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kolbaska11 [484]
2 years ago
15

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?
Mathematics
1 answer:
uysha [10]2 years ago
6 0

Answer:

76075, 76076, 76077

Step-by-step explanation:

<u>There are 3 teams; Team A, Team B and Team C</u>

<em>Team A has most wins</em>

<em>Team C has least wins</em>

<em>Team B is in between</em>

<u>All these will be consecutive numbers.</u>

<em>Team B: x</em>

<em>Team A: x + 1 (most wins)</em>

<em>Team C: x - 1 (least wins)</em>

<em>Team A + Team B + Team C = Total number of wins</em>

x + x + 1 + x - 1 = 228228

3x = 228228

x = 76076

<em>Wins of Team B : x =</em> 76076

<em>Wins of Team A : x + 1 = 76076 + 1 =</em> 76077

<em>Wins of Team C : x - 1 = 76076 - 1 =</em> 76075

Therefore, in the 2010 season, Team A had 76077 wins, Team B had 76076 wins and Team C had 76075 wins.

!!

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