(7 votes, average: 3.43 out of 5)There are 25 horses. We have to find out the fastest 3 horses. In one race maximum 5 horses can run. How many such races are required in minimum to get the result ?
find 3 fastest horses from 25 horses puzzle
Check your answer:-
Click here to See SolutionLets say it looks like this
O1,O2,O3,O4,O5
T1,T2,T3,T4,T5
TH1,TH2,TH3,TH4,TH5
F1,F2,F3,F4,F5
FV1,FV2,FV3,FV4,FV5
Here O1 > O2 > O3 > O4 > O5 same way for others.
Now lets take fastest horse in each race and then have one race between them, so have race in O1, T1, TH1, F1 and FV1.
Lets say O1, TH1 and F1 comes in top three(and O1 > TH1 > F1), then there are no chances that horses slower than T1 and FV1 can come in top 3 and we can also say that O1 is fastest horse and also F2-F5 horses are not among top 3.
Now we need to find second and third fastest horses
they can be from O2,O3,TH1,TH2 and F1, So we will have once more race among them to determine second and third position.
Thus a total of minimum 7 races are needed to find top 3 horses from 25 horses.
We need fastest 3 horses…
First divide the 25 horses in five groups.. Having 5 horses in each group…
Race the 5 groups (each seperately)..
So 5 races…
And we got top 5 horses in this way…
Now race these 5 horses together….
And select the first three horses finishing the finish line…
So total of #six race…
Surely 7 is 100% correct answer.
step1: Compete in 5 random groups and get results as:
a1 a2 a3 a4 a5
b1 b2 b3 b4 b5
c1 c2 c3 c4 c5
d1 d2 d3 d4 d5
e1 e2 e3 e4 e5
step2: Compete the top runners and say results obtained are:
a1 b1 c1
clearly we get a1 is fastest and whole group d and e are eliminated.
step3: Now race is for second position. a2 and a3 can be faster than b1 and c1 as a1 is faster, b1 and b2 can be faster than c1.
We don’t consider other participants as we need three places only.
Race a2 a3 b1 b2 c1 and select top 2.
So, 7 races.
11
Take any 5 horses for 1st race.(1 race)
Now divide the rest 20 horses into groups of 2 i.e. 10 groups.
Take any group and join them with the top 3 of the 1st race.(2 races)
Now, again take the top 3 of the 2nd race and join them with any other of the remaining 9 groups.
Thus, in total 10+1 = 11 races will be required.
step 1: 7 Races, 5 races we will rank top 3 in all Races
step 2: Take all the first horse in 5 races and conduct a race and mark there rankings in 6 Race
step 3: Horse which were ranked 4 and 5 are eliminated and horses which were ranked 2 and 3 in the same group of Horses which are ranked 4 and 5 are also eliminated as they are inefficent
Step 4: Horse which ranked 3 has chance to be in top 3, but horses which got 2 and 3 in step 1 of gorup of rank 3 horse in step 2 has no chance of being in top 3 so they are also eliminated
step 5: Horse which ranked 2 in step 2 has chance to be in top three, and horse which ranked 2 in step 1 of this horse group also chance to be in 3 position, but horse ranked 3 in step 1 of this horse group will be eliminated as it has chance to be in 4 overall, now we got 2 horses of this group
step 6: Horse which ranked 1 in step 2 will always be the top horse whose rank will be 1, horse ranked 2 and 3 in the group which rank 1 horse of step 2 ran will have chances to be in 2 and 3 overall, so we will take all the 3 horses
step 7: now we have total 6 horses
3 horses from group of a horse which came first in step 2(best of best horses)
2 horses from group of a horse which came second in step 2
1 horse from a group of a horse which came third in step 2
now we have total of 6 horses
Horse which came first in step 2 is always best of best horses so it is ranked 1
now we have 5 horses and we will conduct a race to all the five, and select first and second and they will be 2nd rank and 3rd rank(as we already got first rank horse)
Total ************* 7 races *****************
first conduct 25/5=5 races
take 1st position horses in that 5 races and conduct another race
so that top 3 horses is select… i think so it is crt….
finally 6 races….is it crt guys…
No It is not correct because 2 and 3 horse in first race, can be faster than first horse in second race
first make of group of 5 and make 5 race
take top 5 and make one more race…
select which are top three and take two more horse which group has fastest horse…and make one more race too
total race = 7
Lets have a different solution…
Divide the track in two equal halves. Now we can get 10 horses to run in a go i.e from both sides. after first run we keep the top three and replace the other seven . We do this process again until no horse left.
In this way we get only three attempts 🙂 🙂 . ASSUMPTION :THERE IS NO COLLISION/HARM TO HORSES
And you are assuming track is equal, as some horses might perform better than other in particular condition.
if you talkin about condition then …all horses at one go fits most … condition of tracks may change with every race ….
