Download Addition, Subtraction, Multiplication, and Division of Mixed Numbers

HFCC Math Lab
Arithmetic - 4
Addition, Subtraction, Multiplication and Division of Mixed Numbers
Part I: Addition and Subtraction of Mixed Numbers
There are two ways of adding and subtracting mixed numbers. One way is to express each mixed
number as an improper fraction and then add or subtract the fractions.
2 2
Ex 1: Add 3  1
5
3
If possible, express your answer as a mixed number.
First, we write each mixed number as an improper fraction.
2 3  5  2 15  2 17
3 


5
5
5
5
2 1 3  2 3  2 5
1 


3
3
3
3
So,
2 2 17 5
3 1  
5 3 5 3
The LCD of the fractions is 15
Changing each fraction to an equivalent fraction with 15 as the denominator, we have
17 17  3 51
5 5  5 25




5
5  3 15
3 3  5 15
Therefore,
17 5 51 25 51  25 76
  


5 3 15 15
15
15
Now, express
76
as a mixed number.
15
Hence,
2 2 17 5
1
3 1    5
5 3 5 3
15
Ex 2: Subtract 4
5
1
2
12
4
76
1
5
15
15
If possible, express your answer as a mixed number.
First we write each mixed number as an improper fraction.
So,
Revised 03/10
4
5 4 12  5 48  5 53



12
12
12
12
4
5
1 53 9
2  
12
4 12 4
1 2  4 1 8 1 9
2 


4
4
4
4
The LCD of the fractions is 12.
1
Changing
9
to an equivalent fraction with 12 as the denominator, we have
4
9 9  3 27


4 4  3 12
(The denominator is already 12.)
53 9 53 27 53  27 26
  


12 4 12 12
12
12
Therefore,
Reduce
53 53

12 12
26
to the lowest terms and then change to a mixed number.
12
26 26  2 13
1

 2
12 12  2 6
6
Hence,
4
.
5
1 53 9
1
2    2
12
4 12 4
6
Ex 3: Subtract 7  2
1
7
If possible, express your answer as a mixed number.
First we write the whole number and the mixed number as improper fractions.
7
So,
1 2  7  1 14  1 15
2 


7
7
7
7
7
1
1 7 15
72  
7 1 7
Changing
7
to an equivalent fraction with 7 as the denominator, we have
1
7 7  7 49


1 1 7 7
Therefore,
The LCD of the fractions is 7.
15 15

7
7
(The denominator is already 7.)
7 15 49 15 49  15 34
6
 
 

4
1 7
7 7
7
7
7
1 7 15
6
Hence, 7  2    4
7 1 7
7
1
1
6
Note: It is a common mistake to say that 7  2  5 . Be careful! The correct answer is 4 .
7
7
7
Revised 03/10
2
Ex 4: Perform the indicated operations.
1
4
5  2 1
(Remember the Order of Operations: Add and subtract from left to right.)
10 15
First we write the whole number and the mixed numbers as improper fractions.
5
So,
5
1
52
2
1 21

10 10
1
4 5 21 19
1   
10 15 1 10 15
1
4 19
 .
15 15
The LCD of the fractions is 30.
Changing each fraction to an equivalent fraction with 30 as the denominator, we have
5 5  30 150


1 1 30 30
Therefore,
Reduce
52
21 21 3 63


10 10  3 30
19 19  2 38


15 15  2 30
1
4 5 21 19 150 63 38
1    
 
10 15 1 10 15 30 30 30
150  63  38

30
87  38 125


30
30
125
to lowest terms and then change to a mixed number.
30
125 125  5 25
1


4
30 30  5
6
6
Hence,
52
1
4 5 21 19
1
1     4
10 15 1 10 15
6
Note: The above method of adding and subtracting mixed numbers becomes cumbersome and
time consuming when the whole number parts of the mixed numbers are large numbers. For such
problems, it is strongly recommended that you use the following method.
Revised 03/10
3
Another way to add mixed numbers
Step 1:
Add the fractional parts of the mixed numbers separately.
Step 2:
Reduce the fraction obtained in step 1 to lowest terms. If it is an improper
fraction, change it to a mixed number.
Note: If the fraction obtained in step 1 is an improper fraction, you may first change it to a
mixed number and then reduce the fractional part of the mixed number to lowest terms.
Step 3:
Add the whole number parts of the given mixed numbers.
Step 4:
Add the sum from step 3 to the fraction or the mixed number obtained in step 2.
Write your answer as a mixed number.
Ex 5: Add and simplify.
32
5
7
 25
12
16
Step 1: Add the fractions
5
7
and
.
12
16
The LCD of the fractions is 48.
5
5  4 20


