Difference between revisions of "2005 AMC 12A Problems/Problem 5"

m
(See also)
Line 13: Line 13:
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 21:15, 3 July 2013

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?

$(\mathrm {A}) \ 23 \qquad (\mathrm {B}) \ 24 \qquad (\mathrm {C})\ 25 \qquad (\mathrm {D}) \ 10 \qquad (\mathrm {E})\ 27$

Solution

The sum of the first 20 numbers is $20 \cdot 30$ and the sum of the other 30 numbers is $30\cdot 20$. Hence the overall average is $\frac{20 \cdot 30 + 30 \cdot 20}{50} = 24 \ \mathrm{(B)}$.

See also

2005 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png