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Manager  Joined: 10 Jul 2009
Posts: 84
Location: Ukraine, Kyiv
A train traveling at a constant speed down a straight track  [#permalink]

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Difficulty:   15% (low)

Question Stats: 79% (01:35) correct 21% (02:11) wrong based on 780 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) $$(100/60)(5280/2)$$

Originally posted by barakhaiev on 20 Nov 2009, 04:14.
Last edited by Bunuel on 14 Jul 2012, 02:00, edited 1 time in total.
Edited the question and added the OA.
Math Expert V
Joined: 02 Sep 2009
Posts: 61396
Re: A train crosses a certain line on the track  [#permalink]

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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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Intern  Affiliations: University of Florida Alumni
Joined: 25 Oct 2009
Posts: 30
Schools: Wharton, Booth, Stanford, HBS
Re: A train crosses a certain line on the track  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 61396
Re: A train traveling at a constant speed down a straight track  [#permalink]

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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.

Check other Conversion problems to practice in Special Questions Directory.
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Intern  Joined: 02 Feb 2011
Posts: 35
Re: A train traveling at a constant speed down a straight track  [#permalink]

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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.

Non-Human User Joined: 09 Sep 2013
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Re: A train traveling at a constant speed down a straight track  [#permalink]

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_________________ Re: A train traveling at a constant speed down a straight track   [#permalink] 09 Jan 2020, 18:10
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