iic^ mi^ 
 
 y.rr- The Knickeebockee") 20 fQuartcrly, 75c. aYear. 
 
 i;|I: LiBEAEY. ' Y ■{ 
 
 jj^i Vol. I. No. 3.j CENTS (. Maech, 1890. 
 
 KNICKERBOCKER. ^^^ m\ 
 
 NOTES 
 
 AND 
 
 QUERIES 
 
 EDITED BY 
 
 WILL P. HART, M. S. 
 
 J. E. Sheeeill, Pubusher, 
 
 The Noemal Publishing House, 
 
 Danville, Indiana. 
 
 Entered as second-class matter at the post office at 
 i»7^ Danville, Indiana. <^thj 
 
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 No. 1. AMERICAN POETS. By Will P. Hakt, M. S, 
 
 No. 2. OUTLINES OF ARITHMETIC. By Will P. 
 Hart, MS. 
 
 No. 3 NOTES AND QUERIES. By Will P. Hart, 
 
 M. S. 
 The fourth number of the Knickerbocker Library 
 will be issued June 1, 1890, and is entitled: 
 MEMORY GEMS OF POETRY AND PROSE. By 
 
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 Address J. E. SHERRILL, Publisher, 
 
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Notes and Queries. 
 
 SOLUTIONS, ANALYSES, QUESTIONS AND 
 
 ANSWERS ON THE COMMON SCHOOL 
 
 AND HIGHER BRANCHES. 
 
 rOR TEACHERS AND STUDENTS. 
 
 EDIT]p> 
 
 By Will P. Hart, M. S., 
 »• 
 
 Editor of "The Teacher and Examiner," and Author of "American 
 Poets," "English Poets," "Normal Methods of Teaching Arith- 
 metic," "The Practical Examiner and. Teacher's Manual," 
 "Outlines of Arithmetic," "Oxdlines of Grammar," "Out- 
 lines of Physiology," "Outlines of History" "Out- 
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 THE NORMAL PUBLISHING. HOUSE,' 
 
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 1889. ; 
 

 COPYRIGHT BY 
 
 J. E. SHEER ILL, 
 
 1889. 
 

 I 
 
 
 TO THE 
 
 Teachers and Students 
 
 "Who Have Contributed so Liberally to 
 
 "Notes and Queries," this Volume is Kespectfully 
 
 Inscribed 
 
 BY 
 
 The Editor. 
 
PREFACE. 
 
 This little book is made up of queries and 
 answers from tlie department of ^' Notes and 
 Queries " in The Teacher and Examiner, a 
 school journal of which the compiler is ed- 
 itor. 
 
 The department of ^' Notes and Queries ^' 
 contained so many valuable solutions, analy- 
 ses and answers that the editor deemed them 
 worthy of preservation^ hence the publication 
 of this little book. 
 
 The editor is not responsible for the an- 
 swers given, or the methods of solution. 
 They are published just as they were sent in 
 by the persons answering them. Where au- 
 thorized to do so, we have given the names 
 and addresses of all persons contributing an- 
 swers. 
 
 Danville^ Indiana, December, 1888. 
 
NOTES AND QUERIES. 
 
 1. At what time between 5 and 6 o'clock 
 is the minute-hand midway between 12 and 
 the hour-hand? 
 
 12|^ rain.-f^ distance the hour-hand is from 5= 
 distance traveled by minute-hand ; and minute hand 
 travels 12 times as much distance as the hour-hand, 
 
 ]2 dis=12^min. + ^ dis. 
 11^ dis =12j min. 
 1 dis.=l^% min. 
 12 dis.= l3^\ min.=^M5. Jimes 0. 
 
 2. A merchant sold goods and gained 20 fc ; if, 
 however, they had cost him $140 less, and he had sold 
 for same sum, he would have gained 33^ fo . What 
 did goods cost him ? 
 
 Let a=cost price by first prop, 
 then 1.20a=selling price by first prop. 
 
 a — $140=cost price by second prop, 
 then 1.33J(a — $140)=1.20a, the same selling price. 
 Multiplying and transposing 
 
 a= $1,400, the c^st. 
 
 C, K. Seibbrt, Bristol, Ind. 
 
 3. Suppose a man standing on the bank 
 of a river desires to know the distance to 
 any visible object on the other side of the 
 stream,, how can he find the distance, pro- 
 viding he has nothing but a ten-foot pole? 
 
NOTES AND QUERIES. 
 
 Solution: Let it be required to ascertain 
 
 the distance be- 
 tween A and B, A 
 being inaccessible. 
 rro.Iuce tlie line 
 in the direction of 
 AB to any point, 
 asO; draw the line 
 od at any angle 
 to the line AB; bi- 
 sect the line od 
 through which 
 draw the line Bb, 
 making cb equal to Be; draw the 
 line dba; nlso through c in the 
 direction cA, draw the line Aca, 
 intersecting the line dba; then 
 ba equals AB, the distance re- 
 ^ quired. 
 
 Wm. J. Rogers,. Paterson, New Jersey. 
 
 4. A horse is fastened by a rope 100 ft. 
 
 long to the side of a circular corral 150 ft. 
 
 ill diameter; on what surface can he graze? 
 
 Solution: Let A C F B repre- 
 Fent the corral, C E or A C the 
 rope, and A C F E the area on 
 V7hich the horse can graze. 
 
 Since C A B is a right angle,^ 
 
 / 
 
 C 
 
 CB ^ 
 
 ft., the height 
 
 of the segment A C F. 
 
 DE=CE— CD=33^ ft., the 
 height of the segment A E F. 
 
NOTES AND QUERIES. 9 
 
 By means of a table of segments we find the area, 
 of these two segments in the following manner: 
 
 66|^150=.444+, and .444 per table=.33682; then 
 .33682X(150)2=7578.45 sq. ft., the area of segment 
 ACF. 
 
 33^^200=.! 66+. and .166 per table=.08554; then 
 .08554X(20O)2=342l.60 sq. ft, the area of segment 
 A E F. But the sum of these is the area over which 
 the horse can graze; hence, 7578.45 sq. ft. +3421. 60 
 sq. ft =11000.05 sq. ft., the area required. 
 
 This solution is based on the supposition that the 
 horse grazes on the inside of the corral, though the 
 question does not state definitely whether he grazes^ 
 on the inside or on the outside. If the horse grazes 
 on the outside of the corral, the solution will be en- 
 tirely different and much more difficult. Id. 
 
 5. A point iu the tire of a bugs^y wheel 
 is in contact with the ground. W^at dis- 
 tance will the point have moved when it 
 next has the same position, the circumfer- 
 ence of the wheel being 15 feet? Also, 
 what kind of a figure will it describe ? 
 
 Rule — Multiply the diameter of the gen- 
 erating circle by 4. 
 
 Operation : 15X4 
 
 The point will describe a cycloidal curve. 
 
 Id, 
 
 6. Required, the greatest circular segment 
 that could possibly be cut from a sheet-iron 
 plate 64 feet, 4inches long, and 14 inches 
 wide. 
 
10 NOTES AND QUERIES. 
 
 Solution : Let BAG represent 
 the segment. Since B C is 64^ ft. 
 long, B D is ^ of 64^ ft., or 32|^ ft. 
 in length. A D=l^ ft. 
 
 Let x=D E. Since A. B E is a 
 a right angle, A DXD E=FD'^ 
 
 Or, ^=(32^)-^ . 
 
 6X(32i)2 
 x= ^^ ^ — =886f| ft.=rD E. 
 
 886f| ft. + l^ ft.=888^\ ft.=A E or diameter of 
 circle, and 888^^ ^t. -^2=4444^ ft.==the radius of the 
 required circle. Id. 
 
 7. The Hypotenuse of a right-angled 
 triangle is 41 it,, the base 31 ft. longer than 
 the perpendicular; what is area of triangle ? 
 
 Solution: H=4l ft. 
 
 then P''-1-(P+31)2^1681, the square of H. 
 Cempleting square, transposing and collecting : 
 p2_|_3i p=i360. 
 
 Extracting root 
 
 P=9. 
 
 P+31=40 the B. 
 
 40X9 
 then — o — =180 sq. ft., area. 
 
 C. K. Seibert, Bristol, Ind. 
 
NOTES AND QUERIES. 11 
 
 8. If tbe extreme point of the minute 
 hand of a clock move 3 inches in 4 min- 
 utes, over how many square inches will 
 the hour hand move in 3 hours if the length 
 of the hour hand equals } the length of the 
 minute hand ? 
 
 Solution : M. H moves 3 in, in 4 min., or in 1 hr. 
 it moves 15X3 in, =45 in., its entire circumference. 
 
 •g^^\^=l 1.1408 Id., the diameter of circle. 
 
 |otii^U^=4.4278 in., the length of H. hand. 
 
 4.4278 in. X2X3. 1416=27.82 in., or the circum- 
 ference of H hand's revolution. 
 
 27.82 in.X2.2139 in.=59.416 sq. in., the area of H. 
 H. circle of revolution. 
 
 Then since H, H. moves -^-^ as fast as M. H., it 
 travels in 3 hrs. ^^ of its circumference, and there- 
 fore moves over \ of its own area=i of 59.416 sq. 
 in., or 14.854 sq. inches. . Id. 
 
 9. What persons bailed Jeff. Davis ? 
 Ans. Horace Greely and Cornelius Van- 
 
 derbilt. T. S. Price, Marysville, Cal. 
 
 10. What was the most deadly epidemic 
 ever known ? 
 
 Ans, Fourth epidemic of cholera, 1883, 
 known as the "Damietta Outbreak." 
 
 B. F. MOSBLT. 
 
 11. What period is known as the Hun- 
 dred Years' War between France and 
 England ? 
 
 Ans. The time from 1346 to 1453 cov- 
 ered the period of the Hundred Years' 
 War. Ella S., Ballstown, Ind. 
 
12 NOTES AND QUERIES. 
 
 12. What ancient philosopher thought 
 that the sun was the center of the planetary 
 system ? 
 
 Ans. Ptolemy. R. D. Lyon. 
 
 13. What battle was the turning point 
 in the war of 1812? 
 
 Ans. The Battle of the Thames. Id. 
 
 14. What is the effect on the nervous 
 system of the habitual use of alcohol ? 
 Answer: The effect produced upon the 
 nervous system by the habitual use of 
 alcohol may be divided into four stages as 
 follows : 1. The stage of excitement, in 
 which the nerves become paralyzed, thereby 
 throwing the whole system out of order. 
 2. The stage of muscular weakness. Now 
 the spinal cord becomes affected, the con- 
 trol of some of the muscles is lost, the 
 muscles themselves also become feebler aa 
 the power of contraction diminishes. 3. 
 The stage of mental weakness, in which 
 the cerebrum becomes implicated, the cen- 
 ter of thought being overpowered, the 
 mind a chaos. Ideas flock in thick and 
 fast. The tongue is loosened. The judg- 
 ment loses its hold on the acts, and the 
 hidden nature comes to the surface. 4. 
 The stage of unconsciousness. IS'ow pros- 
 tration ensues. The brain and spinal cord 
 
NOTES AND QUERIES. 13 
 
 are both benumbed, and the man is what 
 we call " dead drunk." J. D. C. C, 
 
 Belleville, Ark. 
 
 15. On what philosophic principle is 
 the diving-bell founded ? 
 
 Ans. That of atmospheric pressure. 
 T. S. Price, Maysville, Cal. 
 
 16. How are vacancies in the office of 
 President and Vice-President filled in all 
 the possible contingencies ? 
 
 Ans. On the death, removal or disa- 
 bility of the President the office is filled 
 successively by the Vice-President and 
 Cabinet officers, the latter ranking in the 
 dignity of their office, the Secretary of State 
 first. In no case does any one succeed to 
 the office of Vice-President. Id. 
 
 17. What was the meaning of the cam- 
 paign cry "Fifty-four, forty, or fight?" 
 
 Ans. A cry adopted during the ]N"orth- 
 western Boundarv discussion bv those who 
 disapproved of yielding our claims to the 
 territory short of 54 degrees 40 minutes 
 north latitude, between the Rocky mount- 
 ains and Pacific ocean. 
 
 Lena IS'^oble, West Lebanon, Ind. 
 
 18. Who were the "Barn-burners?" 
 The " Xnow-nothings ?" 
 
 Ans. (1.) A name applied to the follow- 
 
14 NOTES AND QUERIES. 
 
 era of Van Buren, when in 1844 the Demo- 
 cratic party split in two factions. The 
 story of a man who burned his barn in 
 order to free it of rats was told and the 
 case likened to it. 
 
 (2.) About 1852, when the Whig party 
 was breaking asunder, a secret, oath-bound 
 organization, said to have been called "The 
 Sons of '76" or "Order of Star Spangled 
 Banner,'^ was formed. Those of its mem- 
 bers that had not been admitted to the 
 higher degrees werf kept in ignorance of 
 the aims and name of the organization, and 
 their repeated answer of "I don't know" 
 to questions regarding the society gave 
 them the title of " Know-nothings." Id. 
 
 19. I own 10 acres of land in one field ; 
 the j&eld is 9|^ rods lonsrer than wide. Find 
 width and length of field. 
 
 Solution. 10 acre8= 1,600 square rods. Add half 
 the square of the difiFerence of sides to the product 
 of sides, and extract square root i/ 1600+ ( '^^)^'-^^0^^ 
 
 "2 
 =half sum of sides. 80f =^sum of sides 
 
 — ^^=45=length. ^^ ^ =35f=width. 
 
 L. B. Hayward, Bingham, 0. 
 
 20. A boy can split a cord of wood while 
 a man chops it. The man can split two 
 cords while the boy chops one. What has 
 the boy earned when the man has earned $1? 
 
NOTES AND QUERIES. 15 
 
 Ans. AccordiDg to the data of the prob- 
 lem, I understand that the man can do f 
 more than the boy. If the boy does f the 
 man will do f . If the man has earned |1 
 the boy has earned 60 cents. Id, 
 
 21. A bin in quarter cone shape has an 
 altitude of 9 ft. If it holds 20 bu. 2 pk. 4 
 qt. of grain, what is its slant height? 
 
 Solution : If in quarter cone shape it holds 20 bu., 
 2 pk. and 4 qt., in cone shape it would hold 4 times 
 20 bu., 2 pk. and 4 qt., which is 82 bu. and 2 pk., 
 which incu. in.=2150.4 eu. in X82i==177408 cu. in.; 
 al t.=108 in. Then I 77408^^=4928=area of base. 
 
 v/4928sq in.H-.7854=80:in. nearly, or 6|ft. Radius 
 of base=6|-H2=3J, or base of right an^le triangle 
 of which 9 ft. is the perpendicular. Then i/9^4-3§^ 
 =^9^=9 ft. and 7 in.4- ,2c?. 
 
 22. A toor contains 300 sq.ft., and its 
 diagonal is 25 ft. Find the size of floor. 
 
 Solution: If 25 it. is the diag,, then 25^=fhe 
 sum of the squares of the sides, and 300= 
 the product of the sides. Rule—From the sum 
 of squares subtract twice the product and extract 
 square root, which will be the difference of numbers. 
 
 Then v^625— 6U0=i/25=5=:diff. of numbers. Add 
 to the product the square of half the diff. of num- 
 bers and extract square root which will be J the sum 
 
 of numbers |/3U0-(-(2J)^==l 7. 5=^ half the sum of sides 
 or 35=:sum of sides. 35-f 5=40=;twice the length, 
 or 20. 35 — 5=30=--twice the side, or 15. Id. 
 
16 NOTES AND QUERIES. 
 
 23. 
 yl^z"- [^jJTofindxandy. 
 
 Adding (1.) and (2.) x'+y+y^-^x=\9> (3.) 
 Transposing {x'^-{-x)-\-{y'^-\-y)^=\^. (4.) 
 Completing the square of each of these binomials 
 we have 
 
 (x^+a:+i)+(/+3/4-i)=18+i+J=7^. (5) 
 Since the left band member of this equation con- 
 sists of two perfect ►quares, the right hand member 
 must also be separated into two. and we have 
 (a:^+a:+i) + (yH-y+J)=^«-+\5. (6) 
 Now by inspection of (I.) and (2 ) x is greater 
 than y. Then 
 
 ^+^+i---¥- (7.) 
 x+h=l (8.) 
 x=%='i. (9.) 
 
 y+^+i=¥- (10.) 
 
 y+\=l (11.) 
 y=l=2. (12.) 
 
 John H. Carroll, New Amsterdam, Ind. 
 
 24. What is meant by " Carrying coals 
 to Newcastle " ? 
 
 Ans, It is a phrase meaning doing a 
 needless thing — taking something where 
 least of all it is needed. Newcastle is a 
 town in England, the centre of the coal 
 mining region — a place where coals of all 
 things are most abundant, and hence the 
 expression. G. W. Hoffman, 
 
 South Bloomfield, 0. 
 
 25. What city is known as the " Bride 
 of the Sea," and why \ 
 
NOTES AND QUERIES. 17 
 
 Arts. Venice, Italy, in allusion to the 
 marriage of the Adriatic and the Doge. 
 
 C. L. V. B. 
 
 26. If the earth were inclined 40° in- 
 stead of 23J°, how wide would the zones be ? 
 
 Arts, The Frigid zones would be 40°, 
 the Temperate 10°, and the Torrid 80°. 
 
 S. T. Briscoe, DePauw, Ind. 
 
 27. A man bought a farm for $6,000, 
 and agreed to pay principal and interest in 
 three equal annual installments. What 
 were the annual payments ? Solve by 
 arithmetic. 
 
 This is a problem in geometrical progres- 
 sion, the sum (S=$6,000) the ratio (r=1.06) 
 and the number of terms (n=3) being given 
 to find the first term (a). The following is 
 the formula: a=S (r — l)-^(r''— 1), to which 
 the interest for one year must be added. 
 The rule is as follows : Divide the interest 
 of the sum for one year by the amount of 
 $1 at compound interest for the given time, 
 less 1, and add to the result the interest of 
 the sum for one year. 
 
 Solution: $6,OOOX.06=:$360. $360^.191016= 
 $1,884,658+, and $l,884.658+$360=$2,244.658+, 
 Answer. J. E. Bonn ell, 
 
 Isabel, 111. 
 
18 
 
 NOTES AND QUERIES. 
 
 28. If f of the time past noon equals f 
 of the time to midnight, what is the hour? 
 
 A.M. P.M. 
 
 N 
 
 1. 
 
 2. 
 
