that recurs is called the Repetend, and is marked by placing a dot over the first and last figures of the repetend. Thus, 5, .2i84. 76 ADVANCED ARITHMETIC. The first, if written out, would be — .5555555 to infinity. Note. — "To infinity ^^ means that the expression would never termi- nate, but would continue forever as indicated. The second, if written out, would be — .2134134134134 to infinity. 3. Reduce ^ to a circulating decimal. Process. A 3 2, result. 37)16.000 Explanation: After three divisions ^ ^ ^ the remainder is 16, which is the same 12 as at the start. Therefore the repetend 111 consists of 3 figures. 90 74 16 educe -^ to a circulate. Process. .3i8, result. E^P^ Explanation : The same remainder 9 9)7 recurs after the first and third divi- QQ sions; therefore the second and third . ^ divisions will be repeated as long as 2 2 the division is continued, and the repe- TTT^ tend is 18, commencing with the second 1 7 g decimal place. 4 EXERCISE XXXVIII. Reduce to decimals, obtaining complete results : 1. i I,. 12|f 7. 73/J-^ 2. \\ 5. 48H 8. .5f 3. II 6. 3.7^V 9. .064* Reduce to decimals, obtaining results true to four places : 10. A 13. H 16. .07^^^ 11. -H U. 12H 17. 84.9^iA 12. -jV 15. 8.4f 18. 741.07Ji ADVANCED ARITHMETIC. 77 Reduce to circulating decimals : 19. i 22, t\ ^5- 12iH 20. -ii 23. iU ^^. rU 21. \ 2Jt.. |§ 27. 15|f 54. Reducing' Decimals to Commoii Frac- tions, — Express the denominator, which is 1 with as many O's annexed as there are decimal places. Then reduce the re- sult to its simplest form. EXAMPLES. 1. Reduce .15 to a common fraction. Process ; .15 = ii(^= 2^, result. 2. Reduce .0036 to a common fraction. Process : .0036 = iuVots = a ^gu » result. S. Reduce 5.125 to a mixed number. Process: .125 -=^js%% = h .'. 5i is the required result. Note. — Reduce the decimal part to a common fraction and annex it to the integer. 4. Reduce .12^ to a common fraction. Process: .12^ = ^=12J^100=«2^XTh = i. 5. Reduce .6 to a common fraction. Process : (1) .6 = .666666 to infinity. 10 X (1) = (2) 10 X .6 = 6.666666 to infinity. (2)-(l) = (3) 9x.6 = 6. iof(3) = (4) .6 = 1 = 1, result. Note. — Subtracting the repetend from 10 times the repetend leaves 9 times the repetend. Subtracting .666666 to infinity from 6.666666 to in- finity leaves 6. Therefore 9 times the repetend equals 6, and the repe- tend equals |. 78 ADVANCED ARITHMETIC. 6. Reduce .507 to a common fraction. Process : (1) .507 = .507507507 to infinity. 1000x(l) = (2) 1000 X. 507 = 507.507507 to infinity. (2)-(l) = (3) 999 X. 507 = 507. ^of (3) = (4) 507 = m=J||, result. Principle : Any repetend may he expressed as a common frac- tion hy writing the repetend for the numerator and as many 9^s for the denominator as there are figures in the repetend. 7. Reduce 17.54 to a mixed number. Process : .54 = f f = j\ . .', 17.54 = 17i\, result. 8. Reduce .25108 to a common fraction. Process : (1) .25i08 = .25iif = .25iVi (2) .25iVr = ^ = W = iViVo = Wd , result. Note. — In processes where circulating decimals are involved, reduce the circulate to a common fraction before beginning the process. At the conclusion, the result may be expressed as a circulate, if required. EXERCISE XXXIX. Reduce to com,mon fractions or mixed numbers : 1. .875 7. M 13. .63 2. .125 8, .161 u. .063 3. .425 9. .47S 15. .248 4. 5.625 10. .03711 16. .572 5, 8.875 11. 2.88A 17. .1252 6. .025 12, 15.84,%V 18. 7.0i575 ADVANCED ARITHMETIC. 79 II. STUDY OF PROBLEMS. A. THE MEANS OF EXPRESSING SOLUTIONS. 1. EQUATIONS. 55. Definitions. — A statement, in which the sign of equality ( = ) is used ■ between two numerical expressions to show that they are equal, is an Equation. The expression on the left of the sign is called the first or left member; that on the right, the second or right member. When a member of an equation is separated into parts by either of the signs, plus ( + ) or minus ( — ), or both, these parts are called Terms ; when not so separated, the whole member is called a term. See Terms, Article 21. EXAMPLES. 1. 9-4+5=10. 9, 4, 5, 10, are terms. 2, A's money— $50=B'8 money. A's money, $50, and B's money, are terms. S. 4 pecks = 1 bushel. Each member is a term. 4.. Area of a surface 1 foot long, 1 foot wide=l square foot. Each member is a term. 56. Classification as to Source. — Some equations express constant relations, and are always true; other equa- tions express relations depending upon conditions, and are therefore true for particular conditions only. The following are some of the equations which are always true: 1. 6+4-7=3. 2. 5X8=40. 3. 56^8=7. 4.. 4 pecks = 1 bushel. 5. Area of a surface 1 foot long, 1 foot wide=l square foot. 6, 8xB's mouey+3xB's money = 11 X B's money. 80 ADVANCED ARITHMETIC. The following equations are true for particular conditions only: -?. Cost of 5books=:$8.50. 2. -J of A's age=B's age +6 years. 3. Weight of 1 load =2475 pounds. ^. Money earned by 7 men in 2 days = $85. Note. — As equations of the first class are always true, they may be used in the solution of any problem where such use is of service ; but those of the second class only when by the conditions of the problem they are true. 57. Transformation. — An equation may be changed, or transformed from a given to a required form, by subjecting it to one or more of the following processes : 1. To turn an equation around. 2. To transpose the terms of an equation. 3. To simplify the members of an equation. 4. To multiply an equation. 5. To divide an equation. 58. To Turn an Equation Around, — To turn an equation around is to put the left member on the right and the right member on the left of the sign of equality. Principle : Equals are equal in whatever order considered. It follows from this principle that any equation may be turned around and still remain a true equation. EXAMPIiES. 1. 4 gills = l pint; then, 1 pint=4 gills. 2. Cost of 5 hats = $15; then, $15=cost of 5 hats. 3. $740— cost price =$560; then, $560= $740— cost price. Note. — This process is so easy that no exercise is given for practice. ADVANCED ARITHMETIC. 81 59. Transposition of Terms. — Transposing a term is changing it from one member of an equation to the other. Note. — Transposition is looked upon by many pupils as beyond the- sphere of arithmetic. The" following exercise is given with the aim of suggesting that transposition depends upon the most elementary knowl- edge of number and not upon some difRcult mathematical principle. EXERCISE XL. Answer each question, then tell what number has been transposed : i. 5+6=11; then, 5=:ll-( )? 2. 5+6=11; then, 6 = ll-( )? 3. 9-5=4; then, 9=4+( )? 4. 9=6+8; then, 9- ( )=6? 5. 7-4+9=12; then, 7+9=12+( )? 6. 7-4+9 = 12; then, 7-4=12-( )? 7. 7-4+9=12; then, 9-4= 12-( )? 8. 5+4=12-3; then, 5+4+ ( ) = 12? 9. 14-8=4+2; then, 14-8- ( )=2? Principles: 1. If equals be added to equals, the sums will be equal. 2. If equals be subtracted from equals, the remainders will be equal, BXAMFIiES. 1, $75-$20=$25+$80; transpose $20. Process: (1) $75-$?0 = $25+$3O. (2) $?0 = $2O. (l) + (2) = (3) $75 = $25+$30+$20. Explanation : If (2) be added to (1), the sum of the first members will be $75 - $20+ $20. The two 20's, one to be sub- tracted and the other to be added, balance each other and may be canceled. This leaves $75 for the first member of (3). The second members of (1) and (2) added give $25 + $30 +$20. 82 ADVANCED ARITHMETIC. 2, A 's money +$50= B's money ; transpose A's money. Process: (1) A^g money + $50 = B'8 money. (2) A'g money = A's money. (l)-(2) = (3) $50 = B's money -A's money. Note. — Let the pupil explain. 3. $40 -$80 =$60 -$50; transpose the $50. Process : (1) $40 - $30 = $60 - $^0. (2) $50 = $^0. (l)+(2) = (3) $40 -$30+ $50 = $60. Note. — Let the pupil explain. Each of these exercises and examples has illustrated one principle, which may be called the law of transposition. Law. — Any term of one member of an equation may he trans- posed to the other member^ if at the same time its sign is changed ^ the plus to minus or the minus to plus. 4.. 30—15+40=66 — 11; transpose 11 and 15. Result: 30+40+11 = 66+15. Note. — The process should be mental. 5. 51-12+5=25+19. Transpose the 12 and the 19. Eesult ; 51 + 5 - 19 = 25 + 12. 6. 100% of cost+$415=36i% of cost+$1050. Transpose so that only similar terms will be in the same member. Result : IW of cost - 36^^ of cost = $1050 - $415. Note.— The sign, %, is read per cent; and it means hundredth or hun- dredthSf according as it is used with one or more than one. EXERCISE XLI. 1. Cost+$200=3Xcost+$80. Transpose ^'s to the first member and cost to the second. 2. Cost of 15 hats=$70+cost of 1 hat. Leave fs only in second member. ADVANCED ARITHMETIC. 8o 3. 25—17+44—19=86 — 15+12. Transpose so as to leave no minus signs. 4. 75% of No. -200=40% of No. +200. Transpose so that similar terms will be together, per cent in the first member. 5. i of the tens+^ of the units =^ of the units. Put all the units in the second member. 6. A's money — $600= i of A's money— $200. Transpose so as to leave no minus signs. 7. 10 weeks' wages — $12=4 weeks' wages+$24. Transpose so that similar terms will be together, #'s in the second mem- ber. 8. 5Xmy age +24 years =3 X my age +104 years. Transpose so that similar terms will be together, years in the second member. 60 o To Simplify Members. — A member of an equa- tion is a numerical expression. By transposition, the terms of any member may be made similar. A member whose terms are similar may be simplified by the same processes that are employed to simplify other numerical expressions. See Arti- cle 21. EXAMPLE. 1. i of wt.+i of wt.+i of wt.=6 pounds+5 pounds+8 pounds. Process: (1) i of wt. + ^ of wt. +| of wt. =6 pounds+5 pounds + 8 pounds. (1) simplified = (2) H or lTVxwt. = 19 pounds. EXERCISE XLII. Simplify the members of the resulting equations in Exercise XLI. Note.— Before attempting to simplify, be sure that similar terms are in the same member. 61. To Multiply an Equation. — An equation is multiplied by multiplying each of its members by the same number. 84 ADVANCED ARITHMETIC. General Principles : 1. Equals multiplied by the same num- ber give equal iwoducts. 2. Multiplying each term of a member of an equation by the same number multiplies the member by that number. Note. — In addition to these general principles, a number of special principles are illustrated in the following examples. They will be given in connection with their respective illustrations, and should be commit- ted to memory. EXAKPIiES. i. 1 bushel =4 pecks. Multiply by 5. Process: (l)lbu. = 4pk. *5x(l) = (2) 5bu. = 20pk. 2. My age =35 years. Multiply by 12. Process : (1) My age = 35 yr. 12 X (1) = (2) 12 X my age = 420 yr. 3. Cost of 1 hat = $2. Multiply by 8. (2 forms.) Process : (1) Cost of 1 hat = $2. 8 X (1) = (2) 8 X cost of 1 hat = $16 ; or, 8 X (1) = (3) Cost of 8 hats = $16. Note. — Observe that either equation (2) or (3) is the required form, and that 8 x cost of 1 hat = cost of 8 hats. Principles: 1. The cost of any number of articles of the same kind is equal to that number of times the cost of 1 of those articles. 4. 4+9-3+6=16. Multiply by 7. Process: (1)4+9-3 + 6 = 16. 7x(l) = (2) 28+63-21+42 = 112. (Prin.2.) 5. .$50— A's money=i of B's money. Multiply by 8. Process: (1) $50- A's money = J of B's money. 8x(l) = (2) $400-8 xA's money = 4 x B's money. *"5X(l)=-'(2) " Is read 5 times equation 1 equals equation 2. ADVANCED ARITHMETIC. 85 ^.6X5=30. Multiply by 4. (2 forms.) Process: (1) 6x5 = 30. 4x(l) = (2) (4x6)x5 = 24x5 = 120; or, 4x(l) = (3) 6 X (4x5) = 6x20 = 120. (Prin. 1, Art. 27.) 7. Area of a surface 1 foot long, 1 foot wide = l square foot. Multiply by 6. (2 forms.) Process : (1) Area of a surface 1 ft. 1., 1 ft. w. = 1 sq. ft. 6x(l) = (2) Area of a surface 6 ft. 1., 1 ft. w. = 6 sq. ft. ; or, 6 X (1) = (3) Area of a surface 1 ft. 1. , 6 f t. w. = 6 sq. ft Note 1. — In general, length means the longest dimension of a surface or solid, but mathematically it simply means one of the two or three di- mensions, and it is not necessarily the longest. Note 2. — The equation in the above example expresses a certain area — '^Area of a surface {of certain dimensions) = 1 sq. ft." Any process which multiplies the area of this surface, multiplies this equation. This can be done by multiplying either the length or the width, but not both. A surface 6 times as long and of the same width will have 6 times as much area, or a surface of the same length but 6 times as wide will have 6 times as much area. 2. (1) Multiplying the length or width multiplies the area; therefore, (2) the length and loidth bear to the area the same nu- merical relation as that of factors to the product. Note. — '* Numerical relation" means relation as a number without considering its denomination or kind. Two numbers expressing feet (as, 4 feet and 5 feet) could not be factors, could not be multiplied to- gether, but their numerical values (4 and 5) could be multiplied together and could be factors. Question : Does Prin. 1, Art. 27, apply to example 7? 8. The volume of a solid 1 foot long, 1 foot wide, and 1 foot thick = l cubic foot. Multiply by 12. (3 forms.) Process : (1) Vol. of a solid 1 ft. 1. , 1 ft. w., 1 ft. th. = 1 cu. ft. 12x(l) = (2) Vol. of a solid 12 ft. 1., 1ft. w., 1ft. th. = 12 cu. ft. ; or, 12x(l) = (3) Vol. of a solid 1ft. 1., 12 ft. w., 1ft. th. = 12cu. ft.; or, 12 X (1) = (4) Vol. of a solid 1 ft. 1. , 1 ft. w. , 12 ft. th. = 12 cu. ft. 86 ADVANCED ARITHMETIC. 3. (1) Multiplying the lengthy widths or thickness multiplies the volume; therefore^ (2) the lengthy widths and thickness bear to the volume the same numerical relation as that of factors to the product. Question: Does Prin. 1, Art. 27, apply to example 8? 9. The interest on $1 for 1 yr. at 1%=$.01. Multiply by 50. (3 forms.) Process : (1) Int. on $1 for 1 yr. at 1^ = $.01. 50x(l) = (2) Int. on $50 for 1 yr. at 1^ = $.50; or, 50 X (1) = (3) Int. on $1 for 50 yr. at 1% = $.50 ; or, 50x(l) = (4) Int. on $1 for 1 yr. at 50^ = $.50. 4.. (1) Multiplying the principal, time, or rate, multijjlies the in- terest; therefore, (2) the principal, time., and rate hear to the inter- est the same numerical relation as that of factors to the product. Question: Does Prin. 1, Art. 27, apply to example 9? 10. The amount of wood sawed by 1 man in 1 day=l cord. Multiply by 10. (2 forms.) Process : (1) Amt. sawed by 1 man in 1 da. = 1 c. 10 X (1) = (2) Amt. sawed by 10 men in 1 da. = 10 c. ; or, 10 X (1) = (3) Amt. sawed by 1 man in 10 da. = 10 c. 5. (1) Multiplying the number of working units or the time, multiplies the amount of work done; therefore, (2) the working force and the time bear to the work done the same numerical relation as that of factors to the product. Question : Does Prin. 1, Art. 27, apply to example 10? 11. 8-^4 = 2. Multiply by 4. (2 forms.) Process: (1) 8-^4 = 2. • 4x(l) = (2) (4x8) + 4 = 32-h4 = 8; or, 4x(l) = (3) 8^(4^4)-8^1 = 8. (Prin. 3, Art. 27.) ADVANCED ARITHMETIC. 87 12. The length of a surface of 72 sq. ft., 8 ft. wide =9 ft. Multiply by 8. (2 forms.) Process : (1) Length of a surface of 72 sq. ft. , 8 ft. w. = 9 ft. 8x(l) = (2) Length of a surface of 576 sq. ft., 8 ft. w. = 72 ft. ; or, 8x(l) = (3) Length of a surface of 72 sq. ft., 1 ft. w. = 72 ft. Note.— To multiply the length, (1) multiply the area, or (2) divide the width. Since, by principle 2 above, the length and width bear to the area the numerical relation of factors to a product, the numer- ical relation of the area and one dimension to the other dimen- sion may be expressed as follows : 6. The area and one dimension hear to the other dimension the same numerical relation that the dividend and divisor hear to the quotient. Question : Does Prin. 3, Art. 27, apply to example 12? Explain. IS. The width of a solid of 600 cu. ft., 15 ft. long, 5 ft. thick =8 ft. Multiply by 5. (3 forms.) Process : (1) Width of a solid of 600 cu. ft., 15 ft. 1., 5 ft. th. =8 ft. 5x(l) = (2) Width of a solid of 3000 cu. ft., 15 ft. 1., 5 ft. th.=40ft. ; or, 5x(l) = (3) Width of a solid of 600 cu. ft., 3 ft. l.,5ft. th.=40ft. ; or, 5x(l) = (4) Width of a solid of 600 cu. ft., 15 ft. 1., 1 ft. th. =40 ft. Note.— To multiply one dimension, (1) multiply the volume, or (2) di- vide either of the other dimensions. 7. The volume and two dimensions hear to the third dimension the same numerical relation that the dividend and the factors of the divisor bear to the quotient. Question : Does Prin. 3, Art. 27, apply to example 13? 14.. The principal required to gain $36 in 6 years at 6% = $100. Multiply by 6. (3 forms.) 88 ADVANCED ARITHMETIC. Process : (1) Prin. reqd. to gain $36 in 6 yr. at 6^ = $100. 6x(l) = (2) Prin. reqd. to gain $216 in 6 yr. at 6^ = $600; or, 6x(l) = (3) Prin. reqd. to gain $36 in 1 yr. at 6^ = $600; or, 6x(l) = (4) Prin. reqd. to gain $36 in 6 yr. at 1^ = $600. Note.— To multiply the principal, (1) multiply the interest, or (2) di- vide the time, or (3) divide the rate. 8. The interest and any two of the other elements (principal, rate, or time) bear to the third element the same numerical relation that the dividend and the factors of the divisor bear to the quotient. Question: Does Prin. 3, Art. 27, apply to example 14? 15. ■ The time required by 10 men to saw 10 cords of wood = 1 day. Multiply by 10. (2 forms.) Process : (1) Time reqd. by 10 m. to saw 10 c. = 1 da. 10x(l) = (2) Time reqd. by 1 m. to saw 10 c. = 10 da. ; or, 10x(l) = (3) Time reqd. by 10 m. to saw 100 c. = 10 da. Note. — To multiply the time, (1) multiply the work to be done, or (2) divide the working force. 9. The work done and either element (working force or time) bear to the other element the same numerical relation that the dividend and divisor bear to the quotient. EXERCISE XLIII. 1. 17-44-6 = 19. Multiply by 10. 2. 1 bushel =4 pecks. Multiply by 80. 3. 3 quarts =6 pints. Multiply by 12. 4.. 89.87 inches=l meter. Multiply by 200. 5. 1 ton = 20 hundredweight. Multiply by 6. 6. $50-4x A's money = $30. Multiply by 18. 7. Cost of 1 pound=8/. Multiply by 14. (2 forms.) 8. Value of 1 book=75/. Multiply by 11. (2 forms.) ADVANCED ARITHMETIC. 89 9, Wt. of 12 loads = 150000 pounds. MultiplybyS. (2forms.) 10. Price of 15 share8 = $750. Multiply by 20. (2 forms.) 11. 8X9 = 72. Multiply by 7. (2 forms.) 12. Area of a surface 1 ft. long, 1 ft. wide=l sq. ft. Multi- ply by 7. (2 forms.) ' 13. Volume of a solid 1 ft. long, 1 ft. wide, 1 ft. thick = 1 cu. ft. Multiply by 18. (3 forms.) U. 72^8=9. Multiply by 4. (2 forms.) 15. Width of a surface of 30 sq. ft., 15 ft. long=2 ft. Mul- tiply by 15. (2 forms.) 16. Length of a surface of 80 sq. ft., 2 ft. wide = 15 ft. Mul- tiply by 5. (2 forms.) 17. Time required for $500 to gain $80 at 8% =2 yr. Multi- ply by 3. (3 forms.) 18. Length of a wall 12 ft. high, 2 ft. thick, built by 6 men in 8 days = 600 ft. Multiply by 2. (4 forms.) 19. Principal required to produce $80 at 8% in 2 years = $500. Multiply by 12. ( 3 forms.) 20. Rate required for $500 to produce $80 in 2 years=8%. Multiply by 7. (3 forms.) 21. Thickness of a solid of 30 cu. ft., 6 ft. long, 5 ft. wide= 1 ft. Multiply by 10. ( 3 forms.) 62. To Divide an Equation. — The Third Applica- tion of Division is employed in dividing equations. An equa- tion is divided by taking the same part of each of its members. General Principles: 1. Equals divided by the same number give equal quotients. .2. Dividing each term of a member of an equation by a number divides the member by that number. Note. — In addition to these general principles, all but one of the spe- cial principles given in Article 61 will be of service in the following ex- amples and exercise. 90 ADVANCED ARITHMETIC. -RXATVTPTjTSS. i. 21 bushels =84 pecks. Divide by 7. Note. — Remember, that to divide by 7 means to take \ of each member. Process : (1) 21 bu. = 84 pk. iof(l) = (2) 3bu. = 12pk. 2. Cost of 9 hats=$36. Divide by 9. (2 forms.) Process : (1) Cost of 9 hats = $36. i of (1) = (2) i of cost of 9 hats = $4 ; or, i of (1) = (3) Cost of 1 hat = $4. Note. — Observe that either equation (2) or (3) is the correct result, and that ^ of cost of 9 hats = cost of 1 hat. 5.25-15+80=40 Divide by 5. Process: (1)25-15+30 = 40. ^of (1) = {2) 5-3+6 = 8. (Prin. 2.) Jf.. 20Xmy age — 120 years =8 X John's age. Divide by 4. Process : (1) 20 x my age - 120 yr. = 8 x John's age. i of (1) = (2) 5 X my age - 30 yr. = 2 x John's age. 5. 12X18=216. Divide by 6. (2 forms.) Process: (1) 12x18 = 216. lof (1) = (2) -1^x18 = 2x18 = 36; or, ^of(l) = (3) 12xJ/ = 12x3 = 36. (Prin. 2, Art. 27.) 6. Area of a surface 15 feet long and 12 feet wide = 180 square feet. Divide by 3. (2 forms.) Process : (1) Area of a surface 15 ft. 1., 12 ft. w. = 180 sq. ft. \ of (1) = (2) Area of a surface 5 ft. 1., 12 ft. w. = 60 sq. ft. ; or, I of (1) = (3) Area of a surface 15 ft. 1., 4 ft. w. = 60 sq. ft. Note.— Show how Prin. 2, p. 85, and Prin. 2, Art. 27, apply to exam- ple 6. 7. The volume of a solid 18 ft. long, 12 ft. wide, and 6 ft. thick = 1296 cu. ft. Divide by 6. (3 forms.) ADVANCED ARITHMETIC. 91 Process.: (1) Vol. of a solid 18 ft. L, 12 ft. w., 6 ft. th. = 1296 cu. ft. I of (1) = (2) Vol. of a solid 3 ft. 1., 12 ft. w., 6 ft. th. =216 cu. ft. ; or, ^of (1) = (3) Vol. of a solid 18 ft. 1., 2 ft. w., 6 ft. th. = 216ca. ft.; or, i of (1) = (4) Vol. of a solid 18 ft. 1., 12 ft. w., 1 ft. th. =216 cu. ft. < Note.— Show how I*Hii, 3 (2), Art. 61, and Prin. 2, Art. 27, apply to ex- ample 7. <9. 72-^12 = 6. Divide by 2. (2 forms.) Process: (1) 72-12 = 6. iof (1) = (2) ^2^-^12 = 36 + 12 = 3; or, iof(l) = (3) 72-h(2x12) = 72-h24 = 3. (Prin. 4, Art. 27.) 9. Rate required for $500 to produce $750 in 15 years=10%. Divide by 5. (3 forms.) Process : (1) Rate reqd. for $500 to produce $750 in 15 yr. = W. I of (1) = (2) Rate reqd. for $2500 to produce $750 in 15 yr. =2^ ; or, ^ of (1) = (3) Rate reqd. for $500 to produce $150 in 15 yr. = 2% ; or, I of (1) = (4) Rate reqd. for $500 to produce $750 in 75 yr. = 2^. Note. — Show how Prin. 8, Art. 61, and Prin. 4, Art. 27, apply to ex- ample 9. 10. The time required by 10 men to saw 120 cords of wood = 10 days. Divide by 5. (2 forms.) Process : (1) Time reqd. for 10 m. to saw 120 c. = 10 da. J of (1) = (2) Time reqd. for 50 m. to saw 120 c. =2 da. ; or, J of (1) = (3) Time reqd. for 10 m. to saw 24 c. = 2 da. Note. — Show how Prin. 9, Art. 61, and Prin. 4, Art. 27, apply to ex- ample 10. 11. Thickness of a solid of 480 <3U. ft., 10 ft. long, 8 ft. wide =6 ft. Divide by 2. (3 forms.) Process : (1) Thickness of a solid of 480 cu. ft., 10 ft. 1., 8 ft. w. =6 ft. ^ of (1) = (2) Thickness of a solid of 480 cu. ft., 10 ft. 1. , 16 ft. w. i= 3 ft. ; or, i of (,1) = (3) Thickness of a solid of 240 cu. ft., 10ft. 1., 8 ft. w. = 3 ft. ; or, ^ of (1) = (4) Thickness of a solid of 480 cu. f t. , 20 f 1. 1. , 8 ft. w. = 3 ft. Question : Why increase the width in (2) ? the length in (4) ? Note.— Show how Prin. 7, Art. 61, and Prin. 4, Art. 27, apply to exam- ple 11. 92 ADVANCED ARITHMETIC. EXERCISE XLIV. 1. 18-12+24=80. Divide by 6. 2. 420 hundredweight =21 tons. Divide by 7. 3. $560-40% of cost=$280. Divide by 40. 4.18X30=540. Divide by 6. (2 forms.) 5.4X12X10=480. Divide by 2. (Sforms.) 6. 50% of 75% of a number = 250. Divide by 25. ( 2 forms.) 7. Weight of 20 barrels =5000 pounds. Divide by 10. (2 forms.) 8. Cost of 30 oranges =40/. Divide by 80. (2 forms.) 9. The volume of a solid 12 ft. long, 4 ft. high, 8 ft. thick = 884 cu. ft. Divide by 4. (3 forms.) 10. Interest on $6000 at 8% for IJ yr. = $640. Divide by 8. (3 forms.) ii. 24-^8=3. Divide by 3. (2forms.) 12. Time required by 14 women to make 7 dresses =3^ days. Divide by 3^. ( 2 forms.) 13. Width of a surface of 108 sq. ft., 12 ft. long=9 ft. Di- vide by 9. (2 forms.) 1j^. Length of a wall 12 ft. high, 2 ft. thick, built by 6 men in 12 days = 900 ft. Divide by 6. (4 forms.) 15. Principal required to produce $640 in 1^ years at 8% = $6000. Divide by 64. ( 3 forms.) 16. Rate required for $6000 to produce $640 in 1^ years =8% . Divide by 4. (3 forms.) 17. Time required for $6000 to produce $640 at 8% = 1^ years. Divide by 4. (3 forms.) 2. RATIO. 63. Definitions. — A ratio is an expression of the rela- tive magnitude of one number as compared with another of the same kind taken as the standard. The number compared is the antecedent ; the one taken as the standard of comparison is the consequent. These numbers are also the terms of the ratio, and together form a couplet. ADVANCED ARITHMETIC. \)6 15 yd.: 5 yd. =3 is read, **The ratio of 15 yd. to 5 yd. equals 3," and means that 15 yd. is 3 times 5 yd. '- $8: $12 = 1 means that $8 is | of $12. Observe that a ratio is always ah- stract. 15 yd. is S times 5 yd., but not 3 yd. times 5 yd. ; $8 is f of $12, but not $1 of $12. In arithmetic we have the indicated ratio, and the numerical ratio. As, In $32: $16=2 $32: $16 (indicated ratio). 2 ( numerical ratio). For every indicated ratio whose terms are concrete there may be written a corresponding indicated ratio whose terms are ab- stract. As, 70 bu. : 14 bu. (Terms concrete.) 70 : 14 (Terms abstract.) Note. — The respective terms in the two ratios are numerically equal ; the ratios are absolutely equal. Principle : The numerical ratio between any two similar con- crete numbers is the same as the numerical ratio between their cor- responding abstract numbers. 64. Writing the Corresponding Abstract Terms of Ratios. — In solving ratios and proportions, it is best to deal with abstract terms. The following principles will be of service in determining the corresponding abstract terms of ratios whose terms are concrete. Note.— These principles are found in Article 61, but are not grouped there as they are here. 94 ADVANCED ARITHMETIC. I. Principles Relating to Surfaces: (1) The dimensions (length, and width) hear to the area the same numerical relation as that of factors to the product. (2) The area and one dimension hear to the other dimension the same numerical relation that the dividend and divisor bear to the quotient. II. Principles Relating to Solids: (1) The dimensions (length, width, and thickness) hear to the volume the same numeri- cal relation as that of factors to the product. (2) The volume and two dimensions bear to the third dimension the same numerical re- lation that the dividend and the factors of the divisor bear to the quotient. III. Principles Relating to Interest: (1) The principal, time, and rate bear to the interest the same numerical relation as that of factors to the product. (2) The interest and any two of the elements (principal, rate, or time) bear to the third element the same numerical relation that the dividend and the factors of the divisor bear to the quotient. IV. Principles Relating to Work: (1) The working force and time hear to the work done the same numerical relation as that of factors to the product. (2) The work done and either of the ele- ments (working force or time) hear to the other element the same numerical relation that the dividend and divisor hear to the quotient. EXAMFIiES. The following are examples of writing corresponding abstract terms of ratios : 1. 70 bushels: 14 bushels. Abstract: 70:14. 2. 5X20 men: 4x8 men. Abstract: 5x20:4x8. 3. \ of cost of 32 books : f of cost of 21 books. Abstract: ^of 32:|of 21. 8. ADVANCED ARITHMETIC. 95 ^ Area of a surface } { area of a surface ) ^- \l2 ft. long, 9 ft wide ^ * ^ 9 ft. long, 8 ft. wide ] ' Abstract ; 12 x 9 : 9 x 8. ( Prin. I-(l).) C Length of a solid of ) ( length of a solid of 5. \ 72 cu. ft., 4 ft. wide, > : ^ 90 cu. ft., 5 ft. wide, ( 2 ft. thick ) ( 3 ft. thick Abstract : j^ : g^^. ( Prin. II-(2).) { Interest on $500 ) { interest on $400 ) ^' I at 6% for 3i yr. ^ * ^ at 8% for 5 yr. \' Abstract: 500x6x31:400x8x5. (Prin. III-(1).) C Rate required for ) C rate required for ) 7. < $500 to gain $90 > : < $400 to gain $160 > . ( in 8 years ) ( in 5 years ) Abstract: ^^::^^. (Prin. III-(2).) Work done by } , \ work done by 15 men in 12 days ^ ' ^ 11 men in 8 days Abstract: 15x12:11x8. (Prin. IV-(1).) i Number of men re- ) C number of men re- 9. < quired to build 180 rd. > : < quired to build 88 rd. ( of fence in 12 days ) (of fence in 8 days Abstract : ^ : f . ( Prin. IV-(2;.) EXEECISE XLV. Write the corresponding ratios with abstract terms : 1. 25 1b.: 15 1b. 2. 8x9 books: 7X12 books. 3. Cost of 10 shares : cost of 15 shares. 4.. Value of 20 loads : value of 14 loads. 5. 7 X my age : 12 X my age. 6. 9 X cost of 10 hats: 12 X cost of 5 hats. 96 ADVANCED ARITHMETIC, 7. 8. 9. 10. 11, 12. 13. U. 15. 16. 17. 18. 19. Area of a surface 10 rd. long, 8 rd. wide Length of a surface of 160 sq. ft., 8 ft. wide Width of a surface of 380sq. rd.,20rd.long Volume of a solid 20 ft. long, 8 ft. high, 2 ft. thick Length of a solid ) of 140 cu. yd., 7 yd. \ high, 2 yd. thick ) Width of a solid ) of 120 cu. yd., 8 yd. V long, 3 yd. thick ) Thickness of a solid ) of 108 cu. ft., 12 ft. V long, 9 ft. wide ) Interest on $500 \ \ for 6 yr. at 10% ] ' I Time reqd. for $600 to gain $75 at 6% Rate reqd. for $540 to bear $86.40 in 2 yr. Work done by 12 men in 25 days Time reqd. by 6 men to build 40 rd. of fence Number of men reqd. ) to build 140 rd. in 18 [ days ) Principal reqd. to j produce $112 in 2^ yr. at 7% ] area of a surface ) 17 rd. long, 12 rd. wide ) ) length of a surface of ) I 860 sq.ft., 18 ft. wide \ ' width of a surface of I 540sq.rd.,30rd. long \ ' ) C volume of a solid ) } : ^30 ft. long, 10 ft. high, [ ) ( H ft. thick ) C length of a solid ) : ] of 120 cu. yd., 5 yd. S . ( high, 3 yd. thick ) ( width of a solid ) : ^of 140cu. ft., 10 ft. } . ( long, 2 ft. thick ) ( thickness of a solid ) : ] of 120 cu. ft., 10 ft. [ . ( long, 8 ft. wide ) interest on $750 for 3 yr. at 8% < time reqd. for $480 ) ' \ to gain $60 at 10% ) \ rate reqd. for $620 ) ' I to bear $248 in 4 yr. ) work done by 16 men in 20 days time reqd. by 15 men ) to build 120 rd. of fence ) number of men reqd. ) to build 320 rd. in 15 L days ) principal reqd. to ) produce $120 in 3^ yr. y . at 5% ) ADVANCED ARITHMETIC. 97 65. The Process of rinding" tlie Numerical Ratio. — The process employed in determining the numerical ratio between two numbers is division. ( See Second Applica- tion, Art. 16.) The antecedent becomes the dividend; the consequent, the divisor | and the numerical ratio, the quotient. Principles: 1, The antecedent divided by the consequent eqiiah the numerical ratio, 2. When the terms are abstract^ the antecedent divided by the nu- merical ratio equals the consequent. 3. The product of the consequent multiplied by the numerical ra- tio equals the antecedent. EXAMPLES. 1. The antecedent is $20, the consequent $5. Find the nu- merical ratio. Statement: $20:$5 = ( )? Process: 20-^5 = 4. Note. — It is recommended that, in the process , the corresponding ab- stract terms be always used ; then, no trouble will arise in dealing with complicated expressions, such as are found below (see "No. 7). The statement is made for the purpose of showing clearly the relation of the given parts to the required part, 2. 12 is the ratio of 72 days io what ? Statement : 72 da. : ( ) = 12 ? Process: 72+12 = 6. (Prin. 2.) .'. 6 days is the consequent. Question : Why is the consequent days ? 3. 10 is the ratio of what number to 13 men ? Statement : ( ) : 13 men = 10. Process: 10x13 = 130. (Prin. 3.) .*. 130 men is the antecedent. 98 ADVANCED ARITHMETIC. Jf. Find the ratio of the work of 6 men for 15 days to the work of 9 men for 2 days. Qt^in^.,^4 . S Work of 6 men \ . S work of 9 men ? _/ x o statement : j ^^^, ^^ ^^^^ ^ : j ^^^ ^ ^^^^ | =n ? Abstract: 6xl5:9x2 = ( )? 6x15 , Process 9x2 5. One block of marble is 10 ft. long, 5 ft. wide, 2 ft. thick ; another is 15 ft. long, 4 ft. wide, 5 ft. thick. The volume of the first bears what ratio to the volume of the second ? volume ^ statement: (1) < Ift.'w!, \ ■ j Vu.w., [ =< )? Process. ^- . _ 15 X 4 X 5 i 5 ft. th. 15 X 4x H( )? Note. — Employ cancellation wherever convenient. 6. A tract of land 50 rd. long, 30 rd. wide, is 1^ times as large as another tract, 45 rd. long. How wide is the second tract ? ( Tract ) ( tract ) Statement: (1) < 50 rd. 1., } : ] 45 rd. 1., ^=1}? ( 30 rd. w. ) ( ( ) rd. w. ) (2)||0X \..\IY \-m 50x30 __ Process: -rz — 77 = 25. .'. the required width is 25 rd. 7. What is the ratio of the principal required to produce $216 interest in 5 yr. at 6% to the principal required to pro- duce $192 in 4 yr. at iPrin. reqd. to ) ( prin. reqd. to ) produce $216 in V : -^ produce $192 in [• =( )? 5 yr. at 6^ ) ( 4 yr. at 8^ ) 216^.^92^ ^ ''5x6*4x8 ^ ^' ADVANCED ARITHMETIC. 99 216x4x8 . ., ^''^''''•' 5x6x192 = ^==^^- 8, Find the ratio of the length of a surface of 144 sq. ft., 12 ft. wide, to the length "pf a surface of 180 sq. ft., 10 ft. wide ! Length of a surface ) ( length of a surface ) of 144 sq. ft., 12 ft. \ : ] of 180 sq. ft., 10 ft. \ wide ) ( wide \ Statement: (1) ( )? (2) 144,180^ 12 * 10 ^ Process 144x10 12x180 = |. EXERCISE XLVI. Find the numerical ratios in the following FIND THE BATIO. FIND THE CONSEQUENT. FIND THK ANTECEDENT. Antecedent. Consequent. Antecedent. Ratio. Consequent. Ratio. 1. 56 ft. 8 ft. 7. $64. 11. 13. 342 in. 1 a* 2. 540 yd. 18 yd. 8. 16 da. h U. 34 mi. f. 3. 2760 men. 60 men. 9. 98 sq. ft. 10. 15. 72 yr. 13. 4- H- $20. 10. $7000. 140. 16. .003 oz. 440. 5. 34 bu. 16 bu. 1 11. i hr. 72. 17. 64 cords. 9. 6. 12 hr. 54 hr. ! 12. 24 gal. u- 18. til meters. 1852. Find the numerical ratios in Exercise XL V. 8. PROPORTION. 66. Definitions. — A proportion is the expression of equality between two indicated ratios. Every proportion, therefore, has 4 terms, which are numbered in order from left to right. The first and fourth terms are the extrenus; the second and third terms, the means. The double colon (:: ) is usually (but not always) used for the sign of equality between the ratios. 8bu. : 3bu. :: $40: $15, is read, **8 bu. is to 3 bu. as $40 is to $15," and means that the ratio of 8 bu. to 3 bu. is equal to the ratio of $40 to $15. 100 ADVANCED ARITHMETIC. 67. Denominations of Terms. — (1) All the terms of a proportion may be concrete. As, $25 : $12 : : 100 men : 48 men. Remember, that the two terms of an indicated ratio must be similar ; but terms in different ratios need not be similar, even though those ra- tios are in the same proportion. (2) One couplet may be concrete and the other abstract. As, ■ $25: $12:: 100: 48. (3) All the terms may be abstract. As, 25: 12:: 100:48. By dropping the denominations in a proportion, all or a part of whose terms are concrete, a new proportion is formed whose terms are abstract but are numerically equal to those of the first proportion. Thus, (1) 80 yr. : 12 yr. : : 90 bu. : 86 bu. (Terms concrete.) (2) 80 : 12 : : 90 : 86. (Terms abstract.) 68. Solving" a Proj^ortion with Abstract Terms. — Solving a proportion is the process of finding any one of its terms when the other three are known. Principles: 1. In any 'proportion whose terms are abstract^ the product of the extremes is equal to the product oj the means. Thus, (1) 7: 15:: 21: 45. Then, 7x45 = 15x21. (2) 8:3:: 32: 12. Then, 8x12 = 3x32. 2. Either extreme is equal to the product of the means divided by the other extreme. Thus, 20: 34:: 30: 51. „, 34x30 _, ,34x30 ^ Then, ^ =51, and ^^ =20. ADVANCED ARITHMETIC. 101 3. Either mean is equal to the product of the extremes divided by the other mean. Thus, 36: 14:: 18: 7. . mu 36x7 ,o J 36x7 , . Then, .. ^ =18, and .. „ =14. 14 18 EXAlCFIiES. 1. Find the missing term : 12 : 25 : : ( ) : 125 ? „ 12x125 _^ Process : — — — = 60. Note. — Use cancellation where convenient. ?. Find the missing term : ( ) : 82 : : 60 : 8 ? „ 32x60 „.„ Process : —x — = 240. 8 3. Find the missing term : 13 : 12 : : 78 : ( ) ? 12x78 -„ Process : — i-r — = 72. 4. Find the missing term : 9 : ( ) : : 72 : 4 ? D 9x4 , Process : -fj^^h 5. Find the missing factor: (12x6): 35: :( )X8:70? T „ ... 12x6x70 ... I. Process: (1) — = 144. do (2) 144 + 8 = 18. ^ TT r, 12x6x70 ,„ Or, II. Process : g^^g =18. Note.— The plan of the first process is to neglect the 8 and solve for the entire third term (144) ; then, dividing by 8 will give the required factor. The plan of the second process is this : Since the product of all the factors of the extremes equals the product of all the factprs of the means, the product of all the factors of the extremes divided by the product of all but one of the factors of the means, will give that one. 102 ADVANCED ARITHMETIC. ^ 120X100 108x100 ,^ ^„ ^- ( )X6 '' 720X5 ••12:9? p . 108x100x12x6 J_ ^^^^^^* 720x5x120x100x9 "500" .*. the missing factor is 500. Note. — Where the fractions are to be multiplied together, their nu- merators are placed above and their denominators are placed below the line (why ? ) ; when used as divisors, their denominators are placed above and their numerators below the line (why ? ). Special attention should be given to the fact, that if the required factor is in a denominator, it will appear in the result as a denominator with 1 for its numerator. 7. 128 16 12_ ( ) ::25:45? Process : 128x45 = h 16x72x25 ,*. the required factor is 5 (in the denominator) 1. 12:3 2. 12:3::4 3. ( ):72 4. 43: ( ) 5. 221: 13 6. {):li 7. A:( ) ( ):4? ( )? li:3? 20:600? ( ):21? 360:720? 4:9? EXERCISE XLVII. 125 8. 9|:4::360: ( ), 9. 10. 12. 46X( 8X5 ) 86 25X24 11. 27 19x7'9x( )•• 12X6 108 19 14x28" 162 210 540X5* 700x6* • 175* 12 : 72 ? • 49 * :( ):5? 69. Solving a Proi^ortion with Concrete Terms. — In any proportion whose terms are concrete, a miss- ing part may be found as follows : First, form and solve the corresponding proportion whose terms are abstract ; then, determine from the given proportion by inspec- tion the denomination of the required term. ADVANCED ARITHMETIC. 103 EXAICFXiES. Find the missing part : 1. 7 da. : 12 da. : : ( ) mo. : 72 mo„? A bstract tef ms ; 7 : 12 : : ( ) : 72 ? „ ' '7x72 ._ Process > = 42. .*. 42 mo. is the required term. ( Area of a surface ) ( area of a surface ) 2. \ 40 rd. long, 20 [ : ] 80 rd. long, ( ) V : : 5 A. : 30 A.? / rd. wide \ I rd. wide ) Abstract terms : 20x40: 80x( ):: 5: 30 ? 20x40x30 „ 80x5 =^^- 60 rd. is the required part. „ 20x40x30 „ Process: -^^^^ = m. iSOO reams of 16 pages to the sheet ) ( ( ) reams of ) ( 38400 vol. ) ( 27000 vol. ) [: ]24: pages to V :: ^ of 160 pp. ^ M of 320 pp. [ ? ) ( the sheet ) ( each ) ( each ) 800x16x27000x320 „„ ^<"^^'«- 24X38400X16 ^^- .*. 750 reams is the required term. iBody of soldiers ) ( body of soldiers ) ( ) men long, 60 V : -^ 100 men long, 75 [• : : No. of men : itself ? men wide ) ( men wide ) Abstract terms : ( ) x 60 : 100 x 75 : : No. ! No. ? Process- lQQx75xNo. ^ 60 X No. o*. 125 men is the required part. Note. — The ** No," being in both numerator and denominator, cancels. STime reqd. by ) ( time reqd. by ) ( ) men to build > : -^ 8 men to build \ : : 34: 170 ? 102 rd. of fence ) { 680 rd. of fence ) Abstract terms ; ^ : ^: : 34 : 170 ? V ) 8 104 ADVANCED ARITHMETIC. „ 680x34 .'.6 men is the required part. Note. — Do not forget to use the reciprocal of the result, when the re- quired term is a factor of the denominator. C Principal reqd. to ) ( principal reqd. to ) 6. \ gain $112 in ( ) yr. I : -^ gain $120ineSiyr. V :: 8: 9 ? /at 7% \ I at 5% j 112 120 Abstract terms : r-^ — ^ • tt. — ^ : : 8 : 9 ? ( )x7 3^x5 „ 120x8x7x3 ^ ^^^^^^^•' 112X5X9X10 =^- .*. I or 2| yr. is the reqd. part. EXERCISE XLVIII. Find the missing part in each of the following : 1. $120: ( ) : : 48 cords : 20 cords ? Note.— The 2d term must be $'s ; why ? 2. $3.50: $50:: 8 yd. :( ) ? Note. — The 4th term must be yards ; why ? 3. 44 in.: 18 in. : : ( ) : 900 in. ? of a solid ) ( vol. of a solid 'J ( Vol. :. I 20 ft. (wide, I long, 8 ft. > : < 80 ft. long, 10 ft. V : : 82 : 45 ? ( ) ft. thick \ I wide, 1^ ft. thick) ( Length of a solid ') C length of a solid ^ 5. \ 140 cu. yd., 7 yd. f : -^ ( ) cu. yd., 5 yd. ^ : : 20: 16 ? (high, 2 yd. thick J (high, 8 yd. thick ) f Width of a solid ") ( width of a solid ^ 6. \ 150 cu. ft., 9 ft. \ : \ 180 cu. ft., ( ) ft. f : : 25: 18 ? ( high, 2 ft. thick ) ( high, 8 ft. thick ) ( Thickness of a solid) ( thickness of a solid") 7. \ 162 cu. ft., 12 ft. [ '. \ 216 cu. ft., Oft. f::():2? ( long, 9 ft. high ) ( long, 8 ft. high ) ADVANCED ARITHMETIC. 105 5:8? j Interest on $500 ) j interest on $720 ) . ^' I for 6 yr. at 10% J ' ( f or ( ) yr. at 8% J " * ( Time reqd. by 6 ) ( time reqd. by 15 i 9. \ men to biiild;40 [ : I men to build 120 V : : 5 : ( ) ? ( rd. of fence ') { rd. of fence ) {Number of men reqd. ) ( number of men reqd. ) to build 320 rd. of [■ : ^ to build 140 rd. of [• : ; 96 : 35 ? fence in 15 days j ( fence in ( ) days ) B. PROBLEMS OF ONE BASIS —NATURE AND CLASSIFICATION. 70. Nature. — Typically, a problem consists of two parts: (1) A question, proposed for solution; and (2) a statement of a certain condition or relation, from which that solution may be determined. These parts are called Question and Basis. Thus, If 6 books cost $2.40, what cost 10 books ? Parts: (1) What cost 10 books ? (Question.) (2) 6 books cost $2.40. (Basis.) Note. — The above description refers only to such problems as require but one unknown part or term to be found. One basis only is needed for the solution of such a problem. In Part III, particular attention will be given to problems of two or more bases. Comparatively few problems are stated in the form of the type ; yet, all problems have these two essential parts, expressed or implied. The following are some of the forms not typical : 1. Find I of 200 yards. 2. 150 ounces is f of what ? S. Reduce 40000 lb. to tons. ^. $210 is what part of $450 ? 5. Find the area of a floor 20 ft. long and 15 ft. wide. 6. What cost 6 doz. eggs at 20/ per doz.? 106 ADVANCED ARITHMETIC. These may all be reduced to the form of the type. Thus, 1. Since all {|) of 200 yd. is 200 yd., what is | of it ? 2. If I of a number is 150 ounces, what is the number ? 3. Since 2000 lb. make 1 ton, how many tons in 40000 lb.? 4. Since $450 is all of $450, $210 is what part of it ? 5. Since the area of a surface 1 ft. long, 1 ft. wide is 1 sq. ft., what is the area of a floor 20 ft. long and 15 ft. wide ? 6. If 1 doz. eggs cost 20/, what cost 6 doz.? 71. Stating the Parts of a Problem. — Each part of a problem may be stated in the form of an eq^uation. Take for example the parts — (1) What cost 10 books ? (Question.) (2) 6 books cost $2.40. (Basis.) The question may be stated as an equation between cost of 10 books and the required term. Thus, Cost of 10books=( ) ? In like manner, the basis may be stated as an equation be- tween cost of 6 hooks and $2.J^0. Thus, (1) Cost of 6 books = $2.40; or, (2) $2.40=cost of 6 books. Problems often involve the relation of more than two num- bers. The basis may have as many forms as it has different numbers related. The following is an example : If 15 men harvest 200 acres of grain in 10 days, in how many days can 5 men harvest 80 acres ? The relation here is among men (working force), time, and work done ; and its three forms are : (1) Amt. harvested by 15 men in 10 days =200 A.; or, (2) Time reqd. for 15 men to harvest 200 A. = 10 days; or, (3) Force reqd. to harvest 200 A. in 10 days = 15 men. ADVANCED ARITHMETIC. 107 The question is stated as before : Time reqd. for 5 men to harvest 80 A. = ( ) da.? The following principles, which have been illustrated and explained, should be learned : Principles: /. Every problem has at least two parts, a question {or requirement) and a basis. 2. The parts of a problem may always be stated in the form of equations. 3. Every basis has as many forms as it has numbers related. In each of the following examples, the question and all forms of the basis are stated in equations. EXAMPIjES. 1. James has $720, and John has | as much. How much has John ? Parts : (1) | of James's money = $( ) ? ( Question.) (2) I of James's money = $720 j or, $720 = I of James's money. ( Basis.) 2. What will 10 hats cost, at $2.40 each ? Parts : (1) Cost of 10 hats = $( ) ? ( Question.) (2) Cost of 1 hat = $2.40; or, $2.40 = costof Ihat. (Basis.) 3. 15% of a debt is $630. Find the debt. Parts: (1) 100^ of debt = $( )? (Question.) (2) 15^ of debt = $630; or, $630= 15^ of debt. (Basis.) Note. — The sign, ^, is re&d percent and means hundredth or hundredths. 10^ = ^1^0^ , 25^ = i%% , etc. 100% of anything is all of it. 4.. In ^ of a load of corn, there are 6 bushels. How many bushels in 25 such loads ? Parts: (1) Amount of 25 loads = ( ) bu. ? (Question.) (2) Amount of i\ of a load = 6 bu. ; or, 6 bu. = amount of i\ of a load. ( Basis.) 108 ADVANCED ARITHMETIC. 5. f of an article is worth $.75. What is W of it worth ? Parts : (1) Value of \ I of article = $( ) ? ( Question.) (2) Value of | of article = $.75; or, $.75 = value of | of article. (Basis.) 6. If 6 men plow 45 acres in 3 days, how many acres will 1 man plow in 1 day ? Parts: (1) Amt. plowed by 1 man in Ida. = ( ) A.? (Question.) (2) Amt. plowed by 6 men in 3 da. =45 A. ; or, Force reqd. to plow 45 A. in 3 da. = 6 men ; or, Time reqd. for 6 men to plow 45 A. = 3 da. ( Basis.) 7. If 25 boxes of pens are worth $7.50, what is ^ of a box worth ? Parts : (1) Value of J box = $( ) ? (Question.) (2) Value of 25 boxes = $7.50; or, $7.50 = value of 25 boxes. (Basis.) 8. The volume of a solid 7 ft. long, 5 ft. wide and 4 ft. thick is 140 cubic feet. Find the volume of a solid 10 ft. long, 7 ft. wide and 5 ft. thick. Parts : (1) Vol. of a solid 10 ft. 1., 7 ft. w., 5 ft. th. = ( )cu. ft.? (Question.) (2) Vol. of a solid 7 ft. 1., 5 ft. w., 4 ft. th. = 140 cu. ft. ; or, Length of a solid of 140 cu. ft. , 5 f t. w. , 4 ft. th. = 7 ft. ; or. Width of a solid of 140 cu. ft., 7 ft. 1., 4 ft. th. = 5 ft. ; or, Thicknessof asolidof 140cu.ft.,7ft.l.,5ft.w. = 4ft. (Basis.) 9. 21 is what part of 56 ? Parts: (1) 21 = ( ) of 56 ? (Question.) (2) 56 = all of 56 ; or. All of 56 = 56. (Basis.) Note.—** 21 = ( ) of 56 ? " should be read, ** 21 equals what part of 56 ? " 10. 760 is how many times 38 ? Parts: (1) 760 = ( )x38 ? (Question.) (2) 38 = 1x38; or, 1x38 = 38. (Basis.) ADVANCED ARITHMETIC. 109 EXERCISE XLIX. State the question and all forms of the basis in each of the fol- lowing : 1. Find 20% of $600: 2. What cost f of a ton of coal at $3.80 per ton ? 3. Clara is 12 years old ; her mother is four times as old. How old is her mother ? 4. Find 800% of $240. 5. I of my salary is $875. What is my salary ? 6. 25% of a certain debt is $1600. Find the debt. 7. If 14^ of a boat is worth $9900, what is the whole boat worth ? 8. 8i% of a certain sum is $50. What would 400% of that sum be ? 9. If I of a bolt of cloth is worth $8.20, what are 18 such bolts worth ? 10. If J of Mr. Brown's capital is $700, how much is f J of it ? 11. I deposit 48% of my money, which is $960, in the bank, and invest 28% of my money in calves. How much do I spend for calves ? 12. 860% of a number is 252. Find the number. 13. If 20 bu. of apples cost $18, find the price per bushel. 14.. If 5 men can saw 20 cords of wood in 4 days, how many cords can 1 man saw in 1 day ? 15. 120% of a number is 72. Find 85% of it. 16. If 11 men can set 660 pages of type in 6 days, how many pages will 21 men set in 8 days ? 17. 240% of my age is 84 years. Find 880% of it. • 18. 75 bu. of wheat are worth $60. What are 42 bu. worth ? 19. A lawyer charges $85 for collecting $700. What % does he charge ? 20. I bought an article for $5 and sold it so as to lose $8. What part of my investment did I lose ? 110 ADVANCED ARITHMETIC. 21. I bought an article for $5 and sold it for $7.50. The selling price is what % of the cost price ? 22. 5 times my money would be $4620. How much have I ? 23. 50 doz. eggs sell for $4.50. Find the price per dozen. 2J^. Eggs are 9/ per dozen. How many dozen can I buy for $8.60 ? 25. Eggs are 9/ per dozen. What will 80 dozen cost ? 26. Reduce 64 pints to quarts. (Basis omitted.) 27. The area of a surface 1 ft. long, 1 ft. wdde is 1 sq. ft. Find the length of a lot 150 ft. wide, containing 30000 sq. ft. 28. If 5 boys plow 12 acres in IJ days, in how many days will 12 boys plow 60 acres ? 29. Find the volume of a stone 12 ft. long, 6 ft. wide, and 8 ft. thick. (Basis omitted.) Note. — Preserve your work for use in Exercises L and LI. 72. Classification. — The problems of one basis, though varied in their forms of expression, when considered with ref- erence to the relation of the given 'part to the required part, consist of no more than ten different types ; and these may be classed into four classes, as follows : Class 1. Given a number, to find (1) a part of it, or (2) a mul- tiple of it. EXAMPLES. 1. What is the value of f of an article which cost $7.20 ? 2. What will 50 books cost, at $2.50 each ? Class 2. Given a part of a number, to find (1) the mimber, (2) a multiple of it, or (8) apart of it. EXAlttPIiES. 1. 