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About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at |http: //books .google .com/I LIBRARY University of California. GIFT OF- ^' r /: t • : .-7 ^ ^ / / r * W- ECLECTIC EDUCATIONAL SERIES, RAY'S NEW HIGHER ARITHMETIC A Bevised Edition of the Higher Arithmetic BY JOSEPH RAY, M. D. '♦ LaU Ptv/eaaor in Woodvxxrd ColUfie, IVERSITY VAN ANTWERP, BRAGG & CO. CINCINNATI AND NEW YORK. Ray's Mathematical Series. ARITHMETIC. Bay's New Primary Arithmetic. Ray's New Intellectual Arithmetic. Ray's New Practical Arithmetic. Bay's New Higrher Arithmetic. TWO-BOOK SERIES. Bay's New Elementary Arithmetic. Bay's New Practical Arithmetic. AliGEBRA. Bay's New Elementary Algrebra. Bay's New Higrher Algrebra. HIGHER MATHEMATICS Bay's Plane and Solid Oeometry. Bay's Oeometry and Trigronometry. Bay's Analytic Oeometry. Ray's Elements of Astronomy. Bay's Surveyingr and Navigration. Bay's Differential and Integrral Calculus. Copyright, 1880, BY Van Antwerp, Braoo & Co. PREFACE. Ray*s Higher Arithmetic was published nearly twenty-five years ago. Since its publication it has had a more extensive circii- iation than any other similar treatise issued in this country. To adapt it more perfectly to the wants of the present and future, it has been carefully revised. It has been the aim of the revision to make Ray's New Higher Arithmetic thoroughly practical, useful, and teachable. To this end the greatest care has been given to securing concise definitions and explanations, and, at the same time, the systematic and thorough presentation of each subject. The pupil is taught to think for him* self correctly, and to attain his results by the shortest and best methods. Special attention is given to modem business transactions, and all obsolete matter has been discarded. Almost every chapter of the book has been entirely rewritten, without materially changing the general plan of the former edition, although much new, and some original matter has been introduced. Many of the original exercises are retained. Particular attention is called to the rational treatment of the Arithmetical Signs, to the prominence given to the Metric System, and to the comprehensive, yet practical, presentation of Percentage and its various Applications. The method of combining the algebraic and geometric processes in explaining square and cube root will com- mend itself to teachers. The chapter on Mensuration is unusually' full and varied, and contains a vast amount of useful information. (Hi) 11 I 90S IV PREFACE. The Topical Outlines for Keview will prove invaluable to both teachers and pupils in aiding them to analyze and to classify their arithmetical knowledge and to put it together so as to gain a com- prehensive view of it as a whole. Principles and Formulas are copiously interspersed as summaries, to enable pupils to work intelligently. The work, owing to its practical character, logical exactness, and condensation of matter, will be found peculiarly adapted to the wants of classes in High Schools, Academies, Normal Schools, Commercial Schools and Colleges, as well ad to private students. The publishers take this opportunity of expressing their obligations to J. M. Gbeenwood, a. M., Superintendent of Public Schools, Kan- sas City, Mo., who had the work of revision in charge, and also to Rev. Db. U. Jesse Knisely, of Newcomerstown, Ohio, for his valuable assistance in revising the final proof-sheets. Cincinnati, Jvly^ 1880. CONTENTS. PAGE I. Introduction 9 II. Numeration and Notation 15 III. Addition 23 IV. Subtraction 27 Business Tenns and Explanations 29 V. Multiplication 31 When the multiplier does not exceed 12 . . . .32 When the multiplier exceeds 12 . . . . . .34 Business Terms and Explanations 36 Ck)ntractions in Multiplication 38 VI. Division 43 Long Division 45 Short Division 47 Contractions in Division 49 Arithmetical Signs 50 General Principles 52 Contractions in Multiplication and Division ... 53 VII. Properties OP Numbers 59 Factoring 61 Greatest Common Divisor 64 Least Common Multiple 68 Some Properties of the Number Nine ^ . . . 70 Cancellation 72 VIII. Common Fractions 75 Numeration and Notation of Fractions .... 77 Reduction of Fractions 78 Common Denominator 82 Addition of Fractions 85 Subtraction of Fractions 86 Multiplication of Fractions r / ^' vi CONTENTS. FAOB Division of Fractionfl 90 The G. C. D. of Fractions 92 The L. C. M. of Fractions 94 IX. Decimai. Fractions 99 Numeration and Notation of Decimals 100 Eeduction of Decimals 103 Addition of Decimals 105 Subtraction of Decimals 106 Multiplication of Decimals . . . . . . 108 Division of Decimals Ill X. Circulating Decimals 115 Beduction of Circulates 118 Addition of Circulates 120 Subtraction of Circulates 121 Multiplication of Circulates ^ . 122 Division of Circulates 123 XI. Compound Denominate Numbers 125 Measures of Value 125 Measures of Weight 130 Measures of Extension 133 Measures of Capacity 139 Angular Measure 142 Measure of Time 143 Comparison of Time and Longitude .... 146 Miscellaneous Tables 146 The Metric System 147 Mea.sure of Length 149 Measure of Surface 149 Measure of Capacity 150 Measure of Weight . . * 150 Table of Comparative Values 152 Reduction of Comi>ound Numbers 154 Addition of Compound Numbers 160 Subtraction of Compound NumberH ..... 163 Multiplication of Compound Numbers .... 165 Division of Compound Numbers 167 Longitude and Time 169 Aliquot Parts 172 CONTENTS. vii PAOC XII. Ratio 175 XIII. Proportion 177 Simple Proportion 178 Compound Proportion 184 XIV. Percentage 188 Additional Formulas . 197 Applications of Percentage 197 XV. Percentage. — AppLiCATiONa (Without Time,) . . 199 I. Profit and Loss 199 II. Stocks and Bonds 204 ni. Premium and Discount 208 IV. Commission and Brokerage 213 V. Stock Investments 220 VI. Insurance 228 VII. Taxes 232 VIII. United States Revenue 236 XVI. Percentage. — Applications. ( Wilh Time.) . . . 242 I. Interest 242 Common Method 245 Method by Aliquot Parts 246 Six Per Cent Methods 246 Promissory Notes 254 Annual Interest 259 II. Partial Payments 261 U.S. Rule 261 Connecticut Rule 264 Vermont Rule 265 Mercantile Rule 266 in. True Discount 266 IV. Bank Discount . 268 v. Exchange 278 Domestic Exchange 279 Foreign Exchange 281 Arbitration of Exchange 283 VI. Equation of Payments 286 VII. Settlement of Accounts 292 Account Sales 296 Storage Accounts ^7 viii CONTENTS. PAGE vni. Compound Interest . . . . . . . 298 IX. Annuities 308 / Contingent Annuities .317 Personal Insurance 322 XVn. Partnership 327 Bankruptcy 331 XVIII. Alligation . 333 Alligation Medial 333 Alligation Alternate 334 XIX. Involution 342 XX. Evolution 347 Extraction of the Square Root 349 Extraction of the Cube Root 354 Extraction of Any Root 359 Homer's Method 360 Applications of Square Root and Cube Root . . . 363 Parallel Lines and Similar Figures 366 XXI. Series 369 Arithmetical Progression 369 Geometrical Progression 373 XXII. Mensuration 378 Lines 378 Angles 379 Surfaces 379 Areas . . . • 382 Solids 390 Miscellaneous Measurements 395 Masons' and Bricklayers' Work .... 395 Gauging 396 Lumber Measure 398 To Measure Grain or Hay 398 XXIII. Miscellaneous Exercises 401 OF THE UNIVERSITY RAT'S HIGHER ARITHMETIC. I. mTEODUOnOK Article 1. A definition is a concise description of any object of thought, and must be of such a nature as to dis- tinguish the object described from all other objects. 2. Quantity is any thing which can be increased or diminished; it embraces number and magnitude. Number answers the question, **How many?" Magnitude, **How much?" 3. Science is knowledge properly classified. 4. The primary truths of a science are called Prin- ciples. 6. Art is the practical application of a principle or the principles of science. 6. Mathematics is the science of quantity. 7. The elementary branches of mathematics are Arith- metic, Algebra, and Geometry. 8. Arithmetic is the introductory branch of the science of numbers. Arithmetic as a science is composed of defini- (9) 10 BA Y'S HIGHER ARITHMETIC, tions, principles, and processes of calculation ; as an art, it teaches how to apply numbers to theoretical and practical purposes. 9. A Proposition is the statement of a principle, or of something proposed to be done. 10. Propositions are of two kinds, demonstrable and indemonstrable. Demonstrable propositions can be proved by the aid of reason. Indemonstrable propositions can not be made simpler by any attempt at proof. 11. An Axiom is a self-evident truth. 12. A Theorem is a truth to be proved. 13. A Problem is a question proposed for solution. 14. Axioms, theorems, and problems are propositions. 15. A process of reasoning, proving the truth of a prop- osition, is called a Demonstration. 16. A Solution of a problem is an expressed statement showing how the result is obtained. 17. The terra Operation, as used in this book, is applied to illustrations of solutions. 18. A Rule is a general direction for solving all prob- lems of a particular kind. 19. A Formula is the expression of a general rule or principle in algebraic language; that is, by symbols. 20. A Unit is one thing, or one. One tiling is a con- crete unit ; one is an abstract unit. 21. Nuniber is the expression of a definite quantity. Numbers are either abstract or concrete. An abstra^;^ num- ber is one in which the kind of unit is not named ; a concrete number is one in which the kind of unit is named. Con- crete numbers are also called Denominate Numbers. INTRODUCTION. H 22. Kumbers are also divided into Integral, Fractional, and Mixed. An Integral number , or Integer, is a whole number; a Fractimud nmnber is an expression for one or more of the equal parts of a divided whole; a Mixed number is an Integer and Fraction united. 23. A Sign is a character used to show a relation among numbers, or that an operation is to be performed. 24. The signs most used in Arithmetic are + - X ^ v/ () • . •• \ J •••• 25. The sign of Addition is [+], and is called plus. The numbers between which it is placed are to be added. Thus, 3 + 5 equals 8. Plus is described as a perpendicular cross, in which the bisecting lines are equal. 26. The sign of Subtraction is [ — ], and is called minus. When placed between two numbers, the one that follows it is to be taken from the one that precedes it. Thus, 7 — 4 equals 8. Minus is described as a short horizontal line. Plus and Minus are Latin words. I^us means more; minus means less. Michael Steifel, a German mathematician, first introduced -{- and — in a work published in 1544. 27. The sign of Multiplication is [ X ]> and is read mul- iiplied by, or tim^s. Thus, 4 X 5 is to be read, 4 multiplied by d, or 4 times 5. The sign is described as an oblique cross. William Oughtred, an Englishman, born in 1573, first introduced the sign of multiplication. 28. The sign of Division is [ -^ ], and is read divided by. When placed between two numbers, the one on the left is 12 BAY'S HIGHER ARITHMETIC. to be divided by the one on the right. Thus, 20 -=-4 equals 5. The sign is described as a short horizontal line and two dots; one dot directly above the middle of the line, and the other jnst beneath the middle of it. Dr. John Pell, an English analyst, born in 1610, introduced the sign of division. 29. The Radical »ign, [ l/ ]» indicates that some root is to be found. Thus, |/36 indicates that the square root of 36 is required; ^125, that the cube root of 125 is to be found; and v^625 indicates that the fourth root of 625 is to be extracted. The root to be found is shown by the small figure placed between the branches of the Eadical sign. The figure is called the index. 30. The signs, +> — > X, -r-, i/, are symbols of operation. 31. The sign of Equality is [=], two short horizontal parallel lines, and is read eqvxds or is equal to, and sig- nifies that the quantities between which it is placed are equal. Thus, 3 + 5=9 — 1. This is called an equaJthOUy because the quantity 3 + 5 is equal to 9 — 1. 32. Ratio is the relation which one number bears to another of the same kind. The sign of Ratio is [ : ]. Ratio is expressed thus, 6 : 3 = f = 2, and is read, the ratio of 6 to 3 = 2, or is 2. The sign of ratio may be described as the sign of division with the line omitted. It has the same force as the sign of division, and is used in place of it by the French. 33. Proportion is an equality of ratios. The sign of Proportion is [::], and is used thus, 3: 6:; 4: 8; this may be read, 3 is to 6 as 4 is to 8; another reading, the ratio of 3 to 6 is equal to the ratio of 4 to 8. INTRODUCTION. 13 34. The signs [(), ], are signs of Aggregation — the first is the Parenihem, the second the Yinadum, They are used for the same purpose; thus, 24 — (8 -f 7), or 24 — 8 -|- 7, means that the sum of 8 + 7 is to be subtracted from 24. The numbers within the parenthesis, or under the vinculum, are considered as one quantity. 36. The dots [. . . .], used to guide the eye from words at the left to the right, are called Leaders, or the sign of Gontimuitiony and are read, and so on. 36. The sign of Dedvetion is [.'.], and is read therefore^ hence, or consequently. 37. The signs, =, :, : :, ( ), , . . . ., .•., are symbols of relation. 38. Arithmetic depends upon this primary proposition: that any number may be increased or diminished. ** In- creased" comprehends Addition, Multiplication, and Invo- lution; "decreased," Subtraction, Division, and Evolution. 39. The fundamental operations of Arithmetic in the order of their arrangement, are : Numeration and Notation, Addition, Subtraction, Multiplication, and Division. Topical Outline. Introduction. 1. Definition. 2. Quantity. 3. Science. 4. Principles. 5. Art. 6. Mathematics. 7. Proposition. 8. Demonstration. 9. Solution. 10. Operation. 11. Rule. 12. Formula. 13. Unit. 14. Number. 15. Sign. . 16. Signs most used. 17. Primary Proposition. 18. Fundamental Operations. 14 HAY'S HIGHER ARITHMETIC. Topical Outline of Arithmetic. Preliminary Definitions... 1. Definition. 2. Quantity. 3. Science. 4. Mathematics. 5. Proposition. 6. Tlieorem. 7. Axiom. 8. Demonstration. 9. Solution. 10. Rule. l^ 11. Sign. 1. Definitions. 1. As a Science... ^ 2. Classification of Numbers ^ ( Numeration 1. 3. Operations. ^ and , Notation. 2. Addition. 3. Subtraction. 4. Multiplication. 5. Division. 6. Involution. 7. Evolution. {Abstract Ck)ncrete. ' Integral. 2. - Fractional. . Mixed, g / Simple. v^ I Compound. r 1. Terms often used. 2. As an Art < 2. Signs. ^ 3. Applications. /'I. Problem. 2. Operation 3. Solution. 4. Principle. 5. Formula. 6. Rule. ^ 7. Proof. f 1. To Integers. 2: To Fractions. 3. To Compound Numbers, 4. To Ratio and Proportion. 5. To Percentage. 6. To Alligation. 7. To Progression. 8. To Involution and Evolution, l^ 9. To Mensuration. n. IsTUMERATION AKD NOTATIOK 40. Kiuneration is the method of reading numbers. Notation is the method of writing numbers. Numbers are expressed in three ways ; viz., by words, letters, and figures. 41. The first nine numbers are each represented by a single figure, thus: I 23456789 one. two. three, four. five. six. seven, eight, nine. All other numbers are represented by combinations of these and another figure, 0, called zero, naught, or cipher. Kemabk. — The cipher, 0, is used to indicate no value. The other figures are called significant figures, because they indicate some value. 42. The number next higher than 9 is named ten, and is written with two figures, thus, 10 : in which the cipher, 0, merely serves to show that the unit, 1, on its left, is different from the unit, 1, standing alone, which represents a single thing, while this, 10, represents a single group of ten things. The nine numbers succeeding ten are written and named as follows: II 12 13 14 15 16 eleven, twelve. thirteen, fourteen, fifteen. sixteen. 17 18 19 seventeen, eighteen, nineteen. In each of these, the 1 on the left represents a group of ten tilings, while the figure on the right expresses the units or single things additional, required to make up the number. Kemabk. — The words eleven and tivelve are supposed to be derived from the Saxon, meaning one left after ten, and two left after ten. The words thirteen, fourteen, etc., are contractions of three and ten, ftmr and te»,ete. ^^^j 16 HAY'S HIGHER ARITHMETIC, The next number above nineteen {nine and fen), is ten and ten^ or two groups of few, written 20, and called twenty. The next numbers are twenty-oney 21 ; twenty-two, 22 ; etc. , up to three tern, or thirty, 30 ; forty, 40 ; fifty, 50 ; sixty, 60 ; seventy, 70 ; eighty, 80 ; ninety, 90. The highest number that can be written with two figures is 99, called ninety-nine; that is, nine tens and nine units. The next higher number is 9 tens and ten, or ten tens, which is called one hundred, and written with three figures, 100; in which the two ciphers merely show that the unit on their left is neither a single thing, 1, nor a group of ten things, 10, but a group of ten tens, being a unit of a higher order than either of those already known. In like manner, 200, 300, etc., express two hundreds, three hundreds, and so on, up to ten hundreds, called a thousand, and written with four figures, 1000, being a unit of a still higher order. 43. The Order of a figure is the place it occupies in a number. From what has been said, it is clear that a figure in the 1st place, with no others to the right of it, expresses units or single things; but standing on the left of another figure, that is, in the 2d place, expresses groups of tens; and standing at the left of two figures, or in the 3d place, expresses tens of tens, or hundreds; and in the 4th place, expresses tens of hundreds or Hwusands, Hence, counting from the right hand. The order of Units is in the 1st place, 1 The order of Tens is in the 2d place, 10 The order of Hundreds is in the 3d place, 100 The order of ThotLsands is in the 4th place, 1000 By this arrangement, the same figure has differerd values according to the place, or order, in which it stands. Thus, 3 in the first place is 3 units; in the second place 3 tens, or thirty; in the third place 3 hundreds; and so on. NUMERATION AND NOTATION. 17 44i The word Units may be used in naming all the orders, as follows: Simple units are called Units of the Ist order. Tens " " Units of the 2d order. Hundreds " " Uniis of the Sd order. Thousands " " Units ^ the 4th order. etc. etc. 45. The following table shows the place and name of each order up to the fifteenth. Table op Orders. 15th. 14th. 13lh. 12th. 11th. 10th. 9th. 8th. 7th. 6th. 5th. 4th. 8d. 2d. Ist ac o 00 • • • • • • 00 ^3 • • • OD a • 00 G • a 03 • • OD c o ■ o • oo a o O • oo a o OQ S o 00 S C3 00 f^ •4H «4-4 p-^ «4H 9 O H .^4 o 00 O 00 O 00 ^3 00 O OQ a o n3 2 o 3 c ^ O oo S C 1 o 10 21 «iH 'V o TS TJ p o 'V 00 a c a pq S 00 a 1—^ C P W 00 a s p w oo C H 00 ■♦-» 46. For convenience in reading and writing numbers, orders are divided into groups of three each, and each group is called a period. The following table shows the grouping of 'the first fifteen orders into five periods : Table of Periods. 00 a o o S PQ CO C O O Z3 OQ G a w 6 5 4 00 00 -^ a 'S 3 2 1 ac 'V 0) C P w 00 OQ *J (K 6th Period. 4th Period. H. A. 2. 9 8 7 3d Period. 6 5 4 2d Period. 3 2 1 Ist Period. 18 HA Y'S HIGHER ARITHMETIC. 47. It will be observed that each period is composed of units, tens, and hundreds of the same denomination. 48. List of the Periods, according to the common or French method of Numeration. First Period, Units. Second " Thousands. Third " Millions. Fourth " Billions. Fifth " Trillions. Sixth Period, Quadrillions. Seventh " Quiniillions. Eighth '' Sextillions. Ninth " Septillions. Tenth " Ckitillions. The next twelve periods are, Nonillions, Decillions, Undecillions, Duodecillions, Tredecillions, Quatuordecillions, Quindecillions, Sexdecillions, Septendecillions, Octodecillions, Novendecillions, Vigintillions. Principles. — 1. Ten units of any order always make one of ike next higher order, 2. Removing a significant figure one place to the left increases its value tenfold; one pkice to the right, decreases iti value ten- fold, 3. Vojcant orders in a number are filled with ciphers. Problem. — Express in words the number which is repre- sented by 608921045. Solution. — The number, as divided into periods, is 608*921*045; and is read six hundred and eight million nine hundred and twenty-one thousand and forty-five. Bule for Numeration. — 1. Begin at the right , and point the number into periods of three figures each. 2. Commence at the left, and read in succession each period with its name. Remark. — Numbers may also be read by merely naming each figure wiih the name of the place in which it stands. This method, however, is rarely used except in teaching beginners. Thus, the numbers expressed by the figures 205, may be read two hundred and five, or two hundreds no tens and five units. NUMERATION AND NOTATION. 19 Examples in Numeration. 7 4053 204026 4300201 40 7009 500050 29347283 85 12345 730003 45004024 278 70500 1375482 343827544 1345 165247 6030564 830070320 832045682327825000000321 8007006005004003002001000000 60030020090080070050060030070 504630!209J02'80OV0^24d703l2505O7 Problem. — Express in figures the number four million twenty thousand three, hundred and seven. 4020307. SoiiUTiON. — Write 4 in miUyms period; place a dot after it to separate it from the next period : then write 20 in thoiisands period ; pXace another dot: then write 307 in units period. This gives 4*20'307. As there are but <wo places in the thousands period, a cipher must he put before 20 to complete its orders, and the number correcHy written, is 4020307. Note. — Every period, except the highest, must have three figures ; and if any period is not mentioned in the given number, supply its place with three ciphers. Bule for Notation. — Begin at (lie lefty and write each period in its proper place— filling ilie vacant orders with ciphers. Proof. — ^Apply to the number, as written, the rule for Numeration, and see if it agrees with the number given. 1. Seventy-five. 2. One hundred and thirty-four. 3. Two hundred and four. 4. Three hundred and seventy. 5. One thousand two hundred and thirty-four. Examples m Notation. 6. Nine thousand and seven. 7. Forty thousand five hundred and sixty-three. 8. Ninety thousand and nine. 9. Two hundred and seven thou- sand four hundred and one. 20 BAY'S mo HER ARITHMETIC. 10. Six hundred and forty thou- sand and forty. 11. Seven hundred thousand and seven. 12. One million four hundred and twenty-one thousand six hundred and eighty-five. 13. Seven million and seventy. 14. Ten million one hundred thousand and ten. 15. Sixty million seven hundred and five thousand. 16. Eight hundred and seven million forty thousand and thirty-one. 17. Two billion and twenty mill- ion. 18. Nineteen quadrillion twenty trillion and five hundred billion. 19. Ten quadrillion four hun- dred and three trillion ninety billion and six hun- dred million. 20. Eighty octillion sixty sex- tillion three hundred and twenty-five quintillion and thirty-three billion. 21. Nine hundred decillion sev- enty nonillion six octillion forty septillion fifty quad- rillion two hundred and four trillion ten million forty thousand and sixty. English Method of Numeration. 49. In the English Method of Numeration six figures make a period. The first period is wnife, the second millions^ the third biUions, the fourth triUionSy etc. The following table illustrates this method : § H ->w. OQ H OD ^ .2 .-/^ m CD H OD G CO S3 O -a H oo u C CO C c p CO a .13 00 a CO a 00 O -S3 H CO D CO -»J r; "fh OQ w H p tn H 00 CO S O -S3 H 00 00 00 -M § C HP 00 0) w H OD a 00 a a 00 O 00 P 00 ■»- 432109 876543 210987 654321 By this system the twelve figures at the right are read, two hundred and ten thousand nine hundred and eighty-seven NUMERATION AND NOTATION 21 million six hundred and fifty-four thousand three hundred and twenty-one. By the French method they would be read, two hundred and ten billion nine hundred and eighty-seven million six hundred and fifty-four thousand three hundred and twenty-one. Roman Notation. 50. In the Roman Notation, numbers are represented by seven letters. The letter I represents one; V, five; X, ten; L, fifty; C, (yrve hundred; D, five hundred; and M, one thou- sand. The other numbers are represented according to the following principles : 1st. Every time a letter is repeated, its value is repeated. Thus, n denotes tujo; XX denotes tu^enty. 2d. Where a letter of less value is placed before one of greater value, the less is taken from the greater; thus, TV denotes fi)ur, 3d. Where a letter of less value is placed after one of greater value, the less is added to the greater; thus, XI denotes eleven, 4th. Where a letter of less value stands between two letters of greater value, it is taken from the following letter, not added to the preceding; thus, XIV denotes fourteen, not mieen, 5th. A bar [ — 1 placed over a letter increases its value a thousand times. Thus, V denotes five thoumnd; M denotes one million. » Roman Table. I One. II Two. Ill Three. IV Four. V . ^ Five. VI Six. IX Nine. X Ten. XI Eleven. XIV Fourteen. XV Fifteen. XVI Sixteen. XIX Nineteen. XX Twenty. XXI Twenty-one. XXX Thirty. 22 RAY'S HIGHER ARITHMETIC XL L LX XC C cccc D .OMAN TaBT.E. (Continued.) Forty. DC .... Six hundred. Fifty. DCC .... Seven hundred Sixty. DCCC .... Eight hundred Ninety. DCCCC . . . Nine hundred. One hundred. M .... One thousand. Four hundred. MM .... Two thous,and. Five hundred. MDCCOLXXXI 1881. 1. Definitions. Topical Outline. Numeration and Notation. 2. Methods. 1. Arabic. 1. Characters..... 1. Names. 2. Values. {i 1. Significant Zero. Simple. 2. Terms. f 1. Order. I 2. Periods, i ^' ^"^^^^ 2. Roman... g ^ oS 3. Principles. 4. Rules. 1. Names. LocaL Frencl I 2. English. 2. Values... 1. Alone. 2. Repeated. 3. Preceding. 4. Following. 5. Between. 6. Line Above. 3. Ordinary Language. ni. ADDITION. 51. Addition is the process of uniting two or more like numbers into one equivalent number. Sum or Amoimt is the result of Addition. 62. Since a number is a collection of units of the same kind, two or more numbers can he united into one sum, only when their units are of the same kind. Two apples and 3 apples are 5 apples; but 2 apples and 3 peaches can not be united into one number, either of apples or of peaches. Nevertheless, numbers of different names may be added together, if they can be brought under a common denomination. Principles.— 1. Only like numbers can be added. 2. The sum is equal to dU the units of aU Hie parts. 3. The sum is the same in kind as the numbers added. 4. Units of the same order, and only such, can he added direcUy. 5. The sum is the same in whatever suecession the numbers are added. Bemark. — Like numbers, similar numbers, and numbers of the same kind are those having the same unit. Problem. —What is the sum of 639, 82, and 543? Solution. — ^Writing the numbers as in the margin, operation. Bay, 3, 5, 14 units, which are 1 ten and 4 units. Write 63 9 the 4 units beneath, and carry the 1 ten to the next 82 column. Then 1,,5, 13, 16 tens, which are 6 tens to be 5 43 written beneath, and 1 hundred to be carried to the 12 64 next column. Lastly, 1, 6, 12 hundreds, which is set beneath, there being no other columns to carry to or add. Demonstration. — 1. Figures of the same order are written in the (23) 24 RAY'S HIGHER ARITHMETIC. same column for ccnveniencej since none but units of the same name can be added. (Art. 62.) 2. Commence at the right to add, so that when the sum of any column is greater than nine, the tens may be carried to the next column, and, thereby, units of the same name added together. 3. Carry one for every ten, since ten units of each order make one unit of the order next higher. Rule for Adding Simple Numbers. — 1. Wriie the num- bers to be addedy so that figures of the same order mmj stand in the same column, and draw a line directly beneath. 2. Begin at Hie right and add each column separately^ writing the units under the column added, and carrying the tens, if any, tx) Hie next column. At the last column ivrite the last whole amount. Methods of Proof. — 1. Add the figures downward instead of upward; or 2. Separate the numbers into two or more divisions; find the sum of the numbers in each division, and then add the several sums together; or 3. Commence at the left; add each column separately; place each sum under that previously obtained, but extend- ing one figure further to the right, and then add them together. In each of these methods the result should be the same as when the numbers are added upward. Note. — For proof by casting out the 9's, see Art. 105. Examples for Practice. Find the sum, 1. Of 76767 ; 7654 ; 50121 ; 775. Ans. 135317. 2. Of 97674; 686; 7676; 9017. Ans. 115053. 3. Of 971; 7430; 97476; 76734. Ans. 182611. 4. Of 999; 3400; 73; 47; 452; 11000; 193; 97; 9903; 42 ; and 5100. Ans. 31306. 5. Of four hundred and three ; 5025 ; sixty thousand and seven; eighty-seven thousand; two thousand and ninety; and 100. 154625. 6. Of 20050; three hundred and seventy thousand two hundred ; four million and five ; two million ninety thousand seven hundred and eighty; one hundred thousand and seventy; 98002; seven million five thousand and one; and 70070. 13754178. 7. Of 609505 ; 90070 ; 90300420 ; 9890655 ; 789 ; 37599 ; 19962401; 5278; 2109350; 41236; 722; 8764; 29753; and 370247. 123456789. 8. Of two hundred thousand two hundred; three hundred million six thousand and thirty; seventy million seventy thousand and seventy ; nine hundred and four million nine thousand and forty; eighty thousand; ninety mUlion nine thousand; six hundred thousand and sixty; five thousand seven hundred. 1364980100. In each of the 7 following examples, find the sum of the consecutive numbers from A to B, including these numbers : A. B. 9. 119 131. .4*18. 1625. 10. 987 1001. Am. 14910. 11. 3267 3281. Arts. 49110. 12. 4197 4211. Am. 63060. 13. 5397 5416. Am. 108130. 14. 7815 7831. Am. 132991. 15. 31989 32028. Am. 1280340. 16. Paid for cofiee, $375; for tea, $280; for sugar, $564; for molasses, $119; and for spices, $75: what did the whole amount to? $1413. 17. I bought three pieces of cloth: the first cost $87; the second, $25 more than the first; and the third, $47 more than the second: what did all cost? $358. 18. A man bought three bales of cotton. The first cost H. A. 8. 26 ^A Y'S HIGHER ARITHMETIC $325; the second cost $16 more than the first; and the third, as much as both the others: what sum was paid for the three bales? $1332. 19. A has $75; B has $19 more than A; C has as much as A and B, and $23 more ; and D has as much as A, B, and C together: what sum do they all possess? $722. • _ Suggestions. — ^Two things are of the greatest importance in arithmetical operations, — absolute accuracy and rapidity. The figures should always be plain and legible. Frequent exercises in adding long columns of figures are recommended ; also, practice in grouping numbers at sight into iem and iwerUies is a useful exercise. Accountants usually resort to artifices in Addition to save time and extra lahor, such as writing the number to be carried in small figures beneath the column to which it belongs, also writing the whole amount of each column separately, ete. Topical Outline. Addition. 1. Definitions. 2. Sign. 8. Principles. ' 1. Writing the Numbers. 2. Drawing Line Beneath. 3. Addiug, Reducing, etc 4. Operation.. 5. Rule. 6. Methods of Proof. IV. SUBTEAOTIOR 53. Subtraction is the process of finding the difierence between two numbers of the same kind. The larger number is the Minuend; the less, the Subtra- hend; the number left, the Difference or Remainder, Minuend means to be diminished; subtrahend, to be sub- tracted, 64. Subtraction is the reverse of Addition, and since none but numbers of the same kind can be added together (Art. 62), it follows, therefore, that a number can be sub- tracted only from another of the same kind : 2 cents can not be taken from 5 apples^ nor 3 cows from 8 horses. Principles. — 1. Tlie minuend and subtrahend must be of the same kind, 2. The difference is the same in hind as the minuend and subtrahend, 3. The difference equals the minuend minus the subtrahend. 4. The minuend equals the difference plus the subtrahend, 5. The subtrahend equals the minuend minus the difference. Problem. — From 827 dollars take 534 dollars. OPERATION. Solution. — After writing figures of the same $8 2 7 order in the same column, say, 4 units from 7 units 534 leave 3 units. Then, as 3 tens can not be taken $293 Bern, from 2 tens, add 10 tens to the 2 tens, which make 12 tens, and 3 tens from 12 tens leave 9 tens. To compensate for the 10 tens added to the 2 tens, add one hundred (10 tens) to the 5 hundreds, and say, 6 hundreds from 8 hundreds leaves 2 hundreds; and the whole remainder is 2 hundreds 9 tens and 3 units, or 293. DSMONSTRATIOK. — 1. 6inoe a number can be subtracted only from 28 BAY'S HIOHEE ARITHMETIC. another of the Rame kind (Art. 64), figures of the same name are placed in the same column to be convenient to each other, the less number being placed below as a matter of custom. 2. Commence at the right to subtract,, so that if any figure is greater than the one above it, the upper may be increased by 10, and the next in the subtrahend increased by 1, or, as some prefer, the next in the minuend decreased by 1. Bule for Subtracting Simple Numbers. — 1. TFrife ihe less number under the greater, placing units urtder units, tens under tens, etc., and draw a line directly beneath. 2. Begin at the rigid, subtract each figure from iJie one above it, placing the remainder beneatli. 3. If any figure exceeds the one above it, add ten to the upper, subtract the lower from the sum, increa^ by 1 the units if the next order in Hie subtraJiend, and proceed as before. Proof. — Add the remainder to the less number. If the work be correct, the sum will be equal to the greater. Note. — For proof by casting out the 9's, see Art. 105. Examples for Practice. 1. From 30020037 take 50009. OPERATION. Eemark. — When there are no figures in 30020037 the lower number to correspond with those 50009 in the upper, consider the vacant places 29 9 700 2 8 .Re?ii. occupied by zeros. 2. From 79685 take 30253. - Ans. 49432. 3. From 1145906 take 39876. Ans. 1106030. 4. From 2900000 take 777888. Am. 2122112. 5. From 71086540 take 64179730. Ans. 6906810. 6. From 101067800 take 100259063. Ans. 808737. 7. How many years from the discovery of America in 1492, to the Declaration in 1776 ? 284 years. 8. A farm that coat $7253, was sold at a loss of $395 ; for how much was it sold? $6858. SUBTRACTION. 29 9. The difference of two numbers is 19034, and the greater is 75421 : what is the less ? 66387. 10. How many times can the number 285 be subtracted from 1425? 5 times. 11. Which is the nearer number to 920736; 18l6045 or 25427? Neither. Why? 12. From a tract of land containing 10000 acres, the owner sold to A 4750 acres; and to B 875 acres less than A : how many acres had he left ? 1375 acres. 13. A, B, C, D, are 4 places in order in a straight line. From A to D is 1463 miles ; from A to C, 728 miles ; and from B to D, 1317 miles. How fer is it from A to B, from B toC, and from C to D? A to B, 146 miles ; B to C, 582 miles ; C to D, 735 miles. BUSINESS TERMS AND EXPLANATIONS. 55. Book-keeping is the science and art of recording business transactions. 56. Business records are called Accounts, and are kept in bo<^ called Account Books. The books mostly used are the Day-booh and Ledger. In the Day-book are recorded the daily transactions in business, and the Ledger is used to classify and arrange the results of all transactions under distinct heads. 67. Each account has two sides : Dr. — Debits, and Cr. — Credits. Sums a person oioes are his Debits; sums owing to him are his Credits. The difference between the sum of the Debits and the sum of the Credits, is called the Balance. Debits are preceded by **To," and Credits by '* By." 58. Finding the difference between the sum of the Debits and the sum of the Credits, and writing it under the less side as Balance, is called Balancing. 30 HAY'S HIGHER ARITHMETia Practical Exercises. Dr. JAMES CRAIG. Cr. 1878. 1878. July 4 .. 6 » 10 H 25 » 81 To Merchandise . . „ Interest .... „ Sundries . . . ,, Merchandise . . Ditto . . . 560 50 24 90 870 60 320 10 125 40 July 5 M 11 „ 26 „ 31 By Cash „ BiUs Payable . . „ Sundries . . . „ Cash " Balance . . . • 550 50 890 70 310 SO 100 00 49 50 1878. 1901 50 190150 Aug. 1 To Balance .... 49 50 Balance the following account: Dr. THOMAS BALDWIN. Cr. 1879. 1879. Jan. 3 » 16 „ 25 ,. 31 To Merchandise . . „ Sundries . . . „ Cash ..... „ Merchandise . . 810 30 580 20 381 25 60 75 Jan. 7 „ 20 „ 31 By Sundries . • . „ Cash „ Merchandise . . ,, Balance .... 1 1000 00 300 00 225 20 1879. Feb. 1 To Balance .... Topical Outline. Subtraction. 1. Definition. 2. Terms 3. Sign. 4. Principles. 5. Operation 6. Rule. 7. Proof. 8. Applications. 1. Minuend. 2. Subtrahend. . 3. Diflference or Remainder. 1. Writing the Numbers. 2. Drawing Line Beneath. 3. Subtracting and Writing Difference. Y. MULTIPLIOATION. 69. 1. Multiplication is taking one number as many times as there are units in another ; or, 2. Multiplication is a short method of adding numbers that are equal. 60. The number to be taken, is called the Multiplicand; the other number, the Multiplier; and the result obtained, the Product. The Multiplicand and Multiplier are together called Factors (makers), because they make the Product. PROBLEM.-r-How many trees in 3 rows, each containing 42 trees? . Solution. — Since 3 rows contain 3 times oPEBATioir. as many trees as one row, take 42 three First row, 42 trees times. This may be done by writing 42 Second row, 4 2 trees, three times, and then adding. This gives Third row, 42 trees. 126 trees for the whole number of trees. 126 trees. Instead,* however, of writing 42 three times, write it <mee; then placing under it 42 trees, the figure 3, the number of tiirus it is to be 3 taken, say, 3 times 2 are 6, and 3 times 4 126 trees, are 12. This process is MvUipHcatioTi, Principles. — 1. The multiplicand may be eiOier concrete or abstra4±, 2. The multiplier mu^st always be an abstract number. 3. The product is the same in kind as the multiplicand. 4. The product is the same, tvhichever factor is taken as the multiplier. 5. Tlie partial products are the same in kind as the midti- plicand. 6. The sum of the partial products is equal to the total product. SA Y'S HIGHER ASITHMETIU. MOLTTPLICATION TaBLE. 61. Multiplication is divided into two cases: 1, When the m-ultiptter does wit exceed 12, 2. When the midlipHer exeeedt 12. CASE I. 62. When the multiplier does not exceed 13. Problem. — At the rat« of 53 miles an hour, how far will a railroad car run in four hours? Solution. — Here saj, 4 times 3 (units) are 12 operatioh, (units); write the 2 in units' place, and carry the 63 miles. 1 (ten); then, 4 times 5 are 20, and 1 carried 4 makes 21 (tens), and tlie work is complete. 212 miles. DehOnstratiOM. — The multiplier being written under the mul- tiplicand for convenience, begin with units, so that if the product should contain tens, they may be carried to the ten«; and ho on for each successive order. MULTIPLICATION. 88 Since every figure of the multiplicand is multiplied, therefore, the whole multiplicand is multiplied. Bule. — 1. Write the mvUiplicandy and place the multiplier under it, so thai units of tJie same order shall stand in the same column, and draw a line beneath, 2. Begin with units; multiply each figure <^ the midtiplicand by the multiplier y carrying as in Addition, Proof. — Separate the multiplier into any two parts; multiply by these separately. The sum of the products must be equal to the first product. Examples for Practice. 1. 195X3. Am. 585. 2. 3823X4. Ans. 15292. 3. 8765 X 5. Ans. 43825. 4. 98374 X 6. Ans. 590244. 5. 64382 X 7. Am. 450674. 6. 58765X8. Am, 470120. 7. 837941 X 9. Am. 7541469. 8. 645703 X 10. Am. 6457030. 9. 407649 X H. Am. 4484139. 10. If 4 men can perform a certain piece of work in 15 days, how long will it require 1 man? Solution. — One man must work four times as long as four men. 4 X 15 days = 60 days. 11. How many pages in a half-dozen books, each con- taining 336 pages? 2016 pages. 12. How far can an ocean steamer travel in a week, at the rate of 245 miles a day? 1715 miles. 13. What is the yearly expense of a cotton-mill, if $32053 are paid out every month ? $384^36. 14. A receives from his business an average of $45 a day. He pays three clerks $3; three, $9; and three, $12 a week ; other expenses amount to $4 a day ; what are his profits for one week? $174. 84 JfiA yS HIGHEB ABITHMETia CASE II. 63. When the multiplier exceeds 12. Problem.— Multiply 246 by 235. Solution. — First multiply hy 6 operation. (units), and place the first figure of 2 4 6 the product, 1230, under the 5 (units). 235 Then multiply by 3 (tens), and place 12 3 product by 5 the first figure of the product, 738, 738 product by 3 under the 3 (tens). Lastly, multiply 49 2 product by 2 by 2 (hundreds), and place the first STSTo product by 235 figure of the product, 492, under the 2 (hundreds). Then add these several products for the entire product DEMONS-rtiATiON.— The of the first product, 1230, is unU» (Art. 62). The 8 of the second product, 738, is tens, because 3 (tens) times 6 = 6 times 3 (tens) = 18 (tens) ; giving 8 (tens) to be written in the tens' column. The 2 of the third product, 492, is hundreds, because 2 (hundreds) times 6 = 6 times 2 (hundreds) = 12 (hundreds), giving 2 (hundreds) to be written in the hundreds' column. The right- hand figure of each product being in its proper column, the other figures will fall in their proper columns; and each line being the product of the multiplicand by a part of the multiplier, their sum will be the product by aU the parts or the whole of the multiplier. Rule. — 1. Write the mvUiplier under the multiplicand, placing fijurea (^ the tame order in the same cdumuy and draw a line beneath. 2. Multiply each figure of the mvUiplieand by each figure of Hie mxdtiplier successively ; first by the units^ figure, then by the ten^ figure, etc.; phdng the right-hand fi^pire of each product under that fi>gure of the multiplier whidi produces it, then draw a line beneath, 3. Add the several partial products together; their sum will be the rexpdred produ^i. Methods of Proof.— 1. Multiply the multiplier by the multiplicand; this product must be the same as the first product. MULTIPLICATION. 85 2. The same as when the multiplier does not exceed 12. Note. — For proof by casting out the 9*8, see Art, 106. Bemark. — Although it is custom- ary to use the figures of the multi- plier in regular order beginning with units, it will give the same product to use them in any order, observing that the right-hand figure of ea4:h parlicU product must he pku^ under the figure of the multiplier which produces it. OPERATION. 246 235 7 38 product by 30 492 product by 200 1230 product by 6 67 810 product by 235 Examples for Practice. 1. 7198X216. 2. 8862 X 189. 3. 7575X7575. 4. 15607X3094. 5. 93186X4455. 6. 135790X24680. 7. 3523725 X 2583. 8. 4687319 X 1987. 9. 9264397X9584. 10. 9507340X7071. 11. 1644405X7749. 12. 1389294X8900. 13. 2778588X9867. 14. 204265X562402. Am. 1554768. Am. 1674918. Am. 57380625. Am. 48288058. Am. 415143630. Am. 3351297200. Am. 9101781675. Am. 9313702853. Am. 88789980848. Am. 67226401140. Am. 12742494345. Am. 12364716600. Am. 27416327796. Am. 114879044530. Practical Problems. 1. In a mile are 63360 inches: how many inches are there in the circumference of the earth at the equator if the distance be 25000 miles ? 1584000000 inches. 2. The flow of the Mississippi at Memphis is about 434000 cubic feet a second: required the weight of water passing that point in one day of 86400 seconds, if a cubic foot of water weigh 62 pounds? 2324851200000 pounds. 36 BAY'S HIGHER ARITHMETIC. 3. John Sexton sold 25625 bushels of wheat, at $1.20 a bushel, and received in payment 320 acres of land, valued at $50 an acre ; 60 head of horses, valued at $65 a head ; 10 town lots, worth $150 each ; and the remainder in money : how much money did he receive ? $9350, 4. K light comes from the sun to the earth in 495 seconds, what is the distance from the earth to the sun, light moving 192500 miles a second ? 95287500 miles. 5. If 3702754400 cubic feet of solid matter is deposited in the Gulf of Mexico by the Mississippi every year, what is the deposit for 6000 years? 22216526400000 cu. ft. 6. The area of Missouri is 65350 square miles: how many acres are there in the State, allowing 640 acr^ to each square mile? 41824000 acres. 7. In the United States, at the close of 1878, there were 81841 miles of railroad : if the average cost of building be $50000 a mile, what has been the total cost of building the railroads in this country? $4092050000. 8. The number of pounds of tobacco produced in this country in 1870 was 260000000. If this were manufact- ured into plugs one inch wide and six inches long, and four plugs weigh a pound, what would be the length in inches of the entire crop ? 6240000000 inches. BUSINESS TERMS AND EXPLANATIONS. 64. A Bill is an account of goods sold or delivered, services rendered, or work done. Usually the price or value is annexed to each article, and the date of purchase given. It is customary to write the total amount off to the rigtt, and not directly under the column of amounts added. 66. A Receipt is a written acknowledgment of pay- ment. The common form consists in signing the . name after the, words ** Received Payment" written at the foot of the bill. .. . . MVLTIPLICA TION. m \, Joseph Allen bought of Seth Ward, at Springfield, HI, Jan. 2, 1879, 30 barrels of flour, at $3.60 a barrel; 48 barrels of mess pork, at $16.25 a barrel; 16 boxes of candles, at $3.50 a box; 23 barrels of molasses, at $28.75 a barrel; and 64 sacks of coffee, at $47.50 a sack. Place the purchases in bill form. 1879. Solution. Joseph Allen, Springfield, III., Jan. 2, 1879. Bought of Seth Ward. Jan. 2 >i 2 i> 2 j»" 2 >• 2 To 30 bl. flour. @ S 8.60 a bl. It 48 i> mess pork, „ .16.25 ft „ 16 boxes candles, „ 3.50 „ box „ 23 bl. molasses, ,, 28.75 „ bl. „ 64 sacks coffee, „ 47.50 „ sack 1 108 00 780 00, 56 00 661 25 8040 00 ^15 ■ 25 2. At St. Louis, March 1, 1879, Chester Snyder bought of Thomas Glenn, 4 lb. of tea, at 40 ct. ; 21 lb. of butter, at 21 ct.; 58 lb. of bacon, at 13 ct.; 16 lb. of lard, at 9 ct.; 30 lb. of cheese, at 12 ct.; 4 lb. of raisins, at 20 ct.; and 9 doz. of eggs, at 15 ct. . Place these purchases in the form of a receipted bill? $20.74. 66. A Statement of Account is a written form reur dered to a customer, showing his debits and credits as they appear on the books. The following is an example : 1880. John SMrm, Cincinnati, Feb. 2, 1880. 171 Account with Van Antwerp, Bragg & Co. Jau. 2 10 29 81. To 525 McGuffey's Revised First Readers, @ 16c. „ 50 Ray's New Higher Arithmetics, „ 75c. Cr. By Cash , : ,, Merchandise 84 37 50 ■75 »» 20 12 121 32 50 ■75 - . 988 JSf 38 BAY'S HIGHER ARITHMETIC. 3. James Wilson & Co. bought of the Alleghany Coal Co., March 2, 1880, five hundred tons of coal, at $2.75 a ton, and sold the same Company during the month, as follows: March 3d, 14 barrels of flour, at $6.55 a barrel; March 10th, 6123 pounds of sugar, at 8c. a pound; they also paid them on account, on March 15th, cash, $687.50. Make out a statement of account in behalf of the Alleghany Coal Co. under date of April 1, 1880. $105.96. CJONTRACnONS IN MULTIPLICATION. CASE I. 67. When the multiplier is a composite number. A Composite Number is the product of two or more whole numbers, each greater than 1, called its facUyrs. Thus, 10 is a composite number, whose factors are 2 and 5; and 30 is one whose factors are 2, 3, and 5. Problem. — At 7 cents a piece, what will 6 melons cost? Analysis. — Three times 2 times operation. are 6 times. Hence, it is the same 7 cents, cost of 1 melon, to take 2 times 7, and then take 2 this product 3 times, as to take 6 14 cents, cost of 2 melons, times 7. The same may be shown 3 of any other composite number. J^ cents, cost of 6 melons. Bule. — Separate the multiplier into two or more factors. Multiply first by one of the factorSy then this product by another factor, and so on till each factor has been used as a muUipiier. The last prodiLct wiU be the result required. Examples for Practice. 1. At the rate of 37 miles a day, how fer will a man walk in 28 days ? 1036 miles. MULTIPLICATION. 89 2. Sound moves about 1130 feet per second: how fer will it move in 54 seconds? 61020 feet. 3. If an engine travel at an average speed of 25 miles an hour, how far can it travel in a week, or 168 hours? 4200 miles. CASE II. 68. When the multiplier is 1 with ciphers annexed, as 10, 100, 1000, etc. Demonstration. — By the principles of Notation (Art 48), placing one cipher on the right of a number, changes the units into tens, the tens into hundreds, and so on, and, therefore, mtiUiplies the number by 10. Annexing tvoo ciphers to a number changes the units into hun- dreds, the tens into thousands, and so on, and, therefore, multiplies the number by 100. Annexing three ciphers multiplies the number by 1000, etc. Rule. — Anr\£x to the mtdtiplicand as many ciphers as there are in tfie mvltiplier; the resvlt will he the required product. Examples for Practice. 1. Multiply 743 by 10. Ans. 7430. 2. Multiply 375 by 100. Ans. 37500. 3. Multiply 207 by 1000. Ans. 207000. CASE III. 69. When ciphers are on the right in one or both factors. Problem.— Find the product of 5400 by 130. OPERATION. 5400 Solution.— Find the product of 54 by 13, 130 and then annex three ciphers ; that is, as 16 2 many as there are on the right in both the 5J factors. 702000 40 BAY'S JSIGHEM ARITHMETIC. Analysis. — Since 13 times 54 = 702, it follows that 13 times 54 hundreds (5400) = 702 hundreds (70200) ; and 130 times 5400= 10 times 13 times 5400 = 10 times 70200 = 702000. Bule. — Multiply as if there wei'C no ciphers on the right in the numbers; then annex to the product as many ciphers as there are on tJie right in boHi the factors. Examples for Practice. 1. 15460 X 3200. Ans. 494720(K). 2. 30700 X 5904000. Ans. 181252800000. CASE IV. 70. When the multiplier is a little less or a little greater than 10, 100, 1000, etc. Problem.— Multiply 3046 by 997. Analysis. — Since 997 is equal to 1000 operation. diminished by 3, to multiply by it is the same • 3 4 6 as to multiply by 1000 (that is, to annex 3 99 7 ciphers) and by 3, and take the difference of 3046000 the products ; and the same can be shown in 913 8 any similar case. 3036862 Note. — Where the number is a little greater than 10, 100, 1000, etc., the two products must be added, Bnle. — Annex to the mtdtiplicand as many ciphers as there are figures m the multiplier; multiply the multiplicand by the difference between Hie multiplier and 100, 1000, etc, and add or subtract the smaller result as the multiplier is greater or less than 100, 1000, etc. Examples for Practice. 1. 7023 X 99. Ans. 695277. 2. 16642 X 996. Ans. 16575432. 3. 372051 X 1002. Am. 372795102. MULTIPLICATION. 41 CASE V. 71. When one part taken as units, in the multi- plier, is a flBkctor of another part so taken. Problem.— Multiply 387295 by 216324. Solution. — Commence with the 3 of operation. the multiplier, and obtain the first partial 3 8 7 2 9 5 product, 1161885 ; then multiply this prod- 216324 uct by 8, which gives the product of the 11618 8 5 multiplicand by 24 at once (since 8 times 9 2 9 5 8 3 times any number make 24 times it). 83655720 Set the right-hand figure under the right- 83781 203580 hand hgure 4 of the multiplier in use. Multiply the second partial product by 9, which gives the product of the multiplicand by 216 (since 9 times 24 times a number make 216 times that number). Set the right-hand figure of this partial product under the 6 of the multiplicand ; and, finally, add to obtain the total product. Rule. — 1. Multiply the mvUiplieand by some figure or fibres cf the mvUiplierf which are a factor of one or more parts of the mvUiplier. 2. Multiply this partial product by a factor of some other figure or figures of tJie multiplier, and write Uie right-hand figure thus obtained under the tnght-hand fi^re of the mtdtiplier thus used, 3. Continue thus untU the entire multiplier is used, and then odd the partial products. Examples for Practice. 1. 38057 X 48618. Ans, 185025522B. 2. 267388 X 14982. Ans, 4006007016. 3. 481063 X 63721. Ans. 30653815423. 4. 66917 X 849612. Ans. 56853486204. 5. 102735 X 273162. Ans. 28063298070. 6. 536712 X 729981. Ans. 391789562472. H. A. 4. 42 RAY'S HIGHER ARITHMETIC Topical Outline. 1. Definitions. 2. Tenniu. 8. Sign. 4. Principles. 5. Opeiation. 6. Rule. 7. Proof. 8. Applications. 9. Contractions. Multiplication. 1. Multiplicand. 2. Multiplier. 3. Partial Product 4. Product 1. Writing Numbers. 2. Drawing Line Beneath. %. Finding Partial Products. 4. Drawing a Line Beneath Partial Products. 5. Adding the Partial Products. YL DIYISIOK • 72. 1. Division is the process of finding how many times one number is contained in another ; or, 2. Division is a short method of making several sub- tractions of the same number. 3. Division is also an operation in which are given the product of two &ctors, and one of the factors, to find the other factor. 73. The product is the Dividend; the given factor is the Divisor; and the required factor is the Quotient. The Remainder is the number which is sometimes left after dividing. Note. — Dividend signifies to be divided. Quotient is derived from the Latin word quoties, which signifies haw often. Problem. — ^How many times is 24 cents contained in 73 cents? Solution. — Twenty-four cents operation. from 73 cents leaves 49 cents; 24 73 cents, cents from 49 cents leaves 25 cents ; 24 24 cents from 25 cents leaves 1 cent. 49 cents remaining. Here, 24 cents is taken 3 times 24 from {out of) 73 cents, and 1 cent ^ cents remaining, remains ; hence, 24 cents is eontatTied 24 in 73 cents 3 times, with a remainder "J ^^^ remaining, of 1 cent. 74. The divisor and quotient in Division, correspond to the factors in Multiplication, and the dividend corresponds to the product. Thus : Factors / 5 X 3 = 1 5 | p^^^^^,^ I 3X5 = 15 i Dividend I * ' > Divisors and Quotients. ll5-!-3 = 5i (43) 44 BAY'S HIGHER ARITHMETIC. 75, There are three methods of expressing division ; thus, 12-^3, J^, or 3)12. Each indicates that 12 is to , be divided by 3. Principles. — 1. Wlien. iJie dividend and divisor are . lUce numbers, the quotient is abstract. 2. When the divisor is an abstract number, the quotient is like the dividend, 3. The remainder is like tlie dividend. 4. Tlie dividend is equal to the produet of the quotient by the divisor, plus the remainder. , 76. Multiplication is a short method of making several additions of the same number ; Division is a short method of making several subtractions of the same number ; hence, Division is the reverse of Multiplication. 77. All problems in Division are divided into two classes : li To find the number of equal parts of a nujnber. 2. To divide a number into equal parts. 78. Two methods are employed in solving problems in Division : Long Division, when the work is written in full in solving the problem; and Short Division, when the result only is writteil, the work being performed in the mind. The following illustrates the methods : Problem. — Divide 820 by 5. LONG DIVISION. SHORT DIVISION. 5)820(164 Quotient. 5)820 i_ 164 Quotient. 3 2 tens. 30 Both operations are performed on the same 2 units. principle. In the first, the subtraction is writ- 20 ten ; in the second, it is performed mentally. DIVISION. 45 LONG DIVISION. Problem. — Divide $4225 equally among 13 men. Solution. — As 13 is not contained operation. in 4 (thousandR)i therefore, the quo- tient has no thousands. Next, take 42 Sc «^ J "c « J 2 5 ^^ p ^ a (hundreds) as a partial dividend; 13 Susi^S .eSs is contained in it 3 (hundreds) times ; 13)4225)326 after multiplying and subtracting, there 3 9 h undreds, are 3 hundreds left. Then bring down 3 2 2 tens, and 32 tens is the next partial 2 6 t ens, dividend. In this, 13 is contained 2 6 5 (tens) times, with a remainder of 6 tens. 65^ units. Lastly, bringing down the 5 units, 13 is contained in 65 (units) exactly 5 (units) times. The entire quotient is 3 hundreds 2 tens and 5 units. This may be further shown by separating the dividend into parts, each exactly divisible by 13, as follows: DIVISOR. DIVIDEND. QUOTIENT. 13)3900 + 260 + 65(300+204-5 3900 + 260 + 260 + 65 + 65 Biile for Long Dmsion. — 1. Draw curved lines on tke right and left of tke dividend, placing the divisor on the left. 2. Find how ofte^i the divisor is contained in the left-hand figure, or figures, of the dividend, and write the number in the quotient at the right of (he dividend, 3. Multiply the divisor by this quotient figure, and write the produ/^ under that part of the dividetul from which it was obtained, 4. Subtract this product from the figures above it ; to the re- mainder bring down the next figure of tJie dividend, and divide as before, until all the figures of tShC dividend are brought down. 46 RAY'S HIOHEB ARITHMETia b. If at any time after a figure is brought dovm, the number ihui formed is too smaU to contain the divisor, a cipher must be placed in the qv/otient, and anoUier figure brought down, after which divide as before, 6. ^ there is a final remainder after the last division, place the divisor under it and annex it to the quotient. Proof. — Multiply the Divisor by the Quotient, and to this product add the Remainder, if any ; the sum is equal to the Dividend when the work is correct. Notes. — 1. The product must never be greater than the partial dividend from which it is to be subtracted; if so, the quotient figure is too large, and must be diminished. 2. The remainder after each subtraction must be less than the divisor ; if not, the last quotient figure is too smaUy and must be increased. 3. The order of each quotient figure is the same as the lowest order in the partial dividend from which it was derived* Examples for Practice. 1. 1004835 : 33. 2. 5484888 : 67. 3. 4326422 : 961. 4. 1457924651 : 1204. 5. 65358547823 : 2789. 6. 33333333333 : 5299. 7. 245379633477 : 1263. 8. 555555555555 : 123456. 9. 555555555555 : 654321. Ans. 30449^ Ans, 81864 Aw5. 4502 Ans, 1210900^1 Ans, 23434402^Vf Ans. 6290495^V Ans. 194283161j4f| Ans, 4500028^^:^^ Ans. 849056f||||f In the following, multiply A by itself, also B by itself divide the difference of the products by the sum of A and B. Ans. 909. 4ns, 9158. Ans. 104672. A. B. 10. 2856 3765. 11. 33698 42856. 12. 47932 152604 A. B. 13. 4986 5369. 14. 3973 4308. 15. 23798 59635. 16. 47329 65931. DIVISION. 47 In the following, multiply A by itself, also B by itself: divide the difference of the products by the difference of A and B. Am. 10355. Am. 8281. Am. 83433. Am. 113260. 17. K 25 acres produce 1825 bushels of wheat, how much is that per acre ? 73 bushels. 18. How many times 1024 in 1048576 ? 1024 times. 19. How many sacks, each containing 55 pounds, can be filled with 2035 pounds of flour? 37 sacks. 20. How many pages in a book of 7359 lines, each page containing 37 lines? Id8|f pages. 21. In what time will a vat of 10878 gallons be filled, at the rate of 37 gallons an hour? 294 hours. 22. In what time will a vat of 3354 gallons be emptied, at the rate of 43 gallons an hour? 78 hours. 23. The product of two numbers is 212492745; one is 1035; what is the other? 205307. 24. What number multiplied by 109, with 98 added to the product, will give 106700? 978. SHORT DIVISION. Problem. — ^How often is 2 cents contained in 652 cents? Solution. — ^Two in 6 (hundreds) is contained 3 oferatiok. (hundreds) times; 2 in 5 (tens) is contained 2 2 )652 (tens) times, with a remainder of 1 (ten) ; lastly, 1 326 (ten) prefixed to 2 makes 12, and 2 in 12 (units) is contained 6 times, making the entire quotient 326. Remarks. — Commence at the left to divide, so that if there ia a Temainder it may be carried to the next lower order. 48 J^A Y'S HIOffEB ARITHMETIC. By the operation of the rule, the dividend is separated into parts corresponding to the different orders. Having found the number of times the divisor is contained in each of these parts, the sum of these must give the number of times the divisor is contained in the whole dividend. Analyze the preceding dividend thus : 652 = 600 + 40 + 12 2 in 600 is contained 300 times. 2 in 40 is contained 20 times. 2 in 1 2 is contained 6 times. Hence, 2 in 652 is contained 326 times. Rule for Short Division. — 1. Write the divisor on the left of the dividend with a curved line between them, and draw a line directly beneath the dividend. Begin at die left, divide Buecemvely each figure or figures of the dividend by the divisor , and set the quotient beneath. 2. Whenever a remainder occurs, prefix it to Uie figure in the next lower order, and divide as before, 3. If the figure, except the first, in any order does not oon- tain the divisor, place a cipher beneath it, prefix U to the figure in the next lower order, and divide as before. 4. If there is a remainder after dividing the last figure, place the divisor under it and annex it to the quxMent, Proof. — ^The same as in Long Division. Examples for Practice. 1. Divide 512653 by 5. Am. 102530f. 2. Divide 534959 by 7. Ans. 76422f 3. Divide 986028 by 8. Ans. 123253f 4. Divide 986974 by 11. Ans. 897241^. 5. At $6 a head, how many sheep can be bought for $222 ? 37 sheep, 6. At $5 a barrel, how many barrels of flour can be bought for $895? 179 barrels. DIVISION. 48 OONTBACTIONS IN DIVISION. CASE I. 79. When the divisor is a composite number. This ca«e presente no difficulty except when remainders occur. Problem. — ^Divide 217 by 15. Solution. — 15 = 3X5, hence 217^3 = 72 and 1 remainder; 72 -r- 5 =14 and 2 remainder. Dividing 217 by 3, the quotient is 72 ihreeiy and 1 unU remainder. Dividing by 5, the quotient is 14 (ffleen»)y and a remainder of 2 threes; hence the quotient is 14, and the tnia remainder is 2X^ + 1 = 7* Bule. — 1. Divide (he dividend by one factor of the divisor y cmd divide this qiwtiefni by another fadoTy and so on^ tiU each fouior has been used; the last quotient vM be the required result. 2. MvUiply each remainder by aU of the divisors preceding (he one which produced it. The sum of the products, plus the first remainder, toiU be the true remainder, Remark. — ^This rule is not much used. CASE II. 80. When the divisor is 1 with ciphers annexed. This case presents no difficulty. Proceed thus: Problem.— Divide 23543 by 100. OPERATION. SoLxmoN.—l 100 ) 235143 235^ Rule. — Out off as many figures in the dividend' as there are ciphers in the divisor; the figures cut off vnU be the remainder, «nd the other figure or figures ike quotient. H. A. 5. 50 BAY'S HIGHER ARITHMETIC. CASE III. 81. When ciphers are on the right of the divisor. Problem.— Divide 3846 by 400. Solution. — To divide by 400 is the operation. same as to divide by 100 and then by 4 4 100)38| 46 (Art. 79). Dividing by 100 gives 38, and - 9 Quotient, 46 remainder (Art. 80); then, dividing 200 + 46 = 246, Kem. by 4 gives 9, and 2 remainder: the true remainder is 2 X 100 + 46 = 246 (Art. 79). Bule. — 1. Out off the ciphers at the right of the dm8(yry and as many figures from the right of the dividend. 2. Divide the refmaining part of the dividend by the remain- ing part of the divisor. 3. Annex to the remainder the fibres cut off, and thus obtain the true remmnder. AKITHMEnCAL SIGNS. 82. If a number be multiplied, it is simply repeated as many times as there are units in the multiplier; if a num- ber be divided, it is simply decreased by the divisor as many times as there are units in the quotient. It is thus evident, that Addition and Subtraction are the fundamental concep- tions in all the operations of Arithmetic; and, hence, all numbers may be classified as follows : 1. Numbers to be added; or, positive numbers. 2. Numbers to be subtracted; or, negative numbers. 83. Positive numbers are distinguished by the sign +» negative numbers by the sign — ; thus, + 8 is a positive 8, and — 8 a negative 8. Eemark. — When a number is preceded by no sign, as, for example, the number 4 in the first of the following exercises, it is to be considered positive. ARITHMETICAL SIGN& 61 84. The signs X and -7- do not show whether their remtUs are to be added or to be subtracted ; they simply show what operations are to be performed on the positive or negative numbers which they follow. Thus, in the statement, -f -12 — ^ X 2) the sign X shows that 5 is to be taken twice, but it does not show what is to be done with the resulting 10 ; that is shown by the — . We are to take two 5's from 12. So, in 18 + 9 -H 3, the sign -f- shows that 9 is to be divided by 3 ; what is to be done with the quotient, is shown by the + before the 9. 86. In every such numerical statement, the + or the — must be understood to affect the whole restttt of the opera- tions indicated between it and the next + or — , or between it and the close of the expression. Thus, in 5 + 7X2X9 — 2X6, the -f indicates the addition of 126, not of 7 only ; and the — indicates the subtraction of 12. The same meaning is conveyed by5-f(7X2X9) — (2X6). 86. When the signs X and +- occur in succession, they are to have their particular effects in the exact order of their occurrence. Thus, we would indicate by 96 -?- 12 X 4, that the operator is first to divide by 12, and then multiply the quotient by 4. The result intended is 32, not 2; if the latter were intended, we Rhould write 96 -7- (12 X 4). Usage has been divided on this point, however. Remabk.— It will be observed that in no case can the sigti X or -^ affect any number before the preceding -f or — , or beyond the following + or — . Exercises. 1. 4X3 + 7X2 — 9X3+6X4 — 3X3 = ? SoLXjnoN. + 4X3 = 12, 7X2 = 14, -9X3 = — 27, 6X4 = 24, — 3X3 = — 9. Grouping and adding according to the signs, we have, 12 + 14 4- 24 = 50 ; and — 27 — 9 = — 36. Therefore, 50 — 36 =14, Ans, 2. 2X2 — 1X2 — 2X2 — 5X3— 5X3 — 4 X 2 — 4x2 — 8X3 — 5X2 — 9X3 — 7X2 — 12X4 — 7X 2^? 52 RAY'S HIGHER ARITHMETIC. 80JLirA0N.+ 4— 2 — 4 — 15 — 15 — 8 — 8 — 24— 10— 27 — 14 ':8 — 14. Grouping and adding, we have, 4 — 189= — 185, An^ 3. 21^3X7 — 1x14-1 X 4-^2 + 18 -^3x6-i- ■2x2) + (4— 2 + 6 — 7)X4x6-v-8 = ? 59. Bemabk. — ^Whenever several numbers are included within the marks of parenthesis, brackets, or yinculum, the^are regarded as one number. Note the advantage of this in example 3. 4. 16x4^8—7 + 48-^16—3—7x4x0x9X16 + 24X6-r-48 — 4X9-M2 = ? 1. 5. (16-M6x96-^8 — 7 — 5 + 3)X[(27-^-9)-^8— l]+(91-M3x7— 45— 3)X9=? 9. GENERAL PBINCIPLES. 87. The following are the General Principles of Mul- tiplication and Division. Principle I. — MvUiplying either factor of a produd, muUi' plies ihe product by the same number. Thus, 5X4 = 20, and 5X4X2 = 40, whence 20X2=40. n. — Dividing either factor of a product, divides the produd by the same number. Thus, 5X4 = 20, and 5X4-f-2 = 10, whence 20-^^2 = 10. ni. — MvUiplying one factor of a product, and dividing the other factor by the same number, does not alter the product. Thus, 6X4 = 24, and 6X2X4 -^-2 = 24, whence 6X2X2 = 24. TV .— Multiplying the dividend, or dividing the divisor, by any number, multiplies the quotient by that number. If 24 be the dividend and 6 the divisor, then 4 is the quotient; hence 24X2-5-6 = 8, and 24 -i- (6 -h 2) = 8. CONTRACTIONS. 68 V. — Dividing the dividend^ or muUiplying (ke divisor, bg any number, divides the quotient by that nund)er. Thus, if 24 be the divideod and 6 the diyisor, then 24 -i- 2 = 12; and 12-5-6 = 2; whence 24 -^ (6 X 2) = 2. Therefore, 4-1-2 = 2. VL — Multiplying or dividing both dividend and divisor by the same number^ does not change the quotient. Thus, 24X2 = 48, and 6X2 = 12; consequently, 48-1-12 = 4, and 24-^-6 = 4^ CONTBACnONS IN MULTIPLICATION AND DIVISION. CASE I. 88. To multiply by any simple part of 100, 1000, eto. Note. — ^Let the pupil study carefully the following table of equivalent parts: Pabtb of 100. Parts of 1000. 12^ = 1 of 100. 125 =t of 1000. 16| = 1 of 100. 166| = I of 1000. 25 = i of 100. 250 = i of 1000. 33i = i of 100. 333i = i of 1000. 37i = I of 100. 375 = I of 1000. 62^ = I of 100. 625 = I of 1000. 66f = I of 100. 666| = i of 1000. 75 = I of 100. 750 = I of 1000. 87^ = J of 100. 875 = J of 1000. Pboblem. — ^Multiply 246 by 87^. operation. 24600 SoLTJTiON. — Since 87^ is J of 100, 7 annex two ciphers to ^ the multiplicand, g \ 1 72200 which multiplies it by 100, and then . — TTmrZ ♦«L » r ^1- IX -4n«. 21625 take } of the result. Bule. — Multiply by 100, 1000, efc., and take such a part of the residt as the multiplier is of 100, 1000, etc. 64 I^A Y'S HIOHEB ARITHMETia Examples for Practice. 1. 422X33^. Am. 14066f. 2, 6564 X 62f Am. 410250. 8. 10724 X 16*. Am. 178733|. CASE II. 89. To multiply by any number whose digits are all alike. Problem.— Multiply 592643 by 66666. Solution.— Multiply 592643 by 99999 (Art. 70), the product is 59263707357 ; take } of this product, since 6 is | of 9 ; the result is 39509138238. Bnle. — Multiply as if the digits were ffs, and take such a part of the product as the digit is of 9. Examples for Practice. 1. 451402 X 3333. Am. 1504522866. 2. 281257 X 555555. Am. 156253732635. 3. 630224 X 4444000. Am. 2800715456000. case III. 90. To divide by a number ending in any simple part of lOOy lOOOy etc. Problem.— Divide 6903141128 by 21875. Solution. — Multiply both by 8 and 4 successively. The divisor becomes 700000, and the dividend 220900516096, while the quotient remains the same. (Art. 87, vi.) Performing the division as in Art. 81, the quotient is 315572, and remainder 116096. The remain- der heing a part of the dividend, has been made too large by the multiplication by 8 and 4, and is, therefore^ reduced to its true dimensions by dividing by 8 and 4. This gives 3628 for the true remainder. CONTRACTION& 66 Bole. — Multiply both dividend and diviMr by sudi a number 08 will convert the final figures of the divisor into ciphers, and then divide the firnner product by the latter. Notes. — 1. If there be » remainder, it should be divided by the multiplier, to get the true remainder. 2. The multiplier is 3, 4, 6, etc., according as the final portion of the divisor is thirds, fourths, sixths, etc., of 100, 1000. Examples fob Practige. 1. 300521761 — 225. Ans. 1335652^ 2. 1510337264 -T- 43750. Ans. 34521||f|^ 3. 22500712361-7-1406250. Ans. 160003^^^ 4. 620712480 -r- 20833 J. Quoe. 29794. Rem. 4146| 5. 742851692 -r-2916|. QuoL 254692. Bern. 2^ General Problems. Note. — Let the pupil make a special problem under each general problem, and solve it. 1. When the separate cost of several things is given, how is the entire cost found? 2. When the sum of two numbers, and one of them, are given, how is the other found? 3. When the less of two numbers and the difference between them are given, how is the greater found ? 4. When the greater of two numbers and the difference between them are given, how is the less found? 5. When the, cost of one article is given, how do you find the cost of any number at the same price? 6. If the total cost of a given number of articles of equal value is stated, how do you find the value of one article ? 7. When the divisor and quotient are given, how do you find the dividend? 66 BA Y'S HIGHER ARITHMETIC. 8. How do you divide a number into parts, each contain- ing a certain number of units ? 9. How do you divide a number into a given number of equal parts? 10. K the product of two numbers, and one of them, are given, how do you find the other? 11. If the dividend and quotient are given, how do you find the divisor? 12. If you have the product of three numbers, and two of them are given, how do you find the third? 13. K the divisor, quotient, and remainder are given. Low do you find the dividend ? 14. K the dividend, quotient, and remainder are given, how do you find the divisor? MlBOELLANEOUS EXERCISES. 1. A grqper gave 153 biarrels of flour, worth $6 a bar- rel, for 64 barrels of sugar : what did the sugar cost per barrel ? $17. 2. When the divisor is 36, quotient 217, and remainder 25, what is the dividend ? 7620. 3. What number besides 41 will divide 4879 without a remainder? 119. 4. Of what number is 103 both the divisor and the quotient? 10609. 5. What is the nearest number to 53815, that can be divided by 375 without a remainder? 54000. 6. A fiirmer bought 25 acres of land for $2675 : what did 19 acres of it cost? $2033. 7. I bought 15 horses, at $75 a head : at how much per head must I sell them to gain $210 ? $89. 8. A locomotive has 391 miles to run in 11 hours : after running 139 miles in 4 hours, at what rate per hour must the remaining distance be run? 36 miles. MISCELLANEO US EJ^RCISES. 57 9. A merchant bought 235 yards of cloth, at $5 per yard ; after reserving 12 yards, what will he gain by selling the remainder at $7 per yard ? $386. 10. A grocer bought 135 barrels of pork for $2295 ; he sold 83 barrels at the same rate at which he purchased, and the remainder at an advance of $2 per barrel : how much did he gain ? $104. 11. A drover bought 5 horses, at $75 each, and 12 at $68 each; he sold them all at $73 each: what did he gain? $50. At what price per head must he have sold them to have gamed $118? $77. 12. A merchant bought 3 pieces of cloth of equal length, at $4 a yard; he gained $24 on the whole, by selling 2 pieces for $240: how many yards were there in each piece? 18 yards. 13. If 18 men can do a piece of work in 15 days, in how many days will one man do it? 270 days. 14. If 13 men can build a wall in 15 days, in how many days can it be done if 8 men leave? 39 days. 15. If 14 men can perform a job of work in 24 days, in how many days can they perform it with the assistance of 7 more men? 16 days. 16. A company of 45 men have provisions for 30 days : how many men must depart, that the provisions may last the remainder 50 days? 18 men. 17. A horse worth $85, and 3 cows at $18 each, were exchanged for 14 sheep and $41 in money: at how much each were the sheep valued? $7. 