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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. $4:1$
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Let the distance between $\text{A} \& \text{B}$ be $\text{‘x’} \; \text{km}.$

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

DIAGRAM

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.

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

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

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

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

Total time taken by train$1 :$

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

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

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

Total time taken by train$2 :$

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

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

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

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

$ = \left( \frac{\text{5x}}{\text{2s}} \right) \times \left( \frac{\text{2s}}{\text{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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