$$\eqalign{ & {\text{A}}:{\text{B}} = 5:12 \cr & {\text{B}}:{\text{C}} = 4:5.50 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 12:16.5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 12:\frac{{33}}{2} \cr & {\text{A}}:{\text{B}}:{\text{C}} = 5:12:\frac{{33}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 10:24:33 \cr & {\text{Sum of ratio terms}} \cr & = 10 + 24 + 33 \cr & = 67 \cr & {\text{C's share}} = {\text{Rs}}{\text{.}}\left( {2010 \times \frac{{33}}{{67}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{. }}990 \cr & B'{\text{s share}} = {\text{Rs}}{\text{.}}\left( {2010 \times \frac{{24}}{{67}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}.720 \cr & {\text{Required difference}} \cr & = {\text{Rs}}{\text{.}}\left( {990 - 720} \right) \cr & = {\text{Rs}}{\text{. }}270 \cr} $$

$$\eqalign{ & {\text{Gold in C}} \cr & = \left( {\frac{7}{9} + \frac{7}{{18}}} \right)\,{\text{units}} \cr & = \frac{{\text{7}}}{{\text{6}}}\,{\text{units}}{\text{}} \cr & {\text{Copper in C}} \cr & = \left( {\frac{2}{9} + \frac{{11}}{{18}}} \right)\,{\text{units}} \cr & = \frac{5}{6}\,{\text{units}}{\text{}} \cr & \refore {\text{Gold}}:{\text{Copper}} \cr & = \frac{7}{6}:\frac{5}{6} \cr & = 7:5 \cr} $$

Let earnings of A and B Rs. 8x and Rs. 9x respectively.
Then,
$$\eqalign{ & = \frac{{150\% {\text{ of }}8x}}{{75\% {\text{ of }}9x}} = \frac{{16}}{9} \cr & \Rightarrow \frac{{\frac{3}{2} \times 8x}}{{\frac{3}{4} \times 9x}} = \frac{{16}}{9} \cr & \Rightarrow \frac{{16}}{9} = \frac{{16}}{9} \cr} $$
Hence, A's earnings cannot be determined

Milk in 1 litre mix. in A = $$\frac{8}{{13}}$$
Milk in 1 litre mix. in B = $$\frac{5}{{7}}$$
Milk in 1 litre of final mix. = $$\frac{{900}}{{13}} \times \frac{1}{{100}} \times 1 = \frac{9}{{13}}$$

By rule of alligation, we have :

∴ Required ratio $$ = \frac{2}{{91}} : \frac{1}{{13}} = 2 : 7$$

Let weights of three varieties be 12x, 15x and 20x kg respectively
Then,

Total C.P. = Rs. (100 × 12x + 80 × 15x + 60 × 20x)
= Rs. (3600x)
S.P. of first two verities
$$\eqalign{ & = {\text{Rs}}.\left( {\frac{{120}}{{100}} \times 2400x} \right) \cr & = {\text{Rs}}.\left( {2880x} \right) \cr & {\text{S}}{\text{.P}}{\text{. of third variety}} \cr & = {\text{Rs}}.\left( {3600x - 2880x} \right) \cr & = {\text{Rs}}{\text{. }}720x \cr & {\text{Loss on third variety}} \cr & = {\text{Rs}}{\text{.}}\left( {1200x - 720x} \right) \cr & = {\text{Rs}}{\text{. }}480 \cr & {\text{loss }}\% \cr & = \left( {\frac{{480x}}{{1200x}} \times 100} \right)\% \cr & = 40\% \cr} $$

Let required number be x
Then,
$$ = \frac{{7 - x}}{{9 - x}} = \frac{{11 - x}}{{15 - x}}$$
⇒ (7 - x)(15 - x) = (11 - x)(9 - x)
⇒ 105 - 22x + x² = 99 - 20x + x²
⇒ 2x = 6
⇒ x = 3

Let age of A & B = 5x and 6x years
According to question,
$$\eqalign{ & \frac{{5x + 7}}{{6x + 7}} = \frac{6}{7} \cr & 35x + 49 = 36x + 42 \cr & x = 7 \cr & {\text{A's Present age}} = 5x \cr & = 5 \times 7 = {\text{35 years}} \cr} $$

Let 1 bottle of acid is 30litre and consider first bottle as A and second bottle as B.
Ratio of Acid and water in bottle A = 2 : 3
Volume of Acid in bottle A = $$\frac{2}{5} \times 30 = 12$$
Volume of Water in bottle A = $$\frac{3}{5} \times 30 = 18$$

Ratio of Acid and water in bottle A = 1 : 2
Volume of Acid in bottle A = $$\frac{1}{3} \times 30 = 10$$
Volume of Water in bottle A = $$\frac{2}{3} \times 30 = 20$$

Now 1 bottle of A and 3 bottles of B mixed toger

Volume of Acid in new mixture = 12 + 3 × 10 = 42
Volume of water in new mixture = 18 + 3 × 20 = 78
Ratio of Acid and water in new mixture = 42 : 78 = 7 : 13

Sides of a triangle are in ratio of 7 : 9 : 12
i.e. 7x : 9x : 12x
Difference between length of largest and smaller sides = 12x - 7x = 15
⇒ x = 3
∴ largest sides would be 12 × 3 = 36cm

$$\eqalign{ & {\text{First}}:{\text{Second}}:{\text{Third}} \cr & \,\,\,\,8\,\,\,\,\,\,:\,\,\,\,\,\,9 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,:\,\,\,\,\,4 \cr & \frac{{\overline {\,\,24\,\,\,:\,\,\,\,\,27\,\,\,\,\,\,\,\,:\,\,\,\,36} }}{{\underline {\,8\,\,\,\,\,\,\,\,:\,\,\,\,\,\,\,\,9\,\,\,\,\,\,\,\,:\,\,\,\,12} }} \cr & {\text{Let 8x}}:{\text{9x}}:{\text{12x}} \cr & \refore {\text{First}} \times {\text{Third}} \cr & 8x \times 12x = 2400 \cr & \Rightarrow 96{x^2} = 2400 \cr & \Rightarrow {x^2} = \frac{{2400}}{{96}} = 25 \cr & \Rightarrow x = 5 \cr & \refore {\text{Sum of three numbers }} \cr & {\text{First}} + {\text{Second}} + {\text{Third}} \cr & 8x + 9x + 12x = 29x \cr & \refore 29 \times 5 = 145 \cr} $$