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Total 12 individuals—A, B, C, D, E, F, G, H, I, J, K and L—are sitting in concentric circular arrangement such that six of them form the inner circle and the remaining six form the outer circle. Those who form the inner circle face outwards (away from the center) while those who form the outer circle face inwards (towards the center). The seats are arranged on the two circles in such a way that the angle subtended by the adjacent seats at the center is equal. Further, there are three different diameters of the circles such that four persons (two on outer circle and two on inner circle) sit along each of the three diameters.

The following information is known.
– B is facing both H and F.
– A is facing only J.
– C, D, G and K form a straight line (in no particular order).
– C is to the immediate left of E.
– G is two places to the left of B.
– H, E, F and B form a straight line (in no particular order).
– K is facing only D.
– L is facing both I and J.

PS: It cannot be solved completely. Source : imsindia.com

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  • $\begingroup$ Hi, sam! The description is a bit complicated, looks like it could really use a picture. Also, is there some particular reason why we would want to have an incomplete (as in, not possible to solve completely) problem on the site? $\endgroup$ – Bass Nov 7 '18 at 9:04
  • $\begingroup$ I tried solving it. But yes, it is rather confusing so I am stuck here. It would be good if someone can help in it. $\endgroup$ – sam Nov 7 '18 at 9:08
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I got this far:

Taking the set B, E, F, H
B is on the outer as facing F and H
E is on the inner furthest away
The placing of F, H is unknown

       \         /
        \       /

          \   /
           \ /
 B --- FH ----- E ---- HF
           / \
          /   \

        /       \
       /         \
 
G is 2 to the left of B
                  G
       \         /
        \       /

          \   /
           \ /
 B --- FH ----- E ---- HF
           / \
          /   \

        /       \
       /         \
 
Taking the set C, D, G, K
K is facing only D, so K is inner facing away from G
                  G
       \         /
        \       /
               C
          \   /
           \ /
 B --- FH ----- E ---- HF
           / \
          /   \
         K
        /       \
       /         \
      D
 
Taking the set A, I, J, L
L is facing both I and J so L is outer
A is facing only J so A is inner and J is outer
     J L          G
       \         /
        \       /
        A I    C
          \   /
           \ /
 B --- FH ----- E --- HF
           / \
          /   \
         K    I A
        /       \
       /         \
      D          L J
 
C is to the immediate left of E
Is already known!

So there is some redundant information, but not enough to solve the puzzle.

I found four solutions.

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  • $\begingroup$ The OP said it cant be completely solved, so I think youre right. +1! $\endgroup$ – Omega Krypton Nov 7 '18 at 10:03
  • $\begingroup$ @Weather Vane Yes you are absolutely correct.Thanks for solving it. Can you please tell which are the 3 different diameters in your solution ? $\endgroup$ – sam Nov 7 '18 at 10:24
  • 1
    $\begingroup$ The three diameters are formed by BFEH (or BHEF), DKCG and JAIL (one way or the other). $\endgroup$ – Weather Vane Nov 7 '18 at 10:27
  • 1
    $\begingroup$ 2 of the 3 diameters have 2 solutions each, so there are 4 solutions altogether. $\endgroup$ – Weather Vane Nov 7 '18 at 10:39

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