LIBRARV OF THE University of California. GIK'P OK ^Accession 90.2.0.6 Class Manual Higher Digitized by tine Internet Archive in 2008 witin funding from IVIicrosoft Corporation http://www.archive.org/details/forcomatliematicsOOwalsricli Mathematics for Common Schools MANUAL FOR TEACHERS INCLUDING DEFINITIONS, PRINCIPLES, AND RULES AND SOLUTIONS OF THE MORE DIFFICULT PROBLEMS BY JOHN H. WALSH ASSOCIATE SUPERINTENDENT OF PUBLIC INSTRUCTIOlf BROOKLYN, N.Y. HIGHERABIIHMETIC ^B ft A a CAUFOg BOSTON, U.S.A. D. C. HEATH & CO., PUBLISHERS 1898 W3 ' . Copyright, 18»f^ Bt JOHN H. WALSH. J. S. Gushing & Co. — Berwick & Smith. Boston, Mass.. U.S.A. CONTENTS (Higher Arithmetic Manual.) 1 PAGE Introductory 1 Plan and scope of the work — Grammar school algebra — Con- structive geometry. II General Hints 5 Division of the work — Additions and omissions — Oral and written work — Use of books — Conduct of the recitation — Drills and sight work — Definitions, principles, and rules — Language — Analysis — Obj active illustrations — Approximate answers — Indicating operations — Paper vs. slates. XIII Notes on Chapter Ten 87 XIV Notes on Chapter Eleven 94 XV Notes on Chapter Twelve 117 XVI Notes on Chapter Thirteen 149 XVII Notes on Chapter Fourteen 185 iii IV CONTENTS XVIII PAG. Notes on Chaptee Fifteen 225 XIX Notes on Chapteb Sixteen 239 XX Notes on the Appendix 282 SUPPLEMENT. Definitions, Principles, and Rules i General definitions — Additions — Subtraction — Multiplication — Division — Factoring — Cancellation — Greatest common divi- sor — Least common multiple — Fractions — Decimals — Accounts and bills — Denominate numbers — Percentage — Profit and loss — Commercial discount — Commission — Insurance — Taxes — Duties — Interest — Partial payments — Bank discount — Ex- change — Equation of payments — Ratio — Proportion — Partner- ship — Involution — Evolution — Stocks and bonds — Notes, drafts, and checks. Answebs 1 MANUAL FOR TEACHERS I INTRODUCTOEY Plan and Scope of the Work. — In addition to the subjects generally included in the ordinary i^ext-books in arithmetic, Mathematics for Common Schools contains such simple work in algebraic equations and constructive geometry as can be studied to advantage by pupils of the elementary schools. The arithmetical portion is divided into thirteen chapters, each of which, except the first, contains the work of a term of five months. The following extracts from the table of contents will show the arrangement of topics : First and Second Years Chapter I. — Numbers of Three Figures. Addition and Sub- traction. Third Year Chapters 11. and III. — Numbers of Five Figures. Multipli- ers and Divisors of One Figure. Addition and Subtraction of Halves, of Fourths, of Thirds. Multiplication by Mixed Num- bers. Pint, Quart, and Gallon ; Ounce and Pound. Roman Notation. 1 MANUAL FOR TEACHERS Fourth Year Ohapters IV. and V. — Numbers of Six Figures. Multipliers and Divisors of Two or More Figures. Addition and Subtraction of Easy Fractions. Multiplication by Mixed Numbers. Simple Denominate Numbers. Roman Notation. Fifth Year Ohapters VI. and VII. — Fractions. Decimals of Three Places. Bills. Denominate Numbers. Simple Measurements. Sixth Year Ohapters VIII. and IX. — Decimals. Bills. Denominate Num- bers. Surfaces and Volumes. Percentage and Interest. Seventh Year Ohapters XI. and XII. — Percentage and Interest. Commercial and Bank Discount. Cause and Effect. Partnership. Bonds and Stocks. Exchange. Longitude and Time. Surfaces and Volumes. Eighth Year Ohapters XIII. and XIV. — Partial Payments. Equation of Payments. Annual Interest. Metric System. Evolution and Involution. Surfaces and Volumes. INTEODUCTORY 3 While all of the above topics are generally included in an eight years' course, it may be considered advisable to omit some of them, and to take up, instead, during the seventh and eighth years, the constructive geometry work of Chapter XVI. Among the topics that may be dropped without injury to the pupil are Bonds and Stocks, Exchange, Partial Payments, and Equation of Payments. Grammar School Algebra. — Chapter X., consisting of a dozen pages, is devoted to the subject of easy equations of one unknown quantity, as a preliminary to the employment of the equation in so much of the subsequent work in arithmetic as is rendered more simple by this mode of treatment. To tea.chers desirous of dispensing with rules, sample solutions of type examples, etc., the algebraic method of solving the so-called " problems " in per- centage, interest, discount, etc., is strongly recommended. In Chapter XV., intended chiefly for schools having a nine years' course, the algebraic work is extended to cover simple equations containing two or more unknown quantities, and pure and affected quadratic equations of one unknown quantity. No attempt has been made in these two chapters to treat algebra as a science ; the aim has been to make grammar-school pupils acquainted, to some slight extent, with the great instru- ment of mathematical investigation, — the equation. Oonstructive Geometry. — Progressive teachers will appreciate the importance of supplementing the concrete geometrical instruction now given in the drawing and mensuration work. Chapter XVI. contains a series of problems in construction so arranged as to enable pupils to obtain for themselves a working knowledge of all the most important facts of geometry. Applications of the facts thus ascertained, are made to the mensuration of surfaces and volumes, the calculation of heights and distances, etc. No attempt is made to anticipate the work of the high-school by teaching geometry as a science. 4 MANUAL FOR TEACHERS While the construction problems are brought together into a single chapter at the end of the book, it is not intended that instruction in geometry should be delayed until the preceding work is completed. Chapter XVI. should be commenced not later than the seventh year, and should be continued throughout the remainder of the grammar-school course. For the earlier years, suitable exercises in the mensuration of the surfaces of triangles and quadrilaterals, and of the volumes of right parallelopipedons have been incorporated with the arithmetic work. II GENERAL HINTS Division of the "Work. — The five chapters constituting Part I. of Mathematics for Comiinon Schools should be completed by the end of the fourth school year. The remaining eight arithmetic chapters constitute half-yearly divisions for the second four years of school. Chapter I., with the additional oral work needed in the case of young pupils, will occupy about two years; the re- maining four chapters should not take more than half a year each. When the Grube system is used, and the work of the first two years is exclusively oral, it will be possible, by omitting much of the easier portions of the first two chapters, to cover, during the third year, the ground contained in Chapters I., II., and III. Additions and Omissions. — The teacher should freely supple- ment the work of the text-book when she finds it necessary to do so ; and she should not hesitate to leave a topic that her pupils fully understand, even though they may not have worked all the examples given in connection therewith. A very large number of exercises is necessary for such pupils as can devote a half-year to the study of the matter furnished in each chapter. In the case of pupils of greater maturity, it will be possible to make more rapid progress by passing to the next topic as soon as the previous work is fairly well understood. Oral and Written "Work. — The heading "Slate Problems" is merely a general direction, and it should be disregarded by the teacher when the pupils are able to do the work " mentally." The use of the pencil should be demanded only so far as it may 5 6 MANUAL FOR TEACHERS be required. It is a pedagogical mistake to insist that all of the pupils of a class should set down a number of figures that are not needed by the brighter ones. As an occasional exercise, it may be advisable to have scholars give all the work required to solve a problem, and to make a written explanation of each step in the solution ; but it should be the teacher's aim to have the majority of the examples done with as great rapidity as is con- sistent with absolute correctness. It will be found that, as a rule, the quickest workers are the most accurate. Many of the slate problems can be treated by some classes as "sight" examples, each pupil reading the question for himself from the book, and writing the answer at a given signal without putting down any of the work. Use of Books. — It is generally recommended that books be placed in pupils' hands as early as the third school year. Since many children are unable at this stage to read with sufficient intelligence to understand the terms of a problem, this work should be done under the teacher's direction, the latter reading the questions while the pupils follow from their books. In later years, the problems should be solved by the pupils from the books with practically no assistance whatever from the teacher. Conduct of the Eecitation. — Many thoughtful educators consider it advisable to divide an arithmetic class into two sections, for some purposes, even where its members are nearly equal in attainments. The members of one division of such a class may work examples from their books while the others write the answers to oral problems given by the teacher, etc. Where a class is thus taught in two divisions, the members of each should sit in alternate rows, extending from the front of the room to the rear. Seated in this way, a pupil is doing a different kind of work from those on the right and the left, and he would not have the temptation of a neighbor's slate to lead him to compare answers. GENERAL HINTS 7 As an economy of time, explanations of new subjects might be given to the whole class; but much of the arithmetic work should be done in "sections," one of which is under the im- mediate direction of the teacher, the other being employed in "seat" work. In the case of pupils of the more advanced classes, "seat" work should consist largely of "problems" solved without assistance. Especial pains have been taken to so grade the problems as to have none beyond the capacity of the average pupil that is willing to try to understand its terms. It is not necessary that all the members of a division .should work the same problems at a given time, nor the same number of prob- lems, nor that a new topic should be postponed until all of the previous problems have been solved. Whenever it is possible, all of the members of the division working under the teacher's immediate direction should take part in all the work done. In mental arithmetic, for instance, while only a few may be called upon for explanations, all of the pupils should write the answers to each question. The same is true of much of the sight work, the approximations, some of the special drills, etc. Drills and Sight Work. — To secure reasonable rapidity, it is necessary to have regular systematic drills. They should be employed daily, if possible, in the earlier years, but should never last longer than five or ten minutes. Various kinds are sug- gested, such as sight addition drills, in Arts. 3, 11, 24, 26, etc. ; subtraction, in Arts. 19, 50, 53, etc. ; multiplication, in Arts. 71, 109, etc. ; division, in Arts. 199, 202, etc. ; counting by 2's, 3's, etc., in Art. 61 ; carrying, in Art. 53, etc. For the young pupil, those are the most valuable in which the figures are in his sight, and in the position they occupy in an example ; see Arts. 3, 34, 164, etc. Many teachers prepare cards, each of which contains one of the combinations taught in their respective grades. Showing one of these cards, the teacher requires an immediate answer 8 MANUAL FOR TEACHERS from a pupil. If his reply is correct, a new card is shown to the next pupil, and so on. Other teachers write a number of combinations on the blackboard, and point to them at random, requiring prompt answers. When drills remain on the board for any considerable time, some children learn to know the results of a combination by its location on the board, so that frequent changes in the arrangement of the drills are, therefore, advisable. The drills in Arts. Ill, 112, and 115 furnish a great deal of work with the occasional change of a single figure. For the higher classes, each chapter contains appropriate drills, which are subsequently used in oral problems. It happens only too frequently that as children go forward in school they lose much of the readiness in oral and written work they j)ossessed in the lower grades, owing to the neglect of their teachers to continue to require quick, accurate review work in the operations previously taught. These special drills follow the plan of the combinations of the earlier chapters, but gradu- ally grow more difficult. They should first be used as sight exercises, either from the books or from the blackboard. To secure valuable results from drill exercises, the utmost possible promptness in answers should be insisted upon. Definitions, Principles, and Eules. —Young children should not memorize rules or definitions. They should learn to add by adding, after being first shown by the teacher how to perform the operation. Those not previously taught by the Grube method should be given no reason for " carrying." In teaching such children to write numbers of two or three figures, there is nothing gained by discussing the local value of the digits. Dur- ing the earlier years, instruction in the art of arithmetic should be given with the least possible amount of science. While prin- ciples may be incidentally brought to the view of the children at times, there should be no cross-examination thereon. It may be shown, for instance, that subtraction is the reverse of addition, and that multiplication is a short method of combining equal GENERAL HINTS "' " ""^ 9 numbers, etc. ; but care should be taken in the case of pupils below about the fifth school year not to dwell long on this side of the instruction. By that time, pupils should be able to add, subtract, multiply, and divide whole numbers ; to add and sab- tract simple mixed numbers, and to use a mixed number as a multiplier or a multiplicand ; to solve easy problems, with small numbers, involving the foregoing operations and others contain- ing the more commonly used denominate units". Whether or not they can explain the principles underlying the operations is of next to no importance, if they can do the work with reasonable accuracy and rapidity. When decimal fractions are taken up, the principles of Arabic notation should be developed ; and about the same time, or some- what later, the principles upon which are founded the operations in the fundamental processes, can be briefly discussed. Definitions should in all cases be made by the pupils, their mistakes being brought out by the teacher through appropriate questions, criticisms, etc. Systematic work under this head should be deferred until at least the seventh year. The use of unnecessary rules in the higher grades is to be deprecated. When, for instance, a pupil understands that per cent means hundredths, that seven per cent means seven hun- dredths, it should not be necessary to tell him that 7 per cent of 143 is obtained by multiplying 143 by .07. It should be a fair assumption that his previous work in the multiplication of common and of decimal fractions has enabled him to see that 7 per cent of 143 is ^ of 143 or 143 X .07, without information other than the meaning of the term " per cent." When a pupil is able to calculate that 15 % of 120 is 18, he should be allowed to try to work out for himself, without a rule, the solution of this problem : 18 is what per cent of 120 ? or of this: 18 is 15% of what number? These questions should present no more difliculty in the seventh year than the following examples in the fifth : (a) Find the cost of ^ ton of hay at $12 per ton. (b) When hay is worth $12 per ton, what part of a 10 MANUAL FOR TEACHERS ton can be bought for $ 1.80 ? (c) If ^ ton of hay costs $1.80, what is the value of a ton ? When, however, it becomes necessary to assist pupils in the solution of problems of this class, it is more profitable to furnish them with a general method by the use of the equation, than with any special plan suited only to the type under immediate discussion. In the supplement to the Manual will be found the usual defini- tions, principles, and rules, for the teacher to use in such a way as her experience shows to be best for her pupils. The rules given are based somewhat on the older methods, rather than on those recommended by the author. He would prefer to omit entirely those relating to percentage, interest, and the like as being unnecessary, but that they are called for by many success- ful teachers, who prefer to continue the use of methods which they have found to produce satisfactory results. Language. — While the use' of correct language should be insisted upon in all lessons, children should not be required in arithmetic to give all answers in " complete sentences." Espe- cially in the drills, it is important that the results be expressed in the fewest possible words. — Sparing use of analyses is recommended for begin- ners. If a pupil solves a problem correctly, the natural inference should be that his method is correct, even if he be unable to state it in words. When a pupil gives the analysis of a problem, he should be permitted to express himself in his own way. Set forms should not be used under any circumstances. Objective Illustrations. — The chief reason for the use of objects in the study of arithmetic is to enable pupils to work without them. While counters, weights and measures, diagrams, or the like are necessary at the beginning of some topics, it is important to discontinue their use as soon as the scholar is able to proceed without their aid. GENEKAL HINTS 11 Approximate Answers. — An important drill is furnished in the " approximations." (See Arts. 521, 669, 719, etc.) Pupils should be required in much of their written work to estimate the result before beginning to solve a problem with the pencil. Besides preventing an absurd answer, this practice will also have the effect of causing a pupil to see what processes are necessary. In too many instances, work is commenced upon a problem before the conditions are grasped by the youthful scholar ; which will be less likely to occur in the case of one who has carefully " estimated " the answer. The pupil will frequently find, also, that he can obtain the correct result without using his pencil at all. Indicating Operations. — It is a good practice to require pupils to indicate by signs all of the processes necessary to the solution of a problem, before performing any of the operations. This fre- quently enables a scholar to shorten his work by cancellation, etc. In the case of problems whose solution requires tedious processes, some teachers do not require their pupils to do more than to indicate the operations. It is to be feared that much of the lack of facility in adding, multiplying, etc., found in the pupils of the higher classes is due to this desire to make work pleasant. Instead of becoming more expert in the fundamental operations, scholars in their eighth year frequently add, subtract, multiply, and divide more slowly and less accurately than in their fourth year of school. Paper vs. Slates. — To the use of slates may be traced very much of the poor work now done in arithmetic. A child that finds the sum of two or more numbers by drawing on his slate the number of strokes represented by each, and then counting the total, will have to adopt some other method if his work is done on material that does not permit the easy obliteration of the tell-tale marks. When the teacher has an opportunity to see the number of attempts made by some of her pupils to obtain the correct quo- 12 MANUAL FOR TEACHERS tient figures in a long division example, she may realize the importance of such drills as will enable them to arrive more readily at the correct result. The unnecessary work now done by many pupils will be very much lessened if they find themselves compelled to dispense with the " rubbing out " they have an opportunity to indulge in when slates are employed. The additional expense caused by the introduction of paper will almost inevitably lead to better results in arithmetic. The arrangement of the work will be looked after ; pupils will not be required, nor will they be permitted, to waste material in writing out the operations that can be per- formed mentally ; the least common denominator will be deter- mined by inspection ; problems will be shortened by the greater use of cancellation, etc., etc. Better writing of figures and neater arrangement of problems will be likely to accompany the use of material that will be kept by the teacher for the inspection of the school authorities. The endless writing of tables and the long, tedious examples now given to keep troublesome pupils from bothering a teacher that wishes to write up her '•ecords, will, to some extent, be discontinued when slates are n-* longer used. XIII NOTES ON CHAPTER TEN The formal study of algebra belongs to the high-school ; but some so-called arithmetical problems are so much simplified by the use of the equation that it is a mistake for a teacher not to avail herself of this means of lightening her pupils' burdens. In beginning this part of her mathematical instruction, the teacher should not bewilder her scholars with definitions. The necessary terms should be employed as occasion requires, and without any explanation beyond that which is absolutely neces- sary. 849. Very young pupils can give answers to most of these questions ; so that there will be no need, for the present, at least, of introducing a number of axioms to enable the scholar to obtain a result that he can reach without them. 850. Pupils will learn how to work these problems by work- ing a number of them. They may need to be told that x stands for 1 X ; and that, as a rule, only abstract numbers are used in the equations, the denomination — dollars, marbles, etc. — being supplied afterwards. While the scholars should be required to furnish rather full solutions of the earlier problems, they should be permitted to shorten the work by degrees, writing only whatever may be necessary. 4. a:-|-2a: = 54. 8. a: -f 2 a: 4- 6 ar - 27000. 5. x-\-bx=lS. 9. x-\-bx=-12. 6. 7a;-f 5.r=156. 10. rr -f 2 a: -f- 3 a; = 54. 7. ^x~^x = m. XI. a:-}-6^ = 42. 87 88 MANUAL FOR TEACHERS 12. 2a; + 10a: = 96. 13. Let X — the fourth ; then 4 a; = the third, 12 a; = the second, and 24 a; = the first. a: + 4a:-f-12a:4-24a: = 41. 14. X =the second, 2 a; = first, 9 a: = third. 15. 5a: + 4a; = 81. 17. 4a; = 340. 16. 24 a; = 456. 19. 3 a; + 4 a: =175. 20. Let X = each boy's share ; 2 a; = each girl's share. 2a; + 4a; = 240. 21. a; = number of days son worked; 2 a: = number father worked. 3 a; = son's earnings ; 8 a: = father's earnings. 3a;4-8a;=165. 22. a; = number of dimes; 2 a: = number of nickels; 6a: = number of cents. (10xa;) + (5x2a;)+(lx6a:) = 78, or 10a; + 10a:4-6a; = 78. 23. 15a;— 12a; = 75. 24. a; + 4a; + a: + 4a; = 250. 25. Let X = cost of speller ; then 3 a; = cost of reader. 26. Let a; = smaller ; then 5a: = larger. 27. Let a: = Susan's number ; 2 a; = Mary's; 3 a: = Jane's. 851. 10 : 1 a; is the same as -• ^ 3 852. Pupils already know that J means 3 h- 4, so that they can understand that — means 3 a: -h- 4, or 4 of 3a:. When 4- of 4 something (3 a:) is 24, the whole thing (3 a:) must be 4 times 24, |a 4 or 96 ; that is. when ^ = 24, 3 a: = 96. When ^ = 24, 2y = 24 X 3, or 72. o When ~ = 20, 4 2 = 20 X 5, or 100. 5 From these examples can be formulated the rule for disposing of a fraction in one term of an equation, which is, to multiply NOTES ON CHAPTER TEN 89 both terms by the denominator of the fraction. In changing the 3 a; first term of the equation, — = 24, to 3 a;, it has been multi- plied by 4, so that the second term must also be multiplied by 4. 853. In solving these examples by the algebraic method of " clearing of fractions," attention may be called to its similarity to the arithmetical method. To find the value of y in 2, the pupil multiplies 8 by 5 and divides the product by 2 ; as an ex- ample in arithmetic, he would divide 8 by ■§-, that is, he would multiply 8 by -|; the only difference being that by the latter method he would cancel. While -r^ = 8 may be changed to ^ = 4 by dividing both terms o 5 by 2, beginners are usually advised to begin by '* clearing of fractions," short methods being deferred to a later stage. 854. 6 may be written ^ -j- 5£ = 92. 8. 2|-a; should be reduced to an improper fraction, making the 23a; equation, -— - = 115. Make similar changes in 12, 14, 18, and 20. 8 855. 2. a; -1-^ = 100. ^- i + f = ?'2a: + a: = 267. 6. ?^-?^=:15. 9. Let 5 a; = numerator \lx = denominator. 7 a; — 5 a; = 24 ; 2 a; = 24 ; a; — 12. The numerator, 5 a;, will be 5 times 12, or 60 ; the denominator will be 84 ; and the fraction, fj. Arts. 10. Let X — greater ; - = less. a; + ^=:480. 90 MANUAL FOR TEACHERS Clearing of fractions, 7x-\-x = 3360, 8a; = 3360, X = 420, the greater number, ^ = 60, the less. Or, , let a: = less ; 7x = greater, a: + 7a: = 480, 8 a: = 480, X = 60, the less, *lx = 420, the greater. The employment of the latter plan does away with fractions ib the original equation. 11. 30a:-a: = 522, ora:-;^=522. 30 13. Let a: = number of plums; 4a: = number of peaches. Then 2a: will be cost of plums, and 12a: the cost of the peaches. 2a:+12a: = 70. 16. a:-?^=:80. 7 17. a:-?^-? = 24. 8 4 18. X + l^a; -j-(^xx^) = 15. a: + ^ + 5a: = 15. 19. Let X = price per yard of the 48-yard piece ; 2a: = price per yard of the 36-yard piece ; 48a: will be the total cost of one, and 72 a:, of the other. 48a: -I- 72a: = 240. 20. 160 a: -f 120 a: = 840. 856. The pupils should be permitted to give these answers without assistance. In Art. 857 is explained what is meant by " transposing." 2a;- - «=i«+|-| .2a;- -36 = 96 + 3a;-2a; 2x- -3a; 4-2a;=: 96 + 36 11a;- 132 a; = 12 NOTES ON CHAPTER TEN 91 858. While these exercises are so simple that they can be worked without a pencil, they should be used to show the steps generally taken in more complicated equations. , 07 _ e^ In 1, for instance, the work should take the _ form here indicated, only a single step being _ taken at a time. In 19, the first step is to clear the equation of fractions by multiplying by 6 ; the second step is to transpose the unknown quantities to the left side of the equation, and the known quantities to the right ; the third step is to com- bine the unknown quantities into one, and to make a similar combina- tion of the known quantities; the last step is to find the value of x. After a little more familiarity with exercises of this kind, the pupil can take short cuts with less danger of mistakes ; for the present, however, it will be safer to proceed in the slower way. 859. 5. a; + (a;+75)4-a:-f(a; + 75) = 250. a; + a; -f- a; 4- a; = 250 - 75 — 75. Note. — The parentheses used here are unnecessary. They are employed merely to show that a; + 75 is one side of the field. 6. a; + (a; + 8) = 86. 9. a; + a; + 72 = 96. 7. a; + a; + 318 = 2436. 10. a;-f-f =45. o 4 8. a;+ 1+7 = 100. 11. X = one part ; 2a; — 6 = other part. a; + 2a;-6 = 45. 12. a; = John's money ; a; + 5 = William's money. 3a;+15 + 5a;=103. 92 MANUAL FOR TEACHERS 13. Let X = price of a horse ; a; — 80 = price of a cow ; 4a; = cost of four horses ; Sx— 240 = cost of three cows. 4a; 4- 3a; -240 = 635, 7 a; = 635 + 240 = 875, X = 125, price, in dollars, of a horse ; a; — 80 = 45, price, in dollars, of a cow. Other pupils may solve the problems in this way : X = price of a cow ; a; -j- 80 = price of a horse. 3a;+ 4a; + 320 = 635, 7 a; = 635 -320 = 315, X = 45, price, in dollars, of a cow ; a; + 80 = 125, price, in dollars, of a horse. 14. a; = number of dimes; a; -f- H = number of five-cent pieces; 10a; = value of dimes (in cents); 5 a; -}- 55 = value of five-cent pieces. 10 a; + 5a; + 55 = 100. 15. a; = greater ; a; — 48 = less. a; + a;-48 = 100. Or, X = less ; a; + 48 = greater. a; + a; + 48 = 100. 17. a; = share of the first ; X + 2400 = share of the second ; X + 2400 + 2400 = share of the third. a; + a; + 2400 + a; + 2400 + 2400 = 18000. 18. Let X = less ; a; + 33 = greater. a; + 33 -3a: =11. Bringing known quantities to the left side of the equation, and the unknown quantities to the right, 33-ll = 3a;-ar, 22 = 2a;, 11 = a;. ( -CrNiTEKSITT I NOTES ON CHAPTER tt^Wl^CA U fOT^>^ 93 Or, a; -3a; =11 — 33, -2a; = -22. Changing signs of both terms, 2rr=22, a; =11. This problem may also be worked in this way : X — less ; 3 a: + 11 = greater. 3a;+ll-a;=33. 19. X = number of 5-cent stamps ; a; + 15 = number of 2-cent stamps ; x -}- SO = number of postal cards. 5 a; + 2 a; + 30 + a: + 30 = 100. 20. a; = number of horses; a; + 17 = number of cows; 2a;-f-39 = number of sheep. a; + a;+17 + 2a; + 39 = 88. XIV NOTES ON CHAPTER ELEVEN With this chapter begins the regular work in percentage, and it is important that the pupils obtain, as soon as possible, a correct idea of what is meant by the term per cent. Many of the various subdivisions of this topic found in some books, are taken up only incidentally, while others are omitted altogether, the aim being to give the pupils a foundation upon which they can subsequently build, rather than to scatter their energies over too great a diversity of subjects. 860. The reduction of a common fraction to a per cent, con- sists in changing the former to a decimal of two places. In re- ducing ^ to a decimal, the result is .5, or 5 tenths ; in changing ■J- to an equivalent per cent, the result is 50 per cent, or 50 hundredths. In reducing ^ to a decimal, the answer is given in three places, .125, or 125 thousandths ; in changing it to a per cent, the division is stopped at the second place, and the remain- der written as a fraction, 12^ per cent, or 12^ hundredths. The denominator of a per cent being always the same, 100, the comparative value of several per cents is known at sight. To compare f and f as common fractions, they must be changed to }4 and J^; if a further comparison is to be made between these and -f^, a new common denominator must be employed, and the fractions reduced to -^^^ -^^ and ■^. Changing the fractions to decimals, 625 thousandths, 6 tenths, and 58^^ hun- dredths, simplifies the comparison ; but it is still easier to de- termine their relative value when they are expressed as 62J per cent, 60 per cent, and 58J per cent. NOTES ON CHAPTER ELEVEN 95 The teacher must not be discouraged if the pupil fails to grasp at once the full meaning of percentage, Definitions will not help materially ; much practice in working examples is necessary to give the knowledge desired. 863. Many children find it diflicult to distinguish between ^% and 50%. If the former is read in the business way, -J- of one per cent, it may make the distinction plainer. 864. Per cents being generally given in two figures, scholars hesitate to give the correct answers: 300%, 250%, 125%, 16331%, 420%, 910%. 865. While pupils will find 33|% of 81 cows, by dividing 81 by 3, they should understand that they are really multiplying 81 by 33^ hundredths, or 81 by |. In 4, 6% of 150, or ^^ of 150, may be obtained by multiplying 150 by 6 and cutting off two ciphers ; or by dividing 150 by 100, obtaining 1^, and multi- plying this quotient by 6 ; or by reducing 6% to -^, and finding -^ of 150. In 9, the pupil should find 1% of $ 640 and take one- half of the result. The scholars should be permitted to use their own method of solving these problems, the different analyses given by the pupils furnishing their class-mates an opportunity to select a simpler method. Although every pupil may not be able to determine at once the shortest way of calculating a given example, no one should be allowed to work 3, by multiplying by 33^. "When the multiplication by ^ has been performed, the answer has been ob- tained, except as to the location of the decimal point, and the waste of time in multiplying by 3, repeating this product, and adding three columns should not be tolerated. No fault should be found with the average pupil for failing to recognize in 1, that ^i% ^s ttJ ^^ *^^^ ^^ ^^' ^i% i^ "TO"- ^^^^ general method should be to multiply by the figures given to represent the per 96 MANUAL FOR TEACHERS cent, except in such cases as 12J%, 16|%, 25%, 33J%, 37J%, 50%, and possibly a few others. ^ . Where the given numbers are used, they should be made the multipliers and expressed as hun- (uqq a^. '. — ~ dredths. Nothing is gained in 5, by reducing ^\,, i.oD ^jjg -J- to a decimal; although in 13, writing '. — I 5^%, .055, might make it easier for some. 868. The rule generally given of finding the percentage, by multiplying by the rate expressed as hundredths, is here modified to the extent of using the common fraction to express hundredths, instead of the decimal, as being more in conformity with early algebraic methods. The teacher that prefers to ascertain the base or the rate by the older arithmetical method, will omit 30-41. 30. _^x65 = ^. Am. 35. a:-f-f=132; etc. 100 20 5 31. 15^ = 26; etc. 37. a;-f=78; etc. 20 3 ' 32. lof a; = ~ Am. 38. -^of ? = — Am. 4 4 100 3 150 33. - = 42; etc. 39. -£-=^5; etc. 4 150 3 1 . X 41. -^ = 23; etc. 34. . + l-Am. 40. _of. = ^. ^n.. 800 42. Let X = rate. Then -^ of 65 = 26, or -^ X 65 = 26. 100 100 In an equation containing quantities to be multiplied, the mul- tiplication should be performed before the equation is cleared 65ar \Sx of fractions. This equation becomes — — = 26, or -— - = 26, luu ^0 it being immaterial whether canceling be done or not. NOTES ON CHAPTER ELEVEN 97 43. After a little experience with this class of examples, the equation may be written at once, in the order in which the terms are given : 24 = -l|-of a;,ori|^ = 24. 100 ' 100 44. ?^of a: = 180, or^-180. 100 ' 2 45. a: + ? = 85. 4 .-3 X n 4: X 3 46. — = of -, or = — 5 100 5' 125 5 While the algebraic method is of no advantage to the bright scholar, it makes the employment of rules unnecessary in the case of the ordinary pupil. 48. xX-^Xl=^- 49. ii^ = 44. 100 200 200 50. fx|. 51. f|f = 33. 52. i of 800 = 100 ; i% of 800 - 1. 63. $175 + i of $175. 64. 2i% of a; - 12.50 ; that is, ^ = 12.50. 56. 6i X .16. 56. 8i = -^of ?; thatis,-^ = ^. ^100 3 150 3 57. -^ of 389.50 = 124.64 ; §§M2^= 124.64 ; 100 100 389.5a; =12464; 3895 a; = 124640. 58. ?5£= 174.04 ; 95 a; = 17404. 100 59. ^ + f|=1276. 98 MANUAL FOR TEACHERS 60. 984 = ^ of X ; that is, ~ = 984. 100 ' 3 62. ^=386.75. 4 65. X = cost of oats ; x -\- — — = 1071. Divide the cost of the oats by 30^ to find the number of bushels. 68. Assessed value = | of $ 48000 = $ 32000. Taxes on 32 thousand dollars = $18.50 X 32. 869. In giving answers to these and to all other exercises, no " guessing " should be allowed. The pupil should be permitted to obtain the correct result in his own way — that is, no inflexible rule should be given him to follow — but he should be able to get the answer, using the algebraic method if that seems to him the easiest, as it may be in some instances. The examples are not arranged by " cases," so that each will have to be understood before it can be worked. The careless pupil will probably give the wrong answer to 13 ; saying 6, instead of 600 ; he will be likely too, in 14, to use the larger number as a divisor, and to obtain 44f % instead of 225%. These mistakes are less likely to occur if he uses equations — 3 = — — and — - = 20i. Even those scholars that have solved 200 100 * in their arithmetic work of the lower grades, examples similar to 16, will have new light thrown on their method by using the equation, x -{- - = 20. In mental work, however, the first term 5a; should be made — — , to reduce it in size, so that it can be more 4 easily remembered. 24 is simplified by changing the fractions to whole numbers — ^ is what per cent of fj, 9 is what per cent of 10 — before beginning to calculate the rate. In 25, H and 6| become f and ^, f and ^, 9 and 40. 870. 1-5 can be worked by the pupils without any explana- tion ; 6-20 present more difficulty. The beginner in algebra NOTES ON CHAPTER ELEVEN 99 desires to start at once with his x, without any preliminary cal- culations; and the usual method of treating these examples requires him first to ascertain the gain or the loss before com- mencing his equation. The formula employed in the first five examples is : Cost X — — = gain or loss. When the pupil knows any two of these three terms, he can calculate the third ; and 6-15 furnish data from which the necessary two items can be obtained. The pupil must, however, be careful in 11, for instance, not to subtract the loss from the selling price to obtain the cost. In the following equations, cost X —-- is made equal to the gain or the loss. No canceling has been done. 6. ?50^==18;6. = 18. 7. ^^^ == m. 100 100 8. §^20^^ 12.93, or ?5?^= 1293. 100 100 9. ?5^i^= 181.68. 100 10. §i^=5.25; 84 a: = 525. 100 11. §i£=5.25. 12. ^^^-25. 100 100 13. ^^ = 43.75; 875^ = 4375. 100 14. 934.56^^^^ ^^ 1012.500:^^^3^^^^ 100 100 In 16-20, the cost is represented by x. 16. a: + ^ = 468.75. 18. a: + -= 1646.08. 4 3 17. a: -|= 73.84. 19. rr-i^=204. u lUU 20. 100 21. Gain = i of $275. 22. x% of 60 = 15. 23. :r + f=960. 5 100 MANUAL FOR TEACHERS 27. Saj^ + iof 33^^. 28. a:-i^=33.60. 100 29. Gain = 2J% of $8760. 30. x- — --- — = 70. 10 4 10 24. rr%of32 = 16. 31. 6000 = a: % of 16000. 25. x% of 175 - 25. 32. 6000 = x% of 24000. 26. x% of 200 = 25. 33. 1600 + 2J% of 1600. 34. 4200-3^% of 4200. 871. In 1, the 30 cu. yd. are reduced to cubic feet by multi- plying by 27. Instead of performing the different multiplications, they are merely indicated, so that work may be saved by can- celing. Although 2 should be a simple problem for a bright pupil, it is apt to prove puzzling unless an x is introduced. A paste- board box may be used to represent the walls and the ceiling of a room, the sides and the top being then opened out to permit of its representation on the blackboard. 3. The area in square feet = ^ of 132 X 110. This is reduced to acres by dividing by 9 X 30 J X 160. 132x110x4 _ . 2 X 9 X 121 X 160 4. Number of strips == 6 yd. h- 27 in. = 6 yd. -^ f yd. = 6 X f . 7. The " development " of the fence will be represented by four adjoining rectangles, each marked 6 ft. high, the lengths being 25 ft., 100 ft., 25 ft., and 100 ft, respectively, the whole forming a rectangle 6 ft. X 250 ft. 8. A board's area in square feet = 12x^ = 6. Dividing number of square feet in the fence by 6, gives the number of boards. NOTES ON CHAPTER ELEVEN 101 9. The cost of a square foot is obtained by dividing $181.50 by (160 X 30^ X 9) ; this, multiplied by (300 X 200), gives the cost of the plot. $181.50x4x300x200 160 X 121 X 9 The amount received for the lots will be $160 X 6. 10. Number of cakes = (320 X 160) ^ (4 X 2). 11. Number of cubic feet == 320 X 160 X l^-. 12. (320 X 160 X 1^) ^ (15 X 32). 13. Number of square feet originally = 640 X 440. For build- ing purposes, there will be four pieces, each measuring 300 ft. by 200 ft. 14. The difference between the above areas will represent the number of square feet in the streets. 872. Many of these exercises can be used for mental and sight work. For methods of solution, see Art. 870. 