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

(Created page with 'Without loss of generality, let there be 1 fifth grader. It follows that there are 2 fourth graders and 4 third graders. We have <math>\frac{(1)(10)+(2)(15)+(4)(12)}{1+2+4} = \f…')
 
(Redirected page to 2011 AMC 12A Problems/Problem 4)
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Without loss of generality, let there be 1 fifth grader. It follows that there are 2 fourth graders and 4 third graders.
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#redirect [[2011 AMC 12A Problems/Problem 4]]
We have <math>\frac{(1)(10)+(2)(15)+(4)(12)}{1+2+4} = \frac{88}{7}</math>.
 

Latest revision as of 18:55, 27 June 2020