Since Maruthu has 4 bottles of milk and each containing \(1{1\over2}\) litres, he has 4 times of \(1{1\over2}\) litres of milk.
\(1{1\over2}\times4=\left(1+{1\over 2}\right)\times4=4+{4\over2}\)
= 4 + 2 = 6 litres

Here 9 > 3 and \({11\over 4}<{5\over6}\)So we proceed as follows:
We convert the mixed fraction into improper fraction and then subtract
\(9{1\over4}={(9\times4)+1\over4}={37\over 4}\)
and \(3{5\over6}={(3\times4)+5\over 6}={23\over6}\)
Common multiple of 4 and 6 is 12
Now, \({37\over 4}-{23\over 6}={37\times3\over12}-{23\times2\over12}\)
\(={111\over12}-{46\over 12}={65\over 12}=5{5\over12}\)

The quantity of milk left over = \(5{1\over2}-3{1\over4}\)
Here, note that 5 > 3 and \({1\over2}>{1\over4}\)
The whole numbers 5 and 3 and the fractional numbers \(1\over2\) and \(1\over4\)can be subtracted separately
So, \(5{1\over2}-3{1\over4}=(5-3)+\left({1\over 2}-{1\over4}\right)\)
\(=2+\left({2\over 4}-{1\over4}\right)\)
\(=2+{1\over4}={1\over4}\)litres

\(3{2\over4}+7{2\over5}=3+{2\over4}+7+{2\over5}\)
\(=3+7+\left({2\over4}+{2\over5}\right)\)
\(=10+\left({10\over20}+{8\over20}\right)\)
\(=10+{18\over20}=10+{9\over10}=10{9\over10}\)

Total length of the cloth \(\left(2{3\over4},+2{1\over2}+1{1\over4}\right)m\)
First we add whole numbers: 2 + 2 + 1 = 5 m
Then, add the fractions: \(\left({3\over4}+{1\over2}+{1\over4}\right)={3\over4}+{2\over4}+{1\over4}={3+2+1\over4}={6\over4}={3\times2̶\over2\times2̶}={3\over2}=1{1\over2}m\)
Therefore, the total length of the cloth bought \(=5+1+{1\over2}=6{1\over2}m\)

\(5\cfrac { 4 }{ 9 } \) and \(3\cfrac { 1 }{ 6 } \)
\(\Rightarrow \cfrac { 49 }{ 9 } +\cfrac { 19 }{ 6 } \)
LCM of 9 and 6 is 18
= \(\cfrac { 49\times 2 }{ 9\times 2 } +\cfrac { 19\times 3 }{ 6\times 3 } \)
= \(\cfrac { 98 }{ 18 } +\cfrac { 57 }{ 18 } \)
= \(\cfrac { 98+57 }{ 18 } =\cfrac { 155 }{ 18 } \)
= \(8\cfrac { 11 }{ 18 } \)
(ii) \(7\cfrac { 1 }{ 6 } \Rightarrow \cfrac { 43 }{ 6 } \)
\(12\cfrac { 3 }{ 8 } \Rightarrow \cfrac { 99 }{ 8 } \)
Now to find : \(\cfrac { 99 }{ 8 } -\cfrac { 43 }{ 6 } \)
LCM = 24
= \(\cfrac { 99\times 3 }{ 8\times 3 } -\cfrac { 43\times 4 }{ 6\times 4 } \)
= \(\cfrac { 297 }{ 24 } -\cfrac { 172 }{ 24 } \)
= \(\cfrac { 297-172 }{ 24 } =\cfrac { 125 }{ 24 } \)
= \(5\cfrac { 5 }{ 24 } \)
(iii) To find
\(\left( 9\cfrac { 2 }{ 3 } +2\cfrac { 1 }{ 2 } \right) -\left( 6\cfrac { 1 }{ 6 } +3\cfrac { 1 }{ 5 } \right) \)
\(\Rightarrow \left( \cfrac { 29 }{ 3 } +\cfrac { 25 }{ 2 } \right) -\left( \cfrac { 37 }{ 6 } +\cfrac { 16 }{ 5 } \right) \)
\(\Rightarrow \) LCM of 3 and 2 is 6
LCM of 6 and 5 is 30
\(\Rightarrow \left( \cfrac { 29\times 2 }{ 3\times 2 } +\cfrac { 5\times 3 }{ 2\times 3 } \right) -\left( \cfrac { 37\times 5 }{ 6\times 5 } +\cfrac { 16\times 6 }{ 5\times 6 } \right) \)
\(\Rightarrow \left( \cfrac { 58 }{ 6 } +\cfrac { 15 }{ 6 } \right) -\left( \cfrac { 185 }{ 30 } +\cfrac { 96 }{ 30 } \right) \)
\(\Rightarrow \left( \cfrac { 58+15 }{ 6 } \right) -\left( \cfrac { 185+96 }{ 30 } \right) \)
\(\Rightarrow \cfrac { 73 }{ 6 } -\cfrac { 281 }{ 30 } \)
LCM of 6 and 30 is 30
\(\Rightarrow \cfrac { 73\times 5 }{ 6\times 5 } -\cfrac { 281 }{ 30 } \)
\(\cfrac { 365 }{ 30 } -\cfrac { 281 }{ 30 } =\) 
= \(2\cfrac { 4 }{ 5 } \)

