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, Many Problems and Solutions from the Leading Math- ematical Journals of the United States, and Many Original Problems and Solutions.
WITH
NOTES AND EXPLANATIONS
BY
B. F. FINKEL, A. M., M. Sc. ii
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.
THIRD EDITION- REVISED.
KIBLER & COMPANY, PUBLISHERS,
Springfield, Mo.
COPYRIGHT, 1888,
BY
B. F. FINKEL,
IN THE OFFICE OF THE LIBRARIAN OF CONGRESS,
WASHINGTON, D. C.
PREFACE.
This work is the outgrowth of eight 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 accurate 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 ; viz., the devolpment 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
2 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 combine 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 any 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 formulae 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. D., LL. D.; and the Mathematical Messenger, published by G. H. Harvill, by the kind permission of these distinguished gentlemen.
PREFACE. 3
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.
Hoping that the work will, in a measure, meet the object for which it is written, I respectfully submit 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 rne.
THE AUTHOR.
1 -
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. Jp, 1897.
PREFACE TO SECOND EDITION.
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 students.
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
4 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 Phi to sop hie Enseignement :, Paris, 1898, says, "No country has made greater progress in mathematics during the past twenty-five years than the United States. The most of the text-books on Arithmetic, Algebra, Geometry, and the Calculus, written within the last 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, Weirstrass, 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 Bernouillis, became professor of mathematics and natural philosophy at Harvard University. Since that time the mathematical courses in our leading Universities have been enlarged and strengthened until now the opportunity for research work in mathematics as offered, for example, at the University 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 ; Albegra. — 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 the 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 mathematics studied in our Public Schools be increased, but it is desirable that the quality of
PREFACE. 5
the 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 one 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- ceptions 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 opportunity 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 student in his business in after life, is pedagogically and psychologically wrong. Mathematics 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 best 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
6 PREFACE.
a class of men who have contributed much to the advancement of civilization, will find a short sketch of the mathematicians from 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 Mnothly. 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. Drury College, July, 1899.
CONTENTS.
CHAPTER I.
DEFINITIONS.
Mathematics classified 11 |
CHAPTER II. NUMERATION AND NOTATION.
PAGE
Definitions 11-14
Numeration defined 14
French Method defined 14
English Method defined 14
Periods of Notation 15
Notation defined
Arabic Notation defined
15 15
Roman Notation defined 15
Ordinal Numbers 15
Fractions 18
Irrational Numbers. 20
Examples 21
Addition defined
Subtraction defined
CHAPTER III.
ADDITION.
22 | Examples 23
CHAPTER IV. SUBTRACTION.
23 | Examples 24
CHAPTER V. MULTIPLICATION.
Multiplication defined 24 | Examples 25-26
CHAPTER VI. DIVISION.
Division defined 26 | Examples 27
CHAPTER VII. COMPOUND NUMBERS.
Definitions 28
Time Measure 29
Definitions in Time Measure . 23-31
Longitude and Time 31-34
Standard Time 34-35
The International Date Line . 36-37
Examples 37-36
Solutions 38-39
Examples 39-40
Divisor defined 41
Common Divisor defined 41
Multiple defined 42
Common Multiple defined 42
CHAPTER VIII. GREATEST COMMON DIVISOR.
Greatest Com'n Divisor defined. 41 Examples 41-42
CHAPTER IX. LEAST COMMON MULTIPLE.
Least Common Multiple defined 42
Examples 43-44
8
CONTENTS.
CHAPTER X. FRACTIONS.
PAGB
Definitions 44-46
Fractions classified 44
Solutions of Problems. Examples
PAGE 46-49 49-52
CHAPTER XI.
CIRCULATING DECIMALS.
54 55 55
IV.
I. Addition of Circulates
II. Subtraction of Circulates . . III. Multiplication of Circulates
CHAPTER XII. PERCENTAGE.
Division of Circulates . 56
Examples 56-57
.Definitions ....... 57
Solutions 57-69
II. Commission . 69
Definitions ...... .... 69
Solutions 69-71
Examples . 72
Trade Discount 72
Definitions 72
Solutions 73-76
Examples 76
Profit and Loss 77
III.
IV.
Definitions 77
Examples 83-84
V. Stocks and Bonds 84
Definitions . . . 84
Solutions 85-95
Examples 96-97
VI. Insurance 97
Definitions 97-98
Solutions . . 98-101
Examples 102
CHAPTER XIII.
INTEREST.
II.
III.
Simple Interest 103
Definitions 103
Solutions 103-106
True Discount . 106
Definitions 106
Solutions 106-107
Bank Discount 107
Definitions . . .107
Solutions 108-109
IV. Annual Interest 109
Annual Interest defined 109
Solutions 110-112
V, Compound Interest 112
Compound Int. defined. 112 Solutions 113-115
Definitions Solutions .
CHAPTER XIV.
ANNUITIES.
. . . . 115 I Examples .116-125
126
CHAPTER XV.
MISCELLANEOUS PROBLEMS. Solutions . 127-139
CHAPTER XVI. RATIO AND PROPORTION.
Definitions 139-141 I Problems
Solutions. ......... 141-144
... 144-146
Analysis defined. . . •Solutions . ... . .
CHAPTER XVII.
ANALYSIS. . . . . 146 Problems 146-177
177-180
CONTENTS.
CHAPTER XVIII.
ALLIGATION.
PAGE.
I. Alligation Medial 181
II. Alligation Alternate. . 181
Solutions
PAGE
181-186
CHAPTER XIX. SYSTEMS OF NOTATION.
Definitions 187
Names of Systems 187
Solutions 188-191
CHAPTER XX. MENSURATION.
v V,
Definitions 192-197
Geometrical Magnitudes class- ified 192
I. Parallelogram . . 198-200
II. Triangles 200-204
III. Trapezoid 204-205
IV. Trapezium and Irregular
Polygons 205
V. Regular Polygons . . . 205-207
VI. Circles 207-210
VII. Rectification of Plane Curves and Quadrature of
Plane Surfaces 210-213
II. Conic Sections 223
Definitions 223-224
1. Ellipse 224-227
2. Parabola 227-229
3. Hyperbola 229-232
IX. Higher Plane Curves ... 233
1. The Cissoid Diocles. 233-234
2. The Conchoid of Nicom-
edes 234-235
3. The Oval of Cassini 235
• 4. The Lemniscate of Ber-
nouilli 236
5. The Witch of Agnesi 236-237
6. The Limacon 237
7. The Quadratrix 238
8. The Catenary 238-239
9. The Tractrix 240
10. The Syntractrix 240
II. Roulettes 240
(a) Cycloids 240-243
(b) Prolate and Curtate Cycloid 243-244
(c) Epitrochoid and Hy- potrochoid 245-248
X. Plane Spiral 248
1. Spirals of Archimedes 249
2. The Reciprocal Spiral 249
3. The Lituus 250
4. The Logarithmic Spi- ral 250
XI. Mensuration of Sol- ids 251-254
1. Cylinder 254-255
2. CylindricUngulas 255-262
3. Pyramid and Cone 262-266
4. Conical Ungulas. 266-270 XII. Sphere 270-276
XIII. Spheroid 276-278
1. The Prolate Sphe-
roid 276-278
2. The Oblate Sphe-
roid 278-282
XIV. Conoids 282
1. The Parbolic Co-
noid .... 282-285
2. Hyperbolic Co-
noid 285-286
XV. Quadrature and Cuba- ture of Surfaces and Solids of Revolution 286
1. Cycloid 286-287
2. Cissoid 287-288
3. Spindles 288-289
4. Parabolic Spindle 289-290 XVI. Regular Solids 290
1. Tetrahedron .... 291-292
2. Octahedron 292*
3. Dodecahedron . . . 292-293
4. Icosahedron 293-294
XVII. Prismatoid 294-295
XVIII. Cylindric Rings ... 295-297 XIX. Miscellaneous Measure- ments 297
1. Masons' and Bricklayers' work 297
2. Gauging 297-298
3. Lumber Measure . 298
4. Grain and Hay. . , 298-299
10
CONTENTS.
MENSURATION — Concluded.
PAGE
XX. Solutions of Miscellaneous Problems 299-345
Problems 346-354
Examination Tests : . 354-360
Problems 361-366
GEOMETRY.
Definitions 367 (e) Assumption of the Sphere . . 380
On Geometric Reasoning 369 (/) of Motion 380
On the Advantages Derived from On Logic 380
the Study of Geometry and Laws of Thought 381
Mathematics in General 370 Law of Converse 383
Axioms 375 Methods of Reasoning 384
General Axioms 376 How to Prepare a Lesson in
Assumptions . 377 Geometry 388
(a) Assumption of Straight Line 377 Plane Geometry 390
(b) of the Plane 377 The Three Famous Problems of
(c) of Parallel Line 377 Antiquity 408
(d) of the Circle... 379
ALGEBRA.
Definitions 415 Arithmetical Fallacies 424
Solutions of Problems 416-419 Probability 425
The Quadratic Equation 419 Problems 428
Indeterminate forms 421
Biography of Prof. William Hoover 403
Probability Problems 434-435
Biography of Dr. Artemas Martin 436
Biography of Prof. E. B. Seitz 440
Biography of Rene7 Descartes 442
Biography of Leonhard Euler 446
Biography of Spphus Lie 451
Biography of Simon Newcomb 454
Biography of George Bruce Halsted 457
Biography of Prof. Felix Klein 459
Biography of Benjamin Peirce 462
Biography of James Joseph Sylvester s 468
Biography of Arthur Cayley 475
Table I 473
II 478
™ 478
•JV 479
V 479
VI 480
VII 480
Example 4gl
CHAPTER I.
DEFINITIONS.
1. Mathematics (p.aftv)p.a.Ttxrj, science) is that science which treats of quantity.
!(1.) Arithmetic. (2.) Algebra...
*.j Geometry..
rl. Calculus
1 2. Quaternions.
El. Platonic Geometry.. 2. Analytical Geometry. 3. Descriptive Geometry.
Differential.
Integral.
Calculus of Variations.
: a. Pure Geometry. » b. Conic Sections. i\. Plane Trigon'y. c. Trigonometry.. <2. Analytical Trig. (3. Spherical "
'(1.) Mensuration.
(2.) Surveying.
(3.) Navigation.
(4.) Mechanics.
(5.) Astronomy.
(6.) Optics.
(7.) Gunnery. ^(8.) &c., &c.
2>. Pure Mathematics treats of magnitude or quantity without relation to matter.
3. Applied Mathematics treats of magnitude as subsist- ing in material bodies.
4. Arithmetic (api^^nx^, from d^etf/io?, a number) is the science of numbers and the art of computing by them.
5. Alyebra (Ar. al, the, and geber, philosopher) is that method of mathematical computation in which letters and other symbols are employed.
6. G-eOWietry (y£ajfj.£Tpia, from y£U){j.£Tp£iv to measure land, from p£a, yfh the earth, and p.£Tp£~iv, to measure) is the science of position and extension.
7. Calculus ( Calculus, a pebble) is that branch of mathe- matics which commands by one general method, the most diffi- cult problems of geometry and physics.
12 FINKEL'S SOLUTION BOOK.
8. Differential Calculus is that branch of Calculus which 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.
1C. 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 (quaternis, from quaterni four each, from quator, 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 geometrical elements, and as ex- pressed by an algebraic symbol of quadrinominal form.
12. Platonic Geometry is that branch of geometry in which the argument is carried forward 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 argument 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 (rptytovov, triangle, ptrpov, meas- ure) is that branch of Platonic geometry which treats of the re- lations of the angles and sides of triangles.
16. Plane Trigonometry is that branch of trigonom- etry which treats of the relations of the angles and sides of plane triangles.
17. Analytical Trigonometry is that branch of trig- onometry which treats of the general properties and relations of trigonometrical functions.
18. Spherical Trigonometry is that branch of trig- onometry which treats of the solution of spherical triangles.
19. Analytical 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 projections upon auxiliary planes.
DEFINITIONS. 13
21. Mensuration is that branch of applied mathematics which treats of the measurment of geometrical'magnitudes.
22. 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.
23. Navigation is that branch of applied mathematics which treats of the art of conducting ships from one place to another.
24:. Mechanics is that branch of applied mathematics which treats of the laws of equilibrium and motion.
25. Astronomy d.ffrpovop.ia^ from affrpov, star and VO/JLOS law) is that branch of applied mathematics in which mechan- ical principles are used to explain astronomical facts.
26. Optics (oTtTix-ij, from 8</>ts sight,) is that branch of applied mathematics which treats of the laws of light.
27. GrUnnery is that branch of applied mathematics which treats of the theory of projectiles.
28. A Proposition is a statement of something proposed to be done.
/ rm \ !• Lemma.
, T^ I, ( a. Ineorem. \ on
1. Demonstrable. \ ^ pro^jem ( 2- Corollary.
29. Prop't'n. -
, , ( a. Axiom.
2. Indemonstrable. < , ^»
/ £. Postulate.
30. A Demonstrable Proposition is one that can be proved by the aid of reason.
31. A. Theorem is a truth requiring a proof.
32. A Lemma 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. An Axiom is a self-evident truth.
37. 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.
14 i-lNKEL'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 obtained.
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 direction for solving all problems of a particular kind.
42. 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. Numeration 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 general 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 used in Great Britain and the British provinces.. 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. Trillions. Billions. Millions. Units.
"^845 678^904 325^47 434^913
5-3 5§ 5g -SS ga *.! *p *~<§ ^o? * ^ ^EH ^
5. The number expressed in words by the English method, 'eads thus:
NUMERATION AND NOTATION. 15
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's&nd the number will be correctly expressed in words.
6. The following are the names of the Periods, according to the common, or French method:
First Period, Units.
Second " Thousands.
Third " Millions.
Fourth " Billions.
Fifth " Trillions,
Sixth Period, Quadrillions. Seventh " Quintillions. 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-vigintillions, 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, Pr i mo -sexagil lions, Secundo-sexagillions, and so on to Septua- gillions ; Primo-septuagillions, Secundo-septuagillions, and so on to Octogillions ; Primo-octogillions, Secundo-octogillions, 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, 59 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 inay be gained from the observation of children and of
16 FINKEL'S SOLUTION BOOK.
nations in the childhood of civilization*. When children count or add they use their ringers, 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, sufficiently large for purposes of computation and that they are always at hand.
Let us consider the row of objects, XXXXXXXX
XXXXXXXX , with regard to their order, say
from left to right, freeing our minds from all notions of magni- tude. Beginning 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 fH-F -ffH fH4.
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.
f 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 his 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 tell how many he had.
NUMERATION AND NOTATION. 17
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 0 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 left, 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 a' or is inferior to a', and a'
*See Conant's Number Concept for a full treatment of the various systems of notation.
18 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 integer 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. *£- 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 Laws," 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, -1/-, 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, (^-)Xb=Vj. We thus reach the definition of fractions. The concept of fractions may also be established as in the next article.
11. Fractions. Let us now again assume the row of
0123456789
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 1,2,8,4, . . . . We must then invent marks for the objects previously marked 1,3,5,7 The
NUMERATION AND NOTATION. 19
marks invented are shown in Figure 1, above the objects, the old marks being below the objects.
041f2 "|3}4J5
xxxxxxxxxxxx....
0123456789
Fig. i.
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 J, §; f, |; -J , -f ; and so on.
0 4 f 1 | | 2 | | 3
Thus, XXXXXXXXXX
0123
Fig. 2.
-t n o
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,
OJi|lffJ2
Thus, XXXXXXXXXX....^..
0 \ 1 I 2
Fig. 3.
From this we see that \ and f are marks for the same object. Also | and f . Hence, }=4 and |=f .
A row marked 0, \, f , J, 4, f , 1, . . . . is to be understood as arising from the interpolation of objects by fives; that is, by introducing the objects £, f , f, f, f ,...., or £, J, \, f , f, . . . .
As f comes before f , we say f<f , or J<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, /i,/2'/3» • • • Also, given any positive rational mark, r, other than zero, we can interpolate rational marks be-
20 FINKEL'S SOLUTION BOOK.
tween 0 and r. When no object can be made to fall between an assigned object and 0, that assigned object must be 0 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,'. . . .
0 * i HI
XXXX . . to in finityXXX . . to infinityXXXXX . . to infinityXX . . to infinityX 0 Fig. 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.
12. Irrational Numbers. In considering square num- bers from the ordinal point of view, we re-label our natural row as in Fig. 5.
0 1 i f 2 3
xxxxxxxxxxxxxxxx
0 1 i 2. } 3 4 56789
Fig. 5.
where 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 numbers. Bvery 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 and 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 % 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 not closed on the right and the right-hand class not closed on the left, — leads to the introduction of a new object named by an irrational number.
NUMERATION AND NOTATION. 21
For example, V 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 }/~2. Thus, it is clear that the class on the left of i/~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 s and /, — are equal if the rational objects to the left of 5- are the same as the rational objects to the left of /, and the rational objects to the right of s are the same as those to the right of sr. Thus, 4^ and 2^ effect the same sep- aration of the rational numbers. Hence, 4^=2M.
An equivalent condition for the equality of ^ and / is that every rational number to the left of s shall be to the left of /, and every rational number to the left of s' shall be to the left of s.
Between two unequal irrational objects, s and /, there must lie rational objects; for, since s and s' 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 Morley's Introduction to the Theory of Functions, Chapter I., from which this has been adapted.
22 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 56 thousand.
9. Write 457 thousand 341 million.
10. Write 19 trillion trillion billion billion million million.
CHAPTEK III.
ADDITION.
1. Addition, 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")
6780 " " 9=3
8768 " " 9=2 >Excess of 9's=8.
5343 " " 9=6
3987 " " 9=OJ
32723 excess of 9=8
Explanation. — Ad'ding 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.
3.
SUBTRACTION.
Add the following, beginning at the left :
8456 9799 4363 5809 5432
23
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. 6095054 + 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 the 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. Subtraction 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 of 9's=7 795943 " " 9's=ly
188952 " " 9'i
, ^Excess of 9's=7. -6
24 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 equal 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 99999 9+1 1=9 9999 9+1 ) A , , 1=9 9 9+1 ) . ,
90000000 9856342 j-Add- 4326546 \ Add-
85784895 8978567 3214957
4215105 877775 11 11589 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 920864; 1816090 or 27497?
4. 34567—3451 8 + 3— -2 + 3—4 + 7 + 18—567 + 43812 — 1326 4 678=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. Multiplication 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
39275 21 excess of 9's=3.
23565 31420
3416925=excess of 9*8=3.
MULTIPLICATION. 25
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 the left :
75645 765
1st..
2d 3d
57868425
3. From this operation, we see that it is more convenient to begin at the right to multiply.
5. In multiplication, the multiplicand may be abstract, or concrete; but the multiplier is always abstract.
6. The sign of multiplication is X > and is read, multiplied by, or times. When this sign is placed between two numbers it de- notes that one is to be multiplied 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 789627 by 834, beginning at the left to multiply.
4. 1 acre = 43560 sq. in.; how many square inches in a field containing 427 acres?
26 FINKEL'S SOLUTION BOOK.
5. Multiply 6934789643 by 34789. Prove the result by cast- ing out the 9's.
6. 2778588 X 34678=what ?
7. 2X3X4— 3x7+3— 2X2+4+8X2+4— 3 X5+27=what?
8. 5X7+6X7+8X7— 4X6+6 X 6+7 X6=what?
9. 356789 X4876=what?
10. 395076 X 576426=what ?
11. 7733447 X998800=what?
12. 5654321X999880— 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 are 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
536 Un 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 work is correct.
DIVISION. 27
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 -J-, 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. -Met. = 6, and 24ct.-^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-r-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-f-9 X 3-7-9 — 1+3-^3x9— 8-:-4+5X 2—3 X2-T-2---3-- (3X4-J-6+5— 2)+81-r-27x3-f-9Xl8«-6= what? [Hint.— Per- form the operations indicated by the multiplication and division signs in the exact order of their occurrence.]
6. (64-4-32X96-7-12 — 7—5+3) Xj[(27-:-3)-f-9 — 1+2] + 93H-13X7— 45 } X9+45-T-9+3— l=what?.
7. 2x2-r-2-4-2-7-2x2X2-r-2-r-2-r-2=what? Ans. £.
8. 3-r-3-f-3x3x3-7-3-r-OX4x4x5x5=what? Ans. oo.
9. 2x2x2-f-2x2-7-2-r-2X2X2XOX2X2=what? Ans. 0.
28
FINKEL'S SOLUTION BOOK.
CHAPTER VII.
COMPOUND NUMBERS.
1. A Compound Number is a number which expresses several different units of the same kind of quantity.
2. A Denominate Number is a concrete number in which the unit is a measure; as, 5 feet^ 7 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. Reduction 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. Reduction 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.
Ex. Reduce 2 E. Fr. 1 E. En. 2 E. Fl. 3 yd. 2 na. to nails.
E. Fr. E. En. E. Sc. E. Fl. yd. qr. na. 21 232
65 34
12qr. 5qr.
6qr. 12 qr. 6" 5"
35 qr. 4
140 na.
2"
I42na.
TIME MEASURE. 29
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 " 1 hour (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.
7. The Solar Day is the time between two successive appearances of the sun on a given meridian.
8. The Astronomical Day is the solar day, begin- ning and ending at noon.
9. The Civil Day9 or Mean Solar Day9 is the average of all the solar days of the year. It equals 24 hr. 3 min. 56.556 sec.
1O. The Solar Year, or Tropical Year9 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.
30 FINKEL'S SOLUTION BOOK.
14. A Metonic 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 quotient is the number of cycles, and the remainder is the year of the cycle. Should there be no remainder the proposed year is the twenty-eighth, or last of the cycle. The formula for the above
rule is <! \ in which x denotes the date, and r the re-
mainder which arises by dividing x-\-§ by 28, is the number required.
Thus, for 1892, we have (1 892+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 will fall on the same days in any two years which occupy the same place in the cycle; hence, a table of the 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; the quotient is the number of the cycle elapsed and the remainder is the Golden Number.
r*+ii
The formula for the same is <j \ in which r is the re-
mainder after dividing the date-}-l by 19. It is the Golden Number.
17. A Dionysian or Paschal Period is a period of 532 year, after 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 Induction 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
LONGITUDE AND TIME,
31
lays would give thirty-three for the epact of the fourth year; but in consequence of the insertion of 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.
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 .Fand 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 F and F' are called the/0«, 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. Kepi=peri, near, and fytos=halios, sun); when farthest from the sun, it is said to be in aphelion (Gr. dn6=apo, from, and r\kw~-=halios, sun). The points A and B, in the diagram, 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.
32 FINKEL'S SOLUTION BOOK.
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 around 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 28J°. If the axis of the earth were extended indefi- nitely, it would pass very near, 1J°, 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°.
LONGITUDE AND TIME. 33
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, counting from Greenwich around through the west, is said to be 85° east longitude.
From the above discussion, we see that, since the earth turns on its axis once in 24 hours,
24 hrs. =360° of long., or 360° of long.=24 hrs. 1 hr. =^3- of 360°=! 5° of long., or 15° of long.=l hr. 1 min.=gJ0- of 15°=15' of long., or 15' of long.=l min. 1 sec. =fa of 15' =15" of long., or 15" of long.=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" + 30° l&=m° 16' 36". Hence, since 15°=1 hour, 107° 16' 36"= (107° 16' 36")-KL5f 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. 47J sec. Find the difference of longitude.
1 hr. =15°. 1 min. =15'. II. 1 3. 4 min. =60', or 1°.
Isec. =15". 5. 47J- sec. =47JX15"=710"=11' 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. =15°.
2. 2hr. =30°.
3. 1 min.=15'.
II. <! 4. 30 min.=30X15'=450=7° 30'.
5. .'. 37° 30'=difference of longitude.
6. .'. 56°+37° 30'=93° 30', the longitude of the west-
ern place.
34 FINKEL'S SOLUTION BOOK.
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 time 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 7^-° 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 7J° wide on either side of 90th meridian, i. e., the belt of country lying between 82^° and 97-|° 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 m the Union Depot at St. Louis indicates noon, 12 o'clock M., the clocks in the Union Depots at Indian-
STANDARD TIME. 35
apolis and Kansas City also indicate noon, though at Indianap- olis it is a little more than 16 min. past noon 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 miu., 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 Columbus, Ohio, to Springfield, Mo., and carrying standard time of Columbus, would have standard time at St. L,ouis, standard time and local time at St. L<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 67J-0 west longitude and 82^° west longitude is called Eastern Time ; between 82-^-° W. and 97i° W., Central Time; between 97^-° W. and 112£° W., Mountain Time; and between 11 2^° W. and 127J° W., Pacific Time. We might call the time in the belt between 7^° E. and 7£° W,. Greenwich Time; between 7J° W. and 22|° W., East Atlantic Time; between 22^° W. and 37^° W., Central Atlantic Time; between 37J° W. and 52^° W., West Atlantic Time; and between 52^° W. and 67JC 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.
36 FINKEIv'S SOLUTION BOOK.
THE INTERNATIONAL DATE LINE.
The International Date Line 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. MM while just to left, or west of the line, Wednesday is ending, i. e., Wednesday 12 o'clock, P. 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 Ivondon, twenty-four hours later, it is Tuesday noon there, and there has been no intervening sunset. When does Monday noon become Tuesday noon? The convention is that the change of date occurs at the 180tb 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 western side, it counts the same day twice, passing from Tuesday forenoon back to Monday, and having to do its Tuesday over again."
This line is now little used by sailors, the 180th meridian having taken its place.
The consideration of this line in the solution of problems in longitude and time should add no serious difficulty. Solve the problem completely, leaving out 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? (Ray's Higher Arithmetic, p. iji,prob. 7.)
THE INTERNATIONAL DATE LINE.
37
II. <
1.
x HONOLULU
157° 52 W.= longitude of Honolulu.
OREE«WU:M^*-T:- ^3^
151° 11 E.= longitude of Sydney.
309° 3'=differ- ence of long- itude meas- ured from Honolulu through Greenwich to Sidney.
360°— 309° 3' = 50° 57'= difference of longitude measured directly on the equator from the merid- ian through Honolulu to the meridian through Sydney.
15°=1 hr.
1°=-- hr.=4 min.
5. 6.
7. 50° 57"=50fJ°r=50i§0=50^X4 min.=3 hr. 23 min.
48 sec., difference of time.
8. 4 hr. 5 min., Sunday — 3 hr. 23 min. 48 sec. =41 min.
12 sec., Sunday.
9. Regarding the date line, Sunday is changed to Mon-
day, since Honolulu is east of the line, while Syd- ney is west of it.
EXAMPLES.
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° II7 E.?
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° 59X E., what is the hour and day of the week at Honolulu, Hawaiian [slands, longitude 157° 52X 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.
38
FINKEL'S SOLUTION BOOK.
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 at his destination finds that his watch is 1 hr. 20 min. 30 sec. fast. What is the longitude of his destination if the longitude of New York is 74° (X 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° Ox 24" W.?
Ans. 14 min. G^- sec. past 7 A. M.
SOLUTIONS. Ex. 2. — Reduce 2 p. 3 pn. 1 tr. 1 hhd. 1 gal. 1 qt. to pints,
126 84 42 31*
63 4 2 bbl. T. hhd. gal. qt. pt.
1 11
63 63
P- 2 126
pn. 3
84
tr. 1 42
252 gal. 252 gal. 42 gal.
63 gal.
42 252 252
610 gal. 4
2440 qt.
4882 pt.
Ex. 3. I. Reduce 2 bu. 3 pk. 2 qt. 1 pt. to pints. Equation Method.
-1. 1 bu.=4 pk. 2. 2 bu.=2x4 pk.=8pk. 8 pk.+3 pk.=ll pk.
I pk.=8 qt.
II pk.=llx8qt.=88 qt. 88 qt.+2 qt.=90 qt.
90 qt.=90x2 pt.=180 pt. 180 pt.-f 1 pt.=181 pints. .-. 2 bu. 3 pk. 2 qt. 1 pt.=181 pints.
Solution : II.
Conclusion : III.
EXAMPLES.
Solution: II.
Conclusion:
Solution : II.
Ex. 4. I. Reduce 529 pints to bushels. Equation method.
1. 2 pt.=l qt.
2. 529 pt.=529-4-2=264 qt.+l pt.
3. 8 qt.=l pk.
4. 264 qt.=264-7-8=33 pk.
5. 4 pk.— 1 bu.
6. 33 pk.=33-f-4=8 bu.+l pk. III. .-. 529 pints=8 bu. 1 pk. 1 pt.
Ex. 5. How many gallons will a tank 4 ft. long, 3 ft wide, and 1 ft. 8 in. deep contain?
'l. 4 ft.=length,
2. 3 ft.=width, and
3. 1 ft. 8 in.=l£ ft.=depth.
4. 4x3Xlf=20 cubic ft.=contents of tank.
5. 1 cu. ft.=1728 cu. in.
6. 20 cu. ft.=20Xl728 cu. in.=34560 cu. in.
7. 231 cu. in.=l gal.
8. .;. 34560 cu. in.=34560-f-231=149ff gal.
Conclusion : III. .-. The tank will contain 149ff gallons.
(Fish's Comp. Arith., p. 126, prob. 2.)
EXAMPLES.
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 1 3 3 1 3 19 gr. how many grains ?
