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

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The average (mean) of <math>20</math> numbers is <math>30</math>, and the average of <math>30</math> other numbers is <math>20</math>. What is the average of all <math>50</math> numbers?
 
The average (mean) of <math>20</math> numbers is <math>30</math>, and the average of <math>30</math> other numbers is <math>20</math>. What is the average of all <math>50</math> numbers?
  
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<math> \mathrm{(A) \ } 23\qquad \mathrm{(B) \ } 24\qquad \mathrm{(C) \ } 25\qquad \mathrm{(D) \ } 26\qquad \mathrm{(E) \ } 27 </math>
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<math>
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\textbf{(A) } 23\qquad \textbf{(B) } 24\qquad \textbf{(C) } 25\qquad \textbf{(D) } 26\qquad \textbf{(E) } 27
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</math>
  
 
==Solution==
 
==Solution==
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Since the [[arithmetic mean|average]] of the first <math>20</math> numbers is <math>30</math>, their sum is <math>20\cdot30=600</math>.  
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Since the average of the first <math>20</math> numbers is <math>30</math>, their sum is <math>20\cdot30=600</math>.  
  
 
Since the average of <math>30</math> other numbers is <math>20</math>, their sum is <math>30\cdot20=600</math>.  
 
Since the average of <math>30</math> other numbers is <math>20</math>, their sum is <math>30\cdot20=600</math>.  
  
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So the sum of all <math>50</math> numbers is <math>600+600=1200</math>
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So the sum of all <math>50</math> numbers is <math>600+600=1200</math>.
  
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Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=24 \Longrightarrow \mathrm{(B)}</math>
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Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=\boxed{\textbf{(B) } 24}</math>.
  
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==See Also==
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==Video Solution==
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*[[2005 AMC 10A Problems]]
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https://youtu.be/kLZ3sbmfUb4
  
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*[[2005 AMC 10A Problems/Problem 5|Previous Problem]]
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~Charles3829
  
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*[[2005 AMC 10A Problems/Problem 7|Next Problem]]
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==See also==
 +
{{AMC10 box|year=2005|ab=A|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 17:03, 1 July 2025

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?

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

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 $\frac{1200}{50}=\boxed{\textbf{(B) } 24}$.

Video Solution

https://youtu.be/kLZ3sbmfUb4

~Charles3829

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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