UNIVERSITY OF CALIFORNIA AT LOS ANGELES GIFT OF Dr. ERNEST C. MOORE TEACHING ARITHMETIC A HANDBOOK FOR TEACHERS AND A TEXTBOOK FOR NORMAL AND TRAINING SCHOOLS BY MIDDLESEX A. BAILEY, A.M. HEAD OP THE DEPARTMENT OP MATHEMATICS OF THE NEW YORK TRAINING SCHOOL FOR TEACHERS, NEW YORK CITY PUBLISHED BY MIDDLESEX A. BAILEY YONKERS, N. Y. OOPYEIGHT, 1913, BY MIDDLESEX A. BAILEY. COPYRIGHT, 1913, IN GREAT BRITAIN. J. 8. Gushing Co. Berwick & Smith Co. Norwood, Mass., U.S.A. "55 PREFACE THERE is a science of teaching arithmetic. The feeling of needs and the finding of means for their satisfaction is the law of advancement ; logical division is the law of classifications and definitions ; induction, deduction, and the complete method are the laws for the discovery of principles; experimentation and reasoning are the laws for the solution of simple problems ; and the analysis of complex problems into simple problems is the law for the solution of problems in general. In Part I the author has endeavored to develop and state these laws. In Part II he has endeavored to illus- trate their application ; much has been omitted but enough has been given to suggest the rest. In Part III he has endeavored to give an idea of the advance in complexity from grade to grade in elementary schools and of what is required to secure primary and higher licenses for teaching arithmetic. As a handbook, this work should help the superintend- ent to secure uniformity in his schools ; the principal to harmonize the work of the grades ; and the teacher to get a clear view of fundamental laws. As a textbook, this work should help the teacher in normal and training schools to supplement his lectures ; and prospective teachers in school and out of school to prepare for their chosen profession. MIDDLESEX A. BAILEY. NEW YORK TRAINING SCHOOL FOR TEACHERS, NEW YORK CITY. iii 219364 CONTENTS PAET I. INTRODUCTION PAGE LESSON 1. NEEDS SPECIES LOGICAL DIVISION r . 2 LESSON 2. CASES LOGICAL DIVISION MEASUREMENTS . 6 LESSOX 3. ONE-LINE DIAGRAMS. MECHANICAL AIDS 10 LESSON 4. PRINCIPLES INDUCTION DEDUCTION 14 LESSON 5. PRINCIPLES COMPLETE METHOD 18 LESSON 6. SIMPLE PROBLEMS BY EXPERIMENT 22 LESSON 7. SIMPLE PROBLEMS BY REASONS 26 LESSON 8. COMPLEX PROBLEMS BY ARITHMETIC 30 LESSON 9. WRITTEN PROBLEMS ARRANGEMENT 34 LESSON 10. PROBLEMS BY ALGEBRA BY FORMULA BY RULE BY PROPORTION .... 38 LESSON 11. PROBLEMS BY PROPORTION BY VARIATION . 42 LESSON 12. LESSON PLANS 46 LESSON 13. IN GENERAL 50 PART II. SUBJECT MATTER LESSON 14. NOTATION AND NUMERATION .... 55 LESSON 15. NOTATION AND NUMERATION .... 60 LESSON 16. ADDITION 64 LESSON 17. SUBTRACTION 71 LESSON 18. MULTIPLICATION 76 LESSON 19. DIVISION 82 LESSON 20. FACTORING 90 LESSON 21. 96 LESSON 22. DECIMALS 103 iv CONTENTS LESSON 23. DENOMINATE NUMBERS .... PAGE . 110 LESSON 24. MENSURATION . 117 LESSON 25. INVOLUTION, EVOLUTION, LOGARITION . 125 LESSON 26. ALGEBRA IN ARITHMETIC .... . 131 LESSON 27. LESSON 28. PERCENTAGE . . ... . 137 144 LESSON 29. INTEREST . 152 LESSON 30. INTEREST . 160 PAKT III. EXERCISES SECTION 1. ELEMENTARY SCHOOLS . SECTION 2. PRIMARY LICENSE CITY SECTION 3. PRIMARY LICENSE STATE SECTION 4. HIGHER LICENSES 169 181 187 190 TEACHING ARITHMETIC PART I. INTRODUCTION 1. Object and Scope. The office of this book is to dis- cuss means of assisting pupils six years of age to pass in eight years from complete ignorance of mathematics to the knowledge and efficiency outlined in the .course of study. The keynote is, Every step in response to a need and as a means to an end. The race has advanced to its present stage and will ad- vance to other stages in response to needs. The satisfac- tion of one need gives rise to others, their satisfaction to still others, and so on. Pupils must follow the same path but must accomplish in a few years what the race has accomplished in many centuries. 2. The Plan. The subject matter of mathematics is mental products number and the operations upon num- ber do not exist ready formed in nature. Assuming that the pupil, like the race, should be an inventor, we shall discuss the principal means of helping him to create sub- ject matter for himself, and shall apply these means to the divisions of arithmetic. l LESSON 1. NEEDS 3. Discovering Needs. Pupils are to take every step in response to a need. Place them in the actual situations that give rise to a need or cause them to image these situations, and suggest the need by a question. No labored effort should be made ; the suggestion should be natural and simple. ILL. Number. T. " On your desk there is a handful of sticks. How many are there ? " ILL. Interest. T. " A man borrowed $ 1000. With this money he bought goods which he sold for $ 1200. What should he be will- ing to do for the money borrowed ? " 4. Discovering Species. Most terms have varieties. Thus, numbers are abstract or concrete, concrete numbers are denominate or not denominate, denominate numbers are simple or compound. We do not wish pupils to study these varieties ready formed and to memorize set definitions, but we wish them to find the varieties for themselves and to make definitions in terms of the devel- opment. The process of discovering the species of a genus is logical division. Fix upon a basis of classification, find the differences of this basis, unite the genus and the differences, name the resulting species, define the species. Basis. The basis of classification is determined by some need. Thus, we may wish to find what kinds of triangles there may be with reference to the relative lengths of the sides. 2 4 LESSON 1. SPECIES LOGICAL DIVISION 3 Differences. The differences of a basis of classification may be found by experiment or by the law, a thing must be or must not be. Thus, to find the differences in rela- tive lengths of three lines we may draw several sets of three lines and examine all possible cases. Or we may reason, Three are equal or not three equal ; of three not equal, two are equal or not two equal. This gives rise to three equal, two equal, or no two equal. Names. Names are common words selected for their appropriateness, or unusual words from foreign languages. The pupil is helped to fix a name in mind by an explana- tion of its origin provided the explanation is within his comprehension. Thus, when no two sides of a triangle are equal, the sides if placed parallel are like the rungs of a ladder. The triangle is called scalene from the Latin for ladder. Definitions. To define a term logically is to state its genus and differences. There are as many differences as there have been bases of classification. The proximate genus should be used when its meaning is understood because there is then only one difference. Thus, from the proximate genus, A scalene triangle is a triangle hav- ing no two of its sides equal; from a remote genus, A scalene triangle is a polygon having three sides and having no two of its sides equal. Teaching. It would never do to speak to children about bases of classification, differences of the basis, genus, and species. The formal steps are to be in the mind of the teacher as a guide, the thought being that the teacher with such a guide will do a better piece of work than without it. 4 LESSON 1. SPECIES LOGICAL DIVISION 4 ILL. Triangles. T. " Let us find the different kinds of triangles with reference to the relative lengths of the sides. " Draw a triangle with all its sides equal. Draw some other kind. John has a triangle with two sides equal, and James has a triangle with no two sides equal. Is any other kind possible ? Draw a tri- angle with three sides equal and name it equilateral triangle, draw a triangle with two sides equal and name it isosceles triangle, draw a triangle with no two sides equal and name it scalene triangle. Define each kind." A / Equilateral Aj^..A Triangle (- - Isosceles . \ Scalene ILL. Fractions. T. " Let us find the different kinds of fractions with reference to the relative value of numerator and denominator. il Write a fraction. Mary, how does the numerator of your fraction compare with its denominator? The numerator is less than the denominator. This is true of every fraction that has been written. Write some other kind. John has a fraction whose numerator is equal to its denominator, and Henry has a fraction whose numerator is greater than its denominator. See if you can find some other kind. No one succeeds. We will call such fractions as f proper. The name is appropriate because we can separate a unit into 3 equal parts and take 2 of them. Define a proper fraction. We will call such fractions as f improper. After we have separated a unit into 3 equal parts we cannot take 4 of them. Hence, f is not properly a fraction. We will call such fractions as f improper. After we have separated a unit into 3 equal parts we can take the 3 parts, but ' frac- tion ' means a part and ' f ' is a whole. Hence | is appropriately called an improper fraction. Define an improper fraction. An improper fraction is a fraction whose numerator is equal to or greater than its denominator." / Nu. less than den. / Proper Fraction (- - Nu. equal to den. Fraction {- - Improper \ Nu. greater than den. N Improper 5 LESSON 1. SPECIES LOGICAL DIVISION 5 5. Diagrams. At the end of a classification, a diagram should be made to relate the genus and the species. The diagram may emphasize the differences, the finished prod- ucts, or the names. Following are diagrams illustrating the classification of quadrilaterals : /Adjacent sides equal , All angles right < Adj gides not ^ x Both pair sides || / ... / \ /Adjacent sides equal / N Not all right \Adj. sides not equal Quad. < One pair sides || Neither pair sides || /Square /Rectangle < -T, _ .. , / X Oblong Parallelogram ^ \TM, u -j /Rhombus / Rhomboid < ,. T , / \No name Quad . c Trapezoid Trapezium 6. Exercises. 1. Cause pupils to feel the need of addition. 2. Cause pupils to feel the need of bills. 3. Help pupils to classify integers with reference to divisibility by 2. 4. Help pupils to classify number with reference to the expression of the unit. 6. Ex- plain the appropriateness of equilateral as the name of a triangle whose sides are all equal. 6. Criticise this definition : An equilat- eral triangle is a three-sided triangle having its sides all equal. 7. Criticise this definition : A triangle is one which has three sides. 8. Teach the classification of quadrilaterals. 9. Define square, using as genus : (a) rectangle ; (6) quadrilateral. LESSON 2. CASES LOGICAL DIVISION 7. Discovering Cases. Different types of problems may involve the same terms. Thus in interest, the prin- cipal, time, and rate may be given to find the interest ; the principal, time, and interest may be given to find the rate ; and so on. Pupils should discover the different cases for themselves. The cases are not usually named. ILL. Cost of 1, En. Cost, Number. T. "At 3 ^ each the cost of 5 apples is 15^. Let us find the different problems that can be formed by the omission of each term in succession. " State the problem which arises from the omission of 15 j*. At 3 ^ each what is the cost of 5 apples V State the problem which arises from the omission of 3 ?. If the cost of 5 apples is 15^, what is the cost of 1 apple? State the problem which arises from the omission of 5 apples. At 3 p each how many apples can be bought for 15 ^ ? " ILL. Whole, Fraction of Whole, Part. T. " Let us find the differ- ent problems which arise from f of 20 = 8 by the omission of each term in succession. " State the problem which results from the omission of 8. What is f of 20? State the problem which results from the omission of 20. 8 is f of what number ? State the problem which results from the omission of . 8 is what part of 20? State these problems in gen- eral form." 1. To find a fractional part of a number. 2. To find a number when a fractional part is given. 3. To find what fractional part one number is of another. The teacher should carry such developments farther than he proposes to carry them with pupils, for the sake of widening his own horizon, for he should know more than he attempts to teach. 6 7 LESSON 2. CASES LOGICAL DIVISION 7 ILL. Cases in Percentage. Let us find all the different cases in percentage from the formulae, P = B x R, A = B + P, and D = B-P. There are three equations with five quantities. To solve these equations two of the quantities must be known, because it is impossi- ble to solve three equations with more than three unknown quanti- ties. Hence, there will be three cases in percentage for every two known terms. The combinations of twos in the terms A, B, D, P, R are : AB, AD,AP,AR; BD, BP, BR ; DP, DR; PR. That is, there are 10 times 3, or 30, cases in percentage. They are : Given To Find Given To Find A and B; D, P, R fiandP; A, D, R A and D ; B, P, R B and R ; A, D, P A and P ; B, D, R D and P ; A, B, R A and R ; B, D, P D and R ; A, B, P SandD; A, P, R P and R; A, B, D The teacher should form a concrete problem for each case grouped around some industry or activity. ILL. Sheep. Let us take the activity of buying and selling sheep. A represents the number of sheep after a purchase or 53 ; B, the original number or 50; Z>, the number after a sale or 47; P, the number bought or sold or 3 ; R, the per cent of the original number or 6%. Case 1. Given A and B to find D. A man had 50 sheep ; after purchasing a certain number he had 53. If he had sold as many as he bought, how many would he have had left? Case 6. Given A and D to find R. After purchasing a number of sheep a man had 53 ; if he had sold as many as he purchased, he would have had 47 left. What per cent of the original number did he purchase ? Case 10. Given *4 and R to find B. After purchasing a number of sheep a man had 53, which was 6 % more than the original number. What was the original number? 8 LESSON 2. MEASUREMENTS 8 8. Progressive Difficulty. The teacher must classify the examples under each subject in the order of their difficulty that the pupils may advance in the line of least resistance. He may need several bases of classi- fication. ILL. Subtraction of Mixed Numbers. The difficulty depends upon the sameness of the denominators, and upon the relative value of the fractions in the minuend and subtrahend. Classifying according to these bases, we obtain : denominators the same and fraction in the minuend the greater, denominators the same and fraction in the min- uend the smaller ; denominators different and fraction in the minu- end the greater, denominators different and fraction in the minuend the smaller. Thus : 8| - 5 J, 8J - 5| ; 8f - 5|, 8 - 5|. 9. Measurements. The teacher should have in mind the steps in all measurement as a guide to skilful presen- tation. To measure an object, assert that it possesses as much of a quality as a well-known standard. Abbreviate the concept by a name. Thus, the act requires as much time as the rotation of the earth about its axis, or a day. If the object possesses as much of the quality as the standard, as much more, as much more, and so on, intro- duce number. Thus, the act requires as much time as two rotations, or two days. If the measurement requires a large number of repeti- tions of the standard, select a larger standard. Thus, the act requires as much time as the revolution of the earth about the sun, or a year. If the object possesses less of the quality than the standard, select a smaller standard. Thus, the act requires as much time as the 24th of a day, or an hour. 10 LESSON 2. MEASUREMENTS 9 If necessary, use two or more standards. Thus, the act requires 5 years 2 days and 3 hours. ILL. Length, T. " How long is that line on the board ? About so long (the hands are held apart). How far from your desk to the door? 16 steps (the pupil counts his steps). " It is impossible to hold the hands the exact distance apart ; steps are not all of the same length. Here is a rule which is as long as a certain king's foot ; it is called a foot. Mary may measure the line ; it is 2 feet long. " Here is a rule 3 feet long; it is called a yard ; it is more convenient for measuring long distances than a foot rule ; it is used in meas- uring cloth. John, measure the long line ; it is 2 yards long. " With the foot rule Henry may measure the short line ; it is 1 foot and a little more. To measure this little more, what must we have? Yes, a still shorter rule. This foot rule has been divided into 12 equal parts, and each part is called an inch. Measure the line again, Henry ; it is 1 foot 4 inches. Who will give me the table ? 12 inches make 1 foot, 3 feet make 1 yard." 10. Logical Steps. Preparatory to many exercises, the teacher must discover the steps which are taken in com- mon practice to reach the desired end. Perform the exercise and analyze the steps. ILL. Multiplication. Let us discover the logical steps in the mul- tiplication of an integer by a number of two orders. In multiplying 264 by 24, we multiplied 264 by 4, we multiplied 264 by 20, and added the results. The logical steps are to multiply by the number in units' order, to multiply by the number in tens' order and the result by 10, and to add the products. 11. Exercises. 1. Teach pupils to find the problems which grow out of the statement, The interest of $200 for 2 years at 6% is 824. 2. From the simple interest formulae, / = P x T x R and A = P + 7, find the 20 cases. 3. Why are the cases API to find T and R, impos- sible? 4. The following divisors are to be used in long division, 71 and 17. (a) Arrange them in order of difficulty; (6) give reasons for your arrangement. 6. Teach quart, pint, gallon, in liquid measure. 6. State the logical steps in the addition of fractions. LESSON 3. ONE-LINE DIAGRAMS 12. One-Line Diagrams. To get clear notions of the various operations upon fractions pupils should be taught to make and use diagrams for themselves. Diagrams made by teachers do little good to pupils. One-line dia- grams are based upon separating a line into a number of equal parts and then into another number of equal parts in such a way as to show a common measure. ILL. T. " We wish to separate a line into 8 equal parts and then into 6 equal parts so as to show a common measure. " How shall we proceed ? Separate the line by short vertical lines into 8 equal parts. "Then what? The least number which exactly contains 8 and 6 is 24. Separate the line by dots into 24 equal parts. How can we do this? The line is already separated into 8 equal parts and we can separate each small portion into 24 -j- 8 or 3 equal parts. Do so. I ' i - i t I ' I ' t ' I "Now what? We want to separate the line into 6 equal parts. How many dots to a part ? 24 -*- 6 or 4. Draw a short vertical line above at every 4th dot. I I . i I ' i "What does each vertical line below show? . What does each dot show? ^ ? . What does each vertical line above show? J. What is a common measure of and |? -fa. How many 24ths make 1 8th? 3. How many 24ths make 1 6th? 4." 10 13 LESSON 3. ONE-LINE DIAGRAMS 11 Designating Parts. Designate a required portion by placing its value in the center of a dotted line which con- nects the extremities of the portion. ILL. T. " Let us draw a diagram to represent f and f . How shall we proceed? Divide a line into 4 equal parts and then into 3 equal parts. Mark off J and f as I have done." Uses. Pupils should solve examples in addition, sub- traction, multiplication, and division of fractions as a preparation for solutions by rules. ILL. 1. T. " You have just made a diagram to represent and f . Find the value of f + f . The value is & + T 8 j or f J. " Find the value of - f . The value is ^ - T \ or ^. u Find the value of f -4- |. The value is 8 12ths -^ 9 12ths or f ." ILL. 2. T. "Let us find f of f. What must we do? Represent!, divide | into 3 equal parts, and take 2 parts. " f of | is 8 of the 15 equal parts into which the unit is divided or i of | is A-" 13. Mechanical Aids. Pupils should be taught to use mechanical devices especially in mensuration. Paper Folding and Cutting. Let us agree that the pupil shall do the work. ILL. Triangles. T. " What is the sum of the angles of a triangle ? "How shall we proceed? Let us cut a triangle from paper and fold in such a way as to bring the vertices together at the same point. Fold the upper vertex over \ipon the base so as to make the folded edge parallel to the base. Fold the other parts. What seems to be 12 LESSON 3. MECHANICAL AIDS 13 the sum of the angles of a triangle? The sum of the angles of a triangle seems to be the sum of the angles about a point on one side of a straight line or 180." ILL. Parallelograms. T. " We want a rule for finding the area of a parallelogram. We know how to find the area of a rectangle. Let us find how the area of a parallelogram compares with the area of a rectangle. How shall we make this comparison? Cut a parallelo- gram from paper. Starting from one of the vertices cut off a right- angle triangle. Put the triangle first at one end of the second por- tion and then at the other. What seems to be true ? The area of a parallelogram seems to be the same as the area of a rectangle which has the same base and altitude." Crude Measurements. Some things are hard to measure, as the circumference of a circle or the surface of a sphere. Pupils should be asked to exercise their ingenuity on such measurements. ILL. Circumference of a Circle. T. " Here is a circle on the board. How shall we find its circumference? Sometimes the blacksmith wants to cut off a strip of steel long enough to make the tire of a wagon wheel. Do you know how he finds the proper length ? He has a rule made in the form of a circle, with a handle. He finds how many times this small wheel turns around in moving about the cir- cumference of the wagon wheel. " Can we use such a wheel to measure the circumference of this circle on the board ? What shall we use? I have a piece of electric 14 LESSON 3. MECHANICAL AIDS 13 wire here. How can we use it ? John may lay the wire on the cir- cumference and then measure the wire. What is the length of the circumference? 44 in. " Let us see if we can find an easier way. Measure the diameter of the circle. It is 14 in. Divide 44 in. by 14 in. The answer is 3f. What seems to be the circumference of a circle ? The product of its diameter by 3}." ILL. Surface of a Sphere. T. " How shall we measure the surface of this sphere? " I have here a piece of waxed cord and the half of a croquet ball (a hemisphere) into which I have driven two tacks. I propose to wrap the cord about one tack until the string covers the curved surface, and then about the other tack until the cord covers the plane surface, and then to compare the lengths. Henry may do this. The cord about the curved surface is twice the length of that about the plane surface. What does this seem to show ? The surface of a hemi- sphere seems to be twice the surface of a circle which has the same diameter, or the surface of a sphere seems to be 4 times the surface of a circle which has the same diameter." 14. Exercises. 1. Divide a line into 4 equal parts and then into 8 equal parts. 