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# A train traveling at a constant speed down a straight track

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A train traveling at a constant speed down a straight track [#permalink]

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20 Nov 2009, 05:14
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Question Stats:

66% (02:31) correct 34% (01:40) wrong based on 462 sessions

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A train traveling at a constant speed down a straight track crosses a certain line on the track. If the rear wheels of the train cross the line 2 seconds after the front wheels, and the centers of the rear and front wheels are 100 feet apart, which of the following expresses the speed of the train in miles per hour?
1 mile = 5280 feet

(A) $$(100/5280)(60^2/2)$$
(B) $$(100/5280)(60/2)$$
(C) $$(100/5280)(2/60^2)$$
(D) $$(100/60^2)(5280/2)$$
(E) $$(100/60)(5280/2)$$
[Reveal] Spoiler: OA

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Last edited by Bunuel on 14 Jul 2012, 03:00, edited 1 time in total.
Edited the question and added the OA.
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Re: A train crosses a certain line on the track [#permalink]

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20 Nov 2009, 05:32
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barakhaiev wrote:
A train traveling at a constant speed down a straight track crosses a certain line on the track. If the rear wheels of the train cross the line 2 seconds after the front wheels, and the centers of the rear and front wheels are 100 feet apart, which of the following expresses the speed of the train in miles per hour?
1 mile = 5280 feet

(A) $$(100/5280)(60^2/2)$$
(B) $$(100/5280)(60/2)$$
(C) $$(100/5280)(2/60^2)$$
(D) $$(100/60^2)(5280/2)$$
(E) $$(100/60)(5280/2)$$

Speed of the train would be 100/2 feet per second, as it covers the distance of 100 feet in 2 seconds.

We should transform this to miles per hour:

100 feet=100/5280 miles;
2 seconds=2/60^2 hours;

Hence we would have (100/5280)/(2/60^2)=(100/5280)*(60^2/2) miles per hour.

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Re: A train crosses a certain line on the track [#permalink]

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20 Nov 2009, 12:26
agree. 50 feet per second * 60 (convert to minutes) * 60 (convert to hours) this gives you feet per hour. Divide by 5260 = Miles per hour
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Re: A train traveling at a constant speed down a straight track [#permalink]

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21 Aug 2014, 15:45
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Re: A train traveling at a constant speed down a straight track [#permalink]

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12 Sep 2015, 02:38
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A train traveling at a constant speed down a straight track [#permalink]

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13 Sep 2015, 04:08
Bunuel wrote:
barakhaiev wrote:
A train traveling at a constant speed down a straight track crosses a certain line on the track. If the rear wheels of the train cross the line 2 seconds after the front wheels, and the centers of the rear and front wheels are 100 feet apart, which of the following expresses the speed of the train in miles per hour?
1 mile = 5280 feet

(A) $$(100/5280)(60^2/2)$$
(B) $$(100/5280)(60/2)$$
(C) $$(100/5280)(2/60^2)$$
(D) $$(100/60^2)(5280/2)$$
(E) $$(100/60)(5280/2)$$

Speed of the train would be 100/2 feet per second, as it covers the distance of 100 feet in 2 seconds.

We should transform this to miles per hour:

100 feet=100/5280 miles;
2 seconds=2/60^2 hours;

Hence we would have (100/5280)/(2/60^2)=(100/5280)*(60^2/2) miles per hour.

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Re: A train traveling at a constant speed down a straight track [#permalink]

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26 Jan 2016, 05:42
Hi

I am still unable to understand how the distance covered is 100/2.

The front wheel crosses say in 4 mins ( imagine half of the wheel crossed). Now it has to cover some distance till rear wheel crosses.
We can say that rear wheel crosses as soon as some part of wheel crosses line or half rotation ( distance till center of wheel ) crosses the line,

How are we concluding that 100 is the distance being traveled by wheels.
Somehow unable to imagine the problem.

Re: A train traveling at a constant speed down a straight track   [#permalink] 26 Jan 2016, 05:42
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