NCERT Solutions for Class 7 Maths Chapter 2 - Fractions And Decimals

Exercise - 2.1

1. Solve:

(i). $\text{2-}\frac{\text{3}}{\text{5}}$

Ans: The solution is given as,

$2-\frac{3}{5}=\frac{10-3}{5}=\frac{7}{5}=1\frac{2}{5}$

(ii). $\text{4+}\frac{\text{7}}{\text{8}}$

Ans: The solution is given as,

$4+\frac{7}{8}=\frac{32+7}{8}=\frac{39}{8}=4\frac{7}{8}$

(iii). $\frac{\text{3}}{\text{5}}\text{+}\frac{\text{2}}{\text{7}}$ 

Ans: The solution is given as,

$\frac{3}{5}+\frac{2}{7}=\frac{21+10}{35}=\frac{31}{35}$

(iv). $\frac{\text{9}}{\text{11}}\text{-}\frac{\text{4}}{\text{15}}$ 

Ans: The solution is given as,

$\frac{9}{11}-\frac{4}{15}=\frac{135-44}{165}=\frac{91}{165}$

(v) $\frac{\text{7}}{\text{10}}\text{+}\frac{\text{2}}{\text{5}}\text{+}\frac{\text{3}}{\text{2}}$ 

Ans: The solution is given as,

$\frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7+4+15}{10}=\frac{26}{10}=\frac{13}{5}=2\frac{3}{5}$ 

(vi) $\text{2}\frac{\text{2}}{\text{3}}\text{+3}\frac{\text{1}}{\text{2}}$

Ans: The solution is given as,       $2\frac{2}{3}+3\frac{1}{2}=\frac{8}{3}+\frac{7}{2}=\frac{16+21}{6}=\frac{37}{6}=6\frac{1}{6}$

(vii)  $\text{8}\frac{\text{1}}{\text{2}}\text{-3}\frac{\text{5}}{\text{8}}$ 

Ans: The solution is given as,     $8\frac{1}{2}-3\frac{5}{8}=\frac{17}{2}-\frac{29}{8}=\frac{68-29}{8}=\frac{39}{8}=4\frac{7}{8}$

2. Arrange the following in descending order: 

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

Ans: Converting into the fractions with same denominator,

$\frac{2}{9},\frac{2}{3},\frac{8}{21}\,\,\,\,\,\Rightarrow \,\,\,\,\frac{14}{63},\frac{42}{63},\frac{24}{63}$   

Arranging the terms in descending order,         $\frac{42}{63}>\frac{24}{63}>\frac{14}{63}$                           

Converting the fraction into simplest form,

$\frac{2}{3}>\frac{8}{21}>\frac{2}{9}$ 

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

Ans: Converting into the fractions with same denominator, $\frac{1}{5},\frac{3}{7},\frac{7}{10}\,\,\,\,\,\Rightarrow \,\,\,\,\frac{14}{70},\frac{30}{70},\frac{49}{70}$              

Arranging the term in descending order,

$\Rightarrow \,\,\frac{49}{70}>\,\frac{30}{70}>\,\frac{14}{70}$        

Converting the fraction into simplest form,

$\frac{7}{10}>\frac{3}{7}>\frac{1}{5}$ 

3. In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?

$\left( \text{Along the first row}\frac{\text{4}}{\text{11}}\text{+}\frac{\text{9}}{\text{11}}\text{+}\frac{\text{2}}{\text{11}}\text{+=}\frac{\text{15}}{\text{11}} \right)$

Ans: If the sum of fractions in each row, in each column and along the diagonals is same then it is a magic square.

Calculating the sum of,

first row $=\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}$                

second row \[=\frac{3}{11}+\frac{5}{11}+\frac{7}{11}=\frac{3+5+7}{11}=\frac{15}{11}\] 

third row $=\frac{8}{11}+\frac{1}{11}+\frac{6}{11}=\frac{8+1+6}{11}=\frac{15}{11}$ 

first column $=\frac{4}{11}+\frac{3}{11}+\frac{8}{11}=\frac{4+3+8}{11}=\frac{15}{11}$ 

second column $=\frac{9}{11}+\frac{5}{11}+\frac{1}{11}=\frac{9+5+1}{11}=\frac{15}{11}$ 

third column $=\frac{2}{11}+\frac{7}{11}+\frac{6}{11}=\frac{2+7+6}{11}=\frac{15}{11}$ 

first diagonal along top left to bottom right $=\frac{4}{11}+\frac{5}{11}+\frac{6}{11}=\frac{4+5+6}{11}=\frac{15}{11}$ 

second diagonal along top right to bottom left $=\frac{2}{11}+\frac{5}{11}+\frac{8}{11}=\frac{2+5+8}{11}=\frac{15}{11}$ 

Observe that the sum of fractions in each row, in each column and along the    diagonals is same, 

Hence, it’s a magic square.

4. A rectangular sheet of paper is $\text{12}\frac{\text{1}}{\text{2}}$ cm long and $\text{10}\frac{\text{2}}{\text{3}}$ cm wide. Find its Perimeter?

Ans: Given: A rectangular sheet of paper has

 \[\text{Length =12}\frac{1}{2}\] cm

 \[\text{Breadth =10}\frac{2}{3}\] cm

 \[\text{Perimeter of rectangle = 2 (length + breadth)}\] 

 $=2\left( 12\frac{1}{2}+10\frac{2}{3} \right)=2\left( \frac{25}{2}+\frac{32}{3} \right)$ 

   $=2\left( \frac{25\times 3+32\times 2}{6} \right)=2\left( \frac{75+64}{6} \right)$ 

   $=2\times \frac{139}{6}=\frac{139}{3}=46\frac{1}{3}$ cm

   Hence, the perimeter of the rectangular sheet is $46\frac{1}{3}$ cm.

5. Find the perimeter of 

(i) $\text{ }\!\!\Delta\!\!\text{ ABE}$,

(Image will be uploaded soon)

Ans: Given: In $\text{ }\!\!\Delta\!\!\text{ ABE,}\,\text{AB=}\frac{\text{5}}{\text{2}}\text{cm,}\,\text{BE=2}\frac{\text{3}}{\text{4}}\text{cm,}\,\text{AE=3}\frac{\text{3}}{\text{5}}\text{cm}$ 

The perimeter of triangle is equal to the sum of all the sides of the triangle. According to the given figure,

$\text{ }\!\!\Delta\!\!\text{ ABE=AB+BE+AE}$ $=\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5}=\frac{5}{2}+\frac{11}{4}+\frac{18}{5}$ 

$=\frac{50+55+72}{20}=\frac{177}{20}=8\frac{17}{20}$ cm

Hence, the perimeter of $\text{ }\!\!\Delta\!\!\text{ ABE}$ is $8\frac{17}{20}$ cm.

(ii) The rectangle $\text{BCDE}$ in this figure. Whose perimeter is greater?

(Image will be uploaded soon)

Ans: Given: In rectangle $\text{BCDE,}$ $\text{BE=2}\frac{\text{3}}{\text{4}}\text{cm,}\,\text{ED=}\frac{\text{7}}{\text{6}}\text{cm}$

The perimeter of the rectangle is given by,

\[\text{Perimeter of rectangle = 2 (length + breadth)}\] 

 $=2\left( 2\frac{3}{4}+\frac{7}{6} \right)=2\left( \frac{11}{4}+\frac{7}{6} \right)$ 

$=2\left( \frac{33+14}{12} \right)=\frac{47}{6}=7\frac{5}{6}\text{cm}$ 

Hence, the perimeter of rectangle $\text{BCDE}$ is $7\frac{5}{6}$ cm.

Compare the perimeter of triangle with the perimeter of rectangle,

$\text{8}\frac{\text{17}}{\text{20}}\text{cm }>\text{ 7}\frac{\text{5}}{\text{6}}\text{cm}$ 

Hence, the perimeter of triangle \[\text{ABE}\] is greater as compared to the perimeter of rectangle $\text{BCDE}$

6. Salil wants to put a picture in a frame. The picture is $\text{7}\frac{\text{3}}{\text{5}}$ cm wide. To fit in the frame the picture cannot be more than $\text{7}\frac{\text{3}}{\text{10}}$ cm wide. How much should the picture be trimmed?

Ans: Given: The width of the picture$=7\frac{3}{5}$ cm and 

The width of picture frame$=7\frac{3}{10}$ cm

The picture should be   

trimmed$=7\frac{3}{5}-7\frac{3}{10}=\frac{38}{5}-\frac{73}{10}$ $=\frac{76-73}{10}=\frac{3}{10}$ cm

Hence, the original picture should be trimmed by $\frac{3}{10}$ cm.

7. Ritu ate $\frac{\text{3}}{\text{5}}$ part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?

Ans: Given: The part of an apple eaten by Ritu $=\frac{3}{5}$ 

The part of an apple eaten by Somu $=1-\frac{3}{5}=\frac{5-3}{5}=\frac{2}{5}$ 

Compare the parts of apple eaten by Ritu and Somu,

 $\frac{3}{5}>\frac{2}{5}$ 

Observe that Ritu’s part is larger than Somu’s part.

Also, the larger share is more by $\frac{3}{5}-\frac{2}{5}=\frac{1}{5}$ part as compared to the smaller part.

Hence, Ritu’s part is $\frac{1}{5}$ more as compared to Somu’s part.

8. Michael finished colouring a picture in $\frac{\text{7}}{\text{12}}$ hour. Vaibhav finished colouring the same picture in $\frac{\text{3}}{\text{4}}$ hour. Who worked longer? By what fraction was it longer?

Ans: Given: Time taken by Michael for coloring the picture $=\frac{7}{12}$ hour and

Time taken by Vaibhav for coloring the picture $=\frac{3}{4}$ hour

Convert both fractions in the fractions such that their denominator is same,

$\frac{7}{12}$ and 

$\frac{3\times 3}{4\times 3}=\frac{9}{12}$ 

Observe that, 

$\frac{7}{12}<\frac{9}{12}\Rightarrow \frac{7}{12}<\frac{3}{4}$ 

Hence, Vaibhav worked for a longer time.

