ARITHMETIC 
FORMEDICAL 
TECHNICIANS 
COMPILED AND 
PUBLISHED AT 
[DUSTED TECHniCIOriS SCHOOL 
LETTERIMI GEHERRL HOSPITAL 
FOR USE OF STUDENTS 
PUBLISHED FEBRUARY 1941 
REVISED & REPUBLISHED SEPTEMBER. 1942 
REPUBLISHED JANUARY 194 T 
REPUBLISHED JUNE 1943 ■ 
REPUBLISHED OCTOBER 1943 LINEAR MEASURE 
5,280 feet (ft.) 
3 feet 
12 inches (in.) 
1 mile 
1 yard (yd.) 
1 foot 
CAPACITY MEASURE (Apothecaries) 
U quarts (qt.) 
2 pints (pt.) 
1 pint 
1 gallon 
1 quart 
l 6 fluidobnces 
1 fluidounce 
1 fluidrams 
8 fl. drains (fl.dr.) 
6o minims (m) 
WEIGHT MEASURES (Avoirdupois) 
2000 pounds (lb.) 
100 pounds 
1 ton 
1 hundredweight (cwt.) 
1 pound 
16 ounces (oz.) 
METRIC AND ENGLISH EQUIVALENTS 
l inch 
1 lb. 
centimeters (cm.) 
USU grams (Gm.) 
1 meter 
1.09 yards 
0.621 mile 
2.20 pounds 
U 73.18 c.c. 
1 kilometer 
1 kilogram 
1 pint 
1 fluidounce 
1 gram 
30*0 c.c. (approximate value) 
grains (gr.) 
1 grain 
O.OGUg gram. 
1 cubic centimeter of distilled water at U° C. weights 1 gram. TO THE STUDENT 
ARITHMETIC 
You have learned to add, subtract, multiply and divide,using 
integers and common and decimal fractions. You have studied the tables 
of measures commonly used in the United States. You have learned how to 
find areas and volumes, and you know something about percentage. 
Some of you will remember what you have learned; some will have 
rgotten; and some perhaps have never learned the basic principles of 
•.thmetic. 
This course is mainly for those who have forgotten and those who 
e never learned. To the rest, it is a good review. 
You are to study a very interesting part of arithmetic, one that is 
jx great practical value. You will'be interested to learn more of the 
uses of arithmetic in every day life. The knowledge you acquire from 
this course will be of value to you throughout your life. ’Such a know- 
ledge is essential in the store, the shop, the office, the bank, about 
the mine, on the farm, in chemistry, in engineering, in pharmacy, in 
medicine, in nursing, and many other vocations and professions. 
In each of the vocations and professions one who has not acquired 
a high degree of accuracy and a reasonable degree of speed in computation 
is usually at a great disadvantage. Drill exercises, if widely used, 
will be of value to you because they enable you to increase your speed 
and accuracy. Strive to excel your own best record. Realize you are 
working for yourself, and that the knowledge and skill which you acquire 
will be of great value to you in the years to come. 
You will have a difficult time if you do not ask questions about 
procedures you do not understand. Do not be afraid to ask questions 
because someone in the class wants to know if the problems canft be 
worked with calculus. SuCh questions are ridiculous in such a class 
as ours, but only the instructor and the questioner usually know it; the 
other students, becoming discouraged on hearing such awe inspiring 
questions, do not ask sensible questions for fear of appearing too f,dumb. 
This has been a definite handicap to more than one class of students. 
If you are having difficulty, you can improve only by extra hard 
work, which means practice on problems. Everything new is hard; what is 
easy is easy because you have learned it by constant drill. II 
TABLE OF CONTENTS 
To .The Students ii 
Chapter 
Page 
I. Pleading and Writing Large Numbers 1 
11.-’ Integers., U 
111. Saving Time when Making Computations... 7 
IY. Common Fractions and Decimals.. l 6 
V. How to Solve a Problem U 9 
VI. The Metric System., 
VII. The Thermometer. 79 
VIII, Per Cent - Solutions 8U * 
IX. Food as Energy....'. , 95 
X. ■ . Review - Exercises and Problems '.. * 100 
Tables ’ 10S Chkpter 1 
READING AND WRITING LARGE NUMBERS 
When reading the daily papers and magazines you will often find 
it necessary to read large numbers. You will recall that, for con- 
venience in reading, large numbers are pointed off into commas as shown 
in the following example; 
297,956,U50 
This number if read fourteen billion, two hundred ninety-seven 
million, nine hundred fifty-six thousand, four hundred eighty. 
The_ period to the left of billions is called trillions. It is 
used when expressing great magnitudes, like the distance to one of the 
stars. The distance from the earth to the Pole Star is more than 
250,000,000,000,000‘mi1e5. This is read as two hundred fifty trillion 
miles. 
There is a definite way to go about reading such a large number 
as; 
to) 
Note that every three figures are set off by commas. The first group 
to the right is known as hundreds. The next to the left, as thousands; 
the next millions; the last billions. When you see such a number, 
you should say to yourself as you check each group of three figures, 
"hundreds, thousands, millions, billions". The next step is to read 
the whole number. You say fourteen billion, two hundred ninety-seven 
million, nine hundred fifty-six thousand, four hundred eighty. 
Exercise 1 
Read the following: 
1. The land area of the United States is 3,026,789 and the population 
in was 131,669,275* 
Read the following land areas and populations: 
2. Africa . - H,529,U80 sq. miles; 155.U75.000 
3. Canada - 3.69U.863 sq. miles; 11,209,000 
U. China - 3,756,102 " 11 >+22,527,000 
5. Europe - ' 3.773,958 " " 539,800,000 
6. India - 1,575.109 " " 338.i70.632 
7. Soviet Union - 3,170,268 " 180,122,390 
8. Asia - 16.U9U.2i7 " " 1,090,31U,000 
9. Texas - 265,896 " " 6,U1U,82U 
10. Rhode Island - 1,248 " " 713.3U6 Head the following distances iK miles! 
11. New York to San Francisco - 3>173 
12. Earth to Sun - 92,897.^10 
13. Earth to Mercury - 136,000,000 
14. Earth to Mars . 2%,000,000 
15. Earth to Pluto Uflfoo,ooo,ooo 
16. Pluto to Sun - 3*700,000,000 
' .Read the following areas in square miles; 
17. Atlantic Ocean - U 1,321,000 
18. Pacific Ocean - 68,d3U,000 
19. Indian Ocean - 29 , 3^0,000 
20. North Sea - . 220,000 
21. Red Sea - 178,000 
22. Baltic Sea - 160,000 
23. Mount Everest in Indo China is the highest mountain - feet. 
The deepest place in the ocean yet found is feet in the Pacific 
25. The' speed of light is miles per second. .r 
26. The equatorial circumference of the earth is miles 
Accidental deaths in the United States in 19*40: 
27- Motor Vehicles - 3^>500 
28. , Falls - 25,600 
29. All'"burns • 1,300 
30. 'Drowning - ■ 6,300 
31. " 1 Railroad 5,000 
32. Firearms *'■ - 2,^400 
33* Pois6n Gas - • 1,500 
31. Poisons (not gas)- • < 2,100 
35. In 1913, deaths from acute accidents amounted to *4-, 227; from drowning 
9,875. 
Read the melting uoints of the following chemical elements: 
36. Qarbon(diamonds) Centigrade 
37- ~. Iron 1,535° " 
3'B. " Copper • • .... 1,083-° ” 
39- Lead 327° " 
MO. Molybdenum 2,620' M 
Ul. Mercury ; ’ 1....... . 36° 
Read the Boiling point of the following’chemical elements: 
U 2. Arsenic' 615° Centigrade 
Uj. Copper ‘ .....2\300° * UU, Hydrogen .v ,:.i. . -252.8 
U 5. Iron 3;, 000 
Uo, Platinum ‘V ’ 760 ' 
U 7. Thallium 1,650 
*4B. Tin 2,270 
U 9, Tongs ten 5»900 
50. Zinc 930 
- The Use of Round Numbers 
Explanation; 
In many cases where the use of exact numbers if not necessary, 
it is customary to express large amounts in round numbers. Round 
numbers may be expressed to the nearest hundred, thousand, ten thousand 
and so on, as required. 
For example 
Instead of saying $180,306,599, we m&y state this amount to the 
nearest hundred thousand and call it $180,300,000. $306,000 was 
stated as $300,000, because nearer to being $306,000 than 
is $UOOtOOO or $200,000, umpxer rr^ 
Integers 
Addition, subtraction; ;jpdlt indication and 'division of integers. 
Exercise. 2- 
Addition 
1 
2 
3 
u 
5' 
' '6* 
7 
g 
'9 
1 
7 
5 
u 
t 
6 
U 
1 
7 
5 
3 
c 
2 
I 
8 
3 
7 
U 
1 
4- 
s 
7 
3 
s 
a. 
t). 
5 
5. . 
■ 4 
3 
8 
7 
8 
8 
? 5 
7 
, S 
6 - 
i 
' .2' 
2 
U. 
■ 5 , 
3 
9 
7 
1 
7 
5 
2 
0 
9 
c. 
5 
9 
7 
9 
h 
7 
3 
o 
8 
1 
9 
6 
k 
3 
7 
6 
g 
7 
I 
1 
g 
7 
9 
3 
d 
' 6 • 
7 
3 
u 
6 
p 
7 
8 
5 
r 
3 
e 
7 
g 
u 
3 
u 
9 
9 
8 
e 
__J 
, 5 
g 
9 
e. 
I 
6 
g 
9 
9 
6 
_7 
8 
7 
6 
u 
7 
_9 
6 
g 
g 
9 
8 
1 
1 
9 
6 
8 
7 
6 
Subtraction 
u 
75 
7 
6 
8 
1 
39 
g__ 
p 
91 
'6 
3 
U 6 
U 
5 
29 
7 
7 
5 
8 
36 
7_ 
9 
U 5 
9 
a. 
1). 
76 
9 
6s 
Q 
j 
JL 
Ug 
9_ 
U g 
9 
lU 
9_ 
JL 
51 
6 
c. 
39 
8 
27 
3 
Ul 
u_ 
32 
L 
Ug 
6_ 
S 2 
g_ 
67 
_s_ 
69 
s_ 
1_ Subtraction (Cohtinued) 
1 
2 
3 
k 
5 
6 
7 
g 
9 
d. 
17 
8 
16 
36 
; s_ 
19 
9 
13 
8 
19 
s__ 
14 
0 
J 
15 
_6 
e. 
72 
3 
78 
9 
67 
s_ 
S 7 
6 
72 
9 
S 2 
3 
71 
o 
J 
65 
7 
65 
3 
Multiplication 
k 
6 
1 
2 
3 
5 
7 
8 
9 
7^ 
3 
36 
-X 
ks 
2 
U 6 
6 
7* 
a. 
iL 
27 
52 
L 
3S 
9 
b. 
± 
25 
B_ 
U 2 
9 
35 
7 
79 
_2_ 
56 
5 
gg 
u 
7l 
39 
2 
c. 
S 7 
3 
6? 
3 
3_ 
U 2 
6_ 
8U 
_6 
l 6 
l_ 
JL 
gu 
7 
36 
7 
d. 
JL 
16 
s__ 
h 
97 
_l_ 
JL 
JL 
Si 
3 
1_ 
*1 
e, 
14-7 
7 
jg 
6 
29 
3 
Uy 
6 
87 
3 
92 
6 
99 
9 
29 
6 
87 
9_ 
Division 
1 
2 
3 
u 
5 
6 
7 
8 
c 
J 
a. 
U 752 
3W~ 
TW~~ 
sTSS“ 
9772 
8732“ 
UTSS“ 
6) OS 
iW~ 
b. 
Uo 
30 
~E 
66 
11 
Ik 
7 
96 
b 
36 
9 
_§2_ 
7 
6U 
8 
U9__ 
7 Division (Continued)- 
1 
2 
3 
u 
5 
6 
7 
8 
9 
8)96 
c. 
5T50~ 
8)88 
2)Uc 
5755 
liw~ 
UTss- 
SJW 
3710“ 
d. 
36 
6~ 
IS 
~T 
55 
~5 
U 5 
~5~~ 
63 
SI 
9 
63 
~T~ 
36 
1“ 
72 
~S 
e. 
90 
10 
56 
T 
s 
~5 
81 
—a— • 
j 
I+9 
T~ 
33_ 
3 
Uo 
ir~~ 
28 
“IT" 
15 
“T 
f. 
99 
U 9 
~~T 
U 5 
5 
63 
7 
96 
72 
8 
Ug 
5 
su 
7 
12 
~n~ 
g. 
6T30— 
8T72 
9T90 
67^1— 
7)56 
9T31 
8)56 
5730“ 
3755" 
U)2U 
9)90 
8)72 
6)5U 
U)96 
S)6U 
U)su 
h. 
sypr~ 
7)^9 Chapter. 11l 
SAVING TIME WHEN MAKING COMPUTATIONS 
If you were'asked to find the cost of four pounds of butter at 
twenty-five cents a pound, it would be absurd for you*£b make the 
computation on paper, because you can give the answer immediately by 
computing "in your-‘head,as we sometimes say.’ • Some people call this 
kind of computation mental arithmetic. By using mental arithmetic 
many problems are daily solved without paper or pencil. 
You arc already familiar with some of the short methods of com- 
putation in common use. A multiplication table is a short method. A 
knowledge of the tables enables you to make multiplications more 
quickly than you could obtain the results by addition. 
In the lessons that follow you are to learn some new short methods 
of computation, and also how to apply them practically. Learn these 
thoroughly, and accustom yourself to working with them whenever you can 
save time by doing so. You will find that some of these short methods 
have been placed here for review. Even so,, study them carefully and 
learn them well, for they will be of use to you throughout life. 
When an amount is multiplied by 10 or a power of 10 (100; 1000; 
10,000 etc, ) the product is found by adding as many ciphers to the 
multiplicand as there are ciphers in, the multiplier. Thus, 
786 xlO - 7,860. Of dollars and cents are involved, the two right 
hand figures in,;the product are cents, and the figures,at the left of 
the cents are dollars. For'example, $U5,10 x 100 = $^,510.00. 
SHORT MIL THUDS IN MULTIPLICATION 
1. 
To multiply any number by 10 move the decimal point one place 
to the rights annexing a zero if necessary. 
For example: 10 xgU = gUo. 10 x 3-6 U - 
»' 'i ■ ' • / • . 
11. 
To multiply any number by 100, move the decimal point two 
places to the right, annexing two zeros if necessary. 
For example; .100-x-32 - 3200. 100 x r 378.^ 
To multiply any number by 1000, move the decimal point three 
places to the right, annexing three zeros if necessary. 
1111. 
For example: 1000 x 7^= lOOO x .3568 - 35^.8 SHORT METHODS IN MULTIPLICATION tCotttinued) 
IV. 
To multiply an integer "by a number ending in zero, multiply 
the integer by the number or numbers to the left of the zero, 
and then annex a zero to this product * . , 
For example: 325 xbo = 1 First multiply 325 "by U. This 
gives. 130.0«... Annex .a. aero to this number and obtains 13,000, the 
result, "vr '3 ■3 3 • '" ) 
■- 1 '■ ••• - - - ■ - • 
To •multiply .a number by .5, multiply lO, then divide this result 
by 2. - .V:' ' 
Y. 
For example: ‘5 x JSk - 7&bo s 3320 
VI. 
To multiply a number by 25, multiply by 100, then divide this 
result by ,d. 
For example; 25 x 73 s 73QQ r 1825. 
VII 
To multiply .a number by 125, multiply by 1000, then-divide this 
result -by 8. - . ‘ ■ 
For example; 125 x 67 - =. 8375. 
Vlll 
To multiply a number by 9, multiply by 10, then subtract the 
multiplicand. 
For example:- 9 x - - U 92 = UU2S. 
;IX. 
To multiply-a number by 33~1/3» multiply by 100, then divide 
this result by 3. : 
For example; 33-1/3-*.'7s = = 2600, 
X, 
To multiply a number by 66-2/3, multiply by 100, then take 2/3 
rds, of this result. 
/ a . r,.: ■. ■: ■ 
For example;. 66-2/3 x'3& - 3600 x 2/3 f- 2ta). 
XI. 
To multiply a number by first multiply by 100, then divide 
this result by 8, 
For example; 12g- xbB - l/S of UBOO - 600. 
Suggest' a short method for multiplying a number by 50. By 16-2/3 
by 37i* ,By 62J. By 87^., Give three illustrations of each method 
you work out, ■ • ' 
To multiply a number by 250, first jmltiply by 1000, then divide 
.this result by; U. 
Xll, 
For example: 250 x 328 - Jof 328,000 - 82,000. 
Suggest a short method for multiplying a number by 500. By 
333-1/3. By 666-2/3. By 750. Give three illustrations of each 
method you work out.  3 
(For Accuracy and Speed) 
■Without the use of paper or pencil, multiply each of the numbers 
in this exercise by 10, by 100, by 5* "by 125. 
123'-U 5 6 7 8 9 
a. 19 2U 16 8 18 32 U 8 6 30 
b. 73 2U 62 16 lU 23 36 72 Ug 
c. 92 67 U6U IUU 36s 72 Ul* 22 17 
d. 376 Ugo 26.31 U.6U 968 36.8 96S 188 9656 
Exercise U 
(For Accuracy and Speed) 
Without the use of paper or pencil, multiply each of the numbers 
in this exercise by 33-l/3. by 66-2/3 and by 9- 
1 2 3 U 5 6 7 8 9 
a. 33 27 69 U 5 21 gU 99 39 93 
b. 15 9 21 U 2 81 108 135 : ■ 60 
c. 66 90 78 120 IUU 180 36 I6S 
d. 279 881 95U 98U6 ,360 288 : 336 26U 
Exercise 5 
(For,'Accuracy and Speed) 
Multiply the numbers in column 1 by 12-1/2, those in column 2 by 
16-2/3. etc., as designated. • , 
1 2 3 . . u . 5 
12-1/2 16-2/3 250 25 33-1/3 
X X X ' X X 
a. U6U 360 78 UBO 729 
b. 38U 612 i+2o 388 U6B 
c. 9672 Ug6 36U 20U 32U 
d. UgOO 35U 792 760 86U- 
e. 32.08 270 U.U 79232 936 
f. U2UO U 2.6 980 522 You have probably discovered that 37-1/2 Is 3/3 of 100{ that 
62-1/2 is 5/3 Qf 100; and that 87-1/2 is 7/8 of 100. 