18. A drover bought an equal number of sheep and hogs for $1482 : he gave $7 for a sheep, and $6 for a hog : what number of each did he buy? 114. 19. A trader bought a lot of horses and oxen for $1260; the horses cost $50, and the oxen $17, a head; there were twice as many oxen as horses: how many were there of each? 15 horses and 30 oxen. 58 RAY'S HIGHER ARITHMETia 20. In a lot of silver change, worth 1050 cents, one seventh of the value is in 25-cent pieces; the rest is made up of 10-cent, 5-cent, and 3-cent pieces, of each an equal number: how many of each coin are there? Of 25-cent pieces, 6; of the others, 50 each. 21. A speculator had 140 acres *of land, which he might have sold at $210 an acre, and gained $6800; but after holding, he sold at a loss of $5600: how much an acre did the land cost him, and how much an acre did he sell it for? $165, cost; and $125, sold for. Topical Outline. DrVTBION. 1. Definitions. 2. Terms 8. Sign. 4. Principles. ( 1. Dividend. J 2. Divisor. & Opttration ^ 3. Quotient. 4. Remainder. rl. Writing the Numbers. 2. Drawing Curved Lines. 3. Finding Quotient Figure. 4. Multiplying Divisor and Writing Product 5. Drawing Line. 6. Subtracting. 7. Annexing Lower Order. 8. Repeating the Process from 8. ^9. Writing Remainder. 6 Rules.. / 1^^ Division. I Short Division. 7 Prool 8. Applications. d. Contractions / Division. I Multiplication and Division. f 1. .Of Numbers... / ^^^^"f^- 10. Arithmetical Signs. \ ^ Negative. I 2. Of Operation... /Multiplying. I Dividing. 11. General Principles. 12. Applications. Tn. PEOPEETIES OF NUMBEES. DEFINITIONS. 91. 1. The Properties of Numbers are those qualities which belong to them. 2. Kumbers are classified (1), as Integral, Fractional, and Mixed (Art. 22) ; (2), as Abstract and Concrete (Art. 21); (3), Prime and Composite; (4), Even and Odd; (5), Perfect and Imperfect. 3. An integer is a whole number; as, 1, 2, 3, etc. 4. Integers are divided into two classes— prime numbers and composite numbers. 5. A prime number is one that can be exactly divided by no other whole number but itself and unity, (1) ; as, 1, 2, 3, 5, 7, 11, etc. 6. A composite number is one that can be exactly divided by some other whole number besides itself and unity; as, 4, 6, 8, 9, 10, etc. Remark. — Every compoFiite number is the product of two or more other numbers, called its /acUn-a (Art. 60). 7. Two . numbers are prime to each other, when unity is the only number that will exactly divide both; as, 4 and 5. Kemabk. — Two prime numbers are always prime to each other: sometimes, also, two composite numbers ; as, 4 and 9. 8 An even number is one which can be divided by 2 without a remainder ; as, 2, 4, 6, 8, etc. 9. An odd number is one which can not be divided by 2 without a remainder; as, 1, 3, 5, 7, etc. Remabe. — All even numbers except 2 are composite : the odd numbers are partly prime and partly composite. (59) 60 RA Y^S HIQHEB ARITHMETIC. 10. A perfect number is one which is equal to the sum of aU its divisors; as, 6 = 1 + 2+3; 28 = 1 + 2 + 4 + 7 + 14. 11. An imperfect number is one not equal to the sum of all its divisors. Imperfect numbers are Abuniard or Z>e- fective: Abundant when the number is less than the sum of the divisors ; as, 18, less than 1 + 2 + 3 + 6 + 9; and Defective when the number is greater than the sum; as, 16, greater than 1+2 + 4 + 8. 12. A diYisor of a number, is a number that will exactly divide it. 13. One number is divisible by another when it contains that other without a remainder ; 8 is divisible by 2. 14. A multiple of a number is the product obtained by taking it a certain number of times ; 15 is a multiple of 5, being equal to 5 taken 3 times; hence, 1st. A multiple of a number can always be divided by U vrithoid a remainder, 2d. Every multiple is a corrvposile number. 15. Since every composite number is the product of factors, each fiictor must divide it exactly; hence, every lactor of a number is a divisor of it. 16. A prime &ctor of a number is a prime number that will exactly divide it : 5 is a prime &ctor of 20 ; while 4 is a factor of 20, not a prime factor ; hence, 1st. The prime factors of a number are aU the prime numr bers that 'unll exactly divide it. Example. — 1, 2, 3, and 5 are the prime factors of 30. 2d. Every composite number is equal to the product of aU its prime factors. Example. — All the prime factors of 15 are 1, 3, and 5 ; and 1 X 3 X 5 = 15. 17. Any &ctor of a number is called an aliquot part of it. Example. — 1, 2, 3, 4, and 6, are aliquot parts of 12. \ FACTORING. 61 FACTORING. 92. Factoring is resolving composite numbers into fac- tors ; it depends on the following principles and propositions. Pkenceple 1. — A f(uioT of a number is a factor of any mvUiple of thai number. Demonstration. — Since 6 = 2X3, therefore, any maltiple of 6 = 2 X 3 X some number ; hence, every factor of 6 is also a factor of the maltiple. The same May be proved of the multiple of any composite number. Principle 2. — A factor of any two numbers is also a fador of their Bum. Demonstbation. — Since each of the numbers contains the factor a certain number of times, their sum must contain it as often as both the numbers ; 2, which is a factor of 6 and 10, must be a factor of their sum, for 6 is 3 ticos, and 10 is 5 twos, and their sum is 3 twos + 5 twos =S twos, 93. From these principles are derived the six following propositions : Prop. I. — Every number ending udih 0, 2, 4, Q, or S, is dimsible hy 2. Demonstration. — Every number ending with a 0, is either 10 or some number of tens; and since 10 is divisible by 2, therefore, by Principle 1st, Art. 92, any number of tens is divisible by 2. Again, any number ending with 2, 4, 6, or 8, may be considered as a certain number of tens plus the figure in the units' place ; and since each of the two parts of the number is divisible by 2, there- fore, by Principle 2d, Art. 92, the number itself is divisible by 2; tKus, 36 = 30 + 6 = 3 tens -f 6 ; each part is divisible by 2 ; hence, 36 is divisible by 2. Conversely, No number is divisible by 2, unless it ends uHh 0, 2, 4, 6, or 8. Prop. II. — A number is divisible by 4, when the number denoted by its two right-hand digits is divisible by 4. 62 BA Y' S HIGHER ARITHMETIC. Demonsteation.— ^Since 100 is divisible by 4, any namber of hundreds will be divisible by 4 (Art. 92, Principle 1st) ; and any number consisting of more than two places may be regarded as a certain number of hundreds plus the number expressed by the digits in tens' and units' places (thus, 884 is equal to 3 hundreds + 84) ; then, If the latter part (84) is divisible by 4, both parts, or the number itself, will be divisible by 4 (Art. 02, Prin. 2d). Conversely, No number is divisiUe by 4, unless the number denoted by its two right-hand digits is divisible by 4. Prop. III. — A number ending in or 5 is divisible by 5. Demonstration. — ^Ten is divisible by 5, and every number of two or more figures is a certain number of tens, plus the right-hand digit ; if this is 5, both parts of the number are divisible by 5, and, hence, the number itself is divisible by 5 (Art. 92, Prin. 2d). Conversely, No number is divisible by 5, unless it ends in or 5. Prop. IV. — Every number ending in 0, 00, etc., is divis- ible by 10, 100, etc. Demonstration. — If the number ends in 0, it is either 10 or a multiple of 10 ; if it ends in 00, it is either 100, or a multiple of 100, and so on ; hence, by Prin. 1st, Art. 92, the proposition is true. Prop. V. — A composite number is divisible by ike product of any two or Tnore of its prime factors. Demonstration. — Since 2 X 3 X 5 = 30, it follows that 2X3 taken 5 times, makes 30 ; hence, 30 contains 2 X 3 (6) exactly 5 times. In like manner, 30 contains 3X5 (1^) exactly 2 times, and 2X5 (10), exactly 3 times. Hence, if any even number is divisible by S, it is also divisible by 6. Demonstration. — An even number is divisible by 2 ; and if also by 3, it must be divisible by their product 2X3, or 6. Prop. VI. — Every prime number, except 2 and 5, ends with 1, 3, 7, or 9. Demonstration. — This is in consequence of Props. I. and HI. FACTO RING. 63 Prop. VII.— Jny integer w divisible by 9 or by S, if the sum of its digits be thus divmble. 94. To find the prime fkotors of a composite number. Problem. — ^Find the prime factors of 42. SoLunoN.^42 is divisible by 2, and 21 is divisible by 3 or 7, which is found by trial; hence, the prime factors of 42 are 2, 3, 7, .'. 2 X 3X7=42. OPERATION. 2)42 3)21 Bule. — Divide tJie given nuniber by any prime number that wiU eaxicUy divide it; divide Hie quotient in like manner, and so continue until Hie quotient is a prime number; the several divism's and the last quotient are Hie prime fcuiors. Kemabks. — 1. Divide first by the smallest prime factor. 2. The least divisor of any number is a prime number ; for, if it were a composite number, its factors, which are less than itself, would also be divisors (Art. 92), and then it would not be the least divisor. Therefore, the prime factors of any number may be found by dividing it first by the least number that will exactly divide it, then dividing this quotient in like manner, and so on. 3. Since 1 is the factor of every number, either prime or com- posite, it is not usually specified as a factor. Find the prime factors of: 1. 45. Ans. 3, 3, 5. 2. 54. Ans, 2, 3, 3, 3. 3. 72. Ans, 2, 2, 2, 3, 3. 4. 75. Ans, 3, 5, 5 5. 96. Ans. 2, 2, 2, 2, 2, 3 6. 98. Ans, 2, 7, 7 7. Factor 210. 8. Factor 1155. 9. Factor 10010. 10. Factor 36414. 11. Factor 58425. Ans, 2, 3, 5, 7. Ans, 3, 5, 7, 11. Am, 2, 5, 7, 11, 13. Ans, 2, 3, 3, 7, 17, 17. Ans, 3, 5, 5, 19, 41. 64 ^^ y^ HIOHEB ABITHMETia 95. The prime factors common to several numbers may be found by resolving each into its prime factors, then taking the prime factors alike in all. Find the prime factors common to: 1. 42 and 98. Am. 2, 7. 2. 45 and 105. Am. 3, 5. 3. 90 and 210. Am. 2, 3, 5. 4. 210 and 315. Am. 3, 5, 7. 96. To find all the divisors of any composite ntun- ber. Any composite number is divisible, not only by each of its prime factors, but also by the product of any two or more of them (Art. 93, Prop. V.) ; thus, 42 = 2 X 3 X 7 ; and all its divisors are 2, 3, 7, and 2 X 3, 2 X 7, and 3 X 7; or, 2, 3, 7, 6, 14, 21. Hence, Bule. — Resolve Hie number into iU prime fadorSy and then form from these factors all the different products of which they wiU admit; the prims factors and their products wiU be aU Hie divisors of the given number. Find all the divisors: 1. Of 70. Am. 2, 5, 7, and 10, 14, 35. 2. Of 196. Am. 2, 7, and 4, 14, 28, 49, 98. 3. Of 231. Am. 3, 7, 11, and 21, 33, 77. 4. Of 496 ; and name the properties of 496. GREATEST COMMON DIVISOR. 97. A common divisor (C. D.) of two or more num- bers, is a number that exactly divides each of them. 98. The greatest common divisor (G. C. D.) of two or more numbers is the greatest number that exactly divides each of them. GREATEST COMMON DIVI80B. 65 Principles. — 1. Every prime fadot cf a number u a dimor of that number. 2. Every product of two or more prime factora of a number^ is a divisor of that number, 3. Every number is equal to the continued product of all its prime factors. 4. A divisor of a number is a divisor of any number of times that number, 5. A common divisor of two or more numbers is a divisor of their sum, and also of their difference. 6. The product of all the prime factors, common to two or more numbers, is their greatest common divisor. 7. The greatest commen divisor of tujo numbers, is a divisor <f their difference. To Find the Qreatist Common Divisob. CASE I. 99. By simple Motoring. Problem.— Find the G. C. D. of 30 and 105. OPERATIOK. 30 = 2X3X5. I 3X6 = 15, G.C.D. 105 = 3X5X7. ) ' Demonstration. — ^The product 3X5 is a divisor of both the numbers, since each contains it, and it is their greatest common divisor, since it contains all the factors common to both. PROBLEM.-^Find the G. C. D. of 36, 63, 144, and 324. OPERATION. 36, 63, 144, 324 Solution.^ 3 3 12, 21, 48, 108 4, 7, 16, 36 .-.3X3 = 9, G. C. D. Bale. — Resolve the given numbers into their prime factors, and take the product of the faetors common to all the numbers. H. A. 6. 66 BAY','^ mOHER ARITHMETIC. Find the greatest common divisor: 1. Of 30 and 42. 2. Of 42 and 70. 3. Of 63 and 105. 4. Of 66 and 165. 5. Of 90 and 150. 6. Of 60 and 84. 7. Of 90 and 225. 8. Of 112 and 140. 9. Of 30, 45, and 75. 10. Of 84, 126, and 210. 11. Of 16, 40, 88, and 96. 12. Of 21, 42, 63, and 126. ^na. 2X 3 = 6. Am. 2x7 = 14. Am. 3 X 7 = 21. ^rw. 3x11=33. Am. 2 k 3 X 5 = 30. Am. 2 X 2 X 3 = 12. Am. 3 X 3 X 5 = 45. Am. 2 X 2 X 7 = 28. Am. 3 X 5 = 15. Am. 2 X3X7 = 42. Am. 2X2X2=8. Am. 3 X 7 = 21. CASE II. 100. By successive divisions. Problem.— Find the G. C. D. of 348 and 1024. OPERATION. 348) 1024(2 696 328)348(1 328 20 ) 3 2 8 (16 20 128 120 Demonstbation.— If 348 will divide 1024, it is the G. C. D. ; but it will not divide it. If 328 (Art 98, Prin. 5,) will divide 348, it is the G. C. D. of 328, 348, and 1024 ; but it will not divide 348 and 1024 exactly. If 20 will divide 328 (by same process of reasoning), it is the G. C. D.; but there is a remainder of 8; hence, if 8 will divide 20, it is the G. C. D. ; but there is also a remainder of 4. Now, 4 divides 8 without a remainder. Therefore, 4 is the greatest number that will divide 4, 8, 20, 328, 348, and 1024, and is the G. C. D. _8)20(_2^ 16 G. C. D. 4)8 OBEATE8T COMMON DIVISOR. 67 Bule. — Divide tiie greater number by the less, and the divisor by the remainder, and 90 on; always dividing the last divisor by the last remainder, till withing remains; the hut divisor vriU be the greatest ccmvMm divisor sovght. Note. — A condensed form of operation condensed opekation. may be used after the pupils are familiar «.« 1024 2 with the preceding process. ««« Bemabk. — To find the greatest common divisor of more than two numbers, find the ^ " ^^^ ^ " G. C. D. of any two ; then of that G. C. D., 1^ 8 2 and any one of the remaining numbers, 4 8 2 and so on for all of the numbers ; the last C. D. will be the G. C. D. of all the numbers. Remark. — If in any case it be obvious that one of the numbers has a prime factor not found in the other, that factor may be suppressed by division before applying the rule. Thus, let the two numbers be 715 and 11011. It is plain that the prime 5 divides the first but not the second ; and since .that prime can be no factor of any common divisor of the two, their G. C. D. is the same as the G, C. D. of 143 and 11011. Find the greatest common divisor of: 1. 85 and 120. Ans. 5. 2. 91 and 133. Ans, 7. 3. 357 and 525. Ans. 21. 4. 425 and 493. Ans. 17. 5. 324 and 1161. Ans. 27. 6. 589 and 899. Ans. 31. 7. 597 and 897. ■ Ans. 3. 8. 825 and 1287. Ans. 33. 9. 423 and 2313. Ans. 9. 10. 18607 and 24587. Ans. 23. 11. 105, 231, and 1001. Ans. 7. 12. 165, 231, and 385. Am. 11. 13. 816, 1360, 2040, and 4080. Ans. 136. 14. 1274, 2002, 2366, 7007, and 13013. Ans. 91. 68 RAY' 8 HIGHER ARITHMETIC. LEAST COMMON MULTIPLE. 101. A common multiple (C. M.) of two or more numbers, is a number that can be divided by eaufii of them without a remainder. 102. The least common multiple (L. G. M.) of two or more numbers, is the lead number that is divisible by each of them without a remainder. Principles. — 1. A miiUiple of a number is divisiUe by that number. 2. A multiple of a number mmt contain aU of the prime factors of that number. 3. A commxm multiple of two or mjore numbers is divisible by each of those numbers. 4. A common multiple of two or mme numbers contains aU of the prme factors of ecLch of (hose numbers. 5. The hast common multiple of turn or more numbers must contain all of the prim/e fadtms of each <f those numbers, and no other factors. ^ 6. If two or more numbers are prims to each other, their continued product is their least common multiple. To Find the Least Common Multiplk CASE I. 103. By factoring the numbers separately. Problem. — ^Find the L. C. M. of 10, 12, and 15. Solution. — Resolve each operatioit. number into its prime f ac- 10 = 2X5 tors. A multiple of 10 con- 1^=2X2X3 tains the prime factors 2 and 15=3X5 5 ; of 12, the prime factors .'. L. C. M. = 2 X 2 X 3X5 = 60. 2, 2, and 3 ; of 15, the prime factors 3 and 5. But the L. C. M. of 10, 12, and 15 must contain LEAST COMMON MULTIPLK 69 all of the different prime factors of these numbers, and no other factors ; hence, the L. C. M. = 2 X 2 X3 X 5 = 60 (Art. 102, Prin. 2 and 5). Bule. — Resolve each number into its prime factors, and dien take the continued jproduct of aU the different prime factorSy using ecuih factor the greatest number of times it occurs in any one of the given numbers. Remabks. — 1. Each factor must be taken in the least common mul- tiple the greatest number of times it occurs in either of the numbers. In the preceding solution, 2 must be taken twice, because it occurs twice in 12, the number containing it most. 2. To avoid mistakes, after resolving the numbers into their prime factors, strike out the needless factors. Find the L. C. M. of: 1. 8, 10, 15. Ans. 120. 2. 6, 9, 12. Ans. 36. 3. 12, 18, 24. Ans. 72. 4. 8, 14, 21, 28. Ans. 168. 5. 10, 16, 20, 30. Ans. 60. 6. 15, 30, 70, 105. Ans. 210. CASE II. 104. By dividing the numbers suooessively by their common primes. Problem.— Find the L. C. M. of 10, 20, 25, and 30. Solution. — Write the operation. numbers as in the margin. 2 Strike out 10, because it is *. ;p 20 25 30 10 25 15 2 5 3 contained in 20 and 80. Next, divide 20 and 30 by the prime factor 2; write 2X5X2X5X3 = 300 L. CM. the quotients 10 and 15, and the undivided number 25 in a line beneath. Divide these num* bers by the common prime factor 5. The three quotients — 2, 5, 3, are prime to one another ; whence, the L. C. M. is the product of the divisors 2, 5, and the quotients 2, 5, 3. By division, all needless factors are suppressed. 70 RAY'S HIGHER ARITHMETia Bule. — 1. WrUe the numbers in a harizontxd line; strike out any number that will exactly divide any of the others; divide by any prime number that will divide tuHp or more of them uriUumt a remainder; write the quotients and undivided num^ hers in a line beneath, 2. Proceed with this line as before, and continue the opera- tion tUl no number greater than 1 unU exactly divide two or nurre of the numbers. 3. Multiply together the divisors and the numbers in the last line; their product will be the least common multiple required. Remark. — Prime factors not obvious may be found by Art lOO. Find the least common multiple of: 1. 6, 9, 20. Ans. 180. 2. 15, 20, 30. Ans. 60. 3. 7, 11, 13, 5. Ans. 5005. 4. 35, 45, 63, 70. ' Ans. 630. 5. 8, 15, 20, 25, 30. Ans. 600. 6. 30, 45, 48, 80, 120, 135. Ans. 2160. 7. 174, 485, 4611, 14065, 15423. Ans. 4472670, 8. 498, 85988, 235803, 490546. Am. 244291908. 9. 2183, 2479, 3953. Ans. 146261. 10. 1271, 2573, 3403. Ans. 105493. SOME PROPERTIES OF THE NUMBER NINR 105. Addition, Subtraction, Multiplication, and Division may be proved by ** casting out the 9's." To cast the 9's out of any number, is to divide the sum of the digits by 9, and find the excess. Problem.— Find the excess of 9's in 768945. Explanation. — Begin at the left, thus : 7 + 6 are 13 ; drop the 9; 4 + 8 are 12; drop the 9; 3 + 4 + 5 are 12; drop the 9; the excess is 3. The 9 in the number was not counted. CASTING OUT NINEi Principle. — Any number divided by 9, wiU remainder as the sum of its digits divided by 9. ILLUSTRATION. 700000 = 7 X 100000 = 7 X (99999 4- 1) =7 X 99999 + 7 60000 = 6X 10000 = 6X (9999 + l) = 6X 9999-f 6 768945=J 8000 = 8X 1000 = 8X (9994-l) = 8X 999+8 900 = 9X 100 = 9X (994-l) = 9X 99-|-9 40 = 4X 10 = 4X (9 + l) = 4X 9 + 4 5 = 5X 1= =6 Whence, 7X99999 + 6X9999 + 8X999 + 9X99 + 4X9 + 7 + 6 + 8+9 + 4 + 5 = 768945. Solution. — An examination of the above shows that 768945 has been separated into multiples of 9, and the sum of the digits com- posing the number ; the same may be shown of any other number. There qan be no remainder in the multiples, except in the sum of the digits. Proof op Addition. — The sum of the excess of ffs in the several numbers must equal the excess of ffs in their sum. Illustration. — The excesses in the num- bers are 8, 2, 4, and 3, and the excess in the sum of these excesses is 8. The excess in the sum of the numbers is 8, the two excesses being the same, as they ought to be when the work is correct. operation. 7352 5834 6241 7302 26729 8 2 4 3^ 8 Proof of Subtraction. — The excess of 9*s in the minuend must equal the sum of the excess of 9*s in the subtrahend and remainder. Illustration. — As the min- uend is the sum of the subtra- hend and remainder, the reason of this proof is seen from that of Addition. operation. Minuend, 7 6 4 Subtrahend, 1234 Eemainder, 6 40 6 8 1 7 Proof of Multiplication. — Find the excess of 9*8 in ths factors and in the product, TJie excess of 9*8 in the product of the excesses of the factorSy should equal the excess in the product of the factors themselves. 72 -R^ Y'S HIGHER ABITHMETIC. IiiLuaTRATiON. — Multiply 835 by 76; opebation. the product iB 63460. The excess in the 835X76 = 63460 multiplicand is 7, in the multiplier 4, 835, excess =7 and in the product 1; the two former 76, " =4 multiplied, give 28 ; and the excess in 28 7X^=28, " =1 is also 1, as it should be. 63460, '' =1 Proof op Division. — Find the excess of 9'« in each of the terms. To the excess of ffs in the product of the excesses in the divisor and quotient^ add ike excels in the remainder; the excess in the sum shotdd equal the excess in the dividend. OPERATION. Ilhtstbation.— Divide 8915 by 25; the quo- 35 6, excess 5 tient is 356, and the remainder, 15. The excess 2 5, ** 7 of 9's in the divisor is 7 ; in the quotient, 5 ; 35 their product is 35, the excess of which is 8. The excess in the remainder is 6. 6 -f 8 = 14, of which the excess is 5. The excess of 9'b in the dividend is also 5. ^ ^ 3 5, « 8 15, " 6 14, 8915, CANCELLATION. 106. Oanoellation is the process of crossing out equal factors from dividend and divisor. The sign of Cancellation is an oblique line drawn across a figure ; thus, ^, ^, ^. Principles. — 1. Canceling a factor in any number^ divides Hie number by that factor. 2. Canceling a factor in both dividend and divisoTy does not change the quotient. (Art. 129, III.) Problem. — Multiply 75, 153, and 28 together, and divide hy the product of 63 and 36. CA NCELLA TION. 73 SoiiUnoN. — Indicate the operations operation. as in the margin. Cancel 4 out of 25 17 7 28 and 36, leaving 7 above and 9 7^5 v T^il v9il below. Cancel this 7 out of the divi- /f ^ A APP A^p^ dend and out of the 63 in the divisor, P P X p p leaving 9 below. Cancel a 9 out of 9 9 the divisor and out of 153 in the 3 dividend, leaving 17 above. Cancel 2 5x17 3 out of 9 and 75, leaving 25 above ^ = 1 4 1 J and 3 below. No further canceling is possible; the factors remaining in the dividend are 25 and 17, whose product, 425, divided by the 3 in the divisor, gives 141}. Bule for Cancellation. — 1. Indicate Hie mulHplicatwtis which prodtice the dividend, and those, if any, vMch produce the divisor. 2. Gancd equal factors from dividend and divisor ; mvUiply together the factors remaining in the dividend, and divide the product by the product of the factors left in the divisor. Note. — If no factor remains in the divisor, the product of the factors remaining in the dividend will be the quotient ; if only one factor is left in the dividend, it will be the answer. Examples for Practice. 1. How many cows, worth $24 each, can I get for 9 horsQ3, worth $80 each? 30 cows. 2. I exchanged 8 barrels of molasses, each containing 33 gallons, at 40 cents a gallon, for 10 chests of tea, each containing 24 pounds: how much a pound did the tea cost me? 44 cents. 3. How many bales of cotton, of 400 pounds each, at 12 cents a pound, are equivalent to 6 hogsheads of sugar, 900 pounds each, at 8 cents a pound? 9 bales. 4. Divide 15 X 24 X 112 X 40 X 10 by 25 X 36 X 56 X90. ^. H. A. 7. 74 BAY'S HIQHEB ARITHMETIC. Topical Outline. 1. Definitions.. 2. Factoring 3. G. C. D. 4. L. C. M.. Properties of Numbers. Properties. Numbers Glassiiiect. Integer Fraction. L Mixed. Diyisor; Divisible ; •Multiple ; Factors; L Prime Factors; Aliquot Parts. r 1. Definition. 2. Principles. 3. Propositions. 4. Operations. 5. Rules. 6. Applications. Prime. Composite. Even. Odd. Perfect. Imperfect 1. Definitions.... 2. Principles. 3. Operations....; I 2. Common Divisor. Greatest Common Divisor. r Case I J^"l®- \ Applications. Case II... / Rule. I Applications. 1. Definitions..... | ^' Common Multiple. 2. Principles. 3. Operations Least Common Multiple. r Case I / ^^^e- \ Applications. Case II... /Rule- \ Applications. ). Some Properties of the No. 9- / ^' I'rinciple. 1 2. Application to... 6. Cancell:>t;on... , 1. Definition. 2. Sign, /. 3. Prin<'iples. 4. Operation. 5. Rule. 6. Applications. 1. Addition. 2. Subtraction. 3. Multiplicatiou 4. Division. vm. coMMOi^ FRAcrriONs. PEFINinONS. 107. A Fraction is an expression for one or more of the equal parts of a divided whole. 108. Fractions are divided into two classes; viz., com- mon fractions and decimal fractions. 100. A Common Fraction is expressed by two num- bers, one above and one below a horizontal line; thus, J, which is read tioo thirds. HO. The Denominator is the number below the line. It shows the number of parts into which the whole is divided, and thus the size of the parts. 111. The Numerator is the number above the line. It shows how many of the parts are taken. Note. — The denominator denominatesj or names, the parts; the numerator numbers the parts. 112. The Terms of a fraction are the numerator and the denominator. IiiiiUSTRATiON. — The expression i,four fifths, shows that the whole is divided into five equal parts, and that four of those parts are taken. 5 is the denominator, 4 is the numerator, and the terms of the fraction are 4 and 5. 113. Every fraction implies: 1. That a number is di- vided ; 2. That the parts are equal ; 3. That one or more of the parts are taken. 114. There are two ways of considering a fraction whose numerator is greater than 1. Four fifths may be 4 fifths of one thins:, or 1 fifth of four things ; therefore, ^ (75) 76 RAY'S HIGHER ARITHMETIC. The numerator of a fraction may be regarded as showing the number of units to be divided; the denominator, the number of parts into which the numerator is to be divided ; the fraction itself being the value of one of those parts. Hence, a fraction may be considered as an indicated ditrision (Art. 75) in which, 1. The dividend is the numerator, 2. The divisor is the denominaior. 3. The quotient is Hie fraction itself. 116. The Value of a fraction is its relation to a unit. Ue. Fractions are divided into classes with respect to their value and form. (1). As to value, into Proper, Improper, and Mixed. (2). As to form, into Simple, Complex, and Compound. 117. A Proper Fraction is one whose numerator is less than its denominator; as, \. 118. An Improper Fraction is one whose numerator is equal to, or greater than, its denominator; as, f. 119. A Mixed Number is a number composed of an integer and a fraction ; as, 3f . 120. A Simple Fraction is a single fraction whose terms are integral; as, |, |, f. 121. A Complex Fraction is one which has one or 1 1 both of its terms fractional ; as, -^r, -^, or ^. a 5 122. A Compound Fraction is a fraction of a fraction ; as, \ of |. 123. An Integer may be expressed as a fraction by writing 1 under it as a denominator ; thus, ^, which is read seven ones. COMMON FBACTION& 77 124. The Beciprocal of a number is 1 divided by that number ; thus, the reciprocal of 5 is |. 125. Similar Fractions are those that have the same denominator ; as, f and f . 126. Dissimilar Fractions are those that have unlike denominators; as, f and f. Remark. — The word "fraction" is from the Latin, fimngo, I break, and literally means a broken number. In mathematics, however, the word " fraction,'' as a general term, means simply the Indicated quotient of a required division. NUMEBATION AND NOTATION OF FRACTIONa 127. Numeration of Fractions is the art of reading fractional numbers. 128. Notation of Fractions is the art of writing frac- tional numbers. Bule for Beading Common Fractions. — Bead the num- ber of. parts taken as expressed by the nttmerator, and then the size of the parts as expressed by the denominator. Example. — J is read seven ninths, Kemabk. — Seven ninths (J), signifies 7 ninths of one, or i of 7, or 7 divided by 9. Bule for Writing Common Fractions. — Write the num- ber of parts ; plaee a horizontal line below it, under which unite the number whicfi indicates the size of the parts. Fractions to be written in figures: Seven eighths. Four elevenths. Five thirteenths. One seventeenth. Three twenty-ninths. Eight twenty-firsts. Nine forty-seconds. Nineteen ninety-thirds. Thirteen one- hundredths. Twenty-four one-hundred-and-fifteentbs. 78 RA Y' 8 HIGHER ARITHMETIC 129. Since a fetction is an indicated division (Art. 114) ; therefore, Principles. — ^I. A Fradion is mvltipliedf 1st. By multiplying the numerator. 2d. By dividing the denominator, n. A Fraction is divided, 1st. By dividing the numerator. ' 2d. By multiplying the denominator. HI. The value of a Fraction is not changed, 1st. By multiplying both terms by the same number. 2d. By dividing both terms by Uie same number. Remark. — The proof of I is found in Art. 87, Principle IV ; the proof of U is in Principle V ; and of III, in Principle VI. REDUCTION OF FRACTIONS. 130. Beduction of Fractions consists in changing their form without altering their value. CASE I. 131. To reduce a fraction to its lowest terms. Remarks. — 1. Reducing a fraction to lower terms, is changing it to an equivalent fraction whose terms are smaller numbers. 2. A fraction is in its lowettt terms when the numerator and de- nominator are prime to each other; as, f, but not J. Problem. — Reduc3 f^ to its lowest terms. Solution. — Dividing both terms by the first operation. common factor 2, the result is |§ ; dividing 2 ) fg = JJ this by 5 (129, iii\ the result is f , which 5 ) |^ = |, Ans. can not be reduced lower. Or, dividing at once by 10, the greatest second operation. common divisor of both terms, the result is 10 ) |^ = |, Ans. f , as before. i EEDUCTION OF FRACTIONS. 79 Bule. — Reject aU fcustors common to both terms of thefraetwn- Or, divide both terms of the fraction by their greatest common divism'. Reduce to their lowest terms : 1- H- Ans. f. 6. Mf- Ans. ff. 2. If. Ans. f . 7. Ml. Ans. \^. 3. iV». Am. ^. 8--m- Ans. f f . 4. m- Ant. ■^. 9-HH- Ans. i|. 5.,m- Ans. ^. 10. t'W^ > Ans. ^. Express the following in their simplest forms : 11. 923 : 1491. Ans. ^. 13. 2261- =-4123. Ans. ^. 12. 890 : 1691. Ana. ^. 14. 6160- ^- 40480. Ans. :^. CASE II. 132. To reduce a fraction to higher terms. Remabk. — Reducing a fraction to higher terms, is changing it to an equivalent fraction whose terms are larger numbers. Pkoblem. — Reduce f to fortieths. SoiitmoN. — Divide 40 by 8, the quo- operation. tient is 5 ; multiply both terms of the 40-^-8 = 5 given fraction, f , by 5 (129, m), and the . __ 5X5 __. o, ^^^ result is fj, the equivalent fraction re- 8X5 quired. Rule. — Divide ihe required denominator by the denominator of the given fraction ; multiply both terms of the given fraction by this quotient ; the result is the equivalent fraction required. 1. Reduce ^ and -^ to ninety-ninths. Ans. |^, -J^. 2. Reduce |^, f , and ^ to sixty-thirds. Ans. ff > fi» A* 3. Reduce ^, ^, and ^ to equivalent fractions having 6783 for a denominator. Ans. |4^» iHi* iHi' 80 JiA rs HIGHEB ARITHMETIC. CASE III. 133. To reduce a whole or mixed number to an improper fraction. Problem. — Reduce 3f to an improper fraction ; to fourths. Solution. — In 1 (unit), there are 4 fourths; in 3 (units), there are 3 times 4 fourths, = 12 fourths : and 12 fourths -{- 3 fourths = 15 fourths. Bule. — Multiply together tJie whole number and the denom- inator of the fraction: to the product add the numerator , and vrrite the sum over the denominator, 1. In $7f , how many eighths of a dollar? Ans, ^-. 2. In 19f gallons, how many fourths? Ans. ^. 3. In 13f^ hours, how many sixtieths? Ans. ^^, Reduce to improper fractions: 4. 11|. Ans. ^. 5. 15^. Ans. ^. 6. 127||. Ans. -4^. 7. 109^. Ans. ifp. 8. 5|H. Ans. VtV^. 9. 13fJ. Ans. i^fi. Bemark. — To reduce a whole number to a fraction having a given denominator, is a special case under the preceding. Problem. — ^Reduce 8 to a fraction whose denominator is 7. Solution. — Since | equals one, 8 equals 8 times J, or ^. CASE IV. ' 134. To reduce an improper firaetion to a whole or mixed number. Problem. — Reduce ^ of a dollar to dollars. Solution. — Since 5 fifths make 1 dollar, there will be as many dollars in 13 fifths as 5 fifths are contained times in 13 fifths ; that is, 2| dollars. DEDUCTION OF FRACTIONS. 81 BiQe. — Divide the numerator by the denominator; the quotient vnU be the wJwle or mixed number, Kebiark. — If there be a fraction in the answer, reduce it to its lowest terms. 1. In ^ of a dollar, how many dollars? $4|. 2. In -i^l^ of a bushel, how many bushels? 34J bu. 3. In -^^ of an hour, how many hours? I^tV ^o^"^* Reduce to whole or mixed numbers: 4. f|. Ans. 1. 5. i|9A. ^ns, 35. 6. ^^. Am. 88|. 7. HP- ^^' 105fV. 8. 4l^- ^^' 327^. 9. j^p. An*, 509^. CASE V. 135. To reduce compound to simple fractions. Problem. — Reduce f of ^ to a simple fraction. Solution. — J oi ^ = ^j; I oi ^ operation. = JV; andfof ^ = 2X_5 = li. i oi ^ = ^^ = 1^, Am, ^' * ^ 28 28 * ^ 4X7 28 Bule. — Multiply the numerators together for the numerator, and the denominators together for the denominator of the frac- Hon, canceling common faetors if they occur in both terms, Bemark. — Whole or mixed numbers must be reduced to im- proper fractions before applying the rule. Problem. — ^Reduce f of ^ of ^ to a simple fraction. OPERATION. Solution. — ^Indicate the work, ^ and employ cancellation, as ^X^X7 shown in the accompanying 3X10X12 -^'W. operation. n a 82 RAY'S HIGHER ARITHMETIC Reduce to simple fractions : 1. 4 of i off .Ins. \, 2. 1 of f of 2f i4ns. |. 3. i of II of 2|. ^ns. 2. 4. i of i of 3i. Am, \\. 5. f of 1 of ^ of 8f . Am. 3f 6. \ of 1 of f of f of 4|. -4n8. |. 7. A of f of tV ofH of 7i. ui^lS. J|. 8. if of A of A ofM of 1t^5. ^ns. ^^. COMMON DENOMINATOR 136. A common denominator of two or more fi*actions, is a denominator by which they express like parts of a unit. 137. The least common denominator (L. C. D.) of two or more fractions, is the least denominator by which they can express like parts of a unit. Principles — 1. Ordy a common multiple of different de- nominators can become a common denominator, 2. Only a least common mvUiple can become a least common dcno7ninator. CASE VI. 138. To reduce fractions to equivalent fractions having a common denominator. Problem. — Reduce ^, f , and f to a common denominator. OPERATIOK. Solution.— Since 2X^X4 = 24, 24 is a 1X3X4 ^12 common multiple of all the denominators. 2X^X4 24 The terms of ^ must be multiplied by 3 X 4 ; 2X2X4 _ 16 the terms of |, by 2 X 4; and the terms of |, by 3X2X4 24 2X3. The values of the fractions are un- 3X2X3 _ 1_8 altered. (l29, m.) 4X2X3~~24 V -s REDUCTION OF FR ACTIONS. 88 Bule. — Multiply both terms of eadi fraction by the d/enommoir tors of the other fractions. Remark. — Since the denominator of each new fraction is the product of the ^ame numbers — viz., all the denominators of the given fractions — it is unnecessary to find this product more than once. The operation* is generally performed as in the following example : Pboblem. — Beduce ^, f , and f to a common denominator. OPERATION. 2X^X7 = 70, common denominator. 1X5X7 = 35, first numerator. i = fi) 3X2X7 = 42, second numerator. } = ^ C Ana. 6X2X5 = 60, third numerator. ? = ?S 3 Note. — Mixed numbers and compound fractions must first be reduced to simple fractions ; the lowest terms are preferable. Reduce to a common denominator: 1. h I h ^rui. ii I*, H- 2. h h h ^ris. ^, Y^, ^. 3. 1, h I- ^^. Hh 1%, m- 4. h h h h ^^. ni m^ uh m- 5. I, i of a^, I of |. Am. 1^, J^, ff 6. -loff f off,iofioff of2f. An8.m>m^m' "Remark. — When the terms of the fractions are small, and one denominator is a multiple of the others, reduce the fractions to a common denominator, by multiplying both terms of each by such a number as will render its denominator the same as the largest de- nominator. This number wUl be found by dividing the largest denomina- tor by the denominator of the fraction to be redv/xd. Problem. — Reduce ^ and f to a common denominator. OPERATION. Solution. — The largest denominator, 6, is a 1X2 2 multiple of 3; therefore, if we multiply both gw 2^^^ terms of J by 6 divided by 3, which is 2, it is 5 ^ reduced to ^. "^ =~q 84 -R^ Y'S HIGHER ABITHMETia Reduce to a common denominator: 1. \y |, and f. Ans. f, f, |. 2. I, I and ■^. Am. ^, |f , ^. 3. f, t, A and H- ' ^w«. i|, it, ^, H. CASE VII. 139. To reduoe firaotions of different denominators, to equivalent firactions having the least conunon de- nominator. Problem. — Reduce f , ^, and ^ to equivalent fractions, having the least common denominator. OPEKATION. Solution. — Find the L. C. M. of 8, 9, 5 5X9 4 5. and 24, which is 72 ; divide 72 by the given "g = oTTg ^^"70 denominators 8, 9, and 24, respectively ; fj 7 V 8 6 6 multiply both terms of each fraction by —= w = ^ the quotient obtained by dividing 72 by ^ q \/ ^ 07 its denominator ; the L. C. M. is the de- — = — _ — = — 2424V372 nominator of the equivalent fractions. In "^ practice it is not necessary to multiply each denominator in form, Bule. — 1. Find the L. C. M, of the denominators of the given fractions, for the L. C D. 2. Divide this L. G. D. by the demmiinator of each fraction, and midtiply the numerator by the quoti&nt, 3. Write the 'product after each multiplication as a fmrner- ator above the L. C. D. Remabk.— ^All expressions should be in the simplest form. Reduce to the least common denominator : 1. h h f ^»«- A' A» tI- ^' T* s* inr» J* Ans, j^, jj^, -J^, -J-J. 3- ^, I, H. An». H, M. -H- 4- I. I. A. M- ^n»- ■$. f. I. f- 6- i, A. H. A- ^n«- H. J^. M. M- 6. If, 3|, and ^ of 3f ^««. J^, iyj^, |J. ADDITION OF FB ACTIONS. 85 ADDITION OF FRACTIONa 140. Addition of Fractions is the process of uniting two or more fractional numbers in one sum. Bemabk. — As integers to be added must express Hhe units (Art. 52), BO fractions to be added must express like pcais of like units. Problem. — ^What is the sum of f, |, and -j^? Solution. — Beducing the given fractions to operation. equivalent fractions having a common denom- f = H i = if inator, we have if, Jf , and JJ. Since these ^^ = JJ are now of the same kind, they can be added H + M + H = i J by adding their numerators. Their sum is $}=lff,^7M. H=iM. Bnle. — Meduce {he ftactions to a common denominator^ add their numerators, and vnrite the sum over tlie common denomr inator. Kemares. — 1. Each fractional expression should be in its sim- plest form before applying the rule. 2. Mixed numbers and fractions may be added separately and their sums united. S. After adding, reduce the sum to its lowest terms. Examples for Practice. 1. i, I, and ■^. Ans, |f 2. I f , I, and ^. An^. 2^ 3. 1| and 2|. Ans. 4j\ 4. 2^, af, and 4f . Ans, lOf^ 5- T%» A» A» a^id ^' ^^- Hi 6. H, 2i, 3i, and 4^. . Am. 11^ 7. I of f , and ^ of | of 2^. Ans, 1^% ^' i + 4f + ih + U' ^^«- Wt ^- l + H + H + M + l*- ^^. ^3^ 10. ^ of 96J4- f of H of &J-. Ans. 59^ 11- i+i+n+i^+m+m- ^m. dm 86 BAY'S HIOHEE ABITHMETIC. SUBTRACTION OF FRACTIONS. 141. Subtraction of Fractions is the process of finding the difference between two fractional numbers. Remabk. — In subtraction of integers, the numbers must be of like units (Art. 54) ; in subtraction of fractions, the minuend and subtrahend must express like parts of like units. Problem. — ^Find the difference between f and -^. Solution. — Reducing the given frac- operahok. tions to equivalent fractions having a { =}^ common denominator, we have J = fj, ^^ = |f and A = M; the difference is A = ^. }^ — J4 = ^ = ^, ^iw. Bule. — Redv^ce the fractions to a common denominator^ and write the difference of their numerators over the common denowr inator, Remabk. — Before applying the rule, the fractions should be in their simpi st form. The difference should be reduced to its lowest terms. Examples for Practice. 1. i — T^. ^ns. H 2. A-i^off Am, m 3. -H-^off Ans. ii^ 4. A-i^of 4. Ans, ^ 5. H — liV- ^ns. H 6- A-i^. . Am. ^ 7. A — H- Am. ^^ 8. ii-A' Am. ^ Remark. — When the mixed numbers are small, reduce them to improper fractions before subtracting; if they are not small, sub- tract whole numbers and fractions separately, and then unite the results. Thus, MULTIPLICATION OF FRACTIONS. 87 Problem. — Subtract 23|f from 31 J. SoiiUnoN. — Beducing the fractions to a operation. common denominator, } = \\, Bat ff can 31} H + li-=H not be taken from H. Take a unit, \i, 2 3j; iJ — ii = |i from the integer of the minuend and add it 7 U, Am. to«. Then ii+if = H, and fi-H = }J. 30 — 23 = 7. Therefore the answer is 1]\. 9. 12|— lOJf. Ans. l\i. 10. 12|| — 9f|. Am. 3^. 11. 5|f — 2^. Ans. 3^. 12. 7^— Sf Am. ^^. 13. 15 — f Am. l4. 14. 18 — 5|. ^n«. 12f. 15. I of 2| — 3|f. ^n«. ||. 16. ^ — \ of If ^rw. l|. 17. J^ of 4J — ^ of 3^ = what? Am. 13f . 18. ll| + 4--9H = wbat? ^rw. lOHf. 19. A man owned ^ of a ship, and sold | of his share : how much had he left? Am. -f^. 20. After selling ^^ of f + ^ of f of a farm, what part of it remains? Am. ^. 21. 3i + 4f - 5^ + 16| - 7H + 10 - 14f, is equal to what? Am. 6f^. 22.5l — 2i + \^ — ^ + 3^+Si- 16i, is equal to what? Am. ||. 23. 1 — I of f — I of f = what? Am. ^. MULTIPLICATION OF FRACTIONS. 142. Multiplication of Fractions is finding the product when either or when each of the factors is a fractional num- ber. There are three cases: 1. To multiply a fruction by an integer, 2. To mvUiply an integer by a fraction. 3. To mvUiply one fraction by another. «iriH 88 BA Y*S HIGHER ARITHMETIC. Note. — Since any whole number may be expressed in the form of a fraction, the first and second cases are special cases of the third. Problem. — Multiply f by f. Solution. — Once f is J. \ times f is J of operaxion. f = A- i tim^ h ^^^^ is 5 times /5 = i|. f X f = iJ, -47i«. Problem. — ^Multiply \ by 6. Solution. — Six times 3 fourths is 18 fourths. Beducing to its simplest form, we have 4|. Problem. — Multiply 8 by f . Solution. — One fifth times 8 is f , 3 fifths times 8 is 3 times |=^*. Re- ducing to a mixed number, we have 4^. Note. — The three operations are alike, and from them we may derive the rule. operation. iXf = ^ = 4l,^7w. OPERATION. fX| = V = 4f,^t«. or,8X«=¥ = 4f Bule. — MvUiply {lie numerators together for the numercdor of Hie product, and the denominators for the denominator of the product Kemark. — ^Whole numbers may be expressed in the form of fractions. 1. tt X 12. 2. H X 18. 3. If X 24. Examples for Practice. Ans. 9^. Ans, 8 J. Ans. 14J. 4. T^ X 28. 5. H X 30. 6. 3f X 5. Ans. 15f. Ans. 26. Ans. 18 J. OPERATIONS. 3f 3| = V Remark. — In multiplying a mixed number by a whole num- ber, multiply the whole number and the fraction separately, and add the products; or, reduce the mixed number to an im- " "*' proper fraction, and multiply it ; as, in the last example. 15 3* ¥X5 =^ ^^ = 18}, ^ns. MULTIPLICATION OF FEACTIONS 89 7. 45 X f 8. 50 X H- 9. 25 X |. 10. 32x2f. AuB. 35. Am. Z^. Am. 18|. Am. 76. 11. 28 X 3|. 12. if X -h- 13. li X H. 14. H X H. ;ln«. 102|. Am. ^. -4n«. if. 15. What will 3i yards of cloth cost, at $4^ per yard ? teKMAKK. — ^In finding the product of two mixed numberfl, it is generally best to reduce them to improper fractions ; thus, 41 = 1; 3i = Jjf ; $JX ¥== V^ = $l 5, An*. OPERATION. SoiiTJnoN. — The operation may be performed without reducing to improper fractions ; thus, 3 yards will cost $131, And 1 of a yard will cost \ of $41 = $11; henoe, the whole will cost $15. $41 131 ii 16. 6|X41. Am. 30. 17. 4| X 2|. Am. l^. Ann, $15 18. 12f X3^. Am. 40^. 19. 7ii X 3^. Am. 2&i. Am. 4. Am. 7h Am. 4^. Am. 94|. Am. -5%. Am. 4. Ans. 49. 20. Multiply i of 8 by i of 10. 21. Multiply I of 5f by f of aj. 22. Multiply f of | of 5f by f of 3|. 23. Multiply 5, 4^, 2^, and f of 4f . 24. Multiply |, f , ^, i of 2^, and ^ of 3^. 25. Multiply |, i, ^, 3^, and 3|. 26. Multiply 3i, 4|, 5|, | of ^s^^and 6|. 27. At i of a dollar per yard, what will 25 yards of cloth cost? $21|. 28. A quantity of provisions will last 25 men 12f days: how long will the same last one man ? 31 8| days. 29. At 3i cents a yard, what will 2| yards of tape cost? ' 9|- cents. 30. What must be paid for f of -f of a lot of groceries that cost $18f ? $7^. 31. K owns I of a ship, and sells f of his share to L: what part has he left? -fs* H. A. 8. 90 RAY'S HIGHER ARITHMETIC. DIVISION OF FEACriONa 143. Division of Fractions is finding the quotient when the dividend or divisor is fractional, or when both are fractional. There are three cases: 1. To divide a fradion by an integer. 2. Tb divide an integer by a fra^stwn. 3. To divide one fraction by another. Note. — Since any whole number may be expreesed in the form of a fraction, the first and second cases reduce to the third case. Problem.— Divide f by ^^ Solution. — } is contained in 1, seven operation. times ; } is contained in J, J of 7 = J times ; "|-*-f = fXJ = H I is contained in 4, 5X J=V times; f is contained in f , J of -*/■ == ii times. It will be seen that the terms of the dividend have been miUtipliedf and that the terms of the divisor have exchanged places. Writing the terms thus is called " inverting the terms of the divisor," or simply, " inverting the divisor." Problem. — ^Divide 3 by |. OPERATION. Solution. — J is contained in f-f-f = }Xf = ^ = 7i, Ans. 1, five times ; J is contained in Or, 3-^1 = ^ = 7 J. 3, three times 5 times = 15 times ; f is contained in 3, J of 15 times = -^^ = 7^ times. Problem. — Divide f by 2. OPERATION. Solution. — ^Two is contained ^-^f = ^Xi = A = ?r ^^^^ in 1, J times ; 2 is contained in Or, ^ -s- 2 = J. I, J of f = iV times ; 2 is con- tained in f , 4 times ^^ = ^ = ^ times. Note. — From these solutions we may derive the following rule. Bule. — Multiply the dividend by the divisixr vnth its termn inverted. DIVISION OF FRACUONS. 91 Bkmakk. — The terms of the divisor are inyerted because the solution requires it. The same may be shown by a different solu- tion, as below. Problem. — Divide | by f. Solution. — Bednce both' dividend and divisor to a common denominator. The quotient of H "^ H ^^ the same as 10 -r- 27 = ij. The same result is obtained by multiplying the dividend by the divisor, with its terms inverted ; thus, i X I = i?. OFERATIOK* Beharb:. — Mixed nnmbers mnst be rednced to improper frac- tions. Use cancellation when applicable. Examples for Pbacticb. 1. A-^3. 3. 1^8. 4. 6-M. 5. 21 ^t1^. Am. -^ Am. 4s Am. -^ Am. 9 Am. ^^ Am. 1-J-, Am. 261 8- *iy«. 10. lf-^5. 11. ^^i. 12. \^-^^. 13. ^^-^^. 14. 54if-^25|. 15. Divide 1^ by \ of f of 7|. 16. Divide ^ of 3^ of 3^ by ^ of \^. 17. Dividef of^byjof |. 18. Divide I of 3| by H of 7. 19. Divide i of I X ih by ^ of 3^. 20. Divide ^ of 5^^ by | of ^ of 3^. 21. Divide ^ of ^ of ^ by f of ^ of f 22. Divide 1| times 4| by 1^ times 3|. 23. Divide ^ by f of 8} times -j^ of 3^. 24. Divide ^ of f of 27^ by | of ^^ of 5^. 25. What is 2| X I of 19^-^(4^ X tV ^^ ^)? Am. •^. Am, fl- Am. -^jf. Am. 12^. Am. lOf Ans. H. Am. 2\. Am. f . Am. ^. Am. 2J. A718. If. Am. f . -4n3. If. Am. -j^. J.w«. If. Aii8. 34^. ^ns. 3^. 92 BA Y'S HIGHER ARITHMETIC. 9 144. To reduce complex to simple fractions. Problem. — Reduce — to a simple fraction. OPERATION. Explanation.— The mixed If =^V 2 J numbers are reduced to im- V "^ i = V X 4 = If = f f> ^"»- proper fractions^ and the numerator is divided by the denominator. (Art. 114.) Itule. — Divide the numerator by the denominator' , as in division of fractions. Eeduce to simple fractions: 1. f 2 y 2f ^ ii Am. ■^. Ant. |. Ans. If. 4. ^* 18 62 6. 16^- Am, 1\, Ans, W, Ans, 3|. Bemabk. — Complex fractions may be multiplied or divided, by reducing them to simple fractions. The operation may often be shortened by cancellation. 7. ^ X ^. Ans. iH- 31 ^ 25 "^ lOi* ^* ^' Q ^ V ^ Ann 650 94- 244- ' ^sWt"' 2| • 12^ 10. r? 4- ~. ^n«. 4|H- 40f • 73 • ^' 12 ^A-^-i^ A-nit ^ THE GREATEST COMMON DIVISOE OF FRACTIONS. 145. The gpreatest common divisor of two or more fractions is the greatest fraction that will exactly divide each of them. O, a D. OF FRACTIONS. 98 One fraction is divisible by another when th^ numerator of the divisor is a factor of the numerator of the dividend, and the denominator of the divisor is a multiple of the denominator of the dividend. Thus, ^5 is divisible by ,», ; for A = i! ; H "^ A = ^• The greatest common divisor of two or more fractions, must be that fraction whose numerator is the G. C. D. of all the numerators, and whose denominator is the L. C. M. of the denominators. Thus, the G. C. D. of -^ and J| is ^fy. Problem. — ^Find the G. C. D. of |, |f , and -j^. Solution. — Since 5 and 7 are both operation. prime numbers, 1 is the G. C. D. of 1 = G. C. D. of 5, 25, 7 all the numerators ; 96 is the L. C. 9 6 = L. C. M. of 8, 32, 12 M. of 8, 32, and 12 ; therefore, the G. ^j, Ans. C. D. of the fraction is ^5. Bule. — Fhid the G, C. D, of the numerators of the fractionSy and divide it by the L, C, M, of their denominators, Eemark. — The fractions should be in their simplest forms before the rule is applied. Fmd the greatest common divisor: 1. Of 83^ and 268f. Am. 2^. 2. Of 14^ and 95f . Am, ^. 3. Of 59i and 735||. Am. 2||. 4. Of 23^2^ and 213||. Am. 2||. 5. Of 418| and 1772|. Am. ||. 6. Of 261|| and 652||. Am. 4|f. 7. Of 44t, 546|, and 3160. Am, 4f 8. A farmer sells 137^ bushels of yellow com, 478^ bushels of white corn, and 2093f bushels of mixed corn : required the size of the largest sacks that can be used in shipping, so as to keep the com from being mixed ; also the number of sacks for each kind. 3| bushels ; 44, 153, and 670. 94 HA Y'S HIQBER ARITHMETIC, 9. A owris a tract of land, the sides of which ate 134f , 128^, and 115^ feet long: how many rails of the greatest length possible will be needed to fence it in straight lines, the fence to be 6 rails high, and the rails to lap 6 inches at each end? 354 rails. THE LEAST COMMON MULTIPLE OF FRACTIONS. 146, The least common multiple of two or more ^'actions is the least number that each of them will divide exactly. Note. — The G. C. D. of several fractions must be a fraction, but the L. C. M. of several fractions may be an integer or a fraction. A fraction is a mtdtiple of a given fraction when its numerator is a multiple, and its denominator is a divisor, of the corresponding terms of the given fraction. Illustration. — j^ is a multiple of y^. 8 is a multiple of 2, and 11 is a divisor of 33 ; hence, ■A:-J-A = AX^^ = 12. The same result is otherwise obtained ; thus, -^ = f J, and JJ t- ^ = 12. A fraction is a comnum multiple of two or more given fractions when its numerator is a common multiple of the numerators of the given fractions, and its denominator is a common divisor of the denominators of the given fractions. A fraction is the least common multiple of two or more fractions when its numerator is the least common multiple of the given numerators, and its denominator is the greatest common divisor of the given denominators. Problem. — ^Find the L. C. M. of i, f , and f . Solution. — The L. C. M. of operation. the numerators is 15. The G. L. C. M. of 1, 3, 5 = 3 X ^ = 1 5 C. D. of the denominators is 1 ; G. C. D. of 3, 4, 6 = 1 therefore, the L. C. M. of the •"• V, Ans. fractions is ^, or 15. Z. a M. OF FBACTJON& 95 Bule. — Divide the L. C. M. of Oie mmwraJUm by the O. C. D. of ike denominators. Kemabk. — ^The fractions must be in their simplest forms before the rule is applied. Find the least common multiple : 1- Of I, f , I, I, and f Am. 60. 2. Of 4}, 6f , 5|, and 10^. Am. 472^. 3. Of 4> 4» A» 5f » and 12|. Ans. 350. 4. A can walk around an island in 14^ hours ; B, in 9^^ hours ; C, in 16| hours ; and D, in 25 hours. If they start from the same point, and at th^ same time, how many hours afier starting till they are all together again ? 100 hours. Promiscuous Exercises. Note to Teachers. — All problems marked thus [*], are to be solved mentally by the class. In the solution of such problems, the following is earnestly recommended : 1. The teacher will read the problem slowly and distinctly, and not repeat it. 2. The pupil designated by the teacher, will then give the answer to the question. 3. Some pupil, or pupils, will now reproduce the question in the exact language in which it was first given to the class. 4. The pupil, or pupils, called upon by the teacher, will give a short, logical analysis of the problem. 1. What is the sum of 3|, 4^, 5^, f of |, and | of ^ of I? isu- 2. The sum of 1^ and — is equal to how many times their difference? ^ 5 times. 3. What i8(2f + |ofl_M)-=-l^? 6. 96 SAT^ S nrOHEB ABITBMETia 4. Eeduce i^-^LM and 5 x (100- ?^ + 14) to their simplest forms. 16 and 26^^. 5. What is i of 5^ — ^ of 3f ? ^. 6. What is If X -AAf X ifl X If equal to? ^. 7.* |X-^X^? tV- 8. Ixli^x^-ix-^-ix^^i? Vft. 32354 ^^ 9. (2 + i)^(3 + _ ^i^^,^ (2-i)X(4-3f) ^ 10. ii^lililiizil^ what? 4M. 4iX4i-l ^ 11. Add I of I of I, i X I of 1 J, and \. ^. 12. 4 of V' of what number, diminished by — 3& — , leaves ff ? ^. 13.* James's money equals f of Charles's money ; and J of James's money + ^^^ equals Charles's money : how much has each? James, $36; Charles, $60. 14.* A leaves L for N at the same time that B leaves N for L. The two places are exactly 109 miles apart: A travels 1\ miles per hour, and B, 8J miles per hour; in how many hours will they meet, and how fer will each have traveled? 6ff hours. A, 51^ miles; B, 57^ miles. 15. What number multiplied by •§■ of f of 3|^ will produce 2}? 2|f. 16. What, divided by 1|, gives 14f ? 23|. 17. What, added to 14f , gives 29|| ? ISyf^. 18.* I spend ^ of my income in board, ^ of it in clothes, and save $60 a year : what is my income ? $216. 19.* Divide 51 into two such parts that -J of the first is equal to f of the second. 27 and 24. 20.* \ is what part of |? |. COMMON FRACTIONS. 21. Divide ^ of 3f by jj of 7; ^d ^ of of -^ of 54^. 22. Multiply ^ of 2^ by ^ of 19^; and of 14^ by y\ of ^ of 13|. 23. (iMi+i♦H + H!l)-^f =wbat? ^. J 24 * A bequeathed ^ of his estate to his ( rest to his younger, who received $526 less thj What was the estate? 25. Find the sum, difference, and product also, the quotient of their sum by the differen Sum 5^, diff. Iff, prod. 8fJ 26.* A cargo is worth 7 times the ship : wl 4Jf I cargo is -j^ of the ship and cargo ? 27. By what must the sum of ^^^, -^^ multiplied to produce 1000? 2S, Multiply the sum of all the divisors of I ing 1, by the number of its prime fitctors ex< divide by 149^, and I 29. ^ is what part of 10^? Reduce 1 ^'' ^rCHarle^' 1^ 30..Multiply 1, 14f, ?!, 1, 1*, and 6. ^/ that B lea^^^ 5i' ^' 4' 7V 2|' i ^^ ^^AfiS aP^' 31. J of I of what number equals 9||? y ., per bour;^ 32.* A .63-gallon cask is | fuU: 9^ gaUont b\ ^^^ ^eacb b*^ off, how full wiU it be? ^ bo^ ^ g 57^ 0»^ . 33. If a person going 3f miles per hoi ^ tJUl^' i of ^ I I journey in 14f hours, how long would he be, by t ^^ ^ ^^ I 5i mUes per hour? " '■ I 34. A man buys 32f pounds of coffee, a 24}' ^^ pound : if he had got it 4| cents a pound cheaj 29^ t ^ \k\si cl<>*^ I more pounds would he have received ? I \)OSS^9 e I**.' 1 35. Henry spent -j^ of his money and then \jicoTJCi^ ' f tbe ^ i ^^ tlkeii lost f of all his money, and had in uirts tb»^ » ^1^^ , ' than at first. How much had he at first? ^^ f H. A, 9. 98 BAY'S HIGHER ARITHMETIC Topical Outline. Common Fractions. 1. Definition. 2. ClasKS. & Terms.. ' 1. Ab to Kinds < 1. Common. * \2. Decimal. ( 1. Proper. 2. As to Value • • 2. Improper. i .3. Mixed. r 1. Simple. L 3. As to Form. • ' 2. Complex. .8. Compound. f 1. Numerator.. 2. Denominator. 8. Similar. L 4. Dissimilar. 4. Principles. 5. Reduction. 1. Cases.... 2. Principles. 1. Lowest Tprms. 2. Higher Terms. 8. Mired Numbers to Improper Fractions. 4. Improper Fractions to Mixed Numbers. 5. Compound to Simple Fractions. 6. Common Denominator. 7. licast Common Denominator. 1^ 8. Rules. 6. Practical Applications.. 1. Addition 2. Subtraction. 8. Multiplication. 4. Division Is. 18. {^ 18. rl. Deflni J 2. Prlnd \%, Rule, I. Definition. Principles. Rule. 1. Definition. Principles. Rule. I. Definition. Principles. Rule. 1. Definition. Principles. 5. Divisors, Multiples, etc IX DECIMAL FRAOnOIirS. 147. A Decimal Fraction is a fraction whose denoni- iuator is 10, or some product of 10, expressed by 1 with ciphers annexed. Bemask 1. — A decimal fraction is also defined as &Jraetion who9t denominator is some power cf 10. By the ^^potoer" of a quantity, is usually understood, either that quantity itself, or the product arising from taking only that quantity a certain number of times as a factor. Thus, 9 = 3 X 3» or the aeeond power of 3. Remark 2. — Since decimal fractions form only one of the daeees (Art. 108) under the term fixutUmSy the general principles relat- ing to common fractions relate also to decimals, 148. The orders of integers decrease from left to right in a tenfold ratio (Art. 48). The orders may be continued from the place of units toward the right by the same law of decrease. 149. The places at the right of imits are called decimal places, and decimal fractions when so written, without a denominator expressed, are called decimals. 160. The decimal point, or separatriz, is a dot [ . ] placed at the left of decimals to distinguish them from integers. Thus, -^ is written .1 TrAnr " " .001 From this, it is evident that, The denominojlxyr of any decimal is 1 with as many cyphers annexed as there are places in the dednud. 151. A pure decimal consists of decimal places only ; as, .325 ' (99) 100 I^AY'S HIGHER ARITHMETIC. 152. A mixed decimal consists of a whole number and a decimal written together; as, 3.25 Remark. — A mixed decimal may be read as an improper frac- tion, since ^^^H^- 153. A com.plex decimal has a common fraction in its right-hand place; as, .033^ 154. From the general law of notation (Arts. 48 and 148) may be derived the following principles : Principle I. — If, in any decimal^ ilie point be moved to the right, Hie decimal is multiplied by 10 as often as the point is removed one place. Illustration. — If, in the decimal .032, we move the point one place to the right, we have .32. The first has three decimal places, and represents thousandths; while the second has two places, and represents hundredths, (Art; 129, Prin. i.) Principle II. — if, in any decimjol, the point be moved to the left, the decivwl is divided by 10 a^ often as the point is removed one place. Illustration. — If, in the decimal .35 we move the point one place to the left, we have .035. The first represents hundredths; the second, thrascmdihs, while the numerator is not changed. (Art. 129, Prin. n.) Principle III. — Decimal ciphers may be annexed to, or omitted, from, the right of any number without altering its valve. Illustration.— .5 is equal to .500; for i^ox ISo — i^- The reverse may be shown in the same way. (Art. 129, Prin. in.) NUMERATION AND NOTATION OF DECIMALS. 155. Since .6 = -^; .06 = y^ ; and .006 = y^, any figure expresses tenths, hundredths, or thomandths, according as it is in the 1st, 2d, or 3d decimal place; hence, these places are named respectively the tenil\s\ the hundredtiis\ the DECIMAL FRACTIONS. thousmidih^ place ; other places are named in the same way, as seen in the following table: Table op Becdul Orderb. 1st plac« .2 .... read 2 Tenths. 2d " .08 ... . ' 8 Hundredths. 3d " .005 . . . ' 5 Thoueandths. 4th " .0007 . . . ' 7 Ten-thousandths. 6th " .00003 . . ' 3 Hundred-thousandths. 6lh " .000001 . . • 1 Millionth. 7lh ■■ .0000009 . * 9 Ten-millionths. 8th " .00000004. ' 4 Hundred-millionths. 9th '■ .000000006 ' 6 BiUionths. Note:.— The names of the decimal orders are derired from the names of the ordera of whole numbera. Tlie table may therefore be extended to trilliontha, quadrilliooths, etc. Problem.— Eead the decimal .0325 SoLDTiciN. — The numerator is 325; the denomination u ten- IhousandtliB eince there are four decimal places. It is read three liundred and twenty-five ten-thousandths. Bule. — Bead the number expressfd in tAc d^dmeU ptacet a* tiie numerator, give it the denomination exj^tased by Uie riglit- hand fyure. Examples to be Read. 102 BA Y'S HIOHER ARITHMETIC. 5. .7200 13. 2030.0 6. .5060 14. 40.68031 7. 1.008 15. 200.002 8. 9.00J 16. .0900001 9. 105.01 17. 61.010001 10. .0003 18. 31.0200703 11. 00.100 19. .000302501 12. 180.010 20. .03672113 Exercises in Notation. 156. The numerator is written as a simple number; the denomination is then expressed by the use of the decimal point, and, if necessary, by the use of ciphers in vacant places. Problem. — Write eighty-three thousand and one billionths. Explanation. — First write the numerator, operation. 83001. If the point were placed immediately .000083001 at the left of the 8, the denominator would be hundred-thousandths ; it is necessary to fill four places with ciphers so that the final figure may be in billionths' place. Biile. — WrriQ ihe numerator as a vMe number; then place Vie decimal point m (hat ihe rigid-hand figure shall be of ihe same name as ihe decimal. Examples to be Written. 1. Five tenths. 2. Twenty-two hundredths. 3. One hundred and four thou- sandths. 4. Two units and one hun- dredth. 5. One thousand six hundred and five ten-thousandths. 6. Eighty-seven hundred-thou- sandths. 7. Twenty-nine and one half ten-mi Ilionths. SEDUCTION OF DECIMALS, 103 8. Nineteen million and one billionths. 9. Seventy thousand and forty- two units and sixteen hun- dredths. 10. Two thousand units and fifty-six and one third mill- ionths. 11. Four hundred and twenty- one tenths. 12. Six thousand hundredths. 13. Forty-eight thousand three hundred and five thou- sandths. 14. Eight units and one half a hundredth. 15. Thirty-three million ten- millionths. 16. Four hundred thousandths. 17. Four hundred-thousandths. 18. One unit and one half a billionth. 19. Sixty-six thousand and three millionths. 20* Sixty-six million and three thousandths. 21. Thirty-four and one third tenths. REDUCTION OF DEaMALS. 167. Beduction of decimals is changing their form vdthout altering their value. CASE I. 158. To reduce a decimal to a common fraction. Problem. — Beduce .24 to a common fraction. Solution. — .24 is equal to -^j which, operation. reduced, is A. . 2 4 = ^^^ == ,<^, Aiiz. Problem. — ^Reduce .12^ to a common fraction. operation. Solution.— Write 12J as a 221 numerator, and under it place . 1 2 J =? -— - =.^= ^^ = J, Am, 100 as a denominator. Re- duce the complex fraction according to Art. 144. Bule. — FHte tfte decimal as a common fraction; then re- duce the fraction to its lowest terms. 104 BAY'S HIOHER ARITHMETIC, Bemakk. — If the decimal contains many decimal places, an approximate value is sometimes used. For example, 3.14159 = nearly 3|. Reduce to common fractions . 1. .25625 Am. T^ff. 9. ll.Of Am. lliV- 2. .15234375 Am. ^^. 10. .390625 Am. ||. 3. 2.125 Arvi. 2\. 11. .19441 Am. ^j. 4. 19.01750 Am. 19j^. 12. .24| Am. \i. 5. 16.00^ Am. l&^hf. 13. .33i Am. \. 6. 360.028f Am. 350"^. 14. .66$ Am. \. 7. .6666661 Am. ^. 15. .25 Ans. \. 8. .003125 CAS 16. E II. .75 Am. \. 169. To reduce common fractions to decimals. Problem. — Reduce ^ to a decimal. OPERATION. 8oLiiTioN.--The fraction | = J of 7. 7 = 8)7.000 7.0, ^ of 7.0 = .8, with 6 tenths remaining ; .6. .87 5,. Ans, =r.'60, \ of .60 =.07, with 4 hundredths re- maining; .04 = .040, J of .040 = .005. The answer is .875. Note. — Another form of solution may be obtained by multiply- ing both terms of the fraction by 1000; dividing both terms of the resulting fraction by the first denominator and writing the answer as a decimal. Thus, J = }^^, |^J = x^^\ = .875. Both solutions depend on Art. 129, Prin. ni. . 'Rvle.— Annex ciphers to the numerator and divide it by the denominator. Then point off as many decimal places in Hie qmtient as there are ciphers annexed. Bemark. — Any fraction in its lowest terms haying in its de- nominator any prime factor other than 2 and 5, can not be reduced exactly to a decimal. Thus, ^j = .08333 -{-. The sign + is used at ADDITION OF DECIMALS, 105 the end of a decimal to indicate that the result is less than the true quotient. The sign — is also sometimes used to Indicate that the last figure is too great Thus, \ = .1428 -f-, or, by abbreviating, \ = .143—. Reduce to decimals: 1. |. Am, ,1b 2. \. Am, .125 3. ^V Am. .05 4. If. Am, .46875 5. ^^. Am, .005625 6. ^. Am, .8 7. ^^. Am, .495 8. ^j. Am, .078125 9. ^. ^««. .05078125 10. y^. ^«5. .0009765625 Note. — The rule converts a mixed number into a mixed decimal, and a complex into a pure decimal ; thus, 9f = 9.375, since f = .375 ; and .263^5 = .2612, since ^ = .12. 11. l^ Am. 16.5 12. 42^ Am. 42.1875 13. .015^ Am, .01525 14. lOl.Olf Am, 101.0175 15. 751.19^ Am, 75119.0375 16. 2.00^ Am, 2.00003125 ADDITION OF DECIMALS. 160. Addition of Decimals is finding the sum of two or more decimals. Note. — Complex decimals, if there are any, must be made pure, aa far, at least, as the decimal places extend in the other numbers. -Vi.. Problem.— Add 23.8 and 17^ and .0256 and .41|. operation. Solution. — Write the numbers 2 3.8 as in the operation, and add as in 17i = 17.5 simple addition. Write the deci- .0256 mal point in the sum to the left of .412^= .4166f tenths. 41 .74221, Ana, 106 BAY'S HIOHEB ARITHMETIC. Bule. — 1. Wriie Ihe numbers so that figures of the scam order shaU stand in the same column. 2. Then add as in simple numbers, and put ilie decimal point to the left of tenths. Bemark. — ^The proof of each fundamental operation in decimals is the same as in simple numbers. 1. Find the sum of 1 + .9475 1.9475 2. Of 1.33| added to 2.66^ 4. 3. Of 14.034, 25, .000062^, .0034 39.0374625 4. Of 83 thousandths, 2101 hundredths, 25 tenths, and 94i^ units. 118.093 5. Of .16f, .37^, 5, 3.4|, .OOOJ 8.9805«j 6. Of 4 units, 4 tenths, 4 hundredths. 4.44 7. Of .Hi + .6666f + .2222221 1. 8. Of .14f, .018f, 920, .0139^^.. 920.1754 9. Of 16.008J, .00741, .2f, .00019042| 16.299768199| 10. Of 675 thousandths, 2 millionths, 64|, 3.489107, and .00089407 68.29000307 11. Of four times 4.067| and .000^ 16,272 12. Of 216.86301, 48.1057, .029, 1.3, 1000. 1266.29771 13. Add 35 units, 35 tenths, 35 hundredths, 35 thou- sandths. 38.885 14. Add ten thousand and one millionths; four hundred- thousandths ; 96 hundredths ; forty-seven million sixty thou- sand and eight billionths. 1.017101008 SUBTKACTION OF DEaMALS. 161. Subtraction of DecimalB is finding the difference between two decimals. Problem. — From 6.8 subtract 2.057 Solution. — Write the numherR so that units of operation. the same order stand in the same column ; sup- 6 . 8 pose ciphers to be annexed to the 8, and subtract 2. 057 as in whole numbers. 4.743, Ans. SUBTRACTION OF DECIMALS. 107 Problem.— From 13.256f subtract 6.77^ ExPLAKATiON. — In this example, the complex decimals are carried out by division to the same place, and the common fractions treated by Art. 141. OPERATION. ia.256i 6.77i = 6.773t 6.483f, Am. Bule. — 1. WrSe the subtrahend beneath the mnvfind so Viat uiiUs of the same order stand in the same column. 2. Subtract as in simple numbers, and write the decimal pohd as in addition of decimals, NoT£s. — 1. If eitlier or both of the given decimals be complex, proceed as directed in the second problem. 2. If the minuend has not as many decimal places as the subtra- hendy annex decimal ciphers to it^ or suppose them to be annexed, until the deficiency is supplied. Examples for Practice. 1. Subtract 8.00717 from 19.54 2. 3 thousandths from 3000. 3. 72.0001 from 72.01 4. Subtract .93^ from 1.169^ 5. How much is 19 — 8.999^? 6. How much less is .04^ than .4? 7. How much is .65007 — ^? 8. What is 2f — 1^ in decimals ? 9. Subteaet 1 from 1.684 10. f of a millionth from .000| 11. 1^ hundredths from 49f tenths. 12. 10000 thousandths from 10 units. 13. 24^ tenths from 3701 thousandths. 14. 1^ units from 1875 thousandths. 15. f of a hundredth from ^ of a tenth, 16.^64^ hundredths from 100 units. 11.58283 2999.997 .0099 .238^ lO.OOOf .35f .15007 .95 .684 .000443H 4.9225 0. 1.251 0. 0. 99.351 -i» 108 HA rS HIOHEB ARITHMETIC. MULTIPLICATION OF DECIMALS. 162. Multiplication of Decimals is finding the product when either or when each of the factors is a decimal. Principle. — T/ie numbtr of decimal places in the product equaU Uie number of decimal places in both factors. . Problem. ^Multiply 2.56 by .184 OPERATION. Solution.— 2.56 = fJJ, and .184 := ^y^. 2.5 6 Now, f JJ X iVW - AVijW; that 18, the product . .184 of hundredths by thousandths is hundredth- 10 2 4 thousandths. This requires five places of 20 48 decimals, or as many as are found in both 25 6 factors. .47104, Ann. Bule. — 1. Multiply as in whole numbers, 2. Point off as many decimal places in ffie product as there are decimal places in the two faxstors. Remarks. — 1. If the product does not contain as many places as the fac(tors, prefix ciphers till it does contain as many. 2. Ciphers to the right of the product are omitted after pointing. Examples for Practice. 1. IX.l .1 2. 16 X .03^ .b^ 3. .OlX.ll .0015 4. .J080 X 80. 6.4 5. 37.5 X 82| 3093.75 6. 64.01 X .32 . 20.4832 7. 48000x73.. 3504000. 8. 64.66|X18. 1164. 9. .56JX.O33V .0172-H 10.. 738X120.4 98855.2 MULTIPLICATION OF DECIMALS, 109 11. .0001 X 1.006 12. 34 units X. 193 13. 27 tenths X A\ 14. 43. 7004 X. 008 15. 21.0375 X 4.44| 16. 9300.701 X 251. 17. 430.0126X4000. 18. .059 X .059 X .059 19. 42 units X 42 tenths. 20. 21 hundredths X 600. 21. 7100 X i of a miUionth. 22. 26 millions X 26 millionths. 23. 2700 hundredths X 60 tenths. 24. 6.3029 X .03275 25. 135.027 X 1.00327 .0001006 6.562 1.134 .3496032 93.5 2334475.951 1720050.4 .000205379 176.4 13.5 .0008875 676. 162. .206419975 135.46853829 163. Oughtred's Method for abbreviating multiplica- tion, may be used when the product of two decimals is riequired for a definite number of decimal places less than is found in both Actors. Problem.— Multiply.3.8640372 by 1.2479603, retaming only seven decimal places in the product. Explanation. — It is evident that we need regard only that portion of each partial product which affects the figures in and above the seventh decimal place. Beginning with the highest figure of the multiplier, we obtain the first partial product. Taking the second figure of the multiplier, we carry each figure of the partial product one place to the right, so that figures of the same order shall be in Uie same column. This product is carried out one place further than is required, so as to secure accuracy m the seventh place, and we draw a perpendicular line to separate this portion. The product of OPERATION. 3 6 9 7 4 2 3.8640372 1.2479603 38640372 7728074 1545614 270482 34776 2318 11 4.8221650 4 8 G 3 4 5 ,04 by the right-hand 110 MAY'S HIGHER ARITHMETIC. figure in the seventh place, would extend to the ninth place of decimals ; so we may reject the la»t figure, and commence with the 7. With each succeeding figure of the multiplier, we commence to multiply at that figure of the multiplicand which will produce a product in the eighth place. It is also convenient to place each figure of the multiplier directly oVer the first figure of the multipli- cand taken. In multiplying hy .007, we have 7X3 = 21 ; but, if we had been expressing the complete work, we should have 5 to carry to this place ; the corrected product is therefore 21 -|- ^ = 26. The product from the last figure, 3, is carried two places to the right. In the total product, the eighth decimal is dropped ; but the seventh decimal figure is corrected by the amount carried. Bule. — 1. Multiply only mck figures as shall produce one more than the required nuwher of decimal places. 2. Begin wiih Vie highest order of the multiplier; under the rightrhand figure of each partial product, place the riglUrhand figure of the succeeding one. In obtaining su/h right-liand figure, let that number he added which would he carried from multiplying the figure of the next lower order. 3. Add the partial produds, and ryect the right-hand figure. Kemarks.— 1. It will be found convenient to write the multiplier in a reverse order, with its uniti^ figure under that decimal figure of the multiplicand whose order is next lower than the lowest required. Thus, in the fourth example, the 8 would be written under the 1. 2. In carrying the tens for what is left out on the right, carry also one ten for each 5 of units in the omitted part ; thus, 1 ten for 5 or 14 units, 3 tens for 25 or 26 units, etc. Make the same correction for the final figure rejected in the product. ExAMi^^Es FOR Practice. 1. Multiply 27.653 by 9.157, preserving three decimal places. 253.219 2. Multiply 43.2071 by 3.14159, preserving four decimal places. 135.7390 3. Multiply 3.62741 by 1.6432, preserving four decimal places. 5.9606 DIVISION OF DECIMALS. Ill 4. 9.012 X 48.75, preserving one place. 439.3 5. 4.804136 X .010759, preserving six places. .051688 6. Slij^ X 26||, preserving three places. 21813.475 7. 702.61 X 1.258^, preserving three places. 884.020 8. 849.931 X .0424444, preserving three places. 36.075 9. 880.695 X 131.72 true to units. 116005 10. .025381 X .004907, preserving five places. .00012 11. 64.01082 X .03537, preserving six places. 2.264063 12. 1380. 37^ X .234f, preserving two places. 82416 DIVISION OF DEajdALS. 164. Division of Decimals is the process of finding the quotient when either or when each term is a decimal. 165. Since the dividend corresponds to the product in multiplication (Art. 73), and the decimal places in the divi- dend are as many as in both factors (Art. 162, Prin.), we derive the following principles : Principles. — 1. TAe dividend mud contain as many deci- mal places as the divisor ; and when both have the same number, Hie quotient is an integer, 2. The quotierd must contain as many decimal places as th£ number of those in the dividend exceeds the number of those in Hue divisor. Problem.