\(\frac { 1 }{ 3 } \times \frac { 7 }{ 5 } =\frac { 1\times 7 }{ 3\times 5 } =\frac { 7 }{ 15 } \)

\(\frac { 1 }{ 2 } \times \frac { 5 }{ 8 } =\frac { 1\times 5 }{ 2\times 8 } =\frac { 5 }{ 16 } \)

We know that, Rs 1= 100paise or 1 paise=Rs \(\frac{1}{100}\)
∴ 7 paise =\(7 \times \frac{1}{100}=\frac{7}{100}\)=Rs 0.07

We have, \(\frac{2}{3}\)
Now, 
 
Hence, 0.666 is required decimal number.

We have, 0.24 
\(\frac{0.24}{1}\) on removing decimal, \(\frac{024}{100}\), In simplest form =\(\frac{6}{25}\)

\(\frac{13}{11}\times 6=\frac{13\times 6}{11}=\frac{78}{11}=7\frac{1}{11}\)

\(3\times \frac{1}{8}=\frac{3\times 1}{8}=\frac{3}{8}\)

\(\frac{9}{7}\times 6=\frac{9\times 6}{7}=\frac{54}{7}=7\frac{5}{7}\)

\(\frac{2}{7}\times 3=\frac{2\times 3}{7}=\frac{6}{7}\)

\(\frac{1}{15} \times \frac{1}{12}=\frac{1\times 1}{15\times 12}\frac{1}{180}\)

\(\frac{4}{7}\times 3=\frac{4}{7}\times \frac{3}{1}=\frac{4\times 3}{7\times 1}=\frac{12}{7}=1\frac{5}{7}\)

\(\frac{115}{198}\)

\(\frac{2}{9}<\frac{7}{9}<\frac{4}{5}\)

8

We have\(\frac{1}{5},\frac{3}{7},\frac{7}{10}\)

LCM of 5, 7 and 10 = 5\(\times\)7\(\times\)2 = 70
On converting given fractions into like fractions, we get
\(\frac { 1 }{ 5 } =\frac { 1\times 4 }{ 5\times 14 } =\frac { 14 }{ 70 } \)
\(\frac { 3 }{ 7 } =\frac { 3\times 10 }{ 7\times 10 } =\frac { 30 }{ 70 } \)
and  \(\frac { 7 }{ 10 } =\frac { 7\times 7 }{ 10\times 7 } =\frac { 49 }{ 70 } \)
∵ 14 < 30 < 49     [numerators of fractions]
∴  \(\frac { 14 }{ 70 } <\frac { 30 }{ 70 } <\frac { 40 }{ 70 } \)
Thus, 
Hence, descending order of given fractions is \(\frac { 7 }{ 10 } ,\frac { 3 }{ 7 } ,\frac { 1 }{ 5 } \)

\(8\frac { 1 }{ 2 } -3\frac { 5 }{ 8 } =\frac { 17 }{ 2 } -\frac { 29 }{ 8 } \)
 \(\left[ \because 8\frac { 1 }{ 2 } =\frac { 8\times 2+1 }{ 2 } =\frac { 16+1 }{ 2 } =\frac { 17 }{ 2 } and3\frac { 5 }{ 8 } =\frac { 3\times 8+5 }{ 8 } =\frac { 29 }{ 8 } \right] \)
= \(\frac { 17\times 4 }{ 2\times 4 } -\frac { 29\times 1 }{ 8\times 1 } \)
 [L.C.M of 2 and 8 = 8]
= \(\frac { 68 }{ 8 } -\frac { 29 }{ 8 } =\frac { 68-29 }{ 8 } \)
= \(\frac { 39 }{ 8 } =4\frac { 7 }{ 8 } \)

We have, \(2\frac{2}{3}+3\frac{1}{2}\)
On converting given mixed fractions into improper fractions, we get
\(2\frac{2}{3}+3\frac{1}{2}=\frac{2\times 3+2}{3}+\frac{3\times 2+1}{2}\)
=\(\frac{8}{3}+\frac{7}{2}\)
Now, LCM of 3 and 2 = 3 X2 = 6