'A table showing these-relationships might be expressed-in 
this way: - • • 
10 100 1000 
1/8 - - '■ 1-l/U 12-1/2 125 
1/6 - - 16-2/3:: ■■ 166-2/3 
1/U - - 2-1/2 25 250 
1/2 - - 5 • ' 50 ■’• ■ 500 
Since 1/8 of 100 is 12-1/2 
3/8 is 3 x 12-1/2 37-1/2 
SHORT METHODS IN ADDITION 
Method used in adding two numbers 
For example: 
U 6 -f 32 = .? 
Think: U 6 -f 30 —76 
76 -f 2 =. 78 
Another example, which shows a simple way to add three figure 
numbers' that end in zero r 
-j_ $3.90 = 7 
Think: SU.2O f $3.00 $7.20 
$7.20 -f- $.90 $B.lO Exercise 6 
(Eor'Accuracy and Speed) 
Addition. 
.... 3;. 
32 
7*+ 
75 
22 
lU . 
33 
% 
U 2 
12 
2^ 
82 
g 
i 
U 
93 
72 
96 
23 
21 
66 
56 
62 
Is 
16 
12 
S 5 
55 
5 • " ' 
28 ■" • 
ig . 
Ui 
75._ • 
Y- h 
23 
B2 
21 
3i_ 
73 
ls_ 
12 
21 
3IL 
1 
a. 36 
. 27 
b. Ul 
75 
c. 63 
52 
d. 7U 
5^ 
e. 39 
_9s_ 
f. 12 
Ug 
g. S 9 
31 
2 
U 9 
BB 
69 
13 
15 
63 
lU 
19 
28 
65 
__23_ 
25 
■ 5^ 
Method used in adding more than two numbers. 
jTj may 0;?...50 when you wish to verify the accuracy of your 
answer, you should do it rapidly by mental addition. In adding 
columns of figures you can increase your speed by mentally corn- 
birr ug Uu or ro::e figures and adding them to the others in one 
amount, Figures that amount to 10 when combined are recognized 
most easily. 
■11 ( 85 ) 
10 ( ) 10 
( 25 ) 
( 96 ) 
12 ( ) 10 
( 31* ) 
( gu ) 
15 ( ) 7 
( 73 ) 
( 36 ) 
13 ( ) 10 
( ) 
507 
The method of bshibihing the figures in 
each column is shown in the illustration at 
the left. The two figures are combined hy 
brackets and the sum is indicated at the 
right or the left of the brackets. The mental 
process when the first column is added should 
be "10,17*27»37m* The 7 is written and the 
3 is carried to be included in the first 
group in the second column. The'.process in 
the second column is "13, 28, Uo, 50", 
Exercise J 
(For Accuracy and Speed) 
Add the columns in the following problems. So far as 
possible add two figures at a time. Practice on the problems until 
you can add them quickly and accurately 
l 
a. f, l 6 
QP 
85 
J 3 
Id. 2U 
56 
33 
_72 
c. 37 
51 
U 2: 
lU- 
d. 69 
13 
75 
22 
2 
U 6 
83 
37 
83 
38 
72 
75 
35 
33 ' 
72 
28 
3^ 
§ 
29 
UU 
3 
36 
lU 
93 
17 
63 
68 
‘ 15 
g 
33 
31 
82 
•St- 
u 
23 
72 
38 
lU 
26 
78 
U 6 
U 2 
12 
32 
68 
95 
2U 
12 
7^ 
5 
3^ 
P 
67 
23 
*7 
6 8 
97 
ip 
89 
31 
25 
5^ Exercise 7 (fiohtiiiued) 
U 
U 2 . 
25 * 
.<5l 
‘ 3^ 
2k ‘ 
59 • 
2 
16 
12 
7^ 
12 ’ 
1 ; 
e. UU 
2 ■■ 
32 
f. 23 
82 
7^ 
5^ 
g. il*- 
33 
21 ■ 
66 
v, . 2 
2,U. •; 
82 
l - . 
l2 
22 
s 6 
lU ' 
19 
■i- " 
73- . 
15 
' 3 : 
Ul 
p 
. 63 
• 52' 
12 
U 8 
65 
■ 23 
66, 
32 
■ 5 
31 
t . 
28 
13 
-.98 
56 
32 . 
Ui 
' § 
• 52 
'SHORT METHODS IN DIVISION 
Short methods in division can he used in some cases. The 
method is based on the same principle as shown in the short methods 
in multiplication. 
I. To divide by 10, move the decimal point one place to the 
left, prefixing a zero if necessary. 
For example: 79U- 10 = 79. U .373 -10 
.0378. 
11. To divide by 100, move the decimal point two places to the 
left, prefixing two zeros if necessary. 
For example: JgUo -r 100 = 32. U .U 37-100 - .00U37. 
111. To divide by 1000, move the decimal point three places to 
the left, prefixing three zeros if necessary. 
For example: U 2.760-9 1000 = U 2.76 37 1000 = .037. 
IV. To divide by 5, multiply by 2 and divide by 10. 
fox example: -h 5 7&5 x 1570 1570 —lO ~ 157. 
V. To divide by 25, multiply by U and divide by 100. 
For example; U 75 -r 25 " U 75 x 1900 1900-f-100 —l9. VI. To divide by 125, multiply by S< and divide by 1000. 
For example: 1750 “V 125 1750 x 8 ~ IU.OOO IU,OOO - 1000 sr lU 
Vll. To' divide by 12||, multiply by 8 and divide by 100. 
For example: 225*r12j- 225 x 8 = 1800 1800 t 100 —lB 
Vlll. To divide by 33-1/3• multiply by 3 and divide by 100. 
For example; 800 4- 33-1/3. 800 x 3 * 2UOO 2UOO-7100- 2U. 
IX. To divide by 16—2/3* multiply by 6 and divide by 100. 
For example; 250 x 6 =s 1500 1500-4- 100 s 15. 
X. To divide by 66-2/3, multiply by 3 and- divide by 200. 
For example; 12004-66-2/3 1200 x 3 = 3600 3600T* 200=- 18. 
Exercise 8 
(For Accuracy and Speed) 
Do the following divisions without pencil or paper; 
12 3 u 
a. Us-9 33-1/3 S5O -25 ' 95 -f- 25. U 75- 25 
U. 72 T 12-1/2 3Uo- 66-2/3 38 f 125 780 12-1/2 
I 
c. 375 -r 125 U2r 16-2/3 US t 12-1/2 35 T 16-2/3 
d. 360 -f 12-1/2 72t 25 90t 66-2/3 96 -25 
e. 040 - 12-1/2 320 -r 6-l/U 96 7 33-1/3 Ug -7 25 
f. l6-2/- 6UO T 12-1/2 65 -7 250' 500 - 12-1/2 ;• Exercise 9 
Home Work 
1. At $.16-2/3 a yard, how many yards of cloth can he bought for 
$1.50? For $8.00? 
2. An agent sold a for $20,000, at the rate of sl2b an acre. 
How. many acres'did he sell. 
. r , 
3. At $.25 a yard, how many yards of cloth can he bought for 
$7.25? For $37.50? For $18.75? ,F0r.572.25? 
U. .. At $.33-1/3 a dozen, how many eggs can he bought for $7.00? 
For $52.00? For $75.00? 
5. At $.66-2/3 a gross, how many gross of neris can he bought 
for $12,007 For $18.00? 
6. At the price of s,l2§- a yard, how many yards of gingham can 
you buy for $3.00? For $7.00? For $11.00? For $8.00? 
7. A new clerk finds that calico sells at cents a yard. 
He wants to work out a short method to use when finding how 
many yards of this calico are to be sold for a certain number 
of dollars. Can you suggest a method? Chaktefr if 
COMMON TRACTIONS AND DECIMALS 
, ■ iuDSETS 
V< -p'-, •V' . -.(> • *1 v. 
Budgeting simply mbans estimating in advance what you are going 
to spend of the income yotT receive. ' Budgeting means that you plan to 
set aside a certain part for necessities,., savings, recreation, etc. 
It is one of the first steps 'in the wise.,administration of personal 
finances: Income is not the only thing budgeted.' Time is being budget- 
ed more frequently now than in the past. The chart at the left pictures 
the.-distribution of a soldier's 
time at this Special Service 
School-. -It shows how the 
•hours of a single day are used. 
Fig. 1 
BUDGET OF A STUDENT’S DAY 
The whole 2d hours of a day is spoken of as a unit. A part of 
any unit is called a fraction. Thus, the fractions oFUEe 2k hour 
day are: 7/2U; 35/IUU; 7/144; 1/aU; 5/Ug, 13/Ug. 
An hour is a unit. ¥e speak of 1/2 an hour, l/k of an hour. 
A pound is a unit. Think of l/U, 1/8, 3/U, of a pound. 
The earth is a unit. We read that about 1/3 of the area of the 
earth is land and about 2/3 is water. 
Anything is a unit as a matter of fact. We may speak of part 
of a crowd, half a regiment, a tenth of a class, a twentieth of the 
population. To use such terms with exactness it would be of course 
necessary to know the accurate number of individuals in a crowd 
regiment, class, or population. 
Terms used in speaking of fractions. 
3 . • - numerator 
denominator 
-jj— is called a fraction. 
•16- The denominator,, tfills you the .number 6f equal parts into which 
the unit has been- divided. ' : ■' 
The numerator tells you how many of these equal parts are 
taken to form the fraction," Numerator means numberer. 
Proper fraction/ When the numerator is less than the 
denominator, the fraction is called a proper fraction. This 
definition may seem difficult to understand, so think of a proper fraction 
as a part of a unit. 
Improper fraction. When the numerator is greater than the de- 
nominator, the fraction is called an improper fraction. In other words 
an improper fraction is larger than a unit. 
For example: ; 
7/U , 9/3 , • IU/U , 7/7- 
Exercise 10 
( Do on Black- Board) 
1. Draw a line and divide it into 12 equal parts. What is 
each part called? Ppint out -6/h2 -of the line. .8/12 of, the. line. 
2. Draw a circle, then divide it into 8 equal parts. Point 
out -5/8 of the •circle.. 3/8 of .the circle, 
3. Draw a line and show,b£ it, ‘ ]>/h of it. 5/8 of it,' 
.U. Write five .proper fractions and tell what, the numerator and 
denominator of each indicates, 
5. Which .of the. following fractions indicates the greatest,and 
which the least value; 1/3 1/2 l/U 1/5? By properly marking 
off four lines of equal length, show that your answer is correct. 
■6. Arrange the following fractions in the order"of their value, 
placing the largest first; 2/5 2/3 2/15 2/8 2/6. 
7. Arrange the following fractions in the order of their value, 
placing the smallest first: 3/l2 7/12 5/12 9/12 8/12 1/12. 
8. If I write two fractions with equal numerators hut unequal 
denominatorss how can you tell at once which fraction has the greater 
value? Give examples. 
9. If I write two fractions with equal denominators hut unequal 
numerators, how can you tell at once which has the greater value? Give 
examples, 
10. Write three fractions, each 5 for a numerator, and 
each having a value less than 5/8. 11. Write three fractional' kkCh having 6 for a denominator, and 
each having a value greater than 2/6. 
12. ;-Write three fractions, each of which is greater in value than 
5/12 and less in value- than 5/S. 
13. Into how many equal parts is the line in Fig. 2 divided? 
What is each part called? 
Fig. 2 
HOW TO REDUCE A FRACTION TO ITS LOWEST TERMS 
Observe the line shown in Fig, 2, and compare 2/3 of it with 
S/12 of it. 
What is the result if you divide both terms of the fraction 
S/12 by U? 
Draw a line and divide it into 10 equal parts. Compare the 
length of 5/10 of the line with l/2 of it. 
f' ( * * ‘ t t ' r ’ 
What is the result if you divide'both terms of the fraction 
5/10 by 5? 
Draw two lines of equal length* -.Divide one of them into U equal 
parts, the other into S equal parts. How does the length of of 
one line compare with 6/8. of the; other?;-. 
. What is the- result, if. you multiply, both terms of the fraction 
2? . 
The exercises above.should call your attention to the following 
very important .fact; ’ * 
Multiplying or dividing both terms of a fraction by the same 
number does not change the value of the fraction. ~~ 
One 
Fig, 3 
lg. For example, let us multiply l/2 "by 2 
: -1/2x2 i'-2:xl. . 2 Ji. r- I 
2x' 2 " “5" or two jf:.... ths. 
Look at C in Fig. 
For example, let us divide 2/U by 2. 
4" " 2 = 2 I 2/U = 1/2 
We can do another problem to -prove to ourselves this fact. 
When we multiply 2/U by 2, we get U/8. Look at Fig. 3- Arc two 
l/Uths equal to four l/Sths, If we divide U/S by 2, we get two 
l/Uths. This problem- can be uroved by referring to Figure 3* Of 
course, this manner of proving may be objected to as insufficient, 
If so, you could prove the fact, by showing that actual measurement 
the l/Uth rectangle is equal to two 1/8 ths. 
Reduction to lowest terms. .when the numerator and de- 
nominator of a fraction are divided by the same number, the fraction 
is said to be reduced to lower terms. 
For example; (Look at Fig, 3» I and C) 
6/8 divided by 2 - 2 j 6/8 =' 3/U 
8/12 divided by U - 2/3 
10/12 divided by 2 = 5/6 
3/9 divided by 3 - l/3 
Raising: to higher terms. ■ . .When both the numerator and de- 
nominator Of a fraction are multiplied by the same number, the fraction 
is said.to be raised to higher terms. 
For example; (Look at Fig. B and D). 
1/2 x U - = U/g. 
1/2 is equal (in size) to four 1/8 ths. 
2/3 multiplied by 5 ~ 10/15 
19- EXMCI3E 11 
1. Tell which of the following fractions are not in lowest 
terms. Change them to lowest terms. 
_J_ _2_ _s_ 1_ _l2_ _B_ _9_ _6_ _l6_ 
U 3 10 8 15 12 15 •10 32 
2. Change each of the following fractions to lowest terms. 
1+ 6 8 20 3 12 15 
15 32 2U 25 15 16 25 
3. Change each of the following fractions to 12ths. 
To 2Uths. 
"JL 
3 6 U 3 23 
U. Change each of the following fractions to 30ths. 
JL. _1 1_ _2_ J_ J_ _L J_ J L 
2 U 5 10 U 5 10 -• 5 10 . 5 
5. Write in lowest-terms a fraction which equals each of the 
following: 
12 8 30 25 lU 32 12 15 
16 2h 50 75 70 6U 32 12 
Similar Fractions. 
Fractions that have a common denominator are called like 
fractions or similar fractions. In other words those fractions whose 
denominators are the same a~re thought of as having something in common 
namely the denominators. 
For example. 
1 *7 
—q- ancL_J„ are similar fractions. 
Least Common Denominator 
You can make unlike fractions similar by finding the least 
common denominator; that is, the smallest number that is common to 
both. 
For example. 
0■ x R 
, -f- , —, Find the least common denominator. 
3 4b You ask yourself, "What IS a ritbibkt ini6 which 3, and. 6 
will go evenly?’1 By trial and error, ydii digdbVer that 12 is such a 
number. So are }6, Ug and many more. We speak of 12 as the least 
common denominator, since 12 is the smallest (least) of 12, 2U, }6, Ug, 
Once you find the least common denominator you are ready to reduce 
these fractions to twelfths. 
_2 7_ 
3 12 
• 3 _ ?■ 
.12 
'JL r ? 
6 12 
PROBLEM. Turn to Figure 1, Page 16, and find the least common de- 
nominator of the fractions shorn in Figure 1, 
Mixed Number. 
A number that consists of an integer and a fraction is called a 
mixed number. , 
For example; 
5-1/3 is a mixture of an integer and a fraction. 
Reducing mixed numbers to improper' fractions. 
How can 5~1/3 be reduced to an improper fraction? 
Solution in detail 
i=l 
3 
’•f- 
_i§_+_i_V.IL 
3 3 3 
Quick Solution 
1 
5 - 
• V 
sx3+l l 6 
3 = 1 
21‘ EXERCISE 12 
1, Reduce the following to improper fractions. 
a. b. c. d. e. 
1. 2-1/2 3-1/3 5-1/2 **-2/3 12-1/2 
2. 66-2/3 52-1/2 9-1/5 5-7/8 16-2/3 
3. 6-1/U 37-1/2 33-1/3 7-2/5 8-3/U 
U. 7-1/U 9-2/3 10-2/5 6-8/9 U-3/U 
5. 9-1/8 U-9/10 16-1/3 20-2/5 19-3/U 
6. 100-1/3 9&-2/3 50-1/U 61-2/3 18-1/6 
.7. 68-1/9 72-1/3 **l-1/2 36-1/3 13-2/3 
8. Ul-2/5 16-1/8 12-1/2 ’ 13-l/U 16-1/7 
9. 13-2/3 16-1/5 IU-1/2 10-1/10 U-s/9 
10.,.U-7/g 3-9/U 6-U/aO 8-3/25 8-l/U 
EXEHC X3E Ijjj 
1. Reduce the following to integers or mixed numbers. 
a. b. c. d. e. 
