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
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 Bunnel,
Is this a GMAT-kind question?