Subtraction of Mixed Fractions Word Problems Worksheet includes different problems ranging from easy to the most difficult ones. You can practice the Questions from Subtracting Mixed Numbers Word Problems Worksheet and assess your strengths and weaknesses. To Solve Mixed Fractions Subtraction Sums the logic is to first convert given mixed fractions to improper fractions and then change to like fractions and further simplify them. Answer the Word Problems available in the Subtracting Mixed Numbers Worksheet PDF on a frequent basis and increase your problem-solving ability and accuracy.

Do Read:

- Worksheet on Word Problems on Addition of Mixed Fractions
- Worksheet on Word Problems on Multiplication of Mixed Fractions

## Subtraction of Mixed Fractions Word Problems Worksheet

**Example 1.**

Aarush walks 3 \(\frac {1}{ 3} \) km. Arun walks 1 \(\frac {1}{ 6} \) km. Find the difference in the distance they ran?

**Solution:**

Aarush walks= 3 \(\frac {1}{ 3} \) km

Arun walks = 1 \(\frac {1}{ 6} \)

The difference in the distance they ran=3 \(\frac {1}{ 3} \) -1 \(\frac {1}{ 6} \)

=\(\frac {10}{ 3} \)1-\(\frac {7}{ 6} \)

=\(\frac {10 × 2}{ 3 × 2} \)–\(\frac {7}{ 6} \)

=\(\frac {20}{ 6} \)–\(\frac {7}{ 6} \)

=\(\frac {13}{ 6} \)

Therefore, the difference in the distance they walked is \(\frac {13}{ 6} \) km.

**Example 2.
**Raj has 3 \(\frac {1}{ 8} \) acres of agricultural land. He gave 1 \(\frac {1} {2} \) acres of land to his son. Find how many acres of land does Raj has now?

**Solution:**

Raj has agricultural land=3 \(\frac {1}{ 8} \)

Raj gave land to his son=1\(\frac {1}{2} \)

Raj has agricultural land now=3 \(\frac {1}{ 8} \)-1 \(\frac {1}{ 2} \)

=\(\frac {25}{ 8} \)–\(\frac {3} {2} \)

=\(\frac {25}{ 8} \)– \(\frac {3 × 4}{ 2 × 4} \)

=\(\frac {25}{ 8} \)–\(\frac {12}{ 8} \)

=\(\frac {13}{ 8} \)

Therefore, Now Raj has agricultural land is \(\frac {13}{ 8} \) acres.

**Example 3.
**Suma bought 5 \(\frac {1}{ 5} \) liters of grape juice bottles for the function. Suma used 2 \(\frac {1}{ 3} \) liters of grape juice in the function. She used 1 \(\frac {1}{ 3} \) grape juice after the function. Find how many liters of the grape juice is remaining?

**Solution:**

Suma bought grape juice= 5 \(\frac {1}{ 5} \)

Suma used grape juice in the function=2 \(\frac {1}{ 3} \)

Suma used grape juice after the function=1 \(\frac {1}{ 3} \)

Total no. of liters of grape juice used by suma=2 \(\frac {1}{ 3} \) +1 \(\frac {1}{ 3} \)

=\(\frac {7}{ 3} \)+\(\frac {4}{ 3} \)=3\(\frac {2}{ 3} \)

No. of liters of grape juice remaining=5 \(\frac {1}{ 5} \)-3\(\frac {2}{ 3} \)

=\(\frac {26}{ 5} \)–\(\frac {11}{ 3} \)

=\(\frac {26 × 3}{ 5 × 3} \)– \(\frac {11 × 5}{ 3 × 5} \)

= \(\frac {78}{ 15} \) – \(\frac {55}{ 15} \)

=\(\frac {23}{ 15} \)=1 \(\frac {8}{ 15} \)

Therefore, 1 \(\frac {8}{15} \) liters of juice has remained.

**Example 4.
**Santhi has 4 \(\frac {1}{ 7} \) kg of chocolates. She has 2 \(\frac {1}{ 8} \) chocolates remained after distribution to her friends. Find how many chocolates were distributed by Santhi?

**Solution:**

Santhi has chocolates= 4 \(\frac {1}{ 7} \) kg

Chocolates remained after distribution=2 \(\frac {1}{ 8} \)

No. of chocolates distributed by santhi=4 \(\frac {1}{ 7} \) – 2 \(\frac {1}{ 8} \)

=\(\frac {29}{ 7} \)–\(\frac {17}{ 8} \)

=\(\frac {29 × 8}{ 7 × 8} \)– \(\frac {17 × 7}{ 8 × 7} \)

=\(\frac {232}{ 56} \) – \(\frac {119}{ 56} \)

=\(\frac {113}{ 56} \)

=2 \(\frac {1}{ 56} \)

Therefore, No. of chocolates distributed by Swathi is 2 \(\frac {1}{ 56} \).

**Example 5.
**A juice shop prepared 18 \(\frac {1}{ 9} \)liters of juice, of which it sold 5 \(\frac {1}{3} \) liters of juice. Find how many liters of juice remained by?

