A soccer team won two of its first twelve games.The team ended the season winning exactly half of the games it played.What is the fewest number of games the team could have played

20

To find the fewest number of games the team could have played, we need to consider the total number of games they won.

We know that the team won two of its first twelve games and ended the season winning exactly half of the games played. This means that the team won the same number of games in the second half of the season as it did in the first half.

Let's assume the team played "n" games in total. In the first half of the season (first 12 games), they won two games. In the second half of the season (games 13 to n), they also won "2" games.

So, in total, the number of games won by the team is:
Games won in the first half (2) + Games won in the second half (2)
= 2 + 2
= 4.

Given that the team ended the season winning exactly half of the games played, the number of games won is equal to half of the total number of games played:
4 = n/2.

To solve for "n," we multiply both sides of the equation by 2:
4 x 2 = n
8 = n.

Therefore, the fewest number of games the team could have played is 8.