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NCERT Solutions for Class 6 Maths Chapter 7 - Fractions Exercise 7.8

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Class 6 Maths NCERT Solutions for Fractions Chapter 7 Exercise 7.8 - FREE PDF Download

Vedantu provides NCERT Solutions for Class 6 Maths Chapter 7 Exercise 7.8, which focuses on Addition and Subtraction of Fractions. This exercise helps students learn how to add and subtract like and unlike fractions according to the latest Class 6 Maths Syllabus. With clear, step-by-step solutions, students can easily understand the methods and practice solving various problems. These Class 6 Maths NCERT Solutions are designed to improve students’ problem-solving skills and build a strong foundation in fractions, preparing them for future maths topics and exams. Download the free PDF and start practising today!

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Table of Content
1. Class 6 Maths NCERT Solutions for Fractions Chapter 7 Exercise 7.8 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 7 Exercise 7.8 Class 6 | Vedantu
3. Access NCERT Solutions for Maths Class 6 Chapter 7 - Fractions
    3.1Exercise 7.8
    3.2Subtract as indicated:
    3.3Solve the following problems:
4. Benefits of NCERT Solutions for Class 6 Maths Chapter 7 Fractions Exercise 7.8 
5. Class 6 Maths Chapter 7: Exercises Breakdown
6. Important Study Material Links for Class 6 Maths Chapter 7 - Fractions
7. Conclusion
8. Chapter-Specific NCERT Solutions for Class 6 Maths
9. Related Important Links for Class 6  Maths 
FAQs


Glance on NCERT Solutions Maths Chapter 7 Exercise 7.8 Class 6 | Vedantu

  • Exercise 7.8 in Class 6 Maths focuses on the Addition and Subtraction of Fractions, covering both like and unlike fractions.

  • Students learn how to add and subtract fractions by finding common denominators, simplifying fractions, and solving word problems.

  • Vedantu's NCERT Solutions provides detailed explanations and step-by-step guidance, ensuring students understand each process clearly.

  • This exercise helps students strengthen their fraction skills, build problem-solving abilities, and prepare for more advanced mathematical topics in future classes.

Access NCERT Solutions for Maths Class 6 Chapter 7 - Fractions

Exercise 7.8

1. Add the following fractions using Brahmagupta’s method:

a.$\dfrac{2}{7} + \dfrac{5}{7} + \dfrac{6}{7}$

Answer: Since the denominators are the same, add the numerators:
$\dfrac{2+5+6}{7} = \dfrac{13}{7}$


b.$\dfrac{3}{4} + \dfrac{1}{3}$

Answer: Find the least common denominator (LCD) of 4 and 3, which is 12:
$\dfrac{3}{4}$ = $\dfrac{9}{12}$​ and$\dfrac{1}{3}$ = $\dfrac{4}{12}$
$\dfrac{9+4}{12}$ = $\dfrac{13}{12}$


c.$\dfrac{2}{3} + \dfrac{5}{6}$

Answer: The LCD of 3 and 6 is 6:
$\dfrac{2}{3}$ = $\dfrac{4}{6}$​ and $\dfrac{5}{6}$​ stays the same.
$\dfrac{4+5}{6}$ = $\dfrac{9}{6}$= $\dfrac{3}{2}$​


d.$\dfrac{2}{7} + \dfrac{2}{7}$

Answer: Since the denominators are the same:
$\dfrac{2+2}{7}$ = $\dfrac{4}{7}$

e.$\dfrac{3}{4} + \dfrac{1}{3} + \dfrac{1}{5}$

Answer: The LCD of 4, 3, and 5 is 60:
$\dfrac{3}{4}$ = $\dfrac{45}{60}$, $\dfrac{1}{3}$ = $\dfrac{20}{60}$, $\dfrac{1}{5}$ = $\dfrac{12}{60}$​
$\dfrac{45+20+12}{60}$ = $\dfrac{77}{60}$


f. $\dfrac{2}{3} + \dfrac{4}{5}$

Answer: The LCD of 3 and 5 is 15:
$\dfrac{2}{3} $= $\dfrac{10}{15}$ and $\dfrac{4}{5}$ = $\dfrac{12}{15}$
$\dfrac{10+12}{15} $= $\dfrac{22}{15}$​


g. $\dfrac{4}{5} + \dfrac{3}{4}$

Answer: The LCD of 5 and 4 is 20:
$\dfrac{4}{5}$ = $\dfrac{16}{20}$ and $\dfrac{3}{4}$ = $\dfrac{15}{20}$