873. As a preliminary to the formal study of interest, the teacher will need to see that her pupils understand what is meant by the term. She can explain that a person borrowing money should pay for its use, just as a person who rents a house, etc. 874. In changing 4 mo. 10 da. to the fraction of a year, many teachers prefer to reduce the time to days and to write the result over 360, -^f^, leaving the reduction to lowest terms for the subsequent cancellation. In the same way, 1 yr. 5 mo. 15 da. is changed to (360 + 150 + 15) da., or 525 da. = f||- yr. The reduction to days is done very rapidly. 875. 1. $750xTf^x|. 5. $360x^^X3^^^. 2. $84.75 x-r^X it. 6. $ 94.43 x yj^ X #^. 3. $308.25 Xy^X^V 7. $400x^X|fi. 4. $ 464.75 X yf 7 X Iff. etc., etc. 102 MANUAL FOR TEACHERS 877. The teacher should explain that a person that owes money, frequently gives a note as an acknowledgment of the debt, etc. 878. There is no general method applicable to these problems. 1. Interest for a year is $12, or $1 per month, which gives 5? 19 for 1 yr. 7 mo. 2. $3.60 per year is 1^ per day, 33^ for 33 da. 3. $ 6 per year, or $ 9 for 1^ yr. 4. $ 6 per year, or $ 15 in 2 yr. 6 mo. 5. If $ 50 produces $6 in 2 yr., it will produce $3 in 1 yr. ; rate, therefore, is 6%. 6. $ 18 per year is $1 for 20 da., or Jg- yr. 8. 4% per year-- 1% for 90 da. ; 1% of $150 = $1.50. 9. 5% per year = |% for 36 da. ; \% of $240 = ^ of $2.40. 11. $1 is 100% of $1 ; at 5% per year it will take 20 yr. to make 100%. 12. At 6% it will take 16|yr., or 16 yr. 8 mo., to make 100%. ' 13. Disregarding $14.90, it will take 25 yr. at 4% to make 100%. 14. \% per month = 8% for 16 mo. ; 8% of $90 = $ 7.20. 15. 5% for 360 da. = 1% for 72 da. 16. 360 da. -^ 4^ = 720 da. ^ 9 => 80 da. 17. 5% for 1 yr. = 1% for 72 da. ; 1% of $ 75 -= 75 cents. . 18. l%of$63. 20. ^% of $840. 22. 1% of $275. 19. l%of$570. 21. 1% of $150. 23. 2% of $360. 879. 1. 30 rd. 5 yd. 1 ft. = 511 ft. ; 8 rd. 4 yd. 2 ft. = 146 ft.; iH = f ^^• 2. Number of feet deep = (36 X 5) h- (6 X 4). NOTES ON CHAPTER ELEVEN 103 3. 3 mi. 96 rd. = 1056 rd. ; 3 hr. 16 min. = Sj\ hr. ; 1056 rd. X 3^ = 1056 rd. X f f - lif^^ rd. = 3449f rd. = 10 mi. 249f rd. Ans. 4. (1 + J) X (f X ^) ^ (f X ^) = ^ X f X ^ X 4 X /j 8. The first two figures express 1800 ; the second two, 5.4. 22. $48.37 -^8f 880. 2. Provisions that will supply 450 men for 5 months will supply 5 times 450 men for 1 month, and will supply (5 times 450 men) h- 9 for 9 months, or 250 men. The number that must be discharged = 450 men — 250 men — 200 men. Ans. 15. ^-^-1^=19500. 100 100 16. D bought J X f X I X f of the ship. 18. 100 a; + 50^ = 340x75. 19. X = number distributed by each new man ] 2x = number distributed by each experienced man. 16a; + 32a; = 36000. 20. a; ~?= 1972.65. 4 881. 10. 100 cents -^1.13. 883. 3. $1.10+15% of $1.10. 4. 9876 _ 3^ J 45 9876 = 87.2; + 45. 5. 640 is what per cent of (640 + 560), etc. 6. 43 gal. 3 qt. 1 pt. = 43|- gal. ; $70.20 -h-43|- = Ans. 7. (48 X 32) -^ (16 X i). 8. 20 is what per cent of 160? 20 is what per cent of 180? 10. Selling price per bbl. = ^ = ?^ ; 44% = — = — • ^^ ^ 600 4 ^' 600 24 X 23 Let X = cost per bbl. x = — ^ 24 4 104 MANUAL FOR TEACHERS U. 9:075^ =.24.2. 100 884. 425 + 99 is 1 less than 425 + 100 ; 425 + 999 is 1 less than 425 + 1000. 885. 565-99 is 1 more than 565-100; 1424-999 is 1 more than 1424 - 1000. 886. 24 X 21 = (24 x 20) + (24 x 1). See Art. 786. 887. 16-^ .25 = 16 H- J = 16 X 4 ; 36 ^ .75 = 36 ^ J = 36 Xj = 12x4. 888. 7i^| = ^^| = \ll2i£ = 484.15. 200 15. Principal = $97.57 -$7.57 = $90. 90 X y^ X a; = 7.57. 21. Let a; = principal. , a: X 4 X 846 _ 195 92^ ^ 100x360 The interest is then found by subtracting from $ 196.92, the value obtained for x. 22. First find the principal (x). 25. See 15. 934. The recommendation so frequently made, that all writ- ten work be preceded by oral exercises of the same character,, should not be followed without some modifications. Oral work is necessarily accompanied by a number of devices that tend to simplify the task of handling numbers that are not seen ; written work should follow general rules in order to be learned by a majority of the pupils, although later they may adopt some of the short-cuts of their oral exercises. Even the oral addition of two numbers of two figures each, is done in a manner different from the ordinary slate method, the operation in the former case being commenced generally with the tens* figures, and in the NOTES ON CHAPTER TWELVE 119 latter case with the units' figures. The reduction to a common denominator recommended in oral division of fractions, is seldom employed in slate work. The average scholar is able to handle " mental " problems con- taining small numbers in a way that he cannot always explain, although he may endeavor to stultify himself by repeating a prescribed form of analysis. It is next to impossible, with the average teaching, to get the same pupil to work some varieties of " written " problems containing the same conditions. In order to furnish a general method of treating some classes of examples, it has been thought best to commence with written work, leaving the mental exercises with their various devices until the former task is accomplished. The accomparving exercises are so simple as not to need ex- planation by the teacher ; but sufficient time should be given the pupil to work them out in his own way. They differ in this respect from the oral examples of a single operation containing larger figures, but which do not require any effort on the part of the scholar to determine which process is required. 1. Yearly interest is $6; a year and a half will be needed to make the interest $ 9. 2. The yearly interest is $ 8, making the rate 4%. 3. Yearly interest is $4, requiring a principal of f 100, at the given rate. 5. The pupils may remember (Art. 878, No. 15) that 6% for a year is 1 % for 72 da. 6. 4% per year is 1% for 90 da. 11. 2 mo. 12 da. = 72 da. See 5. 12. 1% for 80 da. is (360 -^ 80) % for a year. 17. 2% for 6 mo. 18. $ 3.60 per year is 1 cent per day. 20. See 18. 120 MANUAL FOR TEACHERS 935. First payment = -, leaving — remainder ; second pay- ment = -J- of -— = -, leaving - remainder ; third payment = f of ^ = ^ ; last payment, = $ 2000. The total cost of the house, x = 3 5 the sum of the payments, q + o + H "^ 2000. o o O 936. The books contain many methods of calculating interest, but it is questionable whether it is not time wasted in giving so much attention to this topic. The average person is required to do comparatively little work in this line ; while those called upon to compute interest often, learn short methods of their own or use interest tables. If a second method is to be taught at all, the one by aliquot parts is the most useful, as modifications of this method may be applied to other operations. 6. See Arithmetic, Art. 384. 937. A modification of the so-called " 60-day method." 16. See Art. 901 as to days of grace. 938. 21. 10% gives 2 years' interest; then 1 yr. (J of the foregoing) ; 6 mo. ; 1 mo.; 18 da. (^^ of 6 mo.). 942. 46. Term, 57 da. (54 da.). 47. Term, 92 da. (89 da.). 49. Term, 34 da. (31 da.). 48. Term, 16 da. (13 da.). 50. Term, 187 da. (184 da.). 943. 9. See Table, Arithmetic, Art. 1303. 944. 11. The net price of goods catalogued at x dollars, and sold at a discount of 20 and 10%, will be (^ -^> or ^\ _/. ^80a;\_80a: 8a: _ 72 a; ^ ^ y^^ 100 J 100 100 100 ■ NOTES ON CHAPTER TW! 13. If tlie selling price of the above is $360,^=360; 12 x^ 36000 ; x = 500. Catalogue price = $ 500. Am. 14. 750 - (-^ of 750) = 500 ; 500 - C"^ of 500^ = 500 - 5 a; = net price. 500 -5a; = 450. Transposing, — 5 a: == — 50. Changing signs of both terms, 5 a; = 50, a;=10. 945. 7. Let x — selling price of muslin. (84 X 40) + 105 a: =-- (84 X 55) + (105 X 20). Another way : He loses 15^ per yard on 84 yd., which is a loss of 15^ X 84. This he must make up on 105 yd., which is (15^ X 84) -T- 105 on each yard, or 12^. Selling price of muslin, 20)2^+12^, or 32^. Ans. 8. •J of them brought $120; \ of remainder, or J of them, brought $96; -J- of remainder, or J of them, brought $40; re- mainder, or i of them, brought $30. Total amount received, $286. 9. Proceeds of gas stock, $25 X 165 = $4125. Cost of lots, $4125 - $27 = $4098. Number of square feet in lots, (32 X 115) + (30 X 105) = 3680 + 3150 = 6830. Value per square foot, $4098 -^6830 = $0.60. Ans, 10. Two walls, each 16 X 14, and two others, each 12 X 14, contain (32 + 24) X 14, or (56 X 14) sq. ft. = 784 sq. ft. The ceiling contains (16 X 12) sq. ft. = 192 sq. ft. Adding this to the walls, makes a total of 976 sq. ft. The deductions are (8 X 4) sq. ft. X 2, and (7 X 3) sq. ft. X 3, or 64 sq. ft. + 63 sq. ft. = 127 sq. ft. Number of square feet to be plastered = 976 - 127 = 849. Cost at ^^ per square foot = 2^X849 = $16.98. Ans. 122 MANUAL FOR TEACHERS 946. 1. A can do \ of the work in 1 hr., and B can do | of it in 1 hr. ; together they can do in 1 hr. (| + 1) of the work, or ^ of it ; and to do the whole work it will take as many hours as -^ is contained times in 1. l^« = lxff = ff-2H. Am.2^houT8. 2. Commission of 2^% ^ :^ of amount collected = $1.60. Amount collected = $ 1.60 X 40 = $64. Amount remitted = $64 -$1.60 = $62.40. Ans. 3. i% of (f of $ 12000) = i% of $9000 = J of $90 = $22.50. Ans. Note. — It may be advisable to explain to the pupils that property is seldom insured for its full value, because it is not Hkely that a fire will completely destroy a building, and insurance companies reimburse the per- son insured, only to the extent of his loss. 4. 32xa; = 6x4; 32a; = 24; a; = |f = f. Ans. f yd. or 27 in. 5. 5% for 360 days = 1% for 72 days = 2% for 144 days. 2% of $87 = -4n5. 6. 2% of $176. 7. Let a; = commission ; 40 a: = amount invested; a:-f40a: = 41 a; = 8200 ; x = 200. Ans.^ 200. 8. $500 is -J of cost, $4000. 9. Let a; = loss, or 20% of cost; 5a; = cost; 5a; — a; = 4a; = selling price. X, the loss, is i of selling price, 4 x. 10. Let a; = gain, which is 20%, or ^, of the cost of the goods ; 5 a; = cost ; 5 a; -|- a;, or 6 a;, = selling price. X, the gain, is ^ of selling price, 6a;. Note. — The amount of money given in these two examples, $1200, does not affect either result. It may be used or not, as the pupil prefers. 11. 3 men earn $72-5-8 in one day, or $3 per day each. 5 men earn $15 a day, or $165 in 11 days. NOTES ON CHAPTER TWELVE 123 12. 3 quarters of the cost, or ^, = 225. Cost = |300. By selling for $325, there is a gain of $25, or -^ of the cost. -^^ = 8J%. Arts. 13. 2| yd., orf yd. cost 40^; 1 yd. costs 40 ^-^f, or f of 40^ = Ib^. 4 yd. 1 ft., or 4^ yd., cost 15^ x 4^. 947. The following is the solution without days of grace : Let X = face of the note. Then, ^xAx^ = ^ = dkcount; X = proceeds = 1000. 200 ^ 200 a; -a; = 200000, 199 a; = 200000, Face of note = $ 1005.03. Arts. Proof. Face of note, $ 1005.03 - Deduct 30 days' discount, i%, 5.03 — Proceeds, $1000.00 949. 1. When days of grace are omitted, the term of dis- count is 90 da. 10. Find the term, and add the number of days to March 15. 950. 2. (a) 1 trillion, 500 billions, etc. 5. The first quarter of 1888 contained (31 + 29 + 31) da., or 91 da. The man was employed 60 da., and unemployed 31 da. His $3 additional paid the expenses of the working days. Deducting $ 2 X 31, or $ 62, for the expenses of the other days, his net income = $ 350 - $ 62 = $ 288. Ans. 124 MANUAL FOR TEACHERS 6. Apr. 16, 79 to Mch. 19, '86, 83^ mo., @ $8, $664.80 Mch. 19, '86 to Mch. 4, '87, llj " " 12, 138.00 Apr. 16, '79 to Sept. 1, '80, 1^ " " 2, 33.00 Apr. 16, '79 to Nov. 22, '82, 43^ " " 2, 86.40 Ans. $922.20 10. Last quarter's salary = $287 = 82% of previous quarter's salary = ^ = 2S7; x = 350. 100 Salary of three quarters @ $ 350 = $ 1050 ; add last quarter's, $287. Total for year, $1337. Ans. 9SL See Art. 784. 953. To find 19 times 91, subtract 91 from 20 times 91, or 1820-91. 82 X 19 = (82 X 20) - (82 x 1). 51 X 29 = (51 x 30) - 51. 27 X 99 = 2700 - 27. 954. See Art. 706. 675 -^ 37| = 6f -4- f = 54 -^- 3. 955. 136 X J = 136 - (^ of 136) = 136 - 17. 290 X 1^ = 290-29. 64^^ 5 =12^=12^. 22 Xl9| = (22 X 20) - (22 X ^). 45 X 9|| = (45xl0)-(^of45). 160 ^ 1^ =320h-3. 18^^ 1| = 94-^9 = 10f 956. 3. At 50^ each, the cost would be $8; at 1^ apiece less, the cost is $8.00 - $.16 = Ans. 4. (I of 100 lb.) X 27 = 100 lb. X (f of 27) = 100 lb. x V = 100 lb. X 20^. Note. — The 100 should not be used until the end; even then, 20^ is changed to 2025 without thinking of multiplication, J being considered 25, and annexed to 20. See Art. 649. NOTES ON CHAPTER TWELVE 125 6. 900 ^ 75 = (900 -f- 100) ^ (75 ^ 100) = 9 -^- j. 7. See Art. 955. 9. (10i-x4)+iofl0i = 42 + 5i = 47i. 12. i of (33 X 42) = 33 X 21. 15. $16i^$li = 65-f-5. 20. 16ix2i- = (16ix2) + (iofl6i). 957. 1. -i^5^of(27/x56x37ixf|). 2. [($4,875 X 17350) -^ 196] + [$4.9375 x 122.75] + [$ .0825 X 2240 x 2i]. 3- :r-"^a: = 49739.55f. 12. Duty - [^ of (55 ^ x 45 x 38)] + [20)^ x (45 X 38 X ff )]• 958. In multiplying by 427, the first figure of the product by 42 (7 X 6) is placed under the 2; in multiplying by 832, the first figure of the product by 8 is placed under the 8, and the first figure of the product by 32 (8 X 4) is placed under the 2. 959. These exercises contain some examples worked by short methods explained in previous chapters. See Arts. 650, 714, 791, 792, and 891. 960. See Arithmetic, Art. 384. 964. 2. It won 17 games out of 30. 3. 1600%. 4. "I is what per cent of -i^? ^ is what per cent of ■§- ? y^ is what per cent of ^^ ? 5 is what per cent of 6 ? f = SS^% . 6. The deduction of the first discount leaves 80% of the list price; the deduction from this of 10% of itself leaves 90% of 80%, or 72%. 7. One fills -J- of tank in 1 hr., the other fills ^ in 1 hr. ; both together fill ^ + ^ in 1 hr., or -^ + A' ^^ A-i ^^ fill fl of tank, it will take 24 hr. h- 7 = 34 hr. 126 MANUAL FOR TEACHERS 8. 6% for 60 da. = 80^; for 12 da., ^ of 80^ or 16^; for 72 da., 80^ + 16)^^-96^. Or, $4.80 for year, and ^ of $4.80 for 72 da. 9. 16%= A; 420 = 1^; etc. 965. 5. Selling price = | of $1.50 = $1.12| ; gain = 22J^ = J of 90^. 6. Selling price = $9.60, a reduction of $2.40 from marked price, or i of $12, or 20%. 7. The rug is sold for $24. If this is f of marked price, the latter is $30. 8. See 7. Note. — It is not to be expected that all the pupils' work will be short- ened to this extent, but the majority of the class should be able to give answers at sight to these four examples. 9. Find y^ ^^ ^^^ 2s. 6c?. by compound division ; do not reduce to pence. 21 X 966. 4. Let X = profits first year ; then — — - = profits second 91 T 2^ year; :r 4-^ = 6970. 5. I wish to gain 15% of $.96, or $.14|-, which makes my selling price $.96 + $.14| = $1.10f Let X = marked price. ._15£ = 1.104; etc. Or, writing all the foregoing in one equation : 100 Vioo J 85 a; = .96x115, ^ .96 X 115 ^ = —85— NOTES ON CHAPTER TWELVE 127 7. The average pupil will be able to obtain the meanings of these terms by inquiries of his parents, friends, etc. ; and he will remember much longer what he learns in this way, than if he finds the answer in the text-book. As the penalties for taking usurious interest vary in the different states, the teacher should ascertain the law of her own state in this matter. See Art. 1306. A tax bill or a policy of insurance brought in by a pupil and described, will add to the interest. The teacher should not spend too much time upon details that have no relevancy in her sec- tion of the country. Poll taxes, for instance, should not be dwelt upon in cities in which they are not collected ; etc. 9. The use of the hogshead as a measure of 63 gal. is fast becoming obsolete. The term " barrel," to indicate 31^ gal., is occasionally used in giving the capacity of large tanks, etc. The U. S. authorities require prices to be stated in the currency of the country from which the articles are exported ; but as this would make the problem more complicated, the text-books generally give prices in U.S. money. No allowance is now made for leakage, the quantity actually imported being ascertained by measuring. 10. A port of entry is a place in which there is a custom house, established by the government. 967. 4. Any principal — 1 150, $575, or |343.75 — will double itself at 5% in (100 ^ 5) yr. 12. The pupils will need to obtain a correct idea of the mean- ing of the word " premium " in this connection, as they will find it used differently when they come to the study of Bonds and Stocks. The premium is the amount paid to the company assum- ing the risk. 968. 11. $3500 is raised on property worth $1750000; the rate is $3500 ^ $1750000 = 2 mills on $1. The man's property tax = 2 mills X 24000 = $48; adding to this 1 poll tax, at $2, gives his total tax of $ 50. Ans. 128 MANUAL FOR TEACHERS 969. 2. Multiply the denominators of the first and the third ; divide the numerators of the second and the fourth. 5. While pupils in lower grades may be permitted to reduce both amounts to pence, it is now time to use a shorter method. The sums given may be changed to 38Js. and 5Js., or J-Ps. and -^s. 970. 1. See diagram, Arithmetic, Art. 897, Problem 2. 18 X + 15 a; + 18 :r + 15 a: -f (18 X 15) = 63 X 9, 66 a; + 270 = 567, 66 a; -567 -270 = 297, x = ^. Am. ^ ft. The arithmetical solution is apparent from the foregoing. The sides and the bottom contain 63 sq. yd., or 567 sq. ft. The bot- tom contains (18 X 15) sq. ft., or 270 sq. ft. The sides contain 567 sq. ft. -270 sq. ft. = 297 sq. ft., which is the area of four rectangles, whose bases measure 18 ft., 15 ft., 18 ft., and 15 ft., respectively, the total being 66 ft. 297-^66 gives 4^ as the number of feet in depth. 2. There are 4840 sq. yd. in an acre. (4840 X 3) -?- (| of 242). Cancel. 3. The area of a trapezoid is found by multiplying one-half the sum of the parallel sides by the perpendicular distance between them. See diagrams, Arithmetic, Art. 929, Problems 6 and 9 ; and Art. 1265, Figs. 9 and 10. 4. The area = (^^^^) X 60 = 30 a; + 3000. Am. 30 a; + 3000 = 5400 ; a; = 80. Am. 80 yd. 5. (m^y,=mx.Am. 100 a: = 4000 ; a: = 40, Am, 40 yd. NOTES ON CHAPTER TWELVE 129 6. /'£±|±i5Ax60 = (a; + 20)x60 = 60:r+1200. Am. 60 a; +1200 = 6000; a; = 80; a; + 40 = 120. Am. 80 yd. and 120 yd. 9. The number of square feet in the walls of a room 16^ ft. long, 14f ft. wide, and 13J- ft. high, may be obtained by adding the bases of the four sides, — 16^ + 14f -f 16^ + 14| = 62|, — and multiplying this by their common height, 13^. Dividing by 9 gives the number of square yards. The operation of finding the cost may be indicated as follows : 10^X125X40^25000^^^^^^^ ^n.. $9.26. 9x2x3 27 -^ ^ ^ 10. (20 X 17-J-) -^- 2^ gives the number of feet of carpet. Di- viding this result by 3 gives the number of yards. 11. 41i lb. X (15f X H X i). Cancel. 12. The "development" will be a modification of the one given in problem 20, Arithmetic, Art. 818. In drawing the development, the pupil should be required to approximate the proper proportions, and to place the faces in the proper order. It is not necessary to have the top and bottom faces in the posi- tions shown in Art. 818. The surface of the four vertical faces should be obtained in one operation, as in 9 ; also the surface of the two horizontal faces : [2 X (3f + 2i) X li] -f- [2 X (3i X 2^)]. 13. (135^-^12J)ft. 14. [(128 X 152 X 105) -^ 2150.4] bu. 15. [(77x45x54) ^231] gal. 16. 40 acres = (160 X 40) sq. rd. = 6400 sq. rd. The dimen- sions are 80 rd. by 80 rd., making 320 rd. of fence necessary. There will be 640 posts, at 15^ each; and 5 times 640 rails, or 3200 rails, at 10^ each. 130 MANUAL FOR TEACHERS 17. There will be 16 fields, 4 rows of 4 fields each. Five parallel fences, each a mile long, and five other parallel fences of the same length, and perpendicular to the first, will be required. 19. 1728 cu. in. of water weigh 1000 oz. ; 1 cu. in. weighs (1000 ^ 16) lb. -^ 1728 ; 231 cu. in. weigh [(1000 X 231) ^ (16 X 1728)] lb. 21. Number of square feet = (320 + 210 + 320 + 210) X 6. Divide by 9 to obtain square yards. 22. Area of outer rectangle in square feet = 332 X 222 ; of inner rectangle — (320 X 210) sq. ft. Divide the difference by 9 to obtain square yards. 23. Area of outside plot = (320 X 210) sq. ft. ; of inner plot = (308 X 198) sq. ft. 972. 1. Each yard measured with the short yardstick con- tains If yd. ; the true length = 25 yd. X ff . Or, each so-called yard is ^^ yd. short, and 25 yd. are ff yd. short ; and the piece contains 25 yd. ~ -||- yd. = 24^ yd. Ans. 4. 32 boys = 20 men. If 15 men do the work in 12 da., 20 men do it in 12 da. X \^. 20. Change 6 lb. 14 oz. and 23 lb. 12 oz. to ounces, or to pounds and fractions. 976. 3. A and B can mow 4- of the field in 1 da., all three can mow -J- of the field in 1 da. C mows (^ — ^) of the field in 1 da., or -^ of it ; in 5 da. he does -^ X 5, or -J of the work, for which he should receive ^ of $25. 4. 5bu.(g 80)2^ -400^. 5 bu. (§60)2^-300^. a;bu. @ 30^ = 30 a;^. (x+ 10) bu. @ 50^ = (30a; + 700)^, 50 a: -1-500 = 30 a: + 700, 20 a; = 200, X = 10. Ans. 10 bu. of oats. NOTES ON CHAPTER TWELVE 131 7. Total cost -= (65 ^ X 128) + 80 ^ ; quantity sold = 128 gal. - 16 gal. Selling price per gallon = f of [(65^ X 128) + 80^] -^(128-16). It is advisable to accustom children to understand that a gam of ^ of cost makes the selling price -| of cost. 10. On sofas sold for $1125 there was a loss of -|-, making the selling price -| of cost. On the remaining sofas, the selling price, $1125, represents f of cost. 4 fifths of cost = $1125, selling price of 25 sofas. 1 fifth of cost = $2811, loss on first lot. 6 fifths of cost = $1125, selling price of remaining sofas. 1 fifth of cost = $ 187^ = gain on second lot. Loss on the transaction, $281.25 - $ 187.50. The following is an algebraic solution without fractions : Let X = loss on first lot ; then 5 a; = cost of first lot. ^x-~x = 4:X — selling price = 1125. X = 281^ = loss in dollars. Let X ■= gain on second lot ; 5 a: = cost of second lot. hx-\- x='^x = selling price — 1125. X = 187-|- — gain in dollars, etc. 11. Let X — cost per egg. 18 a; = cost of 18 eggs. — ^ = selling price per egg. 18 X 1 X ' T ' Gain per egg = x = —-; that is, on x cents I gain -^ of X cents. A gain of -^ of cost = -^^^i of cost = 63^^%. Ans. x-^ = 2. 132 MANUAL FOR TEACHERS 12. Let X = marked price. 100 18. Let X == cost of the horse. "^ "^ 20 "" ^^ ^^^ ^^'^® "" "20" He sold it at ^ of asking price ; i.e., ■— of — ^, or — -^. 399^^275. 400 20. Length of field in rods = (1600 + 146) ^ 18. 21. Let 10 ar represent cost, then loss = a:; and selling price -^ 9 a; = $ 117 ; a: = 13 ; cost = $ 130. A gain of 10% would be $13, making the price at which he should have sold it to gain 10% -$130 + $13 = $143. Ans. 22. Wife receives — -; son, « ^^o' °^ ~q"' ^^-ugli^^r, $5000. ^+^+5000 = a:. 977. The ten problems of this section will call for no special treatment. An occasional problem of this kind has already been given, although, perhaps, with smaller numbers. After the pupils understand in 1, that the joint capital, $700, is the basis upon which the profits are distributed, they will have no diflS- culty in understanding that B is entitled to fJJ of $182, and that C is entitled to fJ-J of $182. Cancellation should be em- ployed. See Art. 1121, No. 5. There is no need in this connection of discussing the subject of business partnerships that are continued for a year or longer. The division of profits in these cases is the subject of a special NOTES ON CHAPTER TWELVE 133 agreement, and is rarely made solely on the basis of the amounts invested. A yearly gain of $4000 made in regular business by two partners, one of whom invested $1000, and the other $3000, might be divided in various equitable ways. The partner invest- ing the larger sum, might first take out $ 120 as interest on his excess of capital ; and the remaining $ 3880 might be equally divided, giving one of them $1940, and the other $2060. Another arrangement might permit each partner to withdraw a fixed sum for services, say $1000, leaving $2000 to be divided on the basis of 1 to 3. This would make the shares ($1000 + $500) and ($1000 + $1500), or $1500 and $2500. 10. Cases of this kind are found only in the books. 979. The cost of the first item is given. The second item contains 451 sq. ft. ; the third contains (4 X 42 X 7-J-) sq. ft., or 1204 sq. ft. ; the fourth contains (3 X 43 X 7|) sq. ft., or 989 sq. ft.; the total of the three items being 2644 sq. ft. The cost @ lOd. is £110- 3-4 First item, 24- 7-8 £134-11 Less 2^% (^V). 3- 7-3^ + Ans. £ 131 - 3 - 8| The fraction of a penny in the discount is ■^, which is nearly 1 farthing, written \d. 980. German currency being a decimal one, the bill is com- puted in the ordinary way. The duty is found by reducing the marks to dollars by multiplying by $.238, and taking 35% of the result. 981. Too much stress cannot be laid upon the importance of requiring children to estimate the probable answer to every " written " problem before placing a figure on paper. The mere drill on the " approximations " found in the text-book is of com- 134 MANUAL FOR TEACHERS paratively little value, unless it leads pupils to the employment of this device throughout all of their work. Besides preventing a pupil from making a very serious mistake in an ordinary com- putation, the habit of careful reading that is necessarily formed by the scholar that is not satisfied with a simple guess, will tend to make his methods simpler and more accurate. He will learn to apply to the apparently more difficult numbers of the written problem the processes employed in solving the compara- tively simple " mental " questions. While all of the pupils should not be expected to give the same answer to each of these examples, they should gradually approach more and more closely to the correct result. 1. 480 is what per cent of 960 ? 2. 52 bu. is about 3 times (17 bu. 37 lb.). 3. 500 cu. ft. -^ 128 cu. ft. Less than 4 cords. 4. 120 cu. ft. -^ about 1^ cu. ft. 5. 1500 sq. rd. -^ 160 sq. rd. Less than 10 acres. 6. A little less than 70 X 70. 7. (6 X 4 X 5) X about 7|-. 8. 64 -^ ^-^, or 64.3 -^ ■^. More than 643. 9. About £ 200 @ f 4.80 to £. 10. About 4 marks to $1. Over 400 marks. 982. The teacher that is allowed nny discretion should omit all problems relating to Bonds and Stocks. The average gram- mar-school pupil cannot be made to understand the subject without the expenditure of more time and energy than should be given to a topic that he will learn by himself when he grows older if he gets the proper foundation. If, however, the course of study requires that this topic be taken up, the teacher should aim to interest the pupils by making some local corporation the basis of the work. A certificate of stock should be obtained, or at least a copy made, which might be placed upon the blackboard. NOTES ON CHAPTER TWELVE 135 The scholars should be led to see that large undertakings, such as the construction of a railroad, the building of water-works, and the like, require more money than any individual might have, or would care to risk. It then becomes necessary to interest a number of persons that will be willing to invest more or less money in the new enterprise. It is found, for instance, that $ 50000 will be needed to build and equip the street rail- road mentioned in 1. The projectors divide this amount into 500 shares, each of which represents a one-five-hundredth interest in the profits. It may happen sometimes that it is considered advisable to interest people of small means, and who are unwill- ing to take a $ 100 share. In these cases the original (par) value of the shares may be fixed as low as $10 each. When the par value is not given, it is understood to be $100. In the distribution of profits, the owner of 10 shares is entitled to 3^4 of $2000, or $40. 2. These profits are generally distributed annually, semi- annually, or quarterly. Before the time comes for " declaring " the dividend, the directors of the company meet and determine how much money shall be thus distributed. It may be con- sidered advantageous to reserve a portion of the profits for the purchase of new cars, or for the extension of the road, etc. ; so that the amount distributed at any time does not necessarily include all that has been gained. The dividend is generally announced as a per cent of the capital, which in this case is $50000; so that the semi-annual dividend is 4%, equal to 8% per year. 3. The owner of a $100 share, on which he receives $8 per year, is not likely to be willing to sell it for $100 if he can obtain only $4 per year interest on that sum of money deposited in a savings bank. The person desirous of obtaining stock after it is reasonably certain that the railroad is going to prove suc- cessful, will have to pay more than $100 per share. Mr. H. pays $150 per share, or 150% of the par value. 136 MANUAL FOR TEACHERS 4. Mr. H. receives 4% dividend on $3000, or $120. From the savings bank he would obtain 2% on $4500, or $90. 5. The $120 semi-annual dividend is 2|% on the $4500 invested, equivalent to 5^% per year. 6. The words " per cent " are not used in stating the price of stock. A $100 share at 164^% is worth $164|-; 30 shares are worth $1641x30. 7. Assuming the par value of each to be $ 100, the first pays $6 per year on $150, or 4% ; the second pays $7 per year on $175, or 4%. The par value of $100 is assumed for convenience; a par value of $50 would make the cost of a share of gas stock 150% of $50, or $75, and its annual dividend would be 6% of $50, or $3, the rate on the amount invested being 4%. 8. The buyer of stock at 125 wishes to receive 4% of $125, or $5. As the dividend is based on the par value ($100), the rate must be 5% per year, or 2^% semi-annually. 9. [93|-% of ($50 X 17)] + [102f % of ($10 X 143)]. 10. A person that desires to buy stocks is not always likely to know where he can find any for sale; so he goes to a stock- broker, who makes a business of buying and selling stocks on commission. This commission is a small per cent of the par value, the charge of ^% for buying or selling a share of the par value of $100 being 12^^, whether the actual value of the stock be $150 or $50. This broker receives, therefore, \% of [($50x17) + ($10x143)]. 11. A bond is a note issued by a corporation, and is generally secured by a mortgage on its property. It is a much larger document than the note of an individual, and frequently contains at the bottom a number of " coupons," one for each half-year's interest, upon which is engraved the date when due and the sum NOTES ON CHAPTER TWELVE 137 payable. A 10 years' U. S. 4% coupon bond for $ 1000 would have 20 coupons, each worth $20 when due. At the expiration of each 6 months, the holder of the bond cuts off the proper coupon and presents it for payment. A " registered " bond con- tains no coupons, a check for the interest being mailed to the owner. Although the railroad company in this example receives only $95 for each $100 bond, it promises to pay $4 interest per year, and $ 100 at the end of 20 years. In considering the rate of interest received by the owner of such a bond, it is not customary to complicate the example too much by requiring the pupils to take into account the additional $5 above the cost received when the bond is redeemed at the end of 20 years, although buyers of bonds include it in their calcula- tions. For our purpose, at present, it will be sufficient to assume that the purchaser of one of these $ 100 bonds receives $ 4 on each $95 invested, the rate being 400^95, or 4^^%. 12. Omitting the question of redemption, at which time the purchaser for $116.50 would receive only $100, the rate is 400-116^, or 3ifi%. The holder of a U. S. bond knows that the face value of the bond will be paid in full at maturity, and that the interest pay- ments will be made on the dates when due ; in the case of the bond of a railroad, or the like, there is always the possibility that something may occur to prevent the company from meeting its obligations. . 13. The rate of income from stocks may vary at each dividend period, depending upon the amount of business done, etc. ; the rate of income from bonds is fixed as stated on their face. Bonds are redeemed at the time specified ; there is no reason why a successful company should sell out and divide the proceeds among its stockholders. When, however, the property of a cor- poration is sold, the claims of bondholders and all other obliga- tions must be satisfied before the stockholders receive anything. 138 MANUAL FOR TEACHERS 14. The stock-broker's fee is called brokerage, or commission. 15. A cotton-mill obtains material through a cotton '* factor " ; property is purchased through a real estate agent ; a grocer may- buy butter, eggs, etc., from a commission merchant, the seller in each case remitting the amount received to the owner after deducting his fee, or commission. 16. The hose in insurance is the sum for which the property is insured; in taxes, it is the assessed value of property; in brokerage, it is the par value of stocks or bonds ; in commission, it is the sum for which goods are bought or sold ; the principal is the base upon which interest is calculated ; the face of the note is the base in bank discount ; in commercial discount, the gross price is the base in the first instance, the base for each subsequent discount being the successive remainders left after the deduction of the previous discounts ; in stocks and bonds, the base is the par value. 17. The assessed value of property is the value for purposes of taxation, and is fixed annually by ofl[icers chosen for this duty, generally called assessors. 18. 2J% of assessed value (a;) = $540; assessed value = $24000. This is | of the actual value, or 66|%. For various reasons, the assessed value is placed below the sum that would be realized by the sale of the property under favor- able conditions ; but care is taken that all property is assessed upon the same basis. 19. If all property were assessed at its actual value, the same amount of taxes would be produced by a lower rate. To obtain $540 taxes on property assessed at $36000, the rate would be 21. 1674 ft. = 558 yd. ; 558 yd. ^ 5^ yd. = 1116 half-yd. ^11 half-yd., which gives 101 rd. and 5 half-yd. = 101 rd. 2^ yd. - 101 rd. 2 yd. 1 ft. 6 in. 22. $ 8575 H- $ 245 =r 35 = number of shares. Quarterly divi- dend = 2^% of ($100 X 35). NOTES ON CHAPTER TWELVE 139 983. 27. Each figure of the product is written two places to the right of the corresponding figure of the multiplicand. Principal, $375. '0 11.25 $386.25 Amount I yr. 3% 11.5875 $397.8375 Amount 1 yr. %% 11.9351 $409.7726 Amount 1-J- yr. 12.2931 $422.0657 Amount 2 yr. 1% ) 4.2206 \% i 2.1103 $428.3966 Amount 2 yr. 3 mo. Principal, 375. $53.3966 Interest 2 yr. 3 mo. Am. $428.40, amount; and $53.40, interest. It is not necessary throughout the work to carry the multi- plication beyond four places of decimals. 985. 6. After deducting the first 25%, or i, f of list price remains ; a second discount of \ of this remainder leaves } of this remainder, or J of f of list price, or -^^ of list price = 90^. List price = 90^-^3^ = 90^Xi/ = $ 1.60. 7. All together can do (i + -g- + i) of the work in 1 hr. 8. Selling price, $60 == f of value of cow ; \ of value = $60 -^3 = $20, the loss. 9. Selling price, $60 = | of value of cow ; J of value = $ 60 -^5 = $12, the gain. 986. 6. The wrong weights are ^5^, or |^ of correct weights, so that the customer receives for $352, W of this amount, the 140 MANUAL FOR TEACHERS gain to the grocer being -^^ of $352, or $16.50. By selling 16J oz. to the pound, the grocer gives |-| of the proper amount, his loss being -^ of $ 320, or $ 10. The net gain is $ 16.50 - $ 10 = $6.50. Ans. 7. Cost of alcohol, $2.50 X 42 = $105 ; 3 yr. interest on $105 = $18.90. Amount to be realized, $105 + $18.90 = $123.90. Number of gallons to be sold, 42 — 7 = 35. Selling price per gallon, $123.90 -^ 35 = $3.54. 8. £4500 = $4.85x 4500 = $21825. Income on consols at 3% = $654.75. Selling price of consols, $21825 X. 96; value of U. S. bonds, $21825 X .96 h- 108. Canceling, we obtain $19400; 6% of which gives the income on bonds = $1164. Difference = $1164 - $654.75 = $509.25. Ans. 9. Number of cubic feet =. 9| X 9^ X 6f = 609 ; weighing 609000 oz. ; etc. 988. 11. Agent collected 80% of $4500 = $3600; on this, his commission at 7^% is $270; making the amount to be given me = $ 3600 - $ 270. 13. a; X -I- X I or 5^ = 15.12. 100 2 40 Find the discount for 17 da., the time from June 20 to July 7. 15. Ignoring the price — if he could buy 80 lb. with 81^% of his money, he could buy with 100%, (80 lb. h- 81^) X 100. 17. Let a: = cost of each cow; 6 a; = cost of 6 cows; com- S r n 18a: .1 • • c 1 18a; mission = — — of 6 a; = — - ; cost and commission = 6 a; + — — 100 100 100 = 525.30. 18. The note is discounted 35 da. after it is made, so that it has (93 — 35) da. to run, or 58 da. [Without grace, 55 da.] The interest for a year is $36, which is 10 cents a day, or $5.80 for 58 days ; etc. 19. If the selling price (regardless of its amount) is six-fifths of the cost, the gain is -J- of cost, or 20%. NOTES ON CHAPTER TWELVE 141 989. 6. The walls contain [(20 + 15 + 20 + 15) X 10] sq. ft. ; the ceiling contains (20 X 15) sq. ft. Dividing by \^, the width in feet of the paper, gives the number of feet of paper required, which is then reduced to yards. 991. 7. Dividend is 3^% of ($9562.50^ 1.27^). 993. Mr. Smith wishes to pay Mr. Thompson the exact amount of his bill. A check on a Memphis bank for $3475.86 would not be sufficient, as Mr. Thompson would have to pay a New York bank for collecting the check in Memphis. As the charge may not always be the same, Mr. Smith cannot know how much to add to the amount of his bill to cover this expense. From a banker that has an account in a New York bank, he can obtain a draft, payable in that city, for the exact amount, by giving the Memphis banker $3475.86 + $5.21, or $3481.07. Exchange is at a premium when the cost of a sight draft is greater than its face ; it is at a discouiit when the cost of a siglit draft is less than its face. Bank Check. ^«- . Quogue, JSr.Y., fayyv. S, 189 6. Jfo. /^^i'. SHINNECOCK NATIONAL BANK. Pay to the order of £&w-ia. /C. 3^k^L1tow^, <^6Sf^^ '^i/xXA^-&v^At ^ — .^.^ ^jQQ Dollars HOWELL & PENNIMAN. An examination of the above check will show wherein it differs in form from the draft. A draft may be made payable at a future time, whereas a check is always payable on presentation. 142 MANUAL FOR TEACHERS 8. A%=TATr;^-|^-=632.18. 9. ^+^=1000. 10. $339.66 - (2% of $339.66) = sum remaining for the pur- chase of dry-goods, etc., and the commission. Dividing this by 1.02 gives the cost of the goods. Another method of solving the foregoing is to indicate the money remaining as $339.66 X .98. Using 1.02 as a divisor, and canceling, gives the result. 339.66 X 98 102 994.2. Noon Monday to 6 p.m. Thursday = 78 hr. The loss in time = 35 sec. X 78 = 45 min. 30 sec. The time shown is 6 hr. — 45 min. 30 sec. = 5 hr. 14 min. 30 sec, or 14^ min. past 5. 4. The number of rows, 2 ft. apart in a space 36 ft. — 4 ft., is (32-f- 2)-f 1 = 17. The number of plants, 16 in. apart in a row 60 ft. - 2| ft., or 57| ft., in length, is (57^^ IJ) + 1 = 44. Total number of plants = 44 X 17 == 748. If the rows run crosswise there will be 29 of them, each con- taining 26 plants. 6. Number of revolutions = 14 mi. -^ 13 ft. 4 in. 995. 8. Length of a degree on the equator = 25000 mi. -^- 360. 20° will measure (25000 ^ 360) X 20. Cancel. 9. The circumference = 18 ft. X 3.1416. Divide by 360. Cancel. 10. The difference is 20°, and the distance will be about one- half that found in 8. The teacher should remember that the shortest distance between these two places is not measured on the parallel of 60°. The shortest distance between two points on a sphere is measured by the arc of a great circle joining the points, and the 20° are -^ of a small circle. 