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MATHEMATICS
FRACTIONS
STUDENT TEXT
REV 1
©2003 General Physics Corporation, Elkridge, Maryland
T
M
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TABLE OF CONTENTS
FIGURES AND TABLES ...........................................................................................................ii
OBJECTIVES ............................................................................................................................ iii
DEFINITION OF FRACTIONS .................................................................................................. 1
Addition and Subtraction of Fractions ...................................................................................... 2
Multiplication of Fractions ..................................................................................................... 10
Division of Fractions............................................................................................................... 11
REDUCING FRACTIONS ........................................................................................................ 12
Changing the Form of Fractions ............................................................................................. 12
CALCULATOR EXERCISES ................................................................................................... 14
Addition of Fractions + ........................................................................................................ 14
Subtraction of Fractions  ..................................................................................................... 17
Multiplication of Fractions  ............................................................................................... 19
Division of Fractions  ......................................................................................................... 20
SUMMARY ............................................................................................................................... 22
PRACTICE EXERCISES .......................................................................................................... 23
GLOSSARY ............................................................................................................................... 24
EXAMPLE EXERCISE ANSWERS ......................................................................................... 25
PRACTICE EXERCISE ANSWERS ........................................................................................ 30
MATHEMATICS – CHAPTER 2 FRACTIONS
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FIGURES AND TABLES
Figure 2-1 Example 2/7 + 3/7 .................................................................................................... 2
Figure 2-2 Example 3/4 ............................................................................................................. 2
Figure 2-3 Example 6/8 ............................................................................................................. 3
Figure 2-4 Example 1/2 × 1/3 .................................................................................................. 10
No Tables
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OBJECTIVES
Upon completion of this chapter, the student will be able to perform the following
objectives at a minimum proficiency level of 80%, unless otherwise stated, on an oral
or written exam.
1.
DEFINE and GIVE EXAMPLES of:
a.
proper fractions
b.
improper fractions
c.
mixed numbers.
2.
CONVERT between proper fractions, improper fractions, and mixed numbers.
3.
SOLVE mathematical problems involving fractions by:
a.
reducing the fraction
b.
solving for the Lowest Common Denominator
c.
changing the form of the fraction
4.
Without a calculator; ADD, SUBTRACT, MULTIPLY, and DIVIDE fractions (proper
fractions, improper fractions, and mixed numbers).
5.
With an approved calculator; ADD, SUBTRACT, MULTIPLY, and DIVIDE fractions
(proper fractions, improper fractions, and mixed numbers).
MATHEMATICS – CHAPTER 2 FRACTIONS
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DEFINITION OF
FRACTIONS
What is meant by
Whole numbers are used for counting; that is,
describing the number of objects in a group.
However, the result of a measurement need not
be a whole number, and in fact, rarely is. The
number of pages in this book is by definition a
whole number, but the weight of the book in
pounds is probably not a whole number. We
need a method that can describe the magnitude
of numbers that lie between the whole numbers.
This is achieved through the use of fractions.
Suppose that we have a circle and we divide the
circle into 4 equal parts.
How can we
mathematically describe one of the parts? A
fraction is the ratio of 2 whole numbers, and it
indicates division.
Each part of the circle is called “one fourth,” and
1
is denoted by the fraction
. A fraction
4
indicates division, and the dividend is called the
numerator.
The divisor is called the
denominator.
9
of the area of a circle?
16
Divide the circle into 16 equal parts. Combine 9
9
of these parts. The resulting area will be
of
16
the area of the circle.
Example 2-2
Suppose the circle we had has a total area of 12
square inches. We still divide the circle into 4
equal parts.
What will be the area of one of the parts?
If the total area is divided by 4, the result will be
the area of 1 of the parts.
Total Area = 12 square inches
Number of parts = 4
Area of one part = 12  4 =
12
4
= 3 square inches
Example 2-3
1
The fraction
means 1  4.
4
1
4
There are three different types of fractions. A
proper fraction is one in which the numerator is
less than the denominator, and so has a value less
191
3 7
than 1. The fractions
,
, and
are
4 16
327
proper fractions; their values are less than 1.
NUMERATOR
DENOMINATO R
Example 2-1
The meaning of a fraction, then, is that some
entity is divided into an equal number of parts
indicated by the denominator, and then added as
many times as indicated by the numerator.
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An improper fraction is one in which the
numerator is equal to or greater than the
denominator. Its value will be equal to or greater
23
6 16
than 1. The fractions
,
, and
are
6
7
13
improper fractions; their values are equal to or
greater than 1.
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A mixed number consists of a whole number and
3
a fraction. The mixed number 3 or 3 3/4
4
3
means the whole number 3 plus the fraction .
4
This is the type of number that would arise from
the measurement of the width of this page; 8
1
1
inches plus
inch equals 8 or 8 ½ inches.
2
2
numerators and placing the result over the
common denominator.
Add
2
3
and .
7
7
2 3 23 5
 

7 7
7
7
Subtract
ADDITION AND SUBTRACTION OF
FRACTIONS
It is often necessary to add and subtract fractions.
In adding and subtracting fractions, it is
important to remember first what a fraction
represents. It means a division of a unit into a
number of equal parts as indicated by the
denominator, and then adding these parts as
many times as indicated by the numerator.
2
means divide by 7 and add 2
7
3
times. The fraction
means divide by 7 and
7
add 3 times. If these 2 fractions are added, the
result will be the same as dividing by 7 and
adding 5 times. The circle below is divided into
7 equal parts. Two of the parts are lightly
shaded. Three of the parts are shaded dark. The
5
sum will be 5 of the parts, or
of the area of the
7
circle.
The fraction
7
9
from
11
11
9 7 97 2
 