7. Change 16 Ib. 3 oz. 1 gr., Troy weight to Avoirdupois weight.
8. An apothecary bought by Avoirdupois 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 the 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 between the weight of 42f ft), of iron and 42.375 ft), of gold ? Ans. 52545 gr.
21. How many bushels of corn will a vat hold that holds 5000 gallons of water. Ans. 537 A 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 ? Ans. $22.797+.
23. If a man buys 10 bu. of chestnuts at $5 a bushel, dry meas- ure, and sells the same at 25 cents a quart, liquid measure, how much does he gain? Ans. $43.09-f- 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. How 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 ^ to the weather?
. 20160.
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-J- Ibs. ? Ans. 40-
GREATEST COMMON DIVISOR. 41
CHAPTER VIII.
GREATEST COMMON DIVISOR.
1. A Divisor of a number is a number that will exactly divide it.
2. A Common Divisor of two or more numbers is a number that will exactly divide each of them.
3. The Greatest Common Divisor 9 or Highest Common 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.
II. 60=2X2X3X5. 2. 120=2X2X2X3X5. 3. 150=2X3X5X5. 4. 180=2X2X3X3X5. 5. G. C. D.=2X3X5=30. III. .-. G. C. D.=30.
Explanation. — By inspecting the factors of each number 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 180 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.
(1. 180=2X2X3X3X5. TT J2. 1260=2X2X3X3X5X7. ' ]3. 1980=2X2X3X3X5X11-
U. 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 : the first containing 16 acres, the second 20 acres, and the third 24 acres. What is the largest sized lots
42 FINKEL'S SOLUTION BOOK.
containing each an exact number of acres, into which the whole can be divided? Ans. 4 A. lots.
4. A farmer has 12 bu. of oats, 18 bu. of rye, 24 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 256, 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.
CHAPTER IX.
LEAST COMMON MULTIPLE.
1. A. Multiple of a number is a number that will exactly contain it ; thus, 24 is a multiple of 6.
3. A Common, Multiple of two or more numbers is a number that will exactly contain each of them.
3. The Least Common Multiple of two or more num- bers is the least number that will exactly contain each of them. I. Find the L. C. M. of 30, 40, 50.
II. 30=2X3X5. 2. 40—2X2X2X5. 3. 50=2X5X5. 4. 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 in 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.
II. 2310=2X3X5X7X11- 2. 210=2X3X5X7. 3. 30=2X3X5. 4. 6=2X3. 5. L. C. M.=2X3X5X7X 11=2310. III. .-. L. C. M. of 2310, 210, 3(\ 3
LEAST COMMON MULTIPLE. 43
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 taken 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. of three numbers between 1 and 100 is 6804 ; and the quotient of the L. C. M. divided by the G. C. D. is 84. What are the numbers?
1. L. C. M.xG. C. D.=6804, and
L.C.M
c P TT — 3. .-.'L.'C.'M.XG. C. D.-s-k' C' p=L. C. M.X
G. C. D-X''=(G- C. D. )2=6804-;-84=81.
II.
4. G. C. D.— y'si =9, by extracting the square root.
5. .-. L. C. M.=6804-i-9=756.
6. 9=3x3.
7. 756=2x2x3x3x3x7.
8. 3x3x2x2=36.
9. 3x3x3x2=54. 10. 3x3x7 =63.
III. .-. 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.
44
FINKEL'S SOLUTION BOOK.
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 tho first time?
8. J. S. H. has 5 pieces of land; the first containing 3 A. 2 rd. 1 p.; the second, 5 A. 3 rd. 15 p.; the third 8 A. 29 p.v 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 X.
FRACTIONS. 1, A Fraction is a number of the equal parts of a unit.
1. Simple. 1. As to Form.<! 2. Complex.
2. Fraction. <
'1.
Common, J or Vulgar.
,2.
As to Value."
I
i*
Proper. Improper. Mixed.
2.
Decimal. J
'l. 2. 3.
Pure. Mixed. I Circulating. <^
Pure. Mixed.
3.
Continued
Fractions.
3. A Common 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, •§• expresses 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- ator, or denominator, or both, are fractional; as, — , ^f, -^.
of o^ 5
FRACTIONS. 45
6. A CoMiptttt'fld Fraction is a fraction of a fraction; as, J of f .
7. A jProper Fraction is a simple fraction whose numerator is less than its denominator; as, |.
8. An Improper Fraction is a simple fraction whose numerator is greater than its denominator; as, |.
9. A Mixed Wumber is a whole number and a frac- tion; as, 3|.
1C. A Decimal Fraction is a fraction whose denomi- nator is ten, or some power of ten; as, y3^, T|¥, T|^. The de- nominator of a decimal is usually omitted and the point (.) is used to determine the value of the decimal expression. Thus,
11. A Pure Decimal is one which consists of decimal figures only; as, .375.
13. A Mixed Decimal is one which consists of an integer and a decimal; as, 5.25.
13. A Circulating Decimal, or a Circulate, is a deci- mal in which one or more figures are repeated in the same 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, T^Vir^o^ — 5 — FT wl^5 when reduced to a
L X o X o
decimal, be circulating because the denominator contains the factor 3.
The repeating figure or set of figures is called a Repetend* and is indicated by placing a dot over the first and the last fig- ure repeated.
14. A Pure Circulate is one which contains no figures but those which are repeated; as, .273.
15. A Mixed Circulate is one which contains one or more figures before the repeating part; as, .45342.
16. A Simple Hepetend contains but one figure; as, .3.
17. A Compound Hepetend contains more than one figure; as, 354.
18. Similar Hepetends are those which begin and end at the same decimal places; as, .3467) and .0358-
19. Dissimilar Repetends are those which begin or end at different decimal places ; as, .536. .835, and .3567.
46 FINKEL'S SOLUTION BOOK.
2O. A Perfect Hepetend is one which contains as many decimal places, less 1, as there are units in the denominator of the equivalent common fraction ; thus, 7=.142857-
21. Conterminous Jiepetends are those which end at the same decimal place; as, .4267, -3275, and .0321.
22. Co-originous Hepetends are those which begin at the same decimal place ; as, .378, -5624, and 3-623.
I. Reduce - to its lowest terms.
Explanation. — Dividing the numerator 9, by 3, without changing the denominator, the value of the fraction is dimin- ished as many times as there are units in the divisor 3. Dividing the denominator 12, by 3, without changing the numerator 9, the value of the fraction is increased as many times as there are units in the divisor 3. Hence, if we divide both terms by 3, the in- crease by dividing the denominator will be equal to the decrease by dividing the numerator, and the value of the fraction will remain unchanged.
I. Reduce j- to a higher denomination.
Explanation. — Multiplying the numerator 2, by 4, without changing the denominator, the value of the fraction is increased as many times as there are units in the multiplier 4. Multiply- ing the denominators, by 4, without changing the numerator, the value of the fraction is decreased as many times as there are units in the multiplier 4. Hence, if we multiply both terms by 4, the increase by multiplying the numerator is equal to the de- crease by multiplying the denominator, and the value of the fraction remains unchanged.
I. Reduce 9£ to an improper fraction.
i. 9
oi* TT J 2- l=i=8-eighths.
• 1 3. 9=9Xf =-V-=9X8-eighths=r72-eighths.
Conclusion- III. .'. 9%=^j>-=79-eighths.
I. Reduce f to 24ths.
' 1. f =f f , or 8-eighths=24-twenty-fourths.
2. 4=4 of }J=A. or l-eighth=4 of 24- twenty-fourths=3-twenty-fourths.
f :=5 times /j:^^! , or 5-eighths=5 times 3-twenty-fourths=15-twenty-fourths. |=:||=r!5-twenty-fourths.
II. } 3.
FRACTIONS.
47
I. Reduce f to 8ths.
1. f=|, or 6-sixths=:8-eighths.
9 i — i ~f 8 f .a li, or
*• t— £ 01 f — o— fi-
ll.
III. I.
3. 4=5 times ±?=-
=J- of 8-
eighths.
or 5-sixths=5 times IJ-eighths^Gf- eighths.
Reduce to 3rds.
£Ji! i4off^M*=
I O O
= of 3-thirds.
III. .-. i~
Explanation. — In taking \ of f , 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.
1 . f ^rf^ll-elevenths.
i i _r 11 y~ ?j- i Of 11-elevenths
t-i of ii_n_n_^ _,
II.
-3 times S
2i-elevenths.
-^=3 times 24-elevenths ~ll =6felevenths.
to a mixed number.
3-thirds=l.
29-thirds=as many times 1 as 3-
thirds is contained in 29-thirds,
which is 9f .
III.
I. Reduce f . f , | to their I,. C. Denominator.
1. L. C. D.=12.
2.
3.
4.
II.-
48
FINKEI/S SOLUTION BOOK. I. Reduce |, |, | to their L. C. DenoMrntf-or
fl. L. C. D.=40.
2. !=«.
li. 3. 4=4x1*=!*.
i- •'• i, I, *=-**, H
I. Add |, f, J. 1. L. C. D.=24.
2. I=.
4. |=fxM= 5- J=x=
6. .'.
I. Reduce f to a fraction whose numerator is 15.
III. .-. f=if.
I. Reduce f, f , ^ to equivalent fractions having least common numerators.
1. L. C. N.=12.
2. 1=||.
A 4.
in. .-.
I. Subtract f from - I. L. C. D.=40.
2- 1=4*-
3- f=fX«=it-
4- -
TIL ,. A-t^U, 1. Multiply J by f.
fl-
n.{2.
13.
f X|=5 times
III. .'. X*=£l:
FRACTIONS.
I. Divide -| by ^. II. J 2. |44=7 times f=V-
U. 4-HN4 of ¥=44=H±.
in. ...
Analysis to the last example :
'1. nj- is contained in 1, or J-, 7 times.
2. f is contained in l,orf, ^ of 7 times— |- times.
3. f is contained in l, ^ of J times^-^T times.
A- 7 is contained in -|, 5 times ^ times, or |^J times. Note. — By inverting the divisor, we find how many times it is contained in 1.
EXAMPLES.
1. One-fifth equals how many twelfths?
2. Reduce -|, ^, -|, and \ to fractions having a common de- nominator.
1 ^
3. Reduce I- to a fraction whose numerator is 13. Ans. _ ! •
4.
5. 6.
7. 8. 9.
10. 11.
i 1
1 13. l=what? ^4 ns. 907200.
Reduce £ to a fraction whose denominator is 11. Ans. _
11
Reduce ~|, -$-, f, to fractions having common numerators. Add £, -J, |l, |, and ^. 3 Of 8f— | of 5— what? Multiply f by 8f . Multiply I of 9i of f by f of 17. 197 r;2
1Z¥ . Of
Ans.
a
14. !i=
4
50 FINKEL'S SOLUTION BOOK.
15. (2iX2*+* of TV) X (f)3-H7f-3iXff)=what?
400000 Ans' 407511:
. A+H-A _
"
-7H 62A 4f - ' 4fx5i-200|^ " 3
8
17. 2-~-2-j-2-;-2H-2-r-2-r-2-r-2-: i : | : ^ : j
Ans. 1.
18. Reduce f to thirds. Ans. 2f thirds.
19. What fraction is as much larger than f as f is less than f ?
Ans. if.
20. What is the value in the 13th example if a heavy mark be drawn between ^ and -J-. Ans. If.
21. l--=what?
22. Subtract * of from ^ of -j- Ans. f JJJ.
*«t °f
23. What is the relation of 11 to 3?
(1. I=iof3.
Solution: 4 2. 11=11 times J of 3=Jf of 3=3§ ( times 3.
Conclusion: /. 11 is 3f times 3.
24. What is the relation of 19 to 5? Ans. 3J.
25. What is the relation of T6T to 24? Ans. ^
26. What part of 3 is 2?
Solution:
i 2. 2=2 times J of 3=§ of 3.
Conclusion: .-. 2 is § of 3.
27. What part of 6 is 7? Ans. J..
28. What part of 3 is J? Ans. TV
29. What part of 4 is 3? Ans. -^
30. What part of f of f is f of TV
EXAMPLES. 51
ri. |oft=f
Solution: f f °f Af-|V
( 4. ^ is A times f of f =T72- of f . Conclusion: .'. f of T7^, or ^, is T7^ of f of f .
i + i,!
31. 24i=what? Ans. 1&.
[Note. — This is a continued fraction.]
32. Find the number of which 75 is ^.
33. Find the number of which 180 is f .
34. if is f of what number?
11. f of some number=if . I of that number=i of «=*. o. I oi that number, or the number rer quired, =4 times T%— if .
Conclusion: .-. if is f of if.
35. 27 is .3 of what number? Ans. 90.
36. f of if is \ of ,f of what number?
*' . '' ' f =* of somj nuniber » or
4. ^ of some number=§.
5. f of that number, or that number,
Solution :
Conclusion: .-. § of |f is J- of f of -J¥5-.
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.
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 f of itself and $6 equals $12; what is the sum? Ans.
40. If •£% of a ton of hay is worth $8^, how much is 10 tons worth? Ans. $204.
41. What number is that if of which exceeds -fa of it by 111? Ans. 216.
42. What part of 2J feet is 3J inches? Ans. ^
43. A has $2400; f of his money plus $500 is } of B's; what sum has B? Ans. $1600.
52 FINKEL'S SOLUTION BOOK.
44. What fraction of -^V— TT-^T 1 I -^T+^^T 1
Ans.
45. A pole stands f in the mud, -^ in the water, and the re- mainder, 12f feet, above water. Find the length of the pole?
Ans. 44f feet.
46. If 48 is ^ of some number, what is f of the same num- ber? Ans. 63.
47. A can do a certain piece of work in 8 days, and B can do the same work in 6 days. In what time can both together do the work? Ans. 3^ days.
48. The lesser of two numbers is - ' gQa , and their differ-
7 °* °f
ence is - |-. What is the greater number? Ans. -^-jp--
*r
49. What number multiplied by f of f X3f will produce |f ?
Ans. f .
50. What number divided by If will give a quotient of 9J?
Ans. *$••
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 C in 6 days. In what time can A, B, and C do the work? Ans. 3T% days.
53. If f of 6 bushels of wheat cost $4J, how much will f of 1 bushel cost? Ans. 80 cents.
55. What number diminished by the difference between ^ and £ of itself leaves 1152? Ans. 2268.
56. If a piece of gold is f 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 \ as dense as the earth. How many times denser than water is the sun? Ans. .
CIRCULATING DECIMAL:
53
CHAPTER XI.
CIRCULATING DECIMALS. I. Change .63 to a common fraction.
.63=.636363+ etc., ad injinitum.
.636363+etc.,=.63+.0063+.000063+etc., ad infinitum. 13. This is a geometrical infinite decreasing series whose first term is .63 and ratio .63-7-.0063=Tfo. The sum
of such a series is =.63-r-(l — 5-^)=!-^=^-.
III. .-. .63=T7T.
I. Reduce 1.001 to a common fraction.
1. i.OOi=l.q0110011001100H0011+etc., ad infinitum.
2. 1.001=1.0011.
3. 1.00li= l+.0011+00000011 + 000000000011+etc., ad
infinitum.
II.,
4. .60li=rirst term.
5- TTmnr=-001 l-r-00000011=ratio.
6. .
17. .'.
(Ray's H. A., p. 120, ex. 8.)
III. /. 1.001=1*^.
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 by this fraction, its numerator becomes the denominator of the fraction required. Hence, a circulate may be reduced to a common fraction by writing for the denominator of the repetend as many 9's as there are figures in the repetend. Thus, .63—
I. Reduce .034639 to a common fraction. i 034639 - 034« * »-34ttl.-34X 999 +639_ 34 X 999+639 ~-TO4ttf=3!Tooo VWr 1000X999 '
__ 34 X (1000—1) +639 34000— 34+639 34000+639— 34
9990<X) 999000 999000
34639— 34_ 34605 __ 6921 = 2807 769 999000 ~999000~199800 66600 ~22200'
54 FINKEL'S SOLUTION BCQK.
In case the circulate is mixed, we have the following rule :
1. For the numerator, subtract that part which precedes the repetend from the whole expression, both quantities being consid- ered as units.
2. For the denominator, write as many 9*s as there are figures in the repetend, and annex as many ciphers as there are decimal fgures before each repetend.
I. ADDITION OF CIRCULATES.
I. Add 5.0770, .24, and 7.124943.
(1. 5.0770 = 5.0770 = 5.07707707 etc. 2. .24 = .242 = .24242424 etc. 3. 7.124943= 7.124943J= 7.124943J2 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 , /. 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 in 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-r-3=2. .*. 770 must be repeated twice. 6-j-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.
fl. .946 = .946 = .946666666666666 etc.
2. .248 = .2484 = .2484-84848484848 etc.
3. 5.0770 = 5.07707 = 5.077077077077077 etc. IL<U. 3.4884 = 3.488448 == 3.488448844884488 etc.
5. 7.124943= 7.12494312 = 7.124943124943124 etc,
6. Sum =16.88562056205620+, • =16.885620.
III. .-. Sum=16.885620.
III. I.
II. III.
I.
I.
CIRCULATING DECIMALS.
II. SUBTRACTION OF CIRCULATES. Subtract 190.476 from 199.6428571
1. 199.6428571 = 199.64285714
2. 190.476 = 19Q 47619047
55
I Difference = 9.1666666 = 9.16. .-. Difference=9.16. Subtract 13.637 from 104.1. 1. 104.1 =104.14 =104.1414141 etc. 13.637= 13.637= 13.6376376 etc.
-3.
Difference = 90.503776
.-. Difference=90.503776.
III. MULTIPLICATION OF CIRCULATES.
Multiply .07067 by .9432. .07067=.070677 .9432 = .9^?- Multiply by the fraction thus :
.06§609 .070677
.003056 16
.066665=product. .42406=.424062 .7067 =.706770
37)1.13083(3056=
HI 003056, be- cause the fraction is
208 185
Multiply 1.256784 by 6.42081.
1.256784=1.2567842
6.42081 = 6.420T9T
.02513568 = .0251356851
.5027137 = .502713702 7
7.540705 — 7.5407055407
.001028270 = .001028276°
233 222
~
Multiply by the fraction thus :
1.2567842
8.069583198
11).011311Q57 .001628276-
56 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 products, 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 repetend ; and so when we multiply 4 by 4, thtf first figure of the repetend in the partial product is 7.
IV. DIVISION OF CIRCULATES.
RULE. — Change the terms to common fractions; then divide as 4n division of fractions, and reduce the quotient to a repetend.
I. Divide .75 by .1
fl. .75=H=»
II. 2. .1=4.
[3. f-f -Hr=ifx9=fi=6.8181 etc.=6.81.
III. .-. .75-r-.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, .71, and 6.1234.
Ans. 63.1690670868888.
4. Add 5.16345, 8.6381, and 3.75.
Ans. 17.55919120847374090302.
5. From 315.87 take 78.0378. Ans. 237-838072095497. •6. From 16.1347 take 11.0884. Ans. 5.0462.
7. 18 is .6 of what number? Ans. 27.
8. From ^ take ^. Ans. .1764705882352941.
9. From 5.12345 take 2.3523456.
-4*5.2.7711055821666927777988888599994.
10. Multiply 87.32586 by 4.37. Ans. 381.6140338.
11. Multiply 382.347 by .03. Ans. 13.5169533.
12. Multiply .9625668449197860 by .75. Ans. .72.
13. Divide 234.6 by .7. Ans. 701.714285.
14. Divide 13.5169533 by 3.145. Ans. 4.297.
PERCENTAGE AND ITS VARIOUS APPLICATIONS. 57
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? . Ans. 6.290.
19. 4.956-f-.75=what? Ans. 6.6087542.
20. 7.714285-7-.952386=what? Ans. 8.1.
CHAPTER XII.
I. PERCENTAGE AND ITS VARIOUS APPLICATIONS.
1. Percentage is a method of computation in which 100 is taken as the basis of comparison.
2. Per cent, is an abbreviation from the L,atin, per centum, per, by, and centum, a hundred.
3. The 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. The 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 i(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:
5 per cent. =5 one-hundredths=yf^=.05^5%. In each of these expressions the fractional unit is TJ^. 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 TII)%=TIQ, we can take T^ of a number instead of ^ °f 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
58 FINKEL'S SOLUTION BOOK.
fractions should also be used, as this method, because of its brevity, is more often used in practice.
As an illustration, find 5% of $600.
1. 100 one-hundredths, or {%%, or 1.00, or 100%=$600,
2. 1 one-hundredth, or -j-fr0 , or .01, or 1%,=^ of II. ^j $600=$6,
3. 5 one-hundredths, or T^, or .05, or 5%— 5 times
III. /. 5% of $600=$30. I. What is 8% of 150 yards?
FIRST SOLUTION.
I. m=lW yards. II. 2. T^Tfo of i$r=Tiir of 150 yards=1.5 yards.
3. y^¥=:8 times 1.5 yards=12 yards. III. /.8% of 150 yards=:12 yards. "
Remark. — This solution is by the method of percentage purely.
II. \ 2. (3.
SECOND SOLUTION.
1. 100%=:150 yards.
1%=-^ of 100%=^ of 150 yards=1.5 yards. 8%=8 times 1%— 8 times 1.5 yards=12 yards.
III. /.8% of 150 yards=12 yards.
THIRD SOLUTION.
Briefly, by fractions :
8% of 150 yards^y^ of 15(0 yards=12 yards.
CASE I.
P— tage.
Formula.— BXR=P, where B is the base, R the rate, and P the percentage.
PERCENTAGE AND ITS VARIOUS APPLICATIONS. 59
I. What is 8% of $500?
rl. 100%— $500, II J 2. 1 %— y^ of $500— $5, and
U. 8%— 8 times $5— $40. III. .-. 8% of $500— $40.
1. What is |% of 800 men? -1. 100%— 800 men.
1%— YTFTT °f 800 men— 8 men, and ^%=| times 8 men— 6 men. III. .-. |% of 800 men— 6 men.
II./2. U.
II
I. What is 10% of 20% of $13.50?
1.50.
2. l%=Ti<> of $13.50— $.135, and .3. 20%— 20 times $.135— $2.70.
(2.) 100%— $2.70.
(3.) l%=riir 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?
rl. 100%— $1200, J2. 1%— Tfo of $1200— $12, and ( ') 3. 30%— 30 times $12— $360— son's share. U. $1200— $360— $840— remainder, rl. 100%— $840.
I 2. 1%=1^ of $840— $8.40, and
(2. W3. 20%— 20 times $8.40— $168— daughter's share. j4. $840— $168— $672— amount divided among four
brothers.
(3.) 100%— fourth brother's share; (4.) 100%+$12=--third brother's share. (5.) 100% +$24— second brother's share, and (6.) 100% + $36=first hi other's share. (7.) 100%+(100%+$12)-j-(100%+$24)+(100%+ $36)— 400%+$72— am'tthe four brothers rec'd. (8.) $672— amount the four brothers received. (90 ..400%-f$72— $672. ( 10. ) 400 % —$672— $72— $600. / 1 1 \ ~\ ot i f\f 'tfinri *ki ^n
IXJL.J -I m TlTlf <pOUU •pl.c'v/.
(12.) 100%— 100 times $1.50— $150— fourth brother's share.
III. .-. The last received $150. (R. H. A., p. 191,prob. 25.)
60 FINKEL'S SOLUTION BOOK.
1. What number increased by 20% of 3.5, diminished by 12^% of 9-6, gives 3|?
100%=the number. . 100% =3.5,
;. 1%=T^- of 3.5=.035, and .3. 20% =20 times .0*5=.7. TTJ rl. 100% =9.6,
(3.)<|2. 1%=TF(T of 96=.096, and
12£%=12-J times .096=1.2. (4.) .-. 100%+-7— 1.2=3i, (5.) 100%— .5=3.5, and 1(6.) 100%=4, the number. III. .-. The number=4. (/?. H. A., p. 191, prob. 26.}
CASE II.
Given j the base *nd the I to find the rate per cent. ( percentage j
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?
12000 men=100%,
2. 1 man=T2^o of 100%=Tf ff % , and
,3. 750 men=750 times -&*%*=&\%- III. .'. 750 men is 6^% of 12000 men.
I. A's money is 50% more than B's; then B'sishow many % less than A's?
II. 100%=B.'s money. Then, 2. 100%+50%=150%=A.'s money. 3. 150%=100% of itself. 4. 1 %=Tio of 100%=|% , and 5. 50%=50 times f %=33£%. III. .-. B.'s money is 33^% less than A.'s
(R. H. A.,p. 192, prob. 11.)
I. 30% of the whole of an article is how many % of f of it?
'1. 100%=whole article. 2. 66f %=f of 100%=f of the article. ?=100% of itself.
4. 1%=— of 100%=H%, and
bo 3
5. 30%=30 times U%=45%.
Ill .-. 30%-of the whole of an article is 45% of f of it,
(R. H. A., p. 192, prob. 20.)
I.
PERCENTAGE.
61
If a miller takes 4 quarts for toll from every bushel he grinds, what % does he take for toll?
(1. 1 bu.— 32 qt. TT 1 2. 32 qt.— 100%, '13. 1 t.=
V4. 4 qt.=4 times 3^%— 12|%. III. .'. He takes 12|% for toll.
CASE III.
~ . ( tne percentage and ) . r , , , , Glven to find the base'
I the rate per cent
d ) . . } to
Formula.— P-^R—B, where P is the percentage, R the rate per cent., and B the base.
$24 is f% of what sum?
1. 100%=sum.
2. |%=$24,
3. i%=4 of $24=48,
4. , or 1%,=8 times $8=$64, and
III.
I.
5. 100%— 100 times $64— $6400. .-. $24isf% of $6400.
I drew 48% of my funds in bank, to pay a note of $150; how much had I left?
III.
100%— amount in bank.
48%— amount drawn out. 100%— 48%— 52%— amount left.
48%— $150, !%=& of $150— $3.125, and
52%— 52 times $3.125— $162.50— amount left.
$162.50— amount I had left.
III.
I pay $13 a month for board, which is 20% of my salary; what is my salary?
100%— my monthly salary.
20%— $13,
l%=^r of $13— $.65, and 100%— 100 times $.65— $65, my monthly salary.
.'. $780—12 times $65— my yearly salary.
My salary=$780. (R. H. A,, p. 194, prob. 20.)
I. 2. 3. 4. 5.
62
FINKEL'S SOLUTION BOOK.
CASE IV.
Given
\ the °Unt
(
to find the base.
, rate per cent.
Formula. — A-*-(\+R)=B, where A is the amount, that is, the base and the percentage, R the rate per cent., and B the base.
III. .-. $540 is 8% greater than $500.
I. A sold a horse for $150 and gained 25%; what did the horse cost?
1. 100%=cost of horse.
2. 25%=gain.
3. 100%+25%=125%=selling price of horse, and
4. $150=selling price of horse ;
.'. 125%=$150,
of $150=$1.20, and
III.
5. 6. 7. 100%=100 times $1.20=412Q=cost of horse.
.-. The horse cost $120.
I. I 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, A. <4. $150=selling price of first horse;
5. .'. 125%=$150,
6. 1%=T^. of $150=$1.20, 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%=selling price of 2d horse, and B <J4. $150=selling price of second horse;
5. .-. 75%=$150,
6. 1 %=TV of $150=$2, and
7. 100%=100 times $2— $200=cost of second horse.
TTT 11
J$120=cost of first horse, and '* 1 $200=cost of second horse.
I. A coat cost $32; the trimmings cost 70% less, and the making 50% less than the cloth; what did each cost?
PERCENTAGE.
63
II.
III.
II.
III.
r 1.
2. 3. 4. 5. 6.
8.
9.
10.
100%=cost of cloth. Then 100%— 70%=30%=cost of trimmings, and 100%— 50%=50%=cost of making. 100%+30%+50%=180%=cost of coat. $32=cost of coat;
^- of $32=$.
"
IOO%=tlOO times $.1777£=$17.77£=cost of cloth. 30%=30 times $.1777J=$5.33i=cost of trimming. 50%=50 times $.1777J=$8.88f=cost of making.
r$17.77J=cost of cloth,
<[$ 5.33^=cost of trimmings, and
^$ 8.88|=cost of making.
(R. H. A., p. 196,prob. 12.)
n a company of 87, the children are 37^% of the women, who are 444% of the men; how many of each?
(1.)
(2.)
(4-)
(5.) (6.)
(V.)
(8.)
(9.)
.(10.)
100%— number of men. Then 44f%=number of women.
3. 37^%=37|- times .44f %=16f %=number of chil- dren in terms of the number of men.
10C%+44f%-f-16f% = 161^-%= number in the company,
87— number in the company ;
l%=i of 87=.54,
100%-=100 times .54==54=number of men, 444%==44£ times .54=24=number of women, and 16|%=16| times .54=19=number of children.
54=number of men, 24=number of women, and 19=number of children.
H. A., p. 197,prob. 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?
64 FINKEL'S SOLUTION BOOK.
100%=original stock.
33i%=decrease.
100%— 33£%=66f %=stock after first decrease.
100%=66f%,
!%=TiTof 66J%=f%, and 20%— 20 times f %=13£%=second decrease. 66f %— 13i%=53£%=stock after second decrease.
20%=20 times .53£%=10f %—first increase. II.<| U. 53^%+lOf %=64%=stock after first increase
2. 1%=T^ of 64%=.64%, and I 33£%=33| times .64%=21£%=second increase. [. 64%-f-21£%=85i|-%=stock after second increase.