2. Represent | and J. 3. By diagram find: (a) f + f ; (*) $ 1 5 ( c ) $ -*- f 4 - B y diagram find | of f . 6. By paper cut- ting find the relation of a triangle to a parallelogram which has the same base and altitude. 6. Cut a triangle from paper and fold as in finding the sum of its angles. Show that the area of a triangle seems to be the area of a rectangle which has the same base and half the same altitude. LESSON 4. PRINCIPLES INDUCTION 15. Inductive Method. There are many principles in arithmetic which must be established as guides to methods of procedure. It is better for the pupil to develop these for himself than to study them ready formulated by another. The inductive method is the process of estab- lishing principles by experiment. Five canons of induc- tion are discussed in logic, but only the canon of agree- ment will be considered in this treatise. See logic (J. & H., p. 215). 16. Canon of Agreement. Whatever is true of several individuals of a class is probably true .of all the indi- viduals of that class. Thus, by experiment it is found of several examples in subtraction that adding the same number to both minuend and subtrahend does not affect the remainder ; this principle is probably true of all ex- amples in subtraction. When a proposition is established in regard to individuals by experiment alone, the inference in regard to the class can never be more than probable unless every individual in the class has been examined. The greater the number of individuals the greater the probability. Take several instances in which the phenomenon occurs, examine these instances for a common circumstance, and make the common circumstance the basis of a general- ization. ILL. T. " We wish to discover a rule for the divisibility of a number by 9. U 16 LESSON 4. PRINCIPLES INDUCTION 15 " Let us examine several numbers which are exactly divisible by 9. To get such numbers those who sit in the first row may multiply a number of three figures by 9 ; those in the second row a number of 4 figures ; all others, a number of 5 figures. Read some of the products : 2034, 3861, 23895, 808884. Do you find anything which these numbers have in common? Get the sums of the digits 9, 18, 27, 36. Now do you find anything in common? The sum of the digits is exactly divisible by 9. What seems to be the rule for the divisibility of a number by 9? A number seems to be divisible by 9 if the sum of its digits is divisible by 9. Memorize this rule." Weakened Form. An inference from a single instance can have little weight of itself, but is of value in a few cases where its establishment by other methods is too difficult for the grade. For examples, see 13. It plays an important part also in the complete method ( 20). Its use elsewhere is not recommended. Observe the weakness of the following : ILL. T. "Find by diagram f of f. The value is & (12). What seems to be a rule for multiplying fractions? To multiply fractions multiply the numerators for a new numerator, and the denominators for a new denominator." Use in Arithmetic. Except as a part of the complete method, the inductive method should be rarely used in arithmetic. An examination of enough instances to warrant a conclusion is long and laborious ; no appeal is made to the intelligence ; an experimental method is not suited to an exact science ; and at the best, the conclusion can never be more than probable. Used by itself, it is valuable for finding rules for divisibility in the lower grades, and for rinding a few rules for mensuration in the upper grades, because in both cases the pupils are not sufficiently mature to use more satisfactory methods. 16 LESSON 4. PRINCIPLES DEDUCTION 17 17. Deductive Method. The deductive method is the method of establishing principles by giving reasons for the steps. Several forms of deduction are discussed in logic but only the form, A is B, B is C, . *. A is C, will be con- sidered in this treatise. See logic (J. & H., p. 145). 18. A is B, etc. Whatever is true of a term is true of what is included within that term or of what is identical with that term. The argument may take the form A is B, B is C, . '. A is C, or the form of a chain of such argu- ments, A is B, B is (7, C is D, D is E, .-. A is E. Each premise must be established by a definition, an axiom, or a proposition previously proved. ILL. Multiplying both numerator and denominator of a fraction by the same number first multiplies the fraction by a number and then divides the result by the same number, because multiplying the numerator multiplies a fraction and multiplying the denominator divides a fraction (A is E). Multiplying a fraction by a number and dividing the result by the same number does not change the value of a fraction by axiom (B is C). Therefore, multiplying both numerator and denominator of a frac- tion by the same number does not change the value of a fraction (A is C). Make some predication about the subject of the required proposition based upon a definition, an axiom, or a propo- sition already proved. This gives the form, A is B. Make some predication about the predicate of the last proposition based upon a definition, an axiom, or a propo- sition already proved. This gives the form, B is 0. Continue as before until a serviceable predicate is found, and unite it with the subject of the first proposition. This may give the form, A is E. Individual Subject. In teaching, it is better to use an 19 LESSON 4. PRINCIPLES DEDUCTION 17 individual than a general term because the process is then more vivid. Thus, ' multiplying the numerator and denominator of | by 2 ' is more vivid than ' multiplying both numerator and denominator of a fraction. by the same number.' , Whatever is proved of the individual in this way is proved of the whole class which includes that individual because the principle could be proved of every other individual of the class in the same way. ILL. Multiplication of Decimals. T. " We are going to learn how to multiply a decimal by a decimal. We know how to multiply a decimal by an integer and how to multiply by .1, .01, and so on. .24 .6 .144 Take .24 x .6. What is to multiply by .6? To multiply by 6 and to multiply the result by .1 because .6 is 6*x .1. "You may all multiply by 6; the answer is 1.44; to multiply a decimal by an integer multiply as in integers and point off as many decimal places in the product as there are decimal places in the multiplicand. Multiply the result by .1; the answer is.l x 44; to multiply by .1, .01, .001, and so on, move the decimal point as many places to the left as there are decimal places in the multiplier. " What is to multiply .24 by .6 or to multiply a decimal by a deci- mal ? Multiply as in integers and point off as many decimal places in the product as there are decimal places in both multiplicand and multiplier." 19. Exercises. 1. Using the method of agreement, help pupils to find a rule for the divisibility of a number by 4. 2. Determine whether the formula, x' 2 + x + 41 = a prime, is true for all integral values of x. Suggestion. For x = 0, 1, 2, 3, 4, 5, 6, the values are 41, 43, 47, 53, 61, 71, 83, respectively. 3. Using the form, A is B, B is C, and so on, assist pupils to find how to multiply integers by a number of two orders. 4. Why is the deductive method better than the inductive method for establishing the principles of arithmetic? LESSON 5. PRINCIPLES COMPLETE METHOD 20. The Complete Method. The race has established many important principles in mathematics by inferring some property from an examination of one or more indi- viduals and by discovering why the property must be true of all the individuals of the class. Thus, the race found by measurement that of a right-angle triangle whose legs are 3 in. and 4 in. the hypotenuse is 5 in. From this and other measurements they inferred that the square of the hypotenuse of every right-angle triangle must be the sum of the squares of the other two sides. By the deductive methods of geometry they proved that their surmise was correct. The complete method consists in finding a prin- ciple by experiment (induction) and in proving it by rea- sons (deduction). It is the method of the discoverer. See logic (J. & H. p. 249). Find by experiment a proposition which is true of one or more individuals of a class and discover from defini- tions, axioms, or propositions already proved, why it must be true of the whole class. ILL. Mult. Terms of Fraction. T. " We wish to discover the effect on a fraction of multiplying its numerator, multiplying its denomina- tor, and multiplying both numerator and denominator by the same number. " You may draw a diagram like mine showing f , f , and f . - * . * 18 20 LESSON 5. PRINCIPLES COMPLETE METHOD 19 " Let us discover the effect of multiplying the numerator. Take f and multiply the numerator by 2 ; the result is f . From the diagram, tell me what has been done to the fraction ; it has been multiplied by 2. What seems to be the effect of multiplying the numerator? To mul- tiply the fraction. Who can tell me why? Multiplying the numera- tor multiplies the number of equal parts taken without affecting the size of the parts. Write the rule, multiplying the numerator multi- plies the fraction. "Let us discover the effect of multiplying the denominator. Take f. and multiply the denominator by 2; the result is |. From the dia- gram, tell me what has been done to the fraction ; it has been divided by 2. What seems to be the effect of multiplying the denominator? To divide the fraction. Who can tell 'me why ? What was the size of one of the equal parts before we multiplied the denominator? A fourth. After we multiplied the denominator? An eighth. What did we do to the size of one of the equal parts ? Divided it by 2. Now who can tell why? Multiplying the denominator divides the size of the equal parts without affecting the number of parts taken. Write the rule, multiplying the denominator divides the fraction. " Let us discover the effect of multiplying both terms by the same number. Take \ and multiply both terms by 2 ; the result is f. From the diagram, tell me what has been done to the fraction ; its value has not been changed. What seems to be the effect on a fraction of multi- plying both terms by the same number? It does not change the value of the fraction. Who can tell me why? When we multiplied the numerator by 2 what did we do to the fraction? We multiplied the fraction. When we multiplied the denominator of the result by 2 what did we do to the result? We divided the result by 2. If we multiply a fraction by 2 and divide the result by 2 what do we do to the fraction? Write the rule, multiplying both numerator and de- nominator of a fraction by the same number does not change the value of the fraction." Brief Rules. After several rules of kindred nature have been established it is sometimes possible to make a brief rule which comprehends them all. A valuable illustra- 20 LESSON 5. PRINCIPLES COMPLETE METHOD 20 tion is found in the principles of multiplying just devel- oped, together with the principles of division that may be developed in a similar manner. ILL. T. "Multiplying the numerator multiplies the fraction, di- viding the numerator divides the fraction. Who can express the two rules by one ? In the case of multiplying and dividing, doing either thing to the numerator does the same thing to the fraction. Let us put it shorter. Doing a thing to the numerator does the same thing to the fraction. "Multiplying the denominator divides the fraction, dividing the de- nominator multiplies the fraction. Give me a short rule for these two. Doing a thing to the denominator does the opposite thing to the frac- tion. " Multiplying both numerator and denominator by the same number does not change the value of the fraction, dividing both numerator and denominator by the same number does not change the value of the fraction. Give me a short rule for these two. Doing the same thing to both numerator and denominator does not change the value of the fraction." 1 ' Use in Arithmetic. The complete method should be used in developing nearly all the principles of arithmetic because it is the method of the discoverer. It discovers something by trial that may be true and proves by reasons that it is true. ILL. Multiplying Fractions. T. "To-day we are going to learn how to multiply a fraction by a fraction. Take f of f and find the result by diagram; ^5. (See diagram, p. 11.) " How can 8 be obtained from the numerators 2 and 4 ? How can 15 be obtained from the denominators 3 and 5? What seems to be the rule for multiplying fractions? Multiply the numerators for a new numerator and the denominators for a new denominator. " Let us find why this rule is true. What does f of f mean? That is to be divided by 3 and the result multiplied by 2 because the de- nominator shows into how many equal parts a thing is to be divided, and the numerator shows how many equal parts are taken. 21 LESSON 5. PRINCIPLES COMPLETE METHOD 21 " Divide - by 3 ; , because multiplying the denominator 5 5x3 4x2 divides the fraction. Multiply the result by 2 ; , because rnul- 5x3 tiplying the numerator multiplies the fraction. 9 A 4x2 " Write, - of - = Now we see why the rule is true. 3 55x3 u To multiply fractions, multiply the numerators for a new numera- tor, and the denominators for a new denominator. Write the rule." ILL. Dividing Fractions. T. " We want a rule for dividing a fraction by a fraction. " By diagram, divide f by f ; the result is f . (See diagram, p. 11.) "How could we get f? By inverting the divisor and proceeding as in multiplication ; f x $ = f . What may possibly be a rule for dividing a fraction by a fraction ? Invert the divisor and proceed as in multiplication. " Let us see if we can find reasons for this rule. The other day we learned how to divide 1 by a fraction ; invert the divisor. What is 1 -H |? f If '1 divided byf ' is f, whatis|of '1 divided by '? f of f. The rule must be true. To divide a fraction by a fraction invert the divisor and proceed as in multiplication. Write the rule." 21. Exercises. 1. Using the diagram, p. 18, teach pupils by the complete method to discover the effect on a fraction : (a) of dividing the numerator; (ft) of dividing the denominator; (c) of dividing both numerator and denominator by the same number. 2. Teach pupils by the complete method how to divide 1 by a fraction. 3. Why is the complete method better for No. 1 ; (a) than the induc- tive method alone ? (6) than the deductive method alone ? 4. What is the advantage of the brief rules suggested for the principles in fractions ? LESSON 6. SIMPLE PROBLEMS BY EXPERIMENT 22. Kinds of Problems. An exercise involving num- ber must state the operations directly or indirectly; this gives rise to examples and problems. A problem must involve one operation or more than one ; this gives rise to simple problems and complex problems. A simple prob- lem must involve addition, subtraction, multiplication, the first case in division known as quotition, or the second case in division known as partition. ILL. Example. Multiply 3 1 by 5. ILL. Simple Problem. At 3 ^ each what is the cost of 5 apples ? ILL. Complex Problem. If 2 apples cost 6 f, how much do 5 apples cost? 23. Importance of Simple Problems. The solution of simple problems is the basis of the solution of all prob- lems, because a complex problem can be separated into a chain of simple problems, and can be solved by solving each simple problem in succession. 24. Stages of Progress. In the advancement of the race there have been three stages ; the stage of obtaining results by counting, the stage of obtaining results chiefly by addition, the stage of obtaining results by the most fitting operation. These three stages should be observed in the advancement of the child. ILL. At 3 f each what is the cost of 5 apples ? OOO OOO OOO OOO OOO Counting Stage. Here are 5 apples, and 3 cents for each apple. They cost 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 cents. Addition Stage. They cost 3, 6, 9, 12, 15 cents. Advance Stage. They cost 5 times 3 cents, or 15 cents. 25 LESSON 6. SIMPLE PROBLEMS BY EXPERIMENT 23 25. Counting Stage. As soon as pupils can count they have at command a means for solving the problems of their daily experience. They can represent the terms by objects and find the results by counting. This work ex- ercises their ingenuity, gives them a feeling of power, and lays the foundation for all subsequent work. ILL. Simple Prob. in Add. On each desk there is a bundle of sticks. T. "If John has 3 apples and Joseph has 2 apples, how many apples have both ? Use the sticks for apples and find out. Mary may explain." M. " Here are John's apples, 1, 2, 3 ; here are Joseph's, 1, 2 ; both have 1, 2, 3, 4, 5 apples." ILL. Simple Prob. in Sub. T. "Susan had 6^ and spent 2?. How many cents did she have left? Use the sticks for cents and find out. Henry may explain."

and the selling price 25 ^, what was the gain on 1 Ib. ? 13 p. If the gain on 1 Ib. was 13 ^, what was the gain onlOlb.? $1.30. 37. Finding Component Problems. The chief work in the solution of a complex problem is to find the component problems. It must be done by a study of the situation. There are two principal aids. Analytic Aid. The value of the term required in the complex problem can be found from one or more terms by a single operation. Thus, the entire gain can be found from the gain on 1 Ib. and the number of pounds. The value of each unknown term thus used can be found from one or more terms by a single operation, and so on. Thus, the gain on 1 Ib. can be found from the cost of 1 Ib. and the selling price of 1 Ib. As soon as the values of 30 37 LESSON 8. COMPLEX PROBLEMS -BY ARITHMETIC 31 all the terms are known the simple problems can be formed. Consider from what terms the required term can be found by a single operation, consider from what terms each unknown term thus used can be found by a single opera- tion, and so on. State the simple problems. Make a diagram by joining each term to its dependent terms. i- 1 1K /Cost lib., 12? Gain on 1 lb.< ' \S.P. 1 Ib, 25 j* The above may be read, The entire gain can be found from the gain on 1 Ib. and the no. of pounds; the gain on 1 Ib. can be found from the cost of 1 Ib. and the selling price of 1 Ib. ILL. 1. T. "If the cost of 3 apples is 6^, what is the cost of 5 apples ? " From what can we find the cost of 5 apples? From the cost of 1 apple. From what can we find the cost of 1 apple ? From the cost of 3 apples. Cost 5 ap. < Cost of 1 ap. < Cost of 3 ap., 6 ^. " Solve. If the cost of 3 apples is 6 ^, what is the cost of 1 apple ? 2 p. If the cost of 1 apple is 2 p, what is the cost of 5 apples ? 10 ^." ILL. 2. T. " If A requires 2 da. for a work and B requires 3 da., how many days do both require ? " From what can we find the no. of days both require ? From the part both can do in 1 da. From what can we find the part both' can do in 1 da. ? From the part A can do in 1 da. and the part B can do in 1 da. (and so on). /Part A 1 da. < Days A, 2 Days both < Part both 1 da.< _ J \Part B 1 da. < Days B. 3 " Solve. If A requires 2 da., what part can he do in 1 da. ? \, If B requires 3 da., what part can he do in 1 da. ? \. If A can do \ of it in 1 da. and B can do \ of it in 1 da., what part can they both do in 1 da.? \. If both can do | of it in 1 da., in how many days can they do fofit? lda." 32 LESSON 8. COMPLEX PROBLEMS-BY ARITHMETIC 38 Grraphic Aid. The situation can often be studied to ad- vantage from a drawing. See 12. ILL. 3. T. "An article sold for $8 at a gain of J, what was the cost? " 'Sold at a gain of |' means ' the selling price is the cost plus \ the cost.' Draw a diagram to represent the terms. ....- SP COST '"GAIN"" " Solve. If the gain is \ the cost, what is the selling price? f the cost. If 4 the cost is $8, what is \ the cost? $2. If \ the cost is $ 2, what is | the cost? $6." ILL. 4- T. " A is 6| % taller than B. B is how many per cent shorter than A ? " " What does this mean ? A's height is B's height plus ^ B's height. B's shortage is what part of A's height? Draw a diagram to repre- sent the terms. " Solve. If B's shortage is ^ B's height and A's height is || B's height, what part of A's height is B's shortage ? -fa or 6^ %." When drawings are used, fractions may usually be omitted in the solutions. Thus : In ILL. 3, If 4 parts are $ 8, how much is 1 part ? $2. If 1 part is $ 2, how much are 3 parts ? , the gain on 10 Ib. is $ 1.30. Or, still shorter, the gain on 1 Ib. is 13 p, the gain on 10 Ib. is $ 1.30. It is often best in problems involving only multiplica- tion and division to express the answers to component problems without performing the operation. ILL. What is the simple interest of $ 720 for 157 days at 7 % ? Use cancellation method. P, 9 720 1.57 T, 157 da. 2 14 R ' 7 % X#x-?-x 1x157 628 I, ? 100 Jm 157 I, $21.98 cms. 21.98 Multiplying $ 720 by r ^ gives the interest for 1 yr. at 7 % ; dividing by 360, for 1 da. ; multiplying by 157, for 157 da. 40. Model Analysis. The model analysis of a complex problem consists of the model analyses of the simple prob- lems into which the complex problems may be separated. Its use is not recommended because of verbiage. ILL. Since the cost of 2 apples is 6 p, the cost of 1 apple is of 6 ^ or 3 f. Since the cost of 1 apple is 3 ^, the cost of 5 apples is 5 times 3 f or 15 p. 41. Exercises. 1. A, B, and C eat 8 loaves of bread, each the same amount; A furnishes 3 loaves and B 5 loaves; C pays 24^ for what he eats. How much should A receive ? Make the analytic dia- gram. 2. State and solve the simple problems. 3. Give the model analysis. 4. A buys a chair and a table for $ 35 ; the cost of the chair is f of the cost of the table. What is the cost of each ? Solve by the aid of a drawing : (a) using fractious ; (6) not using fractions. LESSON 9. WRITTEN PROBLEMS ARRANGEMENT 42. Arrangement. In order to grasp the situations involved in a problem, it is helpful to write what is given and what is required. Since the answer to each com- ponent problem is to be used in the solution of another, it is helpful to write each answer as soon as it is found. The work which cannot be performed mentally may appear at the right. No denominations need appear in the scratch work if they are kept in the statements. Write as concisely as possible what is given and what is required in a vertical line with a short horizontal line below, expressing the denominations. Write below the line the answer to each component problem as soon as it is found, expressing the denominations. At the right put all work that is not performed mentally, omitting the denominations. After the answer write ans. 43. Simple Problems. Addition Subtraction ApT, 38 Had, $358 Pr T, 96 Spent, $299 T, ? Left, ? T, 134 ans. Left, $ 59 ans. Multiplication Quotition C 1 H, $216 216 C 1 H, $216 19 C 19 H, ? 19 En C, $4104 216)1I()4 C 19 H, $4104 ans. 1944 H, ? 