11. 46°22'30" = 46f°. The number of miles = 69J x 46f 12. The approximate length of the 45th parallel is 25000 mi. X .7071 ; the length of a degree on this parallel = (25000 mi. NOTES ON CHAPTER TWELVE 143 X .7071) -^ 360; multiplying by 22^ gives the required distance. Cancel. 996. Time drafts are so little used that it is scarcely worth while to spend much time on their study. A sight draft being payable on presentation (except in those states allowing days of grace), there is no need of formal accept- ance. Acceptance is necessary in the case of time drafts, as they are not payable until the specified time after this acceptance. The acceptance of a draft makes the person or corporation accepting it liable to its owner for the amount, a draft being transferable by endorsement just as a check or a note. 997. In calculating the cost of a sight draft, days of grace — even when allowed — do not enter into the result, this being included in the rate charged. Time drafts are allowed days of grace, except in the states given in the Appendix, Art. 1305. The number of states in which days of grace are no longer allowed, increases yearly, there being no good reason for promising to pay in 60 da. when the intention of the signer is to take 63 da. 1000. 1. Although days of grace are not allowed in Cali- fornia, the pupils of other states should not be expected to know this. In states that grant days of grace, they should be allowed in every note or time draft, no matter where payable ; while in the other states, pupils should be taught not to employ them in any case. The premium on the draft = 1 1.75 X .840 = 1 1.47. The interest (with days of grace) = $840 X yf^ X ^\ = $13.02. The cost of the draft = $840 -f$ 1.47- $13.02 =$828.45. Ans. Or, without days of grace : $840 X yf^X -jV^ = $ 12.60, the cost being $840 + $1.47 -$12.60 = $828.87. Ans. Some teachers prefer to find the cost of a draft for $1 at the given premium — in this case, $1.00175 ; from which is deducted the interest on $1 for 93 da., or $.0155; making the cost of a 144 MANUAL FOR TEACHERS 90-day draft for $1, $1.00175 -$.0155 =$.98625. Multiplying this by 840 gives $825.45, the cost of a draft for $840. This method is not so much shorter as to make it advisable to use it. 2. The discount = ^% of $400 = ^ of $400= 50^. The interest for 33 da. = $400 X yf^ X ^^ = $2.20. Cost = $400 -$.50 -$2.20 = $397.30. Ans. Note. — The word "interest" is used instead of "bank discount," to avoid the confusion arising from the use of "discount" with two meanings in the same example. 6. The six remaining examples should be omitted by pupils that do not use algebraic methods of solution. Scholars that have readily worked the first four examples will find no great difficulty in solving 5-10 by means of the equation. The premium is $^ per $1000, or :f7nnr ^^ ^^® ^^^® ^^ ^^® draft, X. Premium = The interest on x dollars for 63 da. 4000 at 6% =a;X— — X— — = -— — • Adding the premium to the ^ 100 360 2000 & f face of the draft, and deducting the interest, gives a:-}- . ^^ 21 X . 4000 — as the cost of the draft. This may be changed to 4000 x + x-^2x _ 3959 x . 4000 4000 ' '^' 6. i% of $1200 = $1.50; interest for (a: + 3) da. = 1200 X — — X ^^t" = ^~l ; cost of the draft (with days of grace), 100 360 5 V J 6 /• 1200 U ^ + ^ - l^QQQ 1^ 2^ 6 _ 11979 -2 a: j^^ * 5 10 10 10 10 10 ■ ■ Note. — Unless the pupil has studied algebraic subtraction in Chap. XV., he may make a mistake in deducting ^ "^ , by failing to change the sign of the second term. By writing ^ » - + -, he may see more clearly that 5 5 5 the cost of the draft is 1200 — 1^ — - — , etc. Without days of grace, cost -1200-11--; etc. ^ ^ 5 NOTES ON CHAPTER TWELVE 145 - 1001. 1. 15°x3i = 50°. Ans. 2. (61 -^ 15) hr. =^ 4^5 lir. = 4 hr. 4 min. Ans. 3. Difference in time — -J^ hr. = 5 hr. London time = 1 p.m. + 5 hr. = 6 P.M. Ans. 4. 2 P.M. — 5 hr. = 9 a.m. Ans. 5. Vienna is -f-l hr. later = 1-| hr. == 1 hr. 40 min. Time at Vienna 40 min. after 1 p.m., or 20 min. to 2 p.m. Ans. 6. 3 hr. 40 min. = 3| hr. Difference in longitude = 15° X 3| = 55°. Ans. 7. Difference in longitude = 75° + 30° -= 105°. Ans. 8. Philadelphia time is ^^- hr. earlier, or 7 hr. 3 p.m. — 7 hr. = 8 a.m. Ans. 9. Correct Washington time is j\ hr., or 8 min. earlier than standard time. 10. A town in 84° west longitude is 6° east of 90°, so that its correct time is -^^ hr., or 24 min., later. Time, 12 : 24 p.m. Ans. 1002. 1. Longitude difference = 15° X 3|-f . The pupil should see that 15 X |^ = 44 ^ 4 = 11 ; so that 15° X 3*^ = 45° + 11° = 56°. Ans. 2. 15 )37 hr. 18 min. 2 hr. 29 min. 12 sec. At 1 hr. to a degree, the difference in time would be 37 hr. 18 min. ; as it requires 15° to make an hour's difference, dividing 37 hr. 18 min. by 15 gives the result. Shorter methods should be deferred for the present. Using multiplication to obtain the difference in degrees, and division to obtain the difference in time, is more easily understood by beginners. 3. Time difference = y^-g- of 87 hr. 35 min. earlier at Chicago, because it is west of Greenwich. Standard Chicago time is the time of 90°, or 90 hr. -;- 15 = 6 hr. earlier than Greenwich. Standard time = 1 P.M. — 6 hr. = 7 a.m. Ans. 146 MANUAL FOR TEACHERS 4. Vessel's time is 2^ hr. earlier, showing that the vessel is 15° X 2i, or 37^° west of Greenwich. Am. 37° 30'. 6. Time difference = H hr. Longitude difference == 15° X 1-^ = 22J°* The latter place, having the later time, is the more easterly ; so that its longitude is 22 J° east of 11° east, or 33° 30' east. Am. 7. 3 da. 12 hr. 17 min. = 84|J hr. ; 3313.5 h- 84fJ = number of miles per hour. 11. 12^ (ft.) X 3f (ft.) X X (ft.) = 730^;% = 730^ (cu. ft.) ; = 1^= 15^^. Am. 15f^ ft. - 15 ft. 7 in. 12. 48% = 237 bu. 3 pk. + 4% = 19 bu. 3 pk. 2 qt. -j^ of 48% Remainder, 52% = 257 bu. 2 pk. 2 qt. Am. 14. Number of degrees -= (34 X 24) -f- 48.96. 1003. 5. 84 half-dollars -84 cents ==$42.00 -$.84. 9. After taking ^, |- are left; when ^ of the remainder is taken, f of remainder are left, or -J of -f = -J = 4 gal. ; etc. 1004. 5. 12 men working 8 hr. daily build 90 rd. in 15 da.; 7 men working 10 hr. daily build 70 rd. in ? days. 15 da. X 12 X 8 X 70 90 X 7 X 10 9. 72 (in.) X 48 (in.) X x (in.) = 2150.4 (cu. in.) X 75. 16. a; + (a: +15) + (5: +15 4- 27) = 320. 17. .64 bu. = 4 pk. X .64 = 2.56 pk. ; .56 pk. = 8 qt. x .56 = 4.48 qt. ; 3.64 bu. = 3 bu. 2 pk. 4.48 qt. ; -^^ bu. = 4 pk. X A n= I pk. = 2i pk. = 2 pk. 2 qt. ; 3 bu. 2 pk. 4.48 qt. + 2 pk. 2 qt. + 1 bu. 3 pk. 6.52 qt. = 6 bu. 5 qt. ; 10 bu. — 6 bu. 5 qt. = 3 bu. 3 pk. 3 qt. Am. 18. Each step takes 7^ in. + 10 in. ^ 17^ in. = ^ yd. Co8t = 90^X-y^X 18.. Cancel. 24. a: + (a: +1211) = 9891. NOTES ON CHAPTER TWELVE 147 1005. Formerly, bills of exchange were issued to purchasers in sets of three bills, two of which were sent by different steamers to the foreign payee, who presented for payment or acceptance the one that reached him first. The third bill was retained by the purchaser, to be sent in case both of the others failed to reach their destination. At present, only two bills of a set are issued. The second will read as follows : Exchange for £180 17s. 6c?. New York, Dec. 14, 1895. Sixty days after sight of this Second of Exchange (First unpaid) pay to the order of John W. Moran & Bro., One Hun- dred Eighty pounds sterling, seventeen shillings, six pence. Value received, and charge the same to account of To James Lennon & Co., ) -r» r^ o oi London. i PeTER CoMERFORD & SON. No. 39. 1. No deduction for interest is made for the 60 da., the quo- tation giving the price per pound for 60-day bills. The method given in the text-book is a form of the aliquot part method used in calculating interest. 1007. 3. Cost in dollars -1000 ^5.1625. 4. Cost in dollars = 1874.35 X .9525 ^ 4. 5. Number of marks = 1000 ^ (.955 ^ 4) = 4000 -^ .955. 6. Number of francs == 1637.5 X 5.185. 8. £437 5s. lOd'. Less 4%, or ^, 17 9s. 10c?. £419 16s. Cost = $4,885 X 419if = $4,885 X 419.8. 9. 18pcs.,44m. each, or 792 m. @ fr. 25 = fr. 19800 Less 7|-%, 1485 fr. 18315 3 pes. 50 m. each, or 150 m. @ fr. 20, fr.3000 Less 5%, 150 2850 Packing charges, 60.50 fr. 21225.50 Cost of bill = 19^^ X 21225.5 = $ 4138.97. Am. 148 MANUAL FOR TEACHERS 10. M. 3598.60 Less 10%, 359.86 M. 3238.74 Less 5%, 161.94 M. 3076.80 Less2|%, 76.92 M. 2999.88 Freight, 165 kilos @ M. 4.80, 792.00 M. 3791.88 95J^X 3791.88 ^4 -$908.87. Am, TJNITEKSITT XVI NOTES ON CHAPTER THIRTEEN 1008. The word "endorsement" means something that is written on the back of a document. As applied to notes, checks, drafts, etc., it generally means the signature of the person in whose favor the note, check, etc., is made out, which is written on the back in order to transfer the ownership. If the payee of the following note sells it to William Simms, he writes his name on the back, as shown below. AccoTiNK, Va., March 4, 1897. Four months after date, I promise to pay to the order of James McWilliams, Two Hundred Dollars, value received, at the Pohick National Bank. I2OO1-V7. Victor Struder. The effect of the " endorsement in blank " is to make it pay- able to any holder; the " endorsement in full" transfers it to I William Simms, who may transfer it to another by either kind of endorsement. 149 150 MANUAL FOR TEACHERS Besides transferring ownership in a note, the effect of an endorsement is generally to bind the signer to pay the note, in case of default by the maker or preceding endorsers. This lia- bility is avoided if the endorser writes after his name the words, " without recourse." The *' endorsements " mentioned in this chapter are a record of the payments received by the holder of the note. This is usually kept on the back of the note, the date and the amount received being written in each instance. 1010. Although the maker of a note is generally supposed to pay the interest at the end of each year, the U. S. Courts, by whom this rule has been formulated, do not permit the collection of interest upon deferred payments of interest. This rule is followed in all the states except Connecticut (see Art. 1307) for computing the amount due on notes that do not expressly provide for the payment of interest annually (Art. 1172). Connecticut pupils should learn only their own rule ; in other states, no reference whatever to the Connecticut rule should be made. See Art. 1307 for Connecticut-rule answers to the partial pay- ment examples of this chapter. 1013. 6. Let 100 X = cost of coal ; 2 a; = commission. 102 X = cost of coal + commission = 7650. 1015. 1. The man expended 30% of 50% of f of his money, or ^ X -J- X f of it ; which was yf^ of his money. -rJ^a; = 1-|- X 728 = 819 ; x = 9100 Two-fifths of $9100 equals the balance in bank. 2. Cost per gallon, $ 1.50; selling price, $1.60; gain per gallon, 10^ on 150)Z^, or yV = 6|%. 5. Let 10a; = cost, then x will represent the loss in one case and the gain in another, making the selling prices 11 a; and 9 a;. 11 a; = 99 ; a; = 9, gain in dollars. 9 a;= 99; a; = 11, loss in dollars, making the net loss $2. NOTES ON CHAPTER THIRTEEN 151 Note. — Some teachers, wishing to avoid fractions as far as possible in equations, assume x for loss or gain, making 10 a; = cost; etc. Solutions of this kind are given occasionally as a suggestion to be followed or not, as may seem most desirable. 8. Arithmetic, Art. 924, 7 and 8. ■: 1016. As there is no such thing as " true discount," it is un- profitable to spend time upon it'. Any problem involNring find- ing the " present worth " can be solved by an intelligent pupil, from his previous work in interest. 1019. 2. To \\% of I of % 25000, add $ 1. 3. Longitude difference = 5° 59' 18". See Art. 1002, 2. 10. Number of yards = (5616 ^ 1.04) ^ \\. Art. 1013, 6. 11. Term of discount 36 (33) days. Yearly interest is |30, which is $3 for 36 da., yV y^'- $500 - $3 = proceeds, $497. Without grace, the term is 33 days, the interest for which time is $2.75. Proceeds, $497.25. 13. Cost of an acre = $21.78 X 5. Cost per square foot = ($21.78x5) -4- (4840x9), the divisor being the number of square feet in an acre. To gain 20% or \ of cost, the selling price must be f of cost. Multiplying the foregoing by f, the selling price per foot will be $21.78x5x6 4840 X 9 X 5 * In getting the number of square feet in an acre, the pupil may use 160 X 30| X 9, unless he remembers that there are 4840 sq. yd. in an acre. 14. Specific duty (duty by weight) -- $i X 700 X 1^. Ad- valorem duty = 30 % of $1^ X 700. The sum of the two gives the total duty. 1020. 7-9. See Arithmetic, Art. 384. 1021. 4. (50 ft. + 38 ft ) — (7 ft. + 2 ft.). 5. f of cost of horse = $ 90. Cost = $80. A selling price of $100 makes a gain of $20 =^ of cost = 25%. Ans. 162 MANUAL FOR TEACHERS 6. Loss = ^ of cost = 20%. Ans. 12. 100%. 15-16. See Art. 878. 19. ^% of $700. 20. $10 for 60 da. + A ^^ ?10 for 5 da. 21. 200%. 22. ^ of $7. 26. 4A^A=^42-^^3. 28. 1^ = 1 of 63 = 49; ^=7; a: = 56. 29. ^ = 8J = ^, 2a: = 25 ; a: = 12^. Aris. 12 yr. 6 mo. o o 30. 5 qt. = A pk. ; I = .625 ; .625 ^ 4 = .156J = .15625. 1022. 1. $fx5i(yd.)x4(yd.)^li(yd.). ?ix¥xtxf Cancel. 2. What sum in 4 years at 6% will amount to $105.71? Arithmetic, Art. 1017. ^ + (^XTf^x4) = 105.71; a: + ?^=^ 105.71; 100a; -f 24a: =10571; 124a; = 10571; a; = 85.25. $85.25. Ans. 3. Term = 27 da. -}- 3 da. = yV yr. a?-(^XTfcXiV) = 95; a;-^=95; etc. 4. Problems of this kind may be solved without finding the cost; 18^ per yard represents -^ of cost; what price will rep- resent f or U of cost ? If A of cost = 18^, ^ig- will equal 2^, and II will equal 24^. Ans. 5. 24i%=3j4j^; ^=1372; a: = 5600. The whole real estate was worth $5600; the part remaining after the sale of $1372 worth = $5600 -$1372 = $4228. $14000 + $4228 = $18228. Ans. 6. 6 times (5 X 5) sq. ft. 7. Mr. Jones paid |% of | of $ 48000 = ^^^ X | X -^ 1 u 4UU ~ $ 300. The company loses $ 40000 less the premium received, $300. NOTES ON CHAPTER THIRTEEN 153 8. This may be solved without finding the cost, although many pupils will prefer the more tedious way. $ 764.40 repre- sents 91% of cost; what per cent does $894.60 represent? 91^. X 89460 ^21M„_^ ,061%; gam 6if.. 10. Cancel. |3| X 25 X 8 X 8 ^ 128. 11. Total cost,$252.50; loss=$252.50-$141.40-$111.10, which is 44% of the total cost. 12. [14 (yd.) + 5 + 14 + 5] X 3. 14. Profit$2perton, Jof cost. Cost = $198x3. 15. Art. 546. A bill is receipted by writing the words, " Received payment" at its foot, followed by the date and the name of the seller : ^ev f. ^. If the money is received by a clerk, he writes his initials underneath, preceded by the word " per " or " by." 16. ^ = 9f ; 9|a; = 8|; etc. 17. Omit 4^ bu. I gave away \ and f = it + A = tt- T^® remainder = ^^ = .26f = 26f %. 21. $3 per yd. = 80% of cost; $?= 115%. 22. Number of gallons sold = (65 gal. X 60) - 80 gal. = 3820 gal. Selling price per gallon = f of $ 1542 ^ 3820. Cancel. 23. % 180 = \ cost of one horse = f cost of other. 24. Number of square yards in walls = (6 + 4 + 6-f4)x3; in ceiling, 6x4. Number of cords = (18 X 12 X 9) ^ 128. 25. Let X = less, x -^\=^ greater ; x -\- x -{- \ = ^\. ' 26. 16 cu. yd. is :f% of (10 X 8 X 2) cu. yd. ? 154 MANUAL FOR TEACHERS 27. 672 yd. @ $2^ = $1512. Discount without grace = $1512 XjhXi = $17.64. Profit is $1 per yard less discount = $672 -$17.64 = $654.36. Ans. The discount for 3 days' additional (grace) = -^j^ of $ 17.64 = $.88, making the profit 88^ less than the above, or $653.48. Ans. 28. 40^ X [(55x600x5^) ^27]. Cancel. 1023. 5. The circumference of the wheel = distance traveled in 1 revolution = 1 mi. 94 rd. 2 yd. 1 ft. h- 526 = 6838 ft. ^ 526 = 13 ft. = 4 yd. 1 ft. 6. The weight in pounds = if f »- X 9| X 9J X 6}. 8. Rate of income received on 6% bonds = 6 -^- 1.18 ; rate on 4J% bonds = 4:^ -. — — . The income being the same, and the same amount being invested, the rates must be equal ; there- fore, §52 == 450 . 600 a; = 118 X 450 = 53100 ; x = 88 J. Price per $100 = $88.50. 9. The shrinkage being 1 lb. in 10, he must sell 9 lb. for the cost of 10 lb. to suffer no loss. 10 lb. cost $1.80 ; by charging 20^ per pound, he receives the cost. To gain 20%, he must sell for I" of 20^, or 24^ per pound. Since he loses 4% of the amount of sales, or ^, he receives only f|- of the price charged per pound. Therefore to receive 24^, he must charge 24 ^-^|4 = 25^ per pound. [(i8^-.A)x|]-M. 1024. 1. 12xa: = 20xf. 2. This may be solved by analysis, or the following method may be employed : The solid contents of first beam in cubic feet = 16 X 2J X f ; of the second = a: X 3J X -J-}. The second weighs -f-^M times the first ; its contents, therefore, = f f ff times the contents of the first. a;X3iX-J^=16x2Jx|X ff|| ; a; = (16x2ixtX«tt)^(3ixM) = 16 X I X J X fMf X 3^ X H. Cancel. NOTES ON CHAPTER THIRTEEN 156 3. The carpet costs 50^ per foot. $-1- X 22-1- x 15|-^2J = Ans. Or, changing all dimensions to yards : $ 1-|- X 7-| X 5J H- J = Ans. 4. As a sight example, some pupils may see that the width of the large box is double that of each small one, and its depth is three times that of each small one, so that with the same length as the small one, it would contain 2 X 3, or 6, small ones. A length twice as great — 8^ ft. — is required to enable it to hold 12 boxes. 6. [i of (ISixllf)] sq.ft. 22 X 14 X 12 X 1728 2150.4 Drop the decimal, point in the denominator, and annex a cipher to one of the numbers in the numerator. Cancel. 9. 49 X 44 X 27 ^ 231 = number of gallons. Cancel. 10. Solve at sight. 7 yd., 6 yd., 4 yd. 11. Number of gallons = 5i X 6 X 7 X 1728 ^ 231. Cancel- ing, we obtain 1728 gal. One empties it in (1728 -h 9) min. = 192 min. ; the other in (1728 ^ 7) min. = 246f min. ; both in (1728 ^ 16) min. = 108 min. 12. The dimensions of the room are 6 yd. and 5 yd., and the carpet is f yd. wide. 6 contains f an exact number of times (8), so that if the carpet runs across the room it will take 8 strips each 5 yd. long. As 5 -^ f = 6|-, to lay the carpet lengthwise would require 6 strips, and -| of a seventh strip, which would have to be cut. Carpet 30 in. wide, | yd., could be laid lengthwise without splitting the breadths, 5-7-|-, or 6, strips being needed, each 6 yd. long. 13. 36 yd. are needed to cover the floor; including 4-J- yd. cut off in matching the pattern'^ 40-|- yd. must be bought at 95^. At 10^ per yard, the sewing and laying should cost 10)^ X 36 = $3.60, but the custom is to charge for the number of yards 156 MANUAL FOR TEACHERS purchased, 40|, making $4.05 ; (5 X 6) sq. yd. or 30 sq. yd. @ 5^ = $1.50 for lining. Total cost, $38.47^ + $4.05 + $ 1.50 = $44.02J, or $44.03. 14. A strip I of (18 — 15) or | of (21 — 18) is left uncovered on each of the four sides, or H ft. The area of the uncovered space in square feet = (21 X 18) — (18 X 15). 15. The number of square feet in the walls = (18 + 24 -f 18 4- 24) X 9. The ceiling contains (18 X 24) sq. ft. Deduct 60 sq. ft. for two doors, 48 sq. ft. for two windows, 25 sq. ft. for the fireplace. The total number of feet around the four walls = 18 + 24 + 18+24 = 84 ft. Baseboard will not be required at the doors, 8 ft. ; nor at the fireplace, 5 ft. — a deduction of 13 ft., making 71 running feet of baseboard, 1 ft. wide, contain- ing, therefore, 71 sq. ft. The total deduction from the area of walls and ceiling, 1188 sq. ft., are 60 sq. ft. +48 sq. ft. + 25 sq. ft. + 71 sq. ft. = 204 sq. ft., leaving 1188 sq. ft. - 204 sq. ft., or 984 sq. ft., to be plastered. 16. The first pile contains (25 X 20 X 10) cu. ft. and costs $ 1400. 1 cu. ft. cost $ 1400 ^ (25 X 20 X 10). Multiplying by (50 X 40 X 20), the number of cubic feet in the second pile, gives the cost : $1400x50x40x20 25x20x10 17. Pupils that endeavor to solve a problem without examin- ing the conditions, will be likely to assume that this example resembles 16. In the latter, the cost of the second pile is 8 times the cost of the first ; in this one, the surface to be painted in th<' second room is 4 times that of the first room, making the cost $56. As they may not be familiar enough with similar surfaces to know the ratio, they should find the surface of each. 1025. In dividing decimals, change the divisor to a whole number. See Arithmetic, Art. 663. 1026. 2. XXV = 25000. NOTES ON CHAPTER THIRTEEN 157 5. The furlong is seldom used. 3.7082 mi. ^ 4 -= .92705 mi., the length of one side. Multiplying by 8 to reduce to furlongs, we obtain 7.4164 fur. Change the decimal part, .4164 fur., to rods by multiplying by 40, obtaining 16.656 rd. ; .656 rd. = 3.608 yd. ; etc. ; etc. 6. See Art. 986, 7. That selling price, $3.54, must be in- creased |, or $ .59, to gain 16|%. $3.54 + $.59 -$4.13. Ans. „ qj 5 ^ 240 X 38 X 9 ^ 1 7- ffX — Cancel. 10. The inventors of the expression "true discount" assume that interest is not payable in advance. They claim that a borrower that promises to pay $380 at the end of 2 yr. 5 mo. should receive as a loan the principal that will amount to $380 in that time. Let X = principal. Interest = a: X — X — = —• 100 12 200 Amounts a; + — = 380; 200 200 a: + 29 a; = 229 a; = 76000 ; a; =331.88-. The principal (** present worth ") = $331.88. The " true discount " = $380 - $331.88 = $48.12. This "true discount" is the interest at 6% on $331.88 for 2 yr. 5 mo. The interest on $380 at 6% for 2 yr. 5 mo. is $55.10; the difference between the interest and the "true discount" being $55.10- $48.12 = $6.98. Ans. 1027. 1. A gain of 25%, or J of cost, makes the selling price ^9, equal |- of cost; \ of cost, the present gain, is $9-^5, or $ 1.80. A gain of 50% would be $ 1.80 X 2 = $ 3.60. N. B. — It is not necessary to find the cost. 158 MANUAL FOR TEACHERS 2. $30 in one case represents f of cost, making the gain $6 ; in the other case, $30 represents f of cost, making the loss $10. Net loss $4. Note. — The thoughtful teacher will recollect that every member of the class does not "see through" an example in the same way, nor with equal rapidity. While quick work should be exacted where the question involves but a single arithmetical operation, time should be given, in problem work, to pupils that do not quickly grasp the conditions. Such as can dispense with unnecessary figures should be encouraged to do so as much as possible; but care should be taken not to injure others by requiring them to adopt a short method whose underlying principles they do not thoroughl}^ under- stand. Each should, to a certain extent, be allowed to use his own mode of "analyzing" oral problems and of setting down his written ones; shorter ways, however, being presented from time to time in the oral and blackboard work of his brighter classmates. The scholar that reaches his results by a circuitous course will, by these models, be led to see the time saved by shorter methods, and he will probably try to master some of them. 3. Together they have $300 = a: + 2 a;. 6. $40 in 3| yr. = $12 per year. This is produced at 6% by $200. Ans. 7. The interest of x dollars for 5 yr. at 6% = •— ; x -{- --, or 8. Yearly interest = $3. To obtain $18 interest will re- quire 6 yr. 9. The interest is $2, \ of principal in 3J yr. ; in 1 yr. it is ^ X -^ of principal, or -^ = 5%. Or, the interest on $12 @ 1% for a year is 12^, or 40^ for 3^ yr. ; to obtain $2.00 interest, which is 5 times as much, the rate must be 5%. 10. I lose $25, or | of cost = 2,^% . 102a 4. 23i-f(23i-f3}) + (23i-f-3f + 3i). 6. Let 3a; represent the amount received by one; and 4a; the amount received by the other. Then, 7a: ^21.63; a; = 3.09; 4a; = 12.36, and 3a; = 9.27. Ans. $9.27 and $12.36. NOTES ON CHAPTER THIRTEEN 159 7. [(26 X 12) X (4 X 12)] ^ (8 X 4). 8. $10-24x2700x890. ^^^^^^^ 1500 X 356 1044. In addition to the details usually given in a bill, an invoice shows the marks and the numbers placed upon each package shipped. In this invoice, the mark is given in the first column, and is the same on each case. The number of each case is written in the second column. Besides informing the receiver in what case a particular article is packed, it is required by the U. S. custom authorities. A certain percentage of cases in each invoice is examined, and their contents must agree in description, quantity, value, etc., with the invoice. 1500 yd. @ IJ^., £11-14- 4i 1500 " " 2 3000 " " IJ • 2889 " " 2^' Less ^V. 12- -10 23- - 8- - 9 29- -11- -lOi £77- - 4- -llf 1- -18- - n £75- - 6- - 4i Duty @ 50% Value in U. S. money, $366.53. Duty @ 50% = i value = $183.27, nearly. English accountants employ to a great extent a method by aliquot parts called "practice." The cost of the fourth item is obtained as here shown. The cost at Id per yard, 28Sdd., is easily reduced to pounds, etc. It is repeated to obtain the cost @ 2d. Y^d is I of Id., ^^d. is \ of ^^d., and y| is ^ of ^^d. As the smallest denomination is the farthing, or ^ of a penny, | of f c?. is called 2 farthings, or ^d., instead of |c?. A similar, method may be used in, reducing the above result to U.S. money. See Arithmetic, Art. 1005. The value of £50 is ascertained by taking ^ of $486.65, or $ 243.325. £ 25 = ^ of £50. 5s. = £|- ; etc. In assessing duties, the government ignores fractions of a dollar in the cost less than J; over 50;* is considered another dollar. The duty that 2889 @lcf. = £12- 0- 9 " 1 " 12- 0- 9 "tS" 4- 0- 3 "tV" 1- 0- Of "ft" 10- 0^ £29-11-10^ 160 MANUAL FOR TEACHERS would be collected upon the foregoing would be 50% of |367, or $183.50. Children should not, of course, be burdened with such details ; their answer should be $183.27. Some zealous teachers fall into the mistake of endeavoring to make their pupils familiar with the methods of calculation peculiar to some callings. The time assigned to the study of arithmetic can be employed more profitably to the scholar by giving him the ability to handle ordinary problems with reasonable readiness, than by dissipating his energies in trying to make him understand a multitude of small matters that are entirely outside of his present experience. The average boy would make a better accountant if he did not hear of taxes, partnership, insurance, bonds, stock, brokerage, commissions, etc., during his school life, provided the time thus misspent were given to elementary algebra and constructive geometry, as well as to better work in what are considered the more elementary portions of arithmetic. 1046. The ratio of 3 to 6 is generally expressed as 1 to 2. 1049. 3. The ratio of the price of coffee to that of sugar is ^, or 5 to 1. 4. A goes 1 J times as fast as B. The ratio of A to B is J, or 5 to 4. 5. E's earnings are to D's as 4 to 3. 6. 3 to 4. 10. 30 to 3 = 10 to 1. 1050. 6. One wheel makes 70 revolutions per second ; the other makes 90 per second. 7. The diameter = twice radius = 4 ft. 8. One goes 48 mi. per hour ; the other goes the same distance in the same time. 9. Area of first = 6| X 4 J ; of second = 4| X 2^. Ratio = %^^- Reduce. 1051. 2. Cost of farm = $75.50 X 156^ X 124.6 ^ 160 = $9201.52. Interest for 1 yr. on one-half of the cost = $230.04 ; $ 4600.76 -f $230.04 = $4830.80, amount of mortgage. Deduct- ing $500 then paid, gives $4330.80, balance due. NOTES ON CHAPTER THIRTEEN 161 4. Number of cubic feet of excavation = 41| X 8 X 33 = 10890. The inner dimensions of the cellar are 38J X 30 X 8, deducting from the length and the breadth 3 ft., which is the thickness of two walls. The number of cubic feet in the cellar = 9180, leaving 10890 cu. ft. - 9180 cu. ft. = 1710 cu. ft. as the contents of the walls. The cost of the excavation is $^ X 10890 -^27 = $201f ; to lay the wall, costs $15 X 17.1 = |256.50. Total cost, $201.66| + $256.50 = $458.16J, or $458.17. Ans. Note. — One of the practices of builders in some sections is to take only the outside measure of walls in ascertaining the contents. The number of cubic feet in this case, according to their calculations, would be (41^ + 33 + 41 J + 33) X 8 X IJ = 1782 cu. ft., or 72 cu. ft. too many. By this method the four corners, measuring each 1 J X 1^ X 8, or 72 cu. ft. in all, are in- cluded twice. As has already been said, children should be expected to obtain only the correct results, leaving later experience to furnish informa- tion as to local usages. 5. For $107.25 there can be obtained, at 3J%, insurance amounting to $107.25 -^ .0325 - $3300. If this is 80% of the value, the flour must have cost him $3300 ^ | = $4125, which is $4125 -^ 500, per barrel, or $8.25. The algebraic method would be : Let X = the cost per barrel. a: X 500 X 80% X 3J% = 107.25 ; or, 500a: X f X ^\\ = 13:c = 107.25 ; X = 8.25. 6. The bank discount is calculated on the sum due at maturity, which is $1250+ interest from June 12 to Dec. 15 (12), 186 da. (183 da.), at 5% = $1250 + $32.29 ($31.77) = $ 1282.29 ($1281.77). The discount for 30 da. (including grace) on $1282.29 at 6% is $6.41, making the proceeds $1275.88. The discount on $1281.77 for 27 da. (no grace) at 6% = $5.77, making the proceeds $1276. 8. A furnished 5 men for 20 da. and 6 men for 15 da., which is the same as 100 men and 90 men for 1 da. B furnished 162 MANUAL FOR TEACHERS the equivalent of 120 men -|- 180 men for 1 da. The money received should be divided on the basis of 190 men for A and 300 for B, 490 in all ; and A should receive ^§ of the sum, and BfiJofit. itt of $857.50 = $332.50, A's share; itt of $857.50 = $ 525.00, B's share. 9. Arrange the work so as to have the required term in the last place : A ditch (403 X 3 X 3) cu. ft. is dug in (62 X 13) hr. by 27 men. A ditch (750 X 4 X 3) cu. ft. is dug in (250 X 12) hr. by ? men. If (403 X 3 X 3) cu. ft. are dug in (62 X 13) hr. by 27 men, 1 cu. ft. will be dug in (62 X 13) hr. by ^^^ °^^^" ^ ; ^ ^ ^ -^ 403 X 3 X 3 1 r^ -11 k J • 1 k k 27 men X 62 X 13 1 cu. ft. will be dug in 1 hr. by -— — - — ; ^ ^ 403x3x3 ' etc. See Arithmetic, Arts. 973, 974. . 27 men X 62 X 13 X 750 X 4 X 3 ^ Ans. = Uancel. 403 X 3 X 3 X 250 X 12 23a; 10. Let a: = price per barrel; — — = cost of flour. 2J% of 23a; 1 .23a; 23a; • . 1 01" t::: 01 — r-» — tt::: — commission. 4 40 4 160 Then ^ + 11^=1508.80. Note. — The words "after deducting his commission" mislead some pupils, who think that this requires them to begin work by deducting a commission of ^ of the amount sent. If the purchasing agent did not receive money in advance, he would render his employer the following account : To 256 bbl. flour @ |5.75, $1472.00 Commission at 2^%, 36.80 $ 1508.80 2| % on the amount sent, $ 1508.80, would be 2^ % of the cost of the flour, 1 1472, and 2^% of the 1 36.80 commission. NOTES ON CHAPTER THIRTEEN 163 A pupil that wishes to deduct the commission first, may work in this way : Let X = commission, then 40a; = cost of flour ; 41 a; = 1508.80; a; = 36.80. $1508.80, amount sent, -| 36.80, commission, -|1472 left for the pur- chase of the flour. The commission to be deducted is :^j of the sum sent. 1052. See Arithmetic, Arts. 936, 937. 1053. Pupils generally find it most convenient to change the time to days, leaving the reduction to lowest terms for the sub- sequent cancellation. 11. 2 yr. 11 mo. 18 da. = 720 da. + 330 da. + 18 da. = W/ yr. 712a; =3204; a: - 4f Note. — The teacher will notice that canceling 100 and 356 by dividing each by 4, and reducing the equation to = 32.04, is of no advantage. The French method of solving an example like the foregoing, is to indi- cate all the work, and then to cancel : ,. 240 X a; X 1068 ^ 3204 . ^ ^ 100 X 360 100 ' (b) 240 X a; X 1068 = ^ X 100 X 360 ; 3 3 (c) ^_ ?imxX99xm ^^_u 2 The first member of equation (^^y^ the terms of this problem ; it is merely intro- ^^ "vT duced here to show the development of the en- U-;^.t«-__j tire surface. As it is difficult to lay off the re- -^V J^ quired number of inches for the arc AB, the '^^J""!^ pupil will appreciate the foregoing method of ^'^^'- 4. determining the number of degrees it should contain. The compasses or the protractor may be employed to construct an angle of 120° at C. The convex surface of a cone is equal to the circumference of the base X \ slant height. An examination of Fig. 4 will show the resemblance between the methods of calculating the surface of a sector and of a triangle. The area of a triangle = \ (base x altitude) ; that of a sector = \ (base X radius). 22. The altitude, one-half the base, and the slant height, form a right-angled triangle ; and the lengths of the two first being 12 in. and 5 in., respectively, the length of the latter is Vi44-j- 25 in., or 13 in. The convex surface = |- (10 X 3.1416 X 13). 23. The entire surface = [|- of (6 X 3.1416 X 10)] + (3' X 3.1416) = (30 X 3.1416) -f (9 X 3.1416) = 39 X 3.1416. Using IT (pi) instead of 3.1416, the circumference of the base = 67r inches. The radius of the sector representing the develop- ment, is 10 in., and the circumference of the whole circle = 207r inches. As the arc of the sector must be 67r inches, it measures in degrees ^ of 360°, or 108°. 202 MANUAL FOR TEACHERS 24. The slant height will be ^ of 6 in. The circumference of the base will equal the arc of the semicircle, 3ir inches; its diameter will, therefore, be 3 in. 1142. 3. Due Sept. (21) 24. Term of discount from July 21 = 65 (62) da. 4. $ 600 yearly interest represents a principal of $ 10000. 5. Length of one fence, (20 + 20 + 20 + 20) rd. ; of the other, (40 + 10 + 40 + 10)rd. 6. The distance between the center of the first and of the last post = 10 ft. X (11 - 1) = 100 ft. Adding ^ of the diameter of each post, gives 100 ft. 6 in. ; and an additional 3 in. at each end to fasten the wire, makes a total of 101 ft. of wire required for each length, or 303 ft. in all. Am. 1143. 7. Troy weight. 8. 43| mo. (g25)i^ per mo. 43 quarter dollars =^$10.75; i of 25^ = 10^. Total $ 10.85. Ans. 11. Without grace, 1% of $400, or $4. Ans. With grace, $4 + ^of $4, or$420. Ans. 12. iof cost = $2; etc. 13. The principal is unimportant. Time = (100 -^ 8) yr. 17. (100-^6) yr. 23.' i% of $1234. 24. 2% of $1234. 25. The number of rings = 60 pwt. -f- 2^ pwt. 26. 2°3'xl5. 27. [Twice (4 + 3) X 10] + [twice (4 X 3)]. 28. $1.12^ + iof $1.12^=-$1.12i + $.28f 30. Selling price = 1 of cost = 1^; cost = -J^-^f = J^. Ans. 1144. 4. 14% profit = 7^; cost = 7^ -^ .14 = 50^; selling price —b*lf. Ans. NOTES ON CHAPTER FOURTEEN 203 17. Cost of 350 tons @ $3.50 = $1225. Selling price = $4.25 X 350 XHU- Note. — The scholars should not use pencils to obtain answers to problems that can be solved at sight, 1145. See Art. 1290. 3. Area of base = 6 sq. in. ; etc. 5. See Art. 1107, Measurements, 7. 8. Changing given dimensions to inches, the number of gal- lons will be 36^ X 3.1416 X 66 -^ 231. 9. 1 cu. ft. = if|^ lb. Cubical contents = (3' X 3.1416 X 51) cu. ft. 10, 11. Careful pupils will be much interested in ascertaining how closely their calculations as to the contents of the measure, agree with the number of cubic inches it is supposed to contain. There should be 231 cu. in. -^ 8, in a quart. The cup used must be cylindrical. Tapering measures should be left until the frustum of a cone has been studied, Art. 1295. The paper box used for ice cream, a frustum of a pyramid, can also be employed at that time. Some of these measurements should be made out of school, and comparisons made as to the results obtained by different pupils and the methods employed by them to secure accuracy. A random measurement will not obtain the correct diameter of a quart measure. After calculating the altitude of an equilateral triangle or the diagonal of a square, the pupil should draw the figure to a scale, measure the altitude or the diagonal, and compare the measured length with the length obtained by calculation. Pupils should ascertain the weight of a cubic foot of water by weighing a quart of water, for instance, etc. 12. Measure the height to which the water rises in the box, etc. If a solid heavier than water, is placed in a rectangular or a cylindrical vessel containing sufficient water to cover it, and the difference in the depth of the water before and after immersion is noted, the volume of the solid 204 MANUAL FOR TEACHERS can be calculated. It will be equal to tbat of a solid whose base is the base of the vessel, and whose altitude is the difference in depth above mentioned. If the water in a rectangular box, whose base measures 5^ by 3 in., is raised l^ in. by the introduction of a piece of marble, the volume of the latter = 5| X 3 x 1| cu. in. This method is useful in determining the contents of a solid of irregular shape. 13. The radius of the base = i of (25.1328 yd. -f- 3.1416) = 4 yd. Volume of cone = (4' X 3.1416 X i of 18) cu. yd. 14. See Art. 1141, 22. 15. The slant height of the pyramid = V24'^ + (^ of 14)'. See Art. 1283, 13. 1146. 6. [(lOf ^ X li) + (3f)2^x If)] X 10840). 7. 1000 grams -^ 279 = weight in grams of a 10-mark piece. Weight in Troy grains = (1000 -^ 279) X 15.432349. Dividing this result by 23f\^ gives the number of U. S. gold dollars. 1000x15 43.2349 279x23 ©22. Note. — 23.22 is changed to a whole number by removing the decimal point two places to the right, and a corresponding change is made in one of the numbers in the dividend. 11. Interest on $237453250 @ 3 %=$ 7123597.50 250000000 @ 4|% = 11250000. 737954700 @ 4. % = 29518188. $1225407950 @x %= $47891785.50 4789178550 -^ 1225407950 - 3.9083 + The interest on the entire amount at 2^% would be $30635198.75, the saving being $17256586.75. Ans. 12. $100 worth of stock costs $85J. The annual dividend is 5% of $100, or $5. This is (5 -f- .8575) per cent on the money invested. 13. $8930-h1.11|. NOTES ON CHAPTER FOURTEEN 205 14. See Art. 1122, 15. The following shows the account as it stands on the books of the Interior Department. The items that appear as debits on Mr. Well's books, here appear as credits, and vice versa. Dr. RICHARD WELLS. Or. 1S82 1882 Jan. 31 To Cash 885 — Jan. 1 By 646 bbl. Flour 9.45 6095 25 Feb. 5 u tt 450 — (I 16 " 1912 bu. Oats .57 1089 84 Apr. 11 " " 615 35 Apr. 4 " 92311b. Bacon .09 830 79 May 30 " " 4162 15 May 3 " 8264 bu. Corn .74 6115 36 June 25 " 345 lb. Bacon .09 81 05 June 20 " 825 bbl. Pork 12.65 4111 25 " 25 " 35 bbl. Vork 12.66 442 75 u 30 " Cartage 65 — " 30 " Penalty 75 — it 80 " CashinfuU 11646 19 1830T 49 18307 49 1148. Multiply the length in feet by the width in feet by the thickness in inches. 16. 3f^ (per ft.)X 15x 16xf X3. 17. The floor contains (36 X 17^) sq. ft., or 630 sq. ft. If 1-inch boards were used, 630 board feet would be required. The number of feet of 2^-inch planks required — 630 ft. X 2^. 18. (150 X 13 X I X 1) board ft. = 1300 board ft. (60 X 14 X f X 2) " " -= 1260 " " (40X15X^X4) " " -1000 " " Total, 3560 " " the duty on which, at $1 per M, is $3.56. Ans. 19. The length of the fence -- 480 ft. + 360 ft. + 480 ft. + 360 ft. = 1680 ft. For a fence 4 boards high, (1680 X 4) running feet of boards will be needed, or 6720 running feet. If the boards are ^ ft. wide and 1 in. thick, the number of board feet = 6720x^x1 = 3360 ft. Cost = $18 X 3.36. 20. The length of the fence = (25 -f 100 + 25 + 100) ft. Surface = (250 X 6) sq. ft. = 1500 sq. ft. As the boards are 1 in. thick, the number of board feet = 1500. Cost = $25 X 1.5 206 MANUAL FOR TEACHERS == $37.50. The number of posts = 250 -^ 64 = 40 ; cost, at 25 i each, $10. The number of running feet of scantling, two strips, = 250 X 2 = 500 ; the number of board feet =: 500 X ^ X 2 - 250 ; cost, $18 X .25 = $4.50. Total cost, $37.50 + $10 + $4.50 - $52. Ans. 1149. 