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THE ELEMENTS
OF
RAILROAD ENGINEERING
Prepared for Students of
The International Correspondence Schools
SCRANTON, PA.
Volume V
ANSWERS TO QUESTIONS
Pint Edition
SCRANTON
THE COLLIERY ENGINEER CO.
1897
967
v. 5
( W
Copyright, 1897, by The Colliery Engineer Company.
Arithmetic, Key : Copyright, 1893, 18W, 1896, 1897, 1898, by The COLLIERY ENGI-
NEER Company.
Algebra, Key : Copyright, 1894, 1896, 1897, 1898, by THE COLLIERY ENGINEER COM-
PANY.
Logarithms, Key: Copyright, 1897, by The Colliery Engineer Company.
Geometry and Trigonometry, Key : Copyright, 1893, 1894, 1895, 1898, by The Col-
liery Engineer Company.
Elementary Mechanics, Key : Copyright, 189;}, 1894, 1895, 1897, by THE COLLIERY
Engineer Company.
Hydromechanics, Key : Copyright, 1893, 1894, 1895, 1897, by THE COLLIERY ENGI-
NEER Company.
Pneumatics, Key : Copyright, 1803, 1895, 1897, by The Colliery Engineer Com-
pany.
Strength of Materials, Key: Copyright, 18M, 1899, by The Colliery Engineer
Company.
Surveying, Key : Copyright, 189.5, by The Colliery Engineer Company.
Land Surveying, Key : Copyright, 1895, by The Colliery Engineer Company.
Railroad Location, Key : Copyright, 1895, by The Colliery Engineer Company.
Railroad Construction, Key : Copyright, 1895, by The Colliery Engineer Com-
pany.
Track Work, Key : Copyright, 1896, by The Colliery Engineer Company.
Railroad Structures, Key : Copyright, 1896, by The COLLIERY ENGINEER COM-
PANY.
burr PRINTING HOUSE,
FRANKFORT AND JACOB STREETS,
NEW YORK.
A KEY
TO ALL THE
QUESTIONS AND EXAMPLES
INCLUDED IN
Vols. I and II,
EXCEPT THE
EXAMPLES FOR PRACTICE.
It will be noticed that the Key is divided into sections
which correspond to the sections containing the questions
and examples at the end of Vols. I and II. The answers
and solutions are so numbered as to be similar to the num-
bers before the questions to which they refer.
157143
CONTENTS
Arithmetic, -
Algebra,
Logarithms,
Geometry and Tri
onometry.
Elementary Me-
chanics,
Hydromechanics,
Pneumatics,
Strength of Ma-
terials,
Surveying, -
Land Surveying, -
Railroad Location,
Railroad Construc-
tion, -
Railroad Construe
tion, -
Track Work,
Railroad Struc
tures.
Answers to Questions
Answers to Questions
Answers to Questions
Answers to Questions
Answers to Questions
Answers to Questions
Answers to Questions
Answers to Questions
Answers to Questions
Answers to Questions
Answers to Questions
- Answers to Questions 813-872 281-289
Answers to Questions 873-94G 291-297
Answers to Questions 947-1016 299-304
NOS.
PAGES.
1-1G7
1-94
1G8-257
95-13G
258-272
137-143
273-354
145-1 64
355-453
165-186
454-503
187-200
504-553
201-208
554-613
209-232
614-705
233-2^5
706-755
257-268
756-812
269-279
- Answers to Questions 1017-1082 305-309
ARITHMETIC.
(QUESTIONS 1-75.)
(1) See Art. 1.
(2) See Art. 3.
(3) See Arts. 5 and 6.
(4) See Arts. lOand 11.
(5) 980 = Nine hundred eighty,
605 = Six hundred five.
28,284 = Twenty-eight thousand, two hundred eighty-four,
9,006,042 = Nine milHon, six thousand and forty-two.
850,317,002= Eight hundred fifty million, three hundred
seventeen thousand and two.
700,004 = Seven hundred thousand and four.
(6) Seven thousand six hundred = 7, GOO.
Eighty-one thousand four hundred two = 81,402.
Five million, four thousand and seven =: 5,004,007.
One hundred and eight million, ten thousand and one =
108,010,001.
Eighteen million and six = 18,000,006.
Thirty thousand and ten = 30,010.
