WORD PROBLEMS ON MIXED FRACTIONS

About "Word problems on mixed fractions"

Here we are going to see, how to solve word problems on mixed fractions through some examples.

Examples

To learn solving word problems on mixed fractions, let us look at some examples.

Example 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 = 2 1/3 + 3 2/5

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

L.C.M of (3, 5) = 15

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

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

= (2+3) (5/15 + 6/15) (by regrouping)

= 5 11/15

Hence, she walked 5 11/15 miles in all

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Example 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 = 2 1/7 + 1 3/14

2 1/7 + 1 3/14 = 2 2/14 + 1 3/14

= (2+1) (2/14 + 3/14) (by regrouping)

= 3 5/14

Hence, initially David had 3 5/14 pizzas.

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Example 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.

L.C.M of (3, 4) = 12

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

3 2/3 - 1 1/4 = 3 8/12 - 1 3/12

= (3 - 1) (8/12 - 3/12) (by regrouping)

= 2 5/12

Now, Mr. A has 2 5/12 acres of land.

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Example 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 = 3 1/3 + 1 1/3

3 1/3 + 1 1/3 = (3+1) (1/3 + 1/3) (by regrouping)

= 4 2/3

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

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Example 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 = 1 1/2 + 1 3/4

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

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

= 2 5/4, but 5/4 = 1 1/4

So, 2 5/4 = 3 1/4

Hence, no. of periods in all he played = 3 1/4

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Example 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 pounds) / (No of pounds per bag)

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

Total no. of pounds of flour = 7 1/2 = 15/2

No. of pounds per bag = 1 1/2 = 3/2

Then, no. of bags = (15/2) / (3/2)

= (15/2) x (2/3)

= (15x2) / (2x3)

= 30 / 6

= 5

Hence, the no. of bags that Mimi can make = 5

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Example 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 = 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

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

= (3 + 2) (15/20 + 4/20) (by regrouping)

= 5 19/20

Hence, the total weight of the fish they caught = 5 19/20 kg.

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Example 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 = 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

3/5 + 1 1/4 = 12/20 + 1 5/20

= 1 (12/20 + 5/20) (by regrouping)

= 1 17/20

No. of bottles of milk used = 1 17/20

No. of bottles remaining = 3 1/2 - 1 17/20

= 3 10/20 - 1 17/20 ---------(1)

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

3 10/20 = 2 (10/20 + 1) = 3 (10/20 + 20/20) = 2 30/20

(1) -------> 3 10/20 - 1 17/20 = 2 30/20 - 1 17/20

= (2 - 1) (30/20 - 17/20)

= 1 (30-17)/20

= 1 13/20

Hence, no. of bottles of milk that Amy has left over = 1 13/20

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Example 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

So, 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

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

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

= 76 (15-16+15)/20

= 76 14/20

= 76 7/10

Hence, final volume of water in the tank = 76 7/10 liters

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Example 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 were sold by the Trader?

Solution :

Initial strock = 21 1/2 liters

Closing stock = 2 5/8 liters

No. of liters sold = Initial stock - closing strock

= 21 1/2 - 2 5/8

= 21 4/8 - 2 5/8 -----------(1)

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

21 4/8 = 20 (4/8 + 1) = 20 (4/8 + 8/8) = 20 12/8

(1) ----------> 21 4/8 - 2 5/8 = 20 12/8 - 2 5/8

= (20-2) (12/8 - 5/8) (by regrouping)

= 18 7/8

Hence, no. of liters of lemonade were sold = 18 7/8 liters