Students will learn how to solve problems on multiplication of mixed fractions by referring to this page. We have covered various models of Mixed Fractions Multiplication Word Problems that range from easy ones to difficult ones. Try to answer all the questions on mixed fractions multiplication on your own and then verify with our solutions provided to test your skills on the concept. Practice as much as you can and enhance your problem-solving ability and speed of attempting the questions too.

Do Check:

- Word Problems on Division of Mixed Fractions
- Word Problems on Subtraction of Mixed Fractions
- Word Problems on Addition of Mixed Fractions

## Mixed Numbers Multiplication Word Problems

**Example 1.
**Sai spent 8 hours selling vegetables in the market. Murali sells for 1\(\frac {1}{ 4} \)Â times as many hours as Sai. How many hours did Murali sell vegetables in the market?

**Solution:**

Time is taken by Sai for selling the vegetables= 8 hours

Murali sells vegetables =1 \(\frac {1}{ 4} \) Ã— 8

=\(\frac {5}{ 4} \) Ã— 8

=\(\frac {5}{ 4} \)Ã— \(\frac {8}{ 1} \)

=\(\frac {40}{ 4} \) =10

Therefore, Murali spent 10 hours selling vegetables in the market.

**Example 2.
**On Doctor’s advice, Rajesh decided to walk every day. He walks \(\frac {4}{ 7} \) kilometers every day. How many kilometers did he walk in a week?

**Solution:**

Rajesh walk every day= \(\frac {4}{ 7} \)

No. of kilometers Rajesh walk in a week=7 Ã— \(\frac {4}{ 7} \)

=4 km

Therefore, Rajesh walks 4 km a week.

**Example 3.
**The heaviest weight in the family of Sarath is 90 kg. Sarath’s weight is \(\frac {1}{ 3} \) of the heaviest weight. Find the weight of Sarath?

**Solution:**

The heaviest weight in the family = 90 kg

Sarath weight= \(\frac {1}{ 3} \)Ã— 90

=30 kg

Hence, Sarath’s weight is 30 kg.

**Example 4.
**Supriya is interested in singing and dancing. Supriya practices 28 hours a week with them. She takes \(\frac {1}{ 3} \) of the time for singing and \(\frac {1}{ 4} \) of the time for dancing. How many hours does she take for dancing?

**Solution:**

Total no. of hours Supriya practice every week= 28

**Supriya takes time for singing=\(\frac {1}{ 3} \)**

Supriya takes time for dancing=\(\frac {1}{ 4} \)

No. of hours she spends in dancing=\(\frac {1}{ 4} \)Ã— 28

=7

Therefore, Supriya takes 7 hours for dancing.

**Example 5.
**In an office, there are 2000 employees. Male employees are \(\frac {1}{ 4} \). Find the Female employees in the office?

**Solution:**

Total no. of employees in the office=2000

No. of male employees=\(\frac {1}{ 4} \)

No. of female employees=1-\(\frac {1}{ 4} \)=\(\frac {3}{4} \)

\(\frac {3}{ 4} \)Ã— 2000=1500.

Therefore, the total no. of female employees in the office is 1500.

**Example 6.
**Games are conducted for the students in class X. \(\frac {3}{ 8} \) of the total no of students are in class X. \(\frac {1}{ 5} \) of the students in class X participate in games. What fraction of all the students participate in the games?

**Solution:**

No. of students in the class X= \(\frac {3}{ 8} \)

No. of students participate in the games=\(\frac {1}{ 5} \)

Fraction of all students participate in the games=\(\frac {3}{ 8} \) Ã— \(\frac {1}{5} \)=\(\frac {3}{40} \)

Therefore, the fraction of all the students participating in the exam is \(\frac {3}{40} \) .

**Example 7.**

Jagadish planted a mango tree and guava tree in the garden. The guava tree is 3 \(\frac {1}{ 5} \) feet tall. The mango tree is 1 \(\frac {1}{ 3} \) as tall as the guava tree. How tall is the mango tree?

