(a) On the number line, it will mean four jumps of \(\frac{6}{5}\) to the left from zero. We reach at \(\frac{-24}{5}.\)

So, \(\frac{-6}{5}\times4=\frac{-24}{5}\)
(b) \(\frac{-6}{5}\times4=\frac{-6\times4}{5}=\frac{-24}{5}\)
We observe that we arrive at the same rational number.

\(-\frac{5}{3}=-\frac{5\times7}{3\times7}=-\frac{35}{21}\)
\(-\frac{8}{7}=-\frac{8\times3}{7\times3}=-\frac{24}{21}\)
Five rational numbers:
\(-\frac{35}{21}<-\frac{33}{21}<-\frac{32}{21}<-\frac{31}{21}<-\frac{30}{21}<\frac{-29}{21}<\frac{-24}{21}\)
\(\Rightarrow -\frac{5}{3}<(-\frac{33}{21})<(-\frac{32}{21})<(-\frac{31}{21})<(-\frac{30}{21})<(-\frac{29}{21})<\frac{-8}{7}\)
So, five rational numbers between \(-\frac{5}{3}\ and-\frac{8}{7}\)
\(-\frac{33}{21},-\frac{32}{21},\frac{-31}{21},\frac{-30}{21}\ and -\frac{29}{21}\)

Given points, P, Q, R, S, T, U, A and B are on the number line such that TR =RS =SU and AP =PQ =QB.
It means that distance between AB is divided into 3 equal parts and similarly UT on the left of zero is divided into 3 equal parts.
Now, point P is on the right of zero on the number line and between 2 and 3.
So, the rational number represented by P = \(\\ 2+\frac { 1 }{ 3 } \) 
[ since, AB is divided into 3 equal parts and each part shows \(\\ \frac { 1 }{ 3 } \)]
\(=\frac { 2\times 3 }{ 1\times 3 } =\frac { 1 }{ 3 } =\frac { 6+1 }{ 3 } =\frac { 7 }{ 3 } \)
Similarly, Q is on the right of zero and between  2 and 3.
So, the rational number represented by Q = \(2+\frac { 1 }{ 3 } +\frac { 1 }{ 3 } \)
[since, Q shows 2 equal parts out of 3 between 2 and 3 ] 
\(\\ =\frac { 2\times 3 }{ 3 } +\frac { 1 }{ 3 } +\frac { 1 }{ 3 } =\frac { 6+1+1 }{ 3 } =\frac { 8 }{ 3 } \)
Also, R and S lie on the left of zero and between -1 and -2 then each part show \(\frac { -1 }{ 3 } \).
So, the rational number represented by 
R=\(-1+\left( \frac { -1 }{ 3 } \right) =\frac { -1\times 3 }{ 3 } -\frac { 1 }{ 3 } =\frac { -3-1 }{ 3 } =\frac { -4 }{ 3 } \)
and the rational number represented by 
S = -1+\(-1+\left( \frac { -1 }{ 3 } \right) +\left( \frac { -1 }{ 3 } \right) \)
[since, S shows 2 equal parts out of 3 parts between -1 and -2 ]
 \(=\frac { -1\times 3 }{ 3 } -\frac { 1 }{ 3 } -\frac { 1 }{ 3 } =\frac { -3-1-1 }{ 3 } =\frac { -5 }{ 3 } \)

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.

We have,( x - y) +z
Put the value of x,y and z 
\(=\left( \frac { -4 }{ 9 } -\frac { 5 }{ 12 } \right) +\frac { 7 }{ 18 } \\ \frac { -4\times 4-5\times 3 }{ 36 } +\frac { 7 }{ 18 } =\frac { -16-15 }{ 36 } +\frac { 7 }{ 18 } \\ =\left( \frac { -31 }{ 36 } +\frac { 7 }{ 18 } \right) =\frac { -31+7\times 2 }{ 36 } =\frac { -31+14 }{ 36 } =\frac { -17 }{ 36 } \)

Given,35%=\(\frac { 35 }{ 100 } \)

\(\because 35=7\times 5\) 
and 100 = 2\(\times\)2\(\times\)5
\(\because \) HCF of 35 and 100 = 5
On dividing numerator and denominator by their HCF, We get 
\(\frac { 35\div 5 }{ 100\div 5 } =\frac { 7 }{ 20 } \)

Reciprocal of 1 =\(\frac { 1 }{ 1 } \)

\(\frac { 6 }{ 5 } \times \frac { 3 }{ 7 } -\frac { 1 }{ 5 } \times \frac { 3 }{ 7 } \)
\(\\ \therefore \) Product of rational numbers 
\(=\frac { Product\quad of\quad numerators }{ Product\quad of\quad denominators\quad } \)
\(\\ \\ =\frac { 6\times 3 }{ 5\times 7 } -\frac { 1\times 3 }{ 5\times 7 } =\frac { 18 }{ 35 } -\frac { 3 }{ 35 } =\frac { 18-3 }{ 35 } =\frac { 15 }{ 35 } =\frac { 3 }{ 7 } \)

Given,\(\frac { 4 }{ 7 } +\frac { -8 }{ 9 } +\frac { -5 }{ 21 } +\frac { 1 }{ 3 } \)
For same denominator 
LCM of 7,9,21 and 3 is 63,\(\)
\(\frac { 4\times 9 }{ 7\times 9 } =\frac { 36 }{ 63 } \),\(\frac { -8\times 7 }{ 9\times 7 } =\frac { -56 }{ 63 } \)
\(\frac { -5\times 3 }{ 21\times 3 } =\frac { -15 }{ 63 } \) , \(\frac { 1\times 21 }{ 3\times 21 } =\frac { 21 }{ 63 } \)
\(\therefore \)\(\frac { 4 }{ 7 } +\frac { -8 }{ 9 } +\frac { -5 }{ 21 } +\frac { 1 }{ 3 } =\frac { 36 }{ 63 } +\frac { -56 }{ 63 } +\frac { -15 }{ 63 } +\frac { 21 }{ 63 } \)
So,\(\frac { 36+(-56)+(-15)+(21) }{ 63 } \)
\(=\frac { 36+21+(-56)+(-15) }{ 63 } =\frac { 57+(-71) }{ 63 } \)
\(=-\frac { 14 }{ 63 } =\frac { -2 }{ 9 } \)

Given, \(\\ \frac { 1 }{ 2 } +\frac { 1 }{ 6 } +\frac { 1 }{ 8 } \)
For common/same denominator
LCM of 2,6 and 8 is 24.
\(\frac { 1 }{ 2 } =\frac { 1\times 12 }{ 2\times 12 } =\frac { 12 }{ 24 } \frac { 1 }{ 6 } =\frac { 1\times 4 }{ 6\times 4 } =\frac { 4 }{ 24 } \\ \frac { 1 }{ 8 } =\frac { 1\times 3 }{ 8\times 3 } =\frac { 3 }{ 24 } \\ \therefore \frac { 1 }{ 2 } +\frac { 1 }{ 6 } +\frac { 1 }{ 8 } =\frac { 12 }{ 24 } +\frac { 4 }{ 24 } +\frac { 3 }{ 24 } \\ =\frac { 12+4+3 }{ 24 } =\frac { 19 }{ 24 } \)