— Divide .50312 by .19 OPERATION. .19). 50312(2. 648, Am. Solution. — The division is per- 38 formed as in integers. The quo- 12 3 tient 18 pointed according to 114 Principle 2. The quotient must 9 1 have 6 — 2 = 3 places. 7*6 152 152 112 RAY'S HIGHER ARITHMETIC. Proof. — By expressing the decimals as common fractions we have; miifi -^ iV^ = imVir X W =!fU = 2.648 Problem.— Divide .36 by .008 Solution. — The dividend has a less num- operation. ber of decimal places than the divisor. .008 ) . 360 Annex one cipher, making the number 45, A-m, equal. The quotient is an integer. Problem.— Divide .002^ by .06| OPERATION. Solution.— Reducing the . 002 1| = . 002475 mixed decimals to equiva- .06 J =.0 66 lent pure decimals, we have .066). 002475 (.037 5, ilf». .002475 and .066. Dividing, 198 we find one more decimal 4 9 5 place necessary to make the 4 6 2 division exact ; and, pointing 330 by Principle 2, we have .0375 • 3 30 Bule. — Divide as in whole numbers, and point off as matiy decimal places in the quotient as those in the dividend exceed tJwse in tJw divisor. Note. — When the division is not exact, annex ciphers to the dividend, and carry the work as far as may be necessary. Examples for Practice. 1. 63 : 4000. .01575 2. 3.15 : 375. .0084 3. 1.008 : 18. .056 4. 4096 : .0S2 128000. 5. 9.7 : 97000. .0001 6. .9^.00075 1200. 7. 13-^78.12^ .1664 8. 12.9-^8.256 1.5625 9. 81.2096-7-1.28 63.445 10. 1-MOO. .01 11. 10.1 -M7. .59412— 12. .001^100. .00001 DIVISION OF DECIMALS, 113 13. 12755 -f- 81632. .15625 14. 2401 -T- 21.4375 112. 15. 21.13212^.916 23.07 16. 36.72672-^.5025 73.088 17. 2483.25-^5.15625 481.6 18. 142.0281^9.2376 15.375 19. .08^-^.121 .66f = §. 20. .0001— .01 .01 21. 95.3 -^. 264 360.984848 -f 22. 1000 -^ .001 1000000. 23. Ten -~ 1 tenth. 100. 24. .000001-^.01 .0001 25. .00001 -MOOO. .00000001 26. 16.275 4-. 41664 39.0625 27. 1 ten-millionth -f- 1 hundredth. .00001 • 166^ Oughtred's Method. — ^If the quotient is not re- quired to contain figures below a certain denomination, the work may sometimes be abridged. Problem.— Divide 84.27 by 1.27395607, securing a quo- tient true to four places of decimals. OPERATION. Solution.— since 1.2)f';{}>jj^) 84.27000 ( 66.1483 — the divisor is greater 7643 736 than 1 and less than 2, the quotient 7 8 3 2 6 4 will contain six places, — two of in- 7 6 43 7 4 tegers and four of decimals. The i ggQQ highest denomination of the divisor, 12 7 40 multiplied by the lowest denomina- g j ^ q tion of the quotient, would obtain a 50 9 6 figure in the fourth place. We take ~10^54~ one place more as in multiplication 1 1 Q (Art. 163), and also cut off two figures ^-r- of the divisor, since these can not affect „ q the quotient above the fourth place. After obtaining the first figure of the quotient, we drop one right- hand figure of the divisor for each figure obtained. To prevent errors, we cancel the figure before each division. H. A. 10. 114 BAY^S HIQHER ABITHMETIC. Bule. — Fini the figure of the dividend thcd would remit firom imdtiplying a unit in the highest devwminajtion of ihe dmsor by a unit of the lowed denomination required in the quotient. Take one more figure of the dividend to secure aceimwy. Oat off any figures of ihe divisor not needed for the abbreviated dividend. Divide as usual until the fibres remaining in the dividend are all divided. At each subsequent division, drop a figure from the divisor , carrying the number necessary from the produd of ihe figure omitted. Continue until ihe divisor is reduced to two figures. Bemabk. — In the quotient, 5 units of an omitted order may be taken as 1 unit of the next higher ordjer. Examples for Practice. 1. 1000 -r .98, preserving two places. 2. 6215.75 -f- .99^, preserving three places. 3. 28012 -=- .993, preserving two places. 4. 52546.35 -r-.99f, preserving three places. 5. 4840 -^ .9875, preserving two places. 6. 2-=- 1.4142136, preserving seven places. 7. 9.869604401-^3.14159265, preserving eight places. 3.14159265 1020.41 6246.985 28209.47 62678.045 4901.27 1.4142135 Topical Outline. Decimal Fractions. Definitions. Decimal Point {Pure. Mixed. Complex. Principles. Numeration, Rule. Notation, Rule. Reduction In. I Dec. to a Com. Fraction. Com. Fraction to a Dec Addition, Rule. Subtraction, Rule. Multiplication, Rule. Abbreviated Multiplication. Diyisiou, Rule. AbbrvviHtod J>lvli»iou. X. OmOULATrKG DECIMAiS. 107. In reduciDg common fractions to decimals, the process, in some cases, does not terminate. Thb gives rise to Circulating Decimals. Principle I. — Ij any prime facUyn other Hian 2 and 5 are found in Hie denominator of a fraction in Us lowed termSy the resulting decimal vnU be interminate. Demonstration. — If the fraction is in its lowest terms, the nnmerator and denominator are prime to each other (131, Bem. 2). In the process of reduction, the numerator is multiplied by 10. By this means the factors 2 and 5 may be introduced into the numera- tor as many times as necessary; but no others are introduce<). Therefore, if any factors other than 2 and 5 are found in the denom- inator, the division can not be made complete, and the resulting decimal will be interminate. Thus, ^2^ = 2X2X2X2X2X2X6 ^^ '^^375 and, -^ = 2X2X2X2 = -^^^^ l>«t, 61> = 2X^^ = -11^^^+ It is evident that the first will terminate if the numerator be mul- tiplied six times by 10, carrying the decimal to the sixth place. In the same way we reduce ^ to sl decimal containing four places. But since the factor 3 is found in the denominator of 7^, the frac- tion can not be exactly reduced, though the numerator be multi- plied by any power of 10. • Principle II. — Every interminate decimal arising from the reducti&n of a comnum fra^ction will, if the division be carried far evuyagh, contain the same figure, or set of figures, repeated in the same order. (115) 116 HAY'S mOHER ARITHMETIC. Demonstration. — Each of the remainders must be less than the denominator which is used as the divisor (Art. 78, Note 2). If the division be carried far enough, some remainder must be found equal to some remainder already found, and the subsequent figures in the quotient must be similar to the figures found from the former remainder. Thus, in reducing |, we find the decimal .142857, and then have the remainder 1, the number we started with; if we annex a cipher, we shall get 1 for the next figure of the quotient, 4 for the next, etc. 168. 1. In terminate decimals, on this account, have received the name of OircuUding or Reimrring Decimals, 2. A Circulate or Circulating Decimal has one or more figures constantly repeated in the same order. 3. A Bepetend is the figure or set of figures repeated, and it is expressed by placing a dot over the first and last figure ; thus, -f = .142857 ; if there be one figure repeated, the dot is placed over it, thus, f = .6666 + = .6 4. A Pure Circulate has no figures but the repetend ; as, .5 and .124 5. A Mixed Circulate has other figures before the repetend; as, .2083 and .31247 6. A Simple Bepetend has one figure; as, .4 7. A Compound Bepetend has two or more figures ; as, M 8. A Perfect Bepetend is one which contains as many decimal places as there are units in the denominator, less 1 ; thus, | = .i42857 9. Similar Bepetends begin and end at the same deci- • • • • mal place; ap, .427 and .536 10. Dissimilar Bepetends begin or end at diflferent deci- • • • • mal places; as, .205 and .312468 CIRCULATINO DECIMALS. 117 11. Conterminous Bepetends end at the same place; as, .50397 and .42618 12. Co-originou8 Bepetends begin at the same place; • • • as, .5 and .124 169. Any terminate decimal may be considered a circu- • • • late, its repe tend being ciphers; as, .35 = .350 = .350000 Any simple repetend may be made compound, and any compound repetend still more compound, by taking in one • • • or more of the succeeding repetends; as, .3 = .33333, and .0562 = .056262, and .257 = .257257257 Kemarks. — 1. When a repetend is thus enlarged, be careful to take in no part of a repetend without taking the whole of it ; thus, if we take in 2 figures iwthe last example, the result, .25725, would be incorrect, for the next figure understood being 7, shows that 25725 is not repeated. 2. A repetend may be made to begin at any lower place by carry- ing its dots forward, each the same distance ; thus, .5 = .555, and .294i = .29414, and 5.1836 = 5.183683 3. Dissimilar repetends can be made similar, by carrying the dots forward till they all begin at the same place as the one farthest from the decimal point. 4. Similar repetends may be made conterminous by enlarging the repetends until they all contain the same number of figures. This number will be the least common multiple of the numbers of figures in the given repetends. For, suppose one of the repetends to have 2, another 3, another 4, and the last 6 figures ; in enlarging the first, figures must be taken in, 2 at a time, and in the others, 3, 4, and 6 at a time. 170. Circulating decimals originate, as has been already shown, in changing some common fractions to decimals. Then, having given a circulate, it can always be changed to an equivalent common fraction. 171. Circulating decimals may be added, subtracted, multiplied, and divided as other fractions. 118 RAY'S HIGHER ARITHMETia CASE I. 172. To reduce a pure circulate to a common Araction. Problem. — Change .53 to a common fraction. OPERAnON. Solution. — Removing the 100 times the repetend = 5 3.5 3 , decimal point one place to the Once the repetend = : .5 3 right, multiplies the repetend /. 99 times the repetend = 5 3. hy 10 ; two places, by 100, and Once the repetend = JJ, Aw, so on. Then, multiplying by 100, and subtracting the repetend from the product, removes all of that part to the right of the decimal point, and, dividing 53 by 99, we have the common fraction whidi produced the given repetend. « ■ Problem. — Change .456 to a common fraction. OPERATION. 1000 times the repetend = 4 5 6.4 5 6 Once the repetend =• .4 56 999 times the repetend = 4 5 6. 1 .*. once the repetend = fff, Aw. Problem. — Change 25.6 to a common fraction. OPERATION. Carry the dot forward thus : 2 5.6 = 2 6.6 2 6 1000 times the repetend = 6 2 5.6 2 5 Once the repetend = .6 25 999 times the repetend = 6 2 5. .*. once the repetend = ff J Whence the 25.625 = 25 f }{, Aw, Note. — From these solutions the following rule is derived. Bule. — WriU the repetend for ihe numerator , omitting Hie decimal point and the dote, and for the denominator write cw many ff8 as there are fibres in the repetend^ and reduce the fraction to Ue lowed terms. CZBCULAUNG DECIMALS. 119 CASE II. 173. To reduce a mixed circulate to a common fraction. Problem. — Change .82143Y to a common fraction. OPEBATION. 8 21 itii Omitting the decimal point, we have : .821437= ^^^ = 821X999 + 437 ^ 821(1000 — 1)4-437 1000X999 "" 999000 ^ 821000 — 821 + 437 820616 102677 999000 999000 124875 Or, briefly, 821437 — 821 102677 , Ans, Ang 999000 124876' Problem. — Change .048 to a common fittction. OPERATION. Omitting the decimal point, we have : 4ft 4 8 4(10 — 1^ 8 100 100 ' 900 900 ' 900 9 00 ■^900"' 9 00 ~"2 26' ^' Or, briefly, 48 — 4 44 11 900 ""900 226' **** Note. — ^The following rule is derived from the preceding solu- tions. Rule. 1. For ihe nvmeftcAor, subtract the part which precedes ihe repetend from ihe whole expression, both quantities being con- siiered as units. 2. For ihe dentyminatory write as inany 9'« as there are fibres in the repetend, and annex as many ciphers as there are decimal figures before each rqpeiend. 120 RAY'S HIGHER ARITHMETIC. Reduce to common fractions: 1. .3 2. .05 3. .123 4. 2.63 5. .31 6. .0216 7. 48.1 i- 8. 1.001 tV- 9. .138 TtS' 10. .2083 2A- 11. 85.7142 H- 12. .063492 lis- 13. .4476190 48iV. 14. .09027 85f ttV ADDITION OF CIECULATES. 174. Addition of Circulates is the process of finding the sum of two or more circulates. Similar circulates only can be added. Problem.— Add .256, 5.3472, 24.815, and .9098 OPERATION. .25666666,^6 5.8472727272 2 4.8158i58158 .9098000000 31.3295552097 Solution. — Make the circulates similar. The first column of figures which would appear, if the circulates were continued, is the same as the first figures of the repe- tends, 6, 7, 1, 0, whose sum, 14, gives 1 to be carried to the right-hand column. Since the last six figures in each number make a repetend, the last six figures of the sum also make a repetend. Bule. — Make the repetends dmilar, if iJiey be rwt 8o; add, and point off as in ordinary decimalsy increasing the rigkt4iand column by the amount, if any, which nnyuM be carried to it if the circulates were continued; then make a repetend in the sum, similar to those above. Bemark. — In finding the amount to be carried to the right-hand column, it may be necessary, sometimes, to use the two gu<lceediDg figures in each repetend. SUBTBACTION OF CIRCULATES. 121 RxAAfPTJRS FOB PRACTICE. 1. Add .453, .068, .327, .946 1.796 2. Add 3.04, 6.456, 23.38, .248 33.1334 3. Add .25, .104, .61, and .5635 1.536 4. Add i.03, .257, 5.04, 28.0445245 34.37 5. Add .6, .138, .05, .0972, .0416 1. 6. Add 9.21107, .65, 5.004, 3.5622 18.45 7. Add .2045, .09, and .25 .54 8. Add 5.0770, .24, and 7.124943 12.4 9. Add 3.4884, 1.637, 130.81, .066 136.00 SUBTRACTION OF CIRCULATES. 176. Subtraction of Circulates is the process of find- ing the difference between two circulates. The two circu- lates must be similar^ Problem.— Subtract 9.3i56 from 12.902i OPERATION. Solution. — Prepare the numbers for sub- 1 2.9 0212121 traction. If the circulates were continued, the 9.3 1 5 6 1 5 6 1 next figure in the subtrahend (5) would be 3.5 8 6 5 5 5 9 larger than the one above it (2) ; therefore, carry 1 to the right-hand figure of the subtrahend. Rule. — Make ike repetends similar^ if they be not so ; mtbtract and point off as m ordinary decimals, carrying 1, Jwwever, to the right-hand figure of the subtrahendy if on continuing the circulates it be found necessary ; then make a repetend in the remainder, simUar to those above, Bemark. — It may be necessary to observe more than one of the succeeding figures in the circulates, to ascertain whether 1 is to be carried to the right-hand figure of the subtrahend or not. H. A. 11. 122 BAY'S HIGHER ARITHMETIC Examples for Practice. 1. Subtract .0074 from .26 .259 2. Subtract 9.09 from 15.35465 6.25 3. Subtract 4.5i from 18.23673 13.72 4. Subtract 37.0128 fit)m 100.73 63.7i 5. Subtract 8.27 from 10.0563 1.7836290 6. Subtract 190.476 from 199.6428571 9.16 7. Subtract 13.637 from 104.i 90.503776 MULTIPLICATION OF CLBCULATES. 176. Mtiltiplication of drculates is the process of finding the product when either or when each of the factors is a circulate. Problem.— Multiply .3754 by 17.43 Solution. — ^In forming the partial operation. products, carry to the right-hand figures .3754 of each respectively, the numbers 1, 3, 0, 1 7.4 3 = 1 7.4 1- arising from the multiplication of the il601777 figures that do not appear. The repetend 2628lili of the multiplier being equal to J, J of 3.7 5 4 4 4 4 4 the multiplicand is 125148, whose figures 1 2 5 i 4 8 are set down under those of the multipli- . . cand from which they were obtained. 6.5 4 5 2 4 8 1 Aw^ Point the several products, carry them forward until their repetends are similar, and add for answer. Bule. — 1. Ij the multiplier contain a repetend^ change Uto a common fraction. 2. Hien multiply as in mvltiplieation of decinuds, and add to the right-lMnd figure of each partial product the amount necessary if the repetend were repeated, 3. Make the partial products similar, and find their sum. DIVISION OF CIRCIJLATE& 123 Examples for Practice. 1. 4.735 X 7.349 34.800liS 2. .07067 X .9432 .066665 3. 714.32X3.456 2469.173814 4. 16.204X32.75 530.810446 f). 19. 072 X. 2083 3.97348 6. 8.7543X4.7157 17.7045082 7. 1.256784X6.42081 8.069583206 DIVISION OF CIRCULATES. 177. Division of Circulates is the process of finding the quotient when either or when each of the terms is a circulate. Problem. — ^Divide .154 by .2 OPERATION. .i54 = HI, .2 = f Rule. — Change the terms to common fractions; then divide (18 in division of fractions, and rediLce the result to a repetend, Kemark. — This is the easiest method of solving problems in di- yision of circulates. The terms may be made similar, however, and the division performed without changing the circulates to common fractions. Examples for Practice. 1. .*75-r-.i 6.8i 2. 51.49i -^ 17. 3.028 3. 681.5598879-^94. 7.2506371 4. 90.5203749 -^ 6.754 13.40i 5. 11.068735402 ~ .245 45.13 6. 9.5^30663997 -^ 6.217 1.53 7. 3.500691358024^7.684 .45 124 BAY'S HIGHER ARITHMETIC. Topical Outline. Circulating Decimai^. 1. Principles. 2. Definitions — 1. Circulate. 2. Repetend. 3. Pure Circulate. 4. Mixed Circulate. 5. Simple Repetend. 6. Compound Repetend. 7. Perfect Repetend. 8. Similar Repetends. 9. Dissimilar Repetends. 10. Conterminous Rei)etends. , 11. Co-originous Repetends. 3. Reduction / Case I. lease II. { { Definition. 4. Addition \ Rule. Applications. Definition. 5. Subtraction..... } Rule. Applications. Definition. 6. Multiplication \ Rule. Applications. Definition. 7. Division \ Rule. Applications. XL COMPOUND DENOMESTATE . NUMBERS. 178. 1. A Measure is a standard unit used in estimating quantity. Standard units are fixed by law or custom. 2. A quantity is measured by finding how many times it contains the unit. 3. Denomination is the name of a unit of measure of a concrete number. 4. A Denominate Number is a concrete number which expresses a particular kind of quantity; as, 3 feet, 7 pounds. 5. A Ck>mpound Denominate Number is one expression of a quantity by different denomincttiom under one kind of measure; as, 5 yards, 2 feet, and 8 inches. 6. All measures of denominate numbers may be embraced under the following divisions: Valuer Weight, Extenshn, and Time. MEASUBES OF VALUE. 179. 1. Value is the worth of one thing as compared with another. 2. Value is of three kinds: Intrinsic^ Commercial, and Nominal. 3. The Intrinsic Value of any thing is measured by the amount of labor and skill required to make it useful. 4. The Commercial Value of any thing is its purchasing power, exchangeability, or its worth in market. (125) 126 RAY'S HIGHER ARITHMETIC. 5. The Nominal Value is the name value of any thing. 6. Value is estimated among civilized people by its price in money. 7. Money is a standard of value, and is the medium of exchange; it is usually stamped metal, called coins, and printed bills or notes, called paper money. 8. The money of a country is its Currency. Currency is national or foreign. United States Money. 180. United States Money is the legal currency of the United States. It is based upon the decimal system; that is, ten units of a lower order make one of the next higher. The Dollar is the unit. The same unit is the standard of Canada, the Sandwich Islands, and Liberia. Table. 10 mills, marked m., make 1 cent, marked ct. 10 cents ** 1 dime, " d. 10 dimes " 1 dollar, " $. 10 dollars " 1 eagle, ** e. Note. — The cent and miUy which are yj^ and j^un of a dollar, derive their names from the Latin centum and mt//e, meaning a hun- dred and o thousand ; the dime^ which is j\y of a dollar, is from the French word disrwe, meaning Un, Bemabks. — 1. United States money was established, by act of Congress, in 1786. The first money coined, by the authority of the United States, was in 1793. The coins first made were copper cents. In 1794 silver dollars were made. Gold eagles were made in 1795; gold dollars, in 1849. Gold and silver are now both legally standard. The trad^ dollar was minted for Asiatic commerce 2. The coins of the United States are classed a^ brmze, nickd, silver. MEASURES OF VALUE. 127 and gold. The name, value, composition, and weight of each coin are shown in the following table : Table. COIN. VALUE. COMPOSITION. 1 WEIGHT. BBONZS. One cent Icent 95 parts copper, 5 parts tin & sine 48 grains Troy. NICKEL. 3-cent piece. 5-cent piece. Scents. 5 cents. 75 parts copper, 25 parts nickel. 75 •* ** 25 " ** 30 grains Troy. 77.16 •♦ SILVER. Dime. Quarter dollar. Half dollar. Dollar. 10 cents. 25 cents. 50 cents. 100 cents. 90 parts silver, 10 parts copper. 90 •* ** 10 ** •• 90 •♦ •• 10 " •• 90, •• •• 10 ** •• 2.5 grams. 6.25 •• 12.5 •• 412.5 grains Troy. GOIiD. Dollar. Quarter eagle. Three dollar. Half eagle. Eagle. Double eagle. 100 cents. 2>^ dollars. 3 dollars. 5 dollars. 10 dollars. 20 dollars. 90 parts gold, 10 parts copper. 90 " " 10 •• 90 " " 10 " 90 •* " 10 •' 90 •* •• 10 •• 90 *• •• 10 " 25.8 grains Troy. 64.5 77.4 •* 129 258 •♦ *' 516 " •• 3. A deviation in weight of ^ a grain to each piece, is allowed by law in the coinage of Double Eagles and Eagles ; of ^ of a grain in the other gold pieces ; of IJ grains in all silver pieces ; of 3 grains in the five-cent piece ; and of 2 grains in the smaller pieces. 4. The mill is not coined. It is used only in calculations. 181. In reading U. S. Money, name (he dollars and all higher denominaiiona together as dollars^ the dimes and cents as cents, and die next figure, if there he one, as mills ; Or, name the whole number as dollars, and the rest as a decimal of a dollar. Thus, $9,124 is read 9 dollars 12 ct. 4 mills, or 9 dollars 124 thousandths of a dollar. 128 BAY'S HIOHER ABITHMETIQ. English or Sterling Money. 182. English or Sterling Money is the currency of the British Empire. The pound sterling (worth $4.8665 in U. S. money) is the unit, and is represented by the sovereign and the £1 bank-note. Table. 4 farthings, marked qr., make 1 penny, marked d. 12 pence " 1 shilling, " s. 20 shillings " 1 pound, ^ " £. Equivalent Table. & 8. d. qr. 1 = 20 = 240 = 960. 1 = 12 = 48. 1 = 4. Bebcarks. — 1. The abbreviations, JS, s., d., q., are the initials of the Latin words l\hra^ 8olidariu8f denariuSy qwtdixinSf signifying, respectively, pound, shilling, penny, and quarter. 2. The coins are gold, silver, and copper. The gold coins are the sovereign ( = £1), half sovereign ( = 108.), guinea ( = 21s.), and half-guinea ( = lOs. 6d.) The silver coins are the crown ( = 58.), the half crown ( =2s. 6d.), the florin ( = 28.), the shilling, and the six-penny, four-penny, and three- penny pieces. The penny, half- penny, and farthing are the capper coins. The guinea, half-guinea, crown, and half-crown are no longer coined, but some of them are in circulation. • 3. The standard for gold coins is J} pure gold and ^ alloy ; the standard for silver is f J pure silver and ^ copper ; pence and half-pence are pure copper. French Money. 188. French Money is the legal currency of France. Its denominations are decimal. MEASURES OF VALUE 129 The franc (worth 19.3 cents in U. S. money) is the unit The franc is also used in Switzerland and Belgium, and. under other names, in Italy, Spain, and Greece. Table. 10 millimes, marked m., make 1 centime, marked c. 10 centimes " 1 decime, ** d. 10 decimes " 1 franc, " fr. Equivalent Table. 1 fr. = 10 d. == 100 c. = 1000 m. 1 d. = 10 c. = 100 m. 1 c. = 10 m. Remark.— The coins are of gold, silver, and bronze. The gM coins are 100, 60, 20, 10, and 5-franc pieces ; they are .9 pure gold. The 20-franc piece weighs 99.55 gr. The siktr coins are 6, 2, and 1 -franc pieces, and 50 and 20-centime pieces ; they are now .835 pure silver. The brome coins are 10, 5, 2, and 1-centime pieces. The decime is not used in practice. All sums are given in francs and centimes, or hundredths. German Money. 184. German Money is the legal currency of the Grer- man Empire. The mark, or reichsmark (worth 23.8 cents in U. S. money), is the unit. The only other denomination is the pfennig (penny). Table. 100 pfennige, marked Pf., make 1 mark, marked EM. Remabk. — The coins are of gold, silver, and copper. The gold coins are of the value of 40, 20, 10, and 5 marks; the silver coins, 3, 2, and 1-mark pieces, and 50, 20, and 10 pfennige. The coppa-, 2 and 1 -pfennig pieces. Gold and silver are .9 fine. 130 BAY'S HIGHER ARITHMETIC. MEASURES OF WEIGHT. 185. Weight is the measure of the force called gravity, which draws bodies toward the center of the earth. The standard unit of weight in the United States is the Troy pound of the Mint. 188. Three kinds of weight are in use, — 2Voy Weight ^ ApotJiecariea^ Weight, and Avoirdupois Weight. Troy Weight 187. Troy Weight is used in weighing gold, silver, platinum, and jewels. Formerly it was used in philosoph- ical and chemical works. Table. 24 grains, marked gr., make 1 pennyweight, marked pwt. 20 pwt. ** 1 ounce, " oz. 12 oz. ** 1 pound, ft). Equivalent Table. ft). oz. pwt. gr. 1 = 12 = 240 = 5760. 1 = 20 = 480. 1 = 24. Hem ARK. — The Troy pound is equal to the weight of 22.7944 cubic inches of pure water at its maximum density, the barometer being at 30 inches. The standard pound weight is identical with the Troy pound of Great Britain. Apothecaries' Weight. 188. Apothecaries' Weight is used by physicians and apothecaries in prescribing and mixing dry medicines. Medicines are bought and sold by Avoirdupois Weight MEASURES OF WEIGHT. 131 Table. 20 grains, marked gr., make 1 scruple, marked 9. 3 9 *' 1 dram, ** 3. 83 "1 ounce, ** 5. 12 § ** 1 pound, " lb. Equivalent Tablr ft). §. 5. 9. gr. 1 = 12 = 96 = 288 = 5760. 1 = 8 = 24 = 480. 1 = 3 = 60. 1 = 20. Kemark. — The pound, ounce, and grain of this weight are the same as those of Troy weight ; the pound in each contains 12 oz. = 5760 gr. Avoirdupois or Commercial Weight. 189. Avoirdupois or Commercial Weight is used for vreighing all ordinary articles. Table. 16 ounces, marked oz., make 1 pound, marked lb. 25 lb. ** 1 quarter, *' qr. 4 qr. "1 hundred-weight, ** cwt. 20 cwt. '' 1 ton, '* T. Equivalent Table. T. cwt. qr. lb. oz. 1 = 20 =^ 80 = 2000 = 32000. 1 = 4 = 100 = 1600. 1 =r 25 = 400. 1 = 16. 132 JiA yS HlOffJSIi AMITHMETia Kemabks. — 1. In Great Britain, the qr. = 28 lb., the cwt. = 112 lb., the ton = 2240 lb. These values are used at the United States custom-houses in invoices of English goods, and are still used in some lines of trade, such as coal and iron. 2. Among other weights sometimes mentioned in books, are : 1 stone, horseman's weight, = 14 lb.; 1 stone of butcher's meat = 8 lb.; 1 clove of wool == 7 lb. 3. The lb. avoirdupois is equal to the weight of 27.7274 cu. in. of distilled water at 62° (Fahr.) ; or 27.7015 cu. in. at its maximum density, the barometer at 30 inches. For ordinary purposes, 1 cubic foot of water can be taken 62^ lb. avoirdupois. 4. The terms gross and net are used in this weight. Gross weight is the weight of the goods, together with the box, cask, or whatever contains them. Net weight is the weight of the goods alone. 5. The word avoirdupois is from the French avoirs, du, poisj signify- ing goods of weight. 6. The ounce is often divided into halves and quarters in weigh- ing. The sixteenth of an ounce is called a dram. COMPARISON OP WEIGHTS. 190. The pound Avoirdupois weighs 7,000 grains Troy, and the Troy pound weighs 5,760 grains, hence there are 1,240 grains more in the Avoirdupois pound than in the Troy pound. The following table exhibits the relation between certain denominations of Avoirdupois, Troy, and Apothecaries^ Weight. Avoirdupois. Troy. Apotheci iries'. 1 lb. — 1t¥t «>. — 1* lb. 1 oz. — •H4 oz. — m S- 1 ft). 1 lb. 1 oz. 1 5- 1 gr. 1 gf- 1 pwt. i 3- 1 pwt. H a. MEASUBES OF EXTENSION. 133 Bemabk. — In addition to the foregoing, the following, called Diamond Weighty is used in weighing diamonds and other precious stones. Table. 16 parts make 1 carat grain = .792 Troy grains. '4 carat grains " 1 carat =3.168 " Note. — ^This carat is entirely different from the assay carat, which has reference to i\i% fineness of gold. The mass of gold is considered as divided into twenty-four parts, called carats, and is said to be so many carats fine, according to the number of twenty-fourths of pure gold which it contains. MEASURES OF EXTENSION. 191. 1. Extension is that property of matter by which it occupies space. It may have one or more of the three dimensions,— length, breadth, and thickness. 2. A lin6 has only one dimension, — length. 3. A sur&ce has two dimensions, length and breadth. 4. A solid or volume has three dimensions, — length, breadth, and thickness. 192. Measures of Extension embrace : ^ Long Measure. 1. Linear Measure. Chain Measure. Mariners' Measure. Cloth Measure. 2. Superficial Measure. | ^^'^ Measure. (Surveyors Measure. 3. Solid Measure. T Liquid Measure. 4. Measures of Capacity. < Apothecaries Measure. 6. Angular Measure. vDry Measure. 134 BA Y'S HIGHER ARITHMETIC. Long or Linear Measure. 193. Linear Measure is used in measuring distances, or length, in any direction. The standard unit for all measures of extension is the yard, which is identical with the Imperial yard of Great Britain. Table. 12 inches, marked in., make 1 foot, marked ft. 3 ft. ** 1 yard, '' yd. 5^ yd. or 16^ ft. *' 1 rod, " rd. 320 rd. ** 1 mile, " mi. Equivalent Table. mi. rd. yd. ft. in. 1 = 320 = 1760 = 5280 = 63360. 1 = 5^ = 16^ = 198. 1 = 3 = 36. 1 = 12. Bemares. — 1. The standard yard of the United States was obtained from England in 1856. It is of bronze, and of due length at 59.8° Fahr. A copy of the former standard is deposited at each state capital : this was about jjhrj ^^ ^^ i^^^h too long. 2. The rod is sometimes called perch or pole. The furlong, equal to 40 rods, is seldom used. 3. The inch may be divided into halves, fourths, eighths, etc., or into tenths, hundredths, etc. 4. The following measures are sometimes used : 12 lines make 1 inch. 3 barleycorns 1 " 3 inches 1 palm. 4 inches 1 hand. 9 inches 1 span. 18 inches 1 cubit. 3 feet 1 pace. MEASURES OF EXTENSION. 135 194. Chain Measure is used by surveyors in measuring land, laying out roads, establishing boundaries, etc. Table. 7.92 inches, marked in., make 1 link, marked li. 100 li. "1 chain, ** ch. 80 ch. "1 mile, ** mi. Equivalent Table. mi. ch. li. in. 1 = 80 = 8000 = 63360. 1 = 100 = 792. 1 = 7.92. Kemarks. — 1. The surveyors* chain, or Gunter's chain, is 4 rods, or 66 feet in length. Since it consists of 100 links, the chains and links may he written as integers and hundredths ; thus, 2 chains 56 links are written 2.56 ch. 2. The engineers' chain is 100 feet long, and consists of 100 links. 3. The engineers' leveling rod is used for measuring vertical dis- taqces. It is divided into feet, tenths, and hundredths, and, hy means of a vernier, may be read to thousandths. 196. Mariners' Measxire is used in measuring the depth of the sea, and also distances on its surface. Table. 6 feet make 1 fathom. 720 feet " 1 cable-length. Kemarks. — 1. A nautical mile is «ne minute of longitude, meas- ured on the equator at the level of the sea. It is equal to 1.152| statute miles. 60 nautical miles = 1 degree on the equator, or 69.16 statute miles. A league is equal to 3 nautical miles, or 3.458 statute miles. 2. Depths at sea are measured in fathoms ; distances are usually measured in nautical miles. 136 liAY'S mOHER ARITHMETIC. 196. Cloth Measure is used in measuring dry-goods. The standard yard is the same as in Linear Measure, but is divided into halves, qtuirterSf eighthsy sixteenthSy etc., in place of feet and inches. Rebiarks. — 1. There was formerly a recognized tahle for Cloth Measure, but it is now obsolete. The denomiuations were as follows : 2i inches, marked in., make 1 nail, marked na. 4 na. or 9 in. " 1 quarter, " qr. 4 qr. " 1 yard, " yd. 2. At the custom-house, the yard is divided decimally. Superficial or Surface Measure. 197. 1. Superficial Measure is used in estimating the numerical value of surfaces ; such as, land, weather-board- ing, plastering, paving, etc. 2. A surfJEtce has length and breadth, but not thickness. 3. The area of a surface is its numerical value; or the number of times it contains the measuring unit, 4. A superficial iinit is an assumed unit of measure Vor surfaces. Usually the square, whose side is the linear unit, is the unit of measxire; as, the square inch, square foot, square yard, 5. A Beetangle may be defined as a surface bounded by four straight lines forming four square corners; as either of the figures A, B. 6. When the four sides are equal the rectangle is called a square; as the figure C. 7. The area of a rectangle is equal to its length mtdtiplied by its breadth. MEASUEES OF EXTENSION, 137 Explanation. — Take a rectangle 4 inches long by 3 inches wide. If upon each of the inches in the length, a square inch he conceived to stand, there will be a row of 4 square inches, extending the whole length of the rectangle, and reaching 1 inch of its width. Ap the rectangle contains as many such rows as there are inches in its width, its area must be equal to the number of square inches in a row (4) multiplied by the number of rows (3), = 12 square inches. This statement (7), as commonly understood, can present no exception to Prin. 2, Art. 60. Table. 144 square inches (sq. in.) make 1 square foot, marked sq. ft. 9 sq. ft. ** 1 square yard, ** sq. yd. 30J sq. yd. " 1 square rod, leOsq. rd. .r -. ___. " 1 acre, (( <( sq. rd. A. Equivalent Table. A. sq. rd. sq. yd. sq. ft. 1 = 160 = 4840 = 43560 1 = 30i = 272i 1 = 9 1 sq. in. 6272640. 39204. 1296. 144. Note. — ^The following, though now seldom used, are often found in records of calculations : 40 perches (P.), or sq. rds., make 1 rood, marked E. 4 roods " 1 acre, " A. 198. Surveyors' Measure is a kind of superficial measure, which is used chiefly in government surveys. Table. 625 square links (sq. li.) make 1 square rod, sq. rd. 16 so. rd. ** 1 sauare chain, so. ch. 16 sq. rd. 10 sq. ch. 640 A. 36 sq. mi. (6 miles square) H. A. 12. (( (( (( 1 square chain, sq. ch. 1 acre, A. 1 square mile, sq. mi. 1 township, , JX^^^ // ' or T^!£ \ UNIVERSITY I MAY'S HIGHEB ABITHMETja Eqdivalent Table. Tp. Bq. mi. A. sq. ch. 8q. rd. sq. li. 1 = 36 = 23040 = 230400 ^ 368.6400 = 2304000000. 1 = 640 ^ 6400 = 102400 = 64000000. 1 = 10 = 160 =. 100000. 1 = 16 = 10000. 1 ^ 625. Solid Measure. IBS. A Solid has length, breadth, and thickness. Solid Meaaure \a used in estimating the Ciontenia or vdwme of solids. A Cube is a solid, bounded by six equal squares, called facei. Its length, breadth, and thickness are all equal. Remark. — The size or name of anj cube, like that oE a square, depends upon i(s side, as cubic iuch, cubic foot, cubic yard. ExPLANATios. — It each side of a cube is 1 inch iong, it ia calied a cubic inch; it each side is 3 feet (I yard) long, aa repre- sented ID the figure, it ia a cubic or solid Wiien the base o£ a cube is 1 square yard, it contains3X3=9 square feet; and I foot high on this base, contains 9 solid feet; 2 teet high contains 9 X 2 = 18 Boiid feet ; 3 teel high = 27 solid tceL Also it may be shown that 1 solid contains 12 X 12 X 12 = 1728 solid or eMe inches. The unit by which all solids are measured is a culje, whose side is a linear inch, foot, etc, and their size or solidity will be the number of times they contain this unit. Bemark.— The simplest solid ia the lecbmyuiir solid, which is bounded by sin rectangles, called its far.ee, each opposite pair being equal, and perpendicular lo the other tour; as, for example, the MEASURES OF EXTENSION, 139 ordinary form of a brick or a box of soap. If the length, breadth, and thickness are. the same, the faces are squares, and the solid is a cabe. Table. 1728 cubic inches (cu. in.) make 1 cubic foot, cu. ft. 27 cu. fk. ** 1 cubic yard, cu. yd. Equivalent Table. cu. yd. cu. ft. cu, in. 1 = 27 = 46656. 1 = 1728. Hemakes. — 1. A perch of stone is a mass 16\ ft. long, IJ ft. wide, and 1 ft. high, and contains 24| cu. ft. 2. Earth, rock-excavations, and embankments are estimated by the cubic yard. 3. Bound timber will lose ^ in being sawed, hence 50 cubic ft. of round timber is said to be equal to 40 cubic ft. of hewn timber, which is a ton, 4. Fire-wood is usually measured by the cord. A pile of wood 4 ft. high, 4 ft. wide, and 8 ft. long, contains 128 cubic feet or one cord. One foot in length of this pile, or 16 cu. ft., is called a cord foot, 5. Planks and scantling are estimated by board measure. In this measure, 1 reduced footy 1 ft. long, 1 ft. wide, and 1 in. thick, contains 12 X 12 X 1 = 144 cu. in. All planks and scantling less than an inch thick, are reckoned at that thickness ; but, if more than an inch thick, allowance must be made for the excess. Measures of Capacity. 200. Capacity means room for things. Measures of Capacity are divided into Measures of Liquids and Measures of Dry Substances, 201. Liquid Measure is used in measuring liquids, and in estimating the capacities of cisterns, reservoirs, etc. The gallon, which contains 231 cu. in., is the unit of measure in liquids. 140 BA Y'S HIGHER ARITHMETIC. Note. — This gallon of 231 cubic inches was the standard in England at the time of Queen Anne. The present imperial gallon of England contains 10 lb. of water at 62° Fahr., or 277.274 cubic inches. Table. 4 gills, marked gi., make 1 pint, marked pt. 2 pt. "1 quart, ** qt. 4 qt. " 1 gallon, ** gal. Equivalent Table. gal. qt. pt. gi. 1 = 4 = 8 = 32. 1=2= 8. 1=4. Note. — Sometimes the barrel is estimated at SIJ gal., and the hogshead at 63 gal.; but usually each package of this description is gauged separately. 202. Apothecaries' Fluid Measure is used for meas- uring all liquids that enter into the composition of medical prescriptions. Table. 60 minims, marked nt-, make 1 fluid drachm, marked f^, 8 fs ** 1 fluid ounce, " f§. 16 ^ * 1 pint, " O. 8 ** 1 gallon, " cong. Equivalent Table. cong. O. f§. f^. Vl, 1 r= 8 ==r 128 = 1024 = 61440. 1 = 16 = 128 = 7680. 1 = 8 = 480. 1 = 60. MEASURES OF EXTENSION. 141 Notes. — 1. Cong, is an abbreviation for eongioLriumy the Latin for gallon ; O. is the initial of octam, the Latin for one eighth^ the pint being one eighth of a gallon. 2. For ordinary purposes, 1 tea-cup = 2 wine-glasses = 8 table- spoons = 32 tea-spoons = 4 f§. 203. Dry Measure b used for measuring grain, fruit, vegetables, coal, salt, etc. The Winchester bushel is the unit; it was formerly used in England, and so called from the town where the standard was kept. It is 8 in. deep, and 18^ in. in diameter, and contains 2150.42 cu. in., or 77.6274 lb. av. of distilled water at maximum density, the barometer at 30 inches. Note. — ^This bushel was discarded by Great Britain in 1826, and the imperial bushel substituted; the latter contains 2218.192 cu. in., or eighty pounds avoirdupois of distilled water. Table. 2 pints, marked pt, make 1 quart, marked qt. 8 qt. ** 1 peck, " pk. 4 pk. "1 bushel, ** bu. Equivalent Table. bu. pk. qt. pt 1 = 4 = 32 = 64. 1 = 8 = 16. 1=2. Kemabks. — 1. 4qt. or J peck = 1 dry gal. = 268 8 cu. in. nearly. 2. The qaarter is still used in England for measuring wheats of which it holds eight bushels, or 480 pounds avoirdupois. 3. When articles usually measured by the above table are sold by weight, the bushel is taken as the unit. The following table gives the legal weight of a bushel of various articles in avoirdupois pounds : 142 BAY'S HIGHER AMITHMETia Table. ARTICLES. LB. 60 80 56 56 32 60 56 50 60 EXCEPTIONS. Beans. Coal. Corn (Indian). Flax Seed. Oats. Potatoes (Irish). Rye. Salt. Wheat. Me., 64 ; N. Y., 62. rOhio, 70 of cannel ; Ind., 70 mined out of ( the state; Ky., 76 of, anthracite. N. Y., 58 ; Cal., 52 ; Arizona, 54. N. Y. and N. J., 55 ; Kan., 54. r Md., 26 ; Me., N. H., N. J., Pa., 30 ; Neb., 34 ; \ Montana, 35; Oregon and Wash., 36. Ohio, 58 ; Wash., 50. La., 32; Cal., 54. Mass., 70 ; Pa., coarse, 85 ; ground, 70 ; fine^ 62 ; Ky. and 111., fine, 55 ; Mich., 56 ; Col. and Dak., 80. Comparative Table of Measures. cu. in. gal. cu. in. qt. cu. in. pt. Liquid Measure, 231 57f 28J Dry Measure (J pk.), 268| 67^ 33f cu. in.gi. Angular or Circular Measure. 201. A plane angle is the difference of direction of two straight lines which meet at a point. Explanation. — Thus, the two lines AB and AC meet at the point A, called the apex. The lines AB and AC are the sides of the angle, and the difference in direction, or the opening of the lines, is the angU itself, . Angular Measxire is used to measure angles, directions, latitude, and longitude, in navigation, astronomy, etc. A circle is a plane surface bounded by a line, all the points of which are equally distant from a point within. ■ MEASURE OF TIME. 143 ExPLANATtONS. — The bounding lino I> ADBEA is a evrcumfertnce. Every point y^ of this line is at the same distance from / the point C, which is called the center, I The cvrde is the area included within the a I _ circumference. Any straight line drawn \ / from the center to the circumference is \ / called a radius; thus, CD and CB are ^^^^^ ^^y^ radii. Any part of the circumference, as E AEB or AD, is an are. A straight line, like AB, drawn through the center, and having its ends in the cir- cumference, is a diameter; it divides the circle into two equal parts. Notes. — 1. Every circumference contains 360 d^^rees ; and, the apex of an angle being taken as the center of a circle, the angle is measured by the number of degrees in the arc included by the sides of the angle. 2. The angle formed by two lines perpendicular to each other, as the radii AC and DC in the above figure, is a right angU, and is measured by the fourth part of a circumference, 90°, called a quadrant. Table. 60 seconds, marked ", make 1 minute, marked. '. 6(K " 1 degree, *' ®. 360° " 1 circumference, '* c Equivalent Table. 1 == 360 = 21600 = 1296000. 1 = 60 = 3600. 1 = 60. Note. — The twelfth part of a circumference, or 30°, is called a siffn. • MEASURE OF TIME. 205. 1. Time is a measured portion of duration. 2. A Year is the time of the revolution of the earth 144 RAY'S HIGHER ARITHMETIC, around the sun ; a Day is the time of the revolution of the earth on its axis. 3. The Solar Day is the interval of time between two successive passages of the sun over the same meridian. 4. The Mean Solar Day is the mean, or average, length of all the solar days in the year. Its duration is 24 hours, and it is the unit of Time Measure. 5. The Civil Day, used for ordinary purposes, commences at midnight and closes at the next midnight. 6. The Astronomical Day commences at noon and closes at the next noon. Table. 60 seconds, marked sec., make 1 minute, marked rain. 1 hour, ** hr. 1 day, ** da. 1 week, ** wk. 1 month, " men. 1 year, ** yr. 1 common year. 1 leap year. 1 century, marked cen. 60 min. 24 hr. 7 da. 4 wk. 12 calendar mon. 365 da. 366 da. 100 yr. Note.— 1 Solar year = 365 da. 5 hr. 48 min. 46.05 sec. = 365 J da., nearly. Equivalent Table. vr. mo. wk. I = 12 = 52 = 1 = da. hr. min. Rec. (365 — 1366 — 8760 — 525600 — 31536000 8784 — 527040 — 31622400 7 — 168 — 10060 — 604800 1 — 24 — 1440 — 86400 1 — 60 - 3600, 1 — 60, MEASUBE OF TIME. 145 Note. — The ancients were unable to find accurately the number of days in a year. They had 10, afterward 12, calendar months, corresponding to the revolutions of the moon around the earth. In the time of Julius Caesar the year contained 365J days ; instead of taking account of the J of a day every year, the common or civil year was reckoned 365 da^s, and every 4th year a day was inserted (called the intercalary day), making the year then have 366 days. The extra day was introduced by repeating the 24th of February, which, with the Komans, was called the sixth day before the kalends of March. The years containing this day twice, were on this account called bissextile, which means having two sixths. By us they are gen- erally called leap years. But 3651^ days (365 days and 6 hours) are a little longer than the true year, which is 365 days 5 hours 48 minutes 46.05 seconds. The difference, 11 minutes 13.95 seconds, though small, produced, in a long course of years, a sensible error, which was corrected by Gregory XIII., who, in 1582, suppressed the 10 days that had been gained, by decreeing that the 5th of October should be the 15th. 206. 'To prevent difficulty in future, it has been decided to adopt the following rule. Hule for Leap Years. — Every year that is divisible by 4 is a leap year, unless it ends with two ciphers; in which case it must be divisible by 400 to be a leap year. Illustration.— Thus, 1832, 1648, 1600, and 2000 are leap years ; but 1857, 1700, 1800, 1918, are not. Notes. — 1. The Gregorian calendar was adopted in England in 1752. The error then being 11 days. Parliament declared the 3d of September to be the 14th, and at the same time made the year begih January 1st, instead of March 25th. Kussia, and all other countries of the Greek Church, still use the Julian calendar; consequently their dates (Old Style) are now 12 days later tlian ours (New Style), The error in the Gregorian calendar is small, amounting to a day in 3600 years. 2. The year formerly began with March instead of January ; con- sequently, September, October, November, and December were the 7th, 8th, 9th, and 10th months, as their names indicate; being derived from the Latin numerals Septem (7), Octo (8), Novem (9), Decern (10). H. A. 13. 146 RAY'S HIGHER ARITHMETIC, COMPARISON OF TIME AND LONGITUDE. 207. The longitude of a place is its distance in degrees, minutes, and seconds, east or west of an established meridian. • Note. — The difference of longitude of two places on the same side of the e!<tablislied meridian, is found by subtracting the Igj^s longitude from the greater ; but, of two places on opposite sides of the meridian, the difference of longitude is found by adding the longitude of one to the longitude of the other. The circumference of the earth, like other circles, is divided into 360 equal parts, called degrees of longitude. The sun appears to pass entirely round the earth, 360°, once in 24 hours, one day; and in 1 hour it passes over 15°. (360° -f- 24 = 15°.) As 15° equal 900', and 1 hour equals 60 minutes of timey therefore, the sun in 1 minute of time passes over 15' of a degree. (900' -^- 60 = 15'.) As 15' equal 900", and 1 minute of time equals 60 seconds of tiine, therefore, in 1 second of time the sum passes over 15" of a degree. (900" -^ 60 = 15".) Table for Comparing Longitude and Time. 15° of longitude ^= 1 hour of time. 15' of longitude = 1 min. of time. 15" of longitude = 1 sec. of time. Note. — If one place has greater east or less west longitude than another, its time must be later; and, conversely, if one place has later time than another, it must have greater east or less west longitude. MISCELLANEOUS TABLES. 208. The words folio, quarto, octavo, etc., used in speak- ing of books, show how many leaves a sheet of paper makes. THE METRIC SYSTEM. 147 A sheet folded into 2 leaves, called a folio, ] make s 4 4 " " a quarto or 4to, 8 8 " ** an octavo or 8vo, 16 12 " ** a duodecimo or 12mo, 24 16 - " a 16mo, 32 32 '' ** a 32mo, 64 Also, . 24 sheets of paper make 1 quire. 20 quires ** 1 ream. 2 reams " 1 bundle. 5 bundles ** 1 bale. pages. 12 things make 1 dozen. 12 dozens or 144 things 1 gross. 12 grass or 144 dozens 1 great gross. 20 things 1 score. 56 lb. 1 firkin of butter. 100 lb. 1 quintal of fish. 196 lb. 1 bbl. of flour. 200 lb. 1 bbl. of pork. THE METRIC SYSTEM. Historical. The Metrio System is an outgrowth of the French Revolution of 1789. At that time there was a general disposition to break away from old customs ; and the revolutionists contended that every thing needed remodeling. A commission was appointed to determine an invariable standard for all measures of length, area^ solidity, capacity, and weight. After due deliberation, an accurate survey was made of that portion of the terrestrial meridian through Paris, between Dunkirk, France, and Barcelona, Spain ; and from this, the distance on that meridian from the equator to the pole was computed. The quadrant thus obtained was divided into ten million equal parts ; one part was called a metefi\ and is the hose of the system. From it all measures are derived. 148 BAY'S HIGHER ARITHMETIC. In 1795 the Metric System was adopted in France. It is now used in nearly all civilized countries. It was author^ ized by an act of Congress in the United States in 1866. 209. The Metric System is a decimal system of weights and measures. The meter is the primary unit upon which the system is based, and is also the imit of length. It is 39.37043 inches long,* which is very nearly one ten-millionth part of the distance on the earth's surface from the equator to the pole, as measured on the meridian through Paris. o • 0) 00 CO • CO -4 tf ^ O Kemabk. — The stoTicfcird meter, a bar of platinum, is kept among the archives in Paris. Duplicates of this bar have been furnished to the United States. 210. The names of the lAmer denominations in each measure of the Metric System are formed by prefixing the Laiin numerals, dec* (.1), ceaii (.01), and imJUi (.001) to the unit of that measure ; those of the higJier denominations, by prefixing the Greek numerals, deka (10), hddo (100), kUo (1000), and mipna (10000), to the same unit. These prefixes may be grouped about the unit of measure, showing the decimal arrangement of the system, as follows: rmiUi = .001 Lower Denominations. \ centi = .01 I deci = .1 Unit of Measure = deka = hekto = kilo = Higher Denominations. 1. 10. 100. 1000. myria= 10000. THE METRIC SYSTEM. 149 211. The units of the various measures, to which these prefixes are attached, are as follows: . The Meter, which is the unit of Length. The Ar, " ** " " " Surface. The Liter, ** '' '' " " Capacity. The Gram, '* '* *• '* " Weight. Bemabk. — The name of each denomination thus derived, imme- diately shows its relation to the unit of measure. Thus, a cenii- tneter is one one-hundredth of a meter; a kilogram is a thousand grams ; a hehtdiier is one hundred liters, etc. Measure of Length. 212. The Meter is the unit of Length, and is the de- nomination used in all ordinary measurements. Table. 10 millimeters, marked mm. , make 1 centimeter, marked cm. 10 centimeters 10 decimeters 10 meters 10 dekameters 10 hektometers 10 kilometers Hemareb. — 1. The figure on page 148 shows the exact length of the decimeter, and its subdivisions the centimeter and millimeter. 2. The centimeter and millimeter are most often used in meas- uring very short distances ; and the kilometer, in measuring roads and long distances. Measure of Surface. 213. The Ar (p-o. ar) is the unit of Land Measure ; it is a square, each side of which is 10 meters (1 dekameter) in length, and hence its area is one square dekameter. 1 decimeter. " dm. 1 meter. " m. ~1 dekameter. *' Dm. 1 hektometer. " Hm. 1 kilometer. " Km. 1 myriameter. " Mm. 150 EAY'S HIGHER ARITHMETIC, Table. 100 centars, marked ca., make 1 ar, marked a. 100 ars *' 1 hektar, *' Ha. Kemare. — The iquare meter (marked m^) and its subdivisions are used for measuring small surfaces. Measure of Capacity. 214. The Liter (p^o. le'ter) is the unit of Capacity. It is equal in volume to a cube whose edge is a decimeter ; that is, one tenth of a meter. Table. 10 milliliters, marked ml., make 1 centiliter, marked cl. 10 centiliters " 1 deciliter, ** dl. 10 deciliters " 1 liter, '* 1. 10 liters " 1 dekaliter, '' Dl. 10 dekaliters '* 1 hektoliter, '' HI. Bemarks. — 1. This measure is used for liquids and for dry sub- stances. The denominations most used are the liter and hektoliter ; the former in measuring milk, wine, etc., in moderate quantities, and the latter in measuring grain, fruit, etc., in large quantities, 2. Instead of the milliliter and the kiloliter, it is customary to use the cubic centimeter and the cubic meter (marked m^), which ar4 their equivalents. 3. For measuring wood the ster (pro. stdr) is used. It is a cubic meter in volume. Measure of Weight. 215. The Gram {pro, gram) is the imit of Weight. It was determined by the weight of a cubic centimeter of dis- tilled water, at the temperature of melting ice. THE METRIC SYSTEM. 151 Table. 10 milligrams, marked mg., make 1 centigram, marked eg. 10 centigrams * 1 decigram. dg. 10 decigrams * 1 gram. g- 10 grams * 1 dekagram. Dg- 10 dekagrams * 1 hektogram, Hg. 10 hektograms ' 1 kilogram. Kg. 10 kilograms * 1 myriagram. Mg. 10 myriagrams, or 100 kilograms * * 1 quintal. . ^ 10 quintals, or 1000 (( ( * 1 metric ton, M.T. Bemarks. — 1. The gram, kilogram {pro, klKo-gram), and metric ton are the weights commonly used. 2. The gram is used in all cases where great exactness is required ; such as, mixing medicines, weighing the precious metals, jewels, letters, etc. 3. The kilogram, or, as it is commonly abbreviated, the "kilo," is used in weighing coarse articles, such as groceries, etc. 4. The metric ton is used in weighing hay and heavy articles generally. 216. Since, in the Metric System, 10, 100, 1000, etc., units of a lower denomination make a unit of the higher denomination, the following principles are derived: Principles. — 1. A number u redvjced to a lower denom- inaiion by removing ihe decimal point as many places to the RIGHT as there are ciphers in the muUiplier, 2. A number is reduced to a higher denomination by remxmng the decimal point as many places to the left as there are ciphers in the divisor. Illustrations. — Thus, 15.03 m. is read 15 meters and 3 centi- meters ; or, 15 and 3 hundredths meters. Again, 15.03 meters = 1.503 dekameters = .1503 hektometer = 150.3 decimeters = 1503 centi- meters. As will be seen, the reduction is effected by changing the decimal point in precisely the same manner as in United Skites Money. 152 BA rS HIGHER ARITHMETIC. 217. The following table presents the legal and approx- imate values of those denominations of the Metric System which are in common use. Table. DENOMINATION. LEGAL VALUE. 1 APPBOX. VALUE. Meter. 39.37 inches. 3 ft 3| inches. Centimeter. .3937 inch. J inch. Millimeter. .03937 inch. ■it inc^- Kilometer. .62137 mile. 1 mile. Ar. 119.6 sq. yards. 4 sq. rods. Hektar. 2.471 acres. 2i acres. Square Meter. 1.196 sq. yards. 10| sq. feet. Liter. 1.0567 quarts. 1 quart Hektoliter. 2.8375 bushels. 2 bu. 3} pecks. Cubic Centimeter. .061 cu. inch. ■jJj cu. inch. Cubic Meter. 1.308 cu. yards. 35J cu. feet. Ster. .2759 cord. \ cord. Gram. 15.432 grains troy. lb\ grains. Kilogram. 2.2046 pounds av. 2J pounds. Metric Ton. 2204.6 pounds av. 1 T. 204 pounds. Topical Outline. Compound Numbers. 1. Preliminary Definitions. 2. Value. 8. Weight. ri. Definitions. 2. United States and Canadian Money. - 3. English Money. 4. French Money. 5. German Money. 1. Troy Weight 2. Apothecaries' Weight .3. Avoirdupois Weight TOPICAL OUTLINE. 153 Compound Numbers. — {Concluded,) 1. Linear... . 4. Extension. . 1. Long Measure. 2. Chain Measure. 3. Mariners' Measure. .4. Cloth Measure. „ . , „ f 1. Surface Measure. 2. Superficial Measure.... -< 3. Solid Measure. 4. Measures of Capacity. / 1- Liqui<i- 2. Surveyors' Measure. 5. Angular Measure. 5. Time. 6. Comparison of Time and Longitude. 7. Miscellaneous Tables. 12. Dry. 8. Metric System. 1. Historical. 2. Terms.. 1. How Derived. 2. Lower Denominations... 3. Higher Denominations.. 1. Meter. 2. Ar. 3. Liter. L 4 Gram. 1. Ivcngth. 2. Surface. 3. Capacity. I 4. Weight. 5. Principles. 6. Table of Legal and Approximate Values. 3. Units.-.. 4. Measures. 154 BAY'S HIGHER ARITHMETIC, REDUCTION OF COMPOUND NUMBEES. 218. Reduction of Compound Numbers is the process of changing them to equivalent numbers of a different de- nomination. Reduction takes place in two ways: From a higlier denomination to a lower, From a lower denomination to a higher. Principles. — 1. Reduction from a higher denomination to a lower, is performed by midtiplication. 2. Reduction from a lower denomination to a higher,, is per- formed by division. Problem.— Reduce 18 bushels to pints. OPERATION. Solution. — Since 1 bu. = 4 pk., 18 bu. i g \y^ = 18 times 4 pk. =: 72 pk., and since 1 pk. 4 =r 8 qt., 72 pk. = 72 times 8 qt. = 576 qt.; y^ 1^ and since 1 qt. = 2 pt., 576 qt. =: 576 times o 2 pt. = 1152 pt. Or, since 1 bu. = 64 pt., ^ -. « . multiply 64 pt. by 18, which gives 1152 pt. o as before. This is sometimes called Beduo lUm De^cendiny. 1 8 bu. = 1 1 5 2 Jt Problem. — Reduce 236 inches to yards. Solution. — Since 12 inches = 1 ft., operation. 236 inches will be as many feet as 12 in. 12)236 in. is contained times in 286 in., which is 3)19-2 ft. 19| ft., and since 3 ft. = 1 yd., l^ ft will ^ ^ be as many yd. as 3 ft. is contained times 2 3 6 in =64 vd in 19| ft., which is 6| yd. Or, since 1 yd. =r 36 in., divide 236 in. by 36 in., which gives 6f yd., as before. This is sometimes called Beductlon Ascending. Note.— In the last example, instead of dividing 236 in. by 36 in. the unit of value of yards, since 1 inch is equal to ^^ yards, 236 inches = 236 X iV = W yd. = ^ yd. The operation by division is generally more convenient. REDUCTION OF COMPOUND NUMBERS. 155 Remark. — Reduction Descending diminishes the size^ and, there- fore, increases the number of units given ; while Reduction Ascend- ing increases the ^ize, and, therefore, diminishes the number of units given. This is further evident from the fact, that the multipliers in Reduction Descending are larger than 1 ; but in Reduction As- cending smaller than 1. Problem. — Reduce f gallons to pints. Solution.— Multiply by 4 to operation. reduce gal. to qt. ; then by 2 to _3 X ^ X ^ = 3 pt. reduce qt. to pt. Indicate the g operation, and cancel. |. gal. = 3 pt. Problem. — ^Reduce 5f gr. to S. OPERATION. Solution. — Al- ? though this is Ee- ^ 1^ y ivi^ J- 5 duction Ascending, ^7&^' y 2038 84 we use Prin. 1, in . ^ multiplying by the successive unit values, ^V> h and }. ^j g^'=TiO' Problem. — Reduce 9.375 acres to square rods. operation. 9.3 7 5 160 562500 9375 1500.00 sq. rd. 9.3 7 5 A. = 1 5 sq. rd. Problem. — Reduce 2000 seconds to hours. OPERATION. 2 sec. = I hr. 156 JiA y » S HIGHER ARITHMETIC. Problem. — Reduce 1238.73 hektograms to grams. OPERATION. 123 8.7 3X100 = 123873 grams. Problem. — How many yards in 880 meters? OPERATION. 3 9.3 7 in. X 8 8 12X3 = 9 6 2.3 7 7 + yd. Bemark. — Abstract factors can not produce a concrete result; sometimes^ however, in the steps of an indicated solution, where the change of denomination is very obvious, the abbreviations m&y be omitted until the result is written. From the preceding exercises, the following rules are derived : 219. For reducing from higher to lower denomina- tions. Bule. — 1. Multiply {he highest denomination given, by that number of (he next lower which makes a unit of the higher, 2. Add to the prodiict the number, if any, of the Uywer denomination, 3. Proceed in like manner vdth the remit thus obtained, tUl the whole is reduced fo the required denomination, 220. For reducing from lower to higher denomina- tions. Bule. — 1. Divide the given quantity by that number of its own denomination which makes a unit of the next higher, 2. Proceed in like manner with the quotient thus obtained, till (he wJwle is reduced to (he required denomination, 3. The last quotient, with the several remainders, if any, annexed, will be the answer. Note. — In the Metric System the operations are performed by removing the point to tlie right or to the left RED UCTION OP COMPO UND NUMBERS. 1 57 Examples for Practice. 1. How many square rods in a rectangular field 18.22 chains long by 4.76 eh. wide? 1387.6352 sq. rd. 2. Beduce 16.02 chains to miles. .20025 mi. 3. How madly bushels of wheat would it take to. fill 750 hektoliters? 2128^ bu. 4. Reduce 35.781 sq. yd. to sq. in. 46372.176 sq. m. 5. Eeduce 10240 sq. rd. to sq. ch. 640 sq. ch. 6. How many perches of masonry in a rectangular solid wall 40 ft. long by 7^ ft. high, and 2| ft. average thickness ? 82f| P. 7. How many ounces troy in the Brazilian Emperor's diamond, which weighs 1680 carats? 11.088 oz. 8. Reduce 75 pwt. to 3. 30 3. 9. Reduce f gr. to §. ^hs l- 10. Reduce 18f 3 to oz. av. 2^ oz. 11. Reduce 96 oz. av. to oz. troy. 87^ oz. troy. 12. How many gal. in a tank 3 ft. long by 2\ ft. wide and 1| ft. deep ? 75^ gal. 13. How many bushels in a bin 9.3 ft. long by 3| ft. wide and 2\ ft. deep? 61 bu., nearly. 14. How many sters in 75 cords of wood? 271.837+ s. 15. Reduce 2\ years to seconds. 70956000 sec. 16. Forty-nine hours is what part of a week? -^-f wk. 17. Reduce 90.12 kUoliters to liters. 90120 1. 18. Reduce 25" to the decimal of a degree. .00694° 19. Reduce 192 sq. in. to sq. yd. ^j sq. yd. 20. Reduce 6| cu. yd. to cu. in. 311040 cu. in. 21. Reduce $117.14 to mills. 117140 mills. 22. Reduce 6.19 cents to dollars. $.0619 23. Reduce 1600 mills to dollars. $1.60 24. Reduce $5f to mills. 5375 mills. 25. Reduce 12 lb. av. to lb. troy. 14^^ lb. 26. How many grams in 6.45 quintals? 645000 g. 27. Reduce .216 gr. to oz. troy. .00045 oz. troy. 158 RAY'S HIGHER ARITHMETIC. 28. Reduce 47.3084 sq. mi. to sq. rd. 4844380.16 sq. rd. 29. Reduce 4j 9 to lb. -^^ lb. 30. Reduce 7^ oz. av. to cwt. yj-g^ cwt. 31. Reduce 99 yd. to miles. yf^ mi. 32. How many acres in a rectangle 24^^ rd. long by 16.02 rd. wide? 2.4530625 a^res. 33. How many cubic yards in a box 6J ft. long by 2^ ft. wide and 3 ft. high? 1^ cu. yd. 34. Reduce 169 ars to square meters. 16900 m^. 35. Reduce 2^ f^ to n^. 1200 n^. 36. If a piece of gold is ^ pure, how many carats fine is it ? 204- carats. 37. In 18| carat gold, what part is pure and what part alloy? ff pure, and -^ alloy. 38. How many square meters of matting are required to cover a floor, the dimensions of which are 6 m., \\ dm. by 5 m., 3 cm.? 30.9345 m^. 39. How many cords of wood in a pile 120 ft. long, 6 J ft. wide, and 8f ft. high ? 53^ C. 40. How many sq. ft. in the four sides of a room 21^ ft. long, 16^ ft. wide, and 13 ft. high ? 988 sq. ft. 41. What will be the cost of 27 T. 18 cwt. 3 qr., 15 lb. 12 oz. of potash, at $48.20 a ton? $1346.97—. 42. What is the value of a pile of wood 16 m., 1 dm., 5 cm. long, 1 m., 2 dm., 2 cm. wide, and 1 m., 6 dm., 8 cm. high, at $2.30 a ster? $76.13+ 43. What is the cost of a field 173 rods long and 84 rods wide, at $25.60 an acre? $2325.12 44. If an open court contain 160 sq. rd. 85 sq. in.; how many stones, each 5 inches square, will be required to pave it ? 250909 stones. 45. A lady had a grass-plot 20 meters long and 15 meters wide; after reserving two plots, one 2 meters square and the other 3 meters square, she paid 51 cents a square meter to have it paved with stones : what did the paving cost? $146.37 BED UCTION OF COMPO UND NUMBERS. 1 59 46. A cubic yard of lead weighs 19,128 lb.: what is the weight of a block 5 ft. 3^ in. long, 3 ft. 2 in. wide, and 1 ft. 8 in. thick? 9 T. 17 cwt. 7 lb. 11.37 oz. 47. A lady bought a dozen silver spoons, weighing 3 oz. 4 pwt. 9 gr., at $2.20 an oz., and a gold chain weighing 13 pwt., at ?1J a pwt.: required the total cost of the spoons and chain. $23,331 48. A wagon-bed is 10| ft. long, 3^ ft. wide, and 1| ft. deep, inside measure : how many bushels of corn will it hold, deducting one half for cobs? 22 bu. 4 qt. 1.5 — pt. 49. K a man weigh 160 lb. avoirdupois, what will he weigh by troy weight ? 194 lb. 5 oz. 6 pwt. 16 gr. 50. The fore-wheel of a wagon is 13 ft. 6 in. in circum- ference, and the hind wheel 18 ft. 4 in.: how many more revolutions will the fore-wheel make than the hind one in 50 miles? 5155.55+ revolutions. 51. An apothecary bought 5 ft. 10 §. of quinine, at $2.20 an ounce, and sold it in doses of 9 gr., at 10 cents a dose: how much did he gain? $219.33^ 52. How many steps must a man take in walking from Kansas City to St. Louis, if the distance be 275 miles, and each step, 2 ft. 9 in.? 528000 steps. 53. The area of Missouri is 65350 sq. mi.: how many hektars does it contain? 16925940.91+ Ha. 54. A school-room is 36 ft. long, 24 ft. wide, and 14 ft. high; required the number of gallons of air it will contain? 90484.36+ gal. 55. Allowing 8 shingles to the square foot, how many shingles will be required to cover the roof of a barn which is 60 feet long, and 15 feet fi-om the comb to the eaves? 14400 shingles. 56. A boy goes to bed 30 minutes later, and gets up 40 minutes earlier than his room-mate : how much time does he gain over his room-mate for work and study in the two years 1884 and 1885, deducting Sundays only? 731 hours, 30 min. 160 HAY'S HIGHER ARITHMETIC. ADDITION OF COMPOUND NUMBEKS. 221. Compoand Numbers may be added, subtracted, multiplied, and divided. The priuciples ujx)n which these operations are performed are the same as in Simple Num- bers, with this variation ; namely, that in Simple Numbers ten units of a lower denomination make one of the next higher, while in Compound Numbers the scales vary. Addition of Compound Numbers is the process of find- ing the sum of two or more similar Compound Numbers. Problem. — Add 3 bu. 2\ pk.; 1 pk. 1^ pt.; 5 qt. 1 pt; 2 bu. 1^ qt; and .125 pt. Solution. — Eeduce the frac- tion in each number to lower denominations, and write units of the same kind in the same column. The right-hand column, when added, gives 3j\ pt. = 1 qt. Wt P^m write the 1/j and add the 1 qt. with the next column, 6 1 1^= AtiB, making 9 qt. = 1 pk. 1 qt.; write the 1 qt. and carry the 1 pk. to the next column, making 4 pk. = 1 bu.; as there are no pk. left, write down a cipher and carry 1 bu. to the next column, making 6 bu. Problem.— Add 2 rd. 9 ft. 1\ in.; 13 ft. 5.78 in.; 4 rd. 11 ft. 6 in.; 1 rd. lOf ft.; 6 rd. 14 ft. 6| in. Solution. — The numbers are prepared, written, and added, as in the last ex- ample; the answer is 16 rd. 9^ ft. 9.655 in. The \ foot is then reduced to 6 inches, and added to the 9.655 in., making 15.655 in. = l ft. 3.655 in. Write the 3.655 in., and carry the 1 ft., which gives 16 rd. 10 ft. 3.655 in. for the final answer. OPERATION. bu. pk. qt. pt. 3 2 2 : - 3 bu. 2ipk. 1 u= — 1 pk. \\ pt. 5 1 : — 5 qt. 1 pt. 2 1 i -2bu. Hqt. \'- -.125 pt U^ III, OPERATION. rd. ft. in. 2 9 7.25 13 5.78 4 11 6 1 10 8 6 14 6.625 16 91 9.655 but \ ft. = 3 6. 16 10 3.655 ADDITION OF COMPOUND NUMBERS. 161 Rule. — 1. Write the numbers to be added, placing units of the same denomination in the same column. 2. Begin with Hie lowest denomination^ add the numbers, and divide^ their sum by the number of units of this denom- ination whidi make a unit of the next higher, 3. Write the remainder under the column added, and carry the quotierd to the next column. 4. Proceed in the same manner with all (he columns to Hie last, under which write its entire sum, Eemark. — The proof of each fundamental operation in Com- pound Numbers is the same as in Simple Numbers. Examples for Practice. 1. Add f mi.; 146^ rd.; 10 mi. 14 rd. 7 fl. 6 in.; 209.6 rd.; 37 rd. 16 ft. 2^ in.; 1 mi. 12 ft. 8.726 in. 12 mi. 180 rd. 9 ft. 4.633| in. 2. Add 6.19 yd.; 2 yd. 2 ft. 9f in.; 1 ft. 4.54 in.; 10 yd. 2.376 ft.; f yd.; 1| ft.; | in. 21 yd. 2 ft. 3.517 in. 3. Add 3 yd. 2 qr. 3 na. 1^ in.; 1 qr. 2| na.; 6 yd. 1 na. 2.175 in.; 1.63 yd.; | qr.; f na. 12 yd. 1 na. 0.755 in. 4. If the volume of the earth is 1; Mercury, .06; Venus, .957; Mars, .14; Jupiter, 1414.2; Saturn, 734.8; Uranus, 82; Neptune, 110.6; the Sun, 1407124; and the Moon, .018, what is the volume of all? 1409467.775 5. James bought a balloon for 9 francs and 76 centimes, a ball for 68 centimes, a hoop for one franc and 37 cen- times, and gave to the poor 2 francs and 65 centimes, and had 3 francs and 4 centimes left. How much money did he have at first? 17^ francs. 6. Add 15 sq. yd. 5 sq. ft. 87 sq. in.; 16^ sq. yd.; 10 sq. yd. 7.22 sq. ft.; 4 sq. ft. 121.6 sq. in. ; ^ sq. yd. 43 sq. yd. 7 sq. ft. 37.78 sq. in. 7. Add 101 A. 98.35 sq. rd.; 66 A. 74^ sq. rd.; 20 A.; 12 A. 113 sq. rd.; 5 A. 13.33^ sq. rd. 205 A. 139.18^ sq. rd. H. A. 14. 162 RAY'S HIGHER ARITHMETIC. 8. Add 23 cu yd. 14 cu. ft. 1216 cu. in.; 41 cu. yd. 6 cu. ft. 642.132 cu. in.; 9 cu. yd. 25.065 cu. ft.; ^ ca yd. 75 cu. yd. 4 cu. ft. 1279.252 cu. in. 9. Add I C; I cu. ft.; 1000 cu. in. 107 cu. ft. 1072 cu. in. 10. Add 2'lb. troy, 6f oz.; If lb.; 12.68 pwt; 11 oz. 13 pwt. 19^ gr.; I K). II oz.; | pwt. 5 lb. troy, 9 oz. 9 pwt. 2. 85 J gr. 11. Add 83 14.6 gr.; 4.18 g; l^z\ 23 29 18 gr.; 1§ 12 gr.; 19. 1 lb. 2 g 4 3 1 9. 12. Add y\ T.; 9 cwt. 1 qr. 22 lb.; 3.06 qr.; 4 T. 8.764 cwt; 3 qr. 6 lb.; -^ cwt. 5 T. 6 cwt. 2 qr. 14^ lb. 13. Add .3 lb. av.; f oz. 5^ oz. 14. Add 6 gal. 3^ qt.; 2 gal. 1 qt. .83 pt. ; 1 gal. 2 qt. ^ pt.; I gal.; I qt.; | pt. 11 gal. 2 qt. .11|| pt. 15. Add 4 gal. .75 pt.; 10 gal. 3 qt. IJ pt.; 8 gal. | pt.; 5.64 gal.; 2.3 qt.; 1.27 pt; -^ pt. 29 gal. 2 qt. .05f pt. 16. Add 1 bu. ^ pk.; -^^ bu.; 3 pk. 5 qt. 1^ pt.; 9 bu. 3.28 pk.; 7 qt. 1.16 pt.; -^ pk. 12 bu. 3 pk. .46^f pt. 17. Add I bu.; | pk.; ^ qt.; | pt. 2 pk. |f pt. 18. Add 6 f3 2 f3 25 nt ; 2J f3; 7 fs 42 nt ; 1 f3 2| fs; 3f3 6f5 51 rn,. 14 f3 7 f^ 38 nt. 19. Add |- wk.; ^ da.; ^ hr.; \ min. ; ^ sec. 4 da. 30 min. 30J sec. 20. Add 3.26 yr. (365 da. each); 118 da. 5 hr. 42 min. 37J sec; 63.4 da.; 7| hr.; 1 yr. 62 da. 19 hr. 24f min.; Y^ da. 4 yr. 340 da. 1 hr. 14 min. 55^ sec. 21. Add 27° 14' 55.24"; 9° 18^"; 1° 15^; 116° 44' 23.8" 154° 14' 57.29" 22. Add ^\ \ct.\ \ m. 50 ct. 2f m. 23. Add 3 dollars 7 m.; 5 dollars 20 ct.; 100 dollars 2 ct. 6 m.; 19 dollars \ ct. $127 23 ct. 4| m. 24. Add £21 6s. 3|d.; £5 17|s.; £9.085; 16s. 7id.; £^. £37 10s. 8.15d. 25. Add I A.; f sq. rd.; i sq. ft. 107 sq. rd. 12 sq. yd. 6 sq. ft. 34^ sq. in. SUBTRACTION OF COMPOUND NUMBERS. 163 SUBTRACTION OF COMPOUND NUMBERS. 222. (Subtraction of Compound Numbers is the process of finding the difierence between two similar Com- pound Numbers. Problem. — From 9 yd. 1 ft. 6^ in. take 1 yd. 2.45 ft. Solution. — Change the J in. to a decima], operation. making the minuend 9 yd. 1 ft. 6.5 in.; reduce yd. ft. in. .45 ft. to inches, making the subtrahend 1 yd. 2 9 1 6.5 ft 5.4 in. The first term of the difference is 1.1 1 2 5.4 in. To subtract the 2 ft., increase the minuend 7 2 1.1 Am, term by 3 feet, and the next term of the subtra- hend by the equivalent, 1 yard. Taking 2 ft. from 4 ft. we have a remainder 2 ft, and 2 yd. from 9 yd. leaves 7 yd., making the answer 7 yd. 2 ft 1.1 in. Problem. — ^From 2 sq. rd. 1 sq. ft. take 1 sq. rd. 30 sq. yd. 2 sq. ft. Solution. — Write the numbers as before. operation. If the 1 sq. ft. be increased by a whole sq. sq.rd. sq.yd. sq. ft. yd. and the next higher part of the subtra- 2 1 hend by the same amount, we shall have to 1 30 2 make an inconvenient reduction of the first Ans. IJ sq.ft. remainder as itself a minuend. Hence, we make the convenient addition of J sq. yd., and, subtracting 2 sq. ft. from Z\ sq. ft, we have 1\ sq. ft. Then, giving the same increase to the 30 sq. yd., we proceed as in the former case, increasing the upper by one of the next higher, and, having no remainder higher than feety the answer is, simply, \\ sq. ft = 1 sq. ft 36 sq. in. Hule. — 1. Place the subtrahend under tlie minuend^ so that numbers of the same denomination stand in the same column. Begin at the lowest denomination y andy subtracting the parts successively from right to lefty write the remainders beneath, 2. Jf any number in the subtrahend be greater than that of the same denomination in the minuendy increase the upper by a unit, or sudi other quantity of the next higher denmnina- 164 BA Y'S JnOHkB ARITHMEIia tion as vnU render the subtraction possible, and give an equal increase to the next higher term of the subtrahend. Remark. — The increase required at any part of the minuend is, commonly, a unit of the next higher denomination. In a few instances it will be convenient to use more, and, if less be required, the tables will show what fraction is most convenient. Sometimes it is an advantage to alter the form of one of the given quantities before subtracting. Examples for Practice. 1. Subtract | mi. from 144.86 rd. 16.86 rd. 2. Subtract 1.35. yd. from 4 yd. 2 qr. 1 na. If in. 3 yd. 1 qr. f in. 3. Subtract 2 sq. rd. 24 sq. yd. 91 sq. in. from 5 sq. rd. 16 sq. yd. 6| sq. ft. 2 sq. rd. 22 sq. yd. 8 sq. ft. 41 sq. in. 4. Subtract 384 A. 43.92 sq. rd. from 1.305 sq. mi. 450 A. 148.08 sq. rd. 5. Subtract 13 cu. yd. 25 cu. ft. 1204.9 cu. in. from 20 cu. yd. 4 cu. ft. 1000 cu. in. 6 cu. yd. 5 cu. ft. 1523.1 cu. in. 6. Subtract 9.362 oz. troy from 1 lb. 15 pwt. 4 gr. 3 oz. 7 pwt. 22.24 gr. 7. Subtract |$ 3 from ^ g. 3 3 1 9 15^ gr. 8. Subtract 56 T. 9 cwt. 1 qr. 23 lb. from 75.004 T. 18 T. 10 cwt. 2 qr. 10 lb. 9. Subtract ^ lb. troy from -^ lb. avoirdupois. 0. 10. Subtract 12 gal. 1 qt. 3 gills from 31 gal. 1| pt. 18 gal. 3 qt. 3 gi. 11. Subtract .0625 bu. from 3 pk. 5 qt. 1 pt. 3 pk. 3 qt. 1 pt. 12. Subtract 1 f§ 4 fs 38 nt from 4 f^ 2 fs. 2"f3 5f3 22 nt. 13. Subtract 275 da. 9 br. 12 min. 59 sec. from 2.4816 yr. (allowing 365 J days to the year.) 1 yr. 265 da. 18 hr. 29 min. 21.16 sec. MULTIPLICATION OF COMPOUND NUMBERS. 166 14. Find the difference of time between Sept. 22d, 1855, and July 1st, 1856. 9 mon. 9 da. 15. Find the difference of time between December Slst, 1814, and April Ist, 1822. 7 yr. 3 mon. 16. Subtract 43° 18' 57.18" from a quadrant. 46° 41' 2.82" 17. Subtract 161° 34' 11.8" from 180°. 18° 25' 48.2" 18. Subtract ^ ct. from $^. 8 ct. 4J m. 19. Subtract 5 dollars 43 ct. 2\ m; from 12 dollars 6 ct. 8^ m. $6.635f 20. Subtract £9 18s. e^d. from £20. £10 Is. 5id. 21. From \ A. 10 sq. in. take 79 sq. rd. 30 sq. yd. 2 sq. ft. 30 sq. in. 16 sq. in. 22. From 3 sq. rd. 1 sq. ft. 1 sq. in. take 1 sq. rd. 30 sq. yd. 1 sq. ft. 140 sq. in. 1 sq. rd. 1 sq. ft. 41 sq. in. 23. From 3 rd. 2 in. take 2 rd. 5 yd. 1 ft. 4 in. 4 in. 24. From 7 mi. 1 in. take 4 mi. 319 rd. 16 ft. 3 in. 2 mi. 4 in. 25. From 13 A. 3 sq. rd. 5 sq. ft. take 11 A. 30 sq. yd. 8 sq ft. 40 sq. in. 2 A. 1 sq. rd. 30 sq. yd. 1 sq. ft. 32 sq. in. 26. From 18 A. 3 sq. ft. 3 sq. in. take 15 A. 3 sq. rd. 30 sq. yd. 1 sq. ft. 142 sq. in. 2 A. 156 sq. rd. 3 sq. ft. 41 sq. in. MULTIPLICATION OF COMPOUND NUMBERS. 223. Compound Multiplication is the process of mul- tiplying a Compound Number by an Abstract Number. Problem. — Multiply 9 hr. 14 min. 8.17 sec. by 10. SoLUTiOK. — Ten times 8.17 sec. = 81.7 sec. = 1 min. 21.7 sec. Write 21.7 sec. and carry 1 min. to the 140 min. obtained by the next multiplication. This gives 141 min. = 2 hr. 21 min. "^ ^ " ^ ^ ^ ^'^ Write 21 min. and carry 2 hr. This gives 92 hr. = 3 da. 20 hr. OPERATION. da. hr. min. sec. 9 14 8.1 7 10 166 RA Y'S HIGHER ARITHMETIC. Problem.