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Bunuel
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Bunuel
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Suppose in Race1 we got:
Horse No1 finished in 1min
#2 in 2min
#3 in 3min
#4 in 4min
#5 in 5min

In Race2 we got:
Horse No1 finished in 6min
#2 in 7min
#3 in 8min
#4 in 9min
#5 in 10min

In Race3 we got:
Horse No1 finished in 11min
#2 in 12min
#3 in 13min
#4 in 14min
#5 in 15min

...and so on. So in the 6th race the candidates finished in:
1min
6min
11min
16min
21min

And indeed the gold medal goes to the 1st one with 1min timing. The fastest after this one is the 6min, and the one after that the 11 min and so on.
However, we can't assume that the 6min beats any of the horses in Race1. So if anything, in Race7 the candidates should be the 2nd, 3rd, 4th, and 5th from the Race1 and the 2nd from Race6. We have no timer, so this may or may not be true, as well as the previous valid solution. What's wrong with this logic?

You assume that horses run as fast as they can in each race, which is not true. In each race the winner runs just as fast to win. For example, in the second race the winner runs in 6 minutes but this does not mean that in some other race, with faster competitors, it cannot run in 1 minute. Check this post too: on-a-race-track-a-maximum-of-5-horses-can-race-together-at-85080-20.html#p1269589

Hope it's clear.

So each horse ran as fast as it could, therefore assuming the minimum time difference between them.
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This question does expand 'the box', but I still seem to think its not definitive enough. I get that the horse that came second in the race with the 3rd place winning horse(of 6th race) need not be considered, but I still don't see how the horse that came third in the race with the 2nd place winning horse (of 6th race) is assumed to be slower than the horse that came third in the race with the 1st place winning horse (and so, excluded from the 7th race).
And though Skag55 illustrated with a wide difference, it is possible that the first place winning horses of first 5 races are in fact slower than the 2nd s of other races. Maybe horses in one race have differences of just a few seconds while others are wider.
I want to get solid here, but if it is not a GMAT Q...
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each entry in the table below represents a different horse.

X1 X2 X3 X4 X5
X6 X7 X8 X9 X10
X11 X12 X13 X14 X15
X16 X17 X18 X19 X20
X21 X22 X23 X24 X25

Let’s say that we have 5 races of 5 horses each, so each row in the table above represents a race. So, “X1 X2 X3 X4 X5 ” represents a race, and “X6 X7 X8 X9 X10 ” represents another race, etc. In each row, the fastest horses are listed in descending order, from the fastest (extreme left) to the slowest (extreme right). The fastest horses in each race are the ones on the left – so in the first race X1 was the fastest and X5 was the slowest. In the second race X6 was the fastest, X7 was the second fastest and so on.

Only 5 horses each race

So, now we ask ourselves: what do we know after these 5 races? Well, we do have the 5 five fastest horses from each race (X1, X6, X11, X16, and X21). But, does that mean we have the 5 fastest horses? Think about that for a second. Well, actually it does not mean that we have the 5 fastest horses. Because, what if the 5 fastest horses just happened to be in the first race – so X1 X2 X3 X4 X5 are the fastest horses. X1, X6, X11, X16, and X21 are all the fastest horses in their individual groups, but there could be one group that just happened to have all of the fastest horses. Remember we haven’t compared all the horses to each othersince we can only run 5 horses in a race, so that is still a possibility. This is very important to understand in this problem.

Work through a process of eliminatio
Well, now that we’ve had 5 different races, we can eliminate the slowest 2 horses in each group since those horses are definitely not in the top 3. This would leave these horses:

X1 X2 X3
X6 X7 X8
X11 X12 X13
X16 X17 X18
X21 X22 X23

We also know the 5 fastest horses from each group – but it’s important to remember that the 5 group leaders are not necessarily the 5 fastest horses. So what can we do with that information?
Well, we can race those 5 horses against each other (X1, X6, X11, X16, and X21) and that would be the 6th race. Let’s say that the 3 fastest in that group are X1, X6, and X11 – automatically we can eliminate X16 and X21 since those 2 are definitely not in the top 3.

What other horses can we eliminate after this 6th race? Well, we can automatically eliminate all the horses that X16 and X21 competed against in the preliminary races – since X16 and X21 are not in the top 3 then we also know that any horse that’s slower than those 2 is definitely not in the top 3 either. This means we can eliminate X17 X18 X22 and X23 along with X16 and X21.

Now, we also know that X1 is the fastest horse in the group since he was the fastest horse out of the 5 group leaders. So, we don’t need to race X1 anymore. Are there any other horses that we can eliminate from further races? Well, actually there are. Think about it – if X6 and X11 are the 2nd and 3rd fastest in the group leaders, then we should be able to eliminate X8 since X6 raced against him and he was in 3rd place in that race. X7 could only possibly be the 3rd fastest, and since X8 is slower than X7, we can safely eliminate X8. We can also eliminate X12 and X13 since X11 was the 3rd fastest in the group leaders, and X12 and X13 were slower than X11.
So, all together we can eliminate these horses after the 6th race: X17 X18 X22 X23 X16 X21, X12, X13, X8 and X1. This leaves us with the following horses to determine the 2nd and 3rd fastest horses:

X2 X3
X6 X7
X11
What is the solution?

This means we only have 5 horses left! Now we race those horses one more time – in the seventh (7th) race – and we can take out the top 2 horses and that would mean we have the 2nd and 3rd place horses! So, we have found our answer! It takes 7 races to find the top 3 horses in this problem.

Answer: 7 races
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what a complicated question , is there a possibility of using combination in this ? I tried to use combination
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jax91
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

Hi Bunuel

Not sure if this is a GMAT Type question. What if we don't have any winner (I mean what if all the 5 Horses finish the line at same time) during the First Five Races or the consecutive races. Its nowhere mentioned in the question that we a clear winner in each of the race.
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Bunuel
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

PFA and if we consider this case. We cannot literally say that "2. #3 from the last sixth race", "3. the second one from the race with the Gold medal winner from the first five races" is true and hence 4th and 5th case.
Correct me if i am wrong.
Thanks
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Bunuel
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itisSheldon
Bunuel
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

PFA and if we consider this case. We cannot literally say that "2. #3 from the last sixth race", "3. the second one from the race with the Gold medal winner from the first five races" is true and hence 4th and 5th case.
Correct me if i am wrong.
Thanks

I tried to explain this question couple of times on previous two pages. Could not add anything new.
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