216 216 H, 19 ans. 1944 4104 1944 34 44 LESSON 9. WRITTEN PROBLEMS ARRANGEMENT 35 Partition C 19 H, $4104 C 1H, ? C1H, 216 19JH01 $216 ans. JJ8_ etc. NOTE. Observe the awk- wardness of retaining the denominations in division. Quolition Partition 19 $516 $216)$4104 19)$4104 Explanations. The author prefers question and an- swer. See 34' Scratch Work. In addition and subtraction it is un- necessary to rewrite the terms. In multiplication and division it is unnecessary to use the denominations because they appear at the left. To require the denominations is to require what is never done in practice, to insist upon distinctions which are wearisome, and to spend energy which may be better applied. See the note above. 44. Complex Problems. 1. A man gave $5760 cash and 128 cows at $56 in exchange for land at $64 an acre. How many acres did he get ? Pd, $ 5760 and 128 202 56 128 C @ $56 1 A, $64 A, 1_ Cows, $ 7168 Land, $12928 Acres, 202 ans. 768 640 7168 5760 12928 64)12928 128 128 128 EXPL. What is the value of 128 cows @$56? $7168. If the value of the cows is $7168 and the cash $5760, what is the .value of the land? $12928. If the value of 1 A. is $64, how many acres for $12928? 202. 36 LESSON 9. WRITTEN PROBLEMS ARRANGEMENT 45 2. What is the reading of the centigrade thermometer when the reading of the Fahrenheit is 50 ? Fr. C, 0; F, 32 BoilC, 100; F, 212 r F c F 50, C ? "H BOIL (-'> 180 F, 100 C 1 1? 5 P *> 9 ^ Ab. Fr., 18 18 F, 10 C ana. so- X- FR EXPL. If the freezing point of F is 32 and the boiling point 212, what is the difference ? 180. If the freezing point of C is and the boiling point 100, what is the difference? 100. If 180 F is 100 C, what is 1 F ? | C. If the freezing point of F is 32 and the reading is 50, how much above freezing is the reading? 18, etc. 45. Teaching. In writing what is given and what is required, pupils are prone to spend too much time in expressing terms fully. To counteract this it is well for the teacher to read a problem at ordinary speed, while no one writes, that the pupils may understand it as a whole ; and then to read it slowly, requiring every one to finish writing as soon as he finishes reading. The object is to be as concise as possible. Another good plan is for the teacher to read the problem just as he wishes the pupil to write it, and then to call upon some one to state it in full. ) 46. Proofs. Pupils should not be allowed to consider a solution complete until they have proved the answer both approximately and exactly. Approximate. Consider whether the answer is rea- sonable. ILL. Problem, 44. 200 A. at $60 would be worth $12000; 202 A. must be approximately correct. 47 LESSON 9. WRITTEN PROBLEMS ARRANGEMENT 37 Exact. There are four methods : First, review the work with care. This is the common method; it has been used in each of the foregoing problems. Second, solve the problem in a different way. This is of value when the problem can be separated into different sets of com- ponent problems. Third, discover whether the answer meets the conditions of the problem. This is of value when the problem is algebraic in nature. Fourth, form and solve a second problem in which the required term of the first is made a given term of the second, and some given term of the first is made the required term of the second. This method is cumbrous for practice, but valua- ble as an exercise for the pupil. Observe how the simple problems of 43 in multiplication, quotition, and partition prove each other. ILL. Fourth Method. T. "From the solution in 44, make another problem in which the number of acres shall be given to find the cash payment. A man gave 128 cows at $ 56, and a sum in cash for 202 A. of land at $64 an acre. How much cash did he pay?" 47. Discussion. For written work, the requirement through all the grades of writing what is given and what is required, is of supreme importance, because it fixes the attention upon the situations. The time lost in making the statements is usually more than gained in determining the operations. 48. Exercises. 1. At 396 Ib. a day, a ship's crew consume 11088 Ib. of beef in 28 da. Solve the problem arising from the omission of 11088 Ib. 2. From the omission of 396 Ib. 3. From the omission of 28 da. 4. From the land problem in 44, make and solve the complex problem by the use of 202 A. as a known term, and the number of cows as the required term. 5. Solve the last problem in 46. 6. Show that your answer is approximately correct 7. Explain your proof that the answer is exactly correct. LESSON 10. PROBLEMS BY ALGEBRA 49. Solutions by Algebra. A solution by algebra differs from a solution by arithmetic in two respects. By algebra, the unknown terms are represented by #, y, 2 ; by arithmetic, they are expressed in full. By algebra, the equations are solved by the laws of algebra ; by arith- metic, they are solved by analysis. ILL. An article is sold for $ 60 at a gain of . What is the cost ? Algebra S, $ 60 Let x = the cost c, ? - = the gain Qx C, $ 50 ans. = 60 5 6* = 300 x = 50 Arithmetic S, $60 C,J?_ C, $50 ans. The difference up to the formation of the equation is the use of x in the one and of cost in the other. By algebra the equation is solved, Clearing of fractions, 6 x = 300 ; dividing by the coefficient of x, x = 50. By arithmetic the equation is solved, If f of the cost is $ 60, what is of the cost? $10. If i of the cost is $10, what is f of the cost? $50. Use in Arithmetic. Solutions by algebra are valuable for the indirect cases in percentage and interest, and for all other cases in which an operation must be performed upon the required term. 38 50 LESSON 10. PROBLEMS BY FORMULA 39 50. Formulas. The relation of the required term of a problem to the given terms may be expressed by an equation in which the required term is the left-hand member and combinations of the given terms the right- hand member. In solutions by formula, the first step is to get the formula ; it must be recalled from memory, it must be taken from a book, or it must be derived. The second step is to substitute the given values and to solve the equation. ILL. 1. What is the circumference of a circle whose radius is 6 in. ? Recall the formula. R, 6 in. 3.1416 C,?_ C = 2 TT fl 12 C, 37.6992 in. ant. 37.6992 ILL. 2. How far will a body fall from rest in 2 sec.? Get the formula from physics. T, 2 sec. D, ?_ D = 16-^ x T D, 64J ft. ans. ILL. 3. Problem of p. 7, case 6. A, 53 sh. (1) P = B x R R = A - D D, 47 sh. (2) A = B + P ~ A +D R,?_ (3)Z>^B-P ^53-47 R, 6%. ans. 2B = A + D 53+47 2P = A- D =6% PROOF. Second Method ( 46). If the original no. plus the no. purchased is 53, and the original no. minus the no. purchased is 47, what is twice the original no. ? 100. What is twice the no. purchased ? 6. If the original no. is 50, and the no. purchased 3, what is the rate? 6%. Use in Arithmetic. Formulas may be used in problems in which the relations of the terms must be found by geometry, by physics, or by involved processes. 40 LESSON 10. PROBLEMS BY RULE 51 51. Rules. A rule is the translation of a formula into language free from algebraic expressions. ILL. 1. C 2 TT r is translated, Given the radius of a circle to find its circumference, multiply twice the radius by 3.1416. ILL. 2. D = 16 fa x T 2 is translated, Given the time of a body fallen from rest to find the distance, multiply 16^ ft. by the square of the number of seconds. A D ILL. 3. R = - is translated, Given the amount and the differ- ence to find the rate, divide the difference between the amount and difference by the sum of the amount and difference. The teacher will find the exercise of translating from formula to rule excellent to strengthen his command of expression. ILL. A = V s(s a)(s b)(s c). Translate. Given the sides of a triangle to get the area, find the continued product of the half sum of the three sides and the remainders found by subtracting each side from the half sum separately, and extract the square root of the result. Use in Arithmetic. Formerly, rules were used in solv- ing most of the problems of arithmetic. At present, their use is restricted for the most part to mensuration. The rule is stated from memory and its directions are followed. ILL. What is the radius of a circle whose area is 78.54 sq. in.? A., 78.54 sq. in. 25. R, _ ? 3 X 1416.)78 X 5400. R, 5 in. ans. 62832 . 157080 157080 Given the area of a circle to find the radius, divide the area by 3.1416 and extract the square root of the quotient. 51 LESSON 10. PROBLEMS BY PROPORTION 41 52. Mult, and Div. Probs. Problems which involve no other operation than multiplication and division are made up of a number of different terms and have two values for each term. ILL. If 3 men can pick 240 bbl. of apples in 8 da., how many men will be required to pick 480 bbl. in 4 da.? The terms are men, barrels, days. The values for men are 3 and x ; for barrels, 240 and 480 ; for days, 8 and 4. 53. Relation of Terms. The relation of two terms is ascertained by multiplying one of them by a number and noting the effect upon the other. Men and Work. What is the effect on work of multiplying no. of men by a number? To multiply the work by that number. Thus, twice the no. of men do twice the work. The no. of men is propor- tional to the work, or the no. of men varies as the work. Men and Time. What is the effect on time of multiplying no. of men by a number? To divide time by that number. Thus, twice the no. of men require half the time. The no. of men is inversely proportional to the time, or the no. of men varies inversely as the time. Area and Radius. What is the effect upon the radius of multiply- ing the area by a number? To multiply the square of the radius by that number. Thus A=TT R* and 2 A = IT x 2 R 2 . The area of a circle is proportional to the square of the radius, or the area of a circle varies as the square of the radius. 54. Exercises. 1. After losing a third of his sheep a man had 166 left. How many did he have at first? Solve by algebra. 2. In what time will a sum of money double at 6 % simple interest? Solve by algebra. 3. On a lever the weight is 54 lb., the power is 12 'lb., and the power's distance from the fulcrum is 9 in. What is the weight's distance? Solve by formula. 4. What is the surface of a sphere whose radius is 6 in. ? Solve by rule. 6. What is the relation of the distance fallen by a body from rest to the time in seconds ? LESSON 11. PROBLEMS BY PROPORTION 55. Two-Term Problems. Find the relation of the terms by multiplying one of them by a number and not- ing the effect on the other, and form the proportion indi- cated. ILL. 1. At 3 for 5 how many apples can be bought for 30^? 3 ap, 5 f> The no. of apples is proportional to x ap, 30? their cost. No. ap, 18 ans. 3 : x = 5 : 30 3 x 30 1ft x = = 18 5 ILL. 2. If 2 men require 10 da. for a job, how many days do 5 men require? 2 men, 10 da. The no. of men is inversely propor- 5 men I _x da. tional to the time. No. days, 4 ans. 2 : 5 = x : 10 .-xlfi.4 ILL. 3. If the area of a circle whose radius is 5 in. is 78.54 sq. in., what is the area of a circle whose radius is 10 in. ? 5 in., 78.54 sq. in. The area of a circle is proportional 10 in., x sq. in. to the square of the radius. Area7314.16 sq. in. ans. 78.54 :x = 5 2 : 10 2 78.54 x 100 25 = 314.16 ILL. 4. If the area of a circle whose radius is 5 in. is 78.54 sq. in., what is the radius of a circle whose area is 314.16 sq. in.? 5 in., 78.54 sq. in. The area of a circle is proportional x in., 314.16 sq. in. to the square of the radius. Radius, 10 in. ans. 78.54 : 314.16 = 5 2 : x 2 314.16 x 25 78.54 = 10 42 56 LESSON 11. PROBLEMS BY PROPORTION 43 56. N-Term Problems. Problems of more than two terms give rise to compound proportion. This subject is usually omitted from arithmetics. Find the relation of the required term to each of the other terms and make a proportion for each relation. ILL. Problem of 52. 3 men, 240 bbl., 8 da. _ f 240 : 480 x men, 480 bbl., 4 da. 3:*_j 4 . g XT- 10 3 x 480 x 8 10 No. men 12 ans. x = = 12 240 x 4 The no. of men is proportional to the no. of barrels and inversely proportional to the no. of days. This means that so far as the no. of barrels is concerned 3 : x = 240 : 480 ; so far as the no. of days is con- cerned, 3 : x = 4 : 8 ; so far as both are concerned, 3 : x 240 x 4 : 480 x 8. 57. Proportional Parts. The statement that a whole is divided into parts proportional to given numbers means that the ratio of the two sums equals the ratio of each part of the first sum to the corresponding part of the sec- ond sum. ILL. 1. Divide 1728 into parts proportional to 3, 4, 5. 12; 3,4,5 12:1728 = 3:z 1728 ; x,_y, z 12 : 1728 = 4 : y s^s-*^^*-^ Parts, 432, 576, 720 ans. 12:1728 = 5:z ILL. 2. Divide 68 into parts proportional to | and |. H;A,A 17:68= 8:x i 68; ar,_y 17:68 = 9:y 'Cialx 1 I Parts, 32, 36 ans. 58. Use in Arithmetic. There is no great need for pro- portion to solve the problems of arithmetic. It seems necessary, however, to teach the subject in the elementary schools because its terms are so often used in common speech. All mechanical methods which avoid a study of relations should be avoided. 44 LESSON 11. PROBLEMS BY VARIATION 59 59. Two-Term Problems. Find what one value of a term has been multiplied by to give the other value, and multiply or divide the given value of the other term as the relation indicates. ILL. 1. 55. 5 ^ has been multiplied by 6 ; multiplying cost by a number multiplies apples by that number ; 3 apples must be multi- plied by 6. Ans. 18 apples. ILL. 2. 55. 2 men has been multiplied by f ; multiplying men by a number divides days by that number ; 10 da. must be divided by f. Ans. 4 da. ILL. 3. 55. 5 in. has been multiplied by 2 ; multiplying the radius of a circle by a number multiplies the area by the square of that num- ber; 78.54 sq. in. must be multiplied by 4. Ans. 314.16 sq. in. ILL. 4. 55. 78.54 sq. in. has been multiplied by 4; multiplying the area of a circle by a number multiplies the radius by the square root of that number; 5 in. must be multiplied by 2. Ans. 10 in. 60. N-Term Problems. Compare each of the terms with the required term as in two-term problems. ILL. 56. 240 bbl. has been multiplied by 2 ; multiplying barrels by a number multiplies men by that number ; 3 men must be multi- plied by 2. .4ns. 6 men. 8 da. has been multiplied by \ ; multiplying days by a number divides no. of men by that number ; 6 men must be divided by \. Ans. 12 men. 61. Proportional Parts. Find what one sum has been multiplied by to make the other sum. ILL. 1. 57. 12 has been multiplied by 144 ; multiplying the sum by a number multiplies each of the parts by that number ; 3, 4, 5, must be multiplied by 144. A ns. 432, 576, 720. ILL. 2. 57. j has been multiplied by 48 ; f and must be mul- tiplied by 48. Ans. 32, 36. 62. Use in Arithmetic. The method of variation is of great value in all problems which involve multiplication and division because its use requires strict attention to the 63 LESSON 11. PROBLEMS BY VARIATION 45 relations of the terms. It is of special value for problems in mensuration which have to do with similarity. Thus, A is 6 ft. tall; it is proposed to make his statue 12 ft. tall. A's little finger is 3 in. long; to paint a statue of A's size costs $2; the weight of a statue of A's size is 1000 Ib. What will be the length of the little finger of the statue ? What will it cost to paint the statue ? What will be the weight of the statue ? A's height has been multiplied by 2 ; multiplying a linear part by a number multiplies a linear part by that number, multiplies a surface part by the square of that number, and multiplies a solid part by the cube of that number. 2 in., the length of the little finger, must be multiplied by 2 ; $2, the cost of painting a statue of A's size, must be multiplied by the square of 2 ; 1000 Ib., the weight of the statue, must be multiplied by the cube of 2. 63. Problems in General. All problems may be solved by stating and solving their simple problems, by algebra, by formula, and by rule. Problems involving multiplication and division may also be solved by proportion and by variation. The secret of success by each method is to grasp the situations. Arrangement of the work is an important factor. 64. Exercises. 1. If a body falls from rest 64| ft. in 2 sec., how far will it fall in 6 sec.? The distance varies as the square of the time in seconds. Solve by proportion. 2. Solve by variation. 3. If 3 boys earn $3 in 3 da., how many boys will earn $ 100 in 100 da. ? Solve by stating simple problems. 4. Solve by algebra. 6. Solve by formula. 6. Solve by rule. 7. Solve by proportion. 8. Solve by variation. 9. State with reasons which method you prefer for No. 3. LESSON 12. LESSON PLANS 65. Preparation. Before conducting a class exercise the teacher should have a definite plan. He should know exactly what he is going to do and exactly how he is going to do it. He will then look forward with pleasure to the exercise and will know at its close what changes to make the next time he presents a similar exercise. In making the plan the principal things to consider are: 1. The object and scope of the exercise. 2. The logical steps demanded by the subject. 3. The knowledge which pupils must have before they are ready to take up the subject. 4. The means which must be used to induce the pupils to take the logical steps. Object and Scope. The object of a class exercise may be to develop a new subject, to drill upon a subject previously developed, or to determine how well a subject has been mastered. This gives rise to development exercises, drill exercises, and test exercises. By scope is meant where a subject shall begin, where it shall end, and how fully it shall be treated. Important factors are the degree of maturity of the child, the time allowed for the exercise, and the requirements of the course of study. Logical Steps. By logical steps are meant the steps that must be taken in the order of their dependence. See 10. Knowledge. The teacher should consider not only what knowledge pupils must have before they are ready to take 46 66 LESSON 12. LESSON PLANS 47 up a subject, but also whether they actually possess this knowledge. Otherwise he will present what has been presented before or he will present what is irrelevant. Means. This topic demands the teacher's principal study. He must consider the teachings of psychology, logic, and experience to determine how the mind acts ; the teachings of school management to fix upon class government and mechanical movements ; works on meth- ods and history of education to test his theories; and in fine, everything which he has studied bearing upon the subject to add to his efficiency. 66. Development Exercises. These exercises have to do with new subject matter. ILL. Combinations of 4's in Addition. First lesson. Object and Scope. To get the pupils to repeat from memory the table of 4's iu addition. Steps. To discover the results of the new combinations by count- ing objects, to repeat the table with the objects in sight, and to repeat the table with the objects out of sight. Knowledge. Counting and the tables of 1's, 2's, and 3's. Means. To show the need, state that there is frequent occasion to find the sum of 4 and each of the digits, and that it saves time to memorize the results. To help the pupils to satisfy this need, have them write the combinations which they already know by numerals and the new combinations by dots so arranged that their number may be recognized by form, have them find the sums by counting the dots, have them repeat the table with the dots in sight, and have them repeat the table with the dots out of sight. I 2 3 444 67. Drill Exercises. Like finger exercises upon the piano, drill exercises in arithmetic are to give ability to do 48 LESSON 12. LESSON PLANS 68 accurately and rapidly what can already be done less accu- rately and less rapidly. ILL. Combinations of 4's in Addition. Second lesson. Object and Scope. To get pupils to call the results of the combina- tions of 4's in addition arranged in miscellaneous order, with accuracy and at the rate of three a second. Steps. To call the results from memory, and in case a result is missed, to repeat the entire table. Knowledge. Ability to repeat the table of 4's from memory. Means. To get all the combinations before the pupils, use the circle device and the device of writing the addends of each combina- tion in a vertical line. To get speed, tap on the board slowly and require the pupils one by one and also in concert to call the results in unison with the sounds, gradually increasing the speed until the rate of three combinations a second is attained. If this speed is not reached the first day, repeat the exercise at intervals for months or even terms. For additional practice, state simple problems requiring the answers instantly without any form of explanation. B, 9 7 7159380426 , Q , 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 68. Tests. The setting of proper tests requires much skill. Not only must the object and scope be determined with great care, but also the means must be studied with unusual attention. An hour spent in the preparation of a test will often save several hours in the grading of papers and will insure a better measurement of the ability of the pupils. ILL. Oral Test. Combinations of 4's in Addition. A lesson after the subject is well mastered. Object and Scope. To determine whether each pupil can call the results of the combinations of 4's in addition accurately and without hesitation. 69 LESSON 12. LESSON PLANS 49 Steps and Knowledge. Same as in the drill exercise. Means. A pupil should be able to read the results as rapidly when they are expressed by the combinations as when they are expressed by the common method. Write a half dozen of the results by the com- mon method, and then all of the combinations. Require each pupil to read all the results at the same rate he reads those expressed by the common method. 8064293157 12, 10, 11, 13, 9, 6, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4. ILL. Written Test. The forty-five Combinations in Addition. Object and Scope. To determine how rapidly pupils can write the results of the 45 combinations in addition. Steps and Knowledge. Same as in the drill exercise. Means. Give each pupil a paper on which the 45 combinations have been written twice in miscellaneous order, and require him to write as many of the answers as possible in one minute. Graphs. Below is a graph showing the records of ten pupils in the written test. IZ34S6789IO This graph shows that pupil No. 1 wrote 60 correct answers in a minute ; pupil No. 2, 65 ; and so on. 69. Exercises. 1. Plan a development exercise on the combina- tions of 4's in multiplication.* 2. A drill exercise. 3. An oral test. 4. A written test on the 45 combinations in multiplication. 5. Prepare a graph showing the records of ten pupils in the multi- plication test. No. 1, 60 answers in a minute; 2, 80; 3,75; 4, 60; 5, 72; 6, 85; 7, 80; 8, 90; 9, 84; 10, 50. 44 44 * Suggestion. 