Question 14.

Let the length of the train be $l$ kms and speed be $s$ kmph.  

Based on the two scenarios presented, we obtain:

$\frac{l}{s-2} = \frac{90}{3600}$ ...(i)  

$\frac{l}{s-4} = \frac{100}{3600}$ ...(ii)

On dividing (ii) by (i) and simplifying we acquire the value of $s$ as $22$ kmph.  

Substituting this value in (i), we have $l = \frac{90}{3600} \times 20$ kms. {keeping it in km and hours for convenience}

Since we need to find $\frac{l}{s}$, let this be equal to $x$.  

Then, $x = 90 \times \frac{20}{22} = 81.81 \approx 82$ seconds.