12 12  4 48
7
7  3 21


16 16  3 48
5 7 20 21 20  21 41
 



12 16 48 48
48
48
Step 2:
41
is already in lowest terms and cannot be expressed as a mixed number.
48
Step 3:
Add the whole numbers.
Step 4:
Add 57 to
57 
32  25  57
41
.
48
41
41
 57
48
48
(Remember, whenever we add a mixed number and
a proper fraction, we omit the “+” and write the sum
as a mixed number.)
Hence, 32
Revised 03/10
5
7
41
 25  57
12
16
48
4
Ex 6: Add and simplify.
Step 1:
1
13
29  37
2
14
Add the fractions
1
13
and
.
2
14
1 1 7 7


2 2  7 14
The LCD of the fractions is 14.
13 13

14 14
1 13 7 13 7  13 20
   

2 14 14 14
14
14
Step 2:
Reduce
20
to lowest terms and then change to a mixed number.
14
20 20  2 10
3


1
14 14  2 7
7
Step 3:
Add the whole numbers.
Step 4:
3
Add 66 to 1 .
7
Hence,
Ex 7: Add and simplify:
29  37  66
3
3
66 1  67 , so 66  1  67 .
7
7
1
13
3
29  37  67
2
14
7
97
6
5
9
 63  102
18
15
27
First, we reduce the fractional part of each mixed number to lowest terms.
6
66 1


18 18  6 3
5
5 5 1


15 15  5 3
9
9 9 1


27 27  9 3
Therefore, 97
6
5
9
1
1
1
 63  102
 97  63  102
18
15
27
3
3
3
Steps 1 & 2:
Add the fractional parts and reduce.
Step 3:
Add the whole numbers.
Step 4:
Add the results of Step 3 and Step 2.
Hence, 97
Revised 03/10
1 1 1 3
   1
3 3 3 3
97  63  102  262
6
5
9
 63  102
 263 .
18
15
27
5
262  1  263
Another way to subtract mixed numbers
Step 1:
Write the fractional part of each mixed number as an equivalent fraction, using
the LCD as the new denominator.
Step 2:
Look at the two numerators obtained in Step 1.
If the first numerator is greater than the second, then
(A)
Subtract the fractional parts separately. Reduce to lowest terms.
(B)
Subtract the whole number parts of the given mixed numbers.
(C)
The final answer is the mixed number obtained by adding the whole
number in (B) and the fraction in (A).
If the first numerator in Step 1 is smaller than the second, we omit Step 2
and go to Step 3.
Step 3:
Borrow 1 from the whole number part of the first mixed number. Decrease the
whole number part by 1. Add this 1 to the fractional part of the first mixed
number and write it as an improper fraction. Then
(A)
Subtract the fractional parts separately. Reduce to lowest terms.
(B)
Subtract the whole number parts.
(C)
The final answer is the mixed number obtained by adding the whole
number in (B) and the fraction in (A).
Ex 8: Subtract and simplify.
5
1
27  10
6
2
Step 1: The LCD of the fractional parts is 6.
1 1 3 3


2 23 6
5 5

6 6
Since the first numerator, 5, is greater than the second numerator, 3,
we complete Step 2.
Step 2:
(A)
Subtract the fractional parts and reduce to lowest terms:
5 3 53 2 22 1
 
 

6 6
6
6 62 3
Revised 03/10
6
Subtract the whole number parts: 27 10  17
(B)
1
1
Add the results obtained in (B) and (A): 17   17
3
3
5
1
1
Hence, 27  10  17
6
2
3
(C)
Ex 9: Subtract and simplify.
Step 1:
3
7
43  36
8
12
The LCD of the fractional parts is 24.
3 33 9