11 11
11
11
Example 2-4
If the fractions to be added or subtracted do not
have the same denominator, they cannot be
added or subtracted directly as shown above.
They must be altered so that they have the same
3
denominator. Consider the fraction and the
4
circle shown in Figure 2-2. If we divide the
circle into 4 equal parts and add 3 of them, the
3
result will be the area of
of the circle.
4
Figure 2-2 Example 3/4
Figure 2-1 Example 2/7 + 3/7
Fractions with the same denominator are added
or subtracted by adding or subtracting their
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On the other hand, if we divide the circle into 8
equal parts and shade 6 of them, the result will
6
be of the area of the circle.
8
Therefore any number (or fraction) multiplied by
1 (or a number divided by itself) is still equal to
the original number.
3
3
x1 
4
4
1
2
2
3 2 6

4 2 8
Figure 2-3 Example 6/8
Notice that this is exactly the same area as was
3
represented by the fraction . Therefore, the
4
6
3
fraction
is equivalent to the fraction . If we
8
4
multiply both the numerator and denominator of
a fraction by the same number, we do not change
the value of the fraction.
Recall that any number multiplied by one is
equal to itself.
51=5
Example 2-7
This is the method that must be used when
adding or subtracting fractions with different
denominators. One of the fractions is altered to
an equivalent form so that its denominator is the
same as the denominator of the fraction to which
it is to be added (or subtracted). The numerators
are then added (or subtracted) and placed over
the common denominator, as before.
Add
3
3
1 
4
4
2 1

3 6
2 2 2 4


3 3 2 6
Example 2-5
Also any number divided by itself is equal to
one.
4 1 4 1 5
 

6 6
6
6
Example 2-8
5
1
5
Note that
2
1
2
MATHEMATICS – CHAPTER 2-
2 1
2 1 3
 .
+
is NOT equal to
3 6
36 9
What we have done in Example 2-8 is to find the
lowest common denominator (LCD). The LCD
is the smallest number that can be divided by all
the denominators in a problem involving several
Example 2-6
FRACTIONS
6 3

8 4
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fractions. In the above example, the number 6 is
the LCD, since it is the smallest number that can
be divided by 6 and 3.
Step 5. For each fraction, create a fraction
equal to one with each factor required to make
the denominator equal to the LCD.
1 2 3
What is   ?
4 3 8
Step 6. Multiply the numerator and denominator
of each fraction by the required factor of the
LCD.
Step 7. With all fractions now with the same
denominator (the LCD) add and subtract the
numerators over the least common denominator.
The lowest common denominator is 24
1 1 6
6


4 4  6 24
Add the fractions
2 2  8 16


3 3  8 24
1
1
and
4
3
Step 1. Determine the smallest factors of each
of the denominators to be added or subtracted.
3 3 3 9


8 8  3 24
4 has factors of 2 × 2
3 has factors of 1 × 3
1 2 3
  
4 3 8
Step 2.
Determine the Least Common
Denominator (LCD) by determining how many
times each factor must be used.
6 16 9



24 24 24
The LCD must have factors of 2 × 2 × 3.
6  16  9 13

24
24
Denominator 1
Number of times
Factors of 4
2×2
2
||
Denominator 2
Number of times
Factors of 3
1×3
3
|
Example 2-9
The easiest way to solve for the LCD is to factor
each denominator to its smallest factors.
Step 1. Determine the smallest factors of each of
the denominators to be added or subtracted.
Step 2.
Determine the Least Common
Denominator (LCD) by determining how many
times each factor must be used.
Step 3. Multiply together all of the factors of the
Least Common Denominator (LCD).
Step 4. For each denominator to be added or
subtracted determine which factor(s) the
denominator must be multiplied by to reach the
LCD.
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The LCD must have 2 twos and 1 three.
Step 3. Multiply together all of the factors of
the Least Common Denominator (LCD).
The LCD is equal to 2 × 2 × 3 = 12
(Cont'd in next Column)
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Step 4. For each denominator to be added or
subtracted determine which factor(s) the
denominator must be multiplied by to reach
the LCD.
Step 6.
Multiply the numerator and
denominator of each fraction by the required
factor of the LCD.
The first fraction becomes
Least Common Denominator
Factors of LCD 12
2×2×3
1 3 3
 .
4  3 12
The second fraction becomes
1 2  2 4

3  2  2 12
Denominator 1
Factors of 4
2×2
Missing factors
3
Factors of LCD 12
2×2×3
Step 7. With all fractions now with the same
denominator (the LCD) add and subtract the
numerators
over
the least
common
denominator.
3
4 3 4 7



