Adding Mixed Fractions Word Problems Worksheet has problems on the addition of mixed numbers. You can find Adding Mixed Fractions Problems both with like and unlike denominators. By practicing the Questions on Adding Mixed Numbers you will understand how to solve a problem rather than simply getting the answer. Test your knowledge and assess your preparation level by answering the addition of mixed fractions word problems available on a frequent basis.
Do Read:
- Word Problems on Addition of Mixed Fractions
- Worksheet on Word Problems on Multiplication of Mixed Fractions
- Word Problems on Division of Mixed Fractions
Addition of Mixed Fractions Word Problems Worksheet
Example 1.
Ram and Shyam have two mango fields and the total area of the mango field is 4 \(\frac {2}{ 5} \). The big field yields 3 \(\frac {2}{ 5} \) tons of mangoes and the small field yields 2 \(\frac {1}{ 8} \) tons of mangoes. Find the total yield of mangoes?
Solution:
The big field yield tons of mangoes=3 \(\frac {2}{ 5} \)
The small field yields tons of mangoes=2 \(\frac {1}{ 8} \)
The total yield of mangoes=3 \(\frac {2}{ 5} \) +2 \(\frac {1}{ 8} \)
=\(\frac {17}{ 5} \)+\(\frac {17}{ 8} \)
= \(\frac {8 × 17}{ 8 ×5 } \) +\(\frac {17 × 5}{ 8 ×5 } \)
=\(\frac {136}{ 40 } \) +\(\frac {85}{ 40 } \)
=\(\frac {221}{ 40 } \)=5 \(\frac {21}{ 40 } \)
Hence, the total yield of mango fields is 5 \(\frac {21}{ 40 } \).
Example 2.
Sagar spends time in the tuition for 1 \(\frac {1}{ 2 } \) hours in the morning and 2 \(\frac {5}{ 8 } \) hours in the evening. How much time did he spend on the tuition?
Solution:
In the morning he spends time in tuition=1 \(\frac {1}{ 2 } \)
In the evening he spends time in tuition=2 \(\frac {5}{ 8} \)
The time he spends on the tuition=1 \(\frac {1}{ 2 } \) + 2 \(\frac {5}{ 8 } \)
= \(\frac {3}{ 2 } \) + \(\frac {21}{ 8 } \)
= \(\frac {12}{ 8 } \) + \(\frac {21}{ 8 } \)
= \(\frac {33}{ 8 } \) =4 \(\frac {1}{ 8 } \)
Hence, Sagar spends 4 \(\frac {1}{ 8 } \) hours on the tuition.
Example 3.
Sameera attended parents meeting for 1 \(\frac {2}{ 8 } \) hours, which is 1 \(\frac {1}{ 7 } \) longer than scheduled. How much time did she spend in the parent’s meeting?
Solution:
Sameera attended parents meeting = 1 \(\frac {2}{ 8 } \) hours
The meeting was longer by=1 \(\frac {1}{ 7 } \)
The time she spent in the meeting= 1 \(\frac {2}{ 8 } \) + 1 \(\frac {1}{ 7 } \)
=\(\frac {10}{ 8 } \) +\(\frac {8}{ 7 } \)
=\(\frac {10 × 7}{ 8 × 7 } \)+ \(\frac {8 ×8}{ 7 × 8 } \)
=\(\frac { 70}{ 56 } \) + \(\frac {64}{ 56 } \)
=\(\frac {134}{ 56 } \)
=\(\frac {67}{ 28 } \)
=2 \(\frac {11}{ 28 } \)
Hence, Sameera attended the parent’s meeting in 2 \(\frac {11}{ 28 } \) hours.
Example 4.
At a birthday party, Rakesh ordered 3 \(\frac {1}{4 } \) orange juice and 4 \(\frac {2}{ 7 } \) pineapple juice. But he got 4 \(\frac {3}{ 8 } \) orange juice and 3 \(\frac {2}{ 7} \) pineapple juice. Did he get more or fewer juices? By how much?
