› en › download › KeytoRaysNewHig… · P REFACE. IN the p repa ra tio n of the follo w i ng page s, aim the chie f ha s bee n to se rve the teache r. Acco rdi ngly, ag th ro

ECLECTI0 EDUCATIONAL SERIES .

KEY

RAY ’S N'

EW HIGHER

R I T H M E T I C

NEW YORK CINCINNATI CHICAGO

AMERICAN BOOK< COMPANY

\ \ C

FROMTh t GIFTOF

CHARLES HiRS '

m’

fiTnw amM » 11, up ?

SPECIAL Noncs.

Ra y’s Arithmetics h a ve recently been th or

ough ly revised ,a nd issued a s

Ra y ’s New Arithmetics .

Ra y’s New Prim a ry Arithmetic, $0 15

Ra y’s New Intellectua l Arithmetic,Ra y ’s New Pra ctica l Arithmetic,

Ra y ’s Tw o-Book Series .

Ra y’s New Elementa ry ArithmeticRa y’s New Pra ctica l Arithmetic ,

For High Schools a nd Colleges .

Ra y’s New High er Arith m etic ,

Th e m a ny ch a nges in bus iness tra nsa ctions ,a s w ell a s th e a dva nce in meth ods of instruc

tion , h a ve m a de such revis ion necessa ry . Th e

New Arithmetics a re sold for th e s a me low pricesa s th e old editions , notw ith sta nding th e pa per,printing,

binding, a nd genera l a ppea ra nce a refa r superior. S pecia l term s

, for th e exch a nge of

th e new series for th e old , ca n be h a d by a pplication to th e publ ish ers .

COPYRIGHT,1881

,

BY VAN ANTWERP, BRAGG Co.

KEY NEW HIGHER A“.

P REFACE .

IN the preparation of the follow ing pages,the chief

aim h a s been to serve the teacher . Accordingly,through

a great part of the w ork w e have merely indicated theoperations required

,a nd have thus saved the space w hich

w ould be demanded for the full statements of analysis,

useful chiefly in recita tion drill .The help afforded by a Key being of very little value

under the fundamental rules,this w ork does not begin

w ith the first exercises of the NEW HIGHER ARITHMETIC.

In most cases,after a short study of a n arithmetical

problem,the operations themselves a s indicated properly

by signs,w ill suggest the reasoning on w hich the solu

tion is based . In the solutions here given this h a s beenkept in view

,especially w here the text-book ' h a s pre

sented a formula . To the teacher a nd to the class,alike

,

it w ill be advantageous to have the blackboard w orkw ritten in accordance w ith this plan ; a nd

,on this a o

count,early in the Arithmetic

,the subject of Arith

metical Signs h a s been formally presented .

To read the arithmetical syntax understandingly,a nd

to w rite it w ith facility in recording solutions, a re a c

quirements w orth fa r more than they cost. What is(31)

iv PREFACE.

here remarked h a s special reference to the written work.

Oral explanation,reaching even to particulars

,is not to

be set aside ; on the contrary, the judicious teacher w illstill require the minute details of a n analysis

,especially

in the examples designated for such exercise in the

Arithmetic .

There a re a few instances in w hich the brevity men

tioned above h a s not been observed . The experiencedteacher know s

,that

,in some cases

,the operations may

be even very few in number a nd very simple in kind ,While the reasons for them a re not correspondinglyobvious ; in such instances

,a s also w here the chief

difficulty of the problem is in the complexity of the

Operation,w e have aimed to give a n extended solution .

Among the articles under w hich this h a s been thoughtadvisable

,w e may mention Compound Subtraction ,

Pro

portion,Commission

,Stock Investments

,Alligation

,the

applications of Evolution,a nd the added Miscellany.

CINCINNATI,Janua ry, 1881.

KEY,

TO

RAY’

S NEW H IGHER

ARITHMETIC .

MULTIPLICATION.— BILLS AND ACCOUNTS .

Art. 65 .

Sr. LOUIs, Ma rch 1, 1879.

CHESTER SNYDER,

Bought of THOMAS GLENN.

4 lb. tea , 40 ct. a lh . ,

21 butter, 21at a lb .,

58 bacon, 13 ct.

16 la rd , 9 ct.

4 “ra isins,

“ m ct.

9 doz . eggs, 16 ch a dos . ,

Received pa yment,

THOMAS GLENN.

6 KEY TO RAY’S NEW

Art. 66.

ALLEGHENY, April 1, 1880.

JAMES WILSON dc 00

In Acc’t with ALLEGHENY COAL 00.

DE.

Ma rch To 500tons coa l , a ton,

CB .

By 14 bbl. flour, a bbl

6123lb . suga r, 8 ct. a lb . ,

ca sh on a ec’t,

Ba lance due Allegh eny Coa l Co. ,

CONTRACTIONS IN MULTIPLICATION.

CASE IV .

481063

63721

3367441

10102323

30306969

30653815423

IHGHER ARITHMETIC.

66917

849612

803004

6424032

5621028

56853486204

,ARITHMETICAL SIGNS.

Art. 86.

21-2— 3X 7 +49— 1X 1 1X 4 — 2

18 —4 +9

1X 4 X 6 8 +3

— 3— 28 X 0: _3 = 1,Ans.

— 4 x 9 12

— 7 5 3

(27 3— 1

91+ 13X 7 — 45 — 3=Then,3X 0+ 1X 9= 9

CONTRACTIONS INMULTIPLICATIONAND DIVISION.

Art. 88.

6)10724OO

Ans. 178733i, Ans.

( 5 ) 102735 536712

273162 729981

308205 4830408

2773845 43473672

16643070 391263048

28063298070 391789562472

8

Art. 89. CASE II.

(14)4514020000

451402

1504522866,Ans.

( 53)6302240000000

630224000

700178864000

4.

2800715456000,Ans.

Art. 90. CASE III.

225) 300521761

4

9 |OO)Quot. 1335652

,Ans.

4 61, Rem.

696224

( 3) 1406250) 22500712361

8 8

11250000) 180005698888

8 8

Quot. 16000,Ans.

45591104 8 8

712361 Rem. Quot. 29794 , Ans.

199040 4 4 3 Rem.

281257000000

281257

281256718743

5

156253732635

( 2 ) 43750) 15103372644 4

175000) 6041349056

4

Quot. 34521,Ans.

4 4 43514,Rem.

208331) 6207124803 3

62500 )18621374404 4

IEHQLHEEB AARUTUZALEIYCZ 9

( 5 ) 29163) 742851692

3 3

8750 2228555076

8 8

Quot. 254692 Ans .

608 8 3 25h Rem.

MISCELLANEOUS ExEECISES .

(1) $6 X 153 $918 ; $918 —s 54 $17, Ans.

217 X 35 25 7620, Ans.

4879 41 119,Ans.

103X 103 10609,Ans.

53815 375 143 a nd 190 mm ; 144 times 37554000

,this is 185 more than 53815

,a nd since the

next low er product is 190 less than 53815,54000 is the

nearest. Ans. 54000.

$2675 25 $107 $107 X 19 $2033, Ans.

$210 15 $14 , gain on each ; $75 $14

$89, Ans .

391mi . — 139 mi . 252 mi . ; 11hr.— 4 hr. 7

hr. 252 mi . 7 36 miles,Ans.

235 yd . 12 yd . 223 yd . $5 X 235$7 X 223 $1561 $1561 $386, Ans.

135 bl . 83 bl . 52 bl . $2 X 52 $104 , Ans.

X $68 X $73X 17:$1241; $375 $816 $1191; $1241 $1191 $50gain

,Ans.

Also $1191 $118 $1309 $1309 17 $77, Ans.

$240 $24 $216, w hole cost ; $216 3 $72,the cost of 1piece ; $72 $4 : 18 . Ans. 18 yards.

13men — 8 men 5 men ; 1m a n ca n do it in13 times 15 days

,w hich is 195 days ; a nd 5 men

,in k

of 195 days, or 39 days, Ans.


14 men 7 men= 21men ; 1m a n ca n do it in14 times 24 days

,that is

,336 days ; a nd 21men

,in

211

of 336 days,or 16 days

,Ans .

For 1day,the provisions w ill support 30 times

45 m en,or 1350 men ; for 50 days, one fiftieth of 1350

m en 27 men 45 m en 27 men 18 m en,Ans .

618 X 3 $54 $85 654 6139, 501331value$139 $41 $98 1sheep cost one fourteenth of $98 , or$7, Ans .

A sheep a nd a hog cost $13 ; therefore , he w illbuy a s many of each a s $13 is contained times in $1482 ;1482 13 114

,Ans .

1horse a nd 2 oxen cost $84 therefore,he bought

a s many horses a s $84 is contained times in $1260;1260 84 15 tw ice 15 equals 30.

Ans. 15 horses a nd 30oxen.

One seventh of 1050 ct. 150 ct. 150

of the 25-cent pieces. 1050 ct.— 150 ct.= 900 ct. ; 1ofeach of the others w ould make 10 ct. 5 ct. 3 ct.

18 ct. 900 18 50,the number of each of the others.

Ans. Of 25-cent pieces,6 of the others

,50each .

$6300 gain per acre $210

$165, the cost ; $5600 loss per acre ; $165$40 $125, sold for per acre .

Ans. $165, cost ; $125, sold for.

PROPERTIES OF NUMBERS .

LEAST COMMON MULTIPLE.

15 = 5005,Ans.

2 3

2 X 5 X 2 X 3= 60,Ans.

E GHER ARITHMETIC'. 11

( 4 ) ( 5.

45 63 70 15 20 25 30

21 70 m5 ,7 70 5 25 15

70x 3x 3= 630 5 3

2 x 2 x 2 x 5 x 5 x 3

600,Ans.

16 9

16 x 9= 2160,Ans.

( 73>

295—

58 1537 14065 5141

97) 2 33 485 5141

See Rem,p . 70

,N. H. A. 5 53

53x 5 x 2 x 97 x 29 x 3= 4472670, Ans.

( 51)85988 235803 490546

83mm 42994 235803 245273

7) 518 2841

35039

2 2841

2841x 2 x 37 x 7 x 83x 2 244291908,Ans.

By Art. 100,find the common divisors 37 a nd 67 ;

then2479

593953x 37 146261

,Ans .


Find G. C . D. 31; a nd again 83. Then,

3403x 31 105493,Ans.

Art. 106. CANCELLATION.

10 3

(L) hence 30 cow s,

5

10X 24cents

,Ans.

5

2 9

hence 9 bales5 2

25 X 55X 55 X 955 5 9

COMMON FRACTIONS —REDUCTION.

Art. 131. CASE I.

( L) ( 30 ( 4)15m 3. 15mg ,

z,

.

5

14m; 43. 2mm H. 2mm

HIGHER ARITHMETI0.

( 9

46>

(1L)

13384123?H’

Art. 132. CASE II.

63,

"

far X “8“ 65" 6X ‘1‘ 63,

99 63

63 21z :3 ;

381“ X 3 6

25 )

6783 17 399

IQ, X W ,A718 .

199 X 367 6788, Ans.

6783 21 323 ;

HX W ,Ans.

Art. 133,CASE III.

127 x l 7+ 11= 2170;

13

14 KEY TO RAY’S IYEW

5 x 211+ 207= 1262 ;

Art. 134. CASE IV.

Ans.

Lil bu. 137 bu . 4 34} bu.,Ans.

13301hr . 785 hr . 60 13115 hr .

,Ans.

1-3-9-5 1295 37 35,Ans.

800 9 88g, Ans.

kHz 1162 11 Ans.

194311 4260 13 327-155 ,Ans.

15780 31 50931” Ans.

Art. 135. CASE V.

3

( zogxgxfl

5 X 5 > < 7 5 X 7 > < 54

3 2

-1—5- X 8 2

2,Ans

5 X15 > < 5 1

M GHER ARITIIMETIC. 15

2 5

fiX éxfiX éé —E: n

4 336 “

2 2

l X é X fix zsxlfi — g Anssx 5 x7x 4x 5

2

8

5 3

4 2 3 7

Art. 138. CASE VI.

(1 ( 2 )3X 5: 15 ) ( 5 X 6 30

2 X 5 : 20 ) ( 4 X 6: 24

2 X 3: 18 X 4 X 5 20

3X 5 = 30Den. x 5 x 6: 120Den.

Ans. 66! 66, 66 Ans. 1

2260" T

2266.

(3

x 7 > < 8= 112

x 3x 8 72

Ans. H2,

X 2 X 5 X 8 = 400

X 2 X 5 X 6= 420

X 5 X 6 X 8 2 480Den.

Ans . M , m , m , m.


3

2

( 3)

Art. 139. CASE VII.

(1) The L . C . M . of 3,4,a nd

X 4 3x 3 5 x 2A

The L . C . M . of 2,5,10

,and 4 is 20

X 10 3X 4 9 x 2

x10 5 x 433°

Ans . H, 3 , H, H.

2 x 4

3x 4

5 x 2

6 x 2

Ans. 191, H

40

105

24

60Den.

20 240

20 280

7 189

20 420Den.

3x 5

4 x 5

4 x 4

5 x 4

9 x 2

10x 2

Ans. fi , fi ,H,

https://www.forgottenbooks.com/join

18 KEY TO RAY ’S NEW

( 4)

a nd 6,is 84

84

712 x 23= 276

614 x 29= 406

871

38141 Ans.

10 ; L . C . M . of 2,3,4,

a nd 5,is 60 °

60

230

320

4 15

5 12

771

60H

10 11g , Ans.

11129““ 8 °

L . C . M . of

24,a nd 88

,is 264

264

( 5 1’s

_ §L ; 2% L. C. M. of

5,7,3,a nd 3

,is 105

105

5 63

7 15 x 2 30

3 35 x 2 70

3 35 x 7= 245

408

133 319035

333, Ans°

km»

00

L. 0. M. of 8 a nd 15,is 120

120

15 8 X8 15 X

139

120

( 9 ) L . C . M . of 8,12

,18

,

24,a nd 27, is 216

216

12 18 X 11= 19818 12 x24 9 X 23= 207

27 8 X 20= 160

958

333z 412393

H GIIER ARITIIMEH C. 19

55

é -8 —55 ;

2s 11 31 341

9X12X¢ 81

3 3

55 5915,

SUBTRACTION OF FRACTIONS .

Art. 141.

(M t— fi = fi = 5

5 fi — s =m — sflz =m.Ans

( 30 12’s 0f 5 — zfié ; fi —

xfie= %

— ffi%

Ans .

( 40 11. fi — 55 — 55 .

Am

( 50H— fi — ss — m — s le z s ,Ans

—115—12525—

T185 — T

1665:An3°

{5 —13235—

235 a Ans.

( 8015 — 5 4 5 — 55 — 55 — 5

Hg, Ans.

Ira —rh

Ans .

—s¢.

L . C . M. of denominators is 729

729

3243x 2 486

27 27 X 26 702

81 9 X 80 720

243 3X 242 726

728


1; 15 Ans .

l — gz g; 18 — 5g Ans .

( 150 21= 3fl i 0f 315 =W ; W — H= 3&Q — 3&

-H,Ans .

(169 1of 1§ =1;

( 170 1811of 1} of I

ii

1

( 200 fr 0f § = 15n 3

n ; H— n= 3 .Ans

( 219 — s + 13 — 3 =m +

m — m +m ~ — m + % — w =m = 6f s6433, Ans .

( 22 > 5 — 3 =m — m+ 1

115o a

9n= fi xing

= s3 s ou33 33=33 m

MULTIPLICATION OF FRACTIONS .

Art. 142.

( 19 1-3 12=W = 9fi Ans.

xlgg

l—“ 8i, Ans.)

11—

1

4

HIGHER ARITHMETIC'. 21

7

w )”x? u > § x$ T

4

5

26,Ans.

ifx 35,Ans.

4

Ans. Ans.

3 3 3

1—2 A 13.1? A(12 )

gflx

’l

g23m “8 )

lgn1911, ”3

6 2

izn fl ,Ans .

5 5

10 3


( 18 ) 2; 232— 81

2

- 405, Ans.

16

( 19 ) 2;33

0

éggx1 > < w Ans.

4 X 1

2 X 2 ? 8 X 10= 2 x g x 2

3 x 5 s 3

( 22 > z

( 23 % X

( 24 ) sxsxs1X §§

( 25

( 26 ) Ans

X hence Ans.

«511X 2

15 11

11-5 318-2 hence 318-2days, A718 .

( 29 ) gx 141z 1

51 9g; hence 95! cents, Ans.

3 x 2 x 115 : l hence $75, Ans.

6 X 8

( 3103 Ans

REMARK.— The examples under this a rticle

,a s a lso those of the

following, a re compa ra tively ea sy,a nd hence the solutions a re briefly

indica ted here. The pupils m a y work with like brevity, when themultiplica tions or divisions a re mere pa rts of a solution requiringva rious processes. But where, as in these a rticles

,the multiplica tion

or division itself 1'

s the solution to be exhibited,the pupil should have

frequent exercise in writing,with pa rticul a rity a nd a ccura cy

,the

very sta tements a nd the steps which follow . Thus,in the next

a rticle, ta ke the 6th a nd the 15th,for illustra tions.

X 4

352 7k, Ans .

1&1,

4216 ,Ans.

2 232 2 232 943 Ans

—33195 ,Ans.

HIGHER ARITHMETICI. 23

DIVISION OF FRACTIONS.

Art. 143.

3. Z_?1_4

7

2X1 2

13A”

2

( 15 )

i Of -gOf Of lzi= -2

Much m a y be ga ined by pra ctice in the proper use of signs a ndpunctua tion m a rks. Accura cy in the use of a rithmetica l syntax

,is

worth a ll the time itm ay cost the cla ss to m a ke the a cquirement.

( L)

125 X Ans.

( 2 )

gX «1 136,Ans.

( 5311X 1

98 = 23}, Ans. 3x i

rQ= 26§5 Ans.

( 8

fl XHz gig—fi ,

11718 . % ,Ans ,

iX Ans.

( 12 )

263Ans .

( 2

H»x 1=225 ,Ans.

Ans.

“r x A}: Ans.

( 13 )

24 KE Y TO RAY’S NEW

Ba rn um—We have before mentioned tha tusage is notwell settledin rega rd to the efl

'

ect of the signs and X ,occurring in succession.

In fra ctions, we seem to ha ve a facility for avoiding ambiguity,which

usage h a s not a fforded in whole numbers. The word,

“of ” is the

convenient connective by which the divisor is shown to be a result,

a nd not a mere factor of a product. Thus,in fr—z—tof i, the opera

tor,recognizing the compound fra ction, is not likely to use ; a s the

divisor ; it seems most na tura l to use 135 .

Where the divisor is a compound fraction,the opera tor m a y save

time a nd space by m a king all the inversions a t once,a s we illustra te in

examples 24th and 25th following.

See solution on page 23.

(160e. of f s 0fr°r=

r§1 5 11. of fi c es ; Ta x 10—H, Ans.

(170i s of t -1 h ? Ans .

3

i of é x u’a’u—

z‘oau ; w

as0f 1£= 1

11¥w fu'ux lfis“

Ans.

( 2a ) 4 01‘

got"

{Pl- of

lg, Ans

( 210s of ; 0f 1’r=

ri' r i s of f of731 x

1—3-5-= Tfir, Ans.

Ans . See Remark.

( 23

( 24

( 25) AM



( 4 ) G. C. D. Of numera tors : 125 ; L . C. M. of denominators= 48 .

Ans.14255 23—3.

G. C . D. of numera tors 13 ; L . C. M . of denominators 15 . Ans. H.

261H= l fii ; 652H= 13271M; G. C . D. of nu

mera tors 193 ; L . C. M . of denominators 42.

Ans.1492l = 4§~gu

-8 G. C. D.

of numerators: 40; L . C . M . of denominators 9.

Ans .$99 43.

The quantities a re 1311bu .,1332-6 bu .

,”lili bu.

The greatest common measure of these is the capacityof sack required . G. C . D. of numerators= 25 ; L . C .

M . of denominators 8 ; hence 285 , or 35, is th e numberof bushels in each sack

,Ans.

Dividing each quantity by 3} w e have, also, 44 , 153,670

,Ans .

The sides a re Lil ft ,ft

,131 ft. The great

est number of feet w hich w ill exactly measure thesesides

,is the G. C. D. of these numbers. G. C. D. of

numerators: 77, a nd L . C. M. of denominators 12 ;hence H it. is the necessa ry length of the equal rails .

The number in one round is (Li-l 3-3-6 131)59 ; hence , in 6 rounds

,354 rails

,Ans.

REMARK.— }Z being the necessa ry rea ch

, H ft. 15, ft., or, 6 ft.

11in. is the adawed length of rails.

LEAST COMMONMULTIPLE OF FRACTIONS.

Art. 146.

(1) L . C . M . of numerators 60 G. C. D. of denominators 1. Ans. 60.

HIGHER ARITHMETIC. 27

The numbers a re 241,

a nd 221; L . C . M. of

numerators 945 ; G. C . D. of denominators 2.

Numbers 181,fig, 1

57 ,

5911,225; L . C . M . of nu

mera tors 350; G. C . D. of denominators 1.

Their times a re 1331, 11951, 11351, in hours,re

spectively . Any common multiple of these w ill expressthe time to elapse till they a re all together

,a nd the lea st

common multiple w ill express the time to elapse tillthey first come together. By the Rule this is found

,

L. C . M. of numerators G . C . D. of denominators100 l hence 100 hr.

,Ans.

P a om scuous Exm wrss s.

Ans.

lfi z fi ; 1 —8 sum z fi ,difference

325 ; a nd 45 contains 9, five times, Ans.

( 3 of {Lg— g— i= M fl

zflzfis‘ Wu

i + fi t5,Ans.

( 4

115 ;

551+ 1

lq= 16

,Ans.

A180,100 W x

7=W = 26¥tom Ans.

fi xes xm x s :es x s > < fe x ss

1xgx3xq z fi ,Ans.


s of (s—z— s> = s of s of s of s:

Ans.

( 8-) sx s of (1—s) x s of (3— s)

> < i (4181 7 -

7239" Ans .

(4 Ans.

—s—u_

—e&1 are — 1

(11 of t 0f % =fi ; sxsliltgs- z fi ,

Ans.

23 31,r fl ; fl 3, the subtrahend ; 4}

remainder g ,the minuend ; this is 3of 13

0,

g of some number ; hence the number of H.

1929s:Ans.

Qof James ’s money 3 of g, or 296 ,of Charles ’s

money ; hence of Charles ’s $33 2 431-3of it, w hichca n be true only if 33of it $33 ; 12

16of it= 1

11 of $33,

or 83 °

12,-3, or the w hole = $60; then 2 of $60 336,

James s, Ans.

They w ill m eet in a s many hours a s the numberof miles they both tra vel in 1hour is contained times in109 miles 109 hr . (73 89 109 x 3

43: hr .

,Ans.

Also,633X 71} mi . ii

lfifil

i mi . 515-9 mi . , A ; a nd

X 8} mi . 23263331 Ans .

REMARK.— The sign X is often read “

times.

g product g of 3, of fl, 3, one factor the

oth er fa ctor g—z 5 fi 2a ,Ans.


lt z fifi x Ans .

293g1573' s:Ans.

157+A=H5 w hole — H 2

155 ; of it $60;

the w hole 2 IBE of $60 $216, Ans.

(19 ) 3of first= § of second ; 3of first=3of gof

second ; w hole of first a} of of second ; hence gof

second w hole,or 1

31of it 51 a nd second

,therefore

,

157Of 51

,24

,Ans.

Also, 51 24 27 Ans.

RULE .— To find wha t p a rt one number is of

a nother,ta ke the number which is the p a rt for a numer

a tor,a nd tha t of which it is the p a rt for a denomina tor.

Ans

( 21 Ans .

9 23 7 - 3453:Ans.

( 220Ax e} fi xs = ss = 7ss Ans .

Also. ixsxs xe xsx rxT — e s

r— ffn.Ans

( 23) Factor the denominators by finding their com

mon divisors . (Art. 100,Rem .)

1 8 5 5 14 2 2,2 (H f ?)

Ans.

(24 ) 3g— H= 29m younger son’

s share ; difference ofshares

, H—sas 1

10

" of estate estate,therefore

, 8525

X 10 $5250, Ans.

33 2753: 517-3, the sum ; 393— 2

71,133, the dzfierence ; product 3

81x 1

59. 8k} ; a nd

x11227

the quotient.

30 .KEY TO RAY’S’NEW

The ship is w orth 3of the cargo ; both g of

cargo ; 15, of both 15, xgz fi ,

Ans .