Calculate the time for which Vaibhav worked longer,

$\frac{3}{4}-\frac{7}{12}=\frac{9-7}{12}=\frac{2}{12}=\frac{1}{6}$ hour 

Therefore, Vaibhav took $\frac{1}{6}$ hour more than Michael.

Exercise - 2.2

1. Which of the drawings $(a)\,to\,(d)$ show:

(i). $\text{2 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{5}}$

(a)  (Image will be uploaded soon)              

Ans: corresponds to $\text{(d)}$  

Because,  $2\times \frac{1}{5}=\frac{1}{5}+\frac{1}{5}$                                                                

(ii). $\text{2 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{2}}$

(b) (Image will be uploaded soon)

Ans: corresponds to $\text{(b)}$

Because, $2\times \frac{1}{2}=\frac{1}{2}+\frac{1}{2}$

(iii). $\text{3 }\!\!\times\!\!\text{ }\frac{\text{2}}{\text{3}}$

(c)    (Image will be uploaded soon)     

Ans:  corresponds to $\text{(a)}$   

Because, $3\times \frac{2}{3}=\frac{2}{3}+\frac{2}{3}+\frac{2}{3}$ 

(iv). $\text{3 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{4}}$

(d) (Image will be uploaded soon)

Ans: Corresponds to $\text{(c)}$

Because, $3\times \frac{1}{4}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}$ 

2. Some pictures $\left( \text{a} \right)\,\text{to}\,\left( \text{c} \right)$ are given below. Tell which of them show: 

(i). $\text{3 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{5}}\text{=}\frac{\text{3}}{\text{5}}$

(a) (Image will be uploaded soon) 

Ans: Corresponds to $\text{(c)}$                    

Because, $3\times \frac{1}{5}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}$  

(ii). $\text{2 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{3}}\text{=}\frac{\text{2}}{\text{3}}$ 

(b) (Image will be uploaded soon)                                                

Ans: Corresponds to $\text{(a)}$                   

Because, $2\times \frac{1}{3}=\frac{1}{3}+\frac{1}{3}$ 

(iii). $\text{3 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{4}}\text{=2}\frac{\text{1}}{\text{4}}$

(c) (Image will be uploaded soon)                                                             

Ans: Corresponds to $\text{(b)}$                  

Because, $3\times \frac{3}{4}=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}$      

3. Multiply and reduce to lowest form and convert into a mixed fraction:

(i). $\text{7 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{5}}$        

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,   

$7\times \frac{3}{5}=\frac{7\times 3}{5}=\frac{21}{5}=4\frac{1}{5}$

(ii). $\text{4 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{3}}$ 

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,

$4\times \frac{1}{3}=\frac{4\times 1}{3}=\frac{4}{3}=1\frac{1}{3}$ 

(iii). $\text{2 }\!\!\times\!\!\text{ }\frac{\text{6}}{\text{7}}$    

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,

$2\times \frac{6}{7}=\frac{2\times 6}{7}=\frac{12}{7}=1\frac{5}{7}$       

(iv). $\text{5 }\!\!\times\!\!\text{ }\frac{\text{2}}{\text{9}}$ 

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,

$5\times \frac{2}{9}=\frac{5\times 2}{9}=\frac{10}{9}=1\frac{1}{9}$

(v). $\frac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ 4}$     

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,

$\frac{2}{3}\times 4=\frac{2\times 4}{3}=\frac{8}{3}=2\frac{2}{3}$   

(vi)  $\frac{\text{5}}{\text{2}}\text{ }\!\!\times\!\!\text{ 6}$    

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,

$\frac{5}{2}\times 6=5\times 3=15$

(vii)   $\text{11 }\!\!\times\!\!\text{ }\frac{\text{4}}{\text{7}}$    

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,

$11\times \frac{4}{7}=\frac{11\times 4}{7}=\frac{44}{7}=6\frac{2}{7}$ 

(viii)  $\text{20 }\!\!\times\!\!\text{ }\frac{\text{4}}{\text{5}}$ 

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,

$20\times \frac{4}{5}=4\times 4=16$

(ix)  $\text{13 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{3}}$      

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,    

$13\times \frac{1}{3}=\frac{13\times 1}{3}=\frac{13}{3}=4\frac{1}{3}$ 

(x)  $\text{15 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{5}}$ 

Ans: Multiplying and reducing to lowest form and converting into a mixed fraction,   

$15\times \frac{3}{5}=3\times 3=9$                

4. Shade: 

(i). $\frac{\text{1}}{\text{2}}$ of the circles in box 

(ii). (Image will be uploaded soon)                

Ans: Half of the circles in the box are,

$\frac{\text{1}}{\text{2}}\,\text{of}\,\text{12}\,\text{circles=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ 12=6}\,\text{circles}$

(Image will be uploaded soon)

(iii). $\frac{\text{2}}{\text{3}}$ of the triangles in box 

(b) (Image will be uploaded soon)

Ans: Two-third of the triangles in the box are,$\frac{\text{2}}{\text{3}}\,\text{of}\,\text{9}\,\text{triangles=}\frac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ 9=2 }\!\!\times\!\!\text{ 3=6}\,\text{triangles}$ 

(Image will be uploaded soon)

(iv). $\frac{\text{3}}{\text{5}}$ of the squares inbox

(v). (c)   (Image will be uploaded soon)                          

Ans: Three-fifth of the squares in the box are,

$\frac{\text{3}}{\text{5}}\,\text{of}\,\text{15}\,\text{squares=}\frac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ 15=3 }\!\!\times\!\!\text{ 3=9}\,\text{squares}$

(Image will be uploaded soon)

5. Find

(a).$\frac{\text{1}}{\text{2}}\,\text{of}\,\text{(i)}\,\text{24}\,\text{(ii)}\,\text{46}$

Ans:

(i) Calculating the value,

$\frac{\text{1}}{\text{2}}\,\text{of}\,\text{24=12}$

(ii) Calculating the value,

$\frac{\text{1}}{\text{2}}\,\text{of}\,\text{46=23}$ 

(b). $\frac{\text{2}}{\text{3}}\,\text{of}\,\text{(i)}\,\text{18}\,\text{(ii)}\,\text{27}$ 

Ans:

(i) Calculating the value, $\frac{\text{2}}{\text{3}}\,\text{of}\,\text{18=}\frac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ 18=2 }\!\!\times\!\!\text{ 6=12}$

(ii) Calculating the value, $\frac{\text{2}}{\text{3}}\,\text{of}\,\text{27=}\frac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ 27=2 }\!\!\times\!\!\text{ 9=18}$ 

(c)  $\frac{\text{3}}{\text{4}}\,\text{of}\,\text{(i)}\,\text{16}\,\text{(ii)}\,\text{36}$          

Ans:

(i) Calculating the value,

$\frac{\text{3}}{\text{4}}\,\text{of}\,\text{16=}\frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ 16=3 }\!\!\times\!\!\text{ 4=12}$

(ii) Calculating the value,

$\frac{\text{3}}{\text{4}}\,\text{of}\,36\text{=}\frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ 36=3 }\!\!\times\!\!\text{ 9=27}$  

(d)  $\frac{\text{4}}{\text{5}}\,\text{of}\,\text{(i)}\,\text{20}\,\text{(ii)}\,\text{35}$ 

Ans:

(i) Calculating the value,

$\frac{\text{4}}{\text{5}}\,\text{of}\,20\text{=}\frac{\text{4}}{\text{5}}\text{ }\!\!\times\!\!\text{ 20=4 }\!\!\times\!\!\text{ 4=16}$

(ii) Calculating the value,

$\frac{\text{4}}{\text{5}}\,\text{of}\,35\text{=}\frac{\text{4}}{\text{5}}\text{ }\!\!\times\!\!\text{ 35=4 }\!\!\times\!\!\text{ 7=28}$ 

6.   Multiply and express as a mixed fraction:

(a)  $\text{3 }\!\!\times\!\!\text{ 5}\frac{\text{1}}{\text{5}}$       

Ans: Multiplying and expressing the term as mixed fraction,

$3\times 5\frac{1}{5}=3\times \frac{26}{5}=\frac{3\times 26}{5}=\frac{78}{5}=15\frac{3}{5}$

(b) $\text{5 }\!\!\times\!\!\text{ 6}\frac{\text{3}}{\text{4}}$          

Ans: Multiplying and expressing the term as mixed fraction,

$5\times 6\frac{3}{4}=5\times \frac{27}{4}=\frac{5\times 27}{4}=\frac{135}{4}=33\frac{3}{4}$ 

(c) $\text{7 }\!\!\times\!\!\text{ 2}\frac{\text{1}}{\text{4}}$

Ans: Multiplying and expressing the term as mixed fraction,

$7\times 2\frac{1}{4}=7\times \frac{9}{4}=\frac{7\times 9}{4}=\frac{63}{4}=15\frac{3}{4}$        

(d)  $\text{4 }\!\!\times\!\!\text{ 6}\frac{\text{1}}{\text{3}}$ 

Ans: Multiplying and expressing the term as mixed fraction,

$4\times 6\frac{1}{3}=4\times \frac{19}{3}=\frac{4\times 19}{3}=\frac{76}{3}=25\frac{1}{3}$ 

(e)  $\text{3}\frac{\text{1}}{\text{4}}\text{ }\!\!\times\!\!\text{ 6}$ 

Ans: Multiplying and expressing the term as mixed fraction,

$3\frac{1}{4}\times 6=\frac{13}{4}\times 6=\frac{13\times 3}{2}=\frac{39}{2}=19\frac{1}{2}$  

(f)  $\text{3}\frac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ 8}$ 

Ans: Multiplying and expressing the term as mixed fraction,

$3\frac{2}{5}\times 8=\frac{17}{5}\times 8=\frac{17\times 8}{5}=\frac{136}{5}=27\frac{1}{5}$

