5 soccer teams participate in a tournament. Each team plays all the other teams exactly once. If a team wins a game, 3 points are awarded. If a team loses a game, no points are awarded. If the game is tie, both teams get 1 point. At the end of all the games, 4 of the 5 teams get 1, 2, 5 and 8 points in total. The total score of the fifth team is  k. Then  $k-4=$

Let W, L and  T represent a win, a lose and a tie respectively. Let A, B, C and  D  be the four teams who gets 1,2,5 and 8 points in total, respectively. Let  E  be the fifth team.
Neither A nor B has a win, since both of them have less than 3 points, i.e. no wins. So A has $\{T,L,L,L\}$  and   B has $\{T,T,L,L\}$  . Moreover, the match between A and B must be a tie. Thus A lost all other matches against C, D, E Since  C  has a win against A and 5 points in total,  C has $\{W,T,T,L\}$  . Then the match between B and C must be a tie, since neither B or C has no other wins. Thus B losses its matches against D and E. Since D has two wins against A and B, and 8 points in total, D has  $\{W,W,T,T\}$ , i.e. its other matches against C or E are tie.
Finally, E has a win against A, B and C, while it has a tie against D. So its total point is $3\cdot 2+1=10$ .