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A train traveling at 72 kmph crosses a platform in 30 seconds and a ma

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A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 14 Mar 2016, 07:18
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A
B
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D
E

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  85% (hard)

Question Stats:

57% (02:09) correct 43% (02:07) wrong based on 304 sessions

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Re: A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 18 Mar 2016, 02:14
Speed of the train in metres/sec = 72000/3600 = 20

Distance travelled by train to cross the platform = 30 * 20 = 600 = Length of train + Length of platform

Distance travelled by train to cross the man = 18 * 20 = 360 = Length of train

Length of platform = 600 - 360 = 240

Answer: A
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A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 18 Mar 2016, 05:43
What if the the position man is standing is not a point but rather say .004 meters(may be a real fat one)
in that case it would
Distance travelled by train to cross the man = 18 * 20 = 360 = Distance occupied by man in platform + Length of train
The answer has to be E
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Re: A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 18 Mar 2016, 06:24
raarun wrote:
What if the the position man is standing is not a point but rather say .004 meters(may be a real fat one)
in that case it would
Distance travelled by train to cross the man = 18 * 20 = 360 = Distance occupied by man in platform + Length of train
The answer has to be E


I do not think that we have to go into such minute detail and include the length of the person's waist as its negligible when compared to the overall length.
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Re: A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 18 Mar 2016, 06:55
raarun wrote:
What if the the position man is standing is not a point but rather say .004 meters(may be a real fat one)
in that case it would
Distance travelled by train to cross the man = 18 * 20 = 360 = Distance occupied by man in platform + Length of train
The answer has to be E


hi,
Vyshak is correct..
Don't complicate the Q, where it is not required..
If you want to have E as the answer, you don't require a fat person of .004m, even a thin guy of .001m will also effect the distance by that small amount..
But that is not required, we take the person as a point.
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Re: A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 22 Mar 2016, 15:16
Why do we have to subtract 18*20 from 30*20?

I don´t really get the connection between the two numbers.
Maybe someone can explain how the fact that the train needs 18 seconds to pass the person is related to the length of the platform?
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A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 23 Jul 2016, 15:17
first find length of train:
18 s=1/200 h
72 kh*1/200 h=9/25 k=length of train
next find length of platform (p):
30 s=1/120 h
72 kh*1/120 h=9/25 k+p
p=6/25 k=240 m
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A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 23 Jul 2016, 15:37
\(Well\; this\; was\; fun.\)

\(We\; have\; to\; take\; into\; consideration\; the\; lenght\; of\; the\; train\; in\; this\; problem,\;\)
\(considering\; the\; time\; it\; takes\; the\; head\; of\; the\; train\;\)
\(to\; go\; from\; the\; start\; of\; the\; platform\; to\; the\; end\; of\; it.\;\)

\(\mbox{S}o\; if\; it\; takes\; take\; train\; 18\; seconds\; to\; pass\; a\; standing\; man,\;\)
\(it\; means\; that\; the\; lenght\; of\; the\; train\; was\; 72*\frac{1000}{3600}\; =\; 20\; *\; 18\; ->\; 360\; metres.\)

\(Found\; the\; lenght\; we\; simply\; need\; to\; do\; 20*30\; ->\; 600\;\)
\(which\; is\; the\; time\; the\; train\; takes\; to\; completely\; leave\; the\; platform\;\)

\(and\; subtract\; the\; lenght\; of\; the\; train:\; 600-360\; =\; 240\)

\(Answer\; A.\)
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Re: A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 15 Dec 2016, 19:01
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In speed-distance questions, always remember-
Pole, a man (stationary) etc are considered as points, when a train crosses these, it means that the train travelled a distance equal to its own length.
while,
Other trains, railway platform etc are considered for their own lengths, it means that if a moving train crosses these, the train travelled a distance equal to its own length + length of the other train/platform

Coming to this question:
Speed of the train: 72 km/h = \(72*\frac{5}{18}\) = 20 meters/sec ----(converting into mt/sec as everything else is in meters and seconds)

Train crosses a man (a stationary point) in 18 secs: Train covers its own length in 18 secs:
Length of train = \(20 * 18 = 360\) meters

Train crosses a platform in 30 secs: Train covers a distance equal to its own length+length of the platform in 30 secs:
i.e. length of the train + Length of the platform = 20 * 30 ----(Distance = Speed * Time)
\(360+x = 600\)
\(x = 240\)

Length of the platform = 240 meters

A
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Re: A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 21 May 2018, 16:40
Bunuel wrote:
A train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 18 seconds. What is the length of the platform in meters?