$\dfrac{16+15}{20}$ = $\dfrac{31}{20}$​


h. $\dfrac{3}{5} + \dfrac{5}{8}$

Answer: The LCD of 5 and 8 is 40:
$\dfrac{3}{5}$ = $\dfrac{24}{40}$​ and $\dfrac{5}{8}$ = $\dfrac{25}{40}$

$\dfrac{24+25}{40}$ = $\dfrac{49}{40}$​


i. $\dfrac{9}{2} + \dfrac{5}{4}$

Answer: The LCD of 2 and 4 is 4:
$\dfrac{9}{2}$ = $\dfrac{18}{4}$ and $\dfrac{5}{4}$​ stays the same.
$\dfrac{18+5}{4} = \dfrac{23}{4}$


j. $\dfrac{8}{3} + \dfrac{2}{7}$

Answer: The LCD of 3 and 7 is 21:
$\dfrac{8}{3}$ = $\dfrac{56}{21}$ and $\dfrac{2}{7} = \dfrac{6}{21}$
$\dfrac{56+6}{21}$ = $\dfrac{62}{21}$​


k. $\dfrac{3}{4} + \dfrac{1}{3} + \dfrac{1}{5}$

Answer: The LCD of 4, 3, and 5 is 60:
$\dfrac{3}{4} $= $\dfrac{45}{60}$, $\dfrac{1}{3}$ = $\dfrac{20}{60}$, $\dfrac{1}{5}$ = $\dfrac{12}{60}$​
$\dfrac{45+20+12}{60} $= $\dfrac{77}{60}$​


l. $\dfrac{2}{3} + \dfrac{4}{5} + \dfrac{7}{7}$

Answer: The LCD of 3, 5, and 7 is 105:
$\dfrac{2}{3}$ = $\dfrac{70}{105}$, $\dfrac{4}{5}$ = $\dfrac{84}{105}$, $\dfrac{7}{7}$ = $\dfrac{105}{105}$​

$\dfrac{70+84+105}{105} $= $\dfrac{259}{105}$


M. $\dfrac{9}{2} + \dfrac{5}{4} + \dfrac{7}{6}$

Answer: The LCD of 2, 4, and 6 is 12:
$\dfrac{9}{2}$ = $\dfrac{54}{12}$, $\dfrac{5}{4}$ = $\dfrac{15}{12}$, $\dfrac{7}{6}$ = $\dfrac{14}{12}$​
$\dfrac{54+15+14}{12} $= $\dfrac{83}{12}$​


2. Rahim mixes $\dfrac{2}{3}$​ liters of yellow paint with $\dfrac{3}{4}$​ liters of blue paint to make green paint. What is the volume of green paint he has made?

Answer: The LCD of 3 and 4 is 12:
$\dfrac{2}{3}$ = $\dfrac{8}{12}$​ and $\dfrac{3}{4}$ = $\dfrac{9}{12}$

$\dfrac{8}{12}$ + $\dfrac{9}{12}$ = $\dfrac{17}{12}$ = 1 $\dfrac{5}{12}$​ liters of green paint.

3. Geeta bought $\dfrac{2}{5}$​ meter of lace and Shamim bought $\dfrac{3}{4}$​ meter of the same lace to put a complete border on a tablecloth whose perimeter is 1 meter long. Find the total length of the lace they both have bought. Will the lace be sufficient to cover the whole border?

Answer: The LCD of 5 and 4 is 20:
$\dfrac{2}{5}$ = $\dfrac{8}{20}$​ and $\dfrac{3}{4} = \dfrac{15}{20}$​
$\dfrac{8}{20}$ + $\dfrac{15}{20}$ = $\dfrac{23}{20}$ = 1 $\dfrac{3}{20}$​ meters.
They have 1 meter and $\dfrac{3}{20}$​ meters of lace, which is more than enough to cover the 1-meter border.


