\(4\frac { 4 }{ 5 } \div \frac { 24 }{ 5 } +8\frac { 1 }{ 4 } =\frac { (5\times 4)+4 }{ 5 } \div \frac { 24 }{ 5 } +\frac { (8\times 4)+1 }{ 4 } \)
\(\frac { 24 }{ 5 } \div \frac { 24 }{ 5 } +\frac { 33 }{ 4 } \)
Since , the reciprocal  of  \(\frac { 24 }{ 5 } \) is \(\frac { 5 }{ 24 } \) 
\(\\ \therefore \quad \frac { 24 }{ 5 } \times \frac { 5 }{ 24 } +\frac { 33 }{ 4 } =1+\frac { 33 }{ 4 } =\frac { 37 }{ 4 } \)

\(\frac { -6 }{ 7 } =\frac { x }{ 42 } \Rightarrow x=\frac { 42\times (-6) }{ 7 } =6\times (-6)=-36\)

\(\frac { 3 }{ 4 } \times \left( \frac { -2 }{ 3 }  \right) =\frac { 3\times (-2) }{ 4\times (3) } =\frac { -6 }{ 12 } =\)\(-\frac { 1 }{ 2 } \)

On number line \(\frac { 4 }{ 3 } \) is to the right of zero (0)

1 is a positive  rational number [ \(\because \) 1 has positive sign]

The four rational numbers equivalent to \(\frac { 4 }{ 9 } \) are \(\frac { 8 }{ 18 } ,\frac { 12 }{ 27 } ,\)
\(\frac { 16 }{ 36 } \) and \(\frac { 20 }{ 45 } \) .

The four rational numbers equivalent to \(\frac { 5 }{ -3 } \) are \(\frac { 10 }{ -6 } ,\)  
\(\frac { 15 }{ 9 } ,\frac { 20 }{ -12 } \) and \(\frac { 25 }{ -15 } \).

Given \(\frac { -4 }{ 5 } \boxed { } \frac { -5 }{ 7 } \)
Here, both rational numbers are negative.
On ignoring the negative sign, we have the rational number \(\\ \frac { 4 }{ 5 } \)and \(\\ \frac { 5 }{ 7 } \).
Now, convert these rational number into the rational number having same denominator.
\(\therefore \) \(\frac { 4 }{ 5 } =\frac { 4\times 7 }{ 5\times 7 } =\frac { 28 }{ 35 } \) and \(\frac { 5 }{ 7 } =\frac { 5\times 5 }{ 7\times 5 } =\frac { 25 }{ 35 } \)
\(\because \) 28 > 25,so \(\\ \frac { 28 }{ 35 } >\frac { 25 }{ 35 } \)
Then, \(\frac { 4 }{ 5 } >\frac { 5 }{ 7 } \) 
Now, reverse the order i.e. \(\frac { -4 }{ 5 } <\frac { -5 }{ 7 } \)
Hence, \(\frac { -4 }{ 5 } <\frac { -5 }{ 7 } \)

Given, \(\frac { -1 }{ 4 } ,\frac { -2 }{ 8 } ,\frac { -3 }{ 12 } ,......\)
Here,\(\frac { -1 }{ 4 } \) is the rational number in standard form.
Now, \(\\ \frac { -1 }{ 4 } =\frac { -1\times 1 }{ 4\times 1 } ,\frac { -2 }{ 8 } =\frac { -1\times 2 }{ 4\times 2 } ,\frac { -3 }{ 12 } =\frac { -1\times 3 }{ 4\times 3 } \)
or \(\frac { -1\times 1 }{ 4\times 1 } =\frac { -1 }{ 4 } ,\frac { -1\times 2 }{ 4\times 2 } =\frac { -2 }{ 8 } ,\frac { -1\times 3 }{ 4\times 3 } =\frac { -3 }{ 12 } \) 
Thus, we observe a pattern in these numbers. The next four numbers are
\(\frac { -1\times 4 }{ 4\times 4 } =\frac { -4 }{ 16 } ,\frac { -1\times 5 }{ 4\times 5 } =\frac { -5 }{ 20 }, \)
\(\frac { -1\times 6 }{ 4\times 6 } =\frac { -6 }{ 24 } ,\frac { -1\times 7 }{ 4\times 7 } =\frac { -7 }{ 28 } \)
Hence, the required four more rational numbers are
\(\\ \\ \frac { -4 }{ 16 } ,\frac { -5 }{ 20 } ,\frac { -6 }{ 24 } \) and \(\\ \frac { -7 }{ 28 } \).

