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1
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2
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5
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pam lUlItr * fin's ilat^tmiitkal .Strtts.
SOLUTIONS
OF THE MOBE DIFFICULT EXERCISES
AMD
EXAMINATION PAPERS
IN THE
Canadian Editi/)r?
or
HAMBLIN SMITH'S ARITHMETIC;
BT
THOMAS KIRKLAND, M.A.,
Science Master, Normal School, Toronto,
AND
WILLIAM SCOTT, B.A.,
Head Master of the Provincial Model School. Toronto.
TORONTO :
ADAM MILLER & CO.,
1879.
U44
PREFACE.
This work has been prepared for the use of teachers
and private students. With the multiplicity of duties
ordinarily devolving on the teacher, time cannot always
be liad to solve all questions that may be presented by
the pupil. Hence a work suc)i as this becomes a great
convenience, if not an actual necessity.
It has not been thought best to solve such exercises
as are comparatively easy or merclv mechanical llcnco
those under the Simple Kules, many of those in Frac-
tions, Extraction of Roots, the Compound Rules
Interest, etc., are omitted. The Examination Papers
have all been solved.
The solutions have been given with striet reference to '
the Unitary Method, thus showing its applicabiiitv to
questions of every variety and every degree of difficulty
They do not exhibit all the calculations at large, biit
they always furnish results which serve to verify the
operations at the successive stages of the process In
tins way all that is necessary has been brouglit within
a narrow compass, and the connection of the different
parts of each solution will be more readily perceived
It has not been the aim of the Authors to make a
mere Key, but to exhibit the best and neatest mode of
working Arithmetical Exercises. Not only are neatness
and method encouraged by the habit of arranging figures
m their exact places, but the accuracy of the answer is
best secured by the same means.
Indications of any errors or obscurities will be thank-
fully received.
Toronto, May, 1879,
\
CONTENTS.
PAOE.
Examination Papers on the Simple Roles i
Examination Papebs on Mkasures and Multiples.. .. 5
HiGUK.ST Common Factob u
ScinUACTION OF FUACTIONS , U
COMPLKX FuaCTIONS ,f.
MiscKLi.A.NKous Examples in Fkactions ii
EXA5IINATI0N PapKHS ON VoLOAU FuACTIONS ]4
EXAMI.NATION PaPKRS ON DliCIMAL FRACTIONS .'. 21
Examination Papi:ks ON Compound Rules 25
PlUCTICE ,.
n "'^
PkOULEMS involving FRACTIONB oj
Complex Problems „„
Problems relating to work an
Examination Papers „„
Simple Interest ' .„
Partial Payments ^.
Compound Interest .-
Discount '4,.
4y
Examination Papers on Tntebesx and Discount 51
Equation op Pavments .. ., «.-
Equation of accounts.* k-
Commission and Brokerage .... " eo
Insurance ,„
Taxes ^^
Duties
Eaxmination Papers on Percentage .. .' **,.** 61
Profit and Loss * ^ ' * ' * „„
Stocks and Shares «(.
Examination Papers on Profit and Loss, and Stocks . . 74
Division into Proportional Parts gi
Partnership • • • .
00
Vh
CONTENTS.
lARTNEnSHIP SkTTLKMKNTS o/
» •... Bj
Allioation Q.
Exchange ' g-
Examination Papeus on Exchange, Pabtnkuship, &c.. 96
""■" »0»
1 BopoanoN ....
„ 1U4
OIMPLE riJOl'OUTION ,nr
. , „ 1"5
compound puopoktion ^07
Measurkment of Area ' " , <,q
Measdkement of Solii>ity ' II q
Miscellaneous Examination Papers * 115
APPENDIX.
In'erebt .^
Discount * '* ,o«r
. loo
Annuities ,00
/
SOLUTIONS
)F THE MORE DIFFICULT EXAMPLES IN THE CANADIAN EDITION
OF
HAMBLIN SMITH'S ARITHMETIC.
a.
EXAMINATION PAPERS.
VI. — Page 41.
76894764
(112)(7)(56)
688268278
4806106224
8612212448
8670844882024
See art. 50.
8. (8876 -h 5684) x (8876- 5684) - 7859 = 4816
and rem. 576.
7859-576 = 7288.
6. If the sum of two numbers is added to the differ-
ence of the two numbers, the result is equal to twice the
greater number.
.". the greater number =
2x43 814-2x8858
= 7684.
VII.— Page 41.
- ^ X ,, 8 X $23 + 7 X $89
1. Cost of 1 a.= — - — ^ -— « $269.
« „,. 18 X 76
2. Time for 19 men = — ^^ da. = 72 da.
2 SOLUTIONS IIAMBWN HMITU's ABITHMETIO.
•<■ Number Of ouch =. ^-^^ o«o
•8.60 + 21.60 ==262.
i. Number = 876x789 + 802 = 200287.
rj. 19+ 17 + 16 = 61.
• ^""^ received by first out of $61 = |19.
$86700 = ^5-^^^^-^ ^1^
61
= 700x$19
= $18800.
«
«i
second - «i700x^$l7
61
=-- $11900.
•^ thi-d ^ 85700X |15
61
= $10600.
Vlll.-Page 42.
last irsir c'^rrr ^"'" "'" '^«
Anth. Co.p. «f 7846 = 100OO-784« = 21.4. Art. 4«.
8479 = 10000-8479=6521
Difference =4867
Number=a(888+48C7).10}^1225]x4
2.61bs.tea=15,b.eoffee=^«,b,3„g„.
• 7Kii^« * 16x15x8
..751bs.tea=— ^— lb,, ^g^^^gon^
8. Number = 18675 + (45209 -27645) =81230.
BOUITIONH HAMBMN SMITH'b ARITHMETIC.
8
4. 9 times the value of a saddle — !8>2<)1 ;
2«1
.*. value of saddle
$-y- = $29
and value of horsd = 8 x |)20 = $282.
6. Cost of cattle per head = 118 + $2 :-- $20.
XT UK I* «4<^^
Number bought =
20
820.
«
sold =«J^»-.= 200.
Hence the difference, 120 cattle, must sell for
(10400 — $8000 -f- $800) = $3600 ;
$8000
.'. selhhg price of 1 head
120^
= $30.
IX. — Fage 42.
1. See art. 61.
2. Number =- 99995 x 99995 = 9999000025.
3. Share of youngest son = $1789.
second " = 5 x $1789.
eldest " = 15 X $1789.
.-. value of property = $(15 x 1789 + 5 x 1789 + 1789>
= $(21x1789)
= $37569.
4 17694
4. Number of steps = 5 x -V-= 11796.
K T J v. ;j 7770 X $100 ^„.^^^
5. Indebtedness = ^ — = $21000.
87
\ -. z'^ 1998 X $100 ^^,_-
Sum due creditor = -~- — = $5400.
\ 87
Paper X. — Page 43.
1. See art. 47.
2. Use (1728) (144) (12) as the multiplier and mul-
tiply by 12 ; then multiply this product by 12 and the
new product by 12- a^A h,^ «,
getter ae i„ art. so. '' ""''»' P'o-J^^ to-
1
t8
4
t2
16
t68
64
T87a
266
t488
< 1024
t79o2
4096
T1808
16884
T67282
655»6
T868928
262144
T475712
1048576
T7902848
4194304
4.
2796203
786464 I 3457
78 ) 6464
78
Quotient ^766543
See art. 51.
^a..a(tor = quotient.
SOLUnONS HAMBLIN SMITH'S ARITHMETIC.
^ 6 times remainder = divisor.
.-. 43 times remainder =516 ;
• J 616 ,^
.-. remainder = —- = 12.
.'. Dividend^ 12 + 72x432=== 81116.
Highest Common Factor.
Examples (xxiv). Page 46.
The following rule will be found much easier in ptac-
tic6 than the one given in the text book. »
Divide all the given numbers by the least of them,
and bring down the remainders.
2. Divide the first divisor and all of the first remain-
ders by the lea^t of them, and bring down the remain-
ders.
3, Proceed in this manner until a remaindsr is found
that will divide all the other remainders, and the divisor
last used, and this will be the highest common factor
required.
3. ■ 365, 511, 803.
365,
146,
73.
We divide by 365, writing down the remainders 146
and 73. 73 will divide the first divisor, 365, and the
other remainders, and is therefore the H. C. F.
4.
6.
232, 290, 493.
232,
492,
58,
1476
29.
1763.
492,
287.
205,
287.
205,
82-
41,
82.
H. C. F. is 29.
H. C. F. is 41.
6
6.
SOLUTIONS HAMBLIN SMITH'S ARITHMETIC.
I — Page 49. \
1. Number = CL C M nf iq ir
{^. ^. M, ot 13, 15 and 17) 4- 12
= 3315 + 12 = 0327 ^ ^ '^•
2. L. C. M. of 33, 27 and 30 = 2070.
Number of times = If^^^SO __
2970 - '*^-
> 3. Length of rail = H. C. F of 2Snoq f, ^ , „r,.
ft. = 11 ft. '^. i^. 01 zdUI3 ft. and 17765
Number of rails = 6 x AiL^52?_^ + 2 x 17765
= 44496.
4. Since H. C. F. of 210 and 330 ^ 30 • ii
lutions of small wheel — 7 v.., w .' •' ^^ ^^^°-
'^^ ^^eel _ 7 revolutions of large one.
o. Ihe prime factors of 2772 — 9 9 q q r, .
The req,m.ed „„,nbe.. must be dWisiMeW 12 "^ ""
their L. C. M. 2 X 2 X 8 X g''™ ^^^ '2. andiiave
.". one number = 12 x 3 = 3q.
*' a second = 12 x 7 = 84*
"a third =12x11 =.132.
II.
2. We must here find the 3 smallest and also the 3
Ia.|est numbers that will exactly divide 600. '
The prime factors of 600 ==2, 2, 2, 3, 5 and 5.
.. *^^e3«oinllcstbagsmustlioldlbu..2bu. or3bu
- • >-. .-.. 01 5, ^^ ama Vo =. 22 x 75 = 1(J50
•*• smallest suai = !|5lG0Q,
SOLUTIONS HAMBLIN SMITH S AKITHMETIC.
4. Time required by first horse to go once round
5280
440
min. = 12 min.
min. = 20 min.
Time required by second horse to go once round
5280
= o ^o • mm. = 15 mm.
Time required by third horse to go once round
5280
"204"
Time required = L. C. M. of 12 min., 15 min.,
and 20 min. = 60 min.
5. Number = (L. C. M. of G75,- 1050, and 43G8) + 82
= 982800 -f 32 = 9828B2.
Ill — Page 50.
1. Resolve the number into its prime factors. Form
as many series as there are different prime factors, mak-
ing 1 the first term of each series ; the first power of
the prime factor the second term ; the second power ot
that factor the third term, &c. Multiply these series
together.
Prime factors of 8100 ~ 2, 2, 3, 8, 8, 8, 5 and 5.
1st series = 1, 2, 4.
2nd '« = 1, 3, 9, 27, 81.
8rd " ==1, 5, 25.
1, 3, 9, 27, 81
1,2, 4
1, 3, 9, 27, 81, 2, 6, 18, 54, 1G2, 4, 12, 86, 108, 324.
1^5, 25
1, 3, 9, 27, 81, 2, 6, 18, 54, 162, 4, 12, 36, 108, 324.
5, 15, 45, 135, 405, 10, 30, 90, 270, 810, 20, 60, 180.
450, 1350, 4050, 100, 300, 900, 2700, 8100.
8 SOLOTIONS HAMBLIN SMITH 's ARITHMETIC.
2. The prime factors of 10440 = 2», 3», 5 and 29.
• • number required = 29.
8. See art. 87.
4. ^---^"-d by^ 12H„ ^ ^^,,,,.^^ ^^^^^
■Li. 0. M. of 12, 15 and 20^60
.-. time required ^gohrs. ^
. . distance talked by ^=. 60 x 5 mi. .= 800 mi.
]] ^ = 60x4 mi. = 240 mi.
, * <^ =^ 60 X 8 mi. = 180 mi.
6. Number of grs. in lib. Avoir. = ^-T^ii^Z^i^^
Numberofgrainsrequired = H.C.F. oilT^MO '
i , =40.
IV.— Page 50.
2. Number required :^i2I^ ono
2 X 8129 = ^"^•
8. Distance gone =. (360 K 11 X 18) feet
^360x11x18
5280-- mi. = 9f mi.
Number required = L^ CM. of 36, 32. 80 and 98
Col* f?.T■^''^'^^^««^^*«•
Cost of 1 firkm = 56 X 23 cents.
No. of firkins - ^-^-^^ ^ « . • -
~" 56x23~^ = 24.
V.-Page 50.
(1097Zlr!l/^r4^ *^. «"d the H. C. F. of
SOLUTIONS HAMBLIN SMITh's ARITHMETIC. 9
The H. C. F. of 10920 and 15300=60.
8. Since 2 is in the units' place the remainder = 3.
So that the subtraction may be completed 1 must be
borrowed from the 7 in the millions' place, thus the re-
mainder in the millions' place = 6.
4. Length of avenue=3 x 5280 ft. = 15840 ft
L. C. M. of 6, 8, 9, 10 and 12 = 860.
Number of times there are 5 trees in a row =^^^^^^
860
= 44.
15840 15840
^-To-
Total number of trees= i^% ^-^^ +
6 ^ 8 ^ 9
15840
+ -T^sr- = 9284
12
5. Number =2x2x2x2x3x3x 5x5x6x11x17
= 3366000.
Subtraction of Fractions.
Page 57.
The following method will often be found much sim-
pler than the rule given in the text-book:
T i. * , C
b ^^^^ d ^® ^^^^ fractions ;
then r - - = "--^^ - a(d-.^-.cih-a)
h d bd ~ bdT ~~ '
The advantage of this method will be great when the
terms of the fractions are large numbers and nearly
equal to each other.
Examples (xxxi^. Page 57.
10
1 1
^' 18 ~ 12 -
^^1^^-~JJl) — ii (18 — 12) 1
13x12
166
10
SOLUTIONS HAMBLIN SMITH'S ARITHMETIC^
859 199
"• 360 — 200"
_ _1^
~ 450*
860 X 200
160 _
860' x~200
7.
9.
10.
5 +
Complex Fractions.
Examples (xxxviii). Page 67
24
6 -=
9 + 1
5
5 +
86 4- 3
5+ 8
TT
20
T3i'
5
4—2
2 ■-
5
1 + —
20 _ 2
1
90
12
1 + -i_
1 + ^
1 +
1 + J^ — tt
TIT
1 + f
Examples (xxxix.) Page 68.
!• 8f^(2,i+l^)=:3|^(2/^ + lif)
= 32-^4 1
^T
1 7 y 2 1
. 01
(4A +2^) -r 35| = {4U + 21^^ 36^
7. foff+3
. 4-
3"
2
TT-
: 1 1 3 w 5
=5 GV+ttV)
V27x28y
fi 8 fj
When the num. or den. has a common factor, it
should be taken out, the operations performed, and the
common facLor introduced at the last.
8.
9.
11.
SOLUTIONS HAMBLIN SMITH*S ARITHMETIC. H
1 7
(g_-H )4-(3+f ) Vx^V
a-i)x(4-8f
^
X
-^ X ^ X y X ij ■
:7 7
u?r'
12 ffi-2i)j.»of|_|^^»^_^
X
1 6
X#X^=3
iVo^e. Two or more fractions connected by of ar*
always considered as one quality.
Miscellaneous Examples in Fractions.
Examples (xl). Page 69.
6. 3|x8^-^(l^xH.^) = VxVXxVxfi=V=4;
^OTE.-Tndicate all operations before performing an
of them. It IS much easier to simplify he fore perform-
ing the multiplication or division tLn after.
8 n + 4x. 20i __ 19
27 19 63
__8_1
19
21
5| + 3
X
36+21
= 3.
9.
m + H- ,V) (4^ - 3i) = 8« X Ij} ^ 1^J9
and 1/v + 2i - (2 A - ^ _ ^1^) _ 3. ,
1 n
Iff
(2/«
')
75
.*. quotient = -Al? -1- lo =_ «* = 71
9
IF-
10- (H.2,) (j^V^,)= ,„ ,
544 + 110
81
77«r20
76
T
= 3.
\
12
SOLUTIONS UAAIULIN SMITIl's AIHTIIiir/riC.
11. (7i + 1;
and 4-1- — ,"
?5
) m-t-) = ^t2x?A =
4x20
1 r
(2i
T<I
_ 19
12.
-.V) = 3^i-2yv^ = l
3.1
I7 7J
quotient
0}— 1
4x29
9
1 9
T»r
1 acfi
10
1 4P>
17 f
!¥
ii<\ + 1^
94^ — 19
30,' 4- 20
75i
_ 4 r. n
(^ Of IW) -
3i
— A V 19 V ! 3
13.
4 —
= f.
4 —
13
1 —
ij
18
4 —
8
T6~l3
4 +
4 +
1 —
16
4 +
23 _
23 - 16
Iff
32-9
14.
IG.
28 + 23 - 'T-
10^
8
24
71
+ 3^\
(^
of2jV
?i
-7?
10
304
;i (I 4- v" o
iVS
3T-
3 04
3ST*
1-
+ 5t-4t-
9A-8'4 + 7|;-G^
X
IT
X4X
1 7 -- 1 2
lA+1
TITT
'Tff
2 1
'15
3 7 7 !) 7
iB^'^ T5008
64
3 3"
X
.'57797
1 1 -Ul 1
8 7f)
=--1^
= 1.
TTTiT ?•
5 —
3 —
-5 ,
of7=
D —
?¥
* V _9 s/ 7
"23-"-'
5X
2n
X:^XtXAx7 = 5
nvnc.
4x20
"IT"
- 2-/' "- — 1 '^ T
■^17 — -^i m
■^ 2 O
4r. n ij
^ - H>
4— _A.
16-18
+
28
23 - 16
-12
1.
. 3
SOLUTIONS UAMBUN S*nTH'H ARITlIMKTIO. Ig
• 7
. r
X Ax 7 = 5.
SOLUTIONS IIAMBMN SMITh's AHITUMETIC.
49 _40
_lTr 4 5-
ft
I 06
TfS
__ BX220
I 9
1 1 1 s
ijj _ n
r IT
_3S8
•»» — = 1 1 3 V
I V 9
I 9
X ^\ X 1^ X «3^
■JJL = 22n
9
?. «
22
1 1 ()
'2 2ff
_ 3 8 8_
1JJM9
97'"
7 7
1 9
T7 ^ if ^ iff ^
7 X
22«
TIT
1 1
= 2.
19
- 2 3'_
1 14-
•sraff
EXAMINATION PAPERS.
I.— Page 71.
2. iei-x$2| + 27ix$,^^V'x$V + Vx$/^
Sum = 12^2 ^8|-il=:21|7.
Diff. =12^-8M=3^-L
T^-
^¥"ff*
And 21f^ ^ 3^^= 8/^7 X iVt=%V
of yS^ of share = $3600 ;
.-. whole = I of V of $3600 = $18860.
6. Since the sum of two numbers added to their diff.
= twice the greater, we have
4^ + 2| = 2^3/ = twice the gi-eater ;
.-. greater =V(/ = 3f7.
And ^■^S^=il =the less.
SOLUTIONS HAMULIN SMITh's ARITHMETIC.
15
II. — Page 71.
1. The first uutnber is to be made the numerator,
nnrl the second number the denominator of the same
fraction.
(n — (Hi'
2. Art. 66. The relative maguHudes will be obvious
wJien the fractions are reduced to the same denomina-
tor.
8. The sum of the fractions is
TT*
Tr — TiVoV = foio"' next less than j%W;
And 1 - ^y^2y = ^7^2j8^^ ^i^Q fraction required.
2±fl
= TV=:U.and-§.^ =J;
j~ is greater than §.
Also? + !>- = ^=4«and«=««;
3 + 7
fJ-J is less than |.
6.
'8
of ship = ^ of cargo
— I'
TIT
.-. ship = f of cargo ;
••• f of cai-go + -^. of cargo = $60000
# ■'• = $60000 ;
^ * :. cargo = ^iLeopoo. = $36000,
Ship = $60000 -$36000 = $24000. ^
III
1. Art. 69. The denominator, i.e., the '^ name-giver"
jecause it gives the name to the parts.
The numerator, i.e., the '' mmberer,'' or *' counter,"
jecause it indicates how mnnv of tho im-f- ^ ^'^
liic denominator are to be taken, ' ^ '
10
BOhUTIONH HAMULIN «MITil\i AUITUMfcTIC.
2. I of 3 of 2^ bbls. = ? bbl.
Vttluc of '^ bbl. =.
I •
«t
bbl. _::
$7i
6
- ^ '.'
value of 5 or 1 bbl.
ja^xll.
2.^.bblH. = $2A^^
- $18:{.
8. i=^
- 3
. :»
~ a
4
ft
10
I J »
then A = _i + ii^:» + 4 + /, + o. __ .
< ^+4 + + 8 + 10+1 a "•
4. Sum of fractioiiH -. J 7 . then 2 - /jj ,^^ ja .
and ^ of IJ of 88 x ;;•{ = ,« x 5'/, x 11 x V-
To find what fraction thin product is of UDD, wo have
•J 7
20
X 1 1 X 4 ■■♦
^ ^^ ^ S -- 4 7;«
6. Cs age is evidently 84 years.
B'b *• = ♦ of O's --= ^ of 84 = 48 yrs.
^'s «' = i«2 of /^'s = j^j of 48 = 2U yrs.
IV. — Page 72.
1. In the operation of addition of integers, the ad-
dends must have the same name, in order that tlieir
sum may be expressed by one number ; so also in frac-
tions, the addends must Jiave the same fractional unit
in order that their sum inay be expressed as <> lo frnc-
tion.
2. 17^ contains '6}, 5 times, with remainder ^} ; if,
therefore, ^ be taken from 17|, the remainder will
contain 3^ an exact number of times, viz., 5 times.
3. If operations the reverse of those indicated in the
quept^. \'. oe performed on 2|, the required number will
be ..ound ; henco,
f .
SOMJTIONS IIAMIII.IN HMITh's AUITMMETIC. 17
5.
= 4,VoX«iV
4. Carriage = J of horso ;
horso + ^ of horse ^ -: }^226 ;
V value of horse : ^ |i22,'> ;
.-. value of horao =^ l^^'-^j^,^--* = $120.
Carriage = $225 - $120 = $105.
Harness ---.:- {^^ of $120 = $26.
Let 1 represent li'a share, theu
since IV s -^ 1
^'s :. 8 -$88,
antlCa ^ 2- $41 + $176;
ancl^'s+Zrs + C^'s = G + $44 -- ij^8888.
.'. C = $8844,
1 = $1474 -: /i^s share.
^'s = 3x$1474-$88 -== $4834.
C's = 2^ $1474 + $132 = $8080.
v.— Page 72.
1. Arts. 80 and 84.
2 %^?ix3J^-l 2^
8« X 3^ -r" = ^* + SRspi
21
10 X 10^
=z 8*4- •' . ti ?
3. Smallest number equals the L. C. j\I. of $U S6'
and $2i, which =^ $2015. Art. 81 ; then ^ " "^ ''
2015 ...^ ,
-y~ = 105 siieep.
= 3^ +
i ii
ll
IS SOLUTIONS HAMBLIN SMITH'S ARITHMETIC.
20] 5
"ST" = 300 calves ;
2015 _ 5j^,. ,
4, After spending $80 less than f of his money ..'ohn
has -^ of his money + 5^80 loft ; then if 1 i-present his
mouey, we have
(i + $80)~{^(-i.f$80) + $40f -$40
f(i-f$80) — $40 = $40,
cr, |(i-f$80) = $20
i + $80 = $140
i = $C0
/. whole of his money = ^ = $180.
6. ^ o.f »^ = ^ij^ ; ^ of remaining ||, or fj is in the
water. Hence in miul and water theije is y'^r + y4 = ?» ;
and in air, l — |« = i.^, which = Sift. ^
.'. whole post, or ,ij« = ?^J-J = 18 ft.
Vl.-Page 73.
1. Art. 78.
2. Denominator must evidently be equal to sum of
numerators. Hence fi-actions are ^*y, /^, and ^^.
3. 2Jx(8f + 4^ — 6f)-4-4f
= 1.
4. ^ cost of watcli to B =. $86
(<
= 4 of $30
Ag«in, 1^ cost of watch to yl = $48
= ^ofi48
= $40,
SOLUTIONS HAMBLIX SMITH S ABITHMETIC.
19
6. Length of rooms y^", Vif', sV 5 and H. C. F. of
780, 676, and 640, is 5 ;
.'. Tf\ ffc., or If in., is the longest ruler.
VII.— Page 73.
1. To obtain the product of the multiplier and mofii-
plicand we perfo'rm the same operation on the muitipii-
cand as we did on unity to obtain the multiplier.
Thus, to multiply J by i, what was done with 1 to
make -J, the same must be done with §. But, to make
:|, 1 is divided into 4 equal parts, and three of them are
taken. Hence, to make | multiplied by i, | must bo
divided into 4 equal parts, and 3 of them must be
taken.
2. Wife and son had ^ + ^ = » ; daughter had, there-
fore, 1 -| = 1. Mother leaves ^ of ^ = ^ to son, and
rest, ^— ^ ==T^5 to daughter • daughter then had ^ + 1-5^
= f'xy. Son's and daughter's shares make the whole,
and of this daiighter gets ^. Hence daughter's gain =
1
TIT — 3^75"*
_ 5 1 of 1 2 1
3. No. that can read
*' " & write
•' "& cipher = ^\il-(i+,Viy)}
The rest = 1 - (^ + ^Vir + &) = i^VA = 243600 ;
.'. the whole population — %"yy> x 243600
= 1000000.
4. The L. C. M. of 5, 6, 7|, or of
10 12
IT » IT »
they will all be together in 80 min.
A will go round it ^^^ = 6 times ;
15 — 6
=30;
(«
II
ao
30
'71'
=■5
= 4
«<
20
SOLUTIONS HAMBLIN SMITH's AlilTHMKTIO.
B
c
D
«
((
» of «
TT
= 1 «
TT*
Vlll.—Page 73.
1. Operations are l ore easily performed. Art. 64.
2. 3i~2fH;fof 2i^l| = |.^3.._i.
=^- ."5/ 7
*2"
^)
^°<3' -S^ h /A
10 4
.01 13.
and ?««
> 4 li
• . A or ,^- .s more nearly equal to ,- or ^ than to if ^
3. Let 1 denote number of sovereigns ; then
or 3^ -^ 5 ;
Hence whole namber of sovereigns ~ m.
4. Let A, B, C represent the horses.
A would go round the island in 2 min
^ " " - 2i " "
3 "
The L. C. M. of 2 2i q io on i
they will be iojlt ' ' ' '""" " '" '"''^"'-
^ goes round 16 times, and, therefore, travels
15 X 800 rods == 4500 rods.
B goes roiUKl 12 times, and, therefore, travels
12 X SOO rods = 8000 rods.
C goes round 10 times, and, therefore, travels
10 X 800 rods = 8000 rods.
SOLUTIONS HAriBLIN SmTIl'b AHl'i'HMETIO.
21
Decimal Fracrtions.
I. — Page gg.
1. Art. 101.
2. Art. 102.
3. IV actual division, ^^^ = 1-1 Ig-^,-^ ; and
111 — "^ UUJj.j J ,
and since ^jr^-^^ is less than i^fj, evidently the first
statement is more nearly correct.
4. Since division is only a short method of perform-
ing subtraction, divide 2*291 by -0087, and quotient =
2G3, with remainder -0029 ; which is i of -0087.
6. Art. 109.
•Jf^ = 3-1415929 ; hence limits of error lies between
•0000006 and -0000009.
1. Art. 110.
II.
S • 17 _ 7 . 1 8 2 __ ^_
^2^ — 2* ' Tl (JOT — Ton > ^912^ — 'ya
91
13
3(F
13
28 X 10 '
the preceding fractions can evidently be reduced to
finite decimals.
Hi = ^^ = 2^- 3; ¥^5=31^^5; the preceding frac-
tions cannot be reduced to finite decimals.
2. Value = -0625 x 16 x 200 x -0093125 x $8
= $14.90.
3. 3-714536 n= 3'714J4-.., which is evidently more
nearly equal to 3-715 than to 3-714.
' 'l4i4-!-4 4 ~
5. ^ = -7142.
16-84
I2"9e
1 1
m
ill!
Hi
22
SOIAITIONS HAMBLIN SMITH's ARITHMETIC.
III. — Page loo.
1. Tho advantages are (1) that the addition, subtrac-
tion, multiphcation and division of decimals can be per-
formed by processes the same as in ordinary whole
numbers, with only additional rules for placing the
decimal points in the results ; and (2) decimals can be
compared with the same ease as in whole numbers,
whereas vulgar fractions have to bo reduced to a com-
mon denominator.
The disadvanta(jes are in recurring decimals, which
are only approximations.
2. -475 ^ T-VA - \l ; and -38 = ^^,
Share of third := {1 — (A + fl/)} of $t>oOO^ |81C.6GJ,
3. Owns (-'-^ of I) =-,^==.13 ;
and I of ? of value := $1400 ;
:. whole value = ^ x $1400 = $3000.
4. Horse = $120
Buggy = $'-F-f $36/^ = $76-3125
Harness = ||f of ($120 + $76-3125)
= $1^^^«^3^« _ $86-35416
.-. entire outlay = $120+ $76-3125 + $36-35416
=z $232-66f .
Note. — For method of dividing by 999 see Art. 51.
5. -63 X -1 36 X third fraction = | ;
oi" gf ^ 1 of X t^ii'd fraction — 4 ;
.-. third fraction = y x -^Y x *
j .! 2
2 9*
IV.
wliich must be XO or some power of 10, sup.pressed ;
SOLUTIONS HAMBLIN SMITh's AElTHMETlC.
23
vulgar fractions may have any number whatever for
denominator.
2. He gains 14 cents a yard on 140 yds. = $19.60
He next gams V^^^yds., which at 60 cents
a yai-d ^ 2.00
3. If f« = 423
1 = ft X 423
Net gain = $21.60
3 4
= Ttxt|x423 = 425.
6. Correct length = (84 — 7 x 12 x -0208^) yds.
= 82|r yds.
v.— Page loi.
1. Arts. 109 and 110.
2. Fraction - Mi'U'iAIl ^ t o o o v n n.a ? i ssji 7 1
12X5280 40X12X5 28 25 3 4 4^0 '
3. 2700 mi. in 280 hrs. = lU-J^ miles an hour ;
and 405 •* 18 " = 22^ " "
' ^lien 22^— 11|7 _ io^| ^ 10-7608685 &c.
4. div. + quot. = 7^
div. = -^ quot.
r6m. = l^div., and
.-. = 1-^ of f quot. = 14 quot.
.-. f quot. + quot. = 7^, or \p quot. = 7| ;
.-. quot. = 5-|.
But dividend = quot. x div. + rem.
= Vx^ofV+fOofV
>5^.
1 Q23
— J- 'fir-.
5. -B's = ^'s
C7 8 = i?s-
$46.70,
$34.59 ^- (^'s- ^46.70) -$34.59
= J^'s -$81.29;
24
SOLtJTTOMS IfAJfBT.lN SJIITh's ARITfTMETIC.
Sum of all the shares == ^I's 4- ^'s 4- C's
= 4's + ^'8-$46-704-v1'b-$81-29
= 3^'8- $127-99.
,-.8i4'8- $127-99 = $448-715
8fi4'8= $576705
^'s = $192,231.
5'8 = $192,231 - $40-70 = $145-53| ;'
C"s = $192-23.^ -$81-29 = $110-94^.
VI. — Page loi.
1. Arts. 114 and 110.
2. Art. 109.
3. Sum left after the first spending = -^f of money
-$2i.
= -J:of money — $21.
As he spent yWr of ih of money- $2|) - $VA, he
had remaining yV^V of (>, of money- $2^) + $lo9» or
= irrr;^ of money- $||«f + $Hi>,
= $__A?.AfllG__ ;
^ 9X110X2 2
,1 dh 1 4 4 1 X 5 X 5 9 4 5 1 6
.-. themoney = $^,^^„^^j,^,,2
= $341.
This example illustrates the utility of merely indi-
cating the multiplication and division until the final re-
sult is required.
1^ 2«
1
,1
V5»
= Fe = ino =-000064;
••• 1 + 1-r, = -000064.
SOLUTIONS HAJIDLIN S.MlTu's ARITHMETIC.
25
Also,
.i_i 2«
1.1 —i.^' 18-2
" 5» "^io'""" '107 — -0000018;
1 1 -L 1 1
^'•5^' "^ 7%^ = 002G085-
Therefore
IGx
= 16x1
5.
11^1^* 1 .
•200004 - •0020085] - -010780
==3-141592.
10
03^]/ 1Q2+IX2 i(f4^rx2x
_ ^ -T ^ 4 X _^_ + 3 X 4 X r, X _A. I
=^ 10«j
L ^ f 1 ^ c 10 )
97001
108
=••00097001.
102^104"^10«)
( 10«--3xl0-* + 0xl0 + l)
10^
EXAMINATION PAPERS.
I.—Page 146.
1. 3 min. 50 sec, or 230 sec. = difference for 1 day.
1 sec. = - ^^^ »
24 hr. or 24 x GO x GO sec. = diff. for ^iillP ^« » rlnv«
= 300/^ days.
2. Time to pass over 91713000 mi. = 8 min. 18 sec.
I
26
SOLUTIONS IIAMBI.IN SMITH S ARITHMETIC.
Time to pass over 592200 x 91718000 mi.
= 592200 X (8 min. 18 sec.)
= 59220 X 83 min.
= 3413d. 9hr. (between 9 and 10 years.)
8. In a period of 400 years there are 97 leap years ;
(Art. 151).
.-. 400 y (5 hr. 48 min. 49-7 sec.) should he 97 days.
But 400 X (5 hr. 48 min. 49-7 sec.) = 90 d. 21 hr. 81
min. 20 sec. ;
.'. in 400 yr. the error = 2 hr. 28 min. 40 sec. ;
.-. in 12000 yr. " = 30(2 hr. 28 min. 40 sec.)
