LIBRARY OF THE UNIVERSITY OF CALIFORNIA. OK ECLECTIC EDUCATIONAL SERIES. RAY'S NEW HIGHER ARITHMETIC A EEVISED EDITION OF THE HIGHER ARITHMETIC BY JOSEPH RAY, M. D. Late Professor in Woodward College. VAN ANTWERP, BRAGG & CO. CINCINNATI AND NEW YORK. RAY'S MATHEMATICAL SERIES. ARITHMETIC. Ray's New Primary Arithmetic. Ray's New Intellectual Arithmetic. Ray's New Practical Arithmetic. Ray's New Higher Arithmetic. TWO-BOOK SERIES. Ray's New Elementary Arithmetic. Ray's New Practical Arithmetic. ALGEBRA. Ray's New Elementary Algebra. Ray's New Higher Algebra. HIGHER MATHEMATICS Ray's Plane and Solid Geometry. Ray's Geometry and Trigonometry. Ray's Analytic Geometry. Ray's Elements of Astronomy. Ray's Surveying and Navigation. Ray's Differential and Integral Calculus. COPYRIGHT, 1880, BY VAN ANTWERP, BRAGG & Co. PREFACE. KAY'S HIGHER ARITHMETIC was published nearly twenty-five years ago. Since its publication it has had a more extensive circu- lation 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 BAY'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 modern 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. (iii) 111908 IV PREFACE. The Topical Outlines for Review 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 as to private students. The publishers take this opportunity of expressing their obligations to J. M. GREENWOOD, A. M., Superintendent of Public Schools, Kan- sas City, Mo., who had the work of revision in charge, and also to REV. DR. U. JESSE KNISEL.Y, of Newcomerstown, Ohio, for his valuable assistance in revising the final proof-sheets. CINCINNATI, July, 1880. CONTENTS. PAGE I. INTRODUCTION 9 II. NUMERATION AND NOTATION 15 III. ADDITION 23 IV. SUBTRACTION . .27 Business Terms 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 Contractions 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 OF 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 Keduction of Fractions 78 Common Denominator 82 Addition of Fractions 85 Subtraction of Fractions 86 Multiplication of Fractions 87 (v) v i CONTENTS. PAGE Division of Fractions 90 The G. C. D. of Fractions . .92 The L. C. M. of Fractions . . . . . . .94 IX. DECIMAL 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 Keduction 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 Measure of Length . 149 Measure of Surface 149 Measure of Capacity 150 Measure of Weight . . * 150 Table of Comparative Values 152 Reduction of Compound Numbers 154 Addition of Compound Numbers ..... 160 Subtraction of Compound Numbers . . . . 163 Multiplication of Compound Numbers .... 165 Division of Compound Numbers 167 Longitude and Time . . . . . . . . 169 Aliquot Parts . . 172 CONTENTS. vii XII. RATIO . . . . . . . . . PAGE 175 XIII . PROPORTION . 177 Simple Proportion . 178 Compound Proportion . . . . . 184 XIV. PERCENTAGE . 188 Additional Formulas . 197 Applications of Percentage . . . 197 XV. PERCENTAGE. APPLICATIONS. ( Without Time.) . 199 I. Profit and Loss ...... . 199 ii. Stocks and Bonds . 204 m. Premium and Discount . 208 IV. Commission and Brokerage .... . 213 V. Stock Investments . 220 vi. Insurance ........ . 228 vn. Taxes . 232 vni. United States Revenue . 236 XVI. PERCENTAGE. APPLICATIONS. ( With 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 vr. Equation of Payments . 286 vir. Settlement of Accounts . 292 Account Sales . . 296 297 viii CONTENTS. PAGE Viii. Compound Interest 298 ix. Annuities 308 Contingent Annuities . . . . . . 317 Personal Insurance ....... 322 XVII. PARTNERSHIP 327 Bankruptcy 331 XVIII. ALLIGATION . -. 333 Alligation Medial 333 Alligation Alternate 334 XIX. INVOLUTION 342 XX. EVOLUTION 347 Extraction of the Square Koot . . . . . . 349 Extraction of the Cube Root 354 Extraction of Any Koot 359 Horner's Method 360 Applications of Square Koot and Cube Koot . . . 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 RAT'S HIGHER ARITHMETIC. I. INTEODUOTIOK 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. &. The primary truths of a science are called Prin- ciples. 5. 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 RAY'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 term 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 thing is a con- crete unit ; one is an abstract unit. 21. Number is the expression of a definite quantity. Numbers are either abstract or concrete. An abstract 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. INTE OD UCTION. 11 22. Numbers are also divided into Integral, Fractional, and Mixed. An Integral number, or Integer, is a whole number; a Fractional number 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 + -r- V = : :: () '. 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 3. Minus is described as a short horizontal line. Plus and Minus are Latin words. Plus 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- tiplied %, or times. Thus, 4 X 5 is to be read, 4 multiplied bij 5, 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 [-T-], and is read divided by. When placed between two numbers, the one on the left is 12 RA Y'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 just beneath the middle of it. Dr. John Pell, an English analyst, born in 1610, introduced the sign of division. 29. The Radical sign, [ j/ ] 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 iX625 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 Radical sign. The figure is called the index. 30. The signs, +, , X, -f-, i/, are symbols of operation. 31. The sign of Equality is [=], two short horizontal parallel lines, and is read equals 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 equation, 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 Parenthesis, the second the Vinculum. They are used for the same purpose; thus, 24 (8 + 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. 35. The dots [. . . .], used to guide the eye from words at the left to the right, are called Leaders, or the sign of Continuation, and are read, and so on. 36. The sign of Deduction 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 MAY'S HIGHER ARITHMETIC. Topical Outline of Arithmetic, Preliminary Definitions.. 1. Definition. 2. Quantity. 3. Science. 4. Mathematics. 5. Proposition. 6. Theorem. 7. Axiom. 8. Demonstration. 9. Solution. 10. Rule. 11. Sign. 1. As a Science. .. J 1. Definitions. 2. Classification of Numbers , f Numeration 1. J and I Notation. 2. Addition. 3. Subtraction. 4. Multiplication. 5. Division. 6. Involution. 7. Evolution. L 3. Operations. ^ f Abstract. \ Concrete. r Integral. 2. 1 Fractional. (_ Mixed. 3 f Simple. I Compound. 2. As an Art 1. Terms often used... 2. Signs. 3. Applications.. 1. Problem. 2. Operation 3. Solution. 4. Principle. 5. Formula. 6. Rule. 7. Proof. 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. ^ 9. To Mensuration. II. NUMEKATKOT AKD NOTATION. 40. Numeration 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 : 123456789 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. REMARK. 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 : 11 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 things, while the figure on the right expresses the units or single things additional, required to make up the number. REMARK. The words eleven and twelve 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, four and ten, etc. (15) 16 RAY'S HIGHER ARITHMETIC. The next number above nineteen (nine and ten), is ten and ten, or two groups of ten, written 20, arid called twenty. The next numbers are twenty-one, 21; twenty-two, 22; etc., up to three tens, 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 thousands. 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 Thousands is in the 4th place, 1000 By this arrangement, the same figure has different values according to ilie 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. NUMERA TION AND NO TA TION. 1 7 44. The word Unite may be used in naming all the orders, as follows: Simple units are called Units of the 1st order. Tens " " Units of the 2d order. Hundreds " " Units of the 3d order. Thousands " " Units of 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. 13th. 12th. llth. 10th. 9th. 8th. 7th. 6th. 5th. 4th. 3d. 2d. 1st. cc a o H o CO Trillions . . . s of Billions. . CO 3 s CO O O CO cc a o s s of Thousands. Thousands . . CO CO CO a CO PJ 2 H3 *-! O CO p o o> t4 T3 *0 co co R O S T3 % ^H o> ?H TS c o CO ed CO p Q 0) n CO 02 S3 B 1 g S B 'I S B H S 2 B H B 1 P 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. o o;; o ^H e^ .0 e^ .2 08 O S H p BHP K^Jp' WHP WHP BHP 654 321 987 654 321 5th Period. 4th Period. 3d Period. 2d Period. 1st Period. H. A. 2. 18 RAY'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 " Quintillions. Eighth " Sextillions. Ninth " Septillions. Tenth " Octillions. 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 the next higher order. 2. Removing a significant figure one place to the left increases its value tenfold; one place to the right , decreases its value ten- fold. 3. Vacant 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. Rule 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 with 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 NOT A TION. 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 504630209102 J 800 v 70324d703'250'207 PROBLEM. Express in figures the number four million .twenty thousand three hundred and seven. 4020307. SOLUTION. Write 4 in millions period ; place a dot after it to separate it from the next period : then write 20 in thousands period ; place another dot : then write 307 in units period. This gives 4*20*307. As there are but two places in the thousands period, a cipher must be put before 20 to complete its orders, and the number correctly 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. Rule for Notation. Begin at the left, and write each period in its proper place -filling the vacant orders with ciphers. PROOF. Apply to the number, as written, the rule for Numeration, and see if it agrees with the number given. EXAMPLES IN NOTATION. 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, 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 RAY'S HIGHER 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- 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 iionillion 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 units, the second millions, the third billions, the fourth trillions, etc. The following table illustrates this method : o *r^ ^ -^ *0 u I i - -s 1 M i \ c g P / ^; / ^ N rfi H F EH H -M ^ S O S H Thousands Hundreds Tens Units o ^; CO H TJ S HIGHER ARITHMETIC. CASE II. 63. When the multiplier exceeds 12. PROBLEM. Multiply 246 by 235. SOLUTION. First multiply by 5 OPERATION. (units), and place the first figure of 246 the product, 1230, under the 5 (units). 235 Then multiply by 3 (tens), and place 1230 product by 5 the first figure of the product, 738, 738 product by 30 under the 3 (tens). Lastly, multiply 492 product by 2 by 2 (hundreds), and place the first 57810 product by 235 figure of the product, 492, under the 2 (hundreds). Then add these several products for the entire product. DEMONSTRATION. The of the first product, 1230, is units (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 all the parts or the whole of the multiplier. Rule. 1. Write the multiplier wider the multiplicand, placing fijures of the same order in the same column, and draw a line beneath. 2. Multiply each figure of the multiplicand by each figure of die multiplier successively; first by the units' figure, then by the tens 9 figure, etc.; placing the right-hand figure of each product under that figure of the multiplier which produces it, then draw a line beneath. 3. Add the several partial products together; their sum will be the required product. METHODS OP PROOF.!. Multiply the multiplier by the multiplicand; this product must be the same as the first product. MULTIPLICATION. 35 2. The same as when the multiplier does not exceed 12. NOTE. For proof by casting out the 9's, see Art. 105. REMARK. Although it is custom- OPERATION. ary to use the figures of the multi- plier in regular order beginning with 235 units, it will give the same product 738 product by 30 to use them in any order, observing 492 product by 2 that the right-hand figure of each partial _ 1230 product by 5 product must be placed under the figure 57810 product by 2 3 5 of the multiplier which produces it. EXAMPLES FOR PRACTICE. 1. 7198X216. Ans. 1554768. 2. 8862 X 189. Ans. 1674918. 3. 7575X7575. Am. 57380625. 4. 15607X3094. Ans. 48288058. 5. 93186X4455. Am. 415143630. 6. 135790X24680. Am. 3351297200. 7. 3523725 X 2583. Am. 9101781675. 8. 4687319 X 1987. Am. 9313702853. 9. 9264397 X 9584. Am. 88789980848. 10. 9507340X7071. Am. 67226401140. 11. 1644405 X 7749. Am. 12742494345. 12. 1389294X8900. Ans. 12364716600. 13. 2778588 X 9867. Am. 27416327796. 14. 204265X562402. 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 RAY'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. If 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 acres 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 right, and not directly under the column of amounts added. 65. 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. M UL TIPLICATTON. 37 1. Joseph Allen bought of Seth Ward, at Springfield, 111., 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. SOLUTION. JOSEPH ALLEN, 1879. SPRINGFIELD, ILL., Jan. 2, 1879. Bought of SETH WARD. Jan. 2 To 30 bl. flour, @ $ 3.60 a bl. 108 00 ,, 2 ,, 48 ,, mess pork, ,, 16.25 ,, 780 00 ,, 2 ,, 16 boxes candles, ,, 3.50 ,, box 56 00 2 23 bl. molasses, 28.75 ,, bl. 661 25 2 ,, 64 sacks coffee, 47.50 ,, sack 8040 00 84645 25 2. At St. Louis, March 1, 1879, Chester Snyder bought of Thomas Glenn, 4 Ib. of tea, at 40 ct. ; 21 Ib. of butter, at 21 ct.; 58 Ib. of bacon, at 13 ct.; 16 Ib. of lard, at 9 ct.; 30 Ib. of cheese, at 12 ct.; 4 Ib. 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 renr dered to a customer, showing his debits and credits as they appear on the books. The following is an example : JOHN SMITH, 1880. CINCINNATI, Feb. 2, 1880. In Account with VAN ANTWERP, BRAGG & Co. Jan. 10 To 525 McGuffey's Revised First Readers, @ 16c. ,, 50 Ray's New Higher Arithmetics, ,, 75c. Cr. By Cash i , f Merchandise 84 37 50 75 50 :75 ~^r 20 12 121 32 . $88 38 JRAY'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. CONTRACTIONS 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 factors. 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. ^ cents? cost o f 6 me l O ns. Rule. Separate the multiplier into two or more factors. Multiply first by one of the factors, then this product by another factor, and so on till each factor has been used as a multiplier. The last product wiU be the result required. EXAMPLES FOR PRACTICE. 1. At the rate of 37 miles a day, how far will a man walk in 28 days? 1036 miles. MUL TIPLICA T1ON. 39 2. Sound moves about 1130 feet per second: how far 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. 43), placing one cipher on the right of a number, changes the units into tens, the tens into hundreds, and so on, and, therefore, multiplies the number by 10. Annexing two 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. Annex to the multiplicand as many ciphers as there are in the multiplier ; the result will be 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, 1 30 and then annex three ciphers ; that is, as 162 many as there are on the right in both the 54 factors. 702000 40 HAY'S HIGHER 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. Rule. Multiply as if there were no ciphers on the right in the numbers; then annex to the product as many ciphers as there are on the right in both the factors. EXAMPLES FOR PRACTICE. 1. 15460 X 3200. Ans. 49472000. 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 3046 as to multiply by 1000 (that is, to annex 3 997 ciphers) and by 3, and take the difference of 3046000 the products ; and the same can be shown in 9138 any similar case. 3036862 NOTE. Where the number is a little greater than 10, 100, 1000, etc., the two products must be added. Rule. Annex to the multiplicand as many ciphers as there are figures in the multiplier; multiply the multiplicand by the difference between the 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. Ans. 372795102. MULTIPLICATION. 41 CASE V. 71. When one part taken as units, in the multi- plier, is a factor of another part so taken. PROBLEM. -Multiply 387295 by 216324. SOLUTION. Commence with the 3 of OPERATION. the multiplier, and obtain the first partial 387295 product, 1161885 ; then multiply this prod- 216324 uct by 8, which gives the product of the 1161885 multiplicand by 24 at once (since 8 times 9295080 3 times any number make 24 times it). 83655720 Set the right-hand figure under the right- 83781203580 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 multiplicand by some figure or figures of the multiplier, which are a factor of one or more parts of the multiplier. 2. Multiply this partial product by a factor of some other figure or figures of the multiplier, and write the right-hand figure thus obtained under the right-hand figure of the multiplier thus used. 3. Continue thus until the entire multiplier is used, and then add the partial products. EXAMPLES FOR PRACTICE. 1. 38057 X 48618. Am. 1850255226. 2. 267388 X 14982. Ans. 4006007016. 3. 481063 X 63721. An*. 30653815423. 4. 66917X849612. Am. 56853486204. 5. 102735 X 273162. Ans. 28063298070. 6. 536712 X 729981. Am. 391789562472. H. A. 4. 42 RAY'S HIGHER ARITHMETIC. 1. Definitions. 2. Terms.... 3. Sign. 4. Principles. 5. Operation 6. Rule. 7. Proof. 8. Applications. 9. Contractions. Topical Outline. MULTIPLICATION. 1. Multiplicand. 2. Multiplier. 3. Partial Product. 4. Product. 1. Writing Numbers. 2. Drawing Line Beneath. 3. Finding Partial Products. 4. Drawing a Line Beneath Partial Products. 5. Adding the Partial Products. YL DIVISION. 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 factors, 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 how 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 rem aining. remains ; hence, 24 cents is contained 24 m 73 cents 3 times, with a remainder -J cent 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 : Dividend { \ Divisors and Quotients. ^ 1 O -f- o := O ) (43) 44 RAY'S HIGHER ARITHMETIC. 75. There are three methods of expressing division ; thus, 12-1-3, \ 2 -, or 3)12. Each indicates that 12 is to t be divided by 3. PRINCIPLES. 1. When the dividend and divisor are like numbers, the quotient is abstract. 2. When the divisor is an abstract number, the quotient is like the dividend. 3. The remainder is like the dividend. 4. The dividend is equal to the product 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 : 1. To find the number of equal parts of a nwnber. 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 written, 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 5 164 Quotient. 32 tens. i Both operations are performed on the same 20 units. principle. In the first, the subtraction is writ- 2J) teii ; 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 (thousands), therefore, the quo- tient has no thousands. Next, take 42 g'g . *g % (hundreds) as a partial dividend; 13 5.23 3 is contained in it 3 (hundreds) times ; 13)4225)325 after multiplying and subtracting, there 3 9 hundreds, are 3 hundreds left. Then bring down 2 tens, and 32 tens is the next partial 26 tens, dividend. In this, 13 is contained 2 65 (tens) times, with a remainder of 6 tens. 6 5 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+20 + 5 3900 ~+260 + 260 + 65 + 65 Rule for Long Division. 1. Draw curved lines on the right and left of the dividend, placing the divisor on the left. 2. Find how often 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 the dividend. 3. Multiply the divisor by this quotient figure, and write the product under that part of the dividend from which it was obtained. 4. Subtract this product from the figures above it ; to the re- mainder bring down the next figure of the dividend, and divide as before y until all the figures of the dividend are brought down. 46 - RAY'S HIGHER ARITHMETIC. 5. If at any time after a figure is brought down, the number thus formed is too small to contain the divisor, a cipher must be placed in the quotient, and another figure brought down, after which divide as before. 6. If there is a final remainder after the last division, place flie 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 small, 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. Am. 30449f|. 2. 5484888 67. Ans. 81864. 3. 4326422 961. Ans. 4502. 4. 1457924651 -1204. Ans. 1210900|{Hft. 5. 65358547823 -f 2789. Am. 234344022 G T %V 6. 33333333333-^-5299. Ans. 629Q495^\\. I. 245379633477 1263. Ans. 194283161fif|-. 8. 555555555555 123456. Am. 4500028 1 *$$ r . 9. 555555555555 654321. Am. 849056ffff In the following, multiply A by itself, also B by itself: divide the difference of the products by the sum of A and B. A. B. 10. 2856 3765. Am. 909. II. 33698 42856. Ans. 9158. 12. 47932 152604. Am. 104672. 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. A. B. 13. 4986 5369. Ans. 10355. 14. 3973 4308. Ans. 8281. 15. 23798 59635. Ans. 83433. 16. 47329 65931. Ans. 113260. 17. If 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? 198ff 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. SHOET DIVISION. PROBLEM. How often is 2 cents contained in 652 cents? SOLUTION. Two in 6 (hundreds) is contained 3 OPERATION. (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 is a remainder it may be carried to the next lower order. 48 RAY'S HIGHER 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 the left, divide successively each figure or figures of the dividend by the divisor, and set the quotient beneath. 2. Whenever a remainder occurs, prefix it to the figure in the next lower order, and divide as before. 3. If the figure, except the first, in any order does not con- tain the divisor, place a cipher beneath it, prefix it 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 quotient. PROOF. The same as in Long Division. EXAMPLES FOR PRACTICE. 1. Divide 512653 by 5. Ans. 102530f. 2. Divide 534959 by 7. Ans. 76422f 3. Divide 986028 by 8. Ans. 123253f. 4. Divide 986974 by 11. Ans. 89724|f 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. 49 CONTRACTIONS IN DIVISION. CASE I. 79. When the divisor is a composite number. This case presents no difficulty except when remainders occur. PROBLEM. Divide 217 by 15. SOLUTION. 15 = 3 X 5, hence 217 -i- 3 = 72 and 1 remainder ; 72 -r- 5 =14 and 2 remainder. Dividing 217 by 3, the quotient is 72 threes, and 1 unit remainder. Dividing by 5, the quotient is 14 (fifteens), and a remainder of 2 threes; hence the quotient is 14, and the true remainder is 2 X 3 + 1 = 7. Rule. 1. Divide the dividend by one factor of the divisor, and divide this quotient by another factor, and so on, till each factor has been used; the last quotient will be the required result. 2. Multiply each remainder by all of the divisors preceding the one which produced it. The sum of the products, plus the first remainder, will be the true remainder. BEMARK. 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. SOLUTION. 1 1 )235|43 Rule. Cut off as many figures in the dividend as there are ciphers in the divisor; the figures cut off will be the remainder, and the other figure or figures the quotient. H. A. 5. 50 RAY'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)00)38146 (Art. 79). Dividing by 100 gives 38, and 9 Quotient, 46 remainder (Art. 80); then, dividing 200 + 46 = 246, Kern, by 4 gives 9, and 2 remainder: the true remainder is 2 X 100 + 46 = 246 (Art. 79). Rule. 1. Cut off the ciphers at the right of the divisor, and as many figures from the right of the dividend. 2. Divide the remaining part of the dividend by the remain- ing part of the divisor. 3. Annex to the remainder the figures cut off, and thus obtain the true remainder. ARITHMETICAL 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 -J-, negative numbers by the sign ; thus, + 8 is a positive 8, and 8 a negative 8. KEMARK. 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 SIGNS. 51 84. The signs X and -f- do not show whether their results 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, + 12 5X2, 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 -f- 3, the sign -=- shows that 9 is to be divided by 3 ; what is to be done with the quotient, is shown by the + before the 9. 85. In every such numerical statement, the + or the must be understood to affect the whole result of the opera- tions indicated between it and the next + or , or between it and the close of the expression. Thus, in 5 + 7 X 2 X 9 2 X 6, the + indicates the addition of 126, not of 7 only ; and the indicates the subtraction of 12. The same meaning is conveyed by5+(7X2X9) (2X6). 86. When the signs X and -f- occur in succession, they are to have their particular effects in the exact order of their occurrence. Thus, we would indicate by 96 -r- 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 should write 96 -r- (12 X 4). Usage has been divided on this point, however. EEMARK. It will be observed that in no case can the sign X or ~ affect any number before the preceding + or , or beyond the following + or . EXERCISES. 1. 4X3 + 7X2 9X3+6X4 3X3 = ? SOLUTION. + 4X3 = 12, 7X2 = 14, 9X3= -27, 6X4 = 24, 3X3 = 9. Grouping and adding according to the signs, we have, 12 + 14 + 24 = 50 ; and 27 9 == 36. Therefore, 50 36 = 14, Ans. 2. 2X2 1X2 2X2 5X35X3 4X2 4x2 8X3 5X2 9X3 7X2 12 X4 7X 52 RAY'S HIGHER ARITHMETIC. Soi>UTioN.+ 4 2 4 1 5 15 8 8 24 10 27 14 1 .g 14. Grouping and adding, we have, 4 189 185, Ans. 3. 21 -f- 3 X 7 1X1-7-1X4-7-2 + 18 -=-3x6-7- (2X2) + (4 2 + 6 7) X 4x6-^8 = ? 59. EEMARK. Whenever several numbers are included within the marks of parenthesis, brackets, or vinculum, they- are regarded as (me number. Note the advantage of this in example 3. 4. .16x4-1-8 7 + 48-^1637x4x0x9X16 + 24x6-f-48 4 x 9-M2 = ? 1. 5. (16-M6X96-T-8 7 5 + 3) X[ (27-J-9 ) -7-8- 1] + (91 -J-13X7 45 3)X9=? 9. GENEEAL PEINCIPLES. 87. The following are the General Principles of Mul- tiplication and Division. PRINCIPLE I. Multiplying either factor of a product, multi- plies the product by the same number. Thus, 5X4 = 20, and 5X4X2 = 40, whence 20X2 = 40. II. Dividing either factor of a product, divides the product by the same number. Thus, 5X4 = 20, and 5X4-5-2 = 10, whence 20-5-2 = 10. III. Multiplying one factor of a product, and dividing the other factor by the same number, does not alter the product. Thus, 6X4 = 24, and 6X2X4-5-2 = 24, whence 6X2X2 = 24. IV. 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 24X2n-6 = 8, and 24 -s- (6 -5- 2) = 8. CONTRACTIONS. 53 V. Dividing the dividend, or multiplying the divisor, by any number, divides the quotient by that number. Thus, if 24 be the dividend and 6 the divisor, then 24 -^ 2 = 12, and 12 -*- 6 = 2 ; whence 24 -r- (6 X 2) = 2. Therefore, 4 + 2 = 2. VI. Multiplying or dividing both dividend and divisor by the same number, does not change the quotient. Thus, 24 X 2 = 48, and 6 X 2 = 12 ; consequently, 48 -*- 12 = 4, and 24-5-6 = 4? CONTRACTIONS IN MULTIPLICATION AND DIVISION. CASE I. 88. To multiply by any simple part of 100, 1000, etc, NOTE. Let the pupil study carefully the following table of equivalent parts : PARTS OF 100. PARTS OF 1000. 121 = | of 100. 125 = of 1000. 16| = | of 100. 166f = of 1000. 25 = J of 100. 250 = of 1000. = t of 100. 3331 = of 1000. = f of 100. 375 = f of 1000. 621 = I of 100. 625 I of 1000. 66f = | of 100. 666| | of 1000. 75 =f of 100. 750 =| of 1000. 871 = of 100. 875 = $ of 1000. PROBLEM. Multiply 246 by 87^. OPERATION. 24600 SOLUTION. Since 87 J is of 100, 7 annex two ciphers to the multiplicand, g \i72200 which multiplies it by 100, and then . _ take J of the result. Rule. Multiply by 100, 1000, etc., and take such a part of the result as the multiplier is of 100, 1000, etc. 54 RA Y> S HIGHER ARITHMETIC. EXAMPLES FOR PRACTICE. 1. *22 X 33i An*. 14066|. 2. 6564 X 62J. ^w*. 410250. 3. 10724 X 16|. Ans. 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 f of this product, since 6 is f of 9 ; the result is 39509138238. Rule. Multiply as if the digits were 9', and take such a part of the product as the digit is of 9. EXAMPLES FOR PRACTICE. 1. 451402 X 3333. Ans. 1504522866. 2. 281257 X 555555. Ans. 156253732635. 3. 630224 X 4444000. Ans. 2800715456000. / CASE III. 90. To divide by a number ending in any simple part of 100, 1000, 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 being 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. CONTRACTIONS. 55 Rule. Multiply both dividend and divisor by such a number as will convert the final figures of the divisor into ciphers, and then divide the former product by the latter. NOTES. -1. If there be a 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 FOR PRACTICE. 1. 300521761^225. Ans. 2. 1510337264^43750. Ans. 3. 22500712361-^1406250. Ans. 4. 620712480 -=-20833$-. Quot. 29794. Bern. 4146|. 5. 742851692 -r-29161- Quot. 254692. Bern. 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? 56 RA 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. If 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. If the divisor, quotient, and remainder are given, Low do you find the dividend ? 14. If the dividend, quotient, and remainder are given, how do you find the divisor? MISCELLANEOUS EXERCISES. 1. A grocer gave 153 barrels of flour, worth $6 a bar- rel, for 54 barrels of sugar : what did the sugar cost per barrel? $17. 2. When the divisor is 35, 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 farmer 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 EXER CISES. 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 gained $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 ARITHMETIC. 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 $6300; 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. DIVISION. 1. Dividend. 2. Divisor. 3. Quotient. 4. Remainder. ' 1. 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 3. te 9. Writing Remainder. 6 Rules.. | Lon S ^vision. 7 Proof. 8. Applications. 1. Definitions. 2. Terms 3. Sign. 4. Principles. 5. Operation I Short Division. 9. Contractions / Division. \ Multiplication and Division. f 1..0f Numbers... / Positive ' 10. Arithmetical Signs. \ * Ne S ative - I 2. Of Operation... /Multiplying. I Dividing. 11. General Principles. 12. Applications. VII. PEOPERTIES OF NUMBERS. DEFINITIONS. 91. 1. The Properties of Numbers are those qualities which belong to them. 2. Numbers 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 composite number is the product of two or more other numbers, called its factors (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. REMARK. 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. REMARK. All even numbers except 2 are composite ; the odd numbers are partly prime and partly composite. (59) 60 RA Y'S HIGHER ARITHMETIC. 10. A perfect number is one which is equal to the sum of all 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 Abundant or De- 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 divisor 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 it 'without a remainder. 2d. Every multiple is a composite number. 15. Since every composite number is the product of factors, each factor must divide it exactly; hence, every lactor of a number is a divisor of it. 16. A prime factor of a number is a prime number that will exactly divide it : 5 is a prime factor of 20 ; while 4 is a factor of 20, not a prime factor ; hence, 1st. The prime factors of a number are all the prime num- bers that urill 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 all its prime factors. EXAMPLE. All the prime factors of 15 are 1, 3, and 5 ; and 1 X 3 X 5 = 15. 17. Any factor of a number is called an aliquot part of it. EXAMPLE. 1, 2, 3, 4, and 6, are aliquot parts of 12. FACTORING. 61 FACTOEING. 92. Factoring is resolving composite numbers into fac- tors ; it depends on the following principles and propositions. PRINCIPLE 1. A factor of a number is a factor of any multiple of that number. DEMONSTRATION. Since 6 = 2 X 3, therefore, any multiple of 6 = 2 X 3 X some number ; hence, every factor of 6 is also a factor of the multiple. The same may be proved of the multiple of any composite number. PRINCIPLE 2. A factor of any two numbers is also a factor of their sum. DEMONSTRATION. 