Factor denominators by finding G. C. D. of eachtwo ; then ,

sis. dams, 1;

1000,Ans.

Divisors,1 2 4 8 16 32

127 254 508 1016 2032 4064

Sum 8128,a perfect number ; 8128 X 7 133

8 12 8 X 7 > < 84 4 3

381,Ans

MW 115351811 3

6019Ans.

4 X 13g X

979- XHX 5 X 1

35 X 3X 1

’1‘ X 6 1

5232

13313,Ans.

x gx g= 20,Ans.

3 (full) (full) g m ,Ans .

3&mi . X X mi . ; 31335

31535 X 1

41 1033, the number of hours

,Ans.

( 340cazsx 17s — 4t) = e l x1—31 x a s

1h . ; lb . lh .

,Ans.

Whole — 12r T

arathe first remainder ; Tar of thefirst amount $65 amount next in hand losing 3of

this,he had remaining 3, w hich w a s 1

91 of th e first

amount $3141; w ith $10 m ore

,this w ould have been

equal to the first amount ; that is, 12; of it 31—2—5 th e

w hole ; hence $2 of it w a s a nd the Whole w a s 33of $121, $33, Ans.


DEOIMAL FRACTIONS .

REDUCTION OF DECIMALS.

Art. 158.

(1) 2 5625 : 5> a n= 5>ns15616,Ans.

.15234375 $ 33,233,90n mm512640301650 9 34231056

wa s. mess an Ans

.017510 a rfi gv Th . Ans. 19131.

_L 100;

100 5 x 100Ans. 16m .

0284, flag 7a, gs. Ans .

6 66666; W Ans.

003125 a ssess. man.

5_i _0; 10 9 X 10Ans. 11115 .

390625 sh am e. OWmam 5)fi %% mm H,

Ans.

$341

4;-l 944s

10000a ns a ss w e.


24s

.66§— m=3, Ans.

( 15 ) 25 = 5)ess= 1, Ans

.75 535 g, Ans.

Art. 159.

( 2 3).75

,Ans. 125

,Ans. .05

,Ans.

46875,Ans. .005625

,Ans.

Ans. .8 Ans. .495 Ans . .078125

05078125,Ans. Ans. .0009765625

Ans.

.1875 ; w hich , annexed to the 42,gives

3 .25 this, annexed to .015,gives .01525

,Ans .

g: .75 ; this, annexed to gives

.0375 ; w hich , annexed to 75119,gives

75119 0375 Ans .


HIGHER ARITHMETIC’. 35

18 .0001

51728 0001006

6466 Ans.

12

Ans.

008

Ans . .3496032

93007 01

4 .4fi k 251 4000.

841500 9300701 17200504000841500 46503505 Ans.

841500 18601402

93500 233447

Ans.

.059

.059

531

295

903481

.059

31329

17405

000205379 ,

Ans . .0008875

738.

2952

1476

738

888552,

4 2. 02}600.

8 4 1200

168 150

Ans. Ans.

( 22 ) 26000000.

000026

156

52

Ans. 676.


( 24 ).03275

315145

189087

1701783

206419975 ,

Art. 163.

7 5 19

11157

248877 0

2765 3

1382 7

1936

6

Ans.

405081

3645729

135027

9 5 14 18

4320710

314159

12962130

43207 1

17282 8

4321

217 0

38 9

9Ans.

5 4 5 4 5 9 7 s 2

8140737134

16281474 3

48844423

5698516

73266 6

40704

325 6

407

33

4

Ans. 21813475 2


4 4 9 1 8 5 21

12 581944

7026100

140522 0

351305

56209

703

632 Ans. 116005 22 8

3

Ans.

7 09 4 00

.02538 1

004907

Ans. .000124 0Ans. 0

Ans .

DIVISION OF DECIMALS .

Art. 165.

1 ( 2

.01575,Ans. .0084

,Ans.

( 3) ( 4 )

W .056,Ans.

4 09

0

6

3300 128000

,Ans.

953-8336 .0001

,Ans. 1200

,Ans.

gggggo

égg-

“

o

s

oOO 1664,Ans.

12 1.5625,Ans.

8 -2 5 6

4 4 4 4 2 4 0

.0424444

33997 5

1699 9

340O

34 0

34

3

Ans.


(11 (12 )1 59412_ Ans.

J-H-g—Q Ans .

(13) (14 ).15625

,Ans . W 112

,A718

(16 )= 2307 Ans s l i

zzgfl = 73088 Ans .

2 4 8 8 2 5 x4 x 4 x4 15 8 9 2 85 15 6 2 5 x4 x4 x4 330

Ans.

REMARK—Such examples a s the 7th a nd the 17th shonild‘

be per

formed both w ays,th a t the pupil m a y compa re the two processes - f

the one a n easy reduction,a nd the other a tedious division.

W Ans.

i° m Ans.

100° 1000 1000x 1000000,001

oogg

O1_ Ans .

Ans.

1000

16 2 7 5 000000 0625 Anso4 16 6 4

W m + fi 0 — m 01m v q = ,00001

40

Ans.

14142136

58578640

56568544

2010096

1414214

595882

565685

30197

28284

1913

Ans .

942477795

444826451

314159265

130667186

125663706

5003480

3141593

1861887

1570796

291091

CIRCULATING DECIMALS .

Art. 173. CASE II.

( 1)A718 .

( 3

123=m M .

( 2

=nfin Ans .

4

266: Ans.



Ans .

CIRCULATES .

Art. 175.

( 1) .2666 1535465 ( 3) 18 23673

.0074

.2592

Ans. Ans. Ans.

100563563 199 64285718 2727272

Ans.

Ans . Ans.

Ans . 76

MULTIPLICATION OF CIRCULATES.

Art. 176.

( 1)

042618189414

.066666

Ans .

.07067

.063609

.003056

Ans.2142 966060

HIGHER ARITHMETIC. 4 5

7684 3500

3073777774704

4269135798200

3842222218380

426913579820

REDUCTION OF COMPOUND NUMBERS .

2200

(1) x x 16 13876 352 sq. rd . Ans.

REMARK.—Strictly

,the steps of a solution require some expres

sions different from the mere numerica l indica tions. The solution

just given is like the first model on page 156,New Higher Arith.,a nd

is quite convenient. Taking the three a s abstra ct numbers,the sta te

ment is not a ccura te. Usage m a y a llow this for the convenience of

the opera tor, but pupils should he often drilled upon the solutions,

thus ch . X ch . sq. ch . a nd

sq. rd . x 16 sq. rd.,Ans.

( 2 ) 80 .20025 mi .,Ans .

(3) 750x bu .,Ans.

sq. Yd. X 9 X 144 :ln .

,Ans.

10240 16 640 sq . ch .

,Ans. (Art.

P .,Ans .

One carat gr.

W 21x .528 on ,Ans .


303, Ans .

c9. > ssr-= 4 x 3 x 4 x1— m 3. 4 ns

18§ 3 X 60: 1125 gr. ; 1125 gr _L l itgn

2; oz .

,Ans .

96 oz . av. x116 xH"?x 12= 87} oz. tr Ans .

Or,at length

,thus:

1oz . tr. _ 112 of m u) . av. of 16 oz. av .

T}:of 96 oz . av. Hence

,96 oz . av.

LP ,or 87} oz . tr.

,Ans .

Ans

nearly,Ans .

75 s.,Ans.

21yr. 2} X 31536000 sec. 70956000 sec

1Wk . = 168 hr. h ence 11695 , or 1271, Ans.

kl . I. X 1000 90120 l.,Ans .

2535517 degrees ; therefore .00694, or

Ans.

1sq. yd . 1296 sq. in. ; hence 3129025 , or 31, sq.

yd.

,Ans.

6§ cu. yd .230 cu . in . X 27 X

cu. in.,Ans.

mills X 1000 117140 mills,Ans.

( 22 ) ct 100 Ans .

HIGHER ARITENEHc. 47

mills .001of $1600 Ans

( 24 ) 354 mills x 1000 5375 mills,

12 lb. av. =H-gof 12 lb. tr. 14115lh .

,Ans.

6 45 x 100 645,the number of kilograms and

645 x 1000 645000g.

,Ans. (See Art.

1OZ . tr. = 112 Of 5760

,01‘ 480 gr . ;

iii-g: .00045

oz. tr . ,Ans.

sq. rd . X 640x 160 484438016 sq. rd.,

44} of 4 of -

l-l2of g lh .

,Ans.

74 oz. of of £911 cw t. cw t.

,Ans.

99 yd . —T7IG

'

O' of 99 mi . mi .

,Ans.

( 32 HA W— 2 2 3 acres acres,Ans.

16 0

2} of fi of 3X 15, cu . yd . 19-3cu . yd.

,Ans .

169 ars 100X 169 m 22:16900m 2

,Ans.

24 f3=g> < 8 x 60m.= 1200 "L, Ans.

43» pure gof 24 carats fine 204 carats, Ans.

184 carat gold is or 4gpure ; and l — fig31;alloy

,Ans .

Width by breadth X m 2

m 2.,Ans.

( 39 ) 120x x c .,Ans .

48 KEY TO RAY’S NEW( 40) 2 x 214 > < 13 sq. x mix 13 sq. a . _

(43 33) x 13, or 988 sq. ft.

,Ans .

18 cw t. 3 qr. 15 lb. 12 oz. or .94543LT. ;

x 27 .945g Ans.

( 42 ) x x times Ans .

( 43 ) x Ans.

160 sq. rd.

,85 sq. in. 6272725 sq. in . 6272725

25 250909 stones, Ans .

Area = 20X 15, or 300m 2. ; 300m

2. (4 + 9)

1112 287 m 2. X 287 Ans.

17 2 8 X 2 7

Weight esxx a s x z oxw m xw oz . 315323374 oz .

,17 2 8 X 2 7

9 T . 17 cw t. 7 lb. oz .

4 pw t. 9 gr.=105 gr. =317 ,or .218Qoz . Weight

oz . cost x w eight13 pw t. ,

and cost 81} X 13 Sum of costsAns .

REMARK.-In ca ses like this it will be a useful review exercise

to give the solutions the bill form,a s in Art. 66

,New Higher Arith.

( 48) Capacity 221X 1} x 1} X 1728 cu. in.

, 4 of w hich21x 7 x s x17 2 8in bu.

” l 5 ,“bu .

bu.= 22 bu. 4 qt. pt.

,Ans.

160 lb. a v. =H-gof 160 lb. tr. 1943lb. 194

lh.

,5 oz .

,6 pw t.

,16 gr .

,Ans.

50mi . 264000 ft.

264000 134 revolutions.

Ans . 5155 . revolutions.


50

205

KEY TO RAY ’S NEW

1409467.775,Ans .

Ans.

HIGIIE'R ARITHMETIC. 51

cu. ft. cu. in.

106 1152

1000

107 1072

Ans.

5 6 2 14816, Ans.

(13) b.

124 dr.

5“dr.,Ans.

REMARK.—The tea cher may present both of these solutions

,and

remind the pupil of the difference between a concrete addition a nd

a compound addition. (Art. 178,

52

gal .6

2

l

12

qt.

3

1

2

2

2

KEY TO RAY’S NEW

pt.

119636

ts

.11H,Ans .

054, Ans .

3

hr.

12

12

0

(19 )

min.

0

30

30

soc.

0

0

30;304,

E GHBE ARITHMETI0.

4 340

(1. sq. yd . sq. ft.

{LA. 20

9. gsq. rd . 22

fisq. ft.

113 8

4 .

Ans.

COMPOUND SUBTRACTION.

Art. 222.

rd . yd . qr. na. in.

4 2 1 12.

g 128 . 1 1 1

Ans. 3 1 0 Q, Ans.

53


( 30sq. yd . sq. ft.

16 6

24 0

22} 6

22 8

cu.it. cu. in.

4 1000

25

5 Ans

( 709 . gr.

I

1 1519155 ,Ans.

f , 15 . tr. gr.

34,lb. a v . gr.

0,Ans

bu. pk. qt. pt.3 5 1

3 3 1,Ans .

(1. hr. min . yr. mo. da .

175 21 42 1856 7 1

275 9 12 1855 9 22

265412 29 9 9,Ans

265 18 29

oz. pw t.

0 15

9 7

3 7

T . cw t. qr.

75 0 0

56 9 1

18 10 2

gal. qt. pt.31 0 1

12 1 0

18 3 0

12.

13. f3.

4 2

1 4

2 5

( 6

gr.

4

Ans.

lb.

8

23

10,Ans.

gi.

2

3

3,Ans.

m,0

38

22,Ans.

HIGHER ARITHMETIO. 55

yr . mo. da.

1822 4 1 90 0 0

1814 12 31 43 18

7 3,Ans. 46 41 Ans.

180 0 0

161 34

18 25 Ans .

( 21)sq. rd. sq. yd . sq. ft. sq. in . EXP LANATION— In

80 0 0 10 creasing the first tern.

79 30 2 30of the minuend by 1»sq.

ft.,we ta ke 30 sq . in.

16,Ans . from 46 sq . in.

,lea ving

16 sq. in. ; then having ca rried 1» to 2, we increase the second term

of the minuend by 1sq. yd.,a nd subtract 2} from 2} lea ving 0; then,

increasing 30by i, we find a unit to be the next necessa ry increa se,

a nd,proceeding then as usua l

,the a nswer is found

,simply

,16 sq. in.

EXPLANATION. In

crea se the first term of

the minuend by 1} sq.

ft.; a nd sa y,“140 sq. in.

from 181sq. in. lea ve.4\

ct. m.

9 3}99;

8 45,Ana


sq. then ca rrying 1} to next term of subtra hend,we have 2} sq.

ft. to be subtra cted ; increa sing the upper by sq. yd.,we ta ke 2}

from lea ving 1; then ca rrying to 30,a nd using a unit increa se

a bove, we find the a nswer,1sq. rd.

,1sq. ft.

,41sq. in.

rd. yd . ft. in. EXPLANATION—Let the first increase

3 O O 2 be ift. ; then 4 in. from 8 in. lea ve 4 in.;

5 1the next term of the subtra hend be

2 4comes 1} ft.; a nd increasing the upper by

4,Ans . yd.

,the rem ainder there is 0; the next

subtra hend term becomes 5} yd.; a unit of increase answers then to

complete the opera tion.

24 .mi . rd. ft. in. EXPLANATION—h t the first increa se

7 0 0 1 of the minuend be ft.; the next 9. unit,

a nd so on, a s usua l.

(25 )A. s rd. 8 d. s ft. 8 in.

EXPLANATION.

) 3

(13

q

()

y q

5

q

0The first minuendincrease is sq. ft.;

a t the secondminu

1 32,Ans end term the in

crea se is sq. yd.; a t the third minuend term the increa se is 2 sq. rd

a nd the a nswer is 2 A. 1sq. rd. 30sq. yd. 1sq. ft. 32 sq. in.

A. sq. rd. sq. yd . sq. ft. sq. in.

18 0 0 3 3

2 156 5,or



0 3 48

12

5 5 36,Ans .

8 . (1. 10° 28’ w hich215 16 24 x hence

,

75 28° 51’ Ans .

175185 14 i, Ans .

COMPOUND DIVISION.

(3)0. cu. ft. C.

664

11

128

cu .it.

83 cu . ft.

692

664

286

249

376

332

440

8 c. cu . ft,Ans.

18 da. 9 hr. 42 m. sec

1590149 3 sec. ; a nd this X16fi =26454302 sec. w hich306 da. 4 hr. 25 m . 2 sec.

nearly. Ans .

( 4sq. rd. sq. yd . sq. ft. sq. in.

10 29 5 94

301sq.yd. 19 sq. yd .

17

161} 2038

153 17

8} 33

9 17

811; sq.ft. (4 sq.ft. 168

68 153

131} 15

144

2038 sq. in. sq. in.

19 sq. yd . 4 sq. ft. 1191-9sq. in.

Ans.

IZHGIZEHR .AUR172116ETHTZ 59

5

sq. mi . A. sq. rd .

6 0 35

640

3840A. (170A.

225

1590

1575

160

2435 sq . rd . (108 sq. rd.

225

1800

221} «3sq. rd.

170A. 108} sq. rd.,Ans .

600T. 7 cw t. 86 lb. 307401216 dr.,w hich

gives 10578156 dr. reduced,brings 20T. 13 cw t. 20 lb.

14 oz . 12 dr.

,Ans.

312 ga l. 2 qt 1pt. gi. 80058 88 eighths gi.divisor £51 ; quotient gi. this,reduced

,

4 gal. 1qt. gi. , Ans .

19302 bu. bu .,nearly ; re

duced, brings 2 pk . 6 qt. 1} pt. Ans.

min . sec.

38 9 ) 152 46 2

55 16 58 26;19 Ans.

1245 cu. yd . 24 cu. ft.

1627 cu . in . 58129819

cu. in . this5142866 4 cu. in . thisreduced

,110cu . yd . 6

cu. ft. cu . in .

,

Ans .

( 7

3 5 9 sr

12 ) 3 7 0 18

2 1 16


LONGITUDE AND TIME.

Art. 226.

(1) Cincinnati being 84 ° 29’ 45” W. a nd New York74

° 24” W. ,

the difference 10° 29’ 21” 15 41min.

sec. difference in time ; hence , the required time a t themore w esterly place is 41min . sec. earlier than 6A.

M.

,or 18 min. sec. after 5 o clock A. M.

,Ans.

The problem presumes that Springfield is the farther w est. Diff. of time 58 min . 1125 sec. X 15 14° 30’

17” this diff. of long . added to Philadelphia long. 75°

9’3 W. ,

makes required long. 89° 39’20

” W.,Ans.

(3) l et. Obviously, noon a t starting ; 2d. to keep noonfor 24 hrs. the rate of travel to the w est must be 15° perhr.

,or 15 X sta t. mi .

,Ans.

Mobile 88° 2’ 28” W., Chicago 87° 35’ W. ; thelatter is farther east 27’ 28” this 15 1min . 49g

sec.,the advance of Chicago time

,Ans.

9 hr. 7 hr. 13 m in. 32155sec 1hr. 46 min.

274} sec.,by w hich Halifax time is faster ; this diff. x 15

26° 36

'the distance of Halifax, east, w hich taken

from St. Louis long. 90° 12’ 14” W. , leaves required long.

,

63° 35

' 18” W.,Ans.

( 6 ) The diff . of time 46 min. 58 sec. X 15 11° 44’

30 diff. of long. a nd this added toDetroit long.,83° 3

’

W.,

required long. 94°47

’30

” W.

,Ans.

Diff. of longitude 309° 3’ 15 20 hr . 36 min.

12 sec. It is this much la ter in Sidney than in Honolulu,or 41min. 12 sec. after 12 o ’clock A. M. Monday.

No'rE.—The first explorers reached Sidney by traveling ea st,

while Honolulu w a s rea ched by sailing west, a nd the time a t the

two pl a ces w a s fixed by reckoning in contra ry directions from

Greenwich. Hence we must take the longer are as our basis.


St. Petersburg 30° 16’ E.

,N . Orleans 90° 3’ 28

W. ; diff. of longitude 120° 19’ 28” hr . 1min.

1744; sec.,Ans .

12 hr. + 1hr.— 6 h r. 54 min . 34 sec. = 6 hr . 5

min. 26 sec. diff . of time ; this X 15 91° 21’ 30 diff . oflongitude ; a nd Buffalo being so much west of Rom ew hich is 12° 28’ ea st

,the long . of the former is 91° 21’

30” 12° 28’ 78° 53

’30

” W.

,Ans.

St. Helena 5° 42’ W.

,Sa n Francisco 122° 27’ 49

W. ; diff. of long . 116° 45’ 49 15 7 h r. 47 min. 3115

sec. hence required time is so much earlier than 6 P. M.,

that is,12 min . sec. after 10A. M.

,Ans.

4 hr. 43min . 12 sec. difference in time X 15 70°

48’ difference in longitude . Hence

,w hen the time a t the

ship is 4 hr . 43min . 12 sec. ea rlier than a t Greenw ich,the

longitude of the former is 70° 48’ W.,Ans.

ALIQUOT PARTS.

Art. 227.

hr.

of 20min .

of 4 min .

7216

71mi.

62 .KEY TO RAY’S NEW

694 A.

16 pw t. =1960z .

10gr. 11,oz.

20P . R.

1 f 10g’ °

True cost 67999344Sup. costError 810 too much .

1gal . 5 6{L 97920

2 qt. 1}gal .1875

1qt. z } .09375

1pt. .04687

Ans .

805494

80rd . 2047.625

40rd . i20rd.

10rd.

3143071796

92 oz.

3092

13914

322+032+

Ans.

4 hr. 1da.

1hr.

20mm hr.

5 min 112hr.

30sec. 110of 5 min

Ans.

per mi. 684480

20rd . 5280

10rd. 2640

618480

143077.796

8161557 80nearly

, Ans.

HIGHER ARITHMETI0. 63

1cw t

20T. 400

10lb 11,

375

1lb. 0375

81504 2375.

30min . hr.

20min . 1} hr.

6 min . kof 3012 sec.

3laof 6 min .

9709 mi .

SIMPLE PROPORTION.

Art. 231.

( 1) Ana lysis — If I ca n w alk 10} mi . or 221 mi . in 3

hr.,I ca n w alk 4 of or 4 mi . in 1hr. ; in 10 hr.

,10

times a.mi . , w hich is 35 mi . , Ans.

Or,thus ; 3,

hr . 10 hr. th e distance in 3 hr. the

distance in 10 hr. ; that is, 3 hr. 10 hr . 10} iof 10times 10} mi . , or 35 mi . , Ans .

11ft. 8 in . ( 7)140in. 98 in. 670 469 times, Ans.

8 min . 15 sec. :1hr . 3min . ( 7 )495 see. :3600see. 3mi . 31

4355801= Zlfi mi.

,Ans .

16 t . Of 4 OZ .

8 gr.

31°OZ .

1gr.

Ans. £ 59

67} rd. rate per hr.

598 hr. 24 da. 22

17.29125

rd.,Ans.


w eek :3w k 5 da. ( 7 )7 da. 26 da.

l i sp Ans .

15 lb. $5.43}i x 132 15 a

Ans .

The smaller the rank the larger the file,for the

same body ; hence ,36 in rank 42 in rank 24 in file 28

,Ans.

16 sec. 60sec. 28 times 28 X 60 16 105

times,Ans.

Shorter shadow :longer smaller height :larger.

2 ft. 2 in . 25 ft. 9 in. 3fi . 4 in. ( 7 )25 in . 309 in. z:40in. 40 X 309 25

111. z 41fia. 2?in .,Ans .

REMARK.—This simple problem should be illustra ted by a diagram,

a nd a number of simila r questions should be prepa red by the teacher.

Attention to this m ay give the pupil a fair idea of simila rity qftria ngles a nd proportiona lity of sides.

160A. :840A. ::s4 .50 Ans.

Supposed measure :true measure sup . w orthtrue w orth

,or

,8 pt. 7} pt. $240 $225 a nd $240

8225 gain,Ans. Or

,

Sup . measure excess sup. w orth gain8 z } :615, Ans .

f

Sup . pound the la ck sup . w orth its w a nt ;

16 oz . 1} oz . Ans.

1° 80° 24’37 365000ft. 7 )

3600 289477 36500ft. 29349751115 ft ,Ans.

1da . 2 hr. 20min. 5 sec. 5000times 7 )86400sec. 8405 see. 5000 times 4861+}L times

Ans.

Ana lysis —If it take 108 days of 8} or 341h r. each,itw ould take 34 times 108 days or 3672 days of hr . each ;a nd for days each 6} hr. long

,or 2

47 hr .

,itw ould take only

1L, so many, or .,k,of 3672 days, w hich is 136 days, Ans.



( 27 Worth of cloth advance w orth of flour :itsadvance ; or,

per yd . per yd. Ans.

6mi . 60 528 ft. a nd,44 11.:528 ft. :9 strokes

108 strokes , Ans.

Or,thus ; If it be row ed 6 miles in a n hour

,it is row ed

315a s fa r, or 528 ft. in 1minute ; hence , it w ill ta ke a smany times 9 strokes

,a s 44 ft

,the advance by 9 strokes,

is contained times in 528 ft,w hich is 12 times ; hence,

12 times 9, or 108 strokes, Ans.

$100trade trade :87.75 Ans.

1cord :the pile cost of cord ( 7 )128 15 X x 12 Ans.

Fahrenheit h a s 180° from freezing to boiling ;hence,

180° F . 1° F . 80° R. g° R.

180° F . :1° F .

-1oo° o. 5°0.

“ 8 ° 15“

100° C . 1° C . 80° R. R.

