A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?

Answer: Option B

Explanation:

Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27x metres,
Length of the second train = 17y metres.
[because distance = speed*time]

\begin{aligned}
\frac{27x+17y}{x+y} = 23 \\
=> 27x + 17y = 23x + 23y \\
=> 4x = 6y \\
=> \frac{x}{y} = \frac{6}{4}
\end{aligned}
So ratio of the speeds of train is 3:2

Answer: Option C

Explanation:

As trains are running in opposite directions so their relative speed will get added
So, Relative speed = 120 +80 = 200 kmph
= 200*(5/18) = 500/9 m/sec

Let the length of other train is x meter then

\begin{aligned}
\frac{x+270}{9} = \frac{500}{9} \\
=> x+270 = 500\\
=> x = 230
\end{aligned}

So the length of the train is 230 meters

Answer: Option C

Explanation:

Relative Speed = 63-3 = 60 Km/hr
= 60 *(5/18) = 50/3 m/sec

Time taken to pass the man will ne
\begin{aligned}
500*\frac{3}{50} = 30 \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

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