NCERT Solutions Class 7 Mathematics Chapter 2

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.4

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

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.5

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

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.6

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

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.7

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

Ans.

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

Q.8

$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{drawing}\left(\mathrm{a}\right)\mathrm{to}\left(\mathrm{d}\right)\mathrm{show}:\\ \left(\mathrm{i}\right)2\times \frac{1}{5}\left(\mathrm{ii}\right)2\times \frac{1}{2}\\ \left(\mathrm{iii}\right)3\times \frac{2}{3}\left(\mathrm{iv}\right)3\times \frac{1}{4}\end{array}$

Ans.

$\begin{array}{l}\text{(i)\hspace{0.17em}}2\times \frac{1}{5}=\frac{2}{5}\\ \left(\mathrm{ii}\right)2\times \frac{1}{2}=1\\ \left(\mathrm{iii}\right)3\times \frac{2}{3}=2\\ \left(\mathrm{iv}\right)3\times \frac{1}{4}=\frac{3}{4}\\ \text{(a)\hspace{0.17em}Here},\text{there are three pictures and each pictures}\\ \text{show\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}part}=\frac{2}{3}+\frac{2}{3}+\frac{2}{3}\\ =\frac{2+2+2}{3}\\ =\frac{6}{3}=2\\ \text{Hence},\left(\text{iii}\right)\text{denotes picture}\left(\text{a}\right).\\ \text{(b)\hspace{0.17em}Here},\text{there are two pictures}.\text{Each pictures show}\frac{1}{2}\\ \text{part.}\\ \text{So},\frac{1}{2}+\frac{1}{2}=\frac{1+1}{2}=\frac{2}{2}=1\\ \text{Hence},\left(\text{ii}\right)\text{denotes picture}\left(\text{b}\right).\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}Here},\text{there are three pictures}.\text{Each pictures show}\frac{1}{4}\mathrm{part}.\\ \text{Therefore},\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{1+1+1}{4}=\frac{3}{4}\\ \text{Hence},\left(\text{iv}\right)\text{denotes picture}\left(\text{c}\right).\\ \left(\mathrm{d}\right)\text{\hspace{0.17em}Here},\text{there are two pictures and each pictures denotes}\\ \frac{1}{5}\text{\hspace{0.17em}part.}\\ \text{Therefore},\frac{1}{5}+\frac{1}{5}=\frac{1+1}{5}=\frac{2}{5}\\ \text{Hence},\left(\text{i}\right)\text{denotes picture}\left(\text{d}\right).\\ \text{Thus}:\\ \boxed{\begin{array}{l}\left(\text{i}\right)\text{denotes picture}\left(\text{d}\right)\\ \left(\text{ii}\right)\text{denotes picture}\left(\text{b}\right)\\ \left(\text{iii}\right)\text{denotes picture}\left(\text{a}\right)\\ \left(\text{iv}\right)\text{denotes picture}\left(\text{c}\right)\end{array}}\end{array}$

Q.9

$\begin{array}{l}\text{Some pictures}\left(\text{a}\right)\text{to}\left(\text{c}\right)\text{are given below.}\\ \text{Tell which of them show}\\ \left(\mathrm{i}\right)3\times \frac{1}{5}=\frac{3}{5}\left(\mathrm{ii}\right)2\times \frac{1}{3}=\frac{2}{3}\left(\mathrm{iii}\right)3\times \frac{3}{4}=2\frac{1}{4}\end{array}$

Ans.

\begin{array}{l}\text{In picture}\left(\text{a}\right):\\ \text{There are two pictures in left hand side},\text{each denotes}\frac{1}{3}\\ \text{and picture in right hand side denotes}\frac{2}{3}.\\ \text{Therefore},\text{it denotes}\frac{1}{3}\times 2=\frac{2}{3}\text{which denotes}\left(\text{ii}\right).\\ \text{In picture}\left(\text{b}\right)\text{:}\\ \text{Here are three pictures in left hand side},\text{each shows}\frac{3}{4}.\\ \text{There are three pictures in right hand side in which each}\\ \text{of first two denotes 1}.\\ \text{Therefore},\frac{3}{4}\times 3=2\frac{1}{4}\\ \text{Thus picture}\left(\text{b}\right)\text{denotes}\left(\text{iii}\right).\\ \text{In picture}\left(\text{c}\right):\\ \text{There are three pictures in left hand side and each}\\ \text{of them denotes}\frac{1}{5}.\\ \text{Hence},\\ \boxed{\begin{array}{l}\left(\text{i}\right)\text{denotes picture}\left(\text{c}\right)\\ \left(\text{ii}\right)\text{denotes picture}\left(\text{a}\right)\\ \left(\text{iii}\right)\text{denotes picture}\left(\text{b}\right)\end{array}}\end{array}

