2014 AMC 8 Problems/Problem 16
Contents
[hide]Problem
The "Middle School Eight" basketball conference has teams. Every season, each team plays every other conference team twice (home and away), and each team also plays games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?
Solution
Within the conference, there are 8 teams, so there are pairings of teams, and each pair must play two games, for a total of games within the conference.
Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of games outside the conference.
Therefore, the total number of games is .
Solution 2
We can also figure out the number of games the 8 teams play within the conference with the formula n(n - 1)/2, which is used for these kind of match questions, where Team A playing Team B is the same as Team B playing Team A. In this formula, n is the number of teams. So we will do (8 X 7 / 2) X 2, as each team plays the other team twice. The 2s cancel out, to yield 8 X 7, which is 56 games. Then, we can just proceed as above to find the number of games the 8 teams play against 4 teams outside the conference, which is 8 X 4 = 32. Summing these values up yields 88, which is answer choice B.
Video Solution 1 by OmegaLearn
https://youtu.be/Zhsb5lv6jCI?t=991
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution 3
https://youtu.be/Zhsb5lv6jCI?t=991
https://youtu.be/w7Y-iq_kEaY ~savannahsolver
See Also
2014 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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All AJHSME/AMC 8 Problems and Solutions |
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