Word Problems on Addition of Mixed Fractions

Word Problems on Addition of Mixed Fractions | Adding Mixed Numbers Word Problems

Mixed Fractions are the ones having a Whole Number and a Fraction. Learn how to solve problems on adding mixed fractions by availing our quick resource on Word Problems on Addition of Mixed Fractions. Try to solve the Questions on Adding Mixed Numbers available on your own before you cross-check with the respective Solutions and explanations. For better understanding, we have provided the Examples on Mixed Numbers Addition step by step. Understand the Problem Solving Strategy Used and Solve the Related Problems on your own.

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Addition of Mixed Fractions Word Problems with Answers

Example 1.
Ram walked 1\(\frac { 3 }{ 5 } \)  km on Monday, 3 \(\frac { 1 }{ 6 } \) km on Tuesday, and 2 \(\frac { 1 }{ 12 } \)  km on Wednesday. Find out the distance he walked in all?
Solution:
Ram walked on monday= 1 \(\frac { 3 }{ 5 } \)
Ram walked on tuesday=3 \(\frac { 1 }{ 6 } \)
Ram walked on wednesday=2 \(\frac { 1 }{ 12 } \)
The distance he walked in all days= 1 \(\frac { 3 }{ 5 } \) + 3\(\frac { 1 }{ 6 } \) + 2 \(\frac { 1 }{ 12 } \)
=\(\frac { 8 }{ 5 } \) +\(\frac { 19 }{ 6 } \) + \(\frac { 25 }{ 12 } \)
=\(\frac {8 × 12  }{ 5 ×12 } \) +\(\frac {19 × 10  }{ 6 ×10 } \) +\(\frac {25 ×5   }{ 12 ×5 } \) ( LCM of 5,6,12 is 60)
= \(\frac { 96 }{ 60 } \)+\(\frac { 190 }{ 60 } \)1 +\(\frac { 125 }{ 60 } \)
=\(\frac { 411 }{ 60 } \)
=\(\frac { 137 }{ 20 } \) =6 \(\frac { 17 }{ 20 } \)
Hence, Ram walked 6 \(\frac { 17 }{ 20 } \) km.

Example 2.
Laxman bought 2 \(\frac { 1 }{ 2 } \) kg wheat, 4\(\frac { 1 }{ 4 } \)  of sugar, 10 \(\frac { 1 }{ 2 } \) kg rice. How many kg of items did he buy?
Solution:
Laxman bought wheat= 2 \(\frac { 1 }{ 2 } \) kg
Laxman bought sugar=4 \(\frac { 1 }{ 4 } \) kg
Laxman bought rice=10 \(\frac { 1 }{ 2 } \) kg
Total no. of kg of items Laxman bought  = 2 \(\frac { 1 }{ 2 } \)  + 4 \(\frac { 1 }{ 4 } \) + 10 \(\frac { 1 }{ 2 } \)
=\(\frac { 5 }{ 2 } \) +\(\frac { 17 }{ 4 } \) +\(\frac { 21 }{ 2 } \)
=\(\frac { 5 ×2}{ 2×2 } \) + \(\frac { 17}{ 4 } \)  +\(\frac { 21 ×2 }{ 2×2 } \)
= \(\frac { 10 }{ 4 } \) +\(\frac { 17 }{ 4 } \) + \(\frac { 42 }{ 4 } \)
=  \(\frac { 69 }{ 4 } \) =17 \(\frac { 1 }{ 4 } \) .
Therefore, Laxman bought 17 \(\frac { 1 }{ 4 } \) kg of items.

Example 3.
Rajesh donated 19 \(\frac { 1 }{ 2 } \) kg of rice, 25 \(\frac { 1 }{ 4 } \) kg of wheat, and 11 \(\frac { 3 }{ 4 } \) of oil. Find how many kgs of groceries he donated?
Solution:
Rajesh donated rice=19 \(\frac { 1 }{ 2 } \)
Rajesh donated wheat=25 \(\frac { 1 }{ 4 } \)
Rajesh donated oil=11\(\frac { 3}{ 4 } \)
Total no. of kg of groceries Rajesh donated =19 \(\frac { 1 }{ 2 } \)+ 25 \(\frac { 1 }{ 4 } \) + 11 \(\frac { 3 }{ 4 } \)
=\(\frac { 39 }{ 2 } \)+\(\frac { 101 }{ 4 } \)+\(\frac { 47 }{ 4 } \)
= \(\frac { 39 × 2 }{ 2 × 2 } \)+\(\frac { 101 }{ 4 } \)+ \(\frac { 47 }{ 4 } \)
=\(\frac { 78 }{ 4 } \)+\(\frac { 101 }{ 4 } \)+ \(\frac { 47 }{ 4 } \)47/4
=\(\frac { 226 }{ 4 } \)=\(\frac { 113 }{ 2 } \)=56\(\frac {1 }{ 2 } \)
Hence, Rajesh donated 56 \(\frac {1 }{ 2 } \) kgs of groceries.

