NCERT Solutions Class 7 Maths Chapter 9

Q.1

$\begin{array}{l}\mathrm{List}\mathrm{five}\mathrm{rational}\mathrm{numbers}\mathrm{between}:\\ \left(\mathrm{i}\right)-1\mathrm{and}0\left(\mathrm{ii}\right)-2\mathrm{and}-1\\ \left(\mathrm{iii}\right)\frac{-4}{5}\mathrm{and}\frac{-2}{3}\left(\mathrm{iv}\right)–\frac{1}{2}\mathrm{and}\frac{2}{3}\end{array}$

Ans

$\begin{array}{l}\text{(i) Five rational numbers between}-\text{1 and 0 are:}\\ \frac{-2}{3},\frac{-1}{2},\frac{-2}{5},\frac{-1}{3},\frac{-2}{7}\\ \text{(ii)}-\text{2 and}-\text{1}\\ -2=\frac{-12}{6}\text{and}\text{-1=}\frac{-6}{6}\\ \text{Five rational numbers between}-\text{2 and}-\text{1 are:}\\ \frac{-11}{6},\frac{-10}{6},\frac{-9}{6},\frac{-8}{6},\frac{-7}{6}\end{array}$ $\begin{array}{l}(\text{iii)}\frac{-4}{5}\text{and}\frac{-2}{3}\\ \frac{-4}{5}=\frac{-4\times 9}{5\times 9}=\frac{-36}{45}\\ \frac{-2}{3}=\frac{-2\times 15}{3\times 15}=\frac{-30}{45}\\ \text{Five rational numbers between}\frac{-4}{5}\text{and}\frac{-2}{3}\text{are:}\\ \frac{-35}{45},\frac{-34}{45},\frac{-33}{45},\frac{-32}{45},\frac{-31}{45}\\ \text{(iv) –}\frac{\text{1}}{2}\text{and}\frac{2}{3}\\ –\frac{1}{2}=–\frac{1\times 18}{2\times 18}=–\frac{18}{36}\\ \frac{2}{3}=\frac{2\times 12}{3\times 12}=\frac{24}{36}\\ \text{Five rational numbers between –}\frac{\text{1}}{2}\text{and}\frac{2}{3}\text{are:}\\ \frac{19}{36},\frac{20}{36},\frac{21}{36},\frac{22}{36},\frac{23}{36}\end{array}$

Q.2

$\begin{array}{l}\text{Write\hspace{0.33em}four\hspace{0.33em}more\hspace{0.33em}rational\hspace{0.33em}numbers\hspace{0.33em}in\hspace{0.33em}each\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}following}\\ \text{patterns:}\\ \left(\text{i}\right)\frac{-\text{3}}{\text{5}}\text{,}\frac{-\text{6}}{\text{10}}\text{,}\frac{-\text{9}}{\text{15}}\text{,}\frac{-\text{12}}{\text{20}}\text{,…\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\left(\text{ii}\right)\frac{-\text{1}}{\text{4}}\text{,}\frac{-\text{2}}{\text{8}}\text{,}\frac{-\text{3}}{\text{12}}\text{,…}\\ \left(\text{iii}\right)\frac{-\text{1}}{\text{6}}\text{,}\frac{\text{2}}{-\text{12}}\text{,}\frac{\text{3}}{-\text{18}}\text{,}\frac{\text{4}}{-\text{24}}\text{,…\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\left(\text{iv}\right)\frac{-\text{2}}{\text{3}}\text{,}\frac{\text{2}}{-\text{3}}\text{,}\frac{\text{4}}{-\text{6}}\text{,}\frac{\text{6}}{-\text{9}}\text{,…}\end{array}$