 TtttoM 
 f t. p. n.=^| t. to m. 
 1 1. p. n.=h of 1 1. to m.= 
 f t. p. n =3Xt% t. to m.= 
 1 t. to m.=t. to m. 
 t. to m.=t. p. n. 
 
 It.toM 
 
 t- 3 
 9 f 
 
 t. to m. 
 to m. 
 
 If t. to. m. 
 
 1 t. to m.=M of 12 hT.=W hr.=7U hr. 
 
 :12 hr. 
 
 -^3" 
 
 ITS 
 
 '!J3" 
 
 12 hr.— 7^f lir.=4^V hr.=t. p. n. 
 
 Explanation: f of the time past noon 
 equals f of the time to midnight. J of the 
 time past noon equals J of f of the time to 
 midnight, ^ of time to M, etc. 1 [Once] 
 the time to midnight equals the time to 
 midnight. -^ of the time to midnight 
 equals the time past noon, and adding gives 
 14 c>f the time to midnight equal to the 
 time past noon plus the time to midnight, or 
 12 hr. 
 
 The pupil should always be able to place 
 the illustration on the board. The advan- 
 tage claimed for this method is that it re- 
 solves each step into an equation, and Olney 
 says, " The equation is the grand patent for 
 solving problems," to which nearly, if not 
 all, mathematicians will assent. 
 
 RoBT. C. LiNDSEY, Washington, D. C. 
 
NOTES AND QUERIES. 
 
 19 
 
 29. A tree is 100 feet high ; it breaks 
 
 and the top strikes the ground 50 feet from 
 
 the base. What is the height of the stump ? 
 
 (AAB=100ft. 
 B D=50 ft. 
 
 B C=perpendicular (p.) 
 D C=hypotenuse (h.) 
 D B=base (b.) 
 
 '])h+p=100 
 
 y 
 
 (2)-^(l)= 
 
 2) h^— p2=b2= 
 
 [2500 
 (3) h— p=25 
 (l)h+p=100 
 
 (3)4-(l)=r(4) 2 h=125 
 (5) 1 h=62^ 
 (6)100—62^=37^ 
 
 B [=p- 
 
 Note : h^ — p^=(h+p) (h — p) and dividing this by 
 h+p, thus, (^+P) (^~P) ^h— p. These steps may 
 
 be difficult to see for one who has not studied alge- 
 bra, yet some one may be benefited. Id. 
 
 30. A board is 12 feet long, 18 in. broad 
 at one end and 12 in. at the other ; where 
 must I saw it to have the same amount at 
 each end. 
 
20 
 
 NOTES AND QUERIES. 
 
 ' In 12 ft. of length there is 
 
 [a decrease of 6 in. 
 
 In 1 ft. of length there is a 
 
 [a decrease of ^ in. 
 
 Starting at the point formed 
 
 by prolonging the sides, we 
 
 have : 
 
 i in. increase^ 1 ft. length. 
 
 18 " " =36 " " 
 H C=36 ft. 
 G C=24 ft. 
 
 Area of A C B=27 sq. ft. 
 " "DC E=12 sq. ft. 
 " "ABE D=15 sq. ft. 
 " " MN ED=7^^q. ft. 
 " " MCN=19j8q.ft. [12 
 [sq. ft.+7^sq. It.] 
 A C B:MC N::H 0^:8 0^ 
 27: 19*:: SC^iSC 
 27SC2=19JX362 
 
 1 s c^^gse 
 
 1 S C =1/936=30.5941 + 
 30.5941— 24=6.5941+An8. Id. 
 
 31. When the new 4i'8 sell at 105, what 
 must I invest in them to secure an annual 
 income of $983.25 in currency, gold being 
 at 1091, brokerage ^ ? 
 $105 J cur =$100 bond. 4^% of $100=r$4J int. in 
 
 ($1 g=$1.09J cur. [gold. 
 
 1 4^ g =4^X$1.09^ cur.=$4.91625 cur. int. 
 
 f $4 91625 cur. int.=$100 B. or $105J cur. 
 
 I of $105ircur=$-15!5L 
 
 4.91625 ^ ^4.91625 
 
 $1 
 
 $983.25 
 
 =:983.25X$ 
 
 105J 
 
 4.91625 
 
 :$21050l^^5J 
 
NOTES AND QUERIES. 
 
 21 
 
 32. What is the perimeter of a rhombus, 
 one diagonal being 10 rods and the area 
 
 86.60J ? 
 
 Solution and Demonstration : Draw the figure as 
 given here. Let ABCD= the rhombus whose area 
 
 ' is86.60iandBDand 
 A C its diagonals, re- 
 spectively. Now, as 
 the area of a rhom- 
 bus is equal to half 
 the product of its di- 
 agonals, if we divide 
 86.60^ the area of 
 the rhombus, by 10, 
 the length of the 
 known diagonal, it 
 will give us half of 
 the length of the 
 O unknown diagonal. 
 
 86.60|-r-10=r8.66025=:halfthelength of the unknown 
 diagonal; .'. 8.66025X2=17.3205=the whole length of 
 the unknown diagonal. .". diagonal A C=10, and di- 
 agonal B D=17.3205. Now, according to a simple geo- 
 metrical theorem, the diagonals of a rhombus mut- 
 ually bisect each other, and are also perpendicular 
 to each other. .. in A A E B, side A E=5 and side 
 B E=8. 66025. Now, because the diagonals of a 
 rhombus are perpendicular to each other .'. A E B is 
 a right A • . '• side A B of the rhombus ABC D^th e 
 
 hypotenuse of the right A A E B=Va E^+B W^ 
 
 4 
 
 52+8.660252=10+ = the length of one of the sides 
 of the rhombus A B C D. .-. 4X10+=40+=the 
 perimeter of the rhombus A B C D, which was to be 
 found. Jas. K. Rivers, 
 
 Ft. White, Fla 
 
22 NOTES AND QUERIES. 
 
 33. On a pole 100 ft. high is a globe 100 
 ft. in diameter ; and on the globe is a man 
 whese eye is 6 ft. above the globe ; a Jine 
 let fall from his eye is perpendicular to the 
 plane of the earth. What is the area of 
 the circle obscured from his View ? 
 
 B 
 
 Solution: A=6 ft , B=100 ft. and C=100 ft. A to 
 B=56, or hyp. of a rt. an, triangle. B to D=50, or perp. 
 A to B^v'AB^—DB^^i/se^— 50^=1/ 636 or 25.+. 
 Then by prop. 2,5+ :206 ::50:412+=half di«m. of 
 obscured cir. 824=diam. to find area. 824^X7854= 
 area. L. B. Hatward, Bingham, O. 
 
 34. When and where was Jefferson 
 Davis captured? 
 
 Ans. Jefferson Davis was captured May 
 10, 1865, near Irwinville, Ga. 
 
 E. L. Hubbard, Hanover, Kan. 
 
NOTES AND QUERIES. 23 
 
 35. "What was the popular vote cast for 
 Harrison in 1840 ? 
 
 Ans. The popular vote cast for Harrison 
 in 1840 was 1,275,017. He received an 
 electoral vote of 234, which has been 
 equaled in but two instances. Those were 
 for Pierce in 1852, and Grant in 1872, and 
 were 254 and 286, respectively. Id. 
 
 36. How many Presidents, if any, were 
 elected without their receiving a majority 
 of the popular vote ? 
 
 Ans. One, R. B. Hayes, in 1876. Hayes's 
 popular vote was 4,033,295 ; Tilden's popu- 
 lar vote was 4,284,265 ; the electoral vote 
 resulted in 185 ballots for Hayes and 184 
 for Tilden. Id. 
 
 37. If 21 cows eat 8 acres of grass in 6 
 weeks, and 18 cows eat the same in 9 
 weeks, how many cows will it maintain for 
 18 weeks, if the grass grows uniformly dur- 
 ing that time? 
 
 Solution: Consider what each cow eats 
 as composed of 6 parts. Then 21 cows re- 
 quire each week 126 parts, and for 9 weeks 
 1134 parts. From these preliminaries we 
 have the following: Eight acres afford 756 
 parts in 6 weeks, so 8 acres afford 972 parts 
 in 9 weeks; hence 1 acre affords 94| parts 
 in 6 weeks, and 1 acre affords 121J parts in 
 9 weeks. Here we see that 1 acre affords 
 
24 NOTES AND QUERIES. 
 
 27 parts more in 9 weeks than in 6 weeks, 
 on account of the 3 weeks' growth in the 
 one case more than in the other. Hence, 
 if 27 parts grow in three weeks, the growth 
 of one week must be 9 parts. Again : 
 Since 1 acre affords 121J parts in 9 weeks, 
 and the growth in 9 weeks is 81 parts, the 
 remainder, 40J parts, must be the quantity 
 of grass on the acre at first. 
 
 Therefore, the given 8 acres had, at first, 
 324 parts, and through 18 weeks' growth 
 will have furnished 18X9X8, or 1296 parts 
 more, in all, 1620 parts. This amount 
 would pasture 270 cows one week, and for 
 18 weeks 15 cows. Ans. 
 
 The above may be put under the following for- 
 mula: Put a=2l, m=n=p, m='^, g=&, 6^18, r=9, 
 5=18. Then, the number of cows is 
 
 ampqr — hmpqr-\-binprs — anpqs 
 
 mnrs — mnqs 
 Substituting 
 
 72576—62208+186624—145152 51840 
 
 10368—6912 ~^456^-^^ 
 
 the number of cows. 
 
 N. B. — This formula will solve all simi- 
 lar problems. T. A. Pugh, 
 
 Dumontville, O. 
 
 38. Two men are 90 miles apart and 
 travel towards each other, A starting 1 
 hour before B ; A goes 9 miles in 2 hours, 
 B 11 miles in 4 hours. How far will each 
 travel before they meet ? 
 
NOTES AND QUERIES. 25 
 
 Solution. — Both go 7^ miles in 1 hr. 90^-7^= 
 12if=W hr. %%«+ V=34^V miles. 34/^+1=38^. 
 90 — 4^=85 J miles, distance apart after A goes 1 hr. 
 85|^7i=ll|f; ll|fXf=53^V 532^9+41=5711, A's 
 distance. 90— 59|f=32|f, B s distance. Id. 
 
 39. If I sell oats at 42 J cts. per bushel, 
 my gain is only f of what it would be if I 
 should sell at 56i cts. per bushel. What 
 did they cost me ? 
 
 Solution. — Let f =the cost. 
 
 (1). 42i-f =f (561-1). 
 
 (2). 42^-1=37-1—1. 
 
 (3). i:r-.5ct. 
 
 (4). 1=15 ct.=Ans. Id. 
 
 40. Divide the number a into three such 
 parts, that the first shall be to the second as 
 m to n, and the second to the third as p to q^ 
 
 Solution. — Let x=the first part, 2/ the sec- 
 ond, and z the third. 
 Then (1) x-f y+z=a. 
 
 (2) x:y::m:n. 
 
 (3) y:z::p:q. 
 
 (1) = (4) x=a-y-z. 
 
 (2) = (5) x=^. 
 
 • • (6) a-ry— z=-^- 
 
 (7) na — ny — nz=my. 
 
 (8) my+ny=na — nz. ^ 
 
 na — nz 
 
 (3)=(10) y=^- 
 
26 
 
 .-. 01) 
 
 NOTES AND QUERIES, 
 na — nz pz 
 
 m+n q 
 
 (12) anq — nqz=^mpz-f-np^. 
 
 (13) mpz+npz+nqz=anq. 
 
 (14) ,=_^^?^. 
 ^ ' mp+np+nq 
 
 anp 
 
 ^™^mp+np-f-nq 
 amp 
 
 mp+np+nq 
 Put a=42, 71^2, w=3, jt?=4, g=5. 
 Then by substituting these values, we shall have: 
 x=n. 3/=14f, 2=18. Id. 
 
 41. Diagram, and parse " far ": 
 
 Our island home is far beyond the 
 
 sea. 
 
 home I is : beyond sea 
 
 Our 
 
 island 
 
 the I 
 far 
 
 " Far " is an adverb, and modifies 
 the phrase "beyond the sea." Id. 
 
 42. Analyze, and parse italicized word: 
 To be idle is wrong. 
 
 To be idle | is : wrong. 
 
 **To be idle" is the subject, "is" is the 
 'Copula, and "wrong" is the predicate. 
 " Idle" is an adjective used abstractly. 
 
 (See remark under R. XII., Harvey.) Id. 
 
 43. A. T. Stewart & Co. imported 10 
 cases of shawls, averaging 216 pounds a case, 
 invoiced at 24,884.10 francs; the duty being 
 
NOTES AND QUERIES. 27 
 
 50 cents a pound 35 per cent, ad valorem. 
 
 The invoice was paid with a bill of exchange 
 
 bought at 5.16 francs to the dollar. What 
 
 was the duty and what did the shawls cost 
 
 after paying other charges to the amount 
 
 of 175. 80? 
 
 Solution. — 10 cases of 216 lbs. each=2160 lbs. 
 5.16franc8=$1.00; 1 franc— $.193. 
 24884. 10X$.193=$4802.63=Invoice. 
 $4802.63X.35=$1680.92=ad. val. duty. 
 2160 lbs. at 50c.=^$1080.00:^sp. duty. $2760. 92= whole 
 duty. Other charges, $75.80. Whole cost, $7639.35. 
 J. E. ISuTTON, Scianton, Miss. 
 
 44. Can the electoral vote of a State be 
 divided? If so, has it ever been done? 
 When, and where ? 
 
 It may be divided. In California, in 
 1880, five Democratic electors for President 
 and one Republican elector were chosen; 
 one of the Democratic electors for the State 
 at large, Judge Terry, running behind the 
 ticket, resulted in the election of Egerton, 
 B-epublican. The other Democratic elec- 
 tor, Wallace, was elected, making five Dem- 
 ocrats and one Republican. Id, 
 
 45. Who was Pompey ? 
 
 A Roman Consul who raised himself to 
 the dictatorship of Rome without being 
 subject to either the popular or senatorial 
 party. Csesar being at the head of the lat- 
 ter party, Pompey, wishing to rule su- 
 
28 NOTES AND QUERIES. 
 
 preme, made a break with Caesar which 
 led to civil war between them. Pompey 
 was defeated at Pharsalia and fled to Cili- 
 cia, where he collected a small body of 
 soldiers, and then sailed to Alexandria. 
 Septimius stabbed him as he was landing. 
 
 Id. 
 
 46. Four balls are lying on the floor. A 
 fifth ball is placed on top of these. What 
 is the distance from center of fifth ball to 
 floor, each ball being five inches in diameter ? 
 
 Lines connecting centers of any three of 
 lower balls form a right-angled triangle of 
 which base and perpendicular are each 5 
 inches, and whose hypotenuse is therefore 
 ^2 (5)2=7.071 inches. This hypotenuse, 
 7.071 inches, divided by 2=3.535 inches, 
 which forms the base of another right-angled 
 triangle, which has for its hypotenuse a lino 
 connecting centers of a lower and the upper^ 
 or fifth ball being 5 inches long. The per- 
 pendicular of this second triangle is a line 
 drawn from center of fifth ball down to 
 hypotenuse of first triangle, and is in 
 length y/ 52— 3.535^=3.536 inches. 3.536 
 inches+2J inche8^6.036 inches, the dis- 
 tance from center of fifth ball to floor. 
 
 C. K. Seibert, 
 
 Bristol, Ind. 
 
NOTES AND QUERIES. 29 
 
 47. In division of fractions why do you 
 invert the terms of the divisor and multi- 
 ply the numerators together for a new num- 
 erator and the denominators together for a 
 new denominator ? 
 
 The object of inverting the divisor is to 
 find how often the divisor is contained in 
 one. Knowing this, we find how often it is 
 contained in the dividend by taking such a 
 part of the inverted divisor as the dividend 
 is part of one. This is done by multiplying 
 the numerators together for a new numera- 
 tor and the denominators for a new denom- 
 inator. To illustrate, we will take the simple 
 example f ^f . i is contained in one four times, 
 then f are contained in one ^ as often as \, 
 that is ^ times. If f be contained in one |- 
 times it is contained in J of one, J of f or -| 
 times. If the divisor is contained in J |- 
 times, it is contained in f twice |^, or f times, 
 
 P. W". Singleton, 
 
 Farmington, Ky. 
 
 48. A, B and C start from the same 
 point and travel in the same direction about 
 an island 73 miles in compass, A at the rate 
 of 6, B at 10, C at 16 miles per day, in 
 what time will they be next together ? 
 
 Solution. — '^-^j ^, ^|=:number of days it 
 will take each to travel around it. 
 
30 NOTES AND QUERIES. 
 
 The L. C. M. of these numbers =: the re- 
 quired number of days, which is 36J. 
 
 John M. Colaw. 
 
 49. This is money that I did not earn. 
 What is the predicate of this sentence? 
 
 The complex pred. is *' money that I did 
 not earn;" *'that 1 did not earn" is an ad- 
 jective element limiting "money." 
 
 L. B. Hayward, 
 Bingham^ 0. 
 
 50. Where is the Gate of Tears? 
 
 The strait of Babelmandeb, the passage 
 from the Persian Gulf to the Red Sea, is 
 called the "Gate of Tears" by the Arabs. 
 The channel is only about twenty miles 
 wide, and is rocky, and very dangerous for 
 passage in rough weather. It received its 
 melancholy name from the number of ship- 
 wrecks that occured there. F. L. D., 
 
 Lima, 0, 
 
 51. AVhat was the " Rosetta Stone ? " 
 Ans. During the expedition of the French 
 
 to Egypt, under Napoleon, at the close of 
 the last century, an engineer, in digging the 
 foundation of a fort near the Rosetta, mouth 
 of the Nile, found a stone tablet about three 
 feet long, on which was an inscription in 
 three different characters. This was the 
 famous " Rosetta Stone." One of the three 
 texts (the lower one) was Greek, and of 
 
NOTES AND QUERIES. 31 
 
 course, was readily translated ; the text at 
 the head was in the mystic hieroglyphic 
 character ; the intermediate was in a char- 
 acter since called Demotic (Demos, the 
 people), that is, the writing of the common 
 people. This inscription was copied and 
 circulated among scholars, and after long 
 and. ingenious efforts, the alphabet of the 
 hieroglyphic was made out, so that now 
 these carvings are read with ease and cer- 
 tainty, and a new flood of light has been 
 thrown on the history of ancient Egypt. 
 The great work of deciphering these char- 
 acters was mainly effected by the French 
 savant, Champollion. P. C. Coy. 
 
