# Word Problems on Fraction with Solutions | Fraction Word Problems with Answers

Are you feeling difficulty in solving the word problems on fractions? Here you will get plenty of information on how to solve word problems and the method used to solve them. You can apply this related knowledge to the problems you encounter on fractions. By going through the article you can also check the solved examples for a better understanding of the concept.

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## What are Fractions?

A fractional number is considered as the ratio between two numbers. Fractions are defined by $$\frac {a}{b}$$ a is called the numerator which means the equal number of parts that are counted. b is called the denominator which means a number of parts in the whole.

### Fraction Word Problems with Answers

Problem 1:
Mickey has read three-fifth of his 75 pages book. How many more pages he needs to read to complete his book?
Solution:
Let us write the given information,
Mickey has read $$\frac {3}{5}$$ of a 75 page book.
Which can be written as $$\frac {3}{5}$$ * 75
$$\frac {3}{5}$$ * 75 = 45
So, Mickey has completed reading 45 pages from his book.
Now to find the number of pages he needs to read to complete his book
Total number of pages = 75
Number of pages Mickey has completed reading = 45
Number of pages he needs to read to complete his book = (75 – 45)
= 30pages
Number of pages mickey needs to read to complete his book = 30 pages.

Problem 2:
Minnie has Rs. 675. She gave $$\frac {13}{15}$$ of the amount to Daisy. Then Daisy spent $$\frac {9}{15}$$ of the amount given to her. How much amount is Daisy left with?
Solution:
Amount Minnie gave to Daisy = $$\frac {13}{15}$$ of 675
Which can be written as $$\frac {13}{15}$$ * 675
$$\frac {13}{15}$$ * 675 = Rs.585
Amount Daisy spent = $$\frac {9}{15}$$ of 585
Which can be written as $$\frac {9}{15}$$ * 585
$$\frac {9}{15}$$ * 585 = 351
Amount Daisy is left with = Amount Minnie gave to Daisy – Amount Daisy spent
Rs 585 – Rs 351 = Rs 234
Amount left with Daisy left = Rs 234

Problem 3:
Tom bought $$\frac {1}{5}$$L milk on Monday and $$\frac {2}{5}$$L on Tuesday. How much milk did he buy in two days?
Solution:
Milk bought on Monday = $$\frac {1}{5}$$L
Milk bought on Tuesday = $$\frac {2}{5}$$L
Total milk he bought = $$\frac {1}{5}$$L + $$\frac {2}{5}$$L
$$\frac {1}{5}$$L + $$\frac {2}{5}$$L = $$\frac {1 + 2}{5}$$L
= $$\frac {3}{5}$$L
Milk bought by Tom in two days = $$\frac {3}{5}$$L

Problem 4:
Jerry bought $$\frac {5}{7}$$Kg of cheese and used $$\frac {1}{7}$$Kg.  How much cheese is left?
Solution:
Cheese bought = $$\frac {5}{7}$$Kg
Cheese used = $$\frac {1}{7}$$Kg
Cheese left = Cheese bought  – Cheese used
$$\frac {5}{7}$$Kg – $$\frac {1}{7}$$Kg = $$\frac {5 – 1}{7}$$Kg
$$\frac {4}{7}$$Kg
Cheese left with Jerry = $$\frac {4}{7}$$Kg

Problem 5:
Jaggu bought $$\frac {4}{7}$$Kg banana on Monday and $$\frac {2}{7}$$Kg of apple on Tuesday. What is the total quantity of fruits Jaggu bought?
Solution:
Quantity of bananas bought by Jaggu on Monday = $$\frac {4}{7}$$Kg
Quantity of apples bought by Jaggu on Tuesday = $$\frac {2}{7}$$Kg
The total quantity of fruits Jaggu bought = Quantity of bananas + Quantity of apples
$$\frac {4}{7}$$Kg + $$\frac {2}{7}$$Kg = $$\frac {4 + 2}{7}$$Kg
$$\frac {6}{7}$$Kg
The total quantity of fruits Jaggu bought = $$\frac {6}{7}$$Kg

Problem 6:
Ben bought $$\frac {4}{7}$$m cloth at the rate of Rs 140 per meter. How much amount did he pay?
Solution:
Cost per meter = Rs 140
Length of cloth ben bought =$$\frac {4}{7}$$m
Amount Ben paid = Length of cloth ben bought * Cost per meter
$$\frac {4}{7}$$m * Rs 140 = $$\frac {4 * 140}{7}$$
$$\frac {4 * 140}{7}$$ = $$\frac {560}{7}$$
= Rs 80
Amount paid by Ben = Rs 80

Problem 7:
What is the difference between $$\frac {3}{5}$$ of 5000 and $$\frac {5}{8}$$ of 4000
Solution:
First, we need to find what is $$\frac {3}{5}$$ of 5000 and $$\frac {5}{8}$$ of 4000
$$\frac {3}{5}$$ of 5000 = $$\frac {3}{5}$$ * 5000
= 3000
$$\frac {5}{8}$$ of 4000 = $$\frac {5}{8}$$ * 4000
= 2500
Now, difference between $$\frac {3}{5}$$ of 5000 and $$\frac {5}{8}$$ of 4000 = (3000 – 2500)
= 500
The difference between $$\frac {3}{5}$$ of 5000 and $$\frac {5}{8}$$ of 4000 = 500