Ans

$\begin{array}{l}\text{(i)}\frac{-3}{5},\frac{-6}{10},\frac{-9}{15},\frac{-12}{20},\dots \\ \frac{-3}{5},\frac{-3\times 2}{5\times 2},\frac{-3\times 3}{5\times 3},\frac{-3\times 4}{5\times 4},\dots \\ \text{Clearly, the numerator is a multiple of 3 and denominator is}\\ \text{a multiple of 5}\text{.}\\ \text{Therefore, next four rational numbers in this pattern are}\\ \frac{-3\times 5}{5\times 5},\frac{-3\times 6}{5\times 6},\frac{-3\times 7}{5\times 7},\frac{-3\times 8}{5\times 8},\dots \\ \frac{-15}{25},\frac{-18}{30},\frac{-21}{35},\frac{-24}{40},\dots \\ \\ \text{(ii)}\frac{-1}{4},\frac{-2}{8},\frac{-3}{12},\dots \\ \frac{-1}{4},\frac{-1\times 2}{4\times 2},\frac{-1\times 3}{4\times 3},\dots \\ \text{The next four rational number in this pattern are}\\ \frac{-1\times 4}{4\times 4},\frac{-1\times 5}{4\times 5},\frac{-1\times 6}{4\times 6},\frac{-1\times 7}{4\times 7},\dots \end{array}$ $\begin{array}{l}\frac{-4}{16},\frac{-5}{20},\frac{-6}{24},\frac{-7}{28},\dots \\ \\ \text{(iii)}\frac{-1}{6},\frac{2}{-12},\frac{3}{-18},\frac{4}{-24},\dots \\ \frac{-1}{6},\frac{1\times 2}{-6\times 2},\frac{1\times 3}{-6\times 3},\frac{1\times 4}{-6\times 4},\dots \\ \text{The next four rational number in this pattern are}\\ \frac{1\times 5}{-6\times 5},\frac{1\times 6}{-6\times 6},\frac{1\times 7}{-6\times 7},\frac{1\times 8}{-6\times 8},\dots \\ \frac{5}{-30},\frac{6}{-36},\frac{7}{-42},\frac{8}{-48},\dots \\ \\ \text{(iv)}\frac{-2}{3},\frac{2}{-3},\frac{4}{-6},\frac{6}{-9}\dots \\ =\frac{-2}{3},\frac{2}{-3},\frac{2\times 2}{-3\times 2},\frac{2\times 3}{-3\times 3}\dots \\ \text{The next four rational number in this pattern are}\\ \frac{2\times 4}{-3\times 4},\frac{2\times 5}{-3\times 5},\frac{2\times 6}{-3\times 6},\frac{2\times 7}{-3\times 7},\dots \\ \frac{8}{-12},\frac{10}{-15},\frac{12}{-18},\frac{14}{-21},\dots \end{array}$

Q.3

$\begin{array}{l}\mathrm{Give}\mathrm{four}\mathrm{rational}\mathrm{numbers}\mathrm{equivalent}\mathrm{to}:\\ \left(\mathrm{i}\right)\frac{-2}{7}\left(\mathrm{ii}\right)\frac{5}{-3}\left(\mathrm{iii}\right)\frac{4}{9}\end{array}$

Ans

\begin{array}{l}\text{(i)}\frac{-2}{7}\\ \text{Four rational numbers are}\\ \frac{-2\times 2}{7\times 2},\frac{-2\times 3}{7\times 3},\frac{-2\times 4}{7\times 4},\frac{-2\times 5}{7\times 5}\\ =\frac{-4}{14},\frac{-6}{21},\frac{-8}{28},\frac{-10}{35}\\ \text{(ii)}\frac{5}{-3}\\ \text{Four rational numbers are}\\ \frac{5\times 2}{-3\times 2},\frac{5\times 3}{-3\times 3},\frac{5\times 4}{-3\times 4},\frac{5\times 5}{-3\times 5}\\ =\frac{10}{-6},\frac{15}{-9},\frac{20}{-12},\frac{25}{-15}\\ \text{(iii)}\frac{4}{9}\\ \text{Four rational numbers are}\\ \frac{4\times 2}{9\times 2},\frac{4\times 3}{9\times 3},\frac{4\times 4}{9\times 4},\frac{4\times 5}{9\times 5}\\ =\frac{8}{18},\frac{12}{27},\frac{16}{36},\frac{20}{45}\end{array}