7. Find:

(a)  $\frac{\text{1}}{\text{2}}\,\text{of}\,\text{(i)}\,\text{2}\frac{\text{3}}{\text{4}}\,\text{(ii)}\,\text{4}\frac{\text{2}}{\text{9}}$             

Ans:

(i) Calculating the value, \[\frac{\text{1}}{\text{2}}\,\text{of}\,\text{2}\frac{\text{3}}{\text{4}}\text{=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ 2}\frac{\text{3}}{\text{4}}\text{=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\frac{\text{11}}{\text{4}}\text{=}\frac{\text{11}}{\text{8}}\text{=1}\frac{\text{3}}{\text{8}}\] 

(ii) Calculating the value, \[\frac{\text{1}}{\text{2}}\,\text{of}\,\text{4}\frac{\text{2}}{\text{9}}\text{=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ 4}\frac{\text{2}}{\text{9}}\text{=}\frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\frac{\text{38}}{\text{9}}\text{=}\frac{\text{19}}{\text{9}}\text{=2}\frac{\text{1}}{\text{9}}\] 

(b)  $\frac{\text{5}}{\text{8}}\,\text{of}\,\text{(i)}\,\text{3}\frac{\text{5}}{\text{6}}\,\text{(ii)}\,\text{9}\frac{\text{2}}{\text{3}}$ 

Ans:

(i) Calculating the value, \[\frac{\text{5}}{\text{8}}\,\text{of}\,\text{3}\frac{\text{5}}{\text{6}}\text{=}\frac{\text{5}}{\text{8}}\text{ }\!\!\times\!\!\text{ 3}\frac{\text{5}}{\text{6}}\text{=}\frac{\text{5}}{\text{8}}\text{ }\!\!\times\!\!\text{ }\frac{\text{23}}{\text{6}}\text{=}\frac{\text{115}}{\text{48}}\text{=2}\frac{\text{19}}{\text{48}}\] 

(ii) Calculating the value, \[\frac{\text{5}}{\text{8}}\,\text{of}\,\text{9}\frac{\text{2}}{\text{3}}\text{=}\frac{\text{5}}{\text{8}}\text{ }\!\!\times\!\!\text{ 9}\frac{\text{2}}{\text{3}}\text{=}\frac{\text{5}}{\text{8}}\text{ }\!\!\times\!\!\text{ }\frac{\text{29}}{\text{3}}\text{=}\frac{\text{145}}{\text{24}}\text{=6}\frac{\text{1}}{\text{24}}\] 

8.  Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained $\text{5}$ liters of water. Vidya consumed $\frac{\text{2}}{\text{5}}$ of the water. Pratap consumed the remaining water.

(i). How much water did Vidya drink?

Ans:   Water consumed by Vidya is,$\text{=}\frac{\text{2}}{\text{5}}\,\text{of}\,\text{5}\,\text{litres=}\frac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ 5=2}\,\text{litres}$ 

Hence, Vidya drank $2$ litres of water from the bottle.

(ii). What fraction of the total quantity of water did Pratap drink?

Ans: Water consumed by Pratap \[\text{= }\left( \text{1-}\frac{\text{2}}{\text{5}} \right)\text{  }\]part of bottle

Pratap consumed $\frac{\text{3}}{\text{5}}\,\text{of}\,\text{5}\,\text{litres}\,\text{water=}\frac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ 5=3}\,\text{lites}$ 

Hence, Pratap drank $\frac{3}{5}$ part of the total quantity of water present in the bottle.

Exercise - 2.3

1. Find:

(i) $\frac{\text{1}}{\text{4}}\,\text{of}$ (a) $\frac{\text{1}}{\text{4}}$ (b) $\frac{\text{3}}{\text{5}}$ (c) $\frac{\text{4}}{\text{3}}$ 

Ans:

(a) Calculating the value,

$\frac{\text{1}}{\text{4}}\,\text{of}\,\frac{\text{1}}{\text{4}}\text{=}\frac{\text{1}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{4}}\text{=}\frac{\text{1 }\!\!\times\!\!\text{ 1}}{\text{4 }\!\!\times\!\!\text{ 4}}\text{=}\frac{\text{1}}{\text{16}}$ 

(b) Calculating the value,

$\frac{\text{1}}{\text{4}}\,\text{of}\,\frac{\text{3}}{\text{5}}\text{=}\frac{\text{1}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{4}}\text{=}\frac{\text{1 }\!\!\times\!\!\text{ 3}}{\text{4 }\!\!\times\!\!\text{ 4}}\text{=}\frac{\text{3}}{\text{16}}$ 

(c) Calculating the value,

$\frac{\text{1}}{\text{4}}\,\text{of}\,\frac{\text{4}}{\text{3}}\text{=}\frac{\text{1}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\frac{\text{4}}{\text{3}}\text{=}\frac{\text{1 }\!\!\times\!\!\text{ 4}}{\text{4 }\!\!\times\!\!\text{ 3}}\text{=}\frac{\text{1}}{\text{3}}$ 

(ii) \[\frac{\text{1}}{\text{7}}\,\text{of}\] (a) \[\frac{\text{2}}{\text{9}}\] (b) \[\frac{\text{6}}{\text{5}}\] (c) $\frac{\text{3}}{\text{10}}$ 

Ans:

(a) Calculating the value,

$\frac{\text{1}}{\text{7}}\,\text{of}\,\frac{\text{2}}{\text{9}}\text{=}\frac{\text{1}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\frac{\text{2}}{\text{9}}\text{=}\frac{\text{1 }\!\!\times\!\!\text{ 2}}{\text{7 }\!\!\times\!\!\text{ 9}}\text{=}\frac{\text{2}}{\text{63}}$ 

(b) Calculating the value,

$\frac{\text{1}}{\text{7}}\,\text{of}\,\frac{\text{2}}{\text{9}}\text{=}\frac{\text{1}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\frac{\text{6}}{\text{5}}\text{=}\frac{\text{1 }\!\!\times\!\!\text{ 6}}{\text{7 }\!\!\times\!\!\text{ 5}}\text{=}\frac{\text{6}}{\text{35}}$ 

(c) Calculating the value,

$\frac{\text{1}}{\text{7}}\,\text{of}\,\frac{\text{2}}{\text{9}}\text{=}\frac{\text{1}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{10}}\text{=}\frac{\text{1 }\!\!\times\!\!\text{ 3}}{\text{7 }\!\!\times\!\!\text{ 10}}\text{=}\frac{3}{70}$ 

2. Multiply and reduce to lowest form (if possible):

(i) $\frac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ 2}\frac{\text{2}}{\text{3}}$ 

Ans: Multiplying and reducing to lowest form,  

$\frac{2}{3}\times 2\frac{2}{3}=\frac{2}{3}\times \frac{8}{3}=\frac{2\times 8}{3\times 3}=\frac{16}{9}=1\frac{7}{9}$ 

(ii) $\frac{\text{2}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\frac{\text{7}}{\text{9}}$

Ans: Multiplying and reducing to lowest form,  

$\frac{2}{7}\times \frac{7}{9}=\frac{2\times 7}{7\times 9}=\frac{2}{9}$

(iii) $\frac{\text{3}}{\text{8}}\text{ }\!\!\times\!\!\text{ }\frac{\text{6}}{\text{4}}$ 

Ans: Multiplying and reducing to lowest form, 

$\frac{3}{8}\times \frac{6}{4}=\frac{3\times 6}{8\times 4}=\frac{3\times 3}{8\times 2}=\frac{9}{16}$

(iv) $\frac{\text{9}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{5}}$ 

Ans: Multiplying and reducing to lowest form, 

$\frac{9}{5}\times \frac{3}{5}=\frac{9\times 3}{5\times 5}=\frac{27}{25}=1\frac{2}{25}$

(v) $\frac{\text{1}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\frac{\text{15}}{\text{8}}$ 

Ans: Multiplying and reducing to lowest form,

$\frac{1}{3}\times \frac{15}{8}=\frac{1\times 15}{3\times 8}=\frac{1\times 5}{1\times 8}=\frac{5}{8}$

(vi) $\frac{\text{11}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{10}}$ 

Ans: Multiplying and reducing to lowest form,  

$\frac{11}{2}\times \frac{3}{10}=\frac{11\times 3}{2\times 10}=\frac{33}{20}=1\frac{3}{20}$ 

(vii) $\frac{\text{4}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\frac{\text{12}}{\text{7}}$ 

Ans: Multiplying and reducing to lowest form,  

$\frac{4}{5}\times \frac{12}{7}=\frac{4\times 12}{5\times 7}=\frac{48}{35}=1\frac{13}{35}$

3. Multiply the following fractions:

(i) $\frac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ 5}\frac{\text{1}}{\text{4}}$

Ans: Performing multiplication,

$\frac{2}{5}\times 5\frac{1}{4}=\frac{2}{5}\times \frac{21}{4}=\frac{2\times 21}{5\times 4}=\frac{1\times 21}{5\times 2}=\frac{21}{10}=2\frac{1}{10}$ 

(ii) $\text{6}\frac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\frac{\text{7}}{\text{9}}$ 

Ans: Performing multiplication,

$6\frac{2}{5}\times \frac{7}{9}=\frac{32}{5}\times \frac{7}{9}=\frac{32\times 7}{5\times 9}=\frac{224}{45}=4\frac{44}{45}$

(iii) $\frac{\text{3}}{\text{2}}\text{ }\!\!\times\!\!\text{ 5}\frac{\text{1}}{\text{3}}$ 

Ans: Performing multiplication,

$\frac{3}{2}\times 5\frac{1}{3}=\frac{3}{2}\times \frac{16}{3}=\frac{48}{6}=8$ 

(iv) $\frac{\text{5}}{\text{6}}\text{ }\!\!\times\!\!\text{ 2}\frac{\text{3}}{\text{7}}$ 

Ans: Performing multiplication,

$\frac{5}{6}\times 2\frac{3}{7}=\frac{5}{6}\times \frac{17}{7}=\frac{85}{42}=2\frac{1}{42}$ 

(v) $\text{3}\frac{\text{2}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\frac{\text{4}}{\text{7}}$