100%— 85^%=14f %=whole loss;
$66=whole loss;
... 14| %= $66;
HO.) I%=:rr5 of $66— $4.50, and \ / / 14
(11.) 100%=100 times $4.50=$450=original stock. III. .-. $450=original stock.
I. 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%=96%=value of the brewery.
'1. 100%=value of the boat. Then [the boat.
2. I00%+16%=116%=value of tannery in terms of
3. 116%=100%, the value of tannery from step (1),
4. l%=nhr of 100%=ff % , and
5. 100%=100 times jfr%=86ff%=v&lue of the boat
in terms of the tannery. . \. 100%=96%, llA (A \J2- 1%= T*irof96%=.96%, and
75%=75 times .96%=72%=what the owner
of the boat received for it.
•'• 862%%— 72%— 14^V%=what the owner of the boat lost in the trade. (6.) $103=what he lost; (7.) ... 14^g-%=:$103,
(8.) l^^L-of $103— $7.25, and
(9.) 100%=100 times $7.25=$725=value of tannery. III. .'. $725=value of the tannery. (R.H.A.,p.W7,prob.2S.)
(3.)
PERCENTAGE.
65
Remark. — The value of the brewery and boat being ex- pressed vi 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 86^69% for 72% of the same value, losing
86/^%— 72%, or 14^%.
SECOND SOLUTION.
(1.) 100%— value of the boat. Then
(2.) 100%+16%— 116%— value of the tannery.
1(1. 100%— 116%, (3.K2. 1%=^ of 116%— 1.16%, and l3. 4%— 4 times 1.16%— 4.64%. (4.) 116%— 4.64%— 111.36%— the value of brewery in terms of the boat, rl. 100%— 111.36%, \\\ ,,. J2. 1%—^ of 111.36%— 1.1136%, and
I') 3. 75%— 75 times 1.1136%— 83.52%— what the owner of the boat received for it. (6.) .-. 100%— 83.52%— 16.48%— what he lost in the trade. (7.) $103— what he lost. (8.) .', 16.48%— $103, (9.) l%=nr?frir of $103— $6.25, and .(10.) 116%— 116 times $6.25— $725— value of tannery. III. .-. $725— value of the tannery.
THIRD SOLUTION.
100%— value of brewery.
100%— value of tannery. Then
100%— 4%— 96%— value .of the tannery.
.-. 96%.— 100%, the value of brewery in step (1),
l%=-fa of 100%— 1.041%, 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), 1%— rle of 104^%— .89^1% , and
100%— 100 times .89if£%— 89 jf£%— value of the boat in terms of the tannery, and consequently in terms of the brewery.
.*« 89-H1 %— 75%— 14iff %— what the owner of the boat lost in the trade.
$103— what the owner of the boat lost ;
(2.)
(3.)
(4.)
(5.) (6.)
(7.) i%=_L^ of $103=16.96, and
I /\ • M
I (8.) 104i%— 104^ times $6.96— $725=value of tannery.
66 FINKEL'S SOLUTION BOOK.
III. .-. $725=value of of the tannery.
Remark. — In 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|ff % for 75% of the same value.
MISCELLANEOUS PROBLEMS.
I. 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%— 12|%— 87i%— selling price. , 4. $175— selling price.
IL<>5. .-. 87i%— $175,
6. 1^=_L Of $175— $2, and
7. 100%— 100 times $2— $200,
III. .-. $200— cost of the horse. (R. 3d p., p. 204, Pr°t>- 5.)
I. A miller takes for toll 6 quarts from every 5 bushels of wheat ground; what % does he take for toll.?
1. 1 bu.— 32 qt.
2. 5bu.— 5 times 32 qt.— 160 qt.
II.
4. 160 qt.— 100%
c,
4. 1 qt.— T^ of 100%— f%, and
5. 6qt.—6 times |%—3f%.
III. " .-. He takes 3f % for toll. (R. 3d p., p. 204, prob. H-)
I. A farmer owning 45% of a tract of land, sold 540 acres, which was 60% of 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%=^. of 540 acres=9 acres, and ll.{ 6. 100%=100 times 9 acres— 900 acres=what he
owned.
(3.) 45%— what he owned.
(4.) .-.45%— 900 acres, (5.) 1%=^ of 900 acres— 20 acres, and
(6.) 100%— 100 times 20 acres— 2000=number of acres
in the tract. III. .-. The tract contained 2000 acres.
(R. 3d p., p. 204,prob. 12.)
MISCELLANEOUS PROBLEMS.
67
I.
!!.<
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.
2- (3.
(4.
100%— asking price of the cow. Then 100%-H50%=250%=asking price of the horse. 100%— 25%=75%=selling price of cow. 100%=250%,
l%=nnr of 250%=2.50%, and 33£%=33l times 2.5%=83£%=reduction on the
asking price of the horse.
250%— 83i%=166f%=selling price of the horse. 75%+166f%=241f%= selling price of both. $2QO=selling price of both. /. 241f%=$290,
°f 290==1-20 and
) 75%— 75 times$ 1.20=$90=selling price of the
cow. ) 166S%=166f times $1.20=$200=selling price of
the horse.
f$ 90=selling price of the cow, and ''•|$200=selling price of the horse.
(Brooks' H. A., p. 2J.3, prob. 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.
100%=$87560.
1%=T^ of $87560=$875.60, and
5%=5 times $875.60=$4378=the deduction.
18/ — 16/=2/— the deduction per cubic foot.
... $4378=the deduction of 4378-7-.02, or 218900 cubic feet. Then
$87560=cost of 218900 cubic feet.
$.40=$87560-r-218900=:cost of one cubic feet.
.-. $.40 — $. 18=$. 22=cost .of dressing per cubic foot.
Ill
r218900=number of cubic feet, and ' '122 cents=cosc of dressing per cuoic foot.
(Brooks' H. A., p.
, prob. 21.)
68 FINKEL'S SOLUTION BOOK.
EXAMPLES.
1. A merchant, having $1728 in the Union Bank, wishes to withdraw 15%; how much will remain? Ans. $1468.80.
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. 6%.
4. What % of I of f off is -J? Ans. Bl±%.
5. If a man owning 45% of a mill, should sell 33-j-% 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 $500. The horse cost 37-^% less than the buggy, and the harness cost 70% less than the horse ; what was the price of each?
Ans. buggy $275ff , horse $172^f, and harness $51f£.
9. I have 20 yards of yard- wide cloth, which will shrink on sponging 4% in length and 5% in width; how much less than 20 square yards will there be after sponging? Ans. l|~f yards.
10. A. found $5; what was his gain %? Ans. oo.
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. /~
13. A laborer had his wages twice reduced 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.
15. A man sold two horses for the same price $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. 69
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 % ? Ans. 50%.
1 ^
II. COMMISSION.
1. Commission is the percentage 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 Net Proceeds is the sum left after the commission and charges have been deducted from the amount of the sales or collections.
4. The Entire Cost is the sum obtained by adding the commission and charges to the amount of a purchase.
I. An agent received $210 with which to buy goods ; after deducting his commission of 5%, what sum must he expend ?
1. 100%=what he must expend.
2. 5%=his commission.
3. 100%+5%=105%=what he receives. II. S 4. $210=what he receives.
6. 1 %=rhr of $210=$2, and
7. 100%=1C9 times $2=$200=what he expends.
III. .-. $200=what he must expend.
(R. 3d p., p. 207,prob. 4.)
Note. — Since the agent's commission is in the $210, we must not take 5% of $210; 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?
70
FINKEL'S SOLUTION BOOK.
II.
in.
i.
a
3.)
(4.)-
100%— cost of the corn. 3%— the commission. 100%— 3%— 97%— net proceeds, which he invested
in corn.
1. 100%— cost of second lot of corn, commission.
entire cost of second lot of-
=co&t of second
= of $12— $2.06, and
3. 100%+3%— 103%:
corn.
4. 97%— entire cost of second lot of corn.
5. .-. 103%— 97%,
6- 1 %=^ of 97%— jfa% , and
7. 100%— 100 times •£$*%=&±£fo%
lot of corn in terms of the first.
8. 3%— 3 times Ty^%— 2T8-^% — com mission on second lot.
(5. 3%+2T8-053-%—5T8TF\%— whole commission.
(6. $12— whole commission.
(7-
(8.)
(9. ) 100%— 100 times $2 06— $206— cost of 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? r (1. 100%— cost of cotton.
(2. 5%— commission.
(3. 100%— 5%— 95%— net
1. 100%'— cost of sugar.
2. 2 %— commission.
3. 102%— entire cost of sugar.
4. 95%— entire cost of sugar.
5. .-. 102%— 95%,
6. 1%=^ of 95%— T9Ti5Y%, and
7. 100%— 100 timesT9^%— 93-^%— cost of sugar in
terms of cotton.
8. 2%— 2 timesY^% —1|4%— commission on the sugar.
(4.) 5%4-l|T%— 6|4 %— whole commission.
(5.) $210=\vhole commission. (6.) .-. 6|{%=$210,
(7.) 1 % =J4 of $210=$30.60, and
(8.) (9.)
TIT • J$3060— cost of cotton, and '''^$2850— cost of sugar.
[vested in sugar, proceeds, which he in-
100%— 100 times $30.60— $3060— cost of cotton.
— 93-^ times $30.60— $2850— cost of sugar.
. H. A., p. 219,prob. 6.)
COMMISSION.
71
II.
III
III 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%=commission. 4. $11.25=commission.
4. .-. 5%=$11.25.
5. l%=i of $11.25=$2.25, and
6. 100%— 100 times $2.25=$225=an)outtt Of the debt. .'. $225=amount of debt.
(7?, 3d p., p. 207, prob. 6.)
Charge $52.50 for collecting a debt of $525; what was the
rate of commission? 1. $525—100% !!.<! 2. $1=^ of 100%—^-%, and.
3. $52.50=52.5 times ^-%=10%=rate of commission. .'. 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?
100%—cost of flour.
4%=commission on flour. 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. i%_T^_0f (96%+$4.20)= .94T2T%+$.0411|f,
7. 100%=100 times (.94T2T%+$.0411|f )=94&%+
$4.11|f=cost of wheat.
8. 2%=2 times (.94T27%+$.0411i-f-)=ll|-%+$.08TV II ^ I —commission on wheat.
[il
(3.)
(4.)
5=3^- times loss on wheat.
4% + Hf % + $-08T4T + 3-gV% + $• $.21ff=whole loss. $5=whole loss.
9JT%=^$5— $.21|f— $4-78^2T. 1%=J- of $4.178^r=$.53, and
III.
(6.)
(7-) (8.) (9.)
(10.)
(11.) 100%=^100 times $.53=$53. .-. $53=cost of flour. ( /?. H. A., p. 219, prob.
72 FINKEL'S SOLUTION BOOK.
EXAMPLES.
1. A broker in New York exchanged $4056 on Canal Bank, Portland, at -f- %; what did he receive for his trouble?
Ans. $26.35.
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 remitted $160; for what were the books sold? Ans. $200.
4. A lawyer charged $80 for collecting $200; what was his rate of commission? Ans. 40%
5. 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 bbl., if I pay bim 2%% commission? Ans. $1601.5625.
7. If an agent's commission was $200, and his rate of com- mission 5% ; 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 6f%, when he sold them at 14f% commission. I lost in all $86; whatdidthe cattle sell for?
Ans. $200.
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.
III. TRADE DISCOUNT.
1. Trade Discount is the discount allowed in the pur- chas and sale of merchandise.
2. A List, or Hegular JPrice, is an established price, as- sumed by the seller as a basis upon which to calculate discount.
3. A Ifet Price is a fixed price from which no discount is allowed.
4. The Discount is the deduction from the list, or regu- .lar price.
TRADE DISCOUNT.
73
II.
III.
Sold 20 doz. feather dusters, giving the purchaser 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.
(.2.1 10% of 100%— 10%— first discount.
(3.) 100%— 10%— 90%— first net proceeds.
100%— 90%,
1%— Tio of 90%— &%, and 10%— 10 times T9^%z=9%— second discount. 90^— 9 %— 81%— second net proceeds, rl. 100%— 81%,
\3. 10%— 10°times ^%==8.i%==third discount. (6.) 10%+9%+8.1%— 27.1%— sum of discounts. (7.) $325.20— sum of discounts. (8.) .-. 27.1%— $325.20, (9.) 1%— ^T of $325.20— $12, and (10.) 100%— 100 times $12— $1200— wholesale price
of 20 dozen. (11.) $60— $1200-5-20= wholesale price of 1 dozen.
.•. $60— wholesale price per dozen.
(.ff. 3d p., p. 209, prob. 5.)
Bought 100 dozen stay bindings at 60 cents per dozen for 40, 10, and 7^% off; what did I pay for them?
(I-) (2.)
(3.)-
(4.)'
(5.).
60/= list price of 1 dozen. $60—100 times $.60— list price of 100 dozen.
1. 100%— $60,
2. 1 %— y^- of$60— $.60, and
3. 40%— 40 times $.60— $24=first discount.
4. $60— $24— $36— first net proceeds.
1. 100%— $36,
2. 1 %— y-^j- of $36— $.36, and
3. 10%— 10 times $.36— $3.60=second discount. $36 — $3»60— $32.40— second net proceeds.
100%— $32.40,
1%=.^. of $32.40=$.324, and 7|%— 7| times $.324— $2.43— third discount.
2. 3. ,4. $32.40— $2.43— $29.97— cost.
I paid $29.97.
3d p., p. 209, prob. 6.)
I. A retail dealer buys a case of slates containing 10 dozen for $50 list, and 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?
74 FTNKEL'S SOLUTION BOOK.
100%— $50.
2. 1 %— Ho of $50— $.50.
3. 50%— 50 times $50— $25— first discount. A. $50— $25— $25— first net proceeds.
100%— $25.
l^=T^rof$25=$.25.
10%=10 times $.25— $2.50— second discount ._ $25— $2.50— $22.50— second net proceeds, rl. 100%— $22.50. 1 2. 1%—TTnr of $22.50=4.225.
*' n s 10%— 10 times $.225=$2.25=third discount. $22.50— $2.25— $20.25=third net proceeds. 100%=-$20.25.
1%=^ of $20.25=$.2025. 2%=2 times $.2025=$.405=fourth discount. $20.25— $.405=$19.845=cost of 10 dozen slates. $1.9845=$19.845-f-10=:cost of 1 dozen slates.
III. .'. $1.9845=cost of 1 dozen slates.
(R. 3d p., p. 209,prob. 9.)
I. 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-JX on each hat ; what was the wholesale merchant's price per case?
(1.) 37!/==profit on one hat.
(2.) $13.50—36 times $.37|—profit on 3 dozen hats.
(3.) 100%— wholesale merchant's price per case.
(4.) 10%=discount.
(5.) 100%— 10%— 90%—my cost.
(6.) 12. 1%=^ of 90%— .9%.
]3. 12|%— 12| times .9%— lli%=profit in terms of wholesale price.
(7.) .'. lli%— $13.50'.
(8.) l%==*iTT of $13.50— $1.20.
(9.) 100%— 100 times $1.20=$120=wholesale- mer- chant's price per case.
HI. .'. $120=wholesale merchant's'price per case.
(7?. 3d p., p. 212,prob. 4.)
I. A bookseller purchased books from the publishers at 20% off the list ; if he retail them at the list what will be his per cent, of profit?
TRADE DISCOUNT.
II.
1. 100%=list price.
2. 20%=discount.
3. 100%— 20%=80%=cost.
4. 100%=bookseller's selling price, because he sold them
at the list price.
5. .-. 100%— 80%=20%=gain.
6. 80%=100% of itself.
7. 1%=-^ of 100%— li%, and
8. 20%=20 times l^%=25%=his gain %.
(R. 3d p., p. 211,prob. 1.)
III. .-. 25%=his % of profit.
Note. — Observe that since his cost is 80%, and his gain we wish to know what % 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.
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%=Hst price. (2.) 25% of !00%=25%=first discount.
100%— 25%=75%=first net proceeds. 100%=75%,
l%=yi¥ of 75%=|%, and 10%=10 times j%=7£%. 75%— 7i%=67|%= second net proceeds
ir
1.
2.
3.
,4. 1.
2. 1%=T^ of 67i%=.67£%, and
3. 5%=5 times .67i%=3.375%=third discount
4. 67t%— 3.375%=64.125%=cost.
1. 100%=64.125%,
2. !%=.64125%,and
3. 12%=12 times ,64125 %=7.695%=gain.
4. .-. 64.125%-f 7.695%=71.82%=selling price. $35.91=selling price.
... 71.82%=$35.91,
1%=^^ of $35.91=$.50, and
100% =100 'times $.50=$50=list price of 50 gross,
$1.00=$50-;-50=list price of one gross.
til. .'. $1.00=list price of one gross.
(R. 3d p., p. 212,prob. 10.)
I A dealer in notions buys 60 gross shoestrings at 70/ per gross, list, 50,, 10, and 5% off; if he sell them at 2d 10, and 5% off list, what will be his profit?
76
II.
FINKEL'S SOLUTION BOOK.
70/=list price of one gross.
$42=60 times $.70=list price of 60 gross.
100%=$42.
50 times $.42=$21s=first discount. $42— $21=$21=first net proceeds.
10$F==10 times $.21=$2.10=second discount. $21— $2.10=$18.90= second net proceeds.
100%=$18.90.
(7.)
5%=$.945=third discount $18.90— $.945=$! 7.955=cost 100%=$42
1 %=T£Tr of $42=$.42. [count.
20%=20 times $.42=$8.40=first conditional dis- 4. $42— $8 40=$33.60=first conditional net proceeds.
1. 100%=$33.60.
2. 1 %=^ of $33.60=$.336. [discount
3. 10%=10 times $.336=$3.36=second conditional
4- $33.60— $3.36=$30.24= second conditional net proceeds.
1. 100%=$30.24.
2. 1 %=rf(r of $30.24=$.3024. [discount
3. 5%=5 times $.3024=$1.512=third conditional
.24— $1.512=$28.728=selling price. $28.728— $17.955=$10.773=his profit.
III. .'. $10.773=his profit
(R. 3d p., p. 212, prob. 9.)
EXAMPLES.
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£% off. If I sell them at $4 a piece making a profit of 20%, what is the wholesale mer- chant's price per case? Ans. $399|Jf§^.
3. If I receive a discount of 20, 10, and 5% off, and sell at a discount of 10, 5, and 2^% off; what is my % of gain?
Ans. 21|%— .
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. 77
IV. PROFIT AND LOSS.
1. Profit and Loss are terms which denote the gain or loss in business transactions.
2. 3.
I.
III.
I.
II.
Profit is the excess of the selling price above the cost. LOSS is the excess of the cost above the selling price.
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? 1. 100%=cost of cloth per yard.
10%=his profit before reduction.
8%=his profit after reduction. 10%— 8%=2%=his reduction. 5/— reduction.
=2 and
100% =100 times 2^/=$2.50=cost per yard. .-. $2.50=cost of cloth per yard.
(R< 3d p., p. 211,prob. 13.)
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.) l%=Tiir of $150=11.20, and
(7.) 100%=100 times $1.20=$120=cost of first horse. (8.) $150— $120=$30=gain on first horse.
1. 100%=cost of second horse.
2. 25%=loss.
3. 100% — 25%=75%=selling price of second horse.
(9.)
4. $150=selling price.
III.
I.
5. .-. 75%=$150,
6. 1%= TV of $150=$2, and
7. 100%=400 times $2=$200=cost of second horse. $200 — $150=$50=loss on second horse.
$50 — $30=$20=loss in the transaction.
He lost $20 in the transaction.
( R. 3d p. , p. 211, prod. 12. )
A speculator in real estate sold a house and lot for $12000, which sale afford him a profit of 33^% on the cost ; he
(11.)
78
FINKEL'S SOLUTION BOOK.
then invested the $12000 in city lots, which he was obliged to sell at a loss of 33ij%; how much did he lose by the two transactions?
II.
(!•) (2.) (3.)
(4.) (5.)
(6.)
100%— cost of the house and lot.
33^%— gain. [lot.
100%+33£%=133i%=selling price of house and
$12000— selling price of the house and lot.
.-. 133i%=$12000.
(7.) (8.)
(9.) (10.)
=^1 of $12000=190. 100$
[lot.
III. I.
II.
100%— 100 times $90— $9000— cost of house and $12000— $9000— $3000— gain on house and lot. 100%— $12000.
1%— -^ of $12000— $120.
33^%— 33J times $120— $4000— loss on city lots. $4000— $3000— $1000— loss by the two transac- tions. .\ $1000— his loss by the two transactions.
(./?. 3d p., p. 211, prob. 15. )
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, 'l. 100%— cost of first horse.
20%— gain on the first horse. 100%+20%— 120%— selling price of first horse. ... 120%— 100%, from (1),
1%— T-b of 100%—f%, and 100%— 100 times |%=83^%— 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%— spelling price of second horse. .-.80%— 100%, from (1),
1%— ¥V of 100%— li%, and 100%— 100 times 1|%=125%— cost of second
horse in terms of the selling price. 125% — 100%— 25%— loss on the second horse. 25%— 16f %— 8J%— whole loss. $25— whole loss.
(7.)
(8.)
(9.)
((10.)
1.
2.
3.
4.
(z. )<
5.
6.
7.
i.
2.
3!
(3.)<
4.
5.
6.
7.
(4.)
(5.)
(6.)
l%:=_of$25=$3, and
o^ [horse.
100%=100 times $3=$300=selling price of each 83£%=83£ times $3=$250= cost of first horse. 125%=125 times $3=$375=cost of'second horse.
PROFIT AND LOSS.
79
. r$250=cost of the first horse, and | $375=cost of second horse.
I. What % is lost if f of cost equals f of selling price ?
1. f of selling price=|- of cost.
2. £ of selling price=£ of f of cost=f of cost.
3. J- of selling price=4 times -J of cost=f of cost.
4. |=cost.
5! !=selling price.
B. J=fr of 100%=1H%, loss. III. .-. Loss=lli%.
II.
I.
II.
Paid $125 for a horse, and traded him for another, giving 60% additional money. For the second horse I received a third and $25. I then sold the third horse for $150 ; what was my % of profit or loss?
(2.) (3.)
(4.)
(7.) (8.
100%=$125,
1%==^ of $125=$1.25, and 60%=60 times $1.25=$75 = additional money paid for the second horse. $125+$75=$200=cost of second horse. $150=selling price of the third horse. $150+$25=$175=selling price of second horse. $200— $175=$25=loss in the transaction. $200=100%,
$l=innr of 100%=!%, ana $25=25 times |%=12|%=my loss.
III. .-. My loss is 12|%.
(./?. H. A., p. 201,prob. 4.)
I. If [ buy at $4 arid sell at $1, how many % do I lose?
'1. $4=cost
2. $l=selling price.
3. $4— $l=$3=loss.
4. $4=100%.
5. $l=i of 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 B's money ; which had the most, and how much?
of
80
FINKEL'S SOLUTION BOOK.
II.
III.
(1.) 100%=A's money.
(2.) 2J%— whathe lost.
(3.) $5— what he lost.
(4.) .-. 2£%— $5,
(5.) 1%=^7 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.
5. 1 %=4- of $5=$1.50, and
3-$
6. 100%— 100 times $1.50=$150=B's money.
(8.) $180— $150— $30— excess of A's money over B's. .% A had $30 more than B. (R. H. A., p. 203, prob, 5.)
I. 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 cost of each?
I.
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%,
l%=TTnr of 200%=2%, and
25% =25 times 2%=50%= gain on the horse. 200%+50%=250%=selling price of the horse.
80%+250%=330%=selling price of both. $577. 50=selling price of both. .-. 330%=$577.50,
l%=irb of $577.50— $1.75, and 100%=1 00 times $1.75=$175=cost of carnage. 200%=200 times $1.75— $350=cost of the horse.
> J$175=cost of the carriage, and •'• |$350— cost of the horse.
(Milne's prac., p. 259, prob. 19. )
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?
II.
II.
PROFIT AND LOSS. 81
(1.) 100%— cost of the horse.
(2.) 100% + 10%— 110%— asking price.
rl. 100%— 110%, (8..K2. l%=Tio of H0%— 1TV%, and [asking price.
13. 10%— 10 times lT17r%=ll% = reduction from
(4.) 110%— 11%— 99%=selling price.
(5.) $198— selling price.
(6.) .-. 99%— $198,
(7.) 1%— gV of $198— $2, and '
(8.) 100%— 100 times $2— $200— cost of the horse.
III. .-. $200— cost of horse. (Milne's prac., p. 259, prob. 23,)
I. What must be asked for apples which cost me $3 per bbln that I may reduce my asking price 20% and still gain 20% on the cost?
(1.) 100%— $3. (2.) l%=rfo of $3— $.03, and
(3.) 20%— 20 times $.03— $.60— gain.
(4.) $3.00+$.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- l%=8i> of $3-60— $.045, and
7. 100%— 100 times $.045— $4.50— asking price.
ill. .'. $4.50— asking price. (Milne's prac., p. 261, prob.38.)
(5.)
I. A merchant sold a quantity of goods at a gain of
If, however, he had purchased them for $60 less than he did, his gain would have been 25%. What did the goods cost him ?
100%— actual cost of goods.
20%— gain.
100%+20%— 120%— actual selling price. 100% — $60— supposed cost. '1. 100%=100%— $60,
2. 1%=T^ of (100%— $60)— 1%— $.60, and !!.<! ^y>']3. 25%— 25 times (1%— $.60)— 25% —$15 = sup- posed gain. [ual selling price. (100%— $60) + (25%— $15)— 125%— $75— act- .•; 125%— $75— 120%, 5%=$75,
l%=|of $75— $15, and 100%— 100 times $15— $1500— cost of the goods.
III. .-. $1500— cost of goods. (Milne's prac., p. 261, prob. 40.)
82 FINKEL'S SOLUTION BOOK.
Note, — The selling price is the same in the last condition of this problem as in the first. 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.
II.
(1- (2.
(3.
100%— actual cost of the article.
20%— actual gain. 100%+20%— 120%— actual selling price.
(4.) 100% +$300— supposed cost. 1. 100%— 100%+$300,
of (100%+$300)— 1%+$3> and
!!.<! vt"']3, 20%— 20 times (1%-f $3)— 20% +$60— supposed loss. [ual selling price.
(100% +$300)— (20% +$60)— 80%+$240=act- .-. 120%— 80%+$240. 40%— $240,
1%— ^o- of $240— $6, and 100%— 100 times $6— $600— cost of the article.
III. .-. -$600— cost of the article.
(7?. H. A., p. 409,prob. 85.)
I. A man wishing to sell a horse and a cow, asked three times as much for the horse as for the 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.
>) 100% — 10%— 90%— selling price of the cow.
(1. 100%— 300%, (5.K2. l%=iio of 300%— 3%, and
13. 20%— 20 times 3%— 60%— reduction on horse.
(6.) 300%— 60%— 240%— selling price ot the horse.
(7.) 90% +240%— 330%— selling price of both.
(8.) $165— selling price of both
(9.) .-. 330%— $165, (10.) l%=idny of $165— $.50, and
(11.) 90%— 90 times $.50— $45— selling price of cow.
(12.) 240%— 240 times $.50— $120— selling price of horse.
Ill • / $45— amount he received for the cow, and ' I $120— amount he received for the horse.
PROFIT AND LOSS.
EXAMPLES.
1. What price must a man ask fora horse that cost him $200, that he may iali 20% on his asking price 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? Ans. $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. 3T7T/.
5 How must cloth, costing $3.50 a yard, be marked that a merchant mavr deduct 15% from the marked price and still gain 15%? Ans.W&fr.
6. Sold a piece of carpeting for $240, and lost 20%; what selling price would have given me a gain of 20% ?
Ans. $360.
7. Sold two carriages for $240 apiece, and gained 20% on one and lo?t 20% on the other; how much did I gain or lose in the 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. 6J%.
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 j 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 ^ of the selling price is gain, what is the profit?
Ans. 80%.
11. If -J- of an article be sold for the cost of -J of it, what is the rate of loss? Ans. 33^%.
12.. I sold two houses for the same sum; 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 10i sq. yd. of cloth to make me a full suit of clothes. The cloth I am about to buy is 1J yards wide and on sponging it will shrink 5% in length and width. How many yards will it take for my new suit?
Ans.
14. A grocer buys coffee at 15/ per ft>. to the amount of $90 worth, and sells it at the same price by Troy weight ; find the of gain or loss. Ans. Gain
84 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. 50 bu.
16 How must cloth costing $3.50 per yard, be marked that the merchant may deduct 15% from the marked price and still make 15% profit? Ans. $4.735.
17. 1 sold goods at a gain of 20%. If they had cost me $250 more than they did, I would have lost 20% by the sale- How much did the goods cost me? Ans. $500.
18. A merchant bought cloth at $3.25 per yard, and after keeping it 6 months sold it at $3.75 per yard. What was his gain %, reckoning 6% per annum for the use of money?
Ans. 12%+.
V. 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 Bond 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 of $100 each.
4. An Assessment is a sum of money required of the stockholders in proportion to their amount of stock.
5. A Dividend is a sum of money to be paid to the stock- holders in proportion to their amounts of stock.
6. The far Value of money, stocks, drafts, etc., is the nominal value on their face,
7. The Market Value 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 market value over their par value.
1C. Brokerage is the sum paid an agent for buying stocks, bonds, etc.
