Answer / love goel

We will only need 7 races to get the best three out of 25.

Let us Name horses as H1, H2, H3...H25. Now,

Race 1 - H1 - H5
Race 2 - H6 - H10
Race 3 - H11 - H15
Race 4 - H16 - H20
Race 5 - H21 - H21

Let us now assume

First Rank of Race 1 - W11
Second Rank of Race 1 - W12
.
.
First Rank of Race 2 - W21
.
Third Rank of Race 2 - W23
.
.
Fifth Rank Of Race 5 - W55

Race 6 - W11 , W21 , W31, W41, W51 ( toppers from first
five races)

Race 7 - 2nd from Race 6, 3rd from Race 6, X, Y , Z where

X= 2nd winner from the team of winner of race 6
Y= 3rd winner from the team of winner of race 6
Z= 2nd winner from the team of 2nd winner of race 6

So the three best horses will be , Winner of Race 6 and
Winner and Second Winner of Race 7.

Answer / hemant kumar

Hi everybody!

Correct answer is 7.

First conduct 5 races, top horse from each group selected.
Now conduct 1 race among these five, select top three from
these. Now discard the groups of the horses who stood 3rd,
4th and 5th in last race because no one of them can make it
to top three. Now take 2nd and 3rd horse from first group
(the group to which the first ranker belongs) and 2nd horse
from second group (the group to which 2nd ranker belongs).
Now we have total of six horses but first rank is already
decided, so we are left with five horses from these, we have
to select 2. Now conduct 1 race of these five horses and
select top 2.

Thanks for reading this long answer!

Answer / diepark

An explanation as to why the answer is 7, assuming no method
to time the races:

Divide the 25 horses into groups of 5 labeled A to E. Hold 5
races, one for each group and label the horses A1, A2, etc.,
where the number designates what place the horse finished.
Hold a 6th race out of the five winners of each group, and
arbitrarily pick A1 as the winner, B1 as second place, etc.
After the 6th race, you'll have the following tree for
fastest horse:

A1
| \
A2 B1
| | \
A3 B2 C1
| | | \
A4 B3 C2 D1
.
.
.

As you can see, A1 is the fastest horse and is faster than
all the horses in the other groups since he beat their
winners, and likewise, B1 is faster than the horses in
groups C to E since he beat their winners. But we don't know
if B1 is faster than the rest of the horses in group A,
since they never raced against each other, and likewise, we
don't know if C1 is faster than the remaining horses in
group B, since they didn't race each other, etc. To
determine the 2nd fastest horse, let's hold a hypothetical
7th race between A2 and B1. There are two possible outcomes
that result in the following trees:

A1 A1
| |
B1 A2
/ | \ / |
A2 B2 C1 A3 B1
| | | | | \
A3 B3 C2 A4 B2 C1

In the first outcome (left tree), B1 has won and is the
second fastest horse. Now we need to find out which horse is
3rd fastest in an 8th race between A2, B2, and C1, since
they have never raced each other.

In the second outcome (right tree), A2 has won and is the
second fastest. Therefore, an 8th race needs to happen
between A3 and B1 to determine the 3rd fastest horse.

If you consider which horses run in the hypothetical 7th and
8th races (horses A2, A3, B1, B2, and C1), you'll notice
that only 5 horses need to race, so all 5 horses can
actually run in the 7th race to determine the 2nd and 3rd
fastest horses.

So, is it possible to find the top 3 fastest horses in only
6 races? Yes, there are certain situations where this is
possible. Take the scenario where the winner always gets to
run in the next race: the first race has 5 horses (Group F)
that haven't raced; the winner goes on to race 4 more horses
that haven't raced (Group E), and that winner gets to race 4
more (Group D), etc. until the 6th and last race (Group A).

As long as the winner from the 5th race does not finish in
1st or 2nd, then you can determine the 3 fastest horses;
otherwise, you won't be able to determine the 2nd and/or 3rd
fastest horses, since the horses in group A have not raced
any of the other horses.

