THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA LOS ANGELES STATE NORMAL SCHOOL LOS ANGELES, CALlFOftNiA SOUTHERN BRANCH UNIVERSITY OF CALIFORNfA ijLlBRARY LOS ANGELES, CALIF. PRACTICAL METHODS IN ARITHMETIC Nr wa BY R. S. GALER, A. M., Principal of the Iowa City Academy, Iowa City, Iowa. SECOND EDITION, REVISED. CHICAGO: A. FLANAGAN, 1892. COPIBI6HT, 1889, BY R. S. GALEB. 103 CONTENTS, PAGE. I. INTKODUCTION - 5 II. SHORT METHODS - - - 9 III. PRACTICAL MEASUREMENTS Gallons and Barrels in Parallelopipedon Gallons and Barrels in Cylindrical Vessel Gallons and Barrels in Jug Cistern Grain in Bin or Wagon Hay in Rick Hay in Bound Stack Perch of Book to Wall a Well Brick to Wall a Well or Cistern Quantity of Fluid that will flow through a Pipe How High an Object must be to be Visible a Certain Distance How Far an Object, whose Height is Known, is Visible How Far a Body will Fall in a Given Time - - 16-25 Lumber Measure - 25 Shingle Measure 26 Lath Measure 27 IV. COMMON FRACTIONS 28 Addition and Subtraction of Fractions, 28 Multiplication and Division of Frac- tions 30 V. DECIMAL FRACTIONS Contracted Method of Multiplication and Division - * - 32 CONTENTS. PAGE. VI. PUBLIC SURVEYS - 35 VII. THE METRIC SYSTEM - 40 VIII. LONGITUDE AND TIME 43 IX. RATIO - 47 X. PROPORTION 48 Simple Proportion - 49 Compound Proportion Cause and Effect 51 XI. GENERAL PRINCIPLES RULES 53 XII. PERCENTAGE 55 XIII. PERCENTAGE APPLICATIONS 55 Commission and Brokerage - CO Profit and Loss - 63 Stocks and Bonds - G6 Premium and Discount - 70 Investments - 72 Contracts 76 Interest - 78 Promissory Notes 82 Partial Payments - - 85 U. S. Rule 85 Mercantile Rule - - True Discount - 89 Bank Discount - - 91 Exchange 96 XIV. HORNER'S METHOD FOR THE EXTP ACTION OF ROOTS - 102 XV. BITS OF PRACTICAL KNOWLEDGE, 105 XVI. PRACTICAL PROBLEMS - - 110 XVII. ANSWERS 112 PREFACE. This work is not designed to be a complete Arithme- tic. In an experience of several years in teaching, the author was able by a little ingenuity to abridge mathe- matical calculations, and to perceive that many princi- ples in Arithmetic have a general application. (For these he has formulated rules which have at once the merit of brevity and scientific accuracy. Moreover these rules have as wide an application as the princi- ples for which they stand, thus greatly simplifying arithmetical operations and reducing the number of rules to be committed to memory. In addition to this he has shortened the work required for many practical measurements. The constant demand in the school room for these methods and formulas 'led the Author to believe they might be acceptable to a larger class of students and teachers than he could reach personally. With this aim in view this little volume is published- It is not claimed that all of it is original. Much that was considered of value has been gathered from other sources. The Author has taken this step with the pur- pose of embodying the results of his own individual experience in a convenient form, and at the same time of furnishing a book of general value to pupils and teachers. PREFACE. Especial attention is called to the manner of treating Percentage and its numerous applications. This method is entirely original, and, it is believed, results in greatly simplifying these subjects. With the hope that it may aid many in saving much time and labor, and stimulate them to think independ- ently on practical subjects, this little volume is given to the public. BLOOMFIELP, IOWA, JAN. 8th, 1889. TO THE STUDENT. The following general rules for work should be con- tinually and faithfully observed : I. Use the shortest method that is consistent with clearness and accuracy. II. In multiplying, dividing, etc., contract wherever possible. III. Burden the mind with few rules, but learn to apply these rules intelligently. IV. Do not guess at methods in the solution of problems. Reason out each step, and have faith in the result. V. Do as much work mentally as possible. VI. Remember that Rapidity and Accuracy are the great aims. VII. Bear in mind that nothing will take the place of Reasoning in mathematical work. SHORT METHODS. Much time and woik may be saved in arithmetical computations, by the use of certain short methods, which are usually of easy application. A number of them are given here which it is intended that students should use whenever possible. It requires a quick perception that will come only with practice, to tell at a glance which one may be used to the greatest advan- tage. FIRST METHOD. To multiply where the left hand figures are the same, and the sum cf the two right hand figures is ten. EXAMPLE. 55 55 3025 Ans. Beginning at the right we say, 5 times 5 are 25. Then adding 1 to the lower left-hand figure and multi- plying it by the upper one, we have 6 times 5 are 30. Hence, the result is 3025. The figures at the right need not be fives. EXAMPLE. 64 66 4224 Ans. Here the rule for multiplying is the same. (9) 10 PRACTICAL METHODS EXAMPLES. (1) Multiply 83 by 87. (2) Multiply 91 by 99. (3) Multiply 28 by 22. (4) Multiply 114 by 116. SECOND METHOD. To multiply where the right-hand figures are the same, and the sum of the left-hand figures is ten: Rule. Multiply the right-hand figures and bring down the product. Multiply the left-hand figures and add the figure on the right to the product. Write this to the left of the former result. The resulting number will be the required product. EXAMPLE. Multiply 46 by 66. 46 66 3036 Am. 6 times 6 are 36. This is written below. 6 times 4 are 24, to which the right-hand figure 6 is added, mak- ing 30. The result is 3036. EXAMPLES. (5) Multiply 34 by 74. (6) Multiply 28 by 88. (7) Multiply 17 by 97. (8) Multiply 45 by 65. IN ARITHMETIC. \\ THIRD METHOD. Where numbers are nearly 100, 1000, etc. Rule. Add to the upper number as many ciphers as there are figures in the number. From 100 or 1000, etc., subtract the two numbers separately. Multiply the two remainders together, and place the product under the ciphers at the right. Then add the numbers, omit- ting the final 1 at the left. EXAMPLE. Multiply 98 by 95. 9800 9510 9310 Ans. Here the two remainders are 2 and 5, and their pro- duct is 10. This is placed under the ciphers as directed, and the numbers then added, omitting the 1 from the 19 at the left. EXAMPLE. Multiply 994 by 982. 994000 982108 97610S Ans. EXAMPLES. ( 9) Multiply 96 by 95. (10) Multiply 92 by 88. (11) Multiply 995 by 990. Multiply 984 by 993. FOURTH METHOD. Where one of the numbers is an aliquot part of 100 or 1000. 12 PRACTICAL METHODS Rule. Multiply by 100 or 1000, etc., by adding two or more ciphers, or by placing the decimal point two places to the right. Take such a fractional part of the result as the multiplier is of 100 or 1000, etc. EXAMPLE. Multiply 336 by 25. 4)33600 8400 Ans. Since 25= J of 100 the product of 336 and 25 is J the product of 336 and 100 as shown in the operation. EXAMPLE. Multiply 436 by 266|. Operation: 266f=2 of 100. Hence 436x266| =43600x2=116,266. Ans. EXAMPLES. (13) Multiply 72 by 125. (14) Multiply 3456 by 33|. (15) Multiply 75.42 by 62. (16) Multiply 31.8 by 250." (17) Multiply 4072 by 833J. (18) Multiply 60.732 by 444f . FIFTH METHOD. To square a number. Rule. Multiply the number next larger or smaljer that ends in cipher by one an equal amount less or greater than the given number. Add to the result the square of the difference between the given number and the one that ends in 0. REMARK. For numbers of two figures this may be done men- tally. IN ARITHMETIC. 13 EXAMPLE. Square 28. The numbers to be multiplied are 30 and 26. Place at the right 4, the square of 2 ; 30 28. Then 3 times 26 x 78. Hence the result is 784. EXAMPLE. Square 62. 2x2=4. 64x60+4=3844. Ans. EXAMPLES. (19) Square 79. (23) Square 103. (20) Square 61. (24) Square 52. (21) Square 48. (25) Square 27. (22) Square 33. (26) Square 99. SIXTH METHOD. To square a number ending in thirds. Rule. Square the figure on the left and add to the result double the number of thirds of the left-hand figure, indicated on the right. Square the thirds and bring down the numerator thus obtained twice with the same number of ninths. EXAMPLE. Square 63^. Operation: 6x6=36. J of 6=2. 2x2=4+36^40. ^ of 15=!- Hence the result is 4011^. Ans. EXAMPLE. Square 36f. Operation: 3x3=9. | of 3=2. 2x2=4+9 = 13. | of | = |, The result is 1344|. Ans. EXAMPLES. (27) Square 93|. (30) Square 126|. (28) Square 96|. (31) Square 153,1. (29) Square 331 (32) Square 86$. U PRACTICAL METHODS SEVENTH METHOD. To multiply any numbers by carrying mentally. Rule. Multiply the figures that produce units. Bring down into the column of products. Multiply all the figures that produce tens, add the products mentally, and bring down the right-hand figure of the result. Multiply next all the figures that produce hundreds, add mentally and bring down the result, carrying as in ordinary multiplication. Proceed as before until the multiplication is complete. EXAMPLE. Multiply 64 by 52. 64 52 3328 Operation: First, 2x4=8 units. Second, 2x6=12 5x4=20 20+12 =32= tens. Write the 2 in column of results. Third, 5x6+3 (carried) =33= hun- dreds. The total product is 3328. Ans. EXAMPLE. Multiply 145 by 326. 145 326 47270 IN ARITHMETIC. 15 Operation: First, 5x6 = 30. Write the in col- umn of products. Second, 4 x 6 = 24 2x 5 = 10. 24+10+ 3 (carried) =37 = tens. Write the 7 in column of products. Third, Ix 6= 6. 2x 4= 8. 3x 5 = 15. 15+ 8+6+3 ( carried ) =32= hundreds. Write the 2 in column of products. Fourth, 2 x 1 =2. 3x 4=12. 12+ 2+ 3 (carried) =17 = thousands. Write the 7 in column of products. Fifth, 3x 1= 3. 3+ 1 (carried) =4= ten thousands. The whole product is 47270. NOTE., With practice tht, student may learn to multiply very rapidly by this method, the work being entirely mental. EXAMPLES. (33) Multiply 45 by 82. (34) Multiply 27 by 64. (35) Multiply 83 by 36. (36) Multiply 264 by 351. (37) Multiply 746 by 825. (38) Multiply 1285 by 3463. 16 PRACTICAL METHODS PRACTICAL MEASUREMENTS. Nearly all of the following rules and formulas have been worked out by the author for his own convenience in the class room. Some of practical value have been taken from other works or from popular custom. They may be relied on as scientifically correct and accurate. FIRST. To find the capacity in gallons of a parallelopipedon. NOTE. A parallelopipeJoa is a prism whose facjs are six par- allelograms. A dry goods box is an example. Formula: LxWxDx 7. 48 = Gallons. Rule. Multiply the length, width, depth together, expressed in feet. Multiply this product by 7.48. The result will be the capacity of the vessel in gallons. The product of length, width and depth would give the solid contents of the vessel in cubic feet. This multiplied by 1728 would give cubic inches, and this divided by 231 would give gallons. But 1728n-231^ 7.48. Hence the rule. EXAMPLE. Find the capacity in gallons of a box 8 feet long, G feet wide, 2^ feet deep. Solution: 8x6x2^x7.48=897.6 gallons. Ans. EXAMPLES. (39) Find the capacity in gallons of a room 32 feet long, 20 feet wide, 12 feet high. (40) Of a bin 10J feet long, 6| feet wide, 4J feet IN ARITHMETIC. 17 SECOND. To find the capacity in barrels of a parallelopipedon. Formula: L X W X D X ffi or ^4.2. Rule. Multiply the product of the length, width and depth, in feet, by |f, or divide their product by 4.2. Since there are 31^ gallons in a barrel the preceding formula divided by 31^ will give the number of barrels. But 7. 48-5-31 J=fcf Multiplying by g- is equivalent to dividing by 4.2. EXAMPLES. (41) Find the capacity in barrels of a vessel 14 feet long, 12^ feet wide, 4 feet deep. (42) Find the number of barrels in a reservoir 84 feet long, 36 feet wide, 18 feet deep. THIRD. To find the number of gallons in a cylindrical vessel. Formula : Di. 2 X De. X V or 5.9. Rule. Square the diameter, multiply by the depth, and this product by 4 ff 7 or 5.9. NOTE. In all these formulas the dimensions must be expressed in feet before performing the required operation. EXPLANATION. The usual way to obtain this result is to find the solid contents of the vessel in cubic inches and then divide by 231. Or the following formula: Di*X.785AxDeXl728. 231 But .7864X1728 = 47 or 5> Hence the rule. 18 PRACTICAL METHODS EXAMPLES. (43) Find the number of gallons in a well 40 feet deep, and 4 feet in diameter. (44) What is the capacity in gallons of a stand pipe 80 feet high, and 10 feet in diameter ? (45) How many gallons will a cistern hold whose depth is 10 feet, and diameter 5 feet? FOURTH. To find the number of barrels in a cylindrical vessel. Formula : Di. 2 X De. X ^ . 47_!_Q11 3 "'~ O1 ~* EXAMPLES. (46) Find the capacity in barrels of a cistern 12 feet deep and 6 feet across. (47) Find the number of barrels in a cistern 14^ feet deep and 7i feet in diameter, filled to within 1 foot of the top. FIFTH. To find the number of gallons or barrels in a jug cistern. NOTE. Many cisterns are curved at the top, leaving only a narrow mouth. These are called "jug " cisterns. Formula: For gallons, Di. 2 xDe. x5.3. For barrels, Di. 2 X De. X J. REMARK. These rules are only approximate. A slight variation from this may be made by the depth of the cistern, the curve at the top, etc. The rules given are derived from many calculations of the actual capacity of jug cisterns. The mean of these calcula- tions indicates a capacity j^ less than cisterns of a cylindrical shape. The formula for those is, Di. s xDe.x5.9; ^ less than this would be Di. s xDe. X5.3. If this be divided by 3H, the quotient will be barrels; but 5.3-*-314=. Hence the above rule for barrels. IN ARITHMETIC. 