12 12
12
12
Denominator 2
Factors of 3
1×3
Missing factors
2×2
Example 2-10
The first fraction’s denominator must be
multiplied by 3 to reach the LCD.
The second fraction’s denominator must be
multiplied by 2 × 2 to reach the LCD.
Step 5. For each fraction, create a fraction
equal to one with each factor required to make
the denominator equal to the LCD.
The first fraction
1
3
must be multiplied by .
4
3
The second fraction
22
22
1
must be multiplied by
3
(Cont'd in next Column)
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Add the fractions
Step 4. For each denominator to be added or
subtracted determine which factor(s) the
denominator must be multiplied by to reach
the LCD.
5
3
and
.
12
42
Step 1. Determine the smallest factors of each
of the denominators to be added or subtracted.
Least Common Denominator
Factors of LCD
Step 2.
Determine the Least Common
Denominator (LCD) by determining how many
times each factor must be used.
Denominator 1
Denominator 1
Missing factors
Factors
Factors of LCD
Factors of
Denominator 2
Factors of
Missing factors
Denominator 2
Factors of
Step 5. For each fraction, create a fraction
equal to one with each factor required to make
the denominator equal to the LCD.
The first fraction
by
must be multiplied
.
Step 3. Multiply together all of the factors of
the Least Common Denominator (LCD).
The second fraction
The LCD is equal to
(Cont'd in next Column)
must be multiplied
by
(Cont'd in next Column)
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Step 6.
Multiply the numerator and
denominator of each fraction by the required
factor of the LCD.
Calculate the sum of
The first fraction becomes
Step 1. Determine the smallest factors of each
of the denominators to be added or subtracted.
The second fraction becomes
1
3 7

 .
12 16 8
Denominator 1 has factors of
Step 7. With all fractions now with the same
denominator (the LCD) add and subtract the
numerators
over
the least
common
denominator.
Denominator 2 has factors of
Denominator 3 has factors of
Step 2. Determine the Least Common
Denominator (LCD) by determining how many
times each factor must be used.
Denominator 1
Example 2-11
Number of times
Factors of
Denominator 2
Number of times
Factors of
Denominator 3
Number of times
Factors of
The LCD must have
Step 3. Multiply together all of the factors of
the Least Common Denominator (LCD).
(Cont'd in next Column)
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Step 4. For each denominator to be added or
subtracted determine which factor(s) the
denominator must be multiplied by to reach
the LCD.
Step 5. For each fraction, create a fraction
equal to one with each factor required to make
the denominator equal to the LCD.
The first fraction
must be multiplied
Least Common Denominator
by
Factors of LCD
Denominator 1
The second fraction
must be multiplied
Factors of
by
Missing factors
Factors of LCD
The third fraction
Denominator 2
must be multiplied
by
Factors of
Step 6.
Multiply the numerator and
denominator of each fraction by the required
factor of the LCD.
Missing factors
Factors of LCD
The first fraction becomes
Denominator 3
The second fraction becomes
Factors of
The third fraction becomes
Missing factors
Step 7. With all fractions now with the same
denominator (the LCD) add and subtract the
numerators
over
the least
common
denominator.
The first fraction’s denominator must be
multiplied by
The second fraction’s denominator must be
multiplied by
The third fraction’s denominator must be
multiplied by
Example 2-12
(Cont'd in next Column)
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Alternate solution
Multiply each fraction by one in the form of the
missing factors divided by the missing factors.
1
3 7
  .
Calculate the sum of
12 16 8
12 missing 2  2 or 4
12 = 2  2  3
16 = 2  2  2  2
1 4
4
 
12 4 48
8=222
16 missing 3
To determine the LCD each denominator must
contain all factors common to all the other
denominators.
3 3 9
 
16 3 48
8 missing 2  3 or 6
A simple way to do this is to multiply all the
denominators together.
7 6 42
 
8 6 48
12  16  8 = 1,536
Now, with all the denominators the same
perform the required mathematical operation.
However this will not give you the smallest
value and would require extra multiplication.
4
9 42 4  9  42 55




48 48 48
48
48
Determine the maximum number of times a
factor appears in a factor. In this example that
means four 2’s and one 3. Thus the LCD is 2 
2  2  2  3 = 48.
Convert to a mixed fraction and simplify.
55
7
1
48
48
To convert each fraction to a fraction with the
LCD determine which factor(s) is missing from
the current denominator
Since 7 has factors of 1  7 and 48 has factors of
2  2  2  2  3 and there are no common
factors in this case the fraction is simplified.
12 = 2  2  3 is missing 22
16 = 2  2  2  2 is missing 3
Example 2-13
8 = 2  2  2 is missing 23
(Cont'd in next Column)
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When fractions are to be added or subtracted,
their LCD must be found.
If we have to remove
1
inch from a piece of
4
7
wood
inches wide, what will the resulting
8
width be? This would be determined by
1
7
subtracting
from .
4
8
7 1
 
8 4
Note that the numerators are multiplied directly,
as are the denominators. There is no need to find
a lowest common denominator, as in the case of
addition or subtraction.
1
1
and can be visualized
2
3
with regard to the area of the circle. We first
divide the area of the circle into two equal parts.
1
This accounts for the fraction . This area is
2
now divided into three equal parts.
This
1
1
accounts for the fraction . The final area is
3
6
the area of the circle.
The multiplication of
Example 2-14
MULTIPLICATION OF FRACTIONS
Multiplying the numerators together to obtain the
numerator of the product, and multiplying the
denominators together to obtain the denominator
of the product multiply fractions.
Multiply
Figure 2-4 Example 1/2 × 1/3
2
5
and .
3
7
2 5 2  5 10
 

3 7 3  7 21
Example 2-15
Multiply
1
1
and .
2
3
Example 2-16
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DIVISION OF FRACTIONS
Fractions may be divided by utilizing the rule
that the value of a fraction is unchanged if both
the numerator and denominator are multiplied by
1
the same number. The operation divided by
2
1
is indicated as:
3
1
2
1
3
or
2
2 3 3 2 4 8
    
