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A treatise on chemistry and chemical analysis : prepared for students of The International Correspondence Schools, Scranton, Pa (Volume 6) online

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A TREATISE



ON



CHEMISTRY AND CHEMICAL
ANALYSIS



PREPARED FOR STUDENTS OF

THE INTERNATIONAL CORRESPONDENCE SCHOOLS

SCRANTON, PA.



Volume VI



ANSWERS TO QUESTIONS



First Edition



SCRANTON

THE COLLIERY ENGINEER CO.
1900



101



L

U



Copyright, 1897, 1898, 1899, by THE COLLIERY EXGIXKKR COMPANY.



. . .Arithmetic, Key : Copyright, 1893, 1894, 1896, 1897, 1898, by THE COLLIERY ENGINEER
; ' ~- \ / COMPANY.

Elementary Alge'bra and Trigonometric Functions, Key : Copyright, 1894, ls%.
J I 'I 't **; ^189^ l98,"by THE COLLIERY ENGINEER COMPANY.
' *nm-gamlcr* diemris&'y,' '.Key : Copyright, 1898, by THE COLLIERY ENGINEER

COMPANY.

Qualitative Analysis, Key: Copyright. 1898, 1899, by THE COLLIERY ENGINEER

COMPANY.



Press of EATON & MAINS

NEW YORK



A KEY

TO ALL THE

QUESTIONS AND EXAMPLES

CONTAINED IN THE

QUESTION PAPERS INCLUDED IN VOLS. I-V.



It will be noticed that the Keys have been given the same
section numbers as the Question Papers to which they refer.
All article references refer to the Instruction Paper bearing
the vSame section number as the Key in which it occurs,
unless the title of some other Instruction Paper is given in
connection with the article number.



991 179



CONTENTS.



ANSWERS TO QUESTIONS.

Section.

Arithmetic, Part 1 1

Arithmetic, Part 2 2

Elementary Algebra and Trigonometric Functions . . 3

Physics 4

Theoretical Chemistry 5

Inorganic Chemistry, Part 1 6

Inorganic Chemistry, Part 2 7

Inorganic Chemistry, Part 3 8

Inorganic Chemistry, Part 4 9

Qualitative Analysis, Part 1 10

Qualitative Analysis, Part 2 . .. 11

Organic Chemistry, Part 1 12

Organic Chemistry, Part 2 13

Organic Chemistry, Part 3 14

Organic Chemistry, Part 4 15

Quantitative Analysis, Part 1 16

Quantitative Analysis, Part 2 17

Quantitative Analysis, Part 3 . . . 18

Quantitative Analysis, Part 4. , . 19



ARITHMETIC.

(QUESTIONS 1-75. SEC. 1.)



(1) See Art. 1.

(2) See Art. 3.

(3) See Arts. 5 and 6.

(4) See Arts. 1C and 11.

(5) 980 = Nine hundred eighty.
605 = Six hundred five.

28, 284 = Twenty-eight thousand two hundred eighty-four.
9,006,042 = Nine million six thousand forty- two.
850,317,002 = Eight hundred fifty million three hundred
seventeen thousand two.

700,004 = Seven hundred thousand four.

(6) Seven thousand six hundred = 7,600.
Eighty-one thousand four hundred two = 81,402.
Five million four thousand seven = 5,004,007.

One hundred eight million ten thousand one = 108,-
010,001.

Eighteen million six = 18,000,006.
Thirty thousand ten = 30,010.

