Stephanie, Regine, and Brian

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Stephanie, Regine, and Brian [#permalink]

15 Mar 2011, 22:47

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Stephanie, Regine, and Brian ran a 20 mile race. Stephanie and Regine's combined times exceeded Brian's time by exactly 2 hours. If nobody ran faster than 8 miles per hour, who could have won the race?

I. Stephanie II. Regine III. Brian
(A) I only (B) II only (C) III only (D) I or II only (E) I, II, or III  

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Re: Stephanie, Regine, and Brian [#permalink]

15 Mar 2011, 23:29

Onell wrote:

We are given that S+R = B+2 (where S, R and B represent Stephanie, Regine, and Brian respectively)

Also, minimum value of S,R and B is 20/8 = 2.5

Clearly if S and R are at minimum value of 2.5, B has to be a minimum of 3 and hence has to be slower than S and/or R. For all value of S and R greater than 2.5, B will be slower than at least one of S or R and hence B cant win the race. Between S and R there is nothing to distinguish as they are interchangeable in the given equation, so any of them can win the race.

Answer D

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Re: Stephanie, Regine, and Brian [#permalink]

15 Mar 2011, 23:39

Minimum value of time for each the three = 20/8= 2.5 hrs
S+R=B+2
2.5+.25=B+2
B (Brian)=3

S and R will have greater value than B. So B could not win the race. Ans. D
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Joined: 05 Jan 2011

Posts: 183

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Kudos [?]: 12 [0], given: 8

Re: Stephanie, Regine, and Brian [#permalink]

16 Mar 2011, 00:00

Baten80 wrote:

Minimum value of time for each the three = 20/8= 2.5 hrs
S+R=B+2
2.5+.25=B+2
B (Brian)=3

S and R will have greater value than B. So B could not win the race. Ans. D

Thanks guys for your help

Re: Stephanie, Regine, and Brian [#permalink] 16 Mar 2011, 00:00

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Stephanie, Regine, and Brian

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Stephanie, Regine, and Brian

Stephanie, Regine, and Brian

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