1 4 , 2 4 -, 3 4- B , < 4 > 4*4 , . . . LESSON 13. IN GENERAL 70. Mental and Written. There is a tendency on the part of both teacher and pupil to use the pencil too freely. The greater part of the computations made by persons in business and persons in the trades and professions are mental. Often, the use of a pencil is a sign of weakness. The first step in the solution of problems is to grasp the situation ; the second, is to perform the operations. In the schoolroom at least 80 % of these computations should be mental. ILL. Pupils should be able to work such examples as the follow- ing mentally : $7.38 + $ 9.64; $10.50 - $6.84; $7.85x12; $57.65-12; 6f + 8f; 9f - 4|; 49 x f ; 49 -4- |; f x f ; 6f x 8; \ to \ of numbers to 100; | - |; 3 + 4J; 6% of 500; |% of $1000; 528 x 25; 1728 - 25; ... 71. Economy of Time. Operations Easy. Usually, drill in the solution of problems should be separate from drill in the performance of the fundamental operations. In the former, the numbers should be such as not to distract the attention from the consideration of the relations. The greatest care must be exercised that problem work shall not degenerate into an exercise in multiplication and division. ILL. If an article is sold for $525.65 at a gain of 27f % what is the cost? The division of 525.65 by 1.27^ is so difficult that the attention is likely to be focused upon the process of division rather than upon the discovery that division is necessary $540 and 8% are better numbers in general. 60 72 LESSON 13. IN GENERAL 51 Operations Omitted. For a quick review of complex problems, pupils may be asked to state component prob- lems and to name the answers by letters without com- putations. ILL. In what time will $530.50 amount to $641.30 at 5%? If the principal is $ 530.50 and the amount is $ 641.30, what is the interest? fa. What is the interest of $530.50 for 1 yr. at 5%? $ b. If the entire interest is $ a and the interest for 1 yr. is $ b what is the number of years ? -. b Component Problems by Diagrams. For the discovery of whether pupils understand the relations, they may be required to prepare analytical diagrams as in the analysis of sentences. See 37. ILL. Problem above. . En Interest Iforlyr. 72. Expression. By the Teacher. Before a thought is expressed it exists in the mind without words as an im- pulse. When words are selected, the result may be satis- fying to the speaker but unintelligible to the hearer. Hence, it is of prime importance that the teacher should use with accuracy the technical terms and forms of phras- ing peculiar to each subject. Following are expressions having a tendency to arrest development which the author has heard from teachers in the schoolroom. It is criminal for a teacher to give long development exercises which are both inaccurate and silly. Hours and hours are wasted by many teachers in this way, and pupils of intelligence come to despise both the subject and the 52 LESSON 13. IN GENERAL 73 teacher. No explanation at all is better than a foolish ex- planation. ILL. 1. " Tf 1 apple costs 3 ^ the cost of 5 apples will be as many cents as 5 multiplied by 3^ or 15^." Such an expression has a ten- dency to make a fool of the pupil. It is impossible to multiply 5 by 3 ^ ; the phrasing is bad. ILL. 2 " Since 12 is f, is \ of 12." 12 = f is a false statement, | = | of 13 is false. ILL. 3. " If the interest of 1 yr. amounts to f 12, the time it gains $ 24 is y^ of 24 which is 2 yr." ' The interest of 1 yr.' is ridiculous, 'amount' in interest problems means technically 'the principal plus the interest,' ' ^ of 24 ' is 2 and not 2 yr., it is incorrect to take -fa of $ 24, for then the denomination of the answer would be dollars. ILL. 4. " To find the least common denominator of two fractions by inspection, compare the successive factors of the largest number with the smaller until a factor of the smaller is found." The correct expression is, " Compare the successive multiples of the larger denomi- nator with the smaller until a multiple of the smaller is found." By the pupil. When a pupil is taking up a new topic, his major effort is to master the thought. He should not be corrected for the use of fragmentary and crude ex- pressions until quite late. For a time it is enough if he hears invariably the correct forms from the teacher. He will adopt them as soon as he gets a stronger grip upon the thought. He should not be required to give the theoretical explanations of the fundamental operations. He will understand such explanations more or less clearly if they are skilfully given by the teacher, but he has not the power of expression to make them himself. 73. Arrangement. The arrangement of the work is of prime importance. 74 LESSON 13. IN GENERAL 53 The teacher should always paragraph properly. The rule is simple. The first word of a paragraph, even when it is a numeral or a letter, should begin three or four letters farther to the right than the first word of each of the other lines. Thus : [1. Draw two vertical lines. Begin the first word of each paragraph on the right-hand vertical. ;2. Begin every other line on the left-hand vertical. If a drawing is used, it should be put in the center of the page from right to left and no writing should appear on either side. 74. Crutches. The teacher makes a serious mistake if he instructs or permits pupils, when they are beginning a topic, to write figures showing how many are to be car- ried, or to use signs of operations when terms are written in a vertical line, because in many cases pupils never dis- card such crutches. Thus : 68 142 The small figures should never be written, even by the teacher for purposes of explanation. Many high school graduates who enter training school use these small fig- ures habitually in performing all operations. Nothing that can be said or done seems to be effective to prevent them from perpetuating, this practice when they become teachers. The illustration at the right shows how one of these graduates divides by 7. The work is eloquent of im- 7 83 -27 56 68 x 9 61'2 7)378 54 7)3,22(46 28 42 42 54 LESSON 13. IN GENERAL 75 proper instruction. Not only does she use crutches, but she also employs long division in dividing by a number of one order. The use of the signs is entirely unnecessary, is opposed to practice, and is confusing in algebra. 75. Records. Each pupil should have one grade and only one for each unit's work in a term, and the name of the unit should be written in connection with the grade. The average of a great number of grades with no state- ment for what each grade was given is of little value. One grade for each unit together with the name of the unit gives more information of what a pupil has done during a term than a hundred miscellaneous unmarked grades. ILL. In the 4 A grade (N. Y. City, 1912), the units are notation, counting, addition, subtraction, multiplication, division, measure- ments, fractions, problems. Below is a good record of what two pupils have accomplished. NAMES NOT. COUN. ADD. SUB. MULT. Div. MEAS. FK. PROB. Jovo^uyn/f /r/n/\At Jo, a. a 8 8 , or - x TrD 2 x \ D, or i irD a ." % 3 124 LESSON 24. MENSURATION 174 174. Similarity. If a form is enlarged and the likeness is preserved in every respect the original and the result are said to be similar. Multiplying a linear part by a number multiplies every linear part by that number ; multiplies every surface part by the square of that number; multiplies every solid part by the cube of that number. ILL. See 68. 175. Right-Angle Triangles. The right angle triangle is the basis of many problems. The square of the hypote- nuse is equal to the sum of the squares of the^ other two sides. The ancients determined this law experimentally ; pupils in the elementary schools must be content to do likewise. ILL. T. " Let us discover the relation of the hypotenuse of a right angle triangle to the other two sides. "Draw two lines at right angles ; lay off 4 equal spaces on one and 3 equal spaces on the other ; find how many spaces there are oil the hypotenuse by measurement. " Do the same, making 12 and 5 equal parts on the lines. " Do you find a common relation between the hypotenuse and the other sides? 3 2 + 4 2 = 5 2 ; 5 2 + 12 2 = 13 2 . What seems to be the relation of the hypotenuse of a right-angle triangle to the other sides ? 176. Exercises. 1. Teach horizontal, vertical, and oblique lines. 2. Draw an oblique triangle and indicate its three altitudes by dotted lines. 3. Construct a 3-in. cube of pasteboard. 4. Construct a regular tetrahedron of pasteboard. 5. Develop the rule for convex surface of a pyramid. 6. Teach the rule for the volume of a rec- tangular prism. LESSON 25. INVOLUTION, EVOLUTION AND LOGA- RITION 177. Needs. A product may be separated into a num- ber of equal factors. Thus, 8 = 2 x 2 x 2, or 8 = 2 3 . Let us classify the needs which arise from this state- ment by the omission of each term in succession ( 7). If the product is wanting, the requirement becomes No. 1 and gives rise to involution ; if the equal factor is want- ing, the requirement becomes No. 2 and gives rise to evo- lution ; if the number of times the equal factor occurs is wanting, the requirement becomes No. 3 and gives rise to logarition.* 1. What = 2 3 ? 2. 8 = what 8 ? or What = v/8? 3. 8 = 2 vhat ? or What = log 8 2 ? 178. Involution. Involution is the process of finding the product from the equal factor and the number of times the factor occurs. The product is the power, the equal factor is the base, the number of times the factor occurs is the index of the power or the exponent. ILL. T. " There is a short way of expressing that the same num- ber is used several times as a factor. " Write 8 = 2x2x2. The number of times 2 is used as a factor is written over and a little to the right of 2. Thus, 8 = 2 8 . 2 is called the base and 3 is called the exponent. 2 8 is read ' 2 to the 3d power ' or ' the cube of 2.' What does 3' 2 mean ? It is read ' 3 to the 2d power ' or ' the square of 3.' How is 2 6 read ? What does it mean? What is its value ? What is 2 called? What is 5 called? * A word suggested for the phrase, the process of finding logarithms. 125 126 LESSON 25. INVOLUTION AND EVOLUTION 179 " The process of finding the product of equal numbers is called invo- lution ; the word itself means rolled up. The product is called the power. In what way is involution performed ? By multiplication." Use. The principal use of involution is to afford an abbreviated form of expression. If a number is to be raised to a high power, the law for multiplying when the bases are the same may be used to advantage. ILL. The amount of $1 for 20 yr. at 6% compound interest is <$(1.06) 20 . To make 20 separate multiplications would be a long process. The result can be found by 5 multiplications. Thus, 1.06 x 1.06 = 1.1236 or (1.06) 2 ; 1.1236 2 = 1.2625 or (1.06) 4 ; 1.2625 2 = = 1.5938 or (1.06) 8 ; 1.5938 2 = 2.5404 or (1.06) 16 ; 2.5404 x 1.2625 = 3.2071 or (1.06)i 6 x (1.06) 4 or (1.06) 20 . 179. Evolution. Evolution is the process of rinding the equal factor from the product and the number of times the factor occurs. The product is not named, the equal factor is the root, the number of times the factor occurs is the index of the root. ILL. T. "It is sometimes necessary to find one of the equal fac- tors of a number. Can you give me an illustration ? If the cube of a number is 8, what is the number ? Yes, or if the volume of a cube is 8 cu. in., what is its edge? "Find the number whose oth power is 32. How did you get 2, Henry? ' I found the prime factors of 32. 32 = 2 6 , or 2 is one of the 5 equal factors of 32.' Excellent. " It is convenient to give names to the process and to the terms. The process is called evolution ; the word itself means unrolled. The equal factor sought is called the root ; why is root a good name for the term? The number of equal factors is called the index of the root. To find the number whose 5th power is 32 is to find the 5th root of 32.' The < fifth root of 32 ' is written >/32. The sign, ^, is a modification of r, the initial of root. When the index is 2 it is not written." The Process for Pupils. A root which can be exactly expressed by the decimal notation can be found by factor- 179 LESSON 25. INVOLUTION AND EVOLUTION 127 ing. Every root can be found by trial. It is recom- mended that the extraction of roots be limited in elemen- tary schools to these two methods. ILL. By Factoring. T. " Find the number whose square is 576 or find the square root of 576. By factoring, we find 576 = 2 2 x 2 2 x 2 2 x 3 2 = 24 2 ; V576 = 24. Find ^9261." ILL. By Trial. T. " Find V231. 231 = 3 x 7 x 11 ; V231 cannot be found by factoring. What shall we do ? " We will find by trial two numbers differing by 1 unit, such that the square of one shall be less than 231 and the square of the other greater than 231. The smaller will be V23~I true to units. 15 2 = 225 ; 16 2 = 256 ; V231 = 15 +. " We will find by trial two numbers each of which is ' 15 + ' differ- ing by 1 tenth, such that the square of one shall be less than 231 and the square of the other greater than 231. The smaller will be \/231 true to tenths. 15.1 2 = 228.01 ; 15.2 2 = 231.04 ; V231 = 15.1 +. " In a similar way we can find the answer true to any required place. Can the answer be expressed exactly by the decimal notation ? " Find V2T5 to one decimal place." 3 9261 1.3 1.3 1.69 1.3 507 169 1.4 1.4 1.96 1.4 784 196 3 3087 3 1029 7|343 7[49 7 2.JL97 3/ 2.744 5 = 1.3+ The Process for Teachers. To extract the nth root of an integer, point off into periods of n figures each begin- ning with units' place, extract the root of the number de- noted by the first two periods, then the root of the number denoted by the first three periods, and so on. As a guide, raise a + b to the wth power and proceed as the formula indicates. 128 LESSON 25. INVOLUTION AND EVOLUTION 179 ILL. Extract the cube root of 1860867. (a + b) 3 = a 3 + 3 a 2 b + 3 ab* + b s = a 3 + (3 a? + 3 ab + b*)b 1'860'867|123 300 60 364 43200 1080 44289 860 728 132867 132867 a = 10 a = 120 b = S Preparation. As a guide we raise a '+ b to the 3d power and factor. We separate the number into periods of three figures each because the cube root of the number denoted by the first period will give the first figure of the root, the cube root of the number denoted by the first two periods will give the first two figures of the root, and so on. (The teacher should satisfy himself of these facts inductively.) First Extraction. We extract the cube root of 1'860 to obtain the first two figures of the root. If we extract the cube root of a 3 , we get a, the first term of the root in the guide; hence, if we extract the cube root of 1, we must get the first figure of the root in this example ; the cube root of 1 is 1, or the first figure is 1 ; a = 10 because 1 is not 1 unit but 1 thousand. Subtracting the value of a 8 , we get 860 which equals 3 a?b + 3 6 2 + b 8 . If we divide 3 a 2 b by 3 a 2 , we get Z>, the second term of the root in the guide ; hence, if we divide what corresponds to 3 a 2 b, or the greater part of 860, by what corresponds to 3 a 2 , we must get the second figure of the root in this example ; 3 a 2 = 300 ; 860 -*- 300 = 2, or b = 2. If we multiply what is within the parenthesis by b, we get the rest of the power in the guide ; hence, if we multiply what corresponds to what is within the parenthesis by what corresponds^ to b, we should get the rest of the power in this example if it is a perfect power ; 3 a 2 = 300 ; 3 ab = 60 ; 6 2 = 4 ; the parenthesis = 364 ; multiplying by b, or 2, and subtracting, we get 132. Hence, the cube root of 1'860 is 12+, and the remainder found by subtracting the cube of 12 from 1860 is 132. LESSON 25. LOGARITION 129 Second Extraction. We extract the cube root of 1860'867 to get the first three figures of the root; we may regard the whole as 1860 thousands 867 units. We know that the new a is 120 because 1860 is not 1860 units but 1860 thousands, and that the remainder after subtracting the cube of 120 from 1860867 is 132867 which = 3 a*b + 3 aW + b s . Hence, we start in at once to find the new b. If we divide 3 a 2 ft by 3 a 2 , we get b, the second term of the root in the guide ; hence, if we divide what corresponds to 3 a 2 b, or the greater part of 132867, by what corresponds to 3 a 2 , we must get the next figure of the root in this example ; 3 a 2 = 43200 ; and so on. The Process for Teachers. To extract the root of a fraction extract the root of the numerator and then the root of the denominator. To extract the root of a deci- mal, point off into periods beginning with the decimal point and proceed as with integers, pointing off as many decimal places in the root as there are decimal periods in the decimal. ILL. The square root of ff = ; the square root of f = V.66'66+. The square root of .00365 = V.00'36'50. The teacher should discover why the decimal is pointed off from the left rather than from the right. Use. The principal use of evolution in arithmetic is in connection with a few problems in mensuration involving right-angle triangles, areas, and volumes. 180. Logarition. Logarition is the process of finding the number of times the equal factor occurs from the product and the equal factor. The product is the antilog- arithm, the number of times the equal factor occurs is the logarithm, and the equal factor is the base. ILL. How many times must 2 be used as a factor to make 8 ? Or, to what power must 2 be raised to make 8? This question is written, what is log 8 2 ? It is read, what is the logarithm of 8 in the system whose base is 2? Ans. 3. 130 LESSON 25. LOGARITION 181 Use. Tables have been prepared stating the powers to which 10 must be raised to produce all numbers. By their aid, numbers may be multiplied, divided, raised to powers, and depressed to roots with marvelous speed and ease. Logarithms were discovered by Napier in the seventeenth century. It is said, " The miraculous powers of modern calculation are due to three inventions : the Hindu notation, decimal fractions, and logarithms." 181. Exercises. 1. Extract the square root of 2 true to two deci- mal places by trial. 2. True to 4 decimal places by the formula method. 3. Show inductively that if an integer is pointed off into periods of two figures each, the square root of the first period gives the first figure of the root, the square root of the first two periods gives the first two figures of the root, and so on. 4. Show deduc- tively that in extracting the cube root, a decimal must be pointed off from the left. 5. Show deductively that there are as many decimal places in a root as there are decimal periods in the power. LESSON 26. ALGEBRA IN ARITHMETIC 182. Needs. The needs arise of a briefer notation, and of a simpler method of solving equations than is afforded by arithmetic. How to satisfy these needs as fully as is desirable in elementary schools is discussed in this lesson. See 49. 183. Numbers by Letters. The first letters of the alphabet, a, 6, c, are used for known numbers, and the last letters, #, y, z, for unknown numbers. ILL. " If you take a certain number, multiply by 12, add 10, divide by 2, and subtract 5, the result is 24 ; find the number. If the result is 24 after 5 is subtracted what is the number ? 29. If the result is 29 after a number is divided by 2 what is the number? 58. And so on ; the answer is 4. ' A shorter way is to represent the required number by x. Think, x, 12x, 12x + 10, Qx + 5, 6x, Qx = 24, x = 4." 184. Signs + and . A quality may be considered in two opposite phases, as heat and cold, up and down, east and west. One of these phases is called positive and rep- resented by ' + ' ; the other is called negative and repre- sented by ' . ' ILL. T. " Let us agree that opposites may be represented by the signs, ' + ' and ' ', and called positive and negative. " Name two opposites. To the right and to the left ; they may be represented by ' + ' and ' '. What would + 5 in. mean? 5 in. to the right. - 3 in.? 3 in. to the left. " Have you had any such use of ' + ' and < ' before ? Are not addition and subtraction opposites? Is not one represented by ' +' and the other by ' - ' ? " 131 132 LESSON 26. ALGEBRA IN ARITHMETIC 185 185. Coefficients. A number may be used several times as an addend or several times as a subtrahend. The num- ber of times is called the coefficient. A positive coefficient shows how many times a base is used as an addend ; a negative coefficient shows how many times a base is used as a subtrahend. . ILL. T. " Write by the use of signs that a is to be added 3 times. a + a + a is written + 3 a or 3 a ; when the sign is '+ ' it is usually omitted ; + 3 is called a positive coefficient ; coefficient means working together with. Write that a is to be subtracted 3 times. a a a is written 3 a ; 3 is called a negative coefficient. " What does + 3 a mean ? What does 3 a mean ? What does a positive coefficient show? A negative coefficient? " Have you had any such use before ? Does not 3x4 mean 4 + 4 + 4 or that 4 is to be added 3 times ? +3 may be regarded as the coefficient of 4." 186. Numbers with Direction. The need arises of per- forming the fundamental operations upon numbers with direction. Addition. To add when the signs are alike write the sum and use the common sign ; to add when the signs are unlike write the difference and use the sign of the greater. ILL. T. " We want a rule for adding numbers with signs. + 5 - 5 +5 -5 j-jj -_2 -2 + 2 " Suppose that ' + ' means to the right and < ' to the left. Find the sums of the numbers on the board in this way. I draw a line and take the point A for the starting point. I will add + 5 and 2. + 5 means 5 places to the right, one, two, three, four, five; 2 means 2 places to the left, one, two; my last mark is 3 places to the right of A or + 3 ; the sum of + 5 and 2 is + 3. " What seems to be the rule for addition when the signs are alike ? When the signs are unlike ? " Let us see why this rule holds. With like signs, a number of 186 LESSON 26. ALGEBRA IN ARITHMETIC 133 operations is to be united with a number of the same operations and the sum must denote the same operation. With unlike signs, a num- ber of additions cancels the same number of subtractions and the sign of the result is the sign of the greater coefficient." Subtraction. To subtract change the sign of the sub- trahend and proceed as in addition. It would be very confusing if a minus sign were written before each subtrahend to indicate subtraction. _;* would be a monstrosity. See 74. ILL. T. " We want a rule for subtracting numbers with signs. + 2 -2 -2 +2 +_5 _-5 +_5 -5 " If you go to the south, what do you do to your position ? You subtract from your north position and add to your south position. If you go to the north, what do you do to your position ? You sub- tract from your south position and add to your north position. It seems, then, that to subtract a number with either sign is to add the number with its sign changed. Apply this rule to the numbers on the board. To subtract -f 5 from + 2, change the subtrahend and proceed as in addition. What is the rule for subtraction? " Let us look at this in another way. To subtract + 5 is the opposite of to subtract 5 ; the opposite of to subtract 5 is to add 5 ; hence to subtract + 5 is to add 5." Multiplication. The product of like signs is plus ; the product of unlike signs is minus. ILL. T. " We want a rule for multiplying numbers with signs. " 3 x 4 means that 4 is to be subtracted 3 times. 