— Multiply 12 A. 148 sq. rd. 28| sq. yd. by 84. Solution.— Since 84 = 7X12, multiply operation. by one of these factors, and this product -^* sq-ro* sq. yd. by the other ; the last product is the one 12 148 28 1 required. The same result can be obtained 7 by multiplying by 84 at once; performing 9 8 2 17 J the work separately, at one side, and trans- 1 2 ferring the results. 1086 30 24 Rule. — 1. Write the mvltiplier under the lowest denomina- tion of the multiplicand. 2. Multiply the lowest denomination first, and divide the product by the number of units of this denomination which make a unit of the next higher; write the remainder under the denomination multipliedy and carry tJie quotient to Hie product of the next high^ denomination, 3. Proceed in like manner with all the denominations, vrriting the entire product at the last. Examples for Practice. 1. Multiply 7 rd. 10 ft. 5 in. by 6. 45 rd. 13 ft. 2. Multiply 1 mi. 14 rd. S\ ft. by 97. 101 mi. 126 rd. 8 ft. 3 in. 3. Multiply 5 sq. yd. 8 sq. ft. 106 sq. in. by 13. 77 sq. yd. 5 sq. ft. 82 sq. in. 4. Multiply 41 A. 146.1087 sq. rd. by 9.046 379 A. 23.4593+ sq. rd. 5. Multiply 10 cu. yd. 3 cu. ft. 428.15 cu. in. by 67. 678 cu. yd. 1 cu. ft. 1038.05 cu. in. 6. Multiply 7 oz. 16 pwt. 5f gr. by 174. 113 K). 3 oz. 5 pwt. 16^ gr. 7. Multiply 2 3 1 9 13 gr. by 20. 6 § 3 5. 8. Multiply 16 cwt. 1 qr. 7.88 lb. by 11. 8 T. 19 cwt. 2 qr. 11.68 lb. DIVISION OF COMPOUND NUMBERS. 167 9. Multiply 5 gal. 3 qt. 1 pt. 2 gills by 35.108 208 gal. 1 qt. 1 pt. 2.52 gills. 10. Multiply 26 bu. 2 pk. 7 qt. .37 pt. by 10. 267 bu. 7 qt. 1.7 pt. 11. Multiply 3 fs 48 nt by 12. 5 f? 5 fs 36 nt. 12. Multiply 18 da. 9 hr. 42 min. 29.3 sec. by 16yV. 306 da. 4 hr. 25 min. 2 sec, nearly. 13. Multiply £215 16s. 2\di. by 75. £16185 14s. |d. 14. Multiply 10° 28' A2^" by 2.754 28° 51' 27.765" DIVISION OF COMPOUND NUMBERS. 224. Division of Compound Numbers is the process of dividing when the dividend is a Compound NTumber. The divisor may be Simple or Compound, hence there are two cases: 1. To divide a Compound Number into a number of equal parts, 2. To divide one Compound Number by another of the name kind. Note. — Problems under the second case are solved by reducing both Compound Numbers to the same denomination, and then dividing as in simple division. Problem. — ^Divide 5 cwt. 3 qr. 24 lb. 14f oz. of sugar equally among 4 men. SoiiUTiON. — 4 into 5 cwt. gives a opekation. quotient 1 cwt., with a remainder 1 cwt., cwt. qr. lb. oz. = 4 qr., to be carried to 3 qr., making 4)5 3 24 14f 7 qr.; 4 into 7 qr. gives 1 qr., with 3 qr., 1 1 2 4 1 5 1| — 75 lb., to be carried to 24 lb., = 99 lb.; 4 into 99 lb. gives 24 lb., with 3 lb., = 48 oz., to be carried to 14| oz., making 62| oz.; 4 into 62f oz. gives 15^1 oz., and the operation is complete. 168 RAY'S HIGHER ARITHMETIC. Problem. — If $42 purchase 67 bu. 2 pk. 5 qt. If pL of meal, how much will $1 purchase? OPERA.TION. Solution. — Since 42 = 6 X 7, divide bu. pk. qt. pt. first by one of these factors, and the 6)67 2 5 If resulting quotient by the other ; the 7)11 1 \l\ last quotient will be the one required. \ 2 3 1 jyt Rule. — 1. Write the quantity to be divided in the order of its denominations, beginning with tJie highest; place the divisor on the Uft, 2. Begin with Hie highest denomination, divide eacl% number separately, and write Hie gv^otient beneatJi. S, If a remmnder occur after any division, reduce it to the next lower denomination, and, before dividing, add to it the number of its denomiyiation. Examples for Practice. 1. Divide 16 mi. 109 rd. by 7. 2 mi. 107 rd. 2. Divide 37 rd. 14 ft. 11.28 in. by 18. 2rd. 1ft. 8.96 in. 3. Divide 675 C. 114.66 cu. ft. by 83. 8 C. 18.3453+ cu. ft. 4. Divide 10 sq. rd. 29 sq. yd. 5 sq. ft. 94 sq. in. by 17. 19 sq. yd. 4 sq. ft. 119|^ sq. in. 5. Divide 6 sq. mi. 35 sq. rd. by 22^. 170 A. 108| sq. rd. 6. Divide 1245 cu. yd. 24 cu. ft. 1627 cu. in. by 11.303 110 cu. yd. 6 cu. ft. 338.4+ cu. in. 7. Divide 3 § 7 3 18 gr. by 12. 2319 l^ gr. 8. Divide 600 T. 7 cwt. 86 lb. by 29.06 20 T. 13 cwt. 20 lb. 14 oz. 12+ dr. 9. Divide 312 gal. 2 qt. 1 pt. 3.36 gills by 72|. 4 gal. 1 qt. 1.79+ gills. 10. Divide 19302 bu. by 6.215 3105 bu. 2 pk. 6 qt. 1.5+ pt. LONGITUDE AND TIME. 169 11. Divide 76 yr. 108 da. 2 hr. 3* min. 26.18 sec. by 45. 1 yr. 254 da. 27 min. 31.25— sec. 12. Divide 152° 46' 2" by 9. 16° 58' 26|". 225. liongitude and Time give rise to two cases: 1. To find the difference of longitude between two places when the difference of time is given, 2, To find ihe difference of time when their longitudes are given. Problem. — The difference of time between two places is 4 hr. 18 min. 26 sec: what is their difference of longitude? Solution. — Every hour of time corre- j X 1 KO f 1 -^ J • * OPERATION. spends to 15° of longitude; every minute , t ^' ^ ic/ * 1 -x J J l»r. mm. sec. of time to 15' of longitude ; every second . of time to 15'''' of longitude (Art. 207). - e Hence, multiplying the hours in the — ~ • 6436"^ 30 '^^ diflference of time by 15 will give the degrees in the difference of longitude, multiplying the minutes of time by 15 will give minutes ('') of longitude, and multiplying the -seconds of time by 15 will give seconds ('''') of longitude. Problem. — The difference of longitude between two places is 81° 39' 22": what is their difference of time? SoLunoN.^-15 into 81° gives 5 (marked hr.), and 6° operation. to be carried. Instead of ^ ^)^l° 39^ 22'' multiplying 6 by 60, add- ^ hr. 2 6 min. 3 7 ^^^ sec. ing the 39^, and then divid- ing, proceed thus ; 15 into 6° is the same as 15 into 6 X ^ = 6 V ^ 4 — -S- — = 24^^, and as 15 into 39^ gives 2' for a quotient and 9^ 15 remainder, the whole quotient is 26'' (marked min,), and remainder 9^ = 9 X 6(/^, which, divided by 15, gives ?X?Pl =, 3^//^ ^hich 15 with 1^, obtained by dividing 22^^ by 15, gives 37xV^ (marked sec.). The ordinary mode of dividing will give the same result, and may be used if preferred. H. A. 15. 170 RAY'S HIGHER ARITHMETIC, 226. From these solutions we may obtain the following rules : CASE I. Rule. — Multiply ihe difference of time by 15, according to the rule for MvUipHccUion of Compound Numbers, and mark the product ° ' " instead of hr. min. sec. CASE II. Rule. — Divide the difference of longitude by 15, a/xording to the rule for Dvvimn of Coinpound Numbers, an^ mark the quotient hr. min. sec, instead of ° ' ff Note. — The following table of longitudes, as given in the records of the U. S. Coast Survey, is to be used for reference in the solution of exercises. "W." indicates longitude West, and "E." longitude East of the meridian of Greenwich, England. Table op Longitudes. Pl.ACE. LONOrrUDE. Portland, Me., .... o 70 t 15 H 18 W. Boston, Mass., .... 71 3 50 •• New Haven, Conn., . 72 55 45 " New York City, . . . 74 24 " Philadelphia, Pa., . 75 9 3 *• Baltimore, Md., . . . 76 86 59 *• Washington, D. C, . 77 36 " Richmond, Va., . . . 77 26 4 " Charleston, S. C , . . 79 55 49 •' Pittsburgh, Pa., . . . 80 2 •• Savannah, Ga., . . . 81 5 26 " Detroit, Mich., . . . &3 3 " Cincinnati, O., ... 84 29 45 •• Ixjuisville, Ky., . . . 85 25 " Indianapolis, Ind., . 86 6 " Nashville, Tenn,, . . 86 49 " Chicago, 111 87 3.-. " Mobile, Ala., . . . 88 2 28 " Madison, Wis., . . . 89 24 3 •• New Orleans, La., . . 90 3 28 " St Louis, Mo., .... 90 12 H " Minneapolis, Minn., 93 11 8 " PLACE. Des Moines, Iowa, . . . Omaha, Neb., Austin, Tex., Denver, Col Salt Lake City, Utah, . San Francisco, Cal., . . Sitka, Alaska, St. Helena Island, . . . Reykjavik, Iceland, . . Rio Janeiro, Brazil, . . St. Johns, N. F., .... Honolulu, Sandwich Is. Greenwich, Eng., .... Paris, France, Rome, Italy Berlin, German Em p., . Vienna, Austria Constantinople,Tu rkey , St. Petersburg, Russia, . Bombay, India, Pekin, China Sydney, Australia, . . . LONGITUDE. o / 93 37 95 56 97 44 104 59 111 53 122 27 135 19 5 42 22 43 20 52 43 157 52 2 20 12 28 13 23 16 20 28 59 30 16 72 48 116 26 151 11 It 16 W. 14 " 12 " 33 " 47 " 49 42 E " ** " " •• " •* •* (< (I (« <4 LONGITUDE AND TIME. 171 Examples for Practice. 1. It is six o'clock A. M. at New York ; what is the time at Cincinnati? 18 min. 2.6 sec. after 5 o'clock A. M. 2. The difference of time between Springfield, HI., and Philadelphia being 58 min. 1^^ sec, what is the longitude of Springfield ? - 89° 39' 20" W. 3. At what hour must a man start, and how fast would he have to travel, at the equator, so that it would be noon for him for twenty-four hours? Noon; 1037.4 statute miles per hour. 4. What is the relative time between Mobile and Chi- cago? Chicago time 1 min. 49|f sec. faster. 5. A man travels from Hali&x to St. Louis ; on arriving, his watch shows 9 A. M. Halifax time. The time in St. Louis being 13 min. 32^ sec. after 7 o'clock A. M., what is the longitude of Halifax? 63° 35' 18" W. 6. Noon occurs 46 min. 58 sec. sooner at Detroit than at Galveston, Texas: what is the longitude of the latter place? 94° 47' 30" W. 7. When it is five minutes after four o'clock on Sunday morning at Honolulu, what is the hour and day of the week at Sydney, Australia? 41 min. 12 sec. after 12 o'clock A. M., Monday. 8. What is the difference in time between St. Petersburg and New Orleans? • 8 hr. 1 min. 17|| sec. 9. When it is one o'clock P. M. at Rome, it is 54 min. 34 sec. after 6 o'clock A. M. at Buffalo, N. Y.: what is the longitude of the latter? 78° 53' 30" W. 10. When it is six o'clock P. M. at St. Helena, what is the time at San Francisco? 12 min. 56|^ sec. after 10 o'clock A. M. 11. A ship's chronometer, set at Greenwich, points to 4 hr. 43 min. 12 sec. P. M.: the sun being on the meridian, what is the ship's longitude? 70° 48' W. 172 BA F'^y HIOHER ARITHMETIC. ALIQUOT PARTS. 227. All aliquot part is an exact divisor of a number. Aliquot parts may be used to advantage in finding a product when either or when each of the factors is a Compound Number. Problem. — Find the cost of 28 A. 145 sq. rd. \b\ sq. yd. of land, at $16 per acre. Solution. — Multiply $16, the operation. price of 1 A., by 28 ; the product, $ 1 6 $448, is the price of 28 A. 145 '2 8 sq. rd. is made up of 120 sq. rd., 12 8 20 sq. rd., and 5 sq. rd. 120 sq. 3 2 rd. = f of an acre ; hence take f of $16, the price per acre, to find the cost of 120 sq. rd. 20 sq. rd. = J of 120 sq. rd., and cost \ as much. 5 sq. rd. = J of 20 sq. rd., and cost \ as much. 15| sq. yd. = ^ of 1 sq. rd., or ^ of 5 sq. rd., and cost y^y as much as the latter. Add the cost of the several aliquot parts to the cost of 28 acres. The result is the cost of the entire tract of land. 120 sq. rd.= f 2 sq. rd. == \ 5 sq. rd. = J 15Jsq.yd. = ^5 $448 12. 2. .50 .0 5 $462.55 Problem. — A man travels 3 mi. 20 rd. 5 yd. in 1 hr.; how far will he go in 6 da. 9 hr. 18 min. 45 sec. (12 hr. to a day)? OPERATION. rd. yd. Solution. — This example is solved like the preceding, except that here the multiplications and divisions are performed on a Compound instead of a Simple Number. mi. 3 6 da. =8X9 hr. 1 5 min. = J of 1 " 3 " =1 of 15 min. 4 5 sec. = J of 3 min. 249 20 80 5 9 27 188 1 220 225 2J 245 U 49 i • 12 lA « ALIQUOT PARTS, 173 Note. — In all questions in aliquot parts, one of the numbers indicates a raie, and the other is a Compound Number whose txi/ue at this rate is to be found. Bule for Aliquot Parts. — Multiply the number indicating the rate by tlie number of thai denomination for whose unit tlie rate is given y and separate the numbers of the otlier denomina- tions into parts whose values can be obtahted directly by a simple division or muUiplieation of one of the preceding values. Add tfiese different values; the residt will be the entire value required. Notes. — 1. Sometimes one of the- values may be obtained by adding or subtracting two pi^eceding values instead of by multiplying or dividing. 2. Aliquot parts are generally used in examples involving U. S. money, and the following table should be memorized for future use. Aliquot Parts of 100. 5 10 tV = iV 1 TTT 12i 16| 1 1 20 ^\ 25 =i 50 =i Kemark. — The following multiples of aliquot parts of 100, are often used: 18f=A, 37^=1, 40 = |, 60 = f, 62^ = f, 75 = i. Examples for Practice. 1. If a man travel 2 mi. 105 rd. 6^ ft. in 1 hour, how far can he travel in 30 hr. 29 min. 52 sec.? 71 mi. 12 rd. i ft. ^ in. 2. An old record says that 694 A. 1 R 22 P. of land, at $11.52 per acre, brought $8009.344 ; what is the error in the calculation ? $10 too much. 3. At $15.46 an oz., what will be the cost of 7 lb. 8 oz. 16 pwt. 11 gr. of gold? $1435.04 1 74 BA Y' jS HIOHUR ARITHMETIC. 4. What is the cost of 88 gal. 3 qt. 1 pt. of vinegar, at 374^ ct. a gallon? $33.33— 5. If the heart should beat 97920 times in each day, how many times would it beat in 8 da. 5 hr. 25 min. 30 sec? 805494 times. 6. At a cost of $8190.50 per mile over the plain, and at a rate of $84480 per mile of tunnel, what is the cost of a rail- way 17 mi. 150 rd. plain, and 70 rd. tunnel? $161557.80 nearly. 7. What is the value of 20 T. 1 cwt. 13 lb. of sugar, at $3f per cwt.? $1504.2375 8. If £3 6s. silver weigh 1 lb. troy, how much will 17 tb. 11 oz. 16 pwt. 9 gr. be worth? £59 7s. + 9. K a steam-ship could make 16 mi. 67f rd. in 1 hr., how far could it go in 24 da. 22 hr. 56 min. 12 sec. ? 9709 mi. 29.09 rd. Topical Outline. Operations with Compound Numbers. 1. Definition. 1. Reduction J 2. nnsps f ^- From Higher to Lower. From Lower to Higher. f 1. Dennitioii. ■J 2. Cases i^' Is. Rules. ^^• 2. Addition / 1- Definition. \ 2. Rule. 3. SubtracUon / 1- Definition. \ 2. Rule. 4. Multiplication / 1- Definition. (2. Rule. {1. Definition. 2. Cases. 3. Rule. rCase I.— Rule. 6. Longitude and Time. -I Case II.— Rule. (. Table of Longitudes. Definition. 7. Aliquot Parts \ 2. Rule. Table. I 3. Xn. RATIO. DEFINITIONS. 228. 1. Batio is a Latin word, signifying rdation or eon- nectuni; in Arithmetic, it is tJie measure of the rdation af one number to another of the same kindy expressed by their quotient, 2. A Ratio is found by dividing the first number by the second ; as, the ratio of 8 to 4 is 2. The ratio is abstract. 8. The Sign of Batio is the colon (:), which is the sign of division, with the horizontal line omitted ; thus, 6 : 4 si^ifies the ratio of 6 to 4 = |. 4. Each number is called a term of the ratio, and both together a couplet or ratio. The first term of a ratio is the antecedent, which means going before; the second term is the consequent, which mesms foUotuing. 5. A Simple Batio is a single ratio consisting of two terms ; as, 3 : 4 = |. 6. A Compound Batio is the product of two or more simple ratios ; as, < ' I = J— - ^ 15:8/ ' " 5 8' 7. The Beciprocal of a Batio is 1 divided by the ratio, or the ratio inverted ; thus, the reciprocal of 2 : 3, or |, is 8. Inverse Batio is the quotient of the consequent di- vided by the antecedent; thus, | is the inverse ratio of 4 to 5. 9. The Value of the Batio depends upon the relative size of the terms. (175) 176 RAY'S HIGHER ARITHMETIC. 220. From the preceding definitions the following prin- ciples are derived; ^ ^ ^ , Antecedent Principles. — 1. licdw = — -• Conseqtient 2. Antecedent = Gomequent X -fi^w). _ -, Antecedent 3. Consequent = — ^-7; Hence, by Art. 87 : 1. The Ratio is multiplied by multiplying t/w Antecedent or dividing the Consequent. 2. The Ratio is divided by dividing the Antecedent or mul- tiplying the Consequent. 3. The Ratio is not changed by multiplying or dividing both terms by the same number. General Law.— J.ny change in the Antecedent produces a like change in the Ratio, but any clw/nge in tJw Consequent produces an opposite cfiange in Hie Ratio. Problem. — What is the ratio of 15 to 36 ? OPERATION. 15 ; 36 = Ji = ^3. Bnle. — Divide the Antecedent by the Consequent.- Examples for Practice. 1. What is the ratio of 2 ft. 6 in. to 3 yd. 1 ft. 10 in.? j\. 2. What is the ratio of 4 mi. 260 rd. to 1 mi. 96 rd. ? f^f . 3. What is the ratio of 13 A. 145 sq. rd. : 6 A. 90 sq. rd.? ff, 4. What is the ratio of 3 lb. 10 oz. 6 pwt. 10^ gr. : 2 lb. 14i pwt? i,^jft^. PBOPaRTION. 177 5. What is the ratio of 10 gal. 1.54 pt. : 7 gal. 2 qt. .98 pt? 1^. 6. What is the ratio of 56 bu. 2 pk. 1 qt. : 35 bu. 3 pk. 6.055 qt.? ft^f. 7. K the antecedent is 7 and the ratio J^, what is the consequent ? 4f . 8. If the consequent is ^ and the ratio f , what is the antecedent ? -^j, 9. What is the ratio of a yard to a meter, and of a meter to a yard? mm*. IMMU- 10. What is the ratio of a pound avoirdupois to a pound troy ? m- 11. Find the difference between the compound ratios 12. Find the difference between the ratio 4| : 1\ and the inverse ratio. WW - 13. If the consequent is 6^, and the ratio is 2^, what is the antecedent, and what is the inverse ratio of the two numbers? Antecedent 14f, inverse ratio |^. XIII. PEOPORTIOl^. DEFINITIONS. 230. i. Proportion is an equality of ratios. Thus, 4 : 6 : : 8 : 12 is a proportion, and is read ^ is to Q as 8 18 to 12. 2. The Sign of Proportion is the double colon ( ; : ). Note. — It is the same in effect as the sign of equality, which is sometimes used in its place. 3. The two ratios compared are called couplets. The first couplet is composed of the first and second terms, and the second couplet of the third and fourth terms. 178 JRAY'S HIOHEB ARITHMETIC. 4. Since each ratio has an antecedent and consequent, every proportion has two antecedents and two consequents, the 1st and 3d terms being the antecedents, and the 2d and 4th the consequents. 5. The first and last terms of a proportion are called the extremes; the middle terms, the means. All the terms are called proportionals, and the last term is said to be a fourth proportional to the other three in their order. 6. When three numbers are proportional, the second num- ber is a mean proportional between the other two. Thus, 4:6 : : 6:9; six is a mean proportional between 4 and 9. 7. Proportion is either Simple or Compound: Simple when both ratios are simple ; Compound when one or both ratios are compound. Principles. — 1. In every proportion the product of the means is equal to the produd of Hie extremes, 2. The product of the extremes divided by either mean, will give the other mean. 3. Tlie product of (he means divided by either extreme, will give Hie other extreme. SIMPLE PROPORTION. 231. 1. Simple Proportion is an expression of equality between two simple ratios, 2. It is employed when three terms are given and we wish to find the fourth. Two of the three terms are alike, and the other is of the same kind as the fourth which is to be found. 3. All proportions must be true according to Principle 1, which is the test. Principles 2 and .3 indicate methods of finding the wanting term. SIMPLE PMOFORTION. 179 4. The Statement is the proper arrangemeut of the terms of the proportion. Problem. — If 6 horses cost $300, what will 15 horses cost? STATEMENT. 6 horfles : 15 horses :: $300 : ($ ). OPERATION. (|« O /\ /\ \y 1 E ~ =$750. Or,($300X15)^6 = $750,^n«. • Solution. — Since 6 horses and 15 horses maybe compared, they form the first couplet; also, $300 and $ — may be compared, as they are of the same unit of value. Notes. — 1. To find the missing extreme, we use Prin. 3. 2. To prove the proportion, we use Prin. 1. Thus, $750 X 6 ^= $300 X 15. Problem. — If 15 men do a piece of work in 9f da., how long will 36 men be in doing the same? STATEMENT. Solution.— Since 36 men will ^^^ ^^^ ^^ ^^ require less time than 15 men to 36*15"94*M do the same work, the answer should be less than 9f da.; make a operation. decreasing ratio, Jf , and multiply 9 f = -^ da. the remaining quantity by it. ^ X M = ^ <!*•> -^^• Bnle. — 1. For the third term, vrrUe that number which is of the 8ame denominatimi as the number required, 2. For the second term, unite the greater of the two re- maining numbers, when the fourth term is to he greater than the third; and the less, when the fourth term is to be less Vmn the third, 3. Divide the product of the second and third terms by the first; the qmtient will be Hie fourth term, or number required. 180 BA Y'S HIGHER ARITHMETIC, Examples FOR Practice. Note.— Problems marked with an asterisk are to be solved mentally. 1.* If I walk 10^ mi. in 3 hr., how far will I go in 10 hr., at the same rate? 35 mi. 2. If the fore-wheel of a carriage is 8 ft. 2 in. in cir- cumference, and turns round 670 times, how often will the hind- wheel, which is 11 ft. 8 in. in circumference, turn round in going the same distance? 469 times. 3. If a horse trot 3 mi. in 8 min. 15 sec, how far caji he trot in an hour, at the same rate? 21^ mi. 4. What is a servant's wages for 3 wk. 5 da., at $1.75 per week? $6.50 5. What should be paid for a barrel of powder, containing 132 lb., if 15 lb. are sold for «5.43|? $47.85 6. A body of soldiers are 42 in rank when they are 24 in file : if they were 36 in rank, how many in file would there be? 28. 7. If a pulse beats 28 times in 16 sec, how many times does it beat in a minute? 105 times. 8. If a cane 3 ft. 4 in. long, held upright, casts a shadow 2 ft. 1 in. long, how high is a tree whose shadow at the same time is 25 ft. 9 in.? 41 ft. 2| in. 9. If a farm of 160 A. rents for $450, how much should be charged for one of 840 A ? $2362.50. 10. A grocer has a false gallon, containing 3 qt. 1^ pt.: what is the worth of the liquor that he sells for $240, and what is his gain by the cheat? $225, and $15 gain. 11. If he uses 14f oz. for a pound, how much does he cheat by selling sugar for $27.52? $2.15 12. An equatorial degree is 365000 ft.: how many ft. in 80° 24' 37" of the same? 29349751^^ ft. 13. If a pendulum beats 5000 times a day, how often does it beat in 2 hr. 20 min. 5 sec ? 486:^^ times. SIMPLE PROPORTION. 181 14.* If it takes 108 days, of 8^ hr. , to do a piece of work, how many days of 6| hr. would it take ? 136 days. 15. A man borrows $1750, and keeps it 1 yr. 8 mon.: how long should he lend $1200 to compensate for the favor? 2 yr. 5 mon. 5 da. 1^ A garrison has food to last 9 mon., giving each man 1 lb. 2 oz. a day : what should be a man's daily allowance, to make the same food last 1 yr. 8 mon.? 8^^ oz. 17. A garrison of 560 men have provisions to last during a siege, at the irate of 1 lb. 4 oz. a day per man ; if the daily allowance is reduced to 14 oz.' per man, how large a reinforcement could be received? 240 men. 18. A shadow of a cloud moves 400 ft. in 18f sec: what was the wind's velocity per hour? 14^ mi. 19. If 1 ft), troy of English standard silver is worth £3 6s., what is 1 lb. av. worth? £4 2^d. 20. If I go a journey in 12f days, at 40 mi. a day, how long would it take me at 29f mi. a day? 17| da. 21.* If |- of a ship is worth $6000, what is the whole of it worth ? . $10800. 22. If A, worth $5840, is taxed $78.14, what is B worth, who is taxed $256.01 ? $19133.59— 23.* What are 4 lb. 6 oz. of butter worth, at 28 ct. a lb.? $1.22i 24. If I gain $160.29 in 2 yr. 3 mon., what would I gain in 5 yr. 6 mon., at that rate? $391.82 25. If I gain $92.54 on $1156.75 worth of sugar, how much must I sell to gain $67.32? $841.50 worth. 26. If coffee costing $255 is now worth $318.75, what did $1285.20 worth cost? $1028.16 27. A has cloth at $3.25 a yd., and B has flour at $5.50 a barrel. K, in trading, A puts his cloth at $3.62^, what should B charge for his flour? $6.13y\ 28.* If a boat is rowed at the rate of 6 miles an hour, and is driven 44 feet in 9 strokes of the oar, how many strokes are made in a minute? 108 strokes. 182 BAY'S HIGHER ARITHMETIC. 29. If I gain $7.75 by trading with $100, how much ought I to gain on $847.56? $65.6859 30. What is a pile of wood, 15 ft. long, 10\ ft. high, and 12 ft. wide, worth, at $4.25 a cord? $62.75 Eemark. — In Fahrenheit's thermometer, the freezing point of water is marked 32°, and the boiling point 212° : in the Centigrade, the freezing point is 0°, and the boiling point 100°: in Keaumer's, the freezing point is 0°, and the boiling point 80°. 31. From the above data, find the value of a degree of each thermometer in the degrees of the other two. V F.=^° R. = F C-; l"" C. =1|° F. = |°R.; l°R.=li°C. = 2i°F. 32. Convert 108° F. to degrees of the other two ther- mometers. 33^° R. and 42f ° C. 33. Convert 25° R. to degrees of the other two thermom- eters. 31i° C. and 88^° F. 34. Convert 46° C. to degrees of the other two ther- mometers. 36^° R. and 114|° F. Kemabks. — 1. In the working of machinery, it is ascertained that the available power is to the weight overcome, inversely as the diatanees they pass over in the same tim£, 2. Inverse variation exists between two numbers when one in- creases as the other decreases. 3. The available power is taken | of the whole power, \ being allowed for friction and other impediments. • 35. If the whole power applied is 180 lb. and moves 4 ft. , how far will it lift a weight of 960 lb. ? 6 in. 36. If 512 lb. be lifted 1 ft. 3 in. by a power moving 6 ft. 8 in. , what is the power ? 144 lb. 37. A lifts a weight of 1410 lb. by a wheel and axle; for every 3 ft. of rope that passes through his hands the weight rises 4^ in.: what power does he exert? 270 lb. 38. A man weighing 198 lb. lets himself down 54 ft. with a uniform motion, by a wheel and axle : if the weight at the hook rises 12 ft., how much is it? 594 lb. SIMPLE PEOPORTION, 183 39. Two bodies free to move, attract each other with forces that vary inversely as their weights. If the weights are 9 lb. and 4 lb., and the smaller is attracted 10 ft., how far will the larger be attracted ? 4 ft. 5^ in. 40. Suppose the earth and moon to approach each other in obedience to this law, their weights being 49147 and 123 respectively, how many miles would the moon move while the earth moved 250 miles ? 99892+ mi. Can the three following questions be solved by pro- portion? 41. If 3 men mow 5 A. of grass in a day, how many men will mow 13^ A. in a day ? 42.* If 6 men build a wall in 7 da., how long would 10 men be in doing the same? 43.* If I gain 15 cents each, by selling books at $4.80 a doz., what is my gain on each at $5.40 a doz.? 44. A clock which loses 5 minutes a day, was set right at 6 in the morning of January 1st : what will be the right time when that clock points to 11 on the 15th? 11 min. 17.35+ sec. past noon. 45. If water begin and continue running at the rate of 80 gal. an hour, into a cellar 12 ft. long, 8 ft. wide, and 6 ft. deep, while it soaks away at the rate of 35 gal. an hour, in what time will the cellar be full? 95.75+ hr. 46. Take the proportion of 4 : 9 : : 252 : a fourth term. If the third and fourth terms each be increased by 7, while the first remains unchanged, what multiplier is needed by the second to make a proportion? fl^. 47. Prove that there is no number which can be added to each term of 6 : 3 : : 18 : 9 so that the resulting num- bers shall stand in proportion. 48. A certain number has been divided by one more than itself, giving a quotient -jL: what is the number? \. 49. If 48 lb. of sea-water contain 1^ lb. of salt, how much fresh water must be added to these 48 lb. so that 40 lb. of the mixture shall contain \ lb. of salt? 72 lb. 184 BAY'S HIGHER ARITHMETIC. CJOMPOUND PROPORTION. 232. Compound Proportion is an expression of equality between two ratios when either or when each ratio is Com- pound. Problem. — If 3 men mow 8 A. of grass in 4 da., how long would 10 men be in mowing 86 A.? STATEMENT. 1 men : 3 men 1 ^ , r \ a Solution. — Since « * - ^ fi A \ ' * v / ^^* the denomination of the required term operation. is days, make the 9 third term 4 da. In T g forming the first and 3X^^X^ ,, \, j . second terms, con- ^ ^ ^ ="%==<> T cla. Ans, sider each denom- A r A p ination separately ; *^ r 10 men can do the same amount of work in less time than 3 men ; hence, the first ratio is, 10 men : 3 men, the less number being the second term. Since it takes 4 da. to mow 8 A., it will take a greater number of days to mow 36 A., and the second ratio is,'8 A. : 36 A., the gii'eater number being the second term. Then dividing the con- tinued product of the means by that of the extremes (Art. 230, Prin. 3), after cancellation, we have 5f da., the required term. Rule. — 1. For the third term, write that nwnher which is of the same denomination as the number required, 2. Arrange each pair of numbers having the sam£ denom- ination in the compound ratio, as if, with the third term, Hiey formed a simple proportion, 3. Divide the product of the numbers in the second and third tenns by the product of the nwinbers in the first term: the quotient wiU be the required term. 233. Problems in Compound Proportion are readily solved by separating all the quantities involved into ttoo causes and two effects. COMPOUND PROPORTION. 185 Problem. — If 6 men, in 10 days of 9 hr. each, build 25 rd. of fence, how many hours a day must 8 men work to build 48 rd. in 12 days? Solution. — 6 men 10 da. and 9 hr. constitute the first cause, whose effect is 25 rd. ; 8 men 12 da. and ( ) hr. constitute the second cause, whose effect is 48 rd. Hence, STATEMENT. 8 men. ^ : 12 da. >:: 25 rd. : 48 rd. ( ) hr. J OPERATION. ? 5 Biile of Cause and Effect. — 1. Separate all the quan- tities contained in the question into two causes and Hmr effects, 2. Write, for the first tefi^m of a propoHion, all the qudn- tities that constitute the first cause ; for the second term, aU tJiat constitute the second cause; for the third, all that comtittUe tJie effect of the first cause ; and for the fourth, all that constitute the effect of the second coMse, 3. The required quantity may be indicated by a bracket, and found by Art. 230, Principles. Note. — The two causes must be exactly alike in the numbei' and kind of their terms ; and so must the two effects. Examples for Practice. 1. If 18 pipes, each delivering 6 gal. per minute, fill a cistern in 2 hr. 16 min., how many pipes, each delivering 20 gal. per minute, will fill a cistern 7^ times as large as the first, in 3 hr. 24 min. ? 27 pipes. H. A. 16. 186 JiA Y'S HIGHER ARITHMETIC. 2. The use of $100 for 1 year is worth $8 : what is the use of $4500 for 2 yr. 8 mon. worth ? $960. 3. If 12 men mow 25 A. of grass in 2 da. of 10^ hr., how many hours a day must 14 men work to mow an 80 A. field in 6 days ? . 9f hr. 4. If 4 horses draw a raiboad car 9 miles an hour, how many miles an hour can a steam engine of 150 horse-power drive a train of 12 such cars, the locomotive and tender being counted 3 cars ? 22^ mi. per hr. 5. If 12 men, working 20 days 10 hours a day, mow 247.114 hektars of timothy, how many men in 30 days, working 8 hours a day, will mow 1976912 centars of timothy of the same quality ? * 8 men. 6. If the use of $3750 for 8 mon. is worth $68.75, what sum is that whose use for 2 yr. 4 mon. is worth $250? $3896.10+ 7. If the use of $1500 for 3 yr. 8 mon. 25 da. is worth $336.25, what is the use of $100 for 1 yr. worth? $6. 8. A garrison of 1800 men has provisions to last 4^ months, at the rate of 1 lb. 4 oz. a day to each : how long will 5 times as much last 3500 men, at the rate of 12 oz. per day to each man? 1 yr. 7^ months. 9. What sum of money is that whose use for 3 yr., at the rate of $4^ for every hundred, is worth as much as the use of $540 for 1 yr. 8 mon., at the rate of $7 for every hundred? $466. 66f 10. A man has a bin 7 ft. long by 2| ft. wide, and 2 ffc. deep, which contains 28 bu. of corn : how deep must he make another, which is to be 18 ft. long by 1^ ft. wide, in order to contain 120 bu. ? 4|^ ft. 11. If it require 4500 bricks, 8 in. long by 4 in wide, to pave a court-yard 40 ft. long by 25 ft. wide, how many tiles, 10 in. square, will be needed to pave a hall 75 ft. long by 16 ft. wide? 1728 tiles. 12. If 150000 bricks are used for a house whose walls average 1^ ft. thick, 30 ft. high, and 216 ft. long, how COMPOUND PBOPORTION. 187 many will build one with walls 2 ft. thick, 24 ft. high, and 324 ft. long? 240000 bricks. 13. If 240 panes of glass 18 in. long, 10 in. wide, glaze a house, how many panes 16 in. long by 12 in. wide will glaze a row of 6 such houses? 1350 panes. 14. If it require 800 reams of paper to publish 5000 volumes of a duodecimo book containing 320 pages, how many reams will be needed to publish 24000 copies of a book, octavo size, of 550 pages ? 9900 reams. 15. If 15 men cut 480 sters of wood in 10 days, of 8 hours each: how many boys will it take to cut 1152 sters of wood, only ^ as hard, in 16 days, of 6 hours each, pro- vided that while working a boy can do only f as much as a man, and that \ of the boys are idle at a time throughout the work ? 24 boys. Topical Outline. Ratio and Proportion. 1. Ratio 2. Proportion. 1. Definitions. 2. Principles. 3. General Law. 1. Definitions. 2. Principles. 3. Kinds 4. Statement 5. Rules. I 1. Simple. 2. Compound. XIY. PEEOEI^rrAGE, DEFINITIONS. 234. 1. Percentage is a term applied to all calcula- tions in which 100 is the basis of comparison; it is also used to denote the result arising from taking so many hundredths of a given number. 2. Per Cent is derived from the Latin phrase pet centumy which means by or on the hundred, 3. The Sign of Per Cent is ^; the expression 4% =; yI^, is read "4 per cent'* equals yj^; or, decimally, .04 4. The elements in Percentage are the BasCy the Bate, the Percentage^ and the Amount or Difference, 5. The Base is the number on which the Percentage is estimated. 6. The Bate is the number of hundredths to be taken. 7. The Percentage is the result arising from taking that part of the Base expressed by the Rate. 8. The Amount is the Base phis the Percentage. 9. The DifTerenoe is the Base minus the Percentage. NOTATION. 236. It is convenient to use the following notation: 1. Base =B. 2. Rate =R. 3. Percentage = P. A f Amount = A. 1 Difference = D. (ISS) PERCENTAGE. 189 Remark. — All problems in Percentage refer to two or more of the above terms. Owing to the relations existing among these terms, any two of them being given the others can be found. These relations give rise to the following cases : CASE I. 236. Given the base and the rate, to find the per- centage. Principle. — The percentage of any number is the rnrne^ part of that number as the given rate is of 100^. Problem. — If I have 160 sheep, and sell 35% of them: bow many do I sell? OPERATIONS. 1. 160 sheep X-3 5 = 56 sheep, Am. >^ 2. 100/o = 160 sheep. Ifo = 1.60 sheep. .-. 35^ = 1.60X35 = 56 sheep, Ans. 3. 1 6 X iV = ^ ^ sheep, Ans. (Art. 218, Bern.) Suggestion.— 3 5^ = yVtt = sV- Solution. — Take .36 of the base ; the result is the percentage. Formula.— B X B = P. Rule 1. — Multiply the hose by the rate ^ expressed deci- mxdly; the product is the percentage. Rule 2. — Find that part of the base which the rate % is qfim. Examples for Practice. 1. Find 62^^ of 1664 men. 1040 men. 2. Find 35% of f yV 3. Find 9f % of 48 mi. 256 rd. 4 mi. 184 rd. 190 BAY'S HIGHER ARITHMETIC. 4. Fiud 11^% of $3283.47 $364.83 5. Find ZZ\% of 127 gal. 3 qt. 1 pt. 42 gal. 2 qt. 1 pt 6. Find 98^ of 14 cwt. 2 qr. 20 lb. 14 cwt. 1 qr. 15f lb. 7. Find 40% of 6 hr. 28 min. 15 sec. 2 hr. 35 min. 18 sec. 8. Find 104^ of 75 A. 75 sq. rd. 78 A. 78 sq. rd. 9. Find 15|^ of a book of 576 pages. 90 pages. 10. Find 56^% of 144 cattle. 81 cattle. 11. Find 16|^ of 1932 hogs. 322 hogs. 12. Find 1000% of $5.43f $54.37^ 13. Find ^% of f -sV- 14. What part is 25% of a farm? \, 15. What part of a quantity is 18f% of it; 31^%; 37i%; 43f%; 56^%; 62|%; 68f%; 81^%; 83J%; 87^%; 93|% ? A, 1^, f , tV. A» I, ii, «, h h H of it. 16. How much is 100% of a quantity; 125% of it; 250%; 675%; 1000%; 9437^%? 1 time, 1J,2^, 6f, 10, 94f times the quantity. 17.* A man owning f of a ship, sold 40% of his share : what part of the ship did he sell, and what part did he still own ? ^ sold ; ^ left. 18.* A owed B a sum of money; at one time he paid him 40% of it; afterward he paid him 25% of what he owed ; and finally he paid him 20% of what he then owed : how much does he still owe? ^^ of it. 19. Out of a cask containing 47 gal. 2 qt. 1 pt., leaked 6f%: how much was that? 3 gal. If pt. 20. A has an income of $1200 a year; he pays 23% of it for board; 10f% for clothing; 6f% for books; ^% for newspapers; 12^% for other expenses: how much does he pay for each item, and how much does he save at the end of the year? $276, bd.; $124.80, cL; $81, bks.; $7, npr. $154.50, other ex.; $556.70 saved. 21. Find 10% of 20% of $13.50 27 ct. 22. Find 40% of 15% of 75% of $133.33^ $6. PERCENTAGE. IM 23. A man coDtracts to supply dressed stone for a court- house for $119449, if the rough stone costs him-- 16 ct. a cu. ft.; but if he can get it for 15 ct. a cu. ft., he will deduct Z^o from his bill; how many cu. ft. would be needed, and what does he charge for dressing a cu. ft. ? 358347 cu. ft., and 17^ ct. a cu. ft. 24. 48% of brandy is alcohol ; how much alcohol does a man swallow in 40 years, if he drinks a gill of brandy 3 times a day? 657 gal. 1 qt. 1 pt. 2.4 gills. 25. A had $1200; he gave 30% to a son, 20% of the remainder to his daughter, and so divided the rest among four brothers that each after the first had $12 less than the preceding : how much did the last receive ? $150. 26. What number increased by 20% of 3.5, diminished by 12|% of 9.6, gives 3i? 4. CASE II. 237. Given tbe base and the percentage, to find the rate* Pkinciple. — The rate equals Uie number of hundredths Uiat the percentage is of the base. Problem. — ^What per cent of 45 is 9? Solution. — 9 is J of 45 ; but | of any operation. number ia equal to 20^ of that number ; ^^ := ^ ^n^ 20^, Arts. hence, 9 is 20^ of 45- P Formula. — — = R. Bnle 1. — Divide the percentage by the base; the qvMient is the rate; Bnle 2. — Find that part of 100^ that the percentage is of the base. 192 RAY'S HIQHER ARITHMETIC. Examples for Practice. 1. 15 ct. is how many ^ of $2? 2. 2 yd. 2 ft. 3 in. is how many ^ of 4 rd.? 3. 3 gal. 3 qt. is what % of 31^ gal.? 4. f is how many ^ of . -| ? '5. I of I of 4^ is what % of 1^? 6, -^ is how many % of -f ? 3 -^ / 10 7. $5.12 is what % of $640? 8. $3.20 is what % of $2000? 9. 750 men is what % of 12000 men? 10. 3 qt. 1^ pt. is what ^ of 5 gal. 2J qt.? 11. A's money is 50^ more than B's; then B's money is 222|% 16|% 2|% 25% how many ^ less than A's ? 12. What % of a number is 8^ of 35^ of it? 13. What % of a number is 21% of 2^% of it? 14. What ^ of a number is 40^ of 62|% of it? 15. 12^ of $75 is what % of $108? 16. U. S. standard gold and silver are 9 parts pure to 1 part alloy: what % of alloy is that? 10% 17. What ^ of a meter is a yard? 91 ^;|fgj| % 18. How many ^ of a township 6 miles square, does a man own who has 9000 acres ? ^^Tt% • 19. How many ^ of a quantity is 40^ of 25^ of it ? also, 16% of 37^% of it? also, 4^^ of 120% of it? also, 2% of 80% of 66|^ of it? also, |% of 36% of 75% of it? also, ^% of 22^% of 96% of it? 10, 6, 5, ItV 7^^, Wt7%- 20. 30% of the whole of an article is how many % of f of it? 45%. 21. 25% of f of an article is how many % of f of it? 22. How many % of his time does a man rest, who sleeps 7 hr. out of every 24 ? 29^% . PERCENTAOK 193 CASE III. 238. Given the rate and the percentage, to find the base. Principle. — The base bears the same ratio to the percerdxige that 100% does to the rate. Problem. — 95 is 5% of what number? OPERATION. 100/^ = H* = 19X100 = 1900, Ana. Or, 95-!-. 05=1900, Am. Formula. — ~ = B. Bnle 1. — Divide the percentage by the rate, and then mul- tiply the qtu>tient by 100; the product is the base. Bnle 2. — Divide the percentage by the rate expressed deci- mally ; the quotient is the base. Examples for Practice. 1. $3.80 is 5^ of what sum? $76. 2. -^ is 80^ of what number? ■^. 3. 16 is 1^% of what number? 1066|. 4. 31^ ct. is 15f^ of what? $2. 5. $10.75 is 3\fo of what? $322.50 6. 162 men is 4|% of how many men? 3375 men. 7. $19.20 is ^fo of what? $3200. 8. $189.80 is 104% of what? $182.50 9. 16 gal. 1 pt. is 6|^ of what? 262 gal. 2 qt. 10. 10 mi. 316 rd. is 75% of what? . 14 mi. 208 rd. H. A. 17. ^ 194 RAY'S HIGHER ARITHMETIC. 11. Thirty-six men of a ship's crew die, which is 42f % of the whole : what was her crew ? 84 men. 12. A stock-farmer sells 144 sheep, which is 12^ % of his flock: how many sheep had he? 1125 sheep. 13. A merchant sells 35% of his stock for $6000: what is it all worth at that rate? $17142.86 14. I shot 12 pigeons, which was 2§% of the flock: how many pigeons escaped ? 438 pigeons. 15. A, owing B, hands him a $10 bill, and says, ** there is 6J^ of your money:" what was the debt? $160. 16. $25 is 62^% of A's money, and 41f^ of B's: how much has each? A $40, B $60. 17. A found $5, which was 13J% of what he had before: how much had he then ? $42.50 18. I drew 48% of my funds in bank, to pay a note of $150: how much had I left? $162.50 19. A farmer gave his daughter at her marriage 65 A. 106 sq. rd. of land, which was 3^ of his farm : how much land did he own? 2188 A. 120 sq. rd. 20. A pays $13 a month for board, which is 20% of his salary : what is his salary ? $780 a year. 21. Paid 40 ct. for putting in 25 bu. of coal, which was llf ^ of its cost: what did it cost a bu. ? 14 ct. 22. 81 men is 5% of 60% of what? 2700 men. 23. A, owning 60% of a ship, sells 7^% of his share for $2500: what is the ship worth? $55555. 55f 24. A father, having a basket of apples, took out 33^% of them ; of these, he gave 37^^ to his son, who gave 20^ of his share to his sister, who thus got 2 apples: how many apples were in the basket at first ? 80 apples. 25. B lost three dollars, which was 31J^ of what he had left: how much had he at first? $12.60 26. Bought 8000 bu. of wheat, which was bl^% of my whole stock: how much had I before? 6000 bu. 27. If 32^ of 75% of 800% of a number is 1539, what is that number? • 801-^. PER CENTA OE. 195 CASE IV. 239. Given the rate and the amount or the differ- ence, to find the base. Principle. — The base is equal to ihe anumnt divided by 1 plus the rate, or the difference divided by 1 minus tlie raJte, Problem. — ^A rents a house for $377, which is an advance of 16^ on the rent of last year: what amount did he pay last year? OPERATION. $377-1- 1.16 = $325.00, Ans. Or, 10 0^ = rental last year ; 1 6 Jl^ = increase this year ; hence, 1 OO^o -f 1 6^o = 1 16 % =$377; l^o = $3.25; .-. 100^ = 3.25X100 = $325.00, ^ns. Problem. — John has $136, which is 20^ less than Joseph's money: how many dollars has Joseph? OPERATION. $136-^(1 — .2)=$170.00, Am. Or, 1 J^ = Joseph's money ; 8 ^0 = John's money = $136; l^o = $1.7; .-. 100^c)=$1.7X 100 = $170.00, ^w^. Or, 80%=i = $136; ^=$34; and | = $34X 5 = $170.00, Ans. (A^(l+K)) F0RMULA.~B=|j)^(j_j^^| Bule. — Divide ike sum by 1 plus (he rate, or divide the difference by 1 minus ihe rate; ihe quotiefiit wUl be ihe base* 196 RAY'S HIGHER ARITHMETIC. Examples for Practice. • 1. $4.80 is 33^^ more than what? $3.60 2. f is 50% more than what? -|- 3. 96 da. is 100^ more than what? 48 da. 4. 2576 bu. is 60% less than what? 6440 bu. 5. 87^ ct. is 87^% less than what? $7. 6. 42 mi. 60 rd. is 55^ less than what ? 93 mi. 240 rd. 7. 2 lb. 9^ oz. is 50% less than what number of pounds? 5f|f lb. 8. -^ is 99|^ less than what? 155|. 9. $920.93f is 337|% more than what ? $210.50 10. $4358.06^ is 233^% more than what? $1307.41^ 11. In 64^ gal. of alcohol, the water is 7^^ of the spirit: how many gal. of each? 60 gal. sp., 4^ gal. w. 12.* A coat cost $32; the trimmings cost 70% less, and the making 50^ less, than the cloth: what did each cost? Cloth $17.77|, trimmings $5.33^, making $8.88| 13.* If a bushel of wheat make 39^ lb. of flour, and the cost of grinding be 4%, how many barrels of flour can a farmer get for 80 bu. of wheat? 15^ barrels. 14.* How many eagles, each containing 9 pwt. 16.2 gr. of pure gold, can I get for 455.6538 oz. pure gold at the mint, allowing 1^% for expense of coinage? 928 eagles. 15. 2047 is 10% of 110% less than what number? 2300. 16. 4246| is 6% of 50^ of 466|% more than what number ? 3725. 17. A drew out of bank 40% of 50% of 60% of 70% of his money, and had left $1557.20 : how much had he at first? $1700. 18. I gave away 42^% of my money, and had left $2 : what had I at first? $3.50 19. In a school, 5% of the pupils are always absent, and the attendance is 570: how many on the roll, and how many absent? 600, enrolled; 30, absent. APPLICATIONS OF PERCENTAGE. 197 20. A man dying, left 33^% of his property to his wife, 60% of the remainder to his son, 75% of the remainder to his daughter, and the balance, ?500, to a servant : what was the whole property, and each share ? Property, $7500 ; wife had »2500 ; son, $3000 ; daughter, $1500. 21. In a company of 87, the children are 37^^ of the women, who are 44J% of the men: how many of each? 54 men, 24 women, 9 children. 22.* Our stock decreased 33^%, and again 20^ ; then it rose 20%, and again 33 J^ ; we have thus lost $66: what was the stock worth at first? $450. 23.* A brewery is worth 4% less than a tannery, and the tannery 16% more than a boat; the owner of the boat has traded it for 75^ of the brewery, losing thus $103 : what is the tannery worth ? $725. ADDITIONAL FORMULAS. 240. The following additional formulas, derived from preceding data, may also be employed to advantage : 1. By definition, A = B + P ; also, D = B — P. r A 2. From Case IV. < A = B + (BxR) = B + P. APPLICATIONS OF PERCENTAGE. 241. The Applications of Percentage may be divided into two classes, those in which time is not an essential element, and those in which it is an essential element, as follows: 198 BAY'S HIOHEB ARITHMETIC, Those in which Time is not an Essential Element Those in which Time is an Essential Element... I \ 1. Profit and Loss. 2. Stocks and Dividends. 3. Premium and Discount. 4. Commission and Brokerage, 5. Stock Investments. 6. Insurance. 7. Taxes. 8. United States Revenue. 1. Simple Interest. 2. Partial Payments. 3. True Discount. 4. Bank Discount. 5. Exchange. 6. Equation of Payments. 7. Settlement of Accounts. 8. Compound Interest. 9. Annuities. Note.— These topics will be presented in the order in which they stand. Topical Outline. Percentage. 1. Definitions. 2. Notation. Ok x>aSCS.M... « u. m. IV. 1^ I 3. I 3. i 3. I 3. 1. Principle. Formula. Rules. 1. Principle. Formula. Rules. 1. Principle. Formula. Rules. 1. Principle. Formula. Rule. 4. Additional Formulas. „ ,, , ^ . ( Time not an Element 5. Applications of Percentage. -{ *_. *^*^ I Time an Element. XY. PEECENTAGE.-APPLIOATIOI^rS. I. PROFIT AND LOSS. DEFINITIONS. 242. 1. Profit and Loss are commercial terms, and pre- suppose a cost price. 2. The Cost is the price paid for any thing. 3. The Selling Price is the price received for whatever is sold. 4. Profit is the excess of the Selling Price above the Cost. 5. Loss is the excess of the Cost above the Selling Price. 243. There are four cases of Profit or Loss, solved like the four corresponding cases of Percentage. The cost corresponds to the Base; the per cent of profit or loss, to the Bate; the profit or loss, to the Percentage; the cost plus the profit, or the selling price, to the Amount; and the cost minus the loss, or the selling price, to the Difference. CASE I. 244. Given the cost and the rate, to find the profit or loss. Problem. — Having invested 84800, my rate of profit is 1S% : what is my $4800 profit? .13 Solution.— Since the coRt is $4800, and 14400 the rate of profit is 13^; the profit is 13^ ^^^^ of $4800, which is $624. $ 6 2 4.0 Profit. (199) 200 BAY'S HIGHER ARITHMETIC. Examples for Practice. 1. If a man invests $1450 so as to gain 14^%, what is his profit? $210.25 2. I bought $1760 worth of grain, and sold it so as to make 26^% profit : what did I receive for it? $2222. 3. If a man invests $42540, and loses 11|^ of his capital : to what does his loss amount, and how much money has he left ? $4963, loss ; $37577, left. 4. A man buys 576 sheep, at $10 a head. K his flock increases 21^ per cent, and he sells it at the same rate per head, how much money does he receive ? $7000. 5. The cost of publishing a book is 50 ct. a copy ; if the expense of sale be 10% of this, and the profit 25% : what does it sell for by the copy? 68f ct. 6. A began business with $5000: the 1st year he gained 14f %, which he added to his capital ; the 2d year he gained 8^, which he added to his capital; the 3d year he lost 12^, and quit: how much better off was he than when he started? $452.92 7. A bought a farm of government land, at $1.25 an acre; it cost him 160% to fence it, 160% to break it up, 80% for seed, 100% to plant it, 100^ to harvest it, 112% for threshing, 100^ for transportation ; each acre produced 35 bu. of wheat, which he sold at 70 ct. a bushel: how much did he gain on every acre above all expenses the first year? $13.10 8. For what must I sell a horse, that cost me $150, to gain 35% ? $202.50 9. Bought hams at 8 ct. a lb.; the wastage is 10% : how must I sell them to gain 30^ ? llf ct. a lb. 10. I started in business with $10000, and gained 20% the first year, and added it to what I had ; the 2d year I gained 20%, and added it to my capital; the 3d year I gained 20% : what had I then? $17280. PROFIT AND LOSS. 201 11. I bought a cask of brandy, containing 46 gal., at $2.50 per gal.; if 6 gal. leak out, how must I sell the rest, so as to gain 25% ? • $3.59|- per gal. CASE II. 246. Given the cost and the profit or loss, to find the rate. Problem. — ^A man bought part of a mine for $45000, and sold it for $165000 : how many per cent profit did he make? OPERATION. $165000 — $45000 = $120000 profit. 100^o=$46000; 1^^=$450; and 1 20000-f-450-=266}/c, Am. Solution. — Here the profit is $120000, which, compared with $45000, the cost, is -WW = i that is, f of 100^^ = 266f ^c- Examples for Practice. 1. If I buy at $1 and sell at $4, how many per cent do I gain? 300%. 2. If I buy at $4 and sell at $1, how many per cent do Hose? 75%. 3. If I sell f of an article for what the whole cost me, how many per cent do I gain ? 80^ . 4.* Paid $125 for a horse, and traded him for another, giving 60^ additional money. For the second horse I re- ceived a third and $25 ; I then sold the third horse for $150 : what was my per cent of profit or loss? 12^^ loss. 5.* A man bought a farm for $1635, which depreciated in value 25%. Selling out, he invested the proceeds so as to make 33^^ profit : what was his per cent of profit or loss on the entire transaction ? 202 RAY'S HIGHER ARITHMETIC. 6. It cost me $1536 to raise my wheat crop : if I sell it for $1728, what per cent profit is that per bushel? 12^%. 7. If I pay for a lb. of sugar, and get a lb. troy, what ^ do I lose, and what % does the grocer gain by the cheat? ll\% loss; 21^% gain. 8. A, having failed, pays B $1750 instead of $2500, which he owed him: what % does B lose? 30^. 9. An article has lost 20^ by wastage, and is sold for 40^ above cost: what is the gain per cent? 12%. 10. If my retail profit is 33 J^, and I sell at whole- sale for 10% less than at retail, what is my wholesale profit? 20%, 11. Bought a lot of glass; lost 15% by breakage: at what ^0 above cost must I sell the remainder, to clear 20^ on the whole ? 41^^ . 12. If a bushel of com is worth 35 ct. and makes 2^ gal. of whisky, which sells at $1.14 a gal., what is the profit of the distiller who pays a tax of 90 ct. a gallon ? ^t^% • 13. I had a horse worth $80 ; sold him for $90 ; bought him back for $100: what % profit or loss? 12 J^ loss. CASE III. 246. Given the profit or loss and the rate, to find the cost. Problem. — By selling a lot for 34f % more than I gave, my profit is $423.50: what did it cost me? Solution. — Since 34f ^ = $423.50, operation. \(fo = $423.50 -f- 34f, = $12.32 ; and 3 4 J ^^ = $ 4 2 3.5 ; 100^0, or the whole cost, = 100 times 1 ^ = $ 1 2.3 2 $12.32=11232, as in Case III of 1 OO/o =$1 2 32, ^?i8. Percentage. Bemark. — After the cost is known, the profit or loss may be added to it, or subtracted from it, to get the selling price (amount or difference). ''\ xi r( A r-: "y" ■^ OK THC PROFIT AND LOI^, UNIVERSlT^foi Examples for Practice. 1. How large sales must I make in a year, at a profit of 8^ , to clear $2000 ? $25000. 2. I lost $50 by selling sugar at 22^^ below cost : what was the cost? $222. 22| 3. If I sell tea at 13^% profit, I make 10 ct. a lb.: how much a pound did I give? 75 ct. 4. I lost a 2| dollar gold coin, which was 1\% of all I had: how much had I? $35. 5. A and B each lost $5, which was 2J^ of A's and 3^% of B's money: which had the most money, and how much ? A had $30 more than B. 6. I gained this year $2400, which is 120^ of my gain last year, and that is 44|^ of my gain the year before : what were my profits the two previous years ? $2000 last year; $4500 year before. 7. The dogs killed 40 of my sheep, which was A\% of my flock: how many had I left? 920 sheep. CASE IV. 247. Given the selling price (amount or difference) and the rate, to find the cost. Problem. — Sold goods for $25.80, by which I gained T\% • what was the cost? Solution.— The cost is 100^ ; operation. the $25.80 being l\(fo more, is 10 0^ = cost price ; 107i/o ; then, l<fo = $25.80 -^ 107 J 1 7 i /o = $ 2 5.8 ; — 24 ct., and 100^, or the cost, = .*. 1^=24 ct. 100 times 24 ct. = $24; as in 100^o=$24, Ans, Case rV of Percentage. Remabk. — After the cost is found, the difference between it and the selling price (amount or difference) will be the profit or loss. 204 RAY'S HIGHER ARITHMETia Examples for Practice. 1. Sold cloth at $3.85 a yard; my profit was 10^ : how much a yard did I pay? $3.50 2. Gold pens, sold at $5 apiece, yield a profit of 33^%: what did they cost apiece? $3.75 3. Sold out for $952.82 and lost 12% : what was the cost, and what would I have received if I had sold out at a profit of \2% ? $1082.75, and $1212.68 4. Sold my horse at 40% profit; with the proceeds I bought another, and sold him for $238, losing 20% : what did each horse cost me ? $212.50 for 1st, $297.50 for 2d. 5. Sold flour at an advance of 13^^ ; invested the proceeds in flour again, and sold this lot at a profit of 24^, realizing $3952.50: how much did each lot cost me? 1st lot, $2812.50; 2d lot, $3187.50 6. An invoice of goods purchased in New York, cost me 8% for transportation, and I sold them at a gain of 16f^ on their total cost on delivery, realizing $1260: at what were they invoiced? $1000. 7. For 6 years my property increased each year, on the previous, 100^, and became worth $100000: what was it worth at first? $1562.50 IT. STOCKS AND BONDS. DEFINITIONS. 248. 1. A Company is an association of persons united for the transaction of business. 2. A company is called a Corporation when authorized by law to transact business as one person. Corporations are regulated by general laws or special acts, called Charters. STOCKS AND BONDS. 205 3. A Charter is the law which defines the powers, rights, and legal obligations of a corporation. 4. Stock is the capital of the corporation invested in business. Those owning the stock are Stockholders. 5. Stock is divided into Shares, usually of $50 or $100 each. 6. Scrip or Certificates of Stock are the papers issued bj a corporation to the stockholders. Each stockholder is entitled to certificates showing the number of shares that he holds. 7. Stocks is a general term applied to bonds, state and national, and to the certificates of stock belonging to corporations. 8. A Bond is a written or printed obligation, under seal, securing the payment of a certain sum of money at or before a specified time. Bonds bear a fixed rate of interest, which is usually payable either annually or semi-annually. The principal classes of bonds are government, state, city, county, and railroad. 9. An Assessment is a sum of money required of the stockholders in proportion to their amounts of stock. Bemark. — Usually, in the formation of a company, the stock subscribed is not all paid for at once ; but assesamenta are made as the needs of the business require. The stock is then said to be paid for in installments. Other assessments may be made to meet losses or to extend the business. 10. A Dividend is a sum of money to be paid to the stockholders in proportion to their amounts of stock. Remark. — The gross earnings of a company are its total receipts in the transaction of the business ; the net earnings are what is left of the receipts after deducting all expenses. The dividends are paid out of the net earnings. 206 liA Y'S HIGHER ABITHMETIC. 249, Problems involving dividends and assessments give rise to four cases, solved like the four corresponding cases- of Percentage. The quantities involved are, the Stock, the Bate, and the Dividend or Assessment. The stock corresponds to the Base; the dividend or assessment, to the Percentage; the stock plus the dividend, to the Amount; and the stock minus the assessment, to the Difference, CASE I. 250. Given the stock and the rate, to find the dividend or assessment. FoBMULA. — Stock X Baie = Dividend or AsseasmenL Examples for Practice. 1. I own 18 shares, of $50 each, in the City Insurance Co., which haj3 declared a dividend of 7^% : what do I receive? $67.50 2. I own 147 shares of railroad stock ($50 each), on which I am entitled to a dividend of 5%, payable in stock: how many additional shares do I receive ? 7 shares, and $17.50 toward another share. 3. The Western Stage Co. declares a dividend of 4J per cent: if their whole stock is $150000, how much is dis- tributed to the stockholders ? $6750. 4. The Cincinnati Gas Co. declares a dividend of 18% : what do I get on 50 shares ($100 each) ? $900. 5. A railroad company, whose stock account is $4256000, declared a dividend of 3^^ : what sum was distributed among the stockholders? $148960. 6. A telegraph company, with a capital of $75000, de- clares a dividend of 7%, and has $6500 surplus: what has it earned? $11750. STOCKS AND BONDS 207 7. I own 24 shares of stock ($25 each) in a fuel company, which declares a dividend of 6% ; I take my dividend in coal, at 8 ct. a bu. : how much do I get ? 450 bu. CASE II. 251. Given the stock and dividend or assessment, to find the rate. ~, Dividerid or Assessment -, Formula. — ; = Rale. Stock Examples for Practice. 1. My dividend on 72 shares of bank stock ($50 each) is $324: what was the rate of dividend? 9^. 2. A turnpike company, whose stock is $225000, earns during the year $16384.50: what rate of dividend can it declare? 7^, and $684.50 surplus. 3. The receipts of a certain canal company, whose stock is $3650000, amount in one year to $256484; the out- lay is $79383: what rate of dividend can the company declare? ^%y and $12851 surplus. 4. I own 500 shares ($100) in a stock company. K I have to pay $250 on an assessment, what is the rate? ^%. CASE III. 252. Given the dividend or assessment and the rate, to find the stock. _ Dividend or Assessment ^i ^i Formula. = St-ock. Rate Examples for Practice. 1. An insurance company earns $18000, and declares a 15^ dividend : what is its stock account? $120000. 208 RAY'S HIGHER ARITHMETIC. 2 A man gets $94.50 as a 7^ dividend: how many shares of stock ($50 each) has he? 27 shares. 3 Received 5 shares ($50 each), and $26 of another share, as an S% dividend on stock: how many shares had I? 69 shares. CASE IV. 253. Given the rate and the stock plus the divi- dend, or the stock minus the assessment, to find the stock. Stock -f- Div idend Formulas. ~ /S^odk = J ^ 1 -f- i^ote. Stock — As sessment l — Bate. Examples for Practice. 1. Received 10^ stock dividend, and then had 102 shares ($50 each), and $15 of another share: how many shares had I before the dividend? 93 shares. 2. Having received two dividends in stock, one of 5%, another of 8^, my stock has increased to 567 shares: how many had I at first? 500 shares. III. PREMIUM AND DISCOUNT. 254. 1. Premium, Discount, and Par are mercantile terms applied to money, stocks, bonds, drafts, etc. 2. Drafts, Bills of Exchange, or Checks are written orders for the payment of money at some definite place and time. 3. The Par Value of money, stocks, drafts, etc., is the nominal value on their face. 4. The Market Value is the sum for which they sell. PREMIUM AND DISCOUNT. 209 5. Discount is the excess of the par value of money, stocks, drafts, etc., over their market value. 6. Fremiuin is the excess of their market value over their par value. 7. Bate of Premium or Bate of Discount is the rate per cent the premium or discount is of the face. 255. Problems involving premium or discount give rise to four cases, corresponding to those of Percentage. The quantities involved are the Par Valvs, the Rate, the Premium or Discount, and the Market Value, The par value corresponds to the Base; the premium or discount, to the Percentage ; and the market value, to the Amourd or Difference, CASE I. 256. Given the par value and the rate, to find the premium or discount. FOBMUIAS. — \ _ . ( Discount = Par Value X i?«te. Par Value X Rotie. Note. — If the result is a premium, it must be added to the par to get the market value ; if it is a discount, it must be mbtracted. Examples for Practice. 1. Bought 54 shares of railroad stock ($100 each) at 4% discount: find the discount and cost. Dis., «216; cost, $5184, 2. Buy 18 shares of stock ($100 each) at 8^ discount: find the discount and cost. $144, and $1656. 3. Sell the same at 4^^ premium : find the premium, the price, and the gain. $81, $1881, and $225. 4. Bought 62 shares of railroad stock ($50 each) at 28^ premium : what did they cost? $3968. H. A. 18. 210 BAY'S HIGHER ABITHMETIC. 5. What is the cost of 47 shares of railroad stock ($50 each) at 30% discount? $1645. 6. Bought $150 in gold, at f% premium: what is the premium and cost? $1-12J, and $151. 12|^ 7. Sold a draft on New "Sork of $2568.45, at ^^ prem- ium : what do I get for it ? $2581.29 8. §old $425 uncurrent money, at 3% discount: what did I get, and lose? - $412.25, and $12.75 9. What is a $5 note worth, at 6^ discount? $4.70 10. Exchanged 32 shares of bank stock ($50 each), 5^ premium, for 40 shares of railroad stock ($50 each), 10^ dis- count, and paid the difference in cash : what was it ? $120. 11. Bought 98 shares of stock ($50 each), at 15% dis- count ; gave in payment a bill of exchange on New Orleans for $4000, at f % premium, and the balance in cash : how much cash did I pay ? $140. 12. Bought 56 shares of turnpike stock ($50 each), at 69%; sold them at 76^% : what did I gain? $210. 13. Bought telegraph stock at 106%; sold it at 91%: what was my loss on 84 shares ($50 each) ? $630. 14. What is the diflerence between a draft on Philadel- phia of $8651.40, at 1 J^ premium, and one on New Orleans for the same amount, at |^% discount? $151.40 CASE II. 257. Qiven the face and the discount or premium, to find the rate. Formulas. — Bate= " Discount or Premium Par Vctlv£. Difference between Ma rket and Par Value Par Value. \ Notes. — 1. If the par and the market value are known, take their difference for the discount or premium. 2. If the rate of profit or loss is required, the market value or cost is the standard of comparison, not the faxic. PREMIUM AND DISCOUNT. 211 Examples for Pbactice. 1. Paid $2401.30 for a draft of $2360 on New York: what was the rate of premium? li%« 2. Bought 112 shares of raibroad stock ($50 each) for $3640 : what was the rate of discount ? 35^ . 3. If the stock in the last example yields 8^ dividend, what is my rate of profit? 12j\%. 4. I sell the same stock for $5936 : what rate of premium is that? what rate of profit? 6%; 63^^. 5. If I count my dividend as part of the profit, what is my rate of profit ? <^^ A^ • 6. Exchanged 12 Ohio bonds ($1000 each), 1% premium, for 280 shares of railroad stock ($50 each) : what rate . of discount were the latter? 8^^. 7. Gave $266. 66| of notes, A% discount, for $250 of gold : what rate of premium was the gold? 2-|%. 8. Bought 58 shares of mining stock ($50 each), at 40^ premium, and gave in payment a draft on Boston for $4000 : what rate of premium was the draft? li%- 9. Keceived $4.60 for an uncurrent $5 note: what was the rate of discount? 8^. 10. Paid $2508.03 for 26 shares of stock ($100 each), and brokerage, $25.03: what is the rate of discount? ^\%- CASE III. * 258. Given the discount or premium and the rate, to find the face. .Discount or Pi'emium FoRMUiA. — Par Value = Bate. Notes. — 1. After the face is obtained, add to it the premium, or subtract the discount, to get the market value or cost. 2. If the profit or loss is given, and the rate per cent of the face corresponding to it, work by Case III, Percentage. 212 BA Y'S HIOHEB ARITHMETia Examples fOr Practice. 1. Paid 36 ct. premium for gold \% above par: how much gold was there? $48. 2. Took stock at par ; sold it for 2\fo discount, and lost $117: how many shares ($50 each) had I? 104 shares. 3. The discount, at 7^^, on stocks, was $93.75: how many shares ($50 each) were sold ? 25 shares. 4. Buy stock at 4^% premium; sell at 8J^ premium; profit, $345 : how many shares ($100 each) ? 92 shares. 5. Buy stocks at 14^ discount; sell at 3^^ premium; profit, $192.50: how many shares ($50 each)? 22 shares. 6. The premium on a draft, at |^^, was $10.36: what was the face? $1184. 7. Buy stocks at 6% discount ; sell at 42^ discount ; loss, $666 : how many shares ($50 each) ? 37 shares. 8. Bought stock at 10% discount, which rose to 5^ premium, and sold for cash; paying a debt of $33, I invested the balance in stock at 2% premium," which, at par, left me $11 less than at first: how much money had I at first? $148.50 CASE IV. 259. Given the market value and the rate, to find the par value. Market Value FOBMUI.AS.— Par Value =^ J 1 + ^'« «/ Premium. ' Market Value 1 — Rate of Discount. Notes. — 1. After the face is known, take the difference between it and the market value, to find the discount or premium. 2. Bear in mind that the rate of premium or discount, and the rate of profit or loss, are entirely different things; the former is referred to the par value or face, as a standard of comparison, the latter to the market value or cost. COMMISSION AND BROKERAGE. 213 Examples for Practice. 1. What is the face of a draft on Baltimore costing $2861.45, at \\<fo premium? $2819.16 2. Invested $1591 in stocks, at 2^6^ discount ; how many shares ($50 each) did I buy ? 43 shares. 3. Bought a draft on New Orleans, &t i% discount, for $6398.30: what was its face? $6430.45 4. Notes at 65% discount, 2% brokerage, cost $881.79: what is their &ce? $2470. 5. Exchanged 17 railroad bonds ($500 each) 25% below par, for bank stock at 6:^% premium: how many shares ($100 each) did I get? 60 shares. 6. How much gold, at f % premium, will pay a check for $7567 ? $7520. 7. How much silver, at 1\% discount, can be bought for $3172.64 of currency ? $3212.80 8. How large a draft, at \^ premium, is worth 54 city bonds ($100 each), at 12% discount? $4740.15 9. Exchanged 72 Ohio State bonds ($1000 each), at 6\fo premium, for Indiana bonds ($500 each), at 2% premium : how many of the latter did I get ? 150 bonds. IV. COMMISSION AND BROKERAGE. DEFINITIONS. 260. 1. A Commission-Merchant, Agent, or Factor is a person who sells property, makes investments, collects debts, or transacts other business for another. 2. The Principal is the person for whom the commission- merchant transacts the business. 8. Commission is the percentage paid to the commission- merchant for doing the business. 214 RAY'S mOHER ARITHMETIC. 4. A Consignment is a quantity of merchandise sent to a commission merchant to be sold. 5. The person sending the merchandise is the Consignor or Shipper, and the commission merchant is the Consignee. When living at a distance from his principal, the consignee is spoken of as the Correspondent. 6. The Net Proceeds is the sum left after all charges have been paid. 7. A Guaranty is a promise to answer for the payment of some debt, or the performance of some duty in the case of the failure of another person, who, in the first instance, is liable. Guaranties are of two kinds : of payment^ and of collection. 8. In a Guaranty of Payment, the guarantor makes an absolute agreement that the instrument shall be paid at maturity. 9. The usual form of a guaranty, written on the back of a note or bill, is: *^For value received^ I her dry guaranty the payment of the vrithin. John Saunders." 10. A Broker is a person who deals in money, bills of credit, stocks, or real estate, etc. 11. The commission paid to a broker is called Brokerage. 261. Commission and Brokerage involve four cases, corresponding to those of Percentage. The quantities involved are the Arrwurd Bought or SM^ the Rate of Gommimon or Brokerage^ the Commission or Bro- kerage, and the Cost or Net Proceeds. The amount of sale or purchase corresponds to the Base; the commission or brokerage, to the Percentage; the cost, to the Amount; and the net proceeds, to the Difference. COMMISSION AND BROKERAOE. 215 CASE I. 262. Given the amount of sale, purchase, or col- lection and the rate, to find the commission. FoBMUiiA. — Amount of Sale or Purchase X R<^ = Commission, Examples for Practice. 1. I collect for A $268.40, and have 5% commission: what does A get? $254.98 2. I sell for B 650 barrels of flour, at $7.50 a barrel, 28 barrels of whisky, 35 gal. each, at $1.25 a gal.: what is my commission, at 2^%? $137.25 3. Received on commission 25 hhd. sugar (36547 lb.), of which I sold 10 hhd. (16875 lb.), at 6 ct. a lb., and 6 hhd. (8246 lb.) at 5 ct. a lb., and the rest at 5^ ct. a lb.: what is my commission, at 3%? $61.60 4. A lawyer charged 8^ for collecting a note of $648.75: what are his fee and the net proceeds? $51.90, and $596.85 5. A lawyer, having a debt of $1346.50 to collect, com- promises by taking 80%, and charges 5% for his fee: what are his fee, and the net proceeds? $53.86, and $1023.34 6. Bought for C, a carriage for $950, a pair of horses for $575, and harness for $120; paid charges for keeping, pack- ing, shipping, etc., $18.25; freight, $36.50: what was my commission, at 3^^ , and what was the whole amount of my bill? $54.83, and $1754.58 7. An architect charges 3^^ for designing and superin- tending a building, which cost $27814.60: to what does his fee amount? $973.51 8. A factor has 2|^ commission, and 3^% for guar- antying payment: if the sales are. $6231.25, what does he get? $389.45 216 RAY'S HIGHER ARITHMETIC, 9. Sold 500000 lb. of pork at 5^ ct. a lb.: what is my commission at 1 J% ? $343.75 10. An architect charges \\% for plans and specifications, and 2J^ for superintending: what does he make, if the building costs $14902.50? ' $614.73 11. A sells a house and lot for me at $3850, and charges \% brokerage: what is his fee? $24.06+ 12. I have a lot of tobacco on commission, and sell it through a broker for $4642.85: my commission is 2^%, the brokerage IJ^^: what do I pay the broker, and what do I keep? I keep $63.84; brokerage, $52.23 CASE II. 263. Given the commission and the amount of the sale, purchase, or collection, to find the rate. _ Commission _ FoBMULA.— —IT-, ;; — ; — = -Bo^e. Amount of oale or riirchase Examples for Practice. 1. An auctioneer's commission for selling a lot was $50, and the sum paid the owner was $1200 : what was the rate of commission ? ^ 4%. 2. A commission-merchant sells 800 barrels of flour, at $6.43f a barrel, and remits the net proceeds, $5021.25: what is his rate of commission? ^i%' 3. The cost of a building was^l9017.92, including the architect's commission, which was $553.92: what rate did the architect charge? S%. 4. Bought flour for A; my whole bill was $5802.57, in- cluding charges, $76.85, and commission, $148.72: find the rate of commission. 2f %. 5. Charged $52.50 for collecting a debt of $1050: wha( was my rate of commission ? 5%^ COMMISSION AND BROKERAOK 217 6. An agent gets $169.20 for selling property for 88460: what was his rate of brokerage? 2%. 7. My commission for selling books was $6.92, and the net proceeds, $62.28: what rate did I charge? 10^. 8. Paid $38.40 for selling goods worth $6400: what was the rate of brokerage? |%. 9. Paid a broker $24.16, and retained as my part of the commission $42.28, for selling a consignment at $2416: what was the rate of brokerage, and my rate of commission? Brok. 1%; com. 2f %• CASE III. 264. Given the commission and rate, to find the sum on which commission is charged. FoRBiuui. = AmourU of Sale or Purchase. Rale Note. — After finding the sum on which commission is charged, subtract the commission to find the net proceeds, or add it to find the whole cost, as the case may be. Examples for Practice. 1. My commissions in 1 year, at 2^%, are $3500: what were the sales, and the whole net proceeds? $140000, and $136500. 2. An insurance agent's income is $1733.45, being 10^ on the sums received for the company: what were the com- pan/s net receipts? $15601.05 3. A packing-house charged 1^% commission, and cleared 52376.15, after paying out $1206.75 for all expenses of packing : how many pounds of pork were packed, if it cost 4\ ct. a pound ? 5308000 lb. 4. Paid $64.05 for selling coffee, which was |^^ broker- age: what are the net proceeds? $7255.95 H. A. 19. 218 BAY'S mOHER ARITHMETIC. 5. An agent purchased, according to order, 10400 bushels of wheat ; his commission, at l^'^o > was $156, and charges for storage, shipping, and freight, $527.10: what did he pay a bushel, and what was the whole cost? $1.20 a bu., and $13163.10, whole cost. 6. Received produce on commission, at 2^%; my surplus commission, after paying |^ brokerage, is $107.03: what was the amount of the sale, the brokerage, and'^net pro- ceeds? Sale, $6116; brok., $30.58; pro., $5978.39 CASE IV. 265.* Given the rate of commission and the net proceeds or the whole cost, to find the sum on which commission is charged. FOBMULAS. — : Amovmi of Sale or Purchase. (1 + Bate) I Net Proceeds . ^ _ , « , 7i dIZT ^^ Amount of Sale or Purchase. Bale) Note. — After the sum on which commission is charged is known, find the commission by subtraction. Examples for Praotice. 1. A lawyer collects a debt for a client, takes 4^ for his fee, and remits the balance, $207.60: what was the debt and the fee? $216.25, and $8.65 2. Sent $1000 to buy a carriage, commission 2^% : what must the carriage cost? $975.61 Suggestion.— lOO^c + 2\fo — 102J^ = $1000 ; find Ifoy then 100^. 