3 4 3 3 3 9
4
5 1
 
6 8
2 5
 
7 6
Example 2-17
1 1

2 3
If we multiply numerator and denominator by 3,
we obtain:
1
3
3
2
 2
1
3 1
3
If we now multiply numerator and denominator
by 2, we obtain:
3
2
3
2

1 2 2
This is the value of the quotient of 1/2  1/3.
Simply inverting the denominator and
multiplying the result by the numerator obtains
the same result.
1
2  13  3
1 2 1 2
3
The process of inverting means make the
numerator the denominator and vice versa. Thus
5
7
if we invert , we obtain .
7
5
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REDUCING FRACTIONS
Reduce the fraction
In dealing with fractions, it is normal procedure
to express the fraction in such a manner that the
numerator and denominator are as small as
possible. This is known as reducing a fraction to
its lowest terms, and a fraction is said to be
reduced to its lowest terms when the numerator
and denominator have no common factor other
than 1.
Reducing a fraction to its lowest terms is made
possible by the fact that the value of the fraction
is unchanged if both the numerator and
denominator are divided by the same number.
6
Although the fraction
is a perfectly proper
8
3
fraction, it can be reduced to the fraction
by
4
dividing the numerator and denominator by 2.
The fraction is now reduced to its lowest terms
since the numerator and denominator have no
common factor other than 1.
Reduce the fraction
27
to its lowest terms.
45
27 27  9 3


49 45  9 5
Example 2-18
Finding the common factors for the purpose of
fraction reduction is normally done by a trialand-error procedure.
To reduce a fraction to its lowest terms factor
both the numerator and the denominator into
their smallest values.
27
to its lowest terms.
45
27 3  3  3

45 3  3  5
Cancel out any factor found in both terms.
27 3 3 3 3
   
45 3 3 5 5
That will result in the fraction being reduced to
its lowest terms.
6

8
Example 2-19
CHANGING THE FORM OF
FRACTIONS
In performing operations with fractions, it is
sometimes necessary to alter them to an
equivalent form for ease of computation. For
example, if we wished to multiply the two mixed
1
1
numbers, 2 and 7 , it becomes much easier if
3
8
we change the mixed numbers to improper
fractions first. A mixed number can be changed
to an improper fraction by recognizing that a
mixed number is just the sum of a whole number
and a fraction.
Change the mixed number 4
2
to an improper
3
fraction.
4
2
2 4  3 2 12 2 14
 4 
 
 
3
3
3
3 3 3 3
Example 2-20
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Notice in this example that we found the LCD of
the whole number and the fraction, and then
added them. Multiplying the whole number by
the denominator of the fraction and then adding
the numerator of the fraction to the result obtains
the same result. The total is then placed over the
denominator of the fraction.
Change the mixed number 5
3
to an improper
4
fraction.
First factor both the numerator and denominator
into the smallest terms
399
3  7  19

24
2 2 23
In this case the only factor in both terms is a
three.
Cancel out terms that are in the
numerator and denominator.
Multiply the
remaining factors together in each term.
7  19
133

2 2 2
8
Example 2-21
The fraction is in its lowest terms. Now it must
be converted into a mixed number.
We can now multiply two mixed numbers
together.
To convert into a mixed number divide the
denominator into the numerator.
1
1
Multiply 2 by 7 .
3
8
Change the improper fraction
1 2  3  1 6  1 7
2 


3
3
3
3
133  8 = 16 with a remainder of 5.
7
133
to a mixed
8
number.
1 7  8  1 56  1 57



8
8
8
8
133
5
5
 16   16
8
8
8
5
So the mixed number becomes 16 .
8
7 57 399


3 8
24
Example 2-22
Reduced to lowest terms.
399
is the answer but it should be checked to
24
determine if it is reduced to lowest terms and
converted into a mixed number.
An improper fraction can be changed to a mixed
number by dividing the numerator by the
denominator, with any remainder placed over the
denominator as the proper fraction.
(Cont'd in next Column)
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Change the improper fraction
CALCULATOR EXERCISES
29
to a mixed
5
The calculator can be used for solving addition,
subtraction, multiplication, and division of
fractions. The following are examples of its use.
number.
ADDITION OF FRACTIONS +
Find the sum of
Example 2-23
2
3
and .
7
7
2 3
 ?
7 7
Since the denominator is the same, add the
numerators and place the sum over the common
denominator.
2 3 23
 
?
7 7
7
Enter Operation
2
+
2
3
=
5
Display
The common denominator is 7.
Therefore,
2 3 5
 
7 7 7
Example 2-24
MATHEMATICS – CHAPTER 2FRACTIONS
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Find the sum of
3
2
4
and and .
5
5
5
Find the sum of
2
3
and .
7
7
Since the denominators are the same, add the
numerators and place the sum over the common
denominator.
Enter Operation
Display
2

