Question

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train taking speed as y and distance as x.

Answer 1

Let the speed of the train be xkm/h and the time taken by train to travel the given distance be t hours and the distance to travel be dkm. We know that,

$= > \: speed \: = \frac{distance \: }{time}$

$= x = \frac{d}{t}$

$∴d \: = xt......(1)$

Case 1

⇒(x+10)×(t−2)=d

⇒xt+10t−2x−20=d

⇒d+10t−2x−20=d

⇒−2x+10t=20...... (ii)

Case 2

⇒(x−10)×(t+3)=d

⇒xt−10t+3x−30=d

⇒d−10t+3x−30=d

⇒3x−10t=30......... (iii)

Adding equations (ii) and (iii), we gets 

⇒x=50

Substitute the value of x in (ii) we gets

⇒(−2)×(50)+10t=20

⇒−100+10t=20

⇒10t=120

⇒t=12 hours

Substitue the value of t  and x in  equation (i), we gets

Distance to travel =d=xt

⇒d=12×50=600Km

 

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