WORD PROBLEMS ON MIXED FRACTIONS

Problem 1 :

Linda walked 2⅓ miles on the first day and 3⅖ miles on the next day. How many miles did she walk in all?

Solution :

Total no. of miles she walked is

= 2⅓ + 3⅖

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

2⅓ + 3⅖ = 2⁵⁄₁₅ + 3⁶⁄₁₅

By regrouping, we have

= (2 + 3) + (⁵⁄₁₅ + ⁶⁄₁₅)

= 5 + ¹¹⁄₁₅

= 5¹¹⁄₁₅

So, Linda walked 5¹¹⁄₁₅ miles in all.

Problem 2 :

David ate 2⅐ pizzas and he gave 1³⁄₁₄  pizzas to his mother. How many pizzas did David have initially?

Solution :

No. of pizzas he had initially is

= 2⅐ + 1³⁄₁₄

= 2²⁄₁₄ + 1³⁄₁₄

By regrouping, we have

= (2 + 1) + (²⁄₁₄ + ³⁄₁₄)

= 3 + ⁵⁄₁₄

= 3⁵⁄₁₄

So, initially David had 3⁵⁄₁₄ pizzas.

Problem 3 :

Mr. A has 3⅔ acres of land. He gave 1¼ 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

= 3⅔ - 1¼

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

3⅔ - 1¼ = 3⁸⁄₁₂ - 1³⁄₁₂

By regrouping, we have

= (3 - 1) + (⁸⁄₁₂ - ³⁄₁₂)

= 2 + ⁵⁄₁₂

= 2⁵⁄₁₂

Now, Mr. A has 2⁵⁄₁₂ acres of land.

Problem 4 :

Lily added 3⅓ cups of walnuts to a batch of trail mix. Later she added 1⅓ 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

= 3⅓ + 1⅓

By regrouping, we have

= (3 + 1) + (3⅓ + 1⅓)

= 4 + ⅔

= 4⅔

So, Lily put 4⅔ cups of nuts in all.

Problem 5 :

In the first hockey games of the year, Rodayo played 1½ periods and 1¾ periods. How many periods in all did he play?

Solution :

No. of periods in all he played is

= 1½ + 1¾

= 1²⁄₄ + 1¾

By regrouping, we have

= (1 + 1) + (1²⁄₄ + 1¾)

= 2 + ⁵⁄₄

= 2 + 1¼

= 2 + (1 + ¼)

= (2 + 1) + ¼

= 3¼

So, Rodayo played 3¼ periods in all.

Problem 6 :

A bag can hold 1½ pounds of flour. If Mimi has 7½ 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

= 7½

= 15/2

No. of pounds per bag is

= 1½

= 3/2

Then,  we have

Number of bags = (15/2) ÷ (3/2)

= (15/2) ⋅ (2/3)

= 30/6

= 5

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

Problem 7 :

Jack and John went fishing Jack caught 3¾ kg of fish and while John  caught 2⅕ kg of fish. What is the total weight of the fish they caught?

Solution :

Total weight of the fish they caught is

=  3¾ + 2⅕

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

3¾ + 2⅕ = 3¹⁵⁄₂₀ + 2⁴⁄₂₀

By regrouping, we have

= (3 + 2) + (¹⁵⁄₂₀ + ⁴⁄₂₀)

= 5 + ¹⁹⁄₂₀

= 5¹⁹⁄₂₀

So, the total weight of the fish they caught is 5¹⁹⁄₂₀ kg.

Problem 8 :

Amy has 3½ bottles in her refrigerator. She used ⅗ bottle in the morning 1¼ bottle in the afternoon. How many bottles of milk does Amy have left over?

Solution :

No. of bottles of milk used is

= ⅗ + 1¼

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

⅗ + 11¼ = ¹²⁄₂₀ + 1⁵⁄₂₀

By regrouping, we have

= 1 + (¹²⁄₂₀ + ⁵⁄₂₀)

= 1 + ¹⁷⁄₂₀

= 1¹⁷⁄₂₀

So, no. of bottles of milk used is 1¹⁷⁄₂₀.

No. of bottles remaining is

= 3½ - 1¹⁷⁄₂₀

= 3¹⁰⁄₂₀ - 1¹⁷⁄₂₀

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

Then, we have

= (3 + ¹⁰⁄₂₀) - 1¹⁷⁄₂₀

= (2 + 1 + ¹⁰⁄₂₀) - 1¹⁷⁄₂₀

= (2 + ²⁰⁄₂₀ + ¹⁰⁄₂₀) - 1¹⁷⁄₂₀

= (2 + ³⁰⁄₂₀) - 1¹⁷⁄₂₀

= 2³⁰⁄₂₀ - 1¹⁷⁄₂₀

By regrouping, we have

= (2 - 1) + (³⁰⁄₂₀ - ¹⁷⁄₂₀)

= 1 + ¹³⁄₂₀

= 1¹³⁄₂₀

So, 1¹³⁄₂₀ bottles of milk Amy has left over.

Problem 9 :

A tank has 82¾ liters of water. 24⅘ 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 82¾ liters.

24⅘ liters were used -----> Subtract

The tank was filled with another 18¾ liters -----> Add

Then, the final volume of the water in tank is

= 82¾ - 24⅘ + 18¾

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

= 82¹⁵⁄₂₀ - 24¹⁶⁄₂₀ + 18¹⁵⁄₂₀

By regrouping, we have

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

= 76 + ¹⁴⁄₂₀

=  76 + ⁷⁄₁₀

= 76⁷⁄₁₀

So, the final volume of water in the tank is 76⁷⁄₁₀ liters.

Problem 10 :

A trader prepared 21½ liters of lemonade. At the end of the day he had 2⅝ liters left over. How many liters of lemonade was sold by the Trader?

Solution :

Initial stock of lemonade is 21½ liters.

Closing stock is 2⅝ liters.

No. of liters sold = Initial stock - closing stock

No. of liters sold = 21½ - 2⅝

No. of liters sold = 21⁴⁄₈ - 2⅝

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

Then, we have

No. of liters sold = (21 + ⁴⁄₈) - 2⅝

= (20 + 1 + ⁴⁄₈) - 2⅝

= (20 + ⁸⁄₈ + ⁴⁄₈) - 2⅝

= (20 + ¹²⁄₈) - 2⅝

= 20¹²⁄₈ - 2⅝

By regrouping, we have

= (20 - 2) + (¹²⁄₈ - ⅝)

= 18 + ⅞

= 18⅞

So, 18⅞ liters of lemonade was sold by the Trader.

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