Ans: Performing multiplication,

$3\frac{2}{5}\times \frac{4}{7}=\frac{17}{7}\times \frac{4}{7}=\frac{68}{35}=1\frac{33}{35}$

(vi) $\text{2}\frac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ 3}$ 

Ans: Performing multiplication,

$2\frac{3}{5}\times 3=\frac{13}{5}\times \frac{3}{1}=\frac{13\times 3}{5\times 1}=\frac{39}{5}=7\frac{4}{5}$

(vii) $\text{3}\frac{\text{4}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{5}}$ 

Ans: Performing multiplication,

$3\frac{4}{7}\times \frac{3}{5}=\frac{25}{7}\times \frac{3}{5}=\frac{5\times 3}{7\times 1}=\frac{15}{7}=2\frac{1}{7}$

4. Which is greater:

(i) $\frac{\text{2}}{\text{7}}\,\text{of}\,\frac{\text{3}}{\text{4}}\,\text{or}\,\frac{\text{3}}{\text{5}}\,\text{of}\,\frac{\text{5}}{\text{8}}$

Ans: Calculating the greater term,

$\frac{\text{2}}{\text{7}}\,\text{of}\,\frac{\text{3}}{\text{4}}\,\text{or}\,\frac{\text{3}}{\text{5}}\,\text{of}\,\frac{\text{5}}{\text{8}}$                

$\Rightarrow \frac{\text{2}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{4}}\,\text{or}\,\frac{\text{3}}{\text{5}}\text{ }\!\!\times\!\!\text{ }\frac{\text{5}}{\text{8}}$ 

$\Rightarrow \frac{\text{3}}{\text{14}}\,\text{or}\,\frac{\text{3}}{\text{8}}$

$\Rightarrow \frac{3}{14}<\frac{3}{8}$ 

Hence, $\frac{\text{3}}{\text{5}}\,\text{of}\,\frac{\text{5}}{\text{8}}$ is greater.

(ii) $\frac{\text{1}}{\text{2}}\,\text{of}\,\frac{\text{6}}{\text{7}}\,\text{or}\,\frac{\text{2}}{\text{3}}\,\text{of}\,\frac{\text{3}}{\text{7}}$ 

Ans:

Calculating the greater term,

$\frac{\text{1}}{\text{2}}\,\text{of}\,\frac{\text{6}}{\text{7}}\,\text{or}\,\frac{\text{2}}{\text{3}}\,\text{of}\,\frac{\text{3}}{\text{7}}$                

$\Rightarrow \frac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\frac{\text{6}}{\text{7}}\,\text{or}\,\frac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{7}}$ 

$\Rightarrow \frac{\text{6}}{\text{14}}\,\text{or}\,\frac{\text{2}}{\text{7}}$ $\Rightarrow \frac{\text{6}}{\text{14}}>\frac{\text{2}}{\text{7}}$ 

Hence, $\frac{\text{1}}{\text{2}}\,\text{of}\,\frac{\text{6}}{\text{7}}$ is greater.

5. Saili plants \[\text{4}\] saplings in a row in her garden. The distance between

two adjacent saplings is \[\frac{\text{3}}{\text{4}}\] m. Find the 

distance between the first and the last sapling. 

Ans: Given: Saili plants \[4\] saplings in a row where the distance between two 

adjacent saplings $=\frac{3}{4}$m.

(Image will be uploaded soon)

The number of gaps in saplings \[=\text{ }3\] 

Hence, 

The distance between the first and the last saplings$\text{=3 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{4}}\text{=}\frac{\text{9}}{\text{4}}\text{m=2}\frac{\text{1}}{\text{4}}\text{m}$

Therefore, the distance between the first and the last saplings is $\text{2}\frac{\text{1}}{\text{4}}\,\text{m}$

6. Lipika reads a book for $\text{1}\frac{\text{3}}{\text{4}}$ hours every day. 

She reads the entire book in \[\text{6}\] days. How many hours in all were 

required by her to read the book?

Ans: Given: Time taken for reading a book by Lipika $=1\frac{3}{4}$ hours.

Lipika reads the entire book in $6$ days

Calculating the Total hours taken by Lipika to read the entire book,

$=1\frac{3}{4}\times 6=\frac{7}{4}\times 6=\frac{21}{2}=10\frac{1}{2}$ hours.

Hence, it would take $10$ hours to read the book.

7. A car runs $\text{16}$ km using \[\text{1}\] litre of petrol. How much 

distance will it cover using   $\text{2}\frac{\text{3}}{\text{4}}$ litres of 

petrol?

Ans: Given: A car covers the distance$\text{=16}\,\text{km}$ in $1$ litre of 

petrol.

Calculating the distance covered by car in $2\frac{3}{4}$ litres of petrol,

Distance$\text{=2}\frac{\text{3}}{\text{4}}\,\text{of}\,\text{16}\,\text{km=}\frac{\text{11}}{\text{4}}\text{ }\!\!\times\!\!\text{ 16=44}\,\text{km}$

Therefore, car will cover a distance of $44$ km in $2\frac{3}{4}$ litres of petrol.

8. (a)

(i) Provide the number in the box , such that 

$\frac{\text{2}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\text{=}\frac{\text{10}}{\text{30}}$ 

Ans: The number inside the box should be $\frac{2}{3}\times =\frac{10}{30}$ 

(ii) The simplest form of the number obtained in $$ is _____.

Ans: The simplest form of the number obtained in 

$\frac{\text{5}}{\text{10}}\,\text{is}\,\frac{\text{1}}{\text{2}}$

(b) 

(i) Provide the number in the box $$ , such that $\frac{3}{5}\times =\frac{24}{75}$.

Ans: The number inside the box should be $\frac{3}{5}\times =\frac{24}{75}$ 

(ii) The simplest form of the number obtained in is______.

Ans: The simplest form of the number obtained in 

$\frac{\text{8}}{\text{15}}\,\text{is}\,\frac{\text{8}}{\text{15}}$

Exercise - 2.4

1. Find:

(i) $\text{12 }\!\!\div\!\!\text{ }\frac{\text{3}}{\text{4}}$ 

Ans: Calculating the value,

$12\div \frac{3}{4}=12\times \frac{4}{3}=16$

(ii) $\text{14 }\!\!\div\!\!\text{ }\frac{\text{5}}{\text{6}}$ 

Ans: Calculating the value,

$14\div \frac{5}{6}=14\times \frac{6}{5}=\frac{84}{5}=16\frac{4}{5}$ 

(iii) $\text{8 }\!\!\div\!\!\text{ }\frac{\text{7}}{\text{3}}$ 

Ans:  Calculating the value,

$8\div \frac{7}{3}=8\times \frac{3}{7}=\frac{24}{7}=3\frac{3}{7}$

(iv) $\text{4 }\!\!\div\!\!\text{ }\frac{\text{8}}{\text{3}}$ 

Ans: Calculating the value,

$4\div \frac{8}{3}=4\times \frac{3}{8}=\frac{3}{2}=1\frac{1}{2}$

(v) $\text{3 }\!\!\div\!\!\text{ 2}\frac{\text{1}}{\text{3}}$ 

Ans: Calculating the value,

$3\div 2\frac{1}{3}=3\div \frac{7}{3}=3\times \frac{3}{7}=\frac{9}{7}=1\frac{2}{7}$           

(vi) \[\text{5 }\!\!\div\!\!\text{ 3}\frac{\text{4}}{\text{7}}\] 

Ans: Calculating the value,

$5\div 3\frac{4}{7}=5\div \frac{25}{7}=5\times \frac{7}{25}=\frac{7}{5}=1\frac{2}{5}$

2. Find the reciprocal of each of the following fractions. Classify the 

reciprocals as proper fraction, improper fractions and whole numbers.

(i) $\frac{\text{3}}{\text{7}}$

Ans: Calculating the reciprocal and stating the type of the fraction,

Reciprocal of $\frac{\text{3}}{\text{7}}\text{=}\frac{\text{7}}{\text{3}}\to \text{Improper}\,\text{fraction}$  

(ii) $\frac{\text{5}}{\text{8}}$

Ans: Calculating the reciprocal and stating the type of the fraction,

Reciprocal of$\frac{\text{5}}{\text{8}}\text{=}\frac{\text{8}}{\text{5}}\to \text{Improper}\,\text{fraction}$

(iii) $\frac{\text{9}}{\text{7}}$

Ans: Calculating the reciprocal and stating the type of the fraction,

Reciprocal of $\frac{\text{9}}{\text{7}}\text{=}\frac{\text{7}}{\text{9}}\to \text{Proper}\,\text{fraction}$

(iv) $\frac{\text{6}}{\text{5}}$ 

Ans: Calculating the reciprocal and stating the type of the fraction,

Reciprocal of $\frac{\text{6}}{\text{5}}\text{=}\frac{\text{5}}{\text{6}}\to \text{Proper}\,\text{fraction}$

(v) $\frac{\text{12}}{\text{7}}$ 

Ans: Calculating the reciprocal and stating the type of the fraction,

Reciprocal of $\frac{\text{12}}{\text{7}}\text{=}\frac{\text{7}}{\text{12}}\to \text{Proper}\,\text{fraction}$  

(vi) $\frac{\text{1}}{\text{8}}$

Ans: Calculating the reciprocal and stating the type of the fraction,

Reciprocal of $\frac{\text{9}}{\text{7}}\text{=8}\to \text{Whole number}$

(vii) $\frac{\text{1}}{\text{11}}$ 

Ans: Calculating the reciprocal and stating the type of the fraction,

Reciprocal of $\frac{\text{1}}{\text{11}}\text{=11}\to \text{Whole number}$

3. Find:

(i) $\frac{\text{7}}{\text{3}}\text{ }\!\!\div\!\!\text{ 2}$ 

Ans: Calculating the value,

$\frac{7}{3}\div 2=\frac{7}{3}\times \frac{1}{2}=\frac{7\times 1}{3\times 2}=\frac{7}{6}=1\frac{1}{6}$