(7) In adding whole numbers, place 3290
the numbers to be added directly under 504
each other so that the extreme right- 865403
hand figures will stand in the same col- 2074
umn, regardless of the position of those 8 1
at the left. Add the first column of fig- 7

ures at the extreme right, which equals 871359 Ans.
19 units, or 1 ten and 9 units. We place



2 ARITHMETIC. 1

9 units under the units column, and reserve 1 ten for the
column of tens. 1+8+7+9 = 25 tens, or 2 hundreds and
5 tens. Place 5 tens under the tens column, and reserve 2
hundreds for the hundreds column. 2+4 + 5 + 2 = 13
hundreds, or 1 thousand and 3 hundreds. Place 3
hundreds under the hundreds column, and reserve the 1
thousand for the thousands column. 1+2 + 5 + 3 = 11
thousands, or 1 ten thousand and 1 thousand. Place the
1 thousand in the column of thousands, and reserve the
1 ten thousand for the column of ten thousands. 1+6 = 7
ten thousands. Place this 7 ten thousands in the ten
thousands column. There is but one figure, 8, in the hun-
dreds of thousands place in the numbers to be added,
so it is placed in the hundreds of thousands column of the
sum'.

A simple (though less scientific) explanation of the same
proMem is the following: 7 + 1+4 + 3 + 4 + = 19; write
the 9 and reserve ' the 1. 1+8 + 7 + + + 9 = 25; write
the 5 and reserve the 2. 2 + + 4 + 5 + 2 = 13; write the

3 and reserve the 1. 1+2 + 5 + 3 = 11; write the 1 and
reserve 1. 1 + 6 = 7; write the 7. Bring down the 8 to
its place in the sum.

(8) 709
8304725

391
100302

300

909

8407336 Ans.

(9) (a) In subtracting whole numbers, place the subtra-
hend, or smaller number, under the minuend, or larger
number, so that the right-hand figures stand directly under
each other. Begin at the right to subtract. We cannot
subtract 8 units from 2 units, so we take 1 ten from the 6
tens and add it to the 2 units. 1 ten = 10 units, so we
have 10 units + 2 units = 12 units. Then 8 units from 12
units leaves 4 units. We took 1 ten from 6 tens, so only 5



1 ARITHMETIC. 3

tens remain. 3 tens from 5 tens leaves 2
tens. In the hundreds column we have 3 5 9 G 2
hundreds from 9 hundreds leaves 6 hun- 3 3 3 8
dreds. We cannot subtract 3 thousands 47624 Ans.
from thousands, so we take 1 ten thou-
sand from 5 ten thousands and add it to the thousands.
1 ten thousand = 10 thousands, and 10 thousands -f- thou-
sands = 10 thousands. Subtracting, we have 3 thousands
from 10 thousands leaves 7 thousands. We took 1 ten thou-
sand from 5 ten thousands and have 4 ten thousands
remaining. Since there are no ten thousands in the subtra-
hend, the 4 in the ten thousands column in the minuend is
brought down into the same column in the remainder,
because from 4 leaves 4.

(b) 15339
10001
5338 Ans.

(10) (a) 70968 (b) 100000

32975 98735

37993 Ans. 1265 Ans.

(11) We have given the minuend or greater number
(1,004) and the difference or remainder (49). Placing these

1004

in the usual form of subtraction, we have in which

49

the dash ( ) represents the number sought. This num-
ber is evidently less than 1,004 by the difference 49, hence,
1,004 49 = 955, the smaller number. For the sum of the

1004 larger
two numbers we then have 955 smaller

1959 sum. Ans.

Or, this problem may be solved as follows : If the greater
of two numbers is 1,004, and the difference between them
is 49, then it is evident that the smaller number must be
equal to the difference between the greater number (1,004)
and the difference (49); or, 1,004 49 = 955, the smaller



4 ARITHMETIC. 1

number. Since the greater number equals 1,004 and the
smaller number equals 955, their sum equals 1,004 + 955
= 1,959. Ans.

(12) The numbers connected by the plus (-f-) sign must
first be added. Performing these operations we have

5902 3874

8471 2039

9023 5913 sum.
23450 sum.

Subtracting the smaller number (5,913) from the greater
(23,456), we have

23456
5913
17543 difference. Ans.

(13) $ 4 4 6 7 5 = amount willed to his son.

26380 = amount willed to his daughter.
$ 7 1 5 5 = amount willed to his two children.
$125000 = amount willed to his wife and two

children.

