Question

Two trains start from stations A and B and travel towards each other at speeds of 50 km/hr and 60 km/hr respectively. At the time of their meetings, the second train had travelled 120 km more than the first. The distance between stations A and B is…
A) 990 km
B) 1200 km
C) 1320 km
D) 1440 km

Answer 1

Hint: Here both trains are running towards each other with their respective speed. Both trains will meet each other after a time of say t hour. As speed $s = \dfrac{d}{t}$ so, distance travel by train is given by $d = s \cdot t$. The second train has travelled 120 km more distance than the first train, so ${d_2} = {d_1} + 120$. Put ${d_1} = {s_1} \cdot t$ and ${d_2} = {s_2} \cdot t$ in ${d_2} = {d_1} + 120$ equation and calculate time t. Total distance between two stations will be $d = {d_1} + {d_2}$. By using values of s1 and s2, calculate d1 and d2 by using ${d_1} = {s_1} \cdot t$ and ${d_2} = {s_2} \cdot t$. After this use $d = {d_1} + {d_2}$ to obtain distance between two stations.

Complete step-by-step answer:
Here, given that two trains are starting simultaneously from stations A and B towards each other with speeds of 50 km/hr and 60 km/hr respectively.
Let’s say train-1 is starting from station A with a speed of 50 km/hr and train-2 is starting from station B with a speed of 60 km/hr.
So, speed ${s_1} = 50km/hr$ and ${s_2} = 60km/hr$.
After a time of t hour, both trains are meeting at a point.
Let’s assume that the total distance between two stations is d. Let’s say distance travelled by train-1 from station A to meeting point is d1 and by train-2 from station B to meeting point is d2. Above things are illustrated in the diagram below. So, we can say that $d = {d_1} + {d_2}$.

As we know that speed $s = \dfrac{d}{t}$ , so distance $d = s \cdot t$.
Here for train-1 distance travelled in time t is given by ${d_1} = {s_1} \cdot t$.
So, putting value of speed of train-1 ${s_1} = 50km/hr$ , ${d_1} = 50 \cdot t$.
And for train-2 distance travelled in time t is given by ${d_2} = {s_2} \cdot t$.
So, putting value of speed of train-2 ${s_2} = 60km/hr$ , ${d_2} = 60 \cdot t$.
At the meeting point it was observed that the second train has travelled 120 km more distance than the first train.
So, we can write that distance travelled by train-2 = distance travelled by train-1 + 120
So, ${d_2} = {d_1} + 120$
Putting ${d_1} = 50 \cdot t$ and ${d_2} = 60 \cdot t$ in the above equation.
$60 \cdot t = 50 \cdot t + 120$
Taking ala terms of t on one side of equation,
$60 \cdot t - 50 \cdot t = 120$.
Take t common in left side terms of equation, $(60 - 50) \cdot t = 120$
So, $10 \cdot t = 120$
Simplifying, $t = \dfrac{{120}}{{10}}$
So, $t = 12hr$.
So, distance travelled by train-1 in time t hour is, ${d_1} = 50 \cdot t$. Put $t = 12hr$
So, ${d_1} = 50 \cdot 12$
So, ${d_1} = 600km$
And, distance travelled by train-2 in time t hour is, ${d_2} = 60 \cdot t$. Put $t = 12hr$
So, ${d_2} = 60 \cdot 12$
So, ${d_2} = 720km$.
Total distance between station A and B is $d = {d_1} + {d_2}$.
Putting ${d_1} = 600km$ and ${d_2} = 720km$,
$d = 600 + 720$
$d = 1320km$
So, Distance between station A and B is 1320km.

Option (C) is the correct answer.

Note: While solving the problems regarding trains, take care of the following points. If both trains are travelling in the same direction then their relative speed will be different speeds of both trains, while for opposite direction travel relative speed will be the sum of speeds of both trains. If it is given that the train of length l traveling at speed s crosses a stationary man/pole in time t, then time t is calculated by using the formula $t = \dfrac{l}{s}$ formula. If a train of length l traveling at speed s crosses a platform/bridge of length m in time t, then time is calculated by using $t = \dfrac{{l + m}}{s}$ formula.