Q.4

$\begin{array}{l}\mathrm{Draw}\mathrm{the}\mathrm{number}\mathrm{line}\mathrm{and}\mathrm{represent}\mathrm{the}\mathrm{following}\mathrm{rational}\\ \mathrm{numbers}\mathrm{on}\mathrm{it}:\\ \left(\mathrm{i}\right)\frac{3}{4}\left(\mathrm{ii}\right)\frac{-5}{8}\left(\mathrm{iii}\right)\frac{-7}{4}\left(\mathrm{iv}\right)\frac{7}{8}\end{array}$

Ans

\begin{array}{l}\text{(i)}\frac{3}{4}\\ \text{This fraction represent 3 parts out of 4 equal parts}\text{.}\\ \text{Therefore each space between two integers on number}\\ \text{line must be divided into 4 equal parts}\text{.}\\ \text{It can be represented on the number line as:}\end{array}

\begin{array}{l}\text{(ii)}\frac{-5}{8}\\ \text{This fraction represent 5 parts out of 8 equal parts}\text{.}\\ \text{Negative sign represents that it is on the negative side of the}\\ \text{number line}\text{.}\\ \text{Therefore each space between two integers on number}\\ \text{line must be divided into 8 equal parts}\text{.}\\ \text{It can be represented on the number line as:}\end{array} \begin{array}{l}\end{array}

$\begin{array}{l}\text{(iii)}\frac{-7}{4}=-1\frac{3}{4}\\ \text{This mixed fraction represent 1 full part and 3 parts out of}\\ \text{4 equal parts}\text{. Negative sign represents that it is on the}\\ \text{negative side of the number line}\text{.}\\ \text{Therefore each space between two integers on number}\\ \text{line must be divided into 4 equal parts}.\\ \text{It can be represented on the number line as:}\end{array}$ \begin{array}{l}\end{array}

\begin{array}{l}\text{(iv)}\frac{7}{8}\\ \text{This fraction represent 7 parts out of 8 equal parts}\text{.}\\ \text{Therefore each space between two integers on number}\\ \text{line must be divided into 8 equal parts}\text{.}\\ \text{It can be represented on the number line as:}\end{array}

Q.5

$\begin{array}{l}\mathrm{The}\mathrm{points}\mathrm{P},\mathrm{Q},\mathrm{R},\mathrm{S},\mathrm{T},\mathrm{U},\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{on}\mathrm{the}\mathrm{number}\mathrm{line}\mathrm{are}\\ \mathrm{such}\mathrm{that},\mathrm{TR}=\mathrm{RS}=\mathrm{S}\mathrm{U}\mathrm{and}\mathrm{AP}=\mathrm{PQ}=\mathrm{QB}.\mathrm{Name}\mathrm{the}\mathrm{rational}\\ \mathrm{numbers}\mathrm{represented}\mathrm{by}\mathrm{P},\mathrm{Q},\mathrm{R}\mathrm{and}\mathrm{S}.\end{array}$

Ans

$\begin{array}{l}\text{Distance between U and T is 1 unit.}\\ \text{It is divided into 3 equal parts.}\\ \text{TR}=\text{RS}=\text{SU=}\frac{1}{3}\\ \mathrm{R}=-1-\frac{1}{3}=\frac{-3}{3}-\frac{1}{3}=\frac{-4}{3}\\ \mathrm{S}=-1-\frac{2}{3}=\frac{-3}{3}-\frac{2}{3}=\frac{-5}{3}\\ \text{Similarly,}\\ \text{AB}=\text{1unit}\\ \text{It is divided into 3 equal parts.}\\ \text{P}=\text{2+}\frac{1}{3}=\frac{6}{3}+\frac{1}{3}=\frac{7}{3}\\ \text{Q}=\text{2+}\frac{2}{3}=\frac{6}{3}+\frac{2}{3}=\frac{8}{3}\end{array}$

