LCM of the denominators of given rational numbers i.e. 5, 11, 7, 3, 6, and 8 = 9240

$\displaystyle \frac{1}{2} = \frac{1 \times 2}{2 \times 2} = \frac{2}{4}$
$\displaystyle \frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}$
$\displaystyle \frac{1}{2} = \frac{1 \times 8}{2 \times 8} = \frac{8}{16}$
So, $\displaystyle \frac{1}{2} = \frac{2}{4} = \frac{3}{6} = \frac{8}{16}$

$\displaystyle \frac{1}{3} = 0.333000,$ $\displaystyle \frac{7}{8} = 0.875$
$\displaystyle \frac{1}{4} = 0.25,$ $\displaystyle \frac{23}{24} = 0.9583000,$ $\displaystyle \frac{11}{12} = 0.9166000$
$\displaystyle \frac{17}{24} = 0.7083000$
Since $0.7083000 \displaystyle \left ( =\frac{17}{24} \right )$ is greater than 
$0.333000 \displaystyle \left ( =\frac{1}{3} \right )$ and less than $0.875 \displaystyle \left ( =\frac{17}{24} \right )$$\displaystyle \left ( =\frac{7}{8} \right )$
Therefore $\displaystyle \frac{17}{24}$ lies between $\displaystyle \frac{1}{3}$ and $\displaystyle \frac{7}{8}$

$\displaystyle \frac{13}{16}=\frac{13\times 5}{16\times 5}=\frac{65}{80}, \frac{7\times 10}{8\times 10}=\frac{70}{80},\frac{31}{40}=\frac{31\times 2}{40\times 2}$
$\displaystyle =\frac{62}{80}$ and last fraction is $\displaystyle =\frac{63}{80}$
Out of these the largest fraction is $\displaystyle \frac{70}{80}$ $\displaystyle =\frac{7}{8}$

Let the number be $x$

Given fractions are

$\because$ all fractions are having same numerator.
So the fraction having smallest denominator is the largest.
Hence $\displaystyle \frac {29}{23}$ is the largest.

$\displaystyle \frac {19}{20}\, -\, \displaystyle \frac {2}{20}\, =\, \displaystyle \frac {19-2}{20}\, =\, \displaystyle \frac {17}{20}$

$12.1280 < 12.129$

While comparing positive numbers greater is greater, smaller is smaller.

$\displaystyle\frac{-5}{11},\,\frac{-5}{12},\,\frac{-5}{17}$

$\because$ All have same numerator. So the rational number having the least denominator is the greatest. But here all have negative sign. So the number having greatest denominator is greater.

Hence, $\displaystyle\frac{-5}{17}$ is greater.

Alter : Take any two given numbers. $\displaystyle\frac{-5}{11},\, \frac{-5}{12}$

$-5\,\times\,12, -5\,\times\, 11$

- 60, - 55 

$\because\, - 55\, >\, - 60$

So, $\displaystyle\frac{-5}{12}$ is greater.

Now compare this with $\displaystyle\frac{-5}{17}$ 

$\displaystyle\frac{-5}{12},\, \frac{-5}{17}$

$-5\,\times\,17, \, -5\,\times\, 12$

- 85, - 60

$\because\,- 60\, >\,- 85$

So, $\displaystyle\frac{-5}{17}$ is greater. 

The numbers are

$\displaystyle {\frac{-5}{11},\, \frac{-5}{12},\, \frac{-5}{17}}$
$\because$ All have same numerator. Sothe rational number having theleast denominator is the greatest.But here all have negative sign.So, the number having greatestdenominator is greater. Hence, $\displaystyle \frac{-5}{17}$ is greater

The given numbers can be arranged in the ascending order as:
$ {\cfrac{1}{5}\, >\, \cfrac{3}{5}\, >\, \cfrac{7}{5}\, >\, \cfrac{9}{5}}$
Greatest number $= \cfrac{9}{5}$ and Least number $= \cfrac{1}{5}$

Let given numbers arranged in the descending order.
$\displaystyle \frac{1}{2},\, \displaystyle \frac{-4}{5},\, \displaystyle \frac{- 7}{8}$
$- 4\, \times\, 8,\, 5\, \times\, - 7$
$- 32,\, - 35$
$\displaystyle \frac{-4}{5}\, >\, \displaystyle \frac{-7}{8}$
The descending order is $\displaystyle \frac{1}{2}\, >\, \displaystyle \frac{-4}{5}\, >\, \displaystyle \frac{- 7}{8}$
So middle number is $\displaystyle \frac{- 4}{5}$.

