Access Worksheet on Division of Fractions here to get acquainted with various problems on Division of Fractions. Refer to Fractions Division’s step-by-step procedure to solve the problems. Follow the study material and guidance to solve various questions and answers. Know the important methods, rules, and formulae of Dividing Fractions. Try to solve the Dividing Fractions Word Problems Worksheet available on your own to test your knowledge of the concept. We have provided detailed solutions to all the Dividing Fractions Problems available.
Dividing Fractions Word Problems Worksheets with Answers
Problem 1:
Jiya thought it would be nice to include \(\frac { 2 }{ 21 } \) of a pound of chocolate in each of the holiday gift bags she made for her friends and family. How many holiday gift bags could jiya make with \(\frac { 2 }{ 3 } \) of a pound of chocolate?
Solution:
As given in the question,
Amount of chocolate = \(\frac { 2 }{ 21 } \)
No of bags = Total lb/lbs per bag
\(\frac { 2 }{ 3 } \) % \(\frac { 2 }{ 21 } \)
= \(\frac { 2 }{ 3 } \) % \(\frac { 2 }{ 21 } \)
= \(\frac { 2 }{ 3 } \) * \(\frac { 21 }{ 2 } \)
= \(\frac { 2 }{ 3 } \) + \(\frac { 2 }{ 21 } \)
= \(\frac { 2 }{ 3 } \) * \(\frac { 21 }{ 21 } \) + \(\frac { 2 }{ 21 } \) * \(\frac { 3 }{ 3 } \)
= \(\frac { 42 }{ 63 } \) + \(\frac { 6 }{ 63 } \)
= \(\frac { 48 }{ 63 } \)
Problem 2:
Lithi has \(\frac { 1 }{ 5 } \) of a bag of dog food left. She is splitting it between her 3 dogs evenly. What fraction of the original bag does each dog get?
Solution:
As given in the question,
Amount of dog food left = \(\frac { 1 }{ 5 } \)
No of dogs = 3
The fraction of the original bag each dog get = \(\frac { 1%5 }{ 3 } \)
= \(\frac { 1 }{ 5 } \) * \(\frac { 1 }{ 3 } \)
= \(\frac { 1 }{ 15 } \)
Therefore, \(\frac { 1 }{ 15 } \) fraction of original bag each dog gets
Thus, the final solution is \(\frac { 1 }{ 15 } \)
Problem 3:
John has \(\frac { 1 }{ 4 } \) of a gallon of saltwater that he is using for an experiment. He needs to evenly separate the saltwater into 3 separate beakers. How much salt water will be in each beaker?
Solution:
As given in the question,
Fraction of salt water he is using for an experiment = \(\frac { 1 }{ 4 } \)
No of beakers = 3
Amount of salt water in each beaker = \(\frac { 1 }{ 4 } \) % 3
= \(\frac { 1 }{ 4 } \) * \(\frac { 1 }{ 3 } \)
= \(\frac { 1 }{ 12 } \)
Therefore, \(\frac { 1 }{ 12 } \) gallons of salt water
Thus, the final solution is \(\frac { 1 }{ 12 } \)
Problem 4:
David has a board that measures 4ft in length. The board is going to be cut into \(\frac { 1 }{ 4 } \) ft pieces. How many pieces will David split the board into?
Solution:
As given in the question,
Length of the board = 4ft
No of pieces the board is going to be cut = \(\frac { 1 }{ 4 } \)
No of pieces David will get = 4 % \(\frac { 1 }{ 4 } \)
= \(\frac { 4 }{ 1 } \) * \(\frac { 4 }{ 1 } \)
= \(\frac { 16 }{ 1 } \)
= 16 pieces
Therefore, Devin split the board into 16 pieces.
Problem 5:
Jasmine has \(\frac { 1 }{ 4 } \) hour left to finish 4 math problems on the test. How much time does she have to spend on each problem?
