What is the length of the platform in meters...............

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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


Help me to solve this.............

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by Rahul@gurome » Fri Sep 24, 2010 7:07 pm
Solution:
Speed of train is 72 kmph = 72*(5/18) or 20 m/s.
When the train is crossing the man, the train is covering its own length.
So length of train is 18*20 = 360 m.
When its crossing the platform its crossing its own length and the platform's length.
The sum of their lengths is 30*20 = 600 m.
Or length of the platform is 600 - 360 = 240 m.

The correct answer is (A).
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by saurabhmahajan » Fri Sep 24, 2010 11:41 pm
Rahul sir,

Will it be ok if i consider this way...

the train covers the man in 18 seconds and the platform in 30 seconds, i.e the train has to travel for more 12 seconds.
so at the speed of 20m/s = 20*12= 240 mts.
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by chendawg » Fri Jan 14, 2011 7:55 pm
Anyone else not understand the wording of this problem whatsoever? What's the source?

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by ankur.agrawal » Sat Jan 15, 2011 2:36 am
Rahul@gurome wrote:Solution:
Speed of train is 72 kmph = 72*(5/18) or 20 m/s.
When the train is crossing the man, the train is covering its own length.
So length of train is 18*20 = 360 m.
When its crossing the platform its crossing its own length and the platform's length.
The sum of their lengths is 30*20 = 600 m.
Or length of the platform is 600 - 360 = 240 m.

The correct answer is (A).
How do u say this. Plz explain.?

For instance say the platform is 100 metres. Man is standing at say suppose in the middle i.e 50 m.

Train crosses 50 m in order to reach the man & then it will travel its own length in order to cross that man. So how can u assume that the length of the train is 50m .

Pls clarify. It may be a very silly doubt. Thanx in advance to clear the doubt.

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by Rahul@gurome » Sat Jan 15, 2011 9:03 am
ankur.agrawal wrote:How do u say this. Plz explain.?

For instance say the platform is 100 metres. Man is standing at say suppose in the middle i.e 50 m.

Train crosses 50 m in order to reach the man & then it will travel its own length in order to cross that man. So how can u assume that the length of the train is 50m .

Pls clarify. It may be a very silly doubt. Thanx in advance to clear the doubt.
I'm not assuming the trains length is 50m.

I've said when the train crosses the man, the train is basically covering its own length, which you've said also. Then the trains length = (Speed of the train)*(Time taken to cross the man) = (72 km/h)*(18 seconds) = (20 m/s)*(18 s) = 360 m

Hope it is clear now.
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by Giotto84 » Wed Jan 19, 2011 11:51 am
Don't be confused by the man on the platform!

Its simply Distance = Rate x Time

We know the trains rate (72 kmph) and the time it takes to cross the platform (30 seconds) so multiply those numbers together (adjusting for measurements)

Distance = (72 kmph or 20 m/s) x (30 s) = 600 meters

Answer D

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by gmatjeet » Wed Jan 19, 2011 12:08 pm
P = platform length
T = train length

P+ T = Speed of train (72kmph) * Time (30s)

P+T = 72 * 30/(60*60)

Train crossing Man standing on the platform -> take train length as distance covered

T = 72*18/(60*60)

solve both eqn and P = 6/25 km = 240 m

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by ankur.agrawal » Wed Jan 19, 2011 8:58 pm
gmatjeet wrote:P = platform length
T = train length

P+ T = Speed of train (72kmph) * Time (30s)

P+T = 72 * 30/(60*60)

Train crossing Man standing on the platform -> take train length as distance covered

T = 72*18/(60*60)

solve both eqn and P = 6/25 km = 240 m
Pls Explain Why?

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by HashGMAT7XX » Thu Jan 20, 2011 2:47 am
Hi Rahul,

Can I solve it this way?

Time taken to cross the man standing on the platform = 18 secs ; This would account for the time taken for the entire length of the train a point.

Time taken to cross the platform = 30 secs

Hence if the train had negligible length it would have taken (30-18) secs = 12 secs. (This time is the time taken to cross the platform alone)

Now calculate Distance traveled by the train for 12 secs = 72000 * (12/3600) = 240 m

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by Zerks87 » Wed Jan 26, 2011 9:46 am
HashGMAT7XX wrote:Hi Rahul,

Can I solve it this way?

Time taken to cross the man standing on the platform = 18 secs ; This would account for the time taken for the entire length of the train a point.

Time taken to cross the platform = 30 secs

Hence if the train had negligible length it would have taken (30-18) secs = 12 secs. (This time is the time taken to cross the platform alone)

Now calculate Distance traveled by the train for 12 secs = 72000 * (12/3600) = 240 m
I dont think we know that the man is standing at the exact start of the platform and even then the question simply states how long the platform is. If we know how many meter/sec the train goes, which is 20m/s and we multiply by the number of seconds it takes the train to cover the platform (30sec) then we can find how many meters it is. This is because the seconds cancel out when you multiple (20m/s)(30s) and you are left with meters, which = 600, I would choose D

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by HashGMAT7XX » Wed Jan 26, 2011 8:14 pm
Zerks87,

Here while calculating the distance, you are not calculating the length of the train. The length should also be incorporated in the calculation.

For e.g The start of a train (Front portion of the train) might take X secs to cross the platform, but since we have to calculate the time taken for the "Entire Train" to cross the platform, Length of the train should also be included.

Hence the solution provided by you above is incorrect.

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by hey_deep » Mon Jan 31, 2011 11:12 pm
I did not understand the wording or significance of the man on the platform. I went and did the math using some prime factorization and was pleased to see that the numbers worked out so well(!).

72 km/h = 8*9 km/h = 2^3 * 3^2 km/h

Now convert km to m => 2^3 * 3^2 * 10^3 m

And convert h to s =>
1h * (60min/1h) = 60min * (60sec/1min) = 60 * 60 sec = or (2*3*10)(2*3*10)sec = 2^2 * 3^2 * 10^2 sec

So with the original rate,

72 km/h = (2^3 * 3^2 * 10^3) m / (2^2 * 3^2 *10^2) sec

Simplify,

2 * 10 m/s = 20 m/s


rate = distance / time


distance = rate * time
d = 20 m/s * 30s = 600m



It's important for me to show the work. Personally, making the jump from 72km/h to 20m/s is not an easy step, and that jump using prime factorization is the key to easily solving this problem.

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by jayavignesh » Thu Feb 17, 2011 2:21 am
They have asked for only the length of the platform.
They didnt ask you the length of the platform from where the man is standing.
So answer is 600 m .

If iam wrong please correct me.

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by maddy2u » Mon Mar 07, 2011 9:44 am
jayavignesh wrote:They have asked for only the length of the platform.
They didnt ask you the length of the platform from where the man is standing.
So answer is 600 m .

If iam wrong please correct me.
The entire length of the platform is asked.

answer is A .. 240 metres

Let me explain..


Speed = 72kmph = 20mps

lp - length of platform
lm - length of man.

time taken to cross man = (lt)/speed => 18 = lt/20

Therefore length of train lt = 20 *18 = 360 metres

Time taken to cross a platform of length lp => (lp +lt) / speed

Therefore , lp + lt = 20 *30 = 600 metres (You have stopped at this position)

But question has asked to find only length of the platform , lp

lp + lt = 600
lp + 360 = 600
lp = 600-360 = 240 metres.

Hence Answer is (A) . I hope you understood..


P.S: I know it is a bit difficult to understand why I am taking lp +lt at some places but lt at another. Draw a diagram and visualize the distance . You will understand.