1. (a) A note made payable to the order of a certain person or to hearer is negotiable ; in the former case, an endorse- ment is necessary to transfer its ownership. A note payable to hearer does not require endorsement. A note payable to Charles Naumann (without the words, " or order," or the like) is not transferable by endorsement. If Charles Naumann wishes to sell the note, he must assign his interest in it by another document. Note. — The above is the general rule; in some states there are special laws bearing on the subject: In Alabama and Kentucky, a note to be negotiable must be payable at a fixed place; in Indiana and Virginia, at a bank; in West Virginia, at a bank or public office. In Pennsylvania, it should contain the words " without defalcation " ; in New Jersey, " without defalcation or discount"; in Missouri, "negotiable and payable without defalcation or discount." (h) A person unable to write hfs name, makes his mark, as shown below, in the presence of a witness : his Witness: William X Devers. Theodore H. Ficklin. '"="'' 3. In old deeds, the contents of a farm are given in acres (A.), roods (R.), and poles (P.), the rood being \ acre, and containing 40 poles, or square rods. In long measure, the word pole is occasionally employed instead of rod. 6. 2a:-a: + | + |+18; etc. 7. The distance between the ships is the hypotenuse of a right-angled triangle, whose other sides are 72 mi. and 128 mi., respectively. NOTES ON CHAPTER FOURTEEN 207 8. The first capital = 1 3500 orf of $2500 = $1500; etc. 9. See Art. 1026, 10. 10. See Supplement. 1.40 -$2500; put in ||o 100. 50, the area of 1150. 3. :r' + a;' = hypotenuse^ the inscribed square. Ans. 4. Area of circle = (5 X 5 X 3.1416) sq. in. = 78.54 sq. in. 5. Arc of 90° -- i (10 X 3.1416) in. = 7.854 in. ; the chord = V50in. 6. Arc of 90° in a circle whose radius is 10 in. = 15.708 in. Area of sector = | of (15.708 X 10) sq. in. = 78.54 sq. in. Ans. Area of triangle = -| of (10 X 10) sq. in. = 50 sq. in. ; area of segment = 78.54 sq. in. — 50 sq. in. = 28.54 sq. in. Ans. 8. i^'x 3.1416. Ans. 9. Area of outer circle in square yards = (15' X 3.1416) ^ 9 ; of inner circle = (10' X 3.1416) -f- 9. 10. (125 X 3.1416) sq. ft. 11. [(6'~3')x3.1416]sq. in. is 6 in. ; of the inner circle, 3 in. 12. 36 X 3.1416 : 9 X 3.1416 = 4:1. 13. [(30 X 30) - (20 X 20)] sq. ft. 14. Dividing the walk into four trapezoids, as in Fig. 5, the area of each will be [\ of (30 + 20) X 5] sq. ft. = 125 sq. ft. The broken line, XY, drawn mid- way between the parallel sides measures 25 ft. ; the whole length of the center lines is 100 ft. The area of the walk is (100 X 5) sq. ft. • The radius of the outer circle so \x Y^ l\ /I is to 30 ai V- \l A «- x\i 30 Fig. .5. 208 MANUAL FOR TEACHERS • The area of the circular frame in 11 can be ascertained in the same way. The center of the frame is 4J in. from the center of the glass. The mid- dle line of the frame = (3.1416 x 4^ x 2) in. = 28.2744 in. The area = (28.2744 X 3) sq. in. 15. The area of the first = 240 sq. ft. ; of the second, 960 sq. ft. U53. 18. The surface of the sphere = (4 TrXi) sq.ft. = 3.1416 sq. ft. ; the convex surface of the cylinder = 3.1416 sq. ft. 19. The entire surface =- 3.1416 sq. ft. + (2 X J X 3.1416) sq. ft. = 1| times 3.1416 sq. ft. 20. 4 times (^ circumference) 10 X (^ diameter) 10 400 . . . , surface in square inches. 3.1416 3.1416 1154. 1. Rate per cent = 18750 -^ 12500 = If 1^% of $6000 = $90. 3. The price of silver is now given by the ounce. 4. The interest on $200 at 4^% = $9. $9 ^ .08 = $112.50. Ans. 5. $2100 + $4400 + $13000 + $7200 (90% of $8000) = $26700 = total assets. Total liabilities = $1625 + $5625 = $7250. $26700 - $7250 = $19450. To this, add the amounts withdrawn, $850 + $1075= $1925, making the sum of the capital and profits $21375. Of this, H is entitled to i, $7125, less the amount withdrawn by him, $1075, or $6050. 1157. 9-11. Find the cube root of the numerator and of the denominator separately. 12-15. Reduce to an improper fraction before extracting the cube root ; then reduce the root to a mixed number. 1159. 2. [f X 3.1416 X (f X f X f)] cu. in. Cancel. 3. The volume of the first in cubic inches = ^ tt X -j- = |- tf. The volume of the second = ^ir xl= ^ir. 4. The volume of the sphere = .5236 cu. in. ; the volume of the cube = 1 cu. in. NOTES ON CHArTER FOURTEEN 209 6. The ball contains .5236 cu. ft. Its weight = (.5236 X 7.5 X 1000 -^ 16) lb. 1160. 1. 40 tons were to be moved; there remain 22 tons to be moved in 3 da. In 6 da. 18 men moved 18 tons, so it requires 22 men to remove 22 tons in 6 da., or 44 men to remove 22 tons in 3 da. 44 men. Ans. 3. A diagonal on the floor measures V40^ -f 30^ ft. = 50 ft. A diagonal on one wall measures a/40^~+T4^ ft. = 42.38 — ft. ; on the other wall it measures V3(F+T¥^ ft. == 33.11- ft. 4. Selling price, |6500 + 15% of $6500 = $6500 + $975 = $7475. As no date is given, the note may be assumed to be for 120 da., or 123 da. including grace. Proceeds of $7475 for 123 da. = $7321.76. Profits = $7321.76 - $6500 = $821.76. Ans. Without grace, the proceeds would be $3.74 more, making the profits $825.50. Ans. 5. Let X = face ; = premium ; bank discount = x X — — 8 11a; 400 ^ 360 X = (includina; grace). 100 1500^ ^^ ^ 13^ _ 11^^4265, ^ 400 1500 6000 x-^195x-4:4:x = 25590000, 6151 X = 25590000, a; = 4160.30-. $4160.30. Ans. Without grace, x + ^-~-:^ = 4265 ; etc. 4UL) ioU 7. Longitude difference = 48° 24'; time diflference = 48 hr. 24 min. ^ 15 = 3 hr. 13 min. 36 sec. As the more easterly- place has the later time, the watch is fast. Ans. The above result is based upon the assumption that the " sun time " of each place is used. The difference in the "standard " time of the two cities is 3 hr. 210 MANUAL FOR TEACHERS 9. Jof|of|ofa; = 6; ^ = 6; a;=192. 10. ^-f-|^a;-|+1040; 40a; + 8a; ==40 a: - 5 a: + 41600 ; 13 a; = 41600; a; = 3200. Ans. $3200. 11. 2050: a;:: 41: 69. Cancel. 14. $ 1450 - i% of $ 1450 - 63 (60) days' interest on $ 1450 at 5% = cost of draft. 17. A can do ^ in 1 da. ; B can do ^; both can do -}^ in 1 da., and can do the whole work in T^- da. Ans. 19. See Art. 1150, 4. 1161. The definitions and principles called for throughout this chapter should be formulated, as far as possible, by the pupils, the latter being led through the teacher's questioning to see their mistakes, and to make the necessary corrections. If this preliminary work is done as it should be, the scholars will be ready to understand the definition finally given by the teacher. Too much time, however, should not be wasted on formal defini- tions, as they are of next to no help to a pupil in his mathe- matical work, and it is very unlikely that he will ever be called upon to use them in after life. See Supplement. 1. (c?) A decimal fraction is frequently defined as one whose denominator is 10 or some power of 10. In this place, however, the expression is used as synonymous with " decimal." The rule asked for, refers to the method of "pointing oflf" the product. 7. If all three are opened, they will fill (yV + i^i) ^^ iV of the cistern in 1 hr. To fill the whole cistern will require 15 hr. Ans. 9. As they meet in 15 hr., they must approach each other at the rate of (105 -4- 15) miles per hour, or 7 mi. As one goes 3 mi. per hour, the other must travel 7 mi. — 3 mi. = 4 mi. per hour. NOTES ON CHAPTER FOUKTEEN 211 18. A's capital at end of year is $8000, B's is $10000, O's, 3000: total, $21000. Profits - $18000 + $12000 - $21000 -$9000. A, 10000 X i (yr.) = 5000 8000x1 -4000 9000 B, 6000 X i = 3000 10000 X -I- =5000 8000 0, 8000 20000 : 9000 20000 : 9000 20000 : 9000 9000 : 4050 8000 : 3600 3000 : 1350 A is entitled to his capital at the end of the year, $8000, and $4050 profits, or $ 12050. As he receives $ 12000 worth of goods, his cash receipts are $ 50. B receives $10000 + $ 3600 = $ 13600. C receives $3000 + $ 1350 = $4350. 19. f + |of ^ + 20 + ^ = :r; f + f + ^ + 20 = a:; etc. 20. He sold 800 bbl. at $7.50, which amounted to $6000. The cost was $7200 + $312 + $350 -= $7862. His profits be- ing $138, he must have received $7862 + $138 = $8000. As the flour realized only $6000, he must have received $2000 from the railroad company. 28. 16% of the person's money = 20% of $ 160000 ; etc. 29. 4 men do -| of the work in 60 hr. ; to do the remainder, they would need 120 hr. ; and 1 man would require 480 hr. There are 8 da. of 10 hr., or 80 hr., in which to finish it ; 6 men, therefore, will be needed to complete it, or 2 men additional. 30. Cost = $100 + $5 + $50 + $20 = $175. Selling price, $ 300 less 10 % ($ 270) - $ 20 = $ 250. Profit, $ 75. 212 MANUAL FOR TEACHERS If the commission merchant had refunded $ 20 before making returns, his commission would have been 10% of $280, or $28; and the gain would have been $ 77. 36. {h) See Arithmetic, Art. 668. 38. A, 5000 X i- = 2500 3000 x\^ 1500 4000 B, 6000 0, 4000 X i = 2000 12000 X I = 6000 8000 18 : 4 ::$6000 : A, etc. 39. Proceeds of $12000 for 93 da. = $11814. Amount of $ 10000 for 6 mo. at 6% = $10300. Sum remaining - $ 11814 -$10300 = $1514. Without grace, the proceeds of 90-days note = $11820; sum remaining = $ 1520. 40. Cost, $40 each. 10 were sold @ $ 44 each = $ 440 ; 10 @ $ 46 each = $ 460 ; 15 were sold for $ 100 : total $ 1000. To obtain $900 for remaining 15, he must charge $60 each. 1164. 1. (30 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30+17) da. = 291 da. Exact interest = $ 400 X ^ X |f f da. X 630 = da. 4. Business men generally find 243 " X 820 = 199260 " the number of days' credit to 274 " X 950 = 260300 " which each item is entitled. $820 2400 )459560 da. ^^ ^"^ ^^^^^ ^' °^ ^^^ ^^' ^^*^^ — ,g, , . July 5; $920 is due April 5, or ^' 274 da. after July 5. The equated time is 191 da. after July 5, or Jan. 12. Ans. Using months instead of days, the average term of credit is found to be 6 mo. 9 da., nearly, making the equated time Jan. 14. 6. 7500 : A : : 1200 : 250 7500 : B : : 1200 : 950. NOTES ON CHAPTER FOURTEEN 213 6. Bank stock pays (25 X 7|- ^ 85) per cent interest semi- annually ; the railroad stock pays (25 X 3 ^ 31) per cent interest semi-annually. 9. The bases are equilateral triangles, each side of which measures 5 in. Art. 1107, 7, Measurements. 1165. 1. See Supplement. Quantity is anything that can be measured. See Arithmetic, Art. 1072. 2. |of iiX^XyVxfxAxixV-- Cancel. 3. Time difference = 2 hr. 37 min. 33 sec. 11 A.M. 1 : 37 : 33 P.M. W. ? 83° 3' 0° Longitude difference = ? The longitude difference = 2° 37' 33" X 15 - 39" 23' i5". Longitude of San Francisco = 83° 3' + 39° 23' 15" = 122° 26' 15" west. 5. a; + 2a: + 5a: =11480. 6. The equated time for the payment of $600 is [(200 X 1) + (400 X 2)] yr. -^ 600 = 1| yr. The present worth of $600 due in If yr. =$600^ 1.1 = $545.45+. Another way is to calculate the present worth of each, and to add the results : ($200 ^ 1.06) + ($400 ^ 1.12) = $ 188.68 + $357.14 = $545.82. Ans. The latter is the more consistent way inasmuch as it employs the " present worth " method throughout. The first solution uses the " present worth " method to calculate the value at date, oi $600 whose equated time has been found by the "interest" (bank discount) method. 7. Selling price = $120 +15% of $120 = $138. Asking price = $138 +$12 = $150. I threw off $12 from $150, or 8%. 8. Arithme tic, Art. 1250, 8. ^i) = 39, AC = 62, BO = V52^ + 39^ = 65 = £0. Height of tree = 52 ft. + 65 ft. = 117 ft. 214 MANUAL FOR TEACHERS 9. \/l0.125000 = ^ns. 10. If I gain = j\ cost, the gain = ^^ cost X f = J cost = 16| per cent. 1166. 4. [1^(^ + A)]weeks. 15. B's gain of $1400 is -^ of total gain ; ^ of total = $200; A's gain, ^ of total = $ 1000. 1167. 4. Let X = cost per barrel. 75% of 500 a; X. 02^ = 80.85. 5. A does ^ in 1 da. ; B does ^^ in 1 da. A does -Jf of the work, or |, and B does -f^ of it, or f , leaving -| to be done by C in 4 da. To do the whole work C would require 4 da. ^ | - 18 da. 6. A ditch 20 yd. X 18 in. X 4 ft. is dug in (3 X 10) hours by 72 men. A ditch 30 yd. X 27 in. X 5 ft. is dug in (9 X 16) hours by ? men. 72 men X 3 x 10 x 30 X 27 X 5 20 X 18 X 4 X 9 X 15 Ans. 7. £2400 income is produced at 3% by bonds whose face value is £80000. Their cost =-- £80000 X .94f = £75500 - $367433J, 12% of which = $44092.00. $4.86| X (2400 ^ .03) x 94| x .12. 8. a; - — = 30 + 30% of 30-39. 100 ^ 100 :c- 40 a; = 3900; etc. 10. [(15 4- 10 + 15 + 10) X 9f] + (15 X 10) = number of square feet in the walls and ceiling = 637^ sq. ft. = 10^ sq. yd. The cost = 21c?. X ^^ -- 1487|c?. ; etc. 116a 1. J + i =1^-1.41666 + ; ^ + ^+^ = 1^=^ 1.41429 ~. NOTES ON CHAPTER FOURTEEN 215 .7409375 -i- 237100 = .007409375 -^ 2371. See Arithmetic, rt. 668. 2 ^i , 2 fb ^^3^U ■H 4'^ 13^3" 10 ~13' 2 "3^ -2A = 2 3"^ 25_2 12~3^ 12 25 8 25' 8i-7f^ =-v- ^¥- 41- -39^|J 10 8 41. _ 750 + 312 -1025 37 13 "^25 '39' 975 975 975 _2. 3' 37 975a: _2 3' Clearing ; of fractions, 37 ^ 650 ^; ^=-/3V- Ans. \i of ^r of I of tJ^ X if X H X -V- a: = smaller number ; ^ + t^^ = larger. a; + :^ + _7^^ij|; 126a:+126a: + 49=:113; 252a:=113- 49^.64; a:=2V2-if;-^ + A^if + A-=AV + T¥^=--T¥^ = t\- ^^5- if ^^^^ t\- 3. 16s. 4id = 196id ^w5.-£(-| of 196^)^240. I mile -= 1000 meters ; 1 mi. = 1000 m. ^ | = 1600 m. 17 mi. = 1600 m. X 17 = 27200 m. ; 6 furlongs = 200 m. X 6 - 1200 m. ; 82iyd.^l550j^^xi^--^75m. ^ ^ 1760 2 27200 m. -f 1200 m. + 75 m. = 28475 m. Ans. 27 yd. 2 ft. 9 in. = 1005 in. ; 17 yd. 1 ft. 11 in. - 635 in. (635x1) sq. in. cost $25.40; 1 sq. in. -= |2540 ^ 635 ; (1005 X i) sq. in. = ($25.40 ^ 635) X 1005 X i = $35.17f 4. Let a: = B's money ; a;+ 17.50 = A's. 2a: _ a: +17.50 5 3 A's rate is ^ of B's; B's time is -i-J of A's. To run the whole distance, A needs 34 min. ^ ij — 36 min. If he runs 2-J- milea 216 MANUAL FOR TEACHERS in 1G| min., in 1 min. he runs 2^ mi. -^- 1G|-, and in 36 min. he runs (2^ min. -^ 16|) X 36 = J mi. X /^ X \«- = 5 mi. Am. 5. x-li% of a; = 96084; ^-^=96084; 800a;-15a: 800 = 76867200 ; 785a; = 76867200 ; x = 97920. 6. For the information of the teacher, the following method is given : V2-1 _ V2-1 ,, V2-1 _ 2-2V2 + 1 _ 3-2V2 _3 q^/q- V2 + 1 V2+1 V2-1 2-1 1 We learn in algebra that the sum of two numbers {x + y) multi- plied by their difference (x — y) gives the difference of their squares (x^ — y'). If the sum of V2 and 1 be multiplied by their difference (V2 — 1), we obtain the difference of their squares (2 — 1). Multiplying the numerator also by V2 — 1, we retain the equality and obtain a divisor that has no decimals. See Art. 1169, 3. It is not expected that this method should be given to the pupils. 7. For 42 da., 50 men were at work. To do the same work, 30 men would have required 70 da., or (70 + 40) da. to do the whole work. 110 da. — 84 da. = number of days the contractor would have been behindhand. 8. The number of square feet in the wall = (23| + 15| + 23| -f 15|) X llj = 928^ sq. ft. The two windows contain (19 X 5) sq. ft. = 95 sq. ft.; the fireplace contains (4^ X 6) sq. ft. =27 sq. ft.; the door contains (7^ X 3^) sq. ft. = 26J sq. ft. ; a total of 95 sq. ft. + 27 sq. ft. + 26J sq. ft. = 148J sq. ft. There remain to be papered 928J sq. ft. - 148J sq. ft. = 780 sq. ft. = ^f^ sq. yd. A roll of paper contains 12 X ff sq. yd.; the number of 11 -11 u 4.1. f "780 /lov. 26\ 780x1x36 , ., rolls will be, therefore, -^ - (12 X -j =. -^_^^-^; and ite $4.08X780X1X36 ^ Q3Q ^,,. 9 X 12 X 26 ^ NOTES ON CHAPTER FOURTEEN 217 9. a; + ^=336. 10. The "present worth" of $365 due in 30 da. = |365^ 1.005 = $ 363.18 + = the cost of the horse in cash. The " pres- ent worth " of the selling price =- $435 ^ 1.02 = $426.47+. Gain= $426.47 - $363.18 - $63.29, which is 17.43% of the cost. If the seller has the note for $435 discounted at a bank, he will receive in cash $435 -$8.70 - $426.30. If he uses this money to buy the note he has given, he should pay, at bank rates, $365-$1.82i = $363.17f The profit would be $426.30- $363,171 = $63,121, which is 17.38% of the cost. 11. £57 Is. Sd. = 13700c^. ; £2 lis. i^d. = eU^d. 13700 x^Xx = interest - ?55^ =. 616|, 2055 a; = 1233, ^ = UH ; UU y^- = 7 mo. 6 da. Ans. 13. A man that does only -f of a day's work, does 14 da. less work in 84 da. than the average. The contractor therefore loses, in 84 da. on three men, 14 da. + 12 da. + 9-^ da. = 35-^ da. He gains on two others 10-^ da. -f 8f da. = 18j^ da. The net loss = 35-|- da. — ISy^Q- da. = 16-g-| da. The extra 17 men have to do the equivalent of 16|^|- days' work ; each has, therefore, to do 16-^1- days' work -v- 17 = -||- of a day's work, or ^ less than the average. 14. Making no allowance for waste, etc., two strips, each 260 ft. long, will be needed for two sides; two strips, each (93 — 3 — 3) ft., or 87 ft. long, will be needed for the other two, or 520 ft. -j- 174 ft. = 694 ft. = 231^ running yards, 1 yd. wide, making 231-|- sq. yd. Cost at 90^ = $208.20. The surface to be carpeted = (260 - 5) ft. by (93 — 5) ft. = 85 yd. X 29^ yd. Cost = $ 2.09 x 85 X 29^ -^ f f = $ 4936.80. Total, $4936.80 + $ 208.20 = $ 5145. Ans. 15. Since the meeting-place is twice as far from A as from B, the first man goes twice as fast as the other ; the latter, there- 218 MANUAL FOR TEACHERS fore, walks 2^ mi. per hour. If x is the distance between A and B, the first will require f hr., and the second ^^ hr. = — hr. 5 2^ 6 — — ^=1; a^ = 5. An8. 5 mi. 5 5 16. Let X = number of miles between A and C, Then x — lb — distance between B and C. i ^i^li^ = time required for ^ 15 B x-15 ^ 15 first train to run from B to C ; — = time required for second train to run from A to 0. As the latter train leaves 3 hr. later and arrives one-half hour later, the running time of the first is 2^ hr. longer. a:— 15 __^ I 5 . 15 ~25'^2' 10ar-150=:6a;-f375; 4 a: = 525; a:=131i. Am. 131 J mi. 1169. 2. The width of the road = (60 -^- 16^) rd. ; its area = (104 X 60 -^ 16-^) sq. rd. -= [104 X (60 ^ 16|) -^ 160] acres. j^ , $154x104x60x2 (tjQft. ^'' '''' '-= 33^n60 = * ^^^• n.1 , c A- $200x104 ^nf. The cost of grading = - — — — = $65. The cost of fencing = $^ X 104 X 5| - $286. 3. See Art. 1168, 6. V5 + V3 _. V5 + V3 ^^ V5 4- V3 _ 5-f 2V15 + 8 V5-V3 V5-V3 V5 + V3 5-3 • ^8 + 2Vl5^^ ^ 2 NOTES ON CHAPTER FOURTEEN 219 V5-V3 _ V5- V3 ,, V5 -V3 _ 5-2Vl5 + 3 V5 + V3 V5 + V3 V5-V3 5-3 = 8^AV15 = 4-VI5. A ■ 4 + Vl5 - (4 - Vl5) = 2 Vl5 = V60. 4. A cubic centimeter = .3937' cu. in. = (.3937'-^ 1728) en. ft. Its weight = weight of 1 gram = (.3937' X 1000 ^ 1728) oz. - [(.3937' X 1000) -^ (1728 X 16)] lb. A kilogram - [(1000 X .3937' X 1000) ^ (1728 X 16)] lb. The weight of the anchor in kilograms = 6500 lb. -^- the number of pounds in a kilogram ; or, 6500 X 1728 X 16 1000 X .3937 X .3937 X .3937 X lOOO' 5. :r-(a;Xyf^X^%\) = 1500; a;- — =1500; 500 a; - 7 a; = 750000 ; 500 493 a; = 750000 ; x = 1521.30 -. A71S. $1521.30. The proceeds of the new note + $200 must pay the note of $2000 ; the proceeds must therefore be $1800. ^-(^XTf^xA'^) = 1800. 1170. 2. (80.005 - .013) -^ 88. Sx 5. Let x= number of cents received by Thomas. Then — 6x . ^ = number received by Henry, and — = number received by Richard. ^^ ,4-^+1^4.14; 25a; + 15a; + 6 a; =103.50; 46 a; =103.50; a; = 2.25. 6. 272 liquid quarts = 231 cu. in. X 272 -^ 4. A dry quart = 2150.4 cu. in. h- 32. 231 X 272 X 32 231 x 272 x 320 Number of dry quarts 4 X 2150.4 4 X 21504 220 MANUAL FOR TEACHERS 7. The tub holds 12^ qt. X 4| = b^ qt. Both pipes dis- charge 12^ qt. + 83 qt. = 95^ qt. The time required to fill it = (54J H- 95^) min. 8. The number of hours that must elapse before all will again be together at the starting point, is the least common multiple of -^, I", -If. The least common multiple of the numerators is 70. The smallest fraction that will contain the above fractions an exact number of times must have 70 for its numerator, and for the denominator the largest number that will divide 36, 9, and 99 without a remainder; i.e. the greatest common divisor of these numbers. The G. C. D. is 9, and the fraction is ^-. In ^ hr., therefore, A, B, and will be at the starting point. A will have walked around the circle (^^s\) 56 times ; B, (^ ^|) 35 times ; C, (^ ^ |f ) 22 times. Note. — The scholars will readily understand that the fraction which is the least common multiple of ■^^, |, and ff, should have 70 for its numerator. The following may make clear to them why 9 should be the denominator: i , or -, or — should be a whole number ; a; 36 9 99 i.e. — X — .or -. or — should be a whole number, a; 5 2 35 An examination of the second line, in which the divisors are inverted, will show that 70 contains the three denominators, 5, 2, and 35, an exact number of times ; the numerators, 36, 9, and 99, should contain x an exact num- ber of times ; x, therefore, must be a divisor of these numbers, etc. 1172. This work may be slightly abbreviated by combining the interest on the annual interest into one item of 6 years' interest instead of the three separate ones of 3 years' interest, 2 years' interest, and 1 year's interest. In beginning a new topic, how- ever, pupils should not be confused by short methods. 2. $ 1200 -f $ 300 -f (4 -f 3 -f 2 -f 1) years' interest on $ 60. For partial payments on notes bearing annual interest, see Art. 1308. The special rules for New Hampshire and Vermont will be found in Arts. 1309 and 1310. NOTES ON CHAPTER FOURTEEN 221 In states in which the collection of annual interest is not allowed, the teacher should omit this topic. 1173. In the older states, time should not be spent upon this topic. 1176. 1. J of I a section = I of 640 A. 2. I" of ^ section measures 80 rd, by 160 rd. 3. A line from the southwest corner of Sec. 1, to the north- east corner of Sec. 30 (see township diagram on the opposite page), is the hypotenuse of a right-angled triangle whose perpendicular, the eastern boundaries of Sees. 11, 14, and 23, is 3 mi. long ; and whose base, the southern boundaries of Sees. 20, 21, 22, and 23, is 4 mi. long. 6. The number of rods of fence = 80 -f 160 -f 80 + 160 = 480. The number of feet = 16^ X 480 = 7920. A fence 4 boards high requires 7920 ft. X 4, or 31680, running feet of boards. If the latter X ^ = 15840. If the latter are ^ ft. wide, the number of board feet ~ 31680 1184. 2. 26.50 X .85. 4. Multiply 135 by 69, and point off two places in the product. 5. Find the base. 6. 8.50 francs X (10 X 1 X 3.25). 7. Each dimension can be expressed in decimeters, 105 X 80 X 65, whose product is the number of liters ; or the product of the dimensions in meters — 10.5 X 8 X 6.5 — may be multiplied by 1000. 8. 0^.75 means .75/., the denomi- nation in France being generally written before the decimal. 9. 1.25 marks X [(68 ^ 10) X 36] ; i.e. 1^ marks X 6.8 X 36. 10. The number of liters = 50 X 40 X 30 = 60000 ; 92% of this tuimber gives the weight in kilograms (kilos). 1186. These problems are given for practice in obtaining the approximate values of the metric units in terms of our weights and measures. The use of 39.37 in. makes the work too tedious. 13. 4 in. by 4 in. by 4 in. A quart = -2-|J- cu. in. 222 MANUAL FOR TEACHERS 14. A hectoliter = 100 liters = 6400 cu. in. - (6400 h- 2150.4) bu. 6400 cu. in. = (6400 ^ 231) gal. 15. A liter of water, 64 cu. in., weighs a kilo. 1 cu. in. ot water = {^^ oz. ; 64 cu. in. = [(64000 -^ 1728) ^ 16] lb. = 4000 lb. -^ 1728. 16. 400000000 in. = J circumference. See Arithmetic, Art. 1177. 17. A square meter = (40 X 40) sq. in. 18. An are = (400 X 400) sq. in. A hectare = [(400 X 400) X 100] sq. in. 19. Hectometer = 100 meters = 4000 in. 20. A stere = (40 X 40 X 40) cu. in. = (64000 -^ 1728) cu. ft. 21. 1000 grams weigh (4000 lb. -?- 1728) ; 1 gram weighs 4 lb. ^ 1728 -= 28000 grains ^ 1728. 22. A kilometer = 40000 in.; a mile = 63360 in. ; a mile = (6.336 ^ 4) Km. 1197. The average pupil should be permitted to use a pencil for his first solution of these problems. 1. Let X = the value of the second suit. Since $12 and the overcoat = 2x, the overcoat = 2 a; — 12. The second suit (x) and the overcoat (2x— 12) = three times the first suit (36). a;-f 2a:-12 = 36; etc. The second suit is worth $ 16 ; the overcoat, $20. Ans. 2. a; -22 + ^^ = ^; etc. 4 3 The arithmetical analysis might assume some such form as this : The remainder + J of the remainder, or -J of the remain- der = ^ of original sum. The remainder = -J- of original sum X |- = ^ of original sum. The sura lost is 1 — -j^, or -J-J- of original sum. As this is $ 22, the original sum = $ 22 X ff = $ 30. Am. NOTES ON CHAPTER FOURTEEN 223 3. Let a; = time past noon; a; +12 = time past midnight. ^^07+12. 5^^^_pi2; 4ar=12; a; = 3. The time is 3 hr. 5 past noon, or 3 p.m. Ans. 4. At 3 o'clock, the hour hand is 15 minute spaces in ad- vance. To be only 5 spaces behind, the minute hand must gain 10 spaces. While the minute hand goes 1 space, the hour hand goes ^ space ; so that each minute, the minute hand gains \^ space. To gain the 10 spaces necessary, the minute hand must travel (10 ^i|-) minutes = 120 min. -^ 11 = 10|^ min. The time is 10^ min. past 3. 5. 1. A = f B; A = 4 B; 5 B -= 30. B's age = 6 yr. ; A's age = 24 yr. Ans. 6. A takes $ 15 less than f of the profits. If his capital is $30 less than f of the whole, the latter must be double the profits, or $ 1440. A's capital = $ 525 X 2 = $ 1050 ; B's = $ 390. Ans, Or, A takes -f-|-|. or jf, of the profits ; he owns, therefore, Jf of the capital. If ff of the capital + $ 30 = f , or jf , of the capital, ^V of the capital = $ 30 ; etc. 80 7 . Let X = the number of sheep ; — = cost of each ; x— 5 X , . . 2x-Vd . 1-1 2a;-10^80 = number remaining ; = number sold ; — X — = ^3 3 a; 160 a; -800 • . .^ = sum received = 40. 3a; 160 a; -800 = 120 a;; 40 a; = 800; a; = 20. ^Im. 20 sheep. Or, if he received $40 for f of the remainder, he would have received $60 for the remaining sheep. $60 being f of $80, f of the sheep remained, and \ of them died, or 5 sheep. The whole number was, therefore, 20 sheep. 8. Let X = A's age ; then a; + 10 = B's age ; X a;+10 , - = — ^- — ; etc. 2 3 224 MANUAL FOR TEACHERS 1198. 3. 7 ) 19 mi. 180 rd. 2 yd. ft. 9 in. 2 mi. 254 rd. 1 yd. 2 ft. 8| in. Xl2 33 mi. 172 rd. yd. 2 ft. 1^^ in. 4. 18 hr. 24 min. 12 sec. -^ 15 = 1 hr. 13 min. 36| sec. 1 hr. 45 min. time difference = 1° 45' X 15 = 26° 15' difference in longitude. As the place has the later time, it is more east- erly. 7. Calling it 1 in. thick, the number of board feet = 16 X f = 12. $ 40 X. 012 = 48^. Ans. $ 40 X (16 X f X 2|) -^ 1000 = $ 1.20. Ans. 9. [96 (in.) X 90 (in.) X 48 (in.)] ^ 2150.4. 10. If the strips run lengthwise, their number will be 8 yd. -^} yd. = 10J. The number purchased must be 11, each 9 yd long, or 99 running yards of carpet. XVIII NOTES ON CHAPTER FIFTEEN While the work contained in this chapter is intended more particularly for use in such schools as extend their instruction beyond the eighth year of the elementary course, it can profitably replace some of the less useful arithmetical topics taught during the eighth school year. 1199. These exercises should be taken up without any prelim- inary explanations. Their previous work in simple equations has so familiarized the pupils with the use of letters to express numbers, etc., that they need no assistance in the first ten exam- ples. The necessary technical terms should be employed as occa- sion requires, and their meanings should be made clear ; but exhaustive treatment of the diflferent operations should be left for the study of the science of algebra in the high school. 1200. The explanation of the meaning xy, ahc, etc., may be deferred until Art. 1238. For the present, the use of the word coefficient may be limited to simple numerical ones, as given in the text-book. The teacher should not yet explain that in the expression bxy,bx may be considered the coefficient of y; nor that in ^abc, 9 a may be considered the coefficient of he, and 9 ah the coefficient of c. 1204. So far, the pupils have been required to add only single columns containing the same letters. When the signs are alike throughout, as in Art. 1199, they have found the sum of the coefficients, annexed the letter or letters, and prefixed the ■ 225 226 MANUAL FOR TEACHERS common sign. When the signs are unlike, the difference between the sums of the coefficients of the positive and of the negative terms is written, preceded by the sign of the greater sum. It will scarcely be necessary to state to pupils that algebraic expressions containing dissimilar terms are added by placing the plus sign between them; thus the sum of 4a6 and 3ac, for instance, is written 4a6 + Sac. 1205. The expressions employed in the preceding exercises are called Tnonomials, or algebraic expressions of one term. Those of more than one term are called polynomials. A polynomial of two terms is called a binomial; one of three terms, a trinomial. 6. The scholar will readily see that in the addition of poly- nomials, each column should contain similar terms; i.e. terms containing the same letters. That the letters should also be afiected by the same exponents, need not be told him for the present. 1207. From some of the preceding examples, may be seen the use of the plus and of the minus sign to indicate direction north and south, and east and west ; past and future time, etc. In 2, is required the difference between —10° and -f-90°. In 6, there is asked the sum of +40^ and - 50^. Calling the dis- tance north of the starting point -\- 50 miles, in 8, and the dis- tance south — 70 miles, the required location will be (+ 60 miles) 4- ( — 70 miles) == — 20 miles, or 20 miles south. 6 and 8 are problems in algebraic addition ; 10, like 2, is a problem in subtraction. The results of a man's transactions dur- ing a month are ascertained by subtracting the value of his pos- sessions at the beginning of the month from their value at the end. In 10, a man is worth —$250 on Feb. 1 ; deducting from this +$150, which represents his condition on Jan. 1, we obtain -$400. The operation maybe indicated thus: (-$250) — (-|-$150)= —$400, the minus sign in the result indicating a loss. NOTES ON CHAPTER FIFTEEN 227 The algebraic analyses of these problems, if asked at all, should not be required until the pupils have solved them in their own way. The main object of teaching subtraction at this stage, is to enable the scholars to understand the reasons for the change of signs that accompanies the removal of a parenthesis preceded by a minus sign. See Art. 1210. 1210. Considering {a) as an example in / n -ri fL\ OA subtraction, it may be made ^ ' ^. {o) 84 . '. J Take 49-25 _ .Q , or one m addition (6) by cnang- ing the signs of the subtrahend. It may then be written 84 -49 + 25. 1211. The pupil will readily ascertain that a parenthesis pre- ceded by a plus sign may be removed without any alteration being required in the signs of the quantities enclosed within it. 57 + (33 - 16) = 74, may be written 57 + 33 -16 = 74. In 2, the signs of the quantities within a paren- ^ 4_ qo thesis must be changed. The first number ^ , i aq i op; within the parenthesis, being without a sign, is QQ positive ; it therefore becomes negative when — 63 — 25 ^^^ parenthesis is taken away. The equation — -— — — then becomes 92 — 63 - 25 = 4. In 5, the 92 — 63 — ^ 25 multiplier 4 affects only the quantity within the parenthesis. The brackets heretofore used, have been omitted in 5 and 6, to make these arithmetical equations resemble more closely the algebraic equations of Art. 1213. 5 becomes 75 4- 60 - 40 = 95 ; 6 becomes 75 - 60 + 40 = 55. 1213. It has not been considered necessary to give any pre- vious practice in multiplying simple algebraic polynomials by an ordinary number. The average pupil will readily understand that 6 times two x = twelve x. 1. 12a:- 30 = 5a; +12. 12a; -5a; = 12 + 30; 7a; = 42; x = 6. Proof, 6 (12 - 5) = 30 + 12 ; i.e., 6 times 7 = 42. 228 MANUAL FOR TEACHERS 2. 7a;+14 = 3a; + 50; etc. 3. 15 + 5a; +16 = 61; etc. 4. 48--3a; = 52-4a;; etc. 7. 2a;-2-4a; + 38 = 3a:-9; etc. 8. 12a; -30 -5a; = 12; etc. 9. 5a7-12a; + 30 = -12; etc. 10. ll-3a;+10a;=38; etc. 1215. 12. 3a;-3-2a; + 4 = 12. 13. 6a;-6-4a; + 8-3a; + 9 + 24 = 0. 15. 14a; -16 -18a; -36 = 12a; +15 -6a; -12. Transposing, 14a; - 18a; - 12a; + 6a; = 15 - 12 + 16 + 36. Combining, — 10a; = 55. Changing the signs of both members, 10a; = -55. Or. X = - -5i. 17. 39 4 ' 5a; .a; 4 "^2 =¥+ 15a; 4 ' etc 18. 2a;: 4 -5- ¥+ 13 6* 19. 3a; 4 f 9=2a; +¥- X ' i 22. X — 20 = ^ + 60. 1216. 16 ^ = 3. Multiplying by 16 l,3}a;- -7 = 48; or - Lla; 3 Clearing of fractions, 11: i;-2] L = : 144; etc. 2. 3a;. 8 = a;-60; ; etc. 7 = 4a NOTES ON CHAPTER FIFTEEN 229 3. r, + | + a; + f + a;-hf + ^ + f-=99. 3 4 5 6 6 8 3 6. a; -|- 12 = son's present age ; 2 a; + 12 = father's present age. rc+12 + 2a:4-12 = 138; 3a; -138 -24 = 114; X = 38, the son's age 12 yr. ago ; 2 a; = 76, the father's age 12 yr. ago. The present age of the son is 50 yr. (x + 12) ; the present age of the father is 88 yr. (2 a; + 12). Ans. 7. (80 + a;) = 2i-(60-a;); etc. 8. 2(ll+a;) = 25 + a;. 9. Let X = the number of gallons originally in the cask. X 3 a;. 3 a; - = amount drawn off, leaving — in the cask. 60 — = num- 4 ^4 4 ber of gallons required to fill the cask. 24 = 60 - 5;^ ; 4 96 = 240 -3a;; etc. 10. 1-40 = 104. 11. a;4-430_ X = 4 + 76 X Clearing of fractions, x + 430 = 4 a; -{- 76. Transposing, a; — 4 a; = 76 — 430. Combining, — 3 a; = — 354. Changing the signs of both members, 3 a; = 354. Or, x= 118, the smaller number; X + 430 = 548, the larger number. 230 MANUAL FOR TEACHERS 12. Let X = the number of $2 bnls. Then 29 - x = the number of $ 5 bills. 2a; + 5 (29 -a;) = 103; etc. 13. 4(a;4-4) = a: + 34. 14. (a; + 3)X — = 225. Multiplying, ' M£±^ = 225. X Clearing of fractions, 180a; + 540 = 225 x ; etc. 15. 33(a;+l) = 40a;+12. 16. The numbers are x and x -\- 17. Then a; + a;+ 17 =47; etc. 17. Let X = number of years. The mother's age will then be 41 + x, and the son's 5 -f a;, 5 + a;=i(41+^) = ^^; 15 + 3a; = 41 + a;; etc. 1218. 1. Substituting the value of 8 a;, the equation becomes 16 + 7y = 44; etc. 2. 9 + 52 = 34; 5z = 25; z = 5. Ans. 1219. 11. 3a;+14y=78 2a;+14y = 66 Subtracting, x =12 Substituting this value of x in the first equation, 36 + 14 2/ = 78 ; etc. 14. Multiplying the first equation by 2, we have 2x-\-2y = 30. Subtract the new equation from the second one, 2 a; -f 3 y = 38, thus finding the value of y. Substitute this value in either of the original equations. NOTES ON CHAPTEH FIFTEEN 231 17. Multiplying the first equation by 3, and the second by 2, we have 6 a; + 9 y = 120 and 6 ^ + 4 y = 70. 18. Multiply the first by 2, and the second by 7. 19. Multiply the first by 9, and the second by 5. 20. Multiply the first by 8, and the second by 3. 1221. 21. If we add, we have 2.t = 22. Or, subtracting, 2?/^= 14. By adding, y is eliminated; by subtracting, x is eliminated. From this example, pupils should see that either addition or subtraction may be employed to eliminate one of the unknown quantities ; and that either of the two may be elimi- nated, as may be found convenient. 27. hx-^y^ 5, 28. 3:r4-5y-=-8, 3a;-5?/--4. 2a;- ?/ = 12. 29. - 10 re + 2/ = - 1, (1) Subtract (1) from (2). - 5a: + 2/= 9. (2) 30. 3a; + 8y-204, 31. 3:r + 2?/-252, 10a: + 5y-160. 7a: + 5y---609. 34. 3.r + 7-15y-20; 3:r - 15?y =- - 27. (1) 7a;- 6 --10?/+ G; 7a;-10y- 12. (2) Multiply (1) by 7 and (2) by 3, to eliminate x\ or (1) by 2 and (2) by 3, to eliminate y. 35. 51 a; + 44 2/ -804, (1) Multiply (1) by 8, and (2) 45a;-32y= 72. (2) by 11. Add. 37. 2:c - 22 - 2 ?/ + 18 = 6, The pupils should be taught 15 re -I- 135 = 32 •?/ — 96 ^*^ indicate the common de- nominator of 2/ — 3 and 15, as 15 {y — 3), which contains 2/ — 3, 15 times ; and 15, (y — 3) times. 39. 2re + 52/ + 3 = 18rc-242/-12; - 16a;+ 29y = - 15. 4a;-73/ + 5 = 5a;-102/ + 10; -x-^Zy = h. 232 MANUAL FOR TEACHERS 1222. 1. a: + 2/ = 37 ; This problem can also be solved 2a;-f-3y = 96. ^7 tbe use of one unknown quan- tity, by calling the numbers x and 37 — X. The equation becomes 2a; + 3 (37 — a;) = 96 ; or, 2a: + 111 -3a; = 96. 2. Using one unknown quantity, the numbers are x and a; + 28. 5 a; - 2 (a; + 28) = 197. 3. 5a; + 3y = 37; 6a;-5y=10. 4. a;-fy = 65; a; — 2/ = 19. By one unknown quantity, a; + (a; -f 19) = 65. 5. a; -f 2/ = 32 ; 2a; + Sy = 103. See Art. 1216, 12. 6. a; + y = 25; 7a; + 5y=145. 7. 10a;-f 42/ = 38; 6a; + 7y = 32. 8. 5a; + 3y = 375 (cents); 8a; 4-y = 505 (cents). 9. 125 (a;) + 45 (4a;) + 10 (8a;) + 5 (i a;) = 1550. 11. a; + y=19; y+10a;-(a; + 10y) = 45. 12. 13a; = 5y; a; + y=126. 13. 15a; = 83/; a; — y = — 147. -^ being 2i, proper fraction, any equivalent fraction must have a denominator exceeding the numerator. 1223. 4. Eliminate 2 by comparing the first and the sec- ond equation, multiplying the latter by 5. Multiply the second by 3 and compare with the third equation, eliminating z. 5. First eliminate y. 6. a; + a; + y=42; 3a; + 3y - a; + y = 96. 7. 15a;-25y + 30 = 4a; + 2y; 11a; — 27y = - 30. 96-3a; + 6y = 6a;-f 4y; -9a; + 2y = — 96. 9. 15a;~9-9a; + 57 = 24-6y + 2a;, 16a; + 8y - 18a; -f 14 = 12y + 36 - 4a; - 5y. Transposing and combining, 4a;-{-6y = — 24; 2a;-j-y = 22. Eliminate x by dividing the first equation by 2 ; etc. NOTES ON CHAPTER FIFTEEN 233 2. Let X = number of B's chestnuts ; a: + 18 = number of As chestnuts. rr+18 + 4 = 4(a; + 4). 3. rt' + y = 8; 23a; + ITy = 166. 4. a; + y = 55; a; + z = 62 ; 3/ + z = 83. Comparing the first two, we get y — z = — l \ adding this to the third eliminates z ; etc. «• ^ = 1+^ + 1+24- 7. .-1 = 510. 8. .-^=147. 10. 4^.= f+16. 7 5 11. Let a: = value of the clothes, a: + 280 = yearly wages. -J {x + 280) = wages for 6 months = x -{- 130. Clearing of fractions, x + 280 = 2x-\- 260 ; etc. 1228. 14. 3 a: + 6 2a; — 3 2a;(3a; + 6) 6a;^4-12a; -3(3a;+6) -9a;-18 6a;"' + 3a; -18 Ans. 1236. 5. 5a;'^ + 85-3a;^ + 63 = 198; etc. 7. a;2 + 2a;+l-a;'^-49; etc. 8. 43/^ + 20 — 63/^+54 = 24; etc. 9. Clearing of fractions. Art, 1221, 37, we have (z -f 7) (z — 9) :(z-5)(z-3). Performing the multiplication indicated, z^-2z-63 = z'-8z+15; 6 z = 78 ; z = 13. Ans. 234 MANUAL FOE TEACHERS 10. 20x(x-{-l) = S0x(x- 1) ; etc. Divide by x. 13. (a; + 4)(a; + 4) = 8a; + 80; etc. 15. 6A-^ + 36 = 5a;^+72. 16. x"^— Qx-}-9 — (x^—lOx-l- 25) = 12 ; removing the paren- thesis. x'-6x-i-d-x'-j-l0x-2b=12; etc. 18. The common denominator is 36 a:. Clearing of fractions, 9a;' +144 = 4a;' + 324; etc. 19. (a:+7)(.'r - 9) = (a;-3)(a; — 5). 20. (2/-9)(2/ + 7) = (2/-3)(y-6). 1237. 1. Let X --= the breadth ; 2x = the length. The area = a;x2a;-2a;' = 1800; etc. 2. Let X = the length of one edge. The area of one face = x^ ; that of six faces is 6a;', and is equal to 96 sq. in. \ 6a;' = 96; etc. 3. Let X =^ one number ; — = the other. 5 xx^=^ = their product = 80 ; etc. 5 5 100 100 25 6. 40% of a number (a;) = ^ ; 30% of ^ = :i of ^' Q Q 5 5 10 5 .|. 1 = 300; etc. , Let X = the length of the perpendicular ; -^ = the length of the base. The area = -I x X —-]— — --• 2V 4y 8 §^=96; 3ar' = 768; a:" = 256; a:=zbl6. 8 Neglecting the negative result, the perpendicular measures NOTES ON CHAPTER FIFTEEN 235 16 rd., and the base 12 rd. The hypotenuse = Vl6"' + 12' rd. - 20 rd. 8. x' + (— )'= 15' ; x' + ^= 225 ; 16a;' + 9^;'--= 3600 ; etc. 9. (.7; + 9)' = a;' + 15'; a;' + 18a; + 81 = a;' + 225 ; etc. 10. (a;+l)'-a;' = 49; etc. 1238. The pupils should be informed that the product of the numbers represented by two letters is represented by writing the letters together ; thus a times b is written ah, m times n is written mn, just as 3 times x is written 3^7. 1242. 1. a;' + 60; + 9. 6. x^-\-2x-\-l. 2. a;'-12a; + 36. 7. :r'-4a; + 4. 3. ^'-8^ + 16. 8. a;' -10 a; + 25. 1244. 1. a:' + 6:t'+9 = 49; ^ + 3 = ±7. Ans. 2. a;' - 12:r + 36 =- 64 ; a; - 6 = ± 8. Ans. 5. a;^ + 18a; + 81 = 19 + 81 = 100; .t + 9 = ±10. Ans. 6. a;' + 2a-+l = 24 + l = 25; a: + I = ± 5. Ans. 7. a;'- 14a; + 49 = 15 + 49; a;— 7 = ±8. Ans. 1246. 1. a;'-6a; + 9 = 7 + 9;a;-3 = ±4; etc. 2. a;' -12a; + 36 = 108 + 36; etc. 3. a;'+2a; + l =48 + 1; etc. 1247. The first member is made a complete square by adding the square of -J- of the coefficient of x. 1248. 