Difference between revisions of "2014 AMC 8 Problems/Problem 16"
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Therefore, the total number of games is <math>56+32 = \boxed{88}</math>, so <math>\boxed{\text{(B)}}</math> is our answer. | Therefore, the total number of games is <math>56+32 = \boxed{88}</math>, so <math>\boxed{\text{(B)}}</math> is our answer. | ||
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+ | ==Video Solution 1 by OmegaLearn== | ||
+ | https://youtu.be/Zhsb5lv6jCI?t=991 | ||
==Video Solution (CREATIVE THINKING)== | ==Video Solution (CREATIVE THINKING)== | ||
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− | ==Video Solution== | ||
https://youtu.be/Zhsb5lv6jCI?t=991 | https://youtu.be/Zhsb5lv6jCI?t=991 | ||
Latest revision as of 14:25, 29 September 2024
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 , so is our answer.
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.