Answer 9

\[ \begin{align*} \frac{-3}{8} &\quad {\Huge \times} \quad \frac{x}{-24} \\ \\ (-3) \times (-24) &\quad = \quad x \times 8 \\ \\ \frac{(-3) \times \cancel{(-24)}^{-3}}{\cancel{8}^1} &\quad = \quad x \\ \\ (-3) \times (-3) &\quad = \quad x \\ x &\quad = \quad \boxed{\color{red}9} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{-5}{7} \)

\[ \begin{align*} \frac{-15}{-4} &= \frac{15}{4} \text{ (Positive rational number) } \\ \\ 0 &= \text{ Neither Negative or Positive rational number } \\ \\ \boxed{\color{red} \frac{-5}{7}} &= \text{ (Negative rational number) } \\ \\ \frac{4}{9} &= \text{ (Positive rational number) } \\ \\ \end{align*} \]

Answer \(\displaystyle \color{red} \frac{6}{5} \)

\[ \begin{align*} & = \frac{\cancel{12}^6}{\cancel{10}_5} \\ \\ & = \frac{6}{5} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{-5}{17} \)

\[ \begin{align*} \frac{-5}{11}, \frac{-5}{12} , \frac{-5}{17} \\ \\ \text{Compare } & \frac{-5}{11} \text{ and } \frac{-5}{12} \\ \\ \frac{-5}{11} &\quad {\Huge \times} \quad \frac{-5}{12} \\ \\ (-5) \times 12 &\quad \text{and} \quad (-5) \times 11 \\ \\ -60 &\quad < \color{green} \quad -55 \\ \\ \implies \frac{-5}{11} &\quad < \quad \color{green} \frac{-5}{12} \\ \\ \\ \text{Compare } & \frac{-5}{12} \text{ and } \frac{-5}{17} \\ \\ \frac{-5}{12} &\quad {\Huge \times} \quad \frac{-5}{17} \\ \\ (-5) \times 17 &\quad \text{and} \quad (-5) \times 12 \\ \\ -85 &\quad < \quad \color{green} -60 \\ \\ \implies \frac{-5}{12} &\quad < \color{green} \quad \frac{-5}{17} \\ \\ \\ \implies \frac{-5}{11} &\quad < \quad \frac{-5}{12} < \quad \boxed{ \color{red} \frac{-5}{17}} \\ \\ &= {\color{red}\frac{-5}{17}} \text{ is the greatest} \end{align*} \]

Answer \( \displaystyle \color{red} \left| \frac{-2}{11} \right| \)

\[ \begin{align*} \left| \frac{7}{11} \right| &= \frac{7}{11} \\ \\ \left| \frac{-8}{11} \right| &= \frac{8}{11} \\ \\ \left| \frac{-2}{11} \right| &= \frac{2}{11} \\ \\ \left| \frac{-9}{-11} \right| &= \frac{9}{11} \\ \\ \implies \left| {\color{red} \frac{-2}{11}} \right| & \text{ is the smallest} \end{align*} \]

\[ \begin{align*} \color{green} \text{Average} &= \color{green}\frac{\text{Sum of all observation}}{\text{Total No. of observation}} \\ \\ \text{Sum} &= \frac{4}{5} + \frac{2}{3} + \frac{5}{6} \\ \\ \end{align*} \] \[ \begin{array}{c|ccc} 2 & 5, & 3, & 6 \\ \hline 3 & 5, & 3, & 3 \\ \hline 5 & 5, & 1, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array} \] \[ \begin{align*} \text{LCM} &= 2 \times 3 \times 5 \\ &= 30 \\ \\ \frac{4 \times 6 }{5 \times 6 } & = \frac{24}{30} \\ \\ \frac{2 \times 10 }{3 \times 10 } & = \frac{20}{30} \\ \\ \frac{5 \times 5 }{6 \times 5 } & = \frac{25}{30} \\ \\ \text{Sum} &= \frac{24}{30} + \frac{20}{30} + \frac{25}{30} \\ \\ &= \frac{24 + 20 + 25}{30} \\ \\ \text{Sum}&= \frac{69}{30} \\ \\ \text{Average}&= \frac{69}{30} \div 3\\ \\ &= \frac{\cancel{69}^{23}}{30} \times \frac{1}{\cancel3_1}\\ \\ &= \frac{23}{30} \\ \\ \end{align*} \]

Answer \(\displaystyle \color{red} \frac{23}{30} \)

\[ \begin{align*} \frac{12}{-18} & = \frac{12 \times (-1)}{-18 \times (-1) } \\ \\ & = \frac{\cancel{-12}^{-2}}{\cancel{18}_3} \\ \\ &= \frac{-2}{3} \end{align*} \]

Answer \(\color{red}\displaystyle \frac{-2}{3} \)

Answer Infinite

Answer Right side

Answer \( \displaystyle \color{red} \frac{35}{40} \)