1. 5/3 5/>+ ' 6/3 2/2 9/U 
2. 2U/8 IU/3 20/5 18/g IJB/12 
3. 32/10 13/5 20/U 18/15 37/3 
U. U7/5 18/3 16/1+ xU/3 is/5 
5. 16/3 17/5 l“+/u 13/8 16/5 
6. 30/U 50/U 100/2 96/8 78/9 
7. 200/10 U5O/U5 130/10 IU6/12 108/8 
2. 9U/U 87/6 92/8 U9/6 7U/I0 
9. 65/6 39/6 I+6/9 5Vs 37/6 
10, 16/3 28/7 91/10 810/8 67/9 You have now studied many new* Serins. Be sure you understand 
them. Study them again if necessary. 
Budgets 
Denominator 
Numerator 
Proper fraction 
Improper fraction 
■ Reduction to lowest terms 
■Raising to higher- terms 
Similar fractions 
Least common denominator 
Mixed numbers 
23- ADDITION AND SUBTRACTION OF FRACTIONS 
Two fractions may be added or.subtracted by first expressing them 
with the same denominator, then adding or subtracting their numerators, 
as the case requires. 
EXERCISE' lU 
1 -V ■ 2 .... 3 U 
a. 1/2+ 1/3 1/9 +l/10 U/5 4- 1/6 3/U + 1/3 
b. 1/5 +■ 1/6 1/8 +l/10 5/6 +7/8 2/3 + 3/U 
0. 1/5 4 1/8 1/8 -1/6 .3/7 +■ l/U 3/5+ 2/7 
d. l/U 4- 1/7 l/U -+l/3 2/5 + 1/3 5/6 + 7/8 
e. 1/6 +l/5 1/3 +3/*+ l/U + 3/8 U/9+3/;o 
f. l/U + i/s 2/5 + 1/6 1/7 + 3/6 2/15 +3/5 
g. 1/7 + 1/3 3/8+ 1/6 3/U ■+s/8 3/U +3/20 
h. 9/iU 4 17/28 H/17+-8/3U 11/15-3/5 6/8-1/3 
1. 13/15 + 13/20 5/72 + 53/56 10/11 - 3/22 7/20 -l/U 
j. 7/2U +l9/36 31/32 +U9/96 3/U 1/2 3/8-i/u 
k. 25/36 4 19/Ug U3/US+ 35/36 2/3 -1/6 3/8-1/16 
l. 11/17 4- 8/3U 16/25 + 3/50 6/7-1/9 3/16 -2/Us 
m. 22/27 -2/9 7/16—3/6U 23/ 28 5/lU 16/30+-6/7 
n. 9/10 4 3/U 1/10+1/100 1/3+2/3 6/9 4-9/18 
0. Add the fractions shown in Figure 1, Page 16. EXERCISE. 1-5 
Add th,e following’.: 
a. 
a. >2/3 
U-i/U 
5-1/8 
n. 7/8 ' 
3/>* 
1/2 
0. 3-7/8 
i)-3/U 
9-1/2 
S-3JB 
d. 5-7/10 
8-2/5 
*4-2/15 
b. 
■3/5 
2/3 
■2/5 
*3/U 
2/3 
?/!2 
5- 
U-5/S 
7-1/2 
6- 
U-2/3 
7- 
6-5/6 
c. 
' V 5 
3/1+ 
3/20 
7/8 
3/16 
9-3/^ 
5-2/3 
9 
3-5/6 
U-7/12 
6-1/9 
3-3/18 
l*—l/3 
. d. 
1/1+ 
1/3 
5/6 
3/^. 
2/3 
6/aU 
9-5/6 
, 7-2/3 
8-1/12 
7-3/16 
I+-2/S 
7-5/8 
25- MULTIPLICATION OF FRACTIONS- ■ 
In working with fractions, it is frequently.necessary to 
multiply a fraction by a fraction. For example ’in Figure 1, Page 
16, it is shown that of the day hours) is apportioned to 
mess. 
srepresents 2-1/,2 hours. As .a rule you spent about 1-lt/U hours 
a day in the mess. Since 1-l/U is one half of 2-1/2, you spent 
1/2 of of a day. How do you find what l/2 of is? 
5/Ug x 1/2 = 5/96. 
The answer 5/9& is obtained by 
multiplying the numerators to find the 
numerator of the product, and the 
denominators are multiplied to find the 
denominator of the product. 
In the example above 5/96 is already in the lowest term, as no number 
can be divided into both numerator and denominator. 
Terms 
OF. In the example l/2 of note that this means 
1/2 x 5/U8 
Cancellation :- 
When in two or more fractions to be multiplied, there are 
factors common to the numerators and the denominators, cancellation 
of these factors may be made before multiplying. Cancellation 
does not change the value of the product. 
2 
5 **l2 « 7 ,5'- Vt '7 2 
—— of —of “~~ X X —L— 
-6 35 9 -35" 9 9 
.... . - 
The numerator 5 of the first fraction, 5/6, is a factor of 
35, the denominator of the second fraction; ■ Therefore the s’s 
are eliminated by division, leaving 7 as the denominator of the 
second fraction. The 7is then cancelled against the numerator 
of the third fraction, 7/9• As 6, the denominator of the first 
fraction, is a factor of 12, the numerator of the second fraction, 
the 6»s are also eliminated, leaving 2 as the numerator of the 
second fraction. The product, therefore, is 2/9 as 2is the only 
numerator, and 9 the only denominator left. If there were two 
or more numerators left, these would be multiplied to obtain the 
numerator.of the product. The same process would apply in the 
case of two or more denominators. 
Proof JL yJU y _7_ _ 5 XI2X 7 _ Ua 
" 6 35 9 6x 35 x 9" ig<& 
2 
' 9 3XERCI3E 16 
Do the following multiplications of fractions. Make use of 
cancellation where it is possible. 
1 2 3 U 5 
a- 5 X 2/3 3/U X 7/8 5/8 X 12/25 4/9 X lb/3 z 3 
4 U x 5/6 2/3 x 7/16 3/4 x 12/15 16/35 x 27/36 20/U x 8/5 
c. 3/sxlo U/5 x 5/8 5/8 x2U■■ . 27/50 x 20/63 8/9 x 5/8 
d. 2/3 X 3/8 5/8 x IU/15 1/2 X 4/9 36 X 19/2U Us x 35/24 
e. U/sof 5/U 5/6 x 66 1/3 x 9/2 3 x U/9 8 x 3/U 
f. 5/6of 7/10 1/6 X 1/5 8/9 X 5/8 2/3 X 18/U 6 x 2/3 
g. 3/8 X 2/5 3/U X 7/8 1/2 X U/9 6/7 x IU/2 35/86 x 27/U3 
h. 3/U x 5/8 1/5 X 3/U 8/9 X 3/5 1/2 of 1/3 63/64 x 65/72 
i. 7/8 X 2/3 1/3 Xl2 2U/29 x 37/38 1/3 Of 1/6 27/56 X 6U/S1 
j. 3/5 X 5/6 2/5 x 7/8 U5/6U X 56/63 1/8 of l/U 72/91 x 53/6U 
DIVISION OF FRACTIONS 
At times it is necessary to divide one fraction "by another. For 
example, suppose you want to know how many quarters there are in $7.50. 
In other words, you wish to find out how many times l/U of a dollar goes 
into 7-1/2 dollars. 
7-1/2 -- l/U 15/2 ~ l/U 
30/U ~r l/U the denominators are now v 
the same, and can there- 
fore he discarded, the 
problem becoming 
=■ 30 ~ 1 = 30 
Also, the problem may be solved if the division fraction is 
inverted and the two fractions multiplied. This method is easier if 
the fractions do not already have a common denominator. Fundamentally 
this method is the same as that explained above but it makes unnecessary 
the finding of the common denominator. For examples ... _ 
7-1/2 = l/U - 15/2 - l/u = 15/2 X h/1 
- 60/2 
= 30 
compare with the following; 
, 7-1/2 X/U = 15/2 i/u, 
.15x2 1/U, 
2x2 ' ' 
= ( 15 x 2 ) -r (1) 
* I I - * 
= 30/1 
= 30 
EXERCISE 17 
Divide the following fractions; 
i 2 3 5 
a. 1/2 - 1/U ■ 8/25---2/5 18/J6-6/18 8/21~l*/5 U/3 3/g 
t. l/U-rl/2 9/26 -3/2 22/3U--11/17 1/3 4-9/8 5/167/S 
0. 3/1-3/it IU/39-*-7/3 27/I*s- 9/15 V 5 - 2/3 
d. U/9 -3- 2/3 6/15-2/3 6/9 -6 2/3 16-1/9 
e. 2/3 -1/6 U/10 1-U/5 **/7 -- 21/8 1/8 60 
f. 5/8 -- 2/3 7/22 -r 7/2 18/75 9/25 3/8 -r 16/U 1/3-21 
g. 1/5-r 1/2 6/32-61/2 l/3-9-6/15 100-1/5 
h. 1/5 - 1/1* 1/7 -T- 7/8 55/66-6-5/11 15/I*-U/9 1/s—l/100 
1. 5/12-5/8 l/S - 3/5 ; U/10 -.- U/5 32/3 - 6/12 7/9 U/9 
J. 9/21-2/7 1/5--1/2 6/15 -9 2/3 7/8- 8/7 1/25-30 FRACTIONAL RELATIONSHIP BETWEEN NUMBERS 
Frequently it is desired to show the fractional relationship of 
one number to another. In other words, what is the fractional re- 
lationship of one hour to hours? This is expressed as l/2*+. Thus 
1 hour is l/2U of hours. 
What is the relationship of $lO to SUo? It is expressed 
as 10/*4O ~ 1/U. 
The mental process of solution applied to problem .-number.'one 
(below) would be: 2is l/20 of because the fraction 2/^0 
reduced to its lowest terms is l/PO. 
In a like manner U is l/lO of Uo because U/Uo reduced to its 
lowest terms is l/10. 
EXERCISE IS 
What fractional part of; 
X. Uo is 2; U; 8; 10; 20 7 
2. ?U is 2; 3; 6; U; g; 12 7 
3. 72 is 2;. 3; 6; gj ■ 9; . 12; is; 2U: 36 7 
U. Us is 2; 3: u: 6; g; 12; 2U 7 
5. 52 is 2; 5: 10; 25 ? 
6. 56 is 2; U; 7: 8; lu: 28 ? 
7. SO is 2; U; 5; g; 10; l 6; 30; Uo 7 
s. 63 is 3; 7: 9: 21 ? 
Example 
If hQ miles? per hour is 2/5 of the speed of a train, the 
speed of the train may he calculated as follows: 
If 2/5 of the speed is Uo miles per hour, then 1/5 of Uo 
is Uo -r 2, or 20 miles per hour. 
If 1/5 of the speed is 20 miles per hour, then 5/5 is 
5 x 20 or 100 miles per hour. EXERCISE 19 
' . 2 . 3 
a. 27 is 3/l4 of?! ' 26 is 13/lU of? IS is 3/U of? 
b. 15 is 3/5 of? lU is 7/8 of ? is 10/11 of 
c. lb is U/5 of? 27 is 3/5 of 7 l 6 is S/9 of ? 
d. 12 is 2/3 of? 2U is 6/7 of ? a is 3/5 of ? 
e. 25 is 5/6 of? 32 is 8/9 of? 28 is IU/15 of? 
f. 30 is 6/8 of ? 15 is 5/8 of ? 6U is S/ll of 7 
g. 39 is 13/16 of? U 9 is 7/10 of ? . Ug is 6/8 of ? 
h. 33 is 3/7 of ? U 5 is 9/10 of ? ' 8U is 6/9 of ? 
i. l 6 is U/9 of ? , 2*+ is 3/7 of ? • ; l 6 is 8/11 of 7 DECIMALS 
There are two ways to express a- part of a whole, either hy 
fraction or by decimals. You have just studied fractions. We now pro- 
ceed to the study of decimals; 
For example: 
The common fraction l/lO is written as the decimal fraction .1 
The " " 7/10 « » '» ” - " ,? 
The " " 27/100 ” " ’’ » » « .27 
The ’’ " 7/100 " " " " » " .07 
In other words it is possible to express a part of a whole, namely one 
tenth, either by a fraction as l/lO or by a deciminal fraction, .1 
We can think of fractions and decimals as tools, if we like. In a 
similar fashion we can think of an axe and a saw as tools. Either can 
be used to cut a log into pieces,.. One man may use an axe very expertly 
and see no reason why he should learn, to use a saw. But his curiosity may 
lead him to attempt to use the new tool. He may become discouraged because 
he finds it difficult to handle. He is apt to say that it can’t compare 
with this axe and that the saw slows him up. Another man may look at this 
matter of a new tool in a different manner. Suppose he has witnessed an 
expert use a saw. If he thought about the difference, he probably compared 
the effects of the two tools. In some such way, perhaps unconsciously, he 
came to believe that the saw created less waste and what’s more, that it cut 
a log a lot faster than an axe. Probably some such thoughts would lead this 
second man to try to learn how to use the new tool. The hardest job would 
be learning. 
Now it is quite possible that when decimal fractions were” first tried, 
many persons had the same ideas as the first man, while still others were 
like the second man. 
New tools are constantly being invented all the time. One of the im- 
portant features of education is to learn new methods and then use them. 
Of course all new methods are not valuable. The better educated a person 
becomes the better position he is in to discover for himself what methods 
are of value. Some of us may have a great deal of education, yet fail to 
recognize useful tools from useless tools. 
When you think of it, we almost always think of a part of an inch 
as a fraction, as 3/16, l/2, l/U, 1/32. On the other band, when we 
write dollars' and cents, we write them as decimal fractions; as: 
$ 1.25 
$ 1.12 
$ 3-ul 
$ 6.0 s A little thought 6n Why we dd "this suggests several possibilities 
We are accustomedr-to speak of x3/l&-of an inch. This is easier to say 
than;one .thousand eight hundred seventy-five ten-thousandths of an inch 
(0.1875 inch). Moreover the ruler shows the inch divided into sixteenths 
and not decimal fractions. Since this is so, we naturally read off what 
we see, as 3A& for instance, and record this distance. .There might be 
much error if we relied on everyone to remember that 3A& equalled 
.01975.- A 
On the other hand,- we ’are not accustomed to think of dollars 
and cents as fractions, as 
$ 1-1/U $ 1-3/U $6-08/100 $ 3-Ul/100 
4 ' M • 
Six dollars and eight cents is a little easier to say perhaps 
than six dollars and eight one.hundredths. 
EXERCISE gO 
Head these fractions and describe the.method of writing them 
both as a common fraction and a decimal fraction. 
C.F. ■ • -D.F, 
a. three tenths 
b. . nine tenths > • 
1 —— : 
\ - .... 
c. one' tenth, . ■ t * • 
d. eight hundredths 
e. ninety nine hundredths . . ■ 
f. four tenths . , 
g. three hundredths • * ■■•.>■ 1 , . ■ 
h. three'thousandths ■ . : 
i. three ten thousandths i__ . 
j. sixteen hundred thousandths 
k. two thousand seventy-five , 
hundred- thousandths 
Terms 
When we speak of three "places” or four "places" etc., we refer 
to the actual number of integers to the right of the decimal point. 
•32> Designate the number of "places” In each example below; 
Humber of places 
a. U. 6713 
I. 36.75 
c. 1U.16 . : ■' 
d. 3.1U16 
e. .007 
f. .175 ' 
g. .00001 i‘;;- 
h. 13.671 
i. 16.082 
J. , 
k. 7.UU 
l. 17.7U6781 
LXSRCI bE 22 
Write the common fraction, -the decimal fraction, and the number of 
"places" of each of the following; 
C.P. D.P.' • No .Places 
a. eight hundredths __ 
h. four thousandths - ' - __ 
c. two tenths •' . _i. r 
d. nine tenths • ' - ■ ’• •’: ; 
e. three ten thousandths • 
f. eight hundred seventy-three 
thousandths 
g. seventy-five hundred 
thousandths 
•33- EXERCISE 22 (Continued) 
C.3T. D.P. No. Places 
. . . ( . s • f ? . ~ 
h. two hundred fifteen 
ten thousandths • • 
i. twenty-seven thousandths 
j. nine thousand one hundred 
eighty-two ten thousandths 
k. nine tenths 
l. twelve hundredths 
m. seven thousand one 
hundred thirty-nine 
ten thousandths 
n. seven thousand one hundred 
thirty-nine hundred 
thousandths 
o. two thousand seventy- 
eight ten thousandths 
p. three hundred twenty-one 
thousandths 
ADDITION OF DECIMAL FRACTIONS 
When decimal fractions are added, the decimal points appear in 
a vertical (straight up and down) column, so that tenth?,hundredths, 
and so on are in the same line. 
Before adding any decimal fractions, it 
is a good practice to first write the 
decimal point below the line at the spot 
where it should be. It should be directly 
beneath the decimal points above the line 
In the example showp, this place is where 
the point of the arrow is. 
ig!u 
LI 
■3^ EXERCISE 23 
Add the following: 
(1) 
a. .7 
__i3 
■fa. .U 
1.2 
c. .07 
.g 9 
d. 2.1 
2.01 
e. .723 
.12 
(2) 
.U 
.6 
U.i 
3.2 
.27 
.62 
3.001 
b.oio 
.902 
.4 
(3) 
.6 
.3 
6.s 
__9A_ 
A 
.321 
.09 
.3 
.8 
6.7 
8.2 
JL 
6.82 
9.QQ 
.9001 
.012 
(5) 
.6 
.7 
9.1 
g.oi 
.02 ' 
.002 
16.27 
U. 72 
.85 
• 9 
SUBTRACTION OF DECIMAL FRACTIONS 
When one decimal fraction is subtracted from another, the figures 
should he so placed that tenths will he under tenths, hundredths under 
hundredths, and so on. The decimal points will thus appear in a 
cvertical column. 