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 twos, and 10 is 5 twos, and their sum is 3 twos + 5 twos = 8 twos. 93. From these principles are derived the six following propositions : PROP. I. Every number ending with 0, 2, 4, 6, or 8, is divisible by 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; thus, 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 with 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 JRA Y> S HIGHER ARITHMETIC. DEMONSTRATION. Since 100 is divisible by 4, any number 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 7 and units' places (thus, 384 is equal to 3 hundreds -f- 84) ; then, if the latter part (84) is divisible by 4, both parts, or the number itself, will be divisible by 4 (Art. 92, Prin. 2d). Conversely, No number is divisible 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 the product of any two or more of its prime factors. DEMONSTRATION. Since 2 X 3 X 5 = 30, it follows that 2 X 3 taken 5 times, makes 30; hence, 30 contains 2X3(6) exactly 5 times. In like manner, 30 contains 3 X 5 (15) exactly 2 times, and 2X5 (10), exactly 3 fimes. Hence, If any even number is divisible by 3, 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 III. FACTORING. 63 PROP. VII. Any integer is divisible by 9 or by 3, if the sum of its digits be thus divisible. 94. To find the prime factors of a composite number. PROBLEM. Find the prime factors of 42. SOLUTION. 42 is divisible by 2, and 21 is OPERATION. divisible by 3 or 7, which is found by trial ; 2)42 hence, the prime factors of 42 are 2, 3, 7, /. 2 X 3)21 3X7-42. Rule. Divide the given number by any prime number that will exactly divide it; divide the quotient in like manner, and so continue until the quotient is a prime number; the several divisors and the last quotient are the prime factors. KEMARKS. 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: L 45. Ans. 3, 3, 5. 2. 54. Ans. 2, 3, 3, 3. 3. 72. Am. 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. Ans. 2, 3, 5, 7. 8. Factor 1155. Ans. 3, 5, 7, 11. 9. Factor 10010. Ans. 2, 5, 7, 11, 13. 10. Factor 36414. Ans. 2, 3, 3, 7, 17, 17. 11. Factor 58425. Ans. 3, 5, 5, 19, 41. 64 RAY'S HIGHER ARITHMETIC. 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. Ans. 2, 3, 5. 4. 210 and 315. Ans. 3, 5, 7. 96. To find all the divisors of any composite num- 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 = 2X3X7; 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, Rule. Resolve the number into its prime factors, and then form from these factors all the different products of which they will admit; the prime factors and their products will be all the divisors of the given number. Find all the divisors: 1. Of 70. Ans. 2, 5, 7, and 10, 14, 35. 2. Of 196. Ans. 2, 7, and 4, 14, 28, 49, 98. 3. Of 231. Ans. 3, 7, 11, and 21, 33, 77. 4. Of 496 ; and name the properties of 496. GKEATEST 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 DIVISOR. 65 PRINCIPLES. 1. Every prime factor of a number is a divisor of that number. 2. Every product of two or more prime factors 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 common divisor of two numbers, is a divisor of their difference. To FIND THE GREATEST COMMON DIVISOR. CASE I. 99. By simple factoring. PROBLEM. Find the G. C. D. of 30 and 105. OPERATION. 30 = 2 X 3 X 5. | 3 x 5 = 15, G. C. D. 105 = 3X5X7. ) DEMONSTRATION. The product 3 X 5 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. SOLUTION. 3 3 36, 63, 1 44, 324 12, 21, 48, 108 4, 7, 16, 36 /. 3X3 = ' }, G. C. D. Rule. Resolve the given numbers into their prime factors, and take the product of the factors common to all the numbers. H. A. 6. 66 RAY* 8 HIGHER ARITHMETIC. Find the greatest common divisor : 1. Of 30 and 42. Ans. 2x3 = 6. 2. Of 42 and 70. Ans. 2x7== 14. 3. Of 63 and 105. Ans. 3 X 7 = 21. 4. Of 66 and 165. Ans., 3 X 1133. 5. Of 90 and 150. Am. 2 X 3 X 5 = 30. 6. Of 60 and 84. Ans. 2 X 2 X 3 = 12. 7. Of 90 and 225. Ans. 3x3x5 = 45. 8. Of 112 and 140. Ans. 2 X 2 X 7 = 28. 9. Of 30, 45, and 75. Ans. 3 X 5 = 15. 10. Of 84, 126, and 210. Ans. 2x3x7 =42. 11. Of 16, 40, 88, and 96. Ans. 2x2x2 = 8. 12. Of 21, 42, 63, and 126. Ans. 3 X 7 = 21. CASE II. 100. By successive divisions. PROBLEM. Find the G. C. D. of 348 and 1024. OPERATION. DEMONSTRATION. If 348 348) 1 2 4 (J2_ will divide 1024, it is the 696 G. C. D. ; but it will not 328 ) 3 4 8 ( 1 divide it. ~ 3 2 8 ~ If 328 (Art 98, Prin. 5,) , 3 2 g fl 6 will divide 348, it is the ' G. C. D. of 328, 348, and 1024 ; but it will not divide 348 and 1024 exactly. * 2 If 20 will divide 328 (by 8)20(2 same process of reasoning), 1 6 it is the G. C. D.; but there G. C. D. 4)8 is a remainder of 8 ; hence, 2 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. GREATEST COMMON DIVISOR. 67 Rule. Divide the greater number by the less, and the divisor by the remainder, and so on; always dividing the last divisor by the last remainder, till nothing remains; the last divisor will be the greatest common divisor sought. NOTE. A condensed form of operation CONDENSED OPERATION. may be used after the pupils are familiar 348 with the preceding process. 328 REMARK. 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., 16 and any one of the remaining numbers, 4 and so on for all of the numbers ; the last C. D. will be the G. C. D. of all the numbers. 1024 696 328 320 8 1 16 2 2 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. 1. 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. Am. 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. Ans. 11. 13. 816, 1360, 2040, and 4080. Ans. 136. 14. 1274, 2002, 2366, 7007, and 13013. Ans. 91. 68 RAY'S HIGHER ARITHMETIC. LEAST COMMON MULTIPLE. 101. A common multiple (C. M.) of two or more numbers, is a number that can be divided by each of them without a remainder. 102. The least common multiple (L. C. M.) of two or more numbers, is the least number that is divisible by each of them without a remainder. PRINCIPLES. 1. A multiple of a number is divisible by that number. 2. A multiple of a number must contain all of the prime factors of that number. 3. A common multiple of two or more numbers is divisible by each of those numbers. 4. A common multiple of two or more numbers contains all of the prime factors of each of those numbers. 5. The least common multiple of two or more numbers must contain all of the prime factors of each of those numbers^ and no other factors. 6. If two or more numbers are prime to each other, tJieir continued product is their least common multiple. To FIND THE LEAST COMMON MULTIPLE. CASE I. 103. By factoring the numbers separately. PROBLEM. Find the L. C. M. of 10, 12, and 15. SOLUTION. Resolve each OPERATION. number into its prime f ac- 10=2X5 tors. A multiple of 10 con- 12=2X2X3 tains the prime factors 2 and 15 = 3X5 5 ; of 12, the prime factors .'. L. C. M. = 2 X 2 X 3 X 5 = 6 0. 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 MULTIPLE. 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). Rule. Resolve each number into its prime factors, and then take the continued product of all the different prime factors, using each factor the greatest number of times it occurs in any one of the given numbers. REMARKS. 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. Am. 120. 2. 6, 9, 12. Ans. 36. 3. 12, 18, 24. Ans. 72. 4. 8, 14, 21, 28. Ans. 168. 5. 10, 15, 20, 30. Ans. 60. 6. 15, 30, 70, 105. Ans. 210. CASE II. 104. By dividing the numbers successively by their common primes. PROBLEM. Find the L. C. M. of 10, 20, 25, and 30. SOLUTION. Write the OPERATION. 20 25 30 10 25 15 numbers as in the margin. 2 Strike out 10, because it is ~ contained in 20 and 30. Next, divide 20 and 30 by 25 the prime factor 2; write 2X5X2X5X3=300 L. C. M. 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 ARITHMETIC. Rule. 1. Write the numbers in a horizontal line; strike out any number that will exactly divide any of the others; divide by any prime number that will divide two or more of them without a remainder; write the quotients and undivided num- bers in a line beneath. 2. Proceed with this line as before, and continue the opera- tion till no number greater than 1 will exactly divide two or more 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. 100. 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 PROPEKTIES OF THE NUMBER NINE. 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 NINE> PRINCIPLE. Any number divided by 9, remainder as the sum of its digits divided by 9. ILLUSTRATION. 700000 = 7 X 100000 = 7 X ( 99999 + 1 ) = 7 X 99999 + 7 60000 = 6 X 10000 = 6 X (9999 + 1) =6 X 9999 + 6 8000 = 8X 1000 = 8X (999 + l) = 8X 999 + 8 900 = 9 X 100 = 9 X (99 + l) = 9X 99 + 9 40 = 4X 10 = 4X (9 + l)=4X 9 + 4 I 5 = 5X 1= 5 Whence, 7 X 99999 + 6 X 9999 + 8X999+ 9 X 99 + 4X 9 + 7 + > = 768945. 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 can be no remainder in the multiples, except in the sum of the digits. PROOF OF ADDITION. The sum of the excess of ffs in the several numbers must equal the excess of 9's 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 8 5834 2 6241 4 7302 JJ_ 8 26729 PROOF OF SUBTRACTION. The excess of 9's in the minuend must equal the sum of the excess of ffs in the subtrahend and remainder. OPEB ATION. ILLUSTRATION. As the min- Minuend, 7640 8 uend is the sum of the subtra- Subtrahend, 1234 1 Kemainder, 6406 7 hend and remainder, the reason of this proof is seen from that of Addition. PROOF OF MULTIPLICATION. Find the excess of 9's in the factors and in the product. The excess of ffs in the product of the excesses of the factors, should equal the excess in the product of the factors themselves. 72 RAY'S HIGHER ARITHMETIC. ILLUSTRATION. Multiply 835 by 76 ; OPERATION. the product is 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 7X4=28, " =1 is also 1, as it should be. 63460, " =1 PROOF OF DIVISION. Find the excess of 9's in each of the terms. To the excess of 9's in the product of the excesses in the divisor and quotient, add the excess in the remainder; the excess in the sum should equal the excess in the dividend. OPERATION. ILLUSTRATION. Divide 8915 by 25; the quo- 356, excess 5 tient is 356, and the remainder, 15. The excess 25, " 7 of 9's in the divisor is 7 ; in the quotient, 5 ; 35 their product is 35, the excess of which is 8. u The excess in the remainder is 6. 6 + 8 = 14, ' lt of which the excess is 5. The excess of 9's in the dividend is also 5. 14, " 5 8915, " 5 CANCELLATION. 106. Cancellation 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, $, 0, $. PRINCIPLES. 1. Canceling a factor in any number, divides the number by that factor. 2. Canceling a factor in both dividend and divisor, does not change the quotient. (Art. 129, III.) PROBLEM. Multiply 75, 153, and 28 together, and divide by the product of 63 and 36. CANCELLATION. 73 SOLUTION. Indicate the operations OPERATION. as in the margin. Cancel 4 out of 2517 V 28 and 36, leaving 7 above and 9 below. Cancel this 7 out of the divi- dend and out of the 63 in the divisor, PP X #0 leaving 9 below. Cancel a 9 out of ^ ^ the divisor and out of 153 in the 3 dividend, leaving 17 above. Cancel 25X17 3 out of 9 and 75, leaving 25 above g -- = 1 4 1 f 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|. Rule for Cancellation. 1. Indicate the multiplications which produce the dividend, and those, if any, which produce the divisor. 2. Cancel equal factors from dividend and divisor ; multiply 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 horsey, 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 X 90. 34. H. A. 7. 74 RA Y'S HIGHER ARITHMETIC. Topical Outline. PROPERTIES OF NUMBERS. Properties. 1. Definitions ...... Numbers Classified ....... Integer ...... - p ract i on Mixed. Divisor; Divisible ; -Multiple ; Factors; Prime Factors; Aliquot Parts. Prime. Composite. Even. Odd. Perfect. Imperfect. T 1. Definition. 2. Principles. 2. Factoring I 3 - Propositions. j 4. Operations. 5. Rules, i 6. Applications. 3. G. C. D.. 1. Definitions.... 2. Principles. 3. Operations..... 1. Common Divisor. 2. Greatest Common Divisor. Case I J Rule - X Applications. Case II... / Rule. I Applications. f 1. Definitions / L Common Multiple. \ 2. ~ 4. L. C. M i 2. Principles. j 3. Operations j Case II.. .\ . Applications. , 1. Addition. Some Properties of the No. 9.. { l * * 1C1 P 16 ' 2 . Subtraction. I 2. Application to... 4 Least Common Multiple. Case I / Rule ' \ Applications. Case II... f 6. Can cell: L t Lou... . 1. Definition. 12. Sign, /. 3. Principles. 4. Operation. 5. Rule. 6. Applications. 3. Multiplication 4. Division. VIII. COMMON FKACTIOm PEFINITIONS. 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. 109. A Common Fraction is expressed by two num- bers, one above and one below a horizontal line; thus, -f, which is read two thirds. 110. 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 denominates, or names, the parts ; the numerator numbers the parts. 112. The Terms of a fraction are the numerator and the denominator. ILLUSTRATION. The expression f, 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 thing, 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 division (Art. 75) in which, 1. The dividend is the numerator. 2. The divisor is the denominator. 3. The quotient is the fraction itself. 115. The value of a fraction is its relation to a unit. 116. 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, 3-| . 120. A Simple Fraction is a single fraction whose terms are integral ; as, f , f , J-. 121. A Complex Fraction is one which has one or i i both of its terms fractional ; as, -| r , , or ff. 2 3 122. A Compound Fraction is a fraction of a fraction ; as, \ of f . 123. An Integer may be expressed as a fraction by writing 1 under it as a denominator ; thus, , which is read seven ones. COMMON FRACTIONS. 77 124. The Reciprocal 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, fmngo, 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. NUMERATION AND NOTATION OF FRACTIONS. 127. Numeration of Fractions is the art of reading fractional numbers. 128. Notation of Fractions is the art of writing frac- tional numbers. Rule for Reading Common Fractions. Read the num- ber of parts taken as expressed by the numerator, and then the size of the parts as expressed by tJie denominator. EXAMPLE. |- is read seven ninths. REMARK. Seven ninths (J), signifies 7 ninths of one, or J of 7, or 7 divided by 9. Rule for Writing Common Fractions. Write the num- ber of parts ; place a horizontal line below it, wider which write the number which 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-fifteenths. 78 RA Y> S HIGHER ARITHMETIC. 129. Since a fraction is an indicated division (Art. 114) ; therefore, PRINCIPLES. I. A Fraction is multiplied, 1st. By multiplying the numerator. 2d. By dividing the denominator. II. A Fraction is divided, 1st. By dividing ilie numerator. 2d. By multiplying the denominator. III. The value of a Fraction is not changed, 1st. By multiplying both terms by the same number. 2d. By dividing both terms by the same number. KEMARK. The proof of I is found in Art. 87, Principle IV ; the proof of II is in Principle V ; and of III, in Principle VI. KEDUCTION OF FKACTIONS. 130. Reduction 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 lowest terms when the numerator and de- nominator are prime to each other; as, f, but not f. PROBLEM. Reduc3 -f-g- to its lowest terms. SOLUTION. Dividing both terms by the FIRST OPERATION. common factor 2, the result is ^f ; dividing 2 ) | = } this by 5 (129, in\ the result is f, which 5 ) -if f, 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 ) f-J = J, Ans. |, as before. REDUCTION OF FRACTIONS. 79 Rule. Reject all factors common to botfi, terms of the fraction. Or, divide both terms of the fraction by their greatest common ditnsor. Reduce to their lowest terms : 2. |f. Ans. 3. T \V Ans. 4. . Ans. Express the following in their simplest forms: 5.. - An*, 6. |ff Am. ff 8. |||. Ans. if. 9. 10. -f 11. 923 -*- 1491. Ans. 12. 890 1691. 13. 2261-^-4123. ^Irw. 14. 6160 -f- 40480. CASE II. 132. To reduce a fraction to higher terms. REMARK. Reducing a fraction to higher terms, is changing it to an equivalent fraction whose terms are larger numbers. PROBLEM. Reduce f to fortieths. SOLUTION, Divide 40 by 8, the quo- OPERATION. tient is 5 ; multiply both terms of the 40-^-8 = 5 given fraction, f, by 5 (129, ra), and the 5__5X5_ 25 ^us result is ff , the equivalent fraction re- 8 X 5 quired. Rule. Divide the 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 ^ an d ^ to ninety-ninths. Ans. |~J, |-f. 2. Reduce f, -f-, and -^ to sixty-thirds. Ans. ||, ff, ^. 3. Reduce T 8 T , ^ and |^ to equivalent fractions having 6783 for a denominator. Ans. f|||, |f||, ffff 80 RA Y'S HIGHER ARITHMETIC. CASE III. 133. To reduce a whole or mixed number to an improper fraction. PROBLEM. Eeduce 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. Rule. Multiply together the whole number and the denom- inator of the fraction: to the product add the numerator, and write the sum over ilie denominator. 1. In $7f, how many eighths of a dollar? Ans. -% 9 -. 2. In 19f gallons, how many fourths? Am. ^ . 3. In 13f-J hours, how many sixtieths? Ans. - 8 ^y-. Reduce to improper fractions: 4. llf. Ans. 5. 15 T 8 T . Ans. 6. 127-H-. Ans. 7. 109 T V Ans. Ans. 9. 13ft. Am. iflp. REMARK. To reduce a whole number to a fraction having a given denominator, is a special case under the preceding. PROBLEM. Eeduce 8 to a fraction whose denominator is 7. SOLUTION. Since ^ equals one, 8 equals 8 times |, or - 5 7 ? . CASE IV. 134. To reduce an improper fraction to a whole or mixed number. PROBLEM. Reduce /- of a dollar to dollars. SOLUTION. Since 5 fifths make 1 dollar, there will he as many dollars in 13 fifths as 5 fifths are contained times in 13 fifths ; that is, 2| dollars. REDUCTION OF FRACTIONS. 81 Rule. Divide the numerator by the denominator; the quotient will be the whole or mixed number. REMARK. If there be a fraction in the answer, reduce it to its lowest terms. 1. In - 3 -g 7 - of a dollar, how many dollars? 2. In i|^ of a bushel, how many bushels? 3. In J g% 5 - of an hour, how many hours? Reduce to whole or mixed numbers: 5. 6. 341 bu. hours. Ans. 1. 7. ifp. Ans. 105 T 7 T . Am. 35. 8. -ff a . Ans. 827ft- Ins. 88f. 9. .l^LG, Am. 509^. CASE V. 135. To reduce compound to simple fractions. PROBLEM. Reduce f of -f- to a simple fraction. SOLUTION. of | = - 2 V > 1 f y ~28' * OPERATION. 28 4X7 28 Rule. Multiply the numerators together for the numerator, and ike denominators together for the denominator of the frac- tion, canceling common factors if they occur in both terms. REMARK. Whole or mixed numbers must be reduced to im- proper fractions before applying the rule. PROBLEM. Reduce of -ft of ^ to a simple fraction. OPERATION. SOLUTION. Indicate the work, and employ cancellation, as % X $ X 7 shown in the accompanying 3X 10X HQ^^ ^' ^ nSt operation. ^ . 82 RAY'S HIGHER ARITHMETIC. Eeduce to simple fractions: 1. i of f of f Am. f 2. I of |. of 2f . Ans. f . 3. | of || of 2|. .Am. 2. 4. i of | of 3f. Ans. 1. 5. | of |- of ^ of 8f. Ans. 3. 6. | of | of -f- of | of 4|. -4ns. |. 7. T 8 T of f of A of II of 7 i- 8. |f of A of A of ^ of 1 V COMMON DENOMINATOR 136. A common denominator of two or more fractions, 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. Only a common multiple of different de- nominators can become a common denominator. 2. Only a least common multiple can become a least common denominator. CASE VI. 138. To reduce fractions to equivalent fractions having a common denominator. PROBLEM. Reduce -^, f , and f to a common denominator. OPERATION. SOLUTION. Since 2 X 3 X 4 = 24, 24 is a 1X3X4^12 common multiple of all the denominators. The terms of J must be multiplied by 3 X 4 ; 2 X2X4 __ 1_6 the terms of f, by 2 X 4 ; and the terms of f , by 3X2X4 24 2X3. The values of the fractions are un- 3X^X3 = 18 altered. (129, in.) 4X2X3 24 DEDUCTION OF FRACTIONS. 83 Rule. Multiply both terms of each fraction by the denomina- tors of the oilier fractions. REMARK. Since the denominator of each new fraction is the product of the same 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 : PROBLEM. Reduce |, f, and ^ to a common denominator. OPERATION. 2X5X7 = 70, common denominator. 1X5X7 = 35, first numerator. i = $H 3X2X7 = 42, second numerator. f = f f ( Am. 6 X 2 X 5 = .6 0, third numerator. f = f J NOTE. Mixed numbers and compound fractions must first be reduced to simple fractions ; the lowest terms are preferable. Reduce to a common denominator: i- t, I, I- A**- ii li U- 2- \, i f ^. Ao> T 2 2\> T\V 3- I, f, I- -4. HI, A 2 *, IM- 4- i, I, I, ! Am. fti, iff, ||o ; i|o. 5. f, i of 3}, | of f. ^ins. |^, -W-, H- 6. | of f , f of |, i of | of f of 2f . Ant. ffft, |f* iff- EEMARK. 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 will be found by dividing the largest denomina- tor by the denominator of the fraction to be reduced. PROBLEM. Reduce J and f to a common denominator. OPERATION. SOLUTION. The largest denominator, 6, is a 1X2 2 multiple of 3; therefore, if we multiply both terms of J by 6 divided by 3, which is 2, it is 5 reduced to ^. "7T == ~r' 84 RA Y'S HI GHEE ARITHMETIC. Reduce to a common denominator: 1. i, |, and f. Am. f, f, |. 2. |, | and T V -4ns. T 8 2, jf , -fa. 3. |, f , ^ and ^. 4n. |f , |, f, ft. CASE VII. 139. To reduce fractions of different denominators, to equivalent fractions having the least common de- nominator. PROBLEM. Keduce -f, f , and -fa to equivalent fractions, having the least common denominator. OPEllATION. SOLUTION. Find the L. C. M. of 8, 9, 5 5X9 45 and 24, which is 72 ; divide 72 by the given "7 g~\7a ~~7~2 denominators 8, 9, and 24, respectively ; 7 7 V 8 5 6 multiply both terms of each fraction by " = the quotient obtained by dividing 72 by q _ its denominator ; the L. C. M. is the de- = = nominator of the equivalent fractions. In practice it is not necessary to multiply each denominator in form. Rule. 1. Find the L. C. M. of the denominators of the given fractions, for the L. C. D. 2. Divide this L. C. D. by the denominator of each fraction, and multiply the nwnerator by the quotient. 3. Write the product after each multiplication as a numer- ator above the L. C. D. REMARK. All expressions should be in the simplest form. Reduce to the least common denominator : 1. i, \, f Am. &, &, ||. 2- i, f, T%, f- Am. ft, if, if, M- 3- f, I, -H> An*. U, n, U- 4- |, |, -h, M- Ans. f, f, 1, f. 5- f, A. H, A- -Ans. !, fi M, ft- 6. If, 3|, and ^ of 3f .Atw. W> W. If- ADDITION OF FRACTIONS. 85 ADDITION OF FKACTIONS. 140. Addition of Fractions is the process of uniting two or more fractional numbers in one sum. REMARK. As integers to be added must express like units (Art. 52), so fractions to be added must express like parts of like units. PROBLEM. What is the sum of f, |, and ^-? SOLUTION. Reducing the given fractions to OPERATION. equivalent fractions having a common denom- J= J| | Jf inator, we have -|-f, |f, and JJ. Since these T 7 2 = i| are now of the same kind, they can be added if + M + if f f by adding their numerators. Their sum is fi lfi> Am. 11= iff- Rule. Reduce the fractions to a common denominator, add their numerators, and write the sum over the common denom- inator. REMARKS. 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. 3. After adding, reduce the sum to its lowest terms. EXAMPLES FOR PRACTICE. 1. |, f , and ^. Ans. ff, 2. |, |, f , and ^. Ans. 2J 3. If and 2f . Ans. 4. 2i 3f and 4f Ans. 6. li, 2|, 3, and 44, 7. | of f , and f of | of 9- 1 + H + + II + If ^s. 4 T \V 10. f of 96^ + f of || of 5f 11- 1 + 1 + If + lf + 111 + HI- 86 RAY'S HIGHER ARITHMETIC. SUBTKACTION OF FRACTIONS. 141. Subtraction of Fractions is the process of finding the difference between two fractional numbers. REMARK. 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 -f%. SOLUTION. Reducing the given frac- OPERATION. tions to equivalent fractions having a f == f& common denominator, we have f = f J, T 8 ^ = ^f and A = t$; the difference is & = A- Rule. Reduce the fractions to a common denominator, and write the difference of their numerators over the common denom- inator. REMARK. Before applying the rule, the fractions should be in their simpl st form. The difference should be reduced to its lowest terms. EXAMPLES FOR PRACTICE. 1. f A. Ans. |f. 2. T 8 T -ft*- of ^. Ans. 4. ^- y 1 ^- of 4. Ans. 5. ii -gV Am. f|. f* 9 1 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 jf from 31f. SOLUTION. Reducing the fractions to a OPERATION. common denominator, = if. But if- can 3 1 j^f + if ft not be taken from if. Take a unit, if, 23j| ft if it from the integer of the minuend and add it 7 .u. to if. Then if + if = ff, and 4* = f|. 30 23 7. Therefore the answer is ?if . 9. 12f-10ff. 10. 12|| 9. Z o O O nR23 O2 3T *f ' -^2 7_J> Ql 13. 15 f 14. 18 5f. Ans. 12|. 15. | of 2% 3f|. Jlns. ff. 16. 31 f of If. 4n. If. 17. - 1 / qf 41 -^ of 3 =*= what? Ans. 13f . 18. 11-| -f 8-J- 9^f = what ? .4ns. 10^|. 19. A man owned f f of a ship, and sold f of his share : how much had he left? Ans. -f$. 20. After selling ^ of f + | of f of a farm, what part of it remains ? Ans. ---. 21. 3i + 4| -- 5 + 16| -- 7Ji + 10 14f, is equal to what? Ans. 6|f. 22. 5J- 2| + -^ 3f^ + 3^ + 8-J- 16|, is equal to what? Ans. -|-|. 23. 1 f of f | of f = what? ^s. ^. MULTIPLICATION OF FEACTIONS. 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 fraction by an integer. 2. To multiply an integer by a fraction. 3. To multiply one fraction by another. 88 RAY' 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 f . J times f is |- of OPERATION. f = &- I times ?> then > is 5 times s 3 o=ii- f X f = i{!-, PROBLEM. Multiply f by 6. SOLUTION. Six times 3 fourths is 18 OPERATION. fourths. Reducing to its simplest form, J X f ~V~ ~ 4 i> we have 4|. or, f X 6 = \ 8 - = 4 J PROBLEM. Multiply 8 by f. SOLUTION. One fifth times 8 is f , 3 OPERATION. filths times 8 is 3 times f= 2 /-. Ee- f- X f * = 4J, Ans. ducing to a mixed number, we have 4f. or, NOTE. The three operations are alike, and from them we may derive the rule. Rule. Multiply the numerators together for the numerator of the product, and the denominators for the denominator of the product REMARK. Whole numbers may be expressed in the form of fractions. EXAMPLES FOR PRACTICE. 4. T VX28. Am. 15}. 5. if X 30. Ans. 26. 6. 3f X 5. Ans. 1. if X 12. Ans. 2. -14 X 18. Ans. 8-k , 3. ffX24. Ans. HI. OPERATIONS. REMARK. In multiplying a 2 o 2 n mixed number by a whole mini- ^ 1 1. vx 5 _ 5^5 ber, multiply the whole number -T-T ,, r % 15 ^/ = 18i and the traction separately, and j add the products ; or, reduce the mixed number to an im- > s ' proper fraction, and multiply it ; as, in the last example. MULTIPLICATION OF FRACTIONS 89 t 102f. Am. A. 7. 45 X f Am. 35. 11. 28X3}. 8. 50XH- Ans. 39f 12- if X T^- 9. 25 Xf. ^1?18. 18|. 13. ifX-ft- 10. 32X2|. Ans. 76. 14 - If x|f- 15. What will 3-|- yards of cloth cost, at $4-J- per yard? BEMARK. In finding the product of two mixed numbers, it is generally best to reduce them to improper fractions ; thus, OPERATION. SOIAJTION. The operation may be performed without $ 4 J reducing to improper fractions ; thus, 3 yards will cost 3 j- $13 J, and J of a yard will cost J of $4J = $1J; hence, 1 3 \ the whole will cost $15. 1 i. Ans, $15 16. 6| X 4J. 4w8. 30. 18. 12f X 3 T 3 T . ^?is. 40J. 17. 4| X 2|. Ans. 124. 19. 7| X 3^. Ans. 26|. 20. Multiply | of 8 by \ of 10. Ans. 4. 21. Multiply | of 5| by f of 3. Ans. 7. 22. Multiply f of | of 5f by f of 3|. Ans. 4^. 23. Multiply 5, 4J, 2^, and f of 4f . ^?is. 94f. 24. Multiply -f, f , T 5 T , | of 2, and -f- of 3|. ulns. ^\%. 25. Multiply f , , ^-, 31, and 3|. Jns. 4. 26. Multiply 3, 4|, 5|, | of T 5 T ^and 6f . ^Lns. 49. 27. At % 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 ? 318| days. 29. At 3|- cents a yard, what will 2| yards of tape cost? 9f cents. 30. What must be paid for f of f of a lot of groceries that cost $18f ? 87|. 31. K owns f of a ship, and sells f of his share to L: what part has he left? A. H. A. 8. 90 RAY'S HIGHER ARITHMETIC. DIVISION OF FKACTIONS. 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 fraction by an integer. 2. To divide an integer by a fraction. 3. To divide one fraction by another. NOTE. Since any whole number may be expressed in the form of a fraction, the first and second cases reduce to the third case. PROBLEM. Divide f by f. SOLUTION. \ is contained in 1, seven OPERATION. times ; \ is contained in -J, of 7 = | times ; ^-r-| = fXi = ff \ is contained in f , 5 X J V times ; ^ is contained in f , J of -% 5 - = f J times. It will be seen that the terms of the dividend have been multiplied, 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. | is contained in f -f- f f X f = V = 7 J, Ans. 1, five times; J is contained in Or, 3 H- f = *-== 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 ^- f = f X i = T\ = f > ^ r '*- in 1, J times; 2 is contained in Or, f -5- 2 = f. } , | of } = ^ times ; 2 is con- tained in f , 4 times j 1 ^ = T 4 = -f- times. NOTE. From these solutions we may derive the following rule. Rule. Multiply the dividend by the divisor with its term* inverted. DIVISION OF FRACTIONS. 91 REMARK. The terms of the divisor are inverted because the solution requires it. The same may be shown by a different solu- tion, as below. PROBLEM. Divide f by f. SOLUTION. Reduce both dividend and OPERATION. divisor to a common denominator. The f = J f = f J quotient of Jf -*- f | is the same as 10 -H 27 Jg -r- f | = i, ^4ns. = ^f. The same result is obtained by multiplying the dividend by the divisor, with its terms inverted ; thus, f X f = i?. REMARK. Mixed numbers must be reduced to improper frac- tions. Use cancellation when applicable. EXAMPLES FOR PRACTICE. Ans. Ans. Ans. ~9. . 23*. 4V 10. If 5. 11. s* f. 12. 19*-*-!*, 13. 73|-|-9f 14. i-JH-3. 2. **-7. 3. |--8. 4 > 6 * 2^ 5. 21-=--ft, 6. -=-*. 15. Divide 1* by * of f of 7*. 16. Divide T %- of T 3 of ^ by ^ of *|. 17. Divide | of * by * of f. 18. Divide-* of 3f by || of 7. 19. Divide * of f X & by ^ of 3*. 20. Divide f of 5* by |- of ^ of 3*. 21. Divide * of * of & by | of * of f O / 11/O O / 22. Divide If times 4| by 1^ times 3|. 23. Divide 3f by f of 8* times T 5 T of 24* Divide T 9 T of f of 27* by | of T 3 T of 5*. 25. What is 2| X f of 19* -T- (4f X f\ of 8) Ans. Ans. Am. 121. . lOf. Am. Am. if A- if- ? 92 RAY'S HIGHER ARITHMETIC. 144. To reduce complex to simple fractions. PROBLEM. Reduce - to a simple fraction. OPERATION. EXPLANATION. The mixed If = -y- 2J- = f numbers are reduced to im- - -5- | = - 1 g 1 - X I = if- = ff> ^ ? ' s - proper fractions, and the numerator is divided by the denominator. (Art. 114.) Rule. Divide the numerator by the denominator, as in division of fractions. Reduce to simple fractions: 1. 2. -. 3. A Ans. -%. Am. f. Ans. If. 4. . 6. Ans. f J-. 16H' REMARK. Complex fractions may be multiplied or divided, by reducing them to simple fractions. The operation may often be shortened by cancellation. 7. X . 8 - X 9 - x 10 -!-T2T Ans - 40$ 73 ' Am. U. o o Ans. 3 THE GREATEST COMMON DIVISOR OF FRACTIONS. 145. The greatest common divisor of two or more fractions is the greatest fraction that will exactly (Jivide each of them. G. C. D. OF FRACTIONS. 93 One fraction is divisible by another when th'e 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, & is divisible by & ; for T \ = if; ff Hh A = 6- 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 7 f^. PROBLEM. Find the G. C. D. of -|, f -|, and -fy. 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. ^, Ans. C. D. of the fraction is -g 1 ^. Rule. Find the G. G. D. of the numerators of the fractions, and divide it by the L. C. M. of their denominators. REMARK. The fractions should be in their simplest forms before the rule is applied. Find the greatest common divisor: 1. Of 83i and 268f. Ans. 2 T V 2. Of 14 T V and 95f. Am. &. 3. Of 591 and 735-}-f. Ans. 2|f. 4. Of 23 T V and 213if. Ana. 2J|. 5. Of 418| and 17721. ns. jf. 6. Of 261 jf and 652^. Ans. 4|f. 7. Of 44|, 546|, and 3160. Ans. 4|. 