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On a race track a maximum of 5 horses can race together at [#permalink]
10 Oct 2009, 02:56

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A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

31% (02:14) correct
69% (01:32) wrong based on 239 sessions

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?

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.

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.

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.

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...

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.

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;"

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.

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! _________________

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. _________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

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.

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.

Re: On a race track a maximum of 5 horses can race together at [#permalink]
16 Jul 2013, 13:55

It is also possible that one group could be made up of all the fastest horses well another group could have the slowest horse. In this case, the selection of #1's from the first five races is distorted.

Re: On a race track a maximum of 5 horses can race together at [#permalink]
21 Sep 2013, 01:50

Isn't it possible to make just one race? All the horses run the same track, and after 5 races we can get the results for all horses. Now we can just choose based on the timing the top 3 fastest horses..... Why is this wrong?

Re: On a race track a maximum of 5 horses can race together at [#permalink]
21 Sep 2013, 01:53

Expert's post

ronr34 wrote:

Isn't it possible to make just one race? All the horses run the same track, and after 5 races we can get the results for all horses. Now we can just choose based on the timing the top 3 fastest horses..... Why is this wrong?

On a race track a maximum of 5 horses can race together at a time... _________________

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