WORD PROBLEMS WITH MIXED NUMBERS WORKSHEET

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Problem 1 :

Linda walked 2β…“ miles on the first day and 3β…– miles on the next day. How many miles did she walk in all?

Problem 2 :

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

Problem 3 :

Mr. A has 3β…” acres of land. He gave 1ΒΌ acres of land to his friend. How many acres of land does Mr. A have now?

Problem 4 :

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

Problem 5 :

In the first hockey games of the year, Rodayo played 1Β½ periods and 1ΒΎ periods. How many periods in all did he play?

Problem 6 :

A bag can hold 1Β½ pounds of flour. If Mimi has 7Β½ pounds of flour, then how many bags of flour can Mimi make?

Problem 7 :

Jack and John went fishing Jack caught 3ΒΎ kg of fish and while John  caught 2β…• kg of fish. What is the total weight of the fish they caught? 

Problem 8 :

Amy has 3Β½ bottles in her refrigerator. She used β…— bottle in the morning 1ΒΌ bottle in the afternoon. How many bottles of milk does Amy have left over?

Problem 9 :

A tank has 82ΒΎ liters of water. 24β…˜ 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?

Problem 10 :

A trader prepared 21Β½ 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?

1. Answer :

Total number of miles Linda walked is

2β…“ 3β…–

In the above mixed numbers, 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 numbers to be 15.

Then, we have

2β…“ 3β…– = 2⁡⁄₁₅ + 3⁢⁄₁₅

By regrouping, we have

= (2 + 3) + (⁡⁄₁₅ + ⁢⁄₁₅)

= 5 + ΒΉΒΉβ„₁₅

= 5¹¹⁄₁₅

So, Linda walked 5¹¹⁄₁₅ miles in all.

2. Answer :

Number of pizzas David had initially is

= 2⅐ + 1³⁄₁₄

= 2²⁄₁₄ + 1³⁄₁₄

By regrouping, we have

= (2 + 1) + (²⁄₁₄ + ³⁄₁₄)

= 3 + β΅β„₁₄

= 3⁡⁄₁₄

So, initially David had 3⁡⁄₁₄ pizzas. 

3. Answer :

Now, number of acres of land that Mr. A has

= 3β…” - 1ΒΌ

In the above mixed numbers, 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 numbers to be 12. 

Then, we have

3β…” - 1ΒΌ = 3⁸⁄₁₂ - 1³⁄₁₂

By regrouping, we have

= (3 - 1) + (⁸⁄₁₂ - ³⁄₁₂)

= 2 + β΅β„₁₂

= 2⁡⁄₁₂

Now, Mr. A has 2⁡⁄₁₂ acres of land. 

4. Answer :

Number of cups of nuts that Lily put in all is

= 3β…“ + 1β…“

By regrouping, we have

= (3 + 1) + (3β…“ + 1β…“)

= 4 + β…”

= 4β…”

So, Lily put 4β…” cups of nuts in all.

5. Answer :

Number of periods in all he played is

= 1Β½ + 1ΒΎ

= 1²⁄₄ + 1ΒΎ

By regrouping, we have

= (1 + 1) + (1²⁄₄ + 1ΒΎ)

= 2 + β΅β„β‚„

= 2 + 1ΒΌ

= 2 + (1 + ΒΌ)

= (2 + 1) + ΒΌ

= 3ΒΌ

So, Rodayo played 3ΒΌ periods in all.

6. Answer :

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

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

Total no. of pounds of flour is

= 7Β½

= 15/2

No. of pounds per bag is

= 1Β½

= 3/2

Then,  we have

Number of bags = (15/2) Γ· (3/2)

= (15/2) β‹… (2/3)

= 30/6

= 5

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

7. Answer :

Total weight of the fish they caught is 

=  3ΒΎ + 2β…•

In the above mixed numbers, 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 numbers to be 20.

Then, we have

3ΒΎ + 2β…• = 3¹⁡⁄₂₀ + 2⁴⁄₂₀

By regrouping, we have

= (3 + 2) + (¹⁡⁄₂₀ + ⁴⁄₂₀)

= 5 + ΒΉβΉβ„β‚‚β‚€

= 5¹⁹⁄₂₀

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

8. Answer :

No. of bottles of milk used is

β…— + 1ΒΌ

In the above mixed numbers, 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 numbers to be 20.

Then, we have

β…— + 11ΒΌ ΒΉΒ²β„β‚‚β‚€ + 1⁡⁄₂₀

By regrouping, we have

= 1 + (¹²⁄₂₀ + ⁡⁄₂₀)

= 1 + ΒΉβ·β„β‚‚β‚€

= 1¹⁷⁄₂₀

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

No. of bottles remaining is

= 3Β½ - 1¹⁷⁄₂₀

= 3¹⁰⁄₂₀ - 1¹⁷⁄₂₀

(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 + ΒΉβ°β„β‚‚β‚€) - 1¹⁷⁄₂₀

= (2 + 1 + ΒΉβ°β„β‚‚β‚€) - 1¹⁷⁄₂₀

= (2 + Β²β°β„β‚‚β‚€ + ΒΉβ°β„β‚‚β‚€) - 1¹⁷⁄₂₀

= (2 + Β³β°β„β‚‚β‚€) - 1¹⁷⁄₂₀

= 2³⁰⁄₂₀ - 1¹⁷⁄₂₀

By regrouping, we have

= (2 - 1) + (³⁰⁄₂₀ - ΒΉβ·β„β‚‚β‚€)

= 1 + ΒΉΒ³β„β‚‚β‚€

= 1¹³⁄₂₀

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

9. Answer :

Initially, the tank has 82ΒΎ liters.

24β…˜ liters were used -----> Subtract

The tank was filled with another 18ΒΎ liters -----> Add

Then, the final volume of the water in tank is

=  82ΒΎ - 24β…˜ + 18ΒΎ

In the above mixed numbers, 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 numbers to be 20.

Then, we have

= 82¹⁡⁄₂₀ - 24¹⁢⁄₂₀ + 18¹⁡⁄₂₀

By regrouping, we have

= (82 - 24 + 18) + (¹⁡⁄₂₀ - ¹⁢⁄₂₀ + ¹⁡⁄₂₀)

= 76 + ΒΉβ΄β„β‚‚β‚€

=  76 + β·β„₁₀

= 76⁷⁄₁₀

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

10. Answer :

Initial stock of lemonade is 21Β½ liters.

Closing stock is 2⅝ liters.

No. of liters sold = Initial stock - closing stock

No. of liters sold = 21Β½ - 2⅝

No. of liters sold = 21β΄β„β‚ˆ - 2⅝

(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 + β΄β„β‚ˆ) - 2⅝

= (20 + 1 + β΄β„β‚ˆ) - 2⅝

= (20 + βΈβ„β‚ˆ + β΄β„β‚ˆ) - 2⅝

= (20 + ΒΉΒ²β„β‚ˆ) - 2⅝

= 20ΒΉΒ²β„β‚ˆ - 2⅝

By regrouping, we have

= (20 - 2) + (ΒΉΒ²β„β‚ˆ - ⅝)

= 18 + β…ž

= 18β…ž

So, 18β…ž liters of lemonade was sold by the Trader.

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