$2 < 2.1=(2+0.1) < 2.2=(2.1+0.1) < 2.3=(2.2+0.1) < 2.4=(2.3+0.1) < ... < 2.9=(2.8+0.1) < 3=(2.9+0.1)$

We know that, $ \sqrt {2} = 1.414

$ and $ \sqrt {3} = 1.732 $

Hence two rational numbers between $ 1.414 $ and $ 1.732

$  can be $ 1.5 (= \frac {3}{2}) $ and $ 1.6 (= \frac {16}{10}= \frac {8}{5}) $

We know that, $ \sqrt {3} = 1.732$ and $ \sqrt {5} = 2.236 $

Hence three rational numbers between $ 1.732 $ and $ 2.236$  can be $ 1.8 \left(= \dfrac {18}{10}\right)  $ , $ 2 $ and $ 2.2 \left(= \dfrac {22}{10}\right) $.

Required rational number $\displaystyle =\frac{1}{2}\left ( \frac{6}{7}+\frac{7}{8} \right )=\frac{1}{2}\left ( \frac{48+49}{56} \right )=\frac{97}{112}$
Hence option (d) is correct

As $\sqrt 6$ is irrational therefore option A is wrong.

In option B,

The rational number between $\dfrac12$ and $\sqrt1$ :

We know $\dfrac{5}{16} =0.3125$

We know $\dfrac{1}{4} = 0.25$

Between any two rational numbers we can find infinitely many rational numbers. 

rational number always lies on the line.

$\dfrac{-3}{2},-1,3,0,\dfrac{1}{2}$

There are infinite rational numbers between any two rational numbers 

For a rational number to lie between $\dfrac{-2}{5}$ and $\dfrac{-1}{5}$,it should be less than $\dfrac{-1}{5}$ and greater than $\dfrac{-2}{5}$.
Now,$\dfrac{3}{10}$ is not less than $\dfrac{-1}{5}$.
So,$\dfrac{3}{10}$ does not lie between $\dfrac{-1}{5}$ and $\dfrac{-2}{5}$.

All the options have denominator $75$. Hence, let us convert into equivalent fractions having denominator $75$. 
$\dfrac{3}{5} $ $=\dfrac{3\times 15}{5\times 15} $ $=\dfrac{45}{75}$

$\dfrac{2}{3}$ $=\dfrac{2\times 25}{3\times 25}$ $=\dfrac{50}{75}$

here, $\dfrac {5}{\sqrt 3-\sqrt 5}$

Length of the first train$=180m$

A rational number between two numbers $ a $ and $ b = \dfrac {(a +

b)}{2} $
So, a rational number between $\dfrac {3}{8} $ and $ \dfrac {7}{12}$

A rational number between two numbers $ a $ and $ b = \dfrac {(a + b)}{2} $

So, a rational number between $ 5 $ and $ - 2 = \dfrac

{( 5 - 2 )}{2} = \dfrac {3}{2} $

A

rational number between two numbers $ a $ and $ b = \frac {(a +

b)}{2} $

So, a

rational number between $\frac {1}{3} $ and $ \frac {1}{4} = \frac {(\frac {1}{3} + \frac {1}{4})}{2} = \frac {7}{24} $

Now, another rational number between $ \frac {7}{24} $ and $ \frac {1}{4} = \frac {(\frac {7}{24} + \frac {1}{4})}{2} = \frac {13}{48} $



Hence, required two rational numbers between $\frac {1}{3} $ and $ \frac {1}{4} $ are $\frac {1}{3} ,\frac {7}{24}, \frac {13}{48}, \frac {1}{4}$

To insert rational numbers between $2$ numbers, we will arrange the options and check if they are in ascending order.
$\dfrac { 1 }{ 3 } { ? }\dfrac { 117 }{ 300 } \ \Longrightarrow 300<351\ \qquad \dfrac { 3 }{ 4 } { ? }\dfrac { 287 }{ 400 } \ \Longrightarrow 1200>1148$
They are in ascending order, i.e., $\dfrac { 1 }{ 3 } ,\dfrac { 117 }{ 300 } ,\dfrac { 287 }{ 400 } ,\dfrac { 3 }{ 4 } $
From the given options only option $D$ satisfies this condition. Hence, $D$ is the answer.

Let $x=1,\;y=\sqrt{2}$
Then $xy=\sqrt{2}\;\in\;Q^c$
Obvious
$x=-1,\;y=\sqrt{2}$ then $\sqrt{2}x+y=0\;\in\;Q$
$x=1,\;y=\sqrt{2}$ then $x/y=\displaystyle\frac{1}{\sqrt{2}}\;\in\;Q^c$

(I) In $5\sqrt{3}$

a³ = a2 x a is a rational number,

a⁴ = a³x a is a rational number,

......

......

∴ aⁿ = a^n-1 x a is a rational number.

So, the option (III) is true

Mean $= \dfrac {\dfrac {3}{8} + \dfrac {5}{24}}{2} = \dfrac {\dfrac {9 + 5}{24}}{2} = \dfrac{\left (\dfrac {14}{24}\right )}{2}$

Mean $= \dfrac {(-1) + (-2)}{2} = \dfrac {-1 -2}{2} = \dfrac {-3}{2}$.

Mean $= \dfrac {(-6) + (-7)}{2} = \dfrac {-6 -7}{2}$.

$Mean = \dfrac{\left (\dfrac {3}{4} + \dfrac {13}{8}\right )}{2} = \dfrac {6 + 13}{8} \times \dfrac {1}{2} = \dfrac {19}{16}$.

$Mean = \dfrac{\left (\dfrac {4}{9} + \dfrac {4}{5}\right )}{2} = \left (\dfrac {20 + 36}{45}\right ) \times \dfrac {1}{2} = \dfrac {56}{45}\times \dfrac {1}{2}$
$= \dfrac {28}{45}$

Consider $3 \times 4 = 12$, so 
$\dfrac 23 = \dfrac{8}{12}$

The given rational numbers $-\dfrac { 2 }{ 3 }$ and $-\dfrac { 1 }{ 5 }$ are negative rational numbers because the numerator and denominator of both the rational numbers are of opposite signs that is the numerator of both the integers is negative while the denominators are positive.