100° C . 1° C . 180° F . 13° F .

Ans. 2d°

80° R. 1° R. 100° C . C.

80° R. 1° R. 180° F . F .

(32 ) 180°

( 108° 100° C . C.

180° ( 108° 80

° R. 334°R.

80° R. 25° R. 180° F . F . this 32°

Ans.

80° R. 25° R. 100° C . C.,Ans.

100° C . 46° C. 80°R. 36 R.

100° C . 46° C. 180° F . this 32°

Ans.

REMARK.—The teacher who prefers it m a y use the short method,

multiplying by the ra tios found in Ex. 31,but the full sta tements will

be good exercise under this a rticle.

g»of 180 lb . 9601b. ( 7 ) 4 ft.

‘5' ft.2 6 in .

,A718 .

Ans. 3d.

HIGHER ARITM ETI0. 67

( 36 ) 6 ft. 8 in. 1ft. 3in . 512 lb . available pow er.

80in . 15 in . lb. 96 1b. ; of 96.1b. 144

lb.,Ans.

36 in . 4} In. 1440 lb. 180 lb. of 1801h .

270 lb.

,Ans.

12 ft. 54 ft. fit of 198 lb . 594 lb.,Ans.

9 lb . 4 lb. :4} ft. = 4 ft. 5} ih .

,Ans.

Moon ’

s w eight Earth ’

s w eight :Earth ’

s dista nceMoon ’

s distance .

123 49147 250miles 99892 mi .,Ans .

It ca n be so solved for,there is a direct ratio of

the amounts,5 A.

,13} A.

,the time being the same 5 A.

:13} A. ::3men Ans.

It ca n be so solved for,there is a n inverse ratio

,

th e greater the number of m en,the shorter the time

for the same labor. It w ill require one m a n w orking 6times a s long a s 6 m en

,that is

,42 days a nd 10men w illbe kept T1} a s long, or, 4} days .

It ca n not be so solved for,the gains a nd prices

a re not in proportion,the cost remaining the same .

While a true clock indicates that 1440min . havepassed,the losing one indicates but 1435 a s having passed

in each w hole day hence,the ratio of the indicated lapse

to the true lapse,in a ny period , is 1435 1440. From the

6 A. M.,on the first

,to 11on the 15th ,

the time elapsed isda. Hence,

1435 min . :1440 min . da . da .

14 da . 6 hr . 11min . sec. Hence,a s the indica

tion is 14 da . 5 hr.,the true time is later than 11

,by 1hr .

11min . sec. ; hence, 11min . sec. pastnoon

,Ans .

Norm—We see tha t,on the 15th P . M

,it could not a ga in indi

ca te 11o’clock . As the opera tor could determine this for himself

from the ca lcula tion, A. M. w a s omitted in the statement.


( 45 ) At the end of 1h r. the cellar h a s reta ined 8M 5

or 45 gal . that is, in cu . ft.

,45 X 231 1728 .

‘Hence,

Am’

t in 1hr. required am ’

t. 1hr. req. time.231x 45 —1” " x fi x l n fl hr.,

Ans.

2 31x 4 5

( 46 ) 4 :9 567. By the directions w e w rite,

4 :574 . By Prin . 2, the missing term is

found 15W ; a nd dividing by 9, the required multiplier18 found

23W fi -g, Ans.

( 47 The proportion sta tes that 6 3 18 9 . The

quotient in each case is 2 a nd if the addition to each of

the four preserves a proportion, the new quotients mustalso be equal . Now ,

1. When a ny part of a dividend contains a part of thedivisor 2 times exactly, the remaining part of the dividendm ust contain the remaining part of the divisor tw o times

,also,if the whole dividend contain the whole divisor two

times. Here,a part

,6,of th e new dividend

,contains apart

,3,of the new divisor

,2 times ; but the remainingpart of the dividend

,being the added number

,is exactlyequal to the remaining part of the divisor

,a nd of course

ca n not contain itself tw ice -hence,w hatever be theadded number the new quotient ca n never be sogreat a s 2.

2. The dividend a nd divisor being increased by the same

number,their difference af ter the increase must be the

same that it is now ; the new dividend on the left w illexceed its divisor by 3,a nd that on the right w ill exceed

its divisor by 9. The new quotients then must be, on theleft

,1a nd a fra ction whose numera tor is 3 ; on the right

,1

a nd a f ra ction whose numera tor is 9.

On the left the denom inator is 3 a nd the increa se,on the

right,9 a nd the same increa se.

3. If these fractions be e ua l, since the right hand

numerator is 3times the left, t e ri ht hand denom inatormust be three times the left ; t a t is, 9 and once the

increase w ould have to equal three times 3 a nd three timesthe increase . But once the increase can not equal threetimes the increase hence

,the fractions a re not equal

,themixed quotients a re not equal , the ratios a re not the same

,

a nd the proportion is not preserved. Q. E. D.


NUMERICAL ILLUSTRATION— Try the addition of' 5 . Then6 3 18 9. By addition the ratios become 11 8 a nd

23 14 ; but 151 13, a nd H 1191. 14 is not equal to3 times 8.

REMARK.—The tea cher will have observed the importa nt difference

between a mere numerica l illustra tion a nd a rigid demonstra tion,a nd

tha t the former,though useful

, ca n not be substituted for the la tter.

The strict proof given a bove is a good exercise for the a rithmeticia n,

a lthough,certainly

,the a lgebraist ca n sa tisfy himself by a shorter

process. Thus:Let a be the added number. Then

, if 6 + a :3a :9 + a

, ( 6+ a ) (9 + a ) must equa l (3+ a ) or,54

15a a2 54 21a a

2,a nd 15a 21a

,

Which is impossible. If,instea d of “

a”we sa y simply “

the

number,

”we m ay tra nsla te this demonstra tion into common language,

so tha t it will be ava ila ble for the class .

One fifth is here the ra tio w here the divisor is aunit more than the dividend this dividend is required . If

the divisor be 5 times the dividend , for each 1part in thatdividend there is a n excess of 4 such parts in the divisor ;then

,4 parts excess 1such part a unit

,thegiven excess

,dividend required ; or, 4 1 1 Ans.

The w hole w ill still contain 1} lb . of‘

salt,a nd

,

the required ratio being lb. salt to 40 lb. mixture ,lb. salt 1} lb. salt 40 lb . mixture (7 )Then,4 95131 120 lh .

,the w hole ; a nd 120

COMPOUND PROPORTION.

Art. 233.

20 gal . 6 gal .204 min. 136 m in.

1size 7} sizes18 pipes ( 7 )

20x 204 x1

70 KEY 70 RAY ’S NEW

19 32mon. mon.

8100 $4500$8

88 x 32 X 45000960, Ans.

( 314 men 12 men6 da . 2 da .

" 102111325 A. 80A.

10} hr . X 12 X 2 X 80 93, hr.

,Ans.

4 horses 150horses x

15 cars l ca r—1 —22§

30da. 20da .

8 hr. 10 hr. 12 menHa . Ha .

20x10x x 12 men“

30x 8 x8 men

’A”

profit $250profit28 mon. 8 mon .

$3750

63750X 250X 8 Ans.

7

Counting 30da . 1mon .

,3yr. 8 mon . 25 da. 1345 da.

$1500 $100w orth

1345 da.:360da.

10°$6 Ans

HIGHER ARITHMEHC. 71

3500men 1800men

1am ’

t 5 am ’

ts mon.

12 oz . 20oz .

42X 1800X 5 X 20mon. 1yr. 7} mon.

,Ans.

84} rate 67 rate3 yr, 1} yr .

18 long 7 long1 w ide 2} w ide2 bu. 120bu.

100sq. in . 32 sq. in.

40X 25 sq. ft. 75 X 16 sq. ft.

4500tiles 7)

4500X 32 X 75 X 16 1728 tiles,Ans.

1} thick 2 thick30high 24 high 150000bricks 7 )216 long 324 long

150000x 2 x 24 X 324240000bricks, An

1house 6 houses16 long 18 long 240panes12 wide 10w ide

240X 6 X 18 X 10 1350panes , Ans .

n

( 9 )

3540 ( 7 )

M $466 6 Ans.

44 x 363’

2 ft. deep ( 7 )


copies 24000copiesfolds 12 folds 800reams ( 7 )pp. 550pp.

800 X 24000X 12 x5000x 8 x 320

1152 sters

2 degrees80hr.

H 15 men

(m en ) 4 ( boys )( a tw ork ) 3 ( hired )

480x 5 x 96 x 3x 2

9900reams. Ans.

24 boys, Ans.

PERCENTAGE.

Art. 236. CASE I.

FORMULA.— B x R P.

REMARK.—These solutions, in the m a in

,follow the brief model

opera tions,1a nd 3

,p. 189, Higher Arithmetic.

1664 m en X .62} 1040men,Ans .

( a ) 48 mi. 256 rd. 15616 rd. ; 15616 xrd 4 mi . 184 rd.

,Ans.

( 4 . x iof Ans .

a nd of 127 gal . 3 qt. 1pt.

2 qt. 1pt.

,Ans.

9870 of 14 cw t. 2 qr . 20 lb . 1470 lb. X .98lb . 14 cw t. 1qr . 15§ 1b.,Ans.

4070 of 6 hr. 28 min . 15 sec 23295 sec. X .40

sec. 2 hr . 35 min. 18 sec.

,Ans .



onefif th of the second remainder ; he now ow es four fifths

of that remainder,or

,of a

'rof the debt, w hich is of

it,Ans.

47 gal . 2 qt. 1pt. 381pt. ; of 381pt.

115of it 3gal . 1} pt.

,Ans .

81200X .23 boa rd ; 81200X .10§clothing; 81200 X 06} $81books ; $1200 X .001

1, $7

newsp apers ; $1200X .12} other expenses ; sum

of these a nd $1200 sa ved,

Ans .

1070 of 2070 of 110of of 27

cts.,Ans.

of of 2=flT ; $133} x 7 3 -Q x 73}$6, Ans.

If he deducted 3 ct. on each 81, the w hole deduction is The of $119449, w hich is a s this ismade by the allow ance of 1 ct. for each cubic foot

,the

number of cents in this w hole deduction is the number ofcubic feet

,i . s . 358347 cu . ft. the w hole cost of ea ch foot

is $119449 358347 Since the rough stone cost16 ct. per foot, the cost of dressing is 33} ct.

— 16 ct.

17} ct.

,Ans.

40yr. 365} X 40 14610 da.

,a nd hence hedrinks 14610 X 3 43830gi. .48 times this 21038 4

gi. 657 gal . 1qt. 1pt. gi. , Ans.

Having given aw ay 3070 , he had 7070 left ; th angiving 2070 ,or of this

,he had leftgof 7070 ,

or 5670 of

the w hole ; a nd, 5670 of $1200 $672. Now,the share

of the last a nd $12 more that of the third ; the lastshare a nd $24 more the second share ; the last sharea nd $36 more the first share hence

,the $672 must beequal to $72 more than four shares like the last ; $600,

therefore,

4 such shares,a nd one of $600 $150,

Ans.


The increase being of'

is simply .7 but thedecrease is of'

or,

hence,the w hole operationdiminishes the number by or

,therefore thedifference being a nd the subtrahend the minuend

3} i 4,A718 .

Art. 237.CASE II.

PFORMULA BRm nm—Excepting the examples complica ted by fra ctions, the

following a re intended to be like the brief opera tion, p. 191, HigherArithmetic.

.15 2 Ans .

2 yd. 2 ft. 3in. = 99 in. a nd 4 rd. 792 in . 7299}

3gal . 3 qt. gal . + 315 11} Ans.

i "3X -2 ‘3‘

- Of ~§ 0f this —2 H: 101119570, Ans.

640 .008 .001. m,Ans.

2000 .0016 {57m Ans.

750 12000 f , .06; Ans.

3qt. 1} pt. pt. 5 gal . 2} qt. 45 pt.

45 165470 , Ans.

A’

s money h a s 150 for each 100 of B’

s ; hence,B

’

s is less than A’

s by 15505 or of A’

s,w hich is 39470 of‘

it,Ans .

( l z-i iie of rs —rnie — ra = 2in.

m

76 KEY TO RAY’S IVEW

l 1

100 157“A”

127, of 375 = 39 ; 9 108 eg g ,Ans .

9 parts a nd1part make 10pa rts,the w hole and

of 10, one is 1070 ,Ans.

One yd . 36 inches ; the meter, w hich is herethe base

,is ( by Art. 209 inches 36

Ans .

A 6 mi . square contains 36 sq. mi . 23040acres ;9000 23040 Ans.

REMARK.—Where the problem states a number of which is a well

known a nd convenient a liquot,a s 2070 i, itis not nec

essa ry,in the written opera tion, to repea tevery such sta tement a s itis

often convenient, a nd a lw ays exa ct,to write for the the equiva lent

common fra ction, the la tter being well known ; where, however, it isnot usua l or well known,

it is better to indica te the steps of the redue~tion. The following is in point

got },3 6

4; 120 500 5 2f8

m0°f100 10000 100

57“ 100 1—0

°f3

10; 3

3000 1000 10011157“ 57”

Of 367” °f

3 36 1 81,6§ 22} f

555 100°f4 500

x100 535 0170 ’ 15 5

”f10—0

"

96 55 45 96 297 297 1

15 5 500x200X100 200

X100

1720267“

Ans.

3070 : 135 ; Ans.

7 24 : .29} Ans.


Art. 238.

CASE III.

PFORMULA.

R

REMARK.

— The solutions under this a rticle follow ,in the m a in

,

the second model, p. 193,Higher Arithmetic. Both methods a re used,

however, a nd the work of the cla ss should not be restricted to either.

.05 $76, Ans .

-

13T X

139-6Q = = fi ,

Ans .

16 .015 106613, Ans .

31} ct. .15g $2, Ans.

( 5 ) .03§ Ans.

162 men .048 3375 men . Ans .

15070 ,

"

11070 170 332, 10070

$3200, Ans.

170 10070Ans .

16 gal . 1pt. 129 pt. 129 pt. .06} 2100pt.

2627} gal . 262 gal . 2 qt.

,Ans .

10mi. 316 rd. 3516 rd. 3516 rd 75 4688

14 mi . 208 rd.

,Ans.

36 men 42570 ; 170 36 m en 429: 36 m en

X 5 515 13545men ; 100% 84 men

,Ans.

144 sheep 170 144 121} 144 X

35;

915 ; 10070 1125 sheep

,Ans.

Or,briefly

,144 sheep —5 .128 1125 sheep

,Ans.

$6000 .35 z:Ans.

12 pigeons of the flock ; 170 12 X g13pigeons ; 10070 450 a nd 450 12 438

,A718 .

( 150 $10 170 $10 100%$160, Ans.

78 KEY TO RAY'S NEW

$25 or, gof A

’

s {r $5 , w hole.= $40;

$25 or,m of B

’

s3h w hole $60, Ans.

$5 .13} this $5 Ans.

$150 48 170 ; 10070a nd, $150 Ans .

65 A. 106 sq. rd. 10506 sq. rd. 10506 sq. rd

03 350200sq. rd. 2188 A. 120sq. rd.

,Ans .

( 813 x 12 ) .20 $780, Ans.

( 40ct. 25 ) .11; gct. x z :14 ct.,Ans.

81men ( fi x:of 15095 ) = 81men X l -gfl_men

,Ans.

7570 of 6070 .045 825000+ 045 855555 554 ,

Ans.

The sister received 2 apples,w hich w ere 2070 of

37570 of 33170 ,or

, iof gof“

i 7ml“of a ll ; if 2 w ere one40th

, fl 80apples,Ans .

$3 .31; 53

8000bu . 170 8000bu. X Th 140

bu . 10070 14000bu . this 8000bu. 6000bu .

,Ans.

3270 of 7570 of =13025of i of

‘

8

1539 801125 , Ans

Art. 239. CASE IV.

FORMULA.— B fi 8ig;

REMARK.— In the ma in

,the short model opera tion, first under each

problem ,h as been followed here. The tea cher will see the advantage

of ha ving the cla ss familia r with both methods.

$480 $360, Ans .

8 Ans .

HIGHER ARITHMETI0. 79

96 da . 1 10070 = s} of 96 da. 48 da .,Ans.

2576 bu. 1 .6 ) 2576 bu. .4 6440bu .

,

Ans.

87; ct. (1 8 75 ) 873} ct. $7, Ans.

42 mi. 60rd. 13500rd. a nd 13500rd. 1

30000rd. 93mi . 240rd.

,Ans.

2 lb . 931?r oz . 41“ oz . ; 41“oz . (1 .5 )

tw ice oz . lb.,Ans .

17, ( 1 x 155g, Ans.

8920934 ( 1 $12

Ans .

3435806; ( 1 3;Ans.

64} gal . spirit a nd w ater 107470 sp. ;gal . 60gal . spirit,Ans.

7570 of 60gal. 4 ; gal . w ater , Ans.

The w hole cost of it is expressed in ratio to thecost of the cloth

Cost of cloth 10070 ,Trimmings 3070 ,

Ma kingof it

5070 . In all figof it. If $32 bem of the cloth cost,

the latter $32 $17 .77; 30% of this50 70 of it $8 883, Ans.

If 1bu. make 391lb . of flour,80 bu. w ill make

80 times 39% lb.,or

,3136 lh . ; but for each 1 lb . this

contains for the farmer , it also contains .04 of 9. lb . for the

miller ; hence , it affords the farmer a s m any lb . a s is

conta ined times in 3136, w hich a re a nd 3114872 lb

The same result w ill be ba d if w e deduct the

miller ’s percenta ge from the 1bu . ,or ascertain first the

80 KEY TO RAY’S JVEWw hole number of

' barrels, a nd make the deduction fromtha t. The pupil should give a n analysis corresponding toea ch of these operationsl st. 39; lb. x 80 3136 1b. ; 3136 196 16 bl. ; 16

b1.,Ans .

2d. ( 1 x 80 x Lgn 196 b1.,Ans .

The number of grains is X 480,a nd

there w ill be , in value, a s m any eagles a s 9 pw t. gr.

,

or, gr . is contained times in the given sum but for

each one eagle it contains, it must also contain of an

eagle a s expense ; hence , it w ill bring a s many eagles a s

is contained times in the w hole number of eaglesifixfifiw 928 eagles

,Ans.

2047 ( 1 of 2047 ( 1

2047 .89 2300,Ans.

670 of 5070 of x s} Xhence,by formula

,4246; 3725

,Ans.

.40 of .50 of .60 of .70 .084 ; if .084 w eretaken out there must have been left .916 of it ; this being

the w hole must have been 1 .916

$1700, Ans.

You had 10070 a t first ; giving aw ay youhad left or,4} of it, w hich is $2 one-seventh i. of

$2 $3 7 sevenths,or the w hole

,Ans.

570 f rom th e w hole leaves 95 % of the w holethis being 570

,170 is 6, 10070 600

,a nd 570 of this

30,Ans .

f rom th e w hole leaves 6613-70 ,or g, of the

w hole ; T60' of the 3} being taken,leaves 1—45 of the or 1

15

of the w hole ; i of this remainder being taken leavesof it

,or 1

15of the w hole . But this w a s $500; if $500

were 113, the w hole w a s $7500 5, of this $2500 ; I“;



842540X .114 $4963, loss, Ans .

342540 $4963 $37577, left, Ans.

810x 576 x $7000, Ans.

The w hole cost is 50 ct. 11,of 50 ct.

,in all 55

ct. the profit being 25% of the investment,is iof 55 ct.

hence,selling price 55 ct. 134 ct. 684 ct.

,Ans .

Following the formulas for amount a nd difference,85000x x x ( 1

this 85000 Ans.

The sum of the costs ( 160 160 80

100 100 112 100) of 81270 1007091270 of The receipts 35 X 70ct.

diff . Ans .

3150x Ans.

Cost 8 ct. But,losing1

16,each lb . I have bought

enables me to sell only .9 of a lb.

‘

But,Imust realize

times 8 ct.

, or ct. on that .9 of a 1b. hence,selling .9

lb. for ct. , the price of one pound must be ct.

.9 114 ct.,Ans.

By the formula for amount ; 810000 X x

X 817280, Ans.

Price to be realized, X 46 X 14

but a s this is to be broughtby 40gal . , the price of one gal .40 $35 94, Ans.

Art. 245.

cm H‘

REMARK.—The shortopera tions here, follow the model

,p. 201.

$4 $1 $3, profit ; 10070 $1, the base hence,

$3 30070 , A718 .

$4 $1 33, loss ; 10070 $4 , the base ; a nd 34 7570 , Ans.

H GHER ARITHMETIC. 83

5 parts selling for 9 of the same,is a gain of 4 ;

hence, 4 , the gain, 5,the base

,7; 8075 , Ans.

The first outlay w a s $125 , the second 196 a s much,making, in all, H of 8125, or $200, the w hole investment.

Of this, 825 came back w ith the third horse , a nd $150bythe sale,

in all, 8175 , leaving a loss of $25. Of the ih

vestment, $200, the $25 is Ans .

A value decreasing becomes3only if this 2beincreased by i of 1tself'

,it becomes 3! i the w hole .

There w a s,therefore

,no gain

,no loss.

81728 $1536 $192 profit ; 192+ 1536 12570 ,Ans.

( 7 The 1b. sugar compares w ith the 1h . Troy a s 175

with 144 ; hence , 175 for 144 is, on the buyer’s side , a lossof or 17370 on the grocer ’s side

,a ga in of 1

3419or

ma y“ Ans.

$2500 81750 $750 less ; 750 2500 136

3070 , Ans.

2070 less leaves 80of each 100; this 80 increasedby 4070 , or 116 of itself, becomes 112 hence,receiving112

for each 100in the cost, is a gain of 12 on each 100 i. e.

,

1270 , Ans .

For each 100 of the cost, you receive 1335 a t

reta il ; 1070 , or I15 ,less tha n this is 196 of it

,or 120;hence

,by w holesale

,receiving 120on each 100 is a gain

of 2070 , Ans .

15 % loss leaves 85 of each 100 but 2070 gain on

the w hole requires the value of 120 for cost of 85 hence,gaining 35 on 85 , the rate is 35 85 41137 70 ,

Ans.

35 ct. amount of tax the totalcost ; ct. profit ; .25 + 25 0=QQ‘OIO , Am .


Va lue in hand at first wa s $80; receipt exceedingoutlay bv $10 the ga in 10, is, of the base 80, Ana .

Art. 246.CASE III.

$2000 s25000+2000=s27000, Ans.

850 .225 Ans.

10ct. .13§ 122 ct. 75 ct.,Ans.

824} .07} 81} X 150? 835, Ans.

$5 .02; 85 X $180, A’

s money85 .03§ 8150 B

’

s ; difi'

. 030, Ans .

82400 170 $20, 10070 $2000, la styear ; this being 44370 , 1% $2000 443 $45 10070$4500, year before , Ans.

( 40 40 960 40 920,the no.

sheep, Ans.

Art. 247.CASE IV’

( 1 Ans .

$5 1331 Ans .

(1 x

Ans.

$238 (1 the second cost,or

first proceeds a nd —5 2 Ans .

l st cost X X 2d proceeds ; hence,“6

‘0 of first cost a nd X 3

first cost ; of this Ans .


Cost on delivery must have been times theinvoice ; a nd proceeds times times

,or

,times invoice then, $1260 $1000, Ans.

Increasing a value 10070 makes tw ice the value ;hence,the value w a s doubled 6 times

,a nd the final value

w a s 2 X 2 X 2 X 2 X 2 x 2 times,or 64 times

,the

first,w hich must

,therefore

,have been

314of $100000, or

Ans .

STOCKS AND BONDS .

Art. 250. CASE I.

350x 18 x .074 Ans.

350X 147 X .05 dividend:50 7+ hence

,7 shares a nd ( 7 X $50)

on another,Ans .

3150000x .045 $6750, Ans.

3100x 50x .18 $900, Ans .

34256000x .034 3148960, Ans .

( 375000 x .07 ) 36500 311750, Ans.

( 825 x 24 x .06 ) 450,the no. of bu.

,

Ans .

Art. 251.CA“ 11°

3324 + ( 350x 72 ) 09 Ol‘ g% ,Ans.

3163845 0 225000 .07+ hence,770 , a nd

8163845 0 ( 225000X .07 surplus, Ans.

8256484 879383 3177101net this 3650000

4190701 nearly

,or leaving 177101 164250

312851over, Ans.


Art. 252.

CASE III .

818000 .15 $120000, Ans .

.07 50 27 shares, Ans.

350X 50 82650 69 shares

,Ans.

Art. 253.