Q.6

$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{pairs}\mathrm{represent}\mathrm{the}\mathrm{same}\\ \mathrm{rationa}\mathrm{lnumber}?\\ \left(\mathrm{i}\right)\frac{-7}{21}\mathrm{and}\frac{3}{9}\left(\mathrm{ii}\right)\frac{-16}{20}\mathrm{and}\frac{20}{-25}\left(\mathrm{iii}\right)\frac{-2}{-3}\mathrm{and}\frac{2}{3}\\ \left(\mathrm{iv}\right)\frac{-3}{5}\mathrm{and}\frac{-12}{20}\left(\mathrm{v}\right)\frac{8}{-5}\mathrm{and}\frac{-24}{15}\left(\mathrm{vi}\right)\frac{1}{3}\mathrm{and}\frac{-1}{9}\\ \left(\mathrm{vii}\right)\frac{-5}{-9}\mathrm{and}\frac{5}{-9}\end{array}$

Ans

\begin{array}{l}(\text{i})\frac{-7}{21}\text{and}\frac{3}{9}\\ \frac{-7}{21}=\frac{-1\times 7}{3\times 7}=\frac{-1}{3}\\ \frac{3}{9}=\frac{1\times 3}{3\times 3}=\frac{1}{3}\\ \text{Since}\frac{1}{3}\ne \frac{-1}{3}\\ \text{So, given pair does not represent same rational number}\text{.}\\ \left(\text{ii}\right)\frac{-16}{20}\text{and}\frac{20}{-25}\\ \frac{-16}{20}=\frac{-4\times 4}{4\times 5}=\frac{-4}{5}\\ \frac{20}{-25}=\frac{4\times 5}{-5\times 5}=\frac{-4}{5}\\ \text{So, given pair represent same rational number}\text{.}\end{array} $\begin{array}{l}\left(\text{iii}\right)\frac{-2}{-3}\text{and}\frac{2}{3}\\ \frac{-2}{-3}=\frac{2}{3}\\ \text{So, given pair represent same rational number}.\\ \left(\text{iv}\right)\frac{-3}{5}\text{and}\frac{-12}{20}\\ \frac{-12}{20}=\frac{-3\times 4}{5\times 4}=\frac{-3}{5}\\ \text{So, given pair represent same rational number}\\ \left(\text{v}\right)\frac{8}{-5}\text{and}\frac{-24}{15}\\ \frac{-24}{15}=\frac{-8\times 3}{5\times 3}=\frac{-8}{5}\\ \text{So, given pair represent same rational number}\text{.}\\ \left(\text{vi}\right)\frac{\text{1}}{3}\text{and}\frac{-1}{9}\\ \text{Since}\frac{1}{3}\ne \frac{-1}{9}\\ \text{So, given pair does not represent same rational number}\text{.}\\ \left(\text{vii}\right)\frac{-5}{-9}\text{and}\frac{5}{-9}\\ \frac{-5}{-9}=\frac{-1\times 5}{-1\times 9}=\frac{5}{9}\\ \frac{5}{-9}=\frac{5}{-9}\\ \text{Since}\frac{5}{9}\ne \frac{5}{-9}\\ \text{So, given pair does not represent same rational number}\text{.}\end{array}$

Q.7

$\begin{array}{l}\mathrm{Rewrite}\mathrm{the}\mathrm{following}\mathrm{rational}\mathrm{numbers}\mathrm{in}\mathrm{the}\\ \mathrm{simplest}\mathrm{form}:\\ \left(\mathrm{i}\right)\frac{-8}{6}\left(\mathrm{ii}\right)\frac{25}{45}\left(\mathrm{iii}\right)\frac{-44}{72}\left(\mathrm{iv}\right)\frac{-8}{10}\end{array}$

Ans

\begin{array}{l}\left(\text{i}\right)\frac{-8}{6}\\ \frac{-8}{6}=\frac{-4\times \cancel{2}}{3\times \cancel{2}}=\frac{-4}{3}\\ \left(\text{ii}\right)\frac{\text{25}}{45}\\ \frac{25}{45}=\frac{\cancel{5}\times 5}{\cancel{5}\times 9}=\frac{5}{9}\\ \left(\text{iii}\right)\frac{-\text{44}}{72}\\ \frac{-44}{72}=\frac{-11\times \cancel{4}}{18\times \cancel{4}}=\frac{-11}{18}\\ \left(\text{iv}\right)\frac{-8}{10}\\ -\frac{8}{10}=\frac{-4\times \cancel{2}}{5\times \cancel{2}}=\frac{-4}{5}\end{array}

