After the close of a soccer season, with four teams (named A, B, C and D), the final table was:

TEAM   Won   Drawn Lost     Goals Goals     Points
                            For   Against
A      3     0     0        7     1         9
B      1     1     1        2     3         4
C      1     1     1        3     3         4
D      0     0     3        1     6         0

Knowing that the four teams played against each other one time, and that team A beat team B 3x0, can you deduce the results of the other five games?

Since there are 4 teams, it means that there can only have been 6 matches.

AvB, AvC, AvD, BvC, BvD & CvD.

Since we know that  team B lost 3-0 against team A, then we can deduce that the draw with team C must be score less. since they have used all of there "against" goals already. this means team B must have beaten Team D 2-0

AvB 3-0
BvC 0-0
BvD 2-0

the above is ok and since A won all of it's games then the only possible outcomes for A, knowing that they beat B 3-0 are.

AvC 2-0 & AvD 2-1 or
AvC 3-1 & AvD 1-0 or
AvC 2-1 & AvD 2-0.

either C or D scored 1 goal against A,
if it were C then that means that they also scored 2 against D.
if that were the case then A must also have scored 2 against D.

AvC 2-1
AvD 2-0
CvD 2-1

this situation seems to be correct, but i need to disprove the others aswell

Posted by Juggler on 2005-06-10 12:56:19