8. A farmer sells 137^ bushels of yellow corn, 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 corn from being mixed ; also the number of sacks for each kind. 3 bushels ; 44, 153, and 670. 94 RAY'S HIGHER ARITHMETIC. 9. A owns a tract of land, the sides of which are 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 fractions 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 multiple 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. T 8 r is a multiple of -fa. 8 is a multiple of 2, and 11 is a divisor of 33; hence, T 8 T -r- -fa = T 8 T X = I 2 - The same result is otherwise obtained ; thus, T 8 T = f f , and f f -5- -fa = 12. A fraction is a common 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 -^, 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 5 = 1 5 C. D. of the denominators is 1 ; G. C. D. of 3, 4, 6 = 1 therefore, the L. C. M. of the .*. y-, Ans. fractions is ^, or 15. L. C. M. OF FRACTIONS. 95 Rule. Divide the L. C. M. of the numerators by the G. C. D. of the denominators. KEMABK. The fractions must be in their simplest forms before the rule is applied. Find the least common multiple : 1- Of f > t> f> f> and f An*. 60. 2. Of 4, 6f , 5f , and 10J. Ans. 4721, 3. Of 31, 4f , -3^-, 5f , and 12. 4wa. 350. 4. A can walk around an island in 14^ hours ; B, in 9^j hours ; C, in 16f hours ; and D, in 25 hours. If they start from the same point, and at the same time, how many hours after 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|, 4J-, 5J, f of |, and of of |? 13f|. 2. The sum of l^- and is equal to how many times their difference? 9" 5 times. 3. What is (2* + 5 of _M)-=-lJLL? 5. 2 3 2 7 96 &A Y> S HIGHER ARITHMETIC. 4. Reduce lM and I X (100 - ^ + Ii) to 5i - 4J- 7 3 h 2i ; their simplest forms. 16 and 26 T 2 ^. 5. What is of 5J -fr of 3? ^V 6. What is |f X TT 5 2 X iff X If equal to? T V 7* 1 vJ nivAni? ,1 ~2~ X ~I~ X ~~3~ g 1 l-i 2-i 3-| 4-| ? -X--X - > (2-i)X(4-3f) 10. jj X 4i X 4^- 1 ^ what? 4 2^ 11. Add | of f of |, | X | of 1 J, and . |f 12. f of -y~ of what number, diminished by H , leaves -|-|? y 9 ^-. 13.* James's money equals f of Charles's money ; and of James's money -f- $33 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 far will each have traveled? 6ff hours. A, 5 Iff- miles; B, 57-2 2 T miles. 15. What number multiplied by of f of 3-fJ- will produce 2J? 2|f. 16. What, divided by If, gives 14f? 23|. 17. What, added to 14f, gives 29|f ? 15^. 18.* I spend f 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 -| of the first is equal to f of the second. 27 and 24. 20.* \ is what part of |? f. COMMON FRA CTIONS. 97 21. Divide $ of 3f by |f of 7 ; and ^ of | of 27 by f of T 3 T of 5. |f and 34|f 22. Multiply T 7 T of 2 by & of 19; and divide $ of f of 14} by T 8 T of f of 13f 7^ and &&. 23. (Mil + iffl + tttt) - f = what? 5f. 24.* A bequeathed -^ of his estate to his elder son; the rest to his younger, who received $525 less than his brother. What was the estate? $5250. 25. Find the sum, difference, and product of 3 and 2^ ; also, the quotient of their sum by the difference. Sum 5f|, diff. Iff, prod. 8}f ; quot. 3^ 7 . 26.* A cargo is worth 7 times the ship : what part of the cargo is T 5 g- of the ship and cargo? -f^. 27. By what must the sum of ^ 7 T > AW and ATT be multiplied to produce 1000? 1000. 28. Multiply the sum of all the divisors of 8128, includ- ing 1, by the number of its prime factors excluding 1, and divide by 14% 381. 29. 6|- is what part of 10 T 7 T ? Keduce to its simplest formf-! + f--li. 30. Multiply 1, 14f, ?i, I, g, and 6. 31. | of | of what number equals 9f|? 20. 32.* A 63-gallon cask is f full : 9| gallons being drawn off, how full will it be? |||. 33. If a person going 3f miles per hour, perform a journey in 14f hours, how long would he be, if he traveled 5^ miles per hour? lO^f hours. 34. A man buys 32f pounds of coffee, at 17f cents a pound: if he had got it 4| cents a pound cheaper, how many more pounds would he have received? UTTI P un ds. 35. Henry spent ^ of his money and then received $65 ; he then lost f of all his money, and had in hand $10 less than at first. How much had he at first? $33. H. A. 9. 98 HAY'S HIGHER ARITHMETIC. Topical Outline. COMMON FRACTIONS. 1. Definition. 2. Classes.. 8. Terms... 1. As to Kinds.... 2. As to Value.. 3. As to Form.... 1. Numerator.. 2. Denominator. 3. Similar. . 4. Dissimilar. {1. Common. 2. Decimal. ' 1. Proper. Improper. Mixed. Simple. Complex. 1 3. Compound. f 1. Pro j 2. Imi I 3. Mb fl. Sir \ 2. Coi 1 3. Co: 4. Principles. 5. Reduction... - 1. Cases 2. Principles. 3. Rules. 1. Lowest Terms. 2. Higher Terms. 3. Mixed Numbers to Improper Fractions. 4. Improper Fractions to Mixed Numbers. 5. Compound to Simple Fractions. 6. Common Denominator. 7. Least Common Denominator. 6. Practical Applications f fl. Definition. 1. Addition -j 2. Principles. 1 3. Rule. fl. Definition. 2. Subtraction -j 2. Principles. I 3. Rule. fl. Definition. 3. Multiplication -j 2. Principles. I 3. Rule. fl. Definition. 4. Division -j 2. Principles. [ 3. Rule. 5. Divisors, Multiples, etc. IX. DECIMAL FRACTIONS. 147. A Decimal Fraction is a fraction whose denom< inator is 10, or some product of 10, expressed by 1 with ciphers annexed. REMARK 1. A decimal fraction is also defined as a, fraction whose denominator is some power of 10. By the "power" 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 second power of 3. KEMARK 2. Since decimal fractions form only one of the classes (Art. 108) under the term fractions^ 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 units are called decimal places, and decimal fractions when so written, without a denominator expressed, are called decimals. 150. The decimal point, or separatrix, is a dot [ . ] placed at the left of decimals to distinguish them from integers. Thus, T V is written .1 T* 5 " " -01 " " -001 " " - 0001 From this, it is evident that, The denominator of any decimal is 1 with as many ciphers annexed as there are places in the decimal. 151. A pure decimal consists of decimal places only ; as, .325 (99) 100 JRAY'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 3 T 2 o^ = ff. 153. A complex 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 \.If, in any decimal, the point be moved to the right, the 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 hundredth*. (Art. 129, Prin. i.) PRINCIPLE II. If, in any decimal, the point be moved to the left, the decimal is divided by 10 as 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, thousandths, 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 value. ILLUSTRATION. .5 is equal to .500; for ^T == fooo- The reverse may be shown in the same way. (Art. 129, Prin. in.) NUMERATION AND NOTATION OF DECIMALS. 155. Since .6 = ^5 .06=^; and .006 = TT ft nr , any figure expresses tenths, hundredths, or thousandths, according as it is in the 1st, 2d, or 3d decimal place; hence, these places are named respectively the tenths', the hundredths', the DECIMAL FRACTIONS. 101 thousandths 9 place ; other places are named in the same way, as seen in the following table : TABLE OF DECIMAL ORDERS. " A A AH a 1 "? s a rf s *3 .2 5 & .3 Islf If 5 T3 2 S 'O o Ill A, PJ -H A s ^3 3 P g S 3 H W ^ S K -W 1st place .2 . . read 2 Tenths. 2d tt .08 . ti 8 Hundredths. 3d It .00 5 . . it 5 Thousandths. 4th It . 00 7 . . it 7 Ten-thousandths. 5th ii .00 03 . 1 1 3 Hundred-thousandths. 6th (i .00 001 . ft 1 Millionth. 7th .00 0009 (t 9 Ten-millionths. 8th " .00000004. " 4 Hundred-millionths. 9th " .000000006 " 6 Billionths. NOTE. The names of the decimal orders are derived from the names of the orders of whole numbers. The table may therefore be extended to trillionths, quadrillionths, etc. PROBLEM. Read the decimal .0325 SOLUTION. The numerator is 325; the denomination is ten- thousandths since there are four decimal places. It is read three hundred and twenty-five ten-thousandths. Rule. Read the number expressed in the decimal places as the numerator, give it the denomination expressed by the right- hand figure. EXAMPLES TO BE READ. 1. .9 3. .Of 4. .035 102 RAY'S HIGHER ARITHMETIC. 5. .7200 6. .5060 7. 1.008 8. 9. 10. 9.00^ 105.0f .0003 11. 00.100 12. 180.010 13. 14. 15. 16. 17. 18. 19. 20. 2030.0 40.68031 200.002 .0900001 61.010001 31.0200703 .000302501 .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. Rule. Write the numerator as a whole number; then place the decimal point so that the right-hand figure shall be of the same name as the 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-millionths. REDUCTION 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 DECIMALS. 157. Reduction of decimals is changing their form without altering their value. CASE I. 158. To reduce a decimal to a common fraction. PROBLEM. Reduce .24 to a common fraction. SOLUTION. .24 is equal to y 2 ^, which, reduced, is -fg. OPERATION. = ^ = A, Am. PROBLEM. Reduce .12^ to a common fraction. OPERATION. SOLUTION. Write 12 as a ^ 2 i numerator, and under it place . 1 2 J = =^-^= fifa = J, Ans. 100 as a denominator. Re- duce the complex fraction according to Art. 144. Rule. Write the decimal as a common fraction; then re- duce the fraction to its lowest terms. 104 KAY'S HIGHER ARITHMETIC. REMARK. 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 Ans. T 4 ^jy. 9. ll.Of 2. .15234375 Ans. --$ 10. .390625 3. 2.125 Ans. 2$. 11. .1944| 4. 19.01750 A via 1 Q 7 - and ' .184 = ^ftfo. 2.56 Now, -f Jf X T V 8 o 4 o = rWAV> that is, the product .184 of Imndredths by thousandths is hundredth- 1024 thousandths. This requires five places of 2048 decimals, or as many as are found in both 256 'factors. .47104, Am. Rule. 1. Multiply as in whole numbers. 2. Point off as many decimal places in the product as there are decimal places in the two factors. REMARKS. 1. If the product does not contain as many places as the factors, prefix ciphers till it does contain as many. 2. Ciphers to the right of the product are omitted after pointing. EXAMPLES FOR PRACTICE. 1. 1X.1 .1 2. 16 X .03^ .53 3. .OlX.li .0015 4. ,080 X 80. 6.4 5. 37.5 X82i -3093.75 6. 64.01 X. 32 20.4832 7. 48000 X 73. 3504000. 8. 64.66f X 18. 1164. 9. .561 x . 03^ .0172-H- 10. 738X120.4 88855.2 MULTIPLICATION OF DECIMALS. 109 11. .0001 X 1.006 .0001006 12. 34 units X- 193- 6.562 13. 27 tenths X.4 1.134 14. 43. 7004 X- 008 '.3496032 15. 21.0375 X 4.44 93.5 16. 9300.701 X 251. . 2334475.951 17. 430.0126X4000. 1720050.4 18. .059 X -059 X .059 .000205379 19. 42 units X 42 tenths. 176,4 20. 21 hundredths X 600. 13.5 21. 7100 X i of a millionth. .0008875 22. 26 millions X 26 millionths. 676. 23. 2700 hundredths X 60 tenths. 162. 24. 6.3029 X .03275 .206419975 25. 135.027 X 1.00327 135.46853829 163. Oughtred's Method for abbreviating multiplica- tion, may be used when the product of two decimals is required for a definite number of decimal places less than is found in both factors. PROBLEM. Multiply 3.8640372 by 1.2479603, retaining only seven decimal places in the product. EXPLANATION. It is evident that we need OPERATION. regard only that portion of each partial 1 i Q 7 9 product which affects the figures in and above * the seventh decimal place. Beginning with the highest figure of the 38640372 multiplier, we obtain the first partial product. 7728074 Taking the second figure of the multiplier, we 1545614 carry each figure of the partial product one 270482 place to the right, so that figures of the same 34776 order shall be in the same column. This product is carried out one place further than is required, so as to secure accuracy in the 4.8221650 seventh place, and we draw a perpendicular line to separate this portion. The product of .04 by the right-hand 110 KAY'S HIGHER ARITHMETIC. figure in the seventh place, would extend to the ninth place of decimals ; so we may reject the last 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 by .007, we have 7 X 3 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 + 5 = 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. Rule. 1. Multiply only such figures as shall produce one more than the required number of decimal places. 2. Begin with the highest order of the multiplier; under the right-hand figure of each partial product, place the right-hand figure of the succeeding one. In obtaining such right-hand figure, let that number be added which would be carried from multiplying the figure of the next lower order. 3. Add the partial products, and reject the right-hand figure. KEMARKS. 1. It will be found convenient to write the multiplier in a reverse order, with its units 1 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. EXAMPLES 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. 9.012X48.75, preserving one place. 439.3 5. 4.804136 X .010759, preserving six places. .051688 6. 814 T 5 T V X 26f f , preserving three places. 21813.475 7. 702.61 X 1.258-J&, preserving three places. 884.020 8. 849.93| 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. 324.16 DIVISION OF DECIMALS. 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. The dividend must contain as many deci- mal places as the divisor ; and when both have the same number, the quotient is an integer. 2. The quotient must contain as many decimal places as the number of those in the dividend exceeds the number of those in the divisor. PROBLEM. Divide .50312 by .19 OPERATION. .19). 50312(2. 64 8, Am. SOLUTION. The division is per- 38 formed as in integers. The quo- 123 tient is pointed according to 114 Principle 2. The quotient must gY have 5 2 = 3 places. 75 152 152 112 RAY'S HIGHER ARITHMETIC. PROOF. By expressing the decimals as common fractions we have : iWoVV rVo rf iWoVo X W =W* = 2.648 PROBLEM. Divide .36 by .008 SOLUTION. The dividend has a less num- OPERATION. her of decimal places than the divisor. .008 ) . 3 60 Annex one cipher, making the number 45, Ans. equal. The quotient is an integer. PROBLEM. Divide .002Jf by .06f OPERATION-. SOLUTION. Keducing the .002 J$ = . 002475 mixed decimals to equiva- . 6 f =.066 -lent pure decimals, we have . 6 6 ) . 00 2 4 7 5 (.037 5, Ans. .002475 and .066. Dividing, 198 we find one more decimal 495 place necessary to make the 462 division exact ; and, pointing 330 by Principle 2, we have .0375 330 Rule. Divide as in whole numbers, and point off as many decimal places in the quotient as those in the dividend exceed those in the 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 -T- 18. .056 4. 4096-^.0.32 128000. 5. 9.7-^-97000. .0001 6. . 9 -f-. 00075 1200. 7. 13-^78.12^ .1664 8. 12.9-^-8.256 1.5625 9. 81. 2096 -~ 1.28 63.445 10. 1-MOO. .01 11. 10.1 -M7. .59412- 12. . 001 -f- 100. .00001 DIVISION OF DECIMALS. 113 13. 12755 -f- 81632. .15625 14. 2401 -f- 21. 4375 112. 15. 21.13212 -J-.916 23.07 16. 36.72672^.5025 73.088 17. 2483.25 -f- 5. 15625 481.6 18. 142.0281 -T- 9.2376 15.375 19. .OSi-^.121 -66f = f. 20. .0001 -f-. 01 .0! 21. 95.3 -f-. 264 360.984848 + 22. 1000 + .001 1000000. 23. Ten +- 1 tenth. 100. 24. .000001 + .01 .0001 25. .00001 + 1000. .00000001 26. 16.275 + .41664 39.0625 27. 1 ten-millionth ~ 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 1.27395807, securing a quo- tient true to four places of decimals. OPERATION. SOLUTION. Since 1.2 jfr jJ ^ $ ) 8 4.2 7 6 ( 6 6.1 4 8 3 the divisor is greater 7643736 than 1 and less than 2, the quotient 783264 will contain six places, two of in- 764374 tegers and four of decimals. The .18890 highest denomination of the divisor, 12740 multiplied by the lowest denornina- 6150 tion of the quotient, would obtain a 5096 figure in the fourth place. We take 1054 one place more as in multiplication 1019 (Art. 163), and also cut off two figures ~~r~ of the divisor, since these can not affect ~ ^ 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 RAY'S HIGHER ARITHMETIC. Rule. Find the figure of the dividend that would result from multiplying a unit in the highest denomination of the divisor by a unit of the lowest denomination required in the quotient. Take one more figure of the dividend to secure accuracy. Cut off any figures of the divisor not needed for the abbreviated dividend. Divide as usual until the figures remaining in the dividend are all divided. At each subsequent division, drop a figure from the divisor, carrying the number necessary from the product of the figure omitted. Continue until the divisor is reduced to two figures. REMARK. In the quotient, 5 units of an omitted order may be taken as 1 unit of the next higher order. EXAMPLES FOR PRACTICE. 1. 1000 -f-. 98, preserving two places. 1020.41 2. 6215.75 -r- .99^, preserving three places. 6246.985 3. 28012 -f- .993, preserving two places. 28209.47 4. 52546. 35 -^.99f, preserving three places. 52678.045 5. 4840 -f- .9875, preserving two places. 4901.27 6. 2 -^ 1.4142136, preserving seven places. 1.4142135 7. 9.869604401 -f- 3. 14159265, preserving eight places. 3.14159265 Topical Outline. DECIMAL FRACTIONS. Definitions. Decimal Point. C Pure. Classes J Mixed. (_ Complex. Principles. Numeration, Rule. Notation, Rule. Reduction. / L Dec - to a Com - I II. Com. Fraction to a Dec. Addition, Rule. Subtraction, Rule. Multiplication, Rule. Abbreviated Multiplication. Division, Rule. Abbreviated J>i vision. X. CIRCULATING DECIMALS. 167. In reducing common fractions to decimals, the process, in some cases, does not terminate. This gives rise to Circulating Decimals. PRINCIPLE I. If any prime factors oilier than 2 and 5 are found in the denominator of a fraction in its lowest terms, the resulting decimal will be interminate. DEMONSTRATION. If the fraction is in its lowest terms, the numerator and denominator are prime to each other (131, Hem. 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 introduced. 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, ^ = 2X2X2X 3 2X2X2X5 - .009375 but > 6^0 = 23^1^5 =-116666+ 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 ffc to a decimal containing four places. But since the factor 3 is found in the denominator of ^, 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 reduction of a common fraction will, if the division be carried far enough, contain the same figure, or set of figures, repeated in the same order. (115) 116 RAY'S HIGHER 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. Interminate decimals, on this account, have received the name of Circulating or Recurring Decimals. 2. A Circulate or Circulating Decimal has one or more figures constantly repeated in the same order. 3. A Repetend 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 Repetend has one figure; as, .4 7. A Compound Repetend has two or more figures ; as, .59 8. A Perfect Repetend is one which contains as many decimal places as there are units in the denominator, less 1 ; thus, | = . 142857 9. Similar Repetends begin and end at the same deci- mal place; as, .427 and .536 10. Dissimilar Repetends begin or end at different deci- mal places; as, .205 and .312468 CIRCULATING DECIMALS. 117 11. Conterminous Repetends end at the same place; as, .50397 and .42618 12. Co-originous Repetends begin at the same place ; as, .5 and .124 169. Any terminate decimal may be considered a circu- late, its repetend 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 REMARKS. 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 iif the 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 .2941 == .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 ARITHMETIC. CASE I. 172. To reduce a pure circulate to a common fraction. PROBLEM. Change .53 to a common fraction. OPERATION. SOLUTION. Kemoving the 100 times the repetend 5 3.5 3 decimal point one place to the Once the repetend = .53 right, multiplies the repetend /. 99 times the repetend = 53. by 10 ; two places, by 100, and Once the repetend = f f , Ans. 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 which 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 .456 999 times the repetend = 456. .*. once the repetend = f f f , .4ns. PROBLEM. Change 25.6 to a common fraction. OPERATION. Carry the dot forward thus : 2 5.6 = 2 5.6 2 5 1000 times the repetend = 6 2 5.6 2 5 Once the repetend = .625 999 times the repetend = 625. .'. once the repetend = f f f Whence the 25.625 = 2 5 fff, Am. NOTE. From these solutions the following rule is derived. Rule. Write the repetend for the numerator, omitting the decimal point and the dots, and for the denominator write as many ffs as there are figures in the repetend, and reduce the fraction to its lowest terms. CIRCULATING DECIMALS. 119 CASE II. 173. To reduce a mixed circulate to a common fraction. PROBLEM. Change .821437 to a common fraction. OPERATION. 821 -- 7 Omitting the decimal point, we have : .821437 = p* = 821X999 + 437 _ 82 1 ( 1 OOP 1 ) + 437 1000X999 999000 8 2 1000 8 21 + 437 820616 102577 999000 999000 124875 Or, briefly, 8 2143 7 821 102577 , Ans. 999000 124875' Ans. PROBLEM. Change .048 to a common fraction. OPERATION. Omitting the decimal point, we have : 048 = -!!- 4 8 _4(10-1) 8 100 100^900 900 r 900 _36_ _8 .M___li 900 + 900~900~225' m ' Or, briefly, 48 4 44 11 900 ~900~225' NOTE. The following rule is derived from the preceding solu- tions. * . Rule. 1. For tlie numerator, subtract the part which precedes the repetend from the whole expression, both quantities being con- sidered as units. 2. For the denominator, write as many 9's as there are figures in the repetend, and annex as many ciphers as there are decimal figures before each repetend. 120 BAY'S HIGHER ARITHMETIC. Keduce to common fractions : 1. 2. 3. 4. 5. 6. .3 .05 .123 2.63 .31 .0216 48.1 i- TV rfs'- 9. 10. 11. 12. 13. 14. 1.001 .138 .2083 85.7142 .063492 .4476190 .09027 1 9 9 ' A- 85f eV T 4 oV ADDITION OF CIRCULATES. 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 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 figures of the sum also make a repetend. OPERATION. .2566666666 5.3 4 7 2 7 2 7 2 7 2 2 4.8 158158158 .9098000000 31.3295552097 repetend, the last six Rule. Make the repetends similar, if they be not so; add, and point off as in ordinary decimals, increasing the right-hand column by the amount, if any, which would be carried to it if the circulates were continued; then make a repetend in the sum, similar to those above. REMARK. In finding the amount to be carried to the right-hand column, it may be necessary, sometimes, to use the two succeeding figures in each repetend. SUBTRACTION OF CIRCULATES. 121 EXAMPLES FOR 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 1.03, .257, 5.04, 28.0445245 34.37 5. Add .6, .138, .05, .0972, .0416 1. 6. Add 9.21107, .65, 5.004, 3.5622 18.43 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 CIECULATES. 175. 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 12.9021 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 358650559 larger than the one above it (2) ; therefore, carry 1 to the right-hand figure of the subtrahend. Rule. Make the repetends similar, if they be not so ; subtract and point off as in ordinary decimals, carrying 1, however, to the right-hand figure of the subtrahend, if on continuing the circulates it be found necessary ; then make a repetend in the remainder, similar to those above. REMARK. 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 HAY'S HIGHER ARITHMETIC. EXAMPLES FOR PRACTICE. 1. Subtract .0074 from .26 .259 2. Subtract 9.09 from 15.35465 6.25 3. Subtract 4.51 from 18.23673 13.72 4. Subtract 37.0128 from 100.73 63.71 5. Subtract 8.27 from 10.0563 1.7836290 6. Subtract 190.476 from 199.6428571 9.16 7. Subtract 13.637 from 104.1 90.503776 MULTIPLICATION OF CIKCULATES. 176. Multiplication of Circulates 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 arising from the multiplication of the 1501777 figures that do not appear. The repetend 2628lili of the multiplier being equal to J, ^ of 37544444 the multiplicand is 125148, whose figures 125148 are set down under those of the multipli- , , i> i. it: ^ j 6.5452481 Ans. cand from which they were obtained. Point the several products, carry them forward until their repetends are similar, and add for answer. Rule. 1. If the multiplier contain a repetend, change it to a common fraction. 2. Then multiply as in multiplication of decimals, and add to the right-lxind 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 CIRCULATES. 123 EXAMPLES FOR PRACTICE. 1. 4.735 X 7.349 34.800H3 2. .07067 X .9432 .066665 3. 714.32X3.456 2469.173814 4. 16.204 X 32.75 530.810446 5. 19.072 X .2083 3.97348 6. 3.7543 X 4.7157 17.7045082 7. 1.256784 X 6.42081 8.069583206 DIVISION OF CIKCULATES. 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. Rule. Change tJie terms to common fractions; then divide as in division of fractions, and reduce the result to a repetend. REMARK. This is the easiest method of solving problems in di- vision 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 + .1 6.81 2. 51.491 -M 7. 3.028 3. 681.5598879-;- 94. 7.2506371 4. 90.5203749-^6.754 13.401 5. 11. 068735402-^.245 45.13 6. 9.5330663997 -f- 6.217 1.53 7. 3.500691358024-^7.684 .45 124 RAY'S HIGHER ARITHMETIC. Topical Outline. CIRCULATING DECIMALS. 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 Repetends. 11. Co-originous Repetends. 3. Reduction f Case I. lease II. f Definition. 4. Addition -j Rule. V Applications. C Definition. 5. Subtraction -j Rule. I Applications. f Definition. 6. Multiplication -j Rule. I Applications. c Definition. 7. Division -j Rule. (. Applications. XL COMPOUND DENOMINATE 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 Compound Denominate Number is one expression of a quantity by different denominations 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 : Value, Weight, Extension, and Time. MEASUEES 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 mitt, which are T Jo and Y^nr f a dollar, derive their names from the Latin centum and mille, meaning a hun- dred and a thousand ; the dime, which is -^ of a dollar, is from the French word disme, meaning ten. REMARKS. 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 trade dollar was minted for Asiatic commerce 2. The coins of the United States are classed a& bronze, nickel, 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. WEIGHT. BRONZE. One cent. 1 cent. 95 parts copper, 5 parts tin & zinc. 48 grains Troy. NICKEL. 3-cent piece. 3 cents. 75 parts copper, 25 parts nickel. 30 grains Troy. 5-cent piece. 5 cents. 75 " " 25 " 77.16 " SILVER. Dime. 10 cents. 90 parts silver, 10 parts copper. 2.5 grams. Quarter dollar. 25 cents. 90 " " 10 " 6.25 " Half doilar. 50 cents. 90 " " 10 " 12.5 Dollar. 100 cents. 90 " " 10 " 412.5 grains Troy. GOLD. Dollar. 100 cents. 90 parts gold, 10 parts copper. 25.8 grains Troy. Quarter eagle. iy z dollars. 90 " " 10 " 64.5 Three dollar. 3 dollars. 90 " " 10 " 77.4 Half eagle. 5 dollars. 90 " " 10 " 129 Eagle. 10 dollars. 90 " " 10 " 258 Double eagle. 20 dollars. 90 " " 10 " 516 3. A deviation in weight of J 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 1| 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. I 181. In reading U. S. Money, name the dollars and all higher denominations together as dollars, the dimes and cents as cents, and the next figure, if there be 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 RAY'S HIGHER ARITHMETIC. English or Sterling Money. 182. English or Sterling Money is the currency of the British Empire. The pound sterling (worth $4.8665 in TL 8. money) is the unit, aud is represented by the sovereign and the i bank-note. TABLE. 4 farthings, marked qr., make 1 penny, marked d. 12 pence " 1 shilling, " s. 20 shillings " 1 pound, ^ " . EQUIVALENT TABLE. < s. d. qr. 1 = 20 = 240 = 960. 1= 12= 48. 1 = 4. REMARKS. 1. The abbreviations, <, s., d., q., are the initials of the Latin words libra, solidarius, denarius, quadrans, signifying, respectively, pound, shilling, penny, and quarter. 2. The coins are gold, silver, and copper. The gold coins are the sovereign ( = . 3- 3- 9- gr. 1 = 12 = 96 = 288 5760. 1 = 8 = 24 = 480. 1-5= 3 = 60. 1 r= 20. REMARK. 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 weighing all ordinary articles. TABLE. 16 ounces, marked oz., make 1 pound, marked Ib. 25 Ib. " i quarter, ". qr. 4 qr. " 1 hundred-weight, " cwt. 20 cwt. " 1 ton, " T. EQUIVALENT TABLE. T. cwt. qr. Ib. oz. I = 20 =^ 80 = 2000 = 32000. 1 = 4 = 100 = 1600. 1 = 25 = 400. 1 = 16. 132 RA Y'S HIGHER ARITHMETIC. KEMARKS. 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 iveight is the weight of the goods alone. 5. The word avoirdupois is from the French avoirs, du, pois, 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 OF 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. Apothecaries'. 1 lb. 1 T V lb. = IfV* . 1 oz. = lif oz. = Hf g. 1 lb. = 1 ib. 1 oz. = 1 . 1 gr. 1 gr. 1 pwt. = I z- 1 pwt. = \\ 3. MEASURES OF EXTENSION. 133 KEMAKK. In addition to the foregoing, the following, called Diamond Weight, 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 the 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. MEASUKES 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 line has only one dimension, length. 3. A surface has two dimensions, length and breadth. 4. A solid or volume has three dimensions, length, breadth, and thickness. 192. Measures of Extension embrace : f Long Measure. 1. Linear Measure. J Chain Measure. Mariners' Measure. I Cloth Measure. 2. Superficial Measure. / ^ uare Mea * ure ' ( burveyors Measure. 3. Solid Measure. ( Liquid Measure. 4. Measures of Capacity. < Apothecaries 1 Measure. 5. Angular Measure. (Dry Measure. 134 HA 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 16J ft. " 1 rod, " rd. 320 rd. " 1 mile, " mi. EQUIVALENT TABLE. mi. rd. yd. ft. in. 1 = 320 = 1760 = 5280 = 63360. 1 = 5i = 161 = 198. l-.-.:= 3 = 36. 1 = 12. REMARKS. 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 y^V o f an i ncn 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. REMARKS. 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- tances. It is divided into feet, tenths, and hundredths, and, by means of a vernier, may be read to thousandths. 195. Mariners' Measure 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. REMARKS. 1. A nautical mile is one minute of longitude, meas- ured on the equator at the level of the sea. It is equal to 1.152J 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 KAY'S HIGHER 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, quarters, eighths, sixteenths, etc., in place of feet and inches. REMARKS. 1. There was formerly a recognized table for Cloth Measure, but it is now obsolete. The denominations were as follows : 2} 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 surface 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 unit is an assumed unit of measure *for surfaces. Usually the square, whose side is the linear unit, is the unit of measure ; as, the square inch, square foot, square yard. 5. A Rectangle 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 multiplied by its breadth. MEASURES 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 be 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. At* 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. 30^ sq. yd. " 1 square rod, " sq. rd. 160 sq. rd. " 1 acre, " A. EQUIVALENT TABLE. A. sq. rd. sq. yd. sq. ft. sq. in. 1 = 160 = 4840 = 43560 = 6272640. 1 = 30J = 2721 = 39204. 1 = 9 = ' 1296. 1 = 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 B. 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 sq. rd. " 1 square chain, sq. ch. 10 sq. ch. " 1 acre, A. 640 A. u 1 square mile, sq. mi. 36 sq. mi. (6 miles square) " 1 township, J?~p~~ H. A. 12. /T ^ OF THE / IIKIIWCTDQITY 138 RAY'S HIGHER ARITHMETIC. EQUIVALENT TABLE. Tp. sq. mi. A. sq. ch. sq. rd. sq. li. 1 = 36 =-23040 = 230400 = 3686400 = 2304000000. 1 = 640 = 6400 = 102400 = 64000000. 1 = 10 = 160 = 100000. 1 = 16 = 10000. 1 = 625. Solid Measure. 