NCERT Solutions For Class 6 Maths Chapter 7

NCERT Solutions Class 6 Maths Fractions

Topics such as Fractions as Decimals, vice versa, operations on Fractions, word problems and more are covered in the NCERT Solutions For Class 6 Maths Chapter 7 Fractions. For students who are looking to give their best for their class 6 maths exam, it is extremely advisable to practise working on NCERT Solutions For Class 6 Maths Chapter 7. Students can download our NCERT Solutions For Class 6 Maths Chapter 7 Fractions pdf, top help them to learn the concepts effectively and learn the concepts properly as well.

Exercise 2.1

NCERT Solutions For Class 6 Maths Chapter 7 Exercises

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.