19 EXAMPLES. (48) Find the capacity in gallons of a jug cistern 1 1 feet deep and 7 feet in diameter. (49) A jug cistern is 6|- feet across and 9^ feet deep . How many gallons will it contain ? How many barrels? SIXTH. To find the quantity of grain or shelled corn in a wagon or bin. Formula : L X B X D X .8. Rule. Multiply the continued product of the length, breadth and depth by .8. To reduce cubic feet to bushels, we must multiply by 1728 and then divide by 2150.4, the number of cubic inches in a bushel. But ^j^=.8, nearly. To find the number of bushels of corn on the cob the formula is: LxBxDx.4. NOTE. An allowance of ^ is made for the ob. EXAMPLES. (50) How many bushels of grain will a wagon con- tain if the depth is 2 feet, the width 3 feet, and length 10 feet? How. many bushels of corn on the cob? NOTE. Notice that a wagon bed 10 feet long and 3 feet wide holds 1 bushel of corn on the cob for each inch in height. (51) How many bushels of corn on the cob in a bin 16^ feet long, 9-f feet wide, 6 feet high? How many bushels of grain? (52) How many bushels of grain in an elevator bin 63 feet long, 14 feet wide, 9 feet high ? 20 PRACTICAL METHODS SEVENTH. To find the quantity of hay in a rick. Formula : I Pis, over + Breadth \ 2 X L, X - = Ton, 500 Rule. Find, by taking the measurement by a rope, the distance over the stack from ground to ground. To this add the width. Divide by 4, square the quo- tient, and multiply by the length of the stack. This divided by 500 will give the quantity of hay in tons. EXPLANATION. By adding the width of the rick to the distance over we get the entire distance around the rick. Dividing this by 4 "squares" the rick. It is then in the shape of a parallelepiped, whose width and height are equal, and whose length is known. Multiplying these dimensions together gives the solid contents of the rick in cubic feet. Dividing this by the number of cubic feet in a ton gives the tons of hay in the rick. NOTE. It is estimated that 550 cubic feet of clover, or 450 cubic feet of timothy are required for a ton. The number here given, 500, is very nearly correct for ordinary hay, which usually consists of both timothy and clover. EXAMPLE. Find the number of tons of hay in a rick 22 feet long, 12 feet wide, 32 feet over. Solution: 32 ft.+12 ft.=44 ft. 44 ft. --4=11 ft. 11x11x22=2662 cu. ft. ^500^5^ tons. Ans. EXAMPLES. (53) Find the tons of hay in a rick 30 feet long, 14 feet wide, and 30 feet over. (54) How many tons of hay in a rick 13 feet wide, 65 feet long, and 27 feet over? (55) How many tons in a rick of hay 42i feet long, 14 feet wide, 31| feet over? IN ARITHMETIC. 21 EIGHTH. To find the quantity o,. hay in a round stack. Formula: (Dis. over -4- A dis. around) 2 Xi dis. around. i- - 2 - , -== Tons. 4000 Rule. Add to the distance over the stack from ground to ground ^ the distance around the stack, and square the sum. Multiply this by \ the distance around the stack, point off 3 places and divide by 4. The result will be the quantity of hay in tons. EXAMPLE. How many tons of hay in a stack 28 feet over, and 42 feet around? Solution: Distance over 28 feet+ of 42=42 feet. 42X42=1764. 1764x4f ($ of 42)=8232. Point off 3 places and divide by 4. 8.232-f-4=2+tons. Ans. EXAMPLES. (56) Find the number of tons in a stack of hay 32 feet over, and 40 feet around. (57) How many tons of hay in a round stack 26 feet over, and 37 feet around. (58) A round stack of hay has the following dimen- sions: Distance over=25 feet. Distance around=42 feet. How many tons does it contain ? NINTH. To find how many perch of rock it requires to wall a well. Formula: Th(Di Th)De_ p ~~~ 22 PRACTICAL METHODS Ride. From the diameter of the well in feet tract the thickness of the wall, multiply this by the thickness of the wall, and this by the depth of the well. The result divided by 8 will give the number of perch. EXAMPLE. How many perch would be required to wall a well 33 feet deep, 5i feet in diameter, if the wall is 1 foot thick? Solution: 5 J 1=4J. 44_ 189 p EXAMPLES. (59) How many perch would be required to wall a well 52 feet deep, 6^- feet in diameter, the wall being 1J feet thick. (60) A well is 36 feet deep and 4^ feet in diameter. If a wall 10 inches thick be put in, how many perch of rock will be required ? (61) How many perch will wall a well 29 feet deep, 5 feet across, if the wall is 1 foot thick ? TENTH. To find how many bricks it requires to wall a well or cistern. Formula: Th(Di Th)DeX75. Rule. From the diameter of the well or cistern sub- tract the thickness of the wall, multiply this by the thickness of the wall, and the depth of the well or cis- tern. Multiply the last product by 75. The result will be the number of bricks required. IN ARITHMETIC. 23 EXAMPLE. Find the number of bricks it will require to wall a cistern whose depth is 12 feet, distance across CA feet, the wall to be 8 inches thick. B Solution: 8 in.= ft. 6J~=5|. 5|xfx 12X75=3500. Ans. EXAMPLES. (62) Find how many bricks it will require to wall a cistern 14 feet deep, 5| feet across, the wall being 4 inches thick. (63) A well is 18 feet deep, 4 feet 8 inches across. If a wall is put in 4 inches thick, how many bricks will it take? ELEVENTH. To find the number of gallons that would flow through a pipe per minute. Formula : Di 2 X V X - 1 /. Rule. Square the diameter of the tube in inches. Multiply this by the velocity of the water per second in feet, and this by *-/- or 2.4. The result will be the amount discharged by the pipe per minute in gallons. EXAMPLES. (64) How much water with a velocity of 25 feet per second would flow through a pipe 2 inches in diameter, per minute? (65) How much water would be furnished in 12 nours by an artesian well 4^ inches in diameter, the water flowing with a velocity of 12 feet per second? (66) How much water would flow in 24 hours through a discharge pipe 1^ inches across, if the veloc- ity is estimated at 74- feet per second? 24 TWELFTH. To find how high an object must be to be visible a certain distance. The curvature of the earth is 8 inches to the mile, and varies as the square of the distance. That is for 2 miles it is 4 times 8 inches or 32 inches. For 3 miles it is 9 X 8 inches or 72 inches, and so on indefinitely. This is ascertained by a geometrical computation which would be out of place here. Supposing then the curva- ture of the ground to be uniform like the sea level, with no hills or valleys, or other obstructions to sight, an object must be 32 inches above the surface to be visible at the distance of 2 miles, 72 inches to be vis- ible 3 miles, etc. From this we derive the following formula: H=D 2 X, in which H represents Ihe height of the object in feet, D the distance in miles, 8 inches being f ft To find how far an object is visible, whose height is known : Dividing the above formula by -|, we have D 3 = Hxf. Extracting the square root, D= /v /HXf = Formula required. Rule: Extract the square root of f the height of the object in feet. The result will be the distance in miles the object is visible. EXAMPLES. (67) How high must an object be to be seen 10 miles? 12 miles? 20 miles? 100 miles? (68) How far is Washington monument visible (555 feet high)? The Bartholdi Statue of Liberty (322 feet) ? Mt. Chimborazo (4 miles) ? IN ARITHMETIC. 25 THIRTEENTH. To find how far a body will fall in a given time : Formula. D=16 t 2 . Rule. Multiply the square of the time in seconds by 16; or to be more exact, by 16.08. The product will be the distance in feet through which the body will fall. EXAMPLES. (69) How far will a body fall freely through the air in 5 seconds ? 10 seconds ? 30 seconds ? 1 minute ? To find how long it will take a body to fall from a given height. Formula: T=J VH. Rule. Take ^ the square root of the height in feet. The result will be the time in seconds. EXAMPLES. (70) How long will it take a body to fall from a height of 36 feet? 100 feet? 2,500 feet? 250 feet? NOTE. These formulas are for bodies falling in a vacuum. Practically, the velocity of a body falling in air is diminished by friction. Hence the distance through which it falls in a given time is somewhat less than that given above. LUMBER MEASURE. In measuring lumber, the thickness of the boards is not taken into account, provided it be one inch or less. The amount of lumber in a board is ascertained, there- fore, by computing the surface of the board in square feet. The formula would be L X B=No. of feet. 26 PRACTICAL METHODS But when the stuff is more than one inch in thick- ness, the surface expressed in feet must be multiplied by the thickness of the board or plank in inches. EXAMPLES. (71) How many feet of lumber in 16 boards, each 14 feet long, 6 inches wide, and 1 inch thick? (72) How many feet of lumber in 36 planks, each 12 feet long, 10 inches wide, and 2^ inches thick ? (73) How many feet in a pile of lumber made up as follows: 42 boards, each 12 feet long, 5 inches wide, ^ inch thick. 24 planks, each 10 feet long, 12 inches wide, 3 inches thick. 56 pieces, 4 inches square and 9 feet long? How much would the above bill of lumber cost at the following rates: The boards, $2.25 a hundred feet? The planks, $1.50 a hundred feet? The pieces, $1.20 a hundred feet? To find the largest square stick that can be sawed from a log. Rule. Multiply the diameter of the log by . 7. To find the number of feet of lumber in the largest square stick of timber that can be sawed from a log. Rule. Square the diameter of the log in inches and divide by 2. The result will be the amount of lumber in a 12-foot log. SHINGLES. Shingles are put up in bunches, each containing 250 shingles, which are reckoned as 4 inches wide, on an average. Shingles are usually laid 4^ inches to the weather, in which case the "Carpenter's Rule" is as follows: IN ARITHMETIC. 27 900 shingles laid 44 inches to the weather will cover a square, or 100 square feet. An allowance of 100 shingles is made for waste in laying, and because not all bunches are full. LATHS. There are 50 laths in a bunch, each being 4 feet long and 1^ inch wide. They are usually laid f of an inch apart. Allowing for the waste in putting on the laths, the "Contractor's Rule" is as follows: A bunch of laths will cover 3 square yards. For rooms 9 feet high, one-third as many bunches would be required as the distance in feet around the room. EXAMPLES. (74) How large a square stick can be sawed from a log whose diameter is 1 foot? 16 inches? 24 inches? (75) How many feet of lumber can be sawed from a log 12 feet long, 10 inches in diameter? 16 feet long? 22 feet long? (76) How many bunches of shingles would be re- quired to cover a roof," each side of which is 40 feet long, 18 feet wide? To cover a roof 34^ feet long, each side being 16 feet wide ? (77) How many bunches of laths would be required to cover the sides and ceiling of a room 17 feet long, 14| feet wide, 9 feet high? (78) How many bunches would be required for four rooms, each 14 feet long, 12 feet 4 inches wide, 8 feet 6 inches high, the walls only being lathed? 28 PRACTICAL ME1HODH COMMON FRACTIONS. We desire to discuss common fractions only par- tially. What we shall say of them will be for older pupils, as are, indeed, all the contractions presented in this book. We recommend the pupil to commit to memory the models given for the explanation of mul- tiplication and division of fractions. The teacher can invent a sufficient number of practical examples for the use of his classes. ADDITION AND SUBTRACTION OF FRACTIONS. Before fractions can be added or subtracted they must be reduced to equivalent fractions having a com- mon denominator. It is taken for granted that the pupil is able to do this. Fractions may usually be added most rapidly in the following manner: EXAMPLE. Add 2^, |, If, 4J. Operation: 2A L 9A Ans. 'jfr First add ^, |, and ^ mentally, as they are readily reduced to the same denominator. The result is 1. Then add and f . The result is ff, or l y * T . This makes 2 to carry. IN ARITHMETIC, 29 EXAMPLE. Add 4f, 1-fr, 9 T \, 7f, Operation : 4 f IA 9 A Add f and f=l. Then A+ T y=tt- EXAMPLE. Subtract If from 3-|. Operation: 3$ 1 2 2J ^TIS. Subtract without reducing to improper fractions. EXAMPLES. Find the sum of: (79) 4|, 61, J ( i 3$. (80) 8i, 5G T V 28|, 10{f (81) 3|, 8, f, 5J, 9J. / (82) ISA, 6|, 41f, 31^. (83) 16^, HI, 41 V 0^, 8^f, 194, Subtract: (85) 2| from 5J. , (86) 3 1 from 6 T V (87) | from |. /(88) 13| 30 MULTIPLICATION AND DIVISION OF FRACTIONS. PRINCIPLES: I. Multiplying the numerator j Mul J. UeB the or rt ,. .,. ,, , . traction, dividing the denominator ) II. Dividing the numerator j Div - des ^ multiplying the denominator ) EXAMPLE. Multiply by -. Model. It is evident that if I multiply by 4 my answer will be f. For according to Principle I. multi- plying the numerator multiplies the value of the frac- tion. But I have not to be multiplied by 4, but by the of 4. Hence, since my multiplier was 5 times too large, my answer is 5 tiroes too large and in order to obtain the true result I must divide my present answer by 5. But according to Principle II. multiplying the denominator divides the value of the fraction. There- fore |-h5= T 8 g-. This is the result obtained by multi- plying the numerators and denominators together respectively. EXAMPLE. Divide f by f. Model. It is evident that if I divide | by 2 my answer will be T 3 y , for according to Principle II. multi- plying the denominator divides the value of the frac- tion. But I have not | to be divided by 2 but by the ^ of 2. Hence, since my divisor was 3 times too large, my answer is 3 times too small, and in order to obtain IN ARITHMETIC. 31 the true result I must multiply my present answer by 3. But according to Principle I. multiplying the numerator multiplies the value of the fraction. Hence, -j^xSrdrr^. Ans. This is equivalent to inverting the divisor and multiplying, since we multiplied by 3, the denominator o the divisor, and divided by 2, the numerator. When it is possible to shorten the work do so, either by cancellation or by aliquot parts. EXAMPLE. Multiply f by . Operation: * f xf=ij. Ans. By cancellation. EXAMPLE. Multiply 48 by 2|. Operation: 2J=J of 10. Hence, 48x24=480^-4 =120. Ans. EXAMPLE. Divide 40 by 3|. Operation; 3|=* of 10. Hence, 40-^31=4x3 =12. Ans. Here is another method which is recommended for most examples in multiplication of fractions. EXAMPLE. Multiply 4 by 24. Operation: 4| _S 8 11 _ 11 1 Ans. Multiply the upper number first by 2, then by Add the results. The product is as given. 