 62. A person being in a boat 3 miles 
 from the nearest point of the beach, wishes 
 to reach in the shortest time a place 5 miles 
 from that point along the shore ; supposing 
 he can walk 5 miles an hour, but row only 
 at the rate of 4 miles an hour, required the^ 
 place where he must land. 
 
82 
 
 NOTES AND QUERIES. 
 
 B 
 
 3 
 
 
 % 
 
 ^ 
 
 -^ vv 
 
 
 "^ vv 
 
 \ 
 
 
 \ 
 
 1 
 
 
 1 
 
 Solution. — From the 
 figure, the distance 
 rowed is ■i/{x'^-\-9) 
 miles, the distance 
 walked 5 — a:mi., and 
 u, the whole time 
 taken, is evidently, 
 
 "- -4 -i— 5- 
 
 hours, and x must 
 have a value that 
 ^ will make u a mini- 
 mum. Hence, by dif- 
 ferentiation Dx u = 
 
 Dx^ u = 
 
 i, and 
 
 9 
 
 4(a;2+9)^ 
 
 Solving 
 
 4i/(.r2 + 9) 
 — 1=0, we get x = 
 =h 4; but on substi- 
 tuting these values 
 of r» in turn in the 
 expression for Dx u, 
 /^ we see that x=4 is 
 the only value which 
 will make Dx u, =0, 
 since we must take the positive value of !/( a;^ -|- 9 ), 
 as it represents a distance traversed. Hence, we 
 find [i)a;2 w] a;=4=5^; and w then is a minimum 
 
 w^hen «=^4, and the landing place must be 1 mil 
 above the point of destination. 
 
 John M. Col aw, Monterey , Va. 
 
NOTES AND QUERIES. 33 
 
 53. I sold a team of horses for $700.00 ; 
 on one I gained 20%, on the other I lost- 
 20% ; my total loss was $50.00. What was 
 the cost of each? 
 
 Solution. — The loss exceeded the gain by 
 $50.00, and to lose $50.00 at 20% would 
 take $250.00 ; therefore, the horse on which 
 he lost must have cost $250.00 more than 
 the other; but they both together cost 
 $750 00 ; hence, one cost $250.00 and the 
 other $500.00. Id. 
 
 54. An express train left New York City, 
 running at the rate of 25 miles an hour, and 
 reached a certain town at 10 in the morning. 
 A freight train left New York City ten min- 
 utes later, running at the rate of 18 miles an 
 hour, and reached the same town at 6 min- 
 utes past 11 A. M., the same day. Required 
 the starting time of the express and the dis- 
 tance of the town reached. 
 
 8olutio7i.- — 11 hr. 6 min. 
 10 " 10 " 
 
 56 minutes, diff. in time, 
 60 - 25=2| ; and 60^18 = 3J. Then 3J— 2|- 
 =^, difference in time for each mile. 56^ 
 ^=60 mi. to make a diff. of 56 min. 10 
 hours less 2^ hrs.=7f hrs.=7 hrs. 36 min. 
 
 Ans. 7 o'clock and 36 min. A. Mi Dis*- 
 tance, 60 miles. Id, ' 
 
84 NOTES AND QUERIES 
 
 55. ic2_w2z= 
 
 x' 
 
 ;_^3_i9 (2) r to find X and y. 
 
 i3) (x — y) (x-\-y)=z5 from (1) by factoring. 
 4) (a; — y) (x^-txy+y^)=^ld from (2) by fac- 
 toriDg. 
 
 (^) \^^Jy^y ^T9 ^y cancelling :i— 7/ in (5) 
 
 Now since an equation is an expression of 
 the equality of ratios, from (6) we have the 
 following proportion (the denominators be- 
 coming antecedents). 
 
 (7) a?-\-xy+y^ : x+y : : 19 : 5 
 
 (8) x^+xy+y^ : 19 : : x+y : 5 by alternation 
 
 (9) x^-^-xy + 'jf :ld :: x'^ + 2xy + y^ : 25, since 
 raising both terms of a fraction to the same 
 power (2d) does not alter its value. 
 
 (10) x^+xy+y^:x^+2xy+f::19:26 by alter- 
 nation, 
 ril) xy : a^+2xy+y^ : : 6 : 25 by division 
 
 (12) 4xy : oi^+2xy^^ : : 24 : 25, since multi- 
 plying the antecedents of a proportion by the 
 same number (4) does not destroy the pro- 
 portion. 
 
 (13) x^—2xy^7f : x^+2xy+y'' : : 1 : 25 by di- 
 vision 
 
 (14) X — y : x-\-y : : 1 : 5, since like roots of the 
 terms of a proportion are also in proportion. 
 
NOTES AND QUERIES. 86 
 
 (15) 5x — by=zx+y, since product of means 
 equals the product of the extremes. 
 
 (16) 4x^6y by subtracting x and adding 5y 
 to each member of (15). 
 
 (17) a?=— ; substitute this in (1) gives 
 
 ^ V 2 f^ 
 
 4 -^ 
 9/_42/2=20 
 by^=z20 
 f=4 
 
 2/=2 ^ . 
 
 x = |^=3, substituting in (17) 
 Philip B. Hays, 
 
 Brandenburg^ Ky, 
 
 56. What sum invested in U. S. 5's of 
 1881, at 118, will yield an annual income of 
 $1,921 in currency when gold is at 113? 
 Solution : 
 
 $1,921-1. 13-^$l,700=Income in gold. 
 $l,700-.05=$34,000=Par value of bonds. 
 134,000X1. 18=$40,120-=C^st of bonds. 
 
 Ella S., 
 
 Ballstowny Ind. 
 
 57. Multiply 18 da. 9 hr. 42 min. 29.3 sec. 
 by 16^. 
 
 Solution: 
 
 18 da. 9 hr. 42 min. 29.3 sec. 
 
 16-7- 
 ^^11 
 
 305y\da. 16y\hr. 46y\min. 7^ sec. 
 
36 NOTES AND QUERIES. 
 
 Reducing fractions, 
 
 305 da. 16 hr. 46 rain. TiVoSec. 
 
 10 " 54 " 32y\ '' 
 
 . 43 " 38-32- " 
 
 43_7_ " 
 
 306 da. 4 hr. 25 niin. 2 sec, nearly. 
 
 L. B. Hayward, 
 
 Bingham^ 0. 
 
 58. What is the side and end of a rect- 
 angular field whose area is three acres and 
 bounded by 104 rods of fence ? 
 
 Answer : Ride. — From the square of 
 half the perimeter, take four times the given 
 area, and extract the square-root, and to one- 
 half of this add J of the perimeter, which 
 gives the length. Take the difference for the 
 width. 
 
 Solution.— I oi lOA=b2 rods; (52)2=2704; 
 3^=480 sq. jrds; 480X4=1920; 2704— 
 
 1920=784; >/ 784=28 ; J of 28=14; J of 
 104=26; 14+26-=40, length; 26—14=12, 
 width. 
 
 Cadmus, Oak Hill, 0. 
 
 Another solution of the above problem ; 
 
 Put y = one side, 
 
 and X = one end. 
 Then xy = 480 sq. rods = area. (1) 
 
 and 2a;+2Y/=104=rods of fence. (2) 
 
 Then Xz= 
 
 y 
 
and x= 
 
 NOTES AND QUERIES. 37 
 
 104—2 y 
 
 2 
 
 Putting values of ic=to each other 
 480 10-4—2 y (3 
 
 ^ Clearing of fractions, transposing and re- 
 ducing 2/' — 52 :y =— 480. (4) 
 Putting y=z(v+26) in (4) 
 Wc have (?;+26)'— 62 (y+26):=— 480 (5) 
 Then v^+b2v-^()7Q—b2v— 1352=— 480 (6) 
 Cancelling, summing and transposing 
 <G), i;^=rl96_ (7) 
 
 Then v^VWQ ^ (8) 
 
 Substituting y in 8th, 2/=i/ 196+26 (9) 
 
 ^Vhence y=:40::=one side. 
 Substituting value of y in either (1) or (2) 
 it'=:12=one end. 
 
 * The 4th is a complete quadratic, and the 
 method here employed to solve it I have 
 never seen in any work on algebra, and for 
 this rea.«on I use it here, as it may be new to 
 Iiundrcds, perhaps thousands. 
 
 J. K. Ellis, Emma, Ky. 
 
 59. What city was a republic nearly seven 
 centuries ? 
 
 Alls. Eome. Manson Williams, 
 
 Smithjield, Va. 
 
 60. Where are the following: Fingal's 
 €ave ; '^ Land of the Midnight Sun." 
 
38 NOTES AND QUERIES. 
 
 FingaPs Cave is iu Staffa, Isle of Wight; 
 " Land of the Midnight Sun '^ is a term alike 
 applied to Norway and Alaska. 
 
 T. S. Price, 
 Marysvillej Cal. 
 
 61. What and where is Alhambra? The 
 Vatican ? 
 
 Ans. The Alhambra was the great Moor- 
 ish Palace iu Granada, Spain, and it is 
 the finest specimen of arabesque architecture 
 in Europe. The w^orld's most complete and 
 valuable gallery of sculpture is at the Vat- 
 ican, in Kome, Italy. 
 
 Manson Williams, 
 ^ Smithfieldj Va. 
 
 62. Prove that the area of a triangle is 
 equal to half the product of its perimeter by 
 the radius of the inscribed circle. 
 
 Demonstration : — Construct the triangle A 
 BC, with the inscribed circle GEF. Draw 
 the radii, DG, DE and DF, and the lines 
 DA, DB and DC. These last three lines 
 give us three triangles whose bases are the 
 lines AB, BC and CA, respectively, which 
 combined give the perimeter of the larger 
 triangle. The radii DE, DF and DG are 
 also the respective perpendiculars of the three 
 triangles. The rule is well established, and 
 hardly need be demonstrated here, that the 
 
NOTES AND QUERIES. 39 
 
 areas of these triangles will be one-half 
 the product of their bases by their perpen- 
 
 diculars, viz: J (ABXDE)-=area of ABD; 
 i (BCXDF)=area of BCD ; J (CAXDG)== 
 area of CAD. By addition, we have J (AB+ 
 BC+CA)XDE [or DF, or DG, the radius of 
 the inscribed circle and a common perpen- 
 dicular]=the area of the triangle ABC. 
 
 T. S. Peice, 
 
 Marysville, Cal. 
 
 63. How wide must a door 7 feet high be so 
 that a circular saw 8 feet in diameter can pass 
 through it? 
 
 Solution : This is a very simple geomet- 
 rical problem based on the Pythagorean 
 problem. Theoretically, the thickness of 
 the saw-blade should be taken into con- 
 sideration, and would make a difference in 
 
40 NOTES AND QUERIES. 
 
 the result. As thickness is not mentioned 
 in the problem, it will not be taken into con- 
 fiideratiou in the solution. Take 7 feet as 
 the base of the right-angled triangle, and 8 
 feet as the hypothenuse, to find the altitude. 
 By a simple application of the rule, we have 
 8^ — 7^=square of altitude (or width of the 
 door); that is 64—49=15, the square root of 
 which is 3.872 + feet, the width of the door. 
 
 Id. 
 
 64. One mule was found left tied. Parse 
 "left" and ^Hied." 
 
 "Left" is a simple past passive participle 
 from the transitive verb leave; principal parts 
 leave, left, having left, performs the office of 
 an adjective and verb, and is used as adject- 
 ive modifier of "mule." 
 
 "Tied," parsed the same as "left." 
 
 C. D. Hill. 
 
 66. It took Rome three hundred years to 
 die. Analyze and parse italicized words. 
 
 " It took Rome three hundred years to die," 
 is a simple declarative sentence, the subject 
 of which is the personal pronoun "It," the 
 predicate the remainder. The principal part 
 of the predicate is the transitive verb " took," 
 which is followed by its object, " years." 
 ^' Took " is modified by the adverbial phrase 
 "[for] Rome," consisting of the noun 
 " Ro;gQ,e " and a. preposition implied, and the 
 
NOTES AND QUERIES. 
 
 -41 
 
 adverbial infinite phrase ^^ to die/^ " Years '' 
 is limited by the definite adjective ^' three 
 hundred/' 
 
 ^' Rome'' is a proper noun, third, sing., 
 iieut., objc, after a preposition implied. 
 
 '^ Years " is a common noun, third, plu., 
 neut., objc, object of '^took." 
 
 " To die" is a present infinitive verb, prin- 
 cipal parts die, died, having died, used as 
 
 adverbial modifier of " took." 
 6Q. Diagram : 
 
 " Place me on Sunium's marble steep 
 Where nothing save the waves and I 
 May hear our mutual murmurs sweep ; 
 There, swanlike, let me sing and die." 
 
 Parse the italicized words. 
 
 Id, 
 
 -Byron. 
 
 X I Place ! 
 
42 NOTES AND QUERIES. 
 
 "Save ^^ is a preposition showing the rela- 
 tion between "waves and I ^' and " nothing/' 
 
 " Waves ^' is a common noun, third, plu., 
 neut., objc., after the preposition " save." 
 
 " I '' is a personal pronoun, declined sin., 
 I, my or mine, me ; pi., they, their or theirs, 
 them; 1st, sin., mas., objc, after "save." 
 The nominative form is used by poetic license 
 to rhyme with "die." 
 
 " Swanlike" is an adverb of manner modi- 
 fying "sing" and "die." Id. 
 
 67. Required to cut from the smaller ex- 
 tremity of a given right-angled-triangle, J of 
 the area of the triangle. At what point on 
 the base will the perpendicular be erected ? 
 
 Solution : The area of the triangle is to 
 the area of the part to be cut off as 3 to 1, 
 and the length of the base of the triangle 
 will be to the length of the part cut off^ as 
 i/3 is to /l ; or as 1.732 is to 1 ; therefore, 
 the part cut off will be |^^— .57735 of the 
 whole base. U. G. Frisbie, 
 
 New Era, Pa. 
 
 68. Where is the geographical land center 
 of tlie earth ? 
 
 Answer: Great Britain is generally so 
 considered. 
 
 T. S. Price, Marysvilhy CaL 
 
NOTES AND QUERIES. 43 
 
 69. What is the diameter of a spherical 
 cannon-ball that can be made of 890 pounds 
 of iron ? 
 
 ^Solution : 890-480 = 1.8541+ 
 1.8541X1728=3203.8848 
 32 03.8848-^-. 5236=6118.954+ 
 
 ir6118.954=:18.2+inches, diameter. 
 *This solution is based upon the theory^ 
 that a cubic foot of iron weighs 480 pounds. 
 C. C. Monroe, Frankfort , Ky. 
 
 70. Two men carry a kettle weighing 200 
 pounds. The kettle is suspended on a pole, 
 the bail being 2 ft. 6 inches from the hands of 
 one, and 3 ft. 4 inches from the hands of the 
 other. How many pounds does each bear ? 
 
 Solution : The load sustained by each is 
 inversely as the distance between them and 
 the load. Each man will bear half the bur- 
 den, if the kettle hangs from the middle of 
 the pole. In this case the one who has the 
 
 2- 
 long end of the pole will bear -|-=|^^ or ^ of 
 
 5-(r 
 31 
 
 the load, and the other man -^z=^-^=z^ of 
 
 5f 
 
 the load. 
 
 T. A. PUGH, 
 Dumontville, 0. 
 
 71. What was the " Eeign of Terror V' 
 Answer : It was inaugurated May 31, 
 
 1793. Marat was assassinated by a young 
 
44 
 
 NOTES AND QUERIES. 
 
 woman named Charlotte Corday. Danton 
 had a short time before been denounced by 
 the Jacobins and executed. Robespierre re- 
 mained undisputed leader of the Jacobins. 
 He kept the guillotine busy till the execu- 
 tions for months averaged from 50 to 80 per 
 day, and 1,285 from June 10 to July 17, 
 1794. R. Belle Tinsley, 
 
 Joseph, Oregon. 
 
 72. A vessel starts from port and sails 
 
 due east 10 miles the first hour, and thence 
 
 northeast 10 miles the second hour; how far 
 
 is it from port at the end of the second hour ? 
 
 Solution : In the accompanying figure, the 
 
 broken line A 
 B C, represents 
 the course o£ 
 the vessel, and 
 the straight line 
 A C, the dis- 
 tance it is from 
 port at the end of the second hour. 
 
 Since the direction of the vessel in the 
 second hour is due northeast, by constructing 
 the square B D C E, so that B C is its diago- 
 nal, we find the side B E, or its equal C E, 
 as follows : 
 
 (B Cy=2 (B E)2, or 2 (C_E)^ whence 
 
 (BEy=i {10y=:bO ; B E=/50, or 7.07+ 
 miles. 
 
NOTES AND QUERIES. 45 
 
 Now, in the right triangle A E C, A E= 
 (10+7.07), or 17.07 miles, and C E=7.07 
 miles; whence (A C)^=(l7.07)2+(7.07)^ or 
 341.38, and A C=i/341.38, or 18.47+miles. 
 
 73. On a level piece of ground stand two 
 towers 200 feet apart; one tower is 120 feet 
 high and the other is 160 feet high. What 
 point on the ground between the two towers 
 is equally distant from the top of either ? 
 
 
 \ 
 
 Ba 
 
 \ 
 
 
 
 \^^" e 
 
 {Solution: Let AE be tower 160 feet 
 high, B C tower 120 feet high, and E C the 
 distance between the two towers, 200 feet. 
 Let D be the point on the ground equally 
 distant from the tops of the towers. Then A 
 D=B D. 
 
 Let x=E D. 
 
 Then 200-x=:D C. 
 
 Then x^-\-im=z{2m-x)mW. 
 
 a;2+25,600r=40,000-400x+x2+14,400. 
 
 Transposing and collecting : 
 
 400x==28,800. 
 
46 NOTES AND QUERIES. 
 
 x=72, distance from the foot of tower 160 
 feet high, or 200—72=128, distance from 
 foot of tower 120 feet high. 
 
 G. C. Baker, Woodville, 0. 
 74. A farmer paid $76 for calves and 
 sheep, paying $3 for calves and $2 for sheep; 
 he sokl i of his calves and ^ of his sheep for 
 $23, and by so doing lost 8 % of their cost. 
 How many of each did he buy ? 
 
 Solution: $23^.92=^25, cost of J calves 
 and |- sheep. 
 
 Calves. Sheep. 
 
 i + f =$25. 
 } + i -=$51. 
 
 1st lotX3=i + f =$75. 
 Subtracting 2d lot f =$24. 
 