Solution:
Rakesh ordered = 3 \(\frac {1}{ 4 } \) +4 \(\frac {2}{ 7} \)
=\(\frac {13}{ 4 } \)+ \(\frac {30}{ 7} \)
=\(\frac {13 × 7}{ 4 × 7 } \) + \(\frac {30 × 4}{ 7 × 4} \)
=\(\frac {91}{28} \) +\(\frac {120}{ 28 } \)
=\(\frac {211}{ 28 } \)
=7 \(\frac {15}{ 28} \) = 7 15/28
He got= 4 \(\frac {3}{ 8 } \) + 3 \(\frac {2}{ 7 } \)
=\(\frac {35}{ 8} \) + \(\frac {23 }{ 7 } \)
=\(\frac {35 × 7}{ 8 × 7 } \) + \(\frac {23 × 8}{ 7× 8 } \)
=\(\frac {245}{ 86 } \) + \(\frac {184}{ 86 } \) = \(\frac {429}{ 86 } \)
=7 \(\frac {37}{ 56 } \)
7 \(\frac {37}{ 56 } \) – 7 \(\frac {30}{ 56 } \)
=\(\frac { 7}{ 56} \) =\(\frac {1}{8 } \)
Therefore, Rakesh got \(\frac {1}{ 8 } \) of juices more.
Example 5.
Sindhu goes to the temple every Friday. The distance between the temple and home is 5 \(\frac {1}{ 3 } \) miles. She goes to the temple by car in 1 \(\frac {5}{ 9 } \) hours. She came back home in 2 \(\frac {1}{ 2 } \) hours due to heavy traffic. Find the amount of time Sindhu spends on driving?
Solution:
Sindhu goes to the temple by car= 1 \(\frac {5}{ 9 } \)
Sindhu come back Home=2 \(\frac {1}{ 2 } \)
The time Sindhu spent in driving =1 \(\frac {5}{ 9 } \) +2 \(\frac {1}{ 2} \)
=\(\frac {14}{ 9 } \) +\(\frac {5}{ 2} \)
=\(\frac {14 × 2}{ 9 × 2 } \) +\(\frac {5 ×9}{ 9 × 2 } \)
=\(\frac {28}{ 18 } \) + \(\frac {45}{ 18 } \)
=\(\frac {73}{ 18} \) =4 \(\frac {1}{ 18 } \)
Therefore, Sindhu spends 4 \(\frac {1}{ 18 } \) time on driving.
Example 6.
Siri baked cakes 3 \(\frac {1}{5 } \) on Monday, 5 \(\frac {2}{ 9 } \) cakes on Tuesday, and 4 1/9 cakes on Wednesday. Find how many cakes Siri baked on all three days?
Solution:
Siri baked cakes on Monday= 3 \(\frac {1}{5 } \)
Siri baked cakes on Tuesday= 5 \(\frac {2}{ 9 } \)
Siri baked cakes on Wednesday= 4 \(\frac {1}{ 9 } \)
Siri baked cakes on all the three days= 3 \(\frac {1}{ 5} \) + 5 \(\frac {2}{9 } \) + 4 \(\frac {1}{ 9 } \)
=\(\frac {16}{5 } \) +\(\frac {47}{ 9 } \) +\(\frac {37}{ 9 } \)
=\(\frac {16 × 9}{ 5 × 9 } \) +\(\frac {47× 5}{ 9 × 5 } \) + \(\frac {37× 5}{ 9× 5 } \)
=\(\frac {144}{ 45 } \) +\(\frac {235}{ 45 } \) +\(\frac {185}{ 4 5 } \)
=\(\frac {564}{ 45 } \)
Divided by 3 we get \(\frac {188}{ 15 } \)
= 12 \(\frac {8}{ 15} \)
Siri baked cakes on all three days is 12 \(\frac {8}{ 15 } \) .
Example 7.
