Difference between revisions of "2015 AMC 8 Problems/Problem 24"

Revision as of 17:42, 22 December 2024

Big balck men

On one team they play $3N$ games in their division and $4M$ games in the other. This gives $3N+4M=76$ .

Since $M>4$ we start by trying M=5. This doesn't work because $56$ is not divisible by $3$ .

Next, $M=6$ does not work because $52$ is not divisible by $3$ .

We try $M=7$ does work by giving $N=16$ , $~M=7$ and thus $3\times 16=\boxed{\textbf{(B)}~48}$ games in their division.

$M=10$ seems to work, until we realize this gives $N=12$ , but $N>2M$ so this will not work.

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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

@@ Line 1: / Line 1: @@
-== Problem ==
+Big balck men
-A baseball league consists of two four-team divisions. Each team plays every other team in its division <math>N</math> games. Each team plays every team in the other division <math>M</math> games with <math>N>2M</math> and <math>M>4</math>. Each team plays a <math>76</math> game schedule. How many games does a team play within its own division?
-<math>\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72 </math>
 ==Solution 1==