8 8  3 24
7
7  2 14


12 12  2 24
Step 2:
Since the first numerator 9, is smaller than the second numerator, 14, and
we cannot subtract 14 from 9, we go to Step 3.
Step 3:
Borrow 1 from 43 and decrease 43 by 1.
So, 43
3
9
9
9
 43
 43 
 42  1 
24
24
24
8
Thus, 43
 42 
24 9

24 24
 42 
33
24
(Write 1 as
(Add
24
)
24
24
9
)

24 24
3
33
 42
24
8
3
7
9
14
33
14
Therefore, 43  36  43  36  42  36 .
8
12
24
24
24
24
(A) Subtract the fractional parts:
33 14 33  14 19



24 24
24
24
(B) Subtract the whole number parts: 42  36  6
(C) Add the results obtained in (B) and (A):
Hence,
Revised 03/10
3
7
19
43  36  6
8
12
24
7
6
19
19
6
24
24
Ex 10: Subtract and simplify.
5
56  42
9
In this problem, we are subtracting a whole number from a mixed number. Since the
0
whole number, 42, does not have a fractional part, we can say: 42  42 .
9
5
5
0
Therefore, 56  42  56  42
9
9
9
A) Subtract the fractional parts:
5 0 50 5
 

9 9
9
9
(B) Subtract the whole number parts: 56  42  14
(C) Add the results obtained in (B) and (A):
Hence,
14 
5
5
 14
9
9
5
5
56  42  14
9
9
Concept: To subtract a whole number from a mixed number, we subtract the whole number
parts. Then, add this difference to the fractional part of the first mixed number to
obtain the final answer.
Ex 11: Subtract and simplify.
56  42
5
9
In this problem, we are subtracting a mixed number from a whole number. Since the
0
whole number, 56, does not have a fractional part, we can say: 56 = 56 . We borrow 1
9
from 56 and decrease 56 by 1. Write 1 as an improper fraction with its denominator = 9,
5
which is the same as the denominator of the fractional part of 42 .
9
So,
56  55  1  55 
Therefore,
Revised 03/10
9
9
 55
9
9
5
9
5
56  42  55  42
9
9
9
9 5 4
 
9 9 9
(A)
Subtract the fractional parts:
(B)
Subtract the whole number parts: 55  42  13
8
(C)
Add the results obtained in (B) and (A) 13 +
Hence, 56  42
Concept:
4
4
= 13
9
9
5
4
 13
9
9
To subtract a mixed number from a whole number, we write the whole number as
a mixed number by borrowing 1 from the whole number, decreasing the whole
number by 1 and writing 1 as an improper fraction with denominator the same as
the denominator of the fractional part in the given mixed number. Then
(A)
Subtract the fractional parts.
(B)
Subtract the whole number parts.
(C)
Add the results obtained in (B) and (A).
Note: It is very common for students to get mixed up between Ex 10 and Ex 11. Be very
very careful! Notice that
5
5
5
5
56  42  14 but, 56  42  14
9
9
9
9
The correct answer for 56  42
5
4
is 13 .
9
9
Exercises:
Perform the indicated operations by first changing mixed numbers to improper fractions.
Simplify your answers and whenever possible, express your answers as mixed numbers.
1.
1
7
3 2
6
9
2.
5 11
2 1
6 12
3.
7
5
1
3
10
4
4.
2
5.
5
7
4 2
9
12
6.
5 3
7.
2
1 7
1 2 
3
4 8
8.
1
7
4  2 3
5 10
9.
3 5
7 1  2
4 6
10.
7
5 9
1 
8 16 24
Revised 03/10
9
1
2
1
14 7
3
8
Add and simplify without changing mixed numbers to improper fractions. Whenever possible,
express your answer as a mixed number.
11.
7
5
12  10
9
18
12.
43
7
 23
19
13.
75  21
8
23
14.
27
9
7
 41
16
12
15.
47
5
17
 28
14
21
16.
73
5 20

11 22
17.
3 15
53   46
6 30
18.
25
1
5
 24  30
35
7
14
Subtract and simplify without changing mixed numbers to improper fractions. Whenever
possible, express your answer as a mixed number.
19.
52
7
1
 40
12
4
20.
2
5
38  23
6
15
21.
17
9
7
6
18
14
22.
3
7
27  16
8
16
23.
5
7
47  12
9
8
24.
65  24
25.
65
7
 24
15
26.
3
79  51
8
27.
79  51
28.
43
3
8
7
15
3
5
 42
16
12
Solutions to odd numbered problems and answers to even numbered problems
1.
1
7 19 25
3 2  
6
9 6 9
Revised 03/10
The LCD is 18
2.