Answer False

\[ \begin{array}{c|c} \displaystyle \frac{p}{q} = \frac{-2}{3} , \frac{r}{s} = \frac{-4}{3} & \displaystyle \left| \frac{p}{q} \right| = \frac{2}{3} , \left| \frac{r}{s} \right| = \frac{4}{3} \\ \\ \displaystyle \frac{-2}{3} > \frac{-4}{3} & \displaystyle \frac{2}{3} < \frac{4}{3} \\ \\ \end{array} \]

Answer False

\[ \begin{align*} \text{Let} \\ x = \frac{-1}{2} \quad &, \quad y = \frac{1}{2} \\ \end{align*} \] \[ \begin{array}{c|c} | x | = | y | \quad & \quad x = y \\ \\ \displaystyle \left| \frac{-1}{2} \right| , \left| \frac{1}{2} \right| & \displaystyle \frac{-1}{2} , \frac{1}{2} \\ \\ \displaystyle \frac{1}{2} = \frac{1}{2} & \displaystyle \frac{-1}{2} \neq \frac{1}{2} \\ \\ \end{array} \]

Answer False

\[ \begin{align*} \frac{p}{q} & = \frac{-3}{8} \\ \\ \frac{q}{p} & = \frac{8}{-3} \text{ (Not in standard form)} \\ \\ \end{align*} \]

Answer

\( \displaystyle 5\frac{1}{3} = \frac{16}{3} \)

\( \implies \displaystyle \frac{-29}{4} \)

\[ \begin{align*} \frac{-3}{10} , \frac{7}{5} , \frac{-9}{15} , \frac{18}{30} \\ \\ \end{align*} \] \[ \begin{array}{c|cccc} 2 & 10, & 5, & 15, & 30 \\ \hline 3 & 5, & 5, & 15, & 15 \\ \hline 5 & 5, & 5, & 5, & 5 \\ \hline & 1, & 1, & 1, & 1 \\ \end{array} \] \[ \begin{align*} \text{LCM } &= 2 \times 3 \times 5 \\ &= 30 \\ \\ \frac{-3 \times 3}{10 \times 3} &= \frac{-9}{30} \\ \\ \frac{7 \times 6}{5 \times 6} &= \frac{42}{30} \\ \\ \frac{-9 \times 2}{15 \times 2} &= \frac{-18}{30} \\ \\ \frac{18 \times 1}{30 \times 4} &= \frac{18}{30} \\ \\ \text{Descending Order} &= \frac{42}{30}, \frac{18}{30}, \frac{-9}{30}, \frac{-18}{30} \\ \\ &= \frac{-7}{-5} , \frac{18}{30} , \frac{-3}{10} , \frac{9}{-15} \\ \end{align*} \]

Answer \( \displaystyle \color{red} \frac{-7}{-5} , \frac{18}{30} , \frac{-3}{10} , \frac{9}{-15} \)

Answer Length of the line-segment = 6 units.

Answer Length of the line-segment = 1 units.

Answer Length of the line-segment = 2 units.

Answer Length of the line-segment = 2 units.

\[ \begin{align*} \frac{23}{x} &\quad {\Huge \times} \quad \frac{2}{-8} \\ \\ 23 \times (-8) &\quad = \quad 2 \times x \\ \\ \frac{23 \times \cancel{(-8)}^{-4}}{\cancel2_1} &\quad = \quad x \\ \\ 23 \times (-4) &\quad = \quad x \\ \color{green} -92 &\quad = \quad x \end{align*} \]

Answer \( x = \boxed{\color{red}-92} \\ \)

\[ \begin{align*} \frac{x}{9} &\quad {\Huge \times} \quad \frac{19}{3} \\ \\ x \times 3 &\quad = \quad 19 \times 9 \\ \\ x &\quad = \quad \frac{19 \times \cancel9^3}{\cancel3_1} \\ \\ x &\quad = \quad 19 \times 3 \\ x &\quad = \quad \color{green} 57 \end{align*} \]

Answer \( x = \boxed{\color{red}57} \\ \)

\[ \begin{align*} \frac{15}{-x} &\quad {\Huge \times} \quad \frac{1}{-7} \\ \\ -x \times 1 &\quad = \quad 15 \times (-7) \\ \\ -x &\quad = \quad -105 \\ x &\quad = \quad \color{green} 105 \end{align*} \]

Answer \( x = \boxed{\color{red}105} \\ \)

\[ \begin{align*} = \frac{-5}{7} &, \frac{-9}{13} \\ \\ \frac{-5}{7} &\quad {\Huge \times} \quad \frac{-9}{13} \\ \\ (-5) \times 13 &\quad \text{and} \quad (-9) \times 7 \\ \\ -65 &\quad < \quad -63 \\ \\ \frac{-5}{7} &\quad < \quad \frac{9}{-13} \\ \end{align*} \]

