Time, speed & distance

Question

Two trains start at the same time from Mumbai and Pune and proceed towards each other at the rate of $60\hspace{0.1cm} km$ and $40\hspace{0.1cm}km$ $\text{per hour}$ respectively. When they meet, it is found that one train has travelled $20\hspace{0.1cm} km$ more than the other. Find the distance between Mumbai and Pune.

Answer 1

We know that,

Time taken to meet/catch = $\dfrac{\text{Distance separating them}}{\text{Relative Speed}(S_1 \pm S_2)}$

Now, 

Relative Speed =$S_1+S_2$, when both the trains are moving in opposite direction.

Relative Speed =$S_1- S_2$, when both the trains are moving in same direction.

Now, assuming,

Distance between both the trains Or Distance between Mumbai to Pune = $x \hspace{0.1cm}km$

& Time taken to meet = $t\hspace{0.1cm} hr.$

$∴ t = \dfrac{x \hspace{0.1cm}km}{(60+40) \hspace{0.1cm}kmph}$

Or, $t = \dfrac{x}{100}hr$      -----------1)

Now, when both the trains meet , One train travelled 20km. more than the other.

As both the train started their journey at same time & train $A's$ speed is greater than the spped of train $B$.

∴ Train $A$ is travelled more.

∴ Distance travelled by $A$ in $t\hspace{0.1cm}hr= 60\times t$

Distance travelled by $B$ in $t\hspace{0.1cm}hr= 40\times t$

Now, given

$\qquad\qquad 60\times t - 40 \times t = 20$

$\qquad\qquad 20\times t = 20$

∴ $t = 1$

∴ Distance between Mumbai and Pune=

$\qquad\qquad t = \dfrac{x}{100}$   ----------- from 1)

Or, $\qquad\qquad x= 100 \times 1 = 100 \hspace{0.1cm}km$

∴ $\color{green}{\text{Distance between Mumbai and Pune is}}$ $\color{gold}{\text{100 km}}$

Answer 2

let t (time be in hrs) be time for which  train travelled for before meeting at a point.

one train covered 20 km more before meeting at point than other. It would be obvious that 60km/hr train will covered more. So, in time t distance covered by train A is 60t km and second will cover 40t km. now accrd to que.    60t=40t+20 , solving we get t=1 hr. 

now, distance b/w station is 60t+40t

so,60+40 =100km.