^gSs*r^t«*2 ^t\a |iork g)tate College of SgtftuWure at Cornell iHnibetsitp Stbaca, B. I9- itiftrarp Cornell University Library QA 43.F45 A mathematical solution book.Containing 3 1924 002 953 119 Cornell University Library The original of tiiis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924002953119 A MATHEMATICAL SOLUTION BOOK. CONTAINING SYSTEMATIC SOLUTIONS OF MANY OF THE MOST DIFFICULT PROBLEMS Taken from the Leading Authors on Arithmetic and Algebra, Many Prob- lems and Solutions from Geometry, Trigonometry, and Calculus, IWany Problems and Solutions from th'e Leading Math- ematical Journals of the United States, and Many Original Problems and Solutions, NOTES AND EXPLANATIONS. BY B. F. FINKEL, A.M., M.Sc.^ Member of the London Mathematical Society, Member of the American Mathematical Society, Editor of the American Mathematical Monthly, and Professor of Mathematics and , Physics in Drury College. FOURTH EDITION— REVISED AND ENLARGED, KIBLER & COMPANY. PUBLISHERS. Springfield, Mo. Copyright, 1S88, BY B. F. FINKEL, in the office op the lrlerarian of congress, Washington, D. C. DEDICATED TO MY FRIKND, H. S. I.EHR, A. M., Ph. D. PRESIDENT OF THE OHIO NORMAL UNIVERSITY. PREFACE TO THE FIRST EDITION. This work is the outgrowth of a number of years' experience in teaching in the Public Schools, during which time I have ob- served that a work presenting a systematic treatment of solutions of problems would be serviceable to both teachers and pupils. It is not intended to serve as a key to any work on mathe- matics ; but the object of its appearance is to present^ for use in the schoolroom, such an abcurate and logical method of solving problems as will best awaken the latent energies of pupils, and ' teach them to be original investigators in the various branches of science. It will not be denied by any intelligent educator that the so- called "Short Cuts" and "Lightning Methods" are positively in- jurious to beginners in mathematics. All the "whys" are cut out by these methods and the student robbed of the very object for which he is studying mathematics ; I'is., the development of the reasoning faculty and the power to express his thoughts in • a forcible and logical manner. By pursuing these methods, • mathematics is made a mere memory drill and when the memory fails, all is lost ; whereas, it should be presented in such a way as to develop the memory, the imagination, and the reasoning fac- ulty. By following out the method pursued in this book, the mind will be strengthened in these three powers, besides a taste for neatness and a love of the beautiful will^ be cultivated. ' Any one who can write out systematic solutions of problems can resort to "Short Cuts" at pleasure ; but, on the other hand, let a student who has done all his work in mathematics by form- ulae, "Short Cuts," and "Lightning Methods" attempt to write out a systematic solution — one in which the work explains itself — and he will soon convince one of his inability to express' his thoughts in a logical manner. These so-called "Short Cuts" should not be used at all, in the schoolroom. After pupils and students have been drilled on the systematic method of solving problems, they will be able to solve more problems by short methods than they could by having been instructed in all the "Short Cuts" and "Lightning Methods" extant. It can not be denied that more time is given to, and more time wasted in the study of arithmetic in the public schools than vi ■ PREFACE. in any other branch of study ; and yet, as a rule, no better results are obtained in this branch than in any other. The reason of this, to my mind, is apparent. Pupils are allowed to conibine the numbers in such a way as "to get the answer" and that is all that is required. They are not required to tell why they do this, or why they do that, but, "did you get the answer?" is the question. The art of "ciphering" is thus developed at the ex- pense of the reasoning faculty. The method of solving problems pursued in this book is often called the "Step Method." But we might, with equal pro- priety, call any orderly manner of doing any thing, the "Step Method." There are only two methods of solving problems — a right method and a wrong method. That is the right method which takes up, in logical order, link by link, the _ chain of rea- soning and arrives at the correct result. Any other method is wrong and hurtful when pursued by those who are beginners in mathematics. One solution, thoroughly analyzed and criticised by a class,- is worth more than a dozen solutions the difficulties of which are ■ seen through a cloud of obscurities. This book can be used to a great advantage in the class- room^the problems at the end of each chapter affording ample exercise for supplementary work. Many of the Formulae in Mensuration have been obtained by the aid of the Calculus, the operation alone being indicated. This feature of the, work will not detract from its merits for those persons who do not understand the Calculus ; for those' who do understand the Calculus it will afford an excellent drill to work out all the steps taken in obtaining the formulae. Many of the formulae can be obtained by elementary geometry and algebra. But the Calculus has been used for the sake of presenting the beauty and accuracy of that powerful instrument of mathematics. In cases in which the^formulse lead to series, as in the case of the circumference of the ellipse, the rule is given for a near approximation. It has been the aim to give a solution of every problem presenting anything peculiar, and of those which go the rounds of the country. Any which have been omitted will receive space in future editions of this work. The limits of this book have compelled me to omit much curious and valuable matter in Higher Mathematics. I have taken some problems and solutions from the School Visitor, published by John S. Royer; the Mathematical Maga- zine, and the Mathematical Visitor, published by Artemas Mar- tin, A.' M., Ph. p., LL. D. ; and the Mathematical Messenger, published by G. H. Harvill, by the kind permission of these distinguished gentlemen. PREFACE. vii It remains to acknowledge my indebtedness to Prof. William Hoover, A. M., Ph. D., of the Department of Mathematics and Astronomy in the Ohio University at Athens, for critically read- ing the manuscript of the part treating on Mensuration, and to William G. Williams, LL. D., Wright — Professor of the Greek Language and Literature in the Ohio Wesleyan Univer«it-v. for his aid in exten4ing the names of polygons one pages 236-237. Hoping that the work will, in a measure, meet the object for which it is written, I respectfully subrnit it to the use of my fellow teachers and co-laborers in the field of mathematics. Any correction or suggestion will be thankfully received by communicating the same to The Author. PREFACE TO SECOND EDITION. In bringing out a second edition of this work, I am greatly indebted to Dr. G. B. M. Zerr for critically reading the work with a view to eliminating all errors. The Author. Drury College, Feb. ip, i8gy. PREFACE TO THIRD EDITION. The hearty reception accorded this book, as is attested by the fact that two editions of 1,200 copies 'each have already been sold,, encouraged me to bring out this third edition. In doing so, I have availed myself of the opportunity of making some important corrections, and such changes and im- provements as experience and the suggestions of teachers using the book have dictated. The very favorable comments on the- work by some of the most eminent mathematicians in this country confirm the opinion that the book is a safe one to put into the hands of teachers and sttidents. While mathematics is the exact science, yet not every book that is written upon it treats of.it as though it were such. In- deed, until quite recently, there were very few books on Arith- metic, Algebra, Geometry, or Calculus that were not mere copies of the works written a century ago, and in this way the method, the spirit, the errors and the solecisms of the past two hundred years were preserved and handed down to the present genera- tion. At the present time the writers on these subjects are breaking away from the beaten paths of tradition, and the re- sult, though not wholly apparent, is a healthier and more vig- orous mathematical philosophy. Within the last twenty-five vui PREFACE. years there has set in, in America, a reaction against the, spirit and the method of previous generations, so that C. A. Laisant, in his La Mathematique; Philosophie. — Enseignement, Paris, 1898, says, "No country has made greater progress in mathe- matics during the past twenty-five years than the United States." Most of the 'text-books on Arithmetic*, Algebra, Geometry, and the Calculus, written within the la^t five years, are evidence of this progress. . ' The reaction spoken of was brought about, to some ex- tent, by the introduction into our higher institutions of learn- ing of, courses of study in mathematics bearing on the wonder- ful researches of Abel, Cauchy, Galois, Riemann, Weierstrass, and others. This reaction, it may be said, started as early as 1832, the time when Benjamin Peirce, the first American worthy to be' ranked with Legendre, Wallis, Abel, and the BernouiHis, became professor of mathematics and natural philosophy at Harvard University. Since that time the mathematical courses in our leading Universities have been enlarged and streng'thened, until now the opportunity for research work in mathematics as offered, for example, at the Universities of Chicago, Harvard, Yale, Cornell, Johns Hopkins, Princeton, Columbia and others, is as good as is to be found anywhere in the world. For ex- ample, the following are the subjects, offered at Harvard for the Academic year 1899- 1900: Logarithms, Plane and Spherical Trigonometry; Plane Analytical Geometry; Plane and Solid Analytical Geometry ; Algebra ; Theory of Equations. — Invar- iants ; Differential and Integral Calculus ; Modern Methods in Geometry. — Determinants ; Elements of Mechanics ; Quater- nions with application to Geometry and Mechanics ; Theory of Curves and Surfaces; Dynamics of, a Rigid Body ; Trigonomet- ric Series. — Introduction to Spherical Harmonics. — Potential Function ; Hydrostatics. — Hydrokinematics. — Force Functions and Velocity-Potential Functions and their uses. — Hydroki- .netics ; Infinite Series and Products ; The Theory of Functions ; Algebra. ^ — Galois's Theory of Equations ; Lie's Theory as ap- plied to Differential Equations ; Riemann's Theory of Func- tions ; The Calculus of Variations ; Functions Defined by Linear Differential Equations ; The Theory of Numbers ; The Theory of Planetary, Motions ; Theory of Surfaces ; Linear Associative Algebra ; the Algebra of Logic ; the Plasticity of the Earth ; Elasticity ; and the Elliptic and the Abelian Transcendants. While great activity and real progress in mathematics 'is going on in our higher institutions of learning, a like degree of activity is not yet being manifested in many of our colleges and academies and the Public Schools in general. It is not desirable that the quantity of matherhatics studied in our Public. Schools be increased, but it is desirable that the quality of ^ PREFACE. ix ihe teaching should be greatly improved. To bring about this result IS the aim of this book. It does not follow, as is too often supposed, that any onfe familiar with the multiplication table, and able, perhaps, to solve a few problems, is quite competent to teach Arithmetic, or "Mathematics," as arithmetic is popularly called. The very first principles of the subject are of the utmost importance, and unless the correct and refined' notions of these principles are presented at the first, quite as much time is lost by the student in unlearning and freeing himself from erroneous con- c'eptions as was required in acquiring them. Moreover, no ad- vance in those higher modern developments in Mathematics is possible by any one having false notions of its first principles. As a branch for mental discipline, mathematics,- when properly taught, has no superior. Other subjects there are that are equally beneficial, but none superior. The idea en- tertained by many teachers, — generally, those who have pre- pared themselves to teach other subjects, but teach mathematics until an opportuijity to teach in their special line presents itself to them, — that mathematics has only commercial value and only so much of it should be studied as is needed by the studerjt in his business in after life, is pedagogically and psychologically wrong. Mathema,tics has not only commercial value, but edu- cational and ethical value as well, and that to a degree not excelled by any other science. /No other science offers such rich opportunity for original investigation and discovery. So far from being a perfected and complete body of doctrine "handed 'down from heaven" and incapable of growth, as many sup- pose, it is a subject which is being developed at such a mar- velous rate that it is impossible for any but the bdst to keep in sight of its ever-increasing and receding boundary. Because, therefore, of the great importance of mathematics as an agent in disciplining and developing the mind, in advancing the ma- terial comforts of man by its application in every department of art and invention, in improving ethical ideas, and in culti- vating a love for the good, the beautiful, and the true, the teachers of mathematics should have the best training possible. If this book contributes to the end, that a more comprehensive view be taken of mathematics, better services rendered in pre- senting its first principles, and greater interest taken in its study, I shall be amply rewarded for my labor in its prepa- ration. In this edition I have added a chapter on Longitude and Time, the biographies of a few more mathematicians, several hundred more problems for solution, an introduction to the study of Geometry, and an introduction to the study of Algebra. The list of biographies could have been extended indefi- nitely, but the student who becomes interested in the lives of s ' PREFACE. a class of men who have contributed much to the advancement of civiHzation, will find a short sketch of the mathematicians froip the earliest times down to the present day in Cajori's History of Mathematics or Ball's A Short History of Mathematics. The biographies which have been added were taken from the American Mathematical Monthly. I have received much aid in my remarks on Geometry from Study and Difficulties of Mathematics, by Augustus De Morgan. It yet remains for me to express my thanks to my colleague and friend, Prof. F. A. Hall, of the Department of Greek, for .making corrections in the Greek terms used in this. edition. The Author. Drj,try College, July, iSpp. PREFACE TO THE FOURTH EDITION. In bringing out this edition, I have been g(:ided some- what by the suggestions of many teachers who have used the book, and, in view of these suggestions, I have inserted Tables of Compound Numbers; Factoring; Arithmetical and Geometrical Progressions; and Specific Gravity," — giving each of these subjects the fullest possible trea^tment compat- ible with the limits of the book. It did not seem wise to include the Extractiqn of Square and Cube Roots as these art fully discussed in most arithmetics. I have taken this opportunity to 'insert the Metric System of Weights and Measures and have called special attention tO' the importance of this simple, beautiful, and convenient .sys- tem. Fractions have been more fully treated and the prob- lems with their solutions in Analysis have been classified under familiar titles, viz.. Age Problems, Fish Problems, Fox and Hound Problems, Time Problems, etc. Also the prob- lems at the end of Mensuration have been placed in their appropriate places in the body of that subject. Another Examination Test with answers and biographies of Newton and Gauss have been inserted. Many other minor changes and additions have been made. I have endeavored to eliminate all errors, and in this I h^ve received much help from W. J. Greenstreet, editor of the ]\Iatheinatical Gazette, Stroud, Eng., to whom I hereby acknowledge my indebtedness. THE AUTHOR. Drury College, July i^ , igo2. CONTENt^S. Mathematics classified CHAPTER I. DEFINITIONS. PAGE. I .... 1 Definitions PAGE, CHAPTER II. NUMERATION AND NOTATION. Numeration defined 4 French Method defined 4 ' English Method defined 4 Periods of Notation S Notation defined 5 Arabic Notation defined 5 •Roman Notation defined 5 Ordinal Numbers 5-8 Fraction's 8-10 Irrational Numbers 10-11 Examples 11-12 / CHAPTER III. ADDITION. Addition defined 12 | Examples 13 Chapter iv. SUBTRACTION. Subtraction defined 13 | Examples 14 CHAPTER V. \ MULTIPLICATION, Multiplication defined 14 | Examples 15-16 CHAPTER VI. DIVISION. Division defined 16 ] Examples 17 CHAPTER VII. COMPOUND NUMBERS. Definitions 18 I. The Metric System. ...18-21 Long Measure 19 Square Measure 19-20 Land Measure 20 Cubic Measure 20 Wood Measure 20 Mea"sure of Capacity. .20-21 Measure of Weight .. . 21 II. English System of Weights and Mea- sures 22-28 United States Money. 22 English Money 22 Avoirdupois Weight. . 23 Troy Weight 23 Apothecaries' Weight 23 Long Measure 24 (xi) CONTENTS. Compound Numbers - III. PAGE. Cloth Measure "... . .. 24 TV. Surveyor's Linear V. Measure ... 25 Square Measure . . ...25-26 Cubic Measure . . . . .. 26 Liquid Measure .. . ... 27 Dry Measure' . . . 27 Angular or C i r c u 1 a r Measure ...28-29 " Concluded. PAGE. Time Measure ., 30-32 Longitude and Time... 32-39 Standard Time 35-36 The International Date Line 37 Examples 38-39 Miscellaneous P r o b- lems 39-40 CHAPTER VIII. PROPERTIES OF NUMBERS. Definitions 41^3 I. Factoring 43 II. Greatest Common Divi- sor 44-45 Definitions 44 Solution of Problems 44 Examples 44-45 III. Least Common Multi- ple 45-47 Definitions 45 Solutions of P r o b- lems 45-46 Examples 46-47 Definitions 47-50 I. Reduction of Fractions. .50-55 II. Addition of Fractions. . 55 III. Subtraction of Frac- tions '. .55-56 IV. Multiplication of Frac- tions 56-57 V. Division of Fractions . . 57-60 CHAPTER IX FRACTIONS. VI. The G. C. D. of Frac- tions 60-61 VII. The L, C. M. of Frac- tions 61-62 Solutions of Miscellaneous Problems in Fractions 62-65 Examples 65-68 CHAPTER X. CIRCULATING DECIMALS.. I. Addition of Circulates. . 70 II. Subtraction of Circu- lates 71 III. Multiplication of Circu- lates 71-72 IV. Division of Circulates. . 72 Examples 72-73 CHAPTER XI. PERCENTAGE. Definitions 73 Solutions 74-83 Problems 84-85 I. Commission 8-5-88 Definitions 85 Solutions ...85-87 Examples 88 II. Trade Discount 88-92 Definitions 88 Solutions 89-92 Problems 92 III. Profit and Loss 93-100 Definitions 93 .Solutions 93-98 Problems 99-100 IV. Stocks and Bonds.... 100-113 Definitions 100 Solutions 101-111 CONTENTS. PERCENTAGE — Concluded. PAGE. Examples 112-113 V. Insurance 113-118 Definitions 113-114 PAGE. Solutions 114-117 Problems 118 CHAPTER XII. INTEREST. I. Simple Interest 119-122 Definitions 119 Solutions 120-122 II. True Discount 122-123 Definitions 122 Solutions 122-123 III. Bank Discount 123-125 Definitions '. 123 Solutions 124-125 IV. Annual Interest 125-128 Annual Int. defined.. 125 Solutions 125-128 V. Compound Interest. .128-131 Compound Int. defined 128 Solutions 129-131 CHAPTER XIII. ANNUITIES. Definitions 131 I Problems Solutions 132-141 | 142 CHAPTER XIV. MISCELLANEOUS PROBLEMS. Soliitions 143-155 CHAPTER XV. RATIO AND PROPORTION. Definitions 155-157 | Problems Solutions 157-160 .160-162 CHAPTER XVI. ANALYSIS. Analysis defined 162 Solutions 162-207 1. Age Problems 164-165 2. Fox and Hound Prob- lems 165-167 3. Fish Problems 167-168 4. Animal Problems 168-17-0 5. Labor Problems 170 6. Work Problems and Pipe Problems 171-174 7. Wine and Water Prob- lems 174-175 8. Sheep and Cow Prob- lems 175-176 9. With and Against the Current Problems ...176-177 10. Time Problems 177-185 11. Will Problems 185-187 12. Coach Problems 187-189 13. Cup and Cover Prob- lems 189-190 14. Dining and Chess Prob- lems . 190-191 15. Partnership Problems. .191-192 16. Combination Problems 192-193 17. Ditch Problems 193-195 18. Pasture Problems 195 19. Involution and Evolu- tion Problems 196 20. Solutions of Miscellan- , eous Problems 196-207 Problems 207-210 CONTENTS. CHAPTER XVII. ALLIGATION. PAGE. Definitions 211 L AlKgation Medial 211 PAGE. II. Alligation Alternate 211-216 Solutions 211-216 CHAPTER XVIII, SERIES. Definitions 217-218 I. Arithmetical Progres- sion 217-220 Solutions 219, 220 Problems 219, 220 II. Geometrical Progres- sion ;220-223 Solutions 221, 222 Problems 221, 222 CHAPTER XIX. SPECIFIC GRAVITY OR SPECIFIC DENSITY. Definitions 223 | Solutions, 224-228 CHAPTER XX. SYSTEMS OF NOTATION. Definitions 229 I Solutions ... Names of Systems 229 | .230-233 CHAPTER XXI. MENSURATION. Definitions 234-239 Geometrical Magnitudes clas- sified ' 234 I. Parallelogram 240-242 Problems 242-243 IL Triangle .243-248 Problems 248-249 III. Trapezoid 248-249 Problems ". 249 IV. Trapezium and Irregu- lar Polygons 249-250 Problems 250 V. Regular Polygons'. . .250-252 Problems 252-253 VL Circled 253-259 Problems 257-259 VII. Rectification of Plane Curves and Quadrature of Plane Surfaces. . .259-272 Definitions 259-260 Solutions 260-272 VIII. Conic Sections 272-281 Definitions 272-273 1. Ellipse 273-276 2. Parabola 276-278 3. Hyperbola 278-281 IX. Higher Plane Curves 282-297 1. The Cissoid of Dio- des 282-283 2. The Conchoid of Nico- medes 283-284 3. The Oval of Cassini. .^84-285 4. The Lemniscate of Ber- nouilli 285 5. The Witch of Agnesi . 285-286 6. The Limacon 286-287 7. The Quadratrix 287 8. The Catenary 287-289 9. The Tractrix 289 10. The Syntractrix 289 11. Roulettes .- 289-297 (a) Cycloids 289-292 (6) The Prolate and Curtate Cycloid 292-293 (c) Epitrochoid and Hy- . potrochoid 294-297 X. Planev Spirals 297-300 1. Spiral of Archimedes... 298 2. The Reciprocal Spiral 298-299 3. The Lituus 299 4. The Logarithmic Spi- ral 299-300 CONTENTS. Mensuration — Concluded. PAGE. XI. Mensuration of S o 1- ids , 300-302 1. Parallelepipeds 300-303 Problems 302-303 2. Prisms 303-304 Problems 304 3. Cylinder 304-306 Problems 306 4. Cylindric Ungulas. . .306-313 5. Pyramid and Cone. . .313-317 Problems 317-318 6. Conical Ungulas ....318-322 XII. Sphere 322-328 Problems 328-329 XIII. Spheroid -. .329-3^5 1. The Prolate Spheroid 329-331 - 2. The Oblate Spheroid 331-385 XIV. Conoids ....: 385-339 1. The Parabolic Conoid 335-338 2. Hyperbolic Conoid... 338-389 XV. Quadrature and Cuba- ture of Surfaces and Solids of Revolution 339-343 1. Cycloid 339-340 2. Cissoid 340-341 3. Spindles 341-343 (a) Circular Spindles. 341-342 [b) Parabolic Spindles,342-343 PAGE. XVI. Regular Solids 343-347 Definitions 843-344 1. Tetrahedron 344-345 2. Octahedron 345 3. Dodecahedron 345-346 4. Icosahedron 346-347 XVII. Pfismatoid 847-348 Problems 348 XVIII. Cylindric Rings 349-850 Problems 350 XIX. Similar Surfaces... .350-352 Problems 351-352 XX, Similar Solids 352 Problems 852 Theorem, of Pappus... 353 XXI. Miscellaneous Measure- ments 353-355 1. Masons' and Bricklayers' work 353 2. Gauging 353-354 3. Lumber Measure 354-355 4. Grain -and Hay 354-355 XXII. Solutions of Miscella- neous Problems. .. .355-407 Examination Tests With and Without Answers 408^15 Problems 416-421 GEOMETRY. L Definitions 422 II. On Geonietric Reason- ing 424-425 III. On the Advantages De- rived from the Study of Geometry and Mathe- matics in General.. .425^80 IV. Axioms 430^32 General Axioms 432 V. Assumptions 432^35 (a) A s s u m p t i on s of the Straight Line 432 (6) Assumptions of the Plane 432 (c) Assumptions of Par- allel Lines 432^34 (d) Assumptions of the Circle 434-485 (e) Assumptions of the Sphere 435 {f) Assumptions of Mo- tion 485 VI. On Logic .^ 485-443 Laws of Thought 436 Laws of Converse 438 Methods of Reasoning 439 The Syllogysm 439 Rules of the sylogism 440 Logical Fallacies 442-44.3 How to Prepare a Lesson in Geometry 448-444 Plane Qeometry 445-469 A Problem in Modern Ge- ometry ■ 456^57 The Nine Point Circle. .459-462 The Three Famous Prob- lems of Antiquity 463^65 Propositions 466-469 Definitions 470 Solutions of Problems 471^92 I. The Quadratic E q u a- tion 474-476 ALGEBRA. II. Indeterminate Forms. .476-480 III. Probability 480-490 Problems 492-495 xvi CONTENTS. Biography of Prof. William Hoover , 458-459' Biography of Dr. Artemas Martin 484-485- Biography of Prof. E. 3. Seitz 488-489 Biography of Rene Descartes 496-499 Biography of Sir Isaac Newton 600-602 Biography of I^eonhard Euler ; . . . . 503-607 Biography of Karl Friedrich Gauss ' 508-512 Biography of Sophus Lie 513-515- Biography of Simon Newcomb 516-518 Biography of George Bruce Halsted '519-620 Biography of Prof. Felix Klein , 621-623 Biography of Benjamin Peirce 624-529 Biography of James Joseph Sylvester 630-536- Biography of Arthur Cayley ^ . . . 637-539 Fallacies ' 540-543 I. Arithmetical Fallacies 540-641 II. Geometrical Fallacies 541-543 Tables 544-548 Table I '. 644 II 644 " III 544 ■• IV 545 V 545 " VI 64ft " VII .• 64& Examples 647-649* CHAPTER I. DEFINITIONS. 1. JKatheWiaticS (fiat^imrtxTJ, science) is that science which treats of quantity. ('(1.) Arithmetic. la. Differeutial. (1. Calculus ]b. Integral. t2.) Algebra... < <e. Calculus of Variation.s. (2. Quaternions. la. Pure Geometry. 1. Platonic Geometry,, h. Conic Sections, il. Plane Trigon'y. <c. Trigonometry.. <2. Analytical Trig. JS.) (ieometry...) 2. Analytical Geometry. (3. Spherical " 3. Descriptive Geometry. '(1.) Mensuration. , (2 ) Surveying. (3.) Navigation. (4.) Mechanics. (.T.) Astronomy. (6.) Optics. (7.) Gunnery. .(8.) &o., &e. 2. Pure mathematics treats of magnitude or quantity without relation to matter. 3. Applied J\£atJiemaiicS treats of magnitude as subsist- ing in material bodies. 4. Arithmetic {api&fi.rjrix:j, from apid-fioi;, a number) is the science of numbers and the art of computing by them. 5. Algebra {Ar. al, the, and geber, philosopher) is thafr method of mathematical computation in which letters and other symbols are employed. 6. G-eOmetry (jewiieTpia^ from yetufisTpelv to measure land, from ^-^a, )-7J, the earth, and /isTpelv, to measure) is the science of position and extension. 11, Calcul'US ( Calculus, a pebble) is that branch of mathe- matics which commands, by one general method, the most diffi- -;ult problems of geometry and physics. 2 FINKEL'S SOLUTION BOOK. 8. Differential Calculus is that branch of Calculus v^hich investigates mathematical questions by measuring the re- lation of certain infinitely small quantities called differentials. 9. Integral Calculus is that branch of Calculus which determines the functions from which a given differential has been derived. 10. Calculus of Variations is that branch of calculus in which the laws of dependence which bind the variable quanti- ties together -are themselves subject to change. 11. Quaternions {quatemis, from quaterni four each, from quattuor, four) is that branch of algebra which treats of the relations of magnitude and position of lines or bodies in space by means of the quotient of two direct lines in space, considered as depending on a system of four geometi-jcal elements, and as ex- pressed by an algebraic svmbol of quadrinominal form. 13. Platonic Geometri/ is that branch of geometry in which the argument is carried i*rward by a direct inspection of the figures themselves, delineated before the eye, or held in the imagination. 13. Pure Geometry is that branch of Platonic geometry in which the arguinent may be practically tested by the aid of the compass and the square only. 14. Conic Sections is that branch of Platonic geometry which treats of the curved lines formed by the intersection of a cone and a plane. 15. Trigonometry/ {rpiymvov, triangle, /lirpov, meas- ure) is that branch of Platonic geometry which treats of the re- lations of the angles and sides of triangles. 16. Plane TfigononietriJ is that' branch of trigonom- etry Which treats of the relations of the angles and sides of plane triangles. 17. Analytical Trigonometl' g is that branch of trig- onometry which treats of the general properties and relations of trigonometrical functions. 18. Sj)herical Trigononietrg is that branch of trig- onometry which treats of the solution of spherical triangles. 19. Analgtical Geometry is that branch of geometry in which the properties and relations of lines and surfaces are in- vestigated by the aid of algebraic analysis. 20. Descriptive Geometry is that branch of geometry which seeks the graphic solution of geometrical problems by means of pi-ojections upon auxiliary planes. DEFINITIONS. 3 31. JHetlSuration is that branch of applied mathematics which treats of the measurment of geometrical magnitudes. 33. Surveying is that branch of applied mathematics which treats of the art of determining and representing distances, areas, and the relative position of points upon the earth's surface. 33. Wavigation is that branch of applied mathematics which treats of the art" of conducting ships from one place to another. 34. IHechanieS is that branch of applied mathematics which treats of the laws of equilibrium and motion. 35. AstrOtlOmy (darpovn/xia, from aarpov, star and vdfioq law) is that branch of applied mathematics in which mechan- ical principles are used to explain astronomical facts. 36. Optics [oTtTixrj, from S4'ig sight,) is that branch of applied mathematics which treats of the laws of light. 37. Chfinnery is that branch of applied mathematics which treats of the theory of projectiles. 38. -4 FvopOSitiO'H is a statement of something proposed to be done. 39. Prop't'n. < , „, ( I. Lemma. Ts ^ i_i \ «■ Iheorem. < ^ ,, I. Demonstrable. i 7 t^ i_i / 2. Corollary. / b. Problem. •' z. Indemonstrable. a. Axiom. b. Postulate. 30. A Demonsti-ahle Proposition is one that can •be proved by the aid of reason. 31. A Theorem is a truth requiring a proof 33. A LemnKl is a theorem demonstrated for the purpose •of using it in the demonstration of another theorem. 33. A Corollary is a subordinate theorem, the truth of which is made evident in the course of the demonstration of a more general theorem. 34. A Problem is a question proposed for solution. 35. An Indemonstrable Proposition can not be proved by any manner of reasoning. 36. All Axiom is a self-evident truth. 3*7. A Postulate is a proposition which states that some- thing can be done, and which is so evidently true as to require no process of reasoning to show that it is possible to be done. 4 FINKEL'S SOLUTION BOOK. 38. A. Demonstration is the process of reasoning, prov- ing the truth of a proposition. 39. A Solution of a problem is an expressed statement showing clearly how the result is obtainedj 40. An Operation is a process of finding, from given, quantities, others that are known, by simply illustrating the solution. 41. A Rule is a general directibn for solving all "problems of a particular kind. 43. A Formula is the expression of a general rule or principle in algebraic language. 43. A Scholium is a remark made at the close of a dis- cussion, and designed to call attention to some particular feature or features of it. CHAPTER II. NUMERATION AND NOTATION. 1. Wumeration is the art of reading numbers. 2. There are two methods of numeration ; the French and the English. 3. The French method is that in gen^eral use. In this method, we begin at the right hand and divide the number into periods of three figures each, and give a distinct name to each period. 4. The English method is that formerly used in Great Britain. In this method, we divide the number (if it consists of more than six figures) into periods of six figures each, and give a distinct name to each period. The following number illustrates the two methods ; the upper division showing how the number is read by the English method, and the lower division showing how it is read by the French method. 4th period, 3d period, 2d period, 1st period. Millions. Billions. Millions. Units. ^ 678^^ 325 147 434 918 ..(/J 13 PI O O •2n ii .2 ■§ . a S.2 Bnglish. French. 5. The number expressed in words by the English method, reads thus : NUMERATION AND NOTATION. Eight hundred forty-five trillion, six hundred seventy -eight thousand nine hundred four billion, three hundred twenty -five thousand one hundred forty-seven million, four hundred thirty- four thousand nine hundred thirteen. Remark. — Use the conjunction and, only in passing over the decimal point. It is incorrect to read 456,734 four hundred and fifty-six thousand, seven hundred and thirty-four. Omit the and'sd^iA the number will be correctly expressed in words. 6. The following are the names of the Periods, according to the common, or French rtiethod: First Period, Second " Third " Fourth " Fifth " Units. Thousands. Millions. Billions. Trillions, Sixth Period, Quadrillions. Seventh " Quintijlions. Eighth " Sextillions. Ninth " Septillions. Tenth " Octillion. Other periods in order are, Nonillions, Decillions, Undecil- lions, Duodecillions, Tredecilions, Quatuordecillions Quindecil- lions, Sexdecillions, Septendecillions, Octodecillions, Novende- cillions, Vigintillions, Primo-Vigintillions, Secundo-vigintillions, Tertio-vigintilHons, Quarto- vigintillions, Quinto-vigintillions, Sexto-vigintillions, Septo-vigintillions, Octo-vigintillions, Nono- vigintillions, Trigillions; Primo-Trigillions, Secundo-Trigillions, and so on to Quadragillions ; Primo-quadragillions, Secundo- quadragillions, and so on to Quinquagillions; Primo-quinqua- gillions, Secundo-quinquagillions, and so on to Sexagillions, Primo-sexagillions, Secunclo-sexagillions, and so on to Septua- gillions ; Primo-septuagillions, Secundo-septuagilHons, and so on to Octogillions ; Primo-octogillions, Secundo-pctogillions, and so on to Nonogillions ; Primo-nonogillions, Secundo-nonogillions, and so to Centillions. 7. Notation is the art of writing numbers. There are three methods of expressing numbers ; by words, by letters, called the Roman method, and by figures, called the Arabic method. 8. The Roman Notation, so called from its having originated with the ancient Romans, uses seven capital letters to express numbers ; viz., I, V, X, L, C, D, M. 9. The Arabic Notation, so called from its having been made known through the Arabs, uses ten characters to express num- bers ; viz., 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. 10. Ordinal Numbers. A logical definition of number is not easy to give, for the reason that the idea it conveys is a simple notion. The clearest idea of what counting and numbers mean may be gained from the observation of children and of 6 FINKEIv'S SOLUTION BOOK. nations in the childhood of civilization*. When children count or add they use their fingers, or small sticks, or pebbles which they adjoin singly to the things to be counted or otherwise to be ordinally associated with them. History informs us that the Greeks and Romans employed their fingers when they counted or added. The reason why the fingers are so universally used as a means of numeration is, that everyone- possesses a definite number, sufi&ciently large for purposes of computation and that they are always at hand. Let us consider the row of objects, XXXXXXXX X X X X X X X X , with regard to their order, say from left to right, freeing our minds from all notions of magni- tude. Beginningi with any one object in this row, we speak of~ the one we begin with as being the first, the next in order to it to the right the second, the next in order to the right of the sec- ond the third, and so on. The name or mark we thus attach to an object to tell its place in the row is called an integer. This process, or operation, of labeling the objects is called counting and it is the fundamental operation of mathematics. To count objects is to label the objects, not primarily to tell how many there aref- In thus labeling the objects, we may replace the objects by the fingers, by sticks, by pebbles, by marks, or by characters. The method of tallying used at the present time is such a method. In counting objects marks are made until four are made, then these are crossed with a fifth mark and so on. Thus, Ufl. LH1. W1- Suppose that in counting the objects in the row, we use our fingers, and for each object in the row beginning with a certain one we bring in correspondence with that object the little fin- ger of the right hand, with the next object to the right the next to the little finger of the right hand, and so on until an object and the thumb of the right hand are brought into correspond- ence. For the group of objects thus counted, let us bring into correspondence the little finger of the left hand. Now continue the counting of the objects of the row as before, and when a second group is reached bring into correspondence with this group the, next to the little finger of the left hand. Continue this process until a group of the objects as represented by the fingers of the right hand is brought into correspondence with the thumb of the left hand. Thus the fingers of the left hand represent a group of groups of objects. Bring this group rep- resented by the fingers of the left hand into correspondence with * Schubert's Mathematical Essays and Recreations. t My friend, Dr. William Rullkoetter, told me of a case coming under his personal observation, where a farmer, unable to count, but when desirous of knowing if any of bis cattle were missing, would have them driven through a gate or past some point where he could see them as they passed singly. He would then say, " You are here," " and you are here," " and you are here," and so on until all had passed by. In this way he was able to tell if any were missing, but not able to teU how many he had. NUMERATION AND NOTATION. 7 the little toe of the right foot. Now continue the process of counting the objects and so on as before until the big toe of the right foot is brought into correspondence with a group corre- sponding to the fingers of the left hand. Thus the toes of the right foot represent a group of a group of a group of objects. In this manner, we could build up the system of numeration called the Quinary, a system in which five objects as represented by the fingers of the right hand make a unit or group as repre- sented by a finger of the left hand, five groups of five objects as represented by the fingers of the left hand make a group as rep- resented by the toes of the right foot, and so on. The decimal system of numeration may be built up in the same way, except that the group of objects corresponding to the fingers of both hands would be represented by a toe. After the fingers and toes have been exhausted in the process of counting the numeration would have to be continued by using small sticks or pebbles. It is very probably due to the fact that we have 10 fingers that the decimal system was invented. There are, how- ever, among the uncivilized nations of the world a number of different systems of numeration*. At the present time, in labeling objects by the process of counting we use the following characters, viz., 1, 2, 3, 4, 5, 6, 7, 8, 9, etc. 123456789 labels. Thus XXXXXXXXXXXXXXX objects. In labeling, we could begin with the object marked 3 and re-label it 1, then re-label 4 as 2 and 5 as 3, and so on. This is expressed by writing 3— 2==:1, 4—2=2, 5—2=3, meaning that if we begin after the object whose old mark was 2, then the object which was third becomes first, the object which was fourth becomes second, and so on. Beginning after an object instead of with it suggests that our original row might begin after an object; this object after which the counting begins is marked and called the origin. If there are objects to the left of the origin, we count them in the same way; except that we prefix the sign, — , to show that they are to the lefit, and we call the marks so changed negative integers, thus distinguishing them from the old marks which we call positive integers. The marks are ... . —4, —3, —2, —1, 0, 1, 2, 3, 4, 5, 6, .■ These marks constitute what is called the natural integer- system. When an object marked a is to the left of another marked a', we say that a comes before d or is inferior to a! , and a' *See Conant'S" Number Concept for a full treatment of the various systems of notation. 8 FINKEL'S SOLUTION BOOK. comes after a or is superior to a. These ideas are expressed symbolically thus a<a' , a'>a. Here a and a' mean integers, positive or negative. Objects considered as a succession from left to right are in positive order; when considered from right to left, in negative order. * Addition and its inverse operation, Subtraction, are algorithms of counting. Multiplication is an algorithm of Addition, and Division is an algorithm of Subtraction. Addition, Subtraction, Multiplication, and Division are only short methods of counting. If we operate on any integer of the natural-integer series by any one of the operations of Addition, Subtraction, or Multipli- cation, no new integei; is produced. With reference to these operations the natural integer-series is closed, that is to say, there are no breaks in the integer-series into which other inte- gers arising from these operations may be inserted. If, how- ever, we operate on any one of the integers of the integer-series by, the operation of Division, the operations in many cases are impossible. Suppose we wish to divide 17 by 5. This opera- tion is absolutely impossible, -y- is a meaningless symbol with reference to the fundamental operation of mathematics. But in this case, as in the case when negative numbers are introduced by the inverse operation, subtraction, we apply a principle called by Hankel, "The Principle of the Permanence of Formal I,aws," and by Schubert, "The Principle of No Exception," viz., That every time a newly introduced concept depends upon operations previously employed^ the propositions holding for these operations are assumed to be valid still when they are applied to the new con- cepts. In accordance with this principle, we invest the symbol, ^-, which has the form of a quotient without its dividend being the product of the divisor and any number yet defined, with a meaning such that we shall be able to reckon with such apparent quotient as with ordinary quotients. This is done by agreeing always to put the product of such a quotient form with its divisor equal to its dividend. Thus, (Jg^-)X5=17. We thus reach the definition oi fractions. The concept of fractions may also be established as in the next article. 11. Fractions. lyet us now again assume the row of 123456789 objects, XXXXXXXXXXXX. . . . attending to only zero, the object from which we begin, and the objects on the right of it. Suppose we re-label the alternate objects 2, 4, 6, 8, .... marking them 1, 2, 3, 4 We must then invent marks for the objects previously marked 1,3,5,7 The NUMERATION AND NOTATION. 9 marks invented are shown in Figure 1, above the objects, the old marks being below the objects. , 0ilf2f3l4|5 xxxxxxxxxxxx.... 01234 6 6789 Fig. ±. From this it is clear that instead of re-labeling the alternate objects in a row of objects, 0, 1, 2, 3, 4,' 5, 6, 7 we can interpolate alternate objects in the row and then mark them J, f, f , and so on. In the same way we can interpolate two objects between every consecutive two of the given row 0, 1, 2, 3, 4, 5, marking the new objects in order ^, f ; f, |; i, f ; and so on. 0J|lf|2ff3 Thus, XXXXXXXXXX 1 2 ,3 ^; Fig. a. ■ . 12 3 In this way we account for the symbols —'—'—' ... where P P _ P p is any positive integer. These we call positive fractional numbers. By interpolating single objects in the row 0, \, 1, f , 2, f , . . . we have the same sequence of objects as if we interpolate ob- jects by threes in the row 0, 1, 2, 3, 4, 5, 6, and the objects are therefore marked 0, i, i, i, 1, i i i 1 I f I- 2 Thus, XXXXXXXXXX + 1*2 Fig. 3. From this we see that \ and f are marks for the same object. Also f and f. Hence, f^J and f =f . A row marked 0; ^, f , Jf, J, f , 1, . . . . is to be understood as arising from the interpolation of objects by fives; that is, by introducing the objects \, %,%,%,% , or \, j, i, f , f , . • . . As f comes before f , we say f <-|, or \<i\. We may interpolate as many objects as we please in the nat- ural row, and, by the principle of the least common denomina- tor, we can interpolate so as to explain any assigned positive fractional marks,yi,y2,/3, . . . Also, given any positive rational mark, r, other than zero, we can interpolate rational marks be- 10 FINKEL'S SOLUTION BOOK. tween and r. When no object can be made to fall between an-, assigned object and 0, that assigned object must be itself. In the same way we may treat the negative numbers. We can think of an infinity of objects as being interpolated, in the natural row, so that each shall bear a distinct rational number and so that we can say which of any two objects comes, first. It is to be noticed that as we approach any of the natural objects there is no last fractional mark; that is, whatever object, we take there are always others between it and the natural object. Thus, if an infinitude of objects be interpolated in the nat- ural row, 0, 1, 2, 3, ... . + \ I? 1 XXXX. .to infinityXXX. .to infinityXXXXX. .to infinityXX. .to infinityX. Sig. 4- 1 then it is clear that whatever object we take there is an infinity- of objects between it and the natural object, thus rendering it. evident that there are no last fractional marks in this case. 13. Irrational Nnxnbers. In coiisidering square num- bers from the ordinal point of view, we re-label our natural row as in Fig. 5. 1112 3" XXXXXXXXXXXXXXXX 01^21 3 456789 Fig- 5. wheire the old names are below and the new above. We have- now to consider how to bring the omitted objects into the scheme of ordinal njimbers. Every object whose new name is fractional had a fractional name, so that the object whose old name was 2' cannot now have a rational name. We give it a name which we call irrational. We call it the positive or chief square root of 2; and mark it V 2 or 2^ As an ordinal number it is perfectly satisfactory, for we know where it comes, whether left or right, of any proposed rational number, by means of the old marking. Hence, it separates all the rational numbers into two Classes, viz., those on its right and those on its left. A rational number separates all other rational numbers into two classes; we put it into one of the classes afid say it closes that class. Take, for example, f. Now, there is no last fractional mark as we approach f from the left or from the right. Hence, with- out ^ neither the class to the left of f nor the class to the right of f is closed. With f, either class is closed. Any process which serves to separate rational numbers into two classes, — those on the left and those on the right, such that the left-hand class is niDt closed on the right and the right-hand class not closed on the left, — leads to the introduction of a ne-w object named by an irrational number. NUMERATION AND NOTATION. II For example', / 2 separates all rational numbers into two classes, viz., those on the left of it and those on the right^ Now if we take any rational object however near to the V 2 as we please we can always interpolate new rational objects between it and V 2. Thus, it is clear that the class on the left of t/ 2 is not closed at the right nor the class on the right closed on the left. Two rational or irrational numbers, — for simplicity take them both irrational and equal to j and s' , — are equal if the rational objects to the left of i- are the same as the rational objects to the left of s' , and the rational objects to the right of s are the same as those to the right of s' Thus, 4*^ and 2^ effect the same sep- aration of the rational numbers. Hence, 4**=2^. An equivalent condition for the equality of s and / is that every rational number to the left of s shall be to the left of s' , and every rational number to the left of / shall be to the left of s. Between two unequal irrational objects, s and s' , there must lie rational objects; for, since J and / are not equal, there must be a rational number which is before one and not before the other. It is very important to notice that we have now a closed number-system. When we seek to separate the irrational objects as lying left or right of an object, either the object is rational or if not it separates rational objects and is irrational; in any case it must "have for its mark a rational or irrational number, and there is no loop-hole left for the introduction of new real num- bers which separate existing numbers. This is often briefly expressed by saying that the whole system of positive and neg- ative integral, fractional, and irrational numbers is continuous, or is a continuum^. In the way indicated above, the number-concept of Arithme- tic is put on a basis consistent with Geometry. If we select any point on a straight line and call it the zero-point, and also a fixed length, measured on this line, be chosen as the unit of length, any real number, a, can be represented by a point on this line at a distance from the zero-point equal to a units of length. Con- versely, each point on the line is at a distance from the origin equal to a units of length, when a is a real number. That is, there is a one to one correspondence between the points of line and the numbers of the real number-system. For every point of the line, there corresponds a number of the real number-sys- tem and for every number of the real number-system there cor- responds a point of the line. EXAMPLES. 1. Write three hundred seventy quadrillion, one hundred one thousand one hundred thirty-four trillion, seven hundred eighty- *See Harkness and Motley's Introduction to the Theory o/ Functions ^ Qi^&^teiT I., from which, Arts. 10, 11, and 12 have been chiefly adapted 12 FINKEL'S SOLUTION BOOK nine thousand six hundred thirty-two billion, two hundred ninety- eight thousand seven hundred sixty -five million, four hundred thirty-seven thousand one hundred fifty-six. 2. Read by the English method, 78943278102345789328903- 24678. \ 3. Write three thousand one hundred forty -one quintillion, five hundred ninety-two billion six hundred fifty-three million five hundred eighty- nine thousand seven hundred ninety-three ■quadrillion, two hundred thirty -eight billion four hundred sixty- two million six hundred forty-three thousand three hundred eighty-three trillion, two hundred seventy-nine billion five hun- dred two million, eight hundred eighty-four thousand one hundred ninety-seven. 4. Read 141421356237309504880168872420969807856971437- 89132. 5. Is a billion, a million million? Explain. ■6. Write 19 billion billion billion. 7. Write 19 trillion billion million million. 8. Write 19 hundred 66 thousand. 9. Write 457 thousand 341 million. 10. Write 19 trillion trillion billion billion milliorl million. CHAPTER III. ADDITION. 1. A.(ld,ition is the process of uniting two or more numbers of the same kind into one sum or amount. 2. Add the following, beginning at the right, and prove the result by casting out the 9's : 7845 excess of 9=6' 'Excess of 9's=8. 6780 ( ( 'J 9=3 8768 ( i " 9=2 5343 (1 " 9=6 3987 (1 " 9=0 32723 excess of 9=8 Explanation. — Adding the digits in the first number, we have 24. Dividing by 9, we have 6 for a remainder, which, is the excess of the 9's. Treating the remaining numbers in the same manner, we obtain the excesses 3, 2, 6, 0. Adding the excesses and taking the excess of their sum, we have 8 ; this being equal to the excess of the sum the work is correct. SUBTRACTION. IS Add the following, beginning at the left : 8456 9799. 4363 5809 5432 31 26 23 29 33859 From this operation, we see that it is more convenient to be- gin at the right ' Remark. — We can not add 8 apples and 5 peaches because we can not express the result in either denomination. Only numbers of the same name can be added. EXAMPLES. 1. Add the numbers comprised between 20980189 and 20980197. 2. 609.5054 + 900703+90300420+9890655+37699+29753 = ■what? 3. Add the following, beginning at the left: 97674; 347- 893; 789356; 98935679; 123456789. 4. Add all the prime numbers between 1 and 107 inclusive. 5. Add 31989, 63060, 132991, 1280340, 987654321, 78903, and prove tjie result by casting out the 9's. 6. Add the consecutive numbers from 100 to 130. 7. Add the numbers from 9897 to 9910 inclusive. 8. Add MDCCCLXXVI, MDCXCVIII, DCCCCXLIX, DCCCLXII. CHAPTER IV. SUBTRACTION. 1. /S'M'6#f<ifCf*OW' is the process of finding the difference be- tween two numbers. 2. Subtract the following and prove the result by casting out the 9's : 984895 excess of9's=7 795943 " " 9's=l 188952 " " 9's=6j Excess of 9's=:7. 14 FINKEL'S SOLUTION BOOK. Explanation. — Adding the digits in the first number, we have 43. Dividing by 9 the remainder is 7, which is the excess of the 9's. Treating the subtrahend and remainder in the same manner, we have the excesses 1 and 6. , But subtraction is the opposite of addition and since the minuend is eqmal to the sum of the subtrahend and remainder, the excess of the sum of the excesses in the subtrahend and remainder is equal to the excess in the minuend. This is the same proof as that required if we were to add the subtrahend and remainder. 3. We begin at the right to subtract, so that if a figure of the subtrahend is greater than that corresponding to it in the minuend, we can borrow one from the next higher denomination and reduce it to the required denomination and then subtract.- 4. Subtract the following and illustrate the process : 1=9 9 9 9 9 9 9+1 1=9 9 9 9 9 9+1 ) . , , 1=9 9 9+1 ) , , , 90000000 9856842 \ ^^^- 4 3 2 6 5 4 6 \ ^'i'^- 85784895 8978567 8214957 4215105 877775 1 11 1589 EXAMPLES. 1. From 9347893987 take 8968935789. Prove the result by ■casting out the 9's. 2. 7847893578— 6759984699=what? Which is the nearer number to 9^0864; 1816090 or 27497? 4. 34567— 34518 + 3— 2 + 8— 4 + 7 + 18— 567 + 43812 — 1826 -f €78=what.? 5. 5 + 6 + 7— 12— 13 + 14— 2— 3 + 7— 8— 6 + 5 + 12— 8=what? 6. 3 + 4— (6 + 7)— 8 + 27— (1 + 3— 2— 3) — (7— 8 + 5) 3 + 7==- what? CHAPTER V. MULTIPLICATION. 1. Mtlltiplication, is the process of taking one number as many times as there are units in another; or it is a short method of addition when the numbers to be added are equal. 2. Multiply the following and prove the result by casting out the 9's : 7855 excess of 9's=7 435 " " 9's=3 89275 ' 21 excess of 9's=3. 23565 81420 8416925=excess of 9's=3. MULTIPLICATION. 15 Explanation. — Adding the digits in the multiplicand and -dividing the sum by 9, the remainder is 7 which is the excess of the 9's. Adding the digits in the multiplier and dividing the sum by 9, we have the remainder 3 which is the excess of the 9's. Now, since multiplication is a short method of addition when the numbers to be added are equal, we multiply the excess in the multiplicand by the excess in the multiplier and find the excess, and this being equal to the excess in the product, the work is •correct. 3. Multiply the following, beginning at^:he left : 7o(i45 765 1st.. ..49 35 42 5d.. . . . 4228 3035 36 3d.. . . . . 8524 2530 30 20 25 57368425 3. From this operation, we see that it is more convenient to 'begin at the right to multiply. 5. In multiplication, the multiplicand rnay be abstract, or concrete; but the multiplier is always abstract. 6. The sign of multiplication is y,,a.nA\s,reeiA, multiplied by, or times. When this sign is placed between two numbers it de- notes that one is to be nvultiplied by the other. In this case, it has not been established which shall be the multiplicand and which the multiplier. Thus 8x5=40, either may be considered the multiplicand and the other the multiplier. If 8 is the mul- tiplicand, we say, 8 multiplied by 5 equals 40, but if 5 is the ■multiplicand we say, 8 times 5 equals 40. EXAMPLES. 1. 562402 X345728=what? 2. 1 mile ^ 63360 inches; how many inches from the earth to the moon the distance being 239000 miles? 3. Multiply 78962^ by 834, beginning at the left to multiply. 4. 1 acre = 43560 sq. in.; how many square inches in a field ^containing 427 acres ? j^g FINKEL'S SOLUTION BOOK. 5. Multiply 6934789643 by 34789. Prove the result by cast- ing out the 9's. 6. 2778588 X34678=what? 7. 2X3X4— 3x7+3— 2x2+4+8x2+4— 3x5+27=what? 8. 5X7+6X7+8X7— 4x6+6x6+7x6=what? 9. 356789 X4876=what? 10. 395076 X576426=what.? 11. 7733447 X998800=what? 12. 5654321 X999880=what? CHAPTER VI. DIVISION. 1. Division is the process of finding' how many times one number is contained in another; or, it is a short method of sub- traction when the numbers to be subtracted ar^ equal. 2. Divide the following and prove the result by casting out the 9's: 67)5484888(81864 536 Dividend ' 124 5484888 excess of 9's=0. 67 Quotient 578 ,81864 excess of 9's=01 Excess of 9's' ^3^ „. . lin this product Divisor I , ^ "428 67 excess of 9's=4j equals 0. 402 ~268 268. Explanation. — Adding the digits in the dividend and di- viding the sum by 9, we have the remainder 0, which is the ex- cess of the 9's. Adding the digits in the quotient and dividing the sum by 9, we have the remainder 0, which is the excess of the 9's in the quotient. Adding the digits in the divisor and dividing the sum by 9, we have the remainder 4, which is the excess of the 9's in the divisor. Since division is the reverse of multiplication, the quotient corresponding to the multiplicand, the divisor to the multiplier, and the dividend to the product, we multiply the excess in the quotient by the excess in the divisor. The excess of this product is 0. This excess being equal to the excess of the 9's in the dividend, the woi-k is correct. DIVISION. 17 If there be a remainder after dividing, find its excess and add it to the excess of the product of the excesses of the quotient and divisor. Take the excess of the sum and if it is equal to the ex- cess of the dividend the work is correct. 3. The sign of division is -^, and is read divided by. 4. When the divisor and dividend are of the same denomina- tion the quotient is abstract ; but when of different denomina- tions, the divisor is abstract and the quotient is the same as the dividend. Thus , 24 ct. -H 4ct. = 6 , and 24 ct. -i- 4 = 6 ct. Remark. — We begin at the left to divide, that after finding how many times the divisor is contained in the fewest left-hand figures of the dividend, if there be a remainder we can reduce it to the next lower denomination and find how many times the divisor is contained in it, and so on. Note. — The proof by casting out the 9's will not rectify errors caused by inserting or omitting a 9 or a 0, or by interchang- ing digits. EXAMPLES. 1. 4326422-f-961=what? Prove the result by casting out the 9's. 2. 245379633477-^1263=what? Prove the result by casting out the 9's. 3. What number multiplied by 109 with 98 added to the product, will give 106700? 4. The product of two numbers is 212492745 ; one of the numbers is 1035; what is the other number? 5. 27-5-9 X 3-7-9 — 1-1-3-1-3 X 9— 8-r4-l-5 x 2 — 8 X 2-^2-i-3 — (3X4-H6-I-5— 2)-|-81-e-27x3-=-9Xl8-r-6=what? [Hint.— Per- form the operations indicated by the multiplication and division, signs in the exact order of their occurrence.] 6. (64-=-32x96-Hl2 — 7— 5-|-3)x|[(27-r-3)-r-9 — l-|-2]-|- 91~13x7— 45| x9-f45-r-9-|-8— l=what? 7. 2x2-r-2-r-2-T-2x2X2-H2-^2-H2=what? Ans. i. 8. 3-5-3^-3 X3X3-f-3-H)X4x4x5X5=what? Ans. en, 9. 2x2x2-r-2x2-f-2-r-2x2X2X0x2X2=what? Ans. 0. 18 FINKEL'S SOLUTION BOOK. CHAPTER VII. COMPOUND NUMBERS. 1, A. Compound, N^wmher is a number which expresses several different units of the same kind of quantity. 3. A. Denominate Number is a concrete number in which the unit is a measure; as,- 5 feet, 1 pints. ^ 3. The Terms of a compound number are the numbers of its different units. Thus, in 4 bu. 3 pk. 7 qt. 1 pt., the terms are 4 bu. and 3 pk. and 7 qt. and 1 pt. 4. Heduction of Compound Numbers is the process of changing a compound number from one denomination to an- other. There are Two Cases, Reduction Descending and Re- duction Ascending. 5. deduction Descending is the process of reducing a number from a higher to a lower denomination. 6. Reduction Ascending is the process of reducing a number from a lower to a higher denomination. I. THE METRIC SYSTEM OF WEIGHTS AND MEASURES. 1. The Metric System is a system of weights and meas- ures based upon the decimal system. In 1795, France adopted a system of weights and measures based upon the decimal scale. A careful measure' of an arc of the meridian passing through' Paris was made and the To-jroTTTTi' part of the distance from earth's north pole to the equator meas- ured on this meridian was taken as the unit of length. -It was called the meter Upon it is based the entire system, all other units being derived units, that is, units derived from the meter. It is greatly to be regretted that this system did not come into general use in the United States and England a half century ago. It is the very acme of simplicity and elegance. Its use would greatly simplify international commercial relations and promote educational interests in all countries using it. The argument commonly advanced against its introduction is that it would entail a great loss of money now invested in ex- pensive machinery; — practically all machinery used in the arts and the various machine constructions would become worthless. ^ THE METRIC SYSTEM. 19 This is true, but the reason that it is not adopted irrespective of the sacrifice of a large sum of money, is the ignorance and prejudice of the common people. If the common people clearly understood how simple and easy this system is, the popular de- mand would overpower every opposition and its general use would be established in a short time. In order that the common people understand its merits and appreciate them, every teacher in our public schools should devote ■considerable time in presenting this system to his pupils, and he should advocate its use whenever an opportunity presents itself. Teachers, why should our old brain-racking system be used, when such a labor-saving and excellent system has been devised to take its place? Transfer its place in the arithmetics from the latter part of the book to that part just preceding our own cumbersome system of weights and measures, disseminate a healthy knowl- edge of this system among your pupils, and when the time does come that the Metric System shall supercede our present clumsy, awkward, and unscientific system, remember you will have^done your full share in contributing to future generations the greatest heritage to which civilization is heir. 2. The following prefixes are used in the Metric System : Deca signifies 10 times the unit. Hecto signifies 100 times the unit. Kilo signifies 1,000 times the unit. Myria signifies 10,000 times the unit. Deci signifies the 10th part, or .1, part of the unit. Centi signifies the 100th part, or .01 part of the unit. Milli signifies the 1,000th part, or .001 part of the unit. UNITS OF LENGTH. 1, The Prime Unit of I/Cngth is the meter. , LONG MEASURE. 10 rhillimeters (mm.)=: 1 centimeter (cm.) 10 centimeters ' = 1 decimeter (dm.) 10 decimeters ^1 meter (m.) 10 meters := 1 dekameter (Dm.) 10 dekameters = 1 hectometer (Hm.) 10 hectometers =^ 1 kilometer (Km.) 10 kilometers ^ 1 myriameter (Mm.) UNITS OF AREA. 1. The Prime Unit of Square Measure is the square meter, =m.^ . 20 FINKEL'S SOLUTION BOOK. ■ SQUARE MEASURE. 100 square millimeters (sq. mm. := mm^.)^ 1 square centimeter (sq. cm. = cm^.) 100 square centimeters = 1 square decimeter (sq. dm. = dm^.) 100 square decimeters =: 1 square meter (sq. m. = m^.) 100 square meters = 1 square dekameter (sq. Dm. = Dm^.) 300 square dekameters = 1 square hectometer (sq. Hm. = Hm^.' 100 square hectometers = 1 square kilometer, ( sq. Km. = Km^.) Note. — A square decameter is called an are when used in measuring^ land. , Remark. — Observe that the abbreviations, sq. mm. and mm.^, sq. cm. and cm.", sq. m. and m.'', etc., are synonymous. The exponential ab- breviation, being more quickly written, is generally used in Physics and other scientific works. Likewise, cu. mm. and mm.', cu. cm. and cm.', cu. m. and m'. are used synonymously. LAND MEASURE. 10 centares (ca.)= 1 deciare (da.) 10 declares =: 1 are ( a. ) 10 ares ^ 1 decares (Da.) • 10 decares ^ 1 hectare (Ha.) UNITS OF VOLUME. 1. The Standard Unit of Volume is the cubic meter, z^ m^ CUBIC MEASURE. 1000 cubic millimeters (cu. mm. ^mm^.)=l cubic centimeter (cu. cm. = cm^. ) 1000 cubic centimeters ^= 1 cubic decimeter (cu. dm. = dm^.) 1000 cubic decimeters =^ 1 cubic meter (cu. m. = m^.) 1000 cubic meters =1 cubic dekameter (cu. Dm. ^ Dm^.) 1000 cubic dekameters = 1 cubic hectometer (cu. Hm. = Hm^.) 1000 cubic hectometers = 1 cubic kilometer (cu. Km. = Km^.) Note.— A cubic meter is called a slere when used in measuring wood. WOOD MEASURE. 10 decisteres (ds.):=.l stere (s.) 10 steres = 1 dekastere (Ds.) UNITS OF CAPACITY. 1. The Unit of Capacity is the liter. The liter is a cubical vessel whose edge is 1 decimeter. Hence, it equals a cubic decimeter. It is used in measuring liquids and dry sub- stances. THE METRIC SYSTEM. 21 MEASURE OF CAPACITY. 10 milliliters (ml.)=l centiliter (cl.) 10 centiliters = 1 deciliter (dl.) 10 deciliters = 1 liter (1.) 10 liters =1 dekaliter (Dl.) 10 dekaliters = 1 hectoliter (HI.) 10 hectoliters = 1 kiloliter (Kl.) 10 kiloliters =^ 1 myrialiter (Ml.) UNITS OF WEIGHT. 1 . The Unit of Weight is the £^ram. It is the weight of a cubic centimeter of distilled water at 4°C. MEASURE OF WEIGHT. 10 milligrams (mg.)= 1 centigram (eg.) 10 centigrams =1 decigram (dg.) 10 decigrams = 1 gram (g.) 10 grams = 1 dekagram (Dg.) 10 dekagrams =1 hectogram (Hg.) 10 hectograms =1 kilogram (Kg.) 10 kilograms =1 myriagram (Mg.) 10 myriagrams = 1 quintal ( Quin. ) 10 quintals = 1 metric ton (M. T.) I. Find the weight of a cubic meter of gold, specific gravity being IQ^. 1. 1 m.3=rl003 cm.3=l,000,000 cm.^ 2. 1 g. =weight of 1 cm.^ of water. II. -i 3. 19J g.^weight of 1 cm.^ of gold, its sp. gr. being 19;^. 4. 19,250,000 g-.=l,000,000Xl9i g.=weight of 100^ cm.^, or 1 cubic meter of gold. III. . .Imfioi gold weighs 19,250,000 g. I. How many liters in a tank 6 meters long, 5 meters wide, and 2 meters deep? 1. G^number of meters in length. 2. 5:=number of meters in width. 3. 2=number of meters in depth. 4. .'. 6X5X2=60=;number of cubic meters in the tank. 5. 1 cu. m.=1003 cm.3— 1,000,000 cm.^ 6. 60 cu. m.=60X 1,000,000 cm.3=60,000,000 cm.^ 7. 1 cm.3=l !. 8. 60,000,000 cm.3=60,000,000 1. III. .-. The tank contains 60,000,000 1. Compare the above problems with similar ones in the common system of weights and measures. 11.-^ 22 FINKEL'S SOLUTION BOOK. II. ENGLISH SYSTEM OF WEIGHTS AND MEASURES." UNITS OF VALUE. 1. The Prime Unit of United States money is the dollar. UNITED STATES MONEY. 10 mills (m.)= 1 cent (ct. or /) 10 cents = 1 dime (d.) 10 dimes * = 1 dollar (|.) 10 dollars = 1 eagle (E.) Note. — The coins of the United States are made of bronze, nickel, silver, and gold. The gold coins are the double eagle, eagle, half-eagle, quarter - eagle, three -dollar, and one -dollar. The silver coins are the trade dollar, half- dollar, quarter - dollar ^ twenty - cent piece, dime, half- dime, and three-cent piece. The nickel coin is the five-cent piece. The bronze coins are the two-cent piece and the one cent. The three-dollar and one dollar gold pieces are no longer coined ; neither are the silver trade dollar, twenty-cent piece, half dim.e, and three-cent piece, and the nickel cent, the old copper cent, the bronze two cent piece. Many of these coins are still in circulation, however. For many interesting notes under this and the following table the student is referred to Brooks^ Higher Arithmetic, one of the very best arithmetics published in America. ' ENGLISH MONEY. 1. Mnglisb, or Sterling Money, is the money of Great Britain and Ireland. 3. Tie Prime Unit is the pound {£), whose value in United States money is |4.8665. The pound, when coined, is called the sovereign. ' ENGLISH MONEY. 4 farthings (far.)=l penny (d.) 12 pence = 1 shilling (s.) 20 shillings := 1 pound (£) 5 shillings = 1 crown. ' 21 shillings = 1 guinea. Note. — The Unit of French Money is the franc, which is worth 19.3^. The Unit of Germ,an Money is the mark, which is worth 23.85^. I. Express 2 guineas in sovereigns and shillings. {1. 1 guinea :=21 shillings. 2. 2 guineas =2 times 21 shillings^42 shillings. 3. 20 shillings=l sovereign. 4. 42 shillings=42-^20=2 sovereigns+2 shillings. III. .-. 2 guineas =2 sovereigns and 2 shillings. THE ENGLISH SYSTEM. 23 I. What is the value of 80 marks in United States money? „ f 1. 1 mark =23.85 cents=|0.2385. ^^- t 2. 80 marks=80 times |0.2385=|19.08. III. . . 80 marks=|19.08. UNITS OF WEIGHT. 1. Avoirdupois Weight is used for weighing everything except precious metals, and medicines when dispensed. 2. The Prime TJnit of Weight is the pound Avoirdupois. AVOIRDUPOIS V^EIGHT. 16 ounces (oz.) = 1 pound (lb.) 100 pounds = 1 hundredweight (cwt.) 20 hundredweight = 1 ton (T.) I Note. — In the United States Custom House, and in weighing iron and coal at the mines, the long hundredweight, containing 112 pounds, and the long ton, containing 2240 pounds, are used. One pound avoirdupois contains 7000 grains. 3. Troy Weight is chiefly used for weighing gold, silver, and jewels. TROY WtlGHT. 24 grains (gr.) =1 pennyweight (pwt.) 20 penny weights ^ 1 ounce (oz.) 12 ounces ^ 1 pound (lb.) Note. — A Troy pound contains 5760 grains. 4. Apothecaries' Weight is used in mixing and sell- ing medicines. Druggists, buy their medicines by Avoirdupois weight, but mix and sell them by Apothecaries' weight. apothecaries' weight. 20 grains (gr.)= 1 scruple (3) 3 scruples = 1 dram ( 5) 8 drams =: 1 ounce (S) 12 ounces ^ 1 pound (lb.) I. Reduce 2 lbs. 16 gr., Troy weight, to grains. 1. 1 lb. =12 oz. 2. 2 lbs. =2 times 12 oz.=24 oz. 3. 1 oz. =20 pwt. II. "(4. 24 oz. =24 times 20 pwt.=480 pwt. 5. 1 pwt. ^24 grains. 6. 480 .pwt.=480 times 24 grains=ll,520 gr. 7. 11,520 gr.-fl6 gr.=ll,536 gr. III. . . 2 lbs. 16 gr.=ll,536 gr. 24 FINKEL'S SOLUTION BOOK. UNITS OF LENGTH. 1. The Standard Unit of length is the yard. LONG MEASURE. 12 inches (in.) =1 foot (ft.) 3 feet = 1 yard (yd.) 5^ yards or 16^ ft. = 1 rod ( rd.) 40 rods = 1 furlong (fur.) 8 furlongs = 1 mile (mi.) Nbie. — A hand, used in measuring the height of horses, = 4 inches; a knot, used in navigation, = 6086 feet = 1.15 mi.; and a fathom, used in measuring depth at sea, = 6 feet. ^ 1. Reduce 63,244 feet to miles. 1. 3 feet —1 yard. 2. 63,244 feet =63,244-=-3=r21,081 yards+1 foot. 3. 5^ yards=l rod. 4. 21,081 yards=21,081-H5i=3,832 rods+5 yards. 5. 320 rods =1 mile. ' ^ 6. 3,832 rods =3,832-^320=11 miles+312 rods. III. ,-. 63,244 feet=ll mi. 312 rd. 5 yd. 1 ft. Explanation. — In step 4, in order to divide by 5J, we reduce 21,081 to halves obtaining 42,162 halves. 5^=11 halves. Di- viding 42,162 halves by 11 halves, gives 3,832 and a remainder of 10 halves or 5. So the remainder being, 10 half yards, gives 5 yards. CLOTH MEASURE. 2. This is an obsolete system formerly used in measuring cloth. I. Reduce 2 E. Fr. 1 E. En. 2 E. Fl. 3 yd. 2 na. to nails. II. < E. Fr. E. En. E. Sc. E. Fl. yd. qr. na. 2 1 2 3 2 6 5 3 4 12qr. 5qr. 6qr. 12 qr. 6 " 5 '• 12 " 35qr. 4 140 na. 2" 142 na. THE ENGLISH SYSTEM. 25 3. Surveyors' I/inear Measure is used by surveyors in measuring land. 4. The ptandard, or Prime, Unit is the ckatn, called Gunter's chain. ■surveyor's linear measure. 100 links (l.)=l chain (ch.) 80 chains =; 1 mi. Note. — 1 chain = 4 rods = C6 feet I. Reduce 3 chains to inches. 1. 1 chain =100 links. 2. 3 chains=8><;i00 links=300 links. 11 792 inches. 1 link = 7.92 inches. III. «. 1 link =7.92 inches. 4. 300 links =300X7.92 inches= .-. 3 chains=2,376 inches. :2,376 inches. UNITS OF SURFACE, OR SQUARE MEASURE. 1.. The Standard Unit of square measure is the square yard. It is a square each side of which is 1 yard in length. It is derived from the corresponding unit of linear measure and is, therefore, called a derived unit. SQUARE MEASURE. 144 square inches (sq. in.)^ 1 square foot (sq. ft.) 9 square feet =: 1 square yard (sq. yd.) 30^ square yards =1 square rod (sq. rd.) 160 square rods ^= 1 acre (A.) 640 acres ^1 square mile (sq. mi.) Remark. — Observe that the table of square measure is de- rived from the table of linear measure by squaring the correspond- ing units in linear measure. 2. 10 square chains:=l acre ; 1 acre:=4,840 sq. yds. ^43,560 sq, ft. I. 11. III. What is the difference between 10 feet square and 10 square feet ? 1. 10 feet square is a square each side of which is 10 feet in length. 2. .-. 100 square feet=10X10 square jfeet,=10X10Xl sq. ft^the surface of the square. 3. 10 square feet=10 units of which one unit is a square foot. 4. 100 square feet — 10 square feet=90 square feet, the numerical difference. . , 90 square feet is the numerical difference. 26 FINKEL'S SOLUTION BOOK. Explanation. — In finding the area of a rectangle, we multiply the number expressing its length by the number expressing its. width to obtain the number expressing its area. Thus, to find the- area of a rectangle 4 feet long and 3 feet wide : ' 4 is the number expressing its length in the unit of length and 3 is the number expressing its width in the same unit of length. Therefore, 4X3, or 12, is the number expressing its area in the correspond- ing unit of area, viz., the square foot. . It is not true that feet, multiplied by feet gives square feet. If it were, we might have circles multiplied by circles giving square circles, bottles multiplied by bottles giving square bottles,, and days multiplied by days giving square days. For further discussion, see Mensuration, page 239. I. What is the area of a ceiling 20 feet long and 16 feet: wide? C 1. 20=:number of feet in the length. II. ■] 2. 16=:number of feet in the width. (3. . . 20X16=320=number of square feet in its area. III. .-. The ceiling contains 320 square feet. UNITS OF VOLUME, OR CUBIC MEASVRE. 1. The Standard Unit of Volume is the cui>tc j/ard. which is a cube each edge of which is one yard in length. . CUBIC MEASURE. "1,728 cubic inches (cu. in)= 1 cubic foot (cu. ft.) 27 cubic feet =1 cubic yard (cu. yd.) Remarks. — 1. Observe that the table of cubic measure is de- rived from that of linear measure by cubing the corresponding- linear units. Thus, 1 cubic foot^l2X 12X12 XI cubic inch= 1,728 cubic inches; 1 cubic yard=3X3X3Xl cubic foot=27 cubic feet. 2. Firewood is measured by the cord, a cord being a pile 8" feet long, 4 feet wide, and 4 feet high and containing 128 cubic feet. In finding the volume, or solid contents, of a rectangular solid, we multiply the numbers expressing its dimensions together to- obtain the number expressing its volume. I. Find the cost of a pile of wood 24 feet long, 5^ feet high,, and 4 feet wide at |3 per cord. 1. 24 =number expressing the length of the pile in feet.. 2. 5|^=number expressing the height of the pile in feet. 3. 4 =number expressing the width of the pile in feet.. 4. 24X5^X4=528=the number expressing the volume II. -^ in cubic feet. THE ENGLISH SYSTEM. 2r III. 5. 128 cubic reet=l cord. 6. 528 cubic feet=528-^128=4^ cords. 7. |3=cost 5f 1 cord. 8. |12§=4-JXf3=cost of 4^ cords. . . The pile of wood will cost |12|. UNITS OF CAPACITY. 1. The Standard Unit of Capacity is the gallon. LIQlJlD MEASURE. 4 gills (gi.)= 1 pint (pt.) 2 pints = 1 quart (qt.) 4 quarts = 1 gallon (gal.) Note. — The capacity of cisterns, reservoirs, etc., is often expressed in. barrels (bbl. ) of 31 J gallons each, or hogsheads (hhd.) of 63 gallons each, \In old tables the following were given: a tierce {tr.)^42 gal.; a puncheon (pn.)=84 gal.; and a pipe (p.)=126 gal. A gallon contains 231 cubic inches. I. — Reduce 2 p. t pn . 1 tr 1 hhd. 1 gal. 1 qt. to pints _,..--- 126 ,^^-^ 84 ,^-^ 42 ,^^ 31i 4-- 63 4 2 p. pn. tr. bbl. T. hhd. gal. qt. pt 2 3 1 1 1 1 126 84 42 63 63 252 gal. 252 gal. 42 gal "63 gal. 42 252 252 610 gal. 4 2440 qt. 1 " 2441 " 2 48,82 pt. DRY MEASURE. 2 pints (pt.)= 1 quart (qt.) 8 quarts ^ 1 peck (pk.) 4 pecks = 1 bushel (bu.) Note. — A bushel is a cylindrical vessel 18 J inches in diameter and 8 inches deep and contains 2150.42 cubic inches. 28 FINKEL'S SOLUTION BOOK. iSolution : II.' Conclusion' Solution: II I. Reduce 2 bu. 3 pk. 2 qt. 1 pt. to pints. Equation Method. 1. 1 bu.=4 pk. 2. 2 bu.=2x4 pk.=8 pk. 3. 8 pk.-f-3 pk.=ll pk. 4. 1 pk.=8 qt. 5. 11 pk.=llx8qt.=88qt. 6. 88 qt.+2 qt.=90 qt. 7. lqt.=2 pt. , 8. 90 qt.=90x2 pt.=180 pt. 19. 180 pt.+l pt.=181 pints. III. .-. 2 bu. 3 pk. 2 qt. 1 pt.=181 pints. I. Reduce 529 pints to bushels. Equation method. fl. 2 pt.-;=l qt. 2. 529 pt.=529-=-2=264 qt.+l pt. 3. 8 qt.=l pk. 4. 264 qt.=264-i-8=33 pk. 5. 4 pk.^1 bu. 6. 33 pk.==38-T-4=:8 bu.+l pk. ill. .-. 529 pints=8 bu. 1 pk. 1 pt. many gallons will a tank ^ ft. long, 3 ft ■Conclusion: I. How wide, and 1 ft. 8 in. deep contain.'' 1. 4 ft.=length. Solution : '11. Conclusion : 111. 3 ft.=width, and 1 ft. 8 in.=lf ft.=depth. 4x3Xlf=20 cubic ft.=contents of tank. 1 cu. ft.=1728 cu. in. 20 cu. ft.=20Xl728 cu. in.=34560 cu. in. 231 cu. in.=:lgal. .-. 34560 cu. in.=34560-^231=149^f gal. •. The tank will contain 149|^^ gallons. (jFisk's Comp. Arith.,p. 126, f rob. 2.) III. ANGULAR OR CIRCULAR MEASURE. 1. Angular Measure is used to measure angles, direc- tions, latitude, longitude, in navigation, astronomy, etc. 3. An Angle is the amount of divergence between two lines which meet in a point. 3. An Angle is measured by the arc of a circle intercepted by the sides, the vertex of the angle being the center of the circle. It is proved in Geometry that. In the same circle, or equal circles, two angles at the center have the same ratio as their inter- cepted arcs. The two intersecting lines are called the sides of the angle. ANGULAR OR CIRCULAR MEASURE. 29' This being true, we may divide the circumference of a circle into any number of equal parts and adopt one of these equal parts as the unit of circular measure. Then, the angle sub- 'tended by this unit of circular measure may be adopted as the unit of angular measure. In Geometry, the circumference of a circle is divided into four equal parts and one of these equal parts is taken as the unit of . circular measure. It is called a quadrant. The angle subtended by the quadrant is the unit of angular measure and it is called a right angle. In Geometry, therefore, the nnit of' angular measure is the right angle. In Trigonometry, the circumference of the circle is divided into 360 equal parts and one of these equal parts is taken as the: unit of circular measure. It is called a degree of arc, or arc- degree. The .angle subtended by an arc-degree is taken as the" unit of angular measure and is called a degree, (°). However, the most common unit of ang,ular measure used in. trigonometry is the angle subtended by an arc of the circum- ference equal in length to the radius of the circle. This angle is called a radian. Since the radius is contained in a semi- circumference T times, it follows that w radians=180°. From. 180° this it follows that 1 radian= =57°. 3 nearly. Also, 1° = -ron'" radians, marked ysTr'^''''- ANGULAR OR CIRCULAR MEASURE. , 60 seconds (")= 1 minute ('). 60 minutes = 1 degree (°). 360 degrees r= l circumference. I. Reduce 72° to radians. 1) 180°=s: radians. 11.^ 2. 1°=YHK- of -^ radians=rj-7TiY radians. 3. 72°=72Xrr^ radians=47r radians. 180 • ^ III. .-. 72°=|ir radians. Note. — T represents 3.14159265 . . . . , 'the ratio of the circumference of ai circle to the diameter. I. Reduce f jt radians to degrees. TT ri- Tradians=180°. ^^- X 2. |r radians=f of 180°= 120°. III. .-. |7rradians=120°. .30 FINKEL'S SOLUTION BOOK. IV. TIME MEASURE. I. Time is a measured portion of duration. 2. The measures of time are fixed by the rotation of the earth ■on its axis and its revolution around the sun. 3. A. Day is the time of one rotation of the earth on its axis. 4. A. Year is the time of one revolution of the earth :around the sun. TABLE. 60 seconds (sec.) make 1 minute (min.) 60 minutes " lhour(hr.) 24 hours . " 1 day (da.) ' 7 days " 1 week (wk.) • 4 weeks " 1 lunar month (mo.) 13 lunar months, 1 da. 6 hr. " 1 year (yr.) 12 calender months •' 1 year. 365 days " 1 common year. 365 da". 5 hr. 48 min. 46.05 sec. " 1 solar year. 365 da. 6 hr. 9 min. 9 sec. " 1 sidereal year. 365 da. 6 hr. 13 min. 45.6 sec. " 1 Anomalistic year. 366 days " 1 leap year, or bissextile year. 354 days " 1 lunar year. 19 years " 1 Metonic cycle. 28 years " 1 solar cycle. 15 years " 1 Cycle of Indiction. 532 years " 1 Dionysian Period. 5. The unit of time is the day. 6. The Sidereal Day is the exact time of one rotation of the earth on its axis. It equals 23 hr. 56 min. 4.09 sec. 11. The Solar Day is the time between two successiv^e .appearances of the sun on a given meridian. 8. The Astronomical Day is the solar day, begin- ning and ending at noon. 9. Ths Civil Day, or Mean Solar Day, is the average of all the solar days of the year. It equals 24 hr. 3 min. ,56.556 sec. 10. The Solar Year, or Tropical Year^ is the time between two successive passages of the sun through the vernal equinox. II. The Sidereal Year is the time of a complete revolu- tion of the earth about the sun, measured by a fixed star. 13. The Anomalistic Year is the time of two suc- cessive passages of the earth through its perihelion. 13. A Lunar Year is 12 lunar months and consists of 354 day. TIME MEASURE. 31 14. Ji ]t£etonic Cycle is a period of 19 solar years, after ■which the new moons again happen on the same days of the year. 15. A. Solar Cycle is a period of 28 solar years, after which the first day of the year is restored to the same day of the week. To find the year of the cycle, we have the fol- lowing rule: Add nine to the date, divide the sum by twenty-eight; .the qiwtiejit is the number of cycles, and the remainder is the year of the cycle. Should there be no remainder the proposed vear is the twenty-eighth, or last of the cycle. The formula for the above rule is ) — rr — > in which x denotes the date, and r the re- l ^^ Jr maindtr which arises by dividing x-\-'i by 28, is the number "required. Thus, for 1892, we have (J892+9)-^28=67ff. . . 1892 is the ■25th year of the 68 cycle. 16. The Lunar. Cycle is a period of 19 years, after which the new moons are restored to the same day of the civil month. The new moon wilt fall on the same days in any two years •which occupy the same place in the cycle; hence, a table of tl>e moon's phases for 19 years will serve for any year whatever when we know its number in the cycle. This number is called the Golden Number. To find the Golden Number • Add 1 to the date, divide the sum by 19; thequotient is the number of the cycle elapsed and the remainder is the Golden JVutnbcr. The formula for the same is < (■ in which r is the re- mainder after divith'ng the date-|-l by 19. It is the Golden Number. , 17. A DionysUm or Paschal JPeriod is a period of .532 year, alter which the new moons again occur on the same day of the month and the same day of the week. It is obtained by multiplying a Lunar Cycle by a Solar Cycle. 18. A Cycle of IndUtion is a period of 15 years, at the end of which certain judicial acts took place under the Greek emperors. 19. JEpaCt is a word employed in the calender to signify the moon's age at the beginning of the year. The common solar year, containing 365 days, and the lunar year only 354, the difference is 11 days; whence, if a new moon fall on the first of January in any year, the moon will be 11 days old on the first day of the following year, and 22 days on the first of the third year. The epact of these years are, therefore, ■eleven and twenty-two respectively. Another addition of eleven 32 FINKEL'S SOLUTION BOOK. /. days would give thirty-three for the efact of the fourth year; but in consequence of the insertion ot the intercalary month in each third year of the lunar cycle, this epact is reduced to three. In like manner the epacts of all the following years of the cycle are obtained by successively adding eleven to the epact of the former 'year, and rejecting thirty as often as the sum exceeds that number. V. LONGITUDE AND TIME. In the diagram, the curve ACBD represents the path of the earth in its journey around the sun. This curve is called an ellipse. The ellipse may be drawn by taking any two points .^and F' and fastening in them the extremities of a thread whose length is greater than the distance F'F. Place the point of a pencil against the thread and slide it so as to keep the thread constantly stretched; the point of the pencil in its motion will describe the ellipse. The points i^and F' are called X\i^foci, the plural of focus. The sun occupies one of these foci. The plane of the earth's orbit, or path, is called the ecliptic. When the earth is at A it is nearer the sun than when it is at B. When the earth is nearest the sun it is said to be in perihelion (Gr. ■Kspt=^peri, near, and fjAKx; =:heHos, sun) ; when farthest from the sun, it is said to be in aphelion (Gr. d7:6=apo, from, and ^ho<; ^helios, sun). The points A and B, in the diagranj, represent the perihelion and aphelion distances, respectively. The earth is nearest the sun about the first of January and farthest from the sun about the first of July. It takes the earth 365 da. 6 hr. 13 min. 45.6 sec. to travel from A, west around through C, B, and D back to A. LONGITUDE AND TIME. 33 This period of time is called the anomalistic year. The west point as here spoken of, may be thought of thus : Suppose you were located at some point on the surface of the sun in a posi- tion enabling you to see the North Star. Then if you should face that star you would be facing north, your right hand would be to the east, and your left hand to the west, and south to your back. While the earth makes one revolution airound the sun, it rotates 366 times on one of its diameters. The diameter upon which it rotates is called its axis. The axis of the earth is in- clined from a perpendicular to the plane of the earth's orbit at an angle of 23J° . If the axis of the earth were extended indefi- nitely, it would pass very near, \\° , from the North Star. The earth's axis and the sun determine a plane, and this plane is of great importance in gaining a thorough understand- ing of Longitude and Time. Suppose you are on the earth's surface, facing the sun and in this plane. Then you will have noon, while just to the east of you it will be after noon and just to the west it will be before noon. The intersection of this plane with the earth's surface is called the trace of the plane on the earth's surface. This trace is called" a meridian. If you could travel in such a way as to remain in this plane for a whole day, that is, 24 hours, you would haVe noon during the whole time. But if you remain stationary on the earth's surface, you will be carried out of this plane eastward by the earth's rotation. You may conceive that you are at Greenwich, formerly a small suburban town of London, Eng., but now in- corporated in that city, with the sun visible and having such a position that you are in the plane formed by the earth's axis and the sun. It will then be noon to you at that place. Suppose we. take the trace of the plane, in this position, on the earth's surface as our standard meridian. Then all places east of this line will have had noon and all places west of it are yet to have noon. As the earth continues to rotate, rotating as it does from west to east, it will bring points west of the plane into coincidence with the plane and thus these points will have noon successively as they come into the plane. Suppose we start when Greenwich is in this plane, and mark the trace .of the plane on the earth's surface and every four minutes we mark the trace of the plane ; in this way, in a complete rotation of the earth, we will have drawn 360 of these traces, which we have agreed to call meridians. The distance of these lines apart, measured on the equator, is called a degree of longitude, better an arc-degree of longitude. Instead of measuring longitude from Greenwich entirely around the earth through the west, we generally measure it east and west to 180°. 34' ■ FINKEL'S SOLUTION BOOK. Thus, a place located on the 70th meridian, west, is said to be 70° west longitude, and a place situated on the 195th merid- ian, countmg from Greenwich around through the west, is said to be 185° east longitude. From the above discussion, we see that, since the earth turns on its axis once in 24 hours, 24hrs.=c=360=of long., or 360° of long.=o=24 hrs. 1 hr. =c^^ of 360°=15° of long., or 15° of long.=ol hr. 1 min.=c=j-ijy of ■'_5°=15' of long., or 15' of long.=c=l min. 1 sec. =^^jV of 15' ^15" of long., or 15" of long.=o=l sec. Hence, if we have the difference of longitude of two places, we can readily find the difference of time between these two places. For example, the longitude of St. Petersburg is 30° 16' E., and the longitude of _ Washington is 77° 0' 36" W. Now the dif- ference of longitude between these two places is 77° 0' 36" -j- 30° 16'=107° 16' 36". Hence, since 15°<>=1 hour, 107° 16' 36"=o= (107° 16' 36")-J-15, or 7 hrs. 9 min. 6.4 sec, which is the differ- ence of time between Washington and St. Petersburg. Conversely, if we know the difference of time between two places, we can easily find the difference of longitude. For example, the difference of time between New York City and St. Louis is 1 hr. 4 min. 47^ sec. Find the difference of longitude. ' 1. 1 hr. =:>15°. (For =o=, read " corresponds to.") 2. 1 min.=o=15'. II. { 3. 4 min.=c=60', or r. 4. 1 sec. =o=15". 5. 47^ sec. =o=47iXl5"=710"=ll' 50". III. Hence, the difference of longitude is 16° 11' 50". In some cases, problems are so proposed that we are, to find the longitude or time of one place, having given the longitude or time of another place and the difference of time or difference of longitude of the two places. Such problems require no prin- ciples beyond those already established. For example, the difference of -time between two places is 2 hr. 30 min. The longitude of -the eastern place is 56° W. Find the longitude of the western place. [1. Ihr. =c=15°. 2. 2hr. =2=30° 3. 1 min.=o=15'. II, { 4. 30 min.^>30X15'=450'=7° 30'. ,5. .-.37° 30'^=:difference of longitude. 6. .-. 66°+37° 30'=98° 30', the longitude of the west- ern place. STANDARD TIME. 35 Had the place whose longitude is given been in east longi- tude, we would have subtracted the difference of longitude to find the longitude of the western place. The following suggestions may prove helpful in the solution of problems in Longitude and Time : 1. When the longitude of a place is required, having given the longitude of some other place and the difference of longitude between the two places. Conceive yourself located at the place whose longitude is given. Then ask yourself this question, Is the place whose longitude is required, east or west? If west,, add the difference of longitude when the given longitude is west, and subtract if the given longitude is east. If the answer to your question is east, subtract the difference of longitude when the given longi- tude is west and add the difference of longitude when the given longitude is east. If the places are on opposite sides of the standard meridian, subtract the given longitude from the difference of longitude and the difference will be the longitude required, and will be oppo- site in name from the given longitude. That is to say, if the given longitude is east, the required will be west, and vice versa. 2. When the tiine of place is required, having given the time at some other place and the difference of time between the two places. Conceive yourself located at the place whose time is given. Then ask yourself this question. Is the place whose time is required, east or west of me? If the answer to your question is west, subtract the difference of time for the required time. If the answer to your question is east, add the difference of time for the required time. STANDARD TIME. In 1883, the railroad officials of the United States and Canada adopted what is called standard time. These officials agreed to adopt the solar time of some standard meridian as the local time of an extended area. The standard meridians thus adopted are 75th, 90th, 105th, and 120th. All stations in the belt of country 1\° wide on either side of these standard meridians have as local time the solar time of the respective meridian. For example, all points or stations in the belt of country 1\° wide on either side of 90th meridian, i. e., the belt of country lying between 82^° and 97-1° west longitude have as local time the solar time, or sun time, of the 90th meridian. In other words, all time-pieces of the various stations in this belt indicate the same time of day as clocks in the depots situated on the 90th meridian. For exam- ple, when the clock in the Union Depot at St. lyouis indicates noon, 12 o'clock M., the clocks in the Union Depots at Indian- 36 FINKEL'S SOLUTION BOOK. apolis and Kansas City also indicate noon, though at Indianap- olis it is a little more than 16 min. past nooii and at Kansas City it is a little more than 17 min. till noon, sun time. That is, the standard time and local time at Indianapolis differ by a little more than 16 min., standard time being about 16 min. slower than local time, and at Kansas City standard time and local time differ by about 17 min., standard time being about 17 min. faster than local time. If one were to set his watch with the railroad clock in the depot at Columbus, Ohio, then take the train for Springfield, Mo., on arriving at Springfield, Mo., one would find that his watch agrees with the clock in the Frisco» depot. This is be- cause Columbus, Ohio, and Springfield, Mo., are located in the belt of country having central time, i. e., having the sun time of the 90th meridian. How about the local time of these two places? The local time at Columbus, O. , is about 28 min. faster than standard time. The sun comes to the meridian of Columbus before it comes to the 90th meridian. When the sun comes to the meridian at Columbus it is noon, local time, but it will not be noon, standard time, until the sun comes to the 90th meridian, which will be about 28 min. later. Hence, local time at Columbus, O. , is about 28 min. faster than standard time. A passenger going from Columbtis, Ohio, to Springfield, Mo., and carrying standard time of Columbus, would have standard time at St. Louis, standard time and local tinie it St. I^ouis being very nearly the same. On arriving at Springfield, Mo., his watch would still indicate stand- ard time and would agree with- the regulator in the depot at Springfield, but his time would be about 8 min. faster than the local time at Springfield. The time in the belt of country between 67|-° west longitude and 82^° west longitude is called Eastern Time ; between 82^° W. and 97J° W., Central Time; between 97^° W. and 112J° W., Mountain Time; and' between 112|° W. and 127^° W., Pacific Time. We might call the time in the belt between 1\° K. and 7i° W,. Greenwich Time; between 1\° W. and 22|° W., East Atlantic Time; between 22^"^ W. and 37J° W., Central Atlantic Time; between 37^° W. and52J°W., West Atlantic Time; and between 62J° W. and 67^' W., Colonial Time. By some appropriate system of nomenclature, the naming of the time in the belt beginning with 127^° W. longitude might be extended. However, these names would have a very limited use and are therefore not worth coining. THE INTERNATIONAL DATE I.TNE. > 37 THE INTERNATIONAI. DATE I.INE. The International Date I^ine is an irregular line pass- ing through- Bering Strait, along the coast of Asia to near Borneo and Philippine Islands, and thence along the northern limits of the East Indian Islands, New Zealand, and New Guinea. It is the line from which every date on the earth is reckoned. At present, however, the 180th meridian is very generally used in its stead. Suppose one were standing on the 180th meridian at the time it is noon, Wednesday (say)^at Greenwich, and facing north. Then, just to right, or east of this line,. Wednesday is beginning, i. e., Wednesday 12 o'clock, A. M., while just to left, or west of the line, Wednesday is ending, i. e., Wednesday 12 o'clock, V. M. The difference of time between places immediately east and immediately west of the line is therefore 24 hours, and just west of the line it is one day later than just east of it. This is made still clearer by considering what takes place as the earth rotates on 'its axis; the places just west of the line will be carried eastward, and since these places had Wednesday ending, they must now have Thursday begin- ning. But these places are west of the line, the places east of the line still having Wednesday. Hence, it is clear that it is one day later just west of the line than just east of the line. In crossing this line, there- fore, from the, east one day must be added, while in crossing it from the west one day must be subtracted. Professor C. A. Young, in his General Astronomy, in answering the question, "Where does the day begin?" says, "If we imagine a traveler starting from Greenwich on Monday noon and traveling westward as swiftly as the earth turns to the east under his feet, he would, of course, keep the sun exactly on the meridian all day long and have continual noon. But what noon? It was Monday when he started, and when he gets back to London, twenty-four hours later, it is Tuesday noon there, and there has been no intervening sunset. Wlieu does Monday noon become Tuesday noon? The convention is that the change of date occurs at the ISOtb meridian from Greenwich. A ship crossing this line from the east skips one day in so doing. If it is Monday forenoon when the ship reaches the line, it becomes Tuesday forenoon the moment it passes it, the intervening twenty-four hours being dropped from the reckoning on the log-book. Vice versa, ■when a vessel crosses the line from the w^tern side, it counts the same day twice, passing from Tuesday forenoon back to Monday, and having to do its Tuesday over again." The consideration of this line in the solution of problems in longitude and time should add no serious .difficulty.- Solve the problem completely, leaving ouf 'of account the date line. Then, if the time of the given place is west of the line, while the place -whose time is required is east, we simply subtract a day, and if the conditions are reversed, we add a day. I. 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? {Raj's Higher Arithmetic, p. iji^prob. 7.) 38 FINKEL'S SOLUTION BOOK. II. 1. 157° 52' W.= longitude of Honolulu. 2. 151° 11' e;.= longitude of Sydney. 3. 309° 3'=differ- ence of long- itude meas- ured from Honolulu through Greenwich to Sydney. 4. 860°— 309* 3' = 50° 57'= difference of longitude measured directly on "the equator from the merid- ian through Honolulu to the meridian through Sydney. 5.15°=c=l hr, 6. 1°=^5=^V hr.:=4 mini 7.50° 57'=50fX°=50i^°=s=50^X4 min.=3 hr. 28 niin. 48 sec, difference of time. 8. 4 hr. 5 min., Sunday — 3 hr. 28 min. 48 sec. ^41 min. 12 sec, Sunday. 9. Regarding the date line, Stmday is changed to Mon- day, since Honolulu is east of the line, while Syd- ney is west of it. EXAMPIvES. 1. When it is 5 o'clock Monday morning at Paris, France, longitude' 2° 20' E., what is the hour and day of the week at Honolulu, Hawaiian Is- lands, longitude 157° 52' W.? ' Ans. 19 min. 12 sec. past 6 o'clock P. M., Sunday^ 2. When it is five minutes after 3 o'clock on Sunday morning at Hon- olulu, Hawaiian Islands, longitude 157° 52' W., whet is the hour and day of the week at Sydney, Australia, longitude 151° 11' B.? Ans. 41 min. and 12 sec. before 12 o'clock P. M., Sunday. 3. When it is 20 minutes past 12 o'clock on Saturday morning at Chi- cago, 111., longitude 87° 35', what is the hour and day of the week at Pekin, China, longitude 116° 26' E.? Ans. 56 min. 4 sec. past 1 o'clock P. M., Saturday.. 4. When it is ten minutes until 12 o'clock, Friday, midnight, at Con- stantinople, Turkey, longitude 28° 59' E., what is the hour and day of the week at Honolulu, Hawaiian Islands, longitude 157° 52' W.? Ans. 22 min. 36 sec. past 11 o'clock A. M., Friday.. 5. At what hour must a man start, and how fast must he travel, at the equator, so that it would be noon for him for twenty-four hours ? Ans. Noon ; 1037.4 statute miles per hr. EXAMPIvES. 39 6. What, is the difference of time between Constantinople, Turkey, and Sydney, Australia? Ans. 8 hr. 10 min. 48 sec. 7. A traveler sets his watch with the time of the sun at New York. He then travels from there and on arriving athis destination finds that his watch is 1 hr. 20 min. 30 sec. fast. WJiat is the longitude of his destination if the longitude of New York is 74° (/ 24" W.? Ans. 94° 7' 54" W. 8. When it is 1 o'clock P. M. at Rome, Italy, longitude 12° 28' E., what is the hour at New York, longitude 74° 0' 24" W.? Ans. 14 min. 6-5% sec. past 7 A. M. 9. When it is 1:20 A. M. at St. Louis, longitude 90° 15' 15" W., it is 8 hr. 35 min. 1| sec. A. M. at the Cape of Good Hope. What is the longitude of the Cape of Good Hope? Ans. 18° 30' 6" E. 10. When it is 3:55 A. M. at Constantinople, longitude 28° 58' 40" E., it is 6'hr. 50 min. 88 sec. A. M. at Bombay. What is the longitude of Bombay? Ans. 72° 53' 10" E. 11. When it is 6:33 P. M. at Jerusalem, longitude 35° 30' 48"^., it is 11 hr. 17 min. 13 sec. A. M. at Montreal. What is the longitude of Montreal? Ans. 73° 25' 57" W. ^ 12. When it is 1 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 Buffalo? Ans. 78° 53' 30" W. MISCELLANEOUS PROBLEMS. 1. How many links in 46 mi. 3 fur. 5 ch. 25 links? 2. How many acres in afield containing 1377 square chains? 3. How many cubic inches in 29 cords of wood? 4. In 1436 nails how many Ell English? 5. How many miles in 3136320 inches? 6. In 47 ft). 2 § 3 3 13 19 gr. how many grains? 7. Change 16 lb. 3 oz. 1 gr., Troy weight to Avoirdupois ■weight. 8. An apothecary bought by 4-voirdupois weight, 2 ft). 8 oz. of quinine at $2.40 per ounce, which he retailed at 20 ct. a scruple. What was his gain on the whole? 9. How many seconds in a Dionysian Period ? 10. How many seconds in th'e month of February, 1892. 11. How many seconds in the circumference of a wagon wheel? 12. How long would it take a body to move from the earth to the moon, moving at the rate of 30 miles per day. 13. If a man travels 4 miles per hour, how far can he travel in 2 weeks and 3 days? 40 FINKEL'S SOLUTION BOOK. 14. How much may be gained by buying 2 hogsheads of mo- lasses, at 40 ct. per gallon, and selling it at 12 cents per quart? Ans. $10.08 15. In 74726807872 seconds, how many solar years? Ans. 2368 years. 16. At |4 per quintal, how many pounds of fish may be bought for $50.24? Ans. 1256 pounds. 17. How many bottles of 3 pints each will it take to fill a hogshead? Ans. 168. 18. What will 73 bushels of meal cost, at 2 cents per quart? Ans. $46.72. 19. How many ounces of gold are equal in weight to 6 ft), of lead? ' Ans. 87|oz. 20. What is the difference betw^een the weight of 42f ft), of iron and 42.875 ft), of gold ? Ans. 52545 gr. 21. How many bushels of corn will a vat hold that holds 5000 gallons of water. , ^^5. 537-^bu. 22. A cellar 40 ft. long, 20 ft. wide and 8 ft. deep is half full of water. What will it cost to pump it out, at 6 cents a hogshead? .^«i-. $22,797+. 23. If a man buys 10 bu. of chestnlits at $5 a bushel, dry meas- ure, and sells the same at 25 cents a quart, liquid measure, how much does he gain? ^?z5. $43.09-j- gain. 24. How many steps, 2 ft. 8 in. each, will a man take in walk- ing a distance of 15 miles? Ans. 29700. 25. Hoi>v many hair's width in a 40 ft. pole, if 48 hair's width equals 1 line? 26. How many chests of tea, weighing 24 pounds each, at 43 cents a pound, can be bought for $1548? Ans. 150 chests. 27. How long will it take to count 6 million, at the rate of 80 a minute, counting 10, hours a day? Ans. 125 days. 28. How long will it take to count a billion, at the rate of 80 a minute, counting 12 hours a day? Ans. 29. What will 15 hogsheads of beer cost, at 3 cents a pint. Ans. $194.40. 30. How many shingles will it take to cover the roof of a building 60 ft. long and 56 ft. wide, allowing each shingle to be 4 inches wide and 18 inches long, and to lie -J- to the weather? Ans.20U0. 31. There are 9 oz. of iron in the blood of 1 man. How many men would "furnish iron enough in their veins to make a plow- share weighing 22^ lbs. ? Ans. 40. PROPERTIES OF NUMBERS. 41 CHAPTER VIII. PROPERTIES OF NUMBERS. DEFINITIONS. 1. The Properties of Numbers are^ those qualities which belong to them. m n >Ti O p: B 3 n •^ ^ g iTl B o 42 FINKEL'S SOLUTION BOOK. 3. An Integer is a whole number ; as, 1, 2, 3, etc. 4. A Prime Number is one that cannot be divided by- any other integers except itself and unity. Thus, 1, 2, 3, 5, 7, 11, 13, etc., are prime numbers. , JVoU. — There is no general expression known for the representation of primes. Thd numbers, 2P— 1, when p is prime, are known as Mersen- ne's numbers and are prime when p is 1, 2, 3, 5, 7, 13, 17, 19, 31, 61. It is- believed that these and 67, 127, and 257 are the only valines of ^ less than 257 for which 2P — 1 is prime. All values of p less than 257 have been tested, except the following twenty-three : 67, 7l, 89, 101, 103, 107, 109, 127,, 137, 139, 149, 157, 163, 167, 173, 181, 193, 197, 199, 227, 229, 241, and 257. The expression, x^-\-x-{-^\, is prime for all values of .*■ less than 41. The number, 2»— 1=2,305,843,009,213,693,951 is the largest number at pres- ent known to be prime. Euclid fcirc. 330 B.C. ) showed that the, number- of primes is infinite. 5. A Composite Number is a number that can be ex- actly divided by some other whole number besides itself and unity; as,, 4, 10, 12, etc. 6. Two numbers are prime to each other, when unity is the: only number that will exactly divide both ; as, 6 and 25. 7. An Mven Number is a number which is divisible by 2 ; as, 2, 4, 6, 8, etc. 8. An Odd Number is a number which cannot be di- vided by 2 without a remainder; as,*l, 3, 5, 7, 9, 25, etc. 9. A Perfect Number is one which is equal to the sum >of all its divisors ; as, C=l-|-2+3 ; 28=1+2+4+7+14. Noie. — It is probable that all perfect numbers are included in the formula 2;*— 1( 2P — 1), where 2P — 1 is prime; Euclid proved that all num- bers of this form are perfect, and Euler that the formula includes all even perfect numbers. W. W. Rouse Ball says, in his History of Mathematics, "There is reason to believe ^-though a rigid demonstration is wanting — that an odd number cannot be perfect." If we assume this statement to be true, then every perfect number is- of the above form. Then, \i p= 2, 3, 5, ?, 13, 17, 19, 31, 61, the correspond- ing perfect numbers are 2/>-M 2/>— 1 ) =B, 28, 496, 81 '8, 33o50336, 8589869056, 137438691328, 2305843008139952128, 265845599 1569831744654692615953842176. 10. An Imperfect Number is a number not equal to the sum of all its divisors ; and is either Abundant or Defective according as the number is less or greater than the sum of its. divisors; as, 12<l+2+3+4+6 and 101+2+5. 11. An Imaginary Number is a number arising from the extraction of an even root of a negative number; afe, / — 4, fl TT y — o, etc. th&unit of pure imaginary numbers is V — 1. For / — 1, we write i, the initial of the word, imaginary, a notation due to Gauss. Remark. — The limits of this work forbid any discussion of imaginary numbers. Such discussion will have to be looked for FACTORING. 4» in treatises on Algebra, oi which among the very best in the English language is Chrystal's, 2 vols. 12. A Divisor oi a number' is a number that exactly di- vides it. Thus, 3 is a divisor of 12. 13. A Multiple of a number is a number that exactly contains that number. Thus, 30 is a multiple of 6. I. FACTORING. ' 1. l<*actoi"iii^ is resolving composite numbers into factors ;. and depends upon the following principles: Principle 1. A factor of a number is a factor of any mul- tiple of that number. Principle S. A factor of any two numbers is also a factor of their sum or their diiference. From these principles are derived the following six propo- sitions : Proposition 1. Every number ending with 0, 2, 4, 6, or 8 is divisible by 2. Proposition 3. A number is divisible hy 4, when the number denoted by its two right-hand digits is divisible by 4. Proposition 3. A number ending in or 5 is divisible by 5. Proposition 4. Every number ending in 0, 00, 000, etc. is divisible by 10, 100, 1,000, etc. Proposition 5. A composite' number is divisible by the product of any two or more of its prime factors. Proposition 6. Every prime number except 2 and 5, ends with 1, 3, 7, or 9. Remark. — The proofs of the above propositions are easily made and are left for the student as exercises. Proposition 7. Any integer is divisible by 9 if the sum. of its digits' be divisible by 9. f 1. Every number is of the form al0''+&10"-i-|-cl0"-* + +1: 2. But al0»-|-&10'^i-|-cl0"-2_j____j_;_(j(9^1),f -|-&(9 + l)«-i-|-<;(9-)-l)'^2+. . . -|-/=aM(9) + Proof. \ o-|-&M(9) +& + cM(9) -t-c + ....+Z, where M(9) means a multiple of 9,=M(9)-l-a-f &-|-c- + ...+/. 3. .-. The number is divisible by 9 when the sum of its digits is divisible by 9. Proposition 8: Any integer is divisible by 11 if the dif- ference of the sums of the digits in the odd places and even places is divisible by 11. The proof of this proposition is the same as that of the last, except that, for 10, we write 11 — 1. ■44 FINKEL'S SOLUTION. BOOK. II. GREATEST COMMON DIVISOR. .1. jL Divisor of a number is a number that -will exactly divide it. 3. A Common Divisor 6f two or more numbers is a Tiumbei" tliat will exactly divide each of them. 3. The Greatest Cornvnon divisor, or Highest Comvion Factor, of two or more numbers is the greatest number that will exactly divide each of them. I. Find the G. C. D. of 60, 120, 150,, 180. 1. 60=2x2x3x6. 2. 120=2X2X2X3X5. II. <i3. 150=2X3X5X5. 4. 180=2X2X3X3X5. l5. G.C. D.=2X3X5=80. III. .-. G. C. D.==30. JExflanation. — By inspecting the factors of each nurtiber we ■observe that 2 is found in each set of factors; hence, each of the numbers can be divided by 2. But only once, since it is found only once in, the factors of 150. We also observe that 3 will divide the numbers only once, since it occurs only once in the factors of 60 and 120- Also, 5 will divide them but once, since 60, 120 and iSO contain it but once. Hence, the numbers, 60, 120, 150, 180, being divisible by 2, 3 and 5, are divisible by their product, 2X3X5=30. I. Find the G. C. D. of 180, 1260, 1980. rl. 180=2X2X3X3X5. .. l2. 1260=2X2X3X3X5X7. )3. 1980=2X2X3X3X5X11. l4. G. C. D.=2X2X3X3X5=180. - ' III. .-. G. C. D. of-,180, 1260, 1980=180. Explanation. — 2 being found twice in each number, they are each divisible by 2X2 or 4 ; also 3 being found twice in each number, they are each divisible by 3x3 or 9. 5 being found in each number, they are each divisible by 5. Hence, they are divisible by the product of these factors, 2x2x3x3x5=180. EXAMPLES. 1. Find the G. C. D. of 78, 234, and 468. 2. What is the G. C. D. of 36, 66, 198, 264, 600 and 720? 3. I have three fields : t:he first containing 16 acres, the second 20 acres, and the third 24 acres. What is the largest sized lots LEAST COMMON MULTIPLE. 45 containing each an exact number of acres, into which the whole can be divided ? vh/s. 4 A. lots. 4. A farmer has 12 bu. of oats, 18 bu. of rye, :^4 bu. of corn and 30 bu. of wheat. What are the largest bins of uniform size, and containing an exact number of bushels, into which the whole can be put, each kind by itself, and all the bins be full? Ans. 6 bu. bins. 5. A has a four-sided field whose sides are 2.56, 292, 384, and 400 feet respectively; what is the length of the rails used to fence it, if they are all of equal length and the longest that can be used? Ans. 4 ft. 6. In a triangular field whose sides are 288,450, and 390 feet respectively, how many rails will it require to fence it, if the fence is 5 rails high, and what must be the length of the rails if they lap over one foot? Ans. Length of rail, 7 ft. No. 940. III. LEAST COMMON MULTIPLE. 1. A. Multiple of a number is a number that will exactly contain it ; thus, 24 is a multiple of 6. 2. A Common Multiple of two or more numbers is a number that will exactly contain each of them. 3. The Least Comnion llultiple of two or more num- bers is the least number that will exactly contain each of them., v L Find the L. C. M. of 30, 40, 50. (1. 30=2X3X5. . j-r 1 2. 40=2X2X2X5. ' ^'- 13. 50=2X5X5. U. L. C. M.==2X2X2X3X5X5=600. III. .-. L. C. M. of 30, 40, 50=600. Explanation. — The L. C. M. must contain 2 three times, or it would not contain 40; it must contain 5 twice, or it would not contain 50; it must contain 3 once, or it would not contain 30. Since all the factors of the numbers, 30, 40, 50, are contained ia th'^ L. C. M., it will contain each of them without a remainder- I. Find the L. C. M. of 2310, 2.10, 30, 6. ri. 2310=2X3X5X7X11. 2. 210=2X3X5X7. IL <^3. 30=2X3X5. 4. 6=2X3. l5. L. C. M.=2X3X5X7X11=2310. III. .-. L. C. M. of 2310, 210, .^f'. ^,-='2310. 46 FINKEL'S SOLUTION BOOK. Explanation. — 2 and 3 must be used, else the L. C. M' would not contain 6. 2, 3, and 5 must be used, else the L. C. M- ■would not contain 30. Hence 5 must be taken with <the factors ■of 6. In like manner 7 must be taiken with the factors already taken, else the L. C. M. would not contain 210. The factor 11 must be taken with those already taken, else the L. C. M. would not contain 2310. Hence 2, 3, 5, 7, and 11 are the factors to be taken and their product 2310 is the L. C. M. I. The product of the L. C. M. by the G. C. D. of three numbers between 1 and 100 is 6,804; and the quotient of the L. C, M. divided by the G. C. D. is 84. What are the numbers? II. 1. L. C. M.xG. C. D.=6804, and 2. L-C.M. _g^ G. C. D. T- r M 3. .-. L. C. M.XG. C. D.^ p p ^ o r n <j. c u. G. C. D.Xt^^^^=f(G. C. D.)^=6804-=-84: =L. C. M.x 81. III. K_C.M 4. G. C. D.=y'81 =9, by extracting the square root, 5. .-. L. C. M.=6804-f-9=756. 6. 9=3x3. 7. 756=2x2x3x3x3x7. 8. 3x3x2x2=36. 9. 3x3x3x2=54. 10. 3x3x7 =63. .-. 36, 54, and 63=the numbers. Explanation. — Since 9 is the G. C. D., each of the numbers •contains the factors of 9. Since there are two 2's in the L. C. M., one of the numbers must contain these factors. In like man- ner one of'the numbers must contain three 3's; one of them must also contain 7. •■• We write two 3's for each of the numbers, two 2's to any set of these 3's, and 3 and 7 with either of the remain- ing sets, observing that the product of the factors in any set does not exceed 100. If we omit 2 in step 9, the product of the fac- tors is 27. Hence 27, 36, 63 are numbers also satisfying the con- ditions of the problem. EXAMPLES. 1. What is the L. C. M. of 13, 14, 28, 39, and 42.? 2. What is the L. C. M. of 6, 8, 10, 18, 20, 36, and 48? 3. What is the L. C. M. of 18, 24, 36, 126, 20, 48, 96, 720, and 84? 4. What is the smallest sum of money with which I can purchase a number of oxen at $50 each, cows at $40 each, or horses at $75 each? Ans. $600 FRACTIONS. 47 5. Find three numbers whose L. C. M. is 840 and G. C. D. •42. ' - Ans. 84, 210, and 420. 6. What three numbers between 30 and 140 having 12 for their G. C. D. and 2772 for their L. C. M. ? Ans. 36, 84, and 132. 7. At noon the second, minute, and hour hands of a clock are together; how long after will they be together again for the first time? 8. J. S. H. has 5 pieces of land; the first containing 3 A. •2 rd. 1 p.; the second, 6 A. 3 rd. 15 p.; the third 8 A. 29 p., the fourth, 12 A. 3 rd. 17 p.; and the fifth, 15 A. 31 p. Re- quired the largest sized house-lots, containing each an exact number of square rods, into which the whole may be divided. Ans. 1 A. 21 p. 9. The product of the L. C. M. of three numbers by their 'G. C. D.=864, and the L. C. M. divided' by the G. C. D.=24; ■find the numbers. Ans. 12, 18, and 48. CHAPTER IX. FRACTIONS. ^ Fraction is a number of the equal parts of a unit. (1. Simple. 2. Complex. 3. Compound. 2. Fraction.. , Common, < or Vulgar. .2. As to Value.' Proper. Improper. Mixed. 2. Decimal. 1. Pure. 2. Mixed. 3. Circulating. 1 1. Pure. 12. Mixed. 3. Continued Fractions. 3. A Cotrnmon, Fraction, or Vulgar Fraction, is one in which the unit is divided into any number of equal parts; and is expressed by two numbers, one written above the other, with a horizontal line between them. Thus, |- expi-esses five- .sixths. 4:. A. Simple Fraction is a fraction having a single integral numerator and denominator; as, f. 5. A Complex Fraction is a fraction whose numer- 4 2| -' i ^ 2* ator, or denominator, or both, are fractional; as, ^, ^, ~. 48 FINKEL'S SOLUTION BOOK. 6. A Con»potitn,d Fraction is a fraction of a fraction; as, f of |. 7. A. IProper F-fClCtion is a simple fraction whose numerator is less'than its denominator; as, -f. 8. An Inipi'OX>&t' Fraction is a simple fraction whose numerator is gi-eater than its denominator; as, -1. 