Difference between revisions of "2014 AMC 10A Problems/Problem 10"
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Revision as of 14:42, 8 February 2014
- The following problem is from both the 2014 AMC 12A #9 and 2014 AMC 10A #10, so both problems redirect to this page.
Contents
Problem
Five positive consecutive integers starting with 
have average 
. What is the average of 
consecutive integers that start with ?
$\textbf{(A)}\ a+3\qquad\textbf{(B)}\ a+4\qquad\textbf{(C)}\ a+5\qquad\textbf{(D)}}\ a+6\qquad\textbf{(E)}\ a+7$ (Error compiling LaTeX. Unknown error_msg)
Solution 1
Let 
. Our list is 
with an average of 
. Our next set starting with 
is 
. Our average is 
.
Therefore, we notice that 
which means that the answer is 
.
Solution 2
We are given that
\begin{aligned}\frac{a+a+1+a+2+a+3+a+4}5 & =b\rightarrow \\
b & =a+2\end{aligned} (Error compiling LaTeX. Unknown error_msg)
We are asked to find the average of the 5 consecutive integers starting from in terms of . By substitution, this is ![\[\frac{a+2+a+3+a+4+a+5+a+6}5=a+4\]](http://latex.artofproblemsolving.com/5/1/6/5169a0e7b4ba46f5065ee9b32f4356bd2a7b90ef.png)
Thus, the answer is
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
| 2014 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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. 