199. A Solid has length, breadth, and thickness. Solid Measure is used in estimating the contents or volume of solids. A Cube is a solid, bounded by six equal squares, called faces. Its length, breadth, and thickness are all equal. REMARK. The size or name of any cube, like that of a square, depends upon its side, as cubic inch, cubic foot, cubic yard. EXPLANATION. If each side of a cube is 1 inch long, it is called a cubic inch ; if each side is 3 feet (1 yard) long, as repre- sented in the figure, it is a cubic or solid yard. When the base of a cube is 1 square yard, it contains 3X3=9 square feet ; and 1 foot high on this base, contains 9 solid feet ; 2 feet high contains 9 X 2 = 18 solid feet ; 3 feet high contains 9X3 =: 27 solid feet. Also it may be shown that 1 solid or cubic foot contains 12 X 12 X 12 1728 solid or cubic inches. The unit by which all solids are measured is a cube, whose side is a linear inch, foot, etc., and their size or solidity will be the number of times they contain this unit. REMARK. The simplest solid is the rectangular solid, which is bounded by six rectangles, called its faces, each opposite pair being equal, and perpendicular to the other four; 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 cube. TABLE. 1728 cubic inches (cu. in.) make 1 cubic foot, cu. ft. 27 cu. ft. "1 cubic yard, cu. yd. EQUIVALENT TABLE. cu. yd. cu. ft. cu. in. 1 = 27 = 46656. 1 == 1728. REMARKS. 1. A perch of stone is a mass 16| ft. long, 1^ ft. wide, and 1 ft. high, and contains 24J cu. ft. 2. Earth, rock-excavations, and embankments are estimated by the cubic yard. 3. Round 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 foot, 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 RA Y J 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 Ib. 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. NOTE. Sometimes the barrel is estimated at 31 J 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 TV, make 1 fluid drachm, marked f%. 8 f 3 " 1 fluid ounce, " f. 16 f * 1 pint, " O. 8 O " 1 gallon, " cong. EQUIVALENT TABLE. cong. O. f. 3. nt. 1 = 8 = 128 = 1024 = 61440. 1 = 16 = 128 = 7680. 1 = 8 = 480. MEASURES OF EXTENSION. 141 NOTES. 1. Cong, is an abbreviation for congiarium, the Latin for gallon ; O. is the initial of octans, 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 is 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 Ib. 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. EEMARKS. 1. 4 qt. or J peck = 1 dry gal. = 268 8 cu. in. nearly. 2. The quarter is still used in England for measuring wheat, 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 RAY'S HIGHER ARITHMETIC. TABLE. ARTICLES. LB. EXCEPTIONS. Beans. 60 Me., 64; N. Y., 62. Coal. 80 f Ohio, 70 of cannel ; Ind., 70 mined out of \ the state ; Ky., 76 of anthracite. Corn (Indian). 56 N. Y., 58 ; CaL, 52 ; Arizona, 54. Flax Seed. 56 N. Y. andN. J., 55; Kan., 54. Oats. 32 | Md., 26 ; Me., N. H., N. J., Pa., 30 ; Neb., 34 ; \ Montana, 35 ; Oregon and Wash., 36. Potatoes (Irish). 60 Ohio, 58 ; Wash., 50. Eye. 56 La., 32 ; Cal., 54. IMass., 70 ; Pa., coarse, 85 ; ground, 70 ; fine. Salt. 50 62; Ky. and 111., fine, 55; Mich., 56; Col. and Dak., 80. Wheat. 60 COMPARATIVE TABLE OF MEASURES. cu. in. gal. cu. in. qt. Liquid Measure, 231 57f Dry Measure (J pk.), 2684 67^ cu.in.pt. cu. in.gi. 28 7^ 33| 8f Angular or Circular Measure. 204. 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 angle itself. Angular Measure 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 EXPLANATIONS. The bounding lino ADBEA is a circumfertnce. Every point of this line is at the same distance from the point C, which is called the center. The circle is the area included within the circumference. Any straight line drawn from the center to the circumference is called a radius; thus, CD and CB are radii. Any part of the circumference, as AEB or AD, is an arc. 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 degrees ; 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 angle, and is measured by the fourth part of a circumference, 90, called a quadrant. TABLE. 60 seconds, marked ", make 1 minute, marked, '. 60' " 1 degree, " . 360 " 1 circumference, " c. EQUIVALENT TABLE. c. o / // 1 = 360 = = 21600 = 1296000. 1 = 60 = 3600. 1 = 60. NOTE. The twelfth part of a circumference, or 30, is called a sign. 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 RA Y>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 min. 60 min. " 1 hour, " hr. 24 hr. " 1 day, " da, 7 da. " 1 week, " wk. 4 wk. " 1 month, " mori. 12 calendar mon. " 1 year, " yr. 365 da. " 1 common year. 366 da. " 1 leap year. 100 yr. " 1 century, marked cen. NOTE. 1 Solar year = 365 da. 5 hr. 48 min. 46.05 sec. = 365 da., nearly. EQUIVALENT TABLE. yr. mo. wk. da. hr. min. sec. 365 = 876 == 525 ^00 == 31536000. . 12 = 52 = " 366 = 8784 527040 == 31622400. 1 = 7 = 168 10080 = 604800. 1 = 24 = 1440 = 86400. 1 = 60 = 3600. 1 = 60. MEASURE 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 365|- days ; instead of taking account of the \ of a day every year, the common or civil year was reckoned 365 days, 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 Romans, 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 365J 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. Rule 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 begin January 1st, instead of March 25th. Eussia, and all other countries of the Greek Church, still use the Julian calendar; consequently their dates (Old /Style) are now 12 days later than 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 Septeui (7), Octo (8), Novem (9), Decem (10). H. A. 13. 146 RAY 1 8 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 established meridian, is found by subtracting the less 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 +- 24 = 15.) As 15 equal 900', and 1 hour equals 60 minutes of time, therefore, the sun in 1 minute of time passes over 15' of a degree. (900' -j- 60 = 15'.) As 15' equal 900", and 1 minute of time equals 60 seconds of time, therefore, in 1 second of time the sum passes over 15" of a degree. (900" -f- 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, conversel} 7 , 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, 12 16 32 a quarto or 4to, an octavo or 8vo, a duodecimo or 12rno, a 16mo, a 32mo, makes 4 pages. 3 *. 1 1 52 pt. tion Descending. 18 bu. = 1 1 5 2 pt. 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 2o6 in., which is 3 ) 1 9 2 ft. 19| ft., and since 3 ft. = 1 yd., 19| ft will ~~^T y d. be as many yd. as 3 ft. is contained times 236 in 6 ^ yd in 19f ft., which is 6f yd. Or, since 1 yd. = 36 in., divide 236 in. by 36 in., which gives 6f yd., as before. This is sometimes called Reduction 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 sV = W yd. 6f 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 size, 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 jg operation, and cancel. |. gal. = 3 pt. PROBLEM. Reduce 5-f- gr. to . OPERATION. SOLUTION. Al- 2 though this is Re- _A^ y _JL ylv =- Z duction Ascending, ^ T S r * j 2038 84 we use Prin. 1, in . multiplying by the successive unit values, -$, J, and J. ^ T > r< "STo- PROBLEM. Reduce 9.375 acres to square rods. OPERATION. 9.375 160 562500 9375 1500.000 sq. rd. 9.3 7 5 A. 1 5 sq. rd. PROBLEM. Reduce 2000 seconds to hours. OPERATION. 2000X^X;^ = ~ =| hr. 6 ft b ft oo 9 2000 sec. = hr. 156 RA Y> S HIGHER ARITHMETIC. PROBLEM. Keduce 1238.73 hektograms to grams. OPERATION. 123 8.7 3X100 = 123873 grams. PROBLEM. How many yards in 880 meters? RATION. -= 9 6 2.3 7 7 -f yd. OPERATION. 3 9.3 7 in. X 8 8 12X3 REMARK. 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 may 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. Rule. 1. Multiply the higJiest denomination given, by that number of the next lower which makes a unit of the higher. 2. Add to the product the number, if any, of the lower denomination. 3. Proceed in like manner with the result thus obtained, till the whole is reduced to the required denomination. 220. For reducing from lower to higher denomina- tions. Rule. 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 the whole is reduced to the 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 the right or to the left. REDUCTION OF COMPOUND NUMBERS. 157 EXAMPLES FOR PRACTICE. 1. How many square rods in a rectangular field 18.22 chains long by 4.76 ch. wide? 1387.6352 sq. rd. 2. Reduce 16.02 chains to miles. .20025 mi. 3. How many bushels of wheat would it take to. fill 750 hektoliters? 2128J bu. 4. Eeduce 35.781 sq. yd. to sq. in. 46372.176 sq. in. 5. Reduce 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 2f ft. average thickness? 32ff 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 $ gr. to . -^ . 10. Reduce 18f 3 to oz. av. 2-f- 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 \\ ft. deep? 75ff gal. 13. How many bushels in a bin 9.3 ft. long by 3-f ft, wide and 2^ ft. deep? 61 bu., nearly. 14. How many sters in 75 cords of wood? 271.837+ s. 15. Reduce 2J years to seconds. 70956000 sec. 16. Forty-nine hours is what part of a week? -^ wk. 17. Reduce 90.12 kiloliters to liters. 90120 1. 18. Reduce 25" to the decimal of a degree. .00694 19. Reduce 192 sq. in. to sq. yd. 2T S( l- J^- 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 Ib. av. to Ib. troy. 14 T 7 T 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 4^ B to ft. fa ft. 30. Reduce 7-J- oz. av. to cwt. -^-g- cwt. 31. Reduce 99 yd. to miles. T |^ mi. 32. How many acres in a rectangle 24 rd. long by 16.02 rd. wide? 2.4530625 acres. 33. How many cubic yards in a box 6^ ft. long by 2^ ft. wide and 3 ft. high? l-f~| cu. yd. 34. Reduce 169 ars to square meters. 16900 m 2 . 35. Reduce 2 f to r^. 1200 n^. 36. If a piece of gold is y pure, how many carats fine is it? 20f 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., 1^ dm. by 5 m., 3 cm.? 30.9345 m 2 . 39. How many cords of wood in a pile 120 ft. long, 6 ft. wide, and 8f ft. high ? 53 T Vg- C. 40. How many sq. ft. in the four sides of a room 21|- ft. long, 16i ft. wide, and 13 ft. high? 988 sq. ft. 41. What will be the cost of 27 T. 18 cwt. 3 qr, 15 Ib. 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 REDUCTION OF COMPOUND NUMBERS. 159 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. If 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 lb. 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 from 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 RAY'S HIGHER ARITHMETIC. ADDITION OF COMPOUND NUMBERS. 221. Compound Numbers may be added, subtracted, multiplied, and divided. The principles upon 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. 2J pk.; 1 pk. 1^ pt.; 5 qt. 1 pt.; 2 bu. 1 qt; and .125 pt. SOLUTION. Reduce the frac- OPERATION. tion in each number to lower bu. pk. qt. pt. denominations, and write units 3 2 2 = 3 bu. 2J pk. of the same kind in the same 1 1 1 = 1 pk. 1 \ pt. column. The right-hand column, 5 1 = 5 qt. 1 pt. when added, gives 3^- pt. = 1 qt. 20 1 f = 2 bu. 1 \ qt. l^j pt.; write the l/^ and add J .125 pt. the 1 qt. with the next column, 601 1-^= Ans. 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. TI in.; 13 ft. 5.78 in.; 4 rd. 11 ft. 6 in.; 1 rd. lOf ft; 6 rd. 14 ft. 6f in. OPERATION. SOLUTION. The numbers are prepared, rd. ft. in. written, and added, as in the last ex- 2 9 7.25 ample; the answer is 16 rd. 9| ft. 9.655 13 5.78 in. The J foot is then reduced to 6 4 11 6 inches, and added to the 9.655 in., making 1 10 8 15.655 in. = l ft. 3.655 in. Write the 6 14 6.625 3.655 in., and carry the 1 ft., which gives 16 9J 9.655 16 rd. 10 ft. 3.655 in. for the final answer. but ft. = 6. 16 10 3.655 ADDITION OF COMPOUND NUMBERS. 161 Rule. 1. Write the members to be added, placing units of the same denomination in the same column. 2. Begin with the 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 quotient to the next column. 4. Proceed in the same manner with all the columns to the last, under which write its entire sum. REMARK. The proof of each fundamental operation in Com- pound Numbers is the same as in Simple Numbers. EXAMPLES FOR PRACTICE. 1. Add % mi.; 146 J- rd.; 10 mi. 14 rd. 7 ft. 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.; If 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.; f qr.; f na, 12yd. 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 J 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. 74J- sq. rd.; 20 A.; 12 A. 113 sq. rd.; 5 A. 13.33-* sq. rd. 205 A. 139.18sq. 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.; ^ cu. yd. 75 cu. yd. 4 cu. ft. 1279.252 cu. in. 9. Add f C.; f cu. ft.; 1000 cu. in. 107 cu. ft. 1072 cu. in. 10. Add 2 lb. troy, 6| oz.; If lb.; 12.68 pwt.; 11 oz. 13 pwt. 19i gr.; | lb. -if oz.; f pwt. 5 lb. troy, 9 oz. 9 pwt. 2.85-J- gr. 11. Add8g 14.6 gr.; 4.18 g; 7 T \ 3 ; 23 29 18 gr.; Ig 12 gr.; i 9. 1 ft. 2 g 4 3 1 9- 12. Add T 5 e 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. 5Jf oz. 14. Add 6 gal. 3 qt.; 2 gal. 1 qt. .83 pt. ; 1 gal. 2 qt. | pt.; | gal.; g qt.; $ pt. 11 gal. 2 qt. .llfj- pt. 15. Add 4 gal. .75 pt.; 10 gal. 3 qt. 1 pt.; 8 gal. f pt.; 5.64 gal.; 2.3 qt.; 1.27 pt.; ^ pt- 29 gal. 2 qt. ,05f pt. 16. Add 1 bu. % pk.; T 6 T bu.; 3 pk. 5 qt. l\ pt.; 9 bu. 3.28 pk.; 7 qt. 1.16 pt.; -f\ pk. 12 bu. 3 pk. .46^1 pt. 17. Add | bu.; | pk.; f qt.; f pt. 2 pk. ff pt. 18. Add 6 f5 2 f g 25 nt ; 2J f; 7 ft 42 nt ; 1 f 2| f s; 3 fa 6fs51 nt. 14 fg 7 % 38 nt. 19. Add -J- wk.; \ da.; 4^ lir.; \ min. ; -|- sec. 4 da. 30 min. 301 sec . 20. Add 3.26 yr. (365 da. each); 118 da. 5 hr. 42 min. 37|- sec.; 63.4 da.; 7| hr.; 1 yr. 62 da. 19 hr. 24f min.; T ^ 3 2- da. 4 yr. 340 da, 1 hr. 14 min. 55|- sec. 21. Add 27 14' 55.24"; 9 18 J"; 1 15f ; 116 44' 23.8" 154 14' 57.29" 22. Add $i; | 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. 4J m. 24. Add 21 6s. 3Jd.; 5 17|s.; 9.085; 16s. 7Jd.; 37 10s. 8.15d. 25. Add | A.; f sq. rd.; i sq. ft. 107 sq. rd. 12 sq. yd. 6 sq. ft. 344- sq. in. SUBTRACTION OF COMPOUND NUMBERS. 163 SUBTKACTION OF COMPOUND NUMBEKS. 222. Subtraction of Compound Numbers is the process of finding the difference 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 decimal, 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 Ans. 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- 201 hend by the same amount, we shall have to 1 30 2 make an inconvenient reduction of the first Ans. 1J sq.ft. remainder as itself a minuend. Hence, we make the convenient addition of J sq. yd., and, subtracting 2 sq. ft. from 3J sq. ft., we have 1 J- 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 feet, the answer is, simply, 1} sq. ft. = 1 sq. ft. 36 sq. in. Rule. 1. Place the subtrahend under the minuend, so that numbers of the same denomination stand in the same column. Begin at the lowest denomination, and, subtracting the parts successively from right to left, write the remainders beneath. 2. If any number in the subtrahend be greater than that of the same denomination in the minuend, increase the upper by a unit, or such other quantity of the next higher denomina- 164 RAY'S HIGHER ARITHMETIC. tion as will render the subtraction possible, and give an equal increase to the next higher term of the subtrahend. KEMARK. 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 f 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 f 3 from T 6 T 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 T \ 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 fg 38 nt from 4 f g 2 fg. 2"f 5fe 22 HL. 13. Subtract 275 da. 9 hr. 12 min. 59 sec. from 2.4816 yr. (allowing 365^ days to the year.) 1 yr. 265 da. 18 hr. 29 min. 21.16 sec. MULTIPLICATION OF COMPO UND NUMBERS. 165 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 31st, 1814, and April 1st, 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. 4| m. 19. Subtract 5 dollars 43 ct. 2^ m; from 12 dollars 6 ct. 8J- m. $6.635f 20. Subtract 9 18s. G^d. from 20. 10 Is. 5|d. 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 NUMBEKS. 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. OPERATION. SOLUTION. Ten times 8.17 sec. = 81.7 sec. = 1 min. 21.7 sec. Write 21,7 da " ^^ ^ 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. 166 RA F># HIGHER ARITHMETIC. PROBLEM. Multiply 12 A. 148 sq. rd. 28f sq. yd. by 84. SOLUTION. Since 84 = 7 X 12, multiply OPERATION. by one of these factors, and this product A. sq. rd. sq. yd. by the other ; the last product is the one required. The same result can be obtained __ 7 by multiplying by 84 at once ; performing 90 82 1 7 \ the work separately, at one side, and trans- _ 1 2 ferring the results. 1086 30 24 Rule. 1. Write the multiplier 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; ivrite the remainder under the denomination multiplied, and carry the quotient to the product of the next higher denomination. 3. Proceed in like manner with all the denominations, writing 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. 8^ 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. 5| gr. by 174. 113 Ib. 3 oz. 5 pwt. 16| gr. 7. Multiply 2 3 1 B 13 gr. by 20. 6 3 3. 8. Multiply 16 cwt. 1 qr. 7.88 Ib. by 11. 8 T. 19 cwt. 2 qr. 11.68 Ib. 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 fg 48 r^ by 12. 5 f 5 % 36 n^ 12. Multiply 18 da. 9 hr. 42 min. 29.3 sec. by 16 T 7 T . 306 da. 4 hr. 25 min. 2 sec., nearly. 13. Multiply 215 16s. 2d. by 75. 16185 14s. |d. 14. Multiply 10 28' 42f ' by 2.754 28 51' 27.765" DIVISION OF COMPOUND NUMBEKS. 224. Division of Compound Numbers is the process of dividing when the dividend is a Compound Number. 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 same 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 Ib. 14f oz. of sugar equally among 4 men. SOLUTION. 4 into 5 cwt. gives a OPERATION. quotient 1 cwt., with a remainder 1 cwt., cwt. qr. Ib. 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 24 1 5 { J =? 75 Ib., to be carried to 24 Ib., = 99 Ib.; 4 into 99 Ib. gives 24 Ib., with 3 Ib., = 48 oz., to be carried to 14f oz., making 62f oz.; 4 into 62f oz. gives 15JJ oz., and the operation is complete. 168 RAY'S HIGHER ARITHMETIC. PROBLEM. If $42 purchase 67 bu. 2 pk. 5 qt. If pt. of meal, how much will $1 purchase? OPERATION. 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 1 3 3 last quotient will be the one required. 1231 - 2 3 - Rule. 1. Write the quantity to be divided in the order of its denominations } beginning with the highest; place the divisor on the left. 2. Begin with the highest denomination, divide each number separately, and write the quotient beneath. 3. If a remainder occur after any division, reduce it to the next lower denomination, and, before dividing, add to it the number of its denomination. 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^-f- sq. in. 5. Divide 6 sq. mi. 35 sq. rd. by 221 17Q 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 g 7 3 18 gr. by 12. 2 3 1 9 16J gr. 8. Divide 600 T. 7 cwt. 86 Ib. by 29.06 20 T. 13 cwt. 20 Ib. 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. 38 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. Longitude 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 the 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 . , . OPERATION. sponds to 15 of longitude ; every minute -r/ -c i -X j hr. mm. sec. of time to 15 of longitude ; every second of time to 15" of longitude (Art. 207). . r Hence, multiplying the hours in the 64 36' 30" difference 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? SOLUTION. 15 into 81 gives 5 (marked hr.), and 6 OPERATION. to be carried. Instead of 15 ) 81 39' 22" multiplying 6 by 60, add- 5 hr. 26 min. 3 7 T V sec. ing the 39', and then divid- ing, proceed thus : 15 into 6 is the same as 15 into 6 X W = = 24', and as 15 into 39 X gives 2 X for a quotient and 9' 15 remainder, the whole quotient is 26 7 (marked min.)j and remainder 9^ 9X60", which, divided by 15, gives == 36", which 15 with 1 T ^, obtained by dividing 22" by 15, gives 37 T 7 / X (marked sec.). The ordinary mode of dividing will give the same result, and may be used if preferred. H. A. 15- 170 ' S HIGHER ARITHMETIC. 226. From these solutions we may obtain the following rules : CASE I. Rule. Multiply the difference of time by 15, according to the rule for Multiplication of Compound Numbers, and mark the product " instead of hr. min. sec. CASE II. Rule. Divide the difference of longitude % 15, according to the rule for Division of Compound Numbers, and mark the quotient hr. min. sec., instead of ''. 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 OF LONGITUDES. PLACE. LONGITUDE. PLACE. LONGITUDE. Portland, Me., .... o t n 70 15 18 W. Des Moines Iowa o r n 93 37 16 W. Boston, Mass.i .... 71 3 50 " Omaha, Neb., 95 56 14 " New Haven, Conn., . New York City, . . . 72 55 45 " 74 24 Austin, Tex., Denver, Col 97 44 12 " 104 59 33 " Philadelphia, Pa., . Baltimore, Md., . . . Washington, D. C., . Richmond, Va., . . . Charleston, S. C , . . Pittsburgh, Pa., ... Savannah, Ga., . . . Detroit, Mich., . . . Cincinnati, O., ... Louisville Ky 75 9 3 76 36 59 77 36 77 2G 4 79 55 49 80 2 81 5 26 83 3 84 29 45 85 25 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 111 53 47 " 122 27 49 " 135 19 42 " 5 42 " 22 " 43 20 " 52 43 " 157 52 " 000 2 20 E Indianapolis Ind 86 6 Rome Italv. ... 12 28 " Nashville, Tenn., . . Chicago 111 86 49 87 35 Berlin, German Em p., . Vienna, Austria 13 23 " 16 20 " Mobile, Ala., . . . Madison, Wis., . . . New Orleans, La., . . St Lonis Mo 88 2 28 89 24 3 90 3 28 90 12 14 Constantinople,Turkey, St. Petersburg, Russia, . Bombay, India, Pekin China 28 59 " 30 16 " 72 48 " 116 9 6 " Minneapolis, Minn., 93 14 8 Sydney, Australia, . . . 151 11 " 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, 111., and Philadelphia being 58 min. 1 T 2 ^ 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 Halifax to St. Louis ; on arriving, his watch shows 9 A. M. Halifax time. The time in St. Louis being 13 min. 32 T ^ 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-j-f 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 RAY'S HIGHER ARITHMETIC. ALIQUOT PAETS. 227c An 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. 15^ sq. yd. of land, at $16 per acre. SOLUTION. Multiply $16, the price of 1 A., by 28 ; the product, $448, is the price of 28 A. 145 sq. rd. is made up of 120 sq. rd., 20 sq. rd., and 5 sq. rd. 120 sq. 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. i of 120 sq. rd., and cost J as much. 5 sq. rd. J of 20 sq. rd., and cost \ as much. 15| sq. yd. = | of 1 sq, rd., or -f$ of 5 sq. rd., and cost -fa 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. 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. mi. rd. vd. OPERATION. $16 28 128 32 120 sq. rd. == f 2 sq. rd. = 5 sq. rd. = | 1 5 i sq. yd. = T V $448 1 2 2 .50 .05 $462.55 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. 20 6 da. =8X9 hr. 1 5 m in. = J- of 1 " 3 "= of 1 5 min. 45 sec. = J of 3 min. 27 188 1 220 225 2J 245 H 49 i 1 2 1 A 249 80 ALIQUOT PARTS. 173 NOTE. In all questions in aliquot parts, one of the numbers indicates a rate, and the other is a Compound Number whose value at this rate is to be found. Rule for Aliquot Parts. Multiply the number indicating tJie rate by the number of that denomination for whose unit the rate is given, and separate the numbers of the other denomina- tions into parts whose values can be obtained directly by a simple division or multiplication of one of the preceding values. Add these different values; the result will be the entire value required. NOTES. 1. Sometimes one of the values may be obtained by adding or subtracting two preceding 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. 5 - 10 = - ALIQUOT PARTS OF 100. 12 = 20 = 25 = 331 = 50 = REMARK. The following multiples of aliquot parts of 100, are often used: 18} = &, 37*= f, 40 = f, 60 f, 62^ = f, 75 = f, 87} =f 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. 4 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 Ib. 8 oz. 16 pwt. 11 gr. of gold? $1435.04 1 74 RA Y> S HIGHER ARITHMETIC. 4. What is the cost of 88 gal. 3 qt. 1 pt. of vinegar, at 37|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 Ib. of sugar, at $3f per cwt.? $1504.2375 8. If 3 6s. silver weigh 1 ft. troy, how much will 17 ft. 11 oz. 16 pwt. 9 gr. be worth? 59 7s. + 9. If 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. f 1. Definition. 1. Reduction J 2 . Cases... / L From Higher to Lower. 1 3 Rules 5, 2 From Lower to Higher. 2. Addition /I- Definition. 12. Rule. 3. Subtraction 1 1- Definition. I 2. Rule. 4. Multiplication / 1- Definition. I 2. Rule. fl. Definition. 5. Division J 2 . Cases. I 3. Rule. Case I. Rule. f< 6. Longitude and Time. 1 Case II. Rule. (. Table of Longitudes. f 1. Definition. 7. Aliquot Parts -J 2. Rule. (3. Table. XII. EATIO. DEFINITIONS. 228. 1. Ratio is a Latin word, signifying relation or con- nection; in Arithmetic, it is the measure of the relation of one number to another of the same kind, 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. 3. The Sign of Ratio is the colon (:), which is the sign of division, with the horizontal line omitted ; thus, 6 : 4 signifies 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 means followii ig. 5. A Simple Ratio is a single ratio consisting of two terms ; as, 3 : 4 = f . 6. A Compound Ratio is the product of two or more (3*7 simple ratios: as, < I 5 : 8 7. The Reciprocal of a Ratio is 1 divided by the ratio, or the ratio inverted ; thus, the reciprocal of 2 : 3, or f , is 12 3 ~ Z 2 ' 8. Inverse Ratio is the quotient of the consequent di- vided by the antecedent ; thus, f is the inverse ratio of 4 to 5. 9. The Value of the Ratio depends upon the relative size of the terms. (175) 176 RAY'S HIGHER ARITHMETIC. 229. From the preceding definitions the following prin- ciples are derived : Antecedent PRINCIPLES. 1. Ratio = -- Consequent 2. Antecedent = Consequent X Ratio. Antecedent 0. Consequent = ; Ratio Hence, by Art. 87 : 1. The Ratio is multiplied by multiplying the 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 I&w*Any change in the Antecedent produces a like change in the Ratio, but any change in the Consequent produces an opposite change in the Ratio. PROBLEM. What is the ratio of 15 to 36 ? OPERATION. 15 : 36 = H = A- Rule. 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.? T %. 2. What is the ratio of 4 mi. 260 rd. to 1 mi. 96 rd.? fff. 3. What is the ratio of 13 A. 145 sq. rd. : 6 A. 90 sq. rd.? f|, 4. What is the ratio of 3 ft. 10 oz. 6 pwt. KH- gr. : 2 ft. 14f pwt.? PROPORTION. 5. What is the ratio of 10 gal. 1.54 pt. : 7 gal. 2 qt. .98 pt.? $|f 6. What is the ratio of 56 bu. 2 pk. 1 qt. : 35 bu. 3 pk. 6.055 qt.? flM- 7. If the antecedent is 7 and the ratio 1^-, what is the consequent ? 4f . 8. If the consequent is f and the ratio f , what is the antecedent ? ^ 2 r . 9. What is the ratio of a yard to a meter, and of a meter to a yard? 10. What is the ratio of a pound avoirdupois to a pound troy? Uf- 11. Find the difference between the compound ratios {\\\}* fill- H- 12. Find the difference between the ratio 4| : 7-| and the inverse ratio. yf M* 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. PROPORTION. DEFINITIONS. 230. 1. Proportion is an equality of ratios. Thus, 4 : 6 : : 8 : 12 is a proportion, and is read 4 is to 6 as 8 is 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 RAY'S HIGHER 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 product of the extremes. 2. The product of the extremes divided by either mean, will give the other mean. 3. The product of the means divided by either extreme, will give the other extreme. SIMPLE PEOPOKTIOK 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 PROPORTION. 179 4. The Statement is the proper arrangement of the terms of the proportion. PROBLEM. If 6 horses cost $300, what will 15 horses cost? STATEMENT. 6 horses : 15 horses :: $300 : ($ ). OPERATION. $300X15 _ 6 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 men men da da> require less time than 15 men to 36*15-'94-() do the same work, the answer should be less than 9-f da.; make a OPERATION. decreasing ratio, Jf, and multiply 9f =-V- da. the remaining quantity by it. -^ X if == 4 da., Ans. Rule. 1. For the third term, write that number which is of the same denomination as the number required. 2. For the second term, write the GREATER of the two re- maining numbers, when the fourth term is to be greater than the third; and the LESS, when the fourth term is to be less than the third. 3. Divide the product of the second and third terms by Hie first; the quotient will be the fourth term, or number required. 1 80 RA Y> S HIGHER ARITHMETIC. EXAMPLES FOR PRACTICE. NOTE. Problems marked with an asterisk are to be solved mentally. 1.* If I walk 101 m i. i n 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 can he trot in an hour, at the same rate ? 21 T 9 T 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 Ib. 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. 2f 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 T ? 8 - ft. 13. If a pendulum beats 5000 times a day, how often does it beat in 2 hr. 20 min. 5 sec. ? 486^-|f times. SIMPLE PROPORTION. 181 14.* If it takes 108 days, of 8|- hr., to do a piece of work, how many days of 6f 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. l(j. A garrison has food to last 9 mon., giving each man 1 Ib. 2 oz. a day : what should be a man's daily allowance, to make the same food last 1 yr. 8 mon.? 8 T ^ oz. 17. A garrison of 560 men have provisions to last during a siege, at the rate of 1 Ib. 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 T 6 T mi. 19. If 1 Ib. troy of English standard silver is worth 3 6s., what is 1 Ib. 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 Ib. 6 oz. of butter worth, at 28 ct. a Ib.? $1.221 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. If, in trading, A puts his cloth at $3.62^, what should B charge for his flour? $6.13 T 6 3 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 RAY'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 12ft. wide, worth, at $4.25 a cord? $62.75 REMARK. 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 Reaumer'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. 1 F. = R=f C.; 1 C. =lf P.= -| R; 1 R. =1J 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. 31^ C. and 88^ F. 34. Convert 46 C. to degrees of the other two ther- mometers. 36f R. and 114f F. REMARKS. 1. In the working of machinery, it is ascertained that the available power is to the weight overcome, inversely as the distances they pass over in the same time. 2. Inverse variation exists between two numbers when one in- creases as the other decreases. 3. The available power is taken f of the whole power, |- being allowed for friction and other impediments. ' 35. If the whole power applied is 180 Ib. and moves 4 ft., how far will it lift a weight of 960 Ib. ? 6 in. 36. If 512 Ib. be lifted 1 ft. 3 in. by a power moving 6 ft. 8 in., what is the power ? 144 Ib. 37. A lifts a weight of 1410 Ib. 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 Ib. 38. A man weighing 198 Ib. 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 Ib. SIMPLE PROPORTION. 183 39. Two bodies free to move, attract each other with forces that vary inversely as their weights. If the weights are 9 Ib. 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? |||. 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 \\ what is the number? J. 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 RA Y'S HIGHER ARITHMETIC. COMPOUND PKOPOKTION. 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 36 A.? STATEMENT. 1 men : 3 men ) SOLUTION. Since Q A Q a \ I : : 4 da ' : ( ) da ' o xx. . O D A.* ) the denomination of the required term OPERATION. is days, make the 9 third term 4 da. In ^ jj forming the first and 3 Xj! X jft second terms, con- 1 v $ = * = ^ ' S ' sider each denom- ^ y/ A p ination separately ; 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 greater 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 number which is of the same denomination as the number required. 2. Arrange each pair of numbers having the same denom- ination in the compound ratio, as if, with the third term, they formed a simple proportion. 3. Divide the product of the numbers in the second and third term,s by the product of the numbers in the first term: the quotient will be the required term. 233. Problems in Compound Proportion are readily solved by separating all the quantities involved into two 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. V:: 25 rd. : 48 rd. ( ) hr.J OPERATION. Rule of Cause and Effect. 1. Separate all the quan- tities contained in the question into two causes and their effects. 