Difference between revisions of "2012 AMC 8 Problems/Problem 14"

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<math> \textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}10 </math>
 
<math> \textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}10 </math>
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==Solution==
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This problem is very similar to a handshake problem. We use the formula <math> \frac{n(n-1)}{2} </math> to usually find the number of games played (or handshakes). Now we have to use the formula in reverse.
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So we have the equation  <math> \frac{n(n-1)}{2} = 21 </math>. Solving, we find that the number of teams in the BIG N conference is  <math> \boxed{\textbf{(B)}\ 7} </math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=13|num-a=15}}
 
{{AMC8 box|year=2012|num-b=13|num-a=15}}

Revision as of 12:04, 24 November 2012

In the BIG N, a middle school football conference, each team plays every other team exactly once. If a total of 21 conference games were played during the 2012 season, how many teams were members of the BIG N conference?

$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}10$

Solution

This problem is very similar to a handshake problem. We use the formula $\frac{n(n-1)}{2}$ to usually find the number of games played (or handshakes). Now we have to use the formula in reverse.

So we have the equation $\frac{n(n-1)}{2} = 21$. Solving, we find that the number of teams in the BIG N conference is $\boxed{\textbf{(B)}\ 7}$.

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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