Example: Subtract . Ugy from ,5 
]Hs7 
In this incident, ciphers to the 
right of .5 are assumed. That is ,5 can 
he visualized as .500 The addition of 
the ciphers have not altered the value of 
.5 
.013 EXERCISE 2k 
Find the difference by subtraction: 
(1) 
a. .25 • 
.125- 
b. .796 
.906 
C. .7 
• 235 
d. .16 
.1236 
e. .27 
■ 167 
(2) 
.2 
.0001 
.5019 
■’o2ol 
• 3,25 
.0^75 
.271 
,OU6 
.1U69 
.1236 
(U) 
.IW+U9 
■12315 
■76552 
.^9736 
.21359 
.09078 
.9201 
.009 
.Ss^ 
.732 
(3) 
•I 
• 699 
.001 
.0001 
.9 SOI 
.5766 
.u 
.003 
.67 
.059 
SUBTRACTION OF MIXED DECIMALS 
The process of subtracting mixed decimals is the same as that 
of subtracting decimal fractions. The decimal points must be placed 
in a vertical column; with the whole number values on the left and the 
fractional value on the right. 
EXERCISE 23 
Subtract the following: 
(1) 
f 
a. 1000,0001 
ia 
b. 2U3.66 
_12§,35 
(2) 
5138. 
2109.0098 
1^3.7786 
(3) 
256.303 
199-5^5 
U 05.001 
31.786 
36- MULTIPLICATION OF DECIMALS 
If you bought .7 lbs. of candy at $.90 a lb., you would 
multiply .7 x .90 to obtain $,63, the cost. 
Hie process used in multiplying decimal fractions is the same 
as that used with whol'p numbers. The fractional value of the product 
is indicated by showing as many figures at the right of t’he decimal 
point as there ar^-figures at the right of the decimal points-,in both 
the multiplier and the multiplicand. 
The illustration below shows the process of multiplying 
.671 by .015; and ;OU7 by .35 .. 
.6yi 
.015 
3355 
_i>7i 
.010065 
The- solution at the left 
showsJ the multiplication of 
two decimals that have the 
same denominator (1,000); 
the one at the. right, the 
multiplication of the 
•■..oUy 
13,5 
• 235 
iki 
.016U3 
decimals that have different 
denominators. Observe that ih each answer there 
are as many figures at the right of the decimal point as there are in. 
"both the multiplier and the multiplicand. If there are not enough 
f'gures in the product, additional ciphers must be prefixed so that the 
decimal point may appear in the correct position. 
EXERCISE P 6 ' 
Do the following multiplication of decimals: 
(1) (2) 
a. .0725 x .072 .3765 x .012 
■x ' 
■b. .0075 x .005 .009965 x. 9 
c. .2653 x .325 * x .87 
d. .009 x .008 .. . . .. . .2653 x .325 
e. .6562 x .59S .2009 x .675 
f. .3671 X .673 X ,899 
■37- SXERCI'SE 26 * (Continued) ■ 
(1) ' ' . , (2) 
g. U9BI X .9009 :iv 337 x .IUI6 
h. 762 X. 635 , 225 X .13 ■. ' ■ 
i. 377 U X .IOUS , 6725 x .7631 
j. X .1U92 , .. 1883 x .005, 
k. 377 U X .IOUS U 752 X .367 
l. 9.887 x 100.6 ' • ■ 321.175 X 17U.25S 
m. 35.1+2 x 625. S 75.390 S x 21U.U32 
n. 18.6 x 3.1U16 625.107 x U. 00807 
0+ 763.25 x 221U.6 * 876.01 x 6.00071 
p. 1+563.2 x 329.1+3 1967.27 x 13U.67 
q. x 1.231+5 87.67 x 87.67 
Multiplication of decimals by 10 or multiples of 10, as 
10, 100, 1000, 10,000., and so on. 
Example; 
3.7 
10 
37.0 
By multiplying By 10 the decimal in 
the multiplicand is moved one 
place to the right. 
3-7 
100 
370.0 
By multiplying by 100 the decimal 
in the multiplicand is 
. moved two places to the right. 
In other words, we can say that for every cipher in the multiplier 
10j 100, 1000 and so on, the decimal is moved one place to the right. EXERCISE 2? 
At sight multiply the following by 10, 100, and 1000 
(1) (2) (3) (U) (5) 
a. . .lU 6.7 .27 0.02 .01 
b. .8 .13 .02 .001 .0276 
c. .017 .267 .001 .067 .035 
d. .U 32 .067 .321 .UBl .867 
e. .9 .82 .071 .67 .031 
f. .5U .05U .09 .371 .0001 
EXERCISE 28 
Multiply the following by 
(1) (2) (3) W (?) 
a. 10 x $,Ol $.OOl 1 mil 6 mils U,6 
h. 1000 x $.Ol $.003' 3 ®ils .5 mils 3»52 
c. 100 x 1 mgm 10 mgm 1000 mgm .5 mm .06 mgm 
d. 10 x Imm 65 mm .03 mm 1 inch $.03 
e. 1000 x 1 mil .1 mil lU*3 6,01 .001 
f. 100 x 2.<p 10$ 6 mils .006 .0321 
g. 1000. x ,02*46 .78U10 7.6U3 267.1 .00001 
h. 10,000 x .0067 .37U2 DIVISIOIi OF dDcMals 
The process of dividing decimal fractions is the same as that 
used with whole numbers. The position of the decimal point in the 
quotient is determined by the number of decimal' places in the 
dividend and the divisor. The number of places in the quotient is 
equal to the number of places in the dividend minus the number in the 
divisor. 
Example; 
_ -390 U + 
.315 r;i23oooo 
9% 
2550 
2835 
1500 
1360 
2*40 
f 1230000 = 7 places in dividend 
.315- 
-3 places in divisor 
U places in quotient 
There is another way to determine where the decimal point goes. 
Example! 
• 315 ('.'1230000 
When the decimal point is moved to the 
right, the number is multiplied by 10, 
or 100, or 1000 and so on, depending 
how far the point is moved. 
.315 x iooo =• 315 
To make .315 a whole number, move 
decimal point three places to the 
right, which is the same as multiplying 
by 1000. 
.123000 x 1000 
•315 X 1000 
If the divisor is multiplied by 1000, 
the dividend must be multiplied by 
1000 also. 
.1230000 x 1000 = 123.0000 
.315 x 1000 = 315 3 
54~5 ' 
*3 ' 
315 fT23.00 • * . 
5 
X39OU + 
315 (TaTIOOfT 
gU 5 
2 350 
2 535 
T5OO- 
- 
Po~ 
The’problem'then resolves itself into 
dividing 315 dnto 123* The rule to follow 
is this; puVthe first figure of the 
quotient (3) over -the last figure of the 
product (5) thsit results from multiplying 
the divisor (.315) "by the first quotient 
figure (3). Next place the decimal point 
directly above the decimal point in the 
dividend. In this case, an xis placed 
directly above the decimal point, so as 
to clearly indicate the location, • 
Then proceed with the problem.,keeping the 
decimal in place. 
Example; 
Divide .012 by 6 
6 fTUI2 
The process is the same. 
2 
6 nm 2 
12 
The first figure of the quotient goes 
over the last figure of the product,l2. 
X 2 
6 prs 
The decimal point in the quotient goes 
over the decimal point in the dividend. 
Its location is marked hy x. 
.002 
6 f^lt 
The gap hetween the decimal point is 
filled in with ciphers, one over each 
figure in the dividend until 2 is reached. . 
In this case there are two ciphers. 
.002 x 6 - ,012 Proof, 
quotient x divisor dividend 
Example; 
Divide .UIU Dy .0375 
i"”*' 
.0375 ./ 
373. 
This problem differs from the other 
examples, in that there are more places 
in the divisor than in the dividend. You 
can go no■further, as there are no more 
figures to "bring down". 
■Ul; IIOU 
.0375 ( .UII4OOO 
375 
390 
375 
1500 
1500 
In order to carry the division out 
further, you add as many ciphers to 
the right of the.last figures, as 
may he needed.. As la rule quotients 
are "carried out"\t9'two "places". 
.kIUOQO - 6 places 
.0375- = U places 
2 places in quotient 
11. oU 
.0375 POTOOO 
EXERCISE 29 
Do the following divisions: 
(l) . (2) • (3) 
a. .35UU-.22 .21 -.000? .26 -f .0076 
b. .2096 -.o*+ ■ ,U 5 -f.1312 .22U5 -- .17 
0. .7853.37 .l+2 .0006 
d. .1+ 4- .2 l+oo 9- .002 36.7 -7- .72 
e. U 6- .2 1+.6 t .016 8.9 -f- .0162 
f. 1+ -V- .02,; 37<2 -r .026 .92 - .1+72 
g. 76*42 - .21 .6875 - .I+3 .1+756 - 1137 
h. 5375 - .55 .31+13 -6- 13 .5837 + 21137 
1. 981*7 -r .87 .8009-r 11 .6S*+7 U5B 
3. 2781.1+ t 76.398 6U3.1U-6 76.371+ 
k. 13.1+87367 23.67 25.761+3 
l. 365.6 -r- 121.9 1+3.006 r 2.51+67 23.1+73 -r'1+7.362 DIVISION OF DECIMALS BY TEN AND MULTIPLES OF 10. as 10.100, 
1000 and so on. 
The process is the exact -reverse of- multiplication by 10 and 
its multiples. The decimal point is moved to the left one place 
for each cipher in the divisor. 
Examples; 
3.7 -T 10 - .37 Move decimal one place to the left. 
3.7 4- 100 - .037 Move decimal two places to left. 
3.7 ~r 1000 - .0037 Move decimal three places to left. 
EXERCISE 30 
Divide the following-by: 
(1) (2) (3) W (5) 
a. 10 1U6.2 13 167 .01 .1 
Id. 100 16.2 2 .08 .010 6.U 
c. 1000 .002 2UI 67.8 $lOOO .$lO, 
d. 10 S3UI l6 mm 381 ragn Ul mils 
e. ' 100 ' 271 lbs. 3.21 6 gras 3 gr. ' oz. 
f. . 1000 '• 500 nils. 65 grn • 87 mm 2.86 ' $1U,2l 
■g. ■ 10 $.35 100 ran 125 m 6.U gr. 68 mgn 
h. 100 .0U gm .068 gr. .92 oz. 16.6 oz. 1.^2 EXERCISE 31 
Reading the following decimal fractions: 
(1) (2) (3) 
a. . U 25 .7*485 ,1U166 
b. .027*4 .3002*4 .7U675 
c. .765*49 .3781 .32681 
a.. .79U .000*48 .*426*43 
e. .00076 .1408 .67677 
f. .37s .0706 .92*431 
g. .5*46 .62*43 .16812 
h. .037 -3701 .000U6 
i. .165 .002 U .10078 
j. .67*4 .0635 .7**65*4 
k. .875 .0785 .92*465' 
l. .. .926 .87365 
Reading to the nearest hundredths, thousandths, ten thousandths, 
and so on. 
Example; 
Read to the nearest hundredths. The question to 
ask yourself is whether is nearer to being ,79 or .80; 
.79U lacks ,00U from being .79; it also lacks .006 from 
being .80 - Hence the nearest hundredth is ,79 
Example; 
Read .U 25 to nearest hundredth. Is nearer .U 3 or ,U 27 
Since .U 25 ends in 5» it is obvious that 5 is midway 
between .U3O and .U2O - When this is the case the rule 
,is that the higher number is taken as the nearest hundredth 
In this case the answer is .U 3 Reading to nearest thousandths 
Example;; 
Head ,■ to the -nearest .thousandth. The choice is . 
between .300 and .301 The answer is determined by noting 
whether .the figure to the Jright of the thousandth place is 
below five or five or above five. The figure is 2. Hence 
the nearest thousandth This can be read as .3 of 
course, since 300/1000 after cancellation is really 3/10. 
EXERCISE 32 
'•Refer to the last exercise. Read the decimals in column l,to the 
nearest hundredth; those in column 2, to the nearest thousandth; and 
tho.se in- column 3». to the nearest ten-thousandths. 
REVIEW 
01 
FRACTIONS m) DECIMALS 
(!) (2) (3) (U). 
a. • 5 [735“ 7 fITW s 13.0 9 (.^257 
t. ■ru fs.lir- 5 jTSo g [To .5 j 1.U75 
0. 5 ITao 5 (0.025 upTo 7 ]”.'063t~ 
s 1 . 
d. 3 I .14227 9 fg.37 3 (52.02 7 
e. 6' 13.0U5~ g J .0716 6 ( 32.1406 9 11.917 
(1) (2) (3) 
f. 6U.U32-- 16 7.686 105 .1161 -6- 27 
g. "2614.312-36 I2UU.U 307 7*43-986 -r S 2 
h. 668.368 -t 7*4 .63 -r 15 16.215 ~t* 23 
-U5- (1) (2) (3) 
i. 88.32 23 , ■' . 3.6851 T- U 3 .0155*+ -r- 37 
j. 91.205 4- 85 .055188 -7" IU6 33.626 -r U 3 
k. .1081 -T 2.3 .06336 4- 1.32 • 378 2.7 
i. 1.3685 U 4 .18 .01625 - .125 .001216 -r .03s 
Oral 
(1) (2) (3) (U) 
m. .U -f .2 iCM-r .2 .OOU -f .2 .OOU -f- .02 
n. .OU .02 -f 1.5 -r .03 . 0036-r .006 
0. .01U4-.2 8.U32 -r OU .015 -f .05 378 2.7 
Convert to Decimal Fractions 
p. 1/2 1/3 i/*+ 1/5 
q. 1/6 :1/10 2/3 3/8 
r. '7/8 ■ .5/6 3/1+ 1/8 
S. 7/16 15/32 3/25 1/50 
t. 7/15 3/7 '■ 1/15 1/11 
U. 8/9 -4- u/3 11/12 - 5/8 7/9 - 3/5 4- 2/3 
v. U/5 x 5/U 3/3 x 5/U ' 3/3 x 3/3 7/6 - U/5 
w. 7/16-9-7/16 9-1/3-3-1-1/16 ■ g-U/5 -f 1-1/20 5/6 -V-3/U 
x. 1/5 + 1/8 9/16x16/9 6-l/Ux 12-1/2 37-1/2662-1/2 
y. 5/6 X 3/U 7/8 of 2/3 3/5 of 5/16 7/8 of 5/6 
Z. 7-1/2 9-1-1/1+ 9-1/24-7-3/5 5/16-1/5 3/^-2/3 Second Series" 
A. Convert the following to common fractions and add 
• 3 
.15 
.275 
.250 
.0250 
B. Arrange the following fractions in order of size, placing 
the largest first; 
.09 .079SU .102 U .03726 .061 
C. If a Bushel of potatoes contains U 7. pounds of water, what is 
the weight of the wdter in a load containing 32.5 bushels? 
D. A train ran 67.75 miles an hour for 2.U hours. How many miles 
did ig run's 
E. Ahoy tcnk 184 steps, each of which was 2.75 feet long. How 
far did the boy walk? 
F. A field containing 17.375 acres yielded 22. U bushels of wheat 
per aero, What was the total yield? 
Cf. A pound of beef contains .166 pounds of fat. How much fat in a 
U. 5 pound roars dr 
H. If a gallon 01 milk weighs 8.6 pounds, what is the weight of the 
milk that may be put into a tank with a capacity of 11U.5 gallons? 
I. The number cj, Inches or rainfall in a city for each of the first 
five iiunj.hr of. the year was as follows; 4.3, U. 7, 3,6, 5-2, 6.1. 
What was the average rainfall for the five months? 
J. A gallon. 01 water weigh*' 2 3 pounds. What is the weight of the 
water in a tank that contains 1U6.25 gallons? 
K. A field of ac'-cs yielded USI-7 bushels of wheat. What was 
the average yield per- core? 
L. A cubic foot of water weighs 62.5 pounds, and a cubic foot of gold 
weighs 1.9.3 times as men. What would be the weight of a cubic foot 
of geld.? Of a cubic inch of gold? 
M. Find the cost of pounds of pork at $9»25 per 100 pounds. 
IT Find the cost of U?5O envelopes at $2,60 per 1000. O. Find the cost of "bricks at per 1000. 
P. Mrs. Johnson paid $3.50 for a turkey which weighed 1P.3 pounds. 
What was.the price per pound? 
Q. The bank deposits and withdrawals of a merchant during a week 
were as follows; 
D epo Sits Withdrawals 
Monday $ $ 18U.30 
Tuesday 347-85 
Wednesday 792.60 U 73.85 
Thur sday 247. S 2 143,60 
Friday. U27.1S 231.^2 
Saturday 527.2>6 ’ 321.^3 
What was the difference between the deposits and the with- 
drawals for the week? 
R. Mr. Adams paid for an automobile tire. At the end of 
6lUp miles the tire was worthless. What was the cost per mile 
for this tire? 
S. The area, of Texas is 265,896 square miles. The area of 
Rhode Island is square miles. How many times is the area 
of Rhode Island contained in the area of Texas? 
T. A boy raised 62.5 bushels of corn on .75 acres. Find the 
yield per acre. 
U. Mr. Adams took a. motor trip of IUU.9 miles. During the first 
hour he traveled'2^.3 mi,, during the second, 23.9 mi., 
during the third hour, 26,8 mi. At what average speed per 
hour was it necessary to travel to complete the trip in the 
next 3 hours? CHAPTER Y 
HOW TO SOLVE A PROBLEM 
You have now studied addition, subtraction, multindication, 
and division of integers, of common fractions, and of decimal 
fractions. You are familiar with the measurement tables in common 
use in this country,and you know some of the practical short methods 
of computation. You can compute with some degree of speed and ac- 
curacy. It is now desirable that you learn further how to apply this 
knowledge to the solution of problems that arise in the Medical Corps. 