$\displaystyle \dfrac { 6 }{ 7 } ,\dfrac { 4 }{ 5 } ,\dfrac { 7 }{ 9 } ,\dfrac { 3 }{ 4 } \Rightarrow 0.85<0.8<0.78<0.75$

$ \dfrac{8}{16}=\dfrac{1}{2}\ and\ \dfrac{8}{4}=2$

The denominator is the total number of parts in the whole. The lesser the number of parts the greater the value of each part.

$\displaystyle \frac{1}{2}=\frac{1\times 2}{2\times 2}=\frac{2}{4}$

Doing the simplest form of all these options..

$\dfrac{1}{2}\ \ \ of \ \ \dfrac{4}{7}$

In option A, denominator of all the fractions are same and numerator are in increasing order

Let the missing number be $x$

$\dfrac{5}{7} , \dfrac{7}{9} , \dfrac{9}{11}, \dfrac{11}{13}$

Fractions with the same denominators (bottom numbers) are called like fractions.

$\begin{array}{l} We\, have \ \frac { 1 }{ a } <\frac { 1 }{ b }  \ by\, reciprocal\, both\, side\, we\; get \ \frac { a }{ 1 } >\frac { b }{ 1 }  \ \therefore a>b \ Hence,\, the\, option\, C\, is\, the\, correct\, answer. \end{array}$

$0.4=\dfrac 4{10}=\dfrac 25\1.5=\dfrac{15}{10}=\dfrac 32\25.75=\dfrac{2575}{100}=\dfrac{103}{4}\0.072=\dfrac{72}{1000}=\dfrac{9}{125}\1.248=\dfrac{1248}{1000}=\dfrac{156}{125}$

Now,

(a) Let us consider the first set of fraction $\dfrac { 7 }{ 10 } ,\dfrac { 3 }{ 4 }$ and another given fraction $\dfrac { 3 }{ 5 }$ 

(a) Let us consider the first set of fraction $-\dfrac { 1 }{ 2 } ,\dfrac { 5 }{ 6 } ,-\dfrac { 4 }{ 9 }$ 

The rational number $\dfrac {4}{-5}$ is same as $\dfrac {-4}{5}$.

$\displaystyle {\frac{4}{7},\, \frac{1}{3},\, \frac{2}{5},\, \frac{5}{9}}$
The above numbers in ascending order are
$\displaystyle {\frac{1}{3}\, <\, \frac{2}{5}\, <\, \frac{5}{9}\, <\, \frac{4}{7}}$
Middle two numbers are $\displaystyle \frac{2}{5}$ and $\displaystyle \frac{5}{9}$
$\therefore$ Average = $\displaystyle {\frac{2/5\, +\, 5/9}{2}\, =\, \frac{43}{90}}$

The rational number $\dfrac {4}{-5}$ is same as $\dfrac {-4}{5}$.

The given numbers can be arranged in the ascending order as $\displaystyle \frac{1}{5}\, <\, \displaystyle \frac{3}{5}\, <\, \displaystyle \frac{7}{5}\, <\, \displaystyle \frac{9}{5}$
Greatest number $=\, \displaystyle \frac{9}{5}$;
Least number $=\, \displaystyle \frac{1}{5}$.
We have, $\displaystyle \frac{9}{5}\, \times\, \displaystyle \frac{x}{100}\, =\, \displaystyle \frac{1}{5}$
$x\, =\, \displaystyle \frac{100}{9}\, =\, 11\, \displaystyle \frac{1}{9}\, \%$

This question is very easy if we solve it by verification
process.

Take (A) i.e. $\displaystyle\frac{-2}{3}\,<\,\frac{4}{-9}\, \frac{-5}{12}\, <\,\frac{7}{-18}$

First take \displaystyle\frac{-2}{3},\,\frac{-4}{9}$

$-2\,\times\,9,\, 4\,\times\,3$

-18, -12

$\because\, -12\,>\,-18$

So, $\displaystyle\frac{-4}{9}, \frac{-5}{12}$

$- 4\,\times\, 12, \, - 5\,\times\, 9$

- 48, - 45

$\because\, -45\, >\, -48$

So, $\displaystyle\frac{-5}{12}\,>\, \frac{-4}{9},\,i.e.,\frac{-4}{9}\, <\, \frac{-5}{12}$

Finally, $\displaystyle\frac{-5}{12}, \frac{-7}{18}$

$5\,\times\, 18, \, -7\,\times\, 12$

- 90, - 84

$\because\, -84\, >\, -90$

So, $\displaystyle\frac{-7}{18}\, >\, \frac{-5}{12}, i.e., \frac{-5}{12}\, <\, \frac{-7}{18}$

$\therefore\, \displaystyle\frac{-2}{3}\, <\, \frac{-4}{9}\,<\,  \frac{-5}{12}\,<\, \frac{-7}{18}$

You can identify the answer by observing the question by practicing this method.

