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2010 AMC 10A Problems/Problem 1

Revision as of 15:27, 20 December 2010 by T90bag (talk | contribs)

Solution

To find the average, we add up the widths $6$, $\dfrac{1}{2}$, $1$, $2.5$, and $10$, to get a total sum of $20$. Since there are $5$ books, the average book width is $\frac{20}{5}=4$ The answer is $\boxed{D}$.

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