\begin{align} & {\text{Ratio of sides}} \cr & = \frac{{\text{1}}}{{\text{2}}}:\frac{{\text{1}}}{{\text{3}}}:\frac{{\text{1}}}{{\text{4}}} \cr & = 6:4:3. \cr\end{align} Let the sides be 6x, 4x and 3x

Then,

⇒ 6x + 4x + 3x = 104

or 13x = 104

or x = 8

∴ Longest side = 6x = (6 $\times$ 8)m = 48m

	Zinc	:	Tin
A	5x	:	2x	=	7x
B	3y	:	4y	=	7y

⇒ A ⇒ 7x = 7 kg

x = 1 kg

∴ Zinc in alloy A ⇒ 5kg

Tin in alloy A ⇒ 2 kg

⇒ B ⇒ 7y = 21 kg

y = 3 kg

Zinc in alloy B ⇒ 3 $\times$ 3 = 9 kg

Tin in alloy B ⇒ 3 $\times$ 4 = 12 kg

∴ After mix - up the ratio of Zinc and Tin in new alloy \begin{align} & {\text{Zinc}}:Tin \cr & \, \, \, {\text{A}}\, 5\, \, :\, \, \, \, \, 2 \cr & \, \, \, {\text{B}}\, 9\, \, \, :\, \, \, 12 \cr & \overline {{\text{A}} + {\text{B}}\, \, \, \, {\text{14}}\, \, \, \, \, {\text{14}}} \cr & \, \, \, \, \, \, \boxed{1\, \, \, \, \, :\, \, \, \, \, 1} \cr\end{align}

\begin{align} & = \frac{{\text{A}}}{{\text{B}}} = \frac{2}{1}, \, \, \frac{{\text{B}}}{{\text{C}}} = \frac{4}{3}, \, \, \frac{{\text{C}}}{{\text{D}}} = \frac{5}{6} \cr & \Rightarrow \frac{{\text{A}}}{{\text{D}}} = \left( {\frac{{\text{A}}}{{\text{B}}} \times \frac{{\text{B}}}{{\text{C}}} \times \frac{{\text{C}}}{{\text{D}}}} \right) \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \left( {\frac{2}{1} \times \frac{4}{3} \times \frac{5}{6}} \right) \cr & \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \frac{{20}}{9} \cr & \Rightarrow {\text{A}}:{\text{D}} = {\text{20}}:{\text{9}} \cr\end{align}

Rs. 1 : 50paise : 25paise

Number of coins 8x : 5x : 3x \begin{align} & {\text{Value of coins}} \cr & {\text{ = }}\frac{{8x}}{1}:\frac{{5x}}{2}:\frac{{3x}}{4} \cr & \therefore 8x + \frac{{5x}}{2} + \frac{{3x}}{4} = {\text{Rs}}.\, 225, \cr & {\text{Given}}\frac{{32x + 10x + 3x}}{4} = 225 \cr & \Rightarrow 45x = 225 \times 4 \cr & \Rightarrow x = \frac{{225 \times 4}}{{45}} \cr & \Rightarrow x = 5 \times 4 = 20 \cr\end{align} ∴ Number of one rupees coins

= 20 $\times$ 8 = 160

	1st	:	2nd
Fare	4x	:	x
Passengers	1	:	40

Total fare = 4x : 40x = 44x

⇒ 44x = 1100

⇒ x = ¹¹⁰⁰/₄₄ = 25

∴ Fare = 1^st class amount received per day

⇒ 4x = 4 $\times$ 25

⇒ 4x = 100