Answer / desinerd

Detailed Solution for Horse Puzzle:

Divide the set of 25 horses into 5 non-overlapping sets of 5
horses each. Have a race each for all the horses in each
set. This makes it a total of 5 races, one for each set.

Now, have a race for the winners of each of the previous 5
races. This makes it a total of 6 races.

Observe the position of each horse in the 6th race and
correspondingly number the sets. i.e. the set of the winner
of 6th race will be said to be set no. 1 while the set of
the loser of the 6th race will be said to be set no. 5.

Now, possible candidates for the first three positions
exclude the followings:
1. Any horse from set 4 or set 5.
2. Any horse except the winner from set 3,.
3. Any horse except the winner and the 2nd position from set 2.
4. Any horse except the winner, 2nd position and 3rd
position from set 1.

So now we have 6 candidates for top 3 positions. However, we
know that the winner of set 1 is the fastest horse in the
whole group of 25 sets.

So now we have 5 candidates for the second and third
position. What better way to find out who's who than to have
a race of these 5 horses. Race them and this will solve our
problem in just 7 races.

Answer / jai ranjan kumar

No of Races : 7

1 RACE(A) A1 A2 A3 A4 A5
2 RACE(B) B1 B2 B3 B4 B5
3 RACE(C) C1 C2 C3 C4 C5
4 RACE(D) D1 D2 D3 D4 D5
5 RACE(E) E1 E2 E3 E4 E5

LAST TWO CAN BE REMOVED FROM EACH GROUPS.
WE HAVE LEFT WITH 9 HORSES.

6 RACE(F) A1->F1 B1->F5 C1->F2 D1->F3 E1->F4

SUPPOSE (CAN BE GENERIC,ANALYS) FROM RACE(F)

NOW F1(A1) IS FASTES HORSE AMONG 25.
ALL HORSES FROM GORUP B AND E CAN BE ELIMINATED(since E1
and B1 is at 4th and 5th position respectivly). AND ANALYS
A BIT, C3, D2 AND D3 ALSO CAN BE ELIMINATED (since in any
senario C3 will max come 4th, D2->4th and D3->5th). NOW
LEFT WITH 5 HORSES.

7 RACE(G) A2 A3 C1 C2 D1

NUMBER OF RACES REQUIRED : 7

Answer / md

5X5 = 25...
make group of 5 and total 5 such groups ...
race each grup at a time... select winner from each group.
race these 5 winners ... and mark them as their positions

best case: the fastest horse and the 2nd coming horse are
in two different group. Then we can simply say the 2nd
coming horse in the winners' race is the answer.

worst case: the fastest horse and the 2nd coming horse are
in same group. So any from the four except the winner from
the fastest horse's group (say 1st group) may be the 2nd
coming horse.

so need another race.... 4 from 1st group (except winner)
and winner of 2nd group .... who wins is 2nd best.

no of races: 7.

Answer / katie

I was asked this question in microsoft.
The correct answer to find out top 3 horses is 7.
Explaination of Desinerd i.e. Answer#1 seems the most
appropriate.

Answer / joel konecny

Run each horse in a race, always keeping the top two to
compete in the next race, until the last race in which the
top two are identified. So run 8 races instead of 7,
sometimes the simple solution is the best.

Obviously a horse can't run twice in a race. Sometimes when
something is too obvious it makes you think it's a trick
statement.

Answer / mikhailfranco

Obviously horses must be allowed to compete in more than one
race, and they are assumed not to tire as they run races, so
their performance is constant.

Round 1: 5 races of 5
Round 2: 5 winners of Round 1
-> winner is overall 1st place (6 races)
Round 3: 2nd and 3rd places from Round 2,
plus horses that came 2nd & 3rd behind Round 2 1st
placer in
Round 1
plus horse that came 2nd behind Round 2 2nd placer in
Round 1
-> winner is 2nd place overall
-> 2nd place is 3rd place overall

So you can find the winner in 6 races (trivial) and top
three in 7 races.

Answer / aditya kalia

The first soln is correct, but I think its not understandable.
and @ Animesh Sonkar, your soln is correct until 6th race.
In the 7th race, u have eliminated the first rank, the fouth
nd the fifth. But u have raced only 4 horses.. that is whr u
missed. Correct Soln.:-

The fifth horse in the seventh race would be rank 2 horse of
the group which has 2nd rank in the fifth (all winners) race.
So, all the scenarios would be taken care of now.

Eg. after 5th race, let the positions be:

A1 B1 C1 D1 E1 (in order of rank)

now A1 is the fastest.--> eliminate it
D1 nd E1 can't be 2nd nd 3rd.(!!!)
Now we remain with B1 nd C1.

The other horses in the race would be A2 A3 and B2.

So, in every possible case, we can get the first three
positions confirmed.
* We don't need B3 because, B1 nd B2 are already faster than
it (evn after leaving A1), therefore, it can't be 3rd.

* We don't need horse C2 because, B1 nd C1 are already
faster than C2, therefore it is not the contender of top
three positions.

:) :) HAPPY PUZZLING :) :)

Answer / jai ranjan kumar

THE ABOVE SOLUTION IS ALSO WORK FOR WHEN WE NEED TO FIND
FIRST 3 FASTEST HORSES.

the 7th races occures to find 2nd and 3rd positions..

Answer / guest

the horse which crosses the finishes line 2 nd is the second horse.

Answer / hacker

Solution: the answer is 7 races.

Step 1: First, we group the horses into groups of 5 and race each group in the race course. This gives us 5 races.



W11 W12 W13 W14 W15

W21 W22 W23 W24 W25

W31 W32 W33 W34 W35

W41 W42 W43 W44 W45

W51 W52 W53 W54 W55


Step 2:we race the 5 level 1 winners(w11,w21,w31,w41,w51) ans assume winning order of this race is w11,w21,w31,w41,w51

Step 3: BECAUSE WE NEED TOP 3 AND W41 HAS COME 4TH Position that is the reason we don't need to consider W41 W42 W43 W44 W45 and also W51 W52 W53 W54 W55
now we have


W11 W12 W13 W14 W15

W21 W22 W23 W24 W25

W31 W32 W33 W34 W35


Step 4: because we need top 3 then dont need W14 W15 W24 W25 W34 W35

now we have

W11 W12 W13

W21 W22 W23

W31 W32 W33


Step 5: because in 6th race W31 has come on 3rd position that is the reason we do not need to consider W32,W33 and also we will not consider W23
now we have

W11 W12 W13

W21 W22

W31

Step 6: top 1 is already achieved which is W11(winner of 6th race)
remaining are

W12 W13

W21 W22

W31

Step 7: now race W12 w13 w21 w22 w31 to get 2nd and 3rd winner



Hence answer is 7 races.

no.1-6 first is added the last 3 are subtracted , arrange the no. so that each sum and difference give the same answer

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