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Bunuel
A Train requires 7 seconds to pass a pole while it requires 25 seconds to cross a stationary train which is 378 mtrs long. Find the speed of the train.

A. 76.2 kmph
B. 75.6 kmph
C. 75.4 kmph
D. 21 kmph
E. 20 kmph


let speed of train be S
then length of train=7S

as it requires to travel 25 sec for cross 378 mts another train so
(7s+378)/s=25
S=21m/s or 75.6kmph

Ans B
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The train crosses a pole in 7 seconds. (pole- no length)
it crosses a stationery train in 25 s.
Thus total time = 25-7 = 18 seconds.
Thus speed= 378/18 = 21m/s
21m/s*3600/1000
75.6 km/h
Therefore B
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7 seconds to pass a pole:
Time taken by the train to cover a distance that equals to its own length = 7 seconds

25 seconds to cross a train 378 mtrs long:
Time taken by the train to cover a distance equal to the sum of its own length and 378 mtrs = 25 seconds

Therefore,
Time taken by the train to cover only 378 mtrs = 25 - 7 seconds = 18 seconds

Speed = \(\frac{378}{18}\) mtrs/second
=> \(\frac{378}{18} * \frac{18}{5}\) km/hour
=> 75.6 km/hour

B

PS: Remeber this for quick conversions:
To convert speed in m/sec to km/h, multiply by 18/5
To convert speed in km/h to m/sec, multiply by 5/18
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Bunuel
A Train requires 7 seconds to pass a pole while it requires 25 seconds to cross a stationary train which is 378 mtrs long. Find the speed of the train.

A. 76.2 kmph
B. 75.6 kmph
C. 75.4 kmph
D. 21 kmph
E. 20 kmph

It can be solved under 1 minute - Let \(l\) be the length of the train., \(v\) be the speed of train.

\(l = 7v\) and \(l + 378 = 25v\).... we get, \(v = 21m/s\)... Converting to \(km/hr\) ---> \(75.6 km/hr\).
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i missed the unit ughhh
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