 Since |=$24, |=$24Xf=$40, cost of sheep. 
 $76— $40=$36, cost of calves. 
 $40-^$2=20=No. of sheep purchased. 
 $36-$3=12= " calves 
 
 R. E. T., 
 Evergreen, La. 
 75. Who were the eminent men of the 
 Dark Ages ? 
 
 Answer: During the five centuries of 
 this age we find but few names of really emi- 
 nent men. Bede, the Englishman ; Alcuin, 
 also an Englishman ; John, surnamed Scotus 
 or Evigena, a native of Ireland; and Pope 
 Sylvester. Id. 
 
NOTES AND QUERIES. 47 
 
 76. What history was written in the 
 Tower of London ? When ? By whom ? 
 
 Answee : History of the World, written 
 1605-18, by Walter Ealeigh, 1552-1618. 
 
 John M. Colaw, Monterey, Va. 
 
 77. Why is the term, ''Land of the Mid- 
 night Sun,'^ applied to Norway and Alaska ? 
 
 Answer : Because in summer the sun 
 shines in the Arctic regions at 12 o^clock at 
 night. For many days the sun does not set 
 there, as in winter for many days it does not 
 rise. F. J. Swehla, 
 
 Wilson, Kan. 
 
 78. What is the size of the largest angle 
 that can be formed by two straight lines ? 
 
 Answer. 180°. It is sometimes called a 
 straight angle. W. D. Mueller, 
 
 Loimll, 0. 
 
 79. Diagram: He was graduated June 
 10, 1887. 
 
 He I was graduated 
 
 June 10, 1887. 
 
 Id. 
 
 80. Diagram : It was so cold as to freeze 
 the mercury. Parse as. 
 
48 
 
 NOTES AND QUERIES. 
 
 It I was \ cold 
 
 \ o 
 
 freeze I mercury \ p 
 
 
 (A) 
 
 As is a conj. adv., connects the subordinate 
 to the principal clause, and modifies so. Id. 
 
 81. Three equal circles just touch each 
 other, and thus enclose, in the triangular 
 space between them, one acre of ground. What 
 is the radius of the circles? 
 
 Solution : Let .T^Radius of circles. 
 
 Then V^x^ — a^=:x V 3 = alt. of 
 triangle ABC. 
 
NOTES AND QUERIES. 49 
 
 Also, xXx l/Tr= J ^ x" + 160. 
 1.7320 x2=r. 1.5708 x^ + ieO. 
 .1612x2=rl60. 
 
 x2= 992.55 + 
 
 X r=31.5 +, the radius. 
 
 Xax. 
 
 82. If tlie diagonal of a cubical room is 
 50 feet, what are the dimensions of the room ? 
 
 Solution : Let x represent the length of 
 one side of the room ; then x^-\-x^=2x^=ihe 
 square of the diagonal of the floor. The 
 diagonal of the floor, the diagonal of the room, 
 and the height of the room form a right-an- 
 gled triangle. Then 
 
 .r2=833^333|- 
 
 ^=l/833.83J=28.867-h Ans. 
 Rule: Divide the square of the diagonal 
 of any cube by three, and extract the square 
 root of the quotient, the result will be the 
 side of the cube. R. L. Laws, 
 
 Ewartsville, Wash. 
 
 83. Name . the longest river ; highest 
 mountain ; longest tunnel; and highest bridge 
 in the world. 
 
 Ans. Missouri river to the gulf; Mt. 
 Everest; St. Gothard tunnel; and Brooklyn 
 bridge. C. C. Morrison, 
 
 Adamston, W. Ya. 
 
60 NOTES AND QUERIES. 
 
 84. What bone of the human skeleton is 
 not in a cartilaginous state at birth? 
 
 Ans. The petrous which contains the 
 organs of hearing. M. M. Joshua, 
 
 Moreauville, La. 
 
 85. A man rows a boat with the tide 7 
 miles in 40 minutes, and returns against a 
 tide i as strong in one hour, what is the rate 
 of the stronger tide ? 
 
 Solution : Let a;=rate of stronger tide. 
 ^a;=rate of weiker tide. 
 and (7 — |a;)H-|=rate of rowing, neglecting tide, 
 also' (7+|a;)-e-i=rate of rowing, neglecting tide. 
 
 hence, (7-|:.)Xf=(7+i^)+i or ?1^^?}^ 
 
 a;— 2| miles per hr. 
 
 John M. Colaw, 
 Monterey,' Va. 
 
 86. A set out from C toward D, and trav- 
 eled 7 miles a day. After he had gone 32 
 miles, B set out from D toward C, and went 
 every day -^ of the whole journey, and after 
 he had traveled as many days as he went 
 miles in a day, he met A. Required the dis- 
 tance from C to D. 
 
 Solution : 
 
 Let a?=Di8t. from C to D. pxr c j u i. i 
 
 [No. of days he travels. 
 
 — =No. of mi. B travels per day, as well as the 
 
 hence og] =No. of mi. traveled by B. 
 
 ^^ . 7a?=No. of mi. traveled by A. 
 32+19 
 
NOTES AND QUERIES. 51 
 
 And ^+32-f-i|==^, whence, a;^— 228a;= —1152 
 
 Therefore .r=114±i/— 1152+12996=114=fc38 
 
 x=1^2 Ans. Id. 
 
 87. A merchant having a 40 lb. weight 
 let it fall and broke it into 4 pieces. The sizes 
 of the pieces were such that he was enabled 
 to weigh any number of pounds with them 
 from 1 to 40. What did each piece weigh? 
 
 Ans. In any geometrical series proceed- 
 ing in a triple ratio each term is one more 
 than twice the sum of all the preceding; 
 hence, the weights of the pieces are 1, 3, 9 
 and 27. Id. 
 
 88. A and B buy 100 acres of land for 
 |600, and pay equal sums of money. A says 
 to B, ^' give me my choice and I will pay 75 
 cents per acre more than you." How many 
 acres has each man, and what does it cost 
 
 him per acre ? 
 
 [=A's price. 
 Solution : Let a;=what B paid per acre and x-\-^ 
 250 250 ^ ^ ^ 
 
 then, — =No. acres B got. — r^=No. acres A got. 
 
 250 250 
 And — +——^=100, whence, 8x2—34^=15 
 
 '409 20.2237 
 
 TWefo.e, .=n,y-+(n)=,^: 
 
 64 8 
 
 ^^1J_^ 202237-^^^529+, B's price. $4.6529+$. 75 
 8 8 r=$5.4029 A's price. 
 
 $250-^$4.6529=53.729+, B's No. of acres. 
 $250h-$5.4029=46.271+, A's No. of acres. 
 
 Id. 
 
62 NOTES AND QUERIES. 
 
 89. What caused the war of 1812? 
 
 Ans. The violation of American commer- 
 cial rights was the principal of several causes 
 which led to the war of 1812. Besides this, 
 England offered other insults to America,, 
 such as the impressment of our seamen and 
 tlie searching of our vessels for deserters. 
 The firing into the Chesapeake, by the Brit- 
 ish ship Leopard, was the immediate cause. 
 A Subscriber, Lowell, O. 
 
 90. Shylock would have struck Jessica 
 dead beside him. Parse italicized words. 
 
 Ans. Dead, descrip. adj. used as objective 
 complement and belongs to "Jessica.^' Beside, 
 prep, shows the relation of "him'' to " dead.'*' 
 Him, sim. per. pro. ante, is " Shylock '' mas. 
 3rd. sing. obj. c. obj. of beside. Id. 
 
 91. A cubical foot of brass is to be drawn 
 into wire j-q of an inch in diameter. What 
 will be the length of the wire ? No loss in 
 the metal. 
 
 A71S. (4V)'X -7854:=. 00049 sq. in.:=area of 
 the end of the wire. 1 cu. ft.=rrl728 cu. in. 
 1728 cu. in-.00049 sq. in.=3,526,530.61224 
 in.=55 mi. 210 rd. 4 yd. 13 ft. 11.76 in. 
 length of wire. Id. 
 
 92. Three persons. A, B and C, bought a 
 
 12-in. ball of yarn; A paid 91 cents, B paid 
 61 cents, and C 64 cents. How many inches 
 
NOTES AND QUERIES. 53 
 
 must each wind off to get his share if they 
 
 take their shares in the order named? 
 
 Avs. 12^X .')236=907.77 cu. m.= volume of the 
 ball of yarn. $.91+$ 61+$ 64^$2.16=cost of the ball, 
 of which A paid $ 91. Then he would get ^W of the 
 thread B, who paid $ 61 would get -^^^ and C -^^^ 
 of the thread. 20? ^*" 904.77 cu. in =38u'.38 cu. in. A's 
 share. ^V^ of 904.77 cu. in ==254.98 cu. in. Bs 
 .share. ^\ \ of 904.77 cu. in —267.52 cu. in.=C's 
 
 share. ^267..^2^.5236=7.99in.=partC. must wind 
 
 •off. ^ ^267 52+254.98)^5236 = 9.99 in. = part B 
 
 rand C together mu=t wind off. 9.99—7.99=2 in.= 
 B's part. 12 in.— (7 99+2)=2.01 in.=A's part. 
 
 Id. 
 
 93. From a unit of the third order sub- 
 tract the sum of 371 and sixty-five ten-thous- 
 :andths, multiply the remainder by three- 
 tenths and divide the product by five mill- 
 ionths. 
 
 Ans. 100 — .0436=99.9564. 99.9564X.3 
 =29.98692. 29.98692-.000005=5997384. 
 
 Id. 
 
 94. The diameter of a sphere is 2 ft. 
 What is the diameter of another sphere con- 
 taining J as much surface ? 
 
 Ans. 2 ft.x3. 1416=6.2832 ft.=circumfer- 
 ence of the sphere. 6.2832 ft.x2 ft.=12.6664 
 sq. ft.=surface of the sphere. J of 12.5664 
 ,«q. ft.=6.2832 sq. ft.=surface of the required 
 sphere. Then 12.5664 sq. ft. : 6.2832 sq. 
 
54 NOTES AND QUERIES. 
 
 , 6.2832X8 , , 
 
 ft.::(2 ft.)^:(.)^=-y2:ggg^=4. f 4 ==1.59 
 
 +ft: Ans. Id. 
 
 95. If the third of 6 be 3, what'll the 
 fourth of 20 be? 
 
 Ans, J of 6=2 ; J of 20--5. Then 2:5: : 
 5X3 
 3:(?). -2~=7*-^^^^- Jd. 
 
 96. A, B and C invest ^5,640 in business. 
 At the end of six months A puts in |500 and 
 C withdraws $640. At the end of ten months 
 B withdraws $1,000. A's gain is $270, B's 
 $220, and C's $158.40. What did each in- 
 vest at the beginning of the year ? 
 
 Solution: $5,640X6= $33,840 
 
 500 
 
 $6,140 
 640 
 
 
 $5,500X4= 
 1,000 
 
 22,000 
 
 $4,500X2=- 
 
 9,000 
 
 $64,840 
 $270+$220+$158.40=$648.40. 
 $648.40-$64,840= 1 % -rate of gain per month 
 $500X.0lX6=$30. 
 $270-$30=$240. 
 $240-^2=$! 20, gain first six months. 
 
NOTES AND QUERIES. 56 
 
 $120-^(.01X6)=$2,000, am't A first put in. 
 ^1,000X.01X2=$20. 
 $220+$20=$240. 
 $240^1 2==$20, gain per month. 
 |20-.01=|2,000, B's investment. 
 A and B each invested $2^000 at first. 
 ,^5,640-($2,000+$2,000)=$l,640, C's invest- 
 ment. 
 
 J. F. E. Miller, New Bremen, O. 
 
 97. In 1880 February had five Sundays* 
 When will this occur again ? 
 
 Ans. It will occur again in 1920. The 
 period between these occurences is generally 
 28 years, but when a year ending with a 
 double cipher is not devisible by 400, the 
 period in which that year falls is twelve years 
 longer. M. D. Taylor, Osceola, Ky. 
 
 98. A boy can split a cord of wood while 
 a man chops it. The man can split two 
 cords while the boy chops one. What has 
 the boy earned when the man has earned one 
 dollar? 
 
 A71S. This query has been answered before, 
 query and answer No. 20, but the answer 
 given is incorrect. The proportion that the 
 boy's work bears to the man's will be the 
 same whether he be chopping or splitting. 
 
 Let a:=what the boy earns while the man 
 earns $1. Then 
 
6G NOTES AND QUERIES. 
 
 a;: 100 cents :: 100 cents : 2a; 
 2a;2= 1.0000 
 
 ^=/.5000= 707-f 
 The boy then earns 70 cents 7 mills while 
 the man is earning |1. 
 
 Olive E. Cheslj:y, Epsom, N. H. 
 
 99. A and B agree to build a wall for 
 $50. A can lay the wall as fast as B can 
 bring the materials, and B can lay the wall 
 one-half as fast as A brings the materials. 
 Divide the money. 
 
 Solution: Let .t— A^s share ; y=Ws share. 
 Thenx+y=50 (1.) 
 
 B's work is proportional to A's, y '• x'-'-x'- 
 
 22/ (2.) 
 
 From (2) we obtain 2y^=oi? (3.) 
 
 From (1) we obtain x—bO—y (4.) 
 
 Squaring (4.) a;2=2500-100i/+3/l 
 
 Substituting in (3.) 23/^=2500— 100!/+2/^ 
 
 Transposing. 3/'''+ 100?/= 2 5 00. 
 
 Completing square. 3/^-}- lOOy-f 2500=5000. 
 
 Extracting root. 3/+ 50=70.71+ 
 
 2/=70.71-50=20.71 B's share. 
 
 07=50-20.71=29.29 A's share. Id. 
 
 100. Diagram the following: 
 
 (a) I Lad heard of it before, so I was not 
 surprised. 
 
NOTES AND QUERIES. 
 
 had heard 
 
 57 
 
 before 
 
 of I it 
 
 so 
 
 was surprised 
 
 not 
 
 So is a co-ordinate conjunction, and con- 
 nects the two clauses. 
 
 (6) I never go this way that I do not think 
 of it. 
 
 go 
 
 way 
 
 never 
 
 this 
 
 that 
 
 do think 
 
 not I I of I it 
 
 That is a subordinate conjunction and con- 
 nects the subordinate clause to the principal 
 clause. 
 
 (d) Now, there is at Jerusalem, by the 
 sheep-market, a pool. 
 
 Now there 
 
 pool 
 
 is 
 
 by I sheep-market 
 
 at I Jerusalem the 
 
 Now is an independent adverb ; There is an 
 adverb of position. 
 
58 
 
 NOTES AND QUERIES. 
 
 (e) A day later began the fatal retreat of 
 the Grand Army from Moscow. 
 
 retreat 
 
 
 began 
 
 
 '■v 
 
 fatal 
 the 
 
 
 1 later 
 
 1 X , day f^^in 
 
 Moscow 
 
 
 A 
 
 of j Army 
 
 
 
 
 
 Grand 
 
 
 
 
 the 
 
 
 Day is common noun, third person, singular 
 number, neuter gender, objective case, gov- 
 erned by a preposition understood, or objec- 
 tive case without a governing word ; Later is 
 an adverb and modifies began. 
 
 A. M. Barr, 
 
 New Paris, Pa. 
 
 101. Give a sentence using "which'^ both 
 as an adjective and a connective.' 
 
 Ans. Ascertain which book he desires. 
 
 Id. 
 
 102. Define Mercator's projection, equa- 
 torial projection and ])o]ar ])rojection. 
 
 Ans. Mercator's projection is a map of 
 the earth so as to represent the meridians as 
 running ])arallel instead of converging, which 
 makes all the outlines in higher latitudes 
 comparatively wider than near the equator; 
 but,. in order to preserve the proportional 
 
NOTES AND QUERIES. 69 
 
 form, tlie parallels of latitude are not equi- 
 distant apart, but the spaces increase as they 
 are farther from the equator in the ratio that 
 the meridians converge. 
 
 Equatorial projection is representing a 
 hemisphere on a circular map with the equator 
 running through the center, meridians con- 
 verging, as if a globe,' on which all the 
 imaginary lines, as well as real, are drawn, 
 was photographed with the equator in the 
 middle. 
 
 Polar projection is a similar picture of the 
 hemisphere with the pole in the center, from 
 which the meridians are radiating in all di- 
 rections, and the circumference of the map is 
 the equator. 
 
 Recapitulation : — By Mercator^s pro- 
 jection the whole surface of the earth can be 
 shown on one single chart or map. By equa- 
 torial projection the eastern and western 
 hemispheres, and by polar projection the 
 northern and southern hemispheres are shown, 
 but each hemisphere requires a separate chart 
 or picture. • F. J. Swehla, 
 
 Wilson, Ka 
 
 103. I bought three notes drawing 8 per 
 cent, interest, one for $50, due in twelve 
 months; one for |50, due in twenty-four 
 months; and one for |100, due in thirty- 
 six months; I paid only $160 for them^ 
 
60 NOTES AND QUERIES. 
 
 What rate of interest do I get on my invest- 
 ment? 
 
 Ans. The proceeds of the notes at maturity=$54 
 
 +$)8+$124=$236. $236— $160 = $76 = gain. 'Ihe 
 equated time of note.=27 months. $160X!1=$360; 
 $76-^$360:=.21^=per cent, gained on the invest- 
 ment. No days of grace being counted. Id, 
 
 104. How do mariners determine their 
 latitude and longitude at sea? 
 
 Ans. Latitude is found at night from 
 the altitude of the polar star, in daytime from 
 the altitude of the sun. Both observations need 
 corrections for refraction and parallax ; also 
 elongation of the polar star and declination 
 of the sun for that day must be taken in ac • 
 count. Longitude is found by means of a 
 correct time-piece called a chronometer giving 
 the time of the place from which thelongi-^ 
 tude is reckoned, as Greenwich, and compar- 
 ing that with the local time of the place of 
 observation, which is found by observing thcf 
 sun in day time and moon or any othei 
 heavenly body whose time of rising, setting 
 or time it crosses the meridian is given by the 
 astronomical tables. Id. 
 
 105. A boy can split a cord of wood 
 while a man chops it. The man can split 
 two cords while the boy chops one ; what has 
 the boy earned when the man has earned $1 ? 
 
 Ans. (You will see that my solution of 
 the above problem does not produce the same 
 
NOTES AND QUERIES. 61 
 
 result as that of L. B. Hay ward in query and 
 answer number 20.) 
 
 To illustrate the principle underlying the 
 accompanying solution, let us make two sup- 
 positions. 
 
 (1) Suppose their rates of working to be 
 as 3 to 5, i. e.y the man can cut a cord of 
 wood in 3 hours, and the boy in 5 hours. 
 