( 1) x 102 15 ) shares.

times first stock stock after first increase.

x times first stock second amount. 567 shares-1 X 500shares

,Ans.

PREMIUM AND DISCOUNT.

Art. 256. CASE I .

Non a—Nea rly a ll the solutions under the a rticles of Premium and

Discount a re in imita tion of the formulas,a nd thus merely indicate

the processes. It will be to the adva ntage of the cla ss, if the solutionsbe pla ced in this m anner on the blackboa rd, a nd, corresponding to thesigns employed, expla na tions be required in the order of the indicatedsteps. The problems a re ea sy but the drill they afi

'

ord is none the

less va luable.

8100x 18 X .08 $144 ; $1800 $144 $1656,Ans.

31800x .045 381 $1800 $81 $1881 31881

$1656 3225, Ans.

350X 62 x 33968, Ans .

350x 47 x (1 .30) $1645 , Ans.

$150X .00i $150

Ans.


32568 45 x Ans.

$425 x .03 $425

Ans .

35 x ( 1 .06 ) Ans.

$50 x 40 x ( 1 .10) x 32 x

8120, Ans.

350x 98 x ( 1 .15 )

$50x 56 x .69 ) $210, Ans.

350 x 84 x 3630, Ans .

x (1.01i .995 ) $151.399Q, or

Ans.

Art. 257.CA“ H’

1 premium + 2360

1270 ,Ans .

$50x 112 $3640 $1960 1960 5600 3570 ,

$5600 x .08 $448 ; $448 , the gain, $3640, the

investment,

12115 70 ,Ans .

$5936 $3640 $2296 ; 2296 3640

35936 $5600 $336 ; 336 5600 670 , Ans.

6311570 Ans .

( 6 )$50x 280 $1000x 12 x

$50 x 280

$266§ x ( 1 .04 ) $250

8370 , Ans.

350x 58 x 34000

34000157“ A“


( 85 35 Ans .

Art. 258.CASE III.

$117 2} X 100 $5200; $5200 $50 104

shares, Ans . See Art. 86.

71} X ,

100 $1250; $1250 $50 25

shares, Ans .

8470 4570 31170 advance ; $345 37} X 100$9200 $9200 $100 92 shares

,Ans.

For each $1in the pa r value , the buyer gains 17}ct. ; then, .175

gx 100 $1184 , Ans.

4270 670 3670 ; $666 - 36 X 100 $1850

stock $1850 $50 37 shares,Ans.

Each 90 ct. of my money bought $1of the stock,a nd w hen the latter became w orth the sale brought1503of the first money. But it required to buy $1

of the second stock ; hence , the f a ce of the stock I couldhave bought, w ould have been {34} of 19905 , orM , of myfirst money. But I did not obtain so much,having paid

out $33, w hich w ould have bought $3216, of the secondstock. If I could now receive I should havem of

my first money,but to receive $2110, less, ( or only

w ould be to havem of it ; this ca n be true only because3252,of the first money= $211

§7

one 153d of it is212of

or m,a nd 166 of it must be 153 times 813-2, or

Ans .

22 shares,Ans .


90 KEY TO RAY 'S NEW

X 650 X 35 X 28) XAns.

36547 lb. w hole.

16875 X .06 X .03 1st. com .

19672

8246 X .05 X .03 2d

11426 x .05; x .03 3d

Ans,

w hole( 4 ) X .08 the fee ;

For each $1of the debt he accepts 80 keeping5 % of this

,or 4 ct. , he pays over 76 ct. for ea ch $1of thedebt ; then, X .76 X .04

Ans .

$950 $575 $120 $1645, purchase of

this com . $1645

the w hole bill,Ans .

32781460x .03; Ans.

x .035 ) 3389 453, Ans.

3500000x .05; x .01; Ans .

314902 50x .023) 6614 725Ans.

$3850x .005 Ans .

The broker must first take 1g% of or

a nd I retain or 11370 of

nearly, Ans .

Art. 263.

CASE II.(1) 350_z ( 31200 350) 495 , Ans.

M GHER ARITHMETIC. 91

x 800

$6 432 x 800

This is the same a samount net proceedsamount

( 3 $18464 , the cost of building, alone $18464 370 , Ans .

S imilar to the 3d,a nd the form is

,

Ans.

31050 570 , Ans.

$8460 Ans .

Ans .

$6400 .006 $70 , Ans.

$2416 170 ,brokerage a nd

$2416 com . ,Ans.

2570 , Ans.

rate .

Art. 264.

CASE III.

(1) 33500 02; 3140000; 3140000 33500

$136500, Ans .

If 1070 or 110of the gross receipts, it

ca n only equal 5“ of the net receipts ; X 9

Ans.

Including the expense of packing, the w hole commission w a s a s this is$ 7” 1% is 10070 is the cost ; a nd therew ere a s many 1b. a s 44 ct . is contained times in this sum

,that is,5308000 lh .

,Ans. Or

,w ritten thus

X .015) 5308000,the number of lb.

.003) Ans.


$156 014 $12-180; $156 812480

81316310, w hole cost ; $12480 10400 costper bu.

,Ans.

2470 1370 .01g $6116, sale ;$6116 X .02} brokerage ; $6116

proceeds, AnsArt. 265.

CASE IV '

1 .04 ) x .04

Ans.

$1000 or, $97 Ans .

the cost commission hence

,-1 Ans .

NUrn—S ometimes the one ra te of the formula,is ma de up of difer

ent ra tes. See the two examples following, a nd those under Art. 270.

Whole cost ( 1 rate ) cost of the sugar ;$1500 ( 1 .02} .025 ) Ans.

Net proceeds ( 1 rate ) receipts ; thus-e ( 1 .022

4 $2584. Priceper lb. $2584 20672 12; ct.

,Ans.

The agent had 5 ct. out of' each $1in the receipts,leaving 95 ct. proceeds . But the proceeds consisted of tw oparts ; one part, a commission,

w a s of the other part,a n investment. Hence

,if the proceeds w ere 102 parts,

the agent w ould have 2 of' them

,that is

,his commission

w a s T62‘ of the proceeds. Therefore,for each $1of the

receipts the agent had ,first, 5 ct.

,secondly, T}, of 95 ct. ;

both commissions making $157 for each $1of' the receipts.

Hence,there w ere a s many dollars in the w hole a s $162 iscontained times in the w hole $210 commission ; $210$3060 a nd w hole cost

, $3060 commission, $210

$2850, cost of sugar , Ans.

HIGHER ARITHMEH C. 93

Out of each $1, th e agent keeps 31} ct. , puts inflour 663:ct.

,a nd keeps also 1470 of 6613ct. ; in all, theoutlay is $ 034 $ 01 leaving out of

each $1, simply hence, .28-3 $1500,

flour ; then 3570 of this commission ; cost ofcoffee 43; of $1500 $1000; a nd 1370 of $1000 $15 ,commission,Ans.

In each $1in the value of the pork, the principalcould claim 96 ct.

,a nd of this , there w ould be to be

expended for brandy. [ See Formula, a nd Ex. Hence,

for each $1632 contained in $2304 , there w a s $1in the

value of the pork32304 32430, Pork,32430x .04

32304 x .01; 2d Com .

( 9 ) X 1400 $8680 X 96

On each $1in the value of the corn ,the agent

received 3 ct.

,a nd there w a s a proceeds of 97 ct . This 97

ct . w a s a sum to be divided into tw o parts, one of w hichw a s 370 of the other , or

,simply fif-g of the proceeds.

Hence,in all

,the agent took 163 of the corn

,a nd

T3? of

13010of it ; making together “273-31, 166 1or 162 of it ; ifthis w ere $12, one103d of it w a s $2, a nd the whole $206,

Ans.

The proceeds being 152050of the flour

,he lost by

commission of that,or 163—617 of the flour . Then since

the investment without the w a s of the proceeds,he lost 13050of of of the flour, or 1565 of it. But

he lost in the first transaction fin, of the flour ; in all , oftheflour, he lost 737, 1

503; l

—n ,or

T666Uof the flour.

Of the he lost,first

, of it, or $1,-é ."

rsecondly fixe


lost M of m of 811047 ; therefore , on his money.he lost

, $+fi l 1104:a : Taking this from $5,w e have the loss

,equal toTB'HT of the flour then

11411 of the flour must be a ndm ,or the w hole

w orth of the flour, $53, Ans.

STOCK INVESTMENTS.

Art. 269. CASE 1.

$28000 70 $40000, pa r value ; $40000X .08

33200, Ans.

$100962 $94800, face value ; 6% of this$5688, income . Premium saved $5688 X .00}

Ans.

$10200 .30 $34000 pa r ; dividend $34000

X .06 $2040, the income,Ans.

836000 .40 $90000pa r 470 of this $3600,A758 .

Each $1invested in the first buy s or 832-3,a nd yields X .065 $5533, income ; each $1in thesecond buys or $393, a nd yields $g-g-gX .045 35

5353income. The former exceeds the latter by w h ich,

of a dollar,is ” 13370 13; Ans.

The brokerage is estimated on the same base w iththe premium or discount ; hence , in one case the dollarcosts 60} ct. in the other 75} ct. Hence,

10000 .605 ) x .05

10000 .755 ) x .06

570 stock is better,Ans.

Art 270CASE II.

$46000 84594350; a nd 4594350

H GHER ARITHMETIC'. 95

$42150, pa r of stock. Income $46000X .05

$1072 $3372 $3372 $42150 .08,or 870 , Ans.

X 04,1E income ;

863000, cost ; 363000 4470 , Ans .

Each $1of stock cost,in all

,hence,the w hole stock $9850

the income, $500, cost, $9090“ 5470 , Ans.

$2075 being 570 , 10070 , or the w hole farm value ,$41500 stock income $4100 $1.02 3005

the cost of $1in stock . Proceeds of farmpa r of stock , a nd $4100 $4502? 103170 , Ans.

8122400 .004) =3120000, pa r of stock,yielding 8120000X .04} $5000 8122500

.005) the income being $2500, the rate32500 Ans.

Art. 271.CASE III.

(1) Ans.

( 2) .05 ) X .75 $2700, the money in hand—1 X cost of state stock ; $3060$1of R. R. stock costs 80ct. 6 ct. 80ct. .075

$390 .075 $5200, Ans.

Each $1invested in the first buys $192, a nd yields$1511X .02} $31;r income each $1invested in the second

buys a nd yields 8139511X .03 $385income. The

incomes a re a s 166§ a nd 100; hence , the former is 3of

the latter ; that is, it contains $5 a s often a s the lattercontains $3. But a s the first income is

3117 of the investment

,for each $5 in that income there must be $180 in


the investment ; a nd, in like manner, for each 83 in thesecond income there a re 885 in the second investment.Then

,the incomes being a s 5 a nd 3

,the investments must

be a s 180 a nd 85, or , the smaller,H of the larger ; that is,the smaller is H of the difference betw een them ,

a nd,therefore,equals H of 811400= 810200. The larger

810200 811400 821600, a nd both 831800, Ans .

Also,income of first =

31, of of

the second, 335 of 810200 a nd the w hole income

$960, Ans.

821600X proceeds . The incomebeing $840, the face of the stock m ust be $840 .06

814000, at 8070 costing 811200, Ans.

Also, 821465 811200 810265, for land hence

,

the no. of acres 810265 $30 3423, Ans.

Each 81invested in Phila . 6’

s,bought 81

.008 ) = 8H a nd yieldedHI $335 ; each 81invested

in U. P. 7’

s bought 81 .005 ) a nd yielded$916 . The latter investment w a s 3 times the former . The

former income w a s383of the investment the latter would

have been510of the sa me investment

,but wa s 3 times

5201

or$10of it hence

,both incomes w ere

38,

orH"?ofthe first investment,w hich

,therefore

,w a s 43} of 89920

834800, Ans. Also, 834800X 3 8104400, Ans.

Each 81paid for 6 per cents bought ”ill-g, a ndyielded $73-5 ; each $1paid for 4 per cents bought 8111-5495 ,a nd yielded $1,-h ; in the former case the investment w a sLP of the y ield , in the latter a n equa l investment w ouldhave been of the yield ; hence, if the yields be equa l,

the costs a re a s iii a nd 131i, or a s 92 a nd 135 therefore,the smaller is $3of their difference , a nd the largerof that difference ; $8 of 8430 $920, a nd 1458

11of 8430

$1350, Ans.


98 KEY no RAY ’S N'

EW

cost $490, a nd the whole cost 4 times 8490, w hich is$1960, Ans .

By formula, 1X 3430 1960 $1960, Ans .

By formula, 83 the market value of 81hence

,a premium of 12870 , Ans.

8500X 2 06353: 816650, the w hole cost.

8500 10 $5000, f a ce of state stock.

8500 04 812500, f a ce of R. R. stock.

The cost of a state stock share or gthecost of a R. R. share of the same face .

Now,had the R. B . shares been just soma ny a s the

state stock shares,the cost of the R. R. stock w ould have

been 3of the other cost ; but a s the R. B . shares w ere 28times a s many a s the others

,the cost of R. R. stock w a s 25times 3of the other cost. Hence , the R. R. investment

w a s {3of the investment in sta te stock ; the sum of thetw o investments must have been H. of the cost of thestate stock ; hence ,

The state stock costHof 816650_ $5400,a R R.

u u

E44 811250:

REMARK.— The teacher should ca ll a ttention to the compound ra tio

presented in this problem .

P a r of sta te stock to pa r of R R. as 2 5 .

In va lue, 81 $1 6:5 .

Whole va lue whole va lue 12 25.

6 By formula,55— 55g% ,

Ans.

INSURANCE.

Art. 276. CASE 1.

(1 3of 324000 315000 315000x 02;

3of $36000 324000 324000x .013 3270

3270 Ans.

H GHER ARITHMETICC 99

(32500 8600) x .006 Ans.

828000x .013 3490, Ans .

832760x .008 Ans.

81800X .00§ X 10 $135 , deposit ; 8135 X05 ) received

,Ans.

( 31275 x .003) Ans .

325000x .009 3225 37; of 310000 $80 35000

x .01 350 3225 350) 395 , Ans.

Art. 277.CASE II .

3of 84800 83200; X 100 83200

$234 X 100 818600

1470 , Ans.

X 100 $2468

018 nearly,or Ans.

818000 315000 333000 ; 342000 333000

89000 318000x 2; 100 3450; 315000x 3; 100

$570 ; 39000x 43 100 3420; $450 3570 $420

31440 31440x 100 342000 Ans.

810000X 3 85000 835000 845000 835000

810000 X 100 810000 2370 , Ans.

270 of g is 2470 of $70 + 3702670 i12% 1— 3 forfi Ireceive13370 on the w hole for

210, 11,of

43070 for the w hole

,1711of

155 70 1

8170 , Ans.

Art. 278.

CASE III.

$118 .003 314750, Ans.

.015 327425 , Ans.


3. .9 ) x 3 37520, Ans .

4 of 2470 —i

826284 , Ans .

( 50132:of of

33; $70 _ fm,—319370, Ans .

310000 a nd137; of

in all, paid out realiz ing my w hole

receipt must have been 8560 ; 8560 828000, Ans.

Mutual paid 4 of value 470 of 13,of it

rec ’d fromUnion 4 a nd from ow nersOf 3Of it, 1

221010"

lost the diff,or

, 1523090of va lue.

Union p a id 4 of value, received 1111 of value ; lost1129616

. This is,then

, 132904, of value less than Mutual lost ;hence

, 1322020of value 849000 value 8150000 own«

ers lost 4of 8150000 470 of 4 of it, 851750, Ans .

TAXIB .

Art. 281. CASE 1.

(1 3486250 100 Ans.

83800X 1209“ 100 81

Ans.

(3

86000. is

800.

a

10. 125

5 . 062

30 004

Ans.

( 4

Ta x on 810000. is

400

20. 25

4 05

006

Ans .



Art. 284.

CASE IV.

( 1) 100 1,0y 98g ; 01251276 6 9844; x 100

0126840, cap. 0126840 0125127 66 01712 34 tax, Ans.

100 2 102 $7599—s 102 X 100 87450, Ans.

Art. 287. CASE 1.

(1 x 24 x 36 ) x 35 Ans.

x 45 x 36 8648 , Ans .

X 25 X .50

83X 25 75.

$28125, Ans.

6 cw t. 2 qr . 18 lb. 746 lb. 36 Such boxes w eigh26856 lb. duty at 2 ct. per lb . ad . val . duty

X .25 X 26856 both Ans .

(02 56 x 575 x .35 ) x 1154 01092 20;

(01472 01092 20 x 575 Ans.

(112 times X 20) X 825 per ton,Ans.

Duty on 3724 lb.

,10ct. pcr lb.

,

Ad val. 1170 on X 37245466 617

9070 of Ans.

( 8 1120X IiX 8322, value ; ad . val . duty4070 of 8322 this

,With 8112, specific duty,makes Whole cost required sale

times or,

a nd -5 1120

per yd.,Ans .

.HIGHER ARITM ETIC’.

Art. 288.

CASE II.

1 0147380 03684 50 40 Ans.

(010285 314 07618 75 07618 75 35 Ans .

X 40x 63 value X 40X 638126 8126 ad. val . duty

Art. 289.

CASE III.

1 .25 0230320 sum 02879, Ans.

02970 1800 .60 02 75, cost ; a nd, (02970x 1800) x 1800 Ans.

invoice 15 X 144+ 35 ) 5

,bottles to ga l. a nd 8(

X ( 15 X 144 ) 50ct. per bottle, Ans.

Art. 290.

CA“ W ‘

(1) X 1200 83000, spec . duty ; 81000X 60

8600 these , w ith 875, make 83675 w ithout invoice a nd ad.

val . duty 813675 83675 810000 this being 125% of

invoice,the latter is 810000 88000 for 100000

cigars,or 880per thousand, Ans.

845 4 810 835, cost, excluding spec. duty ;hence

, 835 times invoice,w hich is 835

828 per ton, Ans.

X 3x cu. ft.= cu. ft. at 50ct. per ft ,

spec. duty making, with ad . val . given,

a nd leaving times invoice ; hence, invoiceprice per cu. Ans.


Art. 295. CASE 1.FORMULA.

—I P x R x T.

Ba u m—The model on p, 245, Higher Arithmetic, is followedhere ; 5 or a ny grea ter figure in mill

’s place is counted 1ct. in the

a nswers .

x .07 x 240 Ans.

x .045 x 14 Ans.

X .09 x 1m Ans .

x .10x g, Ans.

x .06 x R: Ans .

x .10x 61-017, Ans.

x .074 x 4 Ans.

08722 43x .06 x 50 02878 40, Ans.

x .08 x Ans .

x .06 x 4,a,l,r

Ans.

010000x.06 x “I“, Ans.

Art 297.

(1

12

9

28110422 m . 4 m .

6da .:11,T 24 da . 4

1da. 5108 5 da .

213

01689 2707, 033927 72, Ans.


106 H Y TO RAY’S NEW

5 6 m .

:1, 2 d.

1

4Ans .

117 d. 15 m . ; int. of 81 8.015‘

X {a}required amount X

Ans .

Time,16m . 21da. ; int. of 81at 1295 ;

4 of 81883 X .167 31572 305 ; am ’t 81883

Ans.

Int. of $1 X 33 req. am’

t

X Ans.

Find int. at then 12 a nd X 75, thusX

1—3-1X am ’t Ans .

int. for 1mo. at 195 X H, Xg int. at 2570 Ans .

07302 85 x Ans.

01000000 x x Ans.

x .07 x 74 12 0218 609

3218 609 73

Ans.

( 2L) Time= 40m. 4 d. ; int. at 470 of 812500X

Ans .

30

3 3m .

12 d .

10 d ,

Ans .

HIGHER ARITIflVETIC 107

Time 19 m . 19 d. int. X 4 of

0984 am’

t Ans.

Tim e 93m . 20. d. ; int. 813682 45 X .4684X g. Ans .

Received 8876459 50 X X .014 827608 474Paid 81065252 0 X .06

Gain 821216 96

Ans.

( 25 ) Receives 8100X X 02 X 11Pays 0100 x .06 6

Gain Ans .

X .06 X 14 £ 49384 £ 49 7s. 8d.,

£24 932X .06 X 49 3612 4683 £14s . 114d .,

Ans.

£ 25 x .05 x 4; £ 2 3s. 9d.,Ans.

X .05 X H-g £15 1zs.

10d.,Ans.

Art. 298.

CASE II.

(1) Int. of $1200 for 1yr . at 107; : 8120 ; 8180081200 $600 $600 $120 5 . Ans. 5 yr .

m t. of for 1

yr. at 10% 14 yr. 1yr . 4

mo .,Ans.

int. on

for 1yr. at 870yr. 1yr. 4 mo . 28 da.

,Ans .

To double itself it must gain 100% 100 dividedby 44, 6, 74, 9, 10, 12, 20, 25 , a nd 30, gives 2231, 164, 134,114, 10, 84, 5 , 4 , a nd 34 yr.

,Ans.


It must gain 20070 200 divided by 4 , 10 a nd 12,

gives 50, 20 a nd 164 yr.,Ans .

int. of

for 1yr . at 1070 .7732 yr.

9 mon. 8 da . ,Ans .

int. of

for 1y r, at 1270 .8362 yr.

10mon . 1da .,Ans.

Int. of for 1yr . at 670.612 yr. 7 mon . 10 da.

,Ans .

Int. of for 1yr . at 770 z:

.3676 yr.

, X 365 134 da .

,Ans .

Art. 299 .

CASE HI'

Gain 200 on 100; 200 divided by 5, 10, 15 , 20,25 a nd 30

,w ill give 40, 20, 134, 10, 8 a nd 6470 , Ans.

Gain = 300 on 100; 300 divided by 6, 12, 18, 24a nd 30

, gives 50, 25 , 164, 124, 1070 ,Ans.

Int. of 835000 for 1mon. at 170 = 8294 ; 81758294 2:6 hence Ans .

Int. of 829200 for 1da. at 170 $444 80 ct. ;

80ct. 8 hence 870 ,Ans.

Int. of 812624 80 for 3 mon . at 17010 hence 1070 ,

Ans .

100 5 X 2 10,that is 1070 , on whatcost 6070 ; .164: 16470 , Ans.


110 H Y TO RAY ’S NEW

Art. 301.CASE V.

(1 Am’

t of 81for 2 yr. 3mon . 12 da. at 670Ans .

Am’

t of 81for 10mon. 26 da. at 1070-5 Ans.

Am’

t of 81for 3yr. 1mon . 7 da . at 4470Ans .

Am’

t of 01for yr. a t 74, 0101004 ;

MATURITY or NOTES.

Art. 303.

(1 Due,Aug. 5 time

,2 mon . 3 da. int. of for

2mon .3da.,at 770:

Ans.

Due,Aug. 2 time

,6 mon. 3 da. amount of $430at 1295 8430X (1 .061 Ans.

Due,Nov. 13 ; time , 3 mon. 3 da . ; amount of

at 1070 X Ans.

ANNUAL INTEREST.

Art. 304.

81500 on int. 3yr . 5 mon . 26 da .,draw s

$90on int. 4 yr. 5 mon . 18 da .

,draw s

Ans.

( 2 ) 86000 for 6 intervals 370 a n interval,

8180for 15

M GBER'

ARITHME’TIC’. 111

82500at 770 from Ja n. 11,1871

, to March17, 1873 ; 2 yr . 2 mo. 6 da.

, 8382 084$175 a t 770 for 1yr. 4 mo. 12 da.

,

2500.

Ans. 82898 825

PARTIAL PAYMENTS .

Art. 307. U. S . RULE.

1

Principal,due Sept. 1882

, 5

Int. from Sept. 13,1882

,toNov. 3

,1883

,

mo. 21

Amount due Nov . 3,1883

,

PrincipalInt. fromApr.13

,1873, toDec .8

,1873 7 mo . 25 da.

AmountPayments 810 860 70

Balance,

Int. from Dec.8,1873, toJuly17, mo. 9 da.)

Paym ent,

200.

Balance,

Int. from July toJa n . mo . 15 d .)DueJa n. 1

,1875 ,

Ans.

REMARK.—This solution records no surplus interest ; but in the

following example ea ch deficiency of pa yment is noted, a nd the interest found for the days, a s on p . 249

,Higher Arithmetic.


Principal,Int. to Nov. 25

,1874 (2 yr. 3

Amount,Payment, Balance

,

Int . to July 18 , 1875 (7 mo . 23Paym ent,

Surplus int.,

Int. to Sept. 1, 1875 (1.mo. 14

Balance,

Int. to Dec. 28 , 1875 (3mo . 27

Amount,Payment, Balance

,

Int. to Feb. 10,1876 (1mo. 13

Due at settlement,

The periods a re 1yr. ,6 mo.

