Answer :

Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel was d km.

We know that,




Or, d = xt (i)

If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time

⇒ (x + 10) =

⇒ (x + 10)(t - 2) = d

⇒ xt + 10t - 2x - 20 = d

From (i), we have
⇒ d + 10t - 2x - 20 = d

⇒ - 2x + 10t = 20
⇒ x - 5t = -10 
⇒ x = 5t - 10     (ii)


Also, 
if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time
⇒ (x - 10) =

⇒ (x - 10)(t + 3) = d

⇒ xt - 10t + 3x - 30 = d

By using equation (i),
⇒ d - 10t + 3x - 30 = d
⇒ 3x - 10t = 30 (iii)


Substituting the value of x from eq (ii) into eq (iii), we get
⇒ 3(5t - 10) - 10t = 30

⇒ 15t - 30 - 10t = 30
⇒ 5t = 60
⇒ 
t = 12 hours

Putting this in eq(ii)
⇒ x = 5t - 10
= 5(12) - 10
= 50 km/h

From equation (i), we obtain

Distance to travel = d = xt

= 50 × 12

= 600 km


Hence, the distance covered by the train is 600 km.

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