1. a;' + .r + i = 12 + i = 4^. a; + -i- = ±|-; a; = f, or— f = 3or— 4. Ans. 2. a;^-3a; + f = 10 + 1 = ^; .T-f = ±1-; etc. 3. ^. + 5^_|_(5)2__4_^(^5y. etc. 236 MANUAL FOR TEACHERS 1249. 1. x'-x = 6', x^~x + i = Q-\-i = ^; etc. 1250. 1. 12a; -a;' -32. Changing the signs, and rearranging, a:' -12a; = -32. Completing the square, a;' — 12 a; -f- 36 = - 32 + 36 = 4. Extracting the square root, a; — 6 = rb 2. Transposing, a; = -|-2 + 6 = 8; ora; = — 24-6==4. 12 -a; = 12-8 = 4; or 12 - 4 = 8. 8 and 4, or 4 and 8. Ans, 2. a;^ + 50a; + 625 = 2400 + 625 = 3025. a; + 25 = =h55; a; = 30 or - 80. Neglecting the negative result, the altitude is 30 ft. Ans. 3. a;2 + 225 + 30a; + a;' = 5625. Transposing, 2a;' + 30a; = 5625 - 225 = 5400. Dividing both members by 2, x' + 15 a; = 2700. Completing the square, a;' + 15 a; + (^y = 2700 + (-»/)' ; etc. 4. Perpendicular = V^a;' — a;' = V^^ = f a;. Area 4(.xL^) = 3^ = 150. Clearing of fractions, 3 a;' = 1200, a^= 400, a; = 20. Base = 20 yd. ; hypotenuse = 20 yd. x 1 J = 25 yd. Avs. 5. The convex surface = 6 (a; + a; + a; + a;) = 24 a;; the surfnce of the two bases = 2 a;' ; the entire surface = 2a;' + 24a; = 170 ; j 5;" + 12a; = 85; etc. I 6. The area of the walk = area of outside rectangle — area of inner rectangle. NOTES ON CHAPTER FIFTEEN 237 (40 + 2:c)(50+ 2a;) - 40 X 50 - 784 ; 2000 -{-180x + 4:x'~ 2000 = 784 ; 4a;^+ 180 a; = 784; ar'^- 45a; =196; etc. 7. 12 acres = 1920 sq. rd. i^' = 1920 ; 15a;' - 15360 ; x' = 1024 ; a; = ± 32. 8 Base = 32 rd. ; perpendicular = 32 rd. X If = 60 rd. ; hypote- nuse = V32H^~60' rd. = 68 rd. Diagonal = 68 rd. Ans. 8. a;' + 30' =(50 -a;)'. x" + 900 = 2500 - 100 a; + a;* ; 100 a; =1600; a; = 16. AC= 16 ft., CB, the part broken off = 50 ft. - 16 ft. = 34 ft. Ans. Or, making BC= x, AC= 50 — x. Then, (50 -a;)' + 30' = a;'. 2500 - 100 a; + a;' + 900 = a? - 100a; = - 3400 ; X = 34, the length in feet of the part broken off. 9. 60' + (58 - a;)' = 56' + a;' ; 3600 + 3364 - 116 a; + a;' = 3136 + a;' ; a;' - a;' - 116 a; = 3136 - 3600 - 3364 - 116 a; = - 3828, or 116 a; = 3828 ; a; = 33 = AK The length of the ladder in feet = V56' + 33' = V3136 + 1089 = V4225. 65 ft. Ans. 10. From ABB, the square of BB = 13' - (15 - a:)'. From BOB, the square of BB = A'- x\ Therefore 13' -(15 -a;)' = 4' -a;'; or, 169 - (225 - 30 a; + a;') = 16 - x\ 238 MANUAL FOR TEACHERS Removing the parenthesis, 169 — 225 -{-S0x — 3^= 16 — a:*. Transposing and combining, 30 a; = 16 - 169 + 225 = 72 ; ^ = ^. ^i) = V^C-(7Z)-V4'-2f = Vl6-J^ = J^ = 3f Altitude = 3J ft. Ans. 11. AF=- \^AB' - BF" = V1156 - 256 = V9'00 = 30 ; FC= -y/BC - BF' = V400 - 256 = Vl44 - 12 ; AC= AF+ F0= 30 + 12 = 42. 42 ft. Am. Let AE=x] FC=4:2-x; FD' = AD' - AF' = 26' - x' = 676 - x' ; FD' = DC - FC^ = 40' - (42 - xf = 1600 -(1764-8457 + a;'). Therefore 676 - a;' = 1600 - 1 764 + 84 a: - a;* - 84 a; = 1600 - 1764 - 676 = - 840; X = 10. ED == V26' - 10' - V576 - 24 ; or. = V40'-32' = V576 = 24. 24 ft. Ans, XIX NOTES ON CHAPTER SIXTEEN The geometry work contained in this chapter should be com- menced not later than the seventh year of school, and should be continued throughout the remainder of the grammar-school course. 1251. No formal definitions of lines, angles, etc., should be given at the beginning. After drawing angles of various sizes and with lines of different lengths, the pupils will be able to understand that " an angle is the difference in direction of two straight lines that meet in one point, or that would meet if produced." 1255. The semi-circular protractor is better than the com- mon rectangular one for beginners, as they see more clearly by using the former that an angle is measured by the arc of a circle. Two protractors are printed on a fly-leaf in the back of the text- book, for the use of such pupils as cannot procure others. Pro- tractors made of stout manilla paper can be obtained from the Milton Bradley Co., New York, at one cent each in quantities. A large protractor is needed for blackboard use. This can be made of pasteboard ; or wooden ones can be bought of the Keuffel & Esser Co., New York. Many scholars that are able to measure an angle one of who.se sides is horizontal. Fig. 1, find it difficult at first to ascertain the number of degrees in an angle formed by two oblique lines, Figs. 2 and 3. They should be permitted to discover the method 239 240 MANUAL FOR TEACHERS for themselves. All that is necessary, is to place the center (A) of the base of the protractor on the vertex of the angle, Figs. 1-3, and the edge of the protractor on one of the sides, the other side cutting the circumference. In Figs. 2 and 3, the number of degrees in the angle JCAZ is determined by the number of degrees in the arc BM, and the upper row of figures is used, having thv* zero mark at B. In Fig. 1, the number of degrees in the angle is measured by the arc CM, which requires the use of the lower row of figures. 1256. The average class will find the 100 exercises to Art. 1269, inclusive, sufficient for the first year's work. This will give three per week, and leave some time for review. Pupils should work the exercises at home without any preliminary discussion in class. After the exercises are brought in, they should be done on the blackboard, at which time the mistakes made can be pointed out. While first-class drawing cannot be expected from the instruments used by school-children, the teacher should exact the best work possible under the circum- stances. A hard pencil, kept sharp, is necessary to secure the requisite fineness of line. 1. In drawing an angle, commence at the vertex. This exercise is given to remove the impression sometimes formed, that the size of an angle depends upon the length of the lines, instead of their greater or less difference in direction. In this and all other exercises, the pupils should be encouraged NOTES ON CHAPTER SIXTEEN 241 to commence occasionally with an oblique line. No two results should be exactly alike. If two pupils compare notes, it should be for the purpose of producing a different drawing. One pupil's angle may have its vertex at the right, another at the left ; one vertex may be above, another below ; etc. The better the teaching, the greater will be the variety of results in exer- cises that permit of variety. 3. It is expected that the pupils will see for themselves that each arc will contain ^ of 360°. 4. Using the ruler, draw the first line of any convenient length and in any direction. Placing A of the protractor at either end, mark off 45°, being careful to use the proper row of figures. Remove the protractor ; place the ruler so that its edge just touches the end of the line and the 45° point, and draw the second line. This latter should not be of the same length as the first, unless for some good reason ; so that pupils will not consider that the lines forming an angle should be equally long. Write the number of degrees in each angle. 5. The teacher should not inform the pupils in advance how many degrees they will find in the second angle. They should measure it for themselves, using the protractor. In drawing these angles, the figure in the book should not be followed. The second line should be drawn to the left in some cases ; the lower angle may be made 60° ; etc. When two lines meet to form two angles, it is not at all necessary that the point of meeting should be at the center of one line. 1257. Pupils should be taught that horizontal lines are lines parallel to the surface of still water. Floating straws are hori- zontal, and may point in any direction. A spirit level is used by the carpenter to determine whether or not a beam, for instance, is horizontal. A vertical line is one that has the direc- tion of a plumb line, which is used by a mason to ascertain if a wall is perpendicular. 242 MANUAL FOR TEACHERS In drawings, however, lines that will be horizontal when the paper is placed upon the wall, are called horizontal lines ; and lines that will be vertical when the paper is placed upon the wall, are called vertical lines. 6. The perpendicular need not be drawn to the center of either of the others, nor need it always be drawn above. The teacher should encourage variety. 8. The pupils should draw these lines, and mark in each angle the number of degrees it contains. Encourage the greatest possible variety in the size of the angles and the direction of the lines. 9. While pupils may be able by this time to give the result without drawing the angles and measuring the second one, the teacher should not fail to give them the necessary practice in constructing angles of a given number of degrees, and in measuring the contents of others. Many scholars make as ridiculous mistakes in the measure- ment of angles as they do in their work in numbers, frequently reading the wrong figures, and figure? from the wrong row — marking an angle of 45°, for instance, 135° ; etc. They should learn to ** approximate" the size of an angle, as well as to " esti- mate " the probable answer to an arithmetical problem. An acute angle should not be marked as containing over 90° ; etc. 10. Having learned by observation that the sum of two adja- cent angles is 180°, the pupils should now discover that the sum of any number of angles formed on one side of a straight line is 180°. When they have learned this from drawing the first exercise, they may be permitted to calculate the result in the other two, especially as the protractors are not marked for fractions of a degree. The first exercise should show the same variety in the work of the diflferent pupils as has been recommended for previous work. 12. In constructing a square, the protractor is used to erect a perpendicular at each corner. These perpendiculars are made NOTES ON CHAPTER SIXTE*«bi^S.-*-**^ 243 equal to each other and to the original line. A fourth line is drawn. The accuracy of the work may be tested by measuring, with the protractor, the two upper angles. The base lines used by different pupils should be of different lengths. The pupils should be permitted, also, to construct the square in their own way. Some may erect a perpendicular at one end of the given line, and at the extremity of the second line erect another perpendicular. Some pupils may not measure the first line, drawing the second and third lines lightly of indefi- nite length, and using compasses to make them equal to the first. In this case, the light lines should not be erased ; but the square should be marked off by heavier lines. It is a good practice to have the pupils give a written descrip- tion of their method of working one of these exercises, which should be accepted as a regular composition. The language should be correct; the proper technical terms should be em- ployed ; and there should be sufficient detail to enable any one not familiar with the work to understand just how it was done. 13. Pupils should be permitted to learn for themselves from this exercise and from 14, that vertical, or opposite, angles are equal. 17. After drawing the required lines, the scholar should mark in each angle its contents in degrees. 19. This exercise should enable the pupil to see that the sum of all of the angles formed about a point will be 360°. 20. The teacher should not give unnecessary assistance. If the scholars have a few days in which to work out an exercise, they should find no difficulty in managing this. The word " adjacent " in geometry is applied to each of the two angles formed by one straight line meeting another. In 19, the two lower angles are adjacent ; but none of the upper three angles is adjacent to any other, because three straight lines are used to construct two angles in each case. No two angles in 20 are adjacent, and no two are vertical. 244 MANUAL FOR TEACHERS 21. There are no adjacent angles. They are all vertical, because the lines forming each are produced to form an opposite angle. 22. The pupil is not yet ready to do this in the geometrical way. The teacher should be satisfied if he adds 65° and 25°, and uses the protractor to make an angle of 90° ; etc. 24. If the pupil examines a clock, he will see that the num- ber of degrees between 12 and 1 is ^^ of 360°, or 30°. He has learned already that the length of the sides has nothing to do with the magnitude of the angle. 25. The minute hand goes 90° in a quarter of an hour. The hour hand goes 30° in an hour ; 15° in \ hr. ; 7-J-° in \ hr. To ascertain the angle at 12 : 15, the pupils should draw a clock face, locating the hands properly. Some will place the hour hand at 12, forgetting that it has gone 7^° in \ hour; and will give the answer as 90° instead of the correct one of 90° — 7^°, or 82° 30'. At 6 : 30, the minute hand is at 6, and the hour hand is half way between 6 and 7, or -J- of 30° = 15°. At 8 : 20, the minute hand is at 4, and the hour hand J of the way between 8 and 9 — the number of hour spaces being 4-J-, corresponding in degrees to 30° X 4 J = 130°. Pupils should understand that while the angle at 4 o'clock is 30° X 4, or 120°, and while the angle at 5 o'clock is 30° X 5, or 150°, the angle at 7 o'clock is not 30° X 7, or 210°. By making a drawing, they will see that in the last case the angle should be measured on the left, which will make it 30° X 5, or 150°. Note. — Angles of 180° 210°, etc., may be left for more advanced work. 1260. From 26 should be learned that two lines perpendicular to a third line are parallel to each other ; and from 27, that two lines running in the same direction and making the same angle with a third line, are parallel to each other. 29. DE will be drawn parallel to BC by means of the pro- tractor, an angle of 58° being made at the intersection of A£ and NOTES ON CHAPTER SIXTEEN 245 Fig. 4. DE. Six of the twelve angles will contain 58° each ; the remaining six will each measure 180° — 58° — 122°. The pupils should be permitted to ascertain this for themselves. The card suggested in the note is to be used in schools in which small wooden triangles are not obtainable. 32. To draw from P, a line parallel to the oblique line AB, place the perpendicular of the triangle on the line AB, Fig. 4, and place the ruler XY against the base of the triangle. Hold- ing the ruler in position, slide the triangle along it until the perpendicular passes through the point P. A line BE drawn along this side of the triangle will be parallel to AB. Time should be given the pupils to discover this or a similar method of drawing by means of a ruler and a triangle a line parallel to another line. The method may be made the sub- ject of a composition. 33. QR and C/Fare drawn parallel by means of the ruler and the triangle. They may lie in any position, care only being taken to cut them by a line making angles of 50° and 130° with one of them. Three of the remaining six angles will measure 50° each ; and the others, 130° each. 1261. 35. If the work is done as it should be, the angle at C will measure 80°. The line AB does not need to be horizontal ; nor should all the pupils draw AB of the same length. 36. The third angle will measure 60°. 37. There are 68° in a, and 57° in c. In 5, there are 180° — (68° + 57°), or 55°. In d, which is vertical to h, there are 55°. 38. The angle e should measure 28°, and/ 120°. There are 180° in e (28°) + g +/(120°). 246 MANUAL FOR TEACHERS 39. PRQ = 180° - (70° -f 60°) - 50°. PES = 180° - 50° = 130°. PR8 is therefore equal to the sum of the angles P and Q. 40. 180°. Am. 41. 180° -(36° + 65°). 45. In measuring the side of a triangle, use the smallest frac- tion marked on the ruler. When the ruler in use has the denominations of the metric system on one face, that face should be used, and the length of the line given in millimeters. This will not require any teaching of the metric system beyond show- ing pupils how to read their rulers, and it will do away with the need of using fractions. 46. The length of each side should be marked. If two of them are not found exactly equal, the construction is faulty. 47. The three sides should be equal. 50. Each of the oblique angles will contain 45°. 52. Angles 2 and 3, 90° each ; 4, 50° ; 5 and 6, 40° each. 53. Angles 1 and 4, 67^° each ; 2 and 3, 90° each ; 5 and 6, 22^° each. 54. The angle ;? contains 120°; m, 30°; n, 30°. 1266. 56. Construct this parallelogram by drawing two lines of the given lengths, meeting at an angle of 60°. By means of the ruler and the triangle, construct the other two sides. If the work is properly done, these two sides will measure 2 in. and 3 in., respectively; and the remaining angles will measure 120°, 60°, and 120°, respectively. From this exercise, the pupils should learn that the opposite sides and the opposite angles of a parallelogram are equal, and that the sum of the four angles is 360°. The angles being oblique, the figure is a rhomboid. The work of the scholars should show the variety suggested in previous exercises. It is not essential that the longer of the two NOTES ON CHAPTER SIXTEEN 247 given sides should be taken as the base, nor that the base should be parallel to the lower edge of the paper. 57. Different pupils will construct this trapezoid in different ways. Some, seeing that one angle is a right angle, will use the triangle to draw the second side, 3 in. ; and at the extremity of this side, will draw, by the same means, an indefinite perpendic- ular line to form the third side, which is parallel to the base. The fourth side is drawn to make an angle of 60° with the base. The remaining angles will measure 90° and 120°, respectively ; and the sides will measure 5 in., 3 in., nearly 3J in., and nearly 3iin. 58. The triangle cut off will form a rhombus when opened out, unless the base and the perpendicular are equal. In this case, the paper, when opened, will form a square. Making one angle of the triangle 30° (or 40°) will give a rhombus containing 60° (or 80°). 60. When the three sides of a triangle are equal, its three angles are equal ; but the rhombus has four equal sides without having equal angles. 61. A triangle that contains three equal angles, has its sides equal ; but an oblong has four angles of 90° each, with unequal sides. 62. To construct the rhomboid, draw a base of 2^ in. Two inches above, draw a parallel line 2-1- in. long, with the extremi- ties of the latter on the right or the left of the extremities of the base. If the two remaining sides are exactly equal to each other, the work is correctly done. It is not necessary to draw the altitude, though a broken line may be used. While different methods may be employed, the use of an incorrect one should not be permitted. The work done on the board by a pupil, should be criticised by the class if it be faulty ; or a better way may be suggested, if the one employed by the pupil at the board requiie too much time or unnecessary work. 248 MANUAL FOR TEACHERS The lengths of the two remaining sides should be measured, and marked on the papers. Those of different pupils should be different, there being no limit except the size of the paper: they must, however, be longer than 2 in. each ; and they should not be just 2J in., which would make the figure a rhombus. 63. If the line that shows the altitude of the rhomboid can be drawn within the figure, a right-angled triangle can be cut from one side and transferred to the other, making the figure a rectangle. Arithmetic, Art. 929, 5, last parallelogram. In the case of a rhomboid whose altitude does not fall Fio. 6. within the figure, several cuts will be necessary to change it into a rectangle. See Fig. 5. 65. The areas will be equal, because each rhomboid is equal in area to a rectangle 3 in. by 2 in. 66. The three rhomboids in Fig. 6 have their respective sides equal each to each, but their angles are unequal ; hence their altitudes and their areas are unequal. To construct these rhomboids, draw base lines of three inches, and inclined to each, at any angle except one of 90°, a two-inch line. Use the ruler and the triangle to complete the figures. NOTES ON CHAPTER SIXTEEN 249 68. As the altitudes are not given, the areas of the figures drawn by different pupils should vary. The protractor or the triangle should be used in drawing the altitudes, which should then be carefully measured. 69. See 62, making the upper side 2^ in. The fourth trape- zoid in Arithmetic, Art. 929, 6, shows the manner of determining the dimensions of the required rectangle. 1267. 78. The diameters of these circles should be such as not to admit of the use of the protractor in drawing them. To ascertain the extremities of an arc of 120°, two lines are drawn meeting in the center of the circle at an angle of 120°. The por- tion of the circumference intercepted by these lines will constitute an arc of 120°. The remainder of this circumference will form an arc of 240°. 79. The pupil should be permitted to make the first attempts in his own way. He will doubtless soon discover that the distance between the points of his compasses in drawing the cir- cle, is the length of the chord required, and that by placing one point of the compasses on any portion of the circumference of the circle just drawn, the other point will indicate on the circumfer- ence the other extremity of the chord. To measure the length of the arc in degrees, draw radii to its extremities, and use the protractor to determine the angle made by these radii, which may be produced, if necessary. If the work is properly done, the angle should measure 60°, which is the length of the arc. 80. Draw two light lines meeting at an angle of 72°. Using the vertex of the angle as a center, and any radius, draw an arc be- tween the lines. This arc will measure 72°. Darken the lines from the arc to the vertex to show the radii of the circle of which the arc is a part. 250 MANUAL FOR TEACHERS 81, 82. Figs. 8 and 9 show sectors of 90° and 270°, respec- tively ; Figs. 10 and 11, segments of 120° and 240°, respectively. Pig. 8. Fig. 9. Fig. 10. Fig. 11. 1268. 86. A diameter will divide the circle into two equal parts. A second diameter perpendicular to the first, drawn by means of the protractor or the triangle, will divide the circle into four equal parts. To divide the circumference into four equal parts, it will not be necessary to draw the diameters. When he has the ruler in the proper place to draw the first diameter, the pupil needs to mark only the two points where the ruler cuts the circumference. The third point can be indicated when the triangle is placed at the center of the circle and against the ruler ; etc. While the scholars may be permitted in the beginning of this work to draw a number of unnecessary lines, and while it may be an advantage to even require it, they should gradually learn to make as few lines as possible. The construction lines that are employed, should be drawn very lightly and should not be erased. Other lines should be made more conspicuous. Careful pupils may be allowed to use ink for this purpose. 1269. It is not intended that the methods here suggested should be communicated in advance to pupils. Each should be allowed to try the problem in his own way. The discussion of the method employed afterwards on the blackboard, will suggest other and possibly better modes of procedure. 88. To inscribe a regular pentagon in a circle, it will be neces- sary to divide the circumference into five arcs of 72° each. The NOTES ON CHAPTER SIXTEEN 251 protractor should be used to obtain the first arc ; the remain- ing ones can be set off by the compasses, the first being used as a measure. Careful work should be exacted by the teacher. 89. Many pupils will have learned in their drawing lessons the regular method of inscribing a hexagon in a circle. Those unfamiliar with this way, should not be shown it until after they have constructed the hexagon by means of the protractor. It is a pedagogical mistake to suggest " short-cuts " to pupils before they thoroughly understand a general method. For this reason, the teacher should permit the members of her class to use the protractor in the construction of the inscribed triangle, leav- ing it to themselves to discover a simpler way. She should encourage, also, the employment of a variety of methods even if some of them are not very direct. The experiments made by pupils to discover a new mode of constructing a polygon, will help them in their geometrical study. The chord of 60° being equal in length to the radius (79), the shortest method of inscribing a hexagon is to apply the radius as a chord six times. Two of these divisions of the circum- ference will make arcs of 120°, the chords of three of these form- ing the sides of an inscribed equilateral triangle, 90. Each of the six angles at the center contains 60°. Since the two sides AC and AB, enclosing any central angle, are radii of the circle, and therefore equal to each other, the angles oppo- FiG. 12. Fig. 13. site those sides are equal ; that is, angle A = angle B. The angle at C being 60°, A-\-B = 180° - 60° = 120°, and ^ = ^ - 60°. Angles 1 and 2 (see Arithmetic), therefore, measure 60° 252 MANUAL FOR TEACHERS each, and the whole angle contained between two adjoining sides of the hexagon, measures 120°. After determining by this method the number of degrees in each angle of a regular hexa- gon, the pupils should be required to construct one, and to mark in each angle its contents in degrees, as in Fig. 13, verifying the result by using the protractor. 91. A careless scholar, measuring the number of degrees in each angle of a regular pentagon (Arithmetic, Art. 1268), will sometimes read from the wrong row of numbers on the protrac- tor, getting the result 72°, instead of the correct one of 108°. As there are 72° in each division of the circumscribing circle, he will have no doubt of the correctness of his answer, unless he has been trained to estimate the size of an angle. In this case, he will see that each angle of a regular pentagon is obtuse, and, therefore, greater than 90°. One method of calculating the number of degrees, is to divide the pentagon into five equal triangles, one of which is shown in Fig. 14. The angle at Cis 72°. The sides CA and CB, being radii, are equal ; which makes equal ^^' ^*" angles at A and B, each of which is ^ of (180° - 72°), or 54°. Each of these angles is the half of one of the angles of the pen- tagon, so that these latter angles measure 108° each. 92. A circumscribed square touches the circle at four points, each side constituting a tangent. A tangent being perpendicular to the radius drawn to the point of contact, the square may be constructed by drawing perpendiculars to two diameters inter- secting at right angles, using the triangle or the protractor for the purpose. The ingenious pupil will discover other ways ; drawing, for instance, at each extremity of the two diameters, a line parallel to the intersecting diameter, by means of the ruler and the triangle; etc. No method should be permit- ted that merely approximates accuracy, such as determining that a line is parallel or perpendicular by the eye alone. The NOTES ON CHAPTER SIXTEEN 253 Fig. 15. Through 3 and average class will contain many members intelligent enough to pass upon the correctness of a given method, and they should be called upon to give reasons for any criticisms they may have to offer. In circumscribing some polygons, a hex- agon for instance, many pupils prefer locat- ing points X and Fj instead of using the triangle and the ruler to draw a tangent XF. After dividing the circumference (Fig. 15) into six equal parts at 1, 2, 3, 4, 5, and 6, they draw a diameter from 1 to 4. Through 2 and 6 they draw a secant XA, making MX and MA each equal to the radius. 5 a second secant is drawn, and NY and NB are also made equal to the radius. A line drawn through X and Y will form one side of the circumscribed hexagon. One extremity of this side can be determined by producing the diameter 3(76 to F, and the other by a line through 2(75. Note. — A secant is a line that cuts a circle at two points ; a tangent is a line that touches a circle at one point. 93. The smallest number of triangles into which a pentagon can be divided, is three (Fig. 16). The three angles of each triangle contain 180°, making the sum of angles 1-9, 540°. Since 1 = A, 2 + 4: = B, 6 + 7 = C, 8 = I); 3 -f 5 = jE", the sum of the five equal angles (A, B, C, D, and ^) of a regular pentagon = 540°, and each equals 108°. This is the result that was found in 91 by another method. 94. The hexagon is divisible, as above, into four triangles, containing 180° X 4, or 720° ; making each angle 720° h- 6, or 120°. 95. A quadrilateral is divisible into 2 triangles ; a pentagon, 254 MANUAL FOR TEACHERS into 3 ; a hexagon, into 4 ; a heptagon, into' 5 ; an octagon, into 6, beiiig 2 triangles less in each case than the number of sides in the polygon. 96. The number of degrees in each angle of a regular octagon = [180° X (8 -2)] ^8. 97. At each end of the 2-inch line, draw a 2-inch line at an angle of 108°. At the farther extremity of each of those lines draw a line at an angle of 108°. These last lines meet at an angle of 108° if the work is correctly done, and are each two inches long to the point of their intersection. 98. A line drawn to each extremity of the base at an angle of 60° will form an equilateral triangle. Use angles of 90° for the square, 108° for the pentagon, 120° for the hexagon, nearly 129° for the heptagon, 135° for the octagon, 140° for the non- agon, etc. The first line should be placed near the cen- ter of the bottom of the paper to give room for the successive polygons. See Fig. 17. Bright pupils will calculate the necessary angles and continue to construct polygons as far as the space will permit. 99. By drawing the diameters AB and XY (Fig. 18), the inscribed square will be divided into four triangles, while the circumscribed square contains eight, being double the area of the inscribed square. 1270. These problems are given to en- able the pupils to learn how to bisect lines, erect perpendiculars, construct angles, etc., by means of the ruler and the compasses, and incidentally to learn a number of geometrical facte. The use of other instruments is unnecessary, and should, therefore, not be tolerated. Fig. 17. NOTES ON CHAPTER SIXTEEN 255 1. The distance between the centers == 1^ in. + 1 in. 2. 11 in. — 1 in. 4. The line XY joining the two points of intersection of the equal circles (Fig. 19), bisects the line AB connecting the centers. The radii AX, BX, A Y, BY are each 2 inches long. 5. The previous exercise should lead the pupils to see the steps neces- sary to the construction of the re- quired triangle. The 3-inch base AB is first drawn. The next requirement is to find a point X(Fig. 19) 2 inches from A and from B. A circle of 2-inches radius with i> as a center, will contain every point that is 2 inches from B. A similar circle with ^ as a center, contains every point 2 inches from A. The intersections, X and Y, are each 2 inches from both points A and B. Using either one as a vertex, draw lines to A and B, forming the required triangle. Authorities differ somewhat as to the advisability of requiring pupils to employ circles rather than arcs in geometrical problems. While it may be better at first to use circles, the point to be finally reached is the emj^loy- ment of the shortest lines possible. This should not be inconsistent with the acquirement of geometrical knowledge. A scholar should know after a very few exercises that each point of an arc is 2 inches from the center, as well as he does when he draws the entire circle. In his later construction of an isosceles triangle, the pupil should know that the vertex is above (or below) the center of the base. For this reason the first arc need not be a very long one. The second should be still shorter. 6. Fig. 19 will suggest the necessary steps. Using each end of the 3^-incli base as a center, and with a radius of 4 inches, draw two circles. Draw lines corresponding to XA and XB for the required triangle. Placing the ruler on X and Y will give the perpendicular, which should not extend below the base. If arcs are used, the upper intersection determines the position of the vertex. A lower intersection is used to determine the 256 MANUAL FOR TEACHERS direction of the perpendicular. The pupil will gradually learn that while a definite radius, 4 inches, is required to locate the vertex of the triangle, intersecting arcs, each of 3 inches or 5 inches or any other radius, will serve to locate the second point, used with the vertex to determine the direction of the perpendic- ular. The point of their intersection may be above the base or below it, according to convenience. A point below secures greater accuracy, by being probably at a greater distance from the vertex than one above is likely to be. 7. This is a variation of 6, but without directions as to length of sides. If the perpendicular is correctly drawn, it will bisect the base. See Exercise 48, Art. 1263. The compasses should be employed to determine the equality or inequality of the seg- ments of the base. 8. The same procedure is required as in 7, except that the sides of the triangle are not drawn. The bisecting line should be extremely short. 9. With a 2-inch radius, draw intersecting arcs, using as centers the two extremities of a 2-inch line. 11. Either side may be used as the base ; and different pupils should use a different base, although the greatest number will probably take the longest side. Using the 2-inch side as a base, draw from one end, as a center, an arc with a radius of 1 inch ; and from the other, an arc with a radius of 1 J inches. The intersection of the arcs will be the vertex of the required triangle. With the 3-inch line as a base, the radius of the arcs will be 2 inches and 2-J inches, respectively. Besides employing different bases, the pupils should use oblique lines and vertical lines as bases, and the vertex in some instances should be below the base ; etc. 12. A scholar should be permitted to discover for himself that the intersection of the 2-inch arcs will be at the center of the 4-inch base, After endeavoring, also, to make a triangle whose NOTES ON CHAPTER SIXTEEN 257 sides shall measure 1, 2, and 3 inches, respectively, he will understand that the third side of a triangle must be shorter than the combined lengths of the other two. 13. If the pupil, in drawing arcs to locate the bisecting line, employs the radius used in drawing the circle, he will discovGr that one intersection will take place at the center of the circle. This will lead him to see that only one set of intersecting arcs is necessary — the one beyond the circle, the center of the circle serving as a second point. The teacher should be in no hurry to inform the pupil using two sets of arcs, that a single set will be sufficient ; knowledge that comes in some other way than merely by direct telling, is apt to last longer. 14. Before dividing a sector (Figs. 8 and 9) into two equal parts, some scholars may consider it necessary to draw the chord. Permit them to do so at first. 15. The pupils have already learned that a line drawn from the vertex of an isosceles triangle to the center of the base, is per- pendicular to the base. The method employed in bisecting a line is practically to consider it the common base of two isosceles triangles having their vertices on opposite sides of the base, although the equal sides of the triangles are not drawn. This bisecting line is perpendicular to the assumed base. The pupils have probably discovered that the perpendicular line bisecting the chord in 13, would, if produced, pass through q the center of the circle. From this previous knowledge, they will doubtless be able to answer / the last question of 15. / The scholars that have drawn the chord in \II___ dividing the sector in 14 into two equal parts, ^ will be likely to see that the radius CX (Fig. fig. 20. 20) is perpendicular to the bisected chord AB. 16 shows that a perpendicular that bisects a chord, AB (Fig. 20), also bisects the arc AJCB. 258 MANUAL FOR TEACHERS It will readily be seen by the pupils that an arc AXB can be bisected without drawing the chord AB. 17. This problem is the same as 8. The drawing, however, should show a longer line, and one that does not cut the given line, the requirement being that the perpendicular be drawn io the latter. A perpendicular drawn to a horizontal or to an oblique line from below it, would be correct ; although it might not be accepted as satisfactory if the more common wording of the problem were employed : Erect a perpendicular at the mid- dle point of a line. 18 requires no explanation. 19. Bisecting one of the four divisions of a circumference gives the dividing point between two one-eighths. A ruler placed upon this point and on the center of the circle will indicate the location of another. The distance between two points can be ascertained by the compasses, which can then be used to locate the remaining two dividing points. The lines used to divide a circumference should not be too long. The thoughtless pupil sometimes fails to see, when he employs his compasses to measure the distance between two points on a circumference, that he is measuring the chord of an arc, and not the arc itself. It may be necessary for tJie teacher to explain this in connection with 25. 22. Unless the pupils have learned in their drawing work the method of erecting a perpendicular at the end of a line, some of them may experience a little difficulty in solving this problem. One way of drawing the circumscribed square is to construct on AY (Fig. 18) an isosceles triangle A2Y equal to ACY. Produce 2 A to 1, making -41 equal to 2 A. Produce 2Y to 3 in the same manner. Lines from 1 through JT, and 3 through B, will intersect at 4, which completes the square. 24. The object of 23 and 24 is to lead the pupils to see again that the side of an inscribed hexagon is equal in length to the radius of the circle. NOTES ON CHAPTEK SIXTEEN 259 26. The four triangles will together constitute a circumscribed equilateral triangle. 27. Draw lightly an arc less than 90° in length. After cut- ting off 60°, darken this portion. No radii ^ should be drawn. To construct an angle of 60°, draw an in- definite line AB (Fig. 21). With any con- venient radius, as AM, draw an arc Mli. With Jf as a center, and using the same radius, cut the arc at X. This makes 3fX an arc of 60°. From A draw a line through X. 28. An arc of 30° is constructed by the bisection of an arc of 60°. To construct an angle of 30°, draw AB (Fig. 21) and the arc MN; and cut off MX, as in 27. With the same, or any other convenient radius, and using M and X as centers, draw two arcs intersecting on the right of MX. A line drawn from A through this intersecting point will make with AB an angle of 30°. 29. To construct an angle equal to 60° + 30° (or 30° + 60°), bisect the arc XM (Fig. 21), and from the bisecting point, which is 30° from M, lay off an additional 60°. A line drawn from A through this point will make with AB an angle of 90°. 30. One method of erecting a perpendicular at the end of a line is given in 29. A somewhat similar method consists in prolonging the arc, and marking off at Q and B two divisions of 60° each. Using these two points as centers, draw arcs intersecting at F. From I) draw a line through i^. The line BF bisects the arc QB, which makes the angle FDE = 60°(PQ)+30°(QF)-90°. 