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Difference between revisions of "2011 AMC 10A Problems/Problem 3"
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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 18:57, 25 October 2020

Problem 3

Suppose [$a$ $b$] denotes the average of $a$ and $b$, and {$a$ $b$ $c$} denotes the average of $a$, $b$, and $c$. What is {{1 1 0} [0 1] 0}?

$\textbf{(A)}\ \frac{2}{9} \qquad\textbf{(B)}\ \frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3}$

Solution

Average $1$, $1$, and $0$ to get $\frac23$. Average $0$, and $1$, to get $\frac12$. Average $\frac23$, $\frac12$, and $0$. to get $\boxed{\textbf{(D)}\ \frac7{18}}$

Video Solution

https://youtu.be/JKO9YzQULvM

~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

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