Two persons are walking beside a railway track at respective speeds of 2 and 4 km per hour in the same direction. A train came from behind them and crossed them in 90 and 100 seconds, respectively. The time, in seconds, taken by the train to cross an electric post is nearest to

Let the speed of the train be $V$ km/h and the length of the train be $L$ km.

The relative speed of the train with respect to the first person (walking at 2 km/h in the same direction) is:
$V_1=V-2$ km/h

The train crosses this person in 90 seconds. Convert 90 seconds to hours:
$t_1=\frac{90}{3600}=\frac{1}{40}$ hours

Therefore, the length of the train is:
$L=\left(V-2\right)\cdot \frac{1}{40}$     --- (Equation 1)

The relative speed of the train with respect to the second person (walking at 4 km/h in the same direction) is:
$V_2=V-4$ km/h

The train crosses this person in 100 seconds. Convert 100 seconds to hours:
$t_2=\frac{100}{3600}=\frac{1}{36}$ hours

Therefore, the length of the train is:
$L=\left(V-4\right)\cdot \frac{1}{36}$     --- (Equation 2)

Equating Equation 1 and Equation 2:
$\frac{V-2}{40}=\frac{V-4}{36}$

$9\left(V-2\right)=10\left(V-4\right)$

$9V-18=10V-40$

$V=22$ km/h

Substitute $V=22$ back into Equation 1 to find $L$:
$L=\frac{22-2}{40}=\frac{20}{40}=0.5$ km

The time taken by the train to cross an electric post (which is stationary) is:
$t=\frac{L}{V}=\frac{0.5}{22}$ hours

Converting this time into seconds:
$t=\frac{0.5}{22}\cdot 3600=\frac{1800}{22}\approx 81.82$ seconds

The nearest integer to 81.82 is 82 seconds.