STOCKS AND BONDS.
85
II.
I. At |% 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. i%— 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-r-lOO, or 40 shares.
III. •. 40— number of shares.
I. How many shares of railroad stock at 4% premium can be bought for $9360?
rl. 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 %=rht of $9360— $90.
7. 100%— 100 times $90— $9000— par value. $100— par value of one share.
^9. $9000— par value of 9000-^-100, 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 ?
1. 105X ; or 105% in currency— 100/; or 100% in gold.
2. I/; or 1% in currency— .95-^y/ ; or .95-^% in gold.
3. 100X; or 100% in currency— 95^TX 5 or 95^% in gold. .*. $1 in currency is worth 95^X in gold.
In 1864, the "greenback" dollar was worth only 35f/ in gold; what was the price of gold ?
1. 35fX; or 35f% in gold— 100/ ; or 100% in currency.
2. I/; or 1% in gold— -^ of 100X; or 100%— 2.8X ; or 2.8% in currency. }OT
3. 100X; or 100% in gold— 100 times 2.8X; or 2.8%— 280/;
or 280% in currency.
/. $1 in gold was worth $2.80 in currency.
(R. 3d p., p. 217,prob. 8.)
II.
Ill, I.
Bought stock at 10% discount, which rose to 5% premium and sold for cash. Paying a debt of $33, I invested the balance in stock at 2% premium, which at par, left me $11 less than at first; how much money had I at first?
86 FINKEL'S SOLUTION BOOK.
(1.) 100%=my money at first.
(2.) 100%=par value of stock.
(3.) 10%=discount.
(4.) 100%— 10%=90%=market value.
(5.) .'. 90%=100%, my money; because that is the
amount invested.
(6.) l%=rv of 100%=1£%, and
(7.) 100%=100 times l£$,=lll£%=par value of the
stock in terms of my money.
(8.)
II.
/q \
2. 1%— l-|f%,and [terms ot my money.
3. 5%— 5 times 11%— 5-f %=premium on stock in
4. llll%-|-5-t%=116f% = what I received for the
stock.
5. 116f % — $33=amount invested in second stock.
1. 100%=par value of second stock.
2. 2%=premium.
3. 100%-f-2%=102%=:market valueof second stock.
4. 116f% — $33=market value of second stock
5. .-. 102%=116J %— $33,
6. 1%=^ of
7. 100%=100 times
$32T6T=par value of second stock.
(10.) 114T\\% — $32T6T=what 1 received for the second
stock, since I sold them at par. (11.) .'. 114TVij%— $32T6T=100%— $11, by the last con-
dition of the problem. (12.) 1
(13.) i%__L^ of $21T6T=$1.485, and 14T%%
(14.) 100%=100 times $1.485=$148.50. III. .-. I had $148.50 at first. (7?. H. A., p. 212, prob. 8.)
I. Bought $8000 in gold at 110%, brokerage •£% ; what did
I pay for the gold in currency? '1. 100%=par value of gold.
2. 110%=market value.
3. -£%=brokerage.
H.<[4. 110%+4%=110i%= entire cost.
5. 100%— $8000,
6. 1%=!-^ of $8000=$80, and
7. 110l%=110^ times $80=$8810=cost of gold in currency.
III. .'. $8000 in gold costs $8810 in currency.
I. 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?
STOCKS AND BONDS. 87
100%=par value of the bonds.
116%— market value. $5220— market value. .-. H6% =$5220.
1%— T|¥ of $5220— $45.
100%— 100 times $45— $4500=par value of bonds. 100%— $4500.
1%—^ of $4500— $45.
6%— 6 times $45— $270— income in gold. $1.00 in gold— $1.05 in currency. $270 in gold— 270 times $1.05— $283.50 in currency.
$283.50— income in currency.
(/?. 3d p., p. 217,prob. 5.)
What % of income do U. S. 4^-% bonds, at 108, yield when gold is 105%?
(1.) 100%— amount invested in the bonds.
(2.) 100%— par value of bonds.
(3.) 108%=rnarket value.
(4.) .-. 108%— 100%, from (1).
(5.) l%=riir of 100%— ff%, [of amount invested.
TT I (6.) 100%— 100 times |-f-%=92$%— par value in terms 100%— 92-J%.
4£%— 4^ times ff%=4-^%— income in gold. 8.) 100% in gold— 105% in currency. 9.) 1% in gold— T-^¥ of 105%— l-fa% in currency.
^(10.) 4-g-% in gold— ig- times l-2-1^%— 4f % 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%=ThfOf 100%=f%,
6. 4%— 4 times |%=3£%=gain on the investment.
III. .•. He gained 3£% on the investment.
I.
II.
FINKEL'S SOLUTION BOOK.
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?
$1.13 in currency=$l in gold.
$1 in currency^^Vg. of $l=$i^| in gold, and
$1921 in currency=1921 times
come in gold. 100%=par value of the bonds.
5 %— income in gold. $1700=income in gold. .-. 6%= $1700,
l%==\ of $1700=$340, and [bonds.
100%=iOO times $340=$34000=par value of the 100%=$34000,
l%=^f of $34000=1340, and 118%=118 times $340=$40120=market value, or
amount invested.
III. .'. $40120=amount invested.
II.
III.
I.
(I-) (2.
(3.
(4- (5. (6.
(7.)
(8.)
(9.) (10.)
(11.) (12.)
SECOND SOLUTION.
100%=amount invested in currency. 100%=par value. 118%=market value. ... 118%=100%, from (1.)
l%=rh- of 100?fc=HHF%> and
100%=100 times f|%=^84||%=par value in terms of the investment.
and
d%=5 times £f %==4^%==income in gold. 100% in gold— 113% in currency,
in gold=l-r'^ n currency in gold=4i| times n currency. $1921=income in currency.
and
of $1921=$401.20, and
100%=100 times $401.20=-$40120^amount in- vested in currency.
$40120=amount invested. (JR. 3d p., p. 218, prob. 8.)
How many shares of stock bought at 95^%, and sold at 105, brokerage \% on each transaction, will yield an income of $925 ?
STOCKS AND BONDS.
II.
1. 100%=par value of stock.
2. 95J%=market value of stock.
3. ^%==:brokerage.
4. 95i+i%=95i%=entire cost.
5. 105%=selling price-{-brokerage.
6. ^%=brokerage.
7. 105%— i%= 1041%=selling price.
8. 104|%— <
9. $925=gain.
10. .-. 9i%:
11. 1 %=~ of $925=$100, and
ft*
12. 100%=100 times $100=$10000==pai value of stock.
13. $100=par value one share.
14. $10000=par value 10000-f-lOO, or 100 shares.
III. .-. 100=number of shares.
(R. 3d p., p. 218,prob. 9.)
II.
If .1 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 income?
1. 100%=par value of furnace stock.
2. 5%=income.
3. $180=income.
4. .-. 5%=$180,
5. l%=i of $180=$36, and [nace stock.
6. 100% =100 times $36=$3600=par value of fur-
1. 100%=$3600,
2. 1%=T^ Ot $3600=$36, and [nace stock. $. 75%=75 times $36=$2700=market value of fur-
1. 100%=par value of state stock.
2. 6%=income.
(!•)
(8.)
(5.)
3. $180=income.
4. .-. 6%=$180,
5. 1%=^ of $180=$30, and [stock.
6. 100%=100 times $30=$3000=par value of state
1. 100% =$3000,
2. 1%=T^ 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.
(R. H. A., p. 225, prob. 2.
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 ?
90 FINKEL'S SOLUTION BOOK.
(1.) 100%— par value of the bonds. (2.) 106%— market value.
(3.) $4982— market value, or amount invested. I I (4.) .-. 106%— $4982,
(5.) 1 %— yfo of $4982— $47, and
(6.) 100%— 100 times $47— $4700.
rl. 100%— $4700, (7.K2- 1%— Ti<5- of $4700— $47, and
13. 4%— 4 times $47— $188=income in gold. JII. .-. $188— income in gold. (fi. 3p., p. 218, prob. 11.)
I. The sale of my farm cost me $500, but I gave the pro- ceeds to a broker, allowing him ^%, to purchase rail- road stock then in the market at 102%; the farm paid 5% income, equal to $2075, but the stock will pay $2025 more; what is the rate of dividend?
(1.) 100%— value of the farm.
(2.) 5%— income on the farm.
(3.) $2075— income on the farm.
(4.) .-. 5%— $2075,
(5.) l%=i of $2075— $415, and
(6. ) 100%— 100 times $415— $41500— value of farm.
(7.) $41500— $500— $41000— amount invested in stock. 1. 100%— par value of the stock.
2. 102%— market value, or amount invested.
3. ^%— brokerage
4. 102%+^%— 102|%— entire cost of stock.
fc=$41QOQ,
of $41000— $400, and
[railroad stock. 7. 100%— 100 times $400— $40000— par value of the
1. $2075+$2025— $4100— income on railroad stock.
2. $40000—100%,
3. $1%=^^ of 100%—^% , and [dend.
4. $4100—4100 times =W= rate of divi-
III. /. 10^%— rate of dividend. (/?. H. A., p. 224., prob. 4-)
I. What must be paid for 6% bonds to realize an income of 8% on the investment?
1. 100%— amount invested.
2. 6%— income on the par value of the bonds.
3. 8 %— income on the investment.
4. .'. 8% of investment— 6% of the par value,
5. 1% of investment—^ of 6%— f % of the par value, and 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
II.
STOCKS AND BONDS,
91
II.
II.
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 at 25% advance, or 6% stock at 25% discount.
1.) 100%—amount invested in the 9% stock, 100%=par value. 25%==premium.
100%+25%=125%= market value. .-. 125%=100%,
1%=^ of 100%=f %, and 100%=100 times f %=80%=par value in terms of the investment.
1. 100%=80%,
2. I%=TO-H of 80%=|%, and [stock.
3. 9%=9 times |%=:7i#==income of 9% 100%=amount invested in 6% stock. 100%= par value of 6% stock.
25%=discount.
100%— 25%=75%=market value. .-. 75%=100%.
1%=TV of 100%=li%, and
— 100 times l^%=133^%=par value of the 6% stock in terms of the investment.
LB.
2.) (3. (4- (5. (6.) (7-)
(8.)
(2.) (3.) (4-) (5.) (6.) (7.)
(8.)J2. 1%= IS. 6%=
and [stock.
6%=6 times l£%=8%=income of 6% III. .-. The latter is the better investment, since it pays 8% — 7.1.% ? or \cjc more income on the investment.
( Greenleafs N. A., p. 298, prob. 5.)
If I pay 87i% for railroad bonds that yield an annual in- come of 7%, what % do I get on my investment? 100%=investment. 100%=par value.
87^%=market value, or amount invested. .-. 87£%=100%, from (1.)
1%— 1- of 100%=1|%, and
o/'ff
100%=100 times l^%=114^%==par value in terms of the investment.
1. 100%=114f%,
2. 1%==T^15. of 114f %=1^%, and [ment.
3. 7%=7 times l|%=8%=income on the invest-
(I-)
(2.)
It!
(5.) (6.)
(7.)
Ill.
)=income on the investment.
92
FINKEL'S SOLUTION BOOK.
A banker owns 2-J% 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.
II.
!»J
1.
'2.
(3.).
4.
5.
6.
rl.
(4,)I:
L 1.
2.
3.
(5.)<
4. 5.
6.
1.
/A J*
(6-) 3.
(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 (1), 1%=-^ of 100%=!-^%, and
100%=100 times l£%=lll£%=par value of for- mer in terms of the investment.
100%=11H%,
2-J-%=2£ times 11%— 2J%=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)=1T3T%— $134T2T, 100%=100 times (1T3T%— $134T27) = 117jj%—
$13411^-3r=par value of latter in terms of former.
100%=117||%— $13411if, [$134T2T, and
1% = T^ of (H7fi.%— $l3411f3- )= lT3y % — 3%=3 times (lfV% — $13412T)=319T% — $402T6T =income of latter in terms of the investment. 100%=income of the latter. 100%+66|%=166|%=income of the former. 2J%=income of the former. /. 166f %=2^%,
of2|%=Jo%^nd
[terms of income of former.
"166*
100%=100 times ^%=lj%=income of latter in 3T9T% — $402T6T==income oif the latter.
, Iff [former.
100%=100 times $216=$21600=investment in 100%— $11400= $21600 — $11400 =$10200=- in- vestment in latter.
=2^- times $216=$600=income of former.
STOCKS AND BONDS.
III.
I.
II.
(15.) aA-%— $402;fr=3JV times $216— $402-1<y=$36Q=
income of latter.
(16.) $21600+$10200=$31800=whole investment (17.) $600+$360=$960= whole income.
J$31800=whole investment, and '• |$960=whole income. (/?. H. A., p. 225, prob. 4. )
W. F. Baird, through his broker, invested a certain sum of money in Philadelphia 6's at 115^-%, and three times as much in Union Pacific 7's at 89-J%, brokerage -j-% in both cases; how much was invested in each kincf of stock if his annual income is $9920?
(1.) 100%=amount invested in Philadelphia 6's. (2.) 300%=amount invested in Union Pacific 7's.
i. 2.
iuuy0=pai vaiue 01 .rmiauejpma. o s>. 115£%=market value.
3.
-j-%— brokerage.
(3-).
4.
5.
115-J%+|%=ll"6%=entire cost of Phila. 6's. .r. 116%=100%.
6.
1%=^ of 100%=||%, and
7.
100%— 100 timesff %= 86A%=par value of Phil-
adelphia 6's in terms of investment.
rl.
100% =86 2^%,
MS
1 %— .j-i-g. of 86^-% , and 6%=6 times ff %=5T5¥%=income of Philadel-
I
phia 6's in terms of investment.
1.
100%=par value of Union Pacific 7's.
2.
89^%=market value.
3.
^.%— brokerage.
/ r \
4.
89|%-}-£^==90%==entire cost of Union Pacific 7's.
(&r
5.
... 90%=300%,
6.
l%=fa of 300%=3£%, and
7.
100%=100 times 3|%=333i%=par value of
Union Pacific 7's.
rl.
100%=333£%,
(6.)j2-
-j (ji i Q.C QQQ i <T/.— — Q JL cL. and
1
7%=7 times 3J%=23^%=income of Union Pa-
(7.) (8.) (9.)
(10.)
(11.) (12.)
cific 7's in terms of investment.
52^+23i%=28!r%=whole income. $9920=whole income.
- °f ^9920
[in Philadelphia 7's. 100%=100 times $348=$34800=amount invested 300%=300 times $348=1104600=5= amount in- vested in Union Pacific 7's.
FINKEL'S SOLUTION BOOK.
III.
I.
II.
( $34800=a mount invested in Philadelphia 6's, and I $104400— amount invested in Union Pacific 7's.
(R. H. A., p. 225,prob. 6.)
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?
(2.)
(3.)
(4.)
100%=amount invested in mining stock. 100%+$430=amount invested in furnace srock.
1. 100%=par value of mining stock.
2. 114^%=market value.
4. ii4^%^-^%=1ilo%=ent\re cost.
5. /. 115^=100%, from (1),
6. l%=Tl-g of 100%— ff %, and
7. 100%=100 limes |J%=965|^=par vnlue oi
mining stock in terms of investment.
(5.)
(6.)
(7.) (8.)
l%=rb of 96ff %=£$%, and 6%=6 times |^%:==5-2\%
stock in terms of investment. 100%=par value of furnace stock. 112%=market value.
of
mining
=~. Tof (100%+$430)=f%+$3it, and --
100%=100 times (|%-|_$3|l)_88|%-|-$382f= par value of furnace stock in terms uf invrestm't.
2. l%=rJT of (88f%+$382| )=|%+|3|l, and
3. 4%=4 times (|%+$3H)= 3|- %+$15i|=income
of furnace stock in terms of the investment. .-. 5A%==s3f%.-fll5tti by the conditions of the problem,
!±a of $15=
and
(10.) (11.)
[mining stock. 100%=100 times $9,20=$920=amount invested in 100%+|430=^1350=amount invested in furnace stock. (^?. H. A., p. 225,prob. 7.)
Ill * /$920— amount; invested in mining stock, and ' *|$1350=amount invested in furnace stock.
STOCKS AND BONDS.
95
I.
n.
(i.)
Suppose 10% state stock is 20% better in market than 4% railroad stock; if A.'s income be $500 from each, how much money has he paid for each, the whole investment bringing 6-gf 3-% ?
1. 100 %=par value of state stock.
2. 10%=income.
3. $500= in come.
4. .-. 10% =$500,
5. 1%=TL of $500=$50,and [stock.
6. 100%=100 times $50=$5000=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=$125, and [railroad stock.
6. 100%=100 times $125=$12500=par value of $5000=f of $12500, i. e., the face of state stock
is |- of face of railroad stock.
1. 100%=whole investment.
2. G^l-g %=in come of whole investment.
3. $500+$500=$1000=income of whole investment.
(2.)
(3.)
(4.)
(5.)
6.
1%=— L- of $1000=$166.50, and
6^ [ment.
100%=100 times $166.50=$16650=whole invest-
1. 100%=investment in railroad stock.
l'. 40%=| of 100%=investment in state stock,
excluding the 20% excess. 2'. 100% =40%, 3'. l%=T^o of 40%=|%, and 4. 20%=20 times |% =8%=excess of state
stock over same amount of railroad stock.
3. 40%+8%=48%=investment in state stock.
4. 100%+48%=148%=whole investment.
5. $16650=whole investment.
6. .-. 148%=$16650,
7. l%=T^g- of $16650=$112.50, [railroad stock.
8. 100%=100 times $112. 50=$11250=investment in
9. 48%=48 times $112.50=$5400=investment in
state stock
j. ^ J$11250=amount invested in railroad stock, and ' ' ' 1 $5400=amount invested in state stock.
(R. H, A,, p. 227, prob. 5.)
9d FINKEL'S SOLUTION BOOK.
EXAMPLES.
1. What could I afford to pay for bonds yielding an annual 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. & O. 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 ; how 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. 337-^%.
7. How much must I pay for New York 6's so that I may realize an income of 9%? Ans. 66f %.
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 an annual income of $1800 ?r Ans. $35400.
10. Which is the more profitable, and how much, to invest $5000 in 6% stock purchased at 75%, or 5% stock purchased at 60%? Ans, The latter; $16|.
11. If a man who had $5000 U. S. 6's of 1881 should 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. When gold is at 120, what is a "greenback" dollar worth ?
Ans.
13. Suppose the market value of 5% bank stock to be 'll-g-% 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. $2923.07 A in former, and $1576.92T\ in latter.
14. A bought 5% railroad stock at 109|%, and 4|% pike stock at 107-J%, brokerage |% ; the former cost $100 less than the latter but yielded the same income; what did each cost him? „
Ans. $1100 cost of former, and $1200 cos» of latter
INSURANCE. 97
15. What rate % of income shall I receive if I buy U. S. 5's at a premium of 10%, and receive payment at par in 15 years?
Ans. "
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 $24000.
17. I bought 2|% stock at 80%, and 4|% stock at The income on the former was 44f % 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.
Hint. — Find the whole investment, and whole income as in the problem on page 75. Then find what % the whole income is of the whole invest- ment.
18. Invested in U. S. 4£'s at 105, brokerage £% ; f as much in U. P. 6's at 119|, brokerage £%; and 3 times as much in N. Y. 7's, at 87J, brokerage \%. If my entire income is $1702, find my investment. (School Visitor, vol. 12, p. 97.) Ans. $25320.
19. A. paid $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 stock 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.
VII. INSURANCE.
1. Insurance is indemnity against loss or damage.
~ T M. Fire Insurance.
1. Property Insurance, j 2 j.^ Insurance
2. Insurance.
2. Personal Insurance.
1. Life Insurance.
2. Accident Insurance.
3. Health Insurance.
3. Property Insurance is the indemnity against loss or damage of property.
4. Personal Insurance is indemnity against loss of life or health.
5. Fire Insurance is indemnity against loss by fire.
98 FINKEL'S SOLUTION BOOK.
6. Marine Insurance is indemnity against the dangers of navigation
7. Life Insurance is a contract in which a company agrees, in consideration of certain premiums received, to pay a certain sum to the heirs or assigns of the insured at his death, or to himself if he attains a certain age.
8. Accident Insurance is indemnity against loss by accident.
9. Health Insurance is a weekly indemnity in case of sickness.
1C. TJie Insurer, or Underwriter, is the party, or company, that undertakes the risk.
11. The Itisfc is the particular danger against which the insurer undertakes.
12. The Insured is the party protected against loss.
13. The frennium is the sum paid for insurance; and is a certain per cent, of 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.) I%=TT><T of $1800=$18, and
II.
\&. ) j. /o:==^-^ wi tp±ouu=<j>io, anu
(3.) i%=| times $18=$13.50=one annual deposit.
(4.) $135=10 times $13.50=ten annual deposits.
/•I. 100%=$135, (5.){2. 1%=^ of $135=$1.35, and
13. 5%=5 times $1.35=$6.75=deduction. (6.) $135— $6.75=$128.25=what I received.
ry . ( $135=amount deposited, and ' ' ' ( $128.25=amount received.
(7?. H. A., p. 230, prod. 5.}
I. An insurance company having a risk of $25000, at T9^%, reinsured $10000, at \% , with another office, and $5000, at 1%, with another; how much did it clear above what it paid ?
INSURANCE.
99
II.
(1.) 100%— $25000,
(2.) l%=T-b of $25000— $250, and
(3.) T9(r%=T9<r times $250— $225— what the company received for taking the risk.
1. $10000— amount the company reinsured at f%.
2. 100%— $10000,
(4.) 3. 1%= -rfoof $10000— $100, and
4. f%— 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. (5.) 3. 1%— T|-g- of $5000— $50— what the company paid
4. $80+$50— $130— what the company paid out.
5. $225— $130— $95— what it cleared.
III. .-. $95— what the company cleared.
(/?. H. A., p. 230,prob. 7.)
I took a risk at 4-J% ; reinsured -| of it at 2%, and £ of it at 2-^% ; what rate of insurance do I get on what is left?
(1.)
(2.)
100%— whole risk.
— premium.
— | of 100%^amount reinsured at 2%.
?— y^-g- of 40%— 1%, and [suringf of therisk. %— 2 times -| %— 1^%— ampunt I pay out for rein- fl. 25%— i of 100%— second part reinsured. 2. 100%— 25%. (4.K8. 1%— T^r of 25%— i%, and
4. 2^-%— 2-J- times ^%—|%— amount I paid out for ( reinsuring £ of the risk.
(5.) |%-|-f%—l^J%— amount of premiums paid out. (6.) 11%—1^%—^%—amountof premium I had left. (7.) 40%-f-25%— 65%— whole amount reinsured. (8.) 100%— 65%— 35%— risk left on which I received
TQ% premium. fl. 35%— 100% of itself. (9.K2, 1%— ^ of 100%— 2f%, and
(.3. ^ofa—.^ times 2^%=T3-?.%=r=rate of premium
III. ,'. T3?%— rate of insurance I receive.
(R. H. A., p. 281,prob. 6.)
Remark. — 35% is the base and -fa% is the percentage, and we wish to know what per cent, -}$% is of 35%.
I. Took a risk at 2%; reinsured $10000 of it at 2^% and $8000 at 1|%; my share of the premium was $207-50; what sum was insured?
100 FINKEL'S SOLUTION BOOK.
rl. 100%=$10000,
(1.K2. 1%= -r^of $10000=4100, and [$10000 reinsured. 13.
r%=2^ times $100=$212.50=amount paid out on
rl. 100%=$8000,
(2.K2. l%=Ti¥of$8000=$80,and [$8000 reinsured.
13. If %=lf times $80=$140=amount paid out on
(3.) $212.50+$140=$352.50=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=premium.
(9.) .'. 2%=$560,
(10.) l%=i of $560=$280, and
(11.) 100^=100 times $280=$28000=risk.
III. .-. $28000=risk. (R. H. A., p. 232, prob. 6.)
I. I can insure my house for $2500 at T8¥% 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?
r (1.) 100%=$2500. (2.) l%=Tta of $2500=$25, and (3.) -nr%=T8¥ times $25=$20=one annual premium. (4.) $240=12 times $20=twelve annual premiums.
1. 100%=the amount that will produce $20 an-
nually at 6%.
2. 6%=iriterest. !!.<! ft. J3. $20=interest.
l°'^4. /. 6%=$20,
5. l%=iof$20=$3i, and
6. 100%=100 times $3i=$333^=the amount that
will produce $20 annually at 6%.
(6.) $3331H-$20=$3531i=amount j would have to pay down by the former condition. [tion.
L (7.) .', $353i~$240=$113i=gain by the latter condi-
TTT ( 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 of the year.
INSURANCE.
The Mutual Fire Insurance Company insured a building and its stock for f of its value, charging If %. The Union Insurance Company relieved them of £ of the risk, at l-J-%. The building and stock being destroyed by fire, the Union lost $49000 less than the Mutual; what amount of money did the owners of the building and stock lose?
100%=value of the building and stock. 66f%=f of 100%=amount insured. lj%=rate of insurance. 100%=66f%,
nnr of 66f %=f =1| times f%=
from the owners of the building and stock. 16f %=i of 66|%=amount of which the Union relieved the Mutual.
= and
r (i-)
SI
and =what Mutual received
(10.)
(11.)
(12.) (13.)
(14.) (15.)
(16.)
(17.) (18.)
(21.)
of 1-J%=1-J. times -|-%— ^%— what the Mutual paid
the Union for taking the risk of 16|%. 16f %+li%=17-f%=whole amount the Mutual
received. [paid out.
amount the Mutual i^,~= amount the Mutual
lost." '
16f %=amount the Union paid the Mutual. ^%=amount the Union received from the Mutual. .'. 16f % — i%=16T5T%=amount the Union lost. 491V%— 16T\%= 32f%=what the Mutual lost
more than the Union. [Union.
$49000=what the Mutual lost more than the ... 32f %=$49000,
1 %= JL of $49000=$1500, and ' 32f [ing and stock.
100%=100 times $1500=$ 150000= value of build-
66|%=66|- times $1 500=$ 100000=a mount in- sured, [ers lost, it not being insured.
33j-%=33£ times $1500=$50000=what the own-
100% =$100000,
1%=-^ of $100000=$1000, and
l|%=l| times $1000=$1750=what the owners paid the Mutual for insurance.
.-. $50000 + $1750= $51750=whole amount the owners lost.
III. .'. The owners of the building and stock lost $51750.
102 FINKEL'S SOLUTION BOOK.
EXAMPLES.
1. At l-f%, the premium for insuring my stare was $89.10; what was the amount of the insurance? Ans. $6480.
2. The premium for insuring a tannery for f of its value, at , 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 was the rate? Ans. 7%.
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 of 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 f- of its value at 3%; what was the value of the mill? Ans. — .
7. A house which has been insured for $3500 for 10 years, at \°/c & 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 •J of it for 2^%, and ^ of it at 2-J%; in each case the amount cov- ers premium; how much do I gain? Ans. $99.558*
9. Took a risk at If %; reinsured f of it at 2J% ; my share of the premium was $43 ; what was the amount of the risk?
Ans. $17200.
10. Took a risk at 2£% ; reinsured -J of ?.t at a rate equal to 3% of the whole, by which I lost $37.50. What was the value of the risk? Ans. $5000.
SIMPLE INTEREST. 103
CHAPTER XIII.
INTEREST.
I. SIMPLE INTEREST.
1. Interest is money paid by the borrower to the lender for the use of money.
2. The Principal is the sum of money for which interest is paid.
3. The Rate 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 Amount 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.
Usury is interest at a rate greater than that allowed by-
8. law.
Let P=the principal,
r=the interest on $1 for one year, 7?=l-[-r=a mount of $1 for one year, 7Z=the number of years, yl=amount of P for n years, /V=simple interest on P for a year, Pnr— simple interest on P for n years. P -\-Pnr— P (l-|-?zr)=amountof P forn years, of P for n years.
(I.);
~\+nr* *v ' ... Pnr=A—P.
A — P Interest
When any three of the quantities A, P, n, r are given, the fourth may be found.
CASE I. rPrincipal,^
Given< Rate, and >to find the interest. Formula, 7=s/V«. iTime, J
104 FINKEL'S SOLUTION BOOK'.
I. Find the interest of $300 for two years at 6%. By formula,
Interest /V«=$300X-06x2=$36. By 100% method.
rl. 100%=$300,
1 2. l%=Tfc> of $300=$3, and.
13. 6%=6 times $3=$18=interest for one year. U. $36=2 times $18=interest for 2 years.
III. .'. $36=interest on $300 at 6% for 2 years.
CASE II.
(Principal, >> A—/>
Given<Rate, and >to find the time. Formula, n= — = — .
llnterest, ) Pr
I. In what time, at 5%, will $60 amount to $72? By formula,
A—P $72— $60
By 100% method.
1. $72=amount.
2. $60=principal.
3. $72— $60=$12=interest for a certain time. II.<J4. 100% =$60,
5. i%==_|^ of $60=$f, and
6. 5%=5 times $f=$3=interest for one year.
7. $12=interest for 12-4-3, or 4 years.
