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On a race track a maximum of 5 horses can race together at [#permalink]
 10 Oct 2009, 03:56
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On a race track a maximum of 5 horses can race together at a time. There are a total of 25 horses. There is no way of timing the races. What is the minimum number of races we need to conduct to get the top 3 fastest horses? A. 5 B. 7 C. 8 D. 10 E. 11
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On every race we get the top 3 out of 5.
First 5 Races = Take 3 from each race = 15
Next 3 Races = Take 3 from each race = 9
At this point I figured that we only need 3 horses and we have 2 races. So take the top 2.
Next 2 Races = Take 2 from each race = 4
Last race we get the top 3.
Total races = 5+3+2+1 = 11
Answer = E
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Its not E. Bit more complicated then that 
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I came down to 7. I mean that I can do the task in 7 races. First 5 races: all horses by five. We'll have the five winners. Race 6: the winners of previous five races. We'll have the 3 winners. Now it's obvious that #1 here is the fastest one (gold medal). For the silver and bronze we'll have 5 pretenders: 1. #2 from the last sixth race, 2. #3 from the last sixth race, 3. the second one from the race with the Gold medal winner from the first five races, 4. the third one from the race with the Gold medal winner from the first five races, 5. the second one from the race with the one which took the silver in the sixth race Race 7: these five horse: first and second in this one will have the silver and bronze among all 25. Answer B (7). Good Q. +1
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7 is the correct answer. Good solution Buneul. 
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jax91 wrote: On a race track a maximum of 5 horses can race together at a time. There are a total of 25 horses. There is no way of timing the races. What is the minimum number of races we need to conduct to get the top 3 fastest horses?
A)5
B)7
C)8
D)10
E)11 Nice work by both "yangsta8" and "Bunuel". Both solutions are equally logical... The OA could be controversial as the question is not clear about the method how 1/2/3 are selected... yangsta8 wrote: On every race we get the top 3 out of 5.
First 5 Races = Take 3 from each race = 15
Next 3 Races = Take 3 from each race = 9
At this point I figured that we only need 3 horses and we have 2 races. So take the top 2.
Next 2 Races = Take 2 from each race = 4
Last race we get the top 3.
Total races = 5+3+2+1 = 11
Answer = E Bunuel wrote: I came down to 7. I mean that I can do the task in 7 races.
First 5 races: all horses by five. We'll have the five winners.
Race 6: the winners of previous five races. We'll have the 3 winners.
Now it's obvious that #1 here is the fastest one (gold medal).
For the silver and bronze we'll have 5 pretenders:
1. #2 from the last sixth race,
2. #3 from the last sixth race,
3. the second one from the race with the Gold medal winner from the first five races,
4. the third one from the race with the Gold medal winner from the first five races,
5. the second one from the race with the one which took the silver in the sixth race
Race 7: these five horse: first and second in this one will have the silver and bronze among all 25.
Answer B (7).
Good Q. +1
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Why 7 races? I think we can do it in just 6 races, as shown on the picture. The 6th race will determine the best three winners (Gold, Silver, and Bronze) of the previous 5 races... 
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Financier wrote: Why 7 races? I think we can do it in just 6 races, as shown on the picture. The 6th race will determine the best three of winners of the previous 5 races... We want to get the top 3 fastest horses. The trick here is that though the horse #1 from 6th race will be the fastest one, so gold medal owner but the horse #2 and #3 from this race may not be the second and the third fastest horses out of 25. For the silver and bronze we would have 5 pretenders: 1. #2 from the 6th race; 2. #3 from the 6th race; 3. the second horse from the first round race with the Gold medal winner; 4. the third horse from the first round race with the Gold medal winner; 5. the second horse from the first round race with horse #2 in the 6th. So total 7 races are needed.
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Nice question.. Thanks for posting
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Bunuel wrote:
For the silver and bronze we would have 5 pretenders:
1. #2 from the 6th race;
2. #3 from the 6th race;
3. the second horse from the first round race with the Gold medal winner;
4. the third horse from the first round race with the Gold medal winner;
5. the second horse from the first round race with horse #2 in the 6th.
So total 7 races are needed.
Bunuel, It is uncertain to me that how we can be sure to pick horses for 7th race.. Isn't a possibility that second horse from Gold Medal winner group can be slower than the second horse in Bronze Medal winner group? Lets say: The Gold medal horse took 2 mins, second horse in the Gold medal group took 5 mins to cover the race. While Bronze medal horse took 3 mins and second horse from Bronze medal group took 4 mins. In that case, we need to change this: "second horse from the first round race with the Gold medal winner;" with "second horse from the first round race with the Bronze medal winner;"
Please, explain.