32 PRACTICAL METHODS EXAMPLES. Multiply : W Jby *,',. (93) OJ by 4. (90) |, I T V f (94) 15 ft by 64. ' (91) 82, 4, },,6. r (95) 140^ by 3 T 3 T (92) 423 by 4$. (96) 71|byl|by3J. Divide : ( 97) 56 by 2|. (101) by 2f. ( 98) 13 by 3J. (102) 162 by 18 1. ( 99) 138 by 5f. ^(103) 67Abyl01 T V (100) 4| by If. (104) | by 25| DECIMAL FRACTIONS. Students have opportunities of greatly abridging their work in decimal fractions, which are frequently long and tedious. Below we give methods in both multiplication and division. EXAMPLE. The compound amount of $1 for 40 years, at 6 per cent, is $10.2857179. What is the amount for 80 years to three decimal places ? Operation: 10.2857179 97175 8201 10 2857 2057 823 51 7 $105.795 Ans. Solution: The amount will be the amount for 40 years, multiplied by itself. IN ARITHMETIC. 33 Write the units figure of the multiplier under the decimal place we wish to preserve. Write the other figures of the multiplier in reverse order, as shown in the operation. 1st. Commence at the right, beginning to multiply with that figure of the multiplicand imme- diately above the one used as a multiplier. 2d. Mul- tiply by the next figure of the multiplier, beginning with the figure immediately above it, and placing the first product in the right hand column of results. Continue until all the figures possible have been used, placing the results so that the right hand column may be even. Allowance must be made for carrying as in ordinary multiplication. EXAMPLES. Multiply : (105) 10.87563 by 2.34073 to 2 places. (106) 16.1395 by 4.635 to 1 place. (107) .06708 by 31.97 to 4 places. (108) 19.072 by 84.7624 to 3 places. (109) The compound amount of $1, at 8 % for 35 years, is $14.7853443. What is the amount of $240 for 70 years to 3 decimal places. (110) The compound amount of $1, at 10 % for 28 years, is $14.4209936. For 15 years it is $4.1772482. How much would it be for 43 years to 2 decimal places ? EXAMPLE. Divide 18.9642 by 2.607 to three places. Operation: 2.^)18.9642(7.275 Ans. 18_249 716 52J. 194 182 34 PRACTICAL METHODS Draw a perpendicular line between the third and fourth decimal places. Divide as usual till this line is reached, then instead of bringing down another figure or cipher, strike out one from the divisor after each division and proceed as in ordinary division. The place where the line should be dmwn through the dividend is determined as follows: Multiply the decimal place to be reserved by one of the highest denomination in the divisor. This will give the number of places to be left in the dividend. Illustration: Divide 18.9G42 by 12.007 to three places. .001=place to be reserved. 10=highest denomination in divisor. 10 x. 001=.01. Hence the line should be drawn through the dividend after the second decimal place. EXAMPLES. Divide: (111) 4.562 by 8.75 to 3 places. (112) 20 by 3.104 to 2 places. (113) .04078 by .0036 to 1 place. (114) 91.875 by 15.76 to 4 places. (115) 3,723.84 by 96.7081 to 3 places. (116) If $1 amounts in twelve years, compound in- terest at 7 per cent., to $2.2521916, what sum in the same time would amount to $1,200, reserving two places ? IN ARITHMETIC. 35 PUBLIC SURVEYS. In most of the western and southern states a system of public surveys has made the artificial divisions of land uniform and definite. This is of great import- ance in keeping a public record of land areas and their transfer from one owner to another, which is common in the United States. The method which has been devised makes a description of any piece of land easy and expressed in few words. Several Principal Meridians are taken at convenient intervals from which to measure distances East or West. The one from which lands in Iowa, Missouri and Minnesota are surveyed, is the Fifth, which extends north from the mouth of the Arkansas River. For designating distances North or South, a base line is taken, the one corresponding to the Fifth Meridian, running West from the mouth of the St. Francis River. O The place where these two lines cross is the starting point for the description of lands situated in the above named States. The first row of townships north of the base line is called Township I. North; the second, Township II. North, and so on. Those south of the line are called Township I. South ; Township II. South, and so on. The first row of Townships west of the Principal Meridian is called Range I. West ; the second, Range II. West, etc. Those east are numbered in the same way, Range I. East, Range II. East, etc. It is evident that these two terms, Range and Town- ship, locate exactly tlie position of each township, and 86 PRACTICAL METHODS it is only necessary to further designate by appropriate names the precise part of the township occupied by the land area we wish to describe. Each township is six miles square, hence contains 36 square miles. This however is not exactly true, since the meridians which are followed in surveying the land, converge as we go north, thus making the north- ern line of the township shorter than the southern. The farther north we go the greater this difference. In Iowa it is about 8 feet to the mile. If this differ- ence were allowed to accumulate the townships would continually grow smaller towards the north. In order to keep them of uniform size correction lines are estab- lished, and the township corners placed as upon the base line six miles apart. These correction lines are 24 miles apirt north of the base line, and 30 miles south of the base line. The township is divided into sections, each one mile square, which are numbered from 1 to 36. In number- ing these sections we begin in the north-east corner of the township, which is Section 1. Number west 6 sections, then east 6 sections, etc., till all are numbered. A section contains 640 acres (with the exception of those that are fractional), and is divided into quarters, N.E., N. W., S.E., and S W. Each of these, containing 160 acres, is in turn divided into halves, E.-|, and "W.-J, by a line running north and south. They are also divided into quarters: N.E. J, N.W. J, S.E. J, and S.W. , each containing 40 acres. This method of division may be continued indefinitely. In describing a piece of land, we give its position in the section, township, then the number of township and range. 37 LOS ANC Suppose we wish to describe the township marked with a star in Figure 1. COB REO TIO N L/ INK 6 5 K M * 4a M M fXi M O M M h t 3 b 2 VI. V. IV. III. II. I. B A S E !il NE. FIGURE 1. The description would be as follows: Township 4 North, Kange 5 West from the Fifth Principal Meridian. TOWNSHIP. 6 5 4 3 2 1 7 8 9 a 10 11 12 18 17 16 15 14 13 ! 19 20 21 * 22 23 24 30 29 28 27 26 25 31 32 b 33 34 33 36 FlQUBK 2. 88 PRACTICAL METHODS SECTION. SECTION 22. N. W. '.,. 160 A. N. E. ' 4 . S.W. ',. 8. E. }i. = W N. y t . 40 A. * NN' b S. '/,. N.W.N.E. S.W . Vi. sTw^. s. E. *' N.W. h. N.E.k. .^ + S.W .', S.K,, * FlOUHE 3. FlOUKE 4. In figure 4 the part marked with a star would be described as follows : The N. E. J of the N. E. J of Section 22. If we take the Twp. given in Fig. 1, the entire description will be : The N. E. J of the N. E. J of Section 22, Tmvp. 4 North, Kange 5 West of the Fifth Principal Meridun. EXAMPLES. (117) Describe the pieces of land marked in figures 1, 2 and 4, by f, a, b, *. (118) How much land in the W. J of the S. E. J of Section 20, Township 70 North, Range 7 West of the Fifth Principal Meridian? (119) How much land in the following described property : The N. E. A of the N. E. J of N. W. ] of Section 17, Township G7 North, Range () West? Also, the E. of the S. E. J of Section 17, Town- ship 67 North, Range 6 West? IN ARITHMETIC. 39 There are many forms in which conveyances of Real Estate may be expressed. We give here a form common in Iowa: DEED. KNOW ALL MEN BY THESE PRESENTS, 1 That we, Thomas Milton and Mary Milton, husband and wife, of the County of Davis, and State of Iowa, for the consideration of EIGHT HUNDRED DOLLARS, Hereby convey to James Baxter, of tbe County of Davis, a< d State of Iowa, the following described Real Estate, situate in the County of Davis, and State of Iowa, to-wit : The N. E. ^ of the S. W. & of Section 7, Township 66 Nor.h, Range 13 West 40 acres. And we Warrant the title of the same against all persons whom- soever. In Witness Whereof, We have hereunto set our hands this 22d day of December, 1888. THOMAS MILTON. MARY MILTON. Attest : JOHN SMITH, Notary Public. Appended to this a warranty deed is usually placed an acknowledgment, made by the parties sign- ing the deed, before i notary public, and attested by his notarial seal. ACKNO WLEDGEMEN T. STATE OP IOWA, DAVIS COUNTY. SS. Before me, the undersigned, a Notary Public in and for said county, personally appeared Thomas Milton and Mary Milton, who are personally known to me to be the identical persons whose names are affixed to the foregoing Deed as grantors, and acknowl- edged the same to be their voluntary act and deed. Given under my hand and official seal this 22d day of December, 1888. ,) M. B. HORN. SEAL. > (120) Draw up deeds for the separate parcels of land in the above examples. 40 PRACTICAL METHODS THE METTCIC SYSTEM. We have found that the Metric System can be taught most successfully by arranging all the terms used into one table. There are only fourteen or fifteen different terms used in this simple system of weights and measures, and they can be remembered most easily by thus exhibiting them and their relative values in one group. The Metric System, which owes its great merit to its brevity, and the fact that it is a decimal system, was originated in France, about the time of the French Revolution. It is now in use in all the principal countries of Europe, except England, and was legal- ized in the United States in 1867. It will eventually become the standard for weights and measures in all civilized countries, much to the simplification of math- smatical and commercial operations. THE METER. An arc of one degree on a meridian was accurately measured, and from this the polar cir- cumference of the earth determined. The one 40- millionth part of this was taken as the standard of the new system, and called a meter. Its length is 39.37 + inches. This was divided into ten, one hundred, and one thousand parts, which were called respectively, decimeter, centimeter and millimeter. For greater distances the meter was multiplied by 10, 100, 1000 and 10,000, and the respective results designated by the terms dekameter, hectometer, kilometer and myria- meter. For the fractional parts Latin prefixes were IN ARITHMETIC. 41 taken, ending in i. For multiples and Greek prefixes, meaning 10, .100, etc., and ending in a or o. THE GRAM. The weight of a cubic centimeter of distilled water at its maximum density, 39 Fahrenheit, is the unit of weight. It is called the Gram. THE LITEE. The Liter is the measure of capacity. Its volume is that of a cubic decimeter. THE ARE. The Are is the unit for measuring sur- faces. It is a square, each side of which is a deka- meter or 10 meters, and hence contains 100 square meters. THE STERE. The Ste're is used for measuring wood. It is a cubic meter. The Meter is generally used for measuring the solid contents of a body. TABLE. 10 10 10 10 10 10 10 10 10 Myria Kilo Hecto Deka METEB. Deci Centi Milli Tonneau. Quintal. Myria Kilo Hecto Deka GRAM. Deci Centi Milli Kilo Hecto D. ka LITER. Deci Centi Milli Hecto ARE. Centi Deka STERE. Deci 60 8 394 5 It will be noticed that in the fourth line 100 cen- tiares are required to make 1 are, and 100 ares to make 1 hectare. In the others, 10 of one denomina- tion make one of the next higher. Quantities are written in the metric system as in decimal fractions, and can be changed from one denom- ination to another by a change of reading or of placing the decimal point. Thus, the figures below the table may be read as' 60 hectometers, 8 dekameters, 3 meters, 9 decimeters, 4 centimeters, and 5 millimeters. Or, as 42 PRACTICAL METHODS 60.83945 hectometers. Or, as 6083.945 meters. Or, as 60839.45 decimeters, etc. The use of this table will enable the pupil to obtain a mastery of the Metric System more quickly and easily than in any other manner. REMARKS. The centimeter and millimeter are mostly used in measuring very short distances. The kilometer for measuring long distances. The gram is used for weighing where great exact- ness is required; as jewels, precious stones, etc. For larger articles, such as groceries, the kilogram, or kilo, is commonly used. For very heavy articles, such as hay, etc., the tocneau, or metric ton, is used. For measuring moderate quantities of liquids aud solids, the liter is most commonly employed. The hectoliter for large quantities. The approximate values of the denominations of the metric, system most commonly used, are given in the following table : DENOMINATION. APPROX. VALUE. Meter 3 feet 3f inches. Decimeter 4 inches. Centimeter J inches. Kilometer 5 .furlongs. Are 4 square rods. Hectare 2^ acres. Liter 1 quart. Hectoliter 2jj bushels. Cubic Meter 35| cubic feet. Stere ^ cord. Gram 15^ grains. Kilogram 2| pounds. Metric Ton 2204 pounds. IN ARITHMETIC. 43 LONGITUDE AKD TIME. The earth rotates on its axis once in 24 hours. The shape of the earth is that of an oblate spheroid, the polar diameter being -g-J-g- less than the equatorial. But since the places on the earth's surface rotate in a plane perpendicular to its axis, each place describes an exact circle. Every circle, however large or small, contains SCO degrees. Hence, every point on the earth's surface, except at the very poles, describes every 24 hours, a circle or 360 degrees. In one hour, then, any point would pass through jfa of 360, or 15. But 1 hour=60 minutes. Therefore 15, the space passed over in 1 hour, divided by 60, will give the space passed over in 1 minute. This is ^ of a degree, or 15 '. This, divided by 60, since 60 seconds 1 minute, will give the space passed over in each second of time. The result is ^', or 15". From these facts we form the following table: TABLE FOR LONGITUDE AND TIME. 1 hour of time=15 of Longitude. 1 min. of time=15' of Longitude. 