= 3 d. 2 hrp. 20 min.
4. In 8505 days there are 1417 weeks and 3 d. over ;
.*. the first number appeared on a Friday.
8505 working days —- ^''^^~ ordinary days = 27 yrs.
61 da. nearly.
27 yrs. and 61 days from Monday, June 18th, 1877,
is Friday, April 19th, 1850.
5. The time between 9 hr. 13 min. A.M. on June 2G,
1858, and midnight on Dec. 31, 1873, is 5GG7 d. 14 hr.
47 min. Now 29 d. 12 hrs. 47 min. 30 sec. is con-
tained in 5667 d. 14 hr. 47 min. 191 times and 26 d.
19 hr. 34 min. 30 sec. over.
.'. there were 191 full moons, and the last one
occurred 26 d. 19 hr. 34 min. 30 sec. before 12 P.M. of
Dec. 31, or at 4 hr. 25 min. 30 sec. A.M. of Dec. 4.
II. — Page 146.
1. Since 1 ft. 6 in. = ^- a yard ;
.'. 9 mi. 7 fur. 39 per. 5 yd. 1 ft. 9 in.=10 mi. 3 in.
which can easily be changed to inches, and the result-
ing number of inches reduced to 10 mi. 3 in.
2. No. of revolutions of forc-wheel^^-^^^^ 80 _gg^Q .
of hind-wheel =3360 -718 = 2642
SOLUnONS HAMBLLN SMITH'b AKITHMETIC.
27
Hence the circumfer. of hind- wheel = '' ^* '-^Ao a
ao4a •
8. Time in seconds = ^-5^-* ^lo ^20040
G G
= 7 hr. 24 min. ;
.-. It will reach Montreal at (6.25 + 7.24) or 1.49 p.m.
Time the Toronto train has heeu going at 8= 1 hr. 35 min.
Distance it goes in 1 hr. 35 min. = ^'-*'^'iii^ mi.
5 a 8
= 71imi.
Distance between Montreal and Toronto train at 8
a.m. is (333-711) mi., or 261 J mi.
Each second they approach (88 + 66) ft. or 154 ft.
Number of seconds to meet =z^^y-''P-^-'L?3P
1 /5 4
Distance gone by Montreal f.rn\r\~- 2oi7 « x /> 2 so xBg ~:
a 2 8 X 1 6 4 *"*•
4. Average length =
=149^ mi.
16050 X (202 yd. 9 in.)
98
= 19 mi. 1464^7 yd.
5. Number of strokes=^'*-^^^i^^o
= 28160.
III.— Page 147.
The corresponding unit of area is a square each of
whose sides is equal to the lineal unit, and the corres-
ponding unit of volume is a cube each of whose edges is
equal to the lineal unit.
When the lineal unit is twelve inches, the unit of area
is a square each of whose sides is 12 inches, or a square
whose area is 144 sq. in. ; the unit of volume is a cube
each of vvhosu edges is 12 inches, or a cube whose .
volume is 1728 inches.
28
80LUTION8 HAMULIN SMITH 's ARITHIILTIC.
i
2. Length of table = 90 in.
Width of table = 43 in.
Area of table = (90 x 40) sq. in. ;
.*. number of coins = 90 x 40 ^ 8000.
8600 half pence = £7 lOs.
8. If A gets 1, B gets 2, and C ^ of 3, or 2^ ;
l-h2 + 2i = 6|;
••• ^ gets ~~ of 17 a. 2 r. 88 per. 19 yd. 7ft. 45 in.
=- 8 a. 1 r. 20 per. 21 yd. 77| in.
and B gets 2 x (8 a. 1 r. 20 pei. 21 yd. 77| in.)
--6 a. 3r. 1 per. 11 yd\ 7 ft. 118f in.
aud C gets 2^ x (8 a. 1 r. 20 per. 21 yd. 77| in.)
= 7 a. 2 r. 16 per. 17 yd. 1 ft. 29|in.
4. ITunaber of yards in 1 bale = ~.^,
1 piece =
((
67048
34X68
= 29.
5. Number of sq. in. in 16 sq. ft. - 15 x 144,
.-. pressure = (15x144x15) lb.
= 16 t. 4 cwt.
When the barometer is at 29 the pressure will evi-
dently be ^V less than before.
^^ of 16 t. 4 cwt. = 10 cwt. 3 qrs. 5 lb.
IV.— Page 147.
1. 2 bu. 3 pk. 3 qt.=91 qt.
.-. cost =91 X 12^ cts,
=§11.37i.
2. 130 rods 4 yd. 2i ft. =-1 30^ « rods.
.-. cost =-i:J02«x$2.50.
= $327-^.
Part to be paid in wheat =$227i.i)
207 1 .1
,% Number of bushels ='
■87.
=209 bu.2pk.l gal. Uffpt
TIC.
SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
20
in. ;
]00.
2i;
7ft. 45 iu.
77| in.
77| in.)
ft. 118f in.
7| in.)
1ft. 2U|in.
144,
L44 X 15) lb.
: CWt.
ro will evi-
Ib.
8. 29 gal. 3 qt. 1 pt. =29| gal.
.-. cost of brandy =29^ x 48^ cts. ;
,., - 29^x48?,
.'. quantity of rye = — ~~i — - bu.
81;^
=41 bu. 8 pk. 2f qt.
4. Ill bu. 2 pk. 4 qt. =3572 qt.
2 bu. 1 pk. 4 qt. =76 qt.
.-.number of bags --=—^=47.
5. Number of quarts =
.'. value of produce
129x 95x4^
8
129x95x4i X 45
~ 82 X 8
=$96.98
cts.
v.— Page 148.
1. Number of ounces bought = 12 x 16,
it
<i
sold = i-^i^iooo.
20X24
Cost price = 12 x 16 x 37| cents
= $72.
Selling price
12X 7000x40
30X21
cents
.*. he loses $2.
2. Cost of 1 oz. or 480 grs. =
grs. =
costof '""«
ic
8. It is evident the weight
measure of 8 lb. 20 gr. and '8 lb.
8 lb. 20 gr.
8 lb. 11 oz. 16 dwt. 16 gr.
The H. C. F. of 56020 gr. and
15 cents ;
7oooxig (,pnfn
16x480 *'6"''B
18*1- cents.
must be a common
11 oz. 16 dwt. 16 gr.
= 56020 gr.
= 51760 ffr.
51760 gr. is 20 gr.
80
SOLUTIONB HAMBLIN SMITu'h ARITHMKTIO.
4. 89 mi. 1 fur. 1 per. 9 iuch. = 2-470107 in.
.-. tho weight = 2479107 lb. (Art. 188).
= 12 t. 7 cwt. 8 qr. 1(J«7 lb.
6. The thirtieth part of 1 cwt. 8 lb. = ^j^^^ cwt.
The eighty-fourth part of V, cwt. = ,;^^% cwt.
.-. 500 timcB theii- differcuce = 600x a?^ cwt.
= 250 lb.
Examples (Ixxxvii). Page 151.
1
M s. d.
iL ts. d.
40
28x10
=280 =co8tof28rd.
80i
4x 6 ■
=-100 =. « 4 p.
^-0 8{»f= - 4iyd.
^^ l{l\
281 8|i>'- -entire cost.
2. £ 8. d.
A' 8. d.
4
12 X 8 18 2
= 40 18 —cost of 12 cwt
25
8 X 19 0^
= 2 18 7i = " 8qr.
16
22x 9j»
= 17 2,»5. - " 22 lb.
12x |««
— 00 73^- « 12 oz.
60 14 4^J»=.entire cost.
8. £ s. d.
£ 8. d.
4 10 X 2 18 lOj
— 29 8 10| =cost of 10 a.
40 3x 14 8|
= 242 = " 8 ro.
|26x 4t{-^
- 9 6-^ = » 26 V.
82 2 7i =entire cost.
4. £ 8. d.
£ 8. d.
4 132 X 3 14 8i
-492 18 9 =cost of 132 cwt.
25
8x 18 8iV
- 2 10 OA - *' 8 qr.
10^. X 8|J
= 7 10-,Vir=- " 10^ lb.
496 2 7-i-Vo--ciitire cost.
6. £ 8. d.
£ 8. (/.
4
68 X 12 12
= 793 16 = cost of 63 cwt.
25
3x 8 3
= 990 = " 8qr.
m-x 2 6
US'-
= 2 J_li =.
805 9 n
111 lb.
=entue cost.
SOLUTIONS HAMBLm SMITh'a ARITIIMETIO.
81
6. £ a. d. £ 8. d.
4 29x105 =-8045 = cost of 29 a.
40
8x 26 6
6x 18 U = 3 r> 1^ -
=
I
2
78 15 =
I
7
8 ro.
6 per.
7.
£ s. d.
20
24
8127 7.] -- entire cost.
£ 8. d.
lOx 8J7 6 = G2 =co8tofl6oz.
6x0 8 TOf = 1 3 3 = '« G«lvvt.
8 2:{ = " 20 gr.
20 x m =
8. 2 8. d.
4 25x42 2 4
40
1 X 1 10
Wx~0 5
G3 G 5f = entire cost.
£ .y. d.
1052 18 4 =^ cost of 25 a.
7 = 10 10 7 = " 1 ro.
2 12 7:;=« " 10 p.
ft 7 =
9. £ 8.
4 13 X 22 8
d.
=
lOGG 1
£ 8.
291 4
16 16
3 16
(i'i = entire cost.
d.
=^ cost of 13 cwt.
25
3x 5 12
17 X 4
10. £
319x2 12
=
8. d.
6 ^
= '« 3 qr.
1^}1= " 17 1b.
4
311 16
£ i
837 7
1 19
8
IS J = entire cost,
f. 'd.
6 = cost of 319 cwt
26
3x0 13
1^ =
U = " 3 qr.
IGxO
«rV =
4/0 = " 16 lb.
839 15 3j% = entire cost.
Examples (xc.) Page 157.
1. Value of -^ of estate - $7520 ;
i(
5
¥
((
= $ (4X
(,.»)
$7833^.
I
82
b.
SOLUTIONS HAMBLIN SMITH's ABITHMETIC. "
2. Value of f of | of ship = $1260 ;
.'. " whole ship = $(4 ji 1260)
= $6040.
8. Quantity bought for 866 half-pencc=-3f lb.
• (< (( onto 1- iz ^013x8?
2018 half-pence= — Eirr
866
4. Amount of work done in 25 da. = f^ .^^ ' '
^ 25
6. Time he walks 96800 ft. = 830 min!;^*
.-. " " 7020 ft. =^-»JlO-?i3 3 „,-
9 6 800 *""!•
= 27 min.
6. Value of f of ^ of ^^^ of estate - 1608.125 ;
.-. « i of VV of estate = $t«i4^9?:126
^ofVVof^V
^ ^. , =$1182.126.
7. Distance 15.5 cwt. is carried = 60 mi.
.-. " 8.25 cwt. " = IMA^
8.25
= 286T?g. mi.
8. Value of i of ^ of j\ of vessel .-= $1400 ;
" A of tV of vessel - | tt of jV X 1400
^oflof^V
9. Value of 1 lb. of gold = 12 x £8-89 ;
.-. « -04 '' = £(-04 X 12x3-89)
= £1 17s. 4'128d.
10. Cost of 6 in. of the first kind = Id. — Ud
T A O
«<
«(
seconu "
V/x.~ 1 g 738 , , .
-^ow i^ - 17^4 1, ana ff = ^^^ - ,
.-. the first kind is the cheaper.
I
mi.
0;
of 1 of
1400
3
TT
11.
soLUTiOxNS nimBLiN smith's arithmetic. 8d
Weight carried 20 mi. for £14^ ^ 60 cwt. ;
(I
tt
5^eX60
14^
cwt.
1.
2.
3.
= 22^ cwt.
Examples (xci). Page 158.
Amount mowed by 8 men iu 7 da. = 40 a. ;
24 " 28 " =
it
it
4.
5.
6.
I
7.
24x28x40
^ITT— ^^•
= 480a.
Sum earned by 8 men in 5 da. = $G0 ;
" " ,32 " 24 ** 1^3 2X24 X6
= $1152.
Number needed to consume 351 qr. in 108 da.
= 939 men;
1404 qr. in 50 da.
1404X 1C8X939
= 3-niiJ^ ^en
= 11208.
Number of hordes supplied by 8 bu. for 16 da. = 2 ;
" 3000 X 8 bn. for 24 da!
3 00X16X2
^ """2 4
= 4000.
Cost of carriage of 8 cwt., 150 mi. .-= $12 ;
" 7-89 cwt., 60mi. = $'-:i-*-i^?-^L2
' "^ 3 X 1 5
~ $10.62.
Cost of carriage of 2 1. for 6 mi. rr^ $1.60 :
" 12; J t. for 34 mi. = S-iii^-^-iiM9
^" ^ - 2X6
=: $54-61^
Sum earned by 3 men in 4 da. = $15 ;
" 18 " IGdft, = illi'-l-^JLlxj./'
3x4
^ 8360.
84
SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
8. Number required by 6 peopio for 24 da. = 4 bu. ^
72 « 8 da.
7^X8X4
= 16 bu.
0. Time required to travel 150 mi. = 60 hr. ;
72X8X4 1
6 X 24"^ "^^•
(t
<(
((
500 mi. — ---*' ^JL^ hf
1 5 "^*
sooxco
150
= 200 hr.
But 200 hr. =. y>^o da. = 20 da.
10. Cost of carriage of 5^| cwt. = $15.70 ;
4 X 73^5 cwt. ^ $i^7A >U5^0
-= $78.60.
11. Number required to earn $120 in 6 da = 16 •
" $270 in 8 da.
1 20x8
= 27.
12. Number supplied for $1.20 for 50 hr. = 5 •
$21.60 for 60 hr.
__ 3J:^ X 5 X fl
1.2 Ox 00
=^ 75.
13. Time $190 lasts 3 men ^ 4 weeks ;
" $475 «' 5 <«_ 475X3X4
r=r G weeks
14. Cost of 2 horses for 5 mos. =^ $120 ;
3 " 11 « __ «^3xllxi20
((
11 << __ o^Jii 1 X
== $306.
15. Time 5 horses are fed for 7050d. = weeks ;
•*• " ^ " " 4935(Z. := ^JL1^^12<1
3X7060
«= 7 weeks.
SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
S^
16.
17.
18.
19.
20.
21.
22.
■
23.
Time required by o men to reap 12^ a. --= 50 br. ;
" . " 7 " 15 a.
= -7X121- ^^•
= 48 hr. = 4 da.
Quantity required for 858 men for 6 mo. =234 qr. ;
" " 979 " 3imo. .
97 9 X3tx234
-^ "3 5 8 X 6 ^^*
= 155f qr.
Time in -which 5 men earn $315 = 6 weeks ;
" 4 " $231 = ^^-''^^^"w.
^ 4x315
= 5^ weeks.
Time in whicli 7 men mow 22 a. = 88 hr. ; '
" 12 " 360 a.
_ 7x3COX86Jjj.
~^ 12X22 ~
= 840 hr. = 84 da.
Time in which 10 horses eat 7]- bu. = 7 da. ;
" 28 " 30bu. =
__ 10X30X7 -1
2 8 X tT ^^'
= 10 da.
Time SOO bbl. supply (3 x 44 x 30) rounds = 5da •
" 400 " (5x66x40) "
==: ±g^ X ^ X 4 4 X 3 0J< 5 ,
3 X 5 X e e X 4 o~ ^*'
— 2 days.
Number required to earn 19314(?. in 54 da. = 29 •
" " 407x20x12./. in 12 da!
__ 4 07 X 20X12X54X29
I 9 3 1 4 X 12
= 660.
Cost of hiring 3 Jiorses for 1 mo. =. £18 ;
" 4 *« 5 mo. — .£^''"'"'^5
= £12o!
((
!■
C6
SOLUTIONS IIAMCLIN SJIITu's ARITHMETIC.
Examples (xcii). Page i6i.
1. Part of work done by ^ in 1 hr. == V.
^ " =1-
— If >
A and B *' =1 + 1
(I
— 5
Timo required to do -^\ of work = 1 hr. ;
all the work = 1^^ hr.
Q 3 1
2. In 1 hr. A doos ^\ of work ; B,\'^^ ; C ' '
.-. Part done hy A, B and C in 1 hr.
= Ut> + ?V •*- A) of work
rp. . , = tjWtt of work.
Timo required to do -^^^^ ot work = 1 hr. ;
((
all the work — ?J1±^1 i,r
101 "^*
^. In 1 day ^ and B reap i of field ; AandC i- B
and C, i ; * ^ '
.-. twice (4's work + ^'s work + C"s work) daily
„ .-. ^ , ^ and e do T-Vg. of the work daily.
Time required to do ^W of work = 1 da. ;
all the work = i^"^ aa.
= 222 da.
4. Part filled in 1 min. = (-» +| + ^V) of vessel
- f of vessel.
Time required to fill f of vessel = 1 min.
t(
((
the vessel = s^min.
= 2f . min,
6. Part done by A in 1 da. = ^V of ^V of work ;
2 da. = 2XxVof^7^ "
<(
((
=r 1
of work.
SOLUnONS HAMBLIN SMITIl's ARITHMETIC.
87
Part done by 5 in 2 da. = 1 _(^7^ + ^i^) = ^ ;
. .-. " " Ida. =|ofi = ^V;
.'. B would do the whole work in 10 days.
6. In 1 hr. A does \ of the work ; B and C do § ; A
and C, |.
Part done by C in 1 hr. = f ~ ^ r= ^^ ;
Time B requires to do ^^ of work = 1 hr. ;
.-. " B " all the work = ^-^^hr.
= 4 hr.
7. Part done by A in 12 da. = l\ ;
" 5 " 6 da. = /v ;
.-. " • " C" 4da. = l->(^f + VV)
3
Time C requires to do | of work = 4 da. ;
((
C
(I
all fhe work =
9X4
u
= 18 days.
8. Part filled in 10 min. = j« +^o _^o
= 29
EXAMINATION PAPERS.
I. — Page 164
1. Weight carried 36 mi. = 1200 lb. ;
" " 24 mi.
36x1200
= 1800 lb.
2. Value of f of ship = $13056 ;
^txl305«
i(
^
4
= $18360.
3. Value of 12 x 3| oz. of silver =^ $54 ;
((
22 OS. ** =
$26.40.
s
88
S0LUTI0>JS HAMBI.1N SMITh's AEITIIMETIO.
o3eoo X I
4. Expenses in 35 da. = $61.60 ;
3«5da. = $lM'^°'i.«^
3 5
= $642.40 ;
.-. his total income - $1042.40.
6. Wlion the tax is 6d. tJr ->-jie = £1 ;
QQOOd.
= £615.
II. — Page 164.
1. Tax on $2720^- $(2720-2640-66) = $79.34;
■ .-. " $1 =^?|^ cents.
= 2-9V3^^ cents.
3. Since 5 horses =: 84 sheep ;
•"• 10 " =168 sheep;
.*. 10 horses and 132 sheep = (168 + 132) sheep
= 300 sheep.
And 15 horses and 148 " = (252 + 148) sheep
= 400 sheep.
Cost of keeping 300 sheep == $202 ;
^ ^ 300
= $269^.
8. Debt on which he loses 25 cts. = $1 ;
••• " " $602.10 = $6J?Aio_M
= $2408.40.
4. No. required for 1 work in 22 da. =15 men ;
••• '• " 4 works " %^ da. =4 x 5 x 15 moi
=300 men.
5. Time for 72 men to do 1 work =63 days
••• " 42 « 3works=?-^?A'LG3 ^
4 2 ^""
= 324 da.
SOLUTIONS HAMBLIN SMITH S ARITHMETIC.
90
III. — Page 164.
1. ^'s wages for 12^ da. = ^'s wages for 7i da. + £'a
wages for 7* da. ;
.*. -4's wages for (12§^ - 7i) da. = B'q wagesfor 7i da.;
:. A'a wages for 12^ da. = J5's wages for - J 5 ^ da-
^I¥
= 18 days.
2. No. required for 1 work in 80 da. = 100 men ;
.-. " " 3 woi-ks in \? da. = 3 x 4 x 100 men
=1200 men.
3. In working capacity 5 D^en =. 7 women ;
7 men
((
((
7x7
women
Time for 7 women to do 1 work
,= 9| women.
= 37 da. ;
2 X 7 X 37
•• " (9y + 5) women to do 2 works = - -1.4 da.
= 35 da.
4. Part done by A and 5 in 1 day = ^'^ of work,
" " by ^ alone " = -^\ oi work',
.'. " " by ^ alone " = (^«^_^i^) of work
= y^ of work
.-. Time required by A to lo all the work is 33^ da.
Amount done by A in 20 da. = ^-- = -/^^^ of work.
^ in 20 da. = ^-^ = ^Vxr of work ;
.-.A does {^^Q or ^) of work more than B.
5. Part of cistern emptied in one min. = tV~^V
— "B^xr »
.*. time required to empty cistern = 60 min.
IV.— Page 165.
l.'A works 10 da. ; B, 3 da. ; C, 4 da..
Work done by A in 10 da. =: |f of work ;
♦< ^ in 3 da. = ^V *' '*
lii
40
SOLUTIONS HAMBLIN SMITfl's ARITHMETIC.
••• ^^ *i°ie C' does { 1 - (If + v\) f Of work = 4 da. ;
.'. the time C does the entire work = 6 x 4 da.
= 24 da.
2. Time for (9 + 2 x 12 + 3 x7) boys = 250 da. ;
" (18 + 2xl6 + ax9) "
f£+ 2 xJ^jM XJ) X 2 a 1
l' « + 2 X 1 a + 3 X 9 '**•
= 180 da. ;
.-. to do double the work they would be 2 x 180 da.
= 360 da.
8. They approach each other at the rate of 10 miles
per hour ;
.-. they would meet in VV hr., or 10 hr.;
.*. A would have gone 10 x 6 mi., or 60 mi.
When the sum of the distance each walks equals 50
mi., or 150 mi., they will be 50 mi. apart.
This is the case after they have walked f ^hr., or » «»
hr. = 5 hr., or 15 hr., respectively.
4. Between noon Monday and lOj a.m. Saturday,
there are 118^ hr.
Time lost in 24 hr. = S\ min. ;
" 118i hr. = yM><!i
24
= 15 min. 86/^ sec.
As the watch is 10 min. too fast, it will be 5 min.
364^^ sec. too slow.
5. The watch goes 290 min. in 800 min. of exact
time ;
.-. 290 min. on the watch == 300 min. ;
800
(»
n
3^0 X 30
v¥o '
min
= 6 hr. 10' ^ min
55^
e o mm.
SOLUTIONS HAMBLIN SMITh'h ARITHMETIC.
41
In the second case, 310 min. on the watch = 300 min.;
.-. 800
<i
__ 3 0X300
3 10
= 4 hr. 501^ min.
mm
-
3T
V. — Page 165.
1. Wages for 60 da. at $2 = 60 x $2.
= $120;
.'. total loss by idleness = $28.
Sum lost by being idle lday=$(2+1.50);
.-. number of days of idleness = ^%% = 8 ;
.'. he worked (60- 8) days = 52 days.
2. In 1 hr. 1 man does ^'^ of work ; 1 woman, jf 5 ;
1 toy, T-^Y ;
.*. part done hourly by 1 man, 2 women and 3 boys
I4.4 il 2 3
Time required to do ^^^^^^ of work = 1 hr. ;
all the work == ^;^ hr.
(I
{(
33
= 10 hr.
8. By the second part of the question we see that
pipe -4 —pipe C, empties ;j\y of the cistern in 1 hr.;
.*. pipe C— pipe ^4, would fill :j'g^ ** "
We have, therefore,
part of cistern filled in 1 hr. by pipe A + pipe B—-^
»* ** "0
it " ^ _ " c
" B
•(
((
1 .
BO'
<(
t(
({
tr
7
it A—l _ 7 — «
.*. pipe B fills the cistern alone in */ hr.=6f hrs.
A 'NTlimKn'M ^t l-.yv.'.M/^ 'U^i.^^r^.-. 10 ^_ C - i. J * .-1. X
and noon on Tuesday is 60,
42
SOLUTIONS HAMBLIN HMITII H AUITHMETIO.
Number of hours between 12 on Saturday night and
4 p.m. on Thursday is 112.
8G()8 min. on clock =- 8000 min. of true time ;
' 6720 " _67aox3Goo
mm.
3603
=--111 hr. 54^*2Vt iniii-
■= 3 hr. 54^Yo"T'^i°' P-^^* Thursday.
= 8 hr. 54^ min. nearly "
6. Amount which the work falls behind daily
= (1 + 1 + 1-1- vo) o^ a d^iy's work. ;
.*. in 84 days it falls behind
84x(UH-1-1-tV) " "
— 17 "6 days work;
.'. part which 17 men must do more = -G of a day's work;
1 man
• 6
TT
3
(t
Examples (xciv.) Page 169.
7. Time for which interest is to be calculated
186
days.
Interest =
#, B913xl3 8Xlfi
^ 3CfiX200
-,81X27X1»
^ 200
= $164
•025.
8. Time
—
159 days - iff
yr.
Interest
£(204f-U-xi-gl
p491 V 1 X 63
80X7300
^TOo)
•
£4 9^. ^U-ld.
((
Examples xcv. Page 171.
$326 for 15 yr.
$100 for 1 yr.
1. Interest on $326 for 15 yr. = $220.05 ;
dl.l00x22 0.0/5
3 2 6X15
I
= $4i.
SOLUTIONS HAMBLIN SMITu's ARITHMETIC.
48
2. Interost on $700 for 1 yr. = $''-."-'11 __ $42
Tho entire interest on $700 = $(920.50-700)
= $220.60.
Time for which $42 is interest == 1 yr. ;
X 1
" " $220.60 " __ 230.50X1
4a
= H yr.
8. Interest on $100 for 8 mo. at 9 % = $0 ;
.♦. Principal which amounts to $1 = $joo .
" " " $1825 = sLiL^A'^-il'o
jf I 6
== $1250.
4. Sum on which $54.00 is interest = $100 ;
" " $202.50 " = <g20 2 go)o^o
*? T ± 1 ^ ^ ~ $375.
6. Interest ^ 2 Principal - Principal.
Here the interest of Principal for 1 yr. at 5% is $~ ""- ?^
Time to produce $ -j^^-- = 1 yr. ;
•• Pun. = -^ yr. = 20yr.
b. Interest on Principal for 16| yrs. = J of Principal ;
.-. " $1 f6rlyr.=.$A.^^^^iP^L.
•" •* Principal x 16| '
.-. " $100forlyr. = $i~--^l-^J'i^^^^^
'' ^8 X Principal x 16*
7. Sum on which $70 is interest = $100;
.-. principal of which $1 is the amount ^ $|^^^;
" " rfhir^-,., 42.1 275 X 1 no
170
$1275
m. , , . , = $750
lime ior which $52.50 is interest = 1 yr. ;
.Til l^Wi..>;.1 /i.m tt ••■'b-^u.\l
((
((
^(^liUD.ZD — Yi>U) " = -ll^-'-iiiLilivr
'' 52.50 J^'
= 12^ yr.
1
",'
(I
44
BOLUTIONB HAMIiLIN HMI'ili'H ARITHMETIC.
8. The interest on $400 for 8 mo. = the interest on
$100 for 12 mo.
The interest on $100 for 12 mo., at a certain rato%
= the interest on iJ200 for 12 mo. at half that rate ;
.'. the sum horrowed would pay the same interest aa
$(500 + 200) would.
Interest on $700 for 1 yr. $85 ;
.-. " $100 for 1 yr. ^"'"x'"*
700
730
= .$5.
9. Time for which ^^'^a is interest =866 days :
42^ X 865
<( -
£4 '
2T
7 3 O
da.
= 97 da.
10. Interest on £556^ for 125 da. = i'li^^ I
£100 for 865 da. - £100X366XO,'«
5661x125
- £4-752.
= 4^ per cent,
11. Interest on $S000 for 1 dn =^ $2 ;
$100 for865da -$'*'.?-^l"Ail„2
^ ^ 8000
= $0,^.
. 12. Cost of wheat at end of 6 mo = 5000 x $1.25
= $6260.
Sum realized = i^COOO.
Amount of $6000 for 6 mo .^ ijidJJOO.
,% his gain rrr $(6300-6250) rr-. $50.
13. Interest on Principal for 6\ yr. = •{ of Principal ,
f X Principal
$1
$100
for 1 yr. ^
Principal x 6:|
((
,100 V 5 V PHnPinnl
Principal x 6k
BOLUnONS HAMBLIN SMITH'h AllITHMBriC. 45
14. Interest on !j;i()0 for 4 J yr. at 5 % =r $22. 50 ;
.-. the Principal which amounts to $1 = li»rVff'Vff J
" « II $735 = S'A'xi^
*^ iaa.no
= $000.
Time for which $30 is interest on $000 = 1 yr. ;
<i
II
$540
«<
«(
/>40x 1
y.
.io
^ 18 yr.
(18-45) yr.:^ 13^ yr.
15. Time for which $04-7025 is interest = 305 da. ;
** " $37*905
04:7 6 as ^*'"
=1 140 da.
146 days from May 13th is Oct. 6th.
Examples (xcvi ) Page 173.
1. Principal on Interest Jan. 1, 1877 = $1500.00
Interest to March 16, 1877 = 18.25
Amount = $1518.25
First payment = 100.00
Remainder = $1418.25-
Interest from March 16, to June 13, 1877 =. 20.75
Amount = $? 439.00
Second Payment = 400.00
Remainder = $1039.00
Interest from Jane 13, to Sept. 1 = 18.06
Amount = $1052.60
Third Payment — 200.00
46
SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
Remainder = $852.60
Interest from Sept. 1, to Jan. 1, 1878 — 17.09
Amount = $869.75
2. Principal on Interest March 15, 1876 = $3500.00
Interest to June 1, 1876 = 44.87
Amount = $3644.87
First Payment = 800.00
Remainder = $2744.87
Interest from June 1, to Sept. 1 = 41.51
Amount —. $2786.38
Second Payment = 100.00
Remainder ~ $2686.38
Int. from Sept. 1, 1876, to Jan. 1, 1877 = 53.87
Amount = $2740.25
Third Payment = 1500.00
Remainder = $1180.25
Interest from Jan. 1, to March 1, 1877 = 11.44
Amount = $1191.69
Fourth Payment = 300.00
Remainder = $891.69
Interest from March 1, to May 16, 1877 = 11.14
Amount = $902.83
3. Principal on Interest, Oct. 15, 1859 = $1200.00
Tni^vaai: 4-/\ f\n4- IK 1 0HA net nn
SOLUTIONS HAMBLIN SMITh's ARITHMETIC. 47
Amount = $1272.00
First Payment = 1000.00
- , . ^ Remainder = $272 00
.nt Horn Oct. 15, I860, to April 15, 1861 := 8.'i6
Amount = $280.16
Second Payment = 200.00
y , J, Remainder =
Int. from April 15, 1861, to Oct. 15, 1861 =
.16
2.40
Amount = $82.56
Examples (xcviii). Page 177,
1. Amount of $1 = $(1.03)* = $M25509 •
$1000 = 1000 x$M25509
= $1125.509;
.*. interest = $125,509.
2. Amount of $1 = $(1.03)6 = $1.19405 •
$200 = 200 X $1-19405;
= $238.81.
3. Interestof $1 for4 yr. = $(1.06*- 1) = $-26248
" 8 " =$(1,063-1) = $.19102;
= $(19102 + --2 6i48:::^ijB 1 0.2X
= $-22675 ;
••• interest of $675.75 for 3^ yr. ... $(675.75 x '22676)
, , = $153 22.
4. Amount of $1000 for 4 payments
= $(1000x1.03*)
A^ , ... = $1125-508...
Auiuuni 01 f lOOu at simple mt. = $1120,00 ;
.-. his gain = $5-508...
48
SOLUTIONS HAMBLIN SMITH 's ARITHMETIC.
5. Amount of /?5000 half-yearly = .i'(5000 x 1-02'^)
= ^£5412-1608
Amount of £5000 yearly =. de(5000x 1.042)
== £5408 ;
.-. difference = £4-1608
= £4 36-. ^^%d.
0. Interest on ^40000 = $ { 40000 x (1.05* - 1) I
= 18620.25.
Interest on $80000 -^^ $ 1 80000 x (1.052 - 1) j.
= 18200.00 ;
.*. the difference = $420.25.
7. Compound interest of $248 = ${ 248 x (1-0353-1)1
= $26.96...
Simple interest of $248 = .^(248 x 3 x -035)
== $26.04 ;
.'. the difference = 92 cents.
8. Amount of $1 for 3 yr. = $(1.04)3 = $1-124864.
2yr. = $(1.04)2^ $1-0816.
((
(i
Interest during 3id yr. = $-043264 ;
Amount of $1 for 2iyr. =. $(14)816 + -^|^-^
= $1-103232.
Hence the sum of which $1-103232 is amount = $1 ;
$16989-7728
__ (tjj I C 980.7 7 28
•tP 1.103232
= $15400.
9. The sum of which $(1.05)^ is the amount = $]
$27783
(Bi._?_7 7_83__.
«1P I .1 A 7 C O (t
= $24000/"'
(t
((
SOLUTIONS HAMBLIN SMITH's ARITHMETIC. 49
Examples (xcix). Page i8i.
11. The sum of which $(1.05)3 is the Present Worth
(i
u
$6945.75
dt 6945.76
M
^ 1.05s
10 rru =$6000.
12. The amount of$l= $(1.01375). _^1.070^^gg .
.-. the discount off a debt of $1-070668= $-070668 '•' '
$245.25
"^ 107 06C8
iq rr, • . . ==$16-186...
irf. Ihe mterest on $19-3125 for 1 yr.
= $(20f|/-i9^,^)
= ^9li ;
<(
<i
$100forlyr.=:$l?2i5tt^
19 •'
14. The bill is due on May 4. '~ ^^'
Hence the time is 73 days.
The sum of which $1.02 is the present worths
$1127.10
- $ 1 1 2 7. 1j0 X I
10 2 '
= $1105.
15. Interest on $260 for 1 time = $25
$250 for 2 times = $50
••• discount off $300 for 2 times = $50
" dtorrt- -, .275X60
$275
((
= $■
300
A • . , = $454.
