Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is ?

Answer: Option B

Explanation:

Relative Speed = 60+40 = 100 kmph
= 100*(5/18) = 250/9 m/sec

Distance = 140+160 = 300 meters

Time = Distance/Speed

\begin{aligned}
300*\frac{9}{250} = \frac{54}{5} \\
= 10.8 \text{ seconds}
\end{aligned}

So the time trains will take to cross each other will be 10.8 seconds

Answer: Option B

Explanation:

As we need to calculate answer in seconds, so first convert speed into meter/sec.

we know 1 km/hr = 1*(5/18) m/sec
So, Speed = 132* (5/18) = 110/3 m/sec

Distance need to be covered in passing the platform =
Length of train + length of platform = 110 + 165
= 275 meters

Time = Distance/Speed

\begin{aligned}
=> Time = 275* \frac{3}{110} = \frac{15}{2} \\
= 7.5 seconds
\end{aligned}

Answer: Option B

Explanation:

Distance covered = 120+120 = 240 m
Time = 12 s

Let the speed of each train = x.
Then relative velocity = x+x = 2x

2x = distance/time = 240/12 = 20 m/s

Speed of each train = x = 20/2 = 10 m/s

= 10*18/5 km/hr = 36 km/hr

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 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