A. 240 meters
B. 360 meters
C. 420 meters
D. 600 meters
E. Cannot be determined


We can let p = the length of the platform, in meters, and t = the length of the train, in meters.

When we say the train crosses a platform in 30 seconds, it really means it takes 30 seconds for the nose of the train to enter one end of the platform and the rear of the train to exit the other end of the platform. Thus, in 30 seconds, not only does the train’s nose travel the entire length of the platform but also its entire body length does. Thus, we have (notice that 72 kmph = 72000 meters per hour, and 30 sec = 30/3600 = 1/120 hr):

72000 x 1/120 = p + t

600 = p + t

p = 600 - t

We are also given that the train crosses a man (whose body width is negligible) standing on the platform in 18 seconds. So when the train crosses him, it only travels its body length in 18 seconds. Thus we have (notice that 18 sec = 18/3600 = 1/200 hr):

72000 x 1/200 = t

360 = t

Therefore, the platform has a length of 600 - 360 = 240 meters.

Answer: A
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Re: A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 21 May 2018, 18:28
A train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 18 seconds. What is the length of the platform in meters?

A. 240 meters
B. 360 meters
C. 420 meters
D. 600 meters
E. Cannot be determined

Step 1:
To cross the platform = 72 km/h = 72 km/60 minutes = 1.2 km/60 seconds = 0.6 km/30 seconds = 600 meters

Step 2:
To cross the man = 72 km/h = 72 km/60 minutes = 1200 meter/60 seconds , 18 seconds =360 Meters

Step 3:
Length of platform = 600 - 360 = 240

Ans: A
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Re: A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 27 Nov 2018, 19:54
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Bunuel wrote:
A train traveling at 72 kmph crosses a platform in 30 seconds and a man standing on the platform in 18 seconds. What is the length of the platform in meters?

A. 240 meters
B. 360 meters
C. 420 meters
D. 600 meters
E. Cannot be determined

Let's use some graphics to help see what's happening here.

We'll say that the train STARTS crossing the platform when the front nose of the train meets the beginning edge of the platform . . .
Image

. . . and the train FINISHES crossing the platform when the back end of the train cross the end of the platform . . .
Image

Let p = length of platform (in meters)
Let t = length of train (in meters)

We'll also add a blue dot at the front of train to help determine the distance the train travels.
Below, we have the train when it first meets the platform
Image

Below, we have the train when it FINISHES crossing the platform
Image

During this period, the total DISTANCE the train (and the blue dot) travels = p + t meters

IMPORTANT: At this point, we better convert all units to meters and seconds.

GIVEN: A train traveling at 72 kmph crosses a platform in 30 seconds
1 kilometer = 1000 meters
So, 72 kilometers per hour = 72000 meters per hour

There are 3600 seconds in an hour.
So, 72000 meters per hour = 72000 meters per 3600 seconds
= 20 meters per second [after we divide both parts by 3600]
So, the train's speed = 20 meters per second

Distance = (speed)(time)
So, if the train travels for 30 seconds at a speed of 20 meters per second, the distance traveled = (20)(30) = 600 meters

We already know that the train travels = p + t meters
So, we can write: p + t = 600

GIVEN: A train traveling at 72 kmph crosses a man standing on the platform in 18 seconds
NOTE: I'd prefer that the question asked "Which of the following BEST APPROXIMATES is the length of the platform in meters?", since the width of the man will affect the calculations.
So, let's just say the man does not have a width (width = 0)

So, here's the part where the train MEETS the man . . .
Image

And here's the part where the train finishes PASSING the man . . .
Image

In this case, the train travels a total distance of t meters.

Distance = (speed)(time)
So, if the train travels for 18 seconds at a speed of 20 meters per second, the distance traveled = (20)(18) = 360 meters
We already know that the DISTANCE the train travels = t meters
So, we can write: t = 360

At this point, we know that:
p + t = 600
t = 360

Subtract the bottom equation from the top equation to get: p = 240

Answer: A

Cheers,
Brent
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Re: A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 28 Nov 2018, 03:03
Hi Brent GMATPrepNow

This explanation is elegant and magnificent. Thanks for you effort :thumbup:
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Re: A train traveling at 72 kmph crosses a platform in 30 seconds and a ma  [#permalink]

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New post 28 Nov 2018, 09:17
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Mo2men wrote:
Hi Brent GMATPrepNow

This explanation is elegant and magnificent. Thanks for you effort :thumbup:


Thanks for taking the time to say that!

Cheers,
Brent
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Re: A train traveling at 72 kmph crosses a platform in 30 seconds and a ma   [#permalink] 28 Nov 2018, 09:17
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