NCERT Solutions For Class 6 Maths Chapter 7

NCERT Solutions For Class 6 Maths Chapter 7 PDF Free Download

NCERT Solutions For Class 6 Maths Chapter 7, Fractions, are available here. These solutions are available in PDF format so that students can easily download it and make the best use of it offline as well. These are prepared by experts as per the NCERT curriculum and CBSE Syllabus for 6th class so that students could practice them for final exams.

Some of the other learning materials such as notes, question papers, etc, are good materials for the preparation of apart from NCERT Solutions for class 6. Also, the sample papers and previous year question papers will be helpful in giving the idea of types of questions asked from chapter 7, Fractions.

Class 6 Maths NCERT Solutions for Fractions

Here are the list of topics covered in chapter 7, Fractions are;

  • Types of Fractions
  • Operations on Fractions
  • Fractions on the number line
  • Proper Fractions
  • Mixed Fractions
  • Equivalent Fractions
  • Simplest Form of a Fraction
  • Like and Unlike Fractions
  • Comparing Fractions

Key Insights of Fractions

A fraction can be simply defined as a numerical representation indicating the quotient of two numbers. It is also referred to as the fractional number. For example, 4/6, 7/9, 2/4 and so on. A fraction is mainly used to denote a portion of a whole. Fractions comprise of two numbers: a numerator and a denominator.

For example, 4/6 is a fraction, where 4 is the numerator and 6 is the denominator.

The numerator in the given fraction states the number of equal parts into which something is divided and the denominator states the total number of equal parts into which something is divided. Proper, mixed and improper are the different types of fractions.

  1. Proper Fractions in which the numerator value is less than the denominator.
  2. Mixed Fractions is the ratio of the whole number.
  3. Improper Fractions is the fraction in which the numerator value is greater than or equal to the denominator.

NCERT Solutions For Class 6 Maths Chapter 7 Exercises

NCERT Questions For Fractions


Exercise 2.1


 

1) Solve the following

(a) 2-\(\frac{3}{5}\)

Soln:

To solve ,we have to make both numbers in fraction form

\(\Rightarrow \frac{2}{1}-\frac{3}{5}\)

Lets take LCM of 1,5 = 1*5 =5

\(\Rightarrow \frac{2}{1}*\frac{5}{5}=\frac{10}{5}\)

Since both numbers has the denominator as 5, we can solve the equation now,

The solution is

\(\Rightarrow \frac{10}{5}-\frac{3}{5}=\frac{7}{5}\)

 

(b) \(4+\frac{7}{8}\)

To solve we have to make both numbers in fraction form

\(\Rightarrow \frac{4}{1}+\frac{7}{8}\)

Lets take LCM of 1,8 = 1*8 =8

\(\Rightarrow \frac{4}{1}*\frac{8}{8}=\frac{32}{8}\)

Since both numbers has the denominator as 8, we can solve the equation

now,

The solution is

\(\Rightarrow \frac{32}{8}+\frac{7}{8}=\frac{39}{8}\)

 

(c) \( \frac{3}{5}+\frac{2}{7}\)

Since, the numbers are in fraction form,

Lets take LCM of 5,7= 5*7=35

For first number,

\(\Rightarrow \frac{3}{5}*\frac{7}{7}=\frac{21}{35}\)

For second number,

\(\Rightarrow \frac{2}{7}*\frac{5}{5}=\frac{10}{35}\)

The solution is

\(\Rightarrow \frac{21}{35}+\frac{10}{35}=\frac{31}{35}\)

 

(d) \(\frac{9}{11}-\frac{4}{15}\)

Since ,the numbers are in fraction form

Lets take LCM of 11,15= 11*15=165

For first number,

\(\Rightarrow \frac{9}{11}*\frac{15}{15}=\frac{135}{165}\)

For second number,

\(\Rightarrow \frac{4}{15}*\frac{11}{11}=\frac{44}{165}\)

The solution is

\(\Rightarrow \frac{135}{165}-\frac{44}{165}=\frac{91}{165}\)

 

(e) \(\frac{7}{10}+\frac{2}{5}+\frac{3}{2}\)

Since, the numbers are in fraction form,

Lets take LCM of 10,5,2= 2*5=10

For first number,

\(\Rightarrow \frac{7}{10}*\frac{1}{1}=\frac{7}{10}\)

For second number,

\(\Rightarrow \frac{2}{5}*\frac{2}{2}=\frac{4}{10}\)

For third number,

\(\Rightarrow \frac{3}{2}*\frac{5}{5}=\frac{15}{10}\)

The solution is

\(\Rightarrow \frac{7}{10}+\frac{4}{10} + \frac{15}{10} =\frac{26}{10}\)

 

(f) \(2\frac{2}{3}+3\frac{1}{2}\)

Since, the numbers are in mixed fraction form ,we have to convert it into fractional form

