Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively. When 9 litres of mixture are drawn off, quantity of A in mixture left:

\begin{align} & \left[ {7x - \left( {\frac{7}{{12}}} \right) \times 9} \right] = \left[ {7x - \left( {\frac{{21}}{4}} \right)} \right]\, {\text{litres}} \cr & {\text{Similarly quantity of B in mixture left}}, \cr & \left[ {5x - \left( {\frac{5}{{12}}} \right) \times 9} \right] = \left[ {5x - \left( {\frac{{15}}{4}} \right)} \right]\, {\text{litres}} \cr & \therefore \, {\text{ratio becomes}}, \cr & \frac{{\left[ {7x - \left( {\frac{{21}}{4}} \right)} \right]}}{{\left[ {5x - \left( {\frac{{15}}{4}} \right)} \right]}} = \frac{7}{9} \cr & \Rightarrow \frac{{\left( {28x - 21} \right)}}{{\left( {20x + 21} \right)}} = \frac{7}{9} \cr & \Rightarrow \left( {252x - 189} \right) = 140x + 147 \cr & \Rightarrow 112x = 336 \cr & \Rightarrow x = 3 \cr & {\text{So the can contained}}, \cr & 7 \times x \cr & = 7 \times 3 \cr & = 21\, {\text{litres of A initially}}{\text{.}} \cr\end{align}

Given total runs of A, B, C (A + B + C = 361) \begin{align} & \Rightarrow {\text{A}}:{\text{B}}:{\text{C}} \cr & \, \, \, \, \, \, \, \, \, 3:2 \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 3:2 \cr & \underline {\overline {\, \, \, \, \, \, \, \, \, \, 9:6:4{\text{ }}} } \cr & \therefore 9x + 6x + 4x = 361 \cr & \Rightarrow 17x = 361 \cr & \Rightarrow x = 19 \cr & \therefore {\text{Runs scored by A }} \cr & = 9x \cr & = 9 \times 19 \cr & = 171 \cr\end{align}

Let originally were 4x, 6x and 9x student there in classes receptively.

After 12 students increase in each student then students were 7x, 9x and 12x in each class respectively.

Now, Total Students = 7x + 9x + 12x

4x + 6x + 9x + 3*12 = 28x

9x = 3*12

x = 4.

Then total number of student in three classes, = 4x + 6x + 9x = 19x = 19 * 4 = 76.

Given;

7 jumps of Tom = 5 jumps of Jerry.

Or, Tom / Jerry = 5/7;

Let Jerry's 1 leap = 7 meter and Tom's 1 leap = 5 meter.

Then, ratio of speed of Tom and Jerry = 8*5/6*7 = 40/42 = 20 :21.

\begin{align} & \, \, \, \, \, \, \, \, \, \, \, {\text{A}}:{\text{B}} \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, 5:3 \cr & {\text{Let }}5x:3x = 8x \cr & \Rightarrow {\text{New comers}} \cr & 5y:7y = 12y \cr & \therefore 8x + 12 = 1200 \cr & \Rightarrow 2x + 3y = 300.....(i) \cr & {\text{Again, }}\frac{{5x + 5y}}{{3x + 7y}} = \frac{7}{5} \cr & 25x + 25y = 21x + 49y \cr & \Rightarrow 4x - 24y = 0 \cr & \Rightarrow 4x = 24y \cr & \Rightarrow x = 6y.....(ii) \cr & \therefore {\text{From equation }}.....(i) \cr & 12y + 3y = 300 \cr & y = 20 \cr & \therefore x = 6 \times 20 = 120 \cr\end{align} The number of students initially

8x = 8 $\times$ 120 = 960