NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals (EX 2.1) Exercise 2.1

Q.1

\begin{array}{l}\text{Solve}:\\ \text{(i) 2-}\frac{3}{5}\text{(ii) 4+}\frac{7}{8}\text{(iii)}\frac{3}{5}+\frac{2}{7}\text{(iv)}\frac{9}{11}-\frac{4}{15}\\ \text{(v)}\frac{7}{10}\text{+}\frac{2}{5}\text{+}\frac{3}{2}\text{(vi) 2}\frac{2}{3}+3\frac{1}{2}\text{(vii) 8}\frac{1}{2}-3\frac{5}{8}\end{array}

Ans.

\begin{array}{l}(i)2-\frac{3}{5}=\frac{10-3}{5}=\boxed{\frac{7}{5}}\\ (ii)4+\frac{7}{8}=\frac{32+7}{8}=\frac{39}{8}=\boxed{4\frac{7}{8}}\\ (iii)\frac{3}{5}+\frac{2}{7}=\frac{21+10}{35}=\boxed{\frac{31}{35}}\\ (iv)\frac{9}{11}-\frac{4}{15}=\frac{135+44}{165}=\frac{179}{165}=\boxed{1\frac{14}{165}}\\ (v)\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)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)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}\end{array}

Q.2

$\begin{array}{l}\mathrm{Arrange}\mathrm{the}\mathrm{following}\mathrm{in}\mathrm{descending}\mathrm{order}.\\ \left(\mathrm{i}\right)\frac{2}{9},\frac{2}{3},\frac{8}{21}\left(\mathrm{ii}\right)\frac{1}{5},\frac{3}{7},\frac{7}{10}\end{array}$

Ans.

\begin{array}{l}\text{First},\text{we find the LCM of 9},\text{3 and 21}.\\ \text{LCM of 9},\text{3 and 21}=\text{63}\\ =\frac{2\times 7}{9\times 7},\frac{2\times 21}{3\times 21},\frac{8\times 3}{21\times 3}\\ =\frac{14}{63},\frac{42}{63},\frac{24}{63}\\ \text{Since},\text{42}>\text{24}>\text{14}.\\ \text{Therefore},\\ \frac{42}{63}>\frac{24}{63}>\frac{14}{63}\\ \Rightarrow \boxed{\frac{2}{3}>\frac{8}{21}>\frac{2}{9}}\\ (ii)\frac{1}{5},\frac{3}{7},\frac{7}{10}\\ \text{First},\text{find the LCM of 5},\text{7 and 1}0\\ \text{LCM of denominators 5},\text{7 and 1}0\text{is 7}0.\\ \text{So},\\ \frac{1}{5},\frac{3}{7},\frac{7}{10}\\ \Rightarrow \frac{1\times 14}{5\times 14},\frac{3\times 10}{7\times 10},\frac{7\times 7}{10\times 7}\\ \Rightarrow \frac{14}{70},\frac{30}{70},\frac{49}{70}\\ \text{Since},\text{49}>\text{3}0>\text{14}.\\ \text{Therefore},\\ \frac{49}{70}>\frac{30}{70}>\frac{14}{70}\\ \Rightarrow \boxed{\frac{7}{10}>\frac{3}{7}>\frac{1}{5}}\end{array}

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

$\begin{array}{ccc}\frac{4}{11}& \frac{9}{11}& \frac{2}{11}\\ \frac{3}{11}& \frac{5}{11}& \frac{7}{11}\\ \frac{8}{11}& \frac{1}{11}& \frac{6}{11}\end{array}$

Ans.

\begin{array}{l}\text{Finding the sum of first row}:\\ Sumof1strow=\frac{4}{11}+\frac{9}{11}+\frac{2}{11}\\ =\frac{4+9+2}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of second row}:\\ \text{Sum of 2nd row}=\frac{3}{11}+\frac{5}{11}+\frac{7}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of third row}:\\ \text{Sum of 3rd row}=\frac{8}{11}+\frac{1}{11}+\frac{6}{11}\\ =\frac{8+1+6}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of first column}:\\ \text{Sum of 1st column}=\frac{4}{11}+\frac{3}{11}+\frac{8}{11}\\ =\frac{4+3+8}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of second column}:\\ \text{Sum of 2nd column}=\frac{9}{11}+\frac{5}{11}+\frac{1}{11}\\ =\frac{9+5+1}{11}\\ =\frac{15}{11}\\ \text{Finding the sum of third column}:\\ \text{Sum of 3rd column}=\frac{2}{11}+\frac{7}{11}+\frac{6}{11}\\ =\frac{2+7+6}{11}\\ =\frac{15}{11}\\ \text{Now},\text{we must find diagonal sum}.\\ \text{Sum of}diagonalfromleft\text{}bottomtorighttop\\ =\frac{8}{11}+\frac{5}{11}+\frac{2}{11}\\ =\frac{8+5+2}{11}\\ =\frac{15}{11}\\ \text{Sum of}diagonalfromleft\text{}bottomtorighttop\\ =\frac{6}{11}+\frac{5}{11}+\frac{4}{11}\\ =\frac{6+5+4}{11}\\ =\frac{15}{11}\\ \text{Since the sum of the numbers in each row},\text{in each}\\ \text{column along the diagonal is the same},\text{which is equal}\\ \text{to}\frac{15}{11}.\\ \text{Therefore},\text{given square is magical}.\end{array}