(7) In adding whole numbers, place the numbers to be
added directly under each other so that 3 2 9
the extreme right-hand figures will stand kqa
in the same column, regardless of the g 6 5 4 n 3
position of those at the left. Add the first 2 7 4
column of figures at the extreme right, g-*
which equals 19 units, or 1 ten and 9 m
units. We place 9 units under the units
column, and reserve 1 ten for the column ^ 7 1 3 5 9 Ans.
i ARITHMETIC.
of tens. 1 + 8 + 7 + = 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 + G = T ten-thousands.
Place this seven ten-thousands in the ten-thousands column.
There is but one figure 8 in the hundreds of thousands place
in the numbers to be added, so it is placed in the hundreds
of thousands column of the sum.
A simpler (though less scientific) explanation of the same
problem is the following: 7 + 1 + 4+3 + 4+0 = 19; write
the nine and reserve thel. 1 + 8+7 + + 0+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) 7 9
8304725
391
100302
300
909
8407 33 6 Ans.
(9) (a) In subtracting whole numbers, place the sub-
trahend 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 can not
subtract 8 units from 2 units, so we take 1 ten from the
6 tens and add it to the 2 units. As 1 ten = 10 taiits, we
have 10 units + 2 units =12 units. Then, 8 units from
12 units leaves 4 units. We took 1 ten from 6 ten^. so
ARITHMETIC.
only 5 tens remain. 3 tens from 5 tens 59902
leaves 2 tens. In the hundreds column we 3333
have 3 hundreds from hundreds leaves
G hundreds. We can not subtract 3 thou- '^'^^^^ ^ns.
sands from thousands, so we take 1 ten-thousand from
5 ten-thousands and add it to the thousands. 1 /en-
tJiousand = 10 tJiausands, and 10 thousands + thousands
= 10 thousands. Subtracting, we have 3 thousands from
10 thousands leaves 7 thousands. We took 1 ten-thousand
from 5 ten-thousands and have 4 ten-thousands remaining.
Since there are no ten-thousands in the subtrahend, 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) 15 339
10001
5 3 3 8 Ans.
(lO) {a) 70968 {b) 100000
32975 98735
3 7993 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 number
is evidently less than 1,004 by the difference 49, hence,
1,004 — 49 = 955, the smaller number. For the sum of the
10 4 larger
two numbers we then have 9 5 5 smaller
19 5 9 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)
4 ARITHMETIC.
and the difference (49); or, 1,004 — 49 = 955, the smaller
number. Since the greater number equals 1,004 and the
smaller number equals 955, their sum equals 1,004 -f- 955
= 1,959 sum. Ans.
(12) The numbers connected by the plus (+) sign must
first be added. Performing these operations we have
5962 3874
8471 2039
9023 5913 stun.
23 456 sum.
Subtracting the smaller number (5,913) from the greater
(23,456) we have
23456
5913
17 5 4 3 differ eyice. Ans.
(13) 144675 = amount willed to his son.
2 6 3 8 = amount willed to his daughter.
$71055 = amount willed to his two children.
$1 25000 = amount willed to his wife and two
children.
710 5 5 = amount willed to his two children.
$53945 = 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.
{a) 7x7 units = 49 units, or 4 tens and 9
52 6 3 87 units. We write the 9 units and reserve
7 the 4 tens. 7 times 8 tens = 56 tens ;
3 6 8 4 7 9 Ans. 56 tens + 4 tens reserved = 60 tens or
6 hundreds and tens. Write the
tens and reserve the 6 hundreds. 7 X 3 hundreds = 21 hun-
dreds; 21 + 6 hundreds reserved = 27 hundreds, or 2 thou-
sands and 7 hundreds. Write the 7 hundreds and reserve
ARITHMETIC. 6
the 2 thousands. 7X0 thousands = 42 thousands; 42
+ 2 thousands reserved = 44 thousands or 4 ten-thousands
and 4 thousands. Write the 4 thousands and reserve the
4 ten-thousands. 7x2 ten-thousands = 14 ten-thousands;
14 -|- 4 ten-thousands reserved = 18 ten-thousands, or
1 hundred-thousand and 8 ten-thousands. Write the 8 ten-
thousands and reserve this 1 hundred-thousand. 7x5 hun-
dred-thousands = 35 hundred-thousands; 35 -+- 1 hundred-
thousand reserved = 3G hundred-thousands. Since there
are no more figures in the multiplicand to be multiplied,
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 = GO ; write the and reserve the 6.