Q.10

$\begin{array}{l}\mathrm{Multiply}\mathrm{and}\mathrm{reduce}\mathrm{to}\mathrm{lowest}\mathrm{form}:\\ \left(\mathrm{i}\right)7\times \frac{3}{5}\left(\mathrm{ii}\right)4\times \frac{1}{3}\left(\mathrm{iii}\right)2\times \frac{6}{7}\\ \left(\mathrm{iv}\right)5\times \frac{2}{9}\left(\mathrm{v}\right)\frac{2}{3}\times 4\left(\mathrm{vi}\right)\frac{5}{2}\times 6\\ \left(\mathrm{vii}\right)11\times \frac{4}{7}\left(\mathrm{viii}\right)20\times \frac{4}{5}\left(\mathrm{ix}\right)13\times \frac{1}{3}\\ \left(\mathrm{x}\right)15\times \frac{3}{5}\end{array}$

Ans.

$\begin{array}{c}\left(\mathrm{i}\right)7\times \frac{3}{5}=\frac{7}{1}\times \frac{3}{5}=\frac{21}{5}\\ =\boxed{4\frac{1}{5}}\\ \left(\mathrm{ii}\right)4\times \frac{1}{3}=\frac{4}{1}\times \frac{1}{3}=\frac{4}{3}\\ =\boxed{1\frac{1}{3}}\\ \left(\mathrm{iii}\right)2\times \frac{6}{7}=\frac{2}{1}\times \frac{6}{7}=\frac{12}{7}\\ =\boxed{1\frac{5}{7}}\\ \left(\mathrm{iv}\right)5\times \frac{2}{9}=\frac{5}{1}\times \frac{2}{9}=\frac{10}{9}\\ =\boxed{1\frac{1}{9}}\\ \left(\mathrm{v}\right)\frac{2}{3}\times 4=\frac{2}{3}\times \frac{4}{1}=\frac{8}{3}\\ =\boxed{2\frac{2}{3}}\\ \left(\mathrm{vi}\right)\text{\hspace{0.17em}}\frac{5}{2}\times 6=\frac{5\times 6}{2}=\frac{30}{2}\\ =\boxed{15}\\ \left(\mathrm{vii}\right)11\times \frac{4}{7}=\frac{11}{1}\times \frac{4}{7}=\frac{44}{7}\\ =\boxed{6\frac{2}{7}}\\ \left(\mathrm{viii}\right)20\times \frac{4}{7}=\frac{20}{1}\times \frac{4}{5}=\frac{80}{5}\\ =\boxed{16}\\ \left(\mathrm{ix}\right)13\times \frac{1}{3}=\frac{13}{1}\times \frac{1}{3}=\frac{13}{3}\\ =\boxed{4\frac{1}{3}}\\ \left(\mathrm{x}\right)15\times \frac{3}{5}=\frac{15}{1}\times \frac{3}{5}=\frac{45}{5}\\ =\boxed{9}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Shade}:\\ \left(\mathrm{i}\right)\frac{1}{2}\mathrm{of}\mathrm{the}\mathrm{circles}\mathrm{in}\mathrm{box}\left(\mathrm{a}\right)\\ \left(\mathrm{ii}\right)\frac{2}{3}\mathrm{of}\mathrm{the}\mathrm{triangles}\mathrm{in}\mathrm{box}\left(\mathrm{b}\right)\\ \left(\mathrm{iii}\right)\frac{3}{5}\mathrm{of}\mathrm{the}\mathrm{squares}\mathrm{in}\mathrm{box}\left(\mathrm{c}\right)\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\\ \text{Here},\text{there are 12 circles in box}\left(\text{a}\right).\\ \text{Therefore},\\ \frac{1}{2}\mathrm{of}12=\frac{1}{2}\times 12=6\\ \text{Hence 6 circles to be coloured in}\left(\text{a}\right).\\ \left(\mathrm{ii}\right)\\ \text{Here},\text{there are 9 triangles in box}\left(\text{b}\right)\text{and\hspace{0.17em}}\frac{2}{3}\text{\hspace{0.17em}of the}\\ \text{triangles are to be coloured}.\\ \text{So},\frac{2}{3}\mathrm{of}9=\frac{2}{3}\times 9\\ =\frac{18}{3}=6\\ \text{Thus},\text{6 triangles are needed to be coloured}.\\ \left(\text{iii}\right)\\ \text{Here},\text{there are 15 rectangles in box}\left(\text{c}\right)\text{and}\frac{3}{5}\text{of them}\\ \text{are needed to be coloured}.\\ \text{So},\\ \frac{3}{5}\mathrm{of}\text{\hspace{0.17em}}15=\frac{3}{5}\times 15\\ =\frac{45}{5}=9\\ \text{Thus},\text{9 rectangles are needed to be coloured}.\end{array}$