71055 = amount willed to his two children.
$ 5 3 9 4 5 = amount willed to his wife. Ans.

(14) In the multiplication of whole numbers, place the
multiplier under the multiplicand, and multiply each term
of the multiplicand by each term of the multiplier, writing
the right-hand figure of each product obtained under the
term of the multiplier which produces it.

7 times 7 units 49 uni ts or



8 7

4 tens and 9 units. We write

the 9 units and reserve the 4



34709 Ans. = M tens;



tens reserved = 60 tens, or 6 hundreds and tens.
Write the tens and reserve the 6 hundreds. 7x3 hundreds
= 21 hundreds; 21 -f- 6 hundreds reserved = 27 hundreds,
or 2 thousands and 7 hundreds. Write the 7 hundreds and
reserve the 2 thousands. 7x6 thousands = 42 thousands;



ARITHMETIC. 5

thousands reserved = 44 thousands, or 4 ten thou-
sands and 4 thousands. Write the 4 thousands and reserve
the 4 ten thousands. 7x2 ten thousands = 14 ten thou-
sands; 14 + 4 ten thousands reserved = 18 ten thousands,
or 1 hundred thousand and 8 ten thousands. Write the 8
ten thousands and reserve the 1 hundred thousand. 7X5
hundred thousands = 35 hundred thousands; 35 + 1 hun-
dred thousand reserved = 36 hundred thousands. Since
there are no more figures in the multiplicand to .be multi-
plied, we write the 36 hundred thousands in the product.
This completes the multiplication.

A simpler (though less scientific) explanation of the same
problem is the following:

7 times 7 = 49 ; write the 9 and reserve the 4. 7 times
8 = 56 ; 56 + 4 reserved = 60; write the and reserve the 6.
7 times 3 = 21; 21 + 6 reserved = 27; write the 7 and
reserve the 2. 7 X 6 = 42 ; 42 + 2 reserved = 44 ; write the
4 and reserve 4. 7x2 14; 14 + 4 reserved = 18; write
the 8 and reserve the 1. 7 X 5 = 35 ; 35 + 1 reserved = 36 ;
write the 36.

In this case the multiplier is ,^ 700298
17 units, or 1 ten and 7 units, so .

that the product is obtained by

adding two partial products, .

namely, 7x700,298 and 10

X 700, 298. The actual opera- 11905066 Ans.

tion is performed as follows:

7 times 8 = 56 ; write the 6 and reserve the 5. 7 times 9
= 63; 63 + 5 reserved = 68; write the 8 and reserve the 6.
7 times 2 = 14; 14 + 6 reserved = 20; write the and
reserve the 2. 7 times = ; + 2 reserved = 2 ; write
the 2. 7 times = 0; + reserved = 0; write the 0.
7 times 7 = 49; 49 + reserved = 49; write the 49.

To multiply by the 1 ten we say 1 times 700,298 = 700,-
298, and write 700,298 under the first partial product, as
shown, with the right-hand figure 8 under the multiplier 1.
Add the two partial products; their sum equals the entire
product.



ARITHMETIC. 1

(c) 217 Multiply any two of the numbers

103 together and multiply their product by
651 the third number.
2170
22351
67-



156457
134106

1407517 Ans.

(15) If your watch ticks every second, then to find how
many times it ticks in 1 week, it is necessary to find the
mimber of seconds in one week.

6 seconds = 1 minute.
6 minutes = 1 hour.

3600 seconds = 1 hour.

2 4 hours = 1 day.
14400
7200

86400 seconds = 1 day.
7 days = 1 week.



604800 seconds in 1 week, or the mimber of times that
your watch ticks in 1 week. Ans.

(16) If a monthly publication contains 24 pages, a yearly

2 4 volume will contain 12 X 24, or 288 pages,

1 2 since there are 12 months in one year; and

288 eight yearly volumes will contain 8 X 288,

8 or 2,304 pages.