4 from = + 4 ; - 4 from + 4 = + 8 ; - 4 from + 8=+ 12; -3x-4 = + 12. In the same way, find the value of + 3 x -1-4. Finish the sentence ; the product of numbers with like signs is . " In the same way, find the value of + 3 x 4; of 3 x +4. Finish the sentence; the product of numbers with unlike signs is ." Division. The quotient of like signs is plus ; the quo- tient of unlike signs is minus. 134 LESSON 26. ALGEBRA IN ARITHMETIC 187 ILL. T. " We want a rule for dividing numbers with signs. " Either of two factors is equal to their product divided by the other factor. + 3 x 4 = 12. From this law what must be 12 -.. _ 4 ? _ 12 .j. + 3 ? " - 3 x - 4 = + 12. What must be + 12 -4- - 3? + 12 -4- - 4? + 3 x + 4 = + 12. What must be +12 -* + 3 ? +12 + + 4 ? " Finish the sentences ; the quotient of numbers with like signs is ; the quotient of numbers with unlike signs is ." 187. Removing Parentheses. To remove a parenthesis, multiply every term within it by its coefficient. ILL. T. " A parenthesis about an expression means that every term within it is to be subjected to the same operation. Thus, 5(6z 2) means that Qx 2 is to be multiplied by 5. Remove the parenthe- sis; 5(6x 2) = 30x + 10. What is the rule for removing a parenthesis ? " A bar is sometimes used in place of a parenthesis. Find the value of Qx 2. This means that 6x 2 is to be multiplied by - 1. - Qx- 2 = - Qx + 2." 188. From nx to find x. Given the value of nx to find z, divide both members of the equation by the coefficient of x. ILL. T. "-3z = -12. How shall we find the value of x ? " To find x we must divide 3 x by 3. Dividing both members of an equation by the same number cannot affect the balance. Divide both members of the equation by the coefficient of a:, x = 4." 189. From x" to find x. Given the value of x n to find a;, extract the nili root of both members of the equation. ILL. T. "x 2 = 16. How shall we find the value of a;? " To find x we must extract the square root of x 2 . Extracting the same root of both members of an equation cannot affect the balance. Extract the square root of both members, x 4. x 3 = 27 ; find x." 190. Transposing. To transpose a term from one mem- ber of an equation to the other, change its sign. ILL. T. "2x 3 = 9. It is necessary to transpose 3 to the right-hand member of the equation so that 2 x alone will be in the left-hand member. How shall we do it ? 191 LESSON 26. ALGEBRA IN ARITHMETIC 135 "What must be added to 3 to make zero? + 3. Adding the same number to both sides cannot affect the balance. Adding + 3 to both sides what do you get? 2 x 9 + 3. What is the rule?" 191. Clearing of Fractions. To clear of fractions, mul- tiply both members of an equation by the least common denominator. ILL. T. " 10 = . It is necessary to clear of fractions. "What is the least common denominator? Multiplying both members of an equation by the same number cannot affect the balance. Multiply both members by 6. Say, 2 is contained in 6, 3 times ; 3 times 3 x is 9 x ; 6 times 10 is 60 ; 3 is contained in 6, 2 times ; 2 times 2 x is 4 x. Ans. 9 x 60 = 4 x. What is the rule for clearing of fractions ? "Clear of fractions (I) on the board. The L. C. D. is 5; 5 times 3 x is 15 x ; 5 is contained in 5, 1 time ; 1 times 1 is 1 ; 1 times 6 a: is 6 x ; 1 times 2 is + 2 ; 5 times 4 is 20. .4ns. 15 x 6 x 6 x 2 + 2= 20. I want you to see why we say, ' 1 times 1.' -- means that 6 x 2 is to be multiplied by 1 ( 187) ; the 1 is not written but must be understood. " Clear of fractions (II) on the board. Look out for the ' ' before the fraction. 5 is contained in 10, 2 times ; 2 times 3 is 6 ; 6 times x is 6 x ; 6 times - 4 is -f 24." 192. Discussion. Only enough of algebra should be taught in the elementary schools to give pupils something of an idea of its power and to provide for the indirect cases in percentage and interest and for the formulae of mensuration. The latter, for \TT!) Z and ^TrD 3 , require the handling of equations with the 2d power of x and no other power of x, and equations with the 3d power of x and no other power of x. 136 LESSON 26. ALGEBRA IN ARITHMETIC 193 193. Exercises. 1. By counting as in 186, find the sum of 5 and + 2. 2. If income is constant, savings and spendings are opposites ; what is the effect on savings of subtracting from spendings ? 3. What is the effect on spendings of subtracting from earnings? 4. In equation (II) of 191 find the value of x and explain every step. 6. Solve the equation, \ wD 2 = 78.54, and explain every step (count tr as 3.1416). 6. Solve the equation, $*-Z> 8 = 113.0976, and explain every step. LESSON 27. PERCENTAGE 194. Development. In business a part is usually ex- pressed as a number of the hundred equal portions of a whole. Hundredths is called per cent and the fraction is expressed by aid of the symbol, %. ILL. T. " We are going to consider another way of expressing hundredths. " Write 6 hundredths as a common fraction. T ^. Write small circles in place of the numerator and denominator and write 6 to the left. 6 %. This means that the numerator is 6 and the denominator 100; it is read 6 per cent." 195. Reductions. When the denominator of a fraction is 2, 3, 4, 5, 6, or 8, it is customary to use the common fraction in place of its per cent equivalent. Pupils should find the equivalents and master them so thoroughly that an expression in either form will suggest the other in- stantly without any form of computation. 3 83 J% i 62*% * 75 % f 50% I 16| % 5 I 87*% i 12*% I 33^ % 3 60% i 40 % 4 25 % 1 66f% 2 37i% i 20% fr 80 % ILL. T. " I asked you to bring in on paper the per cent equivalents of the business fractions. Here they are on this chart. As I point call the equivalent instantly (he points at the rate of one a second). You fail on , f, and . Give these special attention ; you must not compute their values." 196. Terms. In the full expression of a part there are three terms, the whole, the fraction of the whole, and the 137 138 LESSON 27. PERCENTAGE 197 part. They are called base, rate, and percentage. The base plus the percentage is the amount ; the base minus the percentage is the difference. ILL. T. "3 = 6 % of 50 ; 53 = 50 + 6 % of 50 ; 47 = 50 - 6 % of 50. 3 is the percentage ; 6 %, the rate ; 50, the base ; 53, the amount ; 47, the difference." 197. Direct Cases. The direct cases are to find the part, the whole plus the part, and the. whole minus the part, from the whole and the fraction of the whole P, A, D from B and R (p. 7). At least 99% of all problems in percentage outside of the schoolroom are of these types. Pupils should work with them for a long time before considering the other types, and until they can find results rapidly and accu- rately both without the pencil and with the pencil. It is best both for mental and for written work to find 1 % and to modify the result. It is worth while in busi- ness problems to be able to call the facts in the following table without stopping to compute them. 1 % of f 100 = $ 1 1 % of $ 10,000 = $ 100 1 % of 1 1000 = $10 1 % of $ 100,000 = $ 1000 1 % of $ 1,000,000 = $ 10,000 Mental. Before taking up written exercises pupils should gain facility in mental computations. It should be remembered that the pencil is required only when the operations are difficult. See 70. Exercises like the following are as valuable in prepara- tion for the solution of problems as examples in abstract multiplication and division. ILL. Fractions. T. "What is f of 30? What is the result when 30 is increased by f of itself ? What is the result when 30 is dimin- ished by of itself? What is the result when a no. is increased 197 LESSON 27. PERCENTAGE 139 by of itself ? - no. What is the result when a no. is diminished by | of itself? f no." ILL. Per Cents. T. What is 6 % of $ 500 ? Find 1 % and multiply by 6; think, 1 % of $500 is $5, and say, 830. What is the result when $500 is increased by 6% of itself? "\Vhat is the result when $500 is diminished by 6% of itself? What is the result when a no. is increased by 6 % of itself? 106% no. What is the result when a no. is diminished by 6 % of itself ? 94 % no." ILL. Equivalents. T. " What is 87 % of $ 48 ? Think, \ of $ 48 is $ 6, and say, $ 42. What is the result when $ 48 is increased by 87^% of itself? What is the result when $48 is diminished by 87 % of itself ? What is the result when a no. is increased by 87 % of itself? no. What is the result when a no. is diminished by 87* % of itself? I no." ILL. Frac. Per Cents. T. ',' You were asked to memorize the facts on the board. Clean the board. What is 1 % of $ 10,000 ? of $ 1,000,000 ? (he drills expecting an answer each second). What is $ % of $ 10,000 ? Think, $ 100, and say, $ 12.50." (Such exercises are valuable because brokerage is usually reckoned as | % of the par value.) Written. Examples of the same nature as the foregoing should be given to afford practice when the computations cannot be readily made without the pencil. Attention is called to the value of finding 1 % as the first step. $2.86 X 50 37 20055 8595 $ 106.005 $286.50 .37 20055 $ 106.005 ILL. To find 37 % of $286.50, we move the decimal point to find 1 %, and then multiply by 37. This practice conforms to the plan in mental work and assists in fixing the place of the decimal point 140 LESSON 27. PERCENTAGE 198 Since 30 x 2 is 60, the answer cannot be so small as $ 10 nor so great as $1060 ; it must be $ 106. As a check, the pupil should be required to fix the point as just suggested, and also by counting the places. 198. Indirect Cases. The indirect cases usually pre- sented in arithmetic are the inverse of the direct cases. DlBEOT P from B & R ; P = BR A from B & R; A = B + BR D from B &. R; D = B - BR INDIRECT R from B & P; R = - B B from P & R ; B - B from A & R ; B = B from D & R ; B- R 1 + R D 1- R ILL. T. " You may copy from the board the cases which you have already had. They are called direct. From the first case write a new problem in which the answer to the first is given in the second and the required term in the second is 6 % ; do the same making the re- quired term 50 ( 46). From the 2d and 3d direct cases, do the same, making the answer the given term and 50 the required term. These four new cases are called indirect." DIRECT What is 6% of 50? .4ns. 3 What is 50 + 6% of 50 ? Am. 53 What is 50 - 6 % of 50 ? A ns. 47 INDIRECT f 3 is what % of 50? J3 is 6% of what? 53 is 6 % more than what? 47 is 6 % less than what ? Solutions in General. The indirect cases can be solved by algebra, by formula, by rule, or by analysis. The 2d and 3d cases can also be solved by proportion, or by variation (Lesson 11). 198 LESSON 27. PERCENTAGE 141 By algebra, the values of the known terms may be sub- stituted in the formula for the direct relation and the equation solved, or the required term may be represented by x and the equation formed by reasoning. See 49. By formula, the required term is made the left-hand member of an equation derived from the direct formula before the substitutions are made. See 50. By rule, the translation of the formula obtained by the last method is stated from memory. See 51. By analysis, the problem is separated into its simple problems. See 36. Each method has its advocates. The author prefers the analysis method, or the 2d algebra method, because by them it is not necessary to decide upon the technical name for each term, a decision which is sometimes quite per- plexing. Thus, in the problem below, by these methods it is not necessary to decide that 47 is the difference. ILL. If a number is diminished by 6 % of itself, the result is 47. What is the number? 1ST ALGEBRA 2n ALGEBRA A 47 D = B-BR No. dec., 47 Let x = no. R, 6% 47 = -.06 B Dec., 6% no. .06 x = dec. Z? Q X>) f 47 = .94 B JJ Q v .94 x = no. dec. ' B 47 50 .94x = 47 5 = 50. .4ns. 9* No., 50. ^ns z = 50 FORMULA RULE D, 47 D=B-BR A 47 Given the differ- ence and the rate tf, 6% J-J ^ = Ij (1 Xt ) ft Q 01 to find the base, 5 ' ? _ j^ Jy B,7 divide the differ- ence by 1 minus l R 5, 50. ;lrw. = ^ = 50 B, 50. Ans. the rate. 47 + .94 = 50 142 LESSON 27. PERCENTAGE 198 Analysis in General. It is convenient to designate the three direct cases as Case 1, and the four indirect cases as Case 2, Case 3, Case 4, and Case 5. ILL. Case 2. T. " 12 is what part of 20 ? Ans. ^ of 20 or .f of 20. " 12 is what % of 20 ? Ans. 12 is ^ of 20 or | of 20 or 60 % of 20." ILL. Case 3. 1. "12 is f of what? Ans. If 12 is f 110., what is no. ? 4. If 4 is I no., what is f no. ? #0. "12 is 3% of what? Ans. If 12 is 3% no., what is 1% no.? 4. If 4 is 1 % no., what is 100 % no. ? ^00." ILL. Cases J+fy 5. T. " What no. increased by | of itself becomes 14? Ans. If a number is increased by f of itself, what does it be- come ? I no. If | no. is 14, etc. "What no. increased by 6 % of itself becomes 53? Ans. If a no. is increased by 6 % of itself, what does it become ? 106 % no. If 106 % no. is 53, what is 1 % no. ? . If 1 % no., is , etc. " What no. decreased by f of itself becomes 21 ? " What no. decreased by 6 % of itself becomes 47 ? " Analysis in Particular. Cases 4 and 5 may be regarded as another way of stating Case 3 and may be treated as Case 3. Pupils should practice changing from one form of expression to the other. ILL. T. " Change to an expression without more or less : 16 is J more than 12. Ans. 16 is f of 12. 8 is | less than 12. Ans. 8 is f of 12. A's money is more than B's. A ns. A's money is f as much as B's. In a mixture of water and vinegar the water is \ more than the vinegar. Ans. In a mixture of water and vinegar the water is f as much as the vinegar. " Change to an expression with more or less : A's weight is f of B's. Ans. A's weight is f less than B's. A's salary this year is f as much as last year. Ans. A's salary this year is \ more than last year. " Solve as Case 3. If a no. is increased by 6 % of itself the result is 53. What is the no. ? Ans. If 106 % of a no. is 53, what is 1 % no. ? etc. In a mixture of 14 gal. of water and vinegar the water is \ more than the vinegar. How much is the vinegar? Ans. If the water is f as much as the vinegar, how much is the mixture ? | as much as the vinegar. If | as much as the vinegar is 14 gal., etc." 199 LESSON 27. PERCENTAGE 143 Analysis Written Problems. A problem can be stated in full from the proper expression of what is given and what is required. CASE 2. Had 210 sheep and sold 54. Sold what % flock ? F, 210 sh Sold, 84 sh Sold, ?%F Jft = H Sold, 40 %F ,4ns. = 40% CASE 3. Sold 84 sheep which was 40 % flock. Flock what ? Sold, 84 sh or | F F,? 42 210 1 F, 42 sh F, 210 sh Ans. CASE 4. Bought 40% flock ; theu had 294 sheep. Flock what ? Aft. P, 294 sh or F F,? 42 210 t F, 42 sh F, 210 sh .4ns. CASE 5. Sold 40% flock ; then had 126 sheep. Flock what ? Aft. S, 126 sh or f F F,? 42 210 t F, 42 sh F, 210 sh ^4ns. Case 2. If he had 210 sheep and sold 84 sheep, what per cent of his flock did he sell ? ^ or ^ or | or 40%. Case 4. If he purchased as many sheep as were in the flock, how many sheep did he then have ? ? flock. If \ flock is 294 sheep, etc. 199. Exercises. 1. Try the reductions of 195. Can you call one a second? 2. Can you name the results in 197 in 5 seconds? 3. In 198 show how each of the indirect formulas is derived from its direct form. 4. State the problem of Case 3, p. 143. 6. Solve it by the 1st algebra method. 6. Solve it by the 2d algebra method. 7. Solve it by the formula method. 8. Solve it by rule. 9. State with reasons which method you prefer. 10. Which of the five cases should receive chief attention ? Why ? 11. How many cases are there in percentage ? See p. 7. 12. State a problem for every case. LESSON 28. PERCENTAGE 200. Important Suggestion. In solutions, what the per cent is of should be expressed after every per cent. It is usually omitted in the statement of a problem, and if it is omitted in the solution also, clear thinking is impossible. 201. Profit and Loss. Unless otherwise specified gain or loss is reckoned as some per cent of the cost. ILL. Development. T. " Shall we regard gain and loss as some per cent of the cost or some per cent of the selling price ? As some per cent of the cost because the cost comes first ; a man cannot sell a thing until he gets it. A merchant sells goods at a gain of 40%. Supply the omission. Ans. A merchant sells goods at a gain of 40% of the cost." ILL. Direct Cases. T. " Goods that cost 50 are sold at a gain of 40%. What is the gain? the selling price? Ans. If the cost is 50^ and the gain is 40 % of the cost, what is the gain ? 20 p. Be sure to use after 40 % what 40 % is of. " Butter that costs 24 f> is sold at a loss of 12 %. What is the loss ? the selling price ? " ILL. Indirect Cases. T. " Make the statement without the use of the words, gain or loss. An article is sold for 60^ at a gain of 33|%. Ans. An article is sold for 60^ or for $ cost. An article is sold for 40^ at a loss of 16f%. Ans. An article is sold for 40^ or for f cost. " A book that cost $ 6 was sold for $ 4. What per cent was lost ? " On goods sold at a profit of 66f % the gain was $24. What was the selling price ? A ns. If f cost was $ 24, etc. " Goods are sold for $2.50 at a gain of 25%. What is the cost? " A coat was sold for $ 15 at a loss of 16|%. What was the loss in dollars ? " Written Problems. Be sure to write after each per cent expression what it is of. 144 201 LESSON 28. PERCENTAGE 145 CASE 1. A man bought goods for $ 528.60 and sold them at a gain of 6 %. What was the gain ? C, $ 528.60 G, 6%C G,? 5.28x60 6 G, $ 31.72 Ans. 31.716 CASE 2. A man bought goods for $528.(>0 and sold them at a gain of $31.716. What % did he gain? C, $528.60 G, $31.716 .06 G, ? % C 528x6.)31*7.16 31 716 G, 6% C Ans. CASE 3. A man sold goods at a gain of 6% or at a gain of -$31.716. What was the cost ? G, $31.716 or6% C C,? 6)31.716 C, $528.60 Ans. CASE 4. A man sold goods for $560.316 at a gain of 6%. What was the cost ? S, $560.316 or 106% C C > ' 5.286 106)560.316 C, $528.60 Ans. 530 EXPL. Case4- If 106% cost is $560.316, what is 1% cost? $5.286. If 1 % cost is $ 5.286, what is the cost ? $ 528.60. Difficult Problems. Problems involving two or more bases are often given in examinations of teachers. They should not be taught in the elementary schools. The secret of success in their solution is to state what the per cent is of after each per cent. ILL. 1. At what per cent must a merchant mark goods so that he can make a discount of 10 % and yet make a gain of 26%? As Case 4 and Case 5 this means, At what per cent of the cost must a merchant mark goods so that he can make a discount of 10% of the marked price and yet make a gain of 26 % of the cost ? As Case 3 this means, At what per cent of the cost must a mer- chant mark goods so that he can sell them for 90% of the marked price or for 126 % of the cost ? If 90% marked price is 126% cost, what is 1% marked price? etc. ILL. 2. A man sold two articles at the same price. On one he gained 20% and on the other he lost 20%. What per cent did he gain or lose on the whole? 146 LESSON 28. PERCENTAGE 202 This means, On the first he gained 20 % cost of first and on the second he lost 20% cost of second. What per cent of the cost of both did he gain or lose ? As Case 3 it means, He sold the first for f cost of first and the second for f cost of second, etc. The cost of each, the cost of both, the selling price of each, and the selling price of both must be expressed by the same unit. If SP is f C p what is C^ f SP. If SP is $ C 2 , what is C 2 ? SP. What is cost of both ? f f SP. What is selling price of both ? f f SP. What is the loss on both? ^ SP., etc. 202. Commission. When an agent buys, his commission is some per cent of his buying price ; when he sells, his commission is some per cent of his selling price ; when he collects, his commission is some per cent of his collection. ILL. T. " A person may engage an agent to transact business. In what capacities may the agent serve ? He may buy, or sell, or collect. " Give me an illustration of an agent buying. A grain dealer in N. Y. City employs an agent in Chicago to purchase 1000 bu. of wheat. The agent's commission is 1 %. What is the commission 1 % of ? 1 % of what the agent pays for the wheat. Why ? " Give me an illustration of an agent selling. A farmer sends 100 bbl. of apples to a merchant in N. Y. City, directing him to sell them. The agent's commission is 10%. 10% of what? Of what the mer- chant sells them for. Why? " Give me an illustration of an agent collecting. An owner of an apartment house employs an agent to collect the rents. The agent's commission is 2| %. 2| % of what ? Of what he collects. Why ? " Difficult Problems. Problems in which the commissions are of different bases are often given in examinations of teachers. They should not be taught in the elementary schools. The secret of success in their solution is to state what the commission is of after every per cent. ILL. An agent sells goods for $4800, charging 3% commission. After paying $25 charges he invests the balance in raw material, re- taining a commission of 2 %. How much does the agent pay for raw material ? 203 LESSON 28. PERCENTAGE 147 Agt'sSP, $4800 48.00* 102^)4619 1st com, 3J% Agt's SP _3 Ch, $25 12 45.0634 2d com, 2 % Agt's C 144 205)9238. Agt's C, ? 156 820 4800 1038 1st com, $ 156 4644 1025 Rem, $4644 etc. Af. ch, $ 4619 or 102^ % Agt's C Agt's C, $ 4506.34 Ans. EXPL. If the agt's SP is $4800 and the 1st com. is 3J% of agt's SP, what is the 1st com.? If the agt's SP is $4800 and the com. $156, what is the rern.? If the rem. is $4644 and the charges $25, what is left after the charges V If the 2nd com. is 2\ % agt's C, what is the amount with the com.? If 102 % agt's C is $4619, etc. 203. Commercial Discount. The net price is the con- tinued product of the list price and the remainders found by subtracting each discount from 100%. ILL. Development. T. " Manufacturers and wholesale dealers usu- ally publish a catalogue of their goods with a fixed price after each article. Here is such a list (he passes it around). For the sake of making quick sales or to meet the market, how can they change these prices? It would be very expensive to publish a new catalogue. They offer a single discount or several successive discounts from the list price. " The first discount is some per cent of the list price, the second is some per cent of the 1st remainder, the third is some per cent of the 2d remainder, and so on. " What is the net price with a single discount of 25%? Ans. 75% L ; it is the list minus 25 % list. " What is the net price with two discounts of 25 % and 20 % ? A ns. 60% L. The rem. after the 1st dis. is 75% L; the rem. after the 2d dis. is ' 80 % or $ of 75 % L ' or 60 % L. " What is the net price with three discounts of 25%, 20% and 10%? ,4ns. 54% L. The rem. after the 2d dis. is 60% L; the rem. after the 3d dis. is ' 90% or T 9 5 of 60% L,' or 54 % L." 148 LESSON 28. PERCENTAGE 204 ILL. Direct Cases. T. " What is the net price of a bill of goods listed at $100 with two discounts of 10% and 5%? Ans. The net price is 95 % of 90 % L or 85.5 % L ; 85.5 % of $ 100 is $ 85.50. "Work it in another way. Ans. The 1st dis. is 10% of $100 or $ 10 ; the 1st rem. is $ 100- $ 10 or $ 90 ; the 2nd dis. is 5 % of $ 90 or $4.50; the 2d rem. is $90 -$4.50 or $85.50. " What is the difference in the net price between three discounts of 20 %, 10 %, and 5 %, and three discounts of 5 %, 10 %, and 20 % ? A ns. One is 80 % of 90 % of 95 % L, and the other is 95 % of 90 % of 80 % L. The product is the same in whatever order the multiplication is performed. There is no difference." 