Q.12

$\begin{array}{l}\mathbf{F}\mathbf{i}\mathbf{n}\mathbf{d}:\\ \left(\mathbf{a}\right)\frac{1}{2}\text{of}\left(\mathbf{i}\right)\mathbf{2}\mathbf{4}\left(\mathbf{i}\mathbf{i}\right)\mathbf{4}\mathbf{6}\\ \left(\mathbf{b}\right)\frac{2}{3}\text{of}\left(\mathbf{i}\right)\mathbf{1}\mathbf{8}\left(\mathbf{i}\mathbf{i}\right)\mathbf{2}\mathbf{7}\\ \left(\mathbf{c}\right)\frac{3}{4}\text{of}\left(\mathbf{i}\right)\mathbf{1}\mathbf{6}\left(\mathbf{i}\mathbf{i}\right)\mathbf{3}\mathbf{6}\\ \left(\mathbf{d}\right)\frac{4}{5}\text{of}\left(\mathbf{i}\right)\mathbf{2}\mathbf{0}\left(\mathbf{i}\mathbf{i}\right)\mathbf{3}\mathbf{5}\end{array}$

Ans.

\begin{array}{l}(a)\\ (i)\frac{1}{2}of24=\frac{1}{2}\times 24=\frac{24}{2}=\boxed{12}\\ (ii)\frac{1}{2}of46=\frac{1}{2}\times 46=\frac{46}{2}=\boxed{23}\\ (b)\\ (i)\frac{2}{3}of18=\frac{2}{3}\times 18=\frac{36}{3}=\boxed{12}\\ (ii)\frac{2}{3}of27=\frac{2}{3}\times 27=\frac{54}{3}=\boxed{18}\\ (c)\\ (i)\frac{3}{4}of16=\frac{3}{4}\times 16=\frac{48}{4}=\boxed{12}\\ (ii)\frac{3}{4}of36=\frac{3}{4}\times 36=\frac{108}{4}=\boxed{27}\\ (d)\\ (i)\frac{4}{5}of20=\frac{4}{5}\times 20=\frac{80}{5}=\boxed{16}\end{array}

Q.13

$\begin{array}{l}\mathrm{Multiply}\mathrm{and}\mathrm{express}\mathrm{as}\mathrm{a}\mathrm{mixed}\mathrm{fraction}:\\ \left(a\right)3\times 5\frac{1}{5}\left(b\right)\text{}5\times 6\frac{3}{4}\left(c\right)7\times 2\frac{1}{4}\\ \left(d\right)4\times 6\frac{1}{3}\left(e\right)\text{}3\frac{1}{4}\times 6\left(f\right)3\frac{2}{5}\times 8\end{array}$

Ans.

\begin{array}{l}\left(\text{a}\right)3\times 5\frac{1}{5}=3\times \frac{26}{5}=\frac{78}{5}=\boxed{15\frac{3}{5}}\\ \left(\text{b}\right)5\times 6\frac{3}{4}=5\times \frac{27}{4}=\frac{135}{4}=\boxed{33\frac{3}{4}}\\ \left(\text{c}\right)7\times 2\frac{1}{4}=7\times \frac{9}{4}=\frac{63}{4}=\boxed{15\frac{3}{4}}\\ \left(\text{d}\right)4\times 6\frac{1}{3}=4\times \frac{19}{3}=\frac{76}{3}=\boxed{25\frac{1}{3}}\\ \left(\text{e}\right)3\frac{1}{4}\times 6=\frac{13}{4}\times 6=\frac{78}{7}=\boxed{19\frac{1}{2}}\\ \left(\text{f}\right)3\frac{2}{5}\times 8=\frac{17}{5}\times 8=\frac{136}{5}=\boxed{27\frac{1}{5}}\end{array}

Q.14

$\begin{array}{l}Find\\ \left(a\right)\frac{1}{2}\mathrm{of}\left(\mathrm{i}\right)2\frac{3}{4}\left(\mathrm{ii}\right)4\frac{2}{9}\\ \left(b\right)\frac{5}{8}\mathrm{of}\left(\mathrm{i}\right)3\frac{5}{6}\left(\mathrm{ii}\right)\text{}9\frac{2}{3}\end{array}$

Ans.

\begin{array}{l}\left(\text{a}\right)\\ \left(\text{i}\right)\frac{1}{2}\text{of}2\frac{3}{4}=\frac{1}{2}\times \frac{11}{4}=\frac{11}{8}=\boxed{1\frac{3}{8}}\\ \left(\text{ii}\right)\frac{1}{2}\text{of}4\frac{2}{9}=\frac{1}{2}\times \frac{38}{9}=\frac{19}{9}=\boxed{2\frac{1}{9}}\\ \left(\text{b}\right)\\ \left(\text{i}\right)\frac{5}{8}\text{of}3\frac{5}{6}=\frac{5}{8}\times \frac{23}{6}=\frac{115}{48}=\boxed{2\frac{19}{48}}\\ \left(\text{ii}\right)\frac{5}{8}\text{of}\text{}9\frac{2}{3}=\frac{5}{8}\times \frac{29}{3}=\frac{145}{24}=\boxed{6\frac{1}{24}}\end{array}

