Worksheet on Word Problems on Fractions

Worksheet on Word Problems on Fractional Numbers is here. Know the step-by-step procedure to solve word problems on fractional numbers. Refer to addition, subtraction, multiplication, and division of fractions to know the different model problems. Go through the below sections to know more about the word problems on fractional numbers. Test your knowledge using these word problems.

Problem 1:
A family eats 1\(\frac {2}{3} \)kg of rice at a meal. Will 3kg of rice be enough for two meals?

Solution:

Quantity of rice required for one meal = 1\(\frac {2}{3} \)kg
To find the quantity of rice required for two meal we need to add 1\(\frac {2}{3} \) + 1\(\frac {2}{3} \)
1\(\frac {2}{3} \) is in mixed fractional form lets convert it into improper fractional number
1\(\frac {2}{3} \) = \(\frac {5}{3} \)
Now quantity of rice required for two meal = \(\frac {5}{3} \) + \(\frac {5}{3} \)
=\(\frac {5+5}{3} \)
= \(\frac {10}{3} \)
\(\frac {5}{3} \) is an improper fractional number to know the quantity in kg lets convert it into mixed fractional form.
\(\frac {10}{3} \) = 3\(\frac {1}{3} \)
So quantity of rice required for two meal = 3\(\frac {1}{3} \)kg
3kg of rice will not be enough for two meals.


Problem 2:
Mary and Jane decided to paint a table. Mary painted \(\frac {1}{2} \) part of the table and Jane painted \(\frac {1}{4} \) part of the table. How much did the girls paint?

Solution:

Fraction of table painted by Mary = \(\frac {1}{2} \)
Fraction of table painted by Jane = \(\frac {1}{4} \)
The total fraction of table painted = Fraction of table painted by Mary + Fraction of table painted by Jane
Which is, \(\frac {1}{2} \) + \(\frac {1}{4} \)
LCM is 4
So, \(\frac {2*1 + 1*1}{4} \)
= \(\frac {2 + 1}{4} \)
= \(\frac {3}{4} \)
Mary and Jane painted \(\frac {3}{4} \) of the table


Problem 3:
Minnie wants to prepare a bow with two colors. She has 2\(\frac {1}{3} \)m of red ribbon and 3\(\frac {1}{2} \)m blue ribbon. What is the total length of ribbon she have together?

Solution:

Length of red ribbon Minne has = 2\(\frac {1}{3} \)m
Length of blue ribbon Minne has = 3\(\frac {1}{2} \)m
The total length of ribbon = Length of red ribbon + Length of blue ribbon
= 2\(\frac {1}{3} \)m + 3\(\frac {1}{2} \)m
Let us convert given mixed fractions into improper fractions
2\(\frac {1}{3} \) = \(\frac {7}{3} \)
and 3\(\frac {1}{2} \) = \(\frac {7}{2} \)
Now total length of ribbon = \(\frac {7}{3} \) + \(\frac {7}{2} \)
LCM is 6
\(\frac {7}{3} \) + \(\frac {7}{2} \)  = \(\frac {2*7 + 3*7}{6} \)
= \(\frac {14 + 21}{6} \)
= \(\frac {35}{6} \)
Lets convert \(\frac {35}{6} \) intlo mixed fraction
\(\frac {35}{6} \) = 5\(\frac {5}{6} \)
Total length of ribbion Minnie has = 5\(\frac {5}{6} \)m


Problem 4:
Tommy had 16 books. He completed \(\frac {1}{4} \) of books. How many books does he need to read to complete all books?

Solution:

Total number of books = 16
Number of books Tommy completed reading = \(\frac {1}{4} \)
Which means \(\frac {1}{4} \) of 16
This can be writtne as \(\frac {1}{4} \) * 16
= \(\frac {16}{4} \)
= 4
This means Tommy completed reading 4 books
Number of books more he needs to read = Total number of books – Number of books completed reading
That is 16 – 4
= 12
Number of books Tommy needs to read = 12 books


Problem 5:
Sid has \(\frac {1}{4} \) basket of apples and \(\frac {3}{4} \) mangoes. How many baskets of fruits do Sid have?