204. Stocks. Development. Several persons may form a company to do business as a single individual, and may become incorporated by obtaining a legal charter from the secretary of state. Such a corporation is a stock company, their holdings is stock, the equal parts into which the stock is divided are shares, and the paper show- ing how many shares have been sold at 'one time is a certificate of stock. ILL. July 1, 1911, several individuals living in N". Y. City obtained a charter under the name of the Hygiea Ice Co. for the production of artificial ice. After electing John Smith president and Henry Brown treasurer, they printed 2000 certificates of stock and bound them into books. CKR. No. CER. No. NEW YORK CITY, N.Y. No. STT. -, 19 No. STT. c/rlLQs i& to &L *i///V Mat. DATE V / isjs vfl& OU^Ti&l/ o-i aAa/ ^ea/ ot NAME / 'uq^&d* c?&& (tLo-.f of/LiA 1J !t/u 0si/h1,&Q' ' / 0^i/i/ w^Lu/E' ol- / a-^/., $/00. PRES. TBEAS. 204 LESSON 28. PERCENTAGE 149 They sold all of the stock at $ 80 a share, and thus obtained a cap- ital of $80,000 for the development of the business. A certificate duly made out and signed is given to each pur- chaser, and the stub duly filled in is retained by the company. ILL. Oct. 15, 191i, the company sold 50 shares to George W. Williams, Yonkers, N. Y., and issued to him certificate No. 100. Write the certificate and the stub and detach the stub. If the company is prosperous, it issues dividends at equal intervals, as annually or semiannually, declared as some per cent of the par value, that is, some per cent of the value printed on the certificate. ILL. During the first year the company made $ 10,000 clear of all expenses. They retained $2000, and paid out $8000 in dividends. What per cent of the par value was the dividend? What was the dividend in dollars on 1 share? How much was Williams' dividend? What effect on the market value of the stock should be expected from this large dividend ? If the owner of a certificate of stock desires to sell, he may offer it to different individuals, but usually he takes it to a dealer in stocks (broker) and pays a commission (brokerage) for making the sale. ILL. Oct. 15, 1912, Williams decided to sell his stock. He wrote on the back of his certificate, " Transfer to the order of George W. Williams," and gave it to a broker, who sold it to Henry Wilson for $102 a share. The broker retained $ per share for brokerage and 2 ^ a share for state tax, and sent Williams the bal- ance. He wrote the name of Henry Wilson on the line left for the name of the purchaser, and sent the stock to the home office, where a new certificate was issued and sent to Henry Wilson. Make the transfer indorsement on the certificate. How much did Williams make on his investment ? Consider cost, dividend, selling price, brokerage, and state tax. Preferred and Common. Sometimes a company issues stock of two kinds, preferred and common. The pre- 150 LESSON 28. PERCENTAGE 204 ferred states that a specified dividend will be paid at fixed intervals if the earnings of the company warrant ; the common states that no dividends will be paid until all dividends of the preferred have been satisfied. The pre- ferred is the more secure, but the common often pays the larger dividends. Forms of Expression. The business terms used in stocks are much abbreviated, and are always some per cent of the par value. For clear thinking, pupils should be required to use the unabbreviated forms, and to express each term as dollars a share. BUSINESS FORM UNABBREVIATED FORM A 6 % dividend. A dividend of $6 a share. 5 % stock. Stock that pays an annual divi- dend of <$ 5 a share. Stock @ 90. Stock @ $ 90 a share. Stock @ 90 %. Stock @ $ 90 a share. Brokerage |. Brokerage $ \ a share. Brokerage \ %. Brokerage $ \ a share. Bought at a premium of 20 %. Bought @ $ 120 a share. Sold at a discount of 20 %. Sold @ $ 80 a share. $ 6000 stock. 60 shares of stock. 6 % stock yields an income of 5 %. Stock pays a dividend of $ 6 a share or 5% of the cost of a share. Written Problems. In the statement of what is given and what is required it is best to use the unabbreviated forms. III. 4. If the first cost of 1 sh. is $89f and the br. is $$, what is the entire cost of 1 sh.? 1 90. If the inc. on 1 sh. is $6, on how many shares is the inc. $ 540 ? 90. If the cost of 1 sh. is $ 90, what is the cost of 90 sh. ? $ 8100. III. 6. If 5% cost is $8, what is 1% cost? $1.60. If 1% cost is $1.60, what is the cost? $ 160. 205 LESSON 28. PERCENTAGE 151 ILL. 1. What is the cost of $4800 stock at 110, brokerage i? Mvl sh, $110 Brlsh,$J 110 , No. sh, 48 48 C, ? ~fi~ 880 C 1 sh 110i C, $5286 ILL. 2. How many sh. at 20 % dis. can be purchased for $5128, br. S ? Mvlsh, $80 Brlsh, 9| C, $5128 No. sh, ? C 1 sh, $80 Sh, 64 Ans. 80^)5128 64 641)41024 3846 2564 2564 ILL. 3. What income will be ob- tained from $1600 invested in 5% stock @ 79J, br. J % ? Inv, $1600 Mv 1 sh, $79$ Br 1 sh, 9| D 1 sh, $5 Td, ?_ No. sh, 20 Td, $100 Ans. ILL. 4. How much must be inv. in 6 % stock @ 89|, br. |, to get an annual inc. of $540? C 1 sh, D 1 sh, $ Td, $540 Inv, ? No. sh, 90 Inv, $8100 Ans. ILL. 5. What percent income will I receive if I buy 6 % stock @ 50? C 1 sh, $50 D 1 sh, $6 I), ? % C 10 v D, 12 %C Ans. ILL. 6. How much must I pay for 8 % stock to make 5 % ? Dl sh, $8 or 5%C C 1 sh, ? C 1 sh, $160 Ans. 205. Exercises. 1. Explain the problems above. 2. Explain how each full form on p. 150 is derived from the business form. 3. Translate the statement of each problem on p. 151 from the busi- ness form to the full form. Thus, ILL. 1. What is the cost of 48 shares of stock at $110 a share, brokerage $ J a share? 4. Consider the par value $50 a share and change every business form on p. 150 to its full form. Thus, " A 6 % dividend " is A dividend of '$ 3 a share." LESSON 29. INTEREST 206. Negotiable Notes. Without Interest. In order to facilitate the transaction of business one person frequently gives to another a written promise to pay a given sum of money at a given time and place. Pupils should write and memorize the form of such an agreement ; they should be careful not to sign their names to business papers pre- pared in school. ILL. T. " July 1, 1912, Henry Jones buys a bill of goods of John Smith for $1100. Instead of paying cash he gives a written promise called a note, agreeing to pay the bill 3 mos. after date at the Tenth National Bank. In order that this note may be negotiable Jones makes it payable to the order of John Smith. $//00.00 N. Y. CITY, N. Y., fa /, 19 c5%-t*5' wu>n i&& to a, to- at B&K&. ofo. I "Copy this note on a piece of paper 7" x 3", and mark it No. 1 ; memorize the form. Explain the features of the note. The note must state the date and place where it is drawn ; the date and place where it is to be paid ; that value has been received ; and by whom, to whom, and how much is to be paid. "What are the technical terms? Henry Jones is the drawer or maker or payer ; John Smith is the drawee or payee ; the date of pay- 152 206 LESSON 29. INTEREST 153 ment is the date of maturity ; the sum for which the note is drawn is the face; the amount to be paid at the date of maturity is the amount at maturity. " What is the date of maturity ? Oct. 1. If the note had read ' 90 days after date,' what would have been the date of maturity ? July 91 or Aug. 60 or Sept. 29 ( 164). What is the amount at maturity? The face or $1100." With Interest. If a note without interest is not paid when it becomes due, interest begins with the date of maturity. Sometimes, however, the maker agrees to pay interest from the date of the note. In this case, with in- terest or with interest at a specified rate is added to the note. ILL. T. " By the sale of these goods Jones expected to make 40% of their cost or $440. He was willing, therefore, to pay something, interest, for the use of the money. Suppose he wrote the note for 3 mo. with interest at 4%. Write the note and mark it No. 2. $/fOO.OO N. Y. CITY, N. Y., fufy /, 19 12 a-it&i, to- a- to at the- Bank, with i/nt&ieat at ty- %. cA"o. 2 /"f&nvy "What is the date of maturity? Oct. 1. What is the period for which $1100 is to bear interest? Your answer, 3 mos., is contrary to business usage. The interest period is the actual no. of days from the date of the note, July 1, to its maturity, Oct. 1, or 92 days. See 164. "If 'with interest' had been written instead of 'with interest at 4% ' what would have been the rate? The legal rate of the state of New York, or 6%." 154 LESSON 29. INTEREST 207 207. Interest Computed. The cancellation method is a favorite with teachers because it is easily taught. It should be made the basis because it leads at once to valu- able modifications. Cancellation Method. No rule is necessary. The pupil finds the simple problems and their answers. See 39. ILL. T. "Look at Note No. 2. We must find the interest of $1100 for 92 days at 4%. Before solving this problem we will solve a few others. "Find the interest of $204 for 1 yr. 5 mo. 17 da. at 7%. There are many ways of proceeding. We will find the interest for 1 da. and then multiply by the no. of days. How many days shall we count as a year? In a year there are 674 sec. less than 365 \ da. (p. 112), but un- less otherwise specified 360 da. are counted to the year in computing interest. How many days are there in 1 yr. 5 mo. 17 da. ? 527. 204 x T fa x ,i T x 527 "Multiplying $204 by T ^ gives the interest for 1 yr. at 7% ; divid- ing by 360, for 1 da. ; multiplying by 527, for 527 da. Finish the work; in cancelling, never divide 100 by a common factor. Why not ? The interest is $20.90." Six Per Cent by Days. To find the interest at 6%, move the decimal point of the principal 3 places to the left, multiply by the number of days arid divide by 6. Modify the result for a different rate. ILL. T. " Let us find a rule for computing interest at 6%. Using the cancellation method, find the interest of P dollars for D days at 6%. P x^x-g^xD = .001 P x D x i " Who can give me the rule ? By this rule what is the interest of $204 for 1 yr. 5 mo. 17 da. at 6% ? $17.918. At 7%? $20.90. 17.918 6 2.986 1 20.904 7 207 LESSON 29. INTEREST 155 "What is the best way of finding interest at 7% from interest at 6%? 7 is 1 more than 6; divide by 6 and add. In dividing 17.918 by 6 the exact quotient is 2.986 J. Is it necessary to write ? Why not? In case of 2.986f what would we have done? Written 7 in place of 6. Why? " Give me the complete rule." Six Per Cent Basis SI. Find the interest of 81 for the given time at 6 % and multiply by the number of dollars. After the multiplication, modify the result for a different rate. To find the interest of $1, count the interest for 1 yr. 6^; for 1 month, ^ff ; for 1 day, ^ m. ILL. T. " Let us compute interest by first finding the interest of $1 at 6%. "What is the interest of $1 for 1 yr. at 6%? Gf<. For 1 mo.? \t (^ of 6f). For 1 da.? \ of a mill (-& of 5 m.). Memorize these facts. " Find the interest of $ 1 for 1 yr. 5 mo. 17 da. at 6%. See next page. " The interest of f 1 for 1 yr. at 6 % is $ .06 ; for 5 mo., $ .025 ; for 17 da., $.002f ; for the whole time, $ .087$. "If the interest of $1 is $.087, what is the interest of $204? $ 17.918. If the interest at 6% is f 17.918, what is the interest at 7 % ? $20.90." Aliquot Part Method. Find the interest for a year at the given rate and modify the result for the years, the months and the days. ILL. T. " Let us compute interest by first finding the interest for 1 year. " What is the interest of $204 for 1 yr. at 7%?" 14.28 1 yr. 1.19 1 mo. 5.95 5 mo. .039| 1 da. .674 17 da. 20.904 156 LESSON 29. INTEREST 207 Bankers' Method. To find the interest for 60 days at 6 %, move the decimal point of the principal two places to the left. Modify the result for a different time or rate. ILL. T. "Notes are usually drawn for 30 da., 60 da., 90 da., or 120 da. Let us find a rule for computing interest for 60 da. at 6%. P x rfo x T*T x 60 = .01 P " Who can give me the rule? At 6% what is the interest of $225 for 60 da.? $2.25, Of $ 250 for 90 da. ? $3.75 ($2.50, $1.25, $3.75). Of $300 for 30 da. ? $1.50 ($3, $1.50)." CANCELLATION P, $204 T, 1 yr. 5 mo. ggQ 17 da. 150 R,7% J7 I, ? 527 204x T JoX 3 J B X527 I, $20.90 Ans. 6 % BY DAYS P, $204 T, 1 yr. 5 mo. 17 da, T? 7 o/ **l ' 10 .204 X 527 107.508 17.918 6 2.986 1 20.904 7 I, $20.90 Ans. 6% BASIS $1 P, $204 .06 T, 1 yr. 5 mo. .025 17 da. ^02| B,-7% T ? 17.918 6 2.986 1 I, $20.90 Ans. 20.904 7 ALIQUOT PARTS P, $204 T, 1 yr. 5 mo. 14.28 1 yr. 17 da. L19 J mo - T> 7 o/ 5.95 5 mo. J% 7 /0 .0391 1 da. * ' .674 17 da. 20.904 ).90 Ans. BANKERS' METHOD P, $287.25 T, 92 da. R,6% I,? I, $4.40 Ans. 2.87x25 60 1.43 62 30 .0957 2 4.4044 BANKERS' METHOD P, $287.25 T, 120 da. 2 . 87x25 60 K, 4 1 % 5.74 50 120 I, ? 1.43 62 1J 4.30 88 I, $4.31 Ans. 208 LESSON 29. INTEREST 157 Discussion. The methods to be used with vhe class must be decided from the course of study or by the prin- cipal of the school. It is important that pupils shall gain the power to compute interest mentally. The cancellation method alone is not sufficient. ILL. A graduate of the N. Y. T. S. for Teachers could not com- pute the interest of $ 100 for 1 yr. at 5% mentally. Given a pencil, she produced the following : This young woman learned the cancellation method in the elemen- tary school, failed to master any other in the training school, and was helpless without a pencil. 208. Note with Interest, Continued. ILL. T. " Take note No. 2. What was the interest at maturity? The interest of $1100 for 92 da. at 4%, or $11.24. How did you get it, Mary?" M. "The int. of $1100 for 'l yr. at 4% is $44; for 90 da. or \ yr., $11; for 10 da. or of 90 da., $1.22; for 2 da. or \ of 10 da., $.24; for 92 da., $11.24." T. " How much must Henry Jones pay Oct. 1 ? $1111.24. Give me the complete history of this note." 209. Bank Discount. To find bank discount, compute the interest on the amount at maturity for the term of discount at the rate of discount. ILL. T. " Take note No. 2 again. Let us suppose that on Aug. 1 Smith needed money and obtained it by selling this note to the bank or getting it discounted. He wrote his name across its back or in- dorsed it, agreeing thereby to pay it in case Jones should fail to do so, and gave it to the bank. Indorse the note. " How much did the bank pay for the note? On Oct. 1 it will re- ceive $1111.24. On Aug. 1 it paid $ 1111.24 less the interest at 6% from Aug. 1 to Oct. 1 ; it simply deducted the interest in advance or the bank discount. Bank discount is interest on the amount at maturity for the term of discount at the rate of discount. Memorize this statement. 158 LESSON 29. INTEREST 210 "We must find the interest of $1111.24 for 61 da. at 6%; it is $11.30. What is the amount at maturity less the bank discount or the proceeds or what Smith will receive? "Complete the history of the note. On Oct. 1, Jones pays the bank $1111.24, the bank writes, 'Paid Oct. 1, 1912, The Tenth Na- tional Bank, John Doe, Cashier,' across the face of the note or cancels it, and returns it to Jones. Cancel note No. 2. " Take note No. 1. Suppose that Smith had sold this note to the bank on Aug. 1, how much would he have received for it?" NOTE No. 1 Face, $ 1100 Date of note, July 1, 1912 Date of dis, Aug 1, 1912 R of dis, 6% Note, without int Proceeds, ? Amt at mat, $1100.00 B dis, $11.18 Proceeds, $1088.82 Ans. NOTE No. 2 Face, $1100 Date of note, July 1, 1912 Date of dis, Aug 1, 1912 R of dis, 6 % Note, with int, 4 % Proceeds, ? Amt at mat, $1111.24 Bdis, $11.30 Proceeds, $1099.94 Ans. 210. Exact Interest. Banks of discount sometimes count 360 and sometimes 365 da. to the year, according as it is to their advantage. Thus, they count 360 when they collect interest and 365 when they pay interest. The government counts 365. Unless otherwise directed, count 360. ILL. T. " Sometimes it is necessary to find exact interest or inter- est counting 365 da. to a year. "Find the exact interest of $204 from July 1, 1911, to Dec. 18, 1912, at 7%. Can we count the time as 365 + 150 + 17 days ? No ! In reckoning 365 da. to a year 5 mo. cannot be counted as 150 da. The time is 365 + 170 days or 535 da. Use the cancellation method." 204 x x 535 211 LESSON 29. INTEREST 159 211. Postal Savings System. Teachers should secure from a post office the pamphlet, " Postal Information," should explain its provisions, should give problems based upon it, should exhibit postal savings cards, ten -cent sav- ings stamps, and savings certificates, and should urge pupils to become depositors. ILL. T. " Have any of you deposited money in the postal savings bank at the post office? How old must a person be to open an ac- count ? 10 jr. How much can you deposit at one time ? An exact number of dollars ; but you can buy 10-cent stamps and stick them to a card, and when you have a dollar in stamps you can make a deposit. Here is a postal savings card with 5 stamps. How much is it worth ? " When you make a deposit, you receive a savings certificate which bears interest at 2 % for every full year the money is on deposit, be- ginning with the first day of the month following the one in which it is deposited. A person has as many certificates as he makes de- posits. Here is a postal savings certificate. "Jan 2, 1912, I obtained a savings certificate of $3 ; Oct. 3, 1912, one of $ 4 ; and Apr. 30, 1913, one of <$ 2. How much are these cer- tificates worth to-day, counting interest ? " 212. Exercises. 1. Compute the interest of $511 for 11 mo. 11 da. at 5 % by the cancellation method. 2. By the 6 % method for days. 3. By the 6 % basis <$ 1 method. 4. By the aliquot part method. 6. Find the rule for computing interest at 36 %. 6. By this rule compute the interest in No. 1. 7. State, with reasons, which method you prefer. 8. Discuss the bankers' method. 9. Write note No. 1, making the time 90 days after date, and adding ' with interest.' 10. Find the proceeds of the note July 20, discounting at 6 %. 11. Prove that for less than a year exact interest is the com- mon interest minus of the common interest. LESSON 30. INTEREST 213. Bonds. A note given by a village, town, city, county, state, or nation, secured by faith and credit, and bearing interest, is called a bond. Bonds are usually treated in arithmetics in connection with stocks, but all which they have in common with stocks is the fact that they are usually bought and sold by brokers. Notes given by individuals and corporations secured by mort- gage and bearing interest are also called bonds. ILL. T. " Suppose the city of Yonkers needed $ 500,000 on June 1, 1912, to construct a sewer, and the money in the treasury was in- sufficient, it might issue and sell to the highest bidder 500 notes (bonds) of $ 1000 each, signed by the city officers as provided by law. The bonds in blank might read : CITY OF YONKERS SEWER LOAN, No. _ YONKERS, N. Y., fane, I, 19 IS. $/ 00 0.00 te> o-n fan& / at &a&ti yza-v at -id (mn-k. &ti& fattA, and vucUt o^- the* @-tty o^ to- the* vuwb&nt at tAC& d&kt. .MAYOR _PRE8. OF COUNCIL CITY TREASURER 160 214 LESSON 30. INTEREST 161 " For the convenience of the purchasers, twenty small notes, called coupons (French, cut), might be printed on the same paper, each for f 40, or the payment of one year's interest, and made payable in order, June 1, 1913, June 1, 1914) and so on. In this case the bond would be a coupon bond. The coupons might be arranged as follows : No. 20 No. 19 No. 13 No. IT No. 16 No. 15 No. 14 No. 13 No. 12 No. 11 No. 10 No. 9 No. 8 No. T No. 6 No. 5 No. 4 No. 3 No. 2 No. 1 " Suppose this was a coupon bond. Write coupon No. 1, or the note becoming due June 1, 1913. Explain the procedure of Mr. A., who owns one of these bonds, in getting his interest each year." Bond and Mortgage. A person owning real estate may borrow money by giving to the lender a note called a bond (bound) and a sealed instrument in writing trans- ferring the property to him. This instrument is a deed or mortgage (death-pledge). The lender has this deed re- corded to prevent fraud, but is in no sense the owner of the property. If the borrower fails to fulfill his agree- ment, the property is sold under the direction of the court and the lender is paid. 214. Drafts. If A owes B, B may request A to pay the whole or a part to C. If A honors the request, B may pay C in this way. The paper containing the request is a draft. It may or may not bear interest and may be payable on demand or at a specified time. ILL. T. " George Williams owes the Home Publishing Co. $ 1000 for books. The latter makes a draft on the former for $ 500 in favor of Henry Brown, payable 30 da. after sight. 162 LESSON 30. INTEREST 215 To ^WME, l&LttLa/m&f 300 BROADWAY, N. Y. CITY, 220 BwadwiMf, cA. If. <y. WOA,. 25, iq/3. a/u to tA.& o-'uie.^ o am-cl 100 at want. f PRES. " The draft is sent to Williams, who writes across the face in red ink, ' Accepted, Mar. 26, 1913, George Williams.' It is then sent to Henry Brown, who sells it or collects it when due. Write the draft and its acceptance." Checks. The most common form of draft is a check. A person has money on deposit in bank and pays debts by checks. As a rule, checks are not sent to the bank for acceptance, but sometimes this is necessary when the validity of the check is to be put beyond question. In such an event the check is said to be certified. It be- comes a draft which has been accepted by the bank. A bank officer writes the acceptance in the usual way. ILL. T. " We will suppose that you have money on deposit at the Tenth National Bank, 100 Broadway, N. Y. City. Draw a sight draft on the bank for no dollars to the order of James Thompson. This draft is a check. " Have this check certified. That is, get one of your classmates to accept it as cashier in the name of the bank." 215. Indirect Cases. The direct cases have already been considered ; given the principal time and rate to find the interest, and to find the amount, that is the principal plus the interest. They should receive 90% of the time devoted to interest. 