Question

A train left station a for station b at a certain speed. after travelling for 100 km, the train meets with an accident and could travel at 4 5 45th of the original speed and reaches 45 minutes late at station b. had the accident taken place 50 km further on, it would have reached 30 minutes late at station b. what is the distance between station a and b?

Answer 1

Ans: 250250 km

Solution 1

See the difference between the two cases.
If 5050 km was traveled with 4545 of the original speed, 45−30=1545−30=15 more minutes was needed.  

Traveling with 4545 of the original speed needs 5454 of the original time. 
54−1=1454−1=14

i.e., 1414 of the original time to travel 5050 km =15=15 minutes
⇒ Original time to travel 5050 km =60=60 minutes =1=1 hour
⇒ Original speed =50=50 km/hr.

After traveling 100100 km, train travels with 4545 of the speed and hence takes 4545 minutes more.

Traveling with 4545 of the original speed needs 5454 of the original time (i.e., 1414 of the original time in excess).

i.e., After traveling 100100 km, 1414 of the original time to cover remaining distance is 4545minutes. (or original time is 45×4=18045×4=180 minutes =3=3 hour).

i.e., after travelling 100100 km, train takes 33 hour for its journey and therefore covers a distance of 3×50=1503×50=150 km.

Therefore distance between A and B =100+150=250=100+150=250 km.

Solution 2

Let original speed =s=s km/hr
total distance =d=d km

From first case,
100s+d−100(4s5)=ds+4560⇒100s+5d−5004s=ds+34⇒400+5d−500=4d+3s⇒d−100=3s⋯(1)100s+d−100(4s5)=ds+4560⇒100s+5d−5004s=ds+34⇒400+5d−500=4d+3s⇒d−100=3s⋯(1)

150s+d−150(4s5)=ds+3060⇒150s+5d−7504s=ds+12⇒600+5d−750=4d+2s⇒d−150=2s  ⋯(2)150s+d−150(4s5)=ds+3060⇒150s+5d−7504s=ds+12⇒600+5d−750=4d+2s⇒d−150=2s  ⋯(2)

(1)⟹2d−200=6s(2)⟹3d−450=6s(1)⟹2d−200=6s(2)⟹3d−450=6s

Therefore 
2d−200=3d−450⇒d=250

Answer 2

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