(ii) $\frac{\text{4}}{\text{9}}\text{ }\!\!\div\!\!\text{ 5}$

Ans: Calculating the value,

$\frac{4}{9}\div 5=\frac{4}{9}\times \frac{1}{5}=\frac{4\times 1}{9\times 5}=\frac{4}{45}$ 

(iii) $\frac{\text{6}}{\text{13}}\text{ }\!\!\div\!\!\text{ 7}$ 

Ans: Calculating the value,

$\frac{6}{13}\div 7=\frac{6}{13}\times \frac{1}{7}=\frac{6\times 1}{13\times 7}=\frac{6}{91}$ 

(iv) $\text{4}\frac{\text{1}}{\text{3}}\text{ }\!\!\div\!\!\text{ 3}$

Ans:  Calculating the value,

$4\frac{1}{3}\div 3=\frac{13}{3}\div 3=\frac{13}{3}\times \frac{1}{3}=\frac{13}{9}=1\frac{4}{9}$

(v) $\text{3}\frac{\text{1}}{\text{2}}\text{ }\!\!\div\!\!\text{ 4}$ 

Ans: Calculating the value,

$3\frac{1}{2}\div 4=\frac{7}{2}\div 4=\frac{7}{2}\times \frac{1}{4}=\frac{7}{8}$

(vi) $\text{4}\frac{\text{3}}{\text{7}}\text{ }\!\!\div\!\!\text{ 7}$ 

Ans: Calculating the value,

$4\frac{3}{7}\div 7=\frac{31}{7}\div 7=\frac{31}{7}\times \frac{1}{7}=\frac{31}{49}$

4. Find:

(i) $\frac{\text{2}}{\text{5}}\text{ }\!\!\div\!\!\text{ }\frac{\text{1}}{\text{2}}$

Ans:  Calculating the value,

$\frac{2}{5}\div \frac{1}{2}=\frac{2}{5}\times \frac{2}{1}=\frac{2\times 2}{5\times 1}=\frac{4}{5}$ 

(ii) $\frac{\text{4}}{\text{9}}\text{ }\!\!\div\!\!\text{ }\frac{\text{2}}{\text{3}}$ 

Ans: Calculating the value,

$\frac{4}{9}\div \frac{2}{3}=\frac{4}{9}\times \frac{3}{2}=\frac{2}{3}$

(iii) $\frac{\text{3}}{\text{7}}\text{ }\!\!\div\!\!\text{ }\frac{\text{8}}{\text{7}}$

Ans:  Calculating the value,

$\frac{3}{7}\div \frac{8}{7}=\frac{3}{7}\times \frac{7}{8}=\frac{3}{8}$

(iv) $\text{2}\frac{\text{1}}{\text{3}}\text{ }\!\!\div\!\!\text{ }\frac{\text{3}}{\text{5}}$ 

Ans: Calculating the value,

$2\frac{1}{3}\div \frac{3}{5}=\frac{7}{3}\div \frac{3}{5}=\frac{7}{3}\times \frac{5}{3}=\frac{35}{9}=3\frac{8}{9}$

(v) $\text{3}\frac{\text{1}}{\text{2}}\text{ }\!\!\div\!\!\text{ }\frac{\text{8}}{\text{3}}$ 

Ans: Calculating the value,

$3\frac{1}{2}\div \frac{8}{3}=\frac{7}{2}\div \frac{3}{8}=\frac{7}{2}\times \frac{3}{8}=\frac{7\times 3}{2\times 8}=\frac{21}{16}=1\frac{5}{16}$

(vi) $\frac{\text{2}}{\text{5}}\text{ }\!\!\div\!\!\text{ 1}\frac{\text{1}}{\text{2}}$

Ans: Calculating the value,

$2\frac{1}{3}\div \frac{3}{5}=\frac{2}{5}\div 1\frac{1}{2}=\frac{2}{5}\div \frac{3}{2}=\frac{2}{5}\times \frac{2}{3}=\frac{2\times 2}{5\times 3}=\frac{4}{15}$

(vii) $\text{3}\frac{\text{1}}{\text{5}}\text{ }\!\!\div\!\!\text{ 1}\frac{\text{2}}{\text{3}}$

Ans:  Calculating the value,

$3\frac{1}{5}\div 1\frac{2}{3}=\frac{16}{5}\div \frac{5}{3}=\frac{16}{5}\times \frac{3}{5}=\frac{16\times 3}{5\times 5}=\frac{48}{25}=1\frac{23}{25}$

(viii) $\text{2}\frac{\text{1}}{\text{5}}\text{ }\!\!\div\!\!\text{ 1}\frac{\text{1}}{\text{5}}$ 

Ans: Calculating the value,

$2\frac{1}{5}\div 1\frac{1}{5}=\frac{11}{5}\div \frac{6}{5}=\frac{11}{5}\times \frac{5}{6}=\frac{11}{6}=1\frac{5}{6}$

Exercise - 2.5

1. Which is greater:

(i) $\text{0}\text{.5}\,\text{or}\,\text{0}\text{.05}$

Ans: Finding the greater term,

$0.5>0.05$

(ii) $\text{0}\text{.7}\,\text{or}\,\text{0}\text{.5}$ 

Ans: Finding the greater term,

$0.7>0.5$

(iii) \[\text{7 or 0}\text{.7}\] 

Ans: Finding the greater term,

$7>0.7$

(iv) \[\text{1}\text{.37 or 1}\text{.49}\]

Ans: Finding the greater term,

$1.37<1.49$

(v) \[\text{2}\text{.03 or 2}\text{.30}\]

Ans: Finding the greater term,

$2.03<2.30$

(vi) \[\text{0}\text{.8 or 0}\text{.88}\]

Ans: Finding the greater term,

$0.8<0.88$

2. Express as rupees using decimals:

(i) $\text{7}\,\text{paise}$

Ans:  Expressing the term as rupees,

$\text{7}\,\text{paise=Re}\text{.}\frac{\text{7}}{\text{100}}\text{=Re}\text{.0}\text{.07}$ 

(ii) \[\text{7 rupees 7 paise}\] 

Ans: Expressing the term as rupees,$\text{7}\,\text{rupees}\,\,\text{7}\,\,\text{paise=Rs}\text{.7+Re}\text{.}\frac{\text{7}}{\text{100}}\text{=Rs}\text{.7+Rs}\text{.0}\text{.07=Rs}\text{.7}\text{.07}$

(iii) \[\text{77 rupees 77 paise}\] 

Ans: Expressing the term as rupees,$\text{77}\,\text{rupees}\,\,\text{77}\,\,\text{paise=Rs}\text{.77+Re}\text{.}\frac{\text{77}}{\text{100}}\text{=Rs}\text{.77+Rs}\text{.0}\text{.77=Rs}\text{.77}\text{.77}$

(iv) \[\text{50 paise}\] 

Ans: Expressing the term as rupees,

$\text{50}\,\text{paise=Re}\text{.}\frac{\text{50}}{\text{100}}\text{=Re}\text{.0}\text{.50}$

(v) \[\text{235 paise}\] 

Ans: Expressing the term as rupees,

$\text{235}\,\text{paise=Re}\text{.}\frac{\text{235}}{\text{100}}\text{=Rs}\text{.2}\text{.35}$

3. Find

(i) Express $\text{5}$ cm in metre and kilometer.

Ans: Expressing $5$ cm in meter and kilometer,

$\because \,\,\text{100}\,\text{cm=1}\,\text{meter}$ 

$\therefore\,\,\text{1}\,\text{cm=}\frac{\text{1}}{\text{100}}\,\text{meter}\rightarrow \text{5}\,\text{cm=}\frac{\text{5}}{\text{100}}\text{=0}\text{.05}\,\text{meter}$ 

And, $\because \,\,\text{1000}\,\text{meters=1}\,\text{kilometers}$ 

$\therefore     \,\,\text{1}\,\text{meter=}\frac{\text{1}}{\text{1000}}\,\text{kilometers}$$\Rightarrow \text{0}\text{.05}\,\text{meter=}\frac{\text{0}\text{.05}}{\text{1000}}\text{=0}\text{.00005}\,\,\text{kilometer}$

(ii) Express $\text{35}$ mm in cm, m and km.

Ans: Expressing $35$ mm in cm, m and km.

$\because \,\,\text{10}\,\text{mm=1}\,\text{cm}$ 

$\therefore \,\,\text{1}\,\text{mm=}\frac{\text{1}}{\text{10}}\,\text{cm}\Rightarrow \text{35}\,\text{mm=}\frac{\text{35}}{\text{10}}\text{=3}\text{.5}\,\text{cm}$ 

And, $\because \,\,\text{100}\,\text{cm=1}\,\text{meter}$ 

$\therefore \,\,\,\text{1}\,\text{cm=}\frac{\text{1}}{\text{100}}\,\text{meter}\Rightarrow \text{3}\text{.5}\,\text{cm=}\frac{\text{3}\text{.5}}{\text{100}}\text{=0}\text{.035}\,\text{meter}$ 

Also, $\because \,\,\text{1000}\,\text{meters=1}\,\text{kilometers}$ 

$\therefore \,\,\text{1}\,\text{meter=}\frac{\text{1}}{\text{1000}}\,\text{kilometer}$ $\Rightarrow \,\,\text{0}\text{.035}\,\text{meter=}\frac{\text{0}\text{.035}}{\text{1000}}\text{=0}\text{.000035}\,\text{kilometer}$ 

4. Express in kg:

(i) $\text{200}\,\text{g}$

Ans: Converting from grams to kilograms,

$\text{200}\,\text{g=}\left( \text{200 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{1000}} \right)\,\text{kg=0}\text{.2}\,\text{kg}$ 

(ii) $\text{3470}\,\text{g}$ 

Ans: Converting from grams to kilograms,

$3470\,\text{g=}\left( \text{3470 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{1000}} \right)\,\text{kg=3}\text{.470}\,\text{kg}$

