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Find 3 Fastest horses from 25 horses

(7 votes, average: 3.43 out of 5)

May 16, 2014 by puzzler 46 Comments

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

find 3 fastest horses from 25 horses puzzle

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We can first do 5 races by taking 5-5 horses and get top horse in each race
Lets 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.

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Comments

  1. Vishal Singh Chauhan says

    July 17, 2017 at 12:54 am

    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…

    Reply
  2. Hardik Patel says

    September 16, 2016 at 9:55 am

    Surely 7 is 100% correct answer.

    Reply
  3. Harshit Anand says

    September 12, 2016 at 2:02 am

    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.

    Reply
  4. Aashirwad Khaitan says

    May 24, 2016 at 12:36 pm

    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.

    Reply
  5. Vijay Kumar says

    November 17, 2015 at 12:45 pm

    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 *****************

    Reply
  6. Ganesh VishnuPriya says

    November 14, 2015 at 1:08 pm

    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…

    Reply
    • Vijay Kumar says

      November 17, 2015 at 12:32 pm

      No It is not correct because 2 and 3 horse in first race, can be faster than first horse in second race

      Reply
  7. Ramesh Chandra says

    July 25, 2015 at 1:52 am

    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

    Reply
  8. Anurag Pushpam says

    February 23, 2015 at 12:54 am

    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

    Reply
    • puzzlersworld says

      March 9, 2015 at 9:24 am

      And you are assuming track is equal, as some horses might perform better than other in particular condition.

      Reply
      • Anurag Pushpam says

        March 10, 2015 at 4:21 am

        if you talkin about condition then …all horses at one go fits most … condition of tracks may change with every race ….

        Reply
  9. Shyam Vijay says

    December 31, 2014 at 8:31 pm

    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.

    Reply
  10. Satya says

    October 25, 2014 at 1:53 pm

    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

    Reply
  11. sandeep says

    October 1, 2014 at 8:29 pm

    this is exact answer 11

    Reply
  12. krisN says

    September 14, 2014 at 1:46 pm

    Answer is 7.Without any duobt

    Reply
  13. Anshul Rajvanshi says

    September 9, 2014 at 5:44 pm

    I think its 6. 5 races would give top 5 racers. Sixth race can give the top 3 racers? Isin’t?

    Reply
  14. Tejaswi Tejas says

    September 1, 2014 at 1:46 pm

    6 races gives three top fastest horses….its simple. ..

    Reply
  15. Pankaj Kumar says

    August 4, 2014 at 6:13 pm

    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

    Reply
    • puzzlersworld says

      August 4, 2014 at 6:15 pm

      Think if you can avoid some races, we need to find minimum number of races needed

      Reply
  16. Srivaths says

    July 24, 2014 at 5:17 pm

    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

    Reply
    • dj says

      August 3, 2014 at 4:43 pm

      that’s seems exact..

      Reply
      • puzzlersworld says

        August 3, 2014 at 4:47 pm

        That does not give minimum number of races required

        Reply
        • dj says

          August 4, 2014 at 5:35 pm

          oh, right! Thankyou

          Reply
      • Raizel says

        October 17, 2014 at 1:41 pm

        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 😀

        Reply
    • Jason says

      October 21, 2014 at 1:24 am

      This is a worst case scenario and gives the upper bound on the problem; however, it is a valid solution.

      Reply
  17. Alok Raghuvanshi says

    July 16, 2014 at 6:21 pm

    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.

    Reply
    • puzzlersworld says

      August 4, 2014 at 6:17 pm

      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…

      Reply
    • Amit says

      September 30, 2014 at 8:49 am

      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

      Reply
  18. Saubi says

    July 11, 2014 at 8:05 pm

    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.

    Reply
    • Alok Raghuvanshi says

      July 16, 2014 at 6:19 pm

      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.

      Reply
      • saandy says

        August 13, 2014 at 5:59 am

        that is absolutely correct…

        Reply
      • damii says

        September 23, 2014 at 1:41 pm

        I think this is the best answer

        Reply
    • Zakeer Belthangady says

      August 29, 2014 at 1:35 pm

      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

      Reply
  19. rohit says

    July 6, 2014 at 10:20 pm

    its simple..and ans is 7

    Reply
  20. loki says

    June 29, 2014 at 6:02 pm

    7 races

    Reply
  21. loki says

    June 29, 2014 at 6:02 pm

    5 races

    Reply
    • Giriraj Mulay says

      April 12, 2017 at 4:40 pm

      How???

      Reply
  22. Anonymous says

    June 28, 2014 at 7:06 pm

    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.

    Reply
    • Vijay says

      July 1, 2014 at 9:10 pm

      here u think wrong. because in first race, line 1 -> second horse can be fast from other 4

      Reply
      • naman says

        August 14, 2014 at 3:57 am

        What should be the answer?

        Reply
  23. Harish Kotkar says

    June 27, 2014 at 2:26 pm

    11

    Reply
  24. Saran MK says

    June 27, 2014 at 9:59 am

    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

    Reply
    • puzzlersworld says

      July 2, 2014 at 7:15 am

      There is a chance that horse which is beaten by fastest horse in earlier races can be better than other 4 horses.

      Reply
    • Anil says

      July 5, 2014 at 5:43 am

      correct ans will be 7………………………think again……..

      Reply
      • crazyforstudy it says

        July 13, 2015 at 3:08 pm

        well explained @puzzlesworld…..answer is 7 only

        Reply
  25. lion says

    June 10, 2014 at 6:32 am

    hi dude…

    Reply

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