3. A buys per order a lot of coffee; charges, $56.85; commission, 1\%; the whole cost is $539.61: what did the coffee cost? $476.80 COMMISSION AND BROKERAGE. 219 4. Buy sugar at 2J^ commission, arid 2^% for guaran- teeing payment: if the whole cost is $1500, what was the cost of the sugar? $1431.98 5. Sold 2000 hams (20672 lb.); commission, 2^^, guar- anty, 2|^, net proceeds due consignor, $2448.34: what did the hams sell for a lb.? 12^ ct. 6. Sold cotton on commission, at 5^; invested the net proceeds in sugar; commission, 2%; my whole commission was $210: what was the value of the cotton and sugar? Cotton, $3060; sugar, $2850. Suggestion. — Note carefully the different processes required here for commission in buying and commission in selling. 7. Sold flour at 3^% commission; invested f of its value in coffee, at 1|^^ commission; remitted the balance, $432.50 : what was the value of the flour, the coffee, and my commissions ? Flour, $1500; coffee, $1000, 1st com., $52.50, 2d com., $15. 8. Sold a consignment of pork, and invested the proceeds in brandy, after deducting my commissions, 4% for selling, and l\^o for buying. The brandy cost $2304.00 : what did the pork sell for, and what were my commissions ? Pork, $2430; 1st com., $97.20; 2d com., $28.80 9. Sold 1400 barrels of flour, at $6.20 a barrel; invested the proceeds in sugar, as per order, reserving my commis- sions, A% for selling and 1|^ for buying, and the ex- pense of shipping, $34.16: how much did I invest in sugar? $8176. 10. An agent sold my com; and, after reserving his com- mission, invested all the proceeds in corn at the same price ; his commission, buying and selling, was 3%, and his whole charge $12 : for what was the corn first sold ? $206. 11. My agent sold my flour at 4% commission ; increas- ing the proceeds by $4.20, I ordered the purchase of wheat at 2^ commission; after which, wheat declining 3J^, my whole loss was $5: what was the flour worth? $53. 220 BAY'S HIGHER ARITHMETIC. V. STOCK INVESTMENTS. DEFINITIONS. 266. 1. A Stock Exchange is an association of brokers and dealers in stocks, bonds, and other securities. Remarks. — 1. The name "Stock Exchange" is also applied to the building in which the association meets to transact business. 2. New York city is the commercial center of the United States, and the transactions of the New York Stock Exchange, as tele- graphed throughout the country, determine the market value of nearly all stocks sold. 2. United States Government Bonds are of two kinds, — coupon and registered. Kemarks. — 1. Coupon bonds may be transferred like bank-notes : the interest is represented by certificates, called coupons, printed at the bottom of the bond, which may be presented for payment when due. 2. Regialei'ed bonds are recorded in the name of the owner in the U. S. Treasurer*s office, and the interest is sent directly to the owner. Registered bonds must be indorsed, and the record must be changed, to effect a transfer. 3. The United States also issues legal tender notes, known as " Greenbacks," which are payable in coin on demand, and bear no interest. 3. The various kinds of United States bonds are dis- tinguished, 1st. By the rate of interest; 2d. By the date at which they are payable. Most of the bonds are payable in coin ; a few are payable in currency. Example. — Thus, "U. S. 4J*s, 1891," means bonds bearing in- terest at ^fc1 and payable in 1891. " U. S. cur. 6's, 1899," means bonds bearing interest at 6^, and payable in currency in 1899. Quotations of the principal bonds are given in the leading daily papers. 4. Bonds are also issued by the several states, by cities and towns, by counties, and by corporations. STOCK INVESTMENTS. 221 Bemarks. — Legitimate stock transactions involve the following terms and abbreviations, which need explanation : 1. A person who anticipates a decline, and contracts to deliver stocks at a future day, at a fixed price which is lower than the present market price, expecting to buy in the interval at a still lower price, is said to sell short, Sliort sales are also made for cash, deliverable on the same day, or in the regvJar way, where the certificates are delivered the day after the sale. In these cases the seller borrows the stock from a third party, advancing security equivalent to the market price, and waits a decline; he buys at what he considers the lowest point, returns his borrowed stock, and reclaims his security. 2. A person who buys stock in anticipation of a rise is said to buy long. 3. Those who sell short are interested, of course, in forcing the market price down, and are called hears; while those who buy long endeavor to force the market price up, and are known as htdh. 4. The following are the principal abbreviations met with in stock quotations: c, means coupon; r., registered; prefd, or pf., pre- Jerredf applied to stock which has advantages over common stock of the same company in the way of dividends, etc.; xd., wiihovi dividendj meaning that the buyer is not entitled to the dividend about to be declared ; c, cash; s3, sSO, s60, seller's option., three, thirty, or sixty days, as the case may be, means that the seller has the privi- lege of closing the transaction at any time within the specified limit; bS, b30, b60, buyer'' s option, three ctoys, etc., giving the buyer the privilege; be., between calls, means that the price was fixed between the calls of the whole list of stocks, which takes place in the New York Exchange twice a day ; opg., for delivei-y at the opening of the books of transfer'. 6. The usual rate of brokerage is ifo on the par value of the stock, either for a purchase or a sale. 267. The quantities involved in problems in stock in- vestments are: the Amount Invested, the Market Value, the Rate of Dividend or Interest paid on the par value of the stock, the Rate of Inemne on the amount invested, and the Income itself. These quantities give rise to five eases, all of which may be solved by the principles of Percentage. .222 MAY'S HIGHER ARITHMETia NOTATION. 268. The following notation may' be adopted to ad- vantage in the formulas : Amount Invested = A. I. Market Value = M. V. Rate of Dividend or Interest = R D. Rate of Income = R. I. Income = I. CASE I. 269. Given the amount invested, the market value, and the rate of dividend or interest, to find the in- come. A. I. Formula.— -rrr X ^- D. = ^ M. V. Problem. — A person invests $5652.50 in Mutual In- surance Company stock at 95 cents : what will be his income if .the stock pay 10% dividend annually? OPERATION. $5652.50-^.95=$59 5 0^ par value of stock purchased. $5950X.1=$595= income. Or, 9 5 ^c = J? of the par value = $5652.50 h<fo = ^ " •* " " =$ 297.50 10% = A " " " " =$ 595, ^ns. Solution. — As many dollars* worth of stock can be bought as $.95 is contained times in $5652.50, which iri 5950 times. Therefore $5950 is the par value of the stock purchased ; and 10^ dividend on $5950 is $595, the income on the investment. Examples for Practice. 1. A invests $28000 in Lake Shore Kailroad stock, at 70^. If the stock yields 8^ annually, what is the amount of his income? $5200. STOCK INVESTMENTS, 223 2. B invests $100962 in U. S. cur. 6's, 1899, at 106^%. If gold is at I premium, what does the government save by paying him his interest in greenbacks? $7.11 3. If I invest $10200 in Tennessee 6's, new, at 30^, what is my annual income ? $2040. 4. A broker invested $36000 in quicksilver preferred stock, at 40^ : if the stock pays 4^, what is the income derived? $3600. 5. Which is the better investment, stock paying 6^^ dividend, at a market value of 106^%, or stock paying 4^% dividend, at 104^% ? The former, lffff|^%. 6. Which is the more profitable, to invest $10000 in 6% stock purchased at 75%, or in 5^ stock purchased at 60%, allowing brokerage ^%? 5^ stock is $31.74+ better. CASE II. 270. Given the amount invested, the market valuer and the income, to find the rate of dividend or in- terest. A. I. Formula. — I. -^ ^/ ' = R. D. M. V. Problem. — Invested $10132.50 in railroad stock at 105%, which pays me annually $965 : what is the rate of dividend on the stock? Solution.— By Art. 259, $10132.50 -^ 1.05 = $9650 = par value of the stock ; and by Art. 251, $965 -4- $9650 = .1, or 10^, Am, Examples for Practice. 1. A has a farm, valued at $46000, which pays him 5% on the investment. Through a broker, who charges $56.50 for his services, he exchanges it for insurance stock at 9^ premium, and this increases his annual income by $1072: what dividend does the stock pay? 8%. 224 RAY'S HIGHER ARTTHMETTC. 2. What dividend must stock pay, in order that my rate of income on an investment of $64968.75 shall be 4^%, provided the stock can be bought at 103^% ? 4\%, 3. My investment was $9850, my income is $500, and the market value of the stock 108^^%, brokerage \% : what is the rate of dividend ? 5^^ . t 4. The sale of my farm cost me $500, but I gave the proceeds to a broker, allowing him ^%, to purchase railroad stock then in market at 102% ; the farm paid a 5^ income, equal to $2075, but the stock will pay $2025 more : what is the rate of dividend? * 10^%. 5. Howard has at order $122400, and can allow broker- age ^%,and buy insurance stock at 101|^, yielding 4^%; but if he send to the broker $100 more for investment, and buy rolling-mill stock at 103^^ , the income will only be half 80 large : what rate does the higher stock pay ? 2^^ . CASE III. 271. Oiven the income* rate of dividend, and mar- ket value, to find the amount invested. FORMUIiA. — =r-^ X M. V. — A. I. Problem. — If U. S. bonds,. pa3dng 5% interest, are sell- ing at 108^%, how much must be invested to secure an annual income of $2000 ? OPERATION. $2000^$. 05 = 40000; $40000 X 1.085 = $43400, Ans. Solutions. — 1. To produce an income of $2000 it will require as many dollar's worth of stock at par as 5 ct. is contained times in $2000, which is 40000 ; and, at lOSJ^c, it will require $40000 X 1.08^ = $43400. 2. $1 of stock will give 5 cents income, and $2000 income will require $40000 worth of stock at par. $40000 of stock, at lOSJ^, will cost $40000 X 1.085 = $43400. STOCK INVESTMENTS. 225 Examples for Practice. 1. What amount is invested by A, whose canal stock, yielding 4^ , brings an income of $300, but sells in market for n% ? «6900. 2. If I invest all my money in 5^ furnace stock, salable at 75%, my income will be $180: how much must I borrow to make an investment in 6^ state stock, selling at 102^ , to have that income? $360. 3. If railroad stock be yielding 6^, and is 20% below par, how much would have to be invested to bring an income of $390? $5200. 4. A banker owns 2^% stocks, at 10% below par, and 3% stocks, at 15% below par. The income from the former is 66-|^ more than from the ^atter, and the investment in the latter is $11400 less thi»,ii in the former: required the whole investment Mid income. $31800, and $960. 5. Howard M. Holden sold $21600 U. S. 4's, 1907, reg- istered, at 99f^, and immediately invested a sufficient amount of the proceeds in Illinois Central Railroad stock, at 80%, which pays an annual dividend of 6^; he receives $840 from the railroad investment ; with the remainder of his money he bought a farm at $30 an acre: required the amount invested in railroad stock, and the number of acres in the farm? $11200 in R. R. stock; 342^ acres. 6. W. T. Baird, through his broker, invested a certain sum of money in Philadelphia 6's, at 115i^, and three times as much in Union Pacific 7's, at 89|%, brokerage ^ % in both cases : how much was invested in each kind of stock if his annual income is $9920? $34800 in Phila. 6'8 ; $104400 in U. P. 7's. 7. Thomas Reed, bought 6% mining stock at 114^^, and 4^ furnace stock at 112% , brokerage ^^; the latter cost him $430 more than the former, but yielded the same income: what did each cost him? Mining, $920; furnace, $1350. 226 HA Y' S HIQHEB ARITHMETIC, CASE IV. 272. Oiyen the market value and the rate of divi- dend or interest, to find the rate of income. R. D. „ ^ Formula. — rr^7= R. I. M. V. Problem. — What per cent of his money will a man realize in bu3dng ^% stock at 80%? OPERATION. $.0 6-^-$.8 = /^ = 5% = 7J/c. Solution. — ^The expenditure of 80 ct. buys a dollar's worth of Btock, giving an income of 6 ct. The rate per cent of income on investment is $.06 -4- $.80 = .07 J, or 7J^o. Examples for Practice. 1. What is the rate of income on Pacific Mail 6's, bought at 30^? 20%. 2. What is the rate of income on Union Pacific 6's, bought at 110%? ^-hfor 3. Which is the better investment: U. S. new 4's, regis- tered, at 99f ^ ; or U. S. new 4^'s, coupons, at 106^ ? The latter is ^^f, better. 4. Thomas Sparkler has an opportunity of investing $30000 in North-western preferred stock, at 76%, which pays an annual premium of 5^; in Panama stock, at 125^, which pays a premium annually of 8^%; or he can lend his money, on safe security, at 6^% per annum. Prove which is the best investment for Mr. Sparkler. The Panama stock. * 5. Thomas Jackson bought 500 shares of Adams Express stock, at 105^^, and paid \\fo brokerage : what is the rate of income on his investment per annum if the annual divi- dend is 8%? 7^^%. STOCK INVESTMENTS. 227 CASE V. 273. Giyen the rate of income and the rate of dividend or interest, to find the market value. Formula. — ' = M. V. XV. I. Problem.— What must I pay for Lake Shore 6's ($100 a share), that the investment may yield 10^ ? OPERATION. $.06-j-.l=3% = 60^o = $60 ashare. Solution.— If bought for $100, or par, it will yield 6^ ; to yield \(fo it must be bought for $100 X 6 = $600 ; to yield 10^ it must be bought for T^ of $600 = $60. Examples for Practice. 1. What must I pay for Chicago, Burlington & Quincy Railroad stock that bears 6^, that my annual income on the investment may yield 5^ ? 120^. 2. Which is the best permanent investment : 4's at 70^ , 5's at 80%, 6's at 90%, or lO's at 120^ ? Why? 3. *rhe rate of income being 7% on the investment, and the dividend rate 4%, what is the market value of $3430 of the stock? $1960. 4. In a mutual insurance company one capitalist has an investment paying 8^: what is the premium on the stock, the dividend being 9^ ? 12|^. 5. Suppose 10% state stock 20% better in market than 4% railroad stock; if A's income be $500 from each, how much money has he 'paid for each, the whole investment bringing 6^% ? $11250, railroad ; $5400, state. 6. At what figure must be government 5 per cent's to make my purchase pay 9%? 55f^. 228 BA Y' S. HIOHER ARITHMETIC. VL INSURANCE. DEFINITIONS. 274. 1. Insurance is indemnity against loss or damage. 2. There are two kinds of insurance, viz.: Property In- surance and Personal Insurance. 3. Under Property Insurance the two most important divisions are : Fire Insurance and Marine Insurance. 4. Fire Insurance is indemnity against loss by fire. 5. Marine Insurance is indemnity against the dangers of navigation. Notes. — 1. Transit Insurance is applied to risks which are taken when property is transferred by railroad, or by railroad and water routes combined. 2. There are several minor forms of property insurance, also, such as Live Stock Insurance, Steam Boiler Insurance, Plate Glass In- surance, etc., the special purposes of which are indicated by their names. 6. Personal Insurance is of three kinds, viz.: Life In- surance. Accident Insurance, and Health Insurance. Personal insurance will be discussed in another chapter. 7. The Insurer or Underwriter is the party or company that undertakes to pay in case of loss. 8. The Risk is the particular danger against which the insurer undertakes. 9. The Insured is the party protected against loss. 10. A Contract is an agreement between two or more competent parties, based on a sufficient consideration, each promising to do or not to do some particular thing possi- ble to be done, which thing is not enjoined nor prohibited by law. INSURANCE. 229 11. The Primary Elements of a contract are: the Parties, the Consideration^ the Subject Matter, the Consent of the Parties, and the Tbne. 12. The written contract between the two parties in in- surance is called a Policy. 13. The Premium is the sum paid for insurance. It is a certain per cent of the amount insured. 14. The Rate varies with the nature of the risk. 15. The Amount or Valuation is the sum for which the premium is paid. Notes. — 1. Whoever owns or has an interest in property, may insure it to the full amount of his interest or liability. 2. Only the (ictual loss can be recovered by the insured, whether there be one or several insurers. 3. Usually property is insured for about two thirds of its value. 16. Insurance business is usually transacted by incorpor- ated companies. 17. These companies are either joint-stock companies, or mutual companies. Remakks. — 1. In joint-stock companies the capital is owned by individuals who are the stockholders. They share the profits and losses. 2. In mutual companies the profits and losses are divided among the insured. 276. The operations in insurance are included under the principles of Percentage. The quantities involved are, the Amount insured, the Per Cent of premium, and the Premium, The amount corresponds to the Base, and the premium, to the Percentage. CASE I. 276. Given the rate of insurance and the amoun insured, to find the premium. FomnTLA.— Amount Insured X -Ka^ = Premium, 1^ 230 J^A Y'S HIQHER ARITHMETIC. Examples for Practice. 1. Insured f of a vessel worth $24000, and f of its cargo worth $36000, the former at 2^%, the latter at 1J%: what is the premium ? $607.50 2. Insured a house for $2500, and furniture for $600, at 3^^: what is the premium ? $18.60 3. What is the premium on a cargo of railroad iron worth $28000, at 1|%? $490. 4. Insured goods invoiced at $32760, for three months, at ^%: what is the premium? $262.08 5. My house is permanently insured for $1800, by a de- posit of 10 annual premiums, the rate per year being |^: how much did I deposit, and if, on terminating the insur- ance, I receive my deposit less 5%, how much do I get? $135 deposited; $128.25 received. 6. A shipment^ of pork, costing $1275, is insured at 4^, the policy costing 75 cents: what does the insurance cost? $7.83 7. An insurance company having a risk of $25000, at •j^^, re-insured $10000, at |^, with another office, and $5000, at 1^0 y with another: how much premium did it clear above what it paid ? $95. CASE II. 277. Oiyen the amount insured and premium, to find the rate of insurance. -^ Premium r, . Formula. ^ = Raie, Amount Insured Examples for Practice. 1. Paid $19.20 for insuring | of a house, worth $4800 what was the rate? f^ INSURANCE. 231 2. Paid $234, including cost of policy, $1.50, for insuring a cargo worth $18600: what was the rate? li%' 3. Bought books in England for $2468 ; insured them for the voyage for $46.92, including the cost of the policy, $2.50: what was the rate? ^\%- 4. A vessel is insured for $42000; $18000 at 2|^, $15000 at 3f%, and the rest at 4|%: what is the rate on the whole $42000 ? ^%. 5. I took a risk of $45000; re-insured at the same rate, $10000 each, in three offices, and $5000 in another; my share of the premium was $262.50: what was the rate? 2|%. 6. I took a risk at l^fo ; re-insured | of it at 2^ , and \ of it at 2^% : what rate of insurance do I get on what is left? ^%. CASE III. 278. Oiven the premium and rate of insurance, to find the amount insured. _, PreTniwm, i . r ■» Formula. — = Amou7U Insureds Bate, Examples FOR Practice. 1. Paid $118 for insuring, at f^ : what was the amount insured ? $14750. 2. Paid $411.37^ for insuring goods, at 1^^ : what was their value? $27425. 3. Paid $42.30 for insuring f of my house, at ^^% : what is the house worth ? $7520. 4. Took a risk at 2\% ; re-insured f of it at 2^%; my share of the premium was $197.13: how large was the risk ? $26284. 5. Took a risk at If %; re-insured half of it at the same rate, and ^ of it at 1^^ ; my share of the premium was $58.11 : how large was the risk? $19370. 232 BA Y'S HIQHER ARITHMETIC, 6. Took a risk at 2%; re-insured $10000 of it at 1\%, and $8000 at ^\%\ niy share of the premium was $207.50: what sum was insured ? $28000. 7. The Mutual Fire Insurance Company insured a build- ing and its stock for |- of its value, charging If^. The Union Insurance Company relieved them of \ of the risk, at 1^%. The building and stock being destroyed by fire, the Union lost forty-nine thousand dollars less than the Mutual: what amount of money did the owners of the building and stock lose? $51750. VII. TAXES. DEFINITIONS. 270. 1. A Tax is a sum of money levied on persons, or on persons and property, for public use. 2. Taxes in this country are: (1.) /Stofe and Local Taxes; (2.) Taoces for the National GovemTnent 3. Taxes are further classified as Direct and Indirect. 4. A Direct Tax is one which is levied on the person or property of the individual. 5. An Indirect Tax is a tax levied on articles of con- sumption, for which each person pays in proportion to the quantity or number of such articles consumed. 6. A PoU-Tax, or Capitation Tax, is a direct tax levied on each male citizen liable to taxation. 7. A Property Tax is a direct tax levied on property. Note. — In legal works properly is treated of under two heads ; viz , Real Property^ or Real Estate^ including houses and lands ; and Personal Property, including money, bonds, cattle, horses, furniture, — in short, all kinds of movable property.. TAXES. 233 8. State and Local Taxes are generally direct^ while the United States Taxes are indirect, 9. An Assessor is a public officer elected or appointed to prepare the Assessment RoU. 10. An Assessment Boll is a list of the names of the taxable inhabitants living in the district assessed, and the valuation of each one's property. 11. The Collector is the public officer who receives the taxes. Note. — In some states all the taxes are collected by the counties ; in others, the towns collect; while in others the collections are made by separate collectors. Generally a number of different taxes are aggregated, — such as state, county, road, school, etc. 280. The quantities involved in problems under tax- ation are, the Assessed Value of the property, the Rate of Taocatioriy the Taa, and the Amount Left after taxation. They require the application of the four cases of Percent- age, the assessed value corresponding to the Base; the tax, to the Percentage; and the amount left after taxation, to the Difference, CASE I. 281. Given the taxable property and the rate, to find the property-tax. Formula. — Taxable Property X -Ka^ = Amount of Tax, Note.— If there be a poll-tax, the sum produced by it should be added to the property-tax, to give the whole tax. Examples for Practice. 1. The taxable property of a county is $486250, and the rate of taxation is 78 ct. on $100 ; that is, ^% : what is the tax to be raised ? $3792.75 H. A. 20. 234 BAY'S HIQHEJR ABITHMETIC, Kemark. — The rate of taxation being usually small, is expressed most conveniently as so many cents on $100, or as so many mills on $1. 2. A's property is assessed at $3800 ; the rate of taxation is 96 ct. on $100 {-^^fo) - what is his whole tax, if he pays a poll-tax of $1 ? $37.48 Remarks.^— 1. In making out bills for taxes, a table is used, con- taining the units, tens, hundreds, thousands, etc., of property, with the corresponding tax opposite each. 2. To find the tax on any sum by the table, take out the tax on each figure of the sum, and add the results. In this table, the rate is l\<fo, or 125 ct on $100. Tax Table. Prop. Tax. Prop. Tax. Prop. Tax. Prop. Tax. Prop. Tax. $1 .0125 $10 .125 $100 $1.25 $1000 $12.50 $10000 $125 2 .025 20 .25 200 2.50 2000 25. 20000 250 3 .0375 30 .375 300 3.75 3000 37.50 30000 375 4 .05 40 .50 400 5. 4000 50. 40000 500 5 .0625 50 .625 500 6.25 5000 62.50 50000 625 6 .075 60 .75 600 7.50 6000 75. 60000 750 7 .0875 70 .875 700 8.75 7000 87.50 70000 875 8 .10 80 1. 800 10. 8000 100. 80000 1000 9 .1125 90 1.125 900 11.25 9000 112.50 90000 1125 3. What will be the tax by the table, on property assessed at $25349 ? Solution.— The tax for $20000 is $250; for $5000 is 62.50; for $300 is $3.75; for $40 is .50; for $9 is .1125; which, added, give $316.86, the tax on $25349. 4. Find the tax for $6815.30 $85.19 5. What is the tax on property assessed at $10424.50, and two polls, at $1.50 each? $133.31 6. A's property is assessed at $251350, and B's at $25135. What is the difference in their taxes? $2827.69— TA2LES. 235 CASE II. 282. Given the taxable property and the tax, to find the rate. Tax ^ FoEMUIiA.— 7= 77— =r = Bxiie. laxable Froperly Examples for Practice. 1. Property assessed at $2604, pays 819.53 tax : what is the rate of taxation ? |% — 75 ct. on $100. 2. The taxable property in a town of 1742 polls, is $6814320; a tax of $66913 is proposed: if a poll-tax of $1.25 is levied, what should be the rate of taxation? 3-95^% = 95 ct. on $100. 3. An estate of $350000 pays a tax of $5670 : what is the rate of taxation ? lf^% = $1.62 on $100. 4. A's tax is $50.46 ; he pays a poll-tax of $1.50, and owns $8704 taxable property ; what is the rate of tax- ation? -^fo = 56i ct. on $100. CASE III. 283. Given the tax and the rate, to find the assessed value of the property. Formula. — - — = Taxable Pi-operty. Bdte Note.— If any part of the tax arises from polls, it should be first deducted from the given tax. Examples for Practice. 1. What is the assessed value of property taxed $66.96, at If % ? $3720. 236 RAY'S HIGHER ARITHMETIC. 2. A corporation pays $564.42 tax, at the rate of x*oV^, or 46 ct. on $100 : find its capital. «122700. 3. A is taxed $71.61 more than B ; the rate is 1^%, or $1.32 on $100: how much is A assessed more than B? $5425. 4. A tax of $4000 is raised in a town containing 1024 polls, by a poll-tax of $1, and a property-tax of -^^^0 (24 ct. on $100) : what is the value of the taxable property in it ? $1240000. 5. A's income is 16% of his capital; he is taxed 2^% of his income,. and pays $26.04: what is his capital ? $6510. CASE IV. 284. Given the amount left after payment of tax and the rate, to find the assessed value of the prop- erty. T, AmouTvt Left after Payment m n tl FoBMUiiA. — = Taxable Property, 1 — Bate (fo. Examples for Practice. 1. A pays a tax of l^V^ ($1.35 on $100) on his capital, and has left $125127.66: what was his capital, and his tax? Capital, $126840; tax, $1712.34 2. Sold a lot for $7599, which covered its cost and 2^ beside, paid for tax : what was the cost ? $7450. VIII. UNITED STATES REVENUE. DEFINITIONS. 285. 1. The Ee venue of the United States arises from the Internal Revenue, from the OustomSy and from Sales of PiMic Lands. UNITED STATES BEVENUE. 237 2. The Internal Revenue is derived from taxes on spirits, tobacco, fermented liquors, banks, and from the sale of stamps, etc. 3. The Customs or Duties are taxes imposed by Gov- ernment on imported goods. They are of two kinds, — ad valorem and specific, 4. Ad Valorem Duties are levied at a certain per cent on the cost of the goods as shown by the invoice. 5. Specific Duties are certain sums collected on each gallon, bushel, yard, ton, or pound, whatever may be the cost of the article. Note. — Problems involving specific duty only are not solved, of course, by the principles of Percentage. 6. An Invoice is a detailed statement of the quantities and prices of goods purchased. 7. A Tariff is a schedule of the rates of duties, as fixed by law. Notes. — 1. The collection of duties is made at the custom-houses established at ports of entry and ports of delivery. The principal custom-house officers are collectors, naval officers, surveyors, and appraisers. 2. Ihre is an allowance for the weight of whatever contains the goods. Duty is collected only on the quantities passed through the custom-house. The ton, for custom-house purposes, consists of 20 cwt., of 112 lb. each. 3. On some classes of goods, both specific and ad valorem duties are collected. The cost price, if given in foreign money, must be changed to United States currency, 286. Problems in United States Customs, where the duty is wholly or in part ad valorem, are solved by the principles of Percentage. The quantities involved are: the Invoice Price corre- sponding to the Base; the Duty, corresponding to the Per- centage ; and the Total Cost of the importation, corresponding to the Amount. 238 RAY'S HIGHER ARITHMETIC. CASE I. 287. Given the invoice price and the rate, to find the duty. FoBMUiiA. — Invoke Price ^(^ Rale = Duty. Examples for Practice. 1. Import 24 trunks, at $5.65 each, and 3 doz. leather satchels, at $2.25 each; the rate is 35^ ad valorem: what is the duty? $75.81 2. What is the duty on 45 casks of wine, of 36 gal. each, invoiced at $1.25 a gal., at 40 ct. a gal. specific duty ? $648. 3. There is a specific duty of $3 per gallon, and an ad valorem duty of 50% on cologne-water: what is the total amount of duty paid on 25 gallons, invoiced at $16.50 per gallon? $281.25 4. What is the total duty on 36 boxes of sugar, each weighing 6 cwt. 2 qr. 18 lb., invoiced at 2^ ct. per lb., the specific duty being 2 ct. per lb., and the ad valorem duty 25^? $r04.97 5. What is the duty on 575 yards of broadcloth, weighing 1154 lb., invoiced at $2.56 per yd., the specific duty being 50 ct. per lb. , and the ad valorem duty 35^ ? If the freight, charges, and losses amount to $160.80, how much a yard must I charge to gain 15% ? Duty, $1092.20; price per yard, $5.45 6. A merchant imported a ton of manilla, invoiced at 5 ct. per lb.; he paid a specific duty per ton, which, on this shipment, was equivalent to an ad valorem duty of 22^^ : what is the specific duty ? $25 per ton. 7. Received a shipment of 3724 lb. of wool, invoiced at 23 ct. per lb.; the duty is 10 ct. T)er lb., and 11^ ad valorem, less 10^: what is the total amount of duty? $419.96 UNITED STATES REVENUE. 239 8. A dry-goods merchant imports 1120 yards of dress goods, \\ yd. wide, invoiced at 23 ct. a sq. yd.; there is a specific duty of 8 ct. per sq. yd., and an ad valorem duty of 40^: what must he charge per yard, cloth measure, to clear 25^ on the whole? $.62|f per yd. CASE II. 288. Given the invoice price and the duty^ to find the rate. IhUy _ FoBMUiA.— - -; — : — frr- = Bate, Invoice Pnce Examples for Practice. 1. If goods invoiced at $3684.50 pay a duty of $1473.80, what is the rate of duty? 40%. 2. If laces invoiced at $7618.75, cost, when landed, $10285.31J, what is the rate of duty? 35^. 3. Forty hhd. (63 gal. each) of molasses, invoiced at 52 ct. a gallon, pay $453.60 duty. The specific duty is 5 ct. a gallon: what is the additional ad valorem duty? 25%. CASE III. 289. Given the duty and the rate, to find the in- voice price. Formula. — ^ = Invoice Price, Bate Examples for Practice. 1. Paid 8575.80 duty on watches, at 25%: at what were they invoiced, and what did they cost me in store? Invoiced, $2303.20; cost, $2879. 2. The duty on 1800 yards of silk was $2970, at 60% 240 BA Y'S HIGHER ABITHMETIC. % ad valorem : what was the invoice price per yard, and what must I charge per yard to clear 20% ? Invoiced at $2.75; sell at 85.28 per yd. 3. The duty on 15 gross, qt. bottles of porter, at a tax of 35 ct. a gallon, was $151.20: if this were equivalent to an ad valorem duty of 20^% on the entire purchase, how many bottles were allowed for a gallon, and at how much per bottle must the whole be sold to clear 20% ? 5 bottles to gal. ; 50 ct. per bottle. CASE IV. * 290. Given the entire cost and the rate, to find the invoice price. ^ Whofe Cost y. . „ . Formula. ==:Invovce Frice. l + BcUe Examples for Practice. 1. 1000 boxes (100 each) pf cigars, weighing 1200 lb., net, cost in store $13675. There is a specific duty of $2.50 per lb., an ad valorem duty of 25%, and an internal revenue tax of 60 ct. a box : freight and charges amount to $75 ; find the invoice price per thousand cigars. $80. 2. Supposing No. 1 pig-iron, American manufacture, to be of equal quality with Scotch pig-iron : at what price must the latter be invoiced, to compete in our markets, if Amer- ican iron sells for $45 a ton ; freight and charges amounting to $10 a ton, and the specific duty being equivalent, in this instance, to an ad valorem duty of 25% ? $28 a ton. 3. A marble-cutter imports a block of marble 6 ft. 6 in. long, 3 ft. wide, 2 ft. 9f in. thick ; the whole cost to him being $130 ; he pays a specific duty of 50 ct. per cu. ft. and an ad valorem duty of 20^ : freight and charges being $20.80, what was the invoice price per cu. ft.? $1.25 TOPICAL OUTLINE. 241 Topical Outline. Applications op Percentage. (WUhmt Time.) 1. Profit and Loss. 2. Stocks and Bonds.. 3. Premium and Discount. 4. Commission and Brokerage. 1. Definitions :— Cost, Selling Price, Profit, Loss. 2. Four Cases. 1. Definitions :— Company, Corporation, Char- ter, Stock, Shares, Scrip, Bond, Assess- ment, Dividend. 2. Four Cases. 1. Definitions :— Drafts, Par Value, Market Value, Discount, Premium, Rate. 2. Four Cases. • 1. Definitions:— Commission-Merchant, Prin- cipal, Commission, Consignment, Con- signor, Consignee, Correspondent, Net Proceeds, Guaranty, Broker, Brokerage. 2. Four Cases. 5. Stock Investments •< 6. Insurance ^ 7. Taxes -« 1. Definitions :— Stock Exchange, Government Bonds, State Bonds, etc. 2. Notation. 3. Five Cases. 1. Definitions:— Fire Insurance, Marine In- surance, Personal Insurance, Insurer, Risk, Insured, Contract, Primary Ele- ments, Policy, Premium, Rate, Amount, Joint-Stock and Mutual Companies. 2. Three Cases. 1. Definitions :— Tax, State and Government, Direct and Indirect, Poll-tax, Property- tax, Assessor, Assessment Roll, Collector. 2. Four Cases. 8. United States Revenue. 1. Definitions:— Internal Revenue, Customs, Ad Valorem Duties, Specific Duties, In- voice, Tariff, Tare. 2. Four Cases. H. A. 21. XYI. PEEOEETAGE -APPLIOATIOI^S. I. INTEEEST. DEFINITIONS. 291. 1. Interest is money charged for the use of money. Note. — The profits accruing at regular periods on permanent, investments, such as dividends or rents, are called interest, since they are the increase of capital, unaided by labor. 2. The Principal is the sum of money on which interest is charged. Note. — The principal is either a sum loaned ; money invested to secure an income ; or a debt, which not being paid when due, is allowed by agreement or by law to draw interest. 3. The Rate of Interest is the number of per cent the yearly interest is of the principal. 4. The AiAount is the sum of the principal and interest. 5. Interest is Payable at regular intervals, yearly, Judf- yearly, or quarterly , as may be agreed : if there is no agree- ment, it is understood to be yearly. Note. — If interest is payable half-yearly, or quarterly, the rate is still the rate per annumy or rate per year. In short loans, the rate per month is generally given ; but the rate per year, being 12 times the rate per month, is easily found ; thus, 2^ a month = 24^ a year. 6. The Legal Rate is the highest rate allowed by the law. 7. If interest be charged at a rate higher than the law allows, it is called Usury; and, in some states, the person offending is subject to a penalty. (242) Rehake:. — The per cent of interest that ia legal in the different Btales and territories, is exhibited in the following table. Alabama Ariaona - Arkansa* California Canada Colorado - Connecticut Dakota - Delaware^ District Columbia. Florida Georgia Illinois Indiaaa Kauatis Keutuckjr Louisiana .*. Maine Maryland Massachuaetts Michigan Minnesota Mississippi Missouri Montana Nebraska Nevada New Hampshire.. New Jersey New MeiicO" New York North CaroliuD~.- Ohio Oregon Pennsylvania Rhode Island South Carolina-... Tennessee Texas United States Utah Vermont Virginia Washington Tei ritory, West Virginia---. Wisconsin Wyoming 10^ 12)S Any. Any. 10^ Note. — When the per i lot mentioned in the t, the first column gives the per cet collected by law. If stipulated in the note, a per cer high a« that in the second column may be collected. 8. Interest is either SimpU or Cbrnpowiwi. 9. Simple Interest is interest which, ev<^n 244 RAY'S HIGHER ARITHMETIC. when due, is not convertible into principal, and therefore can not accumulate in the hands of the debtor by drawing interest itself, however long it may be retained. Note. — Compound Interest is interest which, not being paid when due, is convertible into principal, and from that time draws interest itself and accumulates in the hands of the debtor, according to the time it is retained. It will be treated of in a separate chapter. 292. Simple Interest differs from the applications of Percentage in Chapter XV, by taking time as an element in the calculation, which they do not. 293. The five quantities embraced in questions of in- terest are: the Principal^ the Interedy the Ratey the Time, and the Amourd, — or the sum of the principal and interest. Any three of these being given, the others may be found. They give rise to five cases. The principal corresponds to the Bdse^ and the interest to the Percentage. NOTATION. 294. The following notation can be adopted to ad* vantage in the formulas : « Principal = P. Eate =R Interest = I. Amount = A. Time = T. CASE I. 295. Given the principal, the rate, and the time, to find the interest and the amount. Principle. — The irdered is equal to the continued product of the principal, rate, and time. Formulas. -, p _i_ t a INTEREST. 245 PXBXT = I. { KoTE. — The time is expressed in years or parts of years, or both. COMMON METHOD. Problem. — What is the interest of $320 for 3 yr. 5 mo. 18 da., at 4% ? OPERATION. $320X.04X3A = $44.37J, Int. Solution.— The interest of $320 for 1 year, at 4^, is $12.80; and for 3/y years, it is 3^ times as much as it is for 1 year, or $12.80 X 3 A = $44.37i. Remark. — Unless otherwise specified, 30 days are considered to make a month; hence, 5 mo. 18 da. = ^ of a year. T -r - General Rule. — 1. Multiply the principal by the rate, and that product by the time expressed in years; the product is the. intered, ^ 2. Add the principal and interest, to find the amount. -^ Examples fob Practioe. Find the simple interest of: 1. $178.63 for 2 yr. 5 mo. 26 da., at 7^. $31.12 2. $6084.25 for 1 yr. 3 mo., at 4^. $342.24 3. $64.30 for 1 yr. 10 mo. 14 da., at 9^. $10.83 4. $1052.80 for 28 da., at 10^. $8.19 5. $419.10 for 8 mo. 16 da., at 6^. $17.88 6. $1461.85 for 6 yr. 7 mo. 4 da., at 10^. $964.01 7. $2601.50 for 72 da., at 1^%. $39.02 8. $8722.43 for 5| yr., at 6%. $2878.40 9. $326.50 for 1 mo. 8 da., at 8^. $2.76 10. $1106.70 for 4 yr. 1 mo. 1 da., at 6%. $271.33 11. $10000 for 1 da., at 6%. $1.67 1 yr. = 21.2880 3 yr. = 3 63.864 6 mo. = J 10.644 1 mo. = i 1.774 1 8 da. = A 1.064 1 da. = ^ .059 246 RAY'S HIGHER ARITHMETIC. METHOD BY ALIQUOT PARTS. 296. Many persons prefer computing interest by the method of Aliquot Parts. The following illustrates it : Problem. — What is the simple interest of $354.80 for 3 yr. 7 mo. 19 da., at 6%? Solution. — The yearly in- operation. terest, being 6^ of the prin- $354.80 cipal, is found, by Case I of .06 Percentage; the interest for 3 yr. 7 mo. 19 da. is then ob- tained by aliquot parts; each item of interest is carried no lower than mills, the next figure being neglected if less than 5 ; but if 5 or over, it is $77.41 Ans, counted 1 mill. SIX PER CENT METHODS. FiBST Method. 297. The six per cent method possesses many advantages, and is readily and easily applied, as the following will show : At 6^ per annum, the interest on $1 for 1 year is 6 ct., or .06 of the principal. u 2 mo. " 1 " " .01 *' " " it I it a ^ it it QQ5 a it u « 6 da. = ^ mo. " yV *' " -^^ " " " « 1 " = ^ *' '' ^V « « .OOOi " " Problem. — What is the interest of $560 for 3 yr. 8 mo. 12 da., at 6%? Solution. — The interest on opebation. $lfor 1 yr., at 6 ^o, is 6 ct., and $5 60X.222 = $124.32, Ana. for 3 yr. is 18 ct.; the interest on $1 for 8 mo. is 4 ct., and the interest for 12 da. is 2 mills ; henco, the interest for the entire time is $.222 : therefore, the interest on $560 for 3 yr. 8 mo. 12 da. is equal to $.222 X 660 = $124.32 INTEREST, 247 Per Cent Bule. — 1. Tofe «ic cents J(yr every year, J a cent for every motvthf and i of a miU for every day; their mm 18 the interest on $1, at 6%, /or Hie given Utne. 2. MuUiply the interest on $1 for the given time by the prinr cipal; the prodtust is the interest required. Remarks. — 1. To find the interest at any other rate than 6^, in- crease or decrease the interest at 6^ by such a part of it as the given rate is greater or less than 6^. 2. After finding the interest at 6^o, observe that the interest at 5^ == interest at 6^ — J of itself. And the interest at 4J^ = int. at 6^ — J of it. at 4 ^ = int. at 6% — J of it. at 3 fo = i int. at 6^. at 2 ^ = J int at 6^. at 1 J^ = J int. at 6^. at 1 fo=i int. at 6^. at 7 ^ = int. at 6^ -|- J of it. at 7ifo = " Qfc -f J of it at 8 fo= " 6^0 + J of it. &t 9 ^0= " 6^0 + J of it. at 12fof 18^, 24^ = 2, 3, 4 times interest at 6^. at 5^, 10^, 15^, 20^ =xV» h h i interest at 6^, after moving the point one figure to the right. 3. Or, having the interest at 6^, multiply by the given rate and divide by 6. Second Method. Problem. — ^What is the simple interest of $354.80 for 3 yr. 7 mo. 19 da., at 6%? Solution. — The interest of any opebation. sum ($354.80), at 6^, equals the ($354.80-^2 ) X .436 J = interest of half that sum ($177.40), $77.4055 3, Am. at 12^. But 12^ a year equals 1^ a month, and for 3 yr. 7 mo., or 43 mo., the rate is 43^, and for 19 da., which is \^ of a month, the rate is iM] hence, the rate for the whole time is 43-J-f^, and 43 J J^ of the principal will be the interest. To get 43J§^o, multiply by 43JJ hundredths = .43 J| = .436J. Remark. — In the multiplier .436 J, the hundredths (43) are the number of months, and the thousandths (6J) are J of the days (19 da.), in the given time, 3 yr. 7 mo. 19 da. 248 BAY'S mo HER ARITHMETTC, Rule. — Reduce the years, if any, to m(wii/w, and vrrite the whole rmmber of months as decimal hundredtlis ; after which, 'place ^ of die days, if any, as thousandths; multiply half the principal by this number, the product is the interest. Note. — In applying the rule, when the number of days is 1 or 2, place a cipher to the right of the months, and write the } or | ; otherwise, they will not stand in the thousandths' place : thus, if the time is 1 yr. 4 mo. 1 da., the multiplier is .160J Bemark. — Exact or Accurate Interest requires that the common year should be 365, and leap year 366 days ; hence, exaA;t interest is T^y less for common years, aud ^ less for leap years than the ordi- nary interest for 360 days. Examples for Practice. Find the simple interest of: 1. $1532.45 for 9 yr. 2 mo. 7 da., at 12%. $1689.27 2. $78084.50 for 2 yr. 4 mo. 29 da., at 18^. $33927.72 3. $512.60 for 8 mo. 18 da., at 7%. $25.72 4. $1363.20 for 39 da., at 1\% a month. $22.15 5. $402.50 for 100 da., at 2^ a month. $26.83 6. $6919.32 for 7 yr. 6 mo., at 6%. $3113.69 7. $990.73 for 9 mo. 19 da., at 7^. $55.67 8. $4642.68 for 5 mo. 17 da., at 15^. $323.05 9. $13024 for 9 mo. 13 da., at 10^. $1023.83 10. $615.38 for 4 yr. 11 mo. 6 da., at 20%. $607.17 11. $2066.19 for 3 yr. 6 mo. 2 da., at 30^. $2172.94 12. $92.55 for 3 mo. 22 da., at 5%. , $1.44 Find the amount of: 13. $757.35 for 117 da., at 1J% a month. $801.65 14. $1883 for 1 yr. 4 mo. 21 da., at 6%. $2040.23 15. $262.70 for 53 da., at 1% a month $267.34 16. $584.48 for 133 da., at 7^%. $600.67 17. $392.28 for 71 da., at 2J% a month. $415.49 INTEREST, < 249 18. Find the interest of $7302.85 for 365 da., at 6%, counting 360 da. to a year. $444.26 19. If I borrow $1000000 in New York, at 7%, and lend it at Ifc in Ohio, what do I gain m 180 da.? $479.45 Note. — In New York 365 days are counted a year instead of 360. 20. Find the interest of $5064.30 for 7 mo. 12 da., at 7%, in New York. $218.45 21. If I borrow $12500 at 6%, and lend it at 10^, what do I gain in 3 yr. 4 mo. 4 da.? $1672.22 22. If $4603.15 is loaned July 17, 1881, at 7%, what is due March 8, 1883? $5130.34 Note. — When the time is to be found by subtraction, and the days in the subtrahend exceed those in the minuend, disregard days in subtracting, and take the months in the minuend one less than the actual count. Having found the years and months by subtraction, find the days by actual count, beginning in the month preceding the later of the two dates. Thus, in the above example, we find 1 yr. 7 «io. (not 8), and count from Februcui'y 17th to March 8th 19 da. 23. Find the interest, at 8%, of $13682.45, borrowed from a minor 13 yr. 2 mo. 10 da. old, and retained till he is of age (21 years). $8543.93 24. In one year a broker loans $876459.50 for 63 da., at \\^o a mo., and pays 6% on $106525.20 deposits: what is his gain? $21216.96 25. What is a broker's gain in 1 yr., on $100, deposited at 6%, and loaned 11 times for 33 da., at the rate of 2% a month? $18.20 26. Find the simple interest of £493 16s. 8d. for 1 yr. 8 mo.; at 6^. £49 7s. 8d. Note. — In England the actual number of days is counted. 27. Find the simple interest of £24 18s. 9d. for 10 mo., at 6%. £1 4s. 11 Jd. 28. Of £25 for 1 yr. 9 mo., at 5%. £2 3s. 9d. 29. Of £648 15s. 6d. from June 2 to November 25, at 5%. £15 12s. lOd. 250 CRAY'S HIGHER ARITHMETIC CASE II. 298. Given the principal, the rate, and the interest, to find the time. F0RMU1.A.— ~ .^ ~ = T. Problem. — John Thomas loaned 8480, at 5%, till the in- terest was 8150 ; required the time. Solution. — The inter- operation. est on $480 for 1 yr., at $1 50-f- ($480X. 06) = 6^ years. 5^, is $24. If the prin- cipal produce $24 interest in 1 year, it will require as many years to produce $150 interest as $24 is contained times in $150, which is 6J years, or 6 yr. 3 mo. From the preceding, the following rule is derived : Briile. — Divide the given interest by Hue interest of the prin- cvpal for one year; the quotient is the time, Kemark. — If the principal and amount are given, take their difference for the interest. Examples for Practice. In what time will: 1. $1200 amount to $1800, at 10^ ? 5 yr. 2. $415.50 to $470.90, at 10^? 1 yr. 4 mo. 3. $3703.92 to $4122.15, at 8^ ? 1 yr. 4 mo. 28 da. Note — A part of a day, not being recognized in interest, is omitted in the answer, but must not be omitted in the proof, 4. In what time will any sum, as $100, double itself by simple interest, at 4^, 6, 7|, 9, 10, 12, 20, 25, 30% ? 22f, 16?, 13i, IH, 10, 8i, 5, 4, 3^ yr. INTEREST. 251 5. In what time will any sum treble itself by simple in- terest, at 4, 10, 12% ? 50, 20, 16| yr. 6. How long must I keep on deposit 81374.50, at 10%, to pay a debt of $1480.78? 9 mo. 8 da. 7. How long will it take $3642.08 to amount to 84007.54, at 12% ? 10 mo. 1 da. 8. How long would it take $175.12 to produce $6.43 interest, at 6% ? 7 mo. 10 da. 9. How long would it take $415.38 to produce $10.69 interest in New York, at 7^ ? 134 days. CASE III. 299. Given the principal, interest, and time, to find the rate. Formula. — -= ::r^^ B. PXlfe^XT Problem. —At what rate per cent will $4800 gain $840 interest in 2\ years? Solution. — The operation. interest of $4800, Int. of $4800 at 1^ for 2J yr. = $120; at ] ^, for 2} yr. is .'. $840-5-$l 20 = 7; or, $120 ; hence, the 1^X7 = 7^, Amr rate is as many times 1^ as $120 is contained times in $840, which is- 7 times; or, 7^. Hence the rule. Bule. — Divide (he given inierest by Hie interest of the prin- cipal for the given time, at 1%, Examples for Practice. 1. At what rate per annum will any sum treble itself, at simple interest, in 5, 10, 15, 20, 26, 30 years, respectively ? 40, 20, 13i, 10, 8, 6|%. 252 RAY'S mo HER ARITHMETIC. 2. At what rate of interest per annum will any gum quadruple itself, at simple interest, in 6, 12, 18, 24, and 30 years, respectively? 50, 25, 16f, 12^, 10^. 3. What is the rate of interest when $35000 yields an income of 8175 a month? 6%. 4. What is the rate of interest when $29200 produces $6.40 a day? 8^. 5. What is the rate of interest when $12624.80 draws $315.62 interest quarterly. 10%. 6. Find the rate when stock, bought at 40% discount, yields a semi-annual dividend of 5%? 16f % per annum. 7. A house that cost $8250, rents for $750 a year; the insurance is 3^^, and the repairs ^%, every year: what rate of interest does it pay? 8% — . CASE IV. 900. Given the interest, rate, and time, to find the principal. Problem. — A man receives $490.84 interest annually on a mortgage, at 7^: what is the amount of the mortgage? SoLUTiON.r—Since $.07 is opebation. the interest of $1 for one year, $490. 84-j-$. 07 = 7012 $490.84 is the interest of as $1X7012 = $7012, Ans. many dollars as $.07 is con- Or, 7^=$490.84 tained times in $490.84, which 1^=$70.12 is 7012; therefore, $7012 is 100 /o = $7012. the amount of the mortgage. Or, thus: the time being 1 year, $490.84 is 7^ of the principal; 1^ of it is j of $490.84, which is $70.12, and hence 100^ of it, or the whole principal, is $7012. BrUle. — Multiply ihe rate by the time, and divide the intereM by ihe product; the quotient will he ihe principal. f INTEREST. 253 Examples for Practice. What principal will produce: 1. J1500 a year, at 6%? $26000. 2. $1830 in 2 yr. 6 mo., at 5% ? $14640. 3. $45 a mo., at 9^? $6000. 4. $17 in 68 da., at 1% a mouth? $750. 5. $656.25 in 9 mo., at 3^%? $25000. 6. $86.15 in 9 mo. 11 da., at 10^? $1103.70 7. $313.24 in 112 da., at 7%? $14383.47 8. $146.05 in 7 mo. 14 da., at 6^? $3912.05 9. $58.78 in 1 yr. 3 mo. 20 da., at 4% ? $1125.57 10. $79.12 in 5 mo. 25 da., at. 7% in N. Y.? $2357.46 CASE V. 901. Given the amount, rate, and time, to find the principal. FoBMULA.— A -i- ( 1 + R X T ) = p. Problem. — What is the par value of a bond which, in 8 yr. 8 mo., at 6^, will amount to $15200? OPERATION. Solution.— There are as $1 5200-^$l. 52 = 10000; many dollars in the princi- $1 X 10000 = $! 0000, Ans. pal as $1.52 is contained Or, 100^ = bond; times in $15200, which is 1 00^ + 52 ^o = $l 5200; 10000 ; therefore, $10000 is 1^ = $100 the par value of the bond. 100 ^ = $1000 0, Am. Or, the amount is \%^ of the principal ; hence, \^i of the principal is $15200 ; and yj^ of the principal is $100, and JJJ of the principal is $10000, Hole. — Divide ihe amount by the amount of $1 for fJte given time and raJte; ihe quotient unll be ihe principal. 254 BA Y'S mOHER ARITHMETIC. Examples for Practice. 1. What principal in 2 yr. 3 mo. 12 da., at 6^, will amount to $1367.84 ? $1203.03 2. What principal in 10 mo. 26 da., will amount to $2718.96, at 10% interest? $2493.19 3. What principal, at 4^^, will amount to $4613.36 in 3 yr. 1 mo. 7 da.? $4048.14 4. What principal, at 7%, will amount to $562.07 in 79 da. (365 da. to a year) ? $553.68 PKOMISSOBY NOTEa DEFINITIONS. 802. 1. A Promissory Note is a written promise by one party to pay a specified sum to another. 2. The Face of a note is the sum promised to be paid. 3. The Maker is the party who binds himself to pay the note by signing his name to it. 4. The Payee is the party to whom the payment is promised. Note. — The maker and the payee are called the wigiival parties to a note. 5. The Holder of a note is the person who owns it: he may or may not be the payee. 6. The Indorser of a note is the person who writes his name across the back of it. 7. A Time Note is one in which a specified time is set for payment. Eemark. — There are various forms used in time notes, the prin- cipal ones being as follows? PR0MI8S0BY NOTEIS. 255 1. Oedinaby Fokm. <{450.50 Boston, Mass., June 30, 1880. Three months after date, I promise to pay Thomas Ford, or order, Four Hundred and Fifty, and 3^, Dollars, for value received, vdth interest, Edward E. Morris. 2. Joint Note. $350. Denver, Col., Jan. 2, 1881. Twelve months after date, we, or either of us, promise to pay Frank R. Hairis, or bearer. Three Hundred and Fifty Dollars, for value received, with interest from dale. James West. Daniel Tate. Bemarks. — 1. This note is called a "joint note," because James West and Daniel Tate are jointly liable for its payment. 2. If the note read, " We jointly and severally promise to pay," etc., it would then be called a joint and several note, and the makers would be both jointly and singly liable for its payment. 3. Principal and Surety Note. $125.00 St. Louis, Mo., Nov. 20, 1880. Ninety days after date, I promise to pay James Miller, or order, One Hundred and Twenty-five Dollars, for value received, negotiable and payable without defalcation or discount. Surety, James Miller. H. A. White. Bemare. — The maker of this note, H. A. White, is the piirhcipal, and the note is made payable to the order of the surety, James Miller, who becomes security for its payment, indorsing it on the back to the order of Mr. Whitens creditor. 256 BAY'S HIGHER ARITHMETIC, 8. A Demand Note is one in which no time is specified, and the payment must be made whenever demanded by the holder. The following is an example ; Demand Note. $1100.00 Cincinnati, O., Mar. 18, 1881. For value received, I promise to pay David Swinton, or ord/ir, on demand, Eleven Hundred Dollars, witli interest. Henry Rudolph. 9. A promissory note is negotiable — that is, transferable to another party by indorsement, — when the words **or order," or ** or bearer," follow the name of the payee, as in the above examples ; otherwise, the note is not negotiable. 10. If the note is payable to "bearer," no indorsement is required on transferring it to another party, and the maker nnlv IS rpsnonsihlft for its nflvmftnt. only is responsible for its payment 11. If the note is payable to "order," the payee and each holder in his turn must, on transferring it, indorse the note by writing his name on its back, thus becoming liable for its payment, in case the maker fails to meet it when due. Bemarks. — 1. An indorser may free himself from responsibility for payment if he accompany his signature with the words, " with- out recourse ;" in which case his indorsement simply signifies the transfer of ownership. 2. If the indorser simply writes his name on the back of the note, it is called a " blank " indorsement ; but if he precedes his signature by the words, " Pay to the order of John Smith," for example, it is called a " special " indorsement, and John Smith must indorse the note before it can legally pass to a third holder. 12. It is essential to a valid promissory note, that it contain the words "value received," and that the sum of money to be paid should be written in words. PROMISSOBY NOTES. 257 fr 13. If a note contains the words " with interest," it draws interest from date, and, if no rate is mentioned, the legal rate prevails. Kemabks. — 1. The face of such a note is the sum mentioned with its interest from date to the day of payment. 2. If a note does not contain the words " with interest," and is not paid when due, it draws interest from the day of maturity, at the legal rate, till paid. 14. A note matures on the day that it is legally due. However, in many of the states, three days, called Days of GracCy are allowed for the payment of a note after it is nominally due. Bemabks. — 1. The day before "grace" begins, the note is nomi- nally due ; it is legally due on the last day of grace. 2. The days of grace are rejected by writing " without grace " in the note. 3. Notes falling due on Sunday or on a legal holiday, are due tho day before or the day after, according to the special statute of each state or territory. 4. If a note is payable " on demand," it is legally due when pre- sented. 15. When a note in bank is not paid at maturity, it goes to protest — ^that is, a written notice of this fact, made out in legal form by a notary public, is served on the indorsers, or security. Kemaiik. — Some of the states require that certain words shall be inserted in every negotiable note in addition to the usual forms. For instance, in Pennsylvania, the words " without defalcation " are required; in New Jersey, "without defalcation or discount;" and in Missouri, the statute requires the insertion of " negotiable and payable without defalcation or discount." In Indiana, in order to evade certain provisions of the law, the following words are usually inserted, "without any relief from valuation or appraisement laws." This constitutes what is known as the "iron-clad" note. 303. If a note be payable a certain time ** after date," proceed thus to find the day it is legally due: H. A. 22. 258 RA Y'S HIGHER ARITHMETIC. Bule. — Add to the date of ihe note, Hie number of years and months to elapse before payment ; if this gives the day of a month higher than that month contains, take tlie last day in that month; then, count the number of days mentioned in the note and 3 more: this wUl give the day the note is legally due ; but if it is a Sunday or a national holiday, it mud be paid the day previous, Remarks.^-1. When counting the days, do not reckon the one from which the counting begins. (For exceptions, see Art. 313, 16,Kem.2.) 2. The months mentioned in a note are calendar months. Hence, a 3 mo. note would run longer at one time than at another ; one dated Jan. 1st, will run 93 days ; one dated Oct. 1st, will run 95 days : to avoid this irregularity, the time of short notes is generally given in days instead of months ; as, 30, 60, 90 days, etc. 3. When the time is given in days, it is convenient to use the following table in determining the day of maturity of a note : Table Showing the Number of Days from any Day of: • 365 31 59 90 120 151 181 212 243 273 304 334 • 334 365 28 59 89 120 150 181 212 242 273 303 p • 306 > • 275 p • a s o • a > •>< • 153 184 212 243 • 122 153 181 212 9 • 92 123 151 182 212 243 273 304 335 365 31 61 < • 61 92 120 151 181 212 • 31 62 90 121 151 182 To the same day of next 245 276 304 335 365 31 61 214 245 273 304 334 365 30 61 92 122 153 183 1.84 215 243 274 304 335 365 31 62 92 123 153 Jan. 337 365 31 61 92 306 334 365 30 61 Feb. Mar. April. 273 304 334 365 31 61 92 122 242 273 303 334 365 30 61 91 Mav. June. 122 153 184 214 245 275 91 242 , 212 273 243 July. 122 153 183 214 ?!44 92 123 153 184 214 Aug. 304 334 3H5 30 274 304 335 365 Sept. Oct Nov. Dec. Kemark. — In leap years, if the last day of February be included in the time, 1 day must be added to the number obtained from the table. .1 ANNUAL INTEREST. 259 Examples for Practice. Find the maturity and amount of each of the following notes: $560.60 Chattanooga, Tenn., June 3, 1872. 1. For value received, sixty days after date, I promise to pay Madison Wilcox five hundred and sixty -^ dollars, with interest at 7^. James Daily. $567.47— 8430.00 ToPEKA, Kan., Jan. 30, 1879. 2. Six months after date, I promise to pay Christine Ladd four hundred and thirty dollars, for value received, with interest at 12^ per annum. Amos Lyle. $456.23 $4650.80 St. Louis, Mo., August 10, 1875. 3. For value received, three months after date, I promise to pay Oliver Davis, or order, four thousand six hundred and fifty -^^ doyars, with interest at the rate of 10^ per annum, negotiable and payable without defalcation or dis- count. Milton Moore. Surety, Oliver Davis. $4770.95 — ANNUAL INTEREST. 304. Annual Interest is interest on the principal, and on each annual interest after it is due. Annual interest is legal in some of the states. Explanation.— If Charles Heydon gives a note and agrees to pay the interest annually, but fails to do so, and lets the note run for three years before settlement, the first year's interest would draw interest for two years, and the second year's interest would draw interest for one year. The last year's interest would be paid at the time of settlement, or as it fell due. 260 HA Y'S HIOHEB ARITHMETIC. Problem.— Find the amount of $10500 for 4 yr. 6 mo., interest 6^, payable annually. OPERATION. Int. of $ 1 5 for 4 J yr., at 6^ = $ 2 8 3 5; " " $630 " 8 " " " =$ 302.40 Total annual interest, =$3137.40 Amount =$10600 + $3 13 7. 40 = $ 136 3 7. 4 Solution.— The interest of $10500 for 1 yr., at 6^, is $630, and for 4J yr. is $2835. The first anniual interest draws interest 3J yr., the second 2^, the third 1 J, and the fourth \ yr. This is the same as one annual interest for (3 J + 2^ + 1^-1-^ = 8) eight years. But the interest of $630, for 8 yr., at 6^, is $302.40; therefore, the amount = $10500 + $2835 + $302.40 = $13637.40 Bule.— 1. Find the interest on ihe 'principal for the entire time, and on each yearns interest till the time of settleinent. 2. The sum of these interests is the interest required. Note. — In Ohio and several other states, interest on unpaid annual interest is calculated at the legal rate only. Examples for Practice. 1. Find the amount of a note for $1500, interest 6%, payable annually, given Sept. 3, 1870, and not paid till March 1, 1874. $1838.71 2. A gentleman holds six $1000 railroad bonds, due in 3 years, interest 6^, payable semi-annually: no interest having been paid, what amount is owing him when the bonds mature? $7161. 3. $2500.00 St. Paul, Minn., Jan. 11, 1869. For value received, I promise to pay Morgan Stuart, or order, twenty-five hundred dollars, with interest at 7%, payable annually. Leonard Douglas. What was due on this note March 17, 1873, if the interest was paid the first two years? $2898.825 PARTIAL PAYMENTS, 261 II. PARTIAL PAYMENTS. 905. A Partial Payment is a 'payment in part of a note or other obligation. Whenever a partial payment is made, the holder should write the date and amount of the payment on the back of the note. This is called an Indorsement. 908. The following decision by Chancellor Kent, ''John- son's Chancery Rep., Vol. I, p. 17," has been adopted by the Supreme Court of the United States, and, with few excep- tions, by the several states of the Union, and is known as the ** United States Rule;" XJ. S. Bule. — 1. ** The rule for casting interest when par- tial payments have been made, is to apply the payment, in Hie first place, to the disdiarge of the interest due, 2. ^^If the payment exceeds the interest, the surplus goes towards discharging the principal, and the subsequent interest is to be computed on the balance of principal remaining due, 3. **ijr the payment be less than ike interest, the suiplus of interest must not be taken to augment the principal; but in- terest continues on the former principal until Hie period when the payments, taken together, exceed the inta^est due, and then the surplus is to be applied towards discharging the principal, and intei'est is to be computed on the balxmce as aforesaid, dKyi. This rule is based upon the following principles: Principles. — 1. Payments must be applied first to the dis- diarge of interest due, and the balance, if any, toward paying the principal, 2. Interest must, in no case, become part of the principal. 3. Interest or payment must not draw interest, Eemarks. — 1. It is worthy of remark, that the whole aim and tenor of legislative enactments and judicial decisions on questions of 262 EAY'S HIGHER ARITHMETIC, interest, have been to favor the debtor, by disallowing compound interest, and yet ihu veiy rule fails to secure the end in view, and really maintains and enforces the principle of compound interest in a most objectionable shape ; for it makes interest due (not every year as compound interest ordinarily does), bvi as often as a payment is made; by which it happens that the closer the paymerUs are together, the greajler the loss of the debtor, who thus suffers a penalty for his very promptness. To illustrate, suppose the note to be for $2000, drawing interest at 6^, and the debtor pays every month $10, which just meets the interest then due ; at the end of the year he would still owe $2000. But if he had invested the $10 each month, at 6^, he would have had, at the end of the year, $123.30 available for payment, while the debt would have increased only $120, being a difference of $3.30 in his favor, and leaving his debt $1996.70, instead of $2000. 2. To find the difference of time between two dates on the note, reckon by years and months as far as possible, and then count the davs. Problem. ^850. Cincinnati, April 29, 1880. Ninety days after date, I promise to pay Stephen Ludlow, or order, eight hundred and fifty dollars, with in- terest; value received. Charles K. Taylor. Indoi'sements,— Oct, 13, 1880, $40 ; June 9, 1881, $32 ; Aug. 21, 1881, $125 ; Bee. 1, 1881, $10 ; March 16, 1882, $80. What was due Nov. 11, 1882? Solution.— Interest on principal ($850) from April 29 to Oct. 13, 1880, 5 mo. 14 da., at 6^ per annum, $23.233 . 850. Whole sum due Oct. 13, 1880, . . „ . . 873.233 Payment to be deducted, 40. Balance due* Oct. 13, 1880, ..... 833.233 Interest on balance ($833,233) from Oct. 13, 1880, to June 9, 1881, being 7 mo. 27 da., .... 32.913 Payment not enough to meet the interest, . . . 32. ' Surplus interest not paid June 9, 1881, .... MS Interest on former principal ($833,233) from June 9, 1881, to Aug. 21, 1881, being 2 mo. 12 da., . . 9.999 Whole interest due Aug. 21, 1881, 10.912 PARTIAL PAYMENTS. 263 (Brought forward) Int. due Aug. 21, 1881, . . 10.912 833.233 Whole sum due Aug. 21, 1881, 844.145 Payment to be deducted, 125. Balance due Aug. 21, 1881, 719.145 Interest on the above balance ($719,145) from Aug. 21, 1881, to Dec. 1, 1881, being 3 mo. 10 da., . . 11.986 Payment not enough to meet the interest, . , .10. Surplus interest not paid Dec. 1, 1881, .... 1.986 Interest on former principal ($719,145) from Dec. 1, 1881, to March 16, 1882, being 3 mo. 15 da., . . 12.585 Whole interest due March 16, 1882, .... 14.571 719.145 Whole sum due March 16, 1882, . . . . . 733.716 Payment to be deducted, ...... c 80. Balancedue March 16, 1882, 653.716 Interest on balance ($653,716) from March 16, 1882, to Nov. 11, 1882, being 7 mo. 26 da., .... 25.713 Balance due on settlement, Nov. 11, 1882, . . . $679.43 • Examples for Practice. 1. $304^^. Chicago, March 10, 1882. For value received, six months after date, I promise to pay G. Riley, or order, three hundred and four -^^ dollars. No payments. H. McMakin. What was due Nov. 3. 1883? 8325.63 2. $429^^. Indianapolis, April 13, 1873. On demand, I promise to pay W. Morgan, or order, four hundred and twenty-nine 3^ dollars, value received, with interest. r Wilson. Indorsed : Oct. 2, 1873, $10 ; Dec. 8, 1873, $60 ; July 17, 1874, $200. What was due Jan. 1, 1875? $195.06 264 RAY'S HIGHER ARITHMETIC. 3. $1750. New York, Nov. 22, 1872. For value received, two years after date, I promise to pay to the order of Spencer & Ward, seventeen hundred and fifty dollars, with interest at 7 per cent.' Jacob Winston. Indorsed: Nov. 25, 1874, $500; July 18, 1875, $50; Sept. 1, 1875, $600 ; Dec. 28, 1875, $75. What was due Feb. 10, 1876? $879.71 4. A note of $312 given April 1, 1872, 8% from date, was settled July 1, 1874, the exact sum due being $304.98. Indorsed: April 1, 1873, $30.96; Oct 1, 1873,$ ; April 1, 1874, $20.40 Restore the lost figures of the second payment. $11.08 5. There have been two equal annual payments on a 6^ note for $175, given two years ago this day. The balance is $154.40: what was each payment? $20.50 6. A merchant gives his note, 10% from date, foi* $2442.04: what sum paid annually will have discharged the whole at the end of 5 years? $644,204 308. In Connecticut and Vermont the laws provide the following special rules relative to partial payments: Connecticut Bule. — 1. Gompvle ike interest to the time of the first payment, if that be oive year or more from, ihe time tJiat interest commenced; add it to ihe prindpai, and deduct the payment from the sum total. 2. Jff' there be after payments made, compute the interest on Hie balance due to ihe next payment, and then deduct the pay- ment as above; and in like manner from one payment to another, till all the payments are absorbed: Provided, ihe time between one payment and another be one year or more. 3. But if any payment be made before one yearns interest hath accrued, then compute ihe interest- on ihe sum then due on the obligation, for one year, add it to ihe principal, and compute ihe interest on ihe sum paid, from tJie time it vxis PARTIAL PAYMENTS. 265 jpaici, wp to the end of ike year; add it to 1}ie mm paid^ and deduct that mm from the principal and interest added as above. (See Note.) 4. ijf' any payment be made of a less sum tliait the interest arisen at tfie time of such paymenty no interest is to be com- putedy bvt only on the principal mm, for any period. Note. — Jff a year does not extend beyond the time of payment; but. if it doesj then find the amount of the principal renuiining unpaid up to the time of settiementj likevjise the amount of the payment or payments from the time they were paid to the time of settlement, and deduct the sum of these several amounts from the amount of the principal. What is due on the 2d and 3d of the preceding notes, by the Connecticut rule? 2d, $194.88 ; 3d, $877.95 Vermont Bule. — 1. On all notes, etc., payable with in- terest, partial payments are applied, first, to liquidate the interest that has accrued at the time of such payments; and, secondly f to the extinguishment of the principal. 2. When notes, etc., are drawn wiHi interest payable ANNUALLY, the annual interests that remain unpaid are svbjed to simple interest from the time they become due to the time of final settlement. 3. Partial payments, made after annual interest begins to accrue, also draw simple interest from the time such payments are made to the end of the year; and their amounts are then applied, first, to liquidate the simple interest that has accrued from the unpaid annual interests; secondly, to Uqui- date tJie annual interests that have become due; and, thirdly, to the extinguishment of Hie principal. $1480. Woodstock, Vt., April 12, 1879. For value received, I promise to pay John Jay, or order, fourteen hundred and eighty dollars, with interest annually. James Brown. Indorsed : July 25, 1879, $40 ; May 20, 1880, $50 ; June 3, 1 881, $350. What was due April 12, 1882? $1291.95 H. A. 23. 266 RAY'S HIGHER ARITHMETIC, 909. Business men generally settle notes and accounts payable within a year, by the Mercantile Rule. Mercantile Bule. — 1. Fiad the amount of the principal from the date of the note to the date of 8ettlem£nt 2. Find the amount of eojoh payment from its date to the date of setttement 3. From the amount of the principal svhtract the sum of the amounts of the paym£nt8, NoTR — In applying this rule, the time should be found for the exact number of days. Examples for Practice. 1. $950.00 New York, Jan. 25, 1876. For value received, nine months after date, I promise to pay Peter Finley nine hundred and fifty dollars, with 7% interest. Thomas Hunter. The following payments were made on this note: March 2, 1876, $225; May 5, 1876, $174.19; June 29, 1876, $187.50; Aug. 1, 1876, $79.15 Required — the balance at settlement. $312.57 2. A note for $600 was given June 12, 1865, 6% interest, and the following indorsements were made : Aug. 12, 1865, $100; Nov. 12, 1865, $250; Jan. 12, 1866, $120: what was due Feb. 12, 1866, counting by months instead of days ? $146.65 III. TRUE DISCOUNT. DEFINITIONS. 310. 1. Discount on a debt payable by agreement at some future time, is a deduction made for ** cash," or present payment. It arises from the consideration of the present tvorth of the debt. TRUE DISCOUNT, 267 2. Present Worth is that sum of money which, at a given rate of interest, will amount to the same as the debt at its maturity. 3. True Discount, then, is the difference between the present worth and the whole debt. Remarks. — 1. That is, it is the simple interest on the present worthy from the day of discount until the day of maturity. 2. Discount on a debt must be carefully distinguished from Com- mercial Discount, which is simply a deduction from the regular price or value of an article ; the latter is usually expressed as such a "per cent off." 311. The different cases of true discount may be solved like those of simple interest: the present worth correspond- ing to the Principal; the discount, to the Interest; and the face of the debt, to the Amount, The following case is the only one much used : 312. Given the face, time, and rate, to find the present worth and true discount. Problem. — Find the present worth and discount of $5101.75, due in 1 yr. 9 mo. 19 da., at OPERATION. Amount of $ 1 for 1 yr. 9 mo. 1 9 da., at6^ = $1.108J, and $5101. 7 5 -.-$1.108^ = 4603. 775; $1X4603.775 = $4603.77 5, present worth; $5101.75— $4603. 775 = $497. 97 5, true discount. Solution. — The amount of $1 for 1 yr. 9 mo. 19 da., at 6^, is $1.108|^, and $5101.75 is the amount of as many dollars as $1.108 J is contained times in $5101.75, which is 4603.775 times ; therefore, the present worth is $4603.775, and the true discount is $497,975 Bule. — 1. Divide the debt by the ainourd of %\ for the given time and raie ; the quotient is the present tuorih. 2. The difference bet>ween the debt and the preset worth is the true discount, OF THE '^ UNiVERS/T \^ 268 BAY'S mOHER ARITHMETIC. Examples for Practice. 1. Find the true discount on a debt for $5034.15 due in 3 yr. 5 mo, 20 da., without grace, at 7%. . $984.33 2. What is the present worth of a note for $2500, bearing 6% interest, and payable in 2 yr. 6 mo. 15 da., discounted at 8%.. $2394.10 IV. BANK DISCOUNT. DEFINITIONS. 313. 1. A Bank is an institution authorized by law to deal in money. 2. The three chief provinces of banks are : the receiv- ing of deposits, the loaning of money, and the issuing of bank-bills. Any or all of these provinces may be exercised by the same bank. Bemarks. — 1. A Bank of Deposit is one which takes charge of money, stocks, etc., belonging to its customers. Money so intrusted is called a deposity and the customers are known as deposUors. 2. A Bank of Discount is one which loans money by discounting notes, drafts, etc. 3. A Bank of Issue is one which issues notes, called " bank-bills," that circulate as monev. . 3. The banks of the. United States may be divided into two general classes, — ^National Banks and Private Banks. Kemarks. — 1. National Banks are organized under special legis- lation of Congress. They must conform to certain restrictions as to the number of stockholders, amount of capital stock, reserve, circu- lation, etc. In return, they have privileges which give them certain advantages over Private Banks : they are banks of issue, of discount, and of deposit. 