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Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]
 06 Aug 2010, 01:42
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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
Last edited by Bunuel on 02 Dec 2012, 04:19, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Speed and Race Problem [#permalink]
06 Aug 2010, 02:25
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jakolik wrote: 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. Answer: D.
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Re: Speed and Race Problem [#permalink]
08 Aug 2010, 01:15
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ichha148 wrote: 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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Re: Speed and Race Problem [#permalink]
02 Dec 2012, 01:24
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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.
Answer: I and II only
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Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]
29 Dec 2012, 11:15
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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.
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Re: Speed and Race Problem [#permalink]
07 Aug 2010, 21:14
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
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Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]
09 Dec 2012, 20:25
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.
Why not option E? Where am i going wrong?
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Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]
10 Dec 2012, 01:29
maddyboiler wrote: 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.
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]
10 Dec 2012, 06:40
How did i skip that. Thanks!!
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Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie [#permalink]
11 Dec 2012, 14:59
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.
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Re: Stephanie, Regine, and Brian ran a 20 mile race. Stephanie
[#permalink]
11 Dec 2012, 14:59
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