Figure it out:

1. ​$\dfrac{5}{8}$  - $\dfrac{3}{8}$ 

Solution:

$\dfrac{5}{8}$ - $\dfrac{3}{8}$ = $\dfrac{5 - 3}{8}$ = $\dfrac{2}{8}$ = $\dfrac{1}{4}$​


2.$\dfrac{7}{9}$ - $\dfrac{5}{9}$​

$\dfrac{7}{9}$- $\dfrac{5}{9}$ = $\dfrac{7 - 5}{9}$ = $\dfrac{2}{9}$​


3. $\dfrac{10}{27}$ -$ \dfrac{1}{27}$​

$\dfrac{10}{27}$ - $\dfrac{1}{27}$ = $\dfrac{10 - 1}{27}$ = $\dfrac{9}{27}$ = $\dfrac{1}{3}$​


Figure it out:

1. Carry out the following subtractions using Brahmagupta’s method:

a.$\dfrac{8}{15}$ -$\dfrac{3}{15}$​

Solutions: Since the denominators are the same:
$\dfrac{8 - 3}{15} $=$ \dfrac{5}{15}$ =$ \dfrac{1}{3}$​


b.$\dfrac{2}{5}$ - $\dfrac{4}{15}$​
Solutions: The least common denominator (LCD) of 5 and 15 is 15:
$\dfrac{2}{5}$= $\dfrac{6}{15}$, so $\dfrac{6}{15} $- $\dfrac{4}{15}$ = $\dfrac{2}{15}$​


c. $\dfrac{5}{6} - \dfrac{4}{9}$
Solutions: The LCD of 6 and 9 is 18:
$\dfrac{5}{6} $= $\dfrac{15}{18}$​ and$\dfrac{4}{9} $= $\dfrac{8}{18}$
So,$\dfrac{15}{18}$ - $\dfrac{8}{18}$ = $\dfrac{7}{18}$​

d. $\dfrac{2}{3} - \dfrac{1}{2}$
Solutions: The LCD of 3 and 2 is 6:
$\dfrac{2}{3}$ = $\dfrac{4}{6}$​ and $\dfrac{1}{2}$ = $\dfrac{3}{6}$​
So,$\dfrac{4}{6}$ - $\dfrac{3}{6}$ =$\dfrac{1}{6}$


Subtract as indicated:

a.$\dfrac{13}{4}$​ from $\dfrac{10}{3}$​

Solution: The LCD of 4 and 3 is 12:
$\dfrac{13}{4}$ = $\dfrac{39}{12}$​ and $\dfrac{10}{3}$ = $\dfrac{40}{12}$​
So,$\dfrac{40}{12}$ - $\dfrac{39}{12}$ = $\dfrac{1}{12}$​


b. $185\dfrac{18}{5}518​ from 233\dfrac{23}{3}323​$
Solution: The LCD of 5 and 3 is 15:
$\dfrac{18}{5}$ = $\dfrac{54}{15}$​ and $\dfrac{23}{3}$ = $\dfrac{115}{15}$​
So, $\dfrac{115}{15}$ - $\dfrac{54}{15}$ = $\dfrac{61}{15}$​


c. $\dfrac{29}{7}729​ from 457\dfrac{45}{7}$
Solution: Since the denominators are the same:
$\dfrac{45}{7} $-$ \dfrac{29}{7}$ = $\dfrac{16}{7}$


Solve the following problems:

a. Jaya’s school is $\dfrac{7}{10}$​ km from her home. She takes an auto for $\dfrac{1}{2}$​ km and then walks the remaining distance. How much does she walk daily?

The LCD of 10 and 2 is 10:
$\dfrac{7}{10} $- $\dfrac{1}{2}$ = $\dfrac{7}{10}$ -$ \dfrac{5}{10}$ =$ \dfrac{2}{10}$ = $\dfrac{1}{5}$​ km
So, Jaya walks $\dfrac{1}{5}$​ km daily.


b. Jeevika takes $\dfrac{10}{3}$ minutes to complete a round of the park and her friend Namit takes $\dfrac{13}{4}$​ minutes. Who takes less time and by how much?

The LCD of 3 and 4 is 12:
$\dfrac{10}{3}$ = $\dfrac{40}{12}$​ and $\dfrac{13}{4}$ = $\dfrac{39}{12}$​
Jeevika takes$\dfrac{40}{12}$​ minutes and Namit takes $\dfrac{39}{12}$​ minutes.
Jamit takes less time by$\dfrac{1}{12}$​ minute.


Benefits of NCERT Solutions for Class 6 Maths Chapter 7 Fractions Exercise 7.8 

  • The solutions provide simple, step-by-step explanations for easy understanding of adding and subtracting fractions.

  • Plenty of practice problems help students improve their skills and gain confidence in solving fraction-related questions.