Solution:
As given in the question,
No of hours Jasmine left to finish math problems = \(\frac { 1 }{ 4 } \)
No of problems = 4
The time she have to spend on each problem = \(\frac { 1 }{ 4 } \) % 4
= \(\frac { 1 }{ 4 } \) * \(\frac { 1 }{ 4 } \)
= \(\frac { 1 }{ 16 } \)
Therefore, \(\frac { 1 }{ 16 } \) fraction of time she spend on each problem
Thus, the final solution is \(\frac { 1 }{ 16 } \)
Problem 6:
Sandya finished \(\frac { 1 }{ 3 } \) of her homework problems. Which fraction is equivalent to what Sandya finished?
Solution:
As given in the question,
The fraction of homework problems = \(\frac { 1 }{ 3 } \)
The fraction that is equivalent to wht Sandra finished = \(\frac { 1 }{ 3 } \) * \(\frac { 2 }{ 2 } \)
= \(\frac { 2 }{ 6 } \)
= \(\frac { 1 }{ 3 } \) * \(\frac { 3 }{ 3 } \)
= \(\frac { 3 }{ 9 } \)
Therefore, the fractions \(\frac { 2 }{ 6 } \) and \(\frac { 3 }{ 9 } \) are equivalent to Sandra.
Thus, the final solution is \(\frac { 2 }{ 6 } \) and \(\frac { 3 }{ 9 } \)
Problem 7:
Berlin has 9 cups of sugar. If this is \(\frac { 3 }{ 4 } \) of the number he needs to make a cake, how many cups does he need?
Solution:
As given in the question,
No of cups of sugar = 9
The fraction of cake he needs = \(\frac { 3 }{ 4 } \)
No of cups = 9 % \(\frac { 3 }{ 4 } \)
= \(\frac { 9 }{ 1 } \) * \(\frac { 4 }{ 3 } \)
= \(\frac { 36 }{ 3 } \)
= 12
Therefore, Berlin needs 12 cups of sugar.
Thus, the final solution is 12 cups.
Problem 8:
Mahathi is working on projects that require 3\(\frac { 1 }{ 2 } \) yards of ribbon per project. Mahathi has 28 yards of ribbon. What is the greatest number of projects that Mahathi can complete with this ribbon?
Solution:
As given in the question,
Amount of yards of ribbon per project = 3\(\frac { 1 }{ 2 } \)
No of yards = 28
The greatest number of projects = 28 % 3\(\frac { 1 }{ 2 } \)
= 28 % \(\frac { 7 }{ 2 } \)
= \(\frac { 28 }{ 1 } \) * \(\frac { 2 }{ 7 } \)
= \(\frac { 56 }{ 7 } \)
= 8
Therefore, Mahathi can complete 8 projects with this ribbon.
Thus, the final solution is 8 projects
Problem 9:
Arjun made 40 cookies for the upcoming bake sale. He made \(\frac { 3 }{ 5 } \) of the cookies of chocolate chip and \(\frac { 1 }{ 4 } \) of them peanut butter. How many chocolate chip and peanut butter cookies are there in together?
Solution:
As given in the question,
No of cookies = 40
No of cookies of chocolate chip = \(\frac { 3 }{ 5 } \)
No of cookies of peanut butter = \(\frac { 1 }{ 4 } \)
No of chocolate chips = 40 * \(\frac { 3 }{ 5 } \)
= \(\frac { 120 }{ 5 } \)
= 24
No of peanut cookies = 40 * \(\frac { 1 }{ 4 } \)
= \(\frac { 40 }{ 4 } \)
= 10
Therefore, a total of 10 chocolate and peanut cookies are in together.
Problem 10:
Gillian took 6\(\frac { 1 }{ 2 } \) hours to build a dog house. If she worked for 1\(\frac { 3 }{ 4 } \) hours each night. How many nights did it take?
Solution:
As given in the question,
No of hours Gillian take to build a dog house = 6\(\frac { 1 }{ 2 } \)
No of hours she worked each night = 1\(\frac { 3 }{ 4 } \)
No of nights he took = 6\(\frac { 1 }{ 2 } \) % 1\(\frac { 3 }{ 4 } \)
= \(\frac { 13 }{ 2 } \) % \(\frac { 7 }{ 4 } \)
= \(\frac { 13 }{ 2 } \) * \(\frac { 4 }{ 7 } \)
= \(\frac { 52 }{ 14 } \)
= 3\(\frac { 10 }{ 14 } \)
Therefore, 3\(\frac { 10 }{ 14 } \) nights it took to build a dog house.
