We have
\(\frac{4}{5}=\frac{4\times7}{5\times7}=\frac{28}{35}\)
\(\frac{3}{7}=\frac{3\times5}{7\times 5}=\frac{15}{35}\)
\(\frac{28}{35}>\frac{15}{35}\Rightarrow\frac{4}{5}>\frac{3}{7}\)
\(\Rightarrow \frac{-4}{5}<\frac{-3}{5}\Rightarrow\frac{-3}{7}>\frac{-4}{5}\)

Let required number be x.
\(\therefore\) \(\frac{-10}{3}\times x=\frac{-8}{9}\)
\(\Rightarrow \) x=\(\left[ \frac { -8 }{ 9 } \right] \div \left[ \frac { -10 }{ 3 } \right] =\frac { (-8) }{ 9 } \times \frac { 3 }{ (-10) } \)
=\(\frac{4\times1}{3\times5}=\frac{4}{5}\)
Thus, the required number is \(\frac{4}{15}\).

Sum of \(\frac{12}{5}\) and \(\frac{21}{25}\)=\(\frac{12}{5}+\frac{21}{25}\)
=\(\frac{5(12)-21(1)}{25}=\frac{60-21}{25}=\frac{81}{25}\)
[\(\because\)LCM of 5 and 25 is 25.]
Difference of \(\frac{12}{5}\) and \(\frac{21}{25}\)=\(\frac{12}{5}-\frac{21}{25}\)
=\(\frac{5(12)-21(1)}{25}=\frac{60+21}{25}=\frac{39}{25}\)
Now, \(\left[ \frac { 12 }{ 5 } +\frac { 21 }{ 25 } \right] +\left[ \frac { 12 }{ 5 } -\frac { 21 }{ 25 } \right] =\left[ \frac { 81 }{ 25 } \right] \div \left[ \frac { 39 }{ 25 } \right] \)
=\(\frac{81}{25}\times\frac{25}{39}=\frac{81}{39}=\frac{27}{13}\)

\(\because \) \(\frac{-3}{4}+\frac{-13}{8}\)=\(\frac{-3(2)+(-13)(1)}{8}\)
[\(\because \)LCM of 4 and 8 is 8.]
=\(\frac{-6-13}{8}=\frac{-19}{8}\)
\(\therefore\) \(2-(\frac{-19}{8})=\frac{2}{1}+\frac{19}{8}=\frac{2(8)+19(1)}{8}\)
=\(\frac{16+19}{8}=\frac{35}{8}\)
Thus, \(\frac{35}{8}\)is to be added to\((\frac{-3}{4}+\frac{-13}{8})\) to get 2.

The given rational numbers are: \(\frac{-2}{5},\frac{9}{-10}\) and \(\frac{-5}{6}\).
First let us express the given rational numbers with positive denominators.
\(\therefore\) \(\frac{-2}{5}=\frac{-2\times1}{5\times1}=\frac{-2}{5}\)
\(\frac{9}{-10}=\frac{9\times(-1)}{(-10)\times(-1)}=\frac{-9}{10}\)
\(\frac{-5}{6}=\frac{-5\times1}{6\times1}=\frac{-5}{6}\) 
Since, LCM of 5, 10 and 6 is 30.
\(\therefore\)Making the denominators of the given rationals the same, we have
\(\frac{-2}{5}=\frac{-2}{5}\times \frac{6}{6}=\frac{-12}{30}\)
\(\frac{9}{-10}=\frac{-9}{10}\times \frac{3}{3}=\frac{-27}{30}\)
\(\frac{-5}{6}=\frac{-5}{6}\times \frac{5}{5}=\frac{-25}{30}\)
Since, (-27) <(-25) <(-12)
i.e. \(\frac{-27}{30}<\frac{-25}{30}<\frac{-12}{30}\)
Thus, \(\frac{9}{-10}<\frac{-5}{6}<\frac{-2}{5}\)

It is given that: \(\frac{-5}{8}\) =\(\frac{x}{-32}\) 
\(\therefore\) By cross-multiplication, we get
(x \(\times\) 8) = [(-5) \(\times\) (-32)]
\(\Rightarrow \) 8x = 160 [\(\therefore\)(-1) x (-1) = +1]
\(\Rightarrow \) x=\(\frac{160}{8}\)= \(\frac{20}{1}\)=20
\(\therefore\) The required value of x is 20.

