WORD PROBLEMS ON MIXED FRACTIONS

Problem 1 :

Linda walked 2 1/3 miles on the first day and 3 2/5 miles on the next day. How many miles did she walk in all ? 

Solution :

Total no. of miles she walked is

=  2 1/3 + 3 2/5

In the above mixed fractions, we have the denominators 3 and 5.

LCM of (3, 5)  =  15 

To simplify the above expression, we have to make the denominators of both the mixed fractions to be 15.

Then, we have

2 1/3 + 3 2/5  =  2 5/15 + 3 6/15

By regrouping, we have

2 1/3 + 3 2/5  =  (2 + 3) + (5/15 + 6/15)

2 1/3 + 3 2/5  =  (5) + (11/15)

2 1/3 + 3 2/5  =  5 11/15

So, Linda walked 5 11/15 miles in all. 

Problem 2 :

David ate 2 1/7 pizzas and he gave 1 3/14  pizzas to his mother. How many pizzas did David have initially ?

Solution :

No. of pizzas he had initially is

=  2 1/7 + 1 3/14

=  2 2/14 + 1 3/14

By regrouping, we have

=  (2 + 1) + (2/14 + 3/14)

=  (3) + (5/14)

=  3 5/14

So, initially David had 3 5/14 pizzas. 

Problem 3 :

Mr. A has 3 2/3 acres of land. He gave 1 1/4 acres of land to his friend. How many acres of land does Mr. A have now ? 

Solution :

Now, no. of acres of land that Mr. A has

=  3 2/3 - 1 1/4

In the above mixed fractions, we have the denominators 3 and 4.

LCM of (3, 4)  =  12

To simplify the above expression, we have to make the denominators of both the mixed fractions to be 12. 

Then, we have

3 2/3 - 1 1/4  =  3 8/12 - 1 3/12

By regrouping, we have

3 2/3 - 1 1/4  =  (3 - 1) + (8/12 - 3/12)

3 2/3 - 1 1/4  =  (2) + (5/12)

3 2/3 - 1 1/4  =  2 5/12

Now, Mr. A has 2 5/12 acres of land. 

Problem 4 :

Lily added 3 1/3 cups of walnuts to a batch of trail mix. Later she added 1 1/3 cups of almonds. How many cups of nuts did Lily put in the trail mix in all? 

Solution :

No. of cups of nuts that Lily put in all is

=  3 1/3 + 1 1/3

By regrouping, we have

=  (3 + 1) + (1/3 + 1/3)

=  (4) + (2/3)

=  4 2/3

So, Lily put 4 2/3 cups of nuts in all. 

Problem 5 :

In the first hockey games of the year, Rodayo played 1 1/2 periods and 1 3/4 periods. How many periods in all did he play ? 

Solution :

No. of periods in all he played is

=  1 1/2 + 1 3/4

=  1 2/4 + 1 3/4

By regrouping, we have

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

=  (2) + (5/4)

=  (2) + (1 1/4)

=  (2) + (1 + 1/4)

=  3 1/4

So, Rodayo played 3 1/4 periods in all. 

Problem 6 :

A bag can hold 1 1/2 pounds of flour. If Mimi has 7 1/2 pounds of flour, then how many bags of flour can Mimi make ?

Solution :

No. of bags  =  (Total no. of lbs) / (No of lbs per bag)

Because, we use division, we have to convert the given mixed numbers into improper fractions.

Total no. of pounds of flour is

=  7 1/2

=  15/2

No. of pounds per bag is

=  1 1/2

=  3/2

Then,  we have

Number of bags  =  (15/2) / (3/2)

Number of bags  =  (15/2) ⋅ (2/3)

Number of bags  =  5

So, the number of bags that Mimi can make is 5.

Problem 7 :

Jack and John went fishing Jack caught 3 3/4 kg of fish and while John  caught 2 1/5 kg of fish. What is the total weight of the fish they caught?  

Solution :

Total weight of the fish they caught is 

=  3 3/4 + 2 1/5

In the above mixed fractions, we have the denominators 4 and 5.

