Difference between revisions of "2014 AMC 8 Problems/Problem 16"

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(Solution)
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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.
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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.
  
 
==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 21:20, 20 December 2018

Problem

The "Middle School Eight" basketball conference has $8$ teams. Every season, each team plays every other conference team twice (home and away), and each team also plays $4$ games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?

$\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad \textbf{(E) }160$

Solution

Within the conference, there are 8 teams, so there are $\dbinom{8}{2}=28$ pairings of teams, and each pair must play two games, for a total of $28\cdot 2=56$ games within the conference.

Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of $4\cdot 8 =32$ games outside the conference.

Therefore, the total number of games is $56+32 = \boxed{88}$, so $\boxed{\text{(B)}}$ is our answer.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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

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