 Applying the conditions of the problem, 
 we find that if it takes the boy 3 hours to 
 split a cord of wood, it would require the 
 man |^ of 3, or |- hours, and in one hour he 
 would split |- of. a cord, and in 5 hours, the 
 time required for the boy to cut one cord, the 
 man would split |-X5=2|- cords, a quantity 
 greater than that given in the problem ; hence, 
 our supposition is false. 
 
 (2) Suppose their rates of working to be 
 as 5 to 7. Applying the conditions of the 
 problem we find that if it takes the boy 5 
 hours to split a cord of wood, the man would 
 split it in ^ of 5, or ^- h^urs, and in one 
 hour he would split -^^ of a cord, and in 7 
 hours he would split ■^V^^^-'^M cords, a 
 quanty less than that given in the problem, 
 and our supposition again is false. 
 
 Now, by inspection, we find (and the same 
 is true if the number of suppositions be in- 
 definitely increased), that the above results 
 may be obtained by dividing the square of 
 
G2 
 
 NOTES AND QUERIES. 
 
 the boy's rate by the square of the man's 
 rate, i. e, (6)^^(3)^=2|, and (7)^-(5)^=l|| ; 
 hence, to satisfy the conditions of the prob- 
 lem, the squares of their rates of working 
 must bear the ratio of 1 to 2, and their rateSy 
 that of i/T to 1/2; i. e.j their rates are incom- 
 mensurable, and only an approximate result 
 can be obtained which we find as follows : 
 i/r=l; 1/2=1.414+, hence, when the man 
 has earned $1, the boy has earned 70 cents, 
 nearly. 
 
 S. H. Treher, 
 
 Roxbury, Pa. 
 106. Prove that the lines joining the mid- 
 dle points of any quadrilateral, taken in or- 
 der, enclose a parallellogram. 
 
 The proof of 
 the above propo- 
 sition is based 
 upon the follow- 
 ing: 
 
 T H E o R E M — 
 The straight line 
 joining the middle points of two sides of a 
 triangle is parallel to the third side. 
 
 Let ABCD be a quadrilateral, and FG, 
 GH, HE, and EF be the lines joining the 
 middle points of the sides, taken in order. 
 
 To prove that these sides enclose a paral- 
 lelogram. 
 
NOTES AND QUERIES, 63 
 
 Draw the diagonals AC and BD. 
 
 Now, in the triangle ABC, the line FG 
 joins the middle points of the sides AB and 
 BO, and is, therefore, parallel to the third 
 side AC. 
 
 Again in the triangle ACD, the line EH 
 joins the middle points of the sides AD and 
 CD, and is, therefore, parrallel to the third 
 side AC. 
 
 Since the lines FG and EH are, respect- 
 ively parallel to the line AC, they are paral- 
 lel to each other. 
 
 In the same manner it may be shown that 
 the lines EF and GH are parallel to each 
 other, hence EFGH is a parallelogram. 
 
 Id, 
 
 107. A merchant buys a bill of goods 
 amounting to $2,480; he can have four mouths' 
 credit, or 5% oif for cash; if money is worth 
 only 10% to him, what will he gain by paying 
 cash ? 
 
 Solution, hio of $2480=r$124. 
 
 |2,480-$124==f 2,356. 
 Interest on $2,356 for 4 months at 10%=- 
 $78.53. 
 
 $124--$78.53=:$45.47 Am. 
 
 J. F. E. Miller, New Bremen, O. 
 
 108. What was the " O Grab Me '' act ? 
 Ans. It was a fictitious name for the 
 
 '^ Embargo ^' act. It was applied by those 
 
C4 
 
 NOTES an:) queries. 
 
 who opposed Jefferson and the Democratic 
 })arty. By spelling the word ^' Embargo'' 
 l)ackwards, we have '^O grab me.'' 
 
 Ai.T.iE Hutchinson, 
 
 Crosstown, Ohio. 
 
 109. The length of a rectangular field 
 containing 25 acres is 2 J times its width. 
 Find the length and width. 
 
 A 
 
 D 
 
 F B 
 E C 
 
 Solution : ABCB = rectangular field containing 
 25 acres. 
 BCEF = square field whose side is the 
 width of rectangular field, 
 
 and whose area is -^^rr of area. 
 
 of rectangular field. 
 25 A. = 4,000 sq. rd.= area of rectangular field. 
 4 000 sq. rd h- 2J=1,600 sq. rd.=area of square field. 
 
 = side of square field, 
 = width of rectangular field. 
 = length of rectangular field. 
 C. A.M., Yolo, Cal. 
 
 110. What per cent, is made on a book 
 bought at 10 per cent, below the list price, 
 and sold at 45 per cent, above the list 
 price ? 
 
 / 1,600 sq. rd.= 40rd.: 
 40 rd. X 2^ = 100 rd. 
 
NOTES AND QUERIES. 65 
 
 Solution : Let 100% = list price. 
 
 lOO/o — 10% = 90% = cost price. 
 100 " + 45 '' = 145 " = selling price. 
 145" _90" = 55" =gain. 
 55" -f-90" — .61| = 61i% = rate of gain. 
 
 J. H. S. Barlow, 
 
 Howesville, W. Va. 
 
 111. What will it cost to fence a field 
 in the form of an equilateral triangle, alti- 
 tude 20 rods, at 80 cents per rod ? 
 
 D Solution : 
 
 Let ABC be an 
 equilateral triangle. 
 Section ABD is a 
 right-angled triangle 
 whose hypothenuse 
 equals the side of the 
 equilateral triangle, 
 Cand whose base 
 equals half the side 
 
 of the equilateral triangle. 
 
 Let 2x = hypothenuse and x = base. 
 Then 4a;2— a;2=3rc2=202=400 
 a;^=13 3.3333+ 
 • a;=zi/l33.3333=-11.547 
 
 2a;= 2X11.547 = 23.094=side of 
 equilateral triangle. 
 23.094X3X$.80==$55.425 = Cost of fencing. 
 
 Henry B. Bergey, 
 
 Kulpsville, PennJ 
 
 112. A certain number has been divided 
 by one more than itself, giving a quotient 
 
66 
 Off 
 
 NOTES AND QUERIES. 
 
 What is the number ? (Solve by 
 
 proportion.) 
 
 /Solution : 
 
 Number : number + 1 : : 1:5 
 Then, 5 times number = once the number + 1 
 4 times number = 1 
 
 number = ^ Id. 
 
 113. Find the weight of a spherical 
 shell 2 inches thick, outside diameter 14 
 inches, the metal weighing 450 pounds to 
 the cubic foot. 
 
 Solution : 
 
 14 iD.=| ft.:=outside diameter. 
 10 in.=f ft.=insid'i diameter. 
 [(J)3X.5236]— [(f )^X.5236]^.5-'8448==cu. ft. in shell. 
 450 lbs. X.528448r=237tlbs.=weightof shell 
 
 Id, 
 
 114. Three- sevenths of a sale is gain. 
 What is the gain per cent ? 
 
 Solution : \ = whole sale. 
 
 ^ — f = ^ = cost. 
 y-^ y == 75 = 75% = rate of gain. 
 
 115. The diagonal of a 
 
 Id. 
 
 is 100 
 
 square 
 feet. What is the area ? 
 
 Solution: Let ABCD be 
 the square, and BD the di- 
 
 agonal. 
 
 
 
 AB = 
 
 AD 
 
 
 
 2AB2= 
 
 = BD'^= 
 
 100^== 
 
 10,000. 
 
 AB^= 
 
 = 5,000: 
 
 = area 
 
 of square. 
 Id. 
 
NOTES AND QUERIES. 
 
 67 
 
 116. A, B and C do a piece of work for 
 •.35. On the supposition that A. and B 
 
 do three-fourths of the work, A and C 
 nine-tenths, and B and C thirteen-twenti- 
 eths, what should each receive? 
 
 Solution: f + j^y + if = ff whicli is supposed to 
 be twice the work thai ihe three men do. 
 
 23 _^ 9 _ 23 
 
 Y(j . ^ — 20 
 
 |§ — 4 = 1, the part of ihe work C performs. 
 
 2 3 1 " «.< <« "R II 
 
 "Jo 10 4, D 
 
 2 3 13 10 <i " <« A <l 
 
 2 20 ~" 2 0' -"■ 
 
 These fr^.ciions, rfduced to a common denomina- 
 tor, are to eaca other as their numerators. 
 
 2 1 10 — __8_ _5_ 10 s _|_ 5 _4_ in — 9^ 
 ^1 4' 20 — 2ff' 2 2 0- o -ir o -f lyj — ^o. 
 
 $79 35 X 2^ = $27.60, C's share. 
 $79.35X/V==^17 25, B's " 
 $79.35 XM = $34,50. A's " 
 
 A. E. Taylor. 
 
 117. A rectangular lot of land 64 rods 
 long and 36 rods wide, and a square lot of 
 equal area are to be fenced. Which will 
 require more fencing, and how much more 
 will it require? 
 
 Solution : 
 
68 NOTES AND QUERIES. 
 
 The rectangular lot will require the more fencing. 
 64 rd.X36 -d = 2,304 sq. rd.== area. 
 V2, 304 sq. rd = 48 rd = side of square. 
 (2X64rd.)+(2X36rd.)=200rd. 
 
 =^perimeter of rectangle. 
 4 X 48 rd.= 192 rd.= perimeter of square. 
 200 rd. — 192 rd.= 8 rd.= diflference in fencing. 
 
 Xax. 
 118. Two windows, on opposite sides of 
 the street, and opposite each other, are each 
 28 feet from the ground; a ladder reaching 
 from the middle of the street to either win- 
 dow is 53 feet lons^. What is the shortest 
 line that will reach from one window to the 
 other ? 
 
 Solution : 
 
 
 T/Sa-^— 282=l/2809— 784=l/2025=45=half distance 
 45 ft.X2=90 ft. = whole distance. Id. 
 
 119. A double inclined plane is 50 feet 
 long from the top to the foot on each side ; 
 the top is 30 feet from the ground. How far 
 on the ground from one foot to the other ? 
 
NOTES AND QUERIES. 
 
 6^ 
 
 Solution : 
 
 ao 
 
 This double inclined plane is composed 
 of two right-angled trianglevS,each of which 
 has an hypothenuse of 50 i'eet, an altitude of 
 SO feet, and a base equal to one- half the 
 distance on the ground from one foot to 
 the other. 
 
 l/50"^— yu'-^^l/ 2500— 900= l/ 1 6UO=40=base. 
 40ft,X2=80 ft.=distance from one foot to the 
 •other. • Id. 
 
 120. A tree 90 feet high wa^ broken so 
 that the top struck the ground 30 feet from 
 the root. How much of the tree was 
 broken off? 
 
 Solution: If the tree was perpendicular 
 to the ground, after falling, it, with the 
 ground, formed a right-angled triangle 
 whose base w^as 30 feet, and the sum of 
 whose hypothenuse and perpendicular was 
 SO feet, the hypothenuse being the part 
 broken off, and the perpendicular being 
 the part left standing. 
 
70 NOTES AND QUERIES. 
 
 Let x= part broken off. 
 90 — x=^ pa.t left standing. 
 
 900+8100- 1 mx-^x'=x'' 
 9000— J80x-^0 
 
 180x^-9000 
 a:=z:50=part 
 broken offi 
 
 30/7. ■^^• 
 
 121. Find the area of the quadrilateral 
 ABCD, according to dimensions given. 
 
 fi^^ jD 
 
 ^ AB:=:9 chains. 
 
 B Cr=6 
 
 CD=8 '• 
 AD=:7 " 
 AC 11 " 
 
 D 
 
 Solu'i<iii : 
 
 To fitjd the area of ABC, we have: 
 9+6+ 1 l=r26 ch.. sum of the side-, AB BC, and AC. 
 26^2=13 ch , J of the sHes. AB, BC, and AC. 
 13 — 9=4 ch., difference between AB and half sum 
 
 ot sides. 
 13 — 6=7 ch , differerce between BC and half sum 
 
 of sides. 
 13 — 11=2 ch., difference between AC and half sum 
 
 of sides. 
 13X4X7X'2=728, and l/728=27 sq. ch., nearly. 
 27^10=2.7 Hcres in the triangle ABC. 
 
 To find the area of ADC we have: 
 7+s+l 1=26 ch., sum of the sides AD, DC, and AC. 
 26^2=13 ch., i of the sides AD, DC, and AC. 
 13 — 7=6 ch , difference between AD and half sum 
 
 of sides. 
 
NOTES AND QUERIES. 71 
 
 13 — 8=5 ch., difference between DC and half sum 
 
 of sides. 
 13 — 11=:2 ch , difference between AC and half sum 
 
 of sides. 
 
 13X6X5X2-=780,_and v'780=28 sq. ch., nearly, 
 28-i-10=2.8 acres in the triangle ADC. 
 But ABC+ADC=ABCD; then 2.7+2.8=5.5 acres, 
 nearly. 
 
 Though these results are not precise, 
 they illustrate the rule, and are near enough 
 for all practical purposes. T. S. Pkice, 
 
 Kingsbury, Cal. 
 
 122. What is the office of a '^ Poet Lau- 
 reate?" Who is now " Poet Laureate?" 
 
 Ans. A "Poet Laureate" is an officer of 
 the King's household, whose business it is 
 to compose an ode annually for the King's 
 birthday and other suitable occasions. Al- 
 fred Tennyson is the present "Poet Lau- 
 reate." D. W. Bowman, 
 
 E^ew Madison, O. 
 
 123. Who was Southey? Name some 
 of his works. 
 
 Ans. Robert Southey was a distinguished 
 English poet, born in Bristol, England, 
 August, 1774. He wrote on a great variety 
 of subjects, among which Madoc, Thalaba, 
 The Curse of Kehama, Roderick, and The 
 Last of the Goths are the principal. In 
 1813 he was appointed *'Poet Laureate," 
 a post which he retained till his death in 
 March, 1843. G. L. R., Adams, Ind. 
 
72 
 
 NOTES AND QUERIES. 
 
 124. I have a round garden containing 
 75 8q. rd. How large a square garden can 
 be made in it ? 
 
 Now the square of a side of the square 
 garden inscribed in this circle, the area of 
 which is 75sq. rd., will be its area. Con- 
 ceive this square to be divided by the diag- 
 onal AC into two equal right triangles 
 ABC and ADC. The diagonal AC, which 
 is the diameter of the circle, will be the 
 hypothenuse of each of the right triangles. 
 75-.785398=95.4929857+, is the square of 
 the diameter of the circle — the hypothenuse 
 of the right triangles. 
 
 The square on the hypothenuse of a 
 right triangle is equal to the sum of the 
 squares on the other two sides. Therefore, 
 
NOTES AND QUERIES. 73. 
 
 95.4929857+=the sum of the square on 
 two sides of these right triangles, or twice 
 the square on one. Then the square on one 
 8ide=one.half of 95.4929857+=47.7464927 
 8q. rd. — the area of the garden inscribed 
 in the circle. N. T. G-oldsberry, 
 
 Smoky Ordinary, Ya. 
 
 125. What is the plural of "soliloquy?" 
 ^^Statf?" 
 
 Ans. Soliloquies ; staves, except when it 
 means a corps of officers, then staffs. Id. 
 
 ^^^' Given, 1— i V x—2a-'^=^ to find x, 
 
 -a 
 
 Multiplying by x and transposing x and -y/ x — 2a — , 
 
 X X 
 
 _ . , . a ,a . ^ . a 
 Squaring and transposing x — , (- — x) ^-\ xz= 
 
 [—2a. 
 
 a a 1 — 8a 
 
 Completing ( — x^-^ x-{-\= — 1~" • 
 
 Extracting sq. root, - — x^\=±l ^ 1— 8a ^ 
 
 X 2 
 
 Multiplying by — x and transp'g a;^ ^^ ^ 
 
 [=a. 
 
 Completing the square a;^— ^(^^v^^~H 
 
74 NOTES AND QUERIES. 
 
 p( l±i/l— 8a )^__ 2±2/l— 8a-f8ce. 
 '• 16 16 
 
 Extracting sq. root, x— ~ .~ ^ ~=!::'V2-^2v/l^I^ 
 
 [+8a. 
 
 a; 
 
 _lrti/l— 8azb\2±2i/l— 8aH-8a. 
 4 
 Ed. Haas, 
 
 Cory don, Ind. 
 
 127. Given, (a;«+l)y=(y+l):c='(l),and (v«+l)r== 
 
 [9(a:-^+l)j/3(2). 
 
 Dividing both members of (1) by a^y, x^-\ — .•$= 
 
 b/+\ (3). 
 
 Dividing both members of (2) by xy^ and reversing, 
 
 [3(3:r+?)==/+p (4). 
 
 3 
 Multiplying both members of (3) by 3, ^x^-\--^ = 
 
 [3y4|(5)- 
 
 Adding (4) and (5), 3(a^+3x+-+— )=/+3y+ ?+ 
 
 [p (6). 
 
 Extracting cube root, (ic+~)if 3=3/H — (7). 
 
 X y 
 
 Substituting value of t/+- from (3), u? -\-—f=^ 
 
 y ^ 
 
NOTES AND QUERIES. 76 
 
 [(a:+l)f 3. (8). 
 
 DividiDg both members by x+^fX"^ — l-j--__zi=l^3 (9). 
 
 X x'^ 
 
 Adding 3 to both members, a;^+.2+— =3 +]^3"( 10) 
 
 Adding — 1 to both members, x^ — 2+— r,= #^3 — 1 (H) 
 
 x^ 
 
 Extracting sq. root of (10), x-\--=^^/^.^ ..^^ 
 
 Extracting sq. root of (11), a: "^a/t/^ 1 H'W 
 
 Adding (12) and (13) and dividing by 2, 
 
 '2 
 
 In (12), a;+l_y^g_^^^ substituting this value in. 
 
 (7),y+L^3: -^f 3+3 
 Clearing of fractions and completing, 
 
 2/"-(^-^l/^+3)2/+(f'3. V^3/3_^3) .^ 3f9-~-l 
 
 4 ^ 
 
 ExtractingBq.ro ^^3:^" ^3^ 
 
 ' ^ ^- ~^~ " 
 
 1^3:^/3+3^-^31?/ 9-1 
 y= 2 
 
 Id, 
 
76 NOTES AND QUERIES. 
 
 Given wi?= — ^;;;^ to find x. 
 