260 MANUAL FOR TEACHERS 31. The bisection of VF gives an angle of 45°. Do not draw DF. The recommendation so often made as to getting a variety of drawings from the various members of the class, applies to these problems. The erection of a perpendicular at the end of a vertical line should call forth at least four different results. The perpendiculars may be erected at either end, and may run to the right or to the left. In constructing an angle of 45°, a greater variety is possible. The first line may be horizontal, verti- cal, or oblique ; the second line may start from either end ; and it may extend above or below, to the right or to the left; the lines forming the angles need not be of the same length on all papers, nor need both of them be of the same length on any one paper. 22^° = I of 45°; 135° = 90° + 45° ; 15° = | of 30°; 75° = 60° +15°. 33. The preceding problem should give a hint as to the method of solving the present one. In ^ 32, a circle was drawn with ^ as a center and AX as a radius. This circle cut XY (Fig. 23) in M. To erect a perpen- dicular at A, the center of the circle, the ^' a M ' extremities X and M of the diameter ^^' ^" XM were used as centers to draw arcs intersecting at J. A perpendicular was then drawn from A through J. This problem does not furnish a circle, nor is one necessary. The lino XY is given, and the point A at which the perpendic- ular is to be erected. With ^ as a center, and a radius equal to AX, cut off AM. This gives us the diameter of the circle employed in the preceding problem. It will be noticed that a second set of intersecting arcs on the opposite side of XM is not needed, the point A answering instead. This problem amounts to the bisection of an arc of 180°, of which only the chord XM is drawn, A being the center of the circle. NOTES ON CHAPTER SIXTEEN 261 34. The first two divisions of this problem are reviews of parts of 17 and 30. The erection j^ of a perpendicular at a point between the end and the center, is a variation of the preceding problem. Let BS (Fig. 24) be the required line and A the point. With ^ as a center, and any convenient radius, lay off the Fig. 24. '^ points Jl and M, which will be equidistant from A. Using X and M as centers, draw arcs intersecting at tT. Draw the required perpendicular from A through J. In problem 33 the point M was located at a distance from A, equal to AB\ and while greater accuracy is obtained by having the points J/and X as far apart as possible, the present method is suggested to show that the only essential requirement as to their location is to have them equidistant from A. 35. The base need not be a horizontal line. 36. Proceed as in the construction of a right-angled triangle, base 2 inches, perpendicular 2 inches ; but do not B draw the hypotenuse. With B and H as cen- ters (Fig. 25), and a radius of 2 inches (BP or BIT), draw arcs intersecting at 0. BO and ITO will form the remaining sides. To construct the rectangle, BH (Fig. 25) should be made 3 inches long, BB measuring 2 Fig. 25. inches. With ^ as a center, and a radius of 3 inches, draw an arc. Intersect this by a second arc drawn with IT as a center, and a radius of 2 inches. From the point of intersection, draw lines to B and IT. Some scholars may prefer to erect a perpendicular at each end of the base, etc. 38. Draw a base line, WV (Fig. 26), 3 inches long. At any convenient point, M, erect a perpendicular, MB, 2|- inches long. P H 262 MANUAL FOR TEACHERS At its extremity P, draw the perpendicular PO. With TT as a center and a radius of 3 inches, locate the point Xon the line P0\ TP'Xwill be the second side of the rhombus. On PO draw XN 3 inches; and connect NV. This last line should measure 3 inches if the work is properly done. Fig. To construct a 3-inch rhombus containing an angle of 60°, draw WV\ at W construct an angle of 60°, and make WX 3 inches. At X or at F", draw a 3-inch line making an angle of 120° with XW OY VW, etc. ; or, using Xand Fas centers, and with a radius of 3 inches, draw arcs intersecting at iV; draw iVXandATF. 39. Proceed as in the first problems of 38 ; but make TTFand XN each 4 inches long. 40. See PPB (Fig. 25) ; draw PIL Some pupils, preferring to commence with a horizontal base, calculate the number of degrees in each angle at the base, i (180° - 90°), or 45°. To each end of a base line of any length (Fig. 27), they draw a line making with the base an angle of 45°. If one angle is 120°, the angles at the base will be 30° each; if one is 135°, the angles at the base will measure 22^° each. The method shown in Fig. 27 requires the construction of two angles ; and is, therefore, not so direct as drawing two equal lines meeting at the given angle. Fio. 27. NOTES ON CHAPTER SIXTEEN 263 B Y Fig. 28. 41. Erect a 3-inch perpendicular at the middle of a 3-inch line. A A perpendicular, AB (Fig. 28), bisects the angle at A of the equilat- eral triangle. Erect a 3-inch perpen- dicular at By on an indefinite line ^ CD; at A, construct two angles of 30° each, as shown in the figure ; draw AX and A Y, forming the required equilateral triangle AXY. Test the work by meas- uring XY. 42. To construct a scalene triangle of the required dimen- sions, erect the 3-inch perpendicular at a point not in the center of the 3-inch base, nor at either extremity. To construct the obtuse-angled triangle, produce the 3-inch base by a dotted line, and erect the 3-inch altitude at some point outside of the base. 43. This problem may puzzle the pupils at first. If the tri- angle were isosceles, XY3f, for instance, there would be no diffi- culty. The solution of the prob- lem consists practically in making an isosceles triangle, although the line X3f\s not drawn. With X as a center (Fig. 29), and XYas a radius, the point M is located (without drawing the arc shown in the figure). Using If and Y as centers, and with any con- venient radius, arcs are drawn intersecting at B. XB gives the direction of the altitude. Fig. 29. 44. The given triangle in this case will be XMZ (Fig. 29). Produce ZM indefinitely towards Y, and with XM as a radius, cut the base at Y; etc. XFis, of course, not drawn ; il[fF should be a light line or a broken line ; etc. 264 MANUAL FOR TEACHERS 45. This is the same problem in another form. Let YZ (Fig. 29) be the given line, and X the given point, from which a perpendicular is to be let fall upon YZ. In 43-44, the arc described with X as a center, passes through one extremity of the line. The pupils should see that this is not essential, and that it would be very inconvenient in the case of a very long line. All that is necessary is to intersect the line YZ at two points sufficiently far apart to secure accuracy. 47. Each angle will measure 60°; the arc on which it stands measures 120° ; an angle at the circumference, therefore, is measured by one-half the arc intercepted by its sides. Apply this to the regular pentagon (Arithmetic, Art. 1268). The angle at the top of the figure stands on an arc containing 72° + 72° + 72°, or 216° (made up of three arcs) ; it contains, therefore, \ of 216°, or 108°. See Fig. 30. An angle of a square stands on an arc of 180° ; its contents are, therefore, \ of 180°, or 90° ; etc. Fio. 80. Fig. 81. 48. The number of degrees in each angle of a regular hexagon is 120° (94, Art. 1269). From each end of the given line AB (Fig. 31) draw a line equal in length to AB, and making with it an angle of 120° ; etc. While this is not the best way to construct a regular hexagon, it gives the pupils an opportunity to try the general method on a polygon of six sides. 49. For the construction of an octagon, the angles at A, B, C, etc., should measure 135° (Problem 31). NOTES ON CHAPTER SIXTEEN 2G5 By a later problem, the pupils will learn how to find the center of the circumscribing circle after two sides of the octagon are drawn ; for the present, however, they should be required to repeat the construction of the angle of 135° at the required number of corners. The work should be as accurate as possible, considering the necessary imperfections of the tools employed. 50. A right angle whose vertex is at the circumference, sub- tends an arc of 180°, or a semi-circumference. 51. By means of a ruler, placed on the center of the circle, mark points A and B (Fig. 32) the extremi- ties of a diameter (not drawn). The arc A MB is one-half of the circumference, or 180° ; and the angles X, Y, and Z, whose sides XA and XB, YA and YB, ZA and ZB, subtend this arc, are each measured by one-half of it, and contain, therefore, \ of 180°, or 90°. 52. The hypotenuse of an inscribed right- angled triangle is always the diameter of the circle. In Fig. 32 draw the diameter AOB (or any other). Bisect the arc AXYZB, which gives the vertex of the required trian- gle. Connect this point with the points A and B. Note. — It may be necessary to have the scholars understand that all the vertices of an inscribed polygon must lie in the circumference ; an inscribed triangle, therefore, has three vertices in the circumference. 53. The diameter of a circle having a radius of 2 inches, will bu the diagonal of the inscribed square. The location of the remaining two corners is easily determined. The square may also be constructed by drawing 4-inch diagonals bisecting each other at right angles ; etc. The rhombus will consist of two 8-inch equilateral triangles having a common base. 54. Every angle being on the circumference, each will be measured by one-half the intercepted arc, and the sum of the 266 MANUAL FOR TEACHERS three angles will contain one-half the number of the degrees con- tained in the three arcs ; i.e., ^ of 360°. 65. See the rhombus in 53. 56. The two triangles will be equal in all respects, as will those in 57. 58. From 56 and 57 the pupils should learn that two triangles are equal in every respect (a) when two sides and the included angle of one are equal to two sides and the included angle of the other, each to each ; and (b) when two angles and the included side of one are equal to two angles and the included side of the other, each to each. From 59, they should learn that when two triangles have the three sides of the one equal to three sides of the other, each to each, they are equal throughout. From the present problem, they should learn that any number of triangles can be constructed having their corresponding angles, equal each to each, but whose corresponding sides are unequal, each to each. 60. Impossible, because the sum of three angles of 75° each is greater than 180°. 61. The radius of each circle must be one-half of the hypote- nuse, or 1-J- in. See Problem 52. Any inscribed triangle having one side the diameter of the circle will answer the conditions of the first case. The 2-inch base of the second triangle is obtained by taking A as a center (Fig. 32), and with a radius of 2 inches, drawing an arc at, say, F; ^Fwill be the required base, Y£ the per- pendicular, and AB the hypotenuse. Using a radius of 1^ inches, as above, will give a perpendic- ular, AX, of 1^ inches, etc. Note. — The term base does not necessarily imply a horizontal line; nor perpendicular, a vertical one. 63. To construct an angle at F(Fig. 33) equal to the angle at X, take Xas a center and any convenient radius, and draw the arc Aa ; then take F as a center and the same radius, and draw the arc Gc. The angle X is measured by the arc AB, the NOTES ON CHAPTER SIXTEEN 267 length of whose chord is determined by means of the compasses. Lay off the same on Cb, making the arc CD equal to the arc AB. An angle formed by drawing a line from Y through D Fio. 33. will be equal to the angle at JT, as each angle is measured by an arc of the same number of degrees. See note to Problem 19, Art. 1270. 64. Draw the first line at any angle; and by the preceding problem, draw the two intermediate lines, making, with the horizontal line, angles equal to that made by the first line. The pupil will have no difficulty with the oblique line at the other extremity of the horizontal line. 65. Proceed as in the second part of 64, using a horizontal 3-inch line for the diagonal, drawing from one end a 2|-inch oblique line running upward at any angle, and from the other end a similar line running downward at the same angle. The lines divide the 3-inch diagonal into 5 equal parts, each part measuring f inch. 66. The scholars should gradually learn that it is unnecessary to measure the oblique lines, Km , ^t^ and Ln (Fig. 34). In setting off 5. the divisions ATI, 1 2, etc., La, ah, t etc., it will be found unnecessary to K^=— ^ -^^L use a definite measure; all that is . ^^ required is that they be equal. This equality is obtained by opening the n- compasses slightly and marking off the divisions with them. There is no need of completing the Fig. 34. 268 MANUAL FOR TEACHERS rhomboid, nor of drawing the lines 4a, 36, 2c, etc. Place the ruler on 4 and on a, and use a short line on KL to mark off one division; etc. It is expected that the teacher will not give this method at the outset. To draw a line exactly f of an inch in length, a construction line, 3 inches long, KL (Fig. 34), is drawn lightly, also Kw, and Ln of indefinite length. Only the first division is required on Ktn. Placing the ruler on 1 and c?, mark off on KL f inch, and darken this portion One-seventh of a 5-inch line = -f- inch. 67. When the base of a right-angled triangle measures 3 inches, and the perpendicular measures 2 inches, the hypotenuse will measure V3' -f- 2' = V9 + 4 = Vl3 inches. 68. A line Vl3 inches long is drawn by constructing a right- angled triangle having a 3-inch base and a 2-inch perpendicular. The hypotenuse is the required line. A base of 2 inches and a perpendicular of 1 inch give a hypot- enuse of V5 inches. A hypotenuse of 4 inches and another side of 3 inches give a remaining side of VV inches. For the construction of this tri- angle, see Problem 61. 69. The side opposite the angle of 30° is one-half as long as the side opposite the angle of 90°. 71. The chord of an arc of 60° is 2 inches long ; the chord of an arc of 180° is the diameter, and is 4 inches long. The chord of an arc of 300° is 2 inches long. 72. The perpendicular "erected" at one end of the chord should be turned downwards if the chord is drawn above the cen- ter of the circle. The perpendicular and the diameter meet at the circumference. See Problem 61. 73. The triangle will retain its shape, because only one tri- angle can be drawn with sides of a given length. Problem 59. The rectangle will not retain its shape, because an indefinite vari- NOTES ON CHAPTER SIXTEEN 269 ety of parallelograms can be constructed having their corre- sponding sides equal, each to each. See Exercise 66, Art. 1266. 77. From Problem 15, the pupils have learned that a perpen- dicular bisecting a chord, passes through the center. If there are two lines passing through the center, the latter must be located at their intersection. 78. To inscribe a circle in any triangle, draw lines bisecting the three angles. See problem 28. The intersection of these three lines will be the center of the circle. In practice, only two hnes are drawn ; but the third serves to test the accuracy of a pupil's work. 79. The sides of an inscribed triangle are chords of the cir- cumscribing circle. In Problem 77, we have learned that the intersection of two perpendiculars that bisect chords, locates the center of the circle. With this center, and with a radius equal to the distance from the center to the vertex of any angle of the triangle, draw the circle. 80. The larger the circle the better for beginners. The average scholar may consider it necessary to draw two adjacent chords, but every purpose will be served by cutting the circum- ference by three short lines to mark the boundaries of two adjoining arcs. The bisection of these arcs by perpendiculars will locate the center of the circle. 82. Use a cup to draw the arc. Divide it at random into two parts. Bisect each, as above. q JD 83. The altitude is 2 in. ^' ' ^^^^ 4^ Problem 70. 84. Experienced draughts- men save time by changing the radius as infrequently as possi- ble. To construct a square on a 3-inch line AJB (Fig. 35), '''''■ '^• take a radius of 3 inches, and with A as a center draw the arc ^m. With ^ as a center, mark off JBn = 60°. Bisect Bn, 270 MANUAL FOR TEACHERS using the same radius to obtain the arcs intersecting at o, with B and n as centers. BX= 30° ; place the compasses on X, and cut the first arc at (7, which makes XC 60°, and BC 90°. C is the third corner of the square. Using B and (7 as centers, an I with a radius of 3 inches, draw arcs intersecting at D, the fourth corner of the square. ^ m^ 86. We have found in Problem 26, that constructing an equilateral tri- angle on each side of an inscribed equilateral triangle gives a circum- scribed equilateral triangle. Locate m, n, and o, 120° apart. With each as a center, and with a radius equal to mo, draw arcs intersecting at A, B, and C. Do not draw the triangle 7)ino, shown in the figure. 87. See Problem 41, second part. 89. See Arithmetic, Art. 1089 (Fig. 2). 90. Construct a right-angled triangle having a base and a perpendicular measuring 2 inches and 3 inches, respectively. The hypotenuse will be the side of the required square. Do not construct squares on the other two sides. 91. The hypotenuse is 3 inches, etc. See Problem 68. 92. The square constructed on the hypotenuse of a right- angled triangle whose base and perpendicular measure 3 inches each, will be double the area of a 3-inch square. When the 3-inch square is constructed, measure the length of the diagonal (without drawing it) ; and on a line of this length as a base, construct the required square. 93. Four. See Fig. 36. 94. Nine ; sixteen ; twenty-five. 95. 1, Ij^, 2 inches. 96. 3, 4J, 6 inches. 97. The radius of the second circle is 2 inches. NOTES ON CHAPTER SIXTEEN 271 Fig. 37. 98. A square described on the base of an isosceles rigbt- angled triangle will be one-half the size of the square described on the hypotenuse. A = a-\-a^=2a', a = ^A. The area of an equilateral triangle whose base is YZ, is one-half that of one whose base is XY. The area of a circle whose radius is YZ (or XZ), is one-half that of a circle whose radius is XY. Note. — If the square A in Fig. 37 is drawn above XY instead of below it, the ratio of the large square to that of each small one will be more apparent. Construct an isosceles right-angled triangle having a hypote- nuse of 2 inches (Problem 52). The base of this triangle will be the radius of the required circle. The area of a circle of 2 inches radius = 2^ X tt = 4 tt. The radius of the other circle = V2 ; its area is (V2)''' X tt = 2 TT, which is one-half of 4 tt. 99. The side of the required triangle measures V2 inches. See 98. 100. Cab (Fig. 38) is one-sixth of an inscribed regular hex- agon, and CAB is one-sixth of a circum- a X B scribed regular hexagon. Calling the radius of the circle 2 inches, the altitude CX of the large triangle is 2 inches. Since ab is 2 inches, ax='[ inch -, Ca = 2 inches. In the right-angled triangle Cax, Cx -f ax ---O?; CS'-f 1=4; C'x' = 4-1 = 3; Cx = Vs. The areas of the two triangles will be proportional to the squares of their respective altitudes. C^' = 3 and GX^ = 4. That is, the area of the larger triangle is 1-^^ times the area of the smaller ; hence the area of the circum- scribed hexagon is \\ times the area of the inscribed hexagon. The area of the circumscribed square is double the area of the 272 MANUAL FOR TEACHERS inscribed square; and the area of the circumscribed equilateral triangle is four times the area of the inscribed equilateral triangle. 1271. 3. At the middle point Xof the perpendicular of the right-angled triangle (Fig. 39), cut Xtyi parallel to the base. Re-arranging the two parts gives a rectangle whose dimensions are 4 inches and \\ inches. Figs. 40 and 41 show how the A A A Fig. 39. Vva. 40. Fig. 41. other two triangles are divided to make rectangles of the same dimensions. 4-6. See 56-59, Art. 1270. 7. The two triangles have two sides of one, AC and BC, equal to two sides of the other, CE and CD\ and the angle ACB equal to its opposite angle DCE\ hence the third side, DE, of one triangle is equal to the third side, AB, of the other. 9. The area of the larger triangle is four times that of the smaller. 11. Another method of dividing a line into equal parts. See Problem 66, Art. 1270. 1273. Much interest is added to this work by employing the method here given, in calculating heights and distances in the neighborhood of the school. A comparison between the results obtained by calculations and those obtained by actual measure- ments, will be useful in teaching the pupils the necessity of great accuracy in their preliminary work. 2. The hypotenuse of each triangle represents a ray of light from the sun. etc. NOTES ON CHAPTER SIXTEEN 273 ft. + 4^ 1- ft. = 162 ft. Ans. RN'.: FQ.MJST; 6: (120 + 6) hf =157i ft. 4:ifiV. MJY-- 3. Fr=110ft. 4. CD.DE'.: CB.BA; 50 : 90 : : 1200 : ^^ ; ^^ = 2160ft. The width of the river = 2160 ft. - 100 ft. = 2060 ft. Ans. 5. CD=CG-DG = 6 ft.-4 ft. = 2 ft.; EII=lll ft. + 3 ft. = 180 ft ; AH= 120 ft. ; AB = 120 ft. + 4 ft. = 124 ft. 6. The two triangles are similar, because the angles of one are equal to the angles of the other. Angle D = angle A = 90°. The two angles at C are vertical angles, and, therefore, equal to each other. The remaining angle J5 must be equal to the remaining angle U, because the sum of the angles in each triangle is the same. CD.DE:: AC:AB; 3.25: 5 :: (12 -3.25): ^5. 7. lOi- : (lOi + 195 + 15) : : (12 - 41) : A/. ^=15 8. BP 84 ft. Ans, 9. The tree is 3 ft. X 36, or 108 ft. high. Ans. because angle C= angle B = 45°. 10. AC= AB. A line CB makes an angle of 45° with AC, angle ^ = 90° ; angle ABC= 45°. Note. — In the succeeding problems, the triangle and the protractor may be employed when necessary. 1274. 1. The circumference of the circle, or 360° = 4 in. X 3.1416. The arc of 60° is J of the circumference ; the arc of 120° = i of it ; etc. 2. Draw the circle and the various chords. = chord of 300° = radius = 2 inches ; the chord of 180° = diameter = 4 inches. If the pupils draw the chord of 120°, ZX (Fig. 42), they will find that it bisects the ra- dius YC, making MC 1 inch. CX-= 2 inches ; therefore MX = ^¥^^ = V3 = 1.732 + ; ZX= 1.732 + in. X 2 = 3.464 + in. = chord of 120° = chord of 240°. AC=AB, The chord of 60° 274 MANUAL FOR TEACHERS Diagrams should be employed, unless the pupils can do satis- factory work without them. Since the arc of 180° is three times as long as the arc of 60°, the more careless members of the class may jump to the conclusion that the chord of 180° is three times as long as the chord of 60°. This mistake cannot be made if the circle is constructed, and the chords are drawn and measured. 5. See Exercise 98, Art. 1269. 6. Each side of the hexagon measures 1 inch ; the perimeter = 1 inch X 6 = 6 inches. Ans. Circumference = 2 in. X 3.1416 = 6.2832 inches. Ans. 7. The pupil should use the ruler to ascertain the length of the apothem, which is about -J- in. It can be calculated as follows : — Cb (Fig. 38) is the radius = 1 inch ; bx = one-half the side of the inscribed hexagon = ^ inch ; Cx = apothem. Cx = Cb — bx=l-\ = .1b; Cx = -yJTfb = .866+, or nearly J. 8. The base of each triangle measures one inch, so that AB (Fig. 43) = 3 in. ^X (apothem) x = J in. nearly. 9. The base of each triangle measures about f in. so that the A base of the rectangle (the half ^^^' ^' perimeter) will be nearly 3 inches and the apothem about \^ in. Area about 3 X ||- sq. in. = about 2^ sq. in. 10. AB X AX= 3 X J = 2f . Ans. 2f sq. in. 11. See 9. 12. The perimeter of a 16-sided polygon will be greater than that of an octagon. The 16-sided polygon will have the greater apothem. 13. As the number of sides increases, the perimeter approaches more and more closely the circumference, 6.2832 inches; and the apothem approaches the radius, 1 inch. NOTES ON CHAPTEE SIXTEEN 275 14. The base of the rectangle will be 3.1416 inches, one-half the perimeter (circumference) ; the apothem will be 1 inch (radius). Area = 3.1416 sq. in. Ans. 15. One-half the circumference = 2 X 3.1416 = 6.2832, multi- plied by the radius, 2, gives answer in square inches, 12.5664 sq. in. 16. 78.54 sq. in. Ans. 17. 3.1416 sq. in. Ans. 18. 314.16 sq. in. -^6 = Ans. 19. Subtract from the area of the outer circle, 113.0976 sq. in., the area of the inner circle, 28.2744 sq. in. Ans. 84.8232* sq. in. 1282. Bight prisms, cylinders, etc., are meant when the word oblique is not used. 1. See Arithmetic, Art. 818, Problem 20. The upper squares need not be drawn, as only the convex surface is required. 2. Three rectangles and two triangles. See 1. 3. Three rectangles, each 3 inches high, bases 1, 1^, and 2 inches, respectively. 5. A hollow paper cylinder, without bases, can be opened out into a rectangle whose base is the circumference of the base of the cylinder, and whose height is the altitude of the cylinder. 6. The slant height of a square pyramid is the distance from the apex to the center of one side of the base. This problem requires the pupil to draw, side by side, four isosceles triangles, the base of each being 2 inches and the altitude 3 inches. After constructing the first (Fig. 44), by erect- ing a 3-inch perpendicular at the center of a 2-inch base, he should use B" and A as centers, and radii equal to ^"^'and AB" to locate B"\ On AB' construct a B'"^ third triangle, using A and B' as centers and the radii previously given. Upon one ^t '^Ji^ side of this triangle construct a fourth. Fig. 44. 276 MANUAL FOR TEACHERS The pupils can discover for themselves the method of drawing geometrically the required convex surface. The altitude should be carefully measured. It will be equal in length to the perj^pndicular of a right-angled triangle whose hypotenuse is the slant height of a pyramid, 3 inches ; and whose base is the distance from the center of one edge of the base of the pyramid (the foot of the slant height) to the foot of the alti- tude, 1 inch. Ans. V8 inches = 2.83 inches nearly = about 2|J inches. 7. The area of each convex face of a regular pyramid is found by multiplying one side of the base by one-half the slant height; therefore, the area of all the faces forming the convex sur- face, is obtained by multiplying the sum of all the sides of the base, that is, the perimeter of the base, by one-half the slant height. 8. The pupil requires the slant height in order to proceed as in 6 ; and he should obtain it by drawing it rather than by cal- culating it. He should be able to see that the altitude is a line drawn from the vertex to the center of the base, and that its foot is 1 inch distant from the foot of the altitude. Constructing a right-angled triangle whose base is 1 inch, and whose perpen- dicular is 3 inches, will give a hypotenuse equal to the required slant height. Some scholars will bring in a prism whose slant height is 3 inches. The mistake should be pointed out, but not the mode of correcting it; and a pyramid of the required dimensions should be insisted upon. 1283. When the diameter of the base of a cone is 2 inches, the arc BDC, which forms the circumference when folded, measures 27r inches, or 6.2832 inches. Note. — The Greek letter ir (pi) represents 3.1416. 9. The semi-circumference of paper = 3ir inches, which is the circumference of base of cone. The diameter of base of cone — 3 TT inches -^ TT = 3 inches. The radius of base = 1} inches. r NOTES ON CHAPTER SIXTEEN 277 The slant height = radius of paper = 3 inches = diameter of base of cone = twice radius of base of cone. 10. The area of any sector of a circle = radius X i length of arc. The arc when folded becomes the circumference of base, and the radius becomes the slant height ; so that convex surface = i circumference X slant height = circumference X i slant height. 11. Length of arc of 90° = -^ circumference = -^ of Gtt inches = 1^ TT inches. This is the circumference of the base of the cone, which makes the diameter = 1|- tt inches -^ tt = 1|- inches. The slant height is 3 inches. Length of arc of 60° = ^ circumference = |- of 6 tt inches = TT inches. The diameter of the base of the cone = tt inches ^ tt = 1 inch; slant height, 3 inches. 12. The circumference of the base of the cone = 3ir. This equals the length of the arc of the required sector. If the slant height is 6 inches, the circumference of which the sector is a part = 10 TT. The arc of the sector is, therefore, y^^ of the circumfer- ence ; and its length is -^-^ of 360° = 108°. 13. XY represents the base of the pyramid; and AC, its altitude. The slant height of two faces, AM (Fig. 45), is the hypotenuse of a right-angled triangle, base IJ in., perpendicular 4 in. Using this as the perpendicu- lar of a new triangle, with a base MY, gives as the hy- potenuse AY (Figs. 45 and 46), one of the edges. To draw the development, take AY as a radius, and draw an arc Fig. 45. On this lay off succes- 278 MANUAL FOR TEACHERS sive chords of 3 in., 2 in., 3 in., and 2 in., connecting the extremity of each with the center A, These four triangles constitute the convex faces of the pyramid. On one of them, construct a rectangle 3 inches by 2 inches for the base. 14. In this pyramid, AC (Fig. 45) measures 12 inches, C3I measures -^ of 18 inches, or 9 inches, making AM, one slant height, = Vl44 -f 81 inches = 15 inches. The other slant height, drawn to the center of OF, = Vl44 -f- 25 inches = 13 inches. The pupils should be encouraged to construct stout paper pyramids and cones of required dimensions, making the necessary calculations themselves. The previous fourteen problems will present no difficulty whatever to pupils that are interested. As examples in dry calculation, they may prove somewhat tiresome. Many scholars will, of themselves, construct models of solids much more complicated than the foregoing. 1284. If there are no solids at hand, the pupils should con- struct a supply of paper ones, or make them of modeling clay, turnips, etc. Drawings are not sufficient for effective instruc- tioD. NOTES ON CHAPTER SIXTEEN 279 15. The short method of ascertaining the convex surface should not be given until 18. Each of the convex faces is a trapezoid, whose parallel sides measure 4 inches and 8 inches, respectively, the altitude being 10 inches. 16. Two inches apart, draw two parallel lines, AB and CD (Fig. 47), measuring 1 inch and 2 inches, respectively, a 2-inch perpendicular, ^Y, i connecting the middle point of each. Draw /;\ AC and BD, and produce them until they j \ \ meet in M. With this as a center, and MO as a radius, draw an arc ; on which three / other chords equal to CD are laid off, as in ^p Fig. 46, With jif as a center and a radius / MA, lay off another arc, on which chords / are laid off equal to ^^; etc. See Arithme- / tic. Art. 1296. / jir, ^ X 17. The entire surface = convex surface + ■ 4 sq. ft. + 9 sq. ft. ^'"- '"■ 18. The pupils can readily understand this rule, using the frustums of problems 15 and 17 as illustrations. 19. The pupil should first locate the apex of the cone. This he can do by following the method shown in 16 ; making AB (Fig. 47) 2\ inches; CD, 1^ inches; and AC, 2 inches. This makes MA 5 inches, the slant height of the whole cone. The circumference of the base of the cone — 2^7r. The slant height, 5 inches, is the radius of the required sector, its circumference being 10 tt. 2|-7r, the length of the arc of the sector, being one- fourth of 10 TT, shows that the required sector is a quadrant. 20. The number of square inches in the convex surface = [cir- cumference (perimeter) of upper base (9 X 3.1416) + circumfer- ence of lower base (6 X 3.1416)] X ^ slant height (2). Adding to this, the area of the bottom (3') 9 sq. in. X 3.1416 (Art. 1124, 2) gives the number of square inches of material required. 280 MANUAL FOR TEACHERS 22. Circumference of upper base = 6x3. 1416 Circumference of lower base = 10 X 3.1416 One-half sum = 8 X 3.141G Multiplying by slant height gives 48 X 3.1416 Add to this the area of the upper base, 9 X 3.1416 And the area of the lower base, 25 X 3.1416 Total in square yards, 82 X 3.1416 or ( [(3 + 5) X 6] -f 9 + 25) x 3.1416. 23. FA: UC ::Q:S x:x + 9 :: 6:8 8a: = 6a; + 54 2a:=:54 a; -27. Ans. 27 ft. The slant height of the whole cone =-27 ft. +9 ft. =36 ft. 24. The convex surface of the whole cone = ^ (8x3.1416 X 36) sq. ft. ; of the part cut off =i (6 X 3.1416 X 27) sq. ft. 1287. A sphere (a croquet ball, for instance) and a hemisphere should be used to illustrate these problems. On the plane face of the latter can be drawn the lines AD, FG, HI, CI, etc. ; while on the curved face can be drawn HYI, FXQ^ etc. 25. ^ of 25,000 miles. 26. IH^ chord of 60° of the great circle = radius of the <^ieat circle = 4000 miles. Z5 = J of 111= 2000 miles. 27. The diameter, HI, of the small circle is \ diameter FG of the great circle ; the circumference HYI= \ of 25,000 miles, or 12,500 miles. 28. The length of a degree of longitude on the 60th parallel is about one-half of the length of a degree on the equator. (See Art. 995, Problem 10.) NOTES ON CHAPTER SIXTEEN 281 29. On the plane face of the hemisphere suggested above (Art. 1287), draw diameters AD and FQ at right angles ; and 45° from G, a chord NM parallel to FG. (This chord will not bisect AC.) As MOG - 45°, WCM-= 45° ; and the triangle WCM is a right-angled isosceles tri- angle, and WM'^^^CM'') WM= .lOnCM. (See Art. 995, Problem 12.) If CG measures 4000 miles, CM= 4000 miles, an d WM-= V8,000,000 = 2828.4 miles. 1289. The pupils have already learned that the volume of a rectangular prism is equal to the area of the base multiplied by- its altitude ; these problems are intended to show that the same is true of all prisms and of the cylinder (6). 1292. 8. The volume of the frustum is obtained by deduct- ing the volume of the part cut off from the volume of the whole pyramid (Problem 7). The rule is given later. 10. Fig. 49 gives the method of calculating the slant height. The illustration shows a sec- tion of the frustum formed by passing a plane through the center of the frustum perpendicular to the base. The center of the upper base of the frustum of a right pyr- amid is directly above the center of the lower base, so that a perpendicular let fall from A will fall on CD at a point Jl, 5 in. from C. In the right-angled triangle AXC, AX- = ^"X'+^' = 1444-25 = 169; AC- 5 in, Fig. 49. altitude = 12 in. = Vl69 = 13. AC^ XX NOTES ON THE APPENDIX 1306. A person that contracts to receive a rate of interest greater than is permitted by law, is liable to a penalty, except in Connecticut. In Delaware, Minnesota, etc., the penalty is the forfeiture of the contract ; in New York, the forfeiture of the contract, $1000 fine, and 6 months' imprisonment; in Indiana, Kansas, Kentucky, Maryland, Michigan, Mississippi, Ohio, Penn- sylvania, Tennessee, Vermont, Virginia, and West Virginia, the forfeiture of the excess of interest ; in North Carolina, the forfeiture of double the amount of interest ; in Georgia and New Hamp- shire, the forfeiture of three times the excess of interest ; in Ala- bama, Delaware, Florida, Illinois, Iowa, Louisiana, Missouri, Nebraska, New Jersey, South Carolina, Texas, and Wisconsin, the forfeiture of all interest ; in Arkansas and Oregon, the for- feiture of principal and interest. 1307. 3. Amount of $ 1000, June 1, 1896. to June 1, 1897 Amount of $150, Sept. 16, 1896, to June 1, 1897, 8^ mo. Due June 1, 1897 Interest to settlement, Apr. 16, 1898, lOJ mo. . Amount Apr. 16, 1898 .... Amount of $50, Sept. 16, 1897, to Apr. 16, 1898, 7 mo. Balance due 11,060.00 166.37 $903.63 47.44 $951.07 51.75 $899.32 4. Amount of $500, July 25, 1896, to Apr. 1, 1897, 8 mo. 6 da. ^ $520.50 Amount of $100, Sept. 18, 1896, to Apr. 1, 1897, 6 mo. 13 da $103.22 Amount of $200, Feb. 5, 1897. to Apr. 1, 1897, 1 mo. 26 da 201.86 305.08 Balance due 282 $215.42 NOTES ON THE APPENDIX 283 $985.99 235. $750.99 35.54 $786.53 255.33 5. Amount of $870.50, Jan. 2, 1894, to March 18, 1896, 2 yr, 2 mo. 16 da Less payments of $ 35 and $ 200 New principal ....... Interest March 18, 1896, to Jan. 2, 1897, 9 mo. 14 da. . Amount Amount of $250, Aug. 24, 1896, to Jan. 2, 1897, 4 mo. 8 da, Due at maturity $531.20 Below will be found answers to the partial payments examples of Chapters XIII and XIV, according to tlie Connecticut rule. Note. — Although the legal rate in Connecticut is 6%, the rates given in the examples should be used. Art. 1008. $70.51. Art. 1009. 1. $224.22. 2. $261.21. 3. $1278.15. Art. 1011. 4. $771.24. 5. $899.32 (see Art. 1307, 3). Art. 1013. 5. $1088.31. Art. 1015. 7. $700.50. Art. 1023. 7. $1232.26. Art. 1051. 7. $77.07. Art. 1090. 3. $649.13. 4. $224.64. Art. 1107. 2. $1089.64. The time in each of the preceding examples was found by com- pound subtraction, taking 30 days to each month. In the examples in Art. 1110, the Connecticut rule has been followed. In these, the time is taken in days. (See Art. 1111.) 1308. 7. Principal $700.00 Interest to June 15, 1897, 2 yr. . . , . . 84.00 1 year's interest on $42, unpaid interest . . . 2.52 Amount June 15, 1897 $786.52 Amount of $20, Nov. 15, 1895, to June 15, 1897 . . $21.90 Amount of $80, Feb. 15, 1897, to June 15, 1897 . . 81.60 103.50 New principal June 15, 1897 .... $683.02 Interest to Oct. 15, 1899, 2 yr. 4 mo 95.62 Interest for 1 yr. 4 mo. on $ 40.98, unpaid interest . . 3.28 Interest for 4 mo. on $40.98, unpaid interest ... .82 Amount Oct. 15, 1899 $782.74 Amount of $ 15, Sept. 15, 1898, to Oct. 15, 1899 . . 15.98 Due Oct. 15, 1899 $766.76 Note. — If four places of decimals are used, the answer to 6 is $367.6442, or $367.64 + ; to 7, $ 766.7658, or $ 766.77 - . 284 MANUAL FOR TEACHERS The teacher that wishes other examples of this kind can use 9 and 10 of Art. 1309, and 12, 14, and 15 of Art. 1310, the answers to which are as fol- lows: Art. 1309. 9. $1734. 10. $738.29. Art. 1310. 12. $1901.18. 14. $3432.20. 15. $1010.95. 1309. Pupils in New Hampshire should not be taught the pre- ceding method, but should be confined to the rule laid down by the courts of their own state. An examination of 8 will show the manner of ascertaining the balance. The interest for 1 year is $36. As no interest is due except that which accrued during the year, and as the sura paid is less than the interest due, no interest is allowed on the payment of $30, made during the year. This payment being $6 less than the interest due at the end of the year, the in- terest on $6 is added to the interest on the principal at the end of the next year, making a total of $642.36 then due less the amount of $100 for 11 months. Interest is allowed on the $ 100 payment because it is in excess of the interest then due. 9. Principal Annual interest due May 1, 1897 Payment (no interest) Oct. 1, 1896 Balance of interest Annual interest due May 1, 1898 Interest on balance of interest, 1 yr. . Amount May 1, 1898 . Amount of $1000, June 1, 1897, to May 1, New principal May 1, 1898 . Interest on $ 1648 to May 1, 1899 Payment (no interest) Nov. 1, 1898 Balance of interest Interest on $ 1648 May 1 to Oct. 1 Interest on balance of interest, 5 mo. . Due Oct. 1, 1899 . 10. Principal Annual interest due Jan. 3, 1896 Payment (no interest) June 1, 1895 Balance of interest 1898 $2500.00 $ 150.00 100.00 $ 50.00 150.00 3.00 203.00 1 2703.00 1055.00 $1648.00 $98.88 50.00 $48.88 41.20 1.22 91.30 $ 1739.30 $1000.00 $60.00 10.00 $50.00 NOTES ON THE APPENDIX Amounts brought forward Annual interest due Jan. 3, 1897 Interest on balance of interest, 1 yr. . Interest due Jan. 3, 1897 Amount of payment March 14, 1896, 9 mo. 19 da. Balance of interest Annual interest due Jan. 3, 1898 Interest on balance of interest, 1 yr. . Annual interest due Jan. 3, 1899 Interest for 1 year on $ 102.52 and 1 60 Amount Jan. 3, 1899 . Amount of |500, Sept. 30, 1898, to Jan. 3, 1899 New principal Jan. 3, 1899 . interest on $ 730.67 to March 11, 2 mo. 8 da. Due March 11. 