(iii) $\text{4}\,\text{kg}\,\text{8}\,\text{g}$ 

Ans: Converting from grams to kilograms,$\text{4}\,\text{kg}\,\text{8}\,\text{g=4}\,\text{kg}\,\text{+}\left( \text{8 }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{1000}} \right)\,\text{kg=4}\,\text{kg+0}\text{.008}\,\text{kg=4}\text{.008}\,\text{kg}$

5. Write the following decimal numbers in the expanded form:

(i) $\text{20}\text{.03}$ 

Ans: Converting the decimal number in expanded form, 

$20.03=2\times 10+0\times 1+0\times \frac{1}{10}+3\times \frac{1}{100}$ 

(ii) $\text{2}\text{.03}$ 

Ans: Converting the decimal number in expanded form, $2.03=2\times 1+0\times \frac{1}{10}+3\times \frac{1}{100}$

(iii) $\text{200}\text{.03}$ 

Ans: Converting the decimal number in expanded form, $200.03=2\times 100+0\times 10+0\times 1+0\times \frac{1}{10}+3\times \frac{1}{100}$

(iv) $\text{2}\text{.034}$ 

Ans: Converting the decimal number in expanded form, $2.034=2\times 1+0\times \frac{1}{10}+3\times \frac{1}{100}+4\times \frac{1}{1000}$

6. Write the place value of \[\text{2}\] in the following decimal numbers:

(i) $\text{2}\text{.56}$

Ans: The place value of $2$ in $2.56$ $=2\times 1=2\,$ones

(ii) $\text{21}\text{.37}$ 

Ans: The place value of $2$ in $21.37=2\times 10=2$ tens

(iii) $\text{10}\text{.25}$ 

Ans: The place value of $2$ in $10.25=2\times \frac{1}{10}=2$ tenths

(iv) $\text{9}\text{.42}$

Ans:  The place value of $2$ in $9.42=2\times \frac{1}{100}=2$ hundredth

(v) $\text{63}\text{.352}$ 

Ans: The place value of $2$ in $63.352=2\times \frac{1}{1000}=2$ thousandth

7. Dinesh went from place\[\text{ }\!\!~\!\!\text{ A}\]to place \[\text{B}\] and 

from there to place\[\text{C}\]. 

\[\text{A}\] is \[\text{7}\text{.5}\] km from \[\text{B}\] and \[\text{B}\] is \[\text{12}\text{.7}\] km from\[\text{C}\].  Ayub went from

 place \[\text{A}\] to place \[\text{D}\] and from there to place\[\text{C}\]. \[\text{D}\] is \[\text{9}\text{.3}\] km from

 \[\text{A}\]and \[\text{C}\] is \[\text{11}\text{.8}\] km from\[\text{D}\] . Who travelled more and by how much?

(Image will be uploaded soon)

Ans:

Given: The distance travelled by Dinesh when he went from 

place \[\text{A}\] to place \[\text{B = 7}\text{.5 km}\] and from 

place\[\text{B to C = 12}\text{.7 km}\]

(Image will be uploaded soon)

According to the figure,

The total distance covered by Dinesh \[\text{= AB + BC }\]

Substituting the values,

 \[\text{=7}\text{.5 + 12}\text{.7 = 20}\text{.2 km}\] 

The total distance covered by Ayub \[\text{= AD + DC }\]

Substituting the values, 

\[\text{=9}\text{.3 + 11}\text{.8 = 21}\text{.1 km}\]

Comparing the total distance covered by Ayub and Dinesh, 

\[\text{21}\text{.1 km  20}\text{.2 km}\]

Hence, Ayub covered \[\text{21}\text{.1 -- 20}\text{.2 = 0}\text{.9 km = 900m}\] more distance as compared to Dinesh.

8. Shyam bought \[\text{5 kg 300 g}\] apples and \[\text{3 kg 250 g}\] 

mangoes. Sarala bought \[\text{4 kg 800 g}\] oranges and \[\text{4 kg 150 g}\] 

bananas. Who bought more fruits?

Ans: Given: 

The total weight of fruits bought by Shyam\[\text{ = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g}\]

And the total weight of fruits bought by Sarala\[\text{= 4 kg 800 g + 4 kg 150 g = 8 kg 950 g}\] 

Comparing the quantity of fruits bought by Shyam and Sarala,

\[\text{8}\,\text{kg}\,\text{550}\,\text{g8}\,\text{kg}\,\text{950}\,\text{g}\] 

Observe that quantity of fruits bought by Sarala is greater.

Hence, Sarala bought more fruits then Shyam.

9. How much less is \[\text{28 km}\]then \[\text{42}\text{.6 km}\]?

Ans:

Given: The two distances are $\text{42}\text{.6}\,\text{km}\,\text{and}\,\text{28}\,\text{km}$

Finding the difference of $\text{42}\text{.6}\,\text{km}\,\text{and}\,\text{28}\,\text{km}$,

$\text{42}\text{.6-28}\text{.0=14}\text{.6}\,\text{km}$ 

Hence, $\text{14}\text{.6}\,\text{km}$ less is $\text{28}\,\text{km}$then 

$\text{42}\text{.6}\,\text{km}$.

Exercise 2.6

1. Find:

(i) $\text{0}\text{.2 }\!\!\times\!\!\text{ 6}$ 

Ans: Calculating the value,

\[0.2\times 6=1.2\]

(ii) $\text{8 }\!\!\times\!\!\text{ 4}\text{.6}$

Ans:  Calculating the value,

\[8\times 4.6=36.8\]

(iii) $\text{2}\text{.71 }\!\!\times\!\!\text{ 5}$ 

Ans: Calculating the value,

\[2.71\times 5=13.55\]

(iv) $\text{20}\text{.1 }\!\!\times\!\!\text{ 4}$ 

Ans: Calculating the value,

\[20.1\times 4=80.4\]

(v) $\text{0}\text{.05 }\!\!\times\!\!\text{ 7}$ 

Ans: Calculating the value,

\[0.05\times 7=0.35\]

(vi) $\text{211}\text{.02 }\!\!\times\!\!\text{ 4}$ 

Ans: Calculating the value,

\[211.02\times 4=844.08\]

(vii) $\text{2 }\!\!\times\!\!\text{ 0}\text{.86}$ 

Ans: Calculating the value,

\[2\times 0.86=1.72\]

2. Find the area of rectangle whose length is \[\text{5}\text{.7 cm}\] and 

breadth is \[\text{3 cm}\text{.}\]

Ans: Given: The \[\text{Length of rectangle = 5}\text{.7 cm and Breadth of 

rectangle = 3 cm}\] 

Applying the area of rectangle formula,

\[\text{Area of rectangle = Length x Breadth}\] 

\[\text{= 5}\text{.7 x 3 = 17}\text{.1 c}{{\text{m}}^{2}}\]  

Hence, the area of rectangle is $\text{17}\text{.1}\,\text{c}{{\text{m}}^{\text{2}}}$.

3. Find:

(i) \[\text{1}\text{.3 }\!\!\times\!\!\text{ 10}\] 

Ans: Calculating the value,

$1.3\times 10=13.0$

(ii) \[\text{36}\text{.8 }\!\!\times\!\!\text{ 10}\] 

Ans: Calculating the value,

$36.8\times 10=368.0$

(iii) \[\text{153}\text{.7 }\!\!\times\!\!\text{ 10}\] 

Ans: Calculating the value,

$153.7\times 10=1537.0$

(iv) \[\text{168}\text{.07 }\!\!\times\!\!\text{ 10}\]

Ans: Calculating the value,

$168.07\times 10=1680.7$

(v) \[\text{31}\text{.1 }\!\!\times\!\!\text{ 100}\] 

Ans: Calculating the value,

$31.1\times 100=3110.0$

(vi) \[\text{156}\text{.1 }\!\!\times\!\!\text{ 100}\] 

Ans: Calculating the value,

$156.1\times 100=15610.0$

(vii) \[\text{3}\text{.62 }\!\!\times\!\!\text{ 100}\] 

Ans: Calculating the value,

$3.62\times 100=362.0$

(viii) \[\text{43}\text{.07 }\!\!\times\!\!\text{ 100}\] 

Ans: Calculating the value,

$43.07\times 100=4307.0$

(ix) \[\text{0}\text{.5 }\!\!\times\!\!\text{ 10}\] 

Ans: Calculating the value,

$0.5\times 10=5.0$ 

(x) \[\text{0}\text{.08 }\!\!\times\!\!\text{ 10}\] 

Ans: Calculating the value,

$0.08\times 10=0.80$ 

(xi) \[\text{0}\text{.9 }\!\!\times\!\!\text{ 100}\]

Ans: Calculating the value,

$0.9\times 100=90.0$

(xii) \[\text{0}\text{.03 }\!\!\times\!\!\text{ 1000}\] 

Ans: Calculating the value,

$0.03\times 1000=30.0$ 

4. A two-wheeler covers a distance of\[\text{ }\!\!~\!\!\text{ 55}\text{.3 

km}\] 

in one litre of petrol. How much distance will it cover in \[\text{10 litres}\] of 

petrol?

Ans: Given: In one litre a two-wheeler covers a distance\[\text{ = 55}\text{.3 

km}\]

Since distance covered in one litre by a two-wheeler\[\text{ = 55}\text{.3 km}\]

\[\therefore \,\,\text{In 10 litrs, a two- wheeler covers a distance = 55}\text{.3 x 10 = 553}\text{.0 km}\] 

Hence, $553$ km distance will be covered by two-wheeler in $10$ litres of petrol.