2304 Ans.

(17) If an engine and boiler are worth $3,246, and the
building is worth 3 times as much, plus $1,200, then the
building is worth

$ 3 24 6

X 3

9738
+ 1200
$10938 = value of building.



1 ARITHMETIC. 7

If the tools are worth twice as much as the building, plus
$1,875, then the tools are worth

$ 1 9 3 8

X _ 2

21876

+ 1875

$ 2 3 7 5 1 = value of tools.
Value of building = $10938
Value of tools = 23751

$34089 = value of the building

and tools, (a) Ans.
Value of engine and

boiler = $ 3246

Value of building and

tools = 34689

$37935 = value of the whole
plant, (b) Ans.

(18) (a) (72 X48 X 28 X 5) -v- (96X15X7X6).

Placing the numerator over the denominator the problem

becomes

72X48X28X5 _ ?
96X15X7X6

The 5 in the dividend and 15 in the divisor are both divis-
ible by 5, since 5 divided by 5 equals 1, and 15 divided by 5
equals 3. Cross off the 5 and write the 1 over it; also, cross
off the 15 and write the 3 under it. Thus,

1



3

The 5 and 15 are not to be considered any longer, and, in
fact, may be erased entirely and the 1 and 3 placed in their
stead, and treated as if the 5 and the 15 never existed. Thus,

72X48X28X1
96X3X7X6



8 ARITHMETIC. 1

72 in the dividend and 96 in the divisor are divisible by
12, since 72 divided by 12 equals 0, and 96 divided by 12
equals 8. Cross off the 72 and write the 6 over it ; also, cross
off the 96 and write the 8 under it. Thus,

6



$5x3x7x6

8

The 72 and 96 are not to be considered any longer, and, in
fact, may be erased entirely and the 6 and 8 placed in their
stead, and treated as if the 72 and 96 never existed. Thus,

6 X 48 X 28 X 1 m
8X3X7X6

Again, 28 in the dividend and 7 in the divisor are divisible
by 7, since 28 divided by 7 quals 4, and 7 divided by 7 equals
1. Cross off the 28 and write the 4 over it; also, cross off the
7 and write the 1 tinder it. Thus,

4

6X48X^X1 m
8X3X7X6
1

The 28 and 7 are not to be considered any longer, and, in
fact, may be erased entirely and the 4 and 1 placed in their
stead, and treated as if the 28 and 7 never existed. Thus,

6X48X4X1 .
8X3X1X6

Again, 48 in the dividend and 6 in the divisor are divisible
by 6, since 48 divided by 6 equals 8, and 6 divided by 6
equals 1. Cross off the 48 and write the 8 over it; also, cross
off the 6 and write the 1 under it. Thus,

8

6X^X4X1 _
8x3x1X0



1 ARITHMETIC. 9

The 48 and 6 are not to be considered any longer,
and, in fact, may be erased entirely and the 8 and 1 placed
in their stead, and treated as if the 48 and 6 never existed.
Thus,

6X8X4X1 __

8X3X1X1 =

Again, 6 in the dividend and 3 in the divisor are divisible
by 3, since 6 divided by 3 equals 2, and 3 divided by 3
equals 1. Cross off the 6 and write the 2 over it; also, cross
off the 3 and write the 1 under it. Thus,

2
0x8x4x1 _

8x^x1x1 =
1

The 6 and 3 are not to be considered any longer, and,
in fact, may be erased entirely and the 2 and 1 placed
in their stead, and treated as if the 6 and 3 never existed.
Thus,

2X8X4X1 m

8X1X1X1 ~

Canceling the 8 in the dividend and the 8 in the divisor,
the result is

1
2x$X4xl __ 2x1x4x1

l xl =: ixlxlxl'



Since there are no two remaining numbers (one in the
dividend and one in the divisor) divisible by any number
except 1, without a remainder, it is impossible to cancel
further.