Q.15

$$\begin{gathered} \begin{array}{l}\mathrm{Vidya}\mathrm{}\mathrm{and}\mathrm{}\mathrm{Pratab}\mathrm{}\mathrm{went}\mathrm{}\mathrm{for}\mathrm{}\mathrm{a}\mathrm{}\mathrm{picnic}.\mathrm{}\mathrm{Their}\mathrm{}\mathrm{mother}\\ \mathrm{gave}\mathrm{}\mathrm{them}\mathrm{}\mathrm{a}\mathrm{}\mathrm{water}\mathrm{}\mathrm{bag}\mathrm{}\mathrm{that}\mathrm{}\mathrm{contained}\mathrm{}5\mathrm{}\mathrm{liters}\mathrm{}\mathrm{of}\\ \mathrm{water}.\mathrm{}\mathrm{Vidya}\mathrm{}\mathrm{consumed}\frac{2}{5}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{water}.\mathrm{}\mathrm{Pratap}\\ \mathrm{consumed}\mathrm{}\mathrm{the}\mathrm{}\mathrm{remaining}\mathrm{}\mathrm{water}. \\ \left(\mathrm{i}\right)\mathrm{}\mathrm{How}\mathrm{}\mathrm{much}\mathrm{water}\mathrm{}\mathrm{did}\mathrm{}\mathrm{Vidya}\mathrm{}\mathrm{drink}?\\ \\ \left(\mathrm{ii}\right)\mathrm{}\mathrm{What}\mathrm{}\mathrm{fraction}\mathrm{}\mathrm{of}\mathrm{}\mathrm{total}\\ \mathrm{quantity}\mathrm{}\mathrm{of}\mathrm{}\mathrm{water}\mathrm{}\mathrm{did}\mathrm{}\mathrm{Pratab}\mathrm{}\mathrm{drink}?\end{array} \end{gathered}$$

Ans.

$$\begin{gathered} \begin{array}{l}\mathrm{Vidya}\mathrm{}\mathrm{and}\mathrm{}\mathrm{Pratab}\mathrm{}\mathrm{went}\mathrm{}\mathrm{for}\mathrm{}\mathrm{a}\mathrm{}\mathrm{picnic}.\mathrm{}\mathrm{Their}\mathrm{}\mathrm{mother}\\ \mathrm{gave}\mathrm{}\mathrm{them}\mathrm{}\mathrm{a}\mathrm{}\mathrm{water}\mathrm{}\mathrm{bag}\mathrm{}\mathrm{that}\mathrm{}\mathrm{contained}\mathrm{}5\mathrm{}\mathrm{liters}\mathrm{}\mathrm{of}\\ \mathrm{water}.\mathrm{}\mathrm{Vidya}\mathrm{}\mathrm{consumed}\frac{2}{5}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{water}.\mathrm{}\mathrm{Pratap}\\ \mathrm{consumed}\mathrm{}\mathrm{the}\mathrm{}\mathrm{remaining}\mathrm{}\mathrm{water}. \\ \left(\mathrm{i}\right)\mathrm{}\mathrm{How}\mathrm{}\mathrm{much}\mathrm{water}\mathrm{}\mathrm{did}\mathrm{}\mathrm{Vidya}\mathrm{}\mathrm{drink}?\\ \\ \left(\mathrm{ii}\right)\mathrm{}\mathrm{What}\mathrm{}\mathrm{fraction}\mathrm{}\mathrm{of}\mathrm{}\mathrm{total}\\ \mathrm{quantity}\mathrm{}\mathrm{of}\mathrm{}\mathrm{water}\mathrm{}\mathrm{did}\mathrm{}\mathrm{Pratab}\mathrm{}\mathrm{drink}?\end{array} \end{gathered}$$

Q.16

\begin{array}{l}Find:\\ (i)\frac{1}{4}of\left(a\right)\frac{1}{4}\left(b\right)\frac{3}{5}\left(c\right)\frac{4}{3}\\ (ii)\frac{1}{7}of\left(a\right)\frac{2}{9}\left(b\right)\text{}\frac{6}{5}\left(c\right)\text{}\frac{3}{10}\end{array}

Ans.