5.  Find:

(i) $\text{2}\text{.5 }\!\!\times\!\!\text{ 0}\text{.3}$

Ans:  Calculating the value,

\[\text{2}\text{.5 x 0}\text{.3 = 0}\text{.75}\] 

(ii) $\text{0}\text{.1 }\!\!\times\!\!\text{ 51}\text{.7}$ 

Ans:  Calculating the value,

\[\text{0}\text{.1 x 51}\text{.7 = 5}\text{.17}\]

(iii) $\text{0}\text{.2 }\!\!\times\!\!\text{ 316}\text{.8}$ 

Ans: Calculating the value,

\[\text{0}\text{.2 x 316}\text{.8 = 63}\text{.36}\]

(iv) $\text{1}\text{.3 }\!\!\times\!\!\text{ 1}\text{.3}$ 

Ans: Calculating the value,

\[\text{1}\text{.3 x 3}\text{.1 = 4}\text{.03}\]

(v) $\text{0}\text{.5 }\!\!\times\!\!\text{ 0}\text{.05}$ 

Ans: Calculating the value,

\[\text{0}\text{.5 x 0}\text{.05 = 0}\text{.025}\]

(vi) $\text{11}\text{.2 }\!\!\times\!\!\text{ 0}\text{.15}$ 

Ans: Calculating the value,

\[\text{11}\text{.2 x 0}\text{.15 = 1}\text{.680 }\]

(vii) $\text{1}\text{.07 }\!\!\times\!\!\text{ 0}\text{.02}$ 

Ans: Calculating the value,

\[\text{1}\text{.07 x 0}\text{.02 = 0}\text{.0214}\]

(viii) $\text{10}\text{.05 }\!\!\times\!\!\text{ 1}\text{.05}$

Ans: Calculating the value,

\[\text{10}\text{.05 x 1}\text{.05 = 10}\text{.5525}\]

(ix) $\text{101}\text{.01 }\!\!\times\!\!\text{ 0}\text{.01}$ 

Ans: Calculating the value,

\[\text{101}\text{.01 x 0}\text{.01 = 1}\text{.0101}\]

(x) $\text{100}\text{.01 }\!\!\times\!\!\text{ 1}\text{.1}$

Ans: Calculating the value,

\[\text{100}\text{.01 x 1}\text{.1 = 110}\text{.11 }\]

Exercise 2.7

1. Find:

(i) \[\text{0}\text{.4  }\!\!\div\!\!\text{  2}\] 

Ans: Calculating the value,

\[0.4\div 2=\frac{4}{10}\times \frac{1}{2}=\frac{2}{10}=0.2\]

(ii) \[\text{0}\text{.35  }\!\!\div\!\!\text{  5}\] 

Ans: Calculating the value,

\[0.35\div 5=\frac{35}{100}\times \frac{1}{5}=\frac{7}{100}=0.07\]

(iii) \[\text{2}\text{.48  }\!\!\div\!\!\text{  4}\] 

Ans: Calculating the value,

\[2.48\div 4=\frac{248}{100}\times \frac{1}{4}=\frac{62}{100}=0.62\]

(iv) \[\text{65}\text{.4  }\!\!\div\!\!\text{  6}\] 

Ans: Calculating the value,

\[65.4\div 6=\frac{654}{10}\times \frac{1}{6}=\frac{109}{10}=10.9\]

(v) \[\text{651}\text{.2  }\!\!\div\!\!\text{  4}\] 

Ans: Calculating the value,

\[651.2\div 4=\frac{6512}{10}\times \frac{1}{4}=\frac{1628}{10}=162.8\]

(vi) \[\text{14}\text{.49  }\!\!\div\!\!\text{  7 }\] 

Ans: Calculating the value,

\[14.49\div 7=\frac{1449}{100}\times \frac{1}{7}=\frac{207}{100}=2.07\]

(vii) \[\text{3}\text{.96  }\!\!\div\!\!\text{  4}\]

Ans: Calculating the value,

\[3.96\div 4=\frac{396}{100}\times \frac{1}{4}=\frac{99}{100}=0.99\]

(viii) \[\text{0}\text{.80  }\!\!\div\!\!\text{  5}\] 

Ans: Calculating the value,

\[0.80\div 5=\frac{80}{100}\times \frac{1}{5}=\frac{16}{100}=0.16\]

2. Find:

(i) \[\text{4}\text{.8  }\!\!\div\!\!\text{  10}\] 

Ans: Performing the given calculation,

$4.8\div 10=\frac{4.8}{10}=0.48$

(ii) \[\text{52}\text{.5  }\!\!\div\!\!\text{  10}\] 

Ans: Performing the given calculation,

$52.5\div 10=\frac{52.5}{10}=5.25$

(iii) \[\text{0}\text{.7  }\!\!\div\!\!\text{  10}\] 

Ans: Performing the given calculation,

$0.7\div 10=\frac{0.7}{10}=0.07$

(iv) \[\text{33}\text{.1  }\!\!\div\!\!\text{  10}\]

Ans: Performing the given calculation,

$33.1\div 10=\frac{33.1}{10}=3.31$

(v) \[\text{272}\text{.23  }\!\!\div\!\!\text{  10}\] 

Ans: Performing the given calculation,

$272.23\div 10=\frac{272.23}{10}=27.223$

(vi) \[\text{0}\text{.56  }\!\!\div\!\!\text{  10 }\] 

Ans: Performing the given calculation,

$0.56\div 10=\frac{0.56}{10}=0.056$

(vii) \[\text{3}\text{.97  }\!\!\div\!\!\text{  10}\] 

Ans: Performing the given calculation,

$3.97\div 10=\frac{3.97}{10}=0.397$

3. Find:

(i) \[\text{2}\text{.7  }\!\!\div\!\!\text{  100}\] 

Ans: Converting the terms in fraction form and calculating the value,

$2.7\div 100=\frac{27}{10}\times \frac{1}{100}=\frac{27}{1000}=0.027$ 

(ii) \[\text{0}\text{.3  }\!\!\div\!\!\text{  100 }\]

Ans: Converting the terms in fraction form and calculating the value,

$0.3\div 100=\frac{3}{10}\times \frac{1}{100}=\frac{3}{1000}=0.003$

(iii) \[\text{0}\text{.78  }\!\!\div\!\!\text{  100}\] 

Ans: Converting the terms in fraction form and calculating the value,

$0.78\div 100=\frac{78}{10}\times \frac{1}{100}=\frac{78}{1000}=0.0078$

(iv) \[\text{432}\text{.6  }\!\!\div\!\!\text{  100}\] 

Ans: Converting the terms in fraction form and calculating the value,

$432.6\div 100=\frac{4326}{10}\times \frac{1}{100}=\frac{4326}{1000}=4.326$

(v) \[\text{23}\text{.6  }\!\!\div\!\!\text{  100}\] 

Ans: Converting the terms in fraction form and calculating the value,$23.6\div 100=\frac{236}{10}\times \frac{1}{100}=\frac{236}{1000}=0.236$

(vi) \[\text{98}\text{.53  }\!\!\div\!\!\text{  100}\] 

Ans: Converting the terms in fraction form and calculating the value,

$98.53\div 100=\frac{9853}{10}\times \frac{1}{100}=\frac{9853}{1000}=0.9853$

4. Find:

(i) \[\text{7}\text{.9  }\!\!\div\!\!\text{  1000}\] 

Ans: Converting the terms in fraction form and calculating the value,

$7.9\div 1000=\frac{79}{10}\times \frac{1}{1000}=\frac{79}{10000}=0.0079$ 

(ii) \[\text{26}\text{.3  }\!\!\div\!\!\text{  1000}\]

Ans: Converting the terms in fraction form and calculating the value,

$26.3\div 1000=\frac{263}{10}\times \frac{1}{1000}=\frac{263}{10000}=0.0263$

(iii) \[\text{38}\text{.53  }\!\!\div\!\!\text{  1000}\] 

Ans: Converting the terms in fraction form and calculating the value,

$38.53\div 1000=\frac{3853}{10}\times \frac{1}{1000}=\frac{3853}{10000}=0.03853$

(iv) \[\text{128}\text{.9  }\!\!\div\!\!\text{  1000}\] 

Ans: Converting the terms in fraction form and calculating the value,

$128.9\div 1000=\frac{1289}{10}\times \frac{1}{1000}=\frac{1289}{10000}=0.1289$

(v) \[\text{0}\text{.5  }\!\!\div\!\!\text{  1000}\] 

Ans: Converting the terms in fraction form and calculating the value,

$0.5\div 1000=\frac{5}{10}\times \frac{1}{1000}=\frac{5}{10000}=0.0005$

5. Find:

(i) \[\text{7  }\!\!\div\!\!\text{  3}\text{.5}\] 

Ans: Converting the terms in fraction form and calculating the value,

$7\div 3.5=7\div \frac{35}{10}=7\times \frac{10}{35}=\frac{10}{5}=2$

(ii) \[\text{36  }\!\!\div\!\!\text{  0}\text{.2 }\]

Ans: Converting the terms in fraction form and calculating the value,

$36\div 0.2=36\div \frac{2}{10}=36\times \frac{10}{2}=18\times 10=180$

(iii) \[\text{3}\text{.25  }\!\!\div\!\!\text{  0}\text{.5}\]  

Ans: Converting the terms in fraction form and calculating the value,$3.25\div 0.5=\frac{325}{100}\div \frac{5}{10}=\frac{325}{100}\times \frac{10}{5}=\frac{65}{10}=6.5$

(iv) \[\text{30}\text{.94  }\!\!\div\!\!\text{  0}\text{.7}\]

Ans: Converting the terms in fraction form and calculating the value,

$30.94\div 0.7=\frac{3094}{100}\div \frac{7}{10}=\frac{3094}{100}\times \frac{10}{7}=\frac{442}{10}=44.2$

(v) \[\text{0}\text{.5  }\!\!\div\!\!\text{  0}\text{.25 }\] 

Ans: Converting the terms in fraction form and calculating the value,$0.5\div 0.25=\frac{5}{10}\div \frac{25}{100}=\frac{5}{10}\times \frac{100}{25}=\frac{10}{5}=2$

(vi) \[\text{7}\text{.75  }\!\!\div\!\!\text{  0}\text{.25}\] 

Ans: Converting the terms in fraction form and calculating the value,

$7.75\div 0.25=\frac{775}{100}\div \frac{25}{100}=\frac{775}{100}\times \frac{100}{25}=31$

(vii) \[\text{76}\text{.5  }\!\!\div\!\!\text{  0}\text{.15}\] 

Ans: Converting the terms in fraction form and calculating the value,

$76.5\div 0.15=\frac{765}{100}\div \frac{15}{100}=\frac{765}{10}\times \frac{100}{15}=51\times 10=510$

(viii) \[\text{37}\text{.8  }\!\!\div\!\!\text{  1}\text{.4}\]

Ans: Converting the terms in fraction form and calculating the value,

$37.8\div 1.4=\frac{378}{10}\div \frac{14}{10}=\frac{378}{10}\times \frac{10}{14}=27$

(ix) \[\text{2}\text{.73  }\!\!\div\!\!\text{  1}\text{.3 }\] 

Ans: Converting the terms in fraction form and calculating the value,

$2.73\div 1.3=\frac{273}{100}\div \frac{13}{10}=\frac{273}{100}\times \frac{10}{13}=\frac{21}{10}=2.1$

6. A vehicle covers a distance of \[\text{43}\text{.2 km}\] in

\[\text{2}\text{.4}\]litres of petrol. How much distance will it cover in one 

litre 

petrol?

