2005 AMC 10A Problems/Problem 6: Difference between revisions

Revision as of 18:09, 25 December 2022

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?

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

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

~Charles3829

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

@@ Line 12: / Line 12: @@
 Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=\boxed{\textbf{(B) }24}</math>
+==Video Solution==
+https://youtu.be/kLZ3sbmfUb4
+~Charles3829
 ==See also==