Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.

Then, sum of their values \begin{align} & = Rs.\, \left( {\frac{{25x}}{{100}} + \frac{{10 \times 2x}}{{100}} + \frac{{5 \times 3x}}{{100}}} \right) \cr & = Rs.\, \frac{{60x}}{{100}} \cr & \therefore \frac{{60x}}{{100}} = 30 \Leftrightarrow x = \frac{{30 \times 100}}{{60}} = 50 \cr & {\text{Hence, }}\, {\text{the}}\, {\text{number}}\, {\text{of}}\, {\text{5p}}\, {\text{coins}} \cr & = \left( {3 \times 50} \right) \cr & = 150 \cr\end{align}

\begin{align} & {\text{Second son's share}} \cr & = {\text{Rs}}{\text{.}}\left( {10800 \times \frac{4}{{12}}} \right) \cr & = {\text{Rs}}{\text{. }}3600. \cr\end{align} Money distributed between the two daughters

= Rs. [3600 - (1000 + 600)]

= Rs. 2000 \begin{align} & {\text{First daughter's share}} \cr & = {\text{Rs}}{\text{.}}\left( {2000 \times \frac{{11}}{{20}}} \right) \cr & = {\text{Rs}}.1100. \cr & {\text{Second daughter's share}} \cr & = {\text{Rs}}{\text{.}}\left( {2000 \times \frac{9}{{20}}} \right) \cr & = {\text{Rs}}{\text{.9}}00. \cr\end{align}

Let income of A and B be 3x and 2x respectively. Also, their expenditure is 5y and 3y.

Now, according to question,

3x-5y = 1000 ------- (i)*3

2x-3y = 1000 ---------- (ii)*5

9x-15y-10x+15y = 3000-5000;

Or, -x = -2000;

Or, x = 2000;

Then, income of A = 3x = 3*2000 = Rs. 6000.

Clearly, A and B save 20% and 30% of their respective salaries.

Let the salaries of A and B be x and y respectively.

Then, \begin{align} & {\text{ = }}\frac{{{\text{20}}\% {\text{ of }}x}}{{{\text{30}}\% {\text{ of }}y}} = \frac{4}{3} \cr & \Rightarrow \frac{x}{5} \times \frac{{10}}{{3y}} = \frac{4}{3} \cr & \Rightarrow \frac{x}{y} = 2 \cr & \Rightarrow x = 2y. \cr & \therefore x + y = 2100 \cr & \Rightarrow 2y + y = 2100 \cr & \Rightarrow 3y = 2100 \cr & \Rightarrow y = 700. \cr & {\text{A's salary}} = x = 2y \cr & = {\text{Rs}}{\text{. }}\left( {2 \times 700} \right) \cr & = {\text{Rs}}{\text{. }}1400. \cr\end{align}

\begin{align} & = \frac{x}{y} = \frac{7}{3} \cr & = \frac{{xy + {y^2}}}{{{x^2} - {y^2}}} \cr & = \frac{{\left( {\frac{x}{y}} \right) + 1}}{{\left( {\frac{{{x^2}}}{{{y^2}}}} \right) - 1}} \cr & = \frac{{\frac{7}{3} + 1}}{{{{\left( {\frac{7}{3}} \right)}^2} - 1}} \cr & = \frac{{10}}{3} \times \frac{9}{{40}} \cr & = \frac{3}{4}. \cr\end{align}