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Soccer Games

by Andrew
(New York)











































The chart below shows the results of six soccer games between teams A, B, C, and D, in which each team played every other team. What were the scores of each game? For each game list the two teams and the number of goals that each team scored.

Comments for Soccer Games

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Nov 17, 2011
Soccer Games
by: Staff

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Part III

Substituting these values into the equations:


AC_A + 3= 8
. . . AC_A = 8 - 3
. . . AC_A = 5

AD_D + 3= 5
. . . AD_D = 5 - 3
. . . AD_D = 2


1 + AC_A + AD_A = 7
. . . 5 + AD_A = 6
. . . AD_A = 6 - 5
. . . AD_A = 1


BD_B = 1
CD_C = 3

BD_D + CD_C = 3
BD_D + 3 = 3
BD_D = 0


The final answer:

Here are the values of all the variables:


AB = team A plays team B
AB_A = A team goals for game AB = 1 - won
AB_B = B team goals for game AB = 0

AC = team A plays team C
AC_A = A team goals for game AC = 5 - won
AC_C = C team goals for game AC = 0


AD = team A plays team D
AD_A = A team goals for game AD = 1 - won
AD_D = D team goals for game AD = 0

BC = team B plays team C (tied)
BC _B = A team goals for game BC = 0
BC _C = C team goals for game BC = 0

BD = team B plays team D
BD _B = B team goals for game BD = 1 - won
BD _D = D team goals for game BD = 0

CD = team C plays team D (tied )
CD _C = C team goals for game CD = 3
CD _D = D team goals for game CD = 3




Thanks for writing.

Staff
www.solving-math-problems.com


Nov 17, 2011
Soccer Games
by: Staff

------------------------------------------------------------

Part II


The final equations which need to be solved are:


AB_A + BC_C + BD_D = 1
AC_A + BC_B + CD_D = 8
AD_D + BD_D + CD_D = 5

AB_A + AC_A + AD_A = 7
BC_B + BD_B = 1
BC_C + CD_C = 3
BD_D + CD_D = 3



Team C tied for two games, but has a score of 3 goals. Team C did not score at all for game AC (team A plays team C) because team A won without any opposing goals. Therefore, team C tied in games BC and CD.

Therefore,

BC_C = BC_B
CD_C = CD_D




Substituting BC_C for BC_B:

AB_A + BC_C + BD_D = 1
AC_A + BC_C + CD_D = 8
AD_D + BD_D + CD_D = 5

AB_A + AC_A + AD_A = 7

BC_C + BD_B = 1
BC_C + CD_C = 3
BD_D + CD_D = 3



Substituting CD_C for CD_D :

AB_A + BC_C + BD_D = 1
AC_A + BC_C + CD_C = 8
AD_D + BD_D + CD_C = 5

AB_A + AC_A + AD_A = 7

BC_C + BD_B = 1
BC_C + CD_C = 3
BD_D + CD_C = 3

You now have 7 equations, but 8 variables. To complete the solution to this problem, you need to reduce the number of variables to 7 or less.

AB_A
AC_A
AD_A
AD_D
BC_C
BD_B
BD_D
CD_C

Note that:

AB_A + BC_C + BD_D = 1

Therefore, since team A won every game:

AB_A = 1

BC_C = 0

BD_D = 0


Substituting these values into the equations:


1 + 0 + 0 = 1

AC_A + CD_C = 8
AD_D + CD_C = 5

1 + AC_A + AD_A = 7
. . . or, AC_A + AD_A = 6

BD_B = 1
CD_C = 3

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Nov 17, 2011
Soccer Games
by: Staff


Part I

Question:
by Andrew
(New York)

The chart below shows the results of six soccer games between teams A, B, C, and D, in which each team played every other team. What were the scores of each game? For each game list the two teams and the number of goals that each team scored.


Answer:

Only 4 soccer teams can play one another: A, B, C, D

Order is not important
No repetition of the same combination

Permutations possible = n!/[r!(n-r)!]


Combinations without repetition (n=4, r=2)

n = 4

r = 2


Permutations possible = 4!/[2!(4-2)!]

= 4!/[2!(4-2)!]

= (4*3*2*1)/(2*1*2*1)

= 6


The six possibilities for soccer games

{A,B} {A,C} {A,D} {B,C} {B,D} {C,D}


AB = team A plays team B
AB_A = A team goals for game AB
AB_B = B team goals for game AB

AC = team A plays team C
AC_A = A team goals for game AC
AC_C = C team goals for game AC

AD = team A plays team D
AD_A = A team goals for game AD
AD_D = D team goals for game AD

BC = team B plays team C
BC _B = A team goals for game BC
BC _C = C team goals for game BC

BD = team B plays team D
BD _B = B team goals for game BD
BD _D = D team goals for game BD

CD = team C plays team D
CD _C = C team goals for game CD
CD _D = D team goals for game CD


Initial Equations:


Goals Scored Against Each Team

AB_B + AC_C + AD_D = 0

AB_A + BC_C + BD_D = 1

AC_A + BC_B + CD_D = 8

AD_D + BD_D + CD_D = 5


Goals Scored by Each Team

AB_A + AC_A + AD_A = 7

AB_B + BC_B + BD_B = 1

AC_C + BC_C + CD_C = 3

AD_D + BD_D + CD_D = 3



Adjusting the initial equations for known values

Goals Scored Against Each Team

AB_B + AC_C + AD_D = 0
. . . Therefore, since there can be no negative goals:
. . . AB_B = 0
. . . AC_C = 0
. . . AD_D = 0

AB_A + BC_C + BD_D = 1
AC_A + BC_B + CD_D = 8
AD_D + BD_D + CD_D = 5


Goals Scored by Each Team

AB_A + AC_A + AD_A = 7

AB_B + BC_B + BD_B = 1
. . . AB_B = 0
Therefore,
. . . 0 + BC_B + BD_B = 1
. . . BC_B + BD_B = 1

AC_C + BC_C + CD_C = 3
. . . AC_C = 0
Therefore,
. . . 0 + BC_C + CD_C = 3
. . . BC_C + CD_C = 3


AD_D + BD_D + CD_D = 3
. . . AD_D = 0
Therefore,
. . . 0 + BD_D + CD_D = 3
. . . BD_D + CD_D = 3

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