19  3 25  2

63 9 2

57 50 57  50 107
17



5
18 18
18
18
18
10
4
3
4
3.
7
5
1
3
10
4
Reduce
5
.
10
1
1 15 13
 7 3  
2
4 2 4

5.
7.
9.
5 1
 .
10 2
11
14
6.
1
5
8
8.
5
1
2
10.
1
The LCD is 4.
15  2 13 30 13 30  13 43
3
   

 10
22 4
4 4
4
4
4
5
7
4 2
9
12

41 31

9 12

41 4 31 3

9  4 12  3

164 93 164  93 71
35
 

1
36 36
36
36
36
The LCD is 36.
2
1 7 5 9 7
1 2    
3
4 8 3 4 8
The LCD is 24.

58 9  6 7 3


38 4  6 83

40 54 21 40  54  21 115
19
 


4
24 24 24
24
24
24
3 5
31 11 2
7 1  2   
4 6
4 6 1
Revised 03/10
4.
The LCD is 12.

31 3 11 2 2 12


4  3 6  2 112

93 22 24
 
12 12 12

93  22  24 71  24 47
11


3
12
12
12
12
11
13
16
11.
7
5
12  10
9
18
 12
15.
17.
66
7
19
14.
69
7
48
16.
74
4
11
18.
55
3
14
14  5
19
1
1
 22  22  1  23
18
18
18
18
75  21
8
23
75  21
8
8
 96
23
23
47
12.
14
5
 10
18
18
 22
13.
The LCD is 18
Since 75  21  96 , we have
5
17
 28
14
21
The LCD is 42.
 47
15
34
 28
42
42
 75
15  34
49
49
7
1
1
 75  75 
 75  1  75  1  76
42
42
42
42
6
6
3 15
53   46
6 30
Reducing fractions:
3 1
15 1
and

 , we have
6 2
30 2
3 15
1 1
1 1
53   46  53   46  53    46
6 30
2 2
2 2

 53  1  46  54  46 100
19.
52
7
1
 40
12
4
The LCD is 12.
 52
7
3
 40
12
12
 12
7 3
4
1
 12  12
12
12
3
Revised 03/10
20.
12
15
21.
17
9
7
6
18
14
Reducing fractions:
17
23.
24.
40
9 1
7 1
 and
 , we have
18 2
14 2
5
7
47  12
9
8
The LCD is 72.
8
15
40
40
63
63
 46  1 
 12
12
72
72
72
72
72 40
63
112
63
112  63
49
  12  46
 12  34
 34 .
72 72
72
72
72
72
72
 46 
27.
10
9
7
1
1
 6  17  6  17  6  11
18
14
2
2
 47
25.
15
16
22.
65
7
 24
15
65
7
7
 24  41
15
15
Since 65 – 24 = 41, we have
3
3
79  51  78  1  51
8
8
8
3
 78   51
8
8
8
3
83
 78  51  27
8
8
8
 27
Revised 03/10
5
8
13
26.
28
28.
37
48
3
8
Part II:
Multiplication and Division of Mixed Numbers
To multiply or divide mixed numbers, we use the following two steps.
Step 1:
Express each mixed number as an improper fraction.
Step 2:
Multiply or divide, using the procedures for multiplying and dividing fractions.
Ex 1: Multiply and simplify. If possible, express your answer as a mixed number.
1
1
5 2
2
3
First, we write each of the mixed numbers as an improper fraction.
1 5  2  1 10  1 11
5 


2
2
2
2
1 2  3 1 6 1 7
2 


3
3
3
3
1
1 11 7 11 7 77
5
Hence, 5  2   

 12
2
3 2 3 23
6
6
Ex 2: Multiply and simplify. If possible, express your answer as a mixed number.
8 3
1
6
First, we write the whole number and the mixed number as improper fractions.
4
1 8 19 8  19
8 3   
1 6 3
6 1 6

Ex 3:
(Divide 8 and 6 by 2)
4 19 76
1

 25
1 3
3
3
Multiply and simplify. If possible, express your answers as a mixed number.
5
3
First, we write each mixed number and the whole number as an
7  15  3
12
5
improper fraction.
7
5
3 89
15 18
 15  3 


12
5 12
1
5
3
89  15  18
=
12 2 1 5 1

Revised 03/10
3
(Divide 15 and 5 by 5. Divide 18 and 12 by 6)
89  3  3 801
1

 400
2  1 1
2
2
14
Ex 4: Divide and simplify. If possible, express your answer as a mixed number.
4
5
1
2
8
3
4
5
1 37 7
2 

8
3
8
3
First, we write each mixed number as an improper fraction.

