Learn how to solve mixed fractions subtraction by referring to the interactive exercises over here. Make use of the Word Problems on Subtracting Mixed Numbers and understand how different questions are framed on the concept. The Word Problems provided here involve subtraction of mixed fractions having both like and unlike denominators. Try to solve the Questions over here on your own and then verify with our solutions.

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## Mixed Number Subtraction Word Problems

**Example 1:
**Sailaja spent 8 \(\frac {3}{ 4 } \) hr studying and playing shuttle. If she spent 3 \(\frac {1}{ 2 } \) hr in playing shuttle. Find how many hours she studied?

**Solution:**

Sailaja spent time studying and playing shuttle = 8 \(\frac {3}{ 4} \)

Sailaja spent time playing shuttle=3 \(\frac {1}{ 2 } \)

Sailaja spent time studying =8 \(\frac {3}{ 4 } \) – 3 \(\frac {1}{ 2 } \)

= \(\frac {35}{ 4 } \)–\(\frac {7}{ 2 } \)

=\(\frac {35}{ 4 } \)– \(\frac {2 ×7}{ 2 ×2 } \)

=\(\frac {35}{ 4 } \)–\(\frac {14}{ 4 } \)=\(\frac {21}{ 4 } \)=5 \(\frac {1}{ 4 } \)

Hence, Sailaja spent 5 \(\frac {1}{ 4 } \) hours studying.

**Example 2:
**Srinu runs 8 \(\frac {3}{ 4 } \) in the morning. In the evening he runs 5 \(\frac {5}{ 8 } \). Find how much faster he runs in the morning than evening?

**Solution:**

Srinu runs in the morning= 8 \(\frac {3}{ 4 } \)

Srinu runs in the evening= 5 \(\frac {5}{ 8 } \)

Srinu runs faster in the morning than an evening =8 \(\frac {3}{ 4 } \)– 5 \(\frac {5}{ 8 } \)

=\(\frac {35}{ 4 } \)–\(\frac {45}{ 8} \)

= \(\frac {35 × 2}{ 4 × 2 } \)– \(\frac {45}{ 8 } \) (LCM is 8)

= \(\frac {70}{ 8 } \)–\(\frac {45}{ 8 } \)=\(\frac {25}{ 8 } \)=3 \(\frac {1}{ 8 } \)

**Example 3:
**A tailor has a piece of cloth 15 \(\frac {5}{ 12 } \) m in length. He needs only 11 \(\frac {4}{ 12 } \) m in the cloth. Find how much cloth should he cut?

**Solution:**

A tailor has a piece of cloth =15 \(\frac {5}{ 12 } \)m

He needs cloth = 11 \(\frac {4}{ 12 } \) m

The cloth should be cut by the tailor= 15 \(\frac {5}{ 12 } \)-11 \(\frac {4}{ 12 } \)

= 4 1/12

Therefore, the tailor should cut 4 1/12 cloth.

**Example 4:
**The family drove their car 3 \(\frac {3}{ 4 } \) hours to the function hall. To return home they drove for 5 \(\frac {1}{ 2 } \) hours due to heavy traffic. How much longer time did it take to drive home?

**Solution:**

Time taken by the car to the function hall= 3 \(\frac {3}{ 4 } \)

Time is taken by the car to the home=5 \(\frac {1}{ 2 } \)

Longer time is taken by car to drove home= 5 \(\frac {1}{ 2 } \)– 3 \(\frac {3}{ 4 } \)

= \(\frac {11}{ 2 } \)–\(\frac {15}{ 4 } \)= \(\frac {11 ×2}{ 2 × 2} \)– \(\frac {15}{ 4} \)

=\(\frac {7}{ 4} \)=1 \(\frac {3}{ 4} \).

Hence, 1 \(\frac {3}{ 4} \) time taken by car to drove home.

**Example 5:
**Raju has some money. He spent \(\frac {1}{ 5} \) for the loan, \(\frac {3}{10} \) for the car, and \(\frac {1}{ 8} \) for house expenses. What part of the money is left with him?

**Solution:**

Total money spent by raju= \(\frac {1}{ 5} \)+ \(\frac {3}{ 10} \) +\(\frac {1}{ 8} \)

=\(\frac {1 × 8}{ 5 × 8} \) + \(\frac {3× 4}{ 10 × 4} \) +\(\frac {1 × 5}{ 8 × 5} \)

= \(\frac {8}{ 40} \) +\(\frac {12}{ 40} \) +\(\frac {5}{ 40} \)

=\(\frac {25}{ 40} \)=\(\frac {5} {8} \)

The part of money left with him= 1- \(\frac {5} {8} \)

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

=\(\frac {3} {8} \)

Raju has \(\frac {3} {8} \) money with him.

**Example 6: **

A student has 3 \(\frac {5} {8} \) time to complete the exam. He finished the exam in 2 \(\frac {1} {2} \) hours time. Find in how much time left by the student after completing the exam?

**Solution:
**A student has time to complete the exam= 3 \(\frac {5} {8} \)

He finished the exam =2 \(\frac {1} {2} \)

Time left by the student after finishing the exam= 3 \(\frac {5} {8} \) – 2 \(\frac {1} {2} \)

= \(\frac {29} {8} \)–\(\frac {5} {2} \)

= \(\frac {29} {8} \)–\(\frac {5 ×4 } {2 ×4} \)

= \(\frac {29} {8} \)– \(\frac {20} {8} \)

=\(\frac {9} {8} \)

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

The student has 1 \(\frac {1} {8} \) time after completing the exam.

**Example 7:
**Ramana bought 10 \(\frac {3} {4} \) kg of flowers for decoration. He used 8 \(\frac {1} {2} \) kg of flowers. How many kg of flowers are left after decoration?

**Solution:**

Ramana bought flowers= 10 \(\frac {3} {4} \)

Ramana used flowers for decoration= 8 \(\frac {1} {2} \)

Flowers that are leftover =10 \(\frac {3} {4} \) -8 \(\frac {1} {2} \)

=\(\frac {43} {4} \) –\(\frac {17} {2} \)=\(\frac {43} {4} \) – \(\frac {17 × 2} {2 × 2} \)

=\(\frac {43} {4} \)– \(\frac {34} {4} \)=\(\frac {9} {4} \)=2 \(\frac {1} {4} \).

**Example 8:
**Prasanth has some money. He gave his friend \(\frac {3} {4} \) of the money. Find the money he has?

**Solution:**

Prasanth gave his friend= \(\frac {3} {4} \)

Prasanth has money=1-\(\frac {3} {4} \)

=\(\frac {4} {4} \)– \(\frac {3} {4} \)=\(\frac {1} {4} \)

Therefore, Prasanth has \(\frac {1} {4} \) of the money.

**Example 9:
**A plumber has 6 \(\frac {7} {16} \) pipes. He needs only 3 \(\frac {5} {7} \) for his work. How much pipe does he need to cut?

**Solution:**

A plumber has pipe= 6 \(\frac {7} {16} \)

He needs only=3 \(\frac {5} {7} \)

He needs to cut the pipe=

6[Latex]\frac {7}{16} [\latex] – 3 [Latex]\frac {5}{7} [\latex]

=[Latex]\frac {37}{5} [\latex] -[Latex]\frac {26}{7} [\latex]=\(\frac {37 × 7} {5 × 7} \)– \(\frac {26 × 5} {7 × 5} \)

= \(\frac {259} {35} \)–\(\frac {130} {35} \)

=\(\frac {129} {35} \)=3 \(\frac {24} {35} \)