\begin{array}{l}\left(i\right)\\ \left(a\right)\frac{1}{4}\text{of}\frac{1}{4}=\frac{1}{4}\times \frac{1}{4}=\boxed{\frac{1}{16}}\\ \left(b\right)\frac{1}{4}\text{of}\frac{3}{5}=\frac{1}{4}\times \frac{3}{5}=\boxed{\frac{3}{20}}\\ \left(c\right)\frac{1}{4}\text{of}\frac{4}{3}=\frac{1}{\cancel{4}}\times \frac{\cancel{4}}{3}=\boxed{\frac{1}{3}}\\ (ii)\\ (a)\frac{1}{7}\text{of}\frac{2}{9}=\frac{1}{7}\times \frac{2}{9}=\boxed{\frac{2}{63}}\\ (b)\frac{1}{7}\text{of}\frac{6}{5}=\frac{1}{7}\times \frac{6}{5}=\boxed{\frac{6}{35}}\\ (c)\frac{1}{7}\text{of}\frac{3}{10}=\frac{1}{7}\times \frac{3}{10}=\boxed{\frac{3}{70}}\end{array}

Q.17

$\begin{array}{l}\mathrm{Multiply}\mathrm{and}\mathrm{reduce}\mathrm{to}\mathrm{lowest}\mathrm{form}(\mathrm{if}\mathrm{possible}).\\ \left(\mathrm{i}\right)\frac{2}{3}\times 2\frac{2}{3}\left(\mathrm{ii}\right)\frac{2}{7}\times \frac{7}{9}\left(\mathrm{iii}\right)\frac{3}{8}\times \frac{6}{4}\\ \left(\mathrm{iv}\right)\frac{9}{5}\times \frac{3}{5}\left(\mathrm{v}\right)\frac{1}{3}\times \frac{15}{8}\left(\mathrm{vi}\right)\frac{11}{2}\times \frac{3}{10}\\ \left(\mathrm{vii}\right)\frac{4}{5}\times \frac{12}{7}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\frac{2}{3}\times 2\frac{2}{3}=\frac{2}{3}\times \frac{8}{3}=\boxed{\frac{16}{9}}\\ \left(\mathrm{ii}\right)\frac{2}{7}\times \frac{7}{9}=\frac{2}{\cancel{7}}\times \frac{\cancel{7}}{9}=\boxed{\frac{2}{9}}\\ \left(\mathrm{iii}\right)\frac{3}{8}\times \frac{6}{4}=\frac{3}{\cancel{2}\times 4}\times \frac{\cancel{2}\times 3}{4}=\boxed{\frac{9}{16}}\\ \left(\mathrm{iv}\right)\frac{9}{5}\times \frac{3}{5}=\boxed{\frac{27}{25}}\\ \left(\mathrm{v}\right)\frac{1}{3}\times \frac{15}{8}=\frac{1}{\cancel{3}}\times \frac{\cancel{3}\times 5}{8}=\boxed{\frac{5}{8}}\\ \left(\mathrm{vi}\right)\frac{11}{2}\times \frac{3}{10}=\boxed{\frac{33}{20}}\\ \left(\mathrm{vii}\right)\frac{4}{5}\times \frac{12}{7}=\boxed{\frac{48}{35}}\end{array}$

Q.18

$\begin{array}{l}\mathrm{For}\mathrm{the}\mathrm{fractions}\mathrm{given}\mathrm{below}:\\ \left(\mathrm{a}\right)\mathrm{Multiply}\mathrm{and}\mathrm{reduce}\mathrm{the}\mathrm{product}\mathrm{to}\mathrm{lowest}\\ \mathrm{form}\left(\mathrm{if}\mathrm{possible}\right)\\ \left(\mathrm{b}\right)\mathrm{Tell}\mathrm{whether}\mathrm{the}\mathrm{fraction}\mathrm{obtained}\mathrm{is}\mathrm{proper}\\ \mathrm{or}\mathrm{improper}.\\ \left(\mathrm{c}\right)\mathrm{If}\mathrm{the}\mathrm{fraction}\mathrm{obtained}\mathrm{is}\mathrm{improper}\mathrm{then}\\ \mathrm{convert}\mathrm{it}\mathrm{into}\mathrm{a}\mathrm{mixed}\mathrm{fraction}.\\ \text{(i)}\frac{2}{5}\times 5\frac{1}{4}\\ \text{(ii)}6\frac{2}{5}\times \frac{7}{9}\\ \text{(iii)}\frac{3}{2}\times 5\frac{1}{3}\\ \text{(iv)}\frac{5}{6}\times 2\frac{3}{7}\\ \left(\mathrm{v}\right)3\frac{2}{5}\times \frac{4}{7}\\ \text{(vi)}2\frac{3}{5}\times 3\\ \text{(vii)}3\frac{4}{7}\times \frac{3}{5}\end{array}$

Ans.

