Question

Two trains pass each other on parallel lines. Each train is 100 metres long. When they are going in the same direction, the faster one takes 60 seconds to pass the other completely. If they are going in opposite directions they pass each other completely in 10 seconds. Find the speed of the slower train in km/hr. (a) 30 km/hr (b) 42 km/hr (c) 48 km/hr (d) 60 km/hr (e) 56 km/hr

Answer 1

Answer: A) 30 km/h

Step-by-step explanation:

Let x be the speed of faster train and y be the speed of slower train,

According to the question,

Each train is 100 meters long.

Relative distance = 100 + 100 = 200 meters

When they are running in the same direction,

Their relative speed = (x - y) m/s,

So, the time taken by faster train to overtake the slower train,

$t_1=\frac{\text{ Relative distance}}{\text{Relative speed}}$

$t_1=\frac{200}{x-y}$

According to the question,

$t_1=60\text{ sec}$

$\implies 60=\frac{200}{x-y}$

$60 x- 60 y=200\implies 3x-3y=10$ ------(1)

Again, in opposite direction,

Their relative speed = (x + y) m/s

So, the the time taken by faster train to overtake the slower train,

$t_2=\frac{200}{x+y}$

Again according to the question,

$t_2=10\text{ sec}$

$\implies 10 = \frac{200}{x+y}$

$\implies x + y = 20$ -------(2),

3 × Equation (2) - Equation (1),

6y = 60 - 10

⇒ y = 50/6 m/s

⇒ y = ( 50/6 × 18/5 ) km/h = 30 km/h  ( Since, 1 m/s = 18/5 km/h ),

So, the speed of slower train is 30 km/h.

Answer 2

Answer:

A) 30 km/h

Let x be the speed of faster train and y be the speed of slower train,

According to the question,

Each train is 100 meters long.

Relative distance = 100 + 100 = 200 meters

When they are running in the same direction,

Their relative speed = (x - y) m/s,

So, the time taken by faster train to overtake the slower train,

t_1=\frac{\text{ Relative distance}}{\text{Relative speed}}t1=Relative speed Relative distance

t_1=\frac{200}{x-y}t1=x−y200

According to the question,

t_1=60\text{ sec}t1=60 sec

\implies 60=\frac{200}{x-y}⟹60=x−y200

60 x- 60 y=200\implies 3x-3y=1060x−60y=200⟹3x−3y=10 ------(1)

Again, in opposite direction,

Their relative speed = (x + y) m/s

So, the the time taken by faster train to overtake the slower train,

t_2=\frac{200}{x+y}t2=x+y200

Again according to the question,

t_2=10\text{ sec}t2=10 sec

\implies 10 = \frac{200}{x+y}⟹10=x+y200

\implies x + y = 20⟹x+y=20 -------(2),

3 × Equation (2) - Equation (1),

6y = 60 - 10

⇒ y = 50/6 m/s

⇒ y = ( 50/6 × 18/5 ) km/h = 30 km/h  ( Since, 1 m/s = 18/5 km/h ),

So, the speed of slower train is 30 km/h.