7 times 3 = 21; 21 + 6 reserved = 27 ; write the 7 and re-
serve the 2. 7 X 6 = 42 ; 42 + 2 reserved = 44 ; write the
4 and reserve 4. 7 X 2 = 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 17
units^ or 1 ten and 7 units^ so that ^^) 7 2 9 8
the product is obtained by adding 1 7
two partial products, namely, 7x 4902086
700,298 and 10 X 700,298. The 700298
actual operation is performed as 11905066 Ans,
follows:
7 tiriies 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 re-
serve the 2. 7 times = 0; + 2 reserved = 2 ; write the 2.
7 times = ; + reserved = ; write the 0. 7 times 7 =
49 ; 49 + reserved = 49 ; write the 49.
. To multiply by the 1 ten we say 1 times 700298 = 700298,
and write 700298 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.
(c) 217 Multiply any two of the numbers together
10 3 and multiply their product by the third
(5 51 number.
2170
22351
67
156457
134106
14 9 7 517 Ans.
(15) If your watch ticks every second, then to find how
many times it ticks in one week it is necessary to find the
number of seconds in 1 week.
6 seconds = 1 minute.
6 minutes = 1 hour.
3 6 seconds = 1 hour.
2 4 hours = 1 day.
14400
7200
8 6 40 seconds = 1 day.
7 days = 1 week.
6 4 8 seconds in 1 week or the number of times that
Ans. your watch ticks in 1 week.
(16) If a monthly publication contains 24 pages, a yearly
2 4 volume will contain 12x24 or 288 pages, since
12 there are 12 months in one year; and eight
2gg yearly volumes will contain 8x288, or 2,304
ft pages.
2 3 4 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
$3246
3
9738
plus 1200
$10 9 3 8 = value of building.
ARITHMETIC. 7
If the tools are worth twice as much as the building, plus
11,875, then the tools are worth
$10938
2
21876
plus 18 7 5
$23751 = value of tools.
Value of building = $10938
Value of tools = 2 3 7 51
$34689 = value of the building
and tools, {a) Ans.
Value of engine and
boiler = $3246
Value of building
and tools = 3 4 6 8 9
$ 3 7 9 3 5 = value of the whole
plant, (d) Ans.
(18) {a) (72 X 48 X 28 X 5) -i- (96 X 15 X 7 X 6).
Placing the numerator over the denominator the problem
becomes
72 X 48 X 28 X 5 _
96 X 15 X 7 X 6 ~
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 tinder it. Thus,
1
72 X 48 X 28 X ^ _
96 X ;^ X 7 X 6 ~
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 15 never existed. Thus,
72 X 48 X 28 X 1 _
96 X 3 X 7 X 6
8 ARITHMETIC.
72 in the dividend and 96 in the divisor are divisible by 12,
since 72 divided by 12 equals 6, 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
jr^ X 48 X 28 X 1 _
^0x3x7x6
8
The 72 and 96 are 7iot to be considered any longer, and,
in fact, may be £rased 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 _
8X3X7X6 ~
Again, 28 in the dividend and 7 in the divisor are divisible
by 7, since 28 divided by 7 equals 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 under it. Thus,
4
6 X 48 X 2^ X 1 ^
8x3x^x6
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
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,
ARITHMETIC. 9
6x8x4xl _
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 _
8X1X1X1
Canceling the 8 in the dividend and the 8 in the divisor,
the result is
1
2x^4x1 ^ 2x1x4x1
^xlxlxl~lxixlxl*
1
Since there are no two remaining numbers (one in the
dividend and one in the divisor) divisible by any number trs.-
cept 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 2x1x4x1 = 8; the product of all the
uncanceled numbers in the divisor equals 1x1X1X1 = 1.
„ 2X1X4X18 _.
Hence, :; ; :; = - = 8. Ans.
IXlXlXl 1
2
Or, 4><4^44 = | = 8. Ans.
' 96 X ;^ X 7 X ^ 1
1 \
10 ARITHMETIC.
(d) (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
80 X 60 X 50 X 16 X 14 _ ^
70 X 50 X 24 X 20 ~ '
The 50 in the dividctid 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 offxhe 50 and write the 5 over it;
also, cross off the 70 and write the 7 under it. Thus,
5
80 X 60 X ^P X 16 X 14 _
yp X 50 X 24 X 20
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,
80 X 60 X 5 X 16 X 14 _
7 X 50 X 24 X 20 ~
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
^P X 60 X 5 X 16 X 14 ^
7 X 50 X 24 X ^P
1
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,
4X 60X 5X 16 X 14 ^
7 X 50 X 24 X 1
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\.\\& 16 and write the 2 over it; also cross
off the 24 and write the 3 under it. Thus,
ARITHMETIC. U
2
4 X 60 X 5 X ;0 )i< 14 _
7 X 50 X ^^ X 1 ~
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 _
7 X 50 X 3 X 1 ~
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 ^
7 X ^P X 3 X 1
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 dividend and 7 in the divisor are divisible by
7, since 14 divided by 7 equals 2, and 7 divided by 7 equals 1.