This is the true test of your knowledge of arithmetic. 
It is important that you develop the ability to analyze 
problems. Those whose daily work requires the solution of many pro-• 
blems must always think out the analyses, even though they do not 
write than out. In the problems that follow throughout this book you 
are asked to think out the analysis of each, and often to write it out. 
Such work will give you needed practice that will help you to solve the 
problems that arise when you later enter upon your household duties, 
your trade, or your profession. 
Other things besides analysis are necessary if you are to solve 
the problems you meet. Your computations must be rapid and accurate; 
otherwise your work is of no value. Last-, but very important, your 
figures must be neatly made and legible. Figures made carelessly cause 
a loss of time and money to yourself and to your employer. You may be 
willing to stand that loss, but you may be sure you will not find an 
employer who is willing to do so. 
Below you will find a suggested method for attacking, that is, 
for analyzing a problem. You should study this carefully, and learn how 
to apply the method to other problems. Although there is no one best form 
of analysis, the one given here is a very good guide as to what an analysis 
should be. Study the statement of each problem thoroughly, think the 
solution through carefully, and do as much of the work as you can without 
pencil or paper. At the beginning, it is profitable to indicate the solu- 
tion of every problem before you make your computations. 
To Solve A Problem in Arithmetic 
I. Read the problem carefully so that you may understand 
a. Tiie facts that are stated or suggested. 
b. What you are to find. 
11. Decide upon the operations that are to be used. 
111. Make the necessary computations accurately. 
IT. Check the result. 
-U9- THE SOLUTION OF A PROBLM 
Find the cost of U-3/U yards- of. cloth at 56. cents a yard. 
I. (a) The facts stated; The number of yards of cloth bought. 
. .. The price per yard. 
(b) What you are to find; The total cost of the cloth. 
11. The operation to be used; Multiplication. 
111. Computation: U-3 /Us 19/U 
19/U x $.56 $2.66 
IV. Check the result: ... Do this by making the computation 
a second time. 
THE SOLUTION OF A ■ SECOND PROBLEM 
When a dozen lemons sell for cents, how many can he bought 
for 36 cents? 
I. (a) The facts stated; The selling price per dozen. 
(h) What you are to find; The number that can be bought 
for 36 cents. 
11. The operation to be used; Division. 
111. Computation; l/l2 of ~ s.o^s,the cost of 
1 lemon 
$.36 -7 =s St the number bought. 
IV. Check the result; Do this by making the computation 
... a second time. 
The way to solve such a problem is based on analysis or 
finding out the cost of one lemon. Since 12 lemons cost cents, 
one lemon costs 12 into cents or l/l2 of which is s.o^s. 
One more step is necessary to get the answer. Further analysis is 
needed. You have 36 cents. One lemon costs (U-l/2 cents). 
How many lemons can you buy? In other words, how many times will 
cents go into 36 cents? 
Note to Pupil; In most business transactions it is customary 
to consider fractions of a cent amounting to 1/2 or more as-an extra 
cent. Thus, s’Usj- is considered as S.U6. If the fraction is less 
than 1/2, it is usually discarded in the final result. THE SOLUTION OF A THIRD PROBLEM 
If 12 small radio sets cost a merchant $1.68, what will ~ 
8 sets cost him at the same rate? 
Solution: 
r= cost of one radio SIU 
SIU x 8 =» cost of 8 radio sets. 
By the use of cancellation, this, problem may be solved, 
thereby saving time. 
* ;-Ik 
= $ll2, cost of S sets. 
•56 2 
3.68 x g - |ll2 
, .3 ; - •' 
1’ ■ " 
This is an example of cancellation 
in which different figures happen- 
ed to he cancelled. 
EXERCISE 32 
In solving the following problems, use the method of analysis 
suggested on Pages 50 &51 ♦ Use cancellation whenever possible. 
1. If a farm containing acres is worth,s32,'*4oo, how much 
should be paid for 75 acres' at the same rate? 
Solution; 
$32. to) x 75 _ . 
?M0 ’ " 
2. The daily noon for a week were: .58°, 5U0, 
6l0 , 53° » 56° , and 5U0. ' What the average for the week? 
3. A farmer traded ,IS dozen eggs-* at-'2B .cents a dozen for 
sugar at 9 cents a pound. How many pounds of sugar did he receive? U. In a laundry, burners, each using 7 cubic feet of gas per 
hour, burn 7 hours a day for 6'days., What will be the gas bill at 
90 cents a thousand cubic feet? 
5. The'circumference of a driving wheel•of a Tbcbmotive is 22 
feet. How many revolutions will this wheel make when' the locomotive 
goes a mile? 
6. How far will the locomotive mentioned in Problem 5 travel when 
a driving wheel makes 1000 .revolutions? ■ 
7. When sound travels 1090 feet in a second, how far will it 
travel in lU seconds? , ; v - 
8. At the rate given in Problem J, how long will it take the 
report of a-rifle to .travel 1U,170 feet? 
9. If it takes light about U9B seconds to come from the sun to 
the earth, at the rate of 186,300 miles a second, what is the dis- 
tance from the earth to the sun? „ 
10. In a recent year bales of raw cotton were produced 
in the United States. If a bale of cotton weighs 500 pounds, what 
was the production in pounds? 
11. The receipts of a store for a week wore: Monday $268.h0; 
Tuesday, $313.92; Wednesday, $28U.63; Thursday, s3^7-2d; Friday, 
$318.69; Saturday, $U23.72. ; 'What were the average daily receipts? 
12. Harry's mother bought a table for She paid $9 cash and 
the remainder at the rate of $2.50 a week. How long did it take her 
to pay the bill in full? 
13. If your pulse beats 72 times per minute, how many times does 
it beat in a day? In a year? 
IU. How many bushels of'She! led1 corn are in,a load containing 
2352 pounds? ( A-bushel of shelled corn weighs 56 pounds.) 
15. When children's stockings sell for 66-2/3 cents a pair, how 
many pairs can you buy for SU.? 
16. An automobile traveled a mile in 52 seconds. Wnat was the 
speed per hour? 
17. Find the number of seconds in years. 
18. It is thought that the first life to appear on the earth 
occurred about 1„335,000,000 -..years ago; and thatvman-first appeared 
'about sP,ooo,ooo,,years ago.. Man's tipie is what fraction of the 
time that life has been present. In order to obtain a simple 
fraction for an answer, you must approximate. 19. If one gram of uranium generates in one year 
0.000,000,000,125 grams of lead, how many grams of uranium 
would generate 1 gram of lead in one year? 
20* A man skates UUo yards in seconds. Find his 
speed in miles per hour. 
,21. A man runs ‘UUo yards in U6.U seconds. Find his 
speed per hour. 
22. Man 01 War ran ]>/h of a mile in 1 minute, 12 
seconds. Find speed per hour. 
23- Mr. Brown in 1930 walked and'‘ran 26 miles in 
2:2? 1 29*6 Find speed per hour. CHAPTER VI 
Metric System. , - 
Mathematics ‘is a language, though we hardly ever think of it 
as such. You have been-hearing many new words in your classes. A 
.. •11' 
dictionary will tell you the meaning of such words as calorie, infection 
.'* * . > 
‘ # . ' * * * ’ ‘ ■* 
pathogenic, ulna, shock and so on. .... , ' 
Tables, .when, we refer ..to them., tell us the, meaning of various 
words used in the language of mathematics. A foot, we find is 12’ 
inches; a meter is 39*37 inches. ( > ) means greater than; 
( )means less than; ( =■ ) means equal, 
. There was a time long ago, before integers were in use,when 
the first eight numbers were represented by combinations of horizontal 
bars. Thus' stood for 3* stood for 2. The way to 
write 32 would be 
All this was very complicated before a way was found to 
represent zero. Without stome such means how could one write 320 , 302, 
3,200? Before the introduction of the 0, had become .Z 
and had become Z. the original form of our 3 and 2. 
Furthermore, there was a time, long ago, when fractions were 
unknown. For instance, as time went on, it was found that all measure- 
ments nr all weights could not be expressed accurately by whole numbers 
alone. A piece of cloth U—l/3 ycards long could not be accurately 
measured by a 1 yard rule. It was found that the cloth was more than 
four yards and less than five yards. When, in time, man found that he 
could not measure out exactly a pound of flour, he divided it into 
ounces. Likewise, he divided a yard into feet; and later, feet into 
inches. Instead of developing fractions, smaller units were invented 
as an answer to the problem of measuring parts of a whole. Later,.fractions cane mnre;,.:intc> use.' Thus three, fourths was 
written -3/H* minutes came to,'.be spoken of as of an-hour, o h 
Complications arose in some .-instances .. ;At one time 16-2/3 digits 
equalled a foot. This, of course, -created an .inconvenience when one 
wished to. express ]>/k of a foot in a fraction. To get around this 
difficulty a new digit was formed so - that .each digit equalled l/l6 of 
a foot. (There is considerable belief that this is true, but there is 
no proof.) Certainly an inch that is made up of l 6 parts, is more 
easily divided into smaller fractions of an inch, than would be the case 
if an inch had 12, or 10, or 6 parts. 
..Express l/B*of an inch that has twelve equal parts. What pro-. 
blem does this raise? ■ , - • 
There is another factor in the history of weights and measures. 
An example is that of troy weight, - the weight commonly, used by jewelers. 
During the Bth and 9th centuries, great fairs were held at several Trench 
villages, • including TROYE3. Here came also traders from all countries. 
In time.,it was discovered that the coins were so mutilated that it became 
accessary to .sell coins by weight. - The. standard weight that the village 
of Troyes utilized, was later adopted for precious metal .'(as gold) and 
medicines in all parts of Europe., The troy ounce and the1 avoirdupois 
ounce were originally intended to have the same .weight, ..but after the 
revision in Troyes it was found that the avoirdupois ounce was lighter 
by U2-1/2 grains than the troy ounce. As q result of this development, 
today 
the apothecaries of England/must buy medicine by the .avoirdupois system 
and compound (mix) them by the apothecaries or troy system. 
In 1266, Henry 111 of England decreed that "an English silver 
* 3> ' 
penny called the sterling, round and without clipping, shall weigh 
thirty-two grains of wheat, well dried and gathered out of the middle 
of the ear". In England there continued to develop modifications of the 
basic measures, the pound and the yard. We can indicate some of the 
units under the following headings: 
Weights - ton, hundredweight,stone,pound, ounce. 
Length - mile, furlong, rod, chain, yard, foot, inch. 
Capacity - gallon, quart, pint, ounce. 
quarter, bushel, peck, quart, pint. 
, ' All civilised nations have had experience similar to that of 
the English. Over a period of several thousand years, the various 
values of weights and measures have changed. The changes of course 
lead to confusion, an example of which is shorn by the difference be- 
tween a Troy ounce and an avoirdupois ounce. 
What Henry 111 did, of course, was to establish a standard, 
which was what we call a rough standard, since well dried grains of 
wheat are not the same in weight, although they may be very nearly 
exact. A grain may be lighter than another by .01 grain, which is a 
small difference. But think what a difference would be made by 32 
grains? Let's see: 32 x ,01 —.32 grains. That means that one silver 
penny (32 grains) would be too light in weight by ,32 grain. Thus the 
king attempted to create a standard by which disputes might be settled 
and accuracy maintained. 
In time there arose new troubles. When the first bacteria 
were seen under a microscope, the language of mathematics was at a loss 
L ‘ - v'' o . •, : >.■ '*> * ,fc 
to tell of their size. How was one to express the diameter of a human 
red blood cell? Of course, the diameter could be expressed as 
inch. That was one way. Another method was to divise new and 
12f,000- ■ , . • •. t 
smaller units, as had been done for the foot and other weights and 
measures. There was a second possible means, one that lias been -thought 
* - • ■ • - •’ . , .i. ..*j, ‘ 
of before. Instead of having an inch divided into sixteenths; a foot 
divided into twelfths; yards into thirds; miles into rods; instead of 
.■'•i V :* ; 
having land miles and nautical miles; troy ounces and avoirdupois ounces; 
instead of all this why not try to relate in some way all these weights 
and measures? Instead of thirds, twelfths, sixteenths, and so on, why 
not base a new system on 10, 100, and 1000? 
By 1800, the French had introduced and adopted a new system, The 
Metric System which is widely used throughout the world today. The unit 
of length is the meter, which until very recently has been considered one 
,orty Mllionth. P„. < ,t B. 
earth around'the poles, and'is equivalent to 39*37 inches. 
The unit of volume or capacity is the liter, which is the 
volume of the cubic decimeter. It is also defined as the volume occupied 
by the ma.ss of 1 kilogram of distilled water at its maximum density (U° C) 
It is equivalent to pints. 
The unit of weight is the gram, which is the weight of one milli- 
liter of distilled water at its maximum density (U° C). It is equivalent 
to grains. 
The unit of microscopic measurement is the micromillimeter, or 
micron, which is a thousandth part of a millimeter. ■- • 
In Paris, France, there are preserved (at least up to the time of 
World War II) the models or standards of the meter and kilogram from which 
r. : _ .■ I- ' ’ ’ 
all prototypes are made. 
The United States prototypes, made of a platinum-iridium alloy, 
are kept in the Bureau of Standards at Washington, D. C. and are used for 
standardizing all weights and measures used in the United States, The denominations of this system are multiplied by the Greek 
words deka, meaning ten; hecto, meaning hundred; and kilo, meaning 
thousand, and are divided by the Latin words deed; meaning one tenth; 
centl, meaning one hundredth; and milli, meaning one thousandth. 
i deka 
i 
| hecto 
kilo 
deci 
cent! 
’'mill! 
. ■ . 
i ten 
j hundred 
thousand 
tenth 
hundredth 
one 
thousandth 
10 
i 100' 
1000 
1 
1 
1 • 
I i 
1 
• 
1 
10 
100 
1000 
10 
! ioo 
i 
1000 
0.1 
0.01 
0.001 
’’meter11 for length Micron rr .001 mm 
’’gram” for weight 
1 ’’liter” for capacity or volume 
Common Abbreviations; 
m - meter 
mm - millimeter 
gm - gram 
• • < •• ragm - ■ milligram : 
I.' - liter 
c.c.- milliliter 
». i ’ * , 
To return a moment to the problem of expressing the size 
■ . • 36 
of the red blood cell. Our choice is between 127,000 inch or 7*2 
microns. The new word is micron. It can be looked up in mathematical 
tables if you forget its meaning ; just as you can look up the meaning 
of a new word heard in class or read in a book, 7.2 microns - .0072 mm THE METER 
Fig. 3 
Comparison of the English and metric systems of linear measure 
The metric scale shown above represents 12 and a fraction centi- 
meters (cm). Each centimeter is divided into 10 millimeters., A deci- 
meter is 1/10 of a meter; it is also another name for 10 cm. 
When the meter and yard were compared it was found that: 
One meter - 39*37 inches. 
The meter is divided into: 
Tenths, called decimeters (dm.) 
Hundredths, called centimeters (cm) 
Thousandths, called millimeters (ram) 
The multiples of a meter are; 
10 meters, called one decameter (Dm.) 
100 meters,called one hectometer (hm.) 
1000 meters,called one kilometer (km.) 
10,000 meters,called one myriameter (Mm) EXERCISE 33 
( Do in Class) 
1. One cm. equals how many inches? 
2. One mm. equals how many inches? 
3. One decimeter equals how many inches? 
U. One kilometer equals how many miles? 
5. One inch equals how many cm? 
6. Express your height in meters. Carry to three decimal places. 
7. The Olympic races use the Metric System of weights and 
measures. Convert: ' ■ .. ; - 
100 meters to yards. 
UUo yards to meters, 
1 mil© to meters. 
g. In Europe the Metric System is used almost everywhere. Convert 
to Metric System: 
U 6 miles per hour 
3SU miles per hour 
100 miles per hour 
9. A gun has a 75 nun. diameter. Express in inches. 
10. Twelve inches equal how many 
centimeters - 
millimeters - 
meters - 
11. A dive bomber drops about 9000 ft. in lh seconds. Express 
.9000 ft. as kilometers. Express the speed as kilometer 
per hour; as miles per hour. 
12. When sound travels 332 m, per second, how long will it take 
the report of a gun to travel U Km.? 
•6a EXERCISE 3U ’ 
Convert to centimeters 
(1 ) (2) (3) . , (5) (6) 
a. 13-X/U in. Id dm. 8 in. 3 ft- .6 m. IS mm. 
'o. 3-1/3 yd- 2.1 miles l 6 dm lU in. 15,2 m. 3-9 mm. 
c. 3/U ft. 2/3 yd. 1/8 miles 3/l6 in. 7/8 in. .25 ft. 
Convert to meters 
d. 2-1/2 miles 3 in. U ft. ,6 miles 7/5 miles 68 inches 
e. 28 cm. 360 mm. 1.8 dm. IS Dm. mm. 6 cm. 
THE LITER 
In the metric system, we speak of a liter of milk, whereas in 
the English System the comparable volume is the quart. Like the meter, 
the liter can be divided into tenths, hundredths, and thousandths. 
The liter is related to the meter in the following manner. A 
square container, whose sides and height are exactly 10 centimeters 
(l/lO of a meter, or a decimeter) holds exactly one liter of distilled 
water at U° C. You may be surprised that a temperature is mentioned. 
There is a good reason. As water increases in temperature it expands, 
that is, it takes up more space. 
As mentioned the liter is divided into; 
Tenths, called deciliters (dl.) 
Hundredths, called centiliters (cl.) 
Thousands, called milliliters (ml.) The multiples of a liter are; . . 
10 liters, called ... one decaliter (Dl.) 