20% of a sale is $540. What was the whole amount of the sale ? Note. — Any number of % less than 100^ is a part ; any number of ^ greater than 100^ is a multiple. ADVANCED ARITHMETIC. Ill 2. ^^ of a certain farm is 25 acres. How many acres in 8 such farms ? S. I of a certain article is worth $.75. What is f of it worth ? Class 3. Given h'^iultiple of a number, to find (1) the number , (2) a part of the number, or (3) another multiple of it. EXA]SO>I.ES. 1. If 35 hats sell for $105, find the price of 1 hat. 2. At the rate of $120 in 2 months, how much money can be earned in f of a month ? S. 12 pounds troy equal 144 ounces ; how many ounces do 7 pounds troy equal ? Class 4. Given two numbers, to find (1) what part one is of the other, or (2) what multiple one is of the other. EXAMPIiES. 1. A commission merchant charges $96 for making a sale of $3200. What per cent does he charge ? 2. John has $.50 and George has $5. George's money is how many times John's ? It may be ascertained to which of the above 10 types a prob- lem belongs by examining its parts. Thus, in example 2, Class 3, Parts: (1) Earnings for f mo. = $( )? (Qaestion.) (2) Earnings for 2 mo. =$120. (Basis,) Here are given the earnings for 2 months, or 2 times a num- ber; to find the earnings for f month, or f of the number. It gives a multiple of a number, to find a part of it. EXERCISE L. 1. Determine to which of the 10 types each of the examples under Arti<5le 71 belongs. 2. Determine to which of the 10 types each problem of Ex- ercise XLIX belongs. 112 ADVANCED ARITHMETIC. C. PROBLEMS OF ONE BASIS — SOLUTION. 73. Two Metliods of Solution.— Solving a Problem consists in commencing with a given (or known) relation and passing by means of an authorized process to the re- quired relation. There are two methods employed in arithme- tic for solving problems of one basis : 1. The Equation Metliod, in which equations are used throughout to express the steps of the solution. This method includes what by many authors is called ^^Arithmetical Analy- sis.'' 2. The Proportion Metliod, in which the solution is obtained by forming and solving a proportion. 1. THE EQUATION METHOD. 74. Arrang-ing tlie Parts of a Problem. — For the sake of convenience and system in solution, the following rules should be observed in arranging the parts of a problem : Rules: 1. The question, or requirement, of a problem shoidd be stated in the form of an equation with only the blank term on the right of the sign of equality. 2. The basis of the problem shoidd be stated in that form which agrees in arrangement with the question, or requirement, as stated according to Rule 1. Note. — It is very necessary that the pupil should learn to state by the above rules the parts of a problem without hesitation. In the following examples, the parts of the problems are properly arranged for solution. EXAMPIiES. io If 8 books cost $4, find the cost of 3 books. Partsi : (1) Cost of 3 books = $( ) ? ( Question.) (2) Cost of 8 books = $4. ( Basis.) ADVANCED ARITHMETIC. 113 2. If 8 books cost $4, how many books can be bought for $10 ? Paris : (!) $10 = cost of ( ) books ? (Question.) (2) H = cost of 8 books. ( Basis. ) XoTB.— Compare the ari*q,ngement of parts in Nos. 1 and 2, 3. Reduce 9 feet to inches. (1 foot=12 inches.) Parts: (1) 9 ft. = ( ) in.? (Question.) (2) 1ft. = 12 in. (Basis.) 4. Reduce 72 inches to feet. (1 inch^j^^j foot.) Parts : (1) 72 in. = ( ) ft.? (Question.) (2) 1 in. = i'2 ft. (Basis.) Note. — Compare the arrangement of parts in Nos. 3 and 4. 5. I owe a debt of $720, and can pay 45% of it. How much can I pay ? Parts : ( 1 ) 4b% of debt = $( ) ? ( Question . ) (2) 100^ of debt = $720. ( Basis.) Note. — Remember that 100^ of anything is all of it. 6. I owe $120 and pay $40. What % of the debt do I pay ? Parts : (1) $40 = ( )% of debt ? (Question.) (2) $120 = 100^ of debt. ( Basis.) Note. — Compare the arrangement of parts in Nos. 5 and 6. 7. Find the interest on $500 for 2 years at 8% . (Basis omitted.) Parts : (1) Int. on $500 at 8% for 2 yr. = $( ) ? ( Question.) (2) Int. on $1 at 1% for 1 yr. = $.01. ( Basis.) 8. Find the principal required to produce $80 interest in 2 years at 8%. (Basis omitted.) Parts : (1) Prin. reqd. to produce $80 in 2 yr. at 8^ = $( ) ? ( Question.) (2) Prin. reqd. to produce $.01 in 1 yr. at 1^ = $1. (Basis.) 9. Find the rate required for $500 to produce $80 in 2 years. (Basis omitted.) Parts : (1) Rate reqd. for $500 to produce $80 in 2 yr. = ( )^ ? ( Question.) (2) Rate reqd. for $1 to produce $.01 in 1 yr. = 1%. ( Basis.) 114 ADVANCED ARITHMETIC. 10. Find the time required for $500 to produce $80 at 8%. (Basis omitted.) Parts : (1) Time reqd. for $500 to produce $80 at 8^ = ( ) yr.? (Question.) (2) Time reqd. for $1 to produce $.01 at 1^ = 1 yr. (Basis.) Note. — Compare the arrangement of parts in Nos. 7 to 10. 11. Find the area of a floor 12 feet long and 8 feet wide. (Basis omitted.) Parts : (1) Area of a surface 12 ft. 1. , 8 ft. w. = ( ) sq. ft.? ( Question.) (2) Area of a surface 1 ft. 1., 1 ft. w. = l sq. ft. (Basis.) 12. Find the length of a floor 12 feet wide, containing 240 square feet. (Basis omitted.) Parts : (1) Length of a surface of 240 sq. ft., 12 ft. w. = ( ) ft.? (Question.) (2) Length of a surface of 1 sq, ft., 1 ft, w. = 1 ft. ( Basis.) ♦ 13. Find the width of a surface of 108 square feet, 12 feet long. (Basis omitted.) Parts : (1) Width of a surface of 108 sq. ft., 12 ft. 1. = ( ) ft.? (Question.) (2) Width of a surface of 1 sq. ft., 1 ft, 1. = 1 ft. ( Basis.) Note. — Compare the arrangement of parts in Nos. 11 to 13. It should be remembered that, in stating a problem, the basis is omitted only when it is supposed that the pupil al- ready has the knowledge necessary to enable him to state the basis for himself. EXERCISE LI. I. Arrange for solution the parts of each problem in Exercise XLIX. II. Arrange for solution the parts of each of the following : 1. 150 ounces is | of what number ? 2. Reduce 120 feet to inches. 3. Reduce 228 inches to feet. 4.. If 5 books cost $25, find the cost of 12 books. ADVANCED ARITHMETIC. 115 5. Basis same as in No. 4. Find how many oooks can be bought for $60. 6. If a boy earns^ $.50 in 5 hours, what will he earn in 13 hours ? 7. Find how long it will take $620 at 6% to gain $9 interest. S. Find how much interest $740 will gain at 8% in 8^ years. 9. Basis same as in No. 6. How long can I hire him for $8.50? 10. Find the area of a field 20 rods long and 12 rods wide. 11. A field of 240 sq. rd. is 8 rods wide. How long is it ? ^ 12, If 5 men excavate 20 tons of dirt in 2 days, how many tons will 12 men excavate in 24 days ? IS. Basis same as in No. 12. In what time will 8 men exca- vate 120 tons ? H. 25% of my money is $250. How much money have I ? 15. Basis same as in No. 14. What per cent, of my money is $450? 16. Find the volume of a rectangular cistern 8 ft. long, 5 ft. wide, and 7 ft. deep. 17. At what rate will $800 produce $20 interest in 5 months ? 18. What is the time, when f of the time past noon is 4 hours ? 19. The volume of a rectangular solid is 860 cu. ft., the length 15 ft., the width 8 ft. Find the thickness. 20. A can do a piece of work in 20 days. How much of it can he do in 9 days ? 21. Basis same as in No. 20, How long will A be in doing -J-J- of the work ? 22. I bought an article for $5, and sold it for -f of the cost. Find the selling price. 23. 42 is i of what number ? 21^. 400 is how many times 25 ? 116 ADVANCED ARITHMETIC. 75. Tlie Process of Solving Problems. — After the parts of a problem have been arranged according to the rules given in the last article, the process of solving by equa- tions consists of one or both of the following steps : Steps: 1. Perform such operation, or operations, upon the basis as will change it to an equation, the numerical value of whose first member is unity. 2. Perform such operation, or operations, upon the equation, ob- tained in Step 1, as will change it to an equation whose first mem- ber is identical with the first member of the question. If the numerical value of the first member of the question is unity, the first step only is used ; if the numerical value of the first member of the basis is unity, the second step only is used. Example : If 24 hats cost $12, how many hats can be bought for $16? Solution : (1) $16 = cost of ( ) hats ? ( Question.) (2) $12 = cost of 24 hats. ( Basis.) i^a of (2) = (3) $1 = cos t of 2 hats. 16x(3) = (4) $16 = cost of 32 hats, answer. 76. Integral Solutions. — A solution whose equations contain only integers in their first members is an integral solu- tion. Note. — Integral solutions are not always free from fractions (see Ex- ample 4, page 117) ; but fractions may occur in the second members of the equations only. EXAMFIiES. J. Find the cost of 10 mats at $3 each. Solution: (1) Cost of 10 mats = ${ )? (Question.) (2) Cost of 1 mat = $3. ( Basis. ) 10x(2) = (3) Cost of 10 mats = $30, answer. ADVANCED ARITHMETIC. 117 2. If 10 mats cost $30, what will 1 mat cost ? Solution: (1) Cost of lmat = $( )? (Question.) (2) Cost of 10 mats = $30. ( Basis.) ^0 of (2) = (3) Cost of 1 mat = $3, answer. 3. If 10 mats cost $30, what cost 17 mats ? Solution: (1) Cost of 17 mats = $( )? (Question.) (2) Cost of 10 mats = $30. ( Basis. ) ,V of (2) = (3) Cost of 1 mat = $3. 17x(3) = (4) Cost of 17 mats = $51, answer. 4. If 10 mats cost $30, how many can be bought for $63 ? Solution: (1) $63 = cost of () mats? (Question.) (2) $30 = cost of 10 mats. ( Basis. ) ^ of (2) = (3) $1 = cost of i mat. 63x(3) = (4) $63 = cost of *8* or 21 mats, answer. 5. A man has $75^ and loans $35. What part of his money does he loan ? Solution: (1) $35 = ( ) of his money? (Question.) (2) $75 = all of his money. (Basis.) ,»5 of (2) = (3) $1 = A of his money. 35x(3) = (4) $35 = ?|, or {s of his money, answer. 6. 12 is what part of 40 ? Solution : (l)12 = ()of40? ( Question . ) (2) 40 = all of 40. (Basis.) ^of(2) = (3) 1 = 4'^ of 40. 12 X (3) = (4) 12 = H , or 1^0 of 40, answer. 7. 40 is how many times 12 ? Solution: (l)40 = ()xl2? (Question.) (2) 12 = 1x12, (Basis.) ,Vof(2) = (3) 1 = ,', of 12. 40x(3) = (4) 40= 12, or 3^x12, answer. 8. If 5 men can do a piece of work in 42 days, in what time can 14 men do it ? 118 ADVANCED ARITHMETIC. Solution: (1) Time reqd. by 14 men = ( )da.? (Question.) (2) Time reqd. by 5 men = 42 da. ( Basis.) 5x(2) = (3) Time reqd. by 1 man = 210 da. ^ of (3) = (4) Time reqd. by 14 men = 15 da., answer. 9. Find the area of a surface 16 rods long and 10 rods wide. Solution : (1) Area of a surf. 16 rd. 1., 10 rd. w. =( ) sq. rd.? ( Question.) (2) Area of a surf. 1 rd. 1., 1 rd. w. = 1 sq. rd. (Basis.) 16x(2) = (3) Area of a surf. 16 rd. 1., 1 rd. w. = 16 sq. rd. 10x(3) = (4) Area of a surf. 16 rd. 1., 10 rd. w. = 160 sq. rd., answer. 10. Find the length of a surface of 160 square rods, 10 rods wide. Solution : (1) Length of a surf, of 160 sq. rd., lOrd.w. =( )rd.? (Question.) (2) Length of a surf, of 1 sq. rd., 1 rd. w. = 1 rd. ( Basis.) 160 X (2) = (3) Length of a surf . of 160 sq. rd., 1 rd. w. = 160rd. i^jj of (3) = (4) Length of a surf, of 160 sq. rd., 10 rd. w. = 16 rd., answer. 11. Find the width of a surface of 160 square rods, 16 rods long. Solution: (1) Width of a surf, of 160 sq. rd., 16rd. 1. =( ) rd.? (Question.) (2) Width of a surf . of 1 sq. rd., 1 rd. l. = l rd. (Basis.) 160 X (2) = (3) Width of a surf . of 160 sq. rd., 1 rd. l. = 160rd. 1^6 of (3) = (4) Width of a surf, of 160 sq. rd., 16 rd. 1. = 10 rd., answer. 12. If 5 men build 100 rods of fence in 10 days, how many rods can 8 men build in 7 days ? Solution: (1) Amt. built by 3 men in 7 da. =( ) rd.? (Question.) (2) Amt. built by 5 men in 10 da. = 100 rd. ( Basis.) I of (2) = (3) Amt. built by 1 man in 10 da. = 20 rd. 3^ of (3) = (4) Amt. built by 1 man in 1 da. = 2 rd. 3 X (4) = (5) Amt. built by 3 men in 1 da. = 6 rd. 7x(5) = (6) Amt. built by 3 men in 7 da. =42 rd., answer. 13. If 5 men build 100 rods of fence in 10 days, how many men can build 42 rods in 7 days ? Solution : (1) Force reqd. to build 42 rd. in 7 da. =( ) men ? (Question.) (2) Force reqd. to build 100 rd. in 10 da. =5 men. ( Basis.) jhi of (2) = (3) Force reqd. to build 1 rd. in 10 da. = x^, or ^^ man. ADVANCED ARITHMETIC. 119 10x(3) = (4) Force reqd. to build 1 rd. in 1 da. = 1%, or I man. 42x(4) = (5) Force reqd. to build 42 rd. in 1 da. = 21 men. J of (5) = (6) Force reqd. to build 42 rd. in 7 da. = 3 men, answer. 14.. If 5 men can bviild 100 rods of fence in 10 days, how many days will be required for 8 men to build 42 rods ? Solution : (1) Time reqd. for 3 men to build 42 rd. = ( ) da.? (Question.) (2) Time reqd. for 5 men to build 100 rd. = 10 da. ( Basis.) 5 X (2) = (3) Time reqd. for 1 man to build 100 rd. = 50 da. lU of (3) = (4) Time reqd. for 1 man to build 1 rd. = ^ da. I of (4) = (5) Time reqd. for 3 men to build 1 rd. = ^ da. 42x(5) = (6) Time reqd. for 3 men to build 42 rd. = 7 days, answer. 15. Reduce 42 gallons to quarts. Solution: (1) 42gal. = ( ) qt.? (Question.) (2) lgal. = 4qt. (Basis.) 42 X (2) = (3) 42 gal. = 168 qt. , answer. 16. Reduce 172 quarts to gallons. Solution: (1) 172 qt. = ( ) gal.? (Question.) (2) 1 qt. = i gal. (Basis.) 172 X (2) = (3) 172 qt. = 43 gal . , answer. 17. Find the interest on $500 at 6% per annum for 3 years. Solution : (1) Int. on $500 at Q% for 3 yr. = $( ) ? (Question.) (2) Int. on $1 at 1% for 1 yr. = .$.01. (Basis.) 500 X (2) = (3) Int. on .$500 at 1% for 1 yr. = $5. 6 X (3) = (4) Int. on $500 at 6% for 1 yr. = $30. 3 X (4) = (5) Int. on $500 at 6% for 3 yr. = $90, answer. 18. Find the rate required for $5(X) to gain $90 in 8 years. Solution : (1) Rate reqd. for $500 to gain $90 in 3 yr. = ( )^ ? (Question.) (2) Rate reqd. for $1 to gain $.01 in 1 yr. = 1%. ( Basis.) sU of (2) = (3) Rate reqd. for $500 to gain $.01 in 1 yr. = ^Jn^. 9000 X (3) = (4) Rate reqd. for $500 to gain $90 in 1 yr. = 18^. J of (4) = (5) Rate reqd. for $500 to gain $90 in 3 yr. =6^, answer. 19 annum '. Find the time required for $500 to gain $90 at 6% i^er 120 ADVANCED ARITHMETIC. Solution : (1) Time reqd. for $500 to gain $90 at 6$g = ( ) yr.? ((Question.) (2) Time reqd. for $1 to gain $.01 at 1^ = 1 yr. (Basis.) gfo of (2) = (3) Time reqd. for .$500 to gain $.01 at l%=sU jr. 9000 X (3) = (4) Time reqd. for $500 to gain $90 at 1^^ = 18 yr. I of (4) = (5) Time reqd. for $500 to gain $90 at 6^ = 3 yr., answer. 20. What principal will gain $90 interest in 3 years at 6% per annum ? Solution : (1) Prin. reqd. to gain $90 in 3 yr. at 6^ = $( ) ? (Question.) (2) Prin. reqd. to gain $.01 in 1 yr. at 1% = $1. (Basis.) 9000 X (2) = (3) Prin. reqd. to gain $90 in 1 yr. at 1^ = $9000. I of (3) = (4) Prin. reqd. to gain $90 in 3 yr. at 1^ = $3000. i of (4) = (5) Prin. reqd. to gain $90 in 3 yr. at 6^ = $500, answer. EXERCISE LII. 1. What will 20 hats cost at $2.50 each ? 2. 21 is what part of 49 ? 3. What cost 22 yards of muslin at 10/ per yard ? 18 yards of flannel at 40/ per yard ? 4.. What are 40 acres of land worth at $50.20 per acre ? 6. The light from the sun reaches the earth in 8 minutes, and it travels at the rate of 188000 miles per second. Find the distance to the sun. 6. A has $840; B has 7 times as much; C has 5 times as much as B. How much has B, and C ? 7. What is the area of a surface 120 feet long and 30 feet wide? 8. A can travel on a bicycle 4 times as fast as he can walk. If he can walk 2 miles an hour, how fast can he go on a bi- cycle ? 9. 4 acres is what part of 10 acres ? 10. I charge $75 for collecting $750 : what part of the amount collected will pay me for collecting ? 11. If 1 man can build a certain fence in 12 days, how long will it take 4 men to build it ? ADVANCED ARITHMETIC. 121 12. An area of 140 sq. ft. is 7 ft. wide. How long is it ? 13. An area of 150 sq. ft. is 50 ft. long. How wide is it ? IJ,.. A's money, $500, is how many times B's money, $125? 15. Find the volume of a solid 9 ft. long, 7 ft. wide, and 10 ft. high. " '. 16. A volume of 630 cu. ft. is 9 ft. long and 7 ft. wide. How high is it ? 17. A volume of 680 cu. ft. is 10 ft. high and 7 ft. wide. How long is it ? 18. $6800 is how many times $260 ? 19. The distance to the sun is 90240000 miles. Light trav- els from the sun to the earth in 8 minutes : what is the veloc- ity of light per second ? 20. A lady earns $45 per month (20 days). How much is that per day ? 21. If 20 men harvest 180 acres in 5 days, how much does each man harvest per day ? 22. Three leaps of a fox is as far as two leaps of a hound. Reduce 60 fox-leaps to hound-leaps. Question: 60f. l. = ( ) h. 1.? 23. Basis same as in No. 22. Reduce 60 hound-leaps to fox- leaps. 2 J/.. The minute-hand of a clock travels 12 times as fast as the hour-hand. How long will the hour-hand be in traveling 20 minute-spaces ? 25. How long will the minute-hand be in traveling 8 hour- spaces ? Question: Time reqd. by h. h. to travel 20 min. spaces = { ) min.? 26. The sun passes over 15 degrees of longitude in 1 hour : how many hours will it be in passing over 100 degrees ? 27. If $4.86 buys an exchange on London for £1, how large an exchange can I buy for $9720 ? 122 ADVANCED ARITHMETIC. 28. Find the interest on $720 for 5 years at 8% per annum. 29. What principal will produce $168 interest in 4 years at 7% per annum? 30. At what rate will $540 bear $81 in 3 years ? 77. Fractional Solutions. — Fractional solutions are those in which the first members of all equations are fractions or are expressed as fractions. Preparatory Step : Reduce the fractions in the first members of the parts of the problem to a common denominator. EXAMFIiES. io I of A's money is $4500. Find | of it. Solution : (1) Reduced to L. 0. D., | and t = U and M. (2) M of A's money = $( ) ? (Question.) (3) M of A's money = $4500. ( Basis. ) -iV of (3) = (4) ^V of A's money = $300. 16 X (4) = (5) M of A's money = $4800, answer. 2, If 1 ton of hay costs $5, what costs -^q of a ton ? Solution: (1) Cost of t'd T. = $( ) ? (Question.) (2) CostofHT. = $5. (Basis.) yV of (2) = (3) Cost of iV T. = $.50. 7x(3) = (4) Cost of 1^^ T. = $3.50, answer. 3. Find | of $4000. Solution: ( it f of $4000 = $( )? (Question.) (2) I of $4000 = $4000. (Basis.) i of (2) = (3) i of $4000 = $500. 3 X (3) = (4) I of $4000 = $1500, answer. 4-. 342 is I of what number ? Solution : (1) | of No. = ( ) ? (Question.) (2) f of No. = 342. (Basis.) Jof (2) = (3) i of No. = 114. 4 X (3) = (4) I of No. = 456, answer. ADVANCED ARITHMETIC. 128 5. What number added to | of itself will give 560 ? Solution: (1) f of No. = ( ) ? (Question.) (2) f of No. + 1 of No. =560. (Basis.) (2) = (a) Y of No. = 560. x^of (3) = (4) i.of No.=40. 9 X (4) = (5) I of No. = 360, answer. 6. What number less its f leaves 510 ? Solution : (1) ?^ of No. = ( ) ? (Question. ) (2) I of No. - f of No. = 510. ( Basis. ) (2) = (3) I of No. = 510. iof (3) = (4) ^ of No. = 170. 7 X (4) = (5) I of No. = 1190, answer. 7. Find that number whose | added to its f will give 1240. Solution : (1) Reduced to L. C. D., | and | = ^| and 1%. (2) II of No. = ( )? (Question.) (3) \l of No. + il of No. = 1240. ( Basis. ) (3) = (4) f i of No. = 1240. ^Vof (4) = (5) ^of No. = 40. 24x(5) = (6) iiof No. = 960, answer. 8. Find that number whose ^ less its ^ leaves 850. Solution : (1) Reduced to L. C. D., y'r and i = || and H. (2) t|of No. = ( )? (Question.) (3) II of No. - ii of No. = 850. ( Basis.) (3) = (4) H of No. =850. rVof (4) = (5) ^of No. = 50. 44 X (5) = (6) It of No. = 2200, answer. D. 720 is I more than what number ? Solution: (1) fofNo. = ( )? (Question.) (2) I of No. + 1 of No. = 720. ( Basis.) (2) = (3) I of No. = 720. I of (3) = (4) i of No. = 144. 3 X (4) = (5) I of No. = 432, answer. 10. 102 is f less than what number ? Solution: (1) |of No. = ( )? (Question.) (2) I of No. - 1 of No. = 102. ( Basis.) (2) = (3) I of No. = 102. iof (3) = (4) i of No. = 34. 5 X (4) = (5) I of No. = 170, answer. 124 ADVANCED ARITHMETIC. 11, I is what part of J ? Solution : (1) Reduced to L. 0. D., f and | = f and J. (2) f = ( )of J? (Question.) (3) I = all of |. (Basis.) iof(3) = (4)i=Vof ^. 6x(4) = (5) f = f of|. .-. f isf of|. 12. .12 is how many times .08 ? Solution: (1) .12 = ( )x.08? (Question.) (2) .08 = lx.08. (Basis.) iof (2) = (3) .01 = i of .08. 12x(3) = (4) .12 = ^, or lJx.08, answer. EXERCISE LIII. 1. I of a number is 1240. Find | of the number. 2. My salary is $765. If I spend | of it, what amount do I spend ? S. A man can build f of a wall in 12 days. In what time can he build | of it ? J/.. In selling, a merchant lost y\ of the cost on an article which cost him $20.24. How much did he lose ? What did the article sell for ? 5. A has ^j as much money as B. If they both have $6400, how much has B ? How much has A ? Basis : y^ of B's money +i} of B's money = $6400. (Why ?) 6. 48 is y% of what number ? 7. i^ of a number is 1542. Find the number. 8. 3^ of a bolt of muslin costs $5. What will the whole bolt cost ? 9. 1 charge ^V of a debt for collecting it. If I receive $240, what was the whole debt ? 10. $950 is -^-^ more than what number ? 11. $475 is y\ less than what number ? ADVANCED ARITHMETIC. 125 12. Find the number whose f added to its fg- makes $920. 13. Find the number whose W less its -^ leaves $2080. IJ^. What number added to its -f^ gives 136 ? 15. What number legs its | leaves 75 ? 16. What is I of a ton of hay worth at $3.20 per ton ? 17. If T^ of a farm is worth $1750, what is the whole farm worth ? 18. B can do | of a piece of work in f of a day. What part of it can he do in | of a day ? 19. \\ times a number exceeds its y\ by 54. Find 3^ times the number. 20. A loaned B | of his money and had $480 left. How much money had A before making the loan ? 21. The sum of two numbers is 54; the smaller is \ of the larger. Find the numbers. Basis : f of larger + h of larger = 54. (Why ?) 78. Keciprocals and Their Use.— The Recipro- cal of a number is one divided by that number. EXAMPIiSa 1. The reciprocal of 16=^. Note. — The reciprocal of an integer is expressed as a common fraction whose numerator is 1 and whose denominator is the given integer. 2. The reciprocal of .03=;^. 3. The reciprocal of 3.25=^. Note. — The reciprocal of a decimal is expressed as a common fraction whose numerator is 1 and whose denominator is the given decimal. 4. The reciprocal of f=l-7-f=l Xi= J. Note.— The reciprocal of a common fraction is expressed by inverting the fraction. 5. The reciprocal of 3^, or -^=^. Note. — Reduce to an improper fraction and proceed as in No. 4 above. 126 ADVANCED ARITHMETIC. After stating the parts of a problem for solution often the first step is to proceed from what is given, to unity (Step 1, p. 116). This can always be done directly, either in integral or fractional problems, by employing the following principle: Principle : The product of a number by its reciprocal equals 1 . EXAMPIjES. 1. If 20 books cost $50, what cost 1 book? Solution : (1) Cost of 1 book = $( ) ? (Question.) (2) Cost of 20 books = $50. ( Basis.) aVx(2) = (3) Cost of 1 book = $2.50, answer. Note. — Did we multiply by the reciprocal of 20? We have heretofore used "of" instead of " x " ; is '* of " correct ? ( See Article 88.) 2. If 8 J cords of wood are worth $8, what is 1 cord worth ? Solution: (1) Price of 1 cord = $( )? (Question.) (2) Price of 3^^ cords = $8. ( Basis. ) ^ of (2) = (3) Price of 1 cord = $2.40, answer. Note. — 1% is the reciprocal of Sh 3. At $1.25 per bu., how much wheat can be bought for $22.50 ? Solution: (1) $22.50 = cost of () bu.? (Question.) (2) $1.25 = cost of 1 bu. ( Basis. ) j^x(2) = (8) $l = cost of^bu. 22.50 X (3) = (4) $22.50 = cost of f^, or 18 bu. , answer. Note. — j-^g is the reciprocal of 1.25. 4. 12i is what part of 37i ? Solution: (1) 12| = ( ) of 371^? (Question.) (2) 37i = allof37i. (Basis.) ^of(2) = (3) l = ^of37|. 121 X (3) = (4) 12i^ = ^ X ^ , or ^ of 371, answer. 3 ADVANCED ARITHMETIC. 127 EXERCISE LIV. Solve the problems in Exercise LIII by using reciprocals instead of the fractional solution there given. Thus, 1. Solution: (1) |'of No. = ( )? (Question.) (2) I of No. = 1240. (Basis.) f of (2) = (3) Ix No. = 1550. I of (3) = (4) I of No. = 1356i, answer. 79. Solutions Shortened. — While it is strenuously urged that the pupil should always follow out a complete solu- tion for every problem, it will become monotonous and even repulsive to the pupil to be compelled always to go the " long way around," when he can grasp completely a more direct so- lution. Since solving a problem consists in commencing with a known condition and passing by means of a known authorized process from that known condition to the unknown (required) condition^ then it follows that both the known condition and the process employed may depend largely upon the knowledge and ability of the person taught. The following suggestions will aid in shortening many solutions : (1) Common fractions are usually to be preferred to deci- mals. Learn the following table : EaUIVAIiENTS. f.06y4=r/i6 f. 081/3= $yi2 $.10 =$Vio $.i2y2=r/8 $.i6%=r/6 $.20 =r/5 $.25 =$1/4 $.331/3 = $l/3 $.50 =$1/2 128 ADVANCED ARITHMETIC. EXAMPIiES. 1. At 16f/ each, how many knives can be bought for $8.50 ? Solution: (1) $3| =cost of ( ) knives? (Question.) (2) $^ = cost of 1 knife. ( Basis. ) 6x(2) = (3) $l = cost of 6 knives. 31 x(3) = (4) $3J = cost of 21 knives, answer. 2. What will 220 dozen eggs sell for, at 12^/ per dozen ? Solution : (1) Price of 220 doz. = $( ) ? (Question.) (2) Price of 1 doz . = $| . ( Basis . ) 220 X (2) = (3) Price of 220 doz. = $27^, or $27.50, answer. (2) If the first member of the basis is a known factor of the first member of the question, one step only is needed in solving the problem. EXAMPIiES. 1. 4 times A's money is $800. If B has 12 times as much money as A, how much has B ? Solution: (1) 12 x A's money = $( )? (Question.) (2) 4 X A's money = $800. ( Basis. ) 3 X (2) = (3) 12 X A's money = $2400, answer. 2. Reduce 128 ounces avoirdupois to pounds. Solution: (1) 128 oz. = ( ) lb. ? (Question.) (2) 16oz. = llb. (Basis.) 8 X (2) = (3) 128 oz. = 8 lb. , answer. Note.— Here the basis of solution is not given, and may be stated in either of the forms : 16 oz. = 1 lb., or 1 oz. = i\ lb. As the number to be reduced (128 oz.) is a multiple of 16 oz., the first form is as good as the second, if not better. But if the number to be reduced had not been a multiple of 16 oz., the second form would have been the better, since it starts with unity in its first member. (3) If the first member of the basis is a known multiple of the first member of the question, one step only is needed in solving the problem. ADVANCED ARITHMETIC. 129 EXAMPLES. 1. If $85 will buy 30 books, how many books can be bought for $17? Solution: (1) $17'=f price of () books ? (Question.) (2) $85 = price of 30 books. (Basis.) \ of (2) = (3) $17 = price of 6 books, answer. 2. If 5 men can do a piece of work in 45 days, in what time can 15 men do the same work ? Solution: (1) Time reqd. by 15 men = ( ) da.? (Question.) (2) Time reqd . by 5 men = 45 da. ( Basis. ) I of (2) = (3) Time reqd. by 15 men = 15 da., answer. 3. A bought two tracts of land at a uniform price per acre. The first was 80 rods long and 40 rods wide, and cost $4000. What did the other cost, which was 20 rods long and 8 rods wide ? Solution : (1) Cost of a tract 20 rd. 1., 8 rd. w. =$( ) ? (Question.) (2) Cost of a tract 80 rd. 1., 40 rd. w. =$4000. (Basis.) iof (2) = (3) Costof atract20rd.l.,40rd. w. = $1000. \ of (3) = (4) Cost of a tract 20 rd. 1., 8 rd. w. = $200, answer. (4) When the basis of a problem is omitted, the pupil should be permitted to use whatever basis he can readily un- derstand. EXAMPIiE. Find the interest on $5(X) for 2^ years at 6% per annum. Solution : (1) Int. on $500 at Q% for 21 yr. = $( ) ? (Question.) (2) Int. on $500 at Q% for 1 yr. = $30. ( Basis.) 2^x(2) = (3) Int. on $500 at 6^ for 2^ yr. = $75, answer. Note. — Most advanced pupils will readily see that the ^^ Interest on $500 for 1 yr. at 6% = $30'' ; others may be able only to see that ''Interest on $500 for 1 yr. at 1% = $5'' ; while the beginner will have to commence with ''Interest on $1 for 1 yr. at 1% = $.01." 130 ADVANCED ARITHMETIC. EXERCISE LV. i. What will 50 lb. of butter cost, at 20/ per pound ? 2, What are 80 knives worth, at 25/ each ? 3, I bought 15 bu. of corn, at 83^/ per bu. How much did it cost me ? If.. What will 64 yds. of cloth cost, at 6^/ per yd.? 5. I bought 24 quarts of strawberries, at 8 J/ per qt. How much did I pay for them ? 6. At 12-^/ each, how many tin pails can be bought for $4 ? 7. At 25/ each, how many melons can be bought for $5 ? 8. At 50/ per bu., how many bushels of apples can be bought for $12 ? 9. At 33J/ each, how many books can be bought for $11 ? 10. 15 is what part of 75 ? 11 . A bill of exchange on London costs $4.86 per £. How many £'s will the bill call for, if it cost $486 ? Note. — The English pound, or sovereign, £, is worth about $4.86 in American gold. 12. Reduce 315 gallons to barrels (81.5 gal. = l bbl.). 13. A block of marble 10 ft. long, 8 ft. wide, and 6 ft. high, is worth $420. Find the value of a block 5 ft. long, 2 ft. wide, and 2 ft. high. IJf.. A has $1000, which is 3 times as much as B has ; C has 9 times as much as B. How much has C ? 15. If 20 men build 24 miles of fence in 8 days, in how many days can 5 men build 8 miles ? 16. Since 80° Reaumur is as hot as 100° Centigrade, how many degrees Centigrade will correspond to 240° Reaumur ? 17. Reduce 160 ounces avoirdupois to pounds. 18. The sun passes over 15 degrees of longitude in 1 hour. In what time will it pass over 185 degrees ? 19. John, who is 16 years old, is 4 times as old as James; Henry is 12 times as old as James : how old is Henry ? ADVANCED ARITHMETIC. 181 20. If 96 pounds of sea-water contains 3 pounds of salt, how many pounds of salt in 16 pounds of sea-water ? 21. If the interest on $500 for a certain time at a certain rate is $25, what is the interest on $125 for the same time at the same rate ? 22. 15 times A's money is $720. How much has B, who has 5 times as much as A ? 23. The wages of 3 men for 2 days is $9. Find the wages of 12 men for 8 days. 2Jf.. If 20 men in 15 days build 5 miles of fence, how many miles can 5 men build in 45 days ? 80. Percentage Solutions. — In arithmetic, the term Per Cent means liundredtli or liundredtlis. The sign, %, is used for the words per cent. l%=Tb; 12i%=f|*; and 100%=fft«. You should not think of percentage as something new in principle ; it is only a new form of expressing hundredths. Principle: 100% of any number is all of it. To aid in shortening percentage solutions the following ta- bles are given, and should be committed to memory: 2%=y5o of 100% I2y2%=y8 of 100% 4%=y25 of 100% i6%%=y6 of 100% 5%=y2oof 100% 20%=y5of 100% 6y4%=i/i6 of 100% 25%=y4 of 100% 81/3% =1/12 of 100% 33y3%=:y3 of 100% 10%-yio of 100% 50%=y2 of 100#7 . UNIVERSITY 182 ADVANCED ARITHMETIC. 100% = 50x2% -100% =8x121/2% 100% =25x4% 100% =6x16%% 100% =20x5% 100% =5x20% 100% = 16x6V4% 100%=4x25% 100% = 12x81/3% 100% =3x33%% 100% = 10xlO% 100%=2x50% i. Find 12% of 630 pounds. Solution : (1) 12^ of 630 lb. = ( ) lb.? (Question.; (2) IW of 630 lb. = 630 lb. (Basis.) T^ of (2) = (3) 1% of 630 lb. = 6.3 lb. 12 X (3) = (4) 12^ of 630 lb. = 75.6 lb. , answer. 2. Mr. Jones has $1600, of which 25% is in the bank. How much has he in the bank ? Solution: (1) 25^ of .$1600 = ( )? (Question.) (2) 100^ of $1600 = $1600. (Basis.) i of (2) = (3) 25^ of $1600 = $400, answer. Note. — By knowing (from the table) that 25% is I of 100^, we at once take I of the basis, which gives the required answer without having to pass first to unity. 3, I sell goods for Mr. Carr, amounting to $400, on which I charge a commission of 2%. Find my commission. Solution: (1) 2^ of sales = $( ) ? (Question.) (2) 100^ of sales = $400. ( Basis.) 6»5 of (2) = (3) 2% of sales = $8, answer. Note. — Commission for selling is always some % of the selling price. 4.. I bought goods for a merchant, amounting to $80. I charged 5% commission. Find my commission. Solution: (1) 5^ of purchase = $() ? (Question.) (2) 100^ of purchase = $80. ( Basis. ) ^ of (2) = (3) 5% of purchase = $4, answer. Note. — Commission for buying is always some % of the purchase price. ADVANCED ARITHMETIC. 188 5. An article cost $4.20, and was sold at a profit of 16f %. Find the gain. Solution: (1) .16|^ of cost = $( )? (Question.) (2) 100^ of cost = $4.20. ( Basis.) i of (2) = (3) 161^ of cost = $.70, answer. • Note. — The profit or loss is always some % of the cost price. 6. I bought goods for $800, and sold them at 88^% profit. Find the selling price. Solution : (1) 33 J^ of cost = $( ) ? (Question.) (2) 100^ of cost = $300. (Basis.) I of (2) = (3) 33J^ of cost = $100. (4) $300+$100 = $400, answer. Note. — The profit added to the cost gives the selling price. 7. I bought a horse for $72,* and sold him at a loss of 12^%. Find the selling price. Sohition: (1) 12|^ of cost = $( )? (Question.) (2) 100^ of cost = $72. ( Basis. ) i of (2) = (3) 121^ of cost = $9. (4) $72 -$9 = $63, answer. Note. — The loss subtracted from the cost gives the selling price. 8. VI % of a number is 68. Find the number. Solution: (1) 100^ of number = ( )? (Question.) (2) 17^ of number = 68. (Basis.) yJ, of (2) = (3) 1% of number = 4. 100 X (3) = (4) 100^ of number = 400, answer. Note. — When we cannot solve by a shorter plan, we can always go from what we have given to 1%; then from 1% to the required num- ber of %. 9. A lawyer is paid 5% for collecting a debt. If he gets $45, what is the amount of the debt ? Solution: (1) 100^ of debt = $( ) ? (Question.) (2) b% of debt = $45. ( Basis. ) 20 X (2) = (3) 100^ of debt = $900, answer. 184 ADVANCED ARITHMETIC. 10. My gain on a certain article, which I sold at a profit of 25%, is $.75. Find the cost price. Solution: (1) 100^ of cost = $( )? (Question.) (2) 25^ of cost = $.75. (Basis.) 4x(2) = (3) 100^ of cost = $3, answer. 11. My loss on a certain article, which I sold at a loss of 16|%, is $1.25. Find the cost price. Solution: (1) KW of cost = $( )? (Question.) (2) 161^ of cost = $1.25. (Basis.) 6 X (2) = (3) IW of cost = $7.50, answer. 12. In one book-case I have 50 books ; this is 12-^% of my library. How many books have I in my library ? Solution: (1) 100^ of library = ( ) books ? (Question.) (2) 121 ^ of library = 50 books. ( Basis. ) 8 X (2) = (3) 100^ of library = 400 books, answer. 13. 55 is 25% more than what number ? Solution : (1) 100^ of No. = ( ) ? (Question.) (2) 100^ of No. + 25^ of No. = 55. ( Basis.) (2) = (3) 125^ of No. =55. ^of (3) = (4) 25^ of No. = 11. 4 X (4) = (5) 100^ of No. = 44, answer. 14.. An article was sold at 50% above cost for $4.20. What did the article cost ? Solution : (1) 100^ of cost = $( ) ? (Question.) (2) lOO^of cost +50^ of cost = $4.20. (Basis.) (2) = (3) 150^ of cost = $4.20. ^ of (3) = (4) 50^ of cost = $1.40. 2 X (4) = (5) 100^ of cost = $2.80, answer. 15. 70 is 12^% less than what number ? Solution : (1) 100^ of No. = ( ) ? (Question.) (2) 100^ of No. -12|^ of No. = 70. (Basis.) (2) = (3) 871^ of No. = 70. i.of (3) = (4) 121^ of No. = 10. 8 X (4) = (5) 100^ of No. = 80, answer. ADVANCED ARITHMETIC. 135 16. $7 is what % of $20 ? Solution: (1) $? = ( )^ of $20? (Question.) (2) $20= 100^ of $20. (Basis.) aVof (2) = (3) $l = 5^of $20. 7 X (3) = (4) $7 = 35^ of $20, answer. 17. A. lawyer gets $85 commission for collecting a debt of $500. What % commission does he charge ? Solution : (1) $35 = ( )^ of debt ? (Question. ) (2) $500 = 100^ of debt. ( Basis. ) 5^^ of (2) = (3) $l = k^ofdebto 35 X (3)=(4) $35 = 1% of debt, answer. 18. I paid $52 for a buggy, and sold it so as to gain $13. Find the % of profit. Note. — As profit or loss is some per cent of the cost, in this problem we really want to know how many % $13 is of the cost, or $52. Solution: (1) $13 = ( )^ of cost ? (Question.) (2) $52 = 100^ of cost. (Basis . ) i of (2) = (3) $13 = 25^ of cost, answer. 19. Find 8% of 15% of $3500. Solution : (1) 15^ of $3500 = $( ) ? (Question.) (2) 100^ of 100^ of $3500 = $3500. (Basis.) I i^ of (2) = (3) 1% of $3500 = $35. 15 X (3) = (4) 15^ of $3500 = $525. (5) 8% of $525 = $( ) ? (Question.) (6) 100^ of $525 = $525. ( Basis. ) r^TT of (6) = (7) 1% of $525 = $5.25. 8 X (7) = (8) 8% of $525 = $42, answer. Plan.— The plan of this solution is to find 15% of $3500; then, 8% of that result. This gives 8% of 15^ of $3500. Another form: (1) 8^ of 15^ of $3500 = $( )? (Question.) (2) 100^ of $3500 = $3500. ( Basis. ) jU of (2) = (3) 1% of 100^ of $3500 = $35. jU of (3) = (4) 1% of 1% of $3500 = $.35. 8 X (4) = (5) 8^ of 1% of $3500 = $2.80. 15 X (5) = (6) 8% of lb% of $3500 = $42, answer. 136 ADVANCED ARITHMETIC. 20. 75 is how many % more than 60 ? Solution : (1) 75 = ( )^ of 60 ? (Question. ) (2) 60 = 100^ of 60. (Basis.) JffOf(2) = (3) 1 = -V^of60. 75x(3) = (4) 75 = 125^ of 60. .-. 75 is 25^ more than 60. Note.— Since 100^ of 60 is all of 60, 125^ of 60 is 25^ more than 60. 21. 40 is how many % less than 50 ? Solution: (1) 40 = ( )^ of 50? (Question.) (2) 50 = 100^ of 50. (Basis,) sVof (2) = (3) 1 = 2^ of 50. 40x(3) = (4) 40 = 80^ of 50. .-. 40 is 20^ less than 50. 22. A's money is 25% more than B's; then, B's money is how many % less than A's ? Solution : (1) 100^ of B's money = ( )^ of A's money ? (Question.) (2) 100^ of B's money + 25^ of B's money = 100^ of A's money. (2) = (3) 125^ of B's money = 100^ of A's money. T^ of (3) = (4) 1% of B's money = i% of A's money. 100x(4) = (5) 100^ of B's money = 80^ of A's money. .'. B's money is 20^ less than A's. 23. A's money is 20% less than B's; then, B's money is how many % more than A's ? Solution : (1) 100^ of B's money = ( )^ of A's money ? (Question.) (2) 100^ of B's money - 20^ of B's money = 100^ of A's money. ( Basis.) (2) = (3) 80^ of B's money = 100^ of A's money, i of (3) = (4) 20^ of B's money = 25^ of A's money. 5x(4) = (5) 100^ of B's money = 125^ of A's money. .'. B's money is 25^ more than A's. ADVANCED ARITHMETIC. 137 EXERCISE LVI. 1. Find 6% of $800. 2. Find 12i% of $4000. 3. Find 20% of i of $1600. ^. 8% of 80% of $860 is how much ? 5. Find 40% of 60% of $i. ^. I owe A $640; B, $880; C, $720; D, $.25. How much will each get, if I can pay but 20%, or $.20 on the $1 ? 7. A certain tax levy is 1^%. Find A's taxes, if his prop- erty is worth $46420. 8. 9% of a debt is $189.72. What is the debt ? 9. $3200 is 16% of what? 10. $3200 is 40% of what ? 11. $75 is 88^% more than what number ? 12. $80 is 16|% less than what number ? 13. $250 is how many % more than $200 ? IJ^. $750 is how many % less than $900 ? 15. A's money is 12|% more than B's; then, B's money is how many % less than A's ? 16. A's money is 25% less than B's ; then, B's money is how many % more than A's ? 17. A has a flock of 1800 sheep. He. sells 15% to B, who sells 10% of his to C. After these sales how many has each ? 18. I drew from a bank $60, which was 15% of my deposit. What sum had I left in the bank ? 19. B sells a pair of shoes at 10% less than list price; if list price is 38^% above cost, and B sells them for $1.80, find the cost. 20. What is the face value of Government bonds which imy, at 6%, an income of $720 per annum ? 21. 1 bushel is how many per cent more than 3 pecks ? 22. 1 yard is how many per cent less than 1 meter ? Note.— 36 inches = 1 yard; 39.37 inches = 1 meter. 23. 1 meter is how many per cent more than 1 yard ? 188 ADVANCED ARITHMETIC. ^4. 1 pint is what per cent of 1 gallon ? 25. 5 months is what % of 9 months ? 26. 20% of 50% of $600 is what per cent of 80% of 30% of $400? Plan.— (1) Find 80^ of 3(¥ of $400; (2) find 20^ of 60^ of $600 ; (3) find what % the second result is of the first result. Study the definitions and principles (omitting the formulas and relations), and solve by the method here given all the problems in Profit and Loss, Trade Discount, Commission, Stocks and Bonds, Taxes, Duties, and Insurance. ( See Part II, pp. 251-280.) 3. THE PROPORTION METHOD. 81. Process. — The process of solving a problem by the 'proportion method consists in (1) stating a proportion according to the following principle, and (2) solving that proportion. Principle: In two equations of the same nature^ and depending upon the same condition, the numerical ratio between the first mem- bers is equal to the numerical ratio between the second members. HiliTTSTRATION. Suppose, (a) Cost of 1 book =$3, Then, 40 x (1) = (b) Cost of 40 books = $120. 30 X (1) = (c) Cost of 30 books = $90. 20 X (1) = (d) Cost of 20 books = $60. 15 X (1) = (e) Cost of 15 books = $45. 5 X (1) = (f ) Cost of 5 books = $15. Note. — The last five equations were all obtained from the first, and (1) are all of the same nature, having ''cost of books" for their first mem- bers and "^'s" for their second members; (2) they all depend upon the same condition — that 1 book costs $3. 1. Compare (b) (d). (b) Cost of 40 books = $120. (d) Cost of 20 books = $60. (1) 40:20 = 2. (2) 120:60 = 2. .-. 40: 20:: 120: 60. ADVANCED ARITHMETIC. 189 2. Compare (c) and (f). (c) Cost of 30 books = $90. (f) Cost of 5 books = $15. (1) 30:5 = 6. (2) 90:15 = 6. \ .-. 15:5::90:15. 3. Compare (e) and (f). (e) Cost of 15 books = $45. (f) Cost of 5 books = $15. (1) 15:5 = 3. (2)45:15 = 3. /. 15:5::45:15. When the parts of a problem are stated for solution, the equations formed are of the same nature and depend upon the same condition ; therefore, the above principle may be used in stating the proportion. After the proportion is stated it is solved by methods in Articles 68 and 69. EXAMPLES. 1. If 10 books cost $20, what will 85 books cost? Solution : (1) Cost of 35 books = $( ) ? (Question. ) (2) Cost of 10 books = $20. ( Basis.) (3) 35 : 10 : : ( ) : 20 ? ( Proportion. ) (4)?#=70. .*. the required number is $70. 2. If I can buy 15 hogs for $75, how many can I buy for $90 ? Solution : (1) $90 = cost of ( ) hogs ? (Question.) (2) $75 = cost of 15 hogs. ( Basis. ) (3)90:75::( ):15? (Proportion.) 18 .'. the required number is 18 hogs. 140 ADVANCED ARITHMETIC. 3. Reduce 8 bushels to pecks. Solution : (1) 8 bu. = ( ) pk.? (Question.) (2) lbu.=4pk. (Basis.) (3) 8:1::( ):4? (Proportion.) (4) ^ = 32. .*. the required number is 32 pecks. 4-. If 30 acres of laud cost $450, how much was that per acre ? Solution : (1) Cost of 1 acre = $( ) ? (Question.) (2) Cost of 30 acres = $450. ( Basis. ) (3) 1:30::( ):450? (Proportion.) (4)i#=15. .*. the required answer is $15. 5. If 50 shares of stock cost $4000, what will 78 shares cost ? Solution: (!) Price of 78 shares = $( )? (Question.) (2) Price of 50 shares = $4000. ( Basis.) (3) 78 : 50 : : ( ) : 4000 ? ( Proportion. ) (4)18^ = 6240. oO .-.78 shares will cost $6240. 6. A tract of land 40 rods long, 20 rods wide, contains 5 acres. What must be the length of a tract 60 rods wide to con- tain 80 acres ? 1st Solution: (1) Length of a tract of 30 A., 60rd. w. = ( )rd.? (Question.) (2) Length of a tract of 5 A. , 20 rd. w. = 40 rd. ( Basis. ) (3)g:^::( ):40? (Proportion.) .*. the required length is 80 rods. "30 " 5 Note. — — „ and ^,, do not in fact truly represent the numerical values of the two lengths considered in this problem. The "30" and ADVANCED ARITHMETIC. 141 " 5 " are acresj while the " 60 " and " 20 " are rods. There are 160 square J • ,30x160 -5x160 ,, ,. 4.1, ^ rods in an acre, and — ^t. — and — ^^ — would represent the true numer- oO 20 ical length. But, in working out the proportion, one of these 160's would cancel the other, and they may therefore be omitted altogether. It is not necessary in the proportion method that the blank term should always appear alone in the fight member of the question, or in the right member at all for that matter ; it may fall anywhere in the equation, but care must always be taken to arrange the question and basis in the same order. Number 6 may be solved as follows : 2d Solution : (1) Area of a tract ( ) rd. 1., 60 rd. w. = 30 A.? (Question.) (2) Area of a tract 40 rd. L, 20 rd. w. = 5 A. (Basis.) (8) ( )x60:40x20::30:5? (Proportion.) 40X20X30 ^ ' 60x5 .•. the required length is 80 rods. 7. A tract of land 40 rods long, 20 rods wide, is worth $8(X). Find the width of a tract 50 rods long that is worth $1500. Solution : (1) Value of a tract 50 rd. 1., ( ) rd. w. = $1500 ? (Question.) (2) Value of a tract 40 rd. 1., 20 rd. w. = $800. ( Basis.) (3) 50 X ( ) : 40 X 20 : : 1500 : 800 ? ( Proportion ,) . . 40x20x1500 _ ^^^ 50X800 "^^- .*. the required width is 30 rods. 8. Goods that cost me $820, I sell for $416. How should I mark goods that cost me $40 to make a proportional profit in selling ? Solution: (1) Selling price of $40 worth of goods = $() ? (Question.) (2) Selling price of $320 worth of goods = $416. ( Basis. ) (3) 40:320::( ):416? (Proportion.) (4) '-^^ = 52. 320 .*. the goods should be marked to sell for $52. 142 ADVANCED ARITHMETIC. 9. I sell 40 pounds of sugar, 16 ounces to the pound, for $8.20. For how much should I sell 40 pounds, 15 ounces to the pound ? Solution : (1) Value of 40 lb., 15 oz. to the lb. = $( ) ? (Question.) (2) Value of 40 lb. , 16 oz. to the lb. = $3.20. ( Basis.) (3) 40 X 15 : 40 X 16 : : ( ) : 3.2 ? ( Proportion.) ,,, 40x15x3.2 „ (4) =3. ^ ' 40x16 .'. the required value is $8. Note. — When the same factor occurs in each part of the problem, it has no effect upon the result. If the 40's were omitted in the above problem, would that make any change in the result ? 10. A grocer sells a quantity of sugar by a false weight of 15 ounces to the pound for $8.20. How much does he gain by the cheat ? Solution : (1) Cost of *he number of lb. of 15 oz. each = $( ) ? (Question.) (2) Cost of same number of lb. of 16 oz. each = $3.20. ( Basis.) (3) 15 : 16 : ! ( ) : 3.2 ? ( Proportion.) (4) '^^=3. .'. the true selling price is $3, and his gain by cheating is 20^. Note. — The thief gets pay for true weight (16 oz. to the lb.). If he had delivered full weight, the sales would have been worth $3.20 ; what is the amount worth which he does deliver ? 11. A lawyer who collects for 5%, gets $84.60 for collecting a debt. Find the amount of the debt. Solution: (1) 100^ of debt = $( )? (Question.) (2) 5% of debt = $34.50. ( Basis. ) (3) 100:5:: (): 34.5? (Proportion.) (4) ^^?^=m. 5 .*. the amount of the debt is $690. 12. A's property is assessed at $8800. What is his tax at 96/ on the $100 ? ADVANCED ARITHMETIC. 143 Solution: (1) Tax on $3800 = $( ) ? (Question.) (2) Tax on $100 = $.96. (Basis.) (3) 3800 : 100: : ( ) : .96 ? ( Proportion.) (4) ??^-^^ = 36.48. ■ 100 .'. A's tax is $36.48. 13. B owns $2500 of the capital stock of a $200000 stock company. The company has $14000 for distribution among its stockholders. How much does B get ? Solution: (1) The dividend on $2500 = $( ) ? (Question.) (2) The dividend on $200000 = $14000. (Basis.) (3) 2500 : 200000 : t ( ) : 14000 ? ( Proportion. ) ,,, 2500x14000 ,„^ (4) = 170. ^ ' 200000 .-. B's dividend is $175. 14.. A certain body of soldiers, standing 125 in rank and 60 in file, change the file to 75 : what is the change in rank ? Solution : (1) Length of a certain body of men 75 men w. =( ) men ? (Question.) (2) Length of same body of men 60 men w. = 125 men. ( Basis.) No. No. ^2g ^ Proportion .) ^ ' 75 60 ^ NO.X125X60 ^ ' 75xNo. .'. the rank is changed to 100 men. Note. — If there were two different numbers of men, the amount of each would have to be expressed; but as it is, the same '^No." (what- ever its size) falls both above and below, and does not affect the result. 15. If 800 reams of paper are consumed in printing 88400 volumes of a 160-page book, octavo size, how many reams of paper will be required in printing 27000 volumes of a 820-page book, duodecimo size ? Note. — In making a book octavo size each sheet is folded into 8 leaves, making 16 pages ; in a duodecimo size each sheet is folded into 12 leaves, making 24 pages. 144 ADVANCED ARITHMETIC, Solution : (1) Amt. of paper reqd. for 27000 vol. of 320 pp. each = ( ) reams of 24 pp. to the sheet ? (Question.) (2) Amt. of paper reqd. for 38400 vol. of 160 pp. each = 800 reams of 16 pp. to the sheet. ( Basis.) (3) 27000 X 320 : 38400 x 160 : : ( ) x 24 : 800 x 16 ? ( Proportion .) ,^. 27000x320x800x16 ->^ . (4) = 7oU. ' 38400x160x24 .*. 750 reams will be required. 16. What rate is required for $600 to produce $192 interest in 4 years ? Solution : (1) Int. on $600 for 4 yr. at ( )^ = $192 ? (Question.) (2) Int. on $1 for 1 yr. at 1^ = $.01. (Basis.) (3) 600x4x( ): 1x1x1:: 192:. 01? (Proportion.) 1x1x1x192 ^19200^ 600X4X.01 ~ 2400 * .•. the reqd. rate is 8^. 17. What principal will be required to bear $225 at 9% in 3J years ? Solution : (1) Int. on $( ) for 3^ yr. at 9^ = $225 ? (Question.) (2) Int. on $1 for 1 yr. at 1^ = $.01, ( Basis.) (3) ( )x3ix9: 1x1x1:: 225 :.01? (Proportion.) (4) lxlxlx225^y,Q^ 3^x9x.01 .*. the required principal is $750. 18. If 8 men mow 36 acres in 9 days of 9 hours each, how many men can mow 48 acres in 12 days of 12 hours each ? Solution : (!) Amt. mowed by ( )men in 12 da. of 12 hr. each = 48 A. ? (Question.) (2) Amt. mowed by 8 men in 9 da. of 9 hr. each = 36 A. (Basis.) (3) ( )xl2xl2:8x9x9::48:36? (Proportion.) , . 8x9x9x48 _ 12x12x36 ~ .'. the required answer is 6 men. ADVANCED ARITHMETIC. 145 19. A garrison of men have food to last 9 months, giving each man 1 pound 2 ounces per day. What should be the daily allowance to make the same food last 1 year 8 months ? Note.