Naveen completed 2 \(\frac {3 }{ 10 } \) crosswords on Saturday and 3 \(\frac {4}{ 11 } \) crosswords on Sunday. In total, what fraction of these crosswords did Naveen completed?
Solution:
Naveen completed crosswords on saturday= 2 \(\frac {3 }{ 10 } \)
Naveen completed crosswords on saturday= 3 \(\frac {4}{ 11 } \)
No. of crosswords completed by Naveen =2 \(\frac {3 }{ 10 } \) +3 \(\frac {4}{ 11 } \)
=\(\frac {23 }{ 10 } \) + \(\frac {37 }{ 11 } \)
= \(\frac {23 × 11 }{ 10 × 11 } \) +\(\frac {37 × 10 }{ 11 × 10} \)
=\(\frac {253 }{ 110 } \) + \(\frac {370 }{ 110 } \)
=\(\frac {623 }{110 } \) =5 \(\frac {73 }{ 110 } \)
Hence, Naveen completed 5 \(\frac {73 }{ 110 } \) cross words.
Example 8.
Pooja bought 1 \(\frac {2 }{3 } \) kg of tomatoes, Durga bought 2 \(\frac {5 }{8 } \) dozens of tomatoes. How many dozens of tomatoes bought by Pooja and Durga?
Solution:
Pooja bought tomatoes= 1 \(\frac {2 }{ 3 } \)
Durga bought tomatoes= 2 \(\frac {5 }{ 8 } \)
Tomatoes bought by both of them= 1 \(\frac {2 }{ 3 } \) + 2 \(\frac {5 }{8 } \)
=\(\frac {5 }{ 3 } \) + \(\frac {21 }{ 8} \)
=\(\frac {5 × 8 }{3 × 8 } \) + \(\frac {21 × 3}{8 × 3 } \)
=\(\frac {40 }{24} \) + \(\frac {63 }{24 } \)
=\(\frac {103 }{24 } \)
=4 \(\frac {7 }{24 } \)
Hence, Pooja and Durga bought 4 \(\frac {7}{24 } \) kg of tomatoes.
Example 9.
Kavya ate 1 \(\frac {1 }{7 } \) chocolates and Lasya ate 2 \(\frac {1 }{9 } \) chocolates. In total how many chocolates were eaten by both of them?
Solution:
Kavya ate chocolates=1 \(\frac {1 }{7 } \)
Lasya ate chocolates=2 \(\frac {1}{9 } \)
Chocolates are eaten by both of them=1 \(\frac {1 }{7} \) +2 \(\frac {1 }{9 } \)
= \(\frac { 8 }{7 } \) +\(\frac {19}{9 } \)
=\(\frac {8× 9 }{7 × 9} \) + \(\frac {19 × 7}{9 × 7 } \)
=\(\frac {72 }{63 } \) +\(\frac {133 }{63 } \)
=\(\frac {205 }{63 } \)
=3 \(\frac {16 }{63} \)
Therefore, the Chocolates eaten by both of them are 3 \(\frac {16 }{63 } \) .
Example 10.
Teja read books every day. Teja read comic books 2 \(\frac {4 }{9 } \) , fiction books 1 \(\frac {2 }{3 } \) . How many books does Teja read?
Solution:
Teja read comic books = 2 \(\frac {4 }{9 } \)
Teja read fiction books=1 \(\frac {2}{3 } \)
No. of books Teja read= 2 \(\frac {4}{9} \) + 1 \(\frac {2 }{3 } \) 2/3
=\(\frac {22 }{9 } \) +\(\frac {5 }{3 } \)
=\(\frac {22 }{9 } \) +\(\frac {5 × 3 }{3 × 3 } \)
=\(\frac {22 }{9 } \) + \(\frac {15 }{9 } \)
=\(\frac {37 }{9 } \) =4 \(\frac {1}{9 } \)
Hence, no. of books Teja read is 4 \(\frac {1 }{9 } \) .