Again, mterest on $250 for ^ time = $12 50 •
••• discount off $262.50 " = sjiq ^n .'
'« (fl.27/5x 1 2.ao
$27
$132
263. so
'TT'
60
SOLUTIONS H\MBLIN SxMITH's ARITHMETIC.
16. The amount of $1 = $1-0375.
Hence, if $1 is the cash price, $1-0375 should be the
credit price.
Now, $1-0375 = $1^\.
Hence, if the cash price = 80,
the credit price = 83.
The credit price = $33.20 ;
.-. the cash price = |o of $33.20 ;
17. Interest on $98 for 1 time =^ $30
$98 for i time = $15
.-. the discount off $113 " — $16
" ! $128 " = $~n-^
= $1611^.
18. Sum on which $-80 is int. for 8 mo. — $20 ;
.-. . ♦' " $20.80 " " =$2^8«^^?«
= $520.
Interest on $20 for 8 mo. = $-80 ;
forl2mo. ^-t'*'"^'^'^-'"
<(
20x8
Examples (c.) Page 183.
1. Interest on $950 = $(950 xix^^^)
= $16-625.
True discount off $950 = $^-^^« ^Al
= $16-339... ;
.-. the differec^e — $-285...
2. The bill is due on Sept. 20.
Interest on $722.70 for 40 days at 7^ %
= $(722.70 x.V^Xy'A)
= $5.94 ; )
.'. he received $(722.70-5-94) = $716.76.
SOLUTIONS HAMBLIN SMITH's ARITHMETIC. gl
8. The bill is due on Nov. 12.
Interest on $7850 for 146 days at 10 %
= ${7850 X III xA*^^)
= $314.
4. The note is due on Oct. 6.
The interest on $100 for 95 days = $ ' 33 ;
.-. note for which he receives $(100- Vvl == $100 •
". $601.69 =r $521-^^x12.0
7^1 6 7
f T J . ^^ $511,
6. Interest on $6555 = $(5556 x^-^^)
= $333.30.
True discount off $5555 = $ (t^At'lA)
= $314.43...
.*. the difference = $18.86.
EXAMINATION PAPERS.
I.— Page 184
1. Com^oMwd interest on $1 = $(1.043 __i\ .
" $25000 = 25000 X $(1,043-1)
= 2500ax $-124864
„ , = $3121.60.
^. Amount of $1 at compound interest ^ $M24864.
$1 at simple " = *^1 12 •
.-. sum on which $-004864 is difference = $l'; '
$3.80
^•0 048 6 4
= $781.25.
3. Compound interest on $100 for 2 yr
= 100 X $(1.04'' -1)
= $8.16;
.-. the simple interest on $100 for 2 yr. = $8.16 •
" $100 for 1 yr. = $^« '
= $4.08.
^-^-
52
SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
((
((
((
(i
4. Compound interest for 3 yr. = 1000 x $(1,033 - 1)
= $92-727.
2yr. = 1000x$(1.03»-l)
= $60.90 ;
Syr. - $31-827;
195 da. = ^.g^' of $31-827
= $17.00 ;
2 yr. and 195 da. = $77.90.
5. He adds $20 to his capital for each of 4 years.
Amount of the 1st $20 saved = 20 x $(1.04)*.
2nd $20 " =20x$(1.04)»!
. " 3rd $20 " = 20x$(].04)2. •
4th $20 " = 20 X $(1.04);
.■. his capital is increased by
$(20 + 20x1.04 +-20x1.042 + 20x1.043 + 20x1.04*)
= $(20x5-41032...) = $108-32C... ;
.'. his present ca^iital r^ $(8000 + 108-326...)
::== $8108-326...
II.— Page 184.
1. See articles 181, 182.
2. True discount = $--i^~ = $\^^
= $19.04-i^f
Interest on $ Vt = K W X ^^)
Interest on $400 ^ $-'^
= $20.
Now, $(20-19.04^^) = $.95/^.
3. Discount off £120 = £10 ;
<« -PI in piioxio
£110 ==£
120
I
= £9 36'. id.
SOLUTIONS HAMBUN SM.TH's ARITHUETIO.
i. Interest on $(/0^292-872;for !■ yr. = ts72 ,■
__ «. 100x372
■~ '^9926irri
6. Present value of ade >(; of $(1,052) =, $1 .
" ' $110.25
^ 1.05*
58
■ , ni.—Page 185.
1. Amount of $5000 at end of 18 mo. = $5450.
This was the sum he had to return
Amount of $7500 for 1 yr. = $795*0.
This was the sum he reahzed ;
.-. he gained $(7950-5450) J $2500
2. Discount on $7 for 93 days at 6 7 ^ $.10701 •
.•cash selhng price =. $7-$.107oI = $6-89299
Profit per cwi = $689299 - $5.25 - $l-64299
Hence, total profit =. 43f x $1.6'299 I JJl 88
8. Present worth — $— i*^.ii£2.
See Note I., Aft. 181 ^'''•'^'^•
4. Interest to be received each half vear = $250
Interest on $1 for 1 mo. = - "
^"^^^(i^-+W^M,+i,l,+i,
= $250;
.-. sum X $(6^Vt7) = $250 ;
250
^^i^
'3^4 (T ;
" +U1^)
sum =
6 21
2?Tr
= $41-'''3
Note Thp "r*" i < ^
---^- -»-"« ceUvautuu scuaeiii
page 343
6 may refer to Ex. 1
#
54 SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
5. In<}ere8t on £266| for i yr. = £(266f x J X ^|]<y,
Discount off £83 for 4- yr. = £^lJlli
= £3.
IV.— Page 185.
1. Interest on $6400 for 8 mo. = $218^ =. cli«cnn„f
.-. $6400 -$213^ =: $61861 = sum he has to hnu.
Amount of $1 for 8 mo. .-= $1.03^ ;
'• $61861 " =6186;tx'$1.083^
= $6892.881 ;
.-. sum gained = $(6400- 6392.88f)
= $7.11|.
2. Compound interest on = $(1.08^ - 1)
= $-259712.
Simple interest on $1 = $-24 ;
.-. difference = $-019712.
Sum on which $-019712 is difference = $1 ;
*• *' $985-60 « = $?-«^:6o_>^»
^ 1 97 12
= 150000.
3. Since the discount is the present worth of the
interest,
Interest on £68^^- for 2 yr. .-= £7 19s. l\d. ;
.-. • " £68^^ for 1 yr. = £3;^^7 . '
'^ m\.
ITff
Again, sum on which £im is interest = £63
113 3
■•Tffir
(t
£71.' •'' ^
,1 7 .
2^0 '
<(
7 1 ri 7
== £574 18*.
SOLUTIONS HAMBLIN SMITh's ARITHMETIC. 55
4. Amount of $8000 in 4 yr. = 8000 x $(1.06)*
=- $9724.06 ;
V. ^'s is better by $(9724.06-9600)
= $224.06.
6. Suppose he borrows $100,
then the interest he receives amounts to
2x$(l + 1.02 + 1.022 + l.023) -= $8-248216.
Interest he has to pay = $6 ;
.-. sum on which he gains $2-243216 = $100 ;
" " $269-18592 =:|?A»i8»92x 100
2-2432 1
=.- $12000.
6.
Examples (ci.) Page 187.
X ^ =
6 X ^ = -«
9 X i = «
12 X ^ = '/
16 X ^=i/
45
45
.*. the equated time = _^ = 7^ mo^
IS
6. 16 X 450 = 7200
13^ X 250 = 3875
700 )J0575
15^\ ■= equated time.
It is now required to find the present worth of $700
due in 15y\ mo. tp
Present worth of $700 = $Z2^>ll5?
i.\ru
2 8"
- $6661 2|4
liV4T'
na
SOLUTIONS HAMDLIN SMITIl's ARITHMHTIO.
7.
2 X| =
8 X |f =
6 X 1} = V
The whole debt is due in 4:\ mo. ;
.-. if one-half of it is paid now, the other should no
be paid till 2x4;^ mo., or 8.| mo.
8.
Debt.
$80.75
150.00
80.80
40.50
60.30
When Duo.
Jan. 80
Apr. 3
July 1
Aug. 10
Aufr. 25
No. jf (lavs from
Jan. 30.
(J3
152
192
207
Taking Jan. 80 as the date from which to calculate
the equated time, we have
X 80-75 = 000000
08x150.00— 9450.00
152 X 30.80= 4081.60
192 X 40.50 r^ 7776.00
207 X 60.30 = 12422.10
362.85 ) 34329.7
95 almost.
95 da. from Jan. 30 is May 5.
Time between May 5 and June 2 = 28 da.
Interest on $362.35 for 28 da. at 6 %
=-- $(362.35 x^jV^XtU
= $1.60... ;
.-. $(362.35 + 1.60...), or $304.01... will balance the
account.
SOLUTIONS HAMlJI.tN flMITn'S ARITHMRTIO. 5?
9. £140 is due in 50 da.
£1^0 - 74 da.
£880 «' lOGda.
.-. equated time := LO""+888o + 403flo j^
14 + 13 0+^56^
=: 88 da. (nearly).
88 days from the Ut of March 'is 28th of May.
10. '
1.
, Db.
July 4
Aug. 20
Aug. 2fl
Sei)t. 23
Doc. 5
Debt.
When Duo.
24418./.
7204r)r/.
181088f/.
2()n/)8./.
200.58.f/
Feb. 8
March 5
March 18
May in
May 2H
June 5
No. f>f rtavH from
mth .Ian.
20
61
C4
120
185
148
20 X 24418
51 X 84591
04 X 72946
120 X 181088
185 X 29058
118 X 29058:
084808
1704294
4068544
21802500
4003880
4241094
872902 )_87115JL90
100, nearly.
Examples (cii.) Page igo.
J. Hughes in account with S. Adams.
Cn.
X H7.5.nO =
47 X 815..18 = 38.332!2G
50 X 178.25 -= 9082.00
as X 387.20 - .32i;i7.C0
154 X 418.70 = G4479.80
2175.63 )144931 CO
67 days from July 4 is SepteinvTer 9!
iJne September 9 $mr,.(j:i.
Aug.
S'ept.
Sop-.
Nov.
Dee.
10
1
25
20
1
X 310.00 ==
22 X 675.00 = 14850.00
46 X 512.25 -= 2.3i501..50
102 X 161.75 = 16498.50
113 X 100.00 = 11300.00
1766 ) 66212
.,a ,. ^, . nearly 38.
.38 days from Aug. 10 ia Sept 17.
iJue beptember 17 81765
58
BOLIITIONH HAMBMN HMITH'h ARITHMETIO.
i
If $2175.08 gain a certain interest in 8 days (Art. 185)
$410.08 will gain the same interest in
~-^T6~Hr^ ^ays = 42 days.
42 days before Sept. 17 is Aug. 6.
2. The items of the Dr. side fall due Oct. 12, Nov. 14,
Jany. 17, and Dec. 81, respectively.
I^K. A. B. Conron. Cn.
Oct. 12 Ox -J27.30 —
Nov. 14 a'J X 342.75 -= 11910.7fi
Deo. 31 80 X 175.50 -= 14<J4O.0O
Jan. 17 97 x 212.13 20.070.61
1667.68 ) 4.'H»27.:«
nearly 287
28 (lavB from Oct. 12 is Nov 0.
Due Nov. 9 #1«.';7,«8
Oe», 10 Ox '500 -
Nov. 20 41 X 301) -= 12300 00
Nov. 30 51 X 2.00 = 127.0(».(X)
moo )2r)060X)0
nearly 24.
24 <lnyB from Oct. 10 is Nov. 3.
DueNov.3 *I050
If $1050 gain a certain interest in da. the balance
$007.08, will gain the same interest in
-To't^oV ^^y^ = l^H tlays, nearly.
Hence, the balance will be due on the 11th day from
Nov. 9, or on Nov. 20.
8.
Dr. J. Green in acconnt with Adam Miller & Co. Cb.
March 1,
" 20,
April 11,
" 80,
June 15,
July 18,
Aug. 30,
Sept. 25,
184 X
203 X
224 X
243 X
289 X
323 X
364 X
389 X
720.75 =
81.0.30 =
587.80 =
300.(X) =
625.25 =
560.00 =
68».!K) =
36,0.30 =
1.32618.00 I April 1.
16.0,0O5.(MJ I May 30,
13KMJ7.20 I Julv 20,
72!)00.00 «cpt 25,
18065I7.25
180880.00
24;t:*)3.0O
142101.70
30,
Oct. .TO,
Nov. 20,
243
110
177
!103
3«4
414
700.00 =
nm.m -
.OO'i.CO =
100.00 =
7.00.20 =
329.96 =
500.00 =
00(KK)0.00
i:i8483 2?
55(HKJ 00
17700.00
272322.00
i;i0004.24
207000.00
8450.05 )_820510.11
nearly 238da.
238 (lay« from April 1 is Nov. 25.
Due >fov. 25, 8.3450.05.
4659.30 )J25.0673.65
nearly '200 da.
269 days from March 1 ib Nov. 25.
Due Nov. 25, ^059.30.
Since both sides oi tue account fall due on Nov. 25,
the account should be settled on that day.
Examples cv. Page 193.
5. Brokerage on $578 = $26.01 ;
«
^100 = ^-- '"•
= $4i.
578
SOLUTIONS JIAMHMN SMITIIH AKITIJMETIC.
0. CommisBion on $100 invested = $2^ ;
" |l02i Bont = $2i ;'
•' $8y77 "
69
^ 10 2*"
— S77
7. Beady money payment of $100 =^ $97.60 ;
" ** $7080 = $^J'**»^ »'•««>
^ I b
o Ti? u = $7488.
8. If he sell wheat to tiie value of $100 his com-
mission = $2, and he has $!)8 to invest in silk.
Commission on $98 = $'l"-iLJ __ otu, „ .
, , , " 104 — v^ia »
.*. total commission = $5[o .
•*• ^"'^^°v^««^edwhen$5{;'i8thecom. = $943 •
" $000 "'
<(
^|000x94^?3.
= $9800.
9. Sum on which $1.50 is brokerage = $100 ;
" $570is brokerar(e = $''L«^-L?^
° "^ l.fiO
10. Brokerage on $100 invested = $^5;^^^'
• . " " $100.25 given = $-25 ;
" " $20050 " — $"-Qg,^Q^-3g
'^ 10 0.2 a
Sum invested = $(20050-50) ~ $20000.
Examples (cvi). Page 194
8. Premium on £100 at 2 J- % — £2^ •
sum for which goods worth /97f are insured - £100 •
£4384^V ''
,4384^VxlOO
((
<(
=£
97tf
£4488 15s.
n
ft.
i
1 1
II
60
SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
l«
«
4. Annual payment on 2 policies of $100 each
= $(3.75 + 3.80)
= $7.55 ;
Total payment on 2 policies of $5600 each
= $(55x7.55)
= $415.25.
5. Total payment for cargo worth $100
= $(U + 1 + 1)
= $11^ ;
$20400
= $(204 X HI)
= $473.
6. Sum on which $2.20 is premium = $100 ;
" " $80.85 " =$*-!!•" l*^*'^^'
^ 3.30
— *0/ tJK
But $3675 is tVtt of value of 500 bbl. of iiour ;
.-. value of 1 bbl. = i^^f^^^^ = $9.80.
7. 3^4^ of risk = T-^^ of I of risk + $10 ;
.'. ^U of risk = -^Ya of risk + $10 ;
••• (Ajy ~ z'^%) of risk = $j
2/5 — 2 4
10
risk
10 0^- of risk = $10
= $(1000 X 10) V
= $10000.
8. Sum on which $2i} is premium = $100 ;
" " $71.25 " = ^l±ll^ll^
'2ft
But $8000 is only | of the value of the apples ;
.-. value of the apples = | of $3000 = $4800.
Examples (cvii). Page 195.
1. Tax paid on $100000 = $1050 ;
" «• $5400 =$--1"^''':^®
I u uooo
= $56.70
Die.
ach
each
1 + 1)
< HI)
DO;
75.
3ur ;
! ^/» X 1
'ill
10. ,
)les ;
0. •
SOLUTIONS HAMCLIN SMITH's ABITUMETIC. 61
2. Tax on $8500 = $144.50 ;
** Si _ 14 4. 10 .
^^ - 8500 cents
= 1-7 cents.
8. Tax on $80000 = $1400;
" $75000000 = $110^0000x1400
8 0^
=-- $1312500.
4. Of each $100 collected, $90 is snpnf ,« • .
the school-house ; ^ * '° P^^'^S: for
.'. $8400 requires a tax of $*-i^''M^ = ^8750
Tax paid by $700000 = $875o"^
$1
= Unn cents
li cents.
Examples (cviii.) Page 196.
4. Cost of 8400 lb. =. $6.30.
Specific duty = 8400^ 1 cts. == $42 ,
Ad valorem duty = $^";;^ = $157.50 ;
•'• *«*al duty = $199.50.
5. Value of cotton on whiVli 4:i7 ka • i .
on wiucli $17.50 IS duty = $100 :
*' $1662.50 - . * »
= ^ll? 2.60X100
"^ 17.60
= $9500.
EXAMINATION PAPERS.
I — Page 197.
1. Gain on 24G drams
100 drams
10 drams ;
<(
100X10
2 4»J
10 T
7— drams
2. Since ^«« of debt =
— 4'0C5.,. drams.
$228
! il
.,
-^62
SOLUTIONS HAMBLIN SMITh's AKITHMETIO.
9 3|
TO IF
M X 228
, TTJU
= $225.
8. Value of goods on which $17.50 is duty=
** •♦ " $637 *'
__ (»6JW X 1 00
** 17^50
= S3640.
4. Since ^-^^ of population of 1870 = 5975 ;
.*. the " " = ^QQ^597s
12t
= 47800.
The population in 1860 is 47800 - 6975 = 41826.
5. If r represent the rate per cent.,
then 7600 x (1 + ^^^y = 9196
and (1 4- -^^^)^ — 9 1!' 6 =: 1 01 •
1 + T^ = v/r:2i= 1.1 ;
Tinr = J-
and r = 100 X .1 = 10.
II.— Page 198.
1. He had $1339.60 left out of the part on which he
had to pay tax.
Sum from which $98.60 is left = $100;
((
" $1339.00 " = $±3.3^:«-0J<-i?^
^ 08.60
=$1360;
.*. his entire salary was $(1360 + 400) = $1760.
2. Sum expended on bridge .= ^~\ of $7340
= $7119.80.
8. Sum of 10 results = 10 x 17.5 = 175.
Sum of first 3 = 3 x 16.25 =--= 48.75.
Sum of next 4 = 4 x 16.6 = 66 ;
SOLUTIONS HAJfBLIN SMITh's ARITHMETIC.
63
.-. sum of last 3 = 175-114.75.
Ninth == tenth-1. ^ ^'^'^^'
Eighth r= tenth -4;
.-. sum of last 3 ='3 x tenth - 5 ^ 00.25 ;
• •• 3 X tenth ^ 05.25 ;
.'. tenth = 21.75.
4. Since % of gross receipts = $42525 ;
•*. the " __ laj Jji^o X 4 2 .5 2 5
= $708750.
^ow, 31 % of the capital .. ^'o^ of .$708750 ;
•*• *^6 paid up capital =^ % ^-«^^ •'•.4x708 ij
.-Jixioo
=- $10035000.
o« Part A does in 1 hr ~ i
<(
.'L
T(ioo'
and " 3
IT
Time A and i^ take to do all = ^^ k,
TToo T-TT!(T(r
= ~~'^5 I, .
Hence ^ does *-S 00 V 1 ^>. s Vi.! ' ',
1 7^ -^ TTdTT* or /^ of the work,
and B does 4-«JLp v .< o,. ^^ •+
r. , ^ , 17^ T(5 (T(j. 01 ^\ of it.
Costofjf of vvork= .$85;
A " =$(8x5) = .$40,
~-$(9x5) =.|45.
Ill— Page 198.
1. The cost of a policy of $100 = .$(5| + i ^ .^7^
.-. policy Which covers goods worth = $94.1125 --= 1100;
$7905.45
<!!? JL^_0 fl.4 ff V I o />
^ 4.iT!rr" ~
= $8400.
(<
f ■
1
64 SOLUTIONS HAMBLIN SMITH'S ARITHMETIC.
1
^ 2.
V
1
Date.
Received.
Delivered.
Balance.
Days.
Products.
1
. ^ 1877.
I
January 1
2310
2810
15
34650
16
120
2430
16
38880
: February 1
800
2730
21
57330
1 " 22
1000
1730
7
12110
I March 1
600
1130
33
37290
1 April 3
400
730
7
5110
j April 10
T3«l-»^->. TIT
312
418
21
8778
2730
-1 J ^ n
2312
30)104148
\ iii
I
6471.6 X 5 cents = ^323.58.
^' 2i % = ^1^0^ per unit.
71 y = j.o
'2 /o TCO"
G471-6
<(
101 7 — 4 1
The net increase = ^^ff of males + ^\^^ of females.
The decrease of males ^ /^«y of males
The increase of females =: ^\\ which must
.-. = to decrease of males and total net increase ;
.-. 3V<r of females = ^Vtj of males + ^V'^ of females +
5^^ of males ;
or
^Vir of females = ^\\ of males ;
the numbers are as 31 to 40
T^^- of m of single ticket = $10.50;
.-. cost of single ticket = |.io»x loox lo.so
1 5 X 1 2 ,5
6. Paper duty = UxllJ.
Cost of duty to retail dealer
= .18.
t'a of lu of nd.
= 2-97c/.
••♦'
Products.
SOLUTIONS HAMB.IN SMITh's ARITHMETIC.
66
IV — Page 199.
1. Costof87^yd. = 37ix$4.871 = $182.8125.
49|yd. = 49i1x$.93f =$46.5284875.
on- . Total cost = $229-885987';
Selling pnce ^ ^ ,f $229-3859875 = $805 ^
7 lf;"^"^P^-^«^-^toii lesscom. = -^ of 12000 x
o , ^ . = $825.80.
oum to be inv.estefl in sugar == $^S'±^>^Jjyo ^
Number of pouuds bought = ItfLii^iLljo x 1 ,,
Q XT 1 , = lfi222.11.. lb
3. Number that do well ==,y^ of 750 = 165
barely pass = ^^4^ of 750 = 255'
. ^. " ^""^ = iVtt of 750 = 330.
4. First commission = 1 nf „„^ ,. ,
Sum to be invested = 1 '■
Second commission — 1 of 1 «
, I
. . total commission = (' ' 4- 1 \ .«
((
<(
(I
«
H^"ce TT^o^ of ^^^^ realized' 1*" $70 ;
** << _. (J>1_02_0X70
^ 7
Q . = '11020.
bum invested in groceries = $(1020 - 70^ - ^qrh
NoT..-Sce also solution 8, Ex L., Jage 59"
6. Taking ^'s flour as the standard andreducing ^'s
and Ca to this standard, ^ ^
Amouutofflour^hasof 5'sstandard -- UOofi25bbl.
^ 137.5 bbl.
Amount of flourChas of ^'s '« ^ ^^^^of no of 225 bbh
=- 2G1 bbl.
I
r
66
SOLUTIONS HAMBLIN SMITIl's ARITHMETIC.
Selling price of flour = (125 + 150 -f 225) x $7
= $3500
Sum to be remitted .-= ^%\ of $3500
= $3360.
He must pay $8360 to A, B, and C in the proportion
of 137.5, 150, and 261.
Hence A receives f^|:f of $;^360 =^ $842.30 (nearly).
B receives -^^^'£1, of $3360 = $918.87
G receives ^|^^^ of $3360 = $1598.83 " ]
v.— Page 199.
1. Sum gained, had none proved vvorthlesn
Cost of $1 bill = $(-75 + -01i) :^ 76^ cents.
Sum on which $.23| is gained =
i*
$coox 1
= $2535jf
2. Net sum resulting from sale of goods=|^2 of $1910;
.-. value of goods sold = ?j<^.o of f^|of $1910
:= $2040.
8. Sum invested out of $104 received = $100 ;
((
((
$30056
3 0056X100
Cj. 3 5 G X
— ^- ^0 4
.-. No. of bales bought = ■'^«os«xioo
4.
289x104
= 100 bales.
Sum remitted = 300 x $16.15
= $4845 ;
.'. value of goods sold = $
•'^95
__ ,1^ 4 8 4.5 X 1 .
and commission -- « ^^ <i& 4 8 45 x 100
Toff of $-
85
=: azoo.
\ }
SOLUTIONS IIAMBLIN SMITh's ARITHMETIC. G7
5. Cost of $100 insurance = 15 x |2-8674
= $43011.
itain on |100 insurance = $56-989 •
'• i^sui-ance on which $1709.69 is gain '
__ (& i Too.eox 1 00
'"^ 56.089
= $3000 (nearly).
1. Gain
Examples (cix). Page 203.
116^
=- S3^d.
r
68
SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
7. Sum charged = ^%\ of j^ « of cost
= iffo- o^ cost ;
.*. he gains y^^ of cost, and hence his gain per cent.
= 8.
8. Sum charged = ^%\ of ||^ of cost
= j^^ of cost ;
.*. he gains 10 %.
9. Loss on an outlay of £23 5«. 5d, = £1 Ss. 3}d.;
100 X {£1 3s. Sit/.)
«
it
iglOO
=:£5.
^H
10. Loss on an outlay of £15 6«. Sd. = £3 10.y. 5i</.
<(
^ioo='Ao_x(^;V^'^'-^
ISA
= £23.
11. Cost of 54.87 cwt. = 54.87 X $96 = $5267.52
Gain on $5267.52 = $732.48
((
$100 =
rtt. 100X7 3 248
5267. fl2
= «|)lo.a....
12. Costprice ^ |±i25.6ox loo
108
= $3820.
Gain on $3820 = $(4202 - 3820) = $382 ;
" $100 =$i^^«-^-
= $10.
18. 3 % of original price = $9 ;
the original price =^ $---3---'*
= $300.
14. Cost of 12 lb. = 12 X 2s. G^d, = 30*. 6d.
" 4 lb. = 4 X 3s. 2id. = 12s. 9d.
Total cost
Selling price of 1 lb.
43s. Sd.
j X 43s. Sd.
■= Ss. lid.
SOLUWONS HAMBLIN SMITH'S ARITHMETIC.
15. Cost of 1 lb. of mixture = $i±®*'*JL<'o
"^ i33t
Now$(li'i±-o)_^^^_j^^.
.-. he must have the same quantity of each kind
16. Cost of 80 gal. = 80 x lf;3.G0 = $288
" 180 gal. = 180 X $8.00 = $540.'
Selling price of 1 gal. = $«-^-?AL«J* __ ^.q . J
17. Cost of 80 gal. =: 80 x $8 10 = $248.
OGgal. = 90x$:-}4Lt==$328.'
Selling price of 1 gal. = $UJlll}^o __\o pr,
18. Cost of 3 lb. at 611 ct. = $1.85.
1 lb. at 55 ct. = $.55 ;
.'. cost of 1 lb, of mixture = $''•*?-
^ . = CO ct.
tram on an outlay of 60 ct. = 20 ct. ;
" " 100 ct. = V"-OJ<J_? pf
6 ^'"
= 33 i ct.
Examples (ex) Page 212.
36. Sum paid for an income of $6 = $100 ;
69
(( ct
^5^
i/5 X 100
i.
3'
87. An investment of $125 yields $9 income ;
100X9 .
Again an investment of $75 yields $6 income ;
" $100 «♦ .sjiooxe
(C
«
7a
income
.*. the second is more advantageous by* y.
70
SOLUTIONS HAMBLIN SMITH S ARITHMETIC.
88. Income from £1 in »bo Ifit Htock = £j^s
^'l..l ««
r? 9 O 3
Income from £1 '
Sum invested for difference of income of •£(gV~fV
= dei;
«* " " " " £22i
22^ X 1
= iJ4725.
3
77
89. Income from £00 invested in 8 per cents.
t(
(I
deOG " ill E.R. stocks = £» "JU?
100
= €3.
~ ^ il
= £4.8 ;
.*. the income is increased £1.8.
40. Net income en £91 invested = £(8^-^^)
__ i? s n .1
Sum invested for an incomee of &\^ =^ £91 ;
(i
" £952 = &^'"ij<j\
8 3 3
= £24960.
41. Money from sale of £4500 stock = £l^±?Jiiili^.
First income = £*-y„'^- = ^225.
Second " = £(225 + 108f ) == £893^.
Amount of Egyptian stock =. i^l^I^o
^ £5625.
Sum paid for £5625 stock = £(45 x 112.5) ;
" " £100 stock = £-'-^^^il^iillM
6 6 2a
= £90.
42. Money from sale of ^8200 stock = £^-?±iLili
100
=r £2760.
First income = £-i5^iL2 =_. f nfi
1 u — - _ =
i? 3
*T7'
SOLUTIONS HAMBLIN SMiTU's ARITHMETIC.
71
Income from investing £115 =£i;
" *' £2mo — £^Z1»^*
11 a '
.'. no alteration is made. "^ "^^^ '
48. Income from investing $105| = |6;
" $8229 = «**=*» '^ 6
.-. it is increased by $(4G8 - 4II.45J ^^^^ '
AA J n. ^= $56.55.
44. Income from mvesting $108 in Ist stock
$216 in 2nd **
o'b~r
Sum invested for income of $16| « = $3247
$1674 = $l?Zi^x 324
16 10
_ , = $32076.^^
Sum invested in 1st stock = $^^^^li
tt
2nd
= $10692.
* = 2 X $10692
_ ^ • = $21384.
45. Value of $5500 in currency = $^*-?-^?<J2i.
.-. he will gain $,.000-6875), oriJf ircurrency.
46. On an investment of $92 I lose 7.
Income from investing $85 = $ "^ « ^jI
'^ 100
' • s"^ gamed each year = $(41 _8)
hence number of years = -^ yr, = 5.3 y^^
72
80LUTI0NH UAMIiLIN HHITH'h ARITHMETIC.
47. Money received from sale = $ - -..,. ~,
^ . . ,. 4000x14
Original income = $ i/)/)—
= $560.
Newincome = $19^!?.^™2x9
100X126Y"
=- $514.77... ;
.'. my loss = $16.22... .
48. Loss on each $100 of stock bought == $2,875.
He gained the dividend at 8^ % = $3.60 ;
.'. amount of atock from which the net profit is
$-626 ^ $100 ;
• •
«(
«(
" $87.60
__ di. 37.60x 100
- = $0000.
49. Profit from investing $1 in 6 i)er cents
'99 - ^164'
<i
<«
** 6 per cents
6
86^
hence the former is preferable.
Ka n ■ ^. /1376.60X100
60, Gross income = $ |
n7 » '
+ 400
\ 98i
= f 1800.
(See Ex. paper II. ex. 1, page 62.)
Investment to yield an income of $6 = $101 ;
*(
«
" $1800 = $J-10AlJA_i
, dtorfcono
BOLUTIONH HAMBLIN HMITH'8 AKlTHMETIC.
78
,3 9 3.7 ax 100
61. Since 8% of his stock £2400 .
.'. his stock :r.. 4»-L'lO--^ a-tOQ
= £80000.
Cost of .£8000 stock = ^soooo x 9jj_ '
= $(800x94^x4.80.1);
•'• income in Canada = $?-"- ^ "l*ii 4.8oix la
'^ 100 "
= 144092.
62. Money from sale of £4500stock = £*±^^ ' »>•«
loo ~
First income = jpl*9<l^» r.^^^ ' ^'
s««^ 1 • ioo~ = i'225.
Second income = £-(225 ^ 168j) « £393.75
Amount of Egyptian stock = £^»-^-7«
7
Sum paid for£6025 stock = ,£(45 X 112.5)
** £100 stock = ,£*««^*«xua.a
5 c 2 a
Hence the market price of stock = M(m - i)
= £89f .
63. Amount of stock bought in the 6's
^^.Sum^xjOO
Amount of stock bought in the 7's ^^^
__ Sum X 100
- ^ Y02 — " '
^^->li29 ^ sum X 100
9U - * 1Q2 = 13500;
.-. 8umx(^V* -T^^) =$35
and sum = $— ^^
i I
74 SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
Income from investing in the 6's = ^ ^ " ^ ^ ^ o ^^ x «
= $2040.
Income from investing in the T's = <^±JL" 1x102x7
« 12135 ;
.'. difference of income = ^96,
64. Income from $124.5 — $5 ; ♦
$100 =$'^^
Income from $( '^7 x 34) =^ $ 1^ ;
100 X 4 «
- ^$100 =.$ 100x34
^«0~
65. Since 3^ % of the capital
= 3 % of (capital -$1200000) + 5 % of $1200000 ;
.-. f % of the caijital = 5 % of $1200000 - 3 7 of
$1200000,
and i % of the capital = $21000,
and cajjital = ^^00x2 4000
00.
= $3200000.
Sum to Ve invested = $« 1 « x 1
lost
eifioxiooxioo
Value of this sum in gold = % -
° ^ 1021X116
Number of yards bou^'ht — ai'ox loox 100
•^ ^ 1.034X 102ix 1 15
= 5040 1 51 « yd.
EXAMINATION PAPERS.
I. — Page 215.
1. Desired sale of goods worth $91 = $107 ;
" 3x$728
_. ({1.3x^7 2 8x 107
^ 01
= $2568.
((
i(
SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
Entire cost of tea = ^ ^ ^ 9Jl.i_:« o x-j o o
Entire selling price = $ '^o xj.s o*x i oo x 1 1 o
belling price of 50 lb. = $90 ;
1001b. = $180] 2-
lib. = jj5l.8(l|2.''
4. Marked selling price := .1 -. 5 of n.cf •
~ Beal u ^ i^vd of cost price.
__^ I A of f;}^ of cost price
.1. , ~~ TO A* of cost price :
. . 1113 net gain = 2 1^ „ f^
= 21i %.
• ^'•°' miirei to take i,p tl,e bill = | = ".oxi_oo
Interest on |2400 each ,„a.,e,= |(24o*o:f I^o^,
Amount of $27 for S payn,e„ts ^ f'x.S(l,n .7
Total i„te.o.t receiver'"" '^^^^'l-S^)-
^ i \^-^^?J(F) + ( 171-^,^)2 4- n n \3)
= 27x.f;4.0G8... -+10J413..)|
-$109-836...