— (1) 1 lb. 2 oz.'=L8 oz. ; 1 yr. 8 mo. = 20 mo. Solution : W I in 20^Z' iTn ol Srh'^pTr da. | ^ «-*-" ">» ' (Q"-«-) (2) 1 in /r^^tTo'i'i^^lTr da. | — -"unt. (Basis.) (3) 20 X ( ) : 9 X 18 : : Amt. : Amt. ? (Proportion.) ... 9 X 18 X amt. „ , 20 X amt. .*. the required answer is 8^^^ oz. SO. If 50 men in 10 days of 9 hours each, build a wall 250 yards long, 8 feet high, 3 feet thick, how many hours per day must 75 men work to build a wall 650 yards long, 9 feet high, 4 feet thick, in 80 days ? {Amt. built by 75 men) f * ,„oii acn ^;i i working 30 da. of [ = \ ^.^ , f f/^i^V ' /n„^cf ^^ ()hr.each J |9 ft. h., 4 ft. th? (Question.; {Amt. built by 50 men ") f » „.„n oka ^a i ^Z''.7.r-' hl^a'^^TA!- (Basis.) (3) 75x30x( ):50xlOx9::650x9x4:250x8x3? (Proportion.) . . 50xlOx9x650x9x4 _ 75x30x250x8x3 ~ " .•. the required answer is 7.8 hr. EXERCISE LVII. 1. Find the cost of 150 bushels of apples at $.65 per bushel. 2. If 12 tons of hay can be bought for $36, how much must be paid for 50 tons ? 3. A farmer sows 6 bushels of grain on 4f acres. At that rate, how many bushels will be needed to sow 3.6 acres ? 4.. If it requires 42 yd. of carpet | yd. wide to cover a floor, how many yards of carpet 1 y-d. wide will be needed ? 146 ADVANCED ARITHMETIC. 5. In preparing a certain mixture, 12 gills of water were mixed with 7 gills of other ingredients. How many gills of other ingredients should be mixed with 42 gills of water ? 6. 15% of a debt is $16.50. Find 80% of it. 7. A grocer uses false weight whereby Ib^ oz. is sold for a pound. What is the true value of groceries which he sells for $320? 8. If 20 yd. of cloth 1.5 yd. wide sell for $15, how wide must 51 yd. of the same kind of cloth be to sell for $42.50 ? 9. What are a servant's wages for 3 weeks 5 days, at $1.75 per week ? 10. A body of soldiers are 42 in rank and 24 in file. If they were changed to 36 in rank, how many in file would they be ? 11. An equatorial degree is 865000 feet. How many feet in 75° 24'', measured on the equator ? 12. If a troy pound of English standard silver is worth £3^, what is 1 lb. av. worth ? IS. If coffee which cost $225 is now worth $318.75, what was the cost of coffee now worth $1285.20 ? IJ^. A bin 7 ft. long, 2^ ft. wide, 2 ft. high, contains 28 bu. of corn. How deep must one be which is 18 ft. long, 4f ft. wide, to contain 120 bushels ? 15. If 150000 bricks are used in building a house whose walls average 1^ ft. thick, 80 ft. high, and all together are 216 ft. long,/ how many bricks will be required to build a wall 2 ft. thick, 24 ft. high, and 324 ft. long ? 16. It requires 800 reams of paper to print 28800 volumes of a duodecimo book of 320 pages to the volume. How many reams of paper will be required to print 24000 volumes of an octavo book of 520 pages to the volume ? 17. The premium on a draft for $4320 is $21.60: how large a draft can I buy at a premium of $32.10 ? 18. The rate of premium as in No. 17. What will be the premium on a draft for $1284.20 ? ADVANCED ARITHMETIC. 147 19. A property worth $2540 rents for $139.70: at that rate, what should property valued at $4000 rent for ? 20. At the rate in No. 19, what is the value of property which rentsfor $192.70? ; 21. What is the tax on $3450 worth of property at $1.60 on the $100 ? 22. If A charges 5/ on the dollar for collecting a note and receives as his commission $24.50, what is the face of the note ? 23. A man failing is able to pay only $65 on the $100 of his debts. What does E get, if this man owes him $824 ? 24.. A stock company whose capital is $4256000 is ready to distribute $148960 among its stockholders. Smith gets a divi- dend of $157.50: how much stock has he ? 25. If 16 men build 18 rods of fence in 12 days, how many men can build 72 rods in 8 days ? 26. If $100 gain $8 in 12 months, how many dollars will gain $144 in 4 months ? 27. If 6 men spend $150 in 8 months, how much will 15 men spend in 20 months ? 28. If 180 men in 6 days of 10 hours each dig a trench 200 yd. long, 3 yd. wide, 2 yd. deep, in how many days of 8 hr. each can 100 men dig a trench 180 yd. long, 4 yd. wide, 3 yd. deep ? 29. If a regiment of 840 men has food for 60 days, how many days should the same food last a garrison of 1260 men ? 30. If a pipe discharging 4 gallons per minute fill a cistern in 2 hours, in how many minutes may the cistern be filled by a pipe discharging 9 gallons per minute ? 31. If the use of $1500 for 3 yr. 8 mo. 25 da. is worth $336.25, what is the use of $100 for 1 yr. worth ? 32. If 4 horses draw a car 9 miles per hour, how many miles per hour will an engine of 150 horse-power draw a train of 12 such cars, adding its own weight, 3 cars ? PART II. I. STUDY OF NUMBERS. A. DENOMINATE NUMBERS. 82. Definitions. — A denominate number repre- sents quantity as composed of units of a particular denomina- tion or kind. As, 50 bushels, 30 minutes, 8 pounds. Most magnitudes may be expressed in units of two or more sizes. Thus, time may be expressed in seconds, minutes, hours, etc. The process of determining the number of units of one size in a number expressed in units of another size, is called re- duction. As, (1) To find how many quarts in 10 gallons; or, (2) To find how many gallons in 40 quarts. When the change is from a larger (higher) to a smaller (lower) unit, as in (1), the reduction is descending; when from a smaller to a larger unit, as in (2), the reduction is as- cending. In order to change or reduce a number from units of one size to units of another, the relative size of these units must be known. The equation which states this relation is the basis of solution. These equations are given in the tables. The plan and process of solution are the same as learned in Part I. (148) ADVANCED ARITHMETIC. 149 There are two systems of measurement in use in the United States, The English System and The French System. The English System, originated by the English, has come into common use among the people of the United States for all ordinary measurements and computations ; except the Eng- lish units of value, which are not used in the United States. The French System, originated by the French, is used in the United States principally by scientists in making scientific measurements and computations. I. ENQIilSH SYSTEM. 83. Linear Measures. — A Liine is that which has only one dimension, length. The linear measures, or units, are used in measuring heights, lengths, widths, distances, etc. There are two tables, the Conimon Ijinear Table and the Surveyors' Linear Table. COMMON lilNEAR TABIiB. 1 yard (yd.) =3 feet (ft.). 1 ft. = 12 inclies (in.). SURVEYORS' lilNEAR TABLE. 1 mile (mi.) =80 cliains (ch.). ( 4 rods (rd.), or 1 cliain= •< 66 ft., or ( 100 links (1.). 1 rod= I ^^'^' ^t" ^^ I 25 I. Note 1. — A link is 7.92 inches long. It is seldom, if ever, necessary in practice to reduce links to inches or inches to links. Note 2. — In ordinary land-surveying, surveyors use a chain 100 links, or 66 feet, long, — sometimes called Gunter's Chain. Civil engineers gen- erally use a ''steel tape," 100 feet long. 150 ADVANCED ARITHMETIC. EXAMPIiES. 1. Reduce 5 yd. 2 ft. 9 in. to inches. Plan.— (1) Reduce the yards to feet, and add the 2 ft. (2) Reduce the feet to inches, and add the 9 in. Each step contains one problem and one addition. Solution : (1) 5 yd. = ( ) ft.? (Question.) (2) 1yd. = 3 ft. (Basis.) 5x(2) = (3) 5 yd. = 15 ft. (4) 15 ft. +2 ft. = 17 ft. (5) 17ft. = ( )in.? (Question.) (6) 1ft. = 12 in. (Basis.) 17x(6) = (7) 17 ft. =204 in. (8) 204 in. + 9 in . = 213 in . , answer. 2. Reduce 2 mi. 20 ch. 74 1. to links. KoTE. — If, in the judgment of the teacher, the solution is sufficiently clear to the pupil without stating the ^'Question" each time, the '^Ques- tions' may be omitted. Thus, Solution: (1) 1 mi. = 80 ch. (Basis.) 2x(l) = (2) 2mi. = 160ch. (3) 160ch.+20ch. =180ch. (4) lch. = 1001. (Basis.) 180 X (4) = (5) 180 ch. = 180001. (6) 18000 1. + 74 1. = 18074 1., answer. 3. Prepare the surveyors' linear table for use in reduction ascending. Note. — In reduction descending, the order of the process is from higher to lower units, and the larger unit in each equation is on the left. In reduction ascending, this order is reversed, and the smaller unit is placed on the left. Thus, 1 ' 1 1 25 1 1 ^"•= 7:921- =r98i- 1 1 _ i xiiy ch. ADVANCED ARITHMETIC. 151 lrd. = ich. - 1 ch. = ^ mi. If.. Reduce 24848 1. to higher units. Solution: (1) 1 1. =T^Ty ch. (Basis.) 24348 X (1) = (2) 24348 1. = HU- ch. = 243 ch. +48 1. (3) 1 ch. =^ mi. (Basis.) 243 X (3) = (4) 243ch. = 3mi. + 3ch. .-. 243481. =3 mi. 3 ch. 48 1. Another Solution : (1) 11. = ^V rd. ( Basis.) 24348 X (1) = (2) 24348 1. = 973 rd. + 23 1. (3) lrd. = ich. (Basis.) 973 X (3) = (4) 973 rd. = 243 ch. + 1 rd. (5)lch. = ^mi. (Basis.) 243x(5) = (6) 243ch. = 3mi. + 3ch. .-. 24348 1. =3 mi. 3 ch. 1 rd. 23 1. Note. — In the first solution, equation (2), ^U- ch. is simplified by di- viding the 24348 by 100. The quotient is 243 ch. , the remainder is yVif ch. But 1^ ch. is 1 1., therefore iVir ch. =48 1. Thus, the quotient represents the number of chains, and the remainder the number of links. 5. Reduce 444 rd. to miles. Solution: (l)lrd.=ich. (Basis.) 444 X (1) = (2) 444 rd. = 111 ch. (3)lch. = ^mi. (Basis.) Ill X (3) = (4) 111 ch. = l|^ mi., or, 1.3875 mi., answer. Note. — In reducing to ^^ higher units," the remainders are left as inte- gers, and the quotient only is reduced higher. In reducing to a particu- lar denomination, as in No. 5, the exact quotient is obtained each time and all carried to the required denomination. In reducing fractions to 'Hower units,''' when the result is a mixed number, leave the integral part, and reduce only the fraction lower. 152 ADVANCED ARITHMETIC. 6. Reduce f yd. i ft. to inches. Solution: (1)1 yd. =3 ft. (Basis.) |of(l) = (2) I yd. =2 ft. (3) 2ft. + ift. = 2ift. (4) 1ft. = 12 in. (Basis.) 2i X (4) = (5) 24 ft. = 27 in., answer. 7. Reduce ^ it. to the fraction of a mile. Solution: (l)lft. =^ch, (Basis.) ^of(l) = (2) x\ft. = TTrWch. (3) 1 ch. = gi(5 mi. (Basis.) o^ffe of (3) = (4) xTrVe ch. = lefgr mi., answer. 8. Reduce .32 rd. to the decimal of a mile. Solution : (1) 1 rd. = i ch. = .25 ch. .32of (1) = (2) .32rd. = .08ch. (3) 1 ch. =^xr mi. = .0125 mi. .08of (3) = (4) .08 ch. = .001 mi., answer. Note. — When a decimal result is required, it may be obtained as above by expressing all fractions as decimals ; or the solution may be expressed in common fractions and the result only reduced to a decimal. 9. Find in integral units the value of | mi. minus ^ rd. Solution: (1) 1 mi. = 80 ch. (Basis.) f of (1) = (2) f mi. = 17|ch. (3) lch.=4rd. (Basis.) iot (3) = (4) |ch. = -2/rd. (5) -%^ rd. - j\ rd. = %V rd. - U rd. = W rd. = 2^ rd. (6) 1 rd. = 161 ft. =-3/ ft. ( Basis.) if of (6) = (7) i|rd. = ex-¥-ft. = 12Ht. (8) 1ft. = 12 in. (Basis.) iof (8) = (9) i ft. = 2 in. .-. f mi. - /x rd. =17 ch. 2 rd. 12 ft. 2 in. ADVANCED ARITHMETIC. 153 10. Reduce ^ ft. | rd. | ch. to the fraction of a mile. Solution : (1) 1 ft. = i^ rd. ( Basis.) (2) Jft. = s^jrd. ,k. = 8 quarts (qt.). 1 qt. = 2 pints (pt.). Note. — 1 bushel contains 2150.4+ cu. in. Fruits, vegetables, seeds and the like are bought and sold by the bushel ; but the amount of the bushel is often expressed by weight rather than size. .(See p. 161.) 158 ADVANCED ARITHMETIC. EXERCISE LXI. 1. Reduce 20 bu. to pints. 2. Reduce 2 bu. 2 pk. to quarts. 3. Reduce 7 bu. 8 pk. 7 qt. to pints. 4. Reduce 9 bu. 1 pk. 4 qt. 1 pt. to pints. 5. Reduce 768 pt. to bushels. 6. Reduce 679 pt. to higher units. 7. Reduce f bu. to pints. 8. Reduce -J^ bu. to lower units. 9. |bu.+|pk.+|qt.+ipt. = ( )pt.? 10. ibu.+}pk.+|qt.+ipt. = ( )bu.? 11. 5bu.-ipk.-Hqt.-Hpt. = ( )pt.? 12. ibu.-ipk.-|.375qt.-lipt. = ( )bu.? 13. Find the value of 4 bu. 8 pk. of apples at 20/ per peck. IJ/.. A grocer buys ^ bu. of cherries ; he then sells 1 bu. 8 pk. of them; again, he buys 1 bu. 1 pk. 8 qt„ of cherries; finally he sells all the cherries he has at 6/ per quart. Find the value of the last sale. There are two tables of Liquid Measures in use, (1) the Com- mon Liquid Measures, used in measuring such liquids as water and milk, and (2) the Apothecaries^ Liquid Measures^ used by apothecaries in measuring liquid medicines : COMMON TABLE. 1 liog"slieacl (]i]id.)=2 barrels (bbl.)o 1 bbl. = 311/2 gallons (g-al.)o 1 gal. =4 qt. 1 qt. = 2 pt. 1 pt. = 4 gills (gi.). KoTE. — 1 gallon contains 231 cu. in. The dry measure quart contains 9.45 cu. in. more than the liquid quaint. ADVANCED ARITHMETIC. 159 APOTHECAKTES' TABLE. 1 gallon (*Coiig.) = 8 pints (*0.) 1 O. = 10 fluid ounces (fB)o 1 fS = 8 fluid drams (f5). 1 f5 = 60 minims (m.). EXERCISE LXII, 1, Reduce 1 hhd. to gills. 2. Reduce 2 hhd. 1 bbl. 21 gal. to quarts. S, Reduce ^ gal. -J qt. to gills. ^. Reduce 1008 gi. to barrels. 5. Reduce 1008 pt. to higher units. 6. Reduce 5 Cong, to minims. 7. Reduce 5 O. 2 f 5 3 f 5 to minims. 8. Reduce 1260 m. to higher units. 9. A physician wishes to prepare a 5% solution of carbolic acid. How much water and how much acid must be used to make 2 fluid ounces ? Note. — h% of the 2f5 is carbolic acid, and the rest is water. 10. What will 1 bbl. of syrup sell for, at 80/ per gallon? 87. Measures of Mass. — The Mass of a body is the amount of matter it contains. The Weig-lit is the measure of the attraction between that body and the earth. Weights are used to measure the mass of a body. Note. — To be accurate in comparing the masses of bodies by their weights, the weights must be taken at the same altitude, in the same latitude, and under the same conditions. But for ordinary purposes, this accuracy is not observed. * The Latin words are Congius and Octavua. 160 ADVANCED ARITHMETIC. There are three tables of weights : (1) The table of Avoir- dupois Weights, used in weighing all ordinary articles, such as groceries, meats, live stock, etc. 1 ton. (T.)=20 hundred weiglit (cwt.). 1 c\v^t. = 100 i^ounds (lb,)e 1 lb. = 16 ounces (oz.). (2) The table of Troy Weights, used in jeivelry stores ^ and at mints, when the Government weighs gold and silver for making money. TABIiE. 1 pound (lb.) = 12 ounces (oz.). 1 oz. = 20 penny^veiglits (pwt.). 1 oz. = 20 penny^veiglits 1 pwt. = 24 grains (gr.). (3) The table of Apothecaries' Weights, used by druggists in filling prescriptions. TABLE. 1 pound (lb. = 12 ounces (.^)a 1 S = 8 drams (5)- 1 5 = 3 scruples (9) 1 9=20 grains (gr,). Note. — ^The grain, ounce, and pound of the troy weights are identical with the grain, ounce, and pound, respectively, of the apothecaries' weights. The troy or apothecaries' pound contains 5760 gr., but the av- oirdupois pound contains 7000 gr. Grain and vegetables are often bought and sold by weight- ADVANCED ARITHMETIC. 161 bushels instead of the measured bushels. In the majority of States the bhshel is of the following weight : ^ '^ . TABLE. 1 bu. of wheat =60 lb. 1 bu. of corii=56 lb. 1 bu. of oats = 32 lb. 1 bu. of Irish potatoes = 60 lb. 1 bu. of sweet j)otatoes=55 lb. Note. — The pounds of the above table are avoirdupois pounds. These equivalents are not the same in all States, and the teacher should teach the pupil the table corrected for his own State. EXERCISE LXIII. 1. Reduce 1 T. 3 cwt. 4 lb. to ounces. 2. Reduce 7000 oz. to higher units. 3. Reduce 1 lb. 45 55 23 17 gr. to grains. 4. Reduce 6743 gr. troy to higher units. 6. Reduce 6743 gr. apothecaries' to higher units. 6. Reduce 1 lb. avoirdupois to highest integral units of the troy table. 7. Reduce 2 lb. apothecaries' to highest integral units of the avoirdupois table. 8. A man sells 9600 lb. of Irish potatoes at 70/ per bu. How much do the potatoes bring ? 9. A silver dollar weighs 17 pwt. 4^ gr. What is the weight of $80 in silver ? 10. A druggist buys lactopeptine at $8 per lb. and sells it at 12|/ per 5. What is his per cent of gain ? 11. How many bales of 70 lb. each in 14 T. 17 cwt. 50 lb. of hay? 12. A man sold 3 loads of corn of 2352 lb. each. How many bushels did he sell ? 162 ADVANCED ARITHMETIC. 88 • Measures of Time. — These measures are used in giving the time of day, in Longitude and Time problems, and in computing the interest on notes and bills. 1 year (jt.) = 12 montlis (ino.)« 1 mo. = 30 days (da.), 1 da. =24 hours (lir.). 1 lir. = 60 minutes (min.). 1 min. = 60 seconds (sec). Calendar Months: January, 81 days. July, 31 days. ♦February, 28 days. August, 81 days. March, 81 days. September, 80 days. April, 80 days. October, 81 days. May, 81 days. November, 80 days. June, 80 days. December, 81 days. Note. — The above data give three different lengths to the year. (1) From the table, 1 yr. = 12 mo. =360 da. This may be called the common interest year. The custom in the majority of the States is to use 360 da. for a year. (2) By counting the number of days of the calendar months, the length of the year is found to be 365 days. This is called the com- mon year. (3) Leap Year. — The average length of a year (a complete revolution of the earth around the sun), expressed in mean solar time (sun time) is 365 days, 5 hours, 48 minutes, 47.8 seconds. The Julian Calendar called the year 365^ days. For ordinary purposes it is best to use whole days; so the J^ day each year amounted in 4 years to 1 day. In this way, the Julian Calendar had 3 common years of 365 days each, then 1 leap year of 366 days. The extra day was added to February. This calendar, by counting 3653^ days for a year, used 11 minutes 12.2 seconds too much. The Gregorian Cal- endar corrected this error by omitting the leap year on centennial years, except those which are multiples of 400. The act of Parlia- * February has 29 days in leap years. ADVANCED ARITHMETIC. 163 men t which changed from the Julian Calendar (old style) to the Gre- gorian Calendar (new style) dropped 11 days out of the month of Sep- tember, 1752, making September 3d, old style, September 14th, new style. The civilized nations, except Russia, have adopted the Gregorian Cal- endar. Remember that, when the number of a year is divisible by 4, it is a leap year ; except a centennial year, whose number must be divisible by 400 in order to be a leap year. There are 52 weeks 1 day in a common year, and 52 weeks 2 days in a leap year. A week has 7 days, named Sunday, Mon- day, Tuesday, Wednesday, Thursday, Friday, Saturday. EXERCISE LXIV, 1. Reduce 865 da. 5 hr. 48 min. 47.8 sec. to seconds. 2. Reduce 31556986 sec. to higher units. 3. Name the months in order and give the number of days in each. Jf.. Learn the number of each month, as January, 1st ; Sep- tember, 9th; June, 6th; etc. 5. Find which of the following are leap years : 1756, 1885, 1800, 1892, 1876, 1900, 1904, 1906, 1952, 2000. 6. Find the number of days from June 10, 1895, to March 17, 1896. 7. Find the exact time from 10 o'clock a.m., Wednesday, June 12, 1901, to noon the following Monday. 8. Find the exact time 9170 minutes after 8 p. m., Saturday, June 22, 1901. 9. When was a note due if it was dated June 12, 1901, and due in 98 days ? 10. When was a note due if it was dated December 10, 1879, and due in 125 days ? 164 ADVANCED ARITHMETIC. 89. Measures of Value. — The money of a country is its circulating medium and adopted standard of value. The following is the table of English units of value, or denomina- tions of money. TABIiE. 1 pound (£)=20 sliillings (s.). 1 s. = 12 pence* (d.). 1 d.=4 farthings (far.). Note. — The value of the English pound (or sovereign) in United States money is $4.8665. EXERCISE XXV. 1. Reduce £5 to pence. 2. Reduce 7548 far. to £. 3. Reduce 8424 far. to higher units. Jf.. A bill for £200 12 s. is worth how many dollars ? 5. A bill for $4300 is worth how much in English money ? 90. Review and Rapid Drill Work.— -Heretofore you have been required to consider each step in the process of reduction of denominate numbers as a problem, and to solve it as such. Being now familiar with the nature of reduction, you should hereafter give your attention to speed and accuracy in obtain- ing the result. UIRECTIONS. For Reduction Descending : Multiply hy the number of units of the next lower denomination which makes one unit of the denom- ination to he reduced. For Reduction Ascending : Divide by the number of units of the given denomination , which makes one unit of the next higher de- nomination. Note. — Use the corresponding abstract numbers in this work. ♦The singular ot pence is penny. ADVANCED ARITHMETIC. 165 Reduce, orally — 1. 5 bu. to pk. 2. 6 qt. to pt. S. 8 pk. to pt. EXERCISE LXVI. 4.. 14 pt. to qt. 5. 48 pt. to pk. 6. 128 qt. to bu. 7, 48 in. to ft. 8, 20 ch. to 1. 9, 1 mi. to rd. J3. 12 sq. yd. to sq. ft. 14.. 3 sq. ft. to sq. in. J5. 1 mi. to sq. ch. W. 2 cu. yd. to cu. ft. 20. 1 cu. ft. to cu. in. 21, 1 bu. to cu. in. 10, 11. 12. 800 1. to ch. 198 ft. to ch. 72 in. to yd. 16. 20 A. to sq. rd. 17. 70 sq. ch. to A. 18. 480 sq. rd. to A. 22. 1 gal. to cu. in. 23. 462 cu. in. to gal. 24. 54 cu. ft. to cu. yd. 25. 1 day to min. 26. 1 hr. to sec. 27 , 1 yr. to mo. 31. 3 lb. av. to oz. 32. 3 lb. troy to oz. 33. 3 S to 9. 34. 4 pwt. to gr. 35. 5 5 to gr. 28. 50 min. to hr. 29. 6 hr. to da. 80. 3 mo. to yr. 36. 128 oz. av. to lb. 37. 120 oz. troy to lb. 38. 96 S to lb. 39. 249 to S. 4.0, 120 m. to f 5. EXAMPIjSSS. 1. Reduce 5 bu. to pints. Process: 5x4x8x2 = 320. .'. 5bu. = 320pt. Explanation : (1) Multiplying by 4 reduces to pecks ; by 8, to quarts; by 2, to pints. Therefore the required number 166 ADVANCED ARITHMETIC, is the continued product of 5, 4, 8, and 2, or 320. This mul- tiplying should all be done mentally, and no more figures should be used than are given above. 2. Reduce 7 cwt. 5 lb. 4 oz. to ounces. Process : A. B. (1) 100x7 + 5 = 705. ^ 7' (2) 16x705+4 = 11284, ^^^' 100 705 16 4230 705 4 .'o 7 cwt. 5 lb. 4 oz. = 11284 oz. 11284 Explanation : If the pupil can multiply by 16 (and he ought to), use form A ; if not, use form B. In By the 5 is added to the product of 7 and 100 before writing the result. 3. Reduce 1512 pt. to hogsheads. Process : A. B. 2)1512 3 4)756 j^^ 189 ' jm2^i_^. 2 '^xix^^x^ ""• 63)378(6 + 2 = 3 378 .'. 1512pt. = 3hhd. Explanation : Form ^ is a much-used form. Multiplying by 2 and dividing by 63 is the same as dividing by 31|. The author has some preference foi form B, when all or nearly all of the divisors are small. 4.. Reduce 87560 hr. to years. Process : 313 Wm ^313^ 24x30x;^ 72 '" 3 .-. 37560 hr. = 411 yr. ADVANCED ARITHMETIC. 167 5. Reduce 435 pt. (dry) to higher units. Process : 2 )435 8)217+1 .4)27 + 1 '6 + 3 .-. 435 pt. = 6 bu. 3 pk. 1 qt. 1 pt. 6. Express in integral units ^-f^ hhd. Process: (1) 7 x^_ _ m 60 - 6U- 21 (2) 7 x^? ^0x2 20 =W=sih (3) 27 x^ 10 = ?S=2,V (4) 7 y^ ;0 5 = 1|. .-. lb hhd ..=3 gal. 2qt. 1| pt. 7. Reduce .175 bu. to lower integral units Process : .175 4 .7PP 8 5.6 2 1.2 .-. .175 bu. = 5qt. 1.2 pt. 8. Reduce 8.75 lb. troy to lb. av. Process : 8.75x5760 7000 = 7^. .-. 8.75 lb. troy = 7Hb. av. Plan.— (1) Multiplying by 5760 reduces to gr., and (2) dividing by 7000 reduces to lb. av. 168 ADVANCED ARITHMETIC. EXERCISE LXVII. Reduce : 1. 2 mi. 20 ch. to inches. 7. 5 T. 4 cwt. to oz. 2. 8 sq. mi. 5 A. to sq. rd. 8. 1 T. 6 oz. to oz. 3. 7 cu. yd. 5 cu. ft. to cu. in. 9. 8J lb. troy to gr. 4. 7 bu. 5 qt. to pt. 10. 2 S 4 5 2 9 to gr. 5. 5^ gal. 1 pt. to gi. 11. 5 yr. 4 mo. 17 da. to da 6. 5 O. 1 f S to m. 12. £7 11 d. to far. Reduce to higher units : 13. 143759 in. 17. 840 pt. (liquid). IJ^. 73498 cu. in. 18. 842 pt. (dry). 15. 4375 gr. troy. 19. 546304 min. 16. 8470 A. 20. 4444 oz. av. Reduce to lowest denomination of the table : 21. yV yd. 25. I bu. f pk. 8 qt. 22. .05 yr. 26. .5 sq. rd. . 23. -iV rd. 27. i cu. yd. + .7 cu. ft. 2^. .34 hr. 28. .6 lb.+| pwt. Reduce to highest denomination of the table : 29. 5tV gr. troy. 33. 15 2 9- 30. 5pt. (dry). 3If.. 120 da. 31. lOOSsq. ch. 35. .75 qt. (liquid). 32. 649. cu. in. 36. 7 pt. + .8 qt. Reduce : 37. 5 lb. av. to lb. troy. J^O. 456 gal. to cu. in. 38. 7 lb. apoth. to lb. av. J^l. 840 bu. to gal. 39. 60 bu. to cu. in. J^. 576 gal. to bu. ADVANCED ARITHMETIC. 169 2. FRENCH SYSTEM. 91. Linear Measures. TABIiE. 1 Myriameter (Mm.) = 10 Kilometers (Km.). 1 Km. = 10 Hectomieters (Ilm.). 1 Hm. = 10 Deliameters (Dm.). 1 Dm. = 10 meters (m.). 1 m. = 10 decimeters (dm.). 1 dm. = 10 centimeters (cm.). 1 cm. = 10 millimeters (mm.). Note. — 1 meter- 39.31 inches. By means of this equivalent, any num- ber expressed in units ^of this table may be expressed in units of the English linear table, and vice versa. Reduction descending and ascending may be performed here just as in the English System ; but, owing to the fact that we are now dealing with a decimal system — a system in which 10 units of one denomination make one of the next higher — the process of reduction may be much shortened. The shorter processes are developed in the following examples : EXAMPLES. 1. Reduce 5 Hm. 4 Dm. 2 m. to m. Process : 5 Hm. 4 Dm. 2m. =542 m., result. Note. — When there are no denominations omitted between the high- est denomination given and the denomination required, the figures written in order side by side will be the abstract number correspond- ing to the required result. 2. Reduce 7 Mm. 9 Km. 4 Dm. to Dm. Process : 7 Mm, 9 Km. 4 Dm. = 7904 Dm., result. Note. — Put a in place of the denomination omitted. 3. Reduce 5 Km. to dm. Process : 5 Km. = 50000 dm., result. Note. — Put a in place of each denomination omitted, from Km. to dm. inclusive. 170 ADVANCED ARITHMETIC. If. Reduce 84579 dm. to higher units. Process : 34579 dm. = 3 Km. 4 Hm. 5 Dm. 7 m. 9 dm., result. Note. — This is just the reverse of the process used in reduction de- scending. It is plain, that by dividing successively by lO's, the succes- sive remainders would be 9, 7, 5, etc., and that the result would be just as given above. 5. Reduce 1605 mm. to higher units. Process : 1605 mm. = 1 m. 6 dm. 5 mm., result. Note.— The cm. is omitted in the result. 6. Reduce 5.046 Dm. to cm. Process : 5.046 Dm. =5046 cm., result. Note. — To reduce from Dm. to cm. , we must multiply by 10 three times. This is done by moving the decimal point three places to the right. 7. Reduce 54603.25 m. to Hm. Process : 54603.25 m. =546.0325 Hm. Note. — To divide by 10 twice, move the decimal point two places to the left. 8. Reduce 3 Hm. 6 Dm. to yards. Process : (!) 3 Hm. 6 Dm. = 360 m. (2) ?80^MZ = 393.7. .'.3 Hm. 6 Dm. =393.7 yd. Note. — Always reduce to meters^ and then to inches. 9. Reduce 218 yd. 2 ft. 2 in. to Dm. Process : (1) 218 yd. 2 ft. 2 in. = 7874 in. (2) '^'^ =20. 39.37x10 .-. 218 yd. 2 ft. 2 in. = 20 Dm. Note. — Always reduce to inches^ and then to meters. EXERCISE LXVIII. Reduce : 1. 5 Mm. to m. 6. 43675 dm. to Km. 2. 9 Km. 8 Dm. to dm. 7. 4867.3 m. to Hm. S. 7 Hm. to mm. 8. 4785 cm. to higher units. 4. 5 Mm. 7 Hm. to cm. 9. 54843 m. to higher units. 5. 8 m. 5 dm. to mm. 10. 4.003 Dm. to mm. ADVANCED ARITHMETIC. 171 11, 7 Mm. 8 Km. 6 Hm. 5 Dm. 4 m. to mm. 12. 3 Mm. 8 Hm. 7 Dm. 5 dm. 4 cm. to mm, IS. 86 Mm. to miles. U. 286220 ft. to Dm. 15, 640.0632 Dm.' to. inches. 92. Surface Measures. TABIiE. 1 sq. Min. = 100 sq. Km, 1 sq. Km. = 100 sq. Hm. 1 sq. HmL. = 100 sq. Dm. 1 sq. Dm. = 100 sq. m. 1 sq. m. = 100 sq. dm. 1 sq. dm. = 100 sq. cm. 1 sq. cm. = 100 sq. mm. Note. — A square Dekameter is also called an are. The are is the unit of land measure. A square meter is equal to 39.37x39.37 sq. in., or about 1550 sq. in. EXERCISE LXIX. Reduce : 1. 15 sq. Mm. to sq. m. • 2. 75 sq. Km. to sq. dm. 3. 4 sq. Dm. to sq. mm. 4-. 540000 sq. cm. to sq. m. 5. 89506000 sq. dm. to ares. 6. 48675056 sq. Dm. to higher units. 7. 520 sq. m. to sq. in. 8. 25 sq. Dm. 12 sq. m. to sq. in. 9. 17 sq. Km. to sq. yd. 10. 520 sq. rd. to sq. m. 11. 740 A. to ares. 12. 1 sq. mi. to sq. Hm. 172 ADVANCED ARITHMETIC. 93. Solid Measures. TABIiE. 1 cu. Mm. = 1000 cu. Km. 1 cu. Km. = 1000 cu. Hm. 1 cu. Hm. = 1000 cu. Dm. 1 cu. Dm. = 1000 cu. m. 1 cu. m. = 1000 cu. dm. 1 cu. dm. = 1000 cu. cm. 1 cu. cm. = 1000 cu. mm. Note.— A cubic meter is also called a stere. The stere is used in meas- uring wood. A cubic meter is equal to 39.37x39.37x39.37 cu. in., or about 60923.178 cu. in. EXJ]RCISE LXX. 1. 75 cu. Km. to cu. m. 2, 748 cu. Dm. to cu. cm. 5. 10 cu. m. to cu. ft. 4. 4856000000 cu. dm. to cu. Dm. 6. 8000000000 cu. m. to cu. Km. 6. 182769534 cu. ft. to cu. Dm. 7. How many cords of wood in a pile containing 500 cu. m.? 8. How many steres of wood in a pile containing 500 cu. yd.? 94. Measures of Capacity. TABIiE. 1 Myrialiter (Ml.) = 10 Kiloliters (Kl.). 1 Kl. = 10 Hectoliters (HI.). 1 HI. = 10 Deltaliters (Dl.). 1 Dl. = 10 liters (1.). 1 1=10 deciliters (dl.). 1 dl. = 10 centiliters (cl.). 1 cl. = 10 milliliters (ml.). Note, — The liter is equal in volume to a cubic centimeter (about 1.05 liquid quarts, or .9 dry quart). ADVANCED ARITHMETIC. 173 EXERCISE LXXI. Reduce : U 5 Kl, to 1. 7. 40000 cl. to HI. 2. 9 Ml. to DL '.. 8. 45675 dl. to higher units. 3. 7 HI. 5 1. to dl. ' 9. 7043.08 cl. to Dl. ^. 4 Kl. 4 HI. to cl. 10, 48 1. to qt. 5. 5 Dl. 8 1. to qt. 11. 25 gal. to 1. 6. 7 HI. to gal. 12. 75 bu. to Dl. 95 e Measures of Mass. TABLE. 1 Tonneau (T.)^IO quintals (Q.). 1 Q. = 10 Myriagrams (Mg.)» 1 Mg. = 10 Kilograms (Kg.)- 1 Kg. = 10 Ilektograms (Hg.). 1 Hg. = 10 Dekagrams (Dg.). 1 Dg. = 10 grams (g.). 1 g. = 10 decigrams (dg.). 1 dg. = 10 centigrams (eg.). 1 eg. = 10 milligrams (mg.). Note. — The gram is the weight of a cubic centimeter of distilled water (about 15.42 grains troy). EXERCISE LXXII. Reduce : 1. 6 T. to g. 7. 240 dg. to g. 2. 9 Q. to Dg. 8. 4380 eg. to Dg. 8. 7 Mg. to dg. 9. 5863 eg. to higher units. 4. 5 Dg. to mg. 10. 840 dg. to gr. 5. 50 Mg. to lb. troy. 11. 123.36 lb. troy to Mg. 6. 50 Mg. to lb. av, 12. 128.36 lb. av. to Mg. 174 ADVANCED ARITHMETIC. 96. Measures of Value. TABLE. 1 franc (fr.) = 10 declines (d.). 1 d. = 10 centimes (c.)» 1 c. = 10 Hilllimes (m.)- Note. — The franc is equal in value to 19.3i*. EXERCISE LXXIII. Reduce : 1. 5 fr. to m. 6. $5.79 to fr. 2. 15 fr. 7d. to m. 7. 45 fr. to $'s. 3. 50 fr. to d. 8, 8800 c. to /. 4. 240 c. to fr. 9. 5790 m. to mills. 5. 4834 m. to higher units. 10, 3860 dimes to d. ' 3. COMPOUND DENOMINATE NUMBERS. 97. Addition. EXAMPLES. 1. Add 1 bu. 2 pk. 4 qt. 1 pt. ; 5 bu. 1 pk. 2 qt. Process : Explanation : (1) Write the names bu. pk. qt. pt. of the successive units of the table 12 4 1 used, putting them in horizontal or- der, with the largest to the left. 6 3 6 1, result. (2) Write the addends below- each part in its place, as indicated by the table above. (3) Add as we have done heretofore, placing the sum ob- tained by adding each column below. Result, 6 bu. 3 pk. 6 qt. Ipt. 8gi. Add 2 gal. 3 qt. 1 pt. 8 gi. ; 7 gal. 1 pt. 8 gi. ; 8 gal. 2 qt. Explanation : (1) 3 gi. + 3 gi.+3 gi.=9 gi. But 9 gi.=2 pt. 1 gi. Write the 1 below, and carry the 2pt. 1Q o n 1 1^ (2) 2pt. (carried) + l pt. + l pt. 18 3 l,result. ^^^^ But4pt.=2qt. Write Process: gal . qt. pt. gi. 2 3 1 3 7 1 3 8 2 3 ADVANCED ARITHMETIC. 175 below, and carry the 2 qt. We use the to show that there are no pints left. (3) 2 qt. (carried ) + 2 qt. + 3 qt. = 7 qt. = 1 gal. 3 qt. Write the 3 qt. below, and carry the 1 gal. (4) 1 gal. (carried) + 8 gal. + 7 gal. +2 gal. = 18 gal. Write 18 below. Result, 18 gal. 3 qt. 1 gi. Note. — Teacher should point out the similarity between this and Ad- dition of Simple Numbers. 3. Add 5 yd. 2 ft. 8 in. ; 7 yd. 2 ft. 9 Id. ; 17 yd. 2 ft. 11 in. Explanation : (1) Sum 28 in. =2 ft. 4 in. Write the 4 below, and carry the 2. (2) Sum 8 ft. = 2 yd. 2 ft. Write the ^ (ft.) below, and carry the 2 (yd.). (3) Sum 31 yd. Write the 31 below. 31 2 4, result. Result, 31 yd. 2 ft. 4 in. EXERCISE LXXIV. Process : yd. ft. in. 5 2 8 7 2 9 17 2 11 Add: 1. hr. min. sec. 2. yr. mo. da. 3. yd. ft. in. 7 18 86 50 8 12 9 1 11 9 85 40 46 5 25 8 8 11 48 84 2 9 14 11 2 7 J/., cu. yd. cu. ft. cu. in. 5. sq. yd. sq. ft. sq. in. 6. T. cwt. lb. 120 17 540 12 78 7 5 40 242 9 1048 46 8 96 18 12 72 488 21 1486 52 5 100 50 15 00 868 421 87 7 77 18 84 7. lb. 5. 5. 9. gr. 8. hhd. bbl. gal. qt. pt. 5 7 7 2 14 2 1 25 8 1 11 11 5 1 18 4 80 2 1 7 9 2 15 5 1 11 1 Process : T. 15 6 cwt. lb. 6 12 10 8 oz. 5 14 176 ADVANCED ARITHMETIC. 98. Subtraction. EXAMPLES. i. From 15 T. 6 cwt. 12 lb. 5 oz. take 6 T. 10 cwt. 8 lb. 14 oz. Explanation : (1) 5 is smaller than 14. "We must take 1 lb. from the 12 lb. and add it to the 5 oz. 1 lb. +5 oz. = 21 oz. 21-14 = 7. 8 16 3 7, result. Write the 7 below. (2) There are only 11 lb. left in the minuend. (Why?) 11-8 = 3. Write the ^ below. (3) 10 is larger than 6. (What shall we do ?) 1 T. +6 cwt. = 26 cwt. 26-10=16. Write the 16 below. (4) There are 14 T. left in the minuend. (Why?) 14-6 = 8. Write the 8 below. Result, 8 T. 16 civt. 3 lb. 7 oz. Note. — Teacher should point out the similarity between this and Sub- traction of Simple Numbers. 2. A note dated July 10, 1897, was paid September 23, 1899. How long was the time between these dates ? Explanation : Write the latest date above. Write the number of the month: September, 9th month; July, 7th month. The ^ 2 13^ result pupil should learn the number of each month in the year. S. When I looked at my watch last it was 18 min. after 9 o'clock, but it is now 12 min. after 11 o'clock. How long has it been since I looked at my watch ? Form : j^-j* ^2 ' Note. — 18 minutes after 9 o'clock is 9 hr. 18 9 ig min. (from noon or midnight). 1 54, result. EXEKCISE LXXV. 1. From 13 bu. 1 pk. take 8 bu. 2 pk. 2. From 2 cwt. 38 lb. take 1 cwt. 49 lb. 3. From 2 T. 5 cwt. 41 lb. take 1 T. 10 cwt. 74 lb. Process ; 1899 1897 mo. 9 7 da. 23 10 ADVANCED ARITHMETIC. 177 ^. From 5 mi. 65 ch. 1 rd. take 3 mi. 78 eh. 3 rd. 5. From 50 yr. 3 mo. 12 hr. 50 min. 10 sec. take 24 yr. 6 da. 5 min. 6. From £12 3 &. 9 d. 2 far. take £6 17 s. 11 d. 3 far. 7. From 25 cu. yd. 16 cu. ft. 127 cu. in. take 7 cu. yd. 26 cu. ft. 240 cu. in. 8. Find the time from Jan. 11, 1895, to June 1, 1901. 99. Multiplication. EXAMPIiES. 1, Multiply 3 bu. 1 pk. 5 qt. 1 pt. by 7. Process : Explanation : (1) 7x1 pt. = 7 pt. = bu. pk. qt. pt. 3 qt. 1 pt. Write the 1 below, and 3 1-^1 carry the 3. (2) 7x5 qt. = 35 qt. 35 qt.+3 qt. 23 3 6 1, result. (carried) = 38 qt. = 4 pk. 6 qt. Write the 6 below, and carry the 4. (3) 7xlpk = 7pk. 7pk.+4pk. = llpk. = 2bu.3pk. Write the S below, and carry the ^. (4) 7x3 bu. =21 bu. 21 bu. + 2 bu. = 23bu. Write the 23 below. Result J 23 bu. 3 pk. 6 qt. 1 pt. Note.— The teacher should point out the similarity between this and Multiplication of Simple Numbers. 2. Multiply 27 yd. 2 ft. 9 in. by 9. Process: Explanation: (1)9x9 = 81. 81 in. = 6 ft. 9 in. Write the 9 below, and carry the 6. (2) 9x2=18. 18 ft. + 6 ft.=24 ft. = 8 yd. Write below, and carry the 8. (3) 9x27 = 243. 243 yd. +8 yd. =251 yd. Write the 251 below. Result, 251 yd. 9 in. Multiply 5 T. 8 cwt. 54 lb. 6 oz. by 8. Process : T. cwt. lb. oz. 5 8 54 6 Note.— Have the pupil explain 8 this example. 43 8 35 0, result. yd. 27 ft. in. 2 9 9 251 9, result. 178 ADVANCED ARITHMETIC. EXERCISE LXXVI. Multiply : 1. 12 yd. 1 ft. 10 in. by 8. 2. 5 mi. 43 ch. 51 ft. 9 in. by 11. 3. 17 sq. yd. 7 sq. ft. 7 sq. in. by 7. ^. 8 Tp.^21 Sec. 25 A. 9 sq. ch. by 25. 5. 11 cu. yd. 16 cu. ft. 128 cu. in. by 20. 6. 24 bu. 3 qt. by 24. 7. 4 gal. 8 qt. 1 pt. 2 gi. by 8. 8. 6 Cong. 5 0. 7 f S B f 5 35 m. by 12. 9. 7 T. 13 cwt. 46 lb. 10 oz. by 120. 10. 5 lb. 4 oz. 18 pwt. 22 gr. by 9. 11. 8 1b. 9S 7 5 29 5gr. by 24. 12. 5 yr. 7 mo. 3 da. by 40. 100. Division. EXAMPIiES. 1. Divide 27 bu. 3 pk. 7 qt. by 6. Process: Explanation: (1) 27 + 6 = 4, re- bu. pk. qt. pt. mainder 3. Write the 4 below, 6 )27 3 3 and reduce the 3 bu. to pecks. 4 2 4 1, result. (2) 3 bu. + 3 pk. = 15 pk. 15 + 6 = 2, remainder 3. Write the ^ below, and reduce the 3 pk. to quarts. (3) 3 pk. + 3 qt. = 27 qt. 27 + 6 = 4, remainder 3. Write the 4 below, and reduce the 3 qt. to pints. 3 qt. = 6 pt. 6-r-6 = 1. Write the 1 below. Result j 4 bu. 2 pk. 4 qt. 1 pt. Note. — Teacher should point out the similarity between this and Divi- sion of Simple Numbers. 2. Divide 112 T. 16 cwt. 59 lb. by 7. Process : T. cwt. lb. Note.— The pupil should explain 7)112 16 59 this example. 16 2 37, result. ADVANCED ARITHMETIC. 179 EXERCISE LXVII. Divide : 1. 490 bu. 2 pk.- 4 qt. by 100. 2. 161 yd. 1 ft. 6 in. by 17. 3. 3 yr. 11 mo. 4 da. by 14. 4. 65 gal. 2 qt. 1 pt. by 7. 5. 8 lb. apoth. by 120. 6. 1 Tp. 5 sq. mi. 430 A. by 21. B. INVOLUTION. 101. Definitions. — A composite number that can be formed by using the same number as a factor a number of times is called a PoTver of the number used ; and the number used is called the Root of the power. The degree of a power is the number of times the root is used in producing the power. Thus, 8x3=9. 9 is the second power of 3. A small figure placed just above and to the right of a num- ber indicates the degree of the power to which the number is to be raised, and is called an Exponent. The number itself .is the first power ; the number used twice as a factor gives the second power or square ; the number used three times as a factor gives the third power or cube ; used four times gives the fourth power ; five times, the fifth power, and so on. Thus, 8^=3 (First power.) 8^ = 3 X 3 = 9 ( Second power or square.) 88=8X8X8=27 (Third power or cube.) 8*=8X3X3X3 = 81 (Fourth power.) 85=3X3X3X3X3 = 243 (Fifth power.) 36=3X8X8X8X8X8 = 729 (Sixth power.) 180 ADVANCED ARITHMETIC. Involution is the process of finding a required power of a given root. 102. Process. Since by the definition a power is a product, tlie process of finding a power is multiplication. EXAMPIiES. 1. 5*=( )? Process : 5'* = 5x5x5x5 = = 625, result 2. m4^^{ )? Process : 534 534 2136 1602 2670 285156, result. 3, (A)'=( )• ? Process : {^^f = i^s X A x i^? = j\%\j result. Note. — To raise a common fraction to a given power, both the nu- merator and denominator must be raised to that power. 4. .048^= ( )? Process : .0 4 8 .0 4 3 129 172 .001849 .0 4 3 5547 7396 .0 0007950 7, result. Explanation : Proceed as in multiplication of decimals. EXERCISE LXXVIII. Commit to memory the powers : i. 12; 22; 32; 4^; 5^; 6^; 7^; 8^; 9^; 10^. 2. P; 2^; 3^; 4^; 5^; 6^ 7^ 8^ 9^ W. ADVANCED ARITHMETIC. 181 Raise the following numbers to the powers indicated : 1. 171 5. (|)«. 9, 472. 2. ^ 3. 103. Anotlier Metliod. The multiplication may be performed by separating the root into two numbers, and mul- tiplying each part of the multiplicand by each part of the multiplier. BXAMFLSS. 1272. K 6, iuV' 10, mv\ 87^. ■> 7. (mr- 11, 5.062. 13^ 8. (24&i)^ 1^> (.141) 1. 52z = (8+2)'^ = ( )? Process : 3+2 3+2 6+4 9+ 6 9+12+4 = 25, 2. T-- ={6+2y = ( )? Process : 5+2 5 + 2 10 + 22 52 + 10 52+2x10+22 = 25+20+4 = 49, result. These examples are illustrative of the following principle : Principle : The square of the sum of two numbers is equal to the square of the first plus twice the product of the first times the second plus the square of the second. By referring to the two numbers as 1st and 2d, this principle may be stated in an equation as follows : (lst+2cl)2=lst2+2xlstX2cl + 2d^ (Commit.) 182 ADVANCED ARITHMETIC. 62=.( )? Process 62 = (4 + 2)2 = 42+2x4x2+22= 16 + 16+4 = 36; q^, 62 = (3+3)2 = 32 + 2x3x3+32 = 9 + 18+9 = 36; or, 62 = (l+5)2 = 12+2x1x5+52 = 1 + 10+25 = 36, result. Note. — A number may be separated into any two of its addends and this principle will apply. i. (17)^=( )? Process: 17=' = (lO+7)2 = 102+2xlOx7+72 = 10O+140+49 = 289, result. S. 5'=( )? Process : 6. 9«=( )? Process . 2+ 3 2+ 3 6+ 9 4+ 6 4+12+ 9 2+ 3 12+36+27 8 + 24 + 18 8+36 + 54+27 = 125, result. 5+4 5+4- 5x4+42 52+5x4 52+2x5x4+42 5 + 4 52x4+2x5x42+4'" 5^2x52x4+ 5x42 53+3x52x4+3x5x42+4^ = 729, result. Note. — The teacher should carefully explain each step in this multi- plication. This example illustrates the following principle : Principle : The cube of the sum of tivo numbers is equal to the cube of the first plus three times the square of the first times the sec- Olid "plus three times the first times the square of the second plus the cube of the second ; or. ADVANCED ARITHMETIC. 183 (lst+2cl)^= lst^+3 X Ist^X 2(1 + 3 X 1st X 2d2 + 2dK (Commit.) 7. 233=( )? J; Process: 23'^ = ^20+3)3 = 203+3x202x3+3x20x3«+33 = 8000 + 3600 + 540 + 27 = 12167, result. Note.— These two principles have been here presented, not so much for their practical value in involution, as to prepare for their use in evo- lution — extracting square root and cube root. EXERCISE LXXIX. 1. lP=:(7+4)2=( )? 6. 83=(3+5)« = ( )? 2. 102=(3+7)2=( )? 7. 123=(10+2)3=( )? 3. 132=(5+8)2=( )? 8. 143=(10+4)3=( )? .4. 212=(20+1)2=.( )? 9, 233=(20+3)3=( )? S. 342= (30+4)2= ( )^ 2Q^ 46^= (40+6)3=: { )? 0. EVOLUTION. 104. Definition. — Evolution is the process of find- ing a required root of a given power. The number itself is the first root ; one of the two equal fac- tors which compose a number is its second ov square root ; one of the three equal factors which compose a number is its third or cube root; one of the four equal factors which compose a num- ber is its fourth root; and so on. The required root of a given number may be indicated in either of two ways : ( 1 ) By the use of the radical sign, t/, with a small figure or figures placed between the two parts to indicate the degree of the root. Thus, -^121, the square root of 121. -Vrm, the cube root of 1728. 1^^256, the fourth root of 256. 184 ADVANCED ARITHMETIC. (2) By a fractional exponent. Thus, 121^, the square root of 121. 1728^, the cube root of 1728. 256^, the fourth root of 256. Note. — The radical sign without the small 2 is commonly used to in- dicate the square root of a number. Thus, "V 64, the square root of 64. 1. SQXJABE HOOT. 105. First Process. — The square roots of integral squares can usually be obtained by factoring. When a num- ber is separated into two equal factors, one of the factors is the square root of the number. EXAMPLES. 1. -i/225,-:( ) ? Process: 225 = 5x5x3x3 = 15x15. ••• -v/225 = 15, result. 2, -i/l44x64=( ) ? Process : 144 x 64= 12 x 12 x 8 x 8 = 96 x 96. •'• -1/144x64 = 96, result.^ EXERCISE LXXX. 1. Commit to memory the square root ^f 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. Find by factoring the square root of — 2. 900 5. '824 8. 81x25 . 3. 625 6. 576 9. 676x100 4. 256 7. 1296 10. 144x729 106. Second Process. — When numbers are large, or when they are not perfect squares, the process to be employed ADVANCED ARITHMETIC. 185 in extracting the square root is developed from the following equation, which was learned in Involution: (lst-|-2d)^ = lst'+2xlstx2d+2d^ Preparatory to the development of this process, the following principles should be learned : Principles: 1. When an integral number is separated, from right to left, into periods of two figures each, the number of periods thus formed will equal the number of integral places in the square root of the number. Note, — This principle holds, even though there may be but one figure in the left-hand period. This principle may be illustrated as follows : -1/1== 1, T/roo==io, ViWob=ioo, Vroowoo^iooo. Thus, it appears that for integers less than l^'OO, the square root can- not have more than one integral place ; for integers less than l^OO'OO, the square root cannot have more than two integral places; and so on. ^. When an integer is separated from right to left into periods of two figures each, the square root of the largest integral square in the left period will give the first or left figure of the square root of the number ; the square root of the largest integral square of the number formed by the first two periods will give the first two figures of the square root of the number ; and so on. This principle may be illustrated as follows : 2862=55696; then, -i/5'56'96= 236. The largest integral square in 5 is 4. 1/^4 = 2. 186 ADVANCED ARITHMETIC. Then, 2 is the first figure of the square root, 236. The largest integral square in 556 is 529. -^-"529 = 23. Then, 3 is the next figure of the square root, 236. ' EXAMPLE. Find the square root of 288869. Step 1. — To find the first or left figure of the root. Process : Explanation : By separating the number 28^83^6915 into periods of two figures each, it appears 25 that the square root will contain three fig- 3 ures (Prin. 1). The largest integral square in 28 is 25, of which 5 is the square root. There- fore, the first or hundreds figure of the root is 5 (Prin. 2). Step 2. — To find the second figure of the root. By Step 1, it was found that the first figure of the root is 5 ; then, the number represented by the first two figures of the root must lie between 50 and 60, and the square of this number is the largest integral square in 2883, the first two periods of the number whose root is to be found ( Prin. 2). If this num- ber, which lies between 50 and 60, be represented as the sum of two numbers, then its square may be represented by the following equation : (Ist+2d)2 = lst2+2xlstx2d+2d2; or, (lst+2d)2= lst2+2d(2 X lst+2d) Now, since (1st + 2d) represents a number lying between 50 and 60, let 1st = 50, and proceed to find 2d. Process : 28^83^69 1 50 + 3 = 53. Ist2 = 2500 (to be subtracted) 25 00 '' 2x lst = 100 (trial divisor) 100 2d = 3 (found by trial). 3 2 X 1st + 2d = 103 (complete divisor) ... 103 2d(2 X 1st + 2d) = 309 (to be subtracted) 383 309 74 ADVANCED ARITHMETIC. 187 Explanation : After subtracting the 2500 from the first two periods, there is left 383. Now, 383 must contain 100 x 2d+2d^. Since the 2d represents tlie units of the second place, 2d^ is small as compured to 100 xM; finding how many times 383 contains 100 will' practically find the 2d. Using 100 for a trial divisor, proceed just as in finding the quotient figure in long division. The 2d is found to be 3. The complete divisor is formed by adding the 3 to the 100, making 103. This com- plete divisor multiplied by the 3 in the root gives 309, the number to be subtracted. Step 3. — To find the third or units figure of the root. By Steps 1 and 2, the first and second figures of the root are 5 and 3 respectively. Then, the number represented by the three figures of the root must lie between 530 and 540, and the square of this number must not be greater than 288369. As in Step 2, consider this root as made up of two parts, and its square as represented by the equation — (I8t+2d)2 = l8t2+2d(2xlst+2d). Let 1st = 530, and proceed to find the 2d as before. Process : lst2 = 280900. 28^83^69 1 530 +7 28 09 00 537, result. 2x lst = 1060 1060 2d = 7 7 2xlst + 2d = 1067 1067 2d(2xlst+2d) = 7469 74 69 7469 EXERCISE LXXXI. By steps as above, find : 1. -i/l849 2. -1/19044 S. -1/478864 4. -i/950625 107. Second Process Shortened. — The process of extracting square root, as just given, may be much shortened 188 ADVANCED ARITHMETIC. (1) by consolidating the several steps into one continuous pro- cess. In example, p. 186, 25 subtracted in Step 1 amounts to the same as the 2500 subtracted in Step 2, and the 280900 sub- tracted in Step 3 amounts to the same as the 2500 and the 809 subtracted in Step 2; (2) by dropping all O's that do not affect results. EXAMFIiSS. 1. -1/288369 =( ) ? Process 28^83^69 537 25 ' 103 383 309 Explanation: (1) Separate the number as before, into periods of two figures each. (2) 5x5 = 25. Place 5 in the root, sub- tract 25 from 28, and bring down the next period. (3) 2x5 = 10. Place 10 on the left for a trial divisor, and try 10 into 38 (instead of 100 into 383 as before). 38 + 10 = 3 + . Then, 3 is the second figure of the root. Place the 3 on the right of the 5 in the root and on the right of the 10 in the divisor. Multiply 103 by 3, write the product 309 beneath the 383, sub- tract, and bring down the next period. (4) 2x53= 106. Place the 106 on the left for atrial divisor, and proceed as before. 1067 7469 7469 Note. — The teacher should assist in comparing this shortened process with the process by steps, and explain all differences not clear to the pupil. -1/611.5729 = ( ) ? Process : 6^1.57^29124.73 4 ' 441211 1176 48713557 13409 4943 14829 14829 Note. — In decimals, the periods are formed from the decimal point to the right. Otherwise, the process is the same as in integers. When a number is not an exact power, O's may be added for decimal places, and the process continued at will. ADVANCED ARITHMETIC. 189 3. -i/ 5798649 =( ) ? Process : 5^79^36^49 1 2407, result . 44 179 176 4807 33649 33649 Explanation : If, after bringing down a period, the number thus formed is too small to contain the trial divisor, place a in the root, bring down the next period, cancel the old trial divisor, form a new trial divisor, and proceed as before. 4. V.iWi={ )? _ Process: (1) Vq25 = V 25x25 = 25 (2) Vl024 = -i/32x32 = 32 .'. ^^=11, result. Explanation : The square root of a common fraction is the square root of its numerator divided by the square root of its denominator. This follows as a reverse of the process of squaring a common fraction. Com- mon fractions whose terms are not integral squares, should be reduced to decimals, and their square roots should be extracted as in No. 5, below. 5. t/.048=( )? (Result true to 4 decimal places.) .04^30^00^00 1 . 