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

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Latest revision as of 16:29, 8 December 2020

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?

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

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

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
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

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