$\displaystyle \frac{4}{7},\, \displaystyle \frac{1}{3},\, \displaystyle \frac{2}{5},\, \displaystyle \frac{5}{9}$
The above numbers in ascending order are
$\displaystyle \frac{1}{3}\, <\, \displaystyle \frac{2}{5}\, <\, \displaystyle \frac{5}{9}\, <\, \displaystyle \frac{4}{7}$
Middle two numbers are $\displaystyle \frac{2}{5}$ & $\displaystyle \frac{5}{9}$.
$\therefore$ Average = $\displaystyle \frac{\displaystyle \frac{2}{5}\, +\, \displaystyle \frac{5}{9}}{2}\, =\, \displaystyle \frac{43}{90}$.

Take any two given rational numbers
$\displaystyle \frac{-7}{13},\, \displaystyle \frac{-5}{13}$
$- 7\, \times\, 13,\, -5\, \times\, 13$
$- 91,\, -65$
$\because\, - 65\, >\, - 91$
So $\displaystyle \frac{-7}{13}$ is smaller.
Now compare this with $\displaystyle \frac{- 11}{13}$
$\displaystyle \frac{- 7}{13},\, \displaystyle \frac{- 11}{13}$
$-7\, \times\, 13,\, - 11\, \times\, 13$
$- 91\, ,\, - 143$
$\because\, - 91\, >\, - 143$
So $\displaystyle \frac{- 11}{13}$ is smaller.

By verification process, $A\,\rightarrow\, \displaystyle\frac{3}{4},\, \frac{-2}{1}, \, \frac{-11}{20},\, \frac{-4}{5}$

$3\,\times\, 1,\, -2\,\times\, 4$

3, - 8

Correct, $-2\,\times\, 20, \, -11\,\times\, 1$

- 40, - 11

Wrong.

$B\, \rightarrow\, \displaystyle \frac {3}{4},\, \frac{-11}{20}, \frac{-4}{5},\, \frac{-2}{1}$

$3\,\times\, 20, \, -11\,\times\, 4$

60, - 44

$\displaystyle\frac{3}{4}\, >\,\frac{-11}{20}$ $-11\,\times\, 5,\, -4\,\times\, 20$

- 55, - 80

$\because\, \displaystyle\frac{-11}{20}\, >\, \frac{-4}{5}$

$-4\,\times\, 1,\, -2\,\times\, 5$

- 4, - 10

$\because\, \displaystyle\frac{-4}{5}\, >\, -2$

Since $\dfrac {1}{4}=0.25$.

Given, 

Given, 

$\dfrac{2}{5}$ of $\dfrac{4}{7}=\dfrac{2\times 4}{5\times 7}=\dfrac{8}{35}$

Descending order of the fraction of words spelled correctly is,

$\cfrac {\left(-18\cfrac{1}{3}\times 2\cfrac {8}{11}\right)}{\left|\cfrac {3}{5}+\left(\cfrac {-9}{10}\right)\right|+\left|-\left(\cfrac{-3}{5}\right)\right|}$

To check the order, first make the denominator same of all the fractions, then compare the numerator.

To compare fractions, make the denominators equal

Since we need to compare the fraction $\displaystyle \frac{a+c}{b+c}$ with $\displaystyle \frac{a}{b}$, we cross multiply the terms and check since $a,b,c$ are all given to be positive.
We thus get $b(a + c)$ on the L.H.S. and $a(b + c)$ on R.H.S.
Thus, simplifying we are left with $ab + bc$ on the L.H.S. and $ab + ac$ on the R.H.S.
Now, which side is greater depends on $ac$ and $bc$, which in turn depends upon $a$ & $b.$
L.H.S. is greater if $b > a$, which implies $\frac{a + c}{b + c}$ is greater when $b > a.$