**Solution:
**The height of the guava tree=3 \(\frac {1}{5} \)

The height of the mango tree =1\(\frac {1}{ 3} \) Ã— 3 \(\frac {1}{ 5} \)

=\(\frac {4}{ 3} \) Ã— \(\frac {16}{ 5} \)

=\(\frac {4 Ã— 16}{ 3Ã— 5} \)

=\(\frac {64}{15} \)

=4 \(\frac {4}{15} \)

Therefore, the height of the mango tree is 4 \(\frac {4}{15} \).

**Example 8.
**The cake needs 3 \(\frac {1}{2} \) cups of flour. Janci is baking 2 \(\frac {1}{ 2} \) cakes. How many cups of flour will she need?

**Solution:**

No. of cups of flour required for cake=3 \(\frac {1}{ 2} \)

No. of cakes baking= 2 \(\frac {1}{ 2} \)

No. of cups of flour required=2 \(\frac {1}{ 2} \) Ã— 3 \(\frac {1}{ 2} \)

=\(\frac {5}{ 2} \)Ã— \(\frac {7}{ 2} \)

=\(\frac {35}{ 4} \)

=8 \(\frac {3}{ 4} \)

Therefore, 8 \(\frac {3}{ 4} \) cups of flour required for baking 2 \(\frac {1}{ 2} \) cakes.

**Example 9.
**Sirisha uses \(\frac {2}{5} \) liters of petrol in her Scooty for 1 km. Find how many liters of petrol Sirisha use to drive 1 \(\frac {3}{8} \) km?

**Solution:**

No. of liters of petrol for 1 km=\(\frac {2} {5} \)

No. of liters of petrol used for 1 \(\frac {3}{ 8} \) km= 1 \(\frac {3}{ 8} \)Â Ã— \(\frac {2} {5} \)

=\(\frac {11} {8} \)Ã— \(\frac {2} {5} \)

=\(\frac {22} {40} \)

=\(\frac {11} {10} \)

Therefore, Sirisha uses \(\frac {11} {10} \) lt of petrol for 1 \(\frac {3} {8} \) km.

**Example 10.
**A restaurant makes sweet lime juice on Friday and uses 3 \(\frac {1} {8} \) kg of sweet limes. As Saturday is a busy day, the restaurant plans to make 8 \(\frac {1} {9} \) as much as sweet lime juice. How many kg of Sweet lime juice is required?

**Solution:**

No. of kg of sweet limes used= 3 \(\frac {1} {8} \)

No. of kg of sweet limes used to make 8 \(\frac {1} {9} \) lt=3 \(\frac {1} {8} \) Ã—Â 8 \(\frac {1} {9} \)

=\(\frac {25} {8} \) Ã—Â \(\frac {73} {9} \)

=\(\frac {25 Ã—Â 73} {8 Ã— 9} \)

=\(\frac {1825} {72} \)

=25 \(\frac {25} {72} \)

Therefore, Restaurant uses 25 \(\frac {25} {72} \) kg of sweet limes.

**Example 11.
**Rakesh can ride a bike 3 \(\frac {2} {5} \) km in 1 hour. If he plans to ride a bike for 4 \(\frac {1} {7} \) hours, how much distance he can travel?

**Solution:**

No. of km Rakesh can ride a bike= 3 \(\frac {2} {5} \) km

No. of km Rakesh can ride a bike in 4 \(\frac {1} {7} \)hours=4 \(\frac {1} {7} \)Â Ã— 3 \(\frac {2} {5} \)

=\(\frac {29} {7} \) Ã— \(\frac {17} {5} \)

=\(\frac {29 Ã— 17} {7 Ã—Â 5} \)

=\(\frac {493} {35} \)

=14 \(\frac {3} {35} \)

Therefore, Rakesh can ride a bike in 14 3/35 km.