Problem 8:
Jane spent $$\frac {1}{5}$$ of her pockey money on food and $$\frac {3}{4}$$ on books, how much did she spend alltogether?
Solution:
Amount spent on food = $$\frac {1}{5}$$
Amount spent on books = $$\frac {3}{4}$$
Total amount spent = Amount spent on food + Amount spent on books
$$\frac {1}{5}$$ + $$\frac {3}{4}$$
= ( $$\frac {1}{5}$$ * $$\frac {4}{4}$$ ) + ( $$\frac {3}{4}$$ + $$\frac {5}{5}$$ )
= $$\frac {4}{20}$$ + $$\frac {15}{20}$$
= $$\frac {19}{20}$$
Amount spent by Jane = $$\frac {19}{20}$$

Problem 9:
In a high school contest, Ross jumped 3$$\frac {8}{9}$$m and Joye jumped 4$$\frac {1}{3}$$m. Who jumped more height and by how much more?
Solution:
Height Ross jumped = 3$$\frac {8}{9}$$m
Height Joey jumped = 4$$\frac {1}{3}$$m
The given numbers are mixed fractional numbers let’s convert them to improper fractional numbers
So, 3$$\frac {8}{9}$$m = $$\frac {35}{9}$$m
4$$\frac {1}{3}$$m = $$\frac {13}{3}$$m
This means, Ross jumed $$\frac {35}{9}$$m and Joey jumped $$\frac {13}{3}$$m
Now to know who jumped more height we need to compare these numbers by cross multiplication
$$\frac {35}{9}$$ * $$\frac {13}{3}$$
= $$\frac {35 * 3}{9 * 13}$$
= $$\frac {105}{117}$$
We know 117 > 105 so, $$\frac {13}{3}$$ > $$\frac {35}{9}$$
This means Joey jumped more height
To know by how more Joey jumped than Ross we need to subtract $$\frac {13}{3}$$ from $$\frac {35}{9}$$
LCM is 9
So,( $$\frac {13}{3}$$ * $$\frac {3}{3}$$ ) – $$\frac {35}{9}$$
= $$\frac {39}{9}$$ – $$\frac {35}{9}$$
= $$\frac {39 – 35}{9}$$
= $$\frac {4}{9}$$
Joey jumed $$\frac {4}{9}$$ more than Ross.

Problem 10:
Bunny bought 2$$\frac {2}{5}$$kg of strawberry, 2kg of blackberry and 1$$\frac {2}{5}$$kg of blueberry. What is the total weight of berries Bunny bought?
Solution:
Weight of strawberry = 2$$\frac {2}{5}$$kg
Weight of blackberry = 2kg
Weight of blueberry = 1$$\frac {2}{5}$$kg
We can see that weights of berries are in mixed fractional form and whole number form
Now, let’s convert them to fractional numbers so that we can add them and find the weight of the berries
2$$\frac {2}{5}$$kg =$$\frac {12}{5}$$kg
2kg = $$\frac {2}{1}$$kg
1$$\frac {2}{5}$$kg = $$\frac {7}{5}$$kg
Total weight of berries = Sum of (strawberry + blackberry + blueberry)
= $$\frac {12}{5}$$ + $$\frac {2}{1}$$ + $$\frac {7}{5}$$
LCM is 5
= $$\frac {12}{5}$$ + ($$\frac {2}{1}$$ + $$\frac {5}{5}$$) + $$\frac {7}{5}$$
= $$\frac {12}{5}$$ + $$\frac {10}{5}$$ + $$\frac {7}{5}$$
= $$\frac {12 + 10 + 7}{5}$$
= $$\frac {29}{5}$$
$$\frac {12}{5}$$ is a improper fractional number so let us convert it in to mixed fractional number
$$\frac {29}{5}$$  = 5$$\frac {4}{5}$$
The total weight of berries = 5$$\frac {4}{5}$$kg

Problem 11:
Mickey bought $$\frac {7}{8}$$kg of noddels and Minnie bought $$\frac {6}{8}$$kg of noddels. What is the total quantity of noddels they have?
Solution:
Weight of noddles Mickey bought = $$\frac {7}{8}$$kg
Weight of noddels Minnie bought = $$\frac {6}{8}$$kg
Total weight of noddels = Weight of noddles Mickey bought +Weight of noddels Minnie bought
= $$\frac {7}{8}$$kg + $$\frac {6}{8}$$kg
= $$\frac {7 +6}{8}$$kg
= $$\frac {13}{8}$$kg
This is an improper fractional number so it can be converted to a mixed fractional number
$$\frac {13}{8}$$kg = 1$$\frac {5}{8}$$kg
The total quantity of noddles Mickey and Minnie have = 1$$\frac {5}{8}$$kg

Problem 12:
Kitty’s mother bought 1$$\frac {3}{4}$$kg of cookies and her father bought 1$$\frac {1}{2}$$kg of cookies. What is the total weight of cookies that Kitty has?
Solution:
Weight of cookies bought by Kitty’s mother = 1$$\frac {3}{4}$$kg
Weight of cookies bought by Kitty’s father = 1$$\frac {1}{2}$$kg
To know the total weight of cookies we have to add them
We can’t add mixed fractional numbers so let us convert them into improper fractional numbers
1$$\frac {3}{4}$$kg = $$\frac {7}{4}$$kg
1$$\frac {1}{2}$$kg = $$\frac {3}{2}$$kg
Now we can add these two fractional numbers
$$\frac {7}{4}$$kg + $$\frac {3}{2}$$kg
LCM= 4
= $$\frac {1*7 + 2*3}{4}$$
= $$\frac {7 + 6}{4}$$
= $$\frac {13}{4}$$
$$\frac {13}{4}$$  is an improper fractional number So we have to convert it into a mixed fractional number
$$\frac {13}{4}$$ = 3$$\frac {1}{4}$$
The total quantity of cookies Kitty have = 3$$\frac {1}{4}$$kg

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