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Difference between revisions of "2010 AMC 8 Problems/Problem 4"

m (Undo revision 160325 by Raina0708 (talk) I PM'ed Raina0708 for suggesting NOT to erase LaTeX. So, I will restore the LaTeX code. LaTeX makes solutions more professional.)
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Putting the numbers in numerical order we get the list <math>0,0,1,2,3,3,3,4.</math>
 
Putting the numbers in numerical order we get the list <math>0,0,1,2,3,3,3,4.</math>
 
The mode is <math>3.</math> The median is <math>\frac{2+3}{2}=2.5.</math> The average is <math>\frac{0+0+1+2+3+3+3+4}{8}=\frac{16}{8}=2.</math> The sum of all three is <math>3+2.5+2=\boxed{\textbf{(C)}\ 7.5}</math>
 
The mode is <math>3.</math> The median is <math>\frac{2+3}{2}=2.5.</math> The average is <math>\frac{0+0+1+2+3+3+3+4}{8}=\frac{16}{8}=2.</math> The sum of all three is <math>3+2.5+2=\boxed{\textbf{(C)}\ 7.5}</math>
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== Video Solution by OmegaLearn ==
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https://youtu.be/51K3uCzntWs?t=209
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~ pi_is_3.14
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==Video by MathTalks==
 +
 +
https://youtu.be/EEbksvfujhk
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 +
 +
 +
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2010|num-b=3|num-a=5}}
 
{{AMC8 box|year=2010|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:00, 5 February 2023

Problem

What is the sum of the mean, median, and mode of the numbers $2,3,0,3,1,4,0,3$?

$\textbf{(A)}\ 6.5 \qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 7.5\qquad\textbf{(D)}\ 8.5\qquad\textbf{(E)}\ 9$

Solution

Putting the numbers in numerical order we get the list $0,0,1,2,3,3,3,4.$ The mode is $3.$ The median is $\frac{2+3}{2}=2.5.$ The average is $\frac{0+0+1+2+3+3+3+4}{8}=\frac{16}{8}=2.$ The sum of all three is $3+2.5+2=\boxed{\textbf{(C)}\ 7.5}$

Video Solution by OmegaLearn

https://youtu.be/51K3uCzntWs?t=209

~ pi_is_3.14


Video by MathTalks

https://youtu.be/EEbksvfujhk



See Also

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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