Cross off the 14 and write the 2 over it; also, cross off the 7
and write the 1 under it. Thus,
2
4x6x5x2x;^ ^
7x5x3x1
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 ~
12 ARITHMETIC,
The 5 in the dividend and 5 in the divisor are divis-
ible 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 _
1x^x3x1
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,
4X 6 X 1 X 2X2 _
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 2 oiier it ; also, cross off the 3 and
place 1 under it. Thus,
2
4x 0Xlx2x2 _
1x1x^x1 ~
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 their
stead, and treated as if the 6 and 3 never existed. Thus,
4x2x1x2x2 32 „^ .
— = ^ = ^ — = — = 32. Ans.
1x1x1x1 1
2 1
4 ^ ^ 2 2
^0X ^0x^0x;^x;^ _ 4x2x1 X2x2 _ 32 _
"^''^^' ypx^Px^^x2p - ixlxlxl 1 .^r
7 ^ ^ 1
111
(19) 28 acres of land at $133 an acre would cost
28 X $133 = 13,724.
28
1064
266
13724
ARITHMETIC. 13
If a mechanic earns $1,500 a year and his expenses are
per year, then he would save 11500—1968, or $532
per year. 9 6 8
$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)8724(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
(3 X 365 miles) — 246 miles, or 849 miles the second week.
Thus, 3 6 5
3
1095
246
difference 8 4 9 miles. Ans.
(21) The distance from Philadelphia to Pittsburg is 354
miles. Since there are 5,280 feet in one mile, in 354 miles
there are 354 X 5,280 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 -=- 16, or 116,820 revolutions.
16)1869120(116820 rev. Ans.
• 16
26
16
109
96
131
12 8
32
3 2
~0
14 ARITHMETIC.
(22) (a) 576) 589824 ( 1024 Ans.
576
1382
1152
2 3 04
2304
{6) 43911)369730620(8420 Ans.
351288
184426
175644
87822
87822
(c) 605)2527525(5005 Ans.
2525
2525
2525
(d) 1234)4961794302(4020903 Ans
4936
2579
2468
11143
11106
3702
3702
(23) The harness evidently cost the difference between
$444 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
$417 $2 7 Ans.
ARITHMETIC. 15
(24) (a) 1034
5 76
W
6144
7168
6120
6 8 9 8 2 4 Aas.
5005
505
25025
2 5 2 5
2 5 2 7 5 2 5 Ans.
(c) 43911
8420
878220
175644
351288
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 300 x 10 = 3,000 hours. Hence, he receives for his
work 3,000 X 30 = 90,000 cents, or 90,000 -^ 100 = 1900. Ans.
(26) See Art. 71.
(27) See Art. 77.
(28) See Art. 73.
(29) See Art. 73.
(30) See Art. 76.
13
(31) -X- is an improper fraction, since its numerator 13
8
is greater than its denominator 8.
(32) 4;14l;85l.
16 ARITHMETIC.
(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 num-
ber (except 1) which will divide both of these terms without
a remainder.
4
To reduce the fraction — to its lowest terms we divide
o
both numerator and denominator by 4, and obtain as a
1 4—41 4—4
result the fraction — . Thus, - ' = .7 ; similarly, 777 ' =
<i o -7- 4: Xi lo ~T- 4
]_, ^ -^ 4 _2 -T- 2 _1 . 32 ^ 8 _ 4 -4- 4 _1
4 ''32^4~8 -T-2 4' 64^-8 "8 -T-4~2' '
(34) When the denominator of any number is not
expressed, it is understood to be 1, so that — is the same as
6 -^ 1, or 6. To reduce - 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 mul-
tiplying both terms of - by 4 we shall have — , = — ,
f J ^ I J 1X44
which 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 denom-
inator 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 denomitiator is the denominator of the
fractional part of the mixed number. .
7 7 8
75- means the same as 7 + ^- I^^ 1 there are^— , hence in
8 80
7 there are 7X5- = -^; -jr plus the - of the mixed number