2073 + , result. Explanation: (1) When the right-hand period of a decimal is not complete, it should be made so by annexing O's. (2) When there is a remain- der after using the last period 407 4143 3000 2849 15100 12429 in either decimals or integers, periods of O's may be annexed and the process continued at will. EXERCISE LXXXII. Find the value of- 1. 2. 3. i/6889 -i/5625 t/8186 t/8025 5. 6. 7. 8. t/ 62001 T/98G86 -1/4556.25 9. 10. 11. 12. 1/^.1406844881 4. -1/16.096144 (tVA¥9)^ 190 ADVANCED ARITHMETIC. Obtain results true to three decimal places : IS. 2^ 17. t/8. 141592 U. (|)^ U- B^ 18. -i/4b?i 22. (424)^ 15. 5^ 19. -]/ 326.004 ^^. (28111)^ i^. 11^ 20. t/.OOOT ^^. (33.47^)^ 2. CUBE BOOT. 108. First Process. — The cube roots of integral cubes can usually be obtained by factoring. When a number is sep- arated into three equal factors, one of the factors is the cube root of the number. EXAMFIiES. 1. #^1728= ( )? Process : 1728 = 4 x 4 x 4 x 3 x 3 X 3 = 12 x 12 x 12. .-. ^1728 = 12, result. 2. 1^64 X 512 =( )? Process : 64x512 = 4x4x4x4x4x4x2x2x2 = 32x32x32. .'. #"64x512 = 32, result. EXERCISE LXXXIII. 1. Commit to memory the cubes of the integers from 1 to 10. By factoring f find — 2. 1^2744. s. fimi. 4. #^4851. 109. Second Process. — When numbers are large or when they are not perfect cubes, the process to be employed in extracting the cube root is developed from the following equa- tion, which was learned in Involution : (Ist+2d)3=lst^+3xlst2x2d+3xlstx2d2+2d^ 5. #"3375. 8. #"27X729. 6. 1^4096. 9. #"216X1000. 7. 1^8x729. 10. #"512X343. ADVANCED ARITHMETIC. 191 Preparatory to the development of the process, the following principles should be learned : Principles : it^When an integral number is separated from right to left into periods of three figures each, the number of periods thus formed will equal the number of integral places in the cube root of the number. Note. — This principle holds even though there may be but one or two figures in the left-hand period. To illustrate : n=i. #'Pooo=io. f^rooo'ooo^ioo. #^rooo^ooo^ooo= 1000. Thus, it appears that for integers less than 1^000, the cube root can- not have more than one integral place ; for integers less than 1^000^000 the cube root cannot have more than two integral places ; and so on. 2. When an integer is separated from right to left into periods of three figures each, the cube root of the largest integral cube in the left period will give the first or left figure of the cube root of the num- ber ; the cube root of the largest integral cube of the number formed by the first two periods will give the first two figures of the cube root of the number; and so on. To illustrate : 236^=: 13^144^256; then, 1^13^44^256=236. The largest integral cube in 13 is 8. 1^8 = 2. Then, 2 is the first figure of the cube root, 236. The largest integral cube in 13144 is 12167. f 12167 = 23. Then, 3 is the second figure of the root, 236. 192 ADVANCED ARITHMETIC. EXAMPIiE. Find the cube root of 154854158. Step 1. — To find the first or left figure of the root. Process : 154^854^53(5 125 29 Explanation : By separating the number into periods of three figures each, it appears that the cube root will contain three figures (Prin. 1. ) The largest integral cube in 154 is 125, of which 5 is the cube root. Therefore, the first or hun- dreds figure of the root is 5 (Prin. 2). Step 2. — To find the second figure of the root. By step 1 it was found that the first figure of the root is 5 ; then the num- ber represented by the first two figures of the root must lie between 50 and 60, and the cube of this number is the larg- est integral cube in 154854 (Prin. 2). If this number which lies between 50 and 60 be repre- sented as the sum of two numbers, then its cube may be rep- resented by the following equation : (Ist + 2d)3 = lst3+3xlst2x2d+3xlstx2d2+2d3; or, (Ist+2d)3 = lst3+2d(3xlst2+3xlstx2d+2d2). Now, since (lst+2d) represents a number lying between 50 and 60, let 1st = 50, and proceed to find 2d. Process : 154^854^53 1 50+3 = 53. Ist3= 125000 (to be subtracted) 125000 ' 3xlst2 = 7500 (Trial divisor) 7500~ 3 x 1st X 2d = 450 450 2d2 = 9 9 3 X lst2+3 X 1st X 2d+2d2 (Complete divisor) 7959 2d(3 X lst2+3 X 1st X 2d + 2d2) 29854 23877 5977 Explanation : After subtracting the 125,000 from the first two periods, there is left 29854. Now, 29854 must contain M{3xlst^-\-Sxlstx2d-\-2d^). Of this expression, the 2d repre- sents the figure of the root sought, and {3xl8t^-\-3xlstxM-{- M^) represents the complete divisor. Since the M represents ADVANCED ARITHMETIC. 193 the units of the second place in the root, the second and third terms of the complete divisor are small as compared with the first. Therefore, Sxlst^, or 7500, is used as a trial ^ divisor. Finding the root figure is done in the same way as finding the quotient figure in long division ; but since the trial divisor is somewhat smaller than the complete divisor, allowance must be made for this in selecting the figure of the root. The second figure of the root is found to be 3. The complete divisor is found by adding to the 7500 the values of 5 X 1st X M and 2d^, or 450 and 9, making in all 7959. This complete divisor, multiplied by the root figure 3, gives the number to be subtracted, 23877. Step 3. — To find the third figure of the root. The first and second figures of the root have already been found to be 5 and 3 respectively. Then, the number represented by the three figures for the root must lie between 530 and 540, and the cube of this number must not be greater than 154854153. As in Step 2, consider this root as made up of two parts and its cube as represented by the equation, (lst+2d)3= lstH2d(3 X lst2+3 X 1st x2d+2d2). Let 1st = 530, and proceed to find the M as before. 154'854a53|530+7 = 537, result. Ist3 = 148877000 148877000 ' 3 X lst2 = 842700 842700 3 X 1st x2d = 11130 11130 2d2 = 49 49 3 X lst2+3 X 1st X 2d +2d2 853879 2d(3 X lst2+3x 1st x2d + 2d2) 5977153 5977153 EXERCISE LXXXIV. By steps as above, find — 1. f^274625. 3. #^94818816. 2. ^^7189057. 4. f2087336952. 194 ADVANCED ARITHMETIC. 110. Second Process Sliortened. — The process of extracting cube root, as just given, may be much shortened ( 1 ) by consolidating the several steps into one continuous pro- cess and (2) by dropping all O's that do not affect the result- EXAMPXiES. #" 154854153 Process : ( ) 154^854^53 537, result. 125 7500 450 29854 7959 842700 11130 49 853879 23877 Explanation: (1) Separate the number as before into periods of three figures each. (2)53=125. Place the 5 in the root, subtract the 125 from 154, and bring down the next period. (3) The trial divisor is formed by annexing a to the root al- ready found, squaring the re- sult and multiplying the square by 3. 502x3 = 7500. 29854 + 7500 = 3 + . Place the 3 in the root. The complete divisor is obtained by adding together the trial divisor, three times the last figure of the root times the remainder of the root with a an- nexed, and the square of the last figure of the root. 3 x 3 x 50 = 450; 32 = 9. 7500+450+9 = 7959, complete divisor. Multiply the complete divisor by the last figure of the root, subtract the product, and bring down the next period. (4) Proceed as in (3). 5977153 5977153 r8489664=( ) ? Process : 8^489^664 1 204 8 ' — 120000 2400 16 122416 489664 489664 Explanation : If after bringing down a period, the number thus formed is too small to contain the trial divisor, place a in the root, bring down the next period, can- cel the old trial divisor, form a new trial divisor, and proceed as before. Process : #"? |f= |, result. ADVANCED ARITHMETIC. 195 Explanation : The cube root of a common fraction is equal to the cube root of the numerator divided by the cube root of the denominator. Common fractions, whose terms are not integral cubes, should be reduced to decimals, and their cube roots should be extracted as in No. 4 or 5 below. 4. r .074088= ( ) ? Process : .074^088 .42, result. 64 ' 4800 240 4 5044 10088 10088 Explanation : Separate the number from the decimal point to the right into periods of three figures each ; then, proceed as in integers. 5. ri2.8458: Process , ( ) ? (Result true to three decimal places.) 12^345'800'000 2.311 + ,result. 1200 180 9 4345 1389 4167 1587( 6J X) )0 1 178800 159391 159391 160083C 69S K) 10 1 19409000 m)V )2f n 16015231 Explanation : (1) When the right-hand period of a decimal is not complete, it should be made so by annexing O's. (2) When there is a remainder after using the last period in either decimals or integers, periods of O's may be annexed and the process continued at will. 3393769 Find the value of : 1. f^l5625 2. 1^18824 3. #"1225043 EXERCISE LXXXV. ^. ^517781627 5. #"189119224 6. r 1967221277 196 ADVANCED ARITHMETIC, 7. r65804767688 U. f^66485045871501 8, #^997002999 15. ^1225.043 9. #^260917119 16. #"529.475129 10. f^3422470843 17. f/.013312053 11. f/351895816000 18. (tj^'VV)^" 12. #^35778897375 19. {tUU)^ 13. 1^1881242513625 20. (tWA^)^ Obtain results true to three decimal places : 21. 2^ 23. 3.6^ 25. .006^ 22. 4^ 2^. .244^ 26. (247|)^ 3. BOOTS OF SIGETER DEGREES. 111. Process. — If the index of a root has no other fac- tors than 2's or S's, that root may be extracted by applying processes already explained. Principle: Any root may he extracted hy extracting in succes- sion the roots indicated hy the factors of the index of that root. Explanation : The 4th root is the square root of the square root ; the 6th root is the square root of the cube root ; the 8th root is the square root of the square of the square root ; and so on. EXAMPLES. 1. -1^64= ( ) ? Process: (1) ^64 = 4. (2) -y 4 = 2, result. Note.— The order is not material. The square root may be extracted first, and then the cube root. 2. '1^531441= ( ) ? Process: (1) #"531441 = 81. {2)V8i = d. (3) Vg = 3, result. ADVANCED ARITHMETIC. 197 Roots may often be found by factoring and inspecting the result. 3, -^3125= ( ) ? Process: 3125 = 5x5x5x5x5. .'.-^^3125 = 5, result. 4. yi401=( ) ? Process : 1401 = 7x7x7x7. .'. -v^iiol = 7, result. EXERCISE LXXXVI. 1. -i5^256 3. -^^40353607 5. -if^78125 2. -iJ^ 14641 4. 1^531441 6. "1^248832 D. NUMBERS EXPRESSED BY WORDS OR LETTERS. 112. Explanations. — In writing out the solutions of problems, a great many ivords are used to express numbers. Some of the words commonly used in this way are the follow- ing: number area value volume price work cost force gain time loss interest weight principal length rate width sum thickness amount, etc. NoTB.— Suppose this problem is given: "Five times the cost of my horse is $320; find its cost." The relation given in the problem may be properly expressed as follows : 5 X cost = $320. 198 ADVANCED ARITHMETIC. Here the word " cost " refers to a particular number and *' 5 x cost " is 5 times that number. Not only are words used to express numbers, but letters are also thus used. Note. — Suppose that Cost of 1 apple = If ; then Cost of 2 apples = 2^ ; Cost of 3 apples = 3^ ; and in general, Cost of any number of apples = that number of cents. Now, if n be used for the word number, the expression becomes — Cost of n apples = n^. In the same way it is common to use v for value, p for price, c for cost, etc. It is as necessary to add, subtract, multiply, and divide num- bers thus expressed as it is to perform these processes with numbers expressed in figures. 113. Addition. BXAMPI.E8. 1. 2. 3. 4. 5. 6. 40 4X10 4 tens 4Xco8t 4Xc 4:C 50 5x10 5 tens 5 X cost 5Xc 5c 60 6X10 Qtens QXcost 6xc 6c 150, sum. 15 X 10, sum. 15 tenS, sum. 15 X cost, sum. 15 X C, sum. 15 C, sum. Note. — Carefully compare the above examples (1) as to form of ex- pression and (2) as to results. Notice that ''4c " is read four c, and means 4 c's or 4XC. As a number, 4c is the same as ^ x c. 7. Add lOXcost, 7xcost, bXnumber, 9Xcost, and QXnumber. Process : 10 x cost + 5 x number 7 X cost +Qx number 9 x cost 26 X cost + 11 X number J result. Note. — In this example there are five numbers to be added, three of one kind and two of another. Those of the same kind are united. The ADVANCED ARITHMETIC. 199 resulting numbers cannot be united into one number ; but the sign, + , placed between them, shows that they are to be considered together as one sum. ^K. EXERCISE LXXXVn. Add: 1. . 2. 3. ^. 5 X weight 17 X area 3 X time mxioss 7 X weight 21 X area 12 X time loss 9 X weight area 21 X time 12 X loss 15 X weight 11 X area time 27 Xloss 5. 6. 7. 8 9. IBXg 75Xc 12 Xamt. 53 X/ ISXth. 7xg ISXc 96 X ami. 121 X/ 72 xth. 25Xg SXc S7Xamt. 137 X/ 128 xth. ISXg 54Xc 19 Xamt. 86 X/ 94Sxth. 20Xg 75Xc 96 Xamt. 243 X/ th. 10. 11. 12. 13. u. 72 r 85 Z 18 w; 7v 248 Sr 37 « mw 89 1; 178 148 r 128? 172 w 27 1; 12 8 59 r 374 i 94 ty 34 V 98 15. 12 X work, 17 X work, 9Xwork, 7Xsum, 18 X sum, 14Xwork, 2QXsum, QXwork, 34 X sum, and 12 X sum. 16. 5Xc, 7X/, 8Xc, 12Xc, 17x/, 18x/, and9Xc. 17. 24 w, 72 w, 120 w, 74 1, 95 t, 71 w, and 39 t. 200 ADVANCED ARITHMETIC. 114. Subtraction. EXAMPIiEB. i. 25 X value 17 X value Sx value, dif. 25X^7 17X^; 8 X 17, dif. 25 V 17i? 8'y,dif. 4. 25p 17p 8^, dif. 5. From 25 c take 17 n. Process 25c 17 n 25c-17n, result. Note. — The minuend and subtrahend are not similar numbers ; therefore, the difference can only be indicated by plac- ing the sign, - , between them. EXERCISE LXXXVIII. ibtract : 1, 2. 3. 4. mx weight 12b X time 14&Xcost 105 X loss 12 X weight 57 X time cost 49 X loss 5. 6. .7. 8. • 9. 120 X(/ 224.x f 1 75 XamL 365 Xi; 459 Xw; 88 X(/ 189 X/ 4&Xamt. - 175X1; 2m Xw 10. 11. 12. 13. u. U7p 41 1 lllr 215 s 512 i 195p 2St 55 r 77 s 711 15, 16. 17. 18. 19. 75 X cost 157 XiVo. 17 Xw 189 1; 7b I 28 X area imx price 14 Xf 74.P 88 7/; ADVANCED ARITHMETIC. 201 115. Multiplication. EXAMPIiES. 1. 12 tens 6 Explanaiion : 6 x 12 tens are 72 tens. 72 tens 2. ten 3. 12X cost _6 _6 6 X ten^ or 6 tens, product. 72 X cost, product. j^. 12Xw 5, 12n 6 6 72 X w, product. . 72 n, product. 6. Multiply phj n. Explanation: (1) lxp = p. (2) 2xp = 2p. (3) 3xp = 3ja; and so on. (4) nxp = np, resuli. Thus, writing letters together, as nj), indicates the product of n and p. It is common also to use the expression nxp to indicate the product. Both are correct. 7. Multiply bXp by n. Result: nx5x^j = 5xnx/), or 5pn. Note. — If p and n are abstract, by Principle 4, p. 21, they may stand in any order ; it is customary to put the numerical factors before the letters or words. 8. Multiply 7c by 9 a. Result ;9ax7c = 9xax7xc = 63 ac. Note. — Since factors may stand in any order, the 7 and 9 may be put together, making 63. 202 ADVANCED ARITHMETIC. EXERCISE LXXXIX. Multiply : 1. 2. 3. 4' 7 X cost 24 X area 137 tens 56 X rate 12 86 48 17 •5. 6. 7. 8. 9. mxv 45Xw IbS Xth 4&Xl 129 Xp 5 18 11 - 15 29 10. 11. 12. 13. u. bp 17p lb I 12pr Qlw n r 16. w t 18. h 15. 17. 19. pi Ih 6wh Qpt 7wl r w 41 5Xr 7xh 11 6, Division. 1. Divide 12 tens by 6. Process : 61 12 ^ens ' , or, 12 tens +Q = ^ tens = 2 tens, result. 2 tens, result. Principle : Dividing any factor of a continued 'multiplication divides the product. 2. Divide the cost by 6. Process : Cost -«- 6 = — ^— , result. 6 Note. — The result can only be indicated, and the fractional form is usually employed. ADVANCED ARITHMETIC. 203 3. Divide 5Xc by 9. -, _ „ 5xc 5c ,, Process: 5xc-i-9 = —Q-, or, -q, result. 4.. Divide 16 ly by n. Process : lQw'+n= , result. n 5. Divide 15^ by 25w;. Process : 15l + 25w= -^- = ^- , result. Note. — Cancel any common factors found in both numerator and de- nominator. 6. 6ln-^7wn=( ) ? Process : 5ln + 7 Note. — n is common to both numerator and denominator. Process : bin + 7 wn = ^ — = ,;r— , result. 7 wn 7 w EXERCISE XC. ■orm the process indicated : 1. 64Xco8«-^8. 11. wXl^h. 2. 96Xpnce-j-12. 12. ^Xi^lOXp. 3. ^Q>y.value-^Q. 13. i-7-pXr. Jf.. 20Xweight^6. U- 5x«-^75. 5. IXcost-^U. 15. 12 Iw-hmih. 6. 12Xprice-^72. 16. mi^7Xpt. 7. cost-^S. 17. ipt-^prt. 8. 12^^80. 18. 12cn^27wn. 9. 16x^(;-^w. 19. IS rt -^40 pt. 10. w-^l. 20. 120 prt-T- 144: pnt 117. Involution. EXAMPLES. 1. Square j9. Process : p xp =p^, result. Note. — Read "p^" p square. 2. Square the number. Process : Number x number = (number)^, result. 204 ADVANCED ARITHMETIC. 3. Square 5 a . Process : 5 a x 5 a = 25 a^, result. Square a-\-b. Process : a+6 ab + b^ a'+ab a^-\-2ab + b^, result. Questions : Do this process and result agree with the principle on page 181 ? Can you repeat that principle from memory ? S. Cube 7 Z. Process : 7 1x71x7 l = MSl^, result. Note— Read ''P'' I cube. 6. Cube 5a w ' „ ba^5a ba 125 a^ ,^ Process : — x — x — = — ?- , result. www w^ 7. 7c 4n = { ) o 7c 7c 7c 7c 7c 168070^ Process : .— x r- x .— x ;— x ;i— = t^^oT-^ > result. 4n 4n 4n 4n 4n 1024 n* 8. (a+br=( )? Process : a + b ab-^b^ a^-\-ab a2+2a6 + 62 a+b a^b + 2ab^+b^ a^+2a% + ab^ aH3a26+3a62+6^ result. Questions : Do this process and result agree with principle, page 182 ? Can you repeat that prin- ciple from memory ? ADVANCED AEITHMETIC. EXERCISE XCI. 205 Give results orally : 1. Square (1) ^^y .(2) ^n; (3) j>n; (4) ^; (5) |^^. 2. Square (1) a.^h ; (2) r+p; (8) n+r; (4) t(;+^. 5. Cube (1) ..; (2) 5n; (3) nl ; (4) |^'; (5) -g. 4. Cube (1) r/ + /i; (2) r+p; (3) n+r; (4) w+L O^'-ye luritten processes : 7 9« 146 126^ ^149 a 8. (r/2)2. 118. Evolution, 9. 10. 11. 12. 4:rl 9vt 7c?/7 Il71j9 4ar J EXAMPLES. i. -i/a2=( ) ? Process^ : v a^ = V a x « = a, result. Question : What is the square root of a number ? 2. V^a^V'={ ) ? Process : 'V^S nhxSab = S ab, result. 4 ma' >5c2 ^ ^ /36a2 /6 a 6 a 6 a ,^ Process : ^ = ^ — x — = — > result. \25c2 \5c 5 c 5 c 13. (2+n)2 i4. (3 r/ +4 6)2 15. {w-\-6y 16. (4r+3-^i)' 206 ADVANCED ARITHMETIC. 4. VM^={ ) ? Process : ^Q4a^= 1^4 ax4ax4« = 4a, result. Question : What is the cube root of a number ? 5. l^^^l^=^ ). \216/V ^ ^ • si 27aSr3_ slSar T, o , -. .V , a ,„.., 3<7r Sar Sar ,, Process : ^j = ^ — x — x — = — , result. 6hv 6 Iw 6 Iw 6. -^ SI wH''={ ) ? Process : ~y SI wH* = 'V 3 wl xSwlxSwlxS wl = 3 wl, result. 7. Va^-i-2ab-\-b^=( ) ? Process : V a'^-^2ah-\-h'^ = -/(a-t-6)(a + />) = a + h, result. Note. — "(a+6)(« + 6)" = a + & multiplied by a + 6. Process : f a^ +3 a% +3 ab^ -\-h^ = f {a -\-h){a-\-h){a-\-h) = a + h, result. EXERCISE XCII. (T*'?;e results orally : 1. Square root of (1) l^ ; (2) An^ ; (8) 10/^; (4) ^^ ; ;^. Square root of (1) l'^-\-2lr-\-r'^ ; (2) 7i2+27ic+c2. .?. Cube root of (1) r^• (2) 27 />^• (8) 125 i/;3/^^• (4) ^3; ^^^ 2l6^^* ^. Cuberootof (l)y+3i>V+8pr2+?-3- (9) /3+3/2^+3Zw;24_^3^ ADVANCED ARITHMETIC. 207 Give written process : 5. 125 c^ ] i 9. {p^-}-2rp+r^)^ 6. |-^?i^]^ ' 10. (^/;2+4w+4)^ 7. ^ 11. {t^^^t^r+Ztr'^+T^)^ 8. f_^?_]^ i^. (71^+3 7iV+3wr2+r3)^ 119, Equations Containing' Letters. — Formulas, which are equations containing letters, are often used in arith- metic. For example, (1) C=np. Now, there are three numbers in this equation. If any two of them are known, the other may be found. For (1) = (2) ■np = C. (2)^« = (8) p=-^. (2)-i> = (4) » = y. If the values of n and p be substituted (put in place of the letters) in (1), the result will express the value of C ; if the values of C and n be substituted in (3), the result will express the value of p ; and if the values of C and p be substituted in (4), the result will express the value of n. Note. — When an equation, containing letters, is so arranged that one be expressed in terms of the others. Thus, in (1), C is expressed in terms of p and n; in (3), p is expressed in terms of C and n; and in (4), n is ex- pressed in terms of C and />. 208 ADVANCED ARITHMETIC. EXAMPLES. 1. A=lw. (1) Express the value of w in terms of A and I. (2) Express the value of I in terms of A and w. Process : (1) A = lw. (1)=(2) lw = A. (2) H- Z = (3) u' = A , Is t result. (2) -H w = (4) Z = A , 2d result. 2. F=—-. Express the values of m, v, and t, each in terms of the other numbers. Process: (1) F= — . (1) = (2) !^' = i^. tx(2) = (3) mv = Ft. (3)+i,' = (4) w = -^, 1st result. V Ft (3) + m = (5) V = — , 2d result. (3) = (6) Ft="mv, mv i6)^F={7) < = -p-, 3d result. S, 432 = 16(2^-1). Find the value of «. Process : (1) 432 = 16 (2 1 - 1). (1) = (2) 16(2<-1) = 432. (2)-h16 = (3) 2<-1 = 27. Transposing, we get (4) 2t = 28. I- of (4) = (5) t = 14, result. 4- S = \gt^. Express the values of g and t, each in terms of the other numbers. Process: (1) S = ^gt\ (1) = (2) i,gt''=^S. 2 Sf (2) + i<2 = (3) ^ = -_^ Istresult. 2,S (2)-i-f7 = (4) <2 = — 2^ -i/(4) = (5) < = ^^, 2d result ADVANCED ARITHMETIC. 209 5. A=p-\-prt. Express the values of j9, i, and r. Process: {\) A=p-\-prt=p {l-\-rt). (i) = {2)p(l + r0 = ^. (2)^(l+r0 = (3) ^^fv;:^' 1st result. {l) = {'i) p+prt = A. Transposing, we get (5) prt = A-p. A -» 0)+pr^{Q) t = i^, 2d result. pr (5)+p< = (7) r = —^, 3d result. pt EXERCISE XCIII. ._ PJ^ 1. p—-L~-. Express the values of /) and ^. lOU 2. F=Ma. Express the values of M and a. 3. W—Fd. Express the values of i^and d. 4. W=Mda. Express the values of M, d, and a. 5. 25x30=15 M. Find the value of M. 6. 869.84=1 (/ (24-1). Find the value of g. 7. 1302.48 = 16.08 f. Find the value of t. W 8. A=——-. Express the values of IF and t. 9. V=lwh. Express the values of Z, w, and h. 10. H^=B^+P\ Express the values of P and B. 11. C—~^. Express the values of ^ and i2. 12. A = irl^. Express the value of R. Note. — The Greek letter ir (pi) is used for the number 3.1416. 13. Find the value of F in No. 8, if 1^=75 and d = 15. IJ/.. I=prt. Find the value of p, if /, r and t are respectively 102, T«o, 4. 15. Find the value of B in No. 10, if H and P are respectively 50 and 80. . 210 ADVANCED ARITHMETIC. II. STUDY OF PROBLEMS. A. DEVELOPMENT AND USE OF FORMULAS. 120. General Numbers and General Prob- lems. — If I say, " The fence is 1^.0 rods long,^^ I use a, jjarticular number J J^O rods. If I say, " The fence is I rods long," I use a general number, I rods, which may prove to be 40 rods, or some other number of rods. This general form of expressing a num- ber that may vary in its numerical value is of great service in mathematics. Examine the following problems : 1. What will 5 books cost, at $2 each ? 2. What will 25 lb. of meat cost, at 10/ per lb.? 3. Find the value of 140 A. of land, at $40 per A. 4.. What will 12 horses sell for, at $75 each ? These are all different problems, yet they all belong to the same type. Type : What will a given number of articles cost, at a given price each ? By using letters to represent the numbers in this type, it may be expressed in the following form : What will n things cost, at $p {or pf^) each f Such a type, which is a representative of a great class of problems, is often spoken of as a General Problem. It is of great service, in solving particular problems, to know to what type they belong. EXAMPIiES. 1. What is the area of a field 50 rd. long, 80 rd. wide ? Give the type. Type : What is the area of a surface (or field) I rd. long, w rd. wide? Note. — In giving types, it is well to use initial letters, as far as con- venient, to represent the numbers ; as, I for length and w for width. ADVANCED ARITHMETIC. 211 2. Find the interest on $400 for 2^ years at 7% per annum ? Give the type. Type : Find the interest on %p for t yr. at r % per annum. 3. Find the length of a solid of 8600 cu. ft., 20 ft. wide, 6 ft. thick. Type: Find the length of a solid of v cu. ft., w ft. wide, t ft. thick. EXERCISE XCIV. Give the types : 1. Find 20% of $500. 2. How many books will bring $40 at $5 each ? 3. $75 is 15% of what number ? 4. What must be the width of one acre (160 sq. rd.), if it is 40 rd. long ? 5. What force is required to add 10 ft. per second to the ve- locity of 15 lb. of matter ? Note. — Use a (acceleration) for the number of feet, and J/ (mass) for the number of pounds. 6. 75 is what % of 150 ? 7. I sold 20 lb. of butter for $2. Find the price per lb. 8. Find the width of a wall of 2340 cu. ft., 130 ft. long, 1^ ft. thick. 121. Solving tlie General Problem or Type. The same reasoning and plan are used in solving the general problem^ or type^ as are used in solving thQ particular problem. EXAMPLE. Find the cost of n things, at $p each. Solution: (1) Cost of n things, at $p each = $( )? (Question.) (2) Cost of 1 thing, at $1 each = $1. (Basis.) px{2) = (3) Cost of 1 thing, at '^p each = $p. nx(3) = (4) Cost of n things, at $p each = $np, result. 212 ADVANCED ARITHMETIC. If the cost be represented by C and (4) be expressed in its abstract form, it becomes — (5) C=np. This equation is called a rormula, and it expresses an ab- stract numerical relation that is always true. Relation : The cost of a number of things of the same kind is equal to the product of the number of things and the price of one of the things. This formula relates three numbers : the cost, the number of things, and the price of each. When any two of these numbers are given, the third may be found by aid of the formula. EXAMFIjES. 1. Find the cost of 20 bu. of apples, at 30/ per bu. Solution: (1) C = np. (2) = 20x30 = 600. .-. the cost is 600^, or $6. 2. How many bushels of apples at $.30 each can I buy for $18? Solution: (!) C-np. (2) 18=.3n. (2) = (3) .3ri = 18. (3)-H.3 = (4) n = ~ = 60. .'. the reqd. answer is 60 ba. 3. If 50 bu. of apples cost $15, find the price per bushel. Solution: (1) C = np. (2) 15 = 50p. (2) = (3) 50j9 = 15. 5Vof(3) = (4)/> = B = .3. .'. the price is $.3, or 30^. Note. — In examples 2 and 3, the work would be shortened by one equation if the formula had been turned around in the J3rst equation so that the required number would appear in the left member. ADVANCED ARITHMETIC. 213 EXEEOISE XCV. 1. If n bu. of apples cost $C, find the price per bu. Answer: C ^ n - 2. If $C are paid fdr apples at $p per bu., find the number n of bushels bought. Answer: n= — . Use answer to No. 1 as a formula. 3. Find the price per acre, if 120 A. of land sell for $1440. J^. I sold 5000 lb. of bacon for $400. Find the price per lb. Use answer to No. 2 as a formula : 5. I spend $15.60 for turkeys, at $.65 each. How many did I buy ? 6. A dealer invests $975 in buggies, at $75 each. How many did he buy ? B. MENSURATION. 1. LINES AND ANGIiES. 122. Definitions. — Define a line. (See p. 149.) A line whose length has but one direction is a Straiglit Line. A line whose length changes direction at every point is a Curved Line. AB., a line; also, a straight line. CD, a curved line. Two lines that never approach each other are Parallel. Note. — Lines that are parallel have the same di- rection and never meet. If two lines meet, they form one or more angles, and have different di- rections. An Angle is the difference of direction between two lines that meet. AB and CD, parallel lines. AC By an angle. 214 ADVANCED ARITHMETIC. Note 1. — As line BC, starting from the position AB, swings around point C, the direction of line BC is continually changing, and the difference of direction be- tween line AC and line BC is con- tinually growing larger. Angle ^05 is smaller than angle ACB^ ; angle ACB^ is smaller than angle ACB^^; and so on. Note 2.— If two lines, coming together, stop at the point of meeting, they form but one angle. If only one line stops at the point of meeting, they form two angles. If neither line stops at the point of meeting, they form four angles. One angle. Two angles. Four angles. When the directions of two lines that meet are such that, if the lines are continued past the point of meeting, they form equal angles, the lines are Perpendicular to each other. When the angles thus formed are not all equal, the lines are Oblique to each other. B AC and BC are perpendicu- lar to each other. AC and BC are oblique to each other. Note. — The lines forming an angle are called its sides, and the point of meeting of the sides is called the vertex of the angle. ADVANCED ARITHMETIC, 215 ACB^ a right angle. B I / ACB^ an acute angle. ACB, an obtuse angle. A Riglit Angle has its sides perpendicular to each other. An Acute Angle is smaller than a right angle. An Obtuse Angle is larger than a right angle. Note.— The common unit of angular measure is the degree, marked (°). 1 right angle = 90° A line parallel with the regular lines of writing or print is a Hori- zontal Line. . „ A line perpendicular to the regular lines of writing or print is a Ver- tical Line. AB, a horizon- tal line. CD, a vertical line. Note.— The word " horizontal " came to be applied to lines, probably because in writing upon upright surfaces such as blackboards, such lines are parallel with the horizon ; but if the surface on which you write is parallel with the horison, all lines made on that surface are also parallel with the horizon. EXERCISE XOVT. Define and draw (free-hand ) : 1. A line. 2. A straight line. 3. A curved line. 4.. Two parallel lines. 5. An angle. 6. Two perpendicular lines. 7. Two oblique lines. 8. A right angle. 9. An acute angle. 10. An oblique angle. 11. A horizontal line. 12. A vertical line. Note.— Become familiar with these lines and angles before passing to the next subject. 216 ADVANCED ARITHMETIC. 2. PLANE SURFACES. 123. Parallelogram.— Define Surface. (See p. 154.) A Plane is a surface each of whose dimensions has but one direction. A Polygon is a plane bounded by straight lines. The straight lines are called sides, and the points where the sides meet are called angles or vertices. ^ ^ C A Quadrilateral is a polygon of four sides. The lines joining the al- ternate angles are called diagonals. ABCD, a quadrilateral ; AC a diagonal. A Parallelogram is a quadrilateral whose opposite sides are parallel. The line on which a ^parallelogram rests is usually called its base, and the perpendicular distance from the base to the opposite side (or opposite side produced) is called its altitude. A Rectangle is a parallelogram whose angles are right angles. A Square is a rectangle whose sides are equal. DEC L C ^ ^ B ABCD, a parallelogram and rhomboid ; AB, the base ; EF, the altitude. ABCD, a rectangle ; AB, the base ; DA or CB, the al- titude. B ABCD, a square ; AB, base | CB or AD, the altitude. A Rhomboid is a parallelogram whose angles are not right angles. For form of rhomboid, see parallelogram. ADVANCED ARITHMETIC. A Rhombus is a rhomboid whose sides are equal. ABCDy a rhombus; AB, the base ; DE, the altitude. EXERCISE XCVII. Define and draw (free-hand, hut carefully) : 1. A quadrilateral. 2. A parallelogram. 3. A rectangle. Define: 7. A surface. 8. A plane. 9. A side. 10. A vertex. 4. A square. 6. A rhomboid. 6. A rhombus. 11. A diagonal. 12. A polygon. 13. A base. IJ^. An altitude. Note. — These terms must be defined as applied to polygons. Problem 1 : Letting b represent the length and a the width of a rectangle, develop the formula for the area. BY EQUATIONS. Solution: (1) Area of a surface h ft. 1., a ft. w. = ( ) sq. ft.? (Question). (2) Area of a surface 1 ft. 1., 1 ft. w. = l sq. ft. (Basis). 6x(2) = (3) Area of a surface h ft. 1., 1 ft. w. = 6 sq. ft. ax(3) = (4) Area of a surface h ft. 1., a ft. w. =a6 sq. ft., answer. BY PROPORTION. Solution : (1) Area of a surface 6 ft. 1., a ft. w = ( ) sq. ft.? (Question.) (2) Area of a surface 1 ft. 1., 1 ft. w. = 1 sq. ft. (Basis.) (3) 6xa:lxl::( ): 1? (4) ^^^^^ = 6a, answer. ^ ' 1x1 Using A for area, the result may be written as follows : A = ba. if UWiVERSlTV 218 ADVANCED ARITHMETIC. Relation I. Abstractly, the area of a rectangle is equal to the product of the base and altitude. EXAMPLES. 1. Find the area of a wall 20 ft. long and 7^ ft. high. Solution: (1) A = ba. (2) ^ = 20x7H150. .'. the required area is 150 sq. ft. 2. A rectangular field is 80 rd. long and contains 10 A. How wide is it? Solution : (1) 10 A = 1600 sq. rd. (2) ab = A. (3) 80 a = 1600. (3) + 80 = (4) a = 20. .'. the required width is 20 rd. 3. A sidewalk 4 ft. wide contains 960 square feet. How many yards long is it? Solution: (1) ab = A. (2) 4 6 = 960. iof (2) = (3) 6 = 240. .'. the required length is 240 ft. or 80 yd. Note. — Before applying the formula, the dimensions must be ex- pressed in linear units of the same denomination; then, the area will be square units of that denomination. Use only the corresponding abstract numbers in the formula. Problem 2 : Develop the formula for the square. Development : A square is a rectangle, whose base (6) and altitude (a) are equal. If we call one side of the square s, we can substitute s for a and b in the formula, thus : A = sxs; or, A = sK Relation II. Abstractly, the area of a square equals the square of one of its sid^s. ADVANCED ARITHMETIC. 219 EXAMPLES. 1. Find the area of a square one of whose sides is 46 ft. Formula : (1) A=^s^. (2)^ = 462 = 2116. .', the required area is 2116 sq. ft. 2. The area of a square is 102400 sq. rd. Find the length of one side. Formula: (1) s^ = A. __ (2) 82=102400. -i/(2) = (3) s = 320. . •. the required result is 320 rd. Problem 3: Develop the formula for the parallelogram. Development : Let A BCD be a par- F D F C allelogr am, whose base is ^-B. Draw r '/ j ~7 BE from B perpendicular to CD, \ y/^ \ ^ cutting off the part BEC. Place this ! y/^ \ >^ part on the left as shown in AFD, \y^ I/^ forming the rectangle J J5^i^. Thus, /| 5 the parallelogram has been changed to a rectangle, whose hase and altitude are the same as those of the par- allelogram. Then, the area of the parallelogram is equal to that of the rectangle; or, A=ba. Relation III. The same as that of the formula for a rectangle. EXAMPLES. 1. Find the area of a parallelogram whose base and altitude are 25 ft. and 50 ft. respectively. Formula: (1) A = ba. (2) ^ = 25x50=1250. .'. the reqd. area is 1250 sq. ft. 2. The area of a parallelogram is 1440 sq. rd. What is its length, if it is only 16 rd. wide ? Solution : (1) 16 & = 1440. ( Why ?) (2) b = m^ = 90. .•. the reqd. length is 90 rd. Note.— In rapid work, the formula may be omitted, as in example 2; but never until its use is thoroughly understood — not until the pupil knows just what substituting in a formula means. 220 ADVANCED ARITHMETIC. EXERCISE XOVIII. 1. In what respect is a square a parallelogram ? In what respect a rectangle ? 2. What is the difference between a square and a rhombus ? S. Is it true that the formula A=ah, will apply to a rectan- gle ? A square ? A rhomboid ? A rhombus ? Why ? 4-. Show how the formula A=ah is modified to ^=8^ for the square. 5. Carefully draw and cut out of paper a parallelogram ; then, cut it into two parts and put the parts together to form a rec- tangle as is indicated in problem 3. Can every parallelogram (not already a rectangle) be thus reduced to a rectangle ? 6. In figure in problem 8, does the perpendicular BE have to be erected at B, or may it be erected somewhere else along the base ? 7. Commit the relations I to III. Find the area of the following parallelograms : 8. Base 90 ft., altitude 37 ft. 9. Base 3 ft., altitude 12 in. Note. — Dimensions must be expressed in the same denomination be- fore applying the formula. 10. Base 5 mi., altitude 700 rd. 11. Base 5i yd., altitude 16^ ft. In each of the following parallelograms ^ find the dimension not 12. Area 2700 sq. m., altitude 60 m. 13. Area 960 A., base 600 rd. IJ^. Area 1156 sq. ft., base 34 ft. 15. Area 100500 sq. yd., altitude 450 ft. 16. Area 57500 sq. cm., base 46 dm. ADVANCED ARITHMETIC. 221 124. Triangles.— A Triangle is a polygon of three sides. Any side of a triangle may be considered its base, and the perpendicular from any vertex to the opposite base is the altitude cor- responding to that base. ABCf a triangle ; AB, the base ; C£, the altitude. ABC, an equilateral trian- gle ; also, equiangular. ABC, an isosceles tri- angle. ABC, a scalene triangle; AB, the base : CE, the altitude. An Equilateral Triangle has its three sides equal. An Isosceles Triangle has two of its sides equal. A Scalene Triangle has no two of its sides equal. Every triangle has three angles, which together measure just 180°. An Equiangular Triangle has its three angles equal. ( For form of equiangular triangle, see equilateral triangle. ) NoTB.— Each angle of an equiangular triangle measures just 60°. A Riglit Triangle has one of its angles a right angle. The side opposite the right angle is called the hyj}otenuse ; the other two sides are usually called base and perpendicular. ABC, a right triangle; AB, the base ; CA, the perpendicu- lar or altitude ; CB, the hypote- nuse. 222 ADVANCED ARITHMETIC. ABC, an obtuse triangle. ABC, an acute triangle. An Obtuse Triangle has one of its angles an obtuse angle. An Acute Triangle has all of its angles acute angles. EXERCISE XOIX Define and draw : 1. A triangle. 2. An equilateral triangle. 3. An isosceles triangle. 4-. A scalene triangle. 5. An equiangular triangle. 6. A right triangle. 7. An obtuse triangle. 8. An acute triangle. 9. Define (1) base, (2) altitude, (8) vertex, (4) hypote- nuse, (5) perpendicular. 10. How many degrees in all the angles of a triangle ? 11. Can two of the angles of a triangle be right angles ? Can two of them be obtuse ? Why ? 12. Construct an obtuse triangle and draw its three altitudes. Problem 1 : Develop the formula for the triangle. Development : Let ABC be a triangle. Upon the base AB construct the rectangle ABDE, just as high as the triangle. From C drop a perpendicular CF to the base AB. Parts 1 and 2 are equal, and parts 3 and 4 are equal. Therefore, the triangle is just half as large as the rectangle. But the base and al- ^ titude of the triangle are the same as those of the rectangle. Therefore, , ah A = -^. ADVANCED ARITHMETIC. 223 Relation IV. Abstractly, the area of a triangle is equal to the product of its base and altitude, divided by 2. EXAMPLES. 1. Find the area of a triangle whose base and altitude are respectively 40 ft. and 15 ft. Solution: ^ = ^^^^^ = 300. 2 .'. the required area is 300 sq. ft. 2. The base of a triangle is 75 ft., its area is 400 sq. yd. Find its altitude. Solution : (1) 400 sq. yd. =3600 sq. ft. (2) 1^ = 3600. (3) q = 36(X)x2^Qg .*. the required altitude is 96 ft. Relation V. In right triangles, abstractly, the square of the hypotenuse equals the square of the base plus the square of the per- pendicular. Note.— The development of this relation is too difficult to be given before the pupil has studied geometry . It gives us an important formula : EXAMPLES. 1. The base of a right triangle is 86 ft. ; the altitude, 27 ft. Find the hypotenuse. Solution: (!) H^ = B^-\-P^. (2) iy2 = 362+ 272 = 2025. (3) if= v^2025 = 45. .'. the required result is 45 ft. 2. Find the diagonal of a square, one of whose sides is 32 inches. 224 ADVANCED ARITHMETIC. Note. — ABCD is a square. AC, the diagonal, is the hypotenuse of the right triangle CAB. AB = S2, BC = B2. Solution: (1) if 2 = 322+322 = 2048. (2) iy=V2048 = 45.25 + . .*. the diagonal is 45.25 in. 3. Find the base of a right triangle whose perpendicular is 52 rd. and whose hypotenuse is 65 rd. Solution: (1) B^+T^ = H^. (2) ^2+522 = 652. (3) ^2 = 652 _ 522 =1521. (4) B = Vlb21 = S9. .'. the required result is 39 rd. 4-. The diagonal of a square is 60 yd. Find one side. Note.— The diagonal is the hypotenuse of the right triangle whose sides are equal. Solution: (1) 8^+8^ = H^; or, (2) 2s2 = iy2 = 602 = 3600. iof (2) = (3) 82 = 1800. (4) s=Vl800 = 42.426 + . .'. the reqd. side is 42.426 yd. EXERCISE 0. 1. Study the formula for the area of a triangle till you can develop it from memory. 2. Find the base of a triangle whose area is 5700 sq. ft. and an altitude of 76 ft. 3. Find the altitude of a triangle whose base is 40 rd. and whose area is 5 A. 4.. Find the perpendicular of a right triangle when the base and hypotenuse are respectively 80 ft. and 50 ft. 5. Find the base when the perpendicular and hypotenuse are respectively 45 ft. and 90 ft. ADVANCED ARITHMETIC. 225 6. Find the hypotenuse when the base and perpendicular are respectively 25 in. and 85 in. 7. Find the diagonal of a rectangle whose base and altitude are respectively 10 ft. and 86 ft. 8. Find the altitude of a rectangle whose base and diagonal are respectively 87 ft. and 90 ft. 9. A house measures 20 ft. from north to south ; 15 ft. from east to west, and 12 ft. high. Find the distance on the floor from the northwest to the southwest corner. 10. In the house described in No. 9, find the distance from the northwest lower corner to the southeast upper corner. 11. Commit to memory Relations IV and V. 125. The Other Quadrilaterals. — A Trapezoid is a quadrilateral, having only two parallel sides. The parallel sides are called bases. A Trapezium is a quadrilateral, having no two of its sides parallel. Problem: Develop the formula for the trapezoid. Development : Let A BCD be a trap- ezoid ; AB, lower base (B) ; DC, upper base (b); and FE, altitude. The di- agonal BD divides the trapezoid into two triangles. In the triangle DAB, having for its base AB {B) and its al- titude FE (a), we have — Area = aB In the triangle BCD, having for its base DC (b) and for its altitude FE (a), we have — . ab Area = — . 2 226 ADVANCED ARITHMETIC. But the sum of the areas of the two triangles is the area of the trape- zoid. Then, , _aB ah _aB +ah Factoring a out of both terms of the numerator of the fraction, we have the formula : . a{B-\-h) Relation VI. ' Abstractly y the area of a trapezoid is equal to the product of the altitude and the sum of the bases, divided by 2. There is no formula for a trapezium. If one of its diagonals and the altitudes from the other vertices upon that diagonal are known, its area may be easily found by the formula for tri- angles. EXAMPLiES. 1. Find the area of a trapezoid, if its altitude is 26 ft. and its bases are 50 ft. and 75 ft. Solution : A = ^^ ^^"^^^^ = 1625. 2 .*. the reqd. area is 1625 sq. ft. 2. Find the upper base of a trapezoid whose lower base is 35 in., altitude 15 in., and area 860 sq. in. Solution: (1) i5i^±^ = 360. 2x(l) = (2) 15 (35 + 6) = 720.. T^sof (2) = (3) 35 + 6 = 48. (4) 6 = 48-35 = 13. .•. the reqd. base is 13 in. 3. Find the area of a trapezium whose diagonal is 48 ft. and the altitudes upon that diagonal are respectively 10 ft. and 17 ft. ADVANCED ARITHMETIC. 227 Explanation : Let ABCD be the trapezium ; DB, the diag- onal ; and BA and CF, altitudes. In the triangle DBC, we have given the base 1)5, and altitude CF; and in the triangle DAB, we have the base DB, and altitude EA. SOLUTION. . 48x17 1st triangle : A = = 408. 2d triangle : A = = 240. . '. area of trapezium = 408 sq. ft. +240 sq. ft = 648 sq. ft. EXERCISE CI. 1. Define and draw a trapezoid ; draw one diagonal and the altitude. 2. Define and draw a trapezium. 3. Find the area of a trapezoid whose bases are 46 yd. and 80 yd., and whose altitude is 86 yd. 4.. Find the altitude of a trapezoid whose bases are 30 ft. and 50 ft., and whose area is 480 sq. ft. 5. Find unknown base of a trapezoid whose altitude is 24 in., area 912 sq. in., and given base 45 in. 6. Find the area of a trapezium whose diagonal is 78 ft., and altitudes upon that diagonal are 43 ft. and 21 ft. 7. Study until you can solve from memory the problems of this Article. 126. Regular Polygons and Circles. — A Reg- ular Polygon has its sides equal and its angles equal. A regular poly- gon may always be separated into equal isosceles triangles by drawing lines from its center to each vertex. A line from the center, perpendicular to a side, is called the apothem. The lines bounding a polygon are called its perimeter. Note. — A pentagon has 5 sides; a hexagon, 6 sides sides ; an octagon, 8 sides ; a decagon, 10 sides. a heptagon, 7 228 ADVANCED ARITHMETIC. A Circle is a plane bounded by a curved line, all parts of which are equally distant from a point in the plane. The point is called the center and the curved line bounding the circle is called the circumference. A straight line, passing through the center of a circle and terminating at each end in the circumference, is called a diam- eter. A straight line extending from the center to the circumference is called a radius. ADBC, a circumference ; AB, diameter ; OC, a radius. Peoblem 1 : Develop the formula for the regular polygon. Development : Let ABCDFG be a regular polygon. Divide it into equal triangles by drawing lines from the center to each vertex. Draw OE, the apothem, which is the length of the altitude of each triangle. Represent the bases of the triangles by 6, b^, h'^ ^ etc. Then, In 1st triangle : A = -^ ah' In 2d triangle : A = -jr- In 3d triangle : A = — — ^Q , and so on. The whole area = ah+a¥ + a¥'-[-etc ; or, a{h + ¥^¥'^etc.) But h^-y-\-¥' + etc. = the perimeter (P) of the polygon. Then, the for- mula for the polygon is — aP 2 • A=- Relation VII. Abstractly^ the area of a regular polygon is equal to the product of its apothem, and perimeter, divided by 2. ADVANCED ARITHMETIC. 229 Relation VIII. The ratio of the circumference (C) of a circle to its diameter (D) is S.l^ld (t). Note 1. — The character ir is the Greek letter pi, and is used for 3.1416, in order to shorten the work in writing formulas. Note 2. — The development of this relation is too difficult to be given here. It is found in geometry. Formula : IT =-^=8.1416. Since the diameter equals two times the radius (R), we may substitute 2 R for D. C C=2^R. Relation IX. The circumference of a circle is equal to 2 times SAJf-lG times the radius. Problem 2: Develop the formula for the area of the circle. Development : The figures above suggest the idea, that as the number of sides of a regular polygon increase, the polygon approaches the form of a circle. Proceeding upon the suggestion, and considering a circle a regular polygon of an infinite number of sides, we observe that the apothem becomes the radius of the circle and the perimeter becomes the circumference and the formula for the polygon. aP ^: 2 ' A = becomes A = 230 ADVANCED ARITHMETIC. But from the last formula we know that the circumference equals 2 TT /?. Putting this in the place of C, we have-^ A = ; or, 2 ' Relation X. Abstractly j the area of a circle is equal to 3.14.16 times the square of the radius. EXAMPIiEB. 1. Find the area of a regular pentagon whose apothem is 10 ft. and one side 14.58 ft. Solution : (1) P= 5 x 14.53 ft. = 72.65 ft. (2) ^ = 1«^ = 363.25. .'. the area is 363.25 sq. ft. 2. Find the diameter of a circle whose circumference is 219.912 rods. Solution: (1) -^- = 3.1416. - (2) 219.912 ^3 .^^^g (3) 219.912 = 3.1416 1). (4) 2)= 219.912 ^^Q ' 3.1416 ^ .'. the diameter is 70 rd. S. Find the area of a circle whose radius is 200 ft. Solution: (1) ^=7rll2^ (2) A = 3. 1416 X 40000 = 125664. .'. the area is 125664 sq. ft. 4-. Find the area of the square that can be inscribed in a circle whose diameter is 20 ft. Note. — A square is inscribed in a circle, if all of its angles lie upon the circumference. A square is circumscribed about a circle, if it incloses the circle with its sides touching the circumference. ADVANCED ARITHMETIC. 281 Explanation : In the figure, OB and OC are radii of the circle, and are therefore each 10 feet long. But these radii are sides of the right triangle COB, which is one-fourth of the required square ABCD. Find the area of the triangle and multiply by 4. Solution : In the triangle : A = — - — = 50. :. the area of square = 4 x 50 sq. f t. = 200 sq. ft. EXERCISE Gil. 1. Study and give from memory the solutions of the two problems of this Article. 2. Commit to memory Relations VII to X. « S. Define and draw a regular polygon. If. What is a pentagon? Octagon? Decagon? 5. Define a circle ; circumference ; diameter ; radius ; perim- eter; apothem. 6. What is the formula for the area of a circle ? A regular polygon? 7. Find diameter of a circle whose area is 2829.44 sq. in. 8. Find the radius of a circle whose circumference is 376.992 feet. 9. Find the apothem of a regular polygon whose area is 20800 sq. yd. and whose perimeter is 520 yd. 10. Find the area of a circle whose circumference is 87.6992 ft. 11. Upon how much pasture can a horse graze, if he is tied by a rope 60 ft. long ? 12. A horse is tied to the corner of a house 40 ft. wide and 60 ft. long, by a rope 90 feet long. Over how much ground can the horse go ? IS. Find one side of the square that can be inscribed in the circle whose diameter is 50 ft. 282 ADVANCED ARITHMETIC, ^ IJf. Find the area of the circle if one side of its inscribed square is 25 ft. 15. Find the circumference of a circle whose area is 254.4696 sq. ft. 16. The radius of a circle is 34 inches. Find the area of the inscribed square. n. A circle is 62.882 ft. in circumference. Find the area of the circumscribed square. 3. SOLIDS. 127. Polyhedrons.— Define solid. (See p. 156.) A Polyhedron is a solid bounded by planes. The bound- ing planes are called faces. The lines of intersection of the planes are called edges. The points where the lines of inter- section meet are called vertices. A rectangular solid. A cube. A Rectangular Solid is a polyhedron bounded by rec- tangular planes. A Cube is a rectangular solid whose dimensions are all equal. A Prism is a polyhedron two of whose faces called bases are parallel polygons, and the other faces, called lateral faces, intersect in parallel lines. The altitude of a prism is the perpendicular dis- tance between the planes of its bases. If all the lateral faces of a prism are rectangles, the prism is a right prism; if not, it is an oblique prism. A right triangular prism. NoTK 1. — We will study only right prisms in this book. The lateral (or convex) surface of a prism is the sum of the areas of the lateral faces ; ADVANCED ARITHMETIC. 