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 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 D

Explanation:

Speed of train relative to jogger = (45-9) = 36 kmph
= 36*(5/18) = 10 m/sec

Distance to cover = 240 + 120 = 360 metres

Time = Distance/Speed
So,
\begin{align}
Time = \frac{360}{10} = 36 \text{ Seconds}
\end{align}

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 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}