Difference between revisions of "2010 AMC 10A Problems/Problem 1"
m |
|||
| (One intermediate revision by one other user not shown) | |||
| Line 20: | Line 20: | ||
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>. | ||
| + | ==Video Solution== | ||
| + | https://youtu.be/C1VCk_9A2KE | ||
| + | |||
| + | ~IceMatrix | ||
== See Also == | == See Also == | ||
| − | {{AMC10 box|year=2010|ab=A|before= | + | {{AMC10 box|year=2010|ab=A|before=Non-Existent|num-a=2}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 06:14, 26 May 2020
Contents
Problem 1
Mary’s top book shelf holds five books with the following widths, in centimeters: 
, 
, 
, 
, and 
. What is the average book width, in centimeters?
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 
.
Video Solution
~IceMatrix
See Also
| 2010 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Non-Existent |
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