288 A right pentagonal pyramid. the total surface is the sum of the lateral area plus the areas of the two bases. Note 2. — A prism with a triangle for a base is a triangular prism ; with a square base, a square prism; with a hexagon for a base, a hexagonal prism, etc. A Pyramid is a polygon one of whose faces, the base, is a polygon, and the lateral faces triangles meeting in a point called the vertex of the pyramid. If the lateral faces are all equal isosceles triangles, the pyramid is called a right or regular pyramid ; if the lateral faces are not isosceles triangles, the pyramid is an oblique pyramid. The perpendicular distance from the vertex to the base is called the altitude. The slant height of a right pyramid is the distance from the vertex to the middle point of one side of the base. Note. — A pyramid with a triangle for a base is a triangular pyramid ; with a square base, a square pyramid ; with a pentagon for a base, a pen- tagonal pyramid, etc. EXERCISE cm. Define : 1. A solid. 2. A polyhedron. 3. A face of a polyhedron. 4. An edge of a polyhedron. 5. A vertex of a polyhedron. 6. A rectangular solid, 7. A cube. 8. A prism. 9. Bases of a prism. 10. The altitude of a prism. 11. A right prism. 12. An oblique prism. 13. A triangular prism. IJ^. A square prism. 15. An octagonal prism. 16. A pyramid. 17. The altitude of a pyramid. 18. A triangular pyramid. 19. A pentagonal pyramid. 20. A right pyramid. 21. Lateral area. 22. Total area. Problem 1: Letting 1, w and t rcptresent length, loidth and 234 ADVANCED ARITHMETIC. thickness respectively, develop the formula for the volume of a rec- tangular solid. BY EQUATION. Solution : (1) Vol.of asolidZft. l.,i(;ft.w., = ^. (Prin.l.) 252 ADVANCED AKITHMETIC. Relation I. Abstractly, the gain or loss is equal to the product of cost price and the number of per cent of gain or loss, divided by 100. When there is a gain, (1) S=C+G. (Prin. 2.) But, (2) G = ^ Then,(3)^'=C+^ = 150^;or, 0(100+p) ^' ^- — 100^ • When there is a loss, (1) S=C-L. (Prin. 2.) But, (2) L=P^. 100 O(lOO-y) ^- *- — ioo~- Relation II. Abstractly, the selling price is equal to the product of the cost price and 100 plus the per cent of gain, divided by 100 ; or the product of the cost price and 100 minus the per cent of loss, divided by 100. EXAHPIiES. 1. How much does a man gain by selling an article that cost $12.50 at 40% profit ? Solution: ^,,40x12.5^^ (Why?) » 100 Answer, $5. 2. An article cost $4.20 : for how much must it be sold to gain 20% ? Solution: ^ = 4.2x120^^ q^ (Why?) 100 Answer, $5.04. ADVANCED ARITHMETIC. 258 3. A drover paid $1820 for cattle that he afterwards sold at a loss of 16|%. What did he lose ? Solution: L = M^il??2 = 220. (Why?) 100 Answer, $220. Note. — Use cancellation when practicable. 16| is contained in 100 6 times. 4„ Flour that cost $4.50 per barrel sold for $4.95 per barrel. What was the per cent of gain ? • Solution: (1) ^ = ^iM±£) 100 (2) 4.95 = i:511?2±£> (2) = (3)M1100+£) = 4.95. ' ^ ' ' 100 100 X (3) = (4) 450+4.5j9 = 495. Transpose, (5) 4.5 p = 495 -450 = 45. (5) -H 4.5 = (6) p = 10. Answer, W. Another Solution : (!) G = S-C = $4.95 - $4.50 = $.45. ,3)M^ = .45. 100 X (3) = (4) 4.5p = 45. (4) -=-4.5 = (5) p = 10. Answer, W. 5. Find the per cent of loss, if goods that cost $2720 sell for $2380. Solution: {1)S = ^^^^^^ 100 (2) 2380 = ?™aOO-p) ' 100 100 X (2) = (3) 238000 = 272000 -2720 p. Transpose , (4) 2720 p = 272000 - 238000 = 34000. (4) -H 2720 = (5) « = ?155^=12k ^ 2720 ^ Answer, 121^. 254 ADVANCED ARITHMETIC. Another Solution : (1) L = C-S = $2720 - $2380 = $340. (2) ^ = L. 100 (3) 2720^^340, 100 (4) 2720 p = 34000. (5) « = ?^??9 = 12i. ^ ' ^ 2720 Answer, 12^5^. Note.— In No. 4, first solution, equation (2) turned around gives equa- tion (3). If I had transposed, I would have a minus sign before the p. In No. 5, first solution, I transposed in equation (3) to get equation (4). If I had turned the equation around as in No. 4, I would have a minus sign before the p. If the letter has a plus sign in the right member, turn the equation around, to change to tJie left rnemher; if the letter has a minus sign, transpose, to get rid of the minus sign. 6. A merchant bought goods at 20% less than market value and sold them at 20% more than market value. What % did he gain? Solution : (1) C = 80^ of market value. (Why ?) (2) G^ = 40^ of market value. (Why ?) Note. — The problem then becomes, "40^ is what % of 80^ ? " ' ,-, 40x100 „ Answer, 50^. Oral Solution : Since the cost is 80^ of market value, and the gain is 40^ of the market value, the gain is one-half, or 50% of the cost. 7. A man lost 16|% by selling a house for $538 less than it cost. Find the cost. ADVANCED ARITHMETIC. 255 Solution: (1) L = ^--. (2)538=^. (3) ^^538x100^3^ 16| Answer, 8. A merchant sold goods for $980.28, thereby losing 10% . Find the cost. Solution: {I) S=^^^-^. (2)980.28- ^^ -^^. (3) = ^-55:^^=1089.2. Answer, $1089.20. 9. How must I mark a pair of shoes that cost $2.40, to gain 25% ? ( Key : Come and buy.) _, . . _, 2.4 (100+25) _ Solution: S = ^- ^ = 3. 100 Answer, m.yy. Note. — Keys are employed by many merchants to mark the selling price as well as the cost price of their goods, in a manner known to themselves only. The letters of the key, taken in order, represent the figures : come and buy 1234 567 890 Then, ^3.00 = m.yy. A separate character, as x, is often used instead of repeating a letter. Thus m.yy may be written m.yx. EXERCISE ex. 1. What is the base in profit-and-loss problems? 2. Commit to memory relations I and II. 3. Give from memory the formulas under relation I. 4-. Develop all the formulas for profit and loss. 256 ADVANCED ARITHMETIC. No. 5. $C P% $G #L $S 484 25 ? ? 6. 564 25 ? ? 7. 840 33i ? ? 8. 420 m ? ? 9. 248 80 m ? ? 10. 590 ? 59 ? 11. 780 ? 219 ? 12. 48 50 ? ? 58 20 13. 32 40 ? ? 25 92 u. ? ? 8 58 65 78 15. ? ? 2 50 10 16. ? 18i 6 30 ? 17. ? 6i 27 30 ? 18. 1 ? 120 192 ? 19. Jones sells goods costing him $1500 at a profit of 16f %. Find the gain. 20. I sold 100 hogs, averaging 300 lb. each, at a profit of 33^%. If they cost me 1^1200, find the selling price per lb. 21. What was the loss, if a pair of shoes that cost $2.50 sold at 20% loss? 22. How would you mark an article that cost $3.20 to sell at a 20% profit ? ( Key : Buy for cash.) 23. If $187.50 is gained on goods that cost $1250, find the selling price, and per cent of gain. 24.. If $260 is lost on goods that cost $1300, find the selling price and per cent of loss. 25. If goods sold at 16f % profit for $1085, find the cost price. 26. If goods sold at 25% loss for $480, find the cost price. 27. Sold a horse for $80, losing 20% ; with the $80 bought another, and sold him, gaining 20%. Find the loss on the two transactions. 28. A, failing in business, paid B $1512, which is only 60/ on the $1 (60%). How much did A owe B ? ADVANCED ARITHMETIC. 257 29. Mr. Smith bought goods and sold them at a loss of 12^%, losing $187. Find the selling price. 30. I bought a horse, and afterwards sold him at a loss of 20% ; I added 20% to what I then had, and with this amount bought another horse. I sold the second horse at a profit of 20% for $144. What did the first horse cost ? 134. Trade Discount. — Wholesale merchants usually mark the selling price of their goods so that they can make a reduction from this price in selling to their customers. The marked price is often spoken of as list price. The reduction from the list price is called Trade Discount. Note. — Discounts are made for one or more of the following reasons: (1) The goods may be marked in the first place with the expectation of making a uniform reduction to all retail customers, (2) for large purchas- ers, and (3) for cash. " 25% off," or "25 off," means 25% less than the list price. ** 25 and 10 off " means 25% less than list price, and then 10% less than the amount left after the first reduction. " J and 5 off," means the same as " 88^ and 5 off." Principles : 1. The first discount is some number of per cent of the list price. 2. Each succeeding reduction^ or discount^ is some number of per cent of the amount left after making the last preceding reduction. 3. The selling price ^ after any discount^ is equal to the price be- fore the discount J minus the discount. TERMS. L, list price. Z), discount. S, S\ S^\ etc., successive selling prices. Pj number of % off. 258 ADVANCED ARITHMETIC. Development of the formulas for Trade Discount. 1. D= ^. (Prin.l.) 100 Relation I. Abstractly, the discount is equal to the product of the list price and the number of per cent of discount, divided by 100. (1) S = L-D. (Prin. 3.) ^ jPL lOOL-pL 2. ^.^(1^-^) 100 Relation II. When there is one reduction, abstractly, the sell- ing price is equal to the product of the list price and 100 minus the number of per cent of reduction, divided by 100. When there are several reductions, the amount left, or S, of the first reduction becomes the L for the next reduction, and so on. Then, if another per cent {p^) be taken off the new selling price, S.= «ii^). (Prin. 2.) Putting in place of S its value from the formula above, we have: X(100-p)(100^p-) 100X100 In the same way, a formula for any number of reductions may be obtained. Relation III. After any number of reductions, abstractly, the selling price is equal to the fraction whose numerator is the product of the list price and the several remainders found by sub- tracting each number of per cent from 100, and whose denominator is the product of as many lOO^s as there are reductions. ADVANCED ARITHMETIC. 259 EXAMPIiES. 1. Find a discount of 20% off of a list price of $540. Solution: D = ^^^ = m. Answer, $108. 2. If the list price is $460, find the selling price at 30 off. Solution: S = ^^^ = 322. (Why?) Answer, $322. 3. If the list price is $500, find the selling price at 20 and 10 off. ^ - ^. ^, 500x80x90 -^ Solution: S^ = ^^^^^ =360. Answer, $360. 4. I sold an article that cost me $12, at i, 15, and 5 off. Find the selling price. ^ , ^. ^, 12x661x85x95 ^ ,„ Solution: S^ = .^ ' ..^ =6.46. 100 X 100 X 100 Answer, $6.46. NoTB.— 33i will cancel 66| and 100. 5. If an article sells at 80 off for $14, find the list price. Solution: {1)^^^^^^^ = S. Answer, $20. 6. If an article sells at 30 and 10 off for $6.30, find the list price. c 7 ,• ,,. xaoo+£Mioo+|/) ^ ^^^"'*"^'" ^^^ looTioo ='^- 70X90XX ^^^ 100x100 -^•^^- ,„. ^ 6.30x100x100 ,_. ^^^^- 70^^90 -^^• Answer, $10. ^60 ADVANCED ARITHMETIC. 7. An article, listed at $20, sold at ( ) and 10 off for $13.50. Fill the blank. » c 7 4- /l^ lOKA 20(100-p)x90 ^otoon: (I) 13.50= ^^/^^ . 10000x(l) = (2) 135000 = 1800( 100 -p). (3) 135000 = 180000 -1800 p. Transpose, (4) 1800^-180000-135000. (5) 1800^ = 45000. .^. 45000 ^^ ^'^^=1800=^- Answer, 2b%. EXERCISE CXI. 1. What does " 12i and 10 off " mean ? 2. Give the principles governing Trade Discount. 3. Develop the formulas and commit to memory the rela- tions. No. $L 'P% $D $ S ^. 580 10, 10 ? ? 5. 460 ? 69 ? 6. 500 20, ? ? 375 7. ? 15, 10 9 459 8. 720 ? 20 ? 432 9. ? 15 63 81 ? 10. ? 1 ? 304 04 1216 16 ii. A man sold goods listed at $437 at 20% off. What did they sell for ? 12. Goods sold at 25 and 5 off of a list price of $540. Find the amount of the discount. 13. Goods marked at $380 sold at i, 10, and 10 off. Find the selling price. 11^. Merchandise sold at 20 and 15 off for $1360. Find the list price. ADVANCED ARITHMETIC. 261 15. I bought an article, marked it at $20, and sold it at 80 and ( ) off for $12.60. Fill the blank. 16. How must I mark goods that cost me $450 so that I may sell 10 off of list price at a gain of 20%. 17. I sold an article at 20 and 10 off. If the discounts amount to $3.50, for how much did the article sell ? Note. — The 1st discount is 20?^ of the list price, and the 2d discount is \0% of 80^ of list price. 135. Commission. — Very often merchandise, especially such as produce and live stock, is bought and sold in distant markets. Usually, such buying and selling are not done by the parties furnishing the money and the commodities, but by agents employed for that purpose. An agent, or commission merchant, charges for his services a certain per cent on the amount he receives in selling or the amount he pays out in buying. This charge is called Commission. Principles: 1. The commission is always some number of per rent of the price of that which is bought or sold. 2. The proceeds are equal to the selling price minus the commission . S. The amount is equal to the selling price plus the commission. TERMS. S, price of the sale or purchase. p, number of per cent of commission. 0, commission. P, part left after deducting the commission. A, amount of investment and commission. Note. — S is selected to represent the price of both purchases and mles — that on which commission is based. Development of the formulas for commission. C = ^ — . (Prin. 1.) 100 262 ADVANCED ARITHMETIC. Relation I. Abstractly, the commission is equal to the product of the purchase {or sale) and the number of per cent of commission j divided by 100. To find the proceeds : ( Prin. 2.) 100 100 ' p^ s(im-p) 100 • Relation II. Abstractly, the proceeds are equal to the product of the purchase (or sale) and 100 minus the number of per cent of commission, divided by 100. To find the amount : ( Prin. 3.) A = S + C = S+^', or, 100 Relation III. Abstractly, the amount of investment and com- mission is equal to the product of the purchase (or sale) and 100 plus the number of per cent, divided by 100. examples. 1. 1 sell goods for $1430, commission 4%. Find my com- mission. „ - ,. ^ 1430x4 __ _ Solution: C = = 57.2. 100 Answer, $57.20. 2. How much must an agent remit to his principal, if he sell goods to the amount of $1200, commission 2% ? Solution: P= = 1176. 100 Answer, $1176. ADVANCED ARITHMETIC. 263 3. I send my agent $1442 to be invested in goods, commis- sion 3%. What is the price of the goods bought? . ^(100+p). 100 SxlOS Solution: (1) A (2) 1442 = 100 1442X100^^^^^ 103 Answer, $1400. 4.. A lawyer charges $35 for collecting a debt of $700. Find his % of commission. Solution : (1) (2)p = 100 35x100 700 Answer, 5^ = 5. EXERCISE CXII. 1. What is commission ? A commission merchant ? 2. Name and explain the terms used in commission. 3. Develop the formulas for commission. 4-. Commit to memory the principles and relations. No. $S Pfo $c $P $A 6. 724 50 4 ? ? 6. 840 2^ ? ? 7. 920 ? ? 892 40 8. 128 80 ? 6 44 ? 9. ? ? 48 75 1901 25 10. ? 3 ? 1225 11 11. ? 2 26 50 ? 12. ? H 9 588 70 13. ? ? 14 854 264 ADVANCED ARITHMETIC. i^. A commissiou merchant sells 8000 lb. of bacon at 10/ per pound ; he pays freight $50, and drayage $2. If his com- mission is 2%, what are the net proceeds ? Note. — "Net proceeds" are the balance after deducting all expenses. 15. An agent receives $475 with which to purchase goods. If his commission is 1^%, what amount does he invest? 16. My agent sells 50 head of hogs averaging 225 pounds, at 5/ per pound. He invests the net proceeds in wheat at $.50 per bushel. His commission for selling is 8% ; for buying, 2% . How much wheat does he buy ? 17. My commission at 2% for selling 600 hogs, averaging 225 lb. each, is $135. What did I get for the hogs ? how much per hog ? how much per pound ? 18. I received an account sales, stating that my net proceeds were $4301.80; freight, $55.70; drayage, $8.00; commission, $135. Find the rate of commission. Note. — An ''account sales'' is a final statement made by the agent to his principal. It shows (1) the selling price of the goods, (2) all the ex- penses, including commission, and (3) the proceeds to be remitted. 19. An agent, after deducting his commission, 5%, other expenses, $42.24, had $988.76 to remit. Find the amount of sales. 20. Sold flour at 3^% commission ; invested | of the proceeds in coffee at 1^% commission, and remitted the balance, $482.50. What was the value of the flour, the coffee, and each commis- sion ? The commissions are taken out of the remaining I of the sales of the flour. 136. Stocks and Bonds. Note. — Children in the public schools usually know little or nothing about Stock Companies or Corporations. A day or two spent in working ADVANCED ARITHMETIC. 265 out the eight examples below will be of great service to the pupils. Or- ganize the Galena Mining Company, have pupils represent Jones, Smith, Brown, etc., write out some of the certificates of stock, discuss the mean- ing of dividends, surplus, and so on. Pupils cannot intelligently solve problems which they do not understand. CEKTIFICATE OF STOCK. No. I. Capital Stock, $200,000. Shares, $100 each. £00 Shares. GALENA MINING COMPANY, GALENA. KANSAS. This !s to certify that [AyJUiuJuira. J&yrJMx^ is entitled to c3u>o (hL\J^'Y\xL\>ixL Shares in the Capital Stock of the Galena Mining Connpany. Transferable only on the books of the Connpany, in person or by attorney, upon surrender of this certificate. In Witness Whereof, the seal of said Company is (L. S.) hereto affixed, this I Otfv day of 9eiv., i8C|0. Attested by Cuxburu OtAA/\A^, Secretary. The Par Value, or face value, of a share of stock is what it is valued at on the face of the certificate. The Market Value of a share of stock is its selling price in the market. Note. — When a share sells for more than its face value, it is said to be nhove par, or at a premium. When a share sells for less than its face value, it is said to be below par, or at a discount. Stock companies usually divide their earnings (profits or gains) annually. The amount thus divided is called the divi- dend. Whoever owns a share of stock receives the dividend paid on that share. Buying or selling of stocks is sometimes done through a broker (agent), who charges for his services a brokerage (com- mission) of so many % of the par value of the stocks bought or sold. 266 ADVANCED ARITHMETIC. EXAMPI.E8. The Galena Mining Company was organized and incorporated under the laws of Kansas, January 1, 1890, with the following stockholders : William Jones, 200 shares $20000 William Smith, 400 shares 40000 James Brown, 850 shares 35000 Galen Terry, 400 shares 40000 G. W. Jolley, 650 shares 65000 Capital stock (C. S.) .$200000 At the close of the year 1890, the net earnings of the com- pany were $22480.50. 1. If the company reserves a surplus in the treasury of $2480.50, what will be the rate of dividend (R. D.)? Note. — After reserving a surplus to pay running expenses, what is left of the earnings or gains is divided among the stockholders; Solution: (1) $22480.50 -$2480.50 = $20000. Amt. to be divided. (2) $20000 = ( )% of C. S. ? ( Question.) (3) $200000 = 100^ of O.S. (Basis.) tV of (3) = (4) $20000=10^ of 0. S., answer. 2. What is Smith's share of the dividend (his C. S., $40,000)? Solution : (1) 10^ of his 0. S. = $( ) ? (Question ) (2) 100^ of his 0. S. = $40000. (Basis.) iV of (2) = (3) 10^ of his C. S. = $4000, answer. Note. — Brokerage is the commission an agent (broker) charges for buying or selling stocks or bonds. But unlike commission, brokerage is always some per cent of the Par Value of the stocks or boTK^s, and not some per cent of the cost. 3. Jones sells some of his shares to W. W. Miller at 90 (10% below par), brokerage ^ %, for $13518.75. Find the face value of the stocks sold. ADVANCED ARITHMETIC. 267 Solution : (1) 90^ of par value = price of stocks. (2) I ^ of par value = brokerage. (3) 100^ of par value = $( )? (Question.) (4) m% of par value = $13518.75. (Basis.) ylT of (4) = (5) 1% of par value = $150. 100 X (5) = (6) 100^ of par value = $15000, answer. At the beginning of the year 1891, Terry sells his shares to J. O. Bowers; Jolley sells 300 shares to G. W. Greene at 120 (20% above par) ; at the end, of the year 1891, the company finds that, after reserving a surplus of $3124.40, it can declare a dividend of 6%. Note. — For the sake of brevity the ^'question " will be omitted in the remainder of these solutions. 4-. Find the net earnings of the company for 1891. Solution : (1) 100^ of C. S. = $200000. (Basis.) T^ of (1) = (2) 1^ of 0. S. = $2000. 6x(2) = (3) 6^of 0. S. = $1^000. (4) $12000+3124.40. =$15124.40, answer. 5. What did Greene's 800 shares cost him (at 120) ? Solution: (1) 100^ of 300 shares = $30000. (Basis.) T^jy of (1) = (2) 1% of 300 shares = $300. 120 X (2) = (3) 120^ of 300 shares = $36000, answer. 6. What was Greene's dividend ? SoZw^ion : (1) 100^ of 300 shares = $30000. (Basis.) T^^ of (1) = (2) 1% of 300 shares = $300. 6 X (2) = (3) Q% of 300 shares = $1800, answer. 7. What is Greene's rate of income on his investment ? Note. — The *'rate of income " is always some number of per cent of the investment or cost. Solution : (1) $36000 = 100^ of his investment. (Basis.) aV of (1) = (2) $1800 = 5^ of his investment, answer. 268 ADVANCED ARITHMETIC. 8. If Bowers 's rate of income is 8% for 1891, for how much below par did he buy his stock ? (R. D. was 6%). Parts : (a) 8^ of investment = 6^ of his C. S. (6) IW of investment = ( )% of his 0. S.? Solution : (1) 8^ of investment = 6^ of C. S. I of (1) = (2) 1% of investment = 1% of 0. S. • 100 X (2) = (3) 100^ of investment = 75^ of 0. S., or 2b% below par, answer. Bonds are promissory notes issued by a public corporation (nation, state, county, city), or a private corporation (railroad company, banking company, mining company, etc.). Bonds have a face value (par value), a rate of interest, and when bought and sold they have a market value. Brokerage may be charged by an agent for buying or selling bonds. EXAlCPIiES. 1. I own 6's to the amount of $5500; what is my annual in- come, or interest ? Note. — ** 6's " means that the rate of interest is Q%. Solution : (1) 100^ of P. V. = $5500. (Basis.) i^^of (1) = (2) l^of P. V. = $55. 6 X (2) = (3) 6^ of P. V. = $330, answer. 2. A buys bonds amounting to $50000 at 104, brokerage \% ; what do they cost him ? Solution : (1) 100^ of P. V. = $50000. (Basis.) ^U of (1) = (2) 1% of P. V, = $500. 104i X (2) = (3) 104i^ of P. V. = $52125, answer. 3. A invests $7210 in bonds at 90, brokerage ^% ; what is the face value of the bonds ? Solution : (1) 901^ of P. V. = $7210. (Basis.) yfy of (1) = (2) 1^ of p. V. = $80. 100 X (2) = (3) 100^ of P. V. = $8000, answer. ADVANCED ARITHMETIC. 269 !^. B invests $24000 in 4's, at 80; what is his interest ? Solution : (1) 80^ of P. V. =$24000. (Basis.) i^ of (1) = (2) 1^ of P. V. = $300. 4 X (2) = (3) 4% of P. V. = $ 1200, answer. 5. How must C buy G's to yield an income of 5% ? Solution : (!) 5% of M. V. = Q% of P. V. (Basis.) 20x(l) = (2) lOO^of M. V. = 120^of P. V.,answer. 6. Which furnishes the better income on the investment — 5's at 75, or 6's at 80 ? In the First : (1) 75^ of P. V. = 100^ of investment. (Basis.) tV of (1) = (2) 6% of P. V. = 6|^ of investment. In the Second : (1) 80^ of P. V. = 100^ of investment. (Basis.) ^ of (1) = (2) 1% of P. V. = 1% of investment. 6 X (2) = (3) Q% of P. V. = 7i^ of investment. (4) 71% of investment -61^ of investment = |^ of investment, in favor of the second, answer. The problems of stocks and bonds are so nearly alike that one set of principles and formulas applies to both. Principles: 1. The discount, premium ^ dividend (or interest) and brokerage are each some number of per cent of the par value. 2, The per cent {or rate) of income is some number of per cent of the cost. 3. The market value is equal to the par value plus the premium, or minus the discount. If. The cost is equal to the market value plus the brokerage. terms. P, par value. M, market value. C, cost. B, brokerage. D, dividend (stocks). /, interest (bonds). 270 ADVANCED ARITHMETIC. In., income. Pr., premium. Dis., discount. R, number of % of the dividend or interest. r, number of % of the income on the investment. j9, number of % of premiupi or discount. _p', number of % of the brokerage. Develop the formulas for Stocks and Bonds : pP 1. Pr. or Dis.=f-— . (Prin. 1.) ^' 100' ^^'■'''- ^'^ T. r "fP 3. D or I=-Yqq' (Prin. D Since the income on the investment is the same as the dividend (stocks) or the interest (bonds), it follows that — rC 4. D or 1^1^. (Prin. 2.) When there is no brokerage, 5. D or /=^. (Why?) By comparing formulas 3 and 4, we obtain RP rC 100" 100' °^' 6. RP=rC. When there is no brokerage, 7. RP=rM. When stocks or bonds are at a premium, (1) M=P+Pr. (Prin. 3.) (2) C = P4-Pr+5. (Prin. 4.) ADVANCED ARITHMETIC. 271 P(100+p) 8. M- ^^ . (2) -(4) c r\P^\P'^ mp+pp+p'p (2)-(4) C-^+jgg+^^- ^^ ,or, P(100+p+/) -^^ ^~ loo • When stocks or bonds are at a discount, show that — EXAMPLES. Note. — In this book, the par value of a share or a bond is $100, unless it should be otherwise stated. 1. Find the discount on 20 shares of stock sold at 20% be- low par. Solution: Dis. = — — r — =400. 100 Answer, $400, 2. Find the premium on 60 shares of stock sold at 12^% premium. ^ , ^. „ 12^x6000 ^^^ Solution : Pr. = — ^~ = 750. 100 Answer, $750. 3. Find the brokerage at i% on 40 shares of stock. c, , ,. „ ix4000 ^ Solution: ^ = -^-——- = 5. Answer, $5. 4^. Find the dividend on 50 shares of 8% stock. ^ 8x5000 , ^ = -10^ = ' Answer, $400. Solution: D = l^i^ = 400. 272 ADVANCED ARITHMETIC. 5. Find the rate of interest on 50 bonds that yield $450 an- nual interest. Solution: (1) I=^r^. (2)450 = ^. Answer, %. 6. If the annual interest of $920 on bonds pays an income of 8%, find the cost. rC Solution: (1) -^=77^- (2) 920 = ||. (3) C = 8?^ = 11500. O Answer, $11500. " 7. Find the cost of 45 shares of stock bought at 110. ^ . ^. -. 4500x110 .__ Solution: M= r— = 4950. J-UU Answer, $4950. Note.— The "110" = (100+p) = (100 + 10). 8. Find the cost of 15 bonds at 60, brokerage -|. Solution: (7 = -^^^^^ = 901.875. Answer, $901,875. Note.— The 60i = (100 -p+pO = 100- 40+i). 9. I buy 6% bonds at 80; find my rate of income. Note.— Assuming 1 bond as a basis, the P is $100, the M is $80. Solution: (1) RP=rM. (2) 6xlOO = 80r. ,3) rJ-^^-V. Answer, 11%. ADVANCED ARITHMETIC. 278 10. I bought 5's, ^ brokerage, for $903.75. If they pay me ^2TT% Oil my investment, find the par value, the % of dis- count, the brokerage, and the market value. Note. — '' 5's " means bonds that bear b% interest per annum. Solution: (1) RP=rC, (2) 5 P=8/A:X 903.75. (3) p^2000x 903.75^ 241x5 Par value, $1500. Brokerage, $3.75. (5) if= $903.75 -$3.75 = $900. (6) Dis. = $1500 -$900 = $600. Discount, 40^. EXERCISE CXIII. 1. Discuss the organization of a stock company — the capital stock, the issuing of shares, and the distribution of earnings. 2. Define brokerage, par value, market value, premium, dis- count. 3. When a company loses instead of gaining during the year, how is the assessment made to pay this loss ? J^. What is a bond ? The interest ? 5. Commit to memory the principles. 6. Develop the formulas, and write out the relation expressed by each formula. 7. Find the discount on 34 shares of stock, sold in market at 30% below par. 274 ADVANCED ARITHMETIC. 8. Find the premium on 75 shares of 10% stock, sold in mar- ket at 30% premium. 9. I bought bonds at 105 for $7350. Find the par value. 10. 40 shares of stocks sold in market for $3600. Find the per cent, of discount. 11. I bought 80 shares of 6% stock. Find my annual divi- dend. 12. My interest on 6% bonds is $540. Find the face value of the bonds. 13. I receive a dividend of $400, which is 6|% on my invest- ment. Find the cost of the bonds. 14.. I bought 60 shares of 8% stock for 110. Find the rate of income. 15. I bought 5% stock at 70. Find the rate of income. 16. I bought 6% bonds so as to pay me 10% on my invest- ment. Find the market price. 17. Find the cost of 40 shares of stock, sold at 80, brokerage i%. 18. Find the cost of 75 shares of stock, sold at 120, broker- age i%. 19. How many shares of stock at 77 J, brokerage i%, can be bought for $15600 ? 20. Which pays the better income on the investment, and how much — 7's at 90, or 6's at 75 ? 21. A man owns 5% railroad stocks worth 90 and 4% bonds worth 80. If his income from each is $720, find the par value of each. 22. In No. 21, find the rate of income on each investment. 23. How much money must be invested in 4's at 90 to yield an annual income of $120? 21/.. On bonds bought at 70, my rate of income is 5%. This income annually amounts to $1750. Find the amount of the bonds and the rate of interest. ADVANCED ARITHMETIC. 275 137. Taxes. — A tax is a sum of money levied on persons (poll tax) or property (property tax). Principles: 1, A poll tax is a certain sum required usually of those citizens who have the right to vote. 2. A property tax is some number of per cent of the value of the property taxed. TERMS. F, value of property. p, number of % of the taxes. r, taxes. Formula: T=^. (Prin. 2.) EXAMFIiEB. 1. In a certain county there were 7570 polls, at $1.25. Find the poll tax. Solution: r= 7570 x $1.25 = .$9462.50, answer. 2. The property of the county was valued at $5897500. Find the tax raised by a i% (5 mills) levy. Solution: r= ^^^^^^^ = 29487.50. 100 Answer, $29487.50. Note. — In tax levies is almost the only place in business where the denomination mill is used. 1 mill = i^,<*. A 5-mill levy is a levy of 5 mills or J^ on $1 ; or, |^ of the value of the property taxed. EXERCISE CXIV. 1. Define poll tax; property tax. 2. Give the principle governing poll tax ; property tax. 3. Give the relation expressed by the formula above. 4.. If the number of polls be represented by N, and the tax on each poll by t, give the formula for finding poll tax, T. 5. Find the tax on 1200 polls, at $1.85 each. 276 ADVANCED ARITHMETIC. 6. Find the property tax oii a farm valued at $10400, if the levy is 14 mills (lf%). 7. What is the valuation of A's property who pays $51 on a 15 mill m%) levy? 8. A school district has $20000 taxable property, and desires to raise $240 teachers' wages. What % levy must they run ? 9. A tax of $787.50 is raised on 25 polls at $1.50 each, and a property tax of 6 mills. Find the value of the property taxed. 10. Find the % of the tax levy, when $17500 worth of prop- erty, and 40 polls at $1 each, produce a tax of $285. 138. Duties. — Taxes levied by the government upon im- ported merchandise are called Duties. Duties are of two kinds, specific duties and ad valorem duties. A Specific duty is levied upon the amount of the merchandise imported. An Ad valorem duty is levied upon the value of the merchandise imported. Tare, leakage, and breakage are allowances deducted from the gross amount of imported merchandise. These deductions are made before levying duties. Tare is the deduction made for the weight of the boxes, bar- rels, etc., containing the merchandise. Leakage is the deduction made for the loss of liquids shipped in barrels or casks. The amount to be deducted is determined by gauging. Breakage is the deduction made for the loss of liquids shipped in bottles. Principles: 1. Specific duty is a certain sum per pound, yard, gallon, etc., on the merchandise imported. 2. Ad valorem duty is some number of per cent of the net value of the merchandise imported. TERMS. V, net value of merchandise. p, number of % of ad valorem duty. D, the duty. ADVANCED ARITHMETIC. 277 Formula for ad valorem duty : vV D=-^—-. (Prin. 2.) 100 EXAKPIiES. 1. What is the duty on 560 bottles of wine, duty 60^ per dozen, breakage 12^% ? Solution : (1) Breakage = 12J^ of 560 bottles = 70 bottles. (2) 560 bottles -70 bottles = 490 bottles = 40| dozen bottles. (3) i)M«i/ = 40|x 60^ = $24.50, answer. 2. Find the duty on 400 bags of coffee, each weighing 180 lb. gross, invoiced at 8/ per lb., tare 5%, ad valorem duty 20%. Solution : (1) Amount = 400 x 180 lb. = 72000 lb. (2) Tare = b% of 72000 lb. = 3600 lb. (3) 72000 lb. - 3600 lb. = 68400 lb. (4) raZMe = 68400 x8<* = $5472. 20X5472^ 100 Answer, $1094.40. 3. A merchant imports 10 casks of brandy, each containing 80 gallons, worth $1.40 per gallon; duties are 10/ per gallon, and 25% ad valorem; leakage, 10%. Find the whole duty. Solution : (1) Amount = 10 X 30 gal. = 300 gal. (2) Leakage = 10^ of 300 gal. = 30 gal. (3) 300 gal. -30 gal. = 270 gal. (4) Value = 270 X $1.40 = $378. (5) Sp. 2). = 270xlO,<^ = .$27. (6) Ad. D. = 25^ of $378 = $94.50. (7) $27 +$94.50 = $121.50, answer. EXERCISE CXV. 1. Define duty, specific duty, ad valorem duty, tare, leakage, breakage. 2. Commit the principles. 8. Give the relation expressed by the formula. 278 ADVANCED ARITHMETIC. Jf, Write the formula for specific duty (i)), when the amount (^) of the merchandise and the duty (cZ) on the unit of amount is given. 5. Find the duty on 500 yd. of silk invoiced at $1.50, at 20% ad valorem. 6. What is the duty at 25% ad valorem on 400 boxes of raisins of 40 lb. each, invoiced at 9/ per lb., tare 10% ? 7. The ad valorem duty on 1500 yd. Irish linen, valued at 40/ per yd., is $210. Find the % of duty. 8. Find the specific duty on 750 yd. broadcloth at 50/ per yd. P. What is the specific duty per gallon, when a duty of $540 is paid on 50 casks of wine of 30 gallons each, leakage 10% ? 10. Find the whole duty on 4500 lbs. of wool, invoiced at 20/ per lb., specific duty 10/ per lb., ad valorem duty 12%, tare 139. Insurance. — Insurance is indemnity against loss by damage or death. Insurance is of two kinds, 2Jroperty in- surance and personal insurance. Property insurance is indemnity for loss of property, caused by fire, lightning, tornadoes, death of stock, and the like. Personal insurance is indemnity for loss caused by death, sickness, or accident. A Policy is the written contract between the insurance com- pany and the person insured. A policy which insures the payment of a certain sum only at the death of the person insured, is called a Life policy. A pol- icy which insures the payment of a certain sum at a specified time, or at death occurring before that time, is called an En- doivment policy. Premium is the sum paid for insurance. The Risk is the amount insured. Note. — Insurance companies seldom agree to insure property for its full value. They usually make the risk i or | of the full value. ADVANCED ARITHMETIC. 279 Principles: 1. In property insurance, the pfemium, due an- nually, is some number of per cent of the risk. 2. In life insurance, the premium is usually a certain sum per ^1000, payable annually {semi-annually or quarterly) . TERMS. i?, the risk. p, number of % of the premium. P, the premium. Formula for Property Insurance : EXAMPLES. 1. Find the premium at 2% on a risk of $8400. Solution: P=^^^~ = m. Answer, $168. 2. The annual premium is $480, paid on a factory valued at $48000, insured at f of its value. Find the rate of insurance. Solution : (1) | of $48000 = $32000, risk. (2)480 = ^^^^50. (Why?) ._, 480x100 ,, ^^> P= 32000 '^^- Answer, 1|^. , ' 3. A company insured a vessel worth $60000 for f its value ^ , at 2% ; they reinsured | of their risk at If %. Find the pre- .! mium retained by the first company. Solution : (1) f of $60000 = $40000, whole risk. (2) I of $40000 = $30000, second risk. (3) 1st premium = 2% of $40000 = $800. (4) M premium = 1^^ of $30000 = $525. (5) $800 - $525 = $275, answer. ADVANCED ARITHMETIC. If. What will be the annual premium on a life policy ofj $4000 at $25.40 per $1000 ? Solution: Premmw = 4 x $^5.40 = $101.80, answer. o^. If Mr. Jones was insured for $5000, at an annual pre- mium of $86.40 per $1000, what amount did the company pay out more than it received, if Mr. Jones died after making 11 annual payments ? Solution: (1) PremfMm = 11 x5x $36.40 = $2002. (Why?) (2) $5000 -$2002 = $2998, answer. EXERCISE CXVI. ^•^jy^jiliUc. 1. Define insurance, policy, premium. 2. Define a life policy, endowment policy. 3. Give the principles. 4-. Give the relation expressed by the formula. • 6. How do you find the premium on a life insurance policy, when the amount of the policy and the premium per $1000 is given ? f \j,.)^6. Find the premium at 3% on a risk of $6500. ' ', v"* 7. What is the rate of insurance, if the x:)remium on $5420 is ,> $135.50? ^'\-s/''8. The premium at 2}% on a certain risk was $99. Find the risk. 9. A vessel valued at $50400 was insured for | its value at 1^%. The first company re-insured | of the risk in a second company, paying $441 premiuHi. Find the first premium, and the second rate of insurance. 10. Find the premium paid in 15 yr. on a life policy of $6000, at $27.50 per $1000 annually. 11. A man took out an endowment policy for $4000 and died after making 10 annual payments. If his estate received $2682 more than he had paid out, what was the annual premium per thousand ? -V ADVANCED ARITHMETIC. 281 3. PERCENTAGE "WITH TIME. 140. Terms Used in Interest. Interest is a sum paid for the use of money. The amount of money used is the Principal. The per cent of the principal which equals one year's interest is the Rate. If only the principal bear interest, the interest is called Simple Interest. When by the terms of a note interest is payable annually, (semi-annually, or quarterly,) and is not paid when due, this annual interest (interest due annually, semi-annually, or quarterly) bears interest at the stipulated or legal rate until paid. This is called Annual Interest. A note which reads, "and if the interest be not paid annually [semi-annu- ally, or quarterly], it becomes as principal, and bears the same rate of interest," bears Compound Interest. 141. Time. Since the rate in interest problems is ^^per annum, ^^ or by the year, the time should be expressed in years. By the common method of counting time in interest problems, lmo.=:yVyr. 1 da.=^io yr. This method of counting time is called the common interest method. Example: Express in terms of years (1) 3 mo., (2) 17 da., (3) 6 mo. 8 da., (4) 5 mo. 18 da., (5) 2 yr. 4 mo. 12 da. Results : (1) 3 mo. = j\ yr. (2) 17da. = ^«Vyr. (3) 6 mo. 8 da. = iU jr. (4) 5mo. 18da. = y^yr. (5) 2 y r. 4 mo. 12 da. = ^ yr. (6)' 1 yr. 2 mo. 7 da. = iU yr. Note.— In using these expressions in the interest problems to follow, cancellation will be used after the whole problem is stated ; therefore 282 ADVANCED ARITHMETIC. these expressions need not be reduced to lowest terms. When the num- ber of days is a multiple of 3, divide by 3 and express the result as tenths of a month ; as in (4) and (5) above. Thus : 15da. = j^yr. 33da. = ^yr. g 24 da. =^ yr., and so on. EXERCISE CXVII. Express in years : 1. 6 mo. 2. 8 mo. S. 9 mo. 4. 5 mo. 5. 10 mo. 6. 30 da. 7. 12 da. 8. 24 da. 9. 2 yr. 4 mo. 10. 1 yr. 8 mo. 11. 5 mo. 15 da. 12. 6 mo. 24 da. 13. 7 mo. 21 da. U- 1 yr. 6 mo. 12 da. 15. 2 yr. 4 mo. 10 da. 16. 1 yr. 11 mo. 23 da Find the time in years from — 17. Mar. 20, 1900, to Sept. 25, 1901. yr. mo. da. 1901 9 25 1900 3 20 Form : 19C 1 6 5 = myr. Note. — Subtract as in compound numbers (p. 176). i<5..Sept. 10, 1890, to Jan. 20, 1896. 19. June 1, 1897, to Jan. 5, 1898. 20. July 10, 1895, to June 5, 1896. 21. Feb. 1, 1899, to Nov. 1, 1900. 22. Dec. 25, 1884, to July 5, 1889. 23. Nov. 21, 1893, to Oct. 15, 1895. 21^. Aug. 30, 1898, to April 10, 1900. ADVANCED ARITHMETIC. 288 In Bankers^ Interest (bank discount), the exact number of days is counted; but, in expressing the time in years, 860 da. is considered 1 yr. Example: Find the time in years from Jan. 4, 1900, to Feb. 15, 1900 (bankers' interest). Process : (1) From Jan. 4, 1900, to Feb. 15, 1900 = 42 da. (2) 42da. = s»^yr. Note. — After Jan. 4, there are 27 da. in Jan. and 15 da. in Feb. ; or, 42 da. in all. In Exact Interest, the exact number of days is counted; but, in expressing the time in years, 865 days is considered 1 yr. (866 days in leap year). Example : Find the time in years from July 9, 1897, to Oct. 12, 1897 (exact interest). Process : (!) From July 9, to Oct. 12 = 95 days. (2) 95da. = /^yr. EXERCISE CXVIII. Find the time in years {bankers^ interest) from — 1. Feb. 12, 1900, to July 1, 1900. 2. Feb. 4, 1899, to April 10, 1899. 8. Mar. 17, 1896, to June 25, 1896. 4. Jan. 19, 1896, to May 10, 1896. 5. July 29, 1895, to Oct. 11, 1895. 6. Dec. 10, 1898, to Jan. 27, 1899. Find the time in years (exact interest) from — 7. Nov. 12, 1898, to Jan. 1, 1894. 8. Oct. 21, 1900, to Dec. 21, 1900. 9. May 5, 1896, to Aug. 1, 1896. 10. Sept. 28, 1892, to Dec. 1, 1892. 11. Jan. 12, 1896, to Mar. 1, 1896. 12. Dec. 20, 1897, to Feb. 15, 1898. 284 ADVANCED ARITHMETIC. 142. Simple Interest. (See definition, p. 281.) TERMS. P, the principal. T, the time. R, the number of % of the rate. /, the interest. A, the amount of principal and interest. Develop the formulas for simple interest. We know — (1) Int. on $1 for 1 yr. at 1^ = $t^. (2) Int. on $ P for 1 yr. at 1^ = $ ^ (3) Int. on $Pfor Tyr. at !%■■ 100 PxT 100 * PxTxR (4) Int. on $P for T yr. at R%= . Now, expressing (4) abstractly, we have the interest formula- PRT 1. I: 100 • Relation I. Abstractly, the interest is equal to the continued product of the principal, rate, and time, divided by 100. We know — PRT (1) ^ = P+7=P+- 100 ' .,, .^, . 100 P+Pi?r ^1^=^'^^= ^0 '^^' 2. ^^/(lOO+PT) 100 Relation II. Abstractly, the amount of a given principal bear- ing interest at a given rate for a given time is equal to the product ADVANCED ARITHMETIC. 285 of the principal and 100 plus the product of the number of per cent of the interest and the time, divided by 100. Note.— Remember that in using these formulas the time must be ex- pressed in years before applying the formula. EXAiyiPIiES. 1. Find the interest on $800 for 2 yr. 6 mo. at 8%. Solution : (1) 2 yr. 6 mo. =2| yr. = | yr. 300x8x5 ^^^^- 100x2 "^^• Answer, $60. 2. Find the interest on $1000 at 8% for 1 yr. 3 mo. 20 da. Solution : (1) 1 yr. 3 mo. 20 da= tH yr. 1000x8x47 0, Q ^"^^ ^ 100x360 -l^-^^- Answer, $104.44^. 3. At what rate will $800 gain $86 in 1 yr. 4 mo. ? Solution : (1) 1 yr. 4 mo. = i yr. (2)/ = ^^^ (3) 100 300x4xie 3x100 ... ^ 36x3x100 - ^^^^= 300x4 =^' Answer, 9%. Jf.. In what time will $400 gain $87.60, interest at 400x8xr Solution: (1) 37.6 = 100 37.62000 ^^^ ^"400x8 -^•^^^• Answer, 1.175 yr. = 1 yr. 2 mo. 3 da. 5. What principal will gain $39 in 1 yr. 3 mo. at Solution : (1) 1 yr. 3 mo. = f yr. (2)39 = «^^><^ (3)P 4x100 39x4x100 6x5 Answer, $520. = 520. ^, K .-< Y i ^ ci^V^ Of 286 ADVANCED ARITHMETIC. 6. Find the amount of $640 at 9% interest for 1 yr. 5 mo. 15 da. 17 5 Solution : (1) 1 yr. 5 mo. 15 da. = — -^yr. 640x9xl7.5 _ ^"^^ ^ 100x12 "^^• Interest, $84. (3) Amount = $640 + $84 = $724, answer. 7. The amount of a note at interest for 1 yr. 4 mo. at 6% is $432. Find the principal. Solution : (1) 1 yr. 4 mo. = i yr. (2) 432= ^(100+6x1) ^Pxl08 100 100 * (3) P=*^^ = 4O0. Answer, $400. 8. At what rate will any principal double itself in 8 yr. ? Note. — Any principal has doubled itself when the interest equals the principal. Solution: (1) Interest = P. (2)P = ^. (Why?) Answer, 121%. Note. — In (3), the P's cancel. Another solution : (1) Int. for 8 yr. = 100^ of prin. ( Why ?) i of (1) = (2) Int. for 1 yr. = 12^^ of prin., answer. 9. In what time w^ill any principal treble itself at 6% ? Solution : (1) Interest = 2 P. ( Why ? ) (2) 2P=^^^—. (Why?) \ J 100 ^ ^ ^ Answer, 88^ yr. =33 yr. 4 mo. ADVANCED ARITHMETIC. 287 Another solution : (1) Q% of prin. = int. for 1 yr. i of (1) = (2) 1% of prin. = int. for ^ yr. 200 X (2) = (3) 200^ of prin. = int. for 33^ yr., answer. EXERCISE CXIX. 1. Define interest, principal, rate, amount. 2. What is simple interest f 3. How does annual interest differ from simple interest ? J^. How does compound interest differ from annual interest ? 5. Why should the time be expressed in years ? 6. Explain the common interest method of counting time. 7. Explain the hankers^ interest method. 8. Explain the exact interest method. 9. Develop the formulas for simple interest. 10. Commit to memory the relations. Find the simple interest on — 11. $240 for 2 yr. at 6%. 12. $78 for 8 yr. at 8%. 13. $654 for 5 yr. at 5%. U. $356 for 6 yr. at44%. 15. $2400 for 4 yr. at 8%. 16. $468.50 for 3 yr. at 10%. 17. $763.65 for 2 yr. at 7%. 18. $95000 for 12 yr. at 5i%. 19. $86 for 2 mo. at 10%. 20. $1500 for 3 mo. at 8%. 21. $684 for 6 mo. at 6%. 22. $2450.30 for 4 mo. at 7%. 23. $869 for 9 mo. at 10%. 2J^. $1250 for 5 mo. at 12%. 25. $763.40 for 8 mo. at 9%. 26. $1386.60 for 10 mo. at 8%. 27. $79 for 30 da. at 12%. 288 ADVANCED ARITHMETIC. 28. $156 for 45 da. at 10%. 29. $288 for 50 da. at 10%. 30. $855 for 33 da. at 8%. 31. $926.42 for 15 da. at 12%. 32. $1346.29 for 60 da. at 6%. 33. $2400 for 93 da. at 8%. 3J^. $1198.75 for 100 da. at 7%. 35. $248 for 1 yr. 3 mo. at 8%. 36. $192 for 2 yr. 2 mo. at 7%. 37. $486 for 4 yr. 6 mo. at 6 %. 38. $1573.59 for 3 yr. 4 mo. at 8%. 39. $2576.42 for 5 yr. 6 mo. at 5%. 40. $986.54 for 6yr. 8 mo. at 9%. J^l. $3546 for 10 yr. 9 mo. at 4%. J^2. $5684.76 for 15 yr. 8 mo. at 3^%. 43. $384 for 3 mo. 15 da. at 6%. U' $576 for 2 mo. 18 da. at 7%. 45. $945 for 4 mo. 27 da. at 5%. 46. $128.73 for 3 mo. 20 da. at 6%. 47. $894.65 for 6 mo. 12 da. at 8%. 48. $1246.68 for 6 mo. 15 da. at 8%. 49. $967.95 for 8 mo. 24 da. at 7%. 50. $684.88 for 2 mo. 5 da. at 10%. 51. $129 for 1 yr. 3 mo. 15 da. at 5%. 52. $2700 for 1 yr. 4 mo. 18 da. at 6%. Find the amount of — 53. $678 for 3 yr. 3 mo. 24 da. at 6%. 54. $940 for 4 yr. 3 mo. 21 da. at 7%. 55. $1568 for 5 yr. 8 mo. 18 da. at 8%. 56. $2568.72 for 2 yr. 9 mo. 27 da. at 5 %. 57. $5468.46 for 7 yr. 10 mo. 19 da. at 4%. 58. $4698.58 for 6 yr. 7 mo. 25 da. at 5^%. ADVANCED ARITHMETIC. 289 Find the rate required for — 59. $800 to produce $114 interest in 1 yr. 7 mo. 60. $540 to produce $64.80 interest in 2 yr. 61. $276 to produce 11.96 interest in 6 mo. 15 da. 62. $1500 to produce $117.50 interest in 1 yr. 6 mo. 24 da. 63. $450 to produce $12.60 interest in 8 mo. 12 da. 6J^. $780 to produce $142.35 interest in 1 yr. 9 mo. 27 da. Find the time required for — 65. $300 to produce $52.50 interest at 5%. 66. $540 to produce $86.40 interest at 6%. 67. $1000 to produce $110 interest at 8%. 68. $240 to produce $16.80 interest at 10%. 69. $5000 to produce $262.50 interest at 7%. 70. $840 to produce $11.76 interest at 8%. 71. $235 to produce $2,585 interest at 12%. 72. $340 to produce $5.10 interest at 6%. Find the principal required to produce — 73. $30 interest in 2 yr. 6 mo. at 6%. 7Jf. $50.40 interest in 1 yr. 9 mo. at 8%. 75. $208 interest in 4 yr. 4 mo. at 10% . 76. $96.35 interest in 1 yr. 8 mo. 15 da. at 6%. 77. $35 interest in 3 mo. at 7%. 78. $28. 12 J interest in 4 mo. 15 da. at 5%. 79. $6.40 interest in 45 da. at 8%. 80. $13.50 interest in 90 da. at 6%. Find the principal lohich amounts to — 81. $570 in 2 yr. 4 mo. at 6%. 82. $751.20 in 6 mo. 15 da. at 8%. 83. $904.44f in 1 yr. 3 mo. 20 da. at 10%. 84. $2516.661 in 30 da. at 8%. 290 ADVANCED ARITHMETIC. Find the interest {common interest method) on — 85. $350 from Jan. 10, 1875, to June 15, 1876, at 6%. 86. $500 from May 1, 1880, to Dec. 1, 1884, at S%. 87. $490 from March 15, 1897, to Jan. 1, 1901, at 8%. 88. $875 from July 5, 1896, to May 30, 1899, at 10%. 89. $400 from June 20, 1895, to Feb. 1, 1898, at 6%. Find the interest {bankers'' interest method) on — 90. $80 from Jan. 8, 1887, to Feb. 20, 1887, at 10%. 91. $250 from July 15, 1890, to Sept. 20, 1890, at 8%. 92. $385 from Aug. 12, 1891, to Nov. 1, 1891, at 12%. 93. $1200 from Jan. 18, 1892, to April 1, 1892, at 8%. 9^. $840 from May 12, 1893, to Sept. 5, 1893, at 8%. Find the exact interest on — 95. $125 from Feb. 1, 1891, to Dec. 1, 1891, at 6%. 96. $250 from May 18, 1891, to July 31, 1891, at 8%. 97. $500 from June 15, 1893, to Nov. 1, 1893, at 5%." 98. $475 from Dec. 25, 1894, to March 19, 1895, at 6%. 99. $1000 from May 4, 1895, to Oct. 10, 1895, at 7%. 100. $850 from Jan. 25, 1896, to June 30, 1896, at 8%. 101. In what time will any principal double itself at 8%? 102. In what time will any principal treble itself at 10%? 103. At what rate will any principal double itself in 12 yr.? lOJ^. At what rate will any principal treble itself in 18 yr. ? 143. Notes. [FORM.] No. Ii2. Sedan. Kans., lu.l'T^ 10, i8C)S. OyxAAAj '\X'VoJ\A^ after date, A promise to pay O. \0. C^OxAaXL, or order, at THE FIRST NATIONAL BANK OF SEDAN. CJLAM^ c)=vUyTVcUuxL CUTUcL '^/loo Dollars, with interest at the rate of ten per cent, per annunn until paid, for value received. $500.00. A. d. ']UaAX)^T\^'\^. ADVANCED ARITHMETIC. 291 A Promissory Note is a promise to pay to the party designated a certain sum of money, with or without interest, at the time specified in the note. The principal is the amount of money mentioned in the note. The Face of a note is what it calls for on its face. If a note bears interest, the face includes principal and interest ; if the note does not bear interest, the face is the principal. Note. — Most authors define the face of a note as the principal only ; but, if that be so, we should in the interest of simplicity discard the term face. ( See Bank Discount.) An Indorsement is something written on the back of a note. The above note is payable to E. B. Ferrell, or his order. He may order it j)aid to W. R. Ferrell ; (see first indorsement above.) This is called an indorsement in full, because it is complete. A holder of a note may merely sign his name on the back ; thus he orders it paid to any person who may be the holder at the time the note is due. This is called indorsement in blank. Note.