Q.8

$\begin{array}{l}\mathrm{Which}\mathrm{is}\mathrm{greater}\mathrm{in}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}:\\ \left(\mathrm{i}\right)\frac{2}{3},\frac{5}{2}\left(\mathrm{ii}\right)\frac{-5}{6},\frac{-4}{3}\left(\mathrm{iii}\right)\frac{-3}{4},\frac{2}{-3}\\ \left(\mathrm{iv}\right)\frac{-1}{4},\frac{1}{4}\left(\mathrm{v}\right)-3\frac{2}{7},-3\frac{4}{5}\end{array}$

Ans

\begin{array}{l}\left(\text{i}\right)\frac{2}{3},\frac{5}{2}\\ \frac{2}{3}=\frac{2\times 2}{3\times 2}=\frac{4}{6}\\ \frac{5}{2}=\frac{5\times 3}{2\times 3}=\frac{15}{6}\\ \text{Since, 1}5>4\\ \text{So,}\frac{5}{2}\text{is greater than}\frac{2}{3}.\\ \left(\text{ii}\right)\frac{-5}{6},\frac{-4}{3}\\ \frac{-5}{6}=\frac{-5\times 3}{6\times 3}=\frac{-15}{18}\\ \frac{-4}{3}=\frac{-4\times 6}{3\times 6}=\frac{-24}{18}\\ \text{Since,}-15>-24\\ \text{So,}\frac{-5}{6}\text{is greater than}\frac{-4}{3}.\end{array} \begin{array}{l}\left(\text{iii}\right)\frac{-3}{4},\frac{2}{-3}\\ \frac{-3}{4}=\frac{-3\times 3}{4\times 3}=\frac{-9}{12}\\ \frac{2}{-3}=\frac{2\times 4}{-3\times 4}=\frac{-8}{12}\\ \text{Since,}-8>-9\\ \text{So,}\frac{2}{-3}\text{is greater than}\frac{-3}{4}.\\ \left(\text{iv}\right)\frac{-1}{4},\frac{1}{4}\\ \text{Since,}1>-1\\ \text{So,}\frac{1}{4}\text{is greater than}\frac{-1}{4}.\\ \left(\text{v}\right)-3\frac{2}{7},-3\frac{4}{5}\\ -3\frac{2}{7}=\frac{-23}{7}=\frac{-23\times 5}{7\times 5}=\frac{-115}{35}\\ -3\frac{4}{5}=\frac{-19}{5}=\frac{-19\times 7}{5\times 7}=\frac{-133}{35}\\ \text{Since,}-\text{115>}-\text{133}\\ \text{So,}-\text{3}\frac{2}{7}\text{is greater than}-\text{3}\frac{4}{5}.\end{array}

Q.9

$\begin{array}{l}\mathrm{Write}\mathrm{the}\mathrm{following}\mathrm{rational}\mathrm{numbers}\mathrm{in}\mathrm{a}\mathrm{scending}\mathrm{order}:\\ \left(\mathrm{i}\right)\frac{-3}{5},\frac{-2}{5},\frac{-1}{5}\left(\mathrm{ii}\right)\frac{-1}{3},\frac{-2}{9},\frac{-4}{3}\left(\mathrm{iii}\right)\frac{-3}{7},\frac{-3}{2},\frac{-3}{4}\end{array}$