III. V. $60 at 5% will amount to $72 in 4 years.
CASE III.
rPrincipal,^ A—P
Given^ Time, and >to find the rate. Formula, r=— - — .
llnterest, ) I. I borrowed $600 for two years and paid $48 interest; what
rate did I pay? By formula,
A—P I
By 100% method. '1. $48=interest for 2 years. 2. $24=£ of $48=interest for 1 year. II.J3. $600=100%,
4. $l=:^oflOO%=i [5. $24=24 times i-%=4 III. .'. I paid 4% interest.
SIMPLE INTEREST: 105.
rTime, ^
Given<[Rate, and >to find the principal. llnterest. J Fo
CASE IV. rTime,
A — P 1 Formula, P==
I. The interest for 3 years, at 9%, is $21.60; what is the principal ?
By formula,
P_A-P_ 7_$21.60 . ~-~--
II.
By 100% method.
1. $21.60=interest for 3 years.
2. $7.20=4 of $21.60=interest for 1 year.
3. 100%=principal.
4. 9%=interest for 1 year.
5. $7.20=interest for 1 year.
6. .-. 9%=$7.20,
7. l%=i of $7.20=$.80, and 100%=100 times $.80==$80=principal.
III. .'. $80=the principal.
CASE V. Given<(Rate, and>to find the principal. Formula, P=
fTime, i A
LAmount J l+nr'
I. What principal will amount to $936 in 5 years, at 6% ?
By formula,
A $936
-= - 06~
By 100% method.
1. 100%=principal.
2. 6%=interest for 1 year.
3. 3Q%=5 times 6%=interest for 5 years.
4. 100%+30%=130%=amount.
5. $936=amount.
6. .-. 130%=$936,
7. 1%=^ of $936=$7.20, and
8. 100%=100 times $7.20=$720=principal.
III. .'. $720=the principal that will amount to $936 in 5 years at 6%.
106 FJNKEL'S SOLUTION BOOK.
I. In what time will any sum quadruple itself at
1. 100%=principal. Then
2. 400%=the amount
3. .-. 400%— 100%=300%=interest.
4. 8%=interest for 1 year.
5. 300%=interest for 300-=-8, or 37£ years.
III. .-. Any principal will quadruple itself in 37-J years at
II. TRUE DISCOUNT.
1. Discount on a debt payable by agreement at some future time, is a deduction made for "cash," or present payment; and arises from the consideration of the present 'worth of the debt.
2. Present Worth is that sum of money which, put oa 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 A inn years, P may be considered equivalent to A due at the end of n years.
.*. P may be regarded as the present worth of a given future sum A.
• P— *-
~~
I. Find the present worth of $590, due in 3 years, the rate
of interest being 6%.
By formula,
A $590
=-a=
By 100% method.
1. 100%=present worth.
2. 6%=mterest on present worth for 1 year.
3. 18%=3x6%=interest for 3 years.
TT ;4. 100%+18%=118%=amount, or debt. <5. $590=debt.
6. .-. 118%=$590,
7. 1%—yfg- of $590=$5, and
8. 100% =100 times $5=$500=present worth.
HI. .'. $500=present worth of $590 due in 3 years at
BANK DISCOUNT.
107
I. A merchant buys a bill of goods amounting to $2480; he can have 4 months credit, or 5% off for cash : if money is worth only 10% to him, what will he gain by paying cash?
•••i ;,!
)
(2. (3. (4.
(5. (6.
(7.
(8. (9.
(11.)
III. .'. He
100%=present worth of the debt. 10%=interest on present worth for 1 year. 3-j.%=interest for 4 months. 100%+3i%=103i%=amount of present worth,
which equals the debt, by definition. $2480=the debt. ... 103i%=$2480,
1 %=?7^T of $2480=$24, and 106$
100% =100 times $24=$2400=present worth.
$2480— $2400=$80=true discount.
100%=$2480.
1 %=Tfo of $2480=$24.80, [count for cash.
5%=5 times $24.80=$124=trade discount, or dis-
.'. $124 — $80=$44=his gain by paying cash.
would gain $44 by paying cash.
(/?. 3d p., p. 258,prob. 10.)
1.
Remark.— It is clear that $2480— $124,=$2356 would pay for the goods cash. If the merchant had this sum of money on hand, it would, in 4 months, at 10%, produce $78.53-j- interest. But if he pays his debt he will make $124. Hence he will gain $124 — $78.534=$45.46f.
III. BANK DISCOUNT.
1. J$anh 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.
CASE I.
( Face of note, ) Given < Rate, and > to find the discount and proceeds.
f Time. S T^ i { Z?=^X^X»
v ' Formulae, j P==:j7_jr)t
108 FINKEL'S SOLUTION BOOK.
I. What is the bank discount of $770 for 90 days, at 6% ? By formula,
=$770 X -06 X ±^=$1 1 .935.
By 100% method.
II. 100%=$770, 2. l%=T-^of$770=$7.70, and 3. 6%=6 times $7.70=$46.20=discount for 1 year. 4. $11.935=^ of $46.20=discount for 93 days.
III. /. $11.935=bank discount on $770 for 90 days at 6
CASE II. ds, ) and > to fi
( Rate»
C Proceeds, ) Given < Time, and > to find the face of the note.
TT 77
Formula, F=- - . 1— rn
I. For what sum must a note be made, so that when dis- counted at a bank, for 90 days, at 6% the proceeds will be $393.80? Bv formula,
P $393.80
By 100% method.
1. 100%=face of the note.
2. 6%=discount for one year.
3. 1^%=^ of 6%=discount for 93 days.
4. 100%— lH%=98<nr%=proceeds.
5. $393.80=proceeds.
7. 1%=—— of $393.80=$4, and
II.
5. 100% =100 times $4=$400=face of the note. III. /. $400=face of the note.
CASE III.
Given rate of bank discount, to find the corresponding rate of
interest. Formula, rate of /.== .
1 — rn
I. What is the rate of interest when a 60 day note is dis- counted at 8% per annum? By formula,
of 7==
ANNUAL INTEREST. 109
II.
By 100% method. 4. 100%=face of note.
2. 8%=discount for 1 year.
3. 12^= e^. of 8%=discount for 63 days.
4. 100%— If %=98f %—proceeds.
5. 98|%=100% of itself.
6. 1%=—- of 100%— Jff %, and.
~f, o%=8 times £$£%=81yg-%=rate of interest III. .-. The rate of interest on a 60 day note discounted at 8% per annum=8^%%.
CASE IV.
Given the rate of interest, to find the corresponding rate of
,, . rate of /.
discount Formula,
I. What is the rate of discount on a 60 day note which yields 10% interest?
By formula,
By 100% method. . 100%=proceeds.
2. 10%=interest on proceeds for 1 year.
3. 11%=^ Of 10%=interest on proceeds for 63 days.
4. 100%+l|%=101|%=face of note.
5. 101|%=100% of itself.
II.
7. 10%=1 III. " .-. The rate of discount—
Note. — It must be borne in mind that the interest on the pro ceeds is equal to the discount on the face of the note.
IV. ANNUAL INTEREST.
1. A.in/nAJWJ& Interest is the simple interest of the princi- pal and each year's interest from the time of its accruing until settlement.
I. No interest having been paid, find the amount due Sept. 7, 1877, on a note of $500, dated June 1, 1875, with interest at 6%, payable semi -annually.
no
FINKEL'S SOLUTION BOOK.
2 yr. 3 mon.
1 yr. 9 mon. 6 d.
1 yr. 3 mon. 6 d.
9 mon. 6 d.
1st
2d
3m.6d
3d
4th
6d
II.
(6.)
(1.) 100%=$500, 1877—9—7
(2.) 1%=T^ 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=J of $30=semi-annual interest.
1. 100%=$15, '
2. l%=Tfoof$15=$.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.) V. $500+$68+$3.885=$571.885=amount of the note. III. V. $571.885=amount of the note.
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- ment, 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 period ; also one between 1 and 2. By such diagrams, the time for which to compute interest on the simple 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 5 years, with gold at 105?
ANNUAL INTEREST.
Ill
Syr.
$200 for 4 yr. 9 mon. at 6^,
$200 for 4 yr. 6 mon. at 656.
$200 for 4 yr. 3 mon. at 6$.
2d 3d I 4th
&c.
10 11 12
16 17 1
II.
(i-
(2.
{I:
(v.
$1000=par value of one bond.
$20000=par value of 20 bonds.
100%=$20000,
1%=T^ of $20000=$200, and
4%=4 times $200=$800=income for one year.
$4000=5 times $800=income for five years.
$200=i of $800=interest due at the end of first quarter, and which draws interest to time of set- tlement.
100%=$200,
1%=-^ of $200=$2, and [est for one year.
6%=6 times $2=$12=interest on quarterly inter-
$570=47-^ times $12=interest on quarterly inter- est for the sum of 4f yr.-f-4-^- yr.-f-4^ yr.-j-
-|- i yr. , or 47-J years.
/. $4000+$570=$4570=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.
'1.
2.
(8.).
3. 4.
(9.)
(10.)
1(11.)
Explanation. — It must be borne in mind that the quarterly in- terest, $200, is put on interest at 6% as soo'n 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 4| yr.-|-4J yr. yr.+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' were paid ?
interest
112
FINKEL'S SOLUTION BOOK.
$96 for 9 yr. 6 m. 15 d. at 85*.
$96 for 6 yr. 6 m. 15 d.
4yr. 6m. 15 d.
$96, 2yr. 6m. 15 d.
1 yr.6 m. 15 d
15 d
II.,
(8.)
(9.) 1(10)
100%=$1200.
1%=^ of $1200=$12, and
8%=8 times $12=$96=simple int. for one year $480=5 X$96=five simple interests. =J of $96=interest for 6 months.
$48=interest for 15 days. .*. $532=simple interest unpaid.
1. 100%=$96.
2. l%=Ti? of $96=$.96, and [simple interest.
3. 8%=8 times $.96=$7.6S=interest on one year's
4. $193.92=25i times $7.68=interest on year's sim-
ple interest for 9 yr. 6 mon. 15 da.,+6 yr. 6 mon. 15 da.,+4 yr. 6 mon. 15 da.,-f-2yr. 6 mon. 15 da., +1 yr. 6 mon. 15 da., -(-6 mon. 15 da., or 25 yr. 3 mon.
.'. $532+$193.92=$725.92=amount of interest due. [1, 1892.
$1200+725.92 = $1925.92 = amount due July
III. .-. $1925.92=whole amount due Aug. 1,1892. V. COMPOUND INTEREST.
1. Compound Interest is interest on a principal formed by adding interest to a former principal.
Let />=principal on compound interest.
./?:=( l-j-/')=amount of one dollar for 1 year. P (l+r)=/)7?=amount of P dollars for 1 year. P (l-|-r)2=/>^?2=amount of P dollars for 2 years. P (l+/)3==^?3=amount of P dollars for 3 years. P (l-[-r)n=f>l?n=amount of P dollars for n years. Let ^4=amount of P dollars in n years, and
/—the compound interest of P dollars for » years. Then I==P^—P ..... I. . II.
.III.
COMPOUND INTEREST. 113
n
.*. JR=\/T IV. Applying logarithms to
p
n log. /?=log. A — log. P ', whence _log. A— log. P
log./?
When compound interest is payable semi-annually. P (1-f-f )=amount of P dollars for 4- year. p (!-(-!-) 2=amount of P dollars for"l year. P (l-j-f)2n=amount of P dollars for n years. .-. A=P (l-f-J)2n, when payable semi-annually. When compound interest is payable quarterly, P (1-f-f )=amount of P dollars for £ year. P (1-j-f )2=amount of P dollars for -J- year. P (1-j-f )3=amount of P dollars for £ year. P (1-j-f )4=amount of P dollars for 1 year. P (1-j-f )4n— amount of P dollars for n years.
When the interest is payable monthly, A-£(1+A)»««.
When the interest is payable q times a year, A=P (l+|)qn-
CASE I.
( Principal, } Given < Rate, and > to find the compound interest and amount.
( Time, ) ( I=PR*—P,
Formula,,
I. Find the compound interest and amount of $500 for 3
years at 6%. By formulas,
^=/>7?n=:$500X(l+.06)3=$595.508, and I=PR*— />=$500 X ( 1+.06) 3— $500=$95.508. Remark. — In compound interest, the 100% method becomes very tedious.
By 100% method. ' (1.) 100%=$500,
(3.) 6%=6 times $5= $30=interest for 1 year. (4.) $500+$30-=$530=arnount, or principal for the second year.
.. 100%=$530,
2. 1 %=r^r of $530=45.30, [year.
rIL^
(5.)
3. 6%=6 times $5.30=$31.80=interest for second
4. $530+$31.80=$561.80=amount, or principal for
the third year.
114
FINKEL'S SOLUTION BOOK.
III.
1. 100%=$561.80,
2. 1 %=Tfor of $561.80=45.618, and [year. (6.) 3. 6%=6 times $5.618— $33.70S=interest for third
4. 561.80+$33.7.08=$595.508=amount at end of the
third year. (7.)X $595.508— $500=$95.508=compound interest.
( $95.508=compound interest, and '"' ( $595.508=compound amount.
CASE II.
C Principal, Given < Rate, and
( Compound Interest,
to find the time. Formula, n=
log. A— log. P
log./?
I. In what time will $8000 amount to $12000, at 6% com- pound interest? By formula,
log. A— log. JP=log. 12000— log. 8000^ log. R log. 1.06
4.079181—3.903090 _ 1K ....
=6 yr. 11 mon. 15 da. We may solve the
problem thus: $8000( 1.06 )n=$ 12000, whence (1.06)n=12000-j- 8000=1.50. Referring to a table of compound 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 difference 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 g%tiif of the period or, 11 mon. 15 da. /. The whole time=6 yr. 11 mon. 15 da.
CASE III.
( Principal, ^
Given < Compound Intesest or Amount, and > to find the rate. ( Time, )
Formula, r=n l^ 1.
I. At what rate, by compound interest, will $1000 amount to $1593.85 in 8 years? By formula,
ANNUITIES. 115
CASE IV.
C Compound Interest or Amount }
Given ] Time, and V to find the principal
( Rate, \
I. What principal, at compound interest will amount to 27062.85 in 7 years at 4% ? By formula,
CHAPTER XIV.
ANNUITIES.
I. An, Annuity is a sum of money payable at yearly, or other regular intervals.
!1. Perpetual, or I: SSKor 4. Contingent.
3. A Perpetual Annuity is one that continues forever.
4. A Limited Annuity ceases at a certain time.
5. A Certain Annuity begins and ends at fixed times.
6. A Contingent Annuity begins or ends with the happening of a contingent event.
7. An Immediate 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 Vallie of an annuity is the value of a
deferred annuity at the time it commences.
116
FINKEL'S SOLUTION BOOK.
to find the initial value of a perpetuity.
CASE I. C Annuity, ) Given < Time, and
( Rate,
I. What is the initial value of a perpetual annuity of $300 a year, allowing interest at 6% ?
1. iOO%=initial value.
2. 6%=interest for 1 year.
3. $300=interest for 1 year.
4. .-. 6%=$300.
5. \°/c=\ of $300=$50, and
§. 100%=100 times $50=$5000=initial value. III. ^ .-. Initial value=$5000. (R* H. A., p. 310, prob. 1.)
I. 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 5=the annuity. Then 6'=the amount due in
3 months.
2. 6'-)-5'(l-|-^)=namount due in 6 months.
3. /. v4 = 5+6'(l-f-.01i-)=:$625 + $625(1.01|) =
$1259.37i=amount due at the end of 6 months.
4. 100%— initial value.
5. 3%=semi-annual annuity.
6. $1259.37^=semi-annual annuity.
HJ
B.<
8. l%=Jof $1259.37i=$419.7916f , and
9. 100%=100 times $419.7916f=initial value.
1. Let .Sr^amoiint due in 3 months. Then
2. •S-|-5l(l-f-j)==ainount due in 6 months, [and
3. 614-61(l4-T)+^(l+|r)=amount due in 9 months,
4. S+S(l+-J)+S(l+}r)+S(l+Jr )== amount due in 1 year. [(1+'V)^2556.25.
5. .-. ^=$625+$625( l+-\6)+ $625(1+' V2)+ $625-
6. 100%=5nitial value.
7. 6%=annuitv.
8. $2556.25=annuity.
9. .-. Q%= $2556.25.
10. 1%=4 of $2556.25=$426.0416|, and [value.
11. 100%=100 times $426.0416f =$42604. 16f=initial
1. 100%==initial value.
2. l-^%=quarterly annuity.
3. $625=quarterly annuity.
5. 1^= of $625-4416.6666|, and
6. 100%=100 times $416.6666|=$41666.
ANNUITIES.
Initial value of A=$41979.16f, Initial value of B=:$42604.16f , Initial value of C=$41666.66f .
(R.
117
CASE II.
Given
Annuity, Interval, Rate, and
and
H. A., p.S10,prob.5.)
to find the present value of a deferr-
Time the perpetuity is deferred, J ed perpetuity.
Let 6*=the annuity, f— the rate, and R=l-\-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
au ,. , — rr= nu D -^-r- in which / is the time the perpetuity r (\-\-rJ- R'(R — 1)
is deferred.
I. Find the present value of a perpetuity of $250 a year, de- ferred 8 years, allowing 6% interest. By formula,
$250 _ $250
~R*(R— 1) (l-f-.06)8(l+.06— 1)~~.06(1.06)8~~ By 100% method. (1.) 100%=initial value. 2.) 6%=annuity. 3.) $250=annuity. 4.) /. 6%=$250. 5.) l%=i of $250=$41|, and !!.<! (6.) 100%=100 times $41f=$4166.66£=mitial value.
1. 100%=present value of $4166.66| due in 8 years
at 6%.
2. 159.38481 %==( 1.06) 8XlOO%=compound amount
of the present value for 8 yr. at 6%.
3. /. 159.38481 %=$4166.66f,
4. l^^i-^Wr of $4166.66f=$26. 1422, and
III.
I.
5. 100%=100 times $26.1422 = $2614.22 = present
value.
/. The present value of a perpetuity of $250 a year He- ferred 8 years at 6% interest=$2614.22.
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,
5 $325 $325
"[ $2
.0816(1.04)10
690.67.
118 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,
(6.) 1%=^ of $325— $39.828431, and [value.
(7.) 100%— 100 times $39.828431= $3982.8431 = initial
1. 100%— present value of $3982.8431.
2. 148.024428f*:=(1.04)10X100%==compound amount
of 100% for 5 yr. at 8%.
II.
1(8.)
3. .-.148.024428%— $3982.8431,
4. l%=T48.<ji44?8°f $3982.8431— $26.9067, and
5. 100%— 100 times $26.9067 — $2690.67 — present
value. III. .'. $2690.67— present value of the estate.
(JR. H. A., p. 311,prob. 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 $3982.8431 if the payments are to begin at once, but $3982.8431-r- (1.04)10 if the payments are not to begin until the end of 5 years.
CASE III.
Given -< £.nnul,ty> , Uo find the present valve of an an-
I lime to run, and I ., ~. .
T . T nuity certain.
(^Interval, )
(a) Let P denote the present value. The amount of P for n
Let 6*— the payment, or amount due the first year. = the amount due the second year.
- the amount due the third year.
S'-|-S1/?--»S7?2+S7?8==the amount due the fourth year. [due the nth year.
S+ Sfi+Sfi'*+ Sfi*+ ....... + 67?"-1 — amount
.-. A=
...(1) AR= SR
(2), by multiplying (1) by R. AR—A=SR*—S. . . (3), by subtracting (1) from (2).
1) • • • • 4. But PR*=A.
ANNUITIES. 119
•'" (5')> whence
-
-
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 q= the number of years the annuity continues. Then
, S ~Rp+q 1
P=-= — - X T?p+q — =the present value of an annuity S, for
the time (p-\-y) years, and
C Dp 1
P"=— — -X — D^— =the present value of an annuity S, for p
R P+V ) R — 1 R P+V
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,
S R«-\ $250 (1.05)so-l ,0040110* ^7?=rX~7F~ r05~X (1.05)30 : By 100% method. (1.) 100%— initial value. (2.) 5%— annuity. (3.) $250— annuity. (4.) .-. 5%— $250, (5.) l%-=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 deferred 30 years. [ent value for 30 years.
2. 432.19424%— (1.05)30XlOO%— amount of pres-
3. .-. 432.19424%— $5000,
4. l%=¥T7^Trre of $5000— $11.568865, and
5. 100%=100 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. .!II. V. $3843.1135— present value of an annuity of $250, payable annually for 30 years,
i20
FINKEL'S SOLUTION BOOK.
Remark. — Since $5000 is the initial value which, 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 $250 continuing for 30 years
I.
II.
Find the present value of an annuity of $826.50, to com- mence in 3 years and run 13 years, 9 months, interest 6%, payable semi-annually.
Given £=$826.50, r=.06, /=3 years, and ^=13| years. When interest is payable semi-annually, _/?=(l-[-|-)2. By formula (7),
$826.50.. a0609)^1=$6324m
' .0609 ' (1.0609)16^
(8.)
(9.)
(10.)
III.
By 100% method. (1.) 100%=initial value. (2.) 3%=amount due in 6 months. (3.) 3%+3% (1.03)=6.09%=amount due in 1 year. (4.) $826.5Q=amount due in 1 year. (5.) .-. 6.09%=$826.50, (6.) l%=Tfo of $826.50=$135.712643, and (7.) 100% = 100 times $135.712643 = $13571.2643= initial value
1. 100%=present value of a perpetuity of $826.50
deferred 3 years.
2. 119.40523 %=( 1.0609 )2 times 100%=amount of
present value for 3 years.
3. .-. 11940523%=$13571.2643,
4. l%=mr.ArB7Tr°f $13571.2643=$113.6686,
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)16^ times 100% = amount
of present value for 16f years
3. /. 269.212027%=$13571.2643, .
4. 1%=^^^ of $13571.2643=150.4117,
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 annuity of $826.50 deferred 3 years and continuing 13| years.
$6324.69=present value of $826.50, etc.
If the
ANNUITIES.
121
If the annuity is to begin in p years and continue forever, the formula,
S
if ^=00, the
For, since P= quantity
=1 =1 — 0, approaches 1 as its limit,
CO
and we have ^
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.
8 $1000
II,
By 100% method. (1.) 1 00 %=initial value. (2.) 4%=annuity. .) $1000=annuity.
3
(4-)
(5.) l%=i of $1000=$250, and [$1000.
(6.) 100%=100 times $250=$25000=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,
4. l%=nnriimrof $25000=$222.2492, and
5. 100%=100 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%.
f Annuity, J Rate, \ Interval, and
CASE IV. -to find the final or forborne value.
Given
'
(^Tirne to run, Let $=amount due first year.
$_|_l$fjff=amount due second year.
nt due third year.
=:amount due the fourth year.
2+ &ff3 + +o'7?n-1= amount due
the nth year. Let ^4=amount due the ^th year.
-1 ... (1).
122 FINKEL'S SOLUTION BOOK.
. . (2), by multiplying (1) by R. [from (2).
AR— A=Sfi"— S ...... (3), by subtracting (1)
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,
— 1] = $11064.8139.
II.
By 100% method.
1. 100%=initial value.
2. 10%=annuity.
3. $25=annuity.
4. .-. 10%=$25,
5. 1%=TV of$25=$2.50, and
6. 100%=109 times $2.50=$250=initial value.
7. $44.2592556=[(1.10)40—1]X$1— compound interest of
$1 for 40 yr. at 10%. [$250 for 40 yr. at 10%.
I 8. .'. $11064.8139=44.2592556 X$250=compound int. of III. /. He would be $11064.8139 better off.
Remark. — $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.
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 6% ? (1.) 100%=initial value, or the principal that would
produce the annuity. (2.) 6%=annuity for 1 year. (3.) 84%=14x6%=annuity for 14 years.
1. 100%=6%,
2. 1%=T^ of 6%=-^Q-%, and [1 year. (4.) 3. 6%=6 times -^r%=2T%=interest on annuity for
4. 32.76 %=91 times 2T%=interest on annuity for
(1+2+3+ +14), or 91 years.
(5.) 84%+32.76%=116.76%=whole amount of the n.< annuity.
(6.) $116.76=whole amount of the annuity.
(7.) /. 116.76%=$116.76,
(8.) 1%=^^ of $116,76=$!, and
(9.) 100%=100 times $l=$100=initial value.
(10.) 6%=6 times $l=$6=annuity.
ANNUITIES.
122
(11.) $1.260904— [(1.06) I*— l]x$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. tit 6%.
(13.) .-. $126.0904— $116.76— $9.3304— difference. III. .-. The difference=|9.8304.
CASE V. Final Value or Present Value
Given < Rate, and
( Time to run,
c<
Solving ^—
r>n _ -J
to find the annuity.
R— 1
with respect to 6* and we have T (1). If ^l=the final or
forborne value, by the formula in the last case, we have A- ~™ — 1. Solving this with respect to 6", we have.
S=-
(2).
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 o— Ou -=— — — = $4ob.o&
By 100% method.
(1.) 100%— annuity.
(2.) 7%— annuity.
(3.) .-. 7%— 100%,
(4.) l%=y of 100%— 14f%, and
(5.) 100%— 100 times 14f%— 1428f %— initial value.
1. 100%— present value of 1428^% due in 17 years.
2. 315.8815%— amount of present value for 17 years.
4. i%=-jfTT.foTT of 142T8|%=4.522591%, and
5. 100%— 100 times 4.522591%— 452.2591%— pres- II. < ent value.
(7.) .'. 1428^% —452.2591% — 976.3223% — present
value of an annuity running 17 years. (8.) 3.1588152%— (1.07)17 times l%=amount of 1%
for 17 years. (9.) 3084.0217%— (1.07) 17 times 976.3223 %=amount
of 976.3223% for 17 years at 7%. (10. $15000— amount, or final value.
(11. .-• 3084.0217%— $15000. (12. l%=inrreWTT of $15000— $4.8638, and *( 13. 100%=100 X $4.8638— $486.38— annuity.
124 FINKEL'S SOLUTION BOOK.
III. .-. I must pay $486.38.
CASE VI.
( Annuity, ) , ~ , ,. ..
s^. i T» i TT i c J.T- A •*. j i ^o nna time it
Given < Present Value of the Annuity, and }
) T-» * \ runs.
( Rate, )
In formula (6), Case III., we have P=— — -x — ^ — , whence
Pr_S—Pr
S~ ~~S~*
.•.^?n=— — ^— (1). Applying logarithms,
S \ log. 5— log. (S—Pr)
-- - ~
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), ^Aog. s— log. ( S— Pr )==\og. 80000— log. ( 80000—1000000 X .06 )
log. R log. 1.06
log. 80000— log. 20000^4.903090— 4.301030_.6Q2060:= log. 1.06 .025306 ~ 025306"
years.
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 1000000-f-80000, will be found to be between 23 and 24 years.
Note. — A table of present worths may be computed by form- ula (6.), Case III., in which put 6'=$1.
I. In what time will a debt of $10000, drawing interest at 6%, be paid by installments of $1000 a year. By formula,
log. S— log. (S— 7V)_log. 1000— log. ( 1000—10000 X. 06) log. R log. 1.06
3—2.602060 1KI70- 1K
-=15.725 years=15 yr. 8 mo. 21 da.
.025306
By another method.
Assume $10000 to be the present value of an annuity of $1000 a year. Then $10000-r-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 -f-
ANNUITIES. 125
The compound amount of $10000 for 15 yr. at Q%= $23965.58 The final value of $1000 for 15 years 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 \\\e fraction of a year is of $1000.
6% of $689.61 for fae fraction of a year=$41.3766X fraction of a year.
.-. $689.61+$41.3766X fraction of a year=the amount of $689.61 for the fraction of a year. This amount divided by $1000, a yearly payment, will give infraction. $689.61+141.3766 ^fraction
-$1655~ =/'**>* whence
$689.61+141.3766 X fraction=$\(yM X fraction
.«. $1000 yJraction—$±\. 3766 X/™c#o«==$689.61, or
689.61 0
=& months, 19 days.
.•. The whole time=15 yr. 8 mon. 19 da.
CASE VII.
C Annuity, . }
Given < Time to Run, and > to find the rate of in-
( Present Value of an Annuity, ) terest.
From the formula (6), Case III, £*=-& — ?X pn » we ODtain
=-^r .... (1). This is the simplest expression we can ob- tain for the rate as the equation is of the ^th 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,
^n— 1_^__$650 _ or
————^-—=8.125. Solving this equation by the method of
Double Position, we find r=S%-\--
By another method.
$65Or-$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
126 FINKEL'S SOLUTION BOOK.
PROBLEMS.
1. What is the amount of an annuity of $1000, forborne 15 years, at 3£% 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% ? Ans. -
4. What is the present worth of an annuity of $200, for 7 years, at 5% ? Ans. $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. -
8. A perpetual annuity of $1000 beginning at the end of 10 years, is to be purchased. If interest is reckoned at 3-J%, 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- mence 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 for 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.
127
CHAPTER XV.
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. 16f %— loss.
3. 100%— 16|%— 83^%— selling price.
4. $25— selling price.
5. .-. 83-J%- $25,
A.
B.
=- of $25— $.30, and
in.
6.
7. 100%— 100 times $.30— $30=cost of first cow.
8. $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.
5. 1%— TV of $5— $.3125, and [cow.