— There are other forms of indorsing a note for transfer, but they are not so commonly used as the above. When payments are made upon a note, before final settle- ment, their dates and amounts are placed on the back of the note (see "Indorsements of payments " above). 292 ADVANCED ARITHMETIC. In many States, the holder of a note or bill due at a future date must allow 8 days more than the time stated in the note for payment. These are called Days of Grace. The holder gets interest for this extra time, if the bill bears interest. Note. — It is sometimes supposed that days of grace are only allowable in Bank Discount. This is a mistake. It is a law governing the payment of notes and bills, and not a principle of Bank Discount. But remember that it applies only to notes and bills, and not to all debts that may bear interest. The perplexing question to the student is, *'When am I to include datjs of grace and when not f " If a problem reads " What is due Feb. 10, 1897?" just count to that date. If the problem reads "and runs for 2 yr. 4 mo.," just use that time. But, if the wording of the note is given, you must allow for days of grace, in those States where days of grace are allowed. EXERCISE CXX. 1. Write a note that bears interest from date. 2. Why should a note be dated ? 3. Who is the maker of your note ? If.. Who is the payee f 5. Write a note that bears interest after it is due only. 6. What is the law concerning days of grace in your State ? 7. If a note says nothing about interest, what rate of inter- est will it bear after it is due, in your State ? 8. What is the highest rate of interest you can legally charge in your State ? 9. If you charge more than is legal, what is it called ? 10. What does the law say about usury in your State ? 11. What is indorsement in blank f in full f 12. How are payments indorsed ? 144. Partial Payments. — When partial payments are made on a note or debt bearing interest and running one year or less, it is customary to reckon interest on the principal and on each payment to the time of settlement, and subtract ADVANCED ARITHMETIC. 298 the amount of the payments from the amount of the debt. This is called ^ ^ Merchants^ Rule.''^ Principles of the Merchants' Rule : 1. The whole principal bears interest according to the terms of the note until final settle- ment. 2. Each payment hears interest from the time it is made until date of final settlement. When the debt or note bears interest for more than one year, the interest is computed by the " t/. S. Rule.'" Principles of the U. S. Rule: 1. Payments must first dis- charge interest due, and the balance, if any, will be applied to the discharge of the principal. 2. Interest must not bear interest. 3. Payments must not hear interest. EXAMPIiES. 1. A note of $400, dated Jan. 1, 1897, interest 6%, has the following payments: May 1, 1897, $30; Nov. 16, 1897, $100. What was due Jan. 1, 1898 ? (Merchants' rule. Why?) Solution : (1) Interest on $400 for 1 yr. at 6^ = $24. (2) $400+ $24 = $424, amt. of debt at maturity. (3) From May 1, 1897, to Jan. 1, 1898 = 8 mo. = | yr. (4) Int. on $30 for | yr. at 6^ = $1.20. (5) $30 + $1.20 = $31.20, amt. of first payment. (6) From Nov. 16, 1897, to Jan. 1, 1898 = 1 mo. 15 da. = i yr. (7) Int. on $100 for i yr. at 6^ = $.75. (8) $100+$.75 = $100.75, amt. of second payment. (9) $424 -$31.20 -$100.75 = $292.05, answer. Note. — The work required in finding the interest in (1), (4) and (7) need not appear in the solution ; but the teacher should insist upon speed and accuracy in mechanical work. 2. A note of $150 is dated May 10, 1896; interest 6%. It 294 ADVANCED ARITHMETIC. has the following indorsements: Sept. 10, 1897, $32; Sept. 10, 1898, $6.80. What is due Nov. 10, 1898 ? (U. S. Rule. Why?) Dates. 1898—11—10 1898— 9—10 ^, , , . . . ^onj Q_io Note. — Arrange the dates in order, -.ggg 5_io viith. the latest date above. Begin be- Payments. low ; Subtract each date from the one 1_ 4 $32 n^xt above it. 1— 0— $6,80 2— Solution : (1) Int. on $150 for U yr. at 6^ = $12. (2) $150+$12-$32 = $130, new principal. (3) Int. on $130 for 1 yr. at 6^ = $7.80. (4) $7.80 - $6.80 = $1, unpaid interest. (5) Int. on $130 for ^ yr. at 6^ = $1.30. (6) $130+$1+$1. 30 = $132.30, due at maturity. EXERCISE CXXI. 1. A note for $820 was given June 12, 1897 ; interest 6%. It has the following indorsements : Aug. 12, 1897, $100 ; Nov. 12, 1897, $250; Jan. 12, 1898, $120. What was due Feb. 12, 1898 ? 2. A note of $1200, dated April 1, 1890, payable on demand, with interest at 7%, has the following indorsements: May 6, $210; July 5, $210; Oct. 18, $822. What was due Jan. 1, 1891 ? 3. A note of $2000, dated Jan. 4, 1893, interest 6%, has the following indorsements: Feb. 19, 1898, $400; June 29, 1894, $1000; Nov. 14, 1894, $520. What was due Dec. 24, 1896 ? 4.. Find the amount due at maturity on note No. 12 on pages 290-1, including 8 days of grace (Kansas allows days of grace). 5. A note of $500, dated March 1, 1895, interest 6%, has the following indorsements : Sept. 1, 1895, $10; Jan. 1, 1896, $80; July 1, 1896, $11; Sept. 1, 1896, $80. What is due March 1, 1897 ? 145. Annual Interest. ( See definition, p. 281.) Example: A note of $800 drawing interest at 8%, payable ADVANCED ARITHMETIC. 295 annually, runs for 5 years 3 months : find the amount due at maturity, if no payments have been made. Note. — Each year's interest will bear interest after it becomes due, as follows : 1st year's int. for 4 yr. 3 mo. 2d year's int. for 3 yr. 3 mo. 3d year's int. for 2 yr. 3 mo. 4th year's int. for 1 yr. 3 mo. 5th year's int. for yr. 3 mo. Total 11 yr. 3 mo. Therefore, in addition to the simple interest on the principal for 5 yr. 3 mo., we must count the interest on 1 year's interest for the sum of the above periods, or 11 yr. 3 mo. Solution: (1) Int. on $800 for 1 yr. at 8^ = $64, due annually. (2) Int. on $800 for 5^ yr. at 8^ = $336, int. on principal. (3) Int. on $64 for llj yr. at 8^ = $57.60, int. on int. (4) $800+$336 +$57.60 = $1193.60, amount due. EXERCISE OXXII. Find the annual interest — 1. On $700 for 4 yr. 3 mo., at 6%, payable annually. 2. On $840 for 2 yr. 4 mo., at 8%, payable semi-annually. 3. On $1000 for 6 yr. 2 mo. 15 da., at 10%, payable annually. 4-. On $800 for 1 yr. 9 mo., at 5%, payable quarterly. 5. On $400 for 6 yr. 4 mo. 8 da., at 10%, payable annually. 146. Compound Interest. (See definition, p. 281.) EXAMPIiES. 1. A note for $800, drawing interest at 8% , payable annually, runs for 5 yr. 3 mo. : find the amount due at maturity by com- pound interest. Solution: (1) Amt. of $800 at S% for 1 yr. =$64 +$800 = $864. (2) Amt. of $864 at S% for 1 yr. = $69.12+$864 = $933.12. (3) Amt. of $933.12 at 8% for 1 yr. = $74.6496 = $933.12 = $1007.77. (4) Amt. of $1007.77 at S% for 1 yr. = $80.62 + $1007.77 = $1088.39. (5) Amt. of $1088.39 at H% for 1 yr. = $87.07 + $1088.39 = $1175.46. (6) Amt. of $1175.46 at 8% for i yr. = $23.51 +$1175.46 = $1198.97. 296 ADVANCED ARITHMETIC. Note. — You will observe that this is the same problem which is solved by annual interest. Up to the end of the second year, the results ob- tained by annual interest and compound interest would be the same, but after that time the compound interest plan begins to advance faster than the annual interest plan. Why is this so? Solving problems as above would be too tedious for use in business. A person having many problems to solve in com- pound interest should use a table like the following : TABLE, Showing the amount of $1, at 3, 4, 5, 6, 7 and 8 per cent., at Compound In- terest, for any number of years, from 1 to 25. Yr. 3 per cent. 4: per cent. 5 per cent. 6 per cent. 1 per cent. 8 per cent. 1 1.03 1.04 1.05 1.06 1.07 1.08 2 1.060 1.0816 1.1025 1.1236 1.1449 1.1664 3 1.092727 1.124864 1.157625 1 . 191016 1.225043 1.259712 4 1.125509 1.169859 1.215506 1.262477 1.310796 1.360488 5 1.159274 1.216653 1.276282 1.338226 1.402551 1.469328 6 1.194052 1.265319 1.340096 1.418519 1.500730 1.586874 7 1.229874 1.315932 1.407100 1.503630 1.605781 1.713824 8 1.266770 1.368569 1.477455 1.593848 1.718186 1.850930 9 1.304773 1.423312 1.551328 1.689479 1.838459 1.999004 10 1.343916 1.480244 1.628895 1.790848 1.967151 2. 158924 11 1.384234 1.539454 1.710339 1.898299 2.104851 2.331638 12 1.425761 1.601032 1.795856 2.012196 2.252191 2.518170 13 1.4685S4 1.665074 1.885649 2.132928 2.409845 2.719623 14 1.512590 1.731676 1.979932 2.260904 2.578534 2.937193 15 1.557967 1.800944 2.078928 2.396558 2.759031 3.172169 16 1.604706 1.872981 2.182875 2.540352 2.952163 3.425942 17 1.652848 1.947900 2.292018 2.692773 3.158815 3.700018 18 1.702433 2.025817 2.406619 2.854339 3.379932 3.996019 19 1.753506 2.106849 2.526950 3.025600 3.616527 4.315701 20 1.806111 2.191123 2.653298 3.207135 3.869684 4.660957 21 1.860295 2.278768 2.785963 3.399564 4.140562 5.033833 22 1.916103 2.369919 2.925261 3.603537 4.430401 5.436540 23 1.973587 2.464716 3.071524 3.819750 4.740529 5.871463 24 2.032794 2.563304 3.225100 4.048935 5.072366 6.341180 25 2.093778 2.665836 3.386355 4.291871 5. 427432 6.848475 2. Find the compound interest on $1500 for 12 yr. at 7%, payable annually. ADVANCED ARITHMETIC. 297 Solution: (1) Comp. amt. of $1 for 12 yr. at 7^ = $2.252191. 1500x(l) = (2) Comp. amt. of $1500 for 12 yr. at 7^ = $3378.2865. (3) $3378.2865 -$1500 = $1878.2865, answer.. 3. Find the compound amount of $400 for 4 yr. 2 mo. at 6%, payable semi-annually. Note. — Double the time and use lot the rate. Compound interest for 4 yr. 2 mo. at Q%, payable semi-annually, is the same as for 8 yr. 4 mo. at 3^, payable annually. Solution : (1) Comp. amt. of $1 at S% for 8 yr. = $1.26677. (2) Comp. amt. of $400 at 3% for 8 yr. = $506,708. (3) Int. on $506,708 at S% for 4 mo. = $5,067. (4) $506.708+$5.068 = $511,776, answer. EXERCISE CXXIII. Find the compound interest as in example 1 — 1. On $1000 for 4 yr. at 5%, payable annually. 2. On $850 for 2 yr. at 4%, payable semi-annually. 3. On $900 for 5 yr. 4 mo., at 6%, payable annually. Find the compound interest as in examples 2 or 3 — J^. On $700 for 8 yr. at 5%, payable annually. 5. On $1200 for 9 yr. at 8%, payable semi-annually. 6. On $1100 for 6 yr. at 12%, payable quarterly. 7. Find the compound amount on $1200 for 5 yr. 2 mo. at 6%, payable semi-annually. 8. Find the compound amount on $600 for 17 yr. 3 mo. 15 da. at 10%, payable annually. 147. Bank Discount. — The face oi a note is the amount due at maturity. If the note bears interest, the face is the amount of principal and interest due at maturity. If the note does not bear interest, the face is the principal. Bank Discount is the simple interest counted on the face of the note from the date of discounting to the date of maturity. The 298 ADVANCED ARITHMETIC. manner of counting the interest is not different from that al- ready studied. The j^foceeds are the difference between the bank discount and the face of the note. The "^me to run'''' is the actual number of days from the date of discount to the date of maturity; this is sometimes called the '■HerDi of dis- counts^'' which is perhaps a more appropriate name. To find the date of maturity of a note : ( 1 ) When the time for which the note bears interest is expressed in months, as, "due in 4 mo. , ' ' count by months. From a certain day in one month to the same day in the next month is 1 mo., and so on. (2) When the time is expressed in days, count the actual days to find the date of maturity. Principles: 1. The face of the note, the amount due at matur- ity, is the amount to be discounted. 2. The hank discount is the simple interest on the face of the note for the given time at the given rate. 3. The face of the note minus the hank discount equals the pro- ceeds. TERMS. F, face of the note. R, number of % of the discount. T, time to run. D, discount. P, proceeds. Developing the formulas for hank discount, . r. FRT {!) P. = F-D. (Prin. 3.) ,.. r^. r. T. FRT lOOF-FRT il) = i2) P=F-^ = 10^—'°^' F{10 0-RT) ^' ^- 100 Note. — In the following examples and exercise, days of grace are con- sidered. ADVANCED ARITHMETIC. 299 examfi.es. 1. A note of $240, dated Jan. 1, 1888, time 4 months, was discounted March 5, 1888, at 10% : find the time to run, and proceeds. Solution : (1) Date of maturity : 4 mo. 3 da. from Jan. 1, or May 4. (2) Term of discount : From Mar. 5 to May 4, or 60 days. = } yr (3) p jmoo^i-io )^^ Proceeds, $236. 2. A note of $2400, dated July 3, 1891, bearing 5% interest, was discounted Jan. 9, 1892, at 10% ; the note falls due 10 months from date. Find the time to run, bank discount, and proceeds. Solution ; (1) Date of maturity : 10 mo. 3 da. from July 3, May 6, 1892. (2) Int. on $2400 for Ui yr. at 5^ = $101. (3) $2400+$101 = $2501, face of note. (4) Term of discount: From Jan. 9, 1892, to May 6, 1892, or 118 da. (leap year). = y% yr. (K^ n 2501xl0x59 _ ^^^ ^- 100x180 -^^•^^- Discount, $81.98. (6) P= $2400 - $81.98 = $2419.02, proceeds. 3. For what sum must a 60-day note, without interest, be drawn to produce $500, when discounted for the full time at 6% ? F {100 -TR) Solution: (1) P- 100 JP(100-^x6) (2) 500= ^^ . (2) = (3)500 = ^:^. (4) F=^^^^^^^ = ^.SO,. Face, $505,306. Another Solution : (1) Bank discount on any principal for ^V yr. at Q% = laV^ of principal. (2) Proceeds: 100^ of prin.-lgJjj^ of prin. = 98H^ of prin. (3) 98Jg^of prin.=$500. (4) 1^ of prin. = $5.05306. (5) 100^ of prin. =$505,306, answer. 300 ADVANCED ARITHMETIC. EXERCISE CXXIV. 1. Commit to memory the principles of bank discount. 2. Develop the formulas, and give the relations. Find date of maturity, time to run, and hank discount. 3. Prin., $600; date, May 80; time, 5 mo.; rate, 10% ; dis- counted, July 1. 4. Prin., $840; date, Feb. 15; time, 8 mo.; rate, 0%; dis- counted, June 10. 5. Prin., $1000; date, July 10; time, 90 da.; rate, 8% ; dis- counted, Aug. 1. Find date of maturity, time to run, and jiroceeds, if each of the following hears 8% interest from date : 6. Prin., $800; date, Jan. 10; time, 6 mo. ; rate, 10% ; dis- counted, Mar. 6. 7. Prin., $500; date, Apr. 12; time, 9 mo.; rate, 6%; dis- counted, Aug. 18. 8. Prin., $470; date, May 15; time, 120 da.; rate, 10%; discounted, July 1. 9. I desire to use to-day $080, which I can secure by giving a bankable note payable in 80 days, without interest, discounted at 9%. For what sum must I write the note ? 148. True Discount. — The Present Worth of a debt payable at a future date without interest is that sum of money which, when put at interest for the given time, will amount to the debt. The difference between the present worth and the debt is the True Discount. To find the present worth of a debt is the same problem that we had in Interest : to find the prin- cipal when the amount of the principal and the interest are given. (See formula 2, p. 284.) Note. — If the debt discounted be a note, given in a State where days of grace are recognized by law, days of grace should be considered in true discount as well as in any other application of interest. ADVANCED ARITHMETIC. 301 If a debt or note, bearing interest, is discounted, the interest due at maturity must be added to the principal, before dis- counting. EXAMPIiES. 1. What is the present worth and true discount of a debt of $627 if paid 9 months before it is due, discount 6%? P{im+RT) Solution : (1) A = (2) 627= 100 P{100+6Xj%) ^ PxlO^i 100 100 (3) P=?27xlOOx_2^gQo^ 209 (4) Present Worth = ^Q0O. (5) True Discount = ^27. No. 5. Topeka, Kans., KX/TU. 5, '900. Ui/V-V TnxvTbtnA/ after date, J promise to pay CaxiuTlV ^ Co., or order, c3umX-\K^ jiuyTLcLAyexl^ cutixL "-^/wo Dollars, with interest from date at the rate of 5 per cent, per annum, for value received. $ 1200.00. . c^. Q.. (XclcuTn^. 2. Find the true present worth of note No. 5, on April 9, 1900, discount, 6%. Solution : (1) Interest on $1200 at 5% for 5 mo. 3 da. =$25,50. (2) $1200 +$25. 50 =$1225.50. (3) 5 mo. 3 da. from Jan. 5, 1900, is June 8, 1900, the date of maturity. (4) From April 9 to June 8 = 60 da., the term of dis- count. (5) 1225.50 = -P^^Q^+^><^) . (6) p^jjg5.5x 100 ^-^213.366. Answer, $1213.366. 302 ADVANCED ARITHMETIC. EXERCISE OXXV. 1. A debt of $942 is due in 4yr. 6 mo. 6 da., without interest : find its present worth if money is worth 6%. 2. A debt of $273.75 is due in 1 yr. 7 mo., without interest: find its present worth and true discount at 6%. 3. A debt of $866 is due in 8 yr. 8 mo., without interest : find its true discount at 6%. ^. A debt of $888.20 is due in 5 yr. 7 mo. 27 da., without in- terest. Wliat is its present worth, money being worth 10%? 5. A note of $800, dated Jan. 10, time, 6 mo., bearing interest at 8%, is discounted at true discount Mar. 6, at 10%. Find the true discount. 6. The difference between true and bank discount is equal to the simple interest on the true discount for the given time at the given rate. Illustrate this fact by solving and comparing result in the following: What is the difference between the true and bank discount on $510 for 4 mo. at 6%? 149. Exchange. — Exchange is the process of mak- ing payment in distant places without transferring the money. Exchange is effected by means of drafts or bills of exchange. [A DRAFT.] No. C|. FIRST NATIONAL BANK OF SEDAN. Sedan, Kansas, wXl^ b, '8Cj5. Pay to the order of CJruxhlAJi^ QLoAjfO, $i250.00 c5iAM> c^UyTvdyuxl 3"Lf/ta^ arnxl ]^ Dollars. To WESTERN NATIONAL BANK, ^. c3. DVC»xLu/l^, New York, N. Y. Cashier. This is the form of an ordinary draft, or bill of exchange, ADVANCED ARITHMETIC. 303 issued by one bank upon another. This draft purports to have been issued by the cashier of the First National Bank of Sedan upon the Western National Bank of New York. The New York bank is hereby ordered to pay $250 to Charles Clark. To illustrate the use of a draft: Suppose that you owe Charles Clark $250, and that he lives near New York city. One of the easiest as well as the safest ways to pay the debt is to go to your bank and get a draft on New York, just like the above, and send it to Clark. He can take it to the Western National Bank, or any other bank in that country, and get his money. Banks usually charge a premium for issuing drafts, as, i%, i%, 1%, li%. Sometimes a bank will issue drafts at a dis- count; that is, give you something in order to get you to take the drafts. This is when there is a money panic at the money centers, or a scarcity of money at the local bank. The above is a sight draft; it is payable on demand. Occa- sionally a customer wants a draft and does not care to have it paid for 30, 60, or 90 days. Such a draft is called a time draft, and the purchaser gets interest on his money for that time. Domestic, or Inland Exchange is exchange between people in the same nation or country. Foreign exchange is exchange be- tween people in different nations or countries. Principles: 1. The premium or discount is some number of per cent of the face value of the draft. 2. In sight drafts, the cost is equal to the face of the draft plus the premium or minus the discount. 3. In time drafts, the cost is equal to the face of the draft minus the interest, plus the premium or minus the discount. terms. F, the face of the draft. p, number of % of premium or discount. P, the premium. 304" ADVANCED ARITHMETIC. Z), the discount. (7, the cost. /, the interest. jK, number of % of the interest. y, the time. Development of formulas for exchange. 2. D: 100 pF 100 In sight drafts at a premium, 100 100 ' ' 3 ^^ F{m-{-p) 100 In sight drafts at a discount, 4 ^^ F(lOO-p) 100 In time drafts at a premium, (1) C=F-I+P=F-™^^^. ^ 100 100 - g ^^ J^(100-i?r+j^) 100 In time drafts at a discount, ^ i:t r « rr FET pF 100 Note. — Days of grace are usually considered in time drafts. ADVANCED ARITHMETIC. EXAMPIiES. 1. What will a draft on Kansas City for $880 cost, at i% premium? ssOxlOOj: ,,., , Solution: C = ^ = 882.2. 100 Answer, $882.20. 2. What will a draft on St. Louis for $1200 cost, at i% dis- count? 1200x99^ ,,_. Solution : C = = 1194. 100 Answer, $1194. 3. I have $2000 : how large a draft can I purchase with it, if I have to pay i% premium ? Solution: (1) 2000= ^^ ^^^ 100 (2) F=?0«0^ = 1990.05. lOOi Answer, $1990.05. 4.. How large a draft can I buy for $2000, discount i%? Solution: (1) 2000 = ^^^^^ 100 (2)i.= ?2^^ = 2005.01 + . 99 4 Answer, $2005.01+. 5. I want a 60-day draft on New York for $2000 : what will it cost me, if interest is 6% and premium on New York is i%? Solution: a)C = ^^^^^^^^^±^. 100 2000(100-^x6+1) ^^^^" 100 • 2000 X 99^^ ^ 2000 X 1989 ^ ' 100 100x20 Answer, $1989. Another solution : (!) Int. on draft for 1 yr. at 6^ = 6j^of draft. ^ of (1) = (2) Int. on draft for -i^ yr. at 6^ = l^V^ of draft. (3) 100^ of draft +i^ of draft -l^V^ of draft = 99^o3f of draft. (4) 100^ of draft = $2000. (Basis.) jU of (4) = (5) 1% of draft = $20. 99^ X (5) = (6) 99^x5% of draft = $1989, answer. 306 ADVANCED ARITHMETIC. 6. What will the draft in Example 5 cost if the exchange on 9 Note. — This is the same as above, except that you should use formula No. 6 instead of No. 5. Answer, $1969. 7. How large a 30-day draft can I get for $2000, interest be- ing 8% and premium i%? Solution: m 2m =^^^^^^i^±^. (1) ("12000 ^^99» - -^x59n (1)-W2000- ^^ -100x60 (3) i.= ?000^M^ = 2009.71+. 0971 Answer, $2009.71 + . 8. What will be the size of the draft in Example 7, if ex- change is at a discount of i% ? Note.— This is the same as above, except that you should use formula No. 6 instead of No. 5. Answer, $1980.52+. 9. What will an exchange on London for £240 cost, at $4.85? Solution: (1) Cost of £1 = $4.85. (Basis.) 240x(l) = (2) Cost of £240 = $1164. 10. I have $2635.20 with which to buy a draft on Liverpool, at $4.88 : what will be the size of the draft ? Solution: (1) $4.88 = cost of £1. (Basis.) ^-^of(l) = (2)$l = costof£^g. 2635.2 X (2) = (3) $2635.20 = cost of £540. EXERCISE CXXVI. 1. Define exchange, domestic exchange, foreign exchange. 2. Draw a draft, using amount, names, date, etc., as you may choose. 3. Develop all the formulas for exchange. I ADVANCED ARITHMETIC. 807 Jf.. Commit to memory the principles. 5. Give the relations expressed by the formulas. 6. What will be the cost of a draft on Chicago for $8600 at i% premium ? 7. What will be the cost of a draft on New Orleans for $9400 at i% discount? 8. I paid $5420 for a draft on Kansas City at |% premium : what was the size of the draft ? 9. I paid $7240 for a draft on New York at i% discount: what was the face of the draft ? 10. What will be the cost of a 60-day draft on San Francisco for $6540, premium i%, interest 8%. 11. What will be the cost of a 90-day draft on Philadelphia for $2450, discount ^%, and interest 6%? 12. I have $7800 with which I desire to purchase a 60-day draft on New York: if exchange is at 1^% premium, and money is worth 8%, what will be the face of my draft ? 13. What size bill of exchange can I get on London for $12150, if exchange is $4.86? 14-. I bought a bill of exchange on Liverpool for £842 : what did it cost if exchange is $4.83^ ? 15. A draft on Pittsburg cost $1771, exchange |% premium: what was the face of the draft ? 16. The face of a 80-day draft on St. Paul is $4220: what did it cost if exchange is 1^% premium and interest 8% ? 17. A grain merchant in Toledo sold 11875 bu. of corn at 40/ per bushel; deducting 8% commission, he purchased a 60-day draft with the proceeds, interest 6%, premium 2%: required the face of the draft. 18. An agent owed his principal $1011.84. He bought a draft with the sum, and remitted : the principal received $992. Find the rate of exchange. 19. What is the cost of a bill on Amsterdam for 4800 guild- ers, quoted at 41 J/, brokerage i ? 308 ADVANCED ARITHMETIC. 20. Jones, of St. Louis, has a debt of $6000 to pay in New- York. The direct exchange is -1% premium; but exchange on Philadelphia is i% premium, and from Philadelphia to New York is ^% discount. By circular exchange how much will pay the debt, and how much does he gain over the direct ? C. MECHANICS. 150. Force. — Force is that which tends to produce or overcome motion of matter. The several units of force in common use are the poundal^ dyne, pound, and gram, and may be defined as follows : 1. The force required to add 1 ft. in 1 sec. to the velocity of 1 lb. of matter = 1 poundal (pi.). 2. The force required to add 1 cm. in 1 sec. to the velocity of 1 g. of matter = 1 dyne. 3. The force required to add 32.16 ft. in 1 sec. to the velocity of 1 lb. of matter =1 pound (N. Y.). 4-. The force required to add 980 cm. in 1 sec. to the velocity of 1 g. of matter — 1 gram (N. Y.). Note. — It should be remembered that we have pounds of matter and pounds of force, and grams of matter and grams of force. The pupil should distinguish clearly between the two. The units which have "N. Y." after them are called gravity units, be- cause they vary with the attraction of gravitation at different places on the earth. As given above, the units are correct at the sea-level at New York. The dyne and poundal are called absolute units, and are the same for all places. TABLE OF EQTJrVAIiENTS. 1 lb. offeree (N. Y.) = 32.16 pi. 1 g. of force (N. Y.)=980 dynes. 1 pl. = 13826 dynes. EXERCISE CXXVII. 1. Reduce 584 pi. to dynes. 2. Reduce 840 lb. (N. Y.) to pi. 3. Reduce 34 g. (N. Y.) to dynes. ADVANCED ARITHMETIC. 809 4. Reduce 103675 dynes to pi. 5. Reduce 19296 pi. to lb. (N. Y.). 6. Reduce 8859200 dynes to lb. (N. Y.). 7. Reduce 1960 lb. (N. Y.) to g. (N. Y.). 8. Reduce 84 lb. (N. Y.) to dynes. The amount of velocity added per second to the motion of a body is called its Acceleration. TERMS. F, the force. ilf, the mass. -y, the velocity, t^ the time (in sec). a, the acceleration. Development of the formulas for force. If a body starts to move from a state of rest, its velocity at the end of the Ist sec. will be the same as its acceleration ; or (1) Velocity for 1 sec. = a. and (2) Velocity for i sec. = a< ; or, 1. v=at. From the definition of a poundal, (1) Force required to add 1 ft. in 1 sec. to the velocity of 1 lb. of mat- ter =1 pi. (2) Force required to add v ft. in 1 sec. to the velocity of 1 lb. of mat- ter- v pi. (3) Force required to add v ft. in 1 sec. to the velocity of if lb. of mat- ter =3/y pi. (4) Force required to add v ft. in t sec. to the velocity of M lb. of mat- ter = — pi. ; or, stated abstractly, 2. F=^. z From formula 1, (1) a = f V Substituting a for in formula 2, we have — 3. F=Ma. 310 ADVANCED ARITHMETIC. When if is expressed in lb., v or a in ft., and t in seconds, formulas 2 and 3 give the force in poundals. When M is expressed in g., v or a in cm., and t in sec, these formulas give the force in dynes. Since a poundal is g^^g POund of force (N, Y.), and a dyne is -g^gram of force (N.Y,), these f ormulas will represent 22^^ as m&nj pounds as pound- als; or Yh as many grams as dynes. Now, 32.16 ft. =980 cm. If we represent this amount by g, these formulas, when used to find the force in pounds or grams, will be — 4. F= — , and 5. F=^. 9 Note. — Do not forget that g in feet is 32.16; in centimeters it is 980. EXAMFIiES. 1. What force will give 15 lb. of matter an acceleration of 25 ft. per second ? Solution : F=lbx2b = 375. .'. the required force is 375 poundals. 2. What force will give to 21 grams of matter an accelera- tion of 10 centimeters per second ? Solution: 1^=21x10 = 210. .*. the required force is 210 dynes. 3. What force will give 6 cwt. 20 lb. of matter an accelera- tion of 1 ft. 6 inches per second ? Solution : (1) 6 cwt. 20 lb. = 620 lb. (2) 1ft. 6in. = lHt. (3) jP=620x 11 = 930. .*. the required force = 930 poundals. 4-. What is the force in poundals that will in 10 minutes produce a velocity of a mile a minute in 100 lb.? ADVANCED ARITHMETIC. 311 Solution: (1) 1 mi. =5280 ft. (2) 1 min. = 60 sec. (3) 10 min. = 600 sec. (4) Velocity of 1 mi. per min. = velocity of 88 ft. per sec. 100X88^ ' 600 ^ .'. the reqd. force is 14| poundals. 6. A force of 30 dynes, acting for 12 seconds upon a body, gives it a velocity of 120 cm. per second. Find the mass of the body. Solution : (1) 30 = i?^^= 10 x 3f . 12 (2) 3f=fg = 3. .*. the required mass is 3 grams. 6. A force of 40 poundals is acting upon a mass of 16 pounds ; how much velocity is given the body per second ? Note. — Acceleration is wanted. Solution: (1) 40 = 16xa. (2)a = f^ = 2i. .'. the reqd. acceleration is 2| ft. per sec. 7. A body, acted upon by a force of 100 dynes, receives an acceleration of 20 cm. per second. Find its mass. Solution: (1) 100 = 20x.3f. (2) 3/= -1^0- = 5. .'. the reqd. mass is 5 grams. 8. A force of 360 poundals acts upon a mass of 6 pounds for 4 seconds. What velocity did the body receive ? o 7 ^- /1\ nor, 6XV 3X1' Solution : (1) 360 = = . (2) , = §60x2^240. 3 .". the reqd. velocity is 240 ft. per sec. • 312 ADVANCED ARITHMETIC. 9. A body of 30 pounds received a velocity of 480 ft. per sec- ond from a force of 120 poundals. How long did the force act to do this ? Solution: (1) 120 = ^^^^^^ . t (2) 120x< = 480x30. .*. the reqd. time is 120 sec, or 2 min. 10. How many pounds of force will add to 821.6 pounds of matter a velocity of 20 ft. per second ? o , ,. ^ 321.6x20 ,^^ Solution: F= — = 200, 32.16 .*. the reqd. force is 200 lb. 11. A force of 800 pounds gives a body a velocity of 80 ft. per second in 5 seconds. Find its weight. 80x3f Solution: (1) 800 = 32.16x5 3^^800 X32.16X5 ^^^^3^ 80 /. the reqd. weight is 1608 lb. 12. Find the acceleration per second given to a body of 700 grams by a force of 140 grams. Solution: (1) 140 = ^^-. 980 140x980_ ^^^^-~76o~-^^^- .'. the reqd. acceleration is 196 cm. per sec. EXERCISE CXXVIII. 1. What is force ? Velocity? Acceleration? 2. What is a poundal ? 3. What is a dyne ? 4. What is a lb. of force (N. Y.)? 5. What is a gram of force (N. Y.)? ADVANCED ARITHMETIC. 818 6. Repeat the table of equivalents. 7. How would you reduce grams to pL? 8. How would you reduce pounds to dynes ? 9. What is the difference between a pound of force and a pound of matter ? W. Develop the formulas for force. 11. Give the relation expressed by each formula. 12. A constant force acting upon a mass of 15 g. for 4 sec. gives it a velocity of 20 cm. per sec. Find the force in dynes. 13. What is the force in poundals that will give to 1000 lb. of matter in 20 min. a velocity of 2 miles a minute ? Note.— Express the time in sec. and the velocity in ft. What is the velocity per sec, if the mass is traveling at the rate of a mile a minute ? 14' A mass of 15 lb. lying on a smooth, flat table is acted upon by a force of 60 pi. Find the velocity at the end of 2 sec. 15. A force of 1000 dynes acting on a mass for 1 sec. gives a velocity of 20 cm. per sec. Find the mass in grams. 16. What force in dynes will in 1 sec. give to a mass of 12 g. a velocity of 6 cm. per sec? 17. What velocity will be produced by a force of 490 dynes acting on a mass of 70 g. for 10 sec? 18. In what time will a force of 7500 pi. give to a mass of 800 lb. a velocity of 250 ft. per sec? 151. Work. — Work is moving a body ; and the amount of work depends upon the amount of force (resistance) over- come and the distance through which that force is overcome. The several units of work in common use are the foot-poundal (ft. -pi.), erg , foot-pound (ft. -lb.), and gram-centimeter (g. -cm.), and are defined as follows : 1. The ivork done in overcoming 1 poundal of force through a distance of 1 foot — 1 foot-poundal. 314 ADVANCED ARITHMETIC. 2, The work done in overcoming 1 dyne of force through a dis- tance of 1 centimeters^ 1 erg. 3. The work done in overcoming 1 pound of force through a dis- tance of 1 foot = l foot-pound. Jf. The work done in overcoming 1 gram of force through a dis- tance of 1 centimeter ^=1 gram-centimeter. Note. — The foot-poundal and the erg are called absolute units of work ; and the foot-pound and gram-centimeter are called gravity units. This classification agrees with that of the force units (see p. 308). TABIiE or EQITIVAIiENTS. 1 g.-cm. (]Sr. Y.) = 980 erg:s. 1 ft.-lb. (N. Y.) = 32.16 ft.-pl 1 ft.-pl.=421402 ergs. EXERCISE CXXIX. 1. Reduce 16 ft.-pl. to ergs. 2. Reduce 182 ft.-lb. to ft.-pl. 3. Reduce 842 g.-cm. to ergs. 4. Reduce 8428040 g.-cm. to ft.-pl. 6. Reduce 64820 ft.-pl. to ft.-lb. 6. Reduce 69580 ergs, to g.-cm. 7. Reduce 742 ft.-lb. to grgs. 8. Reduce 21070100 ergs, to ft.-pl. TERMS. ir, the work. F, the force (resistance), d, distance. Developing the formula for ivork. From the definition of foot-poundal, ADVANCED ARITHMETIC. 315 (1) Work done in overcoming 1 pi. of force through a distance of 1 ft. = lft.-pl. (2) Work done in overcoming F pi. of force through a distance of 1 ft. = i^ft.-pl. (3) Work done in overcoming F pi. of force through a distance of d ft. = Fd ft.-pl. ; or, stated abstractly, W=Fd. Note. — When F is poundals and cZ, ft., the work 18 foot-poundals ; when F is dynes and d, cm., the work is ergs; when F is pounds and d, ft., the work is foot-pounds ; and when F is grams and d, cm., the work is gram- centimeters. EXERCISE CXXX. 1. Define work, foot-poundal, foot-pound, erg, and gram- centimeter. 2. Does the question of time enter into the work formula ? Is the work in No. 4 the same whether done in 10 min. or in 10 hr.? 3. Develop the work formula, and give the relation which it expresses. 4-. A team of horses draws a wagon 2 miles. If the draft is 1200 pL, how much work is performed ? 5. 240000 ergs of work is performed in overcoming a force of 800 dynes. Through what distance did the body move ? 6. 9640 ft.-pl. of work overcomes a certain force through a distance of 120 ft. 6 in. Find the force. 7. Two horses draw a machine 40 rd., and thus do 105600 ft. -lb. of work. What is the draft of the machine in pounds ? in poundals? 8. How much work is done in lifting 50 Kg. of matter 20 meters high ? (Answer in g.-cm.) 9. 6240 g.-cm of work is performed in overcoming a resist- ance of 120 dynes. Find the distance. 152. Activity or Po^ver. — The rate at which work is 316 ADVANCED ARITHMETIC. done is called Power or Activity, and its amount depends upon the work done and the time in which it is done. There are two units of activity in use, the watt and the horse-power, which may be defined aS follows : 1. The activity required to do 10000000 ergs of ivork in 1 second = 1 ivatt. 2. The activity required to do 550 ft. -lb. of ivork in 1 second = \ horse-power (h.-p.) Note. — The watt is called the absolute unit of activity, and the horse- power the gravity unit. EQUIVAIiENTS. 1 }i.-p. (N. Y.)=T46 watts. TERMS. A, the activity or power. TT, the work. t, the time. Development of formulas for activity. From the definition of watt, (1) Activity reqd. to do 10000000 ergs of work in 1 sec. = 1 watt. (2) Activity reqd. to do 1 erg of work in 1 sec. = j^qqqqqqq ^^^t. (3) Activity reqd. to do W ergs of work in 1 sec. = j^qqqqqqq watts. W (4) Activity reqd. to do W ergs of work in t ^^<^- = ;^qqqqqqq^ ^2itt^ ; or, stated abstractly. This formula is correct for finding the activity in watts. By using the definition for horse-power, the formula for finding horse-powers will be found to be — ^- ^~550xC ADVANCED ARITHMETIC. 317 EXERCISE OXXXI. 1. What is power or activity f How does it differ from workf 2. Define watt^ and horse-power. 3. Reduce 520 h-p. to watts. ^. Reduce 17158 watts to h-p. 5. Develop all the formulas for activity. Give the relation expressed by each. 6. How many watts of activity will do 720000000 ergs of work in 6 sec? 7. How much work will be done by an activity of 12 h-p. in I min.? 8. How much time will be required for an activity of 5 watts to do 150000000 ergs of work ? 9. How many h-p. of activity will do 385000 ft.-lb. of work in 5 sec? 10. What is the horse-power of an engine that can raise 2376 lb. 1000 ft. in 3 min.? 11. How far can a 2-horse-power engine raise 5 tons in ^0 sec? 12. How long will it take a 2-horse-power engine to raise 5 tons 100 ft.? 153. Simple Macliines. — The simple machines are the Lever, Wheel and Axle, Inclined Plane, and Screw. (See Dic- tionary or Physics for definitions.) I. The elements involved in lever problems are power (P), weight (W), the distance from the fulcrum to the power (Z>), and the distance from the fulcrum to the weight (d). Note. — The distances D and d, are often spoken of as power arm and weight arm, respectively. Law of the Lever : The power is to the weight as the weight arm is to the power arm. Proportion : P : W :: d : D. 318 ADVANCED ARITHMETIC. II. The elements involved in the problems of the ivheel and axle are power (P), weight (W), radius (diameter or circum- ference) of the wheel (R), and the radius (diameter or cir- cumference) of the axle (r). Law of the Wheel and Axle : The power is to the weight as the radius {diameter or circumference) of the axle is to the radius {diameter or circumference) of the wheel. Note. — In this statement of the law, the power is supposed to be ap- plied to the wheel. Proportion : P : W ::r : R. Note. — The diameters or the circumferences may be used instead of the radii. III. The elements involved in problems of the inclined plane are 2Jower (P), weight (IF), the height of the plane (/t), the length of the plane {I), and the base of the plane (6). Laws of the Inclined Plane : ( 1 ) When the power acts parallel to the plane, The power is to the weight as the height of the plane is to its length. (2) When the power acts horizontally, The power is to the weight as the height of the plane is to its base. Proportions : ( 1 ) P : W: :h :l. (2) P:W::h:b. IV. The elements involved in problems of the screw are power (P), weight {W), the distance between the threads of screw {d), and the circumference through which the j)ower moves (C). Law of the Screw : The power is to the weight as the distance between the threads is to the circumference through which the power moves. Proportion : P :W : : d : C. Note, — These proportions are formulas, stated in the form of propor- tion instead of that of equations. ADVANCED ARITHMETIC. 819 EXAXFLSS. 1. The pilot-wheel of a boat is 3 feet in diameter; the axle, 6 inches ; the resistance of the rudder 180 pounds. What power applied to the wheel will move the rudder ? 6in. = ift. Proportion : (1) P: 180 : : ^ : 3. .'. the power will be 30 lb. to balance the resistance of the rudder; anything more than 30 lb. will move it. 2. The distance between the threads of a screw is 1^ inches, the circumference through which the power moves is 18^ feet. What weight will a power of 20 pounds support (neglecting friction)? 13Ht. = 162in. Proportion : (1) 20: W:: U : 162. - ^ (2) TF= 20x162^2160. .-. the required weight is 2160 lb. 3. A man exerts 120 pounds of power in supporting a280-lb. barrel upon a 7-ft. plane extending from the ground to a wagon : how high is the wagon (pressure exerted in the direc- tion of plane)? Proportion: (1) 120 : 280 : /i : 7. ^^^''- 280 -^• .*. the height is 3 ft. 4.. The length of a lever is 16 feet; weight at one end, 100 pounds : what power must be put at the other, 4 feet from the fulcrum, to support the weight ? Proportion : (1) P: 100 : : 12 : 4. (2) P=i5^ = 300. .'. the required power is 300 lb. 320 ADVANCED ARITHMETIC. EXEKOISE CXXXII. Name the elements and give the law of — 1. The lever. 2. The wheel and axle. 3. The inclined plane. If. The screw. 5. State the formula (proportion) for each. 6. Suppose a power of 75 pounds be applied to one end of a 12-foot lever to support a load at the other end : what will be the load when the fulcrum is at the center ? when the fulcrum is 3 feet from the weight ? 7. In a nut-cracker the nut is placed 1 inch from the hinge (fulcrum), the hand 6 inches. If I exert a pressure of 10 pounds, how many pounds of resistance does the nut furnish ? 8. Four men hoist an anchor weighing 1 ton ;. the barrel of the capstan is 8 inches in diameter ; the circle described by the handspikes is 6 ft. 8 in. in diameter. How great a force mUst each man exert ? 9. A power of 70 pounds on a wheel 10 feet in diameter bal- ances a weight of 800 pounds : find the diameter of the axle. 10. The base of an inclined plane is 10 feet, the height 3 feet. What force applied parallel to the base will balance a weight of 2 tons ? 11. In No. 10, if the force applied be 500 pounds, what weight will it support ? 12. A weight of 800 pounds rests on an inclined plane 8 feet high, being held in equilibrium by a force of 25 pounds, acting parallel with the plane : find the length of the plane. 13. How great a pressure will be exerted by a power of 15 pounds applied to a screw whose head is 1 inch in circumfer- ence, and whose threads are -J inch apart ? ADVANCED ARITHMETIC. 321 D. MISCELLANEOUS. 154. Partitive Proportion. — Partitive Pro- portion is the process of dividing a given number into parts having a given ratio to one another. EXAMPIiES. 1. .Divide $25 into two parts, in the ratio of 2 to 3. Note. — If the two answers were obtained and the first answer divided into two parts and the second answer into 3 parts, there would then be 5 parts all of the same size. These 5 parts then represent the amount to be divided, or $25; 2 parts, the first answer; and 3 parts the second an- swer. Hence, Proportions : (1) 5 parts : 2 parts :! $25 : $( )? (2) 5 parts : 3 parts : : $25 : $( ) ? . •. the answers are $10 and $15. 2. A lever 33 feet long has a power of 50 pounds at one end and a weight of 60 pounds at the other end. Where is the ful- crum, if the power balances the weight? From the law of the lever, we know that — 50 lb. : 60 lb. : : wt. arm : pr. arm. Then the weight arm must be to the power arm as 50 to 60, or 5 to 6. Now, the problem is to divide 33 feet into 2 parts in the ratio of 5 to 6. Proportions : (1) 11 parts : 5 parts :: 33 ft. : ( )? (2) 11 parts : 6 parts : : 33 ft. : { ) ? .*. the arms are 15 ft. and 18 ft. 3. Divide 65 bushels of wheat into 3 parts, in the ratio of 3 to 2 to 1^. 3pt.+2pt.+l^pt. = 6| pt. Proportions : (1) Ql parts : 3 parts : : 65 bu. : ( ) bu.? (2) 6^ parts : 2 parts : : 65 bu. : ( ) bu.? (3) 6^ parts : U parts : : 65 bu. t ( ) bu.? .-. thereqd. partsare30bu.,20bu.,and 15 bu. 322 ADVANCED ARITHMETIC. EXERCISE CXXXIII. 1. Divide $750 into 2 parts in the ratio of 1 to 4; in the ratio of 4 to 11. 2. Divide $220 in the ratio of 3 to 7 to 12. 8. A and B balance on opposite ends of a pole 10 feet long. A weighs 120 pounds, and B 180 pounds : where is the fulcrum? 4. At the opposite ends of a lever 20 feet long, two forces are acting whose sum is 1200 pounds. The two arms of the lever are as 2 to 3. What are the two forces when the lever is in equilibrium ? 6. A fails, and can pay his three creditors, E, F, and G, but $750. He owes E $700, F $500, and G $300: how much should each receive ? 155. Mixture Problems. — 1. Different kinds of tea sold as follows : 10 lb. for $7, 30 lb. for $18, and 20 lb, for $8. Find the average price per pound. Solution : (1) Price of 10 lb. = $7. (2) Price of 30 lb. = $18. (3) Price of 20 lb. = $8. (l)+(2)+(3) = (4) Price of 60 lb. =$33. eV of (4) = (5) Price of 1 lb. = $.55. 2, A mixture composed of sugar at 2/ per lb. and sugar at 5/ per lb. is worth 3/ per lb. How is it mixed ? Note. — There is a gain on the 2^ sugar when it becomes a part of a mixture valued at 3^ per lb., and a loss on the bf sugar. They must be mixed in such a ratio as will make the gain on the one equal to the loss on the other. Solution : (1) Gain on 1 lb. of 2f sugar at 3^ = If. (2) Loss on 1 lb. of 5^ sugar at 3^ = 2^. N (2) = (3) Amt. of 5f sugar reqd. to lose 2^ = 1 lb. i of (3) = (4) Amt. of 5<* sugar reqd. to lose If = ^ lb. (1) = (5) Amt. of 2f sugar reqd. to gain lf = l lb. Comparing (5) and (4), we see that the mixture must be made in the ratio of 1 lb. of the 2f sugar to | lb. of the 5f sugar, or in the ratio of 2 to 1. ADVANCED ARITHMETIC. 828 3. How many pounds of coffee at 25/ per lb. must I mix with 15 lb. at 80/ per lb. to make a mixture worth 28/ ? Solution : (1) Loss on 1 lb. of 30^ coffee at 28^ = 2^. 15x(l) = (2) Loss on 15 lb. of 30<* coffee at 28<^ = 30^, whole loss. (3) Gain on 1 lb. of 25,^ coffee at 28^ = 3^. (3) = (4) Amt. of 25^ coffee reqd. to gain 3<* = 1 lb. 10x(4) = (5) Amt. of 25^ coffee reqd. to gain 30,^ = 10 lb., answer. 4-. Mix two kinds of coffee at 20/ and 25/ per lb., to make a mixture of 25 lb. worth 24/ per lb. Solution : (1) Gain on 1 lb. of 20^ coffee at 24^ = 4^. (2) Loss on 1 lb. of 25<^ coffee at 24^ = 1^. (1) = (3) Amt. of 20^ coffee reqd. to gain 4^ = 1 lb. i of (3) = (4) Amt. of 20<5 coffee reqd. to gain l^ = i lb. (2) = (5) Amt. of 25^ coffee reqd. to lose 1^ = 1 lb. From (4) and (5), we see the mixture must be in the ratio of i lb. of 20f coffee to 1 lb. of the 25<^ coffee, or the ratio of 1 to 4. The problem now is, to divide 25 lb. in the ratio of 1 to 4. (6) 5 parts : 1 part : : 25 lb. : ( ) ? (7) 5 parts : 4 parts t : 25 lb. : ( ) ? .a n 4- 1 X 25 lb. -. -, . . Smaller part = = 5 lb. o and the larger part = ' = 20 lb. 5 EXERCISE CXXXIV. 1. How many pounds of sugar at 8/ per lb. must I mix with 10 lb. at 11/ per lb. to make a mixture worth 10/ per lb.? 2. How must I mix coffee at 24/ per lb. with coffee at 80/ per lb. to make a mixture worth 26/ per lb.? 3. I mix sugar at 9/ per lb! with 72 lb. at 5/; the whole mixture is worth 7/ per lb. : how much of the 9/ sugar did I use? 4-. A vintner mixed two kinds of wine, one at 40/ per pint and the other at 60^ per pint. He finds that the mixture is worth just 55/ per pint : how did he mix it ? 324 ADVANCED ARITHMETIC. ^ 5. A saloon-keeper watered his 60/ wine, and sold it at 45/ per pint : if there were 6 gallons and 1 quart of water added, how much wine was there ? 6. A man mixed coffee at 26/ per lb. with coffee at 82/ per lb., making a mixture of 48 lb. worth 80/ per lb. : how much of each did he use ? 7. I mix wine at 40/ and 50/ per pint with wine at 60/ per pint; the 50/ and 60/ wine is mixed in the ratio of 2 of the 50 to 1 of the 60 : if the mixture is worth 48/ per pint, how do I mix it ? 8. If quarters and dimes are mixed in such a ratio that the same money might be molded into the same number of 15/ pieces, how are they mixed ? 156. Equation of Payments.— Equation of Payments is the process of finding the date on which one \ payment may be made in settlement of several debts or an ac- * count of several items, falling due at different times, without loss to either debtor or creditor. EXAMPLES. 1. I owe $600 due in 10 months, $800 due in 9 months, and $1000 due in 6 months, without interest. When can I settle all together ? Solution : Suppose that I should settle all now, then I would lose — (1) Use of $600 for 10 mo. = use of $1 for 6000 mo. (2) Use of $800 for 9 mo. = use of $1 for 7200 mo. (3) Use of $1000 for 6 mo. = use of $1 for 6000 mo. $2400, amt. paid. Use of $1 for 19200 mo. , loss. (4) Use of $1 for 19200 mo. = use of $2400 for 8 mo. .'. 8 months from now, $2400 will settle the debt. 2. My account with Jones is as follows : I owe him $600 due in 11 months, $800 due in 4 months, $500 due in 1 year, and $200 due now ; he owes me $700 due in 8 months and $600 due ADVANCED ARITHMETIC. 825 in 9 months. When will $800 settle the account ? and how much will be due 1 year hence, interest being 6%? Solution : Assume now as the date of settlement. (1) Use of $600 for 11 mo. = use of $1 for 6600 mo. (2) Use of $800 for 4 mo. = use of $1 for 3200 mo. (3) Use of $500 for 12 mo. = use of $1 for 6000 mo. (4) $200 due now. $2100, amt. paid. Use of $1 for 15800 mo., loss. (5) Use of $700 for 8 mo. = use of $1 for 5600 mo. (6 ) Use of $600 for 9 mo. = use of $1 for 5400 mo. $1300, amt.recd. Use of $1 for 11000 mo., gain. (7) $2100 -$1300 = $800, balance. (8) Use of $1 for 15800 mo. -use of $1 for 11000 mo. = use of $1 for 4800 mo., net loss. (9) Use of $1 for 4800 mo. = use of $800 for 6 mo. .*. the time of payment is 6 mo. from now. (istans.) (10) Amt. of $800 at Q% for | yr. = $24+ $800 = $824. (2d ane.) 3. I bought goods Jan. 1, 1898, on 80 days' credit, $600 ; Feb. , 10, on 60 days' credit, $5(X). When can I settle the whole by < paying $1100? Solution: Assume Jan. 1st as the date of settlement. The first debt will be paid 30 days before it is due, and the second will be paid 100 days before it is due. (1) Use of $600 for 30 days = use of $1 for 18000 days. (2) Use of $500 for 100 days = use of $1 for 50000 days. $1100, amt. paid. Use of $1 for 68000 days, loss. (3) Use of $1 for 68000 days = use of $1100 for 62 days, nearly. 62 days after Jan. 1, 1898, is March 4, 1898. Note.— A part of a day is always counted as a full day in equating the time of payment. This is plain, for in the above problem, as the time was more than 61 days, it must fall on the 62d day. These solutions are given in full. After the pupil writes out 326 ADVANCED ARITHMETIC. a few in this way, he should shorten the solution. Thus, for Example 1 : Solution: (1) 600x10 = 6000. (2) 800 X 9 = 7200. (3) 1000 X 6 =6000 . 2400 19200. 19200 + 2400 = 8. Answer, $2400, 8 mo. from now. EXERCISE CXXXV. 