\(\therefore \frac{8}{3}+ \frac{7}{2}= \frac{8\times 2+7\times 3}{6}\)
=\( \frac{16+21}{6}= \frac{37}{6}\)

We have,  4+\(\frac{7}{8}\)=\(\frac{4}{1}+\frac{7}{8}=\frac{4\times 8+7\times1}{8}​​\)  [∵ LCM of 8and 1=8]
=\(\frac{10-3}{5}=\frac{7}{5}\)

The given frraacctnions are \(\frac{2}{5},\frac{3}{10},\frac{9}{14} , \frac{16}{35}\)
LCM of 5,10, 14,35 = 70

Now, \(\frac{2}{5}=\frac{2\times 14}{5\times 14}=\frac{21}{70},\frac{9}{14}=\frac{9\times5 }{14\times 5}=\frac{45}{70}\) and \(\frac{16}{35}=\frac{16\times 2}{35 \times 2}=\frac{32}{70}\)
Clearly, \(\frac{21}{70}<\frac{28}{70}<\frac{32}{70}<\frac{45}{70}\) So, \(\frac{3}{10}<\frac{2}{5}<\frac{16}{35}<\frac{9}{14}\)
Hence, thee zgiiven firacn..ons in ascendimg order are \(\frac{3}{10}<\frac{2}{5}<\frac{16}{35}<\frac{9}{14}\)
Now, let us change each of the given fractions into an equivalent fraction having 70 as its denominator.

We have, \(\frac{7}{10}+\frac{2}{5}+\frac{3}{2}\)
\(\therefore \frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7+4+15}{10}\)
=\(\frac{26}{10}=\frac{13}{5}\)
[dividing numerator and denominator by 2]

We have, \(\frac{3}{5}+\frac{2}{7}\)

LCM of 5 and 7 = 5X7 = 35
∴ \(\frac{3}{5}+\frac{2}{7}=\frac{3\times 7+2\times 5}{35}\)
=\(\frac{21+10}{35}=\frac{31}{35}\)

We have, 2-\(\frac{3}{5}\)=\(\frac{2}{1}-\frac{3}{5}=\frac{2\times 5-3\times 1}{5}\)      [∵ LCM of 5 and 1=5]
=\(\frac{10-3}{5}=\frac{7}{5}\)

\(4\frac{2}{5}+6\frac{1}{5}\) LCM of 5 and 5 is 5.
So, \(\frac{(5\times 4)+2}{5}+\frac{(5\times 6)+1}{5}\)Then \(\frac{22}{5}+\frac{31}{5}=\frac{22+31}{5}=\frac{53}{5}=10\frac{3}{5}\)

Solved part of exercise by Rohan = \(\frac{3}{7}\)       [given]
Solved parr of exercise by Ramu = \(\frac{2}{5}\)    [given]
Let us compare \(\frac{3}{7}\)and \(\frac{2}{5}\)
LCM is 7 and 5 is 35.
\(\frac{3}{7}=\frac{3}{7}\times \frac{5}{5}=\frac{15}{35}\) and \(\frac{2}{5}=\frac{2}{5}\times \frac{7}{7}=\frac{14}{35}\). Thus, \(\frac{15}{35}>\frac{14}{35}\) or \(\frac{3}{7}>\frac{2}{5}\)
So, Ramu solved lesser part i.e. \(\frac{2}{5}\).

Th e given fractions are \(\frac{5}{6},\frac{7}{9},\frac{11}{12}\)
LCM of 6, 9,12 =36
Now, \(\frac{5}{6}=\frac{5\times 6}{6 \times 6}=\frac{30}{36}, \frac{7}{9}=\frac{7\times 4}{9\times 4}=\frac{28}{36},\frac{11}{12}=\frac{11\times 3}{12 \times 3}=\frac{33}{36}\)       

Hence, Iike fractoins are \(\frac{30}{36},\frac{28}{36},\frac{33}{36}\)