9. A jyiixed JVumber is a whole number and a frac- ' tion; as, 3|. 10. A. Decimal Fraction is a fraction whose denomi- nator is ten, or some power of ten; as, -j?^, y|^^, -jf^Tr.' The de- nominator of a decimal is usually omitted and the point (.) is used to determine the value of the decimal expression. Thus, 3 Q 2 7 097 11. A Pure Decimal is one which consists of decimal figures only; as, .375. 13. A Wioced Decimal is one which consists of an integer and a decimal; as, 5.26. 13. A Circulating Decimal, or a Circulate, is a deci- mal in which one or more figures are repeated in the saine order; as, .2121 etc. When a common fraction is in its lowest terms and the denominator contains factors other than 2 or powers of 2, and 5 or powers of 5, the equivalent decimal fraction will be 7 circulating. Thus, -x-sw^^k^ — 5 — ?t will, when reduced to a decimal, be circulating because the denominator contains the factor 3. The repeating figure or set of figures is called a Kepetend, and is indicated by placing a dot over the first and the last fig- ure repeated. 14. A JPure Circulate is one which contains no figures but those which are repeated; as, .273. 15. A JiTixed Circulate is one which contains "one or more figures before the repeating part; as, .45342. 16. A Simple Kejietend contains but one figure; as, .3. n. A Compound Repetend contains more than one figure; as, 354. 18. Similar Repetends are those which begin and end at the same decimal places; as, .3467, and .0358- 19. Dissimilar Jtepetends are those which begin or end at different decimal places; as, .536, .835, and .3567. FRACTIONS. 49 30. ^1 Perfect Repetend is one which contains as many decimal places, less 1, as there are units in the denominator of the equivalent common fraction ; thus, |=.]42S57- 31. Contei-iuinous liepetendH are those which end at the same decimal place; as, .4267, -3275! ind .0321. 33. Co-originous Mepetetids are those which begin at the same decimal place ; as, .378, .5624) and 3-623. 23. A Continued Fraction is a fraction whose numera- tor is an integer and whose denominator is an integer plus a fraction whose denominator is also a fraction, and so on. Thus, f 1 , is a continued fraction, and is equal to the simple fraction -f^. For convenience , in printing such fractions, modern writers have adopted the following method of expressing them: f +J+|+|^. The plus sign being placed between the denominators of the several fractions readily distinguishes" the continued fraction from the expression ■§+J+-|+f , where we wish to indicate that the several simple fractions are to be added together. 34. Discussion of De&nitions. From the definition of a fraction, viz., that it is a number of 2 21 1.2 the equal parts of a unit, it follows that -^, -rf , — ^, and ^ are not fractions. When the idea of a fraction originated the above definition incorporated that idea. But this idea being new became the starting point of numerous 'Other new but related ideas, and in the course of time the new ideas diverged so far from the original idea that, the original description of the original idea became in- adequate to accurately comprise all the new ideas which have thus sprung from it. In all such cases in mathematics it is the custom to enlarge the original description of an idea so as to in- clude new ideas.' Thus while fraction originally meant a num- ber of the equal parts of a unit, its meaning has now been en- larged so as to include such expressions as -^rj, ~, ^"'"ij.i, -y^ , etc. So too the idea of number has been enlarged until it comprises the ideas represented by such expressions as ■J2 _ ' f, 5, -=, /7, 3-2, i/-3, 2-H1/-4, etc. V o ' •'50 FINKEL'S SOLUTION BOOK. Is $|- a fraction ? Yes, for it is a number of fourths. From, the primary idea of a fraction, viz., that it is a number of the equal parts of a unit, it, of course, follows that there cannot be a greater number of parts takeil in a unit than the parts into which [t has been divided, and, therefore, f f would not be a fraction. How is f I read? f of a dollar. For a valuable discussion of such questions the reader is referred to Brooks' Philosophy of Arithmetic, a book that ought to be read by every teacher of arithmetic. I. REDACTION OF FRACTIONS. 1. Reduction of Fractions is the process of changing their form without altering their value. There are six cases of reduction, viz., 1st. Integers or mixed num- bers to fractions. 2d. Fractions to integers or to mixed numbers. 3d. Fractions to. lower terms. 4th. Fractions to higher terms. 5th. Compound fractions to simple fractions. 6th. Complex fractions to simple fractions. CASE I. To reduce integers or mixed numbers to improper fractions. I. Reduce 9f to an improper traction. ri. l=|,=5-fifths. TI. ] 2. 9=9 times |— *^',=9 times 5-fifths=45-fifths. ( 3. 9|=9+|=^+|=^,==45-fifths+3-fifths=48-fifths. III. .-. 9|=^,=48-fifths. Remark 1. — Steps 1 and 2 show how to reduce an integer to an improper fraction. Remark 1.- — In the above solution, we have indicated two ways of writing a solution. Thus, for example, in step 2, we may say at once that 9=9 times 5-fifths=45-fifths. Then 45-fifths+3- fifths=48-fifths. In this way, pupils see a similarity between add- ing -%5. and f , and 45 apples and 3 apples. CASE II. To reduce an improper fraction to an integer or a mixed number. I. Reduce ff to a mixed number, fl. ^|=ll-elevenths=l. (2. |f=49-elevenths=49-elevenths-^ll-elevenths=4T^. III. , .-. n='^- Explanation. — If we consider ff as 49-elevenths and not as 49 divided by eleven, we can not divide the numerator, 49, by the FRACTIONS. 51 denominator, 11, in order to reduce it to a mixed number, because the denominator denominates, or names, the parts. From this point of view, 11 in f f , bear the same relation to 49, that pecks do to 25, in 25 pecks. 25 pecks consist of two parts, viz., (1) a unit of measure, the peck; .(2) 25, called the numeric. 25, the nu- meric, shows how many times the unit of measure is contained in the quantity measui'ed with the peck. So too, 49 in -^f is the numeric showing how many times the unit of measure, one- eleventh, is contained in the quantity measured with the one- eleventh. ' By convention, however, a fraction may be considered as an unexecuted division, the numerator being considered the dividend and the denominator the divisor. From this point of view, we may simply divide the numerator by the denominator. From the first point of view, 49-elevenths equals as many times 1 as 11-elevenths is contained in 49-elevenths which is 4-]^ times 1, or 4^^. ' In presenting this principle to classes, every member of a ■class should have a thorough comprehension of it, but it should not be made a hobby by the teacher so that the other method be excluded entirely. CASE III. To reduce a fraction to lower terms. 1. Reducing a fraction to lower terms is the pro- cess of reducing it to an equivalent fraction having a smaller numerator and denominator. ' I. Reduce -^ to its lowest terms. -h A A A A A A A A H H « FIG. {. Let AB be a line of unit length, and let it be divided into 12 ■equal parts. Counting 9 of these parts, we get t'a of AB, or A of 1. or A- Now if we consider the line divided into 4 equal parts and count 3 of these we get fof AB, or f of ly or |. It is thus seen that A of AB=| of AB. Hence, ^=1 In the same way it may be shown that A=*; A=S • i=i, etc. On comparing y\ and |, it is seen that f may be obtained from ^ by dividing both numerator and denominator of A by 3. In like manner, f may be obtained from ^ by dividing both numerator and denominator of ^\ by 4 ; and that | may be ob- 52 FINKEL'S SOLUTION BOOK. tained from f by dividing b6th numerator and denominator of f by 2. , • Hence, in general, to reduce a fraction to lower terms, divide both numerator and denominator by any common divisor, and to reduce a fraction to its lowest terms, divide both nun\erator and denominator by their greatest common divisor, it being in its I^owest Xerms when its numerator and denominator have no common factor. I. Reduce f to its lowest terms. "8-2' f h I ^ Wmfm, "^m, ^ WM. wmm^ B 2 If FIG. 2. Let- the line AB represent a unit ; and let it be divided into 8 equal parts as indicated by the figure. The row of characters above the line numbers the parts beginning at A. The characters below the line number the parts of the line when it is considered as divided into 4 equal parts. It is thus seen that f of AB=f of AB, or I of a unit= :f of the same unit-. Now the fraction | may be obtained from % by dividing both numerator and denominator of I by 2. lience, if both numerator and denominator of | be divided by 2, the value of the fraction is not changed. In the same way, we can show that %-=^\-=\; -^^■^=\\ it=f. etc., it being understood by these equalities that we mean, for example, \^ of a. unit=| of the same unit. We thus arrive at the general Principle: If both numerator and denominator of a fraction he divided by the same number, the value of the fraction remains unchanged. CASE IV. To reduce a fraction to higher terms. I. Reduce f to ninths. 2X3 „ 1- I 'SXS' FIG. 3. FRACTIONS. 53 Let AB represent a unit and let it be divided into three" equal parts of which two parts are taken. The shaded part of the line represents f of the line AB. Now if we divide each one of these three equal parts into three equal parts, the whole line AB will be broken into 9 equal parts of which the shaded part contains 6 of the 9 parts. Hence, f of AB^f of AB, or f of a unit^| of the same unit. Now f may be obtained from f , by multiplying both numera- tor and denominator of f by 3. From the figure, we also see • that i=^. In the same way, we can show that f ^^^ ; i^TS > i=A, etc. We thus arrive at the general Principle. — // both numerator and denominator of a frac- tion be multiplied by the same number, the value of the fraction remains unchanged. • CASE V. To reduce compound fractions to simple fractions. I. Wiiat is f of I ? II. III. ri. ^ of \=^ of one of the 5 equal parts into which a unit , therefore, one of the 15 equal a unit is divided, that is, ^ of -^ .has been divided parts into which — 1 — 15"- J of 1=4 times J of ^=4 times 1^5=1*3-. f of 1=2 times j of t=2 times ^%=^\. Sof|=ft. iV/AB|'?/t FIG. 4. Explanation. — Let AB represent a unit, and let it be divided into 5 equal parts of which 4 are taken as represented by the shaded part of the line. To take t of | requires, us to divide the shaded part of the line AB into three equal parts and take 2 of them. Thus f of the shaded part of AB=AF. This division may be made by dividing each of the fifths of AB into three equal parts. This is equivalent to dividing AB into 15 equal parts, f, therefore, contains 12 of the 15 equal parts, and | of | or AF contains 8 of the 15 equal parts. Hence f of |=x\. Now -^ is obtained from f and f , by multiplying their numer- ators together for the numerator of the result of the operation and their denominators together for the denominator of the result. 54 FINKEL'S SOLUTION BOOK. CASE VI. To reduce complex fractions to simple fractions. I. Reduce -|- to a simple fraction. This means that f is to be divided by J. Now it is clear that f is 2 times ^ or that f contains i^, twice. I. Reduce -4- to a simple fraction, f This means that f is to be divided by | . By division of frac- tions it is found that f contains •§ , If' times. Remark. — It must not be forgotten that a complex fraction is not a fraction according to the primary idea of a fraction. It is ■I impossible to interpret-|-as a fraction, if we take as our definition of a fraction, that it is one or more of the equal parts into which a unit has been divided. But ideas of fractions are extended to include swch expressions under the term, fraction. Reduction of complex fractions to simple fractions properly comes under Division of Fractions. In the complex fraction above, 3 and 4 are called the means and 2 and 5 the extremes. By the principle of division of fractions, it will be seen that the reduction is effected by multiplying the means together for the denominator of the resulting fraction and the extremes together for the numerator. This operation may often be shortened by cancellation. Thus, ^"■ 2 2 Observe that in the process of cancellation, we say that 5 di- vides 5 and 10, going in 5 once and into 10 twice. The neglect of this often leads pupils to conclude that the result of such can- cellations as the following 2X5X7X11 . „ . ^ , . , 11X2X5X7 ^"^'^°^^^^^°^^- It may not be out of place, to remark here, that with a certain class of teachers overcome with the desire to make everything in education easy, the complex fraction has fallen into disrepute. These teachers strongly advocate the omission of complex frac- tions in our arithmetics and omit the presentation of them in their classes. ' This is all done, it is claimed, for the benefit of the student. As a matter of fact it results in the student's eternal injury. The argument is that since such fractions are seldom encountered in business, it is a loss of time and eftergy to give them any attention in the schoolroom. Such a doctrine loses sight of the fundamental object of an education, viz., the libera- FRACTIONS. 55 tion of the powers of the mind. Education should be directed to the development of the mental powers and not to specialization for business. As a result of such bad teaching and bad pedagogy, it is often a pitiable sight to see average students in Analytical Geometry or Calculus struggling with arithmetical operations and complex fractions, simply because they did not receive the proper training in Arithmetic. In the higher branches of mathe- matics complex fractions occur frequently and the student who takes up these subjects needs all his energy to develop the prin- ciples of these subjects and should not be forced, by necessity, to dissipate any of it in struggling with arithmetical principles. II. ADDITION OF FRACTIONS. I. Addf, I, |. FIRST SOLUTION. II. III. 1. 2. 3. 4. I 6. f L. C. D.=24-twenty-fourths. l=24-twenty- fourths. |:=| times 24-twenty-fourths^l8-twenty- fourths. |-=|- times 24-twenty-fourths=20-twenty- fourths. |-==|- times 24-twenty-fourths^21-twenty fourths. Durths. -|-H-|-=18-twenty-fourths + 20-twenty- fourths -)-21-twenty- fourths =59-twenty- fourths=2-^J. SECOND SOLUTION. n. 1. L. C. D.=24. 9 1 2 4 q 3 3 V 24 1 8 J. 5 5 v 24 20 5- i=ixH=ii. III. SUBTRACTION OF FRACTIONS. I. Subtract f from ^. (1. L. C. D.=40. 2. 1=^, =40-fortieths. 3- |=f Xf|=2|^ =|X40-fortieths=25- fortieths. 56 FINKEL'S SOLUTION BOOK. II. -i III. ^=^Xft=4^, =AX40-fortieths=36- fortieths. ,9,-|=ff-|f=tt, =3,6-fortieths-26- fortieths=ll-fortieths . A— l=tt, =ll-fortieths. IV. MULTIPLICATION OF FRACTIONS. To multiply a fraction by an integer. I. Multiply f by 2. i f 1 f I B FIG. 5. Let the line AB represent a unit and let it be divided into 5 equal parts. Then AC=| of AB and AD=f of AB. But AD is twice AC. Hence, the fraction f is twice the fraction f . Now f may be obtained from f by multiplying the numerator, 2, of | by 2. In the same way we can show that f is three times -f ; that ^ is five times f , and so on. Hence, the Principle. — Multiplying, the numerator of a fraction by any integer multiplies the fraction by that integer. I. Multiply f by 4. * I it t H I I f ¥ ¥ ^'- % ^////A//////m/////\///////A'///////, iwj three -f four^ FIG. 6. Let the line AB represent a unit and let AC represent f of AB, or f of a unit. Now if AC be used as a measure and applied 4 times along the line AD, beginning at A, the second applica- tion carries the extremity of AC to f , the third to f, and the fourth to ^-. Hence, ¥"=^ times f . But' we know by Case III. of Reduction of Fractions, that V^-f- Hence, f=4 times f. But ■§ may be obtained from f by dividing the denominator, 8, of f by 4. In like manner, we can show that f ^2 times f , that 1=3 times ^, that f ^4 times -fg, and so on. Hence, the Principle. — Dividing the denominator of a fraction by any integer, multiplies the fraction by that integer. FRACTIONS. 57 Remark.' — To multiply a whole number by a fraction is the same as multiplying the fraction by the whole number, so far as the result is concerned. Thus 3 times f^f times 3. To multiply a fraction by a fraction. I. Multiply t by f . To multiply t by, f is to take f of f. i f ^ FIG. 7. Let the line AB represent a unit and let AC represent f of AB, or f of a unit. Now if AB be divided into 15 eqilial parts, -% of AB, or AC, will contain 10 of the 15 equal parts. Hence, f of AB=}-§^ of AB. By this division, we may consider AC divided into ft equal parts, and AG will represent f of f . But AG also represents -^. Hence, f of -1=^. Now -f^ may be obtained from f by mul- tiplying the numerator by 4 and the denominator by 5, that is, ^ is obtained by multiplying the numerators of f and f for the numerator and their denominators together fori the denominator •of y\. Hence, to multiply a fraction by a fraction, we have the Rule. — Multiply the numerators together for the numerator cf the product and the denominators together for its denominator . I. Multiply I by I . ri. ftimesi=J. II. X 2. \ times l=\ of- 1=^. (.3. I times |=5 times A=|i. • • |X|=||. I. Multiply f by 3. .^y ri. 1 times f:=f. \2. 3timesf=J74. .-. -fX3=J^i. I. Multiply 7 by %. C 1. I times 7=7. II. j 2. i times 7=i of 7=|. (3. f times 7=3 times J=-y-. .-. 7X|=-V-. V. DIVISION OF FRACTIONS. To divide a fraction by an integer. I. Divide | by 3. 58 FINKEL'S SOLUTION BOOK. 7 A 7 f f m////m////mm//m/m^ . B FIG. 8. Let the line, AB, represent a unit and let AC represent f of AB. Now f is 3 times ^. Hence, -f is f ^3. But -f is obtained from f by dividing the numerator, 6, of f by 3. In like man- ner, it may be shown that ^=t^^2, J;hat \=^-^b', and so on. Hence, the " Principle. — Dividing the numerator of a fraction by any integer divides the fraction by that integer. I. Divide f by 2. * * ^ mMm/M mm. B f FIG. 9. Let the line, AB, represent a unit and let AC represent f of AB, or f of a unit. AC=2 times AD, or | =2X1, or |=2Xf ^ . . f =f-f-2. But f may be obtained from |, by multiplying the denominator of f by 2. In like manner, we can show that ■j^^f^2, that -l^f-^S, that -^-^-^^-^b, and so on. Hence, the Principle. — Multiplying the denominator of a fraction by any integer, divides the fraction by that integer. To divide an integer or a fraction by a fraction.. I. Divide 2 by f . .1 ' I I I I i "/AB f *"/BCf FIG. 10. Let the line, AC, represent two units, AB and BC each repre- senting a unit. Then AB contains 4-fourths and BC contains 4-fourths. Hence, AC contains 4-fourths+4-fQurths or 8- fourths. That is, AC or 2 contains AE or \, 8 times, or 2 con- tains I 8 times, or 2 divided by i is 8. AF is J of AC, that is, AF is J of the number of times i is contained in AC, that is f times. Hence, f is contained in 2, f times. Now f is ob- tained from f , by writing it f , that is by inverting it,, and. multi- FRACTIONS. 59> plying the numerator of f thus written by 2. In like manner, we can show that f is contained in 4, ^ times, that f is .con- tained in 3, ^ times, etc. Hence, the Rule. — To divide any number by a fraction, invert the di- visor and multiply by the fraction thus inverted. Remark. — In presenting the subject of fractions to a primary- class, great care should be taken to make the fundamental prin- ciples stand out .prominently. Thus it should be made clear that in a unit there are 2-halves, 3-thirds, 4-fourths, 5-fifths, and so- on. How many times is \ contained 'in 1, how many times is \, how many times •^? How many eighths in 1, is the same ques- tion, so far as result is concerned, as how many times is \ con- tained in 1? By appropriate object-teaching these fundamental facts can be strongly and, firmly impressed upon the mind of the pupil. Thus, take a foot-rule and apply it 3 times to a straight line drawn on the blackboard, how many feet have been measured ?' Take a 6-inch rule, or -J-foot rule, and apply it to the same line, beginning and ending at the. same points as in the previous meas- urement, how many times will you have to apply the ^-f oot rule 1 6 times. Then you have measured 6 {\ ft.), or 3 feet. Hence, 6 times \ ft.=3 feet. In 3 feet there are 6 half-feet. Use a \ ft. rule, and measure the same segment of a line as before. How many times do you apply the unit of measure? 18 times. The result of the measurement or the length of the segment of the line measured is 18 (^ ft. ) The unit of measure is \ ft. In 1 ft., how many {\ ft.)? Ans. 6. 6 what? 6 (^ ft.) By' continu- ing such examples older students can be made to understand thor- oughly the meaning of such limiting processes as 2-^0. Thus measure a line segment 2 feet long with a foot-rule. It fe applied twice. Measure the same line segment with an inch rule. It must be applied 24 times. Measure the same line segment with a -^ in. rule. It must be applied 240 times. Measure the same- line segment which is 24 inches long with a xir^Tnr i^- rule. It must be applied 240,000 times. In this way, the pupil sees that if any quantity is divided or measured by a quantity which is taken as a unit and made very small, the number of times this- measuring unit is contained in the quantity becomes very large. Of coui-se, we can never take as a measure no part of something. But we can take as measuring unit a quantity as small as we please and thus its application to the quantity to be measured will be correspondingly large. The limit towards which such small measuring units tend is 0, and when we take one smaller than any assignable or expressible quantity, the number of times it is con- tained into the finite quantity to be measured is larger than any assignable or expressible quantity. A quantity larger than any assignable quantity is called an infinite quantity and is desig- nated by the symbol oo . Now we cannot divide 2-f-O, because we -60 FINKEL'S SOLUTION BOOK. have no unit of measure. But such operations often arise as Hmiting cases in mathematics and must be properly interpreted. The real meaning is accurately expressed as follows : 2-f-A, as h approaches 0, approaches co . When ,h becomes inexpressibly small 2-^h becomes inexpressibly large, that is, oo . The sym- Jxilic statement of the above expression is 2"! In the same way 0^2, may be made intelligible ; also 2-=-od ; 00 -^3 ; 5Xqo ; 0X6 ; and many other expressions cornmonly met with in Higlie,r Mathematics, but of which the student has n6t the slightest comprehens'iofi until made intelligible by the teacher. The expression 0-^0 should also receive attention. When the student for the first time comes across this in Algebra, his first answer is 1. He reasons thus 2-^2=:! ; 5-^5=1 ; 7-^7^1 ; 10-^10=1, so then must 0-^0=l. While 0^0 and 0X0 are impossible operations, the student might at this stage be told that these operations should be made to obey the same laws as quanti- tive symbols. Hence, 0-hO=1, 3, 7, f , 27, or any other number, :since in the case of quantitive symbols, the dividend ;= the quo- tient X the divisor, so 0^0-Xl, 0X3, or times any other number. VI. THE GREATEST COMMON DIVISOR OF FRACTIONS. 1. The Greatest Common Divisor of two or more fractions is the greatest fraction that will exactly divide each of them. 2. One fraction is divisible by another fraction when the numerator of the divisor is a factor of the numerator of the divi- dend, and the denominator of the divisor is a multiple, of the ■denominator of the dividend. Thus, f is divisible by ^g ; for f^fg ; |f^^^=10. 3. The greatest common divisor of two or more fractions is ■that fraction whose numerator is the G. C. D. of the numerators and whose denominator is the L. C. M. of the denominators. Hence, for finding the G. C. D. of two or more fractions, we have the following Rule. — Find the G. C. D. of the numerators of the frac- tions, and divide it by the L. C. M. of their denomitiators. Remark. — The fractions should be in their lowest terms before the rule is applied. I. Find the G. C. D. of f , f , ■^. [ 1. 3^G. C. D. of the numerators, 3, 6, and 9. I 2. 140=L. C. M. of the denominators, 4, 7, and 10. FRACTIONS. 61 III. II. ^ 3. T-|Ty=3-^140, the G. C. D. of the numerators divided by the L. C. M. of the denominators, =the G. C. D. of the fractions. •. if^ is the G. C. D. of f , f, and t%. A farmer sells 173| bushels of yellow corn, 478^ bushels of white corn, 2,093J bushels of mixed corn : required the size of the largest sacks that can be used in ship- ping, so as to keep the corn from being mixed ; also the number of sacks for each kind. (R. H. A., page 93, problem 8.) 1 iD7i_276 _oX5Xll I. i.6(^ — ~o~> — o 11.^ 2. 4781= III. 2 3825 3. 2093f= 8 ' 8375 3X3X5X5X17 3X3X5X5X37 25=G. C. D. of the numerators. 8^L. C. M. of the denominators. 3i=25-:-8,=the G. C. D. of the fractions. 44^137-|-H3-J,=the number of sacks required for the yellow corn. 153:=478^-H3-J,=the number of sacks required for the white corn. 670=:2,093|-;-3i,=the number of Sacks required for the mixed corn. [ 1. The capacity of the largest sacks required is 3^ bushels ; and, the number of sacks required for each kind of corn is 44 for the yellow, 153 for the white, and 670 for the mixed. A 2. VII. LEAST COMMON MULTIPLE OF FRACTIONS. 1. The I/cast Common Multiple of two or more frac- tions is the least number, integer or fraction, that each of them will divide without a remainder. 2. A fraction is a multiple of a given fraction when its numerator is a multiple, and its denominatpr is a divisor, of the numerator and denominator, respectively, of the given fraction. Thus, j\ is a multiple of /j- ; for the numerator, 8, is a multiple of the numerator, 2, and the denominator, 11, is a divisor of the denominator, 33. 3. 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? com- mon divisor of the given denominators. Hence, for finding the L. C. M. «f two or more fractions, we have the following ^2 FINKEL'S SOLUTION B^OK. Male. — Divide the L. C. M. of the numerators by the G. D. of the denominators. < -^ ' I. Find the least common multiple of f , f , f , and -|- 1. 12=L. C. M. of the numerators, 2, 3, 4, 6. II. \ 2 1=G. C. D. of the denominators, 3, 4, 5, 7. 3. 12^12H-l,=the L. C. M. of the fractions. .-. 12=the L. C. M. of f, f , f , and f . Ill I A can walk around an island in 14f hours; B in 9^;^ 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 agam ; II. 1. 14^= 100 2X2X5X5 9t^= ■ 7 ' .100 11 50 3. 161=-^,=: 2X2X5X5 11 ' 2X5X5 3 1 III. 3 5X5 .1 100=L. C. M. of the numerators, 100, 100, 50, and 25. , 1=G. C. D. of the denominators, 7, 11, 3, and 1. 100=100-^l,=the L. C. M. of the fractions. In 100 hours after starting, the four men will be to- 4. 25 =-^,= 5. 6. 7. gether again at the point of starting. MISCELLANEOUS PROBLEMS. Solution: Conclusion • II. III. I. II. i Reduce 9^ to an improper fraction. ^ 9^=9+|. l=|=8-eighths. n_nN/«_7,_nN/Q _:_t-.u :72-eighths. '1. 9^: 2. l=|=»-eiglitJis. 3. 9=9X|=-V-=9X8-eighths 4. -\?-+f=-\a-=79-eighths. 7 2 _ _ ^ :-V-=79-eighths. Reduce f to 24ths. 1. f=ff, or 8-eighths=24-twenty-fourths. 2. i=i of |f=A- or l-eighth=i of 24- twenty-fourths:^3-twenty-fourths. f =5 times ^^4=if , or 6-eighths=5 times 3-twenty-fourths:=15-twenty-fourths. L^i4^1 .'i-t\iirpn1-v-fniir1->is 3. l^ d-twenty-tourths:=lo-t\ III. .'. |=if=15-twenty-fourths. FRACTIONS. 63 Reduce- -I to 8ths. 1. 1=1, or 6-sixths=8-eighths. o i_i „f 8_f — i— li, or l-sixth=^ of 8- L. t— * 01 t— g— g— g eightlis=li^ II. { eighths. 3. f=5 times ^|=f ' "^t'^^l^r^ tT^ ° 8 8 l^eighths=6f- eighths. ril. .-. f=^=6f -eighths. I. Reduce \ to 3rds. rl. |.=|=3-thirds. n. \ 2 1^1 „f 3-i_lJ=li-thirds=J of III. .■.\. -li- "3^3 :l|-thirds. 3-thirds. I Explanation. — In taking \ of |, we must divide the numer- ator by 2. The denominator must be left unchanged ; for that •is the denomination to which the given fraction is to be reduced. I. Reduce f to llths. . 4^+4-:=ll-elevenths. II. III. 1 1 f 11 ^ ?i \ of 11-elevenths t— t oi rx—\\ 11 =2|-elevenths. 1=3 times f|=f|=^ times 2i-eleveuths ^ 11 11 =:6f-elevenths. |=r-^=r6|-elevenths. [. Reduce ^- to a mixed number. IT 1 1- ^=1- ■'1. 3-thirds=l. p. 2. 29-thirds^as many times 1 as 8- '^' thirds is contained in 29-thirds, which is 9|. III. .■.-V-=9f I. Reduce |. f, | to their I,. C. Denominator. r 1. L. C. D.=12. II. 2. 12 4. |=2Xt^=A. 6. ^iofi|=A. 7. |=3xA=iV- 8. 1=^. 9. i=Tofif=TV LlO. |=5XT\=if- m. ••■|.f.i=iV,TV,H- 64 FINKEL'S SOLUTION BOOK. I. Reduce ^, |, | to their b. C. Denominator '1. L. C. D.==40. II 3. |=jxJS=!^ -1 4 4 Nv 4 3 2 '-•>■ g S->^4 4iJ- TTT A 4 5 2 3 2 2 S II, {^ Reduce f to a fraction whose numerator is 12. l=il.- 2. |--^fXi|=J-i III • 4 — 1^ £xplanaHon: — Since the numerator of the required fraction is to be 12, we must not, in step 2, cancel 3 in the denominator of f and 12 in the numerator of ^|. Were we to do so and then multiply the resulting fac- tors in the numerators for the numerator and the resulting factors in the denominators for the denominator, we would get J for a result. J is the same as f , but f is not the required fraction ; for its ilumerator is not 12 as was required in tlie problem. i I. Reduce ^ to fourths. fi. 1=1 II. 1. 1=1, =4-fourths. 2. ^=jx|=|=^ of 4-fourths, =|-faurth. I III. .-. l=#-fourth.. I. Reilu^c I to a Iruction whose numerator is 15. IT /I- 1=tI- "1 2. |=fxi|=H- V 5 TTT 3 1 .5 IXi. . J ^^. I. Reduce |, |, | to equivalent fractions having least common numerators. 1. L. C. N =12. 9 1 1 2 Q 2 2.V 1 2 12 6 A 3 STvlS 11 ^- I i ^ 12 1 6' 4 f; 4 « v" 1 2 I 2 ■ 3 TTT . 2 3 4 12 12 U. 111. . . J, J, ^ j-g, ff, ^. I. Divide 2 by f. {1. 1 contains J, 4 times; or l-^|-=4. 2. 1 contains f , ij of 4 times=f times ; or l-5-f= 3. 2 contains f , 2 times f times=| times ; or 2- III. .-. 2 divided by 1=1 . II. FRACTIONS. 65 Divide |- by ^. ,-1 5_:_7--_6 fi. -5-:— 7 J- =7 times 4=\^. (1. t-:-t= Is. *-=-#= L3. |--f=iof-V=M=lH- VTT s • a 1 1 1 Analysis to the last example : !1. Tj- is contained in 1, or f,7 times. 2. Y 's contained in 1, or|^, -j of 7 times:^-^ times. 3. -7- is contained in |^, |- of -J times=-5-'^ times. 4. ^ is contained in f, 5 times -^ times, or |-|- times. Note. — By inverting the divisor, we find how many times it is contained in,l. EXAMPLES. 1. One-fifth equals how many twelfths? .2. Reduce -J, |^, ^, and | to fractions having a common de- nominator. 3. Reduce 4 to a fraction whose numerator is 13. Ans. • 4. Reduce I- to a fraction \vhose denominator is II, Ans, z^ V " ir 5. Reduce -|, ^, |, to fractions having common numerators. 6. Add i, f , i^, I, and tV 7. I of 8|— f of 5=what .? 8. Multiply 3. by Sf. 9. Multiply f of 9^ of f by I of 17. 197 K2 lH-6|^ 7f-5f 12. |fX||XM-^/g=what? - - - - >!».?. .J. 2i 3i i- ^ 13. i=what? Ans. 907200.* 5| i- 14. §i=what? Ans. , i 3^ 6J 66 FINKEL'S' SOLUTION BOOK. If,. (2^X21+1 of ^) X (|)«-^-(7i— SlXM)=what? 400000 Ans. 407511. ( 9J7 41 "ill ) ^11 1 -1-28 1 . |« ! flij-^'^ 2l5L 5 ^4 °T^ • ^Tri"?^ 17 what? "^' ' 3ii+7^4 62t-V ^ ^ 4f • 4f X6H-2001^ X 3 8 17. 2-^2-4-2H-2-=-2-^2-r-2-^2- : j - : -i : i : .l : j-=-^=\vhat? yi;2i. 1. 18. Reduce -i to thircls. Ans. 2f thirds. 19. What fraction is as much larger than |- as ^ is less than ^? Ans. W- 20. What is the value in the 13th example if a heavy, mark be dravyn between \ and \. Ans. l-f. .42* 21. 1 ^=wliat? Ans. 2tVA- off olf. 4^T r..^ — 3 ^r l?i. 22. Subtract i of ^ from t% of -g|- - Ans. ff §^. 23. What is the relation of 11 to 3? ri. l=^of3. Solution: ] 2. 11=11 times ^ qf 3=^- of 3=3f ( times 3. Conclusion: .'. 11 is 3f times 3. 24. What is the relation of 19 to 5 ? * Ans. ^. 25. What is the relation of y^ to 24? Ans. Jj 26. What part of 3 is 2? CI.. n. 1=4 of 3. Solution: | g. 2^2, times J of 3=f of 3. Conclusion: .•. 2 is f of 3. 27. What part of 6 is 7? Ans. f . 28. What part of _3 is i? Ans. ■^\. 29. What part of i is 3? -/ ^«j. Jj^. 30. What part of f of f is f of t\? EXAMPLES. 67 ri. fof |=f. J 2. fofT\=/^. 3. 1 is I of f. 1.4. ^^ is /^ timjes f of |==t'^ of f . • f of x^j, or ^%, is t\ of f of #. 31. +^4|=:.what? Ans.-^. {Note. — This is a continued fraction.] 32. Find the number of which 75 is f . 33. Find the nui^ber of which 180 is f . •34. \l is f of what number? !1. f of some numberp=-^f . 2. ' i of that number=i of'it=A. 3. f of that number, or the number r* quired, :=4 times ^^=^f. .'f . 'i Conclusion: .•. -Jf is f of ^f. 35. 27 is .3 of what number? ' - Ans. 90. ■36. f of -Jf is ^ of f of what number? 1- fofif=|. 2. ioff=i. 3. . • . 4= J of some number, or 4. ^ of some number:=f . 5. f of that number, or that number, -3Xf=J#. Conclusion : . • . f of ^f is |^ of f of ^-. 37. A watch cost $30, and this is f of f of the cost of the watch and chain together. What did the chain cost. Ans. $12 38. A lost f of his money and then found f as much as he lost and then had |120; how much money had he at first? 39. A sum of money diminished by ^ of itself and $6 equals |12; what is the sum? Ans. $31J. 40. If y\ of a ton of hay is worth $8^, how much is 10 tons worth? " Ans. $204. 41. What number is that -^f of which exceeds /j- of it by 111? ' Ans. 216. 42. What part of 2^ feet is 3^ inches? Ans. -^^ ^ 43. A has $2400; f of his money plus $500 is | of B's; what sum has B? Ans. $1600. Solution : -G^ FINKElv'S SOLUTION BOOK. ' 44. What fraction of (^-^).(T^+-^)is 45. A pole stands f in the mud, ^ in the water, and the re- mainder, 12f feet, above water. Find the length of the pole? Ans. 44f feet. 46. If 48 is ^ of some number, what is f of the same num- ber? Ans. 63. 47. A can do a certain piece of work in 8 days, and B can do the same work in 6 days. In what time can both together do the work? Ans. 3f days. 54* 48. The lesser of two numbers is • ^„ , and their difiFer- i ot of ence is -jj-. What is the greater number? Ans. -^-f-^. 49. What number multiplied by f of fX3f -v^ill produce |f ? Ans. f . 50. • What number divided by If will give a quotient of 9^? Ans. ^. 51. A post stands ^ in mud, J in water, and 21 feet above the water? What is its length? Ans. 36 feet. 52. A can do a piece of work in 8 days, A and B can do it in 5 days, and B and Oin 6 days. In what time can A, B, and C do the work? ■ Ans. ^-^ days. 53. If f of 6 bushels of wheat cost $4J, howmuch will f of 1 bushel cost? Ans. 80 cents. 56. What number diminished by the difference between f and i of itself leaves 1152? Ans. 2268. 56. If a piece of gold is | pure, how many carats fine is it? Ans. 15 carats. 57. The density of the earth is 5f times that of water, and the sun is J as dense as the earth. How many times denser than water is the sun? . Ans. fj. CIRCULATING DECIMA^LS. CHAPTER X. CIRCULATINQ DECIMALS. I. Change .63 to a common fraction. 11.(2! I3. 1. .63=.636363-(- etc., ad infinitum. 636363+etc.,=.63+.0063+.000063+etc.,«t^ infinitum. 3. This is a geometrical, infinite, decreasing series whose first term is .63 and ratio .63-i-0063=yfj. The sum of such a series iSj-^^=.63-=-(l— TiTr)=ff=A- III. .-. .63=tV. I. Reduce 1.001 to a common fraction. 1. i.00i=1.00110011001100110011+etc., ad infinitum. 2. 1.001=1.0011. 3. l.OOli— l+.OOll+00000011 + OOOOOOOOOOll+etc., ad infinitum. 4. .60li=first term. 5. T-^^^.j=i.0011-=-00000011=ratio. 6. .-. Sum= -^ =.0011---(l-T,rW)=-0011-^AVTrV=^Uir II.J r "5"T5"- 17. .-. 1.0011=1+^=17^5. {Ray s H. A., p. 120, ex. 8.) III. .•. i.ooi=i7i7. Remark. — Since the denominator of the ratio is always ten or some power of ten, the numerator of the fraction resulting from subtracting the ratio from 1, will have as many 9's in it as there are ciphers in the denominator of the ratio. By dividing the first term fjy ,this fraction, its numerator becomes the denominator of the fraction required. Hence, a circulaii may be reduced to a common fraction by writing for the denominator of the repetend as mariy 9's as there are figures in the repetend. Thus, .63^ ei 10 ^3 fii=rI=_L — 19 •Da— b^^Q IQ-^- I. Reduce .034639 to a common fraction. 1 034639 - 034^ « ,_ 34ff|-_34x999+639 _ 34x999+639 ^ 34 X ( 1000—1 ) +639 ___ 34000—34+639 _ 34000+639—34 '^ 9990^^0 ~ 999000 ~ 999000 ~ 34639_-34^ 34605 ^ 6921 ^ 2307 _ 769 999000 999000 199800 66600 ~"22200' 70 , JJ-INKEL'S SOLUTION BOOK. In case the circulate is mixed, we have the following rule : 1. For the numerator, subtract that fart -which f recedes the repetend from the -whole expression, both quantities being' consid- ered as units. 2. For the denominator, -write as many ffs as there are figures in the repetend, and annex as many ciphers as there are decimal ■figures before each repetend. I. ADDITION OF CIRCULATES. , I. Add 5.0770, .24, and 7.124943. f 1. 5.0770 = 5.0770 = 5.07707707 etc. n.]2. ' .24 = .242 = .24242424 etc. [3. 7.124943= 7.1249431 = 7.12494312 etc. III. .-. Sum=12.44 12.4444444 etc.=12.44. Explanation. — The first thing, in the addition and subtrac- tion of circulates, is to make the circulates co-originous , i. e., to make them begin at the same decimal place. That is, if one be- gins at (say) hundredths, make them all begin at hundredths, providing that each circulate has hundredths repeated. It is best to make them all begin with the circulate whose first re- peated figure is farthest from the decimal point, though any order after that may be taken. In the above example we have made them all begin at hundredths. After having made them all begin at hundredths, the next step is to make them con- terminous, i.e., to make them all end at the same place. To do, this, we find the L. C. M. of the numbers of figures repeated ip each circulate, then divide the L. C. M. by the number of fig- ures repeated in each circulate for the number of times the figures as a group must be repeated. Thus, the number of figures in the first repetend is 3; in the second, 2; and in the third, 6. The L. C. M. of 3, 2, and 6 is 6. 6-i-3=2. .-. 770 must be repeated twice. ,6-^2=3. .'. 42 must be repeated three times. 6-=-6=l. .•. 249431 must be taken once. I. Add .946, .248, 5.0770, 3.4884, and 7.124943. 1. .946 = .946 = .'946666666666666 etc. 2. .248 = .2484 = .248484848484848 etc. 3. 5.0770 = 5.07707 = 5.077077077077077 etc. II-J4. 3.4884 = 3.488448 = 3.488448844884488 etc. 5. 7.124943= 7.12494312 = 7.124943124943124 etc. .6. Sum =16.88562056205620+, =16.885620. III. .-. Sum=16.886626. CIRCULATING DECIMALS. 71 "1 III. III. I. II SUBTRAC'JIUN OF CIRClJLA'l'ES Subf-act 190.476 from 199.6428571 1. l?9.642857i = 199.642857i4 2. 190.476 = 190,47619047 l3. . Difference = 9.1666666 .= 9.16. .-. Difference=9.16. Subtract 13.637 from 104.1. , rl. 104.i =104.14 =104.1414141 etc. 12. 13.637= 13.637= 13.6376376 etc. 13. Difference = 90.503776 Difference=90.503776. III. MULTIPLICATION OF CIRCULATES. Multiply .07067 by .9432. .07067=.070677 .9432 = - .9^ ~ Multiply by the fraction thus : ^" ' .070677 .063609 ■003056 .066666=product. 16 .42406=. 424062 .7067 =.706770 37)1.13083(3056= 11 003056, be- 208 cause the 185 fraction is ■Oi^ 233 222 Multiply 1.256784 by 6.4208i. 1.256784=1.2567842 6.42081 = 6.420,^ .02513568 .5027187 7.540705 \ .001028270 = .0251356851 •= .502713702^ =7.540705540' = .00l62827d'> 8.069583198 11 Multiply by the fraction thus : ■ 1.2567842 .000,^ 11 ).011311057 .001028270 72 FINKEL'S SOLUTION BOOK. Remark. — In multiplying by any number, begin sufficiently far beyond the last figure of the repetend, so that if there is any to carry it may be added to the repetends of the partial p>roducts, making them complete. Thus in the above example, when mul- tiplying by 4, we begin at 5, the second decimal place beyond 4, the last figure of the repeterld ; and so when we multiply 4 by 4, ^« first figure .of the repetend in the partial product is 7. IV. DIVISION OF CIRCULATES. Rule. — Change the terms to common fractions; then divide ai in division of fractions, and reduce the quotient to a repetend. I. Divide .75 by .1 fl. .75=M=II- II.<^2. .1 =i. [3. ||-i4=f|x9=!i=6.8181 etc.=6.8i. 'III. .-. .75-T-.i=6.8i. EXAMPLES. 1. Add .87, .8, and 876. , , ' Ans. 2.644553. 2. Add .3, .45, .45, .351, '.6468, .6468, .6468, and 6468. ^ ' Ans. 4.1766345618. 3. Add 27.56, 5.632, 6.7, 16.356, .7i, and 6.i234. f Ans. 63.1690670868888. 4. Add 5.i634'5, 8.638i, and 3-75. Ans. 17.55919120847374090302. 5. , From 315.87 take 78.0378. , Ans. 237.838072095497. 6. "iFrom 16.1347 take 11.0884. Ans. 5.0462. 7. 18 is .6 of what number? Ans. 27. 8. From tV ta'^'^ tV ^^-^^ -1764705882352941. 9. From 5i2345 take 2.3523456. .^w.?. 2.7711055821666927777988888599994. 10. Multiply 87.32686 by 4.37. Ans. 381.6140338. II. Multiply 382.347 by .03. Ans. 13.6169533. 12. Multiply .9625668449197866'by ,75. Ans. .72. 13. Divide 234.6 by .7. Ans. 701.714285. 14. Divide 13.5169533 by 3.145. Ans. A.m. PERCENTAGE AND ITS VARIOUS APPLICATIONS. 73 15. Divide 2.370 by '4.923076. Ans. 481. 16. Divide -36 by .25. Ans. 1.4229249011857707509881. 17. Divide .72 by .75. ' Ans. .9625668449197860. 18. 54.0678132-r-8.594=what? ^«.y. -6.290. 19. 4.9.5D-^-.75=what,? " ' Ans. 6.6087542. 20. 7.7142S.5-r-.952886=v(rhatF Ans. 8.1. CHAPTER XI. PERCENTAQE AND ITS VARIOUS APPLICATIONS. 1. Percentage is a method of computation in which 100 is taken as the basis of comparison. 3. Per cent, is an abbreviation from the I/'atin, per centum, per, by, and centum, a hundred. , 3. TAe Terms used in percentage are the Base, the Rate, the Percentage, and the Amount or Difference. 4. The Base is the number on which the percentage is computed. 5. Tize Rate is the number of hundredths of the base which is to be taken. 6. The Percentage is the result obtained by taking a cer- tain per cent, of the base. 7. The Amount or Difference is the sum or difference of the base and percentage. 8. The sign,'%, is used instead of the words "per cent." and "one-hundredths," following the number expressing the rate. Thus, for example, for 5 per cent,, or 5 one-hundredths, we write 5%. Hence, we^have the following identical expressions: 6 per cent. =5 one-hundredths=y|^^=.05^5%. In each of these expressions the fractional unit is ■^-^. The fundamental principle of percentage is that our computation shall be made on the basis of hundredths. That this principle be not violated, the denominator of the fraction must always be 100. Thus, since -^■^■=^^, we can take ^^ of a number instead of ^^ of it and get the same result; but using fractions whose denomina- tors are numbers other than 100 to express the rate is not the method of percentage, but merely the method of common frac- tions. However, in teaching percentage the method of common 74 FINKEL'S SOLUTION BOOK. fractions should also be used, as this method, because of it& brevity, is more often usedin practice. As an illustration, find 5% of $600. 1 . 100 one-hundredths, or ^, or 1 . 00, or 100%=! 2. 1 one-hundredth, or y^^, or .01, or 1%,=^^ of II. 3. 5 one-hundredths, or y^, or .05, or 5%=:5 times |6=$30. III. .-. 5% of $600=$30. Explanation. — It must be borne in mind that what we mean by say- ing 100%=f 600 is that {%% of |600 is $600, which is certainly a true state- ment. But for brevity we simply say 100%=$600, with the understanding that we mean 100% of f600=|600, 1% of $600=$6, etc. I. What is 8% of 150 yards? FIRST SOlvUTION. ri. -5-^^=150 yards. II. ] 2. ^i^s=^l^ of m=Thf of 150 yards=1.5 yards. ' (.3. xf ^^8 times 1.5 yards^l2 yards. III. .-.8% of 150 yards=12 yards. Remark. — This solution is by the method of percentage purely, SECOND SOLUTION. ri. 100%==150 yards. II. \ 2. l%=T?rir of 100%=T^ of 150 yards=1.5 yards. , ( 3. 8%=8 times 1%=:8 times 1.5 yards=12 yards. III. .-.8% of 150 yafds=12 yards. THIRD SOLUTION. Briefly, by fractions: 8% of 150 yards=Tfj of 150 yards=12 yards. CASE I. r^. f the base and the ) . ^ j . t. ^'^^" 1 rate per cent. } *" ^""^ ^^^ percentage. Pormnla.—BXR^P , where B is the base, R the rate, and P the percentage. PERCENTAGE AND ITS VARIOUS APPLICATIONS. What is 8% of $500? rl. 100%=$50G, l%=T-^-j of $500=$5, and 89^=8 times |5=$40. III. .-. 8% of $500=$40. 75. 11.(2. u. II. I2 13 III. II. What is 1% of 800 men? 100%=800 men. ifo^rrn °^ ^'-'0 nien=8 men, and 3. |^==f'times 8 men=6 men. .-. 1% of 800 men=6 men. • What is 10% of 20% of $13.60? rl. 100%=$13.50. '(I.U2. l'fo=rU of $18.50=$.135, and L3. 20%=20 times $.135==$2.70. (2.) 100% =$2.70. (3.) 1%=tU of $2.70=$.027, and (4.) 10%=10 times $.027=$.27=27 cents. III. .-. 10% of 20% of $13.50=27 cents. I. A. had $1200 ; he gave 30% to a son, 20% of the remain- der 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? (1. 100%=$12aO, 2. 1%=tU of $1200=$12, and 3. 30%=30 times $12=$360=son's share. 4. $1200— $360=$840=remainder. 1. 100% =« r (1-) II.S (2.) (3.) (4.) (5.) (6.) (7.) (8.) (9) (10.) (11.) (12.) 2. 1 % =j^ of $840=$8.40 , and 3. v20%=20 times $8.40=$168=daughter's share. 4. $840— $168=$672=amount divided among four brothers. 100%)=fourth brother's share; 100%+$12=third brother's , share. 100%-|-^24=second brother's share, and 100%-|-$36=first brother's share. 100%+(100%+$12)+(100% + $24)+(100%-|- $36)=400%+$72=am'tthe four brothers rec'd. $672=amount the four brothers received. .-. 400%+$72=$672. 400 % =$672— $72=$600. 1 %=7iT of $600=$1.50. fo=100 times $1.50=$150=fourth brother's share. III. ,-. The last received $150. (Ji. H. A., p. 191, prob. 85.^ II. 7G . '^ FINKEL'S SOLUTION BOOK. \ I. What number increased by 20% of 3.5, diminished by 12^% of 9.6, gives 3^? r(l.) 100%=the number. -1. 100 '/p =3.5, (2.)<{2. l%=x^Tof 3.5=.035, and Is. 20% =20 times .0S5=.7. r-1. 100% =9.6, (3.)<^2. l%=Ti-^of9.6=.096, and Is. 12i%=12| tijnes .096=1.2. (4.) .•.100%+-7-l-2=3i, (5.) 100%— .5=3.5, and, .(6.) 100%=4, the number. III. .-. The number=4. (H. H. A., f. 191, frob. 26.) CASE II. Given { %^^S„^tage ^^ } to find the rate per cent. Formula. — P^B^R, where B is the base, P the percentage and R the rate per cent. I. 750 men is what % of 12000 men ? 12Q00men=100%, 1 man=T-j^Tr of 100%=^^ ^ % , and 750 men=750 times -x^%=<6\%- III. .-. 750 men ,is 6^% of 12000 men. I. A's money is 60% more than B's; then B'sis how many <Jo less than A's? 1. 100%=B.'s moneyr Then, 2. 100%+50%=150%==A.'s money. 11.-^ 3. 150%=100% of itself 4- l%=Tiu of 100%=!%, and 1.5. 60%=50 times |%=83i%. III. .-. B.'s money is 33^% less than A.'s' {R.H.A.,p.l9Z,prob.ll.) I. 30% of the whole of an article is how many % of f of it? 1. 100%=whole article. 2. 66|%=f of 100%=! of the article. 3. 66t%=i00% of itself. 4. l%=g^ of 100%=li%, and U. 30%=30 times 1|%=45 %. Ill .-. 30% of the whole of an article is 45% of f of it. {R. H. A., p. 192,prob. 20.) 11.(2! I3. II. PERCENTAGE. 77- II. III. If a miller takes 4 quarts for toll from every bushel he grinds, what "fo does he take for toll? fl. 1 bu.=;32 qt. . ^ I 2. 32qt'.=100%, 1 3." 1 qt.=-jV of 100%=3i% , and U. 4 qt.=4 times 3^%=12i%. .-. He takes 12|% for 'toll. CASE III , G-- {Th?r"r;e^c?nf}^°>-<i the base. F^ormnla,.—P^R=B , where P is the percentage, R the rate / per cent., and B the base. I. $24 is 1% of what sum? II. III. 100%=sum. |%=^$24, II. III. ^5=iof$24=$8, 1%, or 1%,=8 times $8=$64, and 100%=100 times $64=16400. $24 is lojo of $6400. I drew 48% of my funds in bank, to pay a note of $150 "j how much had I left? 1. 100%=amount in bank. 2. i48%=amount drawn out. 3. 100%— 48%==52%=;amount left. 4. 48%=$150, 5. l%=^iyof $150=$3.125, and 6. 52%=52 tinjes $3.125=$162.50=amount left. .-. $162.60=amount I had left. IIX III. I pay $13 a month for board, which is 20% of my salary; what is my salary ? 1. 100%=my monthjy salary. 2. 20%=$13, 3. 1%=^ of $13=$.65, and 4. 100%=100 times $.65=$65, my monthly salary. ,5. .-. $780=12 times $65=my yearly salary. My salary=$780. {R. H. A., J>. 194, }roh.^20.\ '78 FINKEL'S SOLUTION BOOIC CASE IV. Given 1 the amount and the I to find the base. rate per cent. Formula. — A-^{1+J?)^=B, where A is the amount, that is, the base and the percentage, H the rate per cent., and B the base. A sold a horse for $150 and gained 25%; horse cost? what did ths lU fl- 2. 3. 4. 5. 6. 7. III. I. II. III. 100%=cost of horse. 25%=gain. 100'%+25%==125%=selling price of horse, and $150=selling price of horse ; .-. 125%=$150, 1 %=ji^ of $150=$1.20, and 100%=100 times $1.20=$120=cost of horse. The horse cost $120. sold' two horses for the same price, $150; on one I gained 25% and on the other I lost 25%; what was the cost of each? 1. 100%==cost of first horse. 2. 25%=gain. 3. 100%+25%=125%=selling price of first horse, 4. $150^selling price of'first horse; 5.. .-. 125%=$150, , 6. l%=i-fj of $150=11.20, and . 7. 100%=100 times $1.20=$120=cost of first horse. 1. 100%^cost of second horse. 2. 25%=loss on second horse. 3. 100% — 25%=75%=seiringpriceof 2d horse, an<^ 4. $150=selling price of se'cond horse; 5. .-. 75%=$150, 6. 1%=tV of $150=$2,,and 7. 100%=100 times $2==$200=co£t of second horse. A. B. J$120=cost of first horse, and I $200=cost of second horse. I. A coat cost $82; the trimmings cost 70% less, and the making 50% less than the cloth; what did each cost? 11."^ PERCENTAGE. 79 i. 100%=cost of cloth. Then 2. 100%— 70foF=30%=cost of trimmings, and 3. 100%— 50%=50%=cost of making! . 4. 100%+30%+50%=180%=cost of coat. 5. $32=cost of coat; 6. . . 180%=$32, 7. 1 %=Ttj o.f $32=$.1777i. 8. 100%=100 times $.1777|=$17.77^— cost of cloth. -• 9. 30%=30 times $.1777|=$5.33-^=cost of trimming. UO. 50%=50 times $.1777|=$8.88f=c6st of making. /.77^=cost of cloth, 5.33^=cost of trimmi 1$ 8.88f=cost of making. {? . a $ 5.334=cost of trimmings, and i -■-- - - {R. H. A..f. 196, f rob. 12.) I. In a company of 87, the children are SH% of the women, who are 44f% of the men; how many of each? (1. 100%=44|%, (8.)<^2. l%=^^.of Is. 37ifo=37i- tii: (1.) 100%=number of men. Then (2.) 44f%^number of women. (1. 100%=44|%, - Df44|%=.44f%,and imes .44-|%=16f %=number of chil- dren in terms of the number of men. (4.) _ 10C%+44f%+16f% = 16H% = number in the III ' company, (5.) 87==number in the company; (6.) .-. 161i%=87, (7.) l%=j^of87=.54, (8.) 100%=100 times .54=54=number of men, (9.) 44|-%^44|- times .54=24=number of women, and 1(10.) 16|%=16| times .54=19=numl)er of children. |54^number of men, III. .-. ■;24=number of women, and / 19=number of children. (i?. If. A., p. 19T,frob. 20.) I, Our stock decreased 33|^%, and again 20%; then it rose 20%, and again 33^%; we have thus lost $66; what was the stock at first ? (4.) 80 ^ FINKEL'S SOLUTION BOOK. ; ' 100%=original stock. 33^%=decrease. 100%— 33i%=66|%=stock after first decrease.,'' 100%=66|%, 1%=T^ of 661%=!%, and 20%=20 times f %=13i%=second decrease. 66f % — 18-^%=53i%=stock after second decrease. 100%=53^%, l%=Tkof63i%=.53i%, and 20%=20^ times .53i%=10|%=first mcrease. 11.^ U. 53i%+10|%=64%=r=stock after first i\icrease (1. 100%==64%, ,. J2. l%=Twof64%==.64%,and ^^■nS. 33i%=33i times .64%=21i%=second increase. l4. 64%-f-21-^%=85^%^stock after second increase (7.) 100%— 86i%=14|%=whole loss; (8.) $66=wiiole loss; (9.) .-. 14|%=$66; ; 10. ) 1 %=j^ of $66=14.50, and 11.) 100%=100 times $4.50=$450=original stock. .-. $450=original stock. A brewery is worth 4% less than a tannery, and the tan- nery 16% more than the boat; the owner of the boat has traded it for 75% of the brewery, losing thus $103 ; w'nat is the tannery worth ? FIRST SOLUTION. ^ (1.) 100 %=value of the tannery. Then (2. ) 100%— 4%=9e%=value of the brewery. (1. 100%=valueof the boat. Then [the boat. 2. 100%+16%=116%=value of tannery in terms of ,g \ Is. 116%^100%, the value of tannery from step (1), (^•>i4. l%=T|^oflOO%==f|%,and 5. 100%=100 times ||%=86^%=value of the boat in terms of the tannery. 100%=96%, 1%=^ of 96%=.96%, and 75% =75 times .96%=72%=what the owner of the boat received for it. •■■ 86^%— 72%=14^%=what the owner of the boat lost in the trade. (6.) $103=what he lost; (7.) .-.' 14A%=$103, (8.) l%==:fi^of$103=$7.25, and (9.) 100%=100 times $7.25=$725=value of tannery. III. .-. $725=value of the tannery. {R. H.A.,f.l91,frol.ZS.) II. i PERCENTAGE. 81 Remark. — The value of the brewery and boat being ex- pressed in terms- of the tannery, 75% of the brewery is also ex- pressed in terms of the tannery; hence, it is plain that the owner of the boat has traded 86^9% for 72% of the same value, losing 863^%— 72%, or 14^%. SECOND SOLUTION. 1. II. s III. n. (2.) (3.)ft 13. (4.) (6.) (7.) (8.) (9.) 1(10.) .-. $725: (1-) (2.) (3.) (4.) (5.) (6.) (7-) I (8.) 100%=value of the boat. Then 100%+16%=116%=value of the tannery. 100%=116%, l%=TiTf of 116%=!. 16%, and 4%=4 times 1.16%==4.64%. , . 116%— 4.64%=111.36%=the value of brewery in terms of the boat. 100%=111.36%, , l%=Ti7 of 111.36%=1.1136%, and 75%=75 times 1.1136 %=83.52%=what the owner of the boat received for it. .-. 100%— 83.52%=16.48%=what he lost in the trade. ^ $103=what he lost. .-. i6.48%=$103, l%=.j.^i„ of $103=16.25, and 116%=il6 times $6.25=$725=value of tannery, ^^value of the tannery. THIRD SOLUTION. ^ 100%=value of brewery. 100%=va]ue of tannery. Then 100%— 4%=96%=value of the tannery. .-. 96%.=100%, the value of brewery in step (1), 1%=^ of 100%=1.04^%, and 100%=100 times 1.04-^%=104^%=value the tan- nery in terms of the brewery. 100%=value of boat. Then 100%+16%=116%=value of the tannery in terms of the boat. .-. 116%=104|%, the value of the tannery in step 5 of (2), l%=TiT of 104i%=.89iff % , and 100%=100 times .89i|f%=89if|%=value of the boat in terms of the tannery, and consequently in terms of the brewery. •■• 891f|%— 75%=14|ff %=what the owner of the boat lost in the trade. $103=what the owner of the boat lost ; .-. 14if|%^$103, 1%=T7T¥5- of $103=$6.96, and 241 39 104|%=i04| times! 3.96=$725^value of tannery. n. 8" FINKEL'S SOLUTION BOOK. III. .-. $725=value of of the tannery. Bemark.—ln step. 5 of (3), we have the value of the boat in terms of the tannery ; but the value of the tannery is in terms of the brewery} hence, the value of the boat is also in terms of the brewery. The owner of the boat, therefore, traded 89|f|% for 75 %i of the same value. MISCELLANEOUS PROBLEMS. L A man sold a horse for $175, which was 12^% less than the horse cost; what did the horse cost? 1. 100%==cost of horse. 2. 12|%=loss. 3. 100%— 12i%=87i%=se)ling price. 4. $175=.selling price. 5. .-. 87i%=$175, , 6- l%=;j^ of $17o=$2, and 7. 100%=100 times $2=$200, III. .-. $200=cost of the horse. {R. Sd p., p. 204., prob. 5.) I. A miller takes for toll 6 quarts from every 5 bushels o' wheat ground; what % does he take for toll.? 1 bu.=32 qt. 6 bu.=5 times 32 qt.=160 qt. IL<;4. 160qt.=100%, 1 qt-=TiT of 100%=!%, and 6 qt.=6 times 4%=3|%. III. .-. He takes 3|% for toll. {E. Sd p.,p. 20i,prob. 11.) I. A fatmer owning 45% of a tract of land, sold 540 acres, which was 60% ot what he owned; how many acres were there in the tract? ' (1.) 100%=^number of acres in the tract. 1. 100%^numbers of acres the farmer owned. 2. 60%=number of acres the farmer sold. 3. 540 acres^what he sold. (2.)J4. .. 60% =540 acres, ( 5. 1%=tV of 640 acres=9 acres, and Il.-i ^ 6. 100%=100 times 9 acres=900 acres=what he owned. (3.) 45%^what he owned. (4.) .-. 45%=900 acres, (5.) 1%=TF of 900 acres=20 acres, and (6.) 100%=100 times 20 acres=2000=number of ^icres in the. tract. i III. .■. The tract contained 2000 acres. {R. Sdp.,p. 20Jf,proh. 12.) MISCELLANEOUS PROBLEMS. 83 A, wishing to sell a cow and a horse to B, asked 150% more for the horse than for the cow; he then reduced the price of the cow 25%, and the horse 33^% , at which price B took them, paying $290; what was the price of each? (1.) 100 %=asking price of the cow. Then (2-) 100%+150%=250%=asking price of the horse. (3.) 100%— 25%=75%=selling price of cow. (1. 100%=250%, , . J2. l%=TiT of 250%=2.50%, and ^^■''|3. 33i%=33^ times 2.5%=83i%=reduction on the I, asking price of the horse. (5.) 250%— 83^%=166|%=selling price of the horse, 11..; . (6.) 75%+166|%=241f %=selling price of both. ( 7. ) $290==selling price of both. (8.) .-. 241f%=$290, (9-) l9^"=2iP "^ $290=$1.20, and (10.) 75%=75 times! 1.20=$90=selling price of the' Cow. (11.) 166|%=166| times $1.20=$200=selling price of the horse. . /$ 90^selling price of the cow, and 111. •■•|^200=selling price of the horse. {Brooks' H. A.,p.2U3,froh. 18.) I. A mechanic contracts to supply dressed stone for a church for $87560, if the rough stone cost him 18 cents a cubic foot; but if he can get it for 16 cents a cubic foot, he will deduct 5% from . his .bill; required the number of cubic teet and the charge for dressing the stone. \ ,1. 100%=$87560. 2. l%=TiT of $87560=$875.60, and 3. 5%=5. times $875.60=$4378=the deduction. 4. 18/ — 16/=2/^the deduction per cubic foot. II.-! 5. .-. $4378=the deduction of 4378-^.02, or 218900 cubic feet. Then 6. $87560=cost of 218900 cubic feet. 7. $.40=$87560-=-218900=cost of one cubic feet. l8. .-. $.40— $.18=$.22=cost of dressing per cubic foot. IIL r218900=number of cubic feet, and '122 cents^cost .of dressing per cubic foot. {Brooks'. H. A., p. 2Jfl,frob. 84 FINKEL'S SOLUTION BOOK. PROBLEMS. 1. A merchant, having $1728 in the Union Bank, wishes to withdraw 15%; how much will remain? Ans. $14^8.80. 2. A Colonel whose regiment consisted of 900 men, lost 8% of them in battle, and 50% of the remainder by sickness; how many had he left? Ans. 414 men. 3. What % of .$150 is 25% of $36? Ans. Q(fo. 4. What % of I off of I is ^? . , Ans. 31^%. 5. If a man owning 45% of a mill, should sell 33^% of -his share for $450 ; what would be the value of the mill? Ans. $3000. 6. A. expends in a week $24, which exceeds by 33^% his earnings in the same time. What were his earnings? Ans. $18. 7. Bought a carriage for $123.06, which was 16% less than I paid for a horse; what did I pay for the horse? Ans. $146.50. 8. Bought a horse, buggy, and harness for $500i The horse cost 37^% less tliain the buggy, and the harness cost 70% less than the horse ; what was the price of each ? Ajts. buggy $275f|, horse $172^, and harness $51fi- 9. I have 20 yards of yard-wide cloth, which jvill shrink on spohging 4% in length and 5% in width; how much less than 20 square yards will there be after sponging? Ans. 1^ yards. 10. A. found $5; what was his gain %? A ns. 11. The population of a city whose gain of inhabitants in 5 years has been 25%, is 87500 ; what was it 5 years ago? Ans. 70000. 12. The square root of 2 is what % of the square root of 3 ? Ans. Ve'X^Hfo- 13. A laborer had his wages twice reduce'd 10%; what did he receive before the reduction, if he now receives $2.02-|^ per day? ', Ans. $2.50. 14. The cube root of 2985984 is what % of the square root of the same number? . Ans. 8^%. 15. A man sold twohorses for |210; on one he gained 25%, and on the other he lost" 25% ; how much did he gain, supposing the second horse cost him f as much as the first ? Ans. $10. COMMISSION. 85 16. A merchant sold goods at 20% gain, but had it cost him $49 more he would have lost 15% by selling at the same price; what did the goods cost him? Ans. $119. 17. If an article had cost 20% more, the gain would have been 25% less; what was the gain %? (See page 147.) Ans. 50% I. COMMISSION. V 1. Commission is the percentag£ paid to an agent for the transaction of business. It is computed on the actual amount of the sale. ■ ' , 2\ An Agent, Factor, or Commission Merchant, is a person who transacts business for another. 3. The Ifet Proceeds is the sum left after the commission and charges have been deducte.d from the amount of the sales or x;ollections. , 4. The Entire Cost is the sum obtained by adding the commission and charges to the amount of a purchase. I. An agent received~$^10 with which to buy goods ; after deducting his commission of 5%, what sum must he ■expend? 1. 100%=what he must exppnd. 2. "5%^his commission. 3. 100%+5%=105%=what he receives. ' II. -(4. $210^what he receives. 5. .-. 105%=$210, e: l%=y^Qf $210=$2, and J. 100%=100 times $2=$200=what he expends. III. .-. $200=wh.at he^must expend. ■ {R. Sdp., p. 207, prob. 4..) Note. — Since the agent's commission is in the $210, we must not take 5% of |2'lO; for we w'ould be computing commission on his commission. Thus, 5% of ($200+$10)=$10+$.50. This is $.50 to" much I. An agent sold my corn, and after reserving his com- mission, invested the proceeds in corn at the same price;' his commission, buying and selling was 3%, and his whole charge $12; for what was the corn first sold? 86 FINKEL'S SOLUTION BOOK. r(i-) (2.) (3.) II. III. I. II. (4.) (5.) (6.) (7.) IJ,=cost of tlie corn. 3%=the commission. 100% — 3%=97%==net proceeds, which he invested in corn. 1. 100%=cost of second lot of corn. 2. 3%=the commission. 3. 100%+3%=103%=entire cost of second lot of corn. 4. 97%=entire cost of second lot of corn. 5. .-. 103%=97%, 6. l%=Tk of 97%=r%% , and ,■ 7. 100%=100 times J^'^%=94^%=cost of second lot of corn in terms' of the first. 8. 8%=3 times yW%=2TW% = commission on second 'lot. 3%+2i%\%=6i\\%=whole commission. $12=whole commission. Hh', =$12, ( 8. ) 1 %=^ of $12=$2.06, and ■ottts (9. ) 100%=100 times $2.06=$206=crislyof first lot of corn -. $206=cost of first lot of corn. {R. H. A., p. 219,prob. 10.) Sold cotton on commission, at 5%; invested the net pro- ceeds in sugar, commission, 2%; my whole commission was $210 ; what was the value of the cotton and sugar? [vested in sugar, proceeds, which he in- III. (1.) 100%=cost of cotton. (2.) 5%=commission. (3.) 100%— 5%=95%=net 1. 100%^cost of^ugar. ^ 2. 2%=commission. , 3. 102%=entire cost of stigar. '- 4. 96%=entire cost of sugar. . 5. .-. 102%=95%,' - 6. l%=T^^of96%=^\%, and . 7. 100%=100 timesJT^%=93-/T%=cost of sugar in terms of cotton. * 8. 2%=2 times^s^^) =l|4^%=commission on the ^ 'sugar. (4.) 5%+l-||%=6|f %=whole commission. (5.) $Sl0^whole commission. (6.) ^ .-. 6|f%=$210, (7. ) ^ 1 % = J of $210=$30.60, and (8. ), 100%=100 times $30.60=$3060=cost of cotton. *- (9.) 93^%=933V times $30.60=$2850=cost of sugar. . r$3060=cqst of cotton, and ■"T$2850=cost of sugar. ( R. H. A., p. 219, prob. 6. ) COMMISStON. 87 I. !!.■ Ill I A lawyer. received $11.25 for collecting a debt ; his com- mission being 5%; what was the amountof the debt? 1. 100%=amount of the debt. 2. 5 %^com mission. 4. $11.25=commission. U. •••. 5%=$11.25. • - 5. l%=^of $11.25=$2.25, and 6. 100%=100 times $2.25=$225=amount of the debt. .-. $225=amount of debt. {H, Sdp.,p. ZOT.trob. 6.) Charge $52.50 for collecting a debt of $525; what was the rate of commission? 1. $525=100% $l=7ij of 100%=A%, and. $52.50=52.5 times -^t-^^^io^^rate of commission. 11.(2! I3. III. I. II. •. 10%=rate of commission. My agent sold my flour at 4% commission; increasing the proceeds by $4.20, I ordered the purchase of wheat at 2% commission; after which, wheat declining 3^'%, my whole. loss was $5 ; what was the flour worth? (1.) 100%=cost of flour. (2.) 4%=commission on flour. (3.) 100%— 4%=96%=net proceeds. 1. 100%=cost of wheat. ^ 2. 2%=commission on wheat. 3. 100%+2%=102%=^entire cost of wheat. 4. 96%-|-$4.20=entire cost of wheat. 5. .-. 102%=96%+$4.20, 6. l%=T^of (96%+$4.20)=.94TV%+$.0411if, 7. 100%=100 times (.94T^%+$.0411|f )=94t^%+ $4.11^^cost of wheat. 8. 2%=2 times {M-^<fc-\-%Mn^)=l\^%-\-%m^f =commission on wheat. (4.) rl. 100%=94TV%+$4.11if III. (6.) (7-) (8.) (9.) (10.) 1(11-) $53= a. 100%=! , , ,_ ,^- . , . J 2. l%=if%+$.04-V. and )3. ^%=^ times (|6 9/^+$.04T2^)=33V%+$-13fi= I loss on wheat. 4% + 1H% + $-08t\ + 3^% + $.13-1^: $.21|f=whoIe loss. $5=whole loss. .-. 9^V%+$-21|f=a5, or 9^%=$5— $.21|f=$4.78A. «tV%+ of$4.78A^=$.53, and ^ti 1% = 100%=100 times $.53=$53. =cost of flour. {R.H.A., p. SW, frob. 21. ) FINKEL'S SOLUTION BOOK EXAMPLES. 1. A broker in New York exchanged $4056 on Canal Bank, Portland, at |-%.; what did he receive for his trouble? Ans. $25.35. 2. A sold on commission for B 230 yards of cloth at $1.25 per yard, for which he received a commission of 3|-% ; what was his commission and what sum did he remit? Ans. Commission $10.06^, and Remittance $277.43|. 3. A sold a lot of books on commission of 20%, and reipitted $160; for what were the books sold? Ans. $200- 4. A lawyer charged $80 for collecting $200; what was his rate of commission ? ^ Ans. A0% 6. I sent my agent $1364.76 to be invested in pork at $6 per bbl. after deducting his commission of 2%; how many barrels of pork did he buy? Ans. 223 bbl. 6. How much money must I send my agent, so that he may purchase 250 bbl.> of flour for me at $6.25 per blDl., if I pay him % commission?. ^«.y. $1601.5625. 7. If an agent's commission was $200, and his rate of com- mission 6% ; what amount did he invest? ' Ans. $4000 8. My agent sold cattle at 10% commission, and after I in- creased the proceeds by $18, I ordered him to buy hogs at 20% commission. The hogs had declined 6|-%, w^hen he sold them at 14f% commission. I lost in all $86; what did the cattle sell for? Ans. 9. An agent sells flour on commission of 2%, and purchases goods on true commission of 3%. If he had received 3% for selling and 2% for buying, his whole commission would have been $5 more. Find the value of the goods bought. ' Ans. $9995 II. TRADE DISCOUNT. 1. Trade Discount is the discount allowed in the pur- chas and sale of merchandise. 2. A. List, or Regular Price, is an established price, as- sumed by the seller as a basis upon which to calculate discount. 3. A Wet Price is a fixed price from which no discount is allowed. 4. The JMscount is the deduction from the list, or regu- lar price. (4.) TRADE DISCOUNT 89 Sold, 20 doz. feather dusters, giving the purciiaser a dis- count of 10, 10 and 10% off, his discounts amounting to $325.20; how much was my price per dozen? (1.) 100%=whosesale price. (2.) 10% of 100 %=10%=first discount. (3.) 100%— 10%=90%=first net proceeds, fl. 100%=^90%, 1 %=TW of 90%=tSj% , and 10%=10 times y8j% =9% ^second discount. 90^ — 9%=81%=second net proceeds. 100%=81%, , ll.\ (5.)<;2. l%=TiTof81%=TVT%.and ■ l3. 10%=10i times yVTr%=8.1%=third discount. (6.) . 10%+9%-t-8.1%=27.1%=sum of discounts. (7.) $325. 20^sum of discounts. (8.) .-. 27.1%=$325.20, (9.-) l%=^Tof$325.20=$12, and . (10.)- 100%=i00 tiipes $12=$1200=wholesale price of 20 dozen. (11.) $60=$1200-T-20=wholesale price of 1 dozen. III. .•. $60=wholesale price per dozen. (i?. SdJ>., f. 209, frob. 5.) I. Bought 100 dozen stay bindings at 60 cents per dozen for 40, 10, and 7i% off; what did I pay for them? ■• ' (1.) 60/=list price of 1 dozen. (2.) $60=100 times $.60=list price of 100 dozen, fl. 100%=$60, 1 %=T¥T of$60=$.60, and 40%=40 times $.60=$24=first discount. $60— $24=$36=first net proceeds. ^^■\ fl. 100%=$36, '• I 2. 1%=TOT of $36=$.36. and ^ '\i. 10%=10 times $.36=$3.60=second discount. U. $36— $3.60=$32.40=second net proceeds. fl. 100%=$32.40, f.M. l%==T^of $32.40=$.324, and ^ '\ 3. l^<fo=ll times $.324=$2.43=third discount. U. $32.40— $2.43=$29.97=cost. 311. .-. I paid $29.97. 'R:Sdp.,p.209,prob.6.) I. A retail dealer buys a case of slates containing 10 dozen for $50 list, a'nd gets 50, 10, and 10% off; paying for them in the usual time, he gets an additional 2%; what did he pay per dozen for the slates? (3.) 90 FINKEL'S SOLUTION BOOK II.<^ III. (10 (2.) (4.) (5.) 100%=$60. l%=^of $60=$.50. 50%=50 times $.50=$26=first discount. $50— $25=$25=first net proceeds. ^100%=$25. l%=^^of $25=$.25. 10%=10 times $.25=$2.50=second discount $25— $2.50=$22.50=second net proceeds. 100%=$22.50. 1%=^ of $22.50=$.225.' 10%=10 times $.225=$2.25=third discount $22.50— $2.25=$20.25=third net proceeds. - 100%=$20.25. l<fo=r^is of $20.25=1.2025. 2%=2 times $.2025=$.405=fourth discount. $20.25— $.405=$19.845=cQSt of 10 dozen slates. $1.9845=$19.845-f-10=cost of 1 dozen slates. ' .9845^cost of 1 dozen slates. {R. 3d p., p. 209, prob. S.) II.<^ ni. Sold a case of hats containing 3 dozen, on which I had re- ceived a discount of 10% and made a profit of 12-^% or 37^/ on each hat ; what was the wholesale merchant's; price per case? (1.) 37^/=profit on one hat. (2.) $13.50=36 times $.37i=profit on 3 dozen hats. (3.) 100%==w^holesale merchant's price per case. (4.) 10%=discount. (5.) 100%— 10%=90%=mycost: {6.)J (7.) (8.) (9.) 100%=90%. l%=TffTrof90%=.9%. 12^%=12| times .9%=lli%=]?rofit in terms of wholesale price. .-. lli%=$13.60. 1%— i- of $13.50— $1.20. 100%=100 times $1.20=$120=wholesale mer- chant's price per case. .". $120=wholesale merchant's price per case. ( R. 8d p. , p. 212, p^ob. 4. } A bookseller purchased books from the publishers at 20% ofT the list ; if he retail %hetn at the list what will be his per cent, of profit? \ TRADE DISCOUNT. 91 1. 100%=list price. 2. 20%=discount. 3. 100%— 20%=80%=cost. 4. 100%==bookseller's selling price, because he sold them- 11.^ at the list price. 5. .-. lOOfc— 80%=20%=;gain. 6. 80%=100% of itself. 7. , 1 %=^ of 100%=li% ' and l8. 20%=20 times li%=25%=hi5 gain fe- ll!.. •. 25%=his % of profit. , {R. Sd p., p. 211, proi. l.y ^ iVo/e. — Observe that since his cost is 80%, and his gain 20%^ we wish to know what fo 20% is of 80%. It will become evi- dent if we suppose the list price to be (say) $400, and then pro- ceed to find the % of gain as in the above solution. I. Bought 50 gross of rubber buttons for 25, 10, and 5% off; disposed of the lot for $35.91, at a profit of 12% ; what was the list price of the buttons per gross ? (1.) 100 %=list price. (2.) 25% of I00%=25%=first discount. (3.) 100%— 25%=75%=first net proceeds. f-1. 100%=75%, ,, J 2. lfo=^h of V5%=|% , and (^•^i 3. 10%=10 times |%=7i%. U. 75% — 7i%=^67-^%^second net proceeds rl. 100%=67i%, ,.J2. l%=T-kof67i%=.67i%,and ^°-''l3. 5%=5 times .67i%=3.375%=third discount. 14. 67i%— 3.375%=64.125%=cost. rl. 100%=64.125%, , I2. l%=.64125%,and v°-''l3. 12%=12 times .64125%=7.695%=gain. L4. .-. 64.125 % +7.695 %=71.82%=selling price. (7.) $35.91=sell;ng price. (8.) .-. 71.82%=$35.91, (9.) \cfg^^.^ of $35.91=$.50, and (10.) 100%=100 times |.50=$50=list price of 50 gross». ( 11. ) $1.00=$50-i-50=list price of one gross. m. .-. $1.00=list price of one gross. {R. Sdp.,p. 212,prob. 10.) I A dealer in notions buys 60 gross shoestrings at 70/ per gross, list, 50, 10, and 5% oflT; if he sell them at 20l 10, and 6% off list, what will be his profit? U n..^ 02 FINKEL'S SOLUTION BOOK. !1.) 70/'=list price of one gross. 2.) $42=60 times $.70=list'price of 60 gross. fl. 100%=$42. .oj2. l%=T^of$42=$.42. ^^■''|3. 50%=50 times $.42=$21=first discount. U. $42— $21=$21=first net proceeds. (1. 100%=$21. ., J2. l%=T-Uof$21=$.21. ^ '' I 3. 10%=10 times $.21=$2.10=second discount. U. $21— $2.10=$18.90=second net proceeds. (1. 100%=$18.90. , .12. 1%=$.189. ^^■nS. 5%=$.945=third discount. U. $18.90— $.945=$17.955=cost. rl. 100%=$42. ,^J2. l%=T¥¥of $42=$.42. - [count. ^^■n 3. 20%=20 times $.42=$8.40=ifirst conditional dis- U. $42— $8.40=$33.60=first conditional net proceeds. fl. 100%=$33.60. 12. l%=y^^of$33.60=$.336. [discount. (7.)\ 3. 10%=10 times $.336=F$3.36=second conditional 4. $33.60— $3.36=$30.24 = second conditional net proceeds. rl. 100%=$30.24. 1 2. l%=iiTj:of $30.24=$.3024. -^ [discount. ^ -'|3. 5%=5 times $.3024=$1.512=third conditional U. $30.24— $1.512=$28.728=se.lling price. (9.) .-. $28.728— $17.955=$10.773=his profit. III. .-. $10.773=his profit (Ji. Sdf., f. 212, prob. 9.) PROBLEMS, i 1. Bought a case of slates containing 12 doz. for $80 list, and got 45, 10, and 10% off; getting an additional 2% off for prompt payment, what did I pay per dozen for the slates ? Ans. $2.9106. 2. Bought a case of hats containing 4 doz., on which I re- -ceived a discount of 40, 20, 10, 5, and 2^% '^^^ ■ ^^ ^ ^^^^ them at $4 a piece making a profit of 20%, what is the wholesale mer- ■chant's price per case ? Ans. $3991^0 7_ 3. If I receive a discount of 20, 10, and 5% off, and sell at a •discount of 10, 5, and 2^'fo off; what is my % of gain? Ans. 21f %— . 4. A bill of goods amounted to $2400; 20% off being allowed, what was paid for the goods? Ans. $1920. ■ 5. Bought goods at 25, 20, 15, and 10% off. If the, sum of my discounts amounted -to $162.30, what was the list price of the goods? ' ^ Ans. $300 PROFIT AND LOSS. 93; III. PROFIT AND LOSS. II. 1. JProfit and Loss are terms which denote the gain or loss. in business transactions. 3. Profit is the excess of the selling price above the cost. 3. Loss is the excess of the cost above the selling price. I. A merchant reduced the price of a certain piece of cloth 5 cents per yard, and thereby reduced his profit on the cloth from 10% to 8%; what was the cost of the cloth, per yard ? (\. 100%=cost of cloth per yard. 2. I0%^his profit before reduction. 3. 8%=his profit after reduction. 4. 10%— 8%=2%=his reduction; 5. 5/=reduction. > 6. .-. 2% =5/, 7. 1%=^ of 5/=2i/, and ^8. 100%=100 times 2|/=$2.50=cost per yard. III. .-. $2.50=cost of cloth per yard. {R. Sdf.,f.'211,prob. IS.) I. A dealer sold two horses for $150 each; on one he gained 25% and on the other he lost 25%; how much did he lose in the transaction? (1.) 100%=cost of the first horse; (2.) 25%=gain. (3.) 100%+25%=125%=selling price of first horse. (4.) $150=selling price. (5.) .-. 125%=$150, (6.) 1%=T^ of $150=$1.20,and (7.) 100%=100 times $1.20=$120=cost of first horse. (8.) $150— $120=$30=gain on first horse. fl. 100%=cost of second hoise. 2. 25%=loss. 3. 100% — 25%i=75%^selling price of second horse. 4. $150:=selling price. 5. .-. 75%=$150, ■ 6. 1%=7V of $150=$2, and [7. 100%=100 times $2=$200=cost of second horse. (10.) $200— $150=$50=loss on, second horse. .(11.) $50— $30=$20=loss in the transaction. III. .■. He lost $20 in the transaction. {R. 3d p. .p. 211,prob. 12.} I. ,A speculator in real estate sold a house and lot for $12000, which sale afford him a profit of 33^% on the cast ; he II. (9.) ■94 FINKEL'S SOLUTION BOOK. JU 311. I. II. then invested the $12000 in city lots,- which he was obliged to pell at a loss of 33^%; how much did he lose by the two transactions.'' (1.) 100%=^cost of the house and lot. , (2.) 33i%=gain. [lot. (3.) V 'l'00%+83i%=133i%=sellingpriceof house and (4.) $12000^selling price of the house and lot. (5.) .-. 133i%=$12000. (6.) 1%=-^ of $12000=190. [lot. (7.) 100%i=100 times $90=$9000=cost of house- and (8.) $12000— $9000=$3000,=gain on house and lot. fl. 100%=$12000. (9.)<J2. l%=TiTof$12000=$120. Is. 33l%=33i times $120=$4000=loss on city lots. (10.) $4000— $3000=$1000=loss by The two transac- tions. .■. $1000=his loss by the two transactions. (R.3df.,f.211,prob.l5.) A dealer sold two horses for the same price; on one he gained 20%, and on the other he lost 20%; his whole loss was $25 ; what did each horse cost? (1.) 100%=selling price of each horse. 1. 100%=cost of first horse. 2. 20%=gain on the first horse. 3. 100%^20%=120%=selling price of first horse. 4. .-. 120%=100%, from (1), 5. 1 %=TiT of 100%=!% , and 100%=100 times |.%=83i%=cost of first horse in terms of the selling price. 100%— 83i%=16f %=gain on first horse. , 100%=cost of the second horse. 20%=loss on second horse. 100^ — 20%=80%=pe]ling price of second horse. .-. 80%=100%, from (1), 1%=^!^ of 100%=li%, and 6. 100%=100 times li%=125%=cost of second horse in terms of the selling price. 7. 125%— 100%=25%=loss on the second horse. 25%— 16t%=8i%=whole loss. $25=whole loss. /o=$25, (2.) (3.) (4.) (5.) (6.) (7-0 (8.) (9.) (10.) l%=^of $25=$3, and ru 8^ [horse. 100%=100 times •$3=$300^sening prifce of each 83^%=83i times $3=$250=cost of first horse. 125%=125 times $3=$375=cost of second horse. II. " Profit and loss. 95 ,.__ _ r$250— -cost of the first horse, and ■ ■ ■ I $375==cost of second horse. I. What "Jo is lost iff of cost equals f of selling price? 1. I of selHng'price':=f of cost. 2. \ of selling price=^ of f of cost=|- of cost. 3. \ of selling price=4 times f of cost==|- of cost. 4. I^cost. 5. -|=selHng price. \ 7. 1=100%. 8. \=\ of 100%=^lli% , loss. III. .-. Loss=ll^%. I. Paid $125 for a horse, and traded hiin for another, giving 60% additional money. For the second horse I received a third and $25. I then sold the third horse for $150 ; vv^hat was my % of profit or loss? , (1.) 100%=$125, (2.) l%=Ti-!r of $125=$1.25, and (3.) 60%=60 times $1.25=$75 = additional money paid for the second horse. (4.) $125+$75=$200=cost of second horse. (5.) $150=selling price of the third horse. (6.) $ieO+$25=$175=Selling price of second horse. (7.) $200— $175=$25=loss in the transaction. rl. $200=100%, I (8.)<^2. $l=7fir of 100%=i%, ana (-3. $25=25 times ^%=12-^%=my loss. III. .-. My loss is 12*%. {R. H. A., p. 201, prob. 4.) I. If I buy at $4 and sell at $1, how many % do I lose? Il.i II. '1. $4=cost. 2. $l^selling price. 3. $4— $l=$3=loss. 4. $4=100%. 5. $l=iof 100%=25%. 6. $3=3 times 25%=75%=loss. III. .-. 75%=loss. I. A and B each lost $5, which was 2|% of A's and 3^% of B's money ; which had the most, and how much ? FINKEL'S SOLUTION BOOK U.< ' (1.) 100%=A's money. (2.) 2J%=what he lost. (3.) $5=what he lost. (4.) .-. 2f%=$5, (5.) ' l%=^of$5=$1.80, and . (6. ) 100%=100 times $1.80=$180=A's money. 1. 100%=B's money. 2. 3^%=what he lost. , . 3. $5=what he lost. 4. .-. 3i%=$5, 5. 1 %=!- of $5=$1.50, and 6. 100%=!=100 times $1.50=$150s=B's money. $180— $I50==$30=excess of A's money over B's. III. (7.) |8.) .-. A had $30 more than B (i?. H. A., f. 203, froh. 5.) Mr. A bought a horse and carriage, paying twice as much for the horse as for the carriage. He afterward sold the horse for 25% more than he gave for it, and the carriage for 20% less than he gave for it, receiving $577.50; what was the cqst of each ? 100%=cost of the carriage. 200%=cost of the horse. 20%^=:loss on the carriage. 100%— 20%==80%=selling price of the carriage. 100,% =200%, 1%=TT^ of 20d%=2%, and 25%=25 times 2%=50%=gain on the horse. 200%+50%=250%=selling ■price of the horse. 80%-i-250%=330%=selling price of both. ' $577.50=selling price of both. .•.330%=$577.50, l%=^fj^ of $577.50=$1.75, and 100%=100 times $1.75=$175=cost of carriage. 200%=200 times $1.75=$350=costof the horse. jjy . r$175=cost of the carriage, and ' ■ I $350=cost of the horse. {Milne's frac, p. 259,prob. 19.) I. Mr. A. sold a horse for $198, which was 10% less than he asked for him, and his asking price was 10% more than the horse cost him. What did the horse cost him? PROFIT AND LOSS. 97 II. III. II.<^ (2.) (3.) 4.) (5.) (1.) 100%^cost of the horse. (2.) 100%+10%=110%=asking price. " rl. 100%=110%, (3.)<^2. l%=TiTof 110%=1tV%> and [asking price. Is. 10%=10 times li-V%=ll%=reduction from (4.) 110%— 11 %=99%=selling price. (5.) $198=selling, price. (6.) .-. 99%=$198, (7.) 1%=-^ of $198=$2, and (8.) 100%=100 times $2=$200=cost of the horse. •. $200=cost of horse. {Milne's frac, f. 269, prob. 23.) What must be asked for apples whicli cost me $3 per bbl., that I may reduce my asking price 20% and still gain ' % on the cost? 100%=$3. lfB=ih of $3=$.03, and 20%=2O times $.03=$.60=gain. $3.004-$.60=$3.60=selling price. 1. 100%=asking price. 2. 20%=reduction., 3. 100%— 20%=80%=selling price. 4. $3.60=selling price. 5. .-. 80%=$3.60, 6. 1%=^^ of $3.60=$.045, and 7. 100%=10,0 times $.045=$4.50=asking price. III. .-. $4.50=asking price. {Milne's j>rac.,f. 261, frob.SS.) I. A merchant sold a quantity of goods at a gain of 20%. If, however, he had purchased them for' $60 less than he did, his gain would have been 25%. What did the goods cost him ? (1.) 100%^actual cost of goods. (2.) 20%=gain. (3.) 100%+20%=120%=actual selling price. (4.) 100%— $60=supposed cost. fl. 100%=100%— $60, .. J 2. l%=Tk of (100%— $60)=1%— $.60, and 11.^ ^°-^]3. 25%=25 times (1%— $.60)=25%— $15 = sup- ,^ posed gain. [ual selling price. (6.) (100%— $60) + (25% — $15)=125%— $75=act- (7.) .-. 125%— $75=120%, (8.) 5%=$75, (9.) l%=^of $75=$15, and (10.) 100%=100 times $l5=$1500=cost of the goods. III. .-. $1500=cost of goods. {Milne's frac, p. 261, prob. 4b.) 11. <^ 98 FINKEL'S SOLUTION BOOK. Note. — The selling price is the same in^ the last condition of this problem as in the iirst. Hence we have the selling price in the last condition equal to the selling price in the first as shown in step (7.) I. I sold an article at 20% gain, had it cost me $300 more, I would have lost 20%; find the cost. _, (1.) 100 %^aGtual cost of the article. » (2.) 20%=actual gain. , (3.) 100%+20%=120%==actual selling price. (4.) 100%-|-$300=supposed cost. fl. l00%=^100%+$300, ,, J2. l%=TiTof(100%+$300)=l%+$3, and '^•'')3. 20%=20 times (l%+$3)=20%+$60=supposed V, loss. [_ual selling price. ( 6. ) ( 100 % +$300 )— ( 20 % +$60 ) =8 % +$24'0=ac t- (7.) .-. 120%=80%+$240. V (8.) • 40%=$240, (9.) 1%=7V of $240=$6, and (10.) 100%=100 times $6=$600=cost of the article. III. .-. $600=cost of the article. {R. H. A., p. Jfi9,prob. 85.) I. A man wishing, to sell a horse and a cow. asked three times as much for the horse as for th^ cow, but, finding no purchaser, he reduced the price of the horse 20%, and the price of the cow 10%, and sold them for $165. What did he get for each? (1.) '100%=;asking price of the cow. (2.) 300-%=asking, price of the horse. (3.) 10%=reduction on the price of the cow. (4.) 100%— 10%=90%=selling price of the cow. (1. 100% =300%, (5.)i2. l%==TiTof300%=3%,and 1-3. 20%=20 times 3%=60%=reduction on horse. II.<! (6.) 300%— 60%=240%=selling price ot thehorse. (7.) 90%+240%=330%=selling price of both. (8.) $165=selling price of both (9.) .-. 330%=$165, (10!) 1 %=^ of $165=$.50, and (11.) 90%=90 times $.50=$45=selling price of cow. (12.) 240%=240 times $.50=$120=selling price Sf horse. III. r$45^amount he received for the cow, and ■|$120=amount he received for the horse. h-' .. PROFIT AND LOS: PROBLEMS. 1. What price must a man ask for a horse that cost him $200, that he may fall 20% on his asking grice and still gain 20%'? Ans. $300. 2. A man paid $150 for a horse which he offered in trade at a price he was willing to discount at 40% for cash, as he would then gain 20%. What was his trading price? ^«.s. $300. 3. A man gained 20% by selling his house for $3600. What '■did it cost him ? 'Ans. $3000. 4. A gained 120% by selling sugar at 8^ per pound. What did the sugar cost him per pound ? Ans. 3-^^. 5 How must cloth, costing $3.50 a yard, be marked that a merchant may deduct 15% from the marked price and still gain 15%? " ^^^.$4.73^,-. 6. Sold 3 piece of carpeting for $240, and lost 20%; what selling price would have given me a gain of 20% ? Ans. $360. 7. S,61d two carrifges for $240 apiece, and gained 20% on one and lo?t 20% on the other; how much did I, gain or lose in I th? transaction ? Ans. Lost $20. 8. Sold goods at a gain of 25% and investing the proceeds, sold at a loss of 25% ; what was my % of gain or loss. Ans. Q\<fo- 9. A man sold a horse and carriage for $597, gaining by the sale, 25% on the horse and 10% on the cost of the carriage. If f of the cost of the horse equals f of the the cost of carriage, what was the cost of each? Ans. Carriage $270; horse $240. 10., If 1^ of the selling price is gain, what is the profit? Ans. 80%. 11. li \ of an article be sold for the cost of ^ of it, what is the rate of loss? Ans. 33^%. 12. I sold two houses for the same sutn; on one I gained 25% and on the other I lost' 25%. My whole loss ^was $240; what did each house cost? Ans. First $1440, second $2400. 13. My tailor informs me that it will take 10^ sq. yd. of cloth to make me a full suit of clothes. The cloth I am about to buy is 1|- yards wide and on sponging it will shrink 5% in length and width. How many yards will it take for my new suit? Ans. 6^11^ yd. 14. A grocer buys coffee at 15/ per Sb. to the amount of $90 worth, and sells it at the same p rice by Troy weight ; find the % of gain or loss. Ans. Gain 21-|-|%. 100 FINKEL'S SOLUTION BOOK. 15. I spent $260 for apples at $1.30 per bushel ; after retain- ing a part for my own use, I sold the rest at ^a profit of 40%^ clearing$13on the whole cost. How many bushels did I retain? Ans. 6(5 bu. 16 How must cloth costing $8.60 per yard, be marked that the merchant may deduct 15% from the marked price and still make 15% profit? Ans. $4,735. 17. I sold goods at a gain of 20%. If they ha,d cost me $250' more than they did, I would have lost 20% by the sale* How much did the^goods cost me ? , Ans. $600. 18. A merchant bought cloth at $3l25 per yard, and after keeping it 6 months sold it at $8.75 per yard. What was his< gain %, reckoning 6% per annum for the use of money? Ans. 12%+> IV. STOCKS AND BONDS. 1. Stocks is a general term applied to bonds, state and national, and to certificates of stocks belong to corporations. 3. A Soncl is a written or printed obligation, under seal,, securing the payment of a certain sum of money at or before a specified time. 3. Stock is the capital of the corporation invested in busi- ness; and is divided into Shares, usually or$100 each. 4. Afl Assessment is a sum of money required of the; stockholders in piroportion to their amount of stock. 5. A Diviclend, is a sum of money to be paid to the stock- holders in proportion to their amounts of stock. 6. Hie Par Value of money, stocks, drafts, etc., is the" nominal value on their face. I'. The Market Yalue is the sum for which they selL 8. Discount is the excess of the par value of money,, stocks, drafts, etc., over their market value. 9. Premium is the excess of their ma^-ket value over their par value. 10. Brokerage is the sum paid an agent for buying stocks^ bonds, etc. ll. STOCKS AND BONDS. 101 At i% brokerage, a broker received $10 for making an in- vestment in bank stock ; how many shares did he buy? 1. 100 %=par value of stock. 2. :J%=brokerage. 3. $10=brokerage. 4. .-. i%=$10, 5. 1%=4 times $10=$40, and 6. 100%=100 times $40=$4000=par value of stock. 7. $100=par value of one share. 8. $4000=par value of 4000-f-lOO, or 40 shares. III. •. 40=number of shares. I. How many shares of railroad stock at Afi/o premium can be bought iof, $9360? 1. 100%^par value of stock I can buy. 2. 4%=premium. 3. 104%=price of what I buy. ■4. $9360=price of what I buy. II. <^ 5. .-. 104% =$9360. 6. 1 fo^T^ of $9860=$90. 7. 100% =100 times $90=$9000=par value. 8. $100=par value of one share. 1-9. $9000=par value of 9000-f-lOO, or 90 shares. III. .•-. 90=number of shares that can be bought. I. When gold is at 105, what is the value of a gold dollar in currency ? rl. 105/.; or 105% in currency=100/; or 100% in gold. II.<2. 1/; or 1% in currency=.95-^V/ ; or .95^% in gold. 1.8. 100/; or 100% in currency=95^/; or 95^% in gold. <II. .-. $1 in currency is worth 95^/ in gold. I. In 1864, the "greenback" dollar was worth only 35^/ in gold; what was the price of gold? '1. 35f/; or 35f % in gold=100/ ; or 100% in currency. , 2. 1/; or 1% in gold=^ of 100/; or 100%=2.8/; or 2.8% in currency. ^°T 3. 100/; or 100% in gold=100 times 2.8/; or 2.8%=280/; or 280% in currency. III. .-. $1 in gold was worth $2.80 in currency. {R. Sdp., p. Z17, :prob.'8.) I. Bought stock at 10% discount, which rose to 5% jJremium and sold for cash. Paying a debt of $33, I invested the balance in stock at 2% premium, which at par, left mi $11 less than at first; how much money had I at first? II. 102 FINKEL'S SOLUTION BOOK. II. (2.) (3.) (4.) (5.) (6.) (7.) (8.) (9-) (10.) (11.) (12.) (13-) 1(14.) 100%=my money at first. 100%^par value of stock. 10%==disc9unt. 100%— 10%=90%=market value. .". 90^^100%, my money; .because that is the amount invested. 1 %=^\ of 100%=li%, and^ 100%=100 times l|%=lll^%=par value of the stock in terms of my money. 100%=11H% 1 %=l-j%, and [terms ot my money. 5%=5 times lj%=5|-%=premium on stock in lll|%4-5|%=li6f% = what I received for the stock. ' 116f-% — $33^amount invested in second stock. 100%^par value of second stock. 2%=premiiim. , 100%+2%=^102%=market value of second stock. 116f % — $33=market value of second stock .-. 102%=116|%— $33, l%=Tiir of (116f%^$33)=lT2jV%— $-32tV iqO%=100 times (1^%— $.32A)=^114t*A%— $32j^y=par value of second stock. 114y\%% — $32^=what I received for the second stock, since I sold them at par. .-. 114^^7%— $32^=100%— f 11, by the last con- dition of the probletnt x 14t'A%=$21tV 1 %=tt4i- of $21t9t=$1.485, and Mi KT III. I. n.<^ III. I. 100^^=100 times $1.485=$148.50. I had $148.50 at first. {R. H. A.,f. 212, f rob. 8.) Bought $8000 in gold at 110%, brokerage 1% ; what did I pay for the gold in currency? 1. 100%^par value of gold. 2. 110%=market value. ' 3. ■J%=brokerage. 4., 110%+i%=110i%=entire cost. 5. 100%=$8000,. 6. l%=TiTr of $8000=^$80, and 7. 110|%=110^ times $80=^$8810=costofgold in currency. .-. $8000 in gold costs $88lb in currency. What income in currency would a man receive by invest- ing $5220 in U. S. 5-20, 6% bonds at 116%, when gold is worth 105? II. STOCKS AND BONDS. 103 (1.) 1009?i=par value of the bonds. (2.) 116%=market value. (3.) $5220=market value. (4.) .-. 116%=$5220. (5.) l%=Ti^cJf$5220=$45. (6.) 100%=100 times $45=$4500=par value of bonds. (1. 100%=$450C. (7.)\ 2. l%=Tk of $450{)=$45. Is. 6%=6 times $45=$270=income in gold. (8.) $1.00 in gold=$1.05 in currencv. (9.) $270 in gold=270 times $1.05=$283.50 in currency. III. .'. $283.50=income in currency. (J?. Sdp.,j>. 2n,prob. 6.) What <fo of income do U. S. 4^% bonds, at 108, yield when gold is 105%? (1.) 100%=amount invested in the bonds. (2.) 100 %=par value of bonds. (3.) 108%=market value. (4.), .-. 108 %=10p%, from (1). , (5.)' 1%=^^ of 100%=|f%, [of amount invested, TT I . (6.) 100%=100 times |f %=92|-%=par value in terms ^■^■^ rl. 100%=92|%. , . . (7.)^2. l%=TiTyof92|%=ff%, • l3. 4^%=4| times |f %=4^%=income in gold. (8.) 100% in gold==105% in currency. (9.) 1% in gold=T-^^ of 105%=1^V% 'i currency. (10.) 4|% in gold=4^ titaes l^i.j%=4|% in currency. III. .-. Income in currency^4f %. Note. — This is a general solution of the preceding problem. Since there is no special amount given, we represent the amount invested by 100%. The market value and the amount invested being the same, we have 108%^100% as shown in (4). I. A man bought Michigan Central at 120, and sold at 124%; what % of the investment did he gain? II. 1. 124%=selling price. 2. 120%=cost. 3. 124%— 120%=4%=gain. 4. 120%=100% of itself. 5- l%=TiTrO'f 100%=f%, . s ^ 6. 4%=4 times f %=3-J%=gain on the investment. III. .-. He gained 3^% on the investment. 11.^ 104 > FINKEL'S SOLUTION BOOK. I. What sum invested in U. S. 5's of 1881, at 118, yielded an annual income of $1921 in currency, when gold was at 113? (1.) $1.13 in currericy=$l'in gold. (2.) $1 in currenCy=Tiy^ of .$l=$.^in gold, and (3.) $1921 in currency=1921 times $^|=$i700=in- come in gold. , (4.) 100%=par value of the bonds. (5.) 5%=income in gold. (6.) $1700=income in gold. (7.) .-. 5%=$1700, (8.)' 1%=^ of$1700=$340, and [bonds. (9.) 100%=100 times $840=$34000=par value of the (1. 100%=$34000, (10.)J2. l%=T^of$34000=$340,.and J3. 118%=118 times-$340=$40120=market value, or '- amount invested. III. .-. $40120=am[ount invested. « SECOND SOLUTION. ' (1.) 100%=amount invested in currency., (2!.) 100 %=par -Value. (3.) 118%=market value. (4.) .-. 118 %=100%, from (1.) •<5-) l%=Ti-T of 100%=|f %, and (6.) 100%=100 times ff%=84||%=paf value in terms of the investment. (1. 100%=84||%. (7.){2. 1%=T^ of 84^f%=4f%, and ' l3. 5%=5 times |f %=4^%=income in gold. - 100% in gold=113% in currency, 1% in gold^l^'(p^% in currency, and 4M% in gold=4i|- times l^%=4^%=income in currency. (9.) $1921^income in currency. (10.) .-. 4i%%=$1921, (11.) 1%==44f of $1921=$401.20, and (12.) 100%=^i00 times $401.20=$40120=amount in- vested in currency. III. .-. $40120=amount invested. (Ji. Sd f., p. 218, f rob. 8.) I. How many shares of stock bought at 95:^%, and sold at 105, brokerage ^% on each transaction, will yield an income of $925 ? _. n. (8.)^ STOCKS AND. BONDS. 105 11. 1. 100%=par value of stock. 2. 95i%=market value of stock. 3. |%=brQkerage. 4. 95|-|-i%=95i%=entire cost. 5. 105%=selling price-(-brokerage. ^%=brokerage. 105%— i%=104|%=sening price. 104|%— 95i%=9i%t=gain. $925=gain. .-. 9i%=$925, 11. 1 %=-L of $925=$100, and 12. 100%=100 times $100=$10000=par value of stock. 13. $]00=par value one share. 14. $10000=par value 10000-f-lOO, or 100 shares. III. .-. 100=number of shares. (i?. 3d p., p. 218, f rob. 9.) 1. II. If I invest all my money in 5% furnace stock salable at 75%, my income will be $180; how much must I bor- row to make an investment in 5% state stock selling at 102%, to have that incdme? 1. 100%=par value of furnace stock. •2. 5%^income. 3. $180=)ncome. 4. .-. 5%=^ (1-) 5. .){: (2 (3.) (4.) (5.) 1%=!. of $180=$36, and [nace stock. 100%=100 times $36=$3600=par value of fur- 100% =$3600, 1%,=^ ot $3600=$36, and [nace stock. 75%=75 times $36=$2700=toarket value of fur- 1. 100%=par value of state stock. 2. 6%=income. 3. $180=income. 4. .-. 6%,=$180, 5. ' l%=i of $180=$30, and [stock. 6. 100%=100 times $30=$3000=par value of state 1. •100%=$3000, 2. l%=y^^ of $3000=$30, and [state stock-. 3. 102%=102 times $30=$3060=market value of $3060— $2700=$360=what I must borrow. III. .-. I must borrow $360. {H. H- A., f. Z25, prob. 3.) I. When U. S. 4% bonds are quoted at 106, what yearly in- come will be received in gold from bonds that can be bought for $4982 ? IIJ II-1 (2 (3 (4 (5 (6 (7 (8.) 106 FINKFl'S SOLUTION BOOK. (1.) 10Q%==par value of the bonds. (2.) 106 %=market value. (3.) * $4982^market value, or amount invested. (4.) .-. 106%=$4982, (5.) l%=T^of$4982=$47, and (6.) 100%=100 times $47=$4700. rl. 100%=$4700, (7.){2. l%=rii! of $4700=$47, and 13. 4%=4 times $47=$188=income in gold. III. .-.' $188=income in gold. • (v?. 3;^., p. 218, f rob. 11.} I. The sale of my farm cost me $600, but I gave the pro- ceeds to a broker, allowing him ^%, to purchase rail- road stock then in the market at 102%; the farm ^aid 5% incomej equaL to $2076, but the stock will pay $2025 more; what is the rate of dividend? (1.) 100%=value of the farm. ".) 5%=income on the farm. .) $2075^1 ncome on the farm. .) :•. 5%=$2075, .) l%=^of$2075=$415, and .) 100%=100 times $415=$41500=value of farm. .) $41500— $500=$41000=amount invested in stock. fl. 100%=par value of the stock. 2. 102%=market value, or amount invested. 3. -|^%=hrokerage 4. 102%-|-i%=102-^%=entire cost of stock. 5. .-. 102i%=$41000, 6. 1 ic=T^ of $41000=$400, and ^^^* [railroad stock.. 7. 100,%=100 times $400=$40000=par value of the rl. $2075+$2026==$4100=income on railroad stock. ,„ J2. $40000=100%, ^^■''13. $l%=^^7ofl00%=:^7%,and [dend. U. $4100=4100 times :j^^%=10i%=rate of divi- III. .-. 10i%=rate of dividend. {E. H. A., p. 22^, prob. ^.} I. What must be paid for 6% bonds to realize an income of 8% on the investment? 1. 100%=amoynt invested. ^ 2. 6%=income on the par value of the bonds, ■rj 3. 8%=income on the investment. ■|4. .-. 8% of investment=6% of the par value, 5. 1% of investment=-^ of 6%=|% of the par value, and 6. 100% of investment=100 times f%=75% of par value. III. .-. Must pay 75% to make 8% on the investment. Note. — It must be borne in mind that 100% of any quantity is the quantity itself. .-. 100%- of, the amount invested equals the STOCKS AND BONDS, 107 amount invested. It must also be remembered that the income on the par value is equal to the income on the investment. Sup- pose I buy a 500-dollar 6% bond for $400. The income on the par value, or face of the bond is 6% of $500, or $30. But $30 is 7-^% of $400, the amount invested. Hence, the truth of step 4 in the above solution. Which is the better investment, buying 9% stock 25% advance, or 6%/ stock at 25% discount. at. II.<^ A. III. II. (1- (2. (3.) (4.) (5.) (6.) (7.) (8 4 100%=amount invested in the 9% stock, 100%=par value. ^ 25 %=premium. 100%+25%=125%=market value. .-. 12.5%=100%, 1 %=Tk of 100%=!%, and 100%==100 times |%=80%=par value in terms- of the investment. 1. 100%=80%, 2. , 1 %=rU of 80%=!%, and [stock. III. 3. 9%==9 times |%=7^%=income of 9% (1.)' 100 %==amount invested in 6% stock. (2.) 100%=par value of 6% stock. ,(3.) 25%=discount. (4.) 100%— 25%=75%=market value. l-B.J (5.) .-. 75%=100%. (6.) 1%=tV of 100%=li%, and (7.) 100%=100 times l^%=183i%=par value of the 6% stock in terms of the investment. rl.'100%=133i%, (8.)<^2. l%=x^T0f 133i%=4%, and [stock. 13. 6%=6 times l^%=8%=income of 6% •. The latter is the better investment, since it pays Sfc — 7^%, or ^% more income on the investment. ( Greenleafs N. A., p. 298, prob. 5.)- If I pay 87-^% for railroad bonds tl^at yield an annual in- come oil ''Jo 1 vvhat % do I get on my investment? (1.) 100%=investment. (2.) 100%=par value. (3.) 87|^%^market value, or amount invested. ' (4.) .-. «7^%=100%, from(l.) (5.) l%=8k °f 100%=1|%, and (6.) 100%=100 times l|%=114f %=par value in. terms of the investment. 100%=114f%, and [ment.- ncome on the invest- ?9^=incoine on the investment. rl. 100%=114f%, (7.)<^2. l%=^^ofll4f%=l|%,i 1,3. 7%=7 times l|%=8%=ir 108 FINKEL'S SOLUTION BOOK. A banker owns 2^% stocks at 10% below par, and 3% stocks at 15% below par. The income from the former is 66f% more than from the latter, and the investment in the latter is $11400 less than in the former; required the whole investment and income. Il-i (2.) (3.) \i i^-)\i (5.') 1. (1. 2. 3. 4. 6. 6. ft- (6-)<|i (7-) (8.) (9.) (10.) (11.) (12.) (13.) (14.) 100%^investment in the former. ,100%— $11400=investment in the latter. 100%=par value of the former. 10%=discount of the former, [vested in former. 100% — 10%=90%=market value, or amount in- .-. 90%==100%, from (i), ' , l%=JjOf.l00%=H%,and . 100%=100 times li%=lll^%=par value of for- mer in terms of the investment.^ 100%=llli%, ' l%=TiTr of 11H%=H%, and 2^%==2^ times l^%=2|%=income of former in terms of the Investment. 100^=par value of the latter. 15%=discount. [vested in the latter. 100% — 15%=85%=market value, or'amount in- .-. 85%=100%-^$11400, from (2), 1%=^ of (100%— $11400 )=lT^%—$134T2r, 100%=100 times (1^%— $134J2^)=117J^%— $13411-|-f =par value of latter in terms of former. 100%=117i|%— $13411if, [$134^, and 1% =T^T of (117H%-$13411if)=lT\%- - 3%=3 times.(lTV%— $134T?r)=3iV%— $402j^ =income of latter in, terms of the investment. 100%^income of the latter. 100%+66|%=166f%=income of the former. ^t7o= =income of the former. 166|%=2^%, 1 %=TfL of mo^i^fc, and _ ^ [terms of income of former. 100%=100 times •^%=l|%=income of latter in 3_9j.%_$402iV=income of the latter. 3^%-$402T«T=lt%. f%=$402TiV%, l%=4of$402A=$216, . ^^^^^^^ 100%=100 times $216=$21600=investment in 100 %— $11400= $21600 — $11400 = $10200 = in- vestment, in latter. ' 2^%=2^ times $216=$600=incpme of former. 144 STOCKS- AND BONDS. log- (15.) ' 3tV%— $402^7=31^ times $216— $4023V=$360= income of latter. (16.) $21600+$10200=$31800=whole investment. (17.) $600+$360=$960=whole income. TTT . V $31800=whole investment, arid ^ ■ •|$960=whole income. {R. H. A., p. 2Z5, prob. i- ) I. W. F. Baird, through his broker, invested a certain sum of money in Philadelphia 6's at llb^fb , and three times- as much in Union Pacific 7's at 89^%, brokerage ^<fo in both cases; how much was invested in each kindof stock if his annual income is $9920? II. (2.) (3.) 100%=amount invested in Philadelphia 6's. 300%=iamount invested in Union Pacific 7's. ' 1. 100%^par value of Philadelphia 6's. 2. 1154-%=market value. 3. ^%^brokerage. ' 4. 115ifo+i%=llGfe=exithe cost of Phila. 6's. ,5. .-. 116%=100%. 6. l%=J^oflOO%=f|%, and 7. 100%=100timesff%=86^%=par value of Phil- adelphia 6's in terms of investment. 1. 100%=86^%, (4.)< 1%= (5.) (6.) (7.) (8.) (9.) (10.) 1%=2^ (11.) - - (12.) =T-^of 86^%, and 6%=6 times |f %=5^^%=income of Philadel- phia 6's in terms of investment. 1. 100%=par value of Union Pacific 7's. 2. 89^%^market value. 3. l-^^^brokerage. 4. 89i%+i%=90%=entire cost of Union Pacific 7's. 5. .-. 90%=300%, 6. l'/o=^ of 300%=3i%, and 7. 100%=100 times 3|%=333i%=par value of Union Pacific 7's. ' rl. 100%=333i%, 2. l%=TiTof333i%=3i%,and ' . 3. 7%=7 times 3|%=23^%=income of Union Pa- cific 7's in terms of investment. 5^\%+23^%=28|4%=whole income. $9920=whole income. ... 28||%=$9920, 1%=kjt7a of $9920=$348, ^. _... ,, ,. „„ 28|4- [m Philadelphia 7 s. ^=100 times $348=$34800=amount invested 300%=300 times $348=$10450(^ amount in- vested in Union Pacific 7's. 110 III. FINKEL'S SOLUTION BOOK. $34800=amount invested in Philadelphia 6's, and $104400:=amount invested in Union Pacific 7's. {R.H.A.,p. 225, prob. 6.) I. 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? 11. (2.) (3.) (4.) (5.) (6.) (7.) (8.) (9.) (10.) (11.) 100%=amount invested in mining stock., 100%+$430^amount invested in furnace stock. 1. 100%=par value of mining stock. 2. 114^% =market value. 3. ■^%=brokerage. 4. 114i%+i%=115%=entirecost. 5. .-. 115%=100%, from (1), 6\ 1%=t1^ of 100%=H%,and 7. 100%=100 ' times If %=96||%=par value of mining stock in' terms of investment. .^ 100%=96f|%-, l%=TiTr of 96f|%=ft%, and 6%=6 times |f %=5^^%=income of mining stock in terms of investment. > 1. 100%=par value of furnace stock. 2. 112%=market value. 3. •|%=l3rokerage. 4. 112%-|-i%=112i%=entire tost. 5. .-. 112i%=100%+$430, 6. l%=^of (100%+$430)=f%+$3fi, and 7. 100%=100 times (|%+$3|i)=88f%+$382f= par value of furnace stock in terms of investm't. 1. 100%=88f%+$382|, 2- 1%=T^T of (88|%+$382|)=.|%+$3H, and 3. 4%=4 times (f %-f-$3H)=3f %-l-$l,5i|=income of furnace stock in terms of the investment. .-. 5^^%=3|-%-|-$15^, by the conditions of the problem, 1^7%=$15i|, '^^=iAl °' «15-=^9-20' -d [miningstock. 100%=100 times $9.20=$920=amount invested in 100%+$430=$1850=amount invested in. furnace stock. {R. H. A., p. 225, prob. 7-) „^ r$920=pamount invested in mining stock, and ■ ■ j $1350^amount invested in furnace stock. STOCKS AND BONDS. Ill II. III. Suppose- 10% state stock is 20% better in market than 4% raili'oad stock; if A.'s income be $500 from each, how- much money has. he paid for each, the whole investment bringing 6^5% ? 1. 100%^par value of state stock. 2. 10%^income. 3. $500=income. 4. .-. 10%=$500, 5. 1%=tV of $500=$50,and [stock. 6. 100%=100 times $50=$5p00=par value of state 1. 100%=par value of railroad stock. 2. 4%=income. 3. $500=income. 4. .-. 4%=$500, 5. l%=i of $500=$135, aad [railroad stock. 6. 100%=100 times $12o=$12500=par value of $5000=1 of $12500, i. e., the face of state stock is 1^ of face of ra,ilroad stock. 1. 100%=whole investment. 2. 6-5^^%^income of whole investment. 3. $500-|-$500=^$1000=income of whole investment. 4. . . 6^f^%=$1000, 5. l%=p^of$1000=$166.50,and orf-5- [ment. 6. 100%=100 times $166.50=$16650=whole invest- 1. 100%=investment in railroad stock. 1'. 40%=-| of 100%=invcstment in state stock, excluding the 20'%'excess.> 2'. 100% =40%, 3;- l%=TiTr of 40%=!%, and 4. 20%=20 times |%=8%=excess of state stock over same amount of railroad stock. 40%+8%=48%=investment in state stock. 100%+48%=148%=whole investment. $16650^whole investment. .-. 148%=$16650. l%=^^j of $1B650=$112.50, [railroad stock. 100%=100 times $112 50=$11250=investment in 48%=48 times $112.50=$5400=investment in state stock > J$11250=amount invested in railroad stock, and I $5400=amount invested in state stock. {R. H. A., p. 227,prob. B.) (1-) (2.) (3.) (4.) (5.) 2.{\ 112 FINKEL'S SOLUTION BOOK. PROBLEMS. r 1. What could I afford to pay for bonds yielding an annua} income of 9% to invest my money so as to realize 6% on the in- vestment? ' Ans. 150%. 2. What must I pay for Chicago, Burlington & Quincy Rail- road stock that bears 6% that my annual income on the invest- ment may yield. 5% ? Ans. 120%. 3. Bought 75 shares N. Y., P. &'0. Railroad stock at 105%, and sold them at 108^% ; /how much did I .gain in the transac- tion? Ans. $262.50. 4. How many shares of bank stock at 5% premium, can be bought for $7665? Ans. 73. 5. A broker bought stock at 4% discount, and, selling them at 3% premium, gained $1400; hovv^ many shares did he buy? Ans. 200. 6. At what price must I buy 15% stock that it may yield the same income as 4% stock purchased at 90% ? Ans. SST^fo. 7. How much must I pay for New York 6's so that I may realize an income of 9%? Ans. Q&^fc. 8. At what price must I buy 7% stock so that they may yield an income equivalent to 10% stocks at par? Ans. 70%. 9. What sum must I invest in U. S. 6's at 118% to secure ah annual income of $1800? Ans. $35400. 10. Which is the more profitable, and how muth, to invest 1^5000 in 6% stock purchased at 75%, or 5% stock purchased at Ans, The latter: $16*. 11. If a man who had $5000 U. S. 6's of 1881 shoulc^ sell them' at 115%, and invest in U. S. 10-40's purchased at 105%, would he gain or lose and how much? Ans. Loss $26.19. 12. Whea gold is at 120, what is a "greenback" dollar worth ? Ans. 83^/. 13. Suppose the market value-of 6% bank stock to be 11^% higher than 8% corporation bonds ; I realize 8% on my invest- ment, and my income from each is $180; what did I invest in each? Ans. $2923.P7tV' '" former, and $1576.92i-% in latter. , 14. A bought 5% railroad ^tock at 109|%, and 4|% pike stock at 107-^%, brokerage -1% ; the former cost $100 less than the latter but yielded the same income; what did each cost him? Ans. $ltOO Cost of former, and $1200 cosi of latter- INSURANCE. 113 15. What rate "Jo of income shall I receive if I buy U. S. 6's at a premium of 10%, and receive payment at par in 15 years? Ans. 3H%. 16."' Suppose the market value of 6% corporation stock is 20% less than 5% state stock; if my income be $1200 from each, what did I pay for each if the whole investment brings 6% ? Ans. $16000, and $2400a 17. I bought 2|% stock at 80%, and 4^% stock at 86%. The income on the former was 44|% more than on the latter, but my investment is $22140 less in the latter than in the former; what do I realize on my investment? Ans. 3|f^%. Hint. — Find the whole investment, and whole incomeas in the problem on page 75. Then find what % tlie whole income is of the whole invest- ment. > 18. invested in U. S. 4^'s ai 105, brokerage i% ; f- as much in U. P. G's at 119f, brokerage ^%; and 3 times as much in N. ' Y. 7's, at 87|-, brokerage \%. If my entire income is $1702, find my investment. {School Visitor, vol. is, p. 97.) Ans. $25320. 19. A. Vaid $1075 for U. S. 5-20 6% bonds at 7^% premium, interest payable semi-annually in^gold. When the average pre- mium on gold was 112%, did he make more or less than B. who invested an equal sum in railroad stock at 14% below par, which paid a semi-annual dividend of 4%? Ans. A. makes $16.40 less than B. every six months. 20. I invested $4200 in railroad stoct at 105, and sold it at 80%; how much must I borrow at 4% so that by investing all I have in 6% bonds at 8% interest, payable annually, I may re-- trieve my loss in one year?^ Ans. $18600. V. INSURANCE. 1. Insurance is indemnity against loss or damage. fi Tj 4. T (1- Fire Insurance. 1. Property Insurance. ,< r, th- • t '^ •' 12. Marine Insurance. ( 1. Life Insurance. 2. Personal Insurance. } 2. Accident Insurance. ( 3. Health Insurance. 3. Property Insurance is, the indemnity against loss or damage of property. 4. Personal Instirance is indemnity against loss of life or health. 5. Fire Insurance is indemnity against loss by fire. 114 FINKEL'S SOLUTION BOOK 6. Marine X»*S«*/*aM.ce is indemnity against the dangers of navigation 7. Life InSHra/rVCe 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. 8. Accident Insura/nce is indemnity against loss by accident. • , 9. Health Insurance is a weekly indemnity in case of sickness. 10. The Insurer, or Underwriter, is the party, or company, that undertakes the risk. 11. The MisJe is the particular danger against which the insurer undertakes. 13. The Insured is the party protected against loss. 13. The Premium is the sum paid for insurance; and is a certain per cent, o'f the amount insured. 14. The Amount, or Valuation, is the sum for which the premium is paid. I. My house is permanently insured for $1800, by a deposit of ten annual premiums, the rate per year being |%; how much did I deposit, and if, on terminating the in- surance, I receive my deposit less 5% ', how much do I get? (1.) 100%=$1800, (2.) l%==T-b of $1800=$18, and (3.) |%=| times $18=$13.50=one annual deposit. (4.) $135=10 times $13.50=ten annual deposits. (1. 100%=$135, {5.)h. l%=Tii! of $135— $1.35, and U. 5%=5 times $1.35=$6.75=deduction. (6.) $135— $6.75=$128.25=what I received. jjj . { $135=amount deposited, and ' ' ' ( $128.25=amount received. {R. H. A., p. ZSO, prab. 5.) I. An insurance company having a risk of $25000, at y9^%, ' reinsured $10000, at ^fo , with another office, and $5000, \ ail<fc, with another; how niuch did it clear above what it paid ? II. INSURANCE. 115 II. 2.) (3.) (4.) (5.) III. 100%=$25000, l%=Ti-S of $25000=$250, and j\fo=-^ times $250=$225=what the company received for taking the. risk. 1. $10000=amount the company reinsured at |%. 2. 100%=$10000, 3. 1%=T^^ of $100Q0=$100, and 4. |%=f times $100=$80=what the company paid for reinsuring $10000. 1. $5000^amount reinsured in another office at 1%. 2. 100%=$5000, [for reinsuring $5000. 3. lfo=y^ of $5000=$50=what the company paid 4. $80+$50=$130=w-hat the company paid out. 15. $225— $130=$95=what it cleared. 5^what tlie company cleared. (H. H. A., p. 2S0,prob. 7.) II. I took a risk at 4-|-% ; reinsured -f of it at 2%, and ^ of it at 2-J% ; what rate of insurance do I get on what is left? (2.) III. ^=whole risk. l^%=premium. 40%=| of 100%=amount reinsured at 2%- 100%=40%, 1%=^ of 40%=!%, and [suringfof therisk. 2%=2 times f j^=f %=amount I pay outforrein- 25%=:^ of 100%=:sefcond part reinsured. 100 fo=25fo. 1%==TOT of 25%=i%, and 2^<fo=^'2\ times :i%=|%^anjount I paid 6ut for reinsuring ^ of the risk. %%-\'%'fo=l^%=^^n\ovint of premiums paid out. H% — 14^%=-5\<?i=amount of premium I had left. 40%+25%=65%=whole amount reinsured. 100%— 65fo=35fo=risk left on which I received ■h'fo premium. 35%=100% of itself. ' 1%=^ of 100%=2f %, and i.3. <nr%=^^Tr times 2f %=y\%^rate of premium i^^rate of insuranqe I receive. - {R.H.A.,p.2Sl,prob.6.) Remark. — 35% is the base and :f\% is the percentage, and we wish to know what per cent, ^% is of 35%. I. Took a risk at 2%; reinsured $10000 of it at 2-|% and l$8000 at 1|%; my share of tlie premium was $207.50 ; what sum was insured? (3.) (4.) (5.) (6.) (7.) (8.) (9.)<^2 (? A 116 FINKEL'S SOLUTION BOOK. II. ri. ioo%=$ioqoo, f (l.)<^2. 1 %=T^T of $10000=1100, and [llOOOpreinsured- Is. 2|%=2itimes$100=$212.50=amountpaidouton. rl. 100%=$8000, (2.K2. l%=T|7-of$8G00==$80,and [$8000 reinsured. 13. 1|%=1| times $80=$140=amount paid out on (3.) $212.50+$140=$352.5'0==whole amount paid out. (4.) $207.'50=what I realize. ( 5. ) .-. $352.50+207.50==$560=premium on whole risk. (6.) 100%=risk. (7.) 2%==premium. (8.) $560^pre,mium. (9.) .-. 2%=$560, (10.) l%=i of $560=$280, and (11.) 100%=100 times $280=$28000=risk. III. $28000=risk. {R. H. A.,p. 2S2,prob. 6.} I. I can insure my house for $2500 at i-\% premium annually, or permanently by paying down 12 annual premiums! which should I prefer, and how much will I gain by it if money is worth 6% per annum to me? II (2.) (8.) (4.) (5.) III. 100%=$2500. 1%=^^^ of $2500=$25, and •^%=y'V times $25=$20^one annual prpmium. $240=12 times $20=twelve annual pi-emiums.' 1. 100%=the amount that will produce $20 an- nually at 6%. ' 2. 6%=interest. 3. $20=interest. 4. .-. 6^«=$26, 5. l%=i of $20=$3i, and 6. 100%=100 times $3i=$333i=the amount that "^ will produce $20 annually at 6%. (6.) $333i4-$20=$353i=amount I would have to pay down by, the former condition. [tion. (7.) .-. $353i— $240=$113i=gain by \he latter condi- ( The latter is the better. , I $113i=gain. Remark. — In (6) we add $20, since a payment must be made immediately. $333^ will not produce that sum until the end oi. the year. INSURANCE. 117 II. The Mutual Fire Insurance Company insured a building and its stock for f- of its value, charging If %. Tljie Union Insurance Company relieved them of ^ of the risk, at l|-%. The building and stock being destroyed by fire, the Union lost $49000 less than the Mutual; vyhat amount of money did the owners of the building and stock lose ? (!■) (3.) (3.) (4.) (5.) (6.) (7.) (8.) (9.) (10.) (11.) (12.) (13.) (14.) (15.) (16.) (17.) (18.) (19.) (20.) (21.) III. 100%=value of the building and stock. 66f%=| of 100%=amount insured. lf%=rate of insurance. 100%=66|%, .. 1%=TTTF of 66f%=f%, and li%=lf times f%=l|%^=what Mutual received from the owners of the building and stock. 16f %=i of 66|%=amount of which the Union relieved the Mutual, rl. 100%=16t%, ■2. l%=Tk of 16|%=i%, and 3. 14-%=1| times ^%=:i%=what the Mutual paid the Union for taking the risk of 16f %. 16|%+li%=17f fo=whole amount the Mutual received. [paid out. 66f%+i%=66H%=whole amount the Mutual .-. 66-^%— 171%= 49tV% = amount the Mutual lost. 16f %=amount the Union paid the Mutual. ;J%^amount the Union received from the Mutual. .-. 16f%— i%=16yV%=amount the Union lost. 49jl2-%— 16/2%=32_f%=what the Mutual lost more than the Union. [Union. $49000=what the Mutual lost moie than the .-. 32f %=$49000, l%=Qj¥of$49000=$1500, and ^. ' , , 32f , ' [ing and stock. 100fo=100 times $1500=$150000=value of build- 66|fo=66f times $1500=$100000=amount in- sured, [ers lost, it not being insured. 33i%=33i times $1500=$50000=what the own- 1. 100%=$100000, 2. l.fo=rU of $100000=$1000, and 3. l|%=lf times $1000=$1760=what the owners paid the Mutual for insurance.- .-. $50000 + $1750= $51750=whole amount the owners lost. The owners of the buil'ding and stock lost $51750. 118 tINKEL'S SOLUTION BOOK. PROBLEMS. 1. At l-f%, the premium for insuring my store' was $89.10; what was the amount of the insurance ? A.ns. 2. The premium for insuring a tannery for -f of its value, at, ^i%i was $145.60; what was the value of the tannery? Ans. $11648. 3. A store and its goods are ^worth $6370. What sum must be insured, at 2%, to cover both property and premium? Ans. ■ 4. The premium for insuring $9870 was $690.90 ; what wai the rate? Ans. I'fo^. 5. A merchant whose stock of goods was valued at $30000, insured it for f ot its value, at f %. In a fire he saved $5000 oi the goods. What was his loss? What was the loss of the in' surance companies? " Ans. - 6. A man paid $180 for insuring his saw mill for |- of itt value at 3%; what was the value of the mill ? ' Ans. > 7. A house which has been insured for $3500 for 10 years, af' \<fo a year, was destroyed by fire ; how. much did the money re- ceived from the company exceed the cost of premiums ? ^ Ans. , 8. Took a risk on a house worth $40000, at 2%; reinsured \ of it for 2^%, and -^ of it at 2^% \ in each case the amount cov- ers premium; how much do I gain? Ans. $99,558. 9. Took a risk at 1|%; reinsured f of it at 2J% ; my share of the premium w^as $43 ; what was the amount of the risk? Ans. $17200. 10. Took a risk at 2^% ; reinsured ^ of it at a rate equal to 3% 6f the whole, by which I lost $37.50. What was the value of the risk? Ans. $5000. SIMPLE, INTEREST. 119 CHAPTER XII. INTEREST. I. SIMPLE INTEREST. 1. Interest is money paid by the borrower to the lender for the use of money. 3. Hl^ Principal is the sum of money for which interest is paid. 3. The Mate of interest is the rate per cent, on $1 for a certain .time. 4. The. Time is the period during which the money is on interest. " ' 5. The A/mount is the sum of the principal and interest. 6. Simple Interest is interest on the principal only. 7. Legal Interest is at the rate fixed by law. - 8. TTswvy is interest at a rate greater than that allowed by law. . _ . Let P^^the principal, »-^the interest on $1 for one year, Ii=\-\-r=a.mount of $1 for one year, «=the number of years, ^==:amount of P for n years, Pr=s,\m^\e interest on P for a year, /'«j-=simple interest on P for n years. P-\-Pnr^:=P (l-j-ra^)=amountof P torn years. ^=amount of P for n. years. .-. A=P+Pnr=P{l+nr) .... (L) ; .-. P=T^ ....(IL); ... Pnr=A—P. „ A — P Interest .^^^ . _ ,'. r — — .... fill. I; nr nr .■.-^.•••(IV.);and ... .=^....(V.). When any three of the quantities A, P,n,r are given, the fourth may be found. CASE I. (-Principal ,-1 Given< Rate, and >to find the interest. Formula, I=Pm. iTime, i 120 FINKEL'S SOLUTION BOOK. I. Find the interest of $300 for two years at 6%. By formula, Interest P?'«=$300xO6x2=$36. By 1009?, method. (1. 100%=$300, J 2. 1%=TW of $800=$3, and ' 1 3. 6%=6 times $B=iF$18=intercst for one year. U- $36=i2 times $18=interest for 2 years. III. .-. $36=interest on $300 at 6% for 2 years. CASE 11. rPrincipai,-. A—I* Given-; Rate, andVto find the time. Formula, n= — . , llnterest, J -^'' I. In what time, at 5%, will $60 amount to $72? By formula, A~jP $72— $60 , /=-^r=$6ox:o^=^y^^'^^- \ By 100% methpd. 1. $72=amount. 2. $60=principal." 3. $72— $60=$12=interest for a certain time. 4. 100% =$60, 5. l%=ii¥ of $60=$f , and 6. 5%=5 times $f=$3=interest for one year. ' 7. $12=interest for 12-7-8, or 4 years. III. .-. $60 at 5% will amount to $72 in 4 years. * lU CASE III. (-Principal, -j Given<J Time, and >to find the rate. Formula, ; A—P r=-^Fi — =-w-=s7^K;^TT7i=-04=4,%. llnterest, J ' Pn ^ I borrowed $600 for two years and paid $48 interest; what rate did I pay? By formula, A—P _ I Pn ~Prr'M<^yS By 100% method, fl. .$48=interest for 2 years. 2. $24=^ of $48=interest for 1 year. II.i3. $600=100%, • 4. $l=^ofl00%=i%,and 5. $24=24 times i.%=4%. III. .•. I paid 4% interest. SIMPLE INTEREST. 121 CASE IV. I '*'' 1 'Given-J Rate, and >to find the principal. A — P jrest. J — - _ j-Time, ^Rate, , ^ . llnterest. J Formula, /*= ^^ fif llA I. The interest for 3 years, at '9%, is $21.60; what is the principal ? By formula, ^-P / $21.60 ^-^;^=^-3><:o9='^^"" By 100% method. <-l. $21.60=interest for 3 years. 2. $7.20=1^ of $21.60=interest for 1 year. 3. 100%=principal. 4. 99^=interest for 1 year. 5. $7.20^interest for 1 year. 6. .-. 9%=$7.20, 7. 1%=-^ of $7.20=$.80, and •-8. 100%=100 times $.80=$80=principal. III. .-. $80=the principal. CASE V. rTime,, , 1 ^ <5iven< Rate, and ^to find the principal; Formula, .P==— j . lAmount ) !