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 228228. How many wins did each team have in the 2010 season?
Accepted Solution
A:
Answer:76075, 76076, 76077Step-by-step explanation:There are 3 teams; Team A, Team B and Team CTeam A has most winsTeam C has least winsTeam B is in betweenAll these will be consecutive numbers.Team B: xTeam A: x + 1 (most wins)Team C: x - 1 (least wins)Team A + Team B + Team C = Total number of winsx + x + 1 + x - 1 = 2282283x = 228228x = 76076Wins of Team B : x = 76076Wins of Team A : x + 1 = 76076 + 1 = 76077Wins of Team C : x - 1 = 76076 - 1 = 76075Therefore, in the 2010 season, Team A had 76077 wins, Team B had 76076 wins and Team C had 76075 wins.!!