  • The solutions cover important concepts, ensuring students are well-prepared for their exams.

  • Practising these problems strengthens students' foundation in fractions, which is essential for future maths topics.

  • The solutions are easy to follow, allowing students to study and practice independently.


Class 6 Maths Chapter 7: Exercises Breakdown

Exercise

Topic

Exercise 7.1 

Fractional Units and Equal Shares 

Exercise 7.2

Fractional Units as Parts of a Whole

Exercise 7.3

Measuring Using Fractional Units

Exercise 7.4

Marking Fraction Lengths on the Number Line

Exercise 7.5

Mixed Fractions

Exercise 7.6

Equivalent Fractions

Exercise 7.7

Comparing Fractions

Exercise 7.9

A Pinch of History



Important Study Material Links for Class 6 Maths Chapter 7 - Fractions

S.No. 

Study Material Links for  Chapter 7 - Fractions

1.

Class 6 Maths Fractions Important Questions

2.

Class 6 Maths Fractions Revision Notes

3.

Class 6 Maths Fractions Worksheets



Conclusion

Vedantu's NCERT Solutions for Class 6 Maths Chapter 7 Exercise 7.8 provide valuable support for students learning about the addition and subtraction of fractions. With clear explanations and step-by-step guidance, students can easily understand how to perform these operations with both like and unlike fractions. The exercise includes a variety of practice problems that help reinforce learning and build confidence. By mastering these concepts, students will be well-prepared for future maths topics and assessments, making this exercise an essential resource for their academic success.


Chapter-Specific NCERT Solutions for Class 6 Maths

The chapter-wise NCERT Solutions for Class 6 Maths are given below. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


S.No

Chapterwise Links for Class 6 Maths NCERT Solutions

1

Chapter 1 Patterns In Mathematics NCERT Solutions

2

Chapter 2 Lines and Angles NCERT Solutions 

3

Chapter 3 Number Play NCERT Solutions 

4

Chapter 4 Data Handling and Presentation NCERT Solutions 

5

Chapter 5 Prime Time NCERT Solutions 

6

Chapter 6 Perimeter and Area NCERT Solutions 

7

Chapter 7 Fractions NCERT Solutions 

8

Chapter 8 Playing with Constructions NCERT Solutions 

9

Chapter 9 Symmetry NCERT Solutions 

10

Chapter 10 The Other Side of Zero NCERT Solutions 



Related Important Links for Class 6  Maths 

Along with this, students can also download additional study materials provided by Vedantu for Maths Class 6-


FAQs on NCERT Solutions for Class 6 Maths Chapter 7 - Fractions Exercise 7.8

1. What is covered in Exercise 7.8 of Class 6 Maths?

Exercise 7.8 focuses on the Addition and Subtraction of Fractions, teaching students how to perform these operations with like and unlike fractions.

2. How does Vedantu's NCERT Solution help with Exercise 7.8?

Vedantu provides clear, step-by-step explanations, making it easy for students to understand and solve problems related to adding and subtracting fractions.

3. Are both like and unlike fractions covered in Exercise 7.8?

The exercise includes addition and subtraction of both like fractions (same denominator) and unlike fractions (different denominators).

4. How does Vedantu explain the process of adding unlike fractions?

Vedantu's solutions guide students through finding the least common denominator (LCD) to add unlike fractions and simplify them.

5. Does the Class 6 Maths Exercise 7.8 include word problems?

Yes, the exercise includes word problems to help students apply the concept of fraction addition and subtraction in real-life situations.

6. Can I download the PDF of Class 6 Maths Exercise 7.8  solutions for FREE?

Vedantu offers a free PDF download of the NCERT Solutions for Class 6 Maths Chapter 7 Exercise 7.8.

7. Will Class 6 Maths Exercise 7.8 solutions help me prepare for exams?

These solutions offer thorough practice and explanations, helping students gain confidence for exams.

8. Are visual aids used in the Class 6 Maths Exercise 7.8  solutions for better understanding?

Diagrams and visual aids are used where necessary to help students visualize the addition and subtraction of fractions.

9. How does Vedantu help with solving word problems involving fractions?

Vedantu provides clear explanations and step-by-step solutions to help students easily understand and solve word problems involving fractions.

10. Are there practice questions included in Class 6 Maths Exercise 7.8  Vedantu’s PDF?

The PDF includes plenty of practice questions to help students strengthen their understanding and master the topic.