,6 mo.

,a nd 3mo. The first

pa ym ent,exceedingthe interest due , w a s a pplied according

to th e rule,a nd the principal , Apr . l st

,w a s $306. If the

next w ere applied in like manner it could have beenno less than the interest of $306 for 6 mo. Hence

,

if the payment equaled the interest, the principal w a s nogrea ter than $306, a nd so the a mount a t next time of payment no greater than a nd the deduction of

(exceeding a six months ’ interest) w ould have leftno greater a principal for the la st 3 mo. thanw hich w ould have amounted

,at the close

,toa sum not sufficient . Hence, it ca n not be true that thelast principal w a s so small

,if the payment wa s applied,

a nd consequently it ca n not be true that the amount, pre

viously ,w a s so sma ll a s but a s it cou ld have



Verification—Principa l.One yea r

’s int.

Ba la nce,

6 )It is obvious that the payments must exceed the interest. Each $1w hich is applied on the principal at a nytime diminishes the interest for the next year by 10 ct.

,

a nd hence , 10 ct. more ca n be applied on the principal,next time ; that is , principal is diminished regularly bypayments w hich a re 1070 more each time . Thus

For each $1. paid on principal , l st time,There is 2d

And 3d

4th

5th

In all, paid on principal a s often a sthere is $1in thefirstapplication to the principal hence,there a re a s many dollars in the first such application a s

is contained times in that is, $400is thefirst payment on the princip a l ; a nd a s the first interest is

the required payment is Ans.

Verification.

HIGHER ARITHMETI0. l15

CONNECTICUT RULE.

Art. 308.Principal of 2d,

Int. for 1yr .

Amount of prin .

,Apr. 13

,1874

,

$10 paid Oct. 2,1873

,being less thanthe interest then due

,draw s nointerest“

$60 on interest from Dec. 8,1873

,to

Apr. 13,1874

,4 mo . 5 da.

,Balance,

Int. toJa n . 1,1875

,8 mo . 19 da.

,

Amt. of'

$200from July 17 1874 , toJa n . 1

,1875

,5 mo. 15 da.

,

Balance due,

Principa l of 3d,Int. to 1st paym ent, (2 yr. 3

Nov. 25 , 1874, paid 500,Ba la nce,

Int. toNov. 25th , 1875, 1yr.

,

$50paym ent less than interest then due

Balance,

Am ’

t of $600 from Sept 1, 1875 , to Nov. 25,1875,

2 mo. 24 da .

Ba la nce,Int. to Feb. 10

,1876, 2 mo. 16 da . ,

875 on int. from Dec. 28, 1875 , to Feb. 10, 1876,1mo. 13da .

Balance due,Axes .

Ans.


VERMONT RULE

Principal,Interest to Apr. 12, 1880, 1yr. )

Am’

t of $40 from July 25 , 1879, toApr. 12,

( 8 mo . 18

Interest debt,Apr. 12

,1880

,

Int. on to Apr. 12,1881

, (1

Int. on principal for 1year ,Am ’

t of $50 from May 20, 1880, to Apr.

( 10mo. 23 da.

Interest debt,Apr. 12

,1881

,

1m. on $86.013 to Apr. (1yr

Int. on principal for 1yr .,

$350on int. from June 3,1881

,to Apr. 12

,1882

,

Balance to apply on principal1480.

Balance due,

Ans.

MERCANTILE RULE.

Art. 309 .

Principal , $950, due Oct. runs from Ja n. 25,inleap year. Hence

,

Int. for 277 da. 8950X .07 Xm950.

Amount, 81000329



BANK DISCOUNT.

Art. 315. CASE I.

[The time in da ys being ea sily found, the following solutions giveonly the ca lcula tion of proceeds a nd discount]

Int. of $1for 138 da, X .023

5,discount ; $792. 50 74 . 27 proceeds,

Ans.

Int. of $1for 95 da . $1X .07 X 95 360

X .018472: discount ;proceeds , Ans.

Int. of 31for 14s da .,at $2672.18X

.024§L discount ;proceeds, Ans.

Int. of $1for 32 da . X 31—30

$3886 X .0155 discount ; $3886proceeds, Ans.

$2850 X the face to be discounted int. of $1for 182 X -030$discount a nd 82965 425proceeds

,Ans.

Int. of $1for 54 da . $ 015 ; X .015

proceeds, Ans .

Int. of $1for 144 da. X .02

0,discount proceeds,

Ans.

Art. 316.

CASE H’

( 1) Bank discount of‘ $1for 33 da.

,at 131570 a mo.

,

$ 0165 proceeds of $1 $ 9835 $1650

Ans.

HIGHER ARITHAIETI0. 119

B. disc . of $1for 63da .,at 673 , proceeds

of $1 $800 .9895 Ans.

B . disc. of $1for 93da.

,770 , $.0180g; a nd

0180; Ans.

B . disc . of $1for 4 mo . 3da.,at 170 a mo.

,proceeds $3375 Ans.

B . disc. of $1for 6 mo. 3 da .

,1070 ,

proceeds $4850 .9491§ Ans .

B . disc . of‘

$1for 63 da. $ 042 proceeds:.958 Ans .

( 7 B . disc . of $1for 43 da .,870 $ 0098 ; proceeds

$ 9904 ; -5 9 9034 Ans.

B . disc . of $1for 33 da .

,670 , proceeds

$1000 .9945 Ans.

Also,disc . of

'

$1for 93 da. $ 0155 ; proceeds$1000 .9845 Ans.

Art. 317.CASE III'

(1) Discount of $100 for 33 da . at 1,

11} a nd 270 amon . a nd proceeds

a nd int. for 33 da. at 170 a yr33330of 170 7 4

1210100

of principal ; of

1555355 70 4 .

TMn, of“

2

nlth-

1y oi' 5

196939 24

I§6Q3% ,

Ans .

Discount of $100 for 63 da. at 6,8 a nd 1070 perannum a nd proceeds

a nd int. for 63 da . at 170 per annum= s§03070

“In of principal “

1” of

151517 of

1010-

0of 5

3050811 10

3150570 , Ans.


Discount of $100for 93 da . at 2,2§ a nd 370 a mon.

a nd proceeds a nd

int. for 93 da. at 170 = 11Q63070 of prin

cipa l of W 25H—z “

3010-

6of

‘

l ag-Q 321

1521570 “

3010-

6of

W 398-3470 , Ans.

Discount of $100 for 1yr . (w ithout grace ) , at 5, 6,7, 8, 9, 10 a nd 1270 ,

is $5 , $6, $7, $8 , $9, $10 a nd $12 ;

proceeds= $95, $94 , $93, $92, $91, $90 a nd $88 ; 355

sis— 73370 3 fi = 8lr%% 5

var Hr — l lm ; fi = 13¥r7o a

Ans

The note being lega lly due in that time , 1yr . 4

mon . 20 da . includes the days of' grace . Out of each $1due in that time he takes a discount of' 114 ct.

,a nd

,there

fore, p a ys out for that $1, 88% ct . Hence

,he receives

ct. interest on 88% ct. in 1116 yr .,or $1on $8 in 111Uyr . $8

at one 70 yields in that time hence,to yield $1, the

rate 70 m ust be 1 .114 ; hence, 970 , Ans.

Art. 318.

CA“ W '

(1) Int. of'

$100for 33 da . at 10,15 a nd 2070 , is

$1.37Q, amounts a re $101.37Q,but int. of a ny sum for 33 da .

,at 170 , T2

101“ of

'

thatsum ; a nd

3 of saggy,fi g“ of 3101371, 70 .

$1.83g, of 1934

Int. of $100 for 63 da . at 6,8,1070am ’t int. for 63 da. a t

170 35631770 43

10-

5Of principal



Corn brought $4750 this,less 370 , w a s purchase

money,

discount on $1for 63 da . a nd

proceeds, hence,each $1in face of draft costs

$ 9895

Ans .

820312 50x (1 (1 .009 32010835,

Bank disc . of $250for 93 da. 5 proceedsgain $18 75

$250 1170 discount, Ans.

$992 ) $992 Ans .

FOREIGN EXCHANGE.

Art. 321.

( 2 ) £ 625 108 . 10d. £ 625 .541fi ; X

Ans.

$1 fr. $1485 Ans.

5000roubles .74 675613-9roubles , Ans .

1milreis 54 ct. ; 4500 54

2% of it hence,taking difii

,in milreis,

8166 663, Ans.

1000guilders $1000 X .40}Disc . for 63 days X .0105 4 226+Proceeds,Brokerage to be paid , $70 .499

,sa y .50

Balance,

Ans.

ARBITRATIONOFEXCHANGE.

Art. 322.

$6000X $6030, direct. N. Y. $6000 Ga l.

Ga l. $1N. N. 0. .99ii $1N. Y.

$6000 X X .992 circular,Ans.

$6030 gain, Ans.

M GHER ARITHMETIG. 123

2) direct,Ans.

St. L. ( 8 Balt.Balt. $1 8 .9911 Ha v.

Ha v . $1 .99§ N. O.

N. St. L.

310000x direct.310000 ( s ) L.

$1 N. Y.

31 0.

810000 1-0053100299 4 , circular, Ans.

10012 50

gain, Ans.

( 4Period of discount from April 26 to June 7, 1mon . 12

da . ; a nd bank proceeds X ( 1 .007

$5247 .6773. On each $1in the N. Y. bill there is a com

mercia l discount of ct . a nd int. off for 10da . w ould bea dditional discount in all

,leaving eachN. Y.

dollar w orth Hence,the N. Y. bill $5247 .6773

.99} a nd premium being theamount in Cin . 85296103, Ans.

From Ma y 3,to June 7

,35 da. ; the interest tocome off w ould have been a 1570 discount ; the $70 pre.

mium w ould reduce the discount to H% ; hence, the

circ.

0,Ans.

gain.

Ans.


direct draft on N. 0. w ould have brought onlyX l .OOH) a nd 85296 103

gain,Ans.

-311X 919 810084 033, direct. (Art.

89339 R. ( s 1.995

1 x .995.

5 f‘

. 4 R.

42016 807 x x 5 10294117715

99 x 25 -38 x .99 x 431016538

81016538 810084 03 Am .

EQUATIONOF PAYMENTS.

Art. 326.

A. to B .

1877.

Ma y 15 To invoice a t 4 mon .

June 1 4 mon.

u 10 u u4 mon

July 20 4 mon. ,

Aug. 1 4 mon. ,

15 4 mon. ,

da . after Sept. 15=Oct. 30, Am .

2. DR.E. in a ec

’t currentwith F.

Prodc’ t

Ma y 4 .

Feb . To inv., 3mon. , Ma y By ca sh .

Ma r. 3mon. , remittance,Apr. 3mon . , June 5 ,

3mon note,1mon.

E. owes F. da . beforeMa y 4==Feb. 26, Ana .



the amounts , w e should have multiplied one by 64 and

the other by 34 , a nd divided the sum of the products by375

,obtaining 44 ; that is, 64 times one a nd 34 times the

other 316500; therefore , 34 times both a nd 30 times one316500. But 34 times both is the same a s 34 times $375,

or 312750 ; hence , 30 times one must be $3750, a nd 310of

33750, or 3125, is that one ; $250 the other, Ans .

Counting from Oct. 3, 3840X 0 0 400X 94

37600; 200X 63 12600 37600 12600 50200,in

A’

s favor he still ow es 8840 3400 3200 $240, w hichhe should retain 50200 240 209 da. after due 209 da .

a fter Oct. 3, is April 30th , of next year, Ans.

Counting from Oct. 25, 33200X 0 0 400X 40

16000 800X 25 20000 16000 20000 36000;hence,I should have a discount on $1for 36000 da .

,for

paying part, before due $3200 $400 3800 32000;

36000 2000 18 da . after Oct. 25,w hich isNov. 12

,Ans.

82500 $500 $500 8500 81000 countingfrom Sept. 16

, 82500X 0 0; 500X 46 23000; 500

X 36 18000; 500X 26 13000; 23000 18000

13000 54000 54000 1000 54 da . after Sept. 16,w hich isNov. 9

, Ans.

Debts. Terms . Equiv. Debts . Terms. Equiv.31200X 41 49200 31300X l 1300

1500X 72 108000 1300X 2 2600

2050X 80 164000 1300X 3 3900

1320X 110 145200 1300X 4 5200

1730X 125 216250 1300X 5 6500

682650 1300 6 7800

$1300 each , Ans. 27300

682650 27300 25 da. interval,a nd the notes run

25, 50, 75, 100, 125 and150da. respectively,Ans.


One drew int. for 5 mo .

,the other w a s discounted

7 mo. before due. On each $1of the first th e gain w a s755

of 6 ct.,that is

, 410of $1. Each $1of the second w a s

bought at its true w orth 7 mo. before due,that is

, 31w a s

bought for rm of a dollar ; hence, the true discount ofa dollar w a s 1" h

a ir'

s 207 of a dollar . Hence,since

the interest drawn a nd the discount taken w ere equal, 410

of the first w a s equal to257 of the other ; that is, one w a s

$34 of the other ; both , then ,must have been fi t of thatother

,a nd since both made $487, the first w a s $280, a nd

the other $207. Now,by the common rule

,equating, w e

should have280 X 1 280.

207 X 2 414

487 )694 ( yr . 1yr . 5 mo. 3 da.

Hence,the difference is 3 days

,Ans.

SETTLEMENT OFACCOUNTS .

Art. 329.

1

HENRY HAMMOND.

F0001Da te,Apr. 4.

Products .

XXXXXXX

855°

888

IIIIIIII

III

I

II

8

3

was

OFl

g

Then the exact sum held beyond May15

,to July 7, draw s interest for 53 da .

,a nd the am ’t

X Ans .

due Ma y 15 , settled 15 days earlier,amounts to 3360180 X .9975 Ans .

128 KEY TO RAY’ S NEW

Due. da . 80x 800

Feb. 15, 45 x 180A r. 14. 401a r 25 , 84 x 500

Apr. 1, 91x 800Ma y 7, 127 x 600:July X 700

June x 200Be L la m xmJu y l 2, 193 > < 500

Oct. 4, 277 X IM =

$7681 15361

391170 2320 hence,169 da. after Ja n.

1,1876

,w hich is June 18

,1876

,Ans. Int. of

from June 18 , to Oct. 4, (108 da .) 1070 am

’

t

Ans.

Gannon:Comm a s.

1876. DR.Focal Da te

,Ja n. 1.

Ja n. 1, 0 X 583Ma r. 3, 62 x 40 2480Apr.30, 120 x 130 15600

$753 18080644 13630

109 4450

4450 109 hence,41da. after Ja n. 1,

or,Feb. 11th

,Ans . $109 on int. to Ma r. 31 49 da.,

amounts to $109 X Ans .

A. L. Morris in a rc’twith T.J. Fisher 00.

da . 8 da .

168 181159 120. 146125 500. 10560 85. 30

7

(Rule on p. 293, High er Arith .)

( 2W11q Sm

Foca l Da fe, Ja n. I.

1876.

Ja n. 1028

Feb. 15.

Ma r. 30,Aa

r 14,y15 ,

June 16,July 19 ,AugSept.

Oct.

Ja n . 10,28 ,

Apr. 15,Feb. 28 ,June % ,

Mpr.

114,

June15 ,4 4 16,

a sSept. 1,Oct. 3,

On.

da . 8 prod.

9 x 400= ssw27 m a 5400105 > < 180= 189w

5m179 x 450 = 80550104 x 401= 41704121x 500 60500

167 x 300 50100200 x 700= 140000242 X M = 484W244 x 150 36600276 X1100= 903600

5361 908034



Aocomrr. ”( Thoma s sold on a ec’t of B . F.Jona a

Sa m .

July 6. ToFreigh tpa id 2000 bu . whea t.

11. 300 on 20

11. Dra ya ge11. Insura nce

11. Commission

11. Loss a nd

CALCULATION FROM FOCAL DATE, JULY 6.

July 8. 2 x31. 25 x

2475. 13287.50

Net proceeds, 1284530

Since on the side 01the sa les we find 132875 0 conta ins 2475,five

times,5 da . qflerJuly 6, or July 11, is the da te the commission should

h a ve, in the ca lcula tion. Its entry a nd time due a re of the same da te,

(which w a s not the ca se in the preceding example) .

12845 30 hence,take 6 days after

July 6,or July 12

,1876, Ans.

Saw s. 011m m.

CALCULATION m om rocu . nu s, JULY 15.

Da ys.0 x 0

15 X

24 X a

15 X

fi na ted thus:$235 $105 .5SP: 5625 .

a nd hence,ta ke 35 da . a fter July

15 ; i. e.

,Ang. 19

,Ans.

”On first side ca lcula te th e da te to be given to commission on th e oth er side.th us:5625 235 239 ; h ence, ta ke 24 da . a fterJuly 15, t e. , Aug. 8.

HIGHER ARITHMETI(7. 131

Ca rnes'

s on interest from August 19 toJanuary1 (4 mo. 13 amounts to

net proceeds,Ans.

STORAGE ACCOUNTS.Art. 332.

DELIVERED.

Bat. on ha nd.

on h and

IKE?

v. no. of bbl

RECEIVED. DELIVERED.

Bbi. Ba l. on ha nd.

Februa ry

bbl . on h a nd , 0

no. of stored.


(DMPOUND INTEREST.

CASE I .

813062 50

2

Art. 334 .

6 mo . ; 116093

1da. 115

.650

457

Amount Ans.

3850.

Interest Ans.

130625 0

133237 5

27,13590225

27013862 03

27014139 271

27014422 056

27,14710497

27;15004 707

1mo . 12 da.:

Amount 15144 751 [Ana5

51000

1050

Amt. for 3 yr .

Ans. Prod .



38 intervals at 4570 ; X850074 45 interest of for 4 mo . at 9 %

350074 45 compoundamount, Ans .

100yr .: 50yr. 50yr . 51000x xAns .

60 intervals at 270 ; 60 55 5 ; $3600 X

X am’

t ;

$3600 int.

,Ans .

80 intervals at 2570 80 z :50 30 ; $4000X

X 56758

Ans.

There a re 111intervals at 370 111 55 55

1 $1200X X X 33192370;amount of for 2 mo. 4 da . at 1270 832604 74

832604 74 $1200 $31404 74 , Ans.

Art. 336.

CASE IL( 1 $12000 -3 $8000 in table

,this is betw een 6

a nd 7 yr . .0814809

.0851112 30W yr . 11mo . 15 da.

Ans. 6 yr . 11mo. 15 da .

$5200 $1308 $6508 ; $6508 $5200

12 5153846 ; this is betw een 7 a nd 8 intervals at 370 ;02166459 ;

‘W of 6 me . 3 mo . 16I 0da. ; 1} yr . 3 mo . 16 da . = 3 yr. 9 mo. 16 da .

,Ans.

$18000 $12500 this is over 14 intervalsat 2470 .02702618 ;

.03532435 ; of 3 mo . 2 mo . 9

da. ;114 yr . 2 mo. 9 da . 3yr . 8 mo . 9 da .

,Ans.

.HIGHER ARITHMETIC. 135

$1 $1 $2 ; $2 $1 in table at 670 ,2.

is betw een 11a nd 12 yr . ; 2 .1017014 ;.1138979 ; yr . 10mo.

21da . ; at 870 ,2. is over 9 yr . ; 2.

.1599204 ; ;W yr . 2 da . ;at 1070 , 2. is over 7 yr . ; 2.

.1948717 ; 3343-29 yr . = 3mo .

5 da.

Ans . 11yr . 10mo . 21da . ; 9 yr . 2 da . ; 7 yr . 3mo . 5 da.

$2257 this is over 14 interva ls at 670 ; — 2 260904: .028186 ~

.1356542 ; 393255613452 of 6 mo. 1mo . 7 da.;

121yr . 1mo. 7 da. 7 yr. 1mo . 7 da.

,Ans.

Art. 337.CASE III.

1 $1000 w hich is found in th etable under 670 . Ans. 670 .

$3600 $3600

w hich is in the table,under 770 . Ans. 770 .

$13200 amount of

for 5 mo. 21da . at 570 Ans.

4 71868847 amount of

4 66734781for 1mo . 14 da . at 4570 = 4 71868864 ; 4270semi -annually 970 per annum,

Ans.

There a re 47 intervals ; amountOf for 2 mo . 3 da . at 25-70 a quarter

2570 quarterly 1070 per annum,Ans.


2+ l = 2 ; looking in the ta bles for2 opposite 10, 15 , a nd 20 yr .

,w e find for 10 yr . betw een

7 a nd Ans . ; for 15 yr. not quite 570 , Ans. ; for 20

yr. a little over 3470 , Ans .

CAs IV.

Art. 338.E

Interest of $1 for 25 yr. at 670 332918707 ;852669 93—5 $16000, Ans .

Int. of $1for 6 yr . 2 me . at 7 payable semiannually .5287 —e .5287 Ans .

Int. of $1for 3 yr . 6 mo. 9 da .

,at 1070 , payablequarterly .416506 $8640, Ans.

Int. of 31, in 19 periods at 470Ans .

Amount of $1 in 7 yr . a t 4 %8 Ans .

Amount of $1in 5 yr . 9 mo. at 675 , payablesemi-annually 38 38

Ans.

Amount of $1in 12 yr . 8 mo . 25 da . at 595P . w orth ;$8767.78 Ans.

ANNUITII‘B .

Art. 340.

CASE I'

( 1) 3300 .06 35000, Ans .

( 2 ) .08 39455 , Ans .

315642 90 .07 = 3223470, Ans .



CASE III.

( 1

Pres . val . of perpetuity of 3125 , deferred 12 yr .

Pres . val . of perpetuity of $125, deferred 24 yr .

Ans.

Art. $ 2.

2

Pres . value of imm ediate perpetuity of 3400Pres. val . of perpetuity 0115400, deferred 154} yr .=

Ans .

6 payable semi -annually, per year ;

.0609 initial value of perpetuity ; presentvalue of such perpetuity ,deferred 3 yr .

, 3135714 2

3113658 6 present value of such perpetuity,deferred 162 yr .,

6113658 6 Ans.

4

Pres . val . of perpetuity of $60, deferred 12 yr .

Pres . val. of perpetuity of $60, deferred 21yr .

Ans .

Int. on $120 for 3 me. at 870 32404—62 40value half-yearly hence the lease may be con

sidered a n immediate annuity of per ha lf year, andrunning 8 yr. 9 mo.

Present value of immediate perpetuity, $6060.

Present value of perpetuity, deferred 82 yr.,

Present value of lease,02500 loss, Ans.

.HTGHE'R ARITEMETI0. 139

Art. 343.

CA“ W “

Initial value of a perpetuity of $300 at 57036000 compound interest of $6000 for 18 yr . 36000X

Ans.

Initial value of a perpetuity of $25 at 10%3250 compound interest of $250 for 40 yr. 3250 X

3110648 1, Ans.

Initial value of a perpetuity of $75 at 670$1250; compound interest of 31250 for 9 yr . =31250X

Ans.

In1tia l value of' a perpetuity of 35 at 970 35553;compound int. of 35554} for 50yr . 35553X 52

Ans.

The annuity may be considered a s the yearlyinterest of som e principal at annual interest . Each 31in14 yr . draw s X a nd one annual interestof 6 ct.

,for 91 intervals (Art. yields X .06 x

91 in all,the interest yielded (or, the annuity ’

s

amount for each 31in the principal w hich yields it)84 ct. z: Hence the principal yieldingsuch a n interest is $100 a nd

,there

fore,the annuity itself 36. Amount of a n annuity of 36,compound interest 14 yr .

,find thus °

Initial value, 3100; compound interest of $100 for 14

yr. 3126 0904 . Ans.

There w ill be 13 deposits, a nd the amount is tha tof a n annuity of $35 running 13 yr . ; initial va lue:compound int. of $350 for 13

X 350 Ans .


Art. 345.

CA “ V‘

Present value of 31a year for 17 years at 770final value of the same X3308 40217 315000 -3

Ans .

31per year for 16 years , commencing now,is

w orth commencing 4 years hence,it is

w orth - s 7 $4800

Ans.

Art. 346.

CA “ VI‘

31000000 380000 this falls betw een 23yr. a nd 24 yr.

,in the table

,under 670

Comp . am ’tof 31000000for 23yr . at 6 0

Final value of 380000per yr . for 23yr.= 3759666 27

Bal . due at end of 23 yr. 3600834 3,Ans.

330000000 32000000 15 ; this falls betw een28 a nd 29 yr .

