Let 2A = 3B = 4C = k \begin{align} & {\text{Then, }} \cr & A = \frac{k}{2}, \cr & B = \frac{k}{3}, \cr & C = \frac{k}{{4.}} \cr & \therefore {\text{A}}:{\text{B}}:{\text{C}} = \frac{k}{2}:\frac{k}{3}:\frac{k}{4} \cr & = \frac{1}{2}:\frac{1}{3}:\frac{1}{4} \cr & = 6:4:3. \cr\end{align}

\begin{align} & {\text{Let number be }}x\& y \cr & {\text{Given, }} \cr & \, \, x:y \cr & \, \, \, \, 3:4 \cr & \, 3a:4a \cr & {\text{Now, given that}} \cr & \Rightarrow {\text{8}}{\left( {3a} \right)^2} = {\left( {4a} \right)^2} + 224 \cr & \Rightarrow 72{a^2} = 16{a^2} + 224 \cr & \Rightarrow 56{a^2} = 224 \cr & \Rightarrow {a^2} = 4 \cr & \Rightarrow a = 2 \cr & {\text{Numbers are}} \cr & x = 3 \times 2 = 6 \cr & y = 4 \times 2 = 8 \cr\end{align}

Let the number of girls be x.

Then, \begin{align} & {\text{Number of boys}} \cr & = 116\% {\text{ of }}x = \frac{{29}}{{25}}x. \cr & \therefore {\text{Required ratio}} \cr & = \frac{{29}}{{25}}x:x \cr & = 29:25. \cr\end{align}

\begin{align} & a:b:c \cr & 2:3:4 \cr & {\text{Let }}2x:3x:4x \cr & \Rightarrow 2a - 3b + 4c = 33 \cr & \Rightarrow 2 \times 2x - 3 \times 3x + 4 \times 4x = 33 \cr & \Rightarrow 4x - 9x + 16x = 33 \cr & \Rightarrow 11x = 33 \cr & \Rightarrow x = 3 \cr & \therefore {\text{c}} \Rightarrow 4 \times 3 = 12 \cr\end{align}

Let the quantity of the wine in the cask originally be x litres.

Using formula:

Final Amount of solute that is not replaced = Initial Amount $\times$( Vol. after removal /Vol. after replacing)^N

Where N = No. of operation done.

Then ratio of wine to total solution in cask after 4 operations,

\begin{align} & 1 \times {\left\{ {\left( {\frac{{x - 8}}{x}} \right)} \right\}^4} = \frac{{16}}{{81}} \cr & \Rightarrow 1 \times {\left\{ {\frac{{\left( {x - 8} \right)}}{x}} \right\}^4} = {\left( {\frac{2}{3}} \right)^4} \cr & \Rightarrow \frac{{\left( {x - 8} \right)}}{x} = \frac{2}{3} \cr & \Rightarrow 3x - 24 = 2x \cr & \Rightarrow x = 24\, {\text{litres}} \cr\end{align}