Total distance = \(26\frac { 1 }{ 4 } \) m
= \(\cfrac { 105 }{ 4 } \)
A rabbit covers in 1 jump = \(1\frac { 3 }{ 4 } \)
= \(\cfrac { 7 }{ 4 } \)m
\(\cfrac { 7 }{ 4 } \) m = 1 jump
\(\cfrac { 105 }{ 4 } m=\cfrac { 7 }{ 7/4 } \times \cfrac { 105 }{ 4 } \)
= \(\cfrac { 4 }{ 7 } \times \cfrac { 105 }{ 4 } =15\)

(i) \(\cfrac { 2 }{ 3 } \times 6=2\times =4\)
(ii) \(8\cfrac { 1 }{ 3 } \times 5\)
= \(\cfrac { 25 }{ 3 } \times 5=\cfrac { 25\times 5 }{ 3 } =\cfrac { 125 }{ 3 } =41\cfrac { 2 }{ 3 } \)
(iii) \(\cfrac { 3 }{ 8 } \times \cfrac { 4 }{ 5 } \)
= \(\cfrac { 3 }{ 8 } \times \cfrac { 4 }{ 5 } =\cfrac { 3\times 1 }{ 2\times 5 } =\cfrac { 3 }{ 10 } \)
(iv) \(3\cfrac { 5 }{ 7 } \times 1\cfrac { 1 }{ 13 } \)
= \(\cfrac { 26 }{ 7 } \times \cfrac { 14 }{ 13 } \)
= 2 \(\times\) 2 = 14

(i) \(3\cfrac { 7 }{ 8 } \)
Improper fraction = \(\cfrac { \left( 3\times 18 \right) +7 }{ 18 } \)
= \(\cfrac { 54+7 }{ 18 } =\cfrac { 61 }{ 18 } \)
(ii) \(\cfrac { 99 }{ 7 } \)
\(Mixed\ fraction=Qutient+\frac { Remainder }{ Divisor } \)
= \(14+\cfrac { 1 }{ 7 } \)
= \(14\cfrac { 1 }{ 7 } \)
(iii) \(\cfrac { 47 }{ 6 } \)
Mixed fraction = \(7\cfrac { 5 }{ 6 } \)
(iv) \(12\cfrac { 1 }{ 9 } \)
Mixed fraction = \(\cfrac { \left( 12\times 9 \right) +1 }{ 9 } \)
= \(\cfrac { 108+1 }{ 9 } =\cfrac { 109 }{ 9 }\)