1 sec. of time=15" of Longitude. The marks, , ', ", indicate degrees, minutes and seconds, respectively, of longitude. To determine the position of a place on the earth's surface, only two things are required its distance north or south from the equator, and its distance east or west 44 PRACTICAL METHODS from a given meridian. The former is called Latitude. The latter, Longitude. A meridian is a great circle passing through the poles. It is evident there may be one for each sepa- rate place on the earth's surface. It is also evident that as the meridians approach the poles they converge, since they all pass through the same point. The distance between the meridians varies as the cosine of the latitude. At 40 north or south latitude a degree is only about f as long as at the equator. The length of a degree at the equator is about 69 miles. Hence, at 40 north latitude, it is about 52 miles. A point on the equator rotates with a velocity of 1037 miles per hour. At 40 north latitude this velocity is only 777 miles per hour. NOTE. If the earth rotated with a velocity 17 times as great as its present rate of motion, bodies would lose their weight. The force of revolution would equal gravity. Longitude is measured east and west 180 from what is called the Prime Meridian. The one usually taken is the meridian which passes through Greenwich, though the meridians of Washington, Paris, and Ber- lin are sometimes used in their respective countries. The difference of longitude between two places is ascertained in various ways. Most generally it is found by astronomical observation of the passage of a star across the meridians of the two places. The time of transit at each place is recorded by means of the telegraph at the other. After allowing for the time necessary for the transmission of the electric fluid, the IN ARITHMETIC. 4 > difference of time between two places may be reduced to longitude by multiplying by 15. The earth rotates on its axis toward the east. Hence the farther east a place, the earlier the sun will appear to rise, or, the time will be laier than at a place farther west. On the other hand the farther west a place, the later the sun will appear to rise, hence the time will be earlier than at places farther east. From the foregoing facts we deduce the following rules: I. To reduce time to longitude, multiply by 15. II. To reduce longitude to time, divide by 15. III. To find the time at a place east of a given place, add their difference of time. IV. To find the time at a place west of a given place, subtract their difference of time. The civil day begins at midnight; the astronomical day at noon. There must be some place where the day changes from one to the succeeding day. Suppose a person start at noon Wednesday, on a^ journey to the west, traveling at the same rate as the sun. Wherever he may be it will be noon to him, as the sun will always be on the meridian. In 24 hours he will have passed around the earth, and though it has always been noon to him, when he returns to the place he started it is not Wednesday noon but Thursday noon. When and where did this change occur? By common consent this place is about 180 from Greenwich, in the Pacific Ocean. It is called the "International Date Line." It is not a meridian but an irregular line, so made to suit 46 PRACTICAL METHODS the purposes of trade, as it would be inconvenient if the date line passed through an inhabited country or island. It is always one day later on one side of the date line than on the other. When it is Tuesday in China it is Monday in America. A ship sailing to- ward China would skip a day upon crossing the line, one sailing in the opposite direction would fall back a day. EXAMPLE. The time at New York is 9:15 A.M. At Bloomfield, la., it is 8:23 A.M. What is the differ- ence in longitude between the two places? Solution 9 : 158 : 23==52 m. 52 m xl5=13=Dif. in Longitude. Ans. EXAMPLE. The longitude of St. Petersburg is 30 16 ' East. That of Boston 71 4 ' West. What is their difference of time? Solution: 71 4' 30 16' 15)101 2Q'=Pif. in Longitude. 6hr. 45m. 20sec. Dif. in Time. Ans. When places are on the same side of the prime meridian, subtract to find their difference of longitude. When they are on opposite sides of the prime meridian, add to find their difference in longitude. EXAMPLES. (121) The longitude of New York is 74 24" W. That of San Francisco 122 27' 49" W. When it is noon at San Francisco, what time is it at New York ? (122) The longitude of Des Moines, la., is 93 37' 16 " W . That of St. Petersburg 30 16 ' E. When it IN ARITHMETIC. 47 is 2 o'clock A.M. at St. Petersburg, what time is it at Des Moines? (123) The difference in time between Chicago, 111., and Paris, Fr., is 5hr. 59in. 40 sec. If Chicago is in longitude 87 35' W., what is the longitude of Paris? (124) The longitude of Pittsburg, Pa., is 80 2 'W. If it is midnight at St. Louis when it is 40m. 41 sec. A.M. at Pittsburg, what is the longitude of St. Louis? (125) The longitude of Pekin, China, is 116, 26' E. That of Rio Janeiro, Brazil, 43 20' W. When it is 6 P. M. Tuesday at Pekin, what is the tin e at Rio Janeiro ? (126) The time at Washington, longitude 77 36" W., is 9 o'clock A. M. The time at Berlin "s 1 min. 34| sec. past 3 p. M. What is the longitude of Berlin? (127) Bombay is in longitude 72 48 ' E. Denver, Colo., in longitude 105 W. When it is 7 P. M. at Denver, what is the time at Bombay? RATIO. Ratio is the numerical relation existing between numbers of the same denomination. The ratio of numbers is obtained by dividing one of the numbers by the other. The sign of a ratio is the colon ( : ), which is an abbreviation of the sign of division, the horizontal line being omitted. 48 PRACTICAL METHODS The antecedent is the first terra of the ratio. The consequent is the second term of the ratio. Since the sign of a ratio is really a sign of division, the ratio between two numbers is obtained by dividing the antecedent by the consequent. Thus: 4 : 6=4-=- 6, or . A simple ratio is one of two terms ; as, 2:3, 5:7, etc. A compound ratio is the combination of two or more simple ratios. __ 2 y 3 "*" ' 3 : 5 The ratio can be found of numbers of the same denomination only. EXAMPLES. What is the ratio of (128) 1 ft. to 1 in.? (129) 2 gal. 2 qt. to 3 gal. 1 pi? (130) 5 hr. 24 min. to 1 day? (131) $16.50 to $3.12|? (132) 2 A. 40rds. to 20 A.? (133) JtoJ? (134) 3 T 6 T to2^.? (135) 34to57|? PROPORTION. An equality of ratios is called a Proportion. That is, when the relation existing between two sets of quantities is the same, the quantities are said to be in proportion. IN ARITHMETIC. 49 The double colon ( : : ) is the sign of proportion. Thus, 4:6 : : 6:9; which is read, 4 is to 6 as 6 is to 9, or the ratio of 4 to 6=the ratio of 6 to 9. There are 4 terms to a proportion, designated in numerical order: 1st, 2d, 3d, 4th. The first and fourth terms are called the extremes; the second and third the means. The great rule of proportion is as follows : Rule. The product of the extremes in a true pro- portion is equal to the product of the means. Hence, when any three terms are given, the other one may be easily found. SIMPLE PROPOBTION. Simple Proportion is the equality of two simple ratios. For finding the fourth proportional when three are given the following is the best method: Rule. Write for the Third Term that number which is of the denomination required in the answer. If from the nature of the problem the answer is to be larger than the third term, place the larger of the two remain- ing numbers for the second term, and the smaller for the first. If the answer is to be smaller than the third term place the smaller of the two numbers for the second term, and the larger for the first. Divide the first term into the product of the second and third. The quotient will be the fourth term. 50 PRACTICAL METHODS EXAMPLE. If $240 produces $12 interest in 1 year, Low much will $400 produce? Solution: 240:400: : 12: ( ) Place $12 for the third term because it is of the denomination (interest) required in the answer. If $240 produces $12, will $400 produce more or less than $12? It will produce more. Therefore place $400 for the second term and $240 for' the first. 400 x $12 ^$20. Ans. 240 NOTE. Use cancellation whenever possible. EXAMPLES. (136) A man earns $2.25 by 1^ days work. How much will he earn in 10^ days? (137) If sound travels 6,672 ft. in 6 sec. how far will it travel in 22| sec. ? (138) A building 42 ft. high casts, at noon, a shadow 26 ft. long. How high must a building be to cast a shadow 42 ft. long? (139) If the velocity of the electric fluid is such that it would travel around the earth (24,897 m.) 11^ times in 1 sec., how far would it travel at the same rate in 2 3 5 of a sec. ? (140) Light travels from the sun to the earth (92,- 000,000 m. ) in 498 sec. How long would it bo in traveling from the earth to the moon (238,000 m.) ? (141) If a pipe flows 118 gals, of water in 6 minutes, how many pipes would it take to flow the same amount in 1 minutes? IN ARITHMETIC. 51 COMPOUND PKOPORTION. Compound proportion is an equality of ratios, one or more of which is compound. Rule. Place for the third term that number which is of the denomination required in the answer. Take each pair of numbers and arrange them according to the rule for simple proportion, always asking the question with regard to the third term. Divide the product of the third and second terms by the first term. The quotient will be the fourth term required. EXAMPLE. If 4 men in 6 days mow a field of 25 acres, how long will it take 7 men to mow a field of 80 acres ? Solution: 1 4 ::6 4x6x16 , . ^xs^ = io 3 4 days. Ara. 86- 16 7x5 7x5 Place 6 days for the third term, because days is required in the answer. If it takes 4 men 6 days to mow a field, will it take 7 men longer, or not so long? It will not take so long. Hence, take 4 for the second term and 7 for the first. If 25 acres are mowed in 6 days, will it take more or less than 6 days to mow 80 acres? It will take longer. Therefore place 80 for the second term and 25 for the first. By multiplication and division, the result is 10-f-f days. REMARK. Notice that the question is always asked with regard to the 3d term, by each pair of numbers, without any reference to the other numbers involved. 52 PRACTICAL METHODS SOLUTION BY CAUSE AND EFFECT. Solutions of examples in proportion, by cause and effect, are based on the theory that: I. Like causes produce like effects. II. Causes are uniform in their operation. And, therefore: III. Effects are propor- tioned to the causes that produce them. Rule. Arrange all the numbers that constitute the first cause as the first term of the proportion. Ar- range all the numbers that constitute the second cause for the second term. Place the first effect for the third term, and the second effect for the fourth term. Divide the product of the terms in which the unknown quantity is found into the product of the other terms. The result will be the required number. EXAMPLE. If a bin 12 ft. long, 8 ft. wide, and 7 ft. high contains 268 bu. of corn, how wide must a bin be whose length is 15 ft. and depth 6* ft. to contain 300 bu. ? Solution: The first cause is a bin 12 ft. long, 8 ft. wide, and 7 ft. high. The second cause is a bin 15 ft. long, 6 ft. deep, and (X) ft. wide. The first effect is 268 bu. The second effect 300 bu. Arrange accord- ingly and solve. 12: 15 8 6 ::268:300 7 (X) 20 ' === -- = 8f 4 ft. Ans. * X 67 NOTE. The student must use bis own judgment, which one of the two methods will be the most convenient for each example. IN ARITHMETIC. 53 EXAMPLES. (142) If the use of $450 for 2 yrs. 6 mo., at 6 per cent, is $67.50, what sum would produce the same amount in 3 yr. 8 mo. at 8 per cent. ? (143) If 12 men cut a pile of wood 96 ft. long, 8 ft. high and 4 ft. wide, in 2^ days, how long would it take 6 men to cut a pile of wood 38 ft. long, 12 ft. wide and 6 ft. high? (144) What sum of money in 2^ yrs., at 10 per cent., would produce 3 times as much interest as $640 at 12 per cent, for 5 yrs. ? (145) If 40 yds. of carpeting are required to cover a floor 22 ft. 6 in. long, 18 ft. 4 in. wide, how many would be required of the same width to cover a floor 14 ft. 6 in. wide, and 15 ft. 2 in. long? (146) If 42 men dig a trench 365 ft. long, 4^ ft. wide, 6 ft. deep, in 7f days, by working 10J hours a day, how long a trench 3 ft. wide and 5 ft. deep, can 18 men dig in 14 days, working 9 hours a day? (147) If 6 men in 4| days, of 8 hours each, mow a field of 21-J A., how many hours a day would 5 men require, working 9| days, to mow a field of 50f A. ? GENERAL PRINCIPLES. As we stated in the preface to this little volume, there are some principles in Arithmetic which apply to a great number and variety of subjects. We give some of them below, and shall constantly use them in the remainder of the work. The same conditions con- stantly reappear in problems under widely different subjects. If we apply the same rules to all these prob- 54 PRACTICAL METHODS lems much time will be saved from committing unnec- essary rules, and Arithmetic begins to show to us the unity of a science, instead of being a mass of details with no organic relation existing between them. PRINCIPLE I. To find the fractional part of a number: Multiply by the fractional part. This rule mny be used in Common and Decimal Fractious, Percentage, and its numerous applications, Mensuration, etc. PRINCIPLE II. To find what part one number is of another: Divide the one that's " is " by the one that's " o/." Probably no rule in Arithmetic applies so widely as this. It may be used in all the subjects named above. PRINCIPLE III. To find a number when a fractional part of it is given : Divide by the fractional part. These rules, in addition to a few special ones belong- ing to different subjects, are usually all that are neces- sary for one-half the operations of Arithmetic. They will be referred to hereafter as Rules I, II. and III. IN ARITHMETIC. 