Difference between revisions of "2011 AMC 10A Problems/Problem 3"
Line 8: | Line 8: | ||
Average <math>1</math>, <math>1</math>, and <math>0</math> to get <math>\frac23</math>. Average <math>0</math>, and <math>1</math>, to get <math>\frac12</math>. Average <math>\frac23</math>, <math>\frac12</math>, and <math>0</math>. to get <math>\boxed{\textbf{(D)}\ \frac7{18}}</math> | Average <math>1</math>, <math>1</math>, and <math>0</math> to get <math>\frac23</math>. Average <math>0</math>, and <math>1</math>, to get <math>\frac12</math>. Average <math>\frac23</math>, <math>\frac12</math>, and <math>0</math>. to get <math>\boxed{\textbf{(D)}\ \frac7{18}}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/JKO9YzQULvM | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
== See Also == | == See Also == | ||
{{AMC10 box|year=2011|ab=A|num-b=2|num-a=4}} | {{AMC10 box|year=2011|ab=A|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:57, 25 October 2020
Contents
Problem 3
Suppose [
] denotes the average of
and
, and {
} denotes the average of
,
, and
. What is {{1 1 0} [0 1] 0}?
Solution
Average ,
, and
to get
. Average
, and
, to get
. Average
,
, and
. to get
Video Solution
~savannahsolver
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.