5215 LESSON 30. INTEREST 163 The indirect cases grow out of the direct cases. Six of them are usually considered in the arithmetics. ILL. T. "The interest of $200 for 2 yr. at 6% is $24. We have mastered this case. Find the three others that arise from the omis- sion of each term in succession." 1. In what time will $200 gain $24 at 6%? 2. At what rate will $ 200 gain $ 24 in 2 yr. ? 3. What principal will gain $24 in 2 yr. at 6%? "The amount of $200 for 2 yr. at 6% is $224. Find the three other cases." 4. In what time will $200 amount to $224 at 6%? 5. At what rate will $200 amount to $224 in 2 yr.? 6. What principal will amount to $224 in 2 yr. at 6%? By Analysis. The author prefers this method. The component problems are displayed in the diagrams. See /En ' T \ , I, $24 /P, $200 yr. " /O \R,6% ILL. T. "Consider Case 1. From what can we find the no. of years? From the entire interest and the interest for 1 yr. From what can we find the int. for 1 yr. ? From the principal and the rate." ILL. T. Solve Case 1. What is the interest of $ 200 for 1 yr. at 6 % ? $ 12. If the interest is $ 12 for 1 yr., in how many years will it be $24? 2. 164 LESSON 30. INTEREST 216 " Solve No. 5. If the amount is $ 224 and the principal is $2( what is the interest ? $ 24. What is the interest of $ 200 for 2 yr. at 1 %? $ 4. If the interest is $ 4 at 1 %, at how many per cent will it be $ 24 ? 6." By Algebra. The known terms may be substituted in the direct formulse, or the unknown term may be repre- sented by x. See p. IJ+l. IST ALGEBRA 2u ALGEBKA P, $200 A, $224 I=200x2.X R 224 = 200 + 1 P, $200 A, $224 Let = R 100 T, 2 yr. J=24 T, 2 yr. 200 X 2 x 100 R,? 24 = 200 X 2 X R R,? -T P 24 a m w R,6% Ans. R, 6% A tft O fa, R ^L /*O. J, \J FORMULA ANALYSIS P, $200 1= Px Tx R P, $200 A, $224 A = P + I A, $224 224 T, 2 yr. R,? R- J T, 2 yr. R,? 200 24 200 X .02 = 4 Px T I=A-P R, 6% R A P 24 R, 6% 24 -r 4 = Ans. 6 Ans. Px T 400 216. Different Cases. For the Teacher. Let us find all the different cases in interest from the formulae, 1= P x T x R and A = P + I. See 7. There are two equations with five quantities. To solve these equations three of the quantities must be known. Hence, for every three known terms there will be two cases. "! The combinations of ttavixMn the terms, A, I, P, R, T are A IP, AIR, AIT; APR, APT; ART; IPR, IPT; IRT; PRT. That is, there are 10 times 2, or 20, cases in interest. Write the 20 cases. Two of them are impossible. Why? 217 LESSON 30. INTEREST 165 217. Kinds of Interest. For the Teacher. Let us clas- sify interest with reference to what may bear interest. ILL. What is the interest of $ 100 for 4 yr. @ 6 %? i IST YE. 2D TK. 3D YB. 4TH YR $6 $6 $6 $6 I on P, $24 .36 .36 .36 .36 .36 I on I on P at the end .36 of each year, $2.16 .0216 .0216 .0216 I on all other unpaid .0216 I at the end of each .001296 year, $ .087696 First Conception. The principal alone may bear interest, simple interest ($ 24). Second Conception. In addition to the above, the $ 6 due at the end of the 1st year may bear interest for 3 yr. ; the $ 6 due at the end of the 2d year may bear interest for 2 yr. ; the $ 6 due at the end of the 3rd year may bear interest for 1 yr. That is, the principal may bear interest and the interest on the principal at the end of each year may bear interest, annual interest ($26.16). Third Conception. In addition to the above, the $ .36 due at the end of the 2d year may bear interest for 2 yr. ; each $ .36 due at the end of the 3d year may bear interest for 1 yr. ; the $ .0216 due at the end of the 3d year may bear interest for 1 yr. That is, the prin- cipal may bear interest, the'interest on the principal at the end of each year may bear interest and all other unpaid interest at the end of each year may bear interest, compound interest ($26.247696). 218. Annual Interest. For the Teacher. Annual inter- est is rarely computed. The above development indicates the method of procedure. ILL. What is the annual interest of $ 525.26 for 3 yr. 5 mo. 17 da. 1. I loO.tfS /50. 3|-, f^f, f$. 10. Change to common fractions and ar- range in order of value: .005, .62^, .35, .33^. SIXTH YEAR 233. 6th yr. ist T. Mental. 1. Change f bu. to quarts. 2. Which is greater, ^ or -f^ ? Give the difference. 3. At 4 ^ a pint what will you pay for 10 gal. 2 qt. of milk ? 4. Sold for 8 ^ and lost 2 ^. What was the per cent of loss ? 6. Have 2 quarters 5 dimes 1 nickel. How many apples can I buy at 5 ^ each ? 6. A man sells a carriage for $ 56, which is ^ less than he gets for his horse. What did he receive for the horse ? 7. If of the number of bushels in a bin is 20, how many bushels in the bin ? 8. How many days between March 12 and April 15, 1913? 9. I had $3.20 and spent 25 % of it. How much did I spend ? 10. An article cost $ 8 and sold at a gain of 37| %. What was the selling price ? Written. 1. Reduce .027 of a ton to ounces. 2. If a bin contains 5 bu. 3 pk. 2 qt., how much will 5 bins of the same size contain ? 3. A man was born Dec. 25th, 1852. How 176 SECTION 1. ELEMENTARY SCHOOLS 234 old was he February 12th, 1881 ? 4. I had $ 1200. I de- posited 25 % of this sum in the Bowery Savings Bank, and 33 % of the remainder in the Metropolitan Savings Bank. How much cash have I ? 6. Make a bill for the following and receipt it ; date it to-day ; make yourself the purchaser. Bought of A. J. Cammeyer : 10 pairs of men's shoes @ $4.75 ; 4 pairs of boys' shoes @ $ 1.47.V ; 6 pairs of slippers @ .87^ ^ ; 9 pairs of girls' shoes @ $2.43; 8 pairs of women's shoes @ $ 3.371. 6. Paid $ 450 for a horse and sold it at a profit of 15 %. What was the selling price ? 234. 6th yr. 2d T. Mental. 1. It will cost 40 ^ to send a ten-word telegram to Boston. Every additional word will cost 3 t. What must I pay for 18 words ? 2. If 9 yards of ribbon cost $ 2.50, what will 27 yards cost ? 3. A man sold 25 % of his wheat to one buyer, 12^ % to a second, and had 350 bu. left. How many bushels in the crop? 4. A dealer bought a second-hand sofa for $40 and having spent 10 % on repairs, sold it at a gain of 25 % on the whole cost. For what did he sell it ? 6. What is f % of $ 5600 ? 6. A piano was marked $480 but sold at a discount of 10 %. Find the discount. 7. My selling price for some goods is $30 and my gain is 20%. What is my money gain? 8. Lost $3 by selling a desk for $18. What per cent lost? 9. Find the commission at 5 % on 400 bbl. of fish at $ 5 a barrel. 10. My store is worth $ 4000. I insured it for f of its value. Find the premium at % %. Written. 1. A commission merchant sold 10 bbl. of pears for a farmer at So a barrel. His commission was 10 %, and $ 2.50 was charged for freight. What sum did he return to the farmer ? 2. A farmer raised 517 bu. of potatoes. He sold 20 /o of them. How many bushels, pecks, and quarts did he sell ? 3. Find the value of a rectangular piece of land 75 rd. long and 56 rd. wide at $ 144 per acre. 4. You sell a 100-lb. keg of horseshoes for $14.05 and gain 25%. What 235 SECTION 1. ELEMENTARY SCHOOLS 177 do they cost you per pound ? 6. The bread from a bbl. of flour (196 lb.) weighs 31^% more than the flour. What is the weight of the bread ? 6. You pay $ 200 rent a month for your store. How many dollars of sales must you make each month to raise this rent if you average 20 % profit ? 7. Sold a horse for $250 and lost 20%. What would have been the selling price if I had gained 20 % ? 8. Jan. 1, 1913, John Smith bought of Henry Jones a piece of land 100 ft. by 25 ft. and paid for it at 50 ^ a square foot by a check on the Tenth National Bank, 100 Broadway. Draw the check. SEVENTH YEAR 235. yth yr. istT. Mental. 1. The principal is $600; the rate, 3% 5 the time, 2 yr. 6 mo. What is the interest? 2. 25 qt. are 62 % of how many gallons? 3. $17 are 20% of how many dollars ? 4. A grocer received 60 bbl. of flour and sold 12 of them. What % had he left ? 5. I collected $450 and received a commission of 4%. What was the com- mission ? 6. Philadelphia is 75 West longitude. When it is noon in London, what is the time in Philadelphia ? 7. How many dollars in 100? How many centimeters in 3 m 70 mm ? 8. A piano listed at $400 was sold with discounts of 25% and 10%. What was the selling price? 9. 50^ is what per cent of f of a dollar ? 10. By selling a book for $4, I lost $1. Find the per cent lost. Written. 1. On May 3d, 1910, I borrowed $640 at simple interest at 5%. How much do I owe on Sept. 27th, 1913? Count the exact number of days. 2. Change $ 112 to English money. 3.- What sum of money invested at 5% per annum will give an annual income of $1200? 4. Two boys have $48 between them, one having $18 more than the other. How much has each? 6. A milkman sold 86 qt. of milk. How many liters did he sell ? 6. Cloth costing 1200 francs in France was shipped to America where a duty of 24% was 178 SECTION 1. ELEMENTARY SCHOOLS 236 levied. How much was the total cost in our money ? 7. A man starts from Chicago and some days later finds that his watch is 2 hr. 30 min. slow. In what direction has he been traveling and over how many degrees of longitude has he gone ? 8. A recipe for 1^ Ib. of fudge calls for 3 cups of sugar (6 ^ a pound), 1 tablespoon of butter (32 ^ a pound), f of a cup of milk (9 ^ a quart), 2 oz. of chocolate (36 X a pound), and 1 teaspoon of vanilla (10^_1-|- oz. bottle). Find the cost of a pound ; count 31 tablespoons to a pint or pound, 2 cups to a pint, and 3 teaspoons to a tablespoon. 236. 7th yr. 2d T. Mental. 1. Principal is what when rate is 5% ; time is 2 yr. 6 mo.; interest is $12.50? 2. Principal is $ 600 ; time is 3 yr. 4 mo. ; interest is $ 80. What is the rate? 3. Principal is $ 900 ; rate 5%; interest $60. What is the time ? 4. The simple interest on a certain principal is $ 146. What is the exact interest ? 5. If 11 books cost $ 1.32, what will 100 books cost ? 6. 9, + 12, -=- 3, x 12, - 3, -- 9, + 4, - 11 ? 7. x, x 12, +9, -=-3, +9, - 12, the result is 20, find x. 8. The following recipe is for a dozen biscuits. How much of each ingredient must be used to make 15 biscuits ? For a dozen, 2 cups of flour, % cup of milk, ^ teaspoon salt, 21 teaspoons of baking powder, 2 tablespoons of shortening. 9. Find the area of a triangle whose base is 12 in. and whose altitude is 10 in. 10. What is the ad valorem duty on 100 yd. of silk valued at $4 a yard, duty 25%? Written. 1. A merchant had $11,640; he invested 26|% of it in dry goods and 12 % of the remainder in groceries. How much money had he left ? 2. Find the interest of $320 for 1 yr. 8 mo. 27 days at 5%. 3. Find the amount of $ 750 from Jan. 12th, 1903, to Dec. 16th, 1905, at 1%. Count the exact number of days. 4. Multiply six hundred twenty- five thousandths by twenty-five millionths and divide the product by one hundred twenty-five hundred thousandths. 6. John Smith bought $400 worth of goods from Thomas 237 SECTION 1. ELEMENTARY SCHOOLS 179 Brown on 3-raonths' credit. Write a bank note for the trans- action, dating it to-day. 6. Find the proceeds at bank dis- count @ 6% if the note is discounted 15 days after to-day. 7. If 29 bales of hay are consumed by 58 cows, how many bales will last 46 cows for the same time ? 8. Sold a wagon for $420, which is 16% less than the cost. What should I have sold it for to gain 25% ? 9. What number increased by | of itself equals 402 ? 10. Find a : 3x 5 (a? + 3) = 21. EIGHTH TEAR 237. 8th Yr. ist T. Mental. 1. Reduce 3 dm 7m 5 cm to mm. 2. A house is valued at $6000. It is insured for 76 % of the valuation at 20 / per S 100. What is the premium ? 3. 15, + 17, H- 4, square, - 1, X 11 = ? 4. Eeduce 2 T. 5 cwt. to pounds. 6. Sold a wagon for $72, gaining 12i%. What was the cost ? 6. How many pint bottles can be filled by 3 gal. 3 qt. of wine ? 7. What is the perimeter of a square whose area is 121 sq. yd. ? 8. What is the circumference of a 28-inch wheel ? 9. What is the ratio of 4 ft. 2 in. to 5 ft. ? 10. State which of the following are multiples, which factors, which powers, and which roots of 8 : 16, 4, 64, 24, 2. Written. 1. If .625 of a cord of wood cost $3.75, what will .75 of a cord cost ? 2. How much will it cost to carpet a room 14 ft. by 12 ft. with carpet f of a yard wide @ $ 1.25 a yard? Strips to run lengthwise. 3. I gained 1 / by selling my farm for $1346.70. What was the cost? 4. A railroad passes through a farm taking a strip \\ miles long and 66 ft. wide. What is the value of this land at $ 80 an acre ? 5. A 60-day note of $ 600, dated Aug. 4th, 1913, was discounted Sept. 1st @6%. Find the proceeds. 6. If 24 men in 12 days of 9 hours each can do a certain amount of work, how many days of 8 hours each will it take 36 men to do twice that amount of work ? 7. A tank full of water is 8 ft. x 4 ft. x 3 ft. What x 1 is the weight of the water ? 8. Find x: 2x = 30. 180 SECTION 1. ELEMENTARY SCHOOLS 238. 8th Yr. 2d T. Mental 1. 35, + 40, -=- 5, x 20, -h 4, -4- 10, 5 = ? 2. How many bottles f qt. in capacity can be filled from a demijohn containing 4| qt.? 3. How many lb. in 1000 Kg.? How many tons? 4. Sold a book for 60^, gaining 20%. What % would I have lost had I sold it for 40^ ? 6. The cost of a bicycle is $ 36. What shall the marked price be to allow a gain of 16f %, after falling 33^% from the marked price ? 6. How many miles in 80 Km ? 7. Solve 8 x 3 = 5 x + 21. 8. Sold a horse for $ 225 and gained $ 25. What per cent gained ? 9. + of my money equals $ 140. How much money have I ? 10. If 6 men can do a piece of work in 18 days, how long will it take 4 men to do the same work ? Written. 1. Add 491673, 28674, 847598, 892654, 34567, 67891, 391638, 328695, 64738 and 473876. 2. Solve by short processes : (a) Find the cost of 5400 lb. @ $ 6.50 per ton ; (6) What is 250 % of 400 ? 3. The surface of a sphere is 1963.5 sq. in. What is its diameter? 4. Find the cost of carpeting a room 12 ft. by 9 ft. with carpet f yd. wide at 95^ a yd. The strips are to run lengthwise. 5. A 90-day note for $ 1000 with interest at 7 % was dated Jan. 17th, 1907, and dis- counted March 2d, at 6%. Find the proceeds. 6. The per- imeter of a square piece of land is 16 rods. How much is it worth at 10 cents a square foot? 7. A cylindrical cistern with a diameter of 5 ft. has 27 inches of water in it. How many gallons are there ? 8. When it is 7 A.M. in New York City, what time is it in London ? 9. I bought 120 meters of lace at 4 francs per meter. For what must I sell it per yard, U. S, money, in order to gain 20 % on the investment ? 10. I have $ 6600 with which to make an investment. I am offered 6% stock @20% premium or 5% stock at 10% premium. Which is better and by how much annually ? SECTION 2. PRIMARY LICENSE CITY 239. Scope. This section gives an idea of the prepara- tion necessary for obtaining a certificate to teach arith- metic in elementary schools. With a few exceptions the following tests have been given for License No. 1, a license to teach the requirements of the first six years in the school system of New York City. 240. Algebra. 1. If an automobile was sold for $ 1025 at a profit of 25 % , how much did it cost ? Use x or other alge- braic symbol. 2. Illustrate the correct use of the equation in solving two easy examples in percentage. 3. State and solve an easy problem in simple interest to find the rate. Use x. 241. Apperception. 4. Apply the principle of apperception to a first lesson in decimals. NOTE. This calls for a knowledge of the five formal steps of the Herbartians : preparation, presentation, comparison, generalization, and application. These steps are taken in 149. The reader should find them. 242. Cancellation. 6. State the principle of cancellation and show how it should be taught. 6. After stating two principles upon which the process of cancellation is based in the multiplication of fractions, show how that process should be taught. 181 182 SECTION 2. PRIMARY LICENSE CITY 243 243. Decimals. 7. State the steps in teaching the rule for the multiplication of a decimal by a decimal. 8. State the steps in teaching the rule for the division of a decimal by a decimal. 244. Denominate Numbers. 9. With respect to a single exercise in linear measurement suitable for the second year, state : (a) what is to be measured ; (6) what it is to be meas- ured with ; (c) how. 10. How should the subject of cubic measure be presented ? 245. Developments. 11. Enumerate the 45 combinations of digits (taken two at a time in addition) which must be learned in the first two years of school and state how you would teach these combinations. 12. Explain "Only like numbers can be subtracted"; show how this principle applies in the subtraction of integers, of common fractions, of decimals, and of denominate numbers. 13. (a) Explain the Austrian method of subtraction, using the following example: subtract 58 from 100. (6) State its advantages and its disadvantages. 14. Show the complete form of blackboard demonstration of the process: (a) of adding three-figure numbers; (&) of multiplying by three-figure numbers ; (c) of dividing by a one-figure number. 246. Devices. 15. Suggest a device for helping a pupil to remember what 17 minus 8 is. 16. Suppose a pupil forgets the product of 6 and 7; suggest three devices which may be helpful to him. 247. Diagrams. 17. Show by a diagram that multiplier and multiplicand (when neither is concrete) can be inter- changed without altering the product. NOTE. This is the commutative law in multiplication. 248 SECTION 2. PRIMARY LICENSE CITY 183 18. Solve by use of a diagram : How many yards of cloth at S f a yard can be bought for $ 6 ? 19. Show graphically that f (of 1) is equal to 3 -5- 5. 20. State a problem in profit and loss, given the gain and gain per cent, and requiring the selling price : (a) Solve the problem stated. (6) Show how to illustrate the solution by a drawing. 248. Drills. 21. "Since memory is served by multiple associations quite as well as by repetition, the drills employed should be varied in form, iii content, and in mode of applica- tion." In the light of the foregoing quotation, suggest three distinct types of drill under each of the following heads: (a) counting; (6) multiplication. 22. Describe three devices or modes of procedure, for enabling the teacher to conduct efficiently a drill in rapid addition, and state the advantages of each. 249. Errors. 23. Following, there are common errors. Correct each and explain its nature as to a pupil : (a) The area of a rectangle is 12 inches multiplied by 6 inches or 72 square inches. (6) Since 39 is 1 less than 40, it may be writ- ten IXL by the Roman notation, (c) 3 is contained in 15 ^ 5ft times, (d) 5 gallons may be taken from 45 gallons 8 times and ^- gallons will remain, because 45 -+- 5 = 8^-. (e) 6%= 144, 1% = 24, 100% = 2400. 250. Fractions. 24. Discover the effect on a fraction of dividing both numerator and denominator by the same number. 25. Discover the rule for multiplying a fraction by a fraction. 26. Discover the rule for dividing a fraction by a fraction. 27. Tell in order the steps involved in getting the pupils to understand the reduction of a common fraction to a decimal. Give illustrations. 184 SECTION 2. PRIMARY LICENSE CITY 251 28. State and illustrate the steps in. simplifying a complex fraction. Define a complex fraction. 29. Find the answer to the following problem : ^ of f is 75 v/o of what number ? Show how you would have children understand each step of the process. 30. State a complete problem which involves finding what part one common fraction is of another. State its solution by a diagram. 31. To find a number when the number plus or minus a part of itself is given. State a practical problem of each type designated. 251. Games. 32. Describe two games or two other recrea- tive' exercises appropriate to the first or second year of school and involving counting or other number work. 252. Geometry. 33. How would you teach finding the area : (a) of a rectangle ? (6) of a parallelogram ? (c) of an oblique triangle ? (d) of a trapezoid ? 34. How would you teach finding the circumference of a circle ? 253. Helps. 35. A pupil cannot solve this problem: Iff of a yard costs 24 ^, how much will 1 yard cost ? Help him. 36. At 3 ^ each how many apples can be bought for 12^? A pupil does not understand why the number of apples for 12^ is the number of times 3^ is contained in 12^. Help him. 254. Inductive Exercises. 37. What is the gist of the inductive method ? 38. Show inductively (as in second year) that -adding 10 to both subtrahend and minuend does not change the difference. 39. Discover inductively a rule for the divisibility of a number by 9. 255 SECTION 2. PRIMARY LICENSE CITY 185 255. Interest. 40. State and solve a practical problem in simple interest to find the time. 41. (a) Compose a problem in bank discount involving the discount of a non-interest bearing note. (6) Write the note, (c) Give the steps in its solution. 256. L. C. D. 42. Add |, f, f, |. (a) Find by inspection the least common denominater. (6) Give the rule for finding the least common denominator by inspection. 43. Explain clearly how to find the least common denom- inator when the denominators are too large for the method by inspection, as in the case of ^, -fa, -^j. 257. Logical Division. 44. Discriminate and illustrate logical division ; logical definition. 45. In beginning the subject of long division, if the follow- ing divisors are to be used, in what order should they be used and why ? 24, 17, 29, 27, 80 ? 46. In presenting the subtraction of one mixed number from another, what is the simplest type of this case ? Illustrate this type and three others of increasing difficulty suitable for early lessons in the subject. 47. State in the order in which they should be taught the types of examples in division which involve one or more deci- mal numbers. Give reasons for the order chosen. 258. Longitude and Time. 48. In a first lesson on longi- tude what points should be brought out? What device should be employed? 49. How would yo\i teach that the time of a place farther west is earlier? 60. What is the difference in time between a place whose longitude is 54 east and a place whose longitude is 60 west ? NOTE. The device required is to locate the w places as suggested at the right. 186 SECTION 2. PRIMARY LICENSE CITY 259 259. Percentage. 61. Taking some one activity, industry, or experience as a center, construct about it five practical prob- lems involving different applications of percentage. 62. State and solve a problem in trade discount. 63. A man bought 5 % stock and thereby secured an income of 6 /o on his investment, (a) -How much did he pay for the stock? (6) Explain as to a pupil how to prove the answer. 260. Proofs. 64. What is meant by proving an example ? by proving a problem ? 65. Give reasons why pupils should be taught checking of results in the four operations and tell what these methods are. 66. In the case of the following processes, state and exem- plify modes of verifying or checking results which are suitable to pupils below the seventh year : (a) addition (two modes) ; (6) finding a whole when a fractional part is given ; (c) reduc- tion ascending. 