Race 1(F) 2(E) 3(D) 4(C) 5(B) 6(A)
----------------------------------------------------
1st F1-. E1----->E1-._ C1. B1-. A1
2nd F2 \ E2 D1 `->E1 \ B2 \ A2
3rd F3 `->F1 D2 C2 \ B3 `->B1
. F4 E3 D3 C3 `->C1 A3
. F5 E4 D4 C4 B4 A4

Thus, to guarantee you can determine the fastest 3 horses,
you need 7 races.

Answer / sasi m

It is seven.....
It is already given by some one .....
But I will try to make it more clear.....

First conduct 5 races with the given 25 horses......
so we will get :
Races : 1 2 3 4 5
First : 1 1 1 1 1
Second : 2 2 2 2 2
Third : 3 3 3 3 3
Fourth : 4 4 4 4 4
Fifth : 5 5 5 5 5

So we are sure that the fastest would be from the horses
which stood first in each race........
that is from First: 1 1 1 1 1

So next we must conduct the race among all the horses which
stood first in each race,i.e. race no 6:

Than we will get the fastest......

Next we have to get the second fastest and third fastest..

Now we have the options like this:

Now the second might be the second from the 6th race or the
second from the earlier conducted race where the fastest
was the first.....

Third might be the third from the 6th race or the third
from the earlier conducted race where the fastest was the
first or the second from the earlier conducted race where
the second fastest from the 6th race was the first.....

So we have to find out the second and third from this group
by conducting the race between:

1)Second from the 6th race
2)Second from the earlier conducted race where the fastest
was the first
3)Third from the 6th race
4)Third from the earlier conducted race where the fastest
was the first
5)Second from the earlier conducted race where the second
fastest from the 6th race was the first

So after conducting this 7th race we will get the second
fastest and third fastest....

Hence the result is 7 races.......

Hope u will understand this.....
if not u can mail me

Answer / alok chandra

I m sorry for the wrong answer I posted earlier. So the
answer is definitely not 11. At the same time Anjali, it is
not even 6. It would be wrong to assume that the horses
which came second and third in the first set of five races
cannot come second and third overall. So let me provide the
correct answer now.

We conduct the 5 races. Choose the top 3 from all the 5
races. Now conduct a race of the top horses of all the
races. Choose the top 3 of the last race. They could
probably be the top 3 of the horses. We discard the horses
which came 4th and 5th along with all the horses of their
group. Let me explain it pictorially.
The top represents the standing of the 6th race. The race
among the top horses of each group. Now we can discard all
the horses which came 4th and 5th in their respective
groups. So we are left with the top 3 horses of each
group.Now consider the horses which came 4th and 5th in the
6th race. They cannot fall in the top 3. Nor can any of
their group members. So we get rid of all the remaining
horses of group 4 and group 5. So we are left with 9 horses
to choose from.
Now consider the horse which came 3rd in the 6th race and
its group memebers. We call it group 3. Now the 2nd and 3rd
horses of group 3 cannot come even 3rd overall. Hence they
can be discarded. SO we consider only the horse which came
first(1 selected for the final race). Consider group 2. We
can discard the horse which came 3rd in the group. And we
select the horses which came 1st and 2nd( 1+2 selected for
the final race). Similarly we consider all the top three
horses from group 1. But we can safely ignore the horse
which came first overall.( 1+2+2 selected for the final
race). So we have 5 horses to choose the best two from. We
already have the horse which came 1st overall. No need to
put that in the race. So the minimum number of races
required is 7 and 11 or 6.

1 2 3 4 5

1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5

Answer / aleeza

atleast 7 races are required.

5 races will select top 5 horses.6th race will help out in
finding 4 best horses. 7th race will give us 3 best horses.

Answer / ashish :p

u c ..........v humans have invented sumtin called a stop
watch......


conduct 5 races......note the timings of the horses....&
select the top 3 :P


ps : it is no mentioned anywhere in the Q tat v r not
allowed 2 use timing devices

Answer / shrey sahay

correct answer is 7... why do yiou people keep writing
anything... atleast look at the top explainations....
answer#4 given by alok chandra is perfectly correct..
atleast have a look at it and then write something

Answer / preethi

6

Answer / dinesh

7 required.....
none others right...

wat will be the next number in the series.... 1,3,5,9,12,21,..

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