During the 2010 baseball season, the number of wins for three teams was three consecutive integers. Of these three teams, the first team had the most wins. The last team had the least wins. The total number of wins by these three teams was 243. How many wins did each team have in the 2010 season?

Question

Mathematics

Karson Oliver

1 June, 20:58

During the 2010 baseball season, the number of wins for three teams was three consecutive integers. Of these three teams, the first team had the most wins. The last team had the least wins. The total number of wins by these three teams was 243. How many wins did each team have in the 2010 season?

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Byron · Answer 1

Step-by-step explanation:

Let x represent the number of wins of the least team.

The number of wins for three teams was three consecutive integers. Of these three teams, if the first team had the most wins and the last team had the least wins. It means that the number of wins of the second team would be x - 1

The number if wins on the last team would be x - 2

The total number of wins by these three teams was 243. It means that

x + x - 1 + x - 2 = 243

3x - 3 = 243

3x = 243 + 3 = 246

x = 246/3 = 82

The first team had 82 wins

The second team had 82 - 1 = 81 wins

The last team had 82 - 2 = 80 wins