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

Revision as of 10:40, 2 August 2006

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

@@ Line 5: / Line 5: @@
 ==Solution==
-Since the average of <math>20</math> numbers is <math>30</math>, their sum is <math>20\cdot30=600</math>.
+Since the [[arithmetic mean|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>.
@@ Line 11: / Line 11: @@
 So the sum of all <math>50</math> numbers is <math>600+600=1200</math>
-Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=24 \Rightarrow B</math>
+Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=24 \Longrightarrow \mathrm{(B)}</math>
 ==See Also==