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