55 PERCENTAGE. Percentage is a general term which embraces all calculations in which 100 is taken as the basis of com- parison. The word is taken from two Latin words, per and centum, meaning by the hundred. The Bate is the number of hundredths that is taken of a number. It is written as a decimal fraction of the denomination of hundredths, or is designated by the per cent mark, %. The Percentage is the result obtained by multiply- ing the number on which percentage is to be computed, by the rate. ALIQUOT PARTS. Before proceeding to the discussion of problems, we give a table of Aliquot Parts of a hundred, by using which the work of many problems may be abridged and made easier. TABLE OF ALIQUOT PARTS. 4 %=&. 37J%=4 5 %-sV 40 %= 10 %=rV 50 %=i- iH%=f 574%=*. 12i%=|. 60 %= 20 %=\. 75 %= 22|%=f. Hi%= 25 %=J. 284 %=|. 66 All the problems in Percentage may be solved by one or more of the general principles given above. It is unnecessary and unwise to divide the subject into four different cases, as is common in most arithmetics. The mind learns and remembers with difficulty the artificial divisions, which confuse by their number, without adding clearness to the mental vision. At any rate, no more than three cases should be formed, corresponding to the 3 principles given, which apply to Percentage and nearly all its applications. It is proposed to give an example for each of the four cases usually treated of, in order to show that they can all be solved by the above principles. EXAMPLE. Find 8 per cent, of 240. Solution: Use Rule I. 240 x. 08=19. 2. Ans. REMARK. Express the rate as a common fraction whenever it is a convenient aliquot part of 100. EXAMPLE. Find 12i per cent, of 328. Solution: 12^ per cent=|-. Rule I. 328 X= 41 . Ans. EXAMPLE. Find 44| per cent, of 72. Solution: 44^ per cent.=. -| ^ 72=32. Ans. EXAMPLE. 8 is how many per cent, of 32 ? Solution: 8=^ of 32. ^=25 per cent. Ans. Or, use Kule II, The " is" number is 8. The "o/" number 32. 8-^32=25 per cent. Ans. EXAMPLE. 18 is how many per cent, of 156? Solution: Eule II. 18^-156=.11 T V = 11 T V per cent. -4ns. IN ARITHMETIC. 57 EXAMPLE. $3.80 is 5 per cent, of what sum? Solution: Kule III. $3.80-f-.05=$76. Ans. Or, Rule II. $3.80 is 5 per cent, of what No.? S3. 80 -=-.05 = $76. Ans. EXAMPLE. r 2 r is 80 per cent, of what No. ? Solution: Rules II. or III. 80%=f T 2 T^-f=irV Ans. EXAMPLE. f- of a number is 10. What is the No. ? Solution: Rule III. 10-^|=16. Ans. Enough problems have now been given to show how the three rules may be applied. The student should solve the following problems in the same manner. EXAMPLES. (148) Find: llf % of $4,230. (149) 37^ % of 172. (150) n % of . (151) IQOO" % of 20. (152) 1C6| % of 84. Solution: 166 %=|. f of 84=140. Ans. (153) ,225 % of 92.6 (154) 10 % of 121 % of 22f % of 23.64. (155) 104 % of 150 cattle. (156) Find 40 % of 22-f % of 15 % of 816. (157) $3.20 is what % of $2000? (158) $5.12 is what % of $960? (159) is what % of f ? (160) 625 men is what % of 16,000 men? (161) 10 % of 33J % of 75 % of a number is what per cent, of it ? 58 PRACTICAL METHODS (1G2) How many per cent, of a Twp. 6 miles square does a man own who has 960 A. 9 (163) 12 is H % of what number? (164) 31 cents is 22 1 % of what sum? (165) 184 men are 3^ % of how many men? (166) 4 gal. is 14 f % of what? (167) 3| mi. is 87 A % of what? (168) $4.50 is 6 %"of what sum? (169) 14 horses are 40 2. When you have the face value, Multiply \ ' remum or 1 Discount. The result of Eule 1 will be the Face Value. The result of Eule 2 will be the Amount Invested. REMARK. Sometimes stock is said to be purchased at 80 per cent. In this case 80 per cent, represents 1 Discount, and the given Bum should be at once multiplied or divided, according to the conditions of the problem. EXAMPLE. I invest $34000 in C., B. & Q. B. E. stock at 85 per cent. If the stock yields 6 per cent. per annum, what will be my income ? Solution: The rule just given applies: $34000-^.85 = $40000 = Face Value. 6 % of $40000=$2400. Ans. Or, by Eule II. $34000 is 85 per cent, of what sum? $34000--.85= : $40000. By Eule I. 6 % of $40000=$2400. Ans. EXAMPLE. Which is the more profitable, U. S. 4J's bought at 105, or Panama stock, bearing 4 per cent. interest, and purchased in the market at 90 per cent. ? Solution: or 3 of the amount invested inU. S. Bonds is returned in the form of interest. -g 4 or 4 2 T of the amount invested in Panama stock is returned in the form of interest. Which is the greater, ^ or -/$ ? 2 . Hence the Panama stock is the better investment. IN ARITHMETIC. 75 SECOND METHOD. What % of 105 is 4? 4J-r-105==4| %. What % of 90 is 4? 4-^-90=4$ %. 4 1 4f J^. Hence the Panama stock is the better investment. EXAMPLE. C. & N. W. E. E. stock, quoted in the market at 79, yields 6 per cent, income. If A receives 400 as his share, what is the value of his stock? Solution: Eules I. and II. $400 is 6 per cent, of what sum ? $400 ~ .06| = $6,000 == face of stock. $6,000 x 79 %=$4,740^value of stock. Ans. EXAMPLE. If 7 per cent. Lake Shore Eailroad shares are bought at 84 per cent., what is the rate of income? Solution: 7 T of the amount invested is received as interest 7 T = 8| %. Ans. EXAMPLE. How must I purchase N. Y. Central E. E. stock which yields 5 per cent, dividend, so as to derive an income of 8 per cent? Solution: Eule II. On every dollar, face value, of the stock I receive 5c. By the conditions of the problem this is to be 8 per cent, of the investment. 5c. is 8 per cent, of what sum ? 5c. -s-8 %=.62= 62 J per cent. Ans. EXAMPLES. (208) I invest $18,400 in. A., T. & S. F. E. E. stock at 92 per cent. If the stock yields 5 per cent, per annum, what will be my income? (209) I invest $31,050 in the purchase of U.S. 3|'s at 103^ per cent. What is my income from them ? 76 PRACTICAL METHODS (210) Denver and Rio Grande stock yielding 7i per cent, may be purchased at 98 per cent. Union Pa- cific stock yielding 44 per cent, at 76 per cent. Which should I prefer? (211) A receives $360 as his share of a 5 per cent, dividend on stock purchased at 107. How much did he have invested? (212) A 3 per cent, dividend was declared on stock which I purchased at 92^ per cent. If I received $195 how much did I pay for the stock? (213) U. S. bonds bearing 5 per cent, are bought at 112^. What is the rate of income? (214) Illinois Central Stock was purchased at 64 per cent, and yields 2^ per cent, dividend. What is the rate of income ? (215) What is the market value of stock yielding a 6^ per cent, dividend and 7^ per cent, rate of income ? (216) Stock which bears 4 per cent, dividend, pro- duces an income of 5 per cent. What is its market value? CONTRACTS. KEMABKS. The subject of contracts, though properly belonging to Commercial Law, is so intimately connected with commercial transactions presented in Arithmetic, as to be discussed here. We hardly see how the faithful teacher can avoid giving his Arithme- tic classes the principal facts in connection with this important branch of the law. A contract is an agreement between two or more persons to do or not to do some certain thing. The following four things are requisite for a valid contract: IN ARITHMETIC. 77 1, Parties; 2, consent of parties; 3, consideration; 4, subject matter. I. PARTIES. The parties to a contract must be such as can legally enter into the agreement. The law usually holds that minors, idiots and lunatics cannot make a binding contract. Formerly this list included married women, but at present they can hold property and make contracts in nearly all the States of the Union. Minors are persons under 21 years of age. Their contracts are not void, but voidable. They are binding on the other party to the contract, but may be repu- diated by the minor. He can be held legally respon- sible, however, for the following things: 1, clothing; 2, food ; 3, lodging ; 4, education ; and 5, medicine. II. CONSENT OF PARTIES. The agreement must be entered into voluntarily. Any fraud, deceit or force by either party invalidates the contract with regard to the party injured. The party at fault, however, can- not, by disregarding the contract, take advantage of his own wrong. III. CONSIDERATION. The consideration is the inducement or reason for entering into the contract. It need not be expressed in property. Kelation- ship, friendship, or gratitude may be a sufficient consideration to bind a contract. Nor is the amount of the consideration essential. Any sum or thing, pro- vided it be valuable, may warrant the execution of the contract. IV. SUBJECT MATTER OF CONTRACTS. It is unnec- essary to enter into a contract to do that which the law commands. Such contracts can not be enforced. On the other hand, a contract to do something which 78 PRACTICAL METHODS is prohibited by law, is void. Such contracts are those which are against public policy, those which are for immoral purposes, or are fraudulent. A contract is either express or implied. An express contract is one which is entered into by express agreement. An implied contract is one in which the assent of the parties to the contract, or some part of it, is presumed from circumstances. As, where a man pur- chases a horse, the law will presume a promise to pay a reasonable compensation, if a price has not been agreed on. Or, when one hires another to work for him and nothing is said about wages, there is an implied agreement to pay a just price for the same. INTEREST. Interest may be denned as the remuneration paid for the use of money. This does not include rent, dividends, etc., which are the remuneration paid for the use of capital in the form of property other than money. INTEREST is common in all civilized countries. Formerly the taking of interest in any form was denominated usury. This term is now very much restricted in its meaning. USURY is interest which is taken in excess of the rate allowed by law. A penalty is attached to the prac- tice of usury in many states. In others any rate of interest whatever may be charged, thus doing away entirely with usury. IN ARITHMETIC. 79 THE LEGAL KATE of interest is the highest rate allowed by law. THE PRINCIPAL is the sum on which interest is to be computed. THE BATE is the amount which is charged for the use of $1 for 1 year. THE AMOUNT is the sum of the principal and interest. SIMPLE INTEREST is interest on the principal for the given time, and which itself cannot draw interest. COMPOUND INTEREST is interest on the principal, and also on the interest, which, upon coming due, is im- mediately converted into principal. METHODS OF COMPUTING INTEREST. Out of the many methods of computing interest, two are selected which are considered the best. These are the Aliquot and the Twelve per cent, method. NOTE : In the following examples 30 days will be considered a month. EXAMPLE. What is the interest on $240 for 2 yrs. 4 m., 15 da., at per cent? Solution: By aliquot parts. 6 % of $240 = $14. 40 = Interest for 1 yr. *=J 28.80= " " 2 " 4.80= " " 4 m. .60= " "15 da. $34.20. Ans. EXAMPLE. Find the interest at 8 per cent, on $165 for 9 m. 27 da. 80 PRACTICAL METHODS Solution: Twelve per cent, method. The interest on $1 at 12 % for 9 m.=.09. For '27 da.=.009. Hence for 9 m. 27 da.=.099. 8 = f of 12. Int. at 8 % = of .099=.066. Int. on $165 = 165 x. 066--= $10.89. Ans. EXPLANATION OF 12 PER CENT. RULE. 12 % per annum = 1 % per m., since 1 yr.= 12 m. 1% for 1 m. or 30 da.=.OOOJ for 1 da. Hence the rule. RULE. The number of months expressed in cents, and ^ the number of days expressed in mills equals the interest on $1 at 12% for the given time. EXAMPLE. Find the simple interest on $1348.25 for 1 yr., 7 m., 18 da., at 12%. Solution: 1 yr. 7 m.=19 m. Interest for 19 m.= .19. Interest for 18 da. =.006. Hence interest for 1 yr. 7 m. 18 da. at 12 % on $1=.196. On $1348.25 it would be 1348.25 X.196=$264.257. Ans. EXAMPLE. How long will it take $640 to produce $120, simple interest, at 6% ? Solution: Interest on $640 for 1 yr. at 6% =$38.40. $120-$38.40-=3i yr.=3 yr. 1 m. 15 da. Ans. EXAMPLE. What is the rate of interest when $1665 produces $116.55 annually. Solution: Rule II. $116.55 is what % of $1665? 116.55-- 1665=.07=7%. Ans. EXAMPLE. What principal will produce $298.35 a year at 9%? IN ARITHMETIC. 81 Solution: Bule II. 1298.35 is 9% of what sum? .35-r-.09=$3315. -4ns. EXAMPLE. What principal will amount in 2 yr. 1 m. 6 da. at 1% to $1376.40? Solution: Eule II. Amount of $1 at 7% for 2 yr. 1 m. 6 da.=1.147. $1376.40=114.7% of what sum? $1376.40^-114.7%=:$1200. Ans. EXAMPLES. NOTE. To get exact or accurate interest 365 days should be counted a year. The interest on any sum counting this number of days will be - 7 V less than by using 360 days. To get exact interest, therefore, it is only necessary to diminish the interest obtained in the usual way by -fa of itself. (217) Find the simple interest on $298.77 for 2| yrs. at 6 %. (218) $112.60 for 4 yr. 7 m. 14 da. at 8 %. (219) $2,356 for 9 m. 27 da. at 8 %. (220) $1,749.32 for 1 yr. 2 m. 25 da. at 9 %. (221) $750.70 for 3 yr. 11 m. 29 da. at 10 %. (222) How long will it take $500 to produce $50 at 8 % ? (223) $244 to produce $32 at 6 % ? (224) $974.60 to produce $118.20 at 8 % ? (225) What is the rate when $400 produces $28 in lyr.? (226) $620 produces $28.93 in 8 mo.? (227) $2,350 produces $302.36 in 1 yr. 7 m. 9 da.? (228) What principal will produce $225.63 in 1 yr. at 9 % ? (229) $29.232 in 5 m. 24 da. at 9 % ? (230) $139.725 in 2 yr. 4 m. 18 da. at 8 %? 82 PRACTICAL METHODS (231) What principal will amount to $059.10 in 1 yr. 7 m. 21 da. at 6 % ? (232) To $512.85 in 10 m. 10 da. at 8 % ? (233) To $566.57 in 4 yr. 1 m. 24 da. at 9 % ? (234) To $2614.21 in 3 yr. 15 da. at 6 % ? (235) Find the exact interest on $519.20 at 7 % for 2 yr. (236) $123.80 at 8 % for 9 m. 3 da. (237) $4075 at 8 % for 3 m. PROMISSORY NOTES. A PBOMISSORY NOTE is a written promise to pay money at some future time. THE MAKER of a note is the party who makes him- self responsible for its payment by signing it. THE PAYEE is the party to whom the promise of payment is made. The one who owns the note at any time is the Holder. AN INDOBSER is one who places his signature on the back of the note. INDORSEMENT is for the purpose of transferring the note from one person to another. REMARK. If the name alone is written, this is a blank indorse- ment, and holds the indorser responsible for the payment of the note in case the maker fails to pay it. If the indorser wishes to free himself from such liability, he may write after his name the words, " with- out recourse." Another method of indorsement is, pay to " John J. Brown." This is a "special" indorsement. IN ARITHMETIC. 