Multiply all the uncanceled numbers in the dividend
together, and divide their product by the product of
all the uncanceled numbers in the divisor. The result
will be the quotient. The product of all the uncanceled
numbers in the dividend equals 2xlX4xl = 8; the



10 ARITHMETIC. i

product of all the uncanceled numbers in the divisor
equals iXlXlXl = 1.

2X1X4X1 8

Hence ' ix
2



. i ? ' '

(b) (80 X 60 X 50 X 16 X 14) ^ (70 X 50 X 24 X 20).

Placing the numerator over the denominator, the problem

becomes

80X60X50X16X14



'OX 50X24X20



= ?



The 50 in the dividend and 70 in the divisor are both divis-
ible by 10, since 50 divided by 10 equals 5, and 70 divided by
10 equals 7. Cross off the 50 and write the 5 over it; also,
cross off the 70 and write the 7 under it. Thus,

5

80x60x^x16x14 __
70X50x24x20

7

The 50 and 70 are not to be considered any longer, and, in
fact, may be erased entirely and the 5 and 7 placed in their
stead, and treated as if the 50 and 70 never existed. Thus,

80X60X5X16X14 _
7X50X24X20

Also, 80 in the dividend and 20 in the divisor are divisible
by 20, since 80 divided by 20 equals 4, and 20 divided by 20
equals 1. Cross off the 80 and write the 4 over it ; also, cross
off the 20 and write the 1 under it. Thus,

4

$0x60x5x16x14



1



1 ARITHMETIC. 11

The 80 and 20 are not to be considered any longer, and,
in fact, may be erased entirely and the 4 and 1 placed in
their stead, and treated as if the 80 and 20 never existed
Thus,

4X60X5X16X14 _

7X50X24X1

Again, 16 in the dividend and 24 in the divisor are divisible
by 8, since 16 divided by 8 equals 2, and 24 divided by 8
equals 3. Cross off the 16 and write the 2 over it; also, cross
off the 24 and write the 3 under it. Thus,

2



7x50xx1
3

The 16 and 24 are not to be considered any longer, and, in
fact, may be erased entirely and the 2 and 3 placed in their
stead, and treated as if the 16 and 24 never existed. Thus,

4X60X5X2X14
7X50X3X1

Again, 60 in the dividend and 50 in the divisor are divis-
ible by 10, since 60 divided by 10 equals 6, and 50 divided
by 10 equals 5. Cross off the 60 and write the 6 over it;
also, cross off the 50 and write the 5 under it. Thus,

6
4x00x5x2x14 _

7x00x3x1

5

The 60 and 50 are not to be considered any longer, and, in
fact, may be erased entirely and the 6 and 5 placed in their
stead, and treated as if the 60 and 50 never existed. Thus,

4X6X5X2X14 _
7X5X3X1

The 14 in the div^lend and 7 in the divisor are divisible by
7, since 14 divided by 7 equals 2, and 7 divided by 7 equals 1,
8-2



12 ARITHMETIC. 1

Cross off the 14 and write the 2 over it ; also, cross off the 7
and write the 1 under it. Thus,

2



/TX5X3X1
1

The 14 and 7 are not to be considered any longer, and, in
fact, may be erased entirely and the 2 and 1 placed in their
stead, and treated as if the 14 and 7 never existed. Thus,

4X6X5X2X2 _
1X5X3X1

The 5 in the dividend and the 5 in the divisor are divisible
by 5, since 5 divided by 5 equals 1. Cross off the 5 of the
dividend and write the 1 over it ; also, cross off the 5 of the
divisor and write the 1 under it. Thus,

1

4x6X^x2x2 _

1x0x3x1

1

The 5 in the dividend and 5 in the divisor are not to be
considered any longer, and, in fact, may be erased entirely
and 1 and 1 placed in their stead, and treated as if the 5 and
5 never existed. Thus,