Given rational numbers are \(\frac { -5 }{ 12 } \) and \(\frac { 7 }{ -18 } \) 
For the same /common denominator,
LCM of 12 and 18 is 36,
\(\frac { (-5)\times 3 }{ 12\times 3 }= \frac { -15 }{ 36 } \) and \(\frac { 7\times 2 }{ (-18\times 2) } =\frac { 14 }{ -36 }= \frac { -14 }{ 36 } \)
-14 is greater than -15.
So,\(\frac { -14 }{ 36 } >\frac { -15 }{ 36 } \)  or \(\frac { -7 }{ 18 } >\frac { -15 }{ 12 } \)

Since, the nearest integer to \(-\frac { 3 }{ 5 } \)is -1.
Let x be subtracted from \(-\frac { 3 }{ 5 } \) to obtain -1.
Then , \(-\frac { 3 }{ 5 } \) -x=-1
\(​​​​​​​\Rightarrow \) -x =-1+\(\frac { 3 }{ 5 } \) [by transposing \(-\frac { 3 }{ 5 } \) to RHS]
\(​​​​​​​\Rightarrow \) -x =\(-\frac { 5 }{ 5 } +\frac { 3 }{ 5 } \left[ \because \frac { -1\times 3 }{ 1\times 3 } =\frac { -3 }{ 3 } \right] \)
\(​​​​​​​\Rightarrow \) -x \(=-\frac { 2 }{ 5 } \Rightarrow x=\frac { 2 }{ 5 } \)
Hence, \(\frac { 2 }{ 5 } \)should be subtracted from \(-\frac { 3 }{ 5 } \) to obtain nearest integer.

Length of the rope = 68m
Length of small pieces = \(\\ 4\frac { 1 }{ 4 } \)m= \(\frac { (4\times 4)+1 }{ 4 } =\frac { 17 }{ 4 }\)m
\(\therefore \)Number of pieces = \(\frac { Total\quad length\quad of\quad rope\quad }{ Length\quad of\quad one\quad small\quad piece\quad } \)
\(\\ =\frac { 68 }{ \frac { 17 }{ 4 }  } \)
\(\\ \\ =\frac { 68 }{ 1 } \times \frac { 4 }{ 17 } [ \because \) reciprocal of \(\frac { 17 }{ 4 } \) is  \(\frac { 4 }{ 17 } \) ]
=\(4\times 4\) 
Hence, number of pieces is 16.

Given rational numbers are \(\frac { 49 }{ 3 } \) and \(\frac { 7 }{ 3 } \). 
To find,\(\frac { 49 }{ 3 } \)\(\div \)\(\frac { 7 }{ 3 } \).Since the reciprocal of \(\\ \frac { 7 }{ 3 } \) is \(\\ \frac { 3 }{ 7 } \) \(\Rightarrow \frac { 49 }{ 3 } \div \frac { 7 }{ 3 } =\frac { 49 }{ 3 } \times \frac { 3 }{ 7 } =\frac { 49\times 3 }{ 3\times 7 } =7\)

We have,\(\frac { 3 }{ 13 } \div \left( \frac { -4 }{ 65 } \right) =\frac { 3 }{ 13 } \times \) reciprocal of \(\left( \frac { -4 }{ 65 } \right) \)
\(\\ =\frac { 3 }{ 13 } \times \left( \frac { 65 }{ -4 } \right) =\frac { 3\times 65 }{ 13\times \left( -4 \right) } \\ \frac { \left( 3\times 5 \right) }{ (-4) } =\frac { 15 }{ -4 } =\frac { -15 }{ 4 } \)

\(\frac { -6 }{ 5 } \times (-2)=\frac { (-6)\times (-2) }{ 5 } =\frac { 12 }{ 5 } \)

\(\frac { 4 }{ 6 } \)

Given rational number are \(\frac { 5 }{ 9 } \) and \(\frac { 4 }{ 7 } \) .
\(\because \) LCM of 9 and 7 is 63, \(\frac { 5\times 7 }{ 9\times 7 } =\frac { 35 }{ 63 } \) and \(\\ \frac { 4\times 9 }{ 7\times 9 } =\frac { 36 }{ 63 } \)
\(\because \)36 > 35 .So, \(\\ \frac { 4 }{ 7 } >\frac { 5 }{ 9 } \)