2. Write, for the first term of a proportion, all the quan- tities that constitute the first cause ; for the second term, all that constitute the second cause; for the third, all that constitute the effect of the first cause ; and for the fourth, all that constitute the effect of the second cause. 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 number 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 RA 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 railroad 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 Ib. 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. 66| 10. A man has a bin 7 ft. long by 2^ ft. wide, and 2 ft. 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 9. 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 PROPORTION. 187 many will build one with walls 2 ft. thick, 24 ft. high, and 324ft. 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 f 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. II. Definitions. 2. Principles. 3. General Law. | 1. Definitions. 2. Principles. 2. Proportion... \ 3. Kinds / 1- Simple. 4. Statement. * 2 - Compound. 5. Rules. XIY. PEKCENTAGE. 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 per centum, which means by or on the hundred. 3. The Sign of Per Cent is I, T 7 e > T 9 6 > f > , it, I, *, if of it. 16. How much is 100% of a quantity; 125% of it; 250%; 675%; 1000^; 9437^? 1 time, 1^,2^, 6f, 10, 94| 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 ; -fa 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? -f^ 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 $133.331 $6. PERCENTAGE. 191 23. A man contracts 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 3ft> 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 3? 4. CASE II. 237. Given the base and the percentage, to find the rate. PRINCIPLE. The rate equals the number of hundredths that the percentage is of the base. PROBLEM. What per cent of 45 is 9? SOLUTION. 9 is J of 45 ; but i of any OPERATION. number is equal to 20^ of that number; ^ = 1^20^, Ans. hence, 9 is 20 ft of 45. p FORMULA. = R. JL> Rule 1. Divide the percentage by the base; the quotient is the rate ; Rule 2. Find that part of 100^ that the percentage is of the base. 192 RAY'S HIGHER ARITHMETIC. EXAMPLES FOR PRACTICE. 1. 15 ct. is how many % of $2? 7$%. 2. 2 yd. 2 ft. 3 in. is how many % of 4 rd.? 3. 3 gal. 3 qt. is what % of 31$ gal.? 4. | is how many % of f ? 831%. 5. $ of | of is what % of 1 T V? 6. is how many ~&*? 222f %. O J-V/ 7. $5.12 is what % of $640? f%. 8. $3.20 is what % of $2000? 9. 750 men is what % of 12000 men? 10. 3 qt. 11 pt. is what % of 5 gal. 2-*- qt.? 16f%. 11. A's money is 50% more than B's; then B's money is how many % less than A's ? 12. What % of a number is 8% of 35% of it? 13. What % of a number is 2$% of 2^% of it? %. 14. What % of a number is 40^ of Q2$% of it? 25%. 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? 17. What % of a meter is a yard? 18. How many ^ of a township 6 miles square, does a man own who has 9000 acres ? ^TQ% 19. How many % of a quantity is 40^ of 25% of it ? also, 16% of yi\% of it? also, 4% of 120% of it? also, 2% of 80^ of 66f % of it? also, f % of 36% of 75% of it? also, 6|% of 221% O f 96% of it? 10, 6, 5, 1 T V, 5Vo> Wo^- 20. 30% of the whole of an article is how many % of f of it? 45%. 21. 25% of | of an article is how many % of f of it? 13i%. 22. How many % of his time does a man rest, who sleeps 7 hr. out of every 24 ? 291% . PERCENTAGE. 193 CASE III. 238. Given the rate and the percentage, to find the base. PRINCIPLE. The base bears the same ratio to the percentage that 100% does to the rate. PROBLEM. 95 is 5% of what number? OPERATION. 5#= A =95; i*?tfc=i*5 100 ^ = ^ = 19X100 = 1900, .4ns. Or, 9 5 -J- . 5 = 1 9 0, Ana. p FORMULA. = B. Rule 1. Divide the percentage by the rate, and then mul- tiply the quotient by 100 ; the product is the base. Rule 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. T 2 T is 80^ of what number ? -fa. 3. 16 is \\% of what number? 1066f. 4. 31 ct. is 15f^ of what? $2. 5. $10.75 is 8|$ of what? $322.50 6. 162 men is 4f % of how many men? 3375 men. 7. $19.20 is T 6 ^ of what? $3200 8. $189.80 is 104% of what? $182.50 9. 16 gal. 1 pt. is \% 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 421 L % 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 2f % of the flock: how many pigeons escaped ? 438 pigeons. 15. A, owing B, hands him a $10 bill, and says, " there is \% of your money : " what was the debt ? $160. 16. $25 is 62|% of A's money, and 41|^ 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 19o 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. 23o 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 of them ; of these, he gave 37-^% to his son, who gave 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 31^ of what he had left: how much had he at first? $12.60 26. Bought 8000 bu. of wheat, which was 57|^ 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 IT. PERCENTAGE. 195 CASE IV. 239. Given the rate and the amount or the differ- ence, to find the base. PRINCIPLE. The base is equal to the amount divided by 1 plus the rate y or the difference divided by 1 minus the rate. 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-^-1.16 = $325.00, Ans. Or, 1 ft = rental last year ; 16^) = increase this year ; hence, 100^ + 16^ = 116J&=$377; 1<&=$3.25; Ans. PROBLEM. John has $136, which is 20^ less than Joseph's money: how many dollars has Joseph? OPERATION. $136-i-(l .2)=$170.00, Ans. Or, 10 (jf Joseph's money ; John's money = $136; .-. 1 (fc $ 1.7 X 1 = $ 1 7 0.0 0, Ans. 136; i=$34; and f = $34 X 5 = $1 7 0.00, Ans. Rule. Divide the sum by 1 plus the rate, or divide the difference by 1 minus the rate; the quotient will be the base. 196 RAY'S HIGHER ARITHMETIC. EXAMPLES FOR PRACTICE. 1. $4.80 is 33$^g more than what? $3.60 2. | is 50% more than what? f. 3. 96 da. is lOO^g 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 Ib. 9|f oz. is 50% less than what number of pounds? 5ff Ib. 8. T ^ 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 l^fo of the spirit: how many gal. of each? 60 gal. sp., 4-J- 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 $8.88f 13.* If a bushel of wheat make 391- Ib. 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 \\% 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-f % 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. APPLICA TIONS OF PER CENT A GE. 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 44f% 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^^ ; 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. 2. From Case IV. < 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 RAY'S HIGHER ARITHMETIC. Those in which Time is not an Essential Element 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. Those in which Time is an Essential Element Topical Outline. PERCENTAGE. J. Definitions. 2. Notation. ( 1. Principle. I. -j 2. Formula. I 3. Rules. 3. Cases f 1. Principle. 11. -j 2. Formula. I 3. Rules. f 1. Principle. III. -j 2. Formula. I 3. Rules. C 1. Principle. . IV. J 2. Formula. I 3. Rule. 4. Additional Formulas. 5. Applications of Percentage. / Time I Time not an Element, an Element. XY. PEKCENT AGE. -APPLICATIONS. I. PKOFIT 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 Rate; 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 $4800, r r. , . -I o^y -I t OPERATION. my rate of profit is 13%: what is my $400 profit ? j 3 SOLUTION. Since the cost is $4800, and 14400 the rate of profit is 13$,; the profit is 13$ 4800 of $4800, which is $624. $ 6 2 4.0 Profit. (199) 200 RA Y'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. If his flock increases 21^f 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% ? $3.59f per gal. CASE II. 245. 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. 1 ^ = $ 4 5 0; and 120000-f-450--=266f^, An*. SOLUTION. Here the profit is $120000, which, compared with $45000, the cost, is *$$- = f , that is, f of 100^ = 266f ft. EXAMPLES FOR PRACTICE. 1. If I buy at $1 and sell at $4, how many per cent do I gain? 300%. 2. If I buy at $4 and sell at $1, how many per cent do Hose? 75%. 3. If I sell |- of an article for what the whole cost me, how many per cent do I gain ? 80^ . 4.* Paid 8125 for a horse, and traded him for another, giving 60^ additional money. For the second horse I re- ceived a third and $25 ; I then sold the third horse for $150: what was my per cent of profit or loss ? ^\% loss. 5.* A man bought a farm for $1635, which depreciated in value 25%. Selling out, he invested the proceeds so as to make 33^ profit : what was his per cent of profit or loss on the entire transaction ? 202 RAY'S HIGHER ARITHMETIC. 6. It cost me $1536 to raise my wheat crop: if I sell it for $1728, what per cent profit is that per bushel? 12^%. 7. If I pay for a Ib. of sugar, and get a Ib. troy, w r hat % do I lose, and what % does the grocer gain by the cheat? 17%% loss; 21$$% gain. 8. A, having failed, pays B $1750 instead of $2500, which he owed him: what % does B lose? 30^. 9. An article has lost 20^ by wastage, and is sold for 40% above cost: what is the gain per cent? 12%. 10. If my retail profit is 33^, and I sell at whole- sale for 10% less than at retail, what is my wholesale profit? 20%. 11. Bought a lot of glass; lost 15% by breakage: at what % above cost must I sell the remainder, to clear 20% on the whole? ^TT%- 12. If a bushel of corn is worth 35 ct. and makes 2$ 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 ? 9 T 8 ^% . 13. I had a horse worth $80 ; sold him for $90 ; bought him back for $100: what % profit or loss? ^\% l ss - 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 T 7 / Difference between Market and Jrar 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 face. PREMIUM AND DISCOUNT. 211 EXAMPLES FOR PRACTICE. 1. Paid 82401.30 for a draft of $2360 on New York: what was the rate of premium? ^-\%* 2. Bought 112 shares of railroad 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? 12 T 4 %. 4. I sell the same stock for $5936 : what rate of premium is that? what rate of profit? 6%; 63 T ^. 5. If I count my dividend as part of the profit, what is my rate of profit ? 75 T 5 g^ . 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, 4% 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 ? 1 ^ % . 9. Received $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 tlie rate, to find the face. .Discount or Premium FORMULA. Par Value Rate. 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 RAY'S HIGHER ARITHMETIC, 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 2ffi 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 8^ 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 ft 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 FORMULAS.- Par Value = J l + Eate f Premium - ] Market Value i 1 Eate 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 FOE, PRACTICE. 1. What is the face of a draft on Baltimore costing $2861.45, at \%8 HIGHER 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 hereby guaranty the payment of the within. 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. 281. Commission and Brokerage involve four cases, corresponding to those of Percentage. The quantities involved are the Amount Bought or Sold, the Rate of Commission 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 BROKERAGE. 215 CASE I. 262. Given the amount of sale, purchase, or col- lection and the rate, to find the commission. FORMULA. Amount of Sale or Purchase X Rate = 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 %\% for guar- antying payment: if the sales are $6231.25, what does he get? $389.45 216 RAY'S HIGHER ARITHMETIC. 9. Sold 500000 Ib. of pork at 5 ct. a lb.: what is my commission at \\% ? $343.75 10. An architect charges \\% f r 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 \\%\ 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 FORMULA. = Rate. Amount of oale or Purchase 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 $6.43| a barrel, and remits the net proceeds, $5021.25: what is his rate of commission? ^\%> 3. The cost of a building was J19017.92, including the architect's commission, which was $553.92: what rate did the architect charge? 3 ft. 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: what was my rate of commission ? 5%, COMMISSION AND BROKERAGE. 217 6. An agent gets $169.20 for selling property for $8460: 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 ? f % . 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. Commission . ~ , n . FORMULA. = Amount of Sale or Purchase. Rate NOTE. After finding the sura 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? 8140000, and $136500. 2. An insurance agent's income is $1733.45, being 10^ on the sums received for the company: what were the com- pany's net receipts? $15601.05 3. A packing-house charged \\% commission, and cleared 32376.15, after paying out $1206.75 for all expenses of packing: how many pounds of pork were packed, if it cost 41 ct. a pound? 5308000 Ib. 4. Paid $64.05 for selling coffee, which was \ = $1000 ; find 1^, then 100^ . 3. A buys per order a lot of coffee; charges, $56.85; commission, \\%\ the' whole cost is $539.61: what did the coffee cost? $476.80 COMMISSION AND BROKERAGE. 219 4. Buy sugar at 2|^ commission, arid %^% for guaran- teeing payment: if the whole cost is $1500, what was the cost of the sugar? $1431.98 5. Sold 2000 hams (20672 Ib.) ; commission, 2^, guar- anty, 2|^, net proceeds due consignor, $2448.34: what did the hams sell for a Ib.? 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 l^/^ 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 \\f ,9 5 ==: $ 5 9 5 = par value of stock purchased. $5950X-1=$595 income. Or, 9 5 ft = if of the par value = $5652.50 5^ = ^ " rt " " =$ 297.50 1 $ = & " " " " = $ 5 9 5, Arts. SOLUTION. As many dollars' worth of stock can be bought as $.95 is contained times in $5652.50, which is 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 Railroad stock, at If the stock yields 8^ annually, what is the amount of his income? $3200. STOCK INVESTMENTS. 223 2. B invests $100962 in U. S. cur. 6's, 1899, at 106^%. If gold is at |- 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 dividend, at a market value of 106^%, or stock paying dividend, at 104%% ? The former, l^fJJ^. 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 value, and the income, to find the rate of dividend or in- terest. A. I. FORMULA. I. -*- - -7- K. D. M. V. PROBLEM. Invested $10132.50 in railroad stock at 105%, which pays me annually $965 : what is the rale of dividend on the stock? SOLUTION. By Art. 259, $10132.50 -*- 1.05 = $9650 = par value of the stock ; and by Art. 251, $965 -r- $9650 = .1, or 10^ , Ans. 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 ARITHMETIC. 2. What dividend must stock pay, in order that my rate of income on an investment of $64968.75 shall be provided the stock can be bought at 103^% ? 3. My investment was $9850, my income is $500, and the market value of the stock 108^%, brokerage \% : what is the rate of dividend ? \% 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 k\%\ but if he send to the broker $100 more for investment, and buy rolling-mill stock at 103^ , the income will only be half so large: what rate does the higher stock pay? 2 6 ^. CASE III. 271. Given the income, rate of dividend, and mar- ket value, to find the amount invested. FORMULA. -^- X M. V. - A. I. Jti. 1). PROBLEM. If U. S. bonds,, paying 5% interest, are sell- ing at 108^%, how much must be invested to secure an annual income of $2000 ? OPERATION. $2000 $. 05 = 40000; $40 000 X 1 -08 5 =$43 4 00, 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, atlOSJ^, 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 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 92^ ? $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 T atter, and the investment in the latter is $11400 less than in the former: required the whole investment and income. $31800, and $960. 5. Howard M. Holden sold $21600 U. S. 4's, 1907, reg- istered, at 99|^, 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 115^, 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's ; $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 RA Y' S HIGHER ARITHMETIC. CASE IV. 272. Given the market value and the rate of divi- dend or interest, to find the rate of income. T? T) FORMULA. rr -~=E. I. M. V. PROBLEM. What per cent of his money will a man realize in buying 6% stock at 80%? OPERATION. $.0*4-$.SO = A = & = 71 SOLUTION. The expenditure of 80 ct. buys a dollar's worth of stock, giving an income of 6 ct. The rate per cent of income on investment is $.06 -r- $.80 .07J, or 7J< 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%? 5^. 3. Which is the better investment: U. S. new 4's, regis- tered, at 99|^; or U. S. new 4^'s, coupons, at 106% ? The latter is -f-f-^yo 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 1^ brokerage : what is the rate of income on his investment per annum if the annual divi- dend is 8%? STOCK INVESTMENTS. 227 CASE V. 273. Given the rate of income and the rate of dividend or interest, to find the market value. FOHMULA. - ' M. V. K. I. PROBLEM. What must I pay for Lake Shore 6's ($100 a share), that the investment may yield 10^ ? OPERATION. a share. SOLUTION. If bought for $100, or par, it will yield 6^> ; to yield l 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 &^% ? $11250, railroad ; $5400, state. 6. At what figure must be government 5 per cent's to make my purchase pay 9%? 228 RA Y> # HIGHER ARITHMETIC. VI. 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 Time. 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 actual 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. KEMARKS. 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. 275. 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 amount insured, to find the premium. FORMULA. Amount Insured X Rate Premium. 230 RAY'S HIGHER 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 \\%\ what is the premium ? $607.50 2. Insured a house for $2500, and furniture for $600, at T%/^ : what is the premium? $18.60 3. What is the premium on a cargo of railroad iron worth $28000, at lf%? $490. 4. Insured goods invoiced at $32760, for three months, at T 8 o%: 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 -f^, the policy costing 75 cents: what does the insurance cost? $7.83 7. An insurance company having a risk of $25000, at T 9 o%, re-insured $10000, at |^, with another office, and $5000, at 1^, with another: how much premium did it clear above what it paid? $95. CASE II. 277. Given the amount insured and premium, to find the rate of insurance. Premium FORMULA. = Rate. Amount Insured EXAMPLES FOR PRACTICE. 1. Paid $19.20 for insuring f of a house, worth what was the rate? INSURANCE. 23: 2. Paid $234, including cost of policy, $1.50, for insuring a cargo worth $18600: what was the rate? \\% 3. Bought books in England for $2468 ; insured them fo the voyage for $46.92, including the cost of the policy, $2.50 what was the rate? If ^ 4. A vessel is insured for $42000; $18000 at 2% ft $15000 at 3f%, and the rest at 4f% : what is the rate oj the whole $42000? 3f% 5. I took a risk of $45000; re-insured at the sam rate, $10000 each, in three offices, and $5000 in another my share of the premium was $262.50: what was th rate? 2f% 6. I took a risk at 1^^ ; re-insured f of it at 2^ , an< \ of it at ^\% what rate of insurance do I get on wha is left? ^% CASE III. 278. Given the premium and rate of insurance, t< find the amount insured. Premium T ORMULA. = Amount Insured. Rate. EXAMPLES FOR PRACTICE. 1. Paid $118 for insuring, at \% : what was the amoun insured ? $14750 2. Paid $411.37^ for insuring goods, at \\ S HIGHER ARITHMETIC. 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 $5.28 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 %-f% 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. Whole Cost FORMULA. = Invoice Price. 1 + Rate 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 3.80, what was the invoice price per cu. ft.? $1.25 TOPICAL OUTLINE. 241 Topical Outline. APPLICATIONS OF PERCENTAGE. (Without Time.) 1. Profit and Loss.... 2. Stocks and Bonds.. 3. Premium and Discount 4. Commission and Brokerage. 5. Stock Investments.. 6. Insurance., 7. Taxes.. 8. United States Revenue 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. l. Definitions: Commission-Merchant, Prin- cipal, Commission, Consignment, Con- signor, Consignee, Correspondent, Net Proceeds, Guaranty, Broker, Brokerage. (^ 2. Four Cases. I 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. 1. Definitions: Internal Revenue, Customs, Ad Valorem Duties, Specific Duties, In- voice, Tariff, Tare. 2. Four Cases. H. A. 21. XVI. PEKCEOTAGE -APPLICATIONS. I. INTEKEST. 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 Kate of Interest is the number of per cent the yearly interest is of the principal. 4. The Amount is the sum of the principal and interest. 5. Interest is Payable at regular intervals, yearly, half- 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 annum, 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 Kate 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) INTEREST. 243 REMARK. The per cent of interest that is legal in the different states and territories, is exhibited in the following table. TABLE. NAME OF STATE. RATE. NAME OF STATE. RATE. Alabama 8% 10* 6% 10% 6% 10% 6% 7% 6% 6% 8% 7% 10% 6% 6% 6% 7% 6% 5% 6% 6% 6% 7% 7% 6% 8% Any. Wfo Any. 6% Any. 6% 12% 6% 10% Any. Any. 24% 8% 8% 10% 12% 10% 8% Any. 6% 6% 10% 10% 10% Missouri 6% 10% 10% 10% 6# 6% 6% 6% 6% 6% 10% 6% 6% 7% 6% 8% 6% 10% 6% 6% 10% 6% 7% 12% 10% Any. 12% Any. 6% 6% 12% 6% 8% 8% 12% 6% Any. Any. 6% 12% 6% Any. 6% 8% Any. 6% 10% Any. _ Montana \rkansas Nebraska California Nevada Canada New Hampshire... N"ew Jersev Colorado Connecticut New M^exico Dakota New ^Tork Delaware North Carolina Ohio District Columbia. Florida Oreffon Oeorsria Pennsylvania Idaho Ehode Island South Carolina Tennessee Illinois oo.. Indiana Iowa Texas United States Utah Vermont Maine Virginia Washington Ter- ritory Massachusetts Michigan West Virginia AVisconsin Minnesota Mississippi \Vvomm 01 i NOTE. When the per cent of interest is not mentioned in the note or contract, the first column gives the per cent that may be collected by law. If stipulated in the note, a per cent of interest as high as that in the second column may be collected. 8. Interest is either Simple or Compound. 9. Simple Interest is interest which, even if not paid 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 Interest, the Rate, the Time, and the Amount, 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 Rase, and the interest to the Percentage. NOTATION. 294. The following notation can be adopted to ad- vantage in the formulas : Principal = P. Eate == K. 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 interest is equal to the continued product of the principal, rate, and time. INTEREST. 245 f P V E) \^ T =^: I FORMULAS. pi - A NOTE. 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. r = $44.37i, Int. SOLUTION. The interest of $320 for 1 year, at 4^, is $12.80 ; and for 3^ years, it is 3^ times as much as it is for 1 year, or $12.80 X 3A = $44.37J. REMARK. Unless otherwise specified, 30 days are considered to make a month; hence, 5 mo. 18 da. = xV ^ a ye General Rule. 1. Multiply the principal by the rate, and that product by the time expressed in years; the product is the interest. 2. Add Hie principal and interest, to find the amount. EXAMPLES FOR PRACTICE. Find the simple interest of: 1. $178.63 for 2 yr. 5 mo. 26 da., at lf c . $31.12 2. 86084.25 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 &%. $271.33 11. $10000 for 1 da., at Q%. $1.67 246 RAY'S HIGHER ARITHMETIC. METHOD BY ALIQUOT PAKTS. 296. Many persons prefer computing interest by the method of Aliquot Parts. The following illustrates it : PROBLEM. What is the simple interest of $354.80 for 3 yr. 7 mo. 19 da., at 6%? SOLUTION. The yearly in- terest, being 6$> of the prin- cipal, is found, by Case I of 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 counted 1 mill. OPERATION. $354.80 .06 l yr. = 21.2880 3 yr. = 3 63.864 6 mo. = ^ 10.644 1 mo. = 1 1.774 1 8 da. == A 1.064 1 da. = /o .059 $77.41, Ans. SIX PEE CENT METHODS. FIRST METHOD. 297. The six per cent method possesses many advantages, and is readily and easily applied, as the following will show : At 6$) per annum, the interest on $1 for 1 year is 6 ct., or .06 of the principal. " 2 mo. " 1 " " .01 " " " tt i tt u i tt t0 Q5 u by such a part of it as the given rate is greater or less than 6^. 2. After finding the interest at 6^, observe that the interest at 5^ = interest at Qf c | of itself. And the interest at 4J^ = int. at 6f J of it. at 4 f c = int. at 6$ J of it. at 3 cf = J int. at 6^>. at 2 ff = \ int. at 6^>. at \\ + J of it. at 12^, 18^,, 24^, = 2, 3, 4 times interest at 6^,. at 5^ , 10^,, 15^, 20/ = T V, i, J, J 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 bj 6. SECOND METHOD. PROBLEM. What is the simple interest of $354.80 for 3 yr. 7 mo. 19 da., at SOLUTION. The interest of any OPERATION. sum ($354.80), at 6^ , equals the ($354.80^-2)X.436J = interest of half that sum ($177.40), $77.40553, Ans. at 12^,. But 12^ a year equals l of the principal will be the interest. To get 43J^, multiply by 43-J-j} hundredths == .43|f = .436J. KEMARK. In the multiplier .436J, the hundredths (43) are the number of months, and the thousandths (6J) are of the days (19 da.), in the given time, 3 yr. 7 mo. 19 da. 248 RAY'S HIGHER ARITHMETIC. Rule. Reduce the years, if any, to months, and write the whole number of months as decimal hundredths; after which, place J of the 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 J 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 REMARK. Exact or Accurate Interest requires that the common year should be 365, and leap year 366 days ; hence, exact interest is 7*3 less for common years, and -^ 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 ruo. 29 da., at 18^. $33927.72 3. $512.60 for 8 mo. 18 da., at 1%. $25.72 4. $1363.20 for 39 da., at \\% 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 1%. $55.67 8. $4642.68 for 5 mo. 17 da., at 15%. $323.05 9. $13024 for 9 mo. 13 da., at Wfi. $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 \\ of the principal ; l< T) = P. PROBLEM. What is the par value of a bond which, in 8 yr. 8 mo., at 6^, will amount to $15200? OPERATION. 5200-*-$!. 52 = 10000; Ans. SOLUTION. There are as many dollars in the princi- pal as $1.52 is contained Or, 100^ = bond; times in $15200, which is 10000; therefore, $10000 is the par value of the bond. Or, the amount is -J-^J of the principal ; hence, ^f of the principal is $15200 ; and principal is $100 ; and -j^g of the principal is $10000. 100 $,=r$ 10000, Am. of the Rule. Divide the amount by the amount of $1 for the given time and rate; the quotient will be the principal. 254 RAY'S HIGHER 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 PKOMISSOKY NOTES. DEFINITIONS. 302. 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 original 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. REMARK. There are various forms used in time notes, the prin- cipal ones being as follows: PR OMISSOE Y NOTES. 255 1. ORDINARY FORM. $450.50 BOSTON, MASS., June 30, 1880. Three montlis after date, I promise to pay Thomas Ford, or order, Four Hundred and Fifty, and -ffa, Dollars, for value received, with 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. Harris, or bearer, Three Hundred and Fifty Dollars, for 'value received, with interest from date. JAMES WEST. DANIEL TATE. REMARKS. 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-jive Dollars, for value received, negotiable and payable without defalcation or discount. Surety, JAMES MILLER. H. A. WHITE. REMARK. The maker of this note, H. A. White, is the principal, 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. White's creditor. 256 JRAY'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 order, on demand, Eleven Hundred Dollars, with interest. HENRY KUDOLPH. 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 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. REMARKS. 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 lie 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. PR OMISSOR Y NOTES. 257 13. If a note contains the words "with interest," it draws interest from date, and, if no rate is mentioned, the legal rate prevails. REMARKS. 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 Grace, are allowed for the payment of a note after it is nominally due. REMARKS. 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 the 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. REMARK. 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 RAY'S HIGHER ARITHMETIC. Rule. Add to the date of the note, the number of years and months to elapse before payment ; if this gives the day of a month higher than that month contains, take the last day in that month; then, count the number of days mentioned in the note and 3 more : this will give the day the note is legally due ; but if it is a Sunday or a national holiday, it must be paid the day previous. REMARKS.!. When counting the days, do not reckon the one from which the counting begins. (For exceptions, see Art. 313, 16, Rem.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 lime 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: *H P 3 Or* K P N > *& *t g P *4 H 5 H c CTQ % HI e 4 o , is $302.40; therefore, the amount == $10500 + $2835 + $302.40 = $13637.40 Rule. 1. Find the interest on the principal for the entire time, and on each year's interest till the time of settlement. 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. 305. 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. 308. 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:" U. S. Rule. 1. "The rule for casting interest when par- tial payments have been made, is to apply the payment, in the first place, to the discharge 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. "If the payment be less than the interest, the surplus of interest must not be taken to augment the principal; but in- terest continues on the former principal until the period when the payments, taken together, exceed the interest due, and then the surplus is to be applied towards discharging the principal, and interest is to be computed on the balance as aforesaid. 307. This rule is based upon the following principles: PRINCIPLES. 1. Payments must be applied first to the dis- charge 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. REMARKS. 1. It is worthy of remark, that the whole aim and tenor of legislative enactments and judicial decisions on questions of 262 RAY'S HIGHER ARITHMETIC. interest, have been to favor the debtor, by disallowing ^compound interest, and yet this very 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), but as often as a payment is made; by which it happens that the closer the payments are together, the greater 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 days. PROBLEM. 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. Indorsements. Oct. 13, 1880, $40 ; June 9, 1881, $32 ; Aug. 21, 1881, $125 ; Dec. 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, .... ^913 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, ... t ... 80. Balance due 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 T 7 ^-. CHICAGO, March 10, 1882. For value received, six months after date, I promise to pay G. Riley, or order, three hundred and four T 7 ^- dollars. No payments. H. McMAKIN. What was due Nov. 3. 1883? $325.63 2. S429y\V INDIANAPOLIS, April 13, 1873. On demand, I promise to pay W. Morgan, or order, four hundred and twenty-nine ffa dollars, value received, with interest. R WlLSON . 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, fov $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 Rule. 1. Compute the interest to the time of the first payment, if that be one year or more from the time that interest commenced; add it to the principal, and deduct the payment from the sum total. 