Example 4.
Giri went on walking 3 \(\frac {1 }{ 2 } \)  km, and then cycling by 4 \(\frac {1 }{ 4 } \) km and then he took a lift and traveled 8 \(\frac {3 }{ 8 } \) km. Find out how much distance he traveled?
Solution:
Giri went on walking =3 \(\frac {1 }{ 2 } \)  km
Giri went on cycling=4 \(\frac {1 }{ 4 } \) km
Giri took a lift and travelled=8 \(\frac {3}{ 8 } \) km
The total distance he travelled= 3 \(\frac {1 }{ 2 } \)+4 \(\frac {1 }{ 4 } \) + 8 \(\frac {3}{ 8 } \)
=\(\frac {7}{ 2 } \)+\(\frac {9 }{ 4 } \)+\(\frac {67 }{ 8 } \)
=\(\frac {7 × 4 }{ 2 ×4 } \) + \(\frac {9 × 2 }{ 4 × 2} \)  + \(\frac {67}{ 8 } \) (LCM of 2,4,8 is 8)
=\(\frac {28}{ 8 } \)+\(\frac {18}{ 8 } \)+ \(\frac {67}{ 8 } \)
=\(\frac {113}{ 8 } \)=14 \(\frac {1}{ 8 } \)

Example 5.
Sameera went to a market and bought mangoes 3 \(\frac {3}{ 4 } \) and her sister bought 4 \(\frac {5}{ 8 } \) mangoes. Find how many kgs of mangoes bought by Sameera and her sister?
Solution:
Sameera bought mangoes= 3 \(\frac {3}{ 4 } \)
Sameera sister bought mangoes=4 \(\frac {5}{ 8 } \)
Total no. of kg of mangoes bought by Sameera and her sister=3 \(\frac {3}{ 4 } \)  + 4 \(\frac {5}{ 8 } \)
= \(\frac {15}{ 4 } \) + \(\frac {29}{ 8 } \)
= \(\frac {15 × 2}{ 4× 2} \)+\(\frac {29}{ 8 } \)
=  \(\frac {30}{ 8 } \)+\(\frac {29}{ 8 } \)=\(\frac {59}{ 8 } \)
=7\(\frac {3}{ 8 } \)
Therefore, Sameera and her sister bought 7\(\frac {3}{ 8 } \)  kg of mangoes.

Example 6.
For a birthday party, Akhil distributed 14 \(\frac {1}{ 2 } \) liters of Pepsi, 10 \(\frac {3}{ 4 } \) liters of sprite, and 12 \(\frac {5}{ 8 } \) liters of thum sup. Find how many liters of cool drinks he distributed at the birthday party?
Solution:
Akhil distributed pepsi= 14 \(\frac {1}{ 2 } \)
Akhil distributed sprite=10 \(\frac {3}{ 4 } \)
Akhil distributed thums up=12 \(\frac {5}{ 8 } \)
Total no. of liters of cool drinks Akhil distributed in the party=14 \(\frac {1}{ 2 } \) +10 \(\frac {3}{ 4 } \) +12 \(\frac {5}{ 8 } \)
=\(\frac {29}{ 2 } \)+\(\frac {43}{ 4 } \)+\(\frac {101}{ 8 } \)
=\(\frac {29 ×4}{ 2 ×4 } \)+\(\frac {43 ×2}{ 4 ×2 } \)+\(\frac {101}{ 8 } \)
=\(\frac {116}{ 8 } \)+ \(\frac {86}{ 8 } \)+\(\frac {101}{ 8 } \)
=\(\frac {303}{ 8 } \)=37 \(\frac {7}{ 8 } \)
Akhil distributed 37 \(\frac {7}{ 8 } \) liters of cool drinks at the birthday party.

