Difference between revisions of "2010 AMC 10A Problems/Problem 1"
Mariekitty (talk | contribs) m (→Solution) |
Hashtagmath (talk | contribs) |
||
(3 intermediate revisions by 3 users not shown) | |||
Line 1: | Line 1: | ||
== Problem 1 == | == Problem 1 == | ||
− | + | Mary's top book shelf holds five books with the following widths, in centimeters: <math>6</math>, <math>\dfrac{1}{2}</math>, <math>1</math>, <math>2.5</math>, and <math>10</math>. What is the average book width, in centimeters? | |
<math> | <math> | ||
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}} |
Latest revision as of 17:12, 18 October 2021
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.