2. Private Banks are unable to- issue their own bank-notes to advantage, owing to the heavy tax on their circulation ; they are, therefore, confined to the provinces of deposit and discount. BANK DISCOUNT, 269 4. A Check is a written order on a bank by a depositor for money. The following is a usual form: No. 1032. Cincinnati, Nw, 27, 1851. Pay to. Bank of Cincinnati, Rvfui B, Shaffer, OR Order Four Hundred and Twenty-five. $425^ j^ Dollars. Oeorge Potter HoUister, Kemabks. — 1. Before this check ifl paid, it must be indorsed by Kuf us B. Shaffer. He may either draw the money from the bank itself, or he may transfer the check to another party, who must also indorse it, as in the case of a promissory note. (See Art. 302, 11.) 2. If the words or Bearer are used in place of or Orders no indorse- ment is necessary, and any one holding the check may draw the money for it. 5. A Draft is a written order of one person or company upon another for the payment of money. The following is a customary form: $1453^. New Orleans, La., July 25, 1880. At sight. Pay to the order of. First Ndtional Bank. Fourteen Hundred Fifty-three and. j^ Dollars Value Received and Charge to Acxx)unt of To John R. Williams & Co,, No. 2136. Memphis, Tenn, Robert James. 270 HA Y'S HIGHER ARITHMETIC, 6. Drafts may be divided into two classes, Sight Drafts and Time Drafts, 7. A Sight Draft is one payable "at sight." (See the form above.) 8. A Time Draft is payable a specified time "after sight," or ** afl^r date." 9. The signer of the draft is the maker or drawer. 10. The one to whom the draft is addressed, and who is requested to pay it, is the drawee. 11. The one to whom the money is ordered to be paid, is the payee. 12. The one who has possession of the draft, is called the owner or holder; when he sells it, and becomes an indorser, he is liable for its payment. 13. The Indorsement of a draft is the writing upon the back of it, by which the payee transfers the payment to another. Bemarks. — 1. A special indorsement is an order to pay the draft to a particular person named, who is called the indorsee^ as " Pay to F. H. Lee. — W. Harris," and no one but the indorsee can collect the bill. Drafts are usually collected through banks. 2. When the indorsemeM is in blanks the payee merely writes his name on the back, and any one who has lawful possession of the draft can collect it. 3. If the drawee promises to pay a draft at maturity^ he writes across the face the word "Accepted," with the date, and signs his name, thus: "Accepted, July 11, 1881. — H. Morton." The ncceptor is first responsible for payment, and the draft is called an acceptance. 4. A draft, like a promissory note, may be payable "to order," or " bearer," and is subject to protest in case the payment or accept- ance is refused. 5. A. time draft is universally entitled to the three days grace: but, in about half of the states, no grace is allowed on sight drafts. 14. When a bank loans money, it discounts the time notes^ BANK DISCO UNT. 271 drafts, etc., offered by the borrower at a rate of discount agreed upon, but not in accordance with the principles of true discount. 15. Bank Discount is simple interest on the face of a note, calculated from the day of discount to the day of maturity, and paid in advance. 16. The Proceeds of a note is the amount which remains after deducting the discount from the face. Bemabks. — 1. Sitice the face of every note is a debt due at a future time, its cost ought to be the present worth of that debt, and the bank discount should be the same as the true discount. As it is, the former is greater than the latter ; for, hank discount ia interest on the face of the note, while true discount is the interest on the present worihf which is always less than the face. Hence, their difference is the interest of the difference between the present worth and face ; that is, the interest of the true discount, (See Art. 310, 3.) 2. In discounting notes, the three days of grace are always taken into account ; and in Delaware, the District of Columbia, Maryland, Missouri, and Pennsylvania, the day of discount and the day ofmaivr rity are both included in the time. 3. If the borrower is not the maker of the notes or drafts, he must be the last indorser, or holder of them. 314. Problems in bank discount are solved like those of simple interest. The face of the note corresponds to the Prindpcd; the bank discount, to the Interest; the proceeds, to the Prindpcd less the Interest; and the term of discount, to the Time, CASE I. 315. Given fhe flEkce of the note, the rate, and time, to find the discount and proceeds. Problem. — What is the bank discount of $770 for 90 days, at 6%? 272 BAY'S HIGHER ARITHMETIC. OPERATION. Int. of $1 for 93 da., at 6 ^^ =$.0 1 55 = Rate of Discount. $770X. 0155 = 111. 935, Discount. $770 — $11.935 = $758.065, Proceeds. Solution.— Find the interest of $770 for 93 da., at 6^; this is $11,935, which is the bank discount. The proceeds is the difference between $770 and $11,935, or $758,065 pRTNCiPLES. — 1. The interest on the sum discounted for Uie given time (plus the tiiree days of grace) at the given rate per cent, is Hie bank discount, 2. The proceeds is equal to Uie sum discounted minus ilie discount. Bule. — 1» Find the interest on the sum discounted for three days more Hum the given time, at the given rate; it is the discount 2. Subtract the discount from Hie sum discounted, and the remainder is the proceeds, Kemarks. — 1. As with promissory notes, the three days of grace are not counted on a draft bearing the words " without grace." 2. The following problems present, each, two dates, — one showing when the note is nominally due, and the other when it is legally due. Thus, in an example which reads, " Due May Yg," the first is the nominal, and the second the legal date. Examples for Practice. Find the day of maturity, the time to run, and the pro- ceeds of the following notes : 1. $792.50 QumcY, III., Jan. 3, 1870. Six months after date, I promise to pay to the order of Jones Brothers seven hundred and ninety-two -j^ dollars, at the First National Bank, value received. Discounted Feb. 18, at 6/o. ALBERT L. ToDD. Due July ^; 138 da. to run; proceeds, J774.27 BANK DISCOUNT. 273 2. 81962yV(r N^^ Y^»^» '^"^^ 26, 1879. Value received, four months after date, I promise to pay B. Thoms, or order, one tliousand nine hundred sixty- two -^jj dollars, at the Chemical National Bank. Discounted Aug. 26, at 7^. Due Nov. 2 6/^^; 95 da. to run; pro. $1926.70 3. $2672^. Philadelphia, March 10, 1872. Nine months after date, for value received, I promise to pay Edward H. King, or order, two thousand six hun- dred seventy-two -^^ dollars, without defalcation. T.. * J T 1 in * /!^ Jeremiah Barton. Discounted July 19, at 6^. Due Dec. lyiaJ 1^8 da. to run; pro. $2606.27 4. $3886. St. Louis, Jan. 31, 1879. One month after date, we jointly and severally promise to pay C. McKnight, or order, three thousand eight hun- dred eighty-six dollars, value received, negotiable and pay- able, and without defalcation or discount. T. Monroe, I. Foster. Discounted Jan. 31, at 1J% a month. Due Feb. ^s/^ Mar.; 32 da. to run; pro. $3823.82 Kemark.— A note, drawn by two or more persons, "jointly and severally," may be collected of either of the makers, but if the words "jointly and severally " are not used, it can only be collected of the makers as a firm or company. 5. $2850. Austin, Tex., April 11, 1879. For value received, eight months after date, we promise to pay Henry Hopper, or order, twenty-eight hundred and fifty dollars, with interest from date, at six per cent per ^°^^""- HaNNA & TUTTLE. Discounted June 15, at 6^. Due Dec. ^^,4; 182 da. to run; pro. $2875.47 274 RA Y'S HIGHER ABITHMETIC. 6. $737yV(r- Boston, Feb. 14, 1880. Value received, two months after date, I promise to pay to J. K. Eaton, or order, seven hundred thirty-seven -f^^ dollars, at the Suffolk National Bank. William Allen. Discounted Feb. 23, at 10^^. Due April ^^/xi\ ^4 da. to run; pro. $726.34 7. $4085^. New Orleans, Nov. 20, 1875. Value received, six months after date, I promise to pay John A. Westcott, or order, four thousand eighty-five -^^ dollars, at the Planters' National Bank. E. Waterman. Discounted Dec. 31, 1875, at 5^. Due May 20/33, 1^76; 144 da. to run; pro. J4003.50 CASE II. 316. Given the proceeds, time, and the rate of dis- count, to find the flEU^e of the note. Problem. — For what sum must a 60 da. note be drawn, to yield $1000, when discounted, at 6% per annum? OPERATION. The bank discount of $1 for 6 3 da., at 6^=$. 0105 Proceeds of $1 = $.9895; $1000 -J-.989 5 = $1 1 0. 6 1, the face of the note. Solution. — For every $1 in the face of the note, the proceeds, by Case I, is $.9895 ; hence, there must be as many dollars in the face as tliis sum, $.9895, is contained times in the given proceeds, $1000; this gives $1010.61 for the face of the note. Hule. — Divide the given proceeds by the proceeds of $1 for Uie given time and rate; or, divide the given discount by the discount of $1 for the given time and rate; the quotient w equal to the face of the note. BANK DISCOUNT. 275 Examples for Practice. 1. Find the face of a 30 da. note, which yields $1650, when discounted at 1^^ a mo. $1677.68 2. The face of a 60 da. note, which, discounted at 6% per annum, will yield $800. $808.49 3. The face of a 90 da. note, bought for $22.75 less than its face, discounted at 7^. $1258.06 4. The face of a 4 mo. note, which, discounted at 1 ^ a month, yields $3375. $3519.29 5. The face of a 6 mo. note, which, discounted at 10^ per annum, yields $4850. $5109.75 6. The face of a 60 da. note, discounted at 2% a month, to pay $768.25 $801.93 7. The face of a 40 da. note, which, discounted at 8^, yields $2072.60 . $2092.60 8. The face of a 30 da. and 90 da. note, to net $1000 when discounted at 6%. $1005.53 at 30 da.; $1015.74 at 90 da. CASE III. 317. Given the rate of bank discount, to find the corresponding rate of interest. Problem. — What is the rate of interest when a 60 da. note is discounted at 2^ a month? OPERATION. Assume $ 1 as the face of the note. Then, 6 3 days = time. $4.20 = discount. $95.80= proceeds. $.00175 = interest of $ 1, at 1 ^, for the given time. And $95. 80X. 00175 = $. 1676 5, interest of proceeds, at 1 ^cy for the given time ; then, $4.20-5-$. 16765 = 25^/^, rate. 276 BAY'S lUOHER ARITHMETia Solution. — The discount of $100 for 63 days, at 2^ a month, is $4.20, and the proceeds is $95.80 The interest of $1 for the given time, at \(fo, is $.00175, and of $95.80 is $.16765 To find the rate, we proceed thus ; $4.20 -^ $.16765 = 25^^. Bule. — 1. Find ike discount and proceeds of $100 or $1 for the time the note runs, 2. Divide the discount by Hie interest of the proceeds^ at 1^^ for the same time; the quotierd is ike rale. Examples for Practice. What is the rate of interest : 1. When a 30 da. note is discounted at 1, 1^, 1\, 2% a month ? 12^11, 15^/ IS^^A^ 24,2^% per annum. 2. When a 60 da. note is discounted at 6, 8, 10^ per annum ? ^ iWa > ^A\> ^^A%^ P^^ annum. 3. When a 90 da. note is discounted at 2, 2^, 3% a month ? 25|ff , 32^^, 39f^^ per annum. 4. When a note running 1 yr. without grace, is discounted at 5, 6, 7, 8, 9, 10, 12^ ? 5^, G^f , 74|, 8^, 9||, 11^, 13^%. 5. My note, which will be legally due in 1 yr. 4 mo. 20 da., is discounted by a banker, at 8%: what rate of interest does he receive? 9^. Remark.— It may seem unnecessary to regard the time the note has to run, in determining the rate of interest ; but, a comparison of examples 1 and 3, shows that a 90 da. note, discounted at 2^ a month, yields a higher rate of interest than a 30 da. note of the same face, discounted at 2^ a month. The discount, at the same rate, on all notes of the same face, tm-ies as the time to run, and if in each case, it was referred to the mme prindpcUy the rate of interest would be the same ; but when the discount becomes Inrgerj the proceeds or principal to which it is referred, becomes smaller j and therefore the rate of interest corresponding to any rate of discount increases %oith the time the note has to run. Hence, the profit of the discounter is greater proportionally on long notes than on short ones, at the same rate. BANK DISCOUNT. 277 CASE IV. 318. Given the rate of interest, to find the corre- sponding rate of discount. Problem. — What is the rate of discount when a 60 day note yields 2% interest a month? OPEBATION. Assume $ 1 as the face of the note. Then, 6 3 days = the time ; $4.20= the interest; and $104.20 = the amount ; also, $.18235 = interest of amount, at 1 ^, for the given time; $4.20-i-$.18235 = 23^2V> rate per cent. Bemark. — Note the difference between this solution and that under the preceding case. In that, the interest was found on the proceeds ; in this, it is computed on the amount. Kule.— 1. Find the intered and Uie amount of $100 or $1 for the given time, 2. Divide the interest by the interest of the amount at 1^ for Hie given time; the quotient is (he corresponding rate of discount. Examples for Practice. What rates of discount : 1. On 30 da. notes, yield 10, 15, 20^ interest? 2. On 60 da. notes, yield 6, 8, 10% interest? 3. On 90 da. notes, yield 1, 2, 4% a month interest? lliWr. 22 W. 42114 %• 4. On notes running 1 yr. without grace, yield 5, 6, 7, 8, 9, 10^ interest? 4^, 5||, 6^, 7^, ^M, ^^fc 278 BAY'S HIGHER ARITHMETIC. * V. EXCHANGE. DEFINITIONS. 319. 1. Exchange is the method of paying a debt in a distant place by the transfer of a credit. Bemabk. — Th'is method of transacting business is adopted as a convenience: it avoids the danger and expense of sending the money itself. 2. The payment is effected by means of a B\J(l of Exchxtn^e, 3. A Bill of Exchange is a written order on a person or company in a distant place for the payment of money. Bemark. — The term includes both drafts and checks. 4. Exchange is of two kinds: Domestic, or Inland, and Foreign. 5. Domestic Exchange treats of drafts payable in the country where they are made. (See page 269 for form.) 6. Foreign Exchange treats of drafts made in one country and payable in another. ^ 7. A foreign bill of exchange is usually drawn in trip- licate, called a Set of Exchange; the different copies, termed respectively the first, second, and third of exchange, are then sent by different mails, that miscarriage or delay may be avoided. When one is paid, the others are void. The following is a common form: Foreign Bill op Exchange. 1. Exchange for New York, July 2, 1878. £1000. Thirty days after sight of this First of Exchange (Second and Third of same tenor and date unpaid), pay to the order of James S. RoUinjs One Thxmsand Pounds Sterling, value received, and charge to amount of To John Brown & Co., J. S. Chick. No. 1250. Liverpool, England. ) EXCHANGE. 279 8. The Bate of Exchange is a rate per cent of the face of the draft. 9. The Course of Exchange is the current price paid in one place for bills of exchange on another. 10. The Par of Exchange is the estimated value of the monetary unit of one country compared with that of another country, and is either Intrhmc or (JommerdaL 11. Intrinsic Par of Exchange is based on the com- parative weight and purity of coins. 12. Commercial Par of Exchange is based on com- mercial usage, or the price of coins in different countries. DOMESTIC EXCHANGE. 320. Where time is involved, problems in Domestic Exchange are solved in accordance with the principles of Bank Discount. Examples for Practice. 1. What is paid for $3805.40 sight exchange on Boston, at|% premium? $3824.43 2. What is the cost of a check on St. Louis for $1505.40, at \fo discount? $1501.64 3. What must be the face of a sight draft to cost $2000, at 1^ premium? $1987.58 4. What must be the face of a sight draft to cost $4681.25, at l\fo discount? $4740.51 5. What will a 30 day draft on New Orleans for $7216.85 cost, at 1% discount, interest 6% ? OPERATION. $1— $.00f = $.9962 5, rate of exchange. $.0055 = bank discount of $ 1 for 3 3 da. $.99075 = cost of exchange for $ 1. $7 216.85X.99075 = $7150.09 4+, Ana. 280 BAY'S HIGHER ARITHMETIC. Solution. — From the rate of exchange subtract the bank dis- count of $1 for 33 days, at 6^; the remainder is the cost of exchange for $1. Then multiply the face of the draft by the cost of exchange for $1, which gives $7150.094, the cost of the draft. 6. What will a 60 day draft on New York for $12692.50 cost, at \^o premium, interest 6%? $12654.42 7. What must be the face of an 18 day draft, costing $5264.15, at \% premium, interest 6^ ? $5256.27 8. What must be the face of a 21 day draft, costing $6836.75, at \% discount, interest 6%? $6925.04 9. A commission merchant in St. Louis sold 5560 lb. of bacon, at \\\ ct. a lb.; his commission is 2J^, and the course of exchange 98^^: what is the amount of the draft that he remits to his consignor? $632.91 10. A grain merchant in Toledo sold 11875 bu. of corn, at 40 cents a bushel; deducting 3% as commission, he pur- chased a 60 day draft with the proceeds, at 2% premium; required the face of the draft? $4564.14 11. Sold lumber to the amount of $20312.50, charging \\% commission, and remitted the proceeds to my consignor by draft; required the face of the draft, exchange \% dis- count? $20108.35 12. If a 45 day draft for $5500 costs $5538.50, find the rate of exchange. OPEKATION. The bank discount of $5500 for 48 days, at 6 ^=$44. Then $5538.50 + $44 — $5500=$82. 50 premium; $82.50 And ^_ -^^ = . 1 5 = 1 J ^, rate of exchange. 13. A father sent to his son, at school, a draft for $250, at 3 mo., interest 6%, paying $244.25 for it; find the rate of exchange. \% discount. 14. An agent owing his principal $1011.84, bought a draft with this sum and remitted it ; the principal received $992 : find the rate of exchange. 2^ premium. EXCHANGE. 281 FOREIGN EXCHANGE. 321. 1. In Foreign Exchange it is necessary to find the value of money in one country in terms of the monetary unit of another country. We reduce foreign coins by comparison with U. S. money, to find their value. Remark.- By an Act of Congress, March 3, 1873, the Director of the Mint is authorized to estimate annually the values of the standard coins, in circulation, of the various nations of the world. In compliance with this law, the Secretary of the Treasury issued the following estimate of values of foreign coins, January 1, 1884: COUNTRY. Austria Bolivia Brazil Chili Cuba Denmark, Norway, Sweden... Egypt France, Belgium, Switz Great Britain German Empire India.. Japan Mexico Netherlands Portugal Russia Tripoli Turkey MONETARY UNIT. Florin , Boliviano Milreis of 1000 reis... Peso Peso Crown Piaster Franc Pound sterling Mark Rupee of 16 annas Yen (silver) Dollar Florin Milreis of 1000 reis... Rouble of 100 copecks. Mahbub of 20 piasters. Piaster VALUE IN U. 8. MONEY. .806 .546 .912 .932 .268 .049 .193 4.866J .238 .383 .869 .875 .402 1.08 .645 .727 .044 Note. — The Drachma of Greece, the Lira of Italy, the Peseta (100 centimes) of Spain, and the Bolivar of Venezuela, are of the same value as the Franc. The Dollar , of the same value as our own, is the standard in the British Possessions, N. A., Liberia, and the Sandwich Islands. The Peso of Ecuador and the United States of Colombia, and the Sol of Peru, are the same in value as the Boliviano. H. A. 24. 282 HA Y' S HIGHER ARITHMETIC. 2. Bills of Exchange on England, Ireland, and Scotland are bought and sold without reference to the jxir of exchange, 3. It is customary in foreign exchange to deal in drafts on the various commercial centers. London exchange is the most common, and is received in almost all parts of the civilized world. Kemark. — London exchange is quoted in the newspapers in several grades : as, " Prime bankers' sterling bills," or bills upon banks of the highest standing ; " good bankers," next in rank ; " prime " and " good commercial," or bills on merchants, etc. The prices quoted depend upon the standing of the drawee and upon the demand for the several classes of bills. Usually a double quo- tation, one for 60 day bills, and the other for 3 day bills, is given for each class. Examples for Practice. 1. Find the cost of a bill of exchange on London, at 3 days sight, for £530 12s., exchange being quoted at $4.88 OPERATION. £530 12s. = £530.6 $4.88X530.6 = 12589.328 2. Find the cost of a bill of exchange on London, at sight, for £625 10s. lOd., when exchange is $4.87 pound sterling. $3046.39 3. What is the cost of a bill on Paris for 1485 francs, $1—5.15 francs. $288.35 4. What is the cost of a bill on Amsterdam for 4800 guilders, quoted at 41^ cents, brokerage J^ ? OPERATION. $.415X4800 = $1992. $1992X.00i = $9.9 6, brokerage. $1992 + $9.96 = $2001.9 6, cost of bill. 5. What will a draft in St. Petersburg on New York for $5000 cost, if a rouble be worth $.74? 6756§f roubles. ARBITRATION OF EXCHANGE. 283 6. What will a draft on Charleston for $4500 cost at Eio Janeiro (milreis = $.54), discount 2% ? 8166. 66f milreis. 7. A gentleman sold a 60 day draft, which was drawn on Amsterdam for 1000 guilders; he discounted it at 6%, and brokerage was \%\ what did he get for it, a guilder being valued at 40^ cents? $397.77 -\- AEBITRATION OF EXCHANGE. DEFINITIONS. 322. 1. Owing to the constant variation of exchange, it is sometimes advantageous to draw through an intermediate point, or points, in place of drawing directly. This is called Circular Exchange. 2. The process of finding the proportional exchange be- tween two places by means of Circular Exchange, is called the Arbitration of Exchange. 3. Simple Arbitration is finding the proportional ex- change when there is but one intermediate point. 4. Compound Arbitration is finding the proportional exchange through two or more intermediate points. Problem. — ^A merchant wishes to remit $2240 to Lisbon : is it more profitable to buy a bill directly on Lisbon, at 1 milreis = $1.10; or to remit through London and Paris, at £1 = $4.88 — - 25.20 francs, to Lisbon, 1 milreis = 5.5 francs ? OPERATION. $2240-5-1.10 = 2036.364 milreis, direct exchange. ( ) milreis =$ 2 2 4 0. $4.88 = 25.20 francs. 5 . 5 francs = 1 milreis. 2240X25.20X1 = 2103.129 milreis, circular exchange. 4.88X5.5 Hence, the gain is 6 6 .7 6 5 milreis in buying circular exchange. 284 BA Y'S HIGHER ARITHMETIC. Solution.— By direct exchange $2240 will buy 2036.364 milrels. By circular exchange we have a set of equations, all of whose members are known except one. Multiplying the right-hand members together for a dividend, and dividing the product by the product of the left-hand members, the quotient is the required mem- ber, which, in this case, is 2103.129 milreis. The diflference in favor of the circular method is 66.765 milreis. Problem. — A merchant in Memphis wishes to remit $8400 to New York; exchange on Chicago \&\\% premium, between Chicago and Detroit 1% discount, and between Detroit and New York \% discount: what is the value of the draft in New York, if sent through Chicago and Detroit ? OPERATION. ($ ) N. Y. = $ 8 4 0, Memphis. $ 1 . 1 i Mem. = %\ Chicago. $.9 9 Chi. = $1, Detroit. $.99i Det. = $l, N. Y. 8400X1X1X1 =$8401.46, .In^ 1.015X.99X.995 Solution.— By the problem, $1,015 in Memphis = $!• in Chicago ; $.99 in Chicago = $1 in Detroit; and $.995 in Detroit = $1 in New York. Hence, multiplying and dividing, as in the preceding problem, we obtain the answer, $8401.46 Biile 1. — 1. Form a series of equationSy expressing the con- ditions of the question; i/te first containing the quantity given equal to an unknown number of the qiumtity requiredy and all arranged in suck a way thai the right-hand qvmitity of each equation and tlie lefb-lw/nd quantity in the equation next foUow- ingy shjJl he of the same denominationy and also the righthand quantity of die la^t and the left-hand quantity of the first, 2. Cancel equal factors on opposite sides y and divide Hie product of Hie quantities in the column which is complete by t/ie product of those in the other column. The quotient will be tlie quantity required. ARBITRATION OF EXCHANGE. 285 Rule 2. — Reduce Hie given quardity to iJie denomination vrith which it is compared; reduce this restdt to the denominor tion with which it is compared; and so on, until the required denomination is reached^ indicating the operations by imriting as midtipliers the proper unit values. The compound fraction thus obtained, reduced to its simplest form, uiU be the amount of the required denomination. Notes. — 1. The first is sometimes called the chain rule, because each equation and the one following, as well as the last and first, are connected as in a chain, having the right-hand quantity in one of the same denomination as the left-hand quantity of the next. 2. In applying this rule to exchange, if any currency is at a premium or a discount, its unit value in the currency with which it is compared, must be multiplied by such a mixed number, or decimal, as will increase or diminish it accordingly. 3. If commission or brokerage is charged at any point of the circuit, for effecting the exchange on the next point, the sum to be transmitted must be diminished accordingly, by multiplying it by the proper number of decimal hundredths. Examples for Practice. 1. A, of Galveston, has $6000 to pay in New York. The direct exchange is ^^ premium; but exchange on New Orleans is ^% premium, and from New Orleans to New York is J% discount. By circular exchange, how much will pay his debt, and what is his gain ? $5999.96; $30.04 gain. 2. A merchant of St. Louis wishes to remit $7165.80 to Baltimore. Exchange on Baltimore is ^% premium; but, on New Orleans it is ^% premium ; from New Orleans to Havana, |% discount; from Havana to Baltimore, \^ discount. What will be the value in Baltimore by each method, and how much better is the circular ? Direct, $7147.93; cir., $7183.77; gain, $36.84 286 BAY'S HIOHEE ARITHMETIC. 3. A Louisville merchant has $10000 due him in Charles- ton. Exchange on Charleston is \% premium. Instead of drawing directly, he advises his debtor to remit to his agent in New York at ^% premium, on whom he immediately draws at 12 da., and sells the bill at f % premium, interest off at 6^. What does he realize, and what gain over the direct exchange? Realizes «10029.94; gain, $17.44 Suggestion. — Interest at 6^ per annum is \(fo a month, or J^ for 15 days, which diminishes the rate of premium to \<fo* 4. A Cincinnati manufacturer receives, April 18th, an account of sales from New Orleans; net proceeds $5284.67, due June Yy. He advises his agent to discount the debt at 6%, and invest the proceeds in a 7 day bill on New York, in- terest off at 6%, at \% discount, and remit it to Cincinnati. The agent does this, April 26. The bill reaches Cincinnati May 3, and is sold 2it \% premium. What is the proceeds, and how much greater than if a bill had been drawn May 3, on New Orleans, due June 7, sold at -J^^ premium, and interest off at 6% ? Proceeds, $5296.10; gain, $35.65 5. A merchant in Louisville wishes to pay $10000, which he owes in Berlin. He can buy a bill of exchange in Louis- ville on Berlin at the rate of $.96 for 4 reichmarks; or he is offered a circular bill through London and Paris, brokerage at each. place \%, at the following rates: £1=$4.90 = 25.38 francs, and 5 francs = 4 reichmarks. What is the difference in the cost? $81.35 VI. EQUATION OP PAYMENTS. DEFINITIONS. 323. 1. Equation of Payments is the process of find- ing ihe time when two or more debts, due at different times, may be paid without loss to either debtor or creditor. EQ UA TION OF PA YMENTS. 287 2. The time of payment is called the Equated Time. 3. The Term of Credit is the time to elapse before a debt is due. 4. The Average Term of Credit is the time to elapse before several debts, due at different dates, may be paid at once. 5. To Average a^ Account is to find the equitable time of payment of the balance. 6. Settling or Closing an Account is finding how much is due between the parties at a particular time. It is some- times called Striking a Balance. 7. A Focal Date is a date from which we begin to reckon in averaging an account. 324. Equation of Payments is based upon the following principles : Principles. — 1. Tluit any sum of money paid before it is duCy is balanced by keeping an equal sum of money for an equal time after it is due, 2. Thai the interest is the measure for the use of any sum of money for any time, Kemare. — The first statement is not strictly correct, although it has the sanction of able writers, and is very generally accepted. For short periods, it will make no appreciable difference. Problem. — A buys an invoice of $250 on 3 mo. credit; $800 on 2 mo.; and $1000 on 4 mo.; and gives his note for the whole amount ; how long should the note run ? OPEBATION. Debts. Terms. Equivalents. 250 X 3 = 750 800 X 2 = 1600 1000 X 4 = 4000 2050 6350 ♦•* 6350 _ o A \ time to run = mo. = 3 mo. 3 da., Ans. 2050 ' 288 HA y'/S' HIGHER ARITHMETia Solution. — The use of $1 for 6350 months is balanced by the use of $2050 for 3 months and 3 days. Or, the interest of $1 for 6350 months equals the interest of $2050, 3 months 3 days, at the same rate. Problem. — ^I buy of a wholesale dealer, at 3 mo. credit, as follows: Jan. 7, an invoice of $600; Jan. 11, $240; Jan. 13, $400; Jan. 20, $320; Jan. 28, $1200; I give a note at 3 mo. for the whole amount : when is it dated ? OPERATION. $600X = $1 for days. $240X 4 = $1 " 960 '" $400X 6=$1 " 2400 " $320X13 = $1 " 4160 " $1200 X 2 1 = $ 1 *' 25200 " $2760 $1 " 32720 " Whence 327 20-f- 27 60== 11.8+ days; hence the time is 12 days, nearly, after Jan. 7 ; that is, Jan. 1 9. Solution. — Start at the date of the first purchase, and proceed as in the preceding solution. 325. When the terms of credit begin at the same date, we have the following rule : Rule. — Mtdtiply eadi debt by its term of credit, and divide tJw mm of the products by Hie sum of the debts; the quotierU tvill be the equaled time. 326. If the account has credits as well as debits, it is called a compound eqmdion, and may be averaged on the same principle as the simple equation. The following problem will illustrate the difference between simple and compound equations, as it involves both debits and credits: Problem. — What is the balance in the following account, and when is it due? EQUATION OF PAYMEm-S. A in aa^t «rUh B. SoLirnoN. — Start with March 1, the earliest day upon which a payment is made or falls due. Find tlie producL'', botli on the I>r. and Cr. side of the account, using the dates when the itemii are due, as cash. The balance due, $625, ie found by subtracting the amount of the credits from the amount of the debits. The balance of products, 5S050, divided by the balance of items, 629, determines the mean time to be 93 days after March 1 ^ June 2. Bule. — 1. FiTtd when each item is due, and take the earlUel or laie^ date ae the focal date. 2. Find the difference betioeen tiie focal date and the re- maining dates, and multiply each item, hj its corre^mnding difference. 3. Find the difference bdween the Dr. and Or. Hems, and dUo between the Jh. and Or. produda, and divide tiie difference of the prodvela by the difference of the items. Add t!te quotient to the focal date '^ it be the first date, or subtract if it be the last date; the restdt in either case vnll be the equated tinie, 4. Jj' the two balances be on oppodte sides of the aecowit, the quotient mitgt be eubiraded from the focal date if the first, or added if it be the last date. Remark. — The solutions thus far have been by the prodml method. The same reiult will be oblained by dividing the interest of the product balance for 1 day by the inlerest of the item balance for 1 day. This depends upon the principle that if both dividend and divisor are multiplied or divided by the same number the quotient is nnchBn){ed. 290 RAY'S BJOHER ABITHMETia Examples for Practice. What is the mean time of the following invoices: 1. A to B, Dk. 1877. May June »t July Aug; Dol. Ct. When due Days after 15 1 10 20 1 15 To Invoice at ^ months. Sum total. 800 700 900 600 500 1000 * - 4500 Products. Oct. 30, 1877. Bemabk. — Start with Sept. 15, the day the first debt is due. 2. Dr. E in accH current with F. Cb. Feb. Mar. A^r. 4 20 1 5 To Invoice 3 mo. I< «« 9 <( It tt Q 4t It <l O (t |550 260 150 325 — P'.'od. May June July 8 28 3 1 By cash, '*^ remit, June 5, " note, 1 munth, >« "2 " tl50 420 .340 170 — Prod, 3. E owes F $205, due Feb. 26. H, Wright to Mason & Giles, Db. 1876. Feb. it Mar: Apr. May June 1 20 10 8 10 15 " 3 " 3 " cash, " 3 " 3 - Dol. Ct. WTien due Days after AprUS. lOnths. 900 «( 700 •( 600 500 t( 900 u 400 Sum total. 4000 Prodtiets. Due June 13, 1876, Bemark. — Start with April 8, the time the first payment is due. 4. Dr. . A in acc^t current with B. Cb. Mar. Apr. June May 19 20 15 10 To Invoice cash, tt «t «< <« t« <« U <« M $901) 800 700 600 — Prod. Feb. Mar. June July 20 5 20 10 By cash, " remit, Mar. 15, '• cash, 4t •• MOO 300 2on 500 — Prod, A owes B a balance of $1600, due April 23. J EQ UA TION OF PA YMENTS, 291 Kemark. — Start with Feb. 20, and date the remittance, March 15, the day it is received. 5. Dr. C in qjccH current with D, Cr. Jan. Feb. (I Apr. 4 3 15 2 To invoice 2 mo. 1 " 2 '• cash, << «250 140 450 100 Prod. Mar. Apr. May June 10 21 4 20 16 By cash, " note, 2 mo. " remit., May 25, " accep. 16 da. sight $350 200 240 120 500 Prod. C is Cr. $470, due Aug. 12. 6. I owe $912, due Oct. 16, and $500, due Dec. 20. K I pay the first, Oct. 1, 15 days before due, when should I pay the last? Jan. 16, next, 27 days after due. 7. Oct. 3, I had two accounts, amounting to $375, one due Dec. 6, and the other Nov. 6, but equated for Nov. 16: what was each in amount? $250, $125. 8. A owes $840, due Oct. 3 ; he pays $400, July 1 ; $200, Aug. 1 ; when will the balance be due? April 30, next year. 9. I owe $3200, Oct. 25 ; I pay $400, Sept. 15 ; $800, Sept. 30: when should the balance be paid? Nov. 12. 10. An account of $2500 is due Sept. 16 ; $500 are paid Aug. 1 ; $500, Aug. 11 ; $500, Aug. 21 : when will the balance be due? Nov. 9. 11. Exchange the five following notes for six others, each for the same amount, and payable at equal intervals: one of $1200, due in 41 days; one of $1500, due in 72 days; one of $2050, due in 80 days; one of $1320, due in 110 days; one of $1730, due in 125 days; total, $7800. The notes are $1300 each, and run 25, 50, 75, 100, 125, 150 days respectively. 12. Burt owed in two accounts $487 ; neither was to draw interest till after due, — one standing a year, arid the othei two years. He paid both in 1 yr. 5 mo., finding the true dis- count of the second, at 6%, exactly equal to the interest of the first : what difference of time would the common rule have made? 3 days* 292 RAY'S HIGHER ARITHMETIC. yil. SETTLEMENT OF ACCOUNTS. DEFINITIONS. 327. 1. An Account is a written statement of debit and credit transactions between two parties, giving the date, quantity, and price of each item. 2. An Account Current is a written statement of the business transactions between two parties, for a definite time. 328. In settling an account, the parties may wish to find: (1). When the balance is equitably due. (2); Wliot mm J at a given time, should be paid to balance the accotint The first process is Averaging the Account (Art. 324) ;* the second is Finding the Cask Balance. Dr. Henry Armor in accH with City Bank. Cr. 1875. Dol. a. 1874. Dol. C*. Jan. 5 To check, 300 Dec. 31 By balance old account, 500 it 20 .t tt 600 1875. 4( 21 ti « 100 Jan. 7 " casb. 50 <t 27 tt tt 850 , ti 15 tt tt ' 400 (t 31 ** . *' 75 tt 24 tt tt 1000 Explanation. — The two parties to this account are Henry Armor and City Bank. The left-hand, or Dr. side, shows, with their dates, the sums paid by the bank on the checks of Henry Armor, for which he is their debtor. The right-hand, or Cr. side, shows, with their dates, the sums deposited in the bank by Henry Armor, for which he is their creditor. 329. Generally, in an account current, each item draws interest from its date to the day of settlement. SETTLEMENT OF ACCOUNTS. 293 Problem. — Find the interest due, and balance this ac- count: F. H, WUlis in accH with E. S, Kennedy. Dr. Cte 1879. Jan. 6 t» 13 ti m ti 2-. <4 28 To check. It Dol. a. 1878. 170 Dec. 31 480 1879. 96 Jan. 7 500 t. 20 50 it 30 By balance forward, " cash deposit, tt It K <t •« •« ct. Interest to Jan. 31, at 6 per cent. Db. $170, 6fcy 25 da 480, " 18 96, " 15 5 0, " 6 50, " 3 OPERATION. Cr. it u ti u $1290 = $.708 = 1.44 = .24 = .50 r=- .025 $2,913 $325, 6%, 31 da =$1,679 = 3.20 .32 .04 8 0*0, 17 5, 240, (t u u 24 11 1 (( u u $1540 $1296 $244 $2.913 $2.33 /. Interest due = $2. 33; and cash balance due Willis is $244-f $2.33= $246.33, Ans. Solution.— The interest is found on each item from its date to the day of settlement; then the sum of the items and the sum of the interests are found on each side. The diflference between the sums of the interests is equal to the interest due, which, added to the diflference between the sums of the items, gives the balance due. Bule.— 1. Find the number of days to dapse between (h^ date when each item is due and die date of settlement 2. Find the mm of the items on Hie Dr. side, and then add to each item its interest^ if tJie item is dii^ before the date of settlement, or »id)tract it if due after the date of settlement; do the same loith Or. side. 3. The difference between the anumnts on Hie two sides of the aecouni is Hie cash balance. 294 BAY'S HIGHER ARITHMETIC. 1. Find the equated time and cash balance of the follow- ing account, July 7, 1876 ; also April 30, 1876, interest ^ per annum : Henry Hammond. Db. Or. 1876. Jan. 4 To Mdse, at 3 mo. 1900.10 Feb. 1 ii ti 44 ^ 44 400.00 44 IS 4t 44 44 O 44 700.50 Mar. 7 44 44 (4 A 44 600.40 April May 8 •• Cash, 500.20 10 '♦ Mdse, at 30 da. 400.00 June 1 ' 1 mo. 100.60 Equated time. May 15, 1876. Cash balance July 7, 1876, $3633.62 Cash balance April 30, 1876, $3592.80 Kemarks. — 1. When, in forming the several products, the cents are. 50 or less, reject them; more than 50, increase the dollars bj 1. 2. Tlie cash balance is the sum that Henry Hammond will be required to pay in settling his account in full at any given date. 2. Find the equated time and cash balance of the follow- ing account, Oct. 4, 1876, money being worth 10^ per annum : William Smith. Dr. Cr. 1876. Dol. Ct. 1876. Dol. a. Jan. 1 To Mdse, 800 00 Jan. 10 By Cash, 400 00 41 16 at 30 da. 180 30 44 28 44 44 200 00 Feb. 14 44 44 44 QQ 44 40() 60 Feb. 15 " Bills Rec. at 60 da. 180 30 Mar. 25 4. (1 500 00 4. 28 '• Cash , 100 00 April 1 " Cash, 800 00 Mar. 30 " Bills Rec. at 90 da. 450 00 May 7 " Mdse, 600 00 April May 14 " Cash, 400 60 «i 21 at 60 da. 700 00 1 44 «4 500 00 June 10 4. 44 200 00 .4 15 '• Bills Rec. at 1 mo. 680 OO 44 15 " 90 da. 20(M) 00 June 16 ♦' Cash, 300 00 July 12 ** *fc 500 00 July 19 44 44 700 ft) Aug. 4 •* 2 mo. 1000 10 Aug. 10 " Bills Rec. at 20 da. 200 00 Sept. 1 '* Cash, 150 00 Oct. 3 4* ftk 1100 00 Equated time, June 18, 1876; cash balance, $2389.70 Kemark. — All written obligations, of whatever form, for which a certain amount is to be received, "are called BiUs Receivable. SETTLEMENT OF ACCOUNTS. 295 3. Find the equated time and the cash balance of account, March 31, 1876, interest 10% per annum. Dr. Gewge Cummings, Cr. 1876. Dol. a. 1875. Dol J Ct. Jan. 2 To Cash, 800 00 Dec. 1876. 1 By Mdse, at 1 mo. ' 583 00 « 21 (i K 194 00 Feb. » .t tt 14 •( 40 00 Mar. 4 41 t( 150 00 Mar. 30 «l 41 tt 4t 130 00 Equated time Feb. 11, 1876; cash balance, $110.48 Find the interest due, and balance the following ac- counts : A, L. Morris in acc'l with T, J. Fisher & Co. 4. Dr. Cr. 1871. Dol. Ct. 1870. Dol. a. Jan. 13 To check, 350 Dec. 31 By bal. flrom old acc't. 813 64 44 22 44 44 275 1871. Feb. 25 It It 100 Feb. 4 " cash. 120 May 1 it 44 400 Mar. 17 41 It 500 June 23 • 4 44 108 25 May 31 44 U 84 50 Interest to June 30, at 6 per cent. Int. due Morris, $13.34; bal. due Morris, S298.23 6. Dr. Wm. White in accH with Beach & Beiiy. Cr. 1875. Dot. Ct. 1875. Dol. Ct. July O «« To check. 212 50 June 80 By bal. from old acc't, 1102 50 4t "^ 20 66 July 6 " cash deposit, 50 Aug. 7 2:» 44 15 44 |4 44 95 t» 25 300 Aug. 9 44 t4 tt 168 75 Sept. 5 110 Sept. 18 44 4t 44 32 .t 11 46 40 Oct. 3 44 44 4t 79 90 44 27 454 25 Interest to Oct. 12, at 10 per cent. Int. due White, $19.68; bal. due White, $123.68 296 JIAY'S HIGHER ARITHMETIC. ACCOUNT SALES. 3dO. 1. An Account Sales is a written statement made by an agent or consignee to his employer or consignor, of the quantity and price of goods sold, the charges, and the net proceeds. 2. Quaranty is a charge made to secure the owner against loss when the goods are sold on credit. 3. Storage is a charge made for keeping goods, and is usu- ally reckoned by the week or month on each piece or article. 4. In Account Sales the charges for freight, commission, etc., are the Debits, and the proceeds of sales are the Credits; the Net Proceeds is the difference between the sums of the credits and debits. 331. Account Sales are averaged by the following rule : Bnle. — 1. Average the sales alone; this result wUl he the date to be given to the commission and guaranty, 2. Make the sales the Or, side, and the charges the Dr. side, and find the eguaied time for paying the net proceeds. 1. Find the equated time of paying the proceeds on the following account of Charles Maynard: Charles Maynard^s Consignment, 1876. Aug. 12 ti 14 « 24 tt 29 (( 3i» « 31 1876. Aug. 10 «i 31 ti 31 (1 31 By J. Barnes, at 10 da.^ " Cash *' Bills Receivable, at 30 da. " Cash •• (4eorge Hand , ** Bills Beceivable. at 20 da., <4 (t (( t< ti CHARGES. To Cash paid, Freight .-....$ 75.00 " " •' Storage 10.00 " " " Insurance >6 per cent 10.26 •* " " Commission 2^ per cent „ 205.23 Net proceeds due Maynard. Dol, 50 60 800 00 850 00 210 00 4900 00 1400 00 8210 800 7910 Ct. 60 51 09 Equated time Sept. 4, 1876. SETTLEMENT OF ACCOUNTS. 297 2. Make an account sales, and find the net proceeds and the time the balance is due in the following: William Thomas sold on account of B. F. Jonas 2000 bu. wheat July 8, 1876, for $2112.50; July 11, 300 bu. wheat, and took a 20 day note for $362.50 ; paid freight July 6, 1876, $150.00; July 11, storage, ?6; drayage, $5; insurance, $4; commission, at 2^%, $61.87; loss and gain for his net gain, $11.57 ^ Net proceeds, $2236.56; due July 12, 1876. 3. Md^e. Co. "5." 1870. July 18 tt '24 it 30 1876. July 15 15 30 30 30 30 By Note, at 20 da ti >t «i 25 '* " Cash J!!!"!!!!!!!!i!!!!Z!!!!""!!!l!""!!!!iZ!!!!"Z!!!!;!!! CHARGES. To Cash paid, Freight f 33.00 " *' " Drayage 4.00 •* " " Inaurunce 1.50 •• *• *• Storage 3.00 " " " Commission, at 2'4 per cent 5.29 C. V. Oanies's uet proceeds 62.73 1109.52 Less our \i net loss 3.93 fl05.59 Find C. V. Cames's net proceeds if paid Jan. 1, 1877, money being worth 10^ per annum ; also the equated time. Equated time Aug. 19, 1876; net proceeds, $65.03 STORAGE AC(X)UNTS. 332. Storage Accounts are similar to bank accounts, one side showing how many barrels, packages, etc., are received, and at what times; the other side showing how many have been delivered, and at what times. Storage is generally charged at so much per month of 30 days on each barrel, package, etc. 298 BAY'S HIGHER ARITHMETIC, 1. Storage to Jan. 31, at 5 ct. a bbl. per month. RECEIVED. DELIVERED. 1876. EU. Balance on hand. Day$. Products. 1876. Bbl. January 2 5 7 10 14 17 20 24 28 200 150 30 120 80 150 75 60 200 January 10 13 17 20 25 27 30 31 110 90 20 115 140 72 100 Storage, $19.25; bbl. on hand, 418. 2. Storage to Feb. 20, at 5 ct a bbl. per month. RECEn^ED. DELIVERED. Bbl. Balance on hand. Bayi, Products. Bbl. January 31 418 February 5 100 February 4 9 12 16 250 120 lUO 30 10 12 14 18 20 80 220 140 90 288 Storage, $15.85; bbl. on hand, 0. VIII. COMPOUND' INTEREST. DEFINITIONS. 333. 1. Compound Interest is interest computed both upon the principal and upon each accrued interest as additional principal. 2. Annual Interest is the gain of a principal whose yearly interests have become debts at simple interest; but, distinct from this, Com])Oiuid Interest is Hie whole gain of a prhicipal, increased ai the end of each interval by all the interest drawi during that interval. 3. The final amount in Compound Interest is called tho compound amount. COMPOUND INTERJSST. 299 Example.— Let the principal be $1000, the rate per cent 6. The first year's interest is $60. If this be added as a new debt, the prin- cipal will become $1060, and the second year's interest $63.60 ; in like manner, the third principal is $1123.60, and a third year's interest $67,416; then, the whole amount is $1191.016, and the whole gain, $191,016 Bemark. — If the above debt be, at the same rate, on annvxd intaest (Art. 304), the whole amount will be $1190.80, and the whole gain $190.80; the difference is the interest of $3.60 (an interest upon an interest debt) for one year, $.216 Compound Interest has four cases. CASE I. 334. Given the principal, rate, and time, to find the compound interest and amount. Problem. — Find the compound amount of $1000, in 4 years, at 2^ per annum Solution. — Multiplying the principal, $1000, by 1.02, the num- ber expressing the amount of $1 for a year, we have the first year's amount, $1020. Continuing the use of the factor 1.02 until the fourth product is obtained, we have for the required amount, $1082.43216 The same numerical result would have been obtained by taking 1.02 four times as a factor, and multiplying the product by 1000. Remark. — Compound Interest may be payable semi-annually or quarterly, and in such cases the computation is made by a multi- [>lication similar to the last. Example. — Let the debt be $1000, and let the interest be com- pounded at 2^0 quarterly. In one year there are four intervals, and, as seen in the last process, the year's amount is $1082.43216 The real gain on each dollar is $.0824+, or, a fraction over S^jf(. According to the usual form of statement, this debt 1s compounding 300 JRA Y'S HIGHER ARITHMETIC. " at 8^ per annum^ payable qiMiierly.^^ But this must not be under- stood as an exact statement of the real gain ; for, when the quarterly rate is 2^, the annual rate exceeds 8^, and when the annual rate is exactly 8^, the quarterly rate is 1.943^, nearly. Bemark. — To ascertain the true rate for a smaller interval when the yearly rate is given, requires to separate the yearly multiplier into equal factors. Thus the true half-year rate, when the annual rate is 21^, is found by separating 1.21 into two equal factors, 1.10 X 1.10; and the quarterly rate, when the annual is 8^, is found by separating 1.08 into four factors, each nearly 1.01943 It is true th^t this process is rarely demanded, and that it is very tedious when 'the intervals are small; but, in a proj>er place, it will be a useful exercise. (See Art. 388). BiQe. — Find the amount of Hie principal for the firtt in- tervaly at the rate for that intervaly and in like manner treat the whole debt, at Hie end of each interval, as a principal at maple interest through the following interval or part of an interval; (he result will be the compound amount. To find the compound interest, deduct the orig^inal debt from the compound amount. Examples for Practice. 1. Find the compound amount and interest of S3850, for 4 yr. 7 mo. 16 da., at 5%, payable annually. 64826.59, and $976.59 2. The compound interest of $13062.50, for 1 yr. 10 mo. 12 da., at 8%, payable quarterly. $2082.25 3. The compound amount of SIOOO, for 3 yr., at 10%, payable ^semi-annually. $1340. 10 4. What sum, at simple interest, 6^, for 2 jr. d mo. 27 da., amounts to the same as $2000, at compound interest, for the same time and rate, payable semi-annually? $2016.03 5. What is the difference between the annual and the compound interest of $5000 in 6 years, 6% per annum? $22,596 L COMPOUND INTEEEST. 301 6. Required the amount of $1000, at compound interest, 21%, for 2 yr. 6 mo. $1617.83 Behabks. — 1. In obtaining the answer to the last problem, the compound amount at the end of the second full interval is treated as a sum at simple interest for 6 months. But, calculated at a true half-year rate, the amount is only $1610.51, and this sum continued at the same true rate for the remaining half year will amount to the same sum which the debt would have reached in a full interval ; for, $1464.10 X 1.21 = $1771.561 ; and $1464.10, for 2 intervals, at 10% comp. int. = $1464.10 X 1.10 X 1.10 = $1771.561 2. Let the student carefully note that the interest drawn during a year may be considered as the mm of interests compounded through smaller intervals, at smaller rates. Thus, 6^ a year may be regarded as the sum of interest compounded through quarterly intervals at 1.467^, through monthly intervals at .487^, or daily at .016^, approximately. Suppose 7 months have passed since interest began. The year may be taken as a period of 12 intervals, and the interest as having been compounded through 7 of them, at .487^. If the amount then reached be continued at compound interest through the other 5 intervals, the amount will be the same as that of the debt continued to the end of the year at the full rate. 3. In this view, the statement may be made, general, that the worth of the debt at any point in a year, is the pHndpalj whichy compounding ai a true partial rate for the remaining fraction of a year^ wiU amount to the same sum as the debt continued through that year at the annual rate. This is in strict accordance with the results obtained by Algebra, but in the common calculations of Arithmetic, the interest is added to the debt, and the rate divided, only according to the statements *' payable annually," "payable quarterly," etc. CALCULATION BY TABLES. 836. When the intervals are many, the actual multipli- cations become laborious; and, therefore, tables of compound interest have been prepared to shorten the work. 302 RAY'S HIGHER ARITHMETIC. Amownl of $1 at Compound Interest in any nujnber of yean, not exceeding fifty-Jive. Yrs. 2 per cent. 2H per cent. 3 per cent. 3>6 per cent. 4 per cent. 4^ per cent. 1 1.0200 0000 1.0250 0000 1.0300 0000 1.0350 0000 1.0400 0000 1.0450 0000 2 1.0«M (XMX) 1.0506 2500 1.0609 0000 1.0712 2500 1.0816 0000 1.0920 2500 3 1.U612 0800 1.0768 9062 1.0927 27rt» 1.1087 1787 1.1248 6400 1.1411 6612 4 1.0824 ;{216 1.1U88 1289 1.1255 0881 1.1475 2300 1.1698 .5856 1.1925 i860 5 l.KMO 8080 1.1314 0821 1.1692 7407 1.1876 8631 1.2166 5-290 1.^61 8194 6 1.1261 6242 1.1596 9342 1.1940 5230 L'lm 5533 1.2653 1902 1.3022 6012 7 l.HSfl 8W7 l.lb86 8575 1.2298 7387 1. -27-22 7926 1.3159 3178 1.36H8 618:< 8 M7I6 5938 1.2184 0290 1.2867 70(« 1.3168 0»)4 1.3685 6905 1.4-221 0061 9 1.1950 9257 1.2488 6297 1.3047 7318 1.3628 9735 1.4'2:« 11»1 1.4860 9514 10 1.21t>9 9442 1.2800 8454 1.3439 1638 1.4105 9876 1.4802 4428 1,5529 6942 11 1.24.^ 7431 1.3120 8666 1.3842 3;«7 1.4599 6972 1.5394 5406 1.6228 5305 12 1.2682 4179 1.3448 8882 1.4257 6089 1.5110 6866 1.6010 3222 1.6958 8143 13 1.2936 0663 1.3785 1104 1.4685 a371 1.56.39 56JI6 1.6650 7351 1.7721 9610 14 1.3194 7876 1.4129 7382 1.5125 8972 1.6186 9452 1.7316 7645 1.8519 4492 15 1.3458 6834 1.4482 9817 1.5579 6742 1.6753 4883 1.8009 4351 1.9352 8244 16 1.3727 8570 1.4845 0562 1.6047 0644 l.7:»9 mn 1.8729 8125 2.0223 7015 17 1.4002 4142 1.5216 18-26 1.6528 4763 1.7946 7555 1.9479 «1050 2.1133 7681 18 1.4*282 4625 1.5596 5»72 1.7024 3306 1.8574 8920 2.0258 1652 2.2084 7877 19 1.4568 1117 1.5986 5019 1.7535 0605 1.9-225 01.32 2.1068 4918 2.3078 6031 20 1.4859 4740 1.6386 1644 1.8061 11-23 1.9897 8886 2.1911 2:il4 2.4117 1402 21 1.5156 6&34 1.6795 8185 1.8602 9457 2.0591 3147 2.2787 6807 2.5202 4116 22 1.5459 7967 1.7215 7140 1.9161 0341 2.1315 1158 2.3699 1879 2.6336 5201 23 1.5768 9926 1.7646 1068 1.9735 8651 2.2061 1448 2.4647 1555 2.7521 6635 24 1.6084 3725 1.8087 2595 2.03-27 9411 2.2833 2849 2.56;*3 W17 2.8760 1383 25 1.6406 0599 1.8539 4410 2.0937 7793 2.3632 4498 2.6658 36:» 3.0054 3446 26 1.6734 1811 1.9002 9270 2.1565 9127 2.4459 5856 2.7724 6979 8.1406 7901 27 1.7068 8B48 1.9478 0UO2 -2.2?12 8901 2.5315 6711 2.88:«6858 3.2820 0956 28 1.7410 '2421 1.9964 9502 2.2b. 9 2768 2.6201 7196 2.9987 0832 3.4296 9999 29 1.77)S 4469 2.(M64 07;» •2.3565 6551 2.7118 7798 3.1 lb6 5145 8.5840 3649 30 J.811! fil58 2.0975 6758 2.4272 6247 2.80B7 9370 8.2433 9751 8.7453 1813 31 1.8475 88S2 2.1500 0677 2.5000 8035 2.9050 3148 8.3731 3341 8.9138 5745 32 1.8845 4059 2.'2U37 5694 2.5750 8276 3.0067 0759 3.5080 5875 4.0899 8101 33 1.9222 3140 2.2588 5086 2.65-23 35-24 3.1119 4235 3.6483 8110 4.2740 3018 34 1.9606 7603 2.3153 2213 2.7319 0530 3.2208 ma 8.7943 16:{4 4.4663 6154 35 1.9998 8956 2.3732 0519 2.8138 6245 3.3.3a5 9045 3.9460 8899 4.6673 4781 36 2.0898 8734 2.4.325 3532 2.8982 7833 8.4502 6611 4.1039 3255 4.8773 7846 37 2.0SU6 85«)9 2.49:» 4870 2.9852 2668 3.5710 2543 4.2680 8986 5.0968 6049 38 2.1-222 9879 2.5556 8242 3.0747 8348 3.6960 1132 4.4.388 1:H5 5.:K62 1921 39 2.1647 4477 2.6195 7448 3.1670 2898 3.8253 7171 4.6163 6599 5.5658 991(8 40 2.2080 3966 2.6850 6384 3.-26-20 3779 3.9592 5972 4.8010 2063 5.8163 6454 41 2.2522 0046 2.7521 9(M3 3.3598 9893 4.0978 3381 4.9930 6145 6.0781 0094 42 2.2972 4447 2.8-209 9520 3.4606 9589 4.-2412 5799 5.1927 8391 6.3516 1-548 43 2.3431 89:^6 2,8915 2008 3.5645 1677 4.:te97 0-202 5.40(« 95-27 6.6374 3818 44 2.3900 5:n4 2.9638 0808 3.6714 5-227 4.54:« 4160 5.6165 1506 6.9)61 2290 45 2.4378 5421 3.0379 0328 3.7815 9584 4.70-23 5655 5.8411 7568 7.2482 4843 46 2.4866 J 129 3.1138 50S6 3.8950 4:^72 4.8669 4110 6.0748 2271 7.5744 1961 47 2.5363 4351 3.1916 9713 4.0118 9503 5.0372 84<>4 6.3178 1562 7.9152 6849 48 2.5870 7039 3.-27I4 8956 4.1322 5188 5.2i:t5 8898 6.57<6 28-24 8.2714 5557 49 2.6:188 1179 S.-mi 7680 4.2562 1944 6.3960 6459 6.83:« 4'J:J7 8.6436 7107 50 2.6915 8803 3.4371 0872 4.3839 0602 5.5849 -2686 7.1066 83:35 9.08-26 3627 51 2.7454 1979 3.5230 3644 4.5154 2320 5.78tW 9930 7.3909 5068 9.4391 0490 52 2.800:J 2»19 3.6111 1235 4.6508 8590 5.98-27 i;i27 7.68<>5 8S71 9.8638 6463 53 2.856:1 3475 3.7013 9016 4.7904 1247 6.1921 08-24 7.9940 5226 10.3077 %t&\ 54 2.01 :M 6144 3.79:J9 2491 4.9;M1 2485 6.4(188 3202 8.3138 1435 10.7715 8677 55 2.9717 3067 3.8887 7303 5.0821 4iSo9 6.6331 4114 8.6463 6692 11.2563 0817 Svhtract $1 from the Amount in this Table to find tJte Interest. COMPOUND INTEREST. 303 Amownl of $1 at Compound Interest in any number of years, not exceeding fifty-five. Yrs. 5 per ceut. 6 per cent. 7 per cent. 8 per cent. 9 per cent. 10 per cent. 1 i.(eooooo 1.0600 000 1.0700 000 1.0800 000 1.0900 000 l.KXXl (XXI 2 1.1023 000 i.vim 000 1.1449 000 1.1664 0(N) 1.1881 UN) 1.-210UIHXJ 3 1.1576 "^50 1.1910 160 l.'2-250 4:iO 1.2597 1-20 l.-29.)0 '2JH) i.:»io 0(>o 4 1.2155 063 1.2624 770 1.3107 9«i0 1:MH 890 1.4115 816 1.4641 000 5 ^ 1.2762 816 1.3382 256 1.4025 517 1.4693 281 1. 5:^86 240 1.6105 100 6 1.3400 936 1.4185 191 1.5007 :m 1.5868 743 1.6771 001 1.7715 610 7 1.4071 004 1.5036 30:j 1.5038 481 l.(i057 815 1.7i:« 243 1.8*280 JttJl 1.9487 171 8 1.4774 554 1.7181 862 1.S509 3112 1.99-2.) 6-26 2.14:i5 888 9 1.5513 282 1.6894 790 1.8384 592 l.mn) 046 •2.1718 9;« 2.:i579 477 10 1.6288 tMO 1.7908 477 1.9671 514 2.1589 250 2.3673 637 2.59-37 425 11 1.7103 394 1.8982 986 2.1048 5*20 2.3316 300 2.5801 264 2.8531 167 12 1.7958 563 2.0121 965 2.-2o21 916 2.5181 701 '2.8126 648 'AASM '284 13 1.8856 491 2.1329 28;t 2.4098 450 2.7196 2:^7 3.0658 046 3.45'22 712 14 1.9799 316 2.-2609O40 2.5783 .342 2.9371 936 3.:i417 270 3.7974 9b;j 15 2.0789 282 2.3965 582 2.7590 315 3.1721 601 3.fr4'24 8-25 4.1772 482 16 2.1828 746 2.5401 517 2.9521 638 3.4-259 426 3.970B 059 4.5949 730 17 2.2920 183 2.6927 7-28 3.1588 152 3.7000 181 4.3-276 :«4 5.0544 703 18 2.4066 192 2.8543 :<92 3.3799 323 3.9960 195 4.7171 204 5.5599 173 19 2.5269 502 3.a»5 995 3.6165 -275 4.3157 Oil 5.1416 613 6.1159 000 20 2.6532 977 3.2071 355 3.8696 845 4.6609 571 5.6044 108 6.7275 000 21 2.7859 626 3.3995 636 4.1403 621 5.0338 337 6.1088 077 7.4002 499 23 2.92)2 607 3.6035 374 4.4304 017 6.4:i65 404 6.6581) WW 8.1402 749 23 3.0715 2:« 3.8197 497 4 7405 299 5.8714 637 7.'2678 745 8.9543 024 24 3.2250 993 4.0489 346 5.072J 670 6.3411 807 7.9110 832 9.8497 :r27 25 3.3863 519 4.2918 707 5.4-274 3-26 6.WM 752 8.6'230 807 10.8347 059 26 3.5556 727 4.549;j 830 5.8073 529 7.3963 632 9.3991 579 11.9181 765 27 8.7334 56:j 4.8223 459 6.2138 676 7.9880 615 10.2450 821 13.1099 942 28 3.9201 291 5.1116 867 6.&488 3^ 8.6271 0&4 11.1671 395 14.4200 936 29 4.1161 350 5.4183 879 7.1142 571 9.3172 749 1-2.1721 821 Id.mW 930 30 4.3219 424 5.74:M 912 7.6122 550 10.0626 569 13.2676 785 17.4494 im 31 4.^380 393 6.0881 006 8.1451 1-29 10.8676 694 14.4617 695 19.1943 425 32 4.7649 415 6.45J3 867 8.7152 708 11.7:J70 830 15.76:« 288 21.1137 768 33 5.00:il 885 6.8405 899 9.3-25!) 398 12.6760 496 17. 1 8*20 -284 23.'2251 544 34 5.25:» 480 7.2510 2>3 9.9781 13.) 13.6901 336 18.7284 109 25.5476 699 35 5.5160 IM 7.6860 868 10.6765 815 14.7853 443 20.4139 679 28.1024 369 36 5.7918 161 8.1472 5-20 11.4-239 4-22 15.9681 718 22.2512 -250 30.9126 805 37 6.0814 069 8.6360 871 1-2.-2-236 181 17.-2456 256 24.'i538 im 34.0039 486 38 6.38->4 773 9.1542 524 13.0792 714 18.6-252 756 26.4366 805 37.4043 4:M 39 6 7047 512 9.70t« 075 13.9^8 '204 20.1152 977 28.8159 817 41.1447 778 40 7.0399 887 10.2857 179 14.9744 578 21.7.M5 215 31.4094 -200 45.2592 556 41 7.3919 882 10.9028 610 16.0226 690 23.4624 &32 34.2362 679 49.7851 811 4-2 7.7615 876 11.5570 327 17.1442 568 25.3;i94 819 37.3175 .m 54.7636 992 4} 8.1496 669 1-2.2504 546 18.3443 548 27.3666 404 40.6761 098 60.-2400 692 44 8.5571 503 1-2.9854 819 19.6284 596 29.5559 717 44.;Wi9 697 66.-264U 761 45 8.9850 078 13.7646 108 21.0021 518 31.9204 494 48.3272 861 72.8904 m7 46 9.4342 582 14.5904 875 22.4?26 234 34.4740 853 52.6767 419 80.1795 321 47 9.9059 711 15.4659 167 24.0457 070 37.2320 122 57.4176 486 88.1974 8.5:1 48 10.4012 697 16.3938 717 25.?2S9 065 40.-2105 7:J1 62.5852 370 97.0172 :«J8 49 10.9213 331 17.3775 WO 27.5299 300 43.4274 190 68.2179 08:^ 106.7189 572 50 11.4673 998 18.4201 543 29.4570 -251 46.9016 125 74.;«75 '201 1 17.3908 5'29 51 12.0407 698 19.5233 635 31.5190 168 50.65.37 415 81.0496 969 128.1299 382 52 1-2.6428 063 20.6968 853 :«.?25:t 480 54.7060 408 88.3441 696 142.04'29 320 53 13.2749 487 2I.9:«6 9*5 :i6.0861 -224 59.0825 241 96.2951 449 15«.'2472 252 54 13.9383 961 2:J.-2550 204 38.61-21 509 6:{.80!)1 '260 104.9617 079 171.8719 477 55 14.6356 300 24.650:j 216 41.3150 015 68.9138 561 114.4082 610 lb9.0591 4-25 Subtract $1 from the Amount in this Table to find ifie Interest, 304 BAY'S HIGHER ARITHMETIC. How to use the table in finding the Compound Amount: 1. Observe at what intervals interest is payahky and also Uie rate 'per interval. 2. If die number of full intervals can be found in tlie year column, note Hie sum corresponding to it in Hie column under Hie proper rate; multiply Hiis sum, or its amount for any re- maining fraction of an interval, by the principal, S. If the number of intervals be not found in the table, separate Hie whole time into periods whidi are each udtJiin ihe limits of ihe table; find Hie amount of the principal for one of them, make Hwi a principal for Hie next, and so mi, till the whole time has been taken into the calculation. Examples for Practice. 1. Find the compound amount of $750 for 17 yr., at 6%, payable annually. $2019.58 2. Of $5428 for 33 yr., 5% annually. $27157.31 3. The compound interest of $1800 for 14 yr., at 8%, payable semi-annually. $3597.67 4. If $1000 is deposited for a child, at birth, and draws 7% compound interest, payable semi-annually, till it is of age (21*yr.), what will be the amount? $4241.26 5. Find the amount of $9401.50, at compound interest for 19 yr. 4 mo., 9%, payable semi-annually. $51576.68 6. Find the compound amount of $1000 for 100 yr., at 10%, payable annually. $13780612.34 7. The compound interest of $3600 for 15 yr. , at 8^ , payable quarterly. $8211.71 8. The compound interest of $4000 for 40 yr., at 5%, payable semi-annually. $24838.27 9. The compound interest of $1200 for 27 yr. 11 mo. 4 da., at 12^, payable quarterly. $31404.74 COMPOUND INTEREST. 305 CASE II. 336. Given the principal, rate, and compound in- terest or amount, to find the time. Problem. — Find the time in which $750 will amount to $2000, the interest being 8 %, payable semi-annually. Solution. — Since a compound amount is found by multiplying the principal by the amount of $1, we here reverse that process, and say: $2000 -i- 750 = $2.66666666, the amount of $1, at 4^^. The number next lower, in the 4^ column, is $2.66583633, the amount for 25 intervals, and is less than $2.66666666 by $.00083033 Since the amount for 25 intervals will, according to the table, gain $.10663343 in 1 interval, it will gain $.00083303 in such a frac- tion of an interval as the latter sum is of the former; .00083033-;- .10663346 = j^, nearly; hence, the required period is 25 jVr inter- vals of 6 mo., or, 12 yr. 6 mo. 1 da., j4w«. Bule. — 1. Divide {he amount by the principal. 2. If die quotient he found in the table under the given rate, ihe years opposite wHl he Hie required number of intervals ; but if not found eocacUyy in Hie table, take tJie number next less, noting its deficiency, its number of years, and its gain during a full interval. 3. Divide the deficiency by the interval gain, and annex the qux)tient to Hie number of full intervals ; the result will be ihe required Hine. Examples for Practice. In what time, at compound interest, will: 1. $8000 amount to $12000, at 6%? 6 yr. 11 mo. 15 da. 2. $5200 amount to $6508, 6^, payable semi-annually? 3 yr. 9 mo. 16 da. H. A. 26. 306 RAY'S HIGHER ARITHMETIC. 3. jil2500 gain $5500, 10%, payable quarterly? 3 yr. 8 mo. 9 da. 4. U gain »1, at 6, 8, 10%? 11 yr. 10 mo. 21 da.; 9 yr. 2 da.; 7 yr. 3 mo. 5 da. 5. $9862.50 amount to $22576.15, 12%, payable semi- annually? 7 yr. 1 mo. 7 da. CASE III. 337. Giyen the principal, the compound interest or amount, and the time, to find the rate. Problem.— At what rate will $1000 amount to $2411.714 in 20 years? Solution.— Dividing $2411.714 by 1000, we have $2.411714, which, in the table, corresponds to the amount of $1 for the time, at ^(fo^ Rule. — Divide Hie amount by ihe principal; search in (ke table, opposite ihe given number of full intervals^ for the eicact q^wtient or ilie number nearest in value; if the time contain also a part of an interval, find the amount of the tabular sum for tJiat time, before comparing with the quotient; the rate per cent at the head of the column will be the exact, or ihe approx- imate rate. Examples for Practice. At what rate, by compound interest, 1. Will $1000 amount to $1593.85 in 8 yr.? 6%. 2. $3600 amount to $9932.51 in 15 yr.? 7%. 3. $13200 amount to 48049.58, in 26 yr. 5 mo. 21 da. ? 5%. 4. $2813.50 amount to $13276.03, in 17 yr. 7 mo. 14 da., interest payable semi-annually? 9%. COMPOUND INTEEEST. 307 5. $7652.18 gain $17198.67, interest payable quarterly, in 11 yr. 11 mo. 3 da.? 10^. 6. Any sum double itself in ^0, 15, 20 yr. ? 1st, between 7% and 8^; 2d, nearly 5^; 3d, little over ^%. CASE IV. 338. Given the compound interest or amount, the time, and the rate, to find the principal. Problem. — ^What principal will yield 831086.78 com- pound interest in 40 yr., at 8^? Solution.— In 40 yr. $1 will gain $20.7245215, at the given rate; the required principal must contain as many dollars as this interest is contained times in the given interest; $31086.78-^20.7245215 = $1500, ^718. Bule. — Divide Hie given interest or amount by the interest or amount of $1 for the given time and at ike given rate; the quotient will he the required principal. Kemark. — If the amount be due at some future time, the prin- cipal is the present worth at compound interest, and the difference between the amount and present worth is the compound discount. Examples for Pra^ctice. What principal, at compound interest, 1. Will yield $52669.93 in 25 yr., 6^? $16000. 2. Will gain $1625.75 in 6 yr. 2 mo., 7^, payable semi- annually? $3075. 3. WiU yield $3598.61 in 3 yr. 6 mo. 9 da., 10^, payable quarterly ? $8640. 4. Will yield $31005.76 in 9^ yr., at 8%, payable semi- annually? $28012.63 308 RAY'S HIGHER ARITHMETIC. 5. Will amount to $27062.85 in 7 yr., at 4% ? $20565.54 6. What is the present worth of $14625.70, due in 5 yr. 9 mo., at 6^ compound interest, payable semi-annually ? $10409.77 7. What is the compound discoimt on $8767.78, due in 12 yr. 8 mo. 25 da., 5%? $4058.87 IX. ANNUITIES.* DEFINITIONS. 339. 1. An Annuity is a sum of money payable at yearly or other regular intervals. {1. Perpetual or Limited. 2. Certain or Contingent. 3. Immediate or Deferred. 2. A Perpetual Annuity, or a Perpetuity, is one that continues forever. 3. A Limited Annuity ceases at a certain time. 4. A Certain Annuity begins and ends at fixed times. 5. A Contingent Annuity begins or ends with the hap- pening of a contingent event — as the birth or the death of a person. 6. An Immediate Annuity is one that begins at once. 7. A Deferred Annuity, or an Annuity in Reversion, is one that does not begin immediately ; the term of the re- version may be definite or contingent. * Since the problems in annuities may be classed usder the Applications of Percentage, the subject is presented here, instead of being placed after Progression ; however, those who prefer may omit this chapter until after Progression has been studied. ANNUITIES. 309 8. An annuity is Forborne or in Arrears if not paid when due. 9. The Forborne or Final Value of an annuity is the amount of the whole accumulated debt and interest, at the time the annuity ceases. 10. The Present Value of an annuity is that sum, which, put at interest for the given time and at the given rate, will amount to the final value. 11. The value of a deferred annuity at the time it com- mences, may be called its Initial Value ; its Present Value is the present worth of its mitial value, at an assumed rate of interest. 12. The rules for annuities are of great importance ; their practical applications include leases, life-estates, rents, dowers, pensions, reversions, salaries, life insurance, etc. Kemark. — An annuity begins, not at the time the first payment is made, but one interval before ; if an annuity begin now, its first payment will be a year, half-year, or quarter of a year hence, accord- ing to the interval named. CASE I. « 340. Given the payment, the interval,* and the rate, to find the initial value of a perpetuity. Problem. — What is the initial value of a perpetual lease of $250 a year, allowing 6^ interest ? OPERATIOK. Solution. — The initial value must $ 1 be the principal, which, at 6^, yields . 06 $250 interest every year ; it is found, .06)250.0000 by Art 300, $4166.66 1, Am. Bule. — Divide tke given payment by the interest of $1 for one interval at ike proposed rate. 310 RA Y'S HIGHER ARITHMETIC, Examples for Practice. 1. What is the initial value of a perpetual leasehold of $300 a year, allowing 6^ interest? $5000. 2. What must I pay for a perpetual lease of $756.40 a year, to secure 8% interest? $9455. 3. Ground rents on perpetual lease, yield an income of $15642.90 a year: what is the present value of the estate, aUowing 7% interest? $223470. 4.^ What is the initial value of a perpetual leasehold of $1600 a year, payable semi-annually, allowing 5^ interest, payable annually? $32400. Suggestion. — Here tlie yearly payment is $1620, by allowing 5^ interest on the half-yearly payment first made. 5. What is the initial value of a perpetual leasehold of $2500 a year, payable quarterly, interest 6% payable semi- annually; 6% payable annually: 6% payable quarterly? $41979.161; $42604.16|; $41666. 66f CASE II. 341. To find the present value of a deferred per- petuity, when the payment, the interval, the rate, and the time the perpetuity is deferred are known. Problem. — Find the present value of a perpetuity of $250 a year, deferred 8 yr., allowing 6% interest. Solution. — Initial value of perpetuity of $250 a year, by last rule = $4166.66§ The present value of $4166.66f , due 8 yr. hence, at 6^ compound interest, = $4166.66§ -^ 1.5938481 (Art. 335 ). Use the contracted method, reserving 3 decimal places ; the quotient, $2614.22, is the present value of the perpetuity. Bule. — Find the initial value of tJie perpetuity by tlie last rule; then find the present worili of this mm for the time the ANNUITIES. 311 'perpetuity is deferred, by Case IV of Compound Interest ; iJm wiU be the present vakie required. Examples for Practice. 1. Find the present value of a perpetuity of $780 a year, to commence in 12 yr., int. 5%. $8686.66 2. Of a perpetual lease of $160 a year, to commence iu 3 yr. 4 mo., int. 7%. $1823.28 3. Of the reversion of a perpetuity of $540 a year, de* ferred 10 yr., int. 6%. $5025.55 4. Of an estate which, in 5 yr. , is to pay $325 a year for- ever: int. 8%, payable semi-annually. $2690.67 5. Of a perpetuity of $1000 a year, payable quarterly, to commence in 9 yr. 10 mo. 18 da., int. 10^, payable semi- annually. $3858.88 CASE III. 342. Given the rate, the payment, the interval^ and the time to run, to find the present value of an annuity certain. Problem. — 1. Find the present value of an immediate annuity of $250 continuing 8 years, 6% interest. Solution. Present value of immediate perpetuity of $250, . . . = $4166.67 Present value of perpetuity of $250, deferred 8 yr., . . = 2614.2 2 Pres. val. of immediate annuity of $250, running 8 yr., = $1552.45 Problem. — 2. The present value of an annuity of $680, to commence in 7 yr. and continue 10 yr., 5% int. Solution. Pres. val. of perpetuity of $680, deferred 7 yr., at 5^o, =$9665.27 Pres. val. of perpetuity of $680, deferred 17 yr., at 6^, = 5933.64 Pres. val. of annuity of $680, deferred 7 yr., to run 10 yr. =$3731.63 312 I^A Y'S HIGHER ARITHMETIC. Rule. -^i^ind f/ie 'present valv^e of ttoo perpetuities having the given rate, payment, and interval, one of them commencing when die annuity commences, and the other when the annuity ends. The difference betioeen these values will he the present value of the annuity. Notes. — 1. This rule applies whether the annuity is immediate or deferred ; in the latter, the time the annuity is deferred must be known, and used in getting the values of the perpetuities. 2. By using the initial instead of the present values of the per- petuities, the rule gives the initial value of the deferred annuity, which may be used in finding its final or forborne value. (Rem. 1, Case IV.) Examples for Practice. 1. Find the present value of an annuity of $125, to com- mence in 12 yr. and run 12 yr., int. 7^. $440.83 2. The present value of an immediate annuity of $400, running 15 yr. 6 mo., int. 8%. $3484.41 3. The present value of an annuity of $826.50, to com- mence in 3 yr. and run 13 yr. 9 mo., int. 6%, payable semi-annually. $6324. 69 4. The present value of an annuity of $60, deferred 12 yr. and to run 9 yr., int. 4^%. $257.17 5. Sold a lease of $480 a year, payable quarterly, having 8 yr. 9 mo. to run, for $2500: do I gain or lose, int. S%, payable semi-annually? Lose $509.96 CASE IV. 343. Given the payment, the interval, the rate, and time to run, to find the final or forborne value of an annuity. Problem. — Find the final or forborne value of an annuity of $250, continuing 8 yr., int. 6^. ANNUITIES. 313 Solution.— The initial value of a perpetuity of $250, at 6^, = $4166.66§ ; its compound interest, at 6% for 8 yr., = $4166.66J X .5938481 = $2474.37, the final or forborne value of the annuity. Bule. — Consider iJie annuity a perpetuity, and find its initial value by Case J. The compound interest of this sum, at tJie given rate for the time iJie annuity runs, mil be ike filial or forborne value. Notes. — 1. The final or forborne value of an annuity may be obtained by finding first the initial value, as in Case III, and then the compound amount for the time the annuity runs. 2. The present value of an annuity can be obtained by finding first the forborne value, as in this case, and then the present worth for the time the annuity runs. Examples for Practice. 1. Find the forborne value of an immediate annuity of 8300, running 18 yr., int. b%. $8439.72 2. A pays $25 a year for tobacco: how much better off would he have been in 40 yr. if he had invested it at 10^ per annum? $11064.81 3. Find the forborne value of an annuity of $75, to com- mence in 14 yr., and run 9 yr., int. 6^. $861.85 Suggestion. — The 14 yr. is not used. 4. A pays $5 a year for a newspaper: if invested at 9%, what will his subscription have produced in 50 yr.? $4075.42 5. An annuity, at simple interest 6%, in 14 yr., amounted to $116.76 : what would have been the difference had it been at compound interest 6%? $9.33 6. A boy just 9 yr. old, deposits $35 in a bank : if he deposit the same each year hereafter, and receive 10%, com- pound interest, what will be the entire amount when he is of age? $858.29 H. A. 27. 314 HAY'S HIGHER AMITHMETia Hie present vcdue of $1 per annum in any number of years, not exceeding fifty-jwe. Yre. 4 per cent. 6 per cent. 6 per cent. 7 per ceni. 8 per cent. 10 per cent. 1 .961538 .952381 .943396 .934579 .925928 .909091 2 1.H86005 1.859410 1.833393 1.806018 1.783255 1.735537 3 2.775091 2.723248 2.673012 2.624316 2.577097 2.486852 4 3.629895 3.54S9S1 3.465106 3.387211 8.312127 3.169865 5 4.451822 4.329477 4.212864 4.100197 3.992710 8.790787 6 5.242137 5.075692 4.917321 4.766540 4.622880 4.355261 7 6.002055 5.7863TS 5.582381 5.389289 5.206370 4.868419 8 6.732745 6.46;<213 6.209794 5.971299 5.746039 5.334926 9 7.435332 7.107822 6.801692 6.515232 6.246888 5.759024 10 8.110696 7.721735 7.360087 7.023582 6.710081 6.144567 11 8.760477 8.306414 7.88&S75 7.498674 7J38964 6.495061 12 9.385074 8.86:K52 8.383844 7.942688 7.536078 6.813692 13 9.986<M8 9.3D3573 8.852683 8.357651 7.903776 7.103366 14 10.563123 9.898641 9.294984 8.745468 8.244237 7.366687 15 11.118387 10.379658 9.712249 9.107914 8.559479 7.606080 16 11.652296 10.837770 10.106895 9.446649 8.851369 7.823700 17 12.165669 11.274066 10.477260 9.763223 9.121638 8.021553 18 12.659297 11.689587 10.827603 10.050087 9.371887 8.201412 19 13.133939 12.065321 11.158116 10.335595 9.603599 8.364920 20 13.590326 12.462210 11.469921 10.594014 9.818147 8.513564 21 14.029160 12.821153 11.764077 10.835527 10.016803 8.648694 22 14.451115 13.163003 12.041582 11.061-241 10.200744 8.771540 23 14.856842 13.488574 1*2.303379 11.272187 10.371059 8.883218 24 15.246963 13.798642 1-2.550353 11.469334 10.628758 8.984744 25 15.622080 14.093945 12.783356 11.653583 10.674776 9.077040 26 15.982769 14.375185 13.003166 11.825779 10.809978 9.160945 27 16.329586 14.643031 13.210534 11.986709 10.935165 9.