Comp . am ’t of330000000for 28 yr . at 5 603873.

Final value of32000000per yr . for 28 yr.= 116805164 .

Bal . due at end of 28 yr. 3798709 ,Ans.

322000 32500 this falls betw een 14 a nd

15 yr .

Comp . am ’t of 322000 for 14 yr . at 770 356727 75 .

Final value of $2500per yr . for 14 yr. 356376 22

Bal . due at end of 14 yr . Ans.

In the former case is the principal at theend of 15 yr.

,or a balance w hich ca n only draw simpleinterest for a fraction of a n interval . But the a mount

of that must be exactly equal to such a fraction



Art. 352.

CAsr. II .

3500 Ans .

$1200 Ans.

3840 Ans .

Art. 353.

CA“ 111.

Pres. value of perpetuity of $1, at 570 ,Pres. value of annuity of 31, age 47 yr .

,Pres . value of reversion of 31,

3500X Ans.

( 2. Pres. value of perpetuity of $1, at 670 ,

Pres. va lue of annuity of 31, age 38 yr.,

Pres. value of reversion of 31,

3165 x 4 4273 3730565 , Ans .

( 3. Pres. va lue of 31, perpetua l lease, at 770 ,

Pres. va lue of 31, life estate, age 62 yr. ,Pres. value of reversion of 31,

31600x 6. 8825, 311012 34 Ans

LIFE INSURANCE.

Art. 355.

( 1 Table number for 40yr ., X 5

3156 50, Ans.

From 40 to 53,inclusive

,14 yr . ; x 14

32191, Ans .

32191in all ; one payment on interest 13 yr .,1


for 12,a nd so on to 1for 1yr. all equivalent to interest

on one for 91yr . hence,

Payments,

Int. on for 91yr .

,

Ans.

Endowment policy cost per year lifepolicy difference on each 31000, 365 50 X

120 37860, Ans .

End’

m’

t premiums X 2

Int. on one for 2 yr .,

32210948

Life pol. cost 3214 .80x2=3429.60

Int. for 2 yr. on one, Profit

,Ans.

810000 32210948 877 Ans.

6 ) On each 31000, per yr . ten of them=3313;annual interest for 9 yr .

, X .06 X 45

3313 or on one thousand,hence the

must have been on 310000, Ans.

15 premiums, of 31each , a nd the annual interestamount to 315 interest of 31for 120yr .

,a nd

on 31000the amount is given 31542 648 a nd 31542 648corresponding in table to 40yr .

,Ans.

Annual payment for 38000 X 8

this,in 38000is contained more th a n 50 a nd less

than 51 times ; hence 51st is the payment m a k ing theexcess ; as l st payment made when 20 yrs. old, the 51st

when 70 yrs. old, Ans.

Annual payment x 300—r . 3 payments a nd interest for 33yr . on one, amounts to3297 X395519 97 330000 39552 320448, Ans.


Amount paid: 311635 .80 on 31000,the table show s 10,15

,20 ; dividing by the sums opposite

42 yr.

,at the third w e have exactly 20yr .

,Ans .

CASE 1.

Art. 357.

( 1 33600 $1500_ $2100 gain ; $2500 81875

34375 ; of 32100 31200, A,Ans . H—H of

B,Ans .

37000 3800 31000 35200 to be divided ;15306of A

’

s gain, Ans .

of 35200: 32560, B’

s gain, Ans .

lift)“ of 35200 3640 31000 31640, C

’

s income,Ans.

324000 328000 332000 384000 3 of

384000 314000 384000 314000 370000 ; g of

3112000

384000— 38000 320000gain H»of 320000 35714 283,A

’

s gain,Ans . {a} of 320000 B

’

s gain,Ans.

Hof 320000: 37G19 .04H,C

’

s ga in ,Ans .

C receives 4473-3:3of the gain ,he

,therefore,contributed 3of the capital , a nd A a nd B w hich

,con

sequently 312960; th is is 9 of 318144 , total ca pita l ;2, of C

’

s stock,Ans

, 1551910; of 33920

31244 .443, A’

s gain ,Ans. ; of 33920 31555 .5543,

B’

s gain,Ans.

38000 312800 315200 35600 ; A a nd Btogether have 35600more than C,

a nd gain 31638 more ;hence the gain is 33332 31173of the stock of 38000

32340, A’

s, Ans. of B’

s,Ans. ; {H

of 315200: 34446, C’

s,Ans.



A h a s 32500x 18 345000; B , 31500x 18

327000; C, 35000X 45 27 + 45: 117 ; A

gains of 33250 $1250, Ans. B ga ins {117 of

33250 3750, Ans. ; C ga ins of 33250:31250, Ans.

A contributes 31000 for 3 mo.

, 3600 for 3 mo.,

3200 for 6 mo. ; B, 31000 for 3 mo., 31400 for 3 mo.,

81800 for 6 mo. ; 1000 x 3 3000;‘

600x 3: 1800;200X 6 1200 sum 6000 ; 1000X 3 3000 1400X3 4200 ; 1800 X 6 10800; sum 18000; 6000

{I of A,Ans . 33of 3800

3600, B,Ans.

A makes 37 strokes per minute for 5 m inutes, a nd32 per minute for 5 minutes ; B makes 40 strokes perminute for 5 minutes

,a nd 35 per minute for 17 minutes ;

C,30strokes per. minute for 12 minutes ;

37 x 5: 185 40x 30

32 x 5:160 35 x 17: 595 12

345 795 360 1500;

A receives 135050 of 32: 46 ct. , Ans. ; B , 115955“ of 32

Ans . C, 1856506of 32 48 ct.

,Ans.

B h a s 35600X 12 367200for 1mo. ; 34200x12 350400; A must contribute 367200 350400

316800 for 1mo.,or 316800 8 32100, for 8 mo .

32100, Ans .

34500X 2 39000 capital ; 31500X 3: 34500

3500 X 3 31500 39000 34500 31500 32200

3800 loss ; A h a s 34500 for 3 mo.

, 33000 for 3 mo.,a nd

31500 for 3mo.,in all

, 327000 for 1mo. B h a s 34500for3mo .

, 34000 for 3 mo .

, 33500 for 3mo.,a nd 33000 for 3

mo.

,in all

, 345000for 1mo. stock-equivalents a re 27000

a nd 45000,or 3 a nd 5 ; hence A should suffer a nd B i

of the loss gof 3800 3300 a s A should lose 3300, and


h a s no capital , he , therefore , ow es B 3300, that is, B takesthe 32200, a nd h a s a claim on A for 3300, Ans .

A h a s 12 shares 3 mo.,5,6 mo .

,a nd 9

,3mo .

,in

all,93 shares 1mo . B h a s 8 sha res 3mo .

,10

,2 mo.

,a nd

.7 , 7 mo .

, total , 93 shares 1mo . C h a s 7 shares 3mo .

,8,2

mo .,9,4 mo.

, a nd 7, 3mo.,in all

,94 shares 1mo. D h a s

3 sha res 3 mo.

,7,2 mo .

,9,4 mo .

,a nd 7

,3 mo.

,total

,80

shares 1mo. 93 93 94 80 36039336of 318000

34650 each for A a nd B ; 32646of 31800 34700 for C ;

35596of 318000 34000 for D ,

Ans.

A had 3400 for 3mo .

, 3200 for 9 mo.

,in all

, 33000

for 1mo. B,3500 for 4 mo .

, 3200 for 4 mo.

, 350 for 4 mol,

in all, 33000 for 1mo. C

, 3300 for 6 mo.

, 3200 for 6 mo .

,

in all, 33000 for 1mo. ; hence their shares of the profitmust be equal . S ince A’

s gain w a s 3225 , the entire gainw a s 3 times 3225 3675, Ans.

Art. 359.

( 2

$1200 3720 3600 31080 33600; of this , 32520hence, pa y 70 ct . on 31, Ans.

11,yof 31200 3840, A 1

70of 3720 3504 , B 1

20

of 3600 3420, C 116of 31080 3756, D.

3

316000 less 570 315200 of 347500this sum ismhence, pa y 12

55of 31, or 32 ct. on 31, a nd to A. 1

55 of

33650, or 31168, Ans.

148 KEY TO BAY’S NEW

ALLIGATION.

Art. 361

54 ct.

,

Ans .

Fineness. wt. prod. wt. sp. gr.

16 x 5 80 15

18 X 2 36 8 -1 6520 X 103 .023809

24 x 1 24

14 260084 ca r, Ans. Ans .

ANALs s.— Copper weighs 71times its bulk of w a ter ; hence, the

bulk of 15 lb. copper is the same a s the bulk of 15 7.75, or,

lb. of w a ter. In like m a nner 1h. w a ter h a s the bulk of 8 lb.

zinc ; a nd .023809 lb. w a ter h a s the bulk of 1lb. silver. In a ll,

lb. w a ter h as the bulk of 23} lb. of the combina tion ; hence,the combina tion weighs gg2 times its bulk of w a ter

,and its

specific gra vity

5

( 2 )Qua lity . Cost.

x 40

x 30 300

X 16 200

x 54

140)

6

1branch 50701 25

1 50

1 60

1 55

1 55

6

w hich , being less than50, show s f a ilure.



each pound will h a ve {74235 bulk too grea t; if the whole be takengold, ea ch pound will h a ve fi g; bulk too sma ll ; hence, ba la nce inra tio of 723 lb. silver to 3487 lb. gold, Ans.

REMARK.— If the sp. gr. of a substa nce be 4

,one pound of it will

h a ve a bulk equa l to ithe bulk of a pound of w a ter ; if sp . gr. be it} ,a pound of the substa nce will h a ve the bulk of 2

31 of a pound of

w a ter. By th us inverting the numbers expressing the sp . gra vities ofdifferent things, their bulks m ay be directly compa red

, just a s we compa re the prices per pound in other examples. When we compa re poices,in a common example

,we ba la nce the losses against the gains, ca lling

the units in the ba la ncing, pounds,”though they m ay be tra nsferred

from a column of prices,named in cents.

”

ILLUSTRATIONw—S uppose w here the a verageprice is 5 ct. we find a loss of “3cents on one

kind,a nd ga in on a nother kind of 11lb. 1

353, cents,

we ta ke 714 of the la tter kind of pounds, a nd 24

pounds of the former. The specific gra vity ca se would rea d:We loseon a pound 333bulks, a nd gain on a nother pound 1

2515 bulls ; hence

ba la nce by 714 of la tter to 24 of former in pounds.

( 59 ( 6 )Ba l. rm

3 3 It,

1 24 2 2 1 3

22 20 2 1 1

3:110 3 18 4 1 1Ans.

Art. 364.

CASE II.

(1Bal .

50 10 12 x 8 : 96,

40 25 2 2x50=100Ans. 50 15 2 2

72 12 10 x 8 = 80. 60 5 2A“

2

75 10 5 3 1 4+ (5x50)=254

,Ans.

(3Ba l.

91 35 8 x 239- = 29 pt.


4

gr.

pw t. 9 gr .

pw t. 21gr.,Ans.

a cidity .

0 0

100 750

) 750( 37370 obta ined.

15 pt.

0 X 25gr pt.desired,221} 123

133pt.

1gal. 2 qt. 13pt.,

OPERATION. l st.)1 13 3 xw = l sfl £31 02 ,

combina tion 1 4} g4 T

Ans.

is to displa ce HP 55349

wa terequa l to 12 W + H=W .

once its weight, a nd, hence, I is the a verage. The lead,while in the

wa ter,displaces 111of its own weight ; the copper displa ces z}; of its own

weight ; the cork,when wholly in w a ter

,displaces 4 times its own

weight. Hence, the piece, say 1oz .,of lead kicks displa cingHof a n

oz . ; 1oz . of copper la cks displa cing3; a cork oz . displa ces toomuch,by 3times its weight; hence, ba la ncing, we ta ke 3oz . of lead for ea ch

Hoz . oi cork , a nd 3 oz . of copper for ea ch 3oz . of cork . But the

conditions require only i of 3 oz . of copper ; hence, to ba la nce th a trequires 3of 3, or 55; oz . cork . The conditions a lso require 12 oz . cork

in ad ; therefore, 12 or oz . cork a re yet required, a nd a s this

containsH,times

,there must be

,by first ba lancing, X 3, or,

‘P oz . lead,which is 2 lb. 73oz .

,Ans.

TION.— Th e


En m rnom OPERATION.

First,ba la ncing in

proportion to bulks,

40 0f cork to 3 of i i 10 83814- 55?

lead,32 of cork to 3 l 9 8 27 4}

of cop p er. This 1110 oz .,

m a kes their actua lweights, as 40X } to11x 3

,and 32 X ito 9 X 3. Then the oz . copper requiring 3

1; oz .

cork , the rema iningW oz . cork require 2 lb. 7ioz . lead,Ans.

120X 74 8880 S ince 60is 12070 of that desired,150X 68 10200 60+ 120, or 5070 is the required130 X 54 7020 average400 26100( 65170 . 10 x40=400

610sha res, Ans.

( 9bbl.

400x7.50=3000 .56 100 x20=2000

640x7.25=4640 req. av.

960x6.75=6480 1 56 x20=11202000 )14120( 7.06

,bbl.

,Ans.average now .

0 III.

Art. 365.

“ E

(1

The given lbs. a nd prices make a n average of 5} ct.

Then,

3 it lb. Ans. The ba la ncing re

6 5+ 7 quires 1of the firs t

7 6 6 23“ 2 Si” ih .

,Ans.

for3of the third ; and

7 of the second for 6

of the third. This gives the required 7 ; but a s there a re 16 13,or


154 KEY TO BAY ’S NEW

Int. gr.

15 3 6 2 64 2 pw t. 16 gr , Ans

202 3 24 1pw t.

,Ans

24 6 3 24 1 Ans

9 5 112 14 8.

Ta king 3’s a nd 5 ’s to make 112gr.

,w e proceed thus

to find other answ ers2d. 2 72 68 64 60 56 52 48

3 6 15 24 33 42 51 60

34 29 24 19 14 9 4

3 2 29 30 31 32 33 34 35 36 37 ca lves,

5 68 60 52 44 36 28 20 12 4 hogs,5 3 10 17 24 3138 45 52 59 lambs.

8 7

Take 5ths of 8 a nd 5ths of 7 to make 100 or whole

8’

s a nd whole 7’

s to make 500.

( 7

If a body have a specific gravity of 2, in w ater it displaces 5 its ow n w eight ; if its sp. gr. be fi, it displaces m

9 5

2 2"

1 3

likem a nneri;121' silver.

so the crow n, 7

sp. gr.Lil , 7

17 gold.

displaceda nd thus w ith the tw o metals. Hence , the question is,If gold disp la ce, in wa ter, 71, of its own weight, a nd silver

22,of its own weight, how should these be combined so a s to

displa ce T}, of their weight? The above balancing show stheir actual w eights should combine a s 74 to 121 i . e .

,the

gold should w eigh fi t of the combined w eights. The

Whole w eight being 171} lb.

,the Weight of the gold mustbe 175 lb. XHt 10% lh .

,Ans.

HIGHER ARITHMEH a 155

INV OLUTION.

Art. 370.

Ans.

14 x 14 x Ans.

Ans .

Ans.

Ans.

Ans .

( 8 G)“= § x 5x 5 x 5 x h a m,

Ans

) s .02 x .02 x .02 .000008,Ans .

= 5 x 5 x 5 x or 5 8,

625 x 625 5190625,Ans.

) 3 .046 X .046 X .046 .000097336,Ans.

xtxtxl xtxt— n fim ,Ans

(20562 2056 x 2056 4227136, Ans.

14 ) X A118 .

Art. 371.

19 2 (10 100 x 9) 81= 361,

29 2 (20— 400 x 9) 81

841,Ans.

Ans.

402 (30 900 x 10) 100

1600, Ans.


125 8 (100 10000 x 25)625 15625

,Ans .

592 (50 2500 x 9) 81

3481,Ans.

Art. 372.

( 1 19 8 (10 =1000 x 9) x

81 729 6859,Ans

298 (20 8000 x 9) x

81) 729 24389, Ans .

4 a X 3) + 3(1X 9)Ans.

408 (20 20)8 8000 x 20) x

400) 8000 64000,Ans.

125 8 (120 1728000 x 5)x 25) + 125 1953125

,Ans .

2168 (200 8000000

x 256) 4096 10077696, Ans.

EXTRACTION OF THE SQUARE ROOT.

Art. 375.

1

25

103 309

309

( 51)

11881009 , Ans.

1

1881



V0025 ; .175 x .0025 x .0028

,Ans .

126 x 58 x 604 4414032, Ans .

1/ x

Art. 377.

V345 ; Ans.

V3"

: V§ _ a nd not—

63 hence, 9more nearly than Ans.

1/ Ans.

Ans.

( ti-Mexmr x sex s fi s s35 ; hence ,

X x 2 x31/fi =mv'

21,Ans.

EXTRACTION OFTHE CUBEROOT.

Art. 380.

( L)

4 X 300AM 2 X 7 X 30

11683


.HIGHER ARITIiMETIC. 163

55 2X300 907500

55X 5X30 8250

5X 5

915775 4578875

462453875

555 2X300 92407500

555X 5X30 83250

5X 5 X

92490775 462453875

701i(19.1393267, Ans.

1

60115859

152000

19 30 570

1

108871108871

431290001912X300=10944300191X3X30 17190

1096149932884497

10244503000

1913x30x9 5165109x9 81

10983872919885485619 etc .

,etr .

125

750046416750

25

8275 41375

5041328


V711; f t 24 11,of

2218845,Ans.

EXTRACTION OF ANY

Art. 384.

15625025, Ans.

1

*5625

64

4719476

4417929

301547736

301547736



889866667

227626024

9381865

1944016

Hence root

109721

x 2 Ans.

Ans.

1/13G7631 c be l‘t. V11 Ans.

.HIGIIER ARITHMEHC. 167

APPLICATIONS OF SQUARE AND CUBE ROOT.

1) 32 31+ ft,Ans.

922 115 ; a nd ( 69 92 )115 46 rd.

,Ans.

3602 616+ yd.,Ans .

GENERAL EXERCISES IN EVOLUTION.

Art. 389.

( 1) One side :other :1; i. e.,it is 34 timesthe other

,Ans .

Larger smaller 122 52 a nd 144 25: 5g ,

Ans.

43 503 16 cu . in . 125000X 16 64: 31250

cu. in.,Ans.

By Art. 389,4,

875 cu . in 189 cu . in . cu . of req. no.

Fourth term a nd 10) in .

,

Ans.

Norm—Evolution h a ving been presented, such examples a fford the

teacher an opportunity of showing th a t the same powers, or same roots ,

of four proportiona ls a re in proportion. This would furnish a veryelega nt solution

875 189 cube of no.

125 27

5 3 the no.,103.

As it w a s a perfect pow er,a nd the right hand

period w a s 25 , the last figure of th e root must ha ve been5 hence

,the last trial divisor w a s 4725 5 945. Then,


by the rule , w e know 5 to have been annexed to 94 , w hichmust have been “double the root already found thatportion of the root

,therefore

,must have been «1of 94, or

47 ; the entire root,475 , a nd the pow er, 475 2 225625

,

Ans. Briefly thus:4725 5 945 4} of 94 47 475 2 225625

,Ans.

( 6 504 3of its square ; the square of 504

1764 ; 42,Ans .

( 7 91252 5 1X X 3 times the cube of the small69

,Ans.

First— If the number be a n integer it w ill be thegreatest square in 1332 if w e extract the root of 1332

,to

tw o places, w e shall find the integral remainder equal tothe integral part of the root. Hence , 36, Ans .

Second— Whether the number sought be integer or

fraction,its squa re increa sed by once the number makes the

given sum . Let the diagram for square root (Ray ’

sNew

Higher) be taken for illustration. The large square A,in

the cut, is the square of the number w e seek. The rectangle B equals the number multiplied by Q ; C equals thesame a nd both equal once the number.

Now , a s in the cut,the w idth of the

rectangle being 5, w e w ant yet a smallsquare , D ,

the squa re of «5, if , to makea com “

lete square the exactroot of w hich is but the numberw e sought is the side of A,

w hich is F18 1.

less by .5 than the root just found hence 36,Ans.

The ceiling is a rectangle of a length equal to gofits w idth , a nd, if 3} of it be taken off by a line parallel tothe end

,a square w ould be left

,w hose side is the present



been 4 3 16,the former part, 60480, 300 times the

squa re of the pa rtia l root a nd 4 times 30 times tha t

root; hence, 331 of this, or must equal once thesquare of that root of it. But

,according to Non 3

,

under the Rule for Square Root, it takes more than twice

a n integer added to its square to make the square of the

next higher integer ; consequently the greatest square in201is the square of the partial root already found beforethe last figure , 4 ; a s this square is 196, a nd the sq. rootis 14

,the cube root sought must have been 144

,a nd the

pow er required 1443 9 985984,Ans.

REHABIL—In every such case where the conditions a re th a t the

root a nd the power a re integra l, the squa re of the la st figure beingtaken a w a y, lea ves 300times the squa re of an integer, 30times the

integer X by the la st figure. Now,a s the la st figure ca n never be

grea ter th a n 9, this sum ca n never be more than 300times the squ

of an integer, 270 times the integer ; the 300th pa rt of this can

never exceed the squa re of the integer by more th an 3—38, or 126, of theinteger, a nd hence the squa re of tha t integer MUSEBE the grea test squarein the quotient of this pa rtia l tria l divisor by 300. The student ca n

convince himself th a t the 300th pa rt of the whole divisor contains

no squa re of a grea ter integral root th an the root“a lready found.

”

Hence the following brief opera tion

1. 241984 4 60496,the complete divisor.

2. 60496 300

3. The root of grea test squa re in 201, is 14.

4. The root found w as 144 , a nd 144 3 is the answer.

If one cube differ from another,1inch

,in the

equal dimensions, the difference in solidity w ill be,one

corner cube,1solid inch

,a nd six blocks of the same

length a s the smaller cube,three of them being squa re,

a nd each of the other three 1 inch w ide . Without thecorner cube the 3 square blocks a nd 3 narrow blocks

,in

this case , w ill contain 1656 solid inches ; or, 1square blocka nd 1narrow block contain 552 solid inches. These being


1inch thick,if w e take one side surf a ce in ea ch of the tw o

,

w e have 552 square inches ; that is, one square a nd a n

inch -w ide strip of the same length contain,together, 552

squa re inches . This surface w ill be the same amount, ifthe inch -strip be divided into two ha lf inch strips, a ndthese he placed on adjacent sides of the square ; there isyet required a square 4 sq. inch

,at the corner

,to make

a full square,

sq. in .

,w hose side is «1inch greaterthan the length of the strip ; therefore

,the side of the square,or edge of the cube, before reduction,

is 2381» in . 4 in . 24 inches ; a nd 24 in. X 24 in . X 24

in . 13824 cu. ih .,Ans.

Conceive one of the sides of the room,or one of

the ends,to be turned dow n ( a s upon hinges) , to a level

w ith the floor . The corner to w hich the fly travels, maythen be considered the diagonally opposite corner of arectangle . In the one case the sides w ill be 46 ft. 124ft.

,or 584 ft. length , a nd 22 ft. w idth , a nd the diagonal

62+ ft. in the other,the sides w ill be 22 ft. 124 ft ,

or

344} ft. w idth, a nd 46 ft. length , a nd the diagonal

I. The answ er to the first part w ill be the same a sif w e required the number of stakes w hich ca n be driven

,

one foot apart, on a square of 10ft.

,there being just 10 of the

allow able divisions in the given 2side . If w e put dow n 11on th e 1°

base line,a nd

, perpendicularlyover these,a s vertices of equi

lateral triangles, 10 others, therew ill be

,continuing thus , 11 in 1?the first row

,10 in the second

,


11 in the third, 10 in the fourth , a nd so on. But the

height of such a tria ngle is l/1°

.86602 ; a nd in 1

ft. this is not conta ined more than once,so tha t, on a

space 1foot wide w e should have, by triangula r a rra ngement,only 11 10 21; w hile by squares w e could

have 22. The triangular a rrangement w ill not be a gainin two feet of w idth unless 2 conta in .86602 of tener than2 times ; a nd so, trying, w e find that in 6 feet there w illbe no gain by that method ; but in 7 ft. there w ill be 8strips, a nd consequently, 9 row s, by triangular a rrangement. In the remaining 3 ft. there w ill be no gain ,

a nd

hence w e shall observe the square form on tha t remainder

,having

5 row s of 11 55O 7 ft.

°

dthW1in all

,128

On 3 ft. 11 33 sta kes,Ans.