$\begin{array}{l}\left(\text{i}\right)\\ \left(\text{a}\right)\frac{2}{5}\times 5\frac{1}{4}=\frac{2}{5}\times \frac{21}{4}=\frac{\cancel{2}}{5}\times \frac{21}{\cancel{2}\times 2}=\boxed{\frac{21}{10}}\\ \left(\text{b}\right)\text{It is an improper fraction}.\\ \left(\text{c}\right)\frac{21}{10}=\boxed{2\frac{1}{10}}\\ \left(\mathrm{ii}\right)\\ \left(\mathrm{a}\right)6\frac{2}{5}\times \frac{7}{9}=\frac{32}{5}\times \frac{7}{9}=\boxed{\frac{224}{45}}\\ \left(\mathrm{b}\right)\text{It is an improper fraction}.\\ \left(\mathrm{c}\right)\frac{224}{45}=\boxed{4\frac{44}{45}}\\ \left(\mathrm{iii}\right)\\ \left(\mathrm{a}\right)\frac{3}{2}\times 5\frac{1}{3}=\frac{3}{2}\times \frac{16}{3}=\frac{\cancel{3}}{\cancel{2}}\times \frac{\cancel{2}\times 8}{\cancel{3}}=\frac{8}{1}=\boxed{8}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}It\hspace{0.17em}is whole number.}\\ \text{(iv)}\\ \text{(a)\hspace{0.17em}}\frac{5}{6}\times 2\frac{3}{7}\\ =\frac{5}{6}\times \frac{17}{7}\\ =\boxed{\frac{85}{42}}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}It is an improper fraction}.\\ \left(\mathrm{c}\right)\frac{85}{42}=\boxed{2\frac{1}{42}}\\ \left(\mathrm{v}\right)\\ \left(\mathrm{a}\right)\text{\hspace{0.17em}}3\frac{2}{5}\times \frac{4}{7}\\ =\frac{17}{5}\times \frac{4}{7}\\ =\frac{68}{35}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}It is an improper fraction}.\\ \left(\mathrm{c}\right)\frac{68}{35}=\boxed{1\frac{33}{35}}\\ \left(\mathrm{vi}\right)\\ \left(\mathrm{a}\right)2\frac{3}{5}\times 3\\ =\frac{13}{5}\times \frac{3}{1}\\ =\frac{39}{5}\\ \left(\text{b}\right)\text{It is an improper fraction}.\\ \left(\text{c}\right)\frac{39}{5}=\boxed{7\frac{4}{5}}\\ \left(\mathrm{vii}\right)\\ \left(\mathrm{a}\right)3\frac{4}{7}\times \frac{3}{5}\\ =\frac{25}{7}\times \frac{3}{5}\\ =\frac{5\times \cancel{5}}{7}\times \frac{3}{\cancel{5}}\\ =\frac{15}{7}\\ \text{(b)\hspace{0.17em}It is an improper fraction}.\\ \left(\mathrm{c}\right)\frac{15}{7}=\boxed{2\frac{1}{7}}\end{array}$

Q.19

$\begin{array}{l}\mathrm{Which}\mathrm{is}\mathrm{greater}:\\ \left(i\right)\frac{2}{7}of\frac{3}{4}\mathrm{or}\frac{3}{5}of\frac{5}{8}\\ \left(ii\right)\frac{1}{2}of\frac{6}{7}\mathrm{or}\frac{2}{3}of\frac{3}{7}\end{array}$

Ans.

\begin{array}{l}\left(\text{i}\right)\\ \frac{2}{7}\text{of}\frac{3}{4}\text{or}\frac{3}{5}\text{of}\frac{5}{8}\\ \frac{2}{7}\times \frac{3}{4}or\frac{3}{5}\times \frac{5}{8}\\ =\frac{\cancel{2}}{7}\times \frac{3}{\cancel{2}\times 2}\text{or}\frac{3}{\cancel{5}}\times \frac{\cancel{5}}{8}\\ =\frac{3}{14}\text{or}\frac{3}{8}\\ \text{Now},\text{the LCM of 14 and 8 is 56}.\text{So},\text{we get}\\ \frac{3}{14}=\frac{3\times 4}{14\times 4}=\frac{12}{56}\text{or}\frac{3}{8}=\frac{3\times 7}{8\times 7}=\frac{21}{56}\\ \text{Since},\text{21}>\text{12},\text{so}\frac{3}{8}>\frac{3}{14}\\ \text{Hence},\boxed{\frac{3}{5}\text{of}\frac{5}{8}>\frac{2}{7}\text{of}\frac{3}{4}}\\ \left(\text{ii}\right)\\ \frac{1}{2}\text{of}\frac{6}{7}or\text{}\frac{2}{3}\text{of}\frac{3}{7}\\ \frac{1}{2}\text{of}\frac{6}{7}or\frac{2}{3}\text{of}\frac{3}{7}\\ =\frac{1}{\cancel{2}}\times \frac{\cancel{2}\times 3}{7}\text{or}\frac{2}{\cancel{3}}\times \frac{\cancel{3}}{7}\\ =\frac{3}{7}\text{or}\frac{2}{7}\\ \text{Since},\text{3}>\text{2},\text{so},\frac{3}{7}>\frac{2}{7}\\ \text{Hence},\boxed{\frac{1}{2}\text{of}\frac{6}{7}>\frac{2}{3}\text{of}\frac{3}{7}}\end{array}