1. I owe $400 due in 4 months and $800 due in 7 months, each bearing interest at 8%. Find the equated time and the amount due. 2. I owe Smith $1200 due in 8 months and $600 due in 5 months ; Smith owes me $2000 due in 10 months : when will $200 settle what he owes me ? 3. A merchant owes his creditor for goods bought, as follows : July 3, $640; July 15, $300; Aug. 10, $520: when will $1460 settle his account ? 4.. If, in No. 3, the merchant pays $500 Aug. 1st, when can he settle the balance with $960 ? 5. If, in No. 3, each amount is a note due in 60 days, bearing 10% interest, find the equated time of payment and the amount due. 157. Partner sliip. EXAMPIiES. i. A, B and C form a partnership, A puts in $200, B $300, and C $400. They gain $450. Divide it among them. Solution : The whole capital is $900. Each partner's capital is to the whole capital as his share of the gain is to the whole gain. Hence — (1) For A, $200 : $900 : : $( ) : $450 ? .-. A's gain is $100. (2) For B, $300 : $900 : : $( ) : $450 ? .*. B's gain is $150. (3) For C , $400 : $900 : : $( ) : $450 ? .-. C's gain is $200. ADVANCED ARITHMETIC. 827 2. A invests $100 for 6 months ; B, $200 for 2 months. They gain $50. Divide it between them. Solution : "* (1) Earnings of $100 for 6 mo. = earnings of $600 for 1 mo., A's part. (2) Earnings of $200 for 2 mo. = earnings of $400 for 1 mo., B's part. (3) The whole earnings = earnings of $1000 for 1 mo. m For A \ Earnings of ). j earnings of } . •*/ n -3;.^? (4) J^or A, j ^g(^ j^j. 1 jno. r \ $1000 for 1 mo. T "^^ ''^^^ .'. A's gain is $30. Note. — The " 1 mo." may be omitted, as it is the same in both the 1st and the 2d terms, and the proportion may be shortened as in (5). (5) ForB, $400: $1000:: $( ):$50? .*. B's gain is $20. EXERCISE OXXXVI. 1. Three persons purchase a farm for $2800, of which A pays $1200, B $1000, and C $600. How shall they divide the rent, which pays $224 a year ? ;g. A, B and C owned a cargo of corn valued at $3475.60. A owned i, B ^, and C the remainder. How much did each lose if the cargo, which was insured for $2512, was lost ? S. A, B and C form a copartnership. B puts in $250 for 6 months, C $275 for 8 months, and A $450 for 4 months. Divide their gain, $825. ^. A, B and C form a copartnership for 1 year, and invest respectively $9600, $8400, and $7200. After 4 months, A in- vests $2000 additional, B $1400, and C $800. They gain $12800: what is each man's gain ? 158. Longitude and Time. — I. Longitude. Longitude is the distance east or west of the prime meridian measured in degrees (°), minutes (^), and seconds (^^). TABIjE. 1° = 60\ r - 60 ^ 328 ADVANCED ARITHMETIC. The meridian of Greenwich, England, has been selected as the prime meridian. 180° east from the prime meridian (0°) reaches half around the earth. Any place on that half of the earth is said to be in east longitude. 180° from 0° west reaches around the other half of the earth. Any place on that half of the earth is said to be in west longitude. The longitude of a place is its distance {less than 180°) east or ivest of the prime meridian. The distance measured in one direction (east or west) around the earth is 860°. II. Time. — (1) Sun Time: The sun appears to travel west- ward around the earth (860°) once in 24 hours. It passes over 15° of longitude in 1 hour of time ; 15' of longitude in 1 min- ute of time ; and 15'^ of longitude in 1 second of time. There- fore the difference of longitude between two places, expressed in ^ ' ^\ is numerically 15 times the difference of time ^ expressed in hours, minutes, seconds. Of two places on meridians nearer to each other than 180°, the place west is earlier and the one east is later, except when the I. D. L. (see below) and midnight are both between the two places ; in which case the place west is later by the difference between their difference of time and 24- hours. (2) The Day : Every day begins first at or near the 180th meridian. On this account the 180th meridian is often called the International Date Line ( I. D. L.).* When midnight is on the I. D. L., there is one, and but one, day on all the earth, (say) Sunday. As midnight moves westward, it changes the day on every part of the earth over which it passes to the new day, Monday. When it has gone all the way around and again *The International Date Line used to have a very winding course. After coming south- ward through Bering Strait, it took a southwesterly course, passing west of the Philippine Is- lands, then bending eastward, then southward into the Southern ocean near the 180th meridian. But now, after passing southward through Bering Strait, it swerves a little (20° or 30°) to the westward. Returning to the 180th meridian about 40° N., it follows that meridian to a point about 10° S. Then it swerves eastward about 10° or 15°. Returning to the 180th meridian again about 50° S., it continues southward on that meridian. This description was taken from a diagram given in The Pathfinder in 1899. ADVANCED ARITHMETIC. 829 reaches the I. D. L., it is Monday all over the earth. When any place is between the I. D. L. and the midnight meridian, if it is west of the I. D. L. and east of midnight, it is the new day ; if it is east of the I. D. L. and west of midnight, it is the old day. (S) Standard or Railroad Time : There are five standards of railroad time in the United States and Canada : Intercolonial, Eastern, Central, Mountain, and Pacific. Each covers a lon- gitude of 15°. Intercolonial commences 52J° W., and the Pa- cific ends 1271° W. The time throughout each standard is the sun time of its central meridian : For Intercolonial, 60th; for Eastern, 75th; for Central, 90th; for Mountain, 105th; and for Pacific, 120th. Note. — Railroad companies vary the boundary -lines of these standards in some places. III. Problems: There are two classes of problems usually given in Longitude and Time : ( 1 ) Given the longitudes of two places and the time at one of them, to find the time at the other; and (2) given the times of two places and the longitude of one, to find the longitude of the other. EXAMPIiES. 1. When it is noon, Monday, 150° W., what is the time 120° E.? Direction. — In drawing a figure to repre- sent a longitude problem, the following will help to make your problem clear: (1) Sketch a circular figure to represent a parallel of lat- itude ; (2) place 0° at the top ; (3) 180° (I. D. L.) at the bottom ; (4) locate approximately the two places under consideration and midnight p^ =i; jy l (M.). M. = midnight. o 330 ADVANCED ARITHMETIC. Solution: (1) 180° - 150° = 30°. (2) 180° -120° = 60°. (3) 30°+60° = 90°, difference of longitude. .'.6 hours = difference of time. 6 hours earlier than noon = 6 a. m. 150° W. is in the old day (Monday). 120° E. is in the new day (Tuesday). 2. When it is 6 hr. 40 min. p. m., Friday, 20° E., what is the time 140° E.? Solution: (1) 140° -20° = 120°, difference of longitude. . • . 8 hr. = difference of time. 8 hr. later than 6 hr. 40 min. p. m. =2 hr. 40 min. a. m. of the following day, Saturday. 3. When it is 7 hr. 20 min. a. m., Friday, 160° W., on what meridian is it 3 hr. 20 min. a. m. Saturday? Solution: (1) 7 hr. 20 min. -3 hr. 20 min. = 4 hr., dif- ference of time. . •. 60° = difference of longitude. Since 3 hr. 20 min. a. m. is earlier than 7 hr. 20 min. a. m,, the required place is west. 60" west of 160° W. is 140° E. ; and it is Saturday, because it is in the new day (beyond the I. D. L.). 4.. When it is noon at Greenwich, 0°, what is the time at St. Louis (standard time)? Solution : The time of St. Louis is that of 90° W., or 6 hr. earlier than noon = 6 hr. a. m. of the same day. 5. When it is 3 o'clock a. m. at Kansas City, where is it 2 o'clock a.m.? 4 o'clock a. m.? Solution : (1) Kansas City is in Central Time ; it is 2 o'clock a. m. on all the standard west, or 7^° each side of 105° W. (2) It is 4 o'clock a. m. on all the standard east of Central Time, or 71° each side of 75° W. ADVANCED ARITHMETIC. 331 EXERCISE CXXXVII. 1. When it is 9 o'clock p. m., Wednesday, 20° E., what is the time at Greenwich (0°)? 2. When it is 9 o'clock p. m., Wednesday, 20° E., what is the time at 30° W.? 3. When it is 9 o'clock ]). m., Wednesday, 20° E., what is the time at 120° W.? ^. When it is 9 o'clock p. m., Wednesday, 20° E., what is the time at 80° E.? 5. When it is noon at Greenwich, at what i:)lace is it 2 o'clock a. m.? 6. When it is noon at Greenwich, at what place is it 7 o'clock a. m.? 7. When it is noon at Greenwich, at what place is it 3 o'clock p. m.? 8. When it is noon at Greenwich, at what place is it 8 o'clock p. m.? 9. When it is 1 hr. 20 min. p. m., Saturday, 110° E., what is the time 170° E.? 10. When it is 1 hr. 20 min. p. m., Saturday, 110° E., what is the time 105° W.? 11. When it is 1 hr. 20 min. p. m., Saturday, 110° E., at what place is it 9 hr. p. m.? 12. When it is 1 hr. 20 min. p. m., Saturday, 110° E., at what place is it 11 hr. p. m.? 13. When it is 1 hr. 20 min. p. m., Saturday, 110° E., what is the time at Boston ? at New York ? at St. Louis ? at Cincin- nati ? at Kansas City? at San Francisco (standard time)? IJ^. When it is noon" on the 180th meridian, it is 6 hr. 52 min. 40 sec. p. m. at Harrisburg, Pa. (sun time) . Find the lon- gitude of Harrisburg. 15. A watch is set right at Pekin, China, longitude 116° 26'' E. On what meridian is the watch when, at noon, it shows 9 hr. 20 min. a.m.? 882 ADVANCED ARITHMETIC. 16. When it is 5 minutes after 4 o'clock on Sunday morning at Honolulu, longitude 157° 52^ W., what is the time at Sydney, Australia, longitude 151° 11' E.? 17. The battle of Manila began 41 min. after 5 o'clock, Sun- day morning, May 1. Manila is in longitude 121° 20' E. Wash- ington is in longitude 77° W. What was the time at Washington when the battle began, by sun time ? by standard time ? 18. When it is 10 o'clock p. m. Wednesday, 20° E., what is the time at Sydney, Australia ? at Honolulu ? railroad time at San Francisco, Cal.? at St. Louis, Mo.? at Boston, Mass.? Note. — When time is required, always give the day as well as the hour of the day. PART III. I. STUDY OF NUMBERS. A. INTEGRAL NUMBERS. 159. Positive and Negative Numbers. — The signs phis (-f-) and minus ( — ) are placed before numbers, when written with figures or letters, to indicate opposite mean- ings or conditions. Thus, If +$4 means $4 gain, -$4 means $4 loss. If +$4 means $4 capital, -$4 means $4 debt. If +4 means 4 to he added, -4 means 4 to he suhtracted. If +4° means 4°(temperature) ahove zero, -4° means 4° heloiu zero. If +4 ft. means 4 ft. traveled to the right, -4 ft. means 4 ft. trav- eled to the left. A Positive Number is a number which, when written, is preceded by the plus sign, expressed or implied. Note. — For convenience, a positive number, standing alone or first in an expression, is usually written without the sign, +. A Neg-ative Number is a number which, when written, is preceded by the minus sign. Note. — The sign, -, is never omitted before negative numbers. All numbers are eiiher positive or negative. To illustrate: 5,4,8,2,1,0, -1*, -2, -3, -4, -5. The numbers in this series decrease in absolute value from left to right; each number, beginning with 4, is one less than * Bead, minus one, minus two, and so on. (333) 884 ADVANCED ARITHMETIC. the number to the left of it. Notice that from to the right the absolute value decreases as the numerical value increases ; —5 is less than —4, —4 is less than —8, etc. The question is sometimes asked, "Can a number be less than nothing?" Yes: a number may be less than nothing. Suppose that I have but $5 and owe nothing, and you ask me how much I am worth: I would say, "$5." That means that I am worth $5 more than nothing, +$5. Suppose now, that I have the $5, and in addition I owe some one $8. I am now worth $8 less than before, or but $2. Suppose again that I owe another debt of $3 : my worth is again decreased $8, and I am now worth $1 less than nothing, — $1. If I should have another debt of $8, I would then be worth $4 less than nothing, — $4. Note.— It does not follow that all negative numbers represent value less than nothing. A negative distance is not less than nothing; -10° ( temperature) is not less than nothing. 160. Numbers Expressed by Means of Let- ters. — In expressing numbers by means of letters, or letters and figures, there are four things to be considered : (1) the literal part, (2) the coefficient, (3) the exponent, and (4) the sign. The Literal Part is the letter or letters used as factors in ex- pressing a number. Thus, In -5a^, a is the literal part ; in 5a%, a and b form the literal part. The Coefficient is a factor (expressed by figures in this book), placed on the left of the literal part. Thus, In Wax, 15 is the coefficient. Note. — In such expressions as ax, or -ax^, the coeflRcient is not ex- pressed, but is understood to be 1. An Exponent is a small figure (sometimes a letter), placed ADVANCED ARITHMETIC. 335 just above and to the right of a factor to show what power of the factor is used. (See p. 179.) The Sign of a number is used to indicate whether the number is positive or negative. A number of one term is called a Monomial. A number of more than one term is called a Polynomial. A polynomial of two terms is called a Binomial; of three terms, a Trinomial. EXERCISE CXXXVIII. 1. Give use of the signs plus and minus. 2. What is a positive number f 3. What is a negative number ? If. Explain how some numbers do have values Zess than nothing. 5. How many things are to be considered in expressing num- bers by letters or by letters and figures ? 6. Explain what is meant by literal part. Give example. 7. Define coefficient. Give example. 8. Define exponent. Give example. 9. What does the sign of a number show ? 10. Define monomial. Give example. 11. Define polynomial. Give example. 12. Define binomial. Give example. 13. Define trinomial. Give example. 161. Addition. — Since we now have to consider both positive and negative numbers, our process of addition must be extended to cover the addition of negative as well as positive numbers. In explaining the following examples, let us assume that + means capital and — means debt. ADVANCED ARITHMETIC. BXAMPIiBS. 1. Add +$8 and +$4. Proce.: +,8+(+»4)= +.12. result. J^^^^ 2. Add -$8 and -$4. $12 (capital). d.o / *.s ^.^ , H (debt) added to Process: -$8 + (-$4)= -$12,result. ^g (debt) makes $12 3. Add +$8 and -$4. (debt). *o / *.v ^. , $4 (debt) added to Process: +$8 + (-$4)= +$4, result. ^g (^^p.^^^^ ^.^^^ ,^^ 4. Add -$8 and +$4. (capital). $4 (capital) added Process: -$8+( + $4)= -$4, result. to $8 (debt) gives $4 (debt). Let us write these processes abstractly and compare : (1) +8 + (+4)=+12. (2) -8 + (-4)=-12. (3) +8 + (-4)=+4. (4) -8 + ( + 4)=-4. Observe that in (1) the signs were alike (both plus), and we added 8 and 4 ; in (2) the signs were alike (both minus), and we added 8 and 4 ; in (3) and (4) the signs were unlike (one plus, the other minus), and we sub- tracted 4 from 8. In every case, the sign of the result is the same as that of the larger addend. Rule for Addition : When the signs are alike, add the num- bers; lohen unlike, subtract the smaller from the larger, giving the result the sign of the larger addend. 5. Add 50, +38, -14, -78, -52, +28, +44. . I. Process: (1) 50+38 = 88 ; (2) 88-14 = 74 ; (3) 74-73 = 1 ; (4) 1-52= -51 ; (5) -51 + 23= -28; (6) -28 + 44 = 16, result. Note.— The process may be much shortened by uniting the positive numbers into one amount and the negative numbers into another. The difference of these amounts preceded by the sign of the greater will be the mathematical sum. Thus: II. Process: (1) 50+38+23+44 = 155. (2) -14-73-52= -139. (3) 155-139 = 16, result. ADVANCED ARITHMETIC. 887 Literal numbers are similar, when they are composed of the same letters affected by the same exponents. Thus : ah^ and 25 aW are similar, but a% and ah"^ are dissimilar. Principles: 1. Only similar numbers can he added to form one term. 2. The addition of dissimilar numbers is indicated by placing between them the proper sign. 6. 7. 8, -4 a 6xy 7x^y -7a 4:xy -6xhj +5 a -dxy -Ux'y +15 a -Sxy -{-llx'y — a result. -Sxy -11 xy, result. -hx'^y +8 a, 0, 0, result. 9. Add -76, +96, -\-6xy, +46, -7xy-j-z. Process : -7b^6xy-\-z +Qb-7xy +46 6b -xy+z, result. 10. Add 12xy'^-lbax-c, 7aa;+9c, -bxif+17 ax-12c. Process : 12 xy^ -lb ax - c + 7 ax +9c -5xy^-i-17ax-12c 7xy^+9ax-4c, result. 11. Add5(x-2/), -7ix-y), l^x-y). Process : 5 (.r - y) -7{x-y) 15{x-y) ld{x-y), result. Note. — When the expressions in parentheses are similar, they may be treated as one number. As, bxNo. -7y No. + 15xNo. = 13xNo. 338 ADVANCED ARITHMETIC. EXERCISE CXXXIX. 1. (jiwe rule for addition. Give principles. 2. Explain examples 1 to 4, letting the numbers represent distance, — the positive numbers distance measured to the right, and negative numbers distance measured to the left. Thus, In Example 1, + 8, to the right. +4, to the right. + 12 ; or, 12 to the right of starting-point. In Example 3, + 8, to the right. -4, t o the left. +4 ; or, still 4 to the right of starting-point. Add (orally) : 3. 4- 6. 6. 7. -7 +7 +7 -7 14 -5 +5 -5 +5 -9 8. 9. 10. 11. 12. -5 +9 -17 +4 -13 +7 -6 +4 -18 + 12 -4 +3 _2 +5 -6 IS. u. 15. 16. 17. — a -6x -{-12 xy -3 A -20 xY +5 a +25 a: -16xy +5 A + 18a:y -7a -3a; -\-9xy -12a^x +4a:y + a -11a: -12 xy +4a2a; -14a;Y 18, 19. 20. 7{x- ■22/) -Hx^-y) 9(x+2/)^ -9{x- ■22/) +Hx'-y) - -Hx-^yY -{-(X- -22/) -\2(x^-y) ^(x^-yy ADVANCED ARITHMETIC. 839 Copy and add : 21. 12a-9c, -4a+17c, -7 c+15a, 27ft-c. 22. 42a^-5ab, -SQ a^-\-12ab, 7a^-15ab, -ab-14:a\ 23. a'-2abW. 5a2-9a6+4?>2, -12a2+26a6-12 6^, 36a-^ -24 62+7 a6. 21^. \la%-^nab''-\-a%\ ba'b -21 ab^-^^a%\ lA.aW-bd'W^ 9a^b, -4a262+5a62+3a26. 25. 9xhj-{-4:X-\-Sij-9y\ 4:y^-2x^y-5x^Sy, 12x-14:y-Qy' -2x^y, x^y-x+y-y^. 162. Subtraction. — It will help to explain the process of subtracting positive and negative numbers, if we will note the fact, that any monomial may be expressed as a binomial , one of whose terms is positive and the other negative, and that the numeri- cal value of either term may be increased at will, provided the nu- merical value of the other be correspondingly increased. EXAMPLES. 1. Represent +5 by an expression containing +8, +12, +17, -5, -14, -21. (1) 5= 8- 3. (4) 5 = 10- 5. (2) 5 = 12- 7. (5) 5 = 19-14. (3) 5 = 17-12. (6) 5 = 26-21. 2. Represent —7 by an expression containing +4, +12, +15, -9, -13, -25. (1) -7= -11+ 4. (4) -7=- 9+ 2. (2) -7= -19 + 12. (5) -7= -13+ 6. (3) -7= -22+15. (6) -7= -25 + 18. The process of subtraction consists in taking from the minu- end the number expressed by the subtrahend. KXAMPIiES. 3. From +8 take +4. (1) +8 -(+4)= +4, result. Taking +4 from +8 leaves +4. 340 ADVANCED ARITHMETIC. 4. From -8 take -|-4. (1) -8= -12+4. The -8 is the same as -12 + 4; tak- (2) (-12+4) -(+4)= -12, result. ing away the +4 leaves -12. 5. From +8 take —4. (1) +8= +12-4. The +8 is the same as +12-4; tak- (2) (+12- 4) -(-4)= +12, result. ing away the -4 leaves +12. 6. From —8 take —4. (1) -8-(-4)= -4, result. Taking -4 from -8 leaves -4. Note. — The same result will be obtained by the capital and debt plan of explanation. (See Addition, p. 336.) Let us compare these examples : (1) +8-(+4)=+ 4. (2) -8-(+4)=-12. (3) +8-(-4)=+12. (4) -8-(-4)=- 4. Observe that, when the signs of the subtrahend and minuend w^ere alike (in (1) and (4)), we took the difference between the 8 and the 4; when the signs of the subtrahend and minuend were unlike (in (2) and (3) ), we took the sum of the 8 and the 4. As we might have expected, this is exactly the reverse of our com- parison in addition. Suppose that in the four examples above, we change the sign of each subtrahend and add (instead of subtracting) and see what results we get : (1) +8+(-4)=+4. (2) -8+(-4)=-12. (3) +8 + (+4)=+12. (4) -8+(+4)=-4. These results are just the same as above, and we have sub- tracted by changing the signs of the subtrahends and adding. Rule for Subtraction : Conceive the sign of the subtrahend to he changed, {the -\-to — or the — to -\-) . and proceed as in addi- tion. Note. — Do not actually change the sign of the subtrahend on paper ADVANCED ARITHMETIC. 341 or slate in written work, but do it mentally. An actual change of written signs leads to confusion in reviewing the work. Principles: 1. The result can he expressed in one term only when the minuend and subtrahend are similar. 2. The subtraction of dissimilar numbers is indicated by placing between them the proper sign. 7. From 12-5+4 take 8-8+2. (1) 12-5+4 = 11. Collect the terms of both minuend and (2) 8-3+2= 7. subtrahend; change the sign of the sub- (3) 11-7 = 4. trahend, and proceed as in addition. Or, ^ i2_t5+4-84-S-2 = 4 change all signs of the subtrahend, ' * and then collect. 8. 5+17-3-(4+8-10+2) = ? Process: 5+ 17 -3 -(4+8- 10+2) = 5 + 17-3- 4-8 + 10-2 =15, result. Note. — The -before the parentheses shows that all the numbers within are to be subtracted, and that the expression within the parentheses is a subtrahend. Rule for Clearing of Parentheses Preceded by a Minus Sign: Drop the parentheses with the minus sign, and change the sign of every term within. 9. 10, 11. 12 7 a -xy +3a;2 -17xy' 5 a +5.^2/ —9x^ —^xy"^ 2 a, result. —6 a;?/, result. +12 .t^, result. — 12 a:?/^, result. 13. 7x^-\-4:ax-\-2b^-{-bb^^7xa-5x^) = { )? Process : 7 x^-\-4:ax+2b^ - 5x^+7 a.r-5b^ 12a;2-8ax+7 62, result. Or, 7x^-{-4ax+2h'^-{-bh^ + 7ax-5a^) = 7 a;2+ 4 ax +2b^ + bb^-7 ax + 5 x^ = 12x^-3 ax +71/, result. 342 ADVANCED ARITHMETIC. 1^. From5(a4-:c) take 2(a+a;). 5(a+a:) 3(a + ar), result. EXERCISE CXL. 1. Explain examples 3 to 6, letting positive numbers mean capital and negative numbers mean debt; letting positive num- bers mean distance moved to the right, and negative numbers distance moved to the left. 2. Give the rule for subtraction. 3. Give the 'principles. J/.. Give the rule for removing parentheses preceded by a minus sign. 5. Do you change any signs in removing parentheses pre- ceded by a plus sign ? Subtract (ora^ lly)'. 6. 7. 8. 9. 10. -7 +7 +7 -7 12 ~5 +5 -5 +5 -9 11. 12. 13. u. 15. —a -la -9a -12 a; + 12 a; -6a 4-3 a -4 a -Qx +6 a: 16. 17. 18. 19. 20. — (J?X -dx'^y — 12mn -\-ab + 17^:^ 7 A -\-12xhj —9mn - 20 ab 23. -3a:^ 21 22. -%x- -22/) -\-^{x'-f) l{x^Jr2xy-\-f) +4(x- -22/) -\-9{x'-y') -A(x^+2xy-\-y^) ADVANCED ARITHMETIC. 343 Copy and subtract : 24. From3tt-562-c take 7 a+4 62-10 c. 25. From 7 a^-\-9xy-Qb^ take nb^-\-4.xy-10a\ 26. From —75{x^y—xy^) take —160{x^y—xy^). 27. 17a:-52/+6-(5a;+ll2/-10)=( )? 28. 9xy-{7ax-^12xy-Sz) = { )? 29. 7x^y-Sxy'^-{-y^-{-7i/+Qxy''-9yx^)=( )? 30. a-\-b — c — d—x — {a — b-\-c—d—x) = { )? 31. x^-j-2xy-{-y^-{x^-2xy^y^) = ( )? ^^. x^+Sxhj-i-Bxy^-\-y^-{x^-Bx^y+Bxy^-f) = ( )? 163. Mviltii^lication. — When the multiplier is posi- tive, it shows how many times the multiplicand is to be taken as an addend (see p. 21) ; but, when negative, it shows how many times the multiplicand is to be taken as a subtrahend. EXAMFIiES. 1. Multiply +7 by +3. Note. — Starting with 0, we add +7 three times. Process : (1) 0+( + 7)= +7, used once. (2) +7-t-(+7)= +14, used twice. , (3) + 14 + ( + 7) = + 21 , used three times . .'. +3x+7=+21, result. 2. Multiply -7 by +3. Note. — Starting with 0, add -7 three times. Process: (1) 0+(-7)= -7, used once. (2) -7+(-7)= -14, used twice. (3) - 14 -|-( - 7) = - 21 , used three times. .*. +3x -7= -21, result. 3. Multiply +7 by -3. Note.— Starting with 0, subtract +7 three times. Process : (1) 0-( + 7)= -7, used once. (2) -7 -( + 7)= -14, used twice. (3) - 14 - ( + 7) = - 21 , used three times. .-. -3x+7= -21, result. 344 ADVANCED ARITHMETIC. U. Multiply -7 by -8. Note. — Starting with 0, subtract -7 three times. Process : (1) 0-(-7)= +7, used once. (2) +7 -(-7)= +14, used twice. (3) +14-(-7)= +21, used three times. .-. _3x-7=+21. Let us compare these examples: (1) +3x+7=+21. (2) +3x-7=-21. (3) -3x+7=-21. (4) -3x -7= +21. Observe (1) that the numerical products are the same (each 21, or the product of 3 and 7), and (2) when the signs of the factors are alike (both plus as in (l)or both minus as in (4) ), the product is positive; but when the signs are unlike (as in (2) and (3) ), the product is negative. Law of Signs in Multiplication : Like signs give plus ; un- like signs give minus. 5, 6. 7. 8. + 12 « +12 a -12a -12a +66 -66 +6 6 -66 +72 a6, result. — 72 a6, result. — 72 a6, result. +72 a6, result. 9. -7aX+56x-8c=( )? Process: (1) -7 ax +5 6 = -35a6. (2) -35a6x-3c = +105 a 6c, result. 10. Multiply 4 m— 7 n by 5 71. Process : 4m -In -5n -20mn+35w2; or, S5 n^ -20 mn, result. Note. — In writing polynomials, it is customary (but not necessary) to write a positive term (if there be one) on the left, and omit its sign. ADVANCED ARITHMETIC. 345 11. Multiply ba(x-\-y) by —h. Note. — This multiplication may be performed (1) by multiplying the 5 a by -h, (2) by multiplying {x + y) by -h, or (3) by removing the paren- theses from the multiplicand and multiplying each term by - h. Thus : (1) (2) ba{x+y) 5a {x+y) - b -b -5ab {x+y), result. 6a {-bx-by), result. (3) ba {x+y)=bax+5ay -b_ -babx-haby, result. 12. {a-b)^={a-b){a-b) = ( )? Process : a-b a-b -ab+b^ a^-ab a^-2ab + b^, result. Principle: 1. The square of the difference of two numbers is equal to the square of the first minus twice the product of the first times the second plus the square of the second. (Commit.) Note. — Compare this principle with the one on page 181, noting the points of similarity and difference. 13. Multiply a +6 by a — 6. Process : a + b a-b +ab-b^ a^ — ab a^ - 6^, result. Principle : 2. The product of the sum and difference of two numbers is equal to the square of the first minus the square of the sec- ond. (Commit.) 346 ADVANCED ARITHMETIC. Uf.. Multiply 3 a— 2a: by 5 6— a. Process: 3a -2 x 56 —a 15ab-10bx-Sa^+2 ax, result. EXERCISE OXLI. 1. What change in the definition (p. 21) is necessary to make it include the multiplication of negative numbers ? 2. What does a positive multiplier mean ? 3. What does a negative multiplier mean ? Give the law of signs. If. Repeat principle 1. In what respect does it differ from that on p. 181 ? 5. Repeat principle 2. Multiply (orally) : 6. -5X+12. 10. -aX-\-b. U. (a+6)2=( )? 7. +6X-10. 11. -2aX-3 6. 15. {a-by = ( )? 8. -7x-9. 12. -i-5aX-\-7m\ 16. \x-yY={ )? 9. -8X+11. 13. -4:X^Xl2y'^. 17. {x-ij)(x-\-y) = { )? Copy and multiply : 18. —9 (x—ij) by —5 a. 19. 7x-12y by a — b. 20. 24m2 — 7mnby 6 m— lln. 21. ^Q>xy—bxy'^hy%x—bz. 22. x^-\-2 xy-\-y'^ by x-\-y. 23. x^ — ^ x^y-\-^ xy^—'f by x^—y^. 21f. ^xy — 7am-^n^hj6a — 7b. 164. Division. — By principle 3 in division (page 25), the dividend equals the product of the divisor and quotient. ADVANCED ARITHMETIC. 347 Since (1) +7x+3=+21, +21 Since (2) -7x+3 = -21, -21 Since (3) -7x-3 = +21, +21 Since (4) +7x-3=-21, -21 +7 = + 3. -7== +3. -7=^3. +7=-3. Observe that when the signs of the dividend and divisor are alike the quotient is positive, and when the signs of the dividend and divisor are unlike the quotient is negative. Law of Signb in Division : Like signs give plus ; .unlike signs give minus. EXAMPIiES. 1. , 2. 3. U- 6 )24 -6 ) -24 -6)24 6 ) -24 4, result. 4, result. —4, result. —4, result. 5. 6. 7. 8. 4y)24xy —Qx)—24xy_ -\-Qy )—24:xy —4y) 24xy 6x, result. 42/, result. —4 a:, result. -6 a;, re 9. 10. Qy)12xy-Q 2x -y , result. —^xil^ax — lOxy — 3a+22/; or, 11. Divide a:2- -2xy-\-y 2 by x—y. 2y-Sa, result. Process : x-y) x^- 2xy + y%x- -xy -xy+yl -xy+y^ y, result. Explanation : (1) Find how many times the first term of the dividend contains the first term of the divisor. x^ + x = x. Place x in the quotient. Multiply the whole divisor by the quotient x, and subtract the product from the dividend. (2) Proceed just as before. Find how many times the first term of the remaining dividend contains the first term of the divisor. -xy-*-x= -y. Place -yin the quotient. Multiply the whole divisor by -y&nd sub- tract. Result, a: -2/. 348 ADVANCED ARITHMETIC. 12. Divide a;2 — 2/2 by a;+2/- Process : x+y) x^-y^ i^-y-i result. x^-\-xy Note. — When we go to subtract +xy, there is no similar term in the dividend ; so the remainder is - xy, to which we bring down the next EXERCISE CXLII. 1. What is the laiv of signs in division ? Compare with the law of signs in multiplication. Divide (orally) : 2. 25^5. 7. —26a^x-^—6ax. 3. -25^5. • 8. -36a;22/2^9a;2/. ^. 25-^—5. 9. 17 ax^y^axy. 6. —25-^—5. 10. — 12 ax(7n-\-n) ^12 ax. 6. -10ah^-\-ah. 11. -12ax (y-z)^(y-z). Copy and divide — 12. bw?n— 4:7)171^ by mn. 13. lhx'^-2bxy^hybx. U. x^-\-2xy-\-y^ hj x-\-y. 15. x^—2ax-\-a^ hj x—a. 16. x^-^x'^a^^xa^-a^hj x^-2ax^a^. 17. x^-\-^x'^y-\-%xy'^-\-y^hj x-{-y. 1 65 . Factoring. — The factors of numbers may be found by inspection or by division. (See pp. 40 and 41.) EXAMPLES. 1. Factor bnh. Process : 5 a& = 5 x a x 6. Factors, 5, a, and b. ADVANCED ARITHMETIC. 349 2. Factor 25a '-^c (Inspection). Factors, 5, 5, a, a, and c. 3. Factor bh^^{x-\-y). Factors, 5, 6, h, and x-\-y. ^. Factor 5 a6- 10 62. Process : 5 h ) 5 a6 - 10 6^ a - 26. Factors, 5, h, and a-2h. 5. Factor a'^-\-2ah-\-h'^ . Note. — By the principle (p. 181), a^+2ah + h'^ is the square of the sum of two numbers. Process: a^+2ah + h^ = {a + h) {a + h). Factors, a+h and a + 6. 6. Factor a2-2a.6+62. Note. — Apply prin. 1, p. 345. Factors, a-h and a-h. 7. Factor a^ - 62. Note.— Apply prin. 2, p. 345. Factors, a + 6 and a-h. EXERCISE CXLIII. Factor — 1. 12 a. 5. ^^xHf. 9. 12(x-y). 2. 1562. 6. 51 mnl 10. x(7n-\-7i). 3. 2b ab. 7. 64^3^2. 11. 16:c7/(a4-6). If.. 12mP'n. 8. 121 x^y^z. 12. 75a;2(a+6)2. 13. m2+2 ' 71171 -\-n'^. 16. x^ -2a:+l. i^. a;2-2 xz-\-z\ 17. 2;2 -2/2. 15. a^^2i x+1. 18. x^- -%x\j^'^xf-y^. 350 ADVANCED ARITHMETIC. 166. Greatest Common Divisor. — The G. C. D is usually found by factoring. The plan here is not different from that found on page 47. EXAMPLES. 1. Find the G. C. D. of 5ac, 2b ax, lb an. Process : (1) 5ac = 5xaxc. (2) 25ax = 5x5xaxx. (3) 15an = 3x5xaxn. .-. G. C. D. = 5xa = 5a. 2. Find the G. C. D. of ba{x-y), Ibmx-Wmy. Process: (1) 5a{x-y) = 5xax{x-y). (2) l5mx-15iny = Sxbxmx{x-y). .•. G.G.I>. = 5x{x-y) = 5{x-ij); or, bx-5y. 3. Find the G. C. D. of x'^+^xy+y^, x^-y^, 7x-\-7y. Process: (1) x^+2xy + y^ = {x-\-y) {x+y). (2) x^-y^ = {x+y){x-y). (3) 7x+7y = 7{x+y). .'. G. G. 'D.=x+y. > EXERCISE OXLIV. Find the G. C. D. of— 1. a2, 5ax, Say. 3. 24x^y^, Wxy^, 40x^yz^. 2. 7mx, 14am, Sbmy. 4- 12a'^m'^n^, IS am^n,S6a^m^n^. 5. Sax(m—n)j m^—n^, 7m — In. 6. 2bax{y-z), Ibay^ — lbaz^, lOay-lOaz. 7. x^-\-2xy-{-y^, x'^-ij^, 17ax-\-17ay. 8. (x-y)^, x^—y^, bhx—bby. 167. Least Com^mon Multiple.— The L. C. M. is usually found by factoring. The plan here is not different from that found on page 49. ADVANCED ARITHMETIC. 351 EXAMPXiES. 1. Find the L. C. M. of 5 a, 10 ay, l^a^y'K Process: (1) 5a = 5xa. (2) 10a^ = 2x5xax?/. (3) 15aV = 3x5xaxaxi/Xi/. .'. L. 0. M. = 2x3x5xaxax?/x?/ = 30aV- 2. Find the L. C. M. oi6a{x-y), 7ab, ^^a(x-y)^. Process: (1) 5a{x-y) = 5yax{x-y). (2) 7ab = 7xaxb. (3) 35a{x-yf = 5x7xax{x-y){x-y). .-. L. CM. =5x7 X axbx{x -y){x -y) = 36 ab{x - yY ; or f 3ba{x^-2xy-\-y^). EXERCISE CXLV. Find the L. C. M. of— 1. 7a, Uah, 2S a^b. 2. 9?/2, 18x2/, 24a;22/2. S. 2mn^z, Qm^nz, 12mnz^. 4-' ms^, 5 mns, 20 xns. 5. \.bax(m—n), 25, 10a{m—n), 6x{m^—n^), 6. 12, X2-2/S {x-yy,Q{x-y). 7. 86, 12c, 862c, 4:b{m-n). 8. x^ — 2xy-{-y^,x^-\-2xy-\-y'^,x^—y^. B. FRACTIONAL NUMBERS. 168. Reduction. — The reduction of fractions here does not differ in principle from that found in Articles 35 to 39. EXAMPLES. 5 a 1. Reduce ^^ to a fraction whose denominator is 10 6a;. Z Process: (1) 10 5a: -5-2 6 = 5. r. --,. bxxba 25 ax ,^ 352 ADVANCED ARITHMETIC. 4Sax'^y 2. Reduce -^ „ , — - — r- to its lowest terms. Note.— 24a:c is the G. 0. D. of the numerator and the denominator. Process: -^ » . ^ — ,-—^. — = _ , ^ — r, result. 72 a^x{m -n)-^24ax 3a{m- n) 3. Reduce t>x—a to the form of a fraction. xy r, r. ab xy (5x-a)-ab 6 x^y - axy -ah Process: ox -a = — = — ^ ^ .result. xy xy xy Note.— (1) Multiply the integral part by the denominator, (2) add the numerator, and (3) place the sum over the denominator. A. Reduce -, to a mixed number. a — b 252 Process: a-b)a^-2ah-\-Sb^{a-b-\ 7, result. gg -ab «-^ -a6+3 52 -ab +62 +2 6^, remainder. 5. — ,77-, — ^"1- to least common denominator. a oc a^d Note. — L. 0. D. of a, 6 c, and aH is Qahd. Process : (1) 6 a^ca -^ a = 6 aca; - — r— » o .« ^ ' 6 acd Xa 6 a^cd io\ a 2 ^ a 2^ a ^dx2b _2a^d iS) Qa^cd^aH = 6c;p^ = ^^^. ' ^cxa^d Qa\d EXEKCISE CXLVI. Reduce — 1. 5 to a fraction whose denominator is x. 2. — to a fraction whose denominator is 7 amz. VIZ ADVANCED ARITHMETIC. 353 3. r^ n -i to lowest terms. 50 a^my"^ 12xy(m—n) , , , , ^- iQ 2/ o iT to lowest terms. 5. ^ \ ^y^y ^o lowest terms. /^2 5'»2 ^. ; — — to mixed number. x-\-y ^ x^-\-2xy-2y^ ^ . , , 7. — ' r ^^ to mixed number. x+y 8. a 2 to fractional form. c 9. b^v—m-\ — — to fractional form. ^ x^y 10. Ti—1-L to L. C. D. ac ah n. 7, -, ^, I to L. C. D. x bo ,^ 6x 7b 7z J. T r^ T^ 12. -^, -^, —7 r to L. C. D. a^y ay^ ay(m—n) 169. Addition, — Addition of fractions here does not differ in principle from that found in Article 40. EXASIFIiES. ^ *Ti3a: 5x 7 X „ S.X 5x 7x lSx+15x+Ux 47 ,^ Process : -4- +"8 + ^2 = 24 " 24^' result. Note : —x is the same as-^rj-. 24 24 „ a c , c 5a/^+6c+ac ,^ Process: -r-\ — r+^T = ^-^ , result. b ab 5b 5ab ' 354 ADVANCED ARITHMETIC. S, Add 26, H , — X m—n Process: 2h-\ = X m-n 2bmx — 2bnx 7 am -7 an Sbmx xm — xn xm-xn xm-xn 2bmx-2bnx-{-7 am-7 an-Sbmx xm — xn ~ 7 am - 2 bnx -7an-Q bmx xm-xn Note. — Have the pupil explain each step. , result. EXERCISE OXLVII. Add: ^' 6' ~^4' "^ 8* ^' "5"' 6~' "^TO"' ?^ __^ 4-L ff 5a;-52 / 7a;- 3y ^- a;' 2/' a:2/' ^- 8 ' 12 * ^ a;-2/ a;+2/ ^ S^I^; ^ 3a;-6 7. 2 ' 2 • a ' ' d 5 3 x—y Sx+Sy' 5a:+52/' (a^+2/)^* ^ 5a 7a6_8a^ mc om 4 c 170. Subtraction. — Subtraction of fractions here does not differ in principle from that found in Article 41. EXAMPLES. 3* 3/ 2. From -r take -=. 4 5 ^ X X 5x 4:X X ,^ iVocm.— -g=-- 20 = 20- '•"suit. ADVANCED ARITHMETIC. 855 2. Prom -Y take — -~. Process :(1) 'J -[-*-l]=^-^+*l. (Why?) a bx+Sy]_ 6a j=( )? a [ 6a J 6a [ 6a J ^ — = — ^ — -, result. 6a 6a 6a:— 5, , 6a:— 5 4. From take — ; — . x-y x-\-y p _ 6.r-5 Qa:-5 _ 6x^-bx+Qxy-6y 6ae^--5.r-6y + 5y _ X-y x+y x^-y^ x"-y^ ^x^-hx-\-Qxy-hy-{Qx^-bx-Qxy-\-by )_ x^ — y^ Qx^-hx-^Qxy-hy-Qx^-^bx^-Qxy-hy _ l2xy- lQy o o o o — J rGSUlt. ^2 _y2 x^ — y^ EXERCISE GXLVIII. 1. From -=^take-7r-. 5 9 ^ ^ 12y ^ , 51/ 2. From — ^ take ^^^ a; 11a: 7 10 5. From -take y 132/* a* a*'^ 4-. From — take^r"- 2/ 72/ ^ „ a4-^ ^ 1 CL — b 5. From -tt^ take — r^-, 12 lo 6, From — ; — take . x-\-y x — y 856 ADVANCED ARITHMETIC. 7. From 12x 10, , 6a;-5 take 8. a — x-\-2y Sx-\-Qy 36 {7 a- 10 I ^-h^ = = ( )? 171. Multiplication. — Multiplication of fractions here does not differ in principle from that found in Article 42. i:XAMPIiES. 1. Multiply^ by 1^. Process : — x -^ = h~» result. ax Q 2 .r' Note.— Cancel, when possible. a — b m-\-n a-\-b a — b Process : ={ )? — -r X = -—J- , result. a+b a-b a+b 3. "^X b ^-X — = i )? „ a ^ c ^ a Process : t- x - t- x i;- b b Sc 362 , result. EXERCISE CXLIX. Multiply — '' b ^^ Wc 2. ^ by P-. 32/ 12 ly 5a6 , l^xz ^' Uy ^ 10^' i. Sab hy 5m 7 ax' Saxy llb'^y 22 bh ^ 12 ax^' ^ x-\-y , x—y 6. —r^ by ^. x-j-z "^ x—z x{a-^b) a- b _ '' a-b ^ 12(a;+a)- 5a— 66 5m— 3n ^' 3.T+6 7/^ 5?i-36' ADVANCED ARITHMETIC. 357 172. Division. — Division of fractions here does not dif- fer in principle from that found in Article 43. EXAMFIiES. 1. Divide — by — ^. n 5 a Sb 6 a m^ bam ,, Process : — -^ — « = — x irr = -ttti result. m m' m Sh 3 6 2. Divide — I — by — ; — . m-\-n '' x-\-y _, m-n m-n m-n x+y x-\-y .^ Process : — -- + — ; — = — -- x — = — -^, result. m+n x+y m+n m-n m+n 3. Divide -. — by 56. 4 m ^ Process : -. — + 56= -. — x -— = , result. 4 m 4971 56 206m EXERCISE CL. Divide — 1x, 92 ^ 82-4, 42-2 -5-^^ To- ^- -y^^^^- 252 •^3522- m+?i •^2m+2ri 1422 ^ 52a; 12-9x 21 ay -" ^my 4 ^ 5 a6c 20 a6c2 ' 16-12 a; ^ ^ 7xyz ^ 12a;2m x-\-y II. STUDY OF PROBLEMS. A. PROBLEMS OF ONE BASIS. 173. Solving Equations of One Unknown Number. — We have learned that various operations may be performed on an equation without destroying the equality : (1) The terms may he transposed. (See p. 81.) 358 ADVANCED ARITHMETIC. EXAMPLES. l.ox—lb=4:X—b. Find the value of ic. Process : (1) 5 .r - 15 = 4 c - 5. Transposing, (2) 5x-4x = 16-b. (2) = (3) .T = 10, result. (2) An equation may he multiplied. ( See p. 83.) 2. - 4 X X - =7. Find the value of x. Process: (1) — = 7. 4 5 (1) = (2) S^zif = 7. 20x(2) = (3) 5x-4.r = 140. (3) = (4) a: = 140, result. Principles : 1. Any equation may he cleared of fractions hy he- ing multiplied hy the L. C. M. of the denominators of all the frac- tions. S, 1 'L^ni=is. Find the value of x. 8 7 Process : (1) — - ^^^-^ = 13. 8 7 56x(l) = (2) 49 ^-(48 .1+32) = 727. (2) = (3) 49 .1-48.*: -32 = 727. Transposing, (4) 49.r-48.r = 727 + 32. Collecting, (5) .r = 759, result. 2. The signs of all the terms of an equation may he changed hy multiplying the equation hy —1. 4. 5.c+3=6.c-9. Find the value of a;. Process: (1) 5a:+3 = 6a;-9. Trans., (2) 5a:-6:r= -9-3. Col., (4) -.r=-12. -lx(4)-(5) .r = 12, result. , Note. — In practice, we simply rewrite the equation with the signs changed. ADVANCED ARITHMETIC. 359 (3) An equation may be divided. (See p. 89.) 5. 35a;-15=18x+70. Find the value of x. Process: (1) 35a;- 15 = 18 a;+70. Trans., (2) 35a:- 18:?: = 70 +15. Col., (3) 17 a: = 85. (3) + 17 = (4) a; = 5, result. Plan: In finding the value of the unknown number in an equation, it is best to follow a definite plan. The operations required should usually be performed in the following order : (1) Clear of fractions. (2) Clear of parentheses. (8) Transpose all terms containing the unknown number to the left member of the equation, and all other terms to the right member. (4) Collect the terms of each member into one term. (5) Change the signs.* (6) Divide by the coefficient of the unknown number. Note. — It is not often that all these operations will be required in solving one equation, but it is usually best to perform those that are re- quired in the order indicated. EXERCISE OLI. 1. What is an equation f 2. What principles are involved in the transposition of the terms of an equation (page 81)? Give the law of transposition (page 82). . 3. What is the rule for removing parentheses (page 341) ? 4- Give principle for clearing of fractions. 5. Upon what principle may the signs of all the terms of an equation be changed ? 'The (5) may be dispensed with, by dlTiding by the negative coefficient in the (6). 860 ADVANCED ARITHMETIC. Find the value of the unknown number in — 7. 25-5a;=0. ^' 5 "^ 6 "^ 3 * 8. 72/+12-482/. ^_ 5^_§^-iii ^. 5a;-18-ja;-38=6. 6 11 ~ 11. 5 71+ 164-871= 10 n. 7 4 ^ ^g. _^+i,=51-68. 7^+80-^-^-205 i5. 24a-12i-6ia=5i. ^-'- ' ^+*^"- 4 -'^^»- -1-1=8- .,.8i^-%^=_74. m m_ 5 m+ 5 6m-7_9 7?i-5i ^^' 4+8 -'^^- ^^- ~4 3~ -^6~" 174. Problems. EXAMPLES. i. J of a number, \ of the number, and 10, are together equal to the number. Find the number. Solution : Let .r= the number.* Then, (1) lx + lx+10 = x. (Why?) 12x(l) = (2) 4a;+3a;+120 = 12a;. (2) = (3) 12a;=120+4a;+3a;. Trans., (4) 12x-4x-Sx = 120. Ool., (5) 5x = 120. ^of(5) = (6) « = 24, answer. 2. Find a number whose J exceeds its i by 150. Solution : Let a: = the number. Then, (1) ^.r-i.r = 150. (Why?) 12x(l) = (2) 4;r-3.r = 1800. Col., (3) a:= 1800, answer. * "Let x=the number," means that x Is to be used to represent the number. The sign = is used because it is more convenient than the word represent. ADVANCED ARITHMETIC. 861 3. The sum of two numbers is 25, and their difference is 5. Find the numbers. Solution : Let x= the smaller number. Then, 25 - a; = the larger number and (1) 25-a;-a: = 5. (Why?) Trans., (2) -a;-a; = 5-25. Ool., (3) -2a;=-20. (3)^- -2 = (4) .T = 10, smaller number. (5) 25- a; = 15, larger number. j^. What four successive integers are together equal to 46 ? Solution : Let x= 1st number. Then, a: + l = 2d number, :r+2 = 3d number, a;+3 = 4th number, and (1) :i:+(:r+l)+0r+2) + CT+3) = 46. (Why?) (1) = (2) x+x+l+x^-2+x+^ = 4Q. Trans., (3) a:+a;+a;+:r = 46-l-2-3. Col., (4) 4 a: = 40. \ of (4) = (5) a; = 10, Ist number. Then, the other numbers are 11, 12, and 13, 5. Henry spent yV of his money, and then received $65 ; he then lost -f of all his money, and had in hand $10 less than at first. • How much had he at first ? Solution : Let a; = his money at first. Then, ^^x = v,TnX,. left, T\a;+65 = amt. after receiving $65, and (1) i(3»i^+65) = :r-10. (Why?) (1) = (2) iiX+^ = x-lO. 44X(2) = (3) 9:c + 715 = 44^-440. Trans., (4) ^x-Ux= -440-715. Col., (5) -35a: =-1155. (5)-t--35 = (6) :k = 33. Answer, $33. 6. Goods sold at 10% gain ; if the goods had cost $120 more, the loss would have been 10%. Find the cost price. Note.— The selling price is 110^ or {^ of the real cost, and W or ^^y of the supposed cost. ADVANCED ARITHMETIC. Solution : Let x = real cost. Then, .r+ 120 = supposed cost and (1) H:t' = xU^+120). (Why?) 10x(l) = (2) ll.r = 9a:+1080. Trans., (3) 11 a: - 9 a; = 1080. Col., (4) 2 a: =1080. ^of (4) = (5) a; = 540. Answer, $540. 7. By selling my watch for $36, I lose | of its cost. What did it cost ? Equation: a; -fa: = 36. (Why?) 8. A and B have equal sums of money ; A gains $100 and B loses $150; then, twice A's money is equal to three times B's. What sum had each at first ? Equation: 2 (a: +100) = 3 (a: -150). (Why?) P. In a certain weight of gunpowder the saltpetre is 6 lb. more than -J of the whole, the sulphur is 5 lb. less than i of the whole, and the charcoal is 3 lb. less than i of the whole. Find the entire weight and the weight of each part. ^g'wah'on : Let a: = whole weight. Then, (1) a-x+Q) + {lx-b) + {ix-S) = x. (Why?) 10. 6 times the ratio of ^ to j is equal to how many times the ratio of 3 to 9 ? Solution : Let a: = required number. Then, (1) 6x(^:|) = a;x(3:9). (Why?) (1) = (2) 6xf = .rx^. (2) = (3)ir=|. 3 X (3) = (4) a: = 8, answer. 11. A and B can perform a piece of work in 10 days. They both work three days and B then finishes it in 12 days. How long will it take each to do the work ? ADVANCED ARITHMETIC. 863 Equation : Let :c = time reqd. by A to do the work. Then, — = part done by A in 1 da. , iV = part done by both in 1 da. , f% = part done by both in 3 da. , -^ — = part done by B in 1 da., 12 12 — = part done by B in 12 da. , 10 a: ^ and(l)^+i?-i? = l. (Why?) Note.— The question is sometimes asked, *' Where did you get the i.^'* The concrete form of the equation will explain : 3 12 12 — of work +— of work of work = 1 x the work. 10 10 X 12. If I subtract $20 from ^ of my money, multiply the dif- ference by 7, and subtract this product from $1800, the result will be equal to my money. How much have I ? Equation: 1300-7(ix-20) = :c. (Why?) Note. — Always require the pupil to explain his reasons for forming the equation ; because there is where the pupil does his investigative thinking. Solving the equation is mechanical. EXERCISE CLII. i. "i, ^ and i of a number exceed the number by 14: find the number. 2. A number decreased by its \ and f equals 54 : find the number. 3. A number increased by its ^ and -^ equals 171 : find the number. J,.. I of a number is 80 greater than its ^'. find the number. 5. 5 times a certain number is 27 more than its i : find the number. 6. The sum of 8 successive numbers is 72: find the numbers. 364 ADVANCED ARITHMETIC. 7. The sum of 5 numbers each differing from the next smaller by 4 is 140: find the numbers. 8. The sum of 3 numbers each 3 times the next smaller is 39: find the numbers. 9. A and B had the same amount of money ; A lost $40 and . B gained $60; i of A's money was then equal to | of B's : what did each have at the start ? 10. A and B had the same amount of money ; A gave B $10; B then had 5 times as much as A: what sum had each at first ? as much as A : what sum had each at first ? 11. In the composition of a quantity of gunpowder, the nitre was 10 lb. more than f of the whole, the sulphur was 4^ lb. less than J of the whole, and the charcoal was 2 lb. less than \ of the whole : what was the amount of powder ? 12. A lady spent at one store $1 more than ^ of her money; at another $^ more than ^ of what was left; at a third, $1 more than -J of what was left, and then had $1 : what did she have at first ? 13. Several detachments of artillery divided a certain num- ber of cannon-balls : the first took 72 balls and -J- of the re- mainder; the second took 144 balls and J of the remainder; the third took 216 balls and ^ of the remainder, and so on. The balls were thus divided equally among the detachments : how many balls and how many detachments ? IJ^. Find a number which, when multiplied by 5, 24 taken from the product, the remainder divided by 3, and 28 taken from the quotient, will give the same number. 15. A and B have the same annual income ; A saves J of his, but B spends annually $400 more than A ; at the end of 4 years B finds himself $1100 in debt: what is the income of each ? 16. What number is as much more than f as f is less than | ? 17. Five times | of what number is 4 less than 8 times \ of it? ADVANCED ARITHMETIC. 365 18. Seven times the ratio of what number to If is equal to that number added to ^ of the ratio of 9 to 4 ? 19. Divide the number 54 into two parts such that f of the smaller increased by 12 will equal f of the larger diminished by 3. 20. The sum of two numbers is 160; | of their difference is 24 : what are the numbers ? 21. The greater of two numbers divided by their difference equals 21 ; their sum is 164: what are the numbers ? 22. One-half of the ratio of 4 to 6 is equal to ^ of the ratio of 5 to what number ? 23. Twelve times the ratio of ^ to f is equal to i of the ratio of what number to 16 ? 2Jf.. How many times the ratio of ^ to | is equal to 3 times the ratio of 1 to 3 ? 25. A can do a piece of work in 10 days ; A and B can do it in 6 days. In what time can B do it alone ? 26. A and B can do a piece of work in 16 days ; after work- ing 4 days, A leaves B to finish the work, which he does in 36 days. In what time can each do it alone ? B. PROBLEMS OF TWO BASES. 175. Solving Equations of Two Unknown Numliers, — When there are two unknown numbers to be found, there must be two independent equations to be used in finding the unknown numbers. Note.— Two equations are independent of each other, when one can- not be obtained from the other alone. Thus, .^+2/ = 10 and are independent ; but oc+y = 10 BLud Sx+dy = SO, are dependent. Two independent equations, relating the same two unknown ADVANCED ARITHMETIC. numbers, may be so combined as to form one new equation, containing but one unknown number. This process is called Elimination. There are three plans of elimination : ( 1 ) by substitution, (2) by comparison, and (3) by addition and sub- traction. I. By Substitution. EXAMPIiES. 1. If x-\-y = 10 and y=4:, find the value of x. Process: (1) x+y = 10. (2) y = 4. Putting 4 for the y in{l),we have — - (3).'r+4 = 10. Trans., (4) a: =10-4 = 6, result. 2. 2x—y = l% and 5=i. Find the values of x and y. Process: (1) 2x-y = 13. (2)| = i 9 X (2) = (3) 2/ = I = 3, one result. (1) = (4) 2:t'-3=13. Trans., (5) 2a: = 13+3 = 16. ^ of (5) = (6) x = 8, the other result. S. x—7y= —38 and ^=B. Find the values of x and y. Process : (1) x-7y= -SS. (2) ^ = 3. 3x4 (3) 2/ = — ^ = 6, one result, (1) = (4) a:-42=-38. Trans., (5) a: = 42 -38 = 4, the other result. 4. x-\-y=lb, and x—y—b. Find the value of x and y. Process: (1) x+y = lb. (2) x-y = 5. Trans., (S) x = 5 + y. (1) = (4) (5+2/) + 2/ = 15; or, ADVANCED ARITHMETIC. 867 (5) b + y+y=i5. Trans, and col. , (6) 2 ?/ = 15 - 5 = 10. (7) y = 5, (3) = (8) a; = 5+5 = 10. 5. 3a:— 42/=10 and 5a;+62/=42. Find the values of a; and 2/. Process: (1) Sx-^y = 10. (2) bx+Qy = 42. Trans, in (1), (3) Sx = 10+iy. iof(3) = = (4).:= -3 ^ (2) = = (5) 5(1^;^^) +6: i/ = 42. 3X(5): = (6) 5(10+4 ^) + 18?/ = 126. (6)^ = (7) 50+202/+18?/ = 126. Trans. , (8) 20i/ + 182/ = 126- -50 = 76. Ool., , (9) 38^ = 76. (10) 2/ = 2. (4) = (11) . = ^'.^' = 6. 3^_i_7 7x— 6 6. -^ — —^ =5 and ^ — r-^=2. Find the values of x and y. Sy-4: 52/+3 Process: (1) ^^ 1=5. 32/-4 7:r-6_ ^^^5i/+3-2- (3 2/-4)x(l) = (3) 3.r + 7 = 152/-20. (5 2/ +3) X (2) = (4) 7:r-6 = 10*/+6. Trans, in (3), (5) dx-15y= -27. Trans, in (4), (6) 7 a; = 10 2/ + 12. (5) = (8)3p±i?]-15.= -27. 7x(8) = (9) 3(10?/ + 12)-105?/=-189. (9) = (10) S0y+SQ-10by= -189. Trans, and col., (11) -75 y= -225. (12) 2/ = 8. (7) = (13) x = ^^ = 6. ADVANCED ARITHMETIC. EXERCISE CLIII. Find the values of the unknown numbers :, +2/= 22. 8. 3. 3a:-22/ = 10. 25 a: y—x 8a:=9. = 15. '\x=y. 6. 8 9a: = -7. = 5. I 5 y- ' 4:X—y 7.< Sx-\-iy ^' ^x-4y^^ x-y ix-S-\-7y=71. i x — 1 II. By Addition and Subtraction. EXAMPIiES. 1. x+y=25 and x—y=7. Find the values of x and y. Process: (1) x-\-y = 25. (2) x-y = 7. (l)4-(2) = (3) 2 a: = 32. (4) a: = 16. (1) = (5) 16 + 2/ = 25. (6) 2/ = 25-16 = 9. 2. 5a:— 72/=18 and 4a:+32/=23. Find the values of a: and 3/. Process: (1) 6x-7y = 18. (2) 4:X+Sy = 2S. 3x(l) = (3) 15x-21y = b4. 7x(2) = (4) 28.r+21?