Question

Moving 6/7 of its usual speed a train is 10 min late. Find its usual time to cover the journey

Answer 1

Answer:

60mins.

Explanation:

Let original time taken be = X

and,

Let original speed be = V

Now,

When Speed = 6/7 V

Then, time taken = X+10

We should keep in mind that the distance covered(S) in both the cases are same.

Which gives,

S = VX

and

S = (6/7)V + (X+10)

This clearly gives us VX = (6/7)V + (X+10)

On calculating both sides , we get

=> 60/7 V = (1/7 ) XV

=> 60/7 = X/7

=> X = 60 minutes. (Ans)

Answer 2

The usual time to cover the journey is 60 minutes

It is given that the train on moving with the 6/7th of its usal speed is 10 minute late.

Let us consider the usual speed to be x km/min . The total Distance is D km.

When the train goes at speed x km/min , time taken to cover distance D is

= Distance/Speed

= D/x minutes

When the train goes at a speed 6x/7 km/min, time taken to cover distance D km is

= 7D/6x minutes

The difference in these times is 10 minutes.

So,

7D/6x  - D/x

= D/6x

This is equal to 10.

So, D/6x = 10

D = 60x

So, usual time is = D/x

= 60x/x

= 60 minutes.