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Place Perception (Posted on 2013-06-07) Difficulty: 3 of 5
Three players A, B and C are nominated and three sportswriters respectively from periodicals D, E and F designate a player as first, second or, third place. However, the point awards for the first-, second- and third-place votes are not known precisely. All we know is that a first place vote is worth x points, a second-place vote is worth y points and a third-place vote is worth z points with x > y > z.

When the points have been counted up, Player A has emerged with 20 points, Player B with 10 points and Player C with 9 points.

If Player A received a second place vote from Periodical D, who received a second place vote from Periodical E?

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution A new solution (spoiler) | Comment 6 of 9 |
While I am waiting for an answer to my "revised brain-twister", allow me a different solution to the original problem.

a) The total score is 3x+3y+3z = 39.  The average score (which nobody got) is x+y+z = 13.  

b) A has an above average score.  Because we know that A received a second place vote, then A's score must be either 2x+y or x+2y.  It cannot be x +y+z, because that is an average score.

b) B has a below average score.  It is either x+2z or 2y+z.  It cannot be x+y+z, because that is an average score.  It cannot be y+2z, because that would be lower than C's score.

c) So what does that make C's score?  Let's consider 4 cases:
   1) A = 2x+y, B = x+2z.  Then C = 2y+z 
   2) A = 2x+y, B = 2y+z.  Then C = x+2z.
   3) A = x+2y, B = x+2z.  Then C = x+y+z, 
        which is greater than B (and also average).  Not possible 
   4) A = x+2y, B = 2y+z.  Not possible.  
         There are only 3 2nd=place scores. 
   
d) So let's do the linear algebra 
   Case 1) 20 = 2x+y, 10 = x+2z, 9 = 2y+z
           Then x = 8, y = 4, z = 1
           And C got two 2nd place votes.
           We have seen this solution already.
           
   Case 2) 20 = 2x+y, 10 = 2y+z, 9 = x+2z 
           Then x = 7+2/3, y = 4+2/3, z = 2/3
                (nobody said that x, y and z had to be integers!)
           And D got two 2nd place votes!
           This is a second, allowable solution.
           
 So, we have proved that there are only two solutions.
 We cannot tell who got the second place vote from Periodical D.
 If the votes have integer values then it is C.  If the votes have non-integer values then it is D.
 
 This problem is definitely difficulty 3!

Edited on June 7, 2013, 7:52 pm
  Posted by Steve Herman on 2013-06-07 19:51:45

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