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

m (Solution)
Line 19: Line 19:
  
 
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>.
 
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>.
 +
 +
 +
== See Also ==
 +
 +
{{AMC10 box|year=2010|ab=A|before=First Question|num-a=2}}

Revision as of 19:33, 1 January 2012

Problem 1

Mary’s top book shelf holds five books with the following widths, in centimeters: $6$, $\dfrac{1}{2}$, $1$, $2.5$, and $10$. What is the average book width, in centimeters?

$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$

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 $\boxed{D}$.


See Also

2010 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
Invalid username
Login to AoPS