2. If there be after payments made, compute the interest on the balance due to the 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, the time between one payment and another be one year or more. 3. BUT if any payment be made before one year's interest hath accrued, then compute the interest, on the sum then due on the obligation, for one year, add it to the principal, and compute the interest on the sum paid, from the time it was PARTIAL PAYMENTS. 265 paid, up to the end of the year; add it to the sum paid, and deduct that sum from the principal and interest added as above. (See Note.) 4. If any payment be made of a less sum than the interest arisen at the time of such payment, no interest is to be com- puted, but only on the principal sum, for any period. NOTE. If a year does not extend beyond the time of payment; but if it does, then find the amount of the principal remaining unpaid up to the time of settlement, likewise 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 Rule. 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, to the extinguishment of the principal. 2. When notes, etc., are drawn with INTEREST PAYABLE ANNUALLY, the annual interests that remain unpaid are subject 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 liqui- date the annual interests that have become due; and, thirdly, to the extinguishment of the 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, 1881, $350. What was due April 12, 1882? $1291.95 H. A. 23. 266 RAY'S HIGHER ARITHMETIC. 309. Business men generally settle notes and accounts payable within a year, by the Mercantile Rule. Mercantile Rule. 1. Find the amount of the principal from the date of the note to the date of settlement. 2. Find the amount of each payment from its date to the date of settlement. 3. From the amount of the principal subtract the sum of the amounts of the payments. NOTE. 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 1% 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. TKUE 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 worth 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 worth, 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 regulai 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 6 ft. OPERATION. Amount of $ 1 for 1 yr. 9 mo. 1 9 da., at 6 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 "after 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. REMARKS. 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 indorsement is in blank, 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 acceptor 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. REMARKS. 1. Since 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, bank discount is interest on the face of the note, while true discount is the interest on the present worth, 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 of matu- 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 Principal; the bank discount, to the Interest; the proceeds, to the Principal less the Interest; and the term of discount, to the Time. CASE I. 315. Given the face 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 272 RAY'S HIGHER ARITHMETIC. OPERATION. Int. of $ 1 for 9 3 da., at 6^, $.0155 Eate of Discount. $770X-0155 = $11.935, 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 PRINCIPLES. 1. The interest on the sum discounted for the given time (plus the three days of grace) at the given rate per cent, is the bank discount. 2. The proceeds is equal to the sum discounted minus the discount. Rule. 1* Find the interest on the sum discounted for three days more than the given time, at the given rate; it is the discount. 2. Subtract the discount from the sum discounted, and the remainder is the proceeds. EEMARKS. 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, tl Due May 3 / 6 ," 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 QUINCY, ILL., Jan. 3, 1870. Six months after date, I promise to pay to the order of Jones Brothers seven hundred and ninety-two -ffa dollars, at the First National Bank, value received. Discounted Feb. 18, at Gfi. ALBERT L. TODD. Due July 3 / 6 ; 138 da. to run; proceeds, $774.27 BANK DISCOUNT. 273 2. $1962 T \ 5 - NEW YORK, July 26, 1879. Value received, four months after date, I promise to pay B. Thorns, or order, one thousand nine hundred sixty- two ^Q dollars, at the Chemical National Bank. Discounted Aug. 26, at 7 ft. E< WlLLIAMS - Due Nov. 26 /2 9 ; 95 da - to run ; pro. $1926.70 3. $2672 T ^. 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 y 1 ^- dollars, without defalcation. JEREMIAH BARTON. Discounted July 19, at 6fo. Due Dec. 10 / 13 ; 148 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 \\% a month. Due Feb. 2 ^ Mar.; 32 da. to run; pro. $3823.82 REMARK. 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 annum. TT p ^ HANNA & TUTTLE. Discounted June 15, at 6^. Due Dec. ll / J4 ; 182 da. to run; pro. $2875.47 274 RA Y'S HIGHER ARITHMETIC. 6. $737 T 4 oV BOSTON, Feb. 14, 1880. Value received, two months after date, I promise to pay to J. K. Eaton, or order, seven hundred thirty-seven T 4 (j dollars, at the Suffolk National Bank. WILLIAM ALLEN. Discounted Feb. 23, at 10^. Due April 14 / 17 ; 54 da. to run; pro. $726.34 7. $4085 T 2 ^V NEW ORLEANS, Nov. 20, 1875. Value received, six months after date, I promise to pay John A. Westcott, or order, four thousand eighty-five T 2 ^ dollars, at the Planters' National Bank. E. WATERMAN. Discounted Dec. 31, 1875, at 5^,. Due May g, 1876; 144 da. to run; pro. $4003.50 CASE II. 316. Given the proceeds, time, and the rate of dis- count, to find the face 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 63 da., at 6 ft = $.0105 Proceeds of $1 $.9 895; $1 00 -f- . 98 9 5 = $ 1 1 . 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 this sum, $.9895, is contained times in the given proceeds, $1000; this gives $1010.61 for the face of the note. Rule. Divide the given proceeds by the proceeds of $1 for the given time and rate; or, divide the given discount by the discount of $1 for the given time and rate; the quotient i* equal to the face of the note. BANK DISCOUNT. 275 EXAMPLES FOR PKACTICE. 1. Find the face of a 30 da. note, which yields $1650, when discounted at \\ c f c a mo. $1677.68 2. The face of a 60 da. note, which, discounted at 6% per annum, will yield $800. $808.49 o. The face of a 90 da. note, bought for $22.75 less than its face, discounted at 1%. $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 ft, for the given time. And $95. 80X- 00175 = $. 1676 5, interest of proceeds, at 1 fa for the given time ; then, $ 4 . 2 -f- $ . 1 6 7 6 5 = 2 5 ^, rate. 276 RAY'S HIGHER ARITHMETIC. SOLUTION. The discount of $100 for 63 days, at 2fi a month, is $4.20, and the proceeds is $95.80 The interest of $1 for the given time, at If 0l is $.00175, and of $95.80 is $.16765 To find the rate, we proceed thus : $4.20 -f- $.16765 = 25 2 / . Rule. 1. Find the discount and proceeds of $100 or 81 for the time the note runs. 2. Divide the discount by the interest of the proceeds, at 1^, for the same time; the quotient is the rate. EXAMPLES FOR PRACTICE. What is the rate of interest : 1. AVhen a 30 da. note is discounted at 1, 1^, 1J, 2% a month? 12-J.ff, 15^, 18 T ^ 4 T , 24^% per annum. 2. When a 60 da. note is discounted at 6, 8, 10^ per annum ? 6iV%> $ AV lO^y^ per annum. 3. When a 90 da. note is discounted at 2, 21, 3% a month ? 25ff , 32 T %%, 39f ffify per annum. 4. When a note running 1 yr. without grace, is discounted at 5, 6, 7, 8, 9, 10, \2% ? _5^, 6tf , 7#, 8tf, 9ft, 111 IB&%. 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 '5" 3. On 90 da. notes, yield 1, 2, 4% a month interest? 4. On notes running 1 yr. without grace, yield 5, 6, 7, 8, 9, 10^ interest? 4Jf 5ff, 278 RAY'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. REMARK. This 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 Bill of Exchange. 3. A Bill of Exchange is a written order on a person or company in a distant place for the payment of money. REMARK. 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 OF 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. Rollins One Thousand Pounds Sterling, value received, and charge to account of To JOHN BROWN & Co., J. S. CHICK. No. 1250. LIVERPOOL, ENGLAND. EXCHANGE. 279 8. The Rate 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 Intrinsic or Commercial. 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 \ 2 2 ' cash, 450 100 Apr. May 4 20 " note, 2 mo. " remit., May 25, 240 120 i June Hi " accep. 16 da. sight 500 C is Cr. $470, due Aug. 12. 6. I owe $912, due Oct. 16, and $500, due Dec. 20. If 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, and the other 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. VII. 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). Wliat sum, at a given time, should be paid to balance the account. The first process is Averaging the Account (Art. 324) ; the second is Finding the Cash Balance. DR. Henry Armor in acc't with City Bank. CR. 1875. Dol. a. 1874. Dol. at. Jan. 5 To check, 300 Dec. 31 Bv balance old account, 500 " 20 ' " 500 1875. " a tt 100 Jan. 7 " cash, 50 " 27 ( H 850 15 400 31 75 24 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 : DR. F. H. Willis in acc't with E. S. Kennedy. CR 1879. DoZ. Ct. 1878. Dot. Ct. Jan. 6 To check, 170 Dec. 31 By balance forward, 325 " 13 " < " 480 J879. 16 .. ^gf (i 96 Jan. 7 " cash deposit, 800 " 21 " " 500 ' 20 it 44 44 175 * 50 90 240 Interest to Jan. 31, at 6 per cent. DR. OPERATION. $170, 6^, 25 da.=:$.708 $325, 6 480, " 18 " =1.44 8 0*0, < ; 96, " 15 a __ 2 4 175, " 500, " 6 " = .50 240, l( 50, 3 " ..= .025 $ 1 5 4 $1296 $2.913 $1296 CR. 31 da .=$1.679 24 " = 3.20 11 " = .32 1 $244 = .04 $5.239 $2.913 $2.33~ .'. Interest due = $2. 33; and cash balance due Willis is $244 + $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 difference between the sums of the interests is equal to the interest due, which, added to the difference between the sums of the items ; gives the balance due. Hule. 1. Find the number of days to elapse between the date when each item is due and the date of settlement. 2. Find the sum of the items on the Dr. side, and then add to each item its interest, if the item is due before the date of settlement, or subtract it if due after the date of settlement; do the same with Cr. side. 3. The difference between the amounts on the tivo sides of the account is the cash balance. 294 RAY'S HIGHER ARITHMETIC. 1. Find the equated time and cash balance of the follow- ing account, July 7, 1876; also April 30, 1876, interest 6^ per annum : Henry Hammond. DR. OK. 1876. Jan. 4 To Mdse, at 3 mo. $900.10 Feb. 1 " 4 " 400.00 ih 700.50 Mar. 600.40 April May 8 10 ' Cash, ' Mdse, at 30 da. 500.20 400.00 June 1 4 " "1 mo. 100.60 Equated time, May 15, 1876. Cash balance July 7, 1876, $3633.62 Cash balance April 30, 1876, $3592.80 REMARKS. 1. When, in forming the several products, the cents are 50 or less, reject them; more than 50, increase the dollars by 1. 2. The 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 : DR. William Smith. CR. 1876. Dol. Ct. 1876. Dot. Ct. Jan. 1 To Mdse, 800 00 Jan. Hi By Cash, 400 00 " 16 4 " at 30 da. 180 30 " 28 200 00 Feb. 14 " 60 " 400 60 Feb. 15 Bills Rec. at 60 da. 180 30 Mar. 25 *' 500 00 28 Cash, 100 00 April 1 Cash, 800 00 Mar. 30 Bills Rec. at 90 da. 450 00 May 7 Mdse, 600 00 April 14 Cash, 400 60 * 2\ at 60 da. 700 00 May 1 * 500 00 June 10 200 00 15 Bills Rec. at 1 mo. 680 00 14 15 1 " 90 da. 2000 00 June 10 Cash, 300 00 July 12 500 00 July 1!> 70(3 OJ Aug. 4 14 2 mo. 1000 10 Aug. 10 Bills Rec. at 20 da. 200 00 Sept. 1 Cash, 150 00 Oct. 3 1100 00 Equated time, June 18, 1876; cash balance, $2389.70 REMARK. All written obligations, of whatever form, for which a certain amount is to be received, are called Bills Receivable. SETTLEMENT OF ACCOUNTS. 295 3. Find the equated time and the cash balance of account, March 31, 1876, interest 10% per annum. DR. George Cummings. CR. 1876. Dot. Ct. 1875. Dot. Ct. Jan. 2 To Cash, 300 00 Dec. 1 By Mdse, at 1 mo. ' 583 00 1876. M 21 it 44 194 00 Feb. 3 ' " " " 40 00 Mar. 4 150 00 Mar. 30 130 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't with T. J. Fisher & Co. 4. DR. CR. 1871. Do!. Ct. 1870. Dol. Ct. Jan. 13 To check, 350 Dec. 31 By bal. from old acc't, 813 64 " '22 44 44 275 1871. Feb. 26 (I 4< 100 Feb. 4 " cash, 120 May 1 '" " 400 Mar. 17 " * 500 June 23 108 25 May 31 84 50 Interest to June 30, at 6 per cent. Int. due Morris, $13.34; bal. due Morris, S298.23 5. DR. Wm. White in acc't with Beach & Berry. CR. 1875. Dol. Ct. 1875. Dol. Ct. July 2 lo check, 212 50 June 30 By bal. from old acc't, 1102 50 " 20 66 July 6 ' cash deposit, 50 Aug. 7 235 " 15 ' " 95 25 300 Aug. 9 4 44 (4 168 75 Sept. 5 110 Sept. 18 4 4. 44 32 11 46 40 Oct. 3 41 44 79 90 27 454 25 Interest to Oct. 12, at 10 per cent. Int. due White, $19.68; bal. due White, $123.68 296 It AY'S HIGHER ARITHMETIC. ACCOUNT SALES. 330. 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. Guaranty 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 : Rule. 1. Average the sales alone; this result will be 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 equated time for paying tJie net proceeds. 1. Find the equated time of paying the proceeds on the following account of Charles Maynard: Charles Maynard? s Consignment. 187G. Aug. 191 By J. Barnes , at 10 da Dol. 50 Ct. 60 14 Cash . . .. 800 00 i 24 (C 9 ii ii Cash 210 00 t per cent 205.25 300 ^ Net proceeds due Maynard 7910 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 2J%, $61.87; loss and gain for his net gain, $11.57 Net proceeds, $2236.56; due July 12, 1876. 3. Mdse. Co. 187(5. Jujy 1876. July 18 24 30 15 15 30 30 30 30 By Note at 20 da ... Dot. 120 60 55 Ct. 00 00 (JO " 25 " " Cash .. .. CHARGES. To Cash paid Freight $ 33 00 ' " " Drayage , 4.00 4 " " Insurance 1.50 " " " Storage ' " " Commission, at 2H per cent . 3.00 529 C. V. Carries 's net proceeds 62.73 Less our / net loss $109.52 3 93 105.59 Find C. V. Carnes'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 ACCOUNTS. 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 RAY'S HIGHER ARITHMETIC. 1. Storage to Jan. 81, at 5 ct. a bbl. per month. RECEIVED. DELIVERED. 3876. Bbl. Balance on hand. Days. Products. 1876. Bbl. January 2 200 January 10 110 5 150 13 90 7 30 17 20 10 120 20 115 14 80 25 J i 17 150 27 20 75 30 100 24 60 31 28 200 Storage, $19.25; bbl. on hand, 418. 2. Storage to Feb. 20, at 5 ct a bbl. per month. RECEIVED. DELIVERED. Bbl. Balance on hand. Days. Products. Bbl. January February 31 4 418 250 February 5 10 100 80 9 120 12 220 12 100 14 140 16 30 18 90 20 288 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, Compound Interest is the whole gain of a principal, increased at the end of each interval by all the interest draw^i during that interval. 3. The final amount in Compound Interest is called tho compound amount. COMPOUND INTEREST. 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 REMARK. If the above debt be, at the same rate, on annual interest (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- plication similar to the last. EXAMPLE. Let the debt be $1000, and let the interest be com- pounded at 2^> 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 8 2 6 -^. According to the usual form of statement, this debt 'is compounding 300 RA Y'S HIGHER ARITHMETIC, " at 82 per cent. 2 3 5 1.0200 0000 1.0404 0000 J.0612 0800 1.0824 3216 1.1040 8080 1.0250 0000 1.0506 2oOO 1.0768 9062 1.1038 1289 1.1314 0821 1.0300 0000 1.0609 (XXX) 1.0927 2700 1.1255 0881 1.1592 7407 1.0350 0000 1.0712 2500 1.1087 1787 1.1475 2300 1.1876 8631 1.0400 0000 1.0816 0000 1.1248 6400 1.1698 5856 1.2166 5290 1.04.50 0000 1.0920 2500 1.1411 6612 1.1925 1S60 1.2461 8194 6 7 8 9 10 1.1261 6242 1.1486 8567 1.1716 5938 1.1950 9257 1.2189 9442 1.1596 9342 1.1886 8575 1.2184 0290 1.2488 6297 1.2800 8454 1.1940 5230 1.2298 7387 1.2667 7008 1.3047 7318 1.3439 1638 1.2293 5533 1 .2722 7926 1.3168 0904 1.3628 9735 1.4105 9876 1.2653 1902 1.3159 3178 1.3685 6905 1.4233 1181 1.4802 4428 1.3022 6012 1.360S 61 Si 1.4221 0061 1.4860 9514 1.5529 6942 11 12 13 14 15 1.2433 7431 1.2682 4179 1.2936 0663 1.3194 7876 1.3458 6834 1.3120 8666 1.3448 8882 1.3785 1104 1.4129 7382 1.4482 9817 1.3842 3387 1.4257 6089 1.4685 3371 1.5125 8972 1.5579 6742 1.4599 6972 1.5110 6866 1.5639 5606 1.6186 9452 1.6753 4883 1.5394 5406 1.6010 3222 1.6650 7351 1.7316 7645 1.8009 4351 1.6228 5305 1.6958 8143 1.7721 9610 1.85 19 4492 1.9352 8244 16 17 18 19 20 1.3727 8570 1.4002 4142 1.4282 4625 1.4568 1117 1.4859 4740 1.4845 0562 1.5216 1826 1.5596 5872 1.5986 5019 1.6386 1644 1.6047 0644 1.6528 4763 1.7024 3306 1.7535 0605 1.8061 1123 1.7339 8604 1.79J6 7555 1.8574 8920 1.9225 0132 1.9897 8886 1.8729 8125 1.9479 4 773 6 7047 512 7.0399 887 8.1472 520 8.6360 871 9.1542 524 9.7035 075 10.2857 179 11.4239 422 12.2236 181 13.0792 714 13.9948 204 14.9744 578 15.9681 718 17.2456 256 18.6252 756 20.1152 97? 21.7245 215 22.2512 250 24.2538 353 20.4300 805 28.8159 b!7 31.4094 200 30.9120 805 34.0039 480 37.4043 434 41.1447 778 45.2592 556 41 42 4:j 44 45 7.3919 882 7.7615 876 8.1496 669 8.5571 503 8.9850 078 10.9028 610 11.5570 327 12.2504 546 12.9854 819 13.7646 108 16.0226 699 17.1442 568 18.3443 548 19.0284 596 21.0024 518 23.4024 832 25.3394 819 27.3660 404 29.5559 717 31.9204 494 34.2362 679 37.3175 320 40.6701 098 44.33(59 597 48.3272 861 49.7851 811 54.7636 992 00.2400 692 00.2040 701 72.8904 837 40 47 48 49 50 9.4342 582 9.9059 711 10.4012 697 10.9213 331 11.4673 998 14.5901 875 15.4659 167 16.3938 717 17.3775 040 18.4201 543 22.4726 234 24.0457 070 25.72S9 065 27.5299 300 29.4570 251 34.4740 853 37.2320 122 40.2105 731 43.4274 190 40.9016 125 52.6707 419 57.4176 486 62.5852 370 08.2179 083 74.3575 201 80.1795 321 88.1974 853 97.0172 338 106.7189 572 117.3908 529 51 52 53 54 55 12.0407 698 12.6428 083 13.2749 487 13.9.583 961 14.6356 309 19.5253 635 20.6968 853 21.9386 985 23.2550 204 24.6503 216 31.5190 168 33.7253 480 30.0801 224 38.6121 509 41.3150 015 50.6537 415 54.7000 408 59.0825 241 63.80!)! 200 68.9138 561 81.0496 969 88.3441 090 90.2951 449 104.9617 079 114.4082 616 129.1299 382 142.0429 320 150.2472 252 171.8719 477 189.0591 425 Subtract $1 from the Amount in this Table to find the Interest. 304 RAY'S HIGHER ARITHMETIC. How to use the table in finding the Compound Amount: 1. Observe at what intervals interest is payable, and also the rate per interval. 2. If the number of full intervals can be found in the year column, note the sum corresponding to it in the column under the proper rate; multiply this sum, or its amount for any re- maining fraction of an interval, by the principal. 3. If the number of intervals be not found in the table, separate the whole time into periods which are each within the limits of the table; find the amount of the principal for one of them, make that a principal for the next, and so on, 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 1% 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 S%, 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 -f- 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 -5- .10663346 = T |-g-, nearly ; hence, the required period is 25 j\j inter- vals of 6 mo., or, 12 -yr. 6 mo. 1 da., Ans. Rule. 1. Divide the amount by the principal. 2. If the quotient be found in the table under the given rate, the years opposite will be the required number of intervals; but if not found exactly, in the table, take the 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 quotient to the number of full intervals ; the result will be the required time. 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, Q%, payable semi-annually? 3 yr. 9 mo. 16 da. H. A. 26. 306 RAY'S HIGHER ARITHMETIC. 3. &12500 gain $5500, 10%, payable quarterly? 3 yr. 8 mo. 9 da. 4. $1 gain 91, at 6, 8, 10 ft? 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- annual ly? 7 yr. 1 mo. 7 da. CASE III. 337. Given 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 Rule. Divide the amount by the principal; search in the table, opposite the given number of full intervak, for the exact quotient or the number nearest in value; if the time contain also a part of an interval, find the amount of the tabular sum for that time, before comparing with the quotient; the rate per Gent at the head of the column ivill be the exact, or the 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 INTEREST. 307 5. $7652.18 gain $17198.67, interest payable quarterly, in 11 yr. 11 mo. 3 da.? 10%. 6. Any sum double itself in 10, 15, 20 yr. ? 1st, between 1% and 8^; 2d, nearly 5^; 3d, little over 3|%. CASE IV. 338. Given the compound interest or amount, the time, and the rate, to find the principal. PROBLEM. What principal will yield $31086.78 com- pound interest in 40 yr., at 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, Am. Rule. Divide the given interest or amount by the interest or amount of $1 for the given time and at the given rate; the quotient will be the required principal. REMARK. 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 PRACTICE. 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., 1%, payable semi- annually? $3075. 3. Will yield $3598.61 in 3 yr. 6 mo. 9 da., W%, 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 discount 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. Annuities are < 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. * Si nee the problems in annuities may be classed under 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 initial 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. REMARK. 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, 4 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 ? OPERATION. SOLUTION. The initial value must $1 be the principal, which, at 6^, yields . 6 $250 interest every year ; it is found, .06)250.0000 by Art. 300, $4166.66 f , Am. Rule. Divide the given payment by the interest of $1 for one interval at the 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, allowing 1% 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 the 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? $41979.16f ; $42604. 16|; $41666.66| 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 6/0 compound interest, = $4166.66 -f- 1.5938481 (Art. 335 ). Use the contracted method, reserving 3 decimal places ; the quotient, $2614.22, is the present value of the perpetuity. Rule. Find the initial value of the perpetuity by the last rule ; then find the present worth of this sum for the time the ANNUITIES. 311 perpetuity is deferred, by Case IV of Compound Interest; this wilt be the present value 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 in Syr. 4 mo., int. 1%. $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., . . = 2614.22 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^,, =$9665.27 Pres. val. of perpetuity of $680, deferred 17 yr., at 5f , = 5933.64 Pres. val. of annuity of $680, deferred 7 yr., to run 10 yr. =$3731.63 312 RA Y'S HIGHER ARITHMETIC. Rule. Find the present value of two perpetuities having the given rate, payment, and interval, one of them commencing when the annuity commences, and the other when the annuity ends. The difference between these values will be 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. (Kern. 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. 1ft. $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. 4fo. $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. 8%, 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., = $4166.6Cf X .5938481 = $2474.37, the final or forborne value of the annuity. Rule. Consider the annuity a perpetuity, and find its initial value by Case I. The compound interest of this sum, at the given rate for the time the annuity runs, will be the final 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 $300, running 18 yr., int. 5^. $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 RAY'S HIGHER ARITHMETIC. The present value of $1 per annum in any number of yearSj not exceeding fifty -five. Yrs. 4 per cent. 6 per cent. 6 per cent. ' 7 per cent. 8 per cent. 10 per cent. 1 2 3 4 5 .961538 1.886095 2.775091 3.629895 4.451822 .952381 1.859410 2.723248 3.545951 4.329477 .943396 1.833393 2.673012 3.4651 06 4.212364 .934579 1.808018 2.624316 3.387211 4.100197 .925926 1.783265 2.577097 3.312127 3.992710 .909091 1.735537 2.48(5852 3.169865 3.790787 6 7 8 9 10 5.242137 6.002055 6.732745 7.435332 8.110896 5.075692 5.786373 6.463213 7.107822 7.721735 4.917324 5.582381 6.209794 6.801692 7.360087 4.766540 5.389289 5.971299 6.515232 7.023582 4.622880 5.206370 5.746(539 6.246888 6.710081 4.355261 4.868419 5.334926 5.759024 6.144567 11 12 13 14 15 8.760477 9.385074 9.985648 10.563123 11.118387 8.306414 8.863252 9.303573 9.898641 10.379658 7.886S75 8.383844 8.852683 9.294984 9.712249 7.498674 7.942686 8.357651 8.745468 9.107914 7.138964 7.536078 7.903776 8.244237 8.559479 6.495061 6.813692 7.103356 7.366687 7.606080 16 17 18 19 20 11.652296 12.165669 12.659297 13,133939 13.590326 10.837770 11.274066 11.689587 12.085321 12.462210 10.105895 10.477260 10.827603 11.158116 11.469921 9.446649 9.763223 10.059087 10.335595 10.594014 8.851369 9.121638 9.371887 9.603599 9.818147 7.823709 8.021553 8.201412 8.3(54920 8.513564 21 22 23 24 25 14.029160 14.451115 14.856842 15.246963 15.622080 12.821153 13.163003 13.488574 13.798642 14.093945 11.764077 12.041582 12.303379 12.550358 12.783356 10.835527 11.061241 11.272187 11.469334 11.653583 10.016803 10.200744 10.371059 10.528758 10.674776 8.648694 8.771540 8.883218 8.984744 9.077040 26 27 28 29 30 15.982769 16.329586 16.663063 16.983715 17.292033 14.375185 14.643034 14.898127 15.141074 15.372451 13.003166 13.210534 13.406164 13.590721 13.764831 11.825779 11.986709 12.137111 12.277674 12.409041 10.809978 10.935165 11.051078 11.158406 11.257783 9.160945 9.237223 9.306567 9.369606 9.426914 31 32 33 34 35 17.588494 17.873552 18.147646 18.411198 18.664613 15.592811 15.802677 16.002549 16.192904 16.374194 13.929086 14.084043 14.230230 14.368141 14.498246 12.531814 12.646555 12.753790 12.854009 12.947672 11.349799 11.434999 11.513888 11 .586934 11.654568 9.479013 9.526376 9.569432 9.608575 9.644159 36 37 38 39 40 18.908282 19.142579 19.367864 19.584485 19.792774 16.546852 16.711287 16.867893 17.017041 17.159086 14.620987 14.736780 14.846019 14.949075 15.046297 13.035208 13.117017 13.193473 13.264928 13.331709 11.717193 11.775179 11.828869 11.878582 11.924613 9.676508 9.705917 9.732651 9.756956 9.779051 41 42 43 44 45 19.993052 20.185627 20 370795 20.548841 20.720040 17.294368 17.423208 17.545912 17.662773 17.774070 15.138016 15.224543 15.306173 15.383182 15.455832 13.394120 13.452449 13.506962 13.557908 13.605522 11.967235 12.006699 12.043240 12.077074 12.108402 9.799137 9.817397 9.833998 9.849089 9.862808 46 47 48 49 50 20.884654 21.042936 21.195131 21.341472 21.482185 17.880067 17.981016 18.077158 18.168722 18.255925 15.524370 15.589028 15.650027 15.707572 15.761861 13.650020 13.691608 13.730474 13.766799 13.800746 12.137409 12.164267 12.189136 12.212163 12.233485 9.875280 9.886618 9.896926 9.906296 9.914814 51 52 53 54 55 21.617485 21.747582 21.872675 21.992957 22.108612 18.338977 18.418073 18.493403 18.565146 18.633472 15.813076 15.861393 15.906974 15.949976 15.990543 13.832473 13.862124 13.889836 13.915735 13.939939 12.253227 12.271506 12.288432 12.304103 12.318614 9.92255* 9.929599 9.935999 9.941817 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. 345. Given 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^,, is $7.886875; $6000 divided by this, gives $760.76, the payment required. Rule. Assume $1 for the payment; determine the present or final value on this supposition, 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 find 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 -r- 1000 $10, the 316 RAY'S HIGHER ARITHMETIC. m 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 6f c (Art. 335), . = $23965.58 Final val. of annuity $1000 for 15 yr., at 6^, (Art. 343), = 23275.97 Balance due at end of 15 yr $689.61 Rule. Divide the present value by the payment, and look in the table, under the given rate, for the quotient; the number of years corresponding to the quotient or to the tabular 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 find the rate of interest. Rule. Divide the present value by the payment ; the quotient will be the present value of $1 for the given time and rate; look in tJie table and opposite the given number of years for the quotient or the tabular number of nearest value, and at Hie head of the column will be found the rate, or a number as near the true rate as the table can exhibit. EXAMPLES FOR PRACTICE. 1. If $9000 is paid for an immediate annuity of $750, to run 20 yr., what is the rate? About &%%. 2. If an immediate annuity of $80, running 14 yr., sells for $650, what is the rate? CONTINGENT ANNUITIES. DEFINITIONS. 348. 1. Contingent Annuities comprise Life Annuities, Dowers, Pensions, etc. 2. The value of such annuities depends upon the expecta- tion of life. 3. Expectation 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 RAY'S HIGHER ARITHMETIC. age would die during any one year, and in how many years the whole number would die. KEMARK. 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, based upon Obsemations at Carlisle (Eng.), shmving the Rate of Extinction of 10,000 lives. i < Number of Survivors. Number of Deaths. | < Number of Survivors. Number of Deaths. o b *% Number of Survivors. Number of Deaths. 10000 1539 35 5362 55 70 2401 124 1 8461 682 36 5307 56 71 2277 134 2 7779 505 37 5251 57 72 2143 146 3 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 80 953 116 11 6431 31 46 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 4458 61 84 529 84 15 6300 39 1 50 4397 59 85 445 78 16 6261 42 1 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 54 4143 70 89 181 39 ; 20 6090 43 i 55 4073 73 90 142 37 1 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 5748 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 5585 57 66 2894 123 101 7 2 32 5528 56 67 2771 123 102 5 2 83 5472 55 68 2648 123 103 3 2 34 5417 55 69 2525 124 104 1 1 CONTINGENT ANNUITIES. 319 TABLE Showing the values of Annuities on Single Lives, according to the 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. 7 per ct. 14.28164 12.083 10.439 9.177 52 12.25793 11.154 10.208 9.392 1 16.55455 13.995 12.078 10.605 53 11.94503 10.892 9.988 9.205 2 17.72616 14.983 12.925 11.342 54 11.62673 10.624 9.761 9.011 3 18.71508 15.824 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 19.59203 16.590 14.325 12.574 57 10.62559 9.771 9.027 8.375 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 14.526 12.770 60 9.66333 8.940 8.304 7.743 9 19.69114 16.742 14.500 12.754 61 9.39809 8.712 8.108 7.572 10 J 9.58339 16.669 14.448 12.717 62 9.13676 8.487 7.913 7.403 11 19.45857 16.581 14.384 12.669 63 8.87150 8.258 7.714 7.229 12 19.33493 16.494 14.321 12.621 64 8.59330 8.016 7.502 7.042 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 18.83(536 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 6.336 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 73 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 23 17.95016 15.525 13.621 12.098 75 5.23901 4.989 4.760 4.549 24 17.80058 15.417 13.541 12.037 76 5.02399 4.792 4.579 4.382 25 17.64486 15.303 13.456 11.972 77 4.82473 4.609 4.410 4.227 26 17.48586 15.187 13.368 11.904 78 4.62106 4.422 4.238 4.067 27 17.32023 15.065 13.275 11.832 79 4.39345 4.210 4.040 3.883 28 17.15412 14.942 13.182 11.759 80 4.1828!) 4.015 3.858 3.713 29 16.99683 14.827 13.096 11.693 81 3.95309 3.799 3.656 3.523 30 16.85215 14.723 13.020 11.636 82 3.74634 3.606 3.474 3.352 31 16.70-51 1 14.617 12.942 11.578 83 3.53409 3.406 3.286 3.174 32 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 34 16.21943 14.260 12.675 11.374 86 2.92831 2.830 2.739 2.652 35 16.04123 14.127 12.573 H.