Ans: Given:\[\,\,\,\text{In 2}\text{.4 litres of petrol, distance covered by the vehicle = 43}\text{.2 km}\]

Since,\[\,\,\,\text{In 2}\text{.4 litres of petrol, distance covered by the vehicle = 43}\text{.2 km}\]

\[\therefore \,\,\text{In 1 litre of petrol, distance covered by the vehicle = 43}\text{.2  }\!\!\div\!\!\text{  2}\text{.4}\] 

Performing the required calculations,

$=\frac{432}{10}\div \frac{24}{10}=\frac{432}{10}\times \frac{24}{10}$ 

$\text{=18}\,\text{km}$ 

Hence, the vehicle can cover \[\text{18 km}\] distance in one litre of petrol.

NCERT Solutions for Class 7 Chapter 2 Maths PDF download

It is the best choice to download NCERT Solutions for Class 7 Maths Chapter 2 PDF available on Vedantu. Students can find all the solutions for solving problems in class 7 at their convenience. Several experts gave their best in preparing these solutions to find answers and compiled them all in our NCERT Solutions for Class 7 Maths Chapter 2 for all students to understand the concepts.

Key Concepts Covered in NCERT Solutions for CBSE Class 7 Maths Chapter 2 Fractions and Decimals

Some important concepts discussed in Chapter 2 Fractions and Decimals of NCERT Solutions Class 7 Maths are:

• Addition and Subtraction of Fractions.

• Multiplication of Fractions.

• Multiplication of a Fraction by a Whole Number.

• Multiplication of a Fraction by a Fraction.

• Division of Fraction.

• Division of Whole Number by a Fraction.

• Reciprocal of Fraction.

• Division of a Fraction by a Whole Number.

• Division of Fraction by Another Fraction.

• Multiplication of Decimal Numbers.

• Multiplication of Decimal Numbers by 10, 100 and 1000.

• Division of Decimal Numbers.

• Division of Decimals by 10, 100 and 1000.

• Division of a Decimal Number by a Whole Number.

• Division of a Decimal Number by Another Decimal Number.

NCERT Solutions for Class 7 Chapter 2 Maths PDF Download

It is the best choice to download NCERT Solutions for Class 7 Maths Chapter 2 PDF available on Vedantu. Students can find all the solutions for solving the sums of this chapter at their convenience. Several experts gave their best in preparing these solutions to find answers and compiled them all in our NCERT Solutions for Class 7 Maths Chapter 2 for all students to understand the concepts. 

2.1 Introduction

In NCERT Solutions Class 7 Chapter 2 Maths, students will learn about fractions and decimals. In junior classes, students have learned about what is a fraction and its types: proper, improper, mixed fractions, etc. Now, in class 7, we are going to learn about multiplication and division of fractions. The concept of fractions mainly focuses on the ratios and proportions, how to distribute etc. At the same time, decimals are the accurate values obtained after the division.

2.2 Recollect

In NCERT Solutions Class 7 Maths Chapter 2, students need to think again on the topics they have learned so far in the previous classes. These include representation of fractions on the number line, ordering of fractions, addition and subtraction of fractions, decimals and their additions, how to keep a point, etc. These are reminded in the first two exercises.

2.3 Multiplication of Fractions

In this section, students can understand how to multiply two fractions. If students have values like a and b, they can say ab is the product of a and b. If the values are like p/q, a/b then, how can we multiply? To multiply these fractions, it has two different methods. One is by using a  whole number and the other is by using a portion.

2.3.1 Multiplication of Fractions Using the Whole Number

Here, let us see what Fraction tells us? It explains that a down part is a whole number (except zero) and the upper part is the integer. In a fraction, the down part is known as the denominator whereas the upper part is the numerator. We use a whole number to multiply fractions if they are the same. For instance, let's say we have p/q. Then we can multiply with the whole number as 3*p/q. It is also applicable for improper or mixed fractions. But students need to make them into simpler forms before multiplying.

2.3.1 Multiplication of Fractions Using the Fraction

In this section, students can learn how to multiply two fractions when they are dissimilar. Students use a fraction to multiply them. The formula for multiplying two fractions is,(product of numerators)/(product of denominators).

The resultant product is less than the two fractions if we multiply two proper fractions. On the other hand, the result is greater than the two fractions if we multiply two improper fractions.

2.4 Division of Fractions

Let's discuss the division of fractions. Students can divide a fraction by a whole number and a whole number by a fraction. Here is a particular case to keep in mind. If two fractions for which numerator and denominator are in reverse order, then they are called reciprocals to each other.  Their product is always 1.

In the same way, while dividing mixed fractions with a whole number, students need to change the mixed fraction into improper fractions. Then it is easy to divide and solve. Next, we have to learn to divide a fraction with another fraction by changing one of the fractions into its reciprocal form.

The three concepts are explained differently in the NCERT Solutions for Class 7 Maths Chapter 2 PDF book available on Vedantu for students to go through if necessary.

2.5 Recalling Decimals

Decimals are the proper forms to represent the results obtained from multiplication and division. Placing the point in between numbers plays a vital role. One can express the heights, distances, weights,  measuring values, interest rates, shares, and fractions, also using decimals. To change the place value of the point, we can multiply by 10,100,...... Let's have a glance at the addition and subtraction of decimals.

2.6 Multiplication of Decimals

In this section, students are going to practice the multiplication of decimals. Even though multiplication is easy, doubts might arise in students' minds while keeping a point. For this, we need to count the number of values after a decimal point in both the numbers and then keep the point before that number of places in the result. It plays a crucial role here. Another variation of Multiplication of Decimals is changing the place value of a decimal point by multiplying it with 10 multiples. It was already discussed by us earlier.

2.7 Division of Decimals

In this section, students will learn how to divide decimals and how many variations it has?  Students can understand the first one, which is explained in 2.7.1 in the PDF. Here let's divide the whole number with 10 multiples; it gives decimals. Similarly, let's divide decimals, we will get whole numbers. Next is the division of decimals with a whole number. Here the place value of the decimal point doesn't change in the result also. 

Students can refer to 2.7.2 in the PDF for further information. Finally,  2.7.3 contains the topic of the division of a decimal with other decimals. Here, students need to replace the decimal point to the right side with the same number of places in both. Then students can divide easily as it becomes the whole number.

NCERT Solutions for Class 7 Maths Chapter 2 Exercises

NCERT Solutions for Class 7 Maths

• Chapter 1 - Integers

• Chapter 3 - Data Handling

• Chapter 4 - Simple Equations

• Chapter 5 - Lines and Angles

• Chapter 6 - The Triangle and Its Properties

• Chapter 7 - Congruence of Triangles

• Chapter 8 - Comparing Quantities

• Chapter 9 - Rational Numbers

• Chapter 10 - Practical Geometry

• Chapter 11 - Perimeter and Area

• Chapter 12 - Algebraic Expressions

• Chapter 13 - Exponents and Powers

• Chapter 14 - Symmetry

• Chapter 15 - Visualizing Solid Shapes

Students can also refer to the following study material for Chapter 2 of Class 7 Maths:

• Chapter 2 Fractions and Decimals: Important Questions

• Chapter 2 Fractions and Decimals: Revision Notes

• Chapter 2 Fractions and Decimals: Formulas

• Chapter 2 Fractions and Decimals: RD Sharma Solutions

• Chapter 2 Fractions and Decimals: NCERT Exemplar Solutions

• Chapter 2 Fractions and Decimals: RS Aggarwal Solutions

Key Features of NCERT Solutions for Class 7 Maths Chapter 2

Practice makes a man perfect. It is perfectly apt for Mathematics. As much as students practice several problems, students will become experts and can solve problems easily and quickly. It helps to improve a student's thinking ability also. It also makes students score cent percent. NCERT Solutions for Class 7th Maths Chapter 2 Fractions and Decimals on Vedantu are an add-on for students' practice and goals. Students can choose the NCERT Solutions for Class 7 Maths for various reasons like:

• It has several examples with answers in detail, which helps students to solve smartly.

• An excellent explanation is available for every concept separately.

• It builds confidence to attempt competitive exams at the national level.

• Highly qualified trainers available online to prepare the PDFs.

• Students should clear their doubts through live chats with instructors.

• They provide the questions in the exam pattern so students can learn how to present in the exam also.

Conclusion

The first few exercises in the Class 7 Maths Chapter 2 Fractions and Decimals explain the addition and subtraction of fractions and decimals, as well as a review of fractions and decimals concepts studied in previous classes with appropriate examples.

The sample problems in class 7 NCERT solutions chapter 2 fractions and decimals are sufficient for students to gain a thorough understanding of applying arithmetic operations to fractions and decimals.

So, to score good marks in Maths easily, solve these NCERT Solutions  for CBSE Class 7 Maths Chapter 2 Fractions and Decimals regularly. It will help you understand all the concepts and you’ll be able to solve all the questions on your own. But, along with the NCERT Solutions do not forget to solve the sample papers and previous year's question papers.