/ = 161. (3)+(4) = (5) 43.T = 215. (6) x = 5. (2) = (7) 20+32/ = 23. Trans., (8) 3?/ = 23-20 = 3. (9)2/ = l. Plan : (1) Multiply the equations by such numbers as will make the ADVANCED ARITHMETIC. 869 coefficient of the x's or the y's equal. (2) If the signs be unlike, add; if alike, subtract. Questions: Why did I nj^ltiply by 3 and 7 in No. (2)? Could I have made the coefficients of the x's equal by multiplying by 4 and 5 ? Can the equations be solved in that way ? Try it. 3. 5 a;— 3 2/ =37 and 6 a;= 11 2/. Find the values of a; and y. Process: (1) bx-3y = 37. {2)Qx = ny. Trans., (3) Qx-lly = 0. 5x(3) = (4) 30x-5by = 0. 6x(l) = (5) 30x-lSy = 222. (4)-(5) = (6) -37^= -222. (7) y = Q. (2) = (8) 6:r = 66. (9)a; = ll. > 4' T= 5 and ^ — ,-^=2. Find the values of x and y. Process : (1) ^r -. =5. 3y-4i ■ (2) ^^^^ = 2. (3y-4)x(l) = (3) dx+7 = 15y-20, (5i/+3)x(2)=(4) 7x-6 = 10y-\-Q. \ Trans, in (3), (5) 3x-15y= - 27. [' Trans, in (4). (6) 7 a; -10 2/ = 12. f 2x(5) = (7) 6a:-302/=-54. I 3x(6) = (8) 21x-S0y = SQ. J (7)-(8) = (9) -15x=-90. [ (10) x = Q. (5).= (11) 18-15?/= -27. i (12) -15?/= -18-27= -45. [ (13) 2/ = 3. \ t EXERCISE CLIV. I Find the values of x and y : ! . {x-\-y = 29. ' \Sx-\-y=22. 370 ADVANCED ARITHMETIC 1. 60. 5. . . . 8£+5_ . 4 8 ■ {x+hy 2y+x ^8+8-'- |bx-2/=2. III. By Comparison. EXAMPLES. 1. x-\-y=12 and x—y=Q. Find the values of x and 2/. Process: (1) x+y = 12. (2) a;-t/ = 6. Transpose in (1), (3) x = 12-y. Transpose in (2), (4) x = Q-\-y. (5) 6+2/ = 12-?/. (Why?) Transpose, (6) y-\-y = 12-Q = 6. (7) 22/ = 6. (8) y = S. (4) = (9) a; = 6+3 = 9. 2. ^x-\-2y = 17 and 4:X-{-y=lQ. Find the values of a: and ?/. Process; (1) 3:r+2^ = 17. {2)4x+y = 16. Trans, in (1), (3) 2y = 17-Sx. i of (3) = (4) 2/ = ^-^. Trans, in (2), (b) y = lQ-4x. (6) 16-4:r=^^^^'. (Why?) 2x(6) = (7) 32-8:r = 17-3a:. Trans. , (8) 3 a; - 8 a; = 17 - 32. Col., (9) -5a;=-15. (10) x = 3. (5) = (11) ^ = 16-12 = 4. Plan: Solve each equation for y in terms of a*; then, make those ADVANCED ARITHMETIC. 871 values of y equal. This gives equation (6), containing one unknown number, x. Questions : Could I have solved for values of x in terms of ^ ? Would this be correct ? Try it. 3. -^-{—T-=Ssindx—z=—S. Find the values of a; and z. Process: (1) 4+4 = 8- (2) x-z=-S. Trans, in (1), (3)-|=8-^. 3x(3) = (4) ^ = 24 -?i?. 4 Trans, in (2), (5) x = z-B. (6) 2-3 = 24-^. (Why?) 4x(6) = (7) 4^-12 = 96-3^. Trans., (8) 4 2; +3 2: = 96 + 12. Col., (9) 7z = 108. (10) z = 15}. (5) = (11) a: = 15f-3 = 12|. 4. 5x—Sy=S4siud5y—Sx=2. Find the values of a; and ?/. Process : (1) 5x-3y = M. (2) 5i/-3.c = 2. Trans, in (1), (3) bx = S4+Sy. Trans, in (2), (5) -Sx = 2-by. Changing signs, (6) 3x = by-2. (8)^ 2_S4 + Sy 5 • ^ 15 X (8) = (9) 252/- 10 = 102+9 y. Trans., (10) 25 2/- 9y = = 102 + 10. Col., (11) lQy = (12) y = 7. 112. (7) = = (13) x = ^ -2 = 11. (Why?) 872 ADVANCED ARITHMETIC. 5.^ + ^=4. '- X y X ■ r-^ Find the \ Process . -^H = 4. <- 1^ y = 8|. Trans, in (1) <- = 4- 3 'y' ^of (3): =<^)^ 4 '3 1 y' Trans, in (2) iof (5): 27yx{7) Trans, and col. <- = (6) \- X = (8)36 , (9) 10 26 = 3 26 ''21 _l^ y y- y= y 27 = 26y+3. 30. 1 (4) = (10) y-- (11)^ = 3. 4 3 -h^- (12) X- = 1. Note. — Instead of expressing the values of x in terms of y, I expressed »;^alues of - in terms of y. This is the more convenient plan. EXERCISE CLV. Find the values of x and y. x-\-Sy=b9. '■U: = 1 18. „ j4x+2 7/=15. '^'\Sx-4y=-S. ^' \2x-{-y: 4- x^y 12 X y 6. 7. X y 1+1-1 X y~lb' x—y_Sx—Q x—2y=Q. nx_ 6y-22 5 "" 4 • Bx-4:y =29. ADVANCED ARITHMETIC. 878 176. Problems. EXAXPIiES. 1. The sum of two numbers is 50, their difference is 30: find the numbers. Solution : Let a: = one number and 2/ = the other number. Then, (1) x-\-y = m and (2) x-y = 30. (l)+(2) = (3) 2:r = 80. (4) a: = 40. (1) = (5) i0+y = 50. (6) 2/ = 50^40 = 10. Answer, 40 and 10. 2. A and B together have $1500; A has twice as much as B. How much has each ? Solution : Let x = A's money and 2/ = B's money. Then, (1) x+y = 1500 and (2) x = 2y. (1) = (3) 2y+y = l5O0. (4) 3?/ = 1500. (5) 2/ = 500. (2) = (6) a; = 1000. Answer: A, $1000; and B, $500. 3. If I of the time past noon is equal to f of the time to midnight, what is the hour ? Note. — The time past noon plus the time to midnight is 12 hours. n. hr. m. past noon. to midnight. Solution : Let a: = time past noon and y = time to midnight. Then, (1) x+y = 12 and (2) |a: = f y. |x(2) = (3)a:=ii/. (i) = (4) hy+y = i2. (5) I?/ = 12. (6) y = 9. v3) = (7) .r = 3. As the time past noon is the hour, the answer is 3 p. m. 374 ADVANCED ARITHMETIC. If. Divide 12 into two such parts that 3 times the one and 5 times the other shall be 46. Solution : Let x = the larger part and 2/ = the smaller part. Then, (1) x+y = 12 and (2) 3x-\-5y = 4Q. 3x(l) = (3) 3x+Sy = 3Q. (2) -(3) = (4) 2 2/ = 10. (5) y = 5. (1) = (6) a:+5 = 12. (7) a: = 12-5 = 7. Answer, 7 and 5. 5. A and B had $30 each ; after B paid A a certain debt, A then had twice as much as B. How much has each ? Solution : Let x = A's money at close and 2/ = B's money at close. Then, (1) a; +2/ = 60 and (2) x = 2y. (1) = (3) 2 2/4-2/ = 60. (4) 3 2/ = 60. (5) 2/ = 20. (1) = (6) a: + 20 = 60. (7) a: = 40. Answer, $40 and $20. 6. A certain number is composed of two digits ; the sum of the units and tens is 6; and if 3 times the digit in tens' place be subtracted from 4 times that in units' place, the remainder will he minus 4. Find the number. Solution : Let :r = the units' figure and 2/ = the t^n=!' figure. Then (1) x-hy=^Q and (2) 4a'-3^v= -4. 3x(l) = (3) 3.r+3?/ = 18. (2) + {3) = (4) 7.r = 14. (5) .r = 2. (6) // = 6-2 = 4. Number, 42. ADVANCED ARITHMETIC. 875 7. Thomas Reed bought 6% mining stock at 114^%, and 4% furnace stock at 112%, brokerage i% in each case; the latter cost him $480 more than the former, but yields the same annual income. What did each cost ? Note. — The cost is the market value plus the brokerage. Then, the cost of the furnace stock is 112 J^, or f of furnace stock ; and the cost of the mining stock is 115^, or |f of mining stock. Solution : Let a: = par value of furnace stock and y = par value of the mining stock. Then (1) ^x-Uy = ^SO (why?) 100x(2) = (3) Qy = ix. 40x(l) = (4) 45 .r- 46 2/ = 17200. 2x(4) = (5) 90 rr- 92 2/ = 34400. Trans, in (3), (6) -4x+Qy = 0. 22Jx(6) = (7) -Q0x + lS5y = 0. (5)+(7) = (8) 43 2/ = 34400. (9) y = SOO. (10) M2/ = 920. (3) = (11) 4 a: = 4800. (12) a; =1200. (13) ^=1350. Answer, $1350 and $920. 8. 5 apples and 4 oranges cost 22/, 8 apples and 7 oranges cost 27/. Find the cost of 1 apj)le and 1 orange. Solution : Let x = cost of 1 apple and y = cost of 1 orange. Then, (1) 5x+iy = 22 and (2) 3 07+ 7 2/ = 27. 3x(l) = (3) 15x + 12y = m. 5x(2) = (4) 15 r +351/ = 135. (4)-(3) = (5) 23y = Q9. (6) y = d. (2) = (7) 3a:+21 = 27. (8) 3x = Q. (9) a: = 2. Answer, 2f and Sf. 876 ADVANCED ARITHMETIC. 9. A and B can do a piece of work in 16 days ; after working 4 days, A leaves B to finish the work, which he does in 86 days. In what time can each do it alone ? Solution : Let a: = time reqd, by A to do the work and y = time reqd. by B to do the work. Then, — = part A can do in 1 da., — = part B can do in 1 da., and (2) 1+^+^ = 1. (Why?) 4x(2) = (3)i5+— = 4. X y 144 (3)-(l) = (4) i= = 3. (5) 3j/ = 144. (6) J/ = 48. <2) = (7) 1+1=1. (9) 44. (10) ^ = 24. Answer, 24 da. and 48 da. 10. Divide $600 into two parts, having the ratio of ^ to \. Solution : Let x = larger part and 2/ = smaller part. Then, (1) x+y-=600 and {2) x:y::l:l. (2) = (3)^=|. 2/X(3) = (4)^=?|. Trans, in (1), (5) x = Q0O-y. {Q)^ = QOO-y. (Why?) ADVANCED ARITHMETIC. 377 (7) Qy = 2400-4y. (8) 61/ +4 2/ = 2400. (9) 10i/ = 2400. (10) y = 240. (4) = (ll)a:=^^ = 360. Answer, $360 and $240. 11. At what time between 4 and 5 o'clock are the hour-hand and minute-hand of a clock together ? Note. — At 4 o'clock the minute-hand is at XII and the hour-hand is at nil. Then, when the minute-hand gains 20 minute spaces on the hour- hand they will be together. Solution: Let .r = the minute-hand's travel and y = the hour-hand's travel. Then, (1) sc-y = 20 and (2) x = 12y. (Why?) Trans, in (1), (3) x = 20+y. (4) 12y = 20 + y. (Why?) (5) 12y-y = 20. (6) 111/ = 20. (7) 2/ = !?. (8) . = ^ = 21^, Answer, 21 r\ min. after 4 o'clock. 12. At what time between 4 and 5 o'clock are the hour-hand and the minute-hand equally distant from V ? Note.— There are two answers to this problem: (1) When the two hands are together, the conditions are the same as in Example 4; (2) when the hands are on opposite sides of V. We will solve for the second answer. Draw a clock-face and verify this statement : When the hands reach the required places, the minute-hand will lack as far of being to VI as the hour-hand is past IIII. Solution : Let x = min. -hand's travel (4) 13 y = 30. and y = hour-hand's travel. ,_> _ 30 Then, (1) a: = 30 -2/ (statement ^^^ ^ ~ l3' above) _ /^n ^ _ 12x30 _ _ , And (2) x = 12y. (Why?) {^)-{G) x- ^^ -zt^^. (3) 12^ = 30-?/. (Why ?) Answ^er, 27tV min. after 4 o'clock. 878 ADVANCED ARITHMETIC. 13. The head of a fish weighs 9 pounds, the tail weighs as much as the head and half the body, and the body weighs as much as the head and tail together. Find the weight of the fish. Solution : Let x = weight of body and 2/ = weight of tail. Then, (1) 2/ = 9 + ia: and (2) x = y-]-9. Trans, in (2) (3) y = x-9. {4)x-d = 9 + lx. (Why?) Trans., (5) a:-ia: = 9+9 = 18. (6) :r = 36. ' (7) 2/ = 36-9 = 27. Answer = 36 lb. +27 lb. +9 lb. = 72 lb. 14' A and B rent a field for $27 ; A puts in 4 horses for 5 months, and B puts in 10 cows for 6 months : what ought each to pay if 2 horses eat as much as 8 cows ? Solution : (1) Pasture for 2 h. = pasture for 3 c. 2 X (1) = (2) Pasture for 4 h. = pasture for 6 c. .'. A's 4 horses are equivalent to 6 cows. (3) Pasture of 6 c. for 5 mo, = pasture of 30 c. for 1 mo. (4) Pasture of 10 c. for 6 mo. = pasture of 60 c. for 1 mo. .'. they should pay in the ratio of 30 to 60. Let x = A's part and 2/ = B's part. Then, (5) x+y = 27 and (6) x:y::30:Q0. m = i7)j = U = h (8) 2x = y. (5) = (9) a:+2rr = 27. (10) x = 9. (11) 2/ = 18. Answer, $9 and $18. 15. A fox is 70 leaps in advance of a hound ; the fox takes 8 leaps to the hound's 6, but 2 of the hound's leaps equal 5 of ADVANCED ARITHMETIC. 379 the fox's: how many leaps must the hound take to catch the fox? Solution: (1) 5f.L = 2h.l. iof(l) = (2) lf.l. = |h.l. 8x(2) = (3) 8f.l. = 3ih.l. 70x(2) = (4) 70f.l.=28h.l. Note.— Re-writing the problem, putting for fox-leaps their equiva- lents found in (3) and (4), the leaps will all be of the same length. We have, **A fox is 28 leaps ahead of a hound, and takes 3^ leaps while the hound takes 6. How many leaps will the hound make in overtaking the fox, the leaps being all of the same length ? " 8v Let a: = No. of leaps for the (8) ^= j^g- ^ ^^''^^ f .y. Trans.in(5),(9):r = i/-28. and 2/ = No. of leaps for the o hound. (10) y-28 = ^. Then, (5) y-x = 2S 15x(10) = (ll) lby-420 = Sy. and {Q)x:y::Sl:6. Trans., (12) Iby-Sy = i20. (6) = (7) ^ = U = j\- (13) ?/ = 60. ^ Answer, 60 leaps. Note. — If fox-leaps had been required in the answer, the hound-leaps should have been reduced to fox-leaps. 16. A banker owns 2|% stocks bought at 10% below par, and 8% stocks bought at 15% below par. The income from the former is 66f % more than from the latter, and the investment in the latter is $11400 less than in the former. Find the invest- ment in each. Solution : Let x = par value of 1st (3) -5- 2| = (5) x = 2y. stocks. (6) 180 2/ -85?/ = 1140000. and y = par value of 2d (Why ?) stocks. (7) 951/ = 1140000. (Why?) ^^^100"^^^- 100 100 ^^^j^y^^ (11) ?^ = 21600. 100x(l) = (3) 2lx = 6y. 100 X (2) = (4) 90 X - 85 // = 1140000. 100 Answer, $10200 and $21600. 380 ADVANCED ARITHMETIC. 17. My agent sold my flour at 4% commission; increasing the proceeds by $4.20, I ordered the purchase of wheat at 2% commission ; after which, wheat declining 8^%, my whole loss was $5. What was the selling price of the flour ? Solution : Let a: = selling price of flour and y = cost price of wheat. 96 X Then, —^= proceeds from flour, QQX r— +4.2 = investment in wheat, and(2) ??|| = :r + 4.2-5. (Why?) 100x(l) = (3) 102?/ = 96.r + 420. 3(X)x(2) = (4) 290?/ = 300 a- +1260 -1500. (5) 29i/ = 30.r-24. Jof (3) = (6) 342/ = 32a:+140. 16x(5) = (7) 464 2/ = 480^-384. 15x(6) = (8) 510i/ = 480.r+2100. (8) -(7) = (9) 46 2/ = 2484. (10) 2/ = 54. Trans, in (2), (11) x = ^-^+5-4.2. (12)ar = ^5|^+5-4.2 = $53. 18. Suppose 10% state stock 20% better in market than 4% railroad stock. If A's income be $500 from each, how much has he paid for each, the whole investment bringing 6^f3% ? Note. — The solution to this problem is given in four steps : (1) We ob- tain the par value of the state stock ; (2) the par value of the railroad stock ; (3) the amount of the investment ; and (4) with the results ob- tained from the first three steps we are enabled to obtain the bases for the last step, which, when solved, gives required results. Step 1 : (1) W s. s. = $500, income on s. s. (Basis.) 10x(l) = (2) 100^ s. s. = $5000, par value of s. s. Step 2 : (1) 4% r. r. s. =$500, income on r. r. s. (Basis.) 25x(l) = (2) 100^ r.r. s. = $12500, par value of r. r. s. ADVANCED ARITHMETIC. 881 Step 3 : (1) .Qsh[% inv. = $1000, whole income. (Basis.) i^\% of (1) = (2) 1% inv. = $166.50. 100 X (2) = (3) 100^ inv. =$16650. Note.— There are 50 shares state stock and 125 shares r. r. stock. Step 4 : Let a: = market value of 1 share s. s. and ^ = market value of 1 share r. r. s. Then, (1) 50^ = ^^x50?/ (why ?) and (2) 50 a: + 1251/ = 16650. Trans, in (1), (3) 5Ox-Q0y = 0. (2) -(3) = (4) 185^ = 16650. (5) y = 90. (6) 5 a: = 540. (7) a; = 108. State stock: 50 sh. at 108^ = $5400, answer. R. R. stock: 125 sh. at 90?^ = $11250, answer. EXERCISE GLVI. 1. Divide $500 between A and B, so that A may have $40 more than B. 2. Divide $420 between Charles and Henry, so that Henry will have $25 less than Charles. 3. Two men owe a debt of $54 in the ratio of 4 to 5 : what does each owe ? 4. Two men owe a debt; for every dollar that A owes, B owes $1.50. If the debt is $250, find each man's part. 5. Divide 20 into two such parts that the smaller plus 5 shall equal the larger minus 5. 6. A and B have $500; A says to B, " Give me $50 and then I will have as much as you will have." How much has each ? 7. James and John bought a v/atermelon for $.85 ; they di- vided it so that John got 2| times as much as James : what should each pay ? 8. i of the time past noon is equal to ^ of the time to mid- night : what is the time ? OH^'^^' 382 ADVANCED ARITHMETIC. 9. i of the time past noon plus 8^ hours is equal to | of the time to midnight. Find the hour. 10. -J of the time past midnight plus 2 hours is equal to ^ of the time to noon minus 1^ hours : find the time past midnight. 11. All the time past midnight minus 2 hours is equal to \ of the time to noon. What is the time ? 12. A number is composed of 2 digits ; the sum of the digits is 5, and the number of units is f of the number of tens. Find the number. 13. A number is composed of 2 digits ; the sum of the digits is 12 ; 5 times the number of units is equal to 2^ times the num- ber of tens. Find the number. IJf.. A number has two places ; the sum of the units and tens is 4; if 86 be subtracted, the order will be reversed. Find the number. Note. — Any number expressed by the French system of notation is equal to the sum of its 1st ( units' ) digit, 10 times its 2d ( tens' ) digit, 100 times its 3d (hundreds') digit, and so on. Convince yourself of this truth by applymg it to several numbers. 15. A and B are partners ; A invests a capital of $500, and B a capital of $850; the gain is $540: divide it. 16. A and B are partners ; they invested capital in the ratio of 5 to 8 ; they have gained $650 : divide it. 17. Divide i in the ratio of 6 to 7. 18. Divide .06 in the ratio of ^ to i. 19. Divide 10.6 in the ratio of .1 and .05. 20. A and B are partners ; A puts in $660 for 10 months, and B has in $500 for the first 6 months and only $100 for the other 4 months ; they gain $600 : divide it. 21. A and B have a joint stock of $2000 by which they gain $640, of which A receives $128 more than B: what is each man's share of the stock ? 22. How many minutes does it lack of 4 o'clock, if | of an hour ago it was twice as many minutes past 2 o'clock ? ADVANCED ARITHMETIC. 383 23. A and B start at the same time from two places, M and N, 154 miles apart, and each travels toward the other till they meet ; A travels 3 miles in 2 hours, and B travels 5 miles in 4 hours : where will they meet ? Note. — In 1 hour A travels 1| miles and B travels IJ miles ; therefore the distance is to be divided in the ratio of 1| to 1^. 24.. A house and garden cost $850; 5 times the price of the house equals 12 times the cost of the garden : find the cost of each. 25. The sum of two numbers is 5760, and their difference is equal to i of the greater: find the numbers. 26. A agrees to work for $2 a day and forfeit $1 for every day he is idle ; at the end of 20 days he receives $25 : how many days does he idle ? Note. — To receive $25 he must work 12| days. During the remaining 7i days he forfeited all he earned ; therefore he idled twice as much as he worked. The problem then becomes: Divide 7^ into two parts in the ratio of 1 to 2. 27. A has 7 loaves of bread, B 5, and C none. The three eat all of the bread, each the same amount; C pays to A and B 12 cents : what shoilld each receive ? Note.— Each will eat one-third of the twelve loaves, or 4 loaves. A eats 4, and has 3 left for C ; B eats 4, and has 1 left for 0. C therefore eats 3 of A's and 1 of B's, and the 12 cents should be divided in the ratio of 3 to 1. 28. Two boys run a race : the smaller boy steps 4 feet and the larger steps 6 feet; and the smaller boy takes 5 steps while the larger boy takes 4; the larger boy gives the smaller 120 feet the start, and they come out even : how many steps does the larger boy take, and how far does he run ? 29. In the last number, how many steps does the smaller boy take, and how far does he run ? 884 ADVANCED ARITHMETIC. 30. Two cog-wheels work together; one has 11 cogs and the other 85 cogs : in how many rounds of the large wheel will the smaller gain 72 rounds ? 81. A hare is 80 leaps in advance of a greyhound ; the hare makes 11 leaps while the hound makes 9, and leajis four-fifths as far as the hound : how many more leaps will the hare make before the hound catches him ? 32. There are two numbers, the one twice as large as the other ; one-third of the smaller and one-half of the larger equal 20 : find the numbers. 33. Four-fifths of A's money is equal to one-half of B's; A gains $200 and B loses $100; they now have the same: what had they at first ? 31/.. A number is composed of 2 digits; the number is equal to 9 times the units less 18 ; it is also equal to 12 times the difference between the units and the tens : find the number. 35. A number is composed of two digits : the difference be- tween the units and the tens is 4 ; 5 times the units is equal to 2^ times the tens : find the number. 36. A is 4 times as old as his son ; in 14 years he will be only twice as old as the son : find their ages. 37. At what time between 8 o'clock and 9 o'clock are the hands of a clock together ? 38. At what time between 8 o'clock and 9 o'clock are the hands of a clock opposite each other ? 39. At what times between 5 and 6 is the minute-hand half- way between the hour-hand and XII ? Ji.0. At what times between 5 and 6 will the hands be at right angles to each other ? j^l. If one-half of the time past midnight is equal to one- sixth of the time past noon, what is the hour ? 4.2. If one-fourth of the time past noon is equal to one-six- teenth of the time past midnight, what is the hour ? ADVANCED ARITHMETIC. 385 4.3. If the time to noon equals one-seventh of the time to midnight, what is the hour ? 4.4.. If one-half of the time to midnight is equal to one- eighth of the time to noon, what is the hour ? 4.5. A farmer sold to one man 20 bushels of wheat and 8 bushels of oats for $14 ; to another 12 bushels of wheat and 10 bushels of oats for $9.70: find the price of the wheat and the oats. 46. A man bought 8 cows and 7 calves for $95; again, he bought 5 cows and 4 calves for $120 : find the price of a cow and a calf. 47. A is 30 years old; A's age is equal to B's plus one-half of C's ; and C is as old as A and B : find the ages of B and C. 4.8. The head of a fish is 6 inches long ; its tail is as long as the head and one-fourth of the body ; and the body is as long as the head and the tail : find the length of the fish. 49. A bag contains three times as many dollars as quarters ; if 8 dollars and 8 quarters be taken away, there will be 5 times as many dollars as quarters : find the number of each. 50. A has two horses and a saddle ; the saddle is worth $20 ; if it be placed on one horse it will make him worth as much as the other ; but if it be placed on the second horse, he is then worth twice as much as the first : find the value of each horse. 51. A flag-pole consists of two parts ; the length of the upper is five-sevenths of the length of the lower; and 9 times the upper part added to 13 times the lower part is longer than 11 times the whole pole by 36 feet ; find the length of the pole. - 52. A party was composed of men and women; 6 of the women left ; there were then twice as many men as women ; when the 6 women returned with their husbands, the number of women was only two-thirds of the number of men. What was the size of the original party ? 386 ADVANCED ARITHMETIC. 53. Twenty-six tons can be carried by 15 wagons and 22 carts, or by 18 wagons and 16 carts : what is a load for a cart and for a wagon ? 54.. A and B started from M and N respectively, and trav- eled till they met ; it appeared that A had traveled five-sev- enths as far as B ; but if A had traveled 15 miles farther, he would have traveled twice as far as B : how far did each travel? 55. A cistern has two pipes ; one can fill it in 15 hours and the other can empty it in 18 hours : if both are left open, what time will the cistern be in filling ? 56. A cistern of 860 gallons has two pipes ; one can fill it in 15 hours and the other can empty it in 20 hours : if both are left open, what time will the cistern be in filling ? 57. A, after doing :i of a work in 5 days, calls the assist- ance of B, and they finish the work in 6 days: in what time could each have done the whole work ? 58. A and B could have done a work in 15 days, but after working together 6 days, B was left to finish it, which he did in 80 days : in what time could A have finished it if B had left at the end of the 6 days ? C. PROBLEMS OF THREE OR MORE BASES. 177. Solving Equations of Tliree or More Unknown Numbers. — When there are three or more un- known numbers to be found, there must be as many equations (bases) as there are unknown numbers. The process of elimination here is not different from that given under Problems of Two Bases. ADVANCED ARITHMETIC. 387 EXAMFIjES. /. Sx-i-^y-z = lS, 6x-Qy-\-5z = S, x-\-y-^z = Q. Find the values of x, y^ and z. Process: (1) Sx+by-z = lS. (2) bx-6y+5z = 8. (3) x+y+z = Q. 3x(3) = (4) Sx-{-Sy-^Sz = l8. (l)-(4) = (5)22/-4s = 0. 5x(3) = (6) 6x+by+5z = 30. (6) -(2) = (7) 11 2/ = 22. (8) y = 2. (5) = (9) 4-4^ = 0. (10) z = l. (3) = (11) x-\-2+l = Q, (12) a; = 3. ^. x-^y-{-z=22, y-\-z-{-r=21, x+z-^r=ld, x-\-y-^r=m. Find the values of x, y, z, and r. Process: (!) x-\-y-\-z = 22. (2) y + 2:+r = 21. (3) x+z+r = 19. (4) . T+y+r = 16. Adding, (5) 3x+3?/+32+3r = 78. Jof (5) = (6) ^+2/+2;+r = 26. (6)-(I) = (7) r = 4. (6) -(2) = (8) :r = 5. (6) -(3) = (9) 2/ = 7. (6) -(4) = (10) 2 = 10. 3. Sx+2y=lS, Sy-2z = S, 2^-32=9. Find the values of x^ y, and z. Process: (1) 2x+2y = ld. 9x(3) = (8) 18 a; -27 2 = 81. {2)Sy-2z = 8. (8)-(7) = (9) -35^ = 35. (3) 2.^-3^ = 9. (10) z=-l. 3x(l) = (4) 9:c + 6^ = 39. (2) = (11) 32/+2 = 8. 2x(2) = (5) 62/-40 = 16. (12) ?/ = 2. (4)-(5) = (6) 9.C + 42+23. (1) = (13) 3^+4 = 13. 2x(6) = (7) 18^+8^ = 46. (14) x = S. 888 ADVANCED ARITHMETIC. EXERCISE CLVII. Find the values of the unknown numbers : 5. { 2x-y-\-4:Z=2Q. i Qx-z = -14:. {x-{-Sy-\-2z = 17. ]4x-2y-\-z = 12. (Sx-y = 10. {Qx+by-\-4:Z=S7. ■{ 7x-Sy-\-dz = 2Q. (Sx-{-2y-z=S. x^y+z = 7. y-\-z-{-w=Q. z-\-w-\-x=S. iv-\-x-{-y=9. 6. ■1 ^x-\-y = n. 72+5r=:ll. [3r+2x=4. 8a;-102/+-ls=60. 7a;+3 2/-52=23. x-2y-\-4:Z=m. 6 "^ 8 "^ 3 9. 2;+2/+2=43. 178. Problems. EXAMPLES. 1. A, B and C have $210: twice A's money and 3 times B's money added to C's will make $400; A's money and 4 times B's added to twice C's will make $510. How much has each ? Solution : Let r = A's money, // = B's money, and 2 = C's money. Then, (1) x+y+z = 210, (2) 2x-hSij+z = 400, and (3) x+iy-\-2z = bl0. (3)-(l) = (4) Sy-^z = SOO. 2x(l) = (5) 2x+2y-\-2z = 420. (5)-(2) = {6) -y+z = 20. (4) -(6) = (7) 41/ = 280. (8) 2/ = 70. (6) = (9) -70+2 = 20. (10) 2 = 90. (1) = (11) .t'+70 + 90 = 210. (12) .1=50. Answer, .$50, $70, and $90. ADVANCED ARITHMETIC. 389 2. A and B together have $600; A and C, $700; B and C, $500. How much has each ? Solution: Let :r = A's money, 2/ = B's money, and 2; = G's money. Then, (1) x+y = Q0O, (2) x+z = 700, and (3) y+z = bOO. Adding (4) 2.r+2?/+2s = 1800. (5) x+y-\-z = 900. (5)-(l) = (6) s = 300. (5) -(2) = (7) 2/ = 200. (5) -(3) = (8) a: = 400. Answer, $400, $200, and $300. 3. A number is expressed by 3. digits, the sum of which is 9; the number is 42 times the sum of the third and the second digits, and the first (units') digit is twice the sum of the other two : find the number. Note. — Any number expressed by the French system of notation is equal to the sum of its 1st (units') digit, 10 times its 2d (tens') digit, 100 times its 3d (hundreds') digit, and so on. Convince yourself of this truth by applying it to several numbers. Solution: Let.r = units* figure. Trans, in (3), (7) x-2y-2z = 0. 2/ = tens' figure, (i)-(7) = (8) Sy+Sz = 9. and 2 = hundreds' (9) y+z = S. figure. ^ of (6) = (10) 19z-ny=-S. Then,(l) x-{-y+z = 9, llx(9) = (ll) llz+ny = 3S. (2) 100z-{-10y-\-x= (10) + (11) = (12) 30^ = 30. 42(2/+2), (13) z = l. and (3) x=:2{z+y) = 2z+ (9) = (14) y+l = S. ^y- (15) y = 2, (2) = (4) 100z+10y-{-x= (1) = (16) a:+2+l = 9. 42^ + 420. (17) a: = 6. Col., (5) r-32y+b8z = 0. Number, 126. (5)-(l) = (6) 57^-33^= -9. 390 ADVANCED ARITHMETIC. ^. A, B, C, D and E have money; B gives A :| of his ; C gives B ^ of his ; D gives C i of his ; and E gives D ^ of his : each then has $80. How much had each at first ? Solution : Let x = A's money, (4) = (7) | w; +6 = 30. y = B's money, (8) f ly = 24. 3 = 0'smoney, (9) w = S2. w = I>'s money, and (3) = (10) f s + 8 = 30. r = E's money. (11)3 = 33. Then, (1) a:+i|/ = 30 (2) = (12) i?/ + ll = 30. (2) ^2/+i2 = 30, (13) y = SS. (3) fs + it/; = 30, (1) = (14) rr+19 = 30. (4) |^/; + ir = 30, (15) x = ll. and (5) f r = 30. Answer, $11, $38, $33, $32, $36. gof (5) = (6) r = 36. 5. A and B can do a piece of work in 12 days ; A and C in 15 days ; and B and C in 20 days. In what time can each do it alone ? Solution : Let r = time req'd by A to do the work, y = time req'd by B to do the work, and 2 = time req'd by C to do the work. Then, — =part done by A in 1 da. — = part done by B in 1 da. 2/ — = part done by C in 1 da. (1) '> 12 =1, (2) X 15 z =1, and (3) 20^ y 20 s = 1. -iVof(2) = = (4) 1 + X 1 z 1 15* hofiS)-- = (5) y 1 z 1 20' I ADVANCED ARITHMETIC. 391 Adding, (7) A+A+A = ^. ^of7 = (8)4+^+^ = A = _^ (8)-(6) = (9)^ = -i,and2 = 60. (8) -(4) = (10) -^=-^, and y = dO. (8)-(5) = (ll)^ = ^,anda: = 20. Answer, 20 da., 30 da., and 60 da. EXERCISE OLVIII. 1. Two calves, 5 sheep and 8 hogs cost $84; 8 calves, 6 sheep and 7 hogs cost $108 ; 10 calves, 12 sheep and 8 hogs cost $98 : find the cost of a calf, a' sheep, and a hog. 2. I bought corn, wheat and oats, as follows: 4 bushels of corn and 3 bushels of wheat for $3.70; 6 bushels of wheat and 8 bushels of oats for $6.20; 12 bushels of corn and 9 bushels of oats for $7.05: find the price of each per bushel. 3. A, B and C have $1000; A and B together own $800; B and C together own $630 : what does each own ? 4.. A, B and C have $3600; if B gives A one-half of his, and C gives B one-third of his, they will all have the same amount : what had each at first ? 5. After a battle in which 24000 men were engaged, it was found that the number slain was one-seventh of those who sur- vived, and that the number wounded was equal to one-half of the slain : find the number slain, the number wounded, and the number not hurt. 6. A farmer has 45 head of horses, cows, and sheep; there are six times as many sheep as cows, and one-third as many horses as sheep : find the number of each. 892 ADVANCED ARITHMETIC. 7. A, B and C built a wall 200 feet long ; B built as many feet as C and one third more, and A built three-fourths as much as B : how many feet did each build ? 8. A man divided his estate of $4000 as follows : To his wife $500 more than to his son, and to the son $1000 more than to the daughter: find the share of each. 9. Four persons compare their money ; the first man has one- half as much as all the others ; the second, one third as much as all the others ; the third, one fourth as much as the other three; and the fourth, $14 less than the first: find the amount of each man's money. 10. A says to B, " Give me $100 and I will have as much money as you will have;" and he says to C, " Give me $200 and I will have as much as you will have." B says to C, "Give me $350 and my money will be to yours as 19: 9." Find how much each has. 11. A and B can do a piece of work in 48 days ; A and C can do it in 80 days ; B and C can do it in 26f days : in what time can each do it alone ? 12. A, B and C earn $68 in 14 days ; A and B can earn it in 18 days ; A and C can earn it in 21 days ; in what time can each earn it alone ? 13. A and B can earn $40 in 6 days ; A and C can earn $54 in 9 days ; B and C can earn $80 in 15 days : what can each earn per day ? 1j^. a cistern is filled by 8 pipes ; the first and second fill it in 1 hr. 10 min. ; the first and third in 1 hr. 24 min. ; and the second and third in 2 hr. 20 min. : in what time can each fill it ? 15. Find three numbers, such that the first together with | of the second is equal to 19 ; ^ of the second with | of the third is equal to 28 ; and \ of the third with ^ of the first is equal to the second. 16. A certain number is composed of three digits whose sum is 15. The digit in units' place is 8 times that in the hundreds' ADVANCED ARITHMETIC. place ; and if 896 be added to the number, the digits will be reversed in order. Find the number. 17. A jeweler sold three rings. The price of the first with -J the sum of the second and third was $25 ; the price of the sec- ond with i of the sum of the first and third was $26 ; and the price of the third with \ the sum of the first and second was $29. What was the price of each ? 18. Find four numbers such that, by adding each to twice the sum of the remaining three, you will obtain 46, 43, 41, and 88, respectively. 19. A gives to B and C twice as much money as each of them has ; B then gives to A and C twice as much as each of them has ; and C then gives to A and B twice as much as each of them has ; each then has $27. How much had each at first ? 20. A, B, C and D together have $46. After B gives to A J of his money, C to B i of his, and D to C J of his, each has $10. How much had each at first ? D. PROBLEMS CONTAINING QUADBATIC EQUATIONS. 179. Quadratic Equations. — An equation that con- tains the second power (or square) of the unknown number is called a Quadratic Equation. An equation that contains no form of the unknown number except the square is called an incomplete quadratic equation. An equation that contains both the first and second powers of the unknown number is called a complete quadratic equation. Thus : (1) a:2=64. (Incomplete.) (2) a:2+4a: = 12. (Complete.) When an equation is of the form of (1), the value of the un- known number may be found by extracting the square root of the equation. When an equation is of the form of (2), something must be 394 ADVANCED ARITHMETIC. added to render the first member a perfect square ; then, the value of the unknown number may be found by extracting the square root of the equation. Principle : Extracting the square root of each member of an equation extracts the square root of the equation. EXAICPIiSS. 1. 6x^=54. Find the values of x. Process: (1) Qx^ = 5i. iof (1H(2) ^r2 = 9. t/(2) = (3) x= ±3, result. Note.— +3x+3 = 9 and -3x -3 = 9. Then,.r may equal +3 or -3. When a number is true for either sign, the double sign =h is used. This sign is read plus or minus, and should be used whenever we extract the square root. 2. — -1 — =T7^H — . Find the values of x. 3 a; 12 x Process : ^^' 3+:r 12+a;' 12:rx(l): = (2) 4.r2+36 = .r2+144. (3) 4.r2-.r2 = 144-36. (4) 3x2 = 108. (5) a:2 = 36. -/(5) = = (6) x= zb6, result. We are now to study how to complete the square of the first member in a complete quadratic equation. We know that — (a;-|-a)2=:a:2+2«a:+a2, and (x—ay=x^—2ax+a^. How could we get the third term in these results, if we had the second term ? Answer : Take half of the 2 a and square it. How can we get the third term in x^ + 12x and x^-12x? A ADVANCED ARITHMETIC. 395 Answer : Take half of the 12 and square it. Thus : a;2+12a;+36 a;2-12a;+36. Rule for Completing the Square : Add the square of half the coefficient of the first power of the unknown number, 8. x'^-\'2x-\Z. Find the values of a;. . Process: (1) a:2-12:c = 13. Completing the sq., (2) a:2-12a:+36 = 13 + 36. (3) a:2- 12 0:4-36 = 49. V(3) = (4) rr-6 = ±7. (5) a: = 6±:7 = 13or -1, results. 4. 5a;2+20a:=60. Find the values of x. Note. — Always divide by the coefficient of x^ before completing the square. Process: (1) 5a:2+20a: = 60. (2) cc^+ A x=l2. (3) a;2+4.r+4 = 12+4 = 16. V73r=(4) 3:+2==h4. (5) a: = 2or -6. 5. x^-\-l\ a:=: — 18. Find the values of x. Process: (1) x^-\-llx= -18. (2) a:2+lla'+i|i= -l8+i|i=-<^ ■i/(2) = (3)^+V = ±|. (4) x= -9 or -1. x 44 6. -r — — ^=4. Find the values of x. 4 x — 2 Process: (1) o"** (2) a;2- 2 a; -176 = 16.^-32. (3) a;2_2a;-16aj = 176-32 = 144. (4) a;2-18a; = 144. (5) a;2-18a; + 81 = 144+81 = 225. -v'^=(6) a- -9 = ±15. a- = 24or -6. 396 ADVANCED ARITHMETIC. EXERCISE CLIX. 1. What is a quadratic equation ? 2. Define incomplete quadratic equation. Complete quadratic equation. 3. Give the principle governing extracting the square root of an equation. If.. How do you complete the square in a complete quadratic equation ? Find the values of x in — 5, a;2_^2a;=8. 15. 3a:2+7 = 43+2a:2. 8, a;2-8a:=33. 17. a:2+10a; = ll. 9. x'^-x=i. 18, 5a;2-40a; = 165. 10. x'^-\-x=-\, 19. 4.T2-4a:=3. 11. a;2-14=-24. 20. 4a:2+20.r=-9. 12. a;2+5a;=-2i. a:+2 4-a;_7 i5. a:2=:625. a;-l 2a; ~3' U. ^-15. 180. Problems. EXAMPIiES. 1. A bought a lot of flour for $126. If the number of $'8 per barrel was equal to f of the number of barrels bought, find the number of barrels and the price per barrel. Solution : Let a; = number of barrels. Then, |a: = price per barrel, and (1) xxlx = 12Q. (2) 1^2=126. (3) .t2 = 441. V{S) =(4) .r=zb21. (5) I of 21 = 6. Answer, 21 bbls. @ $6 eac|^ ADVANCED ARITHMETIC. 397 Note. — While the negative answer satisfies the equation it will not satisfy the problem, and the positive answer only is used in the problem. 2. Divide 12 into two parts, such that their product will be 11. Solution : Let x = one part. Then, 12 - a: = the other part and(l) a;(12-:r) = ll. (2) 12:r-.r2 = ll. (3) :c2_i2x=-ll. (4) x^-12x+SQ= -11+36 = 25. (5)x-6 = =b5. (6) :c = ll, or 1. (7) 12-a: = l, orll. Answer, 1 and 11. 3. If we should reduce the length of an edge of a cube 1 inch, the volume would be reduced 397 cu. in. What is the volume of the cube ? Note. — If we commence with the cube left and think of building the 1 inch back on to it, let .r = one edge of that cube. Then, we must add a block x in. 1., x in. w., 1 in. th., to each of three faces of the cube . =Sx^, a rectangular solid :r in. 1., 1 in. w., 1 in. th., to each of three edges =Sx, and a 1-inch cube in the corner =1 The whole addition =3 a.2+3 x-^1 Then, (1) 3a,i+3a:+l = 397. (2) 3x2 + 3 a =397-1 = 396. (3) x^+x = lS2. (4)a:2+:r+i=132+i = ^. V{i) = (b) x+l = ±-%K (6) ^ = 11, -12. The old cube 11 in. + l in. = 12 in., answer. 898 ADVANCED ARITHMETIC. EXERCISE CLX. 1. Find two numbers, such that their diflference is 5 and their product is 66. 2. Find that number whose square plus 6 times the number equals 55. 3. The sum of two numbers is 12, their product is 85. Find the numbers. Jf.. A rectangular field is twice as long as it is wide, and con- tains 20 acres. How wide is it ? 5. The product of two consecutive integers exceeds six times their sum by 6. Find the numbers. 6. If we cut enough from a cubical block to make its dimen- sions 1 inch shorter, it will lose 1657 cu. in. Find the size of the block. 7. If we cut enough from a cubical block to make its dimen- sions 2 inches shorter, it will lose 728 cu. in. Find the size of the block. 8. The hypotenuse of a right triangle is 50 inches, its base equals one-half of its perpendicular. Find the base and per- pendicular. Hint: Let x = base. Then, 2. t = perpendicular, and (1) a;2+(2af = 502, or (2) :i-2 +4^-2 = 2500. 9. Find the altitude of an equilateral triangle if one side is 40 inches. Hint : A perpendicular from the vertex to the base is the altitude. It divides the triangle into two equal right tri- angles. The base of each right triangle is one-half of the hy- potenuse (40 in.). Letting :r = the altitude, we have — 402 = 202+.r2. 10. A merchant sold a piece of cloth for $24, gaining as many % as the number of dollars the cloth cost. Find the cost. ADVANCED ARITHMETIC. 399 11. A tree 90 ft. high was blown over by a storm so that the top touched the ground 40 ft. from the tree, while the other end of the part broken off rested on the stump. How much was broken off ? 12. The base of a rectangle is 11 ft. longer than its altitude. If its area is 900 sq. ft., find the length of each side. 13. Find the altitude and area of a triangle whose sides are 15, 12, and 8. Note. — When you drop the altitude on the base 15, you have two right triangles. Let a: = the base of one triangle and 15 -a: the base of the other. Then, (Altitude)2 = 122-(15-:j:)2 = 82-a:2. (Why?) Find the value of x, and complete the solution. E. PROGRESSIONS. 181. Aritlimetical Progressions. — An Arith- metical Progression is a series of numbers that in- crease or decrease by a common difference. Thus, I, 4, 7, 10, 18, is an increasing arithmetical progression. The common differ- ence is 3. II, 9, 7, 3, 1, is a decreasing arithmetical progression. The common differ- ence is —2. TERMS. a, the first term. I, the last term. n, the number of terms. (Z, the common difference. S^ the sum of the series. 400 ADVANCED ARITHMETIC. Developing the formulas for arithmetical progressions. I. The formula for the last term: Using the terms given above, we may write an arithmetical progression or series. Thus, (1) (2) (3) (4) (5) (n) S = a+{a-\-d) + {a+2d) + {a+3d) + {a-{-4d) ( ). Observe (1) that each term of the series has a, and that each term after the first has d; (2) that the coeflRcient of d is always 1 less than the number of the term. Then, for the nth term, the coefficient of d is (n-l). From these data, we may write the last term. Thus, l=a-{-{n—l) d. Relation : In an arithmetical progression, the last term is equal to the first term, plus the common difference times the number of terms less 1. II. The formula for the sum of the series : An arithmetical series may be written thus : (1) S = a+{a+d)+{a+2d) +(Z-2d) + (Z-rf) + Z; also, (2) S = l + {l-d) + {l-2d) +(a+2rf)-f(a + d) + a. Adding, (3) 2S = {a + l) + {a-\-l) + {a+l) +{a + l) + {a + l)+{a+l). Since there are n terms and each has (a+Z), (3) = (4) 2S = n{a-\-l); or, S=l(a+l). Relation : The sum of an arithmetical series is equal to the product of one-half of the number of terms and the sum of the ex- tremes (the first and last terms). EXAMPLES. 1. Find the 7th term of the series, whose 1st term is 5, and the common difference is 9. Solution: (1) l = a-\-{n-l)d. (2) Z = 5+(7-l)9 = 59, answer. 2. Find the 10th term of the series 2, 6, 10, etc. Solution: (1) a = 2. (Why?) (2) d = 4. (Why?) (3) Z = 2+(10-l)4 = 38, answer. J ADVANCED ARITHMETIC. 401 3. Find the sum of a series of 19 terms, whose first term is 6 and whose last term is 96. Solution: (1) S = ^{a + l). iq (2) ^=^(6+96) = 969, answer. 4.. Find the last term and the sum of the series, 5, 5, etc., to 25 terms. Solution: (1) a = 5. (2) d = 3. (3) n = 25. (4) ? = 5+(25-l)3 = 77, 1st answer. (5) s = V- (5 +77) = 1025, 2d answer. 5. Find the last term and the sum of the series 12^ 9^ 6, etc., to 15 terms. Solution : (1) a = 12. (2)rf=-3. (Why?) (3) n = 15. (4) Z = 12-f(15-l)(-3). (5) ? = 12-42= -30, 1st answer. (6) s = y [12 + (-30)]. (7) s = JgS X - 18 = - 135, 2d answer. Note. — Be careful to handle the signs correctly, and a minus common diflference will give you no trouble. 6. Find the first term of a series whose last term, common difference, and number of terms are respectively 27, 2, and 11. Solution: (1) 27 = a+(ll-l) 2 = a+20. Trans., (2) a = 27 -20 = 7, answer. 7. Find the number of terms of the series whose first term, last term, and sum are respectively —1, 44, 215. Solution: (1) 215 = -|^(-l+44) = -|-x43. (2) ^4^ = 215. ,„, 215x2 ,_ (3) n = —^ — = 10, answer. 402 ADVANCED ARITHMETIC. EXERCISE OLXT. 1. What is an arithmetical i)rogression or series ? 2, Name and explain the terms used. S. Write a series of 8 terms whose first term is 5 and whose common difference is 12. Jf.. Write a series of 12 terms, whose first term is 12 and whose common difference is —5. 5. Develop the formula for finding I. Give the relation. 6. Develop the formula for finding s. Give the relation. No. I a n d s 7. 9 7 9 5 9 8. ? -5 12 -3 ? 9. ? 27 18 -4 ? 10. 2 ? 7 -1 ? 11. 24 -15 14 ? ? 12. 116 1 ? 5 ? 13. 81 ? '7 7 ? n. m 1 27 ? ? 15. 4 ? f ? 16. 24 8 ? ? 40 17. ? 27 9 ? 68 18. H ? 12 ? -99 19. 15i ? ? *l 168 20. 9 ? 17 4 -891 Note.— The last two (Nos. 19 and 20) will require two equations. Sub- stitute in both formulas. This will give you two equations to solve fcr two unknown numbers. 21. 40 potatoes are 2 yd. apart and the first is 2 yd. from a basket. How far will a boy travel, who gathers them and puts them into the basket one at a time ? ADVANCED ARITHMETIC. 403 182. Falling' Bodies. — Falling of bodies furnishes an example of arithmetical series. A body, if left unsupported, will fall by its own weight during the 1st second, 16.08 ft. (N. Y.); during the 2d second, 48.24 ft.; during the 8d second, 80.4 ft. ; and so on. Thus, 16.08, 48.24, 80.4, 112.56, Note. — Observe that in this series the first term {a) is 16.08, and the common difference (d) is 32.16. EXAMPIiES. 1. How far will a body fall during the 10th second ? Solution: (1) a = 16.08. (2) rf = 32.16. (3) n = 10. (4) l = a + {n-l)d. (5) ? = 16.08+9x32.16 = 305.52. Answer, 305.52 ft. 2. How far will a body fall in 8 seconds ? Solution: (!) Z = 16.08 +7x32.16 = 241.2. (2) ^ = 1(16.08 + 241.2) = 1029.12. Answer, 1029.12 ft. 3. A body is thrown downward so that it travels 70 ft. the first second. How far does it fall in 5 seconds ? Note. — Here a = 70. Solution: (1) Z = 70+4x32.16 = 198.64. (2) >S = 1(70+198.64) = 671.6. Answer, 671.6 ft. In the falling of bodies it has been determined that the ve- locity increases 82.16 ft. per second. This is the acceleration. Using 82.16 for a in formula on page 809, we have — F=32.16^ 404 ADVANCED ARITHMETIC. Relation : Abstractly, the velocity of a falling body is 32.16 times the time expressed in seconds. If. A body strikes the ground with a velocity of 160.8 ft. per second. How many seconds has it been falling and how far has it fallen ? Solution: (1) F=32.16f. (2) 160.8 = 32.16 <. 160.8 ^^^ ^ = 32:i6 = ^- Time, 5 seconds. (5) Z = 16.08+4x32.16 = 144.72. (6) « = 1(16.08 + 144.72) = 402. Distance, 402 ft. Note. — A body thrown upward will decrease in its travel just as a falling body increases in its travel. 5. How high will a body rise, if thrown upward with a veloc- ity of 321.6 ft. per second ? Solution: (1) F= 32.16 i. (2) 321.6 = 32.16 <. Time in rising, 10 sec. Note. — If it was 10 sec. rising, the last second it rose 16.08 ft. To find how far it rose the first second, consider from the top downward and get the 10th term of the series. (4) Z = 16.08+9x32.16 = 305.52. (5) ;S' = J^(16.08+305.52) = 1608. Distance, 1608 ft. EXERCISE CLXII. 1. How far will a body fall in 1 sec? In 2 sec? In 8 sec? 2. Do the distances fallen during the successive seconds form an arithmetical progression ? If so, what is the common difference ? 3. What is the formula for the velocity of falling bodies ? ADVANCED ARITHMETIC. 405 Jf. A body thrown upward is gone 6 sec. How much time is taken in rising ? How much in falling ? 5. How fast will a body be falling at the end of the 6th sec- ond of its descent ? Give its fall during the 7th second. 6. A boy throws a stone over a tree ; it strikes the ground in 4 sec. How high is the tree ? 7. A body strikes the ground with a velocity of 96.48 ft. per sec. How long has it been falling ? 8. A body is thrown downward with a velocity of 60 ft. per second. What will its velocity be in 6 sec? How far will it fall? (a =60+ 16.08. Why?) 9. A tower is 321.6 feet high. With what velocity will a stone, let fall from the top, strike the ground ? Plan : (1) Find the time and (2) the velocity. 183. Geometrical Progressions.— A Geometri- cal Progression is a series of numbers that increase or decrease by a common ratio. 1, 4, 16, 64, etc., is an increasing geometrical series. The common ratio is 4. 24, 12, 6, 3, etc., is a decreasing geometrical series. The common ratio is -J. TERMS. a, the first term. ly the last term. r, the common ratio. n, the number of terms. S, the sum of the series. 406 ADVANCED ARITHMETIC. Developing the formulas for geometrical series. I. The formula for the last term : Using the terms given above, we may •ite a geometrical series. (1) (2) (3) (4) (n). S = a+ar+ar^+ar^ ( ). Observe (1) that each term of the series has a factor a, and (2) every term after the first has a factor r whose exponent is 1 less than the number of the term. Then, we may write the last term — l=ar^-'i^ Note. — n -1 is the exponent of r. Relation: In a geometrical series, the last term is equal to the first term times the ratio raised to a power whose exponent is 1 less than the number of terms. II. The formula for the sum of the series : The series may be written in this form : (1) (2) (3) (4) (n) (1) S = a-\-ar-\-a'i^+ar^ I. (1) (2) (3) {n-D (n) rx(l) = (2) >Sr = ar+ar2+ar3 I +Zr (2)-(l) = (3) Sr-S = lr-a. (4) S{r-l) = lr-a',or, r — 1 Note. — When (1) is subtracted from (2), all the terms in the second members disappear, except the first term in (1) and the last term in (2). Relation : The sum of a geometrical series is equal to the frac- tion whose numerator is the last term times the ratio ^ minus the first term, and lohose denominator is the ratio less 1. EXAMPLES. 1. Find the fifth term of the series whose first term and com- mon ratio are respectively 4 and 3. Solution: (1) l = ar'^~'^ (2) Z = 4x 3^ = 324, answer. I ADVANCED ARITHMETIC. 407 2. Find the ratio of the series whose first term, number of terms and last term are respectively 20, 4, and 540. Solution: (1) l = ar'^~'^ (2) 540 = 20r3. (3) 1^ = 21. (4) r= f27 = 3, answer. 3. Find the sum of 8 terms of the series, 3, 6, 12, etc. Solution: (1) r = 2. (2) Z = 3x27 = 384. (3)^=^^«. (4) S = ^""^^^"^ = 765, answer. 4. Find the sum of the infinite series, 1+i+i+i+etc. Note. — An infinite series continues forever. This series is decreasing and its last term approaches infinitely near to — so near that no other value represents it so accurately as 0. Then, we have a = l, r = i, and 1 = 0. Solution : S = ^ — :r = — r = 2, answer. 5. Find the number of terms in the series whose 1st term, last term, and ratio are respectively 5, 1280, 4. Solution: (1) 1280 = 5x4"—^ (2) 4""-^ = 256. (3) 4^ = 256. (4) n-l = 4. (5) n = 5, answer. NoTE.-^To obtain (3), you know that 4 raised to some power equals 256. Proceed by expanding 4, and when you reach the 4th power you will have 256. EXEECISE CLXIII. 1. What is a geometrical progression or series ? 2. Give and explain all terms used in geometrical series. 408 ADVANCED ARITHMETIC. 3. Write a geometrical series of 4 terms, whose 1st term is 5 and whose ratio is 7. J^.. Write a series of 4 terms, whose 1st term is 4 and whose ratio is ^. 5. Write 5 terms of the series whose 1st term is 4 and whose ratio is —3. 6. Develop the formula for I. Give the relation. 7. Develop the formula for S. Give the relation. 8. What is an infinite series ? If the infinite series is de- creasing, what does the I approach ? No. I a n r »S' 9. ? 4 7 5 ? 10. ? -3 10 3 9 11.' ? 1 8 -4 9 12. 2 infinite f ? 13. 64 1 ? 2 ? n- ? 4 ? 3 118096 15. 13122 ? 9 3 ? 16. ? 2 8 i ? 17. 375000 ? 7 5 468769 18. ? infinite ^ n i YC 49538 lujuei