2
3 2 4
  ?
5 5 5
7
+
0.2857143
3

3
3 2 4 3 2 4
  
?
5 5 5
5
7
=
0.7142857
Enter Operation
Display
2  7 + 3  7 = 0.7142857
3
+
3
2
+
5
4
=
9
In a case where the denominator for both
fractions is the same, multiply the resultant
answer by the denominator to obtain the
fractional number.
0.7142857 × 7 = 5
The common denominator is 5.
Then place the answer (5) over the common
5
denominator (7) to obtain the answer .
7
Therefore,
3 2 4 9
4
   1
5 5 5 5
5
Example 2-26
Example 2-25
Solving equations with a fraction on a calculator
does not necessarily require using all the rules
presented in this chapter. Entering the numbers
into the calculator following the proper rules for
the calculator will provide you with the correct
answer but it will probably be in decimal form.
So there are several steps to convert from
decimal to fraction form.
MATHEMATICS – CHAPTER 2FRACTIONS
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To solve addition of fractions without concern
for keeping the answer as a fraction, simply enter
all the mathematical operations into the
calculator in the order that they occur.
3 2
4
Find the sum of , , and .
5
5 5
Enter Operation
Display
3

3
5
+
.6
2

2
5
+
1
4

4
5
=
1.8
Find the sum of
1 2 3
  .
4 3 8
Enter Operation
3  5 + 2  5 + 4  5 = 1.8
Since in this case the denominator for all the
fractions is 5, multiply the resultant answer by 5
to obtain the fractional number.
Display
1

1
4
+
.25
2

2
3

0.9166667
3

3
8
=
0.5416667
Example 2-28
1.8 × 5 = 9
Then place the answer (9) over the common
9
denominator (5) to obtain the answer .
5
Unless told not to, convert all improper fractions
into mixed numbers.
9 5 4
4
4
   1  1
5 5 5
5
5
The sum of
3 2
4
4
, , and
is 1 .
5 5
5
5
Example 2-27
You can solve fractions on the calculator with
fractions that have different denominators.
However, the answer will result in a decimal
number that may not be easily converted into a
fraction.
MATHEMATICS – CHAPTER 2FRACTIONS
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SUBTRACTION OF FRACTIONS 
Find the difference between
Find the difference between
9
7
and .
11
11
14
2
and .
9
3
14 2
 ?
9 3
9 7
 ?
11 11
To find the lowest common denominator:
Since the denominators are the same, subtract
the numerators and place the difference over the
common denominator.
Enter Operation
9

9
9 7 97
 
?
11 11
11
3
=
3
Enter Operation
Display
9

9
7
=
2
For the numerator:
The common denominator is 11.
Therefore,
3 is the common factor of both denominators.
Therefore, the lowest common denominator of
the two fractions is 9. Hence, in order to
2
subtract these two fractions, the fraction
must
3
have both its numerator and denominator
multiplied by 3.
To convert the numerator of
Enter Operation
9 7
2
 
11 11 11
Display
2
:
3
Display
2

2
3
=
6
Example 2-29
To convert the denominator of
Enter
Operation Display
3

3
3
=
9
2
:
3
(Cont'd in next Column)
MATHEMATICS – CHAPTER 2FRACTIONS
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Solving subtraction of fractions using the
calculator follows the same basic rules as
addition.
The result is
2 6

3 9
Solve for the difference between
Therefore,
Enter Operation
14 2 14 6
 

9 3 9 9
To solve the numerator of the subtraction:
Enter Operation
14
2
and .
9
3
Display
14

14
9

1.5555556
2

2
3
=
0.8888889
Display
14

14
6

8
Example 2-31
Place the difference (8) over the common
denominator (9).
Therefore,
14 6 8
 
9 9 9
Example 2-30
MATHEMATICS – CHAPTER 2FRACTIONS
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MULTIPLICATION OF FRACTIONS

Find the product of
Find the product of
2
5
and .
3
7
1
1
and .
2
3
1 1
 ?
2 3
2 5
 ?
3 7
1 1 11
 
?
2 3 23
2 5 25
 
?
3 7 3 7
To solve for the numerator multiply 1 by 1:
Enter Operation
To solve for the numerator multiply 2 by 5:
Enter Operation
2

Display
1

1
1
=
1
Display
2
To solve for the denominator multiply 2 by 3:
5
=
10
Enter Operation
To solve for the denominator multiply 3 by 7:
Enter Operation
3
7

2

2
3
=
6
Display
3
=
Place the new numerator over the new
denominator.
21
Place the new numerator over the new
denominator.
Therefore,
Display
Therefore,
2 5 10
 
3 7 21
1 1 1
 
2 3 6
Example 2-33
Example 2-32
MATHEMATICS – CHAPTER 2FRACTIONS
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You can also solve multiplication of fractions
using the calculator without concerns for
maintaining fraction form.
DIVISION OF FRACTIONS 
Find the quotient of
Find the product of
Enter Operation
2
5
and .
3
7

2
3

0.6666667
5

3.33333…
7
1 1
 ?
2 3
Display
2
=
1
1
divided by .
2
3
1 1
3
2  2
1 1
3
3 3
To solve for the numerator multiply the first
fraction by the denominator of the second
fraction:
0.4761905
Example 2-34
1
1 3
3  
2
2 1
Or you can multiply the numerators together and
then divide through by the denominators.
The numerator becomes:
Find the product of
Enter Operation
2
5
and .
3
7
Display
2

2
5

10
3

3.33333…
7
=
0.4761905
Example 2-35
Enter Operation
Display
1

1
3
=
3
The denominator becomes:
Enter Operation
Display
2

2
1
=
2
Hence, the numerator becomes:
1 3 3
 
2 1 2
(Cont'd in next Column)
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For the denominator:
Find the quotient of
1
1 3
3  
3
3 1
1 1
 