67. Solve and prove an original example in the second or third case of percentage. 261. Proportion. 68. State and solve a practical problem in proportion. 59. (a) Give two examples involving ideas of ratio or of proportion, appropriate for the third or fourth year. Give model analysis. (6) Write a practical problem in direct pro- portion and one in inverse proportion, (c) Define ratio ; define proportion. 262. Unit of Measure. 60. (a) What is meant by unit oi measure? (6) State and solve a problem in which the num- ber 3 may be used as a unit. 263. Use of Text-book. 61. State how the text-book should be used in arithmetic as in the case of fractions. SECTION 3. PRIMARY LICENSE STATE 264. Scope. The scope of this section is about the same as that of the last. These tests have been given by the Department of Education of the State of New York, for Training School Certificates or certificates to teach in any elementary school of the state except in certain of the large cities. 265. Aliquot Parts. 1. Give the aliquot parts that in your judgment should be memorized during the elementary course. When should they be learned ? Make up two examples of different types, involving in their solution a knowledge of aliquot parts; assume that the examples are to be solved men- tally by pupils of the eighth grade. XOTE. An aliquot part is understood to mean one of the equal parts of 100 ; the word means some times. Thus, the aliquot parts (of 100) are 50 (one of two parts), 33 (one of three), 25 (one of four), 20 (one of five), 16| (one of six), and so on. 266. Analysis. 2. Of what value are forms of analysis ? In what respect are forms of analysis serviceable to you per- sonally in the solution of problems? 3. Give a model analysis: A bookseller sold a book for $ L'.25 at a loss of 10 per cent; how much did he lose ? 4. A man sold 4 of his farm for what the whole of it cost ; what percent did he gain on the part sold? Give a model analysis. 5. Give a logical analysis : A man bought stock paying 4 % dividends, at 20% discount; what rate of income did he receive on the investment? 187 188 SECTION 3. PRIMARY LICENSE STATE 267 6. What must be the marked price of a hat costing $ 6, so that after discounting the price 30 Jo the dealer may make a profit of 16| % ? Analyze. 267. Developments. 7. Outline a development lesson to teach the terms multiplicand, multiplier, and product. 8. Show how to present inductively the idea of a common fraction. NOTE. ' Inductively ' is to be interpreted as ' objectively.' Strictly speaking only principles or laws can be presented inductively. 9. When should decimal fractions be first introduced and how should they be presented ? 10. Explain how the difference between a draft and a eheck may be made clear to a class. 11. Explain how the difference between bonds and stocks may be made clear to a class. 268. Proofs. 12. Write check problems to be used to test the correctness of the solution of the following examples : (a) If a train runs 32 miles in one hour, how far will it run in 45 minutes ? (&) What is the interest on $ 425 for 2 years and 4 months at 6 % ? 269. Teaching. 13. Write three examples such as you would use in a first lesson on long division and show how you would teach the subject. 14. Show by an outline how you would present addition of fractions. 16. Show how you would teach by means of a rectangle or a circle that \ x \ = y^. Suggest what more you would do to develop the rule for the multiplication of fractions. NOTE. The complete method ( 20) is required. The one-line diagram ( 12) is superior to a rectangle or a circle ; fractions with numerators other than 1 are to be preferred. 270 SECTION 3. PRIMARY LICENSE STATE 189' 16. Illustrate how you would explain to a class the reason for inverting the divisor in the -division of fractions. 17. State four cases of problems in simple interest and give a plan for teaching one of them. 270. Theory. 18. How and why should the aim of arith- metic teaching in the primary grades differ from the aim in grammar grades ? 19. Discuss the extent to which problems calling for the application of principles should be given in the first three grades. 20. Explain the principal advantages claimed for : (a) the Austrian method of teaching subtraction ; (&) the spiral method of teaching fractions and other topics in arithmetic. NOTE. The spiral plan calls for teaching the leading principles with fractions which have small denominators, and then in teaching the whole subject again with fractions which have larger denomina- tors, and so on, making each spiral more comprehensive than the one before. 21. State arguments against the use of the Grube method. NOTE. The Grube method teaches the fundamental operations with each number before taking up the next higher. Thus, the fol- lowing exercises with 4 are taught before 5: 2 + 2 = 4, 3 + 1 = 4, 1 + 1 + 1 + 1 = 4; 4-1 = 3, 4-3 = 1; 2x2=4, 4x1 = 4, 1x4 = 4; 4 '-5- 1 = 4, 4 -s- 4 = 1, 4 - 2 = 2, J of 4 = 1, \ of 4 = 2. 22. Discuss the merits of the solution of the examples in simple interest by formula, as compared with the value of solving them by the application of the proper process of rea- soning without the use of formula. SECTION 4. HIGHER LICENSES 271. Scope. This section gives an idea of the prepara- tion necessary to secure a higher license to teach in the schools of New York City. These tests have been given for promotion license (license for the 7th and 8th years), license for head of department, license for assistant to principal, or license for principal. 272. Fundamentals. 1. Discuss these definitions : A num- ber is a unit or collection of units. A number is an abstract ratio of one quantity to another of the same kind. NOTE. How many is measured by comparison. There is a coin for this eye and a coin for this eye ; there are as many coins as a man has eyes ; ' as many as a man has eyes ' is named two. Number is an expression of how many ( 77). There is an act of comparison for each individual or there are as many times of comparison as individuals. Thus there are ' 2 coins 1 time ' or ' 1 coin 2 times.' The number of times of comparison is called ratio. A number is a ratio. An ' act of comparison ' or a ' time ' is an individual. Hence, the definition, ' Number is an expression of how many ' includes the defi- nition, 'Number is a ratio'; the first refers to individuals of any kind while the second refers to individuals alone that are acts of comparison. 2. Describe and criticise the Speer method. NOTE. The Speer method is based upon the ratio idea of number. Thus, the teacher shows two lines, two surfaces, or two solids, one of which (.4) contains the other (B) 3 times. A is how many times Bl 3 times. What is the ratio of A to B ? 3. If B is the cost of 1 apple, what is A 1 The cost of 3 apples. B is what part of A ? ^. What 190 272 SECTION 4. HIGHER LICENSES 191 is the ratio of B to A ? \. If A is the cost of 3 apples, what is 5? The cost of 1 apple. What is the cost of 3 apples at 5^ each ? The ratio of 3 apples to T apple is 3 ; the sum which bears to 5 ^ the ratio of 3 is 15 ^ ; the cost of 3 apples is 15 f. 3. Describe and criticise the McLellan and Dewey method. NOTE. This method is based upon the ratio idea of number. A ratio involves the quantity to be measured, the unit of measure, and the times contained (number). Thus, in 6t : 2 $ = 3, 6? is the quantity to be measured, 2 ^ is the unit of measure, and 3 is the num- 'ber. What is the cost of 3 apples at 5? each? Since the unit of measure is 5 ^ and the number is 3, the quantity to be measured is 3 times 5^ or 15^. 4. " The ratio idea of number should be introduced early, and applied to the work with fractions." D. E. Smith, (a) What is meant by the ' ratio idea of number ' ? (6) By giving two typical problems, illustrate the use of the ratio idea in the early teaching of arithmetic, (c) By giving three typical problems, illustrate its use in fractions. 6. Give two meanings of the expression f . Show graphically their equivalency. 6. What is meant by unitary analysis in arithmetic ? Illus- trate by a problem. NOTE. At 3 for 5 ^, what is the cost of 5 apples ? Unitary analysis requires the finding'of the cost of 1 apple. 7. In arithmetic, give and illustrate the laws of association, commutation and distribution. NOTE. The commutative (interchangeable) law applies to addi- tion and to multiplication. Addends may be interchanged without affecting their sum ; factors, without affecting their product. Thus, 6 + 8 = 8 + 6; 6x8 = 8x6. The associative (bringing together) law applies to addition and to multiplication. Addends may be grouped in any way without affecting their sum ; factors, without affecting their product. Thus, 2 + (3 + 4) = 3 + (2 + 4) = 4 + (2 + 3) ; 2 x (3 x 4) = 3 x (2 x4) = 4 x (2 x 3). SECTION 4. HIGHER LICENSES 272 The distributive (taking apart) law applies to addition and multi- plication combined. A product is the sum of the products of the parts of the multiplicand by the multiplier ; a product is the sum of the products of the multiplicand by the parts of the multiplier. Thus, 287 x 3 = 200 x 3 + 80 x 3 + 7 x 3 ; 287 x 23 = 287 x 3 + 287 x 20. 8. Describe how the following units are derived or fixed : meter ; are ; liter ; grain ; yard ; gallon ; pound ; dollar ; leap year. 9. Give the unit of the metric system which most nearly corresponds to the following : inch; ton; mile; gallon; grain. Give the equivalent of each. 10. State the arguments in favor of beginning number work with counting, and against the system of beginning with num- ber pictures. 11. Show briefly how the simple equation may be made part of elementary arithmetic, indicating the topics to which it is applicable. 12. Give reasons for or against the use of cases, rules, and formulas in teaching percentage. 13. Give the rule or formula and explain an objective illus- tration that would make it clear to pupils : (a) for the area of a circle; (6) for the volume of a sphere. 14. Concerning the calculating of interest name one typical legal enactment or business custom which requires that time and time rate be estimated: (a) by compound subtraction and using 360 da. to a year ; (b) by finding the exact number of days and counting 360 da. to a year ; (c) by finding the exact number of days and counting 365 da. to a year. NOTE, (a) U. S. rule for partial payments; (ft) bank discount; (c) transaction with U. S. government* 15. Illustrate short process : (a) to multiply by 25 ; (6) to multiply by 39 ; (c) to divide by 125 ; (d) to divide by 16|; (e) to add fractional units. 272 SECTION 4. HIGHER LICENSES 193 16. (a) State in the form of a syllogism the argument in- volved in the explanation: If 8 gal. cost $2.40, how many gallons can be bought for $ 5.40 ? (6) State two ways in which the argument may be shortened by the omission of a premise. NOTE. The no. gallons is 8 gal. multiplied by the no. of times $2.40 is contained in $5.40. The no. of times $2.40 is contained in & 5.40 is, etc. 17. State the axioms or general laws of number on which rests the process for finding G. C. D. of two numbers by the division of the greater by the less, the divisor by the re- mainder, etc. (Euclidean method). NOTE. A factor of each of two numbers is a factor of their sum, or of their difference. A factor of a number is a factor of any multi- ple of that number. 18. Show how the process of multiplying numbers, involv- ing decimals, may be explained through the fundamental prin- ciples of the decimal notation without referring to common fractions. 19. Name five topics in arithmetic that can be'taught with- out giving all the reasons, and explain in each case what device you would use to justify your action to your class. 20. Deduce the formula for changing given temperature m from F to (7; R to F. The freezing point and the boiling point are as follows : F, 32-212 ; C, 0-100 ; R, 0-80. 21. What is the test of divisibility of a number by 3 ? By 4 ? By 6 ? By 45 ? 22. Show how to find the G. C. D. and L. C. M. of several fractions, as f, f,f, etc. NOTE. The G. C. D. of 8 12&s, 9 12ths and 10 12ths is 1 12th ; the L. C. M. is 360 12ths or 30. 23. What is meant by rate of exchange ? Explain why it varies. 194 SECTION 4. HIGHER LICENSES 273 24. Describe a good method of conducting a recitation in : (a) written arithmetic ; (&) oral arithmetic. 25. Prove that 7, 11, 13 are factors of all numbers composed solely of the first and fourth orders (of the decimal system) taken in equal amounts as 6006, 8008. 273. Problems and Examples. 26. How much water must be added to a 5 % solution of medicine to get a 1 / so- lution ? 27. If the list price is 60 % more than the cost, what addi- tional % of discount, besides the customary discount of 25 % to the trade, may be allowed for cash payment in order to gain 14 % by the sale ? NOTE. L, f C to manufacturer ; 1st rem, L ; 2d dis, x % ; 2d rem or SP, - x f L ; SP, 114 % C to manufacturer; x = ? 28. For what sum must I draw my note of 3 mo. to yield $ 1000 at 5 % bank discount ? 29. "Find the cost of 100 Km of wire, 55 cm in diameter, at 1 franc 24 centimes, per Kg, the specific gravity of the wire being 8.8. 30. (a) Find the contents of a cylindrical tank 8 m long and 9 dm in diameter; (6) express its contents in liters; (c) express in kilograms the weight of water at maximum density which this tank will hold; (d) express in kilograms the weight of the same quantity of oil, specific gravity .7; (e) translate the answers to (a), (&), and (c) into their approxi- mate English equivalents (cubic inches, quarts, pounds). 31. New York is longitude 73 58' 25.5" ^W., Sydney, Aus- tralia, is 151 12' 39" E. When it is 9 A.M.,*March 16, at New York, what is the time at Sydney ? 32. A certain calculating machine can give prodxicts up to ten digits. Explain how to use it to multiply 73,924,583 by 762,343. 273 SECTION 4. HIGHER LICENSES 195 33. 2240 Ibs. of chalk occupy 15.5 cubic ft. What is the specific gravity of the chalk ? NOTE. The specific gravity of a substance is its weight divided by the weight of an equal volume of water. 34. Find the value of the circulating decimal .426 and ex- plain the solution. NOTE. A circulating (moving in a circle) decimal is a decimal whose last figures repeat without end. Dots are placed over the first and last figures of the part which repeats (the repetend). It is equal to a decimal ending in a common fraction whose numerator is the repetend and whose denominator is as many 9's as there are figures in the repetend. Thus, .426 = .4|| = f^. 35. Simplify: (a) I"*""* f * NOTE. See 122. -s- 2^ -f- 6 is ambiguous. Use the signs as they occur. INDEX Abstract nos., 110. Addition develop., 64 ; checks, 68 ; fractions, 98 ; decimals, 105 ; prob- lems, 23, 34; signs, 132. Algebra develop., 131; problems, 38 ; percentage, 141 ; interest, 164 ; tests, 181. Aliquot parts multiplication, 81 ; div., 88; interest, 156; tests, 187. Altitudes, 119. Analysis develop., 26 ; percentage, 142; interest, 163; tests, 187. Analytic aid develop., 30; int., 163. Angles, 117. Annual interest, 165. Apperception, 181. Arabic notation, 55. Arrangement prob., 34; para- graphs, 53. Associative law, 191. Austrian method sub., 72, 75 ; division, 87. Bank discount develop., 158; tests, 192. Bonds develop., 160 ; tests, 188. Cancellation expl., 100 ; tests, 181. Canon of agreement, 14. Cases develop., 6 ; percentage, 7 ; int., 164. Checks, 162. Circle cir., 12, 121 ; area, 121 ; tests, 192. Classifications, 2. Combinations add., 65; sub., 72; mult., 77; div., 83. Commercial discount, 147. Commission, 146. Commutative law, 81, 100, 191. Complete method, 20. Complex fractions, 102, 195. Complex problems, 30. Composite nos., 90. Compound fractions, 102. Compound interest, 165. Compound numbers, 110. Concrete nos., 110. Cones, 119. Constructions, 120. Convex surface, 122. . , Crutches, 53. Cube root, 127, 128. Cylinders, 119. Decimals develop., 103; history, 109 ; roots, 129 ; tests, 182. Deduction, 16. Definitions, 3. Denominate nos., 110, 114, 182. Development exercises, 47, 182, 188. Diagrams, 5, 182. Differences or differentia, 3. Difficult problems, 145, 146, 194. Dimensions, 118. Discount bank, 158 ; commercial, 147. Distributive law, 191. Divisibility, 15, 92, 193. Division develop., 82; fractions, 99; dec., 105; Austrian, 87; remainders, 101 ; by factors, 87, 101, 108 ; signs, 133. Drafts, 161, 188. Drill exercises, 47, 183. Elementary schools scope of the grades, 169-179. English notation, 62. Equations, 134, 192. Eratosthenes, 90. Errors, 52, 53, 183. Euclidean method, 95, 193. Evolution, 125. Exact interest, 158, 159, 178. Examinations el. schools, 169 ; teachers, 181-195. Exercises, 169-195. Explanations, 29. Expression, 51. Factors develop., 90-95 ; division by, 87, 108, 101 ; roots, 127. Formal steps of Herbartians, 181. Formula, 39, 128, 141, 164. 196 INDEX 197 Fractions develop., 96 ; lowest terms, 95, 98 ; remainders, 101 ; roots of, 129 ; clearing of, 135 ; g. c. d. and 1. c. m., 193. French notation, 57. Fundamentals, 190. Games, 184. Geometry, 184. See mensuration. Graphic aids, 32, 183. Greatest common divisor by fac- toring, 94 ; Euclidean, 95, 193 ; of fractions, 193. Grube method, 189. Helps, 184. Herbartians, 181. Hindu notation, 57, 62. Hypotenuse, 124. Induction, 14, 184. Interest develop., 152; exact, 158, 159, 178 ; tests, 185, 189. Involution, 125. Italian subtraction, 73, 75. Least common multiple by in- spection, 93 ; by factoring, 94 ; of fractions, 193. Ledger balances, 74. Lesson plans, 46. Licenses primary, 181 ; higher, 190. Lines, 117. Logarithms, 129. Logarition, 125. Logical definitions, 3, 185. Logical division, 3, 185. Logical steps, 9. Longitude and time, 185. McLellan and Dewey method, 191. Major analysis, 27. Measurements, 8. Mechanical aids, 11. Mensuration, 117, 184. Mental and written, 50. Metric system, 114. Minor analysis, 27. Mixed nos., 100. Model analysis, 27, 33, 187. Mortgages, 161. Multiplication develop., 76; frac- tions, 16, 99; dec., 17, 106; proofs, 80 ; signs, 132. Names, 3. Needs, 2. Nines, proofs by add., 69; mult., 80 ; div., 86. Notation and numeration Arabic, 55; French, 57; English, 62; Hindu, 63. Notes, 152. Number, 55, 190. Operations combined, 88. Paper folding, 11. Paragraphing, 53. Parallelograms, 5, 12, 119, 121. Parentheses, 88, 134. Partition, 22, 76, 82. Percentage develop., 137; cases, 7; tests, 186. Polygons, 118. Postal savings system, 159. Prime nos., 90. Prisms, 119. Problems class, 22 ; by experi- ment, 22; analysis, 26, 30; algebra, 38 ; formula, 39 ; rule, 40 ; proportion, 41 ; variation, 44 ; arrangement, 34 ; proofs, 36 ; graphic aids, 32 ; explanations, 29 ; exercises, 169-195. Profit and loss, 144. Proofs problems, 36: add., 68; sub., 73; mult., 80; div., 86; tests, 186, 188. Proportion, 41, 186. Pyramids, 119. Quadrilaterals, 5, 121. Quotition, 22, 76, 82. Ratio, 43, 186, 190, 191. Records, 54. Rectangles, 5, 110, 121. Regular polygons, 118. Remainders, 101. Rhomboid and rhombus, 5. Right-angle triangle, 118, 124 Roman notation, 60. Rules, 40, 141, 164. Savings banks, 167. Sequence of signs, 88, 195. Short processes, 192. Sieve of Eratosthenes, 90. Signs, 88, 132, 133. Similarity, 45, 124. Simple problems, 22, 26, 34. Solids, 117. 198 INDEX Species, 2. Specific gravity, 194. Speer method, 190. Sphere, 13, 119. Spiral method, 189. Square root, 127. Stocks, 148. Subtraction develop., 64, 71 ; frac- tions, 98; dec., 105; signs, 133. Surface, 117. Syllogism, 193. Tetrahedron, 119. Textbook, 186. Theory, 189. Transposing, 134. Trapezium, 5. Trapezoid, 5, 119, 121. Triangles, 4, 11, 12, 118, 124. Unit of measure, 186. Unitary analysis, 191. Variation, 44. CONSULTATION BY CORRESPONDENCE For Superintendents, Principals, Teachers and Others By Middlesex A. Bailey, Author of " Teaching Arithmetic " ORDINARY CONSULTATION, $1.00 Special Consultation Proportional to the Amount of Labor PAYMENT IN ADVANCE THE establishment of a bureau of consultation for the teaching of arithmetic is something of an innovation. In other lines, laymen consult physicians, and physicians con- sult specialists; contractors consult engineers, and engineers consult specialists ; and so on. Occasions arise when a superintendent wishes to consult in regard to a course of study ; a principal, in regard to grade work ; a teacher,, in regard to class work ; or a prospective teacher, in regard to preparation. It is with diffidence that the writer offers his services, be- cause he realizes his deficiencies. It is in place, however, to state his preparation. Since graduation from college in 1877, he has been engaged in school work ; 3 years as principal of an elementary school in Winsted, Conn. ; 5 years as princi- pal of a high school in Keene, N.H. ; 14 years as head of the department of mathematics at the State Normal School of Kansas, in Emporia ; and 14 years as head of the depart- ment of mathematics at the New York Training School for Teachers in New York City. During these 36 years he has made the subject of teaching arithmetic his major study and has written a series of arithmetics published by the American Book Company. MIDDLESEX A. BAILEY, YONKERS, NEW YORK INSTRUCTION BY CORRESPONDENCE For Teachers and Prospective Teachers of Arithmetic By Middlesex A. Bailey, Author of " Teaching Arithmetic" THIRTY LESSONS, $15.00 Book, Paper, Envelopes, Postage both ways, Included PAYMENT IN ADVANCE A COURSE of training in methods of teaching arithmetic is offered to those who are preparing to obtain a primary license. Any person seventeen years of age who has mastered the requirements in mathematics of an ejght-year course in the elementary schools is eligible to take this work. A thoroughly satisfactory preparation is impossible without some knowl- edge of algebra and geometry. If the candidate has not taken these subjects, he should arrange to study them by correspond- ence or otherwise at the earliest opportunity. He is advised also to purchase Jevon & Hill's Logic and to read it in con- nection with the study of methods. It should be borne in mind that no course by correspondence can equal in value attendance at a normal or training school. A second course is offered to teachers who are preparing for a higher license and who prefer work by correspondence to attendance in summer schools or afternoon classes. The candidate is requested to state in advance his age, the schools from which he has graduated, the schools which he has attended, the purpose for which he wishes to take the work and the course desired. The candidate for a higher license is requested also to state his teaching experience. MIDDLESEX A. BAILEY, YONKERS, NEW YORK UNIVERSITY OF CALIFORNIA AT LOS ANGELES THE UNIVERSITY LIBRARY This book is DUE on the last date stamped below WOV 9 3J8 19 * 1 3 NOV 5 1946 APR 1 4 1947 nc^ teC 13 1950< FEB-2 1952 Form L-9-15m-3,'34 8 , JAN 19513 If AY a 6 1951 < l 1 1 1954 "MAY 101956 APR 2 2 195S DEC 9 REC'D COL UNIVERSITY of CALIFORNIA JUBRABT