 Clearing of fractions x"^ — 5x=124-8]/'fl7 
 Subtracting 12 from each member and squaring 
 x*—l0x^-\-x''-\-120x+Ui=&4x 
 Subtracting 64a;, x*—l0x^-^x''+5&x-\-Ui=O 
 Factoring {x''—lSx+^6) (x^-{-Sx+4)=0 
 Making x^ — 13a;H-36=0 we find x:=9 or 4 
 
 Or making ar'^+3a;+4=0 we find x== — ^ 
 
 Id. 
 Another solution of the same : 
 
 Given x=^ — f^ to find the value of x, 
 
 X — 5 
 
 Solution : 
 ■(1) oc^ — 5x=12+8|/^, by clearing effractions 
 
 (2) X+4rr=X+4 [(2) 
 
 (3) x'—4x+4.=zl6+SVx+x by adding (1) and 
 
 (4) X — 2=4+1/ a; by extracting the square 
 
 [root of (3). 
 
 (5) X — ■/x=Q by transposing the terms in 
 
 [(4) 
 
 (6) X — V x+i=Q+i=^-£- by completing the 
 
 [square 
 
 (7) /^-i=d=f 
 
 (8) Vx=±i+i 
 
 (9) Vx=S or— 2 
 
 (10) x=9 or 4 
 
 See Ray's New Higher Algebra, p. 232. 
 
 P. B. Hays, 
 Brandenburg, Ky. 
 
NOTES AND QUERIES. 
 
 77 
 
 129. Three persons whose residences are 
 in a triangular area, ten, eleven and twelve 
 rods from each other, wish to di^ a well 
 where it will be as near one house as the 
 other. Where must they dig it? What 
 distance from each house ? 
 
 Solution: Rule: To find equal distances 
 from the vertices or 
 angles of a triangle to 
 a point in the triangle, 
 divide the product of 
 the three sides by four 
 times the area of the 
 triangle. 
 
 10X11X12=1,320== 
 product of the three 
 sides. 
 To find the area of a triangle when the 
 three sides are given : 
 
 EuLE : From half the sum of the three 
 sides subtract each side separately ; multi- 
 ply the half sum and these remainders to- 
 gether, and extract the square root. 
 
 10+11+12=33 J of 33=16.5 half sum. 
 
 16.5—10=6.5 
 
 16.5—11=5.5 
 
 1 6.5—12=4.5 
 
 >/l6.5X6.5X5.5X4.5=51.52=area 
 51.52X4=206.08=four times area 
 
78 NOTES AND QUERIES. 
 
 1,320-206. 08=r6.405+rd.=distance from 
 each one to the well? 
 
 Jerome Erdley, 
 Middleburgh, Pa. 
 
 130. A lady, the mother of three daugh- 
 ters, had a farm of 600 acres in the form of 
 a circle, with a residence in the center. 
 Being desirous of having her daughters 
 near her, she gave each of them a farm in 
 the form of a circle with a residence in the 
 center of eacli. How much did her daugh- 
 ters get, and how far apart were they? 
 How much did the lady keep, and how far 
 was each dauirhter from her? 
 
 Solution : ^500X160— 80, 000sq. rd . Di- 
 ameter of large circle^/ 80,000-^.7854=:: 
 319.19. llTow, letting the diameter of one 
 
NOTES AND QUERIES. 79 
 
 of the Small circles=2 and the radius=l, 
 to find distances from center of circles to 
 center of large circle use rule given in 
 question and answer No. 129. 
 
 2X2X2=product of the three sides. 
 
 Area of triangle=i/2'— r=i/3=^1.732 
 
 1.732X4=6.928=four times the area. 
 
 8-6.928=1.15 distance from A, B or C 
 to center of circle; . • . the ratio of the radius 
 to the distance from the center of each cir- 
 cle to the center of the large circie^^l : 
 
 [1.15. 
 
 . • . 1+1.15=2.15 or radius of large circle. 
 
 Diameter of large circle=2 15X2-=4.30. 
 
 4.30 : 2 : : 319.19 : x, or diameter ot one 
 of the daughters' tracts of ]and=rl48.46+rd. 
 
 The distance they are apart=the diam- 
 eter of the circlesr=148.46+rd. 
 
 The distance the daughters are from the 
 lady=(319.19-2)— (JX148.46)=85.365rd. 
 
 Area of each daughter's tract equals 
 
 148.46+2x.7854=17;310.5sq. rd. 
 
 17,310.5 -:-160=108 A. 30.5sq. rd. 
 
 What the lady had left=500A.— (3X108 
 A. 30.58q. r(i.)=^175A. b8.5sq. rds. Id. 
 
 131. What is meant by the X Y Z Mis- 
 sion ? 
 
 Answer: During the Revolution the II. 
 S. secured the valuable aid of France by 
 Treaties in 1778. In 1789 monarchy was 
 
80 NOTES AND (^)UERIES. 
 
 overthrown in France, and that nation soon 
 found herself at war with England and 
 other European nations. She desired the 
 U. S. as an ally, and Genet was sent to 
 accomplish her purpose. His mission failed. 
 Washington persisted firmly in preserving 
 our neutrality, and Jay's Treaty was con- 
 cluded with England. The course of our 
 government angered France. In 1797 the 
 Directory, which then governed that coun- 
 try, gave permission to the French navy to 
 assail our vessels. Following a policy of 
 conciliation, m spite of French insult^ to 
 our minister and the threat to our commerce. 
 President Adams called a special session of 
 Congress in May, 1797, and Charles Cotes- 
 worth Pickney, John Marshall and Elbridge 
 Gerry were sent to France to arrange mat- 
 ters. In the spring of the next year the 
 President submitted to Congress dispatches 
 that had been received from these com- 
 missioners. They had been kept waiting 
 by Talleyrand, the Minister of Foreign 
 Afiairs, and had been approached by three 
 unofficial persons with what was in effect a 
 demand for a bribe and a loan to the Direc- 
 tory before any arrangement could be con- 
 cluded with the U. S. In the dispatches 
 the names of the three persons were indi- 
 cated merely by the letters X Y Z, and 
 
KOTES AND QUERIES. 81 
 
 hence the whole affair came to be termed 
 the X Y Z Mission. Philip B. Hays, 
 
 Brandenburg, Ky. 
 
 132. I have a circular garden containing 
 ■75sq. rd. What must be the side of a square 
 field to contain it ? 
 
 ^ — nB 
 
 ABC D=the square. 
 E F— diameter of cir- 
 cle. E F=D C or A B. 
 
 ^ 7 5-^.7854= 95.492729. 
 1/95.492729=9.772 = E 
 F=A B or D C in rds. 
 Id. 
 
 The evening was glorious, and light through 
 
 the trees, 
 Played in the sunshine the raindrops, the 
 
 birds and the breeze ; 
 The landscape outstretching in loveliness 
 
 lay 
 On the lap of the year, in the beauty of 
 
 May. 
 Diagram, and parse *4ight." 
 
82 NOTES AND QUERIES. 
 
 'evening | the 
 
 {glorious 
 and through trees | the 
 light 
 
 raindrops | the 
 birds j the 
 and 
 
 breeze I the 
 Played in sunshine | the 
 
 [The 
 ' landscape < 
 
 [ outstretching | in loveliness 
 
 , (the 
 on lap •< 
 
 lay 
 
 of year | the 
 
 m 
 
 beauty I *^®j^^y 
 
 Light is an adjective, descriptive, com- 
 mon, in the predicate after the copulative 
 verb "was,'' and modifies the noun "even- 
 ing." Rule X, Rigdon. Id. 
 
 134. A banker owns 2J% stocks at 10 fo 
 below par, and Sfc stocks at 15% below par. 
 The income from the former is 66|^ more 
 than from the latter, and the investment in 
 the latter is $11,400 less than in the former. 
 Required the whole investment and income. 
 Solution : 
 
 (1.) Let 100%=Investment. 
 (2.) 100%— $ll,400=money to be di- 
 [vided into two equal parts, to 
 [find latter investment. 
 
NOTES AND QUERIES. 83 
 
 (3.) 100%-$11,400 ^ • 
 
 ^ ^ ^ =50 f^— $5,700 latter 
 
 [investment. 
 
 (4.) 50/^-?5,700+$ll ,400=50% +$5,700 
 
 [former investment. 
 
 (5.) 50% , $5,700 ^ _ 
 
 -qTt-H — qk — = par value of former 
 
 [stock, 
 
 2i/»of(6)=(6)=J.^,+%^tbr- 
 
 [mer income. 
 
 A • /QN . Qfc rn\ 50% 15,700 1 
 
 Agam (3)-^85=(7)=-g^-^i^-^par val- 
 
 [ne latter stock. 
 8f. of (7)=(8)=A^/„-!g-l= j^. 
 
 [come from latter. 
 By tlie conditions of the prol)lem, 
 former income=latter i 6^%% or 
 I of latter+latter 
 
 K } % |^j^q7» - 85 J 170 85 
 
 (10-) [iTo^^-^ J + 1 170^"- "ST , 
 
 5 $285 , ., . 
 =jpjQ%-^ latter income 
 
84 NOTES AND QUERIES. 
 
 Now (10)=(6W^4-^,_!|P=3-|5.^,^ 
 
 $5,70a 
 
 3600 
 19 $1,007, . . -, 
 
 ^^1224 ^^^^M~ ^ ti'ansposiDg and 
 
 [uniting terms* 
 19 ?6,042 1. ;, • . 
 '^'' 1221^^^1124 ^ ^e^^ci^g ^o a 
 
 [CD. 
 or 19/^=16,042 by omitting the CD's, 
 1%=$318. 
 100 %=:P1,800, whole investment. 
 
 ^^^^ |?li?2tz!H!^=.|10,200, latter 
 
 [investment (from (3).) 
 
 f^^. 110,200.00 ^,^,,, , 
 
 (12.) 'g^ =112,000, latter stock. 
 
 3% of $12,000=$360, income on 
 
 Ratter, 
 t of 360=$240. [mer[ 
 
 |240+$360=$600, income on for- 
 $600+1360=1960, whole income. 
 
 Id. 
 
 135. She wp with her fist Parse "up.* 
 Ans, C//9 is a verb, irregular, intransi 
 tive, active, indicative, past, third, singular, 
 to agree with "She" as its subject, Rule 13, 
 
NOTES AND QUERIES. 86 
 
 Harvey, See Harvey's Gr., old ed., p. 194, 
 rem. 7. Id. 
 
 136. " Pass we, then, to the next circle." 
 ' — Dante's Inferno. 
 
 Parse Pass and we. 
 
 Aas. Pass is a verb, regular, intransitive, 
 active, imperative, present, first, plural, to 
 agree with its subject "we," Kule 13, Har- 
 vey. Authority, Butler's P. and C. Gr., p. 
 94. We is a pronoun, personal, simple, 
 first, plural, common, Eule — A pronoun 
 agrees with antecedent in gender, number 
 and person, nominative, subject of the finite 
 verb "Pass," Rule 1, Harvey. Id. 
 
 137. They had no suspicion of his being 
 John. 
 
 • thad I suspicion | ^f being-John. 
 
 his 
 
 His is a pronoun,personal5simple, singu- 
 lar, masculine, third, to agree with its ante- 
 <iedent "John," Rule 14, Rigdon, possessive, 
 and limits "being," Rule 8, Rigdon. Joh7i 
 IS a noun, proper, singular, masculine, third, 
 predicate nominative after "being/ 'Rule 6, 
 Rigdon. Id. 
 
 138. Where are the Red mountains ? 
 
86 NOTES AND QUERIES. 
 
 Ans. In the south-western part of iN'evada, 
 
 Id. 
 
 139. In the German language how is 
 the diphthong "eu" pronounced? 
 
 Ans. Like the English "oi" in oil. 
 
 Id. 
 
 140. Correct, give reasons : (a) A group 
 of children were strolling homeward. 
 
 Corrected: A group of children was 
 strolling homeward. 
 
 Because (1) "A" is used with nouns in 
 the singular number only; (2) the speaker 
 thinks of the collection as one body. 
 
 (b) I differ with you. If difference in 
 opinion is meant, the expression is correct, 
 otherwise use frojn. ^ See Kerl's Common 
 School Grammar, p. 291 ; also, Webster's 
 Unabridged Dictionary ou the word "differ.'* 
 
 Id, 
 
 141. Burt owed in two accounts |487 ; 
 neither was to draw interest until after 
 due — one standing a year and the other 
 two years. He paid both in one year and 
 five months, finding the true discount of the 
 second, at 6^, exactly equal to the interest 
 of the first. What difference of time would 
 the common rule have made ? 
 
 Solution : Assume $1 to be the amount 
 of each account. 
 
NOTES AND QUERIES. 87 
 
 Int. on fl for 5mon. at 6,^=4ct.= 
 
 40 
 
 Int. on fl for 7mon. @6f^=|ct.= 
 
 rS 035 
 |1.000+$.035=|1.035, amt. on |i 
 [for 7mon. @6fo. 
 
 Present worth on $1 for 
 
 1-^^A [7mon. @6^. 
 
 Since the Int. of one account=the true 
 discount of other: 
 
 Th-|li„t.=JtrueDise. 
 40 207 
 
 40 J . $280 . . A I. 
 40 ^^'-^2-07 ^'' ^''*- 
 
 Then f 207 ^ , , 
 
 -2(j7=2d Aeet. 
 
 $207 |280_|4_8T , ., 
 
 207 207"" 207 
 487_ 
 207 — ^^°'* 
 
 207~^ * 
 
 207 
 2o-7=$207, 2d. 
 
88 NOTES AND QUERIES. 
 
 280 
 2Q7=$280, Ist. 
 
 !N"ow, equating by the common rule gives : | 
 
 $280X1=1280. ' 
 
 $207X2=414 
 487 )694(1.42505yr.= 
 
 r"lyr. 5mo. 3da. 
 
 I lyr. 5mo. 
 
 L 3da. difference. 
 
 Id. 
 
 142. A borrows a sum of money at 6 
 per cent., payable semi-annually, and lends 
 it at 12 per cent., payable quarterly, and 
 clears $2,450.85 a year. How much money 
 does he borrow ? 
 
 Solution. Suppose he borrows $1 ; at 
 the end of 6mo. it amounts to $1.0609, of 
 which he pays Set. for interest, leaving 
 $1.0309, which, in the next 6mo. will amount 
 to $1.09368181, from which, on paying 3ct. 
 int., he will have remaining^ $1.06368181, 
 thus clearing in the year $.06368181 on each 
 one dollar ($1) borrowed; .-.he will have to 
 borrow as many dollars as $.06368181 is 
 contained in $2,450.85, which is 38485.87 
 times, or $38,485.87. M. E. Eagleton. 
 
 143. A flag-staff is 120ft. high and 2Jin. 
 in circumference. How many feet of twine 
 will it require to wind spirally around it 
 
NOTES AND QUERIES. 89 
 
 from the bottom to the top, passing once 
 jiround it in every 3ft.? 
 
 Solution, The length of the line repre" 
 sents the hypotenuse of a triangle, of which 
 120ft. is the perpendicular, and 40 times 
 5}in. the base. 
 
 14400+69|=14469.4444+ ; . 
 
 v'l4469.4444+=120.28+ft. or 120ft. 3J 
 [in., nearly, length of twine. 
 L. P. EosE. 
 
 144. Who were the * 'Round-heads," and 
 why so called ? 
 
 Answer, "Round-head," in English his- 
 tory, is a nick-name given, in the reign of 
 Charles I, to the Puritans, or Parliament- 
 ary party, who were accustomed to wear 
 their hair cut close to the head. They 
 were so called in opposition to the Cava- 
 liers, or Royalists, who wore their hair in 
 long ringlets. The term was soon extended 
 in its application so as to include all the 
 adherents of the Parliament, whether Puri- 
 tans or not. (See Webster's Unabridged.) 
 
 J. E. McMuLLEN. 
 
 145. A stone was dropped into a well. 
 It was observed that after being dropped it 
 was ten seconds before the sound reached 
 the ear. What is the depth of the well ? 
 
90 NOTES AND QUERIES. 
 
 Solution. To find the distance a body 
 falls in a given time, multiply 16ft. by the 
 square of the number of seconds. — Steele. 
 
 16ft.X102=l,600ft.=depth of well. 
 
 H. C. Rogers. 
 
 146. If the length of the hypotenuse of 
 a right-angled triangle be given, and the 
 sum of the base and the altitude ; how, ac- 
 cording to arithmetic, not by algebra (as it 
 is easy enough by algebra), can the lengths 
 of the base and the altitude respectively be 
 found? 
 
 Answer. J (S=ti/2H2— S2)=thelegsof any '• 
 right-angled triangle. S=sum of the base 
 and altitude and ll=hypotenuse. 
 
 A. 11. Kennedy. 
 
 147. A blank paper book containing 48 
 sheets is sold for $3.50, and another con- 
 taining 78 sheets of the same size for $4.75; 
 the binding costs the same in both, and 
 paper was of same quality. What was the 
 price of the binding? 
 
 Solution. Let ic=cost of binding. 
 
 rp, $3.50- ^ $4.75— x; 
 Then ^^-=-^^ 
 
 Clearincf of fractions, 
 
 $273.00-78x=$228.00-48a: ; 
 
NOTES AND QUERIES. 91 
 
 Transposing", 
 
 30x=$45.00; 
 Dividing by coefficient of x, 
 
 x=|1.50, cost ot binding. 
 
 J. A. S., Leesville, O. 
 
 148. A man has a square yard contain- 
 ing ^ig- of an acre ; he makes a gravel walk 
 around it which occupies ^f of the whole 
 yard. What is the width of the walk? 
 
 Solution. i^A=^ of 160rd.==16rd. 
 i/16=4rd.=length of one side of square. 
 ■If of 16rd=8.75rd. 16.— 3.75=12.25rd. 
 =?area of the square within the walk. 
 
 Let x=8ide of square within the walk. 
 
 Then :r2=12.25. 
 
 .• .a::=3.5, by ext. root. 
 
 4 — 3.5=.5rd.=twice the width of walk. 
 
 .5-2=-.25rd.=4i-ft. Ans. 
 
 Mignonette. 
 
 149. What is the side of a cube that 
 contains as many cubic inches as there are 
 square inches in one of its faces ? 
 
 Solution. Let a:=side of the cube. 
 Then :ii?=7?. 
 
 Or x=l, by dividing both sides by x^. 
 Ans. 1 inch. Id, 
 
 150. Why do the inhabitants of a 
 mountainous country always prize liberty 
 so highly ? 
 
92 NOTES AND QUERIES. 
 
 Answer. A mountainous region pos- 
 sesses a magnificence and sublimity of 
 scenery nowhere else to be found ; and 
 magnificence and sublimity know no mean 
 submission. The natural strongholds defy 
 all power, and the very air is spirit-elating. 
 The rushing streams, the deep and awful 
 chasms, the towering forests, ^' Heaven's 
 concentrated lightning, thunder and 
 storms," the savage rocks and precipices, 
 are all proclaimers of liberty. 
 
