Let, the length of the train $ = \text{L meter}$
Speed of the train $ = \text{S m/sec}$
Now, we know that $ \boxed{\text{Speed} = \frac{\text{Distance}}{\text{Time}}} $
$ \Rightarrow {\text{Time} = \frac{\text{Distance}}{\text{Speed}}} $
So, $\frac{\text{L}}{\text{S} – 2 \times \frac{5}{18}} = 90$
$\Rightarrow \frac{\text{L}}{\text{S} - \frac{5}{9}} = 90 \quad \longrightarrow (1)$
And, $\frac{\text{L}}{\text{S} – 4 \times \frac{5}{18}} = 100$
$\Rightarrow \frac{\text{L}}{\text{S} - \frac{10}{9}} = 100 \quad \longrightarrow (2) $
Divide equation $(1)$ by equation $(2),$ we get,
$ \Rightarrow \frac{ \frac{\text{L}}{\text{S} – \frac{5}{9}}} {\frac{\text{L}}{\text{S} – \frac{10}{9}}} = \frac{90}{100}$
$\Rightarrow \frac{\text{L}}{ \left( \frac{9 \text{S}-5}{9} \right)} \times \frac{ \left( \frac{9 \text{S} – 10}{9} \right)}{\text{L}} = \frac{9}{10}$
$ \Rightarrow \frac{9 \text{S} – 10}{9 \text{S} – 5} = \frac{9}{10}$
$ \Rightarrow 90 \text{S} – 100 = 81 \text{S} – 45 $
$ \Rightarrow 9 \text{S} = 55 $
$ \Rightarrow \boxed{\text{S} = \frac{55}{9} \; \text{m/sec}}$
$ \Rightarrow \frac{1}{ \left( \frac{55}{9} – \frac{5}{9} \right)} = 90 \quad [ \because \text{ From equation (1)}]$
$ \Rightarrow \frac{\text{L}}{ \frac{50}{9}} = 90 $
$ \Rightarrow \text{L} = 90 \times \frac{50}{9}$
$ \Rightarrow \boxed{\text{L} = 500 \; \text{m}}$
$\therefore$ The time is taken by the train to cross an electric post $= \frac{\text{L}}{\text{S}}$
$\qquad = \frac{500}{ \left( \frac{55}{9} \right)}$
$\qquad = \frac{500}{55} \times 9$
$\qquad = \frac{4500}{55}$
$\qquad = 81.8181 \; \text{sec}$
$\qquad \simeq 82 \; \text{sec}.$
Correct Answer$: \text{B}$