There are 25 horses and only five tracks in a race.
How do you find the second coming horse of all the 25
horses, provided there is no stop clock? (obviously, a
horse cannot participate more than once in a race).
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.
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.
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
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
plus horse that came 2nd behind Round 2 2nd placer in
-> 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.
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
* 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
You cannot simply take the fastest horse from each group of
five. You have to look at the times of all the horses and
take the five fastest times from all 25 and then select the
top 5. Some would argue length and turf play in, but all
else equal, the fastest horse of one race could be slower
than the slowest of the other 4 races, so the winner of
each race is not a good answer.
At the Party:
1. There were 9 men and children.
2. There were 2 more women than children.
3. The number of different man-woman couples possible was
24. Note that if there were 7 men and 5 women, then there
would have been 35 man-woman couples possible.
Also, of the three groups - men, women and children - at the
4. There were 4 of one group.
5. There were 6 of one group.
6. There were 8 of one group.
Exactly one of the above 6 statements is false.
Can you tell which one is false? Also, how many men, women
and children are there at the party?
Three boxes labeled as red, blue and mixed.These labels are
incorrect.one box contains red balls and another box
contains blue balls and remaining one box contains both red
and blue balls.
Pickup one ball from any box(u should pickup ball only
once) and name the correct labels.
Tic-Tac-Toe is being played. One 'X' has been placed in one
of the corners. No 'O' has been
Where does the player that is playing 'O' has to put his
first 'O' so that 'X' doesn't win?
Assume that both players are very intelligent. Explain your
one man driving a car in a lonely forest .. it was raining
at that moment ...suddenly one of the car tyre got
punchured...and he is having stephiny in his car ...he gets
it and he tried to fix that stephiny...when he removes the
bolts of punchred tire all are gone into mud...all bolts...
fell into mud...and that bolts he never get...then, how can
he reach safely to home... by driving his car..?
My friend collects antique stamps. She purchased two, but
found that she needed to raise money urgently. So she sold
them for Rs. 8000 each. On one she made 20% and on the other
she lost 20%.
How much did she gain or lose in the entire transaction?
1.At the recent web developres bowling match,two ganes were
played.Kiev beat stuart in both games;also Richard beat John
in both games.The winner in game1 second in game2.Richard
won game2 and John beat stuart in game1.No player got the
same placing twice.Can you determine who finished where in