Solution:

Quantity of apples in a basket that Sid have = \(\frac {1}{4} \)
Quantity of mangoes in a basket that Sid have = \(\frac {3}{4} \)
Total number of baskets Sid have = quantity of apples in a basket + Quantity of mangoes in a basket
Which is \(\frac {1}{4} \) + \(\frac {3}{4} \)
= \(\frac {1+3}{4} \)
=\(\frac {4}{4} \)
=1
Total number of baskets Sid have = 1


Problem 6:
Mickey and Minnie went on a road trip. They traveled \(\frac {4}{6} \)Km in one hour and \(\frac {5}{6} \) km in next hour. How much distance did they travel?

Solution:

Mickey and Minnie traveled in one hour = \(\frac {4}{6} \)Km
Mickey and Minnie traveled in next hour = \(\frac {5}{6} \)Km
Total distance traveled by Mickey and Minnie = Distance traveled in one hour+ Distance traveled in next hour
Which is \(\frac {4}{6} \)Km + \(\frac {5}{6} \)Km
= \(\frac {4+5}{6} \)Km
= \(\frac {9}{6} \)Km
Distance traveled by Mickey and Minnie in two hours = \(\frac {9}{6} \)Km


Problem 7:
Sam, Tom, and Ted were having drinks from a large mango juice pack, Sam drank \(\frac {1}{5} \) from the bottle. Tom drank \(\frac {2}{5} \) and ted drank \(\frac {1}{5} \). How much did they drink all together?

Solution:

Fraction of juice drank by Sam = \(\frac {1}{5} \)
Fraction of juice drank by tom = \(\frac {2}{5} \)
Fraction of juice drank by Ted = \(\frac {1}{5} \)
The total fraction of juice complted = Fraction of juice drank by Sam + Fraction of juice drank by Tom + Fraction of juice drank by Ted
Which is \(\frac {1}{5} \) + \(\frac {2}{5} \) + \(\frac {1}{5} \)
= \(\frac {1+ 2 + 1}{5} \)
= \(\frac {4}{5} \)
Fraction of juice drank all together = \(\frac {4}{5} \)


Problem 8:
Bunny and Kitty were at a birthday party. So as a part of a birthday party game Bunny and Kitty starting a cake competition. Bunny ate \(\frac {3}{5} \) of cake and Kitty ate \(\frac {1}{3} \) of cake. Who ate a large piece of cake and won the competition?

Solution:

Fraction of cake ate by Bunny = \(\frac {3}{5} \)
Fraction of cake ate by Kitty = \(\frac {1}{3} \)
To know who ate large piece we need to compare both fractions
Let us subtract the fraction of cake ate by Bunny and the fraction of cake ate by Kitty
Which is, \(\frac {3}{5} \) – \(\frac {1}{3} \)
LCM is 15
= ( \(\frac {3}{5} \) * \(\frac {3}{3} \)) – (\(\frac {1}{3} \) * \(\frac {5}{5} \))
= \(\frac {9}{15} \) – \(\frac {5}{15} \)
= \(\frac {9 – 5}{15} \)
= \(\frac {4}{15} \)
Bunny ate a larger piece of cake and won the competition


Problem 9:
Bob walks \(\frac {4}{5} \)km in one hour. How many kilometers does he walk in seven hours?

Solution:

Fraction of distance covered by Bob in one hour = \(\frac {4}{5} \)km
Fraction of distance covered by Bob in seven hours =  \(\frac {4}{5} \) * 7
= \(\frac {4}{5} \) * \(\frac {7}{1} \)
= \(\frac {4 * 7}{5 * 1} \)
= \(\frac {28}{5} \)
Fraction of distance covered by Bob in seven hours = \(\frac {28}{5} \)km


Problem 10:
Jack ate \(\frac {1}{2} \)kg of sweets and Ame ate \(\frac {3}{10} \)kg of sweets. Who are more?

Solution:

Quantity of sweet ate by Jack = \(\frac {1}{2} \)
Quantity of sweet ate by Ame = \(\frac {3}{10} \)
We can compare two fractional numbers by doing cross multiplication.
So, let us compare our given fractional numners to find out how ate more sweets
Which gives, \(\frac {1}{2} \) * \(\frac {3}{10} \)
= \(\frac {1 * 10 }{2 * 3} \)
= \(\frac {10}{6} \)
We alredy know that 10 > 6
So, \(\frac {1}{2} \) > \(\frac {3}{10} \)
Jack ate more sweets


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