During the 2010 baseball season, the number of wins for three teams was three consecutive integers. Of these three teams, the

Question

3 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] · Answer 1 · 2022-08-22T14:30:12+03:00

Answer:

76075, 76076, 76077

Step-by-step explanation:

There are 3 teams; Team A, Team B and Team C

Team A has most wins

Team C has least wins

Team B is in between

All these will be consecutive numbers.

Team B: x

Team A: x + 1 (most wins)

Team C: x - 1 (least wins)

Team A + Team B + Team C = Total number of wins

x + x + 1 + x - 1 = 228228

3x = 228228

x = 76076

Wins of Team B : x = 76076

Wins of Team A : x + 1 = 76076 + 1 = 76077

Wins of Team C : x - 1 = 76076 - 1 = 76075

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

!!

During the 2010 baseball​ season, the number of wins for three teams was three consecutive integers. Of these three​ teams, the

During the 2010 baseball season, the number of wins for three teams was three consecutive integers. Of these three teams, the