ADDING AND SUBTRACTING MIXED FRACTIONS WITH UNLIKE DENOMINATORS

Problem 3 :

3/4  - 2  1/2

Solution :

= 3/4  - 2  1/2

Converting the mixed fraction to improper fraction, we get

= 3/4 - 5/2

Least common multiple of 4 and 2 is 4.

= [3(1) – 5(2)]/4

= (3 - 10)/4

= -7/4

Converting the improper fraction to mixed fraction, we get

= -1  3/4

Problem 4 :

4  1/3 + 2  1/6

Solution :

=  4  1/3 + 2  1/6

Converting the mixed fraction to improper fraction, we get

= 13/3 + 13/6

Least common multiple of 3 and 6 is 6.

= [13(2) + 13(1)]/6

= (26 + 13)/6

= 39/6

= 13/2

Converting the improper fraction to mixed fraction, we get

= 6  1/2

Problem 5 :

2  2/3  - 5  5/6

Solution :

=  2  2/3  - 5  5/6

Converting the mixed fraction to improper fraction, we get

= 8/3 - 35/6

Least common multiple of 3 and 6 is 6.

= [8(2) – 35(1)]/6

= (16 - 35)/6

= -19/6

Converting the improper fraction to mixed fraction, we get

= - 3 1/6

Problem 5 :

-2  1/4 + 3  1/8

Solution :

= -2  1/4 + 3  1/8

Converting the mixed fraction to improper fraction, we get

= (-9/4) + 25/8

Least common multiple of 4 and 8 is 8.

= [(-9(2) + 25(1)]/8

= (-18 + 25)/8

= 7/8

Problem 7 :

4  1/5  - 2  1/6

Solution :

=  4  1/5 - 2  1/6

Converting the mixed fraction to improper fraction, we get

= 21/5 - 13/6

Least common multiple of 5 and 6 is 30.

= [(21(6) – 13(5)]/30

= (126 - 65)/30

= 61/30

= 13/2

Converting the improper fraction to mixed fraction, we get

= 2  1/30

Problem 8 :

2  1/3 + 1  1/6

Solution :

= 2  1/3 + 1  1/6

Converting the mixed fraction to improper fraction, we get

= 7/3 + 7/6

= [7(2) + 7(1)]/6

Least common multiple of 3 and 6 is 6.

=(14 + 7)/6

= 21/6

Converting the improper fraction to mixed fraction, we get

= 3  1/2

Problem 9 :

1  1/2 + 4  2/3

Solution :

= 1  1/2 + 4  2/3

Converting the mixed fraction to improper fraction, we get

= 3/2 + 14/3

= [3(3) + 14(2)]/6

Least common multiple of 3 and 2 is 6.

= (9 + 28)/6

= 37/6

Converting the improper fraction to mixed fraction, we get

= 6  1/6

Problem 10 :

3  1/3  - 1  1/2

Solution :

=  3  1/3 - 1  1/2

Converting the mixed fraction to improper fraction, we get

= 10/3 - 3/2

Least common multiple of 3 and 2 is 6.

= [10(2) – 3(3)]/6

= (20 - 9)/6

= 11/6

Converting the improper fraction to mixed fraction, we get

= 1 5/6

Problem 11 :

4  3/7  - 2  1/3

Solution :

=  4  3/7 - 2  1/3

Converting the mixed fraction to improper fraction, we get

= 31/7 - 7/3

Least common multiple of 7 and 3 is 21.

= [31(3) – 7(7)]/21

= (93 - 49)/21

= 44/21

Converting the improper fraction to mixed fraction, we get

= 2  2/21

Problem 12 :

There is a flower garden 7  2/5 m² and vegetable garden 8  3/4 m² of  at Petty's school. Which garden is bigger and by how much in m² ?

Solution :

Area of flower garden = 7  2/5 m²

Area of vegetable garden = 8  3/4 m²

Difference = 8  3/4 - 7  2/5

= (8 - 7) + (3/4 - 2/5)

= 1 + (15 - 8)/20

= 1 + 7/20

= 1   7/20 m²

Vegetable garden is 1   7/20 m² bigger.

Problem 13 :

Aiguo, Destiny and Claudio shared a 12-slice pizza. Aiguo ate 1/3 of the pizza and Destiny ate 1/4 of the pizza. Claudio ate the remaining slices.

a) What fraction of the pizza did Aiguo and Destiny eat together?

b) What fraction of the pizza did Claudio eat?

c) How many slices did each person eat?

Solution :

a) Fraction of the pizza ate by Aiguo + Destiny = 1/3 + 1/4

LCM(3, 4) = 12

= (4 + 3)/12

= 7/12

b) Fraction part of pizza did Claudio eat = 1 - (7/12)

= (12 - 7)/12

= 5/12

c) Total number of slices = 12

Number of slices eat by Aiguo = 1/3 of 12

= 1/3 x 12

= 12/3

= 4 slices

Number of slices eat by Destiny = 1/4 of 12

= 1/4 x 12

= 12/4

= 3 slices

Number of slices eat by Claudio = 5/12 of 12

= 5/12 x 12

= 5 slices

Problem 14 :

Of the students in a class, 3/4 take the bus to school. 3/16 of the students in the class walk to school. The remaining students are driven to school in a car.

a) What fraction of the class is driven to school in a car?

b) What fraction of the class does not walk to school?

c) Is it possible that there is a total of 25 students in the class? Explain.

Solution :

a) Fraction part of students driven to school in a car = 1 - (3/4 + 3/16)

= 1 - (12 + 3)/16

= 1 - 15/16

= (16 - 15)/16

= 1/16

b) Fraction part of students does not walk to school = 1 - 3/16

= (16 - 3)/16

= 13/16

c) Number of students driven to school 

= 3/4 + 3/16 + 13/16

= (12 + 3 + 13)/16

= 28/16

No, it is not possible for there to be a total of 25 students because the total number of students must be a multiple of 16.