**Solution:**

No. of liters of juice prepared=18 \(\frac {1}{ 9} \)

No. of liters of juice sold=5 \(\frac {1}{3} \)

No. of liters of juice remained by=18 \(\frac {1}{ 9} \)– 5 \(\frac {1}{3} \)

=\(\frac {163}{ 9} \)–\(\frac {16 }{3} \)

=\(\frac {163}{ 9} \)–\(\frac {16 × 3}{ 3 × 3} \)

=\(\frac {163}{ 9} \)–\(\frac {48}{9} \)

=\(\frac {115}{ 9} \)

=12 \(\frac {7}{ 9} \)

Therefore, No. of liters of juice remained is 12 \(\frac {7}{ 9} \).

**Example 6.
**Aarya is 1 \(\frac {5}{ 7} \) kg and Surya is 3 \(\frac {2}{ 9} \) kg. How much heavier is Surya than Aarya?

**Solution:**

Aarya weight= 1 \(\frac {5}{ 7} \)

Surya weight= 3 \(\frac {2}{ 9} \)

Surya is heavier than Aarya = 3 \(\frac {2}{ 9} \) – 1 \(\frac {5}{ 7} \)

=\(\frac {29}{ 9} \) – \(\frac {12}{ 7} \)

=\(\frac {29 × 7}{ 9 × 7} \)– \(\frac {12 × 9}{ 7× 9} \)

=\(\frac {203}{ 63} \)–\(\frac {108}{ 63} \)

=\(\frac {95}{63} \)

=1 \(\frac {32}{63} \)

Therefore, Surya is heavier than Aarya by 1 \(\frac {32}{ 63} \).

**Example 7.
**Ayan told to practice shuttle 3 \(\frac {1}{ 9} \) hours per day. He already practiced shuttle 1 \(\frac {1}{ 4} \) hour today. How many hours does Ayan still need to practice today?

**Solution:**

Ayan practice shuttle per day=3 \(\frac {1}{9} \)

Ayan practice shuttle today=1 \(\frac {1}{4} \)

Ayan still need to practice today=3 \(\frac {1}{9} \)-1 \(\frac {1}{4} \)

=\(\frac {28}{9} \)–\(\frac {5}{ 4} \)

=\(\frac {28× 4}{ 9 × 4} \)– \(\frac {5 × 9}{ 4 × 9} \)

=\(\frac {112}{36} \)–\(\frac {45}{36} \)

=\(\frac {67}{36} \)= 1 \(\frac {31}{36} \)

Therefore, Ayan still needs to practice today is 1 \(\frac {31}{36} \).

**Example 8.
**Tarun has 2 \(\frac {3}{ 8} \) fruits remained after selling. He sold 1 \(\frac {1}{8} \) fruits on that day. Find How many more fruits does Tarun have to sell?

**Solution:**

Tarun has remained fruits=2 \(\frac {3}{8} \)

Tarun sold fruits =1 \(\frac {1}{8} \)

Tarun has to sell fruits=2 \(\frac {3}{ 8} \)-1 \(\frac {1}{8} \)

=\(\frac {19}{8} \)–\(\frac {9}{8} \)

=\(\frac {10}{8} \)=\(\frac {5}{4} \)=1 \(\frac {1}{4} \)

Hence, Tarun has to sell 1 \(\frac {1}{4} \) fruits.

**Example 9.
**Sailaja spends 2 \(\frac {1}{ 8} \) hours practicing the violin. She also spends 1 \(\frac {1}{ 2} \) hours dancing. How much more time Sailaja spends practicing the violin?

**Solution:**

Sailaja spends practicing the violin=2 \(\frac {1}{8} \)

Sailaja spends dancing=1 \(\frac {1}{2} \)

Sailaja spends more time in violin= 2 \(\frac {1}{ 8} \) – 1 \(\frac {1}{2} \)

=\(\frac {17}{ 8} \) – \(\frac {3}{ 2} \)

=\(\frac {17}{ 8} \) – \(\frac {3 × 4}{ 2 × 4} \)

=\(\frac {17}{ 8} \)– \(\frac {12}{8} \)

=\(\frac {5}{ 8} \)

Therefore, Sailaja spends \(\frac {5}{8} \) more time practicing the violin.

**Example 10.
**For baking a cake, 2 \(\frac {1}{5} \) cups of flour and 1 \(\frac {1}{ 2} \) cups of sugar are required. Find how many more cups of flour are required for baking the cake?

**Solution:**

No. of cups of flour required=2 \(\frac {1}{5} \)

No. of cups of sugar required=1 \(\frac {1}{ 2} \)

No. of more cups of flour required than sugar= 2 \(\frac {1}{5} \)– 1 \(\frac {1}{2} \)

=\(\frac {11}{5} \)– \(\frac {3}{ 2} \)

=\(\frac {11 × 2}{ 5 × 2} \) – \(\frac {3× 5}{ 2 × 5} \)

=\(\frac {22}{10 } \)–\(\frac {15}{10} \)

=\(\frac {7}{ 10} \)

Therefore, \(\frac {7}{10} \) more cups of flour is required.