/+«'■ I. What principal will amount to $936 in 5. years, at 6^^? By formula, P ^ ^^^^ $720 By 100% method. 1. 100%=principal. 2. 6%=interest for 1 year. 3. 30%=5 times 6%=interest for 5 years. 4. 100%+30%=±=130%=amount. 5. $936=a-mount. 6. .-. 130%=$936, 7. l%=j-^of $936=$7.20, and 8. 100%=100 times $7.20=$720=principal. HI. .'. $720=the principal that will amount to $936 in 5 years at 6%. II. 122 FINKEL'S SOLUTION BOOK. I. In what time will any sum quadruple itself at 8% ? II. 1. 100%=principal. Then 2. 400%=the amount. 3. .-. 400%— 100%=300%=interest 4. 8%=interest for 1 year. 5. 300%=interest for 300-i-8, or 37i years. III. .-. Any principal will quadruple itself in 37^ years at 8%. II. TRUE DISCOUNT. 1. Discount on a debt payable by agreement at some future time, is a deduction made for "cash," or present paymentj and arises from the consideration of the present worth of the debt!. 3. Present Worth, is that sum of money which, put on interest for the given time and rate, will amount to the debt at its maturity. 3. True Discount is the difference between the present worth and the whole debt. Since P will amount to ^ in « years, P may be considered equivalent to A due at the end oi n years.y .-. P may be regarded as the present worth of a given future sum A. , . P=-A^ \-\-nr , I. Find the present worth of $590, due in 3 years, the tatc of interest being 6%. By formula. $590 ^$500. II. IJ^r 1+3 X .06" By 100'% method. ^. 100%=present worth. 2. 6%=interest on present worth for 1 year. 3. 18%=3x6%=interest for 3year8. 4. 100%+18%=118,%=amount, or debt. 5. $590=debt. 6. .-. 118%=$590,- 7. l%=TfH-of $590=$5, and 8. 100%=100 times $5=$500=present worth. m. .*. $500=present worth of $590 due in 3 years at 6%. BANK DISCOUNT. 12a- I. A merchant buys a bill of godds amounting to $2480; he can have 4 months credit, or 5% oiT for cash : if money is worth only 10% to him, what will he gain by paying, cash? II. (2.) (3.) (4.) '(5.) (6.) (7.) (8>) (9.)^ (10.)| (11.) iOO%^present worth of the debt. 109^=interest on present worth for 1 year. 3-J-^^interest for 4 months. 100%-|-6i%=103^%=amount of present worth,. which equals the debt, by definition. $2480=the debt. .-. 103i%=$2480, 1 %=T?T5T of $2480=124, and 100%==100 times $24=$2400=present worth. $2480— $2400=$80=true discount. 100%=$2480. 1 fo=iU of $2480=$24.8Q, [count for cash^ 5%=5 times $24.80=$124=trade discount, or dis- .-. $124— $86=$44'=his gain by paying cash. III. .•. He would g^in $44 by paying cash. (H. Sdf.,f. 258, prob. lO.y Remark.— It is clear that $2480— $124 ,=$2356 would pay for the goods cash. I,f the merchant had this sum of money on hand, it would, in 4 months, at 10%, produce $78.53^ interest. But if he pays his debt he will make $124. Hence he will gain $124 — $78.53i=$45.46f. III. BANK DISCOUNT. 1. 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. 2. The Proceeds of a note is the amount which remains after deducting the discount from the face. / CASET I. ( Face of note, ) Given < Rate, and > to find the discount and proceeds. C^'^'^ > Formula, j J=|><0<« 124 FINKEL'S SOLUTION BOOK n I. What is the bank discount of $770 for 90 days, at 6% ? By formula, i?=-^X r X «=$770 X .06 X ^^^±^=$1 1.935. 360 By 100% method. (1. t00%=$770r 1 Tr |2. 1%=T.^ of $770=$7.70, and ; |3. 6%=6 times $7.70=$46.20=discount for 1 year. U. $11,935=^ of $46.20=discount for 93 days. III. .-. $11.985=bank discount on $770 for'90 days at ,6%. CASE II. ( Proceeds, "i Kyiveh < Time, and > to find the face of the note. (Rate, ) „,_,/> Formula, J<=^ . » 1 — rm I. For what sum must a note be made, so that wheiji dis- counted at a bank, for 90 days, at 6% the proceeds will be $893.80? Bv formula, /> $393.80 By 100% method. A. 100%=face of the note. , - 2. 6%^discount for one year. 3. 1^^%=^ of 6%=d|iscount for 93 days; 4. 100%-lH%=98^%=proceeds. ill.<5. $393.80=proceeds. ( 6. .-. 98A%=$393.80, 7. 1%=Q^ of $393.80— $4, and •■S. 100%=100 times $4=$400=face of the note. III. .-. $400=face of the note. CASE III. Given rate of bank discount, to find the corresponding rate of interest. Formula, rate of /.== . 1 — rn I. What is the rate of interest when a 60 day note is dis- counted at 8% per annum? -By formula, ANNUAL TNTEKKST. 125i IIJ III. By 100% method. 1. 100%=face of note. 2. 8%^discount for 1 year. 3. If %=-/-^ of 8fo=discount for 63 days. 4. 100%— lf%=98f%=proceeds. 5. 98f%=100% of itself. 1 ^-%=^,oflOO,%=m%, and. 8%=8 times ^^<fc=8:f^s%%=rate of interest. . The rate ofi interest on a 60 day note discounted at.,{ per annum^8:^^%. CASE IV. Given the rate of interest, to ^nd the corresponding^ rate of ^rate of /. discount. Formula, l+^Xrateof/. I. What is the rate of discount on a 60 day note which yields 10% intei'est? By formula, ', rate of/. .10 a 7_q3 »7 <,/ 1+nX rate of /.~1 +#ff X .10—"^^^—^^^ /" " By 100% method. fj. 100%=proceeds. 2. 10%=interest on proceeds for 1 year. 3. l|%=^6^3^ of 10%=interest on proceeds for 63 days. 4. 100^c+li%=101f %=face of note. 5. 101|%=100% of itself. ^- ^ ^"=1^1 °^ 100%=H^/« , and ' 7. 10%=10 times |^^%=9f|^%. .-. The rateof discount=9ff^%. JVoie. — It must be borne in mind that the interest on the pro- ceeds is equal to the discount on the face of the note. II. III. IV. ANNUAL INTEREST. 1, Annual Interest is the simple intere&t of the princi- pal and each year's interest frorn the time of its accruing until settlement. I. No interest having been paid, find the amount due Sept. 7, 1877, on anote of $500, dated June 1, 1876, with interest at 6%, payable semi-anniially. 126 FINKEL'S SOLUTION BOOK. 2yr. Smon. 6 3. 1 yr. 9 mon. 6 d. 1 yt. 3 mon. 6 d. 9 mon, 6 d. 2d 3 m. 6 d. 3dl 4th II. S III. r(l.) 100%=$500, ' 1877—9—7 (2.) l%=Th of $500=$5, and 1875—6—1 (3.) 6%=6 times $5=$30=simple interest 2—3—6 for 1 year. (4.) $68^2t*^ times $30==simple interest for 2 years, 3 months, 6 days. (5,) $15=1 of $30^semi-annual interest. i ' '1. 100%=$15, 2. 1%=T^^ of $15=$.15, and 3.,6%:=6 times $.15=$.90=interest on one semi- annual interest for 1 year. 4. $3.885=4^f times $.90^interest on one semi-annual interest for the sum of the periods each draws int. (7.) ■.-. $500+$68+$3.885==$571.885=amount of the not^. .-. $571.885=amount of the note. (6.) Explanation. — At the end of six months there is $15 Interest "due ; and, since it was not paid at that time, it drew interest from that time to the time of settlement, which is 1 yr. 9 mon. •6 da. At the end of the next six months, or at the end of the first year, there is another $15 due; and, since it was not paid at that time, it drew interest from\that time to the time of settle- rhent, which is 1 yr. 3 mon. 6 da. In like manner, the third -semi-annual interest drew interest for 9 mon. 6 da., and the fourth for 3. mon. 6 da. This is the same as one semi-annual interest drawing interest for the sum of 1 yr. 9 mon. 6 da., 1 yr. 3 mon. 6 da., 9 mon. 6 da., 3 mon. 6 da. In the dia- gram, the line A B represents 2 yr. 3 mon. 6 day., A 1 repre- sents the first year the note run, and 1-2 represents the second year the note run. Between A and 1 is a small mark that de- notes the semi-annual pdriod ; also one between 1 and 2. By such diagrams, the time for which to corhpute interest on the. isimple interest may be easily found. I. The interest of U. S. 4% bonds is payable quarterly in gold; granting that the income from them might be immediately invested, at 6%, what would the income on 20 1000-dollar bonds amount to in 6 years, with gold at 105? ANNUAL INTEREST. 127 ^'."0 for 4 yr. 9 mon. at 6^, i ?200 for 4 yr. 6 mon. at 6^. 1 1 ?200 for 4 yr. 3 men. at 6ji. 1 ' 1st M SilUth .S 6 71 8 &C. 9 10 11 12 13 14 15 16 17 18 19 20 1 2 4 d II.<^ (2.) (3.) (4.) (5.) (6.) (7.) (8.) (9.) (10.) $1000=par value of one bond. $20000=par value of 20 bonds. 100%=$20000, 1%==^^ of $20000=$200, and 4%=4 times $200=$800=income for one year. $4000=5 times $800=income for five years. $200=i of $1800=interest due at the end of first quarter, and which draws interest lo time of set- tlement. 1. 100% =$200, 2. lfc=rU of I200=:$2, and [est for one year. 3. 6%^6 times $2=$12=interest on quarterly inter- 4. $570=47-|- times $12^interest on quarterly inter- est for the sum of 4f yr.-(-4-|- yr.-f 4:|^ y-^h -\-^ yr. , or 47-| years. .-. $4000-t-$570=$4570r=income of bonds in gold. $1.00 in gold^$1.05 in currency. [rency. $4570 in gold=4570 times $1.05=$4798.50 in cur- III. .-. The bonds yield $4798.50 in currency. Explanation. — It must be borne in mind that the quarterly in- terest, $200, is put on interest at 6% as soon as it is due. At the •end of the first quarter there is $200 due which draws interest at •6% for the remaining time, 4 years, 9 months. The second quarterly interest is due at the end of six months and draws in- terest for the remaining time, 4 years 3 months, and so on with the remaining quarterly payments. This is the same as one quarterly payment drawing interest for the sum of 4f yr.-f-4^ yr. H~4i yr.-f-etc, or 47-^ years. I. What was due on a note of $1200, dated January 16, 1883, and due Aug. 1, ,1892, and bearing interest at 8%, payable annually, if the 2, 3, 5, and 7th years' interest '^were paid ? 128 FINKEL'S SOLUTION BOOK. t96for9yr. 6 m. 16d. at8^. | 1 , (96 £or6yr. 6 m. 15 d. 4 $96, 4 yr. 6im. 15 d. J96, 2 yr. 6 m. 16 d. $96, 1 yr. 6 m. 15 d. ) 3 ) $96, 6 in, 15 d, (1.) 100%=$1200. (2.) 1%=TW of $1200=$12, and (3.) 8%=8 times $12=$&6=simple int. for one year, (4.) $480=5 X$96=five simple interests. (5.) $48=-^ of $96— interest for 6 months. (6.) $4=Jj of $48=interest for 15 days. (7.) .'. $532=simple interest unpaid, fl. 100%=$96. II I 2. l%=Tk of $96=$.96, and [simple interest, 3. 8%=8 limes $.96=$7.68=interest on one year's ,g J 4. $193.92=25i times $7.68=interest on year's sim- ^ ^ ■•' > • pie interest for 9 yr. 6 mon. 16 da.,+6 yr. 6 mon. 15 da.,4-4 yr. 6 mon. 15 da.,+2yr. 6 mon. 15 da., +lyr. 6. mon. 15 da.,-|-6 mon. 15 da., or 25 yr. » 3 mon. (9.) "" .-. $532+$193.92=$725.92=amount of interest due. [1, 1892. (10) $1200+725.92 = $1925,92 = amount due July* III. .-. $1925.92=whole amount due Aug. 1, 1892. V. ^COMPOUND INTEREST. 1. CompQUnd Interest is interest on a principal formed by adding' interest to a tprmer principal. Let /'=principal on compound interest. r=rate, ^=(l-|-?-)=amount of one dollar for 1 year. P {l-\-r)^PR=a.mon'n.t of P dollars for 1 year. P (l-|-'')^=-^-^^='imount of P dollars for 2 years. P (l-j-r)8=/'7?8=amount oi P dollars for 3 years. P (l-i-r)"=/'i?''=amount oi P dollars for n years. Let yl^amount of P dollars in n years, and /=tlie compound interest of P dollars for n years. Then J=PI^—P I. ' A=PR'' II. •••^=4- ^"- COMPOUND INTEREST. 129 ° _ ' A .'. .ff==-\/£ IV. Applying logarithms to Ji''=r—, p ^ n log. i?=log. A — log. P , whence log, ^-log. i^ ""- log.i? •■_ ■ When compound interest is payable semj-annually. P (l-(-^)^amo}jnt of P dollars for \ year. P (l-|-f )^=am6unt of P dollars for 1 year, P ( l+i-) ^"^amourit of P dollars for n years. .'. A=P (l-)-|-)2"', when payable semi-annually. When com_pound interest is payable quarterly, P (14-f )=amount of P dollars for -^ year. P (l-)-f )^=amount of P dollars for -J year. P (l-|-|)'=amount of P dollars for f year. P (l-j-f )*=amount of P dollars for 1 year. ^•(I'-j-f )*°=amount of -P dollars for n years. , -•. A=P (1+f )*». When the interest is payable monthly, When the interest is payable §> times a year, CASE I. ( Principal, ) Given < Rate, and V to find the comp^ound interest and amount. ( Time, S ( I^PR^—P, Formula, A=PP^. I. Find the compound interest and amount of $500 for 3. years at 6%. By formulas, ^=/'^-=$500X(14-.06)3=$595.508, and I=PR^—P=%hm X ( 1 +.06 ) 3— $500=$95.508. Remark. — In compound interest, the lOO'j/o method becomes very tedious. By 100% method. , (1.) 100%=$500, (2.) l%=,^of$500=$5, (3.) Q%=S times $5=$30=interest for 1 year. (4.) $500-|-$30=$530=amount, or principaV for the second year. rl. 100%=|530, 2. l%=T-^^of$530==$5.30, [year. 3. 6%=6 times $5.30=$31.80=interest for second 4. $530+$31.80=$561.80=amount, or principal for the third year. II. (6.) 130 FINKEL'S SOLUTION BOOK. n. 100%=$561.80, i2. l%=j-|^of $561.80=$5.618, and [year. 3. 6%=6 times $5.618=$33.708=interest for third 4. 561.80+$33.708=$595.508=amount at end of the third year. (7.) $5^5.508— $500=$95.508=compound interest. ( $95.508=compound^ interest, and •*■ •'■ I $595.508=compound amount. CASE II. ( Principal, ) Given < Rate, and V to find the time. i Compound Interest, ) ^^^^^^^^ ^ Jog, ^-log. /> • log. -ff , 1. In what time will $8000 amount to $12000, at 6% com- pound interest? By formula, log. ^— log. ^ _ log. 12000— log. 8000 _ - ' ""7 log. a ~ log. 1.06 ~ 4.079181—3.903090 „ ii ik ^ w i *u- zirJj; -1^6 yr. 11 mon. 15 da. We may solve the .025306 ^ ^ problem thus: $8000(1.06)°=$12000, whence (1.06)°=12000-r- 8000=1.50. Referring to a table of cotapound amounts and passing down the column of 6%, we find this amount between 6 years and 7 years. The amount for 6 years is 1.4185191 ; the amount for required time is 1.50. .-. There is a diflTerence of 1.50— 1.4185191, or .0814809- The difference for the year between 6 and 7 is .0851112. .0851112=amount for the whole period between 6 and 7, .0814809=amount for f^fff of the period or, 11 mon. 15 da. .-. The whole time=6 yr. 11 mon. 15 da. CASE III. ( Principal, ) Given < Compound Interest or Amount, and ^ to find the rate, r Time, ' Formula, ''^'l? — 1. I. At what rate, by compound interest, will $1000 amount to $1593.85 in 8 years? By formula,' ■nI-p -nJ $1000 ' ) ANNUITIES. 131 CASE IV. ( Compound Interest or Amount ^ Given } Time, and > to find the principal. r Rate, ) r A ,or Formulae, jP^< / I. What principal, at compound interest will amount to 27062.85 in 7 years at 4% ? By formula, CHAPTER XIII. ANNUITIES. I. A.n Annuity is a sum of money payable at yearly, or other regular intervals. II. Perpetual, or I' CenS'or 4. Contingent. , 3. A Perpetual Annuity is one that continues forever. 4. A Limited Anmiity ceases at a certain time. 5. A Certain Annuity begins and ends at fixed times. 6. A Contingent Annuity begins or ends with the iiappening of a contingent event. 7. An Imtnediate Annuity is one that begins at once. 8. A Deferred Annuity is one that does not begin im- mediately. 9. The Final or Forborne 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 given rate, will amount to the initial value. II. The Initial Yalue of an annuity is the value of a deferred annuity at the time it commences. ^ • i 132 FINKEL'S SOLUTION BOOK II: III. CASE I. C Annuity, ) Given < Time, and > to find the initial value of a perpetuity. ( Rate, ) I. What is the initial value of a perpetual annuity of $300 a year, allowing interest at 6% ? 1. 100%==initial value. 2. 6%=interest for 1 3-ear. 3. $300=interest for 1 year. 4. .-. 6%=$300. 5. lfo=i of $300=$50, and 6. 100%=100 times $50=$5000=initial value. .-. Initial value=$5000. (Ji. H. A., p. 310, f rob. 1.) What is the initial value of a perpetual leasehold of $2500 a year payable quarterly, interest payable semi-annually at 6%; 6% payable annually ; 6% payable quarterly? 1. Let 6'=the annuity. Then 5'=the amount due in 3 months. ' 2. 5'-|-'S'(l-|-f )^amount due in 6 months. 3. .-. ^■=5.+ 5'(l + .01^) = $625 + $625(1.01i) = $1259.37Y=amount due at the end of 6 months. A.'i4. 100%=initial value. 5. 3%=semi-4nnual annuity. 6. $1259.37-j=semi-annual annuity. 7. .-. 3%=$1259.37i. 8. l%=iof $1259.37i=$419.7916|, and 19. 100%=100 times $419.7916|=initial value. 1. Let 5'=amount due in 3 months. Then 2. 5'-(-5'(l-(-i)=amount due in 6 months, [and 3. 5'-|-5'(l-i-J)+5'(l+|')=amount due in 9 months, 4. S'-f 5(l-f-f)+6'(l+f' )+6'(l+f )= amount due in i year. , [(l+-i^8)=$2566.25. p. I 5. .-. ^=$625+$62&(l+•\6)+$625(l+•V2)+$625- ^^ 6. 100%=initial value. 7. 6%=annuity. 8. $2556.26=annuity. 9. .-. 6%=$2556.25. 10. l%=-^of $2556.25=$426.0416f, and [value. 11. 100%==100 times $426.0416f =$42604. 16f=initial 1. 100%^initial value. 2. l^%=quarterly annuity. $625=quarterly annuity. II. 5. 16. 4% =$625. 1 1%=JL of $625=$416.6666|, and, 10d%=100 times $416.6666f=$41666.66f. ANNUITIES. 133 i initial value of A=$41979.16|, III. .-. } Initial value of B=$42604. 16|, and ( Initial value of C=$41666.66|. (Ji. H. A., p. SIO, frob. 5.) CASE II. {Annuity, "^ Interval, 1 to find the present Rate, and T ^value of a deferr- Time the perpetuity is deferred, J ed perpetuity. Let &=the annuity, »'=the rate, and R=\-\-r. Then by Case I., the initial value of S is S-^r. To find the present value of the initial value, we use formula III., compound interest. .'. P S S = — ■r:r-. — r:= 7-.,/ r. — T^ i" vyhich t is the time the perpetuity r{l-\-ry R\R — 1) _ f f J is deferred. I. Find the present value of a perpetuity of $260 a year, de- ferred 8 years, allowing 6% interest. By formula, P S ■- ^250 _ $250 i?'(i?— 1) (l+.06)8(l+.06— 1) .06(1.06)*"^ By 100% ipethod. ' ■ ( 1. ) 100%=initial value. (2.) 6%=annuity. (3.) |250=annuity. (4.) .-. 6%=$250. (5.) 1%=^ of$250=$41|,and ll.{ (6.) 100%=100 times $41|=$4166.66|=initial value. '1. 100%=present value of $4166.661 due in 8 years at 6%. 2. 159.38481%=(1.06)«Xl00%=compound amount ,„ V of the present value for 8 yr. at 6%. S'-'\Z. .: 159.38481 %=$4166.66|, 4. 1%=TT¥T?TST of $4166.66|=$26.1422, and 5. 100%=lb0 times $26.1422 = $2614.22 = present . I value. III. .'. The present value of a perpetuity of $250 a year de- ferred 8 years at 6% interest=$2614.22. I. Find the present value of an estate which, in 5 years, is to pay $325 a year forever ; interest 8%, payable semi- annually. By formula, S^ $325 $325 _ [( l+i) ^—1]^'~[( 1-04) =^—1] (1.04)1° ~.0816(1.04)i» ~ $2690.67. 134 FINKEL'S SOLUTION BOOK. By 100% method. . . . (1.) 100%=initial value. (2.) 4%=amount due in 6 months. (3.) 4%+(1.04)x4%=8.16%=amount due in 1 year. (4.) $325^amount due in 1 year. (5.) .-. 8.16%=$325, , 11.^(6.) l%=^f^ of $325=139.828431, and , [value. (7.) 100%=100- times $39.828431= $3982.8431 = initial ri. 100%=present value of $3982.8431. . 2. 148.024428%=(1.04)i»X100%=compound amount of 100% for 5 yr. at 8%. I (8.)J3. .•.148.024428%=$3982.8431, 4. l%=T„.^i„^5of $3982.8431=$26.9067, and • 5. 100% =100 times $26.9067 = $2690.67 = present value. III. .-. $2690.67=present value of the estate. {R. H. A., p.Sll,pr0b.4.) Explanation. — The initial value is a sum of money which placed on interest at 8% payable semi-annually will produce $325 per year. But 8% payable semi-annually is the same as 8.16% payable annually. Hence 8.16% is the annual payment. But $325 is the annual payment. Hence 8.16%^$325, from which we find that $3982.8431 is the initial value, or the amount that will produce $325 per year. Then the present value of a sum of money that will pay $325 is $3982.843! if the payments are to begin at once, but $3982.8431-=- (1.04)'" if the payments are not to begin until the end of 5 years. CASE III. {Rate, ^ ™. , ^' J >to find the present valve of an an- Time to run, and f .. \ . Interval, J -""'t^ "'^t*^'^- (0) Let P denote the present value. The amount of P for n Let ■S:^the payment, or amount due the first year. 5'-|-5!i?=the amount due the second year. .$■-)— 5'^?+'^-^^=*^'^^ amount due the third year. S-\-S/i-\-SJi^-]-SJi^ = the amount due the fourth year. [due the nth year. 8-{^SJi -\- SP^-\- SP^-}- -I- 5'i?»-i = amount .-. A=S-^SIi-{-SJi^-\-SP^Jr- -|-5^«-i . ...(1) AIi=SP+SjR''+SIi«-\-SJ?i+ SP^ (2), by multiplying (1) by i?. AJi—A=SJ^—S. . . (3), by subtracting (1) from (2). •■• ^= ^^ (4.) But PP-^A. '— ANNUITIES. ]3.j 5(i?°-l) ; S 7?"-l j?-(i?— i; ~i?^i^ i?" • ■ v^^ (^.) When the annuity is to begin at a certain time, and then to continue a certain time. Let /=the number of years the annuity is deferred, and j'= the number of years the annuity continues. Then /"'=-= — - X p I ^the present value of an annuity S, for the time {j)-\-q) years, and O Dp I -=the present value of an annuity S, for p Rp+i—l S 7?"— 1 S l'^ Rp+a JR—V^ Ro R—1 ^ R^'J^n—nJi' R p+i J R- I. Find the present value of an annuity, of $250, payable an- nually, for 30 years at 5%. Given S, n, and r. By formula, P- S i?°-^l $250 (1.05)'"'-1 By 100% method. (1.) 100 %=initial value. (2.) 5%=annuity. (3.) $250=annuity. (4.) .-. 5%=$250, (5. ) 1 %=i of $250=$50, and (6.) 100%=100 times $50%=$5000=initial value of an immediate perpetuity of $250 per year. 1. 100%^present value of an annuity (/e/erret/ 30 years. [ent value for 30 years; 2. 432.19424%=(1.05)3°Xl00%=amount of pres- C7 \h. .-. 432.19424%=$5000, ^ ''4. l%=T^hT^^ of $5000=$11.568,865, and 5. 100%=lbO times $11.568865=$1156.8865=pres- ent value of annuity of $250 deferred 30 years. (8-) .-. $5000— $1156.8865=$3843.1135=present value of an annuity continuing 30 years. III. \-. $3843.1135^present value of an annuity of $250, payable annually for 30 years, II. 136 FINKEL'S SOLUTION BOOK. Remark. — Since $5000 is the initial value whicli, in this case, is also the present value of an immediate perpetual annuity, or perpetuity of $250, and $1156.8865 the present value of an an- nuity of $250 deferred 30 years, $.5000— $1156.8865=$3843.1135= the present value of an annuity of $260 continuing for 30 years at 5%. II. Find the present value of an annuity of $826.50, to. com- mence in 3i years and run 13' years, 9 months, interest 6%, payable semi-annually. Given /S=$826,50, r=.06, ^=3 years, and q=\%% ye^rs. When interest is payable semi-annually, i?=(l-|-^)2. By formula (7), ' 8 Ri—V $826.50^(1.0609)"%— 1 ' "R—X^R'J'+i) By 100% method. .0609 (,1.0609)i<^ =$6324.69. (1- (2. (3.) (4. (5.) (6.) 7-) (8.) lQO%:^initial value. 3%=amount due in 6 months. 3%-f3% (1.03)=6.09%=amount due in 1 year. $826.50=amount due in 1 year. .-. 6.09%=$826.50, 1%=.^^^ of $826.50=$135.712643, and 100^ = 100 times $135.712643 = $13571.2643= initial value (9.) (10.) of a perpetuity of $826.50 100%=amount of 1. 100%=present value deferred 3 years. 2. 119.40523%=(1.0609)2 times present value for 3 years. 3. .-. 119 40'523%=$13571.2643, 4. l%=TTTiTrK^of $13571.2643=$113.6586, 5. 100%=100 times $113.6586=$11365.86=present value of such a perpetuity deferred 3 years 1. 100%=present value of such a perpetuity deferr- ed 16f years. 2. 269.212027%=(1.0609)i6?i times 100% = amount of present value for 16f years 3. .-. 269,212027 %=$13571.2643, 4. 1%=TWYT^YT of $13571.2643=$50.4117, 5. 100%=100 times $50.4117= $5041. 17 = present value of such a perpetuity deferred 16f years. .-. $11365.86— $5041. 17=$6324.69=present value of an annui|:y of $826.50 deferred 3 years and continuing l3-§ years. III. .-. $6324.69=present value of $826.50, etc. ANNUITIES. 137 If the annuity is to bdgin in f years and continue forever, the formula, rX ^- . - becomes P=-, S R—\ ^ RP+1 'Rp{R—l) ■ T-or, since P=-^[ [l_-^] _ [l—l] J if 5^=00 , the quantity ^ L , ^ =1 =1^ — 0, approaches 1 as its limit, s Rp+Q ■=1- and we have P=^£(l_o)-[l_^)]=^^^^^ . ^, I. Find the present value of a perpetual annuity of $1000 to begin in 3 years, at 4% interest. By formula, [value of the annuity. ^-(^=^=.04^r54F=^2222^-92= p-"--* By 100% method. (1.) 100% =initial value. (2.) 4%=annuity. (3.) $1000=annuity. \ (4.) -•. 4%=$1000, (5.) l%=i-of $1000=$2,50, and [$1000. (6.) 100%=100 times $250=$25009==initial value of 1. 100%^present value. 2. 112.4864% =(1.04)3 times 100%= amount of present value for 3 years at 4%. 3. .-. 112.4864 %=$25000, II. (7.) 4. 1« of $25000=$222.2492, and 5. 100%-=lbO times $222.2492=$22224.92=present value. III. .. $22224.92=present value of an annuity of $1000 to be- gin in 3 years at 4%. CASE IV. {Annuity, Interval, and Time to run,^ Let &=amount due first year. iS^-|-5.ff ^amount due second year. ;S-(-6'i?+,S'i?2=amount due third year. ;S-j-,S'/?-i-«S^'^+'S^^^=amount due the fourth year. S+SR+SR^+^R^ + +Ji?°-i= amount due the «th year. Let ^^amount due the «th year. .•. A^S+SR+SR^+SR^+ +5i?n-i ... (1). -to find the final or forborne value. 138 FINKEL'S SOLUTION BOOK. ^i?=5'i?+5i?2-fS'i?8+5i?*+ 4-5'i?" .. (2), by multiplying (1) by i?. ' [from (2). .-. AR—A=SR'—S (3), by subtracting (1). .•.^-^^^=ii (4.) I A pays $25 a year for tobacco ; how much better off would he have been in 40 years if he had invested it at 10%^ per annum? " ' ' By formula, ^=-^X(i?"— 1)=^X[(1.10)''«— 1] = $11064.8139. By 100% method'. 1. 100 %=initial value. 2. 10%^annuity. 3. $25=annuity. 4. .-. 10%=$25, 5. l%=Jjf of$25=$2.60, and 6. 100%=100 times $2.50=$250=initial value, 7. $44.2592556=[(1.10)*''— l]X$l=compound interestof $1 for 40 yr. at 10%. [$250 for 40 yr. at 10%. " 8. .-. $11064.8189=44.2592556 X$250=compound int. of .-. He would be $11064.8>139 better 6ff. Rentark, — $250 placed on interest at 10% will produce $25 per year. If this interest be put on interest at 10%, instead of spending it for tobacco, it will amount' to $11064.8139 in 40 'years. This would be a very sensible and profitable investment for every young man to make, who is a slave to the pernicious habit. II. III. I. An annuity, at simple interest 6%, in 14 years, amounted to $116.76 ; what would have been the difference, had it been at compound interest fi<?^ ? n. (2.) (8.) (4.) (5.) (6.) (V-) (8.) (9.) (10.1 100%=initial value, or the principal that would produce the annuity. 6%^annuity for 1 year. 84%=14xB%=annuity for 14 years. 1. 100%=6%, 2- 1%=TW of 6%=/-i7%. and [1 year.. 3. 6%==6 times ■^%=^%=interest on annuity for 4. 32.76%=91 times ^%==interest on annuity 'for (l+2-(-3+ +14), or 91 years'. 84%+82.76%=116.76%=whole amount of 'the annuity. $116.76=whole amount of the annuity. .-. 116.76%=$116.76, l%==j^^of$116,76=$l, and 100%=100 times $l=$100=initial value. 6%=6 times $l=$6=annuity. ANNUITIES. 139' (11.) $1.260904=[(l.''06)i*—l]X$l= compound inter- est on $1 for 14 yrs. at 6%. (12.) $126.0904=1.260904 X $100= compound interest on $100 for 14 yrs. at 6%. 1(13.) ..-. $126.0904— $116.76=$9.3304=difference. III. .-. The difference=$9. 3304.^ , CASE V. (• Final Value or Present Value ) Given } Rate, and > to find the annuity, ( Time to run, ) Solving P^^ — ^X — sj~with respect to S and we have S^P{^^_^p^_^_ _ (1) If ^_the final or forborne value, by the formula in the last case, we have .^4=^ S y,R^ — 1. Solving this with respect to 5, we have. R—\ ^ {R~-\)A rA (2). i?°— 1 i?°— 1 ■ ■ I. How much a year should I pay, to secure $15000 at the end of 17 years, interest 7% ? By formula (2), „ rA _ .07 X $15000 <.,„„„« ' ^=.ff^^=i-( 1.07) ^ ,_i=$486.38. By 100% method. (- (1.) 100%=annuity. (2.) 7%=anhuity. (3.) .-. 7%=100%, (4.) 1%=4- of 100%=14^%, and (5.) 100%=100 times 14i%=1428f %=initial value. ri. 100%^present value of 1428|^% due in 17 years. 2. 315.8815 %=amount of present value for 17 years. ,r. J 3. .-. 315.8815 %=1428t%, \'°-)U. l%=^TT.kTTof 1428t%=4.522591%, and 5. 100%=100 times 4.522591 %=452.2591%=pres- II.< 1^ ent value. (7.) . .-. 1428^^% —452.2591% = 976.3223% = present V^lue of an annuity running 17 years. (8.) 3.1588152%=(1.07)i' times l%=amount of 1% for 17 years. (9.) 3084.0217 %=( 1.07)1 Mimes 976.3223%=amount of 976.3223% for 17 years at 7%^ (10.) $15000=amount, or final value. (11.) ■•. 3084.0217 %=$15000. '12.) l%=TTr^V2TT of $1500f)=$4.8638, and 13.) 100%=100X$4.8638=$486.38=annuity. II 140 FINKEL'S SOLUTION BOOK. III. .-. I must pay $486.38. CASE VI. ( Annuity, ' ) + a j t- v Given I Present Value of the Annuity, and )■ ^° *'"'* '^""^ " J -D . '^ \ I'uns. ( Rate, ) , In formula (6), Case III., we have ^=d — rX — 5^ — , vsrhence J{^-^\ PjR—t 1 _ _Pr J Pr_S—Pr R^ 5' ' °'' /?»"" 5' ' R'^ S~ S ' (1). Applying logarithms, «log.i?=log.[3^]. i S \ . _, \oz. S—\os,. (S—Pr) ••• -=i°g[:^=pj-^i°g- ^=- — ^fi \ (2). I. In how many years can a debt of $1,000,000, drawing interest at 6%, be discharged by a sinking fund of $80,000 per year? '^ By formula (2), ^_ .og. 5— log. ( 6'— Pr ) _ log. 80000— log. ( 80000— lOpOOOO X -06 ) * log. R log. 1.06 _ log. 80000— log. 20000 ^ 4.903090— 4.301030 _ .602060 _ ~ log. 1.06 ~ .025306 "02,5306" years. 1 By another method. Assume $1,000,000 to be the present value of an annuity of •$80000 a year. Then $12.50 may be considered as the present value of $1 for the same time and rate. By reference to a table of present worths $12.50, which is lOOOOOO-f-SOQOO, will be found to be between 23 and 24 years. Note. — A table of present worths may be computed by form- ula (6.), iCase III., in which put 5'=$1. I. In what time will a debt of $10000, drawing interest at •6%, be paid by installments of $1000 a year. Bv formula, _ log. 5— log. (5— J° r\__ log.l000— log.( 1000— 10060x^06) ^ log. R ~ lo^TTog 3—2.602060 ,^„„^ \, „ „, , = — 025306~^- years=15 yr. 8 mo. 21 da. By another method. Assume $10000 to be the present value of an annuity of $1000 a year. Then $10000-i-1000=$10=the present value of $1 for the same time and rate. By referring to a table of present worth we find this amount between 15 and 16, years. .•. The time is 15 years -\- ANNUITIES. 141- The compound amount of $10000 for 15 yr. at 6%= $23965.58'- The final value of $1000 for 15 o'ears at 6%= $23275.97 Balance= $ 689.61 This balance, $689.61, will require a fraction of a year to dis- charge it. The part of a year required, will be such a fraction of a year as the amount of $689.61 for ihe fraction of a v^ar is of $1000. 6% of $689.61 for t\ie fraction of a year=$41.3766X fraction of a year. i .-. $689. 61+$41. 3766 X fraction of a year=the^ amount of $689:61 for the fraction of a year. This amount divided by 3, a yearly payment, will give \h& fraction. i.61+$41.3766x/'-ac2'2o« ^ ^. , — $1000 — =fractton, whence i.61-(-$41.3766 X fraction=$lOOO X fraction .-. $1000xfraction~H'i-S7&6Xfraction=$68Q.61, or ■ .-. $958,628 X/'-ac/«oi^=$689.61. .•.fraction= aoiT ^^ months, 19 days. .•. The^whole time^l5'yr. 8 mon. 19 da. CASE VII. \ ( Annuity, ) Given < Time to Run, and > to find the rate of in- ( Present Value of an Annuity, ) terest. , ^ ^D 1 From the formula (6), Case III, P — _ X — d^T"' ^^ obtains — = — =— - .... (1). This is the simplest expression we can ob- tain for the rate as the equation is of the nth degree and can not be solved in a general manner. I. If an immediate annuity of $80. running 14 yr., sells' for $650, what is the rate? By formula, i?»_l r $650 „,„. r^rT=8.125. Solving this equation by the method of r (1+^)'* & ^ J DqubW Position, we find r=8'fo-\-- By anothe^- method. ' $650-^-$80=8^125. By referring to a table of present worths of $1, corresponding to 14 years, we find it to be between 8 and 142 FINKEL'S SOLUTION BOOK. PROBLEMS. 1. What IS the amount of an annuity of $1000, forborne 15 years, at S^fo compound interest? Ans. $19295.125 2. What will an annuity of $30 payable semi-annually, amount to, in arrears 3 years at 7% compound interest? Ans. 3. What is the present -worth of an annuity of $500- to con- tinue 40 years at 7 % ? A'ns. 4. What is the present worth of an annuity of $200, for 7 years, at 5% ? Axs. $1152.27. 5. A father presents to his daughter, for 8 years, a rental of ^$70 per annum, payable yearly, and the reversion for 12 years succeeding to his son. What is the present value of the gift to his son, allowing 4% compound interest? Ans. 6. A yearly pension which has been forborne for 6 years, at ■6%, amounts to $279 ; what was the pension? Ans. $480.03. 7. A perpetual annuity of $100 a year is sold for $2000 ; at what rate is the interest reckoned ? Ans'. I 8. A perpetual annuity of $1000 beginning at the end of 10 years, is to be purchased. If interest is reckoned at 3-^%, what •should be paid for it? Ans. 9. If a clergyman's salary of $700 per annum is 6 years in ar- rears, how much is due, allowing compound interest at 6% ? Ans. $4882.72. 10. A soldier's pension of $350 per annum is 5 years in ar- rears; allowing 5% compound interest, what is due him? Ans. $1933.97. 11. What annual payment will meet principal and. interest of -a debt of $2000 due in 4 year a 8% compound interest? Ans, — 12. What is the present worth of a perpetual annuity of $600 at 6% per annum? Ans. $10000. ' 13. What is the present value of an annuity of $1000, to com- taence at the end of 15 years, and continue forever, at 6% per annum? Ans: $6954.40. 14. To what sum will an annuity of $120 fpr 20 years amount at 6% per annum? , Ans. $4414.27- 15. A debt of $8000 at 6% compound interest, is discharged by eight equal annual installments; what was the annual install- ment? Ans. $1288.286- MISCELLANEOUS PROBLEMS. 143 CHAPTER XIV. MISCELLANEOUS PROBLEMS, INVOLVING THE VARIOUS APPLICATIONS OF PERCENTAGE. I. Sold a COW for $25, losing 16f% ; bought another and sold ■it at a gain of 16 % ; I neither gained nor lost on the two ; what -was the cost of each? 1. 100%=cost of the first cow. 2. 16|%=loss. 3. 100%— 16t%=83i%=selling price. 4. $25=selling price. , 5. .•.-83i%=$25, l%=-i- of $25=$.30, and 100%=100 times $.30=$30=cost of first cow. $30 — $25=$5, loss on the first cow, and gain on second cow. 1. 100%^cost of second cow. 2. 16%=gain. 3. $5=gain. 4. .-. 16%=$5. tl. A.<^ B. 5. 1%=^ of $5=$.3125, and [cow. second Jtl. 6. 100% =100 times $.3125=$81.25=cost of ( $30^cost of first cow, and I $31.25=^cost of second cow. • ' Retnark. — Since I lost $5 on the first cow, and neither gained nor lost on the two, I must have gained $5 on the second cow^. .16« I. There have been two equal annual payments on a 6% note of $175, given 2 years ago this day. The balance is $154.40 ; what was each payment? II.<^ (2.) (3.) (4.) (5.) (6.) (7.) 100 %^a payment. 100% =$175, l%=rh of $175=$1.75, and 6%=6 times $1.75==$10.50=interest for 1 year. $175'+$10.50= $185.50,= amount before paying tha payment. [payment. $185.50— 100% = amount left after paying the 1. 100%=$185.50— 100%, 2. l%=TiTj of ($185.50— 100% )=$1.855—1%, and 3. 6%=:6 times ($1.855— 6%)=$11.13— 6%=inter- est for second year. 4. $185.50— 100% +$11.13— 6 %=$196.63— 106% = ■ amount before paying the last payment. 5. $196.63 — 106% — 100% = $196.63 — 206% = amount left after paying the last payment. 144 FINKEL'S SOLUTION BOOK. $154.40=amount aftier paying the last payment. .-. $154.40=$196.63— 206%. 206%=$i96.6^— $164.40=$42.23, 1%=^ of $42.23=$.205, and %=100 times $.205=$20.6G=the payrnent. 8. (9. (10.) (11.) 1(12.) III.. .-. $30.^0=the payment. Remark. — In this solution "we are obliged to use the minus sign, — , which is no obstacle to the student ofalgebra, but to the student of arithmetic '\\ may seem insurmountable. To avoid this sign, we give another solution. Il.i (1-) (2.) (3.) (4.) (5.) (6.) (7.) (8.') (9.) (10-). (11.) (12.) (13.) (14.) (15.) 100%=the payment. Then $154.4047100 %=amount of the debt at the end of of the second year. 100%=principal that produced this amount. 6%=interest. < 106%=amount. .-. 106%=$154.40+100%, [and ' l%=T-^ir of ($154.40+100% )^$1.4566^+|i%, 100,%=100 times ($1.4566A+lf %) = $145.66^ -|-94-^%=amount at end of the first year after paying off the payment. $145.66 A+94A%+100%= $145.66^ + 194^% ^ =amount before paying off the payment = amount at end of first year. f\. 100%i=the principal that produced it. 2. 6%=interest. 3. 106%=amount. 4. .-. 106%=$145.66^2^+194ii%, ^- 1%=T^ of ($145,66^f+194i|%) = $1.37H§ 1.83^%%, apd 6. 100% = 100 times ($1.37iitl + 1.83^»^V%) = $137U^4-183T°A\%=the amount at first. + $175^the amount at first. -183vW^%=$175. $1372-g^-g- — $37jsft9, 183^W¥' l%=$37ii||-=-i83^VA=$.205, and 100%=100 times $.205=$20.50t=the payment. III. $20.50==the payment. {R. H. A., p. 264,prob. 6.) Explanation. — $154.40^the amount after paying off the last payment. .-. $154.40-|-100%^amount before paying of . the last payment, or it equals the debt at the end of the first year plus the interest on this debt for the second year. .-. We let 100%= the debt at the end of the first year, 106%=amount of 100% for 1 year. .-. 106% = $154.40 + 100%. Then proceed as in the solution. I. If a merchant sells |^ of an article for what ^, of it cost, what is his gain % ? MISCELLANEOUS PROBLEMS. 145 ^U III. I. II. III. 1. 100%=cost of whole article. 2. 87i%=J of 100%=cost of I of the article. 3. 87-^%=selling price of -f of the article. 4. 29i%=i of 87|%=selling price of:i^of the article. 5. 116|%=4 times 29^%= selling price of the whole article. 6. .-. lief^'o— 100%=16|%=gain. (Milne's Prac, p. S60,J>ro6. SI.) 16f %=his gain. A merchant sold goods to a certain amount, on a commis- iion of 4%, and having remitted the net proceeds to the owner, received ^fc for prompt payment, which amounted to $15.60. What was his commission? (2.) (3.) (4.) (5.) (6.) (7.) (8.) I III. 100%=cost of goods. 4% ^commission. 100%— 4%==96%=net proceeds. :J%=amount received for prompt payment. $15.60=amount received for prompt pavment. .-. i%=$15.60. 4. 1%=4 times $15.60=$62.40. 6. 100%=100 times $62.40=$6240=net proceeds. .-. 96%=$6240. 1%=^-^ of $6240=$65, and 100% =100 times $65=$6500=cost of goods. 100% =$6500. 1 %=y^^ of $6500=165, and 4% =4 times $65==$260=his commission. . . His commission:^$260. ( Greenleafs N. A., p. ^^l,prob. 11.) If I sell 30 yards of cloth for $182, and gain 10%, how ought I to sell it a yard to lose 25% .' $132=selling price of 30 yards. $4.40=$132-i-30=selling price of one yard. 100%=cost of one yard. 10%=gain. 100%+10%=110%=selling price per yard. $4.40=selling price per yard. .-. 110%=$4.40. l%=^i-^of$4.40=$.04, , 100%=100 times $.04=$4=cost per yard. 100% =$4. l%=Twof$4=$.04, 25%=26 times $.04=$l=loss. .-. $4 — $l=$3=selling priee per yard to lose 25%, I must sellit at $3 per yard to lose 25%. "' {Stoddard's Co?nplete, -p. 206, proh.'9.\ 146 FINKEL'S SOLUTION BOOK. IIA ni. A merchant receives on commission three kinds of iiour ; from A he receives 20 barrels, from B 25 barrels, and from C 40 barrels. He finds Jhat A's flour is 10% better than B's, and that B's is 20% better than C's. He sells the whole at $6 per barrel. What in justice should each man receive? l3 (2.) (3.) (4.) (5.) (6.) (7.) (8.) <9.), (10.) (11.) (12.) (13.) (14.) (15.) (16.) $6=selling price of 1 barrel. $510=selling price, of (20+25+40), or 85 barrels. 100%"=value of C's flour' per barrel. 120%^value of B's flour per barrel. 100%=1205,. l%=TiTi of'mf.^H^, 10%=10 times 1-|%=12%. 120%+12%=132%=value of A's flour per barrel. 4000%=40 times 100%=what C received. 3000%=25 times 120fo=what B received. 2640%=20 times; 132 %=what A received. 9640%=40d0%+3000%+2640%=what all rec'd. $510=what all received. .-. 9640fc=$510. lf„=-^^^ of $510=$.52|i|,,and [received. 4000%=4000 times $.521^1= $211^— what C 3000%=3000 times $.52fH=*158iH=whatB received. [received. 2640%=2640 times $.52|^=$139i|i=what A ( $139if|=A's share, I $158iJ|-=B's share, and ( $211^f=C's share. ( Greenlcaf s National Arifh. p. 44^-) II. III. f of B's money equals A's money. What % is A's money less than B's, and what <fo is B's money more than A's? '1. 100%=B's money. 2. 75%=f of 100%=A's money. 3. 100%— 75%— 25%=excess of B's money over A's. 4. 75% =100% of itself, 5. l%=yV of 100%=li%, and ' [than A's. 6. 25%=25 times l.J%=33^%=the % B's money is more ( A's money is 25% less than B's, and ■ ■ / B's money is 33^% more than A's money. .(Siod. Comf.,f. 20S,prob. 19,) At what price must an article which cost 30 cents be marked, to allow a discount of 12^% and yield a net profit of 16|-% ? MISCELLANEOUS PROBLEMS. 147 iiA III. 10 (2.) (3.) (4.) (5.)i 100%=30/, l%=TiT"f30/=T%/, and 16|-%=16t times ^.j/=5/=profit. 30/-|-5/=35/=the price at which it must sell to gain.l6f%. 1. 100%=marked price. 2. 12-|^%=discount from marked price. 3. 100%— 12i%=87i%=selling price. 4. 35/=selling price. 5. .-. 87i%=35/. ' 6. 1%=^ of 35/=.40/, and 7. 100%=100 times .40/=40/=marked price. 40^=marked price. ( Seymour's Prac, p. 80S, froh. J^. ) Had an article cost 10% less, the number of % gain -would have been 15% more ; what was the gain? 1. JOO%^selling price. 2. 100 %=actual cost price. 3. iOO%— 100%=gain. 4. 100%— 10.%=90%=supposed cost 5. 100% — 90%=conditional gain. 6. 90%=100% of itself. 7. i%=7V of ioo%=H%. I . 8. iOO%— 90%=( 100— ^Q) times li%==V X.Z0O%— 100% ^conditional gain %. [difference. 9. .-. V°X-ZOO%— 100%— (iOO% — 100%)=iX-?00% = 10. 15%=diff"erence. 11. .-. i-X-Z00%=15%. [the actual cost. 12. 100%=^2 times 15%=135%=selling price in terms of 13. .-. 135%— 100%==35%=gain. .-. 35%=gain. - {R. H. A., p. 406, prob. 87.) A literal solution. L-et 6'=selling price and C=the cost. Then >S — C=gain and Il.i III. s—c =rate of gain. S- c 9 c ■± TTr"-_ ^ ^conditional rate of gain. . ^6'=-^-g-C, whence 5^f-JC^1.35C. ,-. Rate of gain=.35C-^C=.35=35%. ijC=conditional gain and V°6^— C 5— C _ ^ c c~ — ^^'' °^ 1.35 C— C=.35 C=gain. In the erection of my house I paid three times as much for material as for labor. Had I paid. 6% more for labor, and 10% rnore for material, my house would have cost $3052. What did it cost me? 148 FINKEL'S SOLUTION BOOK. IIJ (3.) (4.) III. 100%=cost of labor. 300%=3 times 100%=cost of material. 100%=100%, ^ l%=l%,and 6%=6%. 1 100%+6%=106%=supposed cost of labor. 100% =300%, lfc=Tis of 300%=3%, and 10%=10 times 3%=30%. 300%+30%=330%=supposed cost of materiaL 330%-{-106%=436%=supposed cost of houfee, $3052=supposed cost of house. .-. 436%=$3052, 1 % =^^ of $3052=$7 , and 100%=100 times $7=$700=cost of labor. 300%=300 times $7— $2f00=cost of materiaL $2100-H$760=$2800=cost of house. ^cost of the house. I. I invest f as much in 8% canal" stock at 104%, as in 6% gas stock at 117% ; if my income from both is $1200, how jnuch did I pay for each, and ■what was the incomes ■ from each ? i II.1 (2.) (3.) 100%=investment in gas stock. Then 66f %^investment in canal stock. 100%i=par value of the gas stock. 117%=market value of the gas stock. ^ 117%=100%, from (1),- (5.){2 (-3 (6.) (6.) (7.) (8.) (9.) (10.) 1%=- TTT ofl00%=ff^%,and =8511' ^=par value in 100%=100 times J|^%= terms of the investment. 100%=85f|%- l%=TTT%>and 6%=6 times lxT%=5^%=income of gas stock. 100%=par value of canal stock. 104%=market value. .-. 104%=66f%, l%=Tkof66|%=|^%, and , , 100%=100 times f5%=64^%. 100% =64^%, 1%=T^ of 64^%=|f%, and 8%=8 times ||%==5-/^%^income of canal stocks 5/^%+5-/Tf=10|f %=income from both. $1200=income from both. .-. 10^%=$1200, ' 1%=_1^ of $1200=1117, and MISCELLANEOUS PROBLEMS. 149 (11.) 100%=100 times $117=$11700=investment in gas stock. [canal stock. (12.) 66f%=66f times $117=$7800 = investment in (13.) 5^%=5A times $117=$600=income from each. ^ $600=:income from each. Ill, ' .'. < $11700^investment in gas stock, and ' $7800=investment in canal stock. I, A man bought two horses for $300; he sold them for $250 apiete. The gain on one was h'fo more than on the other; what was the gain on each.? 1. $300==cost of both. 2. $500=$2504-$250=selling price of both. 3. $500— $300=$200=gain on both. 4. 100%^gain on first horse. Then 5. 105%=gain on s'econd horse. II <; 6. 205%=100%+105%=gain on both. 7. $200=gain on both. , 8. .-. 205% =$200. 9. ,l%=^'j of $200=$H, and 10. 100%=100 times $|4=$97.56,*T=gain on the first. 11. 105%=105 times $ti=$102.43fj=gain on the second. _ ( $97.56:i*y^gain on the first, and ■'■ / $102.43|^=gain on the sec9nd. Note. — In this solution, it is assumed that the gain on one was S% of the gain on the other more than the other, and this is the usual assumption. But the problem really means that the per cent, of gain on one, computed on its cost, was 5% more than the per cent, of gain on the other, computed on its cost. iBy this assumption, the problem is algebraic. The following is the solution: Let x=the cost of the first horse, aifd $300 — x^ the cost of the second. Then $250 — ^o^gain on first, and $250 — ($300 — x)=x — $50, the gain on the second. ($250 — x)-^x= rate of gain on the fii'st, and {x — $50)-i-($300 — x), the rate of gain on the second. Then (250 — x)-^x — {x — 50)-=-(300 — x)= ^. Whence, by clearing effractions, transposing and, combin- ing, Ar2_10300 ;f=— 1500000, ^=5150 ± 50^/10009= $147.7755, the cost of the first horse. $300— x=$152. 2245, the cost of the «econd horse. $2.50 — x=$102.2245, gain on the first hokse, and ^ — $50=$97.7755, the gain on the second horse. I. An agent sells nroduce at 2% commission, invests the proceeds in flour at 3% commission: his whole commis- sion was $75. How many barrels of flour did he buy at $5 per barrel ? 150 FINKEL'S SOLUTION BOOK. III. I. II (2.) (3.) (4.) II.I (5.) f6.) (7.) (8.) (9.) (10.) (11.) (12.) 100%=value of the produce. 2%=the commission. [vested in the flour. 100% — 2%=98%=net proceeds, or amount in- 1. 100%=cost of the flour. 2. 3%=commission on flour. 3. 100%+3%=103%=whole cost of the flour. 4. .-. 103%=98%, .5- 1 1o=^^ of 98%=^!^% > and 6. 100%=100:><T%V% = 95iV5^% = cost of flour in terms of the value of the produce. 7. 98% — 95TV\%=2T%*^%=commissian on flour. 2%+2T*A%==4Y'V\%=whole commission. whole commission. * l%=$75-f-4T'A=$l6.45, and [produce. 100%=100 times $15.45=$1545 = value of the 96tV^%=95ttf\ times $15.45= $1470 = value of the the flour. , $5=cost of 1 barrel. $1470=cost of 1470-^-5, or 294 barrels. The agent bought 294 barrels of flour. •. , A distiller sold his whisky, losing 4% ; keeping $18 of the proceeds, he gave the remainder to an agent to buy rye at 8% commission; he lost in all $32 ; what was the whisky worth? 4A\^ (2.) (3.) (4.) (5.) III. (6.) (7.) (8.) (9.) (10.) .(11-) 100%=value of the whisky. 4%=loss. 100%— 4%=96%=amount he had l,eft. 96% — $18^amount he invested in rye. * . 1. 100%.=cost of the rye. 2. 8%=commission on the rye. 3. 100% +8 %=108%=whole cost of rye. 4. .- 108%=96%— $18, 5- l%=Tk of (96%-$18)=|%-$.16f, and 6. 100%=100 times (|%— $.16f)=88|%— $16.66f =cost of rye. 7. 8%=8 times (|%— $.16f )=7i%— $1.33i=com- mission on rye. , 4%+(7i%— $1.33i)=lH%— $l.E3i=wholeloss. $32=whole loss. .-. 11^%.— $1.33i=$32 ll^%=$33.33i, 1 %=j^ of $33.33i=$3, and 100%=100 times $3=$3G0=value of the whisky. $300=value of the whisky. (i?. H. A.,f. 4.06, frob 91.} MISCELLANEOUS PROBLEMS. 151 What will be the cost in New Orleans of a draft on New York, payable 60 days after sight, for $5000, exchange being at l^<fo premium? 1. 100%=face-of the draft. 2. l^%=premium. 3. 100%4-li%=101^%=rate of exchange. 4. 5^%^discount for one year. II.-J5. |%=Tf«/7 of 5%=discount for 63 days. 6. .-. 10H%—|%=100|%=cost of the draft 7. 100yo=$5000. 8. l%=T^^of $5000=$50, and 19. 100j%=100f times $50=$5031.25=cost of the draft. III. .-. $5031.25=cost of the draft. Explanation. — Observe that since the draft is not to be paid in New York for 63 da^'s, the banker in New Orleans, who has the use of the money for that time allow^s the drawer discount on the face of the draft for that time, which goes, (1) towards reducing the premium if there be any, and> (2) towards reducing the face of the draft. Note. — The rate of exchange between two places or countries depends upon the course of trade. Suppose the trade between New York and New Orleans is such, that New York owes New Orleans $10,250,000 and New Orleans owes New York $13,000,- 000. There is a "balance qf trade" of $2,750;000 against New Orleans and in favor of New York. Hence, the demand in New Orleans for drafts on New York is greater than the demand in New York for drafts on New Orleans and, therefore, the drafts are at a premium in New Orleans. But if New York owes Nevy^ Orleans $13,000,000 and New Orleans owes New York $10,250,- 000, the "balance of trade," $2,750,000, is againsfi^e-w York and in_/«i'o?' of New Orleans. Hence, the demand in New Orleans for drafts on New York is less than the demand in New York for drafts on New Orleans and, therefore, the drafts are at a dis- count in New Orleans. If the trade between the two places is the same, the rate of ex- change is at par. The reason why the banks in Nev7 York "should charge a pre- mium, when the balance of trade is against them, is that they must be at the expense of actually sending money to the New Orleans banks or be charged interest on their unpaid balance ; the reason why the New Orleans banks will sell at a discount is that they are willing to sell for less than the face of a draft in order to get the m'oney owed them in New York immediately. Exchange is charged from -J to ,^% , and is designed to cover the cost of transporting the funds from one place to another. 152 FINKEL'S SOLUTION BOOK. I. II.S III. I. II. What will a 30 days' draft on New Orleans for $7216.85 cost, at ■§% discount, interest 6% f 1. 100%=face of draft. 2. f%=discount. 3. 100%— |%=99|%=face less the discount. 4. 6%=bank discount for 1 year. 5. ii%=^ of 6%=bank discount for 33 days. 6. 99i%—^f6=99j\%=cost of the draft. 7. 1009^<,=$7216.85, 8. l%=Tiir of $7216.85=$72.1685, and 9. 99/7%=99:iV times $72.1685=$7160.094=cost of the draft. .-. $7150.094=cost of the draft. The aggregate fape value of two notes is $761.70 and each has 1 year 3 months to run; one of the nQtes I had dis- counted at 10% true discount and the other at 10% bank discount, and realized from both notes $671.50. Find the face Value of both notes. (1.) 100%^face of note discounted at bank discount. (2.) $761.70— 100%=face of note discounted at true disfcount. , (3.) 10%=bank discount for 1 year. (4.) 12|-%=bank discount for 1 year 3 months. 1. 100%=present worth of second note. 2. 10%^interest on present worth for 1 year. 3. 124^%=interest for 1 year 3 months. 4.' 100%+12^%=112|%=amount of present worth. 6. $761.70— 100%=amount of the present worth. 6. .•.,112|%=$761.70— 100%, '^- i%=rnn: °^ ($761.70— ioo%)=$6.7706| — 1%, (5.) (6.) (7.) (8.) (9.) (10.) (11.) (12.) (13.) III. 112i 100Yo=100 times ($6.7706| — 1%) =$677.061— * 88f %^present worth. \$761.70— 100%— ($677.061— 88f% ) = $84.63* — ll^%=true discount. [discount. $84.63i— lli%+12^%=$84.63i+ 1tV% = whole $761.70— $671.50=$90.20=whole discount. , .■.$84.63i+lT\%=$90.20, lJj%=$5.56f, 1%=A. of $5.56f=$4.008, and 100%=^100 times $4.008= $400.80=face 'of note discounted at bank discount. $761.70— 100%=$761.70—$400.80=$360.90=face of note discounted at true discount. $400.80=face of note discounted at bank discount, and $360.90=face of note discounted at true discbunt. MISCELLANEOUS PROBLEMS. 153 11.-^ II. A merchant sold part of his goods at a profit of 20%, and the remainder at a loss of 11%. His goods cost $1000 and his gain was $100; how mnch was sold at a profit? (1.) 100%=cost of goods sold at a profit. ' Then (2.) $1000— 100%=cost of goods sold at a loss. (3.) 20%=profit on 100%, the part sold at a profit (-1. 100%=$1000— 100%. ,, J 2. l%=T{ny of ($1000—100% )=$10— 1% , ^*''']3. 11%=11 times ($10— 1%)=$110— ll%=loss on I • the remainder. (5.) .-. 20%— ($110— 11%)=31%— $110=gain. (6.) $100=gain. (7.) .-. 31%— $110=$100. (8.) 81%=$210, (9.) 1%=^ of $210=^614, [profit. (10.)' . 100%=100 times $6|f=677.41|f=part sold at a III. .-. $677.41|-f=value of the part sold at a profit. ' I. By discounting a note at 20% per annum, I get 22^% pfr annum interest; how long does the note run? 1. 22^% of the proceeds=20% of the face of the note. 2. 1% of the proceeds=jr— of 20%=|% of the face of the note. ^^^ 3. 100% of the proceeds^lOO timss f %=88|%'of the face of the note. 4. 100%=face of the note. 5. 88f %=proceeds. 6. 100%— 88|%=lli%=(liscount for a certain time. 7. 20%=discount for 360 days. 8. l,%=discount for -^-^ of 360 days, or 18 days. 9- ]l^%=discount for 111 times 18 days, or 200 days. III. The note was discounted for'200 days. I. A man bought a farm for $5000, agreeing to pay princi- pal and interest in 5 equal annual instalments. What will be the annual payment including interest at 6?i? 1. 100 %=one annual payment. 2. . . 100?^=^mount paid at end of the fifth year since the debt was then discharged. 3. 100 %=principal that drew interest the fifth year. 4. 6 9'o=interest on this principal, jj i 5. .-. 100^+6%=106 9i=amount of this principal. 6. .-. 106fo=100/o=the annual payment. 7. 1 1o=^ of 100 %=|| 1c, and 8. 100/0=100 times ft<=94^f%= principal at the beginning of the fifth year. 9 . 94^1 fo+lO0 10 = 194^ fo=amount at the end of the fourth year. 154 FINKEL'S SOLUTION BOOK. (2.) (3.) 1. 100%=principal at the beginning of the fourth. year. 2. 6%=interest on this principal. 3. 100%+6%=106%=amount. 4. ,.-.^106%=194i|%, 5- l%=Ti^of 194i|%=1.83A¥ir%> and 6. 100fo=100 times 1.83^^%%=183^VA%— Princi- pal at the beginning of the fourth year. 7. 183/^%-|-100%=283^YT^%=amount at the end of the third year. (1. 100%=principal at the beginning of the third year. 2. 6%=interest. [third year. 3. 100%+6%=106%' = amount at the end of the- 4. "" ■ ' 5. 6. (4.) 106%=283^V-!fe%, l%=TiT of 283^VA%=2.67t\VAV%. and 100%=100 times 2.67t\*^V% =267tV^V% = principal at the beginning of third year. 267T*^VfTV%4-100% = 367T^Wr% = amount at the end of second year. 100%=principal at the beginning of second year. 6%^interest. - [year. 100%+6%=106%=amount at the end of second. ■ 106%=367tWs¥7.%' ^ /<A4 028 674 g/, f,„A -- • ■ - • - ■ 574 / (5.)-' A- 2. 3. 4. 5. 6. l%=TkPf367AVAV%= 100%=100 times 3A6^^1stt^7o=o'±vfttttn7e= principal at the beginning of the second year. 346|fff|i4%+100%=446fff|Hf%=amount at the end of first year. 100%=principal at the beginning of the first year, or the cost of farm. , 6%=interest. 100%+6%=106%>=amount at end of first year. 1%=^^ of 446fff||||.%=4.21AVim\V%> and (6.) (7.) (8.) (9.) III. I. 100%=100 times 4.2IA V1W4V8 %=^'^l^-mih. %^cost of the farm. , $5000=cost of the farm. ••••■421tV¥t¥tWj%=$5000, 1 f^=$5000-^421^8^72^V7=$ll-8698+, and 100%=100 times $11.8698=$! 186.98-|-=the an- nual payment. .-. $1186.98+=the dnnual payment. [Milne's Frac., p. S61,frob. 6~ y -^la^ ' 1% A and B have $4700 ; ^% of A's share equals ^% "2" /" B's share; hov(r much has each? of PROPORTION. 155" IIJ ■ t7<'= =4^9'^==T§Tr%=TTr^!r- TTTTJ-/ III. 9. 10. 11. 12. 13. 14. 15.' 16. 17. ^% ...a... 1% wrnnr 60% 2-. TTnr .6 3 TTnr — T- =^f^%=in=l _^%=|%=1^%. .-. 1|%' of A's=l|% of B's, '. Ifc of A's=^ of H%=^% of B's, and 100% of A's=100 times f«%=74^% of B's. 100fo=B's share. 74^'%=A's share. 100%+74^V%=174^%=sum of their shares. $4700^sutn of their shares. .-. 174^%=$4700, 1%=--1^ of $4700=$27, and 100%=100 times $27=$2700=B's share. 74^%=74^ times $27=$2000=A's share. $2700=B's share and $2000= A 's share. CHAPTER XV. 'ratio and proportion. 1. Ratio is the relative magnitude of one quantity as com- pared with another of the same kind; thus, the ratio of 12 apples to 4 apples is 3. The first, quanity, 12 apples, is called the Antecedent, and the secqnd quantity, 4 apples, the consequent. Taken together they are called Terms of the ratio, or a Couplet. 2. The Sign of ratio is the colon, ., the common sign of division with the horizontal line omitted. Note. — Olney says, "There is a common notion among us, that the French. express a ratio by xlividingthe 'consequent by the antecedent, while the En- glish express it by dividing the antecedent by the consequent. Such is not the fact. French, German, and English writers agree in the above defini- tion. In fact, the Germans very generally use the sign : instead of -^; and 156 FINKEL'S SOLUTION BOOK. 'by all, the two signs are used as exact equivalents." Some writers, however, divide the consequent by the antecedent, as a : b= — This is ac- •Cjording to Webster's definition and illustration. To my mind, to divide the antecedent by the consequent is more simple and philosophical and should be' universally adopted by all writers on mathematics. 3. A Direct Ratio \s- the quotient of the antecedent di- vided by the consequent. 4. An Indirect Ratio is the quotieijt of the consequent by the antecedent. 5. A ratio of Greater 'inequality is a ratio greater than unity; as, 7:3. 6. A ratio of Less 'Ifiequality is a ratio, less than ,unity;' ^s, 4 :5. 7. A Compound Ratio is the product of the correspond- ing terms of several simple ratios. Thus, the compound ratio of 1:3, 5:4, and 7:2 is 1x5x7:3x4x2. 8. ' A Duplicate Ratio is the ratio of the squares of two numbers. 9. A Triplicate Ratio is the ratio of the cubes of two numbers ; as, a^ : d^. 10. A 'Subduplicate 'Ratio is the ratio of the squaa-e roots of two numbers; as, y'^: \/b- 11. A Subtriplicate Ratio is the ratio of the cube roots of two numbers; as,-^^: \^j". PROPORTION. 13. Proportion is an-equality of ratios. The equality is indicated by the ordinary sign of equality or by the double colonj ::. Thus, a : b=c •.d,ora:b:ic:d. 13. The JEoctremes of a proportion are the first and fourth terms. 14. The JMLeans are the second and third terms. 15. A Mean Proportional h&tw&en two quantiti&s is a. quantity to which either of the two quantities bears the same ilatio that the mean does to the other of the two. • '' 16.. A Continued Proportion is.a succession of equal ratios, in which each consequent is the antecedent of the next ratio. 1'7. A Compound Proportion .is an expression of ■equality between a compound and a simple ratio. ' PROPORXrON. 15T l8« jL Conjoined iPropoftion is a proportion which- has each antecedent of a compound ratio equal in value to its- consequent. The first of each pair of equivalent terms is an an- tecedent, and the term following, a consequent. This is also- called the "Chain rule." I. What is the ratio of ^ to f ? lH-|-=i'xf = |,% ratio. ^ / I. What is the ratio of 10 bu. to If bu. ? , 10 bu. -^ If bu. = 10 X t\=7, the ratio. ' I. What is the ratio of 25 apples to 75 boxes? , Ans. No ratio ; for no number of times one will produce- the other In a true proportion, we must always have greater : less :: greater : less or less : greater : : less : greater. The test for the- truth of a proportion is that the product of the means equals the, product of the extremes. i I. If a 5-cent loaf weighs 7oz. when flour is $8 per barrel,, how much should it weigh when flour is $7.50 per barrel? It should evidently weigh more. * .-. less : greater : : less : greater. $7.50 : $8.00 : : 7oz : {■! = 'l^oz.). I, If a staff's feet long, casts a shadow 2 feet, how high is- the steeple whose shadow at the same time is 75 feet? Since the steeple casts a longer shadow than the staff, it is evi- dently higher than the staff. .-. less : greater : : less : greater. 2 feet : 75 feet : : 3 feet : (?=112^fe€t.) I. , What number is that which being divided by one more- than itself, gives \ for a quotient? 1. Let l^^number. Then 2. 5^=1 or f : l-fl : : 1 : 7, whence n.J3. 7(f)=l(f+l)or 4. ^^=|-|-1; whence K 12 1 7. f=2 times J-^^^^number. III. .•. 4=the number. I. What number divided by S more than itself gives ^ for a quotient? 158 FINKEL'S SOLUTION BOOK. 1. Let|-=the number. Then UJ III. I. 2 1+3" 2 =^ or, putting this in the form of a proportion, II. ri. 2. 3. =39 lb. . Y"l"3 : : 7 : 9. - [the product of the extremes. . . i^*=\*+21, the product of the means beine equal to ¥—¥=1=21, i=i of 21=5i, and ^=2 times 5i=10^=the' nuftiber. .-. 10^=the number. If 7 lb. of coffee is equal in value to 5 lb. of tea, and 3 lb. of tea to 13 lb. of sugar, 39 Vn. of sugar to 24 lb. of rice, 12 lb. of rice to 7 lb. of butteij 8 lb. of butter to 12 lb. of cheese ; how many lb. of coffee are equal in value to 65 lb. of cheese? 7 lb. of coffee=5 lb. of tea, 3 lb. of tea^lS lb. of sugar, 39 lb. of sugar=24 lb. of rice, 4. 12 lb. of rice=7 lb. of butter, 5. 8 lb. of butter=i=12 lb. of cheese, and 65 lb. of cheese= J=39 lb. of coffee, 7X3X39X12)^8X65 6X13X24X7X12 III. ■" .-. 65 lb. of cheese=39 lb. of coffee. I. I can keep 10 horses or 15 cows on my farm ; how many horses can I keep if I have 9 cows? 15 cows : 9 cows : : 10 horses : ?=6 horses. 10 horses — 6 horses=4 horses. v .•. I can keep 4 horses with the 9 cows. I. If 2 oxen or 3 cows eat one ton of hay in 60 days, how long will it last 4 oxen and 5 cows? 2 oxen : 4 oxen : : 3 cows : ?=6 cows. .•. 4 oxen eat as mnch as 6 cows. If a ton of hay last 3 cows ■60 days, it will last 6 cows, which are equal to 4 oxen, and 5 cows, or 11 cows, not so long. .'. 11 cows : 8 cows : : 60 days : .?=17J-j- days. I. If 24 men. by working 8 hours- a day, can, in 18 days, dig .a ditch 95 rods long, 12 feet wide at the top, 10 feet wide at the bottom, and 9 feet deep; how many men, in 24 days of 12 hours a day, will be required to dig a ditch 380 rods long, 9 feet wide at the top, 5 feet wide at the bottom, and 6 feet deep? 95 rods : 380 rods 24 men : ?=12 men. 24 days : 18 days 12 hours : 8 hours 12 feet : 9 feet 10 feet : 5 feet 9 feet : 6 feet PROPORTION. 159 380X18X8X9,X5X6X24 95X24X12X12X10X9 =12 men. I. A Louisville merchant wishes to pay $10000, which he owes in Berlin. He can buy a bill of exchaijge in Louis- ville on Berlin at the rate of $.96 for 4 reichmarks ; or he is offered a circular bill through London and Paris, brokerage -J^ at each plate, at the following rates: iEl^$4. 90=25.38 francs, and 5 francs=4 reichmarks. What does he gain by direct exchange? 1. $.238=1/ mark. 2. $10000=10000-^.238=42016.807 marks. 3. $.24^1 mark, since this is the rate of exchange. 4. .-. $10084.033=42016.807 times $.24=42016.807 marks ^direct exchange. 5. 42016.807 marks— (?=$10165.38.) II <; 6. $4.90=i;i— J% of£l=£.99f 7. £1=99^ times 25.38 fr. 8. 6 fr.=4 marks. . 42016.807X4.90X5 .imc^oo .k • Lchange. 9. ^r-- — -- — TT^r-^r^ — , =$10165. 38^cost by circular ex- .99iX.99|X25.38x4 ^ 10. $10165.88---$10084.033=$81.35=gain by direct ex- change. III. .-. $81.35=gain by direct exchange. I. A whepl has 35 cogs ; a smaller wheel working in it, 26 cogs; in how many revolutions of the larger wheel will the smaller one gain 10 revolutions? 1. 35 cogs — 26 oogs^9 cogs=what the smaller wheel gains on larger in 1 revolution of larger wheel. • 2. 26 cogs passed through the point of contact=l revolu- tion of smaller wheel. 3. 1 cog passed through the point of contact=-^ revolu- tion of smaller wheel. II.<1 4. 9 cogs passed through the point of contact=^ revolu- tion of smaller wheel. 5. .". In 1 revolution of larger w^heel the smaller gains -^ revolution of smaller wheel. 6. .•. -^ revolution gained : 10 .revolutions gained : : 1 revolution of larger wheel : ?=28|- revolutions of larger w^heel. III. .■. The smaller wheel will gain 10 revolutions in 28|- revo- lutions of larger wheel. By analysis and proportion.^ 26 cogs passed ihrough the point of contact==l revolution of the smaller wheel. 160 FINKEL'S SOLUTION BOOK. 35 cogs passed through the point, of contact=l revolution of the larger wheel. But when the larger wheel has made 1 revo- lution, 35 cogs of the smaller wheel have passed through the point of contact. If 26 cogs having passed through the point of contact make 1 revolution of the smaller wheel, how many rev- olutions will 35 cogs make? By proportion, 26 cogs : 35 cogs : : 1 rev. : ?=l^'^rev. .•. The smaller wheel makes 1^^^ revolutions while the larger wheel makes 1 resolution. .•. The smaller gains 1^ revolutions — 1 revolution^^^^ revolution. If the smaller wheel gains -i^ revoluion in 1 revolution of the larger wheel to gain 10 revolu- tions on the larger wheel, the larger wheel must make more rev- olutions. .•. less : greater : : less : greater. ■^,\rev. : 10 rev. : : 1 rev. of larger : ?=28|- rev. of larger. I. If the velocity, of sound be 1142 feet per second, and the number of pulsations in a person 70 per minute, what is the distance of a cloud, if 20 pulsations are counted betvveen thfe time of seeing the flash and hearing the thunder? / 1. 1142 ft.=distance sound travels in 1 second. 2. 68520 ft.=60Xll42ft.=distance sound travels in-lmin., or the time of 70 pulsations. 3. .•. If it travels 68520 feet while 70 pulsations are count- ^ ) ed, it will travel not so far while 20 pulsations are I counted. ' \ 4. .•. greater : less : : greater : less, i [145 yd. 2-^ ft. 5. 70 pul. : 20 pul. : : 68520 ft. : ?=19577^ ft.=3 mi. 5 fur, III. .-. The cloud is 3 mi. 5 fur. 145 yd. 2^ ft. distant. (i?., Sd p., p. 289, f rob. J^5.) PROBLEMS. 1. If 8 horses , in ^ of a month eat -f of a ton of hay, hovJ' long will -f of a ton last 6 horses? 2. If a 4-cent loaf weighs 9 oz. when flour is $6 a barrel, how much ou'ght a 5-cent loaf weigh when flour is $8 per barrel? 8. A dog is chasing a hare, which is 46 rods ahead of the dog. The dog runs 19 rods while the hare runs 17; how far must the dog run before he catches the hare ? 4. If 52 men can dig a trench 355 feet long, 60 feet wide, and' 8 feet deep in 15 days, how long will a trench be that is 45 feet wide and 10 feet deep, which 45 men can dig in 25 days? 5. If \ of 12 be 3 what will \ of 40 be? Ans. 15. 6. If 3 be ^ of 12 , what will i of 40 be ? ' Ans. 6S. PROBLEMS. ' 161 7. If 18 men or 20 women do a work in 9 days, in what time can 4 men and 9 women do the same work? Ans. 1B^\ days. 8. If 5 oxen or 7 cows eat 3^ tons of hay in 87 days, in what time will 2 oxen and 3 cows eat the same quantity of hay ? Ans. 105 days. 9. Divide $600 between three men, so that the second man shall receive one-third more than the first, and the third f more than the second. ' 10. Two men in Boston hire a carriage for $25, to go to Con- cord, N. H., and back, the distance being 72 miles; with the privilege of taking in three more persons. Having gone 20 miles, they took in A ; at Concord they took in B ; and when within 80 miles of Boston, they took in C. How much shall each pay? Ans. First man, $7.609i-;f; second, $7.609fJ| ; A, $5.873y9T^; B, $2,864tV; and C, $1.0^.13:%. 11. Three men purchased 6750 sheep. The number of A's sheep is to the number of B's sheep as |- is to 3^, and 4 times the number of C's sheep is to the number of A's sheep as -J is to \. Find the number of sheep each had. I A's= Ans. \ B's = (C's== 12. If $500 gain $10 in 4 months, what is the rate per cent? Ans. 6%. 13. If 12 men can do as much work as 25 women, and 5 wo- men do as much as 6 boys ; how many men would it take to do the work of 75 boys ? Ans. ZO men. 14. If 5 experienced compositors in 16 days, 11 hours each, can compose 25 sheets of 24 pages in each sheet, 44 lines on a page, 8 words in a line, and 5 letters to a word ; how many in- experienced compositors in 12 days, 10 hours each, will it take to compose a volume (to be printed with the same kind of type), consisting of 36 sheets, 16 pages to a sheet, 112 lines to the page, 5 words to a line, and 8 letters to a word, provided that while composing an inexperienced compositor can do only ^ as much as an experienced compositor, and that the latter work is only f as hard as the former? Ans. 16. 15. If A can do f as much in a day as B, B can do | as much as C, and C can do ^ as much as D, and D can do f as much as E, and E can do -f- as much as F; in what time can F do as much work as A can do in 28 days ? Ans. 8. 16. A starts on a journey, and travels 27 miles a day; 7 days after, B starts, and travels the same road, 36 miles a day; in how many days will B overtake A ? Ans. 21 days. 16? FINKEL'S SOLUTION BOOK. 17. A whefil has 45 cogs ; a smaller wheel worljiing in it, 36 cogs ; in how many revolutip'ns of the larger wheel -vyiil the smaller gain 10 revolutions? Aks. 40. 18. If the velocity of sound be 1142 feet per second, and the nnmber of pulsations in a person 70 per minute, what is the dis- tance of a cloud, if 30 pulsations are counted between the time of peeing a flash of lightning and hearing the thunder? Ans. 6i mi. 108 yd. If- ft 19. If William's services are worth $15f a month, when he labors 9 hPV's a day, what onght he to receive for 4|- months, when he labors 12 hqurs a day? Ans $9 20. If 300 cats kill 300 rats in 300 minutes how many cats will kill 100 rats in 100 minutes? , Aks. 300 cats. CHAPTER XVI, ANALYSIS. 1. Analysis, in mathematics, is the process of solving problems by tracing the relation of the parts. I. What will 7 lb. of sugar cost at 5 cents a pound? Analysis for primary classes. If one ponnd of sugar costs 6 cents, 7 pounds will Cost 7 times 5 cents, which are 35 cents. I. If 6 lead pencils cost 30 cents, what will one lead pencil cost ? Analysis: If 6 lead pencils cost 30 cents, one lead pencil will cost as many cents as 6 is contained into 30 cents which are 5 cents. I. If 8 oranges cost 48 cents, what will 5 oranges cost? Analysis: If 8 oranges cost 48 cents, one orange will cost as many cents as § is contained into 48 cents which are 6 cents; if one orange costs 6 cents 5 oranges will cost 5 times 6 cents, which are 30 cents. I. If a boy had 7 apples and ate 2 of them, how many had he left? Analysis: If a boy had 7 apples and ate 2 of thefn, he had left the difference between 7 apples and 2 apples which are 5 apples. I. If John had 12 cents and found 5 cents, how many cents did he then have? Analysis: If John had 12 cents and found 5^ cents, he then had the sum of 12 cents and 5 cents which are 17 cents, ANALYSIS. 163 Note. — If teachers in the Primary Departments would see that their •pupils gave the correct analysis to such problems, their pupils would often be better prepared for the higher grades. After they are thoroughly ac- quainted with the analysis of such questions they may be taught to write out neat, accurate solutions with far less trouble than if allowed to give careless analysis to problems in the lower grades. I. If 4 balls cost 36 cents, how many balls can be bought for 81 cents? Analysis: If 4 balls cost 36 cents, one ball will cost as many cents as 4 is contained into 36 cents which are 9 cents; if one ball costs 9 cents for 81 cents there can be bought as many balls as 9 is contained into 81 which are 9 balls. Written solution. ( 1. 36 cents=cost of 4 balls. II. ^ 2. 9 cents=36 cents-i-4=cost of 1 ball. ( 3. 81 cents=cost of 81-i-9, or 9 balls. III. .-. If 4 balls cost 36 cents, for 81 cents there can be bought 9 balls. I. What number divided by \ will give 10 for a quotient? II. III. 1. -^^the number. 2- ■H-|=*Xi=|=quotient 3. 10=quotient. 4. .-. 1=10, 5. 1=1 of 10=2, and 6. 1=3 times 2=6=the number. .-. 6=the number required. II. HI. $24 is \ of thei cost of a barrel of wine; what did it cost? fl. ^=cost of the wine per barrel. 1 2. f of cost=|24, 3. -^ of cost=4 of $24=$8, U. I of cost=5 times $8=$40, .-. $40=cost of w^ine. What number is that from which, if you take \ of itself, the remainder will be 16? II.<^ ri. 2. 3. 4. 5. 6. III. ^=the number. \ — |-=^=remainder after taking away \. 16=remainder. .•. *=16 -|=|. of 16=4, and •^=7 times 4=28=the number 28=the required number. 164 FINKEL'S SOLUTION BOOK. 1. AGE PROBLEMS. I. A is 30 years old, and B is 6 years old ; in how many years will A be only 4 times as old as B? '1. f=B's age at the required time. Then 2. ■|=A's age at the required time. 3. I — |-=f:^difference of their ages. 4. 30 years — 6 years=24 years:=difFerence of their ages. II.< 5. .•. 1=24 years. 6. i=^ of 24 years=4 years. [time. 7. f=2 times 4 years=8 years, B's age at the required 8. .'. 8 years — 6 years=2 years=the number of years hence when A will be only 4 times as old as B. III. .•. In 2 years A will be only 4 times as old as B. I. Jacob is twice as old~as his son who is 20 years of age ; how long since Jacob was 5 times as old as his son? 1. 20 years^son's age at present. Then 2. 40 ye^rs=Jacob's age at present. 3. f:^s9n's age at required time. Then 4. y=Jacob's age at required time. 5. .•. -y — -f^f ^difference of their ages. II. -J 6. 40 years — 20 years=20 years^difference of their ageSi. 7. .-. 1=20 years, 8. -J^-J of 20 year s=2-^ years, and , [time. 9. f=2 times 2^ years=5 years, son's age at the required" 10. .•. 20 years — 5 years=15 years=time since Jacob was &■ times as old as his son. HI. .•. 16 years ago Jacob was 5 times as, old as his son. Remarks. — Observe that the difference between any two persons' ages Is- constant, that is, if the difference between A's and B's ages is 7 years now, it will be the same in any number of years from now; for, as a year is add- ed to one's age, it is likewise added to the other's age. ' But the ratio of their ages is constantly changing as time goes on. If A is 3 years old and B 5 years old, A is now f as old as B; but in 1 year, A's age will be 4 years, and B's 6 years; A is then f as old as B. In 7 years, A will be 10 years old and B 12; A will then be ^f, or f , as old as B, and so on. The ratip of any two persons' ages approaches unity as its limit. I. At the time of marriage a wife's age was f of the age of her husband, and 10 years after marriage her age was y'^ of the age of her husband ; how old was each at the time of marriage ? 1. |=husband's age at the time of marriage. Then 2. |=wife's age at the time of marriage. 3. H-IO years=i=husband's age 10 years after marriage. 4. |-|-lf^ years=wife's age 10 years after marriage. But ANALYSIS. 165 II. ^^54-7 years==^^ of (|+10 years )^wife's age 10 years after marriage, by second condition of the problem. ■ •■• 1^1+7 years=|-|-10 years. Whende -f=10 years — 7 years, or III. I. II. ' • To 8. tV^S years. . [of marriage. 9. T^=10 times 3 years==30 years=husband's age at time 10. f, or ■3fij,==6 times 3 years=18 years=wife's age at the time of marriage. 30 years=husband's age at time of marriage, and 18 years=wife's age at time of marriage. ( White's Comp. A., f. 24.1, prob. S5.) Ten year? ago the age of A was | of the age of B, and ten years hence the age of A will be |- of the age of B ; find the age of each. ■ 1. f=B's age 10 years ago. Then 2. I^^A's age 10 years ago. / 3. 1+10 years=B's age now, and 4. f-|-10 years=A's age now. 6. |-i-20 years=B's age 10 y.ears hence, and 6. |-|-20 years=;=A's age 10 years hence. 7- f of (1+20 years)=f+16t years=A's age 10" 8. .-. f+16f years=|+20 years ; whence . 9. |— 1=20 years— 16f yeal-s, or 10. 1^=3^ years, and 11. -11=12 times 3^ years=40 years=B's age 10 years ago. [hence, years 12. i=- |=t't=9 times 3^ years=30 years=A's age 10 years ago. 13. .-. fl+10 years^50 years^B's age now, and .14. .j^+10 years^40 years=A's age now. Til. .". 60 years=:B's age, and 40 years=A's age. 2. FOX AND HOUND PROBLEMS. I Under this head comes a class of problems which may be variously named as " Hare and Hound Problems," " Step Prob- lems," " Hare and Tortoise Problems," etc. I. A fox is 50 leaps ahead of a hound, and takes 4 leaps in the same time that the hound takes 3 ; but 2 of the hound's leaps equal 3 of the fox's leaps. How many leaps must the hound take to catch the fox? 1. 2 leaps of hound's=3 leaps of fox's. 2. 1 leap of hound's^=^ of 3 leaps=l^ leaps of the fox's. 3. 3 leaps of hound's=3 times 1^ leaps=4|. leaps'of fox's. »j 1 4. .•. 4^ leaps — 4 leaps^^ leap=what the hound gains in ■'^ taking 3 leaps. [ing 6 leaps. 5. .". 1 leap==2 times ^ leap==what the hound gains in tak- '6. .•. 60 leaps=what the .hound gains in taking 50X6 leaps, or 300 leaps. 166 FINKEL'S SOLUTION BOOK. » III. .-. The hound must take 300 leaps to catch the fox. Remark — We see that 3 of the hound's leaps equals 4J leaps of the foic's. But while the hound takes 3 leaps, the fox takes 4 leaps; hence the hound gains 4| — 4, or \, leap of the fox's. But he has 50 leaps of the fox's to gain,^ and since he gains ^ leap of the fox's in 3 leaps, he must take 300 leaps tO' gain 50 leaps. I. A thief is 20 steps before an officer, and takes 6 steps while the officer takes 5, but 5 of the officer's steps equal 8 of the thief's ; how far will the thief run beforie he is over- • , taken ? ' 1. 5 steps of the officer:=8 steps of the thief, 2. .-. 8 steps — 6 steps=2 steps=the distance the officer gains on the thief every time he takes 5 steps. 3. 1^ step=;distance the officer gains on thief in taking- 1 step. II. ^ 4. .-. 20 steps =: distance the officer gains in taking 20-=-f ,. or 50 steps. But, since 5. 5 steps of the officer=8 steps of the thief, 6. 50 steps of the officer=80 steps of the thief. 7. . . 80 steps — ^20 steps=60 steps=:distance thief runs, before he is overtaken. » III. .-. The thief will run 60 steps. Remark. — It should be observed that the thief is 20 of his owm steps ahead of the officer, and thus his step becomes the unit of measure. It requires 50 of the officer's own steps to overtake the thief. But these are equal to 80 of the thief's. Since the officer took 50 steps equal to. 80 of the thief's steps to overtake the thief and the thief was 20 steps, ahead, it follows that the thief took 60 steps before he was overtaken. I. Achilles is 1 mile behind a tortoise and runs 10 times as fast as the tortoise. Will Achilles overtake the tortoise; ?' Proposed by Zeno, the Eliatic. Zeno's argument : -1. While Achilles is running the 1 mile, the tortoise will have ^un ^V o^ ^ "^i'e farther ; 2. While Achilles is running the y\ mile, the tortoise wilt II. \ have advanced xJr of ^ ^lil^ still farther ; 3. While Achilles is running the -j-Jt of a mile, the tortoise will have advanced xcW of ^ ^^^ still farther ; and so on, ad infiMttum. III. .-. Achilles will never overtake the tortoise. Remark. — This problem was proposed by Zeno, one of the most: famous members of the Eliatic School of philosophers. He was born 495 B. C. and was executed, at Elea, a town on the island of Sicily, in 435 B. C. This paradox and the one to the effect that an arrow cannot move where it is not, and since also it cannot move where it is, that is, in the- space it exactly fills, and hence cannot move at all, were put forth by Zeno to disprove mition. We bring this problem up for discussion not because we have any- thing particularly new to add to it;"~for it has been very thoroughly discussed by some of the greatest thinkers that the world has ever pro- ANALYSIS. 167 duced, and in its complete solution are involved subtile metaphysical questions that have not yet been satisfactorily explained, and perhaps may ne,ver be. The fallacy in Zeno's argument concerning the tflotion of the arrow is that he assumes that at every instant the arrow must be resting in a definite point. But if it is resting, it is not moving and cannot move. Only that rests at a point which remains in it for some consecutive instants. Zeno confounds being in a point in the sense of resting in it, with being in a point, with the sense of, p'assitig through it.* On the fundamental notion of motion, the reader should read articles 33, 84, 85, and 86 of W. B. Smith's Infinitesimal Analysis. In the case of Achilles and the tortoise, if it be understood that they traveled with uniform motion, and traveled a finite distance in a finite time, then Zeno's argument is false; for while the time required is divided into an infinite number of parts, these parts diminish in a geometrical progression and the sum of them all is a finite sum. The distance Achilles must run to overtake the tortoise is li miles. For, in running a mile, Achilles gains x'ly of a mile. Hence, to gain 1 mile, he will have to run as many miles as x°ir is contained in 1 mile or 1^ miles. This would be the distance he would have to run in any case; and whether he would or would not overtake the tortoise depends upon the law of his motion If the w6rd, while, in the three steps of Zeno's argument represent the same period of time, then it would take Achilles as long to travel -fs mile as to travel 1 mile, and as long to travel x^ mile, as to travel xV mile, etc. In this case, Achilles and the tortoise are not traveling with uniform motion, and Achilles would never overtake the tortoise. It is contended by some logicians and metaphysicians, that in the use of the ambiguous terra, "while," lies the fallacy of Zeno's argument. 3. FISH PROBLEMS. I. The head of a trout weighs 2 pounds, the tail weighs 2 pounds more than the head, plus J of the body, and the body weighs as much as the head and tail together; required the weight of the fish ? (Brooks' Int. Arith., p. 143, prob. 10.) 1. |=the weight of the tail. 2. 1+2 lbs., the weight of the head and tail,=the weight "of the body. 3. l+f lbs.=J of (-1+2 lbs.)=the weight of J of the body. 4. (l+f lb^.)+2 lbs. +2 lbs.=f+4f lbs.— 2 lbs. more than the weight of the head and J tif the body. 5. .'. f or 1=1+41 lbs., since the tail weighs as much as the head and J the body-|-2 lbs. .■.f=4f lbs. 7. i=iof 4f lbs.=lilbs. 8. I or 1=6 times 1| lbs. =7 lbs., the weight of the tail. 9. 7 lbs.+2 lbs.=9 lbs.=weight of the body. 10. 7 lbs.+2 lbs.+2 lbs.=ll lbs.=the weight of the fish. III. .-. The weight of the fish is 11 lbs. II. < '■* Bowen's Metaphysics, page 81. The entire chapter on motion in this book is very interesting-. II. 168 FINKEL'S SOLUTION BOOK. Remark. — In the solution of "fish problems'' and those that are similar, we should always take that number as the base from which the other numbers are readily obtained when the base is once known. Thus, in the above problem, when the weight of the tail is known, the weight of the body is easily obtained. Hence, we let | represent the weight of the tail. We might as well have taken f or f to represent the weight of the tail. ' L The head of a fish is 8 inches long, the tail is as long as the head and -^ of the body-|-10 inches, and the body is as long as the 'head and tail ; what is the length of the flshPr 1. f=length of body. 2. 8 in.^length of head. 3. i 1. of b.+lO in.+8 in.=^ 1. of b.+18 in.=length of tail. 4. I 1. of b.:=length of head-|-length of tail. " .-. I 1. of b.=(^ 1. of b.+18 in. )+8 in.=^ 1. of b.+26 in. Whence I 1. of b. — i 1. of b.=-^ 1. of b.=26 in. 7. .'. f 1. of b., or length of body ,=2 times 26 in.=52 in. I 1. of b.+18 in.==26 in.-(-18 in.=44 in.=length. of tail. 19. .-. 52 in. +44 in.+8 in.=104 in.=length of the fish. III. .-. The length of the fish is 104 inches. 4. ANIMAlv PROBLEMS. ' I. A man bought a certain number of sheep for $80 ; if he then buys twice as many more, at |2 less each, they will cost |180.; how many did he buy ? (Brooks' Int. Arith., p. 162, proh. 11.) '1. -1= number of sheep bought, and for which he paid $80. 2. "1=2 times -| =number he bought at f2 less per head and for which he paid $180— 180 , or $100. 3. .-. -I at $2 less per head would have cost ^ of $100, ^ or $50. 4. .•. The same number of sheep that he bought for $80 at $2 less per head would cost $50. 5. $2^the reduction on the price of 1 sheep. ,6. $30==$80— $50=:r)sduction on $30-=-$2, or 15 sheep. III. .-. He bought 15 sheep. I. Henry Adams bought a number of pigs for $48 ; and losing 3 of them, he sold ■§• of the remainder, minus 2, for cost, receiving $32 less than all cost him; required the number purchased. 1. |=^remainder after losing 3. Then 2. |^-|-3=number at first. 3. t of r. — 2=number sold. II. <^ II. ANALYSIS 169 4. $48— $32==$16— what was received for f of r.— 2. 5. $8=^ of $16=what was received for | of (fofr.— 2), or ^ of r. — 1. ■ 1 6. $24=3 times $8=what was received for 3 times (^ of r.— l)=|ofr.— 3. 7. .-. $48— $24=$24=whit (f of r.+3)— (f of r.— 3), or 6 pigs cost. 8. $4=|- of $24=what 1 pig cost. 9. •. $48=what 48H-4, or 12, pigs cost. III. .-. He bought 12 pigs. {Brooks' Int. A., >. 16^, prob. 9.) I. A bought some calves for $80; and having lost 10, he sold ' ■ 4 more than f of the remainder for cost and received $32 less than all cost; required the 'number purchased. 1. |=remainder after losing 10. Then 2. l^-j-lO^number purchased. 3. f of r.-|-4=number sold. [cost. 4. $80— $32=$48= cost of f of r.+4, since they sold at jT 1 5. $24=-^ of $48=cost of ^ of (f of r.+4)=f of r.+2. "^6. $72=3 times $24=cost of 3 times (^ of r.+2)=| of r.+e. [cost. 7. .-. $80— $72=$8=what (f ofr.+lO)— (f of r.+6), or 4 '8. $2=i of $8=what 1 cost! L9. $80=what 80-=-2, or 40 cost. III. .•. He bought 40 calves. {\Brook's Int. A., p. 16^, prob. 10.) I. A lost ^ of his sheep; now if he finds 5 and sells |^ of what he then has for cost price, he will receive $18; but if he loses 5 and sells -| of the remainder for cost price, he will receive $6; how many sheep had he at first? {Brook's Int. A., p. 165, prob. IB.) 1. 4= the number of sheep he had at firs^t. 2. \^= the number he lost. 3. -9 — \=\i the number he had after losing \. 4. \-\-h= the number he had after finding 5. 5. .|of (|+5)=y\+3, the number he sold. 6. I — 5= the number, had he lost 5. 7. |of(| — 5)=^-j — 3, the number he would have sold. 8. |l8=what (-^^+3) sheep cost. II."J 9. $6= what (■j'ij— 8) sheep cost. 10. .-. $12=$18— $6=what (2%+3) sheep— (^—3) sheep, or 6 sheep cost. 11. $2=^ of $12= what 1 sheep cost. 12. $18= what 18-7-2. or 9 sheep cost. But iiA 170 FINKEL'S SOLUTION BOOK. 13. $18= what (^+3) sheep cost. 14.. .'. ^\+3 sh'eep = 9 sheep, or 15. '■^=& sheep. 16. X5= J of 6 sheep=l sheep, and 17. ff=25 times 1 sheep =25 sheep. III. .-. He had 25 sheep at first. I. A man bought a certain riumbef of cows for $200; had he- bought 2 nfore at $2 less each, they would have cost him $216; how many did he buy' 1. $200=cost of cows. > 2. $216=cost of oiiginal number of cows-|-2 more. 3. $216— $200=$16=cost of 2 cows at $2 less per head. 4. .-. $8=i of $16=cost of 1 cow at $2 less per head. Then. 5. $8+$2=$10^cost of each cow purchased. ,6. $200=cost of 200-i-lO, or 20 cows. III. .-. He bought 20 cows. {Brook's Int. A., f. 162, fi-Ob. 8.) 5. I„ABOR PROBI^EMS. I. A laborer agreed to work for |2 a day, on condition that for every day he was idle he should forfeit 50 ^; how many days did he labor, if at the end of 25 days he received $35? 1. |2=amount received for 1 day's labor. 2. |50=25X|2=amount he would have received for 25 days' labor. 3. |50 — |35=|15=amount he'forfeited by his idleness. '' II. \ 4. |2, his wages,+f^, his forfeit,=f2J=amount he lost each day by his idleness. 5. . . |15=amount he lost for as many days' idleness, as |2^ is contained in $15, or 6 days. 6. .-. 25 days — 6 days=19 days, the time he labored. III. .-. He labored 19 days. I. A man was engaged for one year at $80 and a suit of clothes; he served 7 months, arid received for his -wages the clothes and $35; what was the value of the clothes? (\, J-|^value of the suit of clothes. 2. ^-t-$80=wages for 1 year or 12 months. 3. .j1^$6|=tV of (||+$80)=wages for 1 month. 4. i^-|-$46f=7 times (tV+^^I )=wages for 7 months. 5. ||+$35=wages for 7 months. n.<^ 6. tt+$S5=xV 7. T^j=$llf, 8. ^1^=1- of $11|=$2|, and .9. 11=12 times $2i=$28=value of suit of clothes. Ill, .-. The suit of clothes is worth $28. ANALYSIS. 171 6. WORK PROBLKMS AND PIPE PROBIvEMS. I. 11. Three'men, A, B, C, can do a piece of work in 6o days.. After working together 10 days, A withdraws and B. and C work together at the same rate for 20 days, then B withdraws and C completes the work in 96 days, working -J- longer each day. Working at his former rate, C could alone do the work in 222 days ; find Jiow long it would take A and B each separately to do the work. 60 days=:time it would take A, B, and C to do the work. ^=^part of the work they do in 1 day, and ■|-=part of the work they do in 10 days, f — ^=:f ^part left for B and C to do. 1 day of C's after B withdrew^f of one of his days before B withdrew. 96 days of C's after B withdrew^96Xt of one day =128 days of C's before B withdrew. .-. 20 days+128 days^l48 days=:time C worked after A withdrew. 222 days=timeC could do the work alone, .jl^j^part he could do in 1 day. |||^|=part he did in 148 days. .'. I — |=.^=part B did in 20 days. .•. ji-^=-^ of -j-=part B does in 1 day. \^%, or the work,=what he can do in r|f"^TiTr) or 120 days. +"iji^ )^=THTr==P^''t A does in 1 day. 10. 11. 12. 13. 14. 15. ^v- -(tItt- iii%, or the work,=:what he can do in ||-f^ or 261iV days. III. 11. III. . f A can do the work in 261 fV days, and ■ ■ I B can do it in 120 days. A cistern has three pipes : the 1st can fill the cistern in IJ hr. ; the 2d, in 3J hr. ; and the 3d can empty it in .5 hours. How long will it take to fill the cistern if all pipes are left running? 1. 1^ hr.=time it takes the first pipe to fill the cistern. 2. .". Impart of cistern filled by 1st pipe in 1 hr. 3. 3^ hr.=time it takes the 2d to fill it. , 4. .-. T»^=part filledby the2din 1 hr. 5. 6 hr.=:time it takes the 3d to empty the cistern. 6. .'. -J-^part emptied by the 3d in 1 hr. 7. .'. f+A'~i=i¥^P^rt filled when the pipes are all running. 8. .•. f-§^, or the whole cistern, ^what is filled in f-g--^-^,. or 1^ hr. . ■ . The cistern can be filled in 1 ^ hr. .172 FINKEL'S SOLUTION BOOK. I. There is coal now on the dock, and coal is running on also from a shoot at a uniform rate. Six men can clear the dock in 1 hour, but 11 men can clear it in 20 min- utes ; how long would it take 4 men? II.J 1. 2. 3. 4. 5. •6. 7. 8. 9. 10. I^what one man removes in 1 hour! Then 6 times l^what 6 men remove in 1 hour. what 1 man removes in 20 min., or -j- hour. ^^*=11 times ^=what 11 men remove in ^ hour. ■ ■ " " hr. |=iof ^/ — 2^2__i^__ -yyjiat runs on in 1 hr. — ^ hr. Then J=i^-;-|=what runs on in 1 hour. III. [commenced. . . 1/ — 1=1= what was on the dock when the work |=;what 4 men remove in 1 hour. .■. 1^ — J=^^part of coal removed every hour, that was on the dock at first. f=coal to be removed in |-i-^==5 hours. {R. H. A.,f. i06,prob. 90.) It will take 4 men, 5 hours to clear .the dock. Explanation. — ^2r=what 6 men remove in 1 hr. and ^^what 11 men re- moved in § hr. In either case the dock was cleared. ,■. V — V^=V=^ amount of coal that ran on the dock from the shoot in 1 hr. — \ hr., .or f hr. Hence in 1 hr. there will run on, i/~|=^.. Since J run on in 1 hr. and "^ =the whole amount of coal removed in 1 hr., '^ — |, or f must be the amount of coal on the dock when the Work began. Since |=the amount 4 men remove in 1 hr. and ^^the amount that runs on the dock in 1 hr., % — 5, or \ is the paft of the original quantity removea each hour. Hence, if ^ is removed in 1 hour | would be removed in f-j-J, or 5 hours. I. A and B perform .^ of a piece of work in 2 days, when, B leaving, A completes it in -J day, in what time can each complete it alone? II-<^ III. Jj5.=part A and B do in 2 days. •j9^=|- of y'^=part A and B do in 1 day. io_ TTT =J^=ipart left after B quits, and which A com* 2 . rrr- pletes in -^ day. :^^part A can do in 1 day. .|=part A can do in ^ '. \ 5 days. ■f^ — ^=^s^=^=part B can do in 1 day. .-. |^=part B can do in | ; \ , or 4, days. SA can do the work in 6 days, and B chn do the work in 4 days. ( White's Comp. A.,f. 280, f rob. 193.) A and B can do a piece of work in 12 days, B and C in 9 days, and A and C in 6 days; how long will it take each alone to do the work? ANALYSIS. 173. II. IIL 1. 12 days=time it takes A and B to do the work. 2. .•. YV=P^*'t they do in 1 day. 3. 9 days=time it takes B and C to do the work. 4. .•. .^^part they do in 1 day. 5. 6 days=time it takes A and C to do the work. 6. .'. ^^part they do in 1 day. 7 .-. TV+¥+i=il=Pa>'t A and B, B and C, and A and C do in 1 day^twice the work A, B,and C do in 1 day. 8. .•. T^=i of ^^part A, B, and C do in 1 day. 9. If— ^L.=y\=part A, B, and C do in 1 d^y— part B and C do in 1 day:^part C does in 1 day. 10. ■f|=part C does in -fj : 7^ , or lOf days. 11. -fl — ^=^5^=part A, B, and.C do in 1 day — part B and; C do in 1 day=part A does in 1 day. 12.^^|=part A does in ^|-^^5^=14| days. 13. -fl — ^=yij=part A, B, and C do in 1 day — part A and C do in 1 day^part B does in 1 day. 14. .||=part B does in .;f|-^!j=72 days 14|- days=time it takes A, 72 days^time it takes B, and lOy days^time it takes C. ( White's Comf. A., p. 19^., prob. 280.} A man and a boy can niow a certain field in 8 hours,, if the boy rests 3f hours, it takes them 9^, hours. In what time can each do it? 2. 3. 4. II.' 5. 6. 7. 8. l9. III. .-. I. S 1. 9^ hr. — 3f hr.^5f hr.=time they both work together in the second case. 8 hr.^time it takes them to do the work. =part they do in 1 hour. *=||^5f times ^^part they do in 5f hours. .'. 1^1 — |^|=^9j=part the man did in 3| hours, while the- boy j-ested. .-. -^^=-—- of ^^j^part the man did in 1 hour, of .-. |^=part the man can do in f^-i~i\ or 13-J- hours. ^ — -^^=jljy=part the .boy does in one hour. |^=part the boy can do in l-j-f-sV or 20 hours. It will take the man 13^ hours, and The boy 20 hours. {R. H. A., p. 1^02, f rob. SO.) 5ix men can do a work in 4^ days; after working 2 days,, how many must join them so> as to complete it in 3| days? 3.74 FINKEL'S SOLUTION BOOK. II. (1. 2. 3. 4. 5. 6. 7. 8. 9. III. 4J- days==time it takes 6 men, 26 days=6 times 4-J days=time it takes 1 man. .-. ^==part 1 man does in 1 day. ^3^=6 times g'5=part 6 men do in 1 day. ^=2 times y\=part 6 men do in 2 days. [days. i| — ^=Jj=part to be done in 3| days — 2 days, or l-f -l=y|.g.=part 1 miLjci does in 1-| days. .-. J^=part T^s-Hj-lir, or 10 men can do in 1-| days. .-. 10 men — 6 men==i4 men, the number that must join them. . Thev must be joined by 4 more men that they may com- plete the work in 3f days. Ji. H. A., p., 4.0%, prob. 34. vl 7. WINE AND WATER PROBLEMS. ^ I. How much water is there in a mixture of 100 gal. of wine and water, worth $1 per gal., if 100 gal. of the wine costs $120? (Robinson's New Higher Arithmetic, p. 502.) 1. 11.20=1120-^ 100=cost of 1 gal. of wine. 2. fl^cost of wine in 1 gal. of the mixture. 3. .-. •§■ gal.^|1.00-=-|1.20=quantity of wine in each gal. of the mixture. .-.^ gal^l gal. — I gal.=quantity of water in each gal. of the mixture. 5. .-. 16| gal.=^ of 100 gal,=:quantity of water in 100 gal. of the mixture. .-. In 100 gal. of the mixture there are 16f gal. of water. , 11.^ III. I, From a ten-gallon keg of wine, one gallon is drawn off and the keg filled with water ; if this is repeated 4 times, what will be the quantity of wine in the keg? II. a. 2. 8. 4. 5. 6. 7. .jlj=part drawn out each time. Jj^part that was pure wine after the first draught. Jjj. of jS^=j-|Tj=:part wine drawn off the second draught. A — rtTr=AV= P^i'* pure wine left after the second 1 nf -8 1 - 81 81 . Tirxr iTTij- draught. 1_ of J-2JL. 81 -TTTTTTr- draught. [draught. = part wine drawn off" at the third :part pure wine left after the third / , [draught. =part wine drawn off" at the fourth ^.j^.j..j=yYF!nr= P^rt pure wine left after fourth draught. [fourth draught. TTinfV °f 10 gal.=6.561 gal.==pure wine left after the III. .•. There will be 6.561 gal. of pure wine in the keg after the fourth draught. 72 9 7 2 iTTnr - 729 . 05C - ANALYSIS. 175 In the above problem, how many draughts are necessary to draw off half the wirte ? 1. -^rrzipart: wine drawn off at the first draught. 2. \^ — j.i^j=^8^=part wine left after the first draught. 3. -^ of y%=y|^=y^==part wine drawn off at the sec- ond draught. gz g 2 4. J^ — -— -=^i^=(— ) ==part wine left after the second 10 10 draught. 9 2 ga 5. tV of (ztt: ) ^:r^=part wine drawn off at the third lu \jo' 10' draught. 6. (— )^— -^=(-^) =part wine left after the third draught. By induction, 7. (^)»=part wine left after the ^th draught. 8. .•. 10(A-)''=number of gal. left after the «th draught. 9. 5^numDer of gal. left after the «th draught. 10. .-. 10(t:V)''=5. whence 11. (-,9^)"=^. Applying logarithms, 12. n log. T:%=log- i- 13. .-. n = log. i-^log. Jj=.30103-j-. T.954243— 301030-^ .045757=6+. Itl. .". In 7 draughts, half and a little more than half of the wine will be drawn off. 8. SHEEP AND COW PROBLEMS. I. A farmer keeps 60 cows on his farm. For every 4 cows, he plows 1 acre of ground and for every 5 cows, he pastures 1 acre. How many acres in the farm? 1. 1 A.=what he plows for 4 cows. 2. \ A.=what he plows for 1 cow. 8. 1 A.^what he pastures for 5 cows. 11.^ 4. ^A.^what he pastures for 1 cow. 5. .-. i A.+^ A.=-^-5- A.=land required for each cow. 6. .-. 60 cows require as many acres as -^ is contained in 60, or 133J acres. III. . . There are 133 J- acres in the farm. Remark. — The "Coach Problems,'' page 187 are solved on the same principle as these problems and might have been grouped with this set. I. For every 10 sheep I keep I plow an acre of land, and allow one acre of pasture for every 4 sheep; how many sheep can I keep on 161 acres ? 176 FINKEL'S SOLUTION BOOK. n. 1. lA.=what I plow for every 10 sheep I keep. 2. ■^A.'=Vhat I plow for each sheep I keep. 3. lA.=what I allow for pasture for every 4 sheep I keep.. 4. ^A.=what I allow for pasture for each sheep I keep. 5. .•. ^i^A.-|-JA.=^A=land required for every sheep. 6. .•. 161A.=land required for 161-=-^, or 460 sheep. m. .'. I can keep 460 sheep on 161 acres. (J?. Alg. I., p. 112, frob. 6i.y Complete analysis. If for every 10 sheep I plow 1 acre, for 1 sheep I plow -^ of an acre ; and if for every 4 sheep I pasture 1 acre, for 1 sheep, I pasture ^ of an acre ; hence 1 sheep requires ^A.-f-^A., or ^A., and on 161 A. I could keep as many sheep as g^A. is contained ia 161 A., which are 460 sheep. 9. WITH AND AGAINST THE CURRENT PROBLEMS. , I. A man can row down stream at the rate 'of 10 miles per hour, and up stream at the rate of 6 miles per hour; find the rate of rowing in still water and the rate of the current. 1. Rate of rowihg-|-rate of current=:10 miles per hour. 2. Rate of rowing — rate of current=6 miles per hour. 3. .-. 2 times rate of rowing=16 miles per hour, by adding^ steps 1 and 2. 4. .-. The rate of rowings |^ of 16 miles per hour^S miles per hour. 5. 2 times the rate of the curr'ent=4 miles per hour, by subtracting step 2 from step 1. 6. .-. The rate of the current=^ of 4 miles per hour=2: miles per hour. jjj . I The rate of rowing is 8 miles per hour, and ■ I the rate of the current is 2 miles per hour. Note. — This problem was sent to the author, for solution, by a teacher in Michigan where considerable discussion arose as to the cor- rect result. I. How far can a man row up stream and return in 14 hours, if the rate of the current is 5 miles per hour and his rate of rowing in still water , 7 miles per hour ? ■ 1. 7 miles — 5 miles=:2 miles^his rate per hour in going up stream. 2. .-. i hour^time it will take him to go 1 mile up stream. 3. 7 miles-|-5 miles=12 milea==his rate per hour in goinj^ II. ^ down stream. II. <^ ANALYSIS. 177 4. .-. ^V hour=time it takes him to go 1 mile down stream. 5. ,i houT4-^ hour=y\ houri^time it takes him to go 1 mile up stream and return.. 6. .-. 14 hoursi=the time it will take him to goli-^^-,^, or 24 miles up stream and return. III. .-. He can go 24 miles up stream and, return in 14 hours. ^I. III. 10. TIME PROBLEMS. A person being asked the hour of day, said, "the time past noon is ^ of the time past midnight;"" what was the hour? I^time past midnight. I^^time past noon. .-. 1^ — -|^=|=time from midnight to noon. 12 hours=time from midnight to noon. .-. ^=12 hours. ^==^ of 12 hours=6 hours=time past noon. It was 6 o'clock, P. M. At what time between 4 and 5 o'clock is the hour hand ^ as far from 4 as the minute hand is from 2 ? 1. f=distance hour hand moves' past 4. Then 2. -^^-^distance minute hand moves past 12 in same time. 10 min- minute \4 or distance, 10 minuted utes hand is from 2 .■.10 minutes — minutes T — the 24 "5" A. 1 or -Vt — 10 11.^ minutes=5X|=-y-- 5. .-. ^= 10 minutes or i/= 10 minutes, 6. i=f'j of 10 minutes or -^-^ of 10 minutes = y\ minute or minute. 7. -^*=24 times -^ minute or 24 times f minute, = Tt^ij^ minutes or 17-f minutes past 4 o'clock. III. .-. At 7yL minutes or 171- minutes past 4 o'clock, the hour hand is ^ as far from 4 as the minute hand is from 2. Explanation. — Locate the hour hand at 4 and the minute hand at ]'2. Now if the hour hand remained stationary at 4, the minute hand would have to travel on to 2, in order that it may be 5 times as far from 2 as the hoiir hand is from 4.' This is so, because the hour hand is at four, by l^ypothesis, and is, therefore, at a distance minutes from it. Hence, that the minute hand be 5 times as far from 2, it must be at 2, because 5 times is 0. But as the minute hand moves past 12, the hour hand moves past 4. The distance the minute hand is to stop short of 2 or move past 2 is shown in the 3d step. The analysis needs no further detailed explanation. 178 FINKEL'S SOLUiION' BOOK. I. Provided the tim^ past 10 o'clock, A. M., equals f of the time to midnight; whiat o'clock is it? 1. |^=time to midnight. , Then 2. f=time past 10 o'clock. 3. |-j-f:^J==time from 10 o'clock to midnight. II. V^' ^^ h<5urs=time from 10 o'clock to midnight. 5. .•. J=14 hours. 6. i=Y of 14 hours==2 hours, and [o'clock P. M. 7. -1=3, times 2 hours=6 hours, time past 10 6'clock=4 IIL .-. It.is 4 o'clock, P. M. I. At what time between 3 and 4 d'clock will the hour and minute hands of a watch be together? 1. ■|=distance the h. h. moves past 3. Then 2. ^^=12X-|^distance the m. h. moves past 12. 3. ^-^ — |='2^^=distance the m. h. gains on the h. h. ' II.< 4. 15 min.=distance the m. h. gains on the h. h 5. ••. V=15 min. 6. i=Y^ of 15 tnin.=^f min. , [past 12. 7. ^^=24 times ^ min.=l,6T;^ min.=distance m. h. moves III. .'. It is'lG^^^min. past 3 o'clock. Remark. — In problems of this kind, locate the minute hand at 12 and the hour hand at the first of the two numbers between which the conditions of the problem are to be satisfied. Thus in the above problem, at 3 o'clock the minute hand is at 12 and the hour hand at 3. The minute hand moves over 60 minute .spaces while the hour hand moves over 5 minute spaces. Hence the minute hand moves 12 times as fast as the ,hour hand. Since at 3 o'clock the minute hand is at 12 and the hour hand at 3, and the minute hand moves 12 times as fast as the hour hand, it is evident that the minute hand will overtake the hour hand between 3 and 4. So we let f =distance the hour hand moves past 3 until it is overtaken by the minute hand. But since the minute hand moves 12 times as fast as the hour hand, while the hour move |, the minute hand moves 12 times |, or V- Now the minute hand has moved from l2 to 3+|, or 15 minutes+|. Hence the minute hand has gained 15 minutes on the hour hand. It has also gained 'j* — |, or ^. .■. V=15 minutes. In solving any problem of this nature, first locate the hands as previously stated, and then ask yourself how far the minute hand must move to meet the conditions of the problem, if the hour hand should retnain stationary. I. At what tiijie between 6 and 7 o'clock will the minute hand be at right angles with the hour hand ? 1. |^=distance h. h. moves past 6. 2. 2^==12 times |=distance m. h. moves past 12. 3. .-. ^T^ — |=2^^=distance m. h. gains on h. h. 4. 15 min. or 45 min.=distance m. h. gains on the h. h. 5. .•. 2^2=15 niin. or 45 min. 6. ■^=^V ^^ 15 min. or -^ of 45 min.=^ min. or 2-^ min. 7. 2^=24 times \^ min. or 24 times 2^ min.=16Y*T '"*"■ or 49x1" ™*'^- III. .-. The minute hand will be at right angles with the hour hand at 16^^ min. or 49^ min. past 6 o'clock. 11. ANALYSIS. 179 Explanation. — Locate the minute hand at 12 and the hour hand at 6. Now if the hour hand had remained stationary at 6, the minute hand would have to move to 3 or 9, ;'. e., it would have to gain 15 min. or 45 min. While the minute hand is moving to 3 the hour hand is moving from 6. So the min- ute hand must move as far past 3 as the hour hand moves past 6. Or while the minute hand is moving to 9 the hour hand is moving past 6. So the minute hand must move as far past 9 as the hour hand is past 6. .'. The minute hand must gain 15 minutes in the first case and 45 minutes in the second. I. At what time between 2 and 3 o'clock are the hour and minute hands opposite? II. 1. 2. 3. 4. 5. 6. 17. Then past 12, f=distance hour hand moves past 2 ^^=distance the minute hand moves same time. .•. ^ — f=V=distance minute hand gained on the hour 40 min.=distance the minute hand gained on the hour in the [hand. III. hand. > .;. 2^2=40 min. ■^=-^'of 40 mim.:^!^^ min., and \*=24 times 1^ min.=43T^ ""i". , It is 43y\- min. past 2 o'clock when the hands are opposite. Explanation. — Locate the minute hand at 12 and the hour hand at 2. Now if the hour hand remained stationary at 2, the minute hand would have to move to 8 or over 40 minutes in order to be opposite the hoi^r han^. But while the minute hand Is moving to 8, the hour hand is moving fj-om 2. So the minute hand must move as far past 8 as the hour hand is paa(; 2. Since \ is the distance the hour hand moves past 2, f must be the 4istspce the minute hand must move past 8. Hence the distance the minute hand moves is |+40 min. But Y=distance the minute hand moves. .■. V^=?~l~ ■40 min. or ^=40 min. as shown in step 5. I. At what time between 3 and 4 o'clock w^ill the minute hand be 5 minutes ahead of the hour hand? II. 1. |=distance hour hand moves while the m. h. IS moving to be 5 min. ahead. [moves \. 2. 2^=12 Xf ^distance minute hand moves while the h. h. 3. .•. 2^— ^f=2^^distance gained by the minute hand. 4. 15 min. -(-5 mim.=20 min.=distance gained by the m. h. 5. .-. V^=20 min. 6. \-=i^ of 20 min.=ii min. 7. V=24 times W min.=2lT2r min. III. .'.It is 21-j8j^ min. past 3 o'clock. Explanation. — Locate the minute hand at 12 and the hour hand at 3. Now if the hour hand remained stationary ^at 3, the minute hand would have to move to 4 in order to be 5 min. ahead. But while the minute hand is moving to 4 the hour hand i§ mo\>-ing from 3. Hence "the minute hand must move as far past 4 as the hour hand moves past 3. But the hour hand 180 FINKEL'S SOLUTION BOOK. moves f past 3; hence, the minute hand must m'ove f +5 min. past 4, in alij |+20inin. Hence, the minute hand gains (|+20 min.) — 1^20 min. on the hour hand. Remark. — We always find ^, the distance the minute hand moves, for it indicates the time' between any two consec^itive hours. The hour hand indicates the hoiir. / I. At what time between 4 and 5 o'clock do the hands of a clock make with each other an angle of 45° ? II. ■|=distance the hour hand moves past 4. ?^=distance the minute hand moves past. 12. ■^^ — ^ — ?-^ — ^distance the minute hand gains 2 22_ 2 "2" "5" - hour hand. on the- Vl^ min. or 27-^ min.=distance gained by minute hand. .■. 2^2=121^, min. or 27^ min. [min. \=-^-^ of 12^ min. or -^ of 27^ min.=f^ min. or \\ 2^=24 times |f min. or 24 times \\ min.=13-j^ xhva.. or 30 min. III. .•. At 13-j^Y tniii' past 4 or 30 min. past 4, the hands make an angle of 45° with each other. Explanation. — Locate the minute hand at 12 and the hour hand at 4. 45° ^=i of 360". J of 60 min.=7^ min. Hence, that the hands make an angle' of ' 45 , the minute hand must be either 't\ minutes behind the hour hand or 7^ min. ahead. Now if the hour hand remained stationary at 4, the minute' hand would have to move over 12^ min. or 2^ , min. past 2. But while the minute hand is moving this distance, the hour hard is moving past 4. Hence, the minute hand must move as far past 2^ min. past 2 as the hour hand moves past 4, i. e., the minute hand moves |+12J min. Hence, it gains (|+124 min.) — 1=12^ min. The reasoning for the second result is the same as for the first. I. At what time between 4 arid 5 o'clock'is the minute hand as far from 8 as the hour hand is from 3 ? II. .|=distance the hour hand moves past 4. 2^=12 times f=distance minute h^nd moves past 12 in the same tiitie. .•. ^/-|-f^\^=distance both move. 35 min.=distance both move. .-. 2^6=35 min. ' .^=^ of 35 min.=l-/^ min. .B. ^^*=24 times 1^ min.=32T% min. 1. f=distance the h. h. moves past 4. 2. \*=distance minute hand moves past 12. 3. .-. 2^ — |^2Y^=distance the minute hand gains. 4. 45 min. ^distance the minute hand gains. 5. .•. 2^2=45 min. -9 1 mm. III. 6. 4=^V °f 45 min.= 7. V=24 times 2^ min'.=49^ min. It is 32-^^ min. or 49jy min. past 4 o'clock. {'R H. A.,f. lfiB,frob. 40.) ANALYSIS. 181 Explanation. — This problem requires two different solutions. Locate the ■minute hand at 12 and the hour hand at 4. The hour hand is now .5 min- utes from 3. If the hour-hand remained stationary, the mjnute hand would have to move to 7 to be 5 minutes from 8. But while the ininute hand is moving to 7, the hour hand is movihg past 4. Hence the minute hand must stop as far from 7 as the hour hand moves past 4; i. e., if the hour hand moves I past 4 the minute hand must stop I' from 7. Then the hour hand will be 5 minutes+l from 3 and the minute hand will be ^+5 minutes from 8. While the hour hand moved -|, the minute hand moved 85 rain — | . . ^ =35 min. — |, whence \'=35 min. .'. 35 fflin.=distance they both move. The second part has been explained in previous problems. I. At what time bet-ween 5 and 6 o'clock is the minute hand midway bet-ween 12 and the hour hand? When is the hour liand midway between 4 and the minute hand? '1. f=distance the hour hand moves past 5. ll-{ A. ^^=distance the minute hand moves in the same time. f-f"25 min.==distance from 12 to the hour hand. |- of (f-|-25 min. )==^-t-12^ min.=distance minute hand moves. 2 4- =i4-12i min. *— i=¥=12i min. IB. 7. \=-i^ of 12-|- min.=|-| min. .8. V*=^24 times If min.=13-jV «'«■ 1. ^^distance the hour hand moves past 5. 2. ^=distance the minute hand moves in the same time 3 f-|-5 min.=distance the hour hand is from 4. 4. f-j-lO min. =2 times (f-|"^ min.)^distance the min- ute hand is from 4, since the hour hand is midway between it and 4. 5. 20 min.+(f+10 ^in.)=f-|-30 min.=distance the the minute hand is from 12. 2/==|+30 min., or ¥— l=V=30 m. 8 ■J=^V °f 30 min.=l^ min. 9. V==24- times 1\ min.^86 min. A. It is 13^V min. past 5 o'clock. B. It is 36 min. past 5 o'clock. {R. H. A., ;p. JfiS, prob. 4I.) Explanation. — Locate the minute hand at 12 and the hour hand at 5. If the hour hand remained stationary, the minute hand would have to moye over -J of 25 minutes, or 12^ minutes. But while it is moving over 12| minutes, the hour hand is moving past 4. Hence, the minute hand will have to move 12| minutes+s of the distance the hour hand moves past 4. Hence V=i-|-12J minutes, as shown by step 5 of A. In B, if the hour hand 'remained stationary, the minute hand would have to move over 30 minutes, i. e., to 6, that the hour hand may be midway between it and 4. But while the minute hand is moving to 6 the hour hand is moving past 4. Hence the minute hand must move twice as far past 6 as the hour hand moves past 482 FINKEL'S SOLUTION. BOOK. 4. But |=distance the hour hand moves past 4; hence, f ^distance the min- ute hand moves past 6. Hence, 4+30 minutes^distance the minute hand moves. .'. ^•'=1+30 minutes, as shown by step 6 oi B. I. At what time between' 3 and 4 o'clock will the minute hand be as far from 12 on the left .side of the dial plate as the hour hand is from 12 on the right side? n. |^:^distance the hour hand moves past 3. ». 2^=12 times |^=distance the minute hand moves in the same time. 2^-|-|=2^=distance they both move. 45 min.=distance they both move. i=-i-^ of 45 min.=l^ min. \*=24 times 1^ min.=4lT^ min. III. .-. It is 41t\ min. past 3. Explanation.— l^ocaie the minute hand at 12 and the hour Jiand at 3. If the hour hand remained stationary, the minute hand would have to move to 9 to be as far from 12 on the left side of the dial plate as the hour hand is from 12 on the right. But while the minute hand is moving to 9, the hour hand is moving past 3. Hence, the minute hand must stop as far from 9 as the hour hand moves past 3. Hence, it is evident, they both move 45 minutes. I. A man looked at his watch and found the time to be be- tween 5 and 6 o'clock Within an hour he looked again, and found the hands had changed places. What was the exact time when he first looked? II. III. (1-) (2.) (3.) (4.) (5.) (6.) (7.) h. was ahead of h. h. , or the dis- h. moved, since it changed place h. [the two observations. h. moved in the time between • ■ "^^ -\-\^^'''^^=^^s,ta.nce they both moved. 60 min.=distance they both moved. .-. \6=60 min. |=distance m. tance the h. with the m. ^^^distance the m. i — -ii (8.) 1. 2. 3. 4. 5. 6. L7. [ahead of h. h.. =2 times 2-j*^ min.=4jS^ min.^distance m. h. was T¥ of 60 min.=2T\ l^^distance vation 2T^==distance h. h. was past 5, at time of first obser- Then [servation. m. h. was past 12 at time of first ob- 25 min.-|-|+4/7 min.=f4-29y\= distance m. h. was past 12 at time of first observation. \*=|+2V^ min.' ■ 29A min. -2 2. 2 'T-J \=^ of 29tV rT.\n.=l^ min. 24- 2 24 times 1-^ min.=32y\ min. It was 32j\ min. past 5 o'clock. ANALYSIS. 183 Exflanation. — It is clear that the minute hand was ahead of the hour hand at the time of the first observation, or else fhey could not have c-x- changed places within an hour. Now, we call the distance from the point where tho hour hand was located at first to the point where the minrte hand was located first, |. But in the mean time the hour hand has moved to the \ position occupied by the minute hand and the minute hand has moved on around the dial to the position occupied by the hour hand, i. c the hour hand has moved -| and the minute 12 times -|, or V. Hence, thej both moved %". They both moved 60 minutes since the band moved on around! the dial to the position occupied by the hour hand an^ the hour hand mov-j ed to the position occupied by the minute hand. ' Y^60 min. as shownj in step (5.) The remaining part of the solution has been explained in pre- vious problems. I. At a certain time between 8 and 9 o'clock a boy stepped into the schoolroom, and noticed the minute hand be- tween 9 and 10. He left, and on returning within an hour, he found the hour hand and minute hand had ex- changed places. What time was it when he first .en- tered, and how long was he gone? II. |=di stance m h. was ahead of the h. h. or dis tance it moved. moved while the h. (2.) (3.) _ (4 ) 60 min ^distance both moved. (5.) .-. V°=60min. (6.) ^=^V of 60 min.=2T\ min. 2^=distance m. h 2^-j-|^\^ ^distance both moved. [f _"5"' moved (7-) 2 9 "2 ^ [was ahead, times 2y\ min.^4y'^ min.==distanee m. h. 1(8.) 1. |.=distance h. h. moved past 8. 2^==distance m. h 40 min 4-| moved in same time. 2 4 -2 2 2 AA 8 ■J -J ^ *^V if ^=^V of 44A ^4=24 times B. III. 4j*Tf min. =|-(-44J^ min. = dis- tance m. h. moved to be 4y^ min. ahead. 2^*^|_-|-44y% min. min. min.=2y|-j min. [past 8. 2y|-j min. ^48ySj^ min.=time ^j^=distance they both moved. 60 min. ^distance they both moved. .-. 2^6,^60 min. 4. ■^=Y^ of 60 min.=2j\ min. [was gone. 5. \*=24 times 2y^ min.^55y-j min.=time he A. It was 48y^:j^ min past 8 o'cldck when he first en- tered school room. B. He was gone 55y\^ min. Suppose the hour, minute, and second hands of a clock turn upon the same center, and are together at 12 o'clock; how long before the second hand, hour hand, and minute hand respectively, will be midway between the other two hands? ^ 184 FINKEL'S SOLUTION BOOK. fA. II. IC. B. i 5. 7. •|=distance the hour hand moves past 12. Then 2^^distance the minute hand m.o^^es past 12, and ti^to==720 times |=distance y Jhe second hand moves past 12. i4^4o_2^4 ^ i_YJ = distance from the minute hand to the second hand. i_4^ _ 2^4^i_4^ ^ distance from the second hand to the hour hand. 2^ — I = %^^= distance from the hour hand to the second hand 9. 10. 11. • 1 2. 3. 4. 5. 6. 7. Tia.L i_4^i_6v|- 1-4^6 _|_2^2=2_^6_4^distknce around tlie dial, indicated 8. 60 seconds=distance around the dial as by one revolution of the s. h. 9. .-. 2_8^5_4==60 sec. 10. i=WoT of 60 sec— t^-|t sec. 11. 1-^-0=1440 times .jff^ sec.=30T\V'V sec. = time when s. h, is midway between the h. h. and m. h. past 12. Then distance the minute hand moves past 12, and 1. ■|=distance the hour hand moves 9 24_ 3. 1-^" = distance the second hand moves past 12. 4. ^^ — |=2^^=distance from h. h. to m. h. , 2^2-^(];gtance from s. h. to h. h., because the h. h. is mid- way betv?een them. [12. 2^ — ^=distance from s. h. to 1-4^0^2^0 ^ ijMLP = distance around the dial. 60 sec.^distance around the dial riG..a. i-- X^BO=QO sec. . 1 -iTB"!!' of 60 sec.=i^^-5 sec, 1^=1440 times ^^ sec.=59H sec.=time when the h. h. is midway between the s. h. and m. h. distance h. h. moves past 12. Then ^^==distance m. ti. moves past 12, and L*^=distance s h. moves T ] past 12. [h. to s. h. 224_|==2^2^jjistance from h. 2^=distance from m h. to s. h. [from 12 to s h. |_|_2^2 _(_ 2^2 = 4^ = distance U^« — 4^6 ^ i_3^4 = distance around the dial. [dial. 60 sec =distance around the Fias. ANALYSIS. 185 III. 9. 10. 11. A. B. C. i8^?*=60sec. oS60 sec.=-ffj sec. i\4 0=1440 times /g<V sec.=61f|f sec.=time past 12 when the m. h. will be midway between the h. h. and s. h. ' ' . The second hand is midwAy between h. h.'and m. 'h. at SOiWt sec. past 12. [at 69^f sec. past 12. The hour hand is midway between s. h. and m. h. The minute hand is midway between h. h. and s. h. at 61f|? sec. past 12. Explanation. — A. We represent the distance moved by the hour hand by I, = the space Th. And since the minute hand moves 12 times as fast as tlie hour hand, it moves \^. ''he second hand moves 60 times as fast as the minute liand or 720 tim^s as fast as the liour liand. From T to h is f and from lio m is \<. .'. From /i to m is 7'm — T/i = ^i — | = \*. From T to f is IV"- .'. From ;« to J = r:!— 2'ot = iv°— ¥ = '¥"• And, by the condi- tion of the problem, the distance from jii to .s = the distance from in to //. .*. Irom 7« to /; = i Y^ + ^ V^ = ^®/^- ^^ have seen, already, that the distance from A to m is '^. .'. The whole distance around the dial is 2\'' + y = ^^*. B. From r to ^ is |. From T to m is V. •'■ From /; to m=Tm—Th= Y — f=V^. By the condition of the problem, the distance from /;.to m^the dSstance from i to /;. :.sT — Th^^ — l=V- From 7' around the dial to the right of j is ^Y°- •'• "^^^ whole distance around the dial='V°+'\''= C' From rto/iisf. From T tomU^. :. From /; to m=\i— |=Y. ' By the condition of the problem, the distance from m to .j^the distance (romAto»2=V- ■.•.From T to j is ^+if+\i=i^. From T around the dial through 2" to j is '^^. .'. The whole distance around the dial is J-Y^ — • 11. WILL PROBLEMS. II. III. A man at his marriage agreed that if at his death he should leave only a daughter, his wife should have f of his estate, and if he should leave only a son she should have ■^. He left a son and a daughter. What fractional part of tlie estate should each receive, and what was each one's portion, if his estate was worth $6591? ^=daughter's share. f=wife's, share. 