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
_________________
SC Butler has resumed!
Get two SC questions to practice, whose links you can find by date, here.