I
'H'
li.
.76 SOLUTIONS HAMBLIN SMITh's ARTTHAfKTlC.
Sum realized from $2400 = $2509-830... ;
.-. his loss would = $(2520-2509-836...)
= $10-163...
II. — Page 2i6.
1. Present worth of $2.45 = $^*-^-il?-2
^ ^ 1 /»
= $2.33i-.
2 Conditional price -= ^-^% of selling price.
Amount of $100 for 3 mo. = $101^ ;
.•. actual selling price 3 mo. before
! 100
~ lOV- ^^ "^^^ ^^ selling price.
88-«
= j-^j^ ol selhng price ;
.', discount allowed = Ynn " "
— Ill 0/
Again, actual selling price 3 mo. after
101:^ 90
'"Tod °^ iQQ of selling price
91^
.'. discount allowed =
Yqq of selling price ;
100
— 8^ °/
— "« /o*
8. In 1 oz. avoir, weight there are '-?A? gr.
Costof 57G0gr. = $1.20;
♦« 7_ooo irr S-A5 _______
1 A
o760
~ 9^^ cents.
w
fiOT.UTtONS HAMBLtN SMITH 's ARITHMETIC. 77
4. See Art. 198.
Price of ^10000 stock = i?9000 ;
••• ^ •' £'100 stock = £90.
6. Money got from sale =:= $ ll^.'L^" «
'^ loo
= $1032.
lucome from 8 per cents = $' ^''*'^\
- |86 ;
.*. price of 8 % stock = $ ' "-l^ ^8
= $229^
III.— Page 2i6.
1. Sum invested = $^*'?-^L?L*'
'^ 10 2
= $3000.
Number of pounds bought = ";":° = 4000 lb
Total cost of 4000 lb. = $(yi)GO + 30 + ^..^of 3000)
-^$3100;
.*. selling price of 4000 lb. — I;'' V^f ^lo .
'" ion *
" " 1 lb. = $ll^.?_>lA4o
"4000 X 1 6
= $1,081.
2. Selling price = $ '^ " ""M »
• ^ 1 00
= $60.
But $G0 is only ^'^ of asking price ;
.*. asking price = $ ?±>112y
= $80.'
3. Present worth of $2.25 = %?^2±'m2
= $2.14|..
Hence A buys at the lower rate.
Marking price of ^'s silk = ift^i^ii??
^ 100
== $2.67^.
!
I.
h 8
78
SOLUTIONS HAMBT.IN SMITH S ARITHMETIC.
Marking price of ^'s silk = ^'-^J^J^^^s
Gain on an outlay of $2.14|- = fl.So^ ;
.-. " " " $100 r=$., :%
*iJ • .1. X. ir
= i|40.
Gain on an outlay of $2.15 = $-85
.: •* " " $100 =$./tV
4. Supposed cost prico --- l^y of cost price.
Supposed selling prico — |^ of cost price.
Then ^^ of cost price — '^^^^ = -j-^^ of |-^ of cost price ;
and {U-U%) ^'>f cost price = f^V ;
•'• ^Ijj ^^ cost price = $3^,
and cost price — %%^^
=-- $10.
5. The first payment of interest is $6, and will be
due in 1 yr. ; its amount for 2 yrs. will be $6(1.05)- ;
similarly, the amount of the second payment will be
$G(1.05) ; and the amount of the third payment will be
$G. Hence, if P represent the present value of the
bonds, we have
P(1.05)3 = 100.4-6(1.05)2 + 6(1.05) + 6
= 1181)150 ;
.-. P==$102 723....
IV. — Page 217.
1. Value of $4 currency in gold = l^"",'^-
Gain on .f 8 ^l?;};
inn V 2.^
S=S <|j>l^.i'U. .<> •
((
IC.
cost price ;
ind will be
$6(1.05)2 ;
3nt will be
lent will be
due of the
-6
2 3
SOLUTIONS HAMBMN SMITh's ARITHMETIC.
2. Selling price of cheese == 24 x $30 = $720.
79
Cost
Cost
iax30xioo
12 cheese =; $^
^ 130
'TVt
12 cheese = $i-?AiL« \19.9
" 7
=^s514f;
.-. total cost ==s70i|!^;
.-. net loss ~ ^T^l'i-
3. Askiitig price -■ |j;« oi cost price.
Selling price .. ,-', oi {^^ of cost price
= -' " ^ "> of cost price.
TOO II
Hence j -•] « « of cost price -~ cost price - $528 ;
cost price- ^ ^1«'>'"<5 28
S K
-- $6000.
Asking price - {;7;; of $60(;') ..= $blG0.
Soiling price - ^^^ of $8160 -^ $6528.
1. If S represent the sum first invested,
every $73 invested will give $3 interest ;
" $1 '* " $.,^- "
" $S " - Sx$v^3 "
_ and this intorest, S x S;/^., invested, will give
S X f'j X $y\ interest.
Thus at the end of 2 years there was on hand the first
investment and its 2 years' interest, also the interest on
the first year's interest, also a second investment of S
mid one year's interest o ^ it to meet the debt of $1085.
Hence(S + 2xSx,VN !Sxy^xA)+(S + Sx/V)
= $1085 ;
••• ^ " V,ViV = $1085 ;
^.^ j^ ^108.-, X5 3 29
^ 1 1 3 ■.' 4
= $510.59....
'<
I II il
80
SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
6. Net capital Jan. 1, 1875 = 1(40000+ 1750-9350)
= $32400.
Net capitalJan. 1, 1H70 = $(39750 + 2850-7550)
= $35050.
Amount of $32400 at 6% == $34020.
Gross profit = $(85060+1500-84020)
= $2530.
v.— Page 2 1 8.
1. Tiie dividend ^ ^^,^ of stock.
Amonut of new stork purchased
^ V^^ of 3 « ^ o: stock
' of stock.
— T(T
Hence f' of stock =^ $18750,
and stock = $12500 ;
/. the dividend = -^f^ of $12500
= $1000.
2. Cost of $100 of stock -= $76f .
Selling price '• «' = |82f ;
•'• gain on " " = $6^i^.
Amount of stock to gain $6yi2 = $100;
((
$121.66-1 = | 121.66| xl00
6,V
= $2000 ;
.-. number of shares = '-^ = 40.
3. Value of $400 U. S. currency in gold
(J; 1 G 00
— v i~
Sum from which $2i is half yearly dividend - $100 ;
tfli 1 600
V 7
((
$
^^ X 100
^ $9142f.
f*!
SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
4. Whole sum to be collected = $1700000.
Sum already " = $1050000.
Sum to be « = $660000.
Percentage which $650000 is of $1600000
__. lJ)0xG30000
1 S ~
6. Amount oi stock bought = <|'-^?-l^-U_i£
Money from $12000 stock = $
Money from $6000 stock = $
91
$18000.
1200 X 9 3. a
1 do
$11220.
6000 X 86
Money from both sales
.*. loss
1 00
$5100.
$16320;
$60.
• Oiignal income = $ L12^>ii
New Income
Honce increase
1 00
$540.
d|.l 6^320 X 4.5
f|720.
$180,
102
Examples (cxi). Page 220.
8
i + ^
i + i
7 7
1st share
2nd share
3rd share
4 oh oharo
= 7^ of$8470 = $3300.
6
1.
= rj of $8470 . . $2200.
= ,- -^f $8470 -=. $1650.
1
= Tr Of $8470 = $1,820
ifo
81
ft
I'l
6'
,1* I
Piu
82
SOLUTIONS HAMBLIN SMITh's AKITHMETIC.
4. 75 -j- 10 + 15 = 100.
Amount of Saltpetre =. j^,. of 1200 lb. = 900 lb.
Amount of Sulphur =-- ^'^o^ of 1200 lb. = 120 lb.
Amount of CJmrcoal = Vo'^ of 1200 lb. = 180 lb.
<<
((
= 12.
= f*2of480yd. = 120 yd.
= A of 480 yd. = 100 yd.
= A of 480 yd. = 200 yd.
^- 8 + 4-1-5
Length of Ist side
2nd "
3rd *«
G. Bepresenting B's share by 1,
^'s share will be 8,
C's share will be 4 ;
.-. all the shares --= 8 times /i's share.
8 times B's share = $640.
^'s " = 8 X 180 = 1240.
C's " = 4 X $80 = $820. •
7. When the second receives 8 apples, the first
receives 7 and the third 10.
7 + 8-1-10 = 25.
. Share of 1st = ^\ of 100 = 28.
2ud = ,2% of 100 = 32.
3rd = I" of 100 = 40.
8. 5450s. + 7085.S-. -f- 9810.v. = 22345.S.
A gets ^VirVTr of £418 19s. 4U = £102 86- 9./
f " sVirV^of £418 19.. 4^ =: £132 16... 10'^
^ " ^Vir'kof £418 19.S-. 4i,;. .^ £183 18s. 9c/.' '
9. 4150 -f 12450 + 24900 -t 29050 =.70550
Share of Ist town = ^^ of 1921 = 113.
• " 2nd - =. ||4.o of 192J _ yy^^
^^d " = f^n^ of 1921 = 678.
^*^ •' = ti}M§ of 1921 == 791
<(
<(
((
<<
((
ti
.(
SOLUnONS HAMnLm SMITH'S AKlTHMETIC.
-^ on. + i^s. + U. + Qd. + id. =. 282./. ;
.'. Number of each — ^±xioj<2a = gQ
11. £(500 + 350 + 800 + 90) .=. £1740
Share of 1st = ,VVV of 200 a. == 574, „
" 2ud :^. VV\^, of 200 a. = 40.|o a'
" ^^-d = A'V^)- of 200 a. = oil" a'
" 4th=^=.o^.of200a.=:io|oa.'
12. If Ji gets Is., A gets 9d., and 2,.
U + dd. + 28. = i5d,
^'s share = /^ of 45*. = 9^,
^'8 " = U of 45*. = 12,.
6^'s " = II of 45*. == 24*.
13. The^pay of 7 women = the pay of 3 men.
. " 14 boys = " ^i OS
-^ 01 -5S women.
= •* 12
5 men + 3 men + ^^ j^^^ _ ,, ^^^^ ^ ^^^- .
Share of the men = ~ of $10.40 =
88
women ^ ^^ of $10.40 ^ $3.
boys = ^^ of $10.40 =. $2.40.
and'i5lTdr:n:" "^ ' ^^"^^' ^^^^ ^^^ ^^ ' -n
But the share of 9 women = share of 6 men
--d '' 15 children = . 5 J°
6 + 6 + 5 = 17.
Share of men = ^«^ of $517.65 = $182 70
•• ^o;^6n ==^V of $517.65 = $182.70.
chUdren = ^y of $517.65 ^ $152.25
■ (
if
I
84
SOLUTIONS HAMBLIN SMITH 's ARITHMETIC.
15. 20 4- 18 -f. 12 = 50
Share of youngest = j.^ of property = $1440 ;
.'. the value of the property = ^^J-lt®
= $0000.
16. Take B's share as the unit,
then C'« " =lof /i% + $800,
and ^ •« " = I of (f of B's + $800) - $300 ;
Sum of all the shares = ^ r. ^f j^^^ ^ ^(J^^ .
.-. I J of B'8 + $044* = $5000 ;
i^ of B'8 = $4055f ;
B's = $2500.
C"«= 2 of $2o00 + $800
= $1800.
A's = g of $1800 -$300
= $700.
17. Take D's share as the unit.
C's - =Aofi)'«-$100.
B'8 " =lof(T«,of />'*-$100)f$200
= U of D'8 + $120.
A's " =;iof(f«ofiy'^+$120) + $250.
Sum of all the shares = i/^« of D8 + $350.
.-. VV of ^'« + $B50 = $5000 ;
U oi D's = ^4:050 ;
D'8 = $1500.
^'« = Ti> of $1500~$100
=^ $1250.
^'s = I of 51)1250 + $200
= $1200.
J'« = t of $1200 + $250
= $1050.
8f LUTIONS HAMULIN HMITh'h ABITHMETIC.
18. Take the first fraction as the unit,
then the second = |§ of the first,'
and the third = 2rg *«
«<
((
- 4 .
Sum of the 8 fractions = ^ « « « ««
••• VaV of tl first = III
audfirst= A».a_>«i8 3 __
1586x242 — T5 . „
8econd^«|of,VA-f'A;
tliird = II of AW = T2^sV
19. Simple interest — $ ' J ' > x > •' x e
^ loo """~"
=- $5)13.38.
aiTTy
Share of 1st = J^ of $913.38 = $175.50.
«
i(
((
({
2nd
7
rS^, of $913.88 = $218.40.
^'^ ^ *XV ^^ ^'^^'^-^8 = $252.72.
3 o O
4th
5th
HTT of $913.38 = $117.00.
o 6
8
r^ of $913.38 = $149.76.
»♦
20. ^+^ + ^ = 1;
.-. the boys over 15 get ^ of $400
= $200 ;
the boys between 10 and 15 get i of $400
= $133i ;
and tne rest get I of $400
= $CGf .
Number of boys over 15 = 200 x 2 = 400 ;
** *♦ in school -= 3 x 400 = 1200.
85
IMAGE EVALUATION
TEST TARGET (MT-3)
A
4MJ..
m
mt
I.I
11.25
M 11121
1.0 f^~ II
.IT 112
1^ lii {III 2.2
- lis 111 10
1111=
U III 1.6
(^
%.^
/]
^/
?y
C?"^^'. 0%
-c^l
y s>
^^ m ^ *"*
#
V ^'■F ^»
-%
/
Photographic
Sciences
Corporation
23 WEST MAIN STREET
WEBSTER, N.Y. 14S80
(716) 872-4503
\
w
'^
'^
<v
#<;iV
^<i^
6^
86 SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
Examples (cxii). Page 222.
1. Kent of a house worth $2592 ^ $132.75 ;
$864 r:. $iiii-fLlHJL£
^ 2692
-$44.25;
<(
__ ^ 1728 X 132-73
^ 2592
= $88.50.
2. 18 cows eat as much as U horses •
and 90 sheep ♦« " 15 horses.'
6 + 9 + 15 == 30.
J 's share = ,«,- of $22.50' = |4.50.
^'s " = A "f $22.50 = ^6.75.
^'8 " = if of $22.50 = $11.25.
8. Profit to be divided =. ^ of $25780 = $5156
A contributed f of capital ; B, ^% of capital and C, ^
of it. ^
.4'8 profit ^ 2 of |gi5(. ^ $2062.40.
^'b - -Aof$5156=- $2320.20.
C's " -i/Vof$5156 = $773.40.
4. .I's money was in the business 287 days and B's
167 days.
287 X 2400 = 688800
167 X 1800 == 300600
989400
••• ^'s share = «f sfgo of $943
and B's share = U'^^U of $943
■ - $286,^,^^.
5. 8x40 = 120
4x75 = 300
420,
3
B's
SOLUTIONS HAMBLIN SMITh's ARITHMETIC. 8?
D's share ^ ^^ o ^f
E's " = |oo of
6. 7 X 500 = 3500
8 X 600 = 4800
9x900 =8100.
= $20.
= $50.
16400.
A'8 share --. ^{'£^^ of $410 =
^'* " =AVAof$410 =
^* " = iVAV of $410 =
= $87.50.
$120.
$202.50.
7.
3x10 = 30
4x12 =-48
2x14 =28
106.
Share of Ist = ^VV of $106 = $30.
2ad = T^*/^ of $106 = $48.
^^^ =-TVffOf$136--= $28. •
8. First works (6x9 + 6x8) hrs. = 102 hrs.
Second " 10x9|hrs. ^ 95 « *
<(
((
197
Share of Ist = joa of $29.55 -= $16.30.
Share of 2nd ^%^^ of $29.55 = $14.25.
9. 12x400 = 4800
10x5U0 = 5000
12x300 = 3600
9 x 600 = 5400
18800.
A'8 share = ,P/^>^ of $470 - $245.
9800 = A'8 capital for 1 mo.
9000 = B'8
u
4<
B'8
((
?^\%% of $470 = $225
88 SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
10. 12x3000 = 36000)
32x4600 = 64000) =9^000.
16x3600 = 62500)
9x2600 = 22600) "= '^^^^^•
12 X 2250 = 27000 = 27000.
192000.
Johnston's share = iVi^V^nr of $1248
= $586.
Wilson's share = ^^^%oo_o^ ^f |j248
= $487.60.
Miller's share = ^VgO^oo^ of $1248
= $176.60.
11.1^x10 = 16 horses = 45 sheep
2 xSO =^ 60 oxen = 120
3ixl00=^325 sheep = 325
2^ X 40 — 100 horses =z= 300 sheep)
<(
({
Hx50 =621 oxen
125
8 X 115 = 346 sheep = 346
<(
(«
I
A'a share =_- ^^^V of $88.20 = ^o4.30.
^'« " =--TyA of $88.20 =$53.90.
= 490.
= 770.
1260.
12. i + i+i + | =
77
As there are only |o i^ j^jg property he could not
possibly leave ^.
1
IT
A's share = -Jj of $1886.50 = $735.00.
1
B'B " = -}j of $1886.50 = $490.00.
O's
Z>'s
77
1
= T7 of $1886.60 = $367.60.
(«
I
3"
=^ T? of $1886.60 = $294.00.
SOLUTIONS HAMBLIN SMITH's ARITHMETIC,
89
13. 36 + 64 + 78 = 168,
A'a = ^%% of 78 gal. =r 16« gal.
^'s = tV^ of 78 gal. = 25^-1,- gal.
14. A uses the whole house for 4 mo. ; half of it for
5^ mo., and ^ of it for 2^ mo,
B uses I the house for 5^ mo., and ^ of it for 2^ mo.
uses } the house for 2^ mo.
4 X 1 =4 ]
5i X i^2f[ =
5^ X I- 2J,
1 _ « [ =
1 82
2¥ ■
2| X
2i X
90
If?-
ot
Sum =: 2^S^8,
1 82
^'8 ^hare= ^g of $187.20 - $118.30
5's
((
(<
8A
= |m of $187.20 = $5.5.90.
30
_2T
-=■^8 of $187.20 =
¥¥
Examples (cxiii ) Page 225.
1. Resouboes and Liabilities. , Ownbbship.
Db. rjD
•3456 |325(i
ml _^6
tfSO $3696
8080 Resources at clo&ine.
3596 Liabilities.
4484 Present worth of firm.
2510 Credit excess of Ownership,
1974 Net gain.
987 A'8 share of net cain.
987 B'a •
Dr.
8175 A withdrew
315 B
Cr.
<(1500
1500
490
Total investment 3000
'■ withdrawn 490
Firm's net investment 2510
Hence A's net capital = $(1500-175 + 987) = $2812..
B's " " = $(1500-815 + 987) - $2172.
90
SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
2. Rbsourobs and Liabilities.
lg2 , 1240
7680 Besources at c'o 1 ic.
3C90 Liabilities.
3990 Present worth of firm.
Dr.
11000
685
1860
3545
OWNEBSHIP,
Cb.
96000
420
4000
280
InveRtment 10700
Withdrawal 3545
Net invBstmeut 7155
Debit excess of R. & L. 3990
Net loss 3165
A' I of net loss 1899
I B\ 8- of " " 1266
Hence A a net capital at closing
= ${(6000 + 420) -(1000 + 685 + 1899)} = $2836;
and ^'s net capital at closing
= $1(4000 + 280) -(I860 + 1266)} = $1154.'
3. Besources AND Liabilities. Ownbrship.
Dr.
»2263
5000
7263
Liabilities at closing
Resources "
Insolvency of firm
Cr.
$3846
4462
676
8983
7263
1720
Dr.
$2860
5560
8420
Investmenft
Withd awal
Net investment
Credit excess of B. & L,
Net loss
.4 's ? of ret loss
B'B
Or.
$6000
4000
251
10251
8420
1830
1720
3560
also
1421
Hence A'b net capital at closing
= $(6000 - 2860 - 2130) = ^1010;
and B'a net capital at closing
= 1(4250 - 5560 - 1420) = - $2730,
i. e., B'b net insolvency = $2730.
Examples (cxiv). Page 230.
1. Diflfs.
30
23
35
5
7
Arranging as in Ex. 2, we see
that 5 lbs. at 23 cents, and 7 lbs. at
35 cents, will form a mixture that
may be sold for 30 cents a pound.
11
SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
91
Cb.
96000
420
4000
280
Or.
$G00O
4000
251
10251
8420
1830
1720
3550
2130
142)
2. Diff8.55,
20
86 3
60
2
25 802
3. DifFs.TO
I
16 65
••• ■ .
575
20i90
2,5
2,3
1,3
We see that 3 bushels of oats, 2
bushels of rye, and 2 bushels of bar-
ley would form the required mixture.
Of course, any multiples of these
quantities would satisfy the condi-
ditions equally well, so that we might
take 30 bushels of oats, 20 bushels of rye, and 20 bush-
els of barley.
We find that 2 lbs. at 65 cents,
2 lbs. at 75 cents, and 1 lb. at 90
cents, may be sold without gain
or loss. But there^are 30 lbs. at
90 cents. Hence we must have
2 X 30 lbs. = 60 lbs. at 55 cents,
and 2 X 30 lbs. = 60 lbs. at 75 cents. Or, we may take
5 lbs. at 55 cents, 3 lbs. at 75 cents, and 3 lbs. at 90
cents ; we will then have 50 lbs. at 56 cents, and 30 lbs.
at 75 cents.
We see that 4 gallons of alcohol
at $1.50 and 1 gallon of water
will form a mixture that may be
sold for $1.20 a gallon. But there
are 15 x 4, or 60 gallons of alco-
There must, therefore, be 16 x 1,
or 15 gallons of water.
6. 12 gals, at 36 cents each = 432 cents.
8 '* 66 *' = 448 "
20 gals. 880 cents.
Hence cost of 1 gal. = %^^ = 44 cents.
The question now is, how many gallons of Kerosene
oil, worth 60 cents per gallon, must be mixed with 20
gallons of another kind worth 44 cents per gallon, so
that the mixture may be sold for 50 cents a ballon.
4. Diffs.|1.20j
80
1.20
1.60
hoi in the mixture.
02 SOLUTIONS HAMBLIN BMITH's ARITHMKTIO,
As before, we have
DifPs
6
• • •
10
50
44
60
But there are 20 gallons, or 4 times
5 5 gallons at 44 cents. We must, there-
... fore, have 4 times 8, or 12 gallons of
3 Kerosene oil.
6. 16 bushels at 48 cents each = 768 cents.
12 " 34 " = 408
((
28 bushels
1176
Therefore the cost of 1 bushel = ~~^ cents
=■- 42 cents.
As in the previous question, we have
Diff-s.
14
4
24
66
42
60
80
2
That is, 2 bushels at 42 cents will
balance 1 bushel at 60 cents and 1 bushel
at 80 cents. But there are 28 bushels at
42 cents. There must, therefore, be 14
bushels of rye and the same quantity of
barley.
7. Diff's.
10
6
24
14
18
30
We find the proportional parts to
form the mixture to be 3 lbs. at 14
cents, 3 lbs. at 18 cents, and 8 lbs.
at 30 cents. Adding these propor-
tional quantities we find that they
form a mixture of 14 lbs. But the
required mixture is to contain 84 lbs. Hence- -fl = 6
= the number of times the proportional quantities
must be increased in order to give the required quan-
tity of the mixture. We shall, therefore, have
6x8 lbs. =1 18 lbs. at 14 cents,
6x8 lbs. =. 18 lbs. at 18
a
and 6 \ 8 lbs. = 48 lbs. at 80 «
SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
98
8. Diffa.
6
2
39
6
38
87
45
If we take the first proportional
parts indicated, we have 1 lb. at 33
1, 8 cents, 3 lbs. at 37 cents, and 2 lbs.
3, 8 at 45 cents. Adding, we find the
proportional parts form a mixture
2, 4 of lbs. But the required mix-
ture must contain 120 lbs. Hence -~° = 20 := the
number of times the proportional parts must be in-
creased in order to give the required quantity of the
mixture. We shall, therefore, have 20 x 1 lb. ^ 20 lbs
at 33 cents, 20 x 3 lbs. ^- 60 lbs. at 37 cents, and 20 x
2 lbs. = 40 lbs. at 45 cents. If we take the second
proportional parts, viz., 8, 3, and 4, we find that they
form a mixture of 10 lbs. Hence VV = 12 == the num-
ber of times the proportional parts must be increased.
Hence, we have
12x8 lbs. = 86 lbs. at 33 cents, 12x3 lbs. = 36 lbs. at
37 cents, and 12 x 4 lbs. = 48 lbs. at 46 cents.
Examples (cxv). Page 237.
!• Since iJ1500 = $7300 ;
••• ^1 = $Hh
Now the advance on $4* = |(4|^ _ 4|) = $|| ;
it
((
100xJ»
4*
¥5
*' $100 =
Hence exchange is at a premium of 9^
quotation would be 109|.
2. Since 5-3 fr. = $1 ;
.-. 286874 fr. = $iil-«iliiL
= $44693.20...
3. Since 12 fl. = 25.56 fr.;
.;. lfl. = -^^^fr. = 2.13fr.
= 2 fr. 13 cent.
= $9h
and the
94
SOLUTIONS HAMBLIN SMITH'S AttlTHMETIC.
4. Siuce 25^ fr. = 2244 copecks ;
y n f .. 3 0X3344 1
zu ir. = — ^^^ — copecks
.— 1700 copecks.
5. Since 25Hr. = ll^g fl.;
20 V 1 T"-*
= fl. 20 kr.
6. Since5.12i.fr. = $1 (gold) ;
.-. 12600 fi% = $'7«J^?Ll (gold)
= $2472 (gold).
Now $100 (gold) = $185^ (currency) ;
$2472 - = .^^JLl^^iiLy (cui-rency)
i 00
= $3845.44.
7.^ Since |^« of $4| = £1 ;
2707.80 X 1
:. $2767.80 = £
= £
2767.80x100x9
108 X 40
= £576. 12s. 6d.
8. Amount of gold in $1 = ^ of ^j%» gr.
= 23.22 gr.
Amount of gold in £1 = U of ~^- ^ ?-*'- gr.
11X40X480
1869
gr.
Now, 23.22 gr. = ^^ ,
11X4 0X480 __ (jh l 1 X40X480X1
6 • <!? 2 3.22 X 1 869
= $4-8665..,.
1869
9. Since 100 fl. = 209-25 fr. ;
.\ 12.16+ fl. = 12.l6 2 5^X2^.jft
* 100
= 26 fr. 45... cents.
SOLUTIONS HAMBLtN SMITh'b ARITnMETlO.
95
10. Since 18 fl. = 20 mar. ban. ;
.-. 80 fr., or 14 fl. = L*'*^" .. .
18'
and £1, or 25.5 fr. = ^±11^^11 mar. ban.
;> 0x18
= 13-- mar. ban.
Since 14 mar. ban. = .£1 ;
11.
20 mar, ban., or 18 fl. = 4.-
20X1
14 '
and 28 fl., or 00 fr ^ fil^liH^'^.A ,
18x14'
4 fr., or 72 cents = i'l^^^Jl^.o ^ ' ;
60X18X14
.• .'^I — x> 100X4x28x20x1
73X60X18X14
=: i>l Oo
Hence £1 = |S;4.8G.
12. Since 20 fr. = 40.5 fl. ;
.'. £1, or 25.7 fr. = '^'•'^^"•^ fl .
2 '*• »
50X25 7 X 4 0..->
and £50 =
2002.125 fl.
fl.
18. Since 25.65 fr. = 240 d. ;
.'. 3 fr., or 525 rees
3X240
<L
26. G5
and 1 ree = -lJil±^.d •
52 BX 2 5.C -> '
.*. 1000 rees = -l-«ioji^JLlU?^
f>25 X 2 5.65 •
= 5'S^(l. nearly.
14. Since 1 oz. Eng. gold = ?,]-^^ oz. of Fr. gold ;
.-. 1 oz. English gold = ill i]! of 31.1 grammes.
Now 10 gram. — 81 fr. ;
315 1x31.1 -,3151x31.1x31
.^ , . „ gram. _
JlOO ° 10x3100
3 16 1x31.1x31
.*. 1 OZ. Eng. gold =
10X3100
10X3100
fr •
•11. ,
fr. Fr.goia
and hence 1 fr. Fr. gold = j^A^x#if«^; oz. Eng, gold
= -0102045 OZ. of Eng. gold.
f
96
SOLUTIONS HAMBLIN SMITH S AfilTHMETIO.
Now 77 J«.
—
1 oz.;
•
• •
^1
=
!* X 1 ^_
7 71- ®^-
And
•0102045
OZ.
■
1 fr. ;
30
• • TTJ
oz.
=
30
77Jx.(» 10304ft
25.17 fr.
fr.
1.
EXAMINATION PAPERS.
I. — Page 238.
8 X .96 *= 2.85.
7 X 1.15 = 8.05.
12 X 1.36 = 16.32.
22
27.22
Hence sp. gr. of mixture = -^•-'
" =f offofC's
^\ of C's.
22
, = 1.2372...
2. Sum of which $6291 is the interest
qt. G291 XlOO
- ^' ' ~^Th —
= $17475.
Take C's money as the unit ; then,
= U of C's.
Sum of all their money = C's 4- -j% of Cs + 1|
2 3 3 of ^'h •
.-. Ytf" of C"s = $17475 ;
, • and C"s=$«-^LL11?
= $6000.
^'s -- ^\ of $6000
^off of/^of 6"s
/
= $5400.
^'s n=. ^ of $6000
= $6075.
/
8.
SOLUTIONS HAMBLIN SMITH 8 ARITHMETia
Since 21 fr. == $4 :
97
.'. 10 mar. ban., or 851V. — $ -jT"
and £7, or 96 mar. ban. — f-"^
and £1200 ^
»fl X 3fl X4
x~2ll
19()0xflflX''<ffX4
•t' 7x10x21 '
= $5774.48...
4. By direct exchange $100J -r: $100 at New York ;
.-. $14831.00 = $Vl^''|;?£?ii2.«
= $14224.91....
$1 Cinn. = $(i oo~ft) ^*' I^^uis
= ^CwTrX-njVft) N. 0.
.-. $14831.60 Cinn. = %:^^i^i^^^,, N. Y.
= $14476.72....
Hence frain = $251.81.
6. Cost of exchange of $2660 = $2570.89 ;
(I
«
$ a ft 7 0.8 9
26 CO
= $-9665.
But the bank had the use of this money for 63 days,
and allowed a deduction for interest.
Bank discount for 63 da. = $(-9665 x /jV x j^^)
= $010009.
Course of exchange ^ $-976509...
1--9765... = 0234 ;
.'. exchange was at a discount of 2*84 %.
II.— Page 238.
1. Cost of 1 lb. of the mixture ^. »' '^ ' «"
116
76 cents.
cents
1^ of the mixture consisted of the good tea and ^^ of
it of the inferior kind.
m
SOLUTIONS HAMBLIN SMITH S ARITHMETIC.
it
Hence 1^ of cost of dear tea +- ^j^ of (cost of dear tea
— 12 cents) = 75 cents ;
.'. cost of dear t^a == (75 + ^^) cents ;
=: 78| cents,
and cost of cheap tea = 66| cents.
2. Sum expended in paying clerks = $1600.
Sura given to J = i^^-oT" ^ 'li^l^OO.
= $1200.
t(
((
D
I 00
3 0X4
1 00
Sum to be apportioned = $(12800 - 1600 - 1200
-1600-120)= $8280.
Part of this given to ^ = | of S8280 = $3312.
" « ^ =. 3 of fiSH280 r~^ $4968.
Net sum received by A =- $(1600 + 3312) = $4912.
" " " " .8 =^ $(12()0 + 4968) = $6168.
3. Since 57^ fl. = 120 fr.;
.-. £1, or'l2.15 fl. = ' '■''r'''' fr.
' o 7 f
=-- 25.85if fr.
4. 3 X 1400 = 4200\
2 X 3400 = 6800 1
8 X 1800 = 54G0|
4 X 4200 = 16800J
4 X 2000 = 8000)
6 X 1400 = 8400:-
2 X 4000 = 8000J
-= 33200.
24400.
57600.
^'s share = ^ i|{}g of $4082 = $2324.
B'a
(<
^t^^ of .'5^4032 $1708.
5. Every gal. of the firsf. mixture contains l^ or i
sTT
gal. of wine, and every gal. of the second mixture con*
taiuti -^-^ or ,^ gal. of wine ;
SOLUTIONS HAMBLIN SMITH 8 ARITHMETIC.
99
tea
>0
2.
8.
3rv
and ^Q
t«
/. f of number from Ist + i of (14 — number from 1st)
= 7 gal. ;
*. •'• (i — f) <^^ number from 1st ==^ (lO^- 7) gal.;
= 3^ gal.,
2 X 3»
7
= 10.
Henco the number from the 2nd =. 4.
III.— Page 239.
1. By Art. 206 it is found that a mixture of 5 lb. at
8 cents, 5 lb. at 10 cents, 5 lb. at 12 cents, and 15 lb.,
at 20 cents, \vould be worth 15 cents per lb.
5 + 5 + 5 + 15 = 30;
.♦. quantity at 8 cents = ^"^5^ of 200 lb.
= 33^ lb.
10 cents = ^% of 200 lb.
= 33^ lb.
12 cents = /^ of 200 lb.
= 33^ lb. '
20 cents -- ^^ of 200 lb.
=- 100 lb.
((
t<
2. Interest for 18 da. at 6 % — -.vi^y of note.
Discount = vTojT "
•. Note-- (^|^«^ + ^^,-) of note = §1190-234,
andjii;^^ - =J?1190-234;
"^ r 1 C89
= $1212, nearly.
3. A gain of 'f>120 in G mo. = a gain of !3520 in 1 mo.
$150 in 5 mo. -=-- " .|;30 "
?;'210 in 1) mu. _ " $23^ "
rr^h
100 SOLUTIONS HAMBLIN SMITHS ARITHMETIC.
Sum from which $23i is gained =
<<
$73i
((
73ix 400
^ 23*
= $1267|.