6. 100%— 100 times $.3125— $31.25— cost of second ( $30=cost of first cow, and
) $31.25=cost of second cow.
Remark. — Since I lost $5 on the first cow, and neither gained »**r lost on the two, I must have gained $5 on the second cow. .• 16%— $5.
I. There have been two equal annual payments on a 6% note of $175, given 2 years ago this day0 The balance is L40 ; what was each payment?
II.
(1.) 100 %=a payment. (2.) 100%— $175, (3.) l%=ri(5- of $175=$1.75, and (4.) 6%= 6 times $1.75=$10.50=interest for 1 year. (5.) $175+$10.50= $185.50 — amount before paying the payment. [payment.
(6.) $185.50 — 100% —amount left after paying the
1. 100%— $185.50— 100%,
2. 1%— Tio of ($185.50— 100%)— $1.855— 1%, and
3. 6%— 6 times ($1.855— 6%)— $11.13— 6%— inter- (7.)J 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.
128
FINKEL'S SOLUTION BOOK.
(8.
(9. (10.) (11.) (12.)
$154.40=amount after paying the last payment
.-. $154.40=$196.63— 206%.
206%— $196.63— $154.40— $42.23,
1%— ^ of $42.23=$.205, and
100%— 100 times $.205— $20.50=the payment.
III. .*. |20.50=the payment.
Remark. — In this solution we are obliged to use the minus sign, — , which is no obstacle to the student of algebra, but to the student of arithmetic it may seem insurmountable. To avoid this sign, we give another solution.
II.
III.
(I-')
(2.)
(3.) (4-) (5.) (6.)
(7.) (8.)
(9.)
(10,)
(11.) (12.) (13.) (14.) (15.)
100%— the payment. Then
$154.40+100%— 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%-Ti¥ of ($154.40+100% )-$1.4566^+M%, 100%— 100 times ($1.4566^+11% ) = $145.66^
-|-94i|%— amount at end of the first year after
paying off the payment. $145.66^+94/3 %+100%— $145.66^ + 194|f %
—amount before paying oft' the payment —
amount at end of first year. 100%— the principal that produced it. 6%=interest. 106%— amount.
l-83^fr%, and 100% = 100 times ($1.3?Hw +
$137JtfH-183^/V%=tibe amount at first. $175=the amount at first.
and 100%=100 times $.205— $20.50=the payment.
.'. $20.50— the payment.
(R. H. A., p. 26^ prob. 5.)
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 f of an article for what J of it cost, what is his gain % ?
MISCELLANEOUS PROBLEMS.
129
1. 100%=cost of whole article.
2. 87i%=J of 100%=cost of £ of the article.
3. 87^-%=selling price of f of the article.
4. 29i%=£ of 87|%=selling price of £ of the article.
5. 116f%=4 times 29^%= selling price of the whole
article.
6. .'. 116|%— 100%=16f%=gain.
.-. 16|%=his gain. (Milne's Prac., p. 360,prob. 51.)
A merchant sold goods to a certain amount, on. a commis- sion of 4%, and having remitted the net proceeds to the owner, received ^% for. prompt payment, which amounted to $15.60. What was his commission?
100%=cost of goods.
4%=commission.
100%— 4%=96%=net proceeds.
;!%— amount received for prompt payment.
2. $15.60=amount received for prompt payment.
3. .-. i%=$15.60.
4. 1%=4 times $15.60=$62.40.
5. 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%=^ of $6500=$65, and
4%=4 times $65=$260=his commission.
His commission=$260.
( Greenleafs N. A., p. J^l.prob. 11.)
(2.) (3.)
(4.)
(5.) (6.) (7-)
(8.)
If I sell 30 yards of cloth for $132, and gain 10%, how ought I to sell it a yard to lose 25% ?
$132=selling price of 30 yards. $4.40=$132-^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. 1 %=TH of $4.40=$.04, 100%=100 times $.04=$4=cost per yard. I. 100%=$4.
HI.
(1.)
(2.)
( 3. )
(4.)
(5.1
(6 )
(7.
(8.
(9.)
3. 25% =25 times $.04=$l=loss.
4. ... $4 — ^i—^^selling price per yard to lose 25%.
.•. I must sell it at $3 per yard to lose 25%.
(StoddarcCs Complete, p. 206, prot>. 9.)
130 FINKEL'S SOLUTION BOOK.
I. A merchant receives on commission three kinds of flour ; from A he receives 20 barrels, from B 25 barrels, and from C 40 barrels. He finds that 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?
(1.) $6=selling price of 1 barrel.
2.) $510=sellmg price of (20+25+40), or 85 barrels.
3.) 100%—value of C's flour per barrel.
(4.) 120%—value of B's flour per barrel.
(1. 100%=120%.
(5.W2. 1%=^ ofl20%=li%,
13. 10%— 10 times 1-J %=12%.
(6. ) 120% +12 %— 132 %=valtie of A's flour per barrel.
II.
(7.) 4000%=40 times 100%— what C received.
i!
(8.) 3000%— 25 times 120%— what B received.
2640%— 20 times 132%— what A received. 9640%— 4000%+3000%+2640%— what all rec'd. $5 10= what all received. 9640^=4510.
(9.
(10. (11. (12.
(13.) 1%=^ of $510— $.52|-J|, and [received.
(14.) 4000%— 4000 times $.52fii — $21H||=what C
(15.) 3000%— 3000 times $. 52-Jii — $158^1— what B
received. [received.
(16.) 2640%— 2640 times $.52fi| — $139iJ|— what A
v Tic»i/2"^y— A's share, ill. .'. I $158.Hf— B's share, and :C's share.
( Greentcafs National Aritk. p. 442.)
I. f of B's money equals A's money. What % is A's money less than B's, and what % is B's money more than A's?
1. 100%— B's money.
2. 75%— | of 100%— A's money.
TT j 3. 100%— 75%— 25%=excess of B's money over A's. 11X4. 75%— 100% of itself,
5. 1%— TV of 100%— H%, and [than A's.
6. 25%— 25 times l£%=33i%:=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.
(Stod. Comp.,p. 203,prob. 19.)
I. At what price must an article which cost 30 cents be marked, to allow a discount of 12^% and yield a net profit of 16f% ?
MISCELLANEOUS PROBLEMS.
131
II.
1.
2. 3.
(4.
100%=30/,
16f%=16£ times ^Xxm5/=profit.
30/+5/=35/=the price at which it must sell to
III.
I.
1. 100%=marked price.
2. 12^%=discount from marked price.
3. 100%— 12|%=87i%=selling price.
4. 35/=selling price. (5.n 5. .-. 87|%=35/.
6. l%=^~ of 35/=.40/, and
7. 100%=100 times .40/=40/=marked price. /. 40^=marked price.
(Seymour's Prac., p. 203; prob. 4.)
Had an article cost 10% less, the number of % gain would have been 15% more ; what was the gain?
1. .7#6>%=selling price.
2. 100%— actual cost price.
3. 100%— 100%=gain.
4. 100% — 10%=90%=supposed cost.
5. 100% — 90%=conditional gain.
6. 90%=100% of itself.
[difference.
8. 100%— 90%=(
=conditional gain %.
9. /. V X100%— 100%— (100% —
10. 15%=difference.
11. /. ix^#%=15%. [the actual cost.
12. 100%=9 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. Let 5=:selling price and C=the cost. Then 5 — C=gain and
C* S~* C1
• — -~r- =rate of gain. S — T9^C=conditional gain and
^ 10 $ Q $ Q
-=conditional rate of gain. .•. — — ^___^.^ or
.'. 1.35C—C=.35C=gain.
III.
whence 5=fJ C=1.35 C. .-. Rate of gain=.35C-7-C==.35=35%.
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% more for material, my house would have cost $3052. What did it cost me?
132 FINKEL'S SOLUTION BOOK.
(1.) 100%=cost of labor.
(2.) 300%— 3 times 100%=cost of material.
rl. 100%=100%, i
,„ J2. !%=!%, and
<6-')3. 6%=6%.
U. 100%+6%=106%=supposed cost of labor.
II. 100% =300%,
2. l%=Ti-o- of300%=3%, and
3. 10%=10 times 3%=30%.
4. 300%+30%=330%=supposed cost of material.
(5.) 330%+106%==436%=:SUPPosed cost of house.
(6.) $3052=supposed cost of house.
(7.) .-.436% =$3052,
(8.) l%=^k of $3052=$7, and
(9.) 100%=100 times $7=$700=cost of labor.
(10.) 300%=300 times $7=$2100=cost of material.
( 11.) $2100+$700=$2800=cost of house.
III. .-. $2800=cost of the house.
I. I invest |- as much in 8% canal stock at 104%, as in 6% gas stock at 117% ; if my income from both is $1200, how much did I pay for each, and what was the income from each ?
(1.) 100%=investment in gas stock. Then
(2.) 66f %=investment in canal stock.
1. 100%=par value of the gas stock.
2. 117%=market value of the gas stock.
3. .-. 117%=100%,from (1),
4. 1%=^-^ of 100%=-fr5-% > and
5. 100%=100 times }fo%=85||%=par value in
terms of the investment. ;i. I00%=85|f%, (5.)<^2. l%=ff-?-%, and
U. 6%=6 times |^%=5A%=income of gas stock,
II. 100%=par value of canal stock. 2. 104%=market value. 3. /. 104%=66|%, 4. l%=T^of66|%=||%,and 100%=100 times
(3.)
(6.K2. l%=zT^ of 64^%=|f%, and
13. 8%=8 times ff %=5^%==mcome of canal stock. (7.) 5^g-%-}-5^=10^%=income from both. (8.) $1200=income from both. (9.) ,.
(10.) 1%=-—- of $1200=$117, and
MISCELLANEOUS PROBLEMS. 133
(11.) 100%=100 times $117=$11700=investment in
gas stock. [canal stock.
(12.) 66f%=66f times $117=$7800 = investment in
(13.) 5A%=5A times $117=$600=income from each.
$600=income from each.
JJf .'. $11700=investment in gas stock, and $7800=investment in canal stock.
I, A man bought two horses for $300; he sold them for $250 apiece. The gain on one was 5% more than on the . other; what was the gain on each?
1. $300=cost of both.
2. $500=$250+$250=selling price of both.
3. $500— $300=$200=gain on both.
4. 100%=gain on first horse. Then
5. 105%=gain on second horse.
n<| 6. 205%=100%+105%=gain on both.
7. $200=gain on both.
8. .-. 205%=$200.
9. l%=?fa of $200=$Jf, and
10. 100%=100 times $|f=$97.56¥4T=gain on the first.
11. 105%=105 times $££=$102.43f£=gain on the second.
Ill • $ $97.56^T=gain on the first, and ' ( $102.43f|=gain on the second.
Note. — In this solution, it is assumed that the gain on one was 5% 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. By this assumption, the problem is algebraic. The following is the solution: Let #=the cost of the first horse, and $300 — #, the cost of the second. Then $250 — #— gain on first, and $250 — ($300— x)=x— $50, the gain on the second. ($250— x) -t-x= rate of gain on the first, and (x — $50)-f-($300— x), the rate of gain on the second. Then (250 — x)-$-x — (x — 50)-r-(300— x)= j1^. Whence, by clearing of fractions, transposing and, combin- ing, *2__ 10300 *=—i500000, ^=5150^50^10009= $147.7755, the cost of the first horse. $300— *=$152.2245, the cost of the second horse. $250 — #=$102.2245, gain on the first horse, and * — $50=$97.7755, the gain on the second horse.
I. An agent sells produce 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 ?
134
FINKEL'S SOLUTION BOOK.
II.
III. I.
II
(1.) 100%=value of the produce.
(2.) 2 %— the commission. [vested in the flour,
(3.) 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.)
(5.) (6.) (7.) (8.) (9.) (10.)
(11.) (12.)
.-. 103%=, l%=rta of 98%=!%%, and lOO%==100XT9-o8*% = 95^% = cost of flour in terms of the value of the produce. — 95TV\%=2T8o%%=commission on flour.
%==whole commission. «$75=whole commission.
l%^$75-MT8oV= $15.45, and [produce.
100%=100 times $15.45=$1545 = value of the 95TV%%==95TVV times $15.45= $1470 = value of
the the flour. $5=costofl barrel. $1470=cost of 1470-=-5, 01 294 barrels.
\ 2.
(3.) (4.)
.*. 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?
(1.) 100%=value of the whisky. >.) 4%=loss.
100% — 4%=96%=amount he had left. 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. 1%=^ of (96%— $18)=|%— $.16|, and
6. 100%=100 times (f% — $.16f)= 88f%— $16.66f
=cost of rye.
7. 8%=8 times (|%— $.16f)=7i%— $1.33i=com-
mission on rve.
— $1.33^-)=! H% — $1.33|=wholeloss. 2=\vhole loss. H^%_$1.33^=:$32 *o= $33.33i,
of $33.33ih=$3, and
(5.)
(6.)
(7.) (8.) (9.)
(10.) l(ll-)
100%-=100 times $3=$300=value of the whisky.
III. .-. $300=value of the whisky.
(Jt. H. A., p. 4Q6, prob 91.)
MISCELLANEOUS PROBLEMS. 135
I. What will be the cost in New Orleans of a draft on New York, payable 60 days after sight, for $5000, exchange being at \\°/o premium?
1. 100%=face of the draft.
2. l|%=premium.
3. 100%+1-J%= 101i%=rate of exchange.
4. 5%=discount for one year.
II.
of 5%=discount for 63 days.
/. 101-4-%— f %=100|%=cost of the draft
7. 100% =$5000.
8. 1%=^ of $5000=$50, and
lOOf %=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 days, the banker in New Orleans, who has the use of the money for that time allows 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 of 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 New Orleans $13,000,000 and New Orleans owes New York $10,250,- 000, the "balance of trade," $2,750,000, is against New York and in favor 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 New 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 money 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.
136
FTNKEL'S SOLUTION BOOK.
I.
II.
III. I.
!!.
What will a 30 days' draft on New Orleans for $7216.85 cost, at •§% discount, interest 6% ?
1. 100 %=face of draft.
2. -f%=discount.
3. 100%— f %=99f %=face less the discount.
4. 6%=bank discount for 1 year.
5. %v%~'ffir °f 6%=bank discount for 33 days.
6. 99f % — -i^%=99¥3^%=cost of the draft.
7. 100%=47216.85,
8. l%=Tfo. of $7216.85=$72.1685, and
9. 99^ ^^99^. times $72.1685=$7150.094=cost of the
draft.
.-. $7150.094=cost of the draft.
The aggregate face value of two notes is $761.70 and each has 1 year 3 months to run; one of the notes 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.
100%=face of note discounted at bank discount. $761.70 — 100%=face of note discounted at true
discount.
10%— bank discount for 1 year. 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. 12£%=interest for 1 year 3 months.
4. 100%+12|%=112i%=amount of present worth.
5. $761.70 — 100%=amount of the present worth.
(1.)
(2.)
(3.) (4.)
(5.)
(6.)
(70
(8.)
(9.)
(10.)
(11.) (12.)
(13.)
6. .-. 112i%=$761.70— 100%,
of ($761.70— 100% )=:$6.7706f— 1%,
100%=100 times ($6.7706| — 1%) = $677.06|—
88f %=present worth. $761.70— 100%— ($677.06|— 88f % ) = $84.63-J —
H^.%— true discount. [discount.
$84.63i— ll^-%t+12i%=$84.63i+ 1TV % = whole $761.70— $671.50=$90.20=whole discount.
l%=jrof $5.56|=$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.
III.
. ( $400.80=face of note discounted at bank discount, and ( $360.90=face of note discounted at true discount.
MISCELLANEOUS PROBLEMS.
137
II.
II.
3.)
(*•)•
(5.) (6.) (V.)
19
10.)
1. A merchant bold 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? 100%— cost of goods sold at a profit. Then $1000 — 100%— cost of goods sold at a loss. 20%— profit on 100%, the part sold at a profit
1. 100%— $1000— 100%.
2. 1%— -j-^of ($1000— 100%)— $10— 1%,
3. 11%— 11 times ($10— 1%)— $110— 11%— loss on
the remainder.
.'. 20%— ($110— 11%)— 31%— $110=gain. $100— gain. /. 31%— $110— $100. 31%— $210,
1 %— ^T of $210— $6ff , [profit.
100%— 100 times $6ff— 677.41ff— part sold at a III. .-. $677.41f-f— value of the part sold at a profit.
I. By discounting a note at 20% per annum, I get 22-J% per annum interest; how long does the note?
1. 22-£% of the proceeds=20% of the face of the note.
2. 1% of the proceeds— —f of 20%=|% of the face of the
note. «**
3. 100% of the proceeds=100 timss |%=88|% of the face
of the note.
4. 100%— face of the note.
5. 88f%— proceeds.
6. 100% — 88f %=ll£%=discount for a certain time.
7. 20%— discount for 360 days.
8. 1%— discount for -fa of 360 days, or 18 days.
9 lli%==discount for 11^ 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 installments. What will be the annual payment including interest at 6%?
1. 100 c/c=on& annual payment.
2. .-. 100%=amount paid at end of the fifth year
since the debt was then discharged.
3. 100 %-principal that drew interest the fifth year.
4. 6 %= interest on this principal.
5. .*. KXH+6%=106%=amount of this principal.
6 . . ' . 1 06 #F=100 % - the annual payment .
7. l %=^ of 100 #=£ f %, and
8. 100^=100 times f|%=94£f %= principal at the
beginning of the fifth year.
9. 94J| %-hlOO $>=194jf ^-amount at the end of the
fourth year.
II
138
FINKEL'S SOLUTION BOOK.
II.
(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%=194££%,
5. 1%=T^ of 194ff%==1.83Y9¥5-^g-%, and
6. 100%=100 times 1.83^r%%=183^\3¥%=princi-
pal at the beginning of the fourth year.
7. lS&f^%+lQO%=285Jffo%==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. .-. 106%=283^ftftF%,
6. 100%3oO time^F67TVAVT%?^67T4^rV% =
principal at the beginning of third year.
7. 267AWTV%+100%— 367T4A88rV%^ amount at
the end of second year.
1. 100%=principal at the beginning of second year.
2. 6%=interest [year.
3. 100%+6%=106%=amount at the end of second
4. .-. 106%=367T\\8-84rr%>
6. 100%=100 times 3.46fff|m%=346|fff|-|-f %= principal at the beginning of the second year.
the end of first year.
1. 100%— principal at the beginning of the first
year, or the cost of farm.
2. 6%= interest.
3. 100%+6%=106%=amount at end of first year.
4. .
5. 1
6. 100%=100 times 4.2;
%=cost of the farm. (6.) $5000=cost of the farm.
(8.) l%=$5000^421^\^%44%Vz:=$ll-8698-|-, and (9.) 100%=100 times $11.8698=$1186.98+=the an- nual payment.
III. .'. $1186.98+ =the annual payment.
(Milne's Prac., p. 361, prob. 63.)
(4.)
(5.)
I. A and B have $4700 ; f
~i%
B's share; how much has each?
of A's share equals
of
PROPORTION.
II.
of B's,
8. 1% of A's= - of !%%=%%% of B's, and
9. 100% of A's=100 times fo%=74X% of B's.
10. 100%— B's share.
11. 74^2T%=A's share.
12. 100%+7422T%=17422T%==sum of their shares.
13. $4700=sum of their shares.
14. .-. 174 22T% =$4700,
15". l==
>f$4700==$27, and
16. lOOftf^lOO times $27=$2700=B's share.
17. 74^2T%=74Y2T times $27— $2000=A's share. ( $2700=B's share and
' "I $2000=A's share.
1. J
CHAPTER XVI.
RATIO AND PROPORTION.
1. Rdtio 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 second quantity, 4 apples, the consequent. Taken together they are called Terms of the ratio, or a. Couplet.
the common sign of
2. The Sign of ratio is the colon, division with the horizontal line omitted.
Note. — Olney says, "There is a common notion among us, that the French express a ratio by divding the 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 -f-; and
140 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- cording 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 JKatio is the quotient of the antecedent di- vided by the consequent.
4. An Indirect Hatio is the quotient 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 Inequality is a ratio less than unity; as, 4 :5.
7. A Compound Hatio 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 Hatio is the ratio of the cubes of two numbers; as, a3 : bz.
1C. A SubduplicateJRatio is the ratio of the square roots of two numbers; as, \/^: \/~&.
11. A Subtriplicate Ratio is the ratio of the cube roots of two numbers; as, $/^~: $/£.
PROPORTION.
%
12. Proportion is an equality of ratios. The equality is indicated by the ordinary sign of equality or by the double colon, : :. Thus . a : b=c \d, or a : b : : c : d.
13. The Extremes of a proportion are the first and fourth terms.
14. The Means are the second and third terms.
15. A Mean Proportional between two quantities is a quantity to which either of the two quantities bears the same ratio 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.
17. A Compound Proportion is an expression of equality between a compound and a simple ratio.
PROPORTION. 141
A Conjoined Proportion 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."
What is the ratio of -J to f ? £-s-f = i X f = |,the ratio.
What is the ratio of 10 bu. to If bu.? 10 bu. -i- If bu. = 10 X TV = 7, the ratio.
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. Ifa5-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 : (? = 7Ty>z.)
I. If a staff 3 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| feet.)
I. What number is that which being divided by one more than itself, gives -ij- for a quotient?
II..
III. I.
1.
2.
3. 4. 5.
6.
7.
M
Let f=number. T
hen . : : 1 : 7, whence
lumber, ed by 3 more than itself gives $ for
f+1 ^01^ + - 7(*)=1(*+1) or Y*=f+l; whence
i=-rV» and |=2 times TL=i=i
i=the number.
/"hat number divid a quotient?
142 FINKEL'S SOLUTION BOOK.
1. Let f±=the number. Then 2
=|- or, putting this in the form of a proportion,
1-+3 : : 7 : 9. [the product of the extremes.
= 1^-j-21> the product of the means being equal to '—¥=4=21,
6. £=±of21=5i, and
7. |=2 times 5i=10i=the number. III. .-. 10£=the number.
I. 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 lb. of sugar to 24 lb. of rice, 12 lb. of rice to 7 lb. of butter, 8 lb. of butter to 12 lb. of cheese ; how many lb. of coffee are equal in value to 65 lb. of cheese?
1. 7 lb. of coffee=5 lb. of tea,
2. 3 lb. of tea=13 lb. of sugar,
3. 39 lb. of sugar=24 lb. of rice,
4. 12 lb. of rice=7 lb. of butter,
'^5. 8 lb. of butter=12 lb. of cheese, and 6. 65 lb. of cheese= ?=39 lb. of coffee,
7X3X39X12X8X65 '' 5X13X24X7X12 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. .-. 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 I 3 cows : : 60 days : ?=17T3T 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 days : 18 days
12 hours : 8 hours _nA men . p__12
12 feet
10 feet
9 feet
9 feet
5 feet
6 feet
PROPORTION. 143
380X18X8X9X5X6X24
95X24X12X12X10X9
=12 men.
A Louisville merchant wishes to pay $10000, which he owes in Berlin. He can buy a bill of exchange in Louis- ville on Berlin at the rate of $.96 for 4 reichmarks ; or he is offered a circular bill through London and Paris, brokerage |% at each place, at the following rates : £1=$4.90=25.38 francs, and 5 francs=4 reichmarks. What 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.)
6. $4.90=£1— 1% of£l=£.99J.
7. JE1=.99|. times 25.38 fr.
8. 5 fr.=4 marks.
42016.807X4.90X5 ,00 . .
Circular
II
10. $10165.38— $10084.033=$81.35 = gain by direct ex- change.
III. .'. $81.35=gain by direct exchange.
I. A wheel 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 cogs=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. ^ 4. 9 cogs passed through the point of contact=-^- revolu- tion of smaller wheel. 5. .'. In 1 revolution of larger wheel the smaller gains -£%
revolution of smaller wheel.
/. 2^- revolution gained : 10 revolutions gained .' : 1 revolution of larger wheel : ?=28f revolutions of larger wheel.
III. .-. The smaller wheel will gain 10 revolutions in 28-f revo- lutions of larger wheel.
By analysis and proportion.
26 cogs passed through the point of contact==l revolution of the smaller wheel.
144 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^9-g-rev.
.-. The smaller wheel makes 1/g- revolutions while the largei wheel makes 1 revolution. .-. The smaller gains 1^- revolutions, — 1 revolutions^- revolution. If the smaller wheel gains -fa revoluion in 1 revolution of the larger wheel to gain 10 revolu- tions on t|ie larger wheel, the larger wheel must make more rev- olutions. /. less : greater : .: less : greater.
•fa rev. ; 10 rev. ; : 1 rev. of larger ; ?=28f rev. of larger.
I. If the velocity of sound be 1142 fee.t 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 between the time of seeing the flash and hearing the thunder?
1. 1142 ft.=distance sound travels in 1 second.
2. 68520 ft=60Xll42 ft.=distance sound travels in 1 min.,
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 counted.
4. .'. greater : less : : greater : less. [145 yd. 2-iJ- ft.
5. 70 pul. : 20pul. : : 68520 ft. : ?=19577| ft.=3 mi. 5 fur.
III. .-. The cloud is 3 mi. 5 fur. 145 yd. 2| ft. distant.
(/?., 3d p., p. 289,prob. 45.)
PROBLEMS.
1. If 3 horses, in ^ of a month eat f of a tori of hay, how long will •§• of a ton last 5 horses?
2. If a 4-cent loaf weighs 9 oz. when flour is $6 a barrel, how much ought a 5-cent loaf weigh when flour is $8 per barrel?
3. 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 -J- of 12 be 3 what will £ of 40 be ? Ans. 15.
6. If 3 be \ of 12, what will £ of 40 be? Ans. 6&
II.
PROBLEMS. 145
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. 13^7T days.
8. If 5 oxen or 7 cows eat 3T4T 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 30 miles of Boston, they took in C. How much shall each pay? Ans. First man, $7.609|£f; second, $7.609 W ; A» $5.873TV¥; B, $2,864^ ; and C, $1.041TV
11. Three men purchased 6750 sheep. The number of A's sheep is to the number of B's sheep as f is to 3^, and 4 times the number of C's sheep is to the number of A's sheep as -J is to -J-. Find the number of sheep each had. C 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. 30 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 f as much as C, and C can do |- as much as D, and D can do -| 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.
146 FINKEL'S SOLUTION BOOK.
17. A wheel has 45 cogs ; a smaller wheel working in it, 36 cogs ; in how many revolutions of the larger wheel will the smaller gain 10 revolutions? Ans. 40.
18. If the velocity of sound be 1142 feet per second, and the number 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 seeing a flash of lightning and hearing the thunder?
Ans. 5| mi. 108 yd. If ft
19. If William's services are worth $15f- a month, when he labors 9 hours a day, what ought he to receive for if months, when he labors 12 hours a day? Ans $91.91^.
20. If 300 cats kill 300 rats in 300 minutes, how many cats will kill 100 rats in 100 minutes? Ans. 300 cats.
CHAPTER XVII.
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 5 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 8 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 them, 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. 147
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-^4=cost of 1 ball. ( 3. 81 cents=cost of 81-=-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.
1. |r=the number.
2- *-H=* X4=f=quotient
3. 10=quotient.
4. /, 1=10,
5. £=£of 10=2, and
6. =3 times 2=6=the number.
III. .'•'. 6=the number required.
I. $24 is f of the cost of a barrel of wine; what did it cost?
II. |=cost of the wine per barrel. 2. \ of cost=$24, 3. \ of cost=£ of $24=$8, 4. | of cost=5 times $8=$40,
(II. .% $40=cost of wine.
t. What number is that from which, if you take \ of itself, the remainder will be 16 ?
1. ^=the number.
2. If — f=4=remainder after taking away ^.
3. 16=remainder. .-.$=16,
5. f=i of 16=4, and
6. ^=7 times 4=28=the number
III. .'. 28=the required number.
148 FINKEL'S SOLUTION BOOK.
I. A boat is worth $900; a merchant owns £ of it, and sells % of his share ; what part has he left, and what is it worth ?
1. ^=part the merchant owned.
2. | of !=^=part he sold.
i , 3. .-. f-AHri— &=M=T\=Part he had left.
-1 $900=value of Tf , or the whole ship. R , *. $75=TV of $900=value of ^ of the ship.
1 3. $375=5 times $75=value of ^ of the ship, or part he had left.
$375=value of it.
I. A and B were playing cards. B lost $14, which was y7^ times f as much as A then had ; and when they com- menced, | of A's money equaled fy of B's. How much had each when they began to play?
( 1. ) | of A's money=4 of B's.
(2. ) J of A's money=| of f =-£% of B's.
II.
=B's money when they began to play. Then
(3.) f of A's money=8 times A=M of B's-
(4.)
(5.) ^|.=A's money when they began.
1. Tf=A's money after winning $14 from B.
2. $14=what B lost.
3. T7V times f=T7^^part A's money is of $14.
5. ^=4 of $14=$2, and [$14 from B.
6. If— 15 times $2=$30=A's money after winning (7.) .'. $30— $14=$ 16= A's money at first.
.)
.)
(8.) .-. if=$16, from (5),
^(10.) H=35 times $l=$35=B's money at first.
$16=A's money at first, and $35=B's money at first.
(Stod. Int. A., p. lll.prob. 30.)
I. A drover being asked how many 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. 4|=the number of sheep.
2. i+9-^=-J of the number-|-9^.