2 3
The numerator becomes:
Enter Operation
Enter Operation
Display
1

1
3
=
3
The denominator becomes:
Enter Operation
Display
3

1
1
=
3
Display
1

2

1

1
3
)
0.333
=
1.5
1
(
0.5
Example 2-37
Alternately, you can follow the rules of dividing
fractions and invert and multiply the divisor
(second) fraction.
Hence, the denominator becomes:
1 3 3
  1
3 1 3
Find the quotient of
The net result is then:
1
3
3
3
2
 2 
1
3 1 2
3
Therefore,
1
1
divided by .
2
3
1
1
divided by .
2
3
1 1 1 3
   
2 3 2 1
Multiply the numerators then divide through by
the denominators.
1 1 3
 
2 3 2
Enter Operation
Display
1

1
3

3
2

1.5
1
=
1.5
Example 2-36
Dividing fractions by fractions on a calculator
typically requires an additional step of grouping
the mathematical operations together using
parentheses.
Example 2-38
MATHEMATICS – CHAPTER 2FRACTIONS
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To multiply fractions, multiply the numerator by
the numerator and multiply the denominator by
the denominator:
SUMMARY
Fractions

Numerator – top number in a fraction

Denominator – bottom number in a
fraction

Proper fraction – numerator is less than
denominator

Improper fraction – numerator is greater
than or equal to denominator

Mixed number – sum of an integer and a
proper fraction

Fractions, like whole numbers can be:
a.
Added
b.
Subtracted
c.
Multiplied
d.
Divided
Multiply
2 5 2  5 10
 

3 7 3  7 21
To divide fractions invert the fraction and
multiply using the rules above.
2
2 3 3 2 4 8
    
3 4 3 3 3 9
4
After solving problems with fractions, reduce all
fractions to lowest terms:
Reduce the fraction
To add or subtract fraction the denominator must
be the same:
Add
2 1
 .
3 6
2 2 2 4


3 3 2 6
4 1 4 1 5
 

6 6
6
6
7 1
Subtract  
8 4
7  1  2 
    
8  4  2 
7 2
 
8 8
Cancel out any factor found in both terms.
27 3 3 3 3
   
45 3 3 5 5
That will result in the fraction being reduced to
its lowest terms.
To solve problems with mixed numbers,
multiply the whole number by the denominator
and ad to the numerator. Place the sum over the
denominator:
Change the mixed number 4
2
to an improper
3
fraction.
4
FRACTIONS
27
to its lowest terms.
45
27 3  3  3

45 3  3  5
5
8
MATHEMATICS – CHAPTER 2-
2
5
and .
3
7
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2
2 4  3 2 12 2 14
 4 
 
 
3
3
3
3 3 3 3
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PRACTICE EXERCISES
1.
Indicate whether the following numbers
are proper fractions, improper fractions,
or mixed numbers.
a.
7
16
b.
d.
11
29
e. 3
g. 11
j.
5.
1
8
16
11
m. 9
12
13
a.
3 8

4 32
6 3
b. 2 
7 5
2
c. 11  15
3
19
3
c.
12
8
d.
8 1

3 6
e.
9 2

12 4
f.
11 1

13 2
1
2
f.
1
2
g.
9 1

16 8
h.
1 1

3 5
i.
42
6
43
j.
2 4

5 7
k.
9 8

8 9
l.
3
1
12  6
4
4
3
4
h.
7
5
i.
3
k.
11
16
l.
321
322
n.
4
5
o.
6
5
2.
Change all of the improper fractions in
exercise 1 to mixed numbers.
3.
Change all of the mixed numbers in
exercise 1 to improper fractions.
4.
Reduce the following fractions:
a.
5
15
b.
8
32
c.
54
9
d.
18
6
e.
7
21
f.
6
24
g.
18
24
h.
100
1000
i.
19
28
j.
10
10
k.
2
10
l.
3
12
m.
18
21
n.
2
2
o.
18
72
MATHEMATICS – CHAPTER 2FRACTIONS
Compute the following:
6.
One branch in a fluid piping system
1
carries
of the total system flow. If the
6
total system flow is 27,000 lbm. per hour,
what is the flow in the branch?
7.
A chemistry lab has a stock of 175 bottles
of chemicals. 60 of these bottles contain
sulfuric acid.
55 of them contain
hydrochloric acid. What fraction of the
bottles contain sulfuric acid?
What
fraction contains hydrochloric acid?
8.
The outside diameter of a pipe is 4
23 of 33
1
8
1
inches; the pipe wall thickness is
12
inch. What is the inside diameter of the
pipe?
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GLOSSARY
Denominator
The divisor of a fraction. The bottom number in a fraction.
Denominator - Down
Fraction
The ratio of two whole numbers. It indicates division.
Improper fraction
A fraction where the numerator is equal to or greater than the denominator.
Lowest common
denominator (LCD)
The smallest number that can be divided by all the denominators in a
problem involving several fractions.
Mixed number
A number consisting of a whole number and a fraction.
Numerator
The dividend of a fraction. The top number in a fraction.
Proper fraction
A fraction where the numerator is less than the denominator, and so has a
value less than 1.
MATHEMATICS – CHAPTER 2FRACTIONS
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EXAMPLE EXERCISE
ANSWERS
Step 3. Multiply together all of the factors of
the Least Common Denominator (LCD).
The LCD is equal to 2 × 2 × 3 ×7 = 84
Add the fractions
Step 4. For each denominator to be added or
subtracted determine which factor(s) the
denominator must be multiplied by to reach
the LCD.
5
3
and
.
12
42
Step 1. Determine the smallest factors of each
of the denominators to be added or subtracted.
Least Common Denominator
12 has factors of 2 × 2 × 3
21 has factors of 2 × 3 × 7
Factors of LCD 84
2×2×3×7
Step 2.
Determine the Least Common
Denominator (LCD) by determining how many
times each factor must be used.
Denominator 1
Factors of 12
2×2×3
Denominator 1
Number of times
Missing factors
7
Factors of 12
2×2×3
Factors of LCD 84
2×2×3×7
2
||
Denominator 2
3
|
Factors of 42
2×3×7
Denominator 2
Number of times
Missing factors
2
Factors of 42
2×3×7
The first fraction’s denominator must be
multiplied by 7 to reach the LCD.
2
|
3
|
The second fraction’s denominator must be
multiplied by an additional 2 to reach the
LCD.
7
|
(Cont'd in next Column)
The LCD must have 2 twos, 1 three, and 1
seven.
(Cont'd in next Column)
MATHEMATICS – CHAPTER 2FRACTIONS
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Step 5. For each fraction, create a fraction
equal to one with each factor required to make
the denominator equal to the LCD.
Calculate the sum of
Step 1. Determine the smallest factors of each
of the denominators to be added or subtracted.
5
7
The first fraction
must be multiplied by .
12
7
12 has factors of 2 × 2 × 3
3
The second fraction
must be multiplied by
42
2
2
16 has factors of 2 × 2 × 2 × 2
8 has factors of 2 × 2 × 2
Step 2. Determine the Least Common
Denominator (LCD) by determining how many
times each factor must be used.
Step 6.
Multiply the numerator and
denominator of each fraction by the required
factor of the LCD.
5  7 35
The first fraction becomes