-287223 28 16.663063 14.898127 13.406164 12.137111 11.051078 9.306567 29 16.98:i715 15.141074 13.fi90721 1-2.277674 11.158406 9.369606 30 17.2921133 15.872451 13.764831 12.409041 11.257783 9.426914 31 17.588494 15.592811 13.929086 J2.531814 11.349799 9.479013 32 17.873552 15.802677 14.084043 12.646555 11.434999 9.526376 33 18.147646 16.002549 14.230230 12.763790 11.613888 9.569432 34 18.411198 16.192»04 14.368141 1-2.854000 11.586934 9.606575 35 18.664613 16.374194 14.498246 1-2.917672 11.654568 9.644159 36 18.908?82 16.546852 14.620987 13.085-208 11.717193 9.676506 37 19.142579 16.711287 14.736780 13.117017 11.776179 9.706917 38 19.367864 16.867893 14.846019 13.193473 11.828869 9.732651 39 19.584485 17.017041 14.949075 13.264928 11.878582 9.756056 9.779051 40 19.792774 17.159086 15.046297 13.331709 11.924613 41 19.993052 17.294368 15.138016 13.394120 U. 967235 9.799137 42 20.185627 17.423208 15.224543 13.45-2449 12.006690 9.817397 43 20 370795 17.545912 15.306173 13.506962 12.043240 9.833998 44 20.548841 17.6627T3 15.383182 13.557908 12.077074 9.849089 45 20.720040 17.774070 15.455832 13.605522 12.108402 9.862806 46 20.884654 17.880067 15.521370 13.650020 12.137400 9.875280 47 21.042936 17.981016 15.589028 13.691608 12.164-267 9.886618 48 21.195131 18.077158 15.650027 13.730474 12.189136 9.890926 49 21.3414T2 18.168?22 15.7075?2 13.766799 12.212163 9.906296 50 21.482185 18.255925 15.761861 13.800746 12.233485 9.914814 51 21.617485 18.338977 15.813076 18.832473 12.253227 9.922550 52 21.747582 18.418073 15.861393 13.86-21-24 12.-271S06 9.929590 53 21.872675 18.493403 15.906974 13.889836 12.288432 9.935900 54 21.992957 18.565146 15.949976 13.915735 12.304103 9.941817 55 22.106612 18.633472 15.990543 13.939939 12.318614 9.947107 ANNUITIES. 315 CALCULATIONS BY TABLE. 344. By the table on page 314, some interesting and important cases in annuities can be solved, among which are the following three : CASE V. 346. Oiven the rate, time to run, and the present or final value of an annuity, to find the payment. Problem. — An immediate annuity running 11 yr., can be purchased for $6000.: what is the payment, int. 6%? Solution. — The present value of an immediate annuity of $1 for 11 yr., at 6^o, is $7x886875; $6000 divided by this, gives $760.76, the payment required. * Bule. — Assume $1 for the payment; determine the present or final value on this suppositiony and divide the given present or final value by it. Examples for Practice. 1. How much a year should I pay, to secure $15000 at the end of 17 yr., int. 7%? $486.38 2. What is the payment of an annuity, deferred 4 yr., running 16 yr., and worth $4800, int. 6%? $599.64 CASE VI. 346. Given the payment, the rate, and present value of an annuity, to And the time it runs. Problem. — In what time will a debt of $10000, drawing interest at 6%, be paid by installments of $1000 a year? Solution. — The $10000 may be considered the present value of an annuity of $1000 a year at 6^©; but $10000 -t- 1000 = $10, the 316 BAY'S HIGHER ARITHMETia present value of an annuity of $1 for the same time and rate ; by reference to the table, the time corresponding to this present value, under the head of 6^, is 15 yr.; the balance then due may be thus found : Comp. amt. of $10000 for 15 yr., at 6/<, (Art. 336), . = $23965.58 Final val. of annuity $1000 for 15 yr., at 6^o (Art. 343), = 23275.97 ^ Balance due at end of 15 yr $689.61 Bule. — Divide the present value by Hie payment, and look in the table, under the given rate, for tlie quotierd; the number of years corresponding to the quotierd or to Hie tabidar number next less, will be the number of full intervals required, » Note. — The difference between the compound amount of the debt, and the forborne value of the annuity, for that number of intervals, will be the unpaid balance. Examples for Practice. 1. In how many years can a debt of $1000000, drawing interest at 6%, be discharged by a sinking fund of $80000 a year ? 23 yr., and $60083.43 then unpaid. 2. In how many years can a debt of $30000000, drawing interest at 5^, be paid by a sinking fund of $2000000? 28 yr., and $798709.00 unpaid. 3. In how many years can a debt of $22000, drawing 7^ interest, be discharged by a sinking fund of $2500 a year ? 14 yr., and $351.53 then unpaid. 4. Let the conditions be the same as those of the illus- trative example, and let each $1000 payment be itself a year's accumulation of simple interest: what would be the whole time required to discharge the debt? 15 yr. 8 mo. 19 da. 5. Suppose the national debt $2000000000, and funded at 4%: how many years would be required to pay it off, by a sinking fund of $100000000 a year? 41 yr., and $3469275 unpaid. CONTINGENT ANNUITIES. 317 CASE VII. 347. Given the payment, time to run, and present value of an annuity, to And the rate of interest. Bule. — Divide Vie preient value by the payment ; Hie quotient will be the present value of $1 for the given time and rate; look in the table and opposite Vie given number of years for t/ie qmtievd or the tabxdar number of nearest value, and at the head of tlie column wUl be found the rale, or a number as near the true rale as tJie table can exhibit. Examples fob Practice. 1. If $9000 is paid for an immediate annuity of $750, to run 20 yr., what is the rate? About 5^%. 2. If an immediate annuity of $80, running 14 yr., sells for $650, what is the rate? 8%+. CONTINGENT ANNUITIES. DEFINITIONS. 348. 1. Ck>ntingent Annuities comprise Life Annuities, DowerSf Pensions, etc. 2. The value of such annuities depends upon the erpecta- tion (f life. 3. Exi>ectation of Life is the average number of years that a person of any age may be expected to live. 4. Tables, called ** Mortality Tables," have been prepared in England and in this country for the purpose of ascertain- ing how many persons of a given number and of a certain 318 BAY'S HIGHER ARITHMETia age would die during any one year, and in how many years the whole number would die. Remark. — ^These tables, though not absolutely accurate, are based upon so large a number of observations that their approx- imation is very close. Legal, medical, and scientific authorities use them in discussing vital statistics, and insurance companies make them a basis for the transaction of business. 349. The following table differs but slightly from those prepared in this country: Carlisle Table Of Mortality y hosed upon Observations at Carlisle (Eng.), showing the Rate of Extinction of lOfiOO lives. *^ ^ •M *^ ^ «p^ **•* _• •M • o < o E S t B 1 4> bo 2g 2 »« £ u B t . ^1 3 3 S ft 3 3 3 3 B 3 W 525 Cfi 52; 52; C« Jz; 5Z; CO ^ 10000 1539 35 5362 55 70 2401 124 1 8461 682 36 5307 66 71 s 2277 134 2 7779 505 37 5251 57 72 2143 146 8 7274 276 38 5194 58 73 1997 156 4 6998 201 39 5136 62 74 1841 166 5 6797 121 40 5075 66 75 1675 160 6 6676 82 41 5009 69 76 1515 156 7 6594 58 42 4940 71 77 1359 146 8 6536 43 43 4869 71 78 1213 132 9 6493 33 44 4798 71 79 1081 128 10 6460 29 45 4727 70 80 953 1J6 11 6431 31 4C 4657 69 81 837 112 12 6400 32 47 4588 67 82 725 102 13 6368 33 48 4521 63 83 623 94 14 6335 35 49 44^ 61 84 529 84 15 6300 39 50 4397 59 85 445 78 16 6-261 42 51 4338 62 86 367 71 17 6219 43 52 4276 65 87 296 64 18 6176 43 53 4211 68 88 232 51 19 6133 43 M 4143 70 89 181 39 1 20 6090 43 55 4073 73 90 142 37 ! 21 6047 42 56 4000 76 91 105 30 22 6005 42 57 3924 82 92 75 21 23 5963 42 58 ' 3842 93 93 54 14 24 5921 42 59 3749 106 94 40 10 25 5879 43 60 3643 122 95 30 7 26 5836 43 61 3521 126 96 23 5 27 5793 45 62 3395 127 97 18 4 28 574S 50 63 3268 125 98 14 3 29 5698 56 64 3143 125 99 11 2 30 5642 57 65 3018 124 100 9 2 31 5-)85 57 66 2894 123 101 7 2 32 5528 56 67 2771 123 102 5 2 33 5472 55 68 2648 123 103 3 2 34 5417 55 69 2525 124 104 1 1 CONTINOENT ANNUITIES. 319 Table Showing the wives of AnnuUies on Single Lives, ojceording to ike Carlisle Table of Mortality. Age. 4 per ct. 5 per ct. 6 per ct. 7 per ct. Age. 4 per ct. 5 per ct. 6 per ct. 7perct. 14.28164 12.083 10.439 9.177 ' 52 12.25793 11.154 10.208 9.392 1 16.55455 13.995 12.078 10.6(» 53 11.94508 10.892 9.988 9.-205 2 17.72616 14.983 12.925 11.342 54 11.62673 10.621 9.761 9.011 3 18.71508 15.821 13.652 11.978 55 11.29961 10..347 9.524 8.807 4 19.23133 16.271 14.042 12.322 56 10.96607 10.063 9.280 8.595 5 19.592(13 16.590 14.325 12.574 57 10.62559 9.771 9.027 8.875 6 19.74502 16.735 14.460 12.698 58 10.28647 9.478 8.772 8.153 7 19.79019 16.790 14.518 12.756 59 9.96331 9.199 8..529 7.940 8 19.76443 16.786 14.626 12.770 60 9.66333 8.940 8.304 7.743 9 19.69114 16.742 14.500 12.7W 61 9.39809 8.712 8.108 7.572 10 J9.98339 16.669 14.448 12.717 62 9.13676 8.487 7.913 7.408 11 19.45857 16.581 14.384 12.669 e:i 8.87150 8.258 7.714 7.229 12 19.334^3 16.494 14.:J21 12.621 04 8.593:iO 8.016 7.502 7.(M2 13 19.20937 16.406 14.257 12.572 65 8.30719 7.765 7.281 6.847 14 19.08182 16.316 14.191 12.522 66 8.00966 7.503 7.049 6.641 15 18.95534 16.227 14.126 12.473 67 7.69980 7.227 6.803 6.421 16 18.83636 16.144 14.067 12.429 68 7.37976 6.941 6.546 6.189 17 18.72111 16.066 14.012 12.389 69 7.04881 6.643 6.277 5.945 18 18.60656 15.987 13.956 12.348 70 6.70936 6.:»6 5.998 5.690 19 18.48649 15.904 13.897 12.305 71 6.35773 6.015 5.704 5.420 20 18.36170 15.817 13.835 12.259 72 6.02548 5.711 5.424 5.162 21 18.23196 15.726 13.769 12.210 7:^ 5.72465 5.435 5.170 4.927 22 18.09386 15.628 13.697 12.156 74 5.45812 5.190 4.944 4.719 2) 17.95016 15.525 13.621 12.098 75 5.23901 4.989 4.760 4.549 24 17.80058 15.417 13.541 1-2.037 76 5.02399 4.792 ' 4.579 4.382 25 17.64486 15.303 13.456 11.9?2 77 4.8247:} 4.609 4.410 4.227 26 17.48586 .15.187 13.368 11.901 78 4.62106 4.422 4.238 4.067 27 17.32023 15.065 13.275 11.832 79 4.;»:J45 4.210 4.010 3.883 28 17.15412 14.942 13.182 11.759 80 4.182»9 4.015 3.858 8.713 29 16.99683 14.827 13.096 11.693 81 3.95309 8.799 3.656 8.523 30 16.85215 14.723 18.020 11.636 82 8.74634 3.606 8.474 3.352 31 16.70511 14.617 12.942 11.578 83 8.5.3409 3.406 8.'286 3.174 82 16.55246 14.506 12.860 11.516 84 3.32856 3.211 3.102 2.999 33 16.39072 14.387 12.771 11.448 85 3.11515 3.009 2.909 2.815 84 16.21943 14.260 12.675 11.374 86 2.92831 2.830 2.?39 2.652 85 16.04123 14.127 12.573 11.295 87 2.77593 •2.685 2.599 2.519 36 15.85577 13.987 12.465 11.211 88 2.68337 2.597 2.515 2.439 37 15.66586 13.843 12.354 11.124 89 2.57704 2.495 2.417 2.344 38 15.47129 1.3.695 r2.2:» 11.0:)3 90 2 41621 2.3.39 2.266 2.198 39 15.27184 13.542 12.120 10.939 91 2.:)9835 2.321 2.248 2.180 40 15.07363 13.390 12.002 10.845 92 2.49199 2.412 2.337 2.266 41 14.88314 13.245 11.890 10.757 93 2.59955 2.518 2.440 2.367 42 14.69466 13.101 11.779 10.671 94 2.64976 2.569 2.492 2.419 4:t 14.50529 12.957 11.668 10.585 95 2.67433 2.596 2.5-22 2.451 44 14.30874 12.806 11.551 10.494 96 2.62779 2.555 2.486 2.420 45 14.10460 12.648 11.428 10.397 97 2.492M 2.428 2.368 2.309 •M 13.88928 12.480 11.296 10.292 98 2.33222 2.278 2.227 2.177 47 13.66208 12.:«)1 11.154 10.178 99 2.08700 2.045 2.004 1.964 48 13.41914 12.107 10.998 10.aV2 100 1.65282 1.624 1.596 1.569 49 13.15312 11.892 10.823 9.908 101 1.210(» 1.192 1.175 1.159 50 12.86902 11.660 10.6.31 9.749 102 0.76183 0.753 0.744 0.7.35 51 12.56581 11.410 10.422 9.573 103 0.3-JOal 0.317 0.314 0.312 320 BAY'S HIGHER ARITHMETIC. CALCULATIONS BY TABLE. 860. The preceding table of life annuities shows the sum to be paid by a person of any age, to secure an an- nuity of SI during the life of the annuitant. CASE I. 851. To find the value of a given annuity on the life of a person whose age is known. Bule. — Find from the table the value of a life annuity of $1, for Hie given a^e and rate of interest, and mvltipLy it by the payinent of Hie given annuity. Remarks. — 1. To find the value of a life-eBtate or widow's dower (which Ib a life-efltate in one third of her husband's real estate): Estimate the value of the property in which the life-estate is held; the yearly interest of this sum, at an a(p eed rate, will be a life-annuity, whose value for the given aye and rate will be the value of the life-estate, 2 The reversion of a life-annuity, life-estate, or dower is found by deducting its value from the value of the property. Examples for Practice. 1. What must be paid for a life-annuity of $650 a year, by a person aged 72 yr., int. 7%? $3355.30 2. What is the life-estate and reversion in $25000, age 55 yr., int. 6%? Life-estate, $14286; rev., $10714. 3. The dower and reversion in $46250, age 21 yr., int. 6%? Dower, $12736.33; rev., $2680.34 CASE II. 352. To find how large a life-annuity can be pur- chased for a given sum, by a person whose age is known. CONTINGENT ANNUITIES. 321 Bule. — Assume $1 a year for Uw annuity; find from the table its value for tlis given age and rate of interest, and divide Hie given cost by it; Hie quotient vnU be Hie 'payment required. Examples for Practice. How large an annuity can be purchased : 1. For $500, age 26 yr., int 6%?. $37.40 2. For $1200, age 43, int. bfo ? $92.61 3. For $840, age b^, int. 7^? $103.03 CASE III. 853. To find the present value of the reversion of a given annuity ; that is, what remains of it, after the death of its possessor, whose age is known. Hule. — Fiud the present value of Hie annuity during its whole continua'nee; find its value during tfve given life; their difference will be Hie value of the reversion. Note. — It will save work, to consider (he annuity as $1 a year^ then apply the rule, using the tables in Art. 335 and Art. 350, and multiply the result by the given payriienL Examples for Practice. 1. Find the present value of the reversion of a perpetuity of $500 a year, after the death of a person aged 47, int. 5%. $3849.50 2. Of the reversion of an annuity of $165 a year for 30 yr., after the death of a person 38 yr. old, int. 6%. $251.76 3. Of the reversion of a lease of $1600 a year, for 40 yr., afl«r the death of A, aged 62, int. 7^. $9485.93 322 BA Y'S HIGHER ARITHMETIC, PERSONAL INSURANCE. DEFINITIONS. 354. 1. Personal Insurance is of two kinds: (1.) lAje [nsurance; (2.) Accident Insurance, 2. life Insurance is a contract in which a company agrees, in consideration of certain premiums received, to pay a certain sum to the heirs or assigns of the insured at his death, or to himself if he attains a certain age. 3. Accident Insurance is indemnity against loss by accidents. 4. The Policies issued by life insurance companies are: (1.) Tenn Policies; (2.) Ordinary Life Policies; (3.) Joint Life Policies; (4.) Endovrment Policies; (5.) Reserved Endoiv- ment Policies; (6.) Tontine Savings Fund Policies. 5. The chief policies are, however, the Ordinary Life and the Endowment. 6. The Ordinary Life Policy secures a certain sum of money at the death of the insured. Premiums may be paid annually for life, semi-annually, quarterly, or in one pay- ment in advance; or the premiums may be paid in 5, 10, 15, or 20 annual payments. 7. An Endowment Policy secures to the person insured a certain sum of money at a specified time, or to his heirs or assigns if he die before that time. Remark. — It will be advantageous for the student to examine an "application" and a "policy" taken from some case of actual insurance ; by a short study of such papers, the nature of the insur- ance contract will be learned more easily than by any mere verbal description; additional light may be had from the reports pub- lished by various companies. 355. 1. The following is a condensed table of one of the leading companies: PERSONAL INSURANCE. 323 Table. Annual JFVemwm Rates for an Insurance of flOOO. LIFE POLICIES. ENDOWMENT POLICIES. Payable at death only. Paynbl e as indicated, or at death, if prior. Age. Annual Payments. Single Payment Age. In 10 years In 15 years In 20 years For life 20 years 10 years • 20 to • 20 to 25 919 89 927 39 $42 56 $326 58 26 $103 91 $66 02 $47 68 26 20 40 27 93 43 37 832 58 26 104 03 66 15 47 82 27 20 93 28 50 44 22 338 83 27 104 16 66 29 47 98 28 21 48 29 09 45 10 345 31 28 104 29 66 44 48 15 29 22 07 29 71 46 02 352 05 29 104 43 66 60 48 33 30 22 70 80 36 46 97 859 05 30 104 58 66 77 48 53 31 23 35 81 03 47 98 366 33 81 104 75 66 96 48 74 32 24 05 81 74 49 02 873 89 82 104 92 67 16 48 97 38 24 78 82 48 50 10 881 73 83 105 11 67 36 49 22 34 25 56 83 26 5122 389 88 84 105 31 67 60 49 49 35 26 38 84 08 52 40 898 34 35 105 53 67 85 49 79 36 27 25 84 93 58 63 407 11 86 105 75 68 12 50 11 87 28 17 85 83 54 91 416 21 87 106 00 68 41 50 47 88 29 15 86 78 56 24 425 64 88 106 28 68 73 50 86 89 30 19 87 78 57 63 435 42 89 106 58 69 09 6130 40 3180 88 &S 59 09 445 55 40 106 90 69 49 - 61 78 41 32 47 89 93 60 60 456 04 41 107 26 C9 92 52 31 42 33 72 41*10 62 19 466 89 42 107 65 70 40 62 89 43 35 05 42 84 68 84 478 11 43 108 08 70 92 63 54 44 36 46 43 64 65 57 489 71 44 108 55 71 50 64 25 45 37 97 45 03 67 37 501 69 45 109 07 72 14 55 04 46 39 58 46 50 69 26 514 04 46 109 65 72 86 55 91 47 41 30 48 07 71 25 526 78 47 110 80 73 66 56 89 48 43 13 49 73 73 32 539 88 48 111 01 74 54 57 96 49 45 09 51 50 75 49 553 33 49 111 81 75 51 69 15 50 47 18 53 38 77 77 567 13 50 112 68 76 59 60 45 51 49 40 55 38 80 14 581 24 51 113 64 77 77 61 90 52 5178 57 51 82 63 595 66 52 114 70 79 07 63 48 53 54 31 59 79 85 22 610 36 53 115 86 80 51 65 22 54 57 02 62 22 87 94 625 33 54 117 14 82 09 67 14 55 59 91 64 82 90 79 640 54 55 118 54 83 82 69 24 56 63 00 67 60 93 78 655 99 56 120 09 85 73 57 66 29 70 59 96 91 671 64 57 121 78 87 84 58 69 82 73 78 100 21 687 48 58 123 64 90 15 59 73 60 77 22 103 68 703 49 59 125 70 92 70 60 77 63 80 91 107 35 719 65 60 127 96 95 50 61 81 96 84 88 111 23 735 92 61 130 45 62 86 58 89 16 115 32 752 26 62 133 19 63 91 54 93 76 119 66 768 67 63 186 20 64 96 86 98 73 12A 28 785 10 64 189 52 65 102 55 104 10 129 18 80152 65 148 16 1 J 324 J^A T'S HIGHER ARITHMETia 2. Quantities considered in Life Insurance are: 1. Premium on-$1000. 2. The Gain or Loss. 3. Amount of the Policy. 4. Age of the Insured. ^ 5. The Term of years of Insurance. These quantities give rise to five classes of problems, but they involve no new principles, and by the aid of the preceding tables they are easily solved. Simple interest is intended where interest is mentioned in the following prob- lems: Examples for PrAlCtice. 1. W. R. Hamilton, aged 40 years, took a life policy for $5000. Required the annual premium ? 8156.50 2. Conditions as above, how much would he have paid out in premiums, his death having occurred after he was 53? »2191. 3. Conditions the same, what did the premiums amount to, interest 6%? $3045.49 4. James Bragg, aged 50 years, took out an endowment policy for $20000, payable in 10 years, and died after making 6 payments: how much less would he have paid out by taking a life policy for the same amount, the premium pay- able annually? $7860. 5. Thomas Winn, 28 years of age, took out an endow- ment policy for $10000, payable in 10 years; he died in 18 months: what was the gain, interest on the premiums at 6^, and how much greater would the profit have been had he taken a life policy, premiums payable annually? $7789.052 ; $1755.57, profit. 6. P. Darling took out a life policy at the age of 40 years, and died just after making the tenth payment; his premiums amounted to $3975.10, interest 6^ ; required the amount of his jjolicy ? $10000. TOPICAL OUTLINE. 325 7. R C. Storey took out an endowment policy for $10000 for 15 years ; he lived to pay all of the premiums ; but had he put them, instead, at 6^ interest as they fell due, they would have amounted to $15426.78: what was his age? 40 years. 8. Allen Wentworth had his life insured at the age of twenty, on the life plan, for $8000, premium payable annu- ally : how old must he be, that the sum of the premiums may exceed the policy? 71 years old. 9. T. B. Bullene, aged 40 years, took out a life policy for $30000, payments to cease in 5 years, the rate being $9,919 on the $100 ; his death occurred two months after he had mad§ the third payment: what was gained over and above the premiums, interest 6%? $20448. 10. , F. M. Harrington took out an endowment policy for $11000 when he was 42; at its maturity he had paid in premiums $635.80 more than the face of the policy: what was the period of the endowment? 20 years. 1. Simple Interest. Topical Outline. Applications op Percentage. {With Time.) ' 1. Definitions:— Interest, Principal, Rate, Amount, Legal Rate, Usury, Notes Promissory, Face, Payee, In- dorser. Demand Note, Time Note, Principal and Surety, Maturity, Protest. f Methods I 2. Five Cases. I Rules. Common. Aliquots. Six and Twelve per ct L Exact Interest. 2. Rule.— Formula. 3. Rule.— Formula. 4. Rule.— Formula, 5. Rule.— Formula. 3. Annual Interest.— Rule 326 BAY'S HIGHER ABITHMETIC. Topical Outline.— (Oo7i<mw€d.) Applications of Pekcentage. {With Time.) 2. Partial Payments.. . 1. Definitions :— Payment, Indorsement 2. Rules. / ^' ^' ^' I^ule.— Principles. Connecticut, ( / Vermont, and Mercantile Rules. r 1 True Discount / ^* I^finitions:— Present Worth,Discount *■ I 2. - ' 3. Discount... - 2. Bank Discount. Rule. 1. Definitions :— Bank, Deposit, Issue, Check, Drafts, Drawee, Payee, In- dorsement, Discount, Proceeds, Days of Grace, Time to Run; 2. Four Cases, Rules. 4. Exchnnge 1. Definitions .-—Domestic and Foreign Exchange, Bill, Set Rate, Course, Par, Intrinsic, and Commercial. 1. Domestic. r 2. Kinds. 2. Forei^.., 1. Direct. Table of Values. 2. Circular. 1. Definitions :— Arbitra- tion, Simple and Compound. 2. Rules. 5. Equation of Payments... 1. Definitions:— Equated Time, Term of Credit, . Average Term, Averaging Account, Clos- ing Account, Focal Date. 2 Principles. 3. Rules. 6. Settlement of Accounts. 1. Definitions. 2. Kinds.. ri.. I 3. 5 1. Accounts Current; Rule. Account Sales; Rule. Storage. /. Compound Interest | ^- ^definitions: — Comp. Int, Comp. Amt I 2. Four Cases, Rules. 8. Annuities. . 9. Personal Insurance ' 1. Definitions :— Perpetual, Limited, Certain, Contingent. Im- mediate, Deferred, Forborne, Final Value, Initial Value, Present Value. 2. Seven Cases. Rules. Table. Definitions. Table. I 2. Xyn. PAETK"EBSHIP. DEFINITIONS. 856. 1. Partnership is the association of two or more persous in a business of which they are to share the profits and the losses. The persons associated are called partners; together they constitute a Fimif Company y or Home. 2. The Capital is the money employed in the business; the Assets or Besources of a firm are its property, and opposed to these are its Liabilities or Debts. 3. Partnership has two cases : (1.) When all the shares of the capital are continued through the same time; (2.) When the full shares are not continued through the same time. The first is called Simple Partnersldp ; the second, Compound PartnerMp, Principle. — Gains and losses are sliared in proportion to the sums invested and Hie periods of investment CASE I. 867. To apportion the gain or loss, when aU of each partner's stock is employed through the same time. Problem. — A, B, and C are partners, with $3000, $4000, and $5000 stock, respectively ; if they gain $5400, what is each one's share? OPERATION. Solution.— The whole 3 ^^^ of $ 5 4 = $ 1 3 5 0, A's share, stock is $12000, of which 4 ^ " 5 4 0= 1 8 0, B's '* A owns T^j, B t\, C ^^] ^ ^5^ « 5 400= 2250, C's ** hence, by the principle 12 $5 4 0, whole, stated, A should have f^ of the gain, or $1350; in like manner, B, $1800; C, $2250. (327) 328 RAY'S HIGHER ARITHMETIC, Rule. — Divide the gain or loss among (he partners in pro- poHion to their shares of tJw stock. Remark. — The division may be made by analysis or by simple proportion. Examples for Practice. 1. A and B gain in one year $3600; their store expenses are $1500. If A's stock is »2500 and B's «1875, how much does each gain ? A $1200, B 8900. 2. A, B, and C are partners ; A puts in $5000, B 6400, C $1600. C is allowed $1000 a year for personal attention to the business ; their store expenses for one year are $800, and their gain, $7000. Find A's and B*s gain, and Cs income. A $2000, B $2560, C $1640. 3. A, B, and C form a partnership ; A puts in $24000, B $28000, C $32000; they lose ^ of their stock by a fire, but sell the remainder at f more than cost: if all expenses are $8000, .what is the gain of each ? A $5714.281, B $6666.66|, C $7619.04|f 4. A, B, and C are partners; A's stock is $5760, B's, $7200; their gain is $3920, of which C has $1120: what is C*s stock, and A's and B*s gain ? C's stock, $5184; A*s gain, $1244.44|; B's gain, $1555.55f 5. A, B, and C are partners; A's stock, $8000; B's, $12800; C% $15200; A and B together gain $1638 more than C : what is the gain of each ? A $2340 ; B $3744 ; C $4446. 6. A, B, and C have a joint capital of $27000 ; none of them draw from the firm, and when thfey quit A has $20000; B, $16000; C, $12000: how much did each con- tribute? A, $11250; B, $9000; C, $6750. 7. A, B, C, and D gain 30% on the stock ; A, B, and C gain $1150; A, B, D, $1650; B, C, D, $1000; A, C, D, $1600 : what was each man's stock? A, $2666f ; B, $666J; C, $500; D, $2166|. PABTNERSHIR 329 CASE II. 358. To apportion the gain or loes when the fUll shares are not continued throagh the same period. Problem. — A, B, and C are partners; A puts in $2500 for 8 mo.; B, $4000 for 6 mo.; C, $3200 for 10 mo.; their net gain is $4750: divide the gain. Solution. — A*s operation. capital ($2500), $2500X 8 = $20000, A's equivalent, used 8 months, is $4000 X 6= 24000, B's " equivalent to 8 X $3200X10= 32000 ,^8 " $2500, or $20000, $7 6000 used 1 month^'« ^ of $4750 = $1 250, A's share, capital ($4000), |5 of $4750 = $! 500, B^s " used 6 months, IS ^^ $47 50 = $2000, C. - equivalent to 6 X $4000, or $24000, used 1 month ; Cs capital ($3200), used 10 months, is equivalent to 10 X $3200, or $32000 used I month. Dividing the gain ($4750) in proportion to the stock equivalents, $20000, $24000, $32000, used for the pame time (1 month), the results Will be the gain of each; A's $1250, B's $1500, Cs $2000. Bule. — MvMply each partnei^s stock by Hie time it is wsed; and divide Uie gain of loss in proportion to the products so obtained. Examples for Practice. 1. A begins business with $6000 ; at the end of 6 mo. he takes in B, with $10000 ; 6 mo. after, their gain is $3300 : what is each share? A's, $1800; B's, $1500. 2. A and B are partners ; A's stock is to B's, as 4 to 5 ; after 3 mo., A withdraws f of his, and B f of his: divide their year's gain, $1675. A, $800 ; B, $875. 3. A. B, and C join capitals, which are as ^, ^, \; after 4 mo. , A takes out ^ of his ; aft«r 9 mo. more, their gain is $1988: divide it. A, $714; B, $728 ; C, $546. H A. 28. 330 BAY'S HIGHER ARITHMETIC. 4. A and B are partners; A puts in $2500; B, $1500; after 9 mo., they take in C with $5000; 9 mo. afl;er, their gain is $3250 : what is each one's gain ? A's, $1250; B's, $750; Cs, $1250. 5. A and B are partners, each contributing $1000; after 3 months, A withdraws $400, which B advances ; the same is done aft«r 3 months more ; their year's gain is $800 : what should each get ? A, $200 ; B, $600. 6. A, B, and C are employed to empty a cistern by two pumps of the same bore ; A and B go to work first, making 37 and 40 strokes respectively a minute; aft«r 5 minutes, each makes 5 strokes less a minute ; after 10 minutes, A gives way to C, who makes 30 strokes a minute until the cistern is emptied, which was in 22 minutes fi-om the start : divide their pay, $2. A, 46 ct.; B, $1.06; C, 48 ct. 7. A and B are partners; A's capital is $4200; B's, $5600 : aft«r 4 months, how much must A put in, to entitle him to ^ the year's gain ? $2100. 8. A and B go into partnership, each with $4500. A draws out $1500, and B $500, at the end of 3 mo., and each the same sum at the end of 6 and 9 mo.; at the end of 1 j£. they quit with $2200 : how must they settle? B takes $2200, and has t claim on A for $300. 9. A, B, C, and D go in partnership ; A owns 12 shares of the stock; B, 8 shares; C, 7 shares; D, 3 shares. After 3 mo., A sells 2 shares to B, 1 to C, and 4 to D; 2 jno. afterward, B sells 1 share to C, and 2 to D ; 4 mo. after- ward, A buys 2 from C and 2 from D. Divide the year's gain ($18000). A, $4650; B, $4650; C, $4700; D, $4000. 10. A, with $400; B, with $500; and C, with $300, joined in business ; at the end of 3 mo. A took out $200 ; at the end of 4 mo. B drew out $300, and after 4 mo. more, he drew out $150; at the end of 6 mo. C drew out $100; at the end of the year they close ; A's gain was $225 : what was the whole gain ? $675. BANKRUPTCY. 331 BANKRUPTCY. DEFINITIONS. 369. 1. Bankruptcy is the inabUity of a person or a firm to pay indebtedness. 2. A Bankrupt is a pei*son unable to pay his debts. 3. The assets of a bankrupt are usually placed in the hands of an Assignee, whose duty it is to convert them into cash, and divide the net proceeds among the creditors in proportion to their claims. Bemarks. — 1. This act on the part of a debtor ia called making an cuisignmeTity and he is said to be able to pay so much on the dollar. 2. All necessary expenses, including assignee's fee (which is generally a certain rate per cent on the whole amount of property), must be deducted, before dividing. 4. The amount paid on a dollar can be found by taking such a part of one dollar as the whole property is of the whole amount of the debts; each creditor's proportion may be then found by multiplying his claim by the amount paid on the dollar. Note. — Laws in regard to bankruptcy differ in the various states ; usually a bankrupt who makes an honest assignment is freed by law of his remaining indebtedness, and is allowed to retain a homestead of from $500 to $5000 in value, and a small amount of personal property. Examples for Practice. 1. A has a lot worth $8000, good notes $2500, and cash $1500; his debts are $20000: what can he pay on $1, and what will A receive, whose claim is $4500? Solution.— $8000 + $2500 + $1500 = $12000, the amount of prop- erty which is J^g^J or f of the whole debts. Hence, | of $1 = (50 ct., the amount paid on $1, and $4500 X .60 = $2700, the sum paid to A 332 RAY'S HIGHER ARITHMETIC. 2. My assets are $2520 ; I owe A $1200 ; B, $720 ; C, 8600 ; D, $1080 : what does each ^^\.^ and what is paid on each dollar ? A, $840 ; B, $504 ; C, $420 ; D, $756 ; 70 ct on $1. 3. A bankrupt's estate is worth $16000; his debts, $47500; the assignee charges 5% : what is paid on $1, and what does A get, whose claim is $3650 ? 32 ct on $1, and $1 168. Topical Outline. Partnership. c 1. Definitious. 1. Partnership 2. f ^^e I.— Applications. \ Case II.— Applications. f 1. Definitions. \ 2. Applications. 2. Bankruptcy. X YIII. ALLIG ATIOl^I . DEFINITIONS. 860. Alligation is the process of taking quantities of different values in a combination of average value. It is of two kinds, Medial and Alternate. ALLIGATION aMEDIAL. 361. Alligation Medial is the process of finding the mean or average value of two or more things of different given values. Problem. — If 3 lb. of sugar, at 5 ct. a lb, and 2 lb., at 4^ ct. a lb., be mixed with 9 lb., at 6 ct. a lb., what per lb. is the mixture worth ? OPERATION. Solution.— The 3 lb. at 5 ct. Price. Quantity. Cost, a lb. = 15 ct.; the 2 lb. at 4| ct. 5 ct. X 3 = 1 5 ct per lb. == 9 ct.; the 9 lb. at 6 ct. 4 J X 2 =9 per lb. = 54 ct.; therefore, the 6 X 9 = 54 whole 14 lb, are worth 78 ct.; 14 ) 7 8 ( 5 f ct. and 78-i- 14 = 5f ct. per lb., Ans, Bule. — Find the vahtes of Hie definite parts , and divide the sum of Hie values hy the sum of Uie parts. Examples for Practice. 1. Find the average price of 6 lb. tea, at 80 ct.; 15 lb., at 50 ct.; 5 lb., at 60 ct.; 9 lb., at 40 ct. 54 ct. per lb. 2. The average price of 40 hogs, at $8 each ; 30, at $10 each ; 16, at $12.50 each ; 54, at $11.75 each, $10.39 each. (.1.33) 334 BAY'S HIGHER ARITHMETIC, 3. How fine is a mixture of 5 pwt. of gold, 16 carats fine ; 2 pwt, 18 carats fine; 6 pwt., 20 carats fine; and 1 pwt. pure gold? 18f carats fine. 4. Find the specific gravity of a compound of 15 lb. of copper, specific gravity, 7|; 8 lb. of zinc, specific gravity, 6J; and \ lb. of silver, specific gravity, lOJ^. 7.445 — Remarks. — 1. By the specific gravity of a body is usually under- stood, its weight compared with the iveight of an equal bulk of tvater; it may be numerically expressed as the quotient of the former by the latter. Thus, a cubic inch of silver weighing 10} times as much as a cubic inch of water, its specific gravity •■= lOj. 2. To find the specific gravity of a body heavier than water: (1.) Find its weight in air , (2.) Suspending it by a light thread, find its weight in water and note the difference; (3.) Divide the first weight by this difference. For example : if a piece of metal weighs If oz. in air, but in water only IJ oz., its specific gravity = 1} -i- (1 J — IJ) = 7. (See Norton*8 Natural Philosophyy p. 152.) 5. What per cent of alcohol in a mixture of 9 gal., S6% strong; 12 gal., 92^ strong; 10 gal, 95% strong; and II gal, 98^ strong? 93%. 6. At a teacher's examination, where the lowest passable average grade was 50, an applicant received the following grades: In Orthography, 50; Reading, 25; Writing, 50; Arithmetic, 60 ; Grammar, 55 ; Geography, 55 : did he suc- ceed, or did he fail ? He failed. ALLIGATION ALTERNATE. 862. Alligation Alternate is the process of finding the proportional quantities at given particular prices or values in a required combination of given average value. CASE I. 363. To proportion the parts, none of the quanti- ties being limited. ALLIGATION. 335 Problem.— 1. What relative quantities of sugar, at 9 ct. a lb. and 5 ct. a lb., must be used for a corapound, at 6 ct. alb.? 9 Solution. — If you put 1 lb. at 9 ct. in the mixture to be sold for 6 ct., you lose 3 ct.; i£ you put 1 lb. at 5 ct. in the mixture to be sold at 6 ct., you gain 1 ct.; 3 such lb. gain 3 ct.; the gain and loss would then be equal if 3 lb. at 5 ct. are mixed with 1 lb. at 9 ct. OPERATION. . 3 lb. at 5 ct. = 1 5 ct. . 1 lb. at 9 ct. = _9 ct. 4 lb. worth 24 ct. which is ^ = 6 ct. a lb. Problem. — 2. What relative numbers of hogs, at $3, $5, $10 per head, can be bought at an average value of $7 per head? OPERATION. Diff. Balance. Am, 1 1 2 Explanation. — Wr iti ng the average price 7, and the particular values 3, 5, 10, 3 4 3 3 as in the margin, we say : 7 5^ _2^ 3 3 3 sold for 7, is a gain of 4, 10 3 4 2 6 which we write opposite ; 5 sold for 7 is a gain of 2 ; 10 sold for 7 is a loss of 3. We wish to make the gains and losses eqwd; hence, each losing sale must be balanced by one which gains. To iose 3 foursy will be balanced by gaining 4 threes^ and a gain of 3 hvos will balance a loss of 2 threes. To indicate this in the operation, we write the deficiency 3, against the excess 4; then the excess 4 against the deficiency 3; and in another column, in the same manner, pair the 3 and 2, writing each opposite the position which the other has in the colun^.n of differ- ences. The answer might be given in two statements of balance, thus : for each 3 of the first kind take 4 of the third, and for each 3 of the second kind take 2 of the third. Since each balance column shows only proportional parts, we may multiply both quantities in any balance column by any numbet^y fractional or integi^aly and thus the final answers be varied indefinitely. For example, had the second balancing column been multiplied by 4, the answer would have read, 1, 4, 4, instead of 1, 1, 2. The principle just stated is of great value in removing fractions from the balance columns, when integral terms are desired. 336 JiA K'aS' higher ARITHMETIC, Bale. — 1. Write tlie jmrticular values or prices in orders in a column^ having tlie smallest at the liead; write the averagfe value in a middle position at the left and separated from iJte others by a vertical line, 2. In another column to tJie right, and opposite Hie respective vcdues, place in order the dijffh'ences between tliem and the average value, 3. Tlien prepare balance columns, giving to each of Hieni two numbers, one an excess and tlie other a deficiency taken from the difference column; write eoc/i of these opposite tJie position which the oilier has in the difference column; so 'proceed until each number in the difference column lias been balanced with anoHier; tlien, The proportional quantity to be taken of each kbid, will be the sum of Hie nuDibers in a horizontal line to the rigid of its excess or deficiency. Note. — The proof of Alligation Alternate is the process of Alli- gation Medial. Examples for Practige. 1. What relative quantities of tea, worth 25, 27, 30, 32, and 45 ct. per lb., must be taken for a mixture worth 28 ct. per lb.? 19, 4, 3, 1, 3 lb. respectively. Remark. — It is evident that other results raay by obtained by making the connections differently ; as, 6, 17, 3, 3, 1 lb.; or, 17, 6, 1, 1, 3 lb. 2. What of sugar, at 5, 5^, 6, 7, and 8 ct. per lb., must be taken for a mixture worth 6f ct. per lb.? 1, 5, 5, 7, 8 lb. respectively; or, 5, 1, 1, 8, 7 lb., etc. 3. What relative quantities of alcohol, 84, 86, 88, 94, and 96% strong, must be taken for a mixture 87^ strong? 10, 7, 3, 1, 3 gal.; or, 7, 10, 1, 3, 1 gal., etc. 4. What of gold and silver, whose specific gravities are ALLIGATION. 337 19 J and 10|, will make a compound whose specific gravity shall be 16.84 ? 723 lb. silver to 3487 lb. gold. 5. What of silver f pure, and -^ pure, will make a mixt- ure \ pure? 1 lb., f pure; 5 lb., ^ pure. 6. What of pure gold, and 18 carats, and 20 carats fine, must, be taken to make 22 carat gold ? 1 part 18 carats, 1 part 20 carats, 3 pure. CASE II. 864. To proportion the parts, one or more of the quantities, but not the amount of the combination, being given. Problem. — How many whole bushels of each of two kinds of wheat, worth respectively $1.20 and $1.40, per bushel, will, with 14 bushels, at $1.90 per bushel, make a combination whose average value is $1.60 per bushel? Diff. Bal. OPERATION. AiiRwerB. 1.60 1.20 .40 3 1.40 .20 3 1.90 .30 4 2 1 19 14 2 17 14 3 15 14 4 13 5 11 14114 6 9 14 7 7 14 8 5 14 9 3 14 10 1 14 Solution. — We find, by Case I, that to have that average value the parts may, in one balancing, stand 3 of the first to 4 of the third, and in another, 3 of the second to 2 of the third. By directly com- bining, we obtain the proportions 1, 1, and 2, and as the third must be 14, we have for one answer 7, 7, 14. But we find the other answers in the following manner. The proportion will not be altered if in any balancing column we multiply both quantities by the same number, hence the answer can be varied as often as we can multiply or divide the columns, ob- serving the other conditions, which are that the answers shall be integral, and that the number of 4'8 and 2's taken shall make 14. As there are more fractions than there are integers between any two limits, we try fractional multipliers in order to obtain the greatest number of answers. Observe that the number of 4's taken will not stand alone in any answer for the third kind of wheat, but will be H. A. 29. 338 liA Y'S HIGHER ARITHMETIC, added to some humber of 2's ; the number of d's taken as any one answer for the first or second kind vyiU not be increased by any other product ; hence, if we use a frcuUional iuultiplier, it must be such that its denominator will disappear in multiplying by 3 ; and this shows that a fractional multiplier will not be convenient unless it can be expressed as thirds. Therefore, the remaining question is, How many thirds of 4 with thirds of 2 will make 14? Since 14 = ■^, the question is the same as to ask, How many whole 4*s with whole 2's will make 42 ? It is plain that there can not be more than ten 4*s. We can take J of 1 four and 19 twos, or J of first column and -^ of second. 2 fours " 17 " " f " " " " V " " Q « (( 1C (C U 3 U t( U it J5 (t it The answers are now obvious : write 1, 2, 3, etc., parallel with 19, 17, 15, etc., and the 14's in the third row. Rule. — Find Hie proportional parts, as in Case J, and ob- serve the term or terms in the balance columns , standing oppo- site Hie value or price corresponding to the limited quantity; tJien find what midtipliers iviU produce tJie given limited quan- tUy in the required place, and of Hiose multipliers v>se only such as will agree with tlie remaining conditions of the problem. Examples for Practice. * 1. How many railroad shares, at 50%, must A buy, who has 80 shares that cost him 72%, in order to reduce bis average to 60%? 96 share.^ 2. How many bushels of tops, worth respectively 50, 60 and 75 ct. per bu., with 100 bu., at 40 ct. per bu., will mak a mixture worth 65 ct. a bu.? 2, 2, and 254 3. How much water (0 per cent) will dilute 3 gal. 2 qt. 1 pt. of acid 91 % strong, to 56^ ? 2 gal. 1 qt. ^ pt. 4. A jeweller has 3 pwt. 9 gr. of old gold, 16 carats fine : how much U. S. gold, 21f carats fine, must he mix with it, to make it 18 carats fine? 1 pwt. 21 gr. ALLIGATION. 3e39 5^ How much water, with 3 pt. of alcohol, 96% strong, and 8 pt., 78^, will make a mixture 60% strong? 4\ pt. 6. I mixed 1 gal. 2 qt. ^ pt. of water with 3 qt. 1 J pt. of pure acid; the mixture has 15% more acid than desired: ho^ much water will reduce it to the required strength? 1 gal. 2 qt. 1^ pt. 7. How much lead, specific gravity 11, with ^ oz. copper, sp. gr. 9, can be put on 12 oz. of cork. sp. gr. ^, so that the three will just float, that is, have a sp. gr. (1) the same as water ? 2 lb. 1\ oz. 8. How many shares of stock, at 40%, must A buy, who hai bought 120 shares, at 74^, 150 shares, at 68^, and 130 shares, at 54^, so that he may sell the whole at 60^, and gain 20% ? 610 shares. 9. A buys 400 bbl. of flour, at $7.50 each, 640 bbl., at $7.25, and 960 bbl., at $6.75: how many must he buy at $5.50, to reduce his average to $6.50 per bbl.? 1120 bbl. CASE III. 865. To proportion the parts, the amount of the whole combination being given. Problem. — If a man pay $16 for each cow, $3 for each hog, and $2 for each sheep, how many of each kind may he purchase so as to have 100 animals for $600 ? Solution.— Pro- operation. ceeding as in Case DifF. Bal. Answers. I, we find that the given average re- $6 quires 5 of the first kind with 2 of the 7 13 third, and 10 of the second kind with 3 of the third ; taken in two parts, 7 are required in (me balancing, and 13 in another ; these being together 20, which is contained five times in the required number, 100, if we multiply all of the terms in the balance columns by 5, we have for one answer 25 sheep, 50 hogs, and 25 cows. 2 3 16 4 3 10 5 2 10 3 12 64 24 25 5 25 38 36 2 6 51 22 27 64 8 28 340 RAY'S HIGHER ARITHMETia As there are other multipliers affording results within the con- ditionS) we leave the student to find the remaining answers by a process similar to that shown under Case 11. Remark. — Suppose that we had to determine 400 how many 7*s and 11*8 would make 400. By 2 22 trial, we find that tm) ll's taken away leave an 3 7 8. ..5 4 exact number of 7*s. It is now unnecessary to 7 7 take single 11*8 or proceed by trial any farther; 9 3 01... 4 3 for, as 378 is an exact number of 7*8, if we take 7 7 away 11*8 and leave 7*s, we must take seven 11*8; j g 2 2 4... 3 2 and thus the law of continuation is obvious : 400 is composed of . 11*8 2, 9, 16, 23, 30; With 7'8 54, 43, 32, 21, 10. Bule. — Proportion the parts as in Case I; Vien, noting the sums of the balancing columns, find, by trial or by direct di- viston, ivhat multipliers will make those columns together equal to the given amount of the combination, and of tiiose multipliers use only such as will agree with the remaining conditions. Examples for Practice. 1. What quantities of sugar, at 3 ct. per lb. and 7 ct. per lb., with 2 lb. at 8 ct, and 5 lb. at 4 ct. per lb., will make 16 lb., worth 6 ct. per lb.? f lb. at 3 ct, 8^ lb. at 7 ct 2. How many bbl. flour, at $8 and $8.50, with 300 bbl. at $7.50, and 800 at 87.80, and 400 at $7.65, will make 2000 bbl. at «7.85 a bbl. ? 200 bbl., at «8 ; 300 bbl., at $8.50 '3. What quantities of tea, at 25 ct and 35 ct. a lb., with 14 lb. at 30 ct, and 20 lb. at 50 ct, and 6 lb. at 60 ct, will make 56 lb. at 40 ct. a lb.? 10 lb. at 25 ct, and 6 lb. at 35 ct TOPICAL OUTLINE. 341 4. How much copper, specific gravity 7|, with silver, specific gravity 10^, will make 1 tb. troy, of specific gravity 8f ? 7f^ oz. copper, 4Jff oz. silver. 5. How much gold 15 carats fine, 20 carats fine, and pure, will make a ring 18 carats fine, weigliing 4 pwt. 16 gr.? 2 pwt. 16 gr.; 1 pwt.; 1 pwt. 6. A dealer in stock can buy 100 animals for 8400, at the following rates, — calves, $9; hogs, $2; lambs, $1 : how many may he take of each kind? 37 calves, 4 hogs, 59 laml)s; {pile of nine different answers.) 7. Hiero's crown, sp. gr. 14f , was of gold, sp. gr. 19J, and silver, sp. gr. 10 J ; it weighed 17 J lb.: how much gold was in it? lOfJ lb. Topical Outline. AlJ^lGATlON. * r 1. Definitions. . Alligation . 2. Kinds.. . 1. Alligation Medial...... / 1- DefiniUons. I 2. Rules. » 2. Alligation Alteniatc.. . 1. Definitions. £ I. Rulel 2. Cases ..J n. Rule. I III. Rule. XrX. mYOLTTTlOK DEFINITIONS. 806. 1. A power of a quantity is either tliat quantity itself, or the product of a number of factors each equal to that quantity. Kemabk. — Regarding unity as a base, we may say, the power of a quantity is the product arising from taking unity once as a mul- tijilicand, with only the given quantity a certain number of times as a factor. The power takes its name from the number of times the quantity is used as factor. Unity is no power of any other positive number. 2. The root of a power is one of the equal factors which |>roduce the power. 3. Powers are of difierent degrees, named from the number of times the root is taken to produce the power. The degree is indicated by a number written to the right of the root, and a little above ; this index number is called an exponent. Thus, 5X5, or the 2d jmver of 5, is written 52. 5X5X5, " 3d " " " " 5». 4. The second power of any numl>er is called the square, because the area of a square is numerically obtained by form- ing a second power. 5. In like manner the €iird power of any number is called its cube, because the solidity of a cube is numerically ob- tained by forming a third power. 367. To find any power of a number, higher than the first. (342) INVOLUTION. 343 Bole. — Multiply the number by itself ^ and cmUinue the mvlr tiplicatwn till Hud number has been used as jo/dor as many times as are indicated by the exponent. Notes. — 1. The nnmber of multiplications will be one fess than the exponent, because the root is used twice in the first multiplica- tion, once as multiplicand and once as multiplier. 2. When the |K)wer to be obtained is of a high degree, multiply by some of the pK)wers instead of by the root continually ; thus, to obtain the 9th power of 2, multiply its 6th power (64) by its 3d power (8) ; or, its 5th power (32) by its 4th power (16) : the rule being, that the product of any two or more powers of a number is thai power whose degree is eqwil to the sum of their degi-eea, 3. Any power of 1 is 1 ; any power of a number greater' than 1 is gieaier than the number itself : any power of a number less than 1, is less than the number itself. 388. From Note 2, 48 X 48 X 4» X 4^ X 4^ = 4^ «; but the expression on the left is the 5th power of 48; hence, (48)5 =415. i}^Q,t is, when the exponent of the power required is a composite number (15), raise tlte root to a power whose exponent is one of its factors (3), and this resuU to a power whose exponent is the other factor (5). Note. — Let the student carefully note the difference between raising a potver to a power, and mtdliplying together different powers of the same root ; thus, 23X22=2*. Here we have multiplied the cube by the square and obtained the 5th power ; but the 5th power is net the square of the cube; this is the sixth power, and we write (22)3 = 2«, or (23)2 =23^2 ^2«. 369. Any power of a fraction is equal to that power of the numerator divided by that power of the denominator. 370. The square of a decimal must contain twice^ and its cubCy three times as many decimal places as the root, etc.; 344 BA yS HIGHER ARITHMETia hence, to obtain any power of a decimal, we have the foUow- ing rule: Rule. — Proceed as if the decimal were a whole number, and point off in the remit a number of dedmxd places equal to the number in the root multiplied by the exponent of Hie power. Examples for Practice. Show by involution, that: 1. (5)* equals 25. 8. (|)« equals iff|J. 2. 14* 2744. 9. (.02)* " .000008 3. 6* 7776. 10. (5*)2 •' 390625. 4. 192" 36864. 11. (.046)8 " .000097336 6. 1»"> 1. 10 (\\1 n I ^^- W 4'Jiiiti- 6. (I)* tMt- 13. 20562 " 4227136. 7. (2i)« iiH- 14. (7.62i)2 " 58. 1406 J 371. Special processes for squaring and cubing numbers. Problem.— Find the square of 64. 64 64 PARALLEL OPERATIONS. 6 +4 = 6 tens + 4 units. 60 + 4 256 3840 60X4H-42 602+60X4 4096= 602 + 2(60X4) + 4a = 3600 + 480+16. The operations illustrate the following principle : Principle. — The square of tlie sum of two numbers is equal to ihe square of the first, plus twice the product of the first by the second, plus the square of the second. Thus : ^ \ - A >, ^ y -^x^ OF THE INVOLUTrON\ ^p 252 = (22 + 3)2 = 484 + 2 (22 X 3)+ & = 625. 252 = (21 + 4)2 = 441 + 2 (21 X 4) + 16 = 625. 252 = (20 + 5)2 = 400 + 2 (20 X 5) + 25 = 625. Remark. — The usual application of this principle in Aritkmetie is, in squaring a number as composed of tens and units. The third statement above illustrates this ; and, if we represent the tens by t, the units by u, we have the following statement : (t + u)*^=l^ + 2tu+u^; or, in common language : The square <^ any number composed of tens and units is equal to the square 0/' the tens, + twice the product 0/ the tens by the units, + the squa:i^ oj the units. Examples for Practice. Square the following numbers, considering each as the sum of two quantities, and applying the principle announced in Art. 371, on page preceding: 1. 19. 361. 4. 40. 1600. 2. 29. 841. 5. 125. 15625. 3. 4. 16. 6. 59. 3481. Illustration. — Draw a square. From points in the sides, at equal distances from one of the comers, draw two straight lines across the figure, each parallel to two sides of the figure. These two lines will divide the square into four parts, two of them being squares and two of them rectangles. The base being composed of two lines, and the square of four parts, we see that 372. The square described on the sum of two lines is ^ual to the sum of the squares described on the lines, plus twice the rectangle of the lines. Remark.— Both the principle employed above and the illustra- tion are frequently used in explaining the method for square root 346 BAY'S HIGHER ARITHMETIC. Problem. — ^Find the cube of 64. PARALLEL OPEBATIONS. 64«= 4096= 60» + 2(60 X4) + 4» 64= 604-4 16384= 60 2X4+2(60X42) + 4» 24576 = 608 + 2(60»X4)4-3(60X4M 64» = 262144— 60 » + 3(602 X4) + 3(60X4 2)4-4 ». The operation illustrates the following principle : Principle. — The cube of any number composed of two parts, is equal to Vie cube of the first part, phis three times the square of the first by tlie second, plus three times tlie first by the square of the second, plus tfie cube of the second. Thus : 258 = (22 + 3)8 =228 ^ 3 (222 x 3) + 3 (22 X 32) 4- 38 = 10648 + 4356 4- 594 4- 27 = 15625. Bemark. — The usual application of this principle in Arithmetic is, in the cubing of a number as composed of tens and units. Bei)re- senting the tens by i, and the units hy u, we have the following state- ment, which the student will express in common language, similar to that of the principle used in squaring numbers : {t 4- uy = e» 4- st^u + Sfu^ 4- ti8. Examples for Practice. Considering the following numbers as made, each, of two parts, cube them by the principle just stated : (1.) 19. 6859. (4.) 40. 64000. (2.) 29. 24389. (5.) 125. 1953125. (3.) 4. 64. (6.) 216. 10077696. XX. EYOLUTIOK DEFINITIONS. 373. 1. Evolution is the process of finding roots of numbers. 2. A root of a number is either the number itself or one of the equal factors which, without any other factor, produce the number. Since a number is the first root, as also the first pouer of itself, no operation is necessary to find either of these ; hence, in evolu- tion, we seek only one of the equal factors which produce a power. Evolution is the reverse of Involution, and is sometimes called the Extraction of Hoots. 3. EootSf like powers, are of different degrees, 2d, 3d, 4th, etc.; the degree of a root is always the same as the degree of the power to which that root must be raised to produce the given number. Thus, the 3d root of 343 is 7, since 7 must be raised to the 3d power, to produce 343 ; the 5th root of 1024 is 4, since 4 must be raised to the 5th jjower, to produce 1024. Since the 2d and 3d powers are called the square and cube, so the 2d and 3d roots are called the square root and cube root. 4. To indicate the root of a number, we use the Radical Sign (]/), or a fractmial exponmt The radical sign is placed before the number; the degree of the root is shown by the small figure between the branches of the radical sign, called the Index of the root. Thus, ^18 signifies the cube root of 18 ; ^^9 signifies the 5th root of 9. The square root is usually indicated without the index 2; thus, 1^10 is the same as f'lO. (347) 348 HAY'S HIGHER ARITHMETIC. 5. The root of a uiimber raay be indicated by a fractional exponent whose numerator in 1, a7id whose denominator is the index of the root to be expressed. Thus, 1^7 = 7% and 1^5 = 5^ ; fiiniilarly, 4* = 16% the numera- tor indicating a power and the denominator a root. 6. A perfect power is a number whose root can be exactly expressed in the ordinary notation; as 32, whose fifth root is 2. 7. An imperfect power is a number whose root can not be exactly expressed in the ordinary notation ; as 10, whose square root is 3.1622-[- 8. The squares and cubes of the first nine numbers are as follows: Numbers, 123456789 Squares, 1 4 9 16 25 36 49 64 81 Cubes, 1 8 27 64 125 216 343 512 729 9. The Square Boot of a number is one of the two equal factors which, without any other factor, produce that num- ber ; thus, 7X7 = 49, and i/49 = 7. 10. The Cube Boot of a number is one of the three equal factoi-s which, without any other, produce that num- ber; thus, 3 X 3 X 3 = 27, and ^27 = 3. 374. Concerning powers and roots in the ordinary deci- mal notation, we state the following principles: Principles. — 1. The square of any nwnber has twice as ynanyy or mie less ilmn twice as many, figures as Hie number itself lias. 2. 2%cre mil be as many figures in the square root of a perfect powei* as there are periods of two figures each in the power, beginning wiHi units, and also a figure in tlie root cor- responding to a part of sucJi period at Uie left in Hue power. EVOLUTIOK 349 3. The cube of a number has three times as many figures, or one or two less than Viree times as viant/y as the number itself has. 4. TJwre wUl be as vmny figures in Hie cube root of a perfect poivery as there are periods of three figures each in the po^ver, beginning with units, and also a figure corresponding to any part of sudi a period at the left hand. Exercises. 1. Ppove that there will be six figures in the cube of the greatest integer of two figures. 2. Prove that there will be twelve figures in the fourth power of the greatest integer of three figures. EXTRACTION OF THE SQUARE ROOT. FiEST Explanation. pROBLE^r. — What is the length of the side of a square containing 576 sq. in.? Solution. — The length required will operation. be expressed by Ihe square root of 576 ; 5 7 6(20 by Principle 2, we know that the root can 4 have no less than two places of figures; 40 400 24 in. Ans, and since the scjuare of 3 tens is greater, 4 ry^ and that of 2 tens less than 576, the root Jl i 7 ft must be less than 30 and greater than 20; hence, 2 is the first figure of the root, and 400 the greatest square of tens contained in 576. Let the first of the accompanying figures represent the square whose side is to be found. We see that the side must be greater than 20, and that the given area exceeds by 176 sq. in. the square whose side is 20 in. long. It is also evident from the figure that the 176 sq. in. may be regarded as made of three parts, two of them being rectangles and one a small square ; these parts are of the same width, and, if that width be ascertained and 350 RAY'S IIIOHEB ARITHMETIC. 20X4=80 20X20=400 16 % II X s s added to 20 in., the required side will be found. The two 20-inch rectangles, with the .small square, may be considered as making one long rectangle of the required width, as shown in the figure on the right ; and, as the exact area of that rectangle is 176 sq. in., if we knew its length, its true width would be found by dividing the area by the length (Art. 197, 7); but we do know that the length is greater than 40 in., and hence, that the width is, in inches, less than the quotient of 176 by 40; and since, in 176, 40 is contained more than four times, but not five times, 4 is the highest number we need try for the width. Now, as the true length of that rectangle is 40 in. increased by the true widthy the proper uxiy to try 4, is to add it to 40 and multiply the sum by 4; thus, 40 in.-f 4 in. = 44 in.; and 44 X 4 = 176. This shows that 4 in. is the width of the rectangle, and hence the required side is 20 in. -f 4 in. = 24 in., Ans, Remarks. — 1. Since 17 contains 4 as often integrally ^ as 176 con- tains 40, it is convenient to use simply the 17 as dividend with 4 as divisor, and then annex the quotient to the divisor and to the first figure of the root. 2. At the first step we ascertained that the whole root was greater than 2 tens and less than 3 tens ; at the next step we learned that the units were not equal to 5, and by trial they were found to be 4. The whole process was a gradual approach to the exact root, — one figure at a time. It is im()ortant for the student to note that in the processes of evolution there must be steps of trial. Even the higher branches will not exempt from all trial work. The most valuable rules pertaining to such numerical opera- tions, simply narrow the trial by making the limits obvious. Thus our device above showed the second part of the root less than 5; an actual trial showed it to be exactly 4. 3. If the power had been 58081, we should have found there were three figures in the root; here, as in the former caSiS, 4 is found the greatest figure 44 58081(241 180 176 481 481 481 EVOLUTION, 351 which can stand in ten*B place, <ind we may treat the 24 tens exactly as we treated the 2 tens in the first illustration. Second Explanation. We learned in Art. 371 that the square of a number composed of tens and units is equal to the square of the tens, i)lus twice the product of the tens by the units, plus the square of the units. The square of (20 + 4), or 24a, jg 202 + 2 X (20 X 4) + 4}. Now, if the square of the tens be taken away, there will remain 2 X (20 X 4) 4- 42 = 40 times 4, and 4 times 4, or, simply (40 + 4) X 4. We see then that if the square of the tens be taken away, the remain- der is a product whose larger factor is the douUe of the tens, increased by the unitSj the smaller factor being simply the units. Suppose, then, in seeking the square root of 1764, we have found the tens of the root to be 4 ; the remainder 164 must be the product of the units by a factor which is equal to the sum of twice the tens and once the operation. units. If we knew the units, that larger 17 6 4(40 factor could be found by doubling the tens 1600 2 and adding the units; if, on the other 80 hand, we knew the larger factor, the units 2 could be found by diject division ; we cfo 32 know that larger factor to be more than 80, and hence that the units factor is less than the exact number of times 164 contains 80. Therefore, the units figure can not be so great as 3, and the largest we need try is 2. The proper uxty to try 2, is to add it to 80, and then multiply by 2 ; this being done, we see that, the product being equal to 164, 2 is the exact number of units, 80 + 2 the larger factor exactly, and 42 the exact root. Note. — These successive steps showed, that the first figure toas the root as nearly as tens could express it ; with the second figure we found the root exactly. Had the power been 1781, the 42 would still have been the true root as far as tens and units could express it; and at the next step, seeking a figure in tenth^s place, we would have found the true root, 42.2, as far as expressible by tenSj unitSy and tenths. Continuing this operation, we find 42.201895 to be the root, ti'ue as far as miUionihs can express it ; so, in any case, when a figure is correctly found, the true root can not differ from the whole root obtained, by so much as a unit in the place of that figure. 16 4 4 2, Ans, 164 1 \ 352 BAY'S mOHER ARITHMETIC. 375. To extract the square root of a number writ- ten in the decimal notation, as integer, fraction, or jtnixed number. * BtQe. — 1. Point off the number into periods of tvx) figures eachf commencing with units. 2. Find the greatest square in the first period on the left; place its root on the right, like a quotierd in division; subtract the square from the period, and to the remainder bring dovm the next period for a dividend. 3. Double the root already found, as if it were units, and uTite it on the left for a trial divisor; find how often this is contained in (he dividend, exclusive of the right-hand figure, annexing the quotient to the root and to the divisor; then mul- tiply the complete divisor by the quotient, subtract the product from the dividend, and to ike remainder bring down the next period as before. 4. Double the root as before, place it on the left as a trial divisor, proceeding as with the former divisor and quotient figure; continue the operation until the remainder is nothing, or until the lowest required decimal order of the root has been obtained. Notes. — 1. It any product be foand too large, the last figure of the root is too large. 2. The number of decimal places in the power must be even; hence the number of decimal periods can be increased only by annexing ciphers in pairs. Contracted division may be used to find the lower orders of an imperfect root. • 3. When a remainder is greater than the previous divisor it does not follow that the last figure of the root is too small, unless that remainder be large enough to contain twice the part of the root already found and 1 more; for this would be the complete divisor, and would be contained in the remainder if the root were increased by 1. Hence, the square of any number mtist be increased by a unit more than twice the number itself, to make the square of the next higher. Thus, 1252 =15625; simply add 250+1, and find 15876=126«. EVOLUTION. 353 Examples for Practice. 1. 1/2809. 2. 1/1444. 53. 38. 109. 13625. 8944.9 2490.74 uare root ( 7. 1/ 3444736. 8. 1/57600. 1856. 240. 3, 1/11881. 9. 1/16499844. 10. 1/49098049. 11. 1/73005. 12. v''386». )f 3, true to the 4062. 4. 1/185640625, 5. 1/80012304. 6. 1/6203794. 13. Find the sqi 7007. 270.194 7583.69 7th decimal place. 1.7320508 14. Find the square root of 9.869604401089358, true to the 7th decimal place. 3.1415926 15. 1/.030625 X 1/40.96 X l/.00000625 = what? .0028 16. 1/(126)2 X (58)2 X (604)2 = 4414032. 17. 1/12.96 X sq. rt. of ^2^^ = 3.2863 Kemaue. — The remainder, at any point, is equal to the square of an unknovm part, plus twice the product of that part by a Arnoim part. The remainder may also be considered as the difference of (too squares^ which is always equal to the product of the mm of the roots by their difference, 376. The square root of the product of any number of quantities is equal to the product of their square roots; thus, i/l6 X .49 = 4 X .7 = 2.8 377. The square root of a common fraction is equal to the square root of the numerator, divided by the square root of the denominator. Remark. — It is advantageous to multiply both terms by what will render the denominator a square. Examples for Practice. 1. V>_^= 2. i/34| = H. A. 30. .92582+ 5.8843+ S. V^^ = ^ nearly. 4. i/272^= 16f 354 JRA Y'S HIGHER ARITHMETIC. 5. i/6f = 2.5298+ 6. V^ ifXi^JVXMX l/3"= .45886+ 7. 1/123.454321 X .81 = 9.9999 8. 1/1.728 X 4.8 X^ = ^^ X l/^T, written also 2^ 1/21. EXTRACTION OF THE CUBE ROOT. First Explanation. Problem. — What is the edge of a cube whose solid inches number 13824? OPERATION. 13824(20 8000 4 20^X3 =1200 20 X3X4 = 240 42= 16 1456 5824 24, ^ns. 5824 Solution. — The length required will be expressed by the cube root of 13824. By Prin. 4, we know the root can have no less than two places of figures; al:?o, since the cube of 3 tens is greater and that of 2 tens is less than 13824, the root must be less than 30 and more than 20; conse<iuently, 2 is the fii'st figure of the root, and 8000 is the greatest cube of tens contained in 13824. Let the first of the accomi»anying figures represent the cube whose edge is to be found. We see that it must be greater than 20 inches, and that the given solidity exceeds by 5824 cu. in. the cube whose edge is 20 inches, and which for convenience we will call A. (See Fig. 2.) It is also evident from the second figure, where the separate parts are shown, that the 5824 cu. in. may be regarded as made up of seven parts, three of them being square blocks ( 5) 20 in. long, three of them being rectangular blocks ( 0) of the same length, and one a small cube ( C). These parts are of the same thickness, and if that thickness be ascertained and added to 20 in., tiie required e<lge will be found. The scpiare blocks and the oblong blocks, with the small cube, may be considered as standing in line Fig. 1. EVOLUTION, 355 ow, as (Fig. 3) and forming one oblong solid of uniform thickneRs. N the exact solidity of that solid is 5824 cu. in., if we knew its side surface, it8 true thickness would be found by dividing the number expressing the solidity by the number expressing the surface. But we do know that side surface to be greater than 3 times 400 sq. in., and hence the thickness must be less than the quotient of 5824 by 1200; and since in 5824, 1200 is con- tained 4 times but not 5 times, 4 is the highest number we need try for the width ; as the exact surface of one side is equal to 1200 sq. in., increased by 3 rectangles 20 in. long, and a small square also, each of a width equal to the required thickness, the proper way to try 4 for that thickness is, to multiply it by 3 times 20, then by itself, and, adding the prod- ucts to 1200, multiply the sum by 4 ; thus, 1200 sq. in. + 80 sq. in. X 3 -f 16 sq. in. = 1456 sq. in., and 1456 X 4 = 5824, which ly Fig. 2. -y N \ \ S 3; B D n Fis. rt. shows that 4 in. is the true thickness, and 20 in. -f 4 in. = the re- quired edge, 24 in., Ans. Note.— Fig. 1 shows also the complete cube with the section lines marked. 356 RAY'S HIGHER ARITHMETIC. Remarks. — 1. Since 3 times the square of 2 tens is equal to 300 times the H<|uare of 2, it is allowable to use simply the square of the first part of the root as units, multiplying by 300 for a trial divisor ; and so, too, in the second part of the trial work, it will answer to multiply the first part by the last found figure and by 30. 2. The steps are trial . , , , OPERATION. i steps, and as remarked ... , under the rule for square 14172488(242 root, our artifices have r simply narrowed the 2^X300 = 1200 6172 range of the trial. 2X^X30= 240 3. Had the power been ^ ^ = ^^ 14172488, there would 1456 _5824 have been three figures in the root; and here, as 24^X300 = 172800 in the former case, the 24X2X30= 1440 second figure is 4 ; and 2 ^ = 4 348488 348488 we may treat the 24 tens 174244 exactly as we treated the 2 tens in the former illustration. Second Explanation. We have seen (page 346) how the cube of a number, of two figures, is composed ; that, for example, 24 3=8000 + a(20 2X4) + 3(20X4 2) + 4». Here "we see that if the cube of the tens be taken away, there will remain (202X3 + 20X4X3 + 42)X4; that is, the remainder may be taken as a product of two factors, of which the smaller is the units, and the larger made up of 3 times the square of the tens, with 3 times the tens by the units, with also the square of the units. Suppose, then, in seeking the cube root of 74088, we find the tens to be 4; the remainder 10088 must be the product of units by a factor comix)sed of three parts, such as we have described. If we knew the larger factor, the units could be obtained by direct division ; but we do know that larger factor to be greater than 3 times the square of 40 ; hence, we know the units must be less than the quotient of 10088 by 4800, and consequently 2 is the largest figure we need try for units. The way to try 2, is to compose a larger factor M EVOLUTION, 357 after the manner just described ; operation. hence, multiply 2 by 3 times 40, add 22 or 4, add the sum to the 4800, and multiply by 2, 42X300 = 4800 which, being done, shows that 4X2X^0= 240 2 is the units figure, and that 2 2 = 4 42 is the root sought. * 5944 74088(42 64 10088 10088 Hem ARK. — Three times the square of the tens is the convenient trial divisor. This is in most instances a greater part of the com- plete divisor; for example, the least number of tens above otw ten is 2, and the greatest figure in unit's place can not exceed 9; the cube of 29 is 24389, the first complete divisor is 1821, the first trial divisor being 1200, a greater part of it. 378. To extract the cube root of a number written in the decimal notation as whole number, fraction, or mixed number. Bule. — 1. Beginning with units , separate tlie number into periods of three figures each; the extreme lefi period niay Irnve hut one or two figures, but the extreme right, whether of units or decimal orders, must have three places, by the annexing of ciphers if necessary, 2. Find tJie greatest cube in the highest period, place its root on the right as a quotient in division, and tlien subtract the cube from the period, bringing down to the right of the remain- der the next period to complete a dividend, 3. Square Hie root found as if it were units, multiply it by 300, and place the product on the left ojs a trial divisor] find how often it is contained in the dividend, and place tlie quo- tient figure to the right of the root ; midtiply the quotient by 30 tiines Hie preceding part of the root as units, square also the quotient, and add tlie two results to the trial divisor; tJien mul- tiply the sum by the quotient, and subtract the product from the dividend, annexing to the remainder another period as before. 4. Square Hie wlwle root as before, multiply by 300, pro* 358 JiAY'S HIGHER ARITHMETIC. deeding as rmth the former trial divisor^ quotient, and addi- tions; continue Uie operaiion until Hie retnainder is notliing, or until Hie lowest required decimal order of Hie rod has been obtained. Notes. — 1. Should any product exceed the dividend, the quotient figure is too large. 2. If any remainder is larger than the previous divisor, it does not follow that the last quotient figure is too small, unl&ss the remainder is ku'ge enough to ccmtain 3 times the square of thai part of the root already found, with 3 times that part of the root, and 1 more / for this is the proper divisor if the root is increased by 1. 3. Should decimal periods be required beyond those with which the operation begins, the operator may annex three ciphers to each new remainder. 4. When the operator has obtained one more than half the re- quired decimal figures of the root, the last complete divisor and the last remainder may be used in the manner of contracted division. 879. The cube root of any product is equal to the prod- uct of the cube roots of the factors. Thus, #^250 X 4 X 648 X 9 = #^125 X 8 X 216 X 27 = 5X2X6X3. 380. The cube root of a common fraction is equal to the quotient of the cube root of the numerator by the cube root of the denominator. Bemarks. — 1. When the terms are not both perfect cubes, multi- ply both by the square of the denominator, or by some smaller factor which will make the denominator a cube. 2. Reduce mixed numbers to improper fractions, or the fractional parts of such numbers to decimals. Examples for Practice. 1. ^512. 8. 2. 1^19683. 27. 3. 1^7301384. 194. 4. f^94818816. 456. EVOLUTION. 359 5. #"1067462648. 1022. 6. 1^5.088448. 1.72 7. ^22188.041 28.1 8. #"32:65 3.196154 9. #':0079 .1991632 10. #"3.0092 1.443724 11. f':^ .315985 12. #"25. 2.924018 13. #"TT. 2.22398 14. #"17 .87358 15. #"^. .64366 16. #"171.416328875 5.555 17. #'70Tr 19.1393267 1^- "^SIt- .2218845 19. #"1 of ^. .6235319 EXTEACrriON OF ANY EOOT. 381. We have seen in the chapter on Involution, that if a 'power be raised to a power, the new exponent is the product of the given exponent by the number of times the power is taken as a factor; that, for example, 2^ raised to the 4th power, is 2^^*=2^2^ Consequently, reversing that process, a power may be separated into equal factors, if the given exponent be a composite number; thus, 2^^ = 24X8^ and consequently \/2^=2^, 0YfV^=2^\ for 2>2 equals either 2^ X 2* X 2*, or 2^ x 2^ x 23 x 2^. It is important to note the distinction between separating a power into factors which are powers, and separating that power into equal factors, having the same roots and exponents. The latter separa- tion would be extracting a root. Thus, 25 — 23X22 because equal to 23+2. But the square root of 2^ is no< 2 3, nor is the cube root of 2^ equal to 2 2. If 23 be taken twice as a faxior we have 2^, and V2^ = 2^ ; similarly, ^26"= 22. 