$\dfrac{(3\frac{2}{3})^2-(2\frac{1}{2})^2}{(4\frac{3}{2})^2-(3\frac{1}{3})^2} \div \dfrac{3\frac{2}{3}-2\frac{1}{2}}{4\frac{3}{2}-3\frac{1}{3}}$

$\left[\dfrac{170}{3} +\dfrac{6}{7}\right] \div \left[\dfrac{2}{7} \times \dfrac{11}{2}\right]$

Difference of  $\dfrac{1}{5}$  and $\dfrac{2}{7}$ is  

Consider the unit digits of $7^{x}$ series which is $7,9,3,1$

$ \dfrac {\dfrac {5}{2.25}}{\dfrac {9}{2.25}} = \dfrac {5}{9}$

Given exp =$\displaystyle =\frac{2+4}{1\times 4}\div \frac{\frac{1}{2}\div \frac{1}{4}}{\frac{1}{2}+\frac{1}{4}}$
$\displaystyle =\frac{6}{2}\div\frac{\frac{1}{2}\times \frac{4}{1}}{\frac{3}{4}}=3\div \frac{2}{\frac{3}{4}} $
$\displaystyle =3\div \frac{8}{3}=3\times \frac{3}{8}=\frac{9}{8}=1\frac{1}{8}$

Given expression

(Using BODMAS)
$\displaystyle \frac{\frac{1}{3}\div \frac{1}{3}\times \frac{1}{3}}{\frac{1}{3}\div \frac{1}{3}of\frac{1}{3}}-\frac{1}{9}=\frac{\frac{1}{3}\times \frac{3}{1}\times \frac{1}{3}}{\frac{1}{3}\div \frac{1}{9}}-\frac{1}{9}$
$\displaystyle =\frac{1\times \frac{1}{3}}{\frac{1}{3}\times \frac{9}{1}}-\frac{1}{9}=\frac{1}{3\times 3}-\frac{1}{9}=\frac{1}{9}-\frac{1}{9}=0$

$Let\quad fraction\quad x\quad then\ x\times x\div \dfrac { 1 }{ x } =18\dfrac { 26 }{ 27 } =\dfrac { 512 }{ 27 }$

$\displaystyle \frac{2}{\displaystyle1+\frac{1}{\displaystyle1-\frac{1}{\displaystyle2}}}\times\frac{3}{\displaystyle\frac{5}{\displaystyle6}of\frac{3}{\displaystyle2}\div 1\frac{1}{\displaystyle4}}$
$=\displaystyle \frac{2}{\displaystyle1+\frac{1}{\displaystyle\frac{1}{\displaystyle2}}}\times\frac{3}{\displaystyle\frac{15}{\displaystyle12}\div \frac{5}{\displaystyle4}}$

Any number  divided by 1 equals that number.
for eg:
$8 \div 1 = 8$
$10 \div 1 = 10$
$\therefore $When any number is divided by 1, the quotient is number itself.
Option D is correct.

Given exp.
$\displaystyle =\left [ \frac{13}{4}\div \left { \frac{5}{4}-\frac{1}{2}\left ( \frac{5}{2}-\frac{3\, \, \,2}{2} \right ) \right } \right ]\div \left ( \frac{1}{2}of\frac{13}{3} \right )$
$\displaystyle =\left [ \frac{13}{4}\div \left { \frac{5}{4}-\frac{1}{2}\left ( \frac{5}{2}-\frac{1}{12} \right ) \right } \right ]\div \frac{13}{6} $
$=\displaystyle \left [ \frac{13}{4}\div \left { \frac{5}{4}-\frac{1}{2}\times \frac{30-1}{12} \right } \right ]\div \frac{13}{6}$
$\displaystyle =\left [ \frac{13}{4}\div \left { \frac{5}{4}-\frac{29}{24} \right } \right ]\div \frac{13}{6}$
$\displaystyle =\left [ \frac{13}{4}\div \frac{30-29}{24} \right ]\div \frac{13}{6}$
$\displaystyle =\left ( \frac{13}{4} \div \frac{1}{24}\right )\div \frac{13}{4}=\frac{13}{4}\times 24\times \frac{6}{13}=36$

$\displaystyle\frac{280.5}{25.5}\,=\, \frac{280.5}{25.5}\,\times\, \frac{10}{10}\,\times\, \frac{10}{10}$

$=\,\displaystyle\frac{2805}{2.55}\,\times\,\frac{1}{100}$

$=\,\displaystyle\frac{1100}{100}\,=\, 11$

$143.6\div 200=71.8\div100=0.0718$

15 of $\displaystyle \frac{1}{5}\, =\, 15\, \times\, \displaystyle \frac{1}{5}\, =\, 3$

values of $\sqrt { 5 } $ and $\sqrt { 10 } $ are given.