4. After each drawing off f of the wine remaining in
the cask is left.
Hence the part finally left
= f of f of f of f of the wine
= uVff of the wine.
6. Since 25.15 fr. = ^1;
.-. 1 rouble, or 1.2 fr. = £^i^2L±
2S.1 5 »
and 920 roubles = ^i'^^'/J-L^^.L
... ^ ' = ^'43 'l7s. lid. (nearly).
Again, since 25.35 fr. = £1 ; ^^
.'. 1 rouble, or 1.15 fr. = je*-^""*.
2 5.;{5 -^
and 920 roubles
jp9 2 0x_l.l5x 1
2 /5.3 s
= ^41 14.V. 8^d. (nearly)
Hence the broker's gain = £2 8s. 2^d.
IV — Page 240.
1. £354 I65. Sd. = 85155«^.
Since 38|rf. = 1 dollar ;
.-. 851551?. = ^^^'^'^xy doUars
3 8 i
= 2211/^ dollars.
2« Since 1 lira = l|;o.22 ;
.-. 7500 Hre = 7500 x $0.22
= $1650.
By circuitous exchange £1 z^-^ $4.95 ;
.-. 26 fr. = $4.95,'
and 1 lira, or 1 1 fr. = $li2i±^ .
26
iih7500v9.x4.0g
= $1606.37.
Hence the difference = $43.63.
SOLUTIONS HAMBLIN SMITH's AKITHMETIO.
101
g m
vme
f
3. Since £200 = $1000;
4. By direct exchauge £1 -^ $4.86^ ;
.-. £3000 = 3000 X $4.86f
= $14600.
Through Paris 5.25 fr. = $1 ;
.-. £1, or 25 fr. - $-^/^' ,
^ 5.25 *
and £3000 == $''"-"±^111
"^ 5.2 5
= $14285f..
Through Amsterdam 1 guild. = $0.40 ;
.-. £1, or 12^ guild. =. 12^, x $0.40,
and £3000 = 3000 x 12i x $0.40
= $14040.
By direct exchange he has to pay $14G00 for the draft;
by Paris, only $14285f , and by Amsterdam $14640.
5. Cost price |o» of 14^ cents = 12^ cents.
Diff.
= 22| gain.
221 loss.
" 13
" 14
8 --=12
1 =: 6
We have, therefore, 1 lb. at 8 cents,
8^ lb.
and 8 lb.
Of course, the above are only a few of the many
answers that might be found to this question.
V. — Page 240,
1. 2456 + 735 + 4361 = 7552.
102 SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
Number to be provided by 1st = f^«f of 182
= 59, nearly.
" " " 2nd = ^T-Y^ of 182
= 17, nearly,
by 3rd = 44|^ of 182
== 106, nearly.
2. Cost of 9 gal. of mixture = 70.?.
" 1 gal. " = V".^. = 7ls.
Selling price of 1 gal. " = 6 x 2p. = 17.s.
Gain on 7 f^v. = Of*-. ;
= 118|s.
3. Since the gain on $2200 = $880 ;
.-. " " $3500 = SiiP-^A"!!
•^^ 2 2 00
= $1400.
But the gain for 2 mo. less = $1120 ;
.-. 2 mo. gain on $3500 — $280.
Since time for which $280 is gain on $3500 = 2 mo. ;
" $1120 " f3500
1120X2
' 280
= 8 mo.
Time for which $280 is gain on $3500 = 2 mo. ;
$880 " $2200
880 y 3 5 a X 3
2 2 00 X 280
= 10 mo.
Time for which $280 is gain on $3500 = 2 mo. ;
" $1200 " $2500.
1_2 X 3 5 X 2
2 .5 6 61<~2^ 0~
= 12 mo.
SOLUTIONS HAMBLIN SMITH's AKITHMETIC.
103
4. Capital at end of Ist year
= ^ of origmal capital -£1200.
Capital at end of 2nd year
= I of (f of original capital - A'1200) - ^21200
= ^ of original capital — £3000.
Capital at end of 3rd year
= I- of (f of original capital -£3000) -£1200
= y of original capital — £5700.
Capital at end of 4th year
= f of (y of original capital - £5700) -£1200
= ^l of orignal capital — £9750.
Hence f^ of original capital— £9750 = 4 x original
capital ;
.-. -j-l of original capital = £9750,
and origipal capital = £-"^^^/°°
=:£9176iV
5. Strength of 1 gal. of the mixture
/34 iax46\ .»
= ItOTT ^ TXTTF ) — O =
Since the gain on -j^j^
100 X ^u
42
TUS"
8 .
Tir¥ >
««
100 =
.•»4
TO IT
= 23tV.
9.
Examples (cxvi). Page 242.
Relation 80 x G40 : 180 = '''\''^l*^ :
= 128 :
10.
7 :
8
12
15
2
I
15-
7
: 4
1
1.
compound ratio.
Hence the 4th ratio = %^
= 9 : 13.
104
\
SOLUTIONS HAMBLIN SMITH'S ARITHMETIC.
11. If f be the given ratio. Then adding any num-
ber, say 5, to each of the terms, we have ~^^ = |.
Comparing this ratio with f, we have ^^ and |^.
Hence we see that this ratio is increased by adding the
same number to each of its terms.
Again, if we take ^ as the ratio and add, say 2, to each
of the terms, we have I'^l- = f. Comparing this with
the original ratio, we have f^ and [f. Hence, we see that
this ratio is diminished by addiug the same number to
each of its terms. A ratio is, therefore, increased or
diminished by adding the same number to each of its
terms according as the antecedent is less or greater than
its consequent.
I
4.
Examples (cxvii.) Page 245.
A = y> of ^%r of c
U> of 5
2' of C;
A : C
25 : 39.
10. ^'s share = ^ of A's.
C'a " = I of i>"8 = l-of « of^'s.
Z>'s " = t of C's = toff of loft's;
.-. 4's+|of /f's + ft of A's-f-Z^of^'s = $1587;
• •• (l-t-8+l- + t'W) of ^'s = $1587,
and If of ^'s = $1587.
.24x1587
^'s = $•
9
= $552.
-B's = g- of $552 = $460.
C's = f of $400 = $340.
i>"s -^ 5 01 |345 - $230.
SOLUTIONS HAMBLIN SMITh's ARITHMETIC. 105
num-
Examples (cxviii.) Page 247.
1. 233c/. : £1247 10s. 5d. :: A'l : gross income ;
.-. gross income = Jtji^l^i^.^
° 2 3 3
= £1286.
2. The number of hours between 12 at noon on Mon-
lay and 10^ a.m. on Saturday = 118^ hr.
24 hr. : 118^ hr. :: 3 m. 10 s. : time gained ;
.-. time gamed =.= 24-- min.
= 15 min. 86/^ s.
As the watch was 10 min. fast on Monday, it is now
25 m. and 36/g s. too fast, and hence it is 10 h. 40 m.
36^\.s.
8. Gain in 6^ rounds r= .^ mi.
6k rounds : 9 rounds :: ^ mi. : ^I's gain. ;
••• ^'s gain = ^ mi.
TU
.mi.
4. The hands of the watch will be together for the
4th time after noon at 16 1\ min. past 3. Art. 173.
The watch will have gone (0^^^ + 18t) f lGy\) min., or
203 min.
But 59|| min. on the watch correspond to 60 min. of
true time ;
.'. 59l{ : 203 :: 60 min. : time required ;
. , 203x60
.'. time required = g^,^ mm. = 205 min.
5.
= 3 hi*. 25 min.
Since 4 men = G wome = b >y8 ;
.'. 1 man = 5- " -- " " •
4 — ■( »
and 5 men — '^ «< _ ■*:/> u
■f
106
SOLUTIONS HAMBLIN SMITH S ARITHMETIC.
(9 + Y) women : 6 women :: 27|-da. : time required ;
• A 2^2X6
.". time required = g , ho da.
= 10 da.
(8+ V) Ws : 9 boys :: 27i da. : time required;
27^x9 ^
.'. time required = QVisr "8"
= 12f da.
6. Ul : 5| :: $116.15 : value required;
. ^ 5^x116.15
.*. value required = $ — ^^3
= $47.18J.
7. 26 in. : (7 X 9) in. :: 32^ yds. : yards required ;
.*. yards required = ^'^g' —
= 78f .
8. The difference in 24 hr. = 7| min.
Their present difference = 5 min.
In how many hours will tlieir difference amount to
25 min. ?
7i min. : 25 min. :: 24 hr. : hr. required;
.'. hours required = ^— J^ —
= 80.
80 hours from noon on Monday is 8 p.m. on Thursday.
9. 2§ : 3i :: 6336 stones : stones required ;
.•. stones required = .^^ ^ ^^^^
= 7722.
10. 7 : 3 : : 22400 people : number fed ;
1 rj 22400X3
.*. number led = — ^
= 9600 ;
.*. number sent away = 22400 — 9600
= 12800.
60
40
8
16
SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
Examples (cxix). Page 249.
1'^ :: 18 men ; men required •
860
8
10
107
• •• men required = iAliA^a e x a x 1
60X4 0X3X16
= 54.
2. In 18 months 1200 men complete f of the work-
how many men will be required to do f of the work in'
10 months.
16 : 18 :: 1200 men : number required ;
f : #
.'. number required = ij^^iiiixi
lexi
= 2250 ;
.'. number additional =: 2250 ~ 1200
= 1050.
8. 6 : 7 :: Omen : number required;
5.: 6
7 : 10
.'. number required = ^^^^ « *< 1
4.
6X3X7
= 18.
185 : 92^ :: 20 men : number required ;
If : 9 1 ,
.-.number required =:??-^-^^i^^
^ 185 X 1*
5
= 50.
5. 4 times work of soldiers + 4 times work of navvies
-- work- necessary to dig the trench in 1 day •
And 7 times work of soldiers + 7 times work of half
tUe navvies
= work necessary to dig the trench in 1 day ;
>.^ -.4.>^' ~i^'-iiii.a&tlt
108
SOLUTIONS HAMBLIN SMITH S ARITHMETIC.
.-. 8 times work of soldiers + 8 times work of navvies
= 14 times work of soldiers + 7 times work of navvies ;
and hence work of navvies = 6 times work of soldiers.
6. 2:10 : : 1 drona : amount required ; •
10 : 12^
9 : Hi
86 : 45
. 7: 8f
amount required =
loxiajx 1 Itx45x8l
3X10X9x36X7
= 12
6 3
7. 470
7
360
8 ■
658 revolutions : number required ;
. . :• 658X360X8
.'. number required = — 470x7 ~
= 576.
8. If 15 men working 15 hours a day do ^ of a piece
of work in 24 days, how many hours a day must 18
men work to do the rest of it in 12 days ?
18 : 15 :: 15 hr. : hours required i
12 : 24
3
f
hours required
15 X 15x 24x|
~18xl2xf
9.
24
9
7'
232i
2k
248
12
4
387^
5i
H
= 161.
5^ da. : days required ;
Bix248Xl2x4x3874x5jx3i
.-. days required = — ^^^ 9 xTxTFoiTTs JlTa*
155.
10.
SOLUTIONS HAMBLIN SMith's ABITHMKTIC.
10 da. : days required;
10«
9
10
25
24
44
40 ;
: 5
11
80
16
50
45
.*. (lays required = ' " ^ •'' >^ " x ■< « x j^o x a o x 4«
__ ^^'.) X I X a s X a 4 xT4~x ?6~
Examples (cxxiv). Page 258.
1. Area of floor = (14 1 X 15 «) sq.ft.
.'. cost ~
= |4.93^».
2. Area = (146^ x gs^) sq. ft. __^ ^^^^. ^
«° 9x16 ^4' ya.;
.•. cost — llli^ -^ "' X 3 G
- ^^ cents = $520.96|.
9 X Ic
3.Sin.4ro.lpo^a a.6Ht. = 44100;.ft,
• ■ ^'^^ = ^/44I00 ft. = 210 ft.
4. Sincelro.26po.28yd.4^ft_18225sq.ft.
••• 8i<^e = ^/18225 ft. = 135 ft. '
^•Area -^ (40 x 3 x 100) sq. ft.
Number of turfs =
.'. cost
4 0X3X100
3x1 ■ »
4 X JiX 10 0x81
loolTsxT
d.
= £18 10 s.
6. Length of room .^VW9 ft. - 17 ft
Area of walls /^ -.„
= (4x17x11^) sq.ft.
=- 782 sq. ft.
Area to be whitewashed ^ (32^-f l^-^) sq yd
cost = 119x5 cents =
= 119 so, vd.
$5.95.
.Mm
110 fiOLUTlONB HAMBLIN HMlTll's ARITHMETIC.
7 Area of room = {S\ k 6^) sq. yd. = 65 sq. yd.
Length of carpet = -7- yd. = j, yd. ;
.-. cost per yard — $^*fe ^ $1.20.
2
8. Since £2 Ids. Hd. = 7\Qd. ;
.'. length of paper = -^ yd.
Area of paper (^i" x tj) aq. yd.
Height of room = JCr X*) - (^^a + 130/ yd.
= 12 ft.
9. Area - 559504 sq. ft.
Breadth = m~^*'it. = 187 ft.
10. Area, = (880 x 880) sq. yd.
= 22^ a.
y't o^ *^^ 8.rea of the walls
11. Breadth — sT
= TT I (42 + 2 X breadth) x lOj \
21
= Tr) 21 4- breath}
= Vt ft' + fVof^'^eadth;
.••W of breadth = \\^ft;
.-. breadth = V oi Vr ft-
= 17^ ft.
12. Length = ^* in. = 1 ft. 9^ in
13. Area of room
= J2x(12/^ + 9|f)xlc}sq.ft.
= (2 X 22^ X 10) sq. ft.
2 X 221 X 10
1l;: 'Is of paper =
= 100.
3xf
yJ.
SOLUnONS HAMBLIN SMITH'r ARITHMETIC.
And 100 yds., at 12 cents a yaid = $12.
14. Number ^ l^i^iil^
2x51
= 5952.
Ill
16. Area of walls
= { (2xl5 + 2xl2)xl0} sq.ft.
= 540 sq. ft.
T« il, r 640
Liength of paper = - -- ft. = 210 ft.
Cost of paper = "-V* x 1 2. i cents.
16 Area of walls
= .j (2x21.f 2xl5)xl2f sq.ft.
n^j .. ^ = ^64 sq. ft.
Deduction = ^21 f 30 + 2x09} sq. ft. = 189 sq. ft.
Length of paper = ^" ft. == oo yd.
Cost of paper = 90 x 15 cents.
= $18.50.
17.\ Since 60 a. 2 r. 82 po. = 245888 sq. yd. ;
.-. breadth = ^//-«-« yd.
= 429 yd.
Diagonal ^ v/(5722 + 429^) yd.
= v/6ll225 yd. =:z 715 yd.
18. Area of each part = (^~^~^^~19 w ilL!LzlJl\ ^ /•>.
= 10511^ sq. yd.
Area covered by trees = (900 x 20 + 430 x 20) sq ft
= 29554 sq. vd.
112
SOLUTIONS HAMBLlN SMITh's ABITITMETIC.
19. Area of walls
= { (2 X length •!- 2 x breadth) x 11 J ?q. ft.
=- { (4 X breadth + 2 x breadth) x 11 } sq. ft.
— (G6 X breadth) sq. ft. ;
.-. 6G X breadth ^. (143 x 3) x 2, .
and breadth — ^-^-^^ ft.
6 6
--- 13 ft.,
and length = 20 It. ;
.-. length ot moulding ^^ "-^'+12^1^. ft.
.-: 25 vds.
20. Area of ceiling =- (27:\ x 20) sq. ft.
:= — jT" sq- ft-
Area of walls - { (2 x 27i + ^ x 20) x 12^ f sq. ft.
z^ -^ sq. it.;
.*. area to be painted = —^ sq. ft. ;
.*. cost = Y^' ^ '^^ cents.
= 169.20
21. Area of room = (15f x 13^) sq. ft.
Length of carpet = --^^^ ft, ;
.'. cost r- -^-^,— X 95 cents.
= $29,551.
22. Area of room = (lOii x 7i) sq. yd.
Length of carpet = ^^^^ " yd.
^ 3JX2 2 X 4 I .
^ 3 X 3 X 3 -^ ■ '
.-.cost ==^^{^r x$1.08
:^ $112.64.
23. Area of room = (11x8) sq. yd.
TOP
Length of carpet = —3-' yd.
= 132 yd. ;
SOLUTIONS HAMBLIN SMITH 's ARITHMETIC. 113
.-. width of carpet = i^--* yd.
= f yd.
24. Since 12.45 ft. = 4.15 yd. ;
.*. length to be paved
= j (2 X 45.77 + 2 X 4.15 + 2 x41.93)yd.
= 192 yd.
Area to be paved = (192 x 4.15) sq. yd. ;
.-. number of stones = 'a2x4.i5x9
5.7 Gx 4.1 a
=: 300.
Examples (cxxv.) Page 262.
6. Content of the wall = (75 x 12 x 6 x 12 x 18) c. in.
Content of one brick = (9 x f x 3) c. in.;
.-. number of bricks = I?iSi|2i^2ii2jiH
9 X I X 3
= 9600.
7. Number of c. ft. of ice = 45 x 4840 x 9 x i.
Weight in lbs. =r 45 x 4840 x 9 x ^ x \9^
= 1408811- tons.
8. Number of men required to dig (800 x 500 x 40) c. yd.
in 1 month = 4 x 500 ;
.-. number of men required to dig (1000 x 400 x 50) c. yd.
in 1 month = i ooox4oox.5 () x 4 x 500
800X500X40
= 2500 ;
henee number of men to do it in 5 mo. = ^^^
= 500.
9. Area of side = ^|^ sq in. = 361 sq. in.
Length of side ~ n/361 in. = 19 in.
10. Content of cistern = (4 x 2] x 3^) c. ft.
Weight of water
4X5X13X1000
203U lb.
114
BOLUTIONS HA^fBTJN SMITH S ABITHMETIC.
i
11. Content of stone = (4 x 12 x 30 x 15) c. in.
Weight of (4 X 12 X 30 x 15) c. in. == 27 cwt.;
'. weight of 100 c. in.
100x27
4X12X3 0X16
= ^ cwt.
12. 4t. 12 cwt. 3qr. 101b. 7oz. =-- 166375 oz. ;
cwt.
content of vessel =
c. ft.
1 e 3 7_5
1000
166-375 eft.
Length of side = yi66-375 ft.
= 5-5 ft.
13. Number of men required to dig
(i x 1760 X 30 X 7) c. yd. in 1 day = 42 x 120 ;
.'. number of men required to dig
(1000 X 36 X V) c. yd. in 1 day = i^'''-J^±^-l^.JlllJ<JlP
\ ^3/ J J ■(xl7GOx3 0x7x3
= 4800 ; •
hence number required to dig it in 80 days
4800
:m)
= 160.
14. Cubic content of cistern holding 2520 lb.
:-: (1) X y < f ) C. ft. ;
" " " -- 3850 lb.
a^8 -.0x0x10x9
2.3 20x3xf
= 165 c. ft. ;
hence depth of cistern — g-^
= 3|i't.
eft.
I't.
15. Cost of excavation = (110 x 6 x ^)s. = 220.s'.
«* rubble = ( 1 1 x 6 x §).s\ = 1 46if .s.
gravel =(110 x 6 x |) x ^s. = 412i.s
(«
Total cost =
770 1 ..
=- £88 Ids. 2,'
SOLUTIONS HAMBLIN SMITH S ARITHMETIC.
EXAMINATION PAPERS.
115
= 113.
Page 263.
1. Number = JLo.2±oz:_37^al ^ 2»99
2 3 2
2. Profit on 1 yard --= $3.35 -$3.20 - 15 ceut«.
" " 500 yards = 500 x 15 cents == |75.
3.372-2x2x3x31, or, 372, 837, 248.
837 = 3x3x3x31,
248 = 2x2x2x31.
H. C. E. (See pages 5 and 6.)
124, 93, 248.
31, 98 62
81 is, therefore, iLc
Since I = ^±^I1J<1 _ 2639 .
'' 13X29x9 13x^9x9'
11 _ 0X29X i ] 2871
^-^ 9X29X13 .13x29x9 >
2 4 _. 9X1 5i< 2 4 2 80^
^ ^ 9 x 13 X 2 9 9X13X29 »
.'. the decreasing order is {I, 2|., 7.
4. Length of pace in inches = liiiia^jox 1 -z
'I a a n
= 30.
V92
5. Number of each = 9366 - (960 + 480 + 120 + 1)
= 6.
6. ^ = -02.
14 14000
•0 7 — 7
2000.
iV6% ^ -000002.
Sum = 2000-020002
— .200 2 00^3
1606000
j_ 10 1 1
600000
116
SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
7.
7.57 X .86 = 7|1 X U
■ 25
— 7* /
Q qAk 23 22
Q« f-A'i'i 2750-2322
^ ¥¥TT = 990
42 8 — 469
V
8.
8
H
40
8
41707796 ft.
18902598 yds. 2 ft.
2527745 po. ^ yd.
63193 fur. 25 po-
7899 mi. 1 fur.
7899 rai. 1 fur. 25 po. 3 ft. 6 in.
9. Distance passed over in 1 sec. = ^y^ yd. = 22 yd.
■|l-„ _ 60X60X22
t(
((
17 6
= 45 mi.
10. Sum earned in 12 mos. = 2 x $420
Sum spent " " = 3 x $210
Sum laid by " =. $840 - $630
11. Number of steps = *Sx52 8oxi2
3 2
= 9405.
12. One gets 4 parts when the other gets 3.
4 + 3c:.= 7.
One gets f of $13230 = $7560.
The other gets ^ of $13230 = $5670.
mi.
$840.
$630.
.$210.
13.
41
9
3 ,
16 9
4 _u 1 _ 3
ftr.
4
'STi
4 I 4 .
7 + T¥ — T3
5 r. X 6 3
KB
4 6
3 » G 9 X 4 6
^ff^=-0189
r,f>
14.
(I
o _1,
itj^ = |x Vx*xf
= I = -7^
1yd.
ni.
$840.
$630.
$210.
SOLUTIONS HAMBLIN SMITh's AEITHMETIC. 117
16. *x(V + V) = |xV = V.
23 4 1 _ 1 G 8 1 2
14.
16. 11 ro. 11 po. llyd. = 451i'o. 11 yd. = 13663^ yd.
= 122883fft. = 17695260 in.
Fraction = ~~^^ — -^^^^^±-. _ 331
3x4840 — 3X4840X4 — TT5^'
17. Wages for 76 days = 75 x $1.25 =. $93.76.
Sum lost by not working = $93.75 - $69.15 = $24.60.
Sum lost by not working 1 day
=: $1.25 + $-80r^ $2.06.
Number of days he was idle = %%%^ =--12.
18. Number of men required in 1 hr. = 10 x 12 x 24.
80 J^j, ^ 10X12X24
t(
80
«
((
= 86;
" " to do 3 times as
much = 3 X 36 = 108.
19. Value of ^\ of estate = $7500 ;
the estate = S
10X7500
3
tV^ of estate = $
48X 10X7600
il)Fx 3
= $12000.
20. Whole sum remaining = $105.03 ;
.-. sum each ought to have = $i^J^'' = $35.01 •
.-. A must hand over to O $37 50- $'35.01 = $2.49 ;
••• ^ " " $60.82 -$35.01 = $16.81.
21.
n +
11
r.
+ S
7x9x5
6x I4x
— 4 5 _ ] 5— 6
2 2 2
3
4X15
"~
\ 5 + 5
+ 1
3
=
1.
3X23
X 38
_
3 8
4. « _ 3 _
3x23x49
I! i
118
22.
SOLUTIONS HAMBLIN SMir:i's ARITHMETIC,
240025
sxa X 9601
5X0X400
5X5X5X/5X13
9601
8 1 2 fi _ __i„'-_ " '^ " " " ^.j*," 1 J»
TTTTTnTTTFTTTT — 5x6x6x6x16000 ~ TBOIHI*
1.12 14 IJJLA 4
5..S4 5.3 4'
1 121.4 1 1 21400
.5 3 4 6.3 4
= .21.
=: 2100.
23. 7 cwt. 4 lb. = 788 lb.
3 t. 1 qr. = 6748 lb.
.*. fraction =
, 10 a. = 10 X 4840 sq. yd. ;
7 98 _ 197
.-. length of side = y 48400 yd. :^ 220 yd.
Length of 4 sides = 4 x 220 yd. = 880 yd.
\ -- ^ mile ;
,-. number of times r= 1-^^ = 2.
24. The shares of all = (15 + 3 + 10) seamen's shares ;
•. 28 seamen's shares — ^2399 7s. ;
. 1 .. .. £399 7s.
28
— £14 5s. 3d. ;
'. 1 gunner's "
— 3 X (£14 5s. 3d.)
— £42 15s. 9d. ;
1 lieutenant's •'
= 10 X (£14 5s. 3d.)
— £142 12s. 6d,
!5. 1 67.
96 16 (12.96
1
22
67
44
/ 529 y 529
\ 246i y 2401
2.3
49
249
2396
2241
2586
15516
15516
t
SOLUTIONS HAMBLIN SMITH 8 ARITHMETIC.
119
lares ;
26. From midnight on Sunday to 6 p.m. on Wednes-
day is 66 hrs.
Time lost by the clock in 66 hrs. — ^^^ min. =: 22tnin.;
.-. taking away the 10 min. already gained, the clock
will indicate 12 min. to 6, or 5 h. 48 min.
27. Shortness in 22 yd. = ~-^ in. = 9^ in. ,
.'. actual distance = 22 yd. — 9^ in.
= 21 yd. 2 ft. 2^ in.
28. See Art. 174, 178, 181.
Interest = $(1900 x If X ^U) = $266.
Discount =:. $i^^44^ - $233.33jL ;
114
, difference = $(266 - 233.330 = $32.66|.
29. The interest = yf ^ of $170 ;
.-. the discount = ^f ^ of $170;
.-. the P. W. = {%l of $170
= $166,661.
30. Interest = $(880 x f X 2^^) = $49.50,
The interest = ^Viy of $929.50 ;
the discount = ^V? of $929.50
= $49.50.
31.
32.
7X3X3
2X14
3X2X7
^ Q
7x8x3X6x7x 14
2x14x3x2X7 x¥
7 31
100005.:. S8 .
"2"
100005
eeG7
990X55
33. Sum paid to produce $1 income in the 3| per
cents
= $-^Jf^ = $26.
Sum ""aid to Tiroduce -$4 income in the 3|- per cents
= 4 X $26 = $104 ;
120 SOLUTIONS HABIBLIN SMITh's ARITHMETIC.
.-. the 4 per cents at 103 is the better investment.
Sum invested to produce a net income of 98 cents = |26 •
* " $4851
= <i« !-?_?_ Li* ^^^
= $128700.
34. Time required by faster vessel = -J-J^hr. = 120 hr.
^ " " slower " = (120 + 36)hr.;
.-. the average rate of slower «' =1200 »„;
T6G" *"*•
= 7/j mi.
35. One gets 3 parts when the other gets 2 parts •
.-. i of $87.50, or |52.50 = one man's share, '
and f of $87.50, or |35 = the other
36. Money realized by sale ^|34 3ox8ai
(<
100 —
First income - $'-±^^^ = $120.05.
343X17 1
20
Second
200
.343X 17 1x4
= $119.70;
^ 20 X as
.-. the difference = 35 cents.
37. Number of cubic feet in the cistern.
._ 93.75X 112X16
1000 '
93.76X 112x16
ft. = 3 ft.
.*. the depth ^
1000xSk7
38. Interest on $330 for 2 mos. =
" $100" 12mos.^$'-±yil^l = ^Q_?i
Again, the interest on $330 for 12 mos. ="" 6 x" $3 =
.'. the discount off $348
((
((
$333 for 12 mos.
89. (Feet in breadth) 2 =782—552
= (78 4- 55) (78- 55) = 133 x23
= 3059 ;
.'. feet in breadth = / 3059
= 65.3....
iili^H _ «ii7i3
3 4 8 ~ V ■*• ' 'SS'
SOLUTIONS HAMBLIN SSIITh's AJllTHMETIO. 12]
40. Area of floor = (27* x 20^) sq. ft.
__ 8 2X121 f,
'. matting requiied =
j
3X6
9^X 8 2 X 1 2 1
2 2 X 3 X C~
225 i ft. = 76^ yd.
ft.
41.
6 XJB 8X6
4 X 5 ~ 3X5
7 , Jl_7 X 3
^"' 3 4^X~4
X y =
l-l
17
7 I 3 '^ 2
2
17
1.
42. Value of | of f , or i estate =
••• " f X V, or 7 estates = |(7 x 2 x 800)
= $4200.
Part of cable on land = 1 — |X| = J ' •
••• tVV of cable = 234 1 yd. ;
.-. length of cable = ^^^ yd. = 2691 »^ yd.
44.
ir x3
16 I 777 I 216
8
6
6
76
1200
325^
1626
26
187500
4536
192036
8777
7625
1152216
1152216
45. Selling price of 100s. = 110s.
((
a
10s,
! « X J 1 o
Too"
s, ~ 16s. 6d,
122
SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
46. 800 + 756 + 404 ^ 1960.
Number sent from Portsmouth = ^Vf''V o^ ^90 = 200.
« " Plymouth = ^VA o^ ^^^ = 189.
» " Sheerness = ^Vff\ of 490 = 101.
47. (a) Compound interest = 416| x $(1.083 -1)
== $69i.
Simple ** = 416§ x $0.16
= $661 ;
.-. difference = $2|.
(h) Interest on $184 for li yr. = $27.60 ;
^^f^rx <• -I ^2X100X27.60
$100 for lyr. = $ ^^y^^ —
' =$10;
.'. the rate per cent. = 10.
48. $180 = the amount of $100 ;
.-. $3213 = " $^W~° = $1785.
49. Area of field = 30378240 sq. in.
Length of field = 9376 in. ;
... breadth of field = -^^H^" in. = 3240 in.
= 270 feet.
50. Area of walls = { (48 -f 40) x 14^} sq. ft.
= 1254 sq. ft.
Deduction
= {(4x8x5i) + (2xlOx6|) + (6|x5)} sq.ft.
= 333^ sq.ft. ;
.'. area of paper
= (1254 - 333^) sq. ft. = 920^ sq. ft. ;
.-. length
_920| _552_1 a.
- 2^ "• " 45 y^-'
^--i. _ 6 52_i ^ /IK far = fJ.'^'^l far.
= £5 15s. Old,
200.
189.
101.
-1)
sq. ft
51.
SOLUXrONH IIAMBLIN SMlTU's AP^THMETlr. 128
S of 3
TT* of 5 1 X a X 1 X ft
8 3
1 of »j»
52. -^l^ of 25s + ^ViT of 100*. -
= Id. + 7*; - 28,
= 58. hi.
53. ^ of I of f =^ ^ = part whose value is required.
Value of 4' of cargo = i{i;ir>000 ;
" i " =;tof.7 ofliJilGOOO
= Jf;3783^.
54. Part mowed in 1 day by ^, B, and C
= (f + ^ + H) a. - ',2^ a.
Time to mow '3V a. == 1 da. ;
1 a. = T-^/^ da. ;
.-. " " 121 a. - 1±'J^1 da.
= 86 da.
55. Time it loses in 6^ da.
= 5 m. 40 sec. - 2 m. 51 sec. =^ 2 m. 49 sec.
Time it loses in 1 da. - ?-"^ IJ^-^' _ 26 sec.
66. Taxes = -j^J, or ^V^ of rent ;
.-. rent and taxes together = (} 00 ^ ^72^) ^^ ^^^^^
= 1^1 of $720 = $828.08.
57. The Ist and 2nd pay i + f of J, or || of the bill •
.-. x\of the bill = $2.50;
the bill = $iAi<^-J> =$9.37i.
58. Tax on .$1200 when it is half as much again = $27 •
" $760 " a
ilj. 7 .5 X 2 7
^1 2 0~
r'
121 SOLUTIONS HAMDLIN SMITHH ATHTHMETIC,
59. A'H income --= $"-'J^ = $(6x8.1) = $10.50.
<i
.079x3
= $(7x9) ==$21;
B's
.'. difference = $21 -$19.50 ..- $1.50.
cents = $1«5J.
CO. Area of room in sq. yd. = — .; — >
3 1 X 1 a5x 4 2J
1u
1 1 1 1 7
1
^
.'. cost —
()1.
\»* + 10ft *^/ Vn^ 20X10/ »
2 1^21 21
/3 2P\
i;< ' HI ~ l*Off
\ rt 10'
12 3+06 Q
Jt = 1 X 3 - 1 = 2^.
^^ T ,1 !• J. :)Sx 5 3 80 X 1 2 -„
62. Length ot step = no x i - ■ ^'^•
= H3-G in.
)Sx5 3 80Xl2-^ _ 7X48X6 in,
n X 1 I
60
03. Length of street + length of column = 8700 ft.;
.-. time = ^Ja^min. = 00 min.
64. Area to be paved ^ { 850 x (2 x 5i) } sq. ft.
= (425x21) sq. ft.
Cost = 425 X 21 X 37i cents
== $3346.871.
65. Part filled by one pipe in 1 hr. = ^ ;
«« «• the other *' = i.
Time to fill i of cistern = 1 hr. ; ^
'• the cistern — 3 hr.
66. 27 men = 54 boys.
Time for 54 boys --= 280 hrs. ;
Number of hours in 1 day = 4,^42
' = 8 hr.
67. Interest on $125 for 1^ yr. = $13.1 2| ;
,, ^ (»2 X 1 00 X 1
" 1 yr. = $ 3-126
54X280 j^j.^
<t
»2 X 1 00 X I 3.1 2 |
n= $7.
I
SOLUTIONS HAMBMN SMITh's ARITMMKTIC. l'2r,
68. H X $6t, or |» •« == interest on $100;
= $1080.
«0. Length cut ofr.J2i - (U < li) ft. - 14 ft
Length remaining __ (18 _ i^^ ft, . ^^^ ^^'
70. Area of room = (20 x 10:1) sq. ft. .. s.S5 ho. It.'
Length bought for 1C8*. = i|^" ya. = 48 yd
Area of carpet = (48 x J) sq. yd. - .SOsq.yd.-
.-. part uncovered = (886 - » x 86jsq. ft. == 1 1 so . ft.