1=3 times J=aon's share. -^=?^8^whole estate. 5. $6591=whole estate. 6. .-. i^=$6591. [estate. 7. i=j\ of $6591=$507=daughter's share,=^ij of whole 8. 1=3 times $507=$1521=wife's share,=Taj of whole es- , tate. [tate. 9. 1=9 times $607=$4563=son's share,=J^ of whole es- ( $507=x-j of whole estate=daughter's share. .•. < $1521=-^^ of whole estate=wife's share. ( $4563=1% °^ whole estate^son's share. (Milne's Prac. A., p. 862, prob. 74..) 186 FINKEL'S SOLUTION BOOK. Note. — This class of problems originated under the Roman laws of inheritance. The following problem is quoted in Cajori's A History of Mathe- matics, p. 79 : A dying man wills that, if his wife, being w,ith»child, gives birth to a son, the son shall receive f and she J of his estate; but, if a daughter is born, she shall receive J and his wife f . It happens that twins are born,- a boy and a girl. How shall the estate be divided so as to satisfy the will? It is said that the celebrated Roman Jurist, Salvianus Julianus, decided that the estate shall be divided into seven equal parts, of which the son receives four, the wife two, and the daugh-| ter one. It is according to this decision that the above solution is made. Should such a will come into court in this age, it would very likely be set aside. For a valuable critique, by Marcus Baker, U. S. Coast Survey, on this class of problems, see School Visitor, Vol IX, p. 186. 11.^ A gentleman, dying, divided f 5,100 among his three sons,^: whose ages were 9, 11, and 17 respectively, so that the different shares, being on interest at 5%, would amount to equal sums when they became of age; what were the shares? (Brooks' Int. Arith., p. id/, prob. j.) 1. -51^=5%, the amount any principal increases per year at 5%. - 2. I^the amount any principal increases in 12 years at 5%. 3. |+|=|=the amount of the youngest son's share when he becomes of age. In like manner, 4. |.+J=^|=the amount of fhe next youngest son's- share,. and 5. |+^=-|^the amount of the oldest -son's share. •- 6. ." . f of youngest son's principal=f of the next young- est son's principal=f of oldest son's principal, since the amourits are all equal. 7. \ of youngest son's principal=|^ of f , or -f^, of the next youngest son's principal=^ of | , or-^, of the oldest son's principal. 8. f, or the youngest son's principal, =5 Xy^, or ff, of the next youngest son's principal=5X^\=f of the oldest son's principal. 9. .•. ^1 of the principal of the next youngest son^the principal of the youngest son. 10. .". The principal of the next youngest son^-^ of the principal of the youngest son. 11. f of the principal of the oldest son^the principal of the youngest son. 12. .". The principal of the oldest sou^| of the principal , of the youngest son. 13. II =the principal of the youngest son, - ANALYSIS. 187" 14. If =the principal of the next youngest son, and 15. If =|^the principal of the oldest son. 16. .■. ||+-^f+|g=f^==the sum of their principals. 17. $5,100^the .sum of their principals. 18. .-. |i=:$5100. 19. tV=5S: of $5100=$1000. 20. i|=15X$100=$1500, the youngest son's share. 21. •i;|=16X|100=|l600, the next youngest son's share, and , ,22. f|-=::20X|100=$2000, the oldest son's share. C |l,500:=the youngest son's share, III. . . -< $l,600^the next youngest son's share, and (. i2,000=the oldest son's share. 12. COACH PROBI^EMS. ' I. How far may a person ride in a coach, going at the rate of 15 miles an hour, provided he is gone only' 10 hours and walks back at the rate of 12 miles per hour ? 1. 15 miles^the distance he can ride in 1 hour. 2. 1 mile =the distance he can ride in -^j- hotir. 3. 12 m'iles=the distance he can walk in 1 hour. 4. 1 mile ^the distance he can walk in -^ hour. 5. .•. T^g- hour+^hour=T^ijhour=the time it takes him to ride 1 mile -and walk back. 6. .-. 10 hours^the time it takes him to ride 10^-^, or 66f miles, and walk back. III. .-. He can ride 66f miles. I. Ten men hire a coach to ride to Columbus, but by taking in 5 more persons the expense of each is diminished by I J ; what did the coach cost them ? ' 1. -Jf =the amount paid for the coach. 2. y\=the amount each man would have paid had only 10 men paid for it. 3. jJg-^the amount each man paid for it since there are II. -^ 15 men. 4. .•. -^-^ — ^=^=the amount each man saved. 6. $-^^the amount each man saved. , 6. .■. -jijf of the cost of the coach^f^. L7. .-. If, or the cost of the coach,=30XlJ=$6.00. III. . . The coach cost |6. I. Two men, A and B, in Circleville, Ohio, hire a coach for |10, to go to Columbus and back, the distance being 30 miles, with the privilege of taking in three more per- sons. Having gone 10 miles, they take in C ; at Co- lumbus they take in D ; and when within 10 miles of II. 188 FINKEL'S SOLUTION BOOK. Circleville they take in E. How much shall each man pay? .- , , 1. 60 miles=distance A rode, 2. 60 miles=distance B rode, 3. 50 miles^=distance C rode, 4i 30 miles=distance D rode, and 5. 10 miles=distance E rode. II. i 6. . 210 miles=total distance ridden by the 5 men. 7. .-..-^'Vof $10— $2|=amouHt A must pay, 8. •j'VVof |l0=$2|=amount B must pay, 9. ^VV of |10=$2^i=amount C must pay, 10. -jajO^ of |lO=$lf=amount D must pay, and .11. ^1% of $10=$^^=:amount E must pay. |2-f =am,ount A must pay, |2f ^amount B must pay, III. .'. ■{ |2-j8jz= amount C must pay, 14 ^amount D must pay, and 1^4 =amount E must pay. Remark. — ■ The above solution is based upon the principle that a ■passenger pajfs in proportion to the distance he rides, and this it seems to me is the proper view to take of the problem. The following solution based on the principle that the expenses should be borne by those pas- sengers who are responsible for the coach at any time during the jour- :ney has its advocates. Second solution. 1. 60 miles=the whole distance traveled. ' 2. $10=the whole cost. ' 3. $^=$10H-60=cost per mile. 4. $JB<i=10X$^=cost for' the 10 miles, which A ^nd B should share equally. 6 . $-V^=20X$^=cost for the next 20 miles, and this should be shared by A, B, and C equally. 6. |-Vi=20X|^=cost for the first 20 miles on the return trip, and should be shared equally by A, B, C, and D. II. ^ 7. '$-V^= 1 OX$^=cost for the last 10 miles, and should be shared equally by A, B, C, D, and E. 8. |3i=$f +$i^+$|+$f=i of IV-+* of f-V-+i of $^+i off-i^^the amount A should pay. 9. $3^^the amount B should pay. 10. $2T5^=$Y+$f+$*H of $-¥^+i of p^+i of $J,|i= amount C should pay. 11. fl^=$f+$f=i of $V-+i of $V=amount D should pay- 12. $J=i of $^=amount E should pay. III. .-. A should pay $3^ ; B, $3^ ; C, $2^8 ; D, $1^ ; and E, $J. ANALYSIS. 189 11.-^ ni. n. III. Eight men hire a coach; by getting 6 more passengers, the expenses of each were diminished $1|; what do they pay for the coach? ' 1. -I^amount paid tor the coach. [been only 8 men. 2. •|=amount 1 man would have had to pay, had there 3. ^Ij^amount 1 man paid since there were 8 men-|-6 men, or 14 men. 4. ••• i — rV— TS — ^=w="what each saved. 5. $l|^what each saved. ^ 6. .-. ^=$lf , , 7'. ^=f of $l|=$i^, and ^8. -11=56 times $^^:=$32|=amount paid for the coach. .•. $32f=amount paid for the coach. (Ji. H. A., p. ^OS,prob. 48,) Second solution. $l^=amount saved by each man. [the six meu. $14^=8 X$l|==amount saved by the 8 men and paid by .•. $2-J-=-^ of $14=amount paid by each of the 14 men. ,•. $32f=14 times $2-J^amount they paid for the coach. . They paid $32f for the coach. 13. CUP AND COVER PROBLEMS. Under this head comes a class of problems that may be called, "Watch and Chain Problems," "Horse and Saddle ProlDlems," etc, L A lad,y has tiiro silver cups, and only one cover. The first cup weighs 8 ounces. The first cup and cover weighs 3 times as much as the second cup; ai^d the sec- ond cup and cover 4 times as much as the first cup.. I What is the weight of the second cup and the cover? \ 1. 3 times Weight of second cup=w eight of cover-[-weight of first cup, or 8 oz. [2|- oz. 2. 1 times weight of second cup=|- of weight of cover-J- 3. ■|=weight of cover. Then 4. ■J-|-2|- oz.=weight of second cup. [cover. |-[-i+2f oz.^|+2t oz. = weight of second cup and 32 oz.=4 times 8 oz.^weight of second cup and cover,. by the conditions of the problem. ••• l+2f oz.=32 oz. 1=32 oz.— 2f oz.=29^ oz. |=i of 29i oz.=i7i oz. __ f=3 times 7-J- oz.=22 oz.=weight of cover. [cup. 11. ■|-|-2f oz.=7| oz.-}-2| oz.= 10 0^;.= weight of second II.J 5. 6. 7. 8. 9. 10. III. and weight of second cup. S22 oz.=weight of cover, 10oz.= 190 ,FINKEL'S SOLUTION BOOK. I. II. III. A man has two watches, and a chain worth $20; if he put the chain on the first watch it will be worth § as much as the second watch, but if he put the chain on the sec- ond watch it will be worth 2f times the first watch what is the value of each watch ? 1. f s.=f f.+$20. . ' ' 2. i s.==i of (I f.+$20)=i f.+$10. f s.=3 timefe {i f.+$10)=| £+$30. [lem. 1^ s.=^-^ f — $20, by the second condition of the prob- .-. V f.— $20=f f.+$30, whence V f.— f f.=$30+$20, or I f.=$50. i f.=| of $50=$10, and f f.=4 times $10=$40=value of first watch. , I s.=f f.+$30=f of $40+$30=$90=value of the sec- ond watch. ' f , :value of first watch, and i=value of second watch. ( White's Comp. Ariih.,f. 2J^3, prob. 60.) 3. 4. 5. 6. 7. 8. 9. 10. II. 311. 1. 14. DINING, AND CHESS PRpBI,EMS. - A, B, and C dine on 8 loaves of bread ; A furnishes 6 I loaves ; B, 3 loaves; C pays the others 8d. for his share; how must A and B divide the money ? ' 1. 8 Ioaves=what they all eat. 2. 2f loaves=what each eats. 3. .•. 5 loaves — 2f loaves=2-J- loaves=what A furnished towards C's dinner. 4. .•. 3 loaves — 2f loaves=-J loaf=what B furnished to- wards C's dinner. 5. 6. ^=|^A's share, and ^=^=B's share. 7. |- of 8d.=7d.^what A should receive, and 8. \ of 8d.=ld.=what B should receive. A should receive 7d., and B should receive Id. {R. H. A.,f. IfiS, frob. Ji2.) B at a game of chess lost |18, and then Won J as much as he had remaining, and then had \ as much as he had at first; how much had he at first? / 1. |=his money at first. 2. ^ — $18=his money after losing \ ANALYSIS. 191 II. III. I— $6=J- of ( I — $18 )=amount won. I — |l8+f — $6=f— $-24=amount he had after winning. f ^amount he had after winning. .■.f-f24=^, i=i of m=4ih f=6X$4f=$28f=amount he had at first. B had $28f at first. 15. PARTNERSHIP PROBLEMS. I. II. III. A and B enter into partnership and gain f of the stock, lacking |10, and gains |175; the whole stock and share of each. A owns required i6i.) (Brooks' Int. Arithmetic, p 1. f =whole stock. 2. I — $10=what A owns. 3. .-. |— (f— $10), or J+$10=what B owns. Now, 4. whole stock : whole gain=A's stock : A's gain ; or 5. I : $240=1— $10 : $175. 6. . . 240( I— $10 )=175Xf , or 7. i|ii_$2400=4^, 8. '2^-=$2400, 9. i=^ of $2400=1120, 10. |=4XH20=$480=whoIe stock. 11. I— $10=$.360— $10=$350=A's stock. tl2. i+$10=$120-|-$10=$130=B's stock. |480=whole stock, |350=A's stock, and |13p=B's stock. A and B hired a pasture for 4 month§ for |39. At the be- giiining of the first month A put in 5 cows and 3 sheep and B put in 3 cows and 5 sheep ;- at the beginning of the second month A put in 5 sheep and took out 2 cows and B put in 5 cows and took out 2 sheep. What part of the cost of'the pasture should each pay, if a cow eats ^s much as two sheep ? 1. 5 cows-|-3 sheep is equivalent to 10 sheep-|-3 sheep, or 13 sheep. 2. 3 cows+5 sheep is equivalent to 6 sheep-f-5 sheep, or 11 sheep. 3. 5 cows — 2 cows+3 sheep-|-5 sheep is equivalent to 14 sheep for 3 months. 4. 14 sheep for 3 months is equivalent to 42 sheep for 1 month. 192 FINKEL'S SOLUTION BOOK.. 3 cows-J-5 cows+5 sheep — 2 sheep is equivalent to Ift II. ^ sheep for 3 months. 6. 19 sheep for 3 months is equivalent to 51 sheep for 1 month. 7. 13 ^heep for 1 month-)-42 sheep for 1 month is equiv- alent to 55 sheep for 1 month. 8. 11 sheep for 1 month-(-51 sheep for alent to 62 sheep for 1 mortth. 9. 55 sheep for 1 month+62 sheep for alent to 117 sheep for 1 month. 10. .-. $18^T¥r of |39=amount A should pay, and n. $20f=T\\ of f39=amount B should pay. III. I. II. III. month is equiv- month is equiv- . I A should pay $18J, and ■ I B should pay |20f . A, B, and C hire a pasture for $63. A puts in 6 cows, B puts in 18 horsey, and C 48 sheep; how much should each pay, if a cow eats as much as 2 horses, and a horse as much as 4 sheep ? / 1. 1 horse eats as much as 4 sheep. 2. 2 horses eat as much as 8 sheep. 3. . . 1 cow eats as much as 8 -sheep, since 1 cow eats as much as 2 horses.' \ 4. 6 cows eat as much as 6X8 sheep, or 48 sheep. ' 5. 18 horses eat as much as 18X4 sheep, ot 72 sheep, since 1 horse eats as much as 4 sheep. 6. .-. They all put in the equivalent of 48 sheep+48 sheep +72 sheep, or 168 sheep. 7. .-. A, who put in the 'equivalent of 48 sheep, should pay tW of $63=f 18, 8. B, who put in the equivalent of 72 sheep, should pay -^W of |63=$27, and 9. C, who put 'in 48 sheep, should pay yW of |63= f 18=amount A should pay, f27:=amount B should pay, and . $18=amount C should pay. .16. COMBINATION PROBLEMS. r If 62 lb. o"f sea-water contain 2 lb. of salt, how much. salt must be added so fhat 42 lb. of sea-water will contain 2 lb. of salt? {Brooks' Int. Arith., p. 144.) 1. 62 lb.— 2 lb.=60 lb. ^the' quantity of water in the first mixture. 2. 42 lb. — 2 lb.=40 lb.=quantity of water in second mix- ture. ANALYSIS. 193 II. < III. .-. 2 lb.:=quantity of salt to be added to 40 lb. of water to make a mixture of 42 lb. 1 lb.=iquantity to be added to 20 lb. of water. 3 lb.=quantity to be added to 60 lb. of water. . . 3 lb.— 2 lb.=l lb added to the (52 lb. contain 2 lb. of salt. 1 lb. of salt tnust be added. quantity of salt' that must be of sea- water so that 42 lb. will II. III. In a mixture of silver and copper, consisting of 60 oz., there are 4 oz. of copper ; how much silver must be added that there may Idc J oz. copper in 6 oz. of the mixture ? '1. J oz.=what 6 oz. of the new mixture contains. 2. 1 oz.=what 18 oz. of the new mixture contains. 3. 4 oz.=:what 72 oz. of the new mixture contains. 4. . . 72 oz — 4 oz.— 68 oz.=the quantity of silver in 72 oz. of the new mixture. 5. .-. 68 oz. — (60 oz. — 4 oz.)=12 oz.=the quantity of sil- ver that must be added, in order that 6 oz. of the mix- ture will contain ;^ oz. of copper. . . There mtist be added 12 oz. of silver. . ' • 17. DITCH PROBLEMS. I. A and B dig a ditch 100 rods long for flOO ; how many rods does each dig, if they each receive f50, and A digs at 1.75 per rod, and B at |1.25 ? There has been a vast amount of quibbling about this problem; but a few moments consideration should suffice to settle all dis- pute, and pronounce upon it the' sentence of absurdity. We have given, the whole amount each received and the amount each received per rod. Hence', if we divide the whole amount each received by the cost per rod, it must give the num- ber of rods he digs. But by doing this we receive 50-i-.75, or 66f rods, what A digs and 50--i-l-25, or 40 rods, what B digs, or 106f rods which is the length of the ditch, and not 100 rods as stated in the problem. The length of the ditch is a function of the cost per rod and the whole cost, and when they are given the length of the -ditch is determined. We might propose a problem just as absurd by requiring the circumference of a circle whose area is 1 acre, and diameter 20 rods. Since the area and circumference are functions of the diameter, when either 194^ FINKEL'S SOLUTION BOOK. of these are given, the other is determined and should not be limited to an inaccurate statement. If, in the original problem, A's price per rod increases at a constant ratio so that when the ditch is completed he is receiving $1 per rod, and B's price constantly decreases until when the ditch is completed he is re^ceiving $1 per rod, then the problem is solvable, and the result is 50 rods each. I. Two men, A and B, undertake to dig a ditch 100 rods long for $100. A's end of the ditch Being more difficult to excavate than B's, it was agreed that A should receive 25 f per rod more than B. How many rods must each dig in oTder that each, may receive $50? ■ 1. I/Ct 2.?=|100, «=100, the number of rods, and ^=.25, the number of dollars in the difference of the price each is to receive per rod. 2. lyCt x^=number of rods A must dig. Then • 3. n — ^;c=number of rods B must dig. 4. Then — ^=price A receives per rod, and s 5. =price B receives per rod. n — X II. 6. . . — '—=d. X n — X ■ d the number of rods A must dig, and +^=50(5-/17), n-^x=n- -v^-l^)\^y^^=\- !/ J Y V « V .r _ 100 ^ )/ 50 \^ V 100 V — — =50( V 17 — 3 ) , the number of rods B must dig. III. .-. A must dig 50(5— i/l7) rods and B must dig 50(i/l7 —3) rods. Note. — From this algebraip solution, we can derive a rule by which all problems of this class may be solved by arithmetic. Rule. — 1. Divide half the whole cost by the difference of the price per object and to the square of the quotient add the square of half the whole number of objects. 2. Subtract the square root of the sum from half the whole number of objects increased by the quotient of half the whole cost divided by the dif- ference of the price per object, for the number of objects at the higher price per object; and the square root of the sum subtracted from half the number of objects diminished by the quotient of half the whole cost gives the » mber of objects at the J-esser price per object. ANALYSIS. 195 No arithmetical solution of such problems without the aid of Algebra 3s possible. 18. PASTURE PROBLEM, I. If 12 oxen eat up 3^ acres of pasture in 4 weeks, and 21 oxen eat up 10 acres of like pasture in 9 weeks ; to find how many oxen will eat up 24 acres in 18 weeks. II. 1. 10 parts (say)=what one ox eats in a week. Then 2. 120 parts=12XlO parts='what 12 oxen eat in 1 week, 3. 480 parts^4xl20 parts^what 12 oxen eat in 4 weeks. 4. .•. 480 parts^original.grass-)-growth of grass on 3^ A. in 4 weeks. 5. 144 parts=^ of 480 parts=original grass-|-growth of grass on 1 A. in 4 weeks. 6. 210 parts=21XlO parts^what 21 oxen eat in 1 week, 7. 1890 parts=9 X210 parts=what 21 oxen eat in 9 weeks. 8. .'. 1890 parts=original grass-j-growth of grass on 10 A. in 9 weeks. 9. 189 parts=jl0^ of 1890 parts^original grass-j-growth on 1 A in 9 weeks 10. ••. 189 parts — 144 parts=45 parts=growth on 1 A. in 9 weeks — 4 weeks, or 5 weeks. 11. 9 parts=^ of 45 parts=growth on 1 A. in 1 week. 12. 36 parts^4x9 parts^growth on 1 A. in 4 weeks. 13. ••. 144 parts — 36 parts=108 parts=original quantity of grass on 1 A. 14 2592 parts=24xl08 parts=original-quantity on 24 A. 15. 216 parts=24;><9 parts=growth on 24 A. in 1 week. 16 3888 parts=18x216 parts=growth on 24 A. in 18 weeks. 17. .-. 2592 parts-|-3888 parts=6480 parts=quantity of grass to be eaten by the required oxen. 18. 180 parts^l8X10 parts^what 1 ox eats in 18 weeks. 19. .- 6480 parts=what 6480-^-180, or 36 oxen eat in 18 weeks. III. .-. It will require 36 oxen to eat the grass on 24 A. in 18 weeks. Note. — This celebrated problem was, very probably, proposed by Sir Isaac Newton and published in his Arithmetica Universalis in 1704. Dr. Artemas Martin says, "I have not been able to trace it to any earlier work." For a full treatment of this problem see Mathematical Magaiine, Vol. 1, No. 2. II. 196 . FINKEL'S SOLUTION BOOK. 19. INVOI^UTION AND EVOIvUTlON PROBI.EMS. I. If f of the number of trees in an orchard be squared, the result will be 144 ; how many trees in the orchard ? ' 1. |=number of trees in the orchard. 2. ( |X| )2=^.X( I )*=the square of f of the number. 3. 144=the square of f of the number. 4. .■.^%X( 1)2=144,. 5. (f)2=144^^V=900, 1 6. %=t/ 900^30, the number of trees in the orcharci. III. .". 30=:number of trees in the orchard. I. Tw/o-thirds of the cube of^a number is 10 more than the- cube of f of the number ; what is the number ? 1. -|=the number. 2. |(|)»=f of the cube of the number. 3. (fXf )^==the cube of f of the number. II. O. . ■ . f ( I r~{ fXf )^=10, or t( I Y-^^{ I )='=10.^ 5. if(|)=*=10. 7. (1)^=10-^=27. .7. .-. (|)=ir27=3, the number. III. .-. The number is 3. 20. SOI.UTIONS OF MISCELLANEOUS PROBLEMS. I. Two men start from two places 495 miles apart, and travel toward each other ; one travels 20 miles a day, and the other 25 miles a day ; in how many days will they meet? 1. |^=number of days. 2. 20 mi. ^distance first travels in 1 day. 3. f X20 mi. ^distance first travels in f days. 4. 25 mi.=distance second travels in 1 day. 5. f X25 mi.=distance second travels in f days. 6. .-. |X20 mi.+^x25 mi.=f X(20 mi.+25 mi. )=distancfr both travel. 7. 495 mi.^distance both travel; 8. .-. (20 mi.+25 mi.)x|=(45 mi.)X|=495 mi. Whence .9. |=495-;-45=ll=number of days. III. .-. They will- meet in 11 days. Second solution. (1. 20 miles=distance first travels in a day. 1 2. 25 miles=distance second travels in a day. 1 3. .•. 45 niiles=distance both travel in a day. [days. 14. .-. 495 miles=distance both travel in 496-7-45, or 11^ II. II. ANALYSIS. 197 III. .-. They will meet in 11 days. Third solution — the one usually given in the schr^-lro^m. 4 20+25=45)495(11 days. 45 45 45 I. Find a niimber whose square root is ife times its cubi II. )e roo'. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. ==square root of the number. Then X|=the number, because the square rootxthe>' uare root equals the number. =the cube root of the number. Then XfXf=the number. But =5x(f)- Hence, squaring both sides, , X|=25x(|Xf). But Xf=the number, and Xf Xf=the number. • ■jXf X5=25x(^X^)- =25. . (3)8=25'=15625. Dividing by (fXf), III. .-. The number is 15625. {R. H. A., p. 367 , prob. H.) I. A man bought a horse, saddle and bridle for $150 ; the cost of the saddle was \ of the cost of the horse, and the cost of the bridle was ^ the cost of the saddle; vyhat was the cost of each? 11. (-1. J-|=cost of the horse. Then 2. ^\ — \ of ■l-|=cost of the saddle, and ■=\ of Y^^^cost of the bridle. 4. i|=|f+T\+TV-=costofall. 5. $150=cost of all. 6. .-. i|=$150, and 7. J5.=Jj of $150=$10=cost of bridle, 8. f|=12 times .$10=$120=cost of horse. l9. •j-\=2 times $10=$20=cost of saddle. ( $10=cost of the bridle, III. .". < $20=cost of the saddle, and ( $120^cost of the horse. ( White's Comp. A., -fi. S^l, proh. S9.) 198 FINKEL'S SOLUTION BOOK. I. A boat is worth $900; a merchant owns f of it, and sells J of his share ; jvhat part has he left, and what is it worth? {(-1. |==part the merchant owned. A. < 2. ^ of |=^5_^^part he sold. » » l3. ... |_^5^_i|_^5^=|o__5^_part he had left, rl- $900= value of |f, or the whole ship. T, J 2. $75=iV of $900=value of J^ of the ship. 1 3. $375=5 times $75=value of ^ of the ship, i.^ part [ he had left. „_ ( y^^=part he had left, and "^" •'■ I $375=value of it. I. A and B were playing cards. B lost $14, which was -J-y. times f as much as A then had ; and when they com- menced, f of A's money equaled ^ of B's. How mucb had each when they began to play? ( 1. ) I of A's money=f of B's. (2.) I of A's money=i of f=-^ of 'B's. (3.) I of A's money=8 times -^=^ of B's. (4.) |-|=B's money when they began to play. Thea (5.) ^|-=A's money when they began. 1. i-|=A's money after winning' $14 from B. 2. $14=what B lost. 3. y\ times ■|=jij=part A's money is of $14. 4. .-. Jj=$14, 5. Jj=i- of $14=$2, and [$14 from B. 6. Yf=15 times $2:^$30=A's money after winning (7.) ' .-. $30— $14=$ 16= A's money at first. (8.) .-. i|=$16, from (5), (9.) -^=tV of $16=$1, and (10.) 11=35 times $l=$35=B's money at first. $16=A's money at first, and $3.5=B's money at first. (Siod. Int. A., p. Ill, f rob. SO.) II. (6.) III. A drover being asked how rnany sheep he had, said, if to ■J of my flock you add, the number 9^, the sum will be 99^; how many sheep had he? 1. ^:^the number of sheep. 2. ^-|-9^=^ of the number-|-9-^. 3. 99^=^ of the number+9-i. ' 4. .-. i+9i=99i or 5. -^=99-1— -9^=90, and 6. f=3 times 90=270=number of sheep. III. .-. He had 270 sheep. II. 11. ANALYSIS. 199 Heman has 6 books more than Hand ford, and both have 26; how many have each? 1. ■|==number Handford has. Then 2. ■l-f'S^Heman's number. 3. f-f l+6=|+6=number both have. i. 26^number both have. 11.^ 6. .-. 1+6=26 or 1=26—6=20. ^=^ of 20=5, and ■|=2 times 5^10^Handford's number. |— |-6=lD==Heman's number. TjT { Handford iiad 10 books, and •'• I Heman had 16 books. (Stod.Int.A.,j>.116,frob.2.) I. A man and his wife can drink a keg of wine in 6 days, and the man alone in 10 days ; how many days will it last the woman ? ' 1. 6 days^time it takes both to drink it. 2. -|-=part they drink in one day. 3. 10 days=time it takes the jnan to drink it. 4. y^^part he drinks in one day. [day. 5. .•. \ — tV^A — tei^^tV^^^P^-''^ ^^ woman drinks in one 6. -^|-=what the woman drinks in -i-|-f-J-j^l5 days. III. .'. It will take the woman 15 days. {R. Alg. /.,/. 112,prob. 69.) I. A man was hired for 80 days, on this condition :_ that for every day he worked he should receive 60 cents, and for every day he was idle he should forfeit 40 cents. At the expiration of the time, he received $40. How many days did he work? (-1. $.60==what he receives a day. 2. $48=80 X$.60^w hat he would have leceived had he worked the whole time. 3. $40=what he received. II.M. .■. $48— $40=$8=what he lost by his idleness. 5. $1^.60, his wages,-|-$.40, what he had to forfeit,= 'what he lost a day. ^ 6. .-. $8^what he lost in 8-Hl, or 8 days. ,7. 80 days — 8 days=72 days, the time he worked. III. .-. He worked 72 days. I. A ship-mast 51 feet high, was broken off in a storm, and •| of the length broken ofT, equaled f of the length re- maining; how much was broken off, and how much re- mained? 2 200 FINKEL'S SOLUTION BOOK. II. III. 1. f of length broken off=f of length remaining, 2. i of length broken off=^ of f=| of length remaining, 3. f of length broken ofr:=3 times %=^ of length remain- ing. 4. f^length remaining. 5. ■|=length broken off. 6. |-f-|=V=who]e length. 7. 51 feet=whole length. 8. .-. V=51 feet, 9. i=iV of 51 feet=3 feet, and 10. -1=8 times 3 feet=24 feet, length remaining. 11. 1=9 times 3 feet=27 feet, length broken off. 24 feet==length remaining, and 27 feet=length broken off. II.<^ III. A boy being asked his age, said," "4 times my age is 24 years more than 2 times my age;" how old was he? ■|^his age. :f^4 times his age. |-=2 times his age. f=f-|-24 years or 5. I — 1=1=24 years. 6. i^i of 24 years==6 years, and 7. -1^2 times 6 years=12 years, his age. .-. He is 12 years old. (S^od. Int. A., f. 113, frob. 16.) 4xf= 2X1= II. <^ III. If 10 men or 18 boys can dig 1 acre in 11 days, find the number of boys whose assistance will enable 5 men to dig 6 acres in 6 days. 1. 1 A.=what 10 men dig in 11 days. 2. yV A.=what 1 man digs in 11 days. 3. YTTT "^-^TT of tV A.^what 1 man digs .in 1 day. 4. Tf\ A.=y^ A.:^5 times ^^^ A.=what 5 men dig in 1 day. [days. ■^ A.^^\ A.^6 times -^ A.=what 6 men dig in 6 5. 6. 7. 8. 9. 10. 11. 6 A.- A.=5i-'i A.=what is to be dug by the boys in 6 days. 1 A.=what 18 boys dig in 11 days. Jj- A.==what 1 boy digs in 11 days, y^ A.^yij- of y\ A.=whkt 1 boy digs in 1 day. ■^ A.=yf^ A.==6 times j-rg- A.^what 1 boy digs in 6 days. h^ A.=what 5 /i ■: ^^ , or 189, boys dig in 6 days. It will take 198 boys. (/?. 3d p., O. E.,p. 318,prob. 66.) , ANALYSIS. 201 i A man after doing |^ of a piece of work in 30 days, calls an assistant; both together complete it in 6 days. Iti what time could the assistant complete it alone? 1. ^=part the man does in 30 days. ' 2. -5V=¥V of f=part he does in I'day. 3. ■5=3 — ^f=part he and the assistant do in 6 days.- 11.^ 4. -1^=^ of f^^P^'^'- ^^ ^"'^ ^^^ assistant do in 1 day. 5. .•. Yi — sV^TTC — rfir^TT'B'^P^^'* ^^^ assistant does in 1 day. 6. ,|;|^=part the assistant does in l |g : ^^^ =21-^ days. III. .'. It w^ill take the assistant 21f days. (H. 3d p., O. B.,p. 318, f rob. 71.) Bxj>lanai- on : — Since the man does % of the work before he called on the -assistant, there remains | — |^, which he and thp assistant do in 6 daj s. Hence they do \ off, or ^ of the work in one day. If the man and his ■assistant do ^ of the work in 1 day and the man does j^ of the work in 1 •day, the assistant does the difference betweenjij and ^ which is j|g of the work in 1 day. Hence It will take JiS^^ioOi or21f days, to do the work. L A person being asked the time of day, replied that it was past noon, and that f of the time past noon was equal to I of the time to midnight. What was the time of day? 1. |- of the time past noon=|- of the time to midnight. 2. 1^ of the time past noon^| of \==^ of the time to mid- night. ' [midnight. 3. f, or the time past noon,=4 times \=\ of the time to 4. -1=11 me to midnight. Then II. <! 5. |^=time past noon. 6. -l-j-l^l^time from noon to midnight. 7. 12 hours=time from noon to midnight. 8. .•'. 1=12 hours, 9. \-=\ of 12 hours=l-J hours, and [past noon. 10. 3=4 times 1^ hours=5^ hours=5 hr. 20 min., time III. .-.It is 20 min. past 5 o'clock, P. M. {Milne's Prac. A., p. 860, prob. 4.7.} Note. — From 3, we have the statement that the time past noon is | of the 'time to midnight. Hence, if | is the time to midnight, ^ is the time past ooon or if \% is the time to midnight, -^ is the time past noon. I. A person being asked the time of day, said that ^ of the time past noon equals the time to midnight. What is the time of day ? 202 FINKEL'S SOLUTION BOOK. '1. ^:^tiine past noon. Then n.<^ III. time to midnight. i^-pi^*/=time from noon to midnight 12 hours^time from noon to midnight. 12 hoprs. Jj of 12 hours=l hour, and 1 2_ 1. Jj-=1 times 1 hour= It is 7 o'clock P. M. =7 hours=time past noon. II. III. L man being asked the hour of day, replied that ^ of the time past 3 o'clock equaled -J- of the time to midnight; what was the hour? * ^ of the time past 3 o'clock=-^ of the time to nnidnight. |, or the time past 3 o'clock,=4 times •J=|- of the time to midnight. f=time to midnight, f^time past 3 o'clock. |^-|-|^=f=time from 3 o'clock to midnight. 9 li'ours=time from 3 o'clock to midnight. '. .-. f=9 hours. .^=1- of 9 hours^l^ hours, and ^■=\ times \^ hours=6 hours=time past 3 o'clock. f-[~3 hours=9 hours, time past noon. It is 9 o'clock, P. M. {Brooks' Int. A.,f. 156, f rob. 17.) I. A r 1. 2. 3. 4. 5. 6. 7. II. <^ III. I. A person being asked the hour of day, replied, f of the time past noon equals | of time from now to midnight -{-2f hours; what was the time? § of time past noon=|^ of time to midnight-j-2f hours. I of time past noon=^ of (f-(-2f hours )=^ of time to midnight-f-H hours. [to midnight-)-4 hours. |, or time past noon,==3 times (^-|-1-J hours )=-J of time ^=time to midnight. |-|-4 hours=time past noon. [night. |-(-^-|~4 hours=|^-(-4 hours=time from noon to mid- 12 hours=time from noon to midnight. .-. f-(-4 hours=12 hours. |==12 hours — 4 hours=8 hours, i=i of 8 hours=2 hours, and |-|-4 hours=6 hours=time past noon. It is 6 o'clock, P. M. ' (Siod. Int. A., p. 128, Proh. 89.) father gave to each of his sons $5 and had $80 remain- ing; had he given them $8 each, it would have taken all his money; required the number of sons. ANALYSIS. 20S lU 1. 2. 3. U. ni. $8=amount each received by the second condition. $5=atnount each received by the first condition. $3==$8 — $5^ excess of second condition over first, on each son. [10 sons. .-. $30^excess of second condition over first, on 30-5-3, or . Tliere were 10 sons. 11. II. II If 50 lb. of sea water contain 2 lb. of salt, how much fresh water must be added to the 50 lb. so that 10 lb. of the new mixture may contain ^ lb. of salt. 1. -J lb. of salt^what 10 lb. of the new mixture contains. 2. f , or 1, lb. of salt=what 3 times 10 lb., or 30 lb., of the n'ew mixture contain. [mixture contain. 3. 2 Ib.-of salt=what 2 times 30 lb., or 60 lb., of the new 4. .-. 60 lb. — 50 lb.=10 lb.=quantity of fresh water that must be added. .■. 10 lb. of fresh water must be added that 10 lb. of the new mixture may contain ^ lb. of salt. A farmer had his sheep in three fields, f of the number in the first field equals f of the number in the second field, and |- of the number in the second field equals J of the number in the third field. If the entire num- ber was 434, how many were in each field? f (10 (2.) (3.) (4.) (5.) (6.) (7-) (8.) (9.) r |- of number in first field=-| of number in second field. [second field. ^ of number in first field^-^- of ■!=§ of number in -|, or number in first field,=3 times f=f of number in second field. I of number in second field=f of number in third field. [in third field. ■| of number in second field=-| of |=f of numbei I", or number in second field,=3 times -|=-| of num- ber in third field. |=number in third field. Then . -|=number in second field, and |^=f of number in second field^number in first field in terms of number in third field. ^i-^M-*l+li=W=n"mber S 1 9 three fields. the 434=number in the three fields. 2^1^=434, 10.) 11.) W12.) fs=^rT of 434=2, and [field. ff=64 times 2=128=number of sheep in third ^==72 times 2^144=number of sheep in second field. [field. 1^=81 times 2=162=number of sheep in first 204 FINKEL'S SOLUTION BOOK. ( 162=number of sheep in first field, III. .•• < 144=number of sheep in second field, and ( 128=number of sheep in third field. {Milne's Prac. A.,f. 362, f rob. 68.) I. In a certain school of 80 pupils there are 32 girls ; how many boys must leave that there may be 5 boys to 4 girls? '1. 80=whole number of pupils. 2. 32^number of girls. 3. 80— 82=48=number of boys. 4. ^^number of girls. Then ,' since the nu.m her of boys are to be to the number of girls as 5 : 4, II.S 5. f^number of boys. But 6 *=32. 7.' |=iof 32=8, and 8. 1=5 times 8^40=number of boys. 9. .'. 48 — 40=8=number that must leave that there maybe 5 boys to 4 girls. III. .•. 8 boys must leave that there may be 5 boys to 4 girls. I. How far may a person ride in a coach, going at the rate of 9 miles per hour, provided he is gone only 10 hours, and walks back at the rate of 6 miles per hour? 1. 9 mi.^distance he can ride in 1 hour. 2. 1 mi.=distance he can ride in \ hour. 3. 6 mi.=distance he can walk in 1 hour. II.J4. 1 mi.=distance he can walk in \ hour. 5. .•. ^ hi^--|~i hr.=Y^^ hr.=time it takes him to ride 1 mi. and walk back. [and walk back. 6. .". 10 hours=time it takes him to ride 10-r-^\, or 36, mi. III. .-. He can ride 36 miles. I. A hound ran 60 rods bjefore he caught the fox, and f- of the distance the fox ran before he was caught, equaled , the distance he was ahead when they started. How far did the fox run, and how far in advance of the hound was he when the chase commenced? 1. f=distance the fox ran before he was caught. Then 2. ■|=distance he was ahead. 3. |-f-|=f=distance the hound ran to catch the fox. 4. 60 rods=distance the hbund ran to catch the fox. ^ II.<!5. .-. 1=60 rods, 6. ■^=\ of 60 rods=12 rods, and [ahead. 7. -§=2 times 12 rods=24 rodsi^distance the fox was 8. -1=3 times 12 rods=36 rods=distance the fox ran be-. fore he was caught. III. ANALYSIS. 205. 24 rods^distance the fox was ahead, and 36 rods^distance he ran before he was caught. I. If i of 12 be 3, what will i of 40 be? 1. i of 12=4. 2. i of 40='10. By supposition n.<^3. 4=3. Then 4. 1=^ of 8=t, and ,5. 10=10 times |=7i. III. .-. ^ of 40=7^, on the supposition that ^ of 12 is 3. I. A lady spent in one store ^ of her money and $1 more;- in another, J of the remainder and $^ more ; in another, ^ of the remainder and f 1 more ; and in another, ^ of the remainder and $^ more ; she then had nothing left. What sum had she at first?. First Solution. 1. I^et f=;her money at first. Then, 2. •^+fl=amount she spent in the 1st store. 3. |— (i+f 1 )=i— $l,=amount she had left. 4. i—i of (J— |l)+$-i-,=amount she spent in the 2d" store. 5. i— 11 — i=i— $l,=the amount she had left after vis- iting the 2d store. 6. J-+f^=Jof (i— $l)+$l,=amount she spent in the 3d store. 7- 4— $f = (i—|l) — (f-"I4), = amount she had left after visiting the 3d store. 8. ii^ij — $i=i of (4— ff )+$J,=amount she spen-;' in the 4tli store. 9- j\ — $i=(F7"$^a-)— (I'ir— li).=amount she had le±t after visiting the 4th store . 10. |0=what she had left after visiting the 4tli store. 11. . 3^—11=0, whence 12. A=$f, and 1^ 13. ^f=16X$f =$20, the amount she had at first. III. .-. She had $20 at first. Second Solution. 1. $l=f4+fJ,=amount she had on entering the 4th store. 2. $4=2($i+$l),=amount she had on entering the 3d store. 3. $10=2($4+$i)+$l,=amount she had on entering the 2d store. , 4. $20=2X$10,=amount she had on entering the first store. Ill . . ■ . She had $20 at first. 11.^ II. <^ III. 206 FINKEL'S SOLUTION BOOK. I. A gold and silver watch were bought for $160; the silvei watch cost only ^ as much as the gold one ; how mucb was the cost of each? 1. ^=cost of the gold watch. Then 2. •f=cost of the silver watch. 3. ^+i-=8==cost of both. II.< 4. $160=cost of both. 5. .-. f=$160, 6- Y=i of $160^$20=cost of the silver watch, and 7. -^=7 times $20=$140=cost of the gold watch. $20=cost of the silver watch, and $140^cost of gold the watch. I. If 6 sheep are worth 2 cows, and 10 cows are worth 5 horses; how many sheep can you buy for 3 horses? 1. Value of 2 cows=value of 6 sheep. 2. Value of 1 cow=value of 3 sheep. 3. Value of 10 cows=value of 30 sheep. But 10 cows are II. ^j worth 5 horses, 4. .'. Value of 5 horses=value of 30 sheep. 5. Value of 1 horse^value of 6 sheep. 6. Value of 3 horses:^value of 18 sheep. III. .•. 3 horses are worth 18 sheep. I. A teacher agreed to teach a certain time upon these con- ditions : if he had 20 scholars he was to receive $25; but if he had 30 scholars, he was to receive but $30. He had 29 scholars. Required his wages. 1. $25=his rate of wages for 20 pupils. 2. $1.25=-^ of $25=his rate of wages for 1 pupil. 3. $30=his rate of wages for 30 pupils. 4. $1=^1^ of $30=his rate of wages for 1 pupil. 6. .-. $1.25— $1.00=$.25=reduction per pupil by the ad- dition of 10 pupils. 6. $.025=$.25-4-10^reduction per pupil by the addition of 1 pupil. 7. $.225=9, times $.025^reduction per pupil by the addi- tion of 9 pupils. 8. .-. $1.25— $.225=$1.025=his rate of wage per pupil. 9. $29,725=29 times $1.025=his wages for 29 pupils. III. .-. His wages were $29,725. , {Mattoon' s Arith., i>. S85,prob. 200.) Note, — This problem is really indeterminate, because ihere is no definite rate of increase of wages given for each additional scholar, '^'^e might say, - since the wages were increased $5 by the addition of 10 scholars, they would be increased $.50 by the addition of one scholar and, consequently, $4.50 by the addition of 9 scholars. Hence, his wages should be $25+$4.50, or $29.50. By assuming different relations between the increase of wages and additional scholars, other results may be obtained. The above solution seems to be the most satisfactory. II. PROBLEMS. 207 I. 11. III. I. A steamboat that can run 15 rrii. per hr. with the current and 10 mi. per hr. against it, requires 25 hr. to go from Cincinnati to Louisville and return ; what is the dis- tance between the cities ? . 15 mi. ^distance the boat can travel down stream in 1 hour. [hour. 1 mi.=distance the boat can travel down stream in 3. 10 mi.=distance the boat can travel up stream in 1 hr, 4. 1 mi.^distance the boat can travel up stream in -^ hr, 2. ^ IS hr.- hr.=^ hr.^time required for the boat to travel 1 "mi. down and return. 5. .'. 25 br.=time required for the -boat to travel 25-J-5-, or 150, miles down and return. .'. The distance between the two places is 150 miles. A sold to B 9 horses and 7 cows for $300; to C, at , the same price, 6 horses and 13 cows, for the same sum: what was the price of each? II.S o. 6. 7. 8. 9. 10. Cost of 9 hofses+cost of 7 c6ws=$300. Then the Cost of 36 horses+cost of 28 cows=$1200, by taking 4 times the number of each. Cost of 6 horses+cost of 13 cows=$300. Then the Cost of 36 horses-|-cost of 78 cows=$1800, by taking 6 times the number of each. But Cost of 36 horses+ccst of 28 cows=$1200. .-. Cost of 50 cows^$600, by subtracting; and Cost of 1 cow=-jV of $600=$12. The Cost of 7 cows=7 times $12=$84. .-. Cost of 9 horses^$300 — cost of 7 cows= =$216. The V Cost of 1 horse=4 of $216=$24. .II. ( The cows cost $12 apiece, and I The horses $24 apiece. PROBLEMS. 1. A man'bought a horse and a cow for $100, and the cow cost §■ as much as the horse; what was the cost of each? Ans. horse, $60; cow, $40. 2. Stephen has 10 cents more than Marthia, and they to- gether have 40 cents; how many have each? Ans. Stephen, 25/; Marthia, 15/. 3. A's fortune added to ^ of B's fortune, equals $2000; what is the fortune of each, provided A's fortune is to B's as 3 to 4? Ans. A's, $1200 ; B's, $1600. 4. If 10 oxen eat 4 acres of grass in 6 days, in how many days will 30 oxen eat 8 acres? Ans. 4 days. 208 FINKEL'S SOLUTION BOOK. 5. If a 5-cent loaf weighs 7 oz. when flour is worth $6 a bar- rel, how much ought it weigh when flour is worth $7 per barrel ? Ans. 6. A lady gave 80 cents to some poor children; to each boy she gave 2 cents, and to each girl 4 cents; how many were there of each, provided there were three times as many boys as girls? Ans. 8 girls; 24 boys. 7. Two men or three boys can plow^ an acre in |- of a day ; how long wijl it take 3 men and 2 boys to plow it? Ans. Jj da. 8. A- agreed to labor-a certain time for $60, on the condition that for each day he was idle he should forfeit $2, at the expira- tion of the t^me he received $30; how many days did he labor, supposing he received $2 per day for his labor? Ans. 22\ days. 9". The head of a fish is 4 inches long, the tail is as long as the head, plus \ of the body, and the body is as long as the head and tail ; what is the length of the fish? Ans. 32 inches. 10. In a school of 80 pupils there are 30 girls; how many boys must leave that there may be 3 boys to 5 girls? Ans. 32. 11. A steamboat, whose rate of sailing in still. water is l'2 miles an hour, descends a river whose current is 4 miles an hour and is gone 6 hours; how far did it go? Ans. 32 miles. 12. A man keeps 72 cows on his farm, and for every 4 cows he plows 1 acre, and keeps 1 acre of pasture for every 6 cows; how many acres in his farm ? Ans. 30 acres. 13. A company of 15 persons engaged a dinner at a hotel, but before paying the bill § of the company withdrew by which each person's bill was augmented $^; what was the bill? Ans. $15. 14. A man sold his horse and sleigh for $200, and ^ of this is- 8 times what his sleigh cost, and the horse cost 10 times as much as the sleigh ; required the cost of each. ' Ans. horse, $200; sleigh, $20. 15. A went to a store and borrowed as much as he had, and spent 4 cents; he then went to another store and did the same, and then had 4 cents remaining; how much mbne!y had he at first? Ans. 4 cents. 16. A lady being asked her age, said that if her age were in- creased by its \, the sum would equal 3 times her age 12 years ago; what was her age? '' Ans. 20. 17. A lady being asked the hour of day, replied that f- of the- time past noon equaled ^ of the time to midnight,, minus -^ of an^ hour; what was the time? Ans 6 o'clock, P. M.. PROBLEMS. 209 IS. What is the hour of day if -J of the time to noon equals the time past midnight? Ans. 9 o'clock, A. M. 19. A person being asked the time of day, said |- of the time to midnight equals the time past midnight ; what was the time? Ans. 9 o'clock, A. M. 20. A traveler on a train notices that 4J times the number of spaces between the telegraph poles that he passes in a minute is the ratef of the train in miles per hour. How far are the poles apart? , Ans. 198 feet. 21. C's age at A's birth was 5-| times B's age, and now is the sum of A's and B's ages, but if A were now 3 years younger and B 4 years older, A's age would be |- of B's age. Find their ages. Ans. A's, 72, years; B's, 88 years; C's, 160 j'ears. 22. In the above problem change the last and to or, and what are their ages? Ans. A's, 36 ; B's, 44, and C's, 80. 23. I have four casks, A, B, C, and D respectively. Find the capacity of each, if ^ of A fills B, -f of B fills C, and C fills ■f-^ of D; but A will fill C and D and 15 quarts remaining. Ans. A 35 gal., B 15, C 11^, and D 20. 24. A man and a boy can do a certain work in 20 days : if the boy rests 5^ days it will take them 22-|- days; in what time can each do it? Aus. The man, 36 da. ; the boy, 45 da. 25. A can do a job of work in 40 days, B in 60 days; after both work 3 days, A leaves ; when must he return that the work may occupy but 30 days? Ans. 10 days. 26. If 8 men or 15 boys plow a field in 15 days of 9^ hr., how many boys must assist 16 mei» to do the work in 5 days of 10 hr. each? Ans. 12 boys. 27- Bought 10 bu. of potatoes and 20 bu. of apples for $11 ; at another time 20 bu. of potatoes and 10 bu. of apples for $13 ; what did I pay for each per bu. ? Ans. Apples 30/, potatoes 50/. 28. A farmer sold 17 bu. of barley and 13 bu. of wheat for $31.55, getting 35/ a bu. more for whe^t than for the barley. Find the price of each per bu. Ans. ' Barley 90/, wheat $1.25. 29. After losing f of my money I earned $12; I then spent f of what I had and found I had $36 less than I lost; how much money had I at first? Ans. $60. 30. In a company of 87, the children are f of the women, and the women -f of the men; how many are there of each? Ans. 54 men, 24 women, and 9 children. 210 FINKEL'S SOLUTION BCOK. ■ 31. If 4 horses or 6 cows can be kept 10 days on a ton of hay, how long will it last 2 horses and 12 cows? Ans. 4 days, 32. A, B, and C buy 4 loaves of bread, A payings 5 cents, B 8 cents, and C 11 cents. They eat 3 loaves and sell the fourth to D for 24 cents. Divide the 24 cents equitably. Ans. A 5 cents, B 8 cents, and C ll^cents. 33. A and B are at opposite points of a field 135 rods in com- pass, and start to go around in the same direction, A at the rate of 11 rods in 2 minutes and B 17 rods in 3 minutes. In how many rounds will one overtake the other? Ans. B 17 rounds. 34. If a piece of work can be finished in 45 days by 35 men and the men drop off 7 at a time every 15 days, how long will it be before the work is completed? Ans. 75 days. 35 A watch which loses 5 min a day w^as set right at 12 M., July 24th. What will be the true time on the 30th, when the hands of that watch point to 12? Ans. 12:30-jW P. M. 36- A seed is planted. Suppose at the end of 3 years it pro- duces a seed, and on each year thereafter each of which when 3 years old produce a seed yearly. All the seeds produced, do likewise ; how many seeds will be produced in 21 years? Ans. 1872. 37. The circumference of a circle is 390 rods. A, B, and C start to go around at the same time. A walks 7 rods per minute, B 13 rods per minute in the same direction ; C walks 19 rods per minute in the opposite direction. In how many minutes will they meet? Ans. 195 min. 38. If 12 men can empty a cistern into which water is run- ning at a uniform rate, in 40 min., and 15 men can empty it in 30 min., how long will it require 18 men to empty it? Ans. 24 min. 39. Four men A, B, C, and D, agree to do a piece of work in 130 days. Agets42d., B 45d., C 48d., and D 51d., for every day they worked, and when they were paid each man has the same amount. How many days did each work? [da. Ans. A 35^111 da., B 33f||| da., C 31^| da., and D 29fff| 40. A fountain has four receiving pipes. A, B, C, and D; A, B, and C will fill it in 6 hours; B, C, and D in 8 hours; C,,D, and A in 10 hours: and D, A, and B in 12 hr. : it also has four dis- charging pipes, W, X, Y, and Z ; W, X, and Y will empty it in 6 hours; X, Y, Z in 5 hours; Y, Z, and W in 4 hours ; and Z, W, and X in 3 hours. Suppose the pipes all open, and the fountain full, in what time will it be emptied? AnS. 6^ hours. ALLIGATION. 211 CHAPTER XVII. ALLIGATION. 1. Alligation is the process employed in the solution of problems relating to the compounding of articles of different va-lues or qualities. 1. Alligation Medial. 2. Alligation Alternate. 2. Alligation • L ALLIGATION MEDIAL. 1. Alligation JM^e^ial is the process of finding the mean, or average, rate of a mixture composed of articles of different values or qualities, the quantity and rate of each being given. I. A grocer mixed 120 lb. of sugar at 5/ a pound, 150 lb. at 6/., and 130 lb. at 10/.; what is the value of a pound of the mixture.'' 1. 120 lb. @.5/=$6.00, 2. 150 lb. @6/=.$9.00, and II.<l3. ISO lb. @1Q^=%13.00. 4. 400 lb. is worth $28.00. l5. 1 lb. is worth $28-h400=$.07=7 cents. III. .'. One pound of the mixture is worth 7 cents. {Stod. Comp. A., p. 2U,pfol>. 3.) II. ALLIGATION ALTERNATE. 1. Alligation Alternate is the process of finding in what ratio, one to another, articles of different rates of quality or value must be taken to compose a mixture of a given mean, or average, rate of quality or value. ^ CASE I. Given the value of several ingredients, to make a compound of a given value. I. What relative quantities of tea, worth 25, 27, 30, 32, and 45 cents per lb. must be taken for a mixture worth 28 cents per lb. Diff. Bal. average, that the Solution. — In the principle is, gains and loses are equal. We write the average price and the particular values 25 28/- r25/ 27/ 30/ 32/ 45/ 3/ 1/ 2/ 27) 30, 32, and 45 as in the margin. 17/ This 2 lb. 17 lb. • 19 lb. 4 lb. 4 lb. 3 lb. 3 lb. 1 lb. 1 lb. 3 lb. j 3 lb. is only a convenient 212 FINKEL'S SOLUTION BOOK. arrangement of the operation. Now one pound bought for 25/ and sold in a mixture worth 28/ there is a gain of 28/ — 25/, or 3/; one pound bought at 27/ and sold in a mixture worth 28/, there isa gain of 28/ — 27/, or 1/; one pound bought at 30/ and sold in a mixture worth 28/ there is a loss of 30/ — 28/, or 2/ ; one pound bought at 32/ and sold in a mixture worth 28/, there is a loss of 32/ — 28/, or 4/i and ope pound bought at 45/ and sold in a mixture worth 28/ there is a loss of 45/ — 28/, or 17/. Since the gains and losses are equal, we must take the ingredi- ents composing this mixture in such a proportion as to make the gains and losses balance. We will first balance 'the 25/ tea and the 30/ tea. Since we gain 3/ a pound on the 25/ tea, and lose 2/ on the 30/ tea, how many pounds of each must we take so that the gain and loss on these two kinds may be equal? Evi- dently, we should gain 6/ and lose 6/. To find this, we simply find the L. C. M. of 3 and 2. Now if we gain 3/ on one pound of the 25/ tea, to gain 6/, we must take as many pounds as 3/ is contained in 6/, which are 2 lb. If we lose 2/ on one pound of the 30/ tea, to lose 6/, we must take as many pounds as 2/ is contained in 6/, which are 3 lb. Next, balance the 25-cent te^ and the 45-cent tea. The L. C. M. of 3/ and 17/ is 51/. Now if we gain 3/ on one pound of the 25-cent tea to gain 51/, we must take as many pounds as 3/ is contained in 51/ which are 17 lb. If we lose 17/ on one pound of the 45-cent tea, to lose 51/, we must take as many pounds as 17/ is contained in 51/ which are 3 lb. Next, balance the 27-cent tea and the 32-cent tea. The L. C. M. of 1/ and 4/ is 4/. If we gain 1/ on one pound of the 27-cent tea, to gain 4/, we must take as many pounds as 1/ is contained in 4/, which are 4 lb. If we lose 4/ on one pound of the 32-cent tea, it balances the gain on the 27- cent tea. Placing the number' of pounds to be taken of each kind as shown above, and then adding horizontally, we have 19 lb. at 25/, 4 lb. at 27/, 3 lb. at 30/, 1 lb. at 32/, and 3 lb. at 45/. It 'is not necessary to balance them in any particnlar order. All that must be observed, is that all the ingredients be used in balancing. Note. — To prove the problem, use Alligation Medial. CASE II. To pioportionate the parts, one or more of the quantities, but not the amount of the combination, being given. I. How many bushels of hops, worth respectively 50, 60., and 75/ per bushel, with 100 bushels at 40/ per bushel, will make a mixture worth 65/ a bushel? ALLIGATION. 213 40/ 50/ ,75/ 25/ 16/ 5/ Bal. 10/ 2bu. 5bu. 2 bu. 3 bu. 2 bu. Ibu. rlOO bu. 100 bu. X50=<ioObu. 1450 bu. B. 65/. z?,/. ( Ba/. 40/ 50/ 60/ 75/ 25/ 15/ 5/ 10/ 2 bu. 5jbu. 2 bu. 3bu. 2 bu. 1 bu. X50=. '100 bu. 2bu. 2 bu. .254 bu. Solution. — In this solution, we proceed as in Case I. In A, we obtain the relative amounts to be used of each Icind, which is 2 bu. at 40/, 2 bu. at 50/, 2 bu. at 60/, and 9 bu. at 75/. But we are to have 100 bu. of the first kind. Hence, we must multi- ply these results by 100-^2, or 50. Doing this, we obtain 100 ibu. at 40/, 100 bu. at 50/, 100 bu. at 60/, and 450 bu. at 75/. Since either or both of the balancing columns, except the first, may be multiplied by any number whatever without affecting the average, it follows that there are an infinite number of re- sults satisfying the conditions of the problem. Since we are to have 100 bu. at 40/, the first column can be multiplied by only 50. In B, we have multiplied the first column by 50 and added in the results in the other two columns. This gives us 100 bu. at 40/, 2 bu. at 50/, 2 bu. at 60/, and 254 bu. at 75/. The second and third columns may be multiplied by any number whatever. But the fir.st must always must be multiplied by 50, because we .are to have 100 bu. at 40 cents per bushel. (B. H. A., p. 338,prob. 2.) I. 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 Tvill just float, that is, have a sp. gr. (1) the same as water? r tV H i t 3 ■xi . X ¥r^=- 4 3 \\ s g L \ 39i oz.=2 lb. 11 oz. U2 oz. SoLtJTiON. — The specific gravity of any body is the ratio which shows how many times heavier the body is than an equal 214 FINKEL'S SOLUTION BOOK. volume of water. Thus, when we say that the ape.ilfiv --gravity of lead is 11, we mean that a cubic inch, a cubic foot, a cubic yard, or any quantity whatever is 11 times as heavy as an equal quantity of water. Now if a cubic inch (say) of lead be immersed in water, it will displace a cubic inch of water ; and since it weighs 11 times- as much as a cubic inch of water, it displaces .-jlj- of its own weight. Hence, to have equal weights of water ^nd lead we must take only -jL- as much lead as water. Now' since a voluma of water and y'^ ^^ much lead have the same weight, and in the proper combination have a volume of 1, since the sp. gr. of the combination is 1, there is a loss of 1 — yYi or \^, in volume, on the part of the lead. For the same reason, there is a loss of |- in volume on the part of the copper, and 3 on the part of the cork. Balancing, we see that we must take 3 volumes of lead with i2- volumes of cork, a unit volume of water being the basis, in order that the two substances will just float, «'. e., have a specific grav- ity (1). In like manner, we must take 3 volumes of copper with 1^ volumes of cork. Now since we must always take 3 vol- umes of lead for e^fery \j- volumes of cork, it is evident that the weights of th'e substances are in the same proportion. Hence, we may say, we must take 3 oz. of lead with every \j- oz. of cork, and 3 oz. of copper with every -I oz. of cork. But we are to have only ^ oz. of copper. Hence, we must multiply the second balancing column by some number that will give us -J oz. of copper, i. e., we must multiply 3 by some number that will give us ^. The number by which we must multiply is ■J-f-3=-|-. But multiplying f by i, we get -^y oz. of cork. But we are to have altogetber 12 oz. of cork. Hence we must yet have 12 oz. — /y oz.=8/y'' oz. To produce this, we must multi- ply -^-^ by some number that will give \^j oz. This number is- ¥7'-J-tt=¥t • But we must also multiply 3 by Vf . This will give us 39-j oz.=2 lb. 7^ oz. of lead. Hence, we must use 2 lb. T-j- oz. of lead, so that the three will just float. (i?. H. A., p. SS9. prob. 7-) I. 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% ? . (1.) 100%^the average cost. (2.) 20%=gain. (3.) 120%=the average selling price. 1 .J (4.) 60%^the average selling price. (5.) • 120%=60%. (6.) l^/n=^U of 60%=i%. (7.) 100%=100 times -^%=50%, the average cost. ALLIGATION. 215 Il.i r(l.) 120 shares @ 74%= 8880%. {■2.) 150 shares @ 68%=10200%. (3.) 130 shares @ 54%= 7 020%. (4.)' .-. 400 shares are worth 26100%, and (5.) 1 share is worth 2G100%-r-400=65i%, the average. 50% 40 % 65i% 10 %;15i shares. 15|-%||10 shares. 1 j-GlO shares. !X40= J UOO shares. III. He must take (>10 shares. (i?. 71. A., p. SS9, prob. 8-) Explanation. — Since 60% is the average selling price, and his gain is 20%, it is evident that his average cost is 60%-;-1.20, or 50%. In step 3, we iind that the-avcrage cost of tile 400 shares is 65^%. Hence, the problem is the same as to find how many shares at 40%, must A buy who has 400 shares at at an average of 65^% so that his average cost will be .50'^^. Balancing, we find that he must take 15| shares at 40% with 10 shares at lioi%. But he has 400 shares at 65^%. Hence, we must multiply the balancing col- umn by 400-;-10, or 40. This gives 610 shares at 40%. ' CASE IIL To proportion the parts, the amount of the whole combination being given. I. How many barrels of flour, at $8, and $8.50, with 300 bbl. at $7.50, 800 bbl. at $7.80, and 400 bbl. at $7.65, will make 2000 bbl. at $7.85 a bbl.v' fl. 300 bbl. @ $7.50 a bbl.=$2250. 2. 800 bbl. @ $7.80 a bbl.=$6240. 3. ^00 bbl. @ .$7.65 a bbl. =$3060. 4. .-. 1500 bbl. are worth $11550. 5. $7.85^the average price per bbl. of.2000 bbl. 6. .-. $15700=2000 x$7.85=the value of 2000 bbl. 7. . . $15700— $11550=$4150=the value of 2000 bbl.- bbl., or 500 bbl. !.30=$4150-^500=the average value of 1 bbl. II. < -1500 9. $8.30 $8.00 $8.50 $-.30 2 bbl. $.20 3 bbl r200 bbl. X( 500^5 )=<; HI. 1. 200 bbl 2. 300 bbl 1.300 bbl. 5 bbl. at $8.00 per bbl. must be taken with at $8.50 per bbl. {R. H. A.,f. 339, proi. ^.) I. A dealer in stock can buy 100 animals for $400, at the fol- lowing rates: calves, $9; hogs, $2; lambs, $1; hojv many may he take of each kind? , 216 FINKEL'S SOLUTION BOOK. $2 Bai $3;5 lambs. $2; $5;3 calves. 5 hogs. 2 calves. 3 68 29 1017 60 62 30 31 24 44 32 31 36 33 38 28 34 45 20 35 62 12 36 69 4 37 Explanation . — A lamb bought for $1 and sold for $4 is a gain of $3; a hog bought for $2 and sold for $4 is a gain of $2; and a calf bought for $9 and sold for $4 is a loss of $5. We must make the gains and loses equal. The L. C. M. of $3 and $5 is $15. If we gain $3 on one lamb to gain $15 we must take as many lambs as $3 is contained in $15, which are 5 Iambs. If we lose $5 on one calf, to lose $15, we must take as many calves as $5 is contained in $15, which are 3 calves. The L. C. M. of $2 and $5 is $10. If we gain $2 on one hog, to gain $10, we must take as many hogs as $2 is contained in $10, which are 5 hogs. If we lose .$5 on one calf, to lose $10, we must take as many calves as $5 is contained in $10, which are 2 calves. Adding the balancing columns, considering them as abstract numbers, we have 8 and 7. 8+7=15. 100+15=6|. .". Multiplying each balancing column by 6|, will give 33^ lambs, 33J hogs, and 33^ calves. But this result is not compatible with the nature of the problem. Hence we must see if we can take a number of 8's and a number of 7's that will make 100. By trial, we find that tiuo 8's and twelve 7's will make 100. Hence, multiplying the first column by 2 and the second by 12, and adding the columns horizon- tally, we have for our result, 10 lambs, 60 hogs, and 30 calves. Again, we find, by trying three 8's, four 8's, and so on, that nine 8's taken from 100, will leave 28 which is foiir 7's. Hence, nine 8's and four 7's will make 100. Then, multiplying the first column by 9 and the second by 4. and adding the columns horizontally, we have for a second result 45 lambs, 20 hogs, and 35 calves. Now these are the only answers that can be obtained by taking an integral number of 8's and integral number of 7's to make 100. But other answers may be obtained by taking 8 a fractional number of times, and 7 a fractional number of times to make 100- Suppose, for illus- tration, we try to take a number of thirds 8 times. We find that 8 taken 6- third times and 7 taken 36 third times will make 100. JVIultiplying the first column by I and the second by ^, and adding^ the columns horizon- tally, we have, for a result, 10 lambs, 60 hogs, and 30 calves — the same as that obtained by taking 8 twice and 7 twelve times. Again, we find, that 8 taken 13 third times and 7 taken 28-third times will make 100. Multiply- ing and adding as before we find that our results are fractional. Hence, we can not take a fraction whose denominator is three. It is clear that we must take a fraction whose denominator will reduce to unity when being multiplied by 5. Hence, if we try to take 8 a number of fifths times and 7 a number of fifths times to make 100, our results will all be integral. By trial, we find that 8 taken 3-fifths times and 7 taken 68-fifths times will make 100. Multiplying and adding as before, we have, for our results, 3 lambs, 68 hogs, and 29 calves. Again, we find that 8 taken 10-fifths times and 7 taken 60-fifths times, will make 100. Multiplying and adding as be- fore, we have, for results, 10 lambs, 60 hogs, and 80 calves. Again, by trial, we find that 8 taken 17-fifths and 7 taken 52-fifths times will make 100. Multiplying the first column by ^ and the second by "^j and adding the col- umns horizontally, we have, for results, 17 lambs, 52 hogs, and 31 calves. Continuing the process, we find nine admissible answers. These are the only answers, satisfying the nature of the problem. •3. Series SERIES. 217 CHAPTER XVIII. SERIES. DEFINITIONS. 1. A Series is any number of quantities, having' a fixed order, and related to each other in value according to a fixed law. Thus, 1, 3, 5, 7, 9, 11 ; 3, 9, 27, 81 ; 1, 2, 3, 5, 8, 13, 21 ; etc., are series. 2. The quantities involved in a series are called 1'ertns; the first and last are called the extremes and the others the means. 1. Arithmetical. 2. Geometrical. 3. Harmonic. 4. Binomial. 5. Logarithmic. 6. Exponential. 7. Hypergeometric. 8. Etc., etc. 4.^ The series usually treated in arithmetic are the Arithmetical and Geometrical ; and these together with the Harmonic are commonly called Progressions. 5. An Arithmetical Progression is a series of num- bers which increase or decrease, by a constant quantity, called the Common Difference. ] Thus, 5, 8, 11, 14, 17 is an increasing arithmetical progression whose common' difference is 3; and, 17, 15, 13, 11, 9 is a de- ■creasing arithmetical progression whose common difference is 2. 6. A Geometrical Progression is a series of numbers which increase or decrease by a constant multiplier, called the Ratio. • . . . Thus, 2, 6, 18, 54 is an increasing geometrical progress'on whose ratio is 3; and, 1, ^, \, -J, ^\, is a decreasing geometrical progression whose ratio is |-. 7. A Harmonic Progression is a series of numbers, the reciprocals of which are in Arithmetical Progression. Thus, -J-, ^, ^-, -I, is a Harmonic Progression. I. ARITHMETICAL PROGRESSION. 1. Every Arithmetical Progression involves five quantities, viz., thejirst term, represented by a ; the /ast term, represented by /; the number of terms, represented by n ; the commoji differ- ence, represented by d ; and tke S2im of all the terms, represented by .y. 2. If any three of these five quantities be given, the others may be found. Hence, every formula involves four quantities, 218 FINKEL'S SOLUTION BOOK. viz., the three given quantities and the required quantity. But 5X4X3X2 4 things can be selected from 5 things in „ ^ ways or 5 ways. From every one of these 5 combinations of 4 letters, 4 formulas may be obtained. Hence, for the solution of every problem in arithmetical progression 4X5, or 20, formulas are required. 3. But these 20 formulas may be easily derived frgni two fundamental formulas which we shall proceed to derive. rthe£rst term, a, ) Problem. — Given -< the common difference, d, and > to' find (the number of terms, fi, ) the last term, /. Consider the series, 2, 5, 8; .... to 15 terms. Here, first term, a,^2 ; common difference, d,=3 ; and number of terms, n,^15. 2=a, the first term. 5, the second term,=2-f-3=a-|-d, the first term+the com- mon difference. 8, the thii'd term,=2+2X3=a+2d, the first term+Hvice the common difference. 11, the fourth term,=2+3X3=a+3rf, the first term+' three times the common difference.. By continuing this, we see that any term is equal to the first term -|- the common difference multiplied by 1 less than the number of the term. Hence, for an increasing series, we have the Formula. — l=a+ ( n — 1 ) d'. Rale. — Multiply the common difference by one less than the number of the term required and add the product to the first term; if the series is decreasing, subtract the product from the first term. I. A bady falls 1(>,V f^et the first second, and in each suc- ceeding second 3"2-j\ feet more than in the next preceding one. How far will it fall in the 16th second? By formula, /=a+(n— l)rf=16iL ft.+ (l()— 1 )32^-ft.=41)8r\ feet. '1. 16 iV feet=the first term. • 2. 32-J- feet^the common difference. 3. 16=the number of terms. 4. 15=16 — l,^one less than the number of terms. 5. 482^ feet=15X32-J- feet,=tlie common difference mul- tiplied by one less than the number of terms. 6. .-. 16J-2 feet -|-482Jfeet=498f'V feet,=the distance the body falls the 16th second, HI. . The body falls lOSi^j feet the Kith second. II. SERIES. , 219 PROBLEMS. 1. Find the 11th term of the progression 2, 9, 16, etc. 2. Find the 99th term of the series 3^-, 3f, 4i, etc. 3. Find the 59th term of the series ^, f, H, etc. i. Insert five arithmetical means between 3 and 15. 5. How many times does a clocit that strikes quarter hours and hours strike in 12 hours? r the first tertn, a, ) Problem. — Given ■< the last term, /, and V to find the ' (_the number of terms, n, ) sum of the terms. Consider the series 5, 11, 17, 23, 29, 35, 41. , The sum=5+ll+17+23+29+35+41. The sum=41+35+29+23+17+ll+5. .-. 2 times tlie sum=46+46+46+46+46+46+46=6X46= 6(5+41). 6(5+41) n(a+/) .'. The surrv^^ — ^ — —^ = — -. Hence, we have for the sum of the series, the 71 Formula. — s^a+l)-^. Rule. — To flic first term add the last term and multiply the sum by half the number of terms. I. Find the sum of all integers less than 100 which are mul- tiples of 7. ' By formula, s=^{ a+l)^^{ 7 + 98 )ii=:735. ' . 1. 7:=the first term, 2. 7=the common difference, ^ 3. 98=the last term. 4. 14=the number of terms. llA •.735=lin±^Uhesum. III. . The sum=735.- « Remark. — The two formulas, /=a;-|-{ ?; — 1 )rf and j= {«-!-/)-=- are all that are needed to solve any problem in arithmetical progression. They should be committed to memory. I. The first term of a decreasing arithmetical series. is 10, the number of terms 10, and the sum of the series 85 ; required the last term and the common difference. {Ray's Elementary Algebra, p. 221, prob. 7.) II. III. 220 FINKEL'S SOLUTION BOOK. 1. lO^the first term^a, 2. 10= the number of terms^w, 3. 85=the sum of the terms=.f. 4. .-. l=^a — (m — l)d, the series being decreasing, ^10 — (10— l)d=10— 9rf, and 5. s={a+l)-^, or85=(10+/)-^=50+5/. 6. .-. 85=50+5(10— 9c?)=100— 45c?; whence, 7. d=(100 — 85) ^45=: J, the common difference. 8. . . ;=10— 9X^=7, from step 4. . f J^the common difference, and ■ \ 7^the last term . Remark. — It is not necessary to remember the two formulas, /=a-|- (n — 1 )d and /=a — ( n — 1 )d, for finding the last term. The first of these is sufficient; for, when the series is decreasing and a, /, n are given, the for- mula gives a negative .value for rf, the common diflFerence, and adding a negative common difference in forming the terms of a series is the same as subtracting a numerically equal positive common difference. PROBLEMS. 1. Find the sum of the series 1, % 3, etc., to 1000 terms. Ans. 500500. 2. Find the sum of the series 1, 3, 5, 7, . . . to 101 terms. Ans. 10201. 3. Find the sum of all the multiples of 11, from 110 to 990, inclusive. Ans. 44550. 4. If a person saves $100 per year, and puts this sum at simple interest at 4J% at the end of each year, to how tnuch will his property amount at the end of 25 years ? Ans. $3850. II. GEOMETRICAL PROGRESSION. 1. The five quantities involved in a Geometrical Pro- gression are the first term, a; the last term, I; the number of terms, n; the common ratio, r; and the sum of the terms, s. 2. Any three of these five quantities being given, the others may be found. In Geometrical Progression as in Arithmetical Progression, 20 formulas are necessary for the solution of all problems. But these 20 formulas are easily obtained from two fundamental formulas which we will now derive. C the first term, a, ) Problem. — Given < the number of terms, /, and >■ to find the ( the ratio, r, . ) last term. Consider the series 2, 6, 18, 54, 162, etc. 2=the first term. 6, the second term,=2X3=an the first term multiplied by the ratio, 3. 18, the third term,=2X3^=0f^, the first term multiplied by the ratio squared. II. <^ SERIES. 221 54, the fourth term,=2X3^^a>'^, the first term multiplied by the ratio cubed. '■ 162, the fifth term,=2X3*=ar*, the first term multiplied by the ratio raised to the fourth power. From this, we see that any term is equal to the first term mul- tiplied by the ratio raised to a power whose exponent is 1 less than the number of the term. Hence, for finding any term of a geometrical progression, we have the Formula. — l=ar'^'^ Rule. — To find any term of a geometrical, progression, raise the ratio to a power whose exponent is 1 less than the number of the required term. I. , Find the 8th term of the series C, 18, 54, etc. By formula, /=flr"-i=6X-">-^'=l-">122. 1. 6^the first term, 2. 3^the ratio , 3. 8=the number of the term required. 4. 2187^3'', the ratio raised- to a power whose exponent is 1 less than the number of the term. 5. .•.''I3122=(;x2187=the 8th term of the progression. III. . . The 8th term of the series is 13122. PROBLEMS. ' 1. Find the 10th term of the series 1. i, }, |, etc. 'i. Find the 9th term of the series 1, J, i, ^V' ^^c. 3. Find the 6th term of the progression f, i, |, etc. Ans.^^^-. 4. Find the 9th term of the progression e|, 4J, 3, etc. Ans. „«/j. ( the first term, a, '\ JProbleta. — Given ] the last term, I, and > to find the sum of' ( the ratio, r, ) the progression. Consider the progression 5, 15, 45, 135, 405. The sum=5-|-15+45-|-135+405. 3 times the sum=:iX5+3X15+3X45-|-3X135+3X405, =15+45+135-1-405+3X405. 1 times the sum=5+15+45+135+405. .-. 2 times the sum:=3X405 — 5, by subtracting the last equation from the one above. .-. The sum= " . But 3X405 is the last term multi- L plied by the ratio, I'X.r; 5 is the first term, a; and, 2 is the ratio, r,-\ . Hence, for finding the sum. Qf: a gepmetrical progression, we; have the 222 FINKEL'S SOLUTION BOOK. Ir — a Formula. — j: r—1 ' Rule. — Multiply the last term by the ratio, from the product subtract. the first term, and divide the result by the ratio minus 1. Remark 1. — The above formula and rule apply whether the series be increasing or decreasing, though when the series is decreasing Ir — a is negative and r — 1 is negative. But a negative quantity divided by a negative quantity gives a positive quantity. In order that the formula a — If applies arithmetically, for decreasing series, we write the formula s=— j Remark 2. — If the series is an infinite decreasing series, the sum, s, approaches the value as n approaches infinity. Tfius, the sum of the series 1, ^, Y^.-Yi, etc. to an infinite number of terms approaches - — r, or 2. . 1 — i Hence, for the limit of the sum of an Infinite Decreasing Geometrical Progression, we have the limit oi s^- To find the limit of the sum of an infinite decreasing geometrical progression, we have the following Rule. — Divide the first term by the ratio minus 1. Note. — It must be borne in mind that we cannot find the sum of an infinite decreasing geometrical progression. We can only find the limit of the sum as more and more terms are taken. Thus, in the series, 1, Yz, 54. Y&! £tc. to infinity, we can make the sum of the series as nearly equal to 2 as we please, though we can never take enough terms to make it exactly equal to 2. Hence, we should not speak of the sum of an infinite decreasing geometrical progression, but of the limit of the sura. I. A man agreed to work for 14 days on the condition that he should receive 1 cent for the first day, 2 cents for the second day, 4 cents for the third day, and so on. How much did he receive mall? ^ ^ , Ir-a ar^-^Xr-l ar^'-a |0.01X2i*-$0.01 By formula, .f= —^ , = zr-^^- ^ — zr-^ r—1 r—1 r—1 2—1 =$163.83. 1. 1 cent=the first term, 2. 2 ^the ratio, and •i. 14 ^the number of terms. 4. 8192 cents=2"— ^Xl cent, the amount he received for the last day. . 2X8192 cents— 1 cent ,^000 ^ *-icq qo <.i. 5. .-. ^r—z ^16388 cents, =|1d3. 83, the amount he received. III. . . He received |163.83. PROBLEMS. 1. Find the limit of the sum of the series 1, ^^j j^, etc., to infinity. Ans. li- 2. Find the limit of the sum of the series 1, J, i, etc., to infinity. Ans. IJ. II. <^ SPECIFIC GRAVITY, OR SPECIFIC DENSITY. 223 CHAPTER XIX. SPECIFIC GRAVITY, OR SPECIFIC DENSITY. 1, The Specific Gravity of any substance is the ratio •of its weight to that of an equal volume of some other substance taken as a standard. In Physics, it is customary to consider specific gravity as the ratio of the weight of a substance to that of an equal volume of distjlled water at 4° C, if the given substance be either solid or liquid ; but" if gaseous, it is compared with either air or hydrogen at 0° C. and 70 cm. pressure. 3. Density is the quantity of matter in a unit of vohime. Thus, if wz^mass, z'=volume, 5=density, then 5= — . V 3. Specific Density is the ratio of the mass of a unit vol- ume of a substance tO' that of a unit of volume of some other substance taken as a standard. The standards for specific densiiyoA^he same as those for spe- cific gravity. Thus, in the C. G. S. (centimeter — gram — second) system, 1 gram of water occupies 1 cubic centimeter, and, there- fore, the density of water is q =-. :=!. More accurately, 1 cubic cm. 1 cu. cm. of water at 3.9° C. weighs 1.000,013 standard grams ■ 1 •, • 1.000,013 grams ^ nnn mo and, hence, its density is ^ ^1.000,013. 1 cu. cm. If m and OTj be masses of equal volumes of water and of the substance, and v be these equal volumes ; the density of water is — and of the substance — - ■ V V .„ , . ^ , density of substance v Specific density of substances — -. r- ? = — = ■^ , density ot water m m Weight of in units of mass of water, W,^=gm, where g is ac- celeration due to gravity. "Weight of m^ units of mass of the substance, W ^,=.'gm^. .-. Specific gravity of the substance= weight of substance H^j m j g m j weight of equal volume of water W mg rn ' From these two equations of specific gravity and specific density, it is seen that specific gravity and 'specific density are mathematically equivalent. These two expressions are, there- fore, used synonymously. 224 FINKEL'S SOLUTION BOOK. It is a fact, discovered by Archimedes, and proved by experi- ment, that , A body submerged in a fluid is buoyed up by a force equal to the' weight of the fluid displaced. 4. Tme and Apparent Weight. — It, therefore, follows that wlien a body is weighed in air it is buoyed up by a force equal to the weight of the air displaced by it. This is the ap- parent weight of the body. The true weight would be obtained by weighing the body in a vacuum. The apparent weight is taken as the true weight in practice. Prob. 1. — To And the Speci&c Density of a solid in- soluble in water. Find the weight of the body in air, and also its weight as it hangs suspended in ice water. I,et w be the weight in air and w' its weight in water. Then by the principle of Archimedes, w — w'=the weight of the water displaced. .'. Density= •.. ^w — w I. A body weighs 02 grams in air and 42 grams in water; find its specific density. 1. 62 grams=weight of substance in' air. 2. 42 grams=weight of substance in water. II. <^ 3. 20 grams=:weight of water displaced, by principle of Archimedes. 4. 62 grams-f-20 grams=3.1^density, by definition. III. . . The specific density of the body is 3.1. Prob. 2. — To find the Specihc Density of a solid soluble in water. Find the weight of the body in air, and also its weight in some ' liquid of known density in which it is not soluble. Let .y be the density of the liquid, w the weight of the body in air, and w' the weight- of the body in the liquid. Then w — w' equals weight w of the liquid displaced. Hence, r=the specific density of 7il W the body as compared with the liquid. But since the density of the liquid as compared with water is s, therefore, the density of w ws the body as compared with water is j- times t= r. w—w w — w I. A solid weighs 100 grams in air and 64 grams in a liquid of density 1.2 ; what is its density ? {Carhart and Chute's Elements of Physics, p. 117.) 1. 100 grams^weight of body in air. 2. 64 grams=weight of body in the liquid. 3. 36 grams=weight of liquid displaced by the body. II. -i 4. . . 100 grams-^36 grams=2f =density of body as com- pared with the Hquid. SPECIFIC GRAVITY, OR SPECIFIC DENSITY. 225 I 5. .". 1.2X2^=:3|=density of the body as compared with. [ water. III. . . Specific density of the body=3f. Prob. B.—To And the SpeciGc Density of a body lighter than water. Weigh the body in air. Then, to make it sink, attach to it a heavy body whose weight in water is known, and find the weight of the bodies in water. Let w be the weight of the body, in air and w the weight of the body and sinker in water, and w^anAw^ the weight of the sinker in air and water, respectively. Then, zez+ze/j— ze''=weight of water displaced by body and sinker, w-^ — ze/2=weight of water displaced by sinker alone. Hence, w-Vw-^ —It/— ( Wj— zeij ), or w — tt/-^ ze'2=weight of water displaced by w body. Hence, r; equals the specific density. I. Find the density of a solid from the following data: Weight of solid in air^O.5 g. Weight of sinker in water=3.5 g. Weight of solid and sinker in water^3.375 g. (Carhart and Chute's Elements of Physics, p. ii^-) 1. Let TO^ weight of sinker in air. 2. 0.5 g.+«;==weight of solid and sinker in air. 3. 3.375 g.^weight of solid and sinker in water. 4. . . w — ^2.875 g.^weight of vvater displaced by solid and sinker. It. -i 5. 3.5 g.= weight of solid in water. 6. w — 3.5=weight of water displaced by sinker alone. 7. (ly— 2.875 g.)— (te^-3.5 g.) =0.625 g.=weight of wa- ter displaced by the solid. III. .-. The specific density=.008. Prob. 4. — To And the Speci&c Density of a liquid. (1.) By means of the Specific Gravity bottle. The specific gravity bottle is a bottle constructed to hold a given weight of distilled water at a specified temperature. Its capacity is usually 500 cu. centimeters at 4° C. To use the bottle, find the weight of the bottle when empty and also when filled with the liquid at 4° C. The difference will be the weight of 500 cu. centimeters of the liquid. This divided by 500 grams, the weight of 500 cu. centimeters of water, will give the specific density of the liquid. Expressed algebraically, d = - ' ,. where zc^ is the weight of the bottle and liquid and w the weight of the bottle. 226 FINKEL'S SOLUTION BOOK. Remark. — Any ordinary bottle may be used in the following way: Weigh the bottle; weigh the bottle filled with water; and, after care- fully drying the bottle, weigh the bottle filled with air. From the weight of bottle and water, subtract the weight of the bottle, this gives the weight of the water; from the weight of the bottle and the liquid sub- tract the weight of the bottle, this gives the weight of an ,equal volume of the liquid. Dividing the weight of the liquid by the weight of the water gives the specific density, — algebraically expressed as follows: 8= ^^ '^ , where w, is the weight of the bottle -when filled with ze/j — a water, Zf 2 the weight of the. bottle when filled with the liquid, and a is the ■weight of the bottle when empty. (2.) Bv weighing a Solid in the Liquid. Weigh tiie solid in air, then in ice water, and finally in the liquid. From the weight of the solid weighed in the liquid sub- tract the weight of the solid in air ; this gives the weight of the liquid displaced. (Why?) From the weight of the solid weighed in water subtract its weight in air; this gives the weight of the water displaced. Divide the weight of the liquid displaced, by the weight of the water displaced, and the result will be the spe- cific density of the liquid, — algebraically expressed as follows : 5= — ' , where w, is the weight of the solid in air, w' „ the W] Wj 10 ' ^ weight of the solid in the liquid, and w^ the weight of the solid in water. (3.) By the Areometer, or Hydrometer. The Hydrometer consists of a straight stem of wood, glass, or metal weighted at one end so as to float in a liquid in a vertical position. By observing the depths to which it sinks in different liquids, the relative weights of these liquids can be determined. The stem is graduated, the zero being taken at the point to which it sinks in water. I. A body weighs 24 grams in air and 20 grams in water ; what will be its apparent weight in alcohol, specific density 0.8? 1. 24 grams = weight in air. 2. 20 grams^weight in water. 3. 4 grams=24 grams — ^20 grams=weight of water dis- placed. 4. . . 24 grams^4 grams^G, the specific density of the body. ' • 5. 6-=-0.8=7.5, the specific density of the body as com- pared with alcohol. 6. . . The body when weighed in alcohol loses 1^7.5, or ^ of its weight. 7. '.Its apparent weight in alcohol=||X24 grams=20f grams. III. The body will weigh 20^ grams in alcohol. 11.-^ SPECIFIC GRAVITY, OR SPECIFIC DENSITY. 227 Second solution. 1. Let ;ir:=number of grams in the weight of the body in alcohol. Then 2. 24 grams — x grams=:weight of alcohol displaced. 11."! 3. 24 grams — 20 grams=4 g., weight of water displaced. 4. .-.^ill^fzi^II^Ef^densityof alcohol=0.8, 4 grams 5. whence .!tr=20.8, the weight of the body in alcohol. III. . . The weight of the body in alcohol is 20.8 grams. "The method of alligation has practical application in problems of specific density. Thus, if it iS desired to show that a liquid unaffected by gravity, assumes the form of a sphere, it may be accomplished by placing a liquid in some other liquid with which it does not mix and having the same specific density. This experiment is performed by placing, with a pipette, a mass of olive oil in a mixture of alcohol and water. I. In what proportion must water, specific density 1, and al- cohol, specific density 0.8, be mixed to make a mixture specific gravity, 0.9176, the same as olive oil ? , 1. ■.■ 0.9176=specific gravity of olive oil, 2. .-. weight of 0.9176 vol. of water=weight of 1 vol. of olive oil, 3. weight of 1. vol. of water=:weight of 1^0.9176, or It'tA vols, of olive oil. 4. 0.8=specific gravity of alcohol, 5. . . weight of 0.8 vol. of water^weight of 1 vol. of II. \ alcohol, 6. weight of 1 vol. of water^weight of 1-^0.8, or \\ vols, of alcohol. Ratio of Ratio of Given Volumes. Dif. of Volumes. vol. Mix. Vol. in mix. 1 vol. of W. li vol. of A. iV'W vol. of W. tWfvoI. of a. 412 735 735 412 per lb. 1147 III. Hence, to obtain any quantity of a mixture of alcohol and water, specific gravity p. 9176, 735 parts, by weight, of water must be taken for every 412 parts, by weight, of alcohol. The same experiment can be made by placing carbon bisulphide {CSa ) colored with iodine in a solution of zinc sulphate ( ZnS04 ) so that the mixture will have a specific gravity, 1.272, the same as carbon bisulphide. I. Find how much water, specific gravity 1, must be added to a strong solution of zinc sulphate, specific gravity, 1.400, to make a mixture whose specific gravity is 1.272, the same as carbon bisulphide. 228 FINKEL'S SOLUTION BOOK. II. 1. '.• 1.4=:specific gravity of ZnS04, 2. .". volume of 1.4 grams of ZnS04=volume of 1 gram of water, 3. volume of 1 gram of ZnS04=volume of f gram of water. 4. •-• 1.272=specific gravity of QSj, 5. .•. volume of 1.272 grams of CS2=volume of 1 gram of water, 6. volume of 1 gram of CS 2 ^volume of -^f^ gram of water, ' 7. volume of 1 gfam of water=volume of 1 gram of water. Dif. of Ratio of Ratio of Given Weights. Weights. Wts. Mix. ' Weight in Mix. f gram of ZnSO^ 1 gram of HjO 40 119 119 40 Iff gram per unit vol. ' III. Hence, 119 parts, by weight, of zinc sulphate, must be taken for every 40 parts, by weight, of water. Remark. — The last two pioblems are exactly alike so far as the prin- ciple involved in their solution is concerned. In the first problem, we used equal weights and comparative volumes; in the last problem we used equal volumes and comparative weights. I. Hiero's Crown, sp. gr., 14f , was of goW, sp. gr., WJ, and of silver, sp. gr., 10^ ; it weighed 17J lbs. : how much gold was in it? 1. •." 19 J=specific gravity of gold, 2. .". volume of 19:J- lbs. of g6ld=volume of 1 lb. of water, and 3. volflme of 1 lb. of gold=volume of ^ lbs. of water. ■.■•10'J=specdfic gravity of silver, .". volume of 10} lbs. of silver^=volume of 1 lb. of water, and 6. volume of 1 lb. of silver^volume of -^ lbs. of water. ." 14^=specific gravity of crown, ■. volume of 14|- lbs. of the crown^volume of 1 lb. II. ■{ ' of water, and 9. volume of 1 lb. of the crown=volume of yf^ IbS. of water. Dif. in Vol. Ratio of Ratio of Given Volumes. Vol. in crown. 4. 5. 7. 8. 10. per lb. Given Volumes. per lb. Vol. Mix. T^, vol. per lb. for G. ^\, vol. per lb. for S. 74 121 121 74 195 III. 11. .-. m of l"^! Vas.= lOfi lbs.= weight of gold in the crown, and 12. ^^ of 17i lbs.= 6|4 lbs.= -weight of silver in the crown. .•. 10f|- lbs.=weight of gold ib the crown. SYSTEMS OF NOTATION. 229 CHAPTER XX. SYSTEMS OF NOTATION. 1. A ■ System of Wotation is a method of expressing numbers by means of a series of powers of some fixed number called the Radix, or Base of the scale in which the different numbers are expressed. 3. The Madix of any system is the number of units of one order which makes one of the next higher. 3. Names of Systems. Radix. Names of Systems. Radix. Unitary 1 Nonary 9 Biliary 2 Decimal, or Denary 10 Ternary 3 Undenary 11 Quaternary 4 Duodenary, 12 ■Quinary , 5 Vigesimal, 20 •Senary 6 Trigesimal, 30 Septenary 7 Sexagesimal, 60 Octonary - 8 Centesimal, 100 4. In writing any number in a uniform scale, as many dis- tinct characters, or symbols, are required as there are units in the radix of the given system. Thus, in the decimal system, 10 char- acters are required; in the ternary, 3; viz., 1, 2, and 0; in the senary, 6; viz., 1, 2, 3, 4, 5, and 0; and so on. « 5. Let r be the radix of any system, then any number, iV, may be expiessed in the form, N=ar^ _|_i^»i-i_j_cr"-2^<f^i-3-(- .... -f-^^s _|_ ^^_^ ^^ j^^ .^iyhich the co-efficients a, b, c, . . . . , are each less than r. To express an integral number in a proposed scale : Divide the number by the radix, then the quotient by the radix, and so on; the successive remainders taken in order -will be the successive digits beginning from units place. 230 FINKEL'S SOLUTION BOOK. I. Express the common number, 75432, in the senary system, ri. 6)75432 IIJ 2. 6 )12572+0 3. 6 )2095+2 4. 6 )349+1 5. 6)68+1 6. 6 )9+4 7. 1+3 III. .-. 75432 in the decimal sys'tem=1341120 expressed in the senary system. I. Transform 3256 from a scale whose radix is 7, to a scale whose radix is 12. fl. 12)3256- II.<! 2. 12)166+4 3. 12)11+1 4. 0+8 III. .-. 3256 in the septenary system=814 in the duodenary system. Exflanation. — In the senary system, 7 units of one order make one of the next higher. Hence, 3 units of the fourth order = 7X3, or 21, units of the third order. 21 units +2 units =23 units. 23-M2 = l, with a remainder 11. 11 units of the third order =77 urvits of the second order. 77 units +5 units=82 units. 82-;-12=6, with a remainder 10. 10 units of the second order^70 units of the first order. 70 units +6 units=:76 units. 7& -rl2=:6,^with a reniainder 4. Hence, the first quotient is 166, with a re- mainder 4. Treiat this quotient in like manner, and so on, until a quo- tient is obtained, that is less than 12. I. What is the sum of 45324502 and 25405534, in the senary system ? 45324502' 25405534 115134440 Explanation. — 4-1- 2=61 6-^6^1, with no remainder. Write the and. carry the 1. 34-1=4. Write the 4. 5-f-5=10. 10-1-6=1, with a remainder 4. Write the 4 and carry the 1. 5+4+1=10., 10-1-6=1, with a remain- der 4. Write the 4 and carry the 1. 0+2+1=3. Write the 3. 4+3=7. 7-h6=1 with a remainder !.• Write 1 and carry 1 5-l-5-(-l^ll. ll-f-O^ 1, with a remainder 5. Write the 5 and carry the 1. 2-t-4-|-1^7. 7-|-6^ 1, with a remainder 1. Write 1 and carry 1. The result is 115134440. I. What is the difference between 24502 and 5534 in the octonary system? 24502 5534 16746 Explanation. — 1 cannot be taken from 2. Hence, borrow one unit from a higher denomination. Then (2-|-8)— 4=6 (8 — 1) — 3=4. 5 from (4+8> =7. 5 from (3-|-8)=6. Hence, the result is 16746. SYSTEMS OF NOTATION. 231 I. Transform 3413 from the scale of 6 to the scale of 7. (1. 7 )3413 I1J2. 7 )310+3 [3. 7 )24+3 2+2 III. .'. 3413 in the senary system=2233 in the septenary sys- tem. I. Multiply 24305 by 34120 in the senary system. 24305 34120 530140 24305 150032 121323 1411103040 Bxplanation. — Multiplying bhy 2 gives 10. 10^6=1, with a remainder 4. Write 4 and carry 1 to the next order. 2 times 0=0. 0+1=1. Write the 1. 2 times 3=6. 6-5-6=1, with a remainder 0. Write the and carry the 1 to the next higher order. 2 times 4^8. 8-|-l:=9. 9-!-6^1, with a remainder 3. Write 3 and carry the 1 to the next higher order. 2 times 2=4. 4-|-l=5. Write 5. Multiply in like manner by 1, 4, and 3! Add the^ partial products, remembering that 6 units of one order, in the sen- ary system, uniformly make one of the next higher. I. Multiply 2483 by 589 in the undenary system, or system whose radix is 11. We must represent 10 by some character. Let it be t. 2483 589 li'985 •1/502 11184 13322i'5 Explanation. — In the undenary system, 11 units of one order uniformly make one of the next higher order. 9 times 3^=27. 27-!-11^2, with a re- mainder 5. Write 5 and cairy the 2 to the next higher order, or second or- der. 9 times 8=72. 72+2=74. 74-^11=6, with a remainder 8. Write 8 and carry the 6 to the next higher order, or third order. 9 times 4=36. 36-1-6=42. 42-^11=3, with a remainder 9. Write 9 and carry the 3 to the next higher order, or the fourth order. 9 times 2:=18. 18-f 3=21. 21-4-11 =1, with a remainder t. Write t ahd carry the 1 to the next higher order. Multiply in like manner by 8 an-J 5. Add the partial products, remember- ingthat 11 units of one order equals one of the next higher. Wherever 10 occurs, it must be represented by a single character t. 232 FINKEL'S SOLUTION BOOK. I. Divide 1184323 by 589 in the duodenary system. In the duodenary system, we must have 12 characters ; viz., 1, 2, 8, 4, 5, 6, 7, 8, 9, t, e, and 0. t represents 10 and e, 11. 589)1184323(2486 ' em 22if3 IteQ 3e32 39ifO 1523 1523 Explanation. — In the duodenary system, 12 units of one order make one of the next higher. 1184 will contain 589, 2 times. Then multiply the divisor, 589, by 2 thus: 2 times 9=:18. 18-v-12:=l, with a remainder 6. Write the 6 and carry the 1. 2 times 8=16. 16+1=17. 17-J-12=1, with a remainder 5. Write the 5 and carry the 1. 2 times b=t. ^-|-l=e. Write the e. Then 'subtract. 6 from (12+4)=^, 5 from 7=2, and e from (12 +1)=2. Hence, the first partial dividend is 22 t. Bring down 3. Then 22t3 will contain 589, 4 times. Multiply as before. By continuing the operation we obtain 2483 for a quotient. I. Divide 95088918 by ttA, in the duodenary system. i?/4)95088918(/4i'ee 9074 4548 3754 9e49 9074. /951 9e58 /e58 ^e58 I. Extract the square root of 11122441 in the senary system. li 122441 (2405 2X2= 4 44 312 304 2X24=52 2X240=520 5 42441 42441 Exfla)iation. — The greatest square in 11 expressed in the senary system is 4. Subtracting and bringing down the next period, we have 312 for the next partial dividend. Doubling the root already found and finding how many times it is contained in 312 expressed in the senary system, we find it is 4. Continuing the process the same as in the decimal system, the re- sult is 2405. SYSTEMS OF NOTATION. 233 1. Extract the square root of 11000000100001 in the binary system. 11000000100001(1101111 1 101 1000 101 11001 110000 . 11001 110101 1011110 110101 1101101 10100100 1101101 11011101 11011101 11011101 II.. III. I. II.- III. ( Todhunter's Algebra, p. ^5, Ex. 2S.) Find in what scale, or system, 95 is denoted by 137. 1. Let ^:^the radix of the system. Then 2. ^3_|_3r+7=95, 3. ^■■i-|-3r==95— 7=88, and \. ?-2-|-3r-f-|^88+|^'fS by completing the_ square. 5. ^"-1-1='^^ 5 by extracting the square root, and 6. r^^-f — |z=y^8, the radix of the system. .'. 95 is denoted by 137 in the octonary system. ( Todhunter's Alg.,p. 266, prob. S6.) Find in what system I'SSl is denoted by 1000. rl. Let r=the radix of the system. Then I 2. ;-4-(-0r3-|-0r2+0^+0=1331, or 1 3. ?-*=1331. Whence 14. r=v'l331 =11, the radix of the system. . . . 1331 is denoted by 1000 in the undenary system. ( Todhunter' s Alg., p. 255, prob. 28.) 234 FINKEL'S SOLUTION BOOK. CHAPTER XXI. MENSURATION. 1. JKenSUration, is that branch of applied mathematics which treats oi geometrical magnitudes. 2. Creometrical JKagnitudes are Hnes, surfaces, and soiids. 3. Geometrical Magnitudes. D. Solid. ST cn O n "13 o r • 3- a ^3 C. Surface. O c .W A. Line n M ST C. 03 < S. (IS. w ba i-J ■ S-. 1 Oi 4*. 03 lO rr ■"• cr ""• o O o ^ Sot Scrq S Q, g r; S- 3- X- a- M - -^ °= c- 3- X- cr (» n> MENSURATION. 235 •4. A. Line is a geometrical magnitude having lengtli, with- out breadth or thickness. 5. A. StTaight Line is a line which pierces space evenly, so that a piece of space from along one side of it will fit any side of any other portion. 6. A. Curved Line is a line no part of which is straight. 7. A Surface is the common boundary of twO' partg of a solid, or of a solid and the remainder of space. 8. A Plane Surface, or ^lane, is a surface which di- vides space evenly, so that a piece of space from along one side of it will fit either side of any other portion of it. 9. A. Curved Surface is a surface no part of which is plane. 10. A Polygon (niMra,vo^, from /7o/ly?, many, and ymvia, angle) is a portion of a plane bounded by straight lines. 11. A Circle (zj'ioxoc, ' circle, ring) is a portion of a plane bounded by a curved line every point of which is equally distant from a point within called the center, 13. All Ellipse (^skXeKpiq) is a portion of a plane bounded by a curved line any point from which, if two straight' lines are drawn to two points within, called the_/bc«', the sum of the two lines" will be constant. 13. A triangle (Lat. Triangulum, from tres, tria, three, and angulus, corner, angle) is a polygon bounded by three straight lines. 14. An Angle is the opening between two lines which meet in a point. rl. Straight Angle. 15. Angles\^-^'^'^'''^f,'- Is. Oblique 1. Acute. 2. Obtuse. 16. '- A Straight Angle has its sides in the same line, and on different sides of the point of meeting, or vertex. 17. A Right Angle is half of a Straight Angle, and is formed by one straight line meeting another so as to make the adjacent angles equal. 18. An Oblique Angle is formed by one line meeting another so as to make the adjacent angles unequal. 19. An Acute Angle is an angle less than a right angle. 30. An Obtuse Angle is an angle greater than a right angle. -236 FINKEL'S SOLUTION BOOK 31. A Might Triangle is a triangle, one of whose angles "is a right angle. 33. An Oblique- Angled Triangle is one whose angles are all oblique. 33. An Isosceles Triangle is one which has two of its sides equal. 34. A Scalene Triangle is one which has no two of its sides equal. 35. An Equilateral Triangle is one which 'has all the sides equal. 36. A Quadrilateral {l-'&t. quadrilaterus , irom quaiuor , four, and latus, lateris, a side) is a polygon bounded by four ^straight lines. ' , 37. A Parallelogram {UafiaXXriX/iYpap.ti.uv, from Uajidllriloz, parallel, and ypaiLix-q, a stroke in writing, a line) is a quadrilateroil having its opposite sides parallel, two and two. 38. A Might Parallelogram is a parallelogram whose angles are all right angles. 39. An Oblique Parallelogram is a parallelogram whose angles are oblique. 30. A Mectangle (Lat. rectus, right, and angulus, an •angle) is a right parallelogram. 31. A Square is an equilateral rectangle. 33. A Mhomboid (fio/i^ustSi^, from fi6[j.^tK, rhomb,, and ■ildoq, shape) is a parallelogram whose angles are oblique. 33. A Mhombus i<5(5^/?o?, from fiifi^eiv, to turn or whirl round) is an eqtiilateral rhomboid. 34. A Pentagon {nevrdywvnv, IUvts, five, and ytuvia, angle) is a polygon bounded by five sides. Polygons are named in reference to the number of sides that bound them. A Hexa- gon has six -sides; Heptagon, seven; Octagon, e.\^\\\.; Nonagon, nine; Decagon, ten; Undecagon, eleven; Dodecagon, twelve Tr (decagon, thirteen ; Tetrddecagon, fourteen; Pentedecagon fifteen; Hcxdecagon, sixteen; Heptadecagon, seventeen; Octa ■decagon, "eighteen; Enneadecagon , nineteen; Jcosagon, twenty Icosaisagon , twenty-one; Icosadoagon,Xv/enty-tvfo;Icosatriagon twenty-three ; Icosatetragon, twenty-four ; Icosapentegon, twenty five; Icosakexagon, twenty-six; Icosaheptagon, twenty-seven Icosaoctagon, t-v/eaty-e\ght; Icosaenneagon, twenty-nine; Tria contagon, thirty;- Tricontaisagon, thirty-one; Tricontadoagon thirty-two; Tricontatriagon, thirty-three; and so on to Tessa racontagon, forty; P entecontagon , fifty; Hexacontagon, sixty MENSURATION. 237" Ilebdomacontagon , seventy \ Og'doaconia^on, eighty; Enenacon- tagon, ninety; Hecatonagon, one hundred; Diacosiagon, two hundred; TVz'aco^z'a^o;?, three hundred; Tetracosiagon, lour hun- dred; Pentecosiagon, five hundred; Hexacosiagon, six hundred; Ileptacosiagon, seven hundred ; Okiacosiagon, eight hundred ; Enacosiagon , nine hundred; Chiliagon, one thousand; &c. -35. A Spherical Surface is the boundary betw^een a., sphere and outer space. » 36. A. Conical Surface is the boundary .between a cone and outer space. 37. A Cylindrical Surface is the boundary between the cylinder and outer space. 38. A, Solid is a part of space occupied by a physical body, or marked out in any other way. 39. A JPolyhedron (IloXOedpo^, from ^oio^, .'many, andl iSpa, seat, base) is a solid bounded by polygons. 40. A Prism is a polyhedron in \vhich two of the faces, are polygons equal in all respects and having theiai honjologous- sides parallel. 41. The Altitude of a prism is the perpendicular distance- between the planes of its bases. 43. A Triangular Prism is one whose bases are trian- gles. 43. A Quadrangular Prism is one whose bases are- quadrilaterals. 44. A ParalleHopipedon is a prism whose bases are- parallelograms. 45. A Right Parallelopipedon is one whose lateral edges are perpendicular to the planes of the bases. 46. A Rectangular Parallelopipedon is one whose faces are all rectangles. 4*7. A Cube x6j9o?, a cube, a cubical die) is a rectangular- parallelopipedon whose faces are squares.. 48. A Right Prism is one whose lateral edges are per- pendicular to the planes of the bases. 49. An Oblique Prism is one whose lateral edges are oblique to the planes of the bases. 50. A Pyramid {napa/uq) is a polyhedron bounded by a^ polygon called the base, and by triangles, meeting at a. common, point called the vertex of the pyramid.. ^38 FINKEL'S SOLUTION BOOK. 51. The Conveac Surface of a pyramid is the sum of the triangles which bound it. 53. A Might Pyramid is one whose base is a regular polygon, and in which the perpendicular, drawn from the vertex to the plane of the base, passes through the center of the base. The perpendicular is called the axis. 53. A Tetrahedron {rirpa four, and idpa, seat, base) is a pyramid whose faces are all equilateral triangles. 54. The Altitude (l^at. AltHudo, from alius, high, and ude denoting state or condition) of a pyramid is the perpendicu- lar distance from the vertex to the plane of the base. 55. The Slant Height of a. pyramid, is the perpendicu- lar distance from the vertex to any side of the base. 56. A Triangular Pyramid is one whose base is a. triangle. , .57. An .Octahedron (SxTdedpo? from dxrco eight, and kdpa seat, base) is a polyhedron bounded by eight equal equi- lateral triarigles. 58. A Dodecahedron (dwSexa, twelve, and idpa, seat, base) is a polyhedron bounded by twelve equal and regular pentagons. 59. An IcosahedrOn {sixoqi, twenty, and e5^a„ seat, base) is a polyhedron bounded by twenty equal equilateral triangles. 60. A Cylinder (xohvdpo^, from xuXivSecv, xuXieiv^ to roll) is a solid bounded by a surface generated by a line so mov- ing that every two of its positions are parallel, and two parallel planes. 61. The Axis (a^tuv) of a cylinder is the line joining the centers of its bases. 63. A Mights Cylinder is one whose axis is perpendicu- lar to the planes of the bases. 63. A Cone (xwvq(;, from Skr. CO, to bring to a point) is" a solid bounded by a surface generated by a straight line moving so as always to pass through a fixed point called the apex,, and a plane. 64. A Might Cone is a solid generated by revolving a right-angled triangle about one perpendicular. 65. An Oblique Cone is one in which the line, called the axis, drawn from, the apex to the center of the base is not perpendicular. 66. The Frustum (Lat. /rustuni, piece, bit) of a pyra- mid or a cone is the portion included between the base and a parallel section. 67. A Sphere {a<paXpa) is a solid bounded by a curved MENSURATION. 239 ■surface, every point of which is equally distant from a point within, called the center. ' Before we enter into the solution of problems in Mensuration, it will be necessary first to explain a difficulty which we en- counter. _ The common way of teaching that feet multiplied hy feet give square feet is wrong ; for there is no rule in mathematics justify- ing the multiplication of one denominate number by another. If it is correct to %a.y feet multiplied hy feet give 'square feet, we might, with equal propriety, say dollars multiplied by dollars give square dollars — a product wholly unintelligible. In all our reasoning, we deal with abstract numbers alone or the symbols of abstract numbers. These do not represent lines, surfaces, or solids, but the relations between these numbers may represent the relations between the magnitudes under consideration. Suppose, for example, that the line AB contains 5 units, and the the line .5 C 4 units. Let a denote the abstract number 5, and b the abstract number 4. Then <z5=20. Now this product ab is not a surface, nor the representation of a surface. It is simply the abstract number 20. But this number is ex- actly the same as the number of square units contained in the rectan- gle whose side's are AB and B C, as may be seen by constructing the I'ec- ta.ng\e AB CD. Hence the surface of the rectangle is measured by 20 squares described on the unit of length. This relation is universal, and we may always pass from the abstract thus obtained by the product of any two letters, to the measure of the corresponding rectangle by simply considering the abstract units as so many concrete or denominate units. In like manner, the product of three 'letters abc is not a solid obtained by multiplying lines together, which is an impossible operation. It is simply the product of three abstract numbers represented by the letters a, b, and c, and is consequently an abstract number. But this number contains precisely as many units as there are solid units in the parallelopipedon whose edges correspond to the lines a, b, and c; hence, we may easily pass from the abstract to the concrete. Hence, if we wish to find the area of a rectangle whose width is 4 feet and length 6 feet, we simply say, 6x4=24 square feet. We pass at once from t"he ab- stract in the first member to the concrete in the second. It is a question whether pupils should be taught a falsehood in order that they may learn a truth. (See Bledsoe^ s Philosophy of Mathematics, f p. 97-106.) A 240 FINKEL'S SOLUTION BOOK. I. PARALLELOGRAMS. Prob. I. To find the area of a parallelogrram ; whether it- he a square, a rectangle, a rhomboid, or a rhombus. Fornvula. — A^=l%b, where ^^area, /:=length, and b= breadth ; or, A=hy,CL, where yl^area,,3^base, and (j!=altitude. Rule — Multiply the length by the breadth; or, the base by t he- altitude. I. What is the area of a parallelogram whose length is 15- feet and breadth 7 feet? By formula, ^=/x^=lengthXbreadth=15X 7=105 sq. feet. 11. 15 feet=length. 2. 7 feet=breadth. 3. .-. 15X7=105 sq. ft. =area. III. .-. The area i s 105 sq. ft. FIG. 4. Note. — The Sa.se is not necessarily the side toward the ground. Thus in, the parallelogram ABCD, BC may be considered the ba^e, ip which case,, the altitude would be theperpendicular distance .Ei^, between the sides .ffC" and AD. If HG and BC-were given, we could not find the area of the par- allelograrti because we have not the base and altitude given. ' I. What is the area of the parallelogram ABCD, ii £C is 26 feet and -e^ 50 feet? By formula, ^=aX^=-£'i?"X^C=50X 26=1300 sq, ft. 26 feet=.S C=base. 50 feet=^^=altitude. ,[3. .-.26X50=1300 sq. ft.=area. III. .-. The area of AB CD=1ZQ0 sq. ft. ll.{ 2. I. A floor containing 132 square feet, is 11 feet wide ; what: is its length? By formula, A=lx&- ■■■ /=^-r-&=132-Jrll=12. ft.. {1. 132 sq. ft.=area. 2. 11 ft.=breadth. 3. 132-^11=12 ft.=length. III. .-. The floor is 12 ft. long. 1 MENSURATION. 241 Prob. II. The diagonal of a square being- given, to find the area. > 'Formula. — A^d ^-=-2. Rule. — Divide the square of the diagonal by 2, and the quotient will be the area. I. What is the area of a square | whose diagonal is 8 chains ? By formula, A=d ^^-2 = 8^-7-] 2=32 sq. chains. l." 8 ch.^length of diagonal= I BD. 2. 64 sq. ch.=8x8=-EGi^Zr=,| II. ,J square described on the di- agonal BD. 32 sq. ch.=64 sq. ch.-^-2= | area of the square AB CD. III. .-. 32 sq. ch.^the area of the FIG. 5. square. Prob. III. The area of a square being given, to find its diagonal. Formula.— d='\/2A. Rule. — Extract the square root of double the area. I. The area of a square is 578 sq. ft. ; what is the diagonal? By formula, o?=-\/234=a/2X578^Vi156=34 feet. !1. 578 sq. ft.=area of the square. 2. 1156 sq. ft.=2X578 sq. ft.=double the area. 3. 34 feet^'\/ii56=^the' diagonal. III. .-. The diagonal is 34 feet. ^ Prob. IV. The diagonal of a square being- given, to find its side. Formula. — 5==\/p^- Kule. — Extract the square root of one-half the square of the I. ■ What is the side of a square whose diagonal is 12 feet? By formula, 5'='\/p^=Vp<T2^=V72=6-\/2=8.4852+ft " rl. 12 ft.^the diagonal. * ^^ I 2. 144 sq. ft.^l2^:=square described on the diagonal. ) 3. 72 sq. ft.=area of square whose side is required. 14. .-. 8.4852 ft.=6V'2=V72^side of the square. III. .-. The side of the square is 8.4852-t-ft. 242 FI^JKEL'S SOLUTION BOOK. Prob. V. To find the side of a square havlngr its area Sriven. . Formula. — 5=^^- Kllle. — Extract the square root of the number denoting its area. I. What is the side of a square field whose area is 2500 square rods? By forn^ula, 5=V']4=V2500==50 rods. jy i 1. 2500 sq. rd.=area of the field. ■ I 2. 50 rd.=V'2500=side of the square field. III. .-. The side of the field is 50 rods. PROBLEMS. 1. Find the area of the parallelogram ABCD; given AC=1 ft. 2 in., and the perpendicular from B on AC, 3 feet. [See Fig. 4, on p. 240.] 2. Find the area of a parallelogram in which one side is 4 ft. 3 in., and the perpendicular distance between this and the opposite side is 4 feet. 3. The area of a parallelogram is Vlyi acres, and each of two parallel sides is 42 chains ; find the perpendicular distance between them. 4. Find the area of a rhombus, a side of which is 10 feet and a diagonal of 12 feet.-^ ~[The diagonals of a rhombus bisect each other at right angles.] 5. Find the area of a rhombus whose diagonals measure 18 feet and 24 feet. 6. A field is in the form of a rhombus, whose diagonals are 2870 links and 1850 links ; ^find the rent of the field at |5 per acre. 7. The diagonals of a parallelogram are 34 feet and 24 feet, and one side is 25 feet ; find its area. 8. Find the cost of carpeting a room, 30 feet long and 21 feet wide, with carpet 2 feet wide at' 80 cents per yard. 9. How many square yards are there ih a path, 4 feet wide, surrounding a lawn 24 yards long and 22'yards wide ? 10. How many yards of paper, 20 inches wide, will be required to paper the walls of a room, 16 feet by 14 feet by 9 feet, allowing 8 inches for a base- board at the floor and 12 inches for border at the ceiling? ^ 11. The perimeter of a rectangle is 56 feet ; find its area, if its length is 3 times its breadth. 12. What is the area in acres of a square whose perimeter is such that it takes 12 minutes to run around the square, at the rate of byi mil^s per hour ? 13. Cut a rectangular board, 16 feet long and 9 feet wide, into two pieces in such a way that they will form a square. 14. How many feet of framing, 4 inches wide, will it take to frame a picture, 3 feet by 2 feet ? Ans: 6 ft. 4 in. 15. A sheet of galvanized iron, 50 inches wide, is placed against the top of a wall, 6 feet high, while the lower edge is 5 feet 5 iilches from the foot of the wall ; find the area of the sheet of iron. Ans. 4850 sq. in. 16. Allowing 8 shingles to the square foot, how many shingles will it take to roof a barn whicb is 40 feet long and 15 feet from the comb to the eaves ? Ans. 9600 shingles MENSURATION. 243 17. The area of a square is 169 sq. ft. ; find its perimeter, in chains. 18. What is the side of a square, of which the number expressing its area in square feet is equal to the number expressing its perimeter in yards? Ans.ly^ feet. 19. What is the area of a path a yard wide, running diagonally across a square lawn whose side is 30 feet? ^«J. 648[20i/ 2 — 1] sq. in. 20. What is the area of a square whose diagonal is 12 feet ? Ans. 72 sq. ft. 21. What is the area of a square whose diagonal is 5 feet longer than its side ? Ans. 25(3+2/"3) sq. ft. 22. In a garden, in the form of a square, there is located a spring whose distances from the three corners. A, B, C, are 30, 40, an d 50 feet, res pect- ively. Find the length of a side of the garden. Ans. 5|/( 81+52 i/S ) ft. 23. Convert a square, whose sides are 20 inches, into an equivalent area "bounded by circular arcs only. Note.— Tilt required figure is called apelicoid. II. TRIANGLES. ' Prob. VI. Given Ithe base and altitude of a rigbt-ang'led triangle, to And the hypotheUuse. Formula. — ^=v'aH^- lillle. — T'o the square of the base add the square of the alti- tude and extract the square root of the sum. I. In the right-angled triangle A CB, the base A C'=56 and the altitude B C=33 ; what is the hypothenuse ? By formula, h =V'^M^=V38^+56^=Vl089+3136=V'4225 =65. f 1. 56=^ C=the base. 2. 3136=562= the square of the base. 3. 33=^Ci=the altitude. 4. 1089=332=the square of the altitude. 5. 4225=3 136+1089=the sum of "<?. 6. the squares of the base and altitude. 6. 65=\/4225=the square root of the sum of the squares of the base and altitude=the hypothenuse. III. .'. The hypothenuse^65. Prob. yil. To find a side, when the hypothenuse and the other side are g-iven. Formulas.— \ f=V^^ 'Rnle.—B'rom the square of the hypothenuse subtract the square of the given side and extract the square root of the remainder. IlJ 244 FINKEL'S SOLUTION BOOK. I.. The hypothenuse of a right-angled triangle is 109, and the altitude 60; what is the base? IIA By formula, ^^-y//''— n^^Vl09'— 602=V8281==91. 1. 109=hyp6thenuse. 2. 1 1881^109 ^=square of the hypothenuse. 3. S0=the altitude. 1. 3600=602==the square of the altitude. 5. 8281=11881— 3600=difference of the squares of the hypothenuse and .altitude. 6. 91=\/8281=the square root of this diffe¥ence=the base.: III. .-. The base is 91. Remark. — When «=*, ^=v 2a^^av 2. From- this, we see that the diag- onal of a square is v 2 times its side. Prob. VIII. To find the area of a triang-le, having- g-iveu the base and the aljfcitude. , .. \-i ' ■■■ - ' . Formula. — A^lay^b., Rule. — Multiply the base by the altitude and take ' half the product, I. What is the area of a triangle whose base is 24 feet and altitude 1.6, feet? By formula, ^=^aXi5=iX 16X24=192 sq. ft.' .1. 24 ft.=base. 2. 16 ft.=altitude. 3. 38-1 sq. ft.=^16 X24 = pro- tl.<J duct of ba,sp and altitude. 4. 192 sq. it.=\ of 384 sq. ft.= half the product of the base and'the altitude^area. III. .;. The area of the triangle is ■ 192 sq. ft. il FIG. 7. Proh. IX. To find the area of a triangle, having- g-iven its three sides. Formula. — A-='\/ s(s—a)(s—b){s—c), where .y^-|- (^a-\-b-\-c) . *Ilule. — Add the three sides together and take fialfihe sum; from the half sum, subtract each side separately ; multiply the half sum and the three remainders together and extract the square root of the product. '''Demonstration. — In Fig. 7, let ^ C=6, BC^=a, and AB^c. In the right-angled triangle ADB, BV^AB'—AD'^, and in the right-angled tT\a.a%\e.,CDB,BD'^—BC^—nC^. .: AB^—AB^=BC^—DC', or c*— II. MENSURATION. 245 I. What is the area of a triaijgle whose sides' are 13, 14, and 15, feet respectively? By formula, A =\/ s(s—a)(s— b){s—f)^sl21->i (21—13) X.(2]— 14)X(21— 15)^21X8X7X6 = ^7056=84 sq. ft. 1. 42 ft.=13 ft.+14 ft.-fl.5 ft.==sum of the three sides. 2. 21 ft.=^ of 42 ft.=half the sum of the three sides. 3. 21 ft.— 13 ft.=8 ft.=first remainder. ,4. 21 ft. — 14 ft.=7 ft.=second remainder. 5. 21 ft. — 15 ft.^6 ft.=third remainder. [mainders 6. 7056=21X6x7X8=product of half sum and three re- 7. 84 sq. ft.^= ;^7056==square root of the product of the hall" sum and three remainders^the area of the triangle. III. .-. The area of the triangle is 84 sq. ft. Prob. X. To find the radius of a circle inscribed in a triangle. Formula.— Ii=2A-T-{a-{-b-\-c ) . *Rule. — Divide twice the area of the triangle by the sum of the three sides. I. Find the radius of a circle inscribed in a triangle whose sides are 8,4, and 5 feet, respectively. 1. 6 sq. ft.= 4s{s — a){s — b)[s — c)=area of the triangle, by formula, Prob. IX. 2. 12 sq. ft.^twice the area of the triangle. 3. 12 ft.=3 ft.+4 ft.+5 ft.=sum of the three sides. 4. .-. 1 ft.=12-T-12=twice the area divided by the sum of the sides=the radius of the inscribed circle. III. .■. The radius of the inscribed circle is 1 ft. AJy^=a^—JDC^, whence c^—a^—AD'^—DC^. But AD^—DC'^^^(^AI> -\-DC\{AD—DC)=b{AD—DC). :. b{AD—I)C)=c^—a^,&ni AD— DC ^=(c^ — «*)-^*. But ^Z'+-OC=*. .-. By adding the last two equations, we have 2A£>= j 1-*= r""^ — ; whence AD= —L — Since 2o B7)^=AB^—A£>^=^^—AZ>^, if we substitute the value of AD just found wehave^^^=c^-(^ ^^ J = ^^ = (2&<r+'^'— "'+*' ) (2»<:— c'+g'— »' ) _ (^' + 2fc-H:'— g' )[g2— (^»— 2&c-t-e')3 4** ■ W~ ~ ■ - 4p .■.BI>=^[{b+cY—a^\[a^—ib-cy],=~»js{s-a)(s—b)is-c). Now the area of ABC—\AC-><.BD. ,. A-\bXBD=i£b-Xy[l.b+cy—a^-\W~(,b-c)^\ ::= sls(s—a)(s—b){,s—c), where 2s={a+b-\-c). Q. B. D. *Note. — For Demonstration, see any geometry. i\.K 246 FINKEL'S SOLUTION BOOK. Prob. XI. To find, the radius of a circle, circimiserlbedl about a triang-le whose sides are given. .„ , r, ^^c abc Formula. — R= — =- AA 4 \ls{s—a){s—b-){s—c). *Rllle. — Divide'the product of the three sides by four times the area of the triang-le. ^ I. What is the radius of a circle circumscribed about a tri- angle whose sides are 13, 14, and 16 feet, respectively? 1. 2730 cu. ft.=1 3><14Xl5=the prod uct of the three sides., 2. 84 sq. ft.= Vj(j — «)(•$■ — l>){s — c)=the area of the tri- angle, by Prob. IX. [angle. II. ■{ 3. 336 sq. ft.=4x84 sq. ft.=four times the area of the tri- 4. 8| ft.=2730-i-336^the product of the three sides divid- ed by four times the area of the triangle=the radius of the circumscribed circle m. .■. The radius of the circumscribed circle is 8^ ft. Prob. XII. To find the area of an equilateral triangrle,. having given the side. Fovm/ula,.- — ^=:^V3.f ^, where j=side.'- This is what Prob.. IX. becomes, when d.=b=c. Rule. — Multi-ply the square of a side by \ \/3,=.433013-f-. I. What is the area of an equilateral triangle whose sides are 20 feet? * Demonstration. — Let ABC lie any triangle, and ABCB the circumscrib- ed circle. Draw the diameter BE^ and draw EC. Draw the altitude BD of the triangle ABC. The triangles ADB and BCE a.re similar, because both are ^ right- langled triangles, and the angle £iAD=:th.e angle BEC. Ueace, AB:EB:.BI). BC. Hence, ABY.BC=BE^BD or ac=2Ry. BD. But, in the demonstration of Prob. o ^_ IX., we found .BZ>:=-^\/«(s— a)(«— »)(«— o). 2 .-. ac = 2/?X-T-v'«(8-a)(«-6)C«-<!). Whence o - abc abc n z^ f 4v'«(«-o) (»-») (»-c) 4^ Fie. 8. MENSURATION. 247 By formula, A=\ VSX 202=100 V3=l'?3.205+sq. ft. rl. 20 ft.^length of a side. 1 2. 400 sq. ft.=202=square of a side. 3. 173.205 sq. ft.=i v'3X400=.433013X400=iV3 times I ' the square of a side,:=the area of the triangle. III. .". The area of the equilateral triangle is 173.2054-sq. ft- Prob. XIII. The area and base/ of a triangle being' given, to cut off a triangle containing a given area, by a line run- ning parallel to one of its sides. II. ■• — ^ = ^'s|~4'' Fovmula, — ^ = 3-J-t-, where ^=area of the given triangle; ^,the base of the given triangle; and^ , the area of the portion to be cut off. Hule. — As the area of the given triangle is to the area of the triangle to be cut off, so is the square of. the given base to the square of the required base. The square root of the result will be. the base of the required triangle. I. The area of the triangle AB C is 250 square chains and the base AB, 20 chains ; what is the base of the. trian- gle, area equal to 60 sq. chains, cut off by ED parallel to BCl __ _ By formula, ^Z>=3'=i5j^=:20.