Ans

$\begin{array}{l}\text{(i)}\frac{-3}{5},\frac{-2}{5},\frac{-1}{5}\\ \text{Since,}-1>-2>-3\\ \text{So,}\frac{-3}{5}<\frac{-2}{5}<\frac{-1}{5}.\\ \\ \text{(ii)}\frac{-1}{3},\frac{-2}{9},\frac{-4}{3}\\ \frac{-1}{3}=\frac{-1\times 3}{3\times 3}=\frac{-3}{9}\\ \frac{-4}{3}=\frac{-4\times 3}{3\times 3}=\frac{-12}{9}\\ \text{Since,}-2>-3>-12\\ \text{So,}\frac{-4}{3}<\frac{-1}{3}<\frac{-2}{9}.\end{array}$ \begin{array}{l}\text{(iii)}\frac{-3}{7},\frac{-3}{2},\frac{-3}{4}\\ \frac{-3}{7}=\frac{-3\times 4}{7\times 4}=\frac{-12}{28}\\ \frac{-3}{2}=\frac{-3\times 14}{2\times 14}=\frac{-42}{28}\\ \frac{-3}{4}=\frac{-3\times 7}{4\times 7}=\frac{-21}{28}\\ \text{Since,}-12>-21>-42\\ \text{So,}\frac{-3}{2}<\frac{-3}{4}<\frac{-3}{7}.\end{array}

Q.10

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{sum}:\\ \left(\mathrm{i}\right)\frac{5}{4}+\left(\frac{-11}{4}\right)\left(\mathrm{ii}\right)\frac{5}{3}+\frac{3}{5}\left(\mathrm{iii}\right)\frac{-9}{10}+\frac{22}{15}\\ \left(\mathrm{iv}\right)\frac{-3}{-11}+\frac{5}{9}\left(\mathrm{v}\right)\frac{-8}{19}+\frac{(-2)}{57}\left(\mathrm{vi}\right)\frac{-2}{3}+0\\ \left(\mathrm{vii}\right)-2\frac{1}{3}+4\frac{3}{5}\end{array}$

Ans

$\begin{array}{l}\text{(i)}\frac{5}{4}+\left(\frac{-11}{4}\right)\\ =\frac{5}{4}-\frac{11}{4}=\frac{-6}{4}=\frac{-3}{2}\\ \\ \text{(ii)}\frac{5}{3}+\frac{3}{5}\\ \text{L.C.M of 3 and 5 is 15.}\\ \text{So,}\frac{5}{3}+\frac{3}{5}=\frac{5\times 5}{3\times 5}+\frac{3\times 3}{5\times 3}\\ =\frac{25}{15}+\frac{9}{15}\\ =\frac{25+9}{15}=\frac{34}{15}\end{array}$ $\begin{array}{l}\text{(iii)}\frac{-9}{10}+\frac{22}{15}\\ \text{L}\text{.C}\text{.M of 10 and 15 is 30.}\\ \frac{-9}{10}+\frac{22}{15}=\frac{-9\times 3}{10\times 3}+\frac{22\times 2}{15\times 2}\\ =\frac{-27}{30}+\frac{44}{30}\\ =\frac{-27+44}{30}=\frac{17}{30}\\ \\ \text{(iv)}\frac{-3}{-11}+\frac{5}{9}\\ \text{L}\text{.C}\text{.M of 9 and 11 is 99}\text{.}\\ \text{So,}\frac{-3}{-11}+\frac{5}{9}=\frac{3}{11}+\frac{5}{9}\\ =\frac{3\times 9}{11\times 9}+\frac{5\times 11}{9\times 11}\\ =\frac{27}{99}+\frac{55}{99}=\frac{82}{99}\\ \\ \text{(v)}\frac{-8}{19}+\frac{(-2)}{57}\\ \text{L}\text{.C}\text{.M of 19 and 57 is 57.}\\ \text{So,}\frac{-8}{19}+\frac{(-2)}{57}=\frac{-8\times 3}{19\times 3}-\frac{2}{57}\\ =\frac{-24}{57}-\frac{2}{57}\\ =\frac{-24-2}{57}=-\frac{26}{57}\\ \end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\text{(vi)}\frac{-2}{3}+0\\ =\frac{-2}{3}+0\times \frac{3}{3}\\ =-\frac{2}{3}+0=\frac{-2}{3}\\ \\ \text{(vii) -2}\frac{1}{3}+4\frac{3}{5}\\ =-\frac{7}{3}+\frac{23}{5}\\ \text{L}\text{.C}\text{.M of 3 and 5 is 15}\text{.}\\ \text{So,}-\frac{7}{3}+\frac{23}{5}=\frac{-7\times 5}{3\times 5}+\frac{23\times 3}{5\times 3}\\ =\frac{-35}{15}+\frac{69}{15}\\ =\frac{-35+69}{15}=\frac{34}{15}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Find}\\ \left(\mathrm{i}\right)\frac{7}{24}-\frac{17}{36}\left(\mathrm{ii}\right)\frac{5}{63}-\left(\frac{-6}{21}\right)\left(\mathrm{iii}\right)\frac{-6}{13}-\left(\frac{-7}{15}\right)\\ \left(\mathrm{iv}\right)\frac{-3}{8}-\frac{7}{11}\left(\mathrm{v}\right)-2\frac{1}{9}-6\end{array}$