Example 7.
In a competition between two friends, Anjana prepared 20 \(\frac {1}{ 2 } \) lit of orange juice and Sanjana prepared 22 \(\frac {5}{ 8 } \) of orange juice. Find how many liters of juice are prepared by both of them?
Solution:
Anjana prepared orange juice= 20 \(\frac {1}{ 2 } \)
Sanjana prepared orange juice=22 \(\frac {5}{ 8} \)
No. of liters of juice prepared by both of them= 20 \(\frac {1}{ 2 } \) + 22 \(\frac {5}{ 8 } \)
= \(\frac {41}{ 2 } \)+ \(\frac {181}{ 8 } \)
=\(\frac {41 ×4}{ 2 ×4 } \) +\(\frac {181}{ 8 } \)
=\(\frac {164}{ 8 } \)+\(\frac {181}{ 8 } \)=\(\frac {345}{ 8 } \)
=43\(\frac {1}{ 8 } \)
Anjana and sanjana both prepared 43 \(\frac {1}{ 8 } \) liters of juice.

Example 8. 
Sriram has wires of length 6 \(\frac {3}{ 4 } \)  m, 10 \(\frac {5}{ 8 } \). Find the length of both the wires?
Solution:
First wire length= 6 \(\frac {3}{ 4 } \)
Second wire length=10 \(\frac {5}{ 8 } \)
Length of both the wires= 6 \(\frac {3}{ 4 } \)  +10 \(\frac {5}{ 8 } \)
=\(\frac {27}{ 4 } \)  +\(\frac {85}{ 8 } \)
= \(\frac {27 × 2}{ 4 × 2 } \) +\(\frac {85}{ 8 } \)
=\(\frac {54}{ 8 } \)  +\(\frac {85}{ 8 } \) =\(\frac {139}{ 8 } \)
=17 \(\frac {3}{ 8 } \)
The length of both wires is 17 \(\frac {3}{ 8 } \) .

Example 9.
Bhaskar spent 8 \(\frac {1}{ 2 } \) hours on work, 2 \(\frac {1}{ 2} \)  hours on the playground. He spent 1 \(\frac {3}{ 4 } \)on  walking. Find how much time he spent on the day?
Solution:
Bhaskar spent on work=8 \(\frac {1}{ 2 } \)
Bhaskar spent on playground=2 \(\frac {1}{ 2 } \)
Bhaskar spent on walking=1 \(\frac {3}{ 4} \)
Bhaskar spent on the day= 8 \(\frac {1}{ 2 } \) + 2 \(\frac {1}{ 2 } \) + 1\(\frac {3}{ 4 } \)
= \(\frac {17}{ 2 } \) +\(\frac {5}{ 2 } \) +\(\frac {7}{ 4 } \)
= \(\frac {17 × 2}{ 2 ×2 } \) + \(\frac { 5× 2}{ 2 ×2 } \)  +\(\frac {7}{ 4 } \)
=\(\frac {34}{ 4 } \) + \(\frac {10}{ 4 } \) +\(\frac {7}{ 4 } \)
=\(\frac {51}{ 4 } \) =12 \(\frac {3}{ 4 } \)
Therefore, Bhaskar spent 12 \(\frac {3}{ 4 } \)  hours a day.

Example 10.
Srikrishna bought 5 \(\frac {1}{ 4 } \) of vegetables, 20 \(\frac {3}{ 4 } \)  kg of groceries, and 1 \(\frac {1}{ 2} \)  kg of chicken. Find how many kgs he bought?
Solution:
Srikrishna bought vegetables= 5 \(\frac {1}{ 4 } \) kg
Srikrishna bought groceries=20 \(\frac {3}{ 4 } \)   kg
Srikrishna bought chicken= 1 \(\frac {1}{ 2 } \)  kg
Srikrishna bought all=5 \(\frac {1}{ 4 } \) +20 \(\frac {3}{ 4 } \) +1 \(\frac {1}{ 2 } \)
= \(\frac {21}{ 4 } \) +\(\frac {83}{ 4 } \) +\(\frac {3}{ 2 } \)
=\(\frac {21}{ 4 } \) +\(\frac {83}{ 4 } \)+ \(\frac {3 ×2}{ 2 × 2 } \)
=\(\frac {21}{ 4 } \) +\(\frac {83}{ 4 } \) + \(\frac {6}{ 4 } \)
=\(\frac {110}{ 4 } \)=\(\frac {55}{ 2 } \)=27 \(\frac {1}{ 2 } \)
Srikrishna bought 27 \(\frac {1}{ 2 } \) kg.

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