>J^=4V6 =9.7979 + ch. 1. 250 sq. ch.=area of the given triangle ABC. 2.-60i. sq.- ch.p=area of the " "triangle, ,:^i5Z>. 3. ,20 ch.=ib;ase 4f the trian- ^.gle j4^C.'-(,;- '4/^ .". 250 sq. c-h.>:,60 sq. ,ch. '^' v.l^t-.AD^. Whence AD^= (400x60) -h^ 250 =96. ,.-. ^Z>=^%=9.7979+ch. ™. 9. III. .-. The base ^Z?:^9.7979+ch. PROBLEMS. 1. A man travels 20 miles north, then 15 miles due east, finally 28 miles due south ; what is the distance from his starting point ? Ans. 17 mi. 2. A ladder, 50 feet long, is placed so as to reach a window 48 feet high, and on turning the ladder over to the other side of the street it reaches a point 14 feet high. Find the breadth of the street. 3. The hypotenuse of a right-angled triangle is 55 feet and the base is % of the altitude. Find the two sides. II. 248 FINKEL'S SOLUTION BOOK. 4. The hypotenuse of a right-angled triangle is 13 feet and the sum of the sides containing the right angle is 17 feet. Find these sides. 5. In a right-angled triangle the area is half an. acre, and one of the sides containing the right angle is 44 yards ; find the other side in yards. 6. Find the area of a triangle whose sides are 21 feet, 20 feet, and 13 feet, respectively. Also 21 feet, 17 feet, and 10 feet. 7. In a right-angled triangle the sides containi;ig ■the right angle are 30 feet and 40 feet. Find the length of a perpendicular drawn from the right angle to the hypotenuse. 8. The perimeter of a triangle is 48 feet. If one side is 10 feet and the area is 84 square feet, find the two remaining sides. 9. The area of an equilateral triangle is 80 square feet. Find the length of a side. What is the side of a square of equal area ? 10. The sides of a triangle are proportional to 3, 4, and 5. If the perim- eter is 84 feet, find the sides and the area. . III. TRAPEZOIDS. Prob. XIV. To find the area of a trapezoid, haTlng; given the parallel sides and the altitude. Formula. — A=^(l>-{-b')a, -where b and l> are the parallel sides and a, the altitude. Rule* — Multiply half the sum of the parallel sides by the al- titude. I. What is the area of a trapezoid whose parallel sides are 15 meters and 7 meters and altftude 6 meters? By formula, ^=^(^-fy) x«=|(15+7)X6=66 m«. 1. 7 m.^Z>C,thelengthof one of the parallel sides, and 2. 15 ra.=AB, the length of the pther side. 3. 22 m.^7 m.-(-15 m.^sum of the parallel sides. 4. 11 m.=^ of 22 m.=half the sum of the parallel sides. f7ff. 10. .5. 66 m2.=6xll=area of the trapezoid, ABCD. III. .•. The area of the trapezoid is 66 m^. PROBLEMS. 1. The parallel sides of a trapezoid are 18 feet and 24 feet, and the alti- tude is 8 feet ; find the area. 2. The parallel sides of an isosceles trapezoid are 16 feet and 20 feet, and the non-parallel sides are 10 feet each ; find the area of the trapezoid. 3. The line joining the middle points of the non-parallel sides of a trapezoid is 12 feet, and the altitude is 8 feet ; find the area of the trape- zoid. Ans. 96 sq. ft. II. MENSURATION. 249 IV. TRAPEZIUM AND IRREGULAR POLYGONS. Prob. XV. To flnd the area of a trapezium or any irreg-- nlar polygon. Rule. — Divide the. figure into triangles , find the area of the triangles and take their sum. I. What is the area of the trapezium ABCD, whose diag- onal ^ C is 84 feet, and the perpendiculars BE and ^jF, 56 and 22 feet, respectively? 84 ft.=^C=lDase of the triangle AD C. 56 ft.=Z>^ = altitude ot ADC. .-. 2352 sq. ft.=i(^CX D -E)=area of the trian- gle ^7) C. II..' 4. 84 ft.=^C=base of the triangle AJ5C. 5. 22 ft.=^^= altitude' of ABC. 6. .■.924sq.ft.=^(^CX^-?^) FI6. J1. =area of the triangle ABC. 7. 3276 sq. ft.=2352 sq. ft.+924 sq. {t.=ADC-\-ABC= area of the trapezium AB CD. III. .-. The area of the trapezium ABCD is 3276 sq. ft. I. What, is the area of the quadrilateral, ABCD, if AB is 30 feet, BC 17 feet, CD 25 feet, DA 30 feet, and the diag- onal BD 26 feet? By formula for the area of a triangle, A^V s{s — a) (s — b) (s—c) +'Vs{ s — a') (s — d') (s — c') where a, 6, c and a', b' , c', are the sides of the triangles ABD and BCD, respectively. 1. 30 feet=^5=a, 2. 26 ite.t=BD=b, and 3 28 fppf — DA — c 4. . . 42 feet=J(30 feet-f26 feet-|-28 feet)=4(a-|-&+c) II. -! 5. . . 336 sq. feet=-i/ 42Xl2X14X16=the area of the tri- angle ABD, and 6. 204 sq. f eet= v'34X8X9X17=the area of the triangle BDC. 1. .-. 336 sq. feet-l-204 sq. ft.=540 sq. ft.=the area of ABCD. , III. .-. The area of ABCD=5M sq. feet £^orollary 1. If a quadrilateral is such that its diagonals 250 FINKEL'S SOLUTION BOOK. AC and BD are at right angles to each other the area is given by the formula area=i(ACXBD). Corollary 2. If the diagonals are inclined to each other at any angle a degrees, the area is expressed by the formula, area=^(ACXBDXsm a ). PROBLEMS. 1. In the trapezium ABCD, AB=30 in., £C= 17 in., CZ'=25 in., I)A= 28 in., and the diagonal BZ>^26 in.; find its area. Ans. 540 sq. in. 2. In the quadrilateral ASCD, the diagonal AC='\S in., and the per- pendicular on it from .5 and D axe 11 inches and 9 inches respectively;, find the area of the trapezoid.- Ans. 180 sq. in. 3. In the trapezium ABCD, the diagonals AC and BD are perpendicii- larto.each other and measure 16 feet and 2^ feet respectively; find the area. , ' , Ans. 2 sq. yds. 4. , Find the area of the trapezium yi^SCZ?, in which the angles .<4.ffC" and CD4,2.x€. right angles, and.AB is 15 feet, BC is 20 feet, and CD is 7 feet. '••'.' ■ . , • . '. r, '. Ans. 23i sq. ft. 5. -".Find the a^ea ^if theJquadrilateral ABCD, having given that the ap^e?SiJ^Cis''S0°,.ApCisf4: tight angle, ^4.5=13 chains, .ffC=13 chains, and CZ?=l2*'chains. ' .<4»j. 10.31 acres.. 6. The area of a trapezium is 4 acres and the two diagonals measure 16 chains and 10 chains respectively ; at what angle are the two diagonals, inclined to each other ? Ans. 30°. Hint. — I^et / be the intersection of the diagonals.' Then Xfrom B on AC=BI sin.. LBIC, J.from D on AC=DI &\-a..LDlA[=LCIBi ■ . area oiABCD—%AC\BI sin. BIC+DI sin. CIB]=]iAC'X.BD-X.si-a. LBIC. 7. ,Find the area of the polygon ABCDEF, if AD=l&lb links, li^/?" from 'F on ^Z^=850 links,i.BQ from B on AD=1W links,! CS from Con on ^Z?=^500 links,! £■/? from E on ^Z»=250 links, y4P=900 links, AQ= 1040 links, ^.ff=1200 links, and ^5=1380 links. Ans. 9.03625 acres. 8.. Find the area of the field ABCDEF, if ^C=2900 links, Cfi'=25D0- links, £'^=3600 links,! .gJf from B on ^C=400 iinks,!^yfrom D on C".£'=400 links, and i_FZ from F on AE=9bO links. Ans. 63 A. 3 r. 24 p. 9. Find the area of the -polygoa ABCDE, if ^4.5=12 inches, /_ABC R. right angle, BC=b in., CZ?=14 in., AD=15 in., l_ADE a right angle, and DE=% in. Ans. 1 sq. ft. 30 sq. in. 10. What is the area of a figure whose sides are 10, 12, 14, and 16 rd. in order, and the distance from starting point to opposite corner is 18 rd j" Ans. 1 A. 3.9— sq. rd. 11. What is the area of a figure made up of three triangles whose bases, are 10, 12, and IP rd., and whose altitudes are 9, 15, and lOJ rd.? Ans. 1 A. 59 sq. rd. V. REGULAR POLYGONS. Prob. XVT. To find the area of a regular polyg-on. FoTmula. — A=^\aXp, vsrhere /) is the perimeter and a, the apothem. Rule.- — Multiply the perimeter by half' the apothem. MENSURATION. 251 The Perimeter of any polygon is the sum of all its sides. The Afothem is the perpendicular drawn from the cente» to any side of the polygon. I. What is the area of a regular heptagon whose svJe is 19.38 and apothem 20? By formula, ^=^flXi>=iX20X(7xl9.38)=1356.3. 1. 19.38^=length of one side. 2. 135 66=length of 7 sides=the perimeter. 3. 20=apothem. 4. 10=^ of 20=half the 'apothem. 5. 1356.6=10 Xl35.66=product of perimeter by half the apothem. ' III. .'. The area of the heptagon is 1356.6. Prob. XVII. To find the area of a regular poiygon. when the side only is given. *Ilule. — Multifly the square of the side of the polygon by the number standing opposite to its name in the following table of areas of regular polygons whose side ts, 1 : \\.\ Name. Sides. Multipliers. Triangle, Tetragon, or square. Pentagon, Hexagon, .Heptagon, Octagon, Nonagon,,,. _ ,, ,- ^ , .:. Pecagoni f, - ■■■ ■^- Undecagon, Dodecagon, > t-S" 3 4 5 6 ■ 7 8 9 10 11 12 .4330127- 1.0000000. iViTl\^' = 1.7204774. 2.5980762. 8.6339124. 4.8284271. 6.1818242. f cot. if»° 2+2V2 ■| cot. 20° fV5+2V5 = 7.6942088. Vcot. yT"°= 9.3656399,. 3(2+V3) =11.1961524. , *Z«e;«o«i^^(rf<o»,-r$ince" a ^Sgijlar polygon can be divided into as many equal isosceles triangles' as it has sides, we may find the area of one trian- fle ^nd'multi'ply this:ar6a by the number of triangles, for the whole area. ,ef.^^ffC beone of these isosceles triangles, taken from a polygon of n s'des, AB and 5Cthe equal sides, and .4C the base. The angle at the ver- tex 5=360°-^». ^=iC180°— 360°-f-«)=C. From B let fall a perpendicu- BD ' ^ 1 S0° lar on AC at D. Then by trigonometry, -— — =tan (90°— ^-^). .: BD= \A G n (180° \ V The area of the triangle ABC^\ACy,BD=\AC'^ zot /180^\ _ . .pj^g ^j.g^ ^j the polygon=^^ C' cot (^— '\=2.,i where .s^side. By placing ^=1, and »=13, 14, 15, &c., respectively, the area, of polygons of 13, 14, 15, &c., side respectively, may be found. (^) •252 FINKEL'S SOLUTION BOOK. Prolb. XVIII. To find the side of an inscribed square of a ■triang-le, having: griven the base and the altitude. Formula. — s=:- — -^, where 5=side, i> the base, and a a-f-o ti\e altitude. *Rule. — Divide the product of the base and altitude by their 'iutn. I. What is the side, of an inscribed square of a triangle^ ■whose base is 14 feet and altitude 8 feet? By formula, j= — t-j==: ab a+b' 14X8 '14+8 -•tt lU 1. 2. 3. 4. 15. 8 feet=the altitude. 14 feet=the base. 112 sq. ft.=14x8=the product of the base and altitude. 22 feet=14 ft. +8 ft.=their sum. 112-T-22=the product divided by the sum. III. StV feet , 6j\ ft.= nthe side of the inscribed square. PROBLEMS. 1. Within a given regular hexagon, drawn on a side of 10 inches, a •second hexagon is inscribed by joining the middle points of the sides taken in order. Find the area of the inscribed figure. Ans. 194.85 sq. in. 2. ,Find the area of a,regular pentagon on a side of 10 inches. , y Ans. 172.04 sq. in. * Demonstration. —l^ct ABC be any triangle whose base is * and altitude a. Produce AC to H,iaakingCIf=,^Z,. At .«; erect the perpendicular HG and make I/G=SD. Draw , "^ ^ A G and at C, erect the perpen- dicular JFC, and draw J^'K. Then KE=FC^=EN, and KN is the re- quired inscribed square. For, in the similar triangles AUG and ACF, we have AH -.GH:: AC : FC, or a-\-b:a::h:FC. By inver- sion, and then by Division, a:h ■.-.a— FC:FC, or BI-.FC. In the similar triangles ABC and KBE, AC:KE::BD:BI, or BD.AC:: BI-.KE. Whence a:b::BI:KE. :.BI:KE::BI:FC. :. KE^FCz-nd the and its angles right angles by construction A more elegant construction, is to construct a square on the altitude, BD, and having one side adjacent to BD, lying in DH. Then join the vertex, A, of the triangle and the vertex of the square opposite the vertex, D. This line will intersect the side, BC, in E, a vertex of the required square. Jn the same way, we can inscribe a rectangle similar to a given rectangle. FIG. 12. igure Kl^ has its sides equal Hence; it is a square. Q.E.Z>. MENSURATION. 253: 3. Find the area of a regular decagon on a side of 4 inches. J , Ans. liS.lsq. in. 4. Find the area of a regular heptagon inscribed in a circle, radius 6: n 360° ^ inches. [Area of a Regular «-side in terms of the radius is-n- sin. /c^.]. 5. , Find, the" area of a regular heptagon circumscribing a circle whose radius, is 12 jnches. ,|Area of a regular «-side circumscribing a circle in * ' ' 1 80° terms, of the tadius is n tan. (r) ^]. ^ 6., A' fegul^^r octagon is formed by cutting off the corners of a square whose side i'3l'l2 inche^s. Find the side of the octagon. [Side of octagoa ^}ia'{^-^y^ •a), where a is the side of the square]. 7. The a^?a tff a dodecagon is 300 square inches; find the radius of the circle 9irctiiiiscribed about it. 8. ~ Find' the area of the circiilar ring formed by the inscribed and cir- cumscribed circles of a regular hexagon whose side is 20 inches. Show that for a given length of side, the area of the ring is the same whatever the number of sides of the regular polygons. . , ■ 9. What is the area of a path 3 feet wide, around a hexagonal enclosure whose side is 14 feet ? ,' ^»j. 283.17 sq. ft. 10. Find the area of the square formed by joining the middle points of the alternate sides of a regular octagon, whose side is 8 inches. Ans. 186,51 sq. in. 11. The difference between the area of a regular octagon and a square' inscribed in the same circle is 82 . 8 square inches. Bind the radius of the- circle. [Take i/~2=1.414]. Ans. 10 inches. 12. In a circle of a radius 10 inches a regular hexagon is described ; in this hexagon a circle is inscribed; in this circle a regular hexagon is, , inscribed : and so ad infinitum. Find the sum of the areas of all the hex- agons thus formed. Ans. 1039.23 sq. in. 13. In the last example, let the radius be r and the number of sides of the polygon n ; find the sum of the areas of all the circles formed. 180° Ans. ^ «'^-cosec.^ • n 14. In a triangle whose base is 15 inches and altitude 10 inches- a square is inscribed. Find its area. 15. A square is formed by joining the mid points of the sides of a square whose side is 10 feet; in the square thus formed, another is formed in the same way, and so on; find the side of the 7th square thus formed. O^k.?. 1} ft. VI. CIRCLE. Prol). XIX. To And the diameter of a circle, having- given tlie heiglit of an arc and a chord of half tlie arc. Formula. — B^k'^-^a, iu which ^=chord of half the arc and (3!==height. f Rule. — Divide the square of the chord of half the arc by ^the height of the chord. '^Demonstration. — Let AB=^k, the chord of hall' the arc ABC and BZ> =B, the height of the arc ABC. Draw the dinmetev BE and. draw the 254 FINKEL'S SOLUTION BOOK. I. What is the diameter of a circle of which the height of an arc is 5 m. and the chord of half the arc 10 m.? By formula, n=i^-r-a=10^-i-5= 20 m. 11. 1. 10 m.=AB, the length of chord of half the arc. 2. 5 m.=BD, the height of arc. 3. 100 m.*=square of chord. 4. .-. 26 m.==iOO-f-5=BE, the diam- eter of the circle. III. .-. The diameter of the circle is 20 meters. f/c. 75. Prob. XX. Tp find the height of an are, having given the chord of the arc and the radius of the circle. Formula, — «=i? — nJR^—c^ , in which, if=radius and c =|- the chord. *Rule. — From the radius, subtract the square root of the dif- ference of the squares of the radius and half the chord. I. The chord of an arc is 12 feet and the radius of the circle is 10 feet. Find the height of the arc. By formula, a^=R — \JR^—c 2=10 — VlO*— 6»=2 ft. 1. 10 ft.==the radius of the circle. 2. 100 sq. ft.^square of the radius. 3. 12 ft.=the chord. 4. 6 ft.=half the chord. 5. 36 sq. ft.=square of half the chord. 6. 8 ft.=VlOO— 36 = square root of the difference of the squares of the radius and half the chord. 7. .-. 10 ft— 8 ft.=2 ft.=height of the arc. III. .■. The height of the chord is 2 feet. Proto. XXI. To find the cliord of half the arc, having given the chord and height of an are. Formula. — i=^a^-\-c^- radius A O. The triangles ADB and BAE are similar, because their an- gles are equal. Hence, BE:AB::AB:BI), or BB:k::k:a. Whence BE r=Z>=A2-i-a. Q.E.D. N. B. — (1) If a and Z> are given, *=V'-OXn; (2) -if D and k are given * Demonstration.— \ti Fig. 13, we have BD =BO—DO. But ^0= )^AO^—DA^=>f\R'^—<^^\ ■'■ «=-ff— V'>g'— c". If a and i? are given, (1) ^c=%^\R^—(R—aY\ = '2^j(2aR—a^)\ if " and 2c are given, (2) R= n.J MENSURATION. 255 *Ilule. — Take the square root of the sum of the squares of the height of arc and half the chord. I. Given the 6hord=48, the height=10, find the chord of half the arc. • By formula, i=y'a2-)-c2 =^io2+24^=V676=26. 1. 48= the chord. 2. 576=i of 48*=square of half the chord. 3. 10=height of chord. ' jx 1 4. 100'=square of height of chord. j5. 676=5764-100=sum of square of half of chord and height. 6. 26=v'676=square root of sum of square of half of chord and height. III. .-. The chord of half the arc is 26. Prob. XXII. To find the chord of half an arc, having Sriven the chord of the arc and the radius of the circle. Formula. — k=V2Ii ^ — /? v''4^^4c2 . f Rule. — Multiply the radius by the square root of the differ- ence of the squares of twice the radius and the chord; subtract this product from twice the square of the radius and extract the square root of the difference. I. Given the chord of an arc=6 arid the. radius of the circle =5, find the chord of half the arc. By formula^/J=v'2i?8— i?'v/4/f^— 4C2 =V2X52— 5V4X5^— 6' =^10. 1. 5=the radius of the circle. 2. 10=twice the radius of the circle. 3. lOO^square of twice the radius. 4. 6=chord of the arc. 5. 36=square of the chord. 6. 100 — 36=64=difrerence of squares of twice the radius and the chord. 8='\/64=square root of the above difference. 40=5X8^the product of the above square root and the radius. 9. 50=2x5^=twice the square of the radius. .10. -v/50_40=-v/io=chord of half the arc. III. .-. The chord of half the arc is -y/io. * Dem onstrati on. — In Fig. 13, we have AB=^ {AD'^-\-BD'^)^\c^-\-a'^'\:=i ija^+c^. .■.^=V a'+c^. If ^and2caregiven,(l)g=V;ir2— c': if k and a are given, (2) 2c=.2\l k''—a'^ . i Demonstration. — Fro m Prob. XXI.,wehave k^=*i a''-\-c''- . From Prob. XX. we have a=R—>^R'—c^. .-. «'= 2i?'— c '— 2i?V.g'— c'. Sub sti- tuting this value of a^ in the above equation, ,t^v 2.ff^ — Ri^iR^ 4c*. II. 256 FINKEL'S SOLUTION BOOK. Remark. — A continued application of the formula of Problem XXII> enables us to find the ratio of the circumference of a circle to the diame- ter. Thus, let .^=1 and Ci=l, the side of a regular inscribed hexagon. No. of Length of sides.. Length of Side. . Perimeter. . 6 C^-\ =1.00000000 /i =6.00000000 12 Ca=v' 2-V4— Ci^ =(2— \/"3r^ =0.51763809 /^ =6. 21165708 24 C3 = V 2— V4 — C;'' = (2— (2+V"3)") ^) ^ ' " =0 . 26105238 p^-%. 26525722 48 ^4= V2— V4 — C32=(2— (2+(2+v"3r)'^)'^)^ . =0.13080626 />4 =6. 27870041 96 Q=V2-V4— C42={2-(2+(2+(2+V 3) . , , =0.06533817 /b=6. 28206396 192 Ce=V2— V4 — Q2=(2— (2+(2+(2+(2+v"3T^)^)^)''^)^ _^ =0.03272346 >, =6. 28290510 384 C,= V2— V4 — Ce2^(2— (2+(2+(2+(2+(2+V~3)^)^)5^)^)^)^» =0.01636228 />, =6. 28311544 768 C^= V2— V4 — (r,2={2— (2-|-(2-|-(2+(2+(2+(2+vT)'^)^)'-^)^)^)^)^ =0.00818121 ^8=6.28316941 As the number of sides are thus continually inci'eased, the perimeter of the polygons continually approach the circumference of the circle and can be made to differ from the circumference as little as we pleese, though it can never be made to exactly equal the circumference. Hence, we 'may consider 6 . 28317 as approximately the circumference of a circle whose radius is unity. Since IT^R = C, the circumference; therefore, t = J (6.28317) = 3.14159, nearly. Students should learn to use the character, f, instead of 3.14159; for much useless labor can be saved thereby. Thus, suppose we wished to find the mean base of a frustum of a cone, the radii of whose upper and lower bases are 2 feet and 3 feet, respectively. By the riile, we multiply the area of the upper and lower bases together and extract the square root of the product. Now the area of the upper base is 4t sq. ft. and the a rea of the lower base is 9?: sq. ft. Hence, the area of the mean base is *)%%Tfi sq ft. or 6t sq. ft. By using t instead of 3.14159, we have avoided the useless labor of multiplying two numbers, consisting of five decimal places each. While this,' of course, could have been avoided by simply indicating "the multiplication, yet students most often perform the actual multiplication ; and when the multiplication is only indicated, it is much easier arid more quickly done to write one character, w, than to write the five or six charac- ters in 3.14159. Note. — Ahmes, an Egyptian priest who lived somewhere between the years 1700 B. C. and 1100 B. C, gives in a papyrus manuscript, forming part of the Rhind collection, the following rule for finding the area of a circle : Cut off \ of the diatnetfr and on the remainder construct a square; the area of the square thus constructed is equal to the area of the circle. According to this rule, if d is the diameter, Wd"^ is the- area of the circle. Since jTrrfs^the area; therefore, i-rf2=f+rf2, 7r=^j«, or (J5S-)2=3.16-f , a fair approximation for that period in the history of mathematics. A less accurate value of tt is given in I Kings, 7: 23, and II Chronicles, 4: 2, where it is stated that the circumference of a circle is 3 times its diameter. The manuscript of Ahmes, referred to above, is called "directions for knowing all dark things," and consists of a collection of problems in arith- metic and geometry. It is believed that this manuscript is itself a copy, iwith emendations, of an older treatise written about 3400 B. C. ^ MENSURATION. 257 Prob. X!XJ.II. To find the side of a circumscribed polygon, having- given the radius of the circle and a side of a simi- lar inscribed polygon. 'Formula, — K'=^ ; , in which K' is the side of the circumscribed polygon and ^the side of a similar inscribed polygon. Rule* — Divide twice the product of the side of the inscribed ■polygon and radius by the square root of the difference of the squq.res of twice the radius and the side of the inscribed polygon. I. When j??=l, find one side of a regular circumscribed do- decagon. By formula, K = — , The formula does not lead to a direct result, since K is not given. But by the formula of Prob. XXI. , if ^ is replaced by K we have K=V'i — ^4—1 for 2c=l, smce it is the side of a regular inscribed hexagon, and K=\2 — ^3, since 2c is a side of a regular inscribed dodecagon. , 2K 2y^^_„V2^ PROBLEMS. 1. The driving-wheel of a locomotive engine 6 feet 3 inches in diam- eter, makes 110 revolutions a minute ; find the rate at which it is traveling. Ans. 24.54 miles per hour. 2. If the driving-wheel of a bicycle makes 560 revolutions in traveling a mile, what is its radius? [Take t=3J^]. Ans. 1% feet 3. Find the area of a walk 7 feet wide, surrounding a circular pond 252 in diameter. [Take t=31]. Ans. 5390 sq. ft. 4. A wire equal to the radius of a circle is bent so as to fit the circum- ference. How many degrees in the angle formed by joining its ends with the center of the circle ? [Take t'=Z. 14159265]. Ans. 57 . 2957795°. Definition. — The angle subtended at the center of a circle by an arc equal in length to the radius is called a radian. • 5. A wire is bent into the form of a circle whose radius is 30 inches. If the same wire be bent into the form of a square, what would be the length of its side ? 6. , A circle and a square have the same perimeter. What is the differ- ence between their areas ? 7. Two tangents drawn from an external point to a circle are 21 inches long and make angles with each other of 90°. Find the area of the circle. 8. A bicycle driving-wheel is 28 inches in diameter, the sprocket-wheel has 17 sprockets, and the rear sprocket-wheel 7 sprockets ; what is the gear of the wheel ? Hint. — One revolution of the sprocket-wheel makes 17-i-7=JJK. revolutions of the rear sprocket-wheel, or JJ. revolutions of the driving wheel, the rear sprocket-wheel and the 258 FINKEL'S SOLUTION BOOK. driving-wheel being rigidly connected. . • . 5rX28Xi^=:rX68 inches, the distance trav- eled in one revolution of the sprocket-wheel. 68 inches is the gear of the wheel. Gear =( n-i-m )D, where n is the number of sprockets in the sprocket-wheel, m the number of sprockets in the rear sprocket-wheel, and Z> the diameter of the driving-wheel in inches, 9. (a) What is the gear of a bicycle whose driving-wheel is 30 inches in diameter, whose sprocket-wheel has 19 sprockets, and whose rear sptocket- wheel has 6 sprockets? (b) How many revolutions of the sprocket-wheel will be required to travel a mile ? Ans. (a) 95 inches. 10. What is the distance from the center of a chord 70 inches long in a circle whose radius is 37 inches ? Ans. 12 inches. 11. In a circle whose radius is 9 inches, the chord of half an arc is 12 inches ; find the chord of the whole arc. Ans<. 17.69 inches. 12. The length of an arc of a circle is 143 inches and its central angle is 9° 6'; find the radius of the circle. Ans. 900 inches. 13. In a circle of a radius of 37 inches, find the length of the minor arc whose chord is 24 inches. Ans. 24.44 inches. 14. The radius Of a circle is 21 inches ; find the length of an arc which subtends an angle of 60° at the center. 15. The radius of a circle is 9 feet 4 inches ; what angle is subtended at the center by an arc of 28 inches ? 16. The chord of an arc is 48 inches and its height' is 7 inches ; find the length of the arc. [Arc=J^(8A — a), where b is the chord of, half the arc and a is the chord of the whole arc] Ans. 50J^ inches. 17. In a circle whose diameter is 72 inches, find the length of the arc whose height is 8 inches. 18. Find the area of a sector of a circle whose radius is 21 inches and the angle between the radii 40°. 19. Find the area of the sector of a circle having given the arc 32 inches and the radius 17 inches. Ans. 272 sq. in. 20. Angle of a sector is 36° and its area is 385 square feet ; find the length of its arc. Ans. 22 feet. 21. Find the area of a segment cut off by a chord whose length is 14 inches from a circle of a radius of 25 inches. Ans. 9.37 inches. 22.* A regular pentagon is inscribed in a circle of a radius 10 inches ; find the area of a minor segment cut off from the circle by one of its sides. Ans. 15.27 sq. in. 23. Find the area of a segment whose chord is 30 inches and height is 8 inches. Ans. 168.16 sq. in. 24. Find the area of a circle inscribed in a sector whose angle is 120° and whose radius is 10 inches. 25. A line AB is 20 inches long, and C is its middle point. On AB, AC, and CB semicircles are described. Find the area of the circle inscribed in the space inclofeed by the three semicircles. Ans. r—Zy^ inches. 26. Two equal circles, each of a radius 9 inches, touch each other exter- nally, and a common tangent (direct) is drawn to them ; find the area of the space inclosed between the circles and the tangent. Ans. 7 . 53 sq. inches. , 27. Three circles of radius 3 feet are placed so that they touch each other ; find the area of the curvilinear space inclosed by them. Ans. 207 sq. in. 28. Prom the angular points of a regular hexagon, whose side is 10 inches, six equal circles, radii 5 inches, are drawn ; find the area of the figure inclosed between the circles. Ans. 50 (3V 3— t) sq. in. MENSURATION. 259 29. Two equal circles of radius 5 inches are described so that the center of each is on the circumference of the other ; find the area of the curvi- linear figure intercepted between the two circumferences. .<4«j. 30.71 sq. in. 30. Two equal circles of radius 6 inches intersect so that their com- mon chord is equal to their radius ; find the area of the curvilinear figure intercepted between the two circumferences. Ans. 4.53 sq. in. 31. Three circles, radii 10, 12, and 16 inches respectively, touch each other; find the radius of a circle touching the three circles. [See Prob. CLXVI.] 32. ' Find the area of the largest semicircle that can be inscribed in a trapezium ABCD whose sides are AB=AD— 30 feet, and CB=CD=40 feet. 83. Find the area of the equilateral triangle formed by the diameters of three equal semicircles drawn with an equilateral triangle, side=60 feet, in such a way that each semicircle is tangent to two sides of the triangle. .... 84. Three equal circles are inscribed in an equilateral triangel, side =30 feet, in such a way that ea,ch circle is tangent to the two others and tangent to two sides of the triangle. Find the area between the circles and the sides of the triangle. VII. RECTIFICATION •OF PLANE CURVES AND QUADRATURES OF PLANE SURFACES. 1. To Rectify a Curve is to find its lengtii. The term arises from the conception that a right line is to be found which has the same length 3. The Quadrature of a surface is finding its area. The term arises from the conception that we find a square whose area is equal to the area of the required surface. The formula for the rectification of plane curves is *= J"Vc/i*: 2 -\-dy^=:J ^ 1_|_|^^J I dx, when the curve is re- ferred to rectangular co-ordinates. ^/J 1 1+'-^ fS) ' i ^'-' ^ ^=/J 1 ''^+ (S) 1 '^^.-hen the curve is referred to polar co-ordinates. are formulae for the rectifica- tion of curves of double cur- vature, when referred to rec- tangular co-ordinates. 260 FINKEL'S SOLUTION BOOK. s= fJ I r2_|_ [^] V^2 sin 2 e f^l ' [ c^fil ] are formula for the •^ ^ ' W<^J l^^J ^ rectification of A= Cydx or Cxdy is the formula for the quadrature of any plane surface referred to rectangular co-ordinates. curves of double- ( curvature, referred to polar co-ordi- ^ C^r^dO is the formula for the quadrature of plane surfaces, referred to polar co-ordinates. 3« A. Suvface of MeVolution is the surface generated by a line (right or curved) revolving around a fixed right line as- an axis, so that sections of the volume generated, made by a plane perpendicular to the axis are circles. S=27t ly-\\ I 1+ 1 ;/ I f '^^ "* ^^ formula for a surface oi revo- lution, referred to rectangular co-ordinates. S'=27t jyds=a,n: jr sin (9 f | ^2-)-|^l dd is the formulafor a surface of revolution, referred to polar co-ordinates. V=7C Cy^dx or x^dy is the formula for the volume of a solid of revolution referred to rectangular co-ordinates. V= C f fdxdydz and V= C Czdxdy are formula for the- cubature of solids, requiring triple and double integration. V=rjzrdedr and V=JJJr^ sin ddtpdOdr are the formula- for cubature of solids referred to polar co-ordinates. From the equation to the surface of the solid, z must be expressed as a function of r and 6. ;v2-|-y2=:_ff2, is tjje rectangular equation of a circle referred to the center. y^h^lJix — x^ \s the rectangular equation of a circle referred to the left hand vertex as origin of co-ordinates. r=2R cos.^ is the equation of the circle referred to polar co-ordi- nates. Prob. XXIV. To find the circumference of a circle, th& radius being given. MENSURATION. 261 i y' J Formuta.-C.=l£ ^ \ 1+(|) " j d.=i£ d-=^J, cig2_,. yr ^Ri 1+2-3+2J5+2IA7+ ^^^^^^+&c)=4i?Xl-570796+=3.141592x2i?=2;r^, in which jr=3.141592-(-.. Since the diameter is twice the radius, we have '^7cR=7iD, in which D is the diameter. .'. Q,.=^1nR=TtD. C C • (1) Ji^ — '—, 11)D^ — , where C is the circumference. ^ ' In ^ ' Tt Kule. — Multiply twice the radius or the diameter by S.H1592. I. What 'is the circumference of a circle whose radius is 17 rods ? By formula, C=27ri?=3.141592 X 34 rods = 106.814128 ' rods. 1. 17 rods=the radius. 34 rods=2Xl7 rods^the diameter. ' 106.814128 rods=3.141592X34rods=the circumference. 11.(2! 13. III. . . The circumference is 106.814128 rods. Note. — The ratio of the circumference to the diameter can not be exactly ascertained. An untold amount of mental energy has been expended upon this probleni; but all attempts to find an exact ratio have ended in utter failure. Many persons not noted along any other line, claimed to have found this clavem imfossihilitihus by which they have unlocked pU the diffi- culties that have encumbered the quadrature of the circle for more than two thousand years. The QLiadrature of the Circle is to find a square whose area shall be exactly equal to that of the circle. This can not be done, since the ratio of the circumference to the diameter can not be exactly ascertained. Persons claiming to have held communion with the "gods" and extorted from them the exact ratio are ranked by mathematicians' in the same cla.ss with the inventors of Perpetual Motion and the discoverers of the Elixir of Life, Alkahest, the Fountain. of Perpetual Youth, and the Philosopher's Stone. Lambert, an Alsacian mathematician, proved, in 1761, that this ratio is incommensurable. In 18S1, Lindemann, a German mathematician, dem- onstrated that this ratio is transcendental, and that the quadrature of the circle by means of the ruler and compass only, or by means of any algebraic curv'e,.is impossible. Its value has been computed to several hundred deci- mal places. Archimedes, in 287 B. C, found it to be between 3^? and 3?^; Metius, in 1640, gave a nearer approximationin the fraction f-fl; and, in 1873, Mr. W. Shank presented to the Royal Society of London a computation ex- tending the decimal to 707 places. The following is its value to 600 deci- mal places: 3. 141, 592, 653 , 589, 793,238 ,462,643,383, 279 , 502,884, 197 , 169 , :399,375,105,82G,974,944,592,307,816,406,286,208,998,628,034,825, .342,117,067,982,148,086,513,282,306,647,093,844,609,550,582,231, 725,359,408,128,481,117,4.50,284,102,701,938,521,105,559,644,622, '948,954,930,381,964,428,810,975,665,933,446,128,475,648,233,7-86, 262 FINKEL'S .SOLUTION BOOK. t 783,165,271,201,909,145,648,566,923,460,348,610,454,326,648,213^ 393,607,260,249,141,273,724,587,006,606,315,588,174,881,520,920^ 962,829,254,091,715,364,367,892,590,360,011,330 530,548,820,466,. 521,384,146,951,941,511,609,433,057,270,365,759,591,953,092,186, 117,381,932,611,793,105,il8,548,074,462,379,834,749 ,567 ,351,885, 752,724,891,227,938,183,011,949,129,833.673,362,441,936,643.086, 021,396,016,092,448,077,230,943,628,553,096,620,275,569,397,986,. 950,222,474,996,206,074,970,304,123,669+. i'^'=T2-+V+p+J5+&'=-* Bernoulli's Formula. i ""^l^ l^^flflZ'Ac l ' •••• Wallis's Formula, 1655. ijr=l-(-l Sylvester's Formula, 1869. T+L2 1+2^ * 1+3^, 1+45 1+ 4 ^-=1+1^ "^ 2+3^ 2+5^ 2+ Buckner's Formula. J^oie.— T'be Greek letter, Jr is the initial letter of the word TtspitpipiiK ^peripheria, meaning periphery or circumference. It was first used to represent the ratio of the circumference to the diameter by William Jones in his Synopsis Palmariorum Matheseos, 1706, and came into gen- eral use through the influence of Euler. Prob. XXV. To. find the length of any arc of a circle, having- given the chord of the arc and the height of the arc, i. e., the versed sine of half the arc. (a). Formula.—s=J-s]l+\-£\ dx= f y ^ | dx = —J {R^—x^Y^ ^ "^ ir^\-R I 2.3.i?«'^2.4.5.i?='^ 2.4.6.7.i?^ +^"J '=2^^'' +' ^ L^M=^+^^K+^+ ^{S^+Ti^S^l +^'-]''"^^"'''=''^' '''''"^' °^ the arc and c^half the chord of the arc. «i7o/e.— This series was discovered by Bernoulli, but he acknowledged his inability to sum it. Euler found the result to be^Jr'. For an inter- esting discussion of the various formulse for tt, see Squaring the Circle^ Brttannica Encyclopedia MENSURATION. ' 263 . (b.) Formula. — j-^arc=^(8i5 — «)*, where a is the chord of the whole arc and b the chord of half the arc. Rtlle from {b) . From eight times the chord of half the arc sub- tract the chord of the whole arc ; one-third of the remainder will be the length of the arc, approximately . I. Find the length of the arc whose chord is 517638 feet and whose half chord is 261053.6 feet. By formula (fi), s=^{^b—a)=\ (8X261053.6—517638)= 52359.88 feet. 1. 261053.6 feet=length of chord of half arc. 2. 2088428.8 feet=8X261053.6 feet=eight times the length of chord of half arc. 3. 517638 feet=length of chord of whole arc. II. \ 4. 1570790.8 feet=2088428.8 feet— 517638 feet=differ- ence between eight times chord' of half arc and chord of whole arc. 5. 52359.69 feet=:J of 1570790.8 feet=length of arc, nearly. III. .-. The length of the arc is 52359.69 feet. Note. — This important approximation is due to Huygens,(he wrote his name Hugens. It is also sometimes spelled Huyghens), a Danish mathe- matician, born at the Hague, April 14, 1629, and died iii the same town in 1695. For a brief biography of this noted mathematician, see Ball's A Short History of Mathematics, pp. 302-306. The following is Newton's demonstration : Let R be the radius of the circle, L the length of the arc, A the chord of the arc, and B the chord of half the arc. A LB L Then ■;^=2 sm.g;^, "^=2 sxn.|^. X x^ x'' . ^ . Since, sin. ■^="T "37+ TT — ^^'^- (^^^ Bowser's Treatise on Trxgonom- etry, or any other good work on the subject), we have 4-=2(2^ (.2^) , \2Ii) -etc. ) and '^ * ~- 3l^+ 5T~ / V 1 4r=2(iJ? \iJ?) , \i/?) -etc. Y ^ \T 3! "^ 5! / _ ^^ ^'' - * ■ A^L 1, r,2 + 04KI D4 CtC, aUQ 8B=iL . ■ . i{8B—A)=L{j.—j^^Q^y nearly,— L, nearly. I^ the problem proposed, the radius is 100000 feet and the arc is 30°. Using 7!-=3. 1415926, i=52359.88 feet. . • . The result by the formula lacks only about 2 inches of being the same. 264 FINKEL'S SOLUTION BOOK ( c. ) Formula.— s=a^rc=2'J { a^-\-c^) x 1- 10«2 ]■ 60c2 4-33«2 This formula is a vei-y close approximation to the true length • of the arc when a and c are small. The first formula may be ex- tended to any desired degree of accuracy. Rule from {c). — Divide ,10 times the square of the height of the arc by 15 times the square of the chord and SS tim.es the height of the chord; multiply this quotient increased by 1, by 2 times the square root of the sum- of the squares of the height and half the chord. I. The chord of an arc is 25, and versed-sine 15, required the length of the arc. By formula (a), arc a^^c^ La i- _^2 \Ar\ (c^—a^) 5 [a^+c-j I +m(^)'+^'-] r 25'— 15^ ■ [152+252 J {2b''— Ib^ 2a La2_|_^5n-^[^^3^^2j 1 40 _ 152+252 r 252-152 1 ~ 2X15 L1524-252+6 \ 5 5 {- 25^-15^ 1 ' -I J +Il2li5H^25i-J +&=J=' X 53.68 [152+252 J ' 4^^152+252 ,+ft. 1. 25 ft.^length of the chord. 2. 15 ft.=Jieight of the arc, or the versed-sine. 8. 2250 sq. ft. =10 times 1521=10 times the square of the height of the arc. ' [chord. 4. 9375 sq. ft.:^15 times 25^^15 times the square of the 5. 7425 sq. ft.=33 times 152^33 times the square of the height of arc. IT <! 6. 17aOO sq. ft.=7425 sq. ft.+9375 sq. ft. 7. ^Vf=2250-=-17800=10 th-nes l.-,=-^(15 times 25^+33 times 152). 8. l+-/A=lfB-=l + 10 times ir,2-i-(15 times 252+33 times 152). 9. 38H sq. ft.=152 + ( 12^)2. _ 10. 53.58 ft.=.mxVl52 + (12^)2=|f^X|V61=length of arc, nearly. *- III. .-. 53.58 ft.=length of the arc. Prob. XXVI. To flntl tlie area of a circle having- given the tadius, diameter, or' circumference. Formula.— A=A jy dx,=4 f {R^—x^)iidx=4r^x (i?2_^2) ^^7?2sin-iy =2^' 2 ' ' ■ L3.5. .V 7 2.4.5.i?« 2.4.6.7i?' +&c=^;r^ : Ji ' 2.37?8 C2 7r7?2=^;rZ>2=-— =|^X C, when the ra- 47r MENSURATION. 265 dius and circumference are given. . •■ (1) Il =\A -^7r, (2)2?= V4^H-7r=2 R.=2»lA-^7r, and (3) C=^47tA=2>J7iA. Rule I. — Tke area of a circle equals the square of the radius multiplied by S.lJi.1692; or (S) the square of the diameter multi- plied by .785398; or (5) the square of the circumference multiplied by .07958/ or (4) the circumference mhltiflied by \ of the diame- ter; or (6) the circumference multiplied by \ of the radius. Rule II. — Having given the area. (1) To find the radius : Divide the area by 3. 14-1592, and extract the square root of the .quotient. (2) To fnd the diameter : Divide the area by S.H1692 .and multiply the square root of the quotient by 2. (3) To, find ■the circutnference : Multiply the area by 3.1^1592 and multiply the square root of the product by 2. I. What is the area of a circle whose radius is 7 feet? By formula, ^=7ri?2=3.141592 X 7 2— 153.93804+ sq. ft. (\. 7 ft.=the radius. II. J 2. 49 sq. ft.^7^=square of the radius. l3. 153.93804 sq. ft.==3.141592X49 sq. ft.=a'rea of the circle. III. .-. 153.93804 sq. ft.=area of the circle. I. What is the area of a circle whose diameter is 4 rods? By formula, ^=i7rZ>2=ix3.141592x42=12.566368 sq.ft. fl. 4 ft.=the diameten T-r J 2. 16 sq. ft.=square of the diameter. •]3. 12.566368 sq. ft.=ix3.141592x42=.785398xl6 sq. ft. I ^area of the circle. III. .-. 12.566368 sq. ft.=area of the circle.- I. What is the area of a circle whose circumference is 5 meters ? C^ OK 2 By formula, ^=-—=+-=1.989 m^. ■' 47r 47r {1. 5 m.^^the circumference. 2. 25 m.^^the square of the circumference. ' 3. 1.989 m.3=.07958x25 m.3=the area of the circle. III. .•. 1.989 m.^=the area of the circle. Remark. — We might have found the radius by formula (1) under Prob. XXIV and then applied the first of Rulel. afcove. We might have found the radius by formula (1) of Prob. XXIV and then applied (5) of Rule I. above. I. What is the circumference of a circle whose area is 10 A. f 266 FINKEL'S SOLUTION BOOK. By formula (3), C=2V^=2V3.141592xl600=80V'^80X 1.7724539=141.796312 rods. 1. 10 A.=1600 sq. rds.=the area of the circle. 2. 1600-r-7if=the' square of the radius. 3-v -= — V7r==the radius. II. *n n 80 ^ . 40 ^ 4- -—yn=7i2 times ~V;r:^the diameter. 7t 71 II. r- 40,- "-S. 80V7r=;rXl7V7r=141.796312rods=the circumference. III. .•. 141.796812 rQds=the circumference of the circle. I. With what length of rope must a horse be tied to a stake- so that he can graze over one acre of grass and no more ? By formula (1), R=»JA^^=^ im-^7c=A^~ =7.1364+ rd. 1. 1 A.=160 sq. rd.=area of the circle over which the- horse can graze. 2. 160-r-'i'==square of the radius. 3. Vl60-i-7s=4v'l0-f-;r=7.1364 rd.=radius or length of rope.. III. .-. 7.1364 rd.=length of the rope. Prob. XXVII. To find the area of a Sector, or that part of a circle which is bounded by any two radii and their included arc, having^ given the chord of the arc and the hei^rht of the arc. *Formula.—A=Jydx=2J(Ji^—x^y^dx=x(jR<'—x'')ii +^^^'" '^'=-^;r-il-2^J -[^^J f +t"i^J "•"[ 2a J \c^-\-a^~^1.2.s[c^-\-a^} +1.2.3.4.5 U^+^'J "*" 1.2.3.4.5.6.7 ' c^+gg i +^''-^ '" ^h'^^l^ ^ >^ ^^^f the chord of the- arc and a the height of arc. £)emonstration. — Let AB^x, SZ>^y, and If=^AU=z^he i adius of the cir-^ cle. Then x^-\-y^=:It^, the equation oj the circle referred to the center.. Now .(4=2 Cydx; \)\xty=^(R^ — ^^)^, from the equation of the circle. .-. A=2C[R^—x^)iidx=x {Jf^—x^)>i+If^ "'"'^W But *=/?—« and >=c.. R — a Henci A— (,R—a)[R^—{R—ay]+R'^ sin'— „- . ]iut,from(2) Prob. XX, if MENSURATION. 267 Rule. — (1) Find the length of the arc by Problem XXV, and then multiply the arc by half the radius -which may be found by Problem XX, in which c and a are known and R is the un- known quantity. (2) If the arc is given in degrees, take such a part of the •whole area of the circle as the number of degrees in the arc is »fS6(y. I. Find the area of the sector, the chord of whose arc is 4(^ feet, and the versed-sine of half the arc 15 feet. % '— . -^-(^i+mi 1 c2-(-a2 I 1.2.3 1.2.3 1.202+162] +1.2.3.4.5120^+15=] +«=— 658.125 sq. ft. l\.\ 1. 58.58 ft.^2(152+202).x [l+ 60x2o" +^3x15= = length of thcarc, by {b), Prob. XXV. 2. 20|- ft.= — 31 =radius of the circle, by solving tha- formula of Prob. XX with respect to R. 3. .-. 558.125 sq. ft.=^ (20|xff3.58)=area of the sector. III. .'. 558.125 sq. ft.=the area of the sector. Trigonometry, we have sin-'ft=8 — t~2^^° ^ 12^4'; ^'' — ^'^' ^^"<=^> ^^^ ^ ct—a^ \ < c''—a* \ » t g»— g« ]_ [ c'— g' S "» ", 1 r e'— g' 1 = i (. 2a j ""'"l 2« J <<:»+«» 1.2.3 [c'+a^J "^1.2.3.4.5 [c'+b»J ~*°-S angle DBA. For x=R—a— ^^ — the altitude and c is half the base of the triangle — - — |c= the area of the triangle DBA. Therefore, if we subtract the area of the tri- angle DBA from the area of the sector,, we shall have the area of the segment DBC. Hence, the area of the segment DEC is I 1(^5"') "lis (SS)'+i:2i4:5 t^)°-H • ™^ '•«^"i* "^y be carried to any desired degree of accuracy. In this formula, I — ;j^— 1 c is theareaof the tii- FIG 14. 268 FINKEL'S SOLUTION BOOK. I. What is the area of a sector whose arc is40°and the radius of the circle 9 feet? 1. 9 ft.=radius of the circle. 2. 7ri?^^;z9^=area of the whole circle. II J 3. 40°=length of the arc. 4. 40°=i of 360°. 5. ;r9=4 of ;r9 ^=28.274328 sq. ft.=area of the sector. III. . . The area of the sector is 28.274328 sq. ft. Prob. XXVIII. To find the area of the segrment of a circle, having- given the chord of the arc and the height of the segment, i. e., the versed-sine of half the arc. 1.2.3.4.5 ,2~1 5 &C. '-fa2 (^) ^=2(2^+^^2^")=4?+T- *Rllle. — Divide the cube of the height hy tinice the base and in- crease the quotient by two-thirds of the product of the height and base. I. What is the area of a segment whose base is 2 feet and al- titude 1 foot? By formula (3), A=^^-\-l{^ca)=^-\-l{^y^\)=\^^^o,St. • 1-^ . ''' Demonstration— In the last figure, let i5C^«^altitude of the segment and X>£'=2c=the base of the segment. Then BD''=BCxBF=a{2R—a)=c'^. f *— I— ^2 (^1-1— (I'i (-2 (^2 , Whence R= „ , and AD=zR—a=—^ a=-^ — ■ D C^k=zV c^J^a^ . 2a ~ 2a 2a ' By Trigonometry,^— =sin/Z)^C,or-^ = s'in/Z>v4C. Now 27ri? = 360°. „ 180° „, , „ _ ■ 180° , arcBC -iSO° ^ ^ .-. It=z Therefore, 7f: arciJC:: : .'^ — X ■■ Let s ^ arc DCE. Then the /Z>ylC=^-^^. •'• o^^^'^^TH- ^" '■'^^ manner, from DC the right angled triangle i^ZlC, -^—:,=sin / Ci^-O, or since the / Ci^Z>=the k s 1 % /_CAD, ;5-H=sin-— . Now since- the sine of any angle 6=# — flTQ'^'"!" 1.2.3.5 • 1.2.3.5 7 ^'-(-&c., the above equation becomes ^ = 27?--1.2.3 (2^) +1:2^ M °-&<="---(l)' -"-l ^=4^li3l4^)°+i:2^b^J ~^^ ^2^-. Multiplying equation (2) by 8 and subtract equation (1) in order to eliminate the term containing S3, we have approximately,— ^=2^—^ I ^2^1 W^\ + &c. Omitting the negative quantity, since it is very small in comparison with j and be- •cause it is still more diminished by a succeeding positive quantity, we have MENSURATION. 269 (-1. 1 ft.=altitude of the segment. 2. 2 ft.^base of the segment. 3. 4 ft.^twice the base of the segment. 4. 1 cu. ft.=cube of the height of the segment. II. J 5. i sq. ft.=l-^4=quotient of the cube of the height and twice the base. 6. 2 sq. ft.^2 X l=P''oduct of the height and base. 7. 1^ sq. ft.=|- of the product of the height and base. .8. .-. li sq. ft.+i sq. ft.=ly',sq. ft.=area of the segment. III. .-. The area of the segment is ly\ sq. ft. Proto. X!XIX. To find the area of a circular zone, or the space included between any two parallel chords and their intercepted arcs. ' , ^ ^ (c^—a' Formula. — (a ) A= I - •&c. "j } c'-+a' 1.2.3 icH«^ J [ 2a' J I c'«-|-«' 1.2.3 1 ( c'^—a^ ] =_ +1.2.3.4.5 U'+«'J [c's+a'^J ' 1.2.3.4.5 U''+a 'J ' (*) ^=2^+*(2-)-[2p^+*(2^'^')] Rule> — Find the area of each segment by Prob. XXVIII., and take the difference between them, if both chords are on the same side of the center; if on opposite sides of the center, subtract the sum of the areas of the segments from the -whole area of the circle. I. What is the area of a zone, one side of which is 96, and tVip other 60, and the distance between them 25? Let ^^=60=2c', CZ'=96=2c, and HK=1h=h. Then AH =30=c' and C.K=48=c. Let OA=R. Then -c'^^OK =MR^—c "-. But OH OK^HK; .: »J R'^—c''^=^R'^—c^ + h. - ^ic^h^ -f (c^—h^—c'^ ) 2. Whence R= ~2k .: ZH=a'=R—OH=R- -^iR^-c^)=- ^4c2_^2_^(c2=/^2_c'-2 ) •'2^2 i_V4/i2 (,. 1h FIG. IS. .y— ^^~^'' = ^ +"' ^'' =^(4N/c'+a'— g). This is the approximate length of an arc in terms of its height and base. Now the area of the seg- ment />C£=>^ A CXarcr)CE—a.Te&oi the tn&n%\eI)EA—}4 RXS—% AB •270 FINKEL'S SOLUTION BOOK. -}-(c^— A2— c'2)2. In like manner, LiC =^ a=R—i^ Ri'—c'^ ,=.— lift V4c2/i2+(c2— A2_c'2y2_JL(c2>%2_c'2).2 . .: By formula {h), ^=2^+t(2^«)-[2(|7y+K2'^«')] >= j^p4.U^+(c2_A2_^-7^2_(^2_A2^2)?jj3^2.2c_^.^2cX^ £V4C2>%2 + (C«— A2— C'2)2_(c2_>42_c/2)q_|' j ^V4^2^^+(^ -_,j2'II72")2_V4>52(c2— c'2)+(c2— y42_c'3)T| j' ^2(2c)+f X2e' X ^[V4^*6-''+(c''— A*— C'^)2— V4A''(C''— C'2)_|_(c2_;^8_tf'2)2-j] ^/2 J =2547— 408i=2138f. 1. 5o=^=^4c2^24-(c2_;42_e/2^^Y4 X 488 x 25^ + IIX (482— 258— 302)=radius_of the circle. 2. OK=^R^—c^ =V502— 48^=14. 3. .-. Z^ =a=5Q— 14=36==altitude of segment CZZ>. 4. OII=!jR-i—c'^ ==V502— 80'*=40. 5. .. ZZr=a'=50— 40=10=altitude of the segment ALB. Qft3 6. .•.;5^^+f(96x36)=2547=area of segment CDBLA. 7. H-^+f (60X10)=408i=area of the segment ABL. 8. .-. 2547 — 408i=2138f=area of the zone CDBA. III. .-. 2138|=area of the zone ABDC. Note. — Thia result is only approximately correct. The radiui of the cir- ; cle may be found by the following rule: Subtract half the difference bet-ween the two half- chords from the greater half-chord, multifly the remainder by said difference, divide the product by the Tuidth of the zone, and add the quotient to half the luidth. To the square of this sum add the square of the less half chord, and take the square root of the sum. This rule is derived from the formula in the above solution, in which ^^ \ Ji-ic^^{c'^—h^—ii''^Y ^ I r (c'+z^^— </" ) 2+4/^ V2 -j '^ 4A^ , "^L 4^8 _)' MENSURATION. 271 Prob. XXX. To find the area of a circular ring, or the space included between the circumference of two concen- tric circles. FormwlSb. — («■) A=7t {R^ — r^), in which H and r are the radii of the circles. {b.) *A=\nc*, in which c is a chord of the larger circle tangent to the smaller circle. I. Required the area of a ring the radii of whose bounding circles are 9 and 7 respectively. By formula {a), A = 7t{R^ —r^) = 7i(^^—l^)=n7i= ' V 100.530944. 1. 9^i?^i"adius of the larger circle, and 2. 7^>"=radius of the smaller circle. 3. ;r9'=7ri?2==area of larger circle, and , 4. ;r7^=7rr^=area ot smaller circle. 5. .•.7id^—n7^=n{9^—7^) = S27r= 100 530944=area of the ring. III. .-. 100.530944=the area of the ring. *Demotistration. — Let ABC be the chord of the large circle, which Is tangent to the smaller circle, and let ABC=c. Then BC=\c= ^(OC^—OB^)=^^ (It^—r^). .: ic^=Ji^— r^ and i7rc«=ir(i?2— 7-2) But 7r(i?s— r^) is the dif- ference of the areas of the two circles or the area of the ring. > .'. ^7rc'-=the area of the "ring. Q. E. D. ^1^' '*• Prob. XXXI. To find the areas of circular lunes, or the spaces between tbe intersecting arcs of two eccentric circles. II. Formula. — A=- -l{2ca ^-[2^+*(2^''')]- 2(2c) ' "' L2(2c) Rule. — Find the area of the two segments of -which the lunes <ire formed, and their difference ■will be the area required. ■ I. The chord AB is 20, and the height Z>C is 10, and DB 2; find the area of the lune AEB C. =J[('-¥^+*^^)+''" ]=J1[ r>--^) Hh']' +</'' If now we find the altitudes of the two segments and then find the length of the arcs of the segments by formula (b), Prob. XXV, and then find the area of the sectors by multiplying the length of the arcs by half the radius, from the areas of the sectors subtract the triangles formed by the radii of the circles and the chord of the arcs, we shall then have the area of the two segments. Taking their difference, we shall have for the area of the zone 213&76, which is a nearer appro»mation to the true area. 272 FINKEL'S SOLUTION BOOK. ■By formula, ^=^+|(2..)_[2p^(2e«')]= -H(20XlO)-[2^Q+f(20X2)]^131,^. 1. ^^=chord=20. 2. DE = height of segment AEBD==1. [ACBJ?=10 3. Z>C = height of segment "•■^■^- 23^0 + *^^^^ ^°^ = ^^^*= area of the segments CSD. area of the segment AEBJJ. III. .-. 158^26i|=13ixV=area of the Inne A CBB. ne. 17. VIII. CONIC SECTIONS. DEFINITIONIS. 1. The Conic Sections are such plane figures as are- formed by the cutting of a cone. 2. ^ If a cone be cut through the vertex^ by a plane which also- cuts the Fase, the sections will be a triangle. 3. If a right cone be cut in two parts, by a plane parallel to ' the base, the section' will be a ciralR. 4. If a cone be cut by a plane which passes through its two slant sides in an oblique direction, the section will be an ellipse. 5. The Transverse Axis' oi an ellipse is its Ibngest diameter. 6. The Conjugate Axis of an ellipse is its shortest diameter. 7 . An Ordinate is a right line drawn from any point of" the curve, perpendicular to either of the axes. 8. An Abscissa is that part of the diameter which is con- tained betw^een the vertex and the ordinate. , 9. A Parabola is a section formed by passing a plane through a cone parallel to any of its slant sides. - , 10. The Axis of a parabola is a right line drawn from the- vertex, so as to divide the figure into two equal parts. 11. The Ordinate is a right line drawa from any point: ^ in the curve perpendicular to the axis. MENSURATION. 273 18. The Abscissa is that part of the axis which is con- tained between the vertex and the ordinate. 13. A/n Hyperbola is a section formed by passing a plane through a cone in a direction to make an angle at the base greater than that made by the slant height. It will thus pass through the symmetrical opposite cone. 14. The Transverse JMameter of an hyperbola, is that part of the axis intercepted between the two opposite cones. 15. The Conjugate Diameter is a line drawn through the center perpendicular to the transverse diameter 16. A.n Ordinate is a line drawn from any point in the curve perpendicular to the axis. ' 17. The Abscissa is the part of the axis intercepted be- tween that ordinate and the vertex. 1. ELWPSE. a^y^-\-6^x^=a^i^ is the equation of an ellipse referred to the center. _y*^— (2ai»f — }i'^) is the equation of the ellipse referred to left hand vertex. In these equations, a is the semi-transverse diameter and 6 the semi-conjugate diameter -jjc is any ordinate and x is the corres- ponding abscissa. When any three of these quantities are given the fourth may be found by solving either of the above equations with reference to the required quantity. a(l—e^) . , , . , , , /3=7— ^ ' is the polar equation referred to the centre, and 1 — e cos ■ ^ ^ ' p=rr-i ^ is the polar equation referred to the left hand l-\-e cos tf . vertex. Prob. XXXII. To ftnd the circimiference ot an ellipse, the transverse and conjugate diameters heing known. Formula.— cir.= C=4j^/dy'+dx^=iJ'^'^( 1—e^p • J y ./ >l fl,2 ^8 r " L e2»2, ,/ . _^x e2 fa2 . ,» 4/ r^ 5'nI ^dx=i\a sin' —- __sin~*- — Xr- 1 «* rSo* U^ . .,x xj— x^ 274 FINKEL'S SOLUTION BOOK. a3L4 ^2'2/J 2.4.6«M 6L4 V^'Tjjy V"" 2.4a ^^i^-2:2~2:2X4- 2.2.4.4.6.6 -^'^'J - which .= V^^ Kule. — Multiply the square rogt of half the sum of the stfuares of the ttoo diameters by S.H1592, and the product will he t%e cir- cumference , nearly. £. What is the circumference of an ellipse whose axes are 24 and 18 feet respectively ? By formula, C^V.= C=2a-Xl2|l^^(l_j|-^y _ 22^4(1- j|i)*—&«^^ =2^X12 X.87947.=66.31056 ft., nearly. 1. 576 sq. ft;=24*=siquare of the transverse diameter. 2. 324 sq. ft.=18^=square of the conjugate diameter. 3. 900 sq. ft.=:sum of the squares of the diameters. 4. 450_sq. ft.=half the sum of the squares of the diameters. 5. 15\^2ft.=V450=square root of half the sum of the squares of the diameters. 6. ;rl6\/2 ft.= 66.6434 ft., nearly, =the circumference of the ellipse. III. .•. The circumference of the ellipse is 66.6434 ft. nearly, by the rule. Prob. XXXIII. To find the length of any arc of an ellipse, having- given the ordinate, abscissa, and either of the diam- eters. II J^ormw?a.-5=2[^^a^l-(^)2l_(|.|)2-l*_(|.|.5)2 &c \ — asm-' 5-sm-' 7:>^a^—x^ — ?A^ — 5 ^ ) J a 2a 1^2 a 2^ J 2.4^8 L 4 — sin~* ^a^ — x^ — j-'<la* — x^ I — &c., in which x is the ab- la* 38 scissa; a the semi-transverse diameter ; and «=-n. =the 'ec- ^ a* centricity of the ellipse. MENSURATION. 275 ^vAs,.— Find the length of the quadrant CB by Prob. XXXI and CF by substituting the value of x in the above series. Twice the difference between these arcs -will give the length of the arc FBG. I What is the length of the arc FB G, if OB=^x=^,EF=y=i, &nd OC=fe=10? Since a^y^-^b'^x''=a'^b'^, we find, by substituting the values of x,y, and b, a=15. Then by the formula, BBG=s=2 U^ra | 1— ^^> 1 12 4j 3 i2 4 6j 5 \ . ,x e^ (a^ . _.x X J— ; 1 — a sin-' „— hr sin '--.^Vfl^—xZ a 2a [2 a A ) -2A^^Lr\2 -'--^V^W^- fli ei5 1 9 FIG. 18. 9 9 -,, ^A e* r3.15\a52 _ , 2.151, 153 . — - sin '-TTT 2 15 •2VI5 =— 9O c V15^ i^— 72) ^.4.153 L 4 I. 2 _| VI53— 9O — &c. =jrl5X-815— 2 j^ '^— 3c 3.152 ( 185 I J 15" s 30 8.15 1= ^;r— 72— 3']— &c. ^=38.406- ProHt. XXXIV. To find the area of an ellipse, the trans- verse and conjug-ate diameters being given. Formula.— A=:AJydx=4, _ / ^^^2 _ ^^2 )dx=nab, in -which a and b are the semi-transverse, and semi-conjugate diame- ters. Rule. — Multiply the -product of the semi-diameters by it^ .S.H1B92, or tnultifly the product of the diameters by \7i=.785S98. I. What is the area of an ellipse whose traverse diameter is 70 feet and conjugate diameter 50 feet? By formula, ^=;rfl:i5=;r35X 25=2748.893 sq. ft. (•1. 35 ft.=-|^of 70 ft.=length of the semi-transverse diameter. II.<) 2. 25 ft.=^ of 50 ft.=length of the semi-conjugate diameter. I3. .-. 2748.893 sq. ft.=;rx35x25=the area ofthe ellipse. III. .-. The area of the ellipse is 2748.893 sq. ft. Note. — Tiab^^/i^a^.-Kb'. .'. The area of an ellipse is a mean propor- tional between the circumscribed and inscribed circles. 276 FINKEL'S SOLUTION BOOK. Prob. XXXY. To find tbe area of an. elliptic segrment, hav- ing" given the base of the seg*nent, its height, and either di- ameter of the ellipse, the base being parallel to either diam*^ etero Formulae. — {a)A= j ydx, or I xdy=^ j x(a^ — x^)^dx^=: ' »'twe)--(i)*-.(£)'— (£)'-2..,[£]--...a.r (i)'--fe]'-^(5]"--"(S)"---i (b) ^= A^a:=^r*(««— ««)i-|-a2 sin-'H. The former formula of («) gives the area of a segment whose base is parallel to the conjugate diameter and the latter the area of a segment whose base is parallel to the transverse diameter. !Rule< — Find the area of the corresponding segment of the circle described u;pon the same axis to -which the base of. the seg- ment ts ferfendtcular. Then this axis is to the other axis as the area of the circular segment is to the area of the elleftic segment, 2. PARABOi,A. y ^=3,fx is the rectangular equation of the parabola. -cos -Tj is the polar equation of the parabola. In the rectangular equation, HG=y, the ordinate; A Cr=x, the abscissa; ^J?"=^.£'=-i^/. If any two of these are given the re- m?ining one may be found from the equation, p is a constant quantity. Prob. XXXVI. To find the length of any arc of a parabola cut off by a double ordinate. For'mula.—s=2jVdy^+dx'^=2 f{p^^y^)^dy= ^ log [Z±^^!^] or |y(/^+y )>< ^y-2 C-H. i|J- MENSURATION.'i 277 Kule. — When the abscissa is less than half the ordinate: To .the square of the ordinate add \ of the square of the abscissa and ituice the square root of the sum will be the length of the arc. I. What is the length of the arc KAH, if^Gis2 and GH lo'i By formula, 5=^V'/2_|_y2^_ ^log [.H:^^2!S!]=|V^i+P+i> log I ~i~^-^ ~r 1. Since _j'2=2^:«, we have r6+3Vl3n 2\^^9 log j(2+Vl3), or •,by series, ^=20'4i.i-^-i-iiijJ+i.i-fi-i-||I-&'^-]=12.7105 i^length of the arc, nearly. 1. 2=/l G^the abscissa. 2. 6= G^==the ordinate. II J 3. 36=the square of the ordinate. 4. '^=f of 2^=f of the square of the abscissa. 5. 2V(i#+36)=12.858=the length of the arc, nearly. III. .. 12. 858=length of the arc, nearly. Prob. XXXVII. To find the area of a parabola, the base :and heig'ht beingf g-iven. Formula.— A=2 J ydx=2 j-y^dy^=^=%xy =f(x. ly), i. e., the area of pa.ra.hola. J^ICA is |- of the circumscribed rectangle. Ii/Ul6* — Multiply the base by the heig'ht and § of the product •will be the area. I. What is the areA of a parabola whose double ordinate is 24m. :and altitude 16fn. ? By formula, ^=f (x 2_y)=|( 16x24 )=266m''. 1. 24m.^.^^(in last figure)^the double ordinate, or base of the parabola. yT 1 2. 16m.^yl G=the altitude of the parabola. 3. .•.384m^=16x24=the area of the rectangle circumscribed about the parabola. 4. f of 384m2=256m2=the area of the parabola. III. .•. The area of the parabola is 256m^. II. 278 ^ FINKEL'S SOLUTION BOOK. Proli. XXXVIII. To find the area of a parabolic frustunr having- g-iven the double ordinates of its ends and the di* tance between them. FoTmula, — -<4^=f«X-D2 — jr- ^^ which a is the distance be tween the double ordinates, B the greater and b the lesser doubk ,of4inate. Rule. — Divide the difference of the cubes of the two ends by the differenceof their, squares and multiply the quotient by \ of the altitude. I. What is the area of a parabolic frustum whose greater base- is 10 feetj lesser base 6 feet, and the altitude 4 feet? By formula, ^=|aX :giZ:^2=tX4X^^53^=32f sq. ft. rl. 10 ft.=the greater base, 2. 6 ft.=the lesser base, and 3. 4 ft.=the altitude. 4. 784 cu. ft.=103— 6*=the diffeirence of the cubes of the two bases. ■)6. 64 sq. ft.=10^ — 6^=the difference of the squares of the two bases. 6. \2\ ft.=784-i-64=the quotient of the difference of the cubes by the difference of the squares. .7. .-. f X (4Xl2i)=32| sq. ft.=the area of the frustum. III. .•. The area of the frustum is 32f sq. ft. 3. HYPERBOLA. ' 1. a^y^ — ^2^2_. — (^232 is the equation of the hyperbola referred" to its axes in terms of its semi-axes. b"^ I 2. y'^= ^(2ax — x^) is the equation of a hyperbola referred. to its trans-verse axis and a tangent at the left hand vertex. a( 1 e^ ) 3. p=:r^ 2 is the polar equation of the hyperbola. 1 — e cost/ Having given any three of the four quantites a, b, x,y, the oth- er may be found by solving the rectangular equation with refe.r- ence to the required quantity. Prob. XXXIX. To find the leng-th of any arc of an hyper- bola, beg-inning- at the vertex. For'mula.—s=^ldy-^-{-dx^=-^\ J- ^. ali^T) ) '^'>'=' r 1 a^x^ 1.1-3 a^-f4a^3 g ^ 1.1-3-3.5 ■^V +1.2.8 ~T^ 1.2.3.4.5 ■ 38 ^ +12.8.4.5.6.7 a6+4a432_l_8a2a* „ . >v li7 J'''-&c.^ MENSURATION. 279 Ullle. — 1. Find the parameter by dividing the square of the conjugate diameter by the transverse diameter. 2. To 19 times the transverse, add 21 times the parameter of the axis, and multiply the sum by the quotient of the abscissa di- vided by the transverse. S. To 9 times the transverse, add 21 times the parameter , and multiply the sum. by the quotient of the abscissa divided by the transverse. Jj.. To each of the products thus found, add 15 tim.es the para- ^neter, and divide the form.er by the latterj then this quotient being multiplied by the ordinate 'mill give the length, nearly. (Bonnycastle^s Rule.) Note. — A parameter is a double ordinate passing througii the focus. I. In the the hyperbola DA C, the transverse diameter GA =80, the conjugate ..^/=60, the ordinate B C^IQ, and the abscissa ^.5=2.16^7 ; what is the length of the arc DA C? By formula, DAC== 1.1.3 a^-^-ia^b'^ *+ 1.2.3.4.5 b^ 1.1 3.^5 1.2.3.4.5.6-7 a»-\-4a*b^+8x^b* 20.658. .^^ttm^^^m FIG. 20. 1. 45=20/^-^0^=25? -f-a=the parameter Z^which, in the figure, should be drawn to this right of DC, to be consistent with the nature of the problem. 2. 1520=19X80=19 times the transverse diameter. 3. 945=21X46=21 times the parameter. 4. 2465^sum of these two products. 5. .02704=2. 1637-=-80^quotient of abscissa and transverse- diameter. 6. 2.465X.02704=66.6536=sum of the ^;ro(/ac^^ multiplied by the said quotient. Also, 7. [(80X9)+(45X21)]X^^=(720+945)X.02704= 45.0216. Whence 8. (15x45-1-66.8536 )-T-(15x45+45.0216)=741.6536-^ ■720.0216=1.03004. ^S. ..1.03004xlO=10.3004=length of the arc ^C, nearly. III. .-. The length of the arc is 10.3004. II. S 280 FINKEL'S SOLUTION BOOK. Prob. Xli. To find the area of an hyperbola,the transverse and conjugate axes and abscissa being given. Fornmlae.—(a)A=2jydx =2- C''\x^—a^)idx=z-x'' Vx'^—a*—ab log j^ ^'+V^"'— '^ ' n ^xy— Cx^-L^x^^ — a^-i I 1 x^ ——a J = «'•' ^^> ^=^*-^ \i-T3l x-^T^- 1 ( x^ Y 1 ( x^ Y 1 3.5.7. [«'+«',! ~5.7.9 i«2+**J '^'i Rule. — 1. To the product of the transverse diameter and abscissa, add ^ of the square oj the abscissa, and multiply the square root of the sum by 21. ' 2. Add 4 times the square root of the -product of the transverse diam.eter and abscissa, to the product last found and divide the sum by 75. S. Divide If times the product of the conjugate diameter and abscissa by the transverse diameter, and this last quotent multi- plied by the former will give the area required, nearly. — Bonny- castle^ s Rule. I. If, in the hyperbola DA C, the transverse axis AGis 30, the < conjugate diameter HI, 18 and the abscissa AB, 10 ; what is the area of the hyperbola DA C ? By formula {a),A^xY—ab log,[^^^^i:^^_^l=25/— 15X9 log.[^^^fEll!]=300-1351og. [ 25-hV400-[ ^ggp —135 loge3=300— 135Xl.09861228=151.687343,y being found from the equation a^j/a — 52^2= — a^b'. in which a=15, 3=9 and »:'=15+10=25. fl. 1. 21v'30xl0+fxl02=2lV300+500-7-7=2lV371.42857 =21 X 19.272= 404.712, by the first part of the rule. 2. 2. (4V30xlO+404.712)-^-75=(4xl7.3205+404.712)-f- II. ■{ 75=6.3199, by the second part of the rule. 18vl0v4 3. S.- ••• 30 X 6.3199=151.6776, by the third part of the rule, =the area of the hyperbola, nearly. III. .-. 151.6776=the area of the hyperbola. MENSURATION. 281 Prob. Xlil. Tofind the area of a zone of an hyperbola. Formula.— A=2- ('"(x'^—a^ )'dx . .^^^j;^^:^^-ab log, p^+^^']-^;.,V^F=:7^4. .«* log, [_ '^ ^' ^=x^^—x^^—ab log, l^-^T— J J_|_ "^^ l°g^ L ^^ J=«^,-^^,-«3 log, |__j-^^J, in which (^tjij-j ), and (jej,_yj) are the co-mdinates of the points C and Z respectively. I. What is the area of a zone of an hyperbola whose transverse •diameter is 2a=10 feet, conjugate diameter 23=6 feet, the lesser double ordinate of the zone being 8 feet and the greater 12 feet? By for™„I., A- « j-,-, j.,-.Jloi:. \ ^iMft-fl j But from the equation, when^==y2=6, x:=x; 2^10^6 and when j'=J>'i=4,^=x'i=13-J. Substituting these values of aTj andjCzj ^e have ^=50V6-66|— 30 log,(?2g^^±^ ) -= I 50V6— 661—30 log,[f(V6 + l)] | sq. ft. Prob. XLIl. To find the area of a sector of an hyperbola. :kal,o. Formula. — A^ab log,f ,--f^ 1. Rule. — Find the area of the segment AKL by Prob. XL., and subtract it fro?n the area of the triangle KOL. I. What is the area of the sector OAL (Fig. 20) if 0^=a=5, 0/=3=3,and LF=y=^-i By formula, A^ab log,(~+-|)=7i log.(^^+t>=7J log«[TV(3*f20)], But when y=4, »:=13i. Hence, 282 FINICEL'S SOLUTION BOOK. IX. HIGHER PLANE CURVES. 1. Higher IPlane Curves are loci whose equations are above the second degree, or which involve transcendental func- tions, i. e., a function whose degree is infinite. 1. THE CISSOID OF DIOCIvES. 1. The Cissoid of Diodes is the curve generated by th& vej-tex of a parabola rolling on an equal parabola. 2. If pairs of equal ordinates be drawn to the diameter of a circle, and through one extremity of this diameter and the point in the circumference through which one of the ordinates is let fall, a line be drawn, the locus of the intersection of this Ime and the rfqual ordinate, or that ordinate produced is the Cissoid of Diodes. 3. 1/2=: is the equation of the cissoid referred to rectan- gi^lar axes. p=2a sin^ tan^ is the polar equation of the curve. Prob. XLIII. To find the length of an arc OAP of the cissoid. Formula. — s= OAP= n+m Ji+ial "- /J'+CgS^)'-- V(2«— x)" —2+3 log, r V2^(V3+2) -I) LV3 V2a— !V-|-V8a— Sa-J ' I. What is the length of the arc OAJSF, in which case x=a} By formula, s=a J y5— Fig. SI. MENSURATION. 283 Prob. XlilY. To find the. area included between tl»* nurve and its asymptote,. BM. Formula. — ^=2 / y dx=2 I ^-—^dx^ — ^ V2flt :v~i "* Va:y2a — x{a^x) — 3sin~^- , — =37ra2, 2. e., 3 times the area of the circle, OEB. Note. — The name Cissoid is from the Greek xiaaosiSl^^ lik^ ivy, from. z(<ro-<!ij, ivy, eJ5o J form. The curve was invented by tne Greek geometer' Diodes, A D. 500, for the purpose of solving two celebrated problems of the higher geometry; viz., to trisect a plane angle, and to construct two geometrical means lietwet-n two given straight lines. The construction of two geometrical^means between two given straight lines is effected by the cissoid. Thus in the figure of the cissoid, BD and OG are the two geometrical means between the straight lines OD and PG: that is, OD:BD::OG.PG. The trisection of a plane angle is effected by the con- choid. The duplication of the cube. z. e., to find; the edge of a cube whose volume shall be twice that of a cube whose. volume is given, may be effected by the cissoid Thus, on ^C lay off CH=2,BC, &nidra.vi BH. Let fall from the point F, where BH cuts the curve, the perpendicular FR. Then RF^zzlBR. Now a cube described on RF is twice one described on ORy OR^ OR^ for, since FR^y,OR7=x, and BR=2a—x, we have RF^ =^^^= , „„ , or BR 9-^R iRF'=OR0. .-. FR'=20R^ C^E. D. 2. THE CONCHOID OF NICOMEDES. 1. The Conchoid, is the locus formed by measuring, on a line which revolves about a fixed point without a given fixed line, a constant length in either direction from the point where it intersects the given fixed line. 2. x^y^:^{i-\-y)^ (a^ — y^) is the equation of the conchoid re- ferred to rectangular axes. f /is. 22. 3. p=b sec ^±ais the polar equation referred to polar co-ordi- nates. In this equation, ^ is the angle PO makes with AO. Prob. XLiT. To find the length of an arc of the conchoid. ForrHula.~s= f^ \ + C^^ ' de=J»Jl-\-'i&a^ 6 secede. 284 FINKEL'S SOLUTION BOOK. Frob. XL VI. To find the whole area of the conchoid between "two radiants each making an angle 6 with OA. Formula.— A=2,j^r^de=b^ J {ssc e±ayde= i« tan e-\-2a^d-\-Zbi^a^—b^ or b^ tan ^+20^ 0, according as a is or is not greater than b. 1. The area above the directrix m m' and the same radiants z=2ai log^tan ( j -|- - j-j-a^^ 2. The area of the loop which exists when a is _^ ^ is fl^ cos-^-2«Mog j!±^g I ^^^^31^7. Note. — The name conchoid is from the Greek, xoy^aecSiq^ from iKOvZ^, shell, anid e75o?j form, and signifies shell -form. ,It was invented by the Greek geometer Nicomedes, about A. D. 100 for the purpose of tri- secting any plane angle. The trisectioq of an angle may be accomplished fey this curve as follows: Let A OH be any angle to be trisected. From any point, G, in one side let fall a perpendicular, G_F, upon the other. Take ^.¥=.100, and with O as the fixM point, w «»' as the fixed line and /'Pas the revolvins; line of which PDz=m is constant, construct the arc of the con- choid, /'^//. Erect ^G perpendicular to mm' and draw BO. Then is BOA one third of HO A. For bisect ^^at /, and draw GI. Also draw IL parallel to GK. Since BI=lk, BL=LG and GI=BI=IK=GO. By reason of the isosceles triangle BIG, -we have the angle G/0=2/G-ffO= ^£BOA. But /^GIO=IIOG. ■•.21IOA=IJOG,ot I0A=)4 /_HOA. Q^ E. F. 3. THB OVAL OP CASSINI OR CASSINIAN. 1. The Oval of Cassini IS the locus of the vertex of the triangle whose base is 2a and theproductof the other sides =m'' . 2. |j2^(a+x)8J. Sy^-\-{a—xy\=m*or{x'-\-y''-\-a'-y— Aa^x'^=m*\s.t\-\e. rectangular equation of the curve, in which ■2a=AB. 3. r* — 2a^r'^ cos 2B-\-a'^ — ««*=0 is the polar equation of the curve. Discussion. — If ahe^ m, there are two ovals, as shown in the figure. In that case, the last equation shows that if ■OPP' meets the curve in P and P', we have OP.OP'= '*Ja^ — m*; and therefore the curve is its own inverse with respect to a circle of radius= FIG. S3. MENSURATION. 285 FIG. 24. a^dr 4. LEMNISCATB OF BERNOUII^LI. 1. This curve is what a caj.sz«?a« becomes when w^a. The above equations then reduce to 2. (x2+j/2)''=2a2(«2_y8)and| 3. r^^la"^ cos 16. Prob. XliVII. To And the length of the arc of the Lemnis- cate. Foi'mula.—s=j A^+Cj^ ^6 J ■>! f.^\^ a J J r Jo V(a*— H) +i-l-i-^+fifT'Tf^ +&<:•} ■ When^a,^=<z[l+i.i+i.|.i -^^.f.|^.^i5-|-&cn= arc BPA. .-. The entire length of the curve is 4«[l+i.i+if-i+&c] Prob. XIiVIII. To find the area of the lemniscate. For'mula.—A=^C\r^de—^a'^ f^"" cos 26 dff= r2a« sin2^^'''=2«i=. 5. THE VERSIERA OR WITCH OP AGNESI. 1. The VeTSieTa is the locus of the extremity of an ordi- nate to a circle, produced until the produced ordinate is to the ordinate itself, as the diameter of a circle is to one of the seg- ments into which the ordinate divides, the diameter, these seg- ments being all taken on the same side. e * 2. Let P be any point of the curve, I^D=y, the ordinate of the point P and OD==x, the abscis- sa. Then, by defini- tion, ^P i^i^^: : AO:EO, or x: EF: :2 a -.y. But =>i{2 a—y)y.' .-. X ■.>i{2a—y)y: : 2a -.y. Whence x^y= FIG. 25. ia^(2a — y) is the equation referred to rectangular co-ordinates. ■i^KII^H^HC 286 FINKEL'S SOLUTION BOOK. 3. r(r^ — r^ sin^ d-\-4a^) smd=Sa* is the polar equation of the curve. Prob. XlilX. To find the length of an are of the Verslera. ■dx. Formula.-s=f^l+(^^ ' dx=f^l+- 4x^ This can be integrated by series and the result obtained ap- proximately. Probi Li. To find the area between the curve and its asymptote. dx "■■• Formula, — Az=2 I ydx=2 X 8a » / — ^V\a'^ tan +4a2 Kale. — Multiply the area of the given circle byi. Note. — This curve was invented by an Italian lady, Dona Maria Agnesi, Professor ot Mathematics at Bologna, 1748. 6. THE LIMACON. 1. The LimaQOW is the locus of a point P on the radius vector' OP, of a circle 0/?!£' from a fixed point, 0,oh the circle and at a constant distance from either side of the circle. 2. {x^-{-y^ —ax)"^ = b^(x^-\-y^) is the rectangular equation of the curve. 3. r=aQOsd±b \i the polar equa- tion.' In these equations, a= OA and b=PP. Prob. Lil. To find the length of an arc of the Liimacon. o Formula. — s= FIG. 26. Jhl{acosd-\-by-\-a'^ sm<^9 d0=f»i[(~a^-\-b'^-2abcos 6)dd= j\ j (a-\-b)^ cos^-^+(a— by sin"^^ I dd. .: The rectification of the Limacon depends on that of an ellipse whose semi-axes are (a^b) and {a — b.) When a=b, the curve is the cardioid, the polar equation of which is r=a(l-f cos^ ), and s=j ■\l'''+(^-^ J ^*= MENSURATION. 287 «z jr;^2+2cos<' de= +2a fcosifi d&=2a jcosi 6dd— '2a I COS \ 6 c^^=8a=the entire length of the cardioid. Prob. I>II. To And tlxe area of the Limajon. Formula.— A=jir^cl0=ir''(acos0+b)^dd=. ■7t(^a^-\-6^-). When a=^, the curve becomes a cardioid, and A= ^Tta'^. When«^i5, thecurve has two loops and is that in the figure. :r=±a cos d-\-6 is the polar equation of the outer loop, and r^a cos 6 — 6 is the polar equationof the inner loop. The area of the inner loop is A=J^r^db=lj "{acose—6)^d6= i(fl2+52) cos-' a— !*>/«•■'— s^.,; Note. — This curve was invented by Blaise Pascal in 1643. When (z= HI), the curve is called the Trisectrix. 7. THE QUADRATRIX. 1. The Quadratrix is the locus of the intersection, P,pi the radius, OD, and the ordinate §iV,when these move uniformally, so that ON : OA :: L B OD:\n. .^ J is the rectangular equation of the curve, in which a^^OA, x=ON, and y=IJV. 3. The curve effects the quad- rature of the circle, for OC-.OB:: OBaxc ADB. Prob. lilll. To find the area enclosed above the x — axis. Formula. — A^fydx^ X tan I .^ \dx=4a^7r'^ log 2. , Note. — This curve vras invented by Dinostratus, in 370 B. C. " 8. THE CATENARY. 1. The Catenary is the line vphich a perfectly flexible chain assumes when its ends are fastened at two points as B and C in the figure. FI6. 27. b\ z// c 1 M V T>J — "- 1 \ M t- 288 FINKEL'S SOLUTION BOOK. 2. y=^a{ea-\-e'-a) is the rectangular equation of the curve, in-, which 0=0-4. ^ is the origin of co-ordinates. BAC is the- catenary. M'AI'M is the evolute of the catenary and is called the True- trix. To find the equa- tion of the curve, let A be the origin of co-ordinates. Let s denote the length of any arc AB; then, if ^ be the weight of a unit of length, of the chain, the verticle tension at E, is 'sp. Let the horizontal f /ff, 28. 'tension at B, be af, the weight of a units of length of the chain. Let BGhe a tangent at B, then, if BG represents the tension of the chain at B, EF and OF will represent respectively its hori- zontal and its vertical tension at B. < dy PQ__sJ_^ s j ids'^—dx^ C ^^ _ alog (j+Va^+-f^)+c. Since x^zo, when s=o, c= — aloga_ .-. ;«=« logf — ^.^1-4 — 2 I" From this equation, we find .s= , —I e 'ar\-e a I v/^hich is the length of the curve measured from A _ ^ dy s dy s , /' ? _?\ Prob. WV. To find the area included between the Catenary, the axis of x, and two ordinates. Forimvla. — A==Cydx=J\a \ ef-'re^H '^^*= gS-\-fa \z=ayy^ — a^. This is the area included between the axis of X, the curve and the two ordinates, _j'i==a,;j/g==j'. jjOTB. The form gf equilibrium of a flexible chain was first investigatedL by Galileo, who pronounced the curve to be a parabola. His error was de- tected experimentally, in 1669, by Joachim Jungius, a German geometer;, but the true form of the Catenary was obtained by James Bernouilli, in 1691- MENSURATION. . 289 9. THE TRACTRIX. 1. The Tractrioc is the involute of the Catenary. -2. x—a log j fl--|-V(a2— j2) | —a log y—Sl [a^—y'^ , is the rect- •angular equation of the curve. ' l*rob. LiV. To find the length of an arc of the tractrix. Formula, — «= a log i~)- Prob. IiVI. To find the area included by the four branches. FormMla. — A^=J'ydx= — ij^ *ia^—y^ dy^na « . 10. THE SYNTRACTRIX. 1. The Syntractriac is the 'locus of a point, Q, on the tan- gent, PT, of the Tractrix. 2. x= a log J c-|-V'(c* — y"^) \ — a log_y — V(c^ — y^) is the rect- angular equation of the Syntractrix, in which c is §7", a con- stant length. 11. ROULETTES. 1. A. Roulette is the locus of a point rigidly connected with a curve which rolls upon a fixed right line or curve. (a). Cycloids. 1. The Cycloid is the roulette generated^by a point in the circumference of a circle which rolls upon a right line. 3. A. Prolate Cycloid is the roulette generated by a point without the circumference of a circle which rolls upon a right line. 3. A Curtate Cycloid is the roulette generated by a point within the circumfere nce of a circle which rolls upon a right line. 4. , x=versin~'/ — ^2ry — y^ is the rettangular equation of the cycloid referred to its base and a perpendicular at the left hand vertex. To produce this equation, let AN=x and PJV=y, P be- ing any point of the curve. Let OC^ r^the radius of the generatrix OPL. Now AN=A O—NO. But] by con- struction A 0=arc P 0=rrvtxBva.~^ FO, or versin^'j' to a radius r. NO^ PP=»JF L X FO=^ly{ 2r—y)= V2>'_y— jf^- •■• «=versin'^j — 4^iry^y^ Or, we may have «==a(S--sin^), and FIG. 29. j/=fl!(l — cos^) in which is the angle, I^CO, through which the generatrix has rolled. 290 , FINKEL'S SOLUTION BOOK. For x^AO—NO. But AO=aLPCO^a6, &n6 ]VO=PF =PC sin IP CF=a smd. .■ . x=ae—d sin e=a(e—s\.n9) , y= OC—CF=a—CF. 'BxxtCF=PCcoslPCF==acosb .■ y=- a — a cosft==a( 1 — cos6'). Prob. liVII. To find the length of an arc of the cycloid. 4^rf{2r~y)~* dy=: — 242r{^r—y)*-\-c. Reckoning the arc from the origin, c=4r; and the corrected integral is j= — 2{2rY (2r— j) +4r. When^/=2r, 5=4r. .-, The whole length of the cycloid is 8>"=4Z?, i e., the length of the cycloid is 4 times the diameter of the generating circle. Rule. — {1) Multiply the corresponding' chord of the geneta- trig, by 2. To find the length of the cycloid : (^) Multiply the diameter of the generating circle by 4- I. Through what distance will a rivet in the tire of a 3-ft buggy wheel pass in three revolutions of the wheel? By formula, j==3(8r)=24Xlift.=36ft. 1. §ft.=the diameter of the wheel. Then 2. 12ft.=4x3ft.=distance through which it moves in 1 11.^ revolution. 3. .•• 36ft.^3Xl2ft.=distance through which it moves in 3 revolutions. III. .•. It will move through a distance of 36 ft. Prob. LVIII. To find the area of a cycloid. r-lr yidv FonmUa. — A=2 fydx=2 I , -^ -^ = , '^ '0 V2ry-^2 Sr^versin-'^2=B7rr^. Rule. — Multiply the area oj the generating circle by S. I I. What is the area of a cycloid generated by a circle v?hose radius is 2ft. ? By formula,^=3;rr«=37r2='=127r=37.6992 sq. ft. C 1. 2ft.=the radius of the generating circle. II. < 2. ;r2^=12.5664 sq. ft.=the area of the generating circle. ( 3. 3;r2«=37.6992 sq. ft.=the area of the cycloid, III. .-. The area of the cycloid is 37.6992 sq. ft. Prob. IiIX. A wheel whose radins is r rolls along a hori- zontal line with a velocity v'; required the velocity of any point, P, in its circumference ; also the velocity of P horizon' tally and vertically. MENSURATION. ■291 Since a point in the circu'm fere nee of a wheel describ,es, in space, a cycloid, let P, Fig. 29, be the point, referred to the axes ^>1' and a perpendicular at A. Let (.jf^y) be the coordi- nates of the point; then will the horizontal and vertical velocities of P be the rates of change of x and y respectively. ■O being the point of contact, A 0=r versin"' -. Since the cen- ter C, is vertically over 0, its velocity is equal to. the rate of in- crease of AO. In an element of time, dt, the center C will move (y^ rdy y versmr'-c J=T ^ ^ . .-. Its velocity w'^ the distance it moves divided by the time it moves, or v'^^ , ^^-^ ^dt=_ "■ % ... %JEy^^= the velocity V2r)'— _j/' 1 ' i^lry—y"^ dt dt »■ ■ •vertically. ... (1). c I ^ ,, , , ■ From the equation of the cycloid, x=/'versin-^-C — Slry — y''-, we y have dy^ - — r dy. Now dx-^dt^=\!si.& velocity of the point V2ry— -y2 - borizontallv. But dx^^di, or —r,,= , ' -4-- Substituting the dt i^2ry—y-^ dt ^ value of -^, we have ^- = -1)' (2). An element of the at at r ^ ' , curve APBA-' is ds and this is the distance the point travels ds in an element of time, c?if. .-. — = the velocity of the point, P. But ds=\/dy-^-\-dx'^==-\ (^ ^^~-^ +"!':?) -"'dt^-^-^ v'dt, since , from (1), (/j»'=-^^2:^^^and, from (2), rf5<:^('^.^^). .-. By dividing by rfif, we have -5-=o=>J—^''= the velocity of the point, P (3). From (1), (2), and (3), we have, dv „ dx „ , ds '^-^■=^'i=^' 5?="' ^"' dt =''■ ■ dy dx , ,ds^ , — - :f^=.,^=.',^=.',and^-=V2.- •c n dy dx , ■ ■, ds . „ , ,f^=2.,^=0, -=.',a„d-=2.', 292 riNKEL'S SOLUTION BOOK. Hence, when a point of the circumference is in contact withi the line, its velocity is 0; when it is in the same horizontal plane as the cenier, its velocity horizontally and verically is the same- as the velocity of the center, and when it is at the highest point, its motion is entirely horizontal, and its velocity is twice that of the center. Since -=-=>. I— »' = — -i/.vfs have by protjortion ,. at ^ r r ^ : v' : : V2n':r. But V^=x^(Pl^+J?0^ )=ro. .-. The velocity of -P is to that of C as the chord PO is to the radius CO; that is,- -f and C are momentarily moving about O" with equal angular velocity. (*). The Prolate and Curtate Cycloid. 1. x=a(ff — m sinff), y=a(l — /«cos^) are the equations in ' every case. 2. The cycloid is prolate when m is >1 as AIP'B'TA', Fig. 30, and curtate when m is <1, as PB. These equations are found thus: Let CP=ma, and LOCP=e. Th^tn.x=AN=AO—ON. But A0=2Mz subtended by lOCP=ta9, and OiV=/'CXsia lNPC=mas\n LJVPC {=lPCL=7t—0)=ma sin (71—6)= ma sin ff. ,-. x=at) — ma sin6=a(d — m sind). y=PN=OC-\- PC cos INPC {=lPCL=7r—f)) =a-\-ma cos {n—d ) =a— ma cosd =a{X — m cos g ). The same reasoning applies whexi we- assume the point to be /". Note. — ^These curves are also called Trochoids. '• Prob. liX. To And the leng-th of a Troelioid. Formula,— s^ Jul dx » -^dy^ . Since" x=o(^ — m sin^), dx=a{l — »« cos f )£^(^ ; and since _y==- a{\ — »i cost) ), dj=am sinOdO. .-. s= Ndx^-\-dy^=a j ^>JI (1 — m cos^) 2-|-/w2 6inS' e~\dB=a f -^i l+m^—2m cos ff )de= 4a p"^ j (l-^m)^—4m cos^ tp I d^)=4a(l+m)J'^''^(l— cos^ cp)dq). {\-\-m) I. If a fly is on the spoke of a carriage wheel 5 feet in diame- ter, 6 inches up from the ground, through what distance will the MENSURATION. 293 fly move while the wheel makes one revolution on a level plane? Let Cbe the center of the wheel, in the figure, and P the position of the fly at any time. Let; OC= the radius of the carriage wheel =a=2i ft., PC= '2ft., arid the angle OCP ^=B. Let {x,y) be the co- ordinates of the point P. Let F, a point at the inter- Wff. 30. section of the curve and ^/ be' the position of the fly when the motion qf the wheel commenced. Then since :«=:«( f — z«sinS) and _y=a(l^ — wcos^), we have <^Ar^a(l — mcoi6)dd, and dy= ' a m sin<9 dB. .: s= j4 {dx^ -^dy^ ^^£ \ , 1 ''' ^ ^~"' cos») 2+ a^wi^sins Q \ de=a J if(\J[-m^— 2m co% e)dd=4a f ^ ^ \ (1+ .m)" — im cos^ ip y d(p, in vyhich 9)=4^^. But PC=2 ft, and ^ PC=»2«=»2X2ift.,»«=2-H2i=|. .-. .r=4x2i/ " (1+1)^— 4x4cosV I rfGJ=10 /'^'"iVrSl— 80cosa9))(/^»= * 18/^Vri-|icos== '^)rf9*=9;r j 1^|«_3(t-.V)^ (M)^- 5(^.t^^)^(f^)«-7(i^.-M,)''(|^)^-&c 1=18.84 ft. II. .-. The fly wijl move 18.84 ft. Prob. LXI. To find the area contained betw^een the tro- «tu)id. and its axis. 1 Formula. — A=F^Jyd,x=2a'^ I ( 1 — m cos 6) ( 1— m cos e)de=2a^ /^"(l— w cos 6)^de=2a'^(e—2m sin(9+^»?2 =2rt:^(7r-f-|»z'=;r). When »2=1, the curve is the cycloid and the area=37ra!2 as it should be. * When 9>is replaced by (\Tt-\-(p), this \^^ti. elliptic integral oi the second kind and may be written 4a^(f^,<p). -J] 294 FINKEUS SOLUTION BOOK. {c). Epitr.ochoid and Hypotrochoid. ^ 1. Ann, MpitVOChO'id is the roulette formed by a circle roll- ing upon the convex circumference of a fixed circle, and carrying a generating point either within or without the rolling circle. 3. A/n Hypotrochoid is the \oulette formed by a circle rolling upon the concave circumference of a fixed circle, and carrying a generating point either within or without the rolling circle. «+^A a+6. 3. x=(a-{-d)cos 6 — ?nbco%"^^^^dy Jfe=(3-|-3) sin ^ — mb sin^^^^^^ are the equations of the epi'trochoids. In the figure, let C be the center of the fixed circle and O the center of the rolling circle. Let FP' Q be a portion of the curve generated by the point P' situated within the rolling chxle, and ■ let CG=x and P' G=y be the co-ordinates of the point, P'. Let A be the position oi P when the rolling commences, and 9>=Z/'OC through which it rolled. Drav^ O^ perpendicular to CG and 7"7 perpendicular to OK; drav/ £>P and DP". Let OP'=mOP=mb and the angle A CD=e. Then x=CG=Ck -^KG=KC-\-P' I. But CK=OC zo^ii={^a-\-b) cohti and/"/== P' Ocos Z OP'I^mh cos \7t—{cp^e)\=:—mb cos(9>-(-^). But arc AD^arcPD. .-. ad= b(p. Whence (p=^d. .: P'I=^-mb cos^^e, and Ar=(a-f-5) cos d— y=P'G 7 'f+^a mo cos — ; — a. =IK=OK—OI. But OK= OCs\nlKCO=, FIG. 31. (a-f3)sin^, and OJ=OP' s.\nlOP'I=mb sm\7t~{<pJf-e)\^ y=={a-\-b)&m 6 — mb sin 'l 6. mb sin( ip-\-d)^=mb sin "7" 0. o u \im=\, the point /"will be on the circumference of-the roll- ing circle and will describe the curve APN which is called the: MENSURATION.^ , 295- Mpicycloid The equations for the Epicycloid are x-={^a-\-b )cos^— 3cos-4— ^, and y=(a+i5) sin d — iJsin — j- d. The equations for the Hyfotrochoid may be obtained by changing the signs of b and mb, in ]the'equations for the Epitrochoid. /. x=(^a — 3 )co& 6'-)-»?6 cos ' —7 — 6, and_)/=(a — ^)sin 6* — »z3 sin — 7—^ are the equations for the .1 Hypotrochoid. If »?=1, the generating point is in the circumfer- ence of the rolling circle arid the curve generated will be a Hypo- „ z cycloid. .". x=(^(L — b) cos^ -\-b cos —7— 6^, and j(/=(a — 3)sin 6 — b sin — 7— B are the equations of the Hypocycloid . Prob. LiXII. To find the leug-th of the arc of an epitro- choid. Formula.— s=J'^dx'+dy'= J ^ |[^_(a+*) sin 0+ tw(a-f3) sin^i-^~j +r(«+^)"s e—m(a-\-b) cos^i-^l | 40 =(a+b)fA \ \-[-m^—2m{%m e sin^ (9+ cos^cos ^-^9) \dd=(a-^b)J>f{l-\-m^—2mcO6j0)de. This may be expressed as an elliptic integral, JS{k, (p), of the second kind, by substituting (tt -| q)) for 0, and then reducing. 2. By making m=l, we have s=(a-\-6)*j2 f'/Ji — cos^ ) dff, the length of the arc of an hypocycloid. b 3. By changing sign of 6, the above formula reduces to s= (a — b) f'^{l-\-9n^-\-2mcos-j )d0, which is the length of the arc of an hypotrochoid. 4. By making m^^., in the last formula, we have s^ (a — 3)^2 r( 1+cos 7 0p d0, which is the length of the arc of an hypocycloid. I I. A circle 2 ft. in diameter rolls upon the convex circmii- ference of a circle whose diameter is 6 feet. What is the length 296 FINKEL'G SOLUTION BOOK. of the curve described by a point 4 inches from the center of the rolling circle, the rolling circle having made a complete revolu- tion about the fixed circle? In Fig. 31, let Obe the center of the rolling circle; C the center of the fixed circle ; CD=Z ft.=a, the radius of fixed circle; 0D=^ lft.=g, the radius of the rolling circle; 0/'==4inches=J ofl2 inc\ies=mb the distance of the point from the center; and P the positionof the point at any time after the rolling begins. Let ^= the angle A CD and 93=the angle POD through which the roll- ing circle has rolled. Then we have, as previously shown, the equations of the locus P, x=={a-\-b)cos 6 — mb cos(9>-|-6')=(fl;-f-^) cos 6 — mb cos 7^ fi, y=^(^a-\-b) sinff — mb s\n(^<p-\-0)^(^a-\-b)s\n d — mb sin —j~ 0. From these equations, we can find dx and dy. .-. By formula, s=Ndx ^-\-dy^ =6.( a+b)J^ " V^l-j-w/*— 2wcosf &)d&=24: r^Va+Ci)^— f cosS0)d0=& p^VOLO— b Jo ^ - %J ft \ ' & cos ^0)de. Let 3 (9=2?&. Then s=8 /" V(10— 6 cos 3 e)de==^ Jo ^nP^ V(l— f cos^f)d^, =^Hp'' [}—i-i C0S2^^.1.(|) 2 COS* f—i.i.^ay cosotp—i i-i-Ui)* cos8^— &c.^<f ^^ . 21* I ^fc— i-l[i(isin 2^+^)]-i.i.(|)^[Kisin4^+2 sin2^+3^)] -i.i-l(f)''[irV(isin6^+f sin4^-}-V sin 2^+10^) ]-&c. ( *",= lOfT I l-(i)^t)-*(i:l)=(l)^-i(i•l•f)^(i)*-4(M■M)'- (l)4_&c. I =10|;rx. 773=26 9 ft,, nearly. Remark. — When the point is on the circumference of the roll- ing wheel, the length of the curve generated by the point is 5^= {a-\-b) J^l-\-m^—2m cos| e)de={a-\-b) J»J{i^l—2 cos 20)de. If we let the conditions of the above problem remain the same. MENSURATION. 297 ■only changing the generating point to the circumference, we have for the length of the curve, ^=6^2(3+1) /" V(l— cos3l9)</^= -4S fij{l—icos^ ep)d(p, where q)=lH. Expanding this by the Binomial Theorem and integrating each term separately, *= :24;rjl-(i)^(i)-i(i.i)^(i)^-Ki-M)^ (*)'-&<:. | I. A circle whose radius is 1 ft. is rolled on the concave cir- cumference of a circle vv^hose radius is 4 ft. What is the length -of the curve generated by a point in the circumference of the rolling circle, the rolling circle having returned to^the point of /starting? .v=(d! — i5)cos &-|-i5 cos — 7— ^, 7 y={.a — ^) sin S— 3 sin —7— S, are the equations of the curve -which is a hypocycloid. . In these equations a=4: and 6=1, .-. x=^Z cos^-)-cos 3^i^4 cos* f), and _j/=3sini9 — sin 3 ^^4 sin? ^. Whence, cos".^= J(0, and sin ^=.^(9 • .-. cos.2 ^+sin2 ^=(fy+(jiy But cos2 e+sm^ 6^1. ^l_j_yl=4i,x,which is the rectangular equation of the .curve. ,', By formula,.f=V[5^^^T^^=4/"'(^^^^i^)'rfx= •4«J jf^x ''dx=4J [i"^'] V6«=6 X 4=24 ft. X. PLAIN SPIRALS. 1. A Plane Spiral is the locus of a point revolving about a fixed point and continually receding from it in such a manner that the radius vector is a function of the variable angle. Such « curve may cut a right line in an infinite number of points. Thjs would render its re.ctilinear equation of an infinite degree. Hence, these loci are transcendental. 3. The M-easuvinff Circle is the circle whose radius is the radius vector of the spiral, at the end of one revolution of the generating point in the positive direction. 298 FINKEL'S SOLUTION BOOK. 3m ^ Spire is the portion of the spiral generated by any- one revolution of the generating point. 1. THE SPIRAL OF ARCHIMEDES. ^ 1. Tlie Spiral of Archirnedes is the locus of a point revolving about and receding from a fixed point so that the ratio- of the radius vector to, the angle through which it has moved from the polar axis, is constant. 2. r=ad is the polar equation of this curve. Prob. LiXIII. To find the length of the spiral of Archimedes. Formula. — .?= ialogj^+V(l+^') j, which is the length of the curve measured from the origin. FIG- 32. 5-=a7rV[l+(27r)2]+^alog \ 2;r+y[l+(27r )2] | is the length. of the curve made by one revolution of the generating ^oint. Prob. LiXIV. To find the area of the spiral of Archimedes.. F.or'mulct.—A=lJr-'de=^a''je^de=la^e^^*—, the area measured from the origin. 2. THE RECIPROCAIv OR HYPERBO]:.IC SPIRAL. 1. The Recf/procal or Hyperbolic Spiral is the locus, of a point revolving around and receding from a fixed point so. that the inverse ratio of the radius vector to the 'angle through, which it has moved from the polar axis, is constant. 2. ''=r; is the polar equation of the Hyperbolic Spiral. o Prob. liXV. To find the length of the Hyperbolic Spiral. Formula. — s= FIG. 33. MENSURATION. 299 « '(1+0^)1— log \ «+V("l+(?2) I _«-Vl4-(?2, is the length of the spiral measured from the origin. Prob. 1.XVI. To find the area of the Hyperbolic Spiral. /dff a^ measured from the origin. This result must be made positive since the radius vector revolves in the negative direction. 3. THE LITUUS. 1. The XiitUUS is the locus of a point revolving arottnd and receding from a fixed point so that the inverse ratio of thp radius vector to the square root of the angle through which it has moved, is constant. a 2. ''=7/0 is the equation of the Lituus. Prob. liXVII. Xo find the length of the liituus. FIG. 34 Formula.— s= fA^^-\-\-^ de=^ae-lj{{ 1+4^2 )dM= Prob. liXVIII. To find the area of the Lituus. Formula.— A=iJr^d0=J^f'^=^a^ log 6. 4. THE LOGARITHMIC SPIRAL. 1. The Logarithmic Spiral is the locus generated by a point revolving around and receding from a fixed point in such a manner that the radius vector increases in a geometrical ratio, while the variable angle increases in an arithmetrical ratio. 2. r=a9 is the polar equation of the Logarithmic Spiral. If <z is the base of a system of logarithms, this equation becomes- ^^log r. Prob. liXIX. To find the length of the Liog-arithmic Spiral. ■■ Formula.-s=f^r^+(^^^'de^f^r-^+Ql.ye^ 300 FINKEL'S SOLUTION BOOK. {m^-\-l)idr=if(m^-\-l)r, where m is the modulus of the system ■of logarithms. '' Prob. LXX. To find the area of the liOgarithmic Spiral. Formula.— A=lJ'r^d.6=— I rdr= ^mr^. Since m=l, in t^e Naperian System of Logarithms, A=\r^ , i. e., the area is ^ of the square of the radius vector. FIG. 85 XI. MENSURATION OF SOLIDS. 1. PARAI,I,EIX)PIPEDS. Pi-Ob, LXXI. , TTo find the solidity of a cube, the length (Q>£ 4ts fed'ge being- given. 2^orwi-«*?ffi.— F=(edge) x(edge)X(edge)=(edge)3. !]Eiule. — Multiply the edge of the cube by itself, and that product again by the edge. I. What is the volume of a cube whose edge is 5 feet? By formula, r=(edge)8=(5)8==i25 cu. ft. - jj j 1. 5.ft.=the edge of the cube. ■ \ 2. 6X5X5=125 cu. ft.=the volume of the cube. III. .-. The volume of the cube is 125 cu. ft. Remark. — Some teachers of mathematics prefer to express the Tolume by saying 5x5X5x1 cu. ft. =125X1 cu. ft.=125 cu. ft. Prob. iiXXII. To find the volume of a cube, having given its diagoi»aI. (d'\} 1^ ) V3>' RuIGc — Divide the diagonal by the square root of S, and the cube of the quotient will be the volume of the cube. I. What is the volume of a cube whose diagonal is 51.9615 inches? MENSURATION. 301 By forn,«la, V^( Q' =C-^^'y =(?ff^y =^ ^ ' Vvs^- V V3 ^^ Vl.73205>' 27,000 cu. in. 1. 51.9615 in.=the diagonal. 2. 30 in.=51.9615 in.-f-V3=51.96I5 in.-Hl.73205=the edge ir. of the cube. ■. 30X30X30=27,000 gu. in.=the volume of the cube. in. .'. The volume of the cube whose diagonal is 51.9615 in., is 27r000 cu. In. Prob. liXXIIl. To find the volume of a cube whose surf ^c^ is g'iven. Formula. — V= <4)' iiA Rule. — Divide the surface of the cube by 6 and extract tk& squae root of the quotient. This will give the edge of the cube,. The cube of the edge will be the volkme of the cube. I. What is the volume of a cube whose surface is 294 squaie; feet.' By formHla, V=(^^0' =(^^^y ={sfA^)»=l^.= 243cu. in. 1. 294 sq. ft.==the surface of the cube: 2. 49 sq. ft. ==294 sq. ft.-^ 6^area of Psne eaue of the cube- 3. V49=7 ft.=length of the edge of the cube^. ,4. .-. 7X7X7=348 cu. ft.=volume of cube- Ill. .•. 343 cu. ft. is the volume of a cube whose surface iss 294 sq. ft. Prob. LXXIV. To find the solidity of ft piiFSIiiielopipedou.. Formula. — V^ly^bXt, where /==len2th, 3=breadth, and /=thickness. Rule. — Multiply the length, breadth and thinness together. I. What is the volume of a pafallelopipetlon whose length ia, 24 feet, breadth 8 ftet, and thickness 5 feet? By formula, K=/x^X ^=24x8x5=960 Q«. ft, rl. 24 ft.=the length, jj J2. 8 ft.=the breadth, and 1 3. 5 ft.^the thickness. U. .-. 24x8X5=960 cu. ft.=the vgIuto^ III. .". 960 cu. ft.^the length of (he parallelopipedon- 302 FINKEL'S SOLUTION BOOK. Prob.,l^XXV. To Had tbe riimenslons of a paralleloplpe- 4on, having' giveii the ratio"of its cUihenslbus and the volume. Formula.— l=yiV-^{mXnXp)'\m; b=yXV^ (»iXnXp)}fi ; and i=y[ F-^(»zX»X/)]/, where m, n, and^ are the ratios of the length, breadth, and thickness respectively. Rule. — Divide the volume of the faralleioflfeddn hy the pro- duct of the ratios of the dimensions, and extract the the cube root cf the quotient. This gtves tht. G. C. D. of the three dimensions. Multiply the ratios of the dimensions by the G. C. D., and the results will be the dimensions respetively. I. What are the dimensions^of a palrallelopipedon whose length, breadth and thickness are in the ratios of 5). 4 and 3; and whose volume is 12960 cu. ft. ? By formula, /=yfl2960-5-(5><4x3)]5=30 ft.; i=t='J^[12960-i- (5x4x3)]4=24ft. ; and ^=y[12960-=-(5x4x3)]3— 18ft. 1. ^=the quotient obtained by dividing the length by the G. iC. D. of the three dimensions. 2. 4=the quotient obtained by dividing tlie breadth by the ^ G. C. D. of the thrpe dimensions. 3. 3=the quotient obtained by dividing the thickness by the G. C. D. of the three dimensions. 4. .-. 5xG. C. D.=the length, 5. 4xG. C. D.=the breadth, and II..-I 6. 3xG-C. D.=the thickness. 7. .-. (5XG. C. D.)X(4XG.C.D.)X(3XG.C.D.)=60X (G. C. D.)*=the volume of the parallelopipedon. 8. .-. 60 (G. C. D.) 3=12960 cu. ft. 9. (G. CD.) 8=12960^60=216. 10. .-. G. C. D.=f'216=6, 11. .-. 5X(G. C D.)=5x6=30ft.=the length, 12. 4X(G. C. D.)=4X6=24 ft.=the breadth, and .13. 3X(G. C. D.)=3X6=18 ft.=the thickness. III. .'. 30 ft, 24 ft, and 18 ft. are the dimensions of the par- allelopipedon. PROBLEMS. 1. Find the surface of a rectangular solid whose length is 12 feet, breadth 5 feet 4 inches, height 5 feet 3 inches. 2. Find the cost of papering the four walls of a room whose length is 20 feet 6 inches, breadth 15 feet 6 inches, and height 11 feet 3 inches, at 8d. a square yard. MENSURATION. 303 3. A rectangular tank is 16 feet long, 8 feet wide, and 7 feet deep ; how many tons of water will it hold, a cubic foot of water weighing 1,000 oz.? 4. The surface of a rectangle is 1,000 sq. in.; if its length and breadth are respectively 1 ft. 3 in. and 1 ft 2 in., find its height. 5. The dimensions of a rectangular solid are proportional to 3, 4, and 5. If the whole surface contains 2,350 sq. in., find the length, breadth, and height. , /Tin/.— 2,350+[2(3X4)+2(3X5)4-2(4X5!]=25, the greatest common divisor of' the three dimensions.- ' ,< ' < 6. The'whole surface of a rectangular solid contains 1,224 square feet, and the four vertical faces together contain 744 square feet. If the height is 12 feet, find the length and breadth. '7. Find; the" surface and volume of a cube whose diagonal is 2 feet 6 iiieljes. ' Ans. \i}i sq. ft.; 3 cu. ft. 12 cu. in., nearly. 8._:- Find the'edge of a cubical block of lead weighing one ton, having giviifl that a cubic foot of lead weighs 709^ lbs. Ans. 17.60+ inches. 9i The edges of a rectangular block of granite are broportional to 2, 3, and 5, and its volume is 101 cu. ft. 432 cu. in. ; find its dimensions. Ans. 3 ft. ; 4 ft. 6 in. ; 7 ft. 6 in. 10. The diagonal of a rectangular solid is 29 iiches, and its volume is 4,032 cu. in. ; if the thickness is one foot, find the length and breadth. Ans. 21 in. and 16 in. 2. PRISMS. Prob. liXXVI. To And the convex surface of a prism. Formula. — 5==^X<'> in which ^ is the perimeter of the base and a the altitude. Rule. — Multiply the perimeter of the base by the altitude. I. What is the convex surface of the prism ABC — D, if the aJtitude AB is 12 feet, AB, 6 feet, AC, h feeit, and BC, 4 feet? By formula, 5=aX/=12X( 9+5+4 )=180 sq. ft. ( 1. 12 ft ^the altitude of the prism. II. ] 2. 6 ft. +5 ft. +4 ft.=15 ft.=the perimeter of the base. ( 3. .■. 12X15^180 sq. ft.=the convex surface of the prism. III. .-. The convex surface of the prism is 180 sq, ft. Remark. — If the entire surface is required; to the convex sur- face, add the area of the two bases. Formula. — 71=5+2^, where lA is the area of the base, 5' the convex surface, and Z'the total surface. Prob. liXXVII. To find the volume of a prism. For inula. — V=ay.A, where A is the area of the base," a, the altitude. 304 FINKEL'S SOLUTION BOOK. Kule. — Multiply the area of the Base by the altitu de. I. What is the volume of the triangular prism -(4^C — Z),_ whose length ^^ is '8 feet, and either of the equal sides AB^ BC,ir AC, 2ifeet? By formula, F=aX^=8x[(2i)'iV3]=12^b=21.6506 cu. ft- n. 8 ft.=the altitude AE. 2; 2^ ft.^the length of one of the equal sides of the base, as AB. Il.ls. (2|)2iV3=the area of the base ABC, by Prob. XI._ ••• 8x(2i)2iV3=12v'3=21.6506cu. ft. ^the volume of the prism. III. .-. 21.6506 cu. ft.=the volume of the prism. FIG. 36. PROBLEMS. 1. A right prism stands upon a triangular base, whose sides are 13, 14,, and 15 inches. If the height is 10 inches, find its volume and whole sur- face. ^ Ans. 840 cu. in. ; 4 sq. ft. 12 sq. in. , 2. The weight of a brass prism standing on a triangular base is 875 lbs.. If the sides of the base are 25 in., 24 in., and 7 in., find the height of the^ prism, supposing that 1 cu. ft. of brass weighs 8,000 oz. Ans. 3 ft. 3. Water flows at the rate of 30 yards per minute through a woodeni pipe whose cross-section is a square on a side of 4 inches. How long will » it take to fill a cubical cistern whose internal edge is 6 feet? Ans. 21f min. 4. Find the volume of a truncated prism ( that is the part of a prism i included between the base an;! a section made by a plane inclined to the base and cutting all fhe lateral edges ), whose base is a right triangle, base 3 feet, and altitude 4 feet, and the three lateral edges 3 feet, 4 feet, and 5 feet respectively. [Formula.— V—)^A(ei+e^-\-e^), where A is the area of the base and ?i, ^s. a°d e^ the lateral edges.] 5. A square right prism, whose base is 6 inches, intersects a right equi- lateral triangular prism, whose base is 10 inches, in such a way that the square prism is perpendicular to one of the faces of the triangular prism, and the axes of the prisms intersect each other, and a lateral side of the square prism is parallel to the edges of the triangular prism. What is the volume common to the prisms? Ans. 128V 3 cu. in. 3. THE CYLINDER. Prob. liXXVIII. To find the convex surface of a cylinder. Formula. — 5=«X C, in which «is^the altitude and C the circumference of the base. V Rule. — Multiply the circumference of: the base by the altitude. I. What is the convex surface of the rig'ht cylinder A GB — C,, whose altitude EF' is 20 feet and the diameter- of its, base AB is- 4 feet ? MENSURATION. 305 By fonnula, S=<rXC=2&x4«-=80«=251.32.736 sq. ft. 1; 20 ft.=the altitude BF. 2. 4 ft.=rtie diameter AB of the base. jj |3. 12.566368 ft.=4;r=4x3.141592=the I circumference of the base. 4 .-. 20X12.666368=251.32736 sq. ft= the convex surface of the cylinder. III. .■. The convex surface of the eylindsr is 251.32736 sq. ft. Remark. — If the entire surface is required; to the convex surface, add the area of the two bases. F0rm,ul€U— 7'==5+2^=2«aif-f 2wi2 ^ F)^ ST. Prob. LiXXIX. To find the solidity of a cylinder. WormitCa. — F==i-!X^j in iwhich A is the area of the base. Rule. — Multiply the area of the base by the altitude. I , What is the solidity of the cylinder AGB — C,, whose all*' tude FE is 8 feet and diameter AB of the base 2 feet? By formula, V^=ay.A- 25.132736 cu. ft. =8 X ( 1 ' 7r)=87r==8 X 3. 141592= II. :i. 8 ft.=the altitude, BF. 2. 2 ft.=the diameter, AB, of the base. 3. 3.141692 sq. ft.=7r7?2=;7rl2=area of the base. 4. .-. 8X3.141592=25.132786 cu. ft. III. .-. 25.132736 cu. ft. is the volume of the cylinder. I. What is the volume of a slab 4 inches thick sawed from a round log 24 inches in diameter and 10 feet long? 1. Let HGB—C be the slab, Fig. 37, 2. FC=FK=12 in., the radius of the log, 3. FE—1(^ ft., the length of the log, and ZC^4 in., the thickness of the slab. 4. Volume of HGB — C=area of base, HGB,X altitude, FE. 5. Area of if G5=area of IKC=a.rea of sector, IFKC, — area of triangle, IKF. 6. Area of sector, IFKC,=i FCXa^rcICK, Rule (1), II. <; Prob. XXVII. 306 FINKEL'S SOLUTION , BOOK. 7. Arc ICK= ^(8 times fchord I C — chord JK) — \{ 8v/Z.^ + LC ^ — I K) = l{^{FP - F U) + LC - . 2^FP — FL 2)_=i(8V(12^— 8V1- 4''^— 2Vl2^ — 8'') = ■ i(32v"6 — 8v6")=f(4V 6 — V 5), Rule {b), Prob. XXV. . _ _ ■ 8. .-..Areaof sector /T'^ = |[ /rXfMV 6-V 5)] = ' i[12X|(4V 6-V 5)]=16(4V 6-V 6) sq.jn. 9. :■: VoKim^f slab i¥56:-<r=120X16(4\/ 6- V5)= 1920(4V 6-V 0) cu. in. =1451. 52 cu. in., nearly. liL .,-. The volume of the slab=1451.52 cu. in. ;,, i: PROBLEMS. 1. How many cubic yards of earth must be removed in constructing a tunnel 100 yards long, whose section is a semi-circle with a radius of 10 feet? 2. Find the convex surface of a cylinder whose height is three times its diameter, and whose volume is 539 cubic inches. 3. The cylinder of a common pump is 6 inches in diameter ; what must be the beat of the piston if 8 beats are needed to raise 10 gallons? Arts. 12^ in. 4. A copper wire ^ inches in diameter is evenly wound about a cylin- der whose length is 6 inches and diameter 9 . 9 inches, so as to cover the convex surface. Find the length and weight of the wire, if 1 cu. in. of copper weighs 5.1 oz. Ans. 1,885 in., nearly; 75.5 oz. 5. A cubic inch of gold is drawn into a wire 1,000 yards long. Find the diameter of the wire. Ans. .006 in. 6. The whole surface of a cylindrical tube is 264 square inches ; if it^ length is 5 incites, and its external radius is 4 inches, find its thickness. (Use7r=^.] ^«.9. lin. 7. If the diameter of a well is 7 feet, and the water is 10 feet deep, how many gallons of water are there, reckoning 1% gallons to the cubic foot? ^».y. 288.6345 gal. 4. CYLINDRIC UNGULAS. 1. A Cylindric Ungnla (Lat. Ungula, a claw, or hoof) IS any portion of a cylinder "cut off by a plane. Prob. liXXX. To find the convex surface of a cyllndrle nng-ula, when the cutting- plane is parallel to the axis of the cylinder. , fv rdy _ y Formvla. — 5=a fds=:2a l / » „, .=2grsin i-=ax ^ Jo (r^—y^)* r arc of the base. Rule. — Multiply the arc of the base by the altitude. I. What i.* the surface of the cylindric ungula API — Q, whose altitude AD'x&ZI feet and height AT oi the arc of the base, 2 feet and cord Ploi the base 12 feet? MENSURATION. 307 By foimula, 5= ay,arc PAI=a2r sitT'^^^ ^aX^f" &i«~\~-) ^"X^V 2Ar >^" U^-^K 2AT )j = 32(^!±i')sin- |=640(g^j7r)=411.84 sq. ft., nearly. The arc corresponding to the sin -I is found from a table of 17731 natural sines and cosines to be (36° 52/:j'-7-360°) of 27r or ■ ^^ ;r. 1. 2ft.=the height .^4 T'of the arc PAL 2. 12 ft.= the length of the chord PI. 3. 12.87 ft.=2V6H^'x(l+ "•^ 60x6"+33x20 -^^-^-g*"°^^^- arc P^/, by Prob. XXV. 4. .. 32X12.87=411.84 sq. ft. = convex surface PAI^D. III. .-. The convex surface of the cylindric ^ ^ lungula PAI—Q is 411.84 sq. ft. Remark. — r is found, by Prob. XX, formula i?==(a2-|-c2)-^2a. Prob. liXXXI. To find the volume of a cylindric ungula, ■wliose cutting' plane is parallel to the axis. Formula.— V=2 P T^'"-^' f dy dx dz—2ay{r^—y^) J J Jo ^^aj'ff^^'dy dx-2ay{r^-y^ ) =2a£' ^{r^-y^ )dy^ 2ay{r^-y') =a\y(r^-^y'y+r-^ sm-^ ^-2y(r^-y'} \f== .c|r^sin"'- — y(r^ — y'') J , in which j)/ is half the chord of the base. In this formula I r^ sin"' - 1 is the area of the sector ' APBIA , and y(r'^—y'^ )* is the area of the triangle PEI formed foy joining the center E with P and /. ^vASi.—rMultiply the area of the base bji the altitude. I. What is the volume of the cylindric ungula PIA — Z>, if ^/ is 12 feet,^r 2 feet, and altitude .^Z>40"feet? FI6 38. 308 FINKEL'S SOLUTION BOOK By formula, V=aA=a\r' iin »-?^-^(r>— y*)*| =40jl0»sm-'t -6(10^-6'' )^t=4C30sin-' f— 1920=4(»o(||jg?;r)— 192f>= 2574.0] 6- 1920=654.016 cu. ft. (1. 40 ft.=the altitude ^Z). ^ 2. 2 ft.=the height AToi the arc of the base. 3. 12 ft.=the chord rio( the base. II.. 4, 16i sq. ft.i=^^+f of(2Xl2)=the area ot the base^ by rule, Prob. XXYHI. is. .-. 40Xl6^=653icu. ft.=tbe volume oi the cyimdrical trngmla PIA — Z>. III. .*. SSS'J cu. ft.=tbe vohnne of the cylindrical ungula. Memark. — A nearer result would have bee» obtained by findingf- the length of the arc PAJ and multiplying it by half the radius. This would give the area of the sector IE PA. From the area of the sector subtract the area of the triangle PIE formed by join- ing P .nd /with E, and the remainder would be the area of the segment PIA. Prob. LXXXII. To find the convex surface of a cylindrlc ang'ula, when the plane passes obliquely througrli the op- posite sides of the cylinder. ForvntHam — S=z^{a-\-'a')^nr, where a and a" are the least and greatest lengths of the ungula and 2 tt/- the circumfer- ence of the base of the cylinder. Rule. — Multiply the circumferenve of the base by half the sum of the greatest and least lengths of the ungula. I. What is the convex surface of the cylindric ungula A KB A —NM, ifAJV is 8 feet, BM 12 feet and the radius BE of the base 3 feet? By formula, S=^{a-\-a')27rr=:7r(a-\-a') r=7t(S-i-12) X.&= 188.49552 sq. ft. 1. 8 ft.=the least length A^ of the ungula, and 2. 12 ft.=the greatest length BM. ^j i3. 10 ft.=J(8ft.+12 ft.)=half the sum of the least ana " "^ greatest lengths. 4. 18.849552 ft.=6»r=the circumfereaic of the base. 5. .-. 10X18.849552=188.49652 sq, ft.=the convex surface. MENSURATION. 309 III. /, 188.49552 sq. ft.=the convex surface of the ungula. Pr«b. liXXXIII. To flud tlie volume of a cylindric ungula, "nrben the plane passes obliquely tbroug'b tbe opposite sides of tbe cylinder. Formula.— V=l(a^a')7rr'^=^7r(a-{-a()r^. Rule. — Multiply the area of the base, by half the least and greatest lengths of the ungula. I. What is the 'volume ot a cylindric ungula whose least length is; 7 feet, greatest length 11 feet, and the radius of the base'a/feet? ' .> > . By,;fohrivls.;r=^|(.«+a')?rr2=.^(7+ll)^^ 1. 7 ft.=the least length of the ungula, and 2. 11 ft.=the greatest length. 3. 9 ft.=|(7 ft.+ll ft.)=half the length of the least and II. <J greatest lengths. 4. 12.566368 sq. tt.=;r22=the area of the base. 5. .-.9X12.566368=113.097312 cu. ft.=the. volume ot the ungula. III. .-. The .volume of the ungula is 113.097312 cu. ft. Prob. LiXXXIV. To find the convex surface of a cylindric iwng-ula, when the plane passes through the base and one of its sides. rdx *Formula.-S=2j^ -(b-.)ds=.lj^ -(^-_.)-^^== ■=2ry I , dx=2rj\ 6 vers -+V2rx — ^2 —r vers '- , I'Jo >j2rx—x^ '>'- ' rJo ^2r-V »Jlrx—x'^—{r—b)wers''^ -~| ='2r~Vsl2rb—b^— z {r — ^)vers""'- j. Rule. — Multiply the sine of half the arc of the base by the diameter of the cylinder, and from the product subtract the prod- rict of the arc and cosine; this difference multiplied by the quo- tient of the height divided by the versed sine •will be the convex surface. J. What is the convex surface of the cylindric ungula A CB— 310 FINKEL'S SOLUTION BOOK E/_ / r^u^ M):jfT. Z>, whose altitude^/? is 28 feet, height -ff J/ of arc of base 4- feet and chord AC 16 feet? By formula, S=2rJ^^/2ri—6^— (r— 3)vers-0, =2xlOX J [V2 X 10x4-42— ( 10— 4) vers-^A^ , = 140[8-6ve.s-^|]=140[8-3^2.], =140[8— 5.5638]=341-068 sq. ft. FI6. 39. 1. 28 ft.=the altitude BZ>. 2. 4 ft.=the height BM of the arc of the base. 3. 16 ft.=the chord ^ C of the arc of the base. 4. 8 ft.=the sine Cil/ of the arc. 5. 10 ft.=(8'* +42 )-=-(2x4)=tbe radius OC=OB of the- base, by Prob. XX, formula Ji={a^-\-c^ )-7-2a. 6. 6 ft.=10 ft— 4 ft.==cosine O^ of the arc. 7. 160sq. ft.=20x8=sine multiplied by the diameter of the base. 8. 18.5438 ft.=2V§^<l+^?gJ^)=the ar. CBA, by formula of Prob. XXV. 9. .-. 111.2628 sq.ft.=6Xl8.5438=^the arc multiplied by the cosine OM. 10. 160 sq. ft.— 111.2628 sq. ft.=48.7372sq.ft.=the differ- ence. 11. .-. 341.1604 sq. ft.=(28-i-4)x48.7372sq. ft.=the con- vex surface. III. .-. The convex surface is 341.1604 sq. ft. nearly. II.J Note. — The difference in the two answers is caused by the- lengtlioftteB" arc CBA, in the solution, only being a near approximation.- * Demonstration. — In the figure, let BK=x, BM==^6, BD=a,. and the angle BMD=B. Then MK=b—x, and IK=^FL=MK tan e=(i— x)tan^. But tan^=^=^=|. .-. FL^=^{h~x}. Now if we take an elemennt of the arc LBH, and from it draw a line parallel to FJL, we will have an element of the sur- MENSURATION. 311 face LBHBGF. This will be a rectangle whose length is FL=y{b — *■) and width an element of the zxc LBH. An ele- ment of the arc is ds=J^(dx'^-\-dy'^). Let HK==y. Then_)'2= irx — x'^, by a property of the circle, from which we find dy= - dx. .■. ds-=- — .•. The area of the element of >illrx — x^ Sirx — x"^ a ^'^■'^ the surface is -r(b — x)j „ =^> and the whole surface of ABC — ■ 0^ 'S2rx — x'' D is 5=2 ri^b-x)^l£L=, =24/'V-.)-i^= Jo o mrx—x^ oJo sl2rx—x^ 2»-| Vj2rx~x'^—{ r—b ) vers-'-l , =| ["2^2^(5— iS^— 2(r— d)r vers-' ^"1 . Q. E. D. Prob, liXXXV. To And the volume of a cylindric ungula, wlien the cutting-plane passes through the base and one of its sides. For'mvla.— V= j {b—x)dA=^j {b—x)2^2rx—x^dxt =2~^(2rx—x^ )^—{r—b)f '>l{2rx—x^)dx~^=2~^2rx—x^)^ +^^'^~^\{r—x)V2rx—x''-\-;^r'' sin"'^^]+c] . When x=0, F=0. .-. C=—^nr^r—b). .: F=?[f(2/-^— i^)'— (r— d) \^7tr^— (r—b)V2rb—b^—r-' sin"'.^ | 1. Kule. — From f of the cube of half the chord of the base, sub- tract the product of the Area of the base and the difference of the radius of trie base and the height of the arc of the base; this dif- ference multiplied by the quotient of the altitude of the ungula by the height {versed sine^ of the arc of the base, will give the vol- ume. I. What is the volume of a cylindric ungula, whose altitude BD is 8 feet, chord ^ C of base 6 feet, and height BM of arc of base 1 foot? By formula, V=- \ U2rb—b^f—(r—b)l'i7rr^— 312 'FINKEL'S SOLUTION BOOK.' (^r-b)V{2rb—b^)^'' sin-' ^^ j =8 \ |(2x5Xl— 1)*— (5— l)£ir52_4^2x5Xl— 1— ^^ sin-' ^J \^=A ^8— 4ri;r25— 12-^25 sin-'ll [ =528+800 sin-^ |— 200;r= 13.20394 cu. ft. 1. 8 ft=tfae altitude ^Z>. ' 2. 1 ft.=the altitude BM oithe arc ^^C of the base. 3. 6 ft=the chord ^ C of the base. 4. 18 cu. ft.= I of 33=1 of the cube of the sine of half the arc of the base. 1^ 11../ 5. 4yi^ sq. ft.^^-—-^-)-f of 6 Xl^area of the base, by form- ula, (b), Prob. XXVIII 6. 16^ cu. ft.=4 X 4yV=the area of the base X OM, the cosine of the arc CHB. 7. .-., 8,(18 cu. ft.— 16i',CH..ft.)=13^ cu. ft.=the volume of the cylindric ungula A CB — D. III. .-. The volume of the cylindric ungula ACB — Z> is 13^ cu. ft., nearly. Prob. LiXXXVI. To find, tlie convex surface of the frustum of a cylindric ung-ula. Formula.— S=-X^rV'irb—b''—fl{r—b)r vers-rf ~| — "i-V^lr\/2rb'—V—1{r—b')r vers"'- 1 . Kule. — (1) Conceive the section' 'to he continued, till it meets the side of the cylinder produced- then say, as the difference of the heights of the arcs of the two ends of the ungula, is to the height of the arc of the less end, so is the height of the cylinder to the part of the side produced. t (2) Find the surface of each of the ungulas, thus formed, by Prob. LXXXIV., and their difference ivill be the convex sur- face of the frustum of the cylindric ungula. Prob. liXXXVII. To find tbe volume of a frustum of a cylindric ungula. Formula.— V=-Jlj{2rb—b'')^—{r—b) \ i^rr^— (^r—l,)V2rb—b^—r'' sin-' ~ | ~\—±^^(2r¥—¥p'— MENSURATION. »I3 ir—V) \ i7rf^—(r—¥)V(2ry—y^)—r r—y n Rule. — Find the volume of the ungula -whose base is the the upper base of the frustutn and altitude that as found by (1) of the ■ last rule. Also the volume of the ungula whose base is the lower base of the frustum and altitude the sum of the less ungula and altitude of the frustum. T'heir difference will be the volume of ■ the frustum. 5. PYRAMID AND CONE. Prob. LXXXVIII. To find the convex surface of a right cone. Formula. — 5= Cxi/«=27rr x ^V a'^^r'^ , where Cis the circumference, h the slant height, r the radius of the base, and a the altitude. Rule. — Alultifly the circum-ference of the base by the slant height and take half the product. Or, if the altitude and radius ■of the base are given, multiply the circumference of the base ly the ■ Square root of the sutn of the. squares of the radius and altitude, >and take half the product. I. What is the convex surface of a right cone whose altitude is 8 inches and the radius of whose base is 6 inches?. By formula, 5'=2;rrXiV^«2+r2=27r6XiV82-|-62=160«= 188.495559 sq. in. 1. 6 in.=the radius ylZ>ofthe base, and 2. 8 in.=the altitude CD. 3. 10 in.=V'&2+6-^=the slant height CA. 11.14. 37.6991118 in.==2n-r=12x 3.14159265=the circumfer- ence of the base. 4. .-. 188.495559 sq. in= i(10x37.6991118)=the con- vex surface of the cone. III. .-. The convex surface of the cone is 188.495559 sq. in. Prob. liXXXIX. To find the convex surface of a pyramid. Formula.— S=^pXh, in which / is the perimeter of the base and h the slant height. . Rule. — Multiply iJie perimeter of the base by the slant height and take half the product. FIG 40. 314 FINKEL'S SOLUTION BOOK. I. What is'the convex surface of a pentagonal pyramid whose- Slant height is 8 inches and one side of the base 3 inches? By formula, 5==|/ X/«=i( 3+3+3+3+3 ) X 8=60 sq. in. (1. 8 in.=the slant height. 2. 3 in.=the length of one side of the base. 3. 5X3 in.^15 in.^the perimeter of the base. 4. .-. ^(15X8)=60 sq. in,=the convex surface of the pyra- mid. III. .-. The convex surface of the pyramid is 60 sq. in. Remark. — If the entire surface of a pyramid or cone is required^ to the convex surface add the area of the base. Formula. — T=S-\-A , where A is the area of the base- and 5" the convex surface. Prob. XC. To find the volume of a pyramid or a cone. Formula. — V=^aA=^ay,7ir^ , where a is the altitude; and A=nr^ the area of the base. Rule. — Multiply the area of the base by the altitude and take one-third of the product. I. What is the volume of a cone whose altitude CD is 1& inches and the radius ^Z? ot the base 3 inches? By formula, F=iaX'rr2=iXl8X »r3«— 54x3.14159265= 169.646 cu. in. ' (1. 18 in.=the altitude CD, and 2. 3 in.=the radius AI^. ll.\Z. 28.27433385 sq. \xi.=nr^=Z'^n=the area of the base. 4. .