Ans

$\begin{array}{l}\text{(i)}\frac{7}{24}-\frac{17}{36}\\ \text{Since, L.C.M of 24 and 36 is 72.}\\ \text{So,}\frac{7}{24}-\frac{17}{36}=\frac{7\times 3}{24\times 3}-\frac{17\times 2}{36\times 2}\\ =\frac{21}{72}-\frac{34}{72}=\frac{21-34}{72}=\frac{-13}{72}\\ \\ \text{(ii)}\frac{5}{63}-\left(\frac{-6}{21}\right)\\ =\frac{5}{63}+\frac{6}{21}\\ \text{Since, L.C.M of 21 and 63 is 63.}\\ \text{So,\hspace{0.17em}}\frac{5}{63}+\frac{6}{21}=\frac{5}{63}+\frac{6\times 3}{21\times 3}\\ =\frac{5}{63}+\frac{18}{63}=\frac{5+18}{63}=\frac{23}{63}\\ \\ \text{(iii)}\frac{-6}{13}-\left(\frac{-7}{15}\right)\\ =\frac{-6}{13}+\frac{7}{15}\\ \text{Since, L.C.M of 13 and 15 is 195.}\\ \text{So,}\frac{-6\times 15}{13\times 15}+\frac{7\times 13}{13\times 15}=\frac{-90}{195}+\frac{91}{195}\\ =\frac{-90+91}{195}=\frac{1}{195}\end{array}$ $\begin{array}{l}\text{(iv)}\frac{-3}{8}-\frac{7}{11}\\ \text{Since, L.C.M of 8 and 11 is 88.}\\ \text{So,}\frac{-3\times 11}{8\times 11}-\frac{7\times 8}{11\times 8}=\frac{-33}{88}-\frac{56}{88}\\ =\frac{-33-56}{88}=\frac{-89}{88}\\ \\ \text{(v)}-2\frac{1}{9}-6\\ =-\frac{19}{9}-6\\ \text{Since, L.C.M of 9 and 1 is 9.}\\ \text{So,}\frac{-19}{9}-6=\frac{-19}{9}-\frac{6\times 9}{1\times 9}\\ =\frac{-19}{9}-\frac{54}{9}\\ =\frac{-73}{9}\end{array}$

Q.12

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{product}:\\ \left(\mathrm{i}\right)\frac{9}{2}\times \left(\frac{-7}{4}\right)\left(\mathrm{ii}\right)\frac{3}{10}\times (-9)\left(\mathrm{iii}\right)\frac{-6}{5}\times \frac{9}{11}\\ \left(\mathrm{iv}\right)\frac{3}{7}\times \left(\frac{-2}{5}\right)\left(\mathrm{v}\right)\frac{3}{11}\times \frac{2}{5}\left(\mathrm{vi}\right)\frac{3}{-5}\times \frac{-5}{3}\end{array}$

