Bunuel wrote:

A soccer team scores x goals the first game then averages twice as many goals for the next 5 games. The team averages two goals less than the average (arithmetic mean) of the previous 5 games over their next four games. How many goals will the team have scored after 10 games?

A. 19x − 8

B. 15x

C. 12x

D. 19x/10

E. 27x

Attachment:

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AlgebraThere are three groups of games.

Each group has

• A, an average number of goals

• n, a number of games and

• S, Sum (A * n), total number of goals for that group

There are 1 + 5 + 4 = 10 games

Total number of goals? Sum the sums for each group

\(A*n = S\)1) first game: team scores x goals

A = x, n = 1, S =

x2) next 5: average is twice that of first game

A = 2x, n = 5, S =

10x3) next 4: 2 goals less than average for the previous 5 games

A = 2x - 2, n = 4

S = 4(2x - 2) =

8x - 8 The total number of goals after 10 games: sum the sums

\(x + 10x + 8x - 8\) =

\(19x - 8\)Answer

Assign values• Let first week = 1 goal

1 game * average of 1 goal = 1 goal total

• Next 5 weeks, team averages twice as many goals as the first week

5 games * average of 2 goals = 10 goals

• Next 4 weeks, team averages 2 less than average (arithmetic mean) of previous 5 games' average

4 games * average of 0 goals = 0 goals total

• Total number of goals scored after 10 games?

Number of games? 1 + 5 + 4 = 10

Total number of goals: (1 + 10 + 0) = 11 goals total

• Using

x = 1, find the answer choice that yields

11 goals

Eliminate D (not an integer) and E immediately (it's huge)

A. 19x − 8: (19)(1) - 8 = 11.

MATCHB. 15x: (15)(1) = 15. NOT A MATCH

C. 12x (12)(1) = 12. NOT A MATCH

Answer

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