-. 169.6460031 cu. in.=^a^=^xl8x3'';r=the volume of the cone. III. .-. The volume of the cone is 169.6460031 cu. in. Prob. XCI. To flnd'the convex surface of a frustum of a. cone. F ormula.— S= ^{ C-\-a)k=\{2nr-\-27ir')h= 'n{r-\-r')^a^-\-[r- — r')^, in which C is the circumference of the lower base, C the circumference of the , upper base, and h,==. Va^+(>-— r')2,the slant height. - Kule. — Multiply half the sum of the circumferences'of the- two bases by the slant height. 1. What is the convex surface of the frustum of a cone whose altitude is 4 feet, radiusof the lower base 4 feet, and the radius of- the upper base 1 foot? MENSURATION. ' 315- =25».=78.639816 sq. ft. 1. 4 ft.= the altitude OB, 4 ft.^the radius AjB of the lower base, and 1 ft.^the radius DO oi the upper base. 4. 3 it= AE—PE{=D O)=r—r'. 5. 5 it. =sl{I>P^-\-A P^)= ■_ =AD, the slant height. 8;r:^the circumference A GBH oi the lower base. - " ■ "' 7. 2jr==^the circumference DIC of the upper base. 8. 5«'=^(8i'-|-2T)=half the sum of the circumferences. 9. .-. 5x5«=257r=78.539816sq. ft.=the convex surface of the frustum. III. .'. The convex surface of the frustum is 78.539816 sq. ft. Remark. — If the entire surface of the frustum is required, tO' the convex surface add the area of the two bases. Formula.— T=S-\-A-\-A '=w( ^+r' )V'a2^(r— r')* + Prob. XCII. To find the convex surface of the frustum of a pyramid. Formula.— S=^i(p-\-p' )A. Bule< — Multiply half the sum of the perimeters, of the two bases hy the slant height. I. What is the convex surface of the frustum of a pentagonal pyramid, if each side of the lower base is 5 feet, each side of the upper base 1 foot, and the altitude of the frustum 10 feet? Before we can apply the formula, we must find the slant height. Produce ^O, till 0^=0^. Divide Opinio extreme and mean ratio at //; UtAVfEH. Then KO -.OH-.-.OH-.KH. .-. OH^=KOxKff=KOx{K0—OH)=K0'^—KOyi OH; whence OIi^-\-KOX OH=KO'^. Completing the square of this equation, OH^-\-KOxOH-\-\KO'^=\KO^ , from which OII{= EH=EK)=.\KO{^l—\). EE^=BK^—KE^={^X0{*11— l)y—mKO—QH)y,=lK0^4E—l)-'-^\ j KO—\KO{^l— 1)[ ]] =iArO='(V5— 1)2— i^02(3— V5)2 =i^<)2r(V5— 1)2— «16 FINKEL'S SOLUTION BOOK. ■10— 2V^ ^•3_v^)2^=^^Os 1^-^-^— 2^=^i^A'C>2(io— 2V6). But BF= \EA=^s. .-. i*«==Jig-^02(10-,2VS), and s=\KO^\()~.%'^. ^5 .-. KO= , where J is a side of the lower base, = VlO— 2^5 , KO may be considered the radius ./? of a ciircuni- Vio— 2V5 ■ scribed circle of the lower base. In like manner, the radius r of the circumscribed circle of the upper base may be found to be Is' . . 2 _. where s' is a side of the upper base,= VlO— 2V5' ' VlQ—Wl OF, the apothem of the lower base, ^W {EO'^ — EF'^)^= + L^Nr^"'" -")T ( =|V650+10v'5=the slant hei-ht. V5 — V5^ Byforhiula, 5'=|(25+5)|V650+10\/5=6V650+10V5= 155.5795 sq. ft. ri. 10 ft.=the altijtud'e oC 5 ft. = EA, one of the equal sides of the lower base. 3. 1 ft.=efl?, one oif the e"qual sides of the upper base. 4. |V650+10v'5=/^> the ?]ant height. 5. 5x5 ft.^25 ft.=the perimeter ' of the lower base. 6. 5X1 ft.=5 ft.=the perimeter of the V, FIG. 42. upper base. 7. .■. ■^(25+5)|V656+10"\75 = 155.5795 sq.ft.=the conreK surface. III. . .■. The convex surface of the frustum is 155.5795 sq. ft. Prob. XCIII. To And the volume of a frustum of as>yiar mid or a cone. II. MENSURATION. 3ir JB'ormulu.—{a)V=ia{A-\-VAA'+-A'), in wbi^b A is the area of the lower base, A' the area of the upper base and VAA'the area of the mean base. When we ha ve a frustum of acone, (*> V=^a{A-{4AA'-{-A')==^a{ 7ri?2+V( rcB'' X w''' )+ Rule. — (1) J^ind the area of the mean base hy multiplying' the. area of the upper and lower bases together amd extracting th^ sqteare root of the product. (2) Add the upper, lower, and mean bases together and multi- ply the sum. by -J the altitude. I. What is the solidity of a frustum of a cone whose altitude IB 8 feet, the radius of the lower base 2 feet, and the radius of the upper base 1 foot ? By formula (3), V=i^ca{E^-{-Rr-\-r'-)=^^t^4,^^■Jp:.^)== ^X56jr=58.6433cu. ft. -fl. 8 ft.=the altitude. 2. 2 ft.^the radius of the lowCT base. 3. 1 ft.=the radius of the upper base. 4. 4 s'=the area of the lower base. II-' 5. jr=the area of the upper base. ;6. 2?r=V4« X TT^the area of the mean base. 7. 4 *-)-ff4-2ff=7*^the sum of the areas of th« three bases. 8. .-. Ti^x8X77r=58.64S3 cu. ft.=the solidity of the frustum. III. /. Th3 solidity of the frustum is 58.6433 cu. ft PROBLEMS.. 1. Find the entire surface of a riglut pytaaidi, trf whicl* tlhe height ia 2 feet and the base a square on a side of 1 ft. 8 im. Ans. 10 sq. ft. 2. Find the convex surface of a right pyramid 1 fopt higln, standing on a rectangular base whose length is 5 feet 10 inches and breadith 10 inches. Ans. 8 sq. ft. 128 sq. in. 3. Find the con-vex surface of a right pyramid having the same base and height as a cube whose edge is 10 inches. Ans. 223.6 sq. in. 4. Find the weight of a granite pyramid 9 feet high, standing on a square base whose side is 3 feet 4 inches, 1 cubic foot of granite weighing 165 lbs. Ans. 2 tons, 9 cwt. 12 lbs. 5. Find the height of a pyramid of which the volume is 623.52 cu. in., and the base a regular hexagon on a side ©f 1 foot.\ Ans. 5 inches. 6. The volume of a regular oetahedron is 4(?i.4tl cubic, feet; find th© length of each edge. Ans. 10 feet. 7. Find the surface of a regular tetrahedron, if the perpendicular from one vertex to the opposite face is 5 inches.. 8. A conical vessel is 5 inches in diameter and 6 imches deep. To what depth will a ball 4 inches in diameter sink in the vessel ? 9. The ends of the frustum of a pyramid are squaires whose sides are 20 inches and 4 inches, respeetively.. If its altitude is 15 inches, what ia its convex surface ? A^S^ 110 sq. in. 518 FINKEL'S SOLUTION BOOK. 10. What is the volume of a frustum, of a pyramid \yhose upper base is 4 inches square, lower -base 28, inches, and the length of the slant edges 15 inches ? 11. The, volume of a frustum of a cone is 407 cubic inches and its thick- ness is 10>^ inches ? If the diameter of one end is 8 inches, find the diam- eter of the other end. [wzzfi^.] .^» J. 6 inches. 12. A frustum of a pyramid' has for its bases squares whose sides are respectively 0.6 m. and 0.5 m. ; the altitude of the frustrum is 4 m. End the volume. 13. The upper and lower bases of a frustum are squares whose sides are 8 inches and 28 inches, respectively, and the edges of the frustum are each 15 inches., Find the volume of the frustum; also the lat-eral .surface. 14. A conical tent of slant height 9 feet covers a circular area 10 feet in diameter. Find the area of the canvas. 15. The radius of the base of a cone is, to the altitude, as 5:12 and its lateral surfa!ce is 420 sq. ft. Find the radius and altitude. j„^ i Radius=10 feet, and ^''•'- Ultitude=24 feet. 16. The nuniber expressing the volume of a circular cone is 7 times the number expressing its surface. Find the radius and altitude if they are in the ratio of 3 to 4. 17. A right circular cone, radius of base 3 feet and altitude 8 feet, is divided into two equal parts by a plane parallel to the b^se. How far does this plane cut the altitude from the base? 18. Find the dimensions of a right circular cylinder, Jf as large as a similar cylinder, whose height is 20 feet and diameter 10 ft. Aiis. Height, 5^~60 ft.; diameter of base, 2i^ 60 ft. 19. Ho w'miny square inches of tin will be required to make a funnel, the diameters of whose top and bottom are to be 28 inches and 14 inches, respectively, and height 24 inches ? Ans. 525 sq. in. 6. CONICAL UNGULAS. 1. A Conical Ungula (Lat. ungula, a claw, hoof, from unguis, a nail, claw, hoof) is a section or part of a cone cut off by a plane oblique to the base and contained between this plane and the base. Prob. XCIV. To flndtbe surface of a conical ung-ula. Formula. — 5=/ s*^ dx"^ -\-dy^=^ 2x cos ' I ^ ~~R~ — ""^ J '^''' where a is the altitude of the ungula, i? the radius of the base, rthe radius of the upper base of the frustum from which the ungula is cut, i the distance the cutting plane cuts the base from the opposite extremity of the base, and x the radius of a section parallel to the base and at a distance k — y from the base. MENSURATION. 31,9 Prob. XCV. To And the volume of a conical ungnla. Formula. — J^= / Ady= /■^C ^ ^(<lR—t)r—{RJ^r-r-t)x, 1 r._„ ,, -ff— n _( /?_)_r_/ ) x1 ^ f— ( 2i?— if ) V2 4-2?-(2i?— /) (i?+r— 0»— {2/?— /;)(2?-— ^fjjcH c/x, -where the letters represent the same value as in the preceding problem and dy=f ^= jdx, »mcej^—~ -. Prob. XCVI. To And the convex surface of a conical un- ^ula, when the cutting: plane passes throug-h the opposite ■extremities of the ends of the frustum. I Formula. — 5^= This formula is obtained by putting i =0, in the formula of Prob. XCIV. , and integrating the result. For, in this problem, the cutting plane ^i^C-A' passes through the opposite point ^, and therefore the distance from A to the cutting plane is 0. . i^O. FtG- 43. Kule. — Multiply half the sum of the radii of the bases by the square root of their product and subtract the result from the square of the radius of the lower base. Multiply this difference by n tirkes the slant height and divide the result thus obtained by the difference of the radii of the bases. Prob. XCVII. To find the volume of a conical ungnla, when the cutting plane- passes through the opposite extremi- ties of the ends of the frustum. Formula. 3( Ri This formula is obtairjed by putting t==0, in the formula of Prob. XCV., and integrating the result. 'RnlG. ^Multiply the difference of the square roots of the cubes of the radii of the bq-ses by the square root of the cube of the radius of the lower b'ase aiid this product by \it times the altitude. 320 FINKEL'S SOLUTION BOOK. Divide this lasi frodiicct by the Mfference lyffRe rad'vv ef the two- bases and the quotierit 'will be the volume of the ungula. I. A cup in the form of a frustum of a cone is 7 in. in diame- ter at the top, 4 in. at the bottom, and 6 in. deep. If, when full, of water, it is tippifid just so that the raised edge of the bottom;, is visible; what is ^he volume of the W^ater poured out? By formula, F— J^-J-^(^i?^_r?^=^;r(49— 8V7)= 102.016989 cu. in. Remark. — Fig. 43 inverted represents the form of the cup and'. APB Q — C the "quantity of water poured out, C being the tiTfpedi edge of the bottom. I. A tank is 6 feet in diameter at the top^ 8 feet at the bot- tom, and 12 feet deep. A plane passes from the top on one sider to the bottom on the other side : into what segments does it: divide the tank? By formula, F=3^(i.l-.i)=3^(8-3V3),= 327r(8-3V8)=281.87 cu ft. 1. 4 ft.= AZ,, the radius of the lower base. 2. 3 ft=DF','the radius of the upper base, and 3. 12 {t=F'L, the altitude. Then ' 4. J[^^]f ( Vr^-VP )=32 ^(8-3V3)=281.87 3(4- S) I \ t sq. ft.=the volume. III. .-. The volume is 281.87 cu. ft. Prob. XCVIII. To find the convex surface ofa conical un— g-ula, when the cutting plane FCE makes an ang-le CIB less- than the ang-le DAB, i. e. when AIt=t) is less thanDC(=2r). Formula.— S^j^^^a 2+(J?— r/ j »i?:2 cos'* Q:^^^ This formula is obtained by integrating the formula of Prob. XCIV, recollecting that the CO- efficient of x:^ is negative.. Prob. XCIX. To find the volume of a conical ungula, when, the cutting plane FCE makes an angle CIB less than the an- gle DAB. 1. e., when AI (=t) is less than. CD (=2r). II. MENSURATION. 321 r Formula.— V=^zz \ i^'<=°s *( • ^ I FIG. 44. ob, C. , ^o.^flnd the convex surface of a conical un- , -when'tiiecutting- plane JCE is parallel to the side AD, This formula is obtained by integrating tlie formula of Prob. XCV, recollecting, that the co- efficient of Af^ vi negative 1 Prob grula, vi , i. e., ■when^I(=t) is equal to DC(=f2r). +2{J?~2r)^{Ji—r)r—^{J?—r)'i/{Ji—r}r\ This formula is obtained by putting i=2r, in the formula of Prob. XCIV., and inte- grating the resulting equation. Poh. CI. To find the volume of a coni- cal ung-ula, when the cutting- plane FCK is parallel to the side DA, i. e., when AI(=t) is equal to CD (=2r). f;g 45. Formula.— V=ia | -^j^i?2cos-'(l^?±?.'')+ 2(Ii—2r)*J(J?—r)rl—\^r^(J?—r)? I . This formula is obtained by putting i=:2r, in the formula of Prob. XCV., and mtegrating the resulting equation. Prob. CII. To find the convex surface of aconcialungula, when the cuttingr plane FCE makes an angle CIB greater than the angle DAB, i. e., when AI (=t) is greater than DO* (=2r). • ' Formula.- 5'=-^— ^Va^-f (/?_r) 2 j H'^ cos-'(" ~^+^ ^ 322 FINKEL'S SOLUTION BOOK. '. .This formula is obtained by integrating 'the formula of Prob. XCIV., remembering that the coefficient of a:*^ ^^jch occurs in process, of integrating, is positive. Prob. Cill. To find the volume of a conical ungula, when the cutting plane FCE makes an angle CIB greater than the angle DAB, i. e., when AI(r=t) is greater than I>C(=3r). Formula.— V=^—- \ \R^coi-^y ~^ j— {R+r-t){R-r) {2R—t){t—2ry 'og[(^-.+(/-2.)J^)-.]. FIG. 46. This formula is obtained by integrating the formula of Prob. XCV., regarding the coefficient of x^ positive. XII. THE SPHERE. - JProb. CIV. To find the convex surface of a sphere. FortmUa. — 5=2x27rj/V^A4l^^=4;r^2_^2?2, where Z> is the diameter. Rule. — Multiply the square of the diameter by S.H1592. I. What is the surface of a sphere whose radius is 5 inch*? By formula, 5=4;ri?2=4;rx25=314.1592 sq. in. "1. 5 in.=the radius. 2. 25 sq. in.= the square of the radius. 3. .-. 4wX25 sq. in.=314.1592 sq. in.==the surface of the sphere. Ill, .". 314.1592 sq. in.^the surface of the sphere. Note. — Since ?ri?*is the area of a circle whose radius "is R, the area ■(4!r/?^) of a sphere is equal to four great circles of the sphere. The sur- face of a sphere is also equal to the convex surface of its circumscribing cylinder'. Prob. CV. To find the volume of a sphere, or a globe. Formula.— V=27ry^dx=^yrR^=^7e(iZ>y=^ tcD^. Rule. — Multiply the cube of the radius by %Tt {=4.188782); or multi-ply the cube of the diameter by ^Tt {=.5285987). I. What is the volume of a sphere whose diameter is 4 feet? II.< II.<^ MENSURATION. 323 By formula, F"=|7ri?8=|7r23=33.510256cu. ft; L 2 ft.=the radius. 8 cu. ft.==2^=the cube of the radius. 3. .-. 4.188782X8 cu.ft.=33.510256cu.ft.=the volume of the sphere. III. .-. 33.510256 cu. ft.=the volume of the sphere. Prob. CVI. To find the area of a zone; ^ Zone is the curved surface of a sphere included between tvyo parallel planes or cut off by one plane. Formula. — S=27iJia, in which a is the altitude of the segment of which the zone is the curved surface. i ' Rule. — Multiply the circumference of a great circle of the sphere by the altitude of the segment. I, What is the area of a zone whose altitude is 2 feet, on a isphere whose radius is 6 feet? By formula, 5=2jri?a=2 «6X 2=24 tt =75.39822 sq. ft. il. 6 ft.=the radius of the sphere. 2. 2 ft.=the altitude. 3. 127r=37.69911 ft.=the circumference of a great circle of . the sphete. «4. .-.2X37.69911=75.39822 sq. ft.=the area of the zone. III. .-. The area of the zone is 75.39822 sq. ft. Note. — This rule is applicable whether thp zone is the curved surface of the frustum of a sphere or the curved surface of a segment of a sphere. Prob. CVII. To find the volume of the seg-ment of a sphei-e. Formula, — V=\7ta{Zrlr\ra'^) where?-; is the radius of -the base of the segment. Rule. — To three times the square of the radius of the base, add the sqicare of the altitude and multiply the sum by ^71:^.5236987 times the altitude. I. What is the volume of a segment whose altitude is 2 inches and the radius of the base 8 inches ? By formula, F=^7ra(3>'J+fl!2)=^;rX2(3x644-4)=205.2406 cu. in. 1. 8 in.=the radius of the base. 2. 2 in.=th''e altitude of the segment. 3. 192 sq. in =:,3x8^=three times the square of the radius. II. J 4. 4 sq.-in.^the square of the altitude. 5. 196 sq. in.=192 sq. in. -(-4 sq. in. ^three times the square of the radius plus the square of the altitude. 6. |;rX2 X 196=205.2406 cu.in.=the volume of the segment. 324 FINKEL'S SOLUTION BOOK. III. .•. 205.2406 cu. in. =the volume of the segment. ' Note. — From the formula K=Jwa(3>'i4-«')i we have V^^irar ^-\-^ira^ But ^''<'ari is the volume of a cylinder whose radius isr^, and altitude ^a, and Jjra* is the volume of a sphere whose diameter is a, ..'..The volume- of a. segment of a sphere is equal to a cylinder whose base is the base of the seg- ment and altitude half the altitude of the segment, plus a sphere whose diameter is the altitude of the segment. Prob. CVIII. To find the volume of a frustum of a. sphere ,or the portion included between two parallel planes. Formula.— V=irra[B{fl-^rl)-\'a^\=ia(nrl-{-7trl)-^ ^Tta^ *, in which ri is the radius of the lower base, ■ r^ the ra- dius of the upper base. Rule> — To three times the sum of the squared radii of the two- ends, add the square j)f the altitude; multiply this sum by .5235987 times the altitude. ' , , ' , I. What is the volume of the frustumof a sphere, the radius of whose upper base is 2 feet and lower base 3 feet and altitude -^ foot? By formula, V=^7[a[3(rl+rl)+a']=ik Xi[S(^+i)+i']= 8.03839 cu. ft. 1. 3 ft.=|:the radius of theloywer base. 2. 2 ft.=the radius of the upper base. •■ 3. 39 sq. ft.=:3(3'-|-2^)=three times the sum of the squares- Il.l of the radii of the two bases. ' , 4. \ sq. ft.=the square of the altitude. 5. .;. ^TT XiX39i=8.03839 cu. ft.=the volume of the. frustum. - III. .-. 8.0383'9 cii. rt.=the volume of the frustum.' Prob. CIX. To find the volume of spherical sec^tor. A Spherical Sector is the volume generated by any sector of a semi-circle which is revolved about its diameter. Formula. — l'=^naJi^, where a is the altitude of the- zone of the sector. Rule. — Multiply its zone by one-third the radius. *NoTE. — ^a(7r/-j-)-7r>-j)i=the volume of two cylinders whose bases are- the upper and lower bases of the segment and whose altitude is hall the alti- tude ol the segment. jTrn' is the volume of a sphere whose diameter, is the- altitude of the segment. Hence the volume of a segment of a sphere of two- bases is equivalent to the volume of two cylinders whose bases are the up- per and lower bases respectively of the segment and whose common altitude- is the altitude of the segment, plus the volume of a sphere whose diameter is the altitude of the segment. For a demonstration of this and the preceding formula, s?e WentwortKs- Plane and Solid Geometry, Bk. IX„ Prob. XXXII. FIG. 47. MENSURATION. 326 I. What is the volume of a spherical sector the altitude of whose zone is 2 meters and the radius of the sphere 6 meters ? By formula, V=\ 7taR^=ln X 2 X 6^= .150.7964m3. 1. 2m.=the altitude BD of the zene gener- -ated by the arc JSF when the semicir- cle is revolved about AB. 2. Bm.^the radius SC of the sphere. . 3. 2 7r6m.=37. 699104 m ^the circumference I'^II.v -J. , ."ofa great circle of the Sphere. 4. 2;r6x2=75.398208 m2.=the area of .the ■^ ' zone generated by ^^, by Prob. CVl. '■ V' ; 5.':-.ix6X.75.3982O8i=il50.79641'6Hi»=the ' VbltiVn^ of the spherical sector. 3. ■ Ijjtp .- r. .- The'volufne cCf the sphe'ricarseetqr is '150:796416 m " . ■ J, Find the diameter of a sphere of which a sector contains ■789k'98 c-u. ff.,'when the "altitudt'dfits 'zdtfe -IS 6 feet. ' "' ' By formula, F=|7rar2=|7rx6X':^.--.|'i!rX6X/'^==. 7853.98 cu. ft., or 4?-2=2500 sq. ft, whence 2r=50 feet, the diameter of the sphere.' ■■''-^;; , ■ ' -' ■■.':' ' ;" •'■ 1. 6 ft.^the altitude of the zone. 2.; ,-. I^XBX ?'2==the yolutne of the Sector., : But 3. 7853.98 cu. ft.=the volume. 4. . . |7rX6X'-^=7853.98 cu. ft. . ,. . ' 5. r^=Q^5 sq,. ft, by dividing by 4*.'' ' ' 6. .: 2r^50 ft, the diameter of the sphere. ,111. .•.,Xhe diameter of the sphere is 50 feet. Prob. CX, To find tlie area of a lune. A. Lwrie is that portion of a' sphere comprised between two great semi-circles, ,.,,,;,. , .:.■•>- ' Fdr'muldi—S^=4:7tIi^(-^-^j=4t7rJ?'^u, where u is •the quotient of the angle of the lune divided by 360°- Rule*^-AfM^^«^^ Me surface of the sphere by the quotient oj ■the angle: of the lune divided byHQOR I. Given the radius of a sphere 10 incbes; find the area of a lune whose angle is 30°. By formula, S=47rif2 ?/=4x '? XIO^ X(30°-r-360° )= _^;rlG2=104-7197,sq. in.. .,, II. II 326 FINKEL'S SOLUTION BOOK. 1. 10 in.=the radius of the sphere. 2. 30°=the angle of the lune. 3. J^j=30°-r-360°=the quotient of the angle of the lune divided by 360°. 4. -4:;r 10^=400 ''■=1256.6368 sq. in.=the suiface of the sphere. 5. .;;. YVX1256.6368 sq. in.=104.7198 sq. in.=the area of the- lune. III. ..r. The area of the lune is 104.7198 sq. in. WentiaorW s New Plane and Solid Geometry, f. 87 1, Ex, 585. Ppob. CXI. To find the volume of a spherical ung^ula. A. Spherical TTtigulcu is a portion of a sphere bunded by » lune and two great semi-circles. Formula. — V=^7tR^u, where u is the same as in the last problem. ' Rule> — Multiply the area of the lune by one-third the radius/ or, tnultifly the volume of the sphere by the quotient of the angle of the lune divided by 360°. I. What is the volume of a spherical ungula the angle of whose lune is 20°, if the radius of the sphere is 3 feet? By formula, F==|«-^3«=4;r x3»X(20O-f.360°) = 6.283184 cu. ft. i. 3 ft.=the radius of the sphere. 2. 4w3''X(20°-4-360°)=6.283184 sq. ft.=the area of the 11.^ lune, by Prob CX 3. .-. iX3x6.283184=6.283184 cu. ft.=the volume of the ungula.' III. .-. 6.283184 cu. ft. is the volume of the Ungula. Prob. CXII. To find, the area of a spherical triangle. Foj-mM^a.— 5=2;ri?2 x(.^+^4-C— 180°').-^-360°, in' which A , £, and C are the angles of the spherical triangle. Rule» — Multiply the area of the hemisphere in -which the tri- angle is situated by the quotient oj the spherical excess {^the ex- cess of the sum of the spherical angles over 180°) divided by 360°. I. What is the area of a spherical triangle on a sphere whose diameter is 12, the angles of the triangle being 82°, 98°, aiid 100°? By formula, 5'==2 7r/?2x(^+^-f-C— 180°)-f-360°=2>r62x: (82°+98°+100°— 180°;-r-360°=2;r6''Xf\=62.83184^area. II. MENSURATION. 327 I •1. 6==the radius of the sphere. 2. 2* 6^=72 ;r=the area of the hemisphere. 3. (82°+98o+100°— 180°)=100°=the spherical excess., ' 4. 100°-T-360°=^\^=the quotient of the spherical excess divided by 360°. .5. .-. T\X72w^62.83184^the area of the spherical triangle. III. .-. The area of the spherical triangle is 62.83184. (Olney's Geometry and Trigonometry, Un. Md.,p.2S8,Ex. 8.) Prol). CXIII. To find the volume of a spherical pyramid. A Spherical Pyramid is the portion of a sphere bounded by a spherical polygon and the planes of its sides. Formula.— V=lnR^X(E-r-2,QQ°), where ^ is the spherical excess. Rul6< — Multiply the area of the base by one-third of the radius of the sphere I. The angles of a triangle, on a sphere whose radius is 9 feet, are 100°, 115°, and 120° ; find the area of the triangle and the volume of the corresponding spherical pyramid. By formula, F=|7ri?3 x(^H-360° )=|jrif3 X(^+^+C— 180 ° )-j-360 ° = f T 9 3 X ( 100 ° +115 ° +120 ° — 180 ° ) H- 360 ? = ^\ff93=657.377126 cu. ft. 1. 9 ft.=the radius of the sphe,re. 2. 2^9^==the area of the hemisphere in which the pyramid is situated. 3 (100°+115°+120°— 180°)=155°==the sperical ex- cess. II. s 4. •f^^l55°-T-36G°^the quotient of the spherical excess divided by 360°. 5. .-. f|^X27r9^=fiXT92=the area of the base of the pyra- mid. 6. .-. lX9XiiX27r92=657.377126cu. ft.=the volume of the pyramid. ' III. .-. The volume of the spherical pyramid is 657.377126 cu. ft. ( Van Amringe's Davies^ Geometry and Trigonometry, f. 278, Ex. 15. I. Find the area of a spherical hexagon whose angles are 96 ° , 110°, 128°, 136°, 140°, and 150°, if the cfrcumference of a great circle of the sphere is 10 inches. Formula.— S=2nR-' ^'^~''"~^}^^^°^ , where T is the sum of the angles of the polygon and n the number of sides. 328 FINKEL'S SOLUTION BOOK. By formula, S^2nJ^^X^-^^=^^^^^^=2nxQy X (96°+110°+128°+136°+140°+150°— (6— 2)X180°)-^ K(\ Kf) 360°=— X(760°— 720°)-T-360°=i— ;=1.7684 sq. in. tt It , 1. 5-i-'''=the radius of the sphere, since 2>rR=10 in.. 2. 760°=96°+110°+128°+136°+140''+150°=the , , , , sum of ifhe angles of the poljgpn. I 8.760°— (6—2 )>< 180° =40° =the spherical excess. 4. -^=40 °-^-360°=the quotient of th'e spherical excess di- ttJ videdby 860°. " : ' 5. Ink — I ^the area of the hemisphere' on which the polygon _is situated. 6. .-. iX2rr(^|^^=ix50-r-7r=1.7684 sq. in. :,, III. ■ , ,;. The area of the polygon is 1.7684 sq. iii. WentiuortKs Geometry, Revised Md., p. S74, Ex. 596. NoTE.—7-A' Spherical Degree may be defined as half of a lune whose measuring angle is 1°, the line of division being the arc of a great circle whose poles are the points of intersection of- the semicircles, bounding the lune. It, therfefore, follow^ thai: the surface of a sphere contains • 720 spljerical -degrees. It is easily proyed in Spherical Geometry that the surface of a spherical' triangle in spherical degrees equals the excess of the sum of its angles over a straight angle. Hence, if the excess of the sum of the anglps of a spherical triangle is, <?. g., 30°, it contains 30 spherical degrees and the surface of the triahgle ,ia j%. or ^ of the surface of the sphere. ^ PROBLEMS. 1. Find the ratio of the surface of a sphere to the surface: (i) of its circumscribed cylinder, (ii) of its circumscribed cube. ^ 2. A cube and a sphere have equal surfaces ; what is the ratio of their volumes? , .<4«.j. 72:100, nearly. 3. From a cubical block of .rubber the largest possible rubber ball is cut. What decimal of the original solid is cut away? 4. Suppose the earth to be a perfect sphere, 8,000 miles in diameter ; to what height would a person have to ascend in a balloon in order to see ir I one-fourth of its surface? [Formula. — h= „ , where r is the radius of the earth, and -— is the part of the earth's surface visible to the observer. If the part of the earti visible to the observer is -£-, or 1/-?-, «=-i-. 5. A paring an inch wide is cut from a smooth, round orange an inch and a half in diameter. What is its volume, if it is cut from the orange on a great circle of the orange ? Ans. \i^. MENSURATION. 329 6. What would be the volume of a paring cut from the earth on the •equator? .^«i. Jfa^, where a is the width of- the paring. Remark. — This is a remarkable fact, since the volume of the paring is Independent 'Of the radius of th% sphere. 7. If, when a sphere of cork floats in the water, the height of the sub- merged segment is }i of the radius, show that the weights of equal vol- umes of cork and water are as 3*:4*. Note. — The weight of a floatiijg body is equal to the weight of the liquid it displaces. 8. A vertical cylindrical vessel whose internal diameter is 4 feet, is coin- pletely filled with water. , If a metal sphere 25 inches in diameter is laid upon the rim of the vessel, find what weight of water will overflow. _ Ans. 699 lbs., nearly. 9. A conical wine-glass 5j^V 3 inches in diameter and 4 inches deep is filled with water. If a metal sphere 5^ inches in diameter is placed in the vessel, what fraction of the whole contents will overflow ? Ans. f . 10. Pour equal spheres are tangent, to each other. What is the radius of a sphere taiigent to each ? 11. To what depth will a sphere of ice, three feet", in diameteri sink in •water, the specific gravity of ice being f ? "■"12! Find the volume removed by>boring a 2>-inGh auger-hole thirpugh a '6-incto globe. .' I ■ ■-'■■'•:..•::■ ,i. ■.„ ! 13.* What is the volume removed by chiseling a, hole an inch square ■through an 8-inch globe ? --•^■., ■ Note. — This problem cannot be solved without the aid of the calculus. 'XIII.' 'SPHEROIDS. ; '■ 1. A Spheroid is a solid'fdfmed by revolving an ellipse ■about one of its axes as an axis pf revolution. '.' i ,1, THE PROtATE SPHEROID. ' .\ -■- ' f 1. The Prolate Spheroid is, the spheroid formed by re- volving an ellipse about its transverse, diameter as an axis of revolution. Prol). CXiV. To find the surface of a prolate spheroid Formulae.— {a) S==2J'2ny ds=2 r27iy~^i-{.^ dx=. *''/^!('^'-*'^')*'^^^2^^=+2v^^^°"'<''=2^^;^+~sin-.), where ifaS—b^ , . . . , s= ^ =the eccentricity of the ellipse which generates the. surface. (3) ^=4-^(l-2:3-2-X5-2:4t7-2-i8:9-^-) 330 FINKEL'S SOLUTION BOOK. 'R\\\e^.^Multi^ly the circumference of a circle whose radius- is the semi-conjugate diameter by the semi-conjugate diameter in- creased by the product of the arc whose sine is the eccentricity into- the quotient of the semi-transverse diameter divided by the eccen- tricity. I. Find the surface of a prolate spheroid whose transverse diameter is 10 feet and conjugate diameter 8 teet. a 5 By formula (a), S=%7th(^h-\ — sin"'^e)=27r4(4-] — sin~*e)^/ e e 4+(^5-r- g jsin-' \^ J=2>r4[4+ (5-=-4)sin-|] ==1^48+100 sin-|]=|;r[48H-100X ^^«-]= f 7448+100 X. 6435053] =236.3064 sq. ft. '1. 25.1527412=2w4==the circumference of a circle whose: r adiub is the semi-conjugate diameter of the ellipse. " 5 " 3. ''■-§ ft.=6 ft-=-|=the quotient of the semi-transverse diame- ter divided by the eccentricity. II. J 4. .6435053(=the arc(to the radius 1) whose sine is f, or the- eccentricity. 5. 5.8625442 ft.=2^ft. X.6435053=\« ft. X the arc whose- sine is \ 6. 9.3625442 ft.=4 ft.+5.3625442 ft.=semi - conjugate di- ameter increased by said product. 7. .-. 235.3064 sq. ft.=9.3625442x25.1327412=the surface. of the prolate spheroid. III. .-. The surface of the prolate spheroid is 235.3064 sq. ft.. Prob. CXV. To And the volume of a prolate spheroid. 32 /•« Formula. — V^Jny^dx=7r— j (a^ — x^ )dx= TT— I a^x — ^x^ I ^^Ttb^ a, in -which b is the semi-conjugate- diameter, and a the semi-transverse diameter. Rule. — Multiply the square of the semi-conjugate diameter by- the semi-transverse diatneter and this product by ^n. I. What is the volume of a prolate spheroid, whose semi-trans- verse diameter is 50 inches, and semi -conjugate diameter 30 inches^ - By formula, F=|;r^2a=f;r30=' X50=l8849'5.559 cu. in. MENSURATION. 331 1. 30 in.=the semi-conjugate diam<;ter, • 2. 50 in.=the semi-transverse diameter. 3. 900 sq. in=the square of the semi-conjug'ate diameter. II.<(4. 45000 cu. in.=50x900=the squareof the semi-conjugate diameter by the semi- transverse diameter. 5. .•.|7r45000=|x3.14159265x45000cu.in.=: 188495.559 cu. in.==the volume of the prolate spheroid. III. .•. The volume of the prolate spheroid is 188495.559 cu. in. 2. THE OBLATE SPHEROID. 1. An Oblate Spheroid is the spheroid formed by revolving an ellipse about its conjugate diameter as an axis of revolution! Prob. CXVI. To find the surfae of an oblate spheroid. fa I J ^2 Formulae. — {a) S=f27txds=!2 I 2rrx^l-\---^ dy== ,„.(,+-■., J g I). Prob. CXVII. To find the volume of an ohlate spheroid. Formula.— V=Jnx^dy=lC n'^^{b'^-^-^)dy=\na-^b. Rule. — Multiply the square of the semi-transverse diameter by the semi. conjugate diameter and this product by \ n- I. What is the volume of an oblate spheroid, whose tra.ns- verse diameter is 100 and conjugate diameter 60? By formula, K=|ffa33=fw502 X30=314159.266. 1. 30=^ of 60= the semi-conjugate diameter. < 2. 50=1^ of 100=the semi-transverse diameter. 3. 2500^50^=the square of the semi-transverse diameter. II./4. 75000=30 X2500=the square of the semi -transverse di- ameter multiplied by the semi-conjugate diameter. 5. .-. |7rX75000=314159.265= the volume of the oblate spheroid. i III. .-. The volume of the oblate spheroid is 314159.265. Note. — Since the volume of a prolate spheroid is ^irb^a. We may write |;r42«:=|(iTi2X2a). But Tri^xSfe is the volume of a cylinder the radius of whose base is b and altitude 2a. .'. The volume of a prolate spheroid is | of the circumscribed cylinder. In like manner, it may be shown that the vol- ume of an oblate spheroid is f of its circumscribed cylinder. The following is a general rule for finding the volume of a spheroid; Multiply the square.of the revolving axis by the fixed axis and this product by \n . ' FINKEL'S SOLUTION BOOK. Prob. CXVm. To find the TOlume of the middle frnstum vf a prolate spheroid, its leng'th, the middle diameter, and that of either of the ends bein^ given. Case I. When the ends are circular, or parallel to the revolving axis. Formula— V=^sfr{2D^r\^d^)lr where D is the middle diameter. CD, £^the diameter Z^/ of an end, and I the length- of the frustum. l&vi\Si.-^To twice the square of tjf.e Tniddle diameter q.dd the ■square of the-dfiameter. of eitjjter end and this sum, multiplied by^ ihe length of the frustum, and the product again by ^-^n, "will ^ive the solidity. " ' -• ' ' -^- ' -'■■■' _I. What is^the volume of the middle frustum HIGF oi 2l, prolate spheroid, if the middle diameter CD is 50 inches, and that of either of the ends HI or FG is 40 inches, and its. length •O^ 18 inches? ;_. ' '■'.-'■ h''-^' By formula, V==j^^7r{2p^-^d^)l=f^7[(2x50^+^0'')l^—, '31101.767265 cu.:in:.,:-: :.;.;..''•.• "' '"" '■-=-;>) 1. 50 in.=the middle diame- ■ ter CD. ■■■" ■ - - ■-' - 2. 40,in.=the diameter of,ei- ■ ther erid.as '.^/. ■ ' 3. 1.8, in.==the length. O^ of the frustum. 4. 5000 sq. in.=2 X 502=tw'ice II. < t ■^^■: the. square of themiddle: diameter, i ,■ ,. ; ' - . 5. 1600;sq>in.T:^402=Fthe .... sqaureofthe diameter of either end. ' ' ''' '• 6. 5000scj:'iA.^-ieOO'sti. in.'^660a''sq:iri. ' ' - ; 7. 18x6600=118800 cii'.. in. '■■ ■ ' .-. jLttX 118800 cu. in.=31101. 767265 eu. in.^the vol- ume. ' " " '""'■','.", ' ' I ' ' ^ m. .■! The volume of the frustum is 31101.767265 cu.in. Case II. \ When the endsare elliptical, or perpendicular to the revolving axis. ,. , Formula. — V==^^nX2Dd-^D'd')l, where D and d are the transverse and conjugate c5ia,meters of the middle section and D' and d'' the transverae and conjugate diaitieter of the ends and 2 the distance between the ends. Rule. — (1) Multiply tiuice the transverse' diameter of the tniddle section by its conjugate diameter, and to this product add II. MENSURATION., i 333; the -product of the transverse and conjugate- dia/ntetev- of either of the ends. (2) Multiply the sum, thus founds by the distance of the ends, or the height of the frustum, and the product again by -^^Tt and!, the result "will be the volume. I. What is the volume of the middle frustum of au oblate- spheroid, the diameter of the middle section being 100' inchesi and 60 inches; those of the end 60, iwclies and 36 inches ; and the length; 80 inches? By formula, F=T-V^(2Z>c^4.i?V*) /^xV'^C 2X100 X 60+60 X 36) 80=296566.44616 cu. in. 1. 100 in.=the transverse diameter ./^Cof the middle section^ 2. 60 in.=the conjugate diameter ms ofthe middle section^ 3.^12000 sq. in.=2xlOOX60^twice the product of the diameters of the middle sectioifl. • 4. 60 iri.=the transverse diameter AB of the end. 5. 36 in.=the conjugate diameter 2(nc) of the end. 6. 2160 sq. in.==the product of th© diameters of the end. 7. 14160 sq. in.=12p00 sq. m.+2160. sq. in. 80X14160=1132800 cu; in. = the. FIG. 49. product of said sum by the height ofthe frustum. •••'tV«X1132800 cu. in.=296566>44616 cu. in.=the vol- ume of the frustum. III. .-. The volume of the frustum is 29e566K.44«16 cu. in. Prob. CXlX. To find the volume of ai segiueht of a prolatet spberoid i Case Ik When, the base is parallel to the revotving axis. Formula.— V=^7rh^ T-^, (3Z?— 3^) » where .Sis the height of the segment, c^ the irevolving axis^ and i3 the fixed axis. Rule. — (1) Divide the squa/ire- of the K-evohiing axis by the square of the fixed axis, and midt.iply the quotient by the differ- ence betiueen three times thefn^d q,xts; and t-wice the heignt of the segment. (2) Multiply the prodiKtiy tht^s faj^ndy by the square of the heightofthe segment, anji thispro/iMct, b^, ^;f, «%d? the result mli be the volume bf the segment.. 334 FINKEL'S- SOLUTION BOOK. I. What is the volume of a segment of a prolate spheroid of ■which the fixed axis is 10 feet and the revolving axis 6 feet and the height of the segment Ifoot? By formula, K=^»'A2^-^V(3Z»— 2A)= ^'rX6-^(^) '(3X10-2X1)= II. ti-ansverse diameter FIG. SO. 5.277875652 cu. ft. (1. 10 ft.=the 2. 6 ft.^the conjugate diameter AS. g2 3. ^V=r7f5 =the square of the conju- gate diameter divided by the square of the transverse diameter. 4. 2Sft.=^XlO ft— 2X1 ft.= the difference between three times the transverse diameter and twrice the height of thesegment. 5. 5^^X28 ft.=10j^^ ft.=the product of said quotient by said difference. 6. 10/^Xl='=10A cu. ft. 7. .-. i^rXlO^ cu. ft.=:5.277865652 cu. ft.=the volume. III. .-. The volume of the segment is 5.277876652 cu. ft. Oise II. When the base is perpendicular to the revolving axis. Formula. — V=k'^Ji^\-T){^d — 2A), where d is the revolving axis, D the fixed axis, and h the height of the segment.^ Rule. — (1) Divide the fixed axis by the t^evolving axis, and multiply the quotient by the difference between three times the revolving axis and twice the heighf of the^segment, (2). Multiply the product, thus found, by the square of the height of the segment, and this product again by \n. I. Required the volume of the segment of a prolate spheroid, its height being 6 inches, and the axes 50 and 30 inches respect- ively. By formula, V=\nh^{^ {^d-1h)=^ny^^^ Q^^y, MENSURATION. 335 (3X30— 2x6)=2450.442267cu. in. C 1. 50 in.^the transverse diameter, or j|, ^^. J!^~^]]3^ axis. ty^-c yto---^ 2. 80 in.^the conjugate diameter 2il/0. \. ^"■— ■-; — "V/ 3.'|=50-^30=the quotient of the trans ""■^--1_L--^ verse diametej divided by the D conjugate diameter. FIG. 51- 4. 78 in.p=3x30 in. — 2 X6in.=the difference between three I'I.<J times the conjugate or revolving axis, and twice the height of the segment. 5. 130 in.=|^x78 in.=the product of said quotient by said difference. 6. 4680 cu. in=130x62=the square of tlie height of the segment by said product. 7. .-. |jrx4680 cu. in.=2450.442269 cu. in.=the volume of segment. III. .•. The volume of the segment is 2450.442269 cu. in. XIV. CONOIDS. 1. A. Conoid-^is a solid formed by the revolution of a conic section about its axis. ' 1. THE PARABOLIC CONOID. 1. A Parabolic Conoid is the solid formed by icvdlving a parabola about its axis of abscissa. Prob. CXX. To find tlie surface of a parabolic iBonoid, op paraboloid. Formulae. — (a) S^j27iyds^= C'i7iy^\-\-—~^,iy= where 2J> is the latus rectum of the parabola and_y the radius of the base of the conoid, or the ordinate of the parabola. -I I 1 (l>) 5=|7rV^-| (/+;<) — / ^, where 2^ is the same as above and :v the altitude of the conoid, or the axis of abscissa of the parabola. Rule. — To the square of half the latus rectum, or princifal parameter , add the square of the radius of the base of the conoid and extract the square root of the cube of the sum; from this re- sult, subtract the cube of half the latus rectum and multi-ply the. 336 FINKEL'S SOLUTION BOOK. difference hyZn, and divide the product by one and one- half times: the latus rectum, I. Determine the convex surface of a paraboloid whose axis is 20, and the diameter of whose base is 60. From the equation of the parabola, j|'*=2/*, wt have 36^=2/ X 20; whence 2/=45. .-. By formula(a),>9=|^|(/*+y)^-^8^ -5&|C(l)"+30.]'-(f>"S- i»r X25X (125— 27)=49 X25 X3.14159265= 3848.45118. 1. 30= the radius ^ O of the base of the conoid. 2. 20=the altitude OD'. Then by a property of the para<- bola, 3. 30''===2/X,2p, whence 4. ^^22^, the principal parameter of the parabola. 5. (^^yxl25—>Jj^(22i)2+302T=the square root of the cube of the sum of the squares of half the latus. rectum and the radius of the base. II. (45x8 •^ 1 =the cube of half the latus reotum. ' 7. (^^)' X125-(f )'=(«)''(,.^^,_,h, differ. ence between said square root and the pube of half the latus rectum. .... 8. 2^x('^V(125— 27)=7rx98x(Y.)'=2* ^''"^^ said difference. 9. .-. 2 >r98x(^V-i-(|X45)=3848.45118=the surface of the conoid. III. .•'. The surface of the conoid is 3848.45118. Prob. CXXI. Toflnd the volume of a parabolic eonoid. Formula. — V= CTty^dx=^Cn2p x dx=icpe^= .J.7r(2/:c);K=^;ry^;if, where _j/ is the radius ©£ the base- and x the altitude. Rule. — Multiply the area of the base by the aUitutde and take . half the product. MENSURATION. 3af7 T. What is the volume g{ paJ-abolic conoid, the radius of whose base is 10 feet and the altitude 14 feet? By formula, V=i7ry^x = i7rl0'^xU=700X^=22(i2.1U855 cu. ft, 1, 10 ft.=the radius of the base. 2. 14 ft.=the altitude. II.<3. 7rl0-=;U4.159265 sq. ft. the area of the base, 4. .-. ixl4x314.159265=2202.114865 cu. ft.=the volume l_ of the conoid. III. .-. The volume of the conoid=2202.114855 cu. ft. Note. — Since the volume of the conoid is {''^y'^x, it is half of its circum- scribed cylinder. t'rob. CXXII. To And the convex surface of a frustum of a parabolic conoid of wliich the radius of the lower base is B and the upper base r. Formula.— S= f 27tyds=-^ \ (/2-f7?2)*_(/2-j_ri^)^ \ . I. What is the volume of the frtisturn of a parabolic conoid of which the radius of the lower base is 12 feet, the radius of the upper base 8 feet, and the altitude of the frustum 5 feet? Since 122=2/^' and 82=2/Ar, 12«— 82=2/(a;'— «•). ' But ;>;'—;«• =5 feet. .-. 12^— 82=2^X5, whence 2/=l6, the Jatus rectum. .-. By formula, 5=?J[(/2+7?2)^-(/2+r2)*j]= 2^^ [^(82+122)^— (8^+8= )^~|=^(832Vr3—1024V2)= i-/;r[13Vl3— 16V2]. 3X8 16 Prob. CXXIII. Xo find the volume of the frustum of a parabolic conoid, when the bases are perpendicular to- the axis of abscissa. Formula.— V=\nR^x'—\nr'^x=\7i{x'—x){R'^-\-r^ ) Rule. — Multiply the sum of the squares of the radii of the Uivo bases by tc and this product by half the altitude. I. What is the volume of the frustum of a parabolic conoid, the diameter of the greater end being 60 feet, and that of the lesser end 48 feet, and the distance of the ends 18 feet? By formula, F=i;ra(i?24-r2)=^Xl8X'>r(3024-242)=9;r(900 -f576)=9Xl476X'r=132847r=41732.9177626 cu. ft. 338 FINKEL'S SOLUTION BOOK. II. Ill 1. 30 ft.=the radius of the larger base. 2. 24 ft.:^the radius of the lesser base. 3. 18 ft.^thje altitude of the frustum. 4. 900 sq. ft.=the square of the radius of the lower b^se. 5. 676 sq. ft.=the square of the radius of the upper base. 6. 1476 sq. ft =900 sq. ft.4-576 sq. ft.=their sum. 7. .-. iXl8x?rXl476=13284X7r=41732.9177626 cu. ft.= the volume ot the frustum of the conoid. .-. The volume of the frustum is 41732.9177626 cu. ft. 2. THE HVPERB0I.1C CONOID. 1. An Hyperbolic Conoid is the solid formed by revolv- ing an hyperbola about its axis of abscissa. Prob. CXXIV. To find the surface of aji hyperbolic con- oid, or hyperbqloid. Formula.~S=J2nyds=27iry^l-\.^'^\'dx= IgSi^z ^'2 /'b ft i a.i. 1 , — lo'gfjif-l — ye'^'x^- e '- e ) a xi^e^> ^—ab-\- iJa^+b -log ab id'-^b-" x-\ — ye^x'^ Prob. CXXV. To find the volume of an hyperbolic conoid. Formula. — V=^7t[R^-\-d'^)h, where R is the radius of the base, rfthe middle diameter, and h the altitude. Rule. — To the' square of the radius of the base add the square of the middle diameter between the base and the vertex; and this sum. multiplied by the altitude, and the product again by \Tt, •will give the soliditv- I. In the hyperboloid A CB, the altitude CO is 10, the radius -(4 O of the base 12; and the middle diameter DE 15.8745 ; w^hat is the volume.? ri. 10=the altitude CO. 2. 12=the radius AO oi the base. 3. 15.8745=1 he middle diameter £>B. 4. 144=12^=the square of the radius II. J of the base. FIG. 53- 5. 251.99975=15.87452=the square of the middle diameter. ' • MENSURATION. '3^9 ' 6. 395.99975=251.99975+144==the sum of the squares of the radius of the base and the middle diameter; 7. .-. ^;rx 10 X 395.99975=2073.45469 l=the volume. in. .-. The volume of the conoid is 2073.454691. Pi'ob. CXXVI 'To find lilie volume of the frustum of an Jiyp'erbolic conoid. Formula. — V==\7ta{Ji'^-\-d^-\-r^), where i? is the ra- dius of the larger ba%'e,"feaira'r1the radius of the lesser base, and (^ the middle diameter of t^eftu^tum. 'RvA&'.T-^AcCdtoffeifieriihe^-Aquares of the greater and lesser seini-diamtiters, arictihe squarp of the whole diameter in the mid- dle\Jihen- this: sum being niiilt^lied by the altitude', and the prod- uct Vgain iy^Tt, tvtllgive the solidity. XV. QUADRATURE AND CUBATURE OF SUR- FACES AND SOI.IDS OF REVOLUTION. 1. CYCLOID. Prob. CXXVII. To find tlie surface g'enerated by the revolution of a cycloid about its base. Formula.— S=:2J2 nyds^ATtJyiJdx ^ \dy^= Rule. — Multiply the area of the generating circle by *^. Prob. CXXVin. To find the volume of the solid formed Ijy revolving the cycloid about its base. Formula.— V=2 f7ry^dx=27r f^'' /^"^y -=f,^2^._ '^ Jo \l2ry—y^ fX7r(2r)2x2wn Rale. — Multiply the cube of the radius of the generating cir- cle by hn'^. Prob. CXXIX'. To find the surface g-enerated by revolv- ing the cycloid about its axis. Form,ula.—S=J27iyds=i7iiil2?Jy~=&nr^ ( 7t—\). Rule. — Multiply eight times the area of the generating circle hy n minus \. ' Prob. CXXX. To find the volume of the solid fonued by revolving the cycloid about its axis. y^B^BS 340 FINKEL'S SOLUTION BOOK. ^\li»,'^-^Multiply\ofthevotmHeof a sphere toAose radius is thai of the generating' circle by | n^ — \. Prob. CXXXI. To find the surface formed by revolvlngr the cycloid about a tangent at the Tcrtex. Let /* be a point on the curve, AE=PB=syf EP=A£=^y A C= C^=r,and the angle A CF =ft Then we Shall have AE=s. y-=-A C — CE=r — r cos ^ ; and AB^ x=:EP-\-EE^AE-{-EE =zrd-{-rsitt 0. An ry^dx'^-\-dy^ =^itr j (r — f/Q. 54, rcos S^r'^(i^coB&)'^-\-r': sin'^e dd =Anr^ /^(l— cos(9)x V2+2cos6'<«9=8;rr^ I {l-^coiB)co^ede=l&7tr-^ i ''(l_cos2+(9> Jo Jo Xtos:ildd9=\^7rr^J{co^e—tti^»^9)d€^\'e,nr^V2,&\-a\H— |sin|(9cos^^(?— Isin-^i^n t=^^nr^. -10 Rule. — Multiply the area i>f the generating circle by ^. Prob. CXXXH To find the Volume formed by revolving- a cycloid about a tangent at the vertex. Formula.— V'=fiCiiy^ dx^r±%it.\ '^(r_^cos^)2^(l-|- cosd)de=27rr.» f''(l—cosep(l+cosfi)d6=27ir^ 1^(1— cose— cos^^-}-cos'^)t;?ft= 7r^/-*=the volume generated between the curve and the tangent. .-. V=7rAI>^X GI/—V=7r{2r}^x27cr—7t''r^=77r^rK Rule* — Multiply the cube of the radius of the gen^ating cir- cle by 7n^. 2. CISSOID. Prob. CXXXIII. To find the volume generated by revolv- ing the cissoid about the axis of abscissa. Formula, — '^= J" ny»dx=j7r—-—dx=7i( — J-*' — ax^ MENSURATION. 341 Ppob. CXXXrV. To find the volume formed by revolving tbe cissoid about its asymptote. Formula»T- V==2jK{AR)''dy{Ftff. 2i)=27t{2a—x'fy, Prob. CXXXV. To find the volume formed by revolving 4he Witch of Agnesl about its asymptote. Formula.-^ V=~ I ny^dx== Vsrjy^ft — i7iaj'{2ay— =4:7r'a^ —1 2a y*)idy\^ Prob. CXXXVI. To find the volume formed by revolving the Conchoid of S'icomedes about its asymptote, or axis ot Abscissa. Formula.— V=j7ty^dx=n fr—r^^L^+y(l>'' -j^)* 3. SPINDLES. 'l^). TAe Circular, Spindle. A Circular Spindle is the solid fprmed by revolving the segment of a circle about its chord. Prob. CXXXVII . To find the volume of a cli-cular spindle. Let AEBD be the circular spindle formed Iby revplving the segment A CJ3E about the -chord A CB. Let AB=^1a, the length of "the spindle, and MD=2b, the middle diame- ter of the spindle. Let CI^KL^x, the ra- ■dius of any right section of the spindle, and KI^CL=y. Then the required volume of ^^^^/e^s^ the spindle is V=27t f arVj)/. ..(1). Let i?=(a = -f-33 )-f-25 . .(2), Tae the radius of the circle and 6 the angle A GE. Then by a property of the circle, KI'^={21i~EI)xEI, or y^=(1Ji~ EJ)xEI. But EI=E G—IG=Jf—(IC-\-CG)=B— <^+i?cos(9). .-. y^=i^2B—[Ji—{x {-RcosB ) Ur—{x-\.Rcos9)1 ={R-\-{Rco%OJ^-xy\X [B—{Jico%d-{.x)^=R^—{Rcose-\.xy ; whence x^R^ — -j^ — Rcosd . . (3). Substituting this value of a in ( 1 ) , we have F=27r fC^RKZ^^—Rcosff) ^ dy=2 n [i?^ ( \J~ ^ 342 FINKEL'.a SOLUTION BOOK. \ 2a]<?—%a^—{R—b) VliHm ^^-Za>^ R^ —a"- ~\X. Rule. — Multiply the area of the generating segment by ihe path of its center of gravity:^— G^ildin's Rule. 0). T^ Parabolic Spindle. A. Pardbolic Spindle is a solid formed by revolving » parabola abput a double "ordinate perpendicular to the axis. Prob. CXXXVIII. To find the volume of a parabolic spindle. ■ ' ^ Formula.— F=2 f 7t{h—xYdy=1-n: C {h-^—2hx-\- J Jo yB~\ =2n]Ji'^y—^hxy-\-\x'^y~\ =2n[h^b~^hbx+\bx^'\. But ;f= h, when ^=3. .-. V=27tlkH—ihH-\-\h''b^=\^nh-'b=:^^X2bX^h\ Kule. — Multfly the volume of its circumscribed cylinder by •^■^.. I. What is the volume. of a parabolic spindle whose length ^ C is 3 feet and height BD 1 foot? By formula, F=f|7r>%2^=j\7rxl^ X 3=4.9945484 cu. ft. fl- 1 ft.=height BD of the spindle. 2. 3 ft.=length A C. 3. ^ Xl=' X 3=9.42477795 cu. ft. the volume of its circum- scribed cylinder. 4. .-. i\x9.42477795 cu. ft.= 4-9945484 cu. ft, the vol- ume of the parabolic spin- dle. FIG. 56. III. .-. , The volume of the spindle is 4.9945484 cu. ft. Prob. CXXXIX. To find the volume grenerated by revolv- ing- the arc of a parabola about the tangent at its vertex. Let ABC be an arc of a parabola revolved about AB, and let B be any point of the curve. Let AE=BF=x, and AB^BF =y. Then the area of the circle described by the line BB is nx^.. , .-. Dormula.~V=27rJx^dy=:^27cJ'^^dy=27tx~X II. MENSURATION. 313 FIGST,-^"' \y^=^nx'^y=^\nh^b, where ^4^the height, and 3^ CZ?, the or- dinate of the curve. Kule. — Multiply the' volume of its circum- scribed cylinder by \. Prob. ■ CXIi. To find the volume generated by revolving the arc APC of the parabola about BC parallel to the axis AD. . The; afea^ oCthe circle, generated by the line GP\%,Tt\b— yY :..,;■'■ ' . ,; : '. ' .•■"-■I - .-. MoT'm'Ulilb.^^'V=-K C{'b—yydx=l7tb^h. ItiU.le< — Multiply the volume of its circumscribed cylinder by \'. Note. — In the last two problems, the volume considered, lies between the curve and the lines ^4.5 and ^C respectively. The volume generated by the segment \A.CD is found by subtracting the volume foundln the twd problems from the volume of the circumscribed cylinders. Prob. CXLiI. To find the volume formed by revolving a semi-cir<;le about a tangent parallel to its diameter. Let the semi-circle be revolved about the -'tangent .4 C Let AC=R,PP==^AG^BC=y,AF=GP=zx. Then the area of the circle generated by the line GP is TtX^. ButV2=2i?2_2i?(7?2_y2)i_yi=; for, PC^=PC^—PP^,oi (P—xy=P^ —y^ ; wh ence x=P—^/ji''—y^, and x^=2P^—2P^P^—y^—y'^ .: Formula.— V=2j7tx^dy^2nJ{2R^— 1R>i~R'^—y'^—y'^ )dy=\Tt R^ ( 10— 3w), which is the entire volume external to the semi-circle- '' FIG. 59. Hule. — Multiply one-fourth of the volum.e of a sphere -whose radius is that of the generating semi-circle by (10 — Sn). XVI. REGULAR SOLIDS. 1. A. JtegulaT Solid is a solid contained under a certain number of similar and equal plane figures. 3. The Tetrahedron, or Hegular pyramid, is a regular solid bounded by four triangular faces. 3. The Hexahedron, or Cube, is a regular solid bounded by six square faces. 4. The Octahedron is a regular solid bounded by eight triangular faces. 5. The Dodecahedron is a regular solid bounded by twelve pentagonal faces. 844 FINKEL'S SOLUTION BQOK. 6. The Icosahe«l^»n ie«regul«ir solid bounded hy twenty eqaUmUxul triangular faces. These are the ooly JM^ular solids tb«t can poeeibljr be formed. l( Cbe following figures are made of pasteboard, and the dotted lines eut half through, so that the parts niay be turned up and glued together, they will represent ^ five regular solids. FIG. 59. 1. TETRAHKURON. Prob. CXm. To find the surface of a tetrahedron. Formula. — 5'=/8V'3, where /is the length of a linear side. Mule, — Multiply the square of a linear side by sfs=^1.7S20 508. > I. What is the surface of a tetrahedron whose linear edge is 2 inches. By formula, S=Psl^=2^^=i»/^=Q.^2%2 sq. in. 1. 2 in.=;the length of a linear side. sq. in.=22^the square of a linear side. 3. .-. VSX 4 sq. in.=1.73205x4 sq. in.=6.9282 sq.'in., the III. .-. The surface of the tetrahedron is 6.9282 sq. in. Prob. CXLIII. To find the volume of a tetrahedron. Fortn/ula. — V=^>^ P , where / is the length of a linear side. 11.(2' [surface. iRule. — MuUifly the mk* «f a linear side by -^^^iz, or .11786. I. Required the solidity of a tetrahedron whose linear side is 6 feet? By formula, F=iijV2 /'=TiyV5x6»=18V8,==a&,455843 ch, ft, ( 1. 6 ft.s=tlie length pf a linear side. II. < 2. 216 cu. ft.=the cube of the linear side. (3. .-. yV^2"x216 cu. ft.=V2xl8c:u. ft.=25.45843 cu. ft. III. .-. The volume of the tetrahedron is 25.45843 qh, ft. 2. OCTAHEDRON. Prob. CXIilV. To fimi tbe surface of an octahedron. Formula.— S=24% P . Rule. — Multifly the square of ct linear side by 2^8, i. e., by ■ttvo times the square root of three. I. What is the surface of an octahedron whose linear side is -4 feet.? By forumla, 5=2V3/2^2V3X ^'=32^3=55.4256 cu. ft. 1. 4 ft.^the length of a linear side. 2. 16 sq, ft.=4^==:the square of the linear side. 3. .-. 2V3X16 sq. ft.=V3x32 sq. ft.=l. 73205x32 sq. ft.= 55.4256 sq. ft. III. . . The surface of the octahedron is 55.4256 sq. ft. Prob. CLXV. Tofiud the volume of an octahedron. Formula.— V=\\^ P Rule. — Multiply the cube of a linear side by -JV^, /. e., by one - Jhird of the square root of two. 1. What is the volume of an octahedron whose linear side is % inches? By formula, I^=^\/2 /'=^v'2x8''=.4714045x512=241.359104 ■cu. in. 1. 8 in.=che length of a linear side. 2. 512 cu. in.=8'^the cube of a linear side-. 8. .-. iV^2x512cu. in =^X] .4142135x512 cu. in.= 241.359104 cu. in. :ii.<^ II. III. .'. The volume of the octahedron is 241.359104 cu. iu. 3. UODtCAHEDRON. Prob. CXIjVI. To fiiid the surface of a dodecahedron. 346 FINKEL'S SOLUTION BOOK. ..-5=15-J0+2V5>,, II. Rule. — Multiply the square of a linear side by 15\/\^\5-\^ SVS}], or 20.6^.57285. ' , ■; I. What is the surface of a dodecahedron whose linear side is. 3 feet? By formula, 5=16>J(^=i^\='=20.6457285x9 =185.8116565 sq. ft. rl. 3 ft.=the length of a linear side. 2. 9 sq. ft.=3^=square of a linear side. 3. .-. 15->J^ ^+^^ ^X9 sq. ft=20.6457285X9 sq. ft. =185.8115565 sq. ft. III. The surface of the dodecahedron is 185.8115565 sq. ft. Ppob. CXLiVII. To find, tlie volume of a dodecahedron. Formula.— V==h^Qt^^^^\^='l.mZllbXl^- Rule. — Multiply the cube of a linear side by 5 ^( — ^^ 1,. or 7.663115. I. The linear side of a dodecahedron is 2 feet; what is. its. volume ? By formula, F=5 J(^^^i^)/3=^7.663115x8 =61.20492 cu. ft. rl. 2 ft.^the length of a linear side. .)2. 8 cu. ft.=2^=cube of a linear side. ^■]S. .-. 5V[^(47+2lV5)]x8cu. ft.=*7.663115x8cu. ft I =61.20492 Cu. -ft., the volume. III. .-. The volume of the dodecahedron is 61 20492 cu. ft^ 4. ICOSAHEDRON. Prob. CXLiVIII. To And the surface of an icosahedron.. Formula.—S=5^3P=8.66025 X /' • Rule. — Multiply the square of a linear side by 5^3, or- 8.66026. MENSURATION.. 34r I. What is the surface of an icosahedro-n whose linear side is 5 feet. By formula, 5=5^3/2=5^3x52=125^3=216.50626 sq.- ft. rl. 5 ft.^length of a linear side. TT J 2." 25 sq. ft.=52^the square of a linear side. ■ j3. .-. 5V§X25 sq. ft.=8.66025x25 sq. ft.=216.50625 sq. ft. I =the surface. III. .•. The surface of the icosahedron is 216.5062o sq. ft. Prob. CXLiIX. To find the solidity of an icosahedron. Formula.— F"=fV[iC7+3V5)]/3=2.18169X^='. Rule. — Multiply the cube of a linear side by |-V[^(7-(-<?V'5)],. orS. 18169 I. What is the volume of an icosahedron whose linear side is- 3 feet? By formula, r=fV[i(7+3V5)]/3=2.18169x33=58.90563cu.ft. rl. 3 ft.^the length of a linear side. 1 2. 27 cu. ft.==3'=the cube of a linear side. 3. .-. 4V[K7+3V5)]X27cu. ft.=2.18169x27cu. ft. I =58.90563 cu. ft.=the volume. III. .-. The volume of the icosahedron is 58.90503 cu. ft. Note. — The surface and volume of any of the five regular sol- ids may be found as follows : Kule ( 1 ). — Multiply the tabular area by the square of a linear side, and the product -will be the surface > Rule (3). — Multiply the tabular volume by the cube of a linear side, and the product will be the volume. Surfaces and volumes of the regular solids, the edge being 1. II. NO. OF SIDES. NAMES. SURFACES. VOLUMES. 4 6 8 12 20 Tetrahedron Hexahedron Octahedron Dodecahedron Icosahedron 1.73205 6.00000 3.46410 20.64578 8.66025 0.11785 1.00000 0.47140 7.66312 2.18169 XVII. PRISMATOID. 1. A. PTismatoicl is a polyhedron whose bases are any two polygons in parallel planes, and whose lateral faces are triangles determined by so joining the vertices of these bases, that each lateral edge, with the preceding, forms a triangle with one side of either base. ^m FINKEL'S SOLUTION BOOK. 2. jA JViawu)id ii a prisrnatoid whose bases have the same inumber of sides, and every corresponding pair parallel. Ppolj, Chk To fl»4 tfte volwne of any pFismatold. Formula (a).— I^=i«(^j+Sylj^)=^«(J9,-|-34'|<,), where •4 is the altitude, -B^ the area of the lower base, Jlx^ the are* of a ^section distant from the lower base twQ's^hirds the altitu4e, ^, s^thearea of the upper base, and A'tf, the area of It seqtipn distant 'two-thirds the altitude from the upper base. Remark. — This simplest Prismoidal Formula is due to Prof. George B. Halated, A. M., Ph. D., Professor of Mathematics in the University of Texas, Austin, Texas, who was the first to 'demonstrate this important truth. The formula universally ap- plies to/all prisms and cylinders ; also to all solids uniformly "twisted, e. g. the square screw; also to the paraboloid, the right •circular cone, the frustum of a paraboloid, the hyperboloid of one nappe, the sphere, prolate spheroid, oblate spheroid, frustum of ■a right cone, or of a sphere, spheroid, or the elliptic paraboloid, the groin, hyperboloid, or their frustums. For a complete demonstration of the Prismoidal Formula, see Halsted's Elements ■of Geometry or Halsted's Mensuration. Rule. — {a) Multiply one-fourth its altitude by the sum of one base and three times a section distant from that base tviQ-thirds the altitude. u Formula {b). V=ia{B^+4:M+B^), where a is the alti- tude, B^ and ^j the areas of the lower and upper bases respect- ively, and ^1/ the area of a section midway between the two bases. Rule. — (b) Add the area of the two bases and four times the m-id cross-section; multiply this sum by one-sixth the altitude, PROBLEMS. 1. Find the weight of a stfeel wedge whose base measures 8 inches by f inches, and the height of the wedge being 6 inches ; if 1 cu. in. of steel ^weighs 4.53 oz. ? 2. Find the volume of a prismatoid of altitude 3,5 cm., the bases being rectangles whose corresponding dimensions are 3 cm. by 2 cm. and 3.5 cm. by 5 cm. 3. The base of a wedge is 4 by 6, the altitude is 5, and the edge, e, is 3. Find the volume. 4. A stick of timber is 12 feet long and is 4 inches by 5 inches at one end and 8 inches by 3 inches at the other. Find the number of cubic feet in the stick. Ans. \\i cu. ft. 5. How far from the larger end must the stick described in the last problem be sawed into two parts in order that the parts shall have equal solidity? MENSURATION. 84d» XVIII. CYUNDRIC RINGS. 1. A Cylindric Ring is a solid generated by a circle- lying wholly on the same side of a line in its own plane and revolving abont that line. Thus, if a circle vs^hose center is be re- volved about Z>C as an axis, it will gener- ate a cylindric ring whose diameter is AB and inner diameter 2 BC. OC will be the radius of the path of the center O. FIG. 60. Prob. CLiI. To find the area of the surface of a solid ring:,. Forinulai^'"S==2nry(,27iR=:A:n^rIi^ whwe r is the ra-. dius of the ring, and R is the distance from the center of the ring^ to the center of the inclosed space. Rule. — Multiply the generating- circumfer«n,<;e by the path off its center. Or, td the thickness of tht ring add the inneii- diante- ier and this sum being muttifhed hy the thickness, and the pro- duct again by 9. 86970 J).!), will give the area of the- surface. I. What is the area of the surface of a ring -whose diameter- is 3 inches and the inner diameter 12 inches. By formula, 5=47rV/?=4«2XHX (l^+6)=w2 ^45=9.8696044x45 =444.132198 sq. in. 1. 1^ in.^^ of 3 in.=the radius r of the ring. 6 in.=-J of 12 in.^the radius of the inclosed space. 6 in.-j-l^ in.^7^ in. = the ra- il.^ dius R of the center of the ^/ft %%. ring. 4. TtA C^TT 3^the circumference of a section,. 5. nIK=2nIO=2n1\=n I5=the path of the center. 6. . . ;r3Xa'15=7rM5=444.13219&sq. in.=the area of the. surface of the ring. III. .-. The area of the sJurface of the ring is 444.132198 sq. in. Prob. Clill. To find the volume of aeyURdric rimg^. Formula. — V=7t ^r'^R^nr''' X wR, wfeere r is the ra- dius -<4/ of the ring, and R the distance feojm tbe center of the ring to the center of the inclosed space. Gule. — Multiply the area of the generaiiimg ciraie by the path of its center. Or, to the thickness of the ring add the inner di- ameter, and this sum. being multiplied by the square of half the thickness, and the product again by 9:8.€9!S0^,. tpill give the' volume. 350 FINKEL'S SOLUTION "BOOK. I. What is the volume of Un anchooring whose inner diame- ter is 8 inches, and thickness in metal 3 inches? By formula, F=;r'2r2^==^7r2 x(riy2 '>c(3-i-8J==24.'75X ' 9.88l66'#m'2!7a7089 cu, in „, X.'l-J^'.iri'Wi of 3 in.==the radius gf the ring. , , Jti, 8 in.=^tie' inner diaaiieter. ... ^'y^'i4.^-%^ in.==5^ in.=?:the radius i? of the, ■ path of its .'ff^IV'^rfi* V?=fe^,t}ie area of the generating circle. S. 27r('5-|)=ff Xll^the path of its center. 6.,,..-. 7rlLX7r(J^)2=«8X2,4.Z5=9i.86044x24.75.. ., ... , =244.2727089 cu. in., the volume of the ring,. I'lIL .-. The volume of the ring is 244.2727089 cu. iii: ' ' PROBLEMS. 1. Find the surface and volume of a ring, the radius of the inher cir- cumference being ]0>^ inches and the diameter of the cross-seclion S}4 incliies. ./^«j. 847 sq. in. ; 741J cu. in. 2. Find the siirface and volume of a ring, the diameters of the inner and outer circuipferences being 9.8 inches and 12.6 inches respectively. " ' .<4«i. 154.88 sq. in.; 54.21 cu. in. 3. An anchor ring, whose inner diameter is 20 inches and the diameter of a crosg^sectionlO inches, is cut by a plane perpendicular to its axis and at a distance of 4 inches from the center. Find the volume and surface of the segment removed. 4^ In the same ring, find the volume and surface of the segment of the ring, if the plane is passed parallel to the axis of the ring, and at a dis- tance of 6 inches frofa the axis. 5*1 What is the volume, if the plane be passed at a distance of 12 inches from the axis? _.x' XIX. SIMILAR SURFACES. Similar Plane Surfaces are surfaces having their homol- ogous sides proportional and their homologous angles equal. Principle, — Similar areas are to each other as the squares of their like dimensions or as the squares of any other homologous lines. ■■-'..:■■ I. The area of a rectangle w^hose length is 20 rods is 120 sq. rods ; what is the area of a sim:la,'i'-'rbctan^le'*Whose length is 30 rods? ■ , ■ , . . . .- .. xvj".- j'l. 20" rods=the length of the given rectangle, and 2. 120 sq. rd.=its area. ■ .. . . , II.<j3. 30 rods==the length of the required rectangle. ^ = ,,~-j; |4. .-. 202:30^ ::1^0sq. rd. : (?). Whence, ■> . [5. ?=( 120^X302 )-=-202=27Qsq.rd. tl. ,,■-";:, MENSURATION!.'' ■ '351 III. .•. The area of the rectangle is 270 sq. rd. I. The area of a rectangle whose width is 7 feet, is 210 sq. ft. ; what is the length of a similar rectangle whose area is 2100 sq. ft. 1. 210 sq. ft.=the area of given rectangle, and 2. 7 ft.=its- width. Then 3. 210-1-7=30 ft.=its length. 4. .-. 210 sq. ft: 2100 sq. ft. : : 30^ :( ?). Whence, 5. ?=(2100X302 )-T-2 10=300 ft.=the length of the re- quired rectangle. III. The length of the required rectangle is 300 feet. PROBLEMS. 1. The sides of a triangle are al, 20, and 13 inches; find the area of a similar triangle whose sides are to the corresponding sides of the first as 25:3. 2. In a survey map an estate of 144 acres is represented by a quadrilat- eral, ABCD. The diagonal, AC, is 6 inches, and the perpendiculars from ^ and Z? on .(4C are 1.8 inches and .9 inches respectively. On what scale ■was the map drawn ? Ans. 6 inches to the mile. 3. A man 6 feet in height, standing 15 feet from a lamp-post, observes that his shadow cast by the light at the top of the post is 8 feet in length ; how long would his shadow be if he were to approach 8 feet nearer to the post? Ans. 2 ft 4 in. 4. A man, wishing to ascertain the width of an impassable canal, takes two rods, 3 feet and 5 feet in length. The shorter he fixes vertically on one bank and then retires at right angles to the canal, until on resting the other rod vertically on the ground he sees the ends of the two rods in a line with the remote bank ; if the distance between the rods is 60 feet, what is the width of the canal ? , Ans. 90 feet. 5. A man wishing to find the height of a tower, fixes a rod 11 feet in length vertically on the ground at a distance of 80 feet from the tower. On retiring 10 feet further from the tower he sees the top of the rod in line with the top of the tower. If the observer's eye is 5j4 feet above the ground, find the height of the tower. Ans. 55 feet. 6. A triangle ABC is divided into two equal parts by a straight line XY, drawn parallel to the base BC. If ^5=100 inches, find AX. 7. In a given triangle a triangle is inscribed by joining the middle points of the sides. In this inscribed triangle another similar triangle is inscribed, and so on. What fraction of the given triangle is the area of the sixth triangle so drawn? 8. (a) In a given square whose side is 16 inches a square is inscribed by joining the mifidle points of the sides of the given square ; in this inscribed square a square is inscribed iu.like manner, and so on ; find the area of the fifth square, (b) If the process be continued ad infinitum what is the sum of the areas of all the squares ? 9. In a circle of a radius of 32 inches an equilateral triangle is in- scribed, and in this triangle a circle. In this circle an equilateral triangle is again inscribed, and in the triangle a circle, and so on. If the process is continued, find the area of the fourth circle and find which of the circles has an area of 3^ sq. in.? Ans. 50f sq. in.; the sixth. 10. A field of 9 acres is represented in a plan by a triangle whose sides are 25, 17, an(i 12 inches. .On what scale is the plan drawn and what length will be represented by 80 inches? Ans. ^-J^; I mile. 952 FINKEL'S SOLUTION BOOK. 11. The following is (Ised by lutublefiAeA in fifldiflg the diaffieter Of trees at aay height iabove the ground : If the tree casts a definite shadow- on a h'OrkoatM ^lafie^ stand on the edge of the sliadow and observe where- the line of liglit ttdtA the sun tt> yoiif eye strikes thfe tree. Then meteut* the shadow of the trte at the point where the shadow of your head strikes, the ground. The width of the shadow is the diameter of the tree at the- point where the line of light from the eye to tie sun strikes it What principle is involved? XX. SIMILAR SOLIDS. Solids bounded by plane surfaces are similar if the homol- ogous edges are proportional and the homologous polyhedral angles are equal and similarly placed. , All spheres are similar solids. Cylinders generated by similar rectangles, and cones' generated by similar rig-ht triangles are- similar. ■^rineiplet — Similar solids wre to ^ach other as the cubes of their like diwensioKS or •as the cttbes of any other ^mohgViii lines. I. If the weight ota. well proportioned man, 5 feet in height,, be 125 lbs., what will be the weight of a similarly proportioned man 6 feet high ? 1. 5 ft.=the height of the first man, and ' 2. 125 lbs.==his "weight. 3. 6 ft.=the height of the second man. 4 /. ■ 5* : 68 : : 125 lbs : ( ?). Whence, 5. ?==( 125 X6*)-r-5'=216 lbs., the weight of the second; man. IIL .-.The weight of the man whose height is 6 feet, is 216 lbs. PROBLEMS. 1. The edges of two cubes are as 4:3; find the ratio of their surfaces- and their volumes. 2. The surfaefes of two spheres are in the ratio of '25:4,- find the ratio- t)f their volumes. 3. At what distance from the base tnust a cOfie, whose height is 1 foot,, be cut by a plane parallel to the base, in order to be divided into two parts- of equal volume ? Ans. 2.47 in. 4. A right circular cone is intersected by two planes parallel to the base- alid trisecting the height. Compare the volumes of the three parts into- Which the cone is divided. ^ Ans. 1:1 '.l^. 5. The dimensions oi a rectangular parallelopiped are 5, 6, and 9 feet Find the dimensions of a similar solid, having 8 tirctes the volume of the given parallelopiped. Ans. 10, 12, and 18 feet. 6. The voluine of a pyramid whose altitude is 'T in^, is 686 cu. in. Piad the volume of a similar pyramid whose altitude is 21 inicliiies.. Anis.yiA^Q cu. in. V. Two bins of similar form contain, respectively, 375 ajcrd 648 bushels. of wheat If tte first bih is 3 ft '9 in. \Bide„ what is. the -width of the sec- ond ? Am. 6, feet. II. MENSURATION. 353^ THEOREM OP PAPPUS. If a plane curve lies wholly on one side of a line in its owrr , plane, and revolV|ing about that line as an axis, it generates thereby a surfate of revolution, the area of whichis equal to the product of the length of the revolving line into the path of its center of mass ; and a solid the volume of which is equal to the revolving area into the length of the path described by its center of mass. XXI. MISCELIvANEOUS MEASUREMENTS. 1. MASONS' AND BRICKLAYERS' WORK. JUaSons' work is sometimes measured by the cubic foot, and sometimes by the- perch. A perch is 16-^ ft. long, 1^ ft. wide, 1 ft. deep, and contains 16|XliXl=24| cu. ft. Prota. CLIII. To find the number of perch in a piece of masonry. Rule. — Find the solidity of the wall in cubic feet by the rules given for the mensuration of solids, and divide the product by 2^\. I. What is the cost of laying a wall 20 feet long, 7 ft. 9 in. high, and 2 feet thick, at 75 cts. a perch. 1. 20 ft =the length of the wall, 2. 7 ft. 9 in.=7| ft.=the height of the wall, and 3. 2 ft.^the thickness. jj |4. .-. 20X7|X2=3rO cu. ft.=the solidity of the wall. '1 5. 24f cu. ft.^1 perch. 6. 310 cu. ft.=310-;-24|=12|f perches. 7. 75 cts.=thecost of laying 1 perch. 8. .-. 12||x75cts.=$9.39if=thecostof laying 12|| perches. ill. .-. It will cost $9.39^1 to layl2ff perches at 75 cts. a perch. 2. GAUGING. Gauging is finding the contents of a vessel, in bushels, gallons, or barrels. Prob. CLIV. To g-aug-e any vessel. Rule. — Find its solidity in cubic feet by rules already given; this multiplied by 1728-^-2160.4-2 or .88. ivill give the contents in bushels; by 1728-i-2Sl. -will give it in wine gallons, which divided by Sl\ will give the contents in barrels. Prob. CLiV. To find the contents in gallons of a cask or barrel. Rule. — [1) When the staves are straight from the bung to each end; consider the cask two equal frustums of equal cones, and find its contents by the rule of Proh. XCIII. II J ■354 FINKEL'S SOLUTION BOOK (2). When the staves are curved; Add to the head diameter ■^inside) two-tenths of the difference between the head and bung- diameter; but if the staves are only slightly curved, add six- tenths of this difference; this gives trie mean diameter; express it in inches, square it, multiply it by the length in inches, and this product by .008 J^. ; the product will be the contents in wine gallons. 3. L.UMBER MEASURE. , Prob. CliVI. To find the amount of square-edgred inch boards that can be sawed from a round log'. J)oyle's 'Mule. — From the diameter in inches subtract four; the square of the remainder will be the number of square feet of inch boards yielded by a log 16 feet long. I. How much square-edged inch lumber can be cut ft-om a log 32 in. in diameter, and 12 feet long? 1. 32 in.^the diameter of the log. 2. 12 ft.=the length. 3. 32 in. — ,4 in;=28 in.=the diameter less 4. 4. 844 ft.^28^=the square of the diameter less 4, which by the rule, is the number of feet in a log 16 ft. long. 5. 12 ft.=f of 16 ft. 6. .'. f of844 ft.=633 ft.=the number of feet of square- edged inch lumber that can be cut from the log. III. .■. The number of square-edged inch lumber that can be cut from a round log 32 inches in diameter and 12 ft. long is 633 ft. 4. GRAIN AND HAY. Prob. CliVII. To find the quantity of grain in a ivag-on bed or in a bin. Kule. — Multiply ike contents in cubic feet by 1728-i-2150.42, or .8j, and the result will be the contents in bushels. I. How many bushels of shelled corn in a bin 40 feet long, 16 feet wide and 10 feet high ? 1. 40 ft.=the length of the bin. 2. 16 ft.=the width of the bin, and 3. 10 ft.=the height of the bin. 4. .■. 40X16X10=6400 cu. ft.=the contents of the bin in cu. ft. ,5. .-. B400 X 83 bu.=5312 bu.=the contents of the bin in bu. III. .-. The bin will hold 5312 bu. of shelled corn. Kule> — {1) For corn on the cob, deduct one-half for cob. (£) For corn not '■'■shucked''^ deduct two-thirds for cob and shuck. II.< 11'.' MENSURATION. 355 I. How many bushels of corn on the cob vyill a wagon bed ihold that is 10^ feet long, 3^ feet wide, and 2 feet deep? 1. 10^ ft.==the length of the wagon bed,. 2. 3|. ft.^its width, and ,3. 2 ft.=its depth. [in cu. ft 4. .-. 10iX3|X2=73^ cu. ft.=contents of the wagon bed 5. '.-.VTS^ X.8 bu=58.8 bu.=number ot bushels of shelled ■ 'corn', the bed will hold. 6.-.-.^of 58.8 bu=29.4 bu.=the number of bushels of ' • ' cdrn.on the cob that it will hold. ' HI. .-. The wagon bed will hold 29.4 bu. of corn on thie cob. Pr.o}}.- CliVIII. To find the quantity of hay in a stack,rick, •or inow;' ' , • Rule. — Divide the cubical contents in feet by 550 for clover or iy 1^50 for timothy; the quotient -will be the number of tons. Prob. CliXIX. To find the volume' of any irregular solid. Rule. — Im.merse the solid in a vessel of -water and determ.ine the quantity of -water displaced. I A being curious to know the solid contents of a brush pile, put the brush into a vat 16 feet long, 10 feet wide, and S feet deep and containing 5 feet of water. He found, after putting in the brush, that the watqr rose 1.^ feet ; what was the contents of the brush pile ? 1. 16 ft.=the length of the vat, 2. 10 ft.=the width, and 3. 1^ ft.^the depth to which the water rose. 4. .-. 16X10X11^240 cu. ft.=the volume of the brush pile. HI. .•. 240 cu. ■ft=the volume of the brush pile. XXII. SOLUTIONS OP MISCELLANEOUS PROBLEMS. Prob. CLiX. To find at what distance from either end, a "trapezoid must he cut in two to have equal areas, the divid- ing line heing- parallel to the parallel sides. For'm ula.—d=A^\^\{b''^b\)^b'\=.\{b^b^ )a -i-[^i(i5^+^i)4~^]' where A is the area of the trapezoid, b the lower base, and i5j , the upper base. '^ \(_b'^ -\-b\) is the length of the dividing line. Rule. — 1- Extract the square root of half the sum of the squares of the parallel sides and the result -will he the length of ike dividing line. H.J 356 FINKEL'S SOLUTION BOOK. 2. Divide half the area of the -whole trapezoid by half the sut» of the dividing line and either end, and the quotient ■will be the distance of the dividing line from that end. I. I have an inch board 5 fe6t long, 17 inches wide at one end and 7 inches at the other; how far from the large end must it be cut straight across so that' the two parts shall be equal ? By formula, d =\{b-\-b^ )a- ^-[V^(3a_|-3^)-^^] =^( 17-1-7 )60-e-[Vi( 17 ^+7^ )+17]=720-^30 =24 in.=2 ft. 1. 2. 3. 4. 5. But Let ^^CZ> be the board, [end, ^^=17 in.=5, the width of the large Z>C=7 in.=3', the width of the small end, and [board. HK=^ ft.=60 in.=a, the length of the Produce HK, AD, and BC till they meet in £!. Then by similar triangles, 6. ABE-.EGL-.EDCwAB^-.LG^-.DC^. 7. EGL=EDC-\-\{ABCD), or 8. IE GL=2ED C^AB CD=ED C-\-ED C-\-AB CD =EDC-^EAB. 9. .-. BGL=^\{EDC-\-EAB), i. e., EGL is an arithme- tic mean between EAB and EDC. . GE''=^AB^-{-DC^)=i(b^-{-y^)=iin arithmetic mean between EAB and ED C, 111. GL=»Ji(b^+V^)=i'^2(b^+.y^). 11-112. Draw Ct^/" perpendicular to ^^. 13. EL=.i GL=\sl2{ b-'+y^). IL=FE—FI{=KC^\D C=\by- CM—HK^a. MB=\{b—i'). Then in the similar triangles CMB and CIL, MB:IL::CM:CI, or :^{b—b'):{^'X{b^^y^)—\b)::a\ CI. Whence C/=a(iV2(l2+F^— i*)-H(*— ^0= a{^4%{b^J^b'-)-b) ^ ^(^2(17^+7^ )-7) „, -7 10. 14. 15. 16. IS. =iV2(^2_|_^^2)_^3_ b—¥ =6(P 17- 36 in. 19. =^ft. . IM=CM—CJ=b ft. — 3 ft.=2 ft., the distance from the large end at which the board must be cut in two- to have equal areas. III. .-. The board must be cut in two, at a distance of 2 feetL from the large end, to have equal areas in both parts. (i?. H. A.,f. 407,freb. 101.} , MENSURATION. 357 Prob. CIiXI. To divide a trapezoid into n equal parts an4 ["RnA the length of each part. Formfla.-h,=^—^, \_^\ 1--^ i J , ^■T^['-j( '-'-'"r'+"'~"'' )]. "'»• *' '• -the width of the small end, 6 the width of the large end, and a -the length of the trapezoid. A^ is the length of the first part at the small end, A^ the length of the second part, and so on. I. A board ABCD whose length BC is 36 inches, width J^B 8 inches and DC ^ inches, is divided into three equal pieces. Pind the length ot each piece. '"^1=3=3; W {n-\)b\Jrb-' -|_ By formula, h^=j^ |_>j '" ^'"^'' *i J = y'\[Vi(3— 1)42+82— 4]^[V32—4]=36(V2—1) =14.911686 in. =11.442114 in. h^ a_r \ in—Z)b\-\W - [ (^— 2)^^+2^ ^1 36[2— V3]=9.6462 in. 1. 4in.=the width DC bi the' small end, 2. 8 in.=the width AB of the large end, and 3. 36 in.=the length ^C of the board. 4. .-. 216 sq. m.=\{AB-\-DC)xBC =^(8+4) X 36= the area of the board. 5. ^ of 216 sq. in.^72 sq. in.^the area of each piece. 6. AK=AB—KB{=D C)=8 in.— 4 in. =4 in. In the similar triangles AKD and D CB, oe 63 ' 7. AK:DKv.AB:BE, or 4 in.:36in.::8 m.:BB. Whence, 8. ^^=(36x8)-^4=72 in. [triangle ABB. jj . 9. .-. i(^5x^^)=i(8x72)=288sq.in.=theareaofthe ■'10. ABB—ABCD=2'm sq. in.— 216 sq- in.=72 sq. in. =arca of the triangle DCB. 358 ' FINKEL'S SOLUTION BOOK.. 11. Z>C£'+Z>CG/^=^72 sq. ■in.+72 sq. in. ^144sq. in, :=the area of the triangle JPGE. 12. DBC+BCGF-\-I^GIH='J2 sq. in.-|-72 sq. in.+72. sq. in. =216 sq. in'.=the area of the, triarigle HIE. 13. EEG:nEC::EG^:BC\or 144 sq. in.:72 sq.,in.:: (3^2:36^ Whence, 14. (?^=V'(144 X362 )-r-72=36V2=50.911686 itiches. 15. .-. GC=G^— C^=50.911686in.— 36in.=14.91168a , in., the length of ^GCZ>. Again, 16. DEC-MIEy.EC^-.EI-^, or 72 sq . in.:216sq.in■::3 6^•^/^■ Whence, 17. ^/=V(216x362)-f-72=36xV3=62.3538 in._ 18. .-. G/=^/— ^G=36V3— 36V2=36(V3— V2) =11.442114 in., the length oiHIGF^ and 19. ^/=^^=^/=72— 36V3=36(2— V3)=9.6462 in., the length oiABIH. r^/=9.6462 in., III. .:{ G/=11.442114 in., and lGC=14.911686in. Prob. CLiXII. To find the edge of tbe largrest cube tb at c an be cut from a spbere. Formula.— e=^^-^=^yVZD=.bnZbxD, where Z> is the diameter of the sphere. Rule. — Divide the square of ike diameter of the sphere by- three and extract the square root of the quotient; or, multiply the diameter by .57735. I. What is the edge of the largest cube that' can be cut from a sphere 6 inches in diameter? By formula, e=J^'=V'^=6x V^=iV'3x6=.57735x6 =3.4641 in. rl. 6 in.=the diameter of the sphere. J2. .-. .57735X6 in.=3.4641 in.=the edge of the largest cube- t that can be cut from the sphere. III. .•. The edge of the largest cube that can be cut from a sphere whose diameter is 6 inches, .is 3.4641 in. Prob. CLXIII. To And the edg-e of the largest cube that can be cut from a hemisphere. Formula.— e=^~=\V&y.D=.AO^M^xD. Rule. — Divide the square of the diameter by 6, and extract t he- square root of the quotient ; or, multiplythe diameter by .^.08248^ ii.js "•{' MENSURATION. " 35^ I. What is the edge of the largest cube th^t can be cut from a hemisphere, the diameter of whose base is 12. inches? By formula, e=\/Z>2-^6=Vi|5=i2A/|=|V6Xl2=.408248 X 12=4.899176 in. fl. 12 ih.=the diameter of the base of the hemisphere. [2. .-. .408248.x 12, in;=^4.899176 in. III. .'.The edge iff the largest cube that can be cut from a hemisphere, the diameter of whose base is 12 feet, is 4.899176 in. Prob, CliXIV. I'd ftnd the diameter or radius of the three largest equal circle's that can he inscribed in a circle of a given diameter oir radius. FormMld.-^d^D-i-(l-\ri^^)=D-i-2.1bb7=A64:lxD or ?-=i?-e-(l+fA/3)=-4641Xi?. l Rule. — Divide the diameter or radius of the given •cirele by ^.1657 and the quotient will he the diameter or radius of the three largest equal circles inscribed in it; or, multiply the diameter or radius by .4.641, and the result will be the diameter or radius re- spectively of the required circles. I. A circular lot 15 rods in diameter is to have three, circular grass beds just, touching each other and the larger boundary ; what must be the distance between their centers, and how much ground is left in the triangular space about the center? By formula, 2i'^2i?-T-(l+tV3)=2i?-^2.1557=,.T^ =6.9615242 rd.=the distance between their centers. Construction. — Let AHE be the circular lot, C the center, and A CE any diameter. With ^ as a center and radius equal to CE describe an arc intersecting the circumference of the lot in H. Draw a, tangent to the Tot at E and produce the radius CH to intersect the tangent at B. Bisect the angle CBE and draw the bisector GB. It will meet the radius CE'm G, the center of one of the grass beds. Draw GP perpendicular to CB. Then GF=GE, the radius of one of the grass beds. Draw EH. Then EH=CH=EC, and CH=HB, because the triangle EHB is isosceles. 1. CE=1\rd..=R, the radius of the lot. i. CB=2CH=1R. 3. EB=V CB^ — CE^=V {2R)^—R'^=RVZ. In the similar triangles CFG and CBE, 4. CF:FG::CE::BB, or CF: GF::R:RVb. But 5. CF=CB—FB(=FB)=2R—RVs=R{2—V%). 6. .■.R(2—Vsy,GF::R:RV3. Whence, 7. Gi5-=#^=if(2V3-3)=7i(2\/3-3)=7iX ^ — V 3 .4641=3.48075 rd.=the radius. A. FINKEL'S SOLUTION BOOK. ll.< Is. G^=2>-=2-ff(2V'3— 3)=6.9615rd., the distance be- l tween their centers. ri. GD=V GK^—DK^=VAr^—ri=rV'E. 2. i(ZSrx GZ>)=|(_2r X r VS )=r V3=the area of the triangle IGK. 3. Area I)KI?= ^B.) W^ 4 of the small circle, bedkuse the angle 'DKF '. is 60^, or I of 360°. Area BKP=\nr^. ^--H: ■^mrr \ V^T^ III. ■^ wr''=3 tirnes^;!rr^ ±=the area of the FIG. 64 three parts of the small circles within the triangle /GK. .: ^^VI— ^7rr2=r2(v'3— i;r)=.16'l25368r2 ==.16125368x[^(2v'3— 3)]2=.16125368X(21 —12v'3)i?2=.16125368X-2153904x^' =.03473265 xi?'=.03473265x (^i)' =1-953712 sq. rd.==:the area of the space inclosed. j-6.9615 rd.:t=the distance between their centers, and-, .■{1.953712 sq. rd.=the area inclosed about the center of the given lot. (H. H.A. p. i07,prob. 100.) Prob. CliXV. Having: ffiven the area inclosed bj; three equal circles to find the radius of a circle that will jiist in- close the three equal circles. J'o^m**?a.-^=J(^2V3=3)W3=i^) =-s]V., Q3473265 y' ""^^^^ "^ '* ^^^ "^'^ inclosed. • Rule. — Divide the area inclosed hy .0S47S266 and extract the square root of the quotient, and the result -will be the radius of the required circle. Prob. CIjXVI. ^Having- g-iven the radius a, 6, o. of tlie three circles tangeat to each other, t'> find the radius of a circle tangent to thethree circles. Formula.— r or r ^ ^ ..- , . — -j-j — ,-,_, , I nr^^ •2y[a6c(a-\-o-\-c) \ + ( ab-]-ac-\-6c) the minus sign giving the radius of a tangent circle circumscrib- ing the three given circles and the plus sign giving the radius of a tangent circle inclosed by the three given circles. Note. — This formula is due to I^rof. E. B. Seiti;, Late Professor of Mathe- matics in the North Missouri State Normal School, Kirksville , Mo., of whom we give a biographical sketch accompanied by his photograph. This fo' mula is taken from the School Visitor, Vol. II. p. 117, with the MENSURATION. 361 .'slight change that the plus sign is introduced for the case in which the ttangent circle is inclosed by the three given circles. The problem of finding !two circles tangent to three mutually tangent circles, is one supposed 'to have been proposed by Archimedes more than 2000 years ago, though the problem he proposed was not so general — the diameter of one of the ;given circles being equal to the sum of the diameters of the other two. The problem of finding all circles that can be drawn within three mu- tually tangent circles and tangent to each of them, has been simply and elegantly solved by D. H. Davison, Minonk, 111. The above formula led ihim to his wonderful solution. For a complete and elegant solution, where he has actually computed and constructed 81 circles tangent to three given circles, see 5c/ioo/ KjsiVof, Vol. VI.,j>.80. Prob. CLXVII. To find the surface common to two equal ■circular cylinders whose axes intersect at rigrht angrles. Formula. — S=16Ji^, where i? is the radius of the cylinders. Rule. — Multiply the square of the radius of the intersecting •cylinders by J6. I. If the radius of two equal circular cylinders, intersecting at right angles is 4 feet, what is the surface common to both ? By formula, 5=16/?= = 16x4^ =256 sq. ft. fl. 4 ft. = the radius of the cylinders. 11. 1 2. 16 sq. ft. =4^ =the square of the radius of the cylinders I 3. .'. 16X16 sq. ft. ^256 sq. ft. = the surface common to the •^ two cylinders. III. .-. 256 sq.ft. = the surface common to the two cylinders. Prob. CIjXVIII. To find the volume common to two equal •circular cylinders whose axes intersect at right angles. Fovmula. — V = \^ R^ , ^hete B. ii the radius of the <;ylinder. Rule. — Multiply the cube of the radius of the cylinders 1 6 7 ■ ]. A man digging a well 3 feet in diameter, came to a log 3 ieet in diameter lying directly across the entire ^well; what was the volume of the part of the log removed.'' By formula, F= ^jm = 1/(3)3 = jg cu. ft. 1. 8 ft. = the diameter of the log and the ■well. > .by L" II.<^ 2. l|^t.= the radius. 3. 3|-cu. ft. = (1|)3 = the cube of the radius. 4. .'. ^.j X 3f cu. ft. = 18 cu.ft. , the volume of the part of the log removed. . III. .■. The volume of the part of the log removed is 18 cu.ft. prob. CIiXIX. To find the height of an object on the earth's surface by knowing- its distance, the top of the ob- ject being visible above the horizon. 362 FINKEL'S SOLUTION BOOK. Let-ff^=:«be any object, AB=t a tangent to the earth's sur- face from the top of the object, and FE^=D \hs diameter of the eanh. .Then by Geometry, AB^=BF{BF-\-FE),ox P=a{cf -j-Z)). .•.«=— J-—. But « is very small as compared with the-. diameter of the earth jind ^^=^.y without appreciable error. .•. Formula. — a= — f. — =77, where c is the distance t6 the object from the point of observation. When c=l mile, g= = f ft., nearly. Rule. — Multiply the square of the distance, in miles hy\, and the result will be the height of the object in feet • ^ I. What is the height of a steeple whose top can be seen at a distance of 10 miles? piG. 65. c2 102 102 By formula, «=_=_^=__ x5280=f XlO^=66f ft. fl. 10 miles=the distance to the steeple. II. {2. 100=102=the square of the distance. I3. .-. f of 100= 66| ft.=the height of the steeple. III. .-. The height of the steeple is 66f ft. Prob. CLXX. To find the distance to an object by know- ing Its height, the top only of the object being visible above- the horizon. Forniula.-c==V^^^^=^a^^^^Vfa. Rule. — Multiply the height of the object in feet hy | and ex- . tract the square root of the product, and the result will be the dis- tancei in niiles. I. At what distance at sea can Mt. Aconcagua be seen, if its- height is known to be 24000 feet? By formula, c=V'fa=V'f x24000=v'36000=190 mi., nearly. fl. 24000 ft.=the height of the mountain II.] 2. 1x2 4000= 36000. (3. .-. V36000=10V360=190 mi., nearly. III. .-. Mt. Aconcagua can be seen at a distance of 190 miles. Prob. CLXXI. Given the sum of the hypotenuse and perpendicular, and the base, to find the perpendicular. . ^2 ^2 Formula. — ■/= -^ — , where .y is the sum of the hy- AS potenuse and perpendicular, and b the base. MENSURATION. 36& Rule. — l- Prom the square of the sum of the, hypotenuse and ptrt>.endicular subtract the square of the base, and'divide the difference by t-wice the sum of the hypotenuse and perpendicular. 2. To find the hypotenuse: , To the square of the sum of the hypotenuse an4 perpendicular, add the square of the basp and di- vide this sum by ttvice the sum of the hypotenuse and perpendicr ular. 1. A tree 120 feet high is broken off but not severed. The top strikes the ground 34' feet from, the foot of the tree; what is the height of the stump ? j3 l"- 1202—34^ [stump. By formula, /=— ^— -==— ^^^^^^=55H ft-' ^^^ height of the 1. 120 ft.^the sum of the hypoteriuse and perpendicular. 2. 34 ft.i^the base, or the distance the^top strikfes from the foot of the tree. II.J 3. 14400 sq. ft.^l20^^the square of said sum, 4. 1156 sq. ft.=34^^the square of the base, and 5. 14400 sq. ft.— 1156 sq. ft.=13244 sq. ft.=the difference. 6. -:..13244-v-(2Xl20)=55|^ ft.=the height of the stump. Ill, .'. The height of the stump is 55^^ feet. Note. — This rule is easily derived from an algebraic solution. Thus: Let »=the perpendicular, s — a=:the hypotenuse, and fc;the base. Then, »*+*2— (i— x)2, or x^-\-b'^=s'>—2sx-\-x^, and v= t ~ . Prob. CliXXII. To lihd. at what distance from the large end of the frustum of a right pyramid,a plane must be passed parallel to the base so that the two parts shall have equal solidities. SV Formula. — h=- . , -where V is the volume of the frustum, J3 the area of the lower base, Bj the area of the "dividing base," andv^^j the area of the mean base be- tween the "dividing base" and and lower base. Rule. — 1- Find the^volumeof the frustum by Prob. XCIII. 2. Find the dim.ensions of the "dividingbase" by extracting the cube root of half the sum of the cubes of the homologous dimensions of the upper and lower bases. Then find the area of the '■'■divid- ing base." S. Divide half the volume of the frustum by one-third of the :um of the areas of the lower base, '■^dividing base," and mean base between them, and the quotient will be the length of the lower part. I. Ho\Y far from the large end must a stick of timber, 20 feet long, 5 inches square at one end and 10 inches square at the other, be sawed in two parts, to have equal solidities? wt li'lNKEL'S SOLUTION BOOK. 3V By formu4a,.;4=^p^;^p^-^^^-g-^ 240(1024-10x5+52 ) +(^^T €<-+»x-J(l^)+^(i^)*] 42000 1680 2(100-H25<^^+-V-^6) 8+2^36+'J^6 1680 ^ 1680 ^83^6883+ii.. ~8+6J60385"5H-5.45l3618 20.0552168 Co«sz!r«c^iA>w.— Let ^.5 CZ>—^ be the piece of timber, ABCD the lower base^ EFGH the upper base, and OL the altitude. Prolong the edges AH, BE, CF, and DG and the altitude CZ till they meet in P. Draw KL to the middle point of AT), OI Tto the middle point, of GH and draw PIK. Let SMNR be the dividing base. 1. ^^=10 in.^i5, the side of the lower | base. 2. HE=h in.=c, the &ide of the upper base, and 3. 0L===f2!f) ft.=240 in.=a, the altitude. 4. KQ=KL-QL{=JO)=^{b—c)=^\{lQ in.— 5 in)=2^in. By similar triangles, 5. KQ: (il::KL:PL,ox\{b—c):a::\b:PL. Whence, 6. PZ=^=40ft. II. ab ac a=T -c 0- 7. .-. PO=PL—OL=-j- . ^ — c b — c =20 ft.' 8. t;=i/'Ox/Z^^=i«c^=iX240x5^ ^1^ gg ^2000 cu. in., the volume of the pyimnid HEFG. 9. V=:\OLyJyAB-^^ABy^HE^HE-^)=\a{b'^^bc _l_c2 ) =14000 cu. in., the volume of the frustum ABCD—E. 10. .-. I F=4 of 14000 cu. in.=7000cu. in., the volume of each part. 11. ■y+^F=2000 cu, in. +7000 cu. in.=9000 cu. in, the volume ot the pyramid, SMNR — /',.and 12. v-\- F=2000 cu. in. +14000 cu. in.=16000 cu. in., the volume of the pyramid ABCD — P By the princi- ble of similar solids, YAB^,ox 13. -HEFG-P ■■ SMNR-P : ABCD-P : : HE^ : SM-' : 14. v:v^\ V:v-\- SJiP:b». But MENSURATiaN 366, 15. v-i-i V=^[v-\-{v+ V)], i. e., v^^r, or SMNRPis. ac- arithmetical mean betwee© V an.fl '^^-^r V, or HEFCr"- —P&n-iAECD—P. 17. .-. 5vlf»=-|(c-''+33), i. e , BW is aa arithmetical meani between -«!E* and AE~', or c^ aadij^. Whence, 18. >S'.ilf=y[Kc''+^')]=y[i(5=+-lO'0,]M*'36= 8.2548188-l-iii. 19. 5J/2=y[^ ((^+^3) ]3=(|y 36) 2=^^6=68. 14202 sq. in=the area of the dividing base. 20 ^{S^P' X-4^2)=5.1/X -4^=1^36X10=25^36= 82.54818 sq. in.^the area of the mean base of the- part cut from the frustum. 21. .-. \LT{AB*^SMy,AB-\.SM'^)=lLT{b-^ +^|;i(^34.c3)]xi+,^B('5'+c^)]2)=fir[10== +|y(36)Xl0+(fy36),2]^Ji;r(100+82.54818 +68. 14202 ^ == iZ T^X 250. 6902 = Z rx 88.5634 = the- volume of the frustum^.5 CZ>—ikf .But \—M. 23. \ F=7000 cu. in.==the volume of the frustum ABCI> 24. .-. ZT'X 83. 5634=7000 cu. in. Whence, 25. Z 71=7000-^83.5634=83-76883 in.=6 ft. 11.76883 in.,, the leng^^th. III. .-.The stick must be cut in. two at a distance of 83.76883. in., or 6 ft. 11.76883 in., from the- large end. Note. — The frustum of a cone may be divided into two equal parts in the same manner. The'frustum of a. pyramid ora cone can be divided into, any number of equal parts on the same principle as that for dividing a trapezoid into n equal parts, Prob. CLXI. The above problem is one that,, like a wandering Jew, goes the rounds, of the country, and few teachers escape having it proposed to them for ' solution. By a careful study of the sojution here given, every teacher ought to be able not only to soJ,v% ^n4 explain, this problem but a great many others of a similar kind. I. Two poles, perpendicttlsir tQ the same plane, are 40 feet and 50 feet High. At what height from the plane will lines drawn from the top of one to the base of the other, cross? Construction. — I^et BC= 50 I feet^a, the height of the longer pole ; AD = 40 feet = b, the height of the shorter; AB^^d, the distance between the poles ; and AC and BD the lines dra^sn ixaxa. the top of .5 C to the base I AD and the top of AD to the I base BC, respectively. From F", \ the point of intersection of these piQ, ^{(a, two lines, let fall the perpead.icuila?-, FE, tfee, kogtli ol which is required. Then 366 FINKEL'S SOLUTION BOOK. II. 1. AB : BC = AE: FE, or d: a— AE: FE, by thte sim- ilar triangles ABC and AEF. 2. BA: AD=EB : FE, or d : b=d—AE : FE, by the sim- ilar triangles, BAD and BEF. 3. .". a'><.AE=^dy.FE, from the first proportion, and 4. by,{d — AE)^^iXFE, from the second proportion. 5. _-.« XAE=b{d—AE)=bd—bXAE; whence 6. AE^ ^'^ 7. .-. aX a^b bd 8. FE= a+^b ab -dy^FE, by substituting in third step. 50X40 a^-b 50 + 40" =22| feet. III. .'. 22| feet is the distance from the plane to the point ' where the lines cross. Remc^rk. — Observe that the result is independent of the distance the poles are apart. I. James Page has a circular garden lO rods in diameter. How many trees can be set in it so that no two shall "^be within 10 feet of each other and no tree within 2^ feet of the fence? Construction. — Let^^ C be the circular garden, ^ C it diameter, and O its, center. Then with O as a center and radius AO ■= \o\ {10Xl6i ft— 2X2^ ft), or 80 ft, discribe the c\xz\& abcdef , and in it describe the regular hexagon abcdef. Then aO =a^=80 ft. Begin at the center of the circle and put 8 trees 10 ft. apart on each radii, aO, bO, cO, dO. eO, undyO. Then joining these points by lines drawn parallel to the diame- ter of the circle as shown in the figure, their points of intersec- tion will mark the position of the trees. Hence, the trees are ar- ranged in hexagonal form about the center. The first hexagonal Tow^ contains 6 trees, the second, 12, the third 18, and so on. Since the radius of the circle on w^hich the trees are placed is 80 feet and the trees 10 feet apart, there will be 8 hexagonal rows. ^ FIG. 67. -1. 6=the number of trees, in the first hexagonal row. 2. 12=the number of trees in the second hexagonal row. 48=the number of trees in the eighth hexaeonal row. .-. 216=1^(6-1-48) X8=the number of trees in the eight hexagonal rows. 5. 24=6 X4=the^ number of trees at the sides of the hexa- gon abcdef. 6. .-. 216-1-244-1, the tree at the center ,=241=the number of trees that can be set in the garden. II.<^ MENSURATION. 367 JI. Then the triangle AFL is right- thc III. .-. There can be set in the garden, 241 trees. ( Greenleafs NaflArith., f. Ui^prob. Z^.) . I. There is a ball 12 feet in diameter on top of a pole 60 feet •high. On the ball stands a man whose eye is 6 feet above the ball; how much ground beneath the ball is invisible to him? Construction.— 'L^t BE be the pole, L the center of ball, and A the position of the man's eye^ Draw AFC tangent to the ball at F and draw LF and B C: angled at F. (1. 60 it.=BF, the length of pole. 2. 12 it.=ED, the diameter of the ball, and 3. 6 ft.=^Z', the height of the man's eye above the ball. 4. 12 ft.=^Z>+Z>Z=^Z. Now 5. AF=^{AL^—LF^) =V(122— 62)=6V3 ft. In the similar triangles ALF and ■ ACB, 6. AF:LF::AB: BC, or 6^/3 ft. :6 ft. ::.(6ft.+12ft. -f-60ft.), or78ft. : BC. 7. .-. ^C=(6X78)-^6V3=78 ^v/3=ix78V3==26V3 ft. fig. 68. 8. .-. 7r^C2 = 5r (26^3)2=6371.1498932 sq. ft.=the area of the circle over which the man can not see. III^ .-. 6371.1498932 sq. ft.=the area of the invisible ground beneath the ball. I. Three women own a ball of yarn 4 inches in diameter. Hov^r much of the diameter of the ball must each wind off, so that the may share equally? II. 10. 4 in.=the diameter of the ball. Then ■|-7r(4)'=\^;r=the volume of the ball. ■J of ^-j^7r=^-g?7r=each woman's share. '^TT — ^■j^;r=^^*7r=the volume of the ball after the first has unwound her share. But ^Tt Z)^=the volume of any sphere whose diameter is B>. .: \7tl)^=^-^7t. Whence, Z>3=«-g*;r-^^^=i|8^ and Z>==,j/i|8=4^|=|f '18=1X2.6207414=3.4943219 in., diameter of the ball after the first unwound her share. .-. 4 in. — 3.4943219 in=.5056781 in., what the diameter w^as reduced by the first w^oman. ■^Tt — ^^7t=^7t, the volume of the ball after the second had unwound her share. 568 FINKEL'S SOLUTION BOOK. 11. .-. y(V^-H'^)=4yi=|y9=|X2.0800837 =2. 5734448 in., the diameter of the ball after the sec- ond woman unwoud her share. t2. .: 3.4943219 in.— 2.5734448 in.=.7208771 in., what *he- diameter was reduced by the second woman. The diameter was diminished .5056781 in. by the first III • i woman, . ' 'I .7208771 in. by the second woman, and I 2.7734448 in. by the third woman. (Milne's Prac. Arith.,f. SS5, froh. 8.} NoTK. — The following are the formulas to divide a sphere into « equaL parts, the parts being concentric: is the diameter of the sphere; D^, the diameter after the first: part is taken off; Z>2,.the diameter after the second part is takea off; and soon. Then Z> — D^, D^ — Djj,&c, are portions of the diameter taken off by each part. I. A park 20 rods square is surrounded by a drive which con- tains -jij^ of the whole park; what is the width of the drive ? 20 rd. ==^Z>=Z)C, a side of the park. 400 sq, rd.=202=the area of the park AB CD. ,3. TW of 400 sq. rd.=76 sq. rd.=the area of the path. 4. 400 sq. rd.— 76 sq. rd-=324 sq. rd =the area ofthe square ^7^ G^. FIG. 5. ^i5^=V(324)=18 rd., the side of the square EFGH. 6. .-. 7^— -£'7?'=20rd.—18rd.=2rd., twice the width of the path. .7. .■• 1 rd.=| of 2 rd.=the width of the path. III. .-. The width of the path is 1 rod. I. My lot contains 135 sq. rd., and the breadth is to the length as 8 to 5 ; what is the width of a road which shall extend from, one corner half around the lot and occupy \ of the ground. Construction.— Let ABC D he: the lot, and DABSNR the road. Produce AB, till BE is equal to AD. Then AE is. equal to AB-\-AD. On AE, construct the square AEFG, and II. MENSURATION. 369 on .ff^and G^ respectively, lay off EJ and /^^ equal to AB. Then construct the rectangles BEIH, ILKF, and KMDG. They will each be equal to A B CD, for their lengths and widths are equal to the length and width of ABCD. Continue the road around the square. Then the area of the road around the square is four times thp area of the road DABSJVJi. II. 10. 11. 12. 13. 14. 15. 16. III. f =the width AB of the lot. Then f=the length .^^. 1x1=135 sq. rd., the area of the lot. iX|=i of 135 sq. rd. =27' sq. rd. , and |X|=(|)''.=^3 times 27 sq.rd.=81 sq. rd. .-. |=V81=9 rd.,tl}e width AjD, ^=i of 9 rd.=3 rd., and ^^5 times 3 rd.=al5 rd.,the length ^^. 15 rd+9 rd.=24 rd.= "e. AJS,the side of the square AEEG. .: 576 sq. rd.=242= the area of the square AEFG. 33f sq. rd.=;J of 135 sq. rd.^the area of the Toad DABSNR. ••. 135 sq. rd.^4x33f sq. rd.=the area of the road around the square. Then 576 sq. rd. — 135 sq. rd.^441 sq. rd., the area of the square iVOP^. .-. 21rd.=V441 ^NO, a side of the square NOPQ. AE—N0=2i rd.— 21 rd.=3rd.=twice the width of the road. .-. 1-^ rd.=24| ft.=| of 3 rd.=the width of the road. The width of the road is 24f ft. {B. H. . A.,p407,prob.99.) I. The length and breadth of a ceiling are as 6 to 5 ; if each dimension were one foot longer, the area would be 304 sq. ft. ;- what are the dimensions? - Construction. — Let ^^CZ> be the ceiling, ^^ its width and BC its length. Let AIGEhe the ceiling when each dimension "is increased one foot. On BC, lay off .ff.ff"equal to^j5 and draw ZA" parallel to ^^. Then ABKL is a square whose side is the width of the ceiling. 370 FINKEL'S SOLUTION BOOK. I G. 7; 8. 9. 10. 11. 12. 13. ilJ 14. 15. 16. 17. 18. 19. 20. 21, 22, 23. 24. ^=AB, the width of the ceil- ing. 'Then ^=BC, the length, and I X|=^-^X-^C=the area of the ceiling. «X1=-BCxjBA 5/ being 1 foot,=the area of the rect- angle £C^/. |X1=^CXC^, CJ^ being 1 foot,^the area of the rect- angle DCFB. 1 sq. ft.=l^^the area of the square CFGH. •■•|Xf+fXl+fXl+l sq.ft. =the area of AIGE. But pjQ_ ji |X 1+f Xl=V Xl=rectahgle^//rC+rectangle DCPE, i.e.yit equals a rectangle whose length is ( f+f ,, or V , and width 1 ft. ' ■■: |X|+fXl+iXl+l sq. ft.=|X*+VXl+l sq. ft. =the area of AIGE. But 304 sq. ft.=the aresioi AIGE. ■■■ f X|+V Xl+l sq. ft.=304sq. ft. Whence, |X|+VXl=303 sq. ft=the area of AIHCFE. But y^^^ Xf) in which y is ^ ft. ; for a rectangle whose length is y , and the width 1 ft, has the same area as a rectangle whose width is y ft. and length 1, ox\. ■•• IXf+y X|=303 sq. ft., in which U is y ft. iXH-HX|=50i sq 303 sq. ft, |X|+VX|=252i sq. ft.=5X(iX|+MXf)=5x50i sq. ft But |-X-|=the area of the square yl.5^Z, and hose length ft ^ of( V X|)=HXi=half" the rectangle AhNP= the rectangle OMNP, which put to the side AB of the square ABKL as in the figure. |X|+V X|=252^ sq. ft.=thearea of SRAOMK. '^ sq. ft.=(|^)2=theareaof the square i?^(9^, since AR is \\ ft .-. fxf+yxi-Hii sq' ft-=252i sq. ft+ifi sq. ft. =8^48i sq. ft=the area oi {SRAOMK^RQOA). =the area of SQMK: W-W £1.=^-^? i|-== W ft.=the side SK of the square SQMK. |=i-j-V ft.— H.ft.^VV it.^^bit.=SK—SB=^BK =AB, the width of the ceiling. 11 ft.=i of («xf+VXf)=i of y X-|=the area of the rectangle ALNP AL is I and width LN y MENSURATION. 371 |25;.i=^ of 15 ft.=3 ft., and [26. -1=6 times 3 ft.=18 ft.=^ C, the length of the ceiling. ..-.- ( 15 ft.^the width of the ceiling, and 111. .-. I ^g ft ^the length. Remari. — In this solution there is but one algebraic operation ; viz., extracting the square root of the trinomial expression, (|X5_|_i^. ^5_j_i.|i sq in.), in step 23. This might have been omitted and then«the solution would have been purely arithmeti- cal ; for, the area of the square SQMK being known, as shbwn by step 22, its side ^'/sfcould'have been found by simply extract- ing the square root of its area,S-f||i sq. ft. Then by subtracting .S^, which is 1^ ft, from 5Z", We would get BK{=AB), the width of the ceiling. The following solution is quite often given in the schoolroom: 304-=-(5x6) = 10+. V10 = 3+. 5 X3= 15, the width and 6 X 3=18, the length. I. A tin vessel, having a circular mouth 9 inches in diameter, a bottom 4^ inches in diameter, and a depth of 10 inches, is \ part full of water ; what is the' dianneter of a ball which can be put in and just be covered by the vsrater? Construction. — Let AB CD be a vertical section of the vessel, AB the top diameter, DC the bottom diameter, and SB' the al- titude. Produce AD, BC, and .ff^till they meet in G. Draw M'C parallel to EB. In the triangle A CB inscribe the largest circle IBP and let Q be its center. Draw the radius IQ. Now 1. AB^^AB^=R=A\ in.=the radius pf the mouth. 2. Ci''=^Z>C=7-^2^ in., the radius of the bottom, and 8. EF=a=\Q in., the altitude of the vessel. 4. MB=EB— BM{=BC}=B—r=4i i"- — 2-J- in. =2^ in. In the sirr)ilar triangles BMC and BBG, 5. MB:MC: -BB-.BG, or R—r -.a-.-.R-.EG. Whence, 6. EG=^^=4^^t=20 in., the altitude R — r i\ — 2i of the triangle A GB. 2/\AGB ABxBG 7. IQ= ~AB+A G+B G A-BJ^A G+B G 2Ra But AB+BG+BG- ,^^72. 8. AG=BG=^(BB^+BG^)=^[R^+(2ay]=J\(4iy +20^]=V(420i)=20|iri: q • rn_ 4ifg 4iX20 ... . ,.■" ^- • • ^^^2^+2V(^^-f a2)^4|-F20i^^* '"'■'^^^ '■^'*''*^ of the largest sphere that can be put in the vessel or in 372 FINKEL'S SOLUTION BOOK. lU the cone .<4 G£. 10. i^(/C)*=t^(^j J!2c =the volume of the largest sphere that can be ^)J -^' in the cone AQB. 11. i EGx^-SB^=i7t2aJi^=i7t X20x(4i)^=1357r, volume of the cone A GB. put the 12. [' ^TyprJ=l35^- -9099, [i?+V(^^+2. .. . the quantity of water in the cone which will just cover the largest ball that can be put in the cone A GB. 13. in'/?Gx^C2=i;rar2=^jrXlOX(21)2=ifA7r, the volume of the cone DGC. 14. .-. ^waz-^+i of the volume of the vessel = J-|4.;r-|-^ of the volume of the vessel=the quantity of water in the cone necessary to cover the required ball. But 15. \na{R^^Rr^r-^ )=iTlO[(4i)^+4ix2i+(2i)^] if^*, the volume of the vessel, by Prob. XCIII. 16. .". i^ar'^-\-i of the volume of the vessel =^>rar24"i **' 7C-\-\ of ^-^71 — ^\ ^7i, the q'uantity necessary to cover- the required ball. 17. .•. The quantity of water necessary to cover the largest ball: the quantity of water necessary to cover the required ball : : (radius)' of largtest ball : (radius)* of" required ball. Hence, 18. 19. 20. InlaR"^ n ]:i;ra[r^+i(7?2 4- [ie+V(^=+4a2)]8 /^>-+^-)]:: 0+V(X4.^ ))^-^^'- Y^y n : iffA^r : : (3f ) » : ZTO* . Whence, f y ( 337 ) : f y ( s/ ) : : 3| : HO. Whence, 21. HO. -^(^ R[r^J^\{ R^-{-Rr-^r^ ) ] la^R ■)=[l^( ¥)x3f] -=-|y337=9y(/T\), and 22. 18^(^)=6.1967-f-in., the diameter of the raquirei ball. III. .-. The diameter of the required ball is 6.1967+ in. L I have a garden in the form of an equilateral triangle wflose sides are 200 feet. At each corner stands a tower; the^ height of the first tower is 30 feet, the secona 40 feet, and the tnird 60 feet. At what distance from the base of each tower MENSURATION 373 Ttiust a ladder be placed, so that without moving it at the base it may just reach the top of each, and what is the length of the ladder ? Construction. — 'Let ABC he th& triangular garden and ^Z>, BE, and C^the towers at the corners. Connect the tops of the towers by the lines ED and DJP. From G and H, the middle points of DB and DF, draw G^and UN perpendicular to £>E and DP, and at Af and N draw perpendiculars to AB and A C in the triangle ABC, meeting at O. Ihen O is equally distant- from D and E. For, since M\% equally distant from D and E, and MO perpendicular to the plane ABED, every point of MO is equally distant frc \ D anAE. For a like reason, every point of NO is ^^g j^ equally distant frc.i D and E; hence, O their- point of intersec- tion, is equally di-.ant from D, E, and F and is, therefore, the point where the ladder must be placed. Draw Z^/and DJ par- allel to AB and AC, G^ and i¥Z perpendicular to ^^ and ^ C, MP perpendicular to A C, and OP parallel to NP Draw the lines OB, OC, and OA, the required distances from the base of the ladder to the bases of the towers. Draw EO, the length of the ladder. 1. AB=B C=A C==200 ft.=j, the side of the triangle. 2. EC=50 ft.=a, the height of the first tower, 3. EB=4:0 {t.=l>, the height of the second tower, and 4. An=m ft.=c, the height of the third tower. Let ^3 ft=the perpendicular from B to the side A C. 6. EI=BE—BI{=AD)=(d—c)=40 ft.— 30 ft. =10 ft. 7. GK=i(EB+AD}=i(6+c)=i {40 ft.+30ft.) =35 ft. In the similar triangles DJEand GKM, 8. DIJE : : GK: KM, or ^ : b~c : : i(b+c) : KM. A2 ,2 4.02 qri2 KMJ'-^^=^^^y-=\\ ft., 10. 11 12. 2^ 2X200 AM=AK-\-KM=^\s-\ *^ 52+^2 Is 2s 5^-|-^2_f2 j2_|_c2_32 =lQli ft, and BM=AB—AM= ^ _ 2s 2s ==98|^ ft. In like manner, HL=^a+c)=:=\{bO ft.+30 ft.)=40 ft, 374 FINKEL'S SOLUTION BOOK. M ii.<! IB. 13. Z7V=^^^^^4 ft, 14. AJV=A£-\-L^N=is-\ a 8_ 52_|_a2_c2 2s .Is j2^c2_a2 104 ft. 1,. NC=AC-AN=s. 2, - .2, - 96 ft. By similar triangles, 16. AB -.AL:: AM: AP, or s:^s: {s^-\-b"—c'- ) -f-25^ ■.AP. Whence, 17. ^P=(52-|-32_c2J_j_ 4^=^507 ft. 18- .■.PL=AI^AP=[^s—(s^-]-b^~c\}-iAs]= ( j2_^c2— ^2 )H-4.j=49^ ft. 19. PO=PJV=PL+LA^—(s^-{-c^—d^)~i-4s-{-(a^ _c2 )-i-2s=(s^+2a^—6^ — c2 )-e-4.y==53V ft. By similar triangles, 20. AB-.BL: ■.AM:MP,oxs:]i4Zs:'.{s-^-\-b-^—c'^)-^2s: MP. Whence, [lar triangles,, 21. J//'=[(52+^2_c2).i^.y]vi^^3j==50|y'3ft. Bysimi- 22. MP : AP ■.:RO: RM, or [ (52+32-^c2 ) -j- As^Z : (.52_|_32_c2 )_i^j : : (i2^2«2— 32— c2 )-i-4s : i?il/: 23. PM={s^+2a^—6^—c^)i^3s=[(s^+2a''—5^— c2)-j-12:r]V3=17HV3ft. Again 24. MP -.MA ■.-.RO: OM, or [(j2_(_32_c2)_^4j]y'-3 .- (524-32— c2)-^25: : ( j2.j_2aB_32— c2 )-i-4.j : OM.. 25. .-. Oil/=(52+2a2--32— c2)-^2V3.f=[ (^2^2^=— 32 — c2)-i-65]V3=35y\V3 ft. 26. ON=R P=MP—RM=[(s^+i^—c^ )_^5]v'3— (52+2a2_32— c2)-L.l25],y/3=[(j3_a2,..|.232— c2> -f-65]V3=33i V3 ft. Then 27. o c==v( Oiv^'^+iv^C2 )= Jr(£!=i^!±?^!z:£!v3)' ^2. ) J=V[(33iV3)^+96^]=V12516TV =111.87964'ft. 28. +(^^^=£=^)']=^L(33iV3)='+1042] =V14116TV=118.8111+ft. 20. 0B^( OM^+MB^ )= Jr C'"^^''ft~^'~'' v^l (c2_l_c2 AS>v2-i ^'-^ "-^ ^ ^2. y ]=V[(35TW3)^+(98i)M 2j =iV214657i=ll5.8278+ft. 1. O^=V(^^2+O.5^)=v'[(iv'214657i)2+402], =V(13416i+1600)=v'15016TV=122.6402+ft.=the length of the ladder. MENSURATION. 375 1. 111.8796-|-ft.^the distance from base of the ladder- to the base of the tower FC, 2. 1 18.81 ll-t-ft.=the distance from the base of the lad- III. .'.< <ler to' the base of the tower AD. '3. 115.8278+ft.=the distance from the base of the lad- der to the base of the tower BE, and ' [4.. 122.5402-t-ft.=the length of the ladder. { Greenleuf s NaiH Arith., f. J^H, prob. S8.) Remark. — When the sides of the triangle are unequal, pro- ceed in the sanie manner as above. In some cases the base of the ladder will fall without the triangle. I. At the extremities of the diameter of a circular garden stands two trees, one 20 feet high and the other 30 feet high. At what point on the circumference must a ladder be placed so that without moving it at the base it will reach to the top of each tree, the diameter of the garden being 40 feet. Construction. — Let ABC be the circular garden and A C its diameter, t.nd let Alf and CZ?be the two trees at the extremi ties of the diameter. Connect the tops of the tiees by the line B'D and from the middle point £ of FD let fall the the per- pendicular RH'. Draw EG perpendicular to PD. Then all points of jEG are equally distant from FD. At G draw BG perpendicular to AC. .Then all points of BG are equally dis- tant from F and D. Hence, B is the required point. 1. A C=2i?=^40ft., the diameter of the garden. 2. CD=a=iO ft, the height of the tree CD, and 3. AB'=:^b=2(j ft, the height of the tree AF. 4. DI=DC—CI{--=AB') = a—b =^4& ft.— 30 ft.=lO ft. 5. ^/y--= ^ ( CD-X-AF)=^ (a^b) =i(40 ft+30' ft ) = 35 ft. By similar triangles, 6. FI: ID : : EH: HG, or 2i? : a—b\ : i(a-\-b} : BG. Whence, 3 2 II. J 7. HG=- AR -=8|ft, GB=>j{BH~HG-^)=^[R^-Q^y^== =W23 ft. 4i? 9. .■.AB^^{AG^+GB^)=,^[{AH+nG)'''A-(GB'')'\^ 16i?2 "l=5V46ft. =34.91165 ft, nearly, and 376 FINKEL'S SOLUTION BOOK. III. 10. BC=^(GC^-\-GB^)=^\_(ji-^^y+ '-T^=^]=4V«2 ft.=11.31942 ft. \ 34.91165 ft. the distance from the smaller tree, and ■ 11- .31942 ft. the distance from the larger tree. I. Seven men bought a grindstone 5 feet in diameter ; what part of the diameter must each grind off so that they may share equally ? Construction. — Let AB be the diameter of the grind stone, O its center, and A O its radius. From A draw any indefinite line ^jVand on it lay offany convenient unit of length seven times, beginning at A. Let/' be the last point of division. Draw OP, and from the other points of division draw, lines parallel to OP, intersect- ing the radius A O, in the points f, e, d, c, b, and a. Then the radius is divided into seven equal parts. On radius ^0, as a diameter describe a semi-circumference A OK,anA a.ta,b, c, d, e, and/, erect per- pendiculars intersecting the semi-circu'^*'''»"ence in M, L, K, J, H, ana G. Then with /r/g_ 75. O as a center and radii equal the chords MO, LO, KO, lO, HO, and GO, describe the circles as shown in the figure. Then each man's share will be the area lying between the circumfer- ences of these circles. For, the chord GO^=Gf'^-\-fO'^ and, by a property of the circle, G/2=4/x/0. .-. GO^=Afy,fO+fO^, ={Af-\-fO)fO=AOxfO:=\A 02. In like manner HO^=A O XeO'=»AO'', KO^=AOxdO=^AO^, &c. 1. A B^D=5 ft., the diameter of the grind stone. ■2. A 0=R=2\ ft., the radius. ?>. .-. 7iP'^==7rxmy-=&\7r=the area of the stone. ^ ^ ;r=|47r=each man's share. |of7ri?2 6\yr — j^A I of 6^5, l-n- — |47r=i4jr=the area of the stone after the first has 47r/?2= -T¥' ground off his share. 6. .•.^(i|7r-^7r)=i\V'42=2.314554-ft., the radius MO. 7. 2(^0— AfO)=2(2ift.— tV42 ft.) =2(2^ rt.--2.31455 ft.)=:.3709 ft, part of the diameter the first grinds off. 8. Q^TT — I of6^7r=y/7r=the area after the second grinds off his share. MENSURATION. 377 9. 10. radius LO. 11. 12. 64? 14. 15. 16. 17. 18. ■•■ V(W'^-^^)=fV|=2.112875 ft., the Then / 2(yI/0— iC>)=2(5VT\— 4V'f)==2(2.3i45Gft."^ 2.112875 ft.)=.40335 ft., the part of the diameter the second grinds off. \n — ^ cf 6^;r=Y'7=the area after the third has ground off his share. ■• V(V'^-^'')=fVI-=5V'i=l-889822-f-ft., the radius KO. Then, 13. 2(ZC>— irO)=2(|v^— 5v'|)=2(2.112875.ft.— 1.8899822 tt)=.44ei()6 ft., the part of the diameter 11.^ the third grinds off 6^7r — ^ of 6^7r=|^|^»=:t^,ie area after the fourth has ground off his his share. •■■ •\/(B'^-r-'J')=-IVf'=i-636634ft.,theradius70. Then 2(KO—JO)=^'2.{b\lh — ;Vf)=2(1.889822ft.— 1.636634 ft.)=.506;!76 ft.,the part of the diameter the fourth grinds off. ^\7t — ^oi Q\-7t=\^Tf=ithe area after the fifth grinds off his share. ■■• V(28'^^'^)=fVf^= 1-336306 ft, the radius- HO. - Then 19. 2(/0— Z^O)=2(|Vf--|\/f)=2(1.636634 ft.- 1.336306 ft.)^.600f.5/i ft.,, the part of the diameter the fifth grinds off. 20. 6;^7r — f of 6^7r=|-|il-^=the area after the sixth grinds ofT his share. , 21. .-. Vdiir— 7r)=|«/^='.949911 ft, theradius GO. Then 22. 2(/rO—GO)=2(|\/f—fV|)=2( 1.336306 ft.— .944911 ft)=.782790 ft, the part of the diameter the sixth grinds of! . 23. 2X-944911ft=l.^>8»822 ft, the diameter of the part belonging to the seventh man. I. J. A. M., having a w<)olen ball 2 feet in diameter, bored a %ole 1 foot in diamer throuj'h the center. What is the vokime bored out? Construction. — 'Let ABCl^EF be a great circle of the ball -and let A CDF be a verticle section of the ■-auger hole. Draw the diameter BOB -and the radius AG. Then the volume bored out consists of a cylinder, of which A CDF is a vertical section, and two •spherical segments, of which ACB and FDB are vertical sections. 1. BE=1 .feet=2-ff, the radius of the ball, and ■7. ^0=1 foot=2r, the radius of the auger hole. . /r/g. jq 378 FINKEL'S SOLUTION BOOK. 3. iAJ^=0G—^/{A0^—AG^)=^(Ji^—r'-)=y3. 4. .-. AI^=2\'{Ii''—ry=j^3, the length of the cylinder. 11.^5. .'. F==;rr2X(V3)=i'^V3jl^'ie volnme of the cylinder, and 6. 2 V'=^2nBGX7rAG^-\-i7iBG^)=^[Ii—s/(J?^—r^)] 7. i7r(l— 1^3 )+M1-^V3)'=tW 16—9^3), the vol- ume ot the two spherical segments. 8. .-. F+2F-''=i7rV3+iV'^(16— 9V3)=i^(8— 3^3), =1.46809 cu. ft.=2536.85952 cu. in, III. .-. The volume bored out is 2536.85952 cu. in. I. What is the diameter of the largest circular ring that can be put in a cubical box whose edge is 1 foot? Construction. — \^&\.ABCD — ^be the cubical box. let /, K, L, M, N, and P, be the middle points ofthe edge CF, GF, GH, HA, AB , and BC respec- tively. Corinect these points by the lines KI, KL, LM, MN, NP, and 'PI. Theii IKLMNP is a regular hexagon, and the largest ring that can be put in the box will be the inscribed cri- cle of the hexagon. II. 8. ^^=12 in.=e, the edge of the cube. AN=AM=\AB= 6 in.=^e. .-. MN=ML=^MO ^yiAN-'+AAn) =\/{2AN-')= ■ ANy/1=Wle, the side of the hexagon, FIG. 77. MQ=\ML=^ of i-\/2e=iA/2e. Then the radius of the circle. .-. 20R^2 X (iV6e)=i'\/6c=^V6 X 12=6-v/6= 14.6969382 in., the diameter. III. .•. The diameter of the largest circular ring that can be- put in a cubical box whose edge is 1 foot, is 14.6969382 in. I. A fly takes the shortest route from a lower to the opposite upper corner of a room '18 feet long, 16 feet wide, and 8 feet high. Find the distance the fly travels and locate the point where the fly leaves the floor. MENSURATION. 379' FIG. 78. Construction.— 'Let FABE—D be the room, of which AB i& the length, AF the width, arid ADtVe. height; and let 7^ be the position of the fly, and C the op- posite upper corner to which it is to travel. Conceive the side ABCD to revolve about AB until it cOmes to a level with the floor and takes the position of | ABCD'. Then the shortest path of the fly is the diagonal FC of the rectangle FD' C'E, and P will be the point where the fly leaves the floor. , 1. AB^a=l?> ft., the length of the room, 2. AF=b=lQ ft, the width, and 3. AD=h=d> ft, the height. FD'=FA-\-A ^'=3+^=16 ft.+8 ft.=24 ft. Then Fa=V{E-D'^-\^D'C'^)=V[(l>+h)^-\-<^^}, =V'[(16+8)2-|-182]=30 feet, the length of the path of the fly. FD'-.D'.C -.-.AFAP, from the similar triangles C'D'F and PAF, or b-{-h -.a:: 6: AP- Whence, AP=^=^^^^^ I (>-\-/i 1d-|-o =12 feet, the distance from A to where the fly leaves the floor. fA. 11..^ P 15. III. 30 feet is the distance the fly travels, and [floor. 12 feet is the distance from A to where it leaves the ■■fiemark. — If we conceive the side BCHE to revolve about EHviU\.