(See Fig.

REHABIL— If there were 13 rows,of only 10 in each row , the

a rra ngement would be a gain over th a t a bove. Let us see if this can

be m ade. The 13rows would require12 strips, ea ch 3ft. wide. If the rows

12 were to rea ch,from the ba se row

,a lter

10 na tely the left h a nd side,a nd from the

8 second one,a lterna tely the right h and

side,there would h ave to be

,for 10

sta kes,a t lea st 94 ba ses, ea ch equa l to

10 33ft. If so,the h a lf of

this, 33ft., would be a ba se

,a nd 3ft.,

“8 ° 4° the perpendicula r to allow a hypothenuse

to1; hence, there can be noga in by such a rra ngement.

II . The answ er to the second part is the same a s if w erequired the number of stakes w hich ca n be driven

,1ft.



this Key , p . 166 hence,

is the

gain on $1for the interval ; hence , Ans .

(See Art. 386,Remark) . The side being a hy

pothenuse, a nd the perpendicular being that of a rightangled triangle w hose smallest angle is the perpendicula r is such that its squa re is {r of the square of the

side. Hence,the perpendicular 41/3 times the side.

Therefore,the requ ired side 4 4 4 .86602 ;

a nd the wh ole boundary,3 times a s much ; hence, wehave X (12 .86602) Ans.

SERIES .

ARITHMETICAL PROGRESSION.

Art. 392. CASE I .

Ans .

145i, Ans.

.037 012 ; .012 x 9 108 ; .025

108 . Ans .

71 74 Ans.

1234} 64 X 5 324 ; 130— 324Ans .

6} 6} X 364 = 2275 ; 6} 2275

228140 Ans.

Art 393CASE II.

850 2 425 425 X 57 24225,Ans.

100 .0001 mam—u x 12345

6172506 1725, Ans.

HIGHER ARITHMETIO. 175

1x 9999 1 10000 last term ; 10000 1

10001; x 10000= 50005000,Ans.

2 X 999 1 1999 last term ; 1999 1 2000;

3320-11X 1000 1000000

,Ans .

111X 8 888 ; 999 888 111last term ; 999

111 1110; 1121-0 X 9 4995

,Ans .

x 249

last term ; x

Ans .

60 20 39 — 15 = 2§v com

mon difference ; 23} X 21 l 0g, 1st term ;

2g X 45 117 ; 117 last term ;

1273 138k ;13

2

“x 46 31789, Ans.

EXAMPLES FOR PRACTICE .

(1) (28 — 8) Ans.

54 — 8 _- 46 ; 46 Ans.

30 adding 4 to 6,w e oh

ta in 10, 14 , 18 , 22, 26, Ans.

40 36 adding 12 to 4 , w e

obta in 16, 28 , Ans.

2g, 2g, 23, 24, Ans.

42 Ans.

10; 7s+ g= 20; Ans.

500 Ans .


The years of interest on one year ’

s interest oneach 81 of the principal, a re equal to the sum of a n

a rithmetical series,1,2,3, etc.

,to 49

,w hich

,by the

rule, (l 49) X 1225 yr. Hence

,the amount

of each 81in the principal, is1. The annual interests X 50

2. The interest of 10 ct. for 1225 yr .

3. The dollar itself, 1

In a ll,

$270, Ans .

GEOMETRICAL PROGRM ION.

Art. 395.

CA“ L

( 1) —n’n ; 64 X fi n

= § p Ans .

(2

(98 =

w'

51u

'

5 5 100X “M L—T557 5:Ans.

19683 4 Y.19683 78732, Ans.

The 3d term,considered the l st

,the 9th w ould be

the 7th hence,6° 46656 16 X 46656 746496

,Ans.

The 40th term becomes the 8th,starting at the

33d hence, (Q) ?:1

25155571 ; 1024 X125185311 136H~

,Ans.

Ratio of tran . series= 2 ; 32 ; 180 - 32

5g, Ans.

Ratio of tran. series319902145 ; H—HX

“12 7 Th , Ans.

Art 396CASE II.

29 512 ; 6 x 512 last term ; 30726144— 6 Ans.


178 KEY TO RAY’S IVEW

.08 l 1( 108

1. So,using comp. int. ta ble

,w e

1

.08

1

EXAMPLES FOB PRACTICE.

(1) z:64 ; V6 = 4,Ans .

10,Ans.

Included terms 7 1000000 729 11998-5111;‘

i/msggon Ans.

No. of terms: 1 112—z- 63:

V195 : fi»; 63 X Ans .

No. of terms= 6 ; 192 mul

tiplying 6 by 2 four times, w e have 12,24

,48

,96

,Ans.

No. 0f term s : 5 ; — 4096 5 1240—

96

8 multiplyingm three times by 8 , w e have{ hp

1717 ,Ans .

No. of terms 4 ; 216 ; V216

6 ; multiplying tw ice by 6,w e have a nd

Ans .

.HIGHER ARITHMETIC’. 179

Art. 404.

(1) 9 ft. 4 in . : 112 in . ; 2 ft. 5 in . : 29 in . ; 112 in . X29 in. 3248 sq. in . 22 sq. ft. 80 sq. in .

,Ans.

5 ft. 8 in . 42 ft. X 53ih

263sq. yd.

,Ans .

48 it. X 10 ft. 480sq. ft. 69120 sq. in. 8 in .

X 8 in. 64 sq. in . 69120 sq. in . 64 sq. in . 1080,

Ans

13 ft. 3in . 159 in . 9 ft. 6 in . 114 in. 159 in .

X 114 in. X 5 9063 sq. in . 62 sq. ft. 135 sq. in .

,Ans.

22 in . 42 42 42 29 29 29

24 in . 22 24 38 15 18 25

38 in . 20 18 4 14 11 4

in . 42x20X18X4

E 60480;

246 sq. in . 1sq. ft.17864 ;

102 sq. in Ans . sq.rd. Ans.

( 8 ) Bzfl ft. X 16 it. : z40 sq. ft

,

Ans .

43rd. 65 rd. 108 rd. rd. X 27 rd. 1458

sq. rd. 9 A. 18 sq. rd.,Ans.

1OX 9 12 X I5 —1 16 X 105

rd. 1A. 59 sq. rd.

,Ans.

180 KEY TO BAY’S NEW

( 114)10 20 20 20 14 24 24 24

12 10 12 18 16 14 16 18

18 10 8 2 18 10 8 6

20x 10x 8 x 2 3200 ;

”

24 x10x8x6=11520;3m sq. rd. 1/11520 sq. rd.

56.568rd. rd. rd. 1A. sq.

rd.

,Ans .

25 ft. X 14} ft. X 2 7124 sq. ft. in sides ; 18 ft.

X 144 ft.X 2 513sq. ft. in ends ; 712} 513 12255 sq.

ft,tota l ; 551t x 3§ fi n > < 2

X ft.,233sq.

39§ sq. ft. 63; sq. ft. one half of 63; sq. ft. 31; sq.

ft. sq. it. 34gsq. ft. 664} sq. ft. 1225-L» sq. ft.

sq. ft. 1158§ sq. ft. 12837L sq. yd.

,Ans .

the other diagonal.These being at rt. angles, 5 of is the base

,a nd 5

48 it. X 27 ft. X 3 3888 sq. ft. 432 sq. yd. ;

12 ft. X 84 it. X 3 9 97 sq. ft. 33 sq. yd. ; 432 sq. yd.

33 sq. yd . 399 sq. yd . between w alls ; 48 fl}. X g2 X 3 216 sq. ft. 24 sq. yd . in the w alls ; X

399 76 ct. x 24

Ans .

The diagonal of a square w hose side 1,is

i. e.

, it exceeds the side by .41421; a nd, a s squaresa re Similar figures,

.41421 1 given excess req. Side.

.4142 1 ch . 50 ch .

50 x 50 2500 ; 2500 sq. ch . 250A, Ans.



THE CIRCLE

Art. 409.

NOTE—Observe th a t is very nea rly the same a sm .

16 X Ans. ; 22} X

Ans . ; XAns. ; 452 yd . Xm: 1420yd.

,Ans .

56 Ans. ; 1821}Ans .

- z

Ans . ; 639 It.—3fi g 113X It . Ans.

2 it. 13yd . 1it. : 40it ; 5 ft. X 5

ft.X sq. ft.

,Ans; 14} in . X 14 } in .X

sq. in .,Ans . ; 20 It. X 20 ft. X

sq. it. : 139 sq. yd . sq. ft,

Ans.

7 It. 3in . : 87 in . ; 6 yd . 1it. 4 in .: 19§ flu ; 23

ft. X 23It. sq. ft,Ans. ; 43} in . X

434} in. sq. in .: 4 sq. ft. sq.

94 ft. X 934 ft. sq. ft,Ans.

X 3 sq. ft,Ans.

The fraction is that part of the log w hich asegment of a circle , height 4 of diameter, is of the w holecircle . Let the diameter be 3 the radius

,the

distance from center to base of segment .5 ; the halfof segment (or w hole chord ) 2if 2 The area of thesegment, by No. 2

,under V

,p . 389

,is

,

1

Area of circle 9 X .7854,a nd ( 9 X

7854 ) .2918,Ans.

gof 2t/‘2

'

x1

The fraction required is the part w hich a squa re


is of the circle in which it is inscribed. The diameterbeing 1, the side of the square the a rea of thecircle .785398 ; the area of the square .5 the fraction sought .5 .785398 .6366

,Ans.

MENSURATION or SOLIDS.

Art. 410.

( 1) X 40in. =

4 50sq. in .

,Ans .

it. 2§ in . IQ in. ; 14g-in, X45 .5531 5531

°

1n. X21: sq. in .

6 sq. ft. sq. in.,Ans.

60 120 120 120 V 5760000: 2400sq. ft ;

80 60 80 100 2400x2: 4800sq. ft ;

100 60 x 40x 20x 120 5760000;

120

60 80 100: 240fi . ; 240it. X 90it. 4800 sq. ft.

26400sq. ft,Ans.

28 It. X 19 it.: 532 sq. fla ; (121)

?l

sq. ft. in one base sq. ft. X 2 sq.

ft. 532 sq. ft,Ans.

( 5 . 640X 4 X 2—2—1 500480 sq. ft. conv. surfi

,Ans. ;

( 640)2: 409600sq. ft. base ; 500480 409600 910080

squ ft,Ans .

66119. 8 in. : 800°

In . ; . 24 ft A in. : 50in.

,50in. X

2 X in. circ800 in . X in .

—5 2 1256637 06 sq. in .

,convex

surface,Ans . ; X sq. in . in

base 125663706 133517.6876 sq. in .,Ans.

184 KEY TO RAY’ S NEW

it. 2 in. : 14 in . 44} in . X Q in . 201sq. in

in ba se 20} sq. in. X 151in . 94; cu . in.

,Ans .

( 89 12 1818 x 6 x 6 x

12 12sq. in. 1of in . in . ; sq.

12 6in. x in. cu . in .

,Ans .

18

9 in . X 9 in. 81sq. in . ; 81sq. in. X 19 in .

1539 cu . in.

,Ans.

61fli. 78 in. 2 ft. 10111. 34 111. 1ft.

20in. 34 in . X 20in. X 78 in . 53040 cu . in . 30cu .

ft. 1200cu . in .

,Ans .

8 in. X 12 in . 2 48 sq. in. 48 sq. in . X 7 in.

336 cu . in.

,Ans .

2 3g 3g 2&24 2 213 “

f?en.

73; ft. Ans.

(13) g X g X 19.635 sq. in. in base ;sq. in. X 104

°

In. : 206.167 cu . in. ,Ans.

24 in. X 4 : 10 in. ; ( 16+10) 13 in . sq. in .

,Ans.

7 in . X in . ; 3 in . X9.42477795 in . ; 21.99115 9 .42478 2

15 .70796°

1n . 70796°

m . X 5 1n . 78 .5398 sq. in .

,convex

surf. , Ans. X sq. in . (1302

X sq. in . ;

7.06858 124 .0929 Sq. in .

,w hole surfa ce

,Ans .

(16 ) % 0f 1ft 415- 111 — 5-5- 111 ( logy— 1135 ” m

in low er ba se ; (4IV: 18,1Br sq. in . in upper base , mean



150 X 84 X 1§ 1700 cu . ft. ; 1700X 120bricks thousand ; 87 X Ans .

40; 100 16 40 156 ; 156 X 351 4472 cu . ft

,w hole

chimney ; 3X 3X 86= 774 cu . ft. in cavity ; 4472774 3698 cu . ft. of brick Work 3698 X 1

20

89861+ bricks,Ans .

Art. 414.

1 99 X 41X 34 bu.,Ans .

x (1,9)2X314159265

m ean x x

x181 cu . in . cu . in .

-e 231

ga l. , Ans.

11 a . 3 in. 7 a . 3 in . 92 x92 x x 133 3971.3033ga l. ; 3971.3033 31.5

bbl.,Ans .

10 fl) . 120 in .,9 fl}. 108 in .

,5 fl. 60 in . ;

14400; 11664 ; mean 120 X 10812960 14400 11664 12960 39024 ; 39024 X 5

89

780480cu . in . 780480 231 bbl .,Ans

Art. 415.

This is a frustum of a cone ; (1211) 2 X201.062 ; X 3.14159265 132.732

,mean

X X 163.363 ;

X 2; cu . in. ;

cu . in . 231 ga l. , Ans.

HIGHER ARITEMETIC'. 187

Difference of the diameters 3 in. 155of 3 m

in . ; 18 X X 34 X .0034

gal .,Ans.

Difi'

erence of the diameters 6 in . gof 6 in.

4 in. ; 3 ft. + 4 in . = 40 in . ; 40X 40X 64 x .0034

ga l. , Ans.

LUMBER MEASURE.

Art. 416.

( 2 (24 X H 300 it,Ans .

(25 X 34: 6614 ft.,Ans.

(50 XH 1587 ft ,Ans .

MEASURING GRAIN AND HAY.

Art. 419.

cu . ft ;

X .8 bu .

X .8 X 5 bu . Ans.

40x 13x 10x .3 5120 bu.

,Ans.

48000 550 8713,tons clover , Ans. 48000

450 106§ tons timothy,Ans.

MISCELLANEOUS EXERCISES .

( l I w ould gain 2 ct. more a doz en, w hich is «bof acent apiece 1» «b .

15, ct.

,Ans.

3 for a dime is 34 ct. each ; 3§ ct. ct. 24ct. cost ; 4 for a dime is 25 ct. apiece ; 23. ct. — 25 ct.ct.

,Ans.


The difference betw een 374 ct. a nd 45 ct.,or 74

ct.,is 4 of the gain at 45 ct. ; 74 ct. is 4 of 182 ct.

,the

gain per bushel at 45 ct. ; 45 ct.— 18§ ct. = 264 ct. a

bu. ,Ans .

The difference of gain is 15 ct. there must be a smany oranges a s 4 ct. is conta ined times in 15 ct . 15

3: 40 oranges, Ans.

By gaining 11ct. I receive 27 ct. more than bylosing 16 ct. ; the difference in price is 3 ct. a doz . ; therea re 27 3 9 doz .

,costing 5 ct. X 9 16 ct. 61ct.

61ct. 9 64 ct.,Ans.

I must charge 4 4 4 ct. more apiece this is4 ct. X 12 16 ct. more a doz en ; 6 ct. 16 ct. 22 ct.

,

Ans.

4 cf a dime= 18Q= -2- ct. ;

-2 Ans .

After losing 4, I have 3; spending 4 of this, therea re 4 of it left ; gof g=H,

Ans.

A’

s is 1?a s large, a nd, being 44 a s good , is w orth44 of ?: 1

20a s much a s B

’

s,Ans.

If 4 of A’

s: § ,of B

’

s, 4 of A

’

s=§ of B’

s,a nd

4 of A’

s gof B’

s hence, 3of A’

s gof g 2 of B’

s,

Ans.

After giving 161, there remain B receives 11,of fi z fi ; hence

,B receives 44

:310“ of

'

it,more than A

,Ans.

B’

s age is a, of 0’

s ; A’

s is 4 of B’

s,a nd

,there

fore, gof g= 2

9§ of 0

’

s ;295 = 24, Ans .

If 4 of m ine =3of yours, a nd all of

m ine 4 of yours ; w e both have 4 of yours ;of this 15

1, Qwill be 14 ,

Ans.


190 .KEY TO RAY’S NEW

8 it. th ere a re 4 above ; hence, 8 it. must beOf the length ; ff “ ; It

,a nd 4: 18?It ,

A118 .

4 yr. must be the difference betw een 4 a nd 136

of B’

s age ; 195 f o= l °f 4 yn = t yr 18

B’

s age= 26§ yr . ; Qof 26gyr.

- 20yr. ,A’

s age,Ans .

The difference betw een {L of B ’

s a nd g. of B’

s,must be 85 + 84 89 ; -

117 ; if B

’

s

money 12 X 89 $108 ; A’

s in of 8108 $4

$76. Ans. A, 876 ; B, $108.

If i of A’

s age of B’s, 1of A

’

s of B’

s,

a nd gor A’

s of B’

s ; difference, 31yr.,must be of

B’

s,a nd yr. X 8 28 yr . ; A

’

s= 28 yr . + 3}

yr. = 31§ yr. Ans. A,31} yr . ; B,

28 yr.

One boy w ill do it in 3X 7 hr. 21hr .

,a nd a

ma n,w orking 4} times a s fast

,in 21 hr. 4} 4a

hr.

,Ans.

4 more men w ould make 10men ; 6 men ca n do

it in da 1m a n,in 5} da. X 6= 33 da.

, a nd 10 men

in 33 da. 10 3186da . ; time saved= 5§ da.

- 3IQ6da. 2§ da.

,Ans .

A m a n a nd 2 boys= 5 boys ; if 5 boys do thew ork in 4 hr.

,1boy w ould require 4 hr. X hr.

,

a nd 1ma n or 3 boys,

of 20 hr. 6§ hr.,Ans.

The boy w orks 9§ — 33 52 hr.,or 21hr . less

than by the 1st supposition ; the m a n,95— 8 L} hr.

longer ; hence, the m a n does a s much in 1} hr .

, a s theboy in 2} hr . ; w hich a re a s 2 to 3 ; h ence, the m a n doesg, a nd the boy, g, in 8 hr. the m a n w ould do th e w holein 8 hr. 131} hr. ; the boy , in 8 hr. hr.,

Ans.


The 3men w ork 44 da. on w hat 2 men couldhave done in the required time. If 3men w ork Q da.

,

1m a n must w ork 45 da. X 3 13; da .

,a nd 2 men would

be employed 5» of da. 64 da .,Ans.

If the w ork of 9 boys that of 4 men,the w ork

of 3 boys: that of gof 4 men 14} men ; 2 men l§men= 3§ men ; 3men w ork 5 days ; 1m a n m ust w ork5 da. X 3:15 da.

,a nd 3§ men

,15 da g da .

,Ans.

The m a n does .2 a s much a s the boy , or gto theboys 3, or 5 parts in 7 parts, or 19; the boy, 4 ; the firstdoes .9 4 of the w ork, more than the boy ; 4 of 10.A. 43A. ,

A728 .

( 34 ) One m a n could do it in 4} da. X 6 26 da.,a nd

w ould do 1215in 1da . ; 6 men w ould do {fit in 2 da.

,a nd

there w ould remain fi to be done in 1gda .,or Q in 1

da.,w hich requires 10men ; 10 6 4 mon.

,Ans.

There a re in all 62 da. X 8 = 54 da. w ork. If

the 2 men w orked from th e beginning, the 10men w oulddo 55 da. X 10: 582L da . w ork ; hence , the 2 men muststay aw ay a s long after th e commencement a s it w ouldrequire them to do 582 da. w ork ; 42 2

23da.,Ans.

10 men w orking constantly,w ould do a s

much a s 7 men,a nd

,therefore

,to do 4 a s much

,m ust

w ork 116 of the time, a nd to do i}, or a s much,they must

w ork 11,rof the time, a nd rest 1 -

,2¢r 1

33,Ans.

( 37 8 men,on coming, must do that part of the

w ork,w hich 3men could do in 84 da .

,or 1m a n in 25 da .

,

a nd the 8 men in 2841 34 da . th ey

,therefore

,stay aw ay

8} 35» 5753da .

,Ans .


They,a nd those brought w ith them ,

must do in5 da. w hat 4 men could do in 7} da.

,or 1m a n in 30da. ;

there must, therefore , be 30-1—5 6 men ; 6 men— 4

men 2 men,Ans.

At 6 o ’clock the min. hand is 30 min. behindthe hr. hand ; to be 20 min . behind it must gain 30

20 10min . to be 20min . ahead,it must gain 30

20 50 min . While the hr . hand passes over 5 spaces,the min . hand traverses 60 spa ces, thus gaining 55 min.

in 60 min.

,or 1min. in HF: lfi min . ; to gain 10 min.

w ill require 1111. min. X 10 min . to gain 50min.

requires lfi min. X 50 54—ififmin.

Ans. min . past 6, a nd 5410, min. past 6.

If the min. hand is a s fa r p a st 8 a s the hr. handis past 3

,it must be 25 min. in advance . At 4 o ’clock

,it

is 20 min . behind , a nd must gain min.,

which it w ill gain in l llr min . X 45 491k min. But if

the min . hand is a s fa r behind 8 a s the hr . hand is inadvance of 3, it must be betw een 6 a nd 7, a nd a s fa r

behind 7 a s the hr . hand is in advance of 4 . Hence,

w hile the hr. hand ha s gone a certain distance , the min.

hand,w hich goes 12 times a s fast

,must have gone 35

min. (to less that certain distance ; therefore, 35 min.

less the distance 12 times the distance, a nd 35 min.

13 times the distance 35 min .-5 13 spaces passedover by hr. hand since 4 o ’clock ; 2125 X 12 32

34,min.

passed over by the min . hand .

Ans . 32145min . past 4 , a nd 4911r min past 4 .

The hr. hand is a certain space from 5,or 25

min .+ that space, from 12 ; the min . h and is half a s fa r,or

121} min . J; that space. While the hr . hand moves thatsp a ce, the min . hand m oves 121} min . i that space ; but



There a re 1. a s many persons ; hence, each one

must pa y a} a s much,thereby saving 4 31: 60 ct.

,

30 ct. ; 30 ct. X 7 Ans .

( 48 ) The 3 lb. w orth 4 ct. less per lb . w ould make adifference of 12 ct. in 8 lb.

,or 14 ct. a lh . ; 84 ct. 15ct. 10 ct. ; 10 ct. — 4 ct. 6 ct.Ans. 10ct. and 6 ct. a lb.

The cost w ould be 1ct. less a lh .,on 16 lh .

,that

is 16 ct. But the difference is of the cost of 6 lb. of

good sugar. 16 ct. is «l of 48 ct. 48 ct. 6 8 ct. f;of 8 ct. 5§ ct. ; 10 lb. at 8 ct. cost 80 ct. ; 6 lb. at 5§ ct.cost 32 ct. ; 16 lb. cost 80 ct. + 32 ct.=112 ct. ; 1lb.

cost f ; of 112 ct. 7 ct.Ans. Ingredients

,8 ct. a nd 51, ct. ; mixt. 7 ct. a lb.

A w ould have $30more than B he w ould,also

,have 1§ times more ; B w ould have ;$221} $10 325, Ans.

Total cost of w hich B pays 85 ct. ; heshould pa y 1} of 87; ct. ; 871} ct. — 85 ct. 21} ct.B ow es A

,Ans .

285 ft.

—3 ft. : 282 ft ; 282 ft.- 3 ft. 94

row s ; 285 ft. 6 ft. : 279 ft,length of each row ; in

stepping from one row to another,he steps 3 ft. betw een

each tw o row s, in all 279 ft ; 279 ft. X 94 + 279 ft

26505 ft. 5 mi . 6 rd. 6 ft,Ans.

1A. : 43560 sq. ft. ; 1070 of 43560 sq. ft.:4356

sq. ft. 43560 sq. ft.

— 4356 sq. ft. 39204 sq. ft ; 39204

90 ft. front ; $1000 Ans.

( 54 ) $1of stock costs 80 ct.

,a nd is sold for so

th a t th e gain is 30 ct. on 80 ct., or a of the cost,

Ans.


10070 1570 1570 of 85 % 12270 ;8570 12270 _ 97170 10070 — 972% loss

,Ans.

10070 870 10870 12470 of 108%10870 470 of 1215-70 121570

116+§7o 1.1664 $1000,Ans .

15670 of $2500 immediate perpetuity of

$20 a year is w orth $20 .06 to w hich add1st payment, $20, in advance

,a nd w e have

820X 12 $240 gain bythe latter,Ans.