Ans

$\begin{array}{l}\text{(i)}\frac{9}{2}\times \left(\frac{-7}{4}\right)\\ =\frac{9}{2}\times \frac{-7}{4}=\frac{-63}{8}\\ \text{(ii)}\frac{3}{10}\times (-9)\\ =-\frac{3}{10}\times 9=\frac{-27}{10}\\ \text{(iii)}\frac{-6}{5}\times \frac{9}{11}\\ =\frac{-6\times 9}{5\times 11}=\frac{-54}{55}\\ \text{(iv)}\frac{3}{7}\times \left(\frac{-2}{5}\right)\\ =\frac{3\times \left(-2\right)}{7\times 5}=\frac{-6}{35}\\ \text{(v)}\frac{3}{11}\times \frac{2}{5}\\ =\frac{3\times 2}{11\times 5}=\frac{6}{55}\\ \text{(vi)}\frac{3}{-5}\times \frac{-5}{3}\\ =\frac{3\times \left(-5\right)}{-5\times 3}=\frac{-15}{-15}=1\end{array}$

Q.13

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}:\\ \left(\mathrm{i}\right)(-4)÷\frac{2}{3}\left(\mathrm{ii}\right)\frac{-3}{5}÷2\left(\mathrm{iii}\right)\frac{-4}{5}÷(-3)\\ \left(\mathrm{iv}\right)\frac{-1}{8}÷\frac{3}{4}\left(\mathrm{v}\right)\frac{-2}{13}÷\frac{1}{7}\left(\mathrm{vi}\right)\frac{-7}{12}÷\left(\frac{-2}{13}\right)\\ \left(\mathrm{vii}\right)\frac{3}{13}÷\left(\frac{-4}{65}\right)\end{array}$

Ans

$\begin{array}{l}\text{(i)}(-4)÷\frac{2}{3}\\ =-4\times \frac{3}{2}=\frac{-12}{2}=-6\\ \text{(ii)}\frac{-3}{5}÷2\\ =\frac{-3}{5}\times \frac{1}{2}=\frac{-3}{10}\\ \text{(iii)}\frac{-4}{5}÷(-3)\\ =\frac{-4}{5}\times \frac{1}{-3}=\frac{-4}{-15}=\frac{4}{15}\end{array}$ $\begin{array}{l}\text{(iv)}\frac{-1}{8}÷\frac{3}{4}\\ =\frac{-1}{8}\times \frac{4}{3}=\frac{-4}{24}=\frac{-1}{6}\\ \text{(v)}\frac{-2}{13}÷\frac{1}{7}\\ =\frac{-2}{13}\times \frac{7}{1}=\frac{-14}{13}\\ \text{(vi)}\frac{-7}{12}÷\left(\frac{-2}{13}\right)\\ =\frac{-7}{12}\times \frac{13}{-2}=\frac{-91}{-24}=\frac{91}{24}\\ \text{(vii)}\frac{3}{13}÷\left(\frac{-4}{65}\right)\\ =\frac{3}{13}\times \frac{65}{-4}=\frac{3\times 65}{13\times \left(-4\right)}\\ =\frac{3\times \cancel{13}\times 5}{\cancel{13}\times \left(-4\right)}=\frac{15}{-4}=\frac{-15}{4}\end{array}$

Q.14

$\begin{array}{l}\mathrm{Fill}\mathrm{in}\mathrm{the}\mathrm{boxes}\mathrm{with}\mathrm{the}\mathrm{correct}\mathrm{symbol}\mathrm{out}\mathrm{of}>,<,\mathrm{and}=\\ \left(\mathrm{i}\right)\frac{-5}{7}\boxed{}\frac{2}{3}\left(\mathrm{ii}\right)\frac{-4}{5}\boxed{}\frac{-5}{7}\left(\mathrm{iii}\right)\frac{-7}{8}\boxed{}\frac{14}{-16}\\ \left(\mathrm{iv}\right)\frac{-8}{5}\boxed{}\frac{-7}{4}\left(\mathrm{v}\right)\frac{1}{-3}\boxed{}\frac{-1}{4}\left(\mathrm{vi}\right)\frac{5}{-11}\boxed{}\frac{-5}{11}\\ \left(\mathrm{vii}\right)0\boxed{}\frac{-7}{6}\end{array}$