1899 1 50.00 60.00 3.00 $113.00 10.48 $ 102.52 60.00 6.15 60.00 9.75 285 1000.00 238.4: P 1238.42 507.75 1 730.67 8.28 $738.95 Note. — The amount of the payment of March 14 canceled the $3 in- terest on interest and $7.48 additional, leaving $ 102.52 interest still unpaid Jan. 3, 1897 on which two years' interest is taken to Jan. 3, 1899. Two years' interest on the principal is also taken, and one year's interest on the annual interest due Jan. 3, 1898 and unpaid. It will be noticed that no interest is taken on the interest upon interest, $3 and $6.15. See 14 of Art. 1310, as calculated by the N.H. rule : — Principal .... Interest to March 17, 1900, 4 yr. Interest on $ 180 (3 + 2 + 1) yr. Amount of $ 20, 10 mo. Unpaid interest on interest . Interest to March 17, 1903, 3 yr. Interest on $ 180 (3 + 3 + 3 + 3 + 2 + 1) yi Total interest due March 17, 1903 Amount of $ 1000, 6 mo. . Due March 17. 1903 $3000.00 $64.80 21.00 $43.80 162.00 $ 720.00 540.00 205.80 $ 1465.80 1030.00 435.80 $ 3435.80 Note. — Interest is not taken on $43.80. As additional examples, the teacher rnay use 7 of Art. 1308, and 12, 14, and 15 of Art. 1310. The answers by the N.H. method are 7, $ 767.60 ; 12, $1901.27; 14, $3435.80; 15, $ 1011.40. 286 MANUAL FOK TEACHERS 1310. Vermont pupils should omit Arts. 1307, 1308, and 1309. 12. Trincipal . Annual interest due June 1, 1897 Amount June 1, 1897 Amount of $ 100 to June 1, 1897 Payment June 1, 1897 New principal Annual interest to June 1, 1899 Amount of $ 50, 7 mo. Unpaid interest Interest on $ 1705 to settlement Interest on $ 158.99 to settlement Due Oct. 1, 1899 . 13. Principal Jan. 3, 1895 Interest due Jan. 3, 1896 Amount of $ 10, 7 mo. 2 da. Balance of interest Interest due Jan. 3, 1897 1 year's interest on ^49.65 . Total interest Amount of $ 10, 9 mo. 19 da. Balance of interest Interest on $ 1000 for 2 years 2 years' interest on $ 102.15 1 year's interest on $ 60 Amount Jan. 3, 1899 Amount of $500, Sept. 30, 1898, to Jan 3, New principal Jan. 3, 1899 Interest to March 11, 1899 . Due March 11, 1899 14. Principal March 17, 1896 Interest to March 17, 1900, 4 yr. Interest on yearly interest (3 + 2 + 1) yr. Amount of $ 20, 10 mo. Unpaid interest on yearly interest 1899 164.80 21.00 143.80 $ 2500.00 309.00 $2809.00 $104.00 1000.00 1104.00 $1705.00 $210.74 51.75 $158.99 34.10 3.18 196.27 $1901.27 $1000.00 $60.00 10.35 $49.65 60.00 2.98 $112.63 10.48 $102.15 120.00 12.26 3.60 238.01 $ 1238.01 507.75 1^30.26 8.28 1 738.54 $3000.00 $720.00 NOTES ON THE ArPENDIX 287 43.80 162.00 Amounts brought forward 3 years' interest to March 17, 1903 Interest on yearly interest, 3 yr. on $720 (2 + 1) yr. on $ 180 Total interest due March 17, 1903 Amount of $ 1000 to March 17, 1903 . Due March 17, 1903 15. Principal Feb. 25, 1893 . Annual interest to Feb. 25, 1897, 4 yr. Amount Feb. 25, 1897 . Amount of |400, 11 mo. 29 da. . New principal Feb. 25, 1897 . Annual interest on $ 1089.99, 2 yr. . Amount of $ 10, 8 mo. 14 da. Unpaid interest Feb. 25, 1899 Annual interest on $ 1089.99, 2 yr. Interest for 2 yr. on $ 124.30 Amount Feb. 25, 1901 . Amount of |400, 4 mo. 29 da. . New principal Feb. 25, 1901 . Annual interest on $953.99, 1 yr. 1 da. As additional examples, 7 and 9 may be used. The Vermont rule is | 766.75 ; to 9, $ 1733.73. . 720.00 $ 3000.00 540.00 205.80 1465.80 1030.00 435,80 134.72 10.42 124.30 134.72 14.91 $3435.80 $ 1200.00 313.92 $1513.92 423.93 $ 1089.99 273.93 $ 1363.92 409.93 $ 953.99 57.41 $1011.40 answer to 7 by the 1311. 1. [($8500 + $600) X. 0155] + $2 -$141.05. Ans. 2. The amount to be raised on property = $2500 — ($2 X 150) mills 275000 -$.008, or 8 $97. Tax = $2.45 $2200. Rate on $1 = $2200 8 mills on $1, or 1%. A72S. 4. Mr. Simmons' grand list = $ 95 + $ 2 X 97 = $237.65. Am. 5. 1500 -f 2a: = grand list. The grand list multiplied by the rate gives the amount to be raised; (1500 + 2a;) X 2 = 3600 ; 3000 -\-ix = 3600 ; 4 a; = 600 ; a: = 150. 150 taxable polls. Ans. 288 MANUAL FOR TEACHERS 6. Mr. Hallock's grand list = $ 120 + $6 = $126. Since his taxes are $252, the rate = $252 ^ 126 = $2. Let X = appraised value of property. _£_ + 400 = grand list. 2 (-^ + 400) = total levy = 6800. - ^ + 800 = 6800. 2a; + 80000 = 680000. X = 300000. Appraised value of property is $300000. Am. SUPPLEMENT DEFINITIONS, PRINCIPLES, AND RULES A Unit is a single thing. A Number is a unit or a collection of units. The Unit of a Number is one of that number. Like Numbers are those that express units of the same kind. Unlike Numbers are those that express units of different kinds, A Ooncrete Number is one in which the unit is named. An Abstract Number is one in which the unit is not named. Notation is expressing numbers by characters. Arabic Notation is expressing numbers by figures, Eoman Notation is expressing numbers by letters. Numeration is reading numbers expressed by characters. The Place of a Figure is its position in a number. A figure standing alone, or in the first place at the right of other figures, expresses ones, or units of the first order. A figure in the second place expresses tens, or units of the second order. A figure in the third place expresses hundreds, or units of the third order ; and so on. A Period is a group of three orders of units, counting from right to left. Rule for Notation. — Begin at the left, and write the hun- dreds, tens, and units of each period in succession, filling vacant places and periods with ciphers. 11 SUPPLEMENT Rule for Numeration. — Beginning at the rights separate the number into periods. Beginning at the left, read the numbers in each period, giving the narne of each period except the last. ADDITION Addition is finding a number equal to two or more given num- bers. Addends are the numbers added. Tlie Snm, or Amount, is the number obtained by addition. Principle. — Only like numbers, and units of the same order can be added. Rule. — Write the numbers so that units of the same order shall f e in the same column. Beginning at the right, add each column separately, and write the su7n, if less than ten, under the column added. When the sum of any column exceeds nine, wHte the units only, and add the ten or tens to the next column. Write the entire su7?i of the last column. SUBTRACTION Subtraction is finding the difference between two numbers. The Subtrahend is the number subtracted. The Minuend is the number from which the subtrahend is taken. The Eemainder, or Difference, is the number left after subtracting one number from another. Principles. — Only like numbers and units of the same order can be subtracted. The sum of the difference and the subtrahend inust equal the minuend. Rules. — I. Write the subtrahend under the minuend, placing units of the same order in the same column. DEFINITIONS, PRINCIPLES, AND RULES 111 Beginning at the right, find the nuinber that 7nust be added to the first figure of the subtrahend to produce the figure in the corre- sponding order of the minuend, and lurite it below. Proceed in this way until the difference is found. If any figure in the subtrahend is greater than the corresponding figure in the minuend, find the number that must be added to the former to produce the latter increased by ten ; then add one to the next order of the subtrahend and proceed as before. II. Beginning at the units column, subtract each figure, of the subtrahend from the corresponding figure of the minuend and write the remainder below. If any figure of the subtrahend is greater than the corresponding figure in the 7ninuend, add ten to the latter and subtract; then, (a) add one to the next order of the subtrahend and proceed as before ; or, (b) subtract one from the next orden^ of the minuend and proceed as before. MULTIPLICATION Multiplication is taking one number as many times as there are units in another number. The Multiplicand is the number taken or multiplied. The Multiplier is the number that shows how many times the multiplicand is taken. The Product is the result obtained by multiplication. Principles. — The multiplier must be an abstract number. The multiplicand and the product are like numbers. The product is the saine in whatever order the numbers are multiplied. Rule. — Write the m,ultiplier under the multiplicand, placing units of the same order in the same colutnn. Beginning at the right, Tnultiply the multiplicand by the number of units in each order of the multiplier in succession. Write the IV SUPPLEMENT figure of the lowest order in each partial product under the figure of the multiplier that produces it. Add the partial products. To multiply by 10, 100, 1000, etc. Rule. — Annex as many ciphers to the multiplicand as there are cipher's in the multiplier. DIVISION Division is finding how many times one number is contained in another, or finding one of the equal parts of a number. The Dividend is the number divided. . The Divisor is the number contained in the dividend. The Quotient is the result obtained by division. Principles. — When the divisor and the dividend are like num- bers, the quotient is an abstract number. When the divisor is an abstract number^ the dividend and the quotient are like numbe)'S. The product of the divisor and the quotient, plus the remainder, if any, is equal to the dividend. Rule. — Write the divisor at the left of the dividend with a line between them. Find how many times the divisor is contained in the fewest fig- ures on the left of the dividend, and write the result over the last figure of the partial dividend. Multiply the divisor by this quotient figure, and write the product under the figures divided. Subtract the product from the partial dividend used, and to the remainder annex the next figure of the dividend for a new dividend. Divide as before until all the figures of the dividend have been used. If any partial dividend will not contain the divisor, write a ciphei' in the quotient, and annex the next figure of the dividend. If there is a remainder aflei' the last division, write it after the quotient with the divisor underneath. DEFINITIONS, PRINCIPLES, AND RULES V FACTORING An Exact Divisor of a number is a number that will divide it without a remainder. An Odd Number is one that cannot be exactly divided by two. An Even Number is one that can be exactly divided by two. The Factors of a number are the numbers that multiplied to- gether produce that number. A Prime Number is a number that has no factors. A Composite Number is a number that has factors. A Prime Pactor is a prime number used as a factor. A Composite Factor is a composite number used as a factor. Factoring is separating a number into its factors. To find the Prime Factors of a Number. Rule. — Divide the number by any prime factor. Divide the quotient, if composite, in like m^anner ; and so continue until a prime quotient is found. The several divisors and the last quotient will be the prime factors. CANCELLATION Cancellation is rejecting equal factors from dividend and divisor. Principle. — Dividing dividend and divisor by the same num^ ber does not affect the quotient. GREATEST COMMON DIVISOR A Common Factor (divisor or measure) is a number that is a factor of each of two or more numbers. A Common Prime Factor is a prime number that is a factor of each of two or more numbers. The Greatest Common Factor (divisor or measure) is the largest number that is a factor of each of two or more numbers. Numbers are prime to each other when they have no common factor. VI SUPPLEMENT The greatest common divisor of two or more numbers is the product of their common prime factors. PRiNCiPiiES. — A common divisor of two numbers is a divisor of their sum, and also of their difference. A divisor of a number is a divisor of eve^y multiple of that nu77iber ; and a common divism' of two or more number's is a divisor of any of their multiples. To find the Common Prime Factors of Two or More Numbers. Rule, — Divide the numbers by any common prime factors, and the quotients in like manner, until they have no common factor; the several divisors are the common prime factors. To find the Greatest Oommon Divisor of Numbers that are Easily Factored. Rule. — Separate the numbers into thdr prime factors ; the product of those that are common is the greatest common divisor. To find the Greatest Oommon Divisor of Numbers that are not Easily Factored. Rule. — Divide the greater number by the less; then divide the last divisor by the last remainder, continuing until there is no remainder. The last divisor is the greatest common divisoi\ If there are more than two numbers, find the greatest common divisor of two of theyn; then of that divisor and another of the numbers until all of the nu7nbers have been used. The last divisor is the greatest common divisor. LEAST COMMON MULTIPLE A Multiple of a number is a number that exactly contains that number. A Oommon Multiple of two or more numbers is a number that is a multiple of each of them. The Least Oommon Multiple of two or more numbers is the smallest number that is a common multiple of them. DEFINITIONS, PRINCIPLE^^MS^IlfiSj^iTY j y^ Principles. — A multiple of a number comSM^IflTthe prime factors of that number. A cortimon multiple of two or m,ore numbers contains each of the prime factors of those numbers. The Least Common Multiple of tivo or more numbers contains only the prime factors of each of the numbers. To find the Least Common Multiple of Two or More Numbers. Rule. — Divide by any prime numbei' that is an exact divisor of two or more of the numbers, and write the quotients and undivided numbers below. Divide these numbers in like manner, contimiing until no two of the remaining numbers have a common factor. The product of the divisors and remaining numbers is the least common multiple. FRACTIONS A Fraction is one or more of the equal parts of anything. The Unit of a Praction is the number or thing that is divided into equal parts. A Fractional Unit is one of the equal parts into which the num- ber or thing is divided. The Terms of a Fraction are its numerator and its denominator. The Denominator of a fraction shows into how many parts the unit is divided. The Numerator of a fraction shows how many of the parts are taken. A fraction indicates division ; the numerator being the divi- dend and the denominator the divisor. The Value of a Fraction is the quotient of the numerator divided by the denominator. Fractions are divided into two classes — Common and Decimal. A Common Fraction is one in which the unit is divided into any number of equal parts. A common fraction is expressed by writing the numerator above the denominator with a dividing line between. VIU SUPPLEMENT Common fractions consist of three principal classes — Simple, Compound, and Complex. A Simple Fraction is one whose terms are whole numbers. A Proper Fraction is a simple fraction whose numerator is less than its denominator. An Improper Fraction is a simple fraction whose numerator equals or exceeds its denominator. A Compound Fraction is a fraction of a fraction. A Complex Fraction is one having a fraction in its numerator, or in its denominator, or in both. A Mixed Number is a whole number and a fraction written together. The Keciprocal of a Number is one divided by that number. The Eeciprocal of a Fraction is one divided by the fraction, or the fraction inverted. Principles. — Multiplying the numerator or dividing the de- nominator multiplies the fraction. Dividing the numerator or multiplying the denominator divides the fraction. Multiplying or dividing both terms of a fraction by the same number does not alter the value of the fraction. Eeduction of fractions is changing their terms without altering their value. To reduce a Fraction to Higher Terms. Rule. — Multiply both numerator and denominator by the same number. To reduce a Fraction to its Lowest Terms. Rule. — Divide both te^-ms of the fraction by their greatest common divisor. A fraction is in its lowest terms wIkmi the numerator and the denominator are prime to each other. DEFINITIONS, PRINCIPLES, AND RULES ix To reduce a Mixed Number to an Improper Fraction. Rule. — Multiply the whole number by the denominator ; to the product add the numerator ; and place the sum over the denom- inator. To reduce an Improper Fraction to a Whole or to a Mixed Number. Rule. — Divide the num^erator by the denominator. A Common Denominator is a denominator common to two or more fractions. The Least Common Denominator is the smallest denominator common to two or more fractions. To reduce Fractions to their Least Common Denominator, Rule. — Find the least common viultiple of all the denomi- nators/or the least common denominator. Divide this multiple by the denominator of each fraction^ and Tnultiply the numerator by the quotient. ADDITION OF FRACTIONS Principle. — Only like fractions can be added. Rule. — Reduce the fractions, if necessary, to a comm,on denom- inator, and over it write the sum, of the numerators. If there are mixed numbers, add the fractions and the whole numbers separately ^ and unite the results. SUBTRACTION OF FRACTIONS Principle. — Only like fractions can be subtracted. Rule. — Reduce the fractions, if necessary, to a common denom- inator, and over it write the difference between the numerators. If there are mixed numbers subtract the fractions and the whole numbers separately, and unite the results. MULTIPLICATION OF FRACTIONS Rule. — Reduce whole and mixed numbers to improper frac- tions ; cancel the factors common to numerators and denomina- tors, and write the product of the remaining factors in the nitmer- ators over the product of the remaining factors in the denominators. SUPPLEMENT DIVISION OF FRACTIONS Rules. — I. Reduce whole and mixed numbers to improper fractions. Reduce the fractions to a common denominator. Divide the numerator of the dividend hy the nume^^ator of the divisor. II. Invert the divisor and proceed as in multiplication of frac- tions. To reduce a Oomplex Fraction to a Simple One. Rules. — I. Multiply the numerator of the complex fraction by its denoTninator inverted. II. Multiply both terms by the least common multiple of the denominators. DECIMALS A Decimal Fraction is one in which the unit is divided into tenths, hundredths, thousandths, etc. A Decimal is a decimal fraction whose denomination is indi- cated by the number of places at the right of the decimal point. The Decimal Point is the mark used to locate units. A Mixed Decimal is a whole number and a decimal written together. A Oomplex Decimal is a decimal with a common fraction written at its right. To write Decimals. Rule. — Write the numerator ; and from the right, point off as many decimal places as there are ciphers in the denominator, prefixing ciphers, if necessary, to make the required nu/mber. To read Decimals. Rule. — Read the numerator, and give the name of the right- hand order. Principles. — Prefixing ciphers to a decimal diminishes its valv£. DEFINITIONS, PRINCIPLES, AND RULES xi Removing ciphers from the left of a decimal increases its value. Annexing ciphers to a decimal or removing ciphers fronn its right does not alter its value. To reduce a Decimal to a Common Fraction. Rule. — Write the figures of the decimal for the numerator, and 1, with as many ciphers as there are places in the decimal, for the denominator, and reduce the fraction to its lowest terms. To reduce a Common Fraction to a Decimal. Rule. — Annex decimal ciphers to the numerator, and divide it by the denominator. To reduce Decimals to a Common Denominator, Rule. — Make their decimal places equal by annexing ciphers. ADDITION AND SUBTRACTION OF DECIMALS Decimals are added and subtracted the same as whole numbers. MULTIPLICATION OF DECIMALS Rule. — Multiply as in whole numbers, and from the right of the product, point off as many decimal places as there are decimal places in both factors. DIVISION OF DECIMALS Rule. — Make the divisor a whole number by removing the decimal point, and make a corresponding change in the dividend. Divide as in whole numbers, and place the decimal point in the quotient under {or over) the new decimal point in the dividend. ACCOUNTS AND BILLS A Debtor is a person who owes another. A Creditor is a person to whom a debt is due. Xll SUPPLEMENT An Account is a record of debits and credits between persons doing business. The Balance of an account is the diflference between the debit and credit sides. A Bill is a written statement of an account. An Invoice is a written statement of items, sent with merchan- dise. A Beceipt is a written acknowledgment of the payment of part or all of a debt. A bill is receipted when the words, " Received Payment," are written at the bottom, signed by the creditor, or by some person duly authorized. DENOMINATE NUMBERS A Measure is a standard established by law or custom, by which distance, capacity, surface, time, or weight is determined. A Denominate Unit is a unit of measure. A Denominate Number is a denominate unit or a collection of denominate units. A Simple Denominate Number consists of denominate units of one kind. A Compound Denominate Number consists of denominate units of two or more kinds. A Denominate Fraction is a fraction of a denominate number. A denominate fraction may be either common or decimal, Eeduction of denominate numbers is changing them from one denomination to another without altering their value. Eeduction Descending is changing a denominate number to one of a lower denomination. -' Rule. — Multiply the highest denomination hy the number re- quired to reduce it to the next lowei' dctioTnination, and to the prod- uct add the units of thai lower denomination, if any. Proceed in this manner until the required denomination is reached. DEFINITIONS, PKINCIPLES, AND RULES xiii Eednction Ascending is changing a denominate number to one of a higher denomination. Rule. — Divide the given denomination successively hy the numbers that will reduce it to the required deno7nination. To this quotient annex the several remainders. To find the Time between Dates. Rule. — When the time is less than one year, find the exact number of days; if greate)' than one year, find the time by com- pound subtraction, talcing 30 days to the month. PERCENTAGE Per Oent means hundredths. Percentage is computing by hundredths. The elements involved in percentage are the Base, Bate, Per- centage, Amount, and Difference. The Base is the number of which a number of hundredths is taken. The Kate indicates the number of hundredths to be taken. The Percentage is one or more hundredths of the base. The Amount is the base increased by the percentage. The Difference is the base diminished by the percentage. To find the Percentage when the Base and Eate are Given. Rule. — Multiply the base by the rate expressed as hundredths. To find the Eate when the Percentage and Base are Given. Rule. — Divide the percentage by the base. To find the Base when the Percentage and Eate are Given. Rule. — Divide the percentage by the rate expressed as hun- dredths. To find the Base when the Amount and Eate are Given. Rule. — Divide the amount by l-{- the rate expressed as hun- dredths. XIV SUPPLEMENT To find the Base when the Difference and Eate are Given. Rule. — Divide the diff cogence by \-~the rate expressed as hun- dredths. PROFIT AND LOSS Profit or Loss is the difference between the buying and selling prices. In Profit and Loss, The buying price, or cost, is the base. The rate per cent profit or loss is the rate. The profit or loss is the percentage. The selling price is the amount or difference, according as it is more or less than the buying price. COMMERCIAL DISCOUNT Oonuneroial Discount is a percentage deducted from the list price of goods, the face of a bill, etc. The Net Price of goods is the sum received for them. In Oommercial Discount, The list price, or ) . •, 7 The face of the bill | ^^ the Josa. The rate per cent discount is the rate. The discount is the percentage. The list price diminished by the discount is the difference. In successive discounts, the first discount is made from the list price or the face of the bill ; the second discount, from the list price or face of the bill diminished by the first discount ; and so on. COMMISSION Commission is a percentage allowed an agent for his services. A Oommission Agent is one who transacts business on com- mission. DEFINITIONS, PRINCIPLES, AND RULES XV A Oonsignment is the merchandise forwarded to a commission agent. The Consignor is the person who sends the merchandise. The Consignee is the person to whom the merchandise is sent. The Net Proceeds is the sum remaining after all charges have been deducted. In buying, the commission is a percentage of the buying price; in selling, a percentage of the selling price; in collecting, a per- centage of the sunfi collected; hence : The sum invested, or >- is the base. The sum collected ) The rate per cent commission is the rate. The commission is the percentage. The sum invested increased by the commission is the amount. The sum collected diminished by the commission is the differ- ence. INSURANCE Insurance is a contract of indemnity. Insurance is of three kinds — Fire, Marine, and Life. Fire Insurance is indemnity against loss of property by fire. Marine Insurance is indemnity against loss of property by the casualities of navigation. Life Insurance is indemnity against loss of life. The Insurance Policy is the contract setting forth the liability of the insurer. The Pohcy Face is the amount of insurance. The Premium is the price paid for insurance. The Insurer, or Underwriter, is the company issuing the policy. The Insured is the person for whose benefit the policy is issued. In Insurance, The policy face is the base. The rate per cent premium is the rate. The premium is the percentage. XVi SUPPLEMENT TAXES A Tax is a sum of money levied on persons or property foi public purposes. A Personal, or Poll Tax, is a tax on the person. A Property Tax is a tax of a certain per cent on the assessed value of property. Property may be either personal or real. Personal Property consists of such things as are movable. Keal Property is that which is fixed, or immovable. In Taxes, The assessed value is the base. The rate of taxation is the rate. The tax is the percentage. DUTIES Duties are taxes on imported goods. Duties are either Specific or Ad Valorem. A Specific Duty is a tax on goods without regard to cost. An Ad Valorem duty is a tax of a certain per cent on the cost of goods. In Ad Valorem Duties, The cost of the goods is the base. The rate per cent duty is the rate. The ad valorem duty is the percentage. INTEREST Interest is the sum paid for the use of money. Tlie Principal is the sum loaned. The Amount is the sum of the principal and interest. The Eate of Interest is the rate per cent for one year. The Legal Rate is the rate fixed by law. Usury is interest at a higher rate than that fixed by law. Simple Interest is interest on the principal only. DEFINITIONS, PRINCIPLES, AND RULES Xvii To find the Interest when the Principal, Time, and Eate are Given. KuLE. — Multiply the principal by the rate expressed as hun- dredths, and this product by the time expressed in years. To find the Time when the Principal, Interest, and Rate are Given. Rule. — Divide the given interest by the interest for one year. To find the Eate when the Principal, Interest, and Time are Given. Rule. — Divide the given interest by the interest at one per cent. To find the Principal when the Interest, Eate, and Time are Given. Rule. — Divide the given interest by the interest on $ 1. To find the Principal when the Amount and Time and Eate are Given. Rule. — Divide the given amount by the a7nount of $1. Interest by Aliquot Parts. To find the Interest for Years, Months, and Days. Rule. — Mnd the interest for one year and take ihis as many times as there are years. Take the greatest number of the given months that equals an aliquot part of a year and find the interest for this time. Take aliquot parts of this for the remaining months. In the same manner find the interest for the days. The sum of these interests will be the interest required. To find the Interest when the Time is Less than a Tear. Rule. — Find the interest for the time in months or days that will gain one per cent of the principal. Find by aliquot parts, as in the first rule, the interest for the remaining time. The sum of these interests will be the int&rest required. XVlll SUPPLEMENT Interest by Six Per Cent Method. To find the Interest at 6%. Rule. — For Years: Multiply the principal hy the rate ex- pressed as hundredths f and that product hy the number of years. For Months : Move the decimal point two places to the left^ and multiply by one-half the number of months. For Days ! Move the decimal point three places to the left, and multiply by one-sixth the number of days. To find the interest at any other rate per cent, divide the in- terest at 6% by 6, and multiply the quotient by the given rate. To find Exact Interest. Rule. — Multiply the principal by the rate expressed as hun- dredths, and thai product by the time expressed in years of 365 days. ANNUAL INTEREST Annual Interest is interest payable annually. If not paid when due, annual interest draws simple interest. To find the Amount Due on a Note with Annual Interest, when the Interest has not been Paid Annually. Rule. — Find the interest on the principal for the entire time, and on each annual interest for the time it remained unpaid. The sum of the principal and all the interest is the amount due. COMPOUND INTEREST Oompound Interest is interest on the principal and on the un- paid interest, which is added to the principal at regular inter- nals. The interest may be compounded annually, semi-annually, quarterly, etc., according to agreement. To find Oompound Interest. Rule. — Find the amount of the given principal for the first periods Considering this as a ncio principal^ find the amount of DEFINITIONS, PRINCIPLES, AND RULES xix it for the next period, continuing in this manner for the given time. Find the difference between the last amount and the given principal, which will be the compound interest. PARTIAL PAYMENTS Partial Payments are part payments of a note or debt. Eacli payment is recorded on the back of the note or the written obligation. United States Rule. — Find the amount of the principal to the time when the payment or the sum of two or more payments equals or exceeds the interest. From this amount deduct the payment or sum of payments. Use the balance then due as a new principal, and proceed as before. Merchants' Rule. — Find the amount of an interest-bearing note at the time of settlement. Fi7id the amount of each credit from its time of payment to the time of settleftnent ; subtract their sum from the amount of the principal. BANK DISCOUNT Bank Discount is a percentage retained by a bank for advanc- ing money on a note before it is due. The Sum Discounted is the face of the note, or if interest-bear- ing, the amount of the note at maturity. The Term of Discount is the number of days from the day of discount to the day of maturity. The Bank Discount is the interest on the sum discounted for the term of discount. The Proceeds of a note is the sum discounted less the bank dis- count. Problems in bank discount are calculated as problems in interest. XX SUPPLEMENT In Bank Discount, The sum discounted is the priiicipal. The rate of discount is the rate of interest. The term of discount is the time. The bank discount is the proceeds. EXCHANGE Exchange is making payments at a distance by means of drafts or bills of exchange. Domestic Exchange is exchange between places in the same country. Foreign Exchange is exchange between different countries. Exchange is at par when a draft, or bill, sells for its face value ; at a premium when it sells for more than its face value ; at a discount when it sells for less. The cost of a sight draft is the face of the draft increased by the premium, or diminished by the discount. The cost of a time draft is the face of the draft increased by the premium, or diminished by the discount, and this result diminished by the bank discount. To find the Oost of a Draft. Rule. ^-jpmc? the cost of $1 of the draft; multiply this hy the face of the draft. To find the Pace of a Draft. Rule. — Divide the cost of the draft hy the cost q/" $ 1 of the draft. EQUATION OF PAYMENTS Equation of Payments is a method of ascertaining at what time several debts due at different times may be settled by a single payment. The Equated Time of payment is the time when the several debts may be equitably settled by one payment. The Term of Credit is the time the debt has to run before it becomes due. DEFINITIONS, PRINCIPLES, AND RULES xxi The Average Term of Credit is the time the debts due at different times have to run, before they may be equitably settled by one payment. To find the Equated Time of Payment when the Terms of Credit begin at the Same Date. Rule. — Multiply each debt by its term of credit, and divide the sum of the products by the sum of the debts. The quotient will be the average term of credit. Add the average term, of credit to the date of the debts, and the result will be the equated time of payment. To find the- Equated Time when the Terms of Credit begin at Different Dates. Rule. — Find the date at which each debt becomes due. Select the earliest date as a standard. Multiply each debt by the number of days between the standard date and the date whe7i the debt becomes due, and divide the sum of the products by the sum of the debts. The quotient will be the average term of credit from the standard date. Add the average term of credit to the standard date, and the result will be the equated time of payment, RATIO Batio is the relation one number bears to another of the same kind. The Terms of the ratio are the numbers compared. The Antecedent is the first term. The Consequent is the second term. The antecedent and consequent form a couplet. Principles. — See Fractions. PROPORTION A Proportion is formed by two equal ratios. The Extremes of a proportion are the first and last terms. The Means of a proportion are the second and third terms. XXU SUPPLEMENT Principles. — The ^product of the means is equal to the prod- uct of the extremes. Either mean equals the product of the extremes divided by the other m,ean. Either extreme equals the product of the means divided hy the other extreme. Rule for Proportion. — Represent the required term by x. Arrange the terms so that the required tei^m and the similar known term may form one couplet, the remaining tet-ms the other. If the required term is in the extremes, divide the product of the means by the given extrefme. If the required term is in the mean^, divide the product of the extremes by the given mean. PARTNERSHIP Partnership is an association of two or more persons for busi- ness purposes. The Partners are the persons associated. The Capital is that which is invested in the business. The Assets are the partnership property. The Liabilities are the partnership debts. To find the Profit, or Loss, of Each Partner when the Capital of Each is Employed for the Same Period of Time. Rule. — Find the part of the entire profit, or bss, that each partner's capital is of the entire capital. To find the Profit, or Loss, of Each Partner when the Capital of Each is Employed for Different Periods of Time. Rule. — Find each partner's capital for one month, by multi- plying the amount he invests by the number of months it is employed; then find the part of the entire profit, or loss, that each partner's capital for one month is of the entire capital for one month. DEFINITIONS, PEINCIPLES, AND RULES Xxiii INVOLUTION A Power of a number is the product obtained by using that number a certain number of times as a factor. The First Power of a number is the number itself. The Second Power of a number, or the Square, is the product of a number taken twice as a factor. The Third Power of a number, or the Cube, is the product of a number taken three times as a factor. An Exponent is a small figure written a little to the right of the upper part of a number to indicate the power. Involution is finding any power of a number. To find the Power of a Number. Rule. — Take the number as a factor- as Triany times as there are units in the exponent. EVOLUTION A Root is one of the equal factors of a number. The Square Eoot of a number is one of its two equal factors. The Cube Eoot of a number is one of its three equal factors. Evolution is finding any root of a number. Evolution may be indicated in two ways : by the Radical Sign, V~, or by sl fractional exponent. The Index of a root is a small figure placed a little to the left of the upper part of the radical sign, to indicate what root is to be found. In expressing square root, the index is omitted. In the fractional exponent, the numerator indicates the power to which the number is to be raised ; the denominator indicates the root to be taken of the number thus raised. To find the Square Root of a Number. Rule. — Point off in periods of two figures, commencing at units. Find the greatest square in the first period and place the root in the quotient. Subtract this square from the first period, and bring down the next period. XXiV SUPPLEMENT Multiply the quotient figure hy two, and use it as a trial divisor. Place the second figure in the quotient, and annex it also to the trial divisor. Then multiply the figures in the tiial divisor by the second quotient figure, and subtract. Bring down the next period, and proceed as before until the square root is found. To find the Square Eoot of a Praction. Rule. — Reduce the fraction to its simplest form, and find the square root of each term separately. To find the Onbe Boot of a Number. Rule. — Point off in periods of three figures each, beginning at units. Find the greatest cube in the first period and place the root in the quotient. Subtract this cube from the first period, and bring down the next period. Multiply the squxire of the first quotient figure by three and annex two ciphers for a trial divisor. Place the second figure in the quotient. Then, to the trial divisor add three times the prod- uct of the first and second figures, also the square of the second. Multiply this sum by the second figure and subtract. Bring down the next period, and proceed as before until the cube root is found. To find the Oube Eoot of a Fraction. Rule. — Reduce the fraction to its simplest form, and find the cube root of each term separately. STOCKS AND BONDS. Capital Stock is the money or property employed by a corpora- tion in its business. A Share is one of the equal divisions of capital stock. The Stockholders are the owners of the capital stock. The Par Value of stock is the face value. The Market Value of stock is the sum for which it may be sold; DEFINITIONS, PRINCIPLES, AND RULES XXV Stock is at a premium when the market value is above the par value ; at a discount, when below par. Bonds are interest-bearing notes issued by a government or -a corporation. A Dividend is a percentage apportioned among the stockholders. A Stock Broker is a person who deals in stocks. Brokerage is a percentage allowed a stock broker for his services. In Stocks and Bonds, The par value is the base. The rate per cent premium, or discount, is the rate. The premium, ^ discount, or > is the percentage. dividend J f amount, or Terence. The market value is the < ,.