71.
18+28
A
r
(I
TU
1
72.
3 8+31)
= 118 3
j^jy Of 240^. + ^3^. of 90./. + ^^3^ of 88./.
= (4-8 + 2-7 + -476^
= 7-97(R
78. '/30712-5626 == 17/)-25 ; y/^y*
and 175.25 of f « ^_- qq^s^i^, ' '^'^ " ^
v^TTjWinj'TjVD'ViJTro = Tiy^rt^DiT ^- -000805.
^^^''* of $7850 =-. ^^/|. of ^7850
$4900,25.
3ff .
74. They lose
100
75. Time for 14 men to mow 35 a. = GO hr. ;
24X14X00
3xn5
24X 14X00
((
8
((
<{
24 a. = '-'"''*^°?i,,.
Number of days = _
'' 1 2 x3 x 3a
= 10.
70. 7 men and 9 women -= 7 m^n «n.i si ..^
iif men.
.
i. j>.i
»^'M,W
1^0 SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
Time for 1 man to do the work = 9 X H4 da. ;
V^^ men " " = -"Y^Ti'^
=■ 107 ^^^V da.
((
da.
77. -7 + -28 + -056 =- 1-036
1 n ;? .
TTJiOO '
.-. A gets tS-, or |5| of $2849 = $1925.
TT>oo
B gets ^7. « T^A of $2849 = $770.
C gets Hlf, or ^h ot $2849 = $154.
7ft $1400-300) ^ the interest on $360 ; (Art. 181)
((
40
= $3600.
Again, the interest on $360 for 2 yr. = $40 ;
((
^-.r^/^ r 1 ^100X40
$100 for 1 yr. = $ sxaeo
= $5'l'
79 Selling price of 100 oranges = '''-^^ cents
== $1.50.
Loss on an outlay of $2.50 = $1 ;
1 00 X I
((
1 00 = $ '
2.60
= $40.
80. Area of walls = (72 x 11) sq. ft. = 792 sq. it.
Area of windows = (2x9x3) sq. ft. = 54 sq. ft.
Area of door = (7 x 3*) sq. ft. = 24^ sq. ft.
Area of fireplace =-- (4 x 4^) sq. ft. = 18 sq. ft.
Area to be papered = (792-54- 24|- 18) sq. ft.
= 695 i sq. ft. ;
.-. length of paper -= (695^ ^ 2^) ft. = ^r ^^-'^
= ^ 9
= £4 Is ^u
SOLUTIONS HAMBLIN SMITIl's ARITHMETIC.
127
81.
3
If; 1 2 !)
(2)
1.802x7.03
2
g
74
= L2_Lf>fi.80 6
2
1 I 4 :i X 2
;»42U
6.33403.
82. Part sold = -i-g«, or /^ of his share,
.-. part remaming = ^s of his share
= e
83. l*art done by ^, 2 Z?'s, and G in 1 da. = ^ + ^\,
" ^,2/, and (7 " =1;
•'- " " 2^'s, 2Z;'s, 2C"8 " =i;
1 .
— "g- >
^ .-. time required by A and C = 8 days.
84. Number of hours between midnight on Sunday
to 4 p.m. Wednesday = 64.
Time gained in 24 hr. = 7^ min. ;
• '< t« UA U%. 64X74
o4 hr. = — -- - mm.
r= 20 min.
Hence the time on Wednesday is 4 hr. 32 min.
85. 33 + 7 + 5 = 45.
Number of lb. of nitre = f | of 30 lb.
= 22 lb.
" " charcoal = /^ of 30 lb.
= 4f lb.
" " sulphur = ^\ of 30 lb.
= 3i lb.
86. Interest - $ [l639 x -^ x jl^] = $39.95^^.
Discount
difference
■^ $(1639 X ^VjVV) = $39
= $.95
JHf'
128 SOLUTIONS HAMBLIN SMITH'S ARITHMETIC-
87. The bank discount = $(10400 x {'o x yf^) = $416.
The true " -$(10400x^^3:) =$400;
.'. difference — $16.
88. Part sold at cost = ^ of ^, or ,V o^ goods.
" i of cost = I of I, or ^'^
Price of goods realized = (^V + ^ of ;jV) of cost
== y^g of cost ;
.'. cost of goods = 12 X $1155,
and loss ^ 11 x $1155
= $12705.
y
89. The gallon contains --}]'{ ^"^ cub. ft. ;
.. „ -I 27 7.2 74 X 1 000
.-. the gallon weighs p^-^ oz. ;
.-. the pint weighs i^x^lfwVs 1^- = l'2.*;35...1b.
90 Area of floor ^ (22^ x 20^) sq. It. - --;;'- sq. ft.;
.-. cost of carpet == i^' x $1.20 =^$60.75.
Area of walls = | (45 4-40^) x lOf [ sq. ft.
^ Ul^ sq. ft. ;
/. cost of paper ^ -^- ^ 20 cents
= $20.42i.
40-5
_ 8 — 1
— -g— I.
91 • ■
2-4282323 + 3-57G5705 + 2-0001911
92. See Ex. paper X., example 5.
98. iJ-60625 = 12s. Ud.
142857 of 14». lOif?. = | of 14.s. lO^f. = 2.s-. V^rf.
1^ of Vt of '781fl?. = 5*'. 9rf.
Sum = 20s.
Also 20f?. = 1? of 27«.
2 3 of /^ of.£3 5*. 1(?.
?T
= •740„
SOLUnONS HAMBLIN SMITh's ARITHMETIC. 120
04. Time gained in 7^ hr. = {7^ x (3i-24)| min.
= I/tj- min. ;
0-. it must be set at 1^^^ miu. to 12.
95. Interest = $(056,25-750) = $206.25 ;
.-. interest on $750 for 8f yr. = $206.25 ;'
$100 ** 1 yr. c^ {||«X 100X20625
<t
11X760
= $7i
96. 1 per cent on $5420 gives $54.20 ;
.-. income at the lower rate on $9970 = $(453 -54.20)
= $398.80; -
* $100
. <ajl 00 X 398.80
<(
9970
, or
hence the rates are 4 % and 5 %.
97. Area of wall = j 540 + 1841) xS^} sq.ft.
1449 X 26 " ,
¥^G — sq. yd. ;
.-. cost of wall = ^'jl-^-^ X $1.20
= $805.
98. Income from £75 invested = £3.
Money got from ;£75 " = £78.
Income from £78
ri 7 8 X 8
208
= £3.
99. L. C. M. of 2, 3, 4, 5, 6 = 60.
We must now find the least multiple of 60 which is a
perfect square. This is 900.
100. The interest on $320 for 8 mo. = $40 ;
" $300 for 12 mo.
<(
-$
!>r«(5O>f1S5<40
320X8
$67.50.
130
SOLUTIONS HAMBLIN SMITH S ARITHMETIC.
101.
57875
(729)(81)(9)
620875 = 9 X 57876.
4687875 =9x520875.
42190875 = 9 X 4687875.
9
42238274625.
123456
13717 and 3 units over.
1959 and 4 groups of 9^units each ovr ■ ;
.'. quotient = 1959H.
102. 1 metre = 1-0936 yd. ;
.*. 1 centimetre = -010936 yd.
=^ (-010936 X 36) in.
= -393696 in.
108. Part done by 2 ^, i? and G daily = i + ^\.
" B and C
it
{(
((
2A
_ 4 .
= i +
27 .
"SVi
.*. A can do the work in y/ da. = 6|f da.
Part done by B daily = i — y^^
_ 19 •
— T8¥ »
,•. B can do the work in VV da. = 9|# da.
Part done by C daily
'TIF
4 _
1 :$ .
— TST »
1 9
T7f*
.•. C can do the work in ^-^^ da. = 14^-^.
104. M has 12 miles start.
N" gains 4 miles per hour, and hence would overtake
M in 3 hours.
When N arrives M has 4x6 miles to go.
SOLUTIONS HAMBLIN SMITH S ARITHMETIC.
r4 X 6
181
It requires JV—— hr., or 6 hr. to gain this distance
on M.
Hence M travels (5 + 6 + 4) hr. and goes 16 x 6 miles,
or 90 miles.
105: Interest = $(2733^ x 3| x t-^)
= $410.
Amount of $1 at compound interest = $1.157625 ;
.•c sum required = liri^'^ "
625
= $800.
106o Discount off $108'
1 .
6 '
(«
" 11622.50 =. I l«??f^i
= $122.50.
Interest on $1760 = $(1760 x | x ^g^)
=- $132;
.'. difference = $9.60,
107. Cost of 1 apple of Ist kind = \d.
2nd " = ^^, ;
2
«
((
.'. average colst of 1 apple = ^—d.
Selling price of 1 apple = ^d. ;
.'. loss on an outlay of ^\d. = [^^^~^)d.
1 d '
((
((
5
= 4d.
108. What he sold for $91 he should sell for $107 ;
$182 "
((
<t
i S 2 X i 7
9 I
^ $214.
i I
132
SOLUTIONS HAMBLIN SMITH S ARITHMETIC.
109. Area of walls - {(28-| + 271) x 12^} sq.ft.
= C86 sq. ft.
Deduction =-- ^48 + 20 + 3 3 x 2{^) sq. ft
= - ' , sq. ft.
Area of paper = jy^g sq. ft.
Cost - =^"^J cents
0x12x5
= $9.38A.
110. Contents of two longer sides
= (2 X 4 X 2 X tV) c. ft. = f c. ft.
Contents of two shorter sides
-= (2 X 2f X 2 X tV) c- ft. = il c. ft.
Contents of bottom
:= (3f X 2-1 X -i-V) c. ft. -- ^l\ c. ft. ;
.-. whole contents = il±^-i||+-ii:^-^ c. ft. = ^^i c. ft. ;
• f»n«5t _ 12 pf l-^^Sxe _ 125, —Is- IS//
111. We are required to find the L. C, M. of 1, 2, 3,
4, 5, 6. 7, 8 ;
L, C. M. = 840.
Hence the bells will be tolling together m 840 sec,
or 14 min.
112.
1
7
i 1 O 1 *
10) ^4
■euf ~ 1
—. 1
IT
1
6
3 2
1 _ 8 I
113. 2 per cent, of ^'s ca.pital = $220 ;
.-. ^'s capital = -^-"-"^Al£=.$iiooO;
.-. B and C's capital =-- f of $11000
=--- §10500; ■
.-. the capital of each ^ %^^~^.=^ $8260.
114. Since the fast train goes as far in 5 hr. as the
slow one does in 6 hr., the rates are as 6 : 6.
SOLUTIONS HAMBLIN SMITH S ARITHMETIO.
188
Since the fast traiu gains 10 miles in 2 hours, it
gains 5 mi. in 1 hr.
Hence f of rate of slow train = rate of slow traiu + 5 mi. ;
.-. i
ik n
i(
= 5 mi.
and
n a
t(
= 5x5
— 25 mi.
mi.
Rate of fast ti
ain
= (25 +
= 30mi.
5) mi.
115. Income on £80 annually = £6 ;
£100 " = £—~ = £7^.
80
6000X 100
80 ■
Amount of Turkish stock = £
money from sale of stock
80X100 "
= £6500.
Income from £90 invested in railway shares
.-. new income = £?-^«-^-l* = ^-^25.
First income = £^^«JiJ! =. £375 .
.*. he has £,' > less income.
116. 30 men and 10 boys reap 180 a. in 4 da.
14 men and 10 boys " 66 a. *« 4 da. ;
16 men reap 64 a. in 4 da. ;
1 man reaps 1 a. in 1 da.
But 6 men and 2 boys reap 13 a. in 2 da. ;
2 boys reap 1 a. in 2 da. ;
1 boy reaps J a. in 1 da.,
and 2 men and 2 boys reap 2} a. in 1 da.;
" 10 a. in i^*
-•I
= 4 days.
((
((
184
SOLUTIONS HAMBLIN SMITH S ARITHMETIC.
117. Ketail price = (f^o 4 ^Vii) o*' cost prico
= j^ of $4.75
= $6.17^.
118. First interest =-- $(625 x A x tU)
= $29.16|,
Second " = $(1093.75 x y% x y^^)
= $29,161.
119. Time when the difference is 6 min. = 12 hr.
«
«(
t(
16i "
16tx 12
hr.
= 33 hr.
83 hr. from noon on Monday is 9 p.m. Tuesday.
Time gained by the fast goer in 33 hr.
12
=i; 11 min. ;
hence it will indicate 9 hr. 11 min.
Time lost by the slow goer in 83 hr.
33X4
= -2T- ^1^-
= 5^ min. ;
hence it will indicate 5^ min. to 9, or 8 hr. 54 min.
30 sec.
120. Area of each grass plot — {6Q x 36) sq. ft. ;
" covered by grass = i^?A^-? sq. yd.
= 1056 sq. yd.
Area of whole court = (50 x 30) sq. yd.
= 1500 sq. yd.
Cost of grass = $(1056 x -70)
= $789.20.
Cost of stones = $(444 y 9 x .12i)
.'. total cost == $(789.20 + 499.50)
:= $1238.70.
SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
186
121. Number ol leap years in 400 consecutive years
= 97. Art. 151.
Number of times the 29th occurs in an ordinary year
= 11 ;
hence in 400 years it will occur 400 x 11 +97
= 4497
122. Since 62^ cents = ^ of dollar, he received § of
the debt ;
.-. ^ of debt = $281.25,
and '• =$iii^lll«
= $450.
128. The shares are in the ratio of 1 x 4 x 10.
2 X 3 X 12, and 8 X 1 X 20,
1x4x10 = 40.
2x8x12 = 72.
8 X 1 x 20 = 60.
172.
^'s share = yl/V of |43 = $10.
B'a " = ^,23 of $43 = $18.
C's " = /v'V o^" $43 = $15.
124. The cubic content of a brick in tlie second case
=■- (1)^ of that of a brick in the first case.
Hence we may leave the exact dimensions of the firsc
brick out of account and find the cost thus :
cost of 1 brick == $(^%\ of ^i~^„^o" ) ;
.-. « 100 bricks .^ $ (100 X ^ X ^^^)
= $1.12.
125. Number of years = 100 -f- 8^
= 30.
126. At 11 o'clock the hands are 5 minute-spaces
If
li I
w
w
I'l I
li;
186
SOLUTIONS IIAMBLTN SMITIl's ARITHMETIC.
-apart, and as the minute hand moves over 12 minute-
spaces while the hour hand moves over 1, they will be
an exact number of minute-spaces apart at 12 min. past
11. For the same reason they will be an exact number
of minute-spaces apart at 11 hr. 24 min., at 11 hr. 86
min.. and at 11 hr. 48 min. Therefore, they will be an
exact number of minute-spaces apart 4 times between
11 and 12.
127. Time A walks 20 mi. = 20 x 11 min.
= 220 min.
« D
CI
^'s rate
= (220 +46) min.
= 265 min.
= %^^ min. per mile
=^ 13i min.
Again, time A walks 5 mi. = 55 min.
" fi " " = 66i min. ;
.'. A wins by 11^ min.
.*. Distance he walks in 11^ min. = y|| mi.
U mi-
128. ;^ X 8 -h i X (number of months when the
remainder should be paid) = 4| ;
.'. i X (number of months) = 4^ - f,
and number of months = 2 x 3f
129.
= 7^
Buying price = |^^ of 99.
==90.
dhl5 3 46x3
Income
99
• = $466.
130. Cubic contents of tank = (8 x 5^ x 4|) c. ft.
== 192 c. ft.
W^eicrlit of water =^
192 XlOOO
" 1 «
lb.
= 120001b.
SOLUTIONS HAMBLIN SMITHS ARITHMKTIC.
187
Number of gallons
... 1 2 0x4
nx~8
= 1200.
181.
182.
3 0X30 3 (;
4X11
-^SJ ^ 1 —
V 117 /
37
3 X 3 X 3
3 xi
37x4
6x63
j^+fof^:4
hence ^V "^ t^ie boys
— 117 'J AX
.1118 — 10.
53^^ BIT — TJ.l >
B,
and number of boys = —^-^
= 46.
133. The L. C. M. of 8 ami 10 = 40.
In 40 ft. one wheel makes 6 revolutions and the other 4 ;
.*. distance requhed — 100 x40 ft.
= 4000 ft.
134. A works (4i + 3|) hr., or 8 hr. ; li works 4^ hr.
Cost of 12^} hr. work = $1.87^ ;
<» tj i« (ij. 8 X 1.3 71
"** 1 24
= $0-88.
(n>4ix 1374
(«
4*
'«
^ lai
= .50-49^.
185. When the minute hand is between 2 and 8 ;
number of min. past 8 + ^^ (number of miu. past 3) =^ 15 ;
.•. If of number of min. past 3 = 15;
.*. number of min. past 3 = ^ ^ x i a
1 3
18||.
Again, when the hands are together between the
figures 8 and 4, the number of min. past 3--^^^ (number
of min. past 3) = 15 ;
.-. ji of number of min. past 8 = 15;
.•, number of miu. past 8 =»
1 1
'TT'
I
188
SOLUTIONS HAMHLIN SMITH S ARITHMETIC.
1
186. Time for $320 to gain $24 interest = 1 yr. ;
" $H20 " = ~^-^ yr.
4t
i(
= m yr.
187.
Present value = £
93S8X 100
Income from i;l44 invested = £9;
108
('
)23g8 X 1 00 ,,
> 10"8
f
3^3«8 >c_l 00x9
T(r8i<"r4 4
= £180 i)s. 2rf.
Also rate per cent. = ^~^^~
188. Since A can make 50 when B makes 45,
andvl •• ' 50 '' C '' 40;
.-. B " «' 90 " C *• 80:
.'. B can give C 10 points.
139. Sum invested for £8, income = £90 •
... <« " " £2000, *' = £12^22112
^ £60000.
140. 14 mi. 6 fur. = (286 x 830) ft.
Amount of water drawn from Canal
= (236 X 880 X 48 x^^) eft.
Amount of water in the lock
= (80 X 12 X 8^) c. ft. ;
... number of barges - ^•''ex^aox*
80X6X17
= 88.
29
141.
J 3
¥ X i
X H of 5555.67
Ha
. 22X8X4Xl3XgflXg.e7
' 7X39X3X28X45
= $1.76.
142. Work done by A in 10 da. = V-
" destroyed by i? " ~ f 5
SOLUTIONS HAMIII.IN BMITh's ARITHMETIC. 189
.*. part of work done
II
" to be done = ^'j.
Time reciuired by vl to finish = {^\ ~ j^) da.
= ^ da.
148. Let 1 represent the quantity of water in each
cistern ; then,
quantity of water which runs out of first cistern m 1 hr
in no. of hr. required
no. of hours
II
It
II
«i
u
By conditions of question
no. of hours
second cistern
__ no. of hours
II
1 -
= 2 1-
no. of hours \
~~J j
rt no. of hours
.= 2 ^ .
, 8 no. of hours
10
and no. of hours =
= 1.
12.
8
144. In 1 day a man does ^i^ of work ; a woman,
ffiir of it ; a boy, ^^ of it ; a girl, ^^^ of it ;
.-. 1 man, 2 women, 8 boys, anil 4 girls do
CxiiF + izTiy + TOior-l-^tTT) of work daily.
Time to do all the work
^ - ^ ' " ' \)fda.
"I ■'■ • (rBTT + TrfTT + TTOTT+irBTrJ
9 a n n -1
T^^
ua.
= 28* da.
140 SOLUTIONS HAMBLIN SMITH's ABITHMETIC.
145. The fast train runs 5 miles while the slow one
runs B miles ;
.-. distance run by the slow train = f of distance run
by quick train.
But distance run by quick train = distance run by
slow one + 100 miles ;
.-. distance run by quick train = | of distance run by
quick one -f 100 miles ;
.'. § of distance run by quick train = 100 miles,
and
it
«<
5X100
mi.
.-=: 250 mi.
slow " = f of 250 mi.
= 150 mi. ;
.*. distance between London and Edinburgh
-- (250 + 150) mi. = 400 mi.
146. Price of 3 per cents. =75.
3iX 75
H
a
= 87.5.
147.
(2-3 + 1-15 4- -524) = ''^VV^ ;
2*3
A gets TJTTa of $198G.50 = $1155.
1*15
B getsTTH of !5198G.50 -^ $572.
a gets .~7 of $1980.90 = $259.50.
148. Ml == 11 guilders 12 kreut. = G72 kreut.
£1 = 25.5 fr. = y^(/' X 5G0 kreut.
= 714 kreut.
Gain on G72 kreut. ==42 kreut ;
- 100 kreut. ^ l^Vi.
=^ 6+ ki'eut
kreut.
SOLUTIONS HAMBLIN SMITH's ABITHMETIC. 141
149. 35 yards = 32 metres ;
.-. 69^ miles = «_iliLiiil^x_3j ^^^^^^^
= 111835f. metres.
150. Area of walls = (2 x 36 x 14) sq. ft. = 1008 sq. ft.
Deduction = (2 x 8 x 4 + 3 x 10 x 5) sq. ft.
= 214 sq. ft.
Area to be painted = 794 sq. ft.
Cost of 50 sq. ft. = £2 16s. 3d. ;
t(
794 sq. ft. = !2iii(£|.i^ii^).
= £U 13s. 3d.
Area painted for 56i«.= 50 sq. ft. ;
tt
98. =
9X50
5ei
= 8 sq. ft. ;
8
sq. ft.
.-. additional height = ~^ ft. ;
'to
= ift.
151.
9
¥
3
12f
-4-4-4- 9
7 7X6
= 27 I 77X6 ,1,9
=z 2 7 1 3 1 8 !)
^2
= 2.
152. 41-06328 ^ -0438 = 937, and -02268 over;
.-. there are 937 lines, and the length of the remain-
der is 02208 in.
153. Distance A runs in 1 min. =. (2^^ -~ 164) mi.
= (^x^)mi. ;
" 34 '' ==(34x^x/yini.,
" B " 84 "
= (i|x34x^x/^)mi.
K «»: .
— O nil. ;
'. length of course = (2^ + 5) mi.
= 7i mi.
and
I'
I
i
142 SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
164. Rate of boat— rate of stream = 6 x li mi. per hr.
^ 7i mi.
Rate of stream = 2 mi.
.♦. rate of boat in still water = 9^ mi.
/. rate of boat in usual state of stream
= 9 mi.
Time to go 9 mi. :== CO min.
limi.
_ ii^L^ min.
— 9
= 8;^ min.
155. 5 % is 12c?. in the £ ;
.-. he has 240^^. - (10 -hl2)c?. left out of £1.
1^^ of original income = 4J545
((
((
/a a 4 X 5 4 5
= ^' -J 1 8
= £600.
156. Net income from $(1071 + I) = ^(^-^V of 6) ;
... " " $14350 = $ — i-^r~
= $760.
157. He may ride for ^ of 5 hours, because he can
then walk back in f of 5 hours.;
.-. he may ride ^ x 10 mi. = 16| mi.
158. Call the place where the trains meet M ;
the distance from L to Jf = 4 x rate of slow train
in miles per hour ;
.-. distance from iV to 1/ = 1 X rate of quick train
in miles per hour ;
4 X rat e of slow train distance^om L Jo il/ _
•*• 1 X rate of quick train = distance from JV to J/ '
rate of slow train distance from JV to M
i»«f
rate of quick tram distance from L io M
compounding the ratios {kxt. 215),
SOLUTIONS HAMBLIN SMITH's ARITHMETIO.
143
4 X (rate of slow train) 2
(rate of quick train) ^ '
/. 2 X (rate of slow train) = rate of quick train,
169. Since £170 = 4233 fr. ;
.-. £1 =r. \2^^? fr.
= 24-9 fr.
Again, £400 = 603 x 20 fr. ;
£1 =
8 3X20
4 ~
26-15 fr.
fr,
160. The cube root of 60G53 = 3-7;
.'. length of outside edge = (12 x 3*7 + 2 x l-3)in.
= 47
JZ
in.
161.
46
X -
62-
162-80
45
X
= - X
= 1.
63
14
X —
10
I
162. Since £3"
Tiy
£423267
1 oz. Troy
4 2 3 2 6 7 X 1
:j.9
OZ,
Troy
423267X1X2 0X24
7000X39
— 7442,2^ lb.
163. Part done by ^ in 1 hr. = ^.
((
" « i< — 3
r> —■ -g.
(<
" C " =^;
• •
" A, B and C in 1 hr.
= H-l+f
— JO.
.'. time
to do the woik = -^jf hr.
= 48 min.
s. avoir.
144
SOLUTIONS HAMBLIN SMITH S ARITHMETIC.
164. Interests- $771. ©9f -$750 = $21.09^.
Time lor which $56.25 is interest on $750 =- 12 mo. ;
.*. time for which $21.09| is interest on $750
2 1.0 93X 1 2
5 0.26
= 4.V mo.
mo.
((
165. 5 parts in 20 parts in the 1st glass are spirit .
4 " " " 2nd "
.'. 9 " 40 of the mixture are spirit ;
.-. the ratio is 9 of spirit to 31 of water.
166. Selling price of $100 = $125 ;
Toff o^ marked price = $125 ;
«, dl; lOOX 125
ct
=-$-
80
= $156i;
he marks his goods at an advance of 56^ %.
167. Income from investing $101^ - $6 ;
$17255 = $l'-?-'-'^«
= $1020.
Income from investing $85 = $5 ;
«
(i
((
$17255 =$
17 2 r, 5X5
85
= $1015;
.-. total income = $1020 + $1015 = $2035.
168. Time required for 15 men working 9 hours a
day to finish the work
= 16 days ;
.'. to do the work in 1 day 1 man must work (15 x
16 X 9) hrs. ;
.-. to do it in 12 days 18 men must work -li^"^A? hrs
*' 12X18'
= 10 hrs.
169. Increase of shorter in 100 yr. = 3,014 in. •
s ••• ** " 125 vr. = 1^5x3:024 •
= 8-7675 :
SOLUTIONS HAMBlilN SillTH S ARITHJIETIC.
145
hence the longer he.6 to increase (8-7675 — 1-02) in.
or 2-7475 in.
Increase in 125 yr. = 2*7475 in. ;
1 00 X 2 747 «
((
" 100 yr.
125
=-. 2.198 in.
170. B walks at the rate of |S mi., or 4| mi. perhr. ;
.-. B walks 20 miles in (20 -r 4^) hr. = 4i hr. ;
A walks 20 miles in 1 hr. -f- 4i hr. = 5^ hr. :
50 K 5*
A walks 50 miles in — ^~- hr. = 14^ hr. ;
A reaches London at 6 hr. 30 min. P.M.
171.
1074;-4689(103-67
1 "^
203
747
609
2066
13846
12396
20727
145089
145089
u
5
7500
1099
189119224
126
7
64119
8599
49
60193
171
974700
6866
8926224
981666
8926224
U6
SOLUTIONS HAMBLIN SMITHS AUITHJIETIC.
i
172. Time to read (220 x 28 x 12) words = 5^ hr.
.-. " " (400x36x14) «*
__ 400 X 3 e X JJh X 6* 1
2 2 Ox 2 8^1^
«= 15 hr.
173. Distance the train goes in 60 x 60 sec.
= 20x1760 yd.
18 sec.
«
<t
tt
.^ 1 8 X 2 0x1760 -1
¥ox"6cr y
= 170 yd. ;
.•. length of bridge =-- (176-120) yd.
i ^ = 56 yd.
174. 20 men =- 30 women, and 50 children = 30
women ;
.-. sum received by (30 + 40 + 30) women
= $000 for 1 week's worlj,
and sum received by 1 woman = Jjjj ',■ ;; }} " ««
= $6.
175. The first strikes the 7th stroke after --?** sec. ;
sec. :
the second strikes the 7 th stroke alter '"' "" ^•'
1 1
.-. the difference = ^-jT^ ^®°- = ti ^i»-
176. Money received = $(43 x lU) ;
annual income = $i^JJiiL"
1 2 H
= $23.17;^.
177. Number of minutes between 9 a. m. Tuesday
and 11 a. m. Wednesday = 1560.
Number of minutes between 9 a. m. Tuesday and 9
p. m. Wednesday = 21 GO.
The slow clock goes 1550 miu. wJiile the fast on
goes 1560 min. ;
goes
SOLUTIONS IIAMBLIN SMITH's ARITHMETIC. 147
.-. the slow clock goes 2160 min., while the fast one
?160xl6eo . r.,-.^„
— ^^^ — mm., or 2173|« mm. ;
it must be put back (2173§» - 2160) min., or
13|9 min.
178.
1 7 I 8 ^f 2 3 B 7 .
f 1^ of the ore = 506 tons ;
7 6 0X506
the ore
2 63
1520 tons.
tons
179. The net annual increase is 1 in 60, and, hence
the population of each year = «^ of the population of
the preceding year ;
.*. population at end of 5 years = (.«^)fi of 10000000
= 10861578, nearly.
180. Length of side = ill '^J 1*1 yd
° 1134 J^'
= 1120 yd.
Area of each - ^^^^ acres.
= 262 /^ftj- acres.
181. -0416 = 3,'/^ and -0375 = ^|.^ ;
•■• -jVk of number of inmates at first = ^^^ of (num-
ber of inmates at first + 40),
and gi^ of number of inmates at first =« ^-»^ of 40 ;
.-. number of inmates at first = 240 x t^^^ of 40
= 360 ;
..•. number of masters i?^ ^^^.^ of 360
- 15,
and number of boys =:^ 360 — 15
.^ 345.
182. The shares are as 1, 3, 6 and 10.
1-I-.3 + 6-U10 - 20.
il's share = .V of $350 = |;i7.50.
I'll
148
SOLUTIONS HAMBLlN SMlTU's ARITHMETIC.
B's " = 8 X $17 50 ^ !S;r)2.50.
C's " = 6 X $17.50 = $105.
Da •* = 10 X $17.50 = $175.
183. What cost $100 I sell for $92i, and should sell
for $112i to gain 12i % ;
.-. selling price of $3700 =■ $''---"|p*--
= $4500.
184. Interest = $'4
^12 6 6X73X0 _
3 S X 10
/1 265 X li
$15.18.
'iZOO X i-g- -^ p
Discount = $ j — iTvil" I ~ r^^ '
.'. difference — 18 cents.
185. On $100 outlay he should get $160, but receives
only three-eighths of $160, that is, $60. Thus he loses
40 per cent.
186. ^^l- of part of income over $400 = $1024.40 ;
$400
$100X 1024.40
= $1040 ;
.*. the gross income — $1440.
Sum invested for income of $6 = $101^ ;
... u w u $1440 = $^^"^"-1-^-'^-'-*
= $24360.
187. His gross receipts are decreased 85 % by the
fall in flour, and 5 % by the lowering of trade expenses,
and, therefore 40 % in all.
Hence he can lower the 15 ct. loaf 40 %, that is by f
of 15 ct. or 6 ct.
188. ^ of profits for 28 mos. = $7890.50;
, , 1 ni. £ 10 7 X 1 '_' X 7 890.5
.•. total pronts tor 12 mos. =: ^^ ^s
^ $11835.75
I
by I
SOLUnONS HAMBLIN SMITH's ARITHMETIC. 149
189. Cost price = ^ {> «^ of $8.82^ = $3.40.
190. Area of whole rectangle = (72 x 46) sq. yd.
= 3240 sq. yd.
Area of grass plots = (4 x 27 x 18^) sq. ft.
= 162 sq. yd. ;
area to be gravelled = (3240 - 1C2 - 36) sq. yd.
-= 3042 sq. yd. ;
.-. co8t==i^^%ts. = $81.12.
Depth of pond = y^^' yd. = 7 yd.
191.
11X2X18
2x9x7
-1-7
22 10
T ~ T
1-^of ]i + ^ofi}
1 3
1— _o
1-T\0f |i+l}
192. 9 boys are equivalent to 6 men ;
since 12 men do | of the work in 6^ hrs. ;
.'. 1 man does ^ of the work in ^-^-^^hrs. ;
.-. 17 men do ^ of the work in ^r^^iH. hrs..
' 17X3X2 "*°'»
or l^g hr.
193. Principal on which $14i is interest = $100 ;
$1008.75 "
rfs I OG 8.7 5 X 1 00
(i
= $7500.
194. Compound interest on $100 = $8.16.
Simple " " r.^ $8 ;
.*. sum on which $0-16 is diiference = $100 ;
IP ''*' ~. 1 fi
-- $3750.
1Q-^ TU-a vo^-irk M* rti-fcofo ici oc. O vy n ^x OK w 1 O i-^ 1 v > O
X 18x10};
160
SOLUTIONS HAMBLIN SMITh's AUITHMETIC.
.*. the cost of second vessel = --^-^""y^^ of S80000
= $8400.
196. ^%% of a child's share = ^' of a brother's
share ;
.-. a brother's share = ^y^ of a child's share ;
hence 6 times a child's shared Yoir ^i"^^^ ^ child's
share = if> 12670 ;
or, \%y times a child's share = $12670 ;
.-.a child's share -^ $1940, and a brother's share =
$990;
and, when the legacy duty has been taken away, each
child will receive $1920.60, and each brother $960.30.
197. Interest on ^'s debt to A = $(5 x 3:^ x 4) = $66.
P. W. of B'a claim onA = \^^ of i^360.6O = $350 ;
.-. B has to pay $566 -$360 =r $216.
198. £1 16s. 8^. -= UOd.
Buying price per lb. = f*^d. = r'l^d.
Gain on an outlay of ^^d. = (f — f |)(i.
= id;
" " lOOrf. = ^^^^^d.
5, 5
T¥
= Uj\d.
199. 20 %, or ^ of the wheat grown in the conn try
= 10000000 quarters ;
.*. wheat grown = (6 x 10000000) quarters
= 60000000 qrs.
200. Rate ivith the stream = ^/ mi. = 4^ mi. per hr.
Kate against " =11 '^i- = l^j i^ai.
.*. rate in still water = fmi. = 3 mi.
.'. rate of stream
4^ mi.— 3 mi. = l^mi.
