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Difference between revisions of "2010 AMC 10A Problems/Problem 1"

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

Revision as of 16:27, 20 December 2010

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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