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

Revision as of 13:26, 13 August 2019

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?

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

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 $\frac{1200}{50}=24 \Longrightarrow \mathrm{(B)}$

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

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 Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=24 \Longrightarrow \mathrm{(B)}</math>
-==See Also==
+==See also==
-*[[2005 AMC 10A Problems]]
+{{AMC10 box|year=2005|ab=A|num-b=5|num-a=7}}
-*[[2005 AMC 10A Problems/Problem 5|Previous Problem]]
+[[Category:Introductory Number Theory Problems]]
-*[[2005 AMC 10A Problems/Problem 7|Next Problem]]
-[[Category:Introductory Algebra Problems]]
 {{MAA Notice}}