2005 AMC 12A Problems/Problem 5: Difference between revisions

Revision as of 20:15, 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 $\frac{20 \cdot 30 + 30 \cdot 20}{50} = 24 \ \mathrm{(B)}$ .

2005 AMC 12A (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Revision as of 21:26, 25 December 2011 view source Minsoens (talk \| contribs) Administrators 803 edits mNo edit summary ← Older edit		Revision as of 20:15, 3 July 2013 view source Nathan wailes (talk \| contribs) 2,960 edits →See also Newer edit →
Line 13:		Line 13:

	[[Category:Introductory Algebra Problems]]		[[Category:Introductory Algebra Problems]]
			{{MAA Notice}}

Page

Toolbox

Search

2005 AMC 12A Problems/Problem 5: Difference between revisions

Revision as of 20:15, 3 July 2013

Problem

Solution

See also