Question

A train passes two persons walking in the same direction at a speed of 3 km/hour and 5 km/hour respectively in 10 seconds and 11 seconds respectively. the speed of the train is

Answer 1

Let t speed be x

Relative speed of 1st person = x-3

2nd on=x-4

X-3*10/60=x-5*11/60

25

Answer 2

Answer:  

The speed of the train is 25 km/hr.

Solution:

Let the speed of the train be x. So with x speed, person one feels the speed as 3 km/hr and person two as 5 km/hr. So, the relative speed will be for person 1 (P1), it will be (x-3) km/hr and for person 2 (P2), it will be (x-5) km/hr. We know that speed is  

$\text { Speed }=\frac{\text { Distance }}{\text { Time }}$

As the distance covered is same, then the relative speed observed by person 1 and 2 will be equal. This is because,

$\text { Relative speed observed by person 1 }=\frac{\text { Distance covered w.r.t person } 1}{\text { Time taken to cross person } 1}$

So

$\text {Distance covered by train with respect to person 1}=\text {Relative speed of train w.r.t to } P 1 \times \text { Time taken to cross } P 1 \rightarrow(1)$

And

$\text { Relative speed observed by person 2}=\frac{\text { Distance covered w.r.t person } 2}{\text { Time taken to cross person } 2}$

$\text {Distance covered by train with respect to person 2}=\text {Relative speed of train w.r.t to } P 2 \times \text { Time taken to cross } P 2 \rightarrow(2)$

As the distance covered by train with respect to both the persons are same, we can equate eqn (1) and eqn (2). We get

$\text {Relative speed of train w.r.t to } P 1 \times \text { Time taken to cross } P 1$$=\text {Relative speed of train w.r.t to } P 2 \times \text { Time taken to cross } P 2$

$(x-3) \times \frac{10}{3600}=(x-5) \times \frac{11}{3600}$

$10 x-30=11 x-55$

$-30+55=11 x-10 x$

$x=25$

Thus, the speed of the train is 25 km/hr.