Let the number of boys be x. \begin{align} & {\text{Then, }} \cr & {\text{Number of girls}} \cr & = 120\% {\text{ of }}x = \frac{{6x}}{5}. \cr & \therefore x + \frac{{6x}}{5} = 66 \cr & \Rightarrow \frac{{11x}}{5} = 66 \cr & \Rightarrow x = \frac{{66 \times 5}}{{11}} \cr & = 30. \cr & {\text{So, number of boys}} \cr & = 30. \cr & {\text{And, new number of girls}} \cr & = \left( {\frac{{6 \times 30}}{5}} \right) + 4 = 40. \cr & \therefore {\text{Required ratio}} \cr & = 30:40 \cr & = 3:4. \cr\end{align}

Quantity of milk $ = \left( {60 \times \frac{2}{3}} \right)\, {\text{litres}} = 40\, {\text{litres}}.$ Quantity of water in it = (60- 40) litres = 20 litres.

New ratio = 1 : 2

Let quantity of water to be added further be x litres. \begin{align} & {\text{Then, }}\, {\text{milk:water}} = \left( {\frac{{40}}{{20 + x}}} \right) \cr & {\text{Now}}, \, \left( {\frac{{40}}{{20 + x}}} \right) = \frac{1}{2} \cr & \Rightarrow 20 + x = 80 \cr & \Rightarrow x = 60. \cr & \therefore {\text{Quantity}}\, {\text{of}}\, {\text{water}}\, {\text{to}}\, {\text{be}}\, {\text{added = 60}}\, {\text{litres}}{\text{.}} \cr\end{align}

7, 16, 43, 79

If a, b, c, d are in proportion

⇒ a : b :: c : d

Then ad = bc

Let x is added to make it a proportion

⇒ (7 + x) : (16 + x) :: (43 + x) : (79 + x)

⇒ (7 + x)(79 + x) = (16 + x)(43 + x)

∴ x² + 86x + 553 = x² + 59x + 688

⇒ 27x = 135

⇒ x = 5

∴ Hence, it 5 is added to make it a proportion

Let the shares of A, B, C and D be Rs. 3x, 4x, 9x and 10x respectively.

= 9x - 4x =2580

⇒ 5x = 2580

⇒ x = 516

∴ Required Amount = 3x + 10x = 13x

= Rs. (13 $\times$ 516)

= Rs. 6708