Q.4

$\begin{array}{l}\mathrm{A}\mathrm{rectangular}\mathrm{sheet}\mathrm{of}\mathrm{paper}\mathrm{is}12\frac{1}{2}\mathrm{cm}\mathrm{long}\mathrm{and}\\ 10\frac{2}{3}\mathrm{cm}\mathrm{wide}.\mathrm{Find}\mathrm{its}\mathrm{perimeter}.\end{array}$

Ans.

\begin{array}{l}\text{We are given}:\\ \text{Length}\text{of}\text{rectangular}\text{sheet}\text{of}\text{paper}=12\frac{1}{2}\text{cm}=\frac{25}{2}\text{cm}\\ \text{Width}\text{of}\text{rectangular}\text{sheet}\text{of}\text{paper=10}\frac{2}{3}\text{cm}=\frac{32}{3}\text{cm}\\ \text{Since},\boxed{\text{Perimeter of a rectangle}=\text{2}\times \left(\text{length}+\text{width}\right)}\\ \text{Therefore},\\ \text{Perimeter of the given rectangular sheet of paper}\\ =2\times \left(\frac{25}{2}+\frac{32}{3}\right)\text{cm}\\ =2\times \left(\frac{75+64}{6}\right)\text{cm}\\ =2\times \frac{139}{6}\text{cm}\\ =\frac{139}{3}\text{cm}=\boxed{46\frac{1}{3}\text{cm}}\end{array}

Q.5

$\begin{array}{l}\mathrm{Find}\mathrm{}\mathrm{the}\mathrm{}\mathrm{perimeters}\mathrm{}\mathrm{of}\mathrm{}\left(\mathrm{i}\right)\mathrm{\Delta ABE}\mathrm{}\left(\mathrm{ii}\right)\mathrm{}\mathrm{the}\mathrm{}\mathrm{rectangle}\\ \mathrm{BCDE}\mathrm{}\mathrm{in}\mathrm{}\mathrm{this}\mathrm{}\mathrm{figure}.\mathrm{}\mathrm{Whose}\mathrm{}\mathrm{perimeter}\mathrm{}\mathrm{is}\mathrm{}\mathrm{greater}?\end{array}$

Ans.

\begin{array}{l}\text{Since},\text{the Perimeter of a triangle}=\text{side}+\text{side}+\text{side}\\ \text{Therefore},\text{Perimeter of triangle ABE}=\text{AB}+\text{BE}+\text{AE}\\ =\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5}\text{cm}\\ =\frac{5}{2}+\frac{11}{4}+\frac{18}{5}\text{cm}\\ =\frac{50+55+72}{20}\text{cm}\\ =\frac{177}{20}\text{cm}\\ \text{In rectangle BCDE},\text{length BE}=2\frac{2}{4}\text{cm}=\frac{11}{4}\text{cm}\\ \text{Width ED}=\frac{7}{6}\text{cm}\\ \text{Since},\boxed{\text{Perimeter of a rectangle}=\text{2}\times \left(\text{length}+\text{width}\right)}\\ \text{Therefore},\text{perimeter of rectangle BCDE}=2\times \left(BE+ED\right)\\ =2\times \left(\frac{11}{4}+\frac{7}{6}\right)\text{cm}\\ =2\times \left(\frac{33+14}{12}\right)\text{cm}\\ =2\times \frac{47}{12}\text{cm}\\ =\frac{47}{6}\text{cm}\\ \text{Hence},\text{the perimeter of}\Delta ABC=\boxed{\frac{177}{20}\text{cm}}\\ \text{and}\\ \text{The perimeter of rectangle BCDE}=\boxed{\frac{47}{6}\text{cm}}\\ \text{Now},\text{comparing both the perimeter to find which is}\\ \text{greater}.\\ \text{Since},\\ \text{The LCM of denominators of both the perimeter}\left(\text{2}0\text{and 6}\right)\text{is}\\ \text{equal to 6}0.\\ \text{Therefore},\\ \frac{177}{20},\frac{47}{6}\Rightarrow \frac{177\times 3}{20\times 3},\frac{47\times 10}{6\times 10}\Rightarrow \frac{531}{60},\frac{470}{10}\\ \text{Since},\text{531 is greater than 47}0.\\ \text{Therefore},\\ \frac{531}{60}>\frac{470}{60}\Rightarrow \frac{177}{20}>\frac{47}{6}\\ \text{Thus},\text{the perimeter of triangle ABE is greater than}\\ \text{the perimeter of rectangle BCDE}.\end{array}