Thus, the final solution is 3\(\frac { 10 }{ 14 } \)
Problem 11:
A class consists of 40 students. Of them \(\frac { 2 }{ 5 } \) of the fraction are boys. Find the number of girls present in the class?
Solution:Â
As given in the question,
No of students in class = 40
No of boys = \(\frac { 2 }{ 5 } \)
No of boys present = \(\frac { 40 }{ 1 } \) * \(\frac { 2 }{ 5 } \)
= \(\frac { 80 }{ 5 } \)
= 16 boys
Therefore, the total no of girls = 40 – 16
= 24 girls
Therefore, there are 24 girls present in the class
Thus, the final solution is 24 girls.
Problem 12:
John had Php 960. He spent \(\frac { 1 }{ 3 } \) of his money on food. How many were left to him?
Solution:Â
As given in the question,
Amount of Php, John had = 960
Amount of money he spent on food = \(\frac { 1 }{ 3 } \)
Amount of money left out after spending on food = \(\frac { 960 }{ 1 } \) * \(\frac { 1 }{ 3 } \)
= \(\frac { 960 }{ 3 } \)
= 320
Total amount of money left = 960 – 320 = 640
Therefore, John has 640 Php left with him.
Thus, the final solution is Php 640.
Problem 13:
Efren had 480 apples for sale. He sold \(\frac { 3 }{ 5 } \) of them. How many were left?
Solution:Â
As given in the question,
No of apples for sale = 480
Fraction of apples he sold = \(\frac { 3 }{ 5 } \)
No of apples left = \(\frac { 480 }{ 1 } \) * \(\frac { 3 }{ 5 } \)
= \(\frac { 1440 }{ 5 } \)
= 288
No of apples left after selling = 480 – 288
= 192
Therefore, there were 192 apples left after selling
Thus, the final solution is 192 apples.
Problem 14:
Rose and Mary were filling the class-raised garden bed with soil. Rose shoveled in \(\frac { 1 }{ 3 } \) of a cubic yard, and Mary shoveled in \(\frac { 1 }{ 2 } \) of a cubic yard. How much of the soil did they put into the garden bed altogether?
Solution:Â
As given in the question,
The fraction of cubic yard Rose shoveled = \(\frac { 1 }{ 3 } \)
The fraction of cubic yard Mary shoveled = \(\frac { 1 }{ 2 } \)
The amount of soil they put into the garden together = \(\frac { 1 }{ 3 } \) + \(\frac { 1 }{ 2 } \)
Multiply the first fraction with \(\frac { 2 }{ 2 } \) and second fraction with \(\frac { 3 }{ 3 } \)
= \(\frac { 2 }{ 6 } \) + \(\frac { 3 }{ 6 } \)
= \(\frac { 5 }{ 6 } \)
Therefore, \(\frac { 5 }{ 6 } \) fraction of soil they put into the garden bed altogether
Thus, the solution is \(\frac { 5 }{ 6 } \)
Problem 15:
Carl’s invited Otto over to his house. Carl’s liked to share with his guests, so he got out his chocolate stash from last Halloween. He still had \(\frac { 4 }{ 5 } \) of a pound of chocolate. Carl’s asked Otto how much chocolate he would like. Otto said that he would like \(\frac { 1 }{ 3 } \) of a pound of chocolate. Carl’s obliged. How much chocolate does Carl’s have left?
Solution:Â
As given in the question,
Amount of chocolate = \(\frac { 4 }{ 5 } \)
Otto would have chocolate = \(\frac { 1 }{ 3 } \)
Amount of chocolate Carl’s have = \(\frac { 4 }{ 5 } \) – \(\frac { 1 }{ 3 } \)
Multiply the first equation with \(\frac { 3 }{ 3 } \) and the second equation with \(\frac { 5 }{ 5 } \)
= \(\frac { 12 }{ 15 } \) – \(\frac { 5 }{ 15 } \)
= \(\frac { 7 }{ 15 } \)
Therefore, Carl’s have \(\frac { 7 }{ 15 } \) fraction of chocolate
Thus, the final solution is \(\frac { 7 }{ 15 } \).