Q.20

$$\begin{gathered} \mathrm{Saili}\mathrm{plants}4\mathrm{saplings},\mathrm{in}\mathrm{a}\mathrm{row},\mathrm{in}\mathrm{her}\mathrm{garden}.\mathrm{The}\mathrm{distance}\mathrm{between}\mathrm{two}\mathrm{adjacent}\mathrm{}\mathrm{saplings}\mathrm{}\mathrm{is}\frac{3}{4}m. \\ \mathrm{Find}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{first}\mathrm{and}\mathrm{the}\mathrm{last}\mathrm{sapling}. \end{gathered}$$

Ans.

\begin{array}{l}\text{Here},\text{the distance between the two saplings is given as}\frac{3}{4}\text{m}\\ \text{So},\text{the distance between first and the last sapling is}\\ \frac{3}{4}\text{m}\times \text{3=}\frac{9}{4}\text{m}\\ \text{=}\boxed{\text{2}\frac{1}{4}\text{m}}\end{array}

Q.21

$\begin{array}{l}\mathrm{Lipika}\mathrm{}\mathrm{reads}\mathrm{}\mathrm{a}\mathrm{}\mathrm{book}\mathrm{}\mathrm{for}\mathrm{\hspace{0.17em}}1\frac{3}{4}\mathrm{\hspace{0.17em}}\mathrm{hours}\mathrm{}\mathrm{every}\mathrm{}\mathrm{day}.\mathrm{}\mathrm{She}\\ \mathrm{reads}\mathrm{}\mathrm{the}\mathrm{}\mathrm{entire}\mathrm{}\mathrm{book}\mathrm{}\mathrm{in}\mathrm{}6\mathrm{}\mathrm{days}.\mathrm{}\mathrm{How}\mathrm{}\mathrm{many}\mathrm{}\mathrm{hours}\\ \mathrm{in}\mathrm{}\mathrm{all}\mathrm{}\mathrm{were}\mathrm{}\mathrm{required}\mathrm{}\mathrm{by}\mathrm{}\mathrm{her}\mathrm{}\mathrm{to}\mathrm{}\mathrm{read}\mathrm{}\mathrm{the}\mathrm{}\mathrm{book}?\end{array}$

Ans.

\begin{array}{l}\text{Lipika reads}1\frac{3}{4}=\frac{7}{4}\text{hours a day}.\\ \text{So},\text{Number of hours required by her to read the book}\\ \text{in 6 days is}\text{}\frac{7}{4}\text{hour}\times 6=\frac{7}{2\times \cancel{2}}\text{hour}\times \left(\cancel{2}\times 3\right)\\ =\frac{21}{2}\text{hour}\\ \text{=}\boxed{\text{10}\frac{1}{2}\text{hour}}\end{array}

Q.22

$\begin{array}{l}\mathrm{A}\mathrm{}\mathrm{car}\mathrm{}\mathrm{runs}\mathrm{}16\mathrm{}\mathrm{km}\mathrm{}\mathrm{using}\mathrm{}1\mathrm{}\mathrm{litre}\mathrm{}\mathrm{of}\mathrm{}\mathrm{petrol}.\mathrm{}\mathrm{How}\\ \mathrm{much}\mathrm{}\mathrm{distance}\mathrm{}\mathrm{will}\mathrm{}\mathrm{it}\mathrm{}\mathrm{cover}\mathrm{}\mathrm{using}\mathrm{\hspace{0.17em}}2\frac{3}{4}\mathrm{litres}\mathrm{}\mathrm{of}\mathrm{}\mathrm{petrol}?\end{array}$

Ans.

\begin{array}{c}\text{In 1 litre of petrol},\text{car runs 16 km}.\\ \text{So},\text{in}2\frac{3}{4}\text{litres of petrol},\text{distance covered by car is}\\ 2\frac{3}{4}\times 16\text{km}\\ =\frac{11}{\cancel{4}}\times \left(\cancel{4}\times 4\right)\text{km}\\ =\boxed{\text{44}\text{km}}\end{array}

Q.23

$\begin{array}{l}\left(\mathrm{a}\right)\\ \left(\mathrm{i}\right)\text{}\mathrm{Provide}\mathrm{the}\mathrm{number}\mathrm{in}\mathrm{the}\mathrm{box}\text{\hspace{0.17em}}\boxed{},\mathrm{such}\mathrm{that}\frac{2}{3}\times \boxed{}=\frac{10}{30}\\ \left(\mathrm{ii}\right)\mathrm{The}\mathrm{simplest}\mathrm{form}\mathrm{of}\mathrm{the}\mathrm{number}\mathrm{obtained}\mathrm{in}\boxed{}\mathrm{is}\_\_\_\_\\ \left(\mathrm{b}\right)\\ \left(\mathrm{i}\right)\mathrm{Provide}\mathrm{the}\mathrm{number}\mathrm{in}\mathrm{the}\mathrm{box}\boxed{},\mathrm{such}\mathrm{that}\frac{3}{5}\times \boxed{}=\frac{24}{75}\\ \left(\mathrm{ii}\right)\\ \mathrm{The}\mathrm{simplest}\mathrm{form}\mathrm{of}\mathrm{the}\mathrm{number}\mathrm{obtained}\mathrm{in}\boxed{}\mathrm{is}\_\_\_\_\end{array}$

Ans.