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 13.695 12.239 11.033 90 241621 2.339 2.266 2.198 39 15.27184 13.542 12.120 10.939 91 2.39835 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:j 14.50529 12.957 11.668 10.585 95 2.67433 2.596 2.522 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 2.49204 2.428 2.368 2.309 4t> 13.88928 12.480 11.296 10.292 98 2.33222 2.278 2.227 2.177 47 13.6(5208 12.301 11.154 10.178 99 2.08700 2.045 2.004 1.964 48 1:5.41914 12.107 10.998 10.052 100 1.65282 1.624 1.596 1.569 49 13.15312 11.892 10.823 9.908 101 1.210U5 1.192 1.175 1.159 50 12.86902 11.660 10.631 9.749 102 0.76183 0.753 0.744 0.735 51 12.56581 11.410 10.422 9.573 103 0.33051 0.317 0.314 0.312 320 RAY'S HIGHER ARITHMETIC. CALCULATIONS BY TABLE. 350. The preceding table of life annuities shows the sura to be paid by a person of any age, to secure an an- nuity of $1 during the life of the annuitant. CASE I. 351. To find the value of a given annuity on the life of a person whose age is known. Rule. Find from the table the value of a life annuity of $1, for the given age and rate of interest, and multiply it by the payment of the given annuity. REMARKS. 1. To find the value of a life-estate or widow's dower (which is a life-estate 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 ac/reed 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 Rule. Assume $1 a year for tlie annuity; find from tlie table its value for the given age and rate of interest, and divide the given cost by it; the quotient will be the 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. 5^? $92.61 3. For $840, age 58, int. Ifi't $103.03 CASE III. 353. 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. Rule. Find the present value of the annuity during its whole continuance; find its value during the given life; their difference will be the value of the reversion. NOTE. It will save work, to consider the 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 payment. 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., after the death of A, aged 62, -int. 1%. $9485.93 322 RA Y J S HIGHER ARITHMETIC. PERSONAL INSURANCE. DEFINITIONS. 354. 1. Personal Insurance is of two kinds: (1.) Life bisurance; (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.) Term Policies; (2.) Ordinary Life Policies; (3.) Joint Life Policies', (4.) Endowment Policies; (5.) Reserved Endow- 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. TABLE. Annual Premium Rates for an Insurance of $1000. LIFE POLICIES. ENDOWMENT POLICIES. Payable at death only. Payable 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 $19 89 $27 39 $42 56 $326 58 25 $103 91 $66 02 $47 68 26 20 40 27 93 43 37 332 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 30 36 46 97 359 05 30 104 58 66 77 48 53 31 23 35 31 03 47 98 366 33 31 104 75 66 96 48 74 32 24 05 31 74 49 02 373 89 32 104 92 67 16 48 97 33 24 78 3248 50 10 381 73 33 105 11 67 36 49 22 34 25 56 33 26 51 22 389 88 34 105 31 67 60 49 49 35 26 38 34 08 52 40 398 34 35 105 53 67 85 49 79 36 27 25 34 93 53 63 407 11 36 105 75 68 12 50 11 37 28 17 35 83 54 91 416 21 37 106 00 68 41 50 47 38 29 15 36 78 56 24 425 64 38 106 28 68 73 50 86 39 30 19 37 78 57 63 435 42 39 106 58 69 09 51 30 40 31 30 38 83 59 09 445 55 40 106 90 69 49 51 78 41 32 47 39 93 60 60 456 04 41 107 26 69 92 52 31 42 33 72 41 10 62 19 466 89 42 107 65 70 40 52 89 43 35 05 42 34 63 84 478 11 43 108 08 70 92 53 54 44 36 46 43 64 65 57 489 71 44 108 55 71 50 54 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 30 73 66 56 89 48 43 13 49 73 73 32 53988 48 111 01 74 54 57 96 49 45 09 51 50 75 49 553 33 49 111 81 75 51 59 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 51 78 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 7.6 119 66 768 67 63 136 20 64 96 86 98 73 124 28 785 10 64 139 52 65 102 55 104 10 129 18 801 52 65 143 16 i 324 HAY'S HIGHER ARITHMETIC. 2. Quantities considered in Life Insurance are: ' 1. Premium on^lOOO. 2. The Gain or Loss. 3. Amount of the Policy. 4. Age of the Insured. 1 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 PRACTICE. 1. W. R. Hamilton, aged 40 years, took a life policy for $5000. Required the annual premium? $156.50 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 policy? $10000. TOPICAL OUTLINE. 325 7. R. C. Storey took out an endowment policy for SI 0000 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 Wentworih 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 made 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. Topical Outline. APPLICATIONS OF 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. ( Methods [M or 1. Simple Interest. J I Rulcs 2. Five Cases... - Common. Aliquots. Six and Twelve per ct. I Exact Interest. 2. Rule. Formula. 3. Rule. Formula. 4. Rule. Formula. 5. Rule. Formula. ^ 3. Annual Interest. Rule. 326 RAY'S HIGHER ARITHMETIC. Topical Outline. (Continued.} APPLICATIONS OF PEKCENTAGE. (With Time.) 2. Partial Payments.. 1. Definitions : Payment, Indorsement. 2. Rules.. 3. Discount... 1. True Discount... 2. Bank Discount. 1. U. S. Rule. Principles. Connecticut, ' Vermont, and Mercantile Rules. 1. Definitions: Present Worth, Discount. . 2. Rule. 1. Definitions :Bank, Deposit, Issue, Check, Drafts, Drawee, Payee, In- dorsement, Discount, Proceeds, Days of Grace, Time to Ruir. 2. Four Cases, Rules. 4. Exchange-j 1. Definitions : Domestic and Foreign Exchange, Bill, Set, Rate, Course, Par, Intrinsic, and Commercial. f 1. Domestic. 2. Kinds J f L Direct Tab i e of Values. L 2 ' Foreisn " fl. Definitions :-ArbIt- [ 2. Circular. I t ion, Simple and Compound. L 2. Rules. 5. Equation of Payments... 6. Settlement of Accounts. 1. Definitions: Equated Time, Term of Credit, Average Term, Averaging Account, Clos- ing Account, Focal Date. 2 Principles. 3. Rules. 1. Definitions. 1. Accounts Current; Rule. Account Sales; Rule. Storage. 7. Compound Interest. I L Definitions :- Comp. Int., Comp. Amt. I 2. 8. Annuities.. 9. Personal Insurance. Four Cases, Rules. f 1. Definitions: Perpetual, Limited, Certain, Contingent. Im- mediate, Deferred, Forborne, Final Value, Initial Value, Present Value. I 2. Seven Cases. Rules. Table. 1. Definitions. 2. Table. XYIL PAETWEESHIP. DEFINITIONS. 356. 1. Partnership is the association of two or more persons in a business of which they are to share the profits and the losses. The persons associated are called partners; together they constitute a Firm, Company, or House. 2. The Capital is the money employed in the business ; the Assets or Resources 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 Partnership; the second, Compound Partnership. PRINCIPLE. Gains and losses are shared in proportion to the sums invested and the periods of investment. CASE I. 357. To apportion the gain or loss, when all 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. SoLimoN.-The whole 3 ^ of $ 5 4 = $ 1 3 5 0, A's share, stock is $12000, of which 4 4 54QO 1800 B's " A owns T %, B A, C.-&; j^ _/T 5400= 2250* C's " hence, by the principle 12 ' $5^00; whole, stated, A should have ^ of the gain, or $1350; in like manner, B, $1800; C, $2250. (327) 328 RAY'S HIGHER ARITHMETIC. Rule. Divide the gain or toss among the partners in pro- portion to their shares of the 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 $900. 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 C's 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 $5714.28, B $6666.66|, C $7619.04if 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's, $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 they 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, $666|; C, $500; D, $2166|. PARTNERSHIP. 329 CASE II. 358. To apportion the gain or loss when the full shares are not continued through 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 = $2 0000, A's equivalent, used 8 months, is $ 4 X 6 = 2 4 0, B's " equivalent to8X $3200X10= 32000, C's " $2500, or $20000, $ 7 6 used 1 month ; B's f f o f $ 4 7 5 = $ 1 2 5 0, A's share, capital ( $4000 ), f o f $ 4 7 5 = $ 1 5 0, B's used 6 months, is W o f $ 4 7 5 = $ 2 0, C* < equivalent to 6 X $4000, or $24000, used 1 month ; C's capital ($3200), used 10 months, is equivalent to 10 X $3200, or $32000 used 1 month. Dividing the gain ($4750) in proportion to the stock equivalents, $20000, $24000, $32000, used for the same time (1 month), the results will be the gain of each; A's $1250, B's $1500, C's $2000. Rule. Multiply each partner's stock by the time it is used; and divide the gain or 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 y, i> i; after 4 mo., A takes out |- of his ; after 9 mo. more, their gain is $1988: divide it. A, $714; B, $728; C, $546. H A. 28. 330 RAY'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. after, their gain is $3250 : what is each one's gain ? A's, $1250; B's, $750; C's, $1250. 5. A and B are partners, each contributing $1000 ; after 3 months, A withdraws $400, which B advances ; the same is done after 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 ; after 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 from 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 : after 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 yr. they quit with $2200 : how must they settle? B takes $2200, and has & 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 mo. 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. 359. 1. Bankruptcy is the inability of a person or a firm to pay indebtedness. 2. A Bankrupt is a person 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. REMARKS. 1. This act on the part of a debtor is called making an assignment, 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 JJgflf or f of the whole debts. Hence, f of $1 = GO 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, $600 ; D, $1080 : what does each get, 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 $1168. Topical Outline. PARTNERSHIP. 1. Definitions. 1. Partnership ........ ^ j ^ ase * Applications. \ Case II. Applications. l ' Definitions. Applications. 2. Bankruptcy ........ / l ' ( 2. XYIIL ALLIGATION. DEFINITIONS. 360. 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 MEDIAL. 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 Ib. of sugar, at 5 ct. a Ib, 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 =15 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 -f- 14 5f ct. per lb., Ans. Rule. Find the values of the definite parts, and divide the sum of the values by the sum of the 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. (333) 334 RAY'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? 18^ carats fine. 4. Find the specific gravity of a compound of 15 Ib. of copper, specific gravity, 7f ; 8 Ib. of zinc, specific gravity, 6J; and ^ Ib. of silver, specific gravity, 10J. 7.445 REMARKS. 1. By the specific gravity of a body is usually under- stood, its weight compared with the weight of an equal bulk of water; it may be numerically expressed as the quotient of the former by the latter. Thus, a cubic inch of silver weighing 10J 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 1J oz., its specific gravity = 1} -r- (If- 1|) = 7. (See Norton'* Natural Philosophy, p. 152.) 5. What per cent of alcohol in a mixture of 9 gal., 86% strong; 12 gal., 92^ strong; 10 gal., 95% strong; and 11 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. 362. 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.!. What relative quantities of sugar, at 9 ct. a Ib. and 5 ct. a lb., must be used for a compound, at 6 ct. alb.? SOLUTION. If you put 1 lb. OPERATION. at 9 ct. in the mixture to be sold for 6 ct., you lose 3 ct.; if you put 1 lb. at 5 ct. in the . 3 lb. at 5 ct. 1 5 ct. . 1 lb. at 9 ct. = _9 ct. 4 lb. worth 24 ct. mixture to be sold at 6 ct., which is - 2 ^ 6 ct. a lb. 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. PROBLEM. 2. What relative numbers of hogs, at $3, $5, $10 per head, can be bought at an average value of $7 per head? EXPLANATION. Writing OPERATION. the average price 7, and the Diff. Balance. Ans. particular values 3, 5, 10, as in the margin, we say : 3 sold for 7, is a gain of 4, 3 1 3 1 6 2 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 equal; hence, each losing sale must be balanced by one which gains. To lose 3 fours, will be balanced by gaining 4 threes, and a gain of 3 tu-os 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 column 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 number, fractional or integral, 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 RA Y'S HIGHER ARITHMETIC. Rule. 1. Write the particular values or prices in order, in a column, having the smallest at the head; write the average value in a middle position at the left and separated from the others by a vertical line. 2. In another column to the right, and opposite the respective values, place in order the differences between them and the average value. 3. Then prepare balance columns, giving to each of them two numbers, one an excess and the other a deficiency taken from the difference column; write each of these opposite the position which the other has in the difference column; so proceed until each number in the difference column has been balanced with another; then, The proportional quantity to be taken of each kind, ivill be the sum of the numbers in a horizontal line to the right of its excess or deficiency. NOTE. The proof of Alligation Alternate is the process of Alli- gation Medial. EXAMPLES FOR PRACTICE. 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 may 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 and 10|, will make a compound whose specific gravity shall be 16.84? 723 Ib. silver to 3487 Ib. gold. 5. What of silver f pure, and ^_. pure, will make a mixt- ure pure? 1 Ib., f pure; 5 Ib., y 9 ^ 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. 364. 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. Answers. 1.60 1.20 1.40 1.90 .40 .20 .30 1 o 3 4 5 6 7 8 9 10 19 17 15 13 11 9 7 5 3 1 14 14 14 14 14 14 14 14 14 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's 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. 20. 338 It A Y'S HIGHER ARITHMETIC. added to some number of 2's; the number of 3's taken as any one answer for the first or second kind will not be increased by any other product ; hence, if we use a fractional multiplier, 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 = - 4 3 2 , 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 " " | " " " " -V " o tt it 15 it it 3 tt it tt it ij>_ a u 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 the proportional parts, as in Case I, and ob- serve the term or terms in the balance columns, standing oppo- site the value or price corresponding to the limited quantity; then find what multipliers will produce the given limited quan- tity in the required place, and of those multipliers use only such as will agree with the 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 his average to 60%? 96 share? 2. How many bushels of hops, 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, 2 If carats fine, must he mix with it, to make it 18 carats fine? 1 pwt. 21 gr. ALLIGATION. 339 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 pt. of pure acid; the mixture has 15% more acid than desired: how much water will reduce it to the required strength? 1 gal. 2 qt. 11 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 Ib. 7| oz. 8. How many shares of stock, at 40%, must A buy, who has 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. 365. 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 2 3 16 4 3 10 12 64 24 25 50 25 38 36 26 51 22 27 64 28 kind with 2 of the 713 third, and 10 of the second kind with 3 of the third ; taken in two parts, 7 are required in one 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. 340 RAY'S HIGHER ARITHMETIC. 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 II. REMARK. Suppose that we had to determine 400 how many 7's and ll's would make 400. By 2 22 trial, we find that two ll's taken away leave an 3 7 8.. .5 4 exact number of 7's. It is now unnecessary to 77 take single ll's or proceed by trial any farther ; 9 3 01... 4 3 for, as 378 is an exact number of 7's, if we take 7 7 away ll's and leave 7's, we must take seven ll's ; ^ g 2 24.. .3 2 and thus the law of continuation is obvious : 400 is composed of ll's 2, 9, 16, 23, 30; With 7's. .....54, 43, 32, 21, 10. Rule. Proportion the parts as in Case I; then, noting the sums of the balancing columns, find, by trial or by direct di- vision, what multipliers ivill make those columns together equal to the given amount of the combination, and of those multipliers use only such as will agree ivith the remaining conditions. EXAMPLES FOR PRACTICE. 1. What quantities of sugar, at 3 ct. per Ib. and 7 ct. per Ib., with 2 Ib. at 8 ct., and 5 Ib. at 4 ct. per Ib., will make 16 Ib., w r orth 6 ct. per Ib.? f Ib. at 3 ct., 8J Ib. at 7 ct. 2. How many bbl. flour, at $8 and $8.50, with 300 bbl. at $7.50, and 800 at $7.80, 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 Ib., with 14 Ib. at 30 ct., and 20 Ib. at 50 ct., and 6 Ib. at 60 ct., will make 56 Ib. at 40 ct. a Ib.? 10 Ib. at 25 ct., and 6 Ib. at 35 ct. TOPICAL OUTLINE. 341 4. He x ;p much copper, specific gravity 7|, with silver, specific gravity 10^, will make 1 Ib. troy, of specific gravity 8f ? 7fff oz. copper, 4|| oz. silver. 5. How much gold 15 carats fine, 20 carats fine, and pure, will make a ring 18 carats fine, weighing 4 pwt. 16 gr.? 2 pwt. 16 gr.; 1 pwt.; 1 pwt. 6. A dealer in stock can buy 100 animals for $400, at the following rates, calves, $9; hogs, $2; lambs, $1: how many may he take of each kind ? 37 calves, 4 hogs, 59 lambs; (one of nine different answers.) 7. Hiero's crown, sp. gr. 14f , was of gold, sp. gr. 19^, and silver, sp. gr. 10^; it weighed 17 J Ib.: how much gold was in it? lOff Ib. Topical Outline. ALLIGATION. f 1. Definitions. 1. Alligation I C L Alligatiou Medial / L Definitions. 2 - Killds " ^ i 2. Rules. 2. Alligation Alternate.. 2. Cases.. 1. Definitions. fl. Rule. J II. Rule. I III. Rule. XIX. IN VOLUTION. DEFINITIONS. 366. 1. A power of a quantity is either that quantity itself, or the product of a number of factors each equal to that quantity. KEMARK. Kegarding unity as a base, we may say, the power of a quantity is the product arising from taking unity once as a mul- tiplicand, 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 produce the power. 3. Powers are of different 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, 5 X 5, or the 2d power of 5, is written 5 2 . 5X5X5, " 3d " " " " 5 3 . 4. The second power of any number is called the square, because the area of a square is numerically obtained by form- ing a second power. 5. In like manner the third 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 Rule. Multiply the number by itself, and continue the mul- tiplication till that number has been used as factor as many times as are indicated by the exponent. NOTES. 1. The number of multiplications will be one less than the exponent, because the root is used twice in the first multiplica- tion, once as multiplicand and once as multiplier. 2. When the power to be obtained is of a high degree, multiply by some of the powers 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 equal to the sum of their degrees. 3. Any power of 1 is 1 ; any power of a number greater than 1 is greater than the number itself : any power of a number less than 1, is less than the number itself. 368. From Note 2, 4 3 X 4 3 X 4 3 X 4 3 X 4 3 = 4 1 ' 5 ; but the expression on the left is the 5th power of 4 3 ; hence, (4 3 ) 5 =4 15 ; that is, when the exponent of the power required is a composite number (15), raise the root to a power whose exponent is one of its factors (3), and this result to a power whose exponent is the other factor (5). NOTE. Let the student carefully note the difference between raising a power to a power 9 and multiplying together different powers of the same root ; thus, 2 3 X2 2 =2 5 . Here we have multiplied the cube by the square and obtained the 5th power ; but the 5th power is not the square of the cube; this is the sixth power, and we write (2 2 ) 3 = 2 6 , or (2 3 ) 2 = 2 3X2 =2 6 . 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 cube, three times as many decimal places as the root, etc.; 344 It A Y'S HIGHER ARITHMETIC. hence, to obtain any power of a decimal, we have the follow- ing rule: Rule. Proceed as if the decimal were a whole number, and point off in the result a number of decimal places equal to the number in the root multiplied by the exponent of the power. EXAMPLES FOR PRACTICE. Show by involution, that: 1. (5) 2 2. 14 3 3. 6 5 4. 192 2 5. I 10 6. (!) 4 7. processes for squaring and cubing equals 25. 8. (|) 5 equals if-flrf- 2744. 9. (.02) s .000008 7776. 10. (5 4 ) 2 390625. " 36864. 11. (.046) s " .000097336 1. 12. (i) 7 4782969- AV 13. 2056 2 " 4227136. " Hfl- 14. (7.621) 2 " 58.1406}- 371. Special numbers. PROBLEM. Find the square of 64. PARALLEL OPERATIONS. tens + 4 units. 64 = 64 60 +4 = 6 t 60 +4 256 = 3840= 60 2 - 60X4 + 4 2 [-60X4 4096= 60 2 + 2(60X4) + 4 2 = 3600 + 480 + 16. The operations illustrate the following principle : PRINCIPLE. The square of the sum of two numbers is equal to the square of the first, plus twice the product of the first by the second, plus the square of the second. Thus : 25 2 = (22 + 3) 2 = 484 + 2 (22 X 3) 25 2 = (21 + 4) 2 =441 + 2 (21 X 4) + 16 = 625. 25 2 = (20 + 5) 2 =-400 + 2 (20 X 5) + 25 = 625. REMARK. The usual application of this principle in Arithmetic 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 w, we have the following statement : (t + u) 2 = t 2 + 2tu + w 2 ; or, in common language : The square of any number composed of tens and units is equal to the square of the tens, + twice the product of the tens by the units, + the square of 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. 2. 29. 841. 3. 4. 16. 4. 40. 1600. 5. 125. 15625. 6. 59. 3481. ILLUSTRATION. Draw a square. From points in the sides, at equal distances from one of the corners, 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 equal 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 RAY'S HIGHER ARITHMETIC. PROBLEM. Find the cube of 64. PARALLEL OPERATIONS. 64 2 = 4096= 60 2 + 2(60 X4) + 4 _ 64= _ 60 + 4 16384= 60 2 X4+2(60X4 2 ) 24576 =60 The operation illustrates the following principle : PRINCIPLE. The cube of any number composed of two parts, is equal to the cube of the first part, plus three times the square of the first by the second, plus three times the first by the square of the second, plus Hie cube of the second. Thus : 25 3 = (22 + 3) 3 = 22 s + 3 (22 2 X 3) + 3 (22 X 3 2 ) + 3 3 = 10648 -f 4356 + 594 + 27 = 15625. REMARK. The usual application of this principle in Arithmetic is, in the cubing of a number as composed of tens and units. Repre- senting the tens by t, and the units by 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 + u y = t 3 EXAMPLES FOR PRACTICE. Considering the following numbers as made, each, of two parts, cube them by the principle just stated : (1.) 19. 6859. (2.) 29. 24389. (3.) 4. 64. (4.) 40. 64000. (5.) 125. 1953125. (6.) 216. 10077696. XX. EVOLUTION. 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 Roots. 3. Roots, 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 power, 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 fractional exponent. 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 ; jX9 signifies the 5th root of 9. The square root is usually indicated without the index 2 ; thus, 1/10 is the same as "10. (347) 348 RAY'S HIGHEfi ARITHMETIC. 5. The root of a number may be indicated by a fractional exponent whose numerator is 1, and ivhose denominator is the index of the root to be expressed. Thus, 1/7 = 7 2 , and T 3// 5 = 5 1 ; similarly, 4 f = 16 T , 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 -f- 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 Root of a number is one of the two equal factors which, without any other factor, produce that num- ber ; thus, 7 X 7 = 49, and 1/49 = 7. 10. The Cube Root of a number is one of the three equal factors which, without any other, produce that num- ber; thus, 3 X 3 X 3 = 27, and 1^27 = 3. 374. Concerning powers and roots in the ordinary deci- mal notation, we state the following principles : PRINCIPLES. 1. The square of any number has tivice as many, or one less than twice as many, figures as the number itself has. 2. There will be as many figures in the square root of a perfect power as there are periods of two figures each in the power, beginning with units, and also a figure in the root cor- responding to a part of such period at the left in the power. EVOLUTION. 349 3. The cube of a number has three times as many figures, or one or tivo less than 'three times as many, as the number itself has. 4. There will be as many figures in the cube root of a perfect power, as there are periods of three figures each in the power, beginning with units, and also a figure corresponding to any part of such a period at the left hand. EXERCISES. 1. Prove 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. FIRST EXPLANATION. PROBLEM. What is the length of the side of a square containing 576 sq. in.? SOLUTION. The length required will OPERATION. be expressed by the square root of 576; 576(20 by Principle 2, we know that the root can 4 400 2 4 in., Ans. 176 176 have no less than two places of figures; 40 and since the square of 3 tens is greater, 4 and that of 2 tens less than 576, the root |~^ 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 HIGHER ARITHMETIC. 20X4 = 80 20X20=400 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 jive 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 width, the proper way to try 4, is to add it to 40 and multiply the sum by 4; thus, 40 in.+ 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 integrallyj 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 important 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 case, 4 is found the greatest figure 44 58081(241 4_ 180 176 481 481 481 EVOLUTION. 351 which can stand in ten's place, and 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, plus twice the product of the tens by the units, plus the square of the units. The square of (20 + 4), or 24 2 , is 20 2 + 2 X (20 X 4) + 4 2 . Now, if the square of the tens be taken away, there will remain 2 X (20 X 4) + 4 2 = 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 double of the tens, increased by the units, 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 1764(40 factor could be found by doubling the tens 1600 and adding the units; if, on the other 80 hand, we knew the larger factor, the units 2 could be found by direct division ; we do # 2 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 way 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. 164 42, An*. \ 64 NOTE. These successive steps showed, that the first figure was 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 tens, units, and tenths. Continuing this operation, we find 42.201895 to be the root, true as far as millionths 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 jigure. 352 RAY'S HIGHER ARITHMETIC. 375. To extract the square root of a number writ- ten in the decimal notation, as integer, fraction, or mixed number. * Rule. 1. Point off the number into periods tf two figures each, commencing with units. 2. Find the greatest square in the first period on the left; place its root on the right, like a quotient in division; subtract the square from the period, and to the remainder bring down the next period for a dividend. 3. Double the root already found, as if it were units, and write it on the left for a trial divisor ; find how often this is contained in the 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 the 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. If. any product be found 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 must be increased by a unit more than twice the number itself, to make the square of the next higher. Thus, 125 2 =15625; simply add 250+1, and find 15876=126 2 . EVOLUTION. 353 EXAMPLES FOR PRACTICE. 1856. 240. 4062. 7007. 270.194 7583.69 13. Find the square root of 3, true to the 7th decimal place. 1.7320508 14. Find the square root of 9.869604401089358, true to the 7th decimal place. 3.1415926 15. i/,030625 X 1/4O96"X 1/.00000625 what? .0028 16. ]/(126) 2 X (58) 2 X 1. i/2809. 53. 38. 109. 13625. 8944.9 2490.74 7. 1/3444736. 2. 1/1444. 8. 1/57600. 3. 1/11881. 9. 1/16499844. 4. i/185640625. 10. 1/49098049. 5. 1/80012304. 11. 1/73005. 6. 1/6203794. 12. 1/386 3 . 17. 1/12.96 X sq. rt. of ^ = 3.2863 KEMARK. The remainder, at any point, is equal to the square of an unknown part, plus twice the product of that part by a known part. The remainder may also be considered as the difference of two squares, which is always equal to the product of the sum 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, 1/16 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. y-f- = .92582+ 2. i/34f*= 5.8843+ H. A. 30. f ne arly. 3. 4. i/272&"= 161 354 RAY'S HIGHER ARITHMETIC. 5. i/6J = 2.5298+ 6. T/tf X A7 Xlf X T/T= .45886+ 7. T/123.454321 X .81 = 9.9999 8. i/1.728x4.8Xfr = - 1/2L X 1/21, written also EXTRACTION OF THE CUBE KOOT. FIRST EXPLANATION. PROBLEM. What is the edge of a cube whose solid inches number 13824? OPERATION. 13824(20 8000 4 20 2 X3 =1200 20 = 16 1456 5824 24, Ans. 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; also, 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 ; consequently, 2 is the first figure of the root, and 8000 is the greatest cube of tens contained in 13824. Let the first of the accompanying 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 ( B) 20 in. long, three of them being rectangular blocks ( ) 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., the required edge will be found. The square blocks and the oblong blocks, with the small cube, may be considered as standing in line Fig. 1. EVOLUTION. 355 (Fig. 3) and forming one oblong solid of uniform thickness. Now, as the exact solidity of that solid is 5824 cu. in., if we knew its side surface, its 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 Fig. 2. by 4 ; thus, 1200 sq. in. + 80 sq. in. X3 + 16 sq. in. = 1456 sq. in., and 1456X4 = 5824, which \ B B B p shows that 4 in. is the true thickness, and 20 in. + 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 square 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 steps, and as remarked under the rule for square root, our artifices have simply narrowed the 2 2 X300 = 12 2X4X30= 240 4 2 = 16 OPERATION. 14172488(242 range of the trial. 3. Had the power been 14172488, there would have been three figures in the root ; and here, as in the former case, the second figure is 4 ; and we may treat the 24 tens exactly as we treated the 2 tens in the former illustration. 1456 6172 5824 24X^X30= 2 2 = 1440 4 174244 348488 348488 SECOND EXPLANATION. We have seen (page 346) how the cube of a number, of two figures, is composed ; that, for example, Here we see that if the cube of the tens be taken away, there will remain 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 composed 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 EVOLUTION. 357 after the manner just described ; OPERATION. hence, multiply 2 by 3 times 74088(42 40, add 2 2 or 4, add the sum to 64 the 4800, and multiply by 2, 4 2 X300 = 4800 10088 which, being done, shows that 4 x 2X30 240 2 is the units figure, and that 2 2 = 4 42 is the root sought. 5044 10088 KEMARK. 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 one 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. Rule. 1. Beginning with units, separate the number into periods of three figures each; the extreme left period may have but 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 the greatest cube in the highest period, place its root on the right as a quotient in division, and then subtract the cube from the period, bringing down to the right of the remain- der the next period to complete a dividend. 3. Square the root found as if it were units, multiply it by 300, and place the product on the left as a trial divisor ; find how often it is contained in the dividend, and place the quo- tient figure to the right of the root ; multiply the quotient by 30 times the preceding part of the root as units, square also the quotient, and add the tivo results to the trial divisor; then mul- tiply the sum by the quotient, and subtract the product from the dividend, annexing to the remainder another period as before. 