12  7 84
The second fraction becomes
1
3 7

 .
12 16 8
3 2
6

42  2 84
Step 7. With all fractions now with the same
denominator (the LCD) add and subtract the
numerators
over
the least
common
denominator.
35 6 35  6 41



84 84
84
84
Example 2-11
Denominator 1
Number of times
Factors of 12
2×2×3
2
||
3
|
Denominator 2
Number of times
Factors of 16
2×2×2×2
2
||||
Denominator 3
Number of times
Factors of 8
2×2×2
2
|||
The LCD must have 4 twos and 1 three.
(Cont'd in next Column)
MATHEMATICS – CHAPTER 2FRACTIONS
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Step 3. Multiply together all of the factors of
the Least Common Denominator (LCD).
The first fraction’s denominator must be
multiplied by 2 × 2 to reach the LCD.
The LCD is equal to 2 × 2 × 2 × 3 = 48
The second fraction’s denominator must be
multiplied by an additional 3 to reach the
LCD.
Step 4. For each denominator to be added or
subtracted determine which factor(s) the
denominator must be multiplied by to reach
the LCD.
The third fraction’s denominator must be
multiplied by an additional 2 × 3 to reach the
LCD.
Least Common Denominator
Factors of LCD 48
Step 5. For each fraction, create a fraction
equal to one with each factor required to make
the denominator equal to the LCD.
2×2×2×2×3
Denominator 1
The first fraction
Factors of 12
2×2×3
Missing factors
2×2
Factors of LCD 48
2×2×2×2×3
The second fraction
The third fraction
2×2×2×2
Missing factors
3
Factors of LCD 48
2×2×2×2×3
3
must be multiplied by
16
3
.
3
Denominator 2
Factors of 16
1
4
must be multiplied by .
12
4
7
6
must be multiplied by .
8
6
Step 6.
Multiply the numerator and
denominator of each fraction by the required
factor of the LCD.
The first fraction becomes
Denominator 3
Factors of 8
2×2×2
Missing factors
2×3
1 4
4
.

12  4 48
The second fraction becomes
The third fraction becomes
(Cont'd in next Column)
33
9

16  3 48
7  6 42

8  6 48
(Cont'd in next Column)
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Step 7. With all fractions now with the same
denominator (the LCD) add and subtract the
numerators
over
the least
common
denominator.
2
2 3 3 2 4 8
    
3 4 3 3 3 9
4
4
9 42 4  9  42 55




48 48 48
48
48
5
5 1 6 5 8 40 20  2 20
    


6 8 1 6 1 6
3 2
3
8
Example 2-12
If we have to remove
1
inch from a piece of
4
7
inches wide, what will the resulting
8
width be? This would be determined by
1
7
subtracting
from .
4
8
wood
2
2 5 7 2 6 12
    
7 6 5 7 5 35
6
Example 2-17
7 1
 
8 4
Reduce the fraction
7  1  2 
    
8  4  2 
27 3  3  3

45 3  3  5
27
to its lowest terms.
45
Cancel out any factor found in both terms.
7 2
 
8 8
27 3 3 3 3
   
45 3 3 5 5
5
8
That will result in the fraction being reduced to
its lowest terms.
Example 2-14
Multiply
6
23
2  3
3



8 2  2  2 2  2  2 4
1
1
and .
2
3
Example 2-19
1 1 11 1
 

2 3 23 6
Example 2-16
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Change the mixed number 5
3
to an improper
4
fraction.
5
3 (5  4)  3 20  3 23