\\ it is. level with the floor, the path of the fly will be FC" and the length of this is \/l{a-\-hy^b-^\ But v'[('z+/0'' -|-52] ^ -\/[(3-(-^)^-j-a^], because, by expandingthe terms under the radicals, it will be seen that the terms aie the same, except lah and Ibh, and since a is greater than b, FC is less than FC" . When 3=6, FC=FC". I. How many acres are there in _ a square tract of land con- taining as many acres as there are boards in the fence inclosing it, if the boards are 11 feet long and the fence is 4 boards high? (side ) ^ 1- =^number>of acres in the tract, the side being cx- 160 pressed in rods. 2. 4Xl6^X'^^''<?fi=iiumber of feet in the perimeter of the field. 380 FINKEL'S SOLUTION BOOK. 3. .-. 4x1—^^ — 5^^^ n=number of boards in the fence inclosing the tract. ^" •'• 160 L 11 \=^4:Xstde. Whence, 5. {sidey=lQ0x2iXside=BSi0Xstde. 6. .-. stde=SS40 rods=12 miles. 7. .-. (3840) 2-7-1 60=92160=number of acres in the tract. Ill, .-. There are 92160 A. in the tract. {Milne's Pract. Arith.,p. S62,frob. 71.) SECOND SOLUTION. 1. 16=number of acres comprised between two panels of fence on opposite sides of the field. 2 1A.=43560 sq. ft. i 3. 16 A.=16X43560 sq. ft.=696960 sq. ft. 4. llft.^the width of this stripfcomprised between the two 11..^ panels. e. . . 12'mi. ==63360 ft.=698960-=-ll, the length of the strip, , which is the width of the fieltj. 6. 144 sq. mi.=( 12 )2=the area of the field. 7. 1 sq. mi.=640 A. 8. 144 sq. mi.=144x640A=92160A. III. .■ There are 92160 A. in the tract. Explanation — Since for every board in the fence there is an acre of land in the tract for 4 boards, or one panel of fence there would be 4 A. Now a panel on the opposite side of the field w^ould also indicate 4 A. Hence, between two panels on oppo- site sides of the field there would be comprised a tract 11 ft. wide and containing 8 A. But this would make boards on the other two sides of the field have no value. Now the boards on the other two sides having as much value as the boards on the first two sides, it follows that we must take twic^ the area of the rectangle included between two opposite panels for the area com- prised between two opposite -panels in the entire tract. Hence, between two opposite panels in the tract there are comprised 16 A. The length of this rectangle is 16X43560-f-1 1=63360 ft.=12 mi., which is the length of the side of the tract. In any problem of this kind, we may find the length of a side in miles, by multiplying the number of boards in the height of the fence by 33 and divitle the product by the length of a board, expressed in feet. MENSURATION. 381 THIRD SOLUTION. Construction. — Let ABCD be the square tract of land; 0, its. center; and EF, a panel of the fence Draw OI perpendicular to AB. Then, by the conditions of the problem, the area of the triangle, EOF, is 4 acres,= 640 sq. rds., or 174240 sq. ft. c 1. 11 = number of feet in the length of the panal, EF. 2. ^(^^X 0/)=t(llXO/) = the number of square feet in the area of the triangle, EOF. 3. 174240:=the number of square feet in the area of the trian- gle, EOF. ll.\ 4. .•.i(llXO/)=174240, and 5. 0/=:tV of 174240=31680=the Draw OE and OR FIG. 78a.. number of feet in the length of the altitude, 01, of the triangle EOF, which is half the length of a side of the field. 6. .-. 63360^the number, of feet ift the length of a side. 7. 63360 feet=12 miles. . 8. .'. 144 sq. mi. ^12^X1 sq. mi.=the area of the tract in sq. miles. 9. Isq. mi.=:640 A. Iff. 144 sq. mi.=144X640 A.=92160 A. III. .-. The tract contains 92160 acres. I. How many acres are there in a rectangular tract of land cbntaining as many acres as there are boards in the fence enclos- mg it, the fence being 6 boards high, the boards b^ing a rod long, and the length of the field being 1^ times its width ? Construction. — Let ABCD be the tract ; AB, its length ; BC, its breadth; O, its center; EF, a panel in its length; and GH, a panel in its width. Draw the lines OE, OF, OG, and OH. Draw 01 and OK per- pendicular to AB and AD, re- spectively. Then, since AB = 1^ times BC, KO, the half of AB,= 1\ times lO, the half of BC. Hence, the area of triangle, GOH,^=^\ times the area of triangle, EOF. Now, if the tract were a square, the area of the triangle, EOF, would=6 A., or, 960 sq. rds. But, since it is a rectangle whose length is IJ times its breadth, the area of FIG. 78b. 382 FINKEL'S SOLUTION BOOK. Hence, 1200 sq. rds. triangle, EOF,= .}^'t^X , of 960 sq. rds.=800 sq. rds the area of triangle, GO/f,^li times 800 sq. rds,: r 1. .-. i(EFX/O)=U^.X/O)—800 sq.'rds. 2. .-. 7(9=2X800 rds.— 1600 rds., and i;i. I 3. therefore, BC=S-200 rds. .4. .-. AB=-l^ times 3200 rds.=4800 rds. 'IS. .-. 3200X4800-^-160=96000 A., the area of the tract. III. .-. 9'3000 A.=the area of the tract. Pemark.^The above solution is derived from the algebraic solution of th« gener- alized problem. Thus, if the length of the tract is to the breadth as ;» ; », the algebraic solution shows that the area of the triangle, EOF,——^ of the same triangle, were the tr^ct a sqii'are. * I. A horse is" tied to one corner of a rectangular barn 40 feet long and 30 feet wide ; what is the area of the surface over which the horse can range if the rope with which he is tied, is 80 feet long? Construction. — Let ABCD be the barn; AB, its length; EC, its width; A, the corner to which the horse is tied; and AK, the length of the rope. With A as a Center and • a radius equal to AK,^= 80 feet, de- scribe f of a circumference, heginning at .^and ending at Q. (All of this is not shown in the figure as it would, make the figure unnecessarily large). With Z> as a center and DK,= 70 feet, describe the quadrant, KPE. With B as a center and BG,=50 feet, as a radius, describe the quad- rant, GPF. Then draw the lines, DP and BP, and pro- long AB to G, AD \.o K, BC to F, and .£)C" to...fi', Draw PH and /"/perpendicular to DE and BFi respectively, ' ' > 1. Then fof a circle whose radius is AK, or %nAK^,-\- ■ sector A7?/'-t-sector Gi?/*-!- triangle , D CP, + triangle , BCP, ^=^the, area over which the horse can range. But 2, f5-^A'2=|.7s802=48007r=the number of square feet in the area^of J. i^f the circle, radius AK. .■-"':8',_Arf? b%seqtorj KDP,=aTca of quadrant, KDE, — area ■of Sectdr,- Pi)E;^ixKD^— area of sector, PDE,— :|-7r50^-— area of sector, PDE,=^&2^n — area of sector, PDE. But 4. area of sector, PDE,=^PDXarc PE. 5. Arc, P£,=J(8Xchord, P£,— 2XPH), by Prob.XXV, Rule b. FIG. 78c. II ^ MENSURATION. 383 6. {DC-\-CHY-\-HP^=DP\ or DC^+WCXCH+CH^-i^, HP^—DP^. Also, 7. {BC+CIY-\-IP^=BP\ or BC^+2BCXCI+CP+IP'^ =BP\ 8. .-. DC^—BC^+2(DCXCH—BCXCI)=DP^—BP\ by subtracting step 6 from step 5. 9. . . CH=l i[(DP^—BP^)—{DC^-^BC^)]+BC X C/|- DC, by solving the equation in step 7, with resfject to C/f,= Ji[900— 700]+30XC/|h-40, by substituting the numbers for DP, BP, DC, and 5C,=2|+f C/. 10. .-. {BC-\-2^-\-lCIY+IP^{=CH^=\_2l+%CIY)=BP\ , by substituting for CH, in step 6, 2J-f-JC7, as found in step 8. 11. .-. (32J+fC/)2+(2i+fC/)2=402, or 12. 1056i+48|C/+TSV C/2+6i+3f C/+ t«j C/2=1600. 13. .-. |C72+52iC/=537J, or 14. C/2+46|C7=477^. 15. C/2+46fC/-f(23|)2=477^+(23J)2=1022f by making the first side of the last equation a perfect square. 16. . . C/+2?^=:d=Vl022|=±6§V23, o£C/=:6|V23— 23J. 17. .-. Ci/=2i+f(6|v23— 23JJ=5v28— 16. _ _ 18. HE=CE—CH=1 0-{ 5/23—1 5 ) =25— 5V28 =5(5— V23). 19. The chord, PE,=>^PH''+HE^=^CP+HE'^=^\21%Q!^— 561iV23 S =lf V i 996— 202v23 i . 2P. The arc, P£,=J(8Xchord, F£,— 2XPH.)=|{8X1| V|996— 20SV23J— 2( 6|V23— 23^)-] ft.=-V-[ VS996— 202 V28 j— v23+3|] ft.=17.773 ft., by Prob. XXV, Rule (b). 21. Area of sector, PBE,—\{kYC, P£ ,XPI))=25Xarc, PR, =444.325 sq. feet. 22. Area of sector, PBF,=\{2Xc, PF,XPB )=20Xarc, PF, =20X14.63=286.6 sq. ft, the arc, PF, being found in the same way as the arc, PE, was found. 23. Areaof triangle, Z>C/',=J[ DCXPH{—CI) ]=i[ SOX ( 6fv 23— 23^)1=207.624 sq. ft. 24. Areaof triangle,, BCP,=\\BCXPI {=CH)\=\\_ 40X 5( V23— 3 ) ]=179-.574 sq. ft. 25. •. Area over which the horse can range = fff80^+ (i^r S02— 444.325) + (^7:402^ 286.6 )+207.624+179.574= (5825::— 730.925 )=16956.093 sq. ft. Ill, ,". The horse can range over 16956.093 sq. ft. Note. — A solution of this problem is given by the author in the American Mathematical Monthly,. Vol. VII. A solution of the generalized problem was given hy Professor G. B. M. Zerr, in Vol. VIII, of the same journal.- When fhe. barn is a square, the solution is much shorter and simpler. The genjerjlized solution may be solved by the sarne method as pursued in the above problem. By using Rule (6) of Prob. XXV for ■finding the length of the arcs, PE and PF , a very close approximation can be secured if the arc is not greater than 30° or 40°. 384 FINKEL'S SOLUTION BOOK. I. What is the area of the largest square that can be inscribet'- in a semi-circle whose diameter is 10 feet ? FormuIa.—Sid& of square, j,=|V 5R. Rule. — Multiply the radius of the given circle, by f of the square root of 5 ; the result will he a side of the inscribed square. Construction. — Let ABDF be the semi-circle, and C, its center. With 5 as a center and AB, the di- . ameter of the given semi-circle, as radius, describe the quadrant AI. Draw BI perpendicular to AB. Draw ' CI. From D, let fall the perpendicular ~DH to AB and draw DF parallel to j AB. From F drop FG perpendicular to AB. Then GHDF is the required square. Proof. FIG. 78i. 1. C I= '^C B'^ -t- BP — >1CB^+ {2CB)^=\/ 5BC. 2.CD:CI=DH:IB, by the' similar triangles CDH and CIB, on 3. CB : {%CB=DH: 2CB, by substituting for CD, its equal, CB ; for CI, its equal, V bCS ; and for BI, its equal, 2BC. .■.DH=^^bBC. '^ _ CH= -slCDi—DH-^^>lBC^—iBC 2=^V bBC. GH=\>^ bBC. GH=DH,—FD=FG. ■. HDFG is a square, the sides being equal and per- pendicular to each other. ni. .-. The side, DH, of the inscribed square = fi/~5CB == |V6X10=4V5feet. I. What is the edge of the largest cube that can be inscribed in a hemisphere whose diameter is 12 feet? Formula.— e^y &R. Rule. — Multiply the radius by \ of the square root of 6. Construction. — Let KLM — N be the hemisphere and ABCD — Fihs. inscribed cube, vertices E, F, G, and H being in the curved surface of the hemisphere. Let O be the center of the hemisphere. Draw the radius, OF, also the Hne, OB^ NG^JUa. MENSURATION. 385 n. II. OB=\OD=^BC^+CD''-- the cube. :^V 2e, where e is the edge of II. 2. 0^^=>^OB'^+BF^=^lle^+e^=l>J Qe, or ^:^Qe=R. 4. .-. e, the edge of the cube,=JV 6i?=JV 6X6:=2v 6 feet. III. .-. The edge of the required cube is 2v 6 feet. I. How many stakes" can be driven down upon a space 15 feet square, allowing no two to be nearer each other than IJ ft.? A. By Rectangular Arrangement. '1. 13^15-5-lJ+l,^the number of stakes that can be^ut ' in each row. 2. 169=( 13 )', the number that can be placed in the given space. By Triangular - Rectangular Ar- rangement. 1. Place 13 stakes on the base line, AE. 2. Place 12 stakes in the second row, setting each of the twelve directly over the cen- ters of the twelve spaces in the base, as in the figure. 3. Then CD, the width of the strip between the first and second II. 6 F ycYYYYYYVvYVvl row,= V^ C2 —.AD'i= V(li) « ^"^ — Mli)P^V 3=1.0825+ ft. ^(^- ^Sf- 4. .-. IJ ft.— Iv 3 ft.=|(2-V 3)ft.=.1674+ft., the gain in width, by using strips |V 3 ft. wide, instead of IJ ft. wide. 5. .-. To gain l\ ft., we must take \\ ft.-i-|(2— vl) ft., or 2^2+V 3) strips,=7.46+ strips. 6. .•. In 7 strips, \\ ft. wide, we can have 8 strips |V~3 ft. wide. ' 7. .". On these 8 strips we can place 9 rows of stakes, — 6 rows of 13 each, and 4 rows of 12 each, or 113 stakes in all. 8. On the remaining 6 strips, IJ ft. wide, we can place 5 rows of 13 stakes in a row, or 65 stakes. 9. .-. 113 stakes+65 stakes=178 stakes. C. By Isosceles- Triangle Method. '1. Place 12 stakes on the base line, -^.fi", placing the first stake at A, and the remaining ll in such a way that the 12 stakes occupy 11^ equal spaces. 386 FINKEL'S SOLUTION BOOK. n.^ 2. Begin at /, and place a row of 12 stakes exactly as the first row was placed. - G F 3. Begin at H, and place a third row of 12 stakes exactly as in the last, and so on. 4. By this method we can place 15 rows of 12 stakes in a row, in all, 180 stakes. This is pos- sible; for 5. AB=\b ft.^ll^— Vj ft. CD=^AC 2 -^Z>2=V(li) 2- ADB III. [MlA)]^=^385ft. . • . 14 X AV 385 ft. , = IfV 386 ft. _ „ . = 14.92+ ft., which is less FlG.ySg. than the side of the given square. Hence, we can have 15 rows of 12 stakes in a row, or 180 stakes. , 169 stakes can be placed on the square by rectangular arrangement; 178, by triangular - rectangular arrange- ment; and 180, by isosfceles-triangle arrangement. I. Two poles, AD and BC, are a certain distance apart; from the top of ^Cto the foot of AC is 100 feet, and from the top of AC to the foot oiBC is 70 feet ; and from the point where these two lines cross to the plane is 20 feet. What is the height of each pole, and how far apart are they? I,et ^C=100 feet=a ; BD=^ 80 feei:=i5 ; 7=^£•— 20 feet=t; BC \ and AD=y. Then II. 1. 2. FE=^=- — ^, from x-\-y problem, page 365. ex ■'■J'= ■ x—c = AB\ -AD^ = 4. 5. 6. ~y A III. AC^ — BC^ ■■ and BD'^—AD'^ = FIG. 78h ABK .-. AC^~BC^=BD^—AD^ , or a^—x^=6^- .-. x*—2,cx^—(a^—b^)x^ + 2c{a^~b'^)x—{a^—b^)c'^ = 0, by substittiting for y in step 5, its value in step 2. •. .;(;*— 40j;S—3600;ir2+144000;t;— 1440000=0; whence jt=66.473+ feet, by solving the last equation by Hor- nei^'s method. The remaining quantities are now easily found. The height of the longer pole is 66.473+ feet.; etc. MENSURATION. 387 1, How many acres in a circular tract of land, containing as ■many acres as there are boards in the fence inclosing it, the fence being 5 boards high, the boards 8 feet long, and bending to the arc of a circle? ' .".■,-, Cohsiruction. — Let C be the center of of the circular tract, AB =A C=S, the radius, and the arc AB=S feet. Then the area of the sector is-5 A.=217800 sq. ft. 1. 5A.=5><;43560 sq. ft.=217800 sq. ft, the area of the sector A^C. ■. 2. ^(4^X..4C').— ■^(8X^C)=44C==area of the sector ABC. '. Ii:>3. .-. 4^ C=2178O0sq. ft. Whence, ■4. A C=2'17800-^4=.54450 ft.=3300 rods, the radius of the circle.. ■. . ' 5. - :?r 'x('3300)2Vl60=68062.57r = number of acres in tratt. ■;' il. .■' Theiearre 68062.5 TT A., in the tract. I..' What is the' length of a thread wrapped spirally around a •cylinder 40 feet high and 2 feet in diameter, the thread passing -around 10 times? '^ ! •1. %7i fi.==ABCA (iJ^i^. 79), the circumference of the cyl- inder, :■ , ,, ' IIA2. 4'ft.=40 ft.-^10^=^i'', the distance between the spires. 3. 'v/[(27r)2-|-42]=2v'[;r2-|-4] i\..=AEF, the length of one spire. 4. .-. 10x2V['!^^+4]ft. ==20V[7r==+4] ft.=74.4838. ft., the entire length of the thread. III. . . The entire length of the thread=74.4838 ft. Remark. — Each spire is equivalent to the hypotenuse of a right angled triangle whose base is ABC A and altitude AF. This may be clearly shown by covering a cylinder with paper and tracing the position of the thread upon it. Then cut the paper along the line AFK z.i\A spread it upon a plane surface. AEF will then be seen to be the hypotenuse of a right-angled triangle whose base is A CBA and altitude AF. I. A thread passes spirally around a cylinder 10 feet high and 1 foot in diameter. How far will a mouse travel in unwind- ing the thread if the distance between the coils is 1 foot? Construction. — Let A CB — K\ie. a portion df the cylinder and ADBFGK a. portioa of the thread. Let .4 be the position of the mouse when the unwinding begins, P its position at any tirne afterwards, jiyTV^ a portion of the path it describes, and PD the portion of the thread unwound. Draw Z?C parallel ^o HB and draw OZ) and OC. Then FINKEL'S SOLUTION BOOK. IIJ diameter Let 1. AB=1R=\ foot, the 2. 0=10 ft, the altitude. 3. 6'=the angle ^OC, 4. s^AN, the length of a por- tion of the curve, 5. x=OL,- and 6. y=PL. Then 7. PC=a.rcAC=R9, 8. GM=R cos (9, 9. ML=I P=C P co% L CPl '^Rd cos (i;r— I PCF) =R sin IP CI=RB sin ^. OM-\-ML=R cos e+Rff of the cylinder. 10. 11. y==Plf=IM=CM—CI sin ^, and III. =7? sin fii— Ci' cos ^== i?sin^— i?l9cos^. dx=Re cos ddd, by differ- entiating in 10, dy^RSsxnede, by differ- entiating in 11. Now s=j'y/\dx-^-\-dy'^\ .: s=h{Rdcos»^ddy-{- (ROsinff dey]?==R fffdff =iR0^. But "^ 6=2Jt, when one spire is unwound, and 5=10X2a'=20;r, when the unwinding- is coir.plete. .-. 5=i-ff92=ixK20'r)='=100;r2 =989.96044 ft, the distance the mouse travels to unwind the thread. The mouse will travel 989.96044 ft. to unwind the thread. I. What is the length of a thread winding spirally round a. cone, whose radius is R and altitude a, the thread passing round n times and intersecting the slant height at equal distances apart ? Let Phe any p(oint of the thread, (x,y, z,) the co-ordinates of the point jand, let the angle PFC'{=/^OQ)=0, BO— a, the alti- tude, Z>0=itr, the radius of the base of the cone, and r=the radius of the cone at the point P. Then the equations of the thread are : x^^r cos^ (l),y=r sin5 ....'. (2), and z= Inn e ... (Z). From the similar triangles DEP and DOBy r=— (« — z)=R{\ — - — J... (4). Now the distance between P and its consecutive position is ^ (dz'''-\-dx^ -\-dy^)^=^ MENSURATION. 389 ^^^V~l<f^ (5). Substituting the value of r in (1) ^nd (2), and differentiating, we Jiave dx= — =; — 1 cos^4- 2.7rn\— i (2ftn—d) siaO~\de and dy= — — fsinfi'— (2 nn—d)cos0~\iie. From (3), we have d'z=>^ — dff. Substituting =these values of dx, dy, AnA dz\a (B)", we have s= Jo .2nrnHi;J,, • FIG. 80. <27r«-e)2]-i?('^^±P)log, [(27r«-^)+V((2'r«-e)2 + i?2 =iV^('^^+4?r2«27?2). /^^ imiR ^^'L h ^J' where h=\/ {a'^^R'^)., the slant height. Note. — This solution was prepared for the School Visitor, by the author. I. A thread makes n equidistant spiral turns around a cone whose slant height is h, an'd radius of the base r. The cone stands on a horizontal plane and the string is unwound with the lower end in contact with the plane, the part unwound being always tense. Find the length of the trace of the end of the iString on the plane. Let MHhe the part unwound at any time, Z^ being the point in contact with the cone, and BM^=u, the trace on the plane up to this time. Put arc BE^x, AH=y, E being the point in the circumference of the base in the line AH. Let iV/be the posi- 390 FINKEL'S SOLUTION BOOK. tion of the string at the next instant, D and /being homologous, points with E and H. Draw /T/ifparalkl to ED. Then h v BE: : AK : HK,6\:^^=~ (1). Now since the arc BE~ ED h ^ =^Xy is proportional to- the distance the point of contact of the thread with the cone has ascended, ;!£■: A — y..ViTtrn.h.,ox-^ — dx Inrn (2). 'Irmt h This is negative since ^y decreases as x increases. It is evi- j .. X- ^-i. -n ii_ ^^^ 'i^ Inrn. dent from the figure that , .^ — i — ; — ^ . IK. , dy h By similar triangles, IK:HK:: HE'.ME, /, ", . ^ > ,^ . ■•'■ , .„., jvIE siK that IS, form (1) and (2), we get- 271 rn h—y IK ..(3). ; ED- • y dx y. ' ^iK^k^dp^h^. P Therefore, ME=.-^^{h—y)y (4) Put ME=t. Then dt Inrn {k-2y) dy h^ . (5). Py similar figures r:ME: '.ED: MP^ MExED , __ 27rrn ^,, -ME X — r- X IK. FIG. 81. From (3), put MP=^, then^^^=J^i/.-y) ^ .... (6). Equation (5) gives the entire addition to the line AfE which.. consists of NP-\-FD, since EE=ME. Consequently, JVP' dt dx 2nrn , ., „ ^ , 2Ttrn Anrn -(k—2y)- -y (7). Nowil/iV^2. dy dy h^ "■" "" ^ h h'- ^.MP'^^NP^ in the limit. Therefore ^^^^16^'^^^^^ h^ (l+^(^-y)0--- (8)- V(8)=(9), J,=^-^X dy h-' .^( 1+-Tg— (/2— j')M, the integral, of which is u, the length of the trace. Put ,4 — y==2:, and— r — ^=a^. Then 2.!=-—- / (h — z) V(.- + .^M.....(10). Or.=^+(|^-g)v(.^+/.^) + 2.1og.[^±Vl^^]. (11). Write for /«, its equal^ . I4ENSURATION. 391' amt. in Cll) and we have (12), u=- ■ + -7r( nrc J Vil + n^n^j + ^r log,[«?r+V(l + «^'r2)]. This result is independent of ^, the cone's slant height, but involves w the number of turns of the thread. Note. — This solution is by Prof. Henry Gunder and is taken from the School Visitor. Vol. 9, p. 199. Prof. Gunder stands in the very front rank ">f Ohio mathematicians. He has contributed some very fine solutions to Jifficult problems proposed in the Scliool Visitor and the Mathematical Messenger. He is of a very retiring disposition and does not make any pre- tentions as a mathematician. But that he possesses superior ability along that line, his solutions to difficult problems will attest. Prof. Gunder was born at Arcanum, O , Sept. 15th, 1837. He passed his boyhood on a farm and it was while following a plow or chopping winter wood, that difficult problems were solved and hitherto unknown fields of thought explored. He became Principal of the Greenville High School in 1867. After seven years' work here, he became Superintendent of the Public Schools of North Man- chester, Ind. After five years' work at this place he became Superintend- ent of schools of New Castle, Ind. In 1890, Prof. Gunder was elected pro- fessor of Pedagogy in the Findlay (Ohio) College. I. A woman printed 10 lbs. of butter in the shape of a right cone whose base is 8 inches and altitude 10 inches. Having com- pany for dinner, she cut offa piece parallel to the altitude and con- taining .|^ of the diameter. What was the weight of the part cut off? Construction. — Let .4.5 C — G be the cone, ^C the diameter and OG the altitude. Let S be the point where the cutting plane intersected the the di- ameter, F the corresponding point in the slant height, and DLFKB the section formed by the intersection of the cone and the cutting plane. Through F pass a plane parallel to the base ABC&rvA anywhere between this plane and the base, pass a plane NLMK. Then, FIG. 82. 1. A C=2i?=8in., the diameter of the base, 2. OG=a=10 in., the altitude, and 3. OE=OC—B C=R~^A C=R—lR=JiR=\^ in.=c,the distance of the cutting plane from the altitude. Let 4 GQ^x, the distance of the plane MLNK from the vertex G. Bv similar triangles, 5. 0C:OG::BC:EF, or R:a::R~-c:BF. Whence, 6 EF=— — — — ^^6f in. By similar triangles, 7. GO:OC::GQ:QM, or a-.R-.-.x-.QM. Whence, Rx 8. qM=Lq=-~. Now, 392 FINKEL'S SOLUTION BOOK. 9. area of LKM=area of LQKM—area of LKq. But 10. area of LQKM=l(^LQ^ cos-^-^)=C— V x s-(^^, and It. RxJ' 11. areaoiLKQ=l{LKxQI)=U2LIXc)=LJXc= - cV(■N^I•>iJM)=c's/URx-^a + c)X{-Rx-i-a—c)^= 12. .;. ^re« of the segment Z^Af=:?--^--cos-ir—^— ment of volume of the part cut off. 14. .-. V. ' R i I /'R^x^ . / ac \ c '\ lac =ia j /?2cos-'(l)-2cv/(7?=!-c2) +^X liV^[4'-( H)^] + .VX421og.[i±Vli!=(li)!]] . =y j 4«cos-'(i)-VV2 + iflog«[2+fV2] I , =y j 19-6938154— 10.0562976 + .6396202 I = 34.223792 cu. in., the volume of the part cut off. 15. ^a;ri?^=^Xl0x4^X»=53^Br cu. in., the volume of the whole cotie. 16. 10 lbs. ^the weight of the whole cone. Hence, by pro- portion, 17. 53i7rcu. in.: 34.223792 cu. in.: :10-lbs.:(?=2.04258 lbs.) III. .-. The weight of the part cut off is 2.04258- lbs. I. After making a circular excavation 10 feet deep and 6 feet in diameter, it was found necessary to move the center 3 feet to one side; the new excavation being made in the form of a right cone having its base 6 feet in diameter and its apex in the surface of the ground' Reqired the total amount of earth removed. MENSURATION. 393 Construction.— Let ABC—.F be the cylindrical excavation 'first made, A C the diame- ter, HO the altitude. Let A be the center of the con- ical excavation, GAH its diameter, and AP, an ele- ment of the cylinder, the altitude. Pass a plane at a distance x from O and parallel to the base of the excavation. Let figure II. represent the section thus formed, the letters in this section corresponding to f/g_ gg the homologous points in the base represented by the same let- ters in the base of the excavation. An element of the earth re- moved in the conical excavation is (area £AKGNB)dx. The •whole volume removed in the conical part of the excavation is C"'{areaBAKGNB)dx. For let HO = a=\Q ft., the altitude of the excavation, HA^=r=2, ft., the radius of the cylindrical and the conical parts. AB-=AN^= — . This is found from the proportion of similar triangles. BI'^^{rx^a—AI){rx-^a-\-AI). Also BI'>'=^\ir—AI)AI. .-. {2r—AI)AI={rx-T-a—AI)(rx-^a-\-AI). Whence, 9/ 10. 11. 12. 13. 14. 15. HI=r -^2), 5='-(l- «)■ Now 23^ ^" 2a2^ area of BDAKGNB=2{ area of BDAN-\-area of NAG). But ^7t(r^x^-i-a'^)= the area of the quadrant JVAG, and area of BDAN=area of sector BAN-\-area of trian- gle HBA — area of sector BDAH. Now a^-ea of sector BAN=.\ABy^AB %\i\-^ [AI-^AB) = (r2^2H-2a2)sin-i(^), area of triangle ABH=\{AIIxB I)=\rX{rx-i-2a'^ ) Xv'(4a2— :v2)=(r2x-^4a2)v^(4a2— x2), and area of sector BDAH=\[AHxAHcoi-^ {HI^BH)\ =\r'^ X cos-i [1— (;f^2a)]. 394 FINKEL'S SOLUTION BOOK ^^•)16. .-. Area o'i BDAKGNB=2 \ i -— ;r4-li^sin-' -r^cos-x(l-ij). =the volume of the conical part of the exca- vation. 18. 7rar2==the volume of the cylindrical part. 1Q ^ 2j /^64-27V3— 27r\ , x-64— 27V3+16;r>v 19. .-. nar-J^{^ )ar-={^ 18^— > «>-^=±=337.500554 cu. ft., the volume of the entire ex- cavation. III. .'. The volume of the excavation^! ^ ' Jar^,. or 337-50055-{-cu. ft., correct to the last decimal place. Note. — This problem was proposed in the School Visitor by Wayland? Dowling, Rome Center, Mich. A solution of the problem, by Henry Guil- der, was published in Vol. 9, No. 6, p. 121. The solution thpre given is by polar coordinates. The editor gives the answers obtained bv the contribu- tors; viz., Mr. Dowling, H.A. Wood, R. A. Leisy, and William Hoover. Their answers differ from Mr. Gunder's and from each other. Mr. Gun- der's answer is 337.5+cu. ft., the same as above. There is a similar problem, in Todhunter's InUoral Calculus, f. 190,prob. S9. I. A tree 74 feet high, standing perpendicularly, on a hill- side, was broken by the wind but not severed, and the top fell di- rectly down the hill, striking the ground 34 feet from the root of the tree, the horizontal distance from the root to the broken part being 18 feet, find the height of the stub. Construction. — Let AD be the hill-side, A£ the stump, BD MENSURATION. , 395 the broken part, and A C the horizontal line from the root of the tree to the broken part. Produce ^^ to E and draw DE paral- lel to A C. « 1. Let AB=x, the height of the stump. Then 2. BD=n4: ft. — x^s — X, the broken part, since AB-\-BI^ =74 feet. 3. Let AU^^a;^&A ft., the distance from the foot of the tree to where the top struck the ground, 4. v4 C=^=18 ft., the horizontal .distance from the foot of the tree to the broken part. 5. x^AB, the height of the stump. Then 6. ^C=V(^^'+-4C2)=V(*''+'5')- • (!)■ Inthesim- lar triaijgles BA C and BED, 7. \/{x''-\-h^ ):x:: s—x : BE. Whence, «• ^^=#^ ■ • ■ •(2)- ^-'^°' 9. V(»;^+*^ ) -b ■■ -s—x : DE. Whence 11. AE=BE—BA= ''/l ^ "J^ . -X (4). 12. AE^-{-ED^=A£>^, or 13 \Afl=^-J'+\^if=^{ FIG. 84. z^a^ . . . '(S). Developing (5), we have 2a2j2+8^2j2_4a232 )^2_4^2j(^2_a2 )^^ _32(.52_a2) (6). 1.5. 1161x*— 91908^3+1959876a:2— 2.58!)4080;c+ 377913600 =0 (7), by substituting the values of a, b, and .r in (6). 16. .-. X =2^ feet, the height of the stump, by solving (7) by Horner's method. III. .-. The height of the stump is 24 feet. Note. — This problem was taken from the Mathematical Magazine, Vol. I., No. 7, prob. 84. In Vol I., p. 184, of the Mathematical Magazine is a so- lution of it, given by C. H. Scharar and Prof. J. F. W. ShefFer The solu- tion there given is different from the one above. I. What is the longest strip of carpet one yard wide that can be laid diagonally in a room 30 feet long and 20 feet wide ? Construction. — Let ABCD represent the xoo'crv axiAEFGH the atrip of carpet one yard wide placed diagonally in the room. 396 FINKEL'S SOLUTION BOOK. II. ■ 1. Let AB=a=30 ft., the length of the room, 2. BC=i=20 ft., the width, and 3. HG=c=2, ft., the width of the carpet. Let 4. BF==HD=x. Then • 6. B'C=AH=20—x=b—x. 7. .AE=GC=AB—BB=a^^{c^—x^) (2). By similar triangles, 8. EF-.BF: : GF: GC, or 9. c:x:: GF : a~^sj{c'^- Whence, 10. GF.^^=^z:^^. . . X Again, we have 11. EF:BE::GF:FC, or 12.. <;:V(c2— «;2 ) : : GF: h—x. c(b x\ ^ '- (4). By equating G^in(3)and(4) FIG. 85. 13. .-.LrF^ (5). u' ■ ^C*^— ^) _ g[^— V(g'— ^' )] ■ tKc'^—x^) X 15. hx—x'^=:=a>^{f—x'^)—c-^-\-x-^ ...(6), by dividing (5) by •' c and clearing of fractions. 16.' cs — bx — 2x2=aY/(c2_;ic2) . . .(7), by transposing in(6). 17. 4A:*-:-4^x8+(a2_|_^2_4c2) x'^-\-1bc'^x=c'^{a'^—c^). . . ,. . ..(8), by squaring (7) and transposing and com- bining. 18. 4^*— 80^8 +1264^2 _|.360x=8019 . . . , (9), by restoring numbers in (8). 19. .-. A;=.2.5571+ft., by solving (9) by Horner's method.. 20. .•.V'(c'— ^')=V(9— ^^)=l-5689ft. Then, 21. GC=30— ^(9— «2)=28.4311 ft, and 22 FC 20 ^^17.4429 ft. 23! .-. GA=V(^C'2+GC2)=V[(28.4311)8+(17.4229)n =33.3554 ft., the length of the carpet. III. .-. The length of the strip of carpet is 33.3554 ft. I. What length of rope, fastened to a point in the circumfer- ence of a circular field whose area is one acre, will allow a horse to graze upon just one acre outside the field ? Construction. — Let ABPC be the circular field and P the point in the circumference to which the horse is fastened. Let £P represent the length of the required rope. Draw the radius BO oi the field and the line B C. Then 1. 1 A.=160 sq. rd.=the area of the field ^^T'C, and 2. ^O=O/'=i?=v'(160-^7r)=4j(.^Y the radius of II. MENSURATION. 397 the circular field. Let 3. ^=the angle BPO=the angle OBP. Hence, 4. ;r—2/^=the angle ^O/'. Now 5. BP^AP co% I APB=2Rc.oid , the length of the required rope. The 6. area BPCD over which the horse grazesii^area BE CDB—areaBB CPB. But 7. area of circle BECD= 7tBP^=TtAR^cos^d= AnR'^cos'^d, and the 8. area BE'CP^2X{area of -sector EPB~^area of seg- ment BPH). Now 9. area of sector £7'^=^^/' X arc BE=\X^Rco^ X 2Pcos9 X 0=2Ji^ffcos^9, and 10. area of Segment ^/'^=area of sector BOP — area oi triangle OBP==\BOXarcBHP^\OPxBF= \lRXR{Tt—W)'\—\RXR&\vi{7i—ie), =iR^.(7i:—2e)--iR^sin2(f. 11. .-. Area BECP=2[2R^ecos^O+iR^(7C—20)— iR^sin 2f)]==R\[4ecos^ e+TT— 2d— sin20 ]== R^[40(^ ^+P^^ ^+^-2e—sm20-\=R%,t+ 2^cos2i9— siH22(9]. 12. .-. AreaBPCI)B=4^7tR2cos^&—R^[7C-}-2ecos2d— sin26/]. But 13. 7ri?2=lA.=160sq. rd.=the areaof^/'CZ>i?, by the conditions of the problem. . 14. .-. 47cR^cos^e—R^[7r-{-20cos2ff—sin2e]=yTR^. Whence, 15. 4»'('i±^^^— [7r+26'cos2^— sin2(9]=;r, or 16. 2;r4-2»cos25 — TT— 2(9 cos26'+sin2 S=;r. 17. .-. 2 6^0082 e—sin2S=2;rcos2^, or 18. 26 — tan2^=2w, by dividing by cos2^. Whence, 19. ^=51° 16' 24", by solving the last equation by the method of Double Position. 20. .-. -5/'=27?cos(9=8.J(^^\os'-'=8.92926+rods. III. .-. The length of the rope is 8.92926+rods. I. If a 2-inch auger hole be bored diagonally through a 4-inch cube, what will be the volume bored out, the axis of the auger hole coinciding with the diagonal of the cube? .' Formula. — F=r2y'3( ne — 2rv'2), where e is the edge. 398 FINKEL'S SOLUTION BOOK. Coustruction. — Let AFGD be the cube and -Z?7^ the diagonal,^ which is also the axis of the auger hole. The volume bored out •will consist of two equal tetrahedrons acd — D and efg — F plus the cylinder acd—f, minus 6 cylindrical ungulas each equal to ace — b. Pass a plane any where between e and ^, perpendicular to the axis of the cylinder, and let x be the distance the plane is from D. Now let II. 1. AB=e=A^ inches, the edge of the cube ; 2. Z>7^=-\/3i==4V3, the diagonal of the cube; and 3. r==l inch, the .ra- . -1 dius of the a:uger, or the radius of the circle acd.. 4. ac=ad= dc = rsjZ =V3, 5. Z»c = ^ry6 = iV6, by the similar tri- angles dDc and HDc. 6. sl{JDc'i—r^)=. V[(|^v'6)2-r2]= |rV2=iV2, the al- titude of the tetra- hedron acd — D. piQ g-[ 7. .-. 2w= Ifarea of base X altitude) =2(^V3X ac^xi^X |rV2)==4v'6r3=jY'6. '^^^ volume of the two tetrahe- drons, 8. 1/^=71 r'^y^{BF-\-2 times the altitude oiacd—D) = 7Tr^(e\/^—ir\/2)=7r{i\/3—i\/2), the volume of the cylinder acd-^—f. . J 6e=^r'\/2, by similar triangles, not shown in the figure. ^rV'2-(-iry'2=/-V'2=distanee from Z> to where the auger begins to cut an entire circle. ^ — ^X'\/2=versine of an arc of the ungulas at a distance X from J?.' 9. 10. 11. 12 l^an arc of the ungulas at a distance 14. X from D. Q^'^-ix^2,i ment at a distahce x from D. r>J2 13. r^cos r^—^x'^).=t\ie area of a seg- .-. Q,v''=Q,r |^2cos -1 C^^~\—^x^2{r-'-—\x^ Y~[dx ^6 ;'''xcos" MENSURATION. 399 iV^i'-'—i" 5-|V2»- is/ir =6^3(1^6— i7rV2)= . r8(|V6— 7rV2). 15. .-. F,the volume bored out,=2v-\'v'—6v''=ii^6r^ + yrr^e^3—ir^/2)~r^y6—7t^/2}=rWQ{7re — 2r>^2) =16.866105 cu. in. III. .-. The volume bored out is 16.866105 cu. in. I. A horse is tethered to the outside of a circular corral. The length of the tether is equal to the circumference of the corral. Required the radius of the corral supposing the horse to have the libert) of grazing an acre of grass. Coustruction. — Let AEFBK be the circular corral, AB the diameter, and A the point where the horse is tethered. Suppose the horse winds the tether around tbe entire corral; he will then be at A. If he un- vsrinds the tether, keeping it stretched, he will describe an involute, APGH', to the corral. From H' to H, he will de- scribe a semi-circle, radius AH'=AH= to the circumference of the corral. From H through G to A, he will again de- scribe an involute. Then the area over which he grazes is the semi-circle HLH'-\- the two equal involute areas AFGHA and AKGH' A-\-\}n.e. area BFGKB. Let C be the center of the corral and also the origin of co-ordi- nates, AG the AT-axis and P any point in the curve APGH'. ( 1. Let ^=:the angle A CE that the radius CR perpendic- ular to PB, the radius of curvature of the curve APGH' , makes with the ;f-axis, 2. l9„=the angle AFEBK that the radius CK makes with the jc-axis when the radius of curvature PE has moved to the position KG\ 3. R^A C, the radius of the corral: 4. p=PE=arc AFE^RO, the radius of curvature of II. 400 FINKEL'S SOLUTION BOOK. the involute ; 5. x= CM and 6. y^=zPM, the co-ordinates of the point P; and 7. *:.= CGand 8. -y-=0, the co-ordinates of the point G. Then we have- 9. x=CM=IE—CD=PE(=^arc AFE) coslIBP, = IPEC—L OEC{=LECD),— CE Q.o%i EGD =zRe cos( i lEP—i ECD)—R cos(ff— (?)=i?6' cos. l^n—(7t—f))Y-R cos{n—e)=Rd cos -{^Tt—S) —Rcos(7r—e)=Recose+R sin i9 .... (1). 10. y=PM=PI-\.IM{=DE)=PE sin^ PEI-\-ECy(. sin jLECD=RB i\n{B—\n)-\-R Bin(7r—f))=R sin(? —Rff sinf) .... (2). When ^=lV,=angle AFEBK 11. x^:=C G=R ca%%-\-Re„%\T\B^. . . . (3), and 12. j„=0=/? s:n(9„— 7?0„cos/y„ (4). Hence, from (4)^. is: 6'.=i?sine„s-i?cosi9o=tan(9,..,. .(5). Then, from (3),. ■ 14. ;«:o=/?cos^„4-i?tan(9„sin^,=i?^cos^o+^^^sin5oJ ^^dT^ 8ec(?.=^V(l+tan^^,)=i?V(l+e.==). . . .. (6). Now 15. £EGJtB=2[iKGX-K'C— sector BCK\=R^e„—R^ {e-n). ...{!), ■• 16. AEGJIA+AKGir=2jdA=2jip''dd= I R^ O^dd ' =i^2(8^,§8)....(8)^and 17. HH'L=\n{AHY=\n{2nRY^27t^R''....i<d). Ad- , ding (7), (8), and (9), 18. ^2,^0— ie2( ,?__>r)4^i?2(8;rl-$|)-f 2w8i?2= R'''\n-\-\^n'^ — ■J-0„')=area over which the horse- grazes. 19. 1 A.=160 sq. rd.=4356t) sq. ift.=the area over which the horse grazes. 20. .-. if2(;r+y;r3— ^(9|)=43560 sq. ft. Whence, oi o \t 43560 A ,,„, 22. 5„=4.494039=264° 37' 18''.35 by solving (5) by the method of Double Position. 23. .-. i?=19.24738ft.,by substituting the valueofe„in(10). III. .-. The radius of the corral is 19.24738 ft. A 20-foot pole stands plumb against a perpendicular wall; A cat starts to clitnb the pole, but for each foot it ascends the pole slides one foot from the wall; so that when the top of the pole is reached, the pole is on the ground at right angles to the wall. Required the equation to the curve the cat dsscribed: and the- distance through which iftraveled- MfENSURATlOM. 4»1 II. FIG. 89. Coi»struciton.='Lei AC he the wall, P the pQS<ttioin of the cat at any time, and B C the position o( the ladder at the same time. Draw AP and to the middle point £> oi AP draw BD. Thpn AB=PB. , 1. Let ^C=20 ft.=a, the length of the ladder, 2. AP=r, the radius vector of the curve the cat describes, and! 3. ft==tbe angle PAB. 4. K — 2^=the angle ABP, becaiase the angle jP^^=the angle BP4 5. ^^=^Ccos^^C=acos(;r— 2^) a cos2^, 6. iAP—^r=AD=zAB cosZ. BAD ==:— a cos2^cos^. 7. -•. r= — 2fl!cos26'cos6' , or 8. r-\-2aco&1d c^s,0=fi, the equation of the curve de- ^ scribed by the cat. 1. Let j^the distance through which the cat traveled. s=Jy/{dr^-\-r^de^)=2aJ ^(1— 12 cos" (9 + 44 cos* e— 32 cos8 6 )de, \3. = — at V(2' — 4eos0 — cos^^-j- 4cos30)a?0. where 0=;r — Id, 4. = — \a. \ V(6 — 4 cos<^> — 2 cos20-f-4 cos30)<^^ Jo =1.1193 a, 5. =22.386 ft., the distance through which the eat travels. ;--(-2aGos 2 Scos ^==0, is the equation or the curve, and 22.386 ft.=rthe distance through which cat traveled. NoTK. — The integration in this problem is performed by Cotes' Method of Approximation. I. Suppose W. A. Snyder builds a coke oven on a circular bottom 10 feet in diameter. While buildirjg it, he keeps one end of a pole 10 feet long, always against the place he is working and the other end in that point of the circumference of the bot- tom opposite him. Require'd the capacity of the oven. Construction. — Let AB be the diameter of the base and CQ the altitude. At a distance x from the base pass a plane inter- secting the oven in F and E. Draw AE and A C. III. 402 n{ FINKEL'S SOLUTXOJsr BOOK. AB=2Ji=10 feet, the diameter of the base. A C=AE=2R=10 feet, by conditions of the probiem CG=')/{AC''—AG^)=-{4:R^—R-')=JislZ, the alti tude. ' EH^=x^={ZA G-{- GH) ( GB— GH)=ZA G^— 2^GX G//=3B^—2J?X GH{=EI), hecanse BUh the ordinate of a semi-circle whose diameter is 2^7>. From tMs, we find EI=i^{4:R^—x^)—Ji. Then 7r^72==»[V(4i?2— *2)— i?]2, the ' area of the circle whose center; is/. Jo —R'Ydx= / IhR^—x-' .Jo . ■ —2Ry('LR^—x^)]dx, 8. =l5R^x-~ix^ — 4/?8sin-i -1 —RxV(4R^—x^)\ , 2R FIG 90- 9. =i7ri?'(9V3— 4;r)=ijr58(9V3— 4;r), 10. =t»rl25(9V3—4;r) =395.690202+ cu. ft. III. .-. The' capacity is 395.590202 cu. ft. I. At each corner of a square field whose sides are 10 rods, a horse is tiefl with a rope 10 rods long; what is the' area of the part common to the four horses? Construction. — Let ABCD be the fiield and EFGH the are.'i common . to the four horses. Join EF, FG, GH, and EH. Draw Z>A'per- peudicular to EF and draw -DE and DF. ^{nc& AF=I>E=I)F=GB = CE, the triangles ADF and EDC are equilateral and, consequently, the angle ADE=LADC— /.EDC = ^Qi° '_60O=B0°- Also the angle FDC= 30°. Hence, EDF=ZO°. Now let FIG. 91. fl. AZ>=.eZ?=a=10 rods. Then 2. Area of sector EDF=:f\ EDX arc EKF=i[crX(27ra} 3. Area of t riangle EDF=\EFy .DK. But, 4 \EF=^a\2a—,^{4.a^—AF^)'\,\>y formula of Prob. XXII., =W(2—V3), and Il.;5. Z>iir=v/[Z?^^— (i^^)']=W(2+V3). Hence, 6. area of triangle .£'Z>/?=^v/(2+V3)XW(2—V3)=ifl'^. MENSURATION. 403 " 7. .: Area o( sefrment BF=-^^^7ra^— ia^=^a^ (tt— 3). The 8. area of squ.ire BI^GI/=BI^^=a^ (2—^/3). Hence, 9. area of the figure, EB'G//=a^(2—/^B)+iXl^■^a^7T—3) ^a^(^7r-\-l — \/3):^31.5147 sq. rd.=the area common to the four horses. III. .•. The area of the part common to the iour horses is 31.5147 sq. rd. Note. — This problem is similar to problem 348, School Visitor, to which -a fine trigonometrical solution is given by Prof. E. B. Seitz. I. What is the length of the longest straight, infleixible stick -of wood that can be thrust up a. chihmey, the arch, being 4 feet high and 2 feet from the arch to the back of the chimney — the •back of the chimney being perpendicular? • Construction. — ^Let J^DE C be a vertical section of the chim- ney, PB the height of the arch, PE the distance from the arch to the back of chimney, and APD the longest stick of wood that <;an be, thrust, up thexhimney. 1. Let PB=a=^'i^ feet, the height of the arch, 2. PE=b=2 feet, the width of the , chimney, 8. ■«=the length of the longest stick . , of wood, and 4. fcithe angle If A C. Then 5. AP=PP cosecd'^a cosec^, 6. PD=PE sec 6=3 sec d. 7. .-. x=^AP-\-PD=a cosecff -\-b sec 6 (1). Differentia- ting (1), S. 0=— a cosfi'-^sin^e+J sin^H- Il.<! ' cos2^ (2)', or a co%^d=tb sin* d (3), by ^/g g^ clearing of fractions and transposing in (2). ^ 10. U. cos^^ tan ^=3 =tan»^=-. Whence, 12. cot<?=!^ 7. From (3), we may also have Now, from trigonometry, 13. V(l+tan2 6l )=sec# , and 14. v^(l-j-cot2^)=cosec^. Hence, by substituting in (1), 15. x=3v/(l-|-cot2^)+3v/(l+tan26l)= +SV(ai+3l)=(3i-f^i)(V';i4:il)=(«i-f5l)i =V'[ («i-(^3i ) ' ]=-V/[ ( 4l+2i) ' ]=8.323876+ft. 404» FXNKEE'S SOJiUTION' BOOK. in. .-. The length of the longest stick ifr 8;323876+ft. I. .^. stnaJl, garden,, situated' ip a level plkne. is surrounded by ai vyiall having twelve ^qufil, sides, in the Cjenter of whichj are- tvyelve gates. Through these and from, the center of the garden. 12 paths lead op" through the plane in a str,^ight direcliion. Fronii a point in the path leading north and' at a distance of 4 furlongs- 47-^-^ yards from the center of the garden, A. and B. start to travel irt opposite directions a,nd at the same rate. A. continues in the direction he first t£ikes; B., after arriving at the first road (lying east of him) by a i straight ling and^at'right angles with it, turns so- as to arr-ive at the next path bya straight line and atrightangles- with it and so on in like manner untiL be arrives at the same- road from which he started, having made a complfetfe revolution; arpund the center of the g^^en. A,t. the moment that B, has- performed the revplution, hoyir-far will J^. and B. be apart?- point iji the pathf g,hJ, FIG. 93. Let Obe the center of the garden, A, the leading north from which A. and B. start,. C, K,L, M; N, P, the points, at which B. striliek; the paths. The triangles OGA', one, OED-, OFE, &c., are raght triapglesj OCA, OBC, OED, OBF, &c., being, the right angles. Let S in the prolongation of AC denote the position of A., when- B., arrives at:^. It is required to find the distance AS. Let OA=a=4: furlongs, 47^*^ yd:, PS =^ C+ CD+I?E-{^ . . . +NP==x, AS =jy, AP=z, «=12, the number of paths ^d lA OC=-/fGOn=ADd£= '.'... Z iV"O/'=360°-^a^30°. Then, from the right triangles we have OC=OAx cosAOC=acosd, OD=OC co&COD=acos^ 6, 0E=ODx cos D OE=a cos»^ , 0^^= OJV cos NOP=a cos»0 ; A C= OA X sin^OC=asin», CJD=OCs\nCOD=a sin^ cos0,£>E= J?Osin£>OE=a sin^Cos^fl, NP=NO&\n.NOP=a sin^cos^-'^. .-. z=OA — OP^=a(l — cos"(9),and «=a sin^+a sin0cos£'-|- a; sin^ cos^ 0-\- +a!,sin<'cos»*~*fl=asin9^1+cos^+cos2e-|- cos3^-|- -|-cos»""^)=asin5(l — cos"^)tt-(1 — cos^)=^ a;cot-J^(l— cos»(9). Hence, since Z/'.<45=(,90°+^), we have- y^'i/[;x^-^z^—2xzXcQs(90'^-\-d )]=« cosec ^6'(1— cos»#) X V(l+sin2^)=2-^^f^X^p:^[l— (f)«]X^=3292 yd., nearly. Note, — This problem was proposed:in the School Visitor, by Dr. N. R. Oliver, Brampton, Ontario. The above elegant splution was given by Prof. E. B. Sei^?,,and was published in the Sckpol. Visitor, Vol. S^piSS. MiE'N'SURATlQN. 1405 1. A fox is 80 rods north of a hound and runs directly east 360 rods before being overtaken. How far will the houtjd run •before catching the fox if he runs towards the fox all the time, ■and upon a level plain? Construction.^-l^tt 'Caftd A be the position of the hound and ToX at the start, P and «i corresponding positions of the hound and fox any time during ehe chase, and P^ and n their positions the next instant, B the point where the hound catches the fox and CPP'!B the curve described by the hound. Join^w and P, .and n and /"; tbey are tan^gents to the curve at /'and /". Draw Pd and P'e perpendicular to AB, mo perpendicular to P'n , land P'f perpendloul«,r to Pd. 1\. 1. Let^C=a=80rds. % 3 4, 5 6, 7. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24: 25. AB^b=%m i-ds., Bd=y, Pm='w, arc CP=s, curve CPB^s^, and I r=ratio of the FIG 94. hound's rate to the fox's. Then we have mn^dx, ed=P'f=dy, PP'=:ds, no—PP'^dtv .... (1), and s=rx .... (2). From (2), we have, by differentiation, ds^rdx. Whence, dx 1 -3-^-. From the similar right triangles PfP' and mon, we have PP': tnnwpP': mo, or ds 'dx ::dy : mo. Whence, dxy,dy dy . dx 1 „ , . . . ,., , ' mo=i= — -J — —=-=^, since-j-=-. substituting in (1), -i — ds^d-w, or r dy — r^dx=rd'W .... (3). Integrating (3), y—r'^x^=riv-\-C....(4L). But, since when ?(=0, /=0, and 'W==a, Q=ra-\-C. Whence, C== — ra. .-. y — r'^x=:r'w-\-C=r'w — ra .... (5). When x=b,y =3, and w=0, and (5) becomes b — r'^b= — ar, or r'''b — ra=b. Whence, r^ — -,r^\, b r" rA =1- a2 a'^-\.^b^ 4b^ 432 W 406 26. 27. 28. 29. V30. FINKEL'S SOLUTION BOOK, r. Va2+4^2 23~23 But ^6=3^^, what (1) becomes when b=x. .-. 5i=.J-(a+Va2+452;^392.2783 rods, the distance the hound runs to catch the fox. Note. — This solution is substantially the same as the one given by the Late Professor E. B. Seitz, and published in the School Visitor, Vol.IV, p.^fff^ The path of the hound is known as the "Curve, of Pursuit." I. A ship starts on the equator and trayels due north-east at all times; how far, has.it traveled -^yhen its longitude, for the first time, is the same as tha't of the point , of departure ? Let B be the point of the ship's departure, BPJVits course, /*■ its position at any time and Nits position at the next instant. Then PJV is an element of the curve of the ship, which is known as the Loxodrome, or Rhumb line. . Let 0==^F'=th.e longitude- of the point B, <p=PP'=the corresponding latitude, (x,y, z) the rectangular co-ordinates of Z', and 9)=^ ;r=the constant angle PJVQ. Then we have for the equations of the curve, x=P Gcosd =rcos(pcos0 .... (1), y=P G sin d=r cos<pX sin^ .... (2), and z=r sin0. . . .(3), where r is the radius of the earth. Now an element of a curve of double curvature, referred to rectangular co-ordinates is Vidz^+dy^-lrdx^). .-. PN=ds :=sl{dz-^-\-dy^-\-dx^) .. . ...(4). Differentiating (1), (2), and (3), a?x= — /-(cos^sin0(/0-|- cos0sin6'(j?^), FIG- 95. dy^ — r(^hix\.dsirx<pd<p — cos0cos^(j?6'), and </2-=rcos 0(3?0. Sub- stituting these values in (4), ^5^ry'[cos^0(/^^-|-(siii ^sin0^0i ■ — COS0COS 6d6 )^-|-(cost^sin<^(/0-j-cos0sin^(/6')2]^r-v/(cos20</0=i ■\-%\aP-4)d(j)^-\^co%'^<pde'^)=ry/{d(p'^-\-zos'^<t>dB^) . . . (4). Novr PQ=GPxBQ=r cos<pde and JVQ=rd<p.. ^=tanlPJVQ MENSURATION. 407 :=tan9>. .•. -Tf^= — ;^r"^^"9'' °' cos0<<?^^tan <pd(^ .... (6). Substituting the value of cos0t?/9 in(5) ,ds^r-\/ (d<p^-\-ian^ q)cl4'^ ) =riJ(14-ta.n^cp)d<i>=^ -rdd). .■. s= / dd>=: ^^ ' -rj -r (-080 ZO&(pj^^ c-£-9,('^'-^»^- • • •(7)- By integrating (6), ^=tan^/^ =tan9) loge[tan(i;r+|0)] or e9='"'«'=tan(iw+^0) . . . (8). Whence, <?i=2tan-'(e9'=''*9') — \n . When 6'=2;r.and9)=i;r, <?i==2tan-i(e2T)_ia.=89° 47' 9".6=:4988i;r. .-. .5=-^x \ / i » cosier (.4988i-?r— 0)=?-V2(.4988irr)=2.21615937r=8775.991093— mi., the distance the ship travels. The rectangular equations of the Loxodrome are 'V/(*'+J*) I eotan-*i_|_g-atan-i| | =2r, and ^^-f-j/^+^^^^a ^ vyhere a=cot9>. The last equations are easily obtained from the figure. The first is obtained as follows: From (1) and (2), we find ^^tan"^''- ; also, ** -\-y'''=r'''zo%'''4> or cos0=-i\/(^^+>'^). From (8), we get efi«'^V^ r =- , , ,,/ ^ „^, . Whence, «fl'=°'9'4-efi<^<"?'X l-l-sin0 1+V(1 — cos2(^) I ^ 1^(1 — cos20)=rcos0. Transposing eO'">*-^, squaring, and reducing, we have cos0(eS«°"P-fe-fl<:»"P)=2. Substituting the value cos0, and d, we have •y/(x2-)-j)/2) \ e"^^"^ J-f-g-otan *- C ^2r. Note. — This solution was prepared by the author for problem 1501, School Visitor, but it was not published because of its difficult composi- tion. «8 FINKEL'S SOLUTION BOOK. CHAPTER XXII. EXAMINATION TESTS WITH ANSWERS. For the benefit of Students preparing for county or state examinations, we write ovit the answers to Xhs following questiolis as a speeiineu of bo* thfe examination paper ought to be prepared : ARITHMETIC. 1. How do you divide one fraction by, another? Why is the fraction thus divided? 2. Divide foru: inillion and four millionths by one ten • thousandth. Write the answer in figures and words. 3. If a liter of air w«ig&8 1 .273 gr., y*at is the weight in kilos., if the air is in a room which contains 7S cu. m. ? 4. The base of a cylinder is 12 inches in diameter and its altitude is 25 inches. Required the solid contents. 5. The edge of a cube is 6 inches ; what is the length of the diagonal of the cube? 6. A broker bought stock at 4>9J, discount, and sold it at 6% premium, atid gained $450. How many shares did he purchase? 7. A ships 500 tons 'Of cheese, to be sold at 9^ cents a lb. He pays his ^ent 3% for selling ; the proceeds are tp be invested in sugar, after a com- mission of 2<fo is deducted for buying. Required the entire commission. 8. Upon what value are dividends declared? Brokerage estimated? Usual rate of brokerage ^ 9. What is the face of a note dated July 5, 1881, and payable in 4 months to produce $811, when discounted at 9%? 10. Upon what principle is the United States rule for partial payments based? The Mercantile rule? How does compound interest differ from annual interest? Ohio State List, 1884. ANSWERS. il Subject : Arithmetic. Name of Applicant , O (a) Invert the terms of the divisor and then multiply the numerators of the fractions together for the numerator of the quotient and the denominators together for the denominator of the quotient. (b\ The fraction is thus divided because inverting the terms of the divisor gives the number of times the divisor is con- tained in 1, as is shown by analysis. The number of times then it is contained in any other number is obtained by multiplying this number by the number of times the divisor is contained inl. r 1. Four million and four millionths=4000000 . 000004= I 40000000000004 i 1000000 PROBLEMS. 408 II. illl. II. an. 2. Doe teii'thousaDdth= .0001= . 40000000000004 1 10000 40000000000004 10000 "*• 1000000 ^10000" aoooooo ^ i ~ 40000000000004 ^ '4 — jqp =:4Oe99e«O«0OOj|j5 =400008900000. 04= .-. The quotient is four htttiftredtulUoiiatid fourliundredttis. /l. lctt.in.=10001. 1'2. t8 cu. m. =78X10001=78000 1. 3. 1 . 273 g.=the weight of 1 1. of air, and 4. 99294 g.=78000Xl-273 g.=weight of 78000 1. 5. 1000 g.=l kilo. 6. 99294 g.=98294«.-*^000=99. 294 kilos. .'. T,he wei$:1it 0f Is :cu. in. of ^ix weighs 99 .294 kilos. ) 4 11.^ III. • 1. 12 in.=the diameter of the cylinder, and 2. 25 in. =the altitude. Then 3. '/tcW ,aq. io.=36t sq. in., the area of the base of the cylinder. 4. (25X361- ««. in.=:90On- cu. in, =900X3-141592X1 cu. in.= 2827.4328 cu. in., the volume of the cylinder. .-. The-wolame of the [cylinder is 2827.4328 cu. in. 5 II. < III. 1. 6 in.=the length of the edge of the cube. 2. 36 sq. in.4-36 sq. in.=72 -sq. in.=the area of the square described on the diagonal of one of the equal faces, ! which is the sum of the areas of the squares de- scribed on two equal edges. 3. 72 sq. in. +36 sq. in.=108 sq. in.=area of square ;de- scribed on the diagonal of the cube, which equals the sum of the areas described on the three edges. 4. 6V3 iJi.=V108Xl in.=10.392+ in., the length of the diag- _ onal of cube. .-. 6j/3 i>n.=10.392+ in.=length of diagonal of cube. '6 II. - III. ■ 1. 1009^=par value of stock. 2. 4^= discount. 3. 969'(,=100%— 4%=market value, or cost of stock. 4. 5^=premium. 5. 105%=1009J,+5fo=selling price of stock. 6. • 9%=1059'<,— 96%=gain. 7. |450=gain. 8. .■.9%=f4.50. 9. lfo=|50, and 10. 100%=f5000=par value of stock. 11. f 100=par value of one share, usually. Then . 12. $5000=par value of $5000-^-$100, or 5 ,share3. .-. He purchased 5 shares. '1. ^J4 cerits=selling price of one lb. ^ 2. $47.50=500X$0.09K=sellingpriceof 500 lb. • ri. 1009!,=$47.50. o , 2. 1%=$0A75. *■ 1 3. 2^„=2Xf0.475=f0.95=com"mission for selling 410 FINKEL'S SOLUTION BOOK. 10. L the cheese. J47.50— $0 95=f46.55=proiieeds, or the amount to be invested in sugar. 100%=cost of sugar. 3^=commission on sugar. 103^=total cost of sugar. $46 . 55 = total cost of sugar, fl. .-. 103f(,=$46.55. 2. l%=Tk of $46 . 55=$0 .45. 3. 100^=100X$0.45=$45=cost of sugar. 4. 2^=i2X$0.45=$0.90=commission on sugar. $0 . 95+$0 . 90=$1 . 85— totals commission. III. .'. $1.85=entire commission. (a) Dividends are declared upon the par valuer ' 0) Brokerage is reckoned upon the selling price or purchas- ing price of bonds in Commission and Brokerage, but in Stock Tnvestments it is, reckoned on the par value. (4 The usual rate of brokerage is j^ on the par value of the stock, either for a purchase or a sale. II. III. 1881- 1. 100%=face of note. 2. 3^^=discount for 126 da. 3. 96JJ%= proceeds. 4. $811=proceeds. 5. .-. 96M/o=$811, 6. 1%=$8. 37372, and 7. 100^ =$837 . 372, the face of the note. .-. The face of the note must be $837,372. 7 — 5 when dated. 4 3 1881—11— 5-8 when due. (a) Upon the principle that payments be applied first to the discharge of interest due, the balance, if any, toward paying the principal and interest. Interest or payment must in no case jQ draw interest. (b) Upon the principle that partial payments shall draw inter- ~ est from time of payment until date of settlement. (c) Compound Interest increases in a geometrical ratio, and Annual Interest in an arithmetical ratio, SECOND LIST. 1. A and B together have" $9,500. Two-thirds of A's money equals f of" B's. How much money has each ? \ 2. A owes a sum equal to f of his yearly income. By saving -^ of his. income annually for 5 y£ars, he can pay his debt and have $1,^00 lelt. What, is his yearly income? 3. Smith and Jones can do a piece of work in 12 days. If Smith can do. only J as much as Jones, how long will it take each of them to do the work ?• 4. I ard oflFered 6% stock at 84, and 5% stock at 72. Which investment, is preferable, and how much? 5. If in selling cloth f of the gain is equal to ^ of the selling price, for- how much will 3J^ yards sell that cost $5 per yard ? 6. The frustum of a cone is 10 feet in diameter at the bottom and 8 feet, at the top, with a slant height of 12 feet. What is the height of the cone.- from which the frustum is cut ? PROBLEMS. 411 7. A, B and C ate eight pies. If they ate equal shares and A and B fur- nish the pies, A furnishing 5 and B, 3, and C pays 16 cents for his share, how should A and B divide the money? 8. Which is the heavier, and how much, an ounce of lead or an ounce of gold ? ' • Pickaway County List, i8gg. ANSWERS. o Subject: Arithmetic. 4J 1 1 (Name of Applicant.}' 1. •.• 54 of A's money=| of B's money, 2. Yi of A's money=;^ of | of B's money=^5 of* B's money, and 3. 1 of A's money, or A's money,=3 times A of B's money =^ of B's money. 4. Let 3:S=B's money. Then 1 II, 5. i'5f=A's money. 6. S+i%=U=the sum of A's and B's money. 7. $9500=the sum of A's and B's money. 8. .-. i§=$9500. 9- •h=h of $9500=1500, 10. ^=^ times f 500=$4500=A's money, and 11. +5=10 times |500=«5000= B's money. iTt . S $4500=A's money, and ^"- ■• \$5000=B's money. 1. Let |=A's yearly income. 2. 54:=his debt 3. •^=the, amount of his income he is to save for 5 years. 4. .•. f §=the amount he is to save in 5 years. 2 II.- 5. .-. |f-$1200=his debt. 6. .-. ||— $1200=^ of his yearly income. 7. .-. h-m:=^=;=$i2oo. 8. 5V of hisincome=5'ff of $1200=|192'j.^ .9. 5I, or his annual income,=76X$19^r=$1447W. III. .•. $1447i|=A.'s annual income. . - j 1. Let 4=part of the work Jones does in 1 day. Then 2. J^=part Smith does in 1 day. 3. i=part they both do in 1 day. 4. J5=part they both do in 1 day, since they can do the work in 12 days. 3 11.. 5. .•. \=^ of the work. 6. )i=\ of t^=B^ of the work, 7. ^=3XVj=A=part Smith does in 1 day, 8. f=4XJij=2't=P^rt Jones does in 1 day. 9. J|=part Smith can do in ff^-gV) o"" 28 days. ^10. |}=part Jones can do in ll-j-^^^, or 21 days. III. .•. Smith can do the work in 28 days and Jones in 21 days. | I-- ■ 1. 10U%=amount invested in the 6fo stock. 2. 1009'o=par value of the O^J, stock. 3. 845/o=market value of the B^i stotk. 4. .■. 849J, of par value=100% of the investment, «12 FINKEL'S SO'LUTttON BOOK. Answers — Concluded. \ 3. III. , The \% of par Yalue=:ls*i9{) of the investment, <e% -of £be f>af vaMe— 7^% oaf tiK Jmi«8tftfeitts=<A>e iacomeiou theiinvesitiibebt, 1009£>— amotmt invefstediin €% <ete>c^. ' 400^={>ar value of .the 5^ stodc fr2%— madci^t value of 5% stock. .•. 72% of par value=100^ of the investment, \<^ri of par valtte=:l^% of the investment, 5^ of the par value :3=6||^% of the investment. '7^% — 6f|9{,=^^9(i=the differencie in income on the 65^, stock and tlie h^ stock. 5% stock at 84 is ^^% better than the 69^ stock at 72. i II. -i III. 1. '." % of i^aill=t'j of the selling price, 2. X of gain= i^ of i^ of the sellteg prixre=^ -of the sen- ing price. 3. I, or the^a:in,=4XA^, or % of the selling price. 4. I, the selling price, — 5^, the gaifl,=^, the cost. 5. f 5=the cost of 1 yard. 6. $16%=3j^X|5=cost of Z\i, yards. 7. .-. %, the cost,=$I6%, 8. J, the selling price,=4><;$16%=$665^=selling price. .-■ The cloth would sell for $66^. ' II.- III. Let AB, Fig. 63,=5 feet, the radius of the lower base of the frustum, CD=^ feet, the radius of the -upper base of the fnistu^, and AD=^\'i feet, the slant height of the frustum. Then AK=AB—DC=hi^e,\.-^ feet=l foot, and DK=\IAD'- .■.AK:DK= AK^=\jli4r-\='>l\ii feet. AB : B£, by similar triangles ; or 1 foot : V143 feet=5 feet : BE ; whence, ^.£'=5V143 feet, the altitude of the cone from which the frustum is cut. The alti tude of the cone from Vyhich the frustum is cut isSVliSfeet. See solution on page 190. II. III. 1. 1 oz. of gold=5760 gr.-^12, or 480 gr , since there are .5760 *gr. in a pound of gold. 2. I oz. of lead— 7000 gr.T-16, or 437j^ gr., since there are 7000 gr. in a pound of lead. 3. .'. 480 gr.— 437^ gr.=42;^ gr., the excess of weight of 1 oz. of gold over 1 oz. of lead. .■. 1 oz. of gold is 42 J^ gr. heavier than 1 oz. of lead. EXAMINATION TESTS WITHOUT ANSWERS. 1. Define bonds, coupons, exchange, tariff. 2. A field of 12 acres and 30 perches jrields 255 bu. 2 qts of wheats tauch will a field of 15 acres and 10 perches 3deld at the same rate ? 3. Find value of 11% of $180 + 22% of $160 + 92% of $63. how PROBLEMS. ' «» 4 4i piiwo -was, soisk for $SS7, at a gain of 35^ ; what woaM hanie beetk the % or gain if it had been sold for $800 i* 5^ 4^ dealjer imported 120 dozen champagne, invoiced at $23 a dozen^ breakage 12j^ ; what was the duty at 2'2%.? 6. I rent a house for $300 per year, the rent t<^ be paid* monthly in ad- vance ;. wh^t amount of cash, at the begin^niog of the year will pay on& year'a, rent i 7. The rafjteFs of a; house are 20 feet long, the width of the gable is 30 leei^ the rafters project! two feet ; what is the height of the gable ?' 8. What the convex surface of the fenstuni of a come whose slant height 19-6 &et,,th^ diameter of itSil^wer base 5 feet, wd of itstupper 4 feet? I 9. To be analyzed : If for every cow a farmer keeps, he allows- J acre for pasture, wd| f o£ 3^: acre for Gorn>. horn many cows can he keep on 3^ acres ^ lOi How m»Gh can I afford to- give for 6's of '81 so that I may realize 8^ per annum, fgiVdl being ata premium of 1^?' Hancock Eounjty List. 1. What is th|e SHrface of q^ pacallelopiped, 8<feet long,, 4- feet wide, and afeetihigh,? %i, A starts onia journ^gy at the rate of S.miiljes per hourr, 6- hour& after- wards B starts after him at the rate of \ miles per hour. H<owi far -vixill B travel before hg overtakes i^? 3< Tl^tinjie sinc^e noon is ^ of the time to 4 o'clock P. M. ; what is the tijgie? % JV- ipatt h^ing oranges, at, 4 cents each, andi applesi at % for 1 cent, g^jedl 2Q^i by, selling 5 dozen- for j^.(14; how many of each did he sell'i^ 6. The first term of a geometric series is 3, the thirdi term 507 ; find the ratio. 6. A merchant sold a quantity of good's at a ^ain of 209{,. If, however, he had purchased the goods for $60 less, his gain would have Ijeen 25^. What did the good* cost .'' 71 There is a park 400' feet square ; a walk 3 feet wide is made in it, along the edges, how many square yards would such a walk contain? 8j A man sold wheat, commission S^Jj and invested the proceeds in corn, commission 2^ his whole commission, $250-;, for how much did the wheat seU and- what was the value of the corn? Licbmg. County List. r. A man had 43| yards of carpeting; costing $26}^ ; he sold f of the pieces gaining %\ on each yard sold. How much did he receive for it? 2. From the product of f J and -^ subtract the difference of their squares. 3. How many acres in a field whose length is 40 rods and diagonal 50 rods? 4; How many trees will be required to plant the above by placing them 1 rod apart? By 2 rods apart ? 5. Bought a lot of glass ; lost Xh'lo by breakage. At what ^, above cost must r sell the remainder to clear 20% on the whole ? 6. After spending 25^ of my money, and 25% of the remainder, I had left $676. How much had I at first ? 7. How many fifths in ^.? Ans. If. 8. A box is 3 J inches, long, %\ inches wide, and 2 inches deep will con- tain how many f -inch cubes ? 9. Change |_ of quart to the decimal of a bushel. liOi A can hoe 16 rows of corn in a day, B 18j C 20, and D 24, What is the smallest number of rows that will keep each employed an exact num- ber of day^s ? Seneca County List. 414 FINKEL'S SOLUTION BOOK. 1. (o) Define: number, integer, fraction, a CQmmon multiple, and the greatest common divisor of two or more numbers. (6) Prove, (do not inerely illustrate) that to divide by a fraction one may multiply by the divisor inverted. (c) Change 74632 from a scale of 8 to a scale of 9. 2. (o) The freezing and boiling temperatures of water are 32° and 212°, respectively, when measured by a Fahrenheit thermometer; meas- ured by a centigrade thermometer they are 0° and 100°, respectively; if a Fahrenheit thermometer records a temperature of 74° what would the centigrade record be at the same time? _ (b) By what per cent must 8° Fahrenheit be increased so as to ■equal '8° centigrade ? 3. Silver weighs 10.45 times as heavy as water, while gold weighs 19.30 times as heavy as water; find, correct to 3 decimal places, the number •of inches in the edge of a cube of gold which is equal in weight to a cube ■of silver whose edge is 4.3 cm. Also express this weight in'(Troy) grains. '4. A 6% bond, which matures in ft years, with interest payable annu- ally, is selling at 104; a 5J% bond, which matures in IJ years, with interest payable semi-annually, is selling at 102. Which is the better investment? And how much better is it? 5. . A water-tank has connected with it 4 pipes; the first can fill it in 30 min., the second in 40 min., the third can empty it. in 50 min., and the fourth can empty it in one hour. If these pipes are so arra,nged that the third is automatically opened when the tank is precisely J filled, and tiie fourth when the tank is f filled, how long will it take to just fill the tank if the second pipe is set running 10 minutes later than the first? T. Cornell University — Scholarship Examination; 1899. 1. A and B run a race, their rates of funning being as 17 to 18. A runs 2J miles in 16 minutes, 48 seconds and B the whole distance in 34 minutes. What is the distance run ? 2. The surface of the six equal faces of a cube is 1350 sq. inches. What is the length of the diagonal of the cube? 8. A man bought 5% stock at 109J, and 4i% pike stock at 107J, broker- age in each case: J%; the former cost him $200 less than the latter, but .yielded the same income. Find the cost of the pike stock. 4. A, B, and C start together and walk around a circle in the same direction. It takes A ^ hours, B f hours, C ff- hours to walk once around the circle. How many times will 'each go around the circle before they will .all be together at the starting point? 5. I hold two notes, each due in two years, the aggregate face value of which is $1020. By discounting both -at 5%, one by bank, the other by true discount, the proceeds will be $923. Find face of bank note. 6. The hour and minute hands of a watch are together at 12 o'clock; -when are they together again? 7. How many cannon balls 12 inches in diameter can be put into a cubical vessel 4 feet on a side; and how many gallons of wine will it contain after it is filled with the balls, allowing the balls to be hollow, the hoUpw being 6 inches in diameter, and the opening leading to it con- taining one cubic inch? 8. An agent : sold a, house at 2% commission. He invested the pro- <;eedsiin city Ipts at, 3<^ (commission. His commissions amounted to $350. For what w^a? tfie.'.hotise sold? ,' / .V' ' ' ■'" " Ohio State List, December, 1898. PROBLEMS. 415 1. A, B, and C can do a piece of work in 84 days; A, B, and D in 72 days; A, C, and D in 63 days; B, C, and D in 56 days. In what time can each do it alone? 2. A banker bought U. S. 4's at 128§% and U. S. 4J's at 106J%, iDrokerage i%. The latter cdst him $1053.75 more than the former, but yielded him $195 more income. How much was invested in each kind of bonds? 3. f bf the cost of A's house increased by t of the cost of his farm for 2 years at 5%, amounts to $4950. What was the cost of each, if | ■of the cost of the house was only f as much as 4 of the cost of the farm? 4. A man desiring to find the height of a tree, places a 12-foot pole upright 54 feet from the base of the tree; he then steps back 6 feet, and looks over the top of the pole at the top of the tree; his eyes are 4 feet above the ground. How high is the tree? 5. I have, as the net proceeds of a consignment of goods sent by -me, $3816.48, which the consignor desires me to remit by draft at 2 months. If the rates of exchange are |% premium, and the rate of interest 6%, what will be the face of the draft? 6. In a certain factory are employed men, women, and boys ; the boys receive 3 cents per hour, the women 4 cents, and the men 6 cents ; the boys work 8 hours per day, the women 9 hours, the men 12 hours; the toys receive $5 as often as the women receive $10, and for every $10 paid to the women, $24 are paid to the men. How many are there of each, the whole number being 59? 7. Chicago is 87° 35' west. Wh^at is the standard time at Chicago when it is 1 P. M. at Greenwich? 8. From the middle of the side of a square lO-acre field, I run a line cutting oflf 3| acres. Find the length of the line. Ohio State List, June, 1899. 1. How would you present to a class the subject of addition of frac- tions ? Take as an illustrative example, I + f + jV 2. A reservoir is 1.50 meters wide, 2.80 meters long, and 1.25 meters deep. Find how many liters it contains when full, and to what height it would be necessary to raise it that it might contain 10 cu. meters. 3. Reduce (o) .4685 T. to integers of lower denominations, and (6) 1.69408 to a common fraction in its lowest terms. ' 4. The boundaries of a square and circle are each 40 feet. Which has the greater area and how much? , 5. Find the date of a note of $760, at 8%, simple interest, which, when it matured December 1, 1891, amounted to $919.60. 6. A gentleman wishes to invest in 4}% bonds, selling at 102, so as to provide for a permanent income of $1620. How much should he invest ? 7. From one-tenth take one-thousandth ; multiply the remainder by 10000; divide the product by one million, and write the answer in words. 8. Bought 50 gross of buttons for 25, 10, and 5% off, and disposed of the lot for $35.91 at a profit of 12%. _ What was the list price of the "buttons per gross? 9. Had an article cost 10% less, the number of per cent gain would have been 15 more. What was the per cent gain? Give analysis. 10. If the volume of two spheres be 100 cu. in. and 1000 cu. in. re- spectively. Find the ratio of their diameters to the nearest thousandth of an inch. ' Ohio State List, December, 1891. FINKEL'S SOLUTION BOOK. PROBLEMS. li. What is. the area of a field in the form of a. paiiajlejlograin,. whosei- iisngthi is. 160 rods.and wid^h 75' rods? Ans^ 75A. 2. Find the area ot a triangle "whose base is 72 rods and altitude 16i. sods. Amit, 3 A, 2 R. 10 P. 3i Two trees whose heights are 4Oaiid'80 feet respectively, sand on op- posite sides of a stream 30 ffc wide. How far does a squirrel leap in jumpittj- from the top of the highsr to the top qf the lower?' Ains^ 50 feet. 4. How many steps of 3 feet each does a man take ia crossing, diagjomajl ly, a square field that contains 20 acres ^ Ans. 440 steps.. 5. Find- the cost ofpaying a, court, 150 feet square; awajk 10 feet around the whole being pa.>(e4, with flagstones at 54 cents a.sqnar.e yard and the rest at 31>^ cents a, square y ^td ? A-ns. ^89.40. 6. What is the area of a triangle, the three sides ofc wiiich are respect^ ively 180 feet, 150 feet, and 80 feet? A»« 5935 85 sq. ft. 7;( What is the area of a trapezium,. tiie diagonal ol*' which is< 110 feet,, and the perpendiculafs. to the diagoiial are 40i feet, aod 60: feet: respectively? Atus, 5500 sq, lb. 8i At 30 cents,. a bushel, find' the cost of a' box- af oats, the box being, 8 feet long, 4 feet wide and 4 feet deep. Ans. $30.-85^. 9. Two trees stand on opposite sides o:6 a stream 40' feet" wide. The height of one tree is to the width of the stream as-S-is-to 4; and the- width of the stream is to the. height ot the other as.^iis to 5i. What, is the distance between their tops? j^s. 50 feet. 10. How. many miles of-farrow 15 in., wide, is turned in plowing a rect- angular field whose width is 30 rods and length 10 rods less than its diagoal? Ans, 49J^ mi. 11. The sides of a certain trapezium measui^e 10, 12, 14, and! 16- rod^ respectively, and the diagonal, which forms a triangle with the first two sides, is 18 rods; whkt is the area? Ans. 163.796 sq, rds. 12. Three circles, each 40' rods in diameter, touch each other externally; what is the area of the space inclosed between the circles ? Ans. 64:5 sq. rds. 13. How many square .ncnes in one face of a cube which contains 2571353- cubic inches? .^jw. 18769 sq. in. 14. Four ladies bought a ball of thread 3 inches in diameter; what por- tion.of the diameter must each wind off to heve equal shares of the thread?.' First, .27431 91 in. . ) Secondj .3445792 in. .Ans. ■< Third, .4912292 in, ^Fourth, 1,8898815 in. 15. A gentleman proposed to plant a vineyard of 10 A. If he placesithe vines 6 feet apart; how many more can he plant by setting them in the quincunx order than in the square order, allowing the plat to lie in the form ofia.sruare, and no vine to be set nearer its edgethan 1 foot in either case?' Ans. 1870. Ifi. Find the volume generated by the revolution of a circle about a tangent. Ans. 2T«i?2. 17. How many feet in aboard 14 feet long and' 16. inches wide atone- 'eitdand 10 inches at the other, and 3 inches thick? Ans. 45^' feet.' 18. If I saw throughj^ of the diameter of a round log, what portion of; the cutis made?' Ans. .196. PROBLEMS. 417 39. What is the surface of the largest cube that can be cut from a sphere which contains 14137.2 cu. ft. ? Ans. 1800 sq. ft. 20. Two boys are flying a kite. The string is 720 feet long. One boy who stood directly under the kite, was 56 feet from the other boy who held the String; how high was the kite? Ans. 717.8+feet. 21. How many pounds of wheat ',n a cylindrical sack whose diameter is m feet, and whose length is 1% yards? (ir=3.1416) ^ns. 447.31 lb. 22. How large asquare can be cut from a circle 50 inches in diameter? Ans. 35.3553391 in. 28. How many bbl. in a tank in the form of the frustum of a pyramid, 5 feet deep, 10 feet square at the bottom and 9 feet square at the top ? \Ans. 107.26 bbl. 24. From a circular farm of 270 acres, a father gives to his sons equal circular farms, touching each other and the boundary of the farm. He takes for himself a circular portion in the center, equal in area to a son's part, and reserves the vacant tracts around his part for pasture lands and gives each son one of the equal spaces left along the boundary. Required the number of sons and the amount of pasture land each has. Ans.G sons; 8.46079 A. 25. At each angle of a triangle being ori a level plain and having sides respectively 40, 50, and 60 feet, stands a tower whose height equals the sum of the two sides including the angle. Required the length of a ladder to reach the top of each tower, without moving at the base. ^«.s. 116.680316+ft. 26. If, the door of a room is 4 feet wide, and is opened to the angle of 90 degrees, through what distance has the outer edge of the door passed ? Ans. 6.2832 feet. 27. A tinner makes two similar rectangular oil cans whose inside dimen- sions afe as 3, 7, and 11. The first hold 8 gallons and the second being largef requires 4 times as much tin as the other. What are the dimensions of the smaller and the contents of the larger ? , ( Dimensions of smaller 6, 14, and 22 inches. ■ ( Capacity of larger 64 gallons. 28. An 8-inch globe is covered with gilt at 8 cents per square inch; find the cost. ^ Mi. $16.08. 29. A hollow cylinder 6 feet long, whose inner diameter is 1 inch and outer diameter two inches, is transformed into a hollow sphere "whose outer diameter is twice its inner diameter; find outer diameter. Ans. 3.59 in. 30. A circular field is 360 rods in circumference; what is the diagonal of a square field containing the same area? Ans. 20.3 rods. 31. What is the volume of a cylinder, whose length is 9 feet and the cir- cumference of whose base is 6 feet? Ans. 25.78 cu. ft. 32. How many acres in a square field, the diagonal being 80 rods? Ans. 20 acres. 33. How many cubical blocks, each ^dge of which is ^ of a foot, will fill a box 8 feet long, 4 feet wide, and 2 feet thick. _ Ans. 1728 blocks. .34. From one corner of a rectangular pyramid 6 by 8 feet, it is 19 feet to the apex; find the diihentions of a rectangular solid whose dimensions are as 2, 3, and 4, that may be equivalent in volume. Ans. 4, 9, and 8 feet. 35.* A solid metal ball, 4 inches radius, weighs 8 lbs.; what is the thick- ness of spherical shell of the same metal weighing 7^ lb., the external di- ameter of which is 10 inches? , Ans. 1 inch. 36. What is the difference between 25 feet square and 25 square feetr Ans. 600 sq. ft. 418 FINKEL'S SOLUTION BOOK. 37.* Find the greatest number of trees that can be planted on a lot XI rods square, no two trees being nearer each other than one rod? Ans. 152 trees. 38.* A straight line 200 feet long, drawn from one point in the outer edge of a circular race track to another point in the same, just touches the inner edge of the track. Find the area of the track and its width. 1 Ans. Area, Ta2^10000" sq. ft.; width, indeterminate. 39. The perimeter of a certain.field in the form of an equilateral.triangle is 360 rods; what is the area of the field ? Ans. 548.552 sq. rd. 40. A room is 18 feet long, 16 feet wide, and 10 feet high. What length of rope will reach from one upper corner to the opposite upper corner and touch the floor? Ans. 35-3 ft. 41. How many bushels of wheat in a box whose length is, twice its width, and whose width is 4 times its height; diagonal being 9 feet? Ans. '25 bu., nearly. 42 Find the area of a circular ring whose breadth is 2 inches and inside diameter 9 inches. Ans. 69.1152 sq. in. 43 * A round stick of timber 12 feet long, 8 inches in diameter at one end and 16 inches at the other, is rolled along till the larger end describes a complete circle. Required the circumference of the circle. Ans. 150.83 feet. 44. A fly traveled by the shortest possible route from the lower corner to the opposite upper corner of a room 18 feet long, 12 feet wide and 10 feet high. Find the distance it traveled. Ans. 28.42534 feet. 45.* From the middle of one side and through the axis perpendicularly of a right triangular prism, sides 12 inches, I cut a hole 4 inches square. Find the volume removed. " Ans. 138.564064 cu. in. 46.* Two isosceles triangles have equal areas and perimeters. The base of one is 24 feet, and one of the equal sides of the other is 29 feet. The area of both is 10, times the area of a triangle whose sides are 18, 14, and 15 foet. Find the perimeters and ajtitudes. Ans. Perimeters, 98 feet; altitudes 35 and 21 feet. 47. A grocer at one straight cut took off a segment of a cheese which had J^ of the circumference, and weighed 3 pounds; what did the whole weigh? .^H.t. 33.028 lb. 48.* A twelve inch ball is in a corner where walls and floor are at right angles; what must be the diameter of another ball which can touch that ball' while both touch the same floor and the same walls? Ans. 8.2154 in. or 44.7846 in. 49. What will it cost to paint a church steeple, the base of which is an octagon, 6 feet on each side, and whose slant height is 80 feet, at 30 cents per square yard? Ans. $64. 50. A'tree 48 feet high breaks off; the top strikes the level ground 24 feet from the bottom of the tree; find the height of the stump.- Ans. 18 feet. 51. How many acres in a square field whose diagonal is 5J^ rods longer than one of its .sides? Ans. 160.6446 sq. rd. 52.* Three poles of equal length are erected on a plane so that their tops meet, while their bases are 90 feet apart, and distance from the point where the poles meet to the center of the triangle below is 65 feet. What is the length of the poles ? Ans. 83.23 feet. 53. A field contains 200 acres and is 5 times as lOng as wide. What will it cost to fence it, at a dollar per rod ? Ans. $960. 54.* What is the greatest number of plants that can be set on a circular piece of ground 100 feet in diameter, no two plants to be nearer each other than 2 feet and none nearer the circumference than 1 foot? Ans. 2173. PROBLEMS. 419 55. The axes of an ellipse are 100 inches and 60 inches; what is the dif- -Jerencein area between the ellipse and a circle having a diameter equal to the conjugate axis ? , Ans. 600 »r=1884.96 sq. in. 56. Find the diameter of a circle of which the altitude of its greatest in- scribed triangle is 25 feet. Ans. 331^ feet. 57. If we cut from a cubical block enough to make each dimension 1 inch shorter, it will lose 1657 cubic inches, what are the dimensions ? '■ 58. Show that the area of a rhombus is one-half the rectangle formed by its diagonals. Nohle Co. Ex. Test. 59. The length and breadth of a rectangular field are in the ratio of 4 to 3. How many acres in the field, if the diagonal is 100 rods? 60. A spherical vessel 30 inches in diameter contains in depth, 1 foot of -water, how many gallons will it take to fill it'? Holmes Co. Ex. Test. Ans. 39 gallons. 61. A field is 40 rods by 80 rods. How long a line from the middle of one end will cut off 1% acres? Ans. 80.6 rd., nearly. 62. A ladder 20 feet long leans against a perpendicular wall at an angle •of 30°. How far is its middle point from the bottom of the wall? Ans. 10 feet. 63. Four towers, A 125 feet high, B 75 feet, C 180 feet, and D 65 feet, stand on the same plane. B due south and 40 rods from A; C east of B and D south of C. The distance from A to C plus the distance from C to B is half a mile, and the distance from D to B is 82% yd. farther than thedis-' tance from C to D. What length of line is required to connect the tops of A and D ? Ans. 240-f rds. 64. Find the volume of the largest sqnare pyramid that can be cut from a cone 9 feet in diameter and 20 feet high? Ans. 270 cu. ft. 65. A rectangular lawn 60 yd. long and 40 yd. wide has a walk 6 ft. wide around it and paths of the same width through it, joining the points of the opposite sides. Find in square yards the area of one of the four plats in- closed by paths. Ans. 459 sq. yd. 66. Which has the greater surface, a cube whose volume is 13.824 cu. ft., •or a rectangular solid of equal volume whose length is twice its width, and its width twice its height? Ans. Rect. 576 sq. ft., more. 67. The volume of a rectangular tin can is 3 cu. ft. 1058 cu. in.; its di- tnensions are in the proportion of 11, 7, and 3. Find the area of tin in the •can. Ans. 16% sq. ft. 68. A conical well has a bottom diameter of 28 ft. 3 in., top diameter .56 ft. 6 in., and depth 28 ft. 1.2 in. Find its capacity in barrels. Ans. 8023 bbl. , 69. A cylindrical vessel 1 foot deep and 8 inches in diameter was Jf full •of water; after aiball was dropped into the vessel it was full. Find the di- ameter of the ball. Ans. 6 inches. 70. Two logs whose diameters are 6 feet lie side by side. What is the di- ameter of a third log placed in the crevice on top of' the two, if the pile is 9 feet high?. Ans. 4 ft. 71. Circles 6 and 10 feet in diameter touch each other; if perpendiculars from the center are let fall to the line tangent to both circles, how far apart will they be ? Ans. 7.756 ft. 72. What are the linear dimensions of a rectangular box whose capacity is 65910 cubic feet; the length, breadth, and depth being to each other as 5, 3, and 2? ^»j. 65, 89, and 26 ft. 73. The perimeter of a piece of land in the form of an equilateral trian- gle is 624 rods; what is the area ? Ans. 117 A. 18 31 P. 420 FINKEL'S SOLUTION BOOK. 74. Four logs 4 feet in diameter lay side by side and touch each other j on these and in the crevices lay three logs 3 feet in diameter; on these three and in the crevices lay two logs 2 feet in diameter; what is the diame- ter of a log that will lay on the top of the pile touching each of the logs S' feet in diameter and the middle one of the logs 3 feet in diameter? Ans. 75. What will it cost to gild a segment of a sphere whose diameter is 6 inches; the altitude of the segment being 2 inches, at 5 ^ per square inch.' Ans . 76. A grocer cut off the segn:>ent of a cheese, and found it took J of the circumference. ^What is the weight of the whole cheese, if the segment weighed \}4 lbs ? Ans. 52.0228+lbs. 77. Two ladders are standing in the street 20 feet apart. They are in- clined equally toward each other at the top, forming an angle of 45°. Find, by arithmetic, the length of the ladders? Ans. 26.13 ft. Union Co. Ex. List. 78. Two trees stand on opposite sides ofastream 120 feet wide; theheight of one tree is to the width of the stream as 5 is to 4, and the width of the stream is to the height of the other as 5 is to 4; what is the distance between , their tops? . ^«.y. 131.58— ft. 79. How many gallons of water will fill a circular cistern 6 feet deep> and 4 feet in diametei" ? A ns. 564.01 62 gal. 80. A cube of silver, whose diagonal is 6 inches, was evenly plated with gold; if 4 cubic inches of gold were used, how thick was the plating? Ans. xV in. 81. Required the distance between the lower corner and the opposite- upper corner of a room 60 feet long, 32 feet wide, and 51 feet high ? ^«j. 85 ft. 82. How deep must be a rectangular boic whose base inside is 4 inches by 4 inches to hold a quart, dry measure? Ans. 4.2 cu. in. . 83. A fly is in the center of the floor of a room 30 feet long, 20 feet wide, and 12 feet high. How far will it travel by the shortest path to one- of the upper corners of the ceiling? Ans. T/709-|-ft. 84. A corn crib 25 feet long holds 125 bushels. How many bushek will one of like shape and 35 feet long hold ? 85. Let a cube be inscribed in a sphere, a second sphere in this cube, a second cube in this sphere, and so on ; find the diameter of the 7th sphere,, if thatof the first is 27 inches. (2). What is the volume of all the spheres- so inscribed including the first? Ans. . . 86. The area of a rectangular building lot is 720 sq. ft.; its sides are as 4 to 5; what will it cost to excavate the earth 7 feet deep at 36;* per cubic yard? ' ^«.r. $67.20. 87. A owns y^ and B the remainder of a field 60 rods long and 30 rods wide at one end and 20 rods wide at the other end, both ends being parallel to the same side of the field They propose to lay out through it, parallel with the ends, a road one rod wide. Isaving A's )/^ of the remainder at the wide end and B's % at the narrow end of the field. Required the location and area of the road. Ans. ; . 88. The diameter of a circular field is 240 rods. How much grass will be left after 7 horses have eaten all they can reach, the ropes which are al- lowed them being of equal lengths and attached to posts so located that each can touch his neighbor's territory and none can reach beyond the boundary of the field? Ans. 62.831853 A. 89. What is the diameter of a circle inclosing three equal tangent circles^ if the area inclosed by the three equal circles is 1 acre? Ans. ' . PROBLEMS. 421^ 90. What is ihe diameter of a circle inclosing four equal tangent circles ■each being tangent to the the required circle, if the area inclosed by the four equal circles is 1 acre? Ans. y?=4/[5(4 — T)](y'2+1)^(4 — t). 91. What is the greatest number of stal<cs thkt can be driven one foot apart on a rectangular lot whose length is 30 feet and width 'M feet.' Ans. . 92. What is the greatest number ot inch balls that can be put in a box 15 inches long, 9 inches wide, and inches high? Ans. ~ . 93. A conical vessel 6 inches in diameter and 10 inches deep is full of water. A heavy ball 8 inches in diameter, is put into the vessel; how much water will flow out? Ans . 94. How far above the surface of the earth would a person have to ascend in orderthat J^ of its surface would be visible? Atis. 8000 mi. 95. Where mus^t a frustum of a cone be sawed in two parts, to have ■equail solidities, if the frustum is 10 feet long, 2 feet in diameter at one end, and 6 feet at the other? Ans . 96. At the three corners of a rectangular field 50 feet long and 40 feet wide, stands three trees whose heights are 60, 80, and 70 feet. Locate the point where a ladder must be placed so that without moving it at the base it will touch the tops of the three trees, and find the length of the lad- der. What must be the height of a tree at the fourth corner so that the same ladder will reach the top,' the foot of the ladder not being moved? Ans. . 97. A horse is tied to a corner of a barn 50 feet long and 30 feet wide; what is the area of the surface over which the horse can graze, if the rope is 80 feet long? Ans. . 98 How many cubic feet in a stone 32 feet high, whose lower base is a rectangle, 10 feet by 4 feet and the upper base 8 feet by IJ^ feet? Ans. 805j/^ cu. ft. 99. To what height above the ground would a platform, 10 feet by 6 feet, have to be elevated so that 720 sq. ft. of surface would be invisible to a man standing at the center of the platform, the man being 5 f'eet high?, Ahs . 100. Required the side of the least equilateral triangle that will cir. cumscribe seven circles, each 20 inches in diameter. Aii.i. 89 28203 in. 101. Required the sides of the least right triangle that will circumscribe seven circles each 2U inches in diameter. Ans 123.9320 in. and 107.3205 in. 102. How long a ladder will be required to reach a window 40 feet from the ground, if the distance of the foot of the ladder from the wall is % of the length of the ladder. Ans. 50 ft. 103. A circular park is crossed by a straight path cutting off J^ of the circumference; the part cut off contains 10 acres Find the diameter of the park. Ans. 150 rd., nearly 104. Find the length of the minute-hand of aclock, whose extreme point moves 5 ft. 5.9735 in., in 1 da. 18 hr. ? Ans. \ in. 105. A, B, and C, own a triangular tract of land. Their houses are located at the vertices of the triangle; where must they locate a well to be used in common so that the distance irom the houses to the well will be the same, the distance from A to B being 120 rods, from B to C 90 rods and A to C 80 rods. Ans. . 106. A ^lorse is tethered from one corner of an equilateral triangular building whose sides are 100 feet, by a rope 175 feet long. Over what area can he graze? Ans. 90021.109181 sq- ft- 107. Find the area of the triangle formed by joining the centers of the .squares constructed on the sides of an equilateral triangle, whose sides are 20 feet ? A ns. 422 FINKEL'S SOLUTION BOOK, GEOMETRY. Ij.Pure Geometry ■ 1 I.Platonic Geometry ■ 2j. Conic Secio "1. Metrical Geometry - • .Sa.Trigon'try- 2^. Analytical Geometry I. DEFINITIONS. I. Geometry is that branch of mathematics which deduces the properties of figures in space from their defining conditions, by means of assumed properties of space. — Century Dictionary. 'Ig Plane Geometry- 23. Solid Geometry IS Ig. Plane Trig. ^. Geometry^ ...xr^gontry, ^a- Analytical 83. Spherical Trig, . 2. Descriptive Geometry, or Projective Geometry 3. Metrical Geometry is that branch of Geometry which treats of the length of Hnes and the magnitudes of angles, areas, and solids. The fundamental operation of metrical relations is MEASUREMENT. The geometry of Euclid and the Ancients is almost entirely metrical. Tht theorem, 7%(? square described on the hypotenuse of a right-angled triangle- is equal to the sum of the squares described on the other two sides, is a theo- rem of metrical geometry. 4. Descriptive Geometry, also called Projective Geom- etry, Modern Synthetic Geometry, and Geometry of Position, is. that branch of Geometry which treats of the positions, the direc- tions, and intersections of lines, the loci of points, and the nature and character of curves and surfaces. The fundamental operations of Descriptive Geometry are projection and section. Many of the theorems of Descriptive Geometry are very- old, dating as far back as the time of Euclid, but the theories and methods which make of these theorems a homogeneous and harmonious whole is modern having been discovered or perfected by mathematicians of an age nearer our own, such as Monge, Carnot, Brianchon, Poncelet, Moebius, Steiner, Chasles, von Staudt, etc., whose works were published in the earlier half of the present century. Of the synonymous terms I have used to designate this geometry of which I am speaking, the term. Modern Syn- thetic Geometry is the most comprehensive. Descriptive Geometry was invented by Gaspard Monge (1746-1818) in 1794 and at that 'time embraced only the theory of making projections of any accurately defined figure such that from these projections can be deduced, not only the projective proper- ties of the figure, but also its metrical properties. Now this term is used to designate the entire theory and development of geometry as embraced, in the above definition. GEOMETRY. 423 The problem, To draw a third straight line through the inaccessible point of intersection of two (converging) straight lines, \s both metrical and descriptive, that is to say, the required line may be found either by metrical or descriptive geometry, but the method by Descriptive Geometry is far the simpler. The following are the solutions by both methods : «{' Metrical. I. Given the two converg- ing lines AB and C£) which do not intersect in . an acces- sible point. II. Required to draw a third line through the inac- . cessible point K. 1. Draw the transversal LM, intersecting AB and CZ? in E and F respec- tively. 2. DrawjV/'parallelto^/' knd intersecting AB and CD in G and // respec- tively. 3. Divide .£'/^ in any ratio, say 1 : 2, and let Q be the point of division. 4. Divide G/T in the same ratio and let 7? be the point of division. 5. The line through QJ? is the line required. 1. Suppose the line join- ing the inaccessible point A' and the point Q to in- tersect NP in H', if not in Ji. CI o ■■a s CO - I Descriptive. < 1. Choose some point /" out- side the two given straight lines AB and CD. 2. Pass through this point any number of transversals, as FP, HP, KP. 3. Draw the diagonals FG, HE, HI, and KG. 4. The points of intersection L and M lie upon the line which passes through the point of intersection of AB and CD. The proof of this follows from the important harmonic prop- erties of a quadrangle. 424 FINKEL'S SOLUTION BOOK. O 2. Then, from similar tri- angles, KR:KQ=R!'G: QE. 3. Also , KK : KQ=R'H : QF- 4. .-.KG: QE=E!tf:QF. 5. '&Vi\.,QE:QF—\:1.'&y Hyph. 6. .■.R'G:R'H=1:2. 7. B\itRG:RH=l:2. By Const. 8. .■.RG:RH=R'G:R'H. 9. .-.RG^R'G and the point J?' coincides with.^. Many of the properties of the Conic Sections which are estab- lished with great labor and diii&culty by Analytical Geometry are easily and elegantly proved by Descriptive Geometry. Descriptive Geometry stands among the first of the branches of pure mathe- matics in point of interest and simplicity of its methods. The best works on this subject are Luigi Cremona's Elements of Pro- jective Geometry, translated by Charles Leudesdorf, and Theodore Reye's Lectures on Geometry of Position, Part I., translated by .Thomas F. Holgate, and Halsted's Projective Geometry. II. ON GEOMETRICAL REASONING. 5. On Geometrical Reasoning. We are accustomed to speak of mathematical reasoning as being above all other, in accuracy and soundness. This is not correct, if we mean by reasoning the comparing together of different ideas and pro- ducing other ideas from the comparison ; for, in this view, mathe- matical reasonings and all other reasonings correspond precisely. The nature of establishing mathematical truths, however, is totally different from that of establishing a truth in history, political economy, or metaphysics, and the difference is this, viz., instead of showing the contrary of the proposition asserted to be only improbable, it proves it at once to be absurd and impos- sible. For example, suppose one were to ask for the proof of the assassination of Caesar, what would be the method of proof? No one living (o-day is absolutely certain that Caesar was assas- sinated, and, in order to establish this truth, we refer to the testi- mony of historians, men of credit, who lived and wrote their accounts in the very time of which they write ; the statements of these historians have been teceived by succeeding ages as true ; and succeeding historians have backed their accounts by a mass of circumstantial evidence which makes it the most improb- able thing in the world that the account or any particular part of it is false, Ir> this way we have proved that the truth of the GEOMETRY. 425 ;statenient rests on a very high degree of probability, though it ■does not rise to absolute certainty. "In mathematics, the case is wholly different. It is true that the facts asserted in these sciences are of a nature totally distinct from those of history; so much so, that a comparison of the evidence of the two may almost excite a smile. But if it be remembered that acute reasoners, in every branch of learning, have acknowledged the use, we might almost say the necessity ■of a mathematical education, it must be admitted that the points •of connection between these pursuits and others are worth attend- ing to. They are the more so, because there is a mistake into which several have fallen, apd have deceived others, and per- haps themselves, by clothing some false reasoning in what they called a mathematical dress, imagining that, by the application of mathematical symbols to their subject they secured mathe- matical argument. This could not have happened if they had possessed a knowledge of the bounds within '^vhich the empire of mathematics is contained. That empire is sufficiently wide, and might have been better known, had the time which has been wasted in aggressions upon the domains of others, been spent in exploring the immense tracts which are yet untrodden."* In ■establishing a mathematical truth, instead of referring to authority, we contirlually refer our statements to more and more evident statements, until at last we come either to definitions or to state- ments so evidently true, that to deny them would prove the un- soundness of him who makes the denial. Geometry must have recourse to the outside world for its first notions and premises, and is, therefore, a natural science. Yet there is a great difference, between it and the other natural :sciences. For example, contrast Geometry and Qhemistry. Both ■derive their constructive materials from sense-perception ; but while Geometry is compelled to draw only its first results from •observation and is then in a position to move forward deductively to other results without being under the necessity of making fresh observations. Chemistry, on the other hand, is still com- pelled to make observations and to have recourse to nature. III. ON THE ADVANTAGES DERIVED FROM THE STUDY OF GEOMETRY, AND MATHEMATICS IN GENERAL. 6. On the Advantages derived from the Study of Geometry and Mathematics in General. Th« story is told of Abraham Lincoln that before he began the study of law, he worked through Euclid in order to give his mind that training in logical thinking so necessary to a successful lawyer ; '■'De Morgan, Study of Mathematics. 426 FINKEL'S SOLUTION BOOK. and his great success as a lawyer and statesman is largjely to be attributed to the discipline he thus received. There should be no conflict between, the sciences, and the classics. , A student taking a college course should give his time to study in both. The study of language enables a person to express his thoughts accurately and clearly while the study of the sciences provides him with thoughts worthy of expression. How far each of these two great departments should be pursued by the student, must be determined by the student himself. But certainly neither should be pursued exclusively. Yet if one were to pursue one or the other of these two great departments of knowledge exclusively, I heartily agree with Professor Earnst Mach who says, "Here I may coiint upon assent when I say that mathematics and the natural sciences pursued alone as means of instruction yield a richer education in matter and form a more general education, an education better adapted to- the needs and spirit of the time, than the philological branches pursued alone would yield."* As to ' mathematics, "It is admitted by all that ' a finished or even a competent reasoner is not the work of nature alone ; the experience of every day makes it evident that education develops faculties which would otherwise never have manifested their existence. It is,, therefore, as necessary to learn to reason before we can expect to be able to reason, as it is to, learn to swim or fence, in order to attain either of these arts. Now, sometTiing must be reasoned, upon, it matters not much what it is, provided it can be reasoned upon with certainty. The properties of mind' or matter, or the study of languages, mathematics, or natural historvy may be chosen for this purpose. Now, of all these, it is desirable to- choose the one which' adr^its of the reasoning being verified, that is, in which we can find out by other means, such as measurement and ocular demonstrations of all sorts, whether the results are true or not. . . . Now the mathematics are peculiarly well adapted for this purpose, on the following grounds : 1°. Every term is distinctly explained, and has but one mean- ing, and it is rarely that two words are employed to mean the same thing. 2°. The first principles are self-evident, and, though derived from observation, do not require more of it than has been made by children in general. 3° The demonstration is strictly logical, taking nothing for granted except the self-evident first principles, resting nothing upon probability, and entirely independent of authority or opinion. 4°. When the conclusion is attained by reasoning, its truth or falsehood can be ascertained, in geometry by actual measure- '■'See Professor Mach's Popular Snipntijic Lectures, " On Instruction in the Classics and Sciences." Also Grant Allen's Article in.the Oct. No. of the LosmopoUtan for 1897. GEOMETRY. 427 ment, in algebra by common arithmetical calculation. This gives- confidence, and is absolutely necessary, if, as was said before, reason is not to be instructor, but pupil. 5°. There are no -words whose meanings are so much alike that the ideas which they stand for may be confounded. Be- tween the meanings of terms there is no distinction, except ab- solute distinction, and all adjectives and adverbs expressing dif- ference of degree are avoided. Thus it may be necessary to say,. "A is greater than B ;" but it is entirely unimportant whether A is very little greater than B or very much greater than B. Any proposition which includes the foregoing assertion will proves its conclusions generally, that is, for all cases in which A is greater thaij B^ whether the difference be great or little. . . . "These are the principal grounds on which, in our opinion, the utility of mathematical studies may be shown to rest, as a dis- cipline for the reasoning powers. But the habits of mind which these studies have a tendency to form are valuable in the highest degree. The most important of all is the power of concen- trating the ideas which a successful study of them increases where it did exist and creates where it did not. A difficult position, or a new method of passing from one proposition tO' another, arrests all the attention and forces the united faculties to use their utmost exertions. The habit of mind thus formed soon extends itself to other pursuits, and is beneficially felt in all the business of life. "As a key to the attainment of other sciences, the use of the- mathematics is too well known to make it necessary that we should dwell on this topic. In fact, there is not in this country any disposition to undervalue them as regards the utility of their applications. But though they are now generally considered as a part, and a necessary one, of a liberal education, the views which are still taken of them as a part of education by a large proportion of the community are still very confined,"* The advantages derived from a study of geometry, though very great, are only part of those to be derived from a thorough course of study in mathematics. The eminent mathematician Cayley, "the central luminary, the Darwin of the English School of Mathematicians," as Sylvester calls him, said once that if he had to make a defence of mathematics he would do it in the manner in which Socrates, in Plato's "RepubHc" defended jus- tice. Justice, according to the Greek sage, was a thing desir- able, in itself and for its own sake, quite irrespective of the worldly advantages which might accompany a life of virtue and justice. So just for the sake of learning the beauties and the purest truths which mathematics, the oldest and the noblest, the grandest and the most profound of all sciences, represents, * De Morgan, The Study of Mathematics, '428 ' FINKEL'S SOLUTION BOOK. would it be worth while to make ourselves acquainted with its uses as an educational medium and the application it finds in other sciences? Sylvester says,- "The world of, ideas which mathematics discloses or illuminates, the contemplation of divine beauty and order which it induces, the harmonious connection of its parts, the infinite hierarchy and absolute evidence of truths with which mathematical science is concerned, these, and such like, are the surest grounds of its title to human regard." Sylvester, twenty-five years ago called the attention of the Royal Society to the parallelism between the mathematical and musical ■ethos : music being the mathematics of the senses, mathematics the music of reason ; the soul of each the same. Music the -dream, mathematics the working life; each to receive its con- summation from the other, when the human intelligence elevated to its perfect type, shall shine forth glorified in some future Beethoven-Gauss. There is surely something in the beauty of the truths them- selves. .They enrich us by our mere contemplation of them. What a charm and what a wealth of delight and self-content- ment does the finding of mathematical truths afford. In this science, of which geometry is one, out of a few postulates and ;germinating truths, the mind of man can gradually unfold a system of new and beautiful truths never dreamt of before. Locke says', "The mathematician from very plain and easy be- ginnings, by gentle degrees, and a continued chain of reason- ings, proceed to the discovery and demonstration of truths that appear at first sight beyond human capacity." Because mathe- matics is a science of pure reason and rigorous logic a mathe- matician may forget all the preceding propositions of his science and still be able to' guide himself with the utmost confidence through the labyrinth of ideas and reach its exit, if he only keeps clearly before him the ends of the threads of thought. "It is due to the peculiarity of Mathematics, which' is a chain of inseparable reasonings, that one part of it can hardly be studied to the exclusion of the others ; that in order to under- stand the whole, only hard and persistent work, the greatest perseverance and the greatest caution, in which all our mental powers and capabilities have to be brought into pflay, can lead us to the great victory of the mind and enable us to comprehend and see the beauties of pure truths which this magnificent branch of Science represents. To all these peculiarities is due the fact that only a limited number of people are capable of, appreciating the beauties of this oldest of all sciences." No fault has ever been found with Mathematics by the true student. He who has the courage to study diligently in any line of work, can obtain the same results when studying Mathematics with the same diligence and care. As the dril^l will not penetrate the granite unless kept to the work hour after hour, so the 'mind GEOMETRY. 429. will not penetrate th'e secrets of Mathematics unless held long and vigorously to the work. As the sun's rays burn only when concentrated,' so the minfl achieves mastery in Mathematics and indeed in every branch of knowledge only when its possessor hurls all his forces upon it. Mathematics, like all the other sciences, opens its' door to those only who knock long and hard. No more damaging evidence can be adduced to prove the weak- ness of character than for one to have aversion to mathematics; for whether one wishes so or not, it is .nevertheless true, that to have aversion for mathematics means to have aversion to accurate, painstaking, and persistent hard study and to have aversion to hard study is to fail to secure a liberal education, and thus fail to compete in that fierce and vigorous struggle for the highest and the truest and the best in life which only the strong can hope to secure. But we do not judge a painting by the pumber of its admirers. It is as a rule the lowest kind of art which attracts the largest number of admirers. In this practical world, in this w'orld of hard struggle for life,, where the guiding principle is "swim who can and those wha can't may drown," it may not, perhaps, be admissable to judge of the value of a science by its inherent beauty, but rather 'by the share it contributes to the education of our mental faculties, and by the applications it finds in the useful arts and sciences and thus in what measure it contributes to the civilization of the world. He who reads history with some critical judgment cannot fail 'to notice that the degree of civilization of a country- is closely connected with the standard of Mathematics in that country, and this fact is attested by the fierce bidding for the- best mathematicians in the world by such countries as France,, Russia, and Prussia during the latter. part of the last century. Prof. H. J. Stephen Smith, of Oxford, says, "I should not wish to use words which may seem to reach too far, but I often find the conviction forced upon me that the increase of rriathematical- knowledge is a necessary condition for the advancement of science, and if so, a no less necessary condition for the improve- ment of mankind. I could not augur well for the enduring in- tellectual strength of any nation of men, whose education was not based on solid foundation of mathematical learning and whose scientific conception, or in other words, whose notions of the world and of things in it, were not braced and girt together with a strong framework 'of mathematical reasoning."' Fourier, one of the greatest mathematicians of France, on the- completion of his great work on Theory of Heat, says, "Mathe- matics develops step by step, but its progrcrss is steady and cer- tain amid the continual fluctuations and mistakes of the human mind. Clearness is its attribute, it combines disconnected facts, and discovers the secret bond that unife^i. them.. When air and 430 FINKEL'S SOLUTION BOOK. light and the vibratory phenomena of electricity- and magnetism seem to elude us, when bodies are removed from us into the infinitude of space, when man wishes to behold the drama of the heavens that has been enacted centuries ago, when he wants to investigate the effects of gravity and heat in the deep, im- penetrable interior of our earth, then he calls to his aid the help of mathematical analysis. Mathematics renders palpable the most, intangible things, it binds the most fleeting phenomena, it calls down the bodies from the infinitude of the heavens and opens up to us the interior of the earth. It seems a power of the human mind conferred upon us for the purpose of recom- pensing us for the imperfection of our' senses and the shortness of our lives. N.ay,, wha,t is still more wonderful, in the study •of the most diverse phenomena it pursues one and the same method, it explains them all in the same language, as if it were to bear witness to the unity and simplicity of the plan of the universe." Mathematics is the very embodiment of truth. No true de- votee of mathematics can be dishonest, untruthful,- unjust. Be- •cause working ever with that which is true, how can one de- velop in himself that which is exactly opposite? It would be as though one who was always doing acts of kindness i should •develop a mean and groveling disposition. Mathematics there- fore has ethical value as well as educational value. Its prac- tical value is seen about us every day. To do away with every one of the many conveniences of this present civilization in which some mathematical principle is applied, would be to turn the finger of time back over the dial of the ages to the time vi'hen man dwelt in caves and crouched over the bodies of wild beasts. The practical applications of mathematics have in all ages re- ■dounded to the highest happiness of the human race. It rears magnificent temples and edifices, it bridges our streams and rivers ; it sends the railroad car with the speed of the wind across the continent ; it builds beautiful ships that sail on every sea; it has constructed telegraph and telephone lines and made a messenger of something known to mathematics alone that bears messages of love and peace around the globe ; and by these marvellous achievements, it has bound all the nations of the earth in one common brotherhood of man. IV. AXIOMS. . 7. The self-evident first principles of which mention was made in the previous section are called axioms. Thus, A can not be both B arid non-B at the same time ; A horse is a horse; Two times two are four; A body in motion will remain in motion, unless icted upon by some external force. The following are the axioms used in mathematics : GEOMETRY. 431 GENERAL AXIOMS. 1. Things equal to the same thing are equal to each other. Thus, if A=B and B=C, then A=C. ■2. If equals are added to equals, the sums are equal. Thus, Ki A=B and C=D, then A-^C=B^D. 3. If equals be taken from equals the remainders are equal. Thus, if A=B and C=D, then A—C=B—D. ■4. If equals be added. to unequals the sums are unequal iii the same order^ or sense. Thus, if A is greater than B and C=D, then A-\^C is greater than B+D. 5. If equals be taken from unequals the remainders are unequal in the same sense. Thus, if A is greater than B. and C=D, then A — C is greater than B—D. ■6. If unequals be taken from equals the remainders are unequal in the opposite sense. Thus, if A is greater than B and C is equal to D, then C — A is less than D—B. 7. If equals be multiplied by equals, the products are equal. Thus, if v4=5 and C=A then ^C=50. 8. If unequals be multiplied by equals, the products are unequal in the same sense. Thus if v4 is greater than B and C=D, then AC is greater than BD. 9. If equals be divided by equals, the quotients are equal. Thus, i{A=B and C=Z>, then -^=-^- 10. If unequals be divided by equals, the quotients are unequal in the same sense. A B Thus, if A is greater than B and C=D, then —=r is greater than -=-. 11. If unequals be a,dded to' unequals, the greater to the greater and the lesser to the lesser, the sums will be unequal in the same sense. Thus, if A is greater than B and C greater than D, then A-\-C is greater than B-\-D. If m is less than « and / less than q, then m-\-p is less than n-{-q. 12 The whole is- greater than any of its parts. . Thus,'if «!, flj, flj, ai are parts oi A, then A is grerter than any of the a's. IS The whole is equal to the sum. of all its parts. Thus, if fli, 32, a,, a^, a^ are the parts of ^, then A=a.^-\-a^-\-a^^ a^+a^. 14 Magnitudes which coincide with one another are equal to one another. Thus, if A coincides with B, then A and B are equal. 15. If of two unequal quantities, the lesser increases continuously and in- definitely while -the other decreases continuously and indefinitely they must become equal once and but once. Thus, if, in the iigure, the line EF moves parallel 6 to itself, keeping its ex- V tremities in AB and MN and the line GH E^ moves parallel to itself Ay\ keeping its extremities in j^TiV and CD, then ^ — i the two lines are equal ^ '^ once and only once, viz., when both are equal to the line IK'. 432 . FINKEL'S SOLUTION BOOK. 16. If of three quantities t fie first is greater than the second and the second' greater than the third, then the first is greater than the third. Thus, if A is greater than B and B greater than C, then A is greater than C. 17. Two straight lines can not iiiclose a [finite] space. V. ASSUMPTIONS. 8. In addition to the definitions of geometrical magnitudes* and the above axioms the following Assumptions, or Postu- lates, are needed: {a.) ASSUMPTIONS OF THE STRAIGHT LINE. (i.) One and only one straight line may he passed through every two points in space; or, briefly, two points determine a straight line. (2.) Two straight lines lying in a plane, determine a point. If the two lines are parallel, we still say, for the sake of generality and in harmony with conventions adopted in modern geometry, that the two- lines intersect in a point, the point infinity. By taking this view of two parallel lines, many theorems are stated and proved without exceptions to either statement or proof. (3.) Through any point in space a line may be drawn and revolved about this point as a center so as to include any assigned' point. (4.) A straight line-segment, or a sect, may be produced so- as to have any desired length. (5.) A straight line is divided into two parts by any one of its points. (b.) ASSUMPTIONS OF THE PLANE. (i.) Three points not in the same line determine a plane. (2.) A straight line through two points in a plane lies wholly in the plane. (3.) A plane may be passed through a straight line and re- volved about it so as to include any assigned point in spac^. (4.) A portion of a plane may be produced to any desired extent. (5.) A plane is divided into two parts by any of its straight lines. (6.) A plane divides space into two parts. (c.) ASSUMPTION OF PARALLEL LINES. ( I. ) Through a point without a straight line, only one straight- line can be drawn parallel to that lin&. This assumption is a substitute for Euclid's famous eleventh (also called the twelfth) axiom which reads, // a straight line meet tzvo straight lines so as to make the two interior angles- *For definitions of geometrical magnitudes, see Mensuration. GEOMETRY. 433 on the same side^ of it taken together less than two right angles, these straight lines being continually produced shall at length- meet on that side, on which are the angles which are less than, two right angles. An axiom must possess the following properties : ( i ) must be self-evident, (2) must be incapable of being proved from other axioms. That the above so-called axiom does not pos- sess the first of these requisites is proved by the fact that there is a dispute among mathematicians as to whether it is an axiom or not. However, it does satisfy the second criterion as, so far, no valid proof of it from other axioms has ever been given. Many proofs have, indeed been given, but it requires very little thought to see that these proofs are all fallacies of Petitio Principii. > The many attempts to give a rigorous and valid proof of this- assumption, for such it is, has redounded to the eternal glory of geometry in that not only is Euclidean Geometry preserved in all its original purity and integrity but other geometries equally cogent and consistent have been created. The subject is too abstruse for my present purpose and so I shall do nothing more than show the point of departure of these geometries. 1. Let AB be a given straight line, and P the given point. 2. Through P draw any number of lines. 