Difference between revisions of "2011 AMC 10A Problems/Problem 5"

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Revision as of 19:05, 10 February 2011

Without loss of generality, let there be 1 fifth grader. It follows that there are 2 fourth graders and 4 third graders. We have $\frac{(1)(10)+(2)(15)+(4)(12)}{1+2+4} = \frac{88}{7}$.