Q.6

$\begin{array}{l}\mathrm{Sali}\mathrm{}\mathrm{wants}\mathrm{}\mathrm{to}\mathrm{}\mathrm{put}\mathrm{}\mathrm{a}\mathrm{}\mathrm{picture}\mathrm{}\mathrm{in}\mathrm{}\mathrm{a}\mathrm{}\mathrm{frame}.\mathrm{}\mathrm{The}\mathrm{}\mathrm{picture}\mathrm{}\mathrm{is}\mathrm{\hspace{0.17em}}7\frac{3}{5}\mathrm{cm}\mathrm{\hspace{0.17em}}\mathrm{wide}.\mathrm{}\\ \mathrm{To}\mathrm{}\mathrm{fit}\mathrm{}\mathrm{in}\mathrm{}\mathrm{the}\mathrm{}\mathrm{frame}\mathrm{}\mathrm{the}\mathrm{}\mathrm{picture}\mathrm{}\mathrm{cannot}\mathrm{}\mathrm{be}\mathrm{}\mathrm{more}\mathrm{}\mathrm{than}\mathrm{\hspace{0.17em}}7\frac{3}{10}\mathrm{cm}\mathrm{wide}.\mathrm{}\\ \mathrm{How}\mathrm{}\mathrm{much}\mathrm{}\mathrm{should}\mathrm{}\mathrm{the}\mathrm{}\mathrm{picture}\mathrm{}\mathrm{be}\mathrm{}\mathrm{trimmed}?\end{array}$

Ans.

\begin{array}{l}\text{We are given}:\\ \text{The}\text{width}\text{of}\text{the}\text{picture=7}\frac{\text{3}}{\text{5}}\text{cm}\\ \text{Required}\text{width}\text{of picture to be fit in frame=7}\frac{3}{10}\text{cm}\\ \text{Therefore},\text{picture to be trim}=7\frac{3}{5}-7\frac{3}{10}\text{cm}\\ =\frac{38}{5}-\frac{73}{10}\text{cm}\\ =\frac{3}{10}\text{cm}\\ \text{Thus picture should be trimmed by}\boxed{\frac{3}{10}\text{cm}}\end{array}

Q.7

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

Ans.

$\begin{array}{l}\text{Since},\text{Ritu ate\hspace{0.17em}}\frac{3}{5}\text{\hspace{0.17em}part of an apple and her brother eaten the}\\ \text{rest part of the apple}.\\ \text{So},\text{the part of apple left after eaten by Ritu}=1-\frac{3}{5}\\ =\frac{5-3}{5}\\ =\frac{2}{5}\\ \text{Therefore},\text{Somu ate\hspace{0.17em}}\frac{2}{5}\text{\hspace{0.17em}parts of apple}.\\ \text{Now comparing the part eaten by them\hspace{0.17em}}\frac{3}{5},\text{\hspace{0.17em}}\frac{2}{5}\\ \text{Since},\text{3}>\text{2}.\\ \text{Therefore}\\ \frac{3}{5}>\frac{2}{5}\\ \text{That means Ritu ate larger part}.\\ \text{The difference in both parts}=\frac{3}{5}-\frac{2}{5}\\ =\frac{3-2}{5}\\ =\frac{1}{5}\\ \text{Thus,}\\ \text{Somu ate}\frac{2}{5}\text{part of apple}.\\ \text{Ritu ate the larger part of apple}\\ \text{Ritu ate}\text{part more apple than her brother Somu}.\end{array}$

Q.8

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

Ans.

$$\begin{gathered} \begin{array}{l}\text{We are given}:\\ \text{Michael\hspace{0.17em}worked\hspace{0.17em}for\hspace{0.17em}}\frac{\text{7}}{\text{12}}\text{hour}\\ \mathrm{Vaibhav}\mathrm{worked}\mathrm{for}\frac{3}{4}\mathrm{hour} \\ \text{Therefore},\\ \text{In order to find the longer hour of work we have to compare}\\ \text{both the fractions\hspace{0.17em}}\frac{7}{12},\frac{3}{4}\\ \text{The LCM of 12 and 4}=\text{12}\\ \mathrm{So},\frac{7}{12},\frac{3}{4}=\frac{7}{12},\frac{9}{12}\\ \text{Therefore},\frac{9}{12}>\frac{7}{12}\\ \text{That means Vaibhav worked for longer hour}.\\ \text{Difference in their working hour\hspace{0.17em}\hspace{0.17em}}=\frac{9}{12}-\frac{7}{12}\\ =\frac{9-7}{12}\\ =\frac{2}{12}\\ =\boxed{\frac{1}{6}}\\ \text{Thus},\text{Vaibhav worked for}\frac{1}{6}\text{\hspace{0.17em}hour more than Michael}.\end{array} \end{gathered}$$