ADDING AND SUBTRACTING MIXED FRACTIONS

Adding and subtraction can be done in two different ways.

(i)  By regrouping method

(ii) Converting into improper fractions and simplify.

Regrouping Method

Problem 1 :

1  2/3 + 3  1/4

Solution :

= 1  2/3 + 3  1/4

= 1 + 2/3 + 3 + 1/4

Grouping the integers part and fractional part.

= (1 + 3) + 2/3 + 1/4

= 4 + (2/3 + 1/4)

By taking the least common multiple, we get

= 4 + (8 + 3)/12

=  4 + 11/12

= 4  11/12

Problem 2 :

 3  3/4 - 1  1/2

Solution :

= 3  3/4 - 1  1/2

= 3 + 3/4 - 1 + 1/2

By grouping,

= (3 – 1) + (3/4 – 1/2)

= 2 + (3/4 – 1/2)

Here the two proper fractions denominators are not the same.

So, we take the least common multiple of 4 and 2.

= 2 + (3 – 2)/4

= 2 + 1/4

= 2  1/4

To see problems on converting into mixed fractions and simplify, click the link given below.

Adding and Subtracting Mixed Fractions with Unlike Denominators

Problem 3 :

4  1/3 + 2  1/6

Solution :

=  4 + 1/3 + 2 + 1/6

= (4 + 2) (1/3 + 1/6)

By taking the least common multiple, we get

= 6 + (2 + 1)/6

= 6 + 3/6

= 6 + 1/2

= 6  1/2

Problem 4 :

2  2/3  - 5  5/6

Solution :

= 2 + 2/3 - 5 - 5/6

= (2 - 5) + 2/3 - 5/6

= -3 + [(4 - 5)/6]

= -3 - 1/6

= -3  1/6

Problem 5 :

-2  1/4 + 3  1/8

Solution :

= -2 + 1/4 + 3 + 1/8

=(-2 + 3) (1/4 + 1/8)

By taking the least common multiple, we get

= 1 [1(2) + 1(1)]/8

= 1 (2 + 1)/8

= 1  3/8

Problem 6 :

4  1/5  - 2  1/6

Solution :

=  4 + 1/5 - 2 - 1/6

= (4 – 2) + (1/5 – 1/6)

By taking the least common multiple, we get

= 2 + [1(6) – 1(5)]/30

= 2 + (6 – 5)/30

= 2 + 1/30

= 2  1/30

Problem 7 :

2  1/3 + 1  1/6

Solution :

= 2 + 1/3 + 1 + 1/6

= (2 + 1) + (1/3 + 1/6)

By taking the least common multiple, we get

= 3 + [1(2) + 1(1)]/6

= 3 + [(2 + 1)/6]

= 3 + 1/2

= 3  1/2

Problem 8 :

1  1/2 + 4  2/3

Solution :

= 1 + 1/2 + 4 + 2/3

= (1 + 4) + (1/2 + 2/3)

By taking the least common multiple, we get

= 5 [1(3) + 2(2)]/6

= 5 (3 + 4)/6

= 5 + 7/6

= 5 + 1 + 1/6

= 6  1/6

Problem 9 :

3  1/3  - 1  1/2

Solution :

= 3 + 1/3  - 1 + 1/2

=  (3 - 1) + (1/3 – 1/2)

By taking the least common multiple, we get

= 2 + [1(2) – 1(3)]/6

= 2 + (2 – 3)/6

= 2 + (-1/6)

= 1 + 1 - 1/6

= 1 + 6/6 - 1/6

= 1 + 5/6

= 1 5/6

Problem 10 :

4  3/7  - 2  1/3

Solution :

= 4 + 3/7  - 2 + 1/3

= (4 – 2) + (3/7 – 1/3)

By taking the least common multiple, we get

= 2 + [3(3) – 1(7)]/21

= 2 + (9 – 7)/21

= 2 + 2/21

= 2  2/21

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