100 liters, called ... one hectoliter ('hi.) 
1000 liters,called ... one kiloliter (kl.) 
Since the sides and height of this square container are each 
capacity 
10 centimeters, we can obtain the volume (cubic measure) by multi- 
plying side x side x height, or: 
10 cm. xlO cm. xlO cm. = .1000 cubic cm. 
1000 cubic cm. of distilled water at U° C. are called one liter. 
1 cubic cm. of distilled water at U° C. is 1/1000 of distilled water, 
hence 1 cubic cm, can be expressed as 1 milliliter, though the common 
expression for 1/1000 of a liter is 1 c.c, which is an abbreviation 
for 1 cubic centimeter. 
One more point is worth explanation at this moment. So far the 
connection between the linear measures and the volume measures has 
been related in the closest manner. Is there any connection with 
weight? Yes. For 1 c.c. of distilled water at U° C. weighs exactly 
one gram. Arid 1000 c.c. weights 1000 grams. But, we will return to 
the study of the metric weights in subsequent pages. 
Fig. h represents a square container whose sides and height are 
10 centimeters each (l decimeter). 1000 cubic centimeters is another 
way of saying 1 cubic decimeter. 
To sum up. From the linear measure of the metric system, we can 
calculate the capacity (or volume) measure, or liter. ONE LITER 
or 
1000 cc 
one cubic centimeter of water 
weighs - 1 gram 
one decimeter 
one decimeter 
Divided into centimeters 
<-1 -> 
cm 
Fig. U 
One cubic decimeter. The volume of one cubic decimeter is 
1 liter. Moreover a liter of distilled water at U° centrigrade 
temperature, (39*2° F) weighs one kilogram. 
The weight of distilled water at U° C contained in a cube 
of 1/1000 of a liter is equal to a GRAM = grains and 
measures 1 milliliter (one cubic centimeter). It is obvious that the metric system is clear cut and not 
difficult to understand. As this new system was introduced it was 
compared to the English System. We have studied these comparisons 
of linear measure. The capacity measures must be compared in a 
similar way. We shallcompare the metric system with our own American 
Apothecary System. 
• Liquid measure, U. S. 
1 gallon - U quarts 
1 qt. = 2 pints 
1 pt. = 16 ounces 
1 oz. z 8 fluidrams 
1 f. dr. - 60 minims 
By actual measurement one fluidram will Be found to be equivalent 
to 3•6962 c.c. 
Next we should find other equivalents. 
How many c.c. in 1 oz.? 
Method; 
1 fluidram = 3. ©62 cc 
8 fluidram = 1 oz. 
Therefore, 
8 X 3.6962 = 29.573 cc 
How many c.c. in 1 pint? 
Method; > ■ ■■ 
1 fluidram = 3,6962 cc > , 
1 oz. - 29.573 cc 
16 oz. = 1 pint 
Therefore, ■ '■’ ■ . r ■ 
16 x 29.573 cc = U 73.168 cc Approximate values. 
Such values a5,.3.69.6.2. cc; 29.573 cc» • cc are exact 
metric equivalents. However in your work in the ward, there is no reason 
to use these figures. ■ Instead,.round numbers are used. - By so doing, the 
chance of error is greatly reduced. Moreover, it is perfectly safe to 
give a patient Ucc of some medicine, instead of 3• 6962 cc. Such a 
round number as 4 cc- is obviously easier to ~handl®n than 3* cc. 
In a like;manner, the following approximate values are in common use. 
They should be learned. 
Apothecary Metric Eg. App. Value 
l 6 1.00 cc 16 minims ■ 1 
i k 
1 OZ. 29.573 cc 30. cc 
S oz. cc pUo qc 
Problem (Do in class) 
If 3*6962 cc equals 1 fluidram (60 minims) find how many minims 
are in 1 cc. What do you think would be a satisfactory approximate 
value? Pill in your answers in the-space'provided above. _ 
Apothecary 
Metric Eqw. 
App. Value 
Household 
Measure 
Symbols 
1- gal' - 4- qts. 
1 qt. - 2 pts. 
94-6*33 c.c. 
. 
♦ 
1 1 pt, = 16 oz. 
4-73.168 c.c. 
,1 
1 oz. - 8 f.dr, - - 
■ 
. 
29.573 c.c. 
24-0,0 c.c. - - 
■ 
- 1 tumblerful 
17- - - 
y Vlll ; 
30.0 c.c. 
Yr 
—i 
' • •. 
{ - * *7 
1 f .dr = 60 M. 4]- 4? - 
— — —. — — 3.6962 c.c. 
16.0 c.c. - - 
- -1 tablespoon - - - 
! 
'Y/ /JyL I 
0 
4.00 C.c. 
1 teaspoon - - - - 
1 
CyJ 
i- 
m - 
------ 1,00 c.c. 
i 
( 
Figure.; 5 
A table showing apothecary measures with metric equivalents and approximate values. You should know the 
approximate values given in this table. Fill in the approximate value of minims in the last row. 
■66- SPECIAL ATTENTION 
Pay close attention to the approximate value of 1 fluidounce. 
• '■ • ’ v-i . “ *)'.'■■ 
In figuring the number of ounces in 3&0 cc. use 30 cc as the amount 
equivalent to an ounce. 
You will become confused and moreover you will obtain a 
wrong answer, if you find the number of fluidrams first, then divide by 
2 to get the number of ounces. 
Right Way 
-4$Lw s 12oz> 
3? cc 
Wrong Say 
90 
360 c9~ = ■‘i&.fl.dr. 
U co. 
9° fl-dr-- 11.2508. 
g 
In the Medical Corps, the metric capacity system is em- 
ployed in the'following practices: 
'I, Liquid medicines, ordered "by the medical officers, are 
usually put in’a ‘bottle of one of the following sizes: 
x i/) 11 
ft _'' d 
* !. 4. xvr 
VTT 0 , . , 
-rrr —vV X~X I i 
Y o '■• 2. The amount of medicine a patient is to receive is ordered 
by the -medical officef ,J 'The following are common doses or 
„ • V* * »*V • 1 
amounts:’'" 
;of one odnce. *■ l< 
?,v or 4^.or one tablespoon or 16.. cc. ~ 
| or one teaspoon or U cc. 
T cc 
(9 •S’ c.c 
'Ov\_t' fe )>\ "xvT 
, v . ; /• , it 
3» The amount of fluid (water, milk, fruit juices, coffee,etc.) 
that a patient takes in a' hour period, must at.times be de~ 
■ ' C * , . w ’ t '* - u i • ■ 
termined. The figuring may be done in ounces or c.c, ~ according-to 
the order given. 
U. The amount of urine that a patient passes in 2*4 hours must 
at times be calculated. The figuring is likewise in ounces or c-.c. 
5. "Fluids”; or blood are frequently given to a patient by way 
of his veins. Glucose (sugar) and saline (salt) solutions are the 
commonly used "fluids". Common amounts are 500 c.c.; 1000 c.c,; 
1500 c.c. 
6. The amount of urine obtained by catheterization, is usually 
measured in ounces or c.c. 7. It is usually a good plan on the part of the corps man to 
measure the amount -of hlood coughed- »up ,-pr 'Vomited/ •because If is * 
■ . • •TT i;.l rr .*» X ) iff ’ .... 
exceedingly difficult to judge (guess) the amount. An accurate measure 
may help the medical officer to make a diagnosis. 
8. In some diseases,- fluid or pus accumulates within body cavities 
(pleural, pericardial or peritoneal) or joints. Fluid or pus is usually 
measured on removal. •••• *.■ . . 
t. . . i s ‘ 
9. To make solutions.. ; 
EXERCISE 39 
(Do in .Class)" 
1. A patient drinks 2720 c.c. of fluid, 
a) Express this amount in liters, 
h) Express this amount in ounces, 
2. A patient voids 9.2 ounces of urine. Express in c.c. 
3. Reduce .8 pints to ounces. 
U. Reduce .8 ounces to drams, 
5. Reduce . U drams to minims. 
6. How many minims in 1.6$ fluid ounces? 
7- Convert 375 c*c* into fluid ounces. 
8. Convert 0.3 c.c. into minims., ■ • 
9. Convert Uf. oz,, .2 f. dr. into c.c. 
10. Convert 8 minims into c.c. .. 
11. Convert 500 c.c. into fluidqunces. 
12. Convert 20 oz. into c.c. _ , ‘ - 
13. Convert 8 fluidounces into fluidrams. 
IU. You are given a U oz. hot tie of cough medicine. How many 
days will it last if you take ~T three times a day? 15. There are normally about 5»000,000 reel.blood cells 
K(R!B.C.) in a-cubic millimeter ( 1 of a c.c.) of blood 
1000 . • 
How many are there in 5 liters of blood? 
16. There are normally about S,OOO white blood colls 
(W.8.,G.) per .cubic millimeters of blood. Find 
number in 5 liters of blood. 
17. A patient developes pneumonia. His W.B.C. count is 
found to be 26,000 per cubic millimeters. Find the 
number of W.B.C. in 5 liters. How many cells above 
the normal are there? 
1000 
THE GRAM 
One c.c. of distilled water at U° C weighs 1 gram (Gm.). There- 
fore 1 liter of distilled water at U° C weighs 1000 gns. Here then 
we comprehend the close inter-relationship that exists between the 
linear, capacity and weight system. 
The gram is divided into; 
Tenths, called decigrams (dg.) 
Hundredths, called centigrams (eg.) 
Thousandths, called milligrams (mg.) 
The multiples of a gram are; .... 
10 grams, called decagrams (Dg..) . 
100 grams, called hectogram (hg.,) • 
1000 grams, called kilogram (kg.) 
10,000 grams, called myriagram (Mg.). 
The metric system when compared to the apothecary weight system, 
shows; y • ■ ' 
1 oz. =8 drams. 
-70- By actual comparison 1 oz. of distilled water at U° C. weighs 
31.1 grams. ' ‘' " r' iG 
cvSjL XttA-cr* ' 1 £ ' , ■ r ■■ 
There ©ne approximate value-rfrequently used, 4 '“t’ 
1 dram = h grams (one teaspoonful) 
i ounce = 16‘ grams (one tahlespoonfui) 
* \ 1 ' 
In contrast to the liquid measures where fluidounces are fre- 
quently used, there is this general custom in the weight system. 
Weights are expressed almost wholly in grams, thus adhering closely 
to the metric system. For example USO Gm. is expressed as UBO Gm, 
No attempt is made to reduce USO Gm. to ounces. 
Moreover, a drain or teaspoonful of dry measure is given the 
‘Approximate value of k Gm. A tablespoonful is l 6 grams. ■ 
In Europe, food is bought by the gram and its multiples;while 
A- 
-9 ’ ' i • ■ ■ I'' - , • ' 
in America and England we buy by the pound and ounce. 
t - ; ’ . ; • » • ; , . . . 
THE GRAIN 
In England, long ago "before there was a metric system* the wheat 
grain "well dried and gathered out of; the middle of the ear", was made 
the standard for the grain, weight. Drugs and fine metals were measured 
in grains. 
' Some drugs were effective in doses of 5-10 grains. To measure 
out this:''amount would not he difficult. Certain drugs, however, produced 
the desired therapeutic effects when.very small amounts were used. But 
how was "an apothecary to measure out 1/100 of grain to he used as a single 
dose? • ■ . , . 
Two-methods were commonly used; one, hy the use of powders, 
another, hy the’use of solutions. For example. ' A patient may "have required T/lOOof grain of some 
drug, to he taken three times a day. Without delicate scales' to 
measure 1/100 of a grain, the apothecary could solve this problem hy 
placing 1 grain of thee drug and 99 grains of milk sugar in a howl. After 
thoroughly mixing, he had 100 grains of powder, in which there was 
1 grain of medicine.- ‘-Since; 
• 100 grains of - powder contained 1 grain of drug .' 
Then 
1 grain of powder contained l/100 grain of drug. 
This method of mixing the drug with some powder, such as milk 
sugar, was a common means of meeting the problem. By dividing the 100 
grain mixture into 100 equal parts, each part would weigh 1 grain, in 
which there would he l/100 grain of medicine. 
Another method employed the use of a solution. That is, 1 grain 
of a drug is placed in a calculated amount of fluid, such as Water, so 
that a teaspoonful will contain l/100 grain of the medicine. In this 
*cas.e, 100 teaspoons of water would he the necessary amount. 
One of,,the very important drugs of today and several hundred 
years ago, is:digitalis,. It is of great value in certain types of 
heart disease. 
- Digitalis extracted from a certain plant, was not always of the 
same strength. This being the case, it was impossible to state the 
amount of digitalis a patient should he given to obtain benefit, with- 
out poisoning, which results when too much of the drug is given. 
Because the drug varied in potency, the effects varied. To be sure 
the^good effects would be obtained, digitalis was given,to the patient 
until certain signs of poisoning occurred;,then after the drug had been 
withheld several days or more, a fraction of the original 
72- amount was given daily. Without a standard of potency, the physicians 
were forced: 'to follow‘stime rule, even though it was broad rather than 
exact. Therefore, the custom was for the patient to take a certain 
number of drops of the solution daily, until he "puked or purged'*; which 
indicated the early effects of poisoning. 
When the metric system was introduced the gram was promptly com- 
pared to the grain. On measuring it was -found that: 
1 gram = grains. 
= 15.0 grains (approximate?value) 
But what did 1 grain equal? 
' ' 1 gm. - grains 
Therefore 
1 grain = 1 Gm. ~ 
15.^32 
= 0.065 Gm. 
- 0.06 Gm (approximate value) 
This equivalent is very important as it is the basis for con- 
verting one system to the other. You will note that 1 grain is roughly 
equal to 0.06 Gm. - a little more than .05 Gm. and not as much as 0.1 Gm. 
which would be an easier figure to deal with, " 
How to find equivalents. 
Example. ■ • ■ . - 
If 1 grain equals 0.06 Gm. , find “‘the’ metric'equivalent of 0.1 grain 1 
Method 
Since 1 grain - 0.06 Gm. 
l/lO grain = l/lO of 0.06 CVq 
.1 grain r 0.006 Grn. . 
■73' Example; 
'> • •». [-■ ' \*fj . ..V, !■■ **s '•_ --.»j * " • »>'_.• *•! • • • >• v* f 
If 1 Gffl. equals 15.,0 grains, find the metric equivalent of 
' 1/2 G-m. ' ’ 
»'s r, lot . 
Method;. 
1 Gin. = 15.0 grains. 
.•‘i.i.rirW.'v*; V .; ': • vi:. ••• 
IGn j 2 - 15.0 grains /r 2. 
i Cm. = 15.0 grains. 
i Gin. z grains ..•■■■r • • » • 
Example; 
Find the metric equivalent of 10 grains j(as 10 grains of 
l. 1 asperin). 
Method: 
1 grain * 0.06 Gm. 
10 grains = 10 x 0.06 Gin. 
= 0.6 Gm. 
Example; 
Find the metric equivalent of 1/150 of a grain 7 
T 
Method; 
1 grain = 0.06 Gm. 
1/150 of one grain = 1/150 grain. 
Therefore 
1/150 grain = 0.06 Gm. 
:i * *‘lso' •• ' r: 
= O.OOOU Gm. 
■7^ Example; 
Find the ap o the caiy- equival ent~ -of—O;OOOH~’Gfn'J ? 
1 gram = • 0.06 Gm. 
’ ‘- ' " 1 = 1 0..000U Gm. - 
O.OOOU Gm. is what part of 0,06 Gn.? 
*. # ' ’y..' 
... 150 
.oook [ 0.060cr 
u.. 
20 ... 
20 
.OOOU Gm. is one hundred and fiftieth (--p*Q- ) 
of 0.06 Gm. 
Therefore 
since O.OOOU Gin. is --—- of 0.06 Gm. 
150 
and 
since 1 grain = 0.06 Gm. 
Then 
.OOOU Gm.~ grains. 
150 
In the hospital the metric system of measures is used to weigh 
the following: 
1. Objects that are apt to weigh a gram or over. 
a. Food. The patient with diabetes mellitus is taught to 
weigh all the food he eats. The exact way to do this is 
to weigh the food, using metric measures. A rough way is 
to use the teaspoon and tablespoon. 
b. Chemicals In the laboratory, various solutions are made 
and used. This involves the very careful measuring of 
chemicals. 
-75- c. Pathological specimens. Stones and tumors are examples of 
specimens weighed after removal. 
d. Organs removed at post-mortem. , The average weight of 
organs is known. Variations in size due to disease or 
arrested development are most accurately ascertained by 
weighing the organs. Examples are: spleen, kidneys, 
adrenals, heart, etc. 
e. Drugs. In the pharmacy, large quantities of certain com- 
monly used solutions, which are called "stock solutions", 
are made. These solutions usually require the use of drugs 
in amounts of a gram or more. 
The same is true for the making of ointments,salves, 
pastes, etc. 
2. Objects that are apt to weigh a gram or less. 
a. In this category are most of the commonly used drugs. 
EXERCISE 36 
(Do in Glass) 
Convert the following to the metric system: 
1. 10 grains. 
2. 7i grains 
3. 1/200 grain. 
U. 1/50 gr&in. 
5- 1/75 grain. 
6. 1/6 grain. 
7. l/U grain. 
8. 1/2 grain. 
-76- Exercise 36 (Continued) 
9. l/S grain. 
10. 1/150 grain. ■ 
11. 15 grains. 
12. 5 grains. 
13. l/lO grain. 
IU. l/l6 grain. 
15. l/1000 grain. 
PSRCISE 37 
Convert the following; 
1. 30 mg. to Cm. 
2. .6 mg. to Cm. 
3. 8 rag. to Cm. 
U. 500 rag. to Cm. 
5. .OOOU Cm, to mg. 