L.C.M of (4, 5) = 20

To simplify the above expression, we have to make the denominators of both the mixed fractions to be 20.

Then, we have

3 3/4 + 2 1/5  =  3 15/20 + 2 4/20

By regrouping, we have

3 3/4 + 2 1/5  =  (3 + 2) + (15/20 + 4/20)

3 3/4 + 2 1/5  =  (5) + (19/20)

3 3/4 + 2 1/5  =  5 19/20

So, the total weight of the fish they caught is 5 19/20 kg. 

Problem 8 :

Amy has 3 1/2 bottles in her refrigerator. She used 3/5 bottle in the morning 1 1/4 bottle in the afternoon. How many bottles of milk does Amy have left over ?  

Solution :

No. of bottles of milk used is 

=  3/5 + 1 1/4

In the above mixed fractions, we have the denominators 5 and 4.

L.C.M of (5, 4) = 20

To simplify the above expression, we have to make the denominators of both the mixed fractions to be 20.

Then, we have

3/5  + 1 1/4  =  12/20 + 1 5/20

By regrouping, we have

3/5  + 1 1/4  =  1 + (12/20 + 5/20)

3/5  + 1 1/4  =  1 + (17/20)

3/5  + 1 1/4  =  1 17/20

So, no. of bottles of milk used is 1 17/20.

No. of bottles remaining is

= 3 1/2 - 1 17/20

= 3 10/20 - 1 17/20

(Numerator of the first fraction is smaller than the second. In subtraction of mixed numbers, always the numerator of the first fraction to be greater)

Then, we have

=  (3 + 10/20) - 1 17/20

=  (2 + 1 + 10/20) - 1 17/20

=  (2 + 20/20 + 10/20) - 1 17/20

=  (2 + 30/20) - 1 17/20

=  2 30/20 - 1 17/20

By regrouping, we have

=  (2 - 1) + (30/20 - 17/20)

=  (1) + (13/20)

=  1 13/20

So, 1 13/20 bottles of milk Amy has left over. 

Problem 9 :

A tank has 82 3/4 liters of water. 24 4/5 liters of water were used and the tank was filled with another 18 3/4 liters. What is the final volume of the water in the tank ?

Solution :

Initially, the tank has 82 3/4 liters.

24 4/5 liters were used -----> Subtract

The tank was filled with another 18 3/4 liters -----> Add

Then, the final volume of the water in tank is

=  82 3/4 - 24 4/5 + 18 3/4

In the above mixed fractions, we have the denominators 4 and 5.

L.C.M of (5, 4) = 20

To simplify the above expression, we have to make the denominators of both the mixed fractions to be 20. 

Then, we have

=  82 15/20  - 24 16/20 + 18 15/20

By regrouping, we have

=  (82 - 24 + 18) + (15/20 - 16/20 + 15/20)

=  (76) + (14/20)

=  (76) + (7/10)

= 76 7/10

So, the final volume of water in the tank is 76 7/10 liters. 

Problem 10 :

A trader prepared 21 1/2 liters of lemonade. At the end of the day he had 2 5/8 liters left over. How many liters of lemonade was sold by the Trader? 

Solution :

Initial stock of lemonade is 21 1/2 liters.

Closing stock is 2 5/8 liters. 

No. of liters sold  =  Initial stock - closing stock

No. of liters sold  =  21 1/2 - 2 5/8

No. of liters sold  =  21 4/8 - 2 5/8

(Numerator of the first fraction is smaller than the second. In subtraction of mixed numbers, always the numerator of the fraction to be greater)

Then, we have

No. of liters sold  =  (21 + 4/8) - 2 5/8

No. of liters sold  =  (20 + 1 + 4/8) - 2 5/8

No. of liters sold  =  (20 + 8/8 + 4/8) - 2 5/8

No. of liters sold  =  (20 + 12/8) - 2 5/8

No. of liters sold  =  20 12/8 - 2 5/8

By regrouping, we have

No. of liters sold  =  (20 - 2) + (12/8 - 5/8)

No. of liters sold  =  (18) + (7/8)

No. of liters sold  =  18 7/8

So, 18 7/8 liters of lemonade was sold by the Trader.

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