\begin{align} & = \frac{x}{y} = \frac{3}{4} \cr & = \frac{{2x + 3y}}{{3y - 2x}} \cr & = \frac{{2\left( {\frac{x}{y}} \right) + 3}}{{3 - 2\left( {\frac{x}{y}} \right)}} \cr & = \frac{{2 \times \frac{3}{4} + 3}}{{3 - 2 \times \frac{3}{4}}} \cr & = \frac{9}{2} \times \frac{2}{3} = 3 \cr & = \left( {2x + 3y} \right):\left( {3y - 2x} \right) \cr & = 3:1 \cr\end{align}

Product of 1^st and 4_th terms (extremes) = product of 2^nd and 3^rd terms (means) ${\text{Product of 1st and 4th terms (extremes) = product of 2nd and 3rd terms (means)}}$

\begin{align} & {\text{A : B}} \cr & \frac{3}{2}:\frac{8}{3} \cr\end{align} take L.C.M. of denominator and multiply) $ \Rightarrow \frac{3}{2} \times 6:\frac{8}{3} \times 6 = 9x:16x$ After adding 15 in each we get \begin{align} & \therefore \frac{{9x + 15}}{{16x + 15}} = \frac{5}{3} \times \frac{2}{5} \cr & \Rightarrow \frac{{9x + 15}}{{16x + 15}} = \frac{2}{3}\left( {{\text{cross multiply}}} \right) \cr & \Rightarrow 27x + 45 = 32x + 30 \cr & \Rightarrow 5x = 15 \cr & \Rightarrow x = 3 \cr & \therefore {\text{Smaller number}} \cr & = {\text{9}} \times {\text{3 = 27}} \cr & {\text{Greater number}} \cr & {\text{ = 16}} \times {\text{3 = 48}} \cr\end{align}

\begin{align} & = \frac{{{{\text{W}}...{\text{2}}}}}{{{{\text{W}}...{\text{1}}}}} = \frac{3}{2}{\text{and}}\frac{{{{\text{W}}...{\text{1}}}}}{{{{\text{W}}...{\text{3}}}}} = \frac{1}{2} \cr & \Rightarrow \frac{{{W...2}}}{{{W...3}}} = \left( {\frac{{{{\text{W}}...{\text{2}}}}}{{{{\text{W}}...{\text{1}}}}} \times \frac{{{{\text{W}}...{\text{1}}}}}{{{{\text{W}}...{\text{3}}}}}} \right) \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{3}{2} \times \frac{1}{2} = \frac{3}{4} \cr & \Rightarrow {W...2}:{W...3} = 3:4 \cr\end{align}

Let the consequent be x. \begin{align} & {\text{Then, }} \cr & \frac{7}{8} = \frac{{35}}{x} \Rightarrow 35 \times 8 \cr & \Rightarrow x = \frac{{35 \times 8}}{7} = 40. \cr\end{align}