For first number in mixed fraction,

\(2\frac{2}{3}=\frac{\left ( 3*2 \right )+2}{3}=\frac{8}{3}\)

For second number in mixed fraction,

\(3\frac{1}{2}=\frac{\left ( 2*3 \right )+1}{2}=\frac{7}{2}\)

Lets take LCM of 3,2=3*2=6

For first number,

\(\Rightarrow \frac{8}{3}*\frac{2}{2}=\frac{16}{6}\)

For second number,

\(\Rightarrow \frac{7}{2}*\frac{3}{3}=\frac{21}{6}\)

The solution is

\(\Rightarrow \frac{16}{6}+\frac{21}{6}=\frac{37}{6}\)

 

(g) \(8\frac{1}{2}-3\frac{5}{8}\)

Since, the numbers are in mixed fraction form we have to convert it into fractional form

For first number in mixed fraction,

\(8\frac{1}{2}=\frac{\left ( 2*8 \right )+1}{}2=\frac{17}{2}\)

For second number in mixed fraction,

\(3\frac{5}{8}=\frac{\left ( 8*3 \right )+5}{}8=\frac{29}{8}\)

Lets take LCM of 2,8=2*8=8

For first number,

\(\Rightarrow \frac{17}{2}*\frac{4}{4}=\frac{68}{8}\)

For second number,

\(\Rightarrow \frac{29}{8}*\frac{1}{1}=\frac{29}{8}\)

The solution is

\(\Rightarrow \frac{68}{8}-\frac{29}{8}=\frac{39}{8}\)

 

2) Arrange the following numbers in descending order:

a)\(\frac{2}{9},\frac{2}{3},\frac{8}{21}\)

Solution:

\(\frac{2}{9},\frac{2}{3},\frac{8}{21}\)

Let’s take LCM of 9,3,21

9 = 3*3

21= 3* 7

3= 3*1

So, we can take two 3’s as a common term since we four 3’s and 7 is coming only once

Therefore ,the required LCM is= 3*3*7=63

For the first number,

\(\Rightarrow \frac{2}{9}*\frac{7}{7}=\frac{14}{63}\)

For the second number,

\(\Rightarrow \frac{2}{3}*\frac{21}{21}=\frac{42}{63}\)

For the third number,

\(\Rightarrow \frac{8}{21}*\frac{3}{3}=\frac{24}{63}\)

Descending order means arranging the numbers from largest to smallest

So,

\(\frac{14}{63}=0.22\) \(\frac{42}{63}=0.66\) \(\frac{24}{63}=0.38\)

Therefore, the decreasing order of rational numbers are

\(\frac{42}{63}> \frac{24}{63}> \frac{14}{63}\)

i.e)

\(\frac{2}{3}>\frac{8}{21}> \frac{2}{9}\)

 

b) \(\frac{1}{5},\frac{3}{7},\frac{7}{10}\)

Solution:

\(\frac{1}{5},\frac{3}{7},\frac{7}{10}\)

Let’s take LCM of 5,7,10

7=7*1

5=5*1

10=5*2

So, we can take one 5 as a common term .Since, we have two 5’s and 7,2 is coming only once

Therefore the required LCM is= 5*7*2= 70

For the first number,

\(\Rightarrow \frac{1}{5}*\frac{14}{14}=\frac{14}{70}\)

For the second number,

\(\Rightarrow \frac{3}{7}*\frac{10}{10}=\frac{30}{70}\)

For the third number,

\(\Rightarrow \frac{7}{10}*\frac{7}{7}=\frac{49}{70}\)

Descending order means arranging the numbers from largest to smallest

So,

\(\frac{14}{70}=0.2\) \(\frac{30}{70}=0.42\) \(\frac{49}{70}=0.70\)

Therefore, the decreasing order of rational numbers are

\(\frac{49}{70}> \frac{30}{70}> \frac{14}{70}\)

i.e)

\(\frac{7}{10}>\frac{3}{7}>\frac{1}{5}\)

 

 

3) If the sum of the numbers are same along both rows and columns, Will it form a magic square?

\(\frac{5}{13}\) \(\frac{7}{13}\) \(\frac{3}{13}\)
\(\frac{3}{13}\) \(\frac{5}{13}\) \(\frac{7}{13}\)
\(\frac{7}{13}\) \(\frac{3}{13}\) \(\frac{5}{13}\)

 Solution:

  • Sum of first row = \(\frac{5}{13}\)+ \(\frac{7}{13}\)+ \(\frac{3}{13}\)= \(\frac{15}{13}\)

 

  • Sum of second row=\(\frac{3}{13}\)+ \(\frac{5}{13}\)+ \(\frac{7}{13}\)= \(\frac{15}{13}\)

 

  • Sum of third row=\(\frac{7}{13}\)+ \(\frac{3}{13}\)+ \(\frac{5}{13}\)= \(\frac{15}{13}\)

