Question

A train leaves Pune at 7:30 a.m. and reaches Mumbai at 11:30 a.m. Another train leaves Mumbai at 9:30 a.m. and reaches Pune at 1:00 p.m. Assuming that the two trains travel at constant speed at what time do the two trains cross each other?
A. 10:20 a.m.
B. 11:30 a.m.
C. 10:26 a.m.
D. data not sufficient

Answer 1

Hint: When the body moves towards each other, then the relative velocity is the sum of two velocities. When the body moves in the same direction, then the relative velocity is the difference of two velocities.

Complete step by step solution:
Let the constant speeds of the two trains are $u$ and $v$
The distance between Pune and Mumbai is $d$
It is given that the first train leaves Pune at 7:30 am.
First train reaches Mumbai at 11:30 am.
Time taken by the first train to cover the distance is 4 hours.
Second train leaves Mumbai at 9:30 am.
Second train reaches Pune at 1:00 am.
Time taken by the second train to cover the distance is 3.5 hours.
Both the trains will meet each other at the same time.
Speed\[=\dfrac{\text{distance}}{\text{time}}\]
Speed of first train $=\dfrac{d}{4}$
Speed of the second train $=\dfrac{d}{\left( \dfrac{7}{2} \right)}=\dfrac{2d}{7}$
Distance travelled by first train till 9:30 am $=\dfrac{d}{2}$
At 9:30 am the relative distance needed to cover to meet is $\dfrac{d}{2}$
Relative speed $=u+v=\dfrac{d}{4}+\dfrac{2d}{7}$
Let them meet after time $t$ when the second train starts.
Distance travelled by first train in time $t$ is ${{l}_{1}}$
$\begin{align}
  & {{l}_{1}}=ut \\
 & =\dfrac{dt}{4}
\end{align}$
Distance travelled by second train is ${{l}_{2}}$
$\begin{align}
  & {{l}_{2}}=vt \\
 & =\dfrac{2dt}{7}
\end{align}$
It is assumed that the distance between Mumbai and Pune is $\dfrac{d}{2}$
Then,
$\begin{align}
  & {{l}_{1}}+{{l}_{2}}=\dfrac{d}{2} \\
 & \dfrac{dt}{4}+\dfrac{2dt}{7}=\dfrac{d}{2} \\
 & \dfrac{t}{4}+\dfrac{2t}{7}=\dfrac{1}{2} \\
 & \dfrac{7t+8t}{28}=\dfrac{1}{2} \\
 & 15t=14 \\
 & t=\dfrac{14}{15}hours
\end{align}$
As we know that, $1hour=60\min $

Therefore, $\dfrac{14}{15}hour=56\min $

Hence, the time at which both will meet is,
9:30 am $+$56 mins$ = $10:26 am.

Note: When two bodies are moving in opposite directions then the relative velocity is the sum of the two velocities.
When two bodies are moving in the same direction then the relative velocity is the difference of the two velocities.