A jogger running at 9 kmph alongside a railway track is 240 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction. In how much time will the train pass the jogger ?

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 C

Explanation:

Speed = 45 Km/hr = 45*(5/18) m/sec
= 25/2 m/sec
Total distance = 360+140 = 500 meter

Time = Distance/speed
\begin{aligned}
= 500*\frac{2}{25} = 40 seconds
\end{aligned}

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 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 A

Explanation:

Speed = 60*(5/18) m/sec = 50/3 m/sec

Length of Train(Distance) = Speed * Time

\begin{aligned}
= \frac{50}{3}*9 = 150 meter
\end{aligned}