Races To Find Horses For Courses

What was the task? 

First we need to make 5 groups of 5 horses each. Let's put each group on race & note down rank of each horse & corresponding group. This requires 5 races.

Taking out winner of each group aside. In a next race, winner of each group run & fastest winner is found. Total 6 races are conducted till now.

Now let's name group of fastest winner as A, second fastest winner as B & so on as C, D, E. Each horse would be identified by it's group name & rank as Group.Rank. For example, B.3 means the horse that came 3rd in group B & D.5 is horse that came last in group D.

For next race, we can eliminate few. Let's remind we have to find 3 fastest horses. Now let's logically eliminate few horses. Those horse for which we are sure that at least other 3 are faster than those will be eliminated.

1. Now A.1, B.1 & C.1 are faster than D.1 & E.1. So D.1 & E.1 & respective groups eliminated straightaway.

2. The C.2 (& other members of C) eliminated as A.1, B.1, C.1 are faster ones.

3. The B.3 (& other member of B) eliminated as at least A.1, B.1, B.2 are faster ones.

4. Similarly, A.4 & A.5 have 3 horses ahead as A.1, A.2 & A.3.

5. A.1 is already proved it's fastest among all, so no need to re race by taking it.
So we are left with A.2, A.3, B.1,  B.2 & C.1 for next race.  That means total 7 races we need to conduct for finding top 3.

This is the solution you would find everywhere; here it is just simplified. But it's based on big assumption.

It's assumed that, B.3 is not faster than C.1 & A.4 not faster than B.1.

Case 1 :

If B.3/B.4/B.5 faster than C.1 then A.2,A.3,B.1,B.2 & B.3 should run in last race.

Case 2 :

If A.4/A.5 faster than B.1 then we would get top 3 in first race itself as A.1, A.2, A.3!