(a) Sequence \(\frac{2}{5},\frac{7}{10},\frac{8}{15},\frac{13}{30}\)
L.C.M.of 5, 10,15,30 = 30
Sequence be
\(\frac{2\times6}{5\times6},\frac{7\times3}{10\times3},\frac{8\times2}{15\times2},\frac{13}{30}\)
or \(\frac{12}{30},\frac{21}{30},\frac{16}{30},\frac{13}{30}\)
Its ascending order is
\(\frac{12}{30}<\frac{13}{30}<\frac{16}{30}<\frac{21}{30}\)
or \(\frac{2}{5}<\frac{13}{30}<\frac{18}{15}<\frac{7}{10}\)
(b) L.C.M: and to find ascending order.
(c) In a class, the students should stand in ascending order of height.

(a)\(\frac{7}{48}-\frac{17}{36}\)=\(\frac{7(3)-17(4)}{144}\)
=\(\frac{21-68}{144}\)
=\(\frac{-47}{144}\)
(b) \(\frac{5}{63}-(\frac{-6}{21})\)=\(\frac{5}{63}+\frac{6}{21}\)
=\(\frac{5+6(3)}{63}\)
=\(\frac{5+18}{63}=\frac{23}{63}\)
(c) \(\frac{-6}{13}-(\frac{-7}{15})\)=\(\frac{-6}{13}+\frac{7}{15}\)
=\(\frac{-6(15)+7(13)}{195}\)
=\(\frac{-90+91}{195}=\frac{1}{195}\)
(d) \(\frac{-3}{8}-\frac{7}{11}\)=\(\frac{-3(11)-7(8)}{88}\)
=\(\frac{-33-56}{88}=\frac{-89}{88}\)
=-1\(\frac{1}{88}\)

Let the required number be x.
\(\therefore\)  \(\frac{-4}{15}\times x=\frac{-18}{9}\)
\(\Rightarrow \)x=\(\frac{-8}{9}\div\frac{-4}{15}\)
\(\Rightarrow \) x=\(\frac{-8}{9}\times\frac{15}{-4}\)
\(\Rightarrow \) x=\(\frac{8\times15}{9\times4}\)
=\(\frac{10}{3}\)
\(\therefore\) Other number is \(\frac{10}{3}\)

The standard form of \(\frac { -12 }{ 18 } \)is\(\frac { -2 }{ 3 } \)

Given rational numbers is \(\frac { -2 }{ 3 } \)
\(\therefore \frac { -2 }{ 3 } +1=\frac { 2 }{ 3 } +\frac { 1 }{ 1 } \) 
LCM of 1 and 3.
\(\therefore \frac { 1\times 3 }{ 1\times 3 } =\frac { 3 }{ 5 } \) and \(\frac { -2\times 1 }{ 3\times 1 } =\frac { -2 }{ 3 } \)
Now, \(\\ \frac { -2 }{ 3 } +\frac { 3 }{ 3 } =\frac { -2+3 }{ 3 } =\frac { 1 }{ 3 } \) 
So, we subtract \(\frac { 1 }{ 3 } \) from \(\frac { -2 }{ 3 } \) to get the nearest integer.

We have,\(\frac { 1 }{ 3 } +\frac { 2 }{ 6 } \)
LCM of 3 and 6 is 6,
\(\frac { 1\times 2 }{ 3\times 2 } =\frac { 2 }{ 6 } \Rightarrow \frac { 2\times 1 }{ 6\times 1 } =\frac { 2 }{ 6 } \)
So,\(\frac { 2 }{ 6 } +\frac { 2 }{ 6 } =\frac { 2+2 }{ 6 } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)

Representation of rational number \(\frac { 3 }{ 4 } \) on number line .

\(\frac { 1 }{ 5 } <\frac { 2 }{ 8 } <\frac { 3 }{ 4 } \)

\(\frac { 137 }{ 120 } \)