11 races are required and the explanation is here :
1st race – any 5 horses — get 3 fastest horses
2nd race – above 3 fastest horses + 2 from remaining 20 — get 3 fastest horses (18 remain)
3rd race – repeat the 2nd step (16 remain)
4th race – repeat the 2nd step (14 remain)
5th race – repeat the 2nd step (12 remain)
6th race – repeat the 2nd step (10 remain)
7th race – repeat the 2nd step (8 remain)
8th race – repeat the 2nd step (6 remain)
9th race – repeat the 2nd step (4 remain)
10th race – repeat the 2nd step (2 remain)
11th race – repeat the 2nd step (0 remain)
We will get the 3 fastest horses among all 25 horses.
Answer is 6
25 horses => 5 groups (5 in each group)
5 group=> 5 race=>5 winner horse=> group no. 6
now give one race and get 3 winner
this is exact answer 11
Answer is 7.Without any duobt
I think its 6. 5 races would give top 5 racers. Sixth race can give the top 3 racers? Isin’t?
6 races gives three top fastest horses….its simple. ..
ans will b 10….devide in 5*5 and get best 3 of each…now again devide into 5*3 and get 9 best of them..now run any 5 of them eleminate last 2..now we have 7 remaining again run 5 n eleminate now we have 5…….run them all n get best 5…..
5+3+1+1+1=11 ans
Think if you can avoid some races, we need to find minimum number of races needed
Ans is 12
Make 5 horses in 5 groups
The fourth and fifth place of each race can’t be the top three… Which eliminates 10 horses… 15 horses remain… Split them into three groups… Again the last two are removed…. So 9 horses remain… Split into 5 and 4… We get 5-2 and 4-1…so 6 horses remain… Let the first four run… And eliminate one…. Let the remaining 5 run….. And u get the top three….
5+3+2+1+1
that’s seems exact..
That does not give minimum number of races required
oh, right! Thankyou
No the answer should be 6 cause if you took a winner from each group of five horses then you again can make a team of five and after their racing you can get top 3 fastest horse 😀
This is a worst case scenario and gives the upper bound on the problem; however, it is a valid solution.
only 5 race required .
1 ) make 5 groups having 5 horse in each group
2) let the 1st group run for certain distance and note down the time taken by all the horses.
3) repeat step 2 with all groups for the same distance and note down the time taken by each horse.
Now you have time taken by all the 25 horses to complete the race and easy to decide the best 3 of them.
you can not note down the times, as the fastest horse might have slowed down as his opponents were slow, but he might have tried harder if required, you know the last minute stint…
Only 6 race will be required to determine fastest horse.
5 to determine fastest from group of five and one to determine fastest from fast five
I think 8 is the answer..
12345 ABCDE LMNOP PQRST UVXYZ = 5 races
Here’s the order:
123 ABC LMN PQR UVX
1 A L P U = 6th race
1 is clear winner.. Now,
Here are the relationships between these horses ( “>” : may be faster, “>>” is definitely faster)
1 >> 2,3,A,B,C,L,M, N
2,3 > A
2 >> 3
A >> B, C
A >> L
B, C > L
So, we have 6 (2,3,A,B,C,L) horses to find the fastest 2 horses now.
2, 3, A, B, C = 7th race
If, 2 comes first then 3 can come 2nd and A is 3rd fastest.
If, A comes 1st and any1 else comes second then we cannot determine if L is faster or othr 4 horses are faster.
Therefore, in case 2 we will need 1 more race, i.e. total 8 races in worst case.
only 5 race required ….
1 ) make 5 groups having 5 horse in each group
2) let the 1st group run for certain distance and note down the time taken by all the horses.
3) repeat step 2 with all groups for the same distance and note down the time taken by each horse.
Now you have time taken by all the 25 horses to complete the race and easy to decide the best 3 of them.
that is absolutely correct…
I think this is the best answer
11 races my dear
First five group of five horse means 5 race
Result 5×3 15
3 group of 5 horse = 3 race
Result 3×3 =9
1 set race = 1 race
Found best three + 4 from above= 7
1 set race 1 race
Found 3
3+ remaining 2 from 7 means 5
Another race = 1 race
now we found finalist means 11 race
its simple..and ans is 7
7 races
5 races
How???
We can do 5-5 race for total 25 horses. Now we have 5 top horses.
Out of 5 we need to find 3.
For 5 top horses, we can have another race.So we found 1st faster running horse.
Now for 4 top horses we can have another race -> Condition is maximum 5 can include but there is no restriction for minimum. -> Found 2nd faster running horse
Now for remaining 3 horses, we can have another race -> Will get the 3rd faster running horse.
So total 8 races required to find the faster running horses.
here u think wrong. because in first race, line 1 -> second horse can be fast from other 4
What should be the answer?
11
This in not fair. After the final race, those came top 3 is correct, as it went throuh. But again making a race for 2nd of every group is disturbing the logic ? so correct answer should be 6 as i think
There is a chance that horse which is beaten by fastest horse in earlier races can be better than other 4 horses.
correct ans will be 7………………………think again……..
well explained @puzzlesworld…..answer is 7 only
hi dude…