6. .015 Cm. to mg; to grains, 
7. .3 rag. to Cm; to grains' 
Complete the following table: 
30 grains equal Gm. mgn. 
qp, ii it h '• 
]_Q ti II h •' 
yl M 11 11 ii 
1 2 
ll 11 11 ii 
1 grain equals " ,f 
i ii 11 11 11 
-77- Complete the following table (Continued) 
l/k grain equals Gm. mgm. 
1/6 « « « " 
1/g it n ii ' it 
1/10 >» ,f " " 
1/lP " » « » 
l/lb " " « »• 
1/50 " " '» '» 
i/Y5 it 11 11 t» 
1/100 " » »• " 
1/150 u " » •» 
1/200 " " « »» 
NOTE; 
In the hospital the doses of drugs are expressed in the 
metric system. However, at times the apothecary system is used. Be- 
cause both "languages" are used in this country one should be familiar 
with both. CHAPTER VII 
The Thermometer 
There is a difference "between heat and temperature, which 
should be clearly understood. A quart of water at 212° Fahrenheit is 
just as hot as a barrel of water at 212° Fahrenheit because the tempera- 
ture is the come, but the quantity or amount of heat in the barrel of 
water is very much greater than in the quart of water. The intensity 
of heat is 'crown as its temperature, the degree of intensity is de- 
termined by Instruments called thermometers and is expressed relatively 
in degrees. 
A thermometer is a glass tube with a very small bore, sealed 
at one end and the other end terminating in a bulb which is usually 
filled with mercury. When subjected to the influence of heat the mercury 
expands and rises in the tube, the degree of heat being read on a graduated 
scale marked on the tube. 
The three scales used in marking thermometric degrees are 
Fahrenheit, Centigrade, and Reamur. In the Fahrenheit scale the freezing 
point of water is 32° , the boiling point is 212°, and the intervening 
space is divided into 180 degrees. In the Reaumur scale the freezing point 
of water is zero, the boiling point is 80°, and the intervening space is 
/ 
divided into 80 degreesi The Reaumur scale is rarely used in English- 
speaking countries. Most thermometers in use in the United States are 
graduated either in the Centigrade or the Fahrenheit scale, and some ther- 
mometers have both scales. 
,flt is often necessary to convert the degrees of one scale into 
those of another, and in converting Centigrade degrees into Fahrenheit 
degrees or vice versa, it should be kept in mind that 100 degrees Centi- 
grade are equal to 180 degrees Fahrenheit, or that 1° C. is equal to I.B° F. 
It should also be kept in mind that on the Fahrenheit scale the freezing 
-79- point of water is 32 degrees above zero or 32 degrees higher than on 
the Centigrade scale, and this difference of 32 degrees must he 
considered'1. ‘ ' ' ' 
' RULES FOR CokmSlOIT 
With temperatures above the freezing point of water, to convert 
Fahrenheit degrees to Centigrade degrees ihe rule is: Subtract 32 
from the number of Fahrenheit degrees and divide the remainder by 
1.8. 
Example; 
Convert 122° F. to Centigrade. 
Method;. 
122 - 32 *=• 90. 
90 divided by 1.8 50 
r-0 
50 s. answer 
With temperatures above the freezing point of water, to convert 
Centigrade degrees to Fahrenheit degrees the rule is; Multiply the 
number of Centigrade degrees by 1,8 and to the product add 32. 
Example; 
Convert 50° C. to E. 
Method: 
50 x i.s * 90 
90+ 32 => 122 
122 = answer 
-go- With temperatures below tSe freezing point of water, to convert 
Fahrenheit degrees to, Centigrade degrees the rule is; Add 32 to the 
’' • I 
number of degrees below the freezing point and divide the sum by 1,8. 
Example; 
Convert to C. 
Method: 
31 f-32 = 63 
63 divided by 1.8 = 35 
- 35° C ~ answer. 
With temperature below the freezi,ng point of water, to convert 
Centigrade degrees to Fahrenheit degrees, the rule is: Multiply 
the number of degrees below the freezing point by 1.8 and from the 
product subtract 32. 
Example; 
Convert -35° C to F. 
Method; 
35 X i.g - 63' 
63 minus 32 = 3^ 
-310 F r answer 
• There, is a simpler method of conversion, which it is well to 
present after having studied the above. The rule is: 
To convert Fahrenheit to Centigrade, subtract 32° and multiply 
the remit by 5/9 • 
To convert Centigrade to Fahrenheit, multiply hy 9/5 and add 32°. TTHERMOMETER - COMPARATIVE SCALES 
Centi- 
grade 
:v —r-fr 
Fahrenheit 
—-7 r-rs—— 
, - * . * * * 
Water boils at 
r. Centi- 
grade 
Fahrenheit 
mV ' ■ ' ‘ t ■ 1 i; * U f r“ 
sea level 
» 
100° 
£ 
ro 
o 
98.6 
Blood heat 
19U 
95. 
185 
9° 
17^ 
86 
167 
Alcohol "boils 
80 
158 
77 
1U9 
68 
1U0 
60 
131 
55 
i 
i 
127 
Tallow Melts 
50 
122 
U5 , 
113 
35 
| 108 
32 
Water freezes 
o 
•—i 
30' 
- ■ 
OJ 
•- 
•20 
i 
i 
Ik 
i 
i . - 
- 
0 
Zejjo., Fahrenheit 
- k 
'-13 
■ ' • • ■ ■ 
, -22 *, ; 
’1 ' 
-31 
. ...... 
-Uo 
Table I 
— EXERCISE 38 
1. Convert the following Fahrenheit degrees to Centigrade, and 
i 
record in the above table. 
167° f 10k° F 6oc F 0° F - 140° F 
IUo F 98.6 F 32 F -22 F 
. V. •,•'5 :; , 
2. Convert the following Centigrade degrees'to Fahrenheit: 
99° c 15° ' -5° C : : ' ' 
39°' c " 16° ' -25°c 
3. The average temperature on the surface of the sun has been 
estimated at 6,000°’ C. Convert t6' F. degrees'. ' - *'v ‘ - 
U. Convert the following C degree* to F and record. 
Boiling points of several chemicals 
, Element 
. • .. * • j 
i i 
c° 
F° 
; v 
j Antimony 
1,380 
"ur " V* , 
j Argon 
-!S5.7 
| ■. i 
Carbon 
u.aoo 
Helium 
-26s. g 
Iron 
3,000 
Lead ’ 
1,620 
j': 
-• • 
Magnesium 
1,100 
- 
Platinum 
U.300 
Tin 
1 
2,270 
. .» 
i 
Tungsten 
3,900 
f 
i CHAPTER Till 
1. Pei* cent' 
2. Solutions 
1. Per cent. 
"Per cent” is a new word for you in the language of arithmetic. 
It means per 100. Suppose a man borrows $lOO. The bank informs him 
that the interest amounts to 3 Per cent. Translated, this means that 
the cost (interest) of borrowing $lOO is 3 per hundred or $3.00. 
$lOO - amount borrowed 
$ 3 cost, at a rate of 3 per cent interest. 
The short hand expression for "per cent1* is $, so that 3 per cent can 
be expressed as 3s• 
"Per cent" can also be expressed as a decimal fraction. 
3fo is the same as ,03 
"Per cent" can, moreover,he expressed as a common fraction 
yk is the'same as 3/100- 
If these equivalents were written down, we could note that 
$3 is a oart of $lOO and can be expressed in a number of ways. 
No. of Common Decimal Per 
dollars Fraction Fraction Cent 
3 -A— -03 3$ 
100 
To he sure this is understood, a few more examples are given 
below; 
10$ = 10/100 .1 
25$ = 25/100 = .25 
150$= 150/100 = 1.5 
-gU- Examples: 
1. Find 6$ of Uo 
Uo x .06 = 2.U 
2. Find 150$ of 30 
30 x 1.50 s U 5 
a vala? expressed as a common fraction may be expressed as a 
dcJmaL ihudlcn by dividing the numerator by the denominator. For 
f ~ 5 
A poc cent may be converted into a decimal fraction by moving 
the dec3 xh point to the left two places and eliminating the per cent 
sign. Thus '(y/o • ,75 
The decimal fraction may be converted into a common fraction 
by placing the indicated denominator under the numerator and reducing 
the result to a common fraction in lowest terms. Thus; 
.75 = 75/100 = 3/*+ 
EXERCISE 39 
Convert the following; 
Common Decimal Per 
Fraction Fraction « ■ Cent 
• * 
a. 1/2 
t>. 1/h 
c. 2/3 
d. 2/5 
e. 3/U 
f. l/g 
g. 3/16 
h. 3/7 
i. 5/s 
-85- EXERCISE UO. 
Convert the following: 
Per Decimal Common 
Cent Fraction Fraction 
a. 30$ 
b. 
0. 25$ 
d. -5$ , - . ... 
e. .66-2/35..., .... 
f. 75$ 
g. ,50$ ...... . . ...... 
h. . &-i/U$ 
1. 37-1/2$ .. 
j. -50$ . .... . . 
k. . Uos . 
*'’‘. • • i • 
l. 87-1/2$ ..... 
EXERCISE Ul 
Do the following: 
1. 6U x 5C$ 
2. U 5 X 20$ 
3. so x 3$ 
u. 120 XBs 
5. 200 x 12/0 
6. 15 xss 
7. 16 x 25^ 
S. !'9 X 10$ 
9. 90 X g$ 
10. 67 x2s 
11. 
12. lU X 18$ 
-g6~ EXERCISE Ur 
1. A man weighs 150 pounds. His body is composed of the following 
» ‘ . ’ ' ' . * * \ * * 1 
chemicals. Find, in grams, the amount of each. 
Per Cent Grams 
Oxygen - - 65' - - 
Carbon - - * lg - - 
Hydrogen 10 - - 
Nitrogen - - 3 - - 
Phosphorus - - 1 - - 
,*l„# f • , • 
Sulfur - - 0.25 - - 
Chlorine - - 0.15 -* - 
Sodium - - 0.15 - - 
Potassiam - - 0.35 - - 
Calcium - - 2.00 - -*> 
Magnesium - - 0.05 - - 
Iron - - O.OOU - - 
Other elements - 0.0’U6 - - 
2. A potato weighing 100 Cm. contains the following; (Express in Gm.) 
'Carbohydrate - - 19% - 
Protein - -- -- £$ 
‘Fat -0% -- - 
3. A piece of tread weighing 30 Gm, contains the following: 
(Express in Gm.) 
Carbohydrate -- - 18$ 
Protein --- - - - 3$ 
Fat 0,7$ 
Find the amounts of each in 100 Gm. of bread. 2. SOLUTIONS 
Absolution is made “by dissolving a solid in a liquid.' Two 
solutions commonly used in the wards are glucose (sugar) solutions and 
saline (salt) solutions. Those patients who are in need of such 
co.7y 
solutions,Areceive them directly into their own veins. 
There are many, many solutions. The laboratory has need of a 
great variety in order to do work "that involves chemical analysis. The 
pharmacy makes many of the solutions used in the wards, such as cough 
medicines, "hose drops", "eye drops", intestinal medicines, and so on. 
Large pharmaceutical houses,as a rule, prepare glucose and saline 
solutions. 
When a solid as ephedrin (chemical) is dissolved in salt water, 
the solution is called a saline solution of ephedrin. 
If the ephedrin were dissolved in water, the solution would be 
referred to as an aqueous solution of ephedrine. 
At times alcohol is used as the liquid, because the solid will 
dissolve more satisfactorily in alcohol than water. 
It is important to understand the language that deals with the 
strength of solution. What is als solution, a 50$ solution, and so 
on? 
A 1$ solution is one which contains 1 Gm of the solid in every 
100 c.c. of the solution. To make a Is-saline solution. 
Place exactly 1 Gm. of salt (A) 
in the container. 
Then add water until the water 
level in the container reaches the 
100 c.c. mark. 
After the water has been shaken 
to aid in the dispersal of the salt, 
100 c.c. of the'solution-contains 
1 Gm of salt. This is a Vfo saline 
solution. 
Fig, U 
The preparation of a 
1% solution 
This method is one way of making a 1 fo saline solution. Another 
way would be to add the water to a special flask, a flask that is ac- 
curately calibrated by the Bureau of Standards. The resulting solution 
would >e very accurate. Such accuracy is needed in some types of work 
as well as in the preparation of some medicinal solutions. Saline 
solutions that are given by vein, are very accurately prepared. 
To make a liter of 1/6 saline solution, requires the addition of 
10 Gm of salt followed by the addition of enough water to come Mup to11 
the 1000 c.c. mark. 
-89- If 1000 c.c. contains 10 Gm salt 
Then 100 c.c, contains 1 Gm salt 
and this is a 1$ solution. 
A 10$ solution of salt contains 10 Gm. of salt.per 100 c.c. 
A solution of glucose contains 50 Gm. of glucose per 100 c.c 
Problem. How much phenol is needed to make 50 c.c. 6f a solution? 
Method: 
50 c.c. of a Jh solution contains how many grams? 
100 c.c, of 5% solution contains 5 Om. 
Therefore, 
50 c.c. or one-half of 100 c.c. contains one-half the amount 
of phenol, namely 2.5 Gm. 
Problem. How much phenol is needed to make 60 c.c. of a 31° solution? 
One Method. 
bO c.c, of a J/o solution contains how many grams? 
100 c.c, of a jfo solution contains 3 Gm. 
60 c.c. is 3/5 of 100 c.c. 
Therefore, 
3/5 of 3 Gm st amount of phenol in §0 c.c. 
1.8 Gm = answer 
Second Method 
100 c.c, of a3s solution contains 3 Gni, 
1 c.c. of a 370 " " .03 Gm. 
60 c.c. of' a3s " " I.S Gm.(6o x .03) Stock Solutions. 
■ The pharmacy usually has on hand large amounts, perhaps 5000 
>i 
c.c. of certain solutions. Such solutions are stronger than those 
prescribed. Suppose that l/U$ ephedrin solution was commonly used in 
the wards. Uhe pharmacy might keep on hand 5000 c.c. of al% solution. 
As requests were made for l/U$ strength, the pharmacists would dilute 
some of the Ij> solution to l/U$ strength and dispense it. 
Suppose you wished, to make a solution of 50$ strength from a 
solution of 100$ strength. The easiest way is to measure out equal 
amounts of water and 100$ solution and mix them together. You then 
have a 10$ solution. This is practical if you always have a 100$ 
solution, which you rarely have. 
To make 100 c.c, of a 25$ solution of phenol from a 50$ stock 
solution of phenol is the type of problem that comes up. 
Mo '.hod 
ICO c.c. of a 2jf° solution contains 25 Gm phenol 
100 c.c. of a sOfo solution contains 50 Gm phenol 
We want 100 c.c. of a solution or 25 Gm. phenol 
In a solution. 
1 c.c. contains l/100 of 50 Gm - 0,5 Gm phenol. 
How many c.c. of 50% solution contains 25 Gm? 
25 Gm r 0.5 Gni 50 c.c. 
or 
( 1 c.c. of a 30)o solution contains 0.5 Gm.) 
(50 c.c. " " " " " 25.0 Gm.) 
Put 50 c.c. of a 50$ solution (25.0 Gm.) into a glass beaker and 
add enough water to make 100 c.c. This gives 100 c.c. of solution 
in which there are 25 Gm., making it a 25% solution of phenol. 
-91- 50 c.c. water is being added 
Fig. 5 
The figure on the left represents 100 c.c. of 50% solution of phenol, 
in which are 50 Gra. of phenol. The figure on the right represents a con- 
tainer with 50 c.c. of a 50$ solution containing 25 Gm. phenol. into it 
are being poured 50 c.c. of water, This will make a 100 c.c. solution, in 
which are 25 Gm. of phenol; thus creating a 2^. 
Problem 
Make 1000 c.c. of a 25% from a 50% stock solution. 
Method 
1000 c.c, of a 25% solution contains 250 Gm. 
100 c.c, of a 50% solution contains 50 Gm, 
We want 1000 c.c. of a 251° solution or 250 Gm. 
-92— In a 50$ solution 
1 c.c. contains 1/100 of SCI Gm - 0.5 Gm.phenol. 
* J-t*8 urli -i fh iw:,- V,v, 
How many c.c. of stock solution contain 50 Gm.? 
250 Gm. -r 0.5 Gm 500 c.c. 
Put 500 c.c. of a 50$ solution into a beaker and add enough 
water to.make 1000 c.c. This gives 1000 c.c. of solution, in which 
there are 50 Gm,, making it a 25$ solution. 
To make a formula. 
°]o desired , , , 
—- x number c.c. desired - number of c.c. of 
ft of stock sol, stock solution to use. 
X 1000= ■ JgL x ieeO"r 500 0.0. 
50$ s®" 30- 
-2T- 
= 500 c.c, stock sol,to use. 
EXERCISE U 3 
1. Make 1000 c.c. of a 2% solution of phenol from a 25% solution. 
2. How many grams of horic acid are there in 500 c.c, of 2% solution? 
How many grains? 
3. How many grams of silver nitrate in 700 c.c. of a 2.5 1° solution? 
U. How many grains of silver nitrate are there in 6 fluidounces of 
a 10% solution? 
5. How much salt is needed to make 250 c.c. of 1I solution? 
6. A man of 150 pounds1 weight has about 5 liters of blood. The blood 
sugar (amount of sugar in the blood) is 0.1 gm. %, Find the amount 
of sugar in'his blood. 
-93- 7, Pow much salt does the blood contain If the blood‘salt is 
’ ‘ "' ' “ :. Or io C'ji\ 
,U 5 ga.fo 7 
8, How much stock solution is needed to make 30 c..c. l/*+ 1° 
■ephedrin solution from a 1 $ stock solution? 
