2011 AMC 10A Problems/Problem 5

Revision as of 09:55, 8 May 2011 by Kacheep (talk | contribs) (Solution)

Problem 5

At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of $12$, $15$, and $10$ minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?

$\textbf{(A)}\ 12 \qquad\textbf{(B)}\  \frac{37}{3} \qquad\textbf{(C)}\  \frac{88}{7} \qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 14$

Solution

Let there be $x$ fifth graders. It follows that there are $2x$ fourth graders and $4x$ third graders. We have $\frac{(1x)(10)+(2x)(15)+(4x)(12)}{1x+2x+4x} = \boxed{\textbf{(C)}\frac{88}{7}}$.

See Also

2011 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AMC 10 Problems and Solutions