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samark wrote: Bunuel wrote:
For the silver and bronze we would have 5 pretenders:
1. #2 from the 6th race;
2. #3 from the 6th race;
3. the second horse from the first round race with the Gold medal winner;
4. the third horse from the first round race with the Gold medal winner;
5. the second horse from the first round race with horse #2 in the 6th.
So total 7 races are needed.
Bunuel, It is uncertain to me that how we can be sure to pick horses for 7th race.. Isn't a possibility that second horse from Gold Medal winner group can be slower than the second horse in Bronze Medal winner group? Lets say: The Gold medal horse took 2 mins, second horse in the Gold medal group took 5 mins to cover the race. While Bronze medal horse took 3 mins and second horse from Bronze medal group took 4 mins. In that case, we need to change this: "second horse from the first round race with the Gold medal winner;" with "second horse from the first round race with the Bronze medal winner;"
Please, explain. The trick here is that we should choose the horses for the last race so that each horse to be pretender for either the Silver or the Bronze medal. Now, second horse from the first round race with horse #3 in the 6th race could be faster than the second horse from the first round race with the Gold medal winner, but if it is so could this horse get any medal? It's obviously slower than horse #3 and also slower that horse #2 (from the last race) so there is no sense to include this horse in the last race. Similar logic works for other horses. Hope it's clear.
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Bunuel wrote: samark wrote: Bunuel wrote:
For the silver and bronze we would have 5 pretenders:
1. #2 from the 6th race;
2. #3 from the 6th race;
3. the second horse from the first round race with the Gold medal winner;
4. the third horse from the first round race with the Gold medal winner;
5. the second horse from the first round race with horse #2 in the 6th.
So total 7 races are needed.
Bunuel, It is uncertain to me that how we can be sure to pick horses for 7th race.. Isn't a possibility that second horse from Gold Medal winner group can be slower than the second horse in Bronze Medal winner group? Lets say: The Gold medal horse took 2 mins, second horse in the Gold medal group took 5 mins to cover the race. While Bronze medal horse took 3 mins and second horse from Bronze medal group took 4 mins. In that case, we need to change this: "second horse from the first round race with the Gold medal winner;" with "second horse from the first round race with the Bronze medal winner;"
Please, explain. The trick here is that we should choose the horses for the last race so that each horse to be pretender for either the Silver or the Bronze medal. Now, second horse from the first round race with horse #3 in the 6th race could be faster than the second horse from the first round race with the Gold medal winner, but if it is so could this horse get any medal? It's obviously slower than horse #3 and also slower that horse #2 (from the last race) so there is no sense to include this horse in the last race. Similar logic works for other horses. Hope it's clear. Yes, after much head banging and heated discussion with my fellow study partner. Thanks!
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if the runner up in the first race is the fastest among all except the winner of the first race.How could you leave that horse?
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nice question
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Bunuel wrote: I came down to 7. I mean that I can do the task in 7 races.
First 5 races: all horses by five. We'll have the five winners.
Race 6: the winners of previous five races. We'll have the 3 winners.
Now it's obvious that #1 here is the fastest one (gold medal).
For the silver and bronze we'll have 5 pretenders:
1. #2 from the last sixth race,
2. #3 from the last sixth race,
3. the second one from the race with the Gold medal winner from the first five races,
4. the third one from the race with the Gold medal winner from the first five races,
5. the second one from the race with the one which took the silver in the sixth race
Race 7: these five horse: first and second in this one will have the silver and bronze among all 25.
Answer B (7).
Good Q. +1 Hi Bunuel, I have a doubt regarding the solution. The question is asking for Fastest 3 horses & not the Top 3 Winners. Let me put it this way. Suppose 2 separate races were conducted & timing (in minutes) of these 10 horses are as follows: Race 1 - 5, 6, 7, 8 , 9, 10 minutes Race 1 - 8 , 9, 10, 11, 12 minutes Now you are picking 5 from race-1 & 8 from race 2 because these 2 horses are winners. But these 2 horses are not the fastest horses. The 2 fastest horses from this scenario would be 5 & 6 min. As per me had the question asked Top 3 horses then your logic would be completely right but instead the question is asking for the fastest 5. Kindly throw some light this logic. Your inputs are always valuable.
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