$1500—z w orth .

$300 Present w orth of balance Xdue in 6 months, Ans .

The true rate is 370 , quarterly. Amount of $1for 4 periods at 370 , is $1.12550881; $1.12550881

.12550881_ 12150509050501673 , Ans .

The length of the 2571353 cu . in. :

137 in . ; surface of 1 face (137 in.) 2— 18769 sq. in .

,

Ans.

The solidity: the cube of the side,: the square

of the side multiplied by the side . The surfaces: thesquare of the side multiplied by 6 if these numbers a re

equal,the side must equal 6 Ans.

i of 20 ft. 5 ft,the side of the square ;

(5 ft)2 25 sq. ft

,the surface of the square the surface

of the circle (Art.

sq. ft. nearly ; from this, subtracting 25 sq. ft. givessq. ft. nearly, more in the circle, Ans.

196 KEY TO RAY ’S’ NEW

This is the same a s a n annuity for 20 paymentsat 2170 , the present value being $1200; present value ofa n immediate perpetuity of $1per quarter, at 247“$40 present value of same perpetuity, deferred 20 quarters

, $40 $40

present value of annuity of $1 $1200

per quarter a yr .,Ans .

( 64 ) 570 on of $3000: $50; by ta ble (Art.

$50X Ans.

Present w orth of $200 in 1mo.

,at 9 0 ,

31985 11; of $200 due in 2 mo .,at of

$200 due in 3 mo.

,at of $200 due in 4

mo.

,at of $200 due in 5 mo.

,at 970 ,

l92.771 total,

Ans.

10070 — 25% 7570 ; 25% of 757, 18270 ;7570 56i $675 ; 170 $12, a nd 10070$1200, Ans.

Int. of $1for 63 da . 021; the true discountwould be -hif 11, of the principal= $4 80, the

n-

n

principal mustbe $4 80 441X 1021000 Ans.

I lost 2070 of $10000 $2000, a nd have left$8000, on w hich 1870 is $1440 $2000 $1440 $560, to

be gained on money borrow ed ; 1870 4% 1470 gain$5601s 14 % of $560 .14= $4000, Ans.

The quantity w ill be equal to the solidity of abody 38 ft. by 52 ft.

,a nd1in . thick 456 in. X 624 in .X

iin . 71136 cu. in . g 71136 cu . in 231 ga l.,

which 31} 76—1 bbl.,Ans.

The sums w hich w ould amount to $1in 15, 13,


198 KEY 10 BAY’S NEW

answ ers ; so, corresponding to the next there could be 29a nsw ers. In a ll

,w e could have answ ers,

288 a nsw ers , Ans.

A’

s ga in is = 32% B’

s 4070 0’

s

“3, or 48 70 C gains 870 more than B,by contributing

his stock 2 mon. longer; hence, the gain is 470 a mon. ;

3275 495 8 mon. A’

s,Ans . ; 4070 475 10mon.

B’

s,Ans. 4895 495 12 mon.

,C

’

s,Ans .

The discount is 2070 of the face,or 22470 of the

20proceeds ; hence the proceeds27 ;

gof the face, a ndthe discount g. of the face 1

2

1

63?yr. g

yr. 200da . ,Ans .

A receives more thanB ; since be contributed $7.83ir more than B,

his invest7.83§ment must h a ve been28 20 1

5,of his gain 1

5,of $57.90

A,Ans.

, 1-55of $29.70 B

,Ans.

Suppose he borrow s $1 at the end of 6 men . itamounts to of w hich he pays 3 ct. for interest

,leaving w hich,in the next 6 mon.

,w ill amount

to from w hich, on paying 3 ct. int.

,h e w ill

have remaining thus clearing, in th e year,

on each $1borrow ed .06368181

Ans.

Discount for 4 yr. at 470 1249 of debt ;

for 4 yr . at 670 , it is ifs of debt ; f r— fi =fiflg ;Of debt 825 uh : $i a nd the debt is A718 .

HIGHER ARITHME’TI0. 199

$1— 25 ct . = 75 ct. ; $2700 .75 870of $3600 income ; $288 .10 par value$2880X .96 Ans.

6970 is 2 of 9270 I w ill,therefore

,receive it a s

much stock at 9270 , a s at 6970 770 on 4, is the same a s

5470 on the w hole ; so that I gain 5470 570 _ $70 ,annually,on $5200; 470 of $5200: $13 ; $13 a year is

w orth $13 .06_ $216.66g, Ans.

2470 of $1500 $1500

received ; the costw ith int. for 3 mon . ; am

’

t of $1 for 3mon. at 670312717 39 Ans.

The 9th term must be multiplied by the ratio 4times, to give the 13th term hence

,11160261 137781

81,is the 4th pow er of the ratio ; 13“81 1/9 3 ;the 9th term must be divided by the ratio 5 times, or bythe 5th pow er of the ratio

,to give the 4th term ;

243 137781 243 567,Ans.

$19487 .171 $13310 1.4641,ratio of increase

for 4 yr. ; by reference to the table (Art. this isfound

o

opposite 4 yr. under 1070 ,the required rate ; am ’

t

of $1 for 3 yr .

,at 1070 $13310

$10000. Ans. Ca pital, $10000 rate

,1070 .

3min . 28 sec. 60min . 4 in . in. , cir

cum ference ; in. 2

in .

,Ans.

( 84 ) Radius of w hole circle 16 in. ra dius of aperture 3 in. area of stone face 256 X 9 X

= 247 X Each is to have 241X


a nd hence this amount aperture circle left by secondm a n ; twice that amount aperture circle first left.

Thus(3

3 111) X 2d circle left.(3

2 x 1st ( 6

But if w e divide the area of a ny circle by theresult is the square of the radius. Hence

,131

radius 2d left Ans.

Ans.; 16. Ans.

The supposed cost the actual cost $300;

2070 or i, of the supposed cost i of the actual cost$60 ; the difference betw een gaining i of the cost a nd

losing 4 of the cost $60, is of the cost $60 hence,

of the cost $60 gof the cost $240 ; 4 is

$120, a nd the cost $120X 5 $600, Ans.

It goes dow n 1m i. in 1354: B’shr.,

a nd up 1mi .in h r. ; 1

16—

345

2715hr. longer in going up 1mi .than in going down the same distance ; 22} + 2

15: 585

mi .,Ans.

( 87 The actual cost w a s 199 of the supposed cost ;therefore

,the selling rate per cent w a s 196 a s much

,a nd

the difference betw een the tw o selling rates is 110 of thesupposed selling rate

,w hich is 4 of “

1203hence , 1570 is 4

of the actual selling rate 1570 X 9 13570 , selling rate ;1357flZ-110070 Ans.

Each $1in the face of the check cost me 55 ct. ;I receive for it $1 in bonds ; 770 of $1§113ct. ; h ence, I receive yearly 11§ ct. on 55 a t w h ic his 213

7370 , Ans.



parts ; one, a n investment,a nd the other

,870 of the in

vestment hence,the investment w a s +33of the 96 ct.

,or

88; ct. Ha d no money been reserved,the loss w ou ld

have been 114 ct. on each $1of the original value ; thatis,he w ould have lost fi of the w hole had he left th e $18

in the proceeds . But the loss on that $18 w ould havebeen simply the commission, $ 733 of $18, or $14. Hence

,had he kept no money, the w hole loss w ould have been$32 $1§ a nd a s th is w ould have been 1} of thevalue of the w h isky

,that must have been 9 times $334,

or $300, Ans.

( 92 ) 1. The gain being 3; minutes each day,the

clock indicates a period of w hen the true periodis 1440. Hence

,since on the 29th the indicated period

w a s 7 days,min. 1440min . 7 days true lapse.

7 X 1440 da .

6 da. 23 hr . 35 min . sec. ; hence,35 min. see. past 11A. M.

,29th

,Ans .

2. To show the sam e time on the clock face,it must

gain a w hole ha lf -da y, or, 720 min. h ence,

gain in 1da y tim e required req. gain req. time .

m in . 1440m in . 720min .

720 1440 “40da . da.

7

da . a fter Feb. 22d (noon) , lea p -year,is 1?da . a fter

the noon of Sept. 14th ; hence,Sept . 15th, 84, m in . past 5 A. M.

,

-Ans .

( 93 ) S ince 144 ba se X 4 h eight, or, 111 this case,

ba se X ba se,the a rea is the squa re of 4 the height, And


a s the triangles formed a re simila r,w e have 62 36

sq. in .

,a nd so

,by squaring half each alt. , 42} sq. in., 49

sq. in .

,521

25sq. in .

,Ans .

1. The case is one of Arith . Prog . (Art.

a nd a s first term is 16,second 48

,com . diff . 32

,the 5th

term is

(144

;16) x 5= 400 a ,

Ans.

2. Observe that it consta ntly increases ; in the latterhalf of a ny period , falling farther than in the first half.Hence

,in the latter h alf of the 5th second it falls more

than the ha lf of 144 ft. The figure w ill illustrate this.

The length being tw ice the breadth , in a ny one of thetriangles, the addition made for each stepfrom the vertex tow ard the base is a

trapez oid , a nd the low er ha lf of the altitude

,a s KN

,in a ny case , bounds a trap

ezoid smaller th an ()M,but greater than

its half. The w hole distances a re a s the

squares of the times,a s the whole a rea s

a re proportional to the squares of thealtitudes

,in the former problem .

Fig' 5°

16 ft. X =324 ft.

,Ans . ; so 4

2 X [6 ft.: 256 ft,Ans.

REMARK— Compound interest presents a ca se of consta nt increase,

in some respects like these.

As in the tw o preceding problems,the distances

a re a s the squares of the times ; hence, 65 X 6§ = 42imiles,Ans.

( 96 ) The side being to the front a s 2 to 3, the w holebody is now 3 of a squa re, w hose side is the presentw idth . The increase w ill be 16 men at the corner

, 4


ranks at the side,a nd 4 ranks at the end. The former 4

being at of the latter 4 , the side rank a dded to ’ an end

rank w ill make 10 ranks of the same length a s the

pr esent w idth . Hence

,w ithout the 16 at the corner

,

the new body w ill consist of 2304 men,a nd w ill equal 7}

of a square body of the present w idth , 10 ranks of

that length .

For convenience,let each m a n occupy the space

of a square yard . Then, 3} of a square 10strips of the

same length , 1yard w ide,w ill m ake 2304 sq. yd. ; or,taking fi' of this space

,one such square 2

311of such a

strip: 1536 sq. yd. ; that is, 1 such a square a nd two

strips 130 yd. w ide,make 1536 sq. yd . If these strips be

put on tw o adjacent sides of the square, w e shall requireyet a corner square of sq. yd . to make one fullsquare 1536 1

13-Q sq. yd . the side of this1411 yd . this being 1

951 yd . longer than the present

w idth,the required w idth: 36 yd. Hence

,there a re

36 men in a n end rank,54 in a s ide rank

,a nd in all

54 X 36 1944 men,Ans.

If he had cut off four such pieces,he w ould have

left a square piece of cheese, the largest possible. Hence,

the four segments w eighing 12 lb.

,the w hole cheese is

to 12 lh .

,a s a circle to the four segm ents left by cutting

out the largest square . The diagonal of that square w illbe the diameter

,a nd it w ill be a hypothenuse in a r ight

angled triangle of equal ba se a nd perpendicular ; hence ,for each one in the diagonal there a re 1/Q of 1, or V3in the side . Hence

,for ea ch 1 inch in the diagonal

,

there is or 7} squa re inch in the square ; but foreach 1 inch in the dia m eter, there a re 344 —15 or

.785398 sq. in. in the circle, a nd the four segments



the length 15 rd. Suppose the road marked out,a nd

suppose also that w e had four suchlots disposed in the form of a square(a s the pupil ca n arrange four Eclectic Third Readers in the form of

a square inclosing a hollow square) ,the end of each,touching the longer

side of the next ; the side of thew hole square w ould be equal to the Fig. 7

sum of the length a nd breadth of each piece , —ih thiscase

,15 rd. 9 rd.

,or 24 rd. The required roa d in one

,

joining the like rea d in another,there w ill be a square

frame around the w hole four . The area of the largesquare is 576 sq. rd. one such road being i of 135 sq. rd.

,the w hole road-fram e contains 135 sq. rd.,a nd incloses

441 sq. rd.,a square w hose side is 21rd. This differs

from 24 rd. by tw ice the w idth of the road,a nd therefore

that w idth: 5 of (24 21) rd. L} rd. 24; ft.

,Ans.

Three equal circles,of a radius 1

,w ill be to

a large circle just touching them a s the required grassareas to the given lot. Let those of the figure be threesuch circles

,a nd let their centers be joined , forming a n

equilateral triangle, w hose side 2.

From 0, th e lines to F,

H,a ndNdi

vide the triangle into 3equal partsthat is,HON is a]; of HFN having

the same ba se,its altitude

,there

fore,must be 1} of FA. Hence

,

FN being 2,a nd AN

,1,FA

1 a nd

13of this 2 :F0 ; also 1,or

Fig’ 8 '

OF FP OP Now,by proportion,

the glven radius being 74} rd.,

m GHER ARITHMETIC. 207

1 rd. rd.

And,

rd. X 2 rd.,dista nce

betw een their centers,Ans.

Again:the triangle [Art 385 , 4 ,Rem ]is composed of thetriangular space w ithin of each small circle ; hence,FHN— Qa small circle:the triangula r space. One smallcircle (ra d. X 621125 2

X 4 of

thisThe side of the equilateral triangle is a nd

the half baseArea X

Ans.

Consider this board a s a trapez oidal piece cut

off from a triangle. The area is (7+ 17) X 30: 720

sq. in . The required area to be cut off below,is 360 sq.

in . Let the figure ABCD represent the face of the

board,APB the triangle from w hich it w a s cut. The

w hole triangle is sim ilar to HBO or to the piece PDO.

[See Art. 389,Rem . 2

,a nd Art. 231

,Prob. Hence

,

a s the piece HBO h a s a base 17 7,or 10

,a nd a height

6 times this,so the w hole triangle h a s a height

of 6 times 17,or 102 inches ; the area of it: %

of 102 X 17: 867 sq. in. Now,th e problem is

the same a s to cut off 360 sq. in. from the tri D 0

angle, by a line such a s ON a nd the tria ngle

lef t w ill have 507 sq. in . Hence,the height X

4 the base: 507 ; or,the height X th e w hole

base 1014 . But the base is of the height ;hence,the height X by 3of itself makes 1014

,Fis. 9.

and the height X by the whole of itself m akes 6084 ;


hence,the height 1/6084 78 in. Then

,PA P0

,

or 102— 78 24 in. 2 ft,Ans.

Four equal circles , of radius 1,w ill be to a

large circle inclosing a nd touching them a s the requ iredpieces to the given plate. Let

DO,in the figure , be 1; then 00

a nd CP

Area of' the large circle is

2 4142135624 2 XThat of small one is 31415926535hence

,large circle is to small one

,

a s 1,a nd

2 4142135624 2 1 $67 w orth of 1circle piece ;1 $67 Ans.

The square,formed by joining the centers, includes 1

of each circle ; hence , the p a rts of the sma ll circles,

a round the square,m ake three whole circles. Hence

,if

the area of the square a nd 3 times the area of one circlebe taken from the large circle, the remainder is the areaof the four outer portions .

Area of the w hole figure 2 4142135624 2 X18 31054383724 the square 3 circles

4 3 times this,taken

from large circle, leaves Then,

$67 Ans.

( 103) The diameter of the given ball is to the distance of its extreme point from the corner , a s the requiredsmaller diameter is to the distance of the nearest pointof the given ball from th e corner.

If the 12- inch ball w ere cut out of a cube,the

length of the diagonal cut off at each end to make thediameter, w ould be the same a s the distance from nearest


210 KEY TO RAY’

S NEW

sq. in. If the strip 141 of 3 in. w ide

,be dividedinto two, each 3

34 in. w ide, and pla ced on adjacent sides

of the square, there w ill be wanting a square of (33302,

or Lg—i—fi sq. in. , at the corner to make a complete squareof sq. in. ; the side of this is 224 ih . , a nd,

tak ing away the added 333 in.

, there a re left 18 in.,which

a re 6 in. less tha n the thickness of the log. Hence , theca pacity(2 x 24 — 6) x (24— 6) x (24 — 3) + 231: 68T

8T ga l. , Ans.

The whole vessel may be considered a s a frustumcut off from a cone , a section of wh ich is shown in the figure.

AH: 9, DC: 4%, EF: IO, and by prop. ,

3 EO being tw ice EF,E0: 20

, a nd so find

OB The mouth of the vessel , or

P ba se of the cone , : 9 2 X . 7854 , and thewhole cone would hold 81X iX X339 _ 135 X The sma ll cone cut off

being 4 of the large one, the capacity of thevessel is of the la rge cone , or fig-i XHence

,the water in the vessel is X

Let NP or NE be the radius of

the largest ball which could be put in thecone . It is the a ltitude of each of three triangles whichmake up the triangle ABO,

° whose a rea 44} X 20 90.

Hence , dividing double of this by the sum of the ba ses,we shall have 180 (41 9) the ra dius NP ; a ndthe solidity of that la rgest ball : 1

112159X Now ,

the wa ter in the vessel,

a cone DOO, would make a cone(93124 5 14351) X a nd the problem 18 the same a s to finda ball which could be put in a cone a nd be just covered by

this amount of wa ter, if such ball do not extend below DC.

But, the cones being similar, the wa ter which would coverthe la rgest ball is to the wa ter covering the required ball a sthe la rgest ball is to the required ball ; the water covering the largest is (135 — 1

172153) X hence , 9

192953 X

Fig. 13.


X cube of large diameter cube ) f

required one [Art. 389,

(3593 cube of req.diam. , which therefore . equals V5832 x 55 1348 ininches, 18 X 3442634 inches, Ans.


McMa ster’

s United Sta tes Histories

BY JOHN BACH McMASTER

Professor of America n History in the University of Pennsylvania .

PRIMARYHISTORYOFTHEUNITED STATES. Cloth , rzmo,

254 pages . With maps a nd illustra tions

SCHOOL HISTORYOFTHEUNITED STATES . Ha lf lea ther,rzmo, 519 pages. With ma ps and illustra tions

This series is ma rked by many origina l and superior fea tures which

will commend it a like to tea chers , students , a nd genera l rea ders . The

na rra tives form a word-picture of the grea t events a nd scenes of America n

history , told in such a w a y a s to a w a ken enthusia sm in the study a nd

ma ke a n indelible im ression on the mind.

The P rima ryHistory conta ins work for one school yea r, a nd givesa good genera l know ledge of so much of our history a s every America n

should lea rn ; while for those w ho a re to pursue the study further,it will la y a thorough founda tion for subsequent work. It is short, a nd

lea ves unnoticed such questions a s a re beyond the understa nding of

children ; in a simple a nd interesting style it a ffords a vigorous na rra tiveof events a nd a n a ccura te portra ya l of the da ily life a nd customs of the

different periods ; a nd it is w ell proportioned, touching on a ll ma tters of

rea l importa nce for the elementa ry study of the founding a nd building ofour country . Our history is grouped a bout a few centra l idea s

, w hich

a re ea sily comprehended by children. The illustra tions , w hich a re

numerous a nd a ttra ctive, a re historica lly a uthentic, a nd show w ell-knownscenes a nd incidents a nd the progress of civiliz a tion. The m a ps a re

rema rka bly clea r a nd w ell executed, a nd give the loca tion of everyimporta nt pla ce mentioned in the text.

In the S chool History from the beginning the a ttention of thestudent is directed to ca uses a nd results , a nd he is thus encouraged to

follow the best methods of studying history a s a connected grow th of

idea s a nd institutions , a nd not a ba re compendium of fa cts a nd da tes.

Specia l prominence is given to the socia l , industria l, a nd economicdevelopment of the country , to the domestic life a nd institutions of the

people , a nd to such topics a s the grow th of inventions , the highw a ys oftra vel a nd commerce, a nd the progress of the people in a rt, science , a nd

litera ture. The numerous ma ps give vivid impressions of the ea rlyvoyages , explora tions , a nd settlements , of the chief milita ry campa igns,of the territoria l growth of the country, a nd of its popula tion a t different

periods, while the pictures on a lmost every page illustra te diflerea t

pha ses in the civil and domestic life of the people.

Copies will 6:m t, prepa id, on receipt of ”reprise by the Publisher}:America n Book Compa ny

New York Cincinnati

Overton’

s Applied Physiology

BY FRANK OVERTON, A.M M.D.

La te House Surgeon in the City Hospita l , New York City.

OVERTON'

S APPLIED PHYSIOLOGY— Prima ry

OVERTON’

S APPLIED PHYSIOLOGY— Intermediate

OVERTON'S APPLIED PHYSIOLOGY—Advanced

These books form a complete series for the study ofphysiology in a ll gra des . The Prima ry Book follow s a

na tura l order of trea tment a nd presents the elementa ryfa cts a nd principles of physiology in so simple a w a ya s to bring them w ithin the comprehension of children.

The Intermedia te Book is designed to be a completeelementa ry text-book in itself a nd a n introduction to

the a dva nced study of a na tomy a nd physiology. The

Adva nced Physiology is designed to meet the requirements of tea chers a nd schools for a text-book of Physiologya nd Hygiene forHigher Gra des, which should combine thela test results of study a nd resea rch in biologica l, medica l ,a nd chemica l science a nd the best peda gogica l methods ofscience tea ching. The book represents a new a nd ra dica ldepa rture from the old-time methods pursued in the

tea ching of physiology.

The fundamenta l principle of the series is tha t thestudy of a na tomy a nd physiology should be the study ofthe cell

,from the most elementa ry structure in orga nic

life,to its highest a nd most complex form s in the huma n

body. The effects of a lcohol a nd other stimula nts a nd

na rcotics a re trea ted very fully in ea ch book of the series .Illustra tions a nd outline diagrams a re insertedwhereverneeded to expla in the text.

Copies of Overton’:P hysiologic:will be sent, prepa id, to a ny address on

receipt of the price.

America n Book Compa nyCincinnati Chm

Webster’s School Dictiona riesREVISED EDITIONS

WEBSTER’S SCHOOL DICTIONARIES in their revised formconstitute a progressive series. ca refully gra ded a nd especia llya da pted for Prim a ry Schools , Common Schools , High Schools ,

Aca demics , a nd priva te students. These Dictiona ries h a ve a ll

been thorough ly revised, entirely reset, a nd m a de to conform in a ll

essentia l respects to th a t grea t sta nda rd a uthority in EnglishWebster

'

s Interna tiona l Dictiona ry .

WEBSTER’S PRIMARY SCHOOL DICTIONARYConta ining over w ords a nd mea nings , w ith over 400

illustra tions .

WEBSTER’S COMMON SCHOOL DICTIONARYO O O O

Conta rm ng over w ords a nd mea nings , w1th over 500

illustra tions .

WEBSTER’S HIGH SCHOOL DICTIONARYConta ining a bout w ords a nd definitions , a nd a n a ppendix

giving a pronouncing voca bula ry of Biblica l , Cla ssica l , Mythologica l

, Historica l, a nd Geogra phica l proper nam es , w ith over 800

illustra tions.

WEBSTER’S ACADEMIC DICTIONARYCloth , Indexed,

Ha lf Ca lf , Indexed,

Abridged directly from the Interna tiona l Dictiona ry , a nd givingthe orthogra ph y , pronuncia tions , definitions , a nd synonyms of the

la rge voca bula ry of w ords in common use , w ith a n a ppendix con

ta ining va rious useful ta bles , w ith over 800 illustra tions.

SPECIAL EDITIONSWebster

’s Countinghouse Dictiona ry Sheep , Indexed,

Webster’s Condensed Dictiona ry Cloth , Indexed,

The S ame Ha lf Ca lf , Indexed,

Webster’s Ha ndy Dictiona ry 15

Webster’s PocketDictiona ry. Cloth 57

The S ame. Roa n Flexible 69

The S ame. Roa n Tucks 78

The Same. Morocco, Indexed 90

Webster’s Pra ctica l Dictiona ry 80

Copier of a ny of Webster'

s Dictiona ries° °will be sent, prepa id, to a nyaddress on receiptof theprice éy the P ublishers:

AMERICAN BOOK COMPANYNEW YORK CINCINNATI CHICAGO

Documents

› en › download › KeytoRaysNewHig… · P REFACE. IN the p repa ra tio n of the follo w i ng page s, aim the chie f ha s bee n to se rve the teache r. Acco rdi ngly, ag th ro