4. Square the whole root as before, multiply by 300, pro- 358 RAY'S HIGHER ARITHMETIC. ceeding as with the former trial divisor, quotient, and addi- tions; continue the operation until the remainder is nothing, or until the lowest required decimal order of the root 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, unless the remainder is large enough to contain 3 times the square of that 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. 3i 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. 379. The cube root of any product is equal to the prod- uct of the cube roots of the factors. Thus, 1^250 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. REMARKS. 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. f 512. 8. 2. 1^19683. 27. 3. i 3 /7301384. 194. 4. if 94818816. 456. EVOLUTION. 359 5. if 1067462648. 1022. 12. if 25. 13. if IT. 2.924018 2.22398 .87358 .64366 6. if 5.088448. 1.72 7. if 22188.041 28.1 14. iff- 8. if 32.65 3.196154 15. if . 9. if .0079 .1991632 16. if 171 .416328875 5.555 10. if 3.0092 1.443724 11. if^ST .315985 17. if 7011. 19.1393267 18. if-^nHhf. .2218845 .6235319 19. if | of T 4 T . EXTKACTION OF ANY BOOT. 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 3 raised to the 4th power, is 2 3X4 2 12 . Consequently, reversing that process, a power may be separated into equal factors, if the given exponent be a composite number; thus, 2 12 = 2 4X3 , and consequently i/2 = 2 3 , orf 2 TT ^=2 4 ; for 2 12 equals either 2 4 X 2 4 X 2 4 , or 2 3 X 2 3 X 2 3 X 2 3 . 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, 2 5 =2 3 X2 2 because equal to 2 3 + 2 . But the square root of 2 5 is not 2 3 , nor is the cube root of 2 5 equal to 2 2 . If 2 3 be taken twice as a factor we have 2 6 , and 1/2 6 = 2 3 ; similarly, 382. A root of any required degree may be extracted, by separating the number denoting the degree, into its factors, and extracting successively the 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 ARITHMETIC. HORNEK'S METHOD. 383. Horner's Method, named from its inventor, Mr. "W. G. Homer, of Bath, England, may be advantageously applied in extracting any root, especially if the degree of the root be not a composite number. Rule for Extracting any Root. 1. Make as many columns as there are units in the index of the root to be 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 3d column, and so on, placing the last product in the right-hand column, under that part of the gwen 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 next 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 ; and so on, till all the periods have been brought down. EVOLUTION. 361 PROBLEM. Extract the fourth root of 68719476736. 5 5 10 5 25 50 125 375 68719476736(512 Am. 625 75 75 * 500000 15201 * 621947 515201 15 5 *15000 201 515201 15403 * 1067466736 1067466736 *200 15201 202 *530G04000 3129368 201 15403 203 533733368 202 1 *1560600 4084 203 1 1564684 *2040 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 this Contracted Method. Obtain one less than half of the figures required in the root as the rule directs; then, instead of annexing ciphers and bringing down a period to the last numbers in the columns, leave the remainder in the right-hand column for a dividend; cut off the right-hand figure from the last number of the previous column, tivo right-hand figures from the last number in the column before that, and so on, always cutting off one more figure for every column to the left. With the number in the right-hand column and the one in the previous column, determine the next figure of the root, and use it as directed in the ride, recollecting that the figures cut off are not used except in carrying the tens they produce. H. A. 31. 362 KAY'S HIGHER ARITHMETIC. This process is continued until the required number of figures is obtained, observing that when all the figures in the last number of any column are cut off, that column ivill 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 the 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 REMARK. The trial divisors may be known by ending in two ciphers; the complete divisors 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. 125. 2. Extract the cube root of 68719476736. 4096. 3. Extract the fifth root of 14348907. 4. The cube root of 151. 5.325074 5. J/9TAL 3.1416 6. ^L08 1.01943 7. T^S- .83938 8. j/35^~ 2.03848 9. ^782757789696. 9.79795897 10. ^1367631. 4.8058955 EVOLUTION. 363 APPLICATIONS OF SQUARE ROOT AND CUBE ROOT. DEFINITIONS 385. 1. A Triangle is a figure which has three sides and 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 equal to 180, or two right angles. The proof of this belongs to Geometry, but a fair illustration 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 edge through O, while the angles cover just twice 90, or EOD + FOD. Any angle less than 90, as HOF, is an acute angle; any angle greater than 90, as HOE, is an obtuse angle. 4. An Equilateral Triangle is a triangle having three equal sides ; as, MNP. REMARK. 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 It AY'S HIGHER ARITHMETIC. EIGHT-ANGLED TRIANGLES. 388. 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 perpendieular. 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 80 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. Rule. Add together the squares of the base and perpendicular, and extract the square root of the sum. 388. To find one side when the hypothenuse and the other side are given. Rule. Subtract the square of the given side from the square of the hypotiienuse, and extract the square root of Hie differ- ence; or, Multiply the square root of ilie sum of the hypotJienuse and side by the square root of their difference. Representing the three sides by the initial letters h, p, b, we have the following FORMULAS. 1. h = Vp* + b^. 2. p = Vh 2 b* ; or, p = Vh + b X v'T^b. 3. b = Vh 2 p 2 ; or, b = Vh-^-p X Vh p. EXAMPLES FOR PRACTICE. 1. Find the length of a ladder reaching 12 ft. into the street, from a window 30 ft, high. 32.31 + ft. 2. What is the diagonal, or line joining the opposite cor- ners, of a square whose side is 10 ft. ? 14.142 -f- 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 RAY'S HIGHER ARITHMETIC. REMARK. Integers expressing the sides of right-angled triangles may be found to any extent in the following manner : Take any two unequal numbers ; the sum of their squares may represent a hypoth- enuse ; the difference of their squares will then stand for one side, and double their product for the remaining side. Thus, from 3 and 2, form 13, 12, 5 ; from 4 and 1, form 17, 15, 8 ; from 5 and 2, form 29, 21, 20. PARALLEL LINES AND SIMILAR FIGURES. DEFINITIONS. 389. 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. REMARKS. 1. Similar figures have their corresponding 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 off, is similar to the whole triangle. An illustration of this has 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 J. 2. The diagonals of two similar rectangles are as 5 to 12 : how many times does the larger contain the smaller? EVOLUTION. 367 3. The lengths of two similar solids are 4 in. and 50 in. ; the 1st contains 16 cu. in.: what 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. 10^ 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 f of itself makes 504? 42. 7. Separate 91252^ into three factors which are as the numbers 1, 2A, and 3*1 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 corner 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 li ft., and how many allowing no two to be nearer than 1J 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 RAY'S HIGHER ARITHMETIC. compounded through the year? Prove the result by con- tracted multiplication. (Art. 334, Kern. 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 1. Involution..., 2. Evolution . Topical Outline. POWERS AND ROOTS. 1. Definitions. 2. Terms.. 3. Squaring and Cubing.... 1. Definitions. 2. Terms, 1. Powers. 2. Root. 3. Degree. . 4. Exponent. 1. Algebraic Statements. 2. Numerical Illustrations. 3. Geometrical Illustration. . 4. Principles. Perfect. Imperfect. 1. First explanation (Geometrical). 3. Square Root -j 2. Second explanation (Algebraic). [ 3. Rule. 1. First explanation (Geometrical). 4. Cube Root -j 2. Second explanation (Algebraic). 3. Rule. 5. Roots in General Homer's Method. 6. Applications. XXI. SEEIES. 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 Geometrical; these series are commonly called Progres- sions. AKITHMETICAL PROGKESSION. 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 first term is the least or the greatest. Thus, 1, 3, 5, 7, 9, is an ascending 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, I. Number of terms, . . . . n. Common difference, d. Sum of all the terms, . . s. (369) 370 RAY'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 as a convenience. 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. din", is 3, and the terms are composed thus: 1, 1 + 3, 1+6, 1 + 9, etc.; and it is obvious that, as the addition of the com. difF. commences in forming the second term, 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, 1 + (20 1)X3 58, Ans. FORMULA. I = a + (n I)c7; or I a (n l)d. Rule. Multiply the number of terms less one by the com- mon difference, add the product to the 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|, 5f , etc. 145J. 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. 97J. 7. Find the first term of the series 18f , 12|, 6, 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 14- (9 _ 1 ) X 3 25 in full, in the common order, and also in a reverse order, we have "j" \/ 9 __ jjy ^^ 2 Sum = 1+4 + 7 + 10 + 13 + 16 + 19 Sum = 25 + 22 + 19 + 16 + 13 + 10 + 7 + 4 + 1. Twice the sum = 26 + 26 + 26 + 26 + 26 + 26 + 26 + 26 + 26 = 9 times the sum of the extremes; /. the sum J of 9X26 117, Am. 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. s (a + /) n. ~2~ "Rule. Multiply the sum of the extremes by the number of terms, 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 HIGHER ARITHMETIC. 2. Extremes, 100 and .0001: number of terms, 12345. 617250.61725 3. What is the sum of the arithmetical series 1, 2, 3, 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? 3178f. EXAMPLES FOR PRACTICE. REMARK. 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. 1J. 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. 21, 2|, 2|, 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 10^, and common differ- ence, ? 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 be known in order to find the others. These five are thus represented : First term, a. Last term, ...'.../. Ratio, 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 7 X 1 = 1 2 8, Ans. may be formed thus: 1, 1X2, 1 X 2 2 , 1X2 3 , etc., the ratio being raised to its second power in forming the 3d term, to its third power in forming the 4th ; and so, similarly, the 8th term = 1X2 7 = 128, Ans. FORMULA. I = ar n -l. Rale. Consider the given extreme as the first term, and multiply it by 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^. 2. In 2, 5, 12 J, etc., having 6 terms. 195 T V 3. In 100, 20, 4, etc., having 9 terms. 15 | 25 . 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. 5f. 8. Of ^_%> IHfc havin n terms - TT> s- CASE II. 396. Prom 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. OPERATION. EXPLANATION. Writing the 4 X3X4 4 3 . - - = 1 u L 6, whole series, we have : 4 1 SERIES. 375 S = 3 + 12 + 48 + 192 + 768. Also, 4 S = 12 + 48 + 192 + 768 + 3072. It is evident that the lower line exceeds the upper by the difference between 3072 and 3; this difference may be written 4X768 3,N)r 4 X 3 X 4 4 3, and as this is 3 times the series, we have, once the series = /1X/OK//14. O -=1023, Ans. 41 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 + 192 + 48 + 12 + 3 = S, 1 Qf) I A O [ 1 f) | O | 3 1 Q . ]L\Ji ~r~ TtO ~T~ -L^ ~r~ O r~ x 2T ^ > and thus 768 f = f of the series ^ hence, we can write, 768 j:X3^ x 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 Rule I. Find the last tervi 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. a rl a Q a rl FOKMULA. 8= ; or, S = . r 1 1 r The first answer, above given, may take another form, thus : - 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, has the following general statement : S = =!) and corresponds to the following rule ; 376 RAY'S HIGHER ARITHMETIC. Rule II. Raise the ratio to a power denoted by the num- ber of terms, subtract 1, divide the remainder by the ratio less 1, and multiply the quotient by the first term. NOTES. 1. The amount of a debt at compound interest for a num- ber of complete intervals, is the last term of a geometrical progression, whose ratio is 1 + the rate per cent. The table (Art. 335) shows the powers of the ratio. For example, the period being 4 yr., the number of terms is five; the first term is the principal, and the power of the ratio required by Case 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 : $20Q(1.06 40 1) $200X9.2857179 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. 32767ff 3. Of f , |, 2- 8 T , etc., to 7 terms. liftf Find the sum of the following infinite geometrical series : 4. Of 1, i i, etc. 2. 5- Of f , A, ^ etc. 1|. 6. Of , f , &, etc. 2. 7. Of , 1, f etc. 81 8. Of .36 = .3636, etc. .= T \ 6 o + Trftfinr, etc - T 4 y 9. Of .349206, of 480, of 6. ff and ff and f. 10. Find the amount of an annuity of $50, 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 necessary to give special rules for finding the ratio, and the number of terms when these are unknown ; so 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 -^-g^, and -J-. TeVs' 5T6"> T2- 7. Two geometric means between 14.08 and 3041.28 84.48 and 506.88 Topical Outline. SERIES. f 1. Definitions. Terms, Law, Extremes. Means. Series.... f L Terms - 1. Arithmetical... J 2. Cases. 2. Classes... j (Formulas.) f 1. Terms. . 2. Geometrical.... J ^ Case& I (Formulas.) H. A. 32. XXII. MENSUEATIOK. DEFINITIONS. 397. 1. Geometry is that branch of mathematics which treats of quantity having extension and form. When a quantity is so considered, it is called Magnitude. 2. There are four kinds of Magnitude known to Geom- 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. (378) 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 oblique angles. (See Art. 385, 3, Rem., and Art. 386.) SURFACES. POLYGONS. 400. 1. A surface 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 side of a limited surface, or the edye 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- r^^\ N gle ; having four sides, a quadrilateral ; five V sides, a pentagon; six sides, a hexagon, etc. P Quadrilateral. 380 RA Y J S HIGHER ARITHMETIC. The six diagrams following represent regular polygons. Pentagon . Hexagon . Heptagon. Octagon ., Nouagon. 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 Triansrles. Obtuse Triangles. Scalene. Isosceles. Equilateral. Isosceles. 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. 381 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. Rhomboid. Rhombus. 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 Rhombus is a parallelogram whose four sides are equal. 3. A Rectangle is a parallelogram whose angles are all right angles ; when the rectangle has four equal sides, it is a square. A square is a rhomhns whose angles are 90 ; it is also the form of the unit for surface measure. It may properly be denned, an equilateral rectangle. Square. 382 RAY'S HIGHER ARITHMETIC. AKEAS. TRIANGLES AND QUADRILATERALS. 404. The general rules depend on the principles stated in the following remarks: EEMAKKS. 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, LGI 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 base, 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 whole area would be the same, but the perimeter would be inweased, and the figure would be a rhomboid. Different quadrilaterals may have equal areas and unequal 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. MENSURATION. 383 GENERAL RULES. 1. To find the area of a parallelogram. Rule. Multiply one of two parallel sides by the perpendicular distance between them. II. To find the area of a triangle. Rule. Take half the product of the base by the altitude. III. To find the area of a trapezoid. Rule. Multiply half the sum of the parallel sides by the altitude. NOTE. The following is demonstrated in Geometry : IV. To find the area of a triangle when the sides are given. % Rule. Add the three sides together and take half the sum ; from the half sum take the sides separately ; multiply the half sum and the three remainders together, and extract the square root of the product. NOTES. 1. The area of a trapezium may be found by applying this rule to the parts when the sides are known and the diagonal is given in length and in special position as between the sides. The area of any polygon 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 its 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 rcl. 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. rcl. 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-|f 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.? 215 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' 19J 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^ 7 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 fU 3 in. high. $60.56 REGULAR 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 half the apothem. All regular polygons of the same number of sides are similar figures. (Art. 389, Rein. 1.) 407. 1. The circle, 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. IT. A. r,. 386 RAY'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, arid 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 composed of six equilateral triangles, and hence if OB be 1, the side AB 1, and it is easy to compute the apothem, V 1 \ ~ .86602540378 MEN8UEA WON. 387 2. If the distance from center to vertices be unchanged, and a regular polygon of twelve sides be formed about the same center, it will differ less 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. 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 8th 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 i instead of 1, the results will be 3.141590+ and 3.141592 + 388 RAY'S HIGHER ARITHMETIC. Hence, if the circle of diameter 1, be taken as a polygon of 1536 sides, and then as 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 ^ and area = \ circum- ference X diameter, we have area = j- X square of diameter. Rep- resenting the circumference by c, area by CJ diameter by (/, radius by R, we have the following formulas: W 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 the area by J of the diameter; or, 3. Extract the square root of 12.56637 times the area. II. To find the diameter: 1. Divide the circumference by 3.1415926; or, 2. Divide the area by .785398, and extract the square root. III. To find the area: 1. Multiply the diameter by J of the circumference; or, 2. Multiply the square of the diameter by .785398; or, 3. Multiply the square of the radius by 3.1415926 MENSURA TION. 389 IV. To find the area of a sector of a circle : 1. Multiply the arc by one half the radius; or, 2. Take such a fraction of the whole area as the arc is of the whole 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, the area of the triangle whose base is the base of the segment, and whose vertex is the center of the circle; or, 2. Divide the cube of the height by twice the base, and increase the quotient by two thirds of the product of height and base. REMARK. Add the triangle to the sector, 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 1/2 is to 1. 2. The side of an inscribed equilateral triangle is to radius as V 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, 1821, 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 RA Y'S HIGHER ARITHMETIC. 5. Circum. 47.124 ft., diameter 15 ft. 176.715 sq. ft. 6. If we saw down through \ 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 is the largest squared stick which can be cut out of it? .6366 SOLIDS. DEFINITIONS. 410. 1. A Solid is that which has length, breadth, and thickness. A solid may have plane surfaces, curved surfaces, or both. A cumed surface is one no part of which is ji plane. 2. The faces of a solid are the polygons formed by the intersections of its bounding planes ; 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 prism is triangular, quadrangular, etc., according to the shape of its base. The first of the figures here given represents a quadran- gular prism, the second a pentagonal prism. 4. A right prism is a prism whose faces are rectangles. 5. A Parallelepiped is a prism whose faces are parallel- ograms. Its bounding surfaces are six parallelograms. The first figure above represents a parallelepiped whose faces are rectangles. MENSURATION. 391 6. A Cube is a parallelepiped whose faces 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. 8. 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 frustum of a pyramid or cone is the solid which remains when a portion having the vertex is cut off by a plane parallel to the base. 11. A Sphere is a solid bounded by a curved surface, every point of which is at the same distance from a point within, called the center. The diameter of a sphere is a straight line passing through the center and having its ends in the surface; the radius is the distance from the center to the surface. A segment, of a sphere is a portion cut off by one plane, or between two planes; its bases are cirdes, and its height is the portion of the diameter which is cut off with it. 12. The slant height 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 the 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 propor- tional, and their corresponding angles equal. GENERAL RULES. I. To find the convex surface of a prism or cyl- inder : Rule. Multiply the perimeter of the base by the altitude. II. To find the volume of a prism or cylinder : Rule. Multiply the base by the altitude. III. To find the convex surface of a pyramid or cone: Rule. Multiply the perimeter of the base by one half the slant height. IV. To find the volume of a pyramid or cone: Rule. Multiply the base by one third of the altitude. V. To find the convex surface of a frustum of a pyramid or cone : "Rale. Multiply half the sum of the perimeters of the bases by the slant height. VI. To find the solidity of a frustum of a pyramid or cone: MENSURATION. 393 Rule. To the sum of the two bases add the square root of their product, and multiply the amount by one third of the alti- tude. VII. To find the surface of a sphere : Rule. Multiply tJie circumference by the diameter. VIII. To find the volume of a sphere: Rule. Multiply Hie cube of the diameter by .5235987 NOTES. 1. Similar solids are to each other as the cubes of their like dimensions. 2 The sphere is regarded as composed of an infinite number of cones whose common altitude is the radius, and the sniii of whose bases is the whole surface of the sphere. 3. The cone is regarded as a pyramid 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^ in., and sides of base 5^, 6^, 8J, 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 KAY'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. 94^ 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^ 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, 2J, 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 3J in., lower base 4 in. square, upper base 2| in. square. 43J sq. in. 15. The convex surface and whole surface of 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. 4| in. ; lower base, lOf in. square ; upper, 4^ in. square. 974^f| 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 -f sq. ft. and 314.16 sq. in. MENSURATION. 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. 755499J 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 of it? .57735 MISCELLANEOUS MEASUEEMENTS. MASONS' AND BRICKLAYERS' WORK. 411. Masons' work is sometimes measured by the cubic foot, and sometimes by the perch. The latter is 16^- ft. long, 11 ft. wide, and 1 ft. deep, and contains 16^- X 1-i- X 1 = 24| cu. ft., or 25 cu. ft. nearly. 412. To find the number of perches in a piece of masonry. Rule. Find the solidity of the 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 T ^ 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, 8J 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. 9.39 396 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, allowing 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 T ^ for mortar ? $289.17 5. How many bricks of ordinary size will build 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. SUGGESTION. Find the solidity of the whole chimney, then of the hollow part ; the difference will be the solid part of the chimney. GAUGING. 413. Gauging 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, frustum of a cone, etc. Rule. Find the solidity of the vessel in cubic inches by the rules already given; this divided by 2150.42, will give the con- tents in bushels; by 231, will give it in wine gallons, which may be reduced to barrels by dividing the uuniber 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 tlw diameter by .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. REMARK. 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: Rule. Add to the head diameter (inside) two thirds of tlie difference between the head and bung diameters ; but if the staves are only slightly curved , add six tenths of this difference; this gives the mean diameter ; express it in inches, square it, multiply it by the length in inches, an$ this product by .0034: the product will be the contents in wine gallons. NOTE. After 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 head, 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 RAY'S HIGHER ARITHMETIC. LUMBER MEASURE. 416. To find the amount of square-edged inch boards that can be sawed from a round log. REMARK. The following is much used by lumber-men, and is sufficiently accurate for practical purposes. It is known as Doyle's Kule. Rule. From the diameter in inches subtract 4; the square, of the remainder wilt be the number of square feet of inch boards yielded by a log 16 feet in length. EXAMPLES FOR PRACTICE. 1. How 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 ft = 080 feet. Or, f X 28 X 28 = 980 feet. 2. In a log 24 in. in diameter, and 12 ft. long? 300 ft. 3. 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. REMARKS. 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. MENSURATION. 399 418. To find the quantity of grain in a wagon or in a bin: Rule. Multiply the contents in cubic feet by .8 EEMABKS. 1. If it be corn on the cob, deduct one half. 2. For corn not u shucked," deduct two thirds for cob and shuck. 419. To find the quantity of hay in a stack, rick, or mow : Rule. Divide the cubical contents in feet by 550 for clover, or by 450 for timothy; the quotient will be the number of tons. EXAMPLES FOR PRACTICE. 1. How many bushels of shelled corn, or corn on the cob, or corn not shucked, will a wagon-bed hold that is 10^ feet long, 3J feet wide, and 2 feet deep? 58.8 bti.; 29.4 bti. ; 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 T 3 T tons clover ; 106-| tons timothy. Topical Outline. MENSURATION. 1. General Definitions : Geometry, Magnitude, Mensuration. f Parallel. Perpendicular. I Straight \ Horizontal. 2 - Lines \ Broken. | Vertical. Curved. Diagonal. 400 JRAY'JS HIGHER ARITHMETIC. MENSURATION. (Continued. ) 3. Angles . f Right. \ Oblique 4. Surfaces. 5. Solids Right. ie / Acute. \ Obtuse. 1. General Definitions: Plane, Plane figure, Area, Polygons,' Regular, Perimeter, Similar, Center, Altitude, Base, Apothem. 1. Right / IsosceU *' I Sc 2. Triangles... (Rules.) 1 2. Oblique- Obtuse , Scalene. Acute { Scalene, j Isosceles. 3. Quadrilateral.. (Rules.) 4. Circle. Equilateral. Scalene. Isosceles. 1. Trapezium. 2. Trapezoid. [" L Rhomboid. 3. Parallelogram. J 2 . Rhombus... \ Square. ! 3. Rectangle.. .j Square. 1. Terms: Circumference, Radius, Chord, Diam- eter, Segment, Sector, Tangent, Secant. 2. Calculations, Value of IT. I 3. Formulas, Rules. 1. General Definitions: Solid, Base, Face, Edge, Similar. 2. Prism (Rules.) Triangular. (Right.) Quadrangular < j Pentagonal, etc. c Altitude. 3. Pyramid J giant Height. (Rules.) (Frustum.) ( Convex Surface. {Altitude. Slant Height. Convex Surface. f Surface. Parallelepiped. c Cube (Right.) 5. Cylinder (Rules.) \ Solidity. 6. Sphere.. 7. General Formulas. 8. Miscellaneous Applications (1. Terms : Radius, Diameter, Seg- ment. 2. Convex Surface, Rule. 3. Solidity, Rule. ' Masons' and Bricklayers' Work. Gauging. Lumber Measure. . Measuring Grain and Hay. XXIII. MISCELLANEOUS EXERCISES. NOTE. Nos. 1 to 50 are to be solved mentally. 1. If I gain J 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 37 \ 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 ct. per bu. 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 1? 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 6J ct. per dozen. 6. If I sell eggs at 6 ct. per dozen, I lose f- ct. apiece; how much per dozen must I charge to gain f ct. apiece? 22 ct. 7. One eighth of a dime is what part of 3 ct.? $. 8. If I lose | of my money, and spend f of the remainder, what part have I left? Jf . 9. A's land is ^- less in quantity than B's, but jV better in quality: how do their farms compare in value? A's = T % of B's. 10. If f of A's money equals f of B's, what part of B's equals | of A's? f. 11. I gave A f\ of my money, and B j 7 ^ 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 f older than C: how many times C's age is A's? 2J. 13. Two thirds of my money equals f of yours; if we put our money together, what part of the whole will I own? T 6 r . 14. How many thirds in J? 1|. 15. Reduce f to thirds; f to ninths; and f to a fraction, whose nu- o merator shall be 8. 2 ? thirds ; 7 A ninths ; ' i o i loj. 16. What fraction is as much larger than f as f is less than f ? -}|. 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 RAY'S HIGHER ARITHMETIC. 18. In 12 yr. I shall be | of my present age: how long since was J 5 of my present age? 8f yr. 19. Four times f of a number is 12 less than the number; what is the number? 108. 20. A man left T 5 T 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 f less than it cost him : what did it cost me? $8. 22. A is f 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 J yr. ; B, 22 yr. 23. A pole was f 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 T 9 ^ as old as B; how old is each? A, 20 yr.; B, 26* yr. 25. A's money is $4 more than f of B's, and $5 less than f 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 3^ yr. the older: how old is each? A, 31 i 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| hr. 28. If 6 men can do a work in 5J days, how much time would be saved by employing 4 more men? 2J 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 3f hr., it takes them 9J hr. ; in what time can each do it? Man, 13Jhr.; boy, 20 hr. 31. Five men were employed to do a work ; two of them failed to come, by which the work was protracted 4J 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? 4J 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 4 days; after working 2 days, how many must join them so as to complete it in 3| da.? 4 men. 35. Eight men can do a certain amount of work in 6f days; after beginning, how soon must they be joined by 2 more so as to complete it in 5J days? In 2f days. MISCELLANEOUS EXERCISES. 403 36. Seven men can build a wall in 5J days; if 10 men are employed, what part of his time can each rest, and the work be done in the same time? T V 37. Nine men can do a work in 8} 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 2^ 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} min. after 6, and 54 T 6 T min. 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 T 4 F min. after 4; and 49 T 1 T 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^ m i n - 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 21A-50m-8,'61 (Cl795slO)476B .General Library University of California Berkeley Id I /<4UO