Answer \( \displaystyle \frac{-5}{7} < \frac{9}{-13} \)

\[ \begin{align*} \frac{-4}{9} &\quad {\Huge \times} \quad \frac{-3}{7} \\ \\ (-4) \times 7 &\quad \text{and} \quad (-3) \times 9 \\ \\ -28 &\quad < \quad -27 \\ \\ \frac{-4}{9} &\quad < \quad \frac{-3}{7} \\ \end{align*} \]

Answer \( \displaystyle \frac{-4}{9} < \frac{-3}{7} \)

\[ \begin{align*} = \frac{3}{5} &,\frac{12}{20} \\ \\ \frac{3}{5} &\quad {\Huge \times} \quad \frac{12}{20} \\ \\ 3 \times 20 &\quad \text{and} \quad 12 \times 5 \\ \\ 60 &\quad = \quad 60 \\ \\ \frac{-3}{-5} &\quad = \quad \frac{12}{20} \\ \end{align*} \]

Answer \( \displaystyle \frac{-3}{-5} = \frac{12}{20} \)

\[ \begin{align*} \left| \frac{-4}{5} \right| & = \frac{4}{5} \\ \\ \left| \frac{-5}{4} \right| & = \frac{5}{4} \\ \\ \frac{4}{5} &\quad {\Huge \times} \quad \frac{5}{4} \\ \\ 4 \times 4 &\quad \text{and} \quad 5 \times 5 \\ \\ 16 &\quad < \quad 25 \\ \\ \left| \frac{-4}{5} \right| &\quad < \quad \left| \frac{-5}{4} \right| \\ \end{align*} \]

Answer \( \displaystyle \left| \frac{-4}{5} \right| < \left| \frac{-5}{4} \right| \)

\[ \begin{align*} \left| \frac{5}{7} \right| & = \frac{5}{7} \\ \\ \left| \frac{-15}{21} \right| & = \frac{\cancel{15}^5}{\cancel{21}_7} \implies \frac{5}{7} \\ \\ \frac{5}{7} &\quad = \quad \frac{5}{7} \\ \\ \left| \frac{5}{7} \right| & = \left| \frac{-15}{21} \right| \\ \end{align*} \]

Answer \( \displaystyle \left| \frac{5}{7} \right| = \left| \frac{-15}{21} \right| \)

\[ \begin{align*} \left| \frac{-8}{-9} \right| & = \frac{8}{9} \\ \\ \left| \frac{-3}{9} \right| & = \frac{3}{9} \\ \\ \frac{8}{9} &\quad > \quad \frac{3}{9} \\ \\ \left| \frac{-8}{-9} \right| &\quad > \quad \left| \frac{-3}{9} \right| \\ \end{align*} \]

Answer \( \displaystyle \left| \frac{-8}{-9} \right| > \left| \frac{-3}{9} \right| \)

\[ \begin{align*} \frac{3}{5} &= \frac{138}{\boxed{x}} \\ \\ 3 \times x & = 138 \times 5 \\ \\ x & = \frac{\cancel{138}^{46} \times 5}{\cancel3_1} \\ \\ x & = 46 \times 5 \\ x & = 230 \end{align*} \]

Answer \( \displaystyle \frac{3}{5} = \frac{138}{\boxed{\color{red}230}} \)

\[ \begin{align*} \frac{7}{9} &= \frac{\boxed{x}}{108} \\ \\ 9 \times x & = 108 \times 7 \\ \\ x & = \frac{\cancel{108}^{12} \times 7}{\cancel9_1} \\ \\ x & = 12 \times 7 \\ x & = 84 \end{align*} \]

Answer \( \displaystyle \frac{7}{9} = \frac{\boxed{\color{red}84}}{108} \)

\[ \begin{align*} \frac{\boxed{x}}{-15} &= \frac{48}{90} \\ \\ 90 \times x & = 48 \times (-15) \\ \\ x & = \frac{\cancel{48}^8 \times \cancel{(-15)}^{-1}}{\cancel{90}_{\cancel{15}_1}} \\ \\ x & = 8 \times (-1) \\ x & = -8 \end{align*} \]

Answer \( \displaystyle \frac{\boxed{\color{red}-8}}{-15} = \frac{48}{90} \)

\[ \begin{align*} \frac{121}{\boxed{x}} &= \frac{-11}{12} \\ \\ -11 \times x & = 121 \times 12 \\ \\ x & = \frac{\cancel{121}^{11} \times 12}{{\cancel{-11}_{-1}}} \\ \\ x & = \frac{11 \times 12}{-1} \\ \\ x & = \frac{132}{-1} \\ \\ x & = -132 \end{align*} \]

Answer \( \displaystyle \frac{121}{\boxed{\color{red}-132}} = \frac{-11}{12} \)