Ascending order:
II beaker \(\cfrac { 1 }{ 4 } <\cfrac { 1 }{ 2 } <\cfrac { 3 }{ 4 } <1\)
\(\therefore\) Ascending order : \(\cfrac { 1 }{ 4 } ,\cfrac { 1 }{ 2 } ,\cfrac { 3 }{ 4 } ,1\)

Let the number be 'x' 
\(X\times 3\cfrac { 1 }{ 16 } =9\cfrac { 3 }{ 16 } \)
\(X\times \cfrac { 49 }{ 16 } =\cfrac { 147 }{ 16 } \)
\(X=\cfrac { 147 }{ 16 } \times \cfrac { 16 }{ 49 } \)
x = 3
\(\therefore\) Number = 3

Let the fractions be 'x'
\(9\cfrac { 3 }{ 7 } -X=3\cfrac { 1 }{ 5 } \)
\(\cfrac { 66 }{ 7 } -X=\cfrac { 16 }{ 5 } \)
-X = \(\cfrac { 16 }{ 5 } -\cfrac { 66 }{ 7 } \)
-X = \(\cfrac { \left( 16\times 7 \right) -\left( 66\times 5 \right) }{ 35 } \)
-X = \(\cfrac { 112-330 }{ 35 } \)
-X = \(\cfrac { -218 }{ 35 } \)
X =  \(6\cfrac { 8 }{ 35 } \)
\(\therefore 6\cfrac { 8 }{ 35 } \)  is to be subtracted from \(9\cfrac { 3 }{ 7 } \)

Gowri purchased the weight of 
vegetables Tomatoes = \(3\cfrac { 1 }{ 2 } kg=\cfrac { 7 }{ 2 } kg\)
Brinjal = \(\cfrac { 3 }{ 4 } \) kg
Onion = \(1\cfrac { 1 }{ 4 } kg=\cfrac { 5 }{ 4 } kg\)
\(\therefore\) Total weight of the Vegetables = \(\left[ \cfrac { 7 }{ 2 } +\cfrac { 3 }{ 4 } +\cfrac { 5 }{ 4 } \right] kg\)
LCM of 2, 4 is 4
= \(\cfrac { \left( 7\times 2 \right) +3+5 }{ 4 } =\cfrac { 14+3+5 }{ 4 } =\cfrac { 22 }{ 4 } =\cfrac { 11 }{ 2 } \)
\(\therefore\) Total weight = \(5\cfrac { 1 }{ 2 } kg\)

The number of bottles required = \({15\over 2}\div{5\over2}={15\over2}\times{2\over5}\) (reciprocal of \(\frac{5}{2} \text { is } \frac{2}{5}\) )
= 3 bottles