 M. A. Gruber. 
 
 151. "Why do we hear sounds so much 
 plainer when the air is damp? 
 
 Answer, Sound travels in water at > 
 velocity of 4,700 feet per second, and in air 
 1,090 feet per second. Hence the more 
 moisture the air contains the better me- 
 dium of sound it is. Id. 
 
 152. Why is the rainbow always seen 
 bent in the shape of a bow? 
 
 Answer. The deviations of the incident 
 and emergent colored rays of the rain-drop 
 of the primary bow are between 42° 2' and 
 40° 17'. The axis of the bow is a straight 
 line passing from the sun through the ob- 
 server's place to the opposite point of the 
 sky. Hence all the emergent rays meeting 
 the eye of the observer must be equally 
 
NOTES AND QUERIES. 93 
 
 inclined to the axis of the bow, which causes 
 the bow to be circular. 
 
 Id. 
 153. What were the names of the "Seven 
 "Wise Men of Greece," and when did they 
 live? 
 
 Answer, Tbe Seven Wise Men of G-reece 
 are supposed to have lived in the fifth cen- 
 tury B. C. Their names were Pittacus, 
 Bias, Solon, Thales, Chilon, Cleobulus, and 
 Periander. Certain strangers from Miletus 
 agreed to buy whatever should be in the 
 nets of some fishermen without seeing it. 
 When the nets were drawn in they were 
 found to contain a golden tripod, which 
 Helen, as she sailed from Troy, is supposed 
 to have thrown there. A dispute arose 
 between the fishermen and the strangers as 
 to whom it belonged, and as they could not 
 agree, they took it to the temple of Apollo, 
 and consulted the priestess as to what should 
 be done with it. She said it must be given 
 to the wisest man of Greece, and it was 
 accordingly sent to Thales, who declared 
 that Bias was wiser than he, and sent it to 
 him. Bias sent it to another, and so on 
 until it had passed through the hands of all 
 the men, afterwards distinguished by the 
 " Seven Wise Men," each one claiming that 
 the others were wiser than he was. It was 
 
94 NOTES AND QUERIES. 
 
 finally sent to the temple of Apollo, where, 
 according to some writers, it still remains, 
 to teach the lesson that the wisest are the 
 most distrustful of their wisdom. 
 
 T. A. PUGH. 
 
 154. I have a three-cornered lot which 
 contains an acre ; each of the three sides is 
 equal. What is the length of one side? 
 If possible, by arithmetic. 
 
 Solution, This admits of two very nice 
 solutions. First, by the rules of Arith. 
 Art. 532 Milne or 397 Ray's Higher. The 
 area of an equilateral whose sides are 
 one yard each, is .0000894654+ of an acre. 
 We may therefore form the proportion 
 thus: t/ [0000894654+ : 1 :: 1yd. : (?), and 
 solving this we get 105.723+yd. for one 
 side. Second : Since the area of an equi- 
 lateral= the (base)^XJv^. 75, because by Art, 
 529 Milne, 420 White or 396 Ray, the alti- 
 tude=the baseXV.75; therefore, the (base)^ 
 =thearea-^Ji^.75 or thebase^-^^area^Ji/.TSy 
 An acre=160 sq. rds.; therefore, we have: 
 
 i/ 160 
 
 vr;7-^=19.22241327+rds. or 105.72327+ 
 
 yds. Mont. F. Vale. 
 
 155. A borrows a sum of money at 4 
 per annum, and pays the interest at the en( 
 
NOTES AND QUERIES. 95 
 
 of the year. He lends it out at the rate of 
 5% per annum, and receives the interest 
 half-yearly. By this means he gains f 100 
 a year. How much does he borrow? 
 
 Solution, Assume $1 as the amount. At 
 the end of the year at 5%, will have drawn 
 f.050625 interest. Deducting |.04, the 
 amount of interest he pays, from $.050625 
 we have a remainder ot |. 01 0625, w^hich is 
 the gain on fl. In order to gain $100 he 
 must borrow as man^y dollars as $.010625 is 
 contained times in $100, which is 9,411. 76-r 
 times. .*. he borrows $9,411.76+. 
 
 John Schurr, Muncie, Ind. 
 
 156. What must gold eell for that an 
 investment in 5-20's at 120, may yield 8 per 
 cent, interest on the income ? 
 
 Solution. Let 100fo=par value of bonds. 
 
 (1) 6%=income on bonds paid in gold. 
 
 (2) 120%=current value of bonds. 
 
 (1) 6% of 100 per cent. =^6 per cent, in- 
 come paid in gold. 
 
 (2) 6%=? per cent, of 120 per cent, cap- 
 ital invested. 
 
 (3) l%=^^of 120/.. 
 
 (4) 6%=6Xyi-g-=YfQ=5 per cent, rate of 
 income capital. 
 
 6 per cent, of 100 per cent, (par value of 
 bonds)=5 per cent, of 120 per cent. (amt. 
 invested). Then gold must be at a pre- 
 
9G NOTES AND QUERIES. 
 
 mium, so as to increase the rate of income 
 on investment from 5 per cent, to 8 per 
 cent., which is 3 per cent. 
 
 (1) 3 per cent, of 120 per cent.=3.6 per 
 cent. 
 
 (2) 3.6 per cent.=? per cent, of 6 per 
 cent, (income in gold). 
 
 (3) 1 per cent.=i of 6 per cent. 
 
 (4) 3.6 per cent.=3.6Xi=%6=.60=60 per 
 cent. 
 
 Therefore, gold must be at a premium of 
 60 per cent., or be quoted as 160. 
 
 C. H. Allen, Winchester, Ind. 
 
 157. It is required to enclose a piece of 
 ground with a fence 10 rails high, 2 pan- 
 els to the rod, and to have just as many 
 acres enclosed as there are rails in the fence. 
 What will be the size of the enclosure, and 
 what number of rails will it require ? 
 
 Solution. Reduce one acre to the lowest 
 given denomination, divide it by the given 
 number, and multiply the quotient by 4, for 
 the side of the square. In this problem the 
 lowest denomination is twentieths of a rod 
 and the given number is one of these twen- 
 tieths; for there are to be as many acres in 
 the field as there are twentieths of a rod in 
 the circumscribing lines. 160X20=3,200 
 twentieths of a rod in one acre and a strip 
 -^ wide and ^f^^ long will equal one acre. 
 
NOTES AND QUERIES. 97 
 
 Then 3,200X4=12,800 rods, or 40 miles, side 
 of square and 12,800X20=256,000 rails, or 
 twentieths of a rod in one side, 4 times 
 which=l, 024,000, number of rails and acres 
 in the field. Wimmer, 
 
 Coshocton, O. 
 
 158. A farmer buys a flock of sheep at 
 the rate of |35 for every 5 sheep ; he af- 
 terwards loses 9, and sells the remainder at 
 $80 for every 11, and the sum for which he 
 sells the flock is $120 more than that which 
 he gave for it. How many sheep were 
 there ? 
 
 Solution. Let x='^o. of sheep in flock 
 when bought; x — 9=Ko. of sheep in flock 
 when sold. 
 
 If he bought 5 sheep for $35, he bought 
 one sheep for i of $35, or $7 ; and if he 
 sold 11 sheep for $80, he sold one sheep for 
 
 „ of $80, or $f^. 
 
 Then 1x=Q,o^t of flock. 
 
 And ff (x— 9)=3eiling price of flock. 
 
 f^ (x-9)-7x=120 
 
 80X-720 • ,^^ 
 —^ —lx=120 
 
 80x-720-77x=l,320 
 3x=2,040 
 a:=680=]Sro. sheep in flock. 
 
 Xax. 
 159. A twelve-inch ball is in the corner 
 
98 
 
 NOTES AND QUERIES. 
 
 where walls and floor are at right angles. 
 What must be the diameter of another ball 
 which can touch that ball while both touch 
 the same floor and the same walls ? 
 
 Solution. Suppose a cube circumscribed 
 about the two balls ; the line A B will rep- 
 resent the diagonal of the cube and also 
 the axis of the two balls. 
 
 The dimensions of a cube circumscribed 
 about a sphere one ft. in diameter are one 
 ft., and the diagonal of the cube=i/l^X3= 
 T/3=1.732ft. 
 
 A M=D B (each being the distance from 
 the sphere to the nearest dihedral angle of 
 the cube). 
 
NOTES AND QUERIES. 99 
 
 A B=1.732ft. diagonal of the cube. 
 M D=lft. diameter of sphere. 
 AM+D B=1.732ft.— lft=.732ft. 
 AM-Jof.732ft.= 366ft. 
 A M=D B. . • . D B=.366ft. 
 M B=ltt.+.366ft-1.366ft. 
 IS^ow form the proportion. 
 MB :MD ::DB: Dx, 
 
 or 1.366:1::. 366: Da;. 
 
 Dx=lX^6_2679 
 1.366 
 
 .-. Dx= 2679ft. or 3.2148in. 
 
 The diameter of the smaller ball is 3.2148 
 inches, W. L. Damkoehler, 
 
 Sturgeon Bay, Wis. 
 
 160. Suppose a man standing on the 
 hank of a river desires to know the distance 
 to any visible object on the other side of 
 the stream, how can he find the distance, 
 providing he has nothing but a ten-foot 
 pole ? 
 
 Solution. This problem has been solved, 
 but while the solution is true as regards the 
 principles of geometry, its practical value 
 would be rendered useless in nine cases out 
 of ten for lack of suitable back-ground 
 •upon which to make the measurement. 
 The following, I think, has no published 
 equal for ease, simplicity and practical 
 utility : 
 
100 
 
 NOTES AND QUERIES. 
 
 Let B be the object on opposite side of 
 the stream, and A or H the point to meas- 
 ure from. Produce BA. to C, any distance, 
 and construct AC FH, any form of the 
 parallelogram, and produce BH to P; also 
 CF. Now measure AH, also HF and FP. 
 Then by similar triangles : 
 
 FP : FH : : AH : AB, or 
 
 FP : AH : : FH : AB. 
 
 Now, if BH is to be found, measure HP^ 
 and FP : PH : : AH : HB. 
 
 J. K. Ellis, 
 
 Emma, Ky. 
 
 161. What will it cost to fence a field in 
 
NOTES AND QUERIES. 
 
 101 
 
 the form of an equilateral triangle, altitude 
 20 rods, at 80 cents per rod ? 
 
 Solution. Before solving the above, per- 
 mit me to explain how and why I solve it 
 thus : 
 
 Let A B C be an equilateral triangle, with 
 C D the altitude. Let 
 the side, as A C be 1, 
 then A D^J. Finding 
 D C by the rule, we 
 have : V \'^—^'^=V\^— 
 1/1= 1/775=. .866=D C, 
 ^or altitude, when the 
 side is 1. 
 
 Dividing the side by this, we have 1-^ 
 .866=1. 15473=the ratio of the side to the 
 altitude. Now, .*. this is as constant as the 
 ratio of the circumference of a circle to its 
 diameter; .*. when the side is given, 
 To lind the altitude: 
 Rule L Divide the side by 1.15473, the 
 result will be the altitude. 
 
 To find the side when the altitude is 
 ^iven : 
 
 Rule II. Multiply the altitude by 
 1.15473, the result will be the side. 
 
 The above problem is of the latter class. 
 . • . 20X1.15473=23.0946=one side. 
 23.0946X3=69.2838=perimeter. 
 69.2838X$.80=$55.427=cost. Id. 
 
102 NOTES AND QUERIES. 
 
 162. A floor contains 300 square feet, 
 and its diagonal is 25 feet. Find the size 
 of the floor. 
 
 Solution. The sum of the squares of any 
 two numbers increased by twice their pro- 
 duct, will equal the square of their sum; 
 but the sum of the squares of any two 
 numbers diminished by twice their product 
 will equal the square of their dift'erence;. 
 and .'. 25^ equals the sum of their squares^ 
 and 300 is their product. 
 
 . •.i/252+300X2=v^ 625+600=i^T;225=35= 
 their sum, but 
 
 V' 625=600= v'SS-S^their difference. 
 We now have the sum and difference of 
 two numbers to find the numbers, 
 .-. 35+5=40 : 40-2=20=one side. 
 35-5=30 : 30-2=l5=one end. Id. 
 
 163. If a heavy sphere, whose diameter 
 is 4 inches, be dropped into a conical glass 
 full of water, whose diameter is 5 inches 
 and altitude 6 inches, how much water will 
 run over ? 
 
 Solution. Let the diagram represent the 
 ball and glass. 
 
 (B E=6 inches. 
 E C=J A C=2.5 inches. 
 Given \ O F=0 s^=0 M=2 inc hes. 
 
 B C=i/E B^+E C2=v/62+2.52=- 
 
 [6.5 inches. 
 
 I 
 
NOTES AND QUERIES. 
 
 103 
 
 The triangles B E 
 C and B F O are 
 similar, for they are 
 right angled at E 
 and F and have the 
 common angle B. 
 Hence E C : O F : : 
 B C :B 0,or2.5 : 2 
 : : 6.5 : 5.2. E 0= 
 B E-B 0=6-5.2= 
 .8. EM=OM+OE 
 =2+.8=2.8 inches= 
 altitude of the seg- 
 ment M F s' s. E s' 
 
 =i/0 s' 2_o E^=t/2^ -.S'=VSM. Now the 
 quantity of water that runs over is equal to 
 the volume of the segment s s' F M. The 
 geometrical formula for finding the solidity 
 of the segment s s' F M may be thus ex- 
 pressed. Multiply the square of the height 
 plus three times the square of the radius of the 
 base, by the height, and this product by .5236. 
 Height=2.8 in., the square of which=7.84. 
 
 Radius of ba8e=T/3.36, the square of which 
 is 8.36, and three times the 8quare=10.08. 
 . • . (7.84+10.08)X2.8X.5286=26. 2721536 cu. 
 in. Ans. H. A. Withee. 
 
 164. What is the area of a large circle 
 enclosing three equal small circles, the 
 
104 
 
 NOTES AND QUERIES. 
 
 curvilineal space between the small circles 
 bein^ one acre ? 
 
 Solution. Let R represent the radius of 
 
 these equal cir- 
 cles ; then it is 
 obvious that 
 each side of this 
 A is equal to 2R. 
 The triangle is 
 therefore equi- 
 lateral, and it en- 
 closes the given 
 area, and three 
 equal sectors. 
 As the angle 
 J of two right angles, 
 are, together, equal to a 
 semi-circle. But the area of a semi-circle, 
 whose radius is K, is expressed 
 
 ttR' 
 
 by -o- ; and the area of the whole triangle 
 
 must be -^ — ("160; but the area of the 
 
 A is also equal to R multiplied bythe per- 
 pendicular altitude, which is Ri/3. 
 
 Therefore RV3=— r- fl60. 
 
 of each sector is 
 the three sectors 
 
 Or, 
 R2 = 
 
 2 
 
 R2(2i/3-^)=320. 
 320 320 
 
 2v/ 3-3.1416 0.3225 
 
 992.248. 
 
NOTES AND QUERIES. 
 
 105 
 
 Hence, E=31.48+rods. 
 To find distance from A, B or C to cen- 
 ter of large circle, use the following rule: 
 Divide the product of the three sides by 4 
 times the area of the A. 
 
 (31.48X31.48X31.48)-(v/ (31.48X2)2+31.482 
 X31.48X4=36 2+=distance from A, B or C 
 to center of large circle. 
 
 Diameter of large circle=(31 48+36. 2)X 
 
 [2=67.68 rods. 
 Area of large circle=67.68'^X.7854= 
 [3597.59— sq. rds. or 22A. 77.59-sq. rds. 
 Jerome Erdley, Middleburgh, Pa. 
 165. How many feet of lumber in a piece 
 of timber 8 inches square at one end and 
 12 inches at the other, and 45 feet long. 
 Solution. 
 
 Rule. — To find the 
 contents of a frustrum, 
 take the sum of the two 
 bases and the square 
 root of their product, 
 and multiply the sum 
 by one-third of the alti- 
 tude of the frustum. 
 
 Area = 8^ + 1 22+/'¥2 
 Xi22-I44sq. in. X(iX 
 45)^64+144+96- 144 X 
 15=31| cubic ft. 
 
 „ Id. 
 
 12 In, 
 
 8 In. 
 
106 NOTES AND (,)L'h:KIES. 
 
 106. A man agreed tu work a year for 
 a horse and $180. At the end of tlie fifth 
 month, by equitable settlement, he received 
 the horse and $5. What was the value of 
 horse? 
 
 Solution. He received the value of the 
 horse and |180 for 12mo. labor. 
 
 He also received the value of the horse 
 and |5, for 5mo. labor. 
 
 .-. for 12 — 5 or 7mo. labor he received 
 $180— $5=^$175, and for 1 mo. labor he re- 
 ceived 4- of |175=$25. 
 
 .-. the value of the horse =(5X$25)— $5 
 =$120. Id. 
 
 167. A farmer lost 10 ^^o of his wheat 
 crop, but if it had copt him $50 more, his 
 loss would have been 20%. What was the 
 cost of his crop ? 
 
 Solution. 10 %=Yo ^^ cost=lo8s. 
 T^—A= Abseiling price. 
 20%=|- ot 2d cost=^second loss. 
 I — ^=|- of second cost=second S. P.= 
 [100—10 or 90 '/o of first cost. 
 i=90</o. 
 
 f=ll2J% of 1st co8t=2d cost. 
 il2J%— 100%-12if'^=-$50. 
 100%=$400. Cost. Id. 
 
 168. A person purchased 100 animals 
 for $100 — sheep at $3J apiece, calves at $1J- 
 
NOTES AND QUERIES. 107 
 
 at $i. How many animals of 
 
 and pigs 
 
 each kind did he buy ? 
 Solution. 
 
 (21 
 
 I 3 
 
 
 
 
 a 
 1 
 
 b 
 
 
 c 
 5 
 
 d 
 
 e 
 
 1 
 
 15 
 
 
 
 5 
 
 
 
 * :o 
 
 8 
 
 
 
 42 
 
 42 
 
 1 
 
 3 
 
 1 
 
 3 
 
 5 
 
 2 
 
 25 
 
 28 
 
 53 
 
 Sheep. 
 Calves. 
 Pigs. 
 