9, ; How many .teaspoons of salt are needed to make a liter of a 
1~/o saline' solution? CHAPTER IX 
FOOD AS ENERGY 
. ”A .man is continually giving out energy. He is never com- 
pletely at rest as long as he is alive. Even during sleep his limbs, 
twitch, his ribs rise and fall with respiration, and his heart continues 
to beat regularly. Besides these movements, there is another way in 
which he loses energy. It takes energy to keep his body warm. A hot 
pfiece of metal or stone will, in a short time, cool down to the temper- 
ature of the surrounding air, but a living body is always warm to the 
touch. A man, like any warm object, loses heat continually to the cooler 
air which surrounds him.” 
”We can measure the amount of food that a man or an animal con- 
sumes over a given period of time, and we can measure the energy yielded 
during the same period., If we burn an equal weight of similar food in a 
suitable apparatus and find out how much energy its combustion yields,and 
if this value is equal to the energy yielded by the experimental subject, 
then evidently the living organisms, so far as its energy- output is 
concerned, is really and precisely a combustion .engine (like a gasoline 
engine).” .. t 
’’Such experiments have been performed, A man has been shut up for 
a time in a small compartment, so constructed that the energy he gave 
out, as heat or otherwise, could be accurately measured. He was fed on 
a weighed amount of food of known composition and the energy actually 
yielded was compared with that which would have been obtained had the 
food been simply burnt.” In a typical experiment, it was found that a 
man gave out 5,682 calories of heat and ate an amount of food which would 
yield, if burnt, 2,688 calories. The two sums differ by less than l/U of 
* v , . V;. ' 
1 j>, which is within the margin of error of .the methods emplpyed. We can 
say, therefore, that man is fundamentally a" machine driven by the energy 
-95- that is produced from the burning of food. . ,!A mouse or a man works 
in much the same manner as a gasoline engine; its fuel is its food; 
and like a gasoline engine, if it lacks either fuel or air, it will 
cease to move, and slowly becomes cold.11 
Food is fuel, as is gasoline. The automobile engine is driven by 
r . s » •: ' - * f : . 
gasoline when the gasoline is combusted. As it burns it yields energy. 
When food is burned it yields heat, which is the same as energy. 
•'? 'Vi.-I ■■■' ‘ • 
Instead of using the word burn when we speak of food, we say that food 
is oxidized. 
We must find some standard to measure heat, Just as we have found 
standards to measure distance, weight, volume. The unit of heat is the 
large calorie. A calorie is the amount of teat required to raise 
1000 c.c. of distilled, water from 15° to l6° C. 
As mentioned above heat is energy. Food produces energy while 
body warmth and movements of muscles use up energy. 
By experiment it has been found that; 
**■ * • * ' * ' * • i ’ * 
1 gram of carbohydrate yields...U calories 
1 gram of protein yields .... U calories 
1 gram of fat yields 9 calories 
This information gives'us the opportunity to discover how many 
calories, we derive from the food we eat'. ‘ ,a • 
Moreover, it has been found, by experiment*, how many calories 
are used up. in various types bf work and play; in sleep , in health 
and In disease, u , ‘ c .. 
For instance.: o j• • ' 
A at’bed' rest needs about ISOO calories pe,r day. . 
A clerk needs about - - - - 2200 " n it 
, ’-y . ■ . ■ ; ■«, • . * „ •• 
A carpenter needs about -- - 3200 " M H 
A stone mason needs about ' UUOO” ,r M 11 
A lumberman needs about - - -r. . 6000 ,? 11 ,r 
-96- Calories required “by a marching soldier. 
Load Road Normal Extra Total 10$ for 
’Needs ‘v Work waste 
0 Level or rolling 206U' $OO -296 - • 3260" 
50 lb. " « " 206 U 1172 3236 3560■ 
50 lb. Ascending 100 ft. '2O6U 2110 UljU- U 590 
per mile 
The above gives a rough estimate of caloric needs. More 
calories are needed in a long march than in a short one. Since no 
distance is stated, these figures represent only a rough idea of how 
calorie needs vary with types of work or exercise. 
Nothing has been said about the bodyTs need of minerals and 
vitamins. Such a discussion is beyond the scope of a study of the energy 
value of food. 
Classification of foods 
1. Carbohydrates. Sugar Is an example of pure carbohydrate. Bread 
is about 50 - 70$ carbohydrate. 
2. Protein. Meat, fish, and poultry are examples. 
3. Fats. Butter, olive oil and meat fat are examples of fat. 
Method of calculating calories 
1. A man takes 2 teaspoonfuls (8 Cm) of sugar in his coffee. How 
many calories does he obtain from the sugar? 
U calories in one (l) Gm. 
Ux 8 - 32 calories 
2. A man eats 150 Gm. of string beans for lunch. The composition 
of string beans is as follows; 
Carbohydrate Protein Fat 
i % I 
7.7 2.U 0.2 
-97- Calculations; 
$ x amount = total Gm x" calories per Gm r Calories 
7,7 x I*s = 11.5 Gm* x U U 6 cal. 
2.U x 1,5 = 3*6 Gm. xU = IU.U cal. 
.2 x 1.5 = .3 x 9 = 2.7 cal. 
- 63.1 calories from 
string 'beans. 
problem 
On the following page is a list of foods that would 
make an adequate diet. 
Calculate the number of calories that each food yields. 
When all caloric equivalents have “been calculated, add them up, 
The sum total will be the total number of calories provided by 
these foods. FOOD 
lift, in •. 
Equivalents 
C H 
Protein 
Fat 
Gms, 
Household measur 
0 ' % 
at 
P 
Calories 
Milk, whole , 
96.0 
1 quart,. 
,5. - 
r 3 * 3 
-3.8 
;; 
Bread, whole wheat . . , •, 
: 150 . 
i 
5 thick slices. 
50 
■9.7, 
.9/ 
Egg 
50 
1 whole .. 
’■ - 
13.4 
10.5 
Butter 
45 
3 tablespoons 
- 
1. 
85 
Meat,fish, or poultry 
120 
1 large serving 
- 
20, 
14 
Cereal, cooked 
30 
1 serving 
67- 
16. 
7.2 
. 
Fruit.- apple 
100 
1 small 
14. 
- .4. 
.5 
Vegetables 
„ •• 
i\ . 
\ * » 
10% string beans 
100 
3/4 cups 
,7.7 
2,4 
.2 
5$ tomatoes 
, 100. 
1 medium \ 
3.9 
•. .9 
' .4 
Potato ( white) 
100 
1 small 'A 
13,4 
2,2 
' .1 
Cream, 20%; 
120 
X/2 cup 
4.5 
2.5 
18.5 
Mayonnaise 
13 
1 tablespoon 
3.0 
1.5 
73.0 
Sugar 
45 
3 tablespoons 
100 
- 
~ 
Pudding (rice custard) 
100 
l/3 cup 
31 
-4.0 
5.0- - 
i 
■« - 
4 
Total 
Calories 
ANALYSIS OF AN ADEQUATE DIET 
-99- REVIEW 
EXERCISES MD PROBIEMS 
EXERCISE W+ 
Use short cuts for the following; 
■ (1 ) - (2) (3) 
'a.' , 96 x 12-1/2 x 20. 72x33-1/3 
■b.' 172 X5O 1,572 x 3-1/3 156 X 2-1/3 
c. 220 x 125 172 X 2-1/2 92 x 250 
A. Ug X 333-1/3 1,266 x 16-2/3 
e. U 5 X 33-1/3 l,gU6 X 500 132 X 166-2/3 
f. : 56 xbo 112 x 62-1/2 93 x 66-2/3 
g. 15 x 6-2/3 96 X 375 160 X 37-1/3 
h. 145 X 75 ' xBo ' ■ ‘ 12 x 7-1/2 
i. 12 x 7-1/2 159 x 23-1/3 2U x 27-1/2 
j. 322 x 3-3/1* ’ ' ' 'lUx'go' ! ‘IU x 625 
k. IS X 666-2/3 152 X 8-3/U 52 X 37-1/2 
l. 28 X;’l2s 20 x 87-1/2 ’ ' 'IUUx 8-1/3 
EXERCISE U 5 
Subtract the following; 
'CD (2) (3) 
; a. 78-7/12 86-9/10 57-7/15 
53-3/5 
36—5/16 83-U/7 68-5/9 
22-l/U 37-1/3 514-l/ 6 
-100- EXERCISE U 5 (Continued,) 
c. 79-11/12 ■ UG-i]/is Ug-5/g 
-.57-5/8 23-5/12 ■ 37-3/^ 
d. 75-15/21* 6/-IU/36 72-3/5 
67-5/6- . lg-U/9 53-5/8 
EXERCISE U 6 
(1) . • >.:-(a.) , 0) 
a. Ug-3/U x 'l2-1/2 16-1/2x16-1/5 130-1/2x150-2/3 
b. ss-2/3 x 16-l/U 125-3/>+ x 35-3/8 IUL-7/15 x 17-U/5 
c. 196-5/6 X 75-1/2 108-2/3 x 9-1/6 80-1/10 x 25-1/5 
d. 65-1/7 x 27-1/3 190-1/12 1 UB-1/6 27-7/12 x 811-3/U 
EXERCISE U 7 
(1) ' (2) (3)' 
a. 13-1/2 r 14-1/2 7-1/2 T 6-3/7 8-1/3 4- 6-l/U 
b. 37-1/3 ~ 6-2/9 ll+-1/U - 14-i/a 11-5/9-r5-2/3 
c. 2S-3/U f 5-3/U 5-3/7 - 1-U/7 'B-8/15 -f 2-6/35 
d. 37-1/2 -r 3-1/8 16-7/8 -f 6-3/7 2S-U/5 -f- 1-13/23 
EXERCISE U 8 
Express the following as fractions of 100. 
(1) (2) (3) 
a. 6-l/U 20 62-1/2 
b. S-l/3 ' '' ’ ‘ 25 ' 66-2/3 
c. 10 ' 33-1/3 75 
d. 12-1/2 37-1/2 83-1/3 
e. 16-2/3 * 50 S7-1/2 
-101- EXERCISE U 9 
Read orally each, number to be added. Then find the sum. 
(1) (2) (3) 
3578.1329 721.001 " 1925.1926 
8a.098 9.1 1898.1001 
U3.1l 85.Ug 21.0001 
6.9 2000.0002 - 10.01 
2.1 52u3.8755 aooo.iUUU 
75.16 2978.3519 2222.1866 
221.133 ' 1795.2901 3198-7152 
EXffilCl 35 30 
Carry to three places. 
(1) (2) 
a. ij.Ugy. 3.67 U 87.7 25.5714 
b. 13U.57 r 32 ;■ ' 569 -r 23^.67 
c. 1333.6 -f- . 8U.5 " ' .39145 -i- 2.3875 
d. 1,3785 ~ 13.89 >4.85712 -f 2.37^4 
EXERCISE 51 
Change the following values to decimals, Do orally. 
A) (2) 0) W 
a. 3/8$ .2$ 1-3/U i 1/1+ $ 
b. 1/12 of 6$ 1/U of 8$ , - .025$ .1 $ 
o. ; .16$ $ .006$ 
—1Q£“ SXSRCISS 51 (Continued) 
■f. V ■ 
4- 1/S IU/50 3/U 1/5 
e. 3/6 1/10 ,3/10. . .lj/100 
EXERCISE 52 
(1) ■ 
a. 3-1/U X iu.76 72-3/5 X IUU.7S 
b. 7-U/9 X 15.6U2 95.59 x 101-7/11 
c. 6.72 x 19-3/8 23-9/lU X IU. 721 
d. U. 928 X 25-15/16 12-37/bo X’106.78 
e. 25.587x12-2/3 105.723x15-17/18 
EXERCISE 53 
(1) ' (2) 
a. 16-U/5 -r U/2 37.33-1/3 -r 
b. iU.S -r 3-5/8 31,U '-1 17-U/9 
o. U7-17/50 -V 5.26 5U.9375 ■-'■'s-3/16' 
d. 19.991* ~ 6-1/2 132.09 15-U/I7 
e. 17-3/U U.U375 16-5/2 -- 2.072125 
~ , / 
EXERCISE 
iiixpress 
1. 1 week in hours. 
2. 3 hours in minutes. 
3. U days in hours. 
U. 2 years in days. 
■ ,r10> EXERCISE 5U (Continued) 
5. 2-1/2 minutes in seconds. 
6. Us hours in days. 
7. lU days in hours. 
8. 360 seconds in minutes. 
9. 96 hours in days. 
10. 720 seconds in minutes, . 
11,. 5,000 years in centuries. 
12. weeks in hours, 
13. 3 centuries in days. 
IU. years in seconds, 
15. c.c, in liters. 
16. B.U meters in millimeters. 
17. .03 meters in centimeters. 
f;. ' 
IS. 3S grams in milligrams, 
19. 100 c.c. as liters. 
* f 
20. 90 c.c. as ounces. 
21. 8 minims as c.c’s. 
22. .06 Gm. as grains. 
23. 3 miles as kilometers 
2U-. 6 feet as meters. 
25. 10 inches as centimeters. 
26. ■ U. 26 liters as c.c. 
27. 100° F. as C° 
28. 90° C. as F. 
29. 196° F. as C° 
30. 3-5 grams in grains. PROBLEMS 
Iv A man sick in bed, drinks 1 quart of milk during a 2*4 hour period. 
In addition he is given 2000 c.c. of a 10$ solution of glucose 
(sugar). How many calories does .he receive? 
2. 59$ of the body’s weight is water. A man weighs 15,0 pounds. How 
many pounds are water. 
3. A patient is to receive three times a day a teaspoonful of cough 
medicine. In each teaspoonful there is l/U grain of codeine. How 
much codeine is there in 8 fluidounces of cough medicine? Express 
* 
ingrains, nigra., grains/ 
*4. How would you prepare 300 c.c. of bOs alcohol from a stock .solution 
of 95$ ? 
5. During a 7-1/2 mile march a soldier perspires 1,2000 c.c, of fluid. 
It requires 0.5 calorie to evaporate 1 c.c. of'perspiration. How 
many calories are needed to.evaporate 1,200 c.c.? 
6. A certain soldier at rest requires 3000 calories uer day, of which 
1/5 actually goes into work. The remainder is' dissipated as heat. 
Express in calories the number needed' for work and the number 
needed for the dissipation of heat. 
7. A soldier weighing 160 pounds, carrying MO pounds, and walking 15 
miles at the rate of 3 miles-per hour on a level surface will per- 
form an amount of work equivalent to 350 foot-tons of work. 
6.3 foot-tons of work require *4.1 calories. 
Find the number of calories needed to do the foot-tons of 
work. 
-105- 8. If the left ventricle pumps out 70 cc. of blood at each beat, 
and the heart rate is JO per minute, find the amount of blood 
pumped by the ventricle per minute. 
9. If during exercise, the rate increases to 120 per minute, what 
is the volume pumped per minute? 
10. How many grams of sugar yield 3000 calories? 
11. The circumference of the earth at the equator is miles. 
Express in meters? 
,12. When it is noon in Greenwich, England, what time is it at: 
a. Kiev, Russia - 30° B 
b. Leningrad, Russia - 6o° E 
c. Calcutta, India - 90° B 
d. Tokio, Japan - lUo° E 
0 
e. Sydney,Australia - 150 E. 
f. Attu Island, Alaska - 173° E. 
g. Wellington, 1\T. Z. - E. 
h. Dutch Harbor - 165° W 
i. Sitka,Alaska - 135° W 
J. San Francisco - 120° W 
k. New Orleans - 90° W 
l. New York - 75° W 
m Iceland - 20° W 
-io6- Fig 6 
At noon the sun lies directly over Greenwich which is at 0° 
longitude. The earth rotates from West to Sast. It goes completely 
around 3CO0 in 2*i hours. When the- earth moves eastward, the 
western longitudes come directly under the sunTs rays. Before doing 
the problems, it is necessary to find how many degrees the earth moves 
in some unit of time, as an hour. 
-107- TABLES 
MULTIPLICATION AND DIVISION TABLE 
Multiplication 
To find the product of 8 x 9, for example, first locate on the 
top line, the number 9» Next locate the number 8 on the left hand 
vertical line. Lastly, follow the' 9 column downward until it meets the 
extension of the 8 line. The 9 column and 8 line meet at 72, which is 
the product of 8 x 9. ’ 
Division 
To find what numbers when multiplied, give 72, first locate 72, 
Next follow the column upward to the end, which is 9- Next follow the 
line from 72, to the extreme left, to 8. 8 and 9 are the answers. 
72 r 9 ? Find 72.; Locate 9at the end of the vertical 
column. Next follow the horizontal line to the extreme left'. The 
answer is 8. 
1 
2 
3 
U 
5 
*. 6 
7 
S'- 
9 
10 
11 
12 
2 
U 
6 
8 
10 
12 
14 
16 
18 
.20 
22 
• • 2U 
3 
6 
9 
12 
15 
is 
21 
2U 
27 
?? 
?6 
u 
8 
12 
16 
20 
2U 
28 
32 
36 
uo 
UU 
% 
5 
10 
15 
20 
25 
30 
35 
% 
50 
55 
60 
6 
12 
IS 
2U 
30 
36 
U2 
US 
5U 
60 
66 
72 
7 
lU 
21 
28 
35 
42 
U9 
56 
63 
70 
77 
SU 
8 
Id 
2U 
32 
4) 
US 
56 
6U 
72 
80 
88 
96 
9 
IS 
27 
36 
u5 
5U 
63 
72 
81 
90 
99 
108 
10 
20 
30 
uo 
50 
60 
70 
SO 
90 
100 
110 
120 
11 
22 
33 
UU 
55 
66 
77 
ss 
99 
110 
121 
132 
12 
2U 
36 
Ug 
60 
72 
sU 
96 
108 
120 
132 
luU 
This table is to be memorized. 
-108-