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  • The Horses Problem - Need Solution

    Engg.Dee

    Member

    Updated: Oct 25, 2024
    Views: 976
    I have a problem, please give the solution of that with explanation..?

    25 horses are there. you have to decide top 3 against them. only 5 can race at a time. so minimum how many races are required to decide top 3..?
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Replies

  • vinod1993

    Member • Sep 6, 2012

    I think the minimum amount of races will be 6. 5 races from which 5 horses are selected and 6th race top 3 are selected. But this solution may lead to many arguments because it is also possible that the horse which came 2nd in the (say) 2nd race may be faster than the horse which came first in some other race. So not sure, but my guess is 6. Waiting for the best explanation(I'll be trying out too).
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  • clZazy

    Member • Sep 6, 2012

    I think the answer is 7 races.

    5 races among 25 horses in 5 groups(5 horses each group) to find out 1st, 2nd and 3rd in each of the 5 groups.
    now 1 more race among all the 1st position holders, it will leave us with 3 groups only.
    also suppose group3's 1st horse came 1st in the 6th race(it means that group3's 1st horse is all over 1st position holder). now we have to find 2nd the 3rd position holders from the rest. how?
    now I can't explain this in theory so just a rough diagram.
    the horses with * are also rejected as they can't hold any of the 2nd or 3rd position.

    group 1 --> 1st, *, *
    group 2 --> 1st, 2nd, *
    group 3 --> 1st, 2nd, 3rd

    now we are left with only 5 horses.
    1 more race and we have our 2nd and 3rd position holders.
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  • Engg.Dee

    Member • Sep 6, 2012

    hey ur ans is right but ur explanation after 5 race is wrong plz correct it.........
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  • clZazy

    Member • Sep 6, 2012

    how is it wrong?
    tell me which step I am doing wrong?
    their could be an explanation problem.

    1. k first the 5 races. now we know which horses are 1st, 2nd and 3rd in each group. last two from each group are rejected.
    2. then 6th race is between all the 1st position holders from the five groups.
    3. now we can reject whole two groups of horses from which 1st position holders are not in any of the 1st, 2nd or 3rd position in the 6th race.
    4. the horse who came 1st in 6th race is also 1st position holder overall. so now we have to find only the 2nd and 3rd position holders.
    5. and then as I tried to explain with my rough diagram that we can reject 3 more horses from the rest of the 9 horses as they can't take any of the 2nd or 3rd position.
    6. finally we are left with 6 horses from which we know the 1st position holder overall already.
    7. so the 7th race is among these 5 horses to find 2nd and 3rd positions overall.

    I am explaining all this because I am quiet confident about my solution.😉
    your explanation would be helpful though.😀
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  • Engg.Dee

    Member • Sep 6, 2012

    let there is 5 group of 5 horses eaach
    A B C D E
    5 5 5 5 5
    AFTER 5 RACES WE CHOOSE TOP 3 OUT OF 5 GROUPS EACH
    1A 1B 1C 1D 1E
    2A 2B 2C 2D 2E
    3A 3B 3C 3D 3E
    now as a results of sixth race we find that 1a is topper then it will show like..
    1A>1B>1C>1D>1E
    2A 2B 2C 2D 2E
    3A 3B 3C 3D 3E

    NOW for 7th race we choose
    2A 1B 1C
    3A 2B
    OUT OF THESE FIVE WE WILL SELECT THE 2ND AND 3RD POSITIONED HORSES..
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  • clZazy

    Member • Sep 6, 2012

    that's exactly what I explained. 😁
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