Difference between revisions of "2014 AMC 8 Problems/Problem 16"
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Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of <math>4\cdot 8 =32</math> games outside the conference. | Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of <math>4\cdot 8 =32</math> games outside the conference. | ||
| − | Therefore, the total number of games is <math>56+32 = \boxed{88}</math>, so <math>\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. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=15|num-a=17}} | {{AMC8 box|year=2014|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:20, 20 December 2018
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.
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. 
