Bunuel wrote:

If a soccer team scores an average of x goals per game for n games, scores an average of z goals for n-1 games, and scores an average of y goals in one game, what is the teams average goals for all the games?

A. (xn + y + zn - z)/(2n)

B. (x + y + x)/(2n)

C. x + (y + z)/(2n)

D. x/n + z/(n + 1) + y/(n - 1)

E. (xn + y + zn)/(2n)

Average goal for all games =

\(\frac{TotalGoals}{TotalGames}\) i.e.,

Overall Average =

\(\frac{(Ave)(Qty) + (Ave)(Qty) + (Ave)(Qty)}{(TotalQty)}\)Average goals scored =

\(Ave = A\)Number of games =

\(Qty = n\)1) average of x goals per game for n games

\(A_1 = x\) \(n_1 = n\)\(A_1*n_1 = S_1\)

\(xn = S_1\)2) average of z goals for n-1 games

\(A_2 = z\)\(n_2 = (n-1)\)\(A_2*n_2 = S_2\)

\(z(n-1) = zn - z = S_2\)3) an average of y goals in one game

\(A_3 = y\) \(n_3 = 1\)\(A_3*n_3 = S_3\)

\(y*1 = y = S_3\)Total Goals:

\(S_1 + S_2 + S_3 =\)

\(xn + (zn-z) + y = (xn + zn -z + y)\)

Total Games: \(n_1 + n_2 + n_3 =\)

\(n + (n-1) + 1 = (2n)\)

Overall Average: \(\frac{xn + zn - z + y}{2n}\)

Answer

Assign valuesThis approach might save time if you have a good grasp of the algebra involved.

Let

n = 2

x = 1

z = 1

y = 1

1) "average of x goals per game for n games"

(1 * 2 ) = 2 goals

2) "average of z goals for (n-1) games"

(1 * 1) = 1 goal

3) "average of y goals for 1 game"

(1 * 1) = 1 goal

Overall total goals: (2 + 1 + 1) = 4

Overall total games: (2 + 1+ 1) = 4

Average goals per game: \(\frac{4}{4}=1\)

With assigned variables, find the answer that yields 1

If you check, the choice that yields 1 is

Answer

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