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Suppose a train travels x miles in y hours and 15 minutes. Its average

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Suppose a train travels x miles in y hours and 15 minutes. Its average [#permalink]

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New post 20 May 2017, 06:06
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  15% (low)

Question Stats:

81% (00:55) correct 19% (00:45) wrong based on 63 sessions

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Suppose a train travels x miles in y hours and 15 minutes. Its average speed in miles per hour is

(A) (y + 15)/x
(B) x(y − 1/4)
(C) x/(y + 1/4)
(D) x/(y + 15)
(E) (y + 1/4)/x
[Reveal] Spoiler: OA

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Kudos [?]: 135309 [0], given: 12686

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Re: Suppose a train travels x miles in y hours and 15 minutes. Its average [#permalink]

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New post 20 May 2017, 06:15
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Bunuel wrote:
Suppose a train travels x miles in y hours and 15 minutes. Its average speed in miles per hour is

(A) (y + 15)/x
(B) x(y − 1/4)
(C) x/(y + 1/4)
(D) x/(y + 15)
(E) (y + 1/4)/x


Average speed in miles per hour = (total distance in miles)/(total time in hours)
y hours and 15 minutes = y + 1/4 hours

So, average speed in miles per hour = x/(y + 1/4)

Answer:
[Reveal] Spoiler:
C


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Re: Suppose a train travels x miles in y hours and 15 minutes. Its average [#permalink]

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New post 26 Nov 2017, 12:15
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Expert's post
Hi All,

This question can be solved in a couple of different ways; since it makes a direct reference to the AVERAGE SPEED formula, you might find it easiest to just set up that formula and solve:

Total Distance = (Average Speed)(Total Time)

Given the information in the prompt, we have....

(X miles ) = (Average Speed)(Y + 1/4 hours)
Average Speed = (X)/(Y + 1/4)

Final Answer:
[Reveal] Spoiler:
C


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Kudos [?]: 3677 [1], given: 173

Re: Suppose a train travels x miles in y hours and 15 minutes. Its average   [#permalink] 26 Nov 2017, 12:15
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