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28 Apr 2021, 07:45
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
9% (01:49) correct
91%
(02:26)
wrong
based on 34
sessions
History
Date
Time
Result
Not Attempted Yet
I am walking on a platform at 8 km/h. A train comes from behind and crosses me in 30 sec. If the length of the train be 300 m, its speed is:
(A) 40 km/h
(B) 44 km/h
(C) 48 km/h
(D) 52 km/h
(E) 53 km/h
(A) 40 km/h
(B) 44 km/h
(C) 48 km/h
(D) 52 km/h
(E) 53 km/h
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28 Apr 2021, 12:11
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Bunuel
Bunuel the platform is traveling at 8 km/h instead of 8 m/h I suppose?
Because the platform is moving at 8 km per hour and the train is coming from behind, the relative speed of the train to the person is 8 m/h less, hence we can set the speed as \(x - 8\) km/h.
It took \(30 sec * \frac{1 h }{ 3600 sec} = \frac{1}{120} hours\) for it to travel 0.3 km so the relative speed is also \(\frac{0.3 km * 120 }{ 1 hour} = 36 km/h = x - 8\).
Hence x = 44.
Ans: B
02 May 2021, 20:50
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Time in hours * speed in kmh = distance in kilometers
I was confounded momentarily because I had to set up the problem to solve for Kph, but the units in time and distance were in seconds and meters.
\(\frac{30}{3600}\) * (x - 8) = \(\frac{300}{1000}\)
x=44
I was confounded momentarily because I had to set up the problem to solve for Kph, but the units in time and distance were in seconds and meters.
\(\frac{30}{3600}\) * (x - 8) = \(\frac{300}{1000}\)
x=44