Changing division to multiplication by the reciprocal, we have
37 3
37  3 111
55



1
8 7
8 7
56
56
Ex 5: Divide and simplify. If possible, express your answer as a mixed number.
35  5
1
4
First, we write the whole number and the mixed number as improper
fractions.
35  5
1
35 21


4
1
4
Changing division to multiplication by the reciprocal, we have
5
35  4
35 4
=
 
1  21 3
1 21

(Divide 35 and 21 by 7)
5  4 20
2

6
1 3 3
3
Ex 6: Divide and simplify. If possible, express your answer as a mixed number.
16
4
 29
7
First, we write the whole number and the mixed number as improper
fractions.
16
4
116 29
 29 

7
7
1
Changing division to multiplication by the reciprocal, we have
4
116 1
116 1

(Divide 116 and 29 by 29)


7  29 1
7
29

Revised 03/10
4 1 4

7 1 7
15
Ex 7: Divide and simplify. If possible, express your answer as a mixed number.
1
5
1
3
5
15
Since a fraction means to divide, rewrite as a division problem.
1
5  15 1  3 1
1
5
5
3
5
15

76 16

5
5
Change to improper fractions.
Rewrite as a multiplication problem.
19
1
76  5
76 5

 
5 1  16 4
5 16

(Divide 76 and 16 by 4. Divide 5 and 5 by 5)
19 1 19
3

4
1 4
4
4
Exercises:
Multiply the following. Simplify your answers and whenever possible, express your answers as a
mixed numbers.
1.
1
1
1 2
7
3
2.
3
3.
7
1
1
1
8
3
4.
16  3
5.
18  2
6.
4
1
 21
7
7.
5
8.
4
2
9
8
3
16
9.
1
1
1
1 1 1
3
3
3
10.
3
3
3
1
5 2
7
5
2
1
9
1
 12
18
Revised 03/10
16
4
1
3
5
3
1
5
Divide the following. Simplify your answers and whenever possible, express your answers as a
mixed numbers.
11.
1
3
1 2
3
4
12.
5
13.
36  4
1
2
14.
49  4
15.
18
1
9
2
16.
9
1
2
1
2
4
19.
2
3
9
 27
11
1
9
2
2
3
7
5
17.
1
1
3
2
4
18.
1
6
2
8
15
1
2
15
20.
Solutions to odd numbered problems and answers to even numbered problems.
1
1.
1
1 8
7
8
2
1 2 

 2
7
3 71
3
3
3
3.
1 1 57
4
19
1
7 1 


9
8 3
82
31
2
2
5.
1 18
19
38
18  2 


 38
9
1
91
1
19
2.
12
2
3
4.
51
1
5
6.
87
1
2
Revised 03/10
17
2
7.
1
91 12
5  12 

18
1
18 3
9.
1
1
1 4 4 4 44 4
1 1 1    
3
3
3 3 3 3 333

182
2
 60
3
3

8.
21
10.
40
12.
1
14.
10
16.
4
11
64
10
2
27
27
11.
1
3
4 11 4 4
4  4 16
1 2  
 


3
4
3 4
3 11 3 11 33
13.
2
1 36 9 36
8


36  4 
 
8
1
91
2
1
2
1
15.
18
9
13
4
1
37 9 37 1 37
1
9 
 
 
2
2
2
1
2 9 18
18
1
2
2  5 1  2 1  11  9  11  4  22  2 4
1
21 9
2
4
2 4
9
9
2
4
1
2
5
17.
19.
1 15 31 15 15 225
8
15
 15  2
 



7
1
15 1 15
1 31 31
31
2
15
Revised 03/10
18
18.
2
2
3
20.
4
1
12