Latest self Aptitude Question SOLUTION: 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 45th of the original speed and reaches 4

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(#M40020901) SELF Numerical Ability question Keep an EYE

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 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?

vikash singh (10 years ago)

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garvit sahni  (8 years ago)

we will solve in simplest way
let its usual time =ut
1.
5/4ut=ut+45(as time is inversaly proportional to speed)
ut=180minutes
let total distance =d..hence..d-100=180*x.....(i)
2.5/4ut=ut+30...ut=120 minutes
d-150=120*x...(ii)
d-100/180=d-150/120
d=250(ans)

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avinash  (8 years ago)

Let, initial speed of the train = 5 km/h
Then, speed after the accident = 45 x 5 = 4 km/h
Time taken to cover 50 km @ 5 km/h = 10 hours
Time taken to cover 50 km @ 4 km/h = 1212 hours
Difference between times taken = 1212 - 10 = 212 hours = 150 minutes
But, actual difference = (45 - 30) minutes = 15 minutes
= 110 of 150 minutes
Therefore, Speed is 10 times of assumed speed.
Therefore, Speed before accident = 10 x 5 = 50 km/h
And, Speed after accident = 10 x 4 = 40 km/h
Distance after accident is covered @ 40 km/h instead of 50 km/h
And, time difference = 45 minutes = 34 hour
Therefore, Distance between place of accident and B = 50×4050−40×34 = 150 km
Therefore, Distance between A and B = 100 + 150 = 250 km.

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