3. These lines, in relation to the given line, divide themselves into two classes, viz., cutting and NON-CUTTING. Now of the class, non-cutting, how many lines are there? On the answer to this q.uestion "hang all the law and the prophets." A priori, three answers are possible, viz., none, one, many. If we say "none," we have Spherical Geometry ; if we say "one," wehave J^Hc/idean Geometry; if we say "more than, one," we have Psendo- Spherical Geometry. It is true that the answer, "one," is the answer that is usually insisted upon as being the only possible answer. But this an- swer is based upon experience and is not, therefore, a priori. In these geometries, the properties of figures are studied, which figures lie in space, or surfaces, possessing the property that the product of the principal radii of curvature at every point of the surfaces shall be constant. If this product is positive, the surface is spherical and the geometry trfeat'ing of the figures of this surface is Spherical Geometry; if this prpduct is 0, the 434 FINKEL'S SOLUTION BOOK. surface is a plane, and the geometry treating of the 'properties of figures lying in this surface is the ordinary Euclidean Geom- etry; if this product is negative, the surface is pseudo-spherical and the geornetry treating of the properties' of this space is Pseudo-Spherical Geometry. In the above discussion, it has been assumed tacitly that the measure of a distance remains everywhere the same. Professor Felix Klein has shown that if this be not the case and if the law of measurement of distance be properly chosen, we can obtain three systems of plane geometry analogous to the three systems mentioned above. These are called respectively MUip- tic, Parabolic, and Hyperbolic Geometries, meaning ■lacking, equaling, and exceeding. Instead of the above terms given by Klein, we often meet Riemannian, Euclidean, and I/obatscbevskian, from Riemann, Euclid, and Lobat- schevsky, — mathematicians who first set forth clearly the prop- erties of the space-forms^ These geometries refer to hyper-space of two dimensions and are called collectively non-Muclidean Geometry. The notion of hyper-space of two dimensions naturally sug- gested the question as to whether there are different kinds of hyper-space of three or more dinjensions. Riemann showed that there are three kinds, of hyper-space of three dimensions having properties analogous to the three kinds of hyper-space of two dimensions already discussed. These hyper-spaces are differ- entiated by the test whether at every point no geodetical surface, or one geodetical surface, or a fasciculus of geodetical surfaces can be drawn parallel to a given surface, a geodetical surface being defined as such that every geodetic Ime joining any two points on it lies wholly on the surface. The student who would pujrsue the subject should read Dr. Halsted's excellent transla- tions of Lobatschevsky and Bolyai, the Lectures and Addresses ■of Clifford and Helmholtz, Ball's article on Measurement in the Encyclopedia Britannica, Professor Schubert's Essay' on the Fourth Dimension, Russell's Foundations of Geometry, and after- wards the monographs of Riemann, Klein, Newcomb, Beltrami, and Killing. For a full bibliography of the literature of the subject up to the time of its publication, see Bibliography of Non- Euclidean Geometry, by Dr. Halsted, American Journal of Mathematics. (rf.) ASSUMPTIONS OF THE CIRCLE. (i.) A circle may be constructed with any point as center, and with a radius equal to any given sect. (2.) A circle has but one center. ,(3). All radii of the same circle are equal, and, hence all .diameters of the same circle are equal. GEOMETRY. 435 (4.) // an unlimited straight line passes through a point within a circle, it must cut the circumference at least tivice. That it can not cut the circumference more than twice is a theorem. ' The region within a circle is defined as that fr.om any point of which no tangents cari be drawn to the circle. (5.) // one circumference intersects another once, it inter- sects it again. (e) ASSUMPTIONS OF THE SPHERE. (i.) A sphere may be constructed with any point as centeY, and with a radius equal to any given sect. (2.) A sphere has but one center. (3.) All radii of the same sphere' are equal, and, hence all diameters of the same sphere are equal. (4.) // an unlimited straight line passes through a point within a sphere, it must cut the surface at least twice. (5.) // an unlimited plane or if a spherical surface, intersects a spherical surface, it must intersect it in a closed line. (/) ASSUMPTIONS OF MOTION. (i.) A figure may be moved from one position in three di- mensional space to any other position in the same space without altering the size or shape of the figure. By this we mean that a figure may be picked up, turned over in any way, and moved to any other position in space without changing the size or shape of the figure. The proof of many theorems in geometry depends upon this assumption. (2.) A figure may he moved- about in space while one of its points remains fixed. Such movement is called " rotation about a center," the center being the fixed point. (3.) A figure may he moved about in space while two of its points remain fixed. Such movement is called " rotation about an axis," the axis being the line determined by two fixed points. In the higher mathematics and in Physics and other natural sciences other assumptions are needed.' VI. ON LOGIC. 9: On I/Ogic. — As a preliminary to the study of geometry a short discussion of the Methods of Reasoning will be of value. In geometry we are concerned with propositions about space relations. Ideas are images of an object formed by the mind. Words are the spoken or written signs of ideas. 10., A judgment is an. act of the mind affirming a relation between two objects of thought by means of their conceptions. 436 FINKEL'S SOLUTION BOOK. 11. A proposition is a judgment expressed in words. For example, take the ideas represented by "all mushrooms" and "things good to eat," posit these ideas in the mind and dis- cerq the agreement or disagreement of these two ideas, then express the agreement or disagreement in words. It comes out thus, "All mushrooms are things good to eat." Our senses are the instruments by which the qualities of a mushrooms are made known to us. Having found this mushroom, good to eat, and this one, and this one, and so on, tbgether with the experience of the race, we arrive at the conclusion, by in- ductive inference, that "all mushrooms are good to eat." It must be borne in mind that by induction )ve gain no certain knowledge. If the observation of a number of cases shows that alloys of metals fuse at lower temperatures than their constituent metals, we may with more or less probability draw the general inference that , All alloys melt at a lower temperature than their constituent metals. But this can never rise to the rank of an absolutely certain law until all possible cases have been examined. Not olie of the inductive truths which men have established, or think they have established, is really safe from exception or reversal. Lavoisier, when laying the foundations of chemistry, met with so many instances tending to show the existence of oxygen in all acids that he adopted the general conclusion that all acids, contain pxygen, yet subsequent experience has shown this to- be false. Like remarks may be made concerning all other in- ductive inferences, the method never leqding to absolute certainty. 12. The Powers of the Mind engaged in knowledge are the following three, viz., ( I ). The Power of Discrimination, (2) The Power of Detecting Identity, and (3) The Power of Retention. 13. The Laws of Thought are the following three, viz., ( 1 ) The Law of Identity ; as, That which is, is. (2) The Law of Contradiction; as, A thing cannot both be and not be at the same time. (3) The Law of Duality; as, A thing must either be or not be. To these some logicians add a fourth called the "Law of Suf- ficient reason ;" Every effect has a cause. 14. When we join terms together we make propositions; when ive join propositions together we make an argument, or piece of reasoning. GEOMETRY. 437 15. Terms. A concrete term has two meanings, viz., (i) things to which the term applies, and (2) the qualities of those ■things in consequence of which the term is applied. The num- ber of different things to which a term is applied is called its extension, while the number of qualities implied is called its intension. For example, "table" has a larger "extension" than "round table" for the former term applies to a larger number of objects; the latter has the greater "intension" for it includes all the quali- ties that the term "table" does and the additional quahty "round." The word "term" comes from the Latin terminus, meaning end and is so called because it' forms one end of a propos^ition. 16. Propositions. Every proposition is composed of a subject, ( Lat., sub, under, and jectum, laid), a copula, and a predicate (Lat. praedicare, to assert). in the proposition,- "All mushrooms are things good to eat," "all mushrooms" is the subject, "are" is the copula, and "things good to eat" is the predicate. Of the kinds of propositions we have (i) Categorical; As A is B. A is not B; (2) Con- ditional; as. If a triangle is equiangular, it is equilateral. Conditionkl Propositions are divided into twc classes, viz., Hypothetical and Disjunctive. The following is a disjunctive proposition: A I is either B or C. Of the Categorical Propositions we have, A. The Universal AMrmative; as, All horses are animals. E. The Particular AMrmative; as. Some animals are horses. I. ' The Universal Negative; as. No horses are cows. O. The Particular Negative; as, Some animals are not horses. Every proposition which expresses accurately a thought, can be reduced to one of the above forms, though the reduction in many cases is not apparent. For example, Parallel lines never meet, reduces to Parallel lines are lines which never meet. The hypothetical proposition, "If gunpowder be damp, it will not explode" reduces to, "Damp gunpowder will not explode." When 'we make a statement about all the objects which can be- included under a term, we use the term Universally, as logicians say, that is to say, the term is distributed. In the proposition, "all men are mortal," the term "men" is distributed, iDecause the little word "all" indicates that the statement applies 438 FINKEL'S SOLUTION BOOK. to any and every man. But the predicate "mortal" is only TAKEN PARTICULARLY AND IS NOT DISTRIBUTED. Therefore, we see that a Universal Affirmative distrib- utes ITS subject but not its predicate. As a universal negative proposition take, "No sea-weed is a flowering plant." The subject "sea-weed" is distributed. If there could be found a single flowering plant which is a sea- weed, then the proposition would not be true. Hence the predi- cate is also distributed. Hence, the universal negative propqsition distributes it-s SUBJECT AND ITS PREDICATE. , No difficulty is experienced in seeing that the particular affirma- tive' distributes neither its subject nor its predicate, a,nd that* the PARTICULAR NEGATIVE DISTRIBUTES ITS PREDICATE BUT NOT .ITS SUBJECT. In the absence of any knowledge to the contrary, the word "some," in the particular affirmative and particular negative, must be taken to mean "some and.it may be all." ■ 17. The Law of Converse. Two propositions are the converse of each other when the subject of one is the predicate, of the other. Thus, , "Equilateral triangles are equiangular, "(direct). , Equiangular triangles are equilateral, (converse). It does not follow, that because a proposition is true its con- verse will also be true. Thus, "AH regular polygons are equi- lateral (direct); -all equilateral (polygons) are regular, (con- verse). This last is not true. The converse of all definitions are true. Whoever three theorems have the ^following relations, their converses ci/e true : ' > 1. If it is known that when A> B, then x> j/, and 2. If it is known that when A = B, then x = y, and I 3. If it is known that when A <. B, then x Ky, then' the converse of each of thesfe is true. For , • ■?> Ij. li x> y, then A cannot equal B and A cannot be less than B without violating 2 or 3; .-. A> B. (Converse of 1.) 2^. U X — y, then A cannot be greater than B and Acantiot be less than B without violating 1 or 3; .". A=B. (Converse of 2..)' 3i. li X <y, then A cannot be greater than B and A cannot be equal to B without violating 1 or 2; .'. A < B. (Converse of 3.) 18. The opposite of a proposition is formed by stating the negative of its hypothesis and conclusions. Thus, li A = B, then C = D (Direct.) If A is ;not equal B, then C is not equal D. (Opposite.) GEOMETRY, , 439 19. // the direct proposition and its converse are true, the opposite proposition is true; and if a direct proposition and its opposite are true, the converse proposition is true. Thus, 1. li A=B, C = D. (Direct.) li C = D, A—B. (Converse.) If A is not equal to B, C is not equal to D (Opposite.) 2. li A=B, C^D. (Direct.) ' If A is not equal to B, C is not equal to D. (Opposite.) Then, if C = D, A = B. (Converse.) 20. Method^ of Reasoning. There are two methods of reasoning, viz., the Inductive and the Deductive. The Inductive Methodh used in ;-eaching a general truth or principle by an examination and comparison of particular facts. Thus, This apple is equal to the sum of all its parts, this piece of crayon is equal to the sum of all its parts, this orange is equal to the sum of all its parts, and so with peaches, pears, balls, pebbles, slates, knives, and chairs. Therefore, the whole of any object is equal to the sum of all its parts, or the whole is equal to the sum of all its parts. This is inductive Veasoning. The Deductive Method is used in reaching a particulai- truth or principle from general truths or principles. Thus. All animals suffer pain. ' Flies are animals. Therefore, flies suffer pain. 21. Tie Syllogism.^ When we compare propositions we reason. Deriving a third proposition from two given proposi- tions is called syllogistic reasoning, or Deductive Rea- soning. Thus, 1. All English silver cc«ns are coitied at Tower Hill. 2. All sixpences are coined at Tower Hill. Therefore, All sixpences are English silver coins. The last proposition is called the conclusion, the other two propositions are called premises, and the three together the syllogism. ' Again, All electors pay rates. A. No paupers pay rates. E. Therefore, no paupers are electors. E. From the examples given, we see that there are only three terms or classes of things reasoned about; in the first example the three terms are "All Enghsh silver coins," "Tower Hill," and "all sixpences." Of these, the 'class, " English silver coins," does not OQcur in the conclusion. It is used to enable us to compare together the other two classes of things. It is called 440 FINKEL'S SOLUTION BOOK. the middle term. (Things) "coined at Tower Hill," is called -the major term, for the reason that it has the larger exten- sion, and "sixpences," the subject of the conclusion, is called the minor term of the syllogism, for the reason that it has a lesser extension than the subject of the conclusion. The premise in which the "major term" is found is called the major premise, and the one in which the minor term is found is called the minor premise. Hence, the middle term is always the term not found in the conclusion ; the major term is the predicate of the conclusion ; and the minor term is the subject of the conclusion. Suppose that the two premises and the conclusion of the last syllogism be varied in every possible way from aMrmaitive to negative, from universal to particular and vice versa. Each proposition can be converted into four different propo- sitions and each one of these four may be compounded with any one of the other two. Hence the number of changes (called moods) is 4 X 4 X 4 = 64- These moods may be still further varied, if instead of the middle term being the subject of the first and the predicate of the second, this ofder may be reversed, or if the middle term the subject of both,' or the predicate of both. In this way we see that for each of the sixty-four moods we get four syllogisms called £gures. Of the sixty-four moods, . there are altogether nineteen moods of the syllogism that are admissible. 22. Rules of the Syllogism. To find out whether an argument is valid or not, we must examine it carefully to ascer- tain whether it agrees with certain rules discovered by Aristotle. Modern logicians have to some extent broken away from these rules. Without going into the matter in detail we state these rules. I. Bvery syllogism has three terms and only three. These terms are called the the major term, the minor term, and the middle term. II. Mvery syllogism contains three and only three propositions. III. The middle term must be distributed once at least in the premises and must not be am biguo us. Some animals are flesh-eating. Some animals have two stomachs. / y^ ^nimat No conclusion can be drawn. But if we say. Some animals are flesh-eating. All animals consume oxygen, we can say Therefore, some animals consuming oxygen are flesh-eating. GEOMETRY. 441 IV. If both premises are negative no conclusion can he (}rawn. For, from the statements that two things disagree with a third, no proof of agreement or disagreement can "be established. Thus the following is incon- clusive, No Americans are slaves. No, Turks are Americans. V. If both premises are particular no conclusion can be drawn. Thus the following are inconclusive : Some Americans are ignorant. Some Europeans are ignorant. Some laws are unjust. Some men are unjust. VI. No term must be distributed in the conclusion ^bicb was not distributed in the premises. From Some animals eat flesh. All animals consume oxygen. We must conclude that some things that consume oxygen eat flesh. VII. If one premise be negative the conclusion must be negative. Thus from Ry\ V^c^ (3 % 9) 0, All negroes are dark. No American is dark. We draw the conclusion No Atiierican is a negro. VIII. If either premise is particu- lar the conclusion must be particular. Thus, All negroes are black. Some horses are black. Therefore, some horses are not negroes. 23. logical Fallacies. Logical Fallacies result from our -neglect to observe the rules of logic. They occur in the mere form of the statement, that is, in dictione, as it is known in logic. 442 . FINKEL'S SOLUTIOI^ BOOK". There are four purely logical fallacies,, viz.,. I. Fallacy of four terms (Quaternio Terminorum), » Violation of Rule I. • 2. Fallacy of undistributed middle, — Violation of Rule- Ill. 3. Fallacy of illicit process, of the major or minor term.. — Violation of Rule VI. 4. Fallacy of negative premises. — Violation of Rule . IV. There are six semi-logical fallacies, viz., 1. Fallacy of Equivocation. 2. Fallacy of Amphibology. '3. Fallacy of Composition. 4. Fallacy of Division. 5. Fallacy of Accent. 6.' Fallacy of Figure of Speech. In addition to these logical fallicies there are seven Material Fallacies {extra dictionem) that is, • fallacy in the matter of thought, viz., I. Fallacy of Accident. The Converse Fallacy of Accident. 3. The Irrelevant Conclusion. 4. The Petitio Principii'. The Fallacy of the Consequent or Non-sequitur. The False Cause. The Fallacy of Many Questions. We will illustrate some of these fallacies. Light is contrary to darkness. Feathers are ligh,t. .-. Feathers are contrary to darkness. The middle term, "light," has two different meanings in the premises. We have, therefore, four terms instead of three, which violates Rule I. When the middle term is airibiguous, the fal- lacy is known as the ambiguous middle. Every country under a tyranny is distressed. This country is distressed. . . This country is under a Tyranny. — Fallacy of Undis- tributed Middle. All moral beings are accountable. No brute is a moral being. .-. No brute is accountable. — Fallacy of the Illicit Process of the Major Terni. Some- men are not just. No angel is a man. .-. Some angels are not just. — Fallacy of Negative Premises... GEOMETRY. 443 .EXAMPLES. Seven is one number. Two and five are seven. .-. Two and five are one number. — Fallacy of Division. Three and four are two numbers, Seven is three and four. • .-. Seven is two numbers. — ■ Fallacy of Coiiiposition. The duke yet Hves that Henry shall depose. — Fallacy of Amphibology. .The conclusion depending upon the interpretation of the mean- ing of this proposition is doubtful. A hero is a lion. A lion is a quadruped. .-. A hero is k quadruped. — Fallacy of Figure of Speech. Thieves are dishonest ; ' But thieves are men ; .-.•All men are dishonest. — Fallacy of Accident. 24. How to Prepare a I^esson in Geometry. In be- ginning the study of geometry, great care should be taken to grasp a correct notion of the definitions and illustrations. The defi- nitions, axioms, and assumptions are the foundation on which rests the magnificent Structure of geometry. The definitions should be committed to memory, only committing them, however, as they occur in the prosecution of the study. Make haste slowly at first; one proposition per- lesson for the first three lessonS' is quite sufficient: and. two propositions may be takfen at a les- son for the next seven or eight lessons. After this, if the work is thoroughly in hand three propositions together with several originals should' constitute a lesson. In the preparation of the lesson, the student should carefully read the proposition so as to get its full meaning. After the meaning of the proposition is understood, carefully follow the demonstration in the book, never leaving a statement made in the demonstration until it is thoroughly understood. At first, it may be necessary to repeat this two or three times, perhaps oftener. After the given demonstration is thoroughly under- stood, close the book, draw a figure on paper or a slate, and write out a demonstration of your own. Compare your demon- stration with the one in the book, and make such corrections as are necessary. By carefully observing this method, it will be a comparatively short time until one reading of the lesson will generally suffice for the necessary preparation. The theorem should always be 444 FINKEL'S SOLUTION BOOK. •committed to memory, the demonstration never. It is not a bad practice to commit the proposition exactly as it is- stated in the book, for, as a general thing the author has put much time on the statement of each proposition endeavoring to reduce it to its siinplest and most elegant form, and upon this work, the student, as a riije, can not improve. • , In conducting the recitations, no books should be allowed to be consulted. The propositions should be assigned by stating them in part or ^n full to the students called upon to recite. The students so called upon, should go to the board and draw as ' neat and accurate figure as possible, accurate figures often sug- gesting truths not revealed by carelessly constructed figures. It is generally best not to require any part of the demonstration to be written out, unless, indeed, it includes long and complidated algebraic equations. In reciting, if it is convenient, the student should step to the board and, using a pointer in referring to the various parts of his figure, observe the following order in the ■discussion of the theorem : I. Statement of the Theorem. Here give an accu- rate statement of the theorem to be demonstrated. II. Given. Here state, with reference to the figure con- structed whatever is given by the theorem. III. Xo Prove. Here state the exact conclusion -to be de- rived from what is given. IV. Proof. Here set forth,' in logical order the statements to prove the conclusion 'just asserted. The validity, limitations, and general application of the the- =orem may then be discussed by the class. Corollaries coming under the various theorems in the lesson may be assigned to students other than those demonstrating the theorems. The proof of a corollary is usually simple, but its proof should be given with the same care and accuracy. We will now illustrate what we have said by a few proposi- tions. The student should have one of the following excellent texts : I Halsted's Elements of Geometry. Bem^ and Smith's Plane and Solid Geometry. Phillips and Fisher's Elements of Geometry." Wentworth's Plane and Solid Geometry. , GEOMETRY. 445- PLANE GEOMETRY. BOOK I. ANGLES AND STRAIGHT LINES. ' PROPOSITION I. I. Theorem. All straight angles are equal. 1 A C B 1 . D E -F ' II. Given any two straight angles ACB and DEF. III. To pro ve / ACB = / DEF. 1. Apply Z ACB to the / DEF, so that the vertex C shall fall on the vertex E. {First as sumption. of motion}) 2. Then revolve CB so that it contains the point F. (Third assumption of the straight line.) 3. Then CA will coincide with ED. {First assumption of a straight line and Law of Identity}) 4. .-. i.ACB = I DEF. Axiom 10. Corollary 1. All right angles are equal. A A IV. Proof. ^ 1. II. III. C Given -B C -B' any two right angles ACB and ACB'. To prove / ACB = / A'C'B'. IV. Proof. All straight angles are equal. Prop. I. / ACB and / A'C'B' are each the half of a straight angle. By definition. . 3, .-. / ACB = / A'C'B'. Axiom 7. 446 FINKEL'S SOLUTION BOOK. c 3. I. Cor. 2. Tke angular units, degree, minute, and sec- ond have cotistant values. II. Given -a degree angle. III. To prove that it is a constant magnitude. 1. A constant piagnitude is a magnitude whose value is always the same. By def . 2. A straight angle is a magnitude whose value is always the same. By* Prop. I. .'.A straight angle is a constant mag- nitude. A degree angle is one hundred eight-, ieth part of a straight angle. By def. ..■. A degree angle is a constant mag- nitude. By Aristotle's Dictum, — Whatever may be predicated of a whole may be predicated of a part. In like manner, we can prove that minute-angles and second- angles are constants. Cor. 3- Complements of equal angles are equal! IV. Proof. \ 1. 2. I. B » ^ — V II. Given the two equal angles CBD and C B' D and their complements ABC and A'B'C , respectively. III. To prove that / ABC = / A'B'C. fl. / ABC = the difference between a rt. / ^ and / CBD. By def. of comp. 2. / A'B'C = the difference between a rt. / and / CB'D'. By def. of comp. 3. But / CBD = / CB'D'. By hypothesis. 4. .-. / ABC = I A'B'C. By Axiom 1. IV. Proof. I. Cor. 4. Supplements of equal angles are equal. (Proof same as above.) I. Cor. 5- At 0, given point. in a given line, one perpen- dicular, and only one, can be erected in the scmie plane. GEOMETRY. 447 IV. Proof. /r '/> II. Given CD perpendicular to AB at P. III. To prove that no other perpendicular can be drawn to AB at P in the same plane. (1. Suppose that another perpendicular £P could be drawn. 2. Then / BPP: would be a rt. /. By def. of perpendicular. (If two lihes meet and form a rt. Li each is said to be per- pendicular to the other.) 3. But / BPC is a rt. angle. (Since CI? is perpendicular to AB.) 4. . • . 2 ^'P'E would equal / BPC. Prop . I . , Cor. 1. (AU right angles are equal!) ^ ' 5. But this is impossible. By Axiom 8. (The whole is gveater than any of its parts.) 6. .•. The supposition of step 1 is absurd, and a second perpendicular is impossi- ble. - Q. E. D. Remark. In this demonstration, we have used what is called "the Indirect Method, or reductio ad ahsurdum which means a reduction to an absurdity, as distinguished from the Direct Method VLsed in the other proofs. Jevons in his Principles of Science, Vol. I, p. 96, says, "Some philosophers, especially those of France, have held that the Indirect Method of Proof has a certain inferiority to a direct method, which should prevent our using it." He goes on to show that the method is not inferior and holds the belief that nearly half our logical conclusions rest upon its employment. In the above case, by the Law of Duality, a second perpen- dicular can or can not be drawn. It was shown that by sup- posing that a second one could be drawn led us to an absurdity. Hence, a second can not be drawn. This method of proof is often used in geometry.. PROPOSITION II. I. Theorem, If two adjacent angles have their exterior sides in a straight line, these angles are supplements of each other. ms FINKEL'S SOLUTION BOOK. 'B IV. Proof J II. Given the exterior sides OA and OB of the adjacent angles AOD and BOD respectively and the straight AB'in which these two sides lie. . III. To prove /_ AOD = /_ DOB. 1. AOB is a straight line. By hypothesis. 2. .-. I AOB is a. St. I. By def. of a st. /. 3. 'But/_AOD + lDOB = lAOB. By Ax. 9. 4. .'. /'s AOD and DOB are supplementa,ry. By dgf. of supl. angles. Cor. I.' The sum of all the angles about a point in a plane is equal to two straight angles. Cor. 2. The sum of all the angles about a point on the same side of a straight line passing through a point, is equal to a straight angle. PROPOSITION III. Theorem. Conversely: // two adjacent angles are supplements of each other, their exterior angles lie in the same straight line. I. II. III. Given that the sum of the adjacent angles AOI> and DOB are supplements of each other, that is, equal to a straight angle. To prove AG and OB in the same straight line. ' 1. Assume OF in the same straight line with. OA. 2. Then IAOD + /_ DOT is a straight angle. By Prop. II. , 3. But / AOD + I DOB is a straight angle. By hypothesis. 4. .-. / AOD + jTDOB = I AOD + IDOF. IV. Proof. { • By Ax. I. 5. I AOD ^ I AOD. By I,aw of Identity. 6. Subtracting step 5 from step 4, / DOB = I DOF. By Ax. 3„ GEOMETRY. 449 7. .•. OB and OF coincide. By con^^erse Ax. 10. 8. .■ . AO and OB are in the same straight line. Q. E. D. Scholium. Since Propositions II. and III. are true, their opposites are true, viz., // the exterior sides of two adjacent angles are not in a straight line, these angles are. not supplements of each other. If two adjacent angles are not supplements of each' other , their exterior sides are not in the same straight line. PROPOSITION IV. I. Tbedrem. // one straight line intersects another straight line, the vertical angles are equal. /4\ ^f II. III. IV. Proof. By By I. Given the two lines AB and DE intersecting in O. To prove / AOE = / DOB. 1. / AOE + / AOn equals a st. /. Prop. I. 2. / AOB + I BOB equals a St. /. Prop. I. 3. . IAOE+ IA0£> = IAOD + IDOB. By Ax. 1. Take away from each of these equals the common / AOD. Then,/ AOE = /_ DOB. By Ax. 3. In like manner we may prove / AOD = l_ EOB. Q. E. D. Cor. If one of the four angles formed by the intersec- tion of tivo straight lines w-a right angle, the other three angles are right angles. PROPOSITION V. Xheorem. From a point without a straight line one perpendicular , and only one, can be drawn to this line. 4. 5. 6. 450 FINKELS SOLUTION BOOK. II. Given the point, P, and the straight line, AB. III. Xo prove that one perpendicular can be drawn from P to AB, and only one. Iv. Proof. 1 . Turn the part of the plane above AB about AB as an axis until it falls upon the part below AB and denote the position of P by P'. By Assumption 3 of the Plane. 2. Turn the revolved plane about AB to its ' original position. By Assumption 3 of the Plane. 3. Draw the straight line PP', cutting AB in C. By Assumptions 1 and 2 of the Straight Line. 4. Take any other point D in AB, and draw PD and P'£>. 5. Since PCP' is a straight line, PDP' is not a straight line. {Between two points only one straight line can be drawn.) 6. Turn the figure PCB about AB until P falls on P'. By Assumption 3 of the Plane. 7. Then CP will coincide with CP' and DP with DP' 8. .-. I PCD = IP'CD, and / PDC = / P'DC. Ax. 15. 9. .-. / PCD, the half of a st. / PCP' is a right /; and / PDC, the half of / PDP', is not a right angle. 10. .'. PC is perpendicular to AB, and PD is not perpendicular to AB. By def. of Perpendicular! 11. .•. one perpendicular, and only one, can be drawn from P to AB. Q. E. D. PARALLEL LINES. DeAnition. Parallel lines are lines lying in the same plane and never meeting however far produced. On this definition and the assumption of parallel lines rests the whole theory of parallel lines in Euclidean geometry. By convention, we say that parallel lines meet at infinity. Why this convention is adopted will become apparent in studying Higher Modern Geometry. PROPOSITION VI. I. 'Tbeorem. Two straight lines in the same plaiie per- pendicular to the same straight line are parallel. GEOMETRY. 451 A C -D II. Given the two straight lines AB and CD each per- pendicular to the straight line AC. III. To prove AB and CD parallel. 1. AB and CD, lying in the .same plane, must either meet or not meet. By Law of Duality. 2. If thdy meet, we shall have two lines from the same point perpendicular to the same line. By hypothesis. IV. Proof, i (The lines AB and CB being perpendicular to AC.) 3. But this is impossible. By Prop. V. (From a given point without a straight line, one perpen- dicular, and only one, can be drawn to a straight line.) 4. . AB and CJD cannot meet, however far produced. 5. ■. y4.5 and (rZ> are parallel. By definition of Parallel Lines. PROPOSITION VII. I. Theorem.. If a straight line is perpendicular to one of two parallel lines, it is perpendicular to the other. II. A- C- D. K Given the parallel lines AB and CD and the line //TT perpendicular to AB. III. To prove that HK is perpendicular to CD. Suppose .A/TV drawn through /^perpendic- ular to ///iT. Then MJVis parallel to AB. By Prop. VI. (Two lines in the same plane perpendicular to the same line are parallel.) But CD is parallel to AB. By hypothesis. . . AfN coincides with CD. By assump- tion 1 of parallel lines. (Through a point without a straight line only one straight line can be drawn parallel to that line.) CD is perpendicular to ff/C; that is, //'A' is perpendicular to CD. Q. E. D. IV. Proof. \ 5. 16. 452 FINKEL'S SOLUTION BOOK. TRANSVERSALS. De&nition. A straight line intersecting two or more straight lines is called a transversal. 'f In the figure EF is a transversal of the two non-parallel lines AB and CD. The angles AHI, BHI, CIH, and DIH are called interior angles, and the angles ARE, EHB, CIF, ancj FID are called exterior angles. ^ The angles AHI and HID, or BHI and HIC are called alter- nate-interior angles. . The angles AHE and Z)/F, or BHE and C/F are called alter- nate-exterior angles. The angles AHE and C/H, ^/// and CIF, EHB and ff/D, or BHI and D/F are called exterior-interior angles. PROPOSITION VIII. I. Xheorem. If two parallel lines are cut by a third straight line , the alternate-interior angles are equal; and conversely. Exercises, I. Find the value of an angle (1) if it is double its compLemetit; (2) if it is one-fourth of its complement. II. Given (1) that / .^ is double its complement. III. To find the value oi /_ A. 1. rt. / — / ^ — complement of / A. By def. of compl. 2.' /_A = 2(rt. l.— /_A). By hypothesis. 3. .-. / ^ = 2 rt. /'s - 2 / ^. By Distribu- tive Law of Multiplication. 4. Adding 1 /_Aio these two equals, we have 3 / ^ = 2 rt. /'s. By Ax. 2. U. .-. /^ = -|rt. /. By Ax. 7. Q. E. F. Let the student give the solution of (2). 2. Find the value of an angle (1) if it is three times its supplement; (2) if it is one-third of its supplement. 3. How many degrees in the' angle formed by the hands of a clock at 2 o'clock? S o'clock? 4 o'clock? 9 o'clock? TV. Solution. GEOMETRY. 453 PROPOSITION IX. I. Tbeorem. If two parallel lines are cut by a third straight line, the exterior angles are equal, and con- versely. Let the student give the demonstration and state and prove the corojlaries, if any, coming under the theorem. PROPOSITION X. I. Xbeorem. — The sum of three interior angles of a tri- angle is equal to two right angles, or a straight angle. \ [I. Given triangle ABC, III. To prove that / A+/_ B+/_ ACB=ist. /_. 1. Draw CE || to AC, and prolong AC io F. 2. Then / ACB+l BC£+l ECF^st. /_. - (The sum of all the angles about a point on the same side of £f straight line=st. angle.) 3. But 2 ECF^l BAC, TV T^mrif -l * (being exterior-interior angles of H lines) and (beipg alternate-interior angles of | lines.) . Substituting / ^ for / ECF and / B for I BCF, in step 2, 16. .-. l^ACB+l_A^l_B=&V. l_. Q.E.D. Note.^ The truth or this theorem was probably discovered by Thales, 640 B. C. Attempt to prove this theorem without the use of Euclid's " Eleventh Axiom," or any of its equivalents, and you will see where non-Euclidean Geometry comes into the field of human thought. It is high time that teachers of geometry endeavor to gfain a little knowledge of this subject, instead of talking about the "visionary speculations" of the non-Euclidean geometricians. As a help to gain an elementary knowledge of this sub- ject, the reader is recommended to study II. P. Manning's Non-Euclidean Geometry ( Ginn & Co.) BOOK II. In this book is considered the equality of polygons. We shall consider only one theorem properly belonging to this book. 454 FINKEL'S SOLUTION BOOK. Problem. — To bisect a given triangle by a line dra-wn front a randonipoint in cne of its sides. Demonstration. --\^t;\^ ABC be tlio given triangle, D a random point in the sitlf BC, uiul E the middle point bi BC. Join A nnd /?, A and E. Hvayf EF parallel to AD. Draw DF. Then DF bisects the triangle ABC. For the triangle ABE is equivalent to the triangle AEC (?). The triangle AFD is equivalent to the triangle f/6. 1- .ADE {'>). Hence, ABDF \^ equivalent to ABE. (?) and, therefore, DF bisects the triangle^^^'C. Q. E. D. Proposition. — 77/e square described upon the hypoteiiuse of a right triangle is equal to t lie sum of the squares of the other iivo sides. I. Demonstration. — Let C^Z> be any right triangle, right an- gled at F and let A C, CP, and DM be the squares described uppn'its sides. Then the square ^C is equal to the sum of the squares CP and DAI. 'llirough F, draw ^/?" per- pendicular to AB and produce it to meet OP produced, in G\ also pro- duce BC to, meet OP in 7 and AD to meet OP produced, in Ji. Draw GH parallel to PD, and BT par- , allcl to CF. Draw AE. Now the triangles COI and D FC are cquLil (.?). Hence, CI=CD=CB, and therefore the square C/'=the parallelogram CO (?)=the paral- lelogram BF (?)==the rectangle FIG. 2. BK {}). In like manrieV, the square Z> J/ can be proved equal to the rectangle AK. Hence, the square A C^the square CP-\- the square DM. Q. E. D. Note. — This theorem is known to geometricians as the "Pythagorean Theorem," so called from a Greek geometrician, Pythagoras, (569 B. C.) who was the first to prove it It is also known as the ''47th Proposition of Euclid," being the 47th proposition, of the first book of Euclid's Elements, a mathematical work written by Euclid, a .Greek geometer of the 2d century B. C. The above proof is the one essentially given by Euclid. For a great number of different demonstrations the reader is referred to The American Mathematical Monthly, Vols. V and VII. The following demonstration is the one supposed to be given by Pythagoras, and on the discovery of which it is said he sacrificed a hecatomb to the muses that inspired him. This, however, is not authentic. GEOMETRY. 455 II. Demonstration.— Le\. EDChe any riglit triangle, right angled at D. On the sides BE and Z> C construct, the squares EDHG and DCBM respectively. Produce GE and BC until they meet in F., form- ing the square F B A G. On EC, the hypotenuse, construct the square ECKI. Then the square ECKI{& equal to the sum of the squares EDHG and D CBM. For, the square GFBA is equal to G E D H-\- £> C B M-\- 2 EDCF {=iECF). The square GFBA is ill so equal to the square ECKl-\-AECF. Hence, ^C^/+ /r/e. ,3 ^ECF^ GEDH^DCBM^A^ECF {■>.). Whence, ECKI = GEDH-^D CBM. Q.E. D. Proposition. — In any triangle, each angle formed by join- ing the feet of the perpendiculars is bisected by the perpc?idicu- larfrom the opposite vertex. Demonstration. — Let ABC be any triangle and AD, BE, and CF the three perpendiculars. ^Join D and E, D and F, and E and F. angle In the right triangles AEB and AFC, the angle BA C is common to both. Therefore, they are similar. Hence, ^^:^C ■=:AE:AF. Now the triangles BA C and FAE have the angle FAE common and the including sides proportional. Therefore, they are similar, and the angle AFE^(iiit angle A CB. In a similar manner we may prove that the angle DFB=th& A CB ; the angle A FE=tbe angle DBB. From this it follows that the angle CFA —the angle EFA=the angle CFB~ the angle DFB. Hence, angle EFC^=a.n- gle CFD and the angle EFD is bisected by the perpendicular C F. In a similar manner, it can be proved that A D bisects the angle FDE and EB bisects the angle FED. Q. E. D. FIG. 4. Problem. — From a given point' in an arc less than a semi- circumference, draw a chord of the circle -which -will be bisected by the chord of the given arc. Demonstration. — Let ABDC be the given circle, AB the ■given arc, AB the chord of the arc, and P any point of the arc 456 FINKEL'S SOLUTION BOOK. FI6. 5. .iPC. Draw the diameter POC and on the radius PO as a di- ameter describe the circle PEO. Then through the points i?, and G, of inter- secti6n draw the chords PD and PF respectively, and they will be bisected at the points E and G. For draw DC and OE. Then the triangles PEOand PZ> Care right triangles(?) and are also similar (?). Since -f^O and PDC ure Similar, the line OE is parallel to DCi and since O is the middle point of P C,^ is the middle point of PZ>(?). In like manner, G is the middle point of /'7^ Q-E. F. Discussion. — There are, in general, two solutions. When arc AB is diminished until B coincides with A, there is no solution. When AB is a semi-circumference, there is one solution and the chord is the diameter POC. Proposition. — If two equal , straight lines intersect each other anywhere alright angles., the quadrilateral formed by join- ing their extremities is equivalent to half the square on, either straight line. Demonstration. — Let AB and CD be two equal straight lines intersecting each other at right angles at E. Join their extremities, form- ing the quadrilateral ^ C^Z>. Then ACBD is equivak'nt to half the. square of AB or CD. For the area of the triangle ^C^' equals \{AB X CD) and the area of the triangle ADB equa]s-^(:ABxED). Hence, the area oi ACBD=h{ABy.ED)^ \{ABxEC) =\AB\EDArEC) = k{ABxCD). But CZ» equals ^^, by hypothesis. Hence, ACBD =--^ABK Q. E. D. FIG. 6. A PROBLEM IN MODERN GEOMETRY. An equilateral hyperbola passes through the middle points /?, ^, and 7^ of the sides BC, AC, and AB of the triangle ABC, and cutting those sides in order in a, /i, and y. Show that the lines ^«, j5yS, and C;)/ intersect in a point the locus of which is the circumscribing circle of the triangle ABC. Solution. — The equation to any conic is tta^-\-v/3^-\-'wy'^-\- ■2u'/3y-\-2v'ay-\-2w'erp=0 .... (1).. D is {0,U s'm C,ia sinB) ; E, (U sin C, 0, UsinA) ; F, (|csin B, ic sin A, 0). These GEOMETRY. . 457 points being on (1), "wie should have c-v-\-6'^-w-\-2icu'=0 . . . (2). ■ c'^u+a''zv+2acv'=0 (3), 6''z.+a'-v + 2adw' = (4). Whence n^=^^(au' — bv'^^c-w') .... (o), ^'= — ibv' — cw' — au') . . c . . . (6), Tt'= — (cw' — au' — bv'\ .... (7). Substituting in the cop •dition u-\-v-\-w — 2u' co&A — Iv' cosB — 2tv cosC=0 . . ... (8) that (1) is an equilateral h3perbola, ' .a2 ( au'— b -J '—cw' )-\-b'^( bv'—c-w' —au' ) -f-c^ ( civ'—au'~bv' ) abc — 2M'cosyl — ^'2x'' cos i? — 2w'cosC=0 (9). Clearing of fiac tions and noticing that 2abc cosA^a{b'^ -\-c'^ — a"^) (10), -2abc cos Jl=b{a^-\-c'^—b'^) . .'.(11), 2abc cos C=c(a^+0^—c'' ) (12), and reducing, «<'cos-<4-(-'z;'cos.Z?-j-w'cosC=0. . . .(13) Substituting (5), (6), and (7) in (1) an clearing of fractions, a'^(au'—bv'—f:'w')a''-\-b^6v'—cw'—au')/3^+c''iczv'—au'—bv') -}-y--\-2u'abcj3y-\-2v'abcay-\-2iv'abcafi=Q . .(14). Where this xuts BC, a=0, and (14) gives bHbv'—civ'—au'y-^^+2abcu'- ~= — c-(cTv' — au' — bv') . . (15), whence for the point a; a^=0. a ' <^"' — bv' — au' ,^„s-^ r , p,:=, ^ , , ■ yv, . . . .(16). By symmetry, for the point. b — cu -\-bv — au ' ^ \ / i i j > ^. r nv' — au' — bv' ,j „ ,,f-\ r,,, .p,a.,^ ^- ; =— ;y,, Po^O . (]'/)■ J-he equation te- ar — CIV -\-au — bv '■ ' • ^ ' Aa is found to be 6( — cw'-\-bv' — atu')/3 — (c-w' — bv' — aze')y^=0 . ■ ■ (18) ; to B0, a( — cw'-)-'^^' — bv')a—c(c-w' — au'—bv'jy^O- . . . .(19) ; and to Cy, b(—au'-\-bv'—cvj')fi — a{au' — bv' — c-w')a=0 .... (20): ^"y two of which meet tn ^ bc{ — cw'-\-bv' — au'){civ' — au' — bv') "^ A ^^ „, ac{c-w' — bv' — au'){ — c'w'-\-au' — bv') li = -^^ , , ab( — cui'-\-bv' — au')( — cw'-\-au' — bv') ' {n-i\ r=- ' -^^ •■ ■•(■^1) The circumscribing circle is aPy-\-bay-\-ca§^0 (22), which is satisfied by (21) on corKlition (13). proving the proposi- tion. Note. — This problem was solved bv Professor William Hoover, A. M., Ph. D , Professor of Mathemaitics and Astronomy in the Ohio University, -Athens, Ohio, who is one of the leading mathematicians in the United ;States, and whose tiiography follows. 458 FINKEL'S SOLUTION BOOK. BIOGI^APHY. PROF. WILLIAM MOOUER, A. M., PH. D. Prole»sur Hoover was born in the village of Smithville, Wayne county^ Ohio, October 17, 1850, and is the oldest of a family of seven children. Both parents are living in the village where he was born, still enjoying: good health. Up to the age of fifteen he attended the public schools, and for two or three years after, a local academy. Owing to needy circumstances he was obliged to work for his living quite early, and almost permanently closed attendance at any kind of school at eighteen years of age, sometime before which, going into a store in the county seat, as clerk. Nothing could have- been farther from his taste than this work, having been thoroughly in love with study and books long before. After spending two or three years in this way, he went to teaching, about the year 1869i and he has been regularly engaged in his favorite profession to the present day. He attended Wittenberg College and Oberlin College one term each, a thing having very little bearing on his education. He studied no mathe.- matics at either place, excepting a little descriptive astronomy at the latter. After teaching three winters of country school, with indifferent success^ he was chosen, in 1871, a teacher in the Bellefontaine, Ohio, High School^ serving one year, when he wa^ given a place in the public schools of South Bend, Ind. Remaining there two years, he was invited to return to Belle- fontaine as superintendent of schools. He afterwards^ served in the same capacity In Wapakoneta, O., two years, and as principal of the second dis- trict school of Dayton, O. In 18S3, he was elected professor of mathemat- ics and astronomy in the Ohio University, Athens, Ohio, where he is stilL in service. Through all his career of teaching. Professor Hoover has been an inces- sant student, devoting himself largely to original investigations in mathe- matics. Although his pretentions in other lines are very modest, he is em- inently proficient In literature, language, and history. Before going into- college work he had collected a good library. He is indebted to no one for any attainments made in the more advanced of these lines, but by indefati- gable energy and perseverance he has made himself the cultured, classic, and. renowned scholar he is. He has always been a thorough teacher, aiming to lead pupils to a mas- tery of subjects under consideration. His habits of mind and preparation for the work show him specially adapted to his present position, where he- has met great success. He studies methods of teaching mathematics, which in the higher parts is supposed to be dry and uninteresting. He sets the example of enthusiasm as a teacher, and rarely fails to impress upon the minds of his students the immense and varied applications of mathe- matics. He is kind and patient in the class-room and is held in the highest esteem by his students. He is ever ready to aid the patient student inquir- ing after truth. It seems to be a characteristic of eminent mathematicians, that they desire to help others to the same heights to which they them- selves have climbed. T^is was true of Prof. Seitz; it is true of Dr. Martin; and it is true of Prof. Hoover. In 1879, Wooster University conferred upon Prof. Hoover the degree of Master of Arts, and, in 1886, the degree of Doctor of Philosophy cum laudgy he submitting a thesis on Cometary Perturbations. In 1889, he was. elected a member of the London Mathematical Society and is the only man in his state enjoying this honor. In 1890, he was elected a member of the New York Mathematical Society. He has been a member of the Asso- GEOMETRY. 459 ciation for the Advancement of Science for several years. Papers accepted bv the association at the meetings at Cleveland, Ohio, and at Washington, D. C, have been presented on "The Preliminary Orbit of the Ninth Comet of 1S86," and ''On the Mean Logarithmic Distance of Pairs of Points in Two Intersecting Lines." He is in charge of the correspondence work in mathematics in the Chautauqua College of Liberal Arts and of the mathe- matical classes in the summer school at Lake Chautauqua, the principal of which is the distinguished Dr. William R. Harper, president of the new Chicago University. The selection of Professor Hoover for this latter po- sition is of the greatest credit, as his work is brought into comparison with some of the best done anywhere. He is a critical readet and student of the best American and European writers, and besides, is a frequent conlributor to various mathematical jour- nals, the principal of which are School Visitor, Mathematical Messenger, Mathematical Magazine, Mathematical Visitor, Analyst, Annals of Math- ematics, American Mathematical Monthly, and Educational Times, of Lon- don, England. His style is concise and his aim is elegance in form of expression of mathematical thought. While greatly interested in the various branches of pure, mathematics, he is specially interested in the applications to the ad- vanced departments of Astronomy, Mechanics, and the Physical Sciences — such as Heat, Optics, Electricity, and Magnetism. The "electives" of- fered in the advanced work for students in his University are among the best mathematics pursued an> where in this country. He is an active member of the Presbyterian church and greatly interested in every branch of church work. He has been an elder for a number of years and was chosen a delegate to the General Assembly meeting at Port-' land, Oregon, in May, 1892, serving the church in this capacity with fidel- ity and intelligencer In this biography of Professor Hoover, there is a val- uable lesson to be learned, it is this: Energy and perseverance will bring a sure reward to earnest effort. We see how the clerk in a county seat store, in embarrassing circumstances and unknovvn to the world of thinkers, became the well known Professor of Mathematics and Astronomy in one of the leading Institutions of learning in the State of Ohio. "Not to know him argues yourself unknown." THE NINE-POINT CIRCLE. Proposition. — If a circle be described about the pedal trian- gle of any triangle, it tvillpass through the middle points of the lines drciiun from the orthocenter to the vertices cf the (riangle, and through the middle points of the sides of the triangle, in all, through nine points. (2) The center of the nine-point cinle is the mid- dle point of the liiie. joining the orthocenter and the center of the circumscribing circle of the triangle. (3) The radius of the nine- point circle is half the radius of the circumscribing circle of the triangle. (4) The centroid of the triangle also lies on the line joining the orthocenter and the center of the circumscribing circle of the triangle , and divides it in the ratio of 2'. 1. (5) The sides of the pedal triangle intersect the sides of the given triangle in the radical axis of the circumscribing and nine-point circles. (S) The nine-point circle touches the inscribed and escribed circles of the triangle. 460 FINKEL'S SOLUTION BOOK. Tlie Pedal Triangle is a triangle formed by joining the feet of the perpendiculars drawn from the vertices of a triangle to the opposite sides. ' The Orthocenter is the point of intersection of these per- pendiculars. Jifedial lAnes, or 3£edians, are lines drawn from the vertices of a triangle' to the middle point of the opposite sides. TJie CentTOid is the point of intersection of the medians. the Radical A.OCis of two circles is the locus of the points whose powers with respect to the' two circles are equal. Demonstration. — Let .^^ Che anj triangle, AD, BF, and CE the perpendiculars from the vertices to the opposite sides of the triangle. O is the orthocenter. Join the points F, E, and D. Then FED is the pedal triangle. About this triangle, .de- scribe the circle FEHDK. It will then pass through the mid- dle points Z, N, and R of the lines, OA, OB, and OC, and the middle points H, G, and ^of the sides AB, BC, and A C, in all, through nine points. Since the angles AFO and AEO of the quadrilateral are both right angles a circle may be described about it. For the same reason circles may be described about the quadrilaterals EBDO and ODCF. Draw the lines Z^./? and EG. Now the angles FEE and FDE are equal, being measured by half the same arc FE. But FDE equals 2EDD, because AD bisects the angle EDF. .-. FRO equals 2-^Z>Z. Both being, measured by the same arc OF', and FRO being two times FDL, FRO is an angle at the center; therefore, since (9 C is the diameter of the circle circumscribing FODC, R is the middle poinit of OC. In like manner it may be proved that OB and OA are bisected in the points iV and Z respectively. Draw the line RG. The angles RGC and RGB are equal to two right angles. Also the angles 7? Gi5 and J?^Z> are equal to two right angles, because they are opposite angles of a quadrilateral inscribed in a circle. Therefore i?GC is equal to RED. But RED is equal to OBD, because both are measured by half the arc OD. .-. The angle .ffGC equals the angle OBD, and consequently the line RG is parallel to the line OB. But, since RG bisects OC in i? and is parallel to OB, it bisects B C in G. In like manner, it may be shown that AB and A C are bisected by the nine-point circle in the points H and .A' respectively. Hence, the circle passes, in all, through nine points. Q. E. D. (2.) Draw the perpendiculars (?/', KP, and HP \^rom the middle points of the sides of the triangle. They all meet in a common point Z" which is the center of the circumscribing circle of the triangle. With T' as a center and radius equal to PB, GEOMETRY. 461 describe the oirciimscribing circle. Draw the perpendiculars S J' SJ, and SZ to the middle points of the chords FIC, EH, and FIG. 6. DG. These all meet in the same point S, which is the center of the nine-point circle. In the trapezoid PHBO, since .SJ bi- sects ^^ in c/ and is parallel to PH, it bisects-O/' in S. Hence, the center of the nine-point circle is the middle point of the line joining the orthocenter and center of the circumscribing circle. Q. E. D. (3.) Draw the lines KN ■d.nA. PB. Since the angle KFN is a right angle, the line A'jV is a diameter of the nine-point cir- cle. KP=\BO=BN. Since KP and BN are equal and parallel, KPBN is a parallelogram, and consequently KN=B P ^ .-. SN=\BP. But SN is the radius of the nine-point circle and BP is the radius of the circumscribing circle of the triangle. Hence, the radius of the nine-point circle is half the radius of the circumscribing circle. Q. E. D. (4.) Draw the medial lines BK, AG, and CH. Draw the \va.& KH. Now the triangles .A'P// and BOC are similai be- 462 FINKEL'S SOLQTION BOOK. cause the sides of the one are respectively parallel to the sides of the other, and the line IIK\b half the line BC, because^ and ^are the middle points of the sides ^^ and ^ C. .-.BO^^KP. The triangles Ay^ and ^(9§ are similar, because theangles of one are respectively equal to the angles of the other. Then we have KP-.KQwBO-.BQ or KP:BO -.-.KQ: BQ. Bnt B0='1KP. .-. BQ=.1KQ. .-. § is the centroid and divides the line joining orthocenter and the center of the circumscribing circle in the ratio of 2:1. Q.E. D. Hence the line joining the centers of the circumscribing and nine-point circles is divided harmonically in the ratio of 2 : 1 by the centroid and orthocenter of the triangles. These two points are therefore centers of similitude of the circumscribing and nine-point circles. *•. Any line dr^wn through either of these points is divided by the circumferences in the ratio of 2 : 1. (5.) Produce /"^ till it meets .5 C in /". Since two opposite an- gles of the quadrilateral BE PC are equal to two right angles, a circle may be circumscribed about it. Then we have P^E. P' F =P'B. P' C ; therefore the tangents from P^ to the circles are equal. Q. E. D. (6.) Let O be the orthocenter, and /and ^ the centers of the in- .scribed and circumscribed circles. Produce AI to bisect the arc BC\n T. Bisect AO in L, and join GZ, cutting ^Z'in S. The nine point circle passes through G, D , and L, and Z? is a right an- gle. Hence, GL is a diameter, and is therefore =R=QA. There- fore GL and QA are parallel. But QA=Q T, ' therefore GS=GT= CTsm^A=2Ri\n'^^A. Also ST =2 GScosd, being the angle G.? 7" = GTS. JV being the center of the nine-point circle, its radius =JVG=:^Ji ; and r being the radius of the inscribed circle, it is required to show that NI=NG—r. NP' =Sm^Sr—2SN.SIzo^6. Sub- stitute SN=\R~GS; SI=TI— FIG. 7. ST=2PsmiA—2GScose\ and G5=27?sin2|^l, to prove, the proposition. If J be the center of the escribed circle touch ing B C, r^ its radius, it is shown in a/' similar way that NJ=NG^r.,. GEOMETRY. 463 THE THREE FAMOUS GEOMETRICAL PROBLEMS OF ANTIQUITY. The limits of this work forbid our carrying the discussion of elementary geometry further. We have given merely an outline of how the subject may be studied by the student and presented by the teacher and that is our chief aim in this work. But before leaving the subject, it will be of interest to briefly speak of three famous problems in geometry, — problems that have profoundly interested the mathematicians from the time of Plato down to the present time. These problems have been referred to before in this book so that, at this point, we shall only bring them to- gether and speak of them more explicitly. The problems re- ferred to are, (i) The Duplication of the Cube; (ii) the Trisection of an Angle; (iii) the Quadrature of the Circle. The first of these problems means to find the edge of a cube whose volume shall be twice that of a given cube; the second means to divide any given angle into three equal parts ; and the third means to find the side of a square whose area shall be equal to that of a given circle. As has been said, constructions in pure geometry or Euclidean Geometry admit of the use of an ungraduated ruler and a pair of compasses only. With this restriction, all three problems are insoluble. This is the important point to be observed. The problems are only impossible, because we are limited in the use of our instruments to a straight edge, or ungraduated ruler, and' a pair of compasses. In this way many problems may be made impossible. For example, it is impossible, at present, to go across the Atlantic Ocean from Boston to Liverpool on a bicycle, but with a steamship the trip is made very easily. So too, if other instruments are used our three problems are easily solved. The solutions of the first and second problems are implicitly involved in the Galois theory as presented to-day in treatises on higher algebra. The impossibility of the solution of the third was demonstrated in 1882 by Lindenlann. The first two problems may be reduced to one, viz., that of finding two means between two given extremes. In the first prob- lem, if we let a be the edge of given cube and x that of the re- quired cube, then we must have ^^ = 20^, or a : x ■= x -.y^y : 20. In the second, if a is the sine of the given angle, and x the sine of one-third the angle ; then 4.^^ = 3^ — a, or 1 : 4*jr = /fix '■ y - y ■ (sx—a), or I : .r>^=.r^ : (3— 4J^^ )'^=:(3— 4,r2 )H «><;. The problem of the duplication of the cube was known in ancient times as the Delian problem, in consequence of a legend that the Delians had consulted Plato on the subject, It is as- serted by Philoponus, tTiat the Athenians in 430 B. C. were snf- 464 FINKEL'S SOLUTION BOOK. fen'ng from the plague of eruptive typhoid fever and in order to- stop it consulted the oracle at Delos as to how it might be done. Apollo replied that they must double the size of the altar of Minerva which was in the form of a cube. This to the un- learned suppliant, was an easy task, and a new altar having" each of its edges double that of the old one was constructed,, in consequence of which the volume was increased eight-fold. This so enraged the god that he made the pestilence worse than before, and informed a fresh deputation that it was useless to- trifle with him as the new ^Itar must be a cube and have a volume exactly double that of the old one. Suspecting a mystery, the Athenians applied to Plato who referred them to the geom- etricians. In an Arab work, it is related that Plato replied to them, saying, Ye have been neglectful of the science of geometry and, therefore, hath God chastised you, since geometry is the- most sublime of all the sciences. Many solutions of this problem have been given, one of which is given on page 234, by means of the Cissoid. We here give another by means of the parabola. I^etjj/S = ax, be the equation of the parabola whose axis coin- cides with axis of abscissas and .r^ = 2ajj/, the equation of the parabola whose axis coincides with the axis of ordinates. Solv- ing these two equations, we findjj/' = 2a*, that \s, PM^ = 2a^. Hence, if a is the edgfe of the given cube PM is the edge of the required cube. To trisect an angle, we proceed as follows: L,et AOB be the given angle. With Cas center and any radius, describe a circle, ^.ffZ?. Draw the secant BDC so that Z'C." shall be equal to the radius OB. (This is impossible un- less a graduated ruler is used.) Then draw OD. Then angle BCA=i angle AOB. For angle. D C 0=angle DOC. Whv? Angle B£>0—2Xangle: PROBLEMS. 465. DCO. Why? Angle DCO+angle C50=angle BOA. Why? .-. Angle ^05=3 X angle BCO. Why ? The following elegant solution is due Clairaut : lyCt AOB be the given angle. With O as center and any radius describe a circle. Draw AB and trisect it in //"and K, so that AH= HK = A'i?. Bisect the angle AOB by OC, cutting ^^ in Z.. Then AH '= 2I/L. With focus A, vertex //^, and disectrix OC, describe a hyperbola. I^et the branch of this hyperbola which passes through // cut the circle in F. Draw PM perpendicular to OC and produce it to cut the circle in Q. Theri by the focus and directrix property, we have AP : PM = AH:HL = 1:\. .• . AP = 2 PM= PQ. Hence, by symmetry, AP= PQ= QR. , Hence, AOP = POQ = QOR. The Quadrature of the Circle is effected by the Quadratrix. See page 238. For a very full treatment of these problems, see Klein's Famous Problems of Elementary Geometry, translated from the German by Professors Beman and Smith, also see Mathematical Recreations and Problems, by W. W. R. Ball. 466 FINKEL'S SOLUTION BOOK PROPOSITIONS. 1. The lines which ioin the middle points of adjacent sides of any quad- rilateral, form a parallelogram. 2. Two medians of a triangle are equal; prove (without assuming that they trisect each other) that the triangle isisoScles. 3. In an indefinite straight line AS find a point equally distant from two given points which are not l>i)i/i on AB. When does this problem not admit of solution? Cpnstruct a right triangle having given: 4. The hypotenuse andjhe difference of the sides. 5. The perimeter and an acute angle. 6. The difference of the sides and an acute angle. 7., Construct a triangle having given the medians. 8. Construct a triangle, having given the base, the vertical angle, and(l) the sum or (2) the difference of the sides. 9. Describe a circle which shall toubh a given circle at a given point, and also touch a given straight line. 10. Describe a circle which shall pass through two given points and be tangent to a given line. H. Find the point inside a given triangle at which the sides subtend equal angles. 12. Describe a circle which shall be tangent to two intersecting straight lines and passing through a given point. 13. Divide a triangle in two equal parts by a line, perpendicular to a side. 14. Inscribe in a triangle, a rectangle similar to a given rectangle. 15. Construct an equilateral triangle equivalent to a given square. 16- Trisect a triangle by straight lines drawn from a given point in one of its sides, . 17. Draw through a given point a straight line, so that the part of it in- tercepted between a given straight line and a given circle may be divided at the given point in a given ratio. 18. Construct a circle equivalent to the sum of three given circles. 19. Find the locus of a point such that the sum of its distances from three given planes is equal to a given straight line. 20. Construct a sphere tangent to three given spheres and passing througn a given point. 21. Draw a circle tangent to three given circles. Note. — This proposition is known as the Taction Problem. It was pro- posed and solved by ApoUonius, of Pergse, A. El. 200. His solution was indi- rect, reducing the problem to ever simpler and simpler problems. It was lost for centuries, but was restored by Vieta. The first direct solution was given by Gergonne, 1813. An elegant solution of this problem is given by Prof. E. B. Seitz, School Visitor, Vol. IV, f. 61. 22. Construct a sphere tangent to four given spheres. Note. — This problem was first solved by Fermat (1601 — 1665). 23. The perpendicular from the center of gravity of a tetrahedron to any plane without the tetrahedron is one-fourth of the sum of the perpendic- ulars from the vertices to the same plane. PROBLEMS. 467 1. Define: a segment of a circle, four proportional magnitudes, two similar polygons, the projection of a segment of a straight line on another straiglit line. 2. The sum of all the plane angles about a point is four right angles. 3. The locus of all points equally distant from two fixed points is the s-raight line that bisects the line joining the two points, at right angles. 4. A straight line that is perpendicular to a radius at its extremity is tangent to the circle; and conversely. 5. Two polygons that are similar to a third polygon ale similar to each other. . 6. If two triangles have an angle of the one equal to an angle of the other, their areas are to each other as the rectangles of the sides includ- ing those angles. 7. The ratio of the circumference of a circle to its diameter is the satne for all circles. . 8. 'Find the side of -the largest square that can be cut from a tree whose circumference is 14 feet. _ Cornell University — Entrance Examination, 1899. [Proofs by limits are not, in general, satisfactory.] 1. Define : a plane, a straight line perpendicular to a plane, a straight line parallel to a plane, two parallel planes, a diedral angle, the plane angle of a diedral. 2. The sum of the face angles of a convex polyderal angle is less than four right angles. 3. The sections of a prismatic surface made by' two parallel planes are equal polygons. 4. The frustum of a triangular pyramid is equal in volume to three pyramids, whose common altitude is the altitude of the frustum, and whose bases arc the two bases of the frustum and a mean proportional between them. 5. To draw a pl^ne tangent to a cylindar with circle base. 6. If two angles of a spherical triangle be equal, the opposite sides are equal. 7. The lateral area of a cone of revolution is half the product of the perimeter of its base and its, slant height. 8. A cylindrical pail is 6 inches deep and 7 inches in diameter : find how much water it holds, and how much tin it takes to make it. Cornell University — Entrance Examination, 1899. 1. Define a straight line (preferably without using, the ideas of dis- tance or direction). Also define: equal, greater, limit of a variable, length of a curve. 2. If two triangles have two sides of one equal to two sides of the , other, and the included ~ angles of the first greater than that of the second, prove that the third side of the first is greater than the third side of the second. Also prove the converse of this theorem. 3. Similar triangles (and similar polygons) are to each other as the squares on homologous sides. 4. Construct a triangle, being given the lengths of the three per- dendiculars from the vertices on the opposite sides. 5. Compute the side of a regular pentagon inscribed in a circle whose radius is given. 468 FINKEL'S SOLUTION BOOK. 6. Two straight lines in space have one and but one common per- pendicular, and it is the shortest line that can be drawn from one to the other. 7. Compute the volume of a regular octahedron whose edge is two units! 8. Show how to find the radius of a given sphere by means of meas- urements made on the surface. 9. Prove that the volume of a cone is equal to the area of the base multiplied by one-third of the altitude. Also state without proof the chain of propositions which lead up to this theorem. 10., Find the locus of a point in space the ratio of whose distances from two given points is equal to a given constant. Cornell University — ' Scholarship Examination, 1899. Time, 3 hours. 1. State and prove the theorem of Menelaus — and its inverse. 2. Prove : Circles described on any three chords from one point of a circle as diameters, have their other three points of intersection co- straight. 3. Explain the Peaucellier Cell. 4. State and prove the dual theorem of: The pole of any straight through a point is on the polar of the point. 5. Prove; The diagonal triangle of a cyclic quadrangle is self-conju- gate (its own reciprocal polar). 6. (o) ' Explain what is meant by a cross ratio of a range of four points. (6) If (PQRS) = 3, what are the othei- distinct cros? ratios of the same range, and what are their magnitudes? (c) Deduce the .distinct values of the cross ratios of a harmonic range. 7. Prove Pascal's theorem concerning a cyclic hexagon, (b) State the corollaries as, the number of sides is diminished. 8. What is the radical axis> of two circles ? (b) Prove: The difference between the squares of the tangents from any point to two circles equals twice the rectangle of the center sect of the circles and the perpendicular sect from the point to the radical axis. Examination in Halsted's Modern Geometry. The University of Texas, 1894. Time, 3 hours. 1. From the common notion "solid" as a starting point, define surface, line, point, plane, straight line. (b) Define angle, and point out the angles determined by a bi- radial. (f) What is meant by the statement that two magnitudes are equivalent? — that one magnitude is greater than another? ' (d) Define the terms, multiple, submultiple, fraction, ratio. (e) What is the direct meaning of the statement that two series of magnitudes are proportional? State the simplest criteria of propor- tionality between tw6 series of magnitudes in which to every one of either series there corresponds one of the other series._ Apply the test to two such series where it is fulfilled ; and again where one criterion fails. ' (f) When are two'figures perspective? — when similar? 2. Discuss the problem : To describe a circle tangent to three given intersecting straights, not all through the same point. — Also, the analogous problem in, a sphere. PROBLEMS 469 3. (o) State the conditions of congruence of two plane triangles. (b) State the corlditions of similarity of two plane triangles. (c) State the conditions of congruence of two spherical triangles. 4. (o) Investigate the form of the quadrilateral made by joining the mid points of consecutive sides of a quadrilateral. (b) Its relative size. 5. Divide a sect internally and externally in a given ratio. 6. (a) Prove: If four sects are proportional the rectangle of the extremes is equivalent to the rectangle of the means. (b) State the inverse.^ 7. (a) Prove the Pythagorean theorem without using any other con- cerning equivalence of figures. (6) Prove: The altitude to the hypothenuse is a mean propor- tional between the segments of the hypothenuse. 8. (a) When is one spherical triangle A'B'C the polar of another, ABC? ■ . . . (6) Prove: The sides of a spherical triangle intersect the cor- responding sides of its polar on the polar of their orthocenter. Examination in Halsted's Elementary Synthetic Geometry. The University of Texas, 1894. Time, 3 hours. 470 FINKEL'S SOLUTION BOOK. ALGEBRA 1 . Algebra is that branch of mathematics which treats of the general theory of operations with numbers, or quantities. The operations of ordinary abstract arithmetic are a particular case of algebra. Thus, algebra is sometimes caW&A generalized ariihnietic and in turn arithmetic is sometimes called specialized algebra. 2. An Operation, in mathematics, is the act of passing from one number to another, the secorid number having a defi- nite relation to the first. 3. An Operator, in mathematics, is a letter or symbol designating the operation to be performed. Thus, +, — , X. ^-- i/i — . or Z) ,,, and i are operators. 4. The Fundamental I/aws of Algebra are the Com- mutative Law; the Associative Law; and the Distribu- tive Law. The Commutative Law staX&s, that additions and stibtrac- tions fnay be performed in any order ; also the factors of a product may be taken in any order. The Associative Law states, that the terms of an expres- sion may be grouped in any order. Thus, a-\-b — c-\-d—e=(a-\-b)— c+{d—e)=a+(b-c)-\-{d-e)'='a+b-(c—d\—e. Also the factors of a product tnay be grouped in any order. The Distributive Law states, that the product of a com- pound e.xpression ' by a single factor is the algebraic sum. of the partial products of each term of the compound expression by that factor. Thus, (a\b)c=ac-^bc. For a very excellent treatment of these laws, the reader is referred to Chrystal's Algebra, Part I. Note. — The establishment of these three great laws was left for the present century, the chief coutributors thereto being De Morgan, Hankel, and Peacock. These men were working at the philosophy of the first principles. Hamilton, Grassmann, and - Pierce threw a flood of light on the subject by conceiving algebras whose laws dififer from those of ordinary algebra. The student who would become piroficient in mathematics should make himself familiar with ordinary algebra, for it is the basis of all advanced mathematical subjects. For example, in ALGEBRA. 471 analytical geometry, the subject matter is geometry while the language is algebraic; in the calculus, the subject matter may be physics, astronomy, or political economy while' the language is algebraic. We shall solve a few problems in algebra, leaving the student to gain a thorough knowledge of the subject by a study of such works as Chrystal's Algebra, 2 vols. I. An estate was divided among three persons in such a way that the share of the first, was three times that of the sec- ond, and the share of the second twice that of the third. The first received $900 more than the third. How much did each receive? [From If all and Knight s College Algebra, p. 6p, prob ^o.] 1. I,et .«:=the number of dollars in the share of the third person. Then 2. 2;ir=the number of dollars in the share of the sec- ond, and 3. 6:i;=3X2.«r=the number of dollars in the share of the first. 4. 6;f— x=5;i;=the number of dollars the first received more than the second. II. ■{ 5. 900=the number of dollars the first received more than the second. 6. .-. 5x=900, 7. x:^i of 900=180, the number of dollars in .the share of the third, 8. 2;r=360, the number of dollars in the share of the second, 9. 6.r=1080, the number of dollars in the share of the first. r f 180=share of the third, III..-. ] $360= " " " second, and (11080= " " " first. I. The length of a room exceeds its breadth by 8 feet; if each had been increased by 2 feet, the area would have been in- creased by 60 square feet; find the original dimensions of the roo'm. [From. Ifall and Knigkt's College Algebra, p. 6p,prob.j^.'\ (1. I^et j;=the number pf feet in the breadth of the room. Then 2. .a;-|-8=the number of feet in the length, and 3. {x+8)x=x^-\-8x=:th.e number of square feet in the area. 4. ;ir+2=the conditional number of feet in the breadth, and 5. .sr4-2+8=;i:+10=the conditional number of feet in the length. Then 472 FINKEL'S SOLUTION BOOK. II. { 6. (;t:+2)(j»r+10)=:.;t:V12;c+20=the conditional num- ber of square feet in the area. 7. (^V12;«r+20)— (.rV8;c)=4j;+2P=the number of square feet in the increase in the area, 8. 60=the number of square feet in increase in area. 9. .-. 4jr+20=60, 10. 4a:=40, by subtracting 20 from both sides. 11. x=l of 40=10, the number of feet in the breadth, and ^12. .ar+8=18, the number of feet in the length. III. •■■I 10 feet=the breadth, and 18 .feet=the length. I. A takes 3 hours longer than B to walk 30 miles; but if he doubles his pace, he takes 2 hours less ^ime than B; find their rates of walking. [From Ha// and Knight's College Alge- bra, p. 164, prob. J 2.] ■ 1. Let j?=A's rate in miles per hour, and 2. j)/=B's " 80 . Then — =number of hours it takes A to travel 30 X miles. '' II. 4. 5. 6. 7. 9. 10. 11. 12. 80 — =number of hours it takes B to travel 30 miles. y 30 30 . • . — — — =3, by the first condition of the problem. X y 2;i;=A's conditional rate in miles per hour. Then jr— = — =numbei' of hours it takes A to travel Zx X 30 miles. .-. — — — -—2, by the second condition of the prob- y X leva. 15 15 3 , ,., — — — =-g-, from (5). — = 3A, by adding (8) and (9). y J._JL ■ ■ y ~ 30" 30 .•.''jj/=-=-=4f =number of miles B travels per hour. 13. — ^— 4l="2"' ^y substituting for 7 in (9). ^^- X 2^2' ALGEBRA. 473 15. 15 10 =5. III. ■17. .'. x=3=Dumber of miles A travels pei home. . f . ,3 miles=A's rate per hour, and ■ \ 4^ miles=B's rate per hour. I. In a mile race A gives B a start of 44 yards and beat? Wm 51 seconds. In the second trial A gives B a start of 1 min- ute and 15 seconds, and is beaten by 88 yards. Find the rate ot •each in miles per hour. \Todhunter' s Algebra, p. loj, prob. 2j. WentwortKs Complete Algebra, p. lyp, prob. ^^.^ II. < 1. 2. 3. 0. t 6. 7. 8 9. 10. 11. 12. I,et ;e=A's rate in yards per second. j/=B's, rate in yards per second. 1 mile =1760 yards. 1760 .. . , . =time It takes A to run a mile. X 1760- y trial. 1716 -44 1716 y 1760 =time B was running in the first y :51 ■ • (1). 1760 ' y 1760- =:time it takes B to run 1 mile. -88 1672 X trial. 1760 y 39 J. 40j/ 20 19y 59 1672 X 51 '1760 75 X x-'wn 531 =time A was running in second .75 ... . (2). (8), by dividing (1) by 1760. . ... (4), by dividing (2) by 1672. 760y"^ 33440 ' ^^ s^ibtracting (3) from (4); whence 44 13. J/ ="^ yards, B's rate per second. ,. ,„' ., 3600 44 „, . ., 14. .-. 10 miles= „ X'-q-=B s rate m miles per hour. 1 15 15. — ="oQ> by substituting the value oi y in (3) and ' changing the signs. 88 16. .". x=^ yards=A's rate in yards per second. 474 FINKEL'S SOLUTION BOOK. !l7. ^„ ., 3600 88 ., ... . 12 miles— -^^^ Xt^^A- s rate m miles per nour> III 1760^15 rate per h 12 miles=A's rate per hour 1 10 miles=B's rate per hour. I. THE QUADRATIC EQUATION. 5. ax''+dx+c=0, is the general quadratic equation. ax'+6x+c=a( x^ +^x+^) =?( ^' +4^+-^+^" 4^) H^+2^+ - -4(2«r I 2a ''+Ta- b Vd^-iac 2a ^=0. 2a b-Vy^-\ac _^ ^ 2a b+Vb^—4:ac 2a b-Vb^-iac 2a In the solution of exercises involving quadratic equations, Students should be thoroughly grounded in the method of com- pleting the square, and this method should not be superseded by the Hindoo Method nor the Method of Factoring, though this latter method should receive due consideration. When the method is thoroughly impressed upon the mind of the student he should be encouraged to solve examples by merely substitut- ing in the general formulae above. Thus, find the values of x satisfying the equation 2x2-f 6j;-33=0. Here, a=2, b=b, and c=-?,\ Then b+Vb''- -\ac 2a b-VB"- -\ac 5+1/25-4-2—83 __5+17 4 5-17 2-2 =-5^. 2a 5-/25-4-2—33 2-2 = 3. I. Find the price of eggs per score when 10 more in %2\ cents' worth lowers the price 31:|^ cents per hundred. \Went- worth's Complete Algebra-, p. 216, prob. 5.] 1. Let j;=price per score. X 2. 20 =price per egg. ALGEBRA. 475 11. 1 3. 5x=price of 100 eggs . ... . ^ 1250 4- ^2*-^20=^ worth. =number of eggs in 62^ cents' 1250 , ,„ 1250+10;t: t, -f in u -+10= =number if 10 more be X added. 6. 62i-l-?^5±^= 25.r 7. 9. 500_)_4^ =P"ce of each egg, if 10 more be added to 62J cents' worth. 625;t; •j^g— ^=price of 100 eggs. _ 625;i; _.., ' ■ ■ ^^-125+^-^1*- x'^—^^= — j^, by clearing of fractions, transpos- ing, combining, and dividing through by the coefficient of x^. 50625 , by completing the square. ,. 2 25 ,625 10. X^——rX-\-—Tir 4 64 64 25 225 11. A— ^=±— TT— , by extracting the square root. 1 12. x=Z\\, or -25. III. .•. 31Jc.=price of the eggs per score. The negative value is inadmissible in an arithmetical sense. x-'-Vy =11 (1)-) \ Find the values of x and r. [From Schuv- X +y2= 7 (2) J -^ '- -^ lev's Complete Algebra, p. j68, prob. 4..^ 1. ;t2— 9=2— _>' (3), by transposing in (1). 2. X — 3=4— y2 (4)^ by transposing in (2). .r+3. II. 3. x-2,-- o y f+3 A=4--^^' ^"^ J' 5- ^^'-^^-4— ^^qrg- (6), by transposing. 6. y- y =4-^ r+:; 1 (7), by .;t+3 4(x+3)2 " '.jr+3 '4(x+3)2 completing the square in (6). • '^' -^~ 2(^+3) ^^~ 2(x+3) ^^^' ^^ extracting the square root of (7). 476 8. FINKEL'S SOLUTION BOOK. 1 y=2, by transposing , in (8). m.{ 9. x+4:=7, by substituting value of y in (2). UO. .•.x=3. x=B. y='2. X and y have three other values in addition to those found. For a number of different solutions giving the four values of x &XL6.y, see The American Mathematical Monthly . x^+xy+y^=a^ x^+xz+z'=d^ y^+yz+z^—c^ Find X, y, and z. II. For a solution of this example, see The Mathematical Mag- azine, edited by Dr. Artemas Martin, Washington, D. C. I. Find two numbers whose product is equal to the differ- ence of their squares, and the sum of whose squares is equal to the difference of their cubes. \^Ray's Higher Algebra, p. 2jo, prob. p.] , - . ' 1. I,et j;=greater number, 2. and jv— lesser number. 3. xy=x'>—y'^ (1). 4. x'^+y^^x^'—y^ (2). 5. Xet x=ay, then ay2=a3j/2-f-y2 (4)^ by substituting the value of x in (1). «2— a=l (5), by dividing (4) byjj/^, and arranging, whence a=^(l±V 5 , by completing the square of (5)', and extracting the square root, and trans'- posing. 5±V~5" — y= ==iV 5 , by substituting the value of a ^ 2(*±V 5) ^ ^ ^ in (2). _ _ _ 110. x=ay^\i:yH 5) (JV 5) =J(5±V 5). Ill ■ l-^~i(^^~^)' ^°d II. INDETERMINATE FORMS. 6. The symbol, 0, is defined by the equation a— a=0. It is not used to denote nothing, but is used to denote the absence of quantity. AH operations upon this symbol are impossible. Thus, 0X5, 6. 7. 8. 9. ALGEBRA. 477 5^0, 5", are all impossible operations. Standing apart from the conditions imposed upon the quantities from which arises by^ certain limitations, the operations above indicated are absolutely meaningless. But such indicated operations, and many others of the same nature, do very frequently. occur in mathematical investigations, and when they do thus arise they must be inter- preted in such a way as to conform to the fundamental laws of mathematics. In conformity to these laws, OXfl!=aXO=0; a+0=0+a=a; 0-^=0. The symbol, oo , is used to represent a quantity that is larger than any assignable quantity, however large. What meaning shall be attached to the following indicated operation, a-^-O? It is impossible to perform this operation. Suppose we divide a by h. This is possible, provided a and h are real numbers, and is indicated thus, -p. Now what happens to the value of the fraction — , if we conceive h to diminish indefinitely? We know that as the denominator of a fraction decreases, the value of the fraction increases. Hence, if the value of the denominator becomes very small, the value of the fraction becomes very large. If the denominator becomes less than any assignable quantity, the value of the fraction becomes larger than any assignable quantity. All this is concisely and A=0 accurately stated as follows : I -i- I ~^ , or I — I /lio L A J L /i J Hence, the inaccurate though common form, a-^0, must be interpreted in the light of the above explanation and, therefore, a-^0=QO , briefly though inaccurately expressed. a''=l, for all finite values of a; but is indeterminate if a is oo . 0"=0, for all finite values of a; but is infinite if a is infinite, a-^oo =0, for all finite values of a; but is indeterminate if a is infinite. 0° is indeterminate, oo— oo, oo^oo, 0-^0, OXoo , and 1" are all indeterminate. But when these forms occur in any mathematical investigation, they usually have a determinate value. Thus, ^^^^1 . =4=^+^1 . =2a. Here,-J=2«. All the other forms may be reduced to the form —^. Thus, CO- CO- -^---^ -, OXao=OX-^--^ = -g-; 478 FINKEL'S SOLUTION BOOK. _« . i _ a '_OXa ,^„ ^_ ^[.-^ =^ ^=^ -T^-Q— ^T-ox^=T '■ ^°^- (^ ^=* ^''^- ^-°° x^-0 • Since the log. (1") is indeterminate, the quantity 1" is indeter- minate. It is important that the student masters the meaning of these forms, as they occur very frequently in the higher mathematics. For example, the Differential Calculus rests largely upon the , proper mterpretation of -^. 1. Find the limiting value of -- — ^ , , ., „ when x^=2. x^-dx+e n ,_ _ (x-2)(x-3) -\ _x-s-\ _-l_i xs-10;i;+16j^=2 {x-2)(x-S)Xj^2 .^t^sJ^^z -6 2. Find the limiting value of ■ ,, ^ , „ when x-^0 and when (l)j£!±?i."l _ _ x(;i:+2) ~] _ ■r+2 "1 2 2x2+3j: J^=0 ;t:(2;f+3)J^=o 2x-|-3j^i=o /2),£M^1 _oo_ x(^+2) -] _^+2-| . ^_5o.= 2;t:2+3;t:J^^„ °° ^(2;f+3)J^„„ x+3j^J,„ ^ 1+A 4-- 3. Find the 'imiting value of — r — r^^ — ttt. when x=2 and ■when x='x . /■i^ x^+6x-l(i ~\ _ _ (;i:+8)(;ir-2) 1 _ 1 ~|- \r3-12x+16J;^2 (;ir+8)(.r-2)2X=2 jr-2J^=2 x^+6x-1 6 ~[ _oo_ (;tr+8)(;t;-2; ~]. _ 1 1 /0\ ^"~r".* ^"1 ^J^ _ \^^ I u/\^ ^f !■ J. I J_=0. 4. Find the limiting value of ^= =. when x=a. f a--^ X i ALGEBRA. 479 ' Let x=a+h, where h approaches as a limit. Then _^ lA . _i -^4^1 EXAMPLES- (1+x ^^ T— — / when x=0 \.-~x ' 2. Find the limiting value of — when x=^a. 1—f'x^ 3. Find the limiting value of ^ when x^ 1 1— l/ X 4. Find the limjting value o± — j^TT when x^^. 5. ^^-^n =what? 6. f - =what? X — a Jjr=a 1— x+log. x-\ , ^, 7 ^ =what? 1— V'2;i;— X^ J;iri=l 8. fl+a^ |"^=what? L J;ir=,0 [i+«.]j=[[i+«x]jj"=[[i+4(..)+ii^^^, +\^ X J\x J(ax)'^+etc. , by the Binomial Theo- rfu- , 1(1—-^) ,, 1(1-^) (1—2^) 3 , , 1 1" 480 FINKEL'S SOLUTION BOOK. rill ~\" ... r=[l+,+^«.+^,3+etc.] = [l+l+^+-^+etc J =[2.71828+]" =tf" , when a=\. Hence, 1°° is indeterminate.. III. PROBABILITY. 7. De£nition. If an event can happen in a ways and fail in 6 ways, and each of these ways is equally likely, the probability, or the chance of its happening is , and the chance of its failing is — r-j-. a+p That is, the probability of an event happening is the number of favorable ways the event can happen divided by the total number of ways it can happen, and the probability of its failing is the number of ways the event can fail divided by the total number of ways it can happen. / For example, if in a lottery there are 6 prizes and 23 blanks, the chance that a person holding 1 ticket will win a prize is -.fg-, and his chance of not winning is ff . 8. , The reason for the above definition may be made clear by the following considerations : ^ , If an event can happen in a ways and fail to happen in 6- ways, and all these ways are equally likely to occur, we can assert that the chance of its happening is to its chance of failing as a to 6. Thus if the chance of its happening is represented by ka, where k is an undetermined constant, then the chance of its failing is y^(5. .". Chatace of happening + chance of failing = k(a+6). Now the event is certain to happen or to fail; therefore, the sum of the chances of happening and failing must represent certainty. If, therefore, we agree to take certainty as our unit, we have ^ ., , .. . . 1 k(a-\-b)—\, or k——r-;. a-TO a .'. the chance that the event will happen is -^r, and the chance the event will not happen is — rr- a+o • 9. The subject of probability, from the mathematical point of view, is a very difficult one. That it is a very important sub- ject, no one will deny after having read Jevons's Principles of Science, 2 vols., in the first volume of which he has given con- siderable attention to its treatment. Our space is too limited to give here more than a passing reference to the subject. Those who desire to study the subject thoroughly, should read Tod- AI/GEBRA. 481 II. hunter's Hisiory of the Theory of Probability ; LaPlace's Theorie Analytique des Probabilities, 1812 (the most exhaustive treat- ment of the subject 'ever written); and Whitworth's Choice and Chance. The last named book has a large number o£ problems worked out in full. EXAMPI^ES. ' " 1. I. What is the chance of throwing a number greater, than 4 with an ordinary die whose faces are numbered from 1 to 6? ■ 1. 6=number of ways the die can fall. 2. 2=number of ways the die can fall so as to give a ' number greater than 4, viz., 5 and 6. ,3. .■ f=J=the required chance, by definition. III. .". the required chance is \. 2. I. Find the chance of throwing at least one ace in a single throw with two dice. 1. 6=number of ways one of the dice may fall. 2. 6X6=36=number of ways the two dice may fall, since with each of the six ways the first may fall, there are six ways in which the second ■ may fall. 3. 6=number of ways one die may be thrown with- out the ace coming up. 4. 25=5X5=number of ways the two dice can be thrown v^ithout either of them being ace. 6. .'. -|f=;the chance of not throwing an ace. 6. . • . 1— f f =^^=the chafice of throwing at least one ace. • III. .'. the chance of throwing at least one ace is ^-. 3. I. If four coins are tossed find the chance that there should be two heads and two tails. II. II. 1. 2. 3. 4. 5. 6. J=chance of head or tail with one coin. T^5-= (J) *= chance of all heads, tossing 4 coins. yV=(4)*=chance of all tails, tossing 4 coins. i=tXi(i)^=chance of 1 head and 3 t^ls. |=fX(J)3(^)=chance of 1 tail and 3 heads. f=^(^)2(i)'=chance of 2 heads and 2 tails. III. .". the chance of throwing 2 heads and 2 tails is f. 4. I. A bag contains 5 white, 7 black, and 4 red balls; find the chance that three balls drawn at random are all white. 482 FINKEI/'S SOLUTION BOOK. r. * 16-1514 II. III. 2. =:number of ways 3 objects can be selected from 16 objects. =numter of ways 3 objects can be selected 1-2-3 from 5 objects, which is the number of ways the 3 white balls may be selected from the 5 white balls. (6-4-3) ^(l-2-3) , ^, . , , ••• (16-15-14)^(l-2-3) =^=^^" "^q"^'^"^ "^^°""- the required chance is ^. I. If three pennies be piled up at random on a horizontal plane, what is the probability that the pile will not fall down? The pile will stand if the common center of gravity of the sec- ond and third coins falls on the surface of the first or bottom coin. Let r be the radius of a penny ; then the center of the second coin may fall anywhere in a cir- cle whose radius is^r and cen- ter the center of the surface of the first or bottom coin, and the center of the third coin may fall anywhere in a circle whose ra- dius is 2r and the center the center of the surface of the sec- ond coin. The number of posi- tions of the center of the second coin is therefore proportional to 4:7Tr^, and for every one of these positions the center of the third coin can have ircr'^; hence the total number of positions of the second and third coins is proportional to IBTT^r*. We must now determine in how many of these IBtt^;"* posi- tions the pile will stand. Let A be the center of the first or bottom coin, and ^ the cen- ter of the second coin. Take A£>=A£, and with Renter Z> and radius 2r describe the arc CHJ. If the center of the third •coin is on the surface CFJH, the second and third coins will re- main on the first, since ^JV==iV^, -571= T'C, and the pile, will not fall down. When AB is not greater than \r, the circle CHJ will not cut the surface of the second coin, and the pile will stand if the cen- ter of the third coin is anywhere on the second. Mi FIG. 8- PROBABILITY PROBIvEMS. Let AB=AZ>=x,S=suvfiice CFJH, and /=the probability required ; then DB=1x, BG= -^ , DG= -^ , 1 I- -i« /^.'V- — 4^^^ CG=^\_l&r^—{hr'^—^x-'YJ ,arc C/=^cos-'^^ ^^^ ), -^2cos-i('^^^^^^)-'i[l6r*-(5r==-4x2)2]^, and 75= ^—--f''"^7Tr^.27rxdx+-fr'^^f S.lnxdx, 1 1 /■'• /' o ,/^3r2— 4;f2>w =J-_( i_ / ( S—Ttr"^ )xdx. ■ / ^^ cos-i I ^ )xdx, +^r^[lGr*—{5r^-4:X^yj , J ir- cos-^{^ ^^ Jxdx — i^^fier*— (S^2_4;^2-)2~j '^ /"|[^i6r*— (5^2— 4^2) = ^ 'xrfx =ir*cos-i (^5j±=if^^_^(5r2-4x2)[.16r*-(5r2-4x2 )2 ]!" ^ 16 87tr*\-.^ \r irx J (3,.2J-4;<:2>. ,/^5r2— 4a:2x , 1,^2 . „.^ Vi6.^-(5.-4.^)^]'^^=fg-A^(AVT5-2 sin-^). Note. — This solution is due to Artemas Martin, M. A., Ph. D. , LL. D. , member of the London Mathematical Society, member of the Edinburgh Mathematical Society, member of the Mathematical Society of France, member of the American Mathematical Society, member of the Philo- sophical Society of Washington and Fellow of the American Association for the Advancement of Science, Washington, D. C. He is one of the peers of mathematical science. 484 FINKEL'S SOLUTION BOOK. BIOGI^APHY. ARTEMAS MARTIN, M- A., PH. D., LL. D. This eminent mathematician was born in Steuben count/, N. Y., Au- gust 3, 1835. Early, his parents moved to Venango county, Pa., where they lived for many years. Dr. Martin had no schooliijg in his early boyhood, except a little primary instruction; but by self-application and indefatiga- ble energy which have told the story of many a great man, he has become familiar to every mathematician and lover of science in every civilized country of the world. He was never a pupil at school, except when quite small, until in his fourteenth year. -He had learned to read and write at home, but knew nothing of Arithmetic. At fourteen he commenced the study of Arithme- tic, and after spending two winters in the district school, he commenced the study of Algebra. At seventeen, he studied Algebra, Geometry, Nat- ural Philosophy, and Chemistry in the Franklin Select School, walking tvyo and one-half miles night and morning. Three years after, he spent two arid one-half months in the Franklin Academy, studying Algebra and Trigonometry. This finished his schooling. He taught district schools four winters, but not in succession. He was raised on a farm, and worked at farming and gardening in the summer; chopped wood in the winter; and after the discovery of oil in Venango county, worked at drilling oil wells a part of his time, always devoting his "spare moments" to study. In the spring.of 1869, the family moved to Erie county. Pa., where he re- sided until he entered the U. S. Coast Survey OflBce in 1885. While in Erie county, after 1871, he was engaged in market-gardening, which he carried on with great care and skill. He began his mathematical career when in his eighteenth year, by contributing solutions to the Pittsburg Almanac, soon after contributing problems to the "Riddler Column" of the Philadelphia Saturday Evening Post, and was one of the leading contribu- tors for twenty years. In the summer of 1864 he commenced contributing problems and solu- tions to Clark's School Visitor, afterward the Schoolday Magazine, pub- lished in Philadelphia. In June, 1870, he took charge of the "Stairway De partment" as editor, the mathematical department of which he had con- ducted for some years before. He continued in charge as mathematical editor till the magazine was sold to Scribner & Co., in the spring of 1875, at which time it was merged into "St. Nicholas." In September, 1875, he was chosen edito? of a department of higher mathematics in the Normal Monthly, published by Prof. Edward Brooks, Millersville, Pa., and held that position till the Monthly was discontinued in August, 1876. He published in the Normal Monthly a series of sixteen articles on the Diophamtine Analysis. In June, 1877, Yale College conferred on him the honorary degree of Master of Arts (M. A.) In April, 1878, he was elected member of the Lon- don Mathematical Society. In June, 1882, Rutgers College conferred on him the honorary degree of Doctorof Philosophy (Ph. D.) March 7, 1884, he was elected a member of the Mathematical Society of France. In April, 1885, he was elected a member of the Edinburgh Mathematical Society. June 10, 1885, Hillsdale College conferred on him the honorary degree of Doctor of Laws (LL. D.) Februai-y 27, 1886, he was elected a member of the Philosophical Society of Washington. In June, 1881, he was elected Professor of Mathematics of the Normal School at Warrensburg, Mo., but did not accept the position. November 14, 1885, Dr. Martin was appointed > PROBABILITY PROBLEMS. -485 Librarian in the office of the U. S. Coast and Geodetic Survey. On August 26, 1S90, he was elected a Fellow of the American Association for the Ad- vancement of Science. On April 3, 1891 he was elected a, member of the New York Mathematical Society. All these honors have been worthily bestowed and the Colleges and So- cieties conferring them have done honor to themselves in recognizing the merits of one who has become such a power in the scientific world through his own efforts. He has contributed fine problems and solutions to the following journals of the United States: School Visitor, Analyst^ Annals of Mailiematics, Mathematical Monthly, Illinois Teacher, Iowa Instructor, National Edu- cator, fates County Chronicle, Barnes^ Educational Monthly, Wittenherger, Maine Farmers'' Alma7iac, Mathe7natical Messenger, and Educational Notes and Queries, American Mathematical Monthly. Besides other con- tributions, he contributed thirteen articles on "Average " to the Mathemat- ical -Department of the Wittenbcrger, edited by Prof. William Hoover. These are believed to be the first articles published on that subject iil America. Dr. Martin has also contributed to the following English mathematical periodicals,: Lady's, and Gentleman's Diary, Messenger of Mathematics, The Educational Times and Reprint and the Mathematical Gaaette. The Reprint contains a large number of his solutions of difficult "Aver- age " and " Probability " problems, which are master-pieces of mathematical thought and skill, and they will be lasting monuments to his memory. His style is direct, clear and elegant. His solutions are neat, accurate and sim- ple. He has that rare faculty of presenting his solution in the simplest mathematical language, so that those who have mastered the elements of the various branches of mathematics are able to understand his reasoning. Dr. Martin is now ( 1 899 ) editor of the Mathematical Magazine, and The Mathematical Visitor, two of the best mathematical periodicals pub- lished in America. These are handsomely arranged and profusely illu.stra- iti with very beautiful diagrams to the solutions, he doing the typesetting with his own hand. The typographical work of these journals is sa i d to be the finest in America. The best mathematicians from all over the world contribute to these two journals. The Mathematical Visitor is devoted to Higher Mathematics, while The Mathematical Magazine is devoted to the solution's of problems of a more elementary nature. All solutions sent to Dr. Martin receive due credit, and if it is possible to find room for them the solutions are all published. He has thus encouraged many young aspirants to higher fields of mental activity. He is always ready to aid any one who is laboring to bring success with his work. He is of a kind and noble dis- position and his generous nature is in full sympathy with every diligent student who is rising to planes cf honor and distinctian by self application and against adverse circumstances. Dr. Martin has a large and valuable mathematical library containing, many rare and interesting works. His collection of American arithmetics and algebras is one of the largest private collections of the kind in this country. 486 FINKEL'S SOLUTION BOOK. I. Find the average or inean distance of every point of a squuic from one corner. Taking the corner from which the mean distance is to be found for the origin of orthogonal co-ordinates, and one of the sides of the square for the axis of abscissa, we have for the element of the surface dx dy, and since this element is at a distance V(.j;^+y 2) from the origin, tlie average distance =-j/ / dxdyt^{x'^-\-y^) =2^ ! -/•^»V(.-+«')+/'»-.^»log.° +'"'f+'"' \ ■ B». .-. Average distance=Ja[\'2+log^ (1+V2)]. Note.— This solution is by Prof. J. W. F. Sheffer, Hagerstown, Md., whose name may be found attached to the solutions of many difficult prob- lems proposed in the leading mathematical journals of the United States. The above solution is taken from the Mathematical Messenger^ published by G. H. Harville, Simsboro, La. I. All that is known concerning the veracities of two witnesses^ A and B, is that B's statements are twice as reliable as A's. What is the probability of the truth of the concurrent testimon}- of these two witnesses? Let ^^the probability of the truth of any one of A's state- ments ; then 2:v^the probability of any one of B's statements. The event did occur if both witnesses tell the truth, the proba- bility ot which is ^X2;e=2^^. The event did not ^pcur if both testify falsely, the probability of which is (1 — x)y^{\ — 2^)=1 — Zx\-1x''- . Hence, the probability of the occurrence of the event on the supposition that x is known is ^ 2^2^(1— ;^)(l—2»:)' Now, as the veracity of B can not exceed unity, the greatest value oi X is found by putting 2ac=l, which gives x=\; hence, x can have any value from to \. Therefore, the probability in the problem is r^ , -, r^ , , /"^ >''^dx /'>i x'-dx Jo {Sx-By+r Let 8^^— 3=J'- Then x=:-l(y — 3), dx^^dy; the limits of jk are 1 and — 3, and 1 r+i(y+Bydy 1 r+Y, , fiy , 2 >i - = [i>'+flog.0'^+7)+^-, tan-(^^)];;; PROBABIUTY PROBLEMS. 487 =*— |loge2+|^tan-V7. Note. — This solution is taken from the Mathematical Magazine, Vol. II, f. tS2. . Thesolution there given is credited to the author, Prof. William Hoover, and Prof. P. H. Philbrick. I. A cube is thrown into the air and a random shot fired through it; find the chance that shot passed through oppo- site faces. Let AH be the cube. Through P , a point in the faefe EFGH , draw Af.Sr parallel to HE^&^d. draw PiV perpendicular to HE. Now if PA represents the direction of the shot, it will pass through the face ABCB, if it strikes the face EFGH anywhere within HMPN. Let^^=l, LKAF=6, lPAK=ct>, and area HMPN=u. Then ^iir=sec^, PK=iec e tan0, .F^=tan<3 , AP= secdsecip, PM=l—sec0ta.n<p, PJV=1 — tan^, ti=(l — sec^tan0)(] — tari9),the area of the projection of HMPN on a plane perpendicular to PA^ucosffcos^, and that of EFGH=cos0cos(p. Since we are to consider all possible FIG. 9. directions of the shot with respect to the cube, the points of in- tersection of PA with the surface of a sphere whose center is A, and radius luiity, must be uniformly distributed. An element of the surface of this sphere is cos<pd6d(p. By reason of the symmetry of the cube, the required chance is obtained by finding the number of ways the shot can pass through the opposite faces -S'^GA'' and ^^CZ> between the limits 6=0, and (9=J;r, and <p^0 and 0^tan-i (cos^)=0', and the number of ways it can pass through the face ^i^Gi^between the limits 6i=0and 6'=-J;r, and (p=0 and 0=|7r ; and then dividing the fonner by the latter. Hence, the chance required is p. / / ucos6co&'^(f>dtld(f> cosfcos ^ (pdddcf) 4 r}iv /"*' - / / ucosdcos^ <pddd<p. Jo Jo =- / (cos(9— sin^)tan-i(cosi9 )fif5, =^r(sin6l+cosi9)tan-i(cos(9)— ^+^2 tan"! (1^2 tail ^) -iMtt 1 -iloge(l+cos'(9)J^ =-[4V2tan-i(iv'2)+log,(|)-7r]. Note. — This solution is due to Professor Enoch Beery Seitz, member of the London Mathematical Society, and late Professor of Mathematics in the North Missouri State Normal School, Kirksville, Mo. 488 FINKEIv'S SOLUTION BOOK. BIOGRAPHY. PROF. E. B. SEITZ, M. L. M. S. Professor Seitz, a distinguislied mathematician of his day, was born in Fairfield Co., O., Aug. 24, 1846,' and died at Kirksville, Mo., Oct. 8, 1883. His father, Daniel Seitz, was born in Rockingham Co., Va., Dec. 17, 1791, and was twice married. His first wife's name was Elizabeth Hite, of Fairfield Co., O., by whom he had eleven children. His second wife's name was Catharine Beery, born in the same county, Apr. 11, 1808, whom he marr.ed Apr. 15, 1832. This woman was blessed by four sons and three daughters. Mr. Seitz followed the occupation of a farmer and was an in- dustrious and substantial citizen. He died near Lancaster, O., Oct. 14, 1864, in his seventy-third year ; having been a resident of Fairfield Co. for sixty-three years. Professor Seitz, the third son by his father's second marriage, passed his boyhood on a farm, and like most men who have become noted, had only the advantages of a common school education. Possessing a great thirst for learning, he applied himself diligently to his books in private and became a very fine scholar in the English branches, especially excelling in arithmetic, ft was told the author, by his nephew, Mr. Huddle, that when Professor Seitz was in the field with a team, he Would solve problems while the horses rested. Often he would go to the house and get in the garret where he had a few algebras upon which he would satiate his intellectual appetite. This was very annoying to his father who did not see the future greatness of his son, and many. and severe were the floggings he received for going to his favorite retreat to gain a victory over some difficult prob- lem upon which he had been studying while following the plow. Though the way seemed obstructed, he completed algebra at the age of fifteen, with- out an instructor. He chose teaching as his profession which he followed with gratifying success until his death. He took a mathematical course in the Ohio Wesleyan University in 1870. In 1872, he was elected one of the teachers in thfe Greenville High School, which position he held till 1879. On the 24th of June, 1875, he married Miss Anna E. Kerlin, one of Darke county's most refined ladiesr In 1879, he was elected to the chair of mathe- ' matics in the Missouri State Normal School, Kirksville, Mo., which posi- tion he held till dea:th called him from die confines of earth, ere his star of fame had reached the zenith of its glory. He was strjcken by that "demon of death," typhoid fever, and passed the mysterious shades, to be numbered with the silent majority, on the 8'th of October, 1883. On the 11th of March, 1880, he was elected a member of the London Mathematical So- ciety, being the fifth American so honored. He began his mathematical career in 1872, by contributing solutions to the problems proposed in the "Stairway" department of the Schoolday Magazine, conducted by Artemas Martin. His masterly and original solu- tions of difficult Average and Probability problems, sooii attracted universal attention among mathematicians!. Dr. Martin being desirous to know what works he had treating on that difficult subject, was greatly surprised to learn that he had no works upon the subject, but had learned what he knew about that difficult department of 'mathematical science by studying the problems and solutions in the Schoolday Magazine. Afterwards, he con- tributed to the Analyst, the Mathematical Visitor, the Mathematical Maga- zine, the School Visitor, and the Educational Times, of London, Eng. In all of these journals, Professor Seitz was second to none, as his logical and classical solutions of Average and Probability problems, rising as so many monuments to his untiring patience and indomitable energy and perseverance will attest. His name first appeared as a contributor to the Educational Times in Vol. XVIII., of Reprint, 1873^ From that time until his death the Reprint is adorned with some of the finest product of his mighty intellect. (^^■^■^'~t-'^^y- ij_ a-iAy^^ , BIOGRAPHY. 489 On page 21, Vol. II.,' he has given the above solution. This problem had been proposed in 1864 by the great English mathematician, Prof. Wool- house, who solved it with great labor. It was said by an eminent mathe- matician of that day that the task of writing out a copy of that solution was worth more than the book in which it was published. No 'other mathematician seeme,d to have the courage to investigate the problem after Prof. Woolhouse gave his solution to the world, till Profes- sor Seitz took it up and demonstrated it so elegantly in half a page of or- dinary type, that he fairly astonished the mathematicians of both Europe and America. Prof. Woolhouse was the best English authority on Proba- bilities even before Professor Seitz was born. It was the solution of this problem that won for Professor Seitz the acknowledgment of his superior ability to solve difficult Probability prob- lems'over any other living man in the world. In studying his solutions, one is struck with the simplicity to which he has reduced the solutions of some of the most intricate problems. When he had grasped a problem in its entirety, he had mastered all problems of that class. He would so vary the conditions in thinking of one special problerfi and in effecting a solution that he had gene'ralized all similar cases, so exhaustive was his analysis. Behind the words he saw all the ideas represented. These he translated into symbols, and then he handled the symbols with a facility that has rarely been surpassed. What he might have accomplished in his maturcr years, had he turned his splendid powers to investigations in higher and more fruitful fields of mathematics, no man may say. The solving of problems alone is not a high form of mathematical research. While problem solving is very bene- ficial and essential at first, yet, if one confines himself to that sort of work exclusively, it becomes a waste of time. He was a man of the most sipgularly blameless life; his disposition was amiable ; his manner gentle and unobtrusive ; and his decision, when circumstances demanded it, was prompt and firm as the rocks. He did nothing from impulse ; he carefully considered his course and came to conclusions which his conscience approved ; and when his decision was made, it was unalterable. Professor Seitz was not only a good mathematician, but he was also proficient in other branches of knowledge. His mind was cast in a large mold. "Being devout in heart as well as great in intellect, 'signs and quan- tities were to him but symbols of God's eternal truth' and 'he looked through nature up to nature's God.' Professor Seitz, in the very appro- priate words of Dr. Peabody regarding Benjamin Pierce, Professor of Mathematics and Astronomy in Harvard University, 'saw things precisely as they are seen by the infinite mind. He held the scales and compasses with which eternal wisdom built the earth, and meted out the heavens. As a mathematician, he was adored by his friends with awe. As a man, he was a Christian in the whole aim and tenor of life.' " Professor Seitz did not gain his knowledge from books, for his library consisted of only a few books and periodicals. He gained such a profound insight into the subtle relations of numbers by close application with which he \vas particularly gifted. He was not a mathematical genius, that is, as usually understood, one who is born with mathematical power fully de- veloped. But he was a genius in that he was especially gifted with the power to concentrate his mind upon any subject he wished to investigate. This happy faculty qI concentrating all his powers of mind upon one topic to the exclusion of all others, and viewing it from all sides, enabled him to proceed with certainty where others would become confused and disheart- ened. Thread by thread and step by step, he took' up and followed out long lines of thought and arrived at correct conclusions. The darker and more subtle the question appeared to the average mind, the more eagerly lie investigated it. No conditions were so complicated as to discourage liira. His logic was overwhelming. 490 FINKEL'S SOLUTION BOOK. "THE POND PROBLEM." I. On a dark night a circular pond is formed in a circular field. A man undertakes to cross the field. Find the chance that he walks into the pond. Note. — This problem was proposed by Dr. Artemas Martin and pub- lisTied by him in the Mathematical Visitor as problem 300. The first pub- lished solution of this problem was given by the author and appeared in the Dec. No., of Vol. VII., of the American Mathematical Monthly, where also a second solution is given by Prefessor G. B. M. Zerr. Prof: Zerr> solves more difBcult problems than any other man in America. PROBLEMS. 1. State as a theorem the fact implied in the following equa- tions : 92 _ 72 — 4.8; 52 — 32 = 4.4; 932 _ 912 =-4.92; 32 — i2 ^ 4.2. Prove it, and then express the theorem in its most general terms. 2. How find the highest common factor of two polynomials that cannot be readily factored? Prove your answer. Illustrate by finding the H. C. F. of 6x^ -\- yx^ — e,x and 15.^* -f- 31X* + '^ox'^. 3. A cistern can be filled in 4 hours by two pipes running together, and in 6{ hours by one 'of the pipes alone. In how many hours can the other pipe fill the; same cistern? 4. If V a and V b are surds, prove that ^ a ^^ b cannot be: a rational number. 5 . Simplify + j>'2 x~y 1 "^ -t -I j;2 — 1/2 + X y Check your work by substituting Ji: = 4 and y ^ 1, both in the given expression and in the simplified form, and comparing results. 6. Given the two simultaneous equations _y3 _^ y-i _ 5J7 and jr + 3; =_ 3 ; find all the pairs of values of x and 3; that satisfy them. Cornell University — Entrance Examination, 1899. T. 8 — 3^ — ;tr2 1. Resolve — ■ — into partial fractions. x{x -|- 2)2 2. At an election there are 4 candidates, and 3 members to- PROBLEMS. 491 be elected, and an elector may vote for any number of candidates not greater than the number to be .elected. In how many ways may an elector vote ? 3. Find, by using logarithms, the value of V41.72X (.054) . 4. Show that for any two quantities, the square of their geo- metric mean is equal to the product of their arithmetic and har- monic means. 5. Draw the graph of the function x^ -\- x'^ -{- x — 100; and' find the root between 4 and 5, correct to three places of decimals, of the equation x^ -\- x'^ -\- x — 100 = 6. 6. If k, k, are the roots of ax'^ -\- bx -\- c = b, find the value 7. The square of x varies as the cube of y, and *■ = 3 when 1 y = 4. Find the value of y when' x := 'TF^- Cornell University — Entrance Examination^ 1899.. I. (rt) Simplify the expression 26c . 2i>c 6+c " b+c ■+-i 1 1 _^ ^_ c b-2c b^c-2b Show how the symmetry of this expression may be made to serve as a partial check upon the result. (&) As jr becomes more and more nearly equal to 2, what .1-2 — 6.r. -(- 8 value does the fraction '■ approach? What is the .r2 —z^x-\-6 value of this fraction when .r ^ 2? 2. The sum of the three digits of a number is 9; the digit in hundreds' place equals \ of the number composed of the two other digits, and the digit in units' place equals \ of the number composed of the two other digits ; what isi the number ? , 3. Prove that : (0) ;»:«■ -f y'^-'xs, exactly divisible by ;ir -|- 3; if n is any odd, positive integer whatever. (&) \i a-\-'b is constant, then a-h is greatest when «:= ;5. 4. (a) State what seem to you to be the chief differences between algebra and arithmetic. (&) Define negative number, subtraction, and multiplica- tion, and show how, from your definition, the following rules may be deduced: (i) "change the sign of the subtrahend and pro- ceed as in addition" ; (2) give the product the positive or the 492 FINKEL'S SOLUTION BOOK. negative sign according as the two factors have like or unlike signs." 5. Given the equation ax'' -\- bxy -\- cy^ := o in which a, b, and c are real, — the ratio x -.y being unknown. Find the sum and also the product of the roots (i. e. of the values of the ratio X -.y) of this equation. If a approaches o relatively to b and c, what happens to these roots? For what relative values of a, b, and c are the two roots equal? One twice the other? Both imaginary? \^ ' T. Cornell University — Scholarship Examination, 1899. 1. Two men,' A and B, had a nfoney-box containing $2ip, from which each drew a certain sum daily ; this sum being fixed for each, but diflferent for the two. After six weeks, the box was empty. Find the sum which each man drew daily from the box; knowing that A alone would have emptied it five weeks earlier than B alone. Obtain two solutions, and interpret the negative answer. 2. Solve the equation ~2^ x^^~ ~6~\ . x — b) 3. Reduce to their lowest common denominator I I and 1 2.1-3 — 2.r2 — 20Jir — 6 4A'^ — 6x^ — 4^1? + 6 and find, and reduce to their lowest terms, the difference and the quotient of these two fractions. ■4. Write out (.r — yy. o g \aVa.b ^ ) ; reducing the an- swer to the simplest possible form, in which it is free from nega- tive and fractional exponents, and has only one radical sign. Harvard University — Entrance Examination, 1893. 1. Resolve into three factors {x^ — xy — 8. 2. Find the greatest common factor of A-3 -\- 5^2 _^ 5_y ajjfj 2^ -{- yx'^ -\- T,x -\- 2. 3. Solve the equation V;r — 4 + Vj; — \\—tlx + 9=0. V.r— Vj/ V.r+V_y Vjc+Vj/ ^ix—i^y 4. Simplyfy j;2 -F-j/^^j_ x-^y 1 y -1/2 ' 1 1 •^ -I- PROBIvEMS. 493 ~ 5. Solve the simultaneous equations ^{x^ + xy) — apy, x — y = 2. 6. What is the geometrical mean between 2X — 3 and 2X^ -j- xf — /\x — 3 ? 7. A and B start at the same time from the same point m the same direction. A goes at the uniform rate of 60 miles per day ; B goes 14 miles the first day, 16 miles the second day, 18 miles the third day and so on. At the end of 50 days who will be ahead, and by how much ? Massachusetts Institute of Technology — Entrance Ex- amination in Advanced Algebra, 1892. I. (a) Show that (m+i) (n + 2) (» + 3) n 1= — {n^ -\- 6 n -\- 11) . 2 + 3 6 (&) Find the algebraic expression which when divided by x'^ — 2x -\- I gives a quotient .r- -\- 2x -\- i and a remainder X — I. ' 2. (o) Reduce to a common denominator (arran