 6 5 30 70100 
 
 Id, 
 
 169. Four men buy a grindstone 32 
 inches in diameter, with a square hole 
 whose diagonal is four inches. If each 
 grinds off his share in turn, how much of 
 the semi-diameter will each one take? 
 
 Solutio7i. 32in.2X. 7851=804.2496 area of. 
 stone 
 
 4in.2X. 7854=12 5664 area of hole. 
 804 2496~12.5664=791.6832sq. in. 
 Each of the four men received i of 791. 
 6832=197.8208sq. inches. 
 
 32in. — (t/197.9208 X3+12. 5664-. 7854)-^ 
 2=2.108in of 8emi-diameter=what the first 
 man grinds oft*. 
 
 27.784— (i/197.9208 X2 +12.5664-^.7854) 
 -^2=2.4905in. of semi-diameter=what the 
 second man grinds off. 
 
 22.803— (1/197.9208X1+12.5664^.7854)-^ 
 2=3.2165in. of semi-diameter=what the 
 third man grinds off. 
 
108 NOTES AND QUERIES. 
 
 16.37— 4-^2=6.185in. of semi-diameter= 
 what the fourth man grinds off. 
 
 Id. I 
 
 170. Where is the "Run of Cutch," 
 and for what is it noted ? 
 
 Answer. Cutch is a principality under 
 the Presidency of Bombay, and stretches 
 along the Gulf of Cutch and the Indian 
 Ocean, between Guzerat and Snide. It has 
 an area of 15,364sq. mi., and a population 
 of more than half a million. It is naturally 
 divided into Cutch Proper and the Runn of 
 Cutch. Cutch Proper is a belt along the 
 sea shore, and contains all the inhabitants ; 
 while the Runn of Cutch, containing about: 
 8,6003q. mi., is an amphibious desert of 
 hard ground in the dry season and a shal- j 
 low lake in the wet season, caused by the ^ 
 heavy rains and pent-up tides of the south- 
 west monsoon. 
 
 It is supposed to have been originally an 
 inlet of the ocean, and to have had its level 
 raised by volcanic action. The periodical 
 disappearance of the water leaves one con- 
 tinuous crust of salt. It contains a few 
 fertile spots, but animal life is almost un- 
 known. RowENA MiLLSAP, Yolo, Cal. 
 
 171. Pluralize dentifrice, cousin-ger- 
 main, miasma, dormouse. 
 
KOTES AND QUERIES. 
 
 109 
 
 Ans. Dentifrices, cousins-germain, mi- 
 asmata, dormice. Id. 
 
 172. My lot contains 135sq. rd., and the 
 length is to the breadth as 5 to 3 ; what is 
 the width of a road extending from one 
 corner half round the lot, and occupying 
 one-fourth of the ground? 
 
 B 
 
 C 
 
 /^S^r^^- 
 
 A 
 
 Prd 
 
 D 
 
 Solution. 3X5=15. 
 
 135sq. rd.-^15=9rd., length of short side. 
 185sq. rd.-^9— 15rd., long side. 
 9rd.+15rd.=24rd., length of road. 
 135sq. rd.^4=33f3q. rd., area of road. 
 33}sq. rd.-24=1.40|rd., width of road. 
 
 Id. 
 
 173. Who was Pestalozzi ? Please give 
 a sketch of his life, and also of Froebel. 
 
 Answer. Johann Heinrich Pestalozzi was 
 born at Zurich, Jan. 12, 1745. His family 
 belonged to the middle class gentry. He 
 was destined for the Christian min- 
 istry, but turned from it to study law, but 
 
110 NOTES AND QUERIES. 
 
 he (lid not remain long in this pursuit. The 
 perusal of Rosseau's Emile and the unsat- 
 isfactory political condition of Europe dis- 
 gusted him with the artificial life of cities, 
 and he accordingly removed to the country 
 and engaged in farming. Ke shortly after 
 married the daughter of a wealthy merchant. 
 Hismind,however, still dwelt upon the un- 
 happy condition of the masses of the people, 
 and he devoted himself, during his leisure 
 moments, to the consideration of means 
 best suited to promote their elevation. He 
 was convinced that a good education might, 
 in a measure, eradicate the evils by which 
 society was infected.; he accordingly con- 
 verted his own home into an orphan asylum, 
 and endeavored, by a judicious blending of 
 intellectual, industrial and moral training, 
 to afford a specimen of sound education, 
 and one so contrived as to be practicable as 
 a national scheme. Meanwhile, the pursuit 
 of his enterprises involved him in bank- 
 ruptcy, and the failure of his plans brought 
 contempt and opposition upon him. 
 
 His consolation, however, lay in the fact 
 that he had saved upwards of one hundred 
 children from neglect and degradation, and 
 the several volumes which he issued on 
 education contained results of his expe- 
 rience. Many subsequent attempts were 
 
NOTES AND QUERIES. Ill 
 
 made to found schools by him, with varied 
 success, but always with pecuniary embar- 
 rassment. His writings, meanwhile, in- 
 creased in number and importance. The 
 great idea which lay at the basis of his 
 methods of intellectual instruction was 
 that nothing should be treated of except in 
 the concrete. In arithmetic he began with 
 the concrete and proceeded to the abstract. 
 Objects themselves became, in his hands, 
 subjects of lessons, tending to the develop- 
 ment of the observing and reasoning facul- 
 ties — not lessons about objects. Into the 
 teaching of writing, he for the first time 
 introduced graduation. His special atten- 
 tion was directed to the moral and religious 
 training of a child, as distinct from the 
 mere instruction. Almost all of Pestalozzi's 
 methods are now substantially used by the 
 instructors of elementary teachers in JSTor- 
 mal schools of Europe, and to no man, 
 perhaps, has primary teaching been so 
 largely indebted. He died, in 1827, over- 
 whelmed by disappointments and embar- 
 rassments. 
 
 Froebel was a prominent German educa- 
 tor, born in 1782, and died in 1852. He 
 was the founder of the Kindergarten system. 
 Cassie a. Millsap, Yolo, Cal. 
 
 174. Give the uses of the barometer. 
 
112 NOTES AND QUERIES. 
 
 Ans. The uses of the barometer may be 
 classified into physical, hypsometrical and 
 meteorological. It is essentially used in phys- 
 ical researches where the mechanical, chem- 
 ical, optical and acoustical properties of air 
 or other gases are dependent upon the press- 
 ure of the atmosphere. It is used in ascer- 
 taining the heights of mountains, and it m 
 also used as a weather glass. Id. 
 
 175. Who was the author of the Mis- 
 souri Compromise ? 
 
 Ans. Henry Clay. Id. 
 
 176. How long is the Erie canal? 
 
 Ans. The Erie canal is 363 miles long, 
 and cost $9,000,000. Id. 
 
 177. What flower is the national em- 
 blem of France ? 
 
 Ans. The fleur-de-lis is the national em- 
 blem of France ? Id. 
 
 178. What is the national hymn of the 
 United States, and by whom composed ? 
 
 Ans. The national hymn of the United 
 States is "'America," composed by Rev. S. 
 F. Smith. Id. 
 
 179. A fathers age is 40yrs. His son's 
 age is 4yrs. In how many years will their 
 ages be as 1 to 2 ? 
 
 Solution. The ratio of their ages are now 
 as 1 to 10. Hence to be as 1 to 2 the rela- 
 tion would be i^^^ or 5 or ^ less than 40 
 
NOTES AND QUERIES. US 
 
 would be the number of years intervening, 
 which is 32yrs. 
 
 L. B. Hayward, Bingham, O. 
 
 180. A man bought a horse for $156 on 
 8mo. time. He sold him immediately for 
 $180. Whatdid he make, money worth 4J %? 
 
 Solution. 1180— 156=^24. |24X.04jXf= 
 72c. $24+ .72=124. 72=gain. Ans. 
 
 Id. 
 
 181. The base and perpendicular of a 
 right-angle triangle are 30 and 40 respect- 
 ively. Find the hypotenuse. 
 
 Solution. V 302+402=50=Hypotenuse. 
 
 Id. 
 
 182. The product of two numbers mul- 
 tiplied by their sum is 84, and the sum of 
 their squares multiplied by the square of 
 theirproductis 3,600. What are the numbers? 
 
 Solution. Let x represent one number 
 and y the other. From the condition of the 
 problem we have the following equations : 
 
 xy (x+2/)=84 (1). 
 
 xy(x2+2/')-3,600 (2). 
 
 84 
 Dividing (1) by xy, ^+2/=^ (3)- 
 
 Dividing (2) by ^c'yK ^-^f=-^ (4)- 
 
114 NOTES AND QUERIES. 
 
 Squaring (3) and transposing. a?-\-y'^^ 
 
 r(84y 
 
 -2xy (5), 
 
 3 600 (84V 
 Equating (4) and (6). -^=-^2 — 
 
 l2xy. (6). 
 Multiplying (6) by a^/. 3,600= (84)2-2i'i/^ 
 
 Transposing terms in (7). 2a;y=(84)^— 
 
 [3,600 (8). 
 Performing operations indicated in (8), and 
 [dividing by 2. rcy-1,728 (9). 
 Extracting cube root of (9). xy—l^ (10). 
 From (1). x+y=1 (11). 
 Squaring (1). x2+2xi/+2/2=49 (12). 
 Multiplying (10) by 4. 4x]/-=48 (13). 
 Subtracting (13) from (12) a?—2xy-\-y^=^l 
 
 Extracting square root of (14). x—y=±l 
 
 [(15). 
 Adding and subtracting (11) and (15) and 
 [dividing by 2 in each case. a?=4 and 2/=3. 
 W. T. Maddox, 
 
 New Boston, Mo. 
 
 183. Explain the arithmetical rule for 
 finding the area of a triangle when the 
 three sides are given. 
 
NOTES AND QUERIES. 
 B 
 
 115 
 
 Solution. Let ABC represent a triangle 
 whose sides a, b and c are given. Draw 
 B P_i to A C; this will be the altitude of 
 the trianfi^le. From Plane Trig., we have 
 
 JBP 
 
 sin. A= . • . B P=e sin. A. Denoting 
 
 the area of the A by M, and applying the 
 rule for finding the area of a A, the base 
 and altitude being given, we have M=J b c 
 sin. A. But sin. A=2 sin. } A cos. J A; 
 whence M^b c sin. J A. cos. J A. 
 
 But sin 
 
 . i^=4^ 
 
 — b) (is— c) 
 
 be 
 
 and cos |^A= 
 
 \Ml3^.:^- in which S=a+b+c. 
 '- ^ be 
 
 Therefore, substituting in the equation, 
 
 M==bc sin. J A cos. J A, the values of sin. 
 
 J A and cos. J A, 
 
 We have M=bc JEIIE^ / (^«-b) (^s- 
 ^ be ^ be 
 
 ••• ^=Vl^ (i-s-a) (^s-b) (^s-c); 
 
 -c) 
 
116 NOTES AND QUERIES, 
 
 Hence, we may write the following rule : 
 Find half the sum of the three sides, and 
 from it subtract the three sides severally. 
 Find the continued product of the half 
 sum and the three remainders, and extract 
 its square root ; the result will be the area 
 required. Id. 
 
 184. What conveniences of life came 
 into use during the Middle Ages ? 
 
 Answer. They were few and simple. 
 Chimneys came into use about the fifteenth 
 century. Glass windows were introduced 
 into England in the twelfth century. Time 
 was measured by means of sun daily. 
 
 M. M. Joshua, 
 Moreauville, La. 
 
 185. What is the largest public building 
 in use? Give dimensions. 
 
 Ansxoer. Coliseum, in the City of Rome ; 
 capacity, 87,000 people. Id. 
 
 186. Diagram: 
 
 The boast ot heraldry, the pomp of power, 
 And all that beauty, all that wealth e'er 
 gave, 
 
 Await alike the inevitable hour : 
 
 The paths of glory lead but to the grave. 
 
NOTES AND QUERIES. 
 
 117 
 
 boast 
 
 Thomas Shaffer, 
 Bard well, Ky. 
 
 187. In what ways are nouns pluralized ? 
 Answer. Generally by adding s to the 
 
 singular, but where the last sound will not 
 unite with 5, add es. There are many other 
 rules which can be found in any good gram- 
 mar. See Harvey's Grammar, revised, 
 page 34. G. S. Metcalf, 
 
 Carpenteria, Cal. 
 
 188. Diagram and parse italicized words : 
 Pupils should always be required to study 
 what is most profitable, all things con- 
 sidered. 
 
118 NOTES AND QUERIES. 
 
 Answer, 
 
 To study 
 
 should be required 
 
 [that] 
 
 always 
 
 [which is most profitable 
 
 [of] pupils 
 
 f things 
 
 I 
 
 all 
 
 1 
 
 [being]consider'd J 
 
 " To study " is a verb, reg., (study, 
 studied, studied) trans., act. pres. inf., cons, 
 of a noun, sub. of "should be required." 
 
 " What "=that which. ' ' That " is a pro- 
 nominal adj. used as a noun, objective case, 
 object of to study; or it may be parsed 
 as an adj., limiting [subject] understood. 
 
 ^* Which" is a rel. pron., ante, "that," 
 with which it agrees in gen., per., and num- 
 ber, nom. case, sub. of is. 
 
 Things is a noun, com., class, 3d, plu., 
 nom., absolute with the part. " being con- 
 sidered." 
 
 " All things [being] considered " is an 
 independent element, relating logically to 
 *' most profitable," but not grammatically. 
 It is equivalent to " when all things are con- 
 
NOTES AND QUERIES. 119 
 
 sidered," which clause would evidently 
 modiiy most profitable. 
 
 This sentence belongs to a class of sen- 
 tences which gome grammarians pronounce 
 faulty, such as, He was asked a question, 
 He was refused admittance, etc., in which 
 a nbun in the objective case seems to fol- 
 low a passive verb. 
 
 The best way out of the difficulty is to 
 supply a preposition when possible ; if one 
 can not be found, arrange the sentence 
 logically as I have done, and analyze. 
 
 Id. 
 
 189. Diagram: He thought to learn to 
 study to be to learn to think. 
 Answer. 
 
 He I thought 
 
 to be — to learn| 
 
 Sub Element. | tO think. 
 
 to learn | 
 
 . I to study 
 
 "To learn" is subject of the infinitive 
 "to be." 
 
 "To study" is object of "to learn." 
 " To be " is the basis of the abridged sen- 
 tence, and should be parsed as an infinitive, 
 with con. of noun, object of "thought." 
 
120 
 
 NOTES AND QUERIES. 
 
 Second " to learn " is attributive object. 
 '' To think" is object of ** to learn." This 
 is a bungling sentence. Id. 
 
 190. Diagram: The pure attar of roses 
 is worth twenty or thirty dollars an ounce. 
 
 Answer. Second reading. The pure 
 attar of roses is worth [to the amount of] 
 thirty dollars [by] the ounce. 
 Attar I is — worth 
 
 The 
 pure 
 
 
 , 1 
 
 .of — roses. 
 
 
 1 by — ounce 
 
 1 the 
 
 
 
 to — amount 
 
 
 
 /twenty 
 
 
 
 
 
 0^ / thirty dollars. 
 
 Or, 
 
 Attar 
 
 I is — worth— dollars 
 
 The 1 
 
 1 of — roses ) 
 
 by ounce 
 
 thirty 
 
 pure 
 
 (or) 
 
 
 
 twenty. 
 
 Remark : In the last diagram " worth 
 dollars" is an attribute of the second class, 
 "worth" being a prep. Id. 
 
 191. Diagram: The setting of a great 
 hope is like the setting of the sun. 
 
 Answer. 
 
 setting I is — like — setting 
 
 The 
 
 the 
 
 of — hope 
 
 a 
 
 great 
 
 of — sun 
 the I 
 
NOTES AND QUERIES. 
 
 121 
 
 Rem. * 'Like-setting" is an attribute of 
 the second class, "like" being a prep. Id, 
 
 192. Diagram and parse italicized words: 
 
 (a.) I warrant him a warrior tried. 
 
 (6.) By becoming a Quaker, Penn came 
 near being disowned by his father. 
 
 Answer: 
 
 (a.) I I warrant 
 
 [to be]- 
 
 -warrior 
 
 him 
 
 tried 
 
 Remark. This 8entence="I warrant that 
 he is a warrior that has been tried," in the 
 complete form. In this "he" is sub. of " is;" 
 in the abridged form him is sub. of "to 
 be" understand. "Warrior" is the attribu- 
 tive object, objective case. 
 
 Tried is a participle, with cons, of an adj. 
 and limits "warrior." 
 
 {b) Penn 
 
 came 
 
 [to] being disowned 
 
 near 
 
 by-becoming^ 
 
 Quaker. 
 
 Becoming is a verb, irreg. (become, be- 
 came, become), trans., act., pres., participial 
 
122 
 
 NOTES AND QUERIES. 
 
 mode, con. of a nouu, object of prep. *'by." 
 Quaker is a noun, prop., 3d, sing., mas., 
 obj. case, object of the verb '' becoming." 
 
 Near is an adv. modifying "being dis- 
 owned." "Being disowned" is a part., pass. , 
 con. of nouu, obj. of prep, "to." Id, 
 
 193. Diagram and parse italicized words: 
 The army surgeon made him limbs ; 
 
 Said he, the 're only pegs, 
 But theWe as wooden members quite 
 
 As represent my legs. 
 Answer. 
 
 
 surgeon made 
 
 The 
 
 (and) 
 
 limbs 
 
 army 
 
 [for] him 
 
 He I said 
 
 they 
 
 (b 
 theyf are — [s 
 
 are — pegs 
 
 only 
 
 ut) 
 uch] 
 
 as — members 
 
 wooden 
 
 as 
 
 represent 
 
 quite 
 
 legs 
 
 my 
 They is a per. pron., sub. of "are." 
 
NOTES AND QUERIES. 123 
 
 " A.re" is an intrans. copulative verb, join- 
 ing the attribute "such" to the sub. "they.'* 
 
 "As" is a prep, showiog the relation be- 
 tween '*are" and "members." 
 
 Second "as" is a rel. prpn., ante, "such," 
 with which it agrees in gen., per. and num., 
 nom. case, sub. of "represent." 
 
 iN'oTE. — *'Such" limits something under- 
 stood, as "things;" hence, "as" is plural 
 number. 
 
 ''Qaite" is an adv. limiting "represent.' 
 It is equivalent to "wholly" or "completely." 
 
 Id. 
 
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 Pi PHq 741 123 5 
 
OnTLINE^OF^RITHim. 
 
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