A train is running with a speed of 60 kmph and its length is 110 metres. Calculate the time by which it will pass a man running opposite with speed of 6 kmph.

Answer: Option B

Explanation:

First person speed = 2*(5/18) = 5/9 m/sec
Second person speed = 4*(5/18) = 10/9 m/sec

Let the length of train is x metre and speed is y m/sec

then,
\begin{aligned}
\frac{x}{y-\frac{5}{9}} = 9 \\
=> 9y-5 = x \\
=> 9y-x = 5 .....(i) \\
Also, \\
\frac{x}{y-\frac{10}{9}} = 10 \\
90y-9x = 100 .....(ii)\\

\text{from (i) and (ii), we get,} \\
x=50
\end{aligned}
So length of train is 50 metre

Answer: Option D

Explanation:

Speed = Distance/time = 300/18 = 50/3 m/sec

Let the length of the platform be x meters
then
\begin{aligned}
Distance = Speed*Time \\
x+300 = \frac{50}{3}*39 \\
=>3(x+300)= 1950 \\
=> x = 350 \text{ meters}
\end{aligned}

Answer: Option C

Explanation:

Let the length of the train is x meter and Speed of the train is y meter/second
Then x/y = 15 [because distance/speed = time]
=> y = 15/x

\begin{aligned}
=> \frac{x+100}{25} = \frac{x}{15} \\
x = 150 \text{ meters}
\end{aligned}

So length of the train is 150 meters

Answer: Option C

Explanation:

From the given question,
we will first calculate the speed of train relative to man,
=> ( 60 + 6 ) = 66 km/hr ( we added 6 because man is running opposite )
Convert it in metre/second
\begin{aligned}
= 66 \times\frac{5}{18} \\
= \frac{55}{3} m/sec\\
\text{Time it will take to pass man }\\
= 110 \times \frac{3}{55} \\
= 6 \text{ seconds}
\end{aligned}

Answer: Option C

Explanation:

Let the length of each train is x meter
Distance will be x+x = 2x

Relative Speed = 46-36 = 10 km/hr
= 10*(5/18) = 25/9 m/sec

Distance = Speed*Time

\begin{aligned}
2x = \frac{25}{9}*36 \\
2x = 100 \\
=> x = 50
\end{aligned}
So length of both the trains are 50 meters