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The distance from $\text{B}$ to $\text{C}$ is thrice that from $\text{A}$ to $\text{B}.$ Two trains travel from $\text{A}$ to $\text{C}$ via $\text{B}.$ The speed of train $2$ is double that of train $1$ while traveling from $\text{A}$ to $\text{B}$ and their speeds are interchanged while traveling from $\text{B}$ to $\text{C}.$ The ratio of the time taken by train $1$ to that taken by train $2$ in traveling from $\text{A}$ to $\text{C}$ is

1. $1:4$
2. $7:5$
3. $5:7$
4. $4:1$

Let the distance between $\text{A}\; \&\; \text{B}$ be $x\text{’} \; \text{km}.$ So, the distance between $\text{B} \;\&\; \text{C}$ be $3x\text{’} \; \text{km}.$

we know that, $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$

$\Rightarrow \boxed{ \text{Time} = \frac{\text{Distance}}{ \text{Speed}}}$

Let total time taken by train $1$ and train $2$ be  $\text{t}_{1} \; \text{hr}$ and $\text{t}_{2} \; \text{hr}$ respectively.

$\text{A} \longrightarrow \text{B} :$

• $\text{Speed of train 1} = \text{s} \; \text{km/hr}$
• $\text{Speed of train 2} = \text{2s} \; \text{km/hr}$

$\text{B} \longrightarrow \text{C} :$

• $\text{Speed of train 1} = \text{2s} \; \text{km/hr}$
• $\text{Speed of train 2} = \text{s} \; \text{km/hr}$

Total time taken by train $1 :$

$\text{t}_{1} = \left( \frac{x}{\text{s}} \right) + \left( \frac{3x}{\text{2s}} \right)$

$\Rightarrow \text{t}_{1} = \frac{2x+3x}{\text{2s}}$

$\Rightarrow \boxed{\text{t}_{1} = \frac{5x}{\text{2s}} \; \text{hr}}$

Total time taken by train $2 :$

$\text{t}_{2} = \left( \frac{x}{\text{2s}} \right) + \left( \frac{3x}{\text{s}} \right)$

$\Rightarrow \text{t}_{2} = \frac{x+6x}{\text{2s}}$

$\Rightarrow \boxed{\text{t}_{2} = \frac{7x}{\text{2s}} \; \text{hr}}$

Now, $\dfrac{\text{t}_{1}}{\text{t}_{2}} = \dfrac{ \left(\frac{5x}{\text{2s}} \right)} {\left( \frac{7x}{\text{2s}} \right)} = \left( \frac{5x}{\text{2s}} \right) \times \left( \frac{\text{2s}}{7x} \right) = \frac{5}{7}$

$\therefore$ The required ratio of time $\text{t}_{1} : \text{t}_{2} = 5 : 7.$

Correct Answer$: \text{C}$

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