Difference between revisions of "2005 AMC 10A Problems/Problem 6"

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Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=24 \Longrightarrow \mathrm{(B)}</math>
 
Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=24 \Longrightarrow \mathrm{(B)}</math>
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==Video Solution==
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CHECK OUT Video Solution: https://youtu.be/WIR8yPLET9Y
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-Education, The Study Of Everything
  
 
==See also==
 
==See also==

Revision as of 21:01, 30 October 2020

Problem

The average (mean) of $20$ numbers is $30$, and the average of $30$ other numbers is $20$. What is the average of all $50$ numbers?

$\mathrm{(A) \ } 23\qquad \mathrm{(B) \ } 24\qquad \mathrm{(C) \ } 25\qquad \mathrm{(D) \ } 26\qquad \mathrm{(E) \ } 27$

Solution

Since the average of the first $20$ numbers is $30$, their sum is $20\cdot30=600$.

Since the average of $30$ other numbers is $20$, their sum is $30\cdot20=600$.

So the sum of all $50$ numbers is $600+600=1200$

Therefore, the average of all $50$ numbers is $\frac{1200}{50}=24 \Longrightarrow \mathrm{(B)}$

Video Solution

CHECK OUT Video Solution: https://youtu.be/WIR8yPLET9Y

-Education, The Study Of Everything

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions

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