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Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]

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06 Aug 2010, 00:42

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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

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

Given that S+R=B+2, where S, R, and B are times in which Stephanie, Regine, and Brian completed the race.

Min time one could complete the race is 20/8=2.5 hours. Let's see if Brian could have won the race: if he ran at the fastest rate, he would complete the race in 2.5 hours, so combined time needed for Stephanie and Regine would be S+R=B+2=4.5 hours, which is not possible as sum of two must be more than or equal the twice the least time: 2*2.5=5. So Brian could not have won the race.

There is no reason to distinguish Stephanie and Regine so if one could have won the race, another also could. So both could have won the race.

which is not possible as sum of two must be more than or equal the twice the least time: 2*2.5=5. So Brian could not have won the race. - I still don't get this line , please explain

The least time one could complete the race is 20/8=2.5 hours, hence \(S+R\geq{5}\). Let's see if Brian could have won the race: best chances to win he would have if he ran at the fastest rate, so he would complete the race in 2.5 hours, so combined time needed for Stephanie and Regine would be S+R=B+2=4.5 hours, but we know that \(S+B\geq{5}\), so even if Brian ran at his fastest rate to win the race, given equation S+R=B+2 can not hold true. Hence Brian could not have won the race.
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A person with a speed of 8 miles per hour will finish the 20 mile race in: \(t=20/8=5/2=2.5hrs\)

There is no person faster than that so the time of S, R and B must be at least 2.5 hours.

Let S, R and B be the time it took each person to finish the race. S + R = B + 2

So the minimum possible value for S+R = 2.5 + 2.5 = 5 hrs Therefore, the minimum possible value for B = 5-2 = 3 hrs.

Let us test: B=3, S=2.5,R=2.5 ==> S and R could win. B=4, S=3,R=3 ==> B still could not win. Even if we force either S or R to exceed B, the balance will still make B lose.

Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]

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29 Dec 2012, 10:15

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I did it this way-

Let ts=time of steph tr=time of reg tb=time of brian

ts+tr=tb+2

From the problem its clear that Brian takes more time. Since brian takes more time he could not have won the race. (speed and time are inverse proportional) We have no info about the other 2.So, anyone could have won the race.
_________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

which is not possible as sum of two must be more than or equal the twice the least time: 2*2.5=5. So Brian could not have won the race. - I still don't get this line , please explain
_________________

Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]

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09 Dec 2012, 19:25

Can someone clarify this.

1. Stephanie, Regine, and Brian ran a 20 mile race - [D = 20miles] 2. Stephanie and Regine's combined times exceeded Brian's time by exactly 2 hours - [S + R = B + 2] 3. nobody ran faster than 8 miles per hour - [Max Speed = 20/8 = 2.5m/h]

What if A. S = 2, R = 1.8 then B = 1.8 so S would win the race. B. S = 1.8, R = 2 then B = 1.8 so B would win the race. C. S = 2.1, R = 2.1 then B = 2.2 so B would win the race.

1. Stephanie, Regine, and Brian ran a 20 mile race - [D = 20miles] 2. Stephanie and Regine's combined times exceeded Brian's time by exactly 2 hours - [S + R = B + 2] 3. nobody ran faster than 8 miles per hour - [Max Speed = 20/8 = 2.5m/h]

What if A. S = 2, R = 1.8 then B = 1.8 so S would win the race. B. S = 1.8, R = 2 then B = 1.8 so B would win the race. C. S = 2.1, R = 2.1 then B = 2.2 so B would win the race.

Why not option E? Where am i going wrong?

We are told that nobody ran faster than 8 miles per hour, thus the maximum speed is 8 miles per hour not 2.5 miles per hour. Also, it would mean that minimum time one could complete the race is 20/8=2.5 hours, thus your examples are not valid.
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Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]

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11 Dec 2012, 13:59

Isn't this a much easier way to solve this? I may be wrong:

There are 3 times, Ts Tr and (Ts+Tr-2). Now, we're told that the distance is 20 miles and the max they could have run is 8 m/h. Let's say Brian beat Stephanie, and we'll start with Brian because Brian has the funkiest time formula. You would have Ts+Tr-2<Ts, picking Stephanie for no reason, we can do this for Regine as well. You will get Tr<2hours. That means Regine is going 20miles/<2hours, which is >8m/h, which is not allowed. So he cannot beat Regine, and if you just look at the formula you can see he can't beat Stephanie for the same reason. So Brian cannot win. A quick comparison of 20/Ts < 20/Tr will show it is perfectly possible for S to beat R and vice versa.

Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]

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08 Aug 2013, 12:17

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

Time (s) + time (r) = time (b) +2

d=20

If no one ran faster than 8 miles/hour than no one completed the race in under 2.5 hours (time = distance/rate ===> time = 20/8 ===> time = 2.5 hours)

If S+R = B+2, Brian's time will always be greater than S or R. Because no one finishes in under 2.5 hours we can rule out maximum speeds. For example, Brian can not finish in 2.5 hours because S and R would have to have a combined time of 4.5 hours which we know is not possible because 4.5 combined would mean both of them ran it in under 2.5 hours. If Brian finished in 3 hours S and R's time could have been 2.5 each meaning their combined time is exactly two hours greater but their times separately are less than Brian's. The only way the equation can hold true is if Brian's time is equal to or greater than three in which case either S or R (or both) always run in less time.

Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]

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07 Sep 2013, 11:40

I almost made a stupid mistake on this one, but quickly realized that it's obviously D.

Brian would need to complete it in 2.5 hours, meaning the other two would need to complete it in 4.5 combined, which would mean that they both would've been faster than the 8m/h limit given.

Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]

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23 Jun 2014, 21:19

We know t1+t2=t3 +2 -- (1)

where t1, t2 and t3 are the time taken by Stephanie, Regine and Brian to complete the race. Since the minimum value for the time taken is 2.5, t1 or t2 has to be lesser than t3 for equation 1 to hold true. So Brian could not have won the race. But we can find values for t1 and t2 such that stephanie and Regine could have won the race satisfying equation (1).
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Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]

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12 Jul 2015, 03:28

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Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]

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27 Mar 2016, 19:24

tricky one...definitely D. i made the mistake to assume that if B won, then the time combined for the girls is 4.5hours..but no one finished faster than in 2.5 hours..so B for sure did not win the race.

gmatclubot

Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie
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27 Mar 2016, 19:24

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