83 Promissory Notes are either Negotiable or Non- negotiable. A negotiable note is one that can be transferred from one person to another. A non-negotiable note is one which can be collected only by the payee. The words "or order" or "or bearer" are necessary to render a note negotiable. When a note is drawn up to bearer, any person into whose possession the note comes legally can collect it, and no indorsement is necessary. If it is drawn up to order it can be transferred only by the indorsement of the party io whom it is made payable, or the holder. If a note is not paid when due, it is protested -that is, a written notice of the fact of non-payment is made out by a notary public and sent to the indorsers and security. If this is done they become liable for its payment. If not, they are relieved from all liability for its discharge. The Maturity of a note is the date on which it falls due. It is nominally due on the expiration of the time mentioned in the note. It is not legally due for three days thereafter. These three days are called "days of grace." To find the time when a note matures: If the time to elapse before payment is given in years and months, add the number of years and calendar months to the date of the note. If the time is given in days, the actual number of days must be counted from the date of the note, to which three days are added for grace. For example : If a note is dated Jan. 3d to run two months, it will be due March 3 /e. If it is to ran 60 84 PRACTICAL METHODS days it will be due 60 days after Jan . 3, or March It is not necessary to insert the words, " for value received," to make a note valid. The amount in the body of the note should be written in words, to prevent alteration or fraud of any kind. If the words " with interest " are found in a note, it draws interest from the date of the note, and a rate fixed by law is allowed if no rate be mentioned. If the note does not contain the words "with inter- est," it draws interest only after maturity. In many promissory notes the rate of interest is not specified. In all states a rate is fixed by law, which shall be charged in such cases upon the maturity of the note, or from the time specified therein. The following are the common forms used for promissory notes: I. 8317.20. BLOOMFIELD, IA., Dec. 27, 1888. Two months after date, I promise to pay to the order of H. E. Hibbard, three hundred and seventeen and fifo dollars, with inter- est at 6 per cent. B. A. GO AN. II. $250. DBS MOINBS, IA., Aug. 1, 1885. On demand, I promise to pay to J. M. Howie, or bearer, two hundred and fifty and A% dollars. B. M. WORTHINGTON. III. 865.00. CHICAGO, ILL., May 2, 1880. Sixty days after date, we, or either of us, promise to pay Jno. S. Woolson, or bearer, sixty-five dollars, for value received, with interest. FRANK NELSON. G. M. SULLIVAN. The above is a joint note, and the makers are jointly liable for its payment IN ARITHMETIC. 85 EXAMPLES. Find the date of maturity and the amount of each of the following notes : (238) $190.60. GALESBCBG, IM,., July 4, 1879. For value received I promise to pay, two months after date, to W. H. Sadler, one hundred and ninety and ^ dollars, with interest at 9 per cent. J. R. WILCOX. (239) $624.00. MADISON, Wis., Oct. 18, 1887. On or before the tenth day of Feb., 1889, 1 promise to pay U. S. Miller, or order, six hundred and twenty-four dollars, with in- terest at 8 per cent. JOHN MARTIN. (240) $2675.80. ST. PATO, MINN., Feb. 10, 1884. Ninety days after date I promise to pay Chas. Claghorn, or bearer, two thousand six hundred and seventy-five and fflg dollars, value received, with interest at 10 per cent. C. C. COCHEAN. PARTIAL PAYMENTS. When payments are made at various intervals on promissory notes, they are called Partial Payments. There are various methods of computing the amount due on a note when partial payments have been made. The only two which we care to mention here are the United States and the Mercantile rules. UNITED STATES RULE. Compute the interest on the principal from the date when the note begins to bear interest till the time of the first payment, add the interest to the principal and subtract the payment. Compute the interest on this sum as a principal until the next payment. Add the interest and subtract the 86 PRACTICAL METHODS payment as before. Continue in this manner till the date of settlement. Provided, that no interest shall be added which is in excess of the payment made at that time. In case the interest at any time exceeds the payment, interest is to be computed on the original sum to such a time that the sum of the payments may exceed the interest. It is the policy of the law to prevent the payment of compound interest, unless a special contract is made that unpaid interest shall become a part of the principal. This, of course, is to the debtor's advantage. For example: If a man borrows $100 at 6 per cent, for five years, by simple interest he should pay just $6 a year or $30 at the end of the five years. It is evi- dent that if, at any time, any of the interest be added to the principal, more than $6 of interest will accrue in a year, and to this extent will be more than simple interest. Now, this is the very thing done by the United States rule. Every time a payment is made the interest which has accrued at that time is added to the principal before the payment is deducted. The oftener payments are made the oftener is interest added to the former principal, thus making a rapid ac- cumulation of the debt and putting an extra burden upon the debtor for his promptness. If payments are made at short intervals compound interest in its most aggravated form will be the result of applying the U.S. rule. EXAMPLE. $640. OMAHA, NEB., Aug. 12, 1883. Three months after date I promise to pay to John W. Graham or order, six hundred and forty dollars, with interest at 6 per cent ROBERT J. BROWN. IN ARITHMETIC. 87 INDORSEMENTS. Dec. 12, 1883, $25; May 30, 1884, $110; Sept. 3, 1884, $200. What was due Jan. 1, 1885 ? Solution : Time from Aug. 12 to Dec. 12, 1883=4 mo. Int. on $ 1 for 4 m. at 12 per ct.=.04. For 6 per ct.=.02. Int.= Payment= Time from Dec. 12, 1883, to May 30, 1884=5 m. 18 da. Int. at 12 per ct. for 5 m. 18 da.=.056. Int.= At 6 per ct.=.028. Payment= Time from May 30, 1884, to Sept. 3, 1884=3 m. 3 da. Int. at 6 per cent, for 3 m. 3 da.=.0155. Int.= 6.40 .02 12.80 640. 652.80 25. 627.80 .028 17.578 627.80 645.378 110. 535.378 .0155 8.297 535.378 543.675 Payment^ 200. Time from Sept. 3, 1884, to Jan. 1, 1885=3 m. 28 da. 343.675 Int. at 6 per cent.=.019f. .019$ Int,= 6.759 343.675 Amount due Jan. 1, 1885. $350.43 Ans. EXAMPLES. ( 241 ) $2205. ST. Louis, Mo., July 5, 1880. Two years after date I promipe to pay to the order of James Stewart, the sum of two thousand two hundred and five dollars, with interest at 8 per cent. CHARLES MILBURN. INDORSEMENTS. Dec. 16, 1880, $250; May 1, 1881, 1881, $65; Feb. 28, 1882, What is due July 7, 1882? .30; July 1, 88 PRACTICAL METHODS ( 242) $134.40. BLOOMFIELD, IA., Jan. 2, 1883. Thirty days after date I promise to pay to Madison Turner, or order, one hundred and thirty-four and ^ dollars, value received, with interest at 6 per cent. THOMAS TAYLOR. INDORSEMENTS. Jan. 20, 1883, $5; Feb. 10, 1883, $40; March 31, 1883, $81. What was due April 2, 1883. COMMEKCIAL EULE. Compute the interest on the principal till the date of settlement. Add the interest to the principal. Find the interest on each payment from the time it was made to the date of settlement. Find the amount of all the payments and subtract from the amount of the principal. The balance will be the sum due. NOTE. In computing interest by this rule the exact number of days should be used. The Commercial rule is based on the following prin- ciple: If a man borrows a sum of money he should pay interest on it until it is paid, or until the date of settlement. If payments are made the creditor should pay interest to the debtor on these sums from the time they are paid till the date of settlement. This rule prevents any compounding of interest, and is absolutely fair to both debtor and creditor, as it makes each party pay interest on the money received from the other from the time of its receipt till the date of settlement The Mercantile rule is generally used in business for periods less than one year. IN ARITHMETIC. 89 EXAMPLES. (243) $850. MEMPHIS, TENN., Oct. 8, 1887. One year after date I promise to pay to Sidney Smith or order, eight hundred and fifty dollars, value received, with interest at 9 per cent. DOUGLAS JERROLD. INDORSEMENTS. Jan 15, 1888, $116; Jan. 30, 1888, $62.50; Sept. 1, 1888, $180. Find the amount due Oct. 1, 1888. (244) Find the amount due on each of the ex- amples in the previous article, by the Mercantile rule. TRUE DISCOUNT. Notes are frequently transferred from one party to another before maturity. The business of cashing notes is one of the sources of revenue to banks, and money lenders in general. It is evident, that if cash is paid for a note before it is due, a sum should be deducted for advancing the money. The sum deducted is called discount. Since the party selling the note has the use of the money paid from the time of payment till the time the note is due, the amount he deducts should be the in- terest for this period on the sum received. Or in other words, the sum he receives for the note should be such that if put on interest it would amount to the face of the note at the date of its maturity. Or, from the standpoint of the creditor, how much could he afford to pay at the present time for a note due one year hence? Just the sum which, put on interest, would 90 amount in one year to the amount he will receive for the note when it is paid at the end of the year. This sum is called the Present Worth. True Discount is the difference between the present worth and the face or amount of the debt. RULE FOE TRUE DISCOUNT. Divide the face or amount of the debt by the amount of $1 for the given time and rate. The result will be the present worth. This subtracted from the face or amount of the debt will give the true discount. Ex AMPLE.- -A note for $325, due in 3 mo., was sold for cash. If money was worth 8 per cent., how much was paid and what was the true discount ? Solution: $1 in 3 mo. at 8 % would amount to $1.020. It would take as many dollars to amount to $325 as 1.020 is contained in $325 = $318.42 P. W. Ans. $325 $318. 42 = $6.58 Dis. Ans. EXAMPLE. A note due in 1 yr. 4 m. and drawing interest at 9 per cent, was discounted at 8 per cent. What was the amount paid ? Solution: Amount of $260 for 1 yr. 4 mo. and 3 da. at 9 % = $291.395. This is the amount to be re- ceived at the maturity of the note, hence is the sum to be discounted. Ami of $1 for 1 yr. 4 mo. 8 da. at 8 % =$1.107j. $291. 395^-1.107i = $263. 15. Ans. PROBLEMS. (245) A note for $450, running 2 yr. 3 mo. 16 da., was discounted at 7 per cent. What was the amount paid? IN ARITHMETIC. 91 (246) A note for $1, 264 80, bearing 6 per cent, interest, dated Jan. 1, 1885, and due Jan. 1, 1880, was sold Aug. 10, 1885, discount off at 8 per cent. How much was realized? BANK DISCOUNT. Banks are institutions which deal in money and securities. They are either National, State or Private banks. Their principal functions are: 1st. To receive de- posits ; 2d. To loan money ; 3d. To these is added a third in the case of national banks, viz. : To issue bank notes which pass as a circulating medium. The law does not prohibit any bank from issuing bills if it chooses. But since there is a tax of 10 per cent, upon the circulation of all banks, except those organized under the provisions of the National Bank- ing act, private banks are practically cut off from this source of profit. National banks must conform to certain conditions imposed by Congress, which are too numerous to be mentioned here. By complying with these conditions they are granted many privileges which private banks do not have. Banks perform a very useful function in society. It would be impossible to transact the enormous volume of business in the country if money were used in every transaction. It is estimated that 95 per cent, of the business of New York City is effected by exchange 92 PRACTICAL METHODS through banks. They present a safe form of deposit from which sums of money may be checked out at any time. And in this way they greatly facilitate local exchanges. The bank is the purse of society and readily responds to every demand made upon it. The manner in which exchanges are made between different places by means of banks will be explained under the head of exchange. Banks are the usual purchasers of notes, and they have a different method for computing discount from the method given under True Discount. If a note is due in three months the bank simply subtracts inter- est on the face of the note for the given time. Bank Discount is interest on the face or amount of the debt paid in advance. It is evident that by this method the discount will be greater than true discount, and hence the rate of profit greater to the creditor. It really raises the rate of interest paid by the debtor. For example, a man presents his note at the bank for $100, due in one year. The bank discounts it at 10 per cent., paying $90 in cash. At the end of the year the bank receives $100, or $10 interest. Only $90 was loaned by the bank. Hence its rate of interest is |^=11^ per cent. EXAMPLE. Find the bank discount and proceeds of a note for $1,140 due in 3 yr. 6 m. 21 da., the rate of discount being 8 per cent. Solution: Int. on $1 at 8 % for 3 yr. 6 m. 21 da.