 

  • Sum of first column=\(\frac{5}{13}\)+ \(\frac{3}{13}\)+ \(\frac{7}{13}\)= \(\frac{15}{13}\)

 

  • Sum of second column=\(\frac{7}{13}\)+ \(\frac{5}{13}\)+ \(\frac{3}{13}\)= \(\frac{15}{13}\)

 

  • Sum of third column=\(\frac{3}{13}\)+ \(\frac{7}{13}\)+ \(\frac{5}{13}\)= \(\frac{15}{13}\)

 

  • Sum of first diagonal (left to right)= \(\frac{5}{13}\)+ \(\frac{5}{13}\)+ \(\frac{5}{13}\)= \(\frac{15}{13}\)

 

  • Sum of second diagonal (right to left)= \(\frac{3}{13}\)+ \(\frac{5}{13}\)+ \(\frac{7}{13}\)= \(\frac{15}{13}\)

 

  • Yes, it forms a magic square. Since, the sum of fractions in each row, each column and along the diagonals are same.

 

4) A rectangular block of length \(6\frac{1}{4}\)cm and \(3\frac{2}{3}\)cm of width is noted. Find the perimeter and area of the rectangular block.

Solution:

As the block is rectangular in shape

W.K.T

Perimeter of rectangle= \(2*\left ( length+breadth \right )\)

Since, both the numbers are in mixed fraction, it is first converted to fractional form

For length,

\(6\frac{1}{4}\)= \(\frac{\left ( 4*6 \right )+1}{4}=\frac{25}{4}\)

For breadth.

\(3\frac{2}{3}\)= \(\frac{\left ( 3*3 \right )+2}{3}=\frac{11}{3}\)

LCM of 3,4= 12

\(\Rightarrow \frac{25}{4}*\frac{3}{3}=\frac{75}{12}\) \(\Rightarrow \frac{11}{3}*\frac{4}{4}=\frac{44}{12}\)

Perimeter of rectangle= \(2*\left ( length+breadth \right )\)

= \(2*\left ( \frac{75}{12}+\frac{44}{12} \right )\)

= \(2*\left ( \frac{119}{12} \right )\)

= \(\frac{119}{6}\)cm

Area of rectangle = Length*breadth

                              = \(\frac{75}{12}*\frac{44}{12}=\frac{3300}{12}\) \( \Rightarrow 275 cm^{2}\)

 

 

5) Find the perimeter of

(i) DXYZ

(ii) The rectangle YMNZ in this figure given below.

Out of two perimeters, which is greater?

Solution:

(i) In DXYZ,

XY = \(\frac{5}{2}\)cm,

YZ = \(2\frac{3}{4}\) cm,

XZ=\(3\frac{3}{5}\)cm

The perimeter of triangle XYZ = XY + YZ + ZX

=(\(\frac{5}{2}\)+\(2\frac{3}{4}\)+ \(3\frac{3}{5}\))

=\(\frac{5}{2}+\frac{11}{4}+\frac{18}{5}\)

LCM of 2,4,5=

2=2*1

4=2*2

5=5*1

We have, 2 as a common term as we have two 2’s and take 2,5 into account because it’s a single term

LCM=2*5*2=20

\(\Rightarrow \frac{5}{2}*\frac{10}{10}=\frac{50}{20}\) \(\Rightarrow \frac{11}{4}*\frac{5}{5}=\frac{55}{20}\) \(\Rightarrow \frac{18}{5}*\frac{4}{4}=\frac{72}{20}\) \(\Rightarrow \left ( \frac{50+55+72}{20} \right )\)

\(\Rightarrow \frac{177}{20}\)cm

(ii) In rectangle YMNZ,

YZ = \(2\frac{3}{4}\) cm,

MN= \(\frac{7}{6}\)

W.K.T

Perimeter of rectangle = 2 (length + breadth)

                                       = \(2\left(2\frac{3}{4}+\frac{7}{6}\right )\)

= \(2\left(\frac{11}{4}+\frac{7}{6}\right )\)

LCM of 4, 6

4=2*2

6=2*3

We have ,2 as a common term as we have two 2’s and take 2,3 into account because it’s a single term

LCM=2*3*2=12

\(\Rightarrow \frac{11}{4}*\frac{3}{3}=\frac{33}{12}\) \(\Rightarrow \frac{7}{6}*\frac{2}{2}=\frac{14}{12}\) \( 2 \times \frac{33+14}{12} = \frac{47}{6}\)

So the greatest perimeter out of this is,

(i)  \(\frac{177}{20}=8.85\)cm

(ii) \(\frac{47}{6}=7.83\)cm

Comparing the perimeter of rectangle and triangle

\(\frac{177}{20}=8.85\)cm > \(\frac{47}{6}=7.83\)cm

Therefore, the perimeter of triangle XYZ is greater than that of rectangle YMNZ.

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