3. 99£=£ of the number+9f
5. 4=99^—9-^=90, and
6. f=3 times 90=270=number of sheep.
III. .'. He had 270 sheep.
ANALYSIS.
149
I.
II.
III.
I.
II.
III.
I.
Heman has 6 books more than Handford, and both have 26; how many have each?
1. |=number Handford has. Then
2. |+6=Heman's number.
3. f +f +6=|+6=number both have.
4. 26=number both have.
6. .-. f+6=26 or
5. 4^26—6=20.
7. f=iof20=6, and
8. f=2 times 5=10=Handford's number.
9. f+6=16=Heman's number.
Handford had 10 books, and
Heman had 16 books. (Stod. Int. A., p. 116,prob. 2.)
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 man to drink it.
4. T1^=part he drinks in one day. [day.
5. .'. ^ — 'fu=='f^ — A==T"5==:Par* *ne woman drinks in one
6. -j~!=what the woman drinks in ±-^-±-^=1.5 days.
II.
III.
.•. It will take the woman 15 days.
(/?. Alg. /.,/. 112,prob. 59.)
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= what he would have icceived had he
worked the whole time.
3. $40=what he received.
4. ... $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-f-l, or 8 days.
. 80 clays — 8 days=72 days, the time he worked.
/. He worked 72 days.
A ship-mast 51 feet high, was broken off in a storm, and | of the length broken off, equaled f of the length re- maining; how much was broken off, and how much re- mained?
150 FINKEL'S SOLUTION BOOK.
1. f of length broken off=f of length remaining,
2. -J of length broken off=£ of f=| of length remaining,
3. f of length broken off=3 times f=f of length remain-
ing.
4. f— length remaining.
II.
II.
j
5. f— length broken off.
6. f-f-f=y=whole length.
7. 51 feet=whole length.
8. /. V=51 feet,
9. 4=TV of 51 feet=3 feet, and
10. f=8 times 3 feet=24 feet, length remaining.
11. |=9 times 3 feet=27 feet, length broken off.
24 feet— length remaining, and 27 feet=le*igth broken off.
I. A boy being asked his age, said, "4 times my age is 24 years more than 2 times my age;" how old was he?
1. !=his age.
2. 4xf=f=4 times his age.
3. 2Xf=|=2 times his age.
4. .-. |=|._|_24 years or
5. | — 4^4^24 years.
6. i=|- of 24 years=6 years, and
L7. f=2 times 6 years=12 years, his age.
III. /. He is 12 years old. (Stod. Int. A., p. 116,prob. 16.)
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. YTO A.=YT °f TTT A.— what 1 man digs in 1 day.
4. Y2" A.=YTTF A.=5 times YYTF A.=what 5 men dig in 1
day. [days.
5. fT ^.=2% A.=6 times ^ A.=what 5 men dig in 6
6. .'. 6 A. — T3T A.=5T8i A.=what is to be dug by the boys
in 6 days.
7. 1 A.=what 18 boys dig in 11 days.
8. Y*g- A.=what 1 boy digs in 11 days.
9. Yi~g" ^•==TT °f Tff A.=what 1 boy digs in 1 day.
10. -g^ A.=Tf'j A.=6 times T^-g- A.=what 1 boy digs in 6
days. -11. 5-fa A.=what 5T8Y-:~g^, or 189, boys dig in 6 days.
III. .'. It will take 198 boys.
(R. 3d p., O. E.,p. 318,prob. 66.)
ANALYSIS. 151
A man after doing | of a piece of work in 30 days, calls an assistant; both together complete it in 6 days. In what time could the assistant complete it alone?
1. -|==part the man does in 30 days.
2. -ffa—^ of f— part he does in 1 day.
3. -f=f — |=part he and the assistant do in 6 days.
4. T15=-g- of -|=part he and the assistant do in 1 day.
&• •'• T*5 — 5Tff=rA — rTr=Ti7r==Part the assistant does in 1
day. 6. ff$=part the assistant does in -J-J^-i-^^lf days.
II.
III. /. It will take the assistant 21f days.
(R. 3d /., O. E.,p. 318,prob. 71.)
Explanation^.- — Since the man does f of the work before he called on the assistant, there remains f — 1=§, which he and the assistant do in 6 daj s. Hence they do £ of §, or ^ of the work in one day. If the man and his^ assistant do ^ of the work in 1 day and the man does -£$ of the work in 1 day, the assistant does the difference between^ and ^ which is r£0 of the work in 1 day. Hence it will take £$8-7-r£o> or21f days, to do the work.
L A person being asked the time of day, replied that it was- past noon, and that f of the time past noon was equal to f of the time to midnight. What was the time of day?
1. | of the time past noon=f of the time to midnight.
2. i of the time past noon=4 of -f=£ of the time to mid-
night. [midnight.
3. |> or the time past noon,=4 times -$=-f of the time to
4. f=time to midnight. Then
II.
5. -|=time past noon.
6. f-j— |— ^=time from noon to midnight.
7. 12 hours=time from noon to midnight.
8. .*. |=12 hours,
9. -j=7f of 12 hours=l^ hours, and [past noon. 10. ^=4 times 1^- hours=5^- hours=5 hr. 20 min., time
III. .-. It is 20 min. past 5 o'clock, P. M.
(Mztne's Prac. A., p. S60, prob. 47.)
Note. — From 3, we have the statement that the time past noon is £ of the time to midnight. Hence, if f is the time to midnight, £ is the time past noon 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?
152
FINKEL'S SOLUTION BOOK.
1. -5-=time past noon. Then
2. ^=time to midnight.
3. 5— [-7.= 1T2=time from noon to midnight. Ut<J 4. 12 hours=time from noon to midnight.
5. .'. J72=12 hours.
6. Jf—Tz of 12 hours=l hour, and
7. -J=7 times 1 hour=7 hours=time*past noon.
III. .-. It is 7 o'clock P. M.
I. A man being asked the hour of day, replied that £ of the time past 3 o'clock equaled \ of the time to midnight; what was the hour?
1. £ of the time past 3 o'clock=-J of the time to midnight.
2. |, or the time past 3 o'clock,=4 times -J=f of the time
to midnight.
3. |=:time to midnight.
4. |-=time past 3 o'clock.
H.-I 5. f+f =£=time from 3 o'clock to midnight.
6. 9 hours=time from 3 o'clock to midnight.
7. .-. |=9 hours.
8. \-=\ of 9 hours=l^ hours, and
9. |=4 times 1£ hours=6 hours=time past 3 o'clock. .10. f+3 hours=9 hours, time past noon.
III. .-. It is 9, o'clock, P. M.
(Brooks1 Int. A., p. 156,prob. 17.)
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?
1. f of time past noon=J of time to midnight-)-2f hours.
2. J of time past noon=-J of (f+2| hours )=^ of time to
midnight-]- 1£ hours. [to midnight-f-4 hours.
3. f, or time past noon,=3 times (^-[-1-j- hours )=-J of time
4. f=time to midnight.
I J 5. £+4 hours=time past noon. [night.
6. f-HH~4 hours=:|-(-4 hours=time from noon to mid-
7. 12 hours=time from noon to midnight.
8. .'. f-f-4 hours=12 hours.
9. £==12 hours — 4 hours=8 hours,
10. |-=i of 8 hours=2 hours, and *
11. |~|-4 hours=6 hours=time past noon.
III. .-. It is 6 o'clock, P. M.
(Stod. Int. A., p. 128, Prob. 29.)
I. A father gave to each of his sons $5 and had $30 remain- ing; had he given them $8 each, it would have taken all his money; required the number of sons,
ANALYSIS.
153
'1. $8=amount each received by the second condition. 2. $5=amount each received by the first condition. II. •{ 3. $3=$8 — $5= excess of second condition over first, on each son. [10 sons.
,4. /. $30=excess of second condition over first, on 30-i-3, or
I. .•. There were 10 sons.
I. If 50 lb. of sea water contain 2 Ib. of salt, how much fresh water must be added to the 50 lb. so that 10 lb. of the new mixture may contain -J 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 ,, i new mixture contain. [mixture contain.
X 3. 2 lb. 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.
III.
I.
II.
.-. 10 lb. of fresh water must be added that 10 lb. of the new mixture may contain -j- 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 f- of the number in the second field equals £ of the number in the third field. If the entire num- ber was 434, how many were in each field?
•§• of number in first field=f of number in second field. [second field.
-J- of number in first field=-J- of f=| of number in -| , or number in first field,=3 times f=f of number
in second field.
| of number in second field=| of number in third field. [in third field.
•J of number in second field=-J- of f =f of numbet f, or number in second field,=3 times f=f of num- ber in third field. (3.) f=number in third field. Then (4.) -|=number in second field, and (5.) £4=t of number in second field=number in first field in terms of number in third field.
three fields.
434=number in the three fields.
... 2^=434,
_T of 434=2, and [field.
[=64 times 2=128=number of sheep in third 5=72 times 2=144=number of sheep in second field. [field.
1^=81 times 2=162=number of sheep in first
154
FINKEL'S SOLUTION BOOK.
III.
II.
I.
II.
C 162=number of sheep in first field,
< 144=number of sheep in second field, and
( 128=number of sheep in third field.
(Milne's Prac. A., p. 362, prob. 68.)
pupils there are 32 girls ; how that there may be 5 boys to 4
In a certain school of 80 many boys must leave girls?
1. 80=whole number of pupils.
2. 32— number of girls.
3. go — 32=48=number of boys.
4. ^=number of girls. Then, since the number of boys are
to be to the number of girls as 5 : 4,
5. J=number of boys. But
6. |=32.
7. £=£ of 32=8, and
8. 4=5 times 8=40=number of boys.
9. .'. 48 — 40=8=number that must leave that there may be
5 boys to 4 girls.
III. .'. 8 bo^s must leave that there may be 5 boys to 4 girls.
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 -J- hour.
3. 6 mi.=distance he can 'walk in 1 hour.
4. 1 mi.=distance he can 'walk in \ hour.
hr.-)-^ hr.=T5g- hr.=time it takes him to ride 1 mi. and walk back. [and walk back.
10 hours=time it takes him to ride 10-7-T5^, or 36, mi.
5.
6.
III. .-. He can ride 36 miles.
I. A hound ran 60 rods before 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?
rl. 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 hound ran to catch the fox.
5. .-. f=60 rods,
6. -J=^ of 60 rods=12 rods, and [ahead.
7. f=2 times 12 rods=24 rods=distance the fox was
8. f=3 times 12 rods=36 rods=distance the fox ran be-
fore he was caught.
ANALYSIS.
155
( 24 rods=distance the fox was ahead, and
( 36 rods=distance he ran before he was caught.
If J of 12 be 3 , what will ± of 40 be ?
1. •§• of 12=4.
2. i of 40=10. By supposition
3. 4=3. Then
4. 1=4 of 3=|, and
5. 10=10 times |=7|.
.-. £ of 40=7 -J, on the supposition that £ of 12 is 3.
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. f— amount paid for the coach. [been only 8 men.
2. -£=amount 1 man would have had to pay, had there
3. T1?=amount 1 man paid since there were 8 men-f-6 men,
or 14 men.
4. .'• i— TT=T6— Tnr=irs=what each saved.
5. $lf=what each saved.
6. .-.
7. = of $i=$7> and
II.
18. ff=56 times $-J2^==$32|=amount paid for the coach.
HI. .% $32f=amount paid for the coach.
(R. H. A., p. Jt.OS.prob. 46.)
Second solution.
'1. $lf=amount saved by each man. [the six meu.
2. $14=8X$lf=amount saved by the 8 men and paid by
3. .*. $2£=J of $14=amount paid by each of the 14 men. ,4. .-. $32f— 14 times $2£=amount they paid for the coach;
.-. They paid $32f for the coach.
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?
1. 1 A.=what I plow for every 10 sheep I keep.
2. T^A.=what I plow for each sheep I keep.
3. 1 A.=what I allow for pasture for every 4 sheep I keep.
4. ^A.=what I allow for pasture for each sheep I keep.
5. .'. T1irA.-|-JA.=^FA.=land required for every sheep.
6. c-. 161A.=land required for 161-r-^r, or 460 sheep.
.-. I can keep 460 sheep on 161 acres.
(./?. Alg. I.) p. 112,prob. 64') Complete analysis.
If for every 10 sheep I plow 1 acre, for 1 sheep I plow -^ of &n acre ; and' if for every 4 sheep I pasture 1 acre, for 1 sheep, I
III. I.
II.
III.
156 FINKEL'S SOLUTION BOOK.
pasture ^ of an acre ; hence 1 sheep requires ^A.-j-^A., or ^.n.., and on 161 A. I could keep as many sheep as /^A. is contained in, 161 A., which are 460 sheep.
I. A man was engaged for one year at $80 and a suit of clothes; he served 7 months, and received for his wages the clothes and $35; what was the value of the clothes?
1. i|=value of the suit of clothes.
2. ff4-$80=wages for 1 year or 12 months.
3. ^^iej^^ of (ff+$80)=:wages for 1 month.
4. ^-|-$46|=7 times (TVf$6|-)== wages for 7 months. !!.<! 5. ff +$ 3 5= wages for 7 months.
6. /. -if +$35=T^+$46|-.
8. T%=-% of $11^=::=$2^-, and
9. ff=12 times $2£=$28=value of suit of clothes.
III. /. The suit of clothes is worth
A lady has two 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; and the sec- ond cup and cover 4 times as much as the first cup. What is the weight of the second cup and the cover?
1. 3 times weight of second cup=weight of cover-)- weight
of first cup, or 8 oz. [2f oz.
2. 1 times weight of second cup=£ of weight of cover-J-
3. |=weight of cover. Then
4. -J+2f oz.— weight of second cup. [cover.
5. I+J+2J oz.=£-|-2f oz. = weight of second cup and !!.<{ 6. 32 oz.— 4 times 8 oz.=weight of second cup and cover,
by the conditions of the problem.
7. .'. i+2f oz.=32 oz.
8. |=32 oz.— 2f oz.=29£ oz.
9. |=i of 29£ oz.=7£ oz.
10. f=3 times 7^- oz.=22 oz.=weight of cover. [cup.
11. l+2f oz.=7l oz.+2| oz.= 10 oz. = weight of second
. ( 22 oz.=weight of cover, and * " ( 10 oz.=weight of second cup.
I. A steamboat that can run 15 mi. 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?
ANALYSIS. 157
1. 15 mi.=distance the boat can travel down stream in
1 hour. [hour.
2. 1 mi.=distance the boat can travel down stream in -^
3. 10 mi.— distance the boat can travel up stream in 1 hr. II. < 4. 1 mi.=distance the boat can travel up stream in -fa hr.
5. /. y^- hr.-l-y1^ hr.=^r hr.=time required for the boat to
travel 1 mi. down and return.
6. .'. 25 br.=time required for the boat to travel 25-r-J-, or
150, milss down and return.
III. /. The distance between the two places is 150 miles.
I. A, B, and C dine on 8 loaves of bread ; A furnishes 5 loaves ; B, 3 loaves; C pays the others 8d. for his share; how must A and B divide the money ?
1. 8 loaves=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=4 loaf=what B furnished to-
wards C's dinner.
5. .'. — |=-|=A's share, and
6. /. ^-— -J— B's share.
7. | of 8d.=7d.=what A should receive, and .8. ^ of 8d.=ld.=what B should receive.
Ill • \ ^ should receive 7d., and
' I B should receive Id. (R. H. A., p. 403, prob. 42.)
I. A and B dig a ditch 100 rods long for $100; how many rods does each dig, if they each receive $50, and A digs at $.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-—.75, or 66f rods, what A digs and 50-^-1.25, 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
158 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 receiving $1 per rod, then the problem is solvable, and the result is 50 rods each.
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?
!=B's age at the required time. Then •|=A's age at the required time.
3. | — f=J=difference of their ages.
4. 30 years — 6 years=24 years=difFerence of their ages. II.<5. .*. f— 24 years.
6. -i=-jr of 24 years=4 years. [time.
7. 1=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 years=Jacob's age at present.
3. |-:=son's age at required time. Then
4. y)=Jacob's age at required time.
5. .'. l-£ — f — f=difference of their ages.
II.<! 6. 40 years — 20 years=20 years=difference of their ages.
7. ... f=20 years,
8. -J=^ of 20 years=2^ years, and [time.
9. |=2 times 2-J years=5 years, son's age at the required 10. /. 20 years — 5 years— 15 years=time since Jacob was 5
times as old as his son. HI. .'. 15 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 £ 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 {§, or f , as old as B, and so on. The ratio of any two persons' ages approaches unity as its limit.
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?
ANALYSIS. 159
1. 2 leaps of hound's=3 leaps of fox's.
2. 1 leap of hound's=J 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-J- 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. .'. 50 leaps=what the hound gains in taking 50x6
leaps, or 300 leaps. III. .-. The hound must take 300 leaps to catch the fox. Remark — We see that 3 of the hound's leaps equals 4| leaps of the fox'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. 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.
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=2^ of $25=his rate of wages for 1 pupil.
3. $30— his rate of wages for 30 pupils.
4. $l=^j- of $30— his rate of wages for 1 pupil.
5. /. $1.25 — $1.00=$.25=reduction per pupil by the ad-
II.
dition of 10 pupils.
6. $.025— $.25-r-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.
/. $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.
(Matt(*i?sArith.,p. 385, f rob. 200.)
Note. — This problem is really indeterminate, because there is no definite rate of increase of wages given for each additional scholar. We 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.
160 FINKEL'S SOLUTION BOOK.
I. A gold and silver watch were bought for $160; the silver watch cost only ^ as much as the gold one ; how mucb was the cost of each ?
1. ^-=cost of the gold watch. Then
2. ^=cost of the silver watch.
3. 7.+i=8=cost of both.
4. $160=cost of both.
5. .'. f=$160,
6. -J-=-J of $160='$20=cost of the silver watch, and
7. 7._7 times $20=$140=cost of the gold watch.
I $20=cost of the silver watch, and ' ( $140=cost of gold the watch.
A man has two watches, and a chain worth $20; if he put the chain on the first watch it will be worth f as much as the second watch, but if he put the chain on the sec- ond watch it will be worth 2£ times the first watch what is the value of each watch?
1. § B.=J f.+$20.
2. -k s.=J- of (f f.+$20)=J- f.+$10.
3. f s.= 3 times (i f.+$10)=£ f +$30. [lem.
4. I s.=1^ f. — $20, by the second condition of the prob-
II.
5. ... 11 f._$20=f f.+$30, whence y f._ 4 f.=f
6. V f-— f f.=$30+$20, or
7. f f.— $50.
8. i f.=J. of $50=$10, and
9. | f.=4 times $10=$40=value of first watch.
10. f s.=f f.+$30t=f of $40+$30=$90=value of the sec- ond watch. TTT . \ $40=value of first watch, and lli< V I $90=value of second watch.
( White's Comp. Arith., p. 248, prob. 60.)
I. At the time of marriage a wife's age was -| of the age of her husband, and 10 years after marriage her age was Y7^ 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. f+10 years=husband's age 10 years after marriage.
4. f +10 years— wife's age 10 years after marriage. But
5. T7u+7 years— y7^ of (f+10 years )=wife's age 10 years j i after marriage, by second condition of the problem.
'• yVK years=f+10 years. Whence f=10 years — 7 years, or
3 years. [of marriage.
=10 times 3 years=30 years=husband's age at time or T6Q,=6 times 3 years=18 years=wife's age at the time of marriage.
ANALYSIS. 161
( 30 years=husband's age at time of marriage, and ' ' ' (18 years=wife's age at time of marriage.
( White's Comp. A., p. 2^1, prob. 35.)
I. Ten years ago the age of A was f of the age of B, and ten years hence the age of A will be f of the age of B ; find the age of each.
1. |— B's age 10 years ago. Then
2. |=A's age 10 years ago.
3. ^+10 years— B's age now, and
4. f+10 years=A's age now.
5. |+20 years— B's age 10 years hence, and
6. f+20 years=A's age 10 years hence. [hence.
7. 5. Of (|_|_20 years)=f+16f years=A's age 10 years
8. /. f +16f years=£+20 years ; whence
9. 5. — 1=20 years — 16| years, or
10. -^=3^ years, and
11. j-f=12 times 3iJ- years=40 years=B's age 10 years ago.
12. |=T9^=9 times 3£ years=30 years=A's age 10 years
ago.
13. ... 11+10 years=50 years=B's age now, and
14. ^2 I 10 years=40 years=A's age now.
TTT i &Q years=B's age, and
Ui' '*• (40 years=A's age.
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. •J=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.
TT )&• f X25 mi.=clistance second travels in f days.
6. .'. f X20 mi.+lX25 mi.=f X(20 mi.+25 mi.)=distance
both travel.
7. 495 mi.=di stance both travel.
8. .-. (20 mi.+25 mi.)x|=(45 mi.)Xf=495 mi. Whence f =495-1-45=1 l=number of days.
III. .-. They will meet in 11 days.
Second solution.
rl. 20 miles=distance first travels in a day.
TT J 2. 25 miles=distance second travels in a day.
3. .'. 45 miles=distance both travel in a day. [days.
4. .-. 495 miles=distance both travel in 495-7-45, or 11,
162 FINKEL'S SOLUTION BOOK.
III. .-. They will meet in 11 days.
Third solution — the one usually given in the schoolroom.
20+25=45)495(11 days. 45 45 45
I. Find a number whose square root is 25 times its cube root.
1. -|=square root of the number. Then
2. !xf=tne number, because the square root X the square
root equals the number.
3. |— the cube root of the number. Then
4. fXf Xf=the number. But
II.
5. |=5X(f)- Hence, squaring both sides,
6. fxf=25X(|Xf). But
7. f xf=the number, and
8. f Xf Xf=the number.
9- .'. f Xf XS=25X(|Xf )• Dividing by (f Xf ), 10. f=25. [11. ... (f )3=253=15625.
III. .-. The number is 15625. (/?. H. A., p. 367,prob. 14.)
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; what was the cost of each?
1. |"f =cost of the horse. Then
2. •ff=^ of ^|=cost of the saddle, and
3. Y^=\ of T2Y=cost of the bridle.
4. H=!f+T22+TV=co IL<5. $150=cost of all.
and
7. J-^ySr of $150=$10=cost of bridle.
8. H=12 times $10=$120=cost of horse. 19. T\=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., p. 241, prol>. S9.)
ANALYSIS.
163
I.
HI.
A and B perform T9^ 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?
1. T9^=part A and B do in 2 days.
2. ^=4 °f A=Part A and B do in 1 day.
3. -fj — rV=TiF— Part left after B quits, and which A com-
pletes in ^ day.
4. j^=4p=part A can do in 1 day.
5. .'. -|=part A can do in -!—~|=5 days.
7. .'• ^=part B can do in | ; |, or 4, days.
SA can do the work in 5 days, and B can do the work in 4 days.
( White's Comp. A., p. 280, prob. 193.)
II.
I. A and B can do a p.iece 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?
1. 12 days— time it takes A and B to do the work.
2. /. T1-2-=part 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 ... _i__|l|_|_i.==:^3=part 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. .*. Tf=| of £f=part A, B, and C do in 1 day.
9. i| — ^— y-^-^part A, B, and C do in 1 day — part B
and C do in 1 day=part C does in 1 day.
10. |f=part C does in f|-:-TV or 10f days.
11. T| — ^=T5^^=part A, B, and C do in 1 day — part B and
C do in 1 day=part A does in 1 day.
12. T|=part A does in -J|-7-y5^=:14-| days.
13. II — i.— ^i.^part A, B, and C do in 1 day — part A and
C do in 1 day=part B does in 1 day.
14. Tf==part B does in ^-^=72 days
( 14|- days=time it takes A, III. .-. < 72 days=time it takes B, and ( 10^- days=time it takes C.
( White's Comp. A., p. W^prob. 280.)
I. The head of a fish is 8 inches long, the tail is as long as the head and £ of the body-flO inches, and the body is as long as the head and tail ; what is the length of the fish?
164 FINKEL'S SOLUTION BOOK.
1. § =length of body.
2. 8 in.=length of head.
3. | 1. of b.+lO in.+8 in.=£ 1. of b.+18 in.=length of tail,
4. | 1. of b.=length of head-f-length of tail.
II.
5. .-. f 1. of b.=(4 1. of b.+18 in.)+8 in.=i 1. of b.+26 in
Whence
6. I 1. of b.— i 1. of b.=J- 1. of b.=26 in.
7. .'. f 1. of b., or length of body ,=2 times 26 in.=52 in.
1. of b.+18 in.=26 in.+18 iri.=44 in.=length of tail. 19. /. 52 in.+44 in.+8 in.s=104 in.=length of the fish.
III. .-. The length of the fish is 104 inches.
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. f=remainder after losing 3. Then
2. f-j-3=number at first.
3. f of r. — 2=number sold.
4. $48— $32=$ 16= what was received for f of r.— 2.
5. $8=£ of $16=what was received for -J- of (fofr.— 2), j • or -| of r. — 1.
6. $24=3 times $8=what was received for 3 times (-£ of
r— l)=|ofr.— 3. . $48— $24=$24=what (f of r.+3)— (f of r.— 3), or <3
pigs cost.
4=£ of $24=what 1 pig cost. . $48=what 48-T-4, or 12, pigs cost.
III. .-. He bought 12 pigs.
(Brooks' Int. A., f. 164, Prob- &•)
I. A bought some calves for $80; and having lost 10, he sold 4 more than § of the remainder for cost and received $32 less than all cost; required the number purchased.
1. |=remainder after losing 10. Then
2. f-|-10=number purchased.
3. f of r.-|-4=number sold. [cost.
4. $80— $32=$48=cost of f of r.+4, since they sold at TT I 5. $24=i of $48=cost of | of (f of r.+4)=J of r.+2.
^6. $72=3 times $24=cost of 3 times (i of r.+2)=-| of r.-f-6. [cost.
7. .'. $80— $72=$8=what (f of r.+10)— (f of r.+6), or 4
8. $2=^ of $8=what 1 cost.
9. $80=what 80-r-2, or 40 cost.
III. .-. He bought 40 calves.
(Brook's Int. A., p. 164., prob. 10.)
ANALYSIS. 165
A lost f of his sheep; now if he finds 5 and sells f of what he then has for cost price, he will receive $18; but if he loses 5 and sells f of the remainder for cost price, he will receive $6; how many sheep had he at first?
1. J= the number of sheep he had at first.
2. f= the number he lost.
3. -| — f=f > the number he had after losing ^.
4. f-f-5= the number he had after finding 5.
5. fof (f+5)=^+3, the number he sold.
6. | — 5= the number, had he lost 5.
7. | of (| — 5)=^ — 3, the number he would have sold.
8. $18=what (-2VI-3) sheep cost.
9. $6= what (-£% — 3) sheep cost.
10. ,-. $12=$18— $6=what (-fg+S) 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
13. $18= what (-2\- (-3) sheep cost.
14. .-. ^\+3 sheep = 9 sheep, or
15. ^=6 sheep.
16. y^= J of 6 sheep=l sheep, and
17. ff =26 times 1 sheep =25 sheep. .-. He had 25 sheep at first.
(Brooks Int. A., p. 165, prob. 15.)
A man bought a certain number of cows for $200; had he bought 2 more at $2 less each, they would have cost him $216; how many did he buy?
1. $200=cost of cows.
2. .$216— cost of oiig'inal number of cows-j-2 more.
3. $216— $200=$16=cost of 2 cows at $2 less per head.
4. .- $8=4 of $16=cost of 1 cow at $2. less per head. Then
5. $S+$2=$10=cost of each cow purchased.
6. $200=cost of 200-r-lO, or 20 cows.
. . He bought 20 cows.
(Brooks Int. A., p. 162, prob. 8.)
A person being asked the hour of day, said, "the time past noon is ^ of the time past midnight;" what was the hour?
1. !=time past midnight.
2. ^=time past noon.
3. ... I — K^i—time from midnight to noon.
4. 12 hours=time from midnight to noon.
5. .-. |=12 hours.
6. -J=^ of 12 hours=6 hours=time past noon. III. It was 6 o'clock, P. M.
166 FINKEL'S SOLUTION BOOK.
I. Provided the time past 10 o'clock, A. M., equals f of the
time to midnight; what o'clock is it? '1. |=time to midnight. Then
2. |=time past 10 o'clock.
3. |-{-f =J=time from 10 o'clock to midnight. II.<J 4. 14 hours=time from 10 o'clock to midnight.
5. .*. ^=14 hours.
6. £=4 of 14 hours=2 hours, and [o'clock P. M.
7. f=3 times 2 hours=6 hours, time past 10 o'clock=4 III. ' /. It is 4 o'clock, P. M.
I. At what time between 3 and 4 o'clock will the hour and minute hands of a watch be together?
1. -f=distance the h. h. moves past 3. Then
2. 2^=12xf— distance the m. h. moves past 12.
3. 2/ — -f=2^2=distance the m. h. gains on the h. h.
II.
4. 15 min.=distance the m. h. gains on the h. h
5. ... 22=15 min
6. -J=^2 °f 15 min.— |f min. [past 12.
7. 2^=24 times -J-| min.— 16^- min.=distance m. h. moves III. * .'. It is 16^i 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 f, the minute hand moves 12 times f , or ^4. Now the minute hand has moved from 12 to 3-f|, or 15 minutes-)-!, Hence the minute hand has gained 15 minutes on the hour hand. It has also gained 224 — $, or %?. .'. \2=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 tke conditions of the problem, if the hour hand should remain stationary.