4
4
4
4
Example 2-21
Change the improper fraction
29
to a mixed
5
number.
29
 29  5  5 and a remainder of 4
5
29
4
4
 5  5
5
5
5
Example 2-23
MATHEMATICS – CHAPTER 2FRACTIONS
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PRACTICE EXERCISE ANSWERS
1.
2.
4.
a.
7
16
proper fraction
b.
19
3
improper fraction
c.
12
8
improper fraction
d.
11
29
proper fraction
e.
1
2
mixed number
f.
1
2
proper fraction
g.
1
8
mixed number
h.
7
5
improper fraction
i.
3
4
mixed number
j.
16
11
improper fraction
k.
11
16
proper fraction
l.
321
322
proper fraction
m.
12
13
mixed number
n.
4
5
proper fraction
o.
6
5
improper fraction
b.
19
1
6
3
3
c.
12
4
1
1 1
8
8
2
g.
h.
7
2
1
5
5
i
j.
16
5
1
11
11
m.
o.
6
1
1
5
5
a.
5
55 1

 or
15 15  5 3
5 1 5 1


15 3  5 3
11
9
MATHEMATICS – CHAPTER 2FRACTIONS
3
3. e.
b.
30 of 33
3
3
1 3  2   1 7


2
2
2
1 11  8  1 89
11 

8
8
8
3
3 3  4   3 15


4
4
4
9
12 9  13  12 129


13
13
13
8
88 1

 or
32 32  8 4
8
2 2 2
1


32 2  2  2  2  2 4
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5.
c.
54 54  9

 6 or
9
99
54 3  3  6

6
9
3 3
d.
18 18  6

 3 or
6
66
18 2  3  3

3
6
23
e.
7

21
7

21
77 1
 or
21  7 3
1 7 1

3 7 3
f.
6
66 1

 or
24 24  6 4
6
23
1


24 2  2  2  3 4
g.
18 18  6 3

 or
24 24  6 4
18
2  3 3
3


24 2  2  2  3 4
h.
100
100  100
1


or
1000 1000  100 10
100
2 255
1


1000 2  2  2  5  5  5 10
i.
19 19

or
28 28
19
1  19
19


28 2  2  7 29
j.
10 10  10

 1 or
10 10  10
10 2  5

1
10 2  5
k.
2
22 1

 or
10 10  2 5
2
2
1


10 2  5 5
l.
3
33 1

 or
12 12  3 4
3
3
1


12 2  2  3 4
m.
18 18  3 6

 or
21 21  3 7
18 2  3  3 6


21
3 7
7
n.
2 22

 1 or
2 22
2 2
 1
2 2
o.
18 18  18 1

 or
72 72  18 4
18
2  3 3
1


72 2  2  2  3  3 4
a.
3 8
38 8
24  8 32 2  2  2  2  2






1
4 32 4  8 32
32
32 2  2  2  2  2
b.
5
6 3 2  7   6 3 20 3 20  3 60 2  2  3  5 12


1
2  
 
 

5 7
7
7
7 5
7
5 7 5 7  5 35
MATHEMATICS – CHAPTER 2FRACTIONS
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11  3  2  1  35  1  35  5  7  7
2
11  15 
3
3
15 3  15 45 3  3  5 9
c.
d.
8 1 8  2 1 16  1 15 3  5 5
1
 
 


 2
3 6 3 2 6
6
6 23 2
2
e.
9 2 9 23 9  6 3
3
1
 





12 4 12 4  3
12
12 2  2  3 4
f.
11 1 11
22
9
  2 
1
13 2 13
13
13
g.
7
7
9 1 9 1 2 9  2 7
 





16 8 16 8  2
16
16 2  2  2  2 16
h.
2 2 2 8
1 1 1 5 1 3 5  3 8
 





3 5 3 5 5 3
15
15
3 5
15
42
42 1 42  1  6
7
6 
 

43
43 6 43  6  6 43
i.
j.
2 4 2  7 4  5 14  20 34 2  17 34
 





5 7 5 7 75
35
35 5  7 35
k.
17
17
9 8 9  9 8  8 81 64 17
 






8 9 8  9 9  8 72 72 72 2  2  2  3  3 72
3
1 12  4  3 6  4  1 51 4
51 3  17 51
1
12  6 

 



2
4
4
4
4
4 25 25
5  5 25
25
l.
1
of the total system flow.
6
1
Flow in the branch =  27,000 lbm per hour.
6
Flow in the branch = 4,500 lbm per hour.
6.
Flow in the branch =
7.
Fraction of Bottles with Sulfuric Acid

Number wit h Sulfuric Acid
Total Number
60
60  5 12


or
175 175  5 35
MATHEMATICS – CHAPTER 2FRACTIONS

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60 2  2  3  5 12


175
55 7
35
Fraction of Bottles with Hydrochloric Acid 

Number wit h Hydrochlor ic Acid
Total Number
55
55  5 11


or
175 175  5 35
55
5  11
11


175 5  5  7 35
8.
Inside Diameter = Outside Diameter – 2  (Wall Thickness)
1
1

Inside Diameter  4 inches  2 inches 
8
 12

Inside Diameter 
33 2 33  3 2  2 95




8 12 8  3 12  2 24
Inside Diameter  3
MATHEMATICS – CHAPTER 2FRACTIONS
23
inches
24
33 of 33
© 2003 GENERAL PHYSICS CORPORATION
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