$\begin{array}{l}\left(\text{a}\right)\\ \left(\text{i}\right)\frac{\text{2}}{\text{3}}\times \boxed{}\text{=}\frac{\text{10}}{\text{30}}\\ =\frac{\text{2×5}}{\text{3×10}}=\frac{\text{2}}{\text{3}}\text{×}\frac{\text{5}}{\text{10}}\\ \text{So,}\frac{\text{2}}{\text{3}}\times \frac{\text{5}}{\text{10}}=\frac{\text{10}}{\text{30}}\\ \text{Therefore, the number in}\boxed{}\text{is}\frac{\text{5}}{\text{10}}\text{.}\\ \text{(ii)}\\ \text{The simplest form of}\frac{\text{5}}{\text{10}}\text{is}\frac{\text{5}}{\text{10}}=\frac{}{}=\boxed{\frac{\text{1}}{\text{2}}}\\ \text{(b)}\\ \left(\text{i}\right)\frac{\text{3}}{\text{5}}\times \boxed{}\text{=}\frac{\text{24}}{\text{75}}\\ =\frac{\text{3×8}}{\text{5×15}}=\frac{\text{3}}{\text{5}}\text{×}\frac{\text{8}}{\text{15}}\\ \text{So,}\frac{\text{3}}{\text{5}}\text{×}\frac{\text{8}}{\text{15}}=\frac{\text{24}}{\text{75}}\\ \text{Therefore, the number in}\boxed{}\text{is}\frac{\text{8}}{\text{15}}\text{.}\\ \text{(ii)}\\ \text{The simplest form of}\frac{\text{8}}{\text{15}}\text{is}\boxed{\frac{\text{8}}{\text{15}}}\end{array}$

Q.24

$\begin{array}{l}\text{Find:}\\ \text{(i)}\text{12÷}\frac{\text{3}}{\text{4}}\text{(ii)}\text{14÷}\frac{\text{5}}{\text{6}}\text{(iii)}\text{8÷}\frac{\text{7}}{\text{3}}\\ \text{(iv)}\text{}\text{4÷}\frac{\text{8}}{\text{3}}\text{(v)}\text{3÷2}\frac{\text{1}}{\text{3}}\text{(vi)}\text{5÷3}\frac{\text{4}}{\text{7}}\end{array}$

Ans.

$\begin{array}{l}\text{(i)\hspace{0.33em}12÷}\frac{\text{3}}{\text{4}}\text{=12×}\frac{\text{4}}{\text{3}}\text{=}\frac{\text{48}}{\text{3}}\text{=}\boxed{\text{16}}\\ \text{(ii)\hspace{0.33em}14÷}\frac{\text{5}}{\text{6}}\text{=14×}\frac{\text{6}}{\text{5}}\text{=}\frac{\text{84}}{\text{5}}\text{=}\boxed{\text{16}\frac{\text{4}}{\text{5}}}\\ \text{(iii)\hspace{0.33em}8÷}\frac{\text{7}}{\text{3}}\text{=8×}\frac{\text{3}}{\text{7}}\text{=}\frac{\text{24}}{\text{7}}\text{=}\boxed{\text{3}\frac{\text{3}}{\text{7}}}\\ \text{(iv)\hspace{0.33em}4÷}\frac{\text{8}}{\text{3}}\text{=4×}\frac{\text{3}}{\text{8}}\frac{\text{3}}{}\text{=}\frac{\text{3}}{\text{2}}\text{=}\boxed{\text{1}\frac{\text{1}}{\text{2}}}\\ \text{(v)\hspace{0.33em}3÷2}\frac{\text{1}}{\text{3}}\text{=3÷}\frac{\text{7}}{\text{3}}\text{=3×}\frac{\text{3}}{\text{7}}\text{=}\frac{\text{9}}{\text{7}}\text{=}\boxed{\text{1}\frac{\text{2}}{\text{7}}}\\ \text{(vi)\hspace{0.33em}5÷3}\frac{\text{4}}{\text{7}}\text{=5÷}\frac{\text{25}}{\text{7}}\text{=5×}\frac{\text{7}}{\text{25}}\frac{\text{7}}{}\text{=}\frac{\text{7}}{\text{5}}\text{=}\boxed{\text{1}\frac{\text{2}}{\text{5}}}\end{array}$