^ {diffe NOTES, DRAFTS, AND CHECKS. A Promissory Note is a written promise to pay a specified sum on demand, or at a specified time. The Face of a note is the sum named in the note. The Maker is the person who signs it. The Payee is the person to whom the sum specified is to be paid. The Indorser is the person who signs his name on the back of the note, thus becoming liable for its payment in case of default of the maker. An Interest-bearing Note is one payable with interest. If the words " with interest " are omitted, interest cannot be collected until after maturity. A Demand Note is one payable when demand of payment is made. A Time Note is one payable at a specified time. A Joint Note is one signed by two or more persons who jointly promise to pay. XXVI SUPPLEMENT A Joint and Several Note is one signed by two or more persons who jointly and severally promise to pay. In a joint note, each person is liable for the whole amount, but they must all be sued together. In the joint and several note, each is liable for the whole amount, and may be sued separately. A Negotiable Note is one that may be transferred or sold. It contains the words " or bearer," or " or order." A Non-negotiable Note is one not payable to the bearer, nor to the payee's order. The Maturity of a note is the day on which it legally falls due. A Draft, or Bill of Exchange, is a written order directing the payment of a specified sum of money. The Face of a draft is the sum named in it. The Drawer is the person who signs the draft. The Drawee is the person ordered to pay the sum specified. The Payee is the person to whom the sum specified is to be paid. A Sight Draft is one payable when presented. A Time Draft is one payable at a specified time. An Acceptance of a time draft is an agreement by the drawee to pay the draft at maturity, which he signifies by writing across the face of the draft the word " accepted " with the date and his name. A Check is an order on a bank or banker to pay a specified sum of money. ANSWERS. — Part III. Page 448. 2. 20 and 80. 3. $2000; $4000; 1 12,000. 4. 18 girls; 36 boys. 5. 13 and 65. 6. 13. 9. 10. 11. 12. 13. 14. 15. 16. 17. Page 449. . 11. i. $3000; $6000; $ 18,000. 12 and 60. 9 marbles ; 18 marbles ; 27 marbles. 36 years ; 6 years. 8. 1; 4; 12; 24. 30; 15; 135. 9 pounds. 19 rods. 85 feet. Page 450. 18. Son, $ 40 ; daughter, $80. 19. 25 days. 20. Girl, $80; boy, $ 40. 21. Father, 30 da.; son, 15 da. 22. 3 dimes, 6 nickels, 18 cents. 23. 25 yards. 24. 25 rods; 100 rods. 25. Speller, 15 f.-, reader, 45 f. 26. 60 and 12. 27. 18 nuts ; 9 nuts ; 27 nuts. Page 452. 1. 24. 2. 24. 3. 42. 4. 84. 5. 24. 6. 70. 7. 72. 8. 40. 9. 360. 10. 160. 11. 18. 12. 18. 13. 8. 14. 16. 15. 12. 16. 20. 17. 900. 18. 60. 19. 60. 20. 32. Page 453. 21. 36. 22. 222. 23. 180. 24. 72. 25. 320. 26. 7. 1. 15 and 75 2. 28f; 71^-. 3. $816. 4. $180. 5. 89. 6. 100. 7. 40; 15. 8. f|. 9. %\. Page 454. 10. 60; 420. 11. 540; 18. 12. 9. 16. $300. 17. 64 marbles. 18. $2; $3; $10. 19. $4; $2. 20. 3 horses; 12 Page 455. 1. 19. 2. 22. 3. 47. 4. 14. 5. 9. 6. 10. 7. 6. 8. 33. Page 456. 9. 27. 10. 11. 12. 96. 13. 144. 14. 18. 15. 24. 16. 6. 17. 32. 18. 18. 13. 20 peaches; 5 19. 12. plums. 20. 20. 14. $200; $600; $700. 2. 15. 15. $60; $140. 3. 9. 2 ANSWERS. 4. 15 marbles; 33 i 7. $ 10.32. 41. 18,400. 4. Selling price, marbles. 8. $1255.80. 42. 40%. $831.25. 9. $9.16. 43. 133f 5. Selling price, Page 457. 10. $3.68. 44. 72. $ 1051.38. 5. 25 ft.; 100 ft. 11. $ 1.60. 45. 68. 6. 39 acres ; 47 12. $ 1.58. 46. 75% Page 465. acres. 13. $1.62^. 6. 3% 7. 1059 votes; 14. $ 326.80. Page 462. 7. 33^%. 1377 votes. 15. 3 centa. 47. 83.i% 8. 15% 8. 62 years. 16. $ 13.09. 48. Ux 9. 20% 9. 84; 12. 17. $19.98. 200 10. %- 13. Cows, $45; 21. $3.21f. 52. 99. 14. 12^% horses, 22. $11.32. 53. $218.75. 15. H>r/o- $125. 23. $63.04. 54. $500. 16. Cost, $375. 14. 3 dimes; 14 24. $ 19.95. 55. 1. 17. Cost, $92.30. half dimes. 25. $550. 56. r>oo% 18. Cost, $1234.56. 15. 74 and 26. 26. $44.40. 57. 32%. 19. Cost, $ 240. 16. 21 boys; 33 27. $323.40. 58. $183.20. 20. Cost, $63.75. girls. 28. $4.65. 59. 1100. 21. $55. 60. 738. 22. 25% Page 458. Page 461. 61. 10,800. 23. $800. 17. $3600; $6000; 30. 13.T 62. $1547. $ 8400. 20 ' 63. $ 764.80. Page 466. 18. 44; 11. 31. 40%. 24. 50% 19. 5 five-cent 32. X Page 463. 25. 14?% stamps ; 20 two- 4" 64. $1120. 26. 12^/0- cent stamps ; 35 33. 168. 66. $170. 27. 37^ cents. postal cards. 34. 6.r 66. 3500 bn. 28. $ 40. 20. 8 horses; 25 5' 67. $1440. 29. $219. cows ; 55 sheep. 35. 110. 68. $592. 30. $200. 36. 2x 31. 371% Page 460. 3 ■ Page 464. 32. 2-^% 1. $6.34. 37. 117. 1. Selling price. 33. $1646. 2. $4.50. 38. X $2157.40. 34. $4053. 3. $9.60. 150 2. Selling price. 35. $113.75+ 4. $1.55. 39. 500 o/„. $ 29.40. 5. $4,158. 40. .T 3. Selling price. Page 467- 6. $415.80. 800 $1181.25. 1. 3 feet. I ANSWERS. r 2. 18 feet. 19. 331%. 6. $1.16-. 4. m- 3. I acre. 20. 12%. 7. $ 19.55. 5. 16o/„. 4. 1430 yards. 21. $ 1.92. 8. $ 2.94. 6. $7.35. 5. 8 strips. 22. $ 24.24. 9. $1,111 7 '. M- 23. $39.20. 10. $67,721 8 •. 33|- times. Page 468. 24. $1.57. 11. $49.77+. 9 '. ^ocent. 6. 3600 sheets. 12. $ 235.751. 7. 250 ft. 1500 Page 470. Page 475. sq. ft. 25. $ 18.33. Page 472. 10. 1.15; .000625 8. 250 boards. 26. 11^2^% 13. $ 18.88. .0040625 ; 9. $710. 27. $200. 14. $7.80. ^^0 ; tAcj- ; 10. 6400 cakes. 28. $ 893.20. 15. $18.74-. sMo- 11. 76,800 cu. ft. ; 29. $1. 16. 1 1050.801 11. ^ %• 2208 tons. 30. 2%. 17. $ 1035.731. 13. 15 times. 12. 160 feet. 31. $18. 18. $33.73.-. 14. 1 530. 13. 281,600 sq. 32. $ 626.05. 19. $49.04-. 15. .0223125 mile. ft.; 240,- 33. $1.57. 20. $ 154.871 16. 2.432013984. 000 sq. ft. 34. $68.18. 21. $884.53+. 17. 3|%. 14. $27731 35. ^m%- 22. $6191.20. 18. $ 2.64-. 15. 96 lots. 36. 14!%. 23. $2841.81-. 19. $ 3999.24. 16. 164sq. yd. 37. mi %• 24. $2344.50+. 20. 162 days. 38. $ 27.20. 25. $161.00+. 21. 16f%. Page 469. 39. 15 cents. 26. $835.31-. 22. $5.69+. 1. 111%. 40. $6300. 27. $886.17-. 23. $753. 2. 10%. 41. $871.83. 28. $411.65+. 24. $11.28-. 3. 111%. 42. $309.14. 25. T%-,j%\- 4. $26.40. 43. 8%- Page 473. 5. $2.80. 44. 7A%- 29. $1550.21-. ] Page 476. 6. $1,221. 45. $21. 30. $118.14-. 1. 3613 flags; 7. 450/0. 46. $677.25. l^- yards re- 8. 1350%. 47. $7692. Page 474. maining. 9. 41%. 48. $2,36. 18. 63 cents. 2. 200 men. 10. A%. 49. n%- 19. $5.70. 3. 317 acres ; 11. 25%. 50. $21.55. 20. $4.20. $ 14,325.591 ; 12. 3310/0. 21. $ 1.50. $45.19JAV 13. 16|o/,. Page 471. 22. $2.75. 4. $288.75. 14. $75. 1. $112.50. 23. $7.20. 5. 7,002,079.- 15. 21f|o/o. 2. 90f cents. 003129. 16. $35. 3. 85f cents. 1. f- 6. 482.9638599. 17. $71.34|. 4. $19.52-. 2. 71 feet. 7. 2.0635+. 18. $19.20. 5. $ 1.65. 3. 10mi.249| rd. 8. $10.38-. ANSWERS. Page 477. 12. Wt- 22. 58,975. 7. 106| yards. 9. 5or,%; 13. m- 23. , 899,100. 8. 86f bundles. i^nr/o; 14. n- 24, , 426,000. 9. 810 sq. yd. 30jV%. 25, , 16,800. 10. 114 rods. 10. $.00703+. Page 480. 26. 1,172,880. $.04648- ; 1. ^jhh- 27 . 4290. Page 488. $.0055 ; 2. 82. 28, . 71,400. 11. 300 panes. $.00675 ; 3. $1.26J. 29 . 67,716. 12. 20,736 gal. $.00475 ; 4. 113. 30 . 293,249|. 13. 72 cords. $.0225 ; 5. 531%; 14. 13,440 bu. $.005359+. 46f%. Page 485. 15. 7000 cu. yd. 12. 8272.08512. 6. $1.60. " All other arti- 16. 2250 cu. yd. 13. 8.522. 14. 3.7857+. 7. 8. 192 planks 12^%; 1 I '$73,327,274; 122,469 ; . cies, $72,1 1. $313.31-. 15. 30,000 men. 1U%- Increase, $26,- 2. $136.50. 16. i 9. $17.76+. 976,455. 3. $500. 17. 15 miles. 10. $6. 4. 8%- 11. 266|%. Page 486. 6. $410.16-. Page 478. 1. ^^^tmi 18. 170 bushels. Page 484. 2. 6l26|Un- Page 489. 19. 1500 letters; 1. 165. 3. ll,202|tHi. 7. 29.623- feet. 750 letters. 2. 252. 4. 11,6993|0|;}- 8. $30.87i 20. $2630.20. 3. 4048. 5. 786?i|i2|. 9. 12} tons. 4, 910. a 49,167[|ff|. 10 *? 1. 1,163,117,- 5. 2594|. 7. X\J. o. 683.002129. 6. 5646|. 8. 963UIUI. 1. $.41+; 2. 69,092.- 7. 1977f 9. ™mmi (|.37i). 80236843. 8. 6373J. 10. 6462i|HH- 2. $.55+; 3. 176.303- ; 9. 5067tV- ($.46-). 2.0247-. 10. 8852}. 1. 630 boards ; 3. $2.89+; 4. $66.45-. 11. 46,018. 140 posts. ($2.75+). 12. 79,520. 2. 10 feet. 4. $.37-; Page 479. 13. 25,554|. 3. $1.50. ($.32). 6. $42.96-. 14. 106,908. 5. $6.89-; 6. X 3, etc. 15. 65,471. Page 487. ($6.72). 18G.C.D. 16. 65,635^. 4. 4840 sq. yd. ; 7. 25%. 17. 86,855. about 70 Page 490. 8. m% 18. 27,702^. yards. 6. $173.59^ 9. m 19. 37,411. 5. 400 rods. ($173.70). 10. 88VW t'-ents. 20. 26.969. 6. 18 sq.ft.; 270 7. $36.23+ 11. 7.625. 21. 41,382. cu. ft. (1 36.26+). ANSWERS. 8. 10. 11. 12. 13. 14. 15. 16. 17. $791.42+; ($791.89-). $ 176.741; ($ 176.85). $ 985.411 ; (.f 986). 13.10; (!?2.80). $2.60; ($ 2.48). $5.64; ($5.46). $ 1.74- ; ($ 1.69+). $ 5.38- ; ($5.29+). $2.52; ($ 2.40). $5.58; ($5.40). 18 Page 491. Proceeds, $87.34-; ($87.38+). 19. Proceeds, $122.66+; ($122.75-). Proceeds, $ 502.05- ; ($ 502.34+). Proceeds, $71.65-; ($ 71.68+). Proceeds, $ 232.99- ; ($233.10+). Proceeds, $95.58+; ($95.63-). 20. 21. 22. 23 24. 25. Proceeds, $162.65+; ($162.76+). Proceeds, $81.91+; ($81.95+). Page 495. .01020201 10.01. $ 24.28. $10.15+. 1 2.52 f. $ 367.42- $ 10.99+. $2.13. $11.23+. $12.78+. Page 496. 2. 3. 4. Page 492. 674.7+ yards. $613.98+. Page 493. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 371|. 5301. 129rV- 1272f. 517^ 289^. 233||. 260ff. 73413. 10781. 85^. 89f. 264if. 361 If. 337H. 674H. 401,V 421H- 434i. 5. 6. 7. 8. 9. 10. 11. 12. $ 4.90. $115.54-. $5.37.1; $224 07-. $728; $950f. $ 25,000. 20%. Tlie lirst ; 5 % more. 1| acres. .0006216. $1200; $ 4608 ; 25%. Page 497. 13. $55.91-. 14. 43V; 6 If days ; $85,3331. 15. $173,668; $201.880618. 17. $1453.76. 18. 4.23+ times ; 64.68+ in- habitants ; 6.07- inhabi- tants ; $ 15,124,032. 19. 300 sheep. 20. $8572.20. Page 498. 21. 257sq. yd. 6i%; $287.50. 52 weeks. 33,750 qt.; 2250 qt. average. $153.75. $ 33,519.20. $85. 70 feet. $ 187.36. 9 J^ 0/ • «^%; 100%. 25. 26. 27. 28. 29. 30. Page 499. $ 46.80. $ 18.40. 17 lb. 11 oz. 5 pwt. 19 5. 6. 7. 8. 9. 10. 11. 12. 13. Page 500. I. 23,220 gr. gold; 2322 gr. silver ; 258 gr. copper. 24 spoons. $ 2800. iff;i09fX $ 105. 27+ cents. 23t\ gal. 704 sq. ft. If cu. ft. 1317U lb. 6 ANSWERS. 14. 13,405^ 2. $175.94-; 17. $45; $135; 19. $870.48. gal-; ($175.94-). $270; 20. A, $50; B, 1440 bu. 3. $ 350.55+ ; $450. $90; C, 15. 180 o/o. ($ 350.55+). 18. $237; $189; $110. 16. ^2}%. $114. 21. $380. 17. $309.42. 19. 7 days. 22. $ 158.40. Page 504. 20. 240 eggs. Page 501. 4. $846.26-; 21. 5 months. 18. 300 pounds. ($845.77+). 22. 24 weeks. Page 509. 19. A, 43^%; 5. $724.85+; 1. $554.23. B. 36o/o; ($724.69-). Page 506. 2. $171.20. 0,20^0/^ 23. $37,033,131. 3. $1782.67i 20. 672 yards. 1. 38 rd. 3 yd. 24. 10 inches. 4. $158.40. 21. $315. 11 in. 25. $360. 5. $ 28.50. o 113 rd. 3 yd. 1 ft. 6 in. a $60.56^. $50.85. Page 502. /i. 1. 125,422,928,- o. 7. 1; 101.5901+. 3. 175 rd. 1yd. 368.01. 8. $13. 2. n\. 1 ft. 6 in. 6. 12 T. 6 cwt. 9. $7.45. 3. 72. 4. £16 13 s. 2 qr. 13 lb. 4. 390. 4d. Page 510. 5. 3 yards. 5. 21,090d. Page 507. 10. $27.84-. 6. $4. 6. £ 26 5 s. 7. 300. 11. 40 and 10%; 7. 20|o/,. 7. $27.37+. 8. 11 f days. $2 diflfer- 8. $81.25. 8. £108 4 s. 9. 3416^1 lb. ence. 6d. £45 188. 8d. 10. £3068 158. 10 d. 12 $60; 40 % discount; 1. llf rolls. 9. X/6. 2. 1.5548—. 10. 773t?^ oz. 60% net. 3. 5333^ bu. 11. $175. 4. 302. 13. 52%. 4. $3.02^f. 6. ll^V 14. $100. Page 505. 7. 9|. 15. 72% Page 503. 12. $123. 9. 1250. 16. 72%. 5. $60.62^. 13. $72; 6. 70 cents. $119.25; Page 508. 1. $81. 7. IW^cta. $92.25. 10. lOf^f. 2. 75%.. 8. 17 spoons. 14. $450; 11. 37^ pieces. 3. 1^%. 9. 48/jbu. $750; 12. 43,200 min. 4. 627.5. $600. 13. 24f inches. 5. .0075 ; ^. 1. $49.01-; 15. 24 days. 14. 362.16 mi. 6. 77r/o- ($48.99-); 16. 2| hours ; 15. $6.72-. 7. $ 2.63f $48.88-; 48 miles 17. $4761.90+. 8. $24. ($48.88-). from A. 18. $8.28J. 9. 600. ANSWERS. 7 Page 511. 25. $4.16f. ^^ 1969 a; . 5. $1.43-. 10. $3575. 26. 80 hours. * 2000 ' 6. $36. 11. 6|%. 27. 250cu. ft. (]^1^\ 7. 1 yr. 6 mo. 12. $65. 28. $7.12^. V 200 )' 8. $42.17+. 13. $101.50. 29. $18. 9. $5000. 14. $375. 30. $77.95, Page 517. 10. 6%. 15. $10.96-. ($79). 1. 121icu. ft. 11. 1 yr. 9 m 16. $23.14-. 2. $162. 2 da. J 9x 12. $144. Page 512. * 80 ' 13. $83.26i. 1. $ 19. 2. $ 1600. Page 518. 14. 3 %. 2. 35. 3. 255 a;. 3. $126. 15. 2 yr. 1 mo. 3. 2624f yards. 4. 3%. 4. $85.50. 7 da. 4. $27.20 gain. 5. $2000. 5. 16.8 tons. 16. $80. 5. $20. 6. 2 pieces, 17. $2181.99-. 6. 1767.5. 12x12; 18. $72. 7. $5.49-. Page 515. 2 pieces, 8. 35|%. 6. 5 years. 12x14; Page 521. 9. $1431.27. 7. 150a;. 2 pieces, 19. 5 mo. 23 da. 8. 6%. 14x14. 20. 6%. 9 1875 4-^2^ '^' 44iflb. 21. $16.92. Page 513. * 32 8. 5280 lb. 22. $ 402.22. 10. $30. 10. 682.90+ 9. Outside di- 23. 37 days. 11. 4/y. 6829 a; mensions, 24. 5%. 12. .00007865. 3200 ' 14x14x14; 25. 11 mo. 29 da. 13. 108.86- bu. jj 21^ . 2744 cii. m. 14. 2bu.lpk.4qt. * 2000' wood and Page 522. 15. $7425. /_x_\ marble; 1. 570,073,438,- 16. 25%. \100J' 1728 cu. in. 098.53. 17. $49.31+. j2 11a;. fx\ marble ^^(f) 18. $594.50; ' 30 ' V3/ 1016 cu. in. Page 523. ($594.80). j3 x+3_ f x\ wood. 6. .37875. 19. 25%;32j%; ■ 20 ' Uoj' 10. 8 times ; 7. $70.20. 42f%. ^4 ^.(ll] iton; 8. $281.25. 20. $1.56|. * 60 'V 3/ 6|tons. 9. $15,000. 21. Loss, $177. ^g 5997-x . 10. $67.71-. 22. $3.60. * 10 ' Page 520. 23. .Olf. (qoO-^\ ^' •'''^'- Page 524. V 10/ 2. 2 years. 2. i? 41.99+. Page 514. ,g 477x 3. $96. 3. $951.13+. 24. 15,203. ' 400 ■ 4. S%. 4. $112.74-. 8 ANSWERS. 5. $119.43-. 34. I6.55+. 7. 4269.22+ 11. |853.27+; 6. $13.91-. 35. $278.16-. francs. ($853.71-). 36. $1.23-. 8. $1563.55-. 12. $56.62^; Page 525. 37. $196.64+. 9. $1563.55-. ($57.50). 7. $147.19-. 38. $1.10. 10. $1547.37. 8. $8.10-. 39. $389.60. ^^ 18^ Page 536. 9. $52.33-. 40. $6.11+. '25' 3. 2^^. 10. $1005.50. 41. $4.09+. 4. 375. 12. $967.78. 42. $56.32-. 5. $288. 13. $21.79-. 43. $16.72-. Page 531. g. $922.20. 14. $16.53-. 44. $95.43-. 12. 400-4a;. 7. $573.47^ 15. $1274.21-. 45. $15.71+. 13. $500. 8. $1700. 46. $594.30; 14. 10%. 9. $1400; Page 526. ($594.60). 16. Same. $1200. 16. ($800.50-). 47. $10.73+; 10. $1337. 17. $24.74-; ($10.38+). 1. 688,965,549,- ($23.94). 48. $793.73+; 176.65. Page 539. 18. $761.06-; ($794.13+). 1. $165.37^. ($761.45). Page 532. 2. $1453.42-. 19. $43.99-. Page 528. 6. .0019. 3. $50,625.50. 20. $786.39-. 49. 45 cents ; 7. 32 cents. 21. $65.40. (nothing). 8. $46 gain. Page 540. 22. $625.03+. 50. $968.83+; 9. 60 cents. 4. $893,615,929. 23. $98.49+. ($970). 10. $16.98. 7. 284,106,409,- 24. $993.27+. 352.02. 25. $61.68-. 1. 470,952. Page 535. 8. 45.29-%. 26. $252.37+; 1. $106.33+; 9. $223,852.8835. ($252.52-). Page 529. ($106.38)-. 10. 107§ cents. 27. $13.09-; 5. 8614.20375. 2. 7%; ($12,871). 6. 2f {ri^%). Page 541. 28. $486.10-; 7. $110.85+ 3. 24 days. 11. $20,000. ($486.43-). gain. 4. $1200. 12. $728.17^. 29. $2.33+; 8. $16.25. 5. $4.68; 15. Profits, ($1.94+). 9. 45T. 5cwt. ($3.90). $4414.10. 30. $989.67-. 2qr. 6. 5%. 16. |; |; t^; f ($990). 10. $43.33^. 7. 72 days. 31. $1938.43-. 8. $1200. Page 542, 32. $8.06+. Page 530. ($1260). 17. $4752. 4. $196.17. 9. $304.26-. 18. 37'.%; 25%; Page 527. 5. $600.01+. ($304.05+). 12^%; 25%. 33. $1473.52-. 6. $1506.12. 10. Apr. 8, 1894. 19. |81.90. ANSWERS. 9 20. $299.88-. 38. 24,130f. 8. $3. 9. AiiV 39. 651, 329 j\. 9. £73 3s. 10. Largest, f ; 1. 1,287,400. 40. 1,932,560. 10. 6f days. smallest. 2. 3,370,185. 41. $277,133.11. foff. 3. 598,969. 42. $60,887.10. 2. $58.93-. 4. 2,883,736. 43. $48,554.08. 3. 6%- Page 550. 5. 816,669. 4. $3400; 11. 2^. 6. 5,127,460. Page 544. $3570. 12. 2 da. 15 hr. 7. 2,455,038. 44. £7 5s. 9d. 50min. 35 8. 42,327,198. 45. 11 yd. 1 ft. Page 547. sec. 9. 2,513,420. 11 in. 5. $1.30-. 14. iViV 10. 22,944,747. 46. 12 bu. 2 pk. 6. $ 10,000. 15. ItV 5qt. 8. $264.25. 16. f^- Page 543. 47. 11.76+. 9. $ 135.40-. 17. 180. 11. 857,712. 48. 13.72+. 18. 3f. 12. 6,482,112. 49. 14.41+. Page 548. 19. t¥^. 13. 1,230,828. 50. 13.34+. 1. $107.65+. 20. 36f. 14. 921,776. 51. 19.05-. 2. 6 yr. 6 mo. 21. ItV 15. 3,460,704. 52. 22.30-. 3. n%- 22. 3hr.22min. 16. 5,888,304. 4. $750. 23. 85i rods. 17. 1,460,025. 1. 79.98 %. 5. $882. 24. iM- 18. 10,563,960. 2. 5.02%. 6. $97. 25. 15f. 19. 3,911,322. 3. 4.28%. 7. $6; $144. 20. 2,982,840. 4. 3.83 %. 8. 33r/o- 1. 4| feet. 21. 714,186. 5. 3.81%. 9. $ 8960. 2. 120 yards. 22. 3,277,719. 6. 2.44%. 10. $360. 23. 456,375. 7. .54%. Page 551. 24. 174,600. 8. •10%. Page 549. 3. 8750 sq. yd. 25. 362,250. Total, $872,- 11. $50. 4. 3000+ 30 ,t; 26. 104,787i 270,283. 12. 20. 80 yards. 27. 128,550. 13. $61.87|. 5. 100 a; ; 40 28. 625,975. Page 545. 14. $1150. yards. 29. 213,966f. 1. 237.49%. 15. ^%- 6. 60 a; +1200; 30. 413,866f. 2. 234.60 %. 80yd. and 31. 86501 3. 563.36 %. 1. 4ff *120 yd. 32. 10,757f. 3. tV- 7. 10,000 sq. 33. 21,873^. Page 546. 4. 1611. yd. 34. 24,292i. 4. 0.04%. 5. m- 8. 1920 flag- 35. 485,072. 5. 25%. 6. m stones. 36. 167,276. 6. 20%. 7. 73; 240. 9. $9.26-. 37. 24,418^. 7. $24; $30. 8. m- 10. 46 1 yards. 10 ANSWERS. 11. 90^1 lb. 12. 35/^ sq. ft. Page 552. 13. 10 ft. 8 in. 14. 950 bushels. 15. 810 gallons. 16. $416. 17. 10 miles. 18. 615 cords. 19. Bj-mib. 20. 27/^ lb. 21. 70G|sq. j^d. 22. 722f8q. yd. Page 553. 1. 24H yards. 2. 28 cents. Page 554. 3. 84 days. 4. 9 days. 5. $49. 6. 180 bushels. Page 555. 8. (i() husliels. 9. 720 pounds. 10. 4G00 sheets. 11. $312. 12. 8 da. 4 hr. 13. !? 2^.80. 14. 9() rods. 15. 50 days. 16. $45. 17. 120 men. Page 556. 18. 360 men. 19. 20. 2. 3. 4. 5. 6. 7. 10 hours. $1.20. 329. 10 bushels. 12 feet. 90 cents. Page 557. 8. 1680. 85,800. Loss, $93.75. $2.35+. 99.99. 14 feet. $69.05-. 1050 acres. 9792f lb. $ 275.69-. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. Page 558. 20. $873; $184; $143. Estate, $ 45,000 ; son's share, $ 10,000. $10.86+. 6^ days. 21. 22. 23. 24. $78; $104. $36; $17 $48. $25; $20. $16.20; $ 13.80. 5. 45 lb. gold ; 4^ lb. silver; I lb. copper. Page 559. 6. ISaltpeter, 54 lb. ; sulphur, 7ilb; charcoal, 10^ lb. $ 20.25 ; $ 18.63 ; $15.12. 864 bales 540 bales ; 396 bales. $12; $20; $60; $72. 7. 8. 20. 5934.47- meters. 21. 6 quarts. 22. .8125 pound 23. .25 rod. 24. 100 links. 25. .1 acre. Page 561. £13138. 8|-d. 368.90 marks ; $30.73-. Page 562. 1. $40. 2. 8o/o. 3. $150; 150% 2. 326.9843. 3. 6408. 5. 4.2633; 1.405712. Page 560. 6. 16.5393. 9. 28.165; 305.36721612. 10. £.3375. 12. Sum, 8.08690625. 14. .019104141. 15. .05; .03965. 16. 1093.3524 17. 2.8. 18. .215625 m. 19. 12 da. 20 hr. 31 min. 12 sec. Page 563. 4. $30. ^%- $435. Same. 5. 6. 7. 8. 9. 10. 11. 12. f) 0/ . •) 1 0/ '-' /o • -2 /o- $2267.26+. $2.85. Page 564. 17. .f 409.37 i. 18. $24,000*"; 19. 20. 21. 22. u%. $ 18,000. 101 rd. 2 yd. 1 ft. 6 in. 35 shares ; $87.60. ANSWERS. 11 23. 37 rd. 4 yd. Page 570. 26. $ 57,600. 2. $1874.771. 11 in. 1. $45. 27. $2,371 3. $2461.76+. 24. $106.12. 2. 10| acres. 28. $387.36. 4. $ 1000.50+. 3. $452,321 5. $ 1632. 4. $ 4.50. Page 573. Page 565. 5. $1.50; 29. $1000; 25. 229 rd. 4 $5.40; 100%. Page 577. yd. 2 ft. 20%; 30. $ 5487. 6. $ 1845.90+. 6 in. 6. $ 6000. 7. $ 946.04-. 27. $428.40-; 7. $3333^. 2. 688,450. 8. $ 632.65-. $53.40-. 3. 3f f ; 18|f . 9. $985.22+. 28. $119.10+. 4. .140625; 10. $326.34. 29. $149.14+. Page 571. 88.088. 8. $212.51-; 5. 5.820068 ; 1. 9 hr. 45 min. $439.79-. 1000. 2. 14| minutes Page 566. 9. $ 5250 ; past 5 P.M. 30. 51 cents. 18%. Page 574. 3. 4 ft. 6 in. 10 $ 402.50 ; 35%. B 400 yards. $2f. 1. 15%. xv. 7. Page 578. 2. $400. 11. $ 3330 ; 8. $ 10,500. 4. 748 plants ; 3. $102.50. $4.25. 9. 7400 inhab- 754 plants. 4. $2.70. 12. $9.30. itants. 5. 20 da. 6 hr. 5. $40. 13. $201.60; 10. $ 10,500. 40 min. 6. $1.60. $125.93-; 6. 5544 revo- 7. 1 hour. $ 722.40+. Page 575. lutions ; 8. $20. 14. 2| pounds. 1. (a) 1575.- 1320 revo- 9. $12. 15. 98t¥3 lb. 355671 ; lutions. 16. $71.25. (6) .028376- 7. 28 mi. Page 567. 604. 130if rd. 10. $20 loss. Page 572. 2. 49,999.- 8. 1388| mi. 17 $85. $ 594.20 ; 74000 Q 15708 f^n^ 4. 1.. X 1 . 18. 3. (a) 73 ; J7. 10. 694f miles. 5. 962 feet. ($594.50). (b) 2016. 11. 3211i| mi. 6. $6.50 gain. 19. 20% 4. $72. 7. $3.54. 20. $11,356,011 5. $53.95+. Page 579. 21. 700. 6. $160; $140; 12. 1104f| mi. Page 568. 22. $38.59+. $240. 8. $509.25. 23. 12. 7. $262.50. 1. $828.45. 9. 19 T. 62 lb. 24. $ 1006.13^. ($828.87). 8 oz. 25. 63 sq. ft. ; Page 576. 2. $397.30; 10. 24 cents. lO^f rods. 1. $3481.07+. ($397.50). 12 ANSWERS. 3. f 554.40; ($554.68). Page 580. 4. $625.33+. g 3959 a; . 4000 ' / 3961a; \ V4000 j' 6. 7. Il979-2a; . 10 / 11985-2a; ^ V 10 > 7956 + 8 a; (1592. 1). 8. $1200; ($1199.39+). 9. 30 days; (33 days). 10. $1.50 discount per $1000. ($2 discount per $ 1000). Page 581. 1. 56°. 2. 2 hr. 29 min. 12 sec. 3. 5 hr. 50 min. 20 sec. ; 7 A.M. 4. 37° 30^. 5. 1 hr. 14 min. 52 sec. Page 582. 6. 33° 30^ east longitude. 7. 39^gf^ miles per hour. 8. 9. 10. 11. 12. 13. 14. 1800 lb. $ 1000.80. 8 yd. 6 in. 15 ft. 7 in. 257bu.2pk. 2 qt. 20 spoons. 16° 40'. Page 583. 1. $67.46; $ 168.65 ; $236.11. 2. 1 lb. 3 oz. 3. .93. 4. 1 yr. 4 mo. 26 da. nearly. 5. 16 days. 7. .V^. 20. /j; ^; 21. 1.299609375. 23. 109|ffeet. Page 585. 24. A, 4340 votes ; B, 5551 votes. 25. $133.32-. 13. 14. 15. 16. 17. 18. 19. C, 119|. 3 bu. 3 3qt. $7.87i. pk. 1. $880.86+. 2. $1229.01-. Page 586. Page 584. 8. $630.76-. 9. 46f inches. 10. .0005207. 11. yW acre. 12. 30 min. 50 sec. $225.67-. Dec. (10) 13, 1888. Dec. 29, 1892. A, 1071 ; Page 587. 10. $908.S7-. Page 589. 1. $224.46+. 2. $261.19-. Page 590. 4. $772.37-. 5. $899.91+. 3. 12. 4. 2H. Page 591. $150; $7500. 15,544,041.4.= francs. 1. $3640. 2. $4; 6|o/,. Page 592. 6. $ 2 loss. 7. $701.53-. 8. $13.65. 10. $200.02^. $ 193.70+. $446.33-. 4188.48+ marks. 8490.44- francs. £ 307 7 s. 6^ d. nearly. $2050.72+. $4138.97+. 1. $137.61+. Page 593. 2. $12.39-. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. $8500. $282.25. 23 min. 53^ sec. $ 17.82. 1 1 feet. 12 rods. $56. l\ cent«i. $300. 3600 yards. Page 594. 11. $497; ($497.25). 5%. IS.f'oCt $700. $375.10-. 12. 14. 15. ANSWERS. 13 11,181,021.50. $1,037,124.44. 1214,854.74. £ 14 7 s. 2 d. 25 bu. 1 pk. 4qt. 23 rd. 2 yd. 1 ft. 6 in. Page 595. 7. $249,981.53. 8. $318,808.78. 9. $202,722.44. Page 597. $11. $85.25. $95.48-. 24 cents. $ 18,228. 150 sq. ft. $ 39,700. 138 feet. Page 598. 10. $48.43|. 11. 44%. 12. $33.12^. 13. 114sq. yd. 14. $594. 15. $2090.25. 16. if. 17. 26|o/,. 18. $363. 19. 64001b.; $11.40. 20. $870.62-; ($ 870.14-). 21. $4.31^. 22. 45315 ^. 23. $150; $225. Page 599. 24. $33.60; 15 A cords. 'T^ 25. 26. 27. "2 > ^10* 10%. $ 653.48- ; ($654.36). 28. $2688|. 29. $3682.19-; ($3684.371). 30. $4940.28." Page 600. 3. 135.62. 4 19 3 5. 13 feet. 6. 16T. 19cwt. 3 qr. 101 lb. 7. $1235.21-. 8. 881. 9. 25 cents. 10. $16.48-. Page 601. 1. 1^ yards. 2. 18.72 feet. 3. $78.75. 4. 8 ft. 4 in. 5. $14,910.75. 6. 108i| sq.ft. 7. 44.17875 sq. ft. 8. 2970 bu. 9. 252 gallons. 10. 146 sq. yd. 11. 3hr. 12min. 4hr. 6f min. 1 hr. 48 min. Page 602. 13. 40^ yards; $44,021. 14. 11 ft.; 108 sq. ft. 15. 984 sq. ft.; 71ft. 16. $11,200. 17. $56. Page 603. 1. .00007722. 2. 1.485135. 3. .450522. 4. 7.70904. 5. .0712111. 6. .0048393. 7. .63672. 8. .0374715. 9. .8220672. 10. .00004768. 1. .68515625. 2. 589.84. 3. 153.6. 4. 3265. 5. 50. 6. .064. 7. .00002375. 8. .0115. 9. 79,000. 10. 219.32. 3. 4.120275. Page 604. 4. li^^. 5. 9. 10. 7 fur. 16 rd. 3 yd. 1 ft. 9.888 in. $4.13. $ 1900. 3T. 7cwt, 2 qr. 17 lb. 4^ oz. $ 15,000. $6.98-. Page 605. 2. 1,345,595. Page 606. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 80 miles. .49184. $9.27; $12.36. 468 bricks. $46.08. 55 feet. $5700. $1700. $ 77.14. $ 44.48. $215.36+. $ 747.44-. Page 609. 1. 14. 2. 16. 3. 18. 4. 24. 5. 26. 6. 36. 7. 35. 8. 42. 9. 44. 10. 51. 14 1 ANSWERS. 11. 55. 2. 1 4330.80+. 21. $4.10; 7. 12:1:20 p.m. 12. 54. 3. 4 lb. tea, 27 (14). 8. 143° 3' West. 13. 61. lb. coffee. 22. $2750. 9. 82° 40' West. 14. 63. 4. |458.16|. 23. 90 da. 10. 11:33:12 15. 72 6. $8.25. 24. 18 da. A.M. 16. 75. 25. $475. 11. 3:37:20 p.m. 17. 83. Page 614. ($474.76-). 12. 88° 4' East. 18. 84. 6. Dec. (12) 15; 13. 32° 34' East. 19. 91. (27) 30 da.; Page 616. 14. 2:45:42 a.m. 20. 95. $1275.88+; ($1276.00+). 1. 567. 2. 915. 15. 23° West. 1. 7568. 7. $77.61-. 3. 144,200. 3. M- 2. 5107. 8. $332.50; 4. 25,758. 6. if. 3. 6008. $525. 5. 2,114,000. 7. if. 4. 6285^^111. 9. 18 men. 6. 86,526. 8. ff. 5. 6098. 10. 256 barrels. 7. 106,000. 9. n- 6. 3007. 8. 374,625. 10. If- 7. 98,640. 1. $2167.09-. 9. 19,800. 11. il 8. 75,064. 2. $543.16-. 10. 7312. 12. M- 9. 70,921. 3. $1335.23+. 11. 466,000. 13. 3h 10. 78,905. 4. $911.66-. 12. 50,133. 14. H- 13. 98,500. 15. If. Page 610. ] Page 615. 14. 4180. 16. 2^. 1. $183.27-. 6. $ 784.54-. 15. 423,000. 17. m- 2. £150 88. 4 d. 6. $ 724.03+. 16. 40,096. 18. 33^. 7. $1586.88-. 17. 419,904. 19. 12^ Page 612. 8. $1008.10+. 18. 310,050. 20. 16^. 1. iff- 9. $2607.48-. 19. 230^7. 21. lOJ. 2. I 10. $718.00-. 20. 227i 3. ,«tV 11. 4^%- Page 619. 4. m- 12. $600. 1. 12:6 p.m. 1. 666,862,394,- 5. Mi 13. 4 yr. 6 mo. 588.21. 6. f 9 da. Page 617 5. m- 14. 2yr.8mo.lda. 2. 9:36 a.m. 6. $1144.73-. Page 613. 15. $ 600. 3. 142° 55' 30'^ 7. $5166.69+; 7. mi 16. 4%. West. $283.31-. 8. Equal. 17. 63 days. 4. 52° 36' East. 9. ^^. 18. $400. 5. 5:16 a.m. Page 622. 10. ^. 19. $ 295.35. Table. ($295.50). Page 618. 1. 63.30 %. 1. 1389.61-. 20. 7%. 6. 8:24 a.m. 2. 13.14 o/„. 3. 8.58%. 7. M.1M O VV Page 629. 18. 50% 4. 3.83%. 8. h 6. 60 cents. 19. 4% 5. 3.41% 9. If- 7. 20% 20. $372. 6. 2.09%. 10. 9 32"- 8. $150. 21. Iff days. 7. 1.11%. 11. 1. 9. $120. 8. .72%. 12. Iff- 10. 20%. 1. nu- 9. 3.820/,. 13. 6. 2. 163 3T5- All others, 1. 254. 126,632,801. Page 626. 2. 27.1. Page 632. 2. $ 625.92. 3. 1.37. 3. 22T-k. Page 623. 3. 3968 revolu- 4. 26.8. 4. ^; 4 s. 6d. 1. 63,805,573.34. tions. 5. 3.76. 5. 13t\V 2. $ 869,109.89. 4. 27 kilo- 6. .838. 6. 8 f acre ; 3. 524,658,551,- meters. 7. 16.27. 218 lb. 12 oz. 760. 8. .4876. 7. A;H- 4. 986ff|f. Page 627. 9. .306. 8. 1 8- 7. (a) $1104.02+. 5. 770 meters ; 10. .069. 9. 42ff. (6) $1101.56+; 42 kilos. 10. 1. $1116.44+. 6. $ 504. Page 630. 11. 10. 7. 8| days. 1. $6.21. 12. £f; j\ da Page 624. 8. $82.50. 2. 4hr. 24 niin. 13. 8. 8. $9862.33+. 9. 14 yards. 3. $ 22.50. 14. 13 s. 3i}d. 9. f 10. $ 2.45. 4. 150 barrels. 15. t¥iV- 10. $ 72.06. 5. $153.60; 16. 3,3,. Page 628. $192; 17. 370. 1. 80. 21. 40 and 10%. $ 230.40. 18. $520. 2. 4. 22. 50 and 10%. 6. 1740 tiles. 19. $918.75. 3. 17i 23. 30 and 10%. 7. $600;000. 20. £1060 18 s 4. 28. 24. 30 and 5 %. 8. 4i%. 9d. 5. 5. 25. 30 and 15%. 9. 5.0004. 6. 7. 26. 50 and 10%. 10. 2880 gal. Page 633. 7. 36 cents. 27. 20 and 50%. 11. .013972. 1. 49|fi cents 8. 70 cents. 28. 40 and 5 %. 12. $2351.25. 65iff cents 9. 10 bottles. 29. 60 and 10%. 2. $ 13,227.50. 10. 4 men. 30. 40 and 15 0/,. Page 631. 3. 104 days. 13. 3\V- 4. $591.09+. Page 625. 1. $135. 14. 1 mi. 95 rd. 5. tV^t- 3. 12. 2. 10% 1 yd. 6 in. 6. $ 782,300. 4. 8. 3. $150. 16. £23. 7. $100. 5. 70. 4. 10%. 17. $6000; 8. $ 1005.50. 6. ism 5. $ 8.96. $ 14,000. 9. eoo/o. 16 J Page 634. Page 637. 5. 18.708+. 10. $171.98; 1. $100.02. 10. 5 acres ; 6. 27.532-. $ 194.01 ; 2. $500. 23^ days. 7. 28.408-. $ 174.08 ; 3. $1.75 dis- 8. 37.202+. $167.87. count per 1. $327.05+. 9. 43.290-. $1000. 2. $154.44+. 10. 63.245-. Page 643. 4. $700. 5. $198.80. 3. $291.08-. 4. $1276.28+. 1 17. 12. 1. .316+. X. 2. 6. 33 days. 5. $874.75+. 2. .632+. 3. 36. 7. Sight. 3. .949-. 4. 29. 8. 7%. Page 638. 4. .316+. 5. 28. 9. Par. 1. $2109. 5. .632+. 6. 33. 10. 50/0. 2. $95.25. 6. .949-. 7. 73. 3. $40. 4. $34.37|. 7. 8. 1.265-. 1.586+. 8. 9. 48. 16. 1. 23 inches. 2. 1 ft. U in. 5. $5106.25. 9. 1.897+. 10. 113. 3. 120 rods ; 6. $1080; 4%. 13. 2.214-. 11. 180 yards. 2640 yards. 7. 100/0; 5fo/^. 14. 2.530-. 12. 10 rods. 8. 3f%. 15. 2.846-. 13. 12rd.;18rd. 9. i%- 16. 3.162+. 14. 1^ acre. Page 635. 17. .348-. 4. 66 feet ; Page 639. 18. .379+. Page 644. 7.92 inches. 10. $62.50; 19. .411+. 15. $420. 5. 15 inches. i%- 20. .443-. 16. 192 rods. 8. 12| inches. 9. 24 inches. 17. 50 rods ; 1430 yd ; 2. Increased, 3. 16,203.03. 10. 5 feet. $68.75. 21^ acres. 3. $357.42. 18. 150 yards. 4. $175. Page 641. 19. 396 yards. Page 636. 5. 6'8. 4. 172,030. 20. 113.14- rd 1. $8.40; 6. East, 15°. 5. 9%. (18). 7. 7 a.m. 6. $1160.32+. 1. 161%- 2. $629.20. 8. $2437.60+. 7. $592.48-. 2. 3 minutes. 3. $650.39-. 9. $4800; 8. $120.76; 3. 9;*; 3^ 4. $225.16+. ($4797.58-). $162.61; 5. $343.61-. $62.79; Page 645. ($343.43+). Page 640. $83.72; 4. 300%. 6. $484.75; 1. 2.646-. $93.38. 5. 7 X 10, etc. ($485). 2. 3.742-. 6. $1.50. 7. $60.63+. 3. 6.164+. Page 642. 7. 100%. 8. $2. 4. 8.602+. 9. $ 126.80. 8. 20. ANSWERS ). 17 9. 12,000. (^)37i Page 650. Page 653. 10. 48 cases. (e) Impossi- 9. $ 1.72-. 7. 15.588 + ble. 10. 66f%. sq. ft. 1. £45 16 s. 11. 11. 8. 1176 sq. ft. 2. 1 lb. 14 oz. 12. 10,000. 9. 672 sq. ft. 3. 10 men. Page 648. 13. M; H- 10. 3.464+ sq. 4. 44 yards. 4. $24. 14. $ 15,000. in. 5. 3hr.l6min. 5. 1 mi. 7 fur. 15. 896 pounds. Diagonals, 2 6. 319 rd. 4 yd. 27 rd. 3 16. 472AV- in. and 1 ft. 6 in. yd. 2 ft. 17. $ 28.841 3.464+ in. Hi in. 19. 637i 11. 21.008+ feet. Page 646. 6. L.C.M. 360 20. $ 166.581. 12. 50.2656 sq. 7. 10 feet. 7. 194.45+ ft. 21. 6000 copies. in. 8. 4 acres ; 16 8. $6500. 22. 183-^ years. 13. 3.1416 a;2. acres. 9. 22f%. 23. 28ff| bbl. 14. 10 inches. 9. 2^ pounds. 10. 1200 24. 80 rods. 15. 19.635 sq. ft. 10. 432 pounds. peaches. 11. 6 days. Page 654. 12. 40/0. 1. $4.45-. Page 651. 1. $215.75. 13. 4^o/o. 2. $3.60+. 2. $1091.66+. 2. $414.33+. 14. 41 0/0. 3. 128f%. 3. $ 125.90. 15. 18000. 4. $34,312.50. 4. $ 104.55. 16. $1000; Page 649. 5. $ 86.44-. $ 1400. 3. $5.76. 6. $1551.27+; Page 655. 4. $ 4.38. ($ 1552.06-). 5. $157.68+. Page 647. 5. $1.17+. 7. $6000. 6. $453.61+. 17. $70,000. 6. $16.72-. 18. 68 rd. 3 yd. 7. $11.94. Page 652. 1. 84 sq. ft. 4 in. 8. $51.40. 8. (a) $ 33,000 ; 2. 234 sq. yd. 19. $165; $210; 9. $ 76.80. (6)1310/,. 3. 264 sq. rd. $225. 10. $ 92.38+. 9. N.Y. &N. E. 4. 84 sq. in. 20. $315. $50.50. 5. 990 sn ft uov hq. ib. 21. Qt\\%%; 1. $1000. 10. 40 acres. 6. 900 sq. ft. 6%. 2. $189.92- 7. 420 sq. yd. 22. 6 hours. 3. $ 31,000. 2. 1470 sq. ft. 4. m%- 3. 294 sq. rd. Page 656. 1. $3.37.68+. 5. .7525 miles. 4. 1764 sq. rd. 8. 330 sq. rd. 2- 1.5625. 6. 10mo.l7da. 5. 300 sq. yd. 9. 744 sq. rd. 3. (a)2j\%. nearly. 6. 42.332+ ft.; 10. 240 sq. ch. (ft) Vf. 7. (lO^perbu.); 2031.94- Cc) 180. H%- sq. ft. 1. 2i5,ni. 18 ANSWERS. 2. 1.975. 3. $100. 4. 152.02. 5. $489.75, 6. 93 lb. 9f oz. 7. 4ffi 8. $653.08-. 10. 2, 5, 7, 13, 23. Page 657. 1. 3 months. Page 658. 2. 9 months. 3. 1 yr. 5 mo. 4. 3| months. Page 659^ 5. 28^ days. 6. 2^ months. 7. 4^j months. 8. 7 mo. 26 da. 9. 4^ months. 10. August 10. 11. 94f cent». 12. $31.50; $21; $31.50. 13. A, $ 3500 ; B, $3600. Page 660. 14. 60bu.;40bu. 20 bushels. A, $36; B, $24. A, $875; B, $ 1458.33+; C, $1166.67-. 18. A. 54 tons : 15. 16. 17. B, 31^ tons ; C, 94^ tons. 19. 6f quarts. 1. 186,441. Page 661. 2. 34,538,549f. 3. 82,739ffi $540. 220%. $ 440. 2| hours. $ 1440. 1350 sq. in. 8. Page 662. 7. $1.29. Feb. 12, 1809. $5 gain. 2025. 9. 10. $9.52. 1.464375. 56.65- feet. $21. A, $2; B,$3; C, $3.50; D, $4.50. Page 663. 8. M; a). 9. 4725 1b. 1. 9600 men. 2. $37.75-. 3. $9.84+. 4. $4000. 2.16603. 7.95^%. 5 yr. 5 mo. 20 da. $3800.47+. 58 feet. Page 664. 10. 11. 12. 13. 14. 15. 8 A feet. 8 7% 1 yr. 10 mo. $9956.86-. $ 759.76. Page 665; 1. 336AV. 2. n^j rods. $ 1533.75. $116.36-. 7 bu ; 5 bu. 1 pk. 3 qt. 10 acres. .66+. 21 lb. 5 oz. 18 pwt. m gr. 7%. $ 16,000. 113.14^ rd. 3. 4. 5. 8. 9. 10. 11. 12. 13. Page 666. 14. if 4096 cu. ft. $331.86^. 223| sq.ft. $28.01+. 20. 5\i tons. 21. $40.50. 22. 56.8+%. 23.-. $ 27.84+. 15. 17. 18. 19. 24. 16- cents. 25. HI; iH; 26. 240 rods. Page 667. 1. 6.2832 .r. 2. 3.1416x2. 3. .7854 .t2. 4. .07958x2. 5. 1017.8784 sq. ft. 6. 7 yards, 7. 50 rods. 8.2!, 2 9. 11,250 sq. rd. 10. 2688 sq. ra. 11. 62.35+ sq. ft. 12. 1469.69- sq, yd, 13. 7.958 sq. ft. 14. 960 sq. rd. Page 668. 15. 4800 sq. yd. 16. 541.27- sq. rd. 17. 3800 sq. rd. 18. 234 sq. rd. 19; 779,42+ sq, yd. 20; 36 feet. 21. 40 rods, 22. 93.53+ sq. in, 23. 113.10- sq, in. 244.^ 50 sq. ft. Page 669. 1. $6224.35. 6. 210. J. £7 1888, $27,307. 25. 62.832sq.in. 2. 2 yr. 4 mo. 7. 1260. 1889, $ 29.499. 3|da. 8. 594. 1890, $ 25.772. 1. 18,990.59+. 9. 3366. 1891, $ 25.680. 3. i%. Page 672. 10. 3060. 1892, $27,801. 4. ? 49.02. 3. $1441.94+. 5. |21. 4. $150; $270. 1. $88.73+; Page 681. 5. lOo/o. $86.27-. 1. 96 sq. in. Page 670. 6. 288 boards. 2. $309.37i; 2. 72 sq. in. 6. $220.54-; 7. 49 rods. $ 464.061 ; 3. 144 sq. in. ($ 220.67-). 8. $1540. $513.56^. 5. 75.3984 sq. 7. $310.85+. 9. 14 rods. 3. $27xV; in. 8. £1,411,734 10. $50,000. $32i;$29i; 7. 294 sq. in. ; 6s. Id. $41i 6x\ 9. $21,050. 1. $69.85-. 8. 6 inches. 10. $1200. Page 679. 9. 216 sq. in. Page 673. 4. $4945.05+; 1. $1400. 2. $ 12,500. $5934.07-; Page 682. 2. $8.63-. 3. 3o/„. $4120.88-. 10. 128 sq. in. 3. $1917. 7. 20. 5. $42.96-; 11. 10 feet. 4. 284 days. 10. 14 weeks. $34.80-; 12. 1200 sq. ih , 5. 357. 11. 21 men. $28.35+; 20 in. 6. 143. 12. 240 miles. $46.39+. 13. 702 sq. in. 7. .8. 13. $96. 6. 6 men. 14. 4.7124 sq.ft. 8. 14 feet. 14. $15,000. 7. 31f days. 15. 48 sq. in. 15. $128. 8. 208 acres. 17. 64 sq. in. Page 671. 9. 240 men. 18. 576 sq. in. 9. 3 mi. 207 rd. Page 677. 10. 20 days ; 19. 6.93+ sq. in. 1 yd. 1 ft. $ 432.G3-. 24 days. 20. 37.6992 sq. 6 in. in. 10. 33mi.225i\ Page 678. Page 680. 21. 122.5224 sq. rd. 1. 306 sq. ft.; 11. $13. in. 11. 1 hr. 2 min. 12 ft. 12. $226.80. 52 sec. P.M. 2. 126 sq. yd. ; 13. 1363^ lb. Page 683. 12. $72. 12 yd. 14. 12i days. 22. 13 inches. 13. .6 week. 3. 1110 sq. rd.; Table. 23. 122.5224 sq. 14. $34.26+. 15 rd. 1883, $ 25.923. in. 15. '421 yards. 4. 210 sq. ch. ; 1884, $ 26.254. 16. 175 sheep. 15 ch. 1885, $ 28.972. 1. 130,548 ses- 18. $425.51. 5. 600 sq. in. ; 1886, $ 26.306. sions. 19. llf rolls. 16 in. 1887, $27,543. 2. 11 poems. 20 ANSWERS. 3. 1237.40; ($ 237.52). 4. $10,000. Page 684. 5. $40. 6. 303 feet. Page 686. 1. $10.34+; $ 505.38- $78,133^. Page 688. 15. lO. 57 cents. 4 lots. $55.15-. $7600. 285 acres ; $213.75 cornniis.sion. $425.52+; $443; ($446.26+). $900; ($899.54+). $1394.05. Page 687. 11. m tons. 12. 2l5 acrsR. 13. .56; .0012. 14. 144 yards. 15. 26j^ cents. 16. l^^x cents. $441. $12,204. 27 sq. yd. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 395,999.- 922186 ; ?H; I- 24| yards. $ 2022.22+ ; $371.44-; ($371.62^). 20%. $ 1280.42+ ; ($1279.54+). $ 8,575,875. $54,368.52|. 3960 inches. ^1 % gain ; 700 sq. in.; 896 sq. in. ; 1568 cu. in, 30. 67,750 acres. Page 689. 1. 18 cu. ft. 2. 400 cu. in. 3. 60 cu. in. 4. 82^ cu. yd. 5. 83.14- cu. ft. 6. 30.56- cu. ft. 7. 169.65- cu. meters. 8. 1163.29- gal. 9. 9719.325 lb. 10. 57| cu. in. Page 690. 12. 3^ inches. 13. 301.59+ cu. yd. 14. 13 inches. 9. 10. 1. $1,468,380,- 830. 2. 1,001,101. 3. 202,100,001. 00006. 6. $1983.38}. Page 691. 7. $2,382+. 8. lOT. 17c\vt. 3 qr. 8 oz. $51,000; $1260; $49,740. $353,369,- 654.14; 94.84- %. 11. $47,891,- 78.^.50 ; 3.9083+ %. 1/5. OgJJ /(,. Page 692. 13. $8000. 14. Due. $11,- 646.19. Page 693. 1. 16 (board) ft. 2. 7 (board) ft. 3. 8 (board) ft. 4. 14 (board) ft. 5. 4 (board) ft. 18 (board) ft. 7 (hoard) ft. 24 (board) ft. 14 (board) ft. 10. 16 (board) ft. 11. 40J (board) ft. 12. 27 (board Jt. 13. 30 (board) ft. 14. 15 (board) ft. 15. 9 (board) ft. 16. $16.20. 17. 1575 (board) ft. 18. $3.56. Page 694. 19. $60.48. 20. $52. m 6 lots. $ 5889f 36 days. 108. 146.86+ mi. B, $1000; C, $1500. Page 695. 9. $5.94— 1. 682.32 sq.ft. 2. 1 17.81 sq. in. 3. 50 sq. in. 4. 28.54 sq. in. 5. 7.854 in ; 7.071+ in. 6. 78.54 sq. in.; 28.54 sq. in. 7. 3.1416 sq. in. ; 12.5664 sq. in. 8. 1 to 9 ; area 14. i\.i.y o vv ItV xuxvo 8. 185.61- sq. (6) .003672; =i22x3.1416. 15. n- ft. (c) 1600. 9. 192 shares. 16. (a)f; Page 696. Page 701. 10. $3200. (c) .125 acre. 9. 43.61 pq_ yd. 1. 113.0976 cu. 11. 3450 copies. 17. 25%. iO. 392.7 sq.ft. in. 12. A, $375; 18. A, $50; 11. 84.8232 sq. in. 2. 14.1372 cu. B, $150; B, $13,600; 12. 3:1. in. C, $100. C, $4350. 13. 500 sq. ft. 3. 1:8. 13. $ 1059.35+ ; 14. 500 sq. ft. 4. .5236:1. ($1059.39-). Page 707. 15. 1:4. 19. 100 miles. Page 704. 20. $2000. Page 698. Page 702. 14. $1434.29+. 22. (a)i; 16. 12.5664 sq. in. 5. 113.0976 in.; 15. 2 : 58 : 48 (c) M- 17. 31 cents. 4071.5136 P.M. 23. (a) If; 18. Equal. sq. in. ; 16. 19i| days. (b) .00375 ; 19. 2:3. 24,429.0816 17. 71 days. (c) .16. 20. 127.328 sq.in. cu. in. 18. $ 13,000. 6. 7. 245 lb. 7 oz. 2:3. 19. 2013.7824 sq. in. Page 708. 25. (6)4000; 1. 11%; |90. 2. $88,922.4231. 8. .4764 ; 20. 452.3904 cu. (c) 13.163; about f in. (d) 1.706. Page 699. 9. .2146; 27. $331 3. !? 13,173.60. about |. 2. («)VA; 28. $200,000. 4. $1121 (&)tV 29 2 men fa 1. 17,000; Page 718. J.. K'fTt'.t'O ir=24ft. 105°; 105°. in. 15. 8. 16. 2 or -38. Page 784. Page 793. 1. 72 cu. in. 9. 1:4. 10. 14.1372 sq. 2. 36 cu. in. Page 752. 10. 90°, 5 in. ; in. 3, 36 cu. in. 1. 3 or -4. liin.,2in., 11. 1| in., 3 in.; 2. 5 or -2. 2i in. ; 37°, 1 in., 3 in. Page 798. 3. - 1 or - 4. 53°, 90°. 4. 36 cu. in. ; 4. 8 or - 1. Page 794. 124.71- cu. in. 5. -4 or -5. Page 785. 12. 108°. 5. 690 cu. in. 6. 7 or 4. 11. ft in. 14. Two, 15 in.; 6. 62.882 cu. in. 7. _6or-7. 8. 19 or -4. two, 13 in. ; 384 sq. in. 1. 60 feet. Page 799. 9. 18or-l. 2. 48 feet. 15. 240 sq. in. 7. 144 cu. in. ; 10. - 18 or - 1. 17. 40 sq. ft. 18 cu. in. Page 786. 8. 126cu. in. 1. 3,-2. 3. 109^3- feet. Page 795. 9. 1350 cu. in.; 2. 4, - 5. 4. 2160 feet; 20. 122.5224 sq. 50 cu. in. 3. 3, - 7. 2060 feet. in. 10. 1300cu.in. ; 4. 8, 2. 5. 124 feet. 21. 175 sq. in. 520 sq. in. ; 5. 9, - 7. 22. 257.6112 sq. in. 770 sq. in. 6. 3, - 6. Page 787. 23. 27 feet. 7. 5,4. 6. 13y% chains. Page 800. 8. 1, -8. 7. 162 feet. Page 796. 11. 1300 cu. in. 9. 2,-9. 8. 84 feet. 24. 452.3904 sq. 12. 65icu. ft. 10. 4, 1. ft. 13. 2232 cu. in. Page 788. 254.4696 sq. ft. Page 753. 9. 108 feet 25. 4166| mi. Page 801. 1. 4 and 8. 10. 119 yards. 26. 2000 miles. 14. 1927.3716 2. 80 feet. 3. 60 yards. 27. 12,500 miles. 28. 1:2. • 1. 2.0944 in.; cu. in. 15. 929.9136 cu. 4. 25 yards. 4.1888 in.; 29. 2828.4+ mi. in. ; 5. 5 feet. 6.2832 in.; 4.0256- gal. 6. 4 feet. 8.3776 in. ; Page 797. 7. 68 rods. 10.4720 in. 30. 113.0976 sq. Page 803. 2. 2 inches ; in. 16. 3053.6352 cu. Page 754. 3.464+ in. ; 31. 56.5488 sq. in. 8. 34 feet. 4 inches ; in. ; 17. 381. 7044 cu. in 9. 65 feet. 3.464+ in.; 28.2744 sq. 190.8522 cu. 10. 3ifeet. 2 inches. / 0F THJC in. \ MATHEMATICS, Bowser's Academic Algebra. A complete treatise through the progressions, indud iag Permutations, Combinations, and the Binomial Theorem. Half leather. $1.25. Bowser's College Algebra, a complete treatise for colleges and scientific schools Hal£ leather, ^1.65. Bowser's Plane and Solid Geometry. Combines the excellences of Euclid with those of the best modern writers. Half leather. $1.35. Bowser's Plane Geometry. Half leather. 85 cts. Bowser's Elements of Plane and Spherical Trigonometry,, A brief course prepared especially for High Schtjols and Academies. Half leather. JSi-oo. Bowser's Treatise on Plane and Spherical Trigonometry. An advanced work which covers the entire course in higher institutions. Half leather, i^i.65, Hanus's Geometry in the Grammar Schools. 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Present a carefully graded course in arithmetic, to begin with the fourth year and continue through the eighth year. Part 1. aoo^ages. Cloth. 40 cts. Part II. 382 pages. Half leather. 75 cts. W^alsh's Mathematics for Common Schools. Special features of this wo>k arc its division into half-yearly chapters instead of the arrangement by topics; the omission, as far as possible, of rules and definitions; the great number and variety of the problems; the use of the equation in solution of arithmetical problems; and the introduction of tho elements of algebra and geometry. Three Book Series — Elementary, 218 pages. 35 cts. Intermediate, 252 pages. 40 cts. Higher, 387 pages. Half leather. 75 cts Two Book Series — Primary, 198 pages, 35 cis. Grammar School, 433 pages. Half leather. 75 cts. Sutton and Kimbrough's Pupils' Series of Arithmetics. Primary Book. Embraces the lour fundamental operations in all their simple relations. 80 pages. Cloth. 25 cts. Intermediate I'.ook. 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Designed to aid in teaching the (..ur fundamental rules in lowest primary grades. 5 cts. each; per hundred I4.00. Sloane's Practical Lessons in Fractions. For elementary grades. p.oar