Question

A train travelling at a speed of \[75\]mph enters a tunnel \[3.5{\text{ }}\]miles long. The train is $\dfrac{1}{4}$ mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?
\[{\mathbf{A}}.2\]min
\[{\mathbf{B}}.3\]min
\[{\mathbf{C}}.4\]min
\[{\mathbf{D}}.5\]min

Answer 1

Hint: In this problem, we need to calculate the total time taken, we will add the length of tunnel and train so that we can get the total distance and then use Speed \[ = \]$\dfrac{{Dis\tan ce}}{{Time}}$


Complete step by step solution:
We are given that :
Speed of the train \[ = {\text{ }}75km/h\]
Length of the tunnel \[ = {\text{ }}3\dfrac{1}{2}miles\]
\[\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \dfrac{7}{2}\]
Also, the length of the train \[ = \dfrac{1}{4}miles\]
Now, we need to find the time taken to pass through the tunnel.
So Total distance \[ = \] Length of the tunnel \[ + \] Length of train
\[\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \dfrac{7}{2} + \dfrac{1}{4} = \dfrac{{15}}{4}miles\]
As we know, Time Taken \[ = \dfrac{{TotalDis\tan ce}}{{Speed}}\]
 \[ = \dfrac{{\left( {\dfrac{{15}}{4}} \right)}}{{75}} = \dfrac{1}{{20}}\]
\[\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \dfrac{1}{{20}} \times 60\]minutes (converting into minutes )
\[\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = {\text{ }}3\] minutes.
So, time is taken to pass the tunnel \[ = {\text{ }}3\]minutes.
$\therefore $Option ‘B’ is correct.


Note: In questions like these,do not forget to add the length of both the tunnel and the train because total distance travelled by train would be both length of tunnel and the length of train.