I. At what time 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. V=12 times f=distance m. h. moves past 12.
3. .*. 2^ — f=2^2=di stance m. h. gains on h. h.
4. 15 min. or 45 min.=distance m. h. gains on the h. h.
5. .•. 2^2=15 min. or 45 min.
6. %==•£% of 15 min. or -^ of 45 min.=!~f min. or 2-^- min.
7. 2^=24 times ^-J min. or 24 times 2-^ min.=16T4T min.
or 49yT min.
III. .-. The minute hand will be at right angles with the hour hand at 16^ min. or 49^ min. past 6 o'clock.
ANALYSIS.
167
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, i. 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 0. 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?
1. |=distance hour hand moves past 2. Then
2. 2^=distance the minute hand moves past 12, in the
same time. [hand.
3. .'. 2-% — f=2or2=distance minute hand gained on the hour !!.<! 4. 40 min.=distance the minute hand gained on the hour
hand.
5. .*. 2Y2=40 min.
6. •J=^£ of 40 mim.=l-j\- min., and
7. 2^=24 times 1T9T min.=43i7y min.
III. .•. It is 43T7T 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 hour hand. But while the minute hand is moving to 8, the hour hand is moving from 2. So the minute hand must move as far past 8 as the hour hand is past 2. Since \ is the distance the hour hand moves past 2, f must be the distance the minute hand must move past 8. Hence the distance the minute hand moves is f+40 min. But \4=distance the minute hand moves. .'. V=l+ 40 min. or ^=40 min. as shown in step 5.
I.
II.
At what time between 3 and 4 o'clock will the minute hand be 5 minutes ahead of the hour hand?
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=2T2=distance gained by the minute hand.
4. 15 min.-}-5 mim.=:20 min.=distance gained by the m. h.
5. .-. %2=20 min.
6. i=^ of 20 min.=4-f min.
7. V==24 times TT min—21-jSr min.
III. /.It is 21T9T 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 is moving from 3. Hence the minute hand must move as far past 4 as the hour hand moves past 3. But the hour hand
168
FINKEL'S SOLUTION BOOK.
moves | past 3; hence, the minute hand must move f +5 min. past 4, in all, |+20 min. Hence, the minute hand gains (|+20 min.) — |=20 min. on the hour hand.
Remark. — We always find \*, the distance the minute hand moves, for it indicates the time between any two consecutive hours. The hour hand indicates the hour.
I.
II.
At what time between 4 and 5 o'clock do the hands of a clock make with each other an angle of 45° ?
1. |=distance the hour hand moves past 4.
2. 2^=distance the minute hand moves past 12.
3. .'. 224 — f=2Y2= distance the minute hand gains on the
hour hand.
III.
4. 12£ min. or 27-J min.=distance gained by minute hand.
5. /. 2Y2=12£ min. or 27£ min. [min.
6. -|=-j-2 of 12-J min. or •£% of 27-J min.=J|- min. or 1J
7. 2^=24 times -|^ min. or 24 times 1^ min.=1313r min.
or 30 min.
.*. At 13T7T min. 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°. & of 60 min.=7$ min. Hence, that the hands make an angle of 45 , the minute hand must be either 7£ 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 hand is moving past 4. Hence, the minute hand must move as far past 2£ min. past 2 as the hour hand moves past 4, z'. e., the minute hand moves f+124 min. Hence, it gains (f-f-12£ min.) — 1=12| min. The reasoning for the second result is the same as for the first.
I. At what time between 4 and 5 o'clock is the minute hand as far from 8 as the hour hand is from 3 ?
1. -|=distance the hour hand moves past 4.
2. 2^=12 times f=distance minute hand moves past
12 in the same time.
3. .'. V+f= V=distance both move.
4. 35 min.— distance both move.
5. /. 2^6=35 min.
6. •^=-5-$ of 35 min.=l^ min.
7. V=?4 times 1-fc min.=32T\ min.
1. f=distance the h. h. moves past 4.
2. 2Y4=distance minute hand moves past 12.
3. .'. 2-£ — 2f:=?22=di stance the minute hand gains.
4. 45 min.=distance the minute hand gains.
5. .•. 2^2=45 min.
6. -J=^- of 45 min.=2^y min.
7. 2Y4=24 times 2^ min.=4911r min.
It is 32T\ min. or 49yT min. past 4 o'clock.
(7?. H. A., p. 403,prob. 40.)
II.
A.
B.
III.
ANALYSIS.
16&
Explanation. — This problem requires two different solutions. Locate the minute hand at 12 and the hour hand at 4. The hour hand is now 6 min- utes from 3. If the hour hand remained stationary, the minute hand would have to move to 7 to be 5 minutes from 8. But while the minute hand is moving to 7, the hour hand is moving past 4. Hence the minute hand must stop as far from 7 as the hour hand moves past 4; «. <?., if the hour hand moves f past 4 the minute hand must stop f from 7. Then the hour hand will be 5 minutes+| from 3 and the minute hand will be f-j-5 minutes from 8. While the hour hand moved f , the minute hand moved 35 min. — $ /. ^ =35 min. — $, whence ^=35 min. .*. 35 min.=distance they both move. The second part has been explained in previous problems.
I.
At what time between 5 and 6 o'clock is the minute hand midway between 12 and the hour hand? When is the hour hand midway between 4 and the minute hand?
rA.
II.
B.
!=distance the hour hand moves past 5. 2^=distance the minute hand moves in the same
time.
f +25 min.=distance from 12 to the hour hand. -J- of (f-f-25 min.)=-J-[-12-J- min.=distance minute
hand moves.
III.
5. 6.
7.
8 9.
A. B.
mn-
i=-^g- of 12-j- min.=Jf min. V=24 times ff min.=13^ min. |-=distance the hour hand moves past 5. 2^=distance the minute hand moves in the same
time
f+5 min.=distance the hour hand is from 4. Y+10 min. =2 times (f+5 min.)=distance the min-
ute hand is from 4, since the hour hand is midway
between it and 4. 20 min.+(f+10 min.)=f-f-30 min.=distance the
the minute hand is from 12.
==__ mm., or
of 30 min.=l£ min. =24 times H min.=36 min. It is 13-^g- min. past 5 o'clock. It is 36 min. past 5 o'clock
. (R. H. A., p. 403,prob. 41.)
Explanation. — Locate the minute hand at 12 and the hour hand at 5. If the hour hand remained stationary, the minute hand would have to move over \ of 25 minutes, or 12£ minutes. But while it is moving over 12$ minutes, the hour hand is moving past 4. Hence, the minute hand will have to move 12£ minutes+^ of the distance the hour hand moves past 4. Hence V=i~l~12| 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
170
FJNKEL'S SOLUTION BOOK.
4. But |=distance the hour hand moves past 4; hence, £=distance the min- ute hand moves past 6. Hence, 1+30 minutes=distance the minute hand moves. .'. ^*— f+30 minutes, as shown by step 6 of B.
I.
II.<
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?
distance the hour hand moves past 3. =12 times |=distance the minute hand moves in the same time.
3. 2^4_j_|==2_6_:^istance they both move.
4. 45 min.=distance they both move.
^=45 min.
-^ of 45 min.=l|-| min.
=24 times 144 min.=41-A- min.
5. 6.
7.
III. .-. It is 41^ min. past 3.
Explanation. — Locate the minute hand at 12 and the hour hand 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.
II.
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?
|-=distance m. h. was ahead of h. h., or the dis- h. moved, since it changed place
(8.)
III. .'.
tance the h.
with the m. h. [the two observations.
2^=distance the m. h. moved in the time between •"• V44-f:=V=distance they both moved. 60 mm.=distance they both moved. .-. 2^=60 min.
-i=Jg. of 60 min.=2-j^ min. [ahead of h. h.
f=2 times 2T\ min.=4T% min.=distance m. h. was 1. |^=distance h. h. was past 5, at time of first obser- vation. Then [servation. 2^=distance m. h. was past 12 at time of first ob- 25 min.-}-f+4T% min.=f+291^= distance m. h.
was past 12 at time of first observation. 4. 5.
6. -j— ^2 of 29T83- rrin.=l^- min. -7. 224— 24 times 1-^- min.=32T% min.
It was 32T43 min. past 5 o'clock.
ANALYSIS.
171
Explanation. — It is clear that the minute hand was ahead of the hour hand at the time of the first observation, or else they could not have ex- changed places \vithin an hour. Now, we call the distance from the point where the hour hand was located at first lothe point where the minrte hand was located first, -|. But in the mean time the hour hand has moved lo the position occupied by the minute hand and the minute hand ha« moved on around the dial to "the position occupied by the hour hand, /. c., the hour hand has moved | and the minute 12 times f, or V- Hence, they both moved ^6. They both moved 60 minutes since the hand moved on around the dial to the position occupied by t'he hour hand and the hour hand mov- «d to the position occupied by the minute hand. .• \6— 60 min. as shown in step (5.) The remaining part of the solution has been explained in pre- vious problems.
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?
((1.) f=distance m. h. was ahead of the h. h. or dis- tance it moved. [J.
2^4=distance m. h. moved while the h. h. moved 2_4_|_|= 2^6 ^13.^^ both moved. (4 ) 60 min =distance both moved. ^=60 min.
YV °f 60 min.=2T\ min. [was ahead
(7.) f=2 times 2T\ min.=4-£% min.=distance m. h. rl. |=distance h. h. moved past 8-
2. 21£=distance m. h. moved in same time.
3. 40 min+|+4T% min.=|+441^ min. = dis-
tance m. h. moved to be 4T% min. ahead.
rA.
II.
(2.) (3.)
(5.) (6.)
(8.)^ 4. 5.
6. -f— 2T of 44T8-g- min.=2-i£-g- min. [past 8.
7. 2T4=24 times 2T|7 min. = 48^ min.=time
1. 2/>— di stance thty both moved.
2. 60 min. —distance they both moved. B. 3. .-. 2T6=6C min.
4. ^=2if of 60 min.=2T\ min. [was gone.
5. \4— 24 times 2T% min.=55T\ min.=time he
( A. It was 48T9¥6¥ min past 8 o'clock when he first en- < tered school room.
( B. He was gone 55T\ 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 ?
172
FINKEL'S SOLUTION BOOK.
'A.
II.
B.
.C.
1. |=distance the hour hand moves past 12. Then
2. ^—distance the minute hand moves past 12, and
3. i_4^4_o=720 times |=distance the second hand moves past 12.
4. i_4_4_o_2_4 ^ i_4_i_6 = distance
from the minute hand to the second hand.
5. i^u — y ==uyj == distance from the second hand to the hour hand.
2^ — |- = 2^2 = distance from the hour hand to the second hand. 1-4Y1-6 + 1^1~H~¥=2~TJ^=distance around the dial. 60 seconds=distance around the dial as indicated by one revolution of the s. h.
6.
7. 8.
9. 10. 11. MP==I440 times
= °f 60 S6C—
sec.=30T\Vr sec. = time
when s. h0 is midway between the h. h. and m. h. ^=distance the hour hand moves past 12. Then ^distance the minute hand moves past 12, and
distance the second hand moves past 12.
4. *-£ — f=2Y2=distance from h. h. to in. h.
5. 2Y2=distance from s. h. to h. h., because the h. h. is mid- way between them. [12.
6. 2^ — f=distance from s. h. to
7. 1-J2-4-°+y) = 1-4#-° = distance around the dial.
8. 60 sec.=distance around the dial.
9. /. 1 46-0=60 sec.
TIQ.A.
of ^0 sec.=^ sec. 11. Uyy^^o times T3¥ sec.=59|f sec.=time when
the h. h. is midway between the s. h. and m. h. 1 |-=distance h. h. moves past 12. Then
2. 2^=distance m. h. moves past 12, and
3. 1-^-°=distance s h. moves past 12. [h. to s. h.
4. 224 — f=2Y2=distance from h. 5: ^—distance from m h. to
s. h. [from 12 to s h.
6. |- + 222 + V = V = distance
7. -JytP— 4_e = i_3_9_4 ^ distance around the dial. [dial.
8. 60 sec =distance around the
ANALYSIS.
173
III. ,\
= Sec.
= sec-
9.
10.
11. 14240= 1440 times ^ sec.=61f|f sec.=time past
12 when the m. h. will be midway between the
h, h. and s. h.
»A. The second hand is midway between h. h. and m. h. at 30T\9Y°7 sec. past 12. [at 59f| sec. past 12.
B. The hour hand is midway between s. h. and m. h.
C. The minute hand is midway between h. h. and s.
h. at
sec. past 12.
•'• From m to .$• = Ts—Tm = * Y°— V = 1 V6- And» b7 the condi- tion of the problem, the distance from m to s = the distance from m to h. :.
Explanation. — A. We represent the distance moved by the hour hand by |, = the space Th. And since the minute hand moves 12 times as fast as the hour hand, it moves ^*. The second hand moves 60 times as fast as the minute hand or 720 times as fast as the hour hand. From T to h is f and from I to m is ^*. .'. From h to m is Tm — 7V/ = *f — f = ^. From T to s is * Y°- of t
from w to ^ = 14§l6 + 1V6-2¥2- We nave seen, already, that the distance from h to m is \2. .'. The whole distance around the dial is 2{y*2 + 222 = 2*g4.
B. From T to k is f. From T to m is V- •'• From // to m=Tm—Th= ^* — i=V' By the condition of the problem, the distance from h to m=the distance from .9 to //. /. sT— T/i==^ — \=*g. From T around the dial to the right of s is 1^°. .'. The whole distance around the dial=1V°+^Q=
1 -CJ5 0
C. From T to h is f . From T to m is ^*. .'. From // to m=Z£—\=*$-. By the condition of the problem, the distance from m to .?— the distance from h to m= ^. .'. From T to s is 1+^+^=^. From T around the dial through T to s is ^s>. .'. The whole distance around the dial
I.
II.
III.
A sold to B 9 horses and 7 same price, 6 horses and
cows for $300; to C, at the 13 cows, for the same sum;
what was the price of each?
1. Cost of 9 horses-f-cost of 7 cows=$300. Then the
2. Cost of 36 horses-j-cost of 28 cows=$1200, by taking
4 times the number of each.
3. Cost of 6 horses-|-cost of 13 cows=$300. Then the
4. Cost of 36 horses+cost of 78 cows=$1800, by taking
6 times the number of each. But
5. Cost of 36 horses+ccst of 28 cows=$1200.
6. .'. Cost of 50 cows=$600, by subtracting; and
7. Cost of 1 cow=-^j- of $600=$12. The
8. Cost of 7 cows=7 times $12=$84.
9. .-. Cost of 9 horses-=$300— cost of 7 cows=$300— $84
=$216. The 10. Cost of 1 horse=! of $216=$24.
The cows cost $12 apiece, and The horses $24 apiece.
II.
174 FINKEL'S SOLUTION BOOK.
I. A man at his marriage agreed that if at his death he should leave onlv a daughter, his wife should have ^ 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 the estate should each receive, and what was each one's portion, if his estate was worth $6591? -1. ^=daughter's share.
2. ^=wife's share.
3. |=3 times f=son's share.
4. J_j_|._|_j_i?3=:i_w]loie estate.
5. $6591=whole estate.
6. /. ^=$6591. [estate.
7. i=T^ of $6591=$507=daughter's share,=^g- of whole |=3 times $507=$1521=wife's share,=T\ of whole es- tate, [tate.
|=9 times $507=$4563=son's share ,=3% of whole es-
( $507=Y*g- of whole estate=daughter's share. III. .-. ] $1521=T3¥ of whole estate=wife's share. ( $4563=T9^ of whole estate=son's share.
(Milne's Prac. A., p. 362, prob. 74.)
Note. — For a valuable critique, by Marcus Baker, U. S. Coast Survey, on this class of problems, see School Visitor, Vol. IX., p. 186.
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 ?
1. |=what one man removes in 1 hour. Then
2. 1g8=6 times |=what 6 men remove in 1 hour.
3. f=-j of |=what 1 man removes in 20 min., or -J- hour.
4. ¥=H times |=what 11 men remove in % hour.
5. /. lf — ¥=y=wnat runs on in 1 hr-~ i hr.=| hr.
Then II. J 6. £=1-^-7-f=what runs on in 1 hour. [commenced.
7. .*. lf — |=|=what was on the dock when the work
8. |=what 4 men remove in 1 hour.
9. .'. | — J=-J=part of coal removed every hour, that was
on the dock at first. 10. j=coal to be removed in |-s-£=5 hours.
(R. H. A., p. 406, prob. 90.) III. .-. It will take 4 men, 5 hours to clear the dock.
Explanation. — ^2— what 6 men remove in 1 hr. and 2e2=what 11 men re- moved in £ hr. In either case the dock was cleared. /. ^ — ^r^g4— amount of coal that ran on the dock from the shoot in 1 hr. — £ hr , or § hr. Hence in 1 hr. there will run on, **---%— 1*. Since \ run on in 1 hr. and ^ —the whole amount of coal removed in 1 hr., ^2 — |, or f must be the amount of coal on the dock when the work began. Since |=the amount 4 men rerrove in 1 hr. and |— the amount that runs on the dock in 1 hr., | — |, or | is the part of the original quantitv removea each hour. Hence, if \ is removed in 1 hour | would be removed in | -±-\, or hours.
ANALYSIS. 175
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.
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=^4x!20 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=TL of 1890 parts=original grass-fgrowth
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— i 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=24x9 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=18X10 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 Magazine, Vol. 1, No. 2.
I. A man and a boy can mow a certain field in 8 hours, if the boy rests 3| hours, it takes them 9£ hours. In what time can each do it?
176 FINKEL'S SOLUTION BOOK.
1. 9J hr. — 3f hr.=5f hr.=time they both work together in
the second case.
2. 8 hr.=time it takes them to do the work.
3. .-. -J=part they do in 1 hour.
4. -^=ff=5f times -J— part they do in 5f hours.
8
IL<[5. .'. || — f|=79a=part the man did in 3| hours, while the boy rested.
6. /. ¥3ff=— of -^j=part the man did in 1 hour.
df
7. .-. |_jj=part the man can do in ^-§-7-^ or 13-^- hours.
8. -£ — ^=£1-5=part the boy does in one hour.
9. .•. |-jj=part the boy can do in f ^-r-^V or 20 hours. TTT ( It will take the man 13-J hours, and
* / The boy 20 hours. (R. H. A., p. 402,prob. 30.)
I. Six men can do a work in 4£ days; after working 2 days, how many must join them so as to complete it in 3| days?
1. 4£ days=time it takes 6 men.
2. 26 days=6 times 4^- days=time it takes 1 man.
3. /. ^g-— part 1 man does in 1 day.
4. Yg=$ times gL=part 6 men do in 1 day.
5. yV=2 times T3^=part 6 men do in 2 days. [days.
6. Tf — T6^=y7^=part to be done in 3f days — 2 days, or If
7. Kj=T3TT:=IPart 1 man does in 1^ days.
II.
or 10 men can do in 1-| days. 9. .*. 10 men — 6 men=4 men, the number that must join
them.
III. /. They must be joined by 4 more men that they may com- plete the work in 3| days. R. H. A., p. 402, prob. 34.
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? '1. J^z^part drawn out each time.
2. T9^=part that was pure wine after the first draught.
3. y1^ of yVr=Tir7=:Part: wine drawn off the second draught.
4. ^ — r|7y=:T8^= part pure wine left, after the second
draught. [draught.
5. yg- of y8^ == T-| i-Q- = part wine drawn off at the third y8^ — yl^^y7^9^— part pure wine left after the third
draught. [draught.
TTT °f T7Tnnr==rJMir===Par^: w^ne drawn off at the fourth TTsV- TW£ff=TWinr= Part Pure wine left after fourth
draught. [fourth draught.
9. /. -j^nnr °^ 10 gal.=6.561 gal.=pure wine left after the
II.
draught.
9292
5. yV of (— ) ==—— ==part wine drawn off at the third
draught.
6. (-^) — ^r=(^} =part wine left after the third
PROBLEMS. 177
III. .*. There will be 6.561 gal. of pure wine in the keg after the fourth draught.
I. In the above problem, how many draughts are necessary to draw off half the wine?
1. T177==part wine drawn off at the first draught.
2. \ —I\=^=part wine left after the first draught.
Q
3. TV of T9¥— T-^— — — =part wine drawn off at the sec-
ond draught.
92 9 2
4> T9iF— ^TV(T=(T) =part wine left after the second
II.
draught. By induction,
7. (T^)n=rpart wine left after the nth draught.
8. /. 10(T9Tr)n=number of gal. left after the ^th draught.
9. 5=number of gal. left after the n\h draught. 10. .'. 10(T9Tr)n=5, whence
H- (T97y)n=i- Applying logarithms,
12. n log. TV=log. f
13. /. n = log. i-r-log. TV=-30103-:-. T.954243=.301030H-
.045757=6+.
III. /. In 7 draughts, half and a little more than half of the wine will be drawn off.
PROBLEMS.
1. A man bought a horse and a cow for $100, and the cow cost f 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 -J 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.
178 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 will it take 3 men and 2 boys to plow it?
Ans. J-j 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 time he received $30; how many days did belabor, supposing he received $2 per day for his labor? Ans. 22-J days.
9. The head of a fish is 4 inches long, the tail is as long as the head, plus -J 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 12 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 5 of the company withdrew by which each person's bill was augmented $-J; what was the bill? Ans. $15.
14. A man sold his horse and sleigh for $200, and f 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 money 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 -J of an hour; what was the time? Ans. 6 o'clock, P. M.
PROBLEMS. 179
18. 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 f 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 rate 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-J 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 f of B's age. Find their ages.
Ans. A's, 72 years; B's, 88 years; C's, 160 years.
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 f of A fills B, f of B fills C, and € fills T9g- 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? Ans. 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-J hr., how many boys must assist 16 men 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 wheat 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; ho\v much money had I at first? Ans. $60.
30. In a company of 87, the children are -J of the women, and the women f of the men; how many are there of each?
Ans. 54 men, 24 women, and 9 children.
180 FINKEL'S SOLUTION BOOK.
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 paying 5 cents, B 8 cen^s, 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 11 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 was 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-^ 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. A gets 42d., B 45d., C 48d., and D 5id., 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 35ffff da., B 33f \\\ da, C 31^H da-> and D 29fftl
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. 6TV hours.
ALLIGATION. 181
CHAPTER XVIII.
ALLIGATION.
1. Alligation is the process employed in the solution of problems relating to the compounding of articles of different values or qualities.
/I77. . . ( 1. Alligation Medial.
2. All^gat^on \ 2 Alli|ation Alternate.
I. ALLIGATION MEDIAL.
1. Alligation Medial 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?
120 lb. @5/=$6.00, 150 lb. @6/=$9.00, and 130 lb.
II
1. 2. 3. 4. 400 lb. is worth $28.00.
III.
5. .-. 1 lb. is worth $28-r-400=$.07=7 cents. .'. One pound of the mixture is worth 7 cents.
, ( Stod. Comp. A., p. 244, p™b. 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.
Dif.
Bal.
SOLUTION. — In average, the principle is, that the gains and loses are equal. ' We write the average price and the particular values 25,
'25X 27/ 30/
32X
45X
3X
I/
2/ 4X 17/
2 lb. 3 lb.
17 lb.
31b.
i
4 lb. 1 lb.
19 lb. 4 lb. 31b. 1 lb. 3 lb.
27, 30, 32, and 45 as in the margin. This is only a convenient
182 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 is a gain of 28/ — 27/, or I/; 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/; and one pound bought at 45/ and sold in a mixture worth 28/ there is a loss of 45/ — 28/, or 17/V 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 Ib. 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 Ib. Next, balance the 25-cent tea 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 ^ea to gain 51/, we must take as many pounds as 3/ is contained in 51/ which are 17 Ib. 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 Ib. Next, balance the 27-cent tea and the 32-cent tea. The L. C. M. of 1^ and 4/ is 4/. If we gain I/ on one pound of the 27-cent tea, to gain 4/, we must take as many pounds as I/ is contained in 4/, which are 4 Ib. If we lose 4/ on one pound of the 32-cenf tea, if 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 Ib. at 25/, 4 Ib. at 27/, 3 Ib. at 30/, 1 Ib. at 32/, and 3 Ib. at 45/. It is not necessary to balance them in any particular 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 proportionate 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.
183
Dif.
40/I25/
65/.
50/ 60X 75>
15/
r>p
W?
Bal.
2 bu.
2 bu.
2 bu.
2bu.
2 bu.
2 bu.
5 bu.
3 bu.
1 bu
9 bu.
-100 bu. 100 bu. 100 bu. .450 bu.
Dif. { Bal.
B. 65/.
40/ 50/ 60/
25/ 15/
2bu.
2 bu. 3bu.
10/
bu.
1100 bu. 2bu. 2bu. 254 bu.
SOLUTION. — In this solution, we proceed as in Case I. In A,. we obtain the relative amounts to be used of each kind, 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-i-2, or 50. Doing this, we obtain 100 bu. 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 first must always must be multiplied by 50, because we are to have 100 bu. at 40 cents per bushel.
(R. 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 will just float, that is, have a sp. gr. (1) the same as water?
r
3
3
x|
8
1 0
9
Ti
I J
oz.=2 Ib. 1 oz.
12 oz.
SOLUTION.— The specific gravity of any body is the ratio which shows how many times heavier the body is than an equal
184 FINKEL'S SOLUTION BOOK.
volume of water. Thus, when we say that the sp^inY 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 Jy of its own weight. Hence, to have equal weights of water and lead we must take only Jy as much lead as water. Now since a volume of water and y^ as much lead have the same weight, and in the proper combination have a volume of 1, since the sp. gr. oi the combination is 1, there is a loss of 1 — Jy, or T^, in volume on the part of the lead. For the same reason, there is a loss of f 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 T§- 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 -| volumes of cork. Now since we must always take 3 vol- umes of lead for every T-2- volumes of cork, it is evident that the weights of the substances are in the same proportion. Hence, we may say, we must take 3 oz. of lead with every y-f- oz. of cork, and 3 oz. of copper with every -f oz. of cork.
But we are to have only -J- oz. of copper. Hence, we must multiply the second balancing column by some number that will give us -J oz. of copper, /. £., we must multiply 3 by some number that will give us -J. The number by which we must multiply is -|~j-3— -g-. But multiplying -| by -^, we get ^4T oz. of cork. But we are to have altogether 12 oz. of cork. Hence we must yet have 12 oz. — 24T oz.= 3^T° oz. To produce this, we must multi- ply T^ by some number that will give 3f2T° oz. This number is S^O^H^SSJJ. But we must also multiply 3 by $fj. This will .give us 39^ oz.=2 lb. 7-§- oz. of lead. Hence, we must use 2 Ib. 7-j- •oz. of lead, so that the three will just float.
(jR. H. A., p. 339,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. (4.) 60%=the average selling price. (5.) / 120%=60%. (6.) 1^=^ of 60%=^%. (7.) 100%=100 times ^%=50%, the average
cost.
ALLIGATION.
185'
II.
120 shares @ 74%==
150 shares @ 68%=10200%.
130 shares @ 54%= 7020%. .-. 400 shares are worth 26100%, and 1 share is worth 26100%-r-400=65i% , the average.
3. 50%
40 % 65i%
10 %
shares.
10 shares.
X40=<
-610 shares. .400 shares.
III. .-. He must take 610 shares. (R. H. A., p. 339, prob. 8.)
Explanation. — Since 60% is the average selling price, and his gain is it is evident that his average cost is 60%-M.20, or 50%. In step 3, we find that the average cost of the 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 65 \%. But he has 400 shares at 65^%. Hence, we must multiply the balancing col- umn by 400-MO, or 40. This gives 610 shares at 40%.
CASE III.
To proportion the parts, the amount of the whole combination being given.
II.
III.
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. ?
300 bbl. 800 bbl. 400 bbl.
$7.50 a bbl.=$2250. $7.80 a bbl.=$6240. $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.— 1500
bbl., or 500 bbl. 8.30=$4150-f-500=the average value of 1 bbl.
8. .
9. $8.30
$8.00
$.302 bbl.l
X( 500^-5 )=
200 bbl. 300 bbl.
$8.50 $.203 bbl.| 5 bbl.
1. 200 bbl. at $8.00 per bbl. must be taken with
2. 300 bbl. at $8.50 per bbl.
(R. H. A., p. 339,prob.
I. A dealer in stock can buy 100 animals for $400, at the fol- lowing rates: calves, $9; hogs, $2; lambs, $1; how many may he take of each kind ?
186
FINKEL'S SOLUTION BOOK.
BaL
$1
$3i5 lambs.
3
10(17
24
31
38
45
62
59
$4
$2
<feoi $z,
5 hogs.
68
60
52
44
36
28
20
12
4
$9
$5:3 calves.
2 calves.
29
30
31
32
33
34
35
36
37
8
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-i-15=6f. .'• Multiplying each balancing column by 6|, will give 33£ lambs, 33£ 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 two 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 four 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. Multiplying the first column by f and the second by 3^, 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 30 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 y and the second by ^, 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.
SYSTEMS OF NOTATION.
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CHAPTER XIX.
SYSTEMS OF NOTATION.
1. A System of Notation is a method