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# A Train requires 7 seconds to pass a pole while it requires 25 seconds

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Math Expert
Joined: 02 Sep 2009
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A Train requires 7 seconds to pass a pole while it requires 25 seconds  [#permalink]

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20 Mar 2016, 09:51
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Difficulty:

95% (hard)

Question Stats:

52% (02:36) correct 48% (02:30) wrong based on 148 sessions

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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

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Re: A Train requires 7 seconds to pass a pole while it requires 25 seconds  [#permalink]

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20 Mar 2016, 11:09
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In 7s the train crosses the pole and in 25 sec the train crosses one more stationary train

In 18 sec the train travels a distance of 378 mtrs

Speed = 378/18 = 21 m/s = 21 (3600/1000) = 21 * 18/5 = 75.6 kmph

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Re: A Train requires 7 seconds to pass a pole while it requires 25 seconds  [#permalink]

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11 Dec 2016, 05:58
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Bunuel wrote:
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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Re: A Train requires 7 seconds to pass a pole while it requires 25 seconds  [#permalink]

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15 Dec 2016, 06:41
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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Re: A Train requires 7 seconds to pass a pole while it requires 25 seconds  [#permalink]

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15 Dec 2016, 08:26
4
Bunuel wrote:
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

$$\frac{t}{7} = \frac{( t + 378)}{25}$$

$$25t = 7t + 7*378$$

Or, 18t = 7*378

Or, t = 7*21

So, Speed in Km/Hr = $$\frac{7*21}{7}*\frac{18}{5}$$ = $$75.6 Km/Hr$$

Hence, speed is (B) 75.6 Km/Hr
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Re: A Train requires 7 seconds to pass a pole while it requires 25 seconds  [#permalink]

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15 Dec 2016, 18:37
1
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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A Train requires 7 seconds to pass a pole while it requires 25 seconds  [#permalink]

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24 Jan 2018, 08:01
Bunuel wrote:
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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Re: A Train requires 7 seconds to pass a pole while it requires 25 seconds  [#permalink]

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15 Apr 2019, 09:29
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Re: A Train requires 7 seconds to pass a pole while it requires 25 seconds   [#permalink] 15 Apr 2019, 09:29
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