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

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

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

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.

Related Important Links for Class 6  Maths 

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