A 300 meter long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform.

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 B

Explanation:

As we need to get answer in seconds, so never forget to convert speed into meter per second.

Speed = 30 km/hr = 30* 5/18 m/sec
= 25/3 m/sec

Distance = length of train = 100 meter

Required time =
\begin{aligned}
\frac{Distance}{Speed} \\
= \frac{100}{\frac{25}{3}} \\
= 100* \frac{3}{25} = 12 sec \\
\end{aligned}

Answer: Option C

Explanation:

First convert speed from km/hr to m/sec

So, Speed = 72*(5/18) = 20 m/sec
Time = 26 seconds

Let the length of the train be x meters.
We know, Distance = Speed*Time.
[you can remember this formula as remembering DUST = D*ST... Distance=Speed*Time]

x+250 = 20*26
=> x = 270 meters

So length of the goods train is 270 meter

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 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