Question

A train left the station A for station B. Having travelled 450 km which contribute 75% of the distance between A and B, the train was stopped due to very bad weather. Half an hour alter the weather cleared a bit and the engine driver having increased the speed by 15 km/hr arrived at station B on time. Find the initial speed of the train.

Answer 1

Answer:

Hey mate here is your answer.....

Solution 1

See the difference between the two cases.

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

Therefore 2d−200=3d−450⇒d=250

hope it's helpful for you...

thanks

Answer 2

Answer:

250km

Explainations:

See the difference of the two cases.

If 50 km was travelled with 4/5

of the original speed,

45 − 30

= 15

more minutes was needed.

Travelling with

4/5 of the original speed needs 5/4

of the original time.

5/4 − 1 = 1/4

i.e.,

1/4 of the original time to travel

50 km =15 minutes

⇒ Original time to travel

50 km

= 60 minutes

=1 hour

⇒ Original speed

= 50 km/hr.

After travelling

100 km,

train travels with 4/5 of the speed and hence takes 45 minutes more.

Travelling with

4/5 of the original speed needs

5/4 of the original time (i.e.,

1/4 of the original time in excess).

i.e., After travelling

100 km,

1/4 of the original time to cover remaining distance is

45 minutes. (or original time is

45 × 4

= 180 minutes

= 3 hour).

i.e., after travelling

100 km, train takes 3 hour for its journey and therefore covers a distance of

3 × 50

=150 km.

Therefore distance between A and B

= 100 + 150

= 250 km.