= .284. $l,140X.284f=$324.52Dis. Ans. $1,140 $324.52-4815.48 Proceeds. Ans. IN ARITHMETIC. 93 EXAMPLES. (247) A man realized from a sale notes amounting to $900, which were due in 9 mo., interest being 8 per cent. He had them cashed on the day they were given at a bank at 10 per cent. What did he realize? NOTE. To find the amount on which the discount is computed, proceed as in second illustrative example under true discount. Find the date of maturity, days to run, proceeds and bank discount on the following notes: (248) $961.75. BLOOMFIELD, Aug. 29, 1881. Three years after date I promise to pay to Nicholas Nickleby, or order, nine hundred and sixty -one and ^ dollars, value received, interest at 7 per cent. MARTIN CHUZZLEWIT. Discounted March 30, 1883, at 8 per cent. ( 249 ) $ 178.80. SAN FKANCISCO, CAL., May 15, 1888. Ninety days after date I promise to pay to B. L. Smith, or bearer, the sum of one hundred and seventy-eight and -ffa dollars, with interest at 10 per cent. J. M. HAWLEY. Discounted July 5, 1888, at 10 per cent. (250) $230. HILLSBOBO, IA., Nov. 20, 1883. One year after date I promise to pay to the order of John Carter two hundred and thirty dollars. B. R. VALE. Discounted Jan. 1, 1884, at 8 per cent. A bank pays out money to a depositor only upon presentation of a certificate of deposit, or a check. A check is the written order of a depositor ordering the bank to pay money to some certain person. ('HECK. No. 23. QUINCY, ILL., May 2, 1887. first HatioRof BQR^ 0f 0uir2e$. Pay to J. C. Harkness or bearer Three hundred and sixteen DOLLARS. $3164. THOMAS HEDGE. NOTE. The bank must cash this check, provided Thos. Hedge has a sufficient amount deposited. RULE. When you have the Face of a note, Multiply \ ^ " " "Proceeds " " Divide \ y The Proceeds of $1. When you have the Discount, Divide by the discount of $1. EXAMPLE. Find the face of a 60 da, note which yields $632 when discounted at 6 per cent. Solution: Proceeds of $1 =.9895. $632 -K9895 = $638.70. Ans. EXAMPLE. The discount on a 4 mo. note at 8 per cent, was $15.65. What was the face of the note? Solution: Eule II. Discount on $1 for 4 m. 3 da. at 8 % ==.027J. $15.65 is 2.7J % of what sum? = $572.56. Ans. To find the rate of interest corresponding to any given rate of discount, the method usually given in arithmetics is long and intricate. la following out the purposes of this little book this operation is much IN ARITHMETIC. 95 simplified and shortened. No objection can surely be brought against any method that saves time and labor without sacrificing principle. The student will also notice a similar abbreviation in the reverse process of finding the rate of discount, corresponding to a given rate of interest. Rule. Divide the rate of discount by the proceeds, of $1 for the given time and rate. EXAMPLE. If a bank discounts a 60 day note at 10 per cent, what is the rate of interest? Solution: Proceeds of $1 for 63 da. at 10 % = .9825. .10-H.9825= lO^Yy %. Ans. EXPLANATION. Every dollar on the face of the note brings, when discounted, only 98^. lOc. is therefore received as interest on. 98^. What per cent, is lOc. of Kule 2. lQ+9S = lQ . Ans. EXAMPLES. (251) What is the rate of interest: When a 90 da. note is discounted at 6, 7, 8, 9, 10 per cent. ? (252) When a 30 da. note is discounted at 8, 9, 10, 11, 12 percent.? (253) When a note due in 1 yr., without grace, is discounted at ^, , , 1, 1^, 1^ per cent, a month? To find the rate of discount corresponding to a given rate of interest. Rule. Divide the rate of interest by the amount of SI for the given time and rate. EXAMPLE. At what rate should a bank discount a 60 da. note so as to receive just 10 per cent, interest? 96 PRACTICAL METHODS Solution: Ami. of $1 at 10 % for 63 da. = $1.0175. EXPLANATION. Every $1 loaned for 63 da. at 10 per cent, will amount to $1.0175 at the end of the time. $1.0175 is therefore the sum discounted. Since $1 is the amount loaned at 10 per cent., the interest will be lOc. What, therefore, must $1.0175 be dis- counted at (multiplied by), so as to produce the lOc. interest? 10c.-r- 1.0175 = 9f|^ %. Ana. EXAMPLES. What rate of discount should a bank charge (254) On a 30 da. note, to receive 8, 9, 10, 12. 15, 18 per cent. ? (255) On a 90 da. note, to receive 8, 10, 12, 15, 18 per cent. ? (256) On a note due in 1 yr., without grace, to re- ceive 6, 8, 9, 10, 12 per cent. ? EXCHANGE. Exchange is the method of paying debts by means of Drafts, or Bills of Exchange, or by a transfer of credits. If every transaction in the country were effected by means of money, the loss resulting therefrom would be very great, and many exchanges would necessarily be prevented. It would not only make enormously in- creased demands on the safe means of transmitting money, but would result in vast risks, which would be a constant source of danger and annoyance. Money JN ARITHMETIC. 97 once lost could not be restored. Transactions would be delayed by the consequent handling and convey- ance of large sums of coin, while persons would be required to keep by them large sums to discharge debts which are constantly accruing. By the present machinery of the business world, if a party in Chicago wishes to discharge a debt in Boston, it is done by sending a Bill of Exchange. A Bill of Exchange is a general term, including drafts, checks, and bills of exchange proper. They are written orders to pay money, on distant persons or companies. Domestic Exchange is the operation of exchange within the limits of a country. Foreign Exchange is the payment of debts by the interchange of obligations between different countries. Drafts are used for domestic exchange; Bills of Exchange are used for foreign exchange. When a bill of exchange is drawn it is generally written in triplicate, called a set of exchange. The three bills are identical, except that they are numbered first, second and third of exchange. These are sent by different routes, to prevent delay by accident of any kind. When one of them has been accepted or paid, the others are void. BILL OF EXCHANGE. ,2,500. BOSTON, MASS., Dec. 31, 1888. At eight of this First of Exchange (second and third of same tenor and date unpaid), pay to L. P. Morton & Co., two thousand five hundred Pounds Sterling, and charge the same to our account. To PALMER BEOS., J. S. OGILVIE & CO. London, England. 98 PRACTICAL METHODS DRAFT. $676.40. DENVER, COL., Aug. 3, 1884. At sight pay to J. W. Martindale, or order, six hundred and seventy six and ^ s dollars, and charge the same to our account. To THIRD NATIONAL BANK, JOHN I. BROWN & SON. St. Paul, Minn. THE DRAWER of a draft is the one who signs it, or the one who orders the money to be paid. THE DRAWEE of a draft is the party who is ordered to pay it. THE PAYEE is the one to whom payment is ordered. Drafts are negotiable under the same conditions as promissory notes. When the holder of a draft sells it, he must indorse it, by which he is held liable for its payment. A SIGHT DRAFT is an order for the payment of money, payable at sight, or when presented for pay- ment. A TIME DRAFT is one which is payable after a specified time has elapsed from the date of the draft, or from sight. If a party in Bloomfield owes $500 in Chicago, he could pay it by registered letter, by money order, or by draft. If he wishes to purchase a draft he does so by a bank which, by having funds deposited in some corre- spondent bank in Chicago, is enabled to draw drafts payable at that place. The draft is sent to the party in Chicago, to whom the debt is owing, who may have it cashed at any bank in the city. In this way the debt is conveniently discharged without remitting money. If the draft is lost or stolen no loss occurs to IN ARITHMETIC. 99 either party, except time, as it can be duplicated and sent again. The bank of Bloomfield must always have funds de- posited in its correspondent bank in Chicago or its drafts will not be honored. If this amount becomes large through various interchanges in financial opera- tions, it may sell drafts on its correspondent at less than their face value, in order to lessen the amount by inducing people to buy drafts. The draft is then said to be at a Discount. If its balance of deposits becomes small, it will charge more than the face value for drafts, in order to check the paying out of this fund, which would eventu- ally necessitate the shipment of coin to take its place, an expensive method of preserving the proper amount of reserve fund. In this case the draft is sold at a Premium. Drafts are frequently cashed at banks other than the one named in the draft. This necessitates a set- tlement between different banks. In the larger cities this settlement is effected by means of the Clearing House. At the close of business hours representatives of the various banks meet with their separate accounts against each other. These accounts are balanced, and the sum due from any bank paid into the clearing house fund. In a few minutes accounts aggregating millions of dollars are thus correctly settled. Sometimes the draft is drawn on private parties. It is usually presented before due, and the drawee, "accepts" it or not, which will be determined by the credit of the drawer of the draft. An acceptance is made by writing across the face of the draft "Ac- 100 PRACTICAL METHODS cepted," together with the date and signature of the drawee. The draft then becomes the promissory note of the drawee. In case the draft is not accepted the payee falls back upon the drawer for the amount. RULES FOR EXCHANGE. When you have the Face of a draft, Multiply \ -o " " " " Cost " " Divide ( y " 1+Premium, or 1 Discount When time is an element substract the bank dis- count on $1 for the given time, before multiplying or dividing. EXAMPLE. What is the cost of a sight draft on Cincinnati for $150 at 1 per cent. Premium? Solution: 1+premium = 1.01. The face of the draft is given ; hence, $150 X 1.01 == $151.50. Ans. EXAMPLE. What is the face of a sight draft on New Orleans which cost $3,200, exchange being ^ per cent. Discount? Solution: 1 % disc. = .995. $3,200 = cost; hence, $3,200-^.995= $3,216. 08. Ans. EXAMPLE. What will be the cost of a 60 da. draft on Louisville for $648.70 at per cent. Premium, interest at 6 per cent ? Solution: l+.00f = 1.0075. Bank disc, on $1 for 63 days at 6 % =-.0105. 1.0075 .0105 = .997. $648.70=Face; hence, $648. 70 x.997=$646.75. Ans. IN ARITHMETIC. 101 EXAMPLE. What was the face of a 12 da. draft on New York at 1 per cent. Discount, interest at 6 per cent., which cost $500? Solution: 1 .01 = .99. Bank disc, on $1 for 15 da. at 6 % ==.0025. .99 .0025 = .9875. $500 = Cost; hence, $500-K9875 = $506.33. Ans. EXAMPLES. Find the cost of (257) A 30 da. draft on Chicago for $240, exchange being ^ per cent. Discount, interest at 6 per cent. (258) A draft for $1,250 on Buffalo, exchange at 1 per cent. Discount. (259) A 90 da. draft on Philadelphia for $672.30, exchange at f per cent. Premium, interest at 6 per cent. (260) A man purchased a car-load of wheat (500 bu. ) at 74|c. per bu., and paid for it by a draft for the amount, on St. Louis, at 1^ per cent. Premium. What did the draft cost? (261) What was the face of a draft costing $200, exchange being 1 per cent. Discount? (262) Find the face of a 60 da. draft on Denver, exchange at 1 per cent. Premium, and costing $1,480, interest at 6 per cent. (263) A draft on New York costing $416.50, was purchased at ^ per cent. Premium. What was its face? 102 PRACTICAL METHODS HORNER'S METHOD FOB THE EX- TRACTION OF ROOTS. Cube root may be extracted much more easily, especially if more than two figures are required in the answer, by Homer's method for the extraction of roots, than by the method generally given in arithmetics. If this method be taught from the first, pupils will readily grasp it, and it soon becomes easy for them to use it. In addition to this, Horner's method may be used for the extraction of roots higher than the third, since it will apply to any root whatever. This method is explained below, and we hope its use may be extended. EXAMPLE. Find the cube root of 100. Operation : 00 100 I 4.641+ Ans. 4 16 64 4800 -^6000 120 5556 33336 126 634800 2684000 1380 64588800 138 4 72656000 1388 13920 EXPLANATION. As many columns are formed as the number of the root to be extracted. At the head of the last column is placed the number rf which we wish to extract the root, and ciphers for the others. The first figure of IN ARITHMETIC. 103 the root is then found. For 100, the nearest cube root was 4. This number is added to the 1st column, multiplied by itself and added to the 2d column, multiplied by itself again and subtracted from the given number. Commencing at the left add the root to the 1st col., multiply the result by the root and add to the 2d col . Add the root to the 1st col. Then add one cipher to the 1st col., 2 to the 2d, and 3 to the 3d. Divide the 2d col. into the 3d for the next figure of the root, making allowance for the sum to be added to the trial divisor. Add this 2d figure of the root to the 1st col., multi- ply the result by it and add to the 2d col., multiply this result by it and substract from the 3d col. Add the 2d figure of the root to the 1st col., multiply the result by it and add to the 2d col. Add it to the 1st col., stopping one place sooner each time. Bring down ciphers and proceed as before. EXAMPLE. Find the 5th root of 36. Operation : 000 248 4 12 32 6 24 8 8 ^.fyty 816 J0.0 832 2.04767+ Ans. 16 80.00015 83264 86591 8718 877t X X 36. 32 ~1. 00000 3 33056 61026 5.918 5266 652 614 38 104 PRACTICAL METHODS Here five columns are formed, and the work is shortened as follows: Instead of adding ciphers strike out one figure from the 4th col., two from the 3d, three from the 2d, and four (if there be that many) from the 1st, then divide as before, and again, when the time comes to add ciphers, strike out figures as before. In this way the work may be much shortened, as in the example given. EXAMPLES. (264) Find the cube root of 19683. ; of 2,803,221. ; of 500, to three decimal places; of 59319; of 5359375. (265) Extract the fifth root of 7962624; of 10; of 50. IN ARITHMETIC. 