382. A root of any required degree may be extracted, by separating the number denoting tlie degree, into its factors, and extraditing successively Hie roots denoted by those factors. Thus, the 9th root is the cube root of the cube root, and the 6th root the square root of the cube root. 360 RAY'S HIGHER ARITHMETia HORNER'S METHOD. 383. Homer's Method, named from its inventor, Mr. W. G. Horner, of Bath, England, may be advantageously a])plied in extracting any root, especially if the degree of the root be not a composite number. Biile for Extracting any Boot. — 1. Make as many columns as there are units in the index of the root to he extracted; place the given number at the head of the right- hand column, and ciphers at the head' of the others, 2. Commencing at the right, separate the given number into periods of as many figures as there are columns ; extract the required root to within unity, of the left-hand period, for the 1st figure of the root. 3. Write this figure in the 1st column, multiply it then by itself, and set it in the 2d column; multiply this again by the same figure, and set it in the Sd column, and so on, placing the last product in the right-hand column, under that part of the given number from which the figure was derived, and subtracting it from the figures above it 4. Add the same figure to the 1st column again, multiply the result by the figure again, adding the product to the 2d column, and so on, stopping at the neoct to the last column. 5. Repeat this process, leaving off one column at the right every time, until all the columns have been thus dropped; then annex one cipher to the number in the 1st column, two to the number in the 2d column, and so on, to the number in the last column, to which the next period of figures from the given number must be brought down, 6. Divide the number in the last column by the number in the previous column as a trial divisor (making allowance for completing the divisor) ; this will give the 2d figure of the root, which must be used precisely as the 1st figure of the root has been ; o.nd so on, till all the periods have been brought doum. EVOLUTION. 361 Problem.— Extract the fourth root of 68719476736. 6 6 25 50 126 375 68719476736(512 Jm 625 10 5 75 75 ♦500000 ' 15201 * 621947 515201 15 5 ♦15000 201 515201 ' 15403 * 1067466736 1067466736 *200 1 15201 202 ♦530G04000 3129368 201 1 15403 203 633733368 • 202 1 ♦1560600 4084 203 1 ♦2040 2 1564684 2042 Note. — It is convenient to denote by ♦ the place where a column is dropped; i, e., reached for the last time by the use of the root figure in hand. 384. The process may often be shortened by thb Contracted Method. — Obtain one less than half of the figures required in the root as Hie rule directs; ^len, instead of annexing dpJiers and bringing down a period to the last numbers in the columns, leave tlie remainder in Hie right-liand column for a dividend; cut off the right-hand figure from tJie last number of the previous column, two ri^it-hand figures from the last number in the column before tJiaty and so on, ahvays cutting off one more figure for every column to the left. With the number in the rightrhand column and the one in the previous column, determine Hie next figure of the root, and use it as directed in the rule, recoUeoting that (he figures cut off are not used except in carrying the tens they produce, H. A. 31. 362 EA Y'S HIGHER ARITHMETIC, This process is continued until the required number of figures is obtained, observvig Hwt when all Vie figures in the last number of any column are cut off, thai column will be no longer used. Remark. — Add to the 1st column mentally; multiply and add to the next column in one operation: multiply and subtract from tlie right-hand column in like manner. Problem. — Extract the cube root of 44.6 to six decimals. 44.600(3.546323 3 9 17600 6 2700 1725000 90 3175 238136 95 367500 12182 100 371716 865 1050 37594^ 111 1054 37659 1058 zi^U %<i>n Remakk.— The trial divisors may be known by ending in two cipliers; the com iplete diviwrs stand just beneath them. After get- ting 3 figures of the root, contract the operation by last rule. Examples for Practice. 1. Extract the square root of 15625. 2. Extract the cube root of 68719476736. 3. Extract the fifth root of 14348907. 4. The cube root of 151. 5. 1/97:41 6. t/TM 8. ^/35.2 12. 9. -(7782757789696. 10. v" 1367631. 125. 4096. 27. 5.325074 3.1416 1.01943 .83938 2.03848 9.79795897 4.8058955 EVOLUTION. 363 APPLICATIONS OF SQUARE ROOT AND CUBE ROOT. DEFINITIONS. 386. 1. A Triangle is a figure which has three sides aud three angles; as, ABC, MNP. 2. The Base of a triangle is that side upon which it is supposed to stand; as, AB, MN. 3. The Altitude of a triangle is the perpendicular dis- tance from the base to the vertex of the angle opposite; as, HP. Remark. — The three angles of a triangle are together eqnal to 180°, or two right angles. The proof of this belongs to Geometry, but a fair iUvMratlon may be made in the manner indicated above. Mark the angles of a card or paper triangle, 1 , 2, 3 ; and by two cuts divide it into three parts. Place the marked angles with their Ver- tices as at O, and it will be seen that the pieces fit a straight ^^^ through O, while the angles cover just twice 90°, or EOD -j- FOD. Any angle less than 90°, as HOF, is an ujcule angle; any angle greater than 90°, as HOE, is an obtuse angle. 4. An Equilateral Triangle is a triangle haying three equal sides; as, MNP. Bemark. — ^The angles of an equilateral triangle are 60° each ; hence, six equilateral triangles can be formed about the same point as a vertex, each angle at the vertex being measured by the sixth of a circumference. (Art, 204.) 364 MAY'S HIGHER AEITHMETIC. RiOHT-ANOLED TRIANGLES. 380. A Right-angled Triangle is a triangle having one right angle ; as IGK, where G = 90°. The side opposite the right angle is called the hypothenuse ; the other two sides are called the base and perpendievlar. It is demonstrated in Geometry that the square described upon the hypothenuse of a right-angled triangle is equal to the sum of the squares described on the other two sides. Illustration. — A practical proof of this may be made in the following manner, especially valuable when the triangle has no equal sides. It will be a useful and entertaining exercise for the pupil. Let the triangle be described upon a card, and let it stand upon the hypothenuse, as AEB does. Make three straight cuts; — one, perpendicular from A, through the smaller square; one, perpendic- ular from B, through the larger square, and one at right angles from the end of the second cut. The two squares are thus divided into five parts, which may be marked, and arranged, as here shown, in a square equal to one described on the hypothenuse. Remark. — The perpendicular in an equilateral triangle divides the base into two equal parts, and also divides the opposite angle into two which are 30° each. From this it follows that if, in a. right- EVOLUTION. 365 angled triangle, one angle is 30°, the side opposite that angle is half the hypothenuse ; and, conversely, if one side be half the hypothenuse, the angle opposite will be 30°. 387. To find the hypothenuse when the other two sides are given. Bule. — Add together Hie squares of the base and perpendicular y and extract the square root of the sum. 388. To find one side when the hypothenuse and the other side are given. Bule. — Subtract the square of the given side from the square of Ike hypothenuse, and extract tJie square root of the differ- ence; or, Multiply the square root of the sum of the hypothenuse and side by the square root of their difference. Representing the three sides by the initial letters h, p, 6, we have the following FOBMULAS. — 1. h = v^p^ -\- 6*. 3. b = Vh^ — j)2 ; or, b=Vh+pX Vh—p. Examples for Practice. 1. Find the length of a ladder reaching 12 ft. into the street, from a window 30 ft. high. 32.81 -J- ft. 2. What is the diagonal, or line joining the opposite cor- ners, of a square whose side is 10 ft. ? 14.142 + ft. 3. What is saved by following the diagonal instead of the sides, 69 rd. and 92 rd., of a rectangle? 46 rd. 4. A boat in crossing a river 500 yd. wide, drifted with the current 360 yd. ; how far did it go? 616 + yd* 366 BAY'S HIGHER ARITHMETia Bkmark. — Integers expressing the sides of right-angle4 triangles may be found to anj extent in the following manner : Take any two unequal numbers ; the mm of their squarei may represent a hypoth- enuse ; the difference of their squares will then stand for one side, and donMe their prodMct for the remaining side. Thus, from 3 and % form 13, 12, 5; from 4 and 1, form 17, 15, 8; from 5 and 2, form 29, 21, 20. PARALLEL LINES AND SIMILAR FIGUREa DEFINITIONS. 889. 1. Parallel Lines are lines which have the same direction. The shortest distance between two straight par- allels is, at all points, equal to the same perpendicular line. 2. Similar Figures are figures having the same number of sides, and their like dimensions proportional. Rebiabks. — 1. Similar figures have their eorreRponding angles equal. 2. If a line be drawn through any triangle parallel to one of the sides, the other two sides are divided proportionally, and the triangle marked ofi*, is similar to the whole triangle. An illustration of this ha8 already been furnished in solving Ex. 8, Art.. 231. 3. All equilateral triangles are similar; the same is true of all squares, all circles, all spheres. 3. The areas of similar figures are to each other as the squares of their like dimensions. 4. The solidities of similar solids are to each other as the cubes of their like dimensions. General Exercises in Evolution and its Applications. 1. One square is 12|^ times another : how many times does the side of the 1st contain the side of the 2d? 3^. 2. The diagonals of two similar rectangles are as 5 to 12 : how many times does the larger contain the smaller? 5^. EVOLUTION. 367 3. The lengths of two similar solids are 4 in. and 50 in. ; the 1st contains 16 eu. in. : wJiat does the 2d contain ? 31250 cu. in. 4. The solidities of two balls are 189 cu. in. and 875 cu. in. ; the diameter of the 2d is 17^ in. ; find the diameter of the 1st. lOJ in. 5. In extracting the square root of a perfect power, the last complete dividend was found 4725: what was the power? 225625. 6. What number multiplied by ^ of itself makes 504? 42. 7. Separate 91 252 J into three &ctors which are as the numbers 1, 2^, and 3. 23, 57.5, and 69. 8. What integer multiplied by the next greater, makes 1332 ? 36. 9. The length and breadth of a ceiling are as 6 and 5 ; if each dimension were one foot longer, the area would be 304 sq. ft. : what are the dimensions? 18 ft., 15 ft. 10. In extracting the cube root of *a perfect integral power, the operator found the last complete dividend 241984: what was the power? 2985984. 11. If we cut from a cubical block enough to make each dimension one inch shorter, it will lose 1657 cubic inches: what is the solidity? 13824 cu. in. 12. A hall standing east and west, is 46 ft. by 22 ft., and 12^ ft. high : what is the length of the shortest path a fly can travel, by walls and floor, from a southeast lower comer to a- northwest upper corner ? 57^ ft. 13. How many stakes can be driven down upon a space 15 ft. square, allowing no two to be nearer each other than 1^ ft., and how many allowing no two to be nearer than \\ ft. ? 128, and 180 stakes. 14. What integer is that whose square root is 5 times its cube root? 15625. 15. If the true annual rate of interest be 10%, what would be the true rate for each 73 days, if the interest be 368 I^A Y'S HIGHER ABITHMETia compounded through the year? Prove the result by con- tracted multiplication. (Art. 334, Rem. 2.) 1.924%. 16. If a field be in the form of an equilateral triangle whose altitude is 4 rods, what would be the cost of fencing it in, at 75 ct. a rod? $10.39 Topical Outline. Powers and Roots. 1. Involution. 1. Definitions. 2. Terms. 2. Evolution . 3. Squaring and Cubing.. 1. Powers. 2. Root 3. Degree. 4. Exponent. ' 1. Algebraic Statements. 2. Numerical Illustrations. 3. Geometrical Illustration. 4. Principles. 1. Definitions. 2. Terms. 1. Root 2. Powers.. 3. Radical. 4. Index. {I Perfect. Imperfect 1. First explanation (Geometrical). 3. Square Root ^ 2. Second explanation (Algebraic). 3. Rule. 4. Cube Root. 1. First explanation (Geometrical). 2. Second explanation (Algebraic). 3. Rule. 5. Roots in General— Horner's Method. 6. Applications. XXL SERIES. DEFINITIONS. 390. 1. A Series is any number of quantities having a fixed order, and related to each other in value according to a fixed law. These quantities are called Terms; the first and last are called Extremes, and the others Means. The Law of a series is a statement by which, from some necessary number of the terms the others may be computed. 2. There are many different kinds of series. Those usu- ally treated in Arithmetic are distinguished as Arithmetical and Oeometrical; these series are commonly called Progres- suyiis* ARITHMETICAL PROGRESSION. 391. 1. An Arithmetical Progression is a series in which any term differs from the preceding or following by a fixed number. That fixed number is called the com- mon difference; and the series is Ascending or Descending, accordingly as the fird term is the least or the greatest. Thus, 1, 3, 5, 7, 9, is an ascendin;» series, whose common differ- ence is 2; but if it were written in a reverse order (or, if we treated 9 as the first term), the series would be descending. 2. Every Arithmetical Progression may be considered under the relations of five quantities, such that any three of them being given, the others may be found. These five are conveniently represented as follows: First term, a. Last term, L Number of terms, . . . . n. Common difference, . . . ei. Sum of all the terms, . . «. (369) 370 BA Y'S HIGHER ARITHMETIC, 3. These give rise to twenty different cases, but all the calculations may be made from the principles stated in the two following cases. Note. — Some of the problems arising under this subject are, prop- erly, Algebraic exercises. Nothing will be presented here, however, which is beyond analysis by means of principles and processes exhib- ited in this book. The formulas given are easily understood, and the student will find it a very simple operation to write the numbers in place of their corresponding letters, and work according to the signs. The formulas are presented a^ a (xmvenieiux, CASE I. 392. One extreme, the common difference, and the number of terms being given, to find the other ex- treme. Problem. — Find the 20th term of the arithmetical series 1, 4, 7, 10, etc. Solution. — Here the series may be considered as made of 20 terms, and we seek the last. The com. diff. is 3, and the terms are composed thus: 1, 1+3, 1+6, 1+9, etc.; and it is obTioiis that, as the addiiion of the com. diff. commences in forming the second lerm, it is taken twice in the third term, three times in the fourth, and so on; similarly therefore it must be taken 19 times in form- ing the 20th, and the simple operation is, l + (20 — 1)X3 = 58, Ans, Formula.—^ = a + (n — 1 )rf; or / = a — (n — l)d Bule. — Multiply the number of terms less one by the com- mon difference^ add the product to Hie given extreme when the larger is sought, subtract it from the given extreme when the smaller is sought. Examples for Practice. 1. Find the 12th term of the series 3, 7, 11, etc. 47. 2. Find the 18th term of the series 100, 96, etc. 32. 3. Find the 64th term of the series 3 J, 5|, etc. 145 J. SERIES. 371 4. Find the 10th term of the series .025, .037, etc. .133 5. Find the 1st term of the series 68, 71, 74, having 19 terms. 20. 6. Find the 1st term of the series 117, 123^, 130, having 6 terms. 97^. 7. Find the first term of the series 18|, 12^, 6J, having 365 terms. 2281J. CASE II. 393. The extremes and the number of terms being given, to find the sum of the series. Problem. — What is the sum of 9 terms of the series 1, 4, 7, 10, etc.? OPERATION. Explanation. — Writing the series j 1/9 l) v 3 = 25. in fuU, in the common order, and also in a reverse order, we have y X ^ = 117 -Ans, Sum = 1 + 4 + 7 4- 10 + 13 -f 16 4- 19 + 22 + 25. Sum =^ 25 + 22 4- 19 4- 16 + 13 + 10 + 7 + 4 + 1. Twice the sum =26 + 26 + 26 + 264-26 + 26 + 26 + 26 + 26 = 9 times the sum of the extremes; .*. the sum =^ J of 9X26 = 117, Ans, If we add a term whose place is a certain distance beyond the first, to another whose place is equally distant from the last, the sum will be the same as that of the extremes, and hence, as the above illustrates, the double of any such series is equal to the product of the number of terms by the sum of the extremes. Formula. — % = (a + /) n. 2 Bule. — Multiply ihe sum of ihe extremes by Hie nunxber of termSy and divide by 2. Examples for Practice. 1. Find the sum of the arithmetical series whose extremes are 850 and 0, and number of terms, 57. 24225, 372 RAY'S UIQHER ARITHMETIC. 2. Extremes, 100 and .0001: number of terms, 12345. 617250.61725 3. What is the sum of the arithmetical series 1, 2, 8, etc., having 10000 terms? 50005000. 4. Of 1, 3, 5, etc., having 1000 terms? 1000000. 5. Of 999, 888, 777, etc., having 9 terms? 4995. 6. Of 4.12, 17.25, 30.38, etc., having 250 terms? 409701.25 7. Whose 5th term is 21; 20th term, 60; number of terms, 46? 3178|. Examples for Practice. Kemabk. — ^It is not deemed necessary to formulate a special rule for each class of examples here introduced. The following are pre- sented as exercises in analysis, each depending on one or more of the principles above stated. 1. Find the common difference of a series whose extremes are 8 and 28, and number of terms, 6. 4. 2. Extremes are 4^ and 20f, and number of terms, 14. \\. 3. Insert one arithmetical mean between 8 and 54. 31. 4. Insert five arithmetical means between 6 and 30. 10, 14, 18, 22, 26. 5. Insert two arithmetical means between 4 and 40. 16, 28. 6. Insert four arithmetical means between 2 and 3. 2i, 2|, 2f , 2f 7. What is the number of terms in a series whose ex- tremes are 9 and 42, and common difference, 3? 12. 8. Whose extremes are 3 and lOJ, and common differ- ence, f ? 21. 9. In the series 10, 15 ... 500? 99. 10. What principal, on annual interest at 10^, will, in 50 yr., amount to $4927.50? $270. SERIES, 373 GEOMETRICAL PROGRESSION. 394. 1. A Geometrical Progression is a series in which any term after the first is the product of the preceding term by a fixed number. That fixed number is called the ratio; and the series is ascending or descending accordingly as the first term is the least or the greatest. Thus, 1, 3, 9, 27, is an ascending progression, and the com- mon multiplier is the ratio of 3 to 1 (Art. 228, 2), or of any term to the preceding; considering 27 as the first term, the same series may be called descending. 2. Any Geometrical Progression may be considered under the relations of five quantities, of which three must he known in order to find the others. These five are thus represented : First term, a. Last term, I. Satio, r. Number of terms, . . . . n. Sum of all the terms, . . s. 3. These quantities give rise to 20 different classes of problems, but all of the necessary calculations depend upon principles set forth in the following cases. Note. — Some of the problems arising from these quantities require such an application of the formulas as can not be understood without a knowledge of Algebra. CASE I. 395. From one extreme, the common ratio and the number of terms, to find the other extreme. Problem. — Find the 8th term of the series 1, 2, 4, 8, etc. 374 RAY'S HIGHER ARITHMETIC. Explanation. — Here, 1 being the first statement. term, and 2 the ratio, we see that the series 2^^ X 1 = 1 2 8, -4?ia. may be formed thus: 1, 1X2, 1 X 2=*, 1X2', etc., the ratio being raised to its ^ecxmd power in forming the 3d term, to its ihitd power in forming the 4th; and so, similarly, the 8th term = lX2^ = 128, Anz. Formula. — I = ar*-i. Rule. — Consider the given extreme aa the first term, and mvltiply it hy that power of the ratio whose degree is denoted by the number of terms less one. Examples for Practice. 1. Find the last term in the series 64, 32, etc., of 12 terms. -g^j. 2. In 2, 5, 12 J, etc., having 6 terms. 195 j^^. 3. In 100, 20, 4, etc., having 9 terms. i&tis - 4. 1st term, 4; common ratio, 3; find the 10th term. 78732. 5. 3d term, 16; common ratio, 6: find the 9th term. 746496. 6. 33d term, 1024; common ratio, f : find the 40th term. 136H. 7. Find the 1st term of the series 90, 180, of 6 terms. 5§. 8- Of j^^y 1^, having 11 terms. xk- CASE II. 396. From one extreme, the ratio, and the number of terms, to find the sum of the terms. Problem. — The first term is 3, the ratio 4, the number of terms 5; required the sum of the series. (H»ERATI0N. Explanation.— Writing the 4X3X4* — 3 __ j q g 3 An$. whole series, we have : 4 — 1 * SERIES. 375 S=:3 + 12 + 48 + 192 + 768. Also, 4 S = 12-1-48 + 192-1- 768 -h 3072. It is evident that the lower line exceeds the upper by the difference between 3072 and 3; this difference maybe written 4X768 — 3,\)r 4X3X4* — 3, and as this is 3 times the series, we have, once the series = 4-^1 Now, observing the form, note that we have multiplied the last term by the ratio, then subtracted the first term, and then divided by the ratio less one. If the series had stood with 768 for first term, and the multiplier }, we should have had 768 -f 192 -h 48 -h 12 + 3 =S, 192 + 48 + 12-f 3 + i = J S; and thus 768 — } = | of the series \ hence, we can write, 768 — ^X3 -^«« , , ^-^ — = 10 2 3, as before. Here we have taken the product of the last term by the ratio from the first term, and have divided by the excess of unity above the ratio. In either case, therefore, we have illustrated Bule I. — Find the last tenn and multiply it by the ratio; then find the difference between this product and the first term, and divide by the difference between the ratio and unity. _ qW — a Q a — rl r — 1 1 — r The first answer, above given, may take another form, thus : 4X4^ X3 — 3 . ^, 3(4^—1) , ,, — IS the same as — ; — ', where appear the 4 — 1 4 — 1 ratio 4, the first term 3, and the number of terms 5. This form, often used when the series is ascending, hai; the following general statement : g^ a(r* — 1) r — 1 and corresponds to the following rule : 376 RAY'S HIGHER ARITHMETIC, Bule II. — Bxme ihe raiio to a 'power denoted by the num- ber of terms y subtract 1, divide Hie remainder by ihe raiio less 1, and mvltiply Hie qiiotieiit by Hie first term. Notes. — 1. The amoiiut of a debt at coinix)und interest for a num- ber of complete intervals, is the last term of a geometrical progression, whose ratio is 1 -J- the rate per cent. The table (Art. 335) shows the powers of the ratio. For example, the jjeriod being 4 yr., the number of terms is five; the first term is the principal, and the power of the ratio required by Cajfe I, is the fourth. 2. The amount of an annuity at compound interest is conveniently found by the Formula corresponding to Kule II. The table (Art. 335) is available, and the work very simple. Thus, if the annuity be $200, the time 40 yr., and the rate 6^, we have a = 200, n = 40, r = 1.06 Then, writing these values in the Formula, we have : , $200(1.06*0—1) $200X9.2857179 ^onnr^oon a Amount = ■■ ^-- = — = $30952.39, Ans. 1.06 — 1 .06 ^ , 3. If the series be a descending one having an infinite number of terms, the last term is 0, and the product required by Rule I is 0. Examples for Practice. 1. Find the sum of 6, 12, 24, etc., to 10 terms. 6138. 2. Of 16384, 8192, etc., to 20 terms. 32767fJ. 3. Of I, I, 3^, etc., to 7 terms. liffl- Find the sum of the following infinite geometrical series : 4. Of 1, ^, J, etc. 2. 5. Of f , /^, 3?/^, etc. 1|. 6. Of ^, f , ^, etc. 2. 7. Of |, 1, f etc. 8 J. 8. Of .36 = .3636, etc. =^%+ yi^, etc. ■^. 9. Of .349206, of 480, of 6. |f and ^ and |. 10. Find the amount of an annuity of S50, the time being 53 yr., the rate per cent 10. $77623.61 11. Applying the formula used in the last example, to any case of the same kind, prove the truth of the rule, in Case IV, of Annuities. SERIES. 377 12. Calculate a table of amounts of an annuity of $1, for any number of years from 1 to 6, at 8%. Remark. — It is not considered neceasary to give special rules for finding the ratio, and the number of terms when these are unknown ; 80 far as these are admissible here, they involve no principles beyond what are presented in the matter already given. Examples for Practice. 1. Find the common ratio: first term, 8; fourth term, 512. 4. 2. First term, 4}|; eleventh term, 49375000000. 10. 3. Sixteenth term, 729 ; twenty-second term, 1000000. 3^. 4. Insert 1 geometric mean between 63 and 112. 84. 5. Four geometric means between 6 and 192. 12, 24, 48, 96. 6. Three geometric means between ^^^^^j, and ^. 7. Two geometric means between 14.08 and 3041.28 84.48 and 506.88 Topical Outline. Series. Series. 1. Definitions— Terms, Law, Extremes. Means. ' 1. Terms. 1. Arithmetical... . 2. Classes... . , 2. Geometrical.... 2. Cases. (Formulas.) ' 1. Terms. 2. Cases. (Formulas.) H. A. 32. XXn. MENSUEATIOK DEFINITIONS. 397. 1. Gteometry is that branch of mathematics which treats of quantity having extension and form. When a quantity is so considered, it is called Sfojpwfttide. 2. There are four kinds of Magnitude known to Greom- etry : Lines, Angles, Surfaces, and Solids. A point has posi- tion, but not magnitude. 3. Mensuration is the application of Arithmetic to Geom- etry ; it may be defined also as the art of computing lengths^ areas, and volumes, LINES. 398. 1. A line is that which has length only. 2. A straight line is the shortest distance between two points. 3. A broken line is a line made of connected straight lines of different directions. 4. A curve, or curved line, is a line having no part straight. The word "line," used without the qualifying word "curve," is un- derstood to mean a straight line. 5. A horizontal line is a line parallel with the horizon, or with the water level. (See Art. 389; 1.) 6. A vertical line is a line perpendicular to a horizontal plane. (878) MENSURATION. 379 ANGLES. 399. An angle is the opening, or inclination, of two lines which meet at a point. (Art. 204.) Remark. — Angles differing from right angles are called oblUjue anyles, (See Art. 885, 3, Bern., and Art. 886.) SURFACES. Polygons. 400. 1. A surftee is that which has length and breadth without thickness. A solid has length, breadth, and thickness. A line is meant when we speak of the »Ule of a limited surface, or the ed<fe of a solid; a surface is meant when we speak of the side or the base of a solid. 2. A Plane is a surface such that any two points in it can be joined by a straight line which lies wholly in the surface. The application of a straight-edge is the test of a plane. 3. A plane figure is any portion of a plane bounded by lines. 4. A polygon is a portion of a plane inclosed by straight lines ; the perimeter of a polygon is the whole boundary. 5. Area is surface defined in amount. For the numerical expression of area, a square is the measuring unit. (Art. 197.) 6. A polygon is regular when it has all its sides equal, and all its angles equal. 7. A polygon having three sides is a trian- f-^y^ gle ; having four sides, a quadrilateral ; five /y^ \ sides, a pentagon ; six sides, a hexagon, etc. Quadrilateral. 380 BA Y'S HIGHER ARITHMETIC. The six diagrams following represent regular polygons. Pentagon. Hexagon. Hel>Ugon. Octagon. Nooiigon. Decagon . 8. The diagonal of a polygon is the straight line joining two angles not adjacent ; as, PN, on the preceding page. 9. The base is the side on which a figure is supposed to stand. 10. The altitude of a polygon is the perpendicular distance from the highest point, or one of the highest points, to the line of the base. 11. The center of a regular polygon is the point within, equally distant from the middle points of the sides ; the apothem of such a polygon is the perpendicular line drawn from the center to the middle of a side ; as, C a center, and CD an apothem. Triangles. 401. Triangles are classified with respect* to their angles, and also with respect to their sides. Acute Triangles. Obtuse Triangles. Scalene. Isosceles. Equilateral. Isosceles. Scalene. 1. A triangle is right-angled when it has one right angle ; it is acute-angled when each angle is acute; it is obtuse- angled when one angle is obtuse. These three classes may be named right triangles, acute triangles, obtuse triangles; the last two classes are sometimes called oblique triangles. MENSURATION. 881 2. A triangle is scalene when it has no equal sides; isosceles, when it has two equal sides; and equilateral, when its three sides are equal. A right triangle can be scalene, as when the sides are 3, 4, 5 ; or, isosceles, as when it is one of the halves into which a diagonal di- vides a square. An obtuse triangle can be scalene or isosceles ; an acute triangle can be scalene, isosceles, or equilateral. Quadrilaterals. 402. Quadrilaterals are of three classes: 1. A Trapezium is a quadrilateral having no two sides parallel. 2. A Trapezoid is a quadrilateral having two and only two sides parallel. 3. A Parallelogram is a quadrilateral having two pairs of parallel sides. * Trapezium « Trapezoid. Riiomboid. Rliombus. Rectangle. 403. Parallelograms are of three classes: 1. A Rhomboid is a parallelogram having one pair of parallel sides greater than the other, and no right angle. 2. A Bhombiis is a parallelogram whose four sides are equal. 3. A Rectangle is a pai'allelogram whose angles are all right angles ; when the rectangle has four equal sides, it is a square. A square is a rhomhus whose angles are 90° ; it is also the form of the unit for surface measure. It may properly be defined, an equilateral rectangle. 382 BA Y'S HIGHER ARITHMETIC. r G n /K E L I AREAS. Triangles and Quadrilaterals. 404. The general rules depend on the prmciples stated in the following remarks: Kemarks. — 1. The area of a rectangle is equal to the product of its length by its breadth. (Art. 197, Ex.) 2. The diagonal of a rectangle divides it into two equal triangles. The accompanying figure illustrates this; EHI = HEF. Observe also that the per- pendicular GL divides the whole into two rectangles; EGL is half of one of them, LXjrl the half of the other, and both these smaller triangles make EGI, which must therefore be half of the whole; EGI and EHI have the same base and equal altitudes. If the triangle EGH be supposed to stand on GH as a 6a»e, its altitude is EF; the perpendicular which represents the height is, in such a case, said to fall on the base produced; i. e., extended. The triangle EGH is equal to the half of GHIL. Any triangle has an area equal to the product of half the base by the altitude. Observe also that the trapezoid EFGI is made of two triangles, EGF and EGI; each of these has the altitude of the trapezoid, and each has one of the parallel sides for a base. Hence, the area of each being J its base X the common altitude, the two areas, or the whole trapezoid, must equal the half of both bases X the altitude. 3. If the piece EGF were taken off and put on the right of EGHI, the line EF being placed on HI, the whok area would be the same, but the perimeter would be increased, and the figure would be a rhomboid. Different quadrilaterals may have equal areas and waeqwd boundaries; also, they may have the sides in the same order and equal, with unequal areas. To find accurately the area of a quadrilateral, more must be known than merely the four sides in order. A regular polygon has a greater area than any other figure of the same perimeter. 4. When triangles have equal bases their areas are to each other as their altitudes; the altitudes being equal, their areas are as their bases. The area of any triangle is equal to half the product of the perimeter by the radius of the inscribed circle. MENSUEATION. 383 General Bules. 1. To find the area of a parallelogram. Bole. — Multiply one of ttoo paraUd sides by the perpendicular ditiance between them. n. To find the area of a triangle. Bole. — Take half Hie product of Hie base by the altitude. m. To find the area of a trapezoid. Bole. — MuUiply half the sum of the parallel sides by the aUiivde. Note. — ^The following is demonstrated in Geometry : IV. To find the area of a triangle when the sides are given. , Bule. — Add the three sides together and take half the sum; from the half sum take the sides separately ; multiply the half sum and die three remainders together, and extras the square rod of the product. Notes. — 1. The area of a trapezium may be found by applying this rule to the parts when tlie sides are known and the diagonal is given in length and in special position as between the sides. The area of any i)olygon may be found by dividing it into triangles and measuring their bases and altitudes. 2. The area of a rhombus is equal to half the product of ite diag- onaLs; these are at right angles. Examples for Practice. 1. Find the area of a parallelogram whose base is 9 ft. 4 in. and altitude 2 ft. 5 in. 22 sq. ft. 80 sq. in. 2. Of an oil cloth 42 ft. by 5 ft. 8 in. 26| sq. yd. 384 RAY'S HIGHER ARITHMETIC. 3. How many tiles 8 in. square in a floor 48 ft. by 10 ft.? 1080. 4. Find the area of a triangle whose base is 72 rd. and altitude 16 rd. 3 A. 96 sq. rd. 5. Base 13 ft. 3 in.; altitude 9 ft. 6 in. 62 sq. ft. 135 sq. in. 6. Sides 1 ft. 10 in.; 2 ft.; 3 ft. 2 in. 1 sq. ft. 102— sq. in. 7. Sides 15 rd.; 18 rd.; 25 rd. 133.66— sq.'rd. 8. What is the area of a trapezoid whose bases are 9 ft, and 21 ft, and altitude 16 ft.? 240 sq. ft. 9. Bases 43 rd. and 65 rd. ; altitude 27 rd. ? 9 A. 18 sq. rd. 10. What is the area of a figure made up of 3 triangles whose bases are 10, 12, 16 rd. and altitudes 9, 15, 10^ rd.? 1 A. 59 sq. rd. 11. Whose sides are 10, 12, 14, 16 rd. in order, and dis- tance from the starting point to the opposite corner, 18 rd. ? 1 A. 3.9 — sq. rd. 12. How much wainscoting in a room 25 ft. long, 18 ft. wide, and 14 ft. 3 in. high, allowing a door 7 ft. 2 in. by 3 ft. 4 in., and two windows, each 5 ft. 8 in. by 3 ft. 6 in., and a chimney 6 ft. 4 in. by 5 ft. 6 in.; charging for the door and windows half-work? 128|^ sq. yd. 13. What is the perimeter of a rhombus, one diagonal be- ing 10 rd., and the area 86.60J sq. rd. ? 40 — rd. 14. Find the cost of flooring and joisting a house of 3 floors, each 48 ft. by 27 ft., deducting from each floor for a stairway 12 ft. by 8 ft. 3 in., allowing 9 in. rests for the joists; estimating the flooring and joisting between the walls at $1.46 a sq. yd., and the joisting in the walls at 76 ct. a sq. yd. ; each row of rests being measured 48 ft. long by 9 in. wide. $600.78 15. What is the area of a square farm whose diagonal is 20.71 ch. longer than a side? 250 acres. 16. How many sq. yd. of plastering in a room 30 ft;. MENSURATION. 385 long, 25 ft. wide, and 12 ft. high, deducting 3 windows, each 8 ft. 2 in. by 5 ft. ; 2 doors each 7 ft. by 3 ft. 6 in. ; and a fire-place 4 ft. 6 in. by 4 ft. 10 in. ; the sides of the windows being plastered 15 in. deep? And what will it cost, at 25 ct a sq. yd.? 215j^ sq. yd.; cost $53.83 17. From a point in the side and 8 ch. from the corner of a square field containing 40 A., a line is run, cutting off 19i A.: how long is the line? One answer, 20^ ch. 18. How much painting on the sides of a room 20 ft. long, 14 ft. 6 in. wide, and 10 ft. 4 in. high, deducting a fire-place 4 ft. 4 in. by 4 ft., and 2 windows each 6 ft. by 3 ft 2 in.? 73^ sq. yd. 19. Find the cost of glazing the windows of a house of 3 stories, at 20 ct. a sq. ft. Each story has 4 windows, 3 ft. 10 in. wide; those in the 1st story are 7 ft. 8 in. high; those in the 2nd, 6 ft. 10 in. high; in the 3d, 5 ft» 3 in. high. * $60.56S Reguiar Polygons and the Circle. 405. Any regular polygon may be divided into equal isosceles triangles, by lines from the center to the vertices; the apothem is their common altitude, and the perimeter the sum of their bases. 406. To find the area of a regular polygon. Rule. — Multiply the perimeter by Iwlf the apotfiem. All regular ])olygoiis of the same nuuiber of sides are similar figures. (Art. 389, Rem. 1.) 407. 1. The Qircle, as already defined (Art. 204), is a figure bounded by a uniform curve. 2. Any line drawn in a circle, having its ends in the curve, is called a chord; as AB, BD. H. A. r.. 386 EAY\S HIGHER ARITHMETIC, 3. The portion of the curve which is cut off by such a line is called an arc, and the space between the chord and the arc is called a segment. Thus, the curve APB is an arc, AB is the chord of that arc, and these inclose a segment whose base is AB, and whose height is OP. 4. If a line be drawn from the middle of a chord to the center, it will be perpendicular to the chord ; so also, a line perpendicular at the middle of a chord, will, if extended, pass through the center, and bisect either of the arcs stand- ing on that chord. Thus, AB is bisected by the perpen- dicular CO, and the arc AP = PB ; so the arc AD = BD. 5. A tangent to a circle as a straight line having only one point in common with the curve; it simply touches the circle; a secant enters the figure from without. If with C as a center, and CO as a radius, a circle were drawn in the equilateral triangle ABD, the sides would be tangent to the circle; the circle would be inscribed in the triangle. The circle of which CB is radius, is circumscribed about ABD. 6. The space inclosed by two radii and an arc, is called a sector; as, ACP. The arc of that sector is the same fraction of the whole cir- cumference that the area of the sector is of the whole circle. Calculations Pertaining to the Circle. 408. The accompanying diagrams present (Fig. 1) a regular polygon of six sides, (Fig. 2) one of twelve sides, and (Fig. 3) a circle divided into twenty-four sectors. Remarks. — 1. The hexagon is com|K)sed of six equilateral triangles, and hence if OB be 1, the side AB-= 1, and it is easy to compute the apolhem, v^ 1 — i — .86602540378 MENSURATION. 387 2. If the distance from center to vertices be unchanged, and a regular polygon of twelve siden be formed about the same center, it will differ jess from a circle whose ra- dius is OB, than the hexagon differs from such a circle. This is evident from the second figure; and if the polygon be made of twenty-four sides (the distance from cen- ter to vertices remaining the same), it will be still nearer the circle in shape and size ; in the space of the diagram, one of the twenty-four triangles forming such a poly- gon would differ very little from one of the twenty-four sectors here shown. The circle is regarded as composed of an in- finite number of triangles whose common altitude is the radius and the sum of whose bases is the circumference. Hence, the area = J the sum of bases X altitude; or, Area of circle -~ J circumference X radius. Area of circle = \ circumference X diameter. Fig. 2. 3. Since the perimeter of the hexagon is 6, it is easy to compute the next perime- ter shown, 'which is 12 times AP or BP. The apothem being found above," subtract it from OP or 1, and obtain .13397459622 the perpendicular of a right-angled tri- angle; then, the base of that triangle be- ing .5, the half of AB, find the hypothe- nuse .517638090205, = PB. Now, if we treat PB as we treated AB we can find Fig. 3. the apothem of the second figure, and then. find one of the 24 sides of another polygon, still more nearly equal to the circle. If these operations be continued, we shall find results $th and 9th as follows: » Perimeter of polygon of 1536 sides = 6.28318092 Perimeter of polygon of 3072 sides — 6.28318420 If the distance from center to vertices be taken J instead of 1, the results will be 3.141590 -f and 3.141592 -f 388 RAY'S HIGHER ARITHMETIC. Hence, if the circle of diameter 1, be taken as a polygon of 1536 sidefl, and then aa a polygon of 3072 sides, the expressions for perimeter do not differ at the fourth decimal place. The number 3.1416 is usually given, although by more expeditious methods than that above illustrated, the calculation has been carried to a great number of decimal places, of which the following correctly shows eighteen : 3.141592653589793238 This important ratio, of circumference to diameter, is represented by the Greek letter tt (pi,), 4. Since circumference == diameter X "^j ^^^ *r®2i = J circum- ference X diameter, we have area =2rX square of diameter. Rep- resenting the circumference by c, area by C, diameter by (/, radius by R, we have the following formula*^: C = d7r; G = d^^; C=R'^r. General Rules. 409. Pertaining to the circle we have the following general rules: I. To find the circumference: 1. Multiply the diameter by 3.1415926; or, 2. Divide iJie area hy \ of the diameter; or, 3. Hiiract the square root of 12.56637 tim^s the area, II. To find the diameter: 1. Divide the circumference by 3.1415926; or, 2. Divide the area 6i/ .785398, and extract the square root, m. To find the area: 1. Multiply the diameter by ^ of the circumference; or, 2. Multiply Hie square of the diameter by .785398; or, 3. Mtilfiply the square of Hie radius by 3.1415926 MENSUBATION, 389 IV. To find the area of a sector of a cirole : 1. MvlUply ilie arc by one half the radius; or, 2. Take such a fraction of the whole area as the arc is of the xxHiole circumference, V. To find the area of a segment less than a semi- circle : 1. Subtract from the area of the sector having the same arc. Vie area (f the triangle whose base is the base of Hw segment, and wlwse vertex is tJie center of the eirde ; or, 2. Divide Vie cube of Vie height by twice the base, and increase the quotient by two thirds cf the product of height and base. Remark. — Add the triangle to the factor, if the segment be greater than a semicircle. The second rule gives an approximate result. Notes. — 1. The side of a square inscribed in a circle is to radius as V^2 is to 1. 2. The side of an inscribed equilateral triangle Ls to radius as l/ 3 is to 1. 3. If radius be 1, the side of the inscribed regular pentagon is 1.1755; heptagon, .8677; nonagon, .6840; undecagon, .5634 Examples for Practice. 1. What are the circumferences whose diameters are 16, 22i, 72.16, and 452 yd.? 50.265482; 69.900436; 226.6973; 1420 yd. 2. What are the diameters whose circumferences are 56, 182 J, 316.24, and 639 ft.? 17.82539; 58.09; 100.66232; and 203.4 ft. 3. Find the areas of the circles with diameters 10 ft. ; 2 ft. 5 in. ; 13 yd. 1 ft. 78.54 sq. ft. ; 660.52 sq. in. ; 139 sq. yd. 5.637 sq. ft. 4. Whose circumferences are 46 ft. ; 7 ft. 3 in. ; 6 yd. 1 ft. 4 in. 168.386 sq. ft. ; 4 sq. ft. 26.322 sq. in. ; 29.7443 sq. ft;. 390 JiA Y'S HIGHER ARITHMETIC. 5. CircuiD. 47.124 ft., diameter 15 ft. 176.715 sq. ft. 6. If we Baw down through J of the diameter of a round log uniformly thick, what portion of the log is cut in two ? .2918 7. What fraction of a round log of uniform thickness ia the largest squared stick which can be cut out of it ? .6366 410. 1. A Solid is that which has length, breadth, and thickness. A Holid may have plane mirtaces, curved surtnees, or both. A. aaved tarfaa ia one no part of which ia n |>lnne. 2. The taoeB of a solid are the polygons formed by the intersections of its bounding pLaues; the lines of those inter- sections are called edges. 3. A Prism is a solid having two bases which are parallel polygons, and faces which are parallelograms. A prixm is triangular, quadrangular, etc., according to the Rhnpe of itH base. The first iif the ligiirea here given represents a qmulran- gular prism, theeecond apenbtipnat prism. 4. A right prism is a prism whose &ce8 are rectangles. 5. A Farallelopiped is a prism whose faces are parallel- ograms. Its bounding surfaces are six parallelograms. The first figure above represents a parallelepiped whose faces are MENSURATION. 391 6. A Cube 13 a parallelopiped whose feces are squares. 7, A Cylinder is a solid having two bases which are equal parallel circles, and having an equal diameter in any parallel plane between them. . A Pyramid is a solid with only one base, and whose faces are triangles with a common vertex. 9. A Cone is a solid whose base is a circle, and whose other surface is convex, terminating above in a point called the vertex. 10. A &ustum of a pyramid or cone is the solid which remaios when a portion having the vertex is cut ofi* by a plane parallel to the base. 11. A Sphere is a solid hounded by a curved surface, every point of which is at the same distance from a point within, called the center. The diamder of a sphere is a straight line passing through the center and having its ends in the surface; the radius is the distjuice from the center to the surface. A segnient of n nphere is a portion cut off by one plane, or lielween two planes; il^ baies nre cirdes, nnd its height is the portion of llie diameter whicJi iK cut off witli it. 12. The slant heigbt of a 'pyramid is the perpendicular distance from the vertex to one of the sides of a base ; the slant height of a cone is the straight line drawn from the vertex to the circumference of the base. 392 RAY'S HIGHER ARITHMETIC. 13. The altitude of any solid is the perpendicular dis- tance between the planes of its bases, or th^ perpendicular distance from its highest point to the plane of the base. 14. The Volume of a solid is the number of solid units it contains ; the assumed unit of measure is a cube. (Art. 199.) 15. Solids are similar when their like lines are proper- tionaly and their corresponding angles e(}ual. General Rui.es. I. To find the convex surface of a prism or cyl- inder : Rule. — Multiply the perimeter of Vie base by tlie altitude, II. To find the volume of a prism or cylinder : Rule. — Multiply tlie base by the altitude, m. To find the convex surface of a pyramid or cone : Rule. — Multiply the perimeter of tlie base by one lialf Vie slant height, TV. To find the volume of a pyramid or cone: Rule. — Multiply the base by one third of Vie altitude, V. To find the convex surface of a Aioistum of a pyramid or cone: • Rule. — Multiply Judf the sum of the perimeters of Vie bases by the slaiii Imgld, VI. To find the solidity of a Arustum of a pyramid or cone: MENSURATION, 393 Rule. — To ike sum of due two hoses add the square root of iJmr product, and multiply tlie amount by mie Hard of Hie aUi- tude, VII. To find the surface of a sphere : Rule. — Multiply Hie circumference by Hie diatnder. VIII. To find the volume of a sphere : Rule. — Multiply the cube of Hie diameter by .5235987 Notes. — 1. Similar solids are to each other as the cubes of their like dimensions. 2 The sphere is regarded as com[K)sed of an infinite number of cones whose common altitude is the radius, and tlie »«im of whose bases is the whole surface of the sphere. 3. The cone is regarded as a ])yramid of an infinite number of faces, and the cylinder as a prism of an infinite number of faces. Examples for Practice. * 1. Find the convex surface of a right prism with altitude 11 J in., and sides of base 5^^, 6^, 8^, 10^, 9 in. 450 sq. in. 2. Of a right cylinder whose altitude is If ft., and the diameter of whose base is 1 ft. 2^ in. 6 sq. ft., 92.6 sq. in, 3. Find the whole surface of a right triangular prism, the sides of the base 60, 80, and 100 ft. ; altitude 90 ft. 26400 sq. ft. 4. The whole surface of a cylinder ; altitude 28 ft. ; cir- cumference of the base 19 ft. 589.455 sq. ft. 5. Find the convex surface and whole surface of a right pyramid whose slant height is 391 ft. ; the base 640 ft. square. Conv. surf 500480 sq. ft. ; whole surf. 910080 sq. ft. 394 RAY'S HIGHER ARITHMETIC. 6. Of a right cone whose slant height is 66 ft. 8 in. ; radius of the base 4 ft. 2 in. 125663.706 sq. in. ; 133517.6876 sq. in. 7. Find the solidity of a pyramid whose altitude is 1 ft. 2 in., and whose base is a square 4J in. to a side. 94J cu. in. 8. Whose altitude is 15.24 in., and whose base is a triangle having each side 1 ft. 316.76 cu. in. 9. What is the solidity of a prism whose bases are squares 9 in. on a side, and whose altitude is 1 ft. 7 in. ? 1539 cu. in. 10. Whose altitude is 6 J ft. , and whose bases are parallel- ograms 2 ft. 10 in. long by 1 ft, 8 in. wide? 30 cu. ft. 1200 cu. in. 11. Whose altitude is 7 in., and whose base is a triangle with a base of 8 in. and an altitude of 1 ft. ? 336 cu. in. 12. Whose altitude is 4 ft. 4 in., and whose base is a tri- angle with sides of 2, 2 J, and 3 ft. ? 10.75 cu. ft. 13. What is the solidity of a cylinder whose altitude is 10^ in., and the diameter of whose base is 5 in. ? 206.167 cu. in. 14. Find the convex surface of a frustum of a pyramid with slant height 3 J in., lower base 4 in. square, upper base 2J in. square. 43J sq. in. 15. The convex surface and whole surface pf the frustum of a cone, the diameters of the bases being 7 in. and 3 in., and the slant height 5 in. Conv. surf. 78.5398 sq. in., whole surf. 124.0929 sq. in. 16. Find the solidity of a frustum of a pyramid whose altitude is 1 ft. 4J in. ; lower base, lOf in. square ; upper, 4|- in. square. 97 4y| cu. in. 17. Of a frustum of a cone, the diameters of the bases being 18 in. and 10 in., and the altitude 16 in. 2530.03 cu. in. 18. What are the surfaces of two spheres whose diameters are 27 ft. and 10 in. ? 2290.221-1- sq. ft. and 314.16 sq. in. MENSUBATION. 395 19. Find the solidity of a sphere whose diameter is 6 mi., and surface 113.097335 sq. mi. 113.097335 cu. mi. 20. Of a sphere whose diameter is 4 ft. 33.5103 cu. ft. 21. Of a sphere whose surface is 40115 sq. mi. 755499i cu. mi. 22. By what must the diameter of a sphere be multiplied to make the edge of the largest cube which can be cut out ofit? • .57735 MISCELLANEOUS MEASUKEMENTS. Masons' and Beicklayers* Work. 411. Masons' work is sometimes measured by the cubic foot, and sometimes by the perch. The latter is 16^ ft. long, 1^ ft. wide, and 1 ft. deep, and contains 16^ X 1^ X 1 = 24| cu. ft., or 25 cu. ft. nearly. 412. To find the number of perches in a piece of masonry. Bule. — Find the solidity of {he wall in cubic feet by the rules given for mensuration of solids, and divide it by 24f . Note. — ^Brick work is generally estimated by the thousand bricks ; the usual size being 8 in. long, 4 in. wide, and 2 in. thick. When bricks are laid in mortar, an allowance of -^ is made for the mortar. Examples for Practice. 1. How many perches of 25 cu. ft. in a pile of building- stone 18 ft. long, 8^ ft. wide, and 6 ft. 2 in. high ? 37.74 = 37f perches nearly. 2. Find the cost of laying a wall 20 ft. long, 7 ft. 9 in. high, and with a mean breadth of 2 ft., at 75 ct. a perch. $9.39 896 HAY'S HIGHER ARITHMETIC. 3. The cost of a foundation wall 1 ft. 10 in. thick, and 9 ft: 4 in. high^ for a building 36 ft long, 22 ft. 5 in. wide outside, at $2.75 a perch, aUowmg for 2 doors 4 ft. wide. $192.98 4. The cost of a brick wall 150 ft. long, 8 ft. 6 in. high, 1 ft. 4 in. thick, at $7 a thousand, allowing -^ for mortar ? $289.17 5. How many bricks of ordinary size will buiid a square chimney 86 ft. high, 10 ft. wide at the bottom, and 4 ft. at the top outside, and 3 ft. wide inside all the way up? 89861+ bricks. SuQGEsnoN. — ^Find the solidity of the whole chimney, then of the hollow part; the difference will be the solid part of the chimney. Gauging. 413. Gtmglng is finding the contents of vessels, in bush- els, gallons, or barrels. 414. To gauge any vessel in the form of a rect- angular solid, cylinder, cone, fhistum of a cone, etc. Rule. — Fmi the solidity of (he vessel in cubic inches hy Hie rules already given; this divided by 2150.42, iviU give five con- tents in bushels; by 231, will give it in wine gallons^ which may be reduced to barrels by dividing Hie uurnber by 31^. Note. — ^In applying the rule to cylinders, cones, and frustums of cones, instead of multiplying the square of half the diameter by 3.14159265, and dividing it by 231, multiply the square of the diounieter hy .0034, which amounts to the same, and is shorter. Examples for Practice. 1. How many bushels in a bin 8 ft. 3 in. long, 3 ft 5 in. high, and 2 ft. 10 in. wide? 64.18 bu. MENSURATION. 397 2. How many wine gallons in a bucket in the form of a frustum of a cone, the diameters at the top and bottom being 13 in and 10 in., and depth 1^ in. ? 5.4264 gal. 3. How many barrels in a cylindrical cistern 11 ft. 6 in. deep and 7 ft. 8 in wide? 126.0733 bbl. 4. In a vat in the form of a frustum of a pyramid, 5 ft. deep, 10 ft. square at top, 9 ft. square at bottom? 107.26 bbL 415. To find the contents in gallons of a cask or barreL Bemark. — ^When the staves are straight from the bung to each end, consider the cask as two frustums of a cone, and calculate its contents by the last rule ; but when the staves are curved, use this rule: Bule. — Add to (he head diameter (inside) tvx) Udrds of (he difference between the head and bung diameters; but if the staves are only slightly curved, add six tenths of this difference; this gives Vie mean diameter; express it in indies ^ square it, multiply it by the length in inches, anj, this product by .0034: Hie product will be the contents in wine gallons. Note. — ^Aiter finding the mean diameter, the contents are found as if the cask were a cylinder. Examples for Practice 1. Find the number of gallons in a cask of beer whose staves are straight from bung to bead, the length being 26 in., the bung diameter 16 inches, and head diameter 13 in. 18.65 gal. 2. In a barrel of whisky, with staves slightly curved, length 2 ft. 10 in., bung diameter 1 ft. 9 in., head 1 ft. 6 in. 45.32 gal. 3. In a cask of wine with curved staves, length 5 ft. 4 in., bung diameter 3 ft. 6 in., head diameter 3 ft. 348.16 gal. 398 BAY'S HIGHER ARITHMETIC. Lumber Measure. 416. To find the amount of square-edged incli boards that can be sawed from a round log. Kemark. — The following is much used by lumbei*-men, and is sufficiently accurate for practical purposes. It is known as Doyle'a Rule. Bule. — From the diameter in inches svbtract 4; the square of the remainder will he the number of square feet of inch boards yielded by a log 16 feet in lengtJi, Examples for Practice. 1. Ho^ much square-edged inch lumber can be cut from a log 32 inches in diameter, and 20 feet long? OPERATION. 32 — 4 = 28; 28 X 28 X H = ^^^ feet. Or, J X 28 X 28 = 980 feet. 2. In a log 24 in. in diameter, and 12 tt. long? 300 ft. 8. In a log 25 in. in diameter, and 24 ft. long? 661i ft. 4. In a log 50 in. in diameter, and 12 ft. long? 1587 ft. To Measure Grain and Hay. 417. Grain is usually estimated by the bushel, and sold by weight; Hay, by the ton. Hem ARKS. — 1. The standard bushel contains 2150.4 cubic inches. A cubic foot is nearly .8 of a bushel. 2. Hay well settled in a mow may be estimated (approximately) at 550 cubic feet for clover, and 450 cubic feet for timothy, per ton. MENSURATIOK 399 418. To find the quantity of grain in a wagon or in a bin: Rule. — Multiply Hie contents in cubic feet by .8 Kemabks. — 1. If it be com on the cob, deduct one half. 2. For corn not ** shucked," deduct two thirds for cob and shuck. 419. To find the quantity of hay in a stack, rick, or mow : Bule. — Divide the cvbical contents in feet by 550 for clover, or by 450 for timotliy; Hie quotient will be Hie number of tons. Examples for Practice. 1. How many bushels of shelled com, or corn on the cob, or corn not shucked, will a wagon-bed hold that is 10^ feet long, 3 J feet wide, and 2 feet deep? 58.8 bu.; 29.4 bu. ; 19.6 bu. 2. In a bin 40 feet long, 16 wide, and 10 feet high? 5120 bu. 3. A hay-mow contains 48000 cubic feet: how many tons of well settled clover or timothy will it hold? 87^ tons clover ; 106f tons timothy. Topical Outline. Mensuration. 1. General Definitions :— Geometry, Magnitude, MensuraUon. Parallel. Perpendicular. I Straight \ Horizontal. 2. Lines . Broken. Curved. Vertical. Diagonal. 400 HAY'S HIGHER ARITHMETia Mensuration. — (Continued. ) 8. Angles } 4. Surfaces. 6. Solids..... Right Oblique / Acute. Obtuse. 1. General Definitions:— Plane, Plane figure. Area, Polygons/ Regular, Perimeter, Similar, Center, Altitude, Base, Apothem. f Isosceles. \ Scalene. Scalene. Acute. 2. Triangles... . (Rules.) 1. Right. 2. Oblique. Obtuse { Isosceles. EquilateraL Scalene. Isosceles. L Trapezium. 3. Quadrilateral- J 2. Trapezoid. (Rules.) [ 8. Parallelogram. 1. Rhomboid. 2. Rhombus... { Square. 3. Rectangle... j Square. ' 1. Terms:— Circumference, Radius, Chord, Diam- eter, Segment, Sector, Tangent, Secant. 2. Calculations, Value of w. 3. Formulas, Rules. 1. General Definitions :— Solid, Base, Face, Edge, Similar. 4. Circle 2. Prism... (Rules.) Triangular. (Right.) _ „ , . , Quadl^lar. | ^"^^^T"- J^"*-- Pentagonal, etc. *- \ S ) r Altitude. 3. Pyramid J slant Height. (Rules.) (Frustum.) ( Convex Surface. f Altitude. 4- Cone J slant Height (Rules.) (Frustum.) ( Convex Surface. 5. Cylinder. / Surface. (Rules.) ^ Solidity. 1. Terms :--Radius, Diameter, Seg- ment. 2. Convex Surface, Rule. . 3. Solidity, Rule. 6. Sphere. 7. General Formulas. 8. Miscellaneous Applications.. ' Masons' and Bricklayers* Work. Gauging. Lumber Measure. . Measuring Grain and Hay. XXni. MISCELLAZ^TEOUS EXERCISES. Note.— Nos. 1 to 50 are to be solved mentally. 1. If I gain I ct. apiece by selling eggs at 7 ct. a dozen, how much apiece will I gain by selling them at 9 ct. a dozen? -f^ ct. 2. If I gain ^ ct. apiece by selling apples at 3 for a dime, how much apiece would I lose by selling them 4 for a dime? \ ct. 3. If I sell potatoes at 37J ct. per bu., my gain is only | of what it would be, if I charged 45 ct. per bu.: what did they cost me? 26 J ct. per hu. 4. If I sell my oranges for 65 ct., I gain f ct. apiece more than if I sold them for 50 ct.: how many oranges have I? 40 oranges. 5. If I sell my pears at 5 ct. a dozen, I lose 16 ct.; if I sell them at 8 ct. a dozen, I gain 11 ct.: how many pears have I, and what did they cost me? 9 dozen at 6 J ct. per dozen. 6. If I sell ^gs at 6 ct. per dozen, I lose f ct. apiece; how much per dozen must I charge to gain | ct. apiece? 22 ct. 7. One eighth of a dime is what part of 3 ct.? y\. 8. If I lose f of my money, and spend f of the remainder, what part have I left? J|. 9. A's land is \ less in quantity than B's, but ^ better in quality: how do their farms compare in value? A's = -j^ of B's. 10. If f of A's money equals | of B's, what part of B's equals | of A's? J. 11. I gave A t\ of my money, and B ^^ of the remainder: who got the most, and what part? B got -^ of it more than A. 12. A is f older than B, and B J older than C: how many times es age is A's? 2 J. 13. Two thirds of my money equals J of yours; if we \)\xt our money together, what part of the whole will I own? ■^^, 14. How many thirds in ^? 1J-. 15. Reduce | to thirds ; f to ninths ; and f to a fraction, whose nu- merator shall be 8. 21 thirds; 74 ninths; ^ ' ^ ' 13J. 16. What fraction is as much larger than | as f is less than ^? If. 17. After paying out \ and \ of my money, I had left $8 more than I had spent: what had I at first? $80. H. A. 34. (401) 402 HA Y'S HIGHER ARITHMETIC. 18. In 12 yr. I shall be J of my present age: how long since was I ^ of my present age? 8f yr. 19. Four times } of a number is 12 less than the number: what is the number? 108. 20. A man left -jfij- of his property to his wife, f of the remainder to his son, and the balance, $4000, to his daughter, what was the estate? $22000. 21. 1 sold an article for \ more than it cost me, to A, who sold it for $6, which was \ less than it cost him : what did it cost me? $8. 22. A is J older than B ; their father, who is as old as both of them, is 50 yr. of age: how old are A and B? A, 27^ yr.; B, 22^ yr. 23. A pole was J under water; the water rose 8 ft., and then there was as much under water as had been above water before: how long is the pole? 18 § ft 24. A is f as old as B; if he were 4 yr. older, he would be ^ as old as B; how old is each? A, 20 yr.; B, 26§ yr. 25. A's money is $4 more than | of B's, and $5 less than | of B's: how much has each? ' A, $76 • B. $108. 26. Two thirds of A's age is f of B's, and A is SJ yr. the older: how old is each? A, 31 J yr.; B, 28 yr. 27. If 3 boys do a work in 7 hr., how long will it take a man who works 4 J times as fast as a boy? 4 J hr. 28. If 6 men can do a work in h\ days, how much time would be saved by employing 4 more men? 2^ days. 29. A man and 2 boys do a work in 4 hr. : how long would it take the man alone if he worked equal to 3 boys? 6f hr. 30. A man and a boy can mow a certain field in 8 hr. ; if the boy rests 3| hr., it takes them 9} hr. ; in what time can each do it? Man, 13J hr.; boy, 20 hr. 31. Five men were employed to do a work ; two of them failed to come, by which the work was protracted 4 J days : in what time could the 5 have done it? 6| days. 32. Three men can do a work in 5 days; in what time can 2 men and 3 boys do it, allowing 4 men to work equal to 9 boys? 4 J da. 33. A man and a boy mow a 10-acre field; how much more does the man mow than the boy, if 2 men work equal to 5 boys? 4| A. 34. Six men can do a work in 4J days ; after working 2 days, how many must join them so as to complete it in 3f da. ? 4 men. 35. Eight men can do a certain amount of work in 6| days; after beginning, how soon must they be joined by 2 more so as to complete it in 5J days? In 2} days. MISCELLANEOUS EXERCISES. 403 36. Seven men can build a wall in 5^ days ; if 10 men are employed, what part of his time can each rest, and the work be done in the same time? jV 37. Nine men can do a work in 8 J days; how many days may 3 remain away, and yet finish the work in the same time by bringing 5 more with them ? 5/j days. 38. Ten men can dig a trench in 7 J days ; if 4 of them are absent the first 2J days, how many other men must they then bring with them to complete the work in the same time? . 2 men. 39. At what times between 6 and 7 o'clock are the hour-hand and minute-hand 20 min. apart? 10}f min. after 6, and 54/^ niin. after 6. 40. At what times between 4 and 5 o'clock is the minute-hand as far from 8 as the hour-hand is from 3? 32 y\ min. after 4; and 49^^ min. after 4. 41. At what time between 5 and 6 o'clock is the minute-hand mid- way between 12 and the hour-hand? when is the hour-hand midway between 4 and the minute-hand? 13^^ Dciin. after 5; and 36 min. after 5. 42. A, B, and C dine on 8 loaves of bread; A furnishes 5 loaves; B, 3 loaves; C pays the others 8d. for his share: how must A and B divide the money? A takes 7d.; B, Id. 43. A boat makes 15 mi. an hour down stream, and 10 mi. an hour up stream: how far can she go and return in 9 hr.? 54 mi. 44. I can pasture 10 horses or 15 cows on my ground ; if I have 9 cows, how many horses can I keep? 4 horses. 45. A's money is 12 ^ of B's, and 16 ^ of C's; B has $100 more than C: how much has A? $48. 46. Eight men hire a coach ; by getting 6 more passengers, the ex- pense to each is diminished $1}: what do they pay for the coach? $32f. 47. A company engage a supper; being joined by ^ as many more, the bill of each is 60 ct. less : what would each have paid if none had joined them? $2.10 48. By mixing 5 lb. of good sugar with 3 lb. worth 4 ct. a lb. less, the mixture is worth 8^ ct. a lb. : find the prices of the ingredients. 10 ct. and 6 ct. a lb. 49. By mixing 10 lb. of good sugar with 6 lb. worth only J as much, the mixture is worth 1 ct. a lb. less than the good sugar: find the prices of the ingredients and of the mixture. Ingredients, 8 ct. and 5J ct. ; Mixt., 7 ct. per lb. 50. A and B have the same amount of money ; if A had $20 more, 404 RA Y'S HIGHER ARITHMETIO. and B $10 less, A would have 2^ times as much as B: what amount has each? $32^. 51. A and B pay $1.75 for a quart of varnish, and 10 ct. for the bottle; A contributes $1, B, the rest: they divide the varnish equally, and A keeps the bottle: which owes the other, and how much? B owes A 2 J ct 52. How far does a man walk while planting a field of <;orn 285 ft. square, the rows being 3 ft. apart and 3 ft. from the fences? 5 mi. 6 rd. 6 ft 53. Land worth $1000 an acre, is worth how much a front foot of 90 ft. depth; reserving ^ for streets? $2.295-f . 54. I buy stocks at 20 <fo discount, and sell them at 10 <fo premium: what per cent, do I gain? 37 J <fo. 55. I invest, and sell at a loss of 15 ^; I invest the proceeds again, and sell at a gain of 15 ^: do I gain or lose on the two speculations, and how many per cent? Lose 2 J ^. 56. I sell at 8 ^ gain, invest the proceeds, and sell at an advance of 12^ <fc\ invest the proceeds again, and sell at 4 ^ loss, and quit with $1166 40: what did I start with? $1000. 57. I can insure my house for $2500, at -^ <fo premium annually, or permanently by paying down 12 annual premiums: which should I prefer, and how much will I gain by it if money is worth 6 ^ per annum to me? The latter; gain, $113.33} 58. A owes B $1500, due in 1 yr. 10 mon. He pays him $300 cash, and a note of 6 mon. for the balance: what is the face of the note, allowing interest at 6 ^c? $1080.56 59. If I charge 12 ^ per annum compound interest, payable quar- terly, what rate per annum is that? 12yVWWt7 ^• 60. How many square inches in one face of a cube which contains 2571353 cu. in.? 18769 sq. in. 61. What is the side of a cube which contains as many cubic inches, as there are square inches in its surface? 6 in. 62. The boundaries of a square and circle are each 20 ft; which is the greater, and how much? Circle; 6.831 sq. ft. nearly. 63. If I pay $1000 for a 5 yr. lease, and $200 for repairs, how much rent payable quarterly is that equal to, allowing 10 ^ interest? $307.92 a year. 64. What is the value of a widow's dower in property worth $3000, lior age l»eingj 40, and interest 5 ^? $669.50 ( MISCELLANEOUS EXERCISES. 405 65. What principal must be loaned Jan. 1, at 9 ^, to be re-paid by 5 installments of $200 each, payable on the first day of each of the five succeeding months? $978.10 66.. After spending 25 ^ of my money, and 25 ^ of the remainder, I had left $675: what had I at first? $1200. 67. I had a 60-day note discounted at 1 ^ a month, and paid $4.80 above true interest: what was the face of the note? $11112.93 68. Invested $10000 ; sold out at a loss of 20^ : how much must I borrow at 4%, so that, by investing all I have at 18%, I may retrieve my loss? $4000. 69. If J of an inch of rain fall, how many bbl. will be caught by a cistern which drains a roof 52 ft. by 38 ft. ? 9.776+ bbl. 70. A father left $20000 to be divided among his 4 sons, aged 6 years, 8 years, 10 years, and 12 years respectively, so that each share, placed at 4J^ compound interest, should amount to the same when its possessor became of age (21 yr.) : what were the shares? $4360.34; $4761.59; $5199.78; $5678.29 71. $30000 of bonds bearing 7% interest, payable semi-annually, and due in 20 yr. are bought so as to yield 89^ payable semi-annually : what is the price ? $27031.08 72. A man wishes to know how many hogs at $9, sheep at $2, lambs at $1, and calves at $9 per head, can be bought for $400, having, of the four kinds, 100 animals in all. How many different answers can be given ? 288 answers. 73. The stocks of 3 partners, A, B, and C, are $350, $220, and $250, and their gains $112, $88, and $120 respectively ; find the time each stock was in trade, B's time being 2 mon. longer than A's. A*8, 8 mon. ; B*8, 10 mon. ; C*s, 12 mon. 74. By discounting a note at 20 9^ per annum, I get 22 W per annum interest : how long does the note run ? 200 days. 75. A receives $57.90, and B $29.70, from a joint speculation : if A invested $7.83J more than B, what did each invest ? A, $16.08} ; B, $8.25 76. A borrows a sum of money at 6^, payable semi-annually, and lends it at 12^^, payable quarterly, and clears $2450.85 a year: what is the sum? $38485.87 77. Find the sum whose true discount by simple interest for 4 yr. is $25 more at 6% than at 4% per annum. $449.50 78. I invested $2700 in stock at 25^ discount, which pays 89^ annual dividends : how much must I invest in stock at 4^^ discount and paying 10^ annual dividends, to secure an equal income? $2764.80 406 -R-4 Y'S HIGHER ARITHMETIC. 79. Exchanged $5200 of stock bearing 5^ interest at 69^, for stock bearing 7% interest at 92%, the interest on each stock having been just paid : what is my cash gain, if money is worth 6^ to me ? $216.66J 80. Bought goods on 4 mon. credit; after 7 mon, I sell them for $1500, 2i% off for cash ; my gain is 15^^, money being worth 6^ : what did I pay for the goods ? $1252.94 81. The 9th term of a geometric series is 137781, and the 13th term 11160261 : what is the 4th term? 567. 82. My capital increases every year by the same per cent. ; at the end of the 3d year it was $13310 ; at the end of the 7th year it was $19487.171 : what was my original capital, and the rate of gain? $10000, and 10%. 83. Find the length of a minute-hand, whose extreme point moves 4 inches in 3 min. 28 sec. 11.02 — in. 84. Three men own a grindstone, 2 ft. 8 in. in diameter : how many inches must each grind off to get an equal share, allowing 6 in. waste for the aperture? 1st, 2.822— in. ; 2d, 3.621+ in. ; 3d, 6.557— in. 85. I sold an article at 20% gain; had it cost me $300 more, I would have lost 20% : find the cost. $600. 86. A boat goes 16^ miles an hour down stream, and 10 mi. an hour up stream : if it is 22} hr. longer in coming up than in going down, how far down did it go? 585 mL 87. Had an article cost 10% less, the number of % gain would have been 15 more : what was the % gain ? 35%. 88. Bought a check on a suspended bank at 55% ; exchanged it for railroad bonds at 60%, which bear 7% interest: what rate of interest do I receive on the amount of money invested? 21j^%. 89. Bought sugar for refinery; 6% is wasted in the process; 30% becomes molasses, which is sold at 40 per cent less than the same weight of sugar cost ; at what per cent advance on the first cost must the clarified sugar be sold, so as to yield a profit of 14% on the invest- ment? 50%. 90. There is coal now on the dock, and coal is running on also, from a shoot, at a uniform rate. Six men can clear the dock in one hour, but 11 men can clear it in 20 minutes : how long would it take 4 men ? 5 hr. 91. A distiller sold his whisky, losing 4% ; keeping $18 of the pro- ceeds, he gave the remainder to an agent to buy rye, 8% commission ; he lost in all $32: what was the whisky worth? $300. 92. A clock gaining 3} min. a day was started right at noon of the 22d of February, 1804: what was the true time when that clock MISCELLANEOUS EXERCISES. 407 Rhowed noon a week afterward ; and, if kept going, when did it next show true time? 35 min. 32.9 sec. after 11 a. M. True, Sept. 15th, 8^ min. past 5 a. m. 93. The number of square inches in one side of a right-angled tri- angular board is 144, and the base is half the height ; required the areas of the different triangles which can be marked off by lines par- allel to the baee, at 12, 13, 14, 14^ inches from the smaller end. 36 sq. in. ; 42|^ sq. in. ; 49 sq. in. ; 52^^^ sq. in. 94. Suppose a body falls 16 ft. the first second, 48 ft. the next, 80 the next, and so on, constanUy increasing, how far will it have fallen in 4 sec; in 4J sec; in 5 sec? 256 ft.; 324 ft.; 400 ft. 95. A man traveling at a constant increase, is observed to have gone 1 mile the first hour, 3 miles the next, 5 the next, and so on : how far will he have gone in 6i hr. ? 42 J mi. 96. The number of men in a side rank of a solid body of militia, is to the number in front as 2 to 3; if the length and breadth be in- creased so as to number each 4 men more, the whole body will contain 2320 men : how many does it now contain ? 1944 men. 97. A grocer at one straight cut took ofi* a segment of a cheese which had J of the circumference, and weighed 3 lb. : what did the whole cheese weigh? 33.0232-f lb. 98. A wooden wheel of uniform thickness, 4 ft. in diameter, stands in mud 1 ft. deep: what fraction of the wheel is out of the mud? .80449-1- of it. 99. My lot contains 135 sq. rd., and the breadth to length is as 3 to 5 : what is the width of a road which shall extend from one corner half round the lot, and occupy J of the ground ? 24 J ft. 100. A circular lot 15 rd. in diameter is to have three circular grass beds just touching each other and the large boundary : what must be the distance between their centers, and how much ground is left in the triangular space about the main center? Distance, 6.9615242-f rd. Space within, 1.9537115-f- sq. rd. 101. I have an inch board 5 ft. long, 17 in. wide at one end, and 7 in. at the other : how far from the larger end must it be cut straight across, so that the solidities of the two parts shall be equal ? 2 ft. 102. Four equal circular pieces of uniform thickness, the largest possible, are to be cut from a circular plate of the same thickness, and worth $67 : supposing there is no waste, what is the worth of each of the four, and what is the worth of the outer portion which is left? Each small circle, $11.49538-|- Outer portion, $17.87747-1- 408 JiA Y'S HIQHEIt ARITHMETIC, 103. A 12-inch ball Ir in the corner where walls and floor are at right angles : what must be the diameter of another ball which can touch that ball while both touch the same floor and the same walls? 3.2154 in., or 44.7846 in. 104. A workman had a squared log twice as long as wide or deep ; he made out of it a water-trough, of sides, ends, and bottom each 8 inches thick, and having 11772 solid inches: what is the capacity of it in gallons? GS^Sj. gal. 105. How many inch balls can be put in a box which measures, inside, 10 in. square, and is 5 in. deep? 568 balls. 106. A tin vessel, having a circular mouth 9 in. in diameter, a bottom 4J in. in diameter, and a depth of 10 in., is \ part full of water : what is the diameter of a ball which can be put in and just be covered by the water? 6.1967 in. Or T . I C UNIVERSITY YB I 74Co I __