<<
BOLU'^IONS HAMBLIN SMITH S ARITHMETIC.
151
201. 1
1 +
54-
2 +
8+ «
UT
4 +
1
8+^
1 +
^ + U
64-
1 + tVt "^6 +it
4 + fxF
202. Part done hy A, £, and C, daily = i.
•' 4 and B
" fi and C
<«
1
ti
li
" G
" A and C
— s »
Tlf
8 .
— Hf »
/. A and (7 can do the work in 2/ da., or 4| days.
203. Cost price :-: Vy> of $38.25
$3826 .
92 »
gam
$(57-^*);
3835>
•. gain per cent. = ^^-^^iMfJMD
9'J
.100 X (5244 - 3825)
3825
$37 •'
T5T
204.. 1 mile = (1760 x 36) in. -^ U%%\ metres
= 1609.306 . . . metres
205. P. W. of $2.05 = T^^4 of $2.05 = $2.
206. The amount of $1 = (1-02)4 ^ $1.08243....
interest of 1 100 -= lOOx $.08243... =$8.243....
162 SOLUTIONS HAMBI.IN SMITH's ARITHMETIC.
207. Money realized by sale = |"' J||^^-?i.
Income from M. B. stock = $"^^l'-^JlJ?J_i
^ 100X178
= $646.80.
Original income = ^^^^-2^^^
= if; 587. 40 ;
.-. difference = $59.40.
208. Cost price of 1 quarter of mixture
= 1^^ of 67s. Gd.
= 46«.
Sum gained on each quarter of the cheap kind is 78.
Sum lost on each quarter of the dear kind is 2*. ;
.'. they must be mixed in the rates of 2 : 7.
8x3x3
1 tTF
c. ft.
209. Cubic content of block =
= ^s C' ft.
Weight of ^V c. ft. of water ^ ^^ of IMo lb. ;
" gV c. ft. of gold = 19.26 X X of i^^^ lb.
3 6 J g
* = 33 lb. 7 oz.
210.. Content of cistern = (1000 + 8) c. in.
= 1008 c. in.
Area of base = ( y^^ - yyo) g^^ fj.^
= 11 sq. ft.
2JL X 1 44
Ti
sq. in. ;
.-. depth = (1008-^^^'-*^) in.
= 27 in.
oil SfiT 1 42 6
f of £10 145. Id. = £9 3s. 6d.
Again, -85714 of £10 14s. Id. = -85714 of 2569d.
= 2201-99266c^.
less than ^hz^-
£9 Ss. Gd.
=: 2202d.
= v/uioirt., wnicn is
SOLUTIONH HAMHLIN HMITh'h AUITIIMKTU;.
158
212. $400 for 8 inos. gives the samo interest aH $100
for 1 year, aud since the rate in double tliat on the
$827, the interest at the end of the year will be the
same as the interest for a year on $827 + $200 at the
smaller rate ;
.-. interest on $527 for 1 yr. = $20.85 ;
$100 " =$'"7/;?-''= $6;
.". the rates are 6 % and 10 %.
818. Cost of a gallon of mixture T^ (8 x |5^)«.-2^».
But 2 J«. is ^ of 4s. ;
.*. i of the mixture is water
i. e. there are 8 pints of water in each gallon.
214. Interest on $550 for 9 mos. == $16.60 ;
6 (» X «
$100 " 12 '« =
216. The broker first offered /^ of the value ;
then ^^ of the value + $870.76 = Voo* of the value ;
.-. $379.75 = (|U-T\^''ftlie value = -^^ of the value ;
.-. the value = ^t of $379.76 = $2460. .
216. Asking price = \l^^ of cost price ;
.'. selling price = ^-^-^ of l^-} of cost price
= \%%^ of cost price;
•'• Tinr of cost price = $6.76 ;
.-. cost price .-^. $i^-5^|±l^ = $92;
and asking price = \l^ of $92 = $115.
217. 15 masons build 200 sq. yd. in 60 hours;
1 A X 60
1 mason builds 1 sq. yd. in
^ '' 200
42 X 1 JSx «0
7 x'a 00
they take 270 hrs., or 30 dayf^.
.'. 7 masons build 420 sq. yd. in
hrs. ;
'^ hrs.
154
SOLUTIONS HAMBIilN SMITH S ARITHMETIC.
218. The average dividend = $
.7fi+.60
= $-67^.
His debts are ^J of $2700 = $4000.
219. Toll on 240 hhd. = 2 hhd. + S90.
150 hhd. = 2 hhd. - $30 ;
90 hhd. --= $120 ;
150 hhd. ^I'""'^'^"
((
((
((
90
= $200 ;
.-. value of ■ 2 hhd. = $200 + $30 = $230 ;
1 hhd. = $^-|-' = $115.
220. Area of walls = (80 x 6) sq. yd. + 2 (80 x 5)sq. yd.
= 1280 sq. yd.
Deduction =(6x8x3) sq. ft. = 16 sq. yd. ;
.-. number ol pictures = ^ ^^*^ ^ ~ ^74.
221.
32 _ 1 65
X yY A
2 13
88 58
31 V 100
TO ^ WWJf
320-165 1 21300
222.
— 880—680 "^ TT ^ " rotr
I55v 3 1 '05
= irS'6 ^ Ti T — ^TT — ^^'
16376-248001 ( 124-001
1
22 53
44
244
248001
976
976
/ 3 1.36
— /4.48 _ / .«4 =
.8 -= 8
.¥ ¥•
223.
* + '
248001
248001
— 7 .
3 ^ ¥
/. f of the army = 2000 men ;
.-. whole army = ?-^-^"--*'-"- men = 9000 men.
224.- The interest on $2000 for 3 mos. = $37.50. ;
/. at the end of 2 years the second would have re-
SOLUTIONS HAMBLIN SMITH's ARITHMETIC. 155
ceived $19000 + 7 x $37.50 + 6 x $37.50 + 5 x
$37.50 + 4 X $37.50 + 3 x $37.50 + 2 x $37.50
-+- $37.50, or $20050. which is more than the first
tender by $50.
^225. An income of $5 is got from an investment of
$91i;
$450 is got from an investment oi
rViT of sum left = $*!' iiJi
TtfTT
sum left _ $^-^^^-* = $9125.
226. Part done by 2 men and 4 boys hourly
'2 " iboy
3 boys
(t
••• " " 1 boy u
hence 1 boy would do the whole in 18 hr.
Part done by 2 men hourly = ^ _ i
— i
— a-
= 1 •
1 .
— F '
= 1 .
(<
((
5 .
5 .
75ir »
" 1 man
hence 1 man would do the whole in Y' br., or 7' hr.
Part done by 1 man and 1 boy hourly =^ .k,^ i^
=z 7 -
hence 1 man and 1 boy would do the whole in •''« hr
or 51 hr. ^ ''
227. Interest on $15840 ^ $(15840 x^^Xy^^)
= $316.80.
Interest = (j.^ x j^^) of the sum
= g\ of the sum ;
.-. discount = ^% of $3696
= |)uio.8(;.
228. See solution of Ex. 168, page 144.
•^
ICO
SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
229. $20 ^ interest on $840. (Ait. 181) ;
.-. $360 =-- " %^^^^
= $6120.
230. (2 X breadth) x breadth x ^!^^ = 4096 o. ft.;
.*. cube of breadth = 4096 c. ft.,
and breadth = v^ioye ft.
= 16 ft. ;
.-. length = 32 ft. and height = 8 ft.
231. 1 lb. tea =
50X84
lemons
= (5 X 12)(^.
= 6«.
232. Number of killed and wounded
= \(A\Qi\ of army
= -^^ of army ;
.'. -^-^ of army = 500 men ;
,*. army == 84 x 500 men
= 42000 men.
233. Cash price in notes = \l of $135 = $128.25.
" gold = |^« of $128.25
= $106.87^.
Change to be received in gold = $(135 - 106.87i)
= $28.12^ ;
«« ' " " notes = « of $28,121 •
= $33.75.
234. Interest = interest on debt.
Discount = interest on present worth ;
,*. interest— discount = interest on (debt— P. W.)
:„x 4. j: i.
(Sqe Art. 181.)
I
SOLUTIONS HAMBLIN SMITh's ARTTSJIETIC.
235. Cost price = 20 x 16 x 55 cents
No. of Troy oz. bought = ^ox 7000
Selling price =
157
"•2
20X24
20 X 7 000X60
2 0X24
.*. loss = $1.
cts. = $175 ;
. 3 2 X 9 1 J
236. $6 is got from investing $91^ ;
.-. $320 " "
Income from $80 = $5 ;
« (( (m320x9U rt£.3 20x91ixa
• o JB JT) .
^ 6 ^ (i x 8
= $305.
237. P. W. of debt ^ %^'-;^''- = $25A ;
.-. difference = $(25/^-25.20) = $1*.
238. ^%% of his property + .^« •} of his property + ^|] ^
of lis property ^=^ £6190 ;
.-. 1^^ of his property = £6190 ;
(( /jGOOx 1 90
239. Sum invested
No. of bbl. bought
010
=: £6000.
(lt,3G81 XlOO
102t
3
7.7VTT
= $3600.
=^ 480.
Total selling price = j^^of $(3681 + 119) =$4560;
.-. selling price of 1 bbl. = $ VsV = $9.50.
240. Content of wall = (60 x 20 x 4) cub. ft.
Space occupied by bricks = |^| of 4800 cub. ft.
= (375x12) cub. ft.;
number of bricks
I
375x12x 1 728
9 X 4ix"4
= 48000.
241. ^ x tV X I of 168*. = *f s. = 3>.
242. A, By and C together do i of (i + 1 + 1) in 1 day ;
j-U^.. J_ 1 n :„ J .
. . Micj uu ^ g 111 tjuv uiiy ,
.". they do the whole in |f days = 8^ days.
lo8
SOIiUTlONS HAMBLIN SMITH S ABJTHMETIC.
248. For every 8 days of the time he earned $(2 x
1.50) and paid 60 cents, or his net earning was $2.40.
Time he took to earn $2.40 — 8 days ;
((
244. '+14-1 =
$72 - \~-^' days = 90 days.
28+^24421
7 3
THIS '
1
IT
.-. Ist gets -7^ of $140000 = $56000.
T?rs-
I
2nd gets -^ of $146000 = $48000.
TTiff
I
3rd gets --^ of $146100 = $42000.
245. Interest on $200 for 3 mo. = $10.
$200 for 1 yr. = $40.
Discount off $240 for 1 yr. = $40 ;
$210forlyr. =$^iil?=$35.
246. Since £S lis. lO^d. = 1869 half-pence,
and 1 sovereign = 480 half-pence,
the least number of sovereigns will be the L. C. M,
of 1869 and 480 half-pence.
L. C. M. of 1869 and 480 =. 3 x 623 x 160.
But (3 X 623 X 160) half-pence
= 623 sovereigns,
and 623 sovereigns weigh 160 oz.
247. Investment to give $7 dividend = $176.
Investment to give $445.50 dividend
= $11137.50;
.*, selling price of flour = VV of $11137.50;
.-. number of bbls. = -—■''"''
> X 99
= 1500.
SOLUTIONS HAMIJTJN SMITHS ARITHMETIC.
150
248.
1 lb. Troy =.. |||. lb. Avoir. (Art. 157).
olc 2 4 lb- Iroy
lb. Avoir.
Weight of riugs — '"^"^'-^s
I •-' X a
Weight of rings and box
14j4X1050x28
= V lb. Avoir.
-~(V + 3^) lb.
= 7tV lb.
Cost of carrying 1 ton, 1 mi. ^ 5s. ;
7rVx 144 X 6
2240 "*
" 7tV lb.. 144 mi.
Value of rings - - 1050 x 22*. ;
cost of insurance = ^^^ x 1050 x 22«.
¥0T)
1 \±.'i
4
6'.
.-. total cost = i^^L*_%.
4 •
= £1 ll6'. 4'2£?.
249. Interest for Ist year ^ $250 ; '
2nd " ^ $275 ;
" 3rd " :^- $802.5 ;
" 4th " .= $332.75 ;
5th " = $366.025 ;
.-. the sum of these is $1526.275 ; so that the inter-
est to be gained after the 5th year is $201.31f, but the
interest for the 6th year = $402. 62f.
.'. 5^ years is the time required.
250. Length = § of breadth, and height = ^ of
breadth ;
.-. t of breadth x breadth x ^ of breadth = 5832 c. ft.,
and cube of breadth = 5832 c. ft. ;
.-. breadth = y.5832"ft.
= 18 ft.
Length = 27 ft.
Height = 12 ft.
160
SOLUTIONS HAMBLIN SMITH S ARITHMETIC.
h
I
3
251.
121711
(32)(8)(56)
973688 = 800 times the multiplicand.
6815816 = 7 times 8 times 121711.
8894752 = 40 times 800 times 121711.
3998936616.
'J8v22.5 40_2«.1
OKO T^-^ ^^ I JL-r.JJL V 2040
•^O-^. 1125 "T 16 I 2(5 ' V ''^ ■3"2'5
28x225, 146X 2 04X t 1 1 2 8 i ^Ji^_L?_JLLL
1 ] 25 ' 421^2 a 2 2 6~ ^ "*" 10 73X46
28if63 = Q
253. Commission on $2480 = $21.70 ;
> 100X21.7
((
$100 = $-
2480
5i;
•. his commission was at the -rate of |^ %.
254. Income from $92 realized =
<«
$25760 "
,25760X6
^ 92
-^ $1680.
" investing $45i = $3^.
((
$25760 = $
25760x3^
46i
= $1840 ;
,*. his income is increased $160.
3 48x100 34800X2
255. P. W. of $348 =
. W. of $292
.-. total gain
P. W. of $292 = $^'^V'-' = ^'lll"^ ;
3 4 8 (• X 2 2 9 2 0x8
):
'.;■ 3 8 3
100/ ^ ' ** 00X2 29200X8^
gain per cent
Bwjr^wvu. \'''vn
!> i X 8
£.'03
^
3
V
SOLUTIONS HAMBLIN S.MITh's AUITIIMKTIC.
IGI
■ T4 8 00X4X803— 20200x8x30;$ "
s= 2 0:rx 8 03"
M 3
3_4 8 X 2 X G^O 3 — 20200X8X203
2 03X20 2ir8
34800X2X11-400X8X2 3 3
> 2 03X4X8
2 1 7 5 xj^l — 1 0x203
" aoTa
_ 362S 170
— 5 era =^ ■'•'T'
256. Since there is a difference of half a day in the
time of completion, according as the boy or man com-
mences first, the man must do twice as much work
each day as the boy
Part of work done in one day by both = (^"3 + ^2^) ;
.-. they will finish the work in ^/ <iays,
or 4^ days.
257. Share of 1st = -i-^^Oy of $80U = $90.
2ad = tVit of $100 _-^. 5g;30.
But as the machine works the same length of time
for each and earns $120, in all, or $60 for each, there-
fore, the latter must give the former the difference
between $(> J and $80, or $30.
258. Since B gets $2750 at the end of two years,
he receives -[^« of $2750, or $2500 ;
A calculates $2500 to be the P. W. of $2725, that is,
that the interest on $2500 for 2 years is $225 ;
$100 " 1 yr. = $l<'"-^-^i^*-
^ "^ ^2 500X2
259. Time 3G men dig (72 x 18 x 12) c. yd.
= (16x8)hr.;
3 C X 1 C X 8 Jjj.
man
7d. =
7 2 X 1 » X 1 2
0x3
hr.
162
SOLUTIONS HAMBLIN SMITH'S ARITHMETIC.
.-. time 82 men dig (64 x 27 x 18) c. yd = ?±J,f '-^-If-^-Mir.
2X3X6X8 ^g^
260. P. W. of bill = £
I so X 100
120
1 '->
= 24 da.
= del 50.
The wine and picture are charged at £(21 + 19), or £40.;
.-. he pays in cash i:(150-40), or £110 ;
.'. the cash price of the bill to the userer is £(110 +
10), or £120 ;
.-. the interest on £120 lor 4 mo. = £60 ;
((
" £100 for 12 mo. = £liL«_li^l.«o
12 0X4
= £150.
261. tV+^V+A+A+^V+ A - in --= nW ;
.'. length of 1^^ of rod =^ 302 in. ;
.*. length of rod =
2 0x302 •
15 1
'■= 400 in.
262. $0.75 is 9 mo. interest on $20 ;
,20.75X 20
m.
.-. $20.75
((
(t
((
or on $553^.
Again, interest on $20 for 9 mo. =z $0.75 ;
((
$100 for 12 mo. = %':1.1^^JL:1A-
263. Distance the first goes
10X3
mi. =
mi.
((
6 — ^
" second " =^ ^^^ mi. = | mi. ;
.'. length of walk = (t + ^) mi. = 1^ mi.
264. A runs 100 yards while B runs 96 yards ;
A runs 100 yards while C runs 95 yards ;
B runs 96 yards while C runs 95 yards ;
hence B, giving C 1 yard start, will overtake him at
the end of 96 yards, and will therefore beat him in a
hundred yards' race.
SOLUTIONS HAMBLIN SMITH's ARITHMETIC. 16iJ
265. The selling price = |^J of cost price ;
•■• lift 0^ tVV of cost price = \^^ of cost price —
••• (i^^-n^a) of cost price- $1;
•*• TffVxr of cost price = $1 ;
and cost price = $200".
266. Compound interest of $1 —{(1.05)2 — 1}
= $0.1025.
Simple " $1 = $0.10 ;
.-. $.0025 = the difference on $1 ;
•■• ^^ = " $.-n.vA= $1200.
267. Cost price == ]«» of $69.55 =. $65 ;
.-. loss is $65 - $61.75 = $8.25;
1 00 X 3.2 5
6 5
.•. loss per cent =
6.
268. Cost of £360 = $1736.10
1 = $4.82i ;
But this includes the commission at |- % ;
hence the course of exchange = |§« of $4.82^
= $4.81 nearly.
209. 3mos.
i year, and } of 8 % -2%.
Discount at 2 % = ^^^ of bill.
Interest '* := ^^ ««
Hence, (^-f^ - ^U) of bill = $16 ;
or Ti7^TTtr •• =$16;
.-. the bill = $L«?^«_li«
= $40800.
270. Cost of 30 lb. = $(18 x .30 + 12 x .05) =
Seliirig price ^ ||^ of $G = $7.50 ;
of 1 ik = $^ = 26 cents.
ij
I
1C4 noLiynoiii. tiamblin smith's arithmetic.
271. 8-14169 ) 1000000000 ( •81881.
94247V
11
2
7-00
«
•68
675280
814159
2610710
2518272
•918181818
•000039772
974880
942i77
•818221590
.*. the difference lies between
•0001 and -0002.
319030
814169
272. See solution of Ex. 4, paper II., page 71.
278. ^'s share = « of D's share ;
B'8 share = * of ^ of /)'s share = || of Z)'s share ;
^"s share = f of || of D'q share = /J of D's share ;
.'. (1 + ^ + II + li) of D's share = |21000 ;
.-. W* oi /)'8 share = $21000 ;
.•. Z)'s share -- $7000;
^ .'. C gets $6000, B gets $4800, A gets $3200.
d). G ■ .T X 1 4
274. Amount of $6.30 at end of 6 mo.
^ 100
= i§0-652.
.'. buying at $6.50 on 6 mo. is the more profitable,
or present value of $6.60 = $
(5.50X100
1 04
25
.*. buying at $6.60 on 6 mo. is the more pioiitablc.
275. Income = $l£:i_';|-LJl^ ^ $700.
Tax on $700 = $12.25;
((
1 2.2S
= $ Yo o" = ■' 4 cents.
m
SOLUTIONS HAMBUN SMITH's ARITHMETIC.
lare ;
lare ;
i 04
90
2.
5 ;
ble.
16£
276. Cost price of iron = $
44
<4
» gBOx 1
1 a
187icwt. = $1050;
1 cwt. = '^i°*«
== $1060;
277. Sum invested in tea = $ —
AO X 100
$6.60.
103
^?f" = 4000.
Number of pounds bought
Selling price = |2« of ${3060 1 30 + ^^^ of 3000)
•= $3875 ;
.-. selling price of 1 lb. = $^« J« =: 90J cents.
278. Interest received each year = $2i -f- $2 i + (in-
terest on $2^ for 6 mo.) = $5/^ ;
.-. on each $100 he gains $(5/o--3i) = $1|^;
4. ■ OJOAnl .1. ^200x100
.'. to gain $200 ho must borrow $ — ^Hi —
= $12800.
279. Cost of tea = 250 x 80 cents = $200.
A.llowance for carriage = y^ of $200 == $5 ;
.', customer has to pay $(9.30 - 5) = $4.30.
280. Cubic content of the plate = f c. in. ;
.-. thickness of new surface = {f-^(7x7x9x 144) }
inches ;
1 . _ 1 .
— 8x7X7Xi44^^* 66448^^*
281. After the sale to A he has left ^y^, or |;^ of the
flock;
after the sale to B he has left || of the flock— 90 ;
after the sale to C he has left
ill of j If of the flock - 90 } ;
••• \U of I If of the flock - 90 } = 579 ;
.-. f f of the flock - 90
.". If of the nock ^^ 690 ;
•^}f^ = 600 ;
the whole flock
= 2_«-«i^ =750.
♦ 3
166
SOLUTIONS HAMDLIN SMITh's ARITHMKTIO.
282. (l8t.)^ X (2ncl.)=^ X (Sv^.y
= 176882 X 270152 x 215496;
1st. X 2nd. X 3id.= v/ (176882 x 279152 x 215496)
= 2 X 8 X 9 X 41 X 73 X 289:
1st number
2nd "
8X 8 XB^Xjt 1 X 7 3 X a 3
3 I 5 4 a G
a x 8 X U X 4 1 X 7 3 X i 3 y
a 7 1) 1 ."i 2
8rd
<(
= 478.
= 369.
2x8 x0x41X73x8 3 ^A.
r^^.ro„ = 584.
176382
283. Sum of rates = llLtl_«« ft. = no ft. per sec.
Difference of rates = ^-iL+^L^ ft. = 22 ft. ♦* ;
.'. twice the rate of faster = (110 + 22) ft. per sec. ;
.-. rate of faster = ''"lll^--'-^ mi. per hr.
= 45 mi. per hr.
Twice the rate of slower = (110 — 22) ft. per sec. ;
.-. the rate of slower = '^'.Hf^ mi. per hr.
= 30 mi. per hr.
284. Cost of tea = | of 72 cents = 60 cents ;
.-. gain per cent. = ^^~ = 50.
285. Total cost =. $(175 x -60 + 225 x -50) =. $217.60 ;
.•. the selling price of the whole = jfg. of $217.60
=: J^261.
He sells 250 lb. for 250 x 55 ots. = $187.50 ;
(175 + 225-250) lb. for $123.50;
he sells 1 lb. for $^f'^^^«, or 82 1 cents.
286. Weight of nitre -= (^-7,^ + tVo ) o^ 10 owt.
= 151 cwt.
sulphur = (^'/u + T^^) of 10 cwt.
— l-j^ cwt.
charcoal = (yV^ + iVir) of 10 owt.
t^ 2^=^ cwt.
((
««
SOLUTIONS HAMBLIN SMITH*k ABITHMRTIO. 167
c.
2H7. Water admitted in 1 hr. = 6 x 8J t. - ISJ t. ;
.-. water gains (I83 - 12) t., or Q\ t. in 1 hr.
Number of hours to gain 60 t. = --^ = 8f hr. ;
.*. rate of sailing = -_ mi. = 4^ m
288. I *^ of * of the capital = $82000 ; •*
.'. J J^ of the whole " = $'»_'* •»*-*_"^,
and
«(
<(
dkl OOX A X 93000
'^ ^ 4x4
= $1000000.
iVff of I of capital = $20000 ;
^Vo of the receipts .- $(32000 + 20000),
and *' ^looxflgooo
= $100000.
289. Interest = ?£^^ 2 '^^^^^j
100
discount = ^^^'^ ^ 2| X rate .
100 + 2,ixrate'
interest _ 100 + 2^ x rate
discount 100
or, — = ^\, or, rate = 8^.
or « 7 — 1 -i_ r^*®
» w*» ^ir — '■^
To'
290.
— flO .
Eate of boat + rate of stream
Rate of boat — rate of stream ~ ^s*
55 X rate of boat -1- 55 x rate of stream
= 60 X rate of boat — 60 x rate of stream ;
115 X rate of stream = 5 x rate of boat ;
23 X rate of p.tream = rat« of boat.
168
SOLUTIONS HAMBLIN SMITH 8 AIIITHMETIO.
291.
(«)
2 +
3-
^r —
1 3
7x20
4x2
2 4- ''I
■—J IX.
1 3
7X29
2 05X7X29
13x87
-36
3J .
3 If '
(^)
6| + 17f - 7i 16
3^ + 2i -~ 4-,V ^^ 1
1 1
2 3 5x8
T4>r8
r_10
as
I TV
292. 9 men and 16 women do f of the work daily
4 men and 14 children do ^
((
13 men, 15 women and 14 children do (g + jr)
of work daily, or ^^ of work ;
.*. they do the work in 1\-^ days.
293. Proceeds of sale = £(100 X 93i) = £9350.
Income from 4 % = £ "YoV* = ^^^^ l^""- ''
Original income = £^-^^fJ^ =.- £300;
100
= £66 13s. M,
294.
1 7
51
<(
.*. increase
5 men in f ^ hr. do y^^'^y of work ;
1 man in 1 hr. does
3 men in 3 hr. do.
hence 7 boys in 3 hr. do yV^
1 boy in 1 hr. does ^^-^
6 boys in 1 hr. do /^
Work to be done by 6 boys = 1
Time for 6 boys to do /^ = 1 hr. ;
xl
901 — 599
1 ffTTO" — TZHU
<(
«
«(
699
TTSHU —
5 99
7
7ir
hr.
= 2-8u25809 hr.
295. See solution of example 5, paper IV., page 165.
296. A does t^^ of work daily ;
/. he does -^ of work in half a day ;
.'. in two days A and 5 do (1 —^—^^) of work ;
.*. in one day A and B do ,'^ -f- 2 = /^ of work ;
.'. in one day B does ^^^ - j\ = -^^ of work ;
.'. B does the work in "/' ^^-ys = 32 days.
SOLUTIONS H/ MBLIN SMITH's ARITHMETIC. 169
297. 30 children + 9 children + 1 child earn $34 ;
'. 40 children earn $34 ;
1 child earns i|^ ;
.'■ (86 + 6 + 5) children earn $~^''^~ = $39.96.
^'-98. Cost price = i oo of $132.50 =, $125 ;
. loss per cent. = 1^^'^'fo _ q
299. Amount of stock bought - $''•'' ^ S'<i 00
= $7500.
Proceeds of first sale =^ ,ff^^^^^»
^100
= $4675.
** second sale = $-'^''Ax««
^' 100
^ = $2125.
Dividend from M. B, S. - ,f ^''-"i*.!?.
= |4CGf .
Original income - $''Al^i21-«*
= $450 ;
.*. increase = $16f.
300. Area of room = (141 x 13 i) sq. ft.
Area of 1 plnnk = (j% x 10) nq. ft. ;
.'. number of planks = "^"^ ' ^ ^^=« == ^sh
VV X 10 ^'
Number of c. in. in 1 plauk =.. (^ x 8 x 120) c. in. ;
.-. weight of 28i planks =^ (28^ x i x 8 x 120 x ^) oz.
= 6840 oz. = 427i lb.
301. 4957.5081 ( 70.41
49
1404
I4nfti
5756
6616
1_2 9.494 V
6 0.7 6
14.38 83 1.69 8 7
3.7 8 — .7 6
1/inQi
14081
= •"2^-^, and the square
root of t
~ i'4u.
root of this fraction is —^
• o
I
170 SOLUTIONS HAMBLIN SMITH'S ARITHMETIC.
802. Interest on $157.50 for 5 yr. = |(189 - 157.nO)
= $31.50;
^100X31. 50
« $100 for 1 yr. =r $ \Tj'^oTrr
= $4.
303. Time from 2nd to 12tli July = 10 days.
Interest on $273.75 for ^^"5 y^'- ~" ^'' ' I'^fxi 00^
= $.375;
.-. value of first bill = $(273.75 + .375)
= $274,125.
Time from 12th to 22nd July = 10 days.
^456.875 X 100
Present value of $456,875 = $-
lOOj^
= $456.25.
304. Suppose tliecask to contain 12 gallons, of wbicli
9 are wine and 3 water,
then 9- i of part drawn — quantity of wine remaining ;
and 3-i of part drawn = quantity of water remaining;
/. 3 + f of part drawn — quantity of water in the
vessel when water is substituted lor tlie part drawn off;
.-. 9 — i of part drawn = 3 + J of part drawn ;
.'. I of part drawn = 6 ;
.'. part drawn = 4 gallons, that is, one-third of the
mixture.
305. Amount of bread each eats = § loaves ;
«♦ *' given by first = 7^ "
and " " given by second == ^ "
.'. he pays the first 7 half-pence and the second 1
half-penny.
306. Interest of $100 - $(100xf X^^Vif) = l^"'
.-. the bill of which $5 is the discount = $105 ;
^48,75 » = pil^.ll^
=- $1023.75.
<(
SOLUTIONS HAMBLIN SJIITu's ARITHMETIC.
171
.)7. Capital at end of 1st year
{«
i<
((
~ Toff of original capital.
2ncl yeai-
Toff "I TO TV 01
3rd year
= -J^J^ of l''^* of 120 of '«
TCFTT "^ TOT) "^ Toff "^
= ^''o^o- o^ original capital ;
original capital - ^^^ of original capital = $200 ;
original capital = 100 x $200 = $20000.
308. Cost of 1st horse =
2nd " = $
100X100
= $80.
((
= $1333^ ;
1 2 r,
fl. 1 X 1
.'. he loses $(80 + 138^-200) = $13^.
309. $2 is the interest of the discount for 6 mo.
$2 Ts also Ihe interest of $100 for 6 mo. ;
.-. discount = $1'00, and interest = $102 ;
.-. the sum ^ $l^^lfL^« -^ $5100.
310. Area of 5 external sides
- U54-f44)x6i + 27x22f sq. ft. = 1231 sq. ft.,
and neglecting the thickness of the material the area
of the inside will also be 1281 sq. ft. ;
.-. cost = =-^-;'^-±* cts. =. $12.31.
311.
(a) 3
2 -
-divided by 1 +
5T
4 +
8-1
= 3 -fl divided by 1-f-^
^ VV - (1 +U) = '-^-P - '-•^-^ _ «8 1
7« . 13
s
1 7
+ tV
¥3' "= iftof
1 ^
1 2 » 3
172 SOLUTIONS HAMBLIN SMITH's ARITHJMETIC.
fl
312. ^ of selling price = ^%% of cost price ;
.-. 16 X selling price = 16 x cost price ;
.-. selling price — || of cost price ; ,
.'. the gaii; = yV ^^ ^^^^ ^^^^ price ;
.-. the gain per cent. -- VV — ^^*
313. $175 has for principal $100
dh787. 50X100
$787.50
({
175
= $450.
Time for which $33.75 is interest on $450 = 1 yr. ;
... " « $540 " 1450 ^/^y^yr.
= 16 yr.
314. In 4 years I save 4 x G|%, or 26| % of my
income ;
.-. (26f % - 25 %) of my income =- $50 ;
.-. my income .-=: $'*-«'^^-'L«
== $3000,
315. $T^|^ =r tax from
.-. $101.40 =
=. $7800.
316. Cost price of 1 lb. - '''^^;' = 66! cents;
225 lb. = 225 x 66f cts. -^ $150.
Sum to be realized = $'''':^' ' ■= $1«5.
Selling price of 45 lb. -- 45 x 72 cents -- $32.40;
" (125-45)lb.---: $132.00:
*« of 1 lb. -- $'\'^o- - '7'^ii cents.
317. Interest on $500 = $(500 x^ x y«-^-)-$13:\.
$1500 remain to be paid, and I ought to keep this sum
beyond the original time of payment till the interest on
it amounts to ijjlS^.
,101.10 xlOOO
' 13
SOLUTIONS HAMBLIN SMxTh's ABITHMF.TIC.
173
$120 = the interest on it for 12 mo. ;
^ ^ 120
= 1^ mo.
----- mo.
318. A rims 880 yd. while B runs 800 yd. ;
then B runs 900 yd. while A rnns -^\ of 960 yd., or,
864 yd. ;
.-. B runs 1760 yd. while A runs (880 + 864) yd.
= 1744 yd.;
.'. B wins by 16 yd.
319. Cost of X egg in one case ^',5.
" other case ^^s. ;
it
' 4- 1
average cost of 1 egg — ^-^-^--s.
20 .,
.*. selling price of 1 egg = ^\s. ;
.-. sum lost on 1 egg = (^,p^_ ,V)s. ^j^\-^8. ;
.*. loss per cent.
100 X ^^Vir
'-' o
h
320. (6 X breadth) x 11 = area to be papered ;
.'. 66 X breadth = 148 x 3 x 2 sq. ft. ;
.-. breadth = '^•'''''^1 ft. =- 13 ft. ;
(! r,
.*. circumference of room == 6 x 13 ft = 26 yd.
821.
^
43 V 7
7 X 7 X !>
+ -%
7
(i ;$
r .
1?T »
3 X 'J X 7
822. (^„Vrt- of SOOfi. + %%?^> of 66.V.) x.^-^
== (i'U 8.S. 3f/.) X 5-^- = ,i^62 5«.
323. From 9 a.m. on Monday to 2 p.m. on Friday
^tn n «• ^ 4-h •■,• r\ vl /) y-I *-» »▼ *^
174
SOLUTIONS HAMBLIN SMITH S AKITHMKTIiJ
Difference between the watches for 1 da. =- 'A\ nuD ;
♦« " " Ig^ da.
= 4^\x3i min.
= 14 min. 48fj sec.
824. 6 men reap 85 a. in 7 x 12 hr. ;
.-. 1 man reaps 1 a. m — Yr — * '
_ . f, • 45X0 X7Xlai„, rut 1,,. .
.-. 9 men reap 45 a. in — .^ ^.j- hr. ^ 72 hr.
/. they will take J) ila. of 8 hr. each.