105 SOME USEFUL FACTS. The following, though not all strictly belonging to mathematics, may be taught in connection with some of the common school branches. They should induce the teacher to make discoveries and calculations for himself, and collect bits of information not found in text books, yet useful in every day life. The Specific Gravity of a body is the weight of a body in air, divided by the difference between its weight in air and in water. The principle of Specific Gravity was discovered by Archimedes, a famous philosopher in Sicily 2000 years ago. The foundations of Geometry were laid by Euclid, an ancient mathematician. Arithmetic is a modern science. The Romans did not carry their computations very far because their system of notation was too cumbersome to admit of this. Arithmetic really began with the introduction of the Arabian method of notation. 106 PRACTICAL METHODS Algebra was introduced into Europe by the Arabs Also the figures 1, 2, 3, etc., which are universally used to-day. The weight of the earth is 6 sextillion tons. The rotation of the earth on its axis tends to hurl bodies on its surface off into space. This centrifugal force increases from the poles to the equator. At the latter place it diminishes the weight of substances If the earth rotated with a velocity 17 times as great as at present, this force would equal gravity, and bodies on the earth's surface would not weigh any- thing. When the earth was thrown off from the sun the impelling force acted at a point 26 miles out of line with the center of the earth, in order to give it a rotary motion once in 24 hours. Every particle of matter attracts every other particle in proportion to the mass, and inversely proportional to the square of the distance. This is termed the law of Gravitation, discovered by Sir Isaac Newton. A body on the sun would weigh 28 times as much as on the earth. A man weighing 150 Ibs. here would weigh over two tons on the sun's surface. IN ARITHMETIC. T07 The North or Pole star is li degrees from the true pole of the heavens, which is continually changing on account of an oscillatory movement of the earth's axis. The stars are composed of much the same materials as the earth. Their composition is ascertained by the use of the spectroscope. TABLE OF SPECIFIC GRAVITIES. The following table shows the weight of a few com- mon substances compared with the weight of an equal bulk of water : FOB GASES Air, 1. Hydrogen, .069 Nitrogen, .972 Oxygen, 1.106 FOB LIQUIDS AND SOLIDS Water, 1. Sea Water, 1.026 Cork, .24 Brick, 1.9 Ice, .92 Limestone, 2. 5 Mercury, 13.6 Milk, 1.032 Glass, 3. Iron, 7.7 Steel, 7.8 Lead, 11.3 Silver, 10.5 Gold, 19.25 108 PRACTICAL METHODS CLASSIFICATION OF FIGURES USED IN MENSITRATION. I Scalene no two sides equal. Triangles: < Isosceles two sides equal. ( Equilateral All three sides equal. ( Trapezium no two sides parallel. Quadrilaterals: < Trapezoid two sides parallel. ( Parallelogram opposite sides parallel. f Rhomboid a parallelogram with obligue angles. Rhomb an equilateral rhomboid. Rectangle a parallelogram with right angles. Square an equilateral rectangle. Parallelograms : -| Solids:-! Cube a solid whose faces are 6 equal parallelograms. Globe a solid whose surface is everywhere equally distant from the center. Pyramid a solid with triangular faces meeting at the apex. Cone a pyramid with an infinite number of sides. RULES FOR MEASUREMENT OF FIGURES. To find the area of a triangle: Base x ^ Altitude. NOTE. The altitude of a triangle is the perpendicular distance from the base to the vertex of the opposite angle. IN ARITHMETIC. 109 To find the area of a trapezium: Divide by diagonals the figure into triangles. Take the dimensions of each and solve by rule for triangles. To find the area of a trapezoid : Multiply half the sum of the parallel sides by the perpendicular distance between them. To find the area of a parallelogram : Multiply any side by the perpendicular distance be- tween it and its opposite side. To find the surface of a cube: Square one side of the cube and multiply by 6. To find the surface of a globe: Multiply the diameter by the circumference, or mul- tiply the Di. s by 8.1416. To find the surface of a pyramid: Find by a preceding rule the area of each face and multiply by the number of faces. To this add the area of the base. To find the surface of a cone: Multiply the circumference of the base by ^ the slant height. 110 PRACTICAL METHODS To find the solidity of a cube: Cube the length of one side. To find the solidity of a globe: Cube the diameter and multiply by .5236. To find the solidity of a pyramid : Multiply the area of the base by | the altitude. To find the solidity of a cone: Multiply the area of the base by | the altitude. NOTE. The altitude of a pyramid or cone is the perpendicular distance from base to apex. PEACTICAL PKOBLEMS. No answers will be given to the following problems, which every teacher should be able to work. They are given as suggestions as to the kind of examples which the teacher should give his pupils. I. How many tons of ice in an icehouse 22x1(3, the ico being packed in solid to a depth of 12 ft. ? II. Find the cost of a board sidewalk 4 ft. wide from the door of the school-house to the street, the price of lumber being given. III. How many scholars are present to-day ? What per cent, is that of the whole number enrolled? IN ARITHMETIC. Ill IV. Write a note payable to the order of some one, and have it properly endorsed. V. Write a receipt, a due bill, a check, a draft, a note with security. VI. What is the total cost price each year of all the books used in this school, and how much money must be placed on interest at 8 % to meet this expense ? VII. Find out by a careful calculation which would be the most profitable ; to invest your money in a farm, or loan it at 4 % ? VIII. Find out what the taxes, interest on money invested, repairs, insurance, etc., on your house are annually, and then ascertain whether it would be cheaper to rent or own property. IX. How many cords of wood can be piled in a wood shed 15 ft. long, 10 ft. wide, 6 ft. high? X. What does a box of water weigh, 3 ft. each way ? NOTE. A cubic foot of water weighs 62^ Ibs. XI. Calculate the weight of the earth, specific gravity 5. XII. How many freight cars would it take to haul the grain that could be put in your school-room ? XIII. If a box contain 3 bu. of potatoes, iiow much would a box hold twice as large each way ? XIV. How many sq. rods in your school yard? How many in a lot 3 times as large each way ? XV. Two notes for $100 each are discounted, one for 30 da., the other for 90 da. Which would yield the greatest rate of interest to the bank ? 112 IN ARITHMETIC. ANSWERS. 1. 7,221. 29. 1,111|. 2. 9,009. 30. 16,044f. 3. 616. 31. 23,51H. 4. 13,224. 32. 7,511-}. 5. 2,516. 33. 3,690. 6. 2,464. 34. 1,728. 7. 1,649. 35. 2,988. 8. 2,925. 36. 92,664. 9. 9,120. 37. 615,450. 10. 8,096. 38. 4,449,955. 11. 985,050. 39. 57,446.4 gal. 12. 977,112. 40. 2,225.3 gal. 13. 9,000. '41. 166J bbl. 14. 115,200. 42. 12,927.6 bbl. 15. 4,713|. 43. 3,776 gal. 16. 7,950. 44. 47,000 gal. 17. 3,393,333|. 45. 1,475 gal. 18. 26,992. 46. 81 bbl. 19. 6,241. 47. 140f bbl. 20. 3,721. 48. 2,856.7 gal. 21. 2,304, 49. 2,127.3 gal. 22. 1,089. 66.9 bbl. 23. 10,609. 50. 48 bu. 24 bu. 24. 2,704. 51. 429 bu. 858 bu. 25. 729. 52. 6,350.4 bu. 26. 9,801. 53. 7.26 tons. 27. 8,7111* 54. 13 tons. 28. 9,3444. 55. 11 tons. PRACTICAL METHODS 1U 56. 2.28 tons. 81. 28^V 57. 1.55 tons. 82. 61^- 58. 1.77 tons. 83. 59. 41+ perch. 84. 60. 13| perch. 85. 61. 14| perch. 86. 62. 1,750 brick. 87. 63. 1,950 brick. 88. 64. 240 gal. 89. ft. 65. 419,904 gal. 90. 1. 66. 40,500 gal. 91. 80. 67. 66| ft. 104^ ft. 92. 1,880. 266| ft. 6,666 ft. 93. 32 jV 68. 28.8 mi. 22 mi. 94. 94 7 \. 178 mi. 95. 452f|. 69. 400ft. 1,600ft. 96. 416. 14,400ft. 57,600ft. 97. 22|. 70. l|sec. 2Jsec. 98. 3 T 9 . 12| sec. 3.95 sec. 99. 24||. 71. 112ft. 100. 3. 72. 900ft. 101. T 5 T . 73. 1,602 ft. $23.59. 102. 8|f 74. 8.4 in. 11.2 in. 103. trAV 16.8 in. 104. T \. 75. 50 ft. 66 ft. 105. 25.45. 91 ft. 106. 74.8. 76. 51.84 bunches. 107. 2.1445. 39 j bunches. 108. 1616.588. 77. 30 T 7 T bunches. 109. $52,465.536. 78. 66| bunches. 110. $60.24. 79. 15 T . 111. .521. 80. 103 T 8 A 5 r- 112. 6.44. 114 PRACTICAL METHODS 113. 12.9. 114. 5.8296. 115. 38.506. 116. $532.82. 117. The N. E. J of the S. E. J of Section 19, Twp. 3 North, Range 3 West of the 5th P. M. The W. of the N. W. JoftheN.W.J of Section 10, Twp. 4 North, Range 1 West of the 5th P. M., containing 20 A. The South of the N. E. 4 of the N. W. ^ of Section 33, Twp. 2 North, Range 6 West, con- taining 20 A. The N. E. J of the N. E. l of Section 22, Twp. 4 North, Range 5 West, con- taining 40 A. 118. 80 A. 119. 20 A. 80 A. 121. 3hr. 13m. 49f sec. P. M. 122. 5hr. 44m. 26f|sec. P. M. of the previ- ous day. 123. 2 20' East. 124. 90 12' 15" West. 125. 7hr. 20 m. 56 sec. j A. M., Tuesday. 126. E. 13 23'. 127. 6 hr. 51 m. 12 sec. A. M. of the follow- ing day. 128. 12. 129. 4.^ 130. TV 131. 5^. 132. 9 ' 133. 4 T- 134. Hi- 135. -III- 136. $18.90. 137. 25,298 ft. 138. 67H ft. 139. 34357. 86 mi. 140. 1 663 1 ftA/~* 141. 5 pipes. 142. $230.11 T 4 T . 143. 4|f da. 144. $5,222.40. 145. 2i-|4| yds. 146. 435 HI ft. 147. 11^ hrs. 148. $470. 149. 64}. 150. VT- 151. 200. 153. 208.35. IN ARITHMETIC. 115 154. .065|. 155. 156 cattle. 156. lOff. 157. A*- 158. A*- 159. 88f%. 160. Q 2 9 of 161. %^-tf 162. 4|%. 163. 800. 164. $1.40f. 165. 5,520 men. 166. 31* gal. 167. 4^- mi. 168. $75. 169. 35 horses. 170. 25% greater. 20% less. 171. 21i| % greater. 17 1% less. 172. 68| % greater. 40ff% less. 173. 61^f-% greater. 38^% less. 174. Com. $129.87. Pro. $3,116.88. 175. $73.96. 176. $39. 177. i%. 178. 3Ji%. 179. 2 2T%' 180. 10 bbls. 181. $884.21. 182. 1st com., $36. 2d com., $33.23. Sugar, $830.77. 183. $1,435. $1,358. 184. $321.43. 185. $103.50. 186. 14|%. 187. ?* 188. $8,500. 189. $65.80. 190. $1,200. 191. $51.20. 192. 6ff|%. 193. 8-^%. 194. $69.75. 195. $50.62J. 196. 4%. 197. 5^%. 198. 39 sharea 199. 105 shares. 200. $700, 201. 500 shares. 202. $1,503.75. 203. $428.875. Sell'g pr.,$2,865.25 204. 1J%. 205. $3,000. 206. $2,000. 207. $850. 208. $1,000. 116 PRACTICAL METHODS 209. $1,050. 210. Denver&RioGrande 239. Stock. 240. 211. $7,704. 212. $6,012.50. 213. <** 241. 214. w.<- 242. 215. 86f%. 243. 216. 80%. 244. 217. $44.82. 245. 218. $41.64 246. 219. $155.50. 247. 220. $194.61. 248. 221. $300.07. 222. 1 yr. 3 m. 223. 2 yr. 2 m. 7 da. 224. 1 yr. 6 m. 6 da. 249. 225. 7%. 226. 7%. 227. 8%. 228. $2,507. 229. $672. 230. $732.82. 250. 231. $600. 232. $479.80. 233. $412.50. 234. $2,210.75. 251. 235. $72.69. 236. $7.41. 237. $80.38. 252. 238. Date of Maturity, Sept. 7, 1879. Amount, $193.60. Amount $689.87. Date of Maturity, May 10 13, 1884. Amount, $2,744.92. $1,221.89. $10.00. $554.03. $1,201.96. $9,97. $387.72. $1,299.93. $882.21. Date of Maturity, Sept. 1, '84. Proceeds, $1032.07. Discount, $132.21. Date of Maturity, July 1316, 1888. Time to run, 42 da. Proceeds, $181.28. Discount, $2.14. Due, Nov. 23, 1884. Time to run, 327 da. Proceeds, $213.29. Discount, $16.71. 2 9 7 r . "O 6 T ?0 , ni 33 1 IN ARITHMETIC. 117 253. 6|$% I 8U; 9||%; 257. $237.48. k; 17-B%; 258. $1,237.60. 254. 26 - . 261 - ^202.02. 255. 7|fff%; 9 T Vfl%; 262. $1,480.74. 5 26a sr (T Q "q" - ( Q q 256. 5ff%;7^ 7 %; 264 ' 27; 141; 7.934; 39; ;9TV%; 175 ' 265. 24; 1,585; 2.187. /AT TEACHING PATRIOTISM YOU WILL BE GREATLY ASSISTED KNOWING //Olf THE LAM'S ARE MADE A.\I> EXECCTJJ >. THERE IS .\O fiOOK ISSTEI) TO COMPARE I.\ THIS RESPECT If/ Til HISTORY AND . . . ^_ CIVIL GOVERNMENT OF IOWA. For the Use of Normal and Public Schools, Teachers' Institutes and Private Instruction. By GEORGE CHANDLER, Superintendent of Schools, Osage, lo-wa. The book is revised to date and contains : HISTORY OF IOWA. It gives a brief history of the settlement and growth of the State. No resident of the State should neglect reading and studying this chapter. It is highly interesting. INSTITUTIONS OF IOWA. An interesting account of the various State insti- tutions is given; when established, where located now, how supported, how governed, in- mates, etc. This chapter is worth the price of the book. CONSTITUTION OF IOWA, The Constitution is given in full with side notes for each article. Any item can be readily found by running the- eye down the side notes. This arrangement will save much time, and is a valuable feature of the book. OFFICERS OF THE STATE. A full list of the various State, District, County, Town, and Township Officers, and how elected, is given. Also the duties of each, the amount of bond required, the salary received, and other information pertaining to each. No intelligent citizen should be ignorant of these, yet I fear that there aie many boys and girls graduated every year with a smattering of Greek and I^atin, that cannot name and give duties of the State officers, and perhaps there are a few teachers likewise deficient. OUTI/INES. Several pages of outlines suitable for Class Room use are given and will be found valuable in Institute work. REVISED TO DATE. The book is up with the times, containing all of the latest laws, changes in salaries and duties of various officers. ADOPTED FOR EXCI/USIVE USB in several counties by the County School Boards for terms of five years. This shows its value as a text book. That the book supplies a long felt want no one can deny. It is written in a pleasant forcible manner. It has already been adopted in many schools and is a regular text book in several County Institutes. Every parent, every pupil, every teacher, should have a copy. BOUND IN CLOTH, 170 PAGES. MAILING PRICE, 60 CENTS. A very low price will be made for Introduction, Institute Use, and to Agents. A. RLANAQAN, Chicago. UNIVERSITY OF CALIFORNIA AT LOS ANGELES THE UNIVERSITY LIBRARY This book is DUE on the last date stamped below JIM >9 1940 M>& wrrnwii OCT N 3 Form L-8 20m-12,'3(3as8) FEB 1 4 1952 MAR t 7953 SEP 1 7 1953 ;i A 000933214 9 103 G13p 1892 :ALli< BRANCA CALIFORfifc LOS ANGELES, CALIF.