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 15:27, 20 December 2010
Solution
To find the average, we add up the widths 
, 
, 
, 
, and 
, to get a total sum of 
. Since there are 
books, the average book width is 
The answer is 
.
