1 \(\times\) 1 = 1
as reciprocal of 1 is 1

Mixed fraction = 19\(\frac{1}{2}\)

(i) We have, \(7\times \frac{3}{5}\)=\( \frac{7\times3}{5}=​​\frac{21}{5}\) 

This is required lowest form.
Now, dividing 21 (numerator) by 5 (denominator),
we get, quotient = 4 and remainder = 1
∴ Mixed fraction =Quotient \(\frac{Reminder}{Divisor}=4 \frac{1}{5}\)

\(8\frac { 1 }{ 2 } -3\frac { 5 }{ 8 } =\frac { 17 }{ 2 } -\frac { 29 }{ 8 } \)
 \(\left[ \because 8\frac { 1 }{ 2 } =\frac { 8\times 2+1 }{ 2 } =\frac { 16+1 }{ 2 } =\frac { 17 }{ 2 } and3\frac { 5 }{ 8 } =\frac { 3\times 8+5 }{ 8 } =\frac { 29 }{ 8 } \right] \)
= \(\frac { 17\times 4 }{ 2\times 4 } -\frac { 29\times 1 }{ 8\times 1 } \)
 [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, \(\frac{7}{10}+\frac{2}{5}+\frac{3}{2}\)
\(\therefore \frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7+4+15}{10}\)
=\(\frac{26}{10}=\frac{13}{5}\)
[dividing numerator and denominator by 2]

∵ Car covers in 1 litre = 8.6 km
Car covers in 36.5 litres
= 8.6\(\times\)36.5
= 313.90
= 313.9 km

In order to know who solved less part of the exerci.se, we will compare \(\frac { 2 }{ 7 } \)and\(\frac { 4 }{ 5 } \)
We have,
LCM of denominators (i.e., 7 and 5), 7 \(\times\)5 = 35.
Converting each fraction into an equivalent fraction having 35 as its denominator, we have
\(\frac { 2 }{ 7 } =\frac { 2\times 5 }{ 7\times 5 } =\frac { 10 }{ 35 } \)and\(\frac { 4 }{ 5 } =\frac { 4\times 7 }{ 5\times 7 } =\frac { 28 }{ 35 } \)
∵10 < 28
∴ \(\frac { 10 }{ 35 } <\frac { 28 }{ 35 } \Rightarrow \frac { 2 }{ 7 } <\frac { 4 }{ 5 } \)
Hence, Mukul solved lesser part than Deeksha.

7÷3.5=\(\frac { 7 }{ 3.5 } =\frac { 7\times 10 }{ 35 } \)
=\(\frac { 70 }{ 35 } \)=2
OR
7÷3.5=\(7\div \frac { 35 }{ 10 } \)
= \(7\times \frac { 10 }{ 35 } \)
= \(\frac { 70 }{ 35 } \)
=2

We have, \(2\times \frac{2}{5}=\frac{4}{5}\)
The pictorial representation is shown as below :

\(\frac{11}{5}\)

We have 
\(4\frac { 2 }{ 3 } -3\frac { 1 }{ 4 } +2\frac { 1 }{ 6 } \)=\(\frac { 14 }{ 3 } -\frac { 13 }{ 4 } +\frac { 13 }{ 6 } \)
= \(\frac { 14\times 4 }{ 3\times 4 } -\frac { 13\times 3 }{ 4\times 3 } +\frac { 13\times 2 }{ 6\times 2 } \)
∵ LCM of 3, 4 and 6 is 12, so we convert each fraction into an equivalent fraction with denominator 12
i,e  \(\frac { 56 }{ 12 } -\frac { 39 }{ 12 } +\frac { 26 }{ 12 } =\frac { 56-39+26 }{ 12 } \)
= \(\frac { 82-39 }{ 12 } =\frac { 43 }{ 12 } =3\frac { 7 }{ 12 } \)

Total hours spend by Ritika for studies =11\(\frac { 1 }{ 3 } \) hours =\(\frac { 34 }{ 3 } \)hours
 Total hours devoted by Ritika for Hindi and Sanskrit 5\(\frac { 3 }{ 5 } \)= \(\frac { 28 }{ 5 } \)hours
Total hours devoted by Ritika for other subjects
= \(\frac { 34 }{ 3 } -\frac { 28 }{ 5 } \)
= \(\frac { 170-84 }{ 15 } =\frac { 86 }{ 15 } \)
=5\(\frac { 11 }{ 15 } \) hours

We have
\(\frac { 1 }{ 2 } of\frac { 6 }{ 7 } =\frac { 1 }{ 2 } \times \frac { 6 }{ 7 } =\frac { 1\times 6 }{ 2\times 7 } \)
= \(\frac { 6 }{ 14 } =\frac { 6\div 2 }{ 14\div 2 } =\frac { 3 }{ 7 } \)
and  \(\frac { 2 }{ 3 } of\frac { 3 }{ 7 } =\frac { 2 }{ 3 } \times \frac { 3 }{ 7 } =\frac { 2\times 3 }{ 3\times 7 } \)
= \(\frac { 6 }{ 21 } =\frac { 6\div 3 }{ 14\div 2 } =\frac { 2 }{ 7 } \)
We know that, if two fractions have the same denominator but different numerators, the fraction with greater numerator is greater
Since 3>2 ,So,\(\frac { 3 }{ 7 } >\frac { 2 }{ 7 } \)
⇒ \(\left( \frac { 1 }{ 2 } of\frac { 6 }{ 7 } \right) \)of greater than \(\left( \frac { 2 }{ 3 } of\frac { 3 }{ 7 } \right) \)

\(\frac{1}{4}\)