325. Cost price of tea sold = "iVo~ ^^^' " ^'^^^'^^' :
.'. he gains 12 cents on each lb. of 48 ct. loa and
loses cents on eaoli lb. of 00 ct. tea ;
;, he must put 1 lb. of the former to 2 lb. of the
latf^^r.
820. i ct., or $7,Vfr — difference of tax on $1 ;
^8.00 X 1
= $1440.
827. 5 cents in the $ is paid by !$500 assets ;
.-. his debts = ^'-^l"^ = $10000.
and his assets = ,^V o^ ^l^^OO = i54()()0.
328. Time 85 men do a work = ;58 da. ;
3 5X38
ixn;
19
K
da.
1 !»
= 70 days.
329. Robert's debt to Charles
= f of Robert's debt to Charles + 10^/. ;
.-. ^ of Robert's debt to Cliarles = 10c/. ;
" " " = 3 X 10(/.
830. Area of surface
= 2 X (4 X 2.1 4- 4 X 8 + 3 X ^) sq. ft.
= 59 sq. ft. ;
cost =
•J X 1 6
8.y. 2.U/.
I,.
SOLUTIONS HAMBUN SMITh's ARITHMETIC. 175
•^^l-Vxj2-f 4,^-3,0, 4 + 4^-8,^ m
" 8 X a C 7 ~ ='
'i> Tff
' J_^ 7 2^!) X 3 9
••'>2 ^ 8x7
ni-..'V
"- 4- 1
7 1
I ■
II )
¥
JJ32. Time lie takes to ride 1 way = -^ hr. = 1} br.;
" walk 1 way = (3]-!^) hr.
— 2.i hr. ;
. " " to walk both ways = 5 hr.
ooQ Distance in miles ,, , . . .
o,iii = the certain time in lir. + ,\y ;
distance in miles ,, ^ . .
^ = tlie certain time in hr. — '. :
distance in miles , _ distance in miles .
6 •"
"TiT-
I >
. distance in miles distance in miles
, distance in miles _ ^
" 20
.•. distance = 5 miles.
334. A and B, and ./ and C contribute .'^(1300 -f-
1500), or $2890.
7i and C contribute $1590 ;
.•, twice ^'s contribution = $1300 ;
hence A contributes .$G50, J] $740, $850, IJ $900.
Now $(050 + 740 + 850 + 900) gain $U52;
.-. ^'s share = ^"^''Mll^ .^^ <i,2U
"^ 32 V^'^^i
.7 40 x 1 1 .I 2 a. .,./. ^/-w
= $200.40,
C's '-'
" 3 -J
__ ^ rt 5 X 1 1 .0 a a> on/.
JD'i
[. rt 5 X 1 1 f) 2
3 2
dr. l! X 1 1 r>
f
3 2
$345.60.
t f.
■J-
'J
170 SOLUTIONS HAMBLIN SMITH's ARITHMETIC.
835. From A to B it takes ^ of 7 hr., or 3| lir. ;
.-. from B to A it takes 5^ hr. - 3^ hr., or 1^ hr. ;
.-. from C to A it takes 2 x If hr., or 3^ hr.
336. 40 X (number of 10 ct. pieces + to)== ^o. of iints ;
50 X (number of 10 ct. pieces — 1) - no. of nuts ;
.-. 40 X no. of 10 ct. pieces -j- 20 = 50 x no. of 10 ct.
pieces — 50 ;
.-. 10 X number of 10 ct. pieces = 70 ;
.'. number of 10 ct. pieces = 7 ;
.-. I have 7 X 10 ct. = 70 ct.
337. Income from 1st investment = $■ — „-— "
(«
2i)a
99
== $2300.
— ^ TTs
.*. the former is better by $50.
338. The note is due on 21st Nov.
Number of days between 18 Aug. and 21 Nov. ^95
Interest on $100 for 95 days = $f |f ;
.-. he gets $(100 - -f |f ) from a note for $100 ;
14315x100
$14315 from a note for $-
((
14 .t 13
I4n
= $14600.
339. 133 oxen to 20 a. = 26-| oxen to 4 a.
28 oxen to 5 a. =-- 22^ oxen to 4 a.
26f oxen eat the original grass and 13 days" growth
in 13 days ;
original grass + 18 days' growth .
.-. 1 ox eats — Y^ x~~^Gj ^^ ^^'*
22f oxen eat the original grass and 16 days' growth
in 16 days ;
original grass I- 16 days' growth . ^ ^
~ in 1 uay;
.*. 1 ox eats
16 X 221
SOLUTIONS HAMBLIN SMITH's ARITHMEXIO. 177
original grass + 13 days' growth
original grass + 16 days' growth
.-. 03 X original grass --=^ 4368 days' growth;
.*. original grass = 69i days' growth.
Quantity of grass to be eaten
=-(69^ + 14) days' growth.
Quantity eaten by 1 ox in 14 days
/69^ 4- 10\ ,
= ^^ il6 ^22l) ^^y'' S''°''*^ '
83^
.'. number of oxen required = 14(69;V + 16)
16x22f
83;V X 16 X 22|,
~" 14 x"85i
= 25.
The following is another solution : —
339. 133 oxen in 13 days eat grass on 20 a. + growth
on 20 a. for 13 days ;
.-. 1 ox in 1 day eats grass on Ysxisa ^' + gi'owth
13x20 p -, -I
TIirTsl a. for 1 day.
Again,
28 oxen in 16 days eat grass on 5 a. + growth on 5 a.
for 16 days ;
.•• ox in 1 day eats grass on yJxTs *• + growth on
Hence,
Il-xT¥5 «" + T%% a- growth = y^^^g a. + ^% a. growth ;
••• 13x133 a. - r?^8a.=-2\a.growth-T-'Aa.growth;
.-. 3 a. - 208 a. growth in 1 day.
hence, 1 a. growth in 1 day == -^^-g a.
178
SOLUTIONS HAMBLIM SMITH S AniTIIMRTIO.
Hence, 183 oxen in 13 days eat grass on 20 a. +-gra?;.5
on 13 X 20 X ^^if a., or 23^ a. ; and it is reqiiiicd to
find how rr?any oxen in 11 days can eat tlie giaa.s un
4 a. + 4 X 14 X 5 ^^ a., or 4|| a.
Oxen which eat 23| a. in 13 days :== 133
H I <« 1 << __ l^x I '']^
i<
((
4^i
14 " =
42JI.X18X 133
'ui^
25.
14 X 23:1
340. i of the constituency vote for A ;
^ of the constitiT'jncy vote for B ;
f ol :|^ of the constituency vote for D and A' ;
now i + i + A = II ;
.«. ^1^ of the constituency = = 540, and .*. number of
electors = C480.
^'8 votes are -*/-» or. 3240 ;
j?'s votes are j\ of 32iO + !, of 3240, or, 2916 ;
0*8 votes are j\ of 3240 + VV o^ ^'^'^^^ oi'. 1944 ;
D's votes are ^% of 3240 + ^V of 3240 + t of 1G20,
or, 2052 ;
^'s votes are ^V of 3240 + Vtt ot 3240 + i^ of 1G20,
or, 1728.
341.
342.
1 4
+
5 T) X 4
r Kl . -773
= 1914. « 7
2 I
1 I 9 2
(i X 9
9 U 6 X 1 i
6X5
4fl X 3x^31
""sTTa X 2
2 X 3 7 X H . ., ^
343. Sum got for $104 ii = .^100 ;
((
$2304.50 = $
.2 304.50 X loo
1 04i
= JII52200.
SOLUTIONS HAMBLlN SMITH's ARITHMETIC.
170
rfl. 4 .') X r> 4
= rooo.
7 .
34 I . Price of stock = ^^
840 Pait B fills in 1 min/-\v-
" Sfmiu. -8|x,,»,
.-. part yl fills = f « ;
.'. time required = (.}!? 4- j^.) min.
= 18 min.
346. In the last 4 years he saves t'2()0 + €120, or £820 ;
.-. his income- ^\y (his income + 4*40) ^^ ,4'8l) ;
.-. yV of Ilia income -£36 = £80 ;
.'. his income = 10 x (,£36 + i'80)
= £1160.
347. 15 men and 30 children get £(177 - 60) = £'117;
.-. 1 man and 2 children get £V/ = ^7 16^.
But 1 man and 1 child get £6 ;
.*. 1 child gets £1 16s.,
and 1 man gets 46 -.£1 16*. = 44 4s.,
and 1 woman gets £3.
348. 1 kilogramme = weight of ,-^Vtt cub. met. of water
(
5 \3
weight of ~j-0(,*y - cub. miles of water
weight of
U'\nTXl760x§)
1000
(^')'x6x20xll2
7 cub. fathoms of water
lb. avoirdupois
1000
~ f IM^ ^^ avoirdupois ;
.-. the ratio is 27951 : 12500.
349. 4285 - (2540 + 980) - 765, the number of
grains of soda and potash that take up 980 grains of
the sulphuric acid ;
49 X number of gr. of soda
82 ~
hence
49 V (765 _ nnmh^v r»f rrr r^f ar^A.-,\
48
= 980 ;
IMAGE EVALUATION
TEST TARGET (MT-S)
/
O
///// / s V^^i^
1.0
I.I
2.5
l^|28
■50 ™^
£ Hi
«.,. I
2.2
1.8
1-25 1.4 III 1.6
■^
— fj>
^ — — ^
Photographic
Sciences
Corporation
s.
4f
^
•>#
\\
^^. ^:\ w^
V
o
23 WEST MA^N STREET
WEBSTER, N.Y. 14580
(716) 872-4503
r^^s
/
^
180
SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
or 3 X number of gr. of soda + 2 x (766 - num-
ber of gr. of soda) = 20 x 96 ;
••. number of gr. of soda L 1920 - 1530 = 390-
.'. number of gr. of potash - 765 - 390 = 375.'
860. Area of sides = { (42 + 31) x 10 J sq.ft. '
= 730 sq. ft. ;
area of windows = (3 x 8^ x 4i) sq. ft.
= 112i sq. ft. ;
area of doors = (4 x 6| x 8^) sq. ft.
= 84| sq.ft. ;
area of fireplace =(6x4) sq. ft. -24 sq ft •
area of skirting =-- (64 x U) sq. ft. -=90 sq. ft. -
.-. area to be papered = (730 - 311) sq. ft.
/ . " = 419 sq. ft. ;
.-. cost ^ 419 X 5 cts. =: $20.95.
AiPPENDIX.
I- — Page 330.
1. Art. 1. 5 ^ '
2. Arts. 2 and 8.
8. Let X be the required sum ; then if ,• be the rate
of interest, we shall have
and P = a;E« ;
whence JL — ^ .
M P '
pa
.*. X = .
M
4. Here, we have
2P = P(l +r)«. and .-. 2 == (1 4-r)".
Also, 2P = P(l + 2r)'", and .v2 = (l + 2r)«;
whence, (l-|-2>-)"» = (l + r)«;
SOLUTIONS HAMBLIN SMITH'S ARITHME JIC. 181
.-. m log (1 + 2r) = n log (1 + r) ;
.-. - = ^og(lH-r)
n logfl-|-2r)
log(l + r)
>
log(l + 2rHhr2)
log(l+r)
log(l+r)3
> h
5. Let P denote the sum of money, and if n be the
required number of years^ we shall have
8P = P(1.05)" ;
whence (1.06)" = 3;
.-. n log (1.06) = log 8 ;
logs
n =
log (1.05)
__ .477 1212
= 22.6 years.
6. Let ic, y, z denote the three shares ; then we shall
have
a? + 2/ + « = P;
also, .tE" = y& = zB% are the equations of condi-
tion ;
whence y = R"" ^f, and z= R^-^a; ; so that
0) + B/'-^x 4- Il"-''a-= P;
whence xB^-^' + xBf'+' + xB"^^ - PE''+'' ;
and .'. X
similarly, y
yc+a
b+a
^c^a _^ ^o+a _^ ^
>b+c»
and z = ^^:pb
PR
a + h
-f R^-^'' -t- K*'^'*
182 SOLUTIONS HAMBI.IN SMITH's ARITHMETIC.
7. Lot r be the interest of $1 for 1 year ; then
amount of $4410 for 2 yr. S. I. ^ 4410(1 + 2r) • and
" $4400 << C.I. =.-4400(1+.).;'
whence 4410(1 + 2/-) = 4400(l+r)2,
or, 440r'2-2»- ^ 1 ;
.-. r = ^?jj or 6 per cent.
8. Let P represent the population ; then, population
at end of nth year
- I' 1 + f ; see Ex. 4 ;
therefore, by the question, we have
' P{l + ^-r=2P,
(i . II ~m
or, n log j 1 +
mn
n =
= log2;
log 2
log(»m + n - nif- log mn '
9. Let P„ represent his property at the ^nd of r - -
m the next year, the (n + l)th, his interest = P„r," '
and expenditure = (n + l)m.P r -
.'. the property left
= P„ + P„r-(«+l)w.P„r
= I*«-|l + r - (n + l)m,r\.
Now putting 2p for n+1, or 2^,-1 for n, we have
Pgp-i. i 1 + r - 2pmr f = ;
.*. 1+r = 2pmr.
Bnt his expenditure in the pth year = jomP,,_,r,
and property left at end of pth year = P^_, { l Irlprnr]
(since 1 + ,- = 2i>m,-) = expenditure in pth years.
nd,
fcion
SOLUTIONS HAMBLIN SMITHS ABITHMETIC. 188
10. From Ex. 3, we see that
M ^ Pe^' ;
by the question
or 6*° = e« ;
••• 20 log 6 ^7tloge;
loge '
11. Let P denote his capital; r the interest of $1 for
one year.
Then the sum he spends every year if? ?P/-.
At the end of the first year he has lefi P(l + r)_2Pr
orP-Pr. ^ ^
At the end of the second year (P _ p^) n + v) - •• 2Pr
or P - 2Pr - P, -^ JK ^ ) ^rr
At the end of the third year (P — 2P/- — P/-2) (l + ,.)
— 2Pr or P :^ 3Pr — 3P/-2 — p^3. ^ ^
By proceeding in this way we may show that the sum
he has left at the end of n years is
-J ii{n — 1)
if — ni^r j--^— P»-^ - ... - Pr" or 2P- P(l +r)».
Thus we have to find n from the equation
2P -- P (1 -|-r)» =. 0, or (1 +,•)» = 2.
Putting for r its value ^-^^ we get (-|§^)'* = 2.
Taking the logarithm of each side
n (log. 13 + log. 8 — log 100) ^ log. 2 ;
'• ^ = iiUj^t = 17.673, nearly
12. Births 62 in 1000
Deaths 27 ♦* 1000
Increase 85 " 1000 or 8i per cent,
184 SOLUTIONS HAMBLIN SMITh's ARITHMETIC.
Population at end of 6 years = 35748 (l + ~ j
= 85743 (l-036)«
= 42461.471 . . .
Hence, increase =.- 42451-471 35743
= 6708-471 . . .
13. The annual increase = _..
45
u
decrease =
60
.*. the net annual increase = -i_— = -^
TT , , 46 60 180"
Hence, by the question,
PJl+-,|^(n = 2P;
• ( } »\'\n o
• • VTffTyy — -^J
or, n|log 181 -log 180} = log 2 ;
• n .30 103
n= 125 nearly.
14. Let P denote the sum borrowed.
p
Then -— . = annual income in the first case ; and
—aP — = second " ;
whence, by the question,
P-600 ^ 2 P ^ P ,
25 8-20" ~ 30 '
.-. P = $3600.
15. If r denote the interest of $1 for one year,
amount of debt in n years •= aE".
Amounc of annual payments
= -^Ir-^+r-h.-.+i} === -^pJiil ;
»» ( J w (R-1 J
SOLUTIONS HAMBLIN SMITh's ARITHMETIC. 186
by the question, these two amouuts must be equal ;
hence, we have '
a(l-fr)'' = _^|(l+r)«- ll ;
mr( )
:. (l-mr)(l + r)" = l.
II.— Page 335.
1. Bank Discount at 5 % = ^'^P = $87.10 ;
.-. P = 1^742.
749
Present worth of $742 = % - -
= $706.66|.
2. Let S represent the sum of money ; then
T^^ S = $536.25 ;
.'. 8 = $18406.26.
If P represent the face of the note,
„ P_ _ P 300P
^ 1 +«r ~ 1 + i.T*^ = 304 = $18406.26 ;
.-. P = $13585.
3. If P represent the sum, then
4 per cent. = -gg-, and
Discount =2Q= $15 ;
.*. P = $390.
Interest on $390 at 6 % = ^\ of $390
= $19.50.
4.
p — J = D; Art. 9.
180P
hence, ^-:^iqq = 150 ;
P
P -1-180
^ »
186 SOLUTIONS HAMBMN SMITH 's ARITHMETIC.
•• 180
or P = $900.
5. Interest on $A for 1 yr. = Ar.
Br ,
Br
(C
Discount on $B
Hence, k.r =
1+r'
A + Ar = B;
_B-A.
A
B-A
.-. lOOr = 100.
A
6. 3 % for 12 mo. = H % for 5 mo. ;
.-. real value of stock = 90-li = 88|.
Income from 88f = 3 ;
tt
100
100X3
8 8i
^Tr-
ie
(«
7. Let X = the amount of the hill due from B to i4 ;
then,
Present Worth of ^a due in m years = aE"^, ^
$6 " n " = 6R-",
$a; " p " =a;R-P;
whence, hy the conditions of the question, we have
a;R-* + 6R-" = aR-'";
.-. xB.-^ = aR-^-iR-";
hence x = «R^-^- 6R^^.
8. Let P represent the sum ;
then $a assures $100 ;
PxlOO
((
a
S0!,UTI0N8 HAMBLIN SMITHH ARrTHSrETlC.
187
A - P is the reduced income ; wiience, by condition
f'f question
P X 100
. at r %,
A - P =.: Int. on
a
r P X 100
100 ®* ^ « ~
a'
Aa
a i-r
9. Let p = price of goods ;
9p
P =
worth
■ To ~ J?"^® 1^*^^ ^^ ^ months; and its present
9??
To"
36p
41 •
1-H
21)
100
A-gain, let x = the rate of discount allowed on pay-
ment at two months, so that p (l-x) is theii paid for
goods.
The present worth of p (1 -a) at 2 mos.= L^Mlz^);
121 '
120jo(l-jr) 36/>
heuce. ^^ = --
.*. X = .11463 . . .
and 100a; = 11.463 . . . rate per c&nt.
Or, we may more briefly reason thus : —
For every $100, B pays $90, if paid in 6 months; ^
and present worth of $90 := ^^^^ of $90 :=. $87.8040.
Amount of $87.8048 for 2 mos. at 5 % = $88.5365 ;
.-. $100-188.5365= $11,468.
188 SOLUTIONS HAMBLIN SMITH's AEITHMETIO.
PI
10. Since D = p^j, Art. 9, we have
1001
^•^^ =100^1'
• T — ft 1 o ft .
and if the interest in 1 year is 6J5», the time in
which 8|g| will be the interest, the rate being the same,
will be
^U - 6|^ = n years.
III.— Page 344.
1. Since
nA 2-Kn~l)r .... ^
^ ^* 1 -\-nr ' ®*"^^^® interest,
We have, in this case
/ 5 X 530 2+(6-l)x.07
P =
f (5-l)x.07\
2 1 + 6X.07 /
6x630 200+28'
200+ 28 \
135 /
- $(630 X V)
= $2237.77 ...
Again, since
^ A ( )
P == - j 1 — il~" [ , Compound Interest
630 ( (1.07) »-;^ I
~ .07 ( (1.07)'
= $2173.10.
A
2. 10000 =
•05
.-. 600 = A
== A
(1.06)-»o|j
1 (i-Q5)=^»-i
(LO'Sp
1^533'!
2.C533/ '
time in
te same,
SOLUTIONS HAMBLIN SMITB's ABITIU|£TI0.
1826.06
;. A ==
8. Since
4. Since
1.6588
= $802.42.
■P = -, Art. 19, in this caso
1000
"" * .06"
= $16666.661.
_A_jR9_.n
'I
Art. 20,
in this case, we have
p^ 400 I (1-06)' 0-1
(106)»M -oeT
= $2199.96...
6. Son's =^"{l~(1.06r^'>(
^ 1000 / (1-06)' »-l )
•o« ( (1-06)10 /
_ l-OOO/.TOOSfi \
— .0 6 \T.7VJifE)
= $7860.08.
Daughter's = J???. ( (l'06)«o-l l
(106)30) -06 /
1 O OP ( 3307 13-1 I
T A.7434887 1 .08 t
= $6404.74.
Institution's =
1000
(106)3 X -06
1000
a.74S4a8 x.o«
$9,901.88.
189
190
SOLUTIONS UAMBLIN SMITH 8 ABITUMETIO.
M»=A
luo
■ 06
R»-l
R:rr
{(106)»''-lf
= !^° (1-86484)
= $8090.56.
7. We may consider the £8769 as the Present Worth
of an Annuity that has 80 years to run, and, therefore,
^3769 = ^|l-(1.08)-3o|;
.-. £160.76 = A |l - (1.04)-3ol ;
A =
£160.76
l-(1.04)-3»
8. The lease is renewed for a years ; $d may, l^re-
fore, be considered as the Present Worth of an Annuity
that has a years to run. Hence,
dR* = amount of annuity = M ;
R"— 1
but
M
-'^- R-
- 1 '
•
••
dB«
= A ^'-
1
1
= ^B..
-1);
•
A
drB"
-•
R*— 1
9. The fine P may be regarded as the Present Worth
of an Annoity, A or exira-rent ;
••• P = ; (1 - B-Jk
r
• A =
Pr
1— B"»*
nt Worth
lerefore,
SOLUnOKS RABTBLTN SMITH's ARTTHHKTIC. 101
The new fine,/, must provide for this extra rent dur-
ing the q — p years, which are to be added to the ori-
ginal term = tine for q years — fine for p years ;
.-. /= A|(l_R-,)_(l_.R-P^j
1 Pr ,
', i^re-
Anuuity
t Worth
10. Each owns
a
2*
to ooutinuo for
Present Worth of a freehold producing $^ per
aunuou
~2r
a
= 2r •
Present Worth of an Annuity "of I
n years
••i=;(i-B-).
or, J = 2(1 - B-")
= 2 (l-g.).
11. Since
A
p = :^(i-K-),
192
We have
iOLUtlOI^a HAr-BLIN smith's a ilTHMBTIO.
lis case.
n
n
p=:i-^i- h+:
n
\
1 -
r^])
Again,
p n
r
n
-(-t)
—mn
= i_ ( l_(l _ mr,':_^^J!^{!!!^^) / M ''_
r i ^ n 1-2 \n I
mn{mn+l){nm~i-*2) / r \ 3 . » \ )
r^B (^) +*°")}
r I n 1-2 ' /
/
w \ r
r ( ^ ' n I 1-2
No«7, as n increases, it is plain that ~ decreases,
n
and tends to zero as its limit , hence, as « increases, the
Limit of the abore series is
r \ 1-2 1-2-8 7
SOLUTIONS HAMBLIN SMITH'S ARITHMETIC.
193
. r 1 ^ 1-2 1-2-3 ^ 7 )
1 - e-""-
r
Hence, limit of P =
12. The Present Torth of an Annuity of $10 per
month for 6 months at i % per month
= Mm I 1 - (i + 10-h) "{•
If P be the sum to be paid at once, P in 19 years
must amount to the preceding present worth ; hence,
we have
5 \ i« 10 ( I 5 \-«)
19
L' |1 -f
Now ( 1
I
H- (fee.
o 19.18 5J
"^ i ■ 103 + i;2 "lOfi^"
10-\' .005
, lOT 3 /
19.18. n 5^
17273" ' io'»
= 1 + . 095 + .004275 + .000121126 + &c.
= 1.099396 . . .
10 r. / , 5 \-«i
10^
And
.005
{l- (l-^l^)
= 2000
1
1 -
6
G.7 52 6.7.8 53
1 *ld3'^1.2'l0« 172.3 • 103 "^
fi.7.8.9 5*
^2-&C
\. 2.3.4 10'
^ 2000 (.03 - .000525 + .000007 - .00000007876 + &c.)
= 58 9G38 ....
Hence P (1.099396) =.- 58.963844 ;
.-. P :^ $53.68.
194
SOLUTIONS HAMBLIN 35IITH S ARITHMETIC.
13. By reference to Ex. 3, we see that
100 { (l-05)-'*«-l }
P(105)3«*=:4000 +
•05
(1.05)36 =. 5-7918149
(l-05)3s* = 5-6064572.
A • ^nnn . 100{ (l-05)3«-l }
Again, 4000 + LA L 1
•05
= 40004-2000x4-7918149
:= 13583-6298.
Hence, P(5-6064572) = 13583-6298 ;
, .-. P = $2422.85.
14. (1) The equation established in Art. 11, which is
the same as that given in the exercise, proves the first
statement.
(2) Multiplying each side of the equation
SjR *i + s^R '■
by R"^, where T is some time subsequent to ^3, we have
T t T t T t
which shows that the amounts are the same at the
subsequent time T.
(3) Again, multiplying each side of
*-,R~'> +.v,R~^'-' =(*•, 4- S2)R~*
by R', and we have
5, + ^2 '
5.R'-''
-S, == §2
S^^
■{t,-t)
ich is
e first
have
it the
SOLUTIONS HAMBLIN SMITIt's ARITHMETIC.
195
Now, the comimmd interest of any sum is found by
subtracting the sum from its amount for the given
time ; therefore, .s,R^~*i - s^ is the C. I. of s^ for the
time it is overdue.
The discount of any sum is found by subtracting the
Present Worth, for tlie given time, from the sum itself;
hence.
.S.R
~(t.-t)
is the discount on Sg for the
time before it is due ; and the equation shov/s that "at
the intermediate time, t, of payment, the interest of tlie
sum overdue is the discount of that not due."
t)
The New Inthorized Elementary Grammar.
MILLER'S SWIKTON'S lAHCUACE LESSONS.
MiLf-ER's Swinton's Language Lessons is used exclu-
sively in nearly all the Principal Public and Model
Schools of Ontario. Among them are
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Slraihroy, Mfaford, Uzbrldge. Brantrord,
Wliidtor, Clinton, St. Thomas, Perth,
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Adopted by the Protestant Schools of llontreal and Levi
College, 4|iiebee, Schools of Winiilpeg, Manitoba,
and St. John's, .%ew FoundIai>d.
Resolution passed unanimously by the Teachers' As
sociation, (North Huron), heldat Brussels, May 17, 1878
" Resolved, That the Teachers at this Convention are of
opmion that « Miller's Swinton Language Lessons,'
by McMillan, is the best introductory work on Grammar
for Public School use, since the definitions, classification
and general treatment are extremely simple and satis
factory."
In my opinion the best introductory Text-book to
Mason's Grammar. All pupils who intend to enter a
High School or to become students for Teachers' Certifi-
Jatea, would save time by using it.
W. J. CARSON, H. M.,
Model School, London.
The definition's in « Miller's Swinton Language Les-
sons" are brief, clear and exact, and leave little to be
unlearned in after years. The arrangement of the sub-
jects is logical and progressive, and the book admirably
helps the judicious teacher in making correct thinkers
and ready readers and writers.
B. W. WOOD,
l3t A Provincial H., P.S., Trenton Falls.
■• carefnl to ask for MILLER'S SWINTON, as other edilions
are in the asarket.
Mental Arithmetic.
By J. A. McLELLAN, M.A., LL.D , Intpectoi
of High Schools, Out.
PART 1. -FUNDAMENTAL RULES, FRACTIONS,
ANALYSIS. PRICE, 80o.
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Provincial Model School, Nova Scotia.
Dr. McLellan's Mental Arithmetic supplies
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ba made in Mathematical work, unless what
we call Mental Arithmetic is thoroughly and
systematically pursued. A boy who is con-
versant with the principles of Mental Arith-
metic, as given m this little text-book, is worth
as a clerk or accoimtant 50 per cent more
than the prodigy who can boast of havixig
" gone " through his written arithmetic half a
dozen times.
J. S. DEACON, Principal Ingersoll Model School.
Dr. McLellan's Mental arithmetic, Part I,, is a-
credit tc Canadians, and it supplies a long-felt
"vrant. It is just what is wanted for "waking up
mind" in the school room. After two weeks use of
the book with my class I am convinced that it is
much superior to any of the American texts that
have been used here both as to the grading of ques-
tions and the style of the problems.
J. A. CLARKE, M. A., H. M. H. S., Picton.
Dr. McLellan's Mental Arithmetic contains a
freat number of useful problems well adapted to
evelop by regular gradations the thinking powers
of the pupil, and to suggest similar examples for the
use of the ieaoher.
D, J. GOCKJIN, Head Master Model and Pablio
Schools, Port Hope,
simple in its arrangement, varied in its types of
{tractical questions and 8uggj;e8tive in its methods,
t is the best book of its l:i*>.l that I have examined.
Prom THE WE8LEYAW, Halifax, Nova Scotda.
The series bids fair to take a good place in saho.
' iMtio work.
NOW BEADY-4th EDITION.
a
o
9
o
•c
B
<
O
a
0)
l-H
.o
o
o
a
o
o t^
03 S
o
8
o
§1
EXAMINATION PAPEE8
m
AEITHMETIC,
%f;^l^^*i"'''^1^^;^•?•.' Inspector High Schools, and
TH08. KiBKLAND. M.A., Science Master, Normal SohdoT
jLoronto.
PBIOB »l.JO.
From the GUELPH MERCURY
Commission wSe^ &c StocS,^ „'nH ^S?**°^,?' ^'»«a«on ;
blems. The Becond^U nn ifiot! * ' "'"i Miscellaneous Pro-
BupplyingVestions ftve tK^JlaBBe^ ThosIwhTr "^^^^ 1°
be teachers cannot have a better Se^in^««l+T^° *^P*''? ^
good a one-on the subject wfihwffihVSt'cupier '' "°* "°
Prom the ADVERTISER.
From the TELESCOPE
One of the mogt popular Text Books ever published.
NEW ELEMENTARY ARITHMETIC
ON THE UNITARY METHOD.
By Thomas Kirkland, M.A., Science Master Normal School,
and William Scott, B.A., Head Master Model School,'
Toronto.
Intended as an Introductory Text-Book to Hamblm Smith's
Arithmetic.
Oloth Bxtra, 176 Pages. Frloa 26 Cents.
Highly recommended by the leading Teachers
of Ontario.
Adopted in many of the best Schools of Quebec.
Adopted in a number of the Schools of New-
foundland.
Authorized by the Council of Public Instruc-
tion. Prince Edward Island.
Authorized by the Conncil of Pnblio Instruction,
Manitoba.
Within, one year the 40th thousand ha* been issued.
ADAM MILLEB & Co., '
TOBONTO,
c5
c
« •• Xpoohi in HUtory mark an j^^^mA In the Study of it "
^ G. W. JoHirioir, H.M.M.S., Hamilton.
s • —
An Acceptable Text-Book on English History
I EPOCHS OF ENGLISH HISTORY,
o
1 REV. M. CREIGHTON, M.A.
a .
Authorized by Abe Kdocatlom Oepaitment,
s* —
1 AdopieJ by the Public Schools of Montreal, and a number of
S the best Schools in Ontario.
OQ
08
A
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»
I
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Cajr™i:^^l'^ ^""'^'"' Comprehensiveaess."-
" Amongst manuals in English History the Epoch
Series is sure to take high rank.''-Daily Globe.
" Nothing was more needed than your excellent
JVimersofEnghshHistory/'-FKED.W.KELlY.Ml'a^^^
Lect. in Enghsh History. High School, Mont eal.
In Eight Volumes, 20 cents each,
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ittillcr ^ (Ep.'s €biu*iitiomil ^tvxts.
HAMBLIN SMITHS
MATHEMATICAL WORKS,
ARB UoCD ALMOar I.XCLIHIVKLV
In the Normal and Model Schools, Torontc ;
Upper Canada College ; Hamilton ani
Brantford Collegiate Institutes ; Bov-
manville, Berlin, Belleville, and a large
number of leadmg High Schools in the
Province.
HAMBLIN SMITH'S ALGEBRA,
With Appendix, by Alfred Baker. B.A., Mathematical Tutor, Univer
■ity College, Toronto. Price, 90 cents.
THOMAS KIRKLAND, M.A., Science Master, Normal School.
"It is the text-book on Algebra for candidates for .sccond-clasti
ceitiflcates, and for the Intermediate Examination. Not the least
valuable part of it is the Appendix by Mr. Baker."
GEO, DICKSON, R.A., Head Master, Collegiate Institute, Hamilton,
" Arrangement of subject" good ; explanations and proofs exhaus*
tlTe, concise and clear ; examples, for the most part from Univerei^
and College Examination Papers, are numerous, easy and progressive.
There is no better Algebra in use in our High Schools and Collegiate
Institutes."
WM, R. RIDDELL, B.A., B.Sc, Mathematical Ma.ster, Normal
School, Ottawa.
Oook"
The Algebra la admirable, and well adapted as a general text.
W, E, TILLET, B.A., Mathematical Master, Bownianville High School.
" I look on the Algebra as decidedly the best Elementary Work on
the subject we have. The examples are excellent and well arranged.
The explanations are easily undcrtitood.
R. DAWSON, B.A.,T.C,D., Head Ma.ster, High SchofW, Bellevilla
"With Mr Baker's admirable Appendix, there would .«eem to he
nothing left to be desired. We have now a first class book, w«l|
adapted in all respects to the wants of pupils of all grades, from the
beginner in our Public Schools to the mdst advanced student in o«ir
Collegiate Institutes and Hijih Schools. Its publication is a great h«Hin
to the over-worked mathematical teachers of the Province.
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-*•*-
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KDITOBIAL COBIMITTEIL
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T?u'J^i^?. ^i^)^^^^^' ^•^' «°'ence Master, Normal School
JAMK8 HUGHES, Public School Inspector, Toronto.
WM HOUSTON^m\^' ^°'^^' '^^'' University College. Toront*.
PROVINCIAL EDITORS.
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Each Number will contain A PORTRAIT ofsomelead-
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NOTES AND NEWS from the diflferent Provinces of
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1