4X6X1X2X2 _
1X1X3X1

The 6 in the dividend and 3 in the divisor are divisible by
3, since 6 divided by 3 equals 2, and 3 divided by 3 equals 1.
Cross off the 6 and place the 2 over it; also, cross off the 3
and place the 1 under it. Thus,

2
4x0x1x2x2

Ixlx^xl
1

The 6 and 3 are not to be considered any longer, and,
in fact, may be erased entirely and 2 and 1 placed in



1 ARITHMETIC. 13

their stead, and treated as if the 6 and 3 never existed.
Thus,

4X2X1X2X2 32

ixixixi T' - 32 - Ans "

Hence,

2 1
4 2 2

4x2x1x2x2 __ 32 _

ixixixi l



(19) 28 acres of land at $133 an acre would cost

28 X$ 1 3 3 = $3,724

2 8

1064
266
$3724

If a mechanic earns $1,500 a year and his expenses are $968
per year, then he would save $ 1 5 $968, or $532 per year.

968

$532

If he saves $532 in 1 year, to save $3,724 it would take as
many years as $532 is contained times in $3,724, or 7 years.

532)3724(7 years. Ans.
3724

(20) If the freight train ran 365 miles in one week, and
3 times as far lacking 246 miles the next week, then it ran
(3X365 miles) 246 miles, or 849 miles the second week.
Thus,

365

3

1095

246

difference 849 -miles. Ans.



14 ARITHMETIC. 1

(21) The distance from Philadelphia to Pittsburg is 354
miles. Since there are 5,280 feet in 1 mile, in 354 miles
there are 354x5,280 feet, or 1,869,120 feet. If the driving
wheel of the locomotive is 16 feet in circumference, then in
going from Philadelphia to Pittsburg, a distance of 1,869,120
feet, it will make 1, 869, 120 -J- 16, or 116,820 revolutions.

16)1869120(116820 rev. Ans.
1




(22) (a) 576)589824(1024 Ans.
576

1382
1152



2304
2304

(b) 43911)369730620(8420 Ans.
351288
184426
175644



87822
87822

(c) 505)2527525(5005 Ans.
2525

2525
2525



1 ARITHMETIC. 15

(d) 1234)4961794302(4020903 Ans.
4936

2579
2468



11143
11106



3702
3702

(23) The harness evidently cost the difference between
1444 and the amount which he paid for the horse and wagon.
Since $264 + $153 $417, the amount paid for the horse and
wagon, $444 $417 = $27, the cost of the harness.

$264 $444

153 417

$ 4 1 7 $ 2 7 Ans.

(24) (a) (b) (c)
1024 5005 43911

576 505 8420



6144 25025 878220

7168 250250 175644

5120 2527525 Ans. 351288

589824 Ans. 369730620 Ans.

(25) Since there are 12 months in a year, the number of
days the man works is 25 X 12 300 days. As he works 10
hours each day, the number of hours that he works in one
year is 300x10 = 3,000 hours. Hence, he receives for his
work 3, 000X30 = 9 0, 000 cents, or 90, 000 -r- 100 = $900. Ans.

(26) See Art. 71.

(27) See Art. 77.

(28) See Art. 73.

(29) See Art. 73.

(30) See Art. 75.

(31) - 1 / is an improper fraction, since its numerator, 13, is
greater than its denominator, 8.

(32) 4J



16 ARITHMETIC. 1

(33) To reduce a fraction to its lowest terms means to
change its form without changing its value. In order to do
this, we must divide both numerator and denominator by the
same number until we can no longer find any number (except 1)
which will divide both of these terms without a remainder.

To reduce the fraction | to its lowest terms, we divide
both numerator and denominator by 4, and obtain as a result

4 4 4-^-4

the fraction i. Thus, - ' = -J-; similarly, T77_!_ 4 = i;

*-j-4 2^-2 32-^-8 4^



(34) When the denominator of any number is not
expressed, it is understood to be 1, so that f is the same as
6 -T- 1, or 6. To reduce -f- to an improper fraction whose
denominator is 4, we must multiply both numerator and
denominator by some number which will make the denomi-
nator of 6 equal to 4. Since this denominator is 1, by multi-

6x4
plying both terms of -f- by 4 we shall have - = - 2 ^, which

1 X 4:

has the same value as 6, but has a different form. Ans.

