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

Latest 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
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

@@ Line 2: / Line 2: @@
 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?
-<math> \mathrm{(A) \ } 23\qquad \mathrm{(B) \ } 24\qquad \mathrm{(C) \ } 25\qquad \mathrm{(D) \ } 26\qquad \mathrm{(E) \ } 27 </math>
+<math> \textbf{(A) } 23\qquad \textbf{(B) } 24\qquad \textbf{(C) } 25\qquad \textbf{(D) } 26\qquad \textbf{(E) } 27 </math>
 ==Solution==
@@ 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 \Longrightarrow \mathrm{(B)}</math>
+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==
 {{AMC10 box|year=2005|ab=A|num-b=5|num-a=7}}
-[[Category:Introductory Number Theory Problems]]
 {{MAA Notice}}