$\frac { 1 }{ \sqrt { 3 } +\sqrt { 2 } -1 } $
$=\frac { \sqrt { 3 } -\left( \sqrt { 2 } -1 \right)  }{ \left( \sqrt { 3 } +\left( \sqrt { 2 } -1 \right)  \right) \left( \sqrt { 3 } -\left( \sqrt { 2 } -1 \right)  \right)  } Multiplying\sqrt { 3 } -\left( \sqrt { 2 } -1 \right) \quad with\quad numerator\quad and\quad denominator\quad $
$=\frac { \sqrt { 3 } -\left( \sqrt { 2 } -1 \right)  }{ 3-{ \left( \sqrt { 2 } -1 \right)  }^{ 2 } } $
 $=\frac { \sqrt { 3 } -\left( \sqrt { 2 } -1 \right)  }{ 3-{ \left( 2+1-2\sqrt { 2 }  \right)  } } $
 $=\frac { \sqrt { 3\quad  } -\sqrt { 2 } +1 }{ 2\sqrt { 2 }  } $
$=\frac { 1.732-1.4142+1 }{ 2.8284 } =0.466$
                                               $(\sqrt { 3 } =\frac { \sqrt { 6 }  }{ \sqrt { 2 }  } =\frac { 2.4495 }{ 1.4142 } =1.732)$

That is the role of the decimal point. The decimal point separates the place values that are whole values on the left from the place values that are fractional parts on the right.

$\displaystyle \frac{280.5}{25.5}=\frac{280.5}{25.5}\times \frac{10}{10}\times \frac{10}{10}$

$\displaystyle =\frac{2805}{2.55}\times \frac{1}{100}$

$\displaystyle =\frac{1100}{100}=11$

$\dfrac{6\dfrac{1}{3}}{9}$ is complex fraction.

Given that 

$\dfrac{4}{5} \div 2 = \dfrac{\dfrac{4}{5}}{2} = \dfrac{4}{10}$

Every number has a reciprocal except 0 $(\dfrac{1}{0}$ is undefined$)$. The reciprocal is shown as $\dfrac{1}{x}$. 

sum of $\dfrac{65}{12}$  and $\dfrac{12}{7} $

$\left(\large{\frac{-5}{3}}\right)^5$ $\div$ $\left(\large{\frac{-5}{3}}\right)^{7}$

$\dfrac {\left(\dfrac {-3}{5}\right)^{3} \times \left(\dfrac {9}{25}\right)^{2} \times \left(\dfrac {-18}{125}\right)^{0}}{\left(\dfrac {-27}{125}\right) \times \left(\dfrac {-3}{5}\right)}=\dfrac{\dfrac{-3^3\times 3^4\times 1}{5^3\times 5^4\times 1}}{\dfrac{3^3\times 3}{5^3\times 5}}=-\dfrac{3^{7-4}}{5^{7-4}}=-\dfrac{27}{125}$

Given, $ 8 \times \displaystyle \frac {\frac {5}{24}}{\frac {7}{12}} $

$ I=\cfrac{3}{4}\times \cfrac{6}{5}=\cfrac{9}{10}$

$\displaystyle 1\frac{1}{3}\div 1\frac{1}{2}=\frac{4}{3}\div \frac{3}{2}\div \frac{10}{9}$
$\displaystyle =\frac{4}{3}\times \frac{2}{3}\div \frac{10}{9}=\frac{8}{9}\times \frac{9}{10}=\frac{4}{5}$
$\displaystyle \therefore \frac{1}{5}$ should be added to $\displaystyle \frac{4}{5}$ to make it an integer

$m, n$ are integer
$1/3 < m/n < 1 $...(i)
$\displaystyle \frac{m + \alpha}{n \alpha} = \frac{m}{n}        n \neq 0$
$\alpha = \displaystyle \frac{m}{m - 1} = +$ve integer
$\therefore m = 2$
Using equation (i)
$n = 3, 4, 5$
$\therefore \displaystyle \frac{m}{n}= \frac{2}{3} , \frac{2}{4}, \frac{2}{5}$

Given $\cfrac {25\%\quad of\quad 50\%\quad of\quad 100\%}{25\quad of\quad 100\times 50\%\quad of\quad 100}$

Given, a=3567, b=10, c=10, d=1000, e=10000
$\frac{a}{b}+\frac{a}{c}-\frac{a}{d}+\frac{a}{e}$put the values a, b, c, d , e
in the above expression, we get
$=356.7+35.67-3.567+0.3597$
$=389.1597$

$\cfrac { \left( a+2 \right) x+{ a }^{ 2 }-1 }{ ax-2a+18 } $ is independent of x.

$\frac{a}{b}=\frac{\frac{7}{4}}{\frac{5}{3}}$
$=\frac{21}{20}$

$\frac{b}{a}=\frac{\frac{5}{3}}{\frac{7}{4}}$
$=\frac{20}{21}$

Now check all the options