(35) In order to reduce a mixed number to an improper
fraction, we must multiply the whole number by the
denominator of the fraction and add the numerator of
the fraction to that product. This result is the numerator
of the improper fraction, of which the denominator is the
denominator of the fractional part of the mixed number.

7-J means the same as 7 + \. In 1 there are |, hence in 7
there are 7 X f = 5 -V" P^ 118 ^ e i ^ ^ e m i xe d number
= ^i _|_ 7. _ e^ which is the required improper fraction.

= w; lot =



(36) The value of a fraction is obtained by dividing the
numerator by the denominator.

To obtain the value of the fraction -^ we divide the
numerator, 13, by the denominator, 2. 2 is contained in 13,
6 times, with 1 remaining. This 1 remaining is written



1 ARITHMETIC. 17

over the denominator, 2, thereby making- the fraction -|-,
which is annexed to the whole number, G, and we obtain G^-
as the mixed number. The reason for performing- this
operation is the following: In 1 there are f (two halves), and
in -^ (thirteen halves) there are as many ones (1) as 2 is
contained times in 13, which is G, and -J (one-half) remain-
ing. Hence, ^ = G + 1- = G J, the required mixed number.

= * A;



= 8; tt =



(37) In division of fractions, invert the divisor (or, in
other words, turn it upside down) and then proceed as in
multiplication.

(a) 35-5-A = XV.= ^ = H 4 = >12. Ans.



(c) V-9 = V-H = xl = = if. Ans.



"28") 113 (4^. Ans.
1 12
1

(^) 15f-f-4f ? Before proceeding with the division,
reduce both of the mixed numbers to improper fractions.



Thus, 15| = 4 = - = V, and 4| =

32 i 3

= - - = -M. The problem is now ^r - -^f = ? As

o

before, invert the divisor and multiply; ^f-^^f = ^-X-fg
_ 63X8 __ 504 ._ 252 __ 126 __ 18 18
" 4X35 " TT * ' ~^~ ~~ " 5 ' 5 ) 1 8 ( 3f. Ans.

L5

3

(38) i + | + 4 = 1 + g + 5 =| = 1. Ans.
When the denominators of the fractions to be added are
alike, we know that the units are divided into the same



18 ARITHMETIC. 1

number of parts (in this case eighths) ; we, therefore, add the
numerators of the fractions to find the number of parts
(eighths) taken or considered, thereby obtaining f or 1 as
the sum.

(39) When the denominators are not alike we know that
the units are divided into unequal parts, so before adding
them we must find a common denominator for the denomi-
nators of all the fractions. Reduce the fractions to fractions
having this common denominator, add the numerators, and
write the sum over the common denominator.

In this case, the least common denominator, or the least
number that will contain all the denominators, is 16; hence,
we must reduce all these fractions to IGths and then add
their numerators.

i + i + A ? To reduce the fraction \ to a fraction
having 16 for a denominator, we must multiply both terms
of the fraction by some number which will make the

1x4

denominator 16. This number evidently is 4; hence,

4 X 4 1

= A>

Similarly, both terms of the fraction - must be multiplied

3x2

by 2 to make the denominator 16, and we have -' = %.

o X A

The fractions now have a common denominator, 16 ; hence,
we find their sum by adding the numerators and placing
their sum over the common denominator, thus : T \ + TG + TQ



(40) When mixed numbers and whole numbers are to be
added, add the fractional parts of the mixed numbers sep-
arately